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**Unformatted text preview: **PROBLEM 1.1KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermalconductivity k and inner temperature, T1.FIND: The outer temperature of the wall, T2.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions,(3) Constant properties.ANALYSIS: The rate equation for conduction through the wall is given by Fouriers law,q cond = q x = q A = -kxT TdT A = kA 1 2 .dxLSolving for T2 givesT2 = T1 q cond L.kASubstituting numerical values, findT2 = 415 C -3000W 0.025m0.2W / m K 10m2T2 = 415 C - 37.5 CT2 = 378 C.COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.<PROBLEM 1.2KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from-15 to 38C.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3)Constant properties, (4) Outside wall temperature is that of the ambient air.ANALYSIS: From Fouriers law, it is evident that the gradient, dT dx = q k , is a constant, andxhence the temperature distribution is linear, if q and k are each constant. The heat flux must bexconstant under one-dimensional, steady-state conditions; and k is approximately constant if it dependsonly weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15Care)(25 C 15 CdTT1 T2q = k=k= 1W m K= 133.3W m 2 .xdxL0.30 mq x = q A = 133.3 W m 2 20 m 2 = 2667 W .x(1)(2)<Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15 T2 38C, with different wall thermal conductivities, k.3500Heat loss, qx (W)25001500500-500-1500-20-10010203040Ambient air temperature, T2 (C)Wall thermal conductivity, k = 1.25 W/m.Kk = 1 W/m.K, concrete wallk = 0.75 W/m.KFor the concrete wall, k = 1 W/mK, the heat loss varies linearily from +2667 W to -867 W and is zerowhen the inside and ambient temperatures are the same. The magnitude of the heat rate increases withincreasing thermal conductivity.COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a planewall would not be linear.PROBLEM 1.3KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiencyof gas furnace and cost of natural gas.FIND: Daily cost of heat loss.SCHEMATIC:ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.ANALYSIS: The rate of heat loss by conduction through the slab isT T7Cq = k ( LW ) 1 2 = 1.4 W / m K (11m 8 m )= 4312 Wt0.20 m<The daily cost of natural gas that must be combusted to compensate for the heat loss isCd =q Cgf( t ) =4312 W $0.01/ MJ0.9 106 J / MJ( 24 h / d 3600s / h ) = $4.14 / d<COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulationbetween it and the concrete.PROBLEM 1.4KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribedthickness.FIND: Thermal conductivity, k, of the wood.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-stateconditions, (3) Constant properties.ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may bedetermined from Fouriers law, Eq. 1.2. Rearranging,k=qxLW= 40T T2m21k = 0.10 W / m K.0.05m( 40-20 ) C<COMMENTS: Note that the C or K temperature units may be used interchangeably whenevaluating a temperature difference.PROBLEM 1.5KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions.FIND: Heat loss through window.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-stateconditions, (3) Constant properties.ANALYSIS: Subject to the foregoing conditions the heat flux may be computed fromFouriers law, Eq. 1.2.T Tq = k 1 2xLW (15-5 ) Cq = 1.4xm K 0.005mq = 2800 W/m 2 .xSince the heat flux is uniform over the surface, the heat loss (rate) isq = q Axq = 2800 W / m2 3m2q = 8400 W.COMMENTS: A linear temperature distribution exists in the glass for the prescribedconditions.<PROBLEM 1.6KNOWN: Width, height, thickness and thermal conductivity of a single pane window andthe air space of a double pane window. Representative winter surface temperatures of singlepane and air space.FIND: Heat loss through single and double pane windows.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-stateconditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy inducedmotion).ANALYSIS: From Fouriers law, the heat losses areSingle Pane:$T1 T22 35 C = 19, 600 Wqg = k g A= 1.4 W/m K 2mL0.005m()()T T25 $CDouble Pane: qa = k a A 1 2 = 0.024 2m2= 120 WL0.010 mCOMMENTS: Losses associated with a single pane are unacceptable and would remainexcessive, even if the thickness of the glass were doubled to match that of the air space. Theprincipal advantage of the double pane construction resides with the low thermal conductivityof air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, useof the double pane construction would also increase the surface temperature of the glassexposed to the room (inside) air.PROBLEM 1.7KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures.FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribedvalue.SCHEMATIC:ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 52walls of area A = 4m , (3) Steady-state conditions, (4) Constant properties.ANALYSIS: Using Fouriers law, Eq. 1.2, the heat rate isq = q A = kTA totalL2Solving for L and recognizing that Atotal = 5W , find5 k T W2L=qL=()5 0.03 W/m K 35 - (-10 ) C 4m 2500 WL = 0.054m = 54mm.<COMMENTS: The corners will cause local departures from one-dimensional conductionand a slightly larger heat loss.PROBLEM 1.8KNOWN: Dimensions and thermal conductivity of food/beverage container. Inner and outersurface temperatures.FIND: Heat flux through container wall and total heat load.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through bottomwall, (3) Uniform surface temperatures and one-dimensional conduction through remainingwalls.ANALYSIS: From Fouriers law, Eq. 1.2, the heat flux is$T T 0.023 W/m K ( 20 2 ) C = k 2 1 =q= 16.6 W/m 2L0.025 m<Since the flux is uniform over each of the five walls through which heat is transferred, theheat load isq = q A total = q H ( 2W1 + 2W2 ) + W1 W2 q = 16.6 W/m2 0.6m (1.6m + 1.2m ) + ( 0.8m 0.6m ) = 35.9 W<COMMENTS: The corners and edges of the container create local departures from onedimensional conduction, which increase the heat load. However, for H, W1, W2 >> L, theeffect is negligible.PROBLEM 1.9KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of thatthrough a composite wall of prescribed thermal conductivity and thickness.FIND: Thickness of masonry wall.SCHEMATIC:ASSUMPTIONS: (1) Both walls subjected to same surface temperatures, (2) Onedimensional conduction, (3) Steady-state conditions, (4) Constant properties.ANALYSIS: For steady-state conditions, the conduction heat flux through a one-dimensionalwall follows from Fouriers law, Eq. 1.2,q = kTLwhere T represents the difference in surface temperatures. Since T is the same for bothwalls, it follows thatL1 = L2k1q 2.k2q1With the heat fluxes related asq1 = 0.8 q 2L1 = 100mm0.75 W / m K1= 375mm.0.25 W / m K0.8<COMMENTS: Not knowing the temperature difference across the walls, we cannot find thevalue of the heat rate.PROBLEM 1.10KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boilwater. Rate of heat transfer to the pan.FIND: Outer surface temperature of pan for an aluminum and a copper bottom.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction through bottom of pan.ANALYSIS: From Fouriers law, the rate of heat transfer by conduction through the bottomof the pan isT Tq = kA 1 2LHence,T1 = T2 +qLkA2where A = D2 / 4 = (0.2m ) / 4 = 0.0314 m 2 .Aluminum:T1 = 110 $C +Copper:T1 = 110 $C +600W ( 0.005 m )(240 W/m K 0.0314 m 2600W (0.005 m )(390 W/m K 0.0314 m2))= 110.40 $C= 110.25 $CCOMMENTS: Although the temperature drop across the bottom is slightly larger foraluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible forboth materials. To a good approximation, the bottom may be considered isothermal at T 110 C, which is a desirable feature of pots and pans.PROBLEM 1.11KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one surface.FIND: Temperature drop across the chip.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heatdissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction inchip.ANALYSIS: All of the electrical power dissipated at the back surface of the chip istransferred by conduction through the chip. Hence, from Fouriers law,P = q = kATtorT =t PkW 2=T = 1.1 C.0.001 m 4 W2150 W/m K ( 0.005 m )<COMMENTS: For fixed P, the temperature drop across the chip decreases with increasing kand W, as well as with decreasing t.PROBLEM 1.12KNOWN: Heat flux gage with thin-film thermocouples on upper and lower surfaces; outputvoltage, calibration constant, thickness and thermal conductivity of gage.FIND: (a) Heat flux, (b) Precaution when sandwiching gage between two materials.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat conduction in gage,(3) Constant properties.ANALYSIS: (a) Fouriers law applied to the gage can be written asq = kTxand the gradient can be expressed asTE / N=xSABtwhere N is the number of differentially connected thermocouple junctions, SAB is the Seebeckcoefficient for type K thermocouples (A-chromel and B-alumel), and x = t is the gagethickness. Hence,q =kENSABtq =1.4 W / m K 350 10-6 V= 9800 W / m2 .-6 V / $ C 0.25 10-3 m5 40 10<(b) The major precaution to be taken with this type of gage is to match its thermalconductivity with that of the material on which it is installed. If the gage is bondedbetween laminates (see sketch above) and its thermal conductivity is significantly differentfrom that of the laminates, one dimensional heat flow will be disturbed and the gage willread incorrectly.COMMENTS: If the thermal conductivity of the gage is lower than that of the laminates,will it indicate heat fluxes that are systematically high or low?PROBLEM 1.13KNOWN: Hand experiencing convection heat transfer with moving air and water.FIND: Determine which condition feels colder. Contrast these results with a heat loss of 30 W/m2 undernormal room conditions.SCHEMATIC:ASSUMPTIONS: (1) Temperature is uniform over the hands surface, (2) Convection coefficient isuniform over the hand, and (3) Negligible radiation exchange between hand and surroundings in the caseof air flow.ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss. The heatloss can be determined from Newtons law of cooling, Eq. 1.3a, written asq = h ( Ts T )For the air stream:q = 40 W m 2 K 30 ( 5) K = 1, 400 W m 2air<For the water stream:22qwater = 900 W m K (30 10 ) K = 18,000 W m<COMMENTS: The heat loss for the hand in the water stream is an order of magnitude larger than whenin the air stream for the given temperature and convection coefficient conditions. In contrast, the heatloss in a normal room environment is only 30 W/m2 which is a factor of 400 times less than the loss inthe air stream. In the room environment, the hand would feel comfortable; in the air and water streams,as you probably know from experience, the hand would feel uncomfortably cold since the heat loss isexcessively high.PROBLEM 1.14KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinderwith an imbedded electrical heater for different air velocities.FIND: (a) Determine the convection coefficient for each of the air velocity conditions and displaythe results graphically, and (b) Assuming that the convection coefficient depends upon air velocity ash = CVn, determine the parameters C and n.SCHEMATIC:V(m/s)Pe (W/m)h (W/m2K)145022.0265832.2498348.18150773.812196396.1ASSUMPTIONS: (1) Temperature is uniform over the cylinder surface, (2) Negligible radiationexchange between the cylinder surface and the surroundings, (3) Steady-state conditions.ANALYSIS: (a) From an overall energy balance on the cylinder, the power dissipated by theelectrical heater is transferred by convection to the air stream. Using Newtons law of cooling on a perunit length basis,Pe = h ( D )(Ts T )where Pe is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/scondition, using the data from the table above, findh = 450 W m 0.025 m 300 40 C = 22.0 W m 2K()<Repeating the calculations, find the convection coefficients for the remaining conditions which aretabulated above and plotted below. Note that h is not linear with respect to the air velocity.(b) To determine the (C,n) parameters, we plotted h vs. V on log-log coordinates. Choosing C =22.12 W/m2K(s/m)n, assuring a match at V = 1, we can readily find the exponent n from the slope ofthe h vs. V curve. From the trials with n = 0.8, 0.6 and 0.5, we recognize that n = 0.6 is a reasonable<10080604020024681012Air velocity, V (m/s)Data, smooth curve, 5-pointsCoefficient, h (W/m^2.K)Coefficient, h (W/m^2.K)choice. Hence, C = 22.12 and n = 0.6.10080604020101246Air velocity, V (m/s)Data , smooth curve, 5 pointsh = C * V^n, C = 22.1, n = 0.5n = 0.6n = 0.8810PROBLEM 1.15KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power requiredto maintain a specified surface temperature for water and air flows.FIND: Convection coefficients for the water and air flow convection processes, hw and ha,respectively.SCHEMATIC:ASSUMPTIONS: (1) Flow is cross-wise over cylinder which is very long in the directionnormal to flow.ANALYSIS: The convection heat rate from the cylinder per unit length of the cylinder hasthe formq = h ( D ) ( Ts T )and solving for the heat transfer convection coefficient, findh=q. D (Ts T )Substituting numerical values for the water and air situations:Waterhw =Airha =28 103 W/m 0.030m (90-25 ) C400 W/m 0.030m (90-25 ) C= 4,570 W/m 2 K= 65 W/m 2 K.<<COMMENTS: Note that the air velocity is 10 times that of the water flow, yethw 70 ha.These values for the convection coefficient are typical for forced convection heat transfer withliquids and gases. See Table 1.1.PROBLEM 1.16KNOWN: Dimensions of a cartridge heater. Heater power. Convection coefficients in airand water at a prescribed temperature.FIND: Heater surface temperatures in water and air.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) All of the electrical power is transferredto the fluid by convection, (3) Negligible heat transfer from ends.ANALYSIS: With P = qconv, Newtons law of cooling yieldsP=hA (Ts T ) = h DL ( Ts T )PTs = T +.h DLIn water,Ts = 20 C +2000 W5000 W / m K 0.02 m 0.200 m2Ts = 20 C + 31.8 C = 51.8 C.<In air,Ts = 20 C +2000 W50 W / m K 0.02 m 0.200 m2Ts = 20 C + 3183 C = 3203 C.<COMMENTS: (1) Air is much less effective than water as a heat transfer fluid. Hence, thecartridge temperature is much higher in air, so high, in fact, that the cartridge would melt.(2) In air, the high cartridge temperature would render radiation significant.PROBLEM 1.17KNOWN: Length, diameter and calibration of a hot wire anemometer. Temperature of airstream. Current, voltage drop and surface temperature of wire for a particular application.FIND: Air velocitySCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from the wire bynatural convection or radiation.ANALYSIS: If all of the electric energy is transferred by convection to the air, the followingequality must be satisfiedPelec = EI = hA (Ts T )where A = DL = ( 0.0005m 0.02m ) = 3.14 105 m 2 .Hence,h=EI5V 0.1A== 318 W/m 2 KA (Ts T ) 3.14 105m 2 50 $C(()V = 6.25 105 h 2 = 6.25 105 318 W/m 2 K)2= 6.3 m/s<COMMENTS: The convection coefficient is sufficiently large to render buoyancy (naturalconvection) and radiation effects negligible.PROBLEM 1.18KNOWN: Chip width and maximum allowable temperature. Coolant conditions.FIND: Maximum allowable chip power for air and liquid coolants.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides andbottom, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer byradiation in air.ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection tothe coolant. Hence,P=qand from Newtons law of cooling,2P = hA(T - T) = h W (T - T).In air,22Pmax = 200 W/m K(0.005 m) (85 - 15) C = 0.35 W.<In the dielectric liquid22Pmax = 3000 W/m K(0.005 m) (85-15) C = 5.25 W.<COMMENTS: Relative to liquids, air is a poor heat transfer fluid. Hence, in air the chip candissipate far less energy than in the dielectric liquid.PROBLEM 1.19KNOWN: Length, diameter and maximum allowable surface temperature of a powertransistor. Temperature and convection coefficient for air cooling.FIND: Maximum allowable power dissipation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through base oftransistor, (3) Negligible heat transfer by radiation from surface of transistor.ANALYSIS: Subject to the foregoing assumptions, the power dissipated by the transistor isequivalent to the rate at which heat is transferred by convection to the air. Hence,Pelec = q conv = hA (Ts T )()2where A = DL + D2 / 4 = 0.012m 0.01m + ( 0.012m ) / 4 = 4.90 104 m 2 .For a maximum allowable surface temperature of 85C, the power is(Pelec = 100 W/m2 K 4.90 104 m 2) (85 25)$ C = 2.94 W<COMMENTS: (1) For the prescribed surface temperature and convection coefficient,radiation will be negligible relative to convection. However, conduction through the basecould be significant, thereby permitting operation at a larger power.(2) The local convection coefficient varies over the surface, and hot spots could exist if thereare locations at which the local value of h is substantially smaller than the prescribed averagevalue.PROBLEM 1.20KNOWN: Air jet impingement is an effective means of cooling logic chips.FIND: Procedure for measuring convection coefficients associated with a 10 mm 10 mm chip.SCHEMATIC:ASSUMPTIONS: Steady-state conditions.ANALYSIS: One approach would be to use the actual chip-substrate system, Case (a), to perform themeasurements. In this case, the electric power dissipated in the chip would be transferred from the chipby radiation and conduction (to the substrate), as well as by convection to the jet. An energy balance forthe chip yields q elec = q conv + q cond + q rad . Hence, with q conv = hA ( Ts T ) , where A = 100mm2 is the surface area of the chip,q q cond q radh = elecA (Ts T )(1)While the electric power ( q elec ) and the jet ( T ) and surface ( Ts ) temperatures may be measured, lossesfrom the chip by conduction and radiation would have to be estimated. Unless the losses are negligible(an unlikely condition), the accuracy of the procedure could be compromised by uncertainties associatedwith determining the conduction and radiation losses.A second approach, Case (b), could involve fabrication of a heater assembly for which theconduction and radiation losses are controlled and minimized. A 10 mm 10 mm copper block (k ~ 400W/mK) could be inserted in a poorly conducting substrate (k < 0.1 W/mK) and a patch heater could beapplied to the back of the block and insulated from below. If conduction to both the substrate andinsulation could thereby be rendered negligible, heat would be transferred almost exclusively through theblock. If radiation were rendered negligible by applying a low emissivity coating ( < 0.1) to the surfaceof the copper block, virtually all of the heat would be transferred by convection to the jet. Hence, q condand q rad may be neglected in equation (1), and the expression may be used to accurately determine hfrom the known (A) and measured ( q elec , Ts , T ) quantities.COMMENTS: Since convection coefficients associated with gas flows are generally small, concurrentheat transfer by radiation and/or conduction must often be considered. However, jet impingement is oneof the more effective means of transferring heat by convection and convection coefficients well in excessof 100 W/m2K may be achieved.PROBLEM 1.21KNOWN: Upper temperature set point, Tset, of a bimetallic switch and convection heattransfer coefficient between clothes dryer air and exposed surface of switch.FIND: Electrical power for heater to maintain Tset when air temperature is T = 50C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Electrical heater is perfectly insulatedfrom dryer wall, (3) Heater and switch are isothermal at Tset, (4) Negligible heat transfer fromsides of heater or switch, (5) Switch surface, As, loses heat only by convection.ANALYSIS: Define a control volume around the bimetallic switch which experiences heatinput from the heater and convection heat transfer to the dryer air. That is,Ein - Eout = 0qelec - hAs ( Tset T ) = 0.The electrical power required is,qelec = hAs ( Tset T )qelec = 25 W/m2 K 30 10-6 m2 ( 70 50 ) K=15 mW.<COMMENTS: (1) This type of controller can achieve variable operating air temperatureswith a single set-point, inexpensive, bimetallic-thermostatic switch by adjusting power levelsto the heater.(2) Will the heater power requirement increase or decrease if the insulation pad is other thanperfect?PROBLEM 1.22KNOWN: Hot vertical plate suspended in cool, still air. Change in plate temperature with time atthe instant when the plate temperature is 225C.FIND: Convection heat transfer coefficient for this condition.SCHEMATIC:ASSUMPTIONS: (1) Plate is isothermal and of uniform temperature, (2) Negligible radiationexchange with surroundings, (3) Negligible heat lost through suspension wires.ANALYSIS: As shown in the cooling curve above, the plate temperature decreases with time. Thecondition of interest is for time to. For a control surface about the plate, the conservation of energyrequirement isE in - E out = E stdT 2hA s ( Ts T ) = M c pdtwhere As is the surface area of one side of the plate. Solving for h, findh=h=McpdT2As (Ts T ) dt3.75 kg 2770 J/kg K2 ( 0.3 0.3) m 2 ( 225 25 ) K 0.022 K/s=6.4 W/m 2 K<COMMENTS: (1) Assuming the plate is very highly polished with emissivity of 0.08, determinewhether radiation exchange with the surroundings at 25C is negligible compared to convection.(2) We will later consider the criterion for determining whether the isothermal plate assumption isreasonable. If the thermal conductivity of the present plate were high (such as aluminum or copper),the criterion would be satisfied.PROBLEM 1.23KNOWN: Width, input power and efficiency of a transmission. Temperature and convectioncoefficient associated with air flow over the casing.FIND: Surface temperature of casing.SCHEMATIC:ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3)Negligible radiation.ANALYSIS: From Newtons law of cooling,q = hAs ( Ts T ) = 6 hW 2 ( Ts T )where the output power is Pi and the heat rate isq = Pi Po = Pi (1 ) = 150 hp 746 W / hp 0.07 = 7833 WHence,Ts = T +q6 hW 2= 30C +7833 W6 200 W / m 2 K (0.3m )2= 102.5CCOMMENTS: There will, in fact, be considerable variability of the local convection coefficientover the transmission case and the prescribed value represents an average over the surface.<PROBLEM 1.24KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficientand emissivity of a person in the room.FIND: Basis for difference in comfort level between summer and winter.SCHEMATIC:ASSUMPTIONS: (1) Person may be approximated as a small object in a large enclosure.ANALYSIS: Thermal comfort is linked to heat loss from the human body, and a chilledfeeling is associated with excessive heat loss. Because the temperature of the room air isfixed, the different summer and winter comfort levels can not be attributed to convection heattransfer from the body. In both cases, the heat flux is2$2Summer and Winter: qconv = h ( Ts T ) = 2 W/m K 12 C = 24 W/mHowever, the heat flux due to radiation will differ, with values ofSummer:)()(844244442q = Ts Tsur = 0.9 5.67 10 W/m K 305 300 K = 28.3 W/mrad())(4Winter: q = Ts4 Tsur = 0.9 5.67 108 W/m 2 K 4 3054 287 4 K 4 = 95.4 W/m 2radThere is a significant difference between winter and summer radiation fluxes, and the chilledcondition is attributable to the effect of the colder walls on radiation.2COMMENTS: For a representative surface area of A = 1.5 m , the heat losses are qconv =36 W, qrad(summer) = 42.5 W and qrad(winter) = 143.1 W. The winter time radiation loss issignificant and if maintained over a 24 h period would amount to 2,950 kcal.PROBLEM 1.25KNOWN: Diameter and emissivity of spherical interplanetary probe. Power dissipationwithin probe.FIND: Probe surface temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible radiation incident on the probe.ANALYSIS: Conservation of energy dictates a balance between energy generation within theprobe and radiation emission from the probe surface. Hence, at any instant-E out + E g = 0A sTs4 = E g EgTs = D21/ 41/ 4150WTs = 0.8 0.5 m 2 5.67 108 W/m2 K 4 ()Ts = 254.7 K.<COMMENTS: Incident radiation, as, for example, from the sun, would increase the surfacetemperature.PROBLEM 1.26KNOWN: Spherical shaped instrumentation package with prescribed surface emissivity within alarge space-simulation chamber having walls at 77 K.FIND: Acceptable power dissipation for operating the package surface temperature in the range Ts =40 to 85C. Show graphically the effect of emissivity variations for 0.2 and 0.3.SCHEMATIC:ASSUMPTIONS: (1) Uniform surface temperature, (2) Chamber walls are large compared to thespherical package, and (3) Steady-state conditions.ANALYSIS: From an overall energy balance on the package, the internal power dissipation Pe willbe transferred by radiation exchange between the package and the chamber walls. From Eq. 1.7,44q rad = Pe = As Ts Tsur)(For the condition when Ts = 40C, with As = D2 the power dissipation will be4Pe = 0.25 ( 0.10 m ) 5.67 108 W m 2 K 4 ( 40 + 273) 77 4 K 4 = 4.3 W<Repeating this calculation for the range 40 Ts 85C, we can obtain the power dissipation as afunction of surface temperature for the = 0.25 condition. Similarly, with 0.2 or 0.3, the family ofcurves shown below has been obtained.Power dissipation, Pe (W)108642405060708090Surface temperature, Ts (C)Surface emissivity, eps = 0.3eps = 0.25eps = 0.2COMMENTS: (1) As expected, the internal power dissipation increases with increasing emissivityand surface temperature. Because the radiation rate equation is non-linear with respect totemperature, the power dissipation will likewise not be linear with surface temperature.(2) What is the maximum power dissipation that is possible if the surface temperature is not to exceed85C? What kind of a coating should be applied to the instrument package in order to approach thislimiting condition?PROBLEM 1.27KNOWN: Area, emissivity and temperature of a surface placed in a large, evacuatedchamber of prescribed temperature.FIND: (a) Rate of surface radiation emission, (b) Net rate of radiation exchange betweensurface and chamber walls.SCHEMATIC:ASSUMPTIONS: (1) Area of the enclosed surface is much less than that of chamber walls.ANALYSIS: (a) From Eq. 1.5, the rate at which radiation is emitted by the surface is4q emit = E A = A Ts)(qemit = 0.8 0.5 m 2 5.67 10-8 W/m 2 K 4 (150 + 273) K 4<q emit = 726 W.(b) From Eq. 1.7, the net rate at which radiation is transferred from the surface to the chamberwalls is(44q = A Ts Tsur())44q = 0.8 0.5 m2 5.67 10-8 W/m 2 K 4 ( 423K ) - ( 298K ) q = 547 W.<COMMENTS: The foregoing result gives the net heat loss from the surface which occurs atthe instant the surface is placed in the chamber. The surface would, of course, cool due to thisheat loss and its temperature, as well as the heat loss, would decrease with increasing time.Steady-state conditions would eventually be achieved when the temperature of the surfacereached that of the surroundings.PROBLEM 1.28KNOWN: Length, diameter, surface temperature and emissivity of steam line. Temperatureand convection coefficient associated with ambient air. Efficiency and fuel cost for gas firedfurnace.FIND: (a) Rate of heat loss, (b) Annual cost of heat loss.SCHEMATIC:ASSUMPTIONS: (1) Steam line operates continuously throughout year, (2) Net radiationtransfer is between small surface (steam line) and large enclosure (plant walls).ANALYSIS: (a) From Eqs. (1.3a) and (1.7), the heat loss is()44q = qconv + q rad = A h ( Ts T ) + Ts Tsur where A = DL = ( 0.1m 25m ) = 7.85m 2 .Hence,()q = 7.85m2 10 W/m2 K (150 25) K + 0.8 5.67 108 W/m2 K 4 4234 2984 K 4 q = 7.85m2 (1, 250 + 1, 095) w/m 2 = (9813 + 8592 ) W = 18, 405 W<(b) The annual energy loss isE = qt = 18, 405 W 3600 s/h 24h/d 365 d/y = 5.80 1011 JWith a furnace energy consumption of Ef = E/f = 6.45 1011 J, the annual cost of the lossisC = Cg Ef = 0.01 $/MJ 6.45 105MJ = $6450<COMMENTS: The heat loss and related costs are unacceptable and should be reduced byinsulating the steam line.PROBLEM 1.29KNOWN: Exact and approximate expressions for the linearized radiation coefficient, hr and hra,respectively.FIND: (a) Comparison of the coefficients with = 0.05 and 0.9 and surface temperatures which mayexceed that of the surroundings (Tsur = 25C) by 10 to 100C; also comparison with a free convectioncoefficient correlation, (b) Plot of the relative error (hr - rra)/hr as a function of the furnace temperatureassociated with a workpiece at Ts = 25C having = 0.05, 0.2 or 0.9.ASSUMPTIONS: (1) Furnace walls are large compared to the workpiece and (2) Steady-stateconditions.ANALYSIS: (a) The linearized radiation coefficient, Eq. 1.9, follows from the radiation exchangerate equation,22h r = (Ts + Tsur ) Ts + Tsur)(If Ts Tsur, the coefficient may be approximated by the simpler expressionh r,a = 4 T3T = ( Ts + Tsur ) 2For the condition of = 0.05, Ts = Tsur + 10 = 35C = 308 K and Tsur = 25C = 298 K, find thath r = 0.05 5.67 108 W m 2 K 4 (308 + 298 ) 3082 + 2982 K 3 = 0.32 W m 2 K)(h r,a = 4 0.05 5.67 108 W m 2 K 4 ((308 + 298 ) 2 ) K3 = 0.32 W m 2 K3<<The free convection coefficient with Ts = 35C and T = Tsur = 25C, find thath = 0.98T1/ 3 = 0.98 (Ts T )1/ 3= 0.98 (308 298 )1/ 3<= 2.1W m 2 KFor the range Ts - Tsur = 10 to 100C with = 0.05 and 0.9, the results for the coefficients aretabulated below. For this range of surface and surroundings temperatures, the radiation and freeconvection coefficients are of comparable magnitude for moderate values of the emissivity, say >0.2. The approximate expression for the linearized radiation coefficient is valid within 2% for theseconditions.(b) The above expressions for the radiation coefficients, hr and hr,a, are used for the workpiece at Ts =25C placed inside a furnace with walls which may vary from 100 to 1000C. The relative error, (hr hra)/hr, will be independent of the surface emissivity and is plotted as a function of Tsur. For Tsur >150C, the approximate expression provides estimates which are in error more than 5%. Theapproximate expression should be used with caution, and only for surface and surroundingtemperature differences of 50 to 100C.Ts (C)351350.050.90.050.9Coefficients (W/m K)hr,ahhr0.320.322.15.75.70.510.504.79.29.0Relative error, (hr-hra)/hr*100 (%)30220100100300500700Surroundings temperature, Tsur (C)900PROBLEM 1.30KNOWN: Chip width, temperature, and heat loss by convection in air. Chip emissivity andtemperature of large surroundings.FIND: Increase in chip power due to radiation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between small surfaceand large enclosure.ANALYSIS: Heat transfer from the chip due to net radiation exchange with the surroundingsis(4q rad = W 2 T 4 - Tsur)()q rad = 0.9 ( 0.005 m ) 5.67 108 W/m 2 K 4 3584 - 2884 K 42q rad = 0.0122 W.The percent increase in chip power is thereforeqP0.0122 W 100 = rad 100 = 100 = 35%..Pq conv0.350 WCOMMENTS: For the prescribed conditions, radiation effects are small. Relative toconvection, the effect of radiation would increase with increasing chip temperature anddecreasing convection coefficient.<PROBLEM 1.31KNOWN: Width, surface emissivity and maximum allowable temperature of an electronic chip.Temperature of air and surroundings. Convection coefficient.21/4FIND: (a) Maximum power dissipation for free convection with h(W/m K) = 4.2(T - T) , (b)2Maximum power dissipation for forced convection with h = 250 W/m K.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between a small surface and alarge enclosure, (3) Negligible heat transfer from sides of chip or from back of chip by conductionthrough the substrate.ANALYSIS: Subject to the foregoing assumptions, electric power dissipation by the chip must bebalanced by convection and radiation heat transfer from the chip. Hence, from Eq. (1.10),(44Pelec = q conv + q rad = hA (Ts T ) + A Ts Tsur)2where A = L2 = (0.015m ) = 2.25 104 m 2 .(a) If heat transfer is by natural convection,()qconv = C A ( Ts T )5 / 4 = 4.2 W/m 2 K5/4 2.25 104 m2 ( 60K )5 / 4 = 0.158 W()()q rad = 0.60 2.25 104 m2 5.67 108 W/m2 K 4 3584 2984 K 4 = 0.065 W<Pelec = 0.158 W + 0.065 W = 0.223 W(b) If heat transfer is by forced convection,()qconv = hA ( Ts T ) = 250 W/m2 K 2.25 104 m2 ( 60K ) = 3.375 WPelec = 3.375 W + 0.065 W = 3.44 W<COMMENTS: Clearly, radiation and natural convection are inefficient mechanisms for transferring2heat from the chip. For Ts = 85C and T = 25C, the natural convection coefficient is 11.7 W/m K.2Even for forced convection with h = 250 W/m K, the power dissipation is well below that associatedwith many of todays processors. To provide acceptable cooling, it is often necessary to attach thechip to a highly conducting substrate and to thereby provide an additional heat transfer mechanismdue to conduction from the back surface.PROBLEM 1.32KNOWN: Vacuum enclosure maintained at 77 K by liquid nitrogen shroud while baseplate ismaintained at 300 K by an electrical heater.FIND: (a) Electrical power required to maintain baseplate, (b) Liquid nitrogen consumption rate, (c)Effect on consumption rate if aluminum foil (p = 0.09) is bonded to baseplate surface.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) No heat losses from backside of heater or sides ofplate, (3) Vacuum enclosure large compared to baseplate, (4) Enclosure is evacuated with negligibleconvection, (5) Liquid nitrogen (LN2) is heated only by heat transfer to the shroud, and (6) Foil isintimately bonded to baseplate.PROPERTIES: Heat of vaporization of liquid nitrogen (given): 125 kJ/kg.ANALYSIS: (a) From an energy balance on the baseplate,E in - E out = 0q elec - q rad = 0and using Eq. 1.7 for radiative exchange between the baseplate and shroud,()4qelec = p A p Tp - T 4 .sh()Substituting numerical values, with A p = D 2 / 4 , findp()2qelec = 0.25 ( 0.3 m ) / 4 5.67 108 W/m2 K 4 3004 - 774 K 4 = 8.1 W.<(b) From an energy balance on the enclosure, radiative transfer heats the liquid nitrogen streamcausing evaporation,E in - E out = 0q rad - m LN2 h fg = 0where m LN2 is the liquid nitrogen consumption rate. Hence,m LN2 = q rad / h fg = 8.1 W / 125 kJ / kg = 6.48 10-5 kg / s = 0.23 kg / h.<(c) If aluminum foil (p = 0.09) were bonded to the upper surface of the baseplate,()q rad,foil = q rad f / p = 8.1 W ( 0.09/0.25 ) = 2.9 Wand the liquid nitrogen consumption rate would be reduced by(0.25 - 0.09)/0.25 = 64% to 0.083 kg/h.<PROBLEM 1.33KNOWN: Width, input power and efficiency of a transmission. Temperature and convectioncoefficient for air flow over the casing. Emissivity of casing and temperature of surroundings.FIND: Surface temperature of casing.SCHEMATIC:ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3)Radiation exchange with large surroundings.ANALYSIS: Heat transfer from the case must balance heat dissipation in the transmission, whichmay be expressed as q = Pi Po = Pi (1 - ) = 150 hp 746 W/hp 0.07 = 7833 W. Heat transferfrom the case is by convection and radiation, in which case()44q = As h ( Ts T ) + Ts Tsur 2where As = 6 W . Hence,()27833 W = 6 ( 0.30 m ) 200 W / m 2 K ( Ts 303K ) + 0.8 5.67 108 W / m 2 K 4 Ts4 3034 K 4 A trial-and-error solution yieldsTs 373K = 100C<COMMENTS: (1) For Ts 373 K, qconv 7,560 W and qrad 270 W, in which case heat transfer isdominated by convection, (2) If radiation is neglected, the corresponding surface temperature is Ts =102.5C.PROBLEM 1.34KNOWN: Resistor connected to a battery operating at a prescribed temperature in air.FIND: (a) Considering the resistor as the system, determine corresponding values for Ein ( W ) ,E g ( W ) , Eout ( W ) and Est ( W ) . If a control surface is placed about the entire system, determine, and E . (b) Determine the volumetric heat generation rate withinthe values for E , E , Eginoutst3the resistor, q (W/m ), (c) Neglecting radiation from the resistor, determine the convectioncoefficient.SCHEMATIC:ASSUMPTIONS: (1) Electrical power is dissipated uniformly within the resistor, (2) Temperatureof the resistor is uniform, (3) Negligible electrical power dissipated in the lead wires, (4) Negligibleradiation exchange between the resistor and the surroundings, (5) No heat transfer occurs from thebattery, (5) Steady-state conditions.ANALYSIS: (a) Referring to Section 1.3.1, the conservation of energy requirement for a controlvolume at an instant of time, Eq 1.11a, isEin + Eg Eout = Estwhere Ein , E out correspond to surface inflow and outflow processes, respectively. The energygeneration term E g is associated with conversion of some other energy form (chemical, electrical,electromagnetic or nuclear) to thermal energy. The energy storage term Est is associated withchanges in the internal, kinetic and/or potential energies of the matter in the control volume. E g ,Est are volumetric phenomena. The electrical power delivered by the battery is P = VI = 24V6A =144 W.Control volume: Resistor.E in = 0<Eout = 144 WEg = 144 WEst = 0The E g term is due to conversion of electrical energy to thermal energy. The term E out is due toconvection from the resistor surface to the air.Continued...PROBLEM 1.34 (Cont.)Control volume: Battery-Resistor System.E in = 0E =0gEout = 144 WE = 144 W<stThe Est term represents the decrease in the chemical energy within the battery. The conversion ofchemical energy to electrical energy and its subsequent conversion to thermal energy are processesinternal to the system which are not associated with Est or E g . The E out term is due to convectionfrom the resistor surface to the air.(b) From the energy balance on the resistor with volume, = (D2/4)L,E g = q()144 W = q (0.06 m ) / 4 0.25 m2q = 2.04 105 W m3<(c) From the energy balance on the resistor and Newton's law of cooling with As = DL + 2(D2/4),Eout = qcv = hAs ( Ts T ))(144 W = h 0.06 m 0.25 m + 2 0.062 m 2 4 (95 25 ) C144 W = h [0.0471 + 0.0057 ] m 2 (95 25 ) Ch = 39.0 W m 2KCOMMENTS: (1) In using the conservation of energy requirement, Eq. 1.11a, it is important torecognize that Ein and E out will always represent surface processes and E g and Est , volumetricprocesses. The generation term E g is associated with a conversion process from some form ofenergy to thermal energy. The storage term Est represents the rate of change of internal energy.(2) From Table 1.1 and the magnitude of the convection coefficient determined from part (c), weconclude that the resistor is experiencing forced, rather than free, convection.<PROBLEM 1.35KNOWN: Thickness and initial temperature of an aluminum plate whose thermal environment ischanged.FIND: (a) Initial rate of temperature change, (b) Steady-state temperature of plate, (c) Effect ofemissivity and absorptivity on steady-state temperature.SCHEMATIC:ASSUMPTIONS: (1) Negligible end effects, (2) Uniform plate temperature at any instant, (3)Constant properties, (4) Adiabatic bottom surface, (5) Negligible radiation from surroundings, (6) Nointernal heat generation.ANALYSIS: (a) Applying an energy balance, Eq. 1.11a, at an instant of time to a control volumeabout the plate, E in E out = E st , it follows for a unit surface area.()()()()22SGS 1m 2 E 1m 2 qconv 1m = ( d dt )( McT ) = 1m L c ( dT dt ) .Rearranging and substituting from Eqs. 1.3 and 1.5, we obtaindT dt = (1 Lc ) SGS Ti4 h ( Ti T ) .(dT dt = 2700 kg m3 0.004 m 900 J kg K)10.8 900 W m 2 0.25 5.67 108 W m 2 K 4 ( 298 K )4 20 W m 2 K ( 25 20 ) C dT dt = 0.052 C s .(b) Under steady-state conditions, E st = 0, and the energy balance reduces toSGS = T 4 + h ( T T )<(2)0.8 900 W m 2 = 0.25 5.67 108 W m 2 K 4 T 4 + 20 W m 2 K ( T 293 K )The solution yields T = 321.4 K = 48.4C.(c) Using the IHT First Law Model for an Isothermal Plane Wall, parametric calculations yield thefollowing results.Plate temperature, T (C)70605040302000.20.40.60.81Coating emissivity, epsSolar absorptivity, alphaS = 1alphaS = 0.8alphaS = 0.5COMMENTS: The surface radiative properties have a significant effect on the plate temperature,which decreases with increasing and decreasing S. If a low temperature is desired, the platecoating should be characterized by a large value of /S. The temperature also decreases withincreasing h.<PROBLEM 1.36KNOWN: Surface area of electronic package and power dissipation by the electronics.Surface emissivity and absorptivity to solar radiation. Solar flux.FIND: Surface temperature without and with incident solar radiation.SCHEMATIC:ASSUMPTIONS: Steady-state conditions.ANALYSIS: Applying conservation of energy to a control surface about the compartment, atany instantE in - E out + E g = 0.It follows that, with the solar input,SAsqS As E + P=04SAsqS As Ts + P=01/ 4 A q + P Ts = S s SAs.In the shade ( qS = 0 ) ,1/ 41000 WTs = 1 m 2 1 5.67 108 W/m 2 K 4 = 364 K.<In the sun,1/ 4 0.25 1 m 2 750 W/m 2 + 1000 W Ts = 1 m 2 1 5.67 108 W/m 2 K 4 = 380 K.<COMMENTS: In orbit, the space station would be continuously cycling between shade andsunshine, and a steady-state condition would not exist.PROBLEM 1.37KNOWN: Daily hot water consumption for a family of four and temperatures associated with groundwater and water storage tank. Unit cost of electric power. Heat pump COP.FIND: Annual heating requirement and costs associated with using electric resistance heating or aheat pump.SCHEMATIC:ASSUMPTIONS: (1) Process may be modelled as one involving heat addition in a closed system,(2) Properties of water are constant.PROPERTIES: Table A-6, Water ( Tave = 308 K): = vf 1 = 993 kg/m3, cp,f = 4.178 kJ/kgK.ANALYSIS: From Eq. 1.11c, the daily heating requirement is Qdaily = U t = McT= Vc (Tf Ti ) . With V = 100 gal/264.17 gal/m3 = 0.379 m3,()()Qdaily = 993kg / m3 0.379 m3 4.178kJ/kg K 40 C = 62,900 kJThe annual heating requirement is then, Qannual = 365days ( 62,900 kJ/day ) = 2.30 107 kJ , or,with 1 kWh = 1 kJ/s (3600 s) = 3600 kJ,Qannual = 6380 kWh<With electric resistance heating, Qannual = Qelec and the associated cost, C, isC = 6380 kWh ($0.08/kWh ) = $510<If a heat pump is used, Qannual = COP ( Welec ). Hence,Welec = Qannual /( COP ) = 6380kWh/(3) = 2130 kWhThe corresponding cost isC = 2130 kWh ($0.08/kWh ) = $170<COMMENTS: Although annual operating costs are significantly lower for a heat pump,corresponding capital costs are much higher. The feasibility of this approach depends on other factorssuch as geography and seasonal variations in COP, as well as the time value of money.PROBLEM 1.38KNOWN: Initial temperature of water and tank volume. Power dissipation, emissivity,length and diameter of submerged heaters. Expressions for convection coefficient associatedwith natural convection in water and air.FIND: (a) Time to raise temperature of water to prescribed value, (b) Heater temperatureshortly after activation and at conclusion of process, (c) Heater temperature if activated in air.SCHEMATIC:ASSUMPTIONS: (1) Negligible heat loss from tank to surroundings, (2) Water is wellmixed (at a uniform, but time varying temperature) during heating, (3) Negligible changes inthermal energy storage for heaters, (4) Constant properties, (5) Surroundings afforded by tankwall are large relative to heaters.ANALYSIS: (a) Application of conservation of energy to a closed system (the water) at aninstant, Eq. (1.11d), yieldsdUdTdT= Mc= c= q = 3q1dtdtdtTftdt = ( c/3q1 ) dT0TiHence,t=()990 kg/m3 10gal 3.79 103m3 / gal 4180J/kg K3 500 W(335 295) K = 4180 s <(b) From Eq. (1.3a), the heat rate by convection from each heater is4/3q1 = Aq1 = Ah ( Ts T ) = ( DL ) 370 ( Ts T )Hence,3/ 43/ 4500 W q1 Ts = T + = T+= ( T + 24 ) K 370 DL 370 W/m 2 K 4/3 0.025 m 0.250 m With water temperatures of Ti 295 K and Tf = 335 K shortly after the start of heating and atthe end of heating, respectively,Ts,i = 319 K<Ts,f = 359 KContinued ..PROBLEM 1.38 (Continued)(c) From Eq. (1.10), the heat rate in air is()44q1 = DL 0.70 ( Ts T )4 / 3 + Ts Tsur Substituting the prescribed values of q1, D, L, T = Tsur and , an iterative solution yieldsTs = 830 K<COMMENTS: In part (c) it is presumed that the heater can be operated at Ts = 830 Kwithout experiencing burnout. The much larger value of Ts for air is due to the smallerconvection coefficient. However, with qconv and qrad equal to 59 W and 441 W, respectively,a significant portion of the heat dissipation is effected by radiation.PROBLEM 1.39KNOWN: Power consumption, diameter, and inlet and discharge temperatures of a hairdryer.FIND: (a) Volumetric flow rate and discharge velocity of heated air, (b) Heat loss from case.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential andkinetic energy changes of air flow, (4) Negligible work done by fan, (5) Negligible heattransfer from casing of dryer to ambient air (Part (a)), (6) Radiation exchange between a smallsurface and a large enclosure (Part (b)).ANALYSIS: (a) For a control surface about the air flow passage through the dryer,conservation of energy for an open system reduces tom ( u + pv ) m ( u + pv ) + q = 0iowhere u + pv = i and q = Pelec. Hence, with m (ii io ) = mcp (Ti To ) ,mcp (To Ti ) = Pelecm=Pelec500 W== 0.0199 kg/scp ( To Ti ) 1007 J/kg K 25$C() m 0.0199 kg/s = 0.0181 m3 / s= = 1.10 kg/m3Vo =<44 0.0181 m3 / s=== 4.7 m/s2Ac D2 (0.07 m )<(b) Heat transfer from the casing is by convection and radiation, and from Eq. (1.10)(44q = hAs ( Ts T ) + As Ts Tsur)Continued ..PROBLEM 1.39 (Continued)where As = DL = (0.07 m 0.15 m ) = 0.033 m 2 . Hence,()( )()q = 4W/m2 K 0.033 m 2 20$ C + 0.8 0.033 m2 5.67 108 W/m2 K 4 3134 2934 K 4q = 2.64 W + 3.33 W = 5.97 W<The heat loss is much less than the electrical power, and the assumption of negligible heat lossis justified.COMMENTS: Although the mass flow rate is invariant, the volumetric flow rate increasesas the air is heated in its passage through the dryer, causing a reduction in the density.However, for the prescribed temperature rise, the change in , and hence the effect on , issmall.PROBLEM 1.40KNOWN: Speed, width, thickness and initial and final temperatures of 304 stainless steel in anannealing process. Dimensions of annealing oven and temperature, emissivity and convectioncoefficient of surfaces exposed to ambient air and large surroundings of equivalent temperatures.Thickness of pad on which oven rests and pad surface temperatures.FIND: Oven operating power.SCHEMATIC:ASSUMPTIONS: (1) steady-state, (2) Constant properties, (3) Negligible changes in kinetic andpotential energy.()PROPERTIES: Table A.1, St.St.304 T = (Ti + To )/2 = 775 K : = 7900 kg/m3, c p = 578J/kgK; Table A.3, Concrete, T = 300 K: k c = 1.4 W/mK.ANALYSIS: The rate of energy addition to the oven must balance the rate of energy transfer to thesteel sheet and the rate of heat loss from the oven. With Ein Eout= 0, it follows thatPelec + m ( u i u o ) q = 0where heat is transferred from the oven. With m = Vs ( Ws t s ) , ( u i u o ) = cp ( Ti To ) , and()44q = ( 2Ho Lo + 2Ho Wo + Wo Lo ) h ( Ts T ) + s Ts Tsur + k c ( Wo Lo )(Ts Tb )/t c ,it follows thatPelec = Vs ( Ws t s ) cp ( To Ti ) + ( 2Ho Lo + 2H o Wo + Wo Lo ) )( h ( T T ) + T 4 T 4 + k ( W L )(T T )/tsosssur bc c oo sPelec = 7900 kg/m3 0.01m/s ( 2 m 0.008 m ) 578J/kg K (1250 300 ) K+ ( 2 2m 25m + 2 2m 2.4m + 2.4m 25m )[10W/m 2 K (350 300 ) K()+0.8 5.67 108 W/m 2 K 4 3504 3004 K 4 ] + 1.4W/m K ( 2.4m 25m )(350 300 ) K/0.5mContinued..PROBLEM 1.40 (Cont.)Pelec = 694, 000W + 169.6m 2 (500 + 313 ) W/m2 + 8400W= ( 694, 000 + 84,800 + 53,100 + 8400 ) W = 840kWCOMMENTS: Of the total energy input, 83% is transferred to the steel while approximately 10%,6% and 1% are lost by convection, radiation and conduction from the oven. The convection andradiation losses can both be reduced by adding insulation to the side and top surfaces, which wouldreduce the corresponding value of Ts .<PROBLEM 1.41KNOWN: Hot plate-type wafer thermal processing tool based upon heat transfer modes byconduction through gas within the gap and by radiation exchange across gap.FIND: (a) Radiative and conduction heat fluxes across gap for specified hot plate and wafertemperatures and gap separation; initial time rate of change in wafer temperature for each mode, and(b) heat fluxes and initial temperature-time change for gap separations of 0.2, 0.5 and 1.0 mm for hotplate temperatures 300 < Th < 1300C. Comment on the relative importance of the modes and theinfluence of the gap distance. Under what conditions could a wafer be heated to 900C in less than 10seconds?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions for flux calculations, (2) Diameter of hot plate andwafer much larger than gap spacing, approximating plane, infinite planes, (3) One-dimensionalconduction through gas, (4) Hot plate and wafer are blackbodies, (5) Negligible heat losses from waferbackside, and (6) Wafer temperature is uniform at the onset of heating.3PROPERTIES: Wafer: = 2700 kg/m , c = 875 J/kgK; Gas in gap: k = 0.0436 W/mK.ANALYSIS: (a) The radiative heat flux between the hot plate and wafer for Th = 600C and Tw =20 C follows from the rate equation,()44q = Th Tw = 5.67 108 W / m 2 K 4rad((600 + 273)4 ( 20 + 273)4)K4= 32.5 kW / m2<The conduction heat flux through the gas in the gap with L = 0.2 mm follows from Fouriers law,(600 20) K = 126 kW / m2Th Twq= 0.0436 W / m Kcond = kL0.0002 m<The initial time rate of change of the wafer can be determined from an energy balance on the wafer atthe instant of time the heating process begins, Ein E = Eoutst dT E = c d w st dt iwhere E ut = 0 and E n = q or q . Substituting numerical values, findoiradconddTwdtq32.5 103 W / m 2= rad == 17.6 K / si,rad cd 2700 kg / m3 875 J / kg K 0.00078 m<dTwdtq= cond = 68.4 K / s cdi,cond<Continued ..PROBLEM 1.41 (Cont.)(b) Using the foregoing equations, the heat fluxes and initial rate of temperature change for each modecan be calculated for selected gap separations L and range of hot plate temperatures Th with Tw =20C.200Initial rate of change, dTw/dt (K.s^-1)400Heat flux (kW/m^2)30020010015010050003 00500700900110013003005007009001100Hot plate temperature, Th (C)Hot plate temperature, Th (C)q''radq''cond, L = 1.0 mmq''cond, L = 0.5 mmq''cond, L = 0.2 mmq''radq''cond, L = 1.0 m mq''cond, L = 0.5 m mq''cond, L = 0.2 m mIn the left-hand graph, the conduction heat flux increases linearly with Th and inversely with L asexpected. The radiative heat flux is independent of L and highly non-linear with Th, but does notapproach that for the highest conduction heat rate until Th approaches 1200C.The general trends for the initial temperature-time change, (dTw/dt)i, follow those for the heat fluxes.To reach 900C in 10 s requires an average temperature-time change rate of 90 K/s. Recognizing that(dTw/dt) will decrease with increasing Tw, this rate could be met only with a very high Th and thesmallest L.1300PROBLEM 1.42KNOWN: Silicon wafer, radiantly heated by lamps, experiencing an annealing process with knownbackside temperature.FIND: Whether temperature difference across the wafer thickness is less than 2C in order to avoiddamaging the wafer.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wafer, (3)Radiation exchange between upper surface of wafer and surroundings is between a small object and alarge enclosure, and (4) Vacuum condition in chamber, no convection.PROPERTIES: Wafer: k = 30 W/mK, = = 0.65.ANALYSIS: Perform a surface energy balance on the upper surface of the wafer to determineTw ,u . The processes include the absorbed radiant flux from the lamps, radiation exchange with thechamber walls, and conduction through the wafer.En E = 0iout q q q = 0sradcd()44 q Tw,u Tsur ksTw,u Tw,L=0(40.65 3.0 105 W / m 2 0.65 5.67 108 W / m 2 K 4 Tw,u ( 27 + 273)4)K430W / m K Tw,u (997 + 273) K / 0.00078 m = 0Tw ,u = 1273K = 1000C<COMMENTS: (1) The temperature difference for this steady-state operating condition,Tw ,u Tw,l , is larger than 2C. Warping of the wafer and inducing slip planes in the crystal structurecould occur.(2) The radiation exchange rate equation requires that temperature must be expressed in kelvin units.Why is it permissible to use kelvin or Celsius temperature units in the conduction rate equation?(3) Note how the surface energy balance, Eq. 1.12, is represented schematically. It is essential toshow the control surfaces, and then identify the rate processes associated with the surfaces. Makesure the directions (in or out) of the process are consistent with the energy balance equation.PROBLEM 1.43KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces exposed to hot and coolzones, respectively.FIND: (a) Initial rate of change of the wafer temperature corresponding to the wafer temperatureTw ,i = 300 K, and (b) Steady-state temperature reached if the wafer remains in this position. Howsignificant is convection for this situation? Sketch how youd expect the wafer temperature to vary asa function of vertical distance.SCHEMATIC:ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Transient conditions when wafer is initiallypositioned, (3) Hot and cool zones have uniform temperatures, (3) Radiation exchange is betweensmall surface (wafer) and large enclosure (chamber, hot or cold zone), and (4) Negligible heat lossesfrom wafer to mounting pin holder.ANALYSIS: The energy balance on the wafer illustrated in the schematic above includes convectionfrom the upper (u) and lower (l) surfaces with the ambient gas, radiation exchange with the hot- andcool-zone (chamber) surroundings, and the rate of energy storage term for the transient condition.En E = Eioutstqrad,h + qrad,c qcv,u q = cdcv,l()(d Twdt)4444 Tsur,h Tw + Tsur,c Tw h u (Tw T ) h l (Tw T ) = cdd Twdt(a) For the initial condition, the time rate of temperature change of the wafer is determined using theenergy balance above with Tw = Tw,i = 300 K,()()0.65 5.67 108 W / m 2 K 4 15004 3004 K 4 + 0.65 5.67 108 W / m 2 K 4 3304 3004 K 48 W / m 2 K (300 700 ) K 4 W / m 2 K (300 700 ) K =2700 kg / m3 875 J / kg K 0.00078 m ( d Tw / dt )i(d Tw / dt )i = 104 K / s<(b) For the steady-state condition, the energy storage term is zero, and the energy balance can besolved for the steady-state wafer temperature, Tw = Tw,ss .Continued ..PROBLEM 1.43 (Cont.))()(440.65 15004 Tw,ss K 4 + 0.65 3304 Tw,ss K 4()()8 W / m 2 K Tw,ss 700 K 4 W / m 2 K Tw,ss 700 K = 0Tw ,ss = 1251 K<To determine the relative importance of the convection processes, re-solve the energy balance aboveignoring those processes to find ( d Tw / dt )i = 101 K / s and Tw,ss = 1262 K. We conclude that theradiation exchange processes control the initial time rate of temperature change and the steady-statetemperature.If the wafer were elevated above the present operating position, its temperature would increase, sincethe lower surface would begin to experience radiant exchange with progressively more of the hot zonechamber. Conversely, by lowering the wafer, the upper surface would experience less radiantexchange with the hot zone chamber, and its temperature would decrease. The temperature-distancetrend might appear as shown in the sketch.PROBLEM 1.44KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactivewastes. Surface convection conditions.FIND: Total energy generation rate and surface temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature drop across thincontainer wall.ANALYSIS: The rate of energy generation isr2Eg = qdV=qo o 1- ( r/ro ) 2 rLdr0 = 2 Lq r 2 / 2 r 2 / 4o oEgo()or per unit length,q rE = o o .g22<Performing an energy balance for a control surface about the container yields, at an instant,E E = 0goutand substituting for the convection heat rate per unit length,2 qo ro= h ( 2 ro )( Ts T )2Ts = T +q o ro.4h<COMMENTS: The temperature within the radioactive wastes increases with decreasing rfrom Ts at ro to a maximum value at the centerline.PROBLEM 1.45KNOWN: Rod of prescribed diameter experiencing electrical dissipation from passage of electricalcurrent and convection under different air velocity conditions. See Example 1.3.FIND: Rod temperature as a function of the electrical current for 0 I 10 A with convection2coefficients of 50, 100 and 250 W/m K. Will variations in the surface emissivity have a significanteffect on the rod temperature?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform rod temperature, (3) Radiation exchangebetween the outer surface of the rod and the surroundings is between a small surface and largeenclosure.ANALYSIS: The energy balance on the rod for steady-state conditions has the form,qconv + q = Eradgen)(4 Dh (T T ) + D T 4 Tsur = I 2R eUsing this equation in the Workspace of IHT, the rod temperature is calculated and plotted as afunction of current for selected convection coefficients.150R o d te m p e ra tu re , T (C )12510075502500246810C u rre n t, I (a m p e re s )h = 5 0 W /m ^2 .Kh = 1 0 0 W /m ^2 .Kh = 2 5 0 W /m ^2 .KCOMMENTS: (1) For forced convection over the cylinder, the convection heat transfer coefficient is0.6dependent upon air velocity approximately as h ~ V . Hence, to achieve a 5-fold change in the2convection coefficient (from 50 to 250 W/m K), the air velocity must be changed by a factor ofnearly 15.Continued ..PROBLEM 1.45 (Cont.)2(2) For the condition of I = 4 A with h = 50 W/m K with T = 63.5C, the convection and radiationexchange rates per unit length are, respectively, q v = 5.7 W / m and q = 0.67 W / m. We concludecradthat convection is the dominate heat transfer mode and that changes in surface emissivity could have2only a minor effect. Will this also be the case if h = 100 or 250 W/m K?(3) What would happen to the rod temperature if there was a loss of coolant condition where the airflow would cease?(4) The Workspace for the IHT program to calculate the heat losses and perform the parametricanalysis to generate the graph is shown below. It is good practice to provide commentary with thecode making your solution logic clear, and to summarize the results. It is also good practice to showplots in customary units, that is, the units used to prescribe the problem. As such the graph of the rodtemperature is shown above with Celsius units, even though the calculations require temperatures inkelvins.// Energy balance; from Ex. 1.3, Comment 1-q'cv - q'rad + Edot'g = 0q'cv = pi*D*h*(T - Tinf)q'rad = pi*D*eps*sigma*(T^4 - Tsur^4)sigma = 5.67e-8// The generation term has the formEdot'g = I^2*R'eqdot = I^2*R'e / (pi*D^2/4)// Input parametersD = 0.001Tsur = 300T_C = T 273eps = 0.8Tinf = 300h = 100//h = 50//h = 250I = 5.2//I = 4R'e = 0.4// Representing temperature in Celsius units using _C subscript// Values of coefficient for parameter study// For graph, sweep over range from 0 to 10 A// For evaluation of heat rates with h = 50 W/m^2.K/* Base case results: I = 5.2 A with h = 100 W/m^2.K, find T = 60 C (Comment 2 case).Edot'gTT_Cq'cvq'radqdotDIR'eTinfTsurepshsigma10.82332.6 59.55 10.23 0.5886 1.377E70.001 5.20.43003000.81005.67E-8 *//* Results: I = 4 A with h = 50 W/m^2.K, find q'cv = 5.7 W/m and q'rad = 0.67 W/mEdot'gTT_Cq'cvq'radqdotDIR'eTinfTsurepshsigma6.4336.5 63.47 5.728 0.6721 8.149E6 0.001 40.43003000.8505.67E-8*/PROBLEM 1.46KNOWN: Long bus bar of prescribed diameter and ambient air and surroundings temperatures.Relations for the electrical resistivity and free convection coefficient as a function of temperature.FIND: (a) Current carrying capacity of the bus bar if its surface temperature is not to exceed 65C;compare relative importance of convection and radiation exchange heat rates, and (b) Showgraphically the operating temperature of the bus bar as a function of current for the range 100 I 5000 A for bus-bar diameters of 10, 20 and 40 mm. Plot the ratio of the heat transfer by convection tothe total heat transfer for these conditions.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Bus bar and conduit are very long in directionnormal to page, (3) Uniform bus-bar temperature, (4) Radiation exchange between the outer surface ofthe bus bar and the conduit is between a small surface and a large enclosure.PROPERTIES: Bus-bar material, e = e,o [1 + ( T To )] , e,o = 0.0171 m, To = 25C, = 0.00396 K 1.ANALYSIS: An energy balance on the bus-bar for a unit length as shown in the schematic above hasthe formE E + E = 0inoutgen2q q radconv + I R = 0e)(4 D T 4 Tsur h D ( T T ) + I2 e / Ac = 0where R = e / A c and A c = D 2 / 4. Using the relations for e ( T ) and h ( T, D ) , and substitutingenumerical values with T = 65C, find(q = 0.85 ( 0.020m ) 5.67 108 W / m 2 K 4 [65 + 273] [30 + 273]rad44)K4= 223 W / m2qconv = 7.83W / m K ( 0.020m )( 65 30 ) K = 17.2 W / m<<0.25where h = 1.21W m 1.75 K 1.25 ( 0.020m )(65 30 )0.25 = 7.83 W / m2 K()I2 R = I2 198.2 106 m / (0.020 ) m 2 / 4 = 6.31 105 I2 W / mewhere2e = 0.0171 106 m 1 + 0.00396 K 1 (65 25) K = 198.2 mThe maximum allowable current capacity and the ratio of the convection to total heat transfer rate areI = 1950 Aq / ( q + q ) = q / qtot = 0.072cvcvradcv<For this operating condition, convection heat transfer is only 7.2% of the total heat transfer.(b) Using these equations in the Workspace of IHT, the bus-bar operating temperature is calculatedand plotted as a function of the current for the range 100 I 5000 A for diameters of 10, 20 and 40mm. Also shown below is the corresponding graph of the ratio (expressed in percentage units) of theheat transfer by convection to the total heat transfer, q v / q tot .cContinued ..PROBLEM 1.46 (Cont.)1311Ratio q'cv / q'tot, (%)Ba r te m p e ra tu re , Ts (C )10080604097532001000200030004000150002040C u rre n t, I (A)6080Bus bar temperature, T (C)D = 10 m mD = 20 m mD = 40 m mD = 10 mmD = 20 mmD = 40 mmCOMMENTS: (1) The trade-off between current-carrying capacity, operating temperature and bardiameter is shown in the first graph. If the surface temperature is not to exceed 65C, the maximumcurrent capacities for the 10, 20 and 40-mm diameter bus bars are 960, 1950, and 4000 A,respectively.(2) From the second graph with q v / q tot vs. T, note that the convection heat transfer rate is always acsmall fraction of the total heat transfer. That is, radiation is the dominant mode of heat transfer. Notealso that the convection contribution increases with increasing diameter.(3) The Workspace for the IHT program to perform the parametric analysis and generate the graphs isshown below. It is good practice to provide commentary with the code making your solution logicclear, and to summarize the results./* Results: base-case conditions, Part (a)IR'ecvovertot hbarq'cvTsur_C eps19506.309E-5 7.1717.82617.21300.85 */q'radrhoeD222.81.982E-8 0.02Tinf_CTs_C3065// Energy balance, on a per unit length basis; steady-state conditions// Edot'in - Edot'out + Edot'gen = 0-q'cv - q'rad + Edot'gen = 0q'cv = hbar * P * (Ts - Tinf)P = pi * Dq'rad = eps * sigma * (Ts^4 - Tsur^4)sigma = 5.67e-8Edot'gen = I^2 * R'eR'e = rhoe / Acrhoe = rhoeo * (1 + alpha * (Ts - To) )To = 25 + 273Ac = pi * D^2 / 4// Convection coefficienthbar = 1.21 * (D^-0.25) * (Ts - Tinf)^0.25// Convection vs. total heat ratescvovertot = q'cv / (q'cv + q'rad) * 100// Input parametersD = 0.020// D = 0.010// D = 0.040// I = 1950rhoeo = 0.01711e-6alpha = 0.00396Tinf_C = 30Tinf = Tinf_C + 273Ts_C = 65Ts = Ts_C + 273Tsur_C = 30Tsur = Tsur_C + 273eps = 0.85// Compact convection coeff. correlation// Values of diameter for parameter study// Base case condition unknown// Base case condition to determine current100PROBLEM 1.47KNOWN: Elapsed times corresponding to a temperature change from 15 to 14C for a referencesphere and test sphere of unknown composition suddenly immersed in a stirred water-ice mixture.Mass and specific heat of reference sphere.FIND: Specific heat of the test sphere of known mass.SCHEMATIC:ASSUMPTIONS: (1) Spheres are of equal diameter, (2) Spheres experience temperature changefrom 15 to 14C, (3) Spheres experience same convection heat transfer rate when the time rates ofsurface temperature are observed, (4) At any time, the temperatures of the spheres are uniform,(5) Negligible heat loss through the thermocouple wires.PROPERTIES: Reference-grade sphere material: cr = 447 J/kg K.ANALYSIS: Apply the conservation of energy requirement at an instant of time, Eq. 1.11a, aftera sphere has been immersed in the ice-water mixture at T.Ein E out = Estq conv = McdTdtwhere q conv = h A s ( T T ). Since the temperatures of the spheres are uniform, the change inenergy storage term can be represented with the time rate of temperature change, dT/dt. Theconvection heat rates are equal at this instant of time, and hence the change in energy storageterms for the reference (r) and test (t) spheres must be equal.M r crdT dT = M t ctdt rdt tApproximating the instantaneous differential change, dT/dt, by the difference change over a shortperiod of time, T/t, the specific heat of the test sphere can be calculated.0.515 kg 447 J / kg Kc t = 132 J / kg K(15 14 ) K6.35s= 1.263kg c t (15 14 ) K4.59s<COMMENTS: Why was it important to perform the experiments with the reference and testspheres over the same temperature range (from 15 to 14C)? Why does the analysis require thatthe spheres have uniform temperatures at all times?PROBLEM 1.48KNOWN: Inner surface heating and new environmental conditions associated with a spherical shell ofprescribed dimensions and material.FIND: (a) Governing equation for variation of wall temperature with time. Initial rate of temperaturechange, (b) Steady-state wall temperature, (c) Effect of convection coefficient on canister temperature.SCHEMATIC:ASSUMPTIONS: (1) Negligible temperature gradients in wall, (2) Constant properties, (3) Uniform,time-independent heat flux at inner surface.PROPERTIES: Table A.1, Stainless Steel, AISI 302: = 8055 kg/m3, c p = 510 J/kgK.ANALYSIS: (a) Performing an energy balance on the shell at an instant of time, Ein E out = Est .Identifying relevant processes and solving for dT/dt,4dT23q 4 ri2 h 4 ro ( T T ) = ro ri3 cpi3dt()()()dT3q r 2 hr 2 ( T T ) .=o3 r3 i idt c rpo i()Substituting numerical values for the initial condition, finddT dt iWW3 105(0.5m )2 500 2 (0.6m )2 (500 300 ) K m2m K=kgJ8055510(0.6 )3 (0.5 )3 m33kg K mdT = 0.089 K/s .dt i<(b) Under steady-state conditions with Est = 0, it follows that2q 4 ri2 = h 4 ro ( T T )i()()Continued ..PROBLEM 1.48 (Cont.)22q ri 105 W/m2 0.5m iT = T + = 300K + = 439Kh ro 500W/m 2 K 0.6m <(c) Parametric calculations were performed using the IHT First Law Model for an Isothermal HollowSphere. As shown below, there is a sharp increase in temperature with decreasing values of h < 1000W/m2K. For T > 380 K, boiling will occur at the canister surface, and for T > 410 K a condition knownas film boiling (Chapter 10) will occur. The condition corresponds to a precipitous reduction in h andincrease in T.1000Temperature, T(K)9008007006005004003001004008002000600010000Convection coefficient, h(W/m^2.K)Although the canister remains well below the melting point of stainless steel for h = 100 W/m2K, boilingshould be avoided, in which case the convection coefficient should be maintained at h > 1000 W/m2K.COMMENTS: The governing equation of part (a) is a first order, nonhomogenous differential equationwith constant coefficients. Its solution is = (S/R ) 1 eRt + i eRt , where T T ,)(323S 3qi ri2 / cp ( ro ri3 ) , R = 3hro / cp ( ro ri3 ) . Note results for t and for S = 0.PROBLEM 1.49KNOWN: Boiling point and latent heat of liquid oxygen. Diameter and emissivity of container.Free convection coefficient and temperature of surrounding air and walls.FIND: Mass evaporation rate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature of container outer surface equalsboiling point of oxygen.ANALYSIS: (a) Applying an energy balance to a control surface about the container, it follows that,at any instant,Ein Eout = 0orqconv + q rad q evap = 0 .The evaporative heat loss is equal to the product of the mass rate of vapor production and the heat ofvaporization. Hence, h ( T T ) + T 4 T 4 A m(1)ssurs s evap h fg = 0(()) h ( T T ) + T 4 T 4 D2ssursmevap = h fg()10 W m 2 K ( 298 263) K + 0.2 5.67 108 W m 2 K 4 2984 2634 K 4 (0.5 m )2mevap = 214 kJ kgmevap =(350 + 35.2 ) W / m2(0.785 m2 ) = 1.41103 kg s .<214 kJ kg(b) Using the energy balance, Eq. (1), the mass rate of vapor production can be determined for therange of emissivity 0.2 to 0.94. The effect of increasing emissivity is to increase the heat rate into thecontainer and, hence, increase the vapor production rate.Evaporation rate, mdot*1000 (kg/s)1.91.81.71.61.51.40.20.40.60.81Surface emissivity, epsCOMMENTS: To reduce the loss of oxygen due to vapor production, insulation should be appliedto the outer surface of the container, in order to reduce qconv and qrad. Note from the calculations inpart (a), that heat transfer by convection is greater than by radiation exchange.PROBLEM 1.50KNOWN: Frost formation of 2-mm thickness on a freezer compartment. Surface exposed toconvection process with ambient air.FIND: Time required for the frost to melt, tm.SCHEMATIC:ASSUMPTIONS: (1) Frost is isothermal at the fusion temperature, Tf, (2) The water melt falls awayfrom the exposed surface, (3) Negligible radiation exchange at the exposed surface, and (4) Backsidesurface of frost formation is adiabatic.PROPERTIES: Frost, f = 770 kg / m3 , h sf = 334 kJ / kg.ANALYSIS: The time tm required to melt a 2-mm thick frost layer may be determined by applyingan energy balance, Eq 1.11b, over the differential time interval dt and to a differential control volumeextending inward from the surface of the layer dx. From the schematic above, the energy in is theconvection heat flux over the time period dt and the change in energy storage is the latent energychange within the control volume, Asdx.Ein E out = Estq onv As dt = dUatch As ( T Tf ) dt = f As h sf dxIntegrating both sides of the equation and defining appropriate limits, findt0h ( T Tf ) m dt = f hsfdx0xotm =tm =f h sf x oh ( T Tf )700 kg / m3 334 103 J / kg 0.002m2 W / m 2 K ( 20 0 ) K= 11, 690 s = 3.2 hour<COMMENTS: (1) The energy balance could be formulated intuitively by recognizing that the totalheat in by convection during the time interval t m ( q t m ) must be equal to the total latent energy forcvmelting the frost layer ( x o h sf ). This equality is directly comparable to the derived expressionabove for tm.(2) Explain why the energy storage term in the analysis has a negative sign, and the limits ofintegration are as shown. Hint: Recall from the formulation of Eq. 1.11b, that the storage termrepresents the change between the final and initial states.PROBLEM 1.51KNOWN: Vertical slab of Woods metal initially at its fusion temperature, Tf, joined to a substrate.()Exposed surface is irradiated with laser source, G W / m 2 .2FIND: Instantaneous rate of melting per unit area, m (kg/sm ), and the material removed in amperiod of 2 s, (a) Neglecting heat transfer from the irradiated surface by convection and radiationexchange, and (b) Allowing for convection and radiation exchange.SCHEMATIC:ASSUMPTIONS: (1) Woods metal slab is isothermal at the fusion temperature, Tf, and (2) The meltruns off the irradiated surface.ANALYSIS: (a) The instantaneous rate of melting per unit area may be determined by applying anenergy balance, Eq 1.11a, on the metal slab at an instant of time neglecting convection and radiationexchange from the irradiated surface.En E = EioutstG =ddM( Mh sf ) = hsfdtdtwhere dM / dt = m is the time rate of change of mass in the control volume. Substituting values,mmm0.4 5000 W / m 2 = 33, 000 J / kg mm = 60.6 103 kg / s m 2The material removed in a 2s period per unit area isM = m t = 121 g / m 22s m(b) The energy balance considering convection and radiation exchange with the surroundings yields<<m G q q = h sf mcvradq = h ( Tf T ) = 15 W / m 2 K ( 72 20 ) K = 780 W / m 2cv()(4q = Tf4 T = 0.4 5.67 108 W / m 2 K [72 + 273] [20 + 273]radmm = 32.3 103 kg / s m 24M 2s = 64 g / m 24)K4= 154 W / m 2<COMMENTS: (1) The effects of heat transfer by convection and radiation reduce the estimate forthe material removal rate by a factor of two. The heat transfer by convection is nearly 5 times largerthan by radiation exchange.(2) Suppose the work piece were horizontal, rather than vertical, and the melt puddled on the surfacerather than ran off. How would this affect the analysis?(3) Lasers are common heating sources for metals processing, including the present application ofmelting (heat transfer with phase change), as well as for heating work pieces during milling andturning (laser-assisted machining).PROBLEM 1.52KNOWN: Hot formed paper egg carton of prescribed mass, surface area and water contentexposed to infrared heater providing known radiant flux.FIND: Whether water content can be reduced from 75% to 65% by weight during the 18speriod carton is on conveyor.SCHEMATIC:ASSUMPTIONS: (1) All the radiant flux from the heater bank is absorbed by the carton, (2)Negligible heat loss from carton by convection and radiation, (3) Negligible mass loss occursfrom bottom side.PROPERTIES: Water (given): hfg = 2400 kJ/kg.ANALYSIS: Define a control surface about the carton, and write the conservation of energyrequirement for an interval of time, t,E in E o ut = E st = 0where Ein is due to the absorbed radiant flux, q , from thehheater and Eout is the energy leaving due to evaporation ofwater from the carton. Hence.q A s t = M h fg .hFor the prescribed radiant flux q ,hq A st 5000 W / m2 0.0625 m2 18sM = h== 0.00234 kg.h fg2400 kJ / kgThe chief engineers requirement was to remove 10% of the water content, orM req = M 0.10 = 0.220 kg 0.10 = 0.022 kgwhich is nearly an order of magnitude larger than the evaporative loss. Considering heatlosses by convection and radiation, the actual water removal from the carton will be less thanM. Hence, the purchase should not be recommended, since the desired water removalcannot be achieved.<PROBLEM 1.53KNOWN: Average heat sink temperature when total dissipation is 20 W with prescribed air andsurroundings temperature, sink surface area and emissivity.FIND: Sink temperature when dissipation is 30 W.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) All dissipated power in devices is transferredto the sink, (3) Sink is isothermal, (4) Surroundings and air temperature remain the same for bothpower levels, (5) Convection coefficient is the same for both power levels, (6) Heat sink is a smallsurface within a large enclosure, the surroundings.ANALYSIS: Define a control volume around the heat sink. Power dissipated within the devicesis transferred into the sink, while the sink loses heat to the ambient air and surroundings byconvection and radiation exchange, respectively.Ein Eout = 0(1)44Pe hAs ( Ts T ) As Ts Tsur = 0.)(Consider the situation when Pe = 20 W for which Ts = 42C; find the value of h.()44h= Pe / As Ts Tsur / ( Ts T )h= 20 W/0.045 m 2 0.8 5.67 108 W/m 2 K 4 3154 3004 K 4 / (315 300 ) K2h = 24.4 W / m K.)(For the situation when Pe = 30 W, using this value for h with Eq. (1), obtain30 W - 24.4 W/m2 K 0.045 m 2 (Ts 300 ) K()40.045 m 2 0.8 5.67 108 W/m 2 K 4 Ts 3004 K 4 = 0()430 = 1.098 (Ts 300 ) + 2.041 109 Ts 3004 .By trial-and-error, findTs 322 K = 49 C.<COMMENTS: (1) It is good practice to express all temperatures in kelvin units when using energybalances involving radiation exchange.(2) Note that we have assumed As is the same for the convection and radiation processes. Since not allportions of the fins are completely exposed to the surroundings, As,rad is less than As,conv = As.(3) Is the assumption that the heat sink is isothermal reasonable?PROBLEM 1.54KNOWN: Number and power dissipation of PCBs in a computer console. Convection coefficientassociated with heat transfer from individual components in a board. Inlet temperature of cooling airand fan power requirement. Maximum allowable temperature rise of air. Heat flux from componentmost susceptible to thermal failure.FIND: (a) Minimum allowable volumetric flow rate of air, (b) Preferred location and correspondingsurface temperature of most thermally sensitive component.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and kineticenergy changes of air flow, (4) Negligible heat transfer from console to ambient air, (5) Uniformconvection coefficient for all components.ANALYSIS: (a) For a control surface about the air space in the console, conservation of energy foran open system, Eq. (1.11e), reduces tom ( u + pv ) m ( u + pv ) + q W = 0iowhere u + pv = i, q = 5Pb , and W = Pf . Hence, with m (ii io ) = mcp (Ti To ) ,mcp (To Ti ) = 5 Pb + PfFor a maximum allowable temperature rise of 15C, the required mass flow rate ism=5 Pb + Pf5 20 W + 25 W== 8.28 103 kg/scp ( To Ti ) 1007 J/kg K 15 $C()The corresponding volumetric flow rate is=m 8.28 103 kg/s== 7.13 103 m3 / s31.161 kg/m<(b) The component which is most susceptible to thermal failure should be mounted at the bottom ofone of the PCBs, where the air is coolest. From the corresponding form of Newtons law of cooling,q = h ( Ts Ti ) , the surface temperature is42q$ C + 1 10 W/m = 70$ CTs = Ti + = 20h200 W/m2 K<COMMENTS: (1) Although the mass flow rate is invariant, the volumetric flow rate increases as theair is heated in its passage through the console, causing a reduction in the density. However, for theprescribed temperature rise, the change in , and hence the effect on , is small. (2) If the thermallysensitive component were located at the top of a PCB, it would be exposed to warmer air (To = 35C)and the surface temperature would be Ts = 85C.PROBLEM 1.55KNOWN: Top surface of car roof absorbs solar flux, qS,abs , and experiences for case (a): convectionwith air at T and for case (b): the same convection process and radiation emission from the roof.FIND: Temperature of the plate, Ts , for the two cases. Effect of airflow on roof temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer to auto interior, (3)Negligible radiation from atmosphere.ANALYSIS: (a) Apply an energy balance to the control surfaces shown on the schematic. For aninstant of time, Ein E out = 0. Neglecting radiation emission, the relevant processes are convectionbetween the plate and the air, qconv , and the absorbed solar flux, qS,abs . Considering the roof to havean area As ,qS,abs As hAs ( Ts T ) = 0Ts = T + qS,abs /hTs = 20 C +800W/m 212W/m 2 K= 20 C + 66.7 C = 86.7 C<(b) With radiation emission from the surface, the energy balance has the formqS,abs As q conv E As = 0qS,abs As hAs ( Ts T ) As Ts4 = 0 .Substituting numerical values, with temperature in absolute units (K),800Wm2 12Wm2 K(Ts 293K ) 0.8 5.67 108WT4 = 02 K4 sm412Ts + 4.536 108 Ts = 4316It follows that Ts = 320 K = 47C.<Continued..PROBLEM 1.55 (Cont.)(c) Parametric calculations were performed using the IHT First Law Model for an Isothermal Plane Wall.As shown below, the roof temperature depends strongly on the velocity of the auto relative to the ambientair. For a convection coefficient of h = 40 W/m2K, which would be typical for a velocity of 55 mph, theroof temperature would exceed the ambient temperature by less than 10C.360Temperature, Ts(K)350340330320310300290020406080100120140160180200Convection coefficient, h(W/m^2.K)COMMENTS: By considering radiation emission, Ts decreases, as expected. Note the manner in whichq is formulated using Newtons law of cooling; since q is shown leaving the control surface, theconvconvrate equation must be h ( Ts T ) and not h ( T Ts ) .PROBLEM 1.56KNOWN: Detector and heater attached to cold finger immersed in liquid nitrogen. Detector surface of = 0.9 is exposed to large vacuum enclosure maintained at 300 K.FIND: (a) Temperature of detector when no power is supplied to heater, (b) Heater power (W) requiredto maintain detector at 195 K, (c) Effect of finger thermal conductivity on heater power.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through cold finger, (3)Detector and heater are very thin and isothermal at Ts , (4) Detector surface is small compared toenclosure surface.PROPERTIES: Cold finger (given): k = 10 W/mK.ANALYSIS: Define a control volume about detector and heater and apply conservation of energyrequirement on a rate basis, Eq. 1.11a,Ein Eout = 0where(1)Ein = q rad + q elec ;E out = q cond(2,3)Combining Eqs. (2,3) with (1), and using the appropriate rate equations,)(44 As Tsur Ts + q elec = kAs (Ts TL )/L .(4)(a) Where q elec = 0, substituting numerical values()40.9 5.67 108 W/m 2 K 4 3004 Ts K 4 = 10W/m K (Ts 77 ) K/0.050 m()5.103 108 3004 Ts4 = 200 (Ts 77 )Ts = 79.1K<Continued..PROBLEM 1.56 (Cont.)(b) When Ts = 195 K, Eq. (4) yields)(0.9 [ ( 0.005 m ) / 4] 5.67 108 W/m 2 K 4 3004 1954 K 4 + q elec2= 10W/m K [ (0.005 m ) /4] (195 77 ) K / 0.050 m2<qelec = 0.457 W = 457 mW(c) Calculations were performed using the First Law Model for a Nonisothermal Plane Wall. With netradiative transfer to the detector fixed by the prescribed values of Ts and Tsur , Eq. (4) indicates thatq elec increases linearly with increasing k.Heater power, qelec(W)191715131197531-10100200300400Thermal conductivity, k(W/m.K)Heat transfer by conduction through the finger material increases with its thermal conductivity. Notethat, for k = 0.1 W/mK, q elec = -2 mW, where the minus sign implies the need for a heat sink, ratherthan a heat source, to maintain the detector at 195 K. In this case q rad exceeds q cond , and a heat sinkwould be needed to dispose of the difference. A conductivity of k = 0.114 W/mK yields a precisebalance between q rad and q cond . Hence to circumvent heaving to use a heat sink, while minimizingthe heater power requirement, k should exceed, but remain as close as possible to the value of 0.114W/mK. Using a graphite fiber composite, with the fibers oriented normal to the direction of conduction,Table A.2 indicates a value of k 0.54 W/mK at an average finger temperature of T = 136 K. For thisvalue, q elec = 18 mWCOMMENTS: The heater power requirement could be further reduced by decreasing .PROBLEM 1.57KNOWN: Conditions at opposite sides of a furnace wall of prescribed thickness, thermalconductivity and surface emissivity.FIND: Effect of wall thickness and outer convection coefficient on surface temperatures.Recommended values of L and h 2 .SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligibleradiation exchange at surface 1, (4) Surface 2 is exposed to large surroundings.ANALYSIS: The unknown temperatures may be obtained by simultaneously solving energy balanceequations for the two surface. At surface 1,q onv,1 = qccond()h1 T,1 T1 = k (T1 T2 )/L(1)At surface 2,qcond = qconv + qrad()(44k ( T1 T2 )/L = h 2 T2 T,2 + T2 Tsur)(2)Surface temperature, T(K)Using the IHT First Law Model for a Nonisothermal Plane Wall, we obtain170015001300110090070050030000.10.20.30.40.5Wall thickness, L(m)Inner surface temperature, T1(K)Outer surface temperature, T2(K)Continued ..PROBLEM 1.57 (Cont.)Both qcond and T2 decrease with increasing wall thickness, and for the prescribed value of h 2 = 102W/m K, a value of L 0.275 m is needed to maintain T2 373 K = 100 C. Note that inner surfacetemperature T1 , and hence the temperature difference, T1 T2 , increases with increasing L.Surface temperature, T(K)Performing the calculations for the prescribed range of h 2 , we obtain170015001300110090070050030001020304050Convection coefficient, h2(W/m^2.K)Inner surface temperature, T(K)Outer surface temperature, T(K)For the prescribed value of L = 0.15 m, a value of h 2 24 W/m2K is needed to maintain T2 373K. The variation has a negligible effect on T1 , causing it to decrease slightly with increasing h 2 , butdoes have a strong influence on T2 .COMMENTS: If one wishes to avoid use of active (forced convection) cooling on side 2, reliancewill have to be placed on free convection, for which h 2 5 W/m2K. The minimum wall thicknesswould then be L = 0.40 m.PROBLEM 1.58KNOWN: Furnace wall with inner surface temperature T1 = 352C and prescribed thermalconductivity experiencing convection and radiation exchange on outer surface. See Example 1.5.FIND: (a) Outer surface temperature T2 resulting from decreasing the wall thermal conductivity k orincreasing the convection coefficient h by a factor of two; benefit of applying a low emissivitycoating ( < 0.8); comment on the effectiveness of these strategies to reduce risk of burn injury when2T2 65C; and (b) Calculate and plot T2 as a function of h for the range 20 h 100 W/m K forthree materials with k = 0.3, 0.6, and 1.2 W/mK; what conditions will provide for safe outer surfacetemperatures.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Radiationexchange is between small surface and large enclosure, (4) Inner surface temperature remains constantfor all conditions.ANALYSIS: (a) The surface (x = L) energy balance is(T T44k 1 2 = h ( T2 T ) + T2 TsurL)With T1 = 352C, the effects of parameters h, k and on the outer surface temperature are calculatedand tabulated below.ConditionsExample 1.5Decrease k by Increase h by 2Change k and hDecrease k (W / m K )1.20.61.20.61.2(h W / m2 K2020404020)T2 (C )0.80.80.80.80.1100697351115(b) Using the energy balance relation in the Workspace of IHT, the outer surface temperature can becalculated and plotted as a function of the convection coefficient for selected values of the wallthermal conductivity.Continued ..O u te r su rfa ce te m p e ra tu re , T2 (C )PROBLEM 1.58 (Cont.)1008060402020406080100C o n ve ctio n co e fficie n t, h (W /m ^2 .K )k = 1 .2 W /m .Kk = 0 .6 W /m .Kk = 0 .3 W /m .KCOMMENTS: (1) From the parameter study of part (a), note that decreasing the thermalconductivity is more effective in reducing T2 than is increasing the convection coefficient. Only ifboth changes are made will T2 be in the safe range.(2) From part (a), note that applying a low emissivity coating is not beneficial. Did you suspect thatbefore you did the analysis? Give a physical explanation for this result.(3) From the parameter study graph we conclude that safe wall conditions (T2 65C) can be2maintained for these conditions: with k = 1.2 W/mK when h > 55 W/m K; with k = 0.6 W/mK2when h > 25 W/m K; and with k = 0.3 W/mK when h > 20 W/mK.PROBLEM 1.59KNOWN: Inner surface temperature, thickness and thermal conductivity of insulationexposed at its outer surface to air of prescribed temperature and convection coefficient.FIND: Outer surface temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in theinsulation, (3) Negligible radiation exchange between outer surface and surroundings.ANALYSIS: From an energy balance at the outer surface at an instant of time,q cond = q conv .Using the appropriate rate equations,k(T1 T2 ) = hL(T2 T ).Solving for T2, find()()0.1 W/m KWk400 C + 50035 CT1 + h T2 K0.025mmT2 = L=kW0.1 W/m Kh+500+L0.025mm2 KT2 = 37.9 C.<COMMENTS: If the temperature of the surroundings is approximately that of the air,radiation exchange between the outer surface and the surroundings will be negligible, since T2is small. In this case convection makes the dominant contribution to heat transfer from theouter surface, and assumption (3) is excellent.PROBLEM 1.60KNOWN: Thickness and thermal conductivity, k, of an oven wall. Temperature and emissivity, , offront surface. Temperature and convection coefficient, h, of air. Temperature of large surroundings.FIND: (a) Temperature of back surface, (b) Effect of variations in k, h and .SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Radiation exchange with largesurroundings.ANALYSIS: (a) Applying an energy balance, Eq. 1.13, at an instant of time to the front surface andsubstituting the appropriate rate equations, Eqs. 1.2, 1.3a and 1.7, find)(T T44k 1 2 = h ( T2 T ) + T2 Tsur .LSubstituting numerical values, findT1 T2 =WW8( 400 K )4 (300 K )4 = 200 K . 20 2 100 K + 0.8 5.67 100.7 W/m K m Km2 K 4 0.05 m<Since T2 = 400 K, it follows that T1 = 600 K.(b) Parametric effects may be evaluated by using the IHT First Law Model for a Nonisothermal PlaneWall. Changes in k strongly influence conditions for k < 20 W/mK, but have a negligible effect forlarger values, as T2 approaches T1 and the heat fluxes approach the corresponding limiting values10000Heat flux, q''(W/m^2)Temperature, T2(K)600500800060004000200040000100200300Thermal conductivity, k(W/m.K)3000100200300Thermal conductivity, k(W/m.K)400Conduction heat flux, q''cond(W/m^2)Convection heat flux, q''conv(W/m^2)Radiation heat flux, q''rad(W/m^2)400PROBLEM 1.60 (Cont.)The implication is that, for k > 20 W/mK, heat transfer by conduction in the wall is extremely efficientrelative to heat transfer by convection and radiation, which become the limiting heat transfer processes.Larger fluxes could be obtained by increasing and h and/or by decreasing T and Tsur .With increasing h, the front surface is cooled more effectively ( T2 decreases), and although qraddecreases, the reduction is exceeded by the increase in qconv . With a reduction in T2 and fixed valuesof k and L, qcond must also increase.30000Heat flux, q''(W/m^2)Temperature, T2(K)600500200001000000100200Convection coefficient, h(W/m^2.K)4000100Conduction heat flux, q''cond(W/m^2)Convection heat flux, q''conv(W/m^2)Radiation heat flux, q''rad(W/m^2)200Convection coefficient, h(W/m^2.K)The surface temperature also decreases with increasing , and the increase in q exceeds the reductionradin q, allowing qto increase with .convcond10000Heat flux, q''(W/m^2)575Temperature, T2(K)5705655608000600040002000055500.20.40.60.8Emissivity55000.20.40.6Emissivity0.81Conduction heat flux, q''cond(W/m^2)Convection heat flux, q''conv(W/m^2)Radiation heat flux, q''rad(W/m^2)COMMENTS: Conservation of energy, of course, dictates that, irrespective of the prescribedconditions, qcond = qconv + q .rad1PROBLEM 1.61KNOWN: Temperatures at 10 mm and 20 mm from the surface and in the adjoining airflow for athick steel casting.FIND: Surface convection coefficient, h.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in the x-direction, (3) Constantproperties, (4) Negligible generation.ANALYSIS: From a surface energy balance, it follows thatq cond = q convwhere the convection rate equation has the formqconv = h ( T T0 ) ,and q cond can be evaluated from the temperatures prescribed at surfaces 1 and 2. That is, fromFouriers law,T1 T2qcond = kx 2 x1(50 40 ) C = 15, 000 W/m2 .Wqcond = 15m K ( 20-10 ) 103 mSince the temperature gradient in the solid must be linear for the prescribed conditions, it follows thatT0 = 60C.Hence, the convection coefficient ish=h=q condT T015,000 W / m240 C= 375 W / m2 K.<COMMENTS: The accuracy of this procedure for measuring h depends strongly on the validity ofthe assumed conditions.PROBLEM 1.62KNOWN: Duct wall of prescribed thickness and thermal conductivity experiences prescribed heat fluxq at outer surface and convection at inner surface with known heat transfer coefficient.oFIND: (a) Heat flux at outer surface required to maintain inner surface of duct at Ti = 85C, (b)Temperature of outer surface, To , (c) Effect of h on To and q .oSCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Constantproperties, (4) Backside of heater perfectly insulated, (5) Negligible radiation.ANALYSIS: (a) By performing an energy balance on the wall, recognize thatbalance on the top surface, it follows thatq = qocond . From an energyq ond = qcconv = q . Hence, using the convection rate equation,o22q = qoconv = h ( Ti T ) = 100 W / m K (85 30 ) C = 5500W /m .(b) Considering the duct wall and applying Fouriers Law,q = ko<T TT=k o iLXq L5500 W/m 2 0.010 m= (85 + 2.8 ) C = 87.8 C .To = Ti + o = 85 C +k20 W/m K<(c) For Ti = 85C, the desired results may be obtained by simultaneously solving the energy balance equationsT TiT Tiq = k oko= h ( Ti T )andoLLUsing the IHT First Law Model for a Nonisothermal Plane Wall, the following results are obtained.91Surface temperature, To(C)Heat flux, q''o(W/m^2)12000100008000600040002000908988878608504080120160Convection coefficient, h(W/m^2.K)20004080120160200Convection coefficient, h(W/m^2.K)Since qconv increases linearly with increasing h, the applied heat flux q must be balanced by anoincrease in q, which, with fixed k, Ti and L, necessitates an increase in To .condCOMMENTS: The temperature difference across the wall is small, amounting to a maximum value of(To Ti ) = 5.5C for h = 200 W/m2K. If the wall were thinner (L < 10 mm) or made from a materialwith higher conductivity (k > 20 W/mK), this difference would be reduced.PROBLEM 1.63KNOWN: Dimensions, average surface temperature and emissivity of heating duct. Duct airinlet temperature and velocity. Temperature of ambient air and surroundings. Convectioncoefficient.FIND: (a) Heat loss from duct, (b) Air outlet temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential andkinetic energy changes of air flow, (4) Radiation exchange between a small surface and a largeenclosure.ANALYSIS: (a) Heat transfer from the surface of the duct to the ambient air and thesurroundings is given by Eq. (1.10)(44q = hAs ( Ts T ) + As Ts Tsur)2where As = L (2W + 2H) = 15 m (0.7 m + 0.5 m) = 16.5 m . Hence,()()q = 4 W/m2 K 16.5 m2 45$ C + 0.5 16.5 m2 5.67 108 W/m2 K 4 3234 2784 K 4q = qconv + q rad = 2970 W + 2298 W = 5268 W<(b) With i = u + pv, W = 0 and the third assumption, Eq. (1.11e) yields,m (ii io ) = mcp (Ti To ) = qwhere the sign on q has been reversed to reflect the fact that heat transfer is from the system.With m = VAc = 1.10 kg/m3 4 m/s (0.35m 0.20m ) = 0.308 kg/s, the outlet temperature isq5268 WTo = Ti = 58$ C = 41$ Cmcp0.308 kg/s 1008 J/kg K<COMMENTS: The temperature drop of the air is large and unacceptable, unless the intent isto use the duct to heat the basement. If not, the duct should be insulated to insure maximumdelivery of thermal energy to the intended space(s).PROBLEM 1.64KNOWN: Uninsulated pipe of prescribed diameter, emissivity, and surface temperature in a roomwith fixed wall and air temperatures. See Example 1.2.FIND: (a) Which option to reduce heat loss to the room is more effective: reduce by a factor of two2the convection coefficient (from 15 to 7.5 W/m K) or the emissivity (from 0.8 to 0.4) and (b) Show2graphically the heat loss as a function of the convection coefficient for the range 5 h 20 W/m Kfor emissivities of 0.2, 0.4 and 0.8. Comment on the relative efficacy of reducing heat lossesassociated with the convection and radiation processes.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between pipe and the room isbetween a small surface in a much larger enclosure, (3) The surface emissivity and absorptivity areequal, and (4) Restriction of the air flow does not alter the radiation exchange process between thepipe and the room.ANALYSIS: (a) The heat rate from the pipe to the room per unit length is(44q = q / L = qconv + q = h ( D )( Ts T ) + ( D ) Ts Tsurrad)Substituting numerical values for the two options, the resulting heat rates are calculated and comparedwith those for the conditions of Example 1.2. We conclude that both options are comparably effective.(h W / m2 KConditionsBase case, Example 1.2Reducing h by factor of 2Reducing by factor of 2)157.5150.80.80.4q ( W / m )998788709(b) Using IHT, the heat loss can be calculated as a function of the convection coefficient for selectedvalues of the surface emissivity.Heat loss, q' (/m)120080040005101520Convection coefficient, h (W/m^2.K)eps = 0.8, bare pipeeps = 0.4, coated pipeeps = 0.2, coated pipeContinued ..PROBLEM 1.64 (Cont.)COMMENTS: (1) In Example 1.2, Comment 3, we read that the heat rates by convection andradiation exchange were comparable for the base case conditions (577 vs. 421 W/m). It follows thatreducing the key transport parameter (h or ) by a factor of two yields comparable reductions in theheat loss. Coating the pipe to reduce the emissivity might to be the more practical option as it may bedifficult to control air movement.(2) For this pipe size and thermal conditions (Ts and T), the minimum possible convection coefficient2is approximately 7.5 W/m K, corresponding to free convection heat transfer to quiescent ambient air.Larger values of h are a consequence of forced air flow conditions.(3) The Workspace for the IHT program to calculate the heat loss and generate the graph for the heatloss as a function of the convection coefficient for selected emissivities is shown below. It is goodpractice to provide commentary with the code making your solution logic clear, and to summarize theresults.// Heat loss per unit pipe length; rate equation from Ex. 1.2q' = q'cv + q'radq'cv = pi*D*h*(Ts - Tinf)q'rad = pi*D*eps*sigma*(Ts^4 - Tsur^4)sigma = 5.67e-8// Input parametersD = 0.07Ts_C = 200// Representing temperatures in Celsius units using _C subscriptingTs = Ts_C +273Tinf_C = 25Tinf = Tinf_C + 273h = 15// For graph, sweep over range from 5 to 20Tsur_C = 25Tsur = Tsur_C + 273eps = 0.8//eps = 0.4// Values of emissivity for parameter study//eps = 0.2/* Base case resultsTinf TsTsurepsh298 4732980.815q'q'cvsigma997.9 577.35.67E-8*/q'radDTinf_CTs_CTsur_C420.60.072520025PROBLEM 1.65KNOWN: Conditions associated with surface cooling of plate glass which is initially at 600C.Maximum allowable temperature gradient in the glass.FIND: Lowest allowable air temperature, TSCHEMATIC:ASSUMPTIONS: (1) Surface of glass exchanges radiation with large surroundings at Tsur = T, (2)One-dimensional conduction in the x-direction.ANALYSIS: The maximum temperature gradient will exist at the surface of the glass and at theinstant that cooling is initiated. From the surface energy balance, Eq. 1.12, and the rate equations,Eqs. 1.1, 1.3a and 1.7, it follows that-k)(dT44 h ( Ts T ) Ts Tsur = 0dxor, with (dT/dx)max = -15C/mm = -15,000C/m and Tsur = T,1.4C WW 15, 000 = 5(873 T ) KmK mm2 K+0.8 5.67 108W8734 T 4 K 4 .m2 K 4 T may be obtained from a trial-and-error solution, from which it follows that, for T = 618K,21,000WWW 1275 2 + 19,730 2 .mmm2Hence the lowest allowable air temperature isT 618K = 345 C.<COMMENTS: (1) Initially, cooling is determined primarily by radiation effects.(2) For fixed T, the surface temperature gradient would decrease with increasing time into thecooling process. Accordingly, T could be decreasing with increasing time and still keep within themaximum allowable temperature gradient.PROBLEM 1.66KNOWN: Hot-wall oven, in lieu of infrared lamps, with temperature Tsur = 200C for heating acoated plate to the cure temperature. See Example 1.6.FIND: (a) The plate temperature Ts for prescribed convection conditions and coating emissivity, and(b) Calculate and plot Ts as a function of Tsur for the range 150 Tsur 250C for ambient airtemperatures of 20, 40 and 60C; identify conditions for which acceptable curing temperaturesbetween 100 and 110C may be maintained.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss from back surface of plate, (3)Plate is small object in large isothermal surroundings (hot oven walls).ANALYSIS: (a) The temperature of the plate can be determined from an energy balance on the plate,considering radiation exchange with the hot oven walls and convection with the ambient air.En E = 0iout(or)q qradconv = 04 Tsur Ts4 h (Ts T ) = 00.5 5.67 108 W / m 2 K 4([200 + 273] T ) K44s4 15 W / m 2 K (Ts [20 + 273]) K = 0<Ts = 357 K = 84C(b) Using the energy balance relation in the Workspace of IHT, the plate temperature can be calculatedand plotted as a function of oven wall temperature for selected ambient air temperatures.Plate temperature, Ts (C)15010050150175200225250Oven wall temperature, Tsur (C)Tinf = 60 CTinf = 40 CTinf = 20 CCOMMENTS: From the graph, acceptable cure temperatures between 100 and 110C can bemaintained for these conditions: with T = 20C when 225 Tsur 240C; with T = 40C when 205 Tsur 220C; and with T = 60C when 175 Tsur 195C.PROBLEM 1.67KNOWN: Operating conditions for an electrical-substitution radiometer having the same receivertemperature, Ts, in electrical and optical modes.FIND: Optical power of a laser beam and corresponding receiver temperature when the indicatedelectrical power is 20.64 mW.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Conduction losses from backside of receivernegligible in optical mode, (3) Chamber walls form large isothermal surroundings; negligible effectsdue to aperture, (4) Radiation exchange between the receiver surface and the chamber walls is betweensmall surface and large enclosure, (5) Negligible convection effects.PROPERTIES: Receiver surface: = 0.95, opt = 0.98.ANALYSIS: The schematic represents the operating conditions for the electrical mode with theoptical beam blocked. The temperature of the receiver surface can be found from an energy balanceon the receiver, considering the electrical power input, conduction loss from the backside of thereceiver, and the radiation exchange between the receiver and the chamber.Ein Eout = 0Pelec q loss q rad = 0()44Pelec 0.05 Pelec As Ts Tsur = 020.64 103()()22444428W (1 0.05 ) 0.95 0.015 / 4 m 5.67 10 W / m K Ts 77 K = 0Ts = 213.9 K<For the optical mode of operation, the optical beam is incident on the receiver surface, there is noelectrical power input, and the receiver temperature is the same as for the electrical mode. The opticalpower of the beam can be found from an energy balance on the receiver considering the absorbedbeam power and radiation exchange between the receiver and the chamber.Ein Eout = 0 opt Popt q rad = 0.98 Popt 19.60 mW = 0Popt = 19.99 mW<where qrad follows from the previous energy balance using Ts = 213.9K.COMMENTS: Recognizing that the receiver temperature, and hence the radiation exchange, is thesame for both modes, an energy balance could be directly written in terms of the absorbed opticalpower and equivalent electrical power, opt Popt = Pelec - qloss.PROBLEM 1.68KNOWN: Surface temperature, diameter and emissivity of a hot plate. Temperature of surroundingsand ambient air. Expression for convection coefficient.FIND: (a) Operating power for prescribed surface temperature, (b) Effect of surface temperature onpower requirement and on the relative contributions of radiation and convection to heat transfer fromthe surface.SCHEMATIC:ASSUMPTIONS: (1) Plate is of uniform surface temperature, (2) Walls of room are large relative toplate, (3) Negligible heat loss from bottom or sides of plate.ANALYSIS: (a) From an energy balance on the hot plate, Pelec = qconv + qrad = Ap ( q conv + q ).rad1/3Substituting for the area of the plate and from Eqs. (1.3a) and (1.7), with h = 0.70 (Ts - T) , itfollows that(2Pelec = D / 4) (444/30.70 ( Ts T )+ Ts Tsur)(44824/3Pelec = ( 0.3m ) / 4 0.70 (175 )473 298+ 0.8 5.67 10Pelec = 0.0707 m)W/m222685 W/m + 1913 W/m = 48.4 W + 135.2 W = 190.6 W2<(b) As shown graphically, both the radiation and convection heat rates, and hence the requisite electricpower, increase with increasing surface temperature.E ffe c t o f s u rfa c e te m p e ra tu re o n e le c tric p o w e r a n d h e a t ra te sH e a t ra te (W )5004003002001000100150200250300S u rfa c e te m p e ra tu re (C )P e le cq ra dq co n vHowever, because of its dependence on the fourth power of the surface temperature, the increase inradiation is more pronounced. The significant relative effect of radiation is due to the small2convection coefficients characteristic of natural convection, with 3.37 h 5.2 W/m K for 100 Ts< 300C.COMMENTS: Radiation losses could be reduced by applying a low emissivity coating to thesurface, which would have to maintain its integrity over the range of operating temperatures.PROBLEM 1.69KNOWN: Long bus bar of rectangular cross-section and ambient air and surroundings temperatures.Relation for the electrical resistivity as a function of temperature.FIND: (a) Temperature of the bar with a current of 60,000 A, and (b) Compute and plot the operatingtemperature of the bus bar as a function of the convection coefficient for the range 10 h 1002W/m K. Minimum convection coefficient required to maintain a safe-operating temperature below120C. Will increasing the emissivity significantly affect this result?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Bus bar is long, (3) Uniform bus-bar temperature,(3) Radiation exchange between the outer surface of the bus bar and its surroundings is between asmall surface and a large enclosure.PROPERTIES: Bus-bar material, e = e,o [1 + ( T To )], e,o = 0.0828 m, To = 25C, = 0.0040 K1.ANALYSIS: (a) An energy balance on the bus-bar for a unit length as shown in the schematic abovehas the formEin E + E = 0outgen2q q radconv + I R = 0e)(4 P T 4 Tsur h P (T T ) + I 2 e / Ac = 0where P = 2 ( H + W ) , R = e / A c and A c = H W. Substituting numerical values,e40.8 2 0.600 + 0.200 m 5.67 108 W / m 2 K 4 T 4 30 + 273 K 4(()[10 W / m 2 K 2 ( 0.600 + 0.200 ) m ( T [30 + 273]) K+ ( 60, 000 A )2{0.0828 106)} m 1 + 0.0040 K 1 ( T [25 + 273]) K / ( 0.600 0.200 ) m 2 = 0Solving for the bus-bar temperature, find<T = 426 K = 153C.(b) Using the energy balance relation in the Workspace of IHT, the bus-bar operating temperature is2calculated as a function of the convection coefficient for the range 10 h 100 W/m K. From thisgraph we can determine that to maintain a safe operating temperature below 120C, the minimumconvection coefficient required ish min = 16 W / m 2 K.<Continued ..PROBLEM 1.69 (Cont.)Using the same equations, we can calculate and plot the heat transfer rates by convection and radiationas a function of the bus-bar temperature.3000H e a t ra te s, q 'cv o r q 'ra d (W /m )175B a r te m p e ra tu re , T (C )1501251007520001000050255075100125150175B u s b a r te m p e ra tu re , T (C )25020406080100C o n ve c tio n h e a t flu x, q 'cvR a d ia tio n e xc h a n g e , q 'ra d , e p s = 0 .8C o n ve c tio n co e fficie n t, h (W /m ^2 .K )Note that convection is the dominant mode for low bus-bar temperatures; that is, for low current flow.As the bus-bar temperature increases toward the safe-operating limit (120C), convection andradiation exchange heat transfer rates become comparable. Notice that the relative importance of theradiation exchange rate increases with increasing bus-bar temperature.COMMENTS: (1) It follows from the second graph that increasing the surface emissivity will beonly significant at higher temperatures, especially beyond the safe-operating limit.(2) The Workspace for the IHT program to perform the parametric analysis and generate the graphs isshown below. It is good practice to provide commentary with the code making your solution logicclear, and to summarize the results./* Results for base case conditions:Ts_C q'cvq'radrhoeHepsh153.3 197317861.253E-7 0.60.810 */ITinf_CTsur_CWalpha6E430300.20.004// Surface energy balance on a per unit length basis-q'cv - q'rad + Edot'gen = 0q'cv = h * P * (Ts - Tinf)P = 2 * (W + H)// perimeter of the bar experiencing surface heat transferq'rad = eps * sigma * (Ts^4 - Tsur^4) * Psigma = 5.67e-8Edot'gen = I^2 * Re'Re' = rhoe / Acrhoe = rhoeo * ( 1 + alpha * (Ts - Teo))Ac = W * H// Input parametersI = 60000alpha = 0.0040rhoeo = 0.0828e-6Teo = 25 + 273W = 0.200H = 0.600Tinf_C = 30Tinf = Tinf_C + 273h = 10eps = 0.8Tsur_C = 30Tsur = Tsur_C + 273Ts_C = Ts - 273// temperature coefficient, K^-1; typical value for cast aluminum// electrical resistivity at the reference temperature, Teo; microohm-m// reference temperature, KPROBLEM 1.70KNOWN: Solar collector designed to heat water operating under prescribed solar irradiation andloss conditions.FIND: (a) Useful heat collected per unit area of the collector, q , (b) Temperature rise of the wateruflow, To Ti , and (c) Collector efficiency.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) No heat losses out sides or back of collector, (3)Collector area is small compared to sky surroundings.PROPERTIES: Table A.6, Water (300K): cp = 4179 J/kgK.ANALYSIS: (a) Defining the collector as the control volume and writing the conservation of energyrequirement on a per unit area basis, find thatE in E out + E gen = E st .Identifying processes as per above right sketch,q q q q = 0solarradconvuwhere q solar = 0.9 q s ; that is, 90% of the solar flux is absorbed in the collector (Eq. 1.6). Using theappropriate rate equations, the useful heat rate per unit area is()44q = 0.9 q Tcp Tsky h ( Ts T )usWWWq = 0.9 7003034 2634 K 4 10 0.94 5.67 108(30 25) Cu22 K42Kmmm)(q = 630 W / m2 194 W / m2 50 W / m2 = 386 W / m2 .u<(b) The total useful heat collected is q A. Defining a control volume about the water tubing, theuuseful heat causes an enthalpy change of the flowing water. That is,q A=mcp ( Ti To )uor(Ti To ) = 386 W/m2 3m2 / 0.01kg/s 4179J/kg K=27.7C.()()(c) The efficiency is = q / qS = 386 W/m 2 / 700 W/m 2 = 0.55 or 55%.u <<COMMENTS: Note how the sky has been treated as large surroundings at a uniform temperatureTsky.PROBLEM 1.71KNOWN: Surface-mount transistor with prescribed dissipation and convection cooling conditions.FIND: (a) Case temperature for mounting arrangement with air-gap and conductive paste between caseand circuit board, (b) Consider options for increasing E g , subject to the constraint that Tc = 40C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Transistor case is isothermal, (3) Upper surfaceexperiences convection; negligible losses from edges, (4) Leads provide conduction path between caseand board, (5) Negligible radiation, (6) Negligible energy generation in leads due to current flow, (7)Negligible convection from surface of leads.PROPERTIES: (Given): Air, k g,a = 0.0263 W/mK; Paste, k g,p = 0.12 W/mK; Metal leads, k =25 W/mK.ANALYSIS: (a) Define the transistor as the system and identify modes of heat transfer.Ein E out + Eg = Est = 0q conv q cond,gap 3q lead + E g = 0T TbT Tb hAs ( Tc T ) k g As c 3k Ac c+ Eg = 0tLwhere As = L1 L 2 = 4 8 mm2 = 32 10-6 m2 and A c = t w = 0.25 1 mm2 = 25 10-8 m2.Rearranging and solving for Tc ,{}Tc = hAs T + k g As /t + 3 ( k Ac /L ) Tb + E g / hAs + k g As /t + 3 ( k Ac /L )Substituting numerical values, with the air-gap condition ( k g,a = 0.0263 W/mK){()Tc = 50W/m 2 K 32 106 m 2 20 C + 0.0263W/m K 32 106 m 2 /0.2 103 m8 23333+3 25 W/m K 25 10 m /4 10 m 35 C / 1.600 10 + 4.208 10 + 4.688 10 W/K(Tc = 47.0 C .)}<Continued..PROBLEM 1.71 (Cont.)With the paste condition ( k g,p = 0.12 W/mK), Tc = 39.9C. As expected, the effect of the conductivepaste is to improve the coupling between the circuit board and the case. Hence, Tc decreases.Power dissipation, Edotg(W)(b) Using the keyboard to enter model equations into the workspace, IHT has been used to perform thedesired calculations. For values of k = 200 and 400 W/mK and convection coefficients in the rangefrom 50 to 250 W/m2K, the energy balance equation may be used to compute the power dissipation for amaximum allowable case temperature of 40C.0.70.60.50.40.350100150200250Convection coefficient, h(W/m^2.K)kl = 400 W/m.Kkl = 200 W/m.KAs indicated by the energy balance, the power dissipation increases linearly with increasing h, as well aswith increasing k . For h = 250 W/m2K (enhanced air cooling) and k = 400 W/mK (copper leads),the transistor may dissipate up to 0.63 W.COMMENTS: Additional benefits may be derived by increasing heat transfer across the gap separatingthe case from the board, perhaps by inserting a highly conductive material in the gap.PROBLEM 1.72(a)KNOWN: Solar radiation is incident on an asphalt paving.FIND: Relevant heat transfer processes.SCHEMATIC:The relevant processes shown on the schematic include:qSIncident solar radiation, a large portion of which q S,abs , is absorbed by the asphaltsurface,q radRadiation emitted by the surface to the air,q conv Convection heat transfer from the surface to the air, andq cond Conduction heat transfer from the surface into the asphalt.Applying the surface energy balance, Eq. 1.12,q S,abs q q conv = q cond .radCOMMENTS: (1) q cond and q conv could be evaluated from Eqs. 1.1 and 1.3, respectively.(2) It has been assumed that the pavement surface temperature is higher than that of theunderlying pavement and the air, in which case heat transfer by conduction and convectionare from the surface.(3) For simplicity, radiation incident on the pavement due to atmospheric emission has beenignored (see Section 12.8 for a discussion). Eq. 1.6 may then be used for the absorbedsolar irradiation and Eq. 1.5 may be used to obtain the emitted radiation q .rad(4) With the rate equations, the energy balance becomes4qS,abs Ts h ( Ts T ) = kdT .dx sPROBLEM 1.72(b)KNOWN: Physical mechanism for microwave heating.FIND: Comparison of (a) cooking in a microwave oven with a conventional radiant orconvection oven and (b) a microwave clothes dryer with a conventional dryer.(a) Microwave cooking occurs as a result of volumetric thermal energy generation throughoutthe food, without heating of the food container or the oven wall. Conventional cooking relieson radiant heat transfer from the oven walls and/or convection heat transfer from the air spaceto the surface of the food and subsequent heat transfer by conduction to the core of the food.Microwave cooking is more efficient and is achieved in less time.(b) In a microwave dryer, the microwave radiation would heat the water, but not the fabric,directly (the fabric would be heated indirectly by energy transfer from the water). By heatingthe water, energy would go directly into evaporation, unlike a conventional dryer where thewalls and air are first heated electrically or by a gas heater, and thermal energy is subsequentlytransferred to the wet clothes. The microwave dryer would still require a rotating drum andair flow to remove the water vapor, but is able to operate more efficiently and at lowertemperatures. For a more detailed description of microwave drying, see MechanicalEngineering, March 1993, page 120.PROBLEM 1.72(c)KNOWN: Surface temperature of exposed arm exceeds that of the room air and walls.FIND: Relevant heat transfer processes.SCHEMATIC:Neglecting evaporation from the surface of the skin, the only relevant heat transfer processesare:q convConvection heat transfer from the skin to the room air, andq radNet radiation exchange between the surface of the skin and the surroundings(walls of the room).You are not imagining things. Even though the room air is maintained at a fixed temperature(T = 15C), the inner surface temperature of the outside walls, Tsur, will decrease withdecreasing outside air temperature. Upon exposure to these walls, body heat loss will belarger due to increased qrad.COMMENTS: The foregoing reasoning assumes that the thermostat measures the true roomair temperature and is shielded from radiation exchange with the outside walls.PROBLEM 1.72(d)KNOWN: Tungsten filament is heated to 2900 K in an air-filled glass bulb.FIND: Relevant heat transfer processes.SCHEMATIC:The relevant processes associated with the filament and bulb include:q rad,fRadiation emitted by the tungsten filament, a portion of which is transmittedthrough the glass,q conv,fFree convection from filament to air of temperature Ta,i < Tf ,q rad,g,iRadiation emitted by inner surface of glass, a small portion of which isintercepted by the filament,q conv,g,iFree convection from air to inner glass surface of temperature Tg,i < Ta,i ,q cond,gConduction through glass wall,q conv,g,oFree convection from outer glass surface to room air of temperatureTa,o < Tg,o , andq rad,g-surNet radiation heat transfer between outer glass surface and surroundings, suchas the walls of a room, of temperature Tsur < Tg,o .COMMENTS: If the glass bulb is evacuated, no convection is present within the bulb; thatis, q conv,f = q conv,g,i = 0.PROBLEM 1.72(e)KNOWN: Geometry of a composite insulation consisting of a honeycomb core.FIND: Relevant heat transfer processes.SCHEMATIC:The above schematic represents the cross section of a single honeycomb cell and surfaceslabs. Assumed direction of gravity field is downward. Assuming that the bottom (inner)surface temperature exceeds the top (outer) surface temperature Ts,i > Ts,o , heat transfer is()in the direction shown.Heat may be transferred to the inner surface by convection and radiation, whereupon it istransferred through the composite byq cond,iConduction through the inner solid slab,q conv,hcFree convection through the cellular airspace,q cond,hcConduction through the honeycomb wall,q rad,hcRadiation between the honeycomb surfaces, andq cond,oConduction through the outer solid slab.Heat may then be transferred from the outer surface by convection and radiation. Note thatfor a single cell under steady state conditions,q rad,i + q conv,i = q cond,i = q conv,hc + q cond,hc+q rad,hc = q cond,o = q rad,o + q conv,o .COMMENTS: Performance would be enhanced by using materials of low thermalconductivity, k, and emissivity, . Evacuating the airspace would enhance performance byeliminating heat transfer due to free convection.PROBLEM 1.72(f)KNOWN: A thermocouple junction is used, with or without a radiation shield, to measurethe temperature of a gas flowing through a channel. The wall of the channel is at atemperature much less than that of the gas.FIND: (a) Relevant heat transfer processes, (b) Temperature of junction relative to that ofgas, (c) Effect of radiation shield.SCHEMATIC:ASSUMPTIONS: (1) Junction is small relative to channel walls, (2) Steady-state conditions,(3) Negligible heat transfer by conduction through the thermocouple leads.ANALYSIS: (a) The relevant heat transfer processes are:q radNet radiation transfer from the junction to the walls, andq convConvection transfer from the gas to the junction.(b) From a surface energy balance on the junction,q conv = q rador from Eqs. 1.3a and 1.7,()()44h A Tj Tg = A Tj Ts .To satisfy this equality, it follows thatTs < Tj < Tg .That is, the junction assumes a temperature between that of the channel wall and the gas,thereby sensing a temperature which is less than that of the gas.()(c) The measurement error Tg Tj is reduced by using a radiation shield as shown in theschematic. The junction now exchanges radiation with the shield, whose temperature mustexceed that of the channel wall. The radiation loss from the junction is therefore reduced, andits temperature more closely approaches that of the gas.PROBLEM 1.72(g)KNOWN: Fireplace cavity is separated from room air by two glass plates, open at both ends.FIND: Relevant heat transfer processes.SCHEMATIC:The relevant heat transfer processes associated with the double-glazed, glass fire screen are:q rad,1Radiation from flames and cavity wall, portions of which are absorbed andtransmitted by the two panes,q rad,2Emission from inner surface of inner pane to cavity,q rad,3Net radiation exchange between outer surface of inner pane and inner surfaceof outer pane,q rad,4Net radiation exchange between outer surface of outer pane and walls of room,q conv,1Convection between cavity gases and inner pane,q conv2Convection across air space between panes,q conv,3Convection from outer surface to room air,q cond,1Conduction across inner pane, andq cond,2Conduction across outer pane.COMMENTS: (1) Much of the luminous portion of the flame radiation is transmitted to theroom interior.(2) All convection processes are buoyancy driven (free convection).PROBLEM 1.73(a)KNOWN: Room air is separated from ambient air by one or two glass panes.FIND: Relevant heat transfer processes.SCHEMATIC:The relevant processes associated with single (above left schematic) and double (above rightschematic) glass panes include.q conv,1Convection from room air to inner surface of first pane,q rad,1Net radiation exchange between room walls and inner surface of first pane,q cond,1Conduction through first pane,q conv,sConvection across airspace between panes,q rad,sNet radiation exchange between outer surface of first pane and inner surface ofsecond pane (across airspace),q cond,2Conduction through a second pane,q conv,2Convection from outer surface of single (or second) pane to ambient air,q rad,2Net radiation exchange between outer surface of single (or second) pane andsurroundings such as the ground, andqSIncident solar radiation during day; fraction transmitted to room is smaller fordouble pane.COMMENTS: Heat loss from the room is significantly reduced by the double paneconstruction.PROBLEM 1.73(b)KNOWN: Configuration of a flat plate solar collector.FIND: Relevant heat transfer processes with and without a cover plate.SCHEMATIC:The relevant processes without (above left schematic) and with (above right schematic)include:qSIncident solar radiation, a large portion of which is absorbed by the absorberplate. Reduced with use of cover plate (primarily due to reflection off coverplate).q rad,Net radiation exchange between absorber plate or cover plate andsurroundings,q conv,Convection from absorber plate or cover plate to ambient air,q rad,a-cNet radiation exchange between absorber and cover plates,q conv,a-cConvection heat transfer across airspace between absorber and cover plates,q condConduction through insulation, andq convConvection to working fluid.COMMENTS: The cover plate acts to significantly reduce heat losses by convection andradiation from the absorber plate to the surroundings.PROBLEM 1.73(c)KNOWN: Configuration of a solar collector used to heat air for agricultural applications.FIND: Relevant heat transfer processes.SCHEMATIC:Assume the temperature of the absorber plates exceeds the ambient air temperature. At thecover plates, the relevant processes are:q conv,a-iConvection from inside air to inner surface,q rad,p-iNet radiation transfer from absorber plates to inner surface,q conv,i-oConvection across airspace between covers,q rad,i-oNet radiation transfer from inner to outer cover,q conv,o-Convection from outer cover to ambient air,q rad,oNet radiation transfer from outer cover to surroundings, andqSIncident solar radiation.Additional processes relevant to the absorber plates and airspace are:q S,tSolar radiation transmitted by cover plates,q conv,p-aConvection from absorber plates to inside air, andq condConduction through insulation.PROBLEM 1.73(d)KNOWN: Features of an evacuated tube solar collector.FIND: Relevant heat transfer processes for one of the tubes.SCHEMATIC:The relevant heat transfer processes for one of the evacuated tube solar collectors includes:qSIncident solar radiation including contribution due to reflection off panel (mostis transmitted),q conv,oConvection heat transfer from outer surface to ambient air,q rad,o-surNet rate of radiation heat exchange between outer surface of outer tube and thesurroundings, including the panel,q S,tSolar radiation transmitted through outer tube and incident on inner tube (mostis absorbed),q rad,i-oNet rate of radiation heat exchange between outer surface of inner tube andinner surface of outer tube, andq conv,iConvection heat transfer to working fluid.There is also conduction heat transfer through the inner and outer tube walls. If the walls arethin, the temperature drop across the walls will be small.PROBLEM 2.1KNOWN: Steady-state, one-dimensional heat conduction through an axisymmetric shape.FIND: Sketch temperature distribution and explain shape of curve.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) Nointernal heat generation.ANALYSIS: Performing an energy balance on the object according to Eq. 1.11a, E in E out = 0, itfollows thatE in E out = q x$and that q x q x x . That is, the heat rate within the object is everywhere constant. From Fourierslaw,q x = kA xdT,dxand since qx and k are both constants, it follows thatAxdT= Constant.dxThat is, the product of the cross-sectional area normal to the heat rate and temperature gradientremains a constant and independent of distance x. It follows that since Ax increases with x, thendT/dx must decrease with increasing x. Hence, the temperature distribution appears as shown above.COMMENTS: (1) Be sure to recognize that dT/dx is the slope of the temperature distribution. (2)What would the distribution be when T2 > T1? (3) How does the heat flux, q , vary with distance?xPROBLEM 2.2KNOWN: Hot water pipe covered with thick layer of insulation.FIND: Sketch temperature distribution and give brief explanation to justify shape.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) Nointernal heat generation, (4) Insulation has uniform properties independent of temperature andposition.ANALYSIS: Fouriers law, Eq. 2.1, for this one-dimensional (cylindrical) radial system has the formq r = kA r16dTdT= k 2rdrdrwhere A r = 2r and is the axial length of the pipe-insulation system. Recognize that for steadystate conditions with no internal heat generation, an energy balance on the system requiresE in = E out since E g = E st = 0. Henceqr = Constant.That is, qr is independent of radius (r). Since the thermal conductivity is also constant, it follows thatr dT "# = Constant.! dr $This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r,remains constant throughout the insulation. For our situation, the temperature distribution mustappear as shown in the sketch.COMMENTS: (1) Note that, while qr is a constant and independent of r, q is not a constant. Howr16does q r vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases withrincreasing radius.PROBLEM 2.3KNOWN: A spherical shell with prescribed geometry and surface temperatures.FIND: Sketch temperature distribution and explain shape of the curve.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial (sphericalcoordinates) direction, (3) No internal generation, (4) Constant properties.ANALYSIS: Fouriers law, Eq. 2.1, for this one-dimensional, radial (spherical coordinate) systemhas the formqr = k Ar()dTdT= k 4 r 2drdrwhere Ar is the surface area of a sphere. For steady-state conditions, an energy balance on the systemyields E in = E out , since E g = E st = 0. Hence,qin = q out = q r q r ( r ) .That is, qr is a constant, independent of the radial coordinate. Since the thermal conductivity isconstant, it follows that dT r 2 = Constant. dr This relation requires that the product of the radial temperature gradient, dT/dr, and the radius2squared, r , remains constant throughout the shell. Hence, the temperature distribution appears asshown in the sketch.COMMENTS: Note that, for the above conditions, q r q r ( r ) ; that is, qr is everywhere constant.How does q vary as a function of radius?rPROBLEM 2.4KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperaturedistribution and heat rate.FIND: Expression for the thermal conductivity, k.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3)No internal heat generation.ANALYSIS: Applying the energy balance, Eq. 1.11a, to the system, it follows that, sinceE in = E out ,q x = Constant f ( x ).Using Fouriers law, Eq. 2.1, with appropriate expressions for Ax and T, yieldsdTdxdK6000W=-k (1-x ) m2 300 1 2x-x3 .mdx q x = k A x)(Solving for k and recognizing its units are W/mK,k=-6000()(1-x ) 300 2 3x 2 =20(1 x )(2 + 3x 2).<COMMENTS: (1) At x = 0, k = 10W/mK and k as x 1. (2) Recognize that the 1-Dassumption is an approximation which becomes more inappropriate as the area change with x, andhence two-dimensional effects, become more pronounced.PROBLEM 2.5KNOWN: End-face temperatures and temperature dependence of k for a truncated cone.FIND: Variation with axial distance along the cone of q x , q , k, and dT / dx.xSCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in x (negligible temperature gradients along y),(2) Steady-state conditions, (3) Adiabatic sides, (4) No internal heat generation.ANALYSIS: For the prescribed conditions, it follows from conservation of energy, Eq. 1.11a, thatfor a differential control volume, E in = E out or q x = q x+dx . Henceqx is independent of x.$Since A(x) increases with increasing x, it follows that q = q x / A x decreases with increasing x.xSince T decreases with increasing x, k increases with increasing x. Hence, from Fouriers law, Eq.2.2,q = kxdT,dxit follows that | dT/dx | decreases with increasing x.PROBLEM 2.6KNOWN: Temperature dependence of the thermal conductivity, k(T), for heat transfer through aplane wall.FIND: Effect of k(T) on temperature distribution, T(x).ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) No internal heatgeneration.ANALYSIS: From Fouriers law and the form of k(T),q = kx16dTdT= k o + aT.dxdx(1)22The shape of the temperature distribution may be inferred from knowledge of d T/dx = d(dT/dx)/dx.Since q is independent of x for the prescribed conditions,x1!6 "#$2d 2 T dT "1 k o + aT6 2 a! dx #$ = 0.dxdq dTx =- dk o + aT=0dxdxdxHence,d 2T "#2!$-adT=2k o + aT dxdx%k o + aT = k > 0Kwhere & dT " 2K! dx #$ > 0'from which it follows that fora > 0: d 2 T / dx 2 < 0a = 0: d 2 T / dx 2 = 0a < 0: d 2 T / dx 2 > 0.COMMENTS: The shape of the distribution could also be inferred from Eq. (1). Since T decreaseswith increasing x,a > 0: k decreases with increasing x = > | dT/dx | increases with increasing xa = 0: k = ko = > dT/dx is constanta < 0: k increases with increasing x = > | dT/dx | decreases with increasing x.PROBLEM 2.7KNOWN: Thermal conductivity and thickness of a one-dimensional system with no internal heatgeneration and steady-state conditions.FIND: Unknown surface temperatures, temperature gradient or heat flux.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional heat flow, (2) No internal heat generation, (3) Steady-stateconditions, (4) Constant properties.ANALYSIS: The rate equation and temperature gradient for this system aredTdT T1 T2=q = kand.xdxdxLUsing Eqs. (1) and (2), the unknown quantities can be determined.dT ( 400 300 ) K(a)== 200 K/mdx0.5mq = 25x(b)WmK 200Km<= 5000 W/m 2 .K= 6250 W/m 2mK mK dT T2 = T1 L = 1000 C-0.5m -250 m dx q = 25xW 250T2 = 225 C.(c) q = 25xWmK 200<Km= 5000 W/m 2T2 = 80 C-0.5m 200(d)K = 20 C.m<q4000 W/m 2K= x == 160dxk25 W/m KmdTT1 = L() dT + T = 0.5m -160 K + 5 C dx 2mT1 = 85 C.(e)(1,2)dT=qx(3000 W/m2 ) = 120 K=25 W/m KKT2 = 30 C-0.5m 120 = 30 C.mdxkm2a +b2<PROBLEM 2.8KNOWN: One-dimensional system with prescribed thermal conductivity and thickness.FIND: Unknowns for various temperature conditions and sketch distribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heatgeneration, (4) Constant properties.ANALYSIS: The rate equation and temperature gradient for this system aredTdT T2 T1q = kand.=xdxdxLUsing Eqs. (1) and (2), the unknown quantities for each case can be determined.(a)(b)(c)dT( 20 50 ) K= 280 K/m0.25mWKq = 50 280 = 14.0 kW/m 2 .xmK mdxdT=q = 50xT2 = L mK WdTdx 160Km<= 8.0 kW/m 2+ T1 = 0.25m 160K + 70 C.m<T2 = 110 C.(d)q = 50xmK WT1 = T2 L 80dTdxKm= 4.0 kW/m 2= 40 C 0.25m 80Km.T1 = 60 C.q = 50x<mK W 200K2= 10.0 kW/mmdTK= 30 C 0.25m 200 = 20 C.T1 = T2 L dxm(e)<( 10 ( 30 )) K= 80 K/m0.25mW K 80 = 4.0 kW/m 2 .q = 50xmK mdx=(1,2)<PROBLEM 2.9KNOWN: Plane wall with prescribed thermal conductivity, thickness, and surface temperatures.FIND: Heat flux, q , and temperature gradient, dT/dx, for the three different coordinate systemsxshown.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional heat flow, (2) Steady-state conditions, (3) No internalgeneration, (4) Constant properties.ANALYSIS: The rate equation for conduction heat transfer isq = kxdT,dx(1)where the temperature gradient is constant throughout the wall and of the form $ $dT T L T 0=.dxL(2)Substituting numerical values, find the temperature gradients,$<400 600 KdT T1 T2=== 2000 K / mdxL0.100m$<600 400 KdT T2 T1=== 2000 K / m.dxL0.100m$<(a)600 400 KdT T2 T1=== 2000 K / mdxL0.100m(b)(c)The heat rates, using Eq. (1) with k = 100 W/mK, are(a)q = 100xW 2000 K / m = -200 kW / m2m K<(b)q = 100xW( 2000 K / m) = +200 kW / m2m K<(c)q = 100xW 2000 K / m = -200 kW / m2m K<PROBLEM 2.10KNOWN: Temperature distribution in solid cylinder and convection coefficient at cylinder surface.FIND: Expressions for heat rate at cylinder surface and fluid temperature.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Steady-state conditions, (3) Constantproperties.ANALYSIS: The heat rate from Fouriers law for the radial (cylindrical) system has the formq r = kA rdT.dr2Substituting for the temperature distribution, T(r) = a + br ,16q r = k 2rL 2br = - 4kbLr 2 .At the outer surface ( r = ro), the conduction heat rate is2q r=ro = 4kbLro .<From a surface energy balance at r = ro,16 16q r=ro = q conv = h 2ro L T ro T ,Substituting for q r=ro and solving for T,16T = T ro +2 kbroh2T = a + bro +!2 kbrohT = a + bro ro +"#$2k.h<PROBLEM 2.11KNOWN: Two-dimensional body with specified thermal conductivity and two isothermal surfacesof prescribed temperatures; one surface, A, has a prescribed temperature gradient.FIND: Temperature gradients, T/x and T/y, at the surface B.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) No heatgeneration, (4) Constant properties.ANALYSIS: At the surface A, the temperature gradient in the x-direction must be zero. That is,(T/x)A = 0. This follows from the requirement that the heat flux vector must be normal to anisothermal surface. The heat rate at the surface A is given by Fouriers law written asq = k wAy,AWKT= 10 2 m 30 = 600W / m.m Kmy AOn the surface B, it follows thatT / y$B = 0<in order to satisfy the requirement that the heat flux vector be normal to the isothermal surface B.Using the conservation of energy requirement, Eq. 1.11a, on the body, findq q = 0y,Ax,BNote that,q = k w Bx,Borq = q .x,By,ATx Band hencey,AT / x$B = qw B = 10 W600 W /m$ = 60 K / m.k/ m K 1m<COMMENTS: Note that, in using the conservation requirement, q = + q iny,A and q = + q .outx,BPROBLEM 2.12KNOWN: Length and thermal conductivity of a shaft. Temperature distribution along shaft.FIND: Temperature and heat rates at ends of shaft.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x, (3) Constantproperties.ANALYSIS: Temperatures at the top and bottom of the shaft are, respectively,T(0) = 100C<T(L) = -40C.Applying Fouriers law, Eq. 2.1,"'$dT= 25 W / m K 0.005 m2 150 + 20x C / mdxq x = 0125 150 - 20x W..q x = kA$Hence,qx(0) = 18.75 Wqx(L) = 16.25 W.<The difference in heat rates, qx(0) > qx(L), is due to heat losses q from the side of the shaft.COMMENTS: Heat loss from the side requires the existence of temperature gradients over the shaftcross-section. Hence, specification of T as a function of only x is an approximation.PROBLEM 2.13KNOWN: A rod of constant thermal conductivity k and variable cross-sectional area Ax(x) = Aoeaxwhere Ao and a are constants.FIND: (a) Expression for the conduction heat rate, qx(x); use this expression to determine thetemperature distribution, T(x); and sketch of the temperature distribution, (b) Considering thepresence of volumetric heat generation rate, q = q o exp ( ax ) , obtain an expression for qx(x) whenthe left face, x = 0, is well insulated.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the rod, (2) Constant properties, (3) Steadystate conditions.ANALYSIS: Perform an energy balance on the control volume, A(x)dx,Ein E out + Eg = 0q x q x + dx + q A ( x ) dx = 0The conduction heat rate terms can be expressed as a Taylor series and substituting expressions for qand A(x),d(q x ) + qo exp ( ax ) Ao exp (ax ) = 0dxq x = k A ( x )(1)dTdx(2)(a) With no internal generation, q o = 0, and from Eq. (1) findd(q x ) = 0dx<indicating that the heat rate is constant with x. By combining Eqs. (1) and (2)ddT k A ( x ) = 0dx dx orA (x )dT= C1dx(3)<Continued...PROBLEM 2.13 (Cont.)That is, the product of the cross-sectional area and the temperature gradient is a constant, independentof x. Hence, with T(0) > T(L), the temperature distribution is exponential, and as shown in the sketchabove. Separating variables and integrating Eq. (3), the general form for the temperature distributioncan be determined,Ao exp ( ax ) dT= C1dxdT = C1Ao 1 exp ( ax ) dxT ( x ) = C1Ao a exp ( ax ) + C2<We could use the two temperature boundary conditions, To = T(0) and TL = T(L), to evaluate C1 andC2 and, hence, obtain the temperature distribution in terms of To and TL.(b) With the internal generation, from Eq. (1),d(q x ) + qo Ao = 0dxorq x = qo Ao x<That is, the heat rate increases linearly with x.COMMENTS: In part (b), you could determine the temperature distribution using Fouriers law andknowledge of the heat rate dependence upon the x-coordinate. Give it a try!PROBLEM 2.14KNOWN: Dimensions and end temperatures of a cylindrical rod which is insulated on its side.FIND: Rate of heat transfer associated with different rod materials.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction along cylinder axis, (2) Steady-state conditions,(3) Constant properties.PROPERTIES: The properties may be evaluated from Tables A-1 to A-3 at a mean temperature of50C = 323K and are summarized below.ANALYSIS: The heat transfer rate may be obtained from Fouriers law. Since the axial temperaturegradient is linear, this expression reduces to$ 100 0$ C = 0.491m C$ k 0.025mT Tq = kA 1 2 = kL420.1mCuAlSt.St.SiNOak MagnesiaPyrex(pure)(2024)(302)(85%)_______________________________________________________________k(W/mK)40117716.314.90.190.0521.4q(W)197878.07.30.0930.0260.69<COMMENTS: The k values of Cu and Al were obtained by linear interpolation; the k value of St.St.was obtained by linear extrapolation, as was the value for SiN; the value for magnesia was obtainedby linear interpolation; and the values for oak and pyrex are for 300 K.PROBLEM 2.15KNOWN: One-dimensional system with prescribed surface temperatures and thickness.FIND: Heat flux through system constructed of these materials: (a) pure aluminum, (b) plain carbonsteel, (c) AISI 316, stainless steel, (d) pyroceram, (e) teflon and (f) concrete.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No heatgeneration, (4) Constant thermal properties.PROPERTIES: The thermal conductivity is evaluated at the average temperature of the system, T =(T1+T2)/2 = (325+275)K/2 = 300K. Property values and table identification are shown below.ANALYSIS: For this system, Fouriers law can be written asq = kxdTT T= k 2 1 .dxLSubstituting numerical values, the heat flux isq = kx275- 325$K = +2500 K k20 10-3 mm2where q will have units W/m if k has units W/mK. The heat fluxes for each system follow.xThermal conductivityMaterial(a)(b)(c)(d)(e)(f)Pure AluminumPlain carbon steelAISI 316, S.S.PyroceramTeflonConcreteTableA-1A-1A-1A-2A-3A-3k(W/mK)23760.513.43.980.351.4Heat flux"q kW / m2x59315133.59.950.883.5'<COMMENTS: Recognize that the thermal conductivity of these solid materials varies by more thantwo orders of magnitude.PROBLEM 2.16KNOWN: Different thicknesses of three materials: rock, 18 ft; wood, 15 in; and fiberglassinsulation, 6 in.FIND: The insulating quality of the materials as measured by the R-value.PROPERTIES: Table A-3 (300K):MaterialThermalconductivity, W/mKLimestoneSoftwoodBlanket (glass, fiber 10 kg/m3)2.150.120.048ANALYSIS: The R-value, a quantity commonly used in the construction industry and buildingtechnology, is defined asR$L in"'k Btu in / h ft 2 F.2The R-value can be interpreted as the thermal resistance of a 1 ft cross section of the material. Usingthe conversion factor for thermal conductivity between the SI and English systems, the R-values are:Rock, Limestone, 18 ft:in1ftR== 14.5 Btu/h ft 2 FWBtu/h ft Fin2.15 0.577812mKW/m Kft18 ft 12()"'Wood, Softwood, 15 in:15 inR=0.12WBtu / h ft Fin 0.5778 12m KW / m Kft= 18 Btu / h ft 2 F1Insulation, Blanket, 6 in:6 inR=0.048WBtu / h ft Fin 0.5778 12m KW / m Kft"'= 18 Btu / h ft 2 FCOMMENTS: The R-value of 19 given in the advertisement is reasonable.1PROBLEM 2.17KNOWN: Electrical heater sandwiched between two identical cylindrical (30 mm dia. 60 mmlength) samples whose opposite ends contact plates maintained at To.FIND: (a) Thermal conductivity of SS316 samples for the prescribed conditions (A) and theiraverage temperature, (b) Thermal conductivity of Armco iron sample for the prescribed conditions(B), (c) Comment on advantages of experimental arrangement, lateral heat losses, and conditions forwhich T1 T2.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional heat transfer in samples, (2) Steady-state conditions, (3)Negligible contact resistance between materials.%PROPERTIES: Table A.2, Stainless steel 316 T = 400 K : k ss = 15.2 W / m K; Armco iron%T = 380 K : k iron = 716 W / m K..ANALYSIS: (a) For Case A recognize that half the heater power will pass through each of thesamples which are presumed identical. Apply Fouriers law to a sampleq = kA ck=Tx$0.5 100V 0.353A 0.015 mqx== 15.0 W / m K.2A c T 0.030 m / 4 25.0 C$<The total temperature drop across the length of the sample is T1(L/x) = 25C (60 mm/15 mm) =100C. Hence, the heater temperature is Th = 177C. Thus the average temperature of the sample is$T = To + Th / 2 = 127 C = 400 K<.We compare the calculated value of k with the tabulated value (see above) at 400 K and note the goodagreement.(b) For Case B, we assume that the thermal conductivity of the SS316 sample is the same as thatfound in Part (a). The heat rate through the Armco iron sample isContinued ..PROBLEM 2.17 (CONT.)q ironq iron$ 0.030 m15.0 C= q heater q ss = 100V 0.601A 15.0 W / m K 40.015 m= 601 10.6 W = 49.5 W.2$whereq ss = k ssA c T2 / x2 .Applying Fouriers law to the iron sample,k iron =49.5 W 0.015 mq iron x2== 70.0 W / m K.2A c T2 0.030 m / 4 15.0 C$<The total drop across the iron sample is 15C(60/15) = 60C; the heater temperature is (77 + 60)C =137C. Hence the average temperature of the iron sample is$T = 137 + 77 C / 2 = 107 C = 380 K.<We compare the computed value of k with the tabulated value (see above) at 380 K and note the goodagreement.(c) The principal advantage of having two identical samples is the assurance that all the electricalpower dissipated in the heater will appear as equivalent heat flows through the samples. With onlyone sample, heat can flow from the backside of the heater even though insulated.Heat leakage out the lateral surfaces of the cylindrically shaped samples will become significant whenthe sample thermal conductivity is comparable to that of the insulating material. Hence, the method issuitable for metallics, but must be used with caution on nonmetallic materials.For any combination of materials in the upper and lower position, we expect T1 = T2. However, ifthe insulation were improperly applied along the lateral surfaces, it is possible that heat leakage willoccur, causing T1 T2.PROBLEM 2.18KNOWN: Comparative method for measuring thermal conductivity involving two identical samplesstacked with a reference material.FIND: (a) Thermal conductivity of test material and associated temperature, (b) Conditions forwhich Tt,1 Tt,2SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer through samplesand reference material, (3) Negligible thermal contact resistance between materials.%PROPERTIES: Table A.2, Armco iron T = 350 K : k r = 69.2 W / m K.ANALYSIS: (a) Recognizing that the heat rate through the samples and reference material, all of thesame diameter, is the same, it follows from Fouriers law thatktTt,1xkt = kr= krTt,2Tr= ktxxTr2.49 C= 69.2 W / m K= 519 W / m K..Tt3.32 CWe should assign this value a temperature of 350 K.<<(b) If the test samples are identical in every respect, Tt,1 Tt,2 if the thermal conductivity is highlydependent upon temperature. Also, if there is heat leakage out the lateral surface, we can expectTt,2 < Tt,1. Leakage could be influential, if the thermal conductivity of the test material were lessthan an order of magnitude larger than that of the insulating material.PROBLEM 2.19KNOWN: Identical samples of prescribed diameter, length and density initially at a uniformtemperature Ti, sandwich an electric heater which provides a uniform heat flux q for a period ofotime to. Conditions shortly after energizing and a long time after de-energizing heater areprescribed.FIND: Specific heat and thermal conductivity of the test sample material. From these properties,identify type of material using Table A.1 or A.2.SCHEMATIC:ASSUMPTIONS: (1) One dimensional heat transfer in samples, (2) Constant properties, (3)Negligible heat loss through insulation, (4) Negligible heater mass.ANALYSIS: Consider a control volume about the samplesand heater, and apply conservation of energy over the timeinterval from t = 0 to E in E out = E = E f E i16Pt o 0 = Mc p T Tiwhere energy inflow is prescribed by the Case A power condition and the final temperature Tf byCase B. Solving for cp,cp =Pt o15 W 120 s=3M T Ti2 3965 kg / m 0.0602 / 4 m2 0.010 m 33.50 - 23.00 C1627<c p = 765 J / kg K2where M = V = 2(D /4)L is the mass of both samples. For Case A, the transient thermal responseof the heater is given byContinued ..PROBLEM 2.19 (Cont.) t "#1/ 2To 1 t 6 Ti = 2q o!cp k #$2t 2q o "k=#c p ! To 1 t 6 Ti $30 sk= 3965 kg / m3 765 J / kg K 2 2653 W / m "#!124.57 - 23.006 C $22= 36.0 W / m K<whereq =oPP15 W=== 2653 W / m2 .2222A s 2 D / 42 0.060 / 4 m4949With the following properties now known, = 3965 kg/m3cp = 765 J/kgKk = 36 W/mKentries in Table A.1 are scanned to determine whether these values are typical of a metallic material.Consider the following,metallics with low generally have higher thermal conductivities,specific heats of both types of materials are of similar magnitude,the low k value of the sample is typical of poor metallic conductors which generally havemuch higher specific heats,more than likely, the material is nonmetallic.From Table A.2, the second entry, polycrystalline aluminum oxide, has properties at 300 Kcorresponding to those found for the samples.<PROBLEM 2.20KNOWN: Temperature distribution, T(x,y,z), within an infinite, homogeneous body at a giveninstant of time.FIND: Regions where the temperature changes with time.SCHEMATIC:ASSUMPTIONS: (1) Constant properties of infinite medium and (2) No internal heat generation.ANALYSIS: The temperature distribution throughout the medium, at any instant of time, mustsatisfy the heat equation. For the three-dimensional cartesian coordinate system, with constantproperties and no internal heat generation, the heat equation, Eq. 2.15, has the form 2Tx2+ 2Ty2+ 2Tz2=1 T. t(1)If T(x,y,z) satisfies this relation, conservation of energy is satisfied at every point in the medium.Substituting T(x,y,z) into the Eq. (1), first find the gradients, T/x, T/y, and T/z.1616161 T2x - y +2 z + 2y =4 y - x + 2z +.xyz tPerforming the differentiations,24+2 =1 T. tHence,T=0twhich implies that, at the prescribed instant, the temperature is everywhere independent of time.COMMENTS: Since we do not know the initial and boundary conditions, we cannot determine thetemperature distribution, T(x,y,z), at any future time. We can only determine that, for this specialinstant of time, the temperature will not change.PROBLEM 2.21KNOWN: Diameter D, thickness L and initial temperature Ti of pan. Heat rate from stove to bottomof pan. Convection coefficient h and variation of water temperature T(t) during Stage 1.Temperature TL of pan surface in contact with water during Stage 2.FIND: Form of heat equation and boundary conditions associated with the two stages.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in pan bottom, (2) Heat transfer from stove isuniformly distributed over surface of pan in contact with the stove, (3) Constant properties.ANALYSIS:Stage 1 2THeat Equation:x 2=1 T tInitial Condition:kqoT= q =ox x = 0 D2 / 4kBoundary Conditions:T= h T ( L, t ) T ( t )x x = L()T ( x, 0 ) = TiStage 2Heat Equation:Boundary Conditions:d 2Tdx 2k=0dT= qodx x = 0T ( L ) = TLCOMMENTS: Stage 1 is a transient process for which T(t) must be determined separately. As afirst approximation, it could be estimated by neglecting changes in thermal energy storage by the panbottom and assuming that all of the heat transferred from the stove acted to increase thermal energystorage within the water. Hence, with q Mcp d T/dt, where M and cp are the mass and specificheat of the water in the pan, T(t) (q/Mcp) t.PROBLEM 2.22KNOWN: Steady-state temperature distribution in a cylindrical rod having uniform heat generationof q1 = 5 107 W / m3 .FIND: (a) Steady-state centerline and surface heat transfer rates per unit length, q . (b) Initial timerrate of change of the centerline and surface temperatures in response to a change in the generation ratefrom q1 to q 2 = 108 W / m3 .SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the r direction, (2) Uniform generation, and(3) Steady-state for q1 = 5 107 W / m3 .ANALYSIS: (a) From the rate equations for cylindrical coordinates,q = krTrq = -kA rT.rHence,1 6 Trq r = k 2rLorq = 2krrTrwhere T/r may be evaluated from the prescribed temperature distribution, T(r).At r = 0, the gradient is (T/r) = 0. Hence, from Eq. (1) the heat rate is16<q 0 = 0.rAt r = ro, the temperature gradient is"#$"#$!"#1 6$4916TK= 2 4.167 105 2 ro = 2 4.167 105 0.025m r r=rmoT= 0.208 105 K / m. r r=roContinued ..PROBLEM 2.22(Cont.)Hence, the heat rate at the outer surface (r = ro) per unit length is161q 1 ro 6 = 0.980 105 W / m.r6q ro = 2 30 W / m K 0.025m 0.208 105 K / mr<(b) Transient (time-dependent) conditions will exist when the generation is changed, and for theprescribed assumptions, the temperature is determined by the following form of the heat equation, Eq.2.20!"#$TT1kr+ q 2 = c prtr rHence!!"#$"#$TT1 1kr=+ q2 . t c p r rrHowever, initially (at t = 0), the temperature distribution is given by the prescribed form, T(r) = 800 524.16710 r , and!"#$4T1kkrr -8.334 105 r=rr rr r=4k16.668 105 rr99= 30 W / m K -16.668 105 K / m216= 5 107 W / m3 the original q = q1 .Hence, everywhere in the wall,T1=5 107 + 108 W / m33 t 1100 kg / m 800 J / kg KorT= 56.82 K / s.t<COMMENTS: (1) The value of (T/t) will decrease with increasing time, until a new steady-statecondition is reached and once again (T/t) = 0.(2) By applying the energy conservation requirement, Eq. 1.11a, to a unit length of the rod for the16 1 6492 in gensteady-state condition, E E + E = 0. Hence q 0 q ro = q1 ro .outrrPROBLEM 2.23KNOWN: Temperature distribution in a one-dimensional wall with prescribed thickness and thermalconductivity.FIND: (a) The heat generation rate, q, in the wall, (b) Heat fluxes at the wall faces and relation toq.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) Constantproperties.ANALYSIS: (a) The appropriate form of the heat equation for steady-state, one-dimensionalconditions with constant properties is Eq. 2.15 re-written as "#!$Substituting the prescribed temperature distribution,d ddq = -k4a + bx2 9"#$ = k dx 2bx = 2bkdx ! dxq = -24-2000 C / m2 9 50 W / m K = 2.0 105 W / m3 .(b) The heat fluxes at the wall faces can be evaluated from Fouriers law,dT "q 1 x6 = kxdx # x$.q = -kd dTdx dx<Using the temperature distribution T(x) to evaluate the gradient, find16q x = kxda + bx 2 = 2 kbx.dxThe fluxes at x = 0 and x = L are then16q 1 L6 = 2 kbL = -2 50W / m K4-2000 C / m2 9 0.050mxq 1 L6 = 10,000 W / m2 .xq 0 = 0x<COMMENTS: From an overall energy balance on the wall, it follows that, for a unit area,E in E out + E g = 016 1616 1 6q 0 q L + qL = 0xxq L q 0 10,000 W / m2 0xq= x== 2.0 105 W / m3.L0.050m<PROBLEM 2.24KNOWN: Wall thickness, thermal conductivity, temperature distribution, and fluid temperature.FIND: (a) Surface heat rates and rate of change of wall energy storage per unit area, and (b)Convection coefficient.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant k.ANALYSIS: (a) From Fouriers law,q = kx$T= 200 60x kxq = q = 200inx=0CW1= 200 W / m2mm K$q = q = 200 60 0.3 C / m 1 W / m K = 182 W / m2 .outx=L<<Applying an energy balance to a control volume about the wall, Eq. 1.11a, E E = E stinoutE = q q = 18 W / m2 .stinout<(b) Applying a surface energy balance at x = L,$q = h T L Toutq 182 W / m2outh==T L T142.7 -100 C$$<h = 4.3 W / m2 K.COMMENTS: (1) From the heat equation,22(T/t) = (k/cp) T/x = 60(k/cp),it follows that the temperature is increasing with time at every point in the wall.(2) The value of h is small and is typical of free convection in a gas.PROBLEM 2.25KNOWN: Analytical expression for the steady-state temperature distribution of a plane wallexperiencing uniform volumetric heat generation q while convection occurs at both of its surfaces.FIND: (a) Sketch the temperature distribution, T(x), and identify significant physical features, (b)Determine q , (c) Determine the surface heat fluxes, q ( L ) and q ( + L ) ; how are these fluxesxxrelated to the generation rate; (d) Calculate the convection coefficients at the surfaces x = L and x =+L, (e) Obtain an expression for the heat flux distribution, q ( x ) ; explain significant features of thexdistribution; (f) If the source of heat generation is suddenly deactivated ( q = 0), what is the rate ofchange of energy stored at this instant; (g) Determine the temperature that the wall will reacheventually with q = 0; determine the energy that must be removed by the fluid per unit area of the wallto reach this state.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform volumetric heat generation, (3) Constantproperties.ANALYSIS: (a) Using the analytical expression in the Workspace of IHT, the temperaturedistribution appears as shown below. The significant features include (1) parabolic shape, (2)maximum does not occur at the mid-plane, T(-5.25 mm) = 83.3C, (3) the gradient at the x = +Lsurface is greater than at x = -L. Find also that T(-L) = 78.2C and T(+L) = 69.8C for use in part (d).Temperature distribution90Temperature, T(x) (C)85807570-20-1001020x-coordinate, x (mm)(b) Substituting the temperature distribution expression into the appropriate form of the heat diffusionequation, Eq. 2.15, the rate of volumetric heat generation can be determined.d dT q+ =0dx dx kwhereT ( x ) = a + bx + cx 2dqq(0 + b + 2cx ) + = (0 + 2c ) + = 0dxkkContinued ..PROBLEM 2.25 (Cont.))(q = 2ck = 2 2 104C / m 2 5 W / m K = 2 105 W / m3<(c) The heat fluxes at the two boundaries can be determined using Fouriers law and the temperaturedistribution expression.q ( x ) = kxdTdxT ( x ) = a + bx + cx 2whereq ( L ) = k [0 + b + 2cx ]xx= L= [b 2cL] k()q ( L ) = 210C / m 2 2 104C / m 2 0.020m 5 W / m K = 2950 W / m 2x<q ( + L ) = ( b + 2cL ) k = +5050 W / m 2x<From an overall energy balance on the wall as shown in the sketch below, E in E out + E gen = 0,?+ q ( L ) q ( + L ) + 2qL = 0xxor 2950 W / m 2 5050 W / m 2 + 8000 W / m 2 = 0where 2qL = 2 2 105 W / m 3 0.020 m = 8000 W / m 2 , so the equality is satisfied(d) The convection coefficients, hl and hr, for the left- and right-hand boundaries (x = -L and x= +L,respectively), can be determined from the convection heat fluxes that are equal to the conductionfluxes at the boundaries. See the surface energy balances in the sketch above. See also part (a) resultfor T(-L) and T(+L).q = q ( L )cv,xh l T T ( L ) = h l [20 78.2] K = 2950 W / m 2h l = 51W / m 2 K<q = q ( + L )cv,rxh r T ( + L ) T = h r [69.8 20] K = +5050 W / m 2h r = 101W / m 2 K<(e) The expression for the heat flux distribution can be obtained from Fouriers law with thetemperature distributionq ( x ) = kxdT= k [0 + b + 2cx ]dx()q ( x ) = 5 W / m K 210C / m + 2 2 104C / m 2 x = 1050 + 2 105 xxContinued ..<PROBLEM 2.25 (Cont.)The distribution is linear with the x-coordinate. The maximum temperature will occur at the locationwhere q ( x max ) = 0,xx max = 1050 W / m 22 105 W / m3= 5.25 103 m = 5.25 mm<(f) If the source of the heat generation is suddenly deactivated so that q = 0, the appropriate form ofthe heat diffusion equation for the ensuing transient conduction isk T T = cpx x t2At the instant this occurs, the temperature distribution is still T(x) = a + bx + cx . The right-hand termrepresents the rate of energy storage per unit volume,E = kstx[0 + b + 2cx ] = k [0 + 2c] = 5 W / m K 2(2 104C / m2 ) = 2 105 W / m3 <(g) With no heat generation, the wall will eventually (t ) come to equilibrium with the fluid,T(x,) = T = 20C. To determine the energy that must be removed from the wall to reach this state,apply the conservation of energy requirement over an interval basis, Eq. 1.11b. The initial state isthat corresponding to the steady-state temperature distribution, Ti, and the final state has Tf = 20C.Weve used T as the reference condition for the energy terms.E E = E = E Einoutstfi E = cp 2L ( Tf T ) cp outE = cp outwithE = 0.in+L(T T ) dxL i+L+L a + bx + cx 2 T dx = cp ax + bx 2 / 2 + cx 3 / 3 T x LL E = cp 2aL + 0 + 2cx 3 / 3 2T L out(E = 2600 kg / m3 800 J / kg K 2 82C 0.020m + 2 2 104C / m2out)(0.020m )3 / 3 2 (20C ) 0.020m E = 4.94 106 J / m 2out<COMMENTS: (1) In part (a), note that the temperature gradient is larger at x = + L than at x= - L. This is consistent with the results of part (c) in which the conduction heat fluxes areevaluated.Continued ..PROBLEM 2.25 (Cont.)(2) In evaluating the conduction heat fluxes, q ( x ) , it is important to recognize that this fluxxis in the positive x-direction. See how this convention is used in formulating the energybalance in part (c).(3) It is good practice to represent energy balances with a schematic, clearly defining thesystem or surface, showing the CV or CS with dashed lines, and labeling the processes.Review again the features in the schematics for the energy balances of parts (c & d).(4) Re-writing the heat diffusion equation introduced in part (b) asd dT k+q =0dx dx recognize that the term in parenthesis is the heat flux. From the differential equation, notethat if the differential of this term is a constant ( q / k ) , then the term must be a linear functionof the x-coordinate. This agrees with the analysis of part (e).(5) In part (f), we evaluated Est , the rate of energy change stored in the wall at the instant thevolumetric heat generation was deactivated. Did you notice that Est = 2 105 W / m3 is thesame value of the deactivated q ? How do you explain this?PROBLEM 2.26KNOWN: Steady-state conduction with uniform internal energy generation in a plane wall;temperature distribution has quadratic form. Surface at x=0 is prescribed and boundary at x = L isinsulated.FIND: (a) Calculate the internal energy generation rate, q , by applying an overall energy balance tothe wall, (b) Determine the coefficients a, b, and c, by applying the boundary conditions to theprescribed form of the temperature distribution; plot the temperature distribution and label as Case 1,(c) Determine new values for a, b, and c for conditions when the convection coefficient is halved, andthe generation rate remains unchanged; plot the temperature distribution and label as Case 2; (d)Determine new values for a, b, and c for conditions when the generation rate is doubled, and the2convection coefficient remains unchanged (h = 500 W/m K); plot the temperature distribution andlabel as Case 3.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with constantproperties and uniform internal generation, and (3) Boundary at x = L is adiabatic.ANALYSIS: (a) The internal energy generation rate can be calculated from an overall energy balanceon the wall as shown in the schematic below.E E + E = 0inoutgen Ein = qconvwhereh ( T To ) + q L = 0(1)q = h ( T To ) / L = 500 W / m 2 K ( 20 120 ) C / 0.050 m = 1.0 106 W / m3<2(b) The coefficients of the temperature distribution, T(x) = a + bx + cx , can be evaluated by applyingthe boundary conditions at x = 0 and x = L. See Table 2.1 for representation of the boundaryconditions, and the schematic above for the relevant surface energy balances.Boundary condition at x = 0, convection surface conditionE E = q inoutconv q ( 0 ) = 0xwhereq (0 ) = kxdT dx x = 0h ( T To ) k (0 + b + 2cx )x = 0 = 0Continued ..PROBLEM 2.26 (Cont.)b = h (T To ) / k = 500 W / m 2 K ( 20 120 ) C / 5 W / m K = 1.0 104 K / m<Boundary condition at x = L, adiabatic or insulated surfaceEin E out = q ( L ) = 0xq ( L ) = kxwheredT dx x = Lk [0 + b + 2cx ]x = L = 0(3)c = b / 2L = 1.0 104 K / m / ( 2 0.050m ) = 1.0 105 K / m 2<Since the surface temperature at x = 0 is known, T(0) = To = 120C, findT ( 0 ) = 120C = a + b 0 + c 0ora = 120C(4)<Using the foregoing coefficients with the expression for T(x) in the Workspace of IHT, thetemperature distribution can be determined and is plotted as Case 1 in the graph below.2(c) Consider Case 2 when the convection coefficient is halved, h2 = h/2 = 250 W/m K, q = 1 1063W/m and other parameters remain unchanged except that To 120C. We can determine a, b, and cfor the temperature distribution expression by repeating the analyses of parts (a) and (b).Overall energy balance on the wall, see Eqs. (1,4)a = To = q L / h + T = 1 106 W / m3 0.050m / 250 W / m 2 K + 20C = 220C<Surface energy balance at x = 0, see Eq. (2)b = h (T To ) / k = 250 W / m 2 K ( 20 220 ) C / 5 W / m K = 1.0 104 K / m<Surface energy balance at x = L, see Eq. (3)c = b / 2L = 1.0 104 K / m / ( 2 0.050m ) = 1.0 105 K / m 2<The new temperature distribution, T2 (x), is plotted as Case 2 below.(d) Consider Case 3 when the internal energy volumetric generation rate is doubled,632q 3 = 2q = 2 10 W / m , h = 500 W/m K, and other parameters remain unchanged except thatTo 120C. Following the same analysis as part (c), the coefficients for the new temperaturedistribution, T (x), area = 220Cb = 2 104 K / mc = 2 105 K / m2<and the distribution is plotted as Case 3 below.Continued ..PROBLEM 2.26 (Cont.)800700Te m p e ra tu re , T (C )60050040030020010005101520253035404550W a ll p o s itio n , x (m m )1 . h = 5 0 0 W /m ^2 .K , q d o t = 1 e 6 W /m ^32 . h = 2 5 0 W /m ^2 .K , q d o t = 1 e 6 W /m ^33 . h = 5 0 0 W /m ^2 .K , q d o t = 2 e 6 W /m ^3COMMENTS: Note the following features in the family of temperature distributions plotted above.The temperature gradients at x = L are zero since the boundary is insulated (adiabatic) for all cases.The shapes of the distributions are all quadratic, with the maximum temperatures at the insulatedboundary.By halving the convection coefficient for Case 2, we expect the surface temperature To to increaserelative to the Case 1 value, since the same heat flux is removed from the wall ( qL ) but theconvection resistance has increased.By doubling the generation rate for Case 3, we expect the surface temperature To to increase relativeto the Case 1 value, since double the amount of heat flux is removed from the wall ( 2qL ) .Can you explain why To is the same for Cases 2 and 3, yet the insulated boundary temperatures arequite different? Can you explain the relative magnitudes of T(L) for the three cases?PROBLEM 2.27KNOWN: Temperature distribution and distribution of heat generation in central layer of a solarpond.FIND: (a) Heat fluxes at lower and upper surfaces of the central layer, (b) Whether conditions aresteady or transient, (c) Rate of thermal energy generation for the entire central layer.SCHEMATIC:ASSUMPTIONS: (1) Central layer is stagnant, (2) One-dimensional conduction, (3) ConstantpropertiesANALYSIS: (a) The desired fluxes correspond to conduction fluxes in the central layer at the lowerand upper surfaces. A general form for the conduction flux is"#!$Hence, A e-aL + B"#q = q cond 1 x=L6 = kl! ka$q cond = kTA -axe +B .= kxka16(b) Conditions are steady if T/t = 0. Applying the heat equation, 2Tx2+q 1 T=k t-!"#$Aq = q +B .ucond x=0 = kka<A -ax A -ax 1 Te+e= tkkHence conditions are steady sinceT/t = 0<(for all 0 L).(c) For the central layer, the energy generation isIILLE = q dx = A e-ax dxg0AE g = e -axa0L0=4949A -aLAe 1 =1 e -aL .aaAlternatively, from an overall energy balance,4 1 69 4 A + B"# k A e-aL + B"# = A 41 e-aL 9.Eg = k! ka $ ! ka$aq q1 + E g = 0 2<1 69gE = q1 q = q 2cond x=0 q cond x=LCOMMENTS: Conduction is in the negative x-direction, necessitating use of minus signs in theabove energy balance.PROBLEM 2.28KNOWN: Temperature distribution in a semi-transparent medium subjected to radiative flux.16FIND: (a) Expressions for the heat flux at the front and rear surfaces, (b) Heat generation rate q x ,(c) Expression for absorbed radiation per unit surface area in terms of A, a, B, C, L, and k.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in medium, (3)Constant properties, (4) All laser irradiation is absorbed and can be characterized by an internalvolumetric heat generation term q x .16ANALYSIS: (a) Knowing the temperature distribution, the surface heat fluxes are found usingFouriers law, dT "# = k - A 1a6e-ax + B"#! dx $ ! ka 2$ A " A + kB"#Front Surface, x=0:q 106 = k + 1 + B# = x! ka $ ! a $A" A"Rear Surface, x=L:q 1 L6 = k + e-aL + B# = e-aL + kB#.x! ka$ !a$q = kx<<(b) The heat diffusion equation for the medium isd dT or q = -k dx dx d A -ax"q1 x6 = k+ e + B# = Ae-ax .dx ! ka$d dTq+ =0dx dxk<(c) Performing an energy balance on the medium,E in E out + E g = 0recognize that E g represents the absorbed irradiation. On a unit area basis16 1 649Ag E = E in + E out = q 0 + q L = + 1 e-aL .xxagAlternatively, evaluate E by integration over the volume of the medium,I16I49LALLAE = q x dx = Ae -ax dx = - e-ax =1 e-aL .g000aa<PROBLEM 2.29KNOWN: Steady-state temperature distribution in a one-dimensional wall of thermal32conductivity, T(x) = Ax + Bx + Cx + D.FIND: Expressions for the heat generation rate in the wall and the heat fluxes at the two wallfaces (x = 0,L).ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3)Homogeneous medium.ANALYSIS: The appropriate form of the heat diffusion equation for these conditions isd 2Tq+ =0dx 2 korq = -kd 2Tdx 2.Hence, the generation rate isq = -k "#!$d dTd= k3Ax2 + 2 Bx + C + 0dx dxdx<q = -k 6Ax + 2Bwhich is linear with the coordinate x. The heat fluxes at the wall faces can be evaluated fromFouriers law,q = kxdT= k 3Ax 2 + 2Bx + Cdxusing the expression for the temperature gradient derived above. Hence, the heat fluxes are:Surface x=0:16<q 0 = kCxSurface x=L:16<q L = k 3AL2 + 2BL + C .xCOMMENTS: (1) From an overall energy balance on the wall, findE E + E = 0inoutg16 1 61 61 6 q 0 q L + E g = kC k 3AL2 + 2 BL + C + E g = 0xx E g = 3AkL2 2 BkL.From integration of the volumetric heat rate, we can also find E asgI16ILLLE = q x dx = -k 6Ax + 2B dx = -k 3Ax 2 + 2 Bxg00E = 3AkL 2 BkL.g20PROBLEM 2.30KNOWN: Plane wall with no internal energy generation.FIND: Determine whether the prescribed temperature distribution is possible; explain yourreasoning. With the temperatures T(0) = 0C and T = 20C fixed, compute and plot thetemperature T(L) as a function of the convection coefficient for the range 10 h 100 W/m2K.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal energy generation, (3) Constantproperties, (4) No radiation exchange at the surface x = L, and (5) Steady-state conditions.ANALYSIS: (a) Is the prescribed temperature distribution possible? If so, the energy balance at thesurface x = L as shown above in the Schematic, must be satisfied.E in E out ? = ? 0q ( L ) q ? = ? 0(1,2)xcvwhere the conduction and convection heat fluxes are, respectively,T ( L ) T (0 )dT q ( L ) = k= k= 4.5 W m K (120 0 ) C 0.18 m = 3000 W m 2xdx x = LLq = h [T ( L ) T ] = 30 W m 2 K (120 20 ) C = 3000 W m2cvSubstituting the heat flux values into Eq. (2), find (-3000) - (3000) 0 and therefore, the temperaturedistribution is not possible.(b) With T(0) = 0C and T = 20C, the temperature at the surface x = L, T(L), can be determinedfrom an overall energy balance on the wall as shown above in the Schematic,T ( L ) T (0 )k h [T ( L ) T ] = 0E in E out = 0q (0) q = 0xcvL4.5 W m K T ( L ) 0 C 0.18 m 30 W m 2 K T ( L ) 20 C = 0<T(L) = 10.9C20Surface temperature, T(L) (C)Using this same analysis, T(L) as a function ofthe convection coefficient can be determinedand plotted. We dont expect T(L) to belinearly dependent upon h. Note that as hincreases to larger values, T(L) approachesT . To what value will T(L) approach as hdecreases?16128400204060Convection cofficient, h (W/m^2.K)80100PROBLEM 2.31KNOWN: Coal pile of prescribed depth experiencing uniform volumetric generation withconvection, absorbed irradiation and emission on its upper surface.FIND: (a) The appropriate form of the heat diffusion equation (HDE) and whether the prescribedtemperature distribution satisfies this HDE; conditions at the bottom of the pile, x = 0; sketch of thetemperature distribution with labeling of key features; (b) Expression for the conduction heat rate atthe location x = L; expression for the surface temperature Ts based upon a surface energy balance at x= L; evaluate Ts and T(0) for the prescribed conditions; (c) Based upon typical daily averages for GSand h, compute and plot Ts and T(0) for (1) h = 5 W/m2K with 50 GS 500 W/m2, (2) GS = 400W/m2 with 5 h 50 W/m2K.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Uniform volumetric heat generation, (3)Constant properties, (4) Negligible irradiation from the surroundings, and (5) Steady-state conditions.PROPERTIES: Table A.3, Coal (300K): k = 0.26 W/m.KANALYSIS: (a) For one-dimensional, steady-state conduction with uniform volumetric heatgeneration and constant properties the heat diffusion equation (HDE) follows from Eq. 2.16,d dT q+ =0dx dx k(1)Substituting the temperature distribution into the HDE, Eq. (1),qL2 x 2 d qL2 2x q1 T ( x ) = Ts +0 + 0 2 + ? = ?022kLdx 2k L k<(2,3)<we find that it does indeed satisfy the HDE for all values of x.From Eq. (2), note that the temperature distribution must be quadratic, with maximum value at x = 0.At x = 0, the heat flux isq ( 0 ) = kx qL2 dT 2x = k 0 +=0 0 2 dx x = 02k L x =0so that the gradient at x = 0 is zero. Hence, thebottom is insulated.(b) From an overall energy balance on the pile, the conduction heat flux at the surface must beq ( L ) = E = qLxg<Continued...PROBLEM 2.31 (Cont.)From a surface energy balance per unit area shown in the Schematic above,q ( L ) q + GS,abs E = 0xcvE in E out + E g = 0qL h ( Ts T ) + 0.95GS Ts4 = 0(4)20 W m 1m 5 W m K ( Ts 298 K ) + 0.95 400 W m 0.95 5.67 103228244W m K Ts = 0<Ts = 295.7 K =22.7CFrom Eq. (2) with x = 0, find30 W m 2 (1m )qL2T ( 0 ) = Ts += 22.7 C += 61.1 C2k2 0.26 W m K2(5)<where the thermal conductivity for coal was obtained from Table A.3.(c) Two plots are generated using Eq. (4) and (5) for Ts and T(0), respectively; (1) with h = 5 W/m2Kfor 50 GS 500 W/m2 and (2) with GS = 400 W/m2 for 5 h 50 W/m2K.Solar irradiation, GS = 400 W/m^280Convection coefficient, h = 5 W/m^2.KTemperature, Ts or T(0) (C)Temperature, Ts or T(0) (C)80604020604020001020304050Convection coefficient, h (W/m^2.K)-200100200300400500T0_CTs_CSolar irradiation, GS (W/m^2)T0_CTs_CFrom the T vs. h plot with GS = 400 W/m2, note that the convection coefficient does not have a majorinfluence on the surface or bottom coal pile temperatures. From the T vs. GS plot with h = 5 W/m2K,note that the solar irradiation has a very significant effect on the temperatures. The fact that Ts is lessthan the ambient air temperature, T , and, in the case of very low values of GS, below freezing, is aconsequence of the large magnitude of the emissive power E.COMMENTS: In our analysis we ignored irradiation from the sky, an environmental radiation effect4youll consider in Chapter 12. Treated as large isothermal surroundings, Gsky = Tsky where Tsky = 30C for very clear conditions and nearly air temperature for cloudy conditions. For low GSconditions we should consider Gsky, the effect of which will be to predict higher values for Ts andT(0).PROBLEM 2.32KNOWN: Cylindrical system with negligible temperature variation in the r,z directions.FIND: (a) Heat equation beginning with a properly defined control volume, (b) Temperaturedistribution T() for steady-state conditions with no internal heat generation and constant properties,(c) Heat rate for Part (b) conditions.SCHEMATIC:ASSUMPTIONS: (1) T is independent of r,z, (2) r = (ro - ri) << ri.ANALYSIS: (a) Define the control volume as V = ridrL where L is length normal to page.Apply the conservation of energy requirement, Eq. 1.11a,E in E out + E g = E stwhereq q +d + qV = Vc16 rTiq = k r LTtq +d = q +(1,2)38q d . (3,4)Eqs. (3) and (4) follow from Fouriers law, Eq. 2.1, and from Eq. 2.7, respectively. Combining Eqs.(3) and (4) with Eq. (2) and canceling like terms, find1TT+ q = c.ktri2 (5) <Since temperature is independent of r and z, this form agrees with Eq. 2.20.(b) For steady-state conditions with q = 0, the heat equation, (5), becomes "#!$ddT= 0.kd d(6)With constant properties, it follows that dT/d is constant which implies T() is linear in . That is,1dT T2 T11== + T2 T1d 2 16or16T = T1 +161T T .21(7,8) <(c) The heat rate for the conditions of Part (b) follows from Fouriers law, Eq. (3), using thetemperature gradient of Eq. (7). That is,116 r1i !+ 1T2 T16"#$ = k ! roi ri "#$L1T2 T16.rq = k r L(9) <COMMENTS: Note the expression for the temperature gradient in Fouriers law, Eq. (3), isT/ri not T/. For the conditions of Parts (b) and (c), note that q is independent of ;this is first indicated by Eq. (6) and confirmed by Eq. (9).PROBLEM 2.33KNOWN: Heat diffusion with internal heat generation for one-dimensional cylindrical,radial coordinate system.FIND: Heat diffusion equation.SCHEMATIC:ASSUMPTIONS: (1) Homogeneous medium.ANALYSIS: Control volume has volume, V = A r dr = 2r dr 1, with unit thicknessnormal to page. Using the conservation of energy requirement, Eq. 1.11a,E in E out + E gen = E stq r q r +dr + qV = Vc pT.tFouriers law, Eq. 2.1, for this one-dimensional coordinate system isq r = kA rTT= k 2r 1 .rrAt the outer surface, r+dr, the conduction rate isq r+dr = q r +T(q r ) dr=q r + k 2 r dr.r r rHence, the energy balance becomes T Tq r q r + k2 rdr + q 2 rdr= 2 rdr cp r r tDividing by the factor 2r dr, we obtainT1 T kr r + q= cp t .r r<COMMENTS: (1) Note how the result compares with Eq. 2.20 when the terms for the ,zcoordinates are eliminated. (2) Recognize that we did not require q and k to be independentof r.PROBLEM 2.34KNOWN: Heat diffusion with internal heat generation for one-dimensional spherical, radialcoordinate system.FIND: Heat diffusion equation.SCHEMATIC:ASSUMPTIONS: (1) Homogeneous medium.2ANALYSIS: Control volume has the volume, V = Ar dr = 4r dr. Using the conservationof energy requirement, Eq. 1.11a,E in E out + E gen = E stq r q r +dr + qV = Vc pT.tFouriers law, Eq. 2.1, for this coordinate system has the formq r = kA rTT.= k 4r 2 rrAt the outer surface, r+dr, the conduction rate isq r+dr = q r +T(q r ) dr = q r + k 4 r 2 dr.r r rHence, the energy balance becomesT2 T 22q r q r + k 4 r r dr + q 4 r dr= 4 r dr cp t . rDividing by the factor 4r 2 dr, we obtain1 2 TTkr.+ q= cp rtr2 r <COMMENTS: (1) Note how the result compares with Eq. 2.23 when the terms for the ,directions are eliminated.(2) Recognize that we did not require q and k to be independent of the coordinate r.PROBLEM 2.35KNOWN: Three-dimensional system described by cylindrical coordinates (r,,z) experiences transient conduction and internal heat generation.FIND: Heat diffusion equation.SCHEMATIC: See also Fig. 2.9.ASSUMPTIONS: (1) Homogeneous medium.ANALYSIS: Consider the differential control volume identified above having a volumegiven as V = drrddz. From the conservation of energy requirement,(1)q r q r +dr + q q +d + q z q z+dz + E g = E st .The generation and storage terms, both representing volumetric phenomena, are1E g = qV = q dr rd dz616E g = Vc T / t = dr rd dz c T / t.Using a Taylor series expansion, we can writeq r +dr = q r +q r dr, q +d = q +q d ,r1638q z+dz = q z +(2,3)16q z dz.z(4,5,6)Using Fouriers law, the expressions for the conduction heat rates are16q = kA T / r = k 1dr dz6 T / rq z = kA z T / z = k1dr rd 6 T / z.q r = kA r T / r = k rd dz T / r(7)(8)(9)Note from the above, right schematic that the gradient in the -direction is T/r and notT/. Substituting Eqs. (2), (3) and (4), (5), (6) into Eq. (1),16163816T.q r dr q d q z dz + q dr rd dz = dr rd dz crztSubstituting Eqs. (7), (8) and (9) for the conduction rates, find 1 6 "#1!!$T+ q dr rd dz = 1dr rd dz6c.t1!6 "#$(10)6 "#$TTT k rd dzdr k drdzd k dr rddzrrzzr(11)Dividing Eq. (11) by the volume of the CV, Eq. 2.20 is obtained.!"#$!"#$!"#$TTTT11kr+2k+k+ q = crz ztrrr <PROBLEM 2.36KNOWN: Three-dimensional system described by cylindrical coordinates (r,,) experiencestransient conduction and internal heat generation.FIND: Heat diffusion equation.SCHEMATIC: See Figure 2.10.ASSUMPTIONS: (1) Homogeneous medium.ANALYSIS: The differential control volume is V = drrsindrd, and the conduction terms areidentified in Figure 2.10. Conservation of energy requiresq r q r +dr + q q +d + q q +d + E g = E st .(1)The generation and storage terms, both representing volumetric phenomena, areTTE st = Vc= dr r sind rd c.ttE g = qV = q dr r sind rd(2,3)Using a Taylor series expansion, we can writeq r +dr = q r +16q r dr,rq +d = q +38q d ,q +d = q +16q d .(4,5,6)From Fouriers law, the conduction heat rates have the following forms.q r = kA r T / r = k r sind rd T / r(7)q = kA T / r sin = k dr rd T / r sin(8)q = kA T / r = k dr r sind T / r .(9)Substituting Eqs. (2), (3) and (4), (5), (6) into Eq. (1), the energy balance becomes161638Tq r dr q d q d + q dr r sind rd = dr r sind rd ctrSubstituting Eqs. (7), (8) and (9) for the conduction rates, find!!!"#$(10)"#$TTdr d k r sind rd k dr rdr sinr"#$TTd + q dr r sind rd = dr r sind rd c k dr r sindtr(11)Dividing Eq. (11) by the volume of the control volume, V, Eq. 2.23 is obtained.!"#$!"#$!"#$TTTT111kr 2kk sin.+2 2+2+ q = c2 rrtrr sin r sin <COMMENTS: Note how the temperature gradients in Eqs. (7) - (9) are formulated. The numeratoris always T while the denominator is the dimension of the control volume in the specified coordinatedirection.PROBLEM 2.37KNOWN: Temperature distribution in steam pipe insulation.FIND: Whether conditions are steady-state or transient. Manner in which heat flux and heat ratevary with radius.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties.ANALYSIS: From Equation 2.20, the heat equation reduces to1T1 Tr=.rr r tSubstituting for T(r), 1 T 1 Cr 1 = 0.= t rrr<Hence, steady-state conditions exist.From Equation 2.19, the radial component of the heat flux isq = krTC= k 1 .rr16Hence, q decreases with increasing r q 1/ r .rr<At any radial location, the heat rate isq r = 2rLq = 2kC1LrHence, qr is independent of r.<COMMENTS: The requirement that qr is invariant with r is consistent with the energy conservationrequirement. If qr is constant, the flux must vary inversely with the area perpendicular to the directionof heat flow. Hence, q varies inversely with r.rPROBLEM 2.38KNOWN: Inner and outer radii and surface temperatures of a long circular tube with internal energygeneration.FIND: Conditions for which a linear radial temperature distribution may be maintained.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant properties.ANALYSIS: For the assumed conditions, Eq. 2.20 reduces tok d dT r+q =0r dr dr If q = 0 or q = constant, it is clearly impossible to have a linear radial temperature distribution.However, we may use the heat equation to infer a special form of q (r) for which dT/dr is a constant (callit C1). It follows thatkd( r C1 ) + q = 0r drCkq= 1r<where C1 = (T2 - T1)/(r2 - r1). Hence, if the generation rate varies inversely with radial location, the radialtemperature distribution is linear.COMMENTS: Conditions for which q (1/r) would be unusual.PROBLEM 2.39KNOWN: Radii and thermal conductivity of conducting rod and cladding material. Volumetric rateof thermal energy generation in the rod. Convection conditions at outer surface.FIND: Heat equations and boundary conditions for rod and cladding.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3) Constantproperties.ANALYSIS: From Equation 2.20, the appropriate forms of the heat equation areConducting Rod:kr ddTr r +q = 0r drdtCladding:<ddTr c = 0.drdr<Appropriate boundary conditions are:<(a)dTr / dr|r =0 = 0(b)Tr ri = Tc ri(c)krdTrdT|ri = k c c |ridrdr<(d)kcdTc|r = h Tc ro Tdr o16<16 16<COMMENTS: Condition (a) corresponds to symmetry at the centerline, while the interfaceconditions at r = ri (b,c) correspond to requirements of thermal equilibrium and conservation ofenergy. Condition (d) results from conservation of energy at the outer surface.PROBLEM 2.40KNOWN: Steady-state temperature distribution for hollow cylindrical solid with volumetric heatgeneration.FIND: (a) Determine the inner radius of the cylinder, ri, (b) Obtain an expression for the volumetricrate of heat generation, q, (c) Determine the axial distribution of the heat flux at the outer surface,q ( ro , z ) , and the heat rate at this outer surface; is the heat rate in or out of the cylinder; (d)rDetermine the radial distribution of the heat flux at the end faces of the cylinder, q ( r, + z o ) andzq ( r, z o ) , and the corresponding heat rates; are the heat rates in or out of the cylinder; (e)zDetermine the relationship of the surface heat rates to the heat generation rate; is an overall energybalance satisfied?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction with constantproperties and volumetric heat generation.ANALYSIS: (a) Since the inner boundary, r = ri, is adiabatic, then q ( ri , z ) = 0. Hence thertemperature gradient in the r-direction must be zero.T = 0 + 2bri + c / ri + 0 = 0r ri1/ 2 cri = + 2b 1/ 212C= 2 150C / m 2 <= 0.2 m(b) To determine q, substitute the temperature distribution into the heat diffusion equation, Eq. 2.20,for two-dimensional (r,z), steady-state conduction1 T T qr + + = 0r r r z z k1q(r [0 + 2br + c / r + 0]) + z (0 + 0 + 0 + 2dz ) + k = 0r r1q[4br + 0] + 2d + = 0rkq = k [4b 2d ] = 16 W / m K 4 150C / m 2 2 300C / m2 ()q = 0 W / m3<(c) The heat flux and the heat rate at the outer surface, r = ro, may be calculated using Fouriers law.Note that the sign of the heat flux in the positive r-direction is negative, and hence the heat flow is intothe cylinder.()q ro, z = krT = k [0 + 2bro + c / ro + 0 ]r roContinued ..PROBLEM 2.40 (Cont.)22q ( ro, z ) = 16 W / m K 2 150C / m 1 m 12C / 1 m = 4608 W / mr()q r ( ro ) = A r q ro, zr<A r = 2 ro ( 2z o )whereq r ( ro ) = 4 1 m 2.5 m 4608 W / m = 144, 765 W<2(d) The heat fluxes and the heat rates at end faces, z = + zo and zo, may be calculated using Fourierslaw. The direction of the heat rate in or out of the end face is determined by the sign of the heat flux inthe positive z-direction.<At the upper end face, z = + zo: heat rate is out of the cylinderq ( r, + z o ) = kzT z zo= k [0 + 0 + 0 + 2dz o ](q ( r, + z o ) = 16 W / m K 2 300C / mzq z ( + z o ) = A z q ( r, + z o )z(q z ( + z o ) = 1 0.222)m) 2.5 m = +24, 000 W / mwhereA = (r2z2<222o ri)<<2 24, 000 W / m = +72, 382 WAt the lower end face, z = - zo: heat rate is out of the cylinderq ( r, z o ) = kzT z zo= k [0 + 0 + 0 + 2dz o ]q ( r, z o ) = 16 W / m K 2 ( 300C / m )( 2.5 m ) = 24, 000 W / mz22q z ( z o ) = 72, 382 W<<(e) The heat rates from the surfaces and the volumetric heat generation can be related through anoverall energy balance on the cylinder as shown in the sketch.E in E out + E gen = 0whereE gen = q = 0E in = q r ( ro ) = ( 144, 765 W ) = +144, 765 WE out = +q z ( z o ) q z ( z o ) = [72, 382 ( 72, 382 )] W = +144, 764 W<<The overall energy balance is satisfied.COMMENTS: When using Fouriers law, the heat flux q denotes the heat flux in the positive zzdirection. At a boundary, the sign of the numerical value will determine whether heat is flowing intoor out of the boundary.PROBLEM 2.41KNOWN: An electric cable with an insulating sleeve experiences convection with adjoining air andradiation exchange with large surroundings.FIND: (a) Verify that prescribed temperature distributions for the cable and insulating sleeve satisfytheir appropriate heat diffusion equations; sketch temperature distributions labeling key features; (b)Applying Fourier's law, verify the conduction heat rate expression for the sleeve, q , in terms of Ts,1rand Ts,2; apply a surface energy balance to the cable to obtain an alternative expression for q inrterms of q and r1; (c) Apply surface energy balance around the outer surface of the sleeve to obtain anexpression for which Ts,2 can be evaluated; (d) Determine Ts,1, Ts,2, and To for the specified geometryand operating conditions; and (e) Plot Ts,1, Ts,2, and To as a function of the outer radius for the range15.5 r2 20 mm.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Uniform volumetric heat generationin cable, (3) Negligible thermal contact resistance between the cable and sleeve, (4) Constantproperties in cable and sleeve, (5) Surroundings large compared to the sleeve, and (6) Steady-stateconditions.ANALYSIS: (a) The appropriate forms of the heat diffusion equation (HDE) for the insulation andcable are identified. The temperature distributions are valid if they satisfy the relevant HDE.Insulation: The temperature distribution is given as(T ( r ) = Ts,2 + Ts,1 Ts,2ln r r) ln ((r r2 ))(1)12and the appropriate HDE (radial coordinates, SS, q = 0), Eq. 2.20,d dT r=0dr dr d 1 r d Ts,1 Ts,2 r 0 + Ts,1 Ts,2? = ?0 = dr ln ( r1 r2 ) dr ln ( r1 r2 ) ()Hence, the temperature distribution satisfies the HDE.<Cable: The temperature distribution is given asqr 2 r 2 T ( r ) = Ts,1 + 1 1 4k c r 2 1and the appropriate HDE (radial coordinates, SS, q uniform), Eq. 2.20,(2)Continued...PROBLEM 2.41 (Cont.)1 d dT q=0r +r dr dr k c1d qr 2 r 0 + 1r dr 4k c2r q 0 + ? = ? 02 kr1 c1dqr 2 2r 2 q 1 + ? = ?0r dr 4k c r 2 k c121 qr1 4r q + ? = ?0r 4k c r 2 k c1<Hence the temperature distribution satisfies the HDE.The temperature distributions in the cable, 0 r r1, and sleeve, r1 r r2, and their key features areas follows:(1) Zero gradient, symmetry condition,(2) Increasing gradient with increasing radius,r, because of q ,(3) Discontinuous T(r) across cable-sleeveinterface because of different thermalconductivities,(4) Decreasing gradient with increasing radius,r, since heat rate is constant.(b) Using Fouriers law for the radial-cylindrical coordinate, the heat rate through the insulation(sleeve) per unit length isq = kArrdTdr= k2 rdT<drand substituting for the temperature distribution, Eq. (1),(q = k s 2 r 0 + Ts,1 Ts,2r) ln (1 r r ) = 2 ksr1 2 (Ts,1 Ts,2 )ln ( r2 r1 )(3)<Applying an energy balance to a control surface placed around the cable,Ein Eout = 0cq q = 0rwhere qc represents the dissipated electrical power in the cableContinued...PROBLEM 2.41 (Cont.)()2q r1 q = 0r(4)<(5)2q = qr1ror<(c) Applying an energy balance to a control surface placed around the outer surface of the sleeve,E in E out = 0q q q = 0rcvrad(())244 qr1 h ( 2 r2 ) Ts,2 T ( 2 r2 ) Ts,2 Tsur = 0This relation can be used to determine Ts,2 in terms of the variables q , r1, r2, h, T, and Tsur.(d) Consider a cable-sleeve system with the following prescribed conditions:kc = 200 W/mKks = 0.15 W/mKr1 = 15 mmr2 = 15.5 mmh = 25 W/m2KT = 25C = 0.9Tsur = 35CFor 250 A with R = 0.005 /m, the volumetric heat generation rate ise( r12 )2q = ( 250 A ) 0.005 / m ( 0.0152 m 2 ) = 4.42 105 W m3q = I 2 R = I 2 R eceSubstituting numerical values in appropriate equations, we can evaluate Ts,1, Ts,2 and To.Sleeve outer surface temperature, Ts,2: Using Eq. (5),( 4.42 105 W m3 ( 0.015m ) 25 W m 2 K ( 2 0.0155m ) Ts,2 298K2())40.9 ( 2 0.0155m ) 5.67 108 W m 2 K 4 Ts,2 3084 K 4 = 0<Ts,2 = 395 K = 122 CSleeve-cable interface temperature, Ts,1: Using Eqs. (3) and (4), with Ts,2 = 395 K,2 qr1 = 2 k s(Ts,1 Ts,2 )ln ( r2 r1 ) 4.42 105 W m3 ( 0.015 m ) = 2 0.15 W m K2Ts,1 = 406 K = 133 C(Ts,1 395 K )ln (15.5 15.0 )<Continued...PROBLEM 2.41 (Cont.)Cable centerline temperature, To: Using Eq. (2) with Ts,1 = 133C,To = T(0) = Ts,1 +2qr14k cTo = 133 C + 4.42 105 W m3 ( 0.015 m )2( 4 200 Wm K ) = 133.1 C<(e) With all other conditions remaining the same, the relations of part (d) can be used to calculate To,Ts,1 and Ts,2 as a function of the sleeve outer radius r2 for the range 15.5 r2 20 mm.Temperature, Ts1 or Ts2 (C)200180160140120100151617181920Sleeve outer radius, r2 (mm)Inner sleeve, r1Outer sleeve, r2On the plot above To would show the same behavior as Ts,1 since the temperature rise between cablecenter and its surface is 0.12C. With increasing r2, we expect Ts,2 to decrease since the heat fluxdecreases with increasing r2. We expect Ts,1 to increase with increasing r2 since the thermal resistanceof the sleeve increases.PROBLEM 2.42KNOWN: Temperature distribution in a spherical shell.FIND: Whether conditions are steady-state or transient. Manner in which heat flux and heat ratevary with radius.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties.ANALYSIS: From Equation 2.23, the heat equation reduces to1 2T1 Tr=.r tr2 rSubstituting for T(r),1 T1 2 C1= 2r 2 = 0. tr rr<Hence, steady-state conditions exist.From Equation 2.22, the radial component of the heat flux isq = krTC= k 1 .rr249Hence, q decreases with increasing r 2 q 1/ r 2 .rr<At any radial location, the heat rate isq r = 4r 2q = 4kC1.rHence, qr is independent of r.<COMMENTS: The fact that qr is independent of r is consistent with the energy conservationrequirement. If qr is constant, the flux must vary inversely with the area perpendicular to the direction2of heat flow. Hence, q varies inversely with r .rPROBLEM 2.43KNOWN: Spherical container with an exothermic reaction enclosed by an insulating material whoseouter surface experiences convection with adjoining air and radiation exchange with largesurroundings.FIND: (a) Verify that the prescribed temperature distribution for the insulation satisfies theappropriate form of the heat diffusion equation; sketch the temperature distribution and label keyfeatures; (b) Applying Fourier's law, verify the conduction heat rate expression for the insulationlayer, qr, in terms of Ts,1 and Ts,2; apply a surface energy balance to the container and obtain analternative expression for qr in terms of q and r1; (c) Apply a surface energy balance around the outersurface of the insulation to obtain an expression to evaluate Ts,2; (d) Determine Ts,2 for the specifiedgeometry and operating conditions; (e) Compute and plot the variation of Ts,2 as a function of theouter radius for the range 201 r2 210 mm; explore approaches for reducing Ts,2 45C toeliminate potential risk for burn injuries to personnel.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial spherical conduction, (2) Isothermal reaction incontainer so that To = Ts,1, (2) Negligible thermal contact resistance between the container andinsulation, (3) Constant properties in the insulation, (4) Surroundings large compared to the insulatedvessel, and (5) Steady-state conditions.ANALYSIS: The appropriate form of the heat diffusion equation (HDE) for the insulation followsfrom Eq. 2.23,1 d 2 dT r=0r 2 dr dr (1)<The temperature distribution is given as() 1 ( r1 r ) 1 ( r1 r2 ) T ( r ) = Ts,1 Ts,1 Ts,2 (2)Substitute T(r) into the HDE to see if it is satisfied:() ? = ? 00 + r1 r 21 d 2 r 0 Ts,1 Ts,2 1 ( r r )r 2 dr 12()1 dr1 + ( Ts,1 Ts,2 )? = ?01 ( r1 r2 ) r 2 dr <and since the expression in parenthesis is independent of r, T(r) does indeed satisfy the HDE. Thetemperature distribution in the insulation and its key features are as follows:Continued...PROBLEM 2.43 (Cont.)(1) Ts,1 > Ts,2(2) Decreasing gradient with increasing radius,r, since the heat rate is constant throughthe insulation.(b) Using Fouriers law for the radial-spherical coordinate, the heat rate through the insulation isdTq r = kA rdr( ) dTdr<= k 4 r 2and substituting for the temperature distribution, Eq. (2),q r = k rqr =2(0 Ts,1 Ts,2(4 k Ts,1 Ts,2()0 + r1 r 2 ) 1 (r1 r2 ))(3)<(4)(1 r1 ) (1 r2 )<Applying an energy balance to a control surface about the container at r = r1,E in E out = 0q q r = 0where q represents the generated heat in the container,3q r = ( 4 3 ) r1 q(c) Applying an energy balance to a control surface placed around the outer surface of the insulation,E in E out = 0q r q cv q rad = 0()()44q r hAs Ts,2 T As Ts,2 Tsur = 0(5)<Continued...PROBLEM 2.43 (Cont.)where2As = 4 r2(6)These relations can be used to determine Ts,2 in terms of the variables q , r1, r2, h, T , and Tsur.(d) Consider the reactor system operating under the following conditions:h = 5 W/m2KT = 25Cr1 = 200 mmr2 = 208 mmk = 0.05 W/mK = 0.9Tsur = 35CThe heat generated by the exothermic reaction provides for a volumetric heat generation rate,q = q o exp ( A To )q o = 5000 W m3A = 75 K(7)where the temperature of the reaction is that of the inner surface of the insulation, To = Ts,1. Thefollowing system of equations will determine the operating conditions for the reactor.Conduction rate equation, insulation, Eq. (3),qr =(4 0.05 W m K Ts,1 Ts,2(1 0.200 m 1 0.208 m ))(8)Heat generated in the reactor, Eqs. (4) and (7),q r = 4 3 ( 0.200 m ) q3((9)q = 5000 W m3 exp 75 K Ts,1)(10)Surface energy balance, insulation, Eqs. (5) and (6),()(4q r 5 W m 2 K As Ts,2 298 K 0.9As 5.67 108 W m 2K 4 Ts,2 (308 K )4)=0As = 4 (0.208 m )2(11)(12)Solving these equations simultaneously, find thatTs,1 = 94.3 C<Ts,2 = 52.5 CThat is, the reactor will be operating at To = Ts,1 = 94.3C, very close to the desired 95C operatingcondition.(e) From the above analysis, we found the outer surface temperature Ts,2 = 52.5C represents apotential burn risk to plant personnel. Using the above system of equations, Eqs. (8)-(12), we haveexplored the effects of changes in the convection coefficient, h, and the insulation thermalconductivity, k, as a function of insulation thickness, t = r2 - r1.Continued...PROBLEM 2.43 (Cont.)12010050Reaction temperature, To (C)Outer surface temperature, Ts2 (C)5545403502468Insulation thickness, (r2 - r1) (mm)k = 0.05 W/m.K, h = 5 W/m^2.Kk = 0.01 W/m.K, h = 5 W/m^2.Kk = 0.05 W/m.K, h = 15 W/m^2.K10806040200246810Insulation thickness, (r2-r1) (mm)k = 0.05 W/m.K, h = 5 W/m^2.Kk = 0.01 W/m.K, h = 5 W/m^2.Kk = 0.05 W/m.K, h = 15 W/m^2.KIn the Ts,2 vs. (r2 - r1) plot, note that decreasing the thermal conductivity from 0.05 to 0.01 W/mKslightly increases Ts,2 while increasing the convection coefficient from 5 to 15 W/m2K markedlydecreases Ts,2. Insulation thickness only has a minor effect on Ts,2 for either option. In the To vs. (r2 r1) plot, note that, for all the options, the effect of increased insulation is to increase the reactiontemperature. With k = 0.01 W/mK, the reaction temperature increases beyond 95C with less than 2mm insulation. For the case with h = 15 W/m2K, the reaction temperature begins to approach 95Cwith insulation thickness around 10 mm. We conclude that by selecting the proper insulationthickness and controlling the convection coefficient, the reaction could be operated around 95C suchthat the outer surface temperature would not exceed 45C.PROBLEM 2.44KNOWN: One-dimensional system, initially at a uniform temperature Ti, is suddenlyexposed to a uniform heat flux at one boundary, while the other boundary is insulated.FIND: (a) Proper form of heat equation and boundary and initial conditions, (b) Temperaturedistributions for following conditions: initial condition (t 0), and several times after heateris energized; will a steady-state condition be reached; (c) Heat flux at x = 0, L/2, L as afunction of time; (d) Expression for uniform temperature, Tf, reached after heater has beenswitched off following an elapsed time, te, with the heater on.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal heat generation, (3)Constant properties.ANALYSIS: (a) The appropriate form of the heat equation follows from Eq. 2.15. Also, theappropriate boundary and initial conditions are:16Initial condition: 2Tx2=Boundary conditions:x=0q o = k T / x) 0x=L1 T tT x,0 = Ti T / x) L = 0Uniform temperature(b) The temperature distributions are as follows:<No steady-state condition will be reached since E in = E st and E in is constant.(c)The heat flux as a function of time for positions x = 0, L/2 and L is as follows:<(d) If the heater is energized until t = te and then switched off, the system will eventuallyreach a uniform temperature, Tf. Perform an energy balance on the system, Eq. 1.11b, foran interval of time t = te,E in = E stIt follows thatE in = Q in =Ite0q o A sdt = q A s t eo1q A s t e = Mc Tf Tio6or1E st = Mc Tf Tiq A tTf = Ti + o s e .Mc6<PROBLEM 2.45KNOWN: Plate of thickness 2L, initially at a uniform temperature of Ti = 200C, is suddenly2quenched in a liquid bath of T = 20C with a convection coefficient of 100 W/m K.FIND: (a) On T-x coordinates, sketch the temperature distributions for the initial condition (t 0), thesteady-state condition (t ), and two intermediate times; (b) On q t coordinates, sketch thexvariation with time of the heat flux at x = L, (c) Determine the heat flux at x = L and for t = 0; what isthe temperature gradient for this condition; (d) By performing an energy balance on the plate,2determine the amount of energy per unit surface area of the plate (J/m ) that is transferred to the bathover the time required to reach steady-state conditions; and (e) Determine the energy transferred to thebath during the quenching process using the exponential-decay relation for the surface heat flux.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, and (3) No internal heatgeneration.ANALYSIS: (a) The temperature distributions are shown in the sketch below.(b) The heat flux at the surface x = L, q ( L, t ) , is initially a maximum value, and decreases withxincreasing time as shown in the sketch above.(c) The heat flux at the surface x = L at time t = 0, q ( L, 0 ) , is equal to the convection heat flux withxthe surface temperature as T(L,0) = Ti.22q ( L, 0 ) = qxconv ( t = 0 ) = h ( Ti T ) = 100 W / m K ( 200 20 ) C = 18.0 kW / mFrom a surface energy balance as shown in the sketch considering the conduction and convectionfluxes at the surface, the temperature gradient can be calculated.Continued ..<PROBLEM 2.45 (Cont.)Ein Eout = 0q ( L, 0 ) qxconv ( t = 0 ) = 0withq ( L, 0 ) = kxT x x = LT 32= qconv ( t = 0 ) / k = 18 10 W / m / 50 W / m K = 360 K / mx L,0<(d) The energy transferred from the plate to the bath over the time required to reach steady-stateconditions can be determined from an energy balance on a time interval basis, Eq. 1.11b. For theinitial state, the plate has a uniform temperature Ti; for the final state, the plate is at the temperature ofthe bath, T.E E = E = E EinoutstfiwithE = 0,in E = cp ( 2L )[T Ti ]outE = 2770 kg / m3 875 J / kg K ( 2 0.010 m )[20 200] K = +8.73 106 J / m 2out<(e) The energy transfer from the plate to the bath during the quenching process can be evaluated fromknowledge of the surface heat flux as a function of time. The area under the curve in the q ( L, t ) vs.xtime plot (see schematic above) represents the energy transferred during the quench process.E = 2 outq ( L, t ) dt = 2Ae Bt dtt =0 xt =011E = 2A e Bt = 2A (0 1) = 2A / BoutB0BE = 2 1.80 104 W / m 2 / 4.126 103 s 1 = 8.73 106 J / m 2out<COMMENTS: (1) Can you identify and explain the important features in the temperaturedistributions of part (a)?(2) The maximum heat flux from the plate occurs at the instant the quench process begins and is equalto the convection heat flux. At this instant, the gradient in the plate at the surface is a maximum. Ifthe gradient is too large, excessive thermal stresses could be induced and cracking could occur.(3) In this thermodynamic analysis, we were able to determine the energy transferred during thequenching process. We cannot determine the rate at which cooling of the plate occurs without solvingthe heat diffusion equation.PROBLEM 2.46KNOWN: Plane wall, initially at a uniform temperature, is suddenly exposed to convective heating.FIND: (a) Differential equation and initial and boundary conditions which may be used to find thetemperature distribution, T(x,t); (b) Sketch T(x,t) for these conditions: initial (t 0), steady-state, t , and two intermediate times; (c) Sketch heat fluxes as a function of time for surface locations; (d)3Expression for total energy transferred to wall per unit volume (J/m ).SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal heatgeneration.ANALYSIS: (a) For one-dimensional conduction with constant properties, the heat equation has theform, 2Tx2=and theconditions are:1 T t%Initial, t 0: T1x,06 = TiKBoundaries: x = 0 T / x) = 0&0K x = L k T / x) L = h T1L, t6 TK'uniformadiabaticconvection(b) The temperature distributions are shown on the sketch.Note that the gradient at x = 0 is always zero, since this boundary is adiabatic. Note also that thegradient at x = L decreases with time.16(c) The heat flux, q x, t , as a function of time, is shown on the sketch for the surfaces x = 0 and xx= L.Continued ..PROBLEM 2.46 (Cont.)1616For the surface at x = 0, q 0, t = 0 since it is adiabatic. At x = L and t = 0, q L,0 is axxmaximum1616q L,0 = h T L,0 Txwhere T(L,0) = Ti. The gradient, and hence the flux, decrease with time.(d) The total energy transferred to the wall may be expressed asE in =I0q conv A sdtE in = hA sI201 67T T L, t dtDividing both sides by AsL, the energy transferred per unit volume isI16E in h T T L, t dt=VL0J / m3COMMENTS: Note that the heat flux at x = L is into the wall and is hence in the negative xdirection.PROBLEM 2.47KNOWN: Plane wall, initially at a uniform temperature Ti, is suddenly exposed to convection with afluid at T at one surface, while the other surface is exposed to a constant heat flux q .oFIND: (a) Temperature distributions, T(x,t), for initial, steady-state and two intermediate times, (b)Corresponding heat fluxes on q x coordinates, (c) Heat flux at locations x = 0 and x = L as axfunction of time, (d) Expression for the steady-state temperature of the heater, T(0,), in terms ofq , T , k, h and L.oSCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) No heat generation, (3) Constant properties.ANALYSIS: (a) For Ti < T , the temperature distributions are$Note the constant gradient at x = 0 since q 0 = q o .x$(b) The heat flux distribution, q x, t , is determined from knowledge of the temperature gradients,xevident from Part (a), and Fouriers law.$(c) On q x, t t coordinates, the heat fluxes at the boundaries are shown above.x(d) Perform a surface energy balance at x = L and an energy balance on the wall:$q cond = q conv = h T L, T(1),q cond = q .o(2)For the wall, under steady-state conditions, Fouriers law givesq = ko$$T 0, T L, dT=k.dxLCombine Eqs. (1), (2), (3) to find:$T 0, = T +qo.1/ h + L / k(3)PROBLEM 2.48KNOWN: Plane wall, initially at a uniform temperature To, has one surface (x = L) suddenlyexposed to a convection process (T > To,h), while the other surface (x = 0) is maintained at To.Also, wall experiences uniform volumetric heating q such that the maximum steady-state temperaturewill exceed T.FIND: (a) Sketch temperature distribution (T vs. X) for following conditions: initial (t 0), steadystate (t ), and two intermediate times; also show distribution when there is no heat flow at the x =L boundary, (b) Sketch the heat flux q vs. t at the boundaries x = 0 and L.x16SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Uniform volumetricgeneration, (4) To < T and q large enough that T(x,) > T.ANALYSIS: (a) The initial and boundary conditions for the wall can be written asInitial (t 0):T(x,0) = ToUniform temperatureBoundary:x = 0 T(0,t) = ToConstant temperaturex=Lk16T= h T L, t T x x=LConvection process.The temperature distributions are shown on the T-x coordinates below. Note the special conditionwhen the heat flux at (x = L) is zero.1616(b) The heat flux as a function of time at the boundaries, q 0, t and q L, t , can be inferred fromxxthe temperature distributions using Fouriers law.COMMENTS: Since T ( x, ) > T and T > To , heat transfer at both boundaries must be out of thewall. Hence, it follows from an overall energy balance on the wall that + q ( 0, ) q ( L, ) + qL = 0.xxPROBLEM 2.49KNOWN: Plane wall, initially at a uniform temperature To, has one surface (x = L) suddenly exposedto a convection process (T < To, h), while the other surface (x = 0) is maintained at To. Also, wallexperiences uniform volumetric heating q such that the maximum steady-state temperature willexceed T.FIND: (a) Sketch temperature distribution (T vs. x) for following conditions: initial (t 0), steadystate (t ), and two intermediate times; identify key features of the distributions, (b) Sketch the heatflux q vs. t at the boundaries x = 0 and L; identify key features of the distributions.x16SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Uniform volumetricgeneration, (4) T < To and q large enough that T(x,) > To.ANALYSIS: (a) The initial and boundary conditions for the wall can be written asInitial (t 0):T(x,0) = ToBoundary:x=LUniform temperatureConstant temperaturex = 0 T(0,t) = ToTk= h T L, t T x x=L16Convection process.The temperature distributions are shown on the T-x coordinates below. Note that the maximumtemperature occurs under steady-state conditions not at the midplane, but to the right toward thesurface experiencing convection. The temperature gradients at x = L increase for t > 0 since theconvection heat rate from the surface increases as the surface temperature increases.1616(b) The heat flux as a function of time at the boundaries, q 0, t and q L, t , can be inferred fromxxthe temperature distributions using Fouriers law. At the surface x = L, the convection heat flux at t =0 is q ( L, 0 ) = h ( To T ). Because the surface temperature dips slightly at early times, thexconvection heat flux decreases slightly, and then increases until the steady-state condition is reached.For the steady-state condition, heat transfer at both boundaries must be out of the wall. It follows froman overall energy balance on the wall that + q ( 0, ) q ( L, ) + qL = 0.xxPROBLEM 2.50KNOWN: Interfacial heat flux and outer surface temperature of adjoining, equivalent plane walls.FIND: (a) Form of temperature distribution at representative times during the heating process, (b)Variation of heat flux with time at the interface and outer surface.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.ANALYSIS: (a) With symmetry about theinterface, consideration of the temperaturedistribution may be restricted to 0 x L.During early stages of the process, heat transferis into the material from the outer surface, aswell as from the interface. During later stagesand the eventual steady state, heat is transferredfrom the material at the outer surface. Atsteady-state, dT/dx = ( q 2 ) k = const . andoT(0,t) = To + ( q 2 ) L k .o(b) At the outer surface, the heat flux is initiallynegative, but increases with time, approachingq /2. It is zero when dT dx x = L = 0 .oPROBLEM 2.51KNOWN: Temperature distribution in a plane wall of thickness L experiencing uniform volumetricheating q having one surface (x = 0) insulated and the other exposed to a convection processcharacterized by T and h. Suddenly the volumetric heat generation is deactivated while convectioncontinues to occur.FIND: (a) Determine the magnitude of the volumetric energy generation rate associated with theinitial condition, (b) On T-x coordinates, sketch the temperature distributions for the initial condition(T 0), the steady-state condition (t ), and two intermediate times; (c) On q - t coordinates,xsketch the variation with time of the heat flux at the boundary exposed to the convection process,q ( L, t ) ; calculate the corresponding value of the heat flux at t = 0; and (d) Determine the amount ofx2energy removed from the wall per unit area (J/m ) by the fluid stream as the wall cools from its initialto steady-state condition.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, and (3) Uniform internalvolumetric heat generation for t < 0.ANALYSIS: (a) The volumetric heating rate can be determined by substituting the temperaturedistribution for the initial condition into the appropriate form of the heat diffusion equation.d dT q+ =0dx dx kwheredqq(0 + 2bx ) + = 0 + 2b + = 0dxkk(T ( x, 0 ) = a + bx 2)q = 2kb = 2 90 W / m K 1.0 104C / m 2 = 1.8 106 W / m3<(b) The temperature distributions are shown in the sketch below.Continued ..PROBLEM 2.51 (Cont.)(c) The heat flux at the exposed surface x = L, q ( L, 0 ) , is initially a maximum value and decreasesxwith increasing time as shown in the sketch above. The heat flux at t = 0 is equal to the convectionheat flux with the surface temperature T(L,0). See the surface energy balance represented in theschematic.252q ( L, 0 ) = qxconv ( t = 0 ) = h ( T ( L, 0 ) T ) = 1000 W / m K ( 200 20 ) C = 1.80 10 W / m<2where T ( L, 0 ) = a + bL2 = 300C 1.0 104C / m 2 ( 0.1m ) = 200C.(d) The energy removed from the wall to the fluid as it cools from its initial to steady-state conditioncan be determined from an energy balance on a time interval basis, Eq. 1.11b. For the initial state, the2wall has the temperature distribution T(x,0) = a + bx ; for the final state, the wall is at the temperatureof the fluid, Tf = T. We have used T as the reference condition for the energy terms.Ein E = E = E EoutstfiwithE = 0inx =LT ( x, 0 ) T dx E = cp L [Tf T ] cp outx =0 E = cp outLx =L a + bx 2 T dx = cp ax + bx 3 / 3 T x 0x =0 3E = 7000 kg / m3 450 J / kg K 300 0.1 1.0 104 (0.1) / 3 20 0.1 K moutE = 7.77 107 J / m 2out<COMMENTS: (1) In the temperature distributions of part (a), note these features: initial conditionhas quadratic form with zero gradient at the adiabatic boundary; for the steady-state condition, the wallhas reached the temperature of the fluid; for all distributions, the gradient at the adiabatic boundary iszero; and, the gradient at the exposed boundary decreases with increasing time.(2) In this thermodynamic analysis, we were able to determine the energy transferred during thecooling process. However, we cannot determine the rate at which cooling of the wall occurs withoutsolving the heat diffusion equation.PROBLEM 2.52KNOWN: Temperature as a function of position and time in a plane wall suddenly subjected to achange in surface temperature, while the other surface is insulated.FIND: (a) Validate the temperature distribution, (b) Heat fluxes at x = 0 and x = L, (c) Sketch oftemperature distribution at selected times and surface heat flux variation with time, (d) Effect ofthermal diffusivity on system response.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant properties.ANALYSIS: (a) To be valid, the temperature distribution must satisfy the appropriate forms of theheat equation and boundary conditions. Substituting the distribution into Equation 2.15, it followsthat 2Tx2=1 T t 2 t 2 cos x 1 6 4 L2 2L 2 L 2 exp 2 t cos x .C= 1 1Ti Ts 6 4 L2 4 L2 2 L C1 Ti Ts exp <Hence, the heat equation is satisfied. Applying boundary conditions at x = 0 and x = L, it follows that16 16 < <T 2 txC|x=0 = 1 Ti Ts exp sin|x=0 = 0x2L4 L22Land16T L, t = Ts + C1 Ti Ts exp 2 txcos|x=L = Ts .24L2LHence, the boundary conditions are also satisfied.(b) The heat flux has the form16 T 2 txkC1q = k=+Ti Ts exp sin.x2x2L4L2LContinued ..PROBLEM 2.52 (Cont.)16<q 0 = 0,xHence,16q L = +x6 1 2 tkC1Ti Ts exp .2L4 L2<(c) The temperature distribution and surface heat flux variations are:(d) For materials A and B of different ,1616!6 "##$2T x, t TsA = exp A B tT x, t Ts4 L2B16116Hence, if A > B , T x, t Ts more rapidly for Material A. If A < B , T x, t Ts morerapidly for Material B.<COMMENTS: Note that the prescribed function for T(x,t) does not reduce to Ti for t 0. Fortimes at or close to zero, the function is not a valid solution of the problem. At such times, thesolution for T(x,t) must include additional terms. The solution is consideed in Section 5.5.1 of thetext.PROBLEM 2.532KNOWN: Thin electrical heater dissipating 4000 W/m sandwiched between two 25-mm thick plateswhose surfaces experience convection.FIND: (a) On T-x coordinates, sketch the steady-state temperature distribution for -L +L;calculate values for the surfaces x = L and the mid-point, x = 0; label this distribution as Case 1 andexplain key features; (b) Case 2: sudden loss of coolant causing existence of adiabatic condition onthe x = +L surface; sketch temperature distribution on same T-x coordinates as part (a) and calculatevalues for x = 0, L; explain key features; (c) Case 3: further loss of coolant and existence ofadiabatic condition on the x = - L surface; situation goes undetected for 15 minutes at which timepower to the heater is deactivated; determine the eventual (t ) uniform, steady-state temperaturedistribution; sketch temperature distribution on same T-x coordinates as parts (a,b); and (d) On T-tcoordinates, sketch the temperature-time history at the plate locations x = 0, L during the transientperiod between the steady-state distributions for Case 2 and Case 3; at what location and when will thetemperature in the system achieve a maximum value?SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internalvolumetric generation in plates, and (3) Negligible thermal resistance between the heater surfaces andthe plates.ANALYSIS: (a) Since the system is symmetrical, the heater power results in equal conduction fluxesthrough the plates. By applying a surface energy balance on the surface x = +L as shown in theschematic, determine the temperatures at the mid-point, x = 0, and the exposed surface, x + L.Ein Eout = 0q ( + L ) qxconv = 0q ( + L ) = q / 2xowhereq / 2 h T ( + L ) T = 0o()T1 ( + L ) = q / 2h + T = 4000 W / m 2 / 2 400 W / m 2 K + 20C = 25Co<From Fouriers law for the conduction flux through the plate, find T(0).q = q / 2 = k T ( 0 ) T ( + L ) / LxoT1 ( 0 ) = T1 ( + L ) + q L / 2k = 25C + 4000 W / m 2 K 0.025m / ( 2 5 W / m K ) = 35CoThe temperature distribution is shown on the T-x coordinates below and labeled Case 1. The keyfeatures of the distribution are its symmetry about the heater plane and its linear dependence withdistance.Continued ..<PROBLEM 2.53 (Cont.)(b) Case 2: sudden loss of coolant with the existence of an adiabatic condition on surface x = +L. Forthis situation, all the heater power will be conducted to the coolant through the left-hand plate. From asurface energy balance and application of Fouriers law as done for part (a), findT2 ( L ) = q / h + T = 4000 W / m 2 / 400 W / m 2 K + 20C = 30CoT2 ( 0 ) = T2 ( L ) + q L / k = 30C + 4000 W / m 2 0.025 m / 5 W / m K = 50CoThe temperature distribution is shown on the T-x coordinates above and labeled Case 2. Thedistribution is linear in the left-hand plate, with the maximum value at the mid-point. Since no heatflows through the right-hand plate, the gradient must zero and this plate is at the maximumtemperature as well. The maximum temperature is higher than for Case 1 because the heat fluxthrough the left-hand plate has increased two-fold.<<(c) Case 3: sudden loss of coolant occurs at the x = -L surface also. For this situation, there is no heat2transfer out of either plate, so that for a 15-minute period, to, the heater dissipates 4000 W/m andthen is deactivated. To determine the eventual, uniform steady-state temperature distribution, applythe conservation of energy requirement on a time-interval basis, Eq. 1.11b. The initial conditioncorresponds to the temperature distribution of Case 2, and the final condition will be a uniform,elevated temperature Tf = T3 representing Case 3. We have used T as the reference condition for theenergy terms.E E + E = E = E E(1)inoutgenstfiNote that E n E = 0 , and the dissipated electrical energy isioutE = q t o = 4000 W / m 2 (15 60 ) s = 3.600 106 J / m 2genoFor the final condition,E = c ( 2L )[Tf T ] = 2500 kg / m3 700 J / kg K ( 2 0.025m ) [Tf 20 ]CfE = 8.75 104 [Tf 20] J / m2fwhere Tf = T3, the final uniform temperature, Case 3. For the initial condition,E = c i+LT2 ( x ) T dx = cL[{0+LT2 ( x ) T dx +T2 (0 ) T dx0L[[where T2 ( x ) is linear for L x 0 and constant at T2 ( 0 ) for 0 x +L.T2 ( x ) = T2 ( 0 ) + T2 ( 0 ) T2 ( L ) x / L}(2)(3)(4)L x 0T2 ( x ) = 50C + [50 30]Cx / 0.025mT2 ( x ) = 50C + 800x(5)Substituting for T2 ( x ) , Eq. (5), into Eq. (4)Continued ..PROBLEM 2.53 (Cont.)0E = c [50 + 800x T ] dx + T2 ( 0 ) T L i L0E = c 50x + 400x 2 T x + T2 (0 ) T L i L{}Ei = c 50L + 400L2 + T L + T2 ( 0 ) T LE = cL {+50 400L T + T2 (0 ) T }iE = 2500 kg / m3 700 J / kg K 0.025 m {+50 400 0.025 20 + 50 20}KiE = 2.188 106 J / m 2i(6)Returning to the energy balance, Eq. (1), and substituting Eqs. (2), (3) and (6), find Tf = T3.3.600 106 J / m 2 = 8.75 104 [T3 20] 2.188 106 J / m 2T3 = ( 66.1 + 20 ) C = 86.1C<The temperature distribution is shown on the T-x coordinates above and labeled Case 3. Thedistribution is uniform, and considerably higher than the maximum value for Case 2.(d) The temperature-time history at the plate locations x = 0, L during the transient period betweenthe distributions for Case 2 and Case 3 are shown on the T-t coordinates below.Note the temperatures for the locations at time t = 0 corresponding to the instant when the surfacex = - L becomes adiabatic. These temperatures correspond to the distribution for Case 2. The heaterremains energized for yet another 15 minutes and then is deactivated. The midpoint temperature,T(0,t), is always the hottest location and the maximum value slightly exceeds the final temperature T3.PROBLEM 2.54KNOWN: Radius and length of coiled wire in hair dryer. Electric power dissipation in the wire, andtemperature and convection coefficient associated with air flow over the wire.FIND: (a) Form of heat equation and conditions governing transient, thermal behavior of wire duringstart-up, (b) Volumetric rate of thermal energy generation in the wire, (c) Sketch of temperaturedistribution at selected times during start-up, (d) Variation with time of heat flux at r = 0 and r = ro.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties, (3) Uniformvolumetric heating, (4) Negligible radiation from surface of wire.ANALYSIS: (a) The general form of the heat equation for cylindrical coordinates is given by Eq.2.20. For one-dimensional, radial conduction and constant properties, the equation reduces to r T + q = c p T = 1 Tr r r k t t1The initial condition isT ( r, 0 ) = TiThe boundary conditions are:<T / r r = 0 = 0kTr r = ro<<<= h [T ( ro , t ) T ](b) The volumetric rate of thermal energy generation isq=EgP500 W83= elec == 3.18 10 W / m22 ro L ( 0.001m ) ( 0.5m )<Under steady-state conditions, all of the thermal energy generated within the wire is transferred to theair by convection. Performing an energy balance for a control surface about the wire, E out + E g = 0,it follows that 2 ro L q ( ro , t ) + Pelec = 0. Hence,q ( ro , t ) =Pelec2 ro L=500 W2 ( 0.001m ) 0.5m5= 1.59 10 W / m2<COMMENTS: The symmetry condition at r = 0 imposes the requirement that T / r r = 0 = 0, andhence q ( 0, t ) = 0 throughout the process. The temperature at ro, and hence the convection heat flux,increases steadily during the start-up, and since conduction to the surface must be balanced byconvection from the surface at all times, T / r r = r also increases during the start-up.oPROBLEM 3.1KNOWN: One-dimensional, plane wall separating hot and cold fluids at T,1 and T ,2 ,respectively.FIND: Temperature distribution, T(x), and heat flux, q , in terms of T,1 , T,2 , h1 , h 2 , kxand L.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constantproperties, (4) Negligible radiation, (5) No generation.ANALYSIS: For the foregoing conditions, the general solution to the heat diffusion equationis of the form, Equation 3.2,T ( x ) = C1x + C2 .(1)The constants of integration, C1 and C2, are determined by using surface energy balanceconditions at x = 0 and x = L, Equation 2.23, and as illustrated above,dT dT (2,3)k = h1 T,1 T ( 0 )k = h 2 T ( L ) T,2 .dt x=0dx x=LFor the BC at x = 0, Equation (2), use Equation (1) to find k ( C1 + 0 ) = h1 T,1 (C1 0 + C2 )(4)and for the BC at x = L to find k ( C1 + 0 ) = h 2 ( C1L + C2 ) T,2 .(5)Multiply Eq. (4) by h2 and Eq. (5) by h1, and add the equations to obtain C1. Then substituteC1 into Eq. (4) to obtain C2. The results areT,1 T,2T,1 T,2C1 = C2 = + T,1111 L1 Lk +h1 +++ h1 h 2 k h1 h 2 k T,1 T,2 x 1 T (x ) = < + + T,1.11 L k h1 h + h + k 21()(())From Fouriers law, the heat flux is a constant and of the formT,1 T,2dTq = k.= k C1 = +xdx11 Lh + h + k 21()<PROBLEM 3.2KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfacesof a rear window.FIND: (a) Inner and outer window surface temperatures, Ts,i and Ts,o, and (b) Ts,i and Ts,o as a function ofthe outside air temperature T,o and for selected values of outer convection coefficient, ho.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiationeffects, (4) Constant properties.PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/mK.ANALYSIS: (a) The heat flux may be obtained from Eqs. 3.11 and 3.12,)(40 C 10 CT,i T,oq ==1L110.004 m1++++2ho k hi65 W m K 1.4 W m K 30 W m2 Kq =50 C(0.0154 + 0.0029 + 0.0333) m(2K W= 968 W m 2 .)Hence, with q = h i T,i T,o , the inner surface temperature isTs,i = T ,i qhi= 40 C 968 W m 2230 W m K<= 7.7 C()Similarly for the outer surface temperature with q = h o Ts,o T,o findTs,o = T,o q= 10 C 968 W m 22= 4.9 C<65 W m K(b) Using the same analysis, Ts,i and Ts,o have been computed and plotted as a function of the outside airtemperature, T,o, for outer convection coefficients of ho = 2, 65, and 100 W/m2K. As expected, Ts,i andTs,o are linear with changes in the outside air temperature. The difference between Ts,i and Ts,o increaseswith increasing convection coefficient, since the heat flux through the window likewise increases. Thisdifference is larger at lower outside air temperatures for the same reason. Note that with ho = 2 W/m2K,Ts,i - Ts,o, is too small to show on the plot.hoContinued ..Surface temperatures, Tsi or Tso (C)PROBLEM 3.2 (Cont.)403020100-10-20-30-30-25-20-15-10-50Outside air temperature, Tinfo (C)Tsi; ho = 100 W/m^2.KTso; ho = 100 W/m^2.KTsi; ho = 65 W/m^2.KTso; ho = 65 W/m^2.KTsi or Tso; ho = 2 W/m^.KCOMMENTS: (1) The largest resistance is that associated with convection at the inner surface. Thevalues of Ts,i and Ts,o could be increased by increasing the value of hi.(2) The IHT Thermal Resistance Network Model was used to create a model of the window and generatethe above plot. The Workspace is shown below.// Thermal Resistance Network Model:// The Network:// Heat rates into node j,qij, through thermal resistance Rijq21 = (T2 - T1) / R21q32 = (T3 - T2) / R32q43 = (T4 - T3) / R43// Nodal energy balancesq1 + q21 = 0q2 - q21 + q32 = 0q3 - q32 + q43 = 0q4 - q43 = 0/* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal pointsat which there is no external source of heat. */T1 = Tinfo// Outside air temperature, C//q1 =// Heat rate, WT2 = Tso// Outer surface temperature, Cq2 = 0// Heat rate, W; node 2, no external heat sourceT3 = Tsi// Inner surface temperature, Cq3 = 0// Heat rate, W; node 2, no external heat sourceT4 = Tinfi// Inside air temperature, C//q4 =// Heat rate, W// Thermal Resistances:R21 = 1 / ( ho * As )R32 = L / ( k * As )R43 = 1 / ( hi * As )// Convection thermal resistance, K/W; outer surface// Conduction thermal resistance, K/W; glass// Convection thermal resistance, K/W; inner surface// Other Assigned Variables:Tinfo = -10// Outside air temperature, Cho = 65// Convection coefficient, W/m^2.K; outer surfaceL = 0.004// Thickness, m; glassk = 1.4// Thermal conductivity, W/m.K; glassTinfi = 40// Inside air temperature, Chi = 30// Convection coefficient, W/m^2.K; inner surfaceAs = 1// Cross-sectional area, m^2; unit areaPROBLEM 3.3KNOWN: Desired inner surface temperature of rear window with prescribed inside and outside airconditions.FIND: (a) Heater power per unit area required to maintain the desired temperature, and (b) Compute andplot the electrical power requirement as a function of T,o for the range -30 T,o 0C with ho of 2,20, 65 and 100 W/m2K. Comment on heater operation needs for low ho. If h ~ Vn, where V is thevehicle speed and n is a positive exponent, how does the vehicle speed affect the need for heateroperation?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Uniform heaterflux, q , (4) Constant properties, (5) Negligible radiation effects, (6) Negligible film resistance.hPROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/mK.ANALYSIS: (a) From an energy balance at the inner surface and the thermal circuit, it follows that for aunit surface area,T ,i Ts,i1 hiq =hTs,i T ,o+ q =hL k + 1 hoTs,i T ,oL k + 1 hoT ,i Ts,i1 hi(=0.004 m1.4 W m Kq = (1370 100 ) W m = 1270 W mh215 C 10 C+)1265 W m K25 C 15 C1210 W m K<2(b) The heater electrical power requirement as a function of the exterior air temperature for differentexterior convection coefficients is shown in the plot. When ho = 2 W/m2K, the heater is unecessary,since the glass is maintained at 15C by the interior air. If h ~ Vn, we conclude that, with higher vehiclespeeds, the exterior convection will increase, requiring increased heat power to maintain the 15Ccondition.Heater power (W/m^2)3500300025002000150010005000-30-20-100Exterior air temperature, Tinfo (C)h = 20 W/m^2.Kh = 65 W/m^2.Kh = 100 W/m^2.KCOMMENTS: With q = 0, the inner surface temperature with T, o = -10C would be given byhT ,i Ts,iT ,i T ,o=1 hi1 hi + L k + 1 ho=0.100.118= 0.846,or()Ts,i = 25 C 0.846 35 C = 4.6 C .PROBLEM 3.4KNOWN: Curing of a transparent film by radiant heating with substrate and film surface subjected toknown thermal conditions.FIND: (a) Thermal circuit for this situation, (b) Radiant heat flux, q (W/m2), to maintain bond atocuring temperature, To, (c) Compute and plot q as a function of the film thickness for 0 Lf 1 mm,oand (d) If the film is not transparent, determine q required to achieve bonding; plot results as a functionoof Lf.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) All the radiant heatflux q is absorbed at the bond, (4) Negligible contact resistance.oANALYSIS: (a) The thermal circuitfor this situation is shown at the right.Note that terms are written on a per unitarea basis.(b) Using this circuit and performing an energy balance on the film-substrate interface,q = q1 + qo2q =oTo TT T+o 1R + R Rscvfwhere the thermal resistances areR = 1 h = 1 50 W m 2 K = 0.020 m 2 K WcvR = L f k f = 0.00025 m 0.025 W m K = 0.010 m 2 K WfR = Ls k s = 0.001m 0.05 W m K = 0.020 m 2 K Wsq =o(60 20 ) C[0.020 + 0.010] m2 K+W(60 30 ) C20.020 m K W= (133 + 1500 ) W m 2 = 2833 W m 2<(c) For the transparent film, the radiant flux required to achieve bonding as a function of film thickness Lfis shown in the plot below.(d) If the film is opaque (not transparent), the thermal circuit is shown below. In order to find q , it isonecessary to write two energy balances, one around the Ts node and the second about the To node..The results of the analyses are plotted below.Continued...PROBLEM 3.4 (Cont.)Radiant heat flux, q''o (W/m^2)70006000500040003000200000.20.40.60.81Film thickness, Lf (mm)Opaque filmTransparent filmCOMMENTS: (1) When the film is transparent, the radiant flux is absorbed on the bond. The fluxrequired decreases with increasing film thickness. Physically, how do you explain this? Why is therelationship not linear?(2) When the film is opaque, the radiant flux is absorbed on the surface, and the flux required increaseswith increasing thickness of the film. Physically, how do you explain this? Why is the relationshiplinear?(3) The IHT Thermal Resistance Network Model was used to create a model of the film-substrate systemand generate the above plot. The Workspace is shown below.// Thermal Resistance NetworkModel:// The Network:// Heat rates into node j,qij, through thermal resistance Rijq21 = (T2 - T1) / R21q32 = (T3 - T2) / R32q43 = (T4 - T3) / R43// Nodal energy balancesq1 + q21 = 0q2 - q21 + q32 = 0q3 - q32 + q43 = 0q4 - q43 = 0/* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal pointsat which there is no external source of heat. */T1 = Tinf// Ambient air temperature, C//q1 =// Heat rate, W; film sideT2 = Ts// Film surface temperature, Cq2 = 0// Radiant flux, W/m^2; zero for part (a)T3 = To// Bond temperature, Cq3 = qo// Radiant flux, W/m^2; part (a)T4 = Tsub// Substrate temperature, C//q4 =// Heat rate, W; substrate side// Thermal Resistances:R21 = 1 / ( h * As )R32 = Lf / (kf * As)R43 = Ls / (ks * As)// Convection resistance, K/W// Conduction resistance, K/W; film// Conduction resistance, K/W; substrate// Other Assigned Variables:Tinf = 20// Ambient air temperature, Ch = 50// Convection coefficient, W/m^2.KLf = 0.00025// Thickness, m; filmkf = 0.025// Thermal conductivity, W/m.K; filmTo = 60// Cure temperature, CLs = 0.001// Thickness, m; substrateks = 0.05// Thermal conductivity, W/m.K; substrateTsub = 30// Substrate temperature, CAs = 1// Cross-sectional area, m^2; unit areaPROBLEM 3.5KNOWN: Thicknesses and thermal conductivities of refrigerator wall materials. Inner and outer airtemperatures and convection coefficients.FIND: Heat gain per surface area.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Negligiblecontact resistance, (4) Negligible radiation, (5) Constant properties.ANALYSIS: From the thermal circuit, the heat gain per unit surface area isq =q =q =T,o T,i(1/ h i ) + (Lp / k p ) + ( Li / k i ) + (Lp / k p ) + (1/ h o )()( 25 4 ) C2 1/ 5 W / m 2 K + 2 (0.003m / 60 W / m K ) + (0.050m / 0.046 W / m K )21C(0.4 + 0.0001 + 1.087 ) m2 K / W= 14.1 W / m 2<COMMENTS: Although the contribution of the panels to the total thermal resistance is negligible,that due to convection is not inconsequential and is comparable to the thermal resistance of theinsulation.PROBLEM 3.6KNOWN: Design and operating conditions of a heat flux gage.FIND: (a) Convection coefficient for water flow (Ts = 27C) and error associated with neglectingconduction in the insulation, (b) Convection coefficient for air flow (Ts = 125C) and error associatedwith neglecting conduction and radiation, (c) Effect of convection coefficient on error associated withneglecting conduction for Ts = 27C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant k.ANALYSIS: (a) The electric power dissipation is balanced by convection to the water and conductionthrough the insulation. An energy balance applied to a control surface about the foil therefore yieldsPelec = q conv + q cond = h ( Ts T ) + k (Ts Tb ) LHence,h=h=Pelec k ( Ts Tb ) LTs T( 2000 8) W2Km2=2000 W m 2 0.04 W m K ( 2 K ) 0.01m2K<= 996 W m 2 KIf conduction is neglected, a value of h = 1000 W/m2K is obtained, with an attendant error of (1000 996)/996 = 0.40%(b) In air, energy may also be transferred from the foil surface by radiation, and the energy balanceyields()44Pelec = q conv + q + qradcond = h (Ts T ) + Ts Tsur + k ( Ts Tb ) LHence,h=()4Pelec Ts4 Tsur k ( Ts T ) LTs T2==2000 W m 0.15 5.67 1082W m K4(3984 2984)K4 0.04 W m K (100 K) / 0.01m100 K( 2000 146 400 ) W100 Km2= 14.5 W m 2 K<Continued...PROBLEM 3.6 (Cont.)If conduction, radiation, or conduction and radiation are neglected, the corresponding values of h and thepercentage errors are 18.5 W/m2K (27.6%), 16 W/m2K (10.3%), and 20 W/m2K (37.9%).2(c) For a fixed value of Ts = 27C, the conduction loss remains at qcond = 8 W/m , which is also thefixed difference between Pelec and q onv . Although this difference is not clearly shown in the plot forc10 h 1000 W/m2K, it is revealed in the subplot for 10 100 W/m2K.200Power dissipation, P''elec(W/m^2)Power dissipation, P''elec(W/m^2)20001600120080040000200400600800Convection coefficient, h(W/m^2.K)No conductionWith conduction100016012080400020406080100Convection coefficient, h(W/m^2.K)No conductionWith conductionErrors associated with neglecting conduction decrease with increasing h from values which aresignificant for small h (h < 100 W/m2K) to values which are negligible for large h.COMMENTS: In liquids (large h), it is an excellent approximation to neglect conduction and assumethat all of the dissipated power is transferred to the fluid.PROBLEM 3.7KNOWN: A layer of fatty tissue with fixed inside temperature can experience differentoutside convection conditions.FIND: (a) Ratio of heat loss for different convection conditions, (b) Outer surfacetemperature for different convection conditions, and (c) Temperature of still air whichachieves same cooling as moving air (wind chill effect).SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-stateconditions, (3) Homogeneous medium with constant properties, (4) No internal heatgeneration (metabolic effects are negligible), (5) Negligible radiation effects.PROPERTIES: Table A-3, Tissue, fat layer: k = 0.2 W/mK.ANALYSIS: The thermal circuit for this situation isHence, the heat rate isq=Ts,1 TR tot=Ts,1 TL/kA + 1/ hA.Therefore,L 1k + h windyqcalm =.qL 1windyk + h calmApplying a surface energy balance to the outer surface, it also follows thatqcond = qconv .Continued ..PROBLEM 3.7 (Cont.)Hence,()(kTs,1 Ts,2 = h Ts,2 TLkT +Ts,1hLTs,2 =.k1+hL)To determine the wind chill effect, we must determine the heat loss for the windy day and useit to evaluate the hypothetical ambient air temperature, T , which would provide the sameheat loss on a calm day, Hence,q =Ts,1 TTs,1 T=L 1L 1+k + h windy k h calmFrom these relations, we can now find the results sought:(a)0.003 m1+qcalm = 0.2 W/m K 65 W/m 2 K = 0.015 + 0.01540.003 m1q0.015 + 0.04windy+0.2 W/m K 25 W/m 2 Kqcalm = 0.553qwindy<15 C +(b)Ts,2 calm=1+(0.2 W/m K(c)windy=36 C= 22.1 C0.2 W/m K<= 10.8 C<(25 W/m2 K )(0.003 m)15 C +Ts,2 )25 W/m 2 K ( 0.003 m )1+0.2 W/m K(65 W/m2 K )(0.003m )(65 W/m2 K )(0.003m )T = 36 C (36 + 15 ) C0.2 W/m K36 C(0.003/0.2 + 1/ 25 ) = 56.3 C(0.003 / 0.2 + 1/ 65)COMMENTS: The wind chill effect is equivalent to a decrease of Ts,2 by 11.3C and-1increase in the heat loss by a factor of (0.553) = 1.81.<PROBLEM 3.8KNOWN: Dimensions of a thermopane window. Room and ambient air conditions.FIND: (a) Heat loss through window, (b) Effect of variation in outside convection coefficient fordouble and triple pane construction.SCHEMATIC (Double Pane):ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constantproperties, (4) Negligible radiation effects, (5) Air between glass is stagnant.PROPERTIES: Table A-3, Glass (300 K): kg = 1.4 W/mK; Table A-4, Air (T = 278 K): ka =0.0245 W/mK.ANALYSIS: (a) From the thermal circuit, the heat loss isT,i T,oq=1 1LLL1++++A hi kg ka kg ho q=(20 C 10 C) 1 10.007 m0.007 m0.007 m1++++ 0.4 m 2 10 W m 2 K 1.4 W m K 0.0245 W m K 1.4 W m K 80 W m 2 K q=30 C(0.25 + 0.0125 + 0.715 + 0.0125 + 0.03125 ) KW=30 C1.021K W= 29.4 W<(b) For the triple pane window, the additional pane and airspace increase the total resistance from1.021 K/W to 1.749 K/W, thereby reducing the heat loss from 29.4 to 17.2 W. The effect of ho on theheat loss is plotted as follows.30Heat loss, q(W)27242118151028466482100Outside convection coefficient, ho(W/m^2.K)Double paneTriple paneContinued...PROBLEM 3.8 (Cont.)Changes in ho influence the heat loss at small values of ho, for which the outside convection resistanceis not negligible relative to the total resistance. However, the resistance becomes negligible withincreasing ho, particularly for the triple pane window, and changes in ho have little effect on the heatloss.COMMENTS: The largest contribution to the thermal resistance is due to conduction across theenclosed air. Note that this air could be in motion due to free convection currents. If thecorresponding convection coefficient exceeded 3.5 W/m2K, the thermal resistance would be less thanthat predicted by assuming conduction across stagnant air.PROBLEM 3.9KNOWN: Thicknesses of three materials which form a composite wall and thermalconductivities of two of the materials. Inner and outer surface temperatures of the composite;also, temperature and convection coefficient associated with adjoining gas.FIND: Value of unknown thermal conductivity, kB.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties, (4) Negligible contact resistance, (5) Negligible radiation effects.ANALYSIS: Referring to the thermal circuit, the heat flux may be expressed asTs,i Ts,o(600 20 ) Cq ==0.3 m0.15 m0.15 mL A L B LC++++kB50 W/m Kk A k B k C 20 W/m Kq=580W/m 2 .0.018+0.15/k B(1)The heat flux may be obtained from()q=h T Ts,i = 25 W/m 2 K (800-600 ) C(2)q=5000 W/m 2 .Substituting for the heat flux from Eq. (2) into Eq. (1), find0.15 580580= 0.018 = 0.018 = 0.098kBq5000k B = 1.53 W/m K.COMMENTS: Radiation effects are likely to have a significant influence on the net heatflux at the inner surface of the oven.<PROBLEM 3.10KNOWN: Properties and dimensions of a composite oven window providing an outer surface safe2to-touch temperature Ts,o = 43C with outer convection coefficient ho = 30 W/m K and = 0.9 whenthe oven wall air temperatures are Tw = Ta = 400C. See Example 3.1.FIND: Values of the outer convection coefficient ho required to maintain the safe-to-touch conditionwhen the oven wall-air temperature is raised to 500C or 600C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in window with nocontact resistance and constant properties, (3) Negligible absorption in window material, (4)Radiation exchange processes are between small surface and large isothermal surroundings.ANALYSIS: From the analysis in the Ex. 3.1 Comment 2, the surface energy balances at the innerand outer surfaces are used to determine the required value of ho when Ts,o = 43C and Tw,i = Ta =500 or 600C.)(44 Tw,i Ts,i + hi (Ta Ts,i ) =Ts,i Ts,o(LA / k A ) + (LB / k B )(Ts,i Ts,o(LA / k A ) + (LB / k B ))(44= Ts,o Tw,o + h o Ts,o T)Using these relations in IHT, the following results were calculated:Tw,i, Ts(C)400500600Ts,i(C)3924935942ho(W/m K)3040.450.7COMMENTS: Note that the window inner surface temperature is closer to the oven air-walltemperature as the outer convection coefficient increases. Why is this so?PROBLEM 3.11KNOWN: Drying oven wall having material with known thermal conductivity sandwiched between thinmetal sheets. Radiation and convection conditions prescribed on inner surface; convection conditions onouter surface.FIND: (a) Thermal circuit representing wall and processes and (b) Insulation thickness required tomaintain outer wall surface at To = 40C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Thermalresistance of metal sheets negligible.ANALYSIS: (a) The thermal circuit is shown above. Note labels for the temperatures, thermalresistances and the relevant heat fluxes.(b) Perform energy balances on the i- and o- nodes findingT,i TiR cv,iT T+ o i + q = 0radR cd(1)Ti To T,o To+=0R R cdcv,o(2)where the thermal resistances areR = 1/ h i = 0.0333 m 2 K / Wcv,i(3)R = L / k = L / 0.05 m 2 K / Wcd(4)2R cv,o = 1/ h o = 0.0100 m K / W(5)Substituting numerical values, and solving Eqs. (1) and (2) simultaneously, find<L = 86 mmCOMMENTS: (1) The temperature at the inner surface can be found from an energy balance on thei-node using the value found for L.T,i TiR cv,o+T,o TiR + R cdcv,i+ q = 0radTi = 298.3CIt follows that Ti is close to T,i since the wall represents the dominant resistance of the system.(2) Verify that q = 50 W / m 2 and q = 150 W / m 2 . Is the overall energy balance on the systemoisatisfied?PROBLEM 3.12KNOWN: Configurations of exterior wall. Inner and outer surface conditions.FIND: Heating load for each of the three cases.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties, (4)Negligible radiation effects.PROPERTIES: (T = 300 K): Table A.3: plaster board, kp = 0.17 W/mK; urethane, kf = 0.026 W/mK;wood, kw = 0.12 W/mK; glass, kg = 1.4 W/mK. Table A.4: air, ka = 0.0263 W/mK.ANALYSIS: (a) The heat loss may be obtained by dividing the overall temperature difference by thetotal thermal resistance. For the composite wall of unit surface area, A = 1 m2,q=T ,i T ,o(1 h i ) + ( L p k p ) + ( Lf k f ) + ( L w k w ) + (1 h o ) A(q=20 C 15 C)( 0.2 + 0.059 + 1.92 + 0.083 + 0.067 ) m 2 K W 1m 2q=35 C2.33 K W<= 15.0 W(b) For the single pane of glass,q=T ,i T ,o(1 h i ) + ( Lg k g ) + (1 h o ) Aq=35 C( 0.2 + 0.002 + 0.067 ) m 2 K W 1m 2=35 C0.269 K W= 130.3 W<(c) For the double pane window,q=T ,i T ,o(1 h i ) + 2 ( Lg k g ) + ( La k a ) + (1 h o ) Aq=35 C( 0.2 + 0.004 + 0.190 + 0.067 ) m 2 K W 1m 2=35 C0.461K W= 75.9 W<COMMENTS: The composite wall is clearly superior from the standpoint of reducing heat loss, and thedominant contribution to its total thermal resistance (82%) is associated with the foam insulation. Evenwith double pane construction, heat loss through the window is significantly larger than that for thecomposite wall.PROBLEM 3.13KNOWN: Composite wall of a house with prescribed convection processes at inner andouter surfaces.FIND: (a) Expression for thermal resistance of house wall, Rtot; (b) Total heat loss, q(W); (c)Effect on heat loss due to increase in outside heat transfer convection coefficient, ho; and (d)Controlling resistance for heat loss from house.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3)Negligible contact resistance.()PROPERTIES: Table A-3, T = ( Ti + To ) / 2 = ( 20 15 ) C/2=2.5C 300K : Fiberglass3blanket, 28 kg/m , kb = 0.038 W/mK; Plywood siding, ks = 0.12 W/mK; Plasterboard, kp =0.17 W/mK.ANALYSIS: (a) The expression for the total thermal resistance of the house wall followsfrom Eq. 3.18.LpLL11R tot =++ b+ s+.<hiA k pA k bA ksA hoA(b) The total heat loss through the house wall isq = T/R tot = ( Ti To ) / R tot .Substituting numerical values, find1R tot =0.01m++0.10m30W/m 2 K 350m 2 0.17W/m K 350m 2 0.038W/m K 350m 20.02m1++2 60W/m 2 K 350m 20.12W/m K 350mR tot = [9.52 + 16.8 + 752 + 47.6 + 4.76] 105 C/W = 831 105 C/WThe heat loss is then,q= 20- (-15 ) C/83110-5 C/W=4.21 kW.2<-5(c) If ho changes from 60 to 300 W/m K, Ro = 1/hoA changes from 4.76 10 C/W to 0.95-5-5 10 C/W. This reduces Rtot to 826 10 C/W, which is a 0.5% decrease and hence a0.5% increase in q.(d) From the expression for Rtot in part (b), note that the insulation resistance, Lb/kbA, is752/830 90% of the total resistance. Hence, this material layer controls the resistance of thewall. From part (c) note that a 5-fold decrease in the outer convection resistance due to anincrease in the wind velocity has a negligible effect on the heat loss.PROBLEM 3.14KNOWN: Composite wall of a house with prescribed convection processes at inner andouter surfaces.FIND: Daily heat loss for prescribed diurnal variation in ambient air temperature.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wallthermal energy storage over 24h period), (2) Negligible contact resistance.3PROPERTIES: Table A-3, T 300 K: Fiberglass blanket (28 kg/m ), kb = 0.038 W/mK;Plywood, ks = 0.12 W/mK; Plasterboard, kp = 0.17 W/mK.ANALYSIS: The heat loss may be approximated as Q =24h T T,i,o0R totdt whereL p L b Ls1++++A hi k p k b ks h o 110.01m0.1m0.02m1R tot =++++222200m 30 W/m K 0.17 W/m K 0.038 W/m K 0.12 W/m K 60 W/m K R tot =11R tot = 0.01454 K/W.Hence the heat rate is12h12Q= 293 273 + 5 sinR tot 240Q = 68.8t dt +24h122 293 273 + 11 sin 24 t dt W 2 t 12 2 t 24 24 24 20t+5 cos 0 + 20t+11 2 cos 24 12 K hK 24 2 60132Q = 68.8 240 + ( 1 1) + 480 240 +(1 + 1) W hQ = 68.8 {480-38.2+84.03} W hQ=36.18 kW h=1.302 108J.COMMENTS: From knowledge of the fuel cost, the total daily heating bill could bedetermined. For example, at a cost of 0.10$/kWh, the heating bill would be $3.62/day.<PROBLEM 3.15KNOWN: Dimensions and materials associated with a composite wall (2.5m 6.5m, 10 studs each2.5m high).FIND: Wall thermal resistance.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature of composite depends only on x(surfaces normal to x are isothermal), (3) Constant properties, (4) Negligible contact resistance.PROPERTIES: Table A-3 (T 300K): Hardwood siding, kA = 0.094 W/mK; Hardwood,3kB = 0.16 W/mK; Gypsum, kC = 0.17 W/mK; Insulation (glass fiber paper faced, 28 kg/m ),kD = 0.038 W/mK.ANALYSIS: Using the isothermal surface assumption, the thermal circuit associated with a singleunit (enclosed by dashed lines) of the wall is( LA / k A A A ) =0.008m= 0.0524 K/W0.094 W/m K (0.65m 2.5m )( LB / k BA B ) =0.13m= 8.125 K/W0.16 W/m K ( 0.04m 2.5m )( LD /k D A D ) =0.13m= 2.243 K/W0.038 W/m K ( 0.61m 2.5m )( LC / k C A C ) =0.012m= 0.0434 K/W.0.17 W/m K ( 0.65m 2.5m )The equivalent resistance of the core is1R eq = (1/ R B + 1/ R D )1= (1/ 8.125 + 1/ 2.243)= 1.758 K/Wand the total unit resistance isR tot,1 = R A + R eq + R C = 1.854 K/W.With 10 such units in parallel, the total wall resistance is(R tot = 10 1/ R tot,1)1 = 0.1854 K/W.COMMENTS: If surfaces parallel to the heat flow direction are assumed adiabatic, the thermalcircuit and the value of Rtot will differ.<PROBLEM 3.16KNOWN: Conditions associated with maintaining heated and cooled conditions within a refrigeratorcompartment.FIND: Coefficient of performance (COP).SCHEMATIC:ASSUMPTIONS: (1) Steady-state operating conditions, (2) Negligible radiation, (3) Compartmentcompletely sealed from ambient air.ANALYSIS: The Case (a) experiment is performed to determine the overall thermal resistance to heattransfer between the interior of the refrigerator and the ambient air. Applying an energy balance to acontrol surface about the refrigerator, it follows from Eq. 1.11a that, at any instant,E g E out = 0Hence,q elec q out = 0(where q out = T,i T,o)R t . It follows thatT,i T,o (90 25 ) C== 3.25 C/Wq elec20 WFor Case (b), heat transfer from the ambient air to the compartment (the heat load) is balanced by heattransfer to the refrigerant (qin = qout). Hence, the thermal energy transferred from the refrigerator over the12 hour period isT,i T,oQout = q out t = qin t =tRtRt =( 25 5 ) C(12 h 3600 s h ) = 266, 000 J3.25 C WThe coefficient of performance (COP) is thereforeQ266, 000COP = out == 2.13Win125, 000COMMENTS: The ideal (Carnot) COP isTc278 KCOP )ideal === 13.9Th Tc ( 298 278 ) KQout =and the system is operating well below its peak possible performance.<PROBLEM 3.17KNOWN: Total floor space and vertical distance between floors for a square, flat roof building.FIND: (a) Expression for width of building which minimizes heat loss, (b) Width and number of floorswhich minimize heat loss for a prescribed floor space and distance between floors. Corresponding heatloss, percent heat loss reduction from 2 floors.SCHEMATIC:ASSUMPTIONS: Negligible heat loss to ground.ANALYSIS: (a) To minimize the heat loss q, the exterior surface area, As, must be minimized. FromFig. (a)As = W 2 + 4WH = W 2 + 4WNf HfwhereNf = Af W 2Hence,As = W 2 + 4WAf Hf W 2 = W 2 + 4A f Hf WThe optimum value of W corresponds todAs4Af Hf= 2W =0dWW2orWop = ( 2Af Hf )1/ 3<The competing effects of W on the areas of the roof and sidewalls, and hence the basis for an optimum, isshown schematically in Fig. (b).(b) For Af = 32,768 m2 and Hf = 4 m,(Wop = 2 32,768 m 2 4 m)1/ 3= 64 m<Continued ..PROBLEM 3.17 (Cont.)Hence,Nf =AfW2=32, 768 m 2(64 m )2=8<and22 4 32, 768 m 4 m q = UAs T = 1W m 2 K ( 64 m ) + 25 C = 307, 200 W64 m<For Nf = 2,W = (Af/Nf)1/2 = (32,768 m2/2)1/2 = 128 m22 4 32, 768 m 4 m q = 1W m 2 K (128 m ) + 25 C = 512,000 W128 m% reduction in q = (512,000 - 307,200)/512,000 = 40%COMMENTS: Even the minimum heat loss is excessive and could be reduced by reducing U.<PROBLEM 3.18KNOWN: Concrete wall of 150 mm thickness experiences a flash-over fire with prescribed radiantflux and hot-gas convection on the fire-side of the wall. Exterior surface condition is 300C, typicalignition temperature for most household and office materials.FIND: (a) Thermal circuit representing wall and processes and (b) Temperature at the fire-side of thewall; comment on whether wall is likely to experience structural collapse for these conditions.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Constantproperties.PROPERTIES: Table A-3, Concrete (stone mix, 300 K): k = 1.4 W/mK.ANALYSIS: (a) The thermal cirucit is shown above. Note labels for the temperatures, thermalresistances and the relevant heat fluxes.(b) To determine the fire-side wall surface temperatures, perform an energy balance on the o-node.T ToT T+ q = L oradR R cvcdwhere the thermal resistances areR = 1/ h i = 1/ 200 W / m 2 K = 0.00500 m 2 K / WcvR = L / k = 0.150 m /1.4 W / m K = 0.107 m 2 K / WcdSubstituting numerical values,( 400 To ) K0.005 m 2 K / WTo = 515C+ 25, 000 W / m 2(300 To ) K0.107 m 2 K / W=0<COMMENTS: (1) The fire-side wall surface temperature is within the 350 to 600C range for whichexplosive spalling could occur. It is likely the wall will experience structural collapse for theseconditions.(2) This steady-state condition is an extreme condition, as the wall may fail before near steady-stateconditions can be met.PROBLEM 3.19KNOWN: Representative dimensions and thermal conductivities for the layers of fire-fightersprotective clothing, a turnout coat.FIND: (a) Thermal circuit representing the turnout coat; tabulate thermal resistances of the layersand processes; and (b) For a prescribed radiant heat flux on the fire-side surface and temperature ofTi =.60C at the inner surface, calculate the fire-side surface temperature, To.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through the layers,(3) Heat is transferred by conduction and radiation exchange across the stagnant air gaps, (3) Constantproperties.PROPERTIES: Table A-4, Air (470 K, 1 atm): kab = kcd = 0.0387 W/mK.ANALYSIS: (a) The thermal circuit is shown with labels for the temperatures and thermalresistances.The conduction thermal resistances have the form R = L / k while the radiation thermalcdresistances across the air gaps have the formR =rad1h rad=134 TavgThe linearized radiation coefficient follows from Eqs. 1.8 and 1.9 with = 1 where Tavg representsthe average temperature of the surfaces comprising the gap()223h rad = ( T1 + T2 ) T1 + T2 4 TavgFor the radiation thermal resistances tabulated below, we used Tavg = 470 K.Continued ..PROBLEM 3.19 (Cont.)Shell(s)Air gap(a-b)Barrier(mb)Air gap(c-d)Liner(tl)R m K / Wcd0.02590.045830.02590.00921R rad20.04264--0.04264----R gap20.01611--0.01611------0.1043() 0.01702(m K / W ) -(m K / W ) -2R total--------Total(tot)--From the thermal circuit, the resistance across the gap for the conduction and radiation processes is111=+R gap R cd R radand the total thermal resistance of the turn coat isR = R + R totcd,sgap,a b + R cd,mb + R gap,c d + R cd,tl2(b) If the heat flux through the coat is 0.25 W/cm , the fire-side surface temperature To can becalculated from the rate equation written in terms of the overall thermal resistance.q = ( To Ti ) / R tot(To = 66C + 0.25 W / cm 2 102 cm / m) 0.1043 m2 K / W2To = 327CCOMMENTS: (1) From the tabulated results, note that the thermal resistance of the moisture barrier(mb) is nearly 3 times larger than that for the shell or air gap layers, and 4.5 times larger than thethermal liner layer.(2) The air gap conduction and radiation resistances were calculated based upon the averagetemperature of 470 K. This value was determined by setting Tavg = (To + Ti)/2 and solving theequation set using IHT with kair = kair (Tavg).PROBLEM 3.20KNOWN: Materials and dimensions of a composite wall separating a combustion gas from aliquid coolant.FIND: (a) Heat loss per unit area, and (b) Temperature distribution.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3)Constant properties, (4) Negligible radiation effects.PROPERTIES: Table A-1, St. St. (304) ( T 1000K ) : k = 25.4 W/mK; Table A-2,Beryllium Oxide (T 1500K): k = 21.5 W/mK.ANALYSIS: (a) The desired heat flux may be expressed asq=T,1 T,2( 2600 100 ) C=1 LAL10.021 m 2 .K 1 0.01++ R t,c + B +++ 0.05 ++h1 k Ak B h 2 50 21.525.4 1000 Wq=34,600 W/m 2 .<(b) The composite surface temperatures may be obtained by applying appropriate rateequations. From the fact that q=h1 T,1 Ts,1 , it follows that(Ts,1 = T,1 ()q34, 600 W/m 21908 C.= 2600 C 2 Kh150 W/m)With q= ( k A / LA ) Ts,1 Tc,1 , it also follows thatL q0.01m 34,600 W/m 2Tc,1 = Ts,1 A = 1908 C = 1892 C.kA21.5 W/m K()Similarly, with q= Tc,1 Tc,2 / R t,cTc,2 = Tc,1 R t,c q=1892C 0.05m2 KW 34, 600= 162 C2WmContinued ..PROBLEM 3.20 (Cont.)()and with q= ( k B / LB ) Tc,2 Ts,2 ,L q0.02m 34,600 W/m 2Ts,2 = Tc,2 B = 162 C = 134.6 C.kB25.4 W/m KThe temperature distribution is therefore of the following form:<COMMENTS: (1) The calculations may be checked by recomputing q from()q=h 2 Ts,2 T,2 = 1000W/m2 K (134.6-100 ) C=34,600W/m2(2) The initial estimates of the mean material temperatures are in error, particularly for thestainless steel. For improved accuracy the calculations should be repeated using k valuescorresponding to T 1900C for the oxide and T 115C for the steel.(3) The major contributions to the total resistance are made by the combustion gas boundarylayer and the contact, where the temperature drops are largest.PROBLEM 3.21KNOWN: Thickness, overall temperature difference, and pressure for two stainless steelplates.FIND: (a) Heat flux and (b) Contact plane temperature drop.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3)Constant properties.PROPERTIES: Table A-1, Stainless Steel (T 400K): k = 16.6 W/mK.ANALYSIS: (a) With R 15 104 m 2 K/W from Table 3.1 andt,cL0.01m== 6.02 104 m 2 K/W,k 16.6 W/m Kit follows thatR = 2 ( L/k ) + R 27 104 m 2 K/W;tott,chenceq=100 CT== 3.70 104 W/m 2 .-4 m 2 K/WR tot 27 10<(b) From the thermal circuit,4 2R Tct,c 15 10 m K/W=== 0.556.Ts,1 Ts,2 R 27 10-4 m 2 K/WtotHence,()()Tc = 0.556 Ts,1 Ts,2 = 0.556 100 C = 55.6 C.<COMMENTS: The contact resistance is significant relative to the conduction resistances.The value of R ,c would diminish, however, with increasing pressure.tPROBLEM 3.22KNOWN: Temperatures and convection coefficients associated with fluids at inner and outersurfaces of a composite wall. Contact resistance, dimensions, and thermal conductivitiesassociated with wall materials.FIND: (a) Rate of heat transfer through the wall, (b) Temperature distribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3)Negligible radiation, (4) Constant properties.ANALYSIS: (a) Calculate the total resistance to find the heat rate,1LL1+ A + R t,c + B +h1A k A Ak BA h 2 A0.01 0.30.021 K1R tot = ++++10 5 0.1 5 5 0.04 5 20 5 WKKR tot = [0.02 + 0.02 + 0.06 + 0.10 + 0.01] = 0.21WWR tot =q=T,1 T,2R tot( 200 40 ) C = 762 W.=0.21 K/W(b) It follows thatTs,1 = T,1 TA = Ts,1 q762 W= 200 C = 184.8 Ch1A50 W/KqL A= 184.8 C kAA762W 0.01m0.1WmK 5m= 169.6 C2KTB = TA qR t,c = 169.6 C 762W 0.06Ts,2 = TB qL Bk BAT,2 = Ts,2 = 123.8 C qh 2AW762W 0.02m0.04= 47.6 C WmK 5m762W100W/K= 123.8 C= 47.6 C2= 40 C<PROBLEM 3.23KNOWN: Outer and inner surface convection conditions associated with zirconia-coated, Inconelturbine blade. Thicknesses, thermal conductivities, and interfacial resistance of the blade materials.Maximum allowable temperature of Inconel.FIND: Whether blade operates below maximum temperature. Temperature distribution in blade, withand without the TBC.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constantproperties, (3) Negligible radiation.ANALYSIS: For a unit area, the total thermal resistance with the TBC is11R tot,w = h o + ( L k )Zr + R + ( L k )In + h it,c()34443R m 2 K W = 3.69 103 m 2 K Wtot,w = 10 + 3.85 10 + 10 + 2 10 + 2 10With a heat flux ofT,o T,i1300 Kq === 3.52 105 W m 2w3 2R 3.69 10 m K Wtot,wthe inner and outer surface temperatures of the Inconel are()Ts,i(w) = T ,i + ( q h i ) = 400 K + 3.52 105 W m 2 500 W m 2 K = 1104 Kw(Ts,o(w) = T ,i + (1 h i ) + ( L k )In q = 400 K + 2 10w3+ 2 104)m2(5 K W 3.52 10 W m2) = 1174 K113 2Without the TBC, R tot,wo = h o + ( L k )In + h i = 3.20 10 m K W , and q wo = ( T ,o T ,i ) R tot,wo =(1300 K)/3.2010-3 m2K/W = 4.06105 W/m2. The inner and outer surface temperatures of the Inconelare then()Ts,i(wo) = T ,i + ( q h i ) = 400 K + 4.06 105 W m2 500 W m 2 K = 1212 KwoTs,o(wo) = T ,i +[(1 h i ) + ( L k )In ] qwo = 400 K + ( 2 103 + 2 104 ) m 2 K(5W 4.06 10 W m2)= 1293 KContinued...PROBLEM 3.23 (Cont.)Temperature, T(K)13001260122011801140110000.0010.0020.0030.0040.005Inconel location, x(m)With TBCWithout TBCUse of the TBC facilitates operation of the Inconel below Tmax = 1250 K.COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increaseswith increasing thickness, limits to the thickness are associated with reliability considerations.PROBLEM 3.24KNOWN: Size and surface temperatures of a cubical freezer. Materials, thicknesses and interfaceresistances of freezer wall.FIND: Cooling load.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties.PROPERTIES: Table A-1, Aluminum 2024 (~267K): kal = 173 W/mK. Table A-1, Carbon steelAISI 1010 (~295K): kst = 64 W/mK. Table A-3 (~300K): kins = 0.039 W/mK.ANALYSIS: For a unit wall surface area, the total thermal resistance of the composite wall isLLLR = al + R + ins + R + sttott,ct,ck alk inskstR =tot0.00635mm2 K0.100mm2 K0.00635m+ 2.5 104++ 2.5 104+173 W / m KW0.039 W / m KW64 W / m K)(R = 3.7 105 + 2.5 104 + 2.56 + 2.5 104 + 9.9 105 m 2 K / WtotHence, the heat flux isq =Ts,o Ts,iR tot 22 ( 6 ) CW== 10.92.56 m 2 K / Wm2and the cooling load isq = As q = 6 W 2 q = 54m 2 10.9 W / m 2 = 590 WCOMMENTS: Thermal resistances associated with the cladding and the adhesive joints arenegligible compared to that of the insulation.<PROBLEM 3.25KNOWN: Thicknesses and thermal conductivity of window glass and insulation. Contact resistance.Environmental temperatures and convection coefficients. Furnace efficiency and fuel cost.FIND: (a) Reduction in heat loss associated with the insulation, (b) Heat losses for prescribedconditions, (c) Savings in fuel costs for 12 hour period.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Constant properties.ANALYSIS: (a) The percentage reduction in heat loss isq qwith 100% = 1 qwithR q = woqqwowoR tot,wo 100% = 1 100% R tot, with where the total thermal resistances without and with the insulation, respectively, areR tot,wo = R cnv,o + R cnd,w + R cnv,i =1 Lw 1++ho k w hi22R tot,wo = ( 0.050 + 0.004 + 0.200 ) m K / W = 0.254 m K / WR tot,with = R cnv,o + R cnd,w + R + R t,ccnd,ins + R cnv,i =L1 Lw1++ R + ins +t,cho k wkins h i22R tot,with = (0.050 + 0.004 + 0.002 + 0.926 + 0.500 ) m K / W = 1.482 m K / WR q = (1 0.254 /1.482 ) 100% = 82.9%<2(b) With As = 12 m , the heat losses without and with the insulation are22q wo = As T,i T,o / R tot,wo = 12 m 32C / 0.254 m K / W = 1512 W()()22q with = As T,i T,o / R tot,with = 12 m 32C /1.482 m K / W = 259 W<<(c) With the windows covered for 12 hours per day, the daily savings areS=( q wo q with )ft C g 106MJ / J =(1512 259 ) W0.812h 3600 s / h $0.01 / MJ 106MJ / J = $0.677COMMENTS: (1) The savings may be insufficient to justify the cost of the insulation, as well as thedaily tedium of applying and removing the insulation. However, the losses are significant andunacceptable. The owner of the building should install double pane windows. (2) The dominantcontributions to the total thermal resistance are made by the insulation and convection at the innersurface.PROBLEM 3.26KNOWN: Surface area and maximum temperature of a chip. Thickness of aluminum coverand chip/cover contact resistance. Fluid convection conditions.FIND: Maximum chip power.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3)Negligible heat loss from sides and bottom, (4) Chip is isothermal.PROPERTIES: Table A.1, Aluminum (T 325 K): k = 238 W/mK.ANALYSIS: For a control surface about the chip, conservation of energy yieldsEg Eout = 0orPc (Tc T ) A=0( L/k ) + R + (1/ h )t,c(85 25 ) CPc,max =Pc,max =(10-4m2 )( 0.002 / 238 ) + 0.5 104 + (1/1000 ) m 2 K/W4 C m 260 10(8.4 10-6 + 0.5 104 + 103 ) m2 K/W<Pc,max = 5.7 W.!&COMMENTS: The dominant resistance is that due to convection R conv > R t,c >> R cond .PROBLEM 3.27KNOWN: Operating conditions for a board mounted chip.FIND: (a) Equivalent thermal circuit, (b) Chip temperature, (c) Maximum allowable heat dissipation fordielectric liquid (ho = 1000 W/m2K) and air (ho = 100 W/m2K). Effect of changes in circuit boardtemperature and contact resistance.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible chipthermal resistance, (4) Negligible radiation, (5) Constant properties.PROPERTIES: Table A-3, Aluminum oxide (polycrystalline, 358 K): kb = 32.4 W/mK.ANALYSIS: (a)(b) Applying conservation of energy to a control surface about the chip ( E in E out = 0 ) , oq qi q = 0cq =cTc T,i1 h i + ( L k )b + R t,c+Tc T ,o1 hoWith q TT 3 ` 10 4 W m 2 , ho = 1000 W/m2K, kb = 1 W/mK and R = 10 4 m 2 K W ,ct,c3 104 W m 2 =Tc 20 C(1 40 + 0.005 1 + 104 ) m2 K W+Tc 20 C(1 1000 ) m 2 KW3 104 W m 2 = (33.2Tc 664 + 1000Tc 20, 000 ) W m 2 K1003Tc = 50,664<Tc = 49C.(c) For Tc = 85C and ho = 1000 W/m2K, the foregoing energy balance yields<q = 67,160 W m 2cwith q = 65,000 W/m and q = 2160 W/m . Replacing the dielectric with air (ho = 100 W/m K), theoifollowing results are obtained for different combinations of kb and R ,c .t222Continued...PROBLEM 3.27 (Cont.)kb (W/mK)R t,cq (W/m )iq (W/m )oq (W/m )c215925742166258365006500650065008659907486669083222(m2K/W)132.4132.410-410-410-510-5<COMMENTS: 1. For the conditions of part (b), the total internal resistance is 0.0301 m2K/W, whilethe outer resistance is 0.001 m2K/W. Hence((T T,oqo= cqTc T,ii) R = 0.0301 = 30 .o) R 0.001iand only approximately 3% of the heat is dissipated through the board. o2. With ho = 100 W/m2K, the outer resistance increases to 0.01 m2K/W, in which case q q i = R i R o = 0.0301/0.01 = 3.1 and now almost 25% of the heat is dissipated through the board. Hence, althoughmeasures to reduce R i would have a negligible effect on q for the liquid coolant, some improvementcmay be gained for air-cooled conditions. As shown in the table of part (b), use of an aluminum oxideboard increase q by 19% (from 2159 to 2574 W/m2) by reducing R i from 0.0301 to 0.0253 m2K/W.iBecause the initial contact resistance ( R = 10 4 m 2 K W ) is already much less than R i , any reductiont,cin its value would have a negligible effect on q . The largest gain would be realized by increasing hi,isince the inside convection resistance makes the dominant contribution to the total internal resistance.PROBLEM 3.28KNOWN: Dimensions, thermal conductivity and emissivity of base plate. Temperature andconvection coefficient of adjoining air. Temperature of surroundings. Maximum allowabletemperature of transistor case. Case-plate interface conditions.FIND: (a) Maximum allowable power dissipation for an air-filled interface, (b) Effect of convectioncoefficient on maximum allowable power dissipation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Negligible heat transfer from the enclosure, to thesurroundings. (3) One-dimensional conduction in the base plate, (4) Radiation exchange at surface ofbase plate is with large surroundings, (5) Constant thermal conductivity.52PROPERTIES: Aluminum-aluminum interface, air-filled, 10 m roughness, 10 N/m contactpressure (Table 3.1): R = 2.75 104 m 2 K / W.t,cANALYSIS: (a) With all of the heat dissipation transferred through the base plate,Pelec = q =Ts,c T(1)R totwhere R tot = R t,c + R cnd + (1/ R cnv ) + (1/ R rad )R tot =andR t,cAc+LkW 2(h r = Ts,p + Tsur+11 1 W2 h + hr (2)22) (Ts,p + Tsur )(3)To obtain Ts,p, the following energy balance must be performed on the plate surface,q=Ts,c Ts,pR t,c + R cnd-42()(= qcnv + q rad = hW 2 Ts,p T + h r W 2 Ts,p Tsur-4)2(4)-42With Rt,c = 2.75 10 m K/W/210 m = 1.375 K/W, Rcnd = 0.006 m/(240 W/mK 4 10 m )= 0.0625 K/W, and the prescribed values of h, W, T = Tsur and , Eq. (4) yields a surfacetemperature of Ts,p = 357.6 K = 84.6C and a power dissipation ofContinued ..PROBLEM 3.28 (Cont.)<Pelec = q = 0.268 WThe convection and radiation resistances are Rcnv = 625 mK/W and Rrad = 345 mK/W, where hr =27.25 W/m K.P o w e r d is s ip a tio n , P e le c (W )(b) With the major contribution to the total resistance made by convection, significant benefit may bederived by increasing the value of h.4 .543 .532 .521 .510 .50020406080100 120 140 160 180 200C o n ve ctio n c o e ffic ie n t, h (W /m ^2 .K )2For h = 200 W/m K, Rcnv = 12.5 mK/W and Ts,p = 351.6 K, yielding Rrad = 355 mK/W. The effectof radiation is then negligible.2COMMENTS: (1) The plate conduction resistance is negligible, and even for h = 200 W/m K, thecontact resistance is small relative to the convection resistance. However, Rt,c could be renderednegligible by using indium foil, instead of an air gap, at the interface. From Table 3.1,R = 0.07 104 m 2 K / W, in which case Rt,c = 0.035 mK/W.t,c2(2) Because Ac < W , heat transfer by conduction in the plate is actually two-dimensional, renderingthe conduction resistance even smaller.PROBLEM 3.29KNOWN: Conduction in a conical section with prescribed diameter, D, as a function of x in1/2the form D = ax .FIND: (a) Temperature distribution, T(x), (b) Heat transfer rate, qx.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in xdirection, (3) No internal heat generation, (4) Constant properties.PROPERTIES: Table A-2, Pure Aluminum (500K): k= 236 W/mK.ANALYSIS: (a) Based upon the assumptions, and following the same methodology ofExample 3.3, qx is a constant independent of x. Accordingly,q x = kA2using A = D /4 where D = ax4q xx dx a 2k)(2 dTdT= k ax1/2 / 4 dx dxx1x1/2TT1= (1). Separating variables and identifying limits,(2)dT.Integrating and solving for T(x) and then for T2,T ( x ) = T1 4q x a 2klnxx1T2 = T1 4q xxln 2 . a 2k x1(3,4)Solving Eq. (4) for qx and then substituting into Eq. (3) gives the results,qx = 2a k ( T1 T2 ) /1n ( x1 / x 2 )4T ( x ) = T1 + ( T1 T2 )ln ( x/x1 )ln ( x1 / x 2 ).(5)<From Eq. (1) note that (dT/dx)x = Constant. It follows that T(x) has the distribution shownabove.(b) The heat rate follows from Eq. (5),qx =W25 0.52 m 236(600 400 ) K/ln = 5.76kW.4mK125<PROBLEM 3.30KNOWN: Geometry and surface conditions of a truncated solid cone.FIND: (a) Temperature distribution, (b) Rate of heat transfer across the cone.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x, (3)Constant properties.PROPERTIES: Table A-1, Aluminum (333K): k = 238 W/mK.()ANALYSIS: (a) From Fouriers law, Eq. (2.1), with A= D 2 / 4 = a 2 / 4 x 3 , it follows that4q x dx a 2 x3= kdT.Hence, since qx is independent of x,T4q x x dx= k dT2 x1 x 3T1aorx4q x 1 = k ( T T1 ). a 2 2x 2 x1Hence1 1 .2 a 2 k x 2 x1 (b) From the foregoing expression, it also follows thatT = T1 +2q x<T2 T1 a 2k21/x 2 1/ x 2 12-1 238 W/m K 1m( 20 100 ) Cqx =2( 0.225 )2 ( 0.075)2 m-2qx =()q x = 189 W.<COMMENTS: The foregoing results are approximate due to use of a one-dimensional modelin treating what is inherently a two-dimensional problem.PROBLEM 3.31KNOWN: Temperature dependence of the thermal conductivity, k.FIND: Heat flux and form of temperature distribution for a plane wall.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-stateconditions, (3) No internal heat generation.ANALYSIS: For the assumed conditions, qx and A(x) are constant and Eq. 3.21 givesLTq dx = 1 ( k o + aT )dTx0q =xTo()1a22 k o ( To T1 ) + 2 To T1 .LFrom Fouriers law,q = ( k o + aT ) dT/dx.xHence, since the product of (ko+aT) and dT/dx) is constant, decreasing T with increasing ximplies,a > 0: decreasing (ko+aT) and increasing |dT/dx| with increasing xa = 0: k = ko => constant (dT/dx)a < 0: increasing (ko+aT) and decreasing |dT/dx| with increasing x.The temperature distributions appear as shown in the above sketch.PROBLEM 3.32KNOWN: Temperature dependence of tube wall thermal conductivity.FIND: Expressions for heat transfer per unit length and tube wall thermal (conduction)resistance.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)No internal heat generation.ANALYSIS: From Eq. 3.24, the appropriate form of Fouriers law isdTdT= k ( 2 rL )drdrdTq = 2 krrdrdTq = 2 rk o (1 + aT ) .rdrq r = kA rSeparating variables,q drr= k o (1 + aT ) dT2 rand integrating across the wall, findqr2qr2qr2ro drrirln orirln orirT= k o o (1+aT ) dTTiaT 2 To= ko T +2 Tia2= k o ( To Ti ) + To Ti2 2()a (T T )q = 2 k o 1 + ( To + Ti ) o i .r2 ln ( ro / ri )It follows that the overall thermal resistance per unit length isln ( ro / ri )TR t =.=qar2 k o 1 + ( To + Ti )2COMMENTS: Note the necessity of the stated assumptions to treating q as independent of r.r<<PROBLEM 3.33KNOWN: Steady-state temperature distribution of convex shape for material with k = ko(1 +T) where is a constant and the mid-point temperature is To higher than expected for alinear temperature distribution.FIND: Relationship to evaluate in terms of To and T1, T2 (the temperatures at theboundaries).SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Nointernal heat generation, (4) is positive and constant.ANALYSIS: At any location in the wall, Fouriers law has the formdTq = k o (1 + T ) .(1)xdxSince q is a constant, we can separate Eq. (1), identify appropriate integration limits, andxintegrate to obtainLT2x0 q dx = T1k o (1 + T )dT2 T2 T2 k o T2 + T1 + 1 .q = xL 22 We could perform the same integration, but with the upper limits at x = L/2, to obtain2 T2 T1 2k o TL/2 + L/2 T1 +q = xL 2 2 whereT +TTL/2 = T ( L/2 ) = 1 2 + To .2Setting Eq. (3) equal to Eq. (4), substituting from Eq. (5) for TL/2, and solving for , itfollows that2To=.222T2 + T1 / 2 ( T1 + T2 ) / 2 + To ()(2)(3)(4)(5)<PROBLEM 3.34KNOWN: Hollow cylinder of thermal conductivity k, inner and outer radii, ri and ro,respectively, and length L.FIND: Thermal resistance using the alternative conduction analysis method.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)No internal volumetric generation, (4) Constant properties.ANALYSIS: For the differential control volume, energy conservation requires that qr = qr+drfor steady-state, one-dimensional conditions with no heat generation. With Fouriers law,q r = kAdTdT= k ( 2 rL )drdr(1)where A = 2rL is the area normal to the direction of heat transfer. Since qr is constant, Eq.(1) may be separated and expressed in integral form,Toq r ro drri r = Ti k (T ) dT.2 LAssuming k is constant, the heat rate isqr =2 Lk ( Ti To )ln ( ro / ri ).Remembering that the thermal resistance is defined asR t T/qit follows that for the hollow cylinder,Rt =ln ( ro / ri )2 LK.COMMENTS: Compare the alternative method used in this analysis with the standardmethod employed in Section 3.3.1 to obtain the same result.<PROBLEM 3.35KNOWN: Thickness and inner surface temperature of calcium silicate insulation on a steam pipe.Convection and radiation conditions at outer surface.FIND: (a) Heat loss per unit pipe length for prescribed insulation thickness and outer surfacetemperature. (b) Heat loss and radial temperature distribution as a function of insulation thickness.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties.PROPERTIES: Table A-3, Calcium Silicate (T = 645 K): k = 0.089 W/mK.ANALYSIS: (a) From Eq. 3.27 with Ts,2 = 490 K, the heat rate per unit length is(2 k Ts,1 Ts,2q = q r L =q =ln ( r2 r1 ))2 ( 0.089 W m K )(800 490 ) Kln ( 0.08 m 0.06 m )<q = 603 W m .(b) Performing an energy for a control surface around the outer surface of the insulation, it follows thatqcond = qconv + qradTs,1 Ts,2ln ( r2 r1 ) 2 k(=Ts,2 T+Ts,2 Tsur1 ( 2 r2 h ) 1 ( 2 r2 h r )where h r = Ts,2 + Tsur22) (Ts,2 + Tsur ) .Solving this equation for Ts,2, the heat rate may bedetermined from()()q = 2 r2 h Ts,2 T + h r Ts,2 Tsur Continued...PROBLEM 3.35 (Cont.)and from Eq. 3.26 the temperature distribution isT(r) =Ts,1 Ts,2ln ( r1 r2 )r + Ts,2 r2 ln As shown below, the outer surface temperature of the insulation Ts,2 and the heat loss q decayprecipitously with increasing insulation thickness from values of Ts,2 = Ts,1 = 800 K and q = 11,600W/m, respectively, at r2 = r1 (no insulation).80010000Heat loss, qprime(W/m)Temperature, Ts2(K)700600500400100030010000.040.080.1200.04Insulation thickness, (r2-r1) (m)0.080.12Insulation thickness, (r2-r1) (m)Outer surface temperatureHeat loss, qprimeWhen plotted as a function of a dimensionless radius, (r - r1)/(r2 - r1), the temperature decay becomesmore pronounced with increasing r2.Temperature, T(r) (K)80070060050040030000.20.40.60.81Dimensionless radius, (r-r1)/(r2-r1)r2 = 0.20mr2 = 0.14mr2= 0.10mNote that T(r2) = Ts,2 increases with decreasing r2 and a linear temperature distribution is approached as r2approaches r1.COMMENTS: An insulation layer thickness of 20 mm is sufficient to maintain the outer surfacetemperature and heat rate below 350 K and 1000 W/m, respectively.PROBLEM 3.36KNOWN: Temperature and volume of hot water heater. Nature of heater insulating material. Ambientair temperature and convection coefficient. Unit cost of electric power.FIND: Heater dimensions and insulation thickness for which annual cost of heat loss is less than $50.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction through side and end walls, (2)Conduction resistance dominated by insulation, (3) Inner surface temperature is approximately that of thewater (Ts,1 = 55C), (4) Constant properties, (5) Negligible radiation.PROPERTIES: Table A.3, Urethane Foam (T = 300 K): k = 0.026 W/mK.ANALYSIS: To minimize heat loss, tank dimensions which minimize the total surface area, As,t, should()be selected. With L = 4/D2, As,t = DL + 2 D 2 4 = 4 D + D 2 2 , and the tank diameter forwhich As,t is an extremum is determined from the requirementdAs,t dD = 4 D2 + D = 0It follows thatD = ( 4 )1/ 3L = ( 4 )1/ 3andWith d 2 As,t dD 2 = 8 D3 + > 0 , the foregoing conditions yield the desired minimum in As,t.Hence, for = 100 gal 0.00379 m3/gal = 0.379 m3,<Dop = L op = 0.784 mThe total heat loss through the side and end walls isq=Ts,1 Tln ( r2 r1 )2 kLop+1h2 r2 Lop(2 Ts,1 T+(2k Dop 4+)12) h ( Dop 4)We begin by estimating the heat loss associated with a 25 mm thick layer of insulation. With r1 = Dop/2 =0.392 m and r2 = r1 + = 0.417 m, it follows thatContinued...PROBLEM 3.36 (Cont.)q=ln ( 0.417 0.392 )(55 20 ) C2 ( 0.026 W m K ) 0.784 m(2 W m K ) 2 (0.417 m ) 0.784 m22 (55 20 ) C+0.025 m(0.026 Wq=1+(2 W m2 K ) 4 (0.784 m )22 (35 C )= ( 48.2 + 23.1) W = 71.3 Wm K ) 4 ( 0.784 m )235 C(0.483 + 0.243) K1+W+(1.992 + 1.036 ) KWThe annual energy loss is therefore()Qannual = 71.3 W (365 days ) ( 24 h day ) 103 kW W = 625 kWhWith a unit electric power cost of $0.08/kWh, the annual cost of the heat loss isC = ($0.08/kWh)625 kWh = $50.00Hence, an insulation thickness of = 25 mm<will satisfy the prescribed cost requirement.COMMENTS: Cylindrical containers of aspect ratio L/D = 1 are seldom used because of floor spaceconstraints. Choosing L/D = 2, = D3/2 and D = (2/)1/3 = 0.623 m. Hence, L = 1.245 m, r1 =0.312m and r2 = 0.337 m. It follows that q = 76.1 W and C = $53.37. The 6.7% increase in the annualcost of the heat loss is small, providing little justification for using the optimal heater dimensions.PROBLEM 3.37KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surfaceand is exposed to a fluid of prescribed h and T. Thermal contact resistance between heaterand tube wall and wall inner surface temperature.FIND: Heater power per unit length required to maintain a heater temperature of 25C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties, (4) Negligible temperature drop across heater.ANALYSIS: The thermal circuit has the formApplying an energy balance to a control surface about the heater,q = q + qabTo TiT Tq =+o ln ( ro / ri )(1/h Do )+ R t,c2 k 25 ( 10 ) C( 25-5 ) Cq=+ln (75mm/25mm )m K 1/ 100 W/m 2 K 0.15m + 0.012 10 W/m KW()q = ( 728 + 1649 ) W/mq=2377 W/m.COMMENTS: The conduction, contact and convection resistances are 0.0175, 0.01 and0.021 m K/W, respectively,<PROBLEM 3.38KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface. Inner andouter wall temperatures. Temperature of fluid adjoining outer wall.FIND: Effect of wall thermal conductivity, thermal contact resistance, and convection coefficient ontotal heater power and heat rates to outer fluid and inner surface.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,(4) Negligible temperature drop across heater, (5) Negligible radiation.ANALYSIS: Applying an energy balance to a control surface about the heater,q = q + qioq =To TiT T+oln ( ro ri )(1 2 ro h )+ Rt,c2 kSelecting nominal values of k = 10 W/mK, R ,c = 0.01 mK/W and h = 100 W/m2K, the followingtparametric variations are obtained3500300030002500Heat rate(W/m)Heat rate (W/m)25002000150010002000150010005005000005010015020000.02Thermal conductivity, k(W/m.K)qiqqo0.040.060.080.1Contact resistance, Rtc(m.K/W)qiqqoContinued...PROBLEM 3.38 (Cont.)20000Heat rate(W/m)160001200080004000002004006008001000Convection coefficient, h(W/m^2.K)qiqqoFor a prescribed value of h, q is fixed, while qi , and hence q , increase and decrease, respectively,owith increasing k and R ,c . These trends are attributable to the effects of k and R ,c on the totaltt(conduction plus contact) resistance separating the heater from the inner surface. For fixed k and R ,c ,tqi is fixed, while q , and hence q , increase with increasing h due to a reduction in the convectionoresistance.COMMENTS: For the prescribed nominal values of k, R ,c and h, the electric power requirement istq = 2377 W/m. To maintain the prescribed heater temperature, q would increase with any changeswhich reduce the conduction, contact and/or convection resistances.PROBLEM 3.39KNOWN: Wall thickness and diameter of stainless steel tube. Inner and outer fluid temperaturesand convection coefficients.FIND: (a) Heat gain per unit length of tube, (b) Effect of adding a 10 mm thick layer of insulation toouter surface of tube.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constantproperties, (4) Negligible contact resistance between tube and insulation, (5) Negligible effect ofradiation.PROPERTIES: Table A-1, St. St. 304 (~280K): kst = 14.4 W/mK.ANALYSIS: (a) Without the insulation, the total thermal resistance per unit length isR tot = R conv,i + R cond,st + R conv,o =R tot =12 (0.018m ) 400 W / m 2 K+ln ( r2 / ri )11++2 ri h i2 k st2 r2 h oln ( 20 /18 )2 (14.4 W / m K )+12 (0.020m ) 6 W / m 2 K)(R tot = 0.0221 + 1.16 103 + 1.33 m K / W = 1.35 m K / WThe heat gain per unit length is thenq =T,o T,iRtot=( 23 6 ) C1.35 m K / W<= 12.6 W / m(b) With the insulation, the total resistance per unit length is now R tot = R conv,i + R cond,st + R cond,ins+Rconv,o , where R conv,i and R cond,st remain the same. The thermal resistance of the insulation isRcond,ins =ln ( r3 / r2 )2 k ins=ln (30 / 20 )2 (0.05 W / m K )= 1.29 m K / Wand the outer convection resistance is nowRconv,o =11== 0.88 m K / W2 r3h o 2 ( 0.03m ) 6 W / m 2 KThe total resistance is now()R tot = 0.0221 + 1.16 103 + 1.29 + 0.88 m K / W = 2.20 m K / WContinued ..PROBLEM 3.39 (Cont.)and the heat gain per unit length isq =T,o T,iR tot=17C= 7.7 W / m2.20 m K / WCOMMENTS: (1) The validity of assuming negligible radiation may be assessed for the worst casecondition corresponding to the bare tube. Assuming a tube outer surface temperature of Ts = T,i =279K, large surroundings at Tsur = T,o = 296K, and an emissivity of = 0.7, the heat gain due to net()4radiation exchange with the surroundings is q = ( 2 r2 ) Tsur Ts4 = 7.7 W / m. Hence, the netradrate of heat transfer by radiation to the tube surface is comparable to that by convection, and theassumption of negligible radiation is inappropriate.(2) If heat transfer from the air is by natural convection, the value of ho with the insulation wouldactually be less than the value for the bare tube, thereby further reducing the heat gain. Use of theinsulation would also increase the outer surface temperature, thereby reducing net radiation transferfrom the surroundings.(3) The critical radius is rcr = kins/h 8 mm < r2. Hence, as indicated by the calculations, heattransfer is reduced by the insulation.PROBLEM 3.40KNOWN: Diameter, wall thickness and thermal conductivity of steel tubes. Temperature of steamflowing through the tubes. Thermal conductivity of insulation and emissivity of aluminum sheath.Temperature of ambient air and surroundings. Convection coefficient at outer surface and maximumallowable surface temperature.FIND: (a) Minimum required insulation thickness (r3 r2) and corresponding heat loss per unitlength, (b) Effect of insulation thickness on outer surface temperature and heat loss.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction, (3) Negligible contactresistances at the material interfaces, (4) Negligible steam side convection resistance (T,i = Ts,i), (5)Negligible conduction resistance for aluminum sheath, (6) Constant properties, (7) Largesurroundings.ANALYSIS: (a) To determine the insulation thickness, an energy balance must be performed at theouter surface, where q = q onv,o + q rad . With q cconv,o = 2 r3h o ( Ts,o T ,o ) , q ad = 2 r3 r(T)()()44s,o Tsur , q = Ts,i Ts,o / R cond,st + R cond,ins , R cond,st = n ( r2 / r1 ) / 2 k st ,and R ond,insc= n ( r3 / r2 ) / 2 k ins , it follows that(2 Ts,i Ts,o)()()44= 2 r3 h o Ts,o T,o + Ts,o Tsur n ( r2 / r1 ) n ( r3 / r2 )+k stk ins2 (848 323 ) K(n 0.18 / 0.15)35 W / m K+(n r3 / 0.18)= 2 r3 6 W / m 2 K ( 323 300 ) K + 0.20 5.67 108 W / m 2 K 4(4323 3004)K40.10 W / m KA trial-and-error solution yields r3 = 0.394 m = 394 mm, in which case the insulation thickness ist ins = r3 r2 = 214 mm<The heat rate is thenq =2 (848 323) K= 420 W / mn ( 0.18 / 0.15 ) n (0.394 / 0.18 )+35 W / m K0.10 W / m K<(b) The effects of r3 on Ts,o and q have been computed and are shown below.Conditioned ..PROBLEM 3.40 (Cont.)Ou te r s u rfa ce te m p e ra tu re , C24020016012080400 .20 .2 60 .3 20 .3 80 .4 40 .5O u te r ra d iu s o f in s u la tio n , mTs ,oH e a t ra te s , W /m250020001500100050000 .20 .2 60 .3 20 .3 80 .4 40 .5O u te r ra d iu s o f in s u la tio n , mTo ta l h e a t ra teC o n ve ctio n h e a t ra teR a d ia tio n h e a t ra teBeyond r3 0.40m, there are rapidly diminishing benefits associated with increasing the insulationthickness.COMMENTS: Note that the thermal resistance of the insulation is much larger than that for the tubewall. For the conditions of Part (a), the radiation coefficient is hr = 1.37 W/m, and the heat loss byradiation is less than 25% of that due to natural convection ( q = 78 W / m, q radconv,o = 342 W / m ) .PROBLEM 3.41KNOWN: Thin electrical heater fitted between two concentric cylinders, the outer surface of whichexperiences convection.FIND: (a) Electrical power required to maintain outer surface at a specified temperature, (b)Temperature at the centerSCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Steady-state conditions, (3) Heaterelement has negligible thickness, (4) Negligible contact resistance between cylinders and heater, (5)Constant properties, (6) No generation.ANALYSIS: (a) Perform an energy balance on thecomposite system to determine the power requiredto maintain T(r2) = Ts = 5C.E E + E gen = Estinout+ q qelecconv = 0.Using Newtons law of cooling,qelec = qconv = h 2 r2 ( Ts T )qelec = 50Wm2 K 2 ( 0.040m ) 5 ( 15 ) C=251 W/m.<(b) From a control volume about Cylinder A, we recognize that the cylinder must be isothermal, thatis,T(0) = T(r1).Represent Cylinder B by a thermal circuit:q=T ( r1 ) TsRBFor the cylinder, from Eq. 3.28,R = ln r2 / r1 / 2 k BBgivingT ( r1 ) = Ts + qR = 5 C+253.1BWln 40/20= 23.5 C 1.5 W/m Km2Hence, T(0) = T(r1) = 23.5C.Note that kA has no influence on the temperature T(0).<PROBLEM 3.42KNOWN: Electric current and resistance of wire. Wire diameter and emissivity. Thickness,emissivity and thermal conductivity of coating. Temperature of ambient air and surroundings.Expression for heat transfer coefficient at surface of the wire or coating.FIND: (a) Heat generation per unit length and volume of wire, (b) Temperature of uninsulated wire,(c) Inner and outer surface temperatures of insulation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction through insulation, (3)Constant properties, (4) Negligible contact resistance between insulation and wire, (5) Negligibleradial temperature gradients in wire, (6) Large surroundings.ANALYSIS: (a) The rates of energy generation per unit length and volume are, respectively,2E = I 2 R gelec = ( 20 A ) (0.01 / m ) = 4 W / m<2 gq = E / Ac = 4 E / D2 = 16 W / m / ( 0.002m ) = 1.27 106 W / m3g<(b) Without the insulation, an energy balance at the surface of the wire yields44E = q = q gconv + q = D h ( T T ) + D w T Tsurrad)(where h = 1.25 [( T T ) / D ]1/ 4 . Substituting,4 W / m = 1.25 ( 0.002m )3/ 4(T 293 )5 / 4 + (0.002m ) 0.3 5.67 108 W / m 2 K 4(T4) 2934 K 4and a trial-and-error solution yields<T = 331K = 58C(c) Performing an energy balance at the outer surface,()(44E = q = q gconv + q = D h Ts,2 T + D i Ts,2 Tsurrad4 W / m = 1.25 ( 0.006m )3/ 4)4(Ts,2 293)5 / 4 + (0.006m ) 0.9 5.67 108 W / m 2 K 4 (Ts,2 2934 ) K 4and an iterative solution yields the following value of the surface temperature<Ts,2 = 307.8 K = 34.8CThe inner surface temperature may then be obtained from the following expression for heat transferby conduction in the insulation.Continued ..PROBLEM 3.42 (Cont.)q =Ts,i T2Rcond4W ==Ts,i Ts,2n ( r2 / r1 ) / 2 k i(2 ( 0.25 W / m K ) Ts,i 307.8 K)n 3<Ts,i = 310.6 K = 37.6CAs shown below, the effect of increasing the insulation thickness is to reduce, not increase, thesurface temperatures.S u rfa ce te m p e ra tu re s , C504540353001234In s u la tio n th ickn e s s , m mIn n e r s u rfa ce te m p e ra tu re , CO u te r s u rfa ce te m p e ra tu re , CThis behavior is due to a reduction in the total resistance to heat transfer with increasing r2. Although()22the convection, h, and radiation, h r = (Ts,2 + Tsur ) Ts,2 + Tsur , coefficients decrease withincreasing r2, the corresponding increase in the surface area is more than sufficient to provide for areduction in the total resistance. Even for an insulation thickness of t = 4 mm, h = h + hr = (7.1 + 5.4)222W/m K = 12.5 W/m K, and rcr = k/h = 0.25 W/mK/12.5 W/m K = 0.020m = 20 mm > r2 = 5 mm.The outer radius of the insulation is therefore well below the critical radius.PROBLEM 3.43KNOWN: Diameter of electrical wire. Thickness and thermal conductivity of rubberized sheath.Contact resistance between sheath and wire. Convection coefficient and ambient air temperature.Maximum allowable sheath temperature.FIND: Maximum allowable power dissipation per unit length of wire. Critical radius of insulation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction through insulation, (3)Constant properties, (4) Negligible radiation exchange with surroundings.ANALYSIS: The maximum insulation temperature corresponds to its inner surface and isindependent of the contact resistance. From the thermal circuit, we may writeE = q =gTin,i TRcond + R conv=Tin,i T()(n rin,o / rin,i / 2 k + 1/ 2 rin,o h)where rin,i = D / 2 = 0.001m, rin,o = rin,i + t = 0.003m, and Tin,i = Tmax = 50C yields the maximumallowable power dissipation. Hence,(50 20 ) CEg,max =n 32 0.13 W / m K+=130C(1.35 + 5.31) m K / W= 4.51 W / m<2 ( 0.003m )10 W / m K2The critical insulation radius is also unaffected by the contact resistance and is given byrcr =k 0.13 W / m K== 0.013m = 13mmh 10 W / m 2 K<Hence, rin,o < rcr and E ,max could be increased by increasing rin,o up to a value of 13 mm (t = 12gmm).COMMENTS: The contact resistance affects the temperature of the wire, and for q = E ,maxg= 4.51 W / m, the outer surface temperature of the wire is Tw,o = Tin,i + q R = 50C + ( 4.51 W / m )t,c(3 104)m K / W / ( 0.002m ) = 50.2C. Hence, the temperature change across the contact2resistance is negligible.PROBLEM 3.44KNOWN: Long rod experiencing uniform volumetric generation of thermal energy, q, concentricwith a hollow ceramic cylinder creating an enclosure filled with air. Thermal resistance per unitlength due to radiation exchange between enclosure surfaces is R . The free convectionrad2coefficient for the enclosure surfaces is h = 20 W/m K.FIND: (a) Thermal circuit of the system that can be used to calculate the surface temperature of therod, Tr; label all temperatures, heat rates and thermal resistances; evaluate the thermal resistances; and(b) Calculate the surface temperature of the rod.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial conduction through thehollow cylinder, (3) The enclosure surfaces experience free convection and radiation exchange.ANALYSIS: (a) The thermal circuit is shown below. Note labels for the temperatures, thermalresistances and the relevant heat fluxes.Enclosure, radiation exchange (given):R = 0.30 m K / WradEnclosure, free convection:11== 0.80 m K / W2 K 0.020mh Dr 20 W / m11R== 0.40 m K / Wcv,cer =2 K 0.040mh Di 20 W / mRcv,rod =Ceramic cylinder, conduction:R =cdn ( Do / Di ) n (0.120 / 0.040 )== 0.10 m K / W2 k2 1.75 W / m KThe thermal resistance between the enclosure surfaces (r-i) due to convection and radiation exchangeis111=+Renc R rad R cv,rod + R cv,cer11R = +enc 0.30 0.80 + 0.40 1m K / W = 0.24 m K / WThe total resistance between the rod surface (r) and the outer surface of the cylinder (o) isR tot = R + R = ( 0.24 + 0.1) m K / W = 0.34 m K / WenccdContinued ..PROBLEM 3.44 (Cont.)(b) From an energy balance on the rod (see schematic) find Tr.Ein E + E = 0outgenq + q = 0() (Tr Ti ) / R tot + q D 2 / 4 = 0r() (Tr 25 ) K / 0.34 m K / W + 2 106 W / m3 0.020m 2 / 4 = 0Tr = 239C<COMMENTS: In evaluating the convection resistance of the air space, it was necessary to define anaverage air temperature (T) and consider the convection coefficients for each of the space surfaces.As youll learn later in Chapter 9, correlations are available for directly estimating the convectioncoefficient (henc) for the enclosure so that qcv = henc (Tr T1).PROBLEM 3.45KNOWN: Tube diameter and refrigerant temperature for evaporator of a refrigerant system.Convection coefficient and temperature of outside air.FIND: (a) Rate of heat extraction without frost formation, (b) Effect of frost formation on heat rate, (c)Time required for a 2 mm thick frost layer to melt in ambient air for which h = 2 W/m2K and TW = 20C.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible convection resistancefor refrigerant flow T,i = Ts,1 , (3) Negligible tube wall conduction resistance, (4) Negligibleradiation exchange at outer surface.()ANALYSIS: (a) The cooling capacity in the defrosted condition ( = 0) corresponds to the rate of heatextraction from the airflow. Hence,()q = h2 r1 T,o Ts,1 = 100 W m 2 K ( 2 0.005 m )( 3 + 18 ) C<q = 47.1W m(b) With the frost layer, there is an additional (conduction) resistance to heat transfer, and the extractionrate isT,o Ts,1T,o Ts,1q ==Rconv + R cond 1 ( h2 r2 ) + ln ( r2 r1 ) 2 kFor 5 r2 9 mm and k = 0.4 W/mK, this expression yieldsThermal resistance, Rt(m.K/W)Heat extraction, qprime(W/m)5045400.40.30.20.103500.0010.0020.003Frost layer thickness, delta(m)Heat extraction, qprime(W/m)0.00400.0010.0020.0030.004Frost layer thickness, delta(m)Conduction resistance, Rtcond(m.K/W)Convection resistance, Rtconv(m.K/W)Continued...PROBLEM 3.45 (Cont.)The heat extraction, and hence the performance of the evaporator coil, decreases with increasing frostlayer thickness due to an increase in the total resistance to heat transfer. Although the convectionresistance decreases with increasing , the reduction is exceeded by the increase in the conductionresistance.(c) The time tm required to melt a 2 mm thick frost layer may be determined by applying an energybalance, Eq. 1.11b, over the differential time interval dt and to a differential control volume extendinginward from the surface of the layer.E in dt = dEst = dU lat()h ( 2 rL ) T,o Tf dt = h sf d = h sf ( 2 rL ) dr(h T,o Tftm =1) 0t m dt = hsf rr2 dr h sf ( r2 r1 )(h T,o Tf)=t m = 11, 690 s = 3.25 h()700 kg m3 3.34 105 J kg (0.002 m )2 W m 2 K ( 20 0 ) C<COMMENTS: The tube radius r1 exceeds the critical radius rcr = k/h = 0.4 W/mK/100 W/m2K = 0.004m, in which case any frost formation will reduce the performance of the coil.PROBLEM 3.46KNOWN: Conditions associated with a composite wall and a thin electric heater.FIND: (a) Equivalent thermal circuit, (b) Expression for heater temperature, (c) Ratio of outer and innerheat flows and conditions for which ratio is minimized.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant properties, (3) Isothermalheater, (4) Negligible contact resistance(s).ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are asshown in the schematic.(b) Performing an energy balance for the heater, Ein = E out , it follows thatq ( 2 r2 ) = q + q =hioTh T,i1( h i 2 r1 )ln ( r2 r1 )+2 k B+Th T,o1(h o 2 r3 )ln ( r3 r2 )+2 k A<(c) From the circuit,ln ( r2 r1 )q2 k Bo=ln ( r r )qTh T,ii( h o 2 r3 )1 + 3 22 k A((Th T,o))( h i 2 r1 )1 +To reduce q q , one could increase kB, hi, and r3/r2, while reducing kA, ho and r2/r1.oiCOMMENTS: Contact resistances between the heater and materials A and B could be important.<PROBLEM 3.47KNOWN: Electric current flow, resistance, diameter and environmental conditionsassociated with a cable.FIND: (a) Surface temperature of bare cable, (b) Cable surface and insulation temperaturesfor a thin coating of insulation, (c) Insulation thickness which provides the lowest value of themaximum insulation temperature. Corresponding value of this temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3)Constant properties.ANALYSIS: (a) The rate at which heat is transferred to the surroundings is fixed by the rateof heat generation in the cable. Performing an energy balance for a control surface about thecable, it follows that E = q or, for the bare cable, I2 R L=h ( D L )(T T ). Withgq=I2 R = ( 700A )e2Ts = T +e(6 104 / m) = 294 W/m, it follows thatisq294 W/m= 30 C+h D i25 W/m 2 K (0.005m )()Ts = 778.7 C.<(b) With a thin coating of insulation, there exist contact and convection resistances to heattransfer from the cable. The heat transfer rate is determined by heating within the cable,however, and therefore remains the same.Ts TTs Tq==1R 1t,cR t,c ++h Di L Di L h Di L Di (Ts T )q=R + 1/ ht,cand solving for the surface temperature, findqTs = Di1294 W/m m2 Km2 K R t,c + h + T = ( 0.005m ) 0.02 W + 0.04 W + 30 CTs = 1153 C.<Continued ..PROBLEM 3.47 (Cont.)The insulation temperature is then obtained fromT Tq= s iR t,corTi = Ts qR t,c = 1153 C qR t,c Di L= 1153 C 294Wm2 K 0.02mW (0.005m )Ti = 778.7 C.<(c) The maximum insulation temperature could be reduced by reducing the resistance to heat transferfrom the outer surface of the insulation. Such a reduction is possible if Di < Dcr. From Example 3.4,rcr =k 0.5 W/m K== 0.02m.h 25 W/m 2 KHence, Dcr = 0.04m > Di = 0.005m. To minimize the maximum temperature, which exists atthe inner surface of the insulation, add insulation in the amountD Di Dcr Di (0.04 0.005 ) mt= o==222<t = 0.0175m.The cable surface temperature may then be obtained fromq=R t,c Di+Ts Tln ( Dcr / Di )2 k+1h Dcr=Ts 30 Cln ( 0.04/0.005 )0.02 m 2 K/W++2 ( 0.5 W/m K ) ( 0.005m )125W2m K ( 0.04m )Hence,Ts 30 CTs 30 CW294==m (1.27+0.66+0.32 ) m K/W 2.25 m K/WTs = 692.5 CRecognizing that q = (Ts - Ti)/Rt,c, findTi = Ts qR t,c = Ts qTi = 318.2 C.R t,c Di L= 692.5 C 294Wm2 K 0.02mW (0.005m )<COMMENTS: Use of the critical insulation thickness in lieu of a thin coating has the effect ofreducing the maximum insulation temperature from 778.7C to 318.2C. Use of the critical insulationthickness also reduces the cable surface temperature to 692.5C from 778.7C with no insulation orfrom 1153C with a thin coating.PROBLEM 3.48KNOWN: Saturated steam conditions in a pipe with prescribed surroundings.FIND: (a) Heat loss per unit length from bare pipe and from insulated pipe, (b) Pay backperiod for insulation.SCHEMATIC:Steam Costs:9$4 for 10 JInsulation Cost:$100 per meterOperation time:7500 h/yrASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3)Constant properties, (4) Negligible pipe wall resistance, (5) Negligible steam side convectionresistance (pipe inner surface temperature is equal to steam temperature), (6) Negligiblecontact resistance, (7) Tsur = T.PROPERTIES: Table A-6, Saturated water (p = 20 bar): Tsat = Ts = 486K; Table A-3,Magnesia, 85% (T 392K): k = 0.058 W/mK.ANALYSIS: (a) Without the insulation, the heat loss may be expressed in terms of radiationand convection rates,()44q= D Ts Tsur + h ( D )( Ts T )Wq=0.8 ( 0.2m ) 5.67 1084864 2984 K 42 K4mW+20( 0.2m ) ( 486-298) Km2 K()q= (1365+2362 ) W/m=3727 W/m.<With the insulation, the thermal circuit is of the formContinued ..PROBLEM 3.48 (Cont.)From an energy balance at the outer surface of the insulation,qcond = qconv + qradTs,i Ts,o44= h Do Ts,o T + Do Ts,o Tsurln ( Do / Di ) / 2 k486 Ts,o KW (0.3m ) Ts,o 298K= 202 Kln ( 0.3m/0.2m )m2 ( 0.058 W/m K )W4 (0.3m ) Ts,o 2984 K 4 .+0.8 5.67 10-82 K4m(())()()()By trial and error, we obtainTs,o 305Kin which caseq=( 486-305) K = 163 W/m.ln ( 0.3m/0.2m )2 (0.055 W/m K )<(b) The yearly energy savings per unit length of pipe due to use of the insulation isSavings Energy Savings Cost=Yr mYr.EnergySavingsJsh$4= (3727 163) 3600 7500 Yr msmhYr 109 JSavings= $385 / Yr m.Yr mThe pay back period is thenPay Back Period =Insulation Costs$100 / m=Savings/Yr. m $385/Yr mPay Back Period = 0.26 Yr = 3.1 mo.<COMMENTS: Such a low pay back period is more than sufficient to justify investing in theinsulation.PROBLEM 3.49KNOWN: Temperature and convection coefficient associated with steam flow through a pipeof prescribed inner and outer diameters. Outer surface emissivity and convection coefficient.Temperature of ambient air and surroundings.FIND: Heat loss per unit length.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3)Constant properties, (4) Surroundings form a large enclosure about pipe.PROPERTIES: Table A-1, Steel, AISI 1010 (T 450 K): k = 56.5 W/mK.ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outersurface thatT,i Ts,oTs,o T,o Ts,o Tsur=+R conv,i + R condR conv,oR rador from Eqs. 3.9, 3.28 and 1.7,(1/T,i Ts,oDi hi ) + ln ( Do / Di ) / 2 k523K Ts,o=Ts,o T,o(1/Do h o )(44+ Do Ts,o Tsur=)Ts,o 293K1ln (75/60 ) 0.075m 25 W/m 2 K2 56.5 W/m K4+0.8 ( 0.075m ) 5.67 108 W/m 2 K 4 Ts,o 2934 K 4523 Ts,oTs,o 2938 T 4 2934 .=+ 1.07 10 s,o0.0106+0.00060.170( 0.6m 500 W/m 2 K)1+()From a trial-and-error solution, Ts,o 502K. Hence the heat loss is()(44q= Do h o Ts,o T,o + Do Ts,o Tsurq= ( 0.075m ) 25 W/m 2 K (502-293) + 0.8 ( 0.075m ) 5.67 108)5024 2434 K 4m K W24q=1231 W/m+600 W/m=1831 W/m.COMMENTS: The thermal resistance between the outer surface and the surroundings ismuch larger than that between the outer surface and the steam.<PROBLEM 3.50KNOWN: Temperature and convection coefficient associated with steam flow through a pipe ofprescribed inner and outer radii. Emissivity of outer surface magnesia insulation, and convectioncoefficient. Temperature of ambient air and surroundings.FIND: Heat loss per unit length q and outer surface temperature Ts,o as a function of insulationthickness. Recommended insulation thickness. Corresponding annual savings and temperaturedistribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constantproperties, (4) Surroundings form a large enclosure about pipe.PROPERTIES: Table A-1, Steel, AISI 1010 (T 450 K): ks = 56.5 W/mK. Table A-3, Magnesia,85% (T 365 K): km = 0.055 W/mK.ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outer surface thatT,i Ts,oTs,o T,o Ts,o Tsur=+RRRconv,i + R cond,s + R cond,mconv,orador from Eqs. 3.9, 3.28 and 1.7,(1 2 r1h i ) + ln ( r2T ,i Ts,or1 ) 2 k s + ln ( r3 r2 ) 2 k m=Ts,o T ,oTs,o Tsur+(1 2 r3h o ) ( 2 r3 ) (Ts,o + Tsur )(T22s,o + Tsur)1This expression may be solved for Ts,o as a function of r3, and the heat loss may then be determined byevaluating either the left-or right-hand side of the energy balance equation. The results are plotted asfollows.Continued...PROBLEM 3.50 (Cont.)2000Thermal resistance, Rprime(K/m.W)2Heat loss, qprime(W/m)160012008004001.510.500.03500.0350.0450.0550.0650.0750.0450.0550.0650.075Outer radius of insulation, r3(m)Outer radius of insulation, r3(m)Insulation conduction resistance, Rcond,mOuter convection resistance, Rconv,oRadiation resistance, Rradq1The rapid decay in q with increasing r3 is attributable to the dominant contribution which the insulationbegins to make to the total thermal resistance. The inside convection and tube wall conductionresistances are fixed at 0.0106 mK/W and 6.2910-4 mK/W, respectively, while the resistance of theinsulation increases to approximately 2 mK/W at r3 = 0.075 m.The heat loss may be reduced by almost 91% from a value of approximately 1830 W/m at r3 = r2= 0.0375 m (no insulation) to 172 W/m at r3 = 0.0575 m and by only an additional 3% if the insulationthickness is increased to r3 = 0.0775 m. Hence, an insulation thickness of (r3 - r2) = 0.020 m isrecommended, for which q = 172 W/m. The corresponding annual savings (AS) in energy costs istherefore$4hs 7000 3600 = $167 / mAS = [(1830 172 ) W m ]yh109 J<The corresponding temperature distribution isLocal temperature, T(K)5004604203803403000.0380.0420.0460.050.0540.058Radial location in insulation, r(m)TrThe temperature in the insulation decreases from T(r) = T2 = 521 K at r = r2 = 0.0375 m to T(r) = T3 =309 K at r = r3 = 0.0575 m.Continued...PROBLEM 3.50 (Cont.)COMMENTS: 1. The annual energy and costs savings associated with insulating the steam line aresubstantial, as is the reduction in the outer surface temperature (from Ts,o 502 K for r3 = r2, to 309 K forr3 = 0.0575 m).2. The increase in R to a maximum value of 0.63 mK/W at r3 = 0.0455 m and the subsequent decayradis due to the competing effects of hrad and A = (1 2 r3 ) . Because the initial decay in T3 = Ts,o with3increasing r3, and hence, the reduction in hrad, is more pronounced than the increase in A , R 3radincreases with r3. However, as the decay in Ts,o, and hence hrad, becomes less pronounced, the increase inA becomes more pronounced and R decreases with increasing r3.rad3PROBLEM 3.51KNOWN: Pipe wall temperature and convection conditions associated with water flow through the pipeand ice layer formation on the inner surface.FIND: Ice layer thickness .SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Negligible pipe wall thermalresistance, (3) negligible ice/wall contact resistance, (4) Constant k.PROPERTIES: Table A.3, Ice (T = 265 K): k 1.94 W/mK.ANALYSIS: Performing an energy balance for a control surface about the ice/water interface, it followsthat, for a unit length of pipe,qconv = qcond()h i ( 2 r1 ) T,i Ts,i =Ts,i Ts,oln ( r2 r1 ) 2 kDividing both sides of the equation by r2,ln ( r2 r1 )( r2 r1 )=Ts,i Ts,ok1.94 W m K15 C== 0.0972 K 0.05 mh i r2 T,i Ts,i3 C2000 W m()()The equation is satisfied by r2/r1 = 1.114, in which case r1 = 0.050 m/1.114 = 0.045 m, and the ice layerthickness is = r2 r1 = 0.005 m = 5 mm<COMMENTS: With no flow, hi 0, in which case r1 0 and complete blockage could occur. Thepipe should be insulated.PROBLEM 3.52KNOWN: Inner surface temperature of insulation blanket comprised of two semi-cylindrical shells of differentmaterials. Ambient air conditions.FIND: (a) Equivalent thermal circuit, (b) Total heat loss and material outer surface temperatures.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial conduction, (3) Infinite contactresistance between materials, (4) Constant properties.ANALYSIS: (a) The thermal circuit is,Rconv,A = R conv,B = 1 / r2 hRcond ( A ) =Rcond ( B ) =ln ( r2 / r1 )< kAln ( r2 / ri ) kBThe conduction resistances follow from Section 3.3.1 and Eq. 3.28. Each resistance is larger by a factor of 2 thanthe result of Eq. 3.28 due to the reduced area.(b) Evaluating the thermal resistances and the heat rate ( q=q + q ) ,AB(2Rconv = 0.1m 25 W/m KRcond ( A ) =q=q=ln ( 0.1m/0.05m ) 2 W/m KTs,1 TRconvcond ( A ) + R +)1= 0.1273 m K/W= 0.1103 m K/WRcond ( B ) = 8 R cond ( A ) = 0.8825 m K/WTs,1 TRconvcond ( B) + R (500 300 ) K(500 300 ) K+= (842 + 198 ) W/m=1040 W/m.(0.1103+0.1273) m K/W (0.8825+0.1273) m K/W<Hence, the temperatures areWmK 0.1103= 407KmWWmKTs,2( B) = Ts,1 q R = 325K.B cond ( B ) = 500K 198 0.8825mWTs,2( A ) = Ts,1 q R A cond ( A ) = 500K 842(<<)COMMENTS: The total heat loss can also be computed from q= Ts,1 T / R equiv ,111 where R equiv = R cond ( A ) + R conv,A+ ( R cond(B) + R conv,B ) = 0.1923 m K/W.Hence q= (500 300 ) K/0.1923 m K/W=1040 W/m.()PROBLEM 3.53KNOWN: Surface temperature of a circular rod coated with bakelite and adjoining fluidconditions.FIND: (a) Critical insulation radius, (b) Heat transfer per unit length for bare rod and forinsulation at critical radius, (c) Insulation thickness needed for 25% heat rate reduction.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3)Constant properties, (4) Negligible radiation and contact resistance.PROPERTIES: Table A-3, Bakelite (300K): k = 1.4 W/mK.ANALYSIS: (a) From Example 3.4, the critical radius isk1.4 W/m Krcr = == 0.01m.h 140 W/m 2 K<(b) For the bare rod,q=h ( Di ) ( Ti T )q=140Wm2 K( 0.01m ) ( 200 25) C=770 W/m<For the critical insulation thickness,( 200 25) CTi Tq==ln ( rcr / ri )ln (0.01m/0.005m )11++2 rcr h2 k2 1.4 W/m K2 (0.01m ) 140 W/m 2 Kq=175C= 909 W/m(0.1137+0.0788) m K/W<(c) The insulation thickness needed to reduce the heat rate to 577 W/m is obtained from( 200 25) CTi TWq=== 577ln ( r/ri )ln ( r/0.005m )m11++2 K 2 1.4 W/m K2 rh2 k2 ( r )140 W/mFrom a trial-and-error solution, findr 0.06 m.The desired insulation thickness is then = ( r ri ) (0.06 0.005 ) m=55 mm.<PROBLEM 3.54KNOWN: Geometry of an oil storage tank. Temperature of stored oil and environmentalconditions.FIND: Heater power required to maintain a prescribed inner surface temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radialdirection, (3) Constant properties, (4) Negligible radiation.PROPERTIES: Table A-3, Pyrex (300K): k = 1.4 W/mK.ANALYSIS: The rate at which heat must be supplied is equal to the loss through thecylindrical and hemispherical sections. Hence,q=qcyl + 2q hemi = qcyl + qspheror, from Eqs. 3.28 and 3.36,q=q=Ts,i Tln ( Do / Di )1+2 Lk Do Lh+Ts,i T1 111D D +22 k io Do h( 400 300 ) Kln 1.041+2 ( 2m )1.4 W/m K (1.04m ) 2m 10 W/m 2 K(+( 400 300 ) K)11(1 0.962 ) m-1 +22 (1.4 W/m K ) (1.04m ) 10 W/m 2 K100K100Kq=+-3 K/W + 15.30 10-3 K/W 4.32 10-3 K/W + 29.43 10-32.23 10q = 5705W + 2963W = 8668W.<PROBLEM 3.55KNOWN: Diameter of a spherical container used to store liquid oxygen and properties of insulatingmaterial. Environmental conditions.FIND: (a) Reduction in evaporative oxygen loss associated with a prescribed insulation thickness, (b)Effect of insulation thickness on evaporation rate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Negligible conduction resistance ofcontainer wall and contact resistance between wall and insulation, (3) Container wall at boiling point ofliquid oxygen.ANALYSIS: (a) Applying an energy balance to a control surface about the insulation, E in E out = 0, itfollows that q conv + q rad = q cond = q . Hence,T Ts,2 Tsur Ts,2 Ts,2 Ts,1+==qR t,convR t,radR t,cond(2where R t,conv = 4 r2 h)1(2, R t,rad = 4 r2 h r((1))11.9, the radiation coefficient is h r = Ts,2 + Tsur, R t,cond = (1 4 k )[(1 r1 ) (1 r2 )] , and, from Eq.22) (Ts,2 + Tsur ) .With t = 10 mm (r2 = 260 mm), =0.2 and T = Tsur = 298 K, an iterative solution of the energy balance equation yields Ts,2 297.7 K,where Rt,conv = 0.118 K/W, Rt,rad = 0.982 K/W and Rt,cond = 76.5 K/W. With the insulation, it follows thatthe heat gain isqw 2.72 WWithout the insulation, the heat gain isq wo =T Ts,1 Tsur Ts,1+R t,convR t,radwhere, with r2 = r1, Ts,1 = 90 K, Rt,conv = 0.127 K/W and Rt,rad = 3.14 K/W. Hence,qwo = 1702 WWith the oxygen mass evaporation rate given by m = q/hfg, the percent reduction in evaporated oxygen is% Re duction =Hence,% Re duction =m wo m wm woq qw 100% = wo 100%q wo(1702 2.7 ) W1702 W 100% = 99.8%<Continued...PROBLEM 3.55 (Cont.)(b) Using Equation (1) to compute Ts,2 and q as a function of r2, the corresponding evaporation rate, m =q/hfg, may be determined. Variations of q and m with r2 are plotted as follows.100000.01Evaporation rate, mdot(kg/s)Heat gain, q(W)10001001010.10.0010.00011E-51E-60.250.260.270.280.29Outer radius of insulation, r2(m)0.30.250.260.270.280.290.3Outer radius of insulation, r2(m)Because of its extremely low thermal conductivity, significant benefits are associated with using even athin layer of insulation. Nearly three-order magnitude reductions in q and m are achieved with r2 = 0.26-3m. With increasing r2, q and m decrease from values of 1702 W and 810 kg/s at r2 = 0.25 m to 0.627W and 2.910-6 kg/s at r2 = 0.30 m.COMMENTS: Laminated metallic-foil/glass-mat insulations are extremely effective and correspondingconduction resistances are typically much larger than those normally associated with surface convectionand radiation.PROBLEM 3.56KNOWN: Sphere of radius ri, covered with insulation whose outer surface is exposed to aconvection process.FIND: Critical insulation radius, rcr.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (spherical)conduction, (3) Constant properties, (4) Negligible radiation at surface.ANALYSIS: The heat rate follows from the thermal circuit shown in the schematic,q= ( Ti T ) / R totwhere R tot = R t,conv + R t,cond andR t,conv =11=hAs 4 hr 2(3.9)R t,cond =1 1 14 k ri r (3.36)If q is a maximum or minimum, we need to find the condition for whichd R tot= 0.drIt follows thatd 1 1 11111 1 +=0 = +dr 4 k ri r 4 hr 2 4 k r 2 2 h r3 givingkhThe second derivative, evaluated at r = rcr, isd dR tot 1131=+dr dr 2 k r3 2 h r 4 r=rrcr = 2cr=11( 2k/h )3 2 k+311131 + > 0=3 2 k 2 h 2k/h ( 2k/h )2Hence, it follows no optimum Rtot exists. We refer to this condition as the critical insulationradius. See Example 3.4 which considers this situation for a cylindrical system.PROBLEM 3.57KNOWN: Thickness of hollow aluminum sphere and insulation layer. Heat rate and innersurface temperature. Ambient air temperature and convection coefficient.FIND: Thermal conductivity of insulation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)Constant properties, (4) Negligible contact resistance, (5) Negligible radiation exchange atouter surface.PROPERTIES: Table A-1, Aluminum (523K): k 230 W/mK.ANALYSIS: From the thermal circuit,T TT1 Tq= 1 =1/r1 1/ r2 1/ r2 1/ r31R tot++24 k A14 k Ih4 r3q=( 250 20 ) C1/0.15 1/ 0.18 1/ 0.18 1/ 0.30K1++24 k I 4 ( 230 )30 ( 4 )( 0.3) W= 80 Wor3.84 104 +0.177230+ 0.029 == 2.875.kI80Solving for the unknown thermal conductivity, findkI = 0.062 W/mK.COMMENTS: The dominant contribution to the total thermal resistance is made by theinsulation. Hence uncertainties in knowledge of h or kA1 have a negligible effect on theaccuracy of the kI measurement.<PROBLEM 3.58KNOWN: Dimensions of spherical, stainless steel liquid oxygen (LOX) storage container. Boilingpoint and latent heat of fusion of LOX. Environmental temperature.FIND: Thermal isolation system which maintains boil-off below 1 kg/day.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible thermal resistancesassociated with internal and external convection, conduction in the container wall, and contact betweenwall and insulation, (3) Negligible radiation at exterior surface, (4) Constant insulation thermalconductivity.PROPERTIES: Table A.1, 304 Stainless steel (T = 100 K): ks = 9.2 W/mK; Table A.3, Reflective,aluminum foil-glass paper insulation (T = 150 K): ki = 0.000017 W/mK.ANALYSIS: The heat gain associated with a loss of 1 kg/day isq = mh fg =1kg day86, 400 s day(2.13 105 J kg ) = 2.47 W()With an overall temperature difference of T Tbp = 150 K, the corresponding total thermalresistance isT 150 KR tot === 60.7 K Wq2.47 WSince the conduction resistance of the steel wall isR t,cond,s =1 11113 = 0.35 m 0.40 m = 2.4 10 K W4 k s r1 r2 4 (9.2 W m K ) 1it is clear that exclusive reliance must be placed on the insulation and that a special insulation of very lowthermal conductivity should be selected. The best choice is a highly reflective foil/glass mattedinsulation which was developed for cryogenic applications. It follows thatR t,cond,i = 60.7 K W =1 11111 =4 k i r2 r3 4 (0.000017 W m K ) 0.40 m r3 which yields r3 = 0.4021 m. The minimum insulation thickness is therefore = (r3 - r2) = 2.1 mm.COMMENTS: The heat loss could be reduced well below the maximum allowable by adding moreinsulation. Also, in view of weight restrictions associated with launching space vehicles, considerationshould be given to fabricating the LOX container from a lighter material.PROBLEM 3.59KNOWN: Diameter and surface temperature of a spherical cryoprobe. Temperature of surroundingtissue and effective convection coefficient at interface between frozen and normal tissue.FIND: Thickness of frozen tissue layer.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible contact resistancebetween probe and frozen tissue, (3) Constant properties.ANALYSIS: Performing an energy balance for a control surface about the phase front, it follows thatq conv q cond = 0Hence,( )(T Ts,2 ) = [(1 r1Ts,2(1rTs,1] 4 k) 2 )2h 4 r22r2 [(1 r1 ) (1 r2 )] =()h (T Ts,2 )k Ts,2 Ts,1 r2 r2 k ( Ts,2 Ts,1 )1.5 W m K 30 = 1 = r1 r1 hr1 ( T Ts,2 )50 W m 2 K ( 0.0015 m ) 37 () r2 r2 1 = 16.2 r1 r1 ( r2 r1 ) = 4.56It follows that r2 = 6.84 mm and the thickness of the frozen tissue is = r2 r1 = 5.34 mm<PROBLEM 3.60KNOWN: Inner diameter, wall thickness and thermal conductivity of spherical vessel containingheat generating medium. Inner surface temperature without insulation. Thickness and thermalconductivity of insulation. Ambient air temperature and convection coefficient.FIND: (a) Thermal energy generated within vessel, (b) Inner surface temperature of vessel withinsulation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional, radial conduction, (3) Constant properties,(4) Negligible contact resistance, (5) Negligible radiation.ANALYSIS: (a) From an energy balance performed at an instant for a control surface about thepharmaceuticals, E g = q, in which case, without the insulationEg = q =Ts,1 T11 1 r r +4 k w 1 2 4 r 2 h21Eg = q =(=(50 25 ) C1 1 1 +4 (17 W / m K ) 0.50m 0.51m 4 ( 0.51m )2 6 W / m 2 K25C1)1.84 104 + 5.10 102 K / W= 489 W<(b) With the insulation, 1 1 1 1 1 11Ts,1 = T + q + + 4 k w r1 r2 4 k i r2 r3 4 r 2 h 3K11111.84 104 +Ts,1 = 25C + 489 W+4 ( 0.04 ) 0.51 0.53 4 ( 0.53)2 6 WKTs,1 = 25C + 489 W 1.84 104 + 0.147 + 0.047 W = 120C<COMMENTS: The thermal resistance associated with the vessel wall is negligible, and without theinsulation the dominant resistance is due to convection. The thermal resistance of the insulation isapproximately three times that due to convection.PROBLEM 3.61KNOWN: Spherical tank of 1-m diameter containing an exothermic reaction and is at 200C when2the ambient air is at 25C. Convection coefficient on outer surface is 20 W/m K.FIND: Determine the thickness of urethane foam required to reduce the exterior temperature to 40C.Determine the percentage reduction in the heat rate achieved using the insulation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial (spherical) conductionthrough the insulation, (3) Convection coefficient is the same for bare and insulated exterior surface,and (3) Negligible radiation exchange between the insulation outer surface and the ambientsurroundings.PROPERTIES: Table A-3, urethane, rigid foam (300 K): k = 0.026 W/mK.ANALYSIS: (a) The heat transfer situation for the heat rate from the tank can be represented by thethermal circuit shown above. The heat rate from the tank isT Tq= tR cd + R cvwhere the thermal resistances associated with conduction within the insulation (Eq. 3.35) andconvection for the exterior surface, respectively, areR cd =(1/ rt 1/ ro ) =(1/ 0.5 1/ ro )4 k=(1/ 0.5 1/ ro ) K / W4 0.026 W / m K0.3267111R cv ==== 3.979 103 ro 2 K / W2 4 20 W / m 2 K r 2hAs 4 hrooTo determine the required insulation thickness so that To = 40C, perform an energy balance on the onode.Tt To T To+=0R cdR cv( 200 40 ) Kro = 0.5135 m( 25 40 ) K=023.979 103 ro K / Wt = ro ri = (0.5135 0.5000 ) m = 13.5 mm(1/ 0.5 1/ ro ) / 0.3267 K / W+<From the rate equation, for the bare and insulated surfaces, respectively,qo =( 200 25 ) K = 10.99 kWTt T=1/ 4 hrt2 0.01592 K / Wqins =( 200 25 )Tt T== 0.994 kWR cd + R cv (0.161 + 0.01592 ) K / WHence, the percentage reduction in heat loss achieved with the insulation is,qins qo0.994 10.99100 = 100 = 91%qo10.99<PROBLEM 3.62KNOWN: Dimensions and materials used for composite spherical shell. Heat generationassociated with stored material.FIND: Inner surface temperature, T1, of lead (proposal is flawed if this temperature exceedsthe melting point).SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constantproperties at 300K, (4) Negligible contact resistance.PROPERTIES: Table A-1, Lead: k = 35.3 W/mK, MP = 601K; St.St.: 15.1 W/mK.ANALYSIS: From the thermal circuit, it follows thatT T4 3q= 1 = q r1 R tot3Evaluate the thermal resistances,11R Pb = 1/ ( 4 35.3 W/m K ) 0.25m 0.30m = 0.00150 K/W11R St.St. = 1/ ( 4 15.1 W/m K ) 0.30m 0.31m = 0.000567 K/W)(R conv = 1/ 4 0.312 m 2 500 W/m 2 K = 0.00166 K/WR tot = 0.00372 K/W.The heat rate is q=5 105 W/m3 ( 4 / 3)( 0.25m ) = 32, 725 W. The inner surface3temperature isT1 = T + R tot q=283K+0.00372K/W (32,725 W )T1 = 405 K < MP = 601K.<Hence, from the thermal standpoint, the proposal is adequate.COMMENTS: In fabrication, attention should be given to maintaining a good thermalcontact. A protective outer coating should be applied to prevent long term corrosion of thestainless steel.PROBLEM 3.63KNOWN: Dimensions and materials of composite (lead and stainless steel) spherical shell used to storeradioactive wastes with constant heat generation. Range of convection coefficients h available forcooling.FIND: (a) Variation of maximum lead temperature with h. Minimum allowable value of h to maintainmaximum lead temperature at or below 500 K. (b) Effect of outer radius of stainless steel shell onmaximum lead temperature for h = 300, 500 and 1000 W/m2K.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant propertiesat 300 K, (4) Negligible contact resistance.PROPERTIES: Table A-1, Lead: k = 35.3 W/mK, St. St.: 15.1 W/mK.ANALYSIS: (a) From the schematic, the maximum lead temperature T1 corresponds to r = r1, and fromthe thermal circuit, it may be expressed asT1 = T + R tot q3where q = q ( 4 3 ) r1 = 5 105 W m3 ( 4 3)( 0.25 m ) = 32, 725 W . The total thermal resistance is3R tot = R cond,Pb + R cond,St.St + R convwhere expressions for the component resistances are provided in the schematic. Using the ResistanceNetwork model and Thermal Resistance tool pad of IHT, the following result is obtained for the variationof T1 with h.Maximum Pb Temperature, T(r1) (K)7006005004003001002003004005006007008009001000Convection coefficient, h(W/m^2.K)T_1Continued...PROBLEM 3.63 (Cont.)To maintain T1 below 500 K, the convection coefficient must be maintained at<h 181 W/m2KMaximum Pb temperature, T(r1) (K)(b) The effect of varying the outer shell radius over the range 0.3 r3 0.5 m is shown below.6005505004504003500.30.350.40.450.5Outer radius of steel shell, r3(m)h = 300 W/m^2.Kh = 500 W/m^2.Kh = 1000 W/m^2.kFor h = 300, 500 and 1000 W/m2K, the maximum allowable values of the outer radius are r3 = 0.365,0.391 and 0.408 m, respectively.COMMENTS: For a maximum allowable value of T1 = 500 K, the maximum allowable value of thetotal thermal resistance is Rtot = (T1 - T)/q, or Rtot = (500 - 283)K/32,725 W = 0.00663 K/W. Hence, anyincrease in Rcond,St.St due to increasing r3 must be accompanied by an equivalent reduction in Rconv.PROBLEM 3.64KNOWN: Representation of the eye with a contact lens as a composite spherical system subjected toconvection processes at the boundaries.FIND: (a) Thermal circuits with and without contact lens in place, (b) Heat loss from anteriorchamber for both cases, and (c) Implications of the heat loss calculations.SCHEMATIC:r1=10.2mmk1=0.35 W/mKr2=12.7mmk2=0.80 W/mKr3=16.5mm2T,i=37Chi=12 W/m KT,o=21Cho=6 W/m K2ASSUMPTIONS: (1) Steady-state conditions, (2) Eye is represented as 1/3 sphere, (3) Convectioncoefficient, ho, unchanged with or without lens present, (4) Negligible contact resistance.ANALYSIS: (a) Using Eqs. 3.9 and 3.36 to express the resistance terms, the thermal circuits are:Without lens:<With lens:<(b) The heat losses for both cases can be determined as q = (T,i - T,o)/Rt, where Rt is thethermal resistance from the above circuits.Without lens: R t,wo =3(12W/m 2 K4 10.2 10-3m3+2(-36 W/m K4 12.7 10 m)2R t,w = 191.2 K/W+13.2 K/W++3(-36W/m K4 16.5 10 m)221 11m4 0.35 W/m K 10.2 12.7 1033= 191.2 K/W+13.2 K/W+246.7 K/W=451.1 K/WWith lens:2)+1 1112.7 16.5 3 m4 0.80 W/m K 103= 191.2 K/W+13.2 K/W+5.41 K/W+146.2 K/W=356.0 K/WHence the heat loss rates from the anterior chamber areWithout lens: q wo = (37 21) C/451.1 K/W=35.5mWWith lens:q w = (37 21) C/356.0 K/W=44.9mW<<(c) The heat loss from the anterior chamber increases by approximately 20% when the contactlens is in place, implying that the outer radius, r3, is less than the critical radius.PROBLEM 3.65KNOWN: Thermal conductivity and inner and outer radii of a hollow sphere subjected to auniform heat flux at its outer surface and maintained at a uniform temperature on the innersurface.FIND: (a) Expression for radial temperature distribution, (b) Heat flux required to maintainprescribed surface temperatures.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)No generation, (4) Constant properties.ANALYSIS: (a) For the assumptions, the temperature distribution may be obtained byintegrating Fouriers law, Eq. 3.33. That is,Tq r r drq1rdT or r= k = k T Ts,1 .Ts,14 r1 r 24 r r1Hence,q 1 1 T ( r ) = Ts,1 + r 4 k r r1 ()2or, with q q r / 4 r2 ,22q r2 1 1 2T ( r ) = Ts,1 + k r r1 <(b) Applying the above result at r2,q =2(k Ts,2 Ts,12 1 1r2 r2 r1 ) = 10 W/m K (50 20 ) C = 3000 W/m2 .2 1(0.1m )11 0.1 0.05 m<COMMENTS: (1) The desired temperature distribution could also be obtained by solvingthe appropriate form of the heat equation,d 2 dT r=0dr dr and applying the boundary conditions T ( r1 ) = Ts,1 and kdT = q .2dr r2(2) The negative sign on q implies heat transfer in the negative r direction.2PROBLEM 3.66KNOWN: Volumetric heat generation occurring within the cavity of a spherical shell ofprescribed dimensions. Convection conditions at outer surface.FIND: Expression for steady-state temperature distribution in shell.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Steady-state conditions, (3)Constant properties, (4) Uniform generation within the shell cavity, (5) Negligible radiation.ANALYSIS: For the prescribed conditions, the appropriate form of the heat equation isd 2 dT r=0dr dr Integrate twice to obtain,dTr2and= C1drCT = 1 + C2 .r(1,2)The boundary conditions may be obtained from energy balances at the inner and outersurfaces. At the inner surface (ri),E = q 4/3 r 3 = qdT/dr) = qr / 3k.= k 4 r 2 dT/dr)g(i)cond,i(i)ririi(3)At the outer surface (ro),22q cond,o = k4 ro dT/dr)ro = q conv = h4 ro T ( ro ) T dT/dr)ro = ( h/k ) T ( ro ) T .(4)From Eqs. (1) and (3), C1 = qri3 / 3k. From Eqs. (1), (2) and (4)3 h qri= + C2 T 2 k 3ro k3kro33qrqrC2 = i i + T .23hro 3ro kHence, the temperature distribution isqri3qri3 1 1 qri3T= ++ T .23k r ro 3hroCOMMENTS: Note that E g = q cond,i = q cond,o = q conv .<PROBLEM 3.67KNOWN: Spherical tank of 3-m diameter containing LP gas at -60C with 250 mm thickness ofinsulation having thermal conductivity of 0.06 W/mK. Ambient air temperature and convection2coefficient on the outer surface are 20C and 6 W/m K, respectively.FIND: (a) Determine the radial position in the insulation at which the temperature is 0C and (b) Ifthe insulation is pervious to moisture, what conclusions can be reached about ice formation? Whateffect will ice formation have on the heat gain? How can this situation be avoided?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial (spherical) conductionthrough the insulation, and (3) Negligible radiation exchange between the insulation outer surface andthe ambient surroundings.ANALYSIS: (a) The heat transfer situation can be represented by the thermal circuit shown above.The heat gain to the tank isq= 20 ( 60 ) KT Tt== 612.4 WR ins + R cv0.1263 + 4.33 103 K / W()where the thermal resistances for the insulation (see Table 3.3) and the convection process on theouter surface are, respectively,11/ r 1/ ro (1/1.50 1/1.75 ) mR ins = i== 0.1263 K / W4 k4 0.06 W / m K111R cv ==== 4.33 103 K / W2hAs h4 ro 6 W / m 2 K 4 (1.75 m )2To determine the location within the insulation where Too (roo) = 0C, use the conduction rateequation, Eq. 3.35,1 1 4 k (Too Tt ) 4 k (Too Tt )q=roo = q(1/ ri 1/ roo ) riand substituting numerical values, find114 0.06 W / m K (0 ( 60 )) K roo = 612.4 W1.5 m= 1.687 m<(b) With roo = 1.687 m, wed expect the region of the insulation ri r roo to be filled with iceformations if the insulation is pervious to water vapor. The effect of the ice formation is tosubstantially increase the heat gain since kice is nearly twice that of kins, and the ice region is ofthickness (1.687 1.50)m = 187 mm. To avoid ice formation, a vapor barrier should be installed at aradius larger than roo.PROBLEM 3.68KNOWN: Radius and heat dissipation of a hemispherical source embedded in a substrate ofprescribed thermal conductivity. Source and substrate boundary conditions.FIND: Substrate temperature distribution and surface temperature of heat source.SCHEMATIC:ASSUMPTIONS: (1) Top surface is adiabatic. Hence, hemispherical source in semi-infinitemedium is equivalent to spherical source in infinite medium (with q = 8 W) and heat transferis one-dimensional in the radial direction, (2) Steady-state conditions, (3) Constant properties,(4) No generation.ANALYSIS: Heat equation reduces to1 d 2 dT rr 2dT/dr=C1=02 dr dr rT ( r ) = C1 / r+C2 .Boundary conditions:T ( ) = TT ( ro ) = TsHence, C2 = T andTs = C1 / ro + TandC1 = ro ( T Ts ) .The temperature distribution has the formT ( r ) = T + ( Ts T ) ro / r<and the heat rate isq=-kAdT/dr = k2 r 2 ( Ts T ) ro / r 2 = k2 ro ( Ts T )It follows thatTs T =q4W== 50.9 C-4 mk2 ro 125 W/m K 2 10()Ts = 77.9 C.COMMENTS: For the semi-infinite (or infinite) medium approximation to be valid, thesubstrate dimensions must be much larger than those of the transistor.<PROBLEM 3.69KNOWN: Critical and normal tissue temperatures. Radius of spherical heat source and radius of tissueto be maintained above the critical temperature. Tissue thermal conductivity.FIND: General expression for radial temperature distribution in tissue. Heat rate required to maintainprescribed thermal conditions.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant k.ANALYSIS: The appropriate form of the heat equation is1 d dT r=0r 2 dr dr Integrating twice,dT C1=dr r 2CT ( r ) = 1 + C2r)(222Since T Tb as r , C2 = Tb. At r = ro, q = k 4 ro dT dr = 4 kro C1 ro = -4kC1.roHence, C1 = -q/4k and the temperature distribution isT (r ) =q+ Tb4 kr<It follows thatq = 4 kr T ( r ) Tb Applying this result at r = rc,q = 4 ( 0.5 W m K )( 0.005 m )( 42 37 ) C = 0.157 W<COMMENTS: At ro = 0.0005 m, T(ro) = q ( 4 kro ) + Tb = 92C. Proximity of this temperature tothe boiling point of water suggests the need to operate at a lower power dissipation level.PROBLEM 3.70KNOWN: Cylindrical and spherical shells with uniform heat generation and surface temperatures.FIND: Radial distributions of temperature, heat flux and heat rate.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Uniform heat generation, (3)Constant k.ANALYSIS: (a) For the cylindrical shell, the appropriate form of the heat equation is1 d dT qr + = 0r dr dr kThe general solution isT (r ) = q2r + C1 ln r + C24kApplying the boundary conditions, it follows thatT ( r1 ) = Ts,1 = q2r1 + C1 ln r1 + C24kT ( r2 ) = Ts,2 = q2r2 + C1 ln r2 + C24kwhich may be solved for()()22C1 = ( q/4k ) r2 r1 + Ts,2 Ts,1 ln ( r2 /r1 )2C2 = Ts,2 + ( q 4k ) r2 C1 ln r2Hence,ln ( r/r2 )222T ( r ) = Ts,2 + ( q 4k ) r2 r 2 + ( q 4k ) r2 r1 + Ts,2 Ts,1 ln ( r /r )21)()(()<With q = k dT/dr , the heat flux distribution isq ( r ) =q2()()22k ( q 4k ) r2 r1 + Ts,2 Ts,1 r r ln ( r2 /r1 )<Continued...PROBLEM 3.70 (Cont.)Similarly, with q = q A(r) = q (2rL), the heat rate distribution isq ( r ) = Lqr 2)(()222 Lk ( q 4k ) r2 r1 + Ts,2 Ts,1 <ln ( r2 /r1 )(b) For the spherical shell, the heat equation and general solution are1 d 2 dT qr+ = 02 dr dr krT(r) = ( q 6k ) r 2 C1/r + C2Applying the boundary conditions, it follows that2T ( r1 ) = Ts,1 = ( q 6k ) r1 C1/r1 + C22T ( r2 ) = Ts,2 = ( q 6k ) r2 C1/r2 + C2Hence,()() [(1 r1 ) (1 r2 )]22C1 = ( q 6k ) r2 r1 + Ts,2 Ts,1 2C2 = Ts,2 + ( q 6k ) r2 + C1/r2and(1 r ) (1 r2 )222T ( r ) = Ts,2 + ( q 6k ) r2 r 2 ( q 6k ) r2 r1 + Ts,2 Ts,1 (1 r ) (1 r )12)(()()<With q (r) = - k dT/dr, the heat flux distribution isq ( r ) =q3(()( q 6 ) r 2 r 2 + k ( T T )21s,2s,1 1r(1 r1 ) (1 r2 )<r2)and, with q = q 4 r 2 , the heat rate distribution is4 q 3q (r ) =r3()()224 (q 6 ) r2 r1 + k Ts,2 Ts,1 (1 r1 ) (1 r2 )<PROBLEM 3.71KNOWN: Temperature distribution in a composite wall.FIND: (a) Relative magnitudes of interfacial heat fluxes, (b) Relative magnitudes of thermalconductivities, and (c) Heat flux as a function of distance x.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties.ANALYSIS: (a) For the prescribed conditions (one-dimensional, steady-state, constant k),the parabolic temperature distribution in C implies the existence of heat generation. Hence,since dT/dx increases with decreasing x, the heat flux in C increases with decreasing x.Hence,q > q34However, the linear temperature distributions in A and B indicate no generation, in which caseq = q32(b) Since conservation of energy requires that q ,B = q and dT/dx)B < dT/dx)C , it follows33,Cfrom Fouriers law thatk B > kC.Similarly, since q ,A = q and dT/dx)A > dT/dx) B , it follows that22,Bk A < k B.(c) It follows that the flux distribution appears as shown below.COMMENTS: Note that, with dT/dx)4,C = 0, the interface at 4 is adiabatic.PROBLEM 3.72KNOWN: Plane wall with internal heat generation which is insulated at the inner surface andsubjected to a convection process at the outer surface.FIND: Maximum temperature in the wall.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with uniformvolumetric heat generation, (3) Inner surface is adiabatic.ANALYSIS: From Eq. 3.42, the temperature at the inner surface is given by Eq. 3.43 and isthe maximum temperature within the wall,To = qL2 / 2k+Ts .The outer surface temperature follows from Eq. 3.46,Ts = T + qL/hWTs = 92 C+0.3 106 0.1m/500W/m2 K=92C+60C=152C.3mIt follows thatTo = 0.3 106 W/m3 ( 0.1m ) / 2 25W/m K+152C2To = 60 C+152C=212C.<COMMENTS: The heat flux leaving the wall can be determined from knowledge of h, Tsand T using Newtons law of cooling.22qconv = h ( Ts T ) = 500W/m K (152 92 ) C=30kW/m .This same result can be determined from an energy balance on the entire wall, which has theformEg Eout = 0whereEg = qALandEout = qconv A.Hence,632qconv = qL=0.3 10 W/m 0.1m=30kW/m .PROBLEM 3.73KNOWN: Composite wall with outer surfaces exposed to convection process.FIND: (a) Volumetric heat generation and thermal conductivity for material B required for specialconditions, (b) Plot of temperature distribution, (c) T1 and T2, as well as temperature distributionscorresponding to loss of coolant condition where h = 0 on surface A.SCHEMATIC:LA = 30 mmLB = 30 mmLC = 20 mmkA = 25 W/mKkC = 50 W/mKASSUMPTIONS: (1) Steady-state, one-dimensional heat transfer, (2) Negligible contact resistance atinterfaces, (3) Uniform generation in B; zero in A and C.ANALYSIS: (a) From an energy balance on wall B,E in E out + E g = E st q1 q + 2qL B = 02q B = ( q1 + q ) 2L B .2To determine the heat fluxes, q1T and q TT , construct thermal circuits for A and C:T2q1 = ( T1 T ) (1 h + L A k A )q = ( T2 T ) ( L C k C + 1 h )210.030 m + 1000 W m 2 K 25 W m K q1 = ( 261 25 ) Cq1 = 236 C ( 0.001 + 0.0012 ) m K Wq = 186 C ( 0.0004 + 0.001) m K W22q1 = 107, 273 W m 0.020 m1+ 50 W m K 1000 W m 2 K q = ( 211 25 ) C222q = 132, 857 W m22Using the values for q1 and q in Eq. (1), find2(q B = 106, 818 + 132,143 W m2) 2 0.030 m = 4.00 106<3Wm.To determine kB, use the general form of the temperature and heat flux distributions in wall B,T(x) = qB2k B2x + C1x + C 2q (x) = k B xq kBx + C1 (1,2)there are 3 unknowns, C1, C2 and kB, which can be evaluated using three conditions,Continued...PROBLEM 3.73 (Cont.)T ( L B ) = T1 = qB2k BT ( + L B ) = T2 = qB2k B( L B )2 C1L B + C 2where T1 = 261C(3)( + L B )2 + C1L B + C2where T2 = 211C(4)where q1 = 107,273 W/m2(5)qq ( L B ) = q1 = k B B ( L B ) + C1 x kBUsing IHT to solve Eqs. (3), (4) and (5) simultaneously with q B = 4.00 106 W/m3, find<k B = 15.3 W m K(b) Following the method of analysis in the IHT Example 3.6, User-Defined Functions, the temperaturedistribution is shown in the plot below. The important features are (1) Distribution is quadratic in B, butnon-symmetrical; linear in A and C; (2) Because thermal conductivities of the materials are different,discontinuities exist at each interface; (3) By comparison of gradients at x = -LB and +LB, find q > q1 .2(c) Using the same method of analysis as for Part (c), the temperature distribution is shown in the plotbelow when h = 0 on the surface of A. Since the left boundary is adiabatic, material A will be isothermalat T1. FindT1 = 835C<T2 = 360CLoss of coolant on surface A400Temperature, T (C)Temperature, T (C)800300200100600400200-60-40-20020Wall position, x-coordinate (mm)T_xA, kA = 25 W/m.KT_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3T_x, kC = 50 W/m.K40-60-40-20020Wall position, x-coordinate (mm)T_xA, kA = 25 W/m.K; adiabatic surfaceT_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3T_x, kC = 50 W/m.K40PROBLEM 3.74KNOWN: Composite wall exposed to convection process; inside wall experiences a uniform heatgeneration.FIND: (a) Neglecting interfacial thermal resistances, determine T1 and T2, as well as the heat fluxesthrough walls A and C, and (b) Determine the same parameters, but consider the interfacial contactresistances. Plot temperature distributions.SCHEMATIC:k A = 25 W m KL A = 30 mmk B = 15 W m KL B = 30 mmk C = 50 W m KL C = 20 mmq B 4 ` 10 6 W m 3ASSUMPTIONS: (1) One-dimensional, steady-state heat flow, (2) Negligible contact resistancebetween walls, part (a), (3) Uniform heat generation in B, zero in A and C, (4) Uniform properties, (5)Negligible radiation at outer surfaces.ANALYSIS: (a) The temperature distribution in wall B follows from Eq. 3.41,qL x T2 T1 xT T2.+1T ( x ) = B B 1 +22k B 2LB2LB 22(1)The heat fluxes to the neighboring walls are found using Fouriers law,q = kxdTdx.qT T At x = L B : q ( L B ) k B + B ( L B ) + 2 1 = q1 (2)x2L B kBqT T At x = + L B : q ( L B ) k B B ( L B ) + 2 1 = q (3)x22L B kBThe heat fluxes, q1 and q , can be evaluated by thermal circuits.2Substituting numerical values, find(q1 = ( T T1 ) C (1 h + LA k A ) = ( 25 T1 ) C 1 1000 W m 2 K + 0.03 m 25 W m K)q1 = ( 25 T1 ) C ( 0.001 + 0.0012 ) K W = 454.6 ( 25 T1 )((4)q = ( T2 T ) C (1 h + LC k C ) = ( T2 25 ) C 1 1000 W m 2 K + 0.02 m 50 W m K2q = ( T2 25 ) C2( 0.001 + 0.0004 ) KW = 714.3 ( T2 25 ) .)(5)Continued...PROBLEM 3.74 (Cont.)Substituting the expressions for the heat fluxes, Eqs. (4) and (5), into Eqs. (2) and (3), a system of twoequations with two unknowns is obtained.T TEq. (2):4 106 W m3 0.03 m + 15 W m K 2 1 = q12 0.03 m1.2 105 W m 2 2.5 102 ( T2 T1 ) W m 2 = 454.6 ( 25 T1 )704.6 T1 250 T2 = 131, 365Eq. (3):(6)+4 106 W m3 0.03 m 15 W m KT2 T12 0.03 m= q2+1.2 105 W m 2 2.5 102 (T2 T1 ) W m 2 = 714.3 ( T2 25 )250 T1 964 T2 = 137,857Solving Eqs. (6) and (7) simultaneously, find(7)T2 = 210.0CT1 = 260.9CFrom Eqs. (4) and (5), the heat fluxes at the interfaces and through walls A and C are, respectively,<q1 = 454.6 ( 25 260.9 ) = 107, 240 W m 2<<q = 714.3 ( 210 25 ) = +132,146 W m 2 .2Note directions of the heat fluxes.(b) Considering interfacial contact resistances, we will use a different approach. The general solution forthe temperature and heat flux distributions in each of the materials isTA ( x ) = C1x + C2TB ( x ) = qB2k Bq = k A C1xx 2 + C3 x + C4TC ( x ) = C5 x + C6 (LA + LB ) x LB(1,2)qq = B x + C3xkBLB x LB(3,4)+ L B x ( L B + LC )(5,6)q = k CC5xTo determine C1 ... C6 and the distributions, we need to identify boundary conditions using surfaceenergy balances.At x = -(LA + LB):(7)q ( L A L B ) + q = 0xcv ( k A C1 ) + h [T TA ( L A L B )] (8)At x = -LB: The heat flux must be continuous, but the temperature will be discontinuous across thecontact resistance.q ( L B ) = q ( L B )x,Ax,B(9)q ( L B ) = [T1A ( L B ) T1B ( L B )] R x,Atc,AB(10)Continued...PROBLEM 3.74 (Cont.)At x = + LB: The same conditions apply as for x = -LB,q ( + L B ) = q ( + L B )x,Bx,C(11)q ( + L B ) = [T2B ( + L B ) T2C ( + L B )] R x,Btc,BC(12)At x = +(LB + LC): q x,C ( L B + LC ) q = 0cv(13) ( k C C5 ) h [TC ( L B + L C ) T ] = 0(14)Following the method of analysis in IHT Example 3.6, User-Defined Functions, we solve the system ofequations above for the constants C1 ... C6 for conditions with negligible and prescribed values for theinterfacial constant resistances. The results are tabulated and plotted below; q1 and q represent heat2fluxes leaving surfaces A and C, respectively.T1B (C)260T2B (C)210T2C (C)210q1 (kW/m )q (kW/m )2R = 0tcT1A (C)260106.8132.0R 0tc23347037122794.6144.2Conditions2500Temperature, T (C)Temperature, T (C)5002300100300100-60-40-2002040-60-40Wall position, x-coordinate (mm)-2002040Wall position, x-coordinate (mm)T_xA, kA = 25 W/m.KT_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3T_x, kC = 50 W/m.KT_xA, kA = 25 W/m.KT_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3T_x, kC = 50 W/m.KCOMMENTS: (1) The results for part (a) can be checked using an energy balance on wall B,E in E out = E gq1 q = q B 2L B2where2q1 q = 107, 240 132,146 = 239, 386 W m2 q B L B = 4 10 W m 2 ( 0.03 m ) = 240, 000 W m .632Hence, we have confirmed proper solution of Eqs. (6) and (7).(2) Note that the effect of the interfacial contact resistance is to increase the temperature at all locations.The total heat flux leaving the composite wall (q1 + q2) will of course be the same for both cases.PROBLEM 3.75KNOWN: Composite wall of materials A and B. Wall of material A has uniform generation, whilewall B has no generation. The inner wall of material A is insulated, while the outer surface ofmaterial B experiences convection cooling. Thermal contact resistance between the materials isR = 10t,c42m K / W . See Ex. 3.6 that considers the case without contact resistance.FIND: Compute and plot the temperature distribution in the composite wall.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with constantproperties, and (3) Inner surface of material A is adiabatic.ANALYSIS: From the analysis of Ex. 3.6, we know the temperature distribution in material A isparabolic with zero slope at the inner boundary, and that the distribution in material B is linear. Atthe interface between the two materials, x = LA, the temperature distribution will show adiscontinuity.TA ( x ) =q L2 Ax2 1 +T2 1A LA 0 x LA2 kA TB ( x ) = T1B ( T1B T2 )x LALA x LA + LBLBConsidering the thermal circuit above (see also Ex. 3.6) including the thermal contact resistance,T TT1B TT T==2q = q L A = 1AR R R totcond,B + R convconvfind TA(0) = 147.5C, T1A = 122.5C, T1B = 115C, and T2 = 105C. Using the foregoing equationsin IHT, the temperature distributions for each of the materials can be calculated and are plotted on theEffect of therm al contact res is tance on tem perature distributiongraph below.150140T (C )130120110100010203040506070x (m m )COMMENTS: (1) The effect of the thermal contact resistance between the materials is to increasethe maximum temperature of the system.(2) Can you explain why the temperature distribution in the material B is not affected by the presenceof the thermal contact resistance at the materials interface?PROBLEM 3.76KNOWN: Plane wall of thickness 2L, thermal conductivity k with uniform energy generation q.For case 1, boundary at x = -L is perfectly insulated, while boundary at x = +L is maintained at To =50C. For case 2, the boundary conditions are the same, but a thin dielectric strip with thermalresistance R = 0.0005 m 2 K / W is inserted at the mid-plane.tFIND: (a) Sketch the temperature distribution for case 1 on T-x coordinates and describe keyfeatures; identify and calculate the maximum temperature in the wall, (b) Sketch the temperaturedistribution for case 2 on the same T-x coordinates and describe the key features; (c) What is thetemperature difference between the two walls at x = 0 for case 2? And (d) What is the location of themaximum temperature of the composite wall in case 2; calculate this temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the plane andcomposite walls, and (3) Constant properties.ANALYSIS: (a) For case 1, the temperature distribution, T1(x) vs. x, is parabolic as shown in theschematic below and the gradient is zero at the insulated boundary, x = -L. From Eq. 3.43,22q 2L5 106 W / m3 2 0.020 mT1 ( L ) T1 ( + L ) =()2k(=2 50 W / m K)= 80Cand since T1(+L) = To = 50C, the maximum temperature occurs at x = -L,T1 ( L ) = T1 ( + L ) + 80C = 130C(b) For case 2, the temperature distribution, T2(x) vs. x, is piece-wise parabolic, with zero gradient atx = -L and a drop across the dielectric strip, TAB. The temperature gradients at either side of thedielectric strip are equal.(c) For case 2, the temperature drop across the thin dielectric strip follows from the surface energybalance shown above.q ( 0 ) = TAB / R xtq ( 0 ) = qLxTAB = R qL = 0.0005 m 2 K / W 5 106 W / m3 0.020 m = 50C.t(d) For case 2, the maximum temperature in the composite wall occurs at x = -L, with the value,T2 ( L ) = T1 ( L ) + TAB = 130C + 50C = 180C<PROBLEM 3.77KNOWN: Geometry and boundary conditions of a nuclear fuel element.FIND: (a) Expression for the temperature distribution in the fuel, (b) Form of temperaturedistribution for the entire system.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3)Uniform generation, (4) Constant properties, (5) Negligible contact resistance between fueland cladding.ANALYSIS: (a) The general solution to the heat equation, Eq. 3.39,d 2Tdx 2is+T=q=0kf( L x +L )q2x + C1x+C2 .2k fThe insulated wall at x = - (L+b) dictates that the heat flux at x = - L is zero (for an energybalance applied to a control volume about the wall, Ein = Eout = 0). HencedT q= ( L ) + C1 = 0dx x = LkfT=C1 = orqLkfq 2 qLxx+C2 .2k fkfThe value of Ts,1 may be determined from the energy conservation requirement thatEg = qcond = qconv , or on a unit area basis.()()kq ( 2L ) = s Ts,1 Ts,2 = h Ts,2 T .bHence,Ts,1 =Ts,1 =q ( 2 Lb )ksq ( 2 Lb )ks+ Ts,2+q ( 2L )hwhereTs,2 =q ( 2L )h+ T+ T .Continued ..PROBLEM 3.77 (Cont.)Hence from Eq. (1),T ( L ) = Ts,1 =q ( 2 Lb )ks+q (2 L )h()23 qL+ T = + C22 kfwhich yields 2b 2 3 L C2 = T + qL + + ks h 2 k f Hence, the temperature distribution for ( L x +L ) isT=q 2 qLxx+qL2k fkf 2b 2 3 L ++ + T ks h 2 k f (b) For the temperature distribution shown below,( L b ) x L: L x +L:+L x L+b:dT/dx=0, T=Tmax| dT/dx | with x(dT/dx ) is const.<PROBLEM 3.78KNOWN: Thermal conductivity, heat generation and thickness of fuel element. Thickness andthermal conductivity of cladding. Surface convection conditions.FIND: (a) Temperature distribution in fuel element with one surface insulated and the other cooledby convection. Largest and smallest temperatures and corresponding locations. (b) Same as part (a)but with equivalent convection conditions at both surfaces, (c) Plot of temperature distributions.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state, (3) Uniform generation, (4)Constant properties, (5) Negligible contact resistance.ANALYSIS: (a) From Eq. C.1,T (x ) =q L2 x 2 Ts,2 Ts,1 x Ts,1 + Ts,21 ++2k f L2 2L2(1)With an insulated surface at x = -L, Eq. C.10 yieldsTs,1 Ts,2 =2 q L2kf(2)and with convection at x = L + b, Eq. C.13 yields()(kU Ts,2 T = q L f Ts,2 Ts,12LTs,1 Ts,2 =)2 LU2 q L2Ts,2 T kfkf()(3)Substracting Eq. (2) from Eq. (3),0=2LU4 q L2Ts,2 T kfkf(Ts,2 = T +)2 qLU(4)Continued ..PROBLEM 3.78 (Cont.)and substituting into Eq. (2) L 1+Ts,1 = T + 2 qL kf U (5)Substituting Eqs. (4) and (5) into Eq. (1),T (x ) = -12 3 Lq 2 qLxx + qL +2 kfkf U 2 kf + T-1or, with U = h + b/ks,T (x ) = 2b 2 3 Lq 2 qLxx + qL + +2 kfkf ks h 2 k f + T(6)<The maximum temperature occurs at x = - L and isb 1 LT ( L ) = 2 qL + + ks h k f + T 0.003m10.015 m ++ + 200C = 530C 15 W / m K210, 000 W / m K 60 W / m K T ( L ) = 2 2 10 W / m 0.015 m 73<The lowest temperature is at x = + L and isT (+L ) = 2b 2 3 L3 qL2+ qL + +2 kf ks h 2 k f + T = 380C<(b) If a convection condition is maintained at x = - L, Eq. C.12 reduces to()(kU T Ts,1 = qL f Ts,2 Ts,12L)2 LU2 qL2Ts,1 Ts,2 =Ts,1 T kfkf()(7)Subtracting Eq. (7) from Eq. (3),0=(2 LUTs,2 T Ts,1 + Tkf)orTs,1 = Ts,2Hence, from Eq. (7)Continued ..PROBLEM 3.78 (Cont.)Ts,1 = Ts,2 =1 b qL+ T = qL + + TU h ks (8)Substituting into Eq. (1), the temperature distribution isT (x ) =1 b qL2 x 2 1 + qL + + T2 k f L2 h ks (9)<The maximum temperature is at x = 0 and isT (0 ) =72 10 W / m3(0.015 m )22 60 W / m K7310.003 m ++ 200C 10, 000 W / m 2 K 15 W / m K + 2 10 W / m 0.015 m T ( 0 ) = 37.5C + 90C + 200C = 327.5C<The minimum temperature at x = L is10.003m + 200C = 290CTs,1 = Ts,2 = 2 107 W / m3 ( 0.015 m ) + 10, 000 W / m 2 K 15 W / m K (c) The temperature distributions are as shown.550Te m p e ra tu re , T(C )500450400350300250200-0 .0 1 5-0 .0 0 9-0 .0 0 30 .0 0 30 .0 0 90 .0 1 5Fu e l e le m e n t lo c a tio n , x(m )In s u la te d s u rfa c eS ym m e trica l co n ve c tio n c o n d itio n sThe amount of heat generation is the same for both cases, but the ability to transfer heat from bothsurfaces for case (b) results in lower temperatures throughout the fuel element.COMMENTS: Note that for case (a), the temperature in the insulated cladding is constant andequivalent to Ts,1 = 530C.<PROBLEM 3.79KNOWN: Wall of thermal conductivity k and thickness L with uniform generation q ; strip heaterwith uniform heat flux q ; prescribed inside and outside air conditions (hi, T,i, ho, T,o).oFIND: (a) Sketch temperature distribution in wall if none of the heat generated within the wall is lostto the outside air, (b) Temperatures at the wall boundaries T(0) and T(L) for the prescribed condition,(c) Value of q required to maintain this condition, (d) Temperature of the outer surface, T(L), ifo corresponds to the value calculated in (c).q=0 but q oSCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniformvolumetric generation, (4) Constant properties.ANALYSIS: (a) If none of the heat generated within the wall islost to the outside of the chamber, the gradient at x = 0 must be zero.Since q is uniform, the temperature distribution is parabolic, withT(L) > T,i.(b) To find temperatures at the boundaries of wall, begin with thegeneral solution to the appropriate form of the heat equation (Eq.3.40).T (x ) = q2x + C1x+C22k(1)From the first boundary condition,dT=0dx x=oC1 = 0.(2)Two approaches are possible using different forms for the second boundary condition.Approach No. 1: With boundary condition T (0 ) = T1T (x ) = q2x + T12k(3)To find T1, perform an overall energy balance on the wallEin E out + Eg = 0 h T ( L ) T,i + qL=0T ( L ) = T2 = T,i +qLh(4)Continued ..PROBLEM 3.79 (Cont.)and from Eq. (3) with x = L and T(L) = T2,T (L) = q2L + T12korT1 = T2 +q2qL qL2L = T,i ++2kh2k(5,6)Substituting numerical values into Eqs. (4) and (6), findT2 = 50 C+1000 W/m3 0.200 m/20 W/m 2 K=50C+10C=60C<T1 = 60 C+1000 W/m3 ( 0.200 m ) / 2 4 W/m K=65C.<2Approach No. 2: Using the boundary conditionkdT= h T ( L ) T,i dx x=Lyields the following temperature distribution which can be evaluated at x = 0,L for the requiredtemperatures,T (x ) = )(q 2 2 qLx L ++ T,i .2kh(c) The value of q when T(0) = T1 = 65Cofollows from the circuitq =oT1 T,o1/ h oq = 5 W/m2 K ( 65-25) C=200 W/m 2 .o<(d) With q=0, the situation is representedby the thermal circuit shown. Hence,q = q + qoabq =oT1 T ,o1/ h o+T1 T,iL/k+1/h iwhich yieldsT1 = 55 C.<PROBLEM 3.80KNOWN: Wall of thermal conductivity k and thickness L with uniform generation and strip heaterwith uniform heat flux q ; prescribed inside and outside air conditions ( T,i , hi, T, o , ho). Strip heateroacts to guard against heat losses from the wall to the outside.FIND: Compute and plot q and T(0) as a function of q for 200 q 2000 W/m3 and T,i = 30, 50oand 70C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform volumetricgeneration, (4) Constant properties.ANALYSIS: If no heat generated within thewall will be lost to the outside of the chamber,the gradient at the position x = 0 must be zero.Since q is uniform, the temperature distributionmust be parabolic as shown in the sketch.To determine the required heater flux q as a function of the operation conditions q and T,i , theoanalysis begins by considering the temperature distribution in the wall and then surface energy balancesat the two wall surfaces. The analysis is organized for easy treatment with equation-solving software.Temperature distribution in the wall, T(x): The general solution for the temperature distribution in thewall is, Eq. 3.40,T(x) = q2kx 2 + C1x + C2and the guard condition at the outer wall, x = 0, requires that the conduction heat flux be zero. UsingFouriers law,dT (1)q (0) = k= kC1 = 0(C1 = 0 )xdx x = 0At the outer wall, x = 0,T(0) = C2(2)Surface energy balance, x = 0:E in E out = 0q q ocv,o q ( 0 ) = 0x()q cv,o = h T(0) T ,o , q x ( 0 ) = 0(3)(4a,b)Continued...PROBLEM 3.80 (Cont.)Surface energy balance, x = L:E in E out = 0q (L) q = 0xcv,iq (L) = kxdT dx x = L(5)= + qL(6)q v,i = h T(L) T ,i c q L2 + T 0 T ( ) ,i 2kq = h cv,i(7)40030020010000500100015002000Volumetric generation rate, qdot (W/m^3)Tinfi = 30 CTinfi = 50 CTinfi = 70 CWall temperature, T(0) (C)Heater flux, q''o (W/m^2)Solving Eqs. (1) through (7) simultaneously with appropriate numerical values and performing theparametric analysis, the results are plotted below.120100806040200500100015002000Volumetric generation rate, qdot (W/m^3)Tinfi = 30 CTinfi = 50 CTinfi = 70 CFrom the first plot, the heater flux q is a linear function of the volumetric generation rate q . Asoexpected, the higher q and T,i , the higher the heat flux required to maintain the guard condition( q (0) = 0). Notice that for any q condition, equal changes in T,i result in equal changes in thexrequired q . The outer wall temperature T(0) is also linearly dependent upon q . From our knowledgeoof the temperature distribution, it follows that for any q condition, the outer wall temperature T(0) willtrack changes in T,i .PROBLEM 3.81KNOWN: Plane wall with prescribed nonuniform volumetric generation having oneboundary insulated and the other isothermal.FIND: Temperature distribution, T(x), in terms of x, L, k, q o and To .SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in xdirection, (3) Constant properties.ANALYSIS: The appropriate form the heat diffusion equation isd dT q dx + k = 0.dx Noting that q = q ( x ) = qo (1 x/L ) , substitute for q ( x ) into the above equation, separatevariables and then integrate,x2 x + C1.2L Separate variables and integrate again to obtain the general form of the temperaturedistribution in the wall,qq x 2 x3 x2 dT = o x T (x ) = o dx+C1dx + C1x+C2 .k2L k 2 6L q dT d = ok dx x1 L dxqdT= odxkIdentify the boundary conditions at x = 0 and x = L to evaluate C1 and C2. At x = 0,qT ( 0 ) = To = o ( 0 0 ) + C1 0 + C2hence, C2 = TokAt x = L,qqLdT L2 = 0 = o L hence, C1 = o + C1dx x=Lk2L 2kThe temperature distribution isqo x 2 x 3 qo LT (x ) = x+To .+k 2 6L 2k<COMMENTS: It is good practice to test the final result for satisfying BCs. The heat flux atx = 0 can be found using Fouriers law or from an overall energy balanceLEout = Eg = qdV0to obtainq = q o L/2.out PROBLEM 3.82KNOWN: Distribution of volumetric heating and surface conditions associated with a quartzwindow.FIND: Temperature distribution in the quartz.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3)Negligible radiation emission and convection at inner surface (x = 0) and negligible emissionfrom outer surface, (4) Constant properties.ANALYSIS: The appropriate form of the heat equation for the quartz is obtained bysubstituting the prescribed form of q into Eq. 3.39.d 2T (1 ) q - xoe+=02kdxIntegrating,(1 ) q e- x + CdTo=+1dxkT=(1 ) q e- x + C x+Cok12 k dT/dx) x=o = qo k dT/dx) x=L = h T ( L ) T Boundary Conditions: (1- )k q + C1 = qookC1 = q / koHence, at x = 0:At x = L: (1- ) (1- )k q e- L + C1 = h q e- L + C1L+C2 T ook kSubstituting for C1 and solving for C2,qC2 = ohHence,qo1 (1 ) e- L + q + o(1- ) e- L + T .kkT (x ) =(1 ) q e- L e- x + qookk(L x ) +q o 1 (1 ) e- L + T . <hCOMMENTS: The temperature distribution depends strongly on the radiative coefficients, and . For or = 1, the heating occurs entirely at x = 0 (no volumetric heating).PROBLEM 3.83KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactivewastes. Surface convection conditions.FIND: Radial temperature distribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties, (4) Negligible temperature drop across container wall.ANALYSIS: The appropriate form of the heat equation isq r2 1 d dT q= = o 1 r2r dr dr kk ro rq r 2 qr 4dT= o ++ C12dr2k4kroq r2 q r4T = o + o + C1 ln r+C2 .4k 16kr 2oFrom the boundary conditions,dT|r=0 = 0 C1 = 0drkdT|r=ro = h T ( ro ) T )dr q r2 q r2q o ro q o ro+= h o o + o o + C2 T 2416k 4kqr3q r 2C2 = o o + o o + T .4h16kHenceqrq r2 3 1 rT ( r ) = T + o o + o o 4hk 16 4 ro241r + . 16 ro <COMMENTS: Applying the above result at ro yieldsTs = T ( ro ) = T + ( qo ro ) / 4hThe same result may be obtained by applying an energy balance to a control surface about thecontainer, where Eg = qconv . The maximum temperature exists at r = 0.PROBLEM 3.84KNOWN: Cylindrical shell with uniform volumetric generation is insulated at inner surfaceand exposed to convection on the outer surface.FIND: (a) Temperature distribution in the shell in terms of ri , ro , q, h, T and k, (b)Expression for the heat rate per unit length at the outer radius, q ( ro ).SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (cylindrical)conduction in shell, (3) Uniform generation, (4) Constant properties.ANALYSIS: (a) The general form of the temperature distribution and boundary conditionsareT (r ) = q2r + C1 ln r+C24kdT q1 = 0 = ri + C1 + 0dr r2kriat r = ri:C1 =idT = h T ( ro ) T dr rkat r = ro:q2r2k isurface energy balanceoqq q 2 q1 k ro + ri2 = h ro + ri2 ln ro + C2 T ro 2k 2k 4k 2kqrC2 = o2h 2 21 + ri + qro ro 2k2 1 ri ln ro + T 2 ro Hence,() 2 r qroq 2 2 qriT (r ) =ro r +ln 4k2k ro 2h 2 1 + ri + T . ro <(b) From an overall energy balance on the shell,q ( r ) = E = q r 2 r 2 .rog(o i )<Alternatively, the heat rate may be found using Fouriers law and the temperature distribution,q ( r ) = k ( 2 ro )rqqr 2dT 2 ro + i 1 + 0 + 0 = q ro r 2 = 2 kro 2kidr r2k roo()PROBLEM 3.85KNOWN: The solid tube of Example 3.7 with inner and outer radii, 50 and 100 mm, and a thermalconductivity of 5 W/mK. The inner surface is cooled by a fluid at 30C with a convection coefficient2of 1000 W/m K.5FIND: Calculate and plot the temperature distributions for volumetric generation rates of 1 10 , 5563 10 , and 1 10 W/m . Use Eq. (7) with Eq. (10) of the Example 3.7 in the IHT Workspace.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constantproperties and (4) Uniform volumetric generation.ANALYSIS: From Example 3.7, the temperature distribution in the tube is given by Eq. (7),T ( r ) = Ts,2 +)(q22q 2 r2 r2 r r "n 4k2k 2 r r1 r r2The temperature at the inner boundary, Ts,1, follows from the surface energy balance, Eq. (10),2 2 q r2 r1 = h2 r1 Ts,1 T()()(1)(2)For the conditions prescribed in the schematic with q = 1 105 W / m3 , Eqs. (1) and (2), with r = r1and T(r) = Ts,1, are solved simultaneously to find Ts,2 = 69.3C. Eq. (1), with Ts,2 now a knownparameter, can be used to determine the temperature distribution, T(r). The results for differentvalues of the generation rate are shown in the graph.Effect of generation rate on temperature distributionsTemperature, T(C)50040030020010005060708090100Radial location, r (mm)qdot = 1e5 W/m^3qdot = 5e5 W/m^3qdot = 1e6 W/m^3COMMENTS: (1) The temperature distributions are parabolic with a zero gradient at the insulatedouter boundary, r = r2. The effect of increasing q is to increase the maximum temperature in thetube, which always occurs at the outer boundary.(2) The equations used to generate the graphical result in the IHT Workspace are shown below.// The temperature distribution, from Eq. 7, Example 3.7T_r = Ts2 + qdot/(4*k) * (r2^2 r^2) qgot / (2*k) * r2^2*ln (r2/r)// The temperature at the inner surface, from Eq. 7Ts1 = Ts2 + qdot / (4*k) * (r2^2 r1^2) qdot / (2*k) * r2^2 * ln (r2/r1)// The energy balance on the surface, from Eq. 10pi * qdot * (r2^2 r1^2) = h * 2 * pi * r1 * (Ts1 Tinf)PROBLEM 3.86KNOWN: Diameter, resistivity, thermal conductivity, emissivity, voltage, and maximum temperatureof heater wire. Convection coefficient and air exit temperature. Temperature of surroundings.FIND: Maximum operating current, heater length and power rating.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Uniform wire temperature, (3) Constant properties, (4)Radiation exchange with large surroundings.ANALYSIS: Assuming a uniform wire temperature, Tmax = T(r = 0) To Ts, the maximumvolumetric heat generation may be obtained from Eq. (3.55), but with the total heat transfercoefficient, ht = h + hr, used in lieu of the convection coefficient h. With()h r = ( Ts + Tsur ) Ts + Tsur = 0.20 5.67 102282W/m K4(1473 + 323 ) K(21473 + 323)222K = 46.3 W / m Kh t = ( 250 + 46.3) W / m 2 K = 296.3 W / m 2 K)(2 296.3 W / m 2 K2 htq max =(Ts T ) =(1150C ) = 1.36 109 W / m3ro0.0005mHence, with2I 2 R e I ( e L / A c ) I 2 eI2 eq====22LAcAc D2 / 4)(1/ 2qImax = max e 1/ 2 D 2 1.36 109 W / m3 = 106 m4 (0.001m )= 29.0 A42<Also, with E = I Re = I (eL/Ac),E AcL==Imax e2110 V ( 0.001m ) / 4 = 2.98m6 m29.0 A 10()<and the power rating isPelec = E Imax = 110 V ( 29 A ) = 3190 W = 3.19 kW<COMMENTS: To assess the validity of assuming a uniform wire temperature, Eq. (3.53) may beused to compute the centerline temperature corresponding to q max and a surface temperature of1200C. It follows that To =2q ro4k( 0.0005m )24 ( 25 W / m K )9+ Ts =1.36 10 W / m3+ 1200C = 1203C. With only a3C temperature difference between the centerline and surface of the wire, the assumption isexcellent.PROBLEM 3.87KNOWN: Energy generation in an aluminum-clad, thorium fuel rod under specified operatingconditions.FIND: (a) Whether prescribed operating conditions are acceptable, (b) Effect of q and h on acceptableoperating conditions.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in r-direction, (2) Steady-state conditions, (3)Constant properties, (4) Negligible temperature gradients in aluminum and contact resistance betweenaluminum and thorium.PROPERTIES: Table A-1, Aluminum, pure: M.P. = 933 K; Table A-1, Thorium: M.P. = 2023 K, k 60 W/mK.ANALYSIS: (a) System failure would occur if the melting point of either the thorium or the aluminumwere exceeded. From Eq. 3.53, the maximum thorium temperature, which exists at r = 0, isT(0) =2qro4k+ Ts = TTh,maxwhere, from the energy balance equation, Eq. 3.55, the surface temperature, which is also the aluminumtemperature, isTs = T +qro2h= TAlHence,TAl = Ts = 95 C +7 108 W m3 0.0125 m214, 000 W m K7 108 W m3 ( 0.0125m )= 720 C = 993 K2TTh,max =4 60 W m K+ 993 K = 1449 K<Although TTh,max < M.P.Th and the thorium would not melt, Tal > M.P.Al and the cladding would meltunder the proposed operating conditions. The problem could be eliminated by decreasing q , increasingh or using a cladding material with a higher melting point.(b) Using the one-dimensional, steady-state conduction model (solid cylinder) of the IHT software, thefollowing radial temperature distributions were obtained for parametric variations in q and h.Continued...PROBLEM 3.87 (Cont.)16001200Temperature, T(K)Temperature, T(K)1500140013001200110010001000800600900400800000.002 0.004 0.006 0.008 0.01 0.012 0.0140.002 0.004 0.006 0.008 0.01 0.012 0.014Radius, r(m)Radius, r(m)qdot = 2E8, h = 2000 W/m^2.Kqdot = 2E8, h = 3000 W/m^2.Kqdot = 2E8, h = 5000 W/m^2.Kqdot = 2E8, h = 10000 W/m^2.Kh = 10000 W/m^2.K, qdot = 7E8 W/m^3h = 10000 W/m^2.K, qdot = 8E8 W/m^3h = 10000 W/m^2.K, qdot = 9E9 W/m^3For h = 10,000 W/m2K, which represents a reasonable upper limit with water cooling, the temperature ofthe aluminum would be well below its melting point for q = 7 108 W/m3, but would be close to the83melting point for q = 8 10 W/m and would exceed it for q = 9 108 W/m3. Hence, under the best of83conditions, q 7 10 W/m corresponds to the maximum allowable energy generation. However, ifcoolant flow conditions are constrained to provide values of h < 10,000 W/m2K, volumetric heatingwould have to be reduced. Even for q as low as 2 108 W/m3, operation could not be sustained for h =22000 W/m K.The effects of q and h on the centerline and surface temperatures are shown below.Surface temperature, Ts (K)Centerline temperature, T(0) (K)200020001600120080040001600120080040001E81E82.8E84.6E86.4E88.2E8Energy generation, qdot (W/m^3)h = 2000 W/m^2.Kh = 5000 W/m^2.Kh = 10000 W/m^2.K1E92.8E84.6E86.4E88.2E81E9Energy generation, qdot (W/m^3)h = 2000 W/m^2.Kh = 5000 W/m^2.Kh = 10000 W/m^2.KFor h = 2000 and 5000 W/m2K, the melting point of thorium would be approached for q 4.4 108 and8328.5 10 W/m , respectively. For h = 2000, 5000 and 10,000 W/m K, the melting point of aluminumwould be approached for q 1.6 108, 4.3 108 and 8.7 108 W/m3. Hence, the envelope ofacceptable operating conditions must call for a reduction in q with decreasing h, from a maximum of q832 7 10 W/m for h = 10,000 W/m K.COMMENTS: Note the problem which would arise in the event of a loss of coolant, for which case hwould decrease drastically.PROBLEM 3.88KNOWN: Radii and thermal conductivities of reactor fuel element and cladding. Fuel heat generationrate. Temperature and convection coefficient of coolant.FIND: (a) Expressions for temperature distributions in fuel and cladding, (b) Maximum fuel elementtemperature for prescribed conditions, (c) Effect of h on temperature distribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible contactresistance, (4) Constant properties.ANALYSIS: (a) From Eqs. 3.49 and 3.23, the heat equations for the fuel (f) and cladding (c) are1 d dTf q=kfr dr dr 1 d dTc r=0r dr dr (0 r r1 )( r1 r r2 )Hence, integrating both equations twice,dTfdrdTcdr==qr2k f+C1kf rC3k crTf = qr 2C+ 1 ln r + C 24k f k fCTc = 3 ln r + C4kc(1,2)(3,4)The corresponding boundary conditions are:dTf dr )r = 0 = 0k fdTfdrdT = k c c dr r = rr = r11Tf ( r1 ) = Tc ( r1 ) kcdTc = h [Tc ( r2 ) T ]dr r = r(5,6)(7,8)2Note that Eqs. (7) and (8) are obtained from surface energy balances at r1 and r2, respectively. ApplyingEq. (5) to Eq. (1), it follows that C1 = 0. Hence,Tf = qr 2+ C24k fFrom Eq. (6), it follows thatC ln rqr 2 1 + C2 = 3 1 + C44k fkc(9)(10)Continued...PROBLEM 3.88 (Cont.)Also, from Eq. (7),qr1C= 32r1qr 2C3 = 12or(11)CCFinally, from Eq. (8), 3 = h 3 ln r2 + C4 T or, substituting for C3 and solving for C4r2 kcC4 =2qr12qr1ln r2 + T+2r2 h 2k cSubstituting Eqs. (11) and (12) into (10), it follows thatC2 =2qr14k f2k c+2qr12r2 h+2qr12k cln r2 + Trqr 2ln 2 + 1 T4k f 2k c r1 2r2 hSubstituting Eq. (13) into (9),C2 =2qr12qr1 ln r1(12)+(2qr1)(13)rqr 2ln 2 + 1 + T4k f2k c r1 2r2 hSubstituting Eqs. (11) and (12) into (4),Tf =Tc =q2r1 r 2 +2qr1(14)(15)2qr1rqr 2ln 2 + 1 + T .2k cr 2r2 h<<(b) Applying Eq. (14) at r = 0, the maximum fuel temperature for h = 2000 W/m2K isTf ( 0 ) =2 108 W m3 ( 0.006 m )2 108 W m3 (0.006 m )24 2 W m K2+2 25 W m K2 108 W m3 ( 0.006 m )ln0.009 m0.006 m2+2 ( 0.09 m ) 2000 W m 2 K+ 300 KTf ( 0 ) = (900 + 58.4 + 200 + 300 ) K = 1458 K .(c) Temperature distributions for the prescribed values of h are as follows:6001300Temperature, Tc(K)Temperature, Tf(K)1500<110090070050030000.0010.0020.0040.005Radius in fuel element, r(m)h = 2000 W/m^2.Kh = 5000 W/m^2.Kh = 10000 W/m^2.K0.0065004003000.0060.0070.0080.009Radius in cladding, r(m)h = 2000 W/m^2.Kh = 5000 W/m^2.Kh = 10000 W/m^2.KContinued...PROBLEM 3.88 (Cont.)Clearly, the ability to control the maximum fuel temperature by increasing h is limited, and even for h , Tf(0) exceeds 1000 K. The overall temperature drop, Tf(0) - T, is influenced principally by the lowthermal conductivity of the fuel material.2COMMENTS: For the prescribed conditions, Eq. (14) yields, Tf(0) - Tf(r1) = qr1 4k f = (2108W/m3)(0.006 m)3/8 W/mK = 900 K, in which case, with no cladding and h , Tf(0) = 1200 K. Toreduce Tf(0) below 1000 K for the prescribed material, it is necessary to reduce q .PROBLEM 3.89KNOWN: Dimensions and properties of tubular heater and external insulation. Internal and externalconvection conditions. Maximum allowable tube temperature.FIND: (a) Maximum allowable heater current for adiabatic outer surface, (3) Effect of internalconvection coefficient on heater temperature distribution, (c) Extent of heat loss at outer surface.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Uniformheat generation, (4) Negligible radiation at outer surface, (5) Negligible contact resistance.ANALYSIS: (a) From Eqs. 7 and 10, respectively, of Example 3.7, we know that(q 2 r2 q 2 2r2 ln r2 r12kr1 4kTs,2 Ts,1 =andTs,1 = T ,1 +(22q r2 r1)(1))(2)2h1r1Hence, eliminating Ts,1, we obtainTs,2 T ,1 =)(2qr2 )(r2 1k222 2 ln r 2 1 r1 r2 + h r 1 r1 r2 2k 111Substituting the prescribed conditions (h1 = 100 W/m2K),()(Ts,2 T ,1 = 1.237 104 m3 K W q W m3)Hence, with Tmax corresponding to Ts,2, the maximum allowable value of q is1400 400q max == 8.084 106 W m341.237 10withq=I 2 Re=I2 e L Ace I2=2LA c r 2 r 21(22I max = r2 r1)1/ 2q e )((2= 0.035 0.0252)63 1/ 22 8.084 10 W mm= 6406 A6 0.7 10 m <Continued ..PROBLEM 3.89 (Cont.)(b) Using the one-dimensional, steady-state conduction model of IHT (hollow cylinder; convection atinner surface and adiabatic outer surface), the following temperature distributions were obtained.Temperature, T(K)1500130011009007005003000.0250.0270.0290.0310.0330.035Radius, r(m)h = 100 W/m^2.Kh = 500 W/m^2.Kh = 1000 W/m^2.KThe results are consistent with key implications of Eqs. (1) and (2), namely that the value of h1 has noeffect on the temperature drop across the tube (Ts,2 - Ts,1 = 30 K, irrespective of h1), while Ts,1 decreaseswith increasing h1. For h1 = 100, 500 and 1000 W/m2K, respectively, the ratio of the temperature dropbetween the inner surface and the air to the temperature drop across the tube, (Ts,1 - T,1)/(Ts,2 - Ts,1),decreases from 970/30 = 32.3 to 194/30 = 6.5 and 97/30 = 3.2. Because the outer surface is insulated, theheat rate to the airflow is fixed by the value of q and, irrespective of h1,()22q ( r1 ) = r2 r1 q = 15, 240 W<(c) Heat loss from the outer surface of the tube to the surroundings depends on the total thermalresistanceln ( r3 r2 )1R tot =+2 Lk i2 r3Lh 2or, for a unit area on surface 2,r2 ln ( r3 r2 )rR +2tot,2 = ( 2 r2 L ) R tot =kir3h 2Again using the capabilities of IHT (hollow cylinder; convection at inner surface and heat transfer fromouter surface through R ot,2 ), the following temperature distributions were determined for the tube andtinsulation.Insulation Temperature, T(K)Tube temperature, T(K)1200116011201080104010000.0251200110010009008007006005000.0270.0290.031Radius, r(m)delta =0.025 mdelta = 0.050 m0.0330.03500.20.40.60.81Dimensionless radius, (r-r2)/(r3-r2)r3 = 0.060 mr3 = 0.085 mContinued...PROBLEM 3.89 (Cont.)Heat losses through the insulation, q ( r2 ) , are 4250 and 3890 W/m for = 25 and 50 mm, respectively,with corresponding values of q ( r1 ) equal to -10,990 and -11,350 W/m. Comparing the tube temperaturedistributions with those predicted for an adiabatic outer surface, it is evident that the losses reduce tubewall temperatures predicted for the adiabatic surface and also shift the maximum temperature from r =0.035 m to r 0.033 m. Although the tube outer and insulation inner surface temperatures, Ts,2 = T(r2),increase with increasing insulation thickness, Fig. (c), the insulation outer surface temperature decreases.COMMENTS: If the intent is to maximize heat transfer to the airflow, heat losses to the ambient shouldbe reduced by selecting an insulation material with a significantly smaller thermal conductivity.PROBLEM 3.90KNOWN: Electric current I is passed through a pipe of resistance R to melt ice underesteady-state conditions.FIND: (a) Temperature distribution in the pipe wall, (b) Time to completely melt the ice.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)Constant properties, (4) Uniform heat generation in the pipe wall, (5) Outer surface of the pipeis adiabatic, (6) Inner surface is at a constant temperature, Tm.3PROPERTIES: Table A-3, Ice (273K): = 920 kg/m ; Handbook Chem. & Physics, Ice:5Latent heat of fusion, hsf = 3.3410 J/kg.ANALYSIS: (a) The appropriate form of the heat equation is Eq. 3.49, and the generalsolution, Eq. 3.51 isT (r ) = q2r + C1lnr+C24kwhereq=(I2R e22 r2 r1).Applying the boundary condition ( dT/dr )r = 0, it follows that20=qr2 C1+2k r22qr2HenceC1 =andqrqT ( r ) = r 2 + 2 lnr+C2 .4k2k2k2Continued ..PROBLEM 3.90 (Cont.)Applying the second boundary condition, T ( r1 ) = Tm , it follows that2q 2 qrTm = r1 + 2 lnr1 + C2 .4k2kSolving for C2 and substituting into the expression for T(r), findT ( r ) = Tm +2qr22kln)(rq22r r1 .r1 4k<(b) Conservation of energy dictates that the energy required to completely melt the ice, Em,must equal the energy which reaches the inner surface of the pipe by conduction through thewall during the melt period. Hence from Eq. 1.11bEst = Ein Eout + EgenEst = E m = t m qcond,r1or, for a unit length of pipe, dT 2 r1 h sf = t m k ( 2 r1 ) dr r1 () qr 2 qr 2 r1 h sf = 2 r1kt m 2 1 2kr1 2k ()())(222 r1 h sf = t m q r2 r1 .Dropping the minus sign, which simply results from the fact that conduction is in the negativer direction, it follows thattm =2 h sf r1(2 2q r2 r1)=2 h sf r1I2R e.With r1 = 0.05m, I = 100 A and R = 0.30 /m, it follows thate920kg/m3 3.34 105 J/kg (0.05m )2tm =or(100A )2 0.30 / mt m = 804s.<COMMENTS: The foregoing expression for tm could also be obtained by recognizing thatall of the energy which is generated by electrical heating in the pipe wall must be transferredto the ice. Hence,2I2 R t m = h sf r1 .ePROBLEM 3.91KNOWN: Materials, dimensions, properties and operating conditions of a gas-cooled nuclear reactor.FIND: (a) Inner and outer surface temperatures of fuel element, (b) Temperature distributions fordifferent heat generation rates and maximum allowable generation rate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,(4) Negligible contact resistance, (5) Negligible radiation.PROPERTIES: Table A.1, Thoriun: Tmp 2000 K; Table A.2, Graphite: Tmp 2300 K.ANALYSIS: (a) The outer surface temperature of the fuel, T2, may be determined from the rateequationT Tq = 2RtotwhereR =totln ( r3 r2 )2 k g+12 r3h=ln (14 11)2 (3 W m K )+(12 ( 0.014 m ) 2000 W m K2)= 0.0185 m K Wand the heat rate per unit length may be determined by applying an energy balance to a control surfaceabout the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that)(()22q = q r2 r1 = 108 W m3 0.0112 0.0082 m 2 = 17, 907 W mHence,T2 = qR tot + T = 17, 907 W m ( 0.0185 m K W ) + 600 K = 931K<With zero heat flux at the inner surface of the fuel element, Eq. C.14 yieldsT1 = T2 +2qr2 r 2 qr 2 r 1 1 1 ln 2 4k t r 2 2k t r1 2T1 = 931K +22108 W m3 ( 0.011m ) 0.008 2 108 W m3 ( 0.008 m ) 0.011 ln 1 4 57 W m K2 57 W m K 0.008 0.011 Continued...PROBLEM 3.91 (Cont.)<T1 = 931K + 25 K 18 K = 938 K(b) The temperature distributions may be obtained by using the IHT model for one-dimensional, steadystate conduction in a hollow tube. For the fuel element ( q > 0), an adiabatic surface condition isprescribed at r1, while heat transfer from the outer surface at r2 to the coolant is governed by the thermalresistance R ot,2 = 2 r2 R = 2(0.011 m)0.0185 mK/W = 0.00128 m2K/W. For the graphite ( q =ttot0), the value of T2 obtained from the foregoing solution is prescribed as an inner boundary condition at r2,while a convection condition is prescribed at the outer surface (r3). For 1 108 q 5 108 W/m3, thefollowing distributions are obtained.250021002100Temperature, T(K)Temperature, T(K)2500170013009005000.0080.0090.01Radial location in fuel, r(m)qdot = 5E8qdot = 3E8qdot = 1E80.011170013009005000.0110.0120.0130.014Radial location in graphite, r(m)qdot = 5E8qdot = 3E8qdot = 1E8The comparatively large value of kt yields small temperature variations across the fuel element,while the small value of kg results in large temperature variations across the graphite. Operationat q = 5 108 W/m3 is clearly unacceptable, since the melting points of thorium and graphite areexceeded and approached, respectively. To prevent softening of the materials, which would occur belowtheir melting points, the reactor should not be operated much above q = 3 108 W/m3.COMMENTS: A contact resistance at the thorium/graphite interface would increase temperatures in thefuel element, thereby reducing the maximum allowable value of q .PROBLEM 3.92KNOWN: Long rod experiencing uniform volumetric generation encapsulated by a circularsleeve exposed to convection.FIND: (a) Temperature at the interface between rod and sleeve and on the outer surface, (b)Temperature at center of rod.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction in rod and sleeve, (2) Steady-stateconditions, (3) Uniform volumetric generation in rod, (4) Negligible contact resistancebetween rod and sleeve.ANALYSIS: (a) Construct a thermal circuit for the sleeve,where22q=E = q D1 / 4 = 24, 000 W/m3 ( 0.20 m ) / 4 = 754.0 W/mgen Rs =ln ( r2 / r1 )2 k sR conv ==ln ( 400/200 )2 4 W/m K= 2.758 102 m K/W11== 3.183 102 m K/Wh D 2 25 W/m 2 K 0.400 mThe rate equation can be written asq=T1 TT T=2 R + RRsconvconv)(-22T1 = T + q ( R + R K/W m=71.8Csconv ) = 27 C+754 W/m 2.758 10 + 3.183 10-2T2 = T + qR conv = 27 C+754 W/m 3.183 10 m K/W=51.0 C.<<(b) The temperature at the center of the rod isT ( 0 ) = To =2qr14k r24, 000 W/m3 ( 0.100 m )2+ T1 =4 0.5 W/m K+ 71.8 C=192C.<COMMENTS: The thermal resistances due to conduction in the sleeve and convection arecomparable. Will increasing the sleeve outer diameter cause the surface temperature T2 toincrease or decrease?PROBLEM 3.93KNOWN: Radius, thermal conductivity, heat generation and convection conditionsassociated with a solid sphere.FIND: Temperature distribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)Constant properties, (4) Uniform heat generation.ANALYSIS: Integrating the appropriate form of the heat diffusion equation,d 2 dT kr dr + q=0r 2 dr 1r2dTqr3=+ C1dr3kT (r ) = ordTqr C= + 1dr3k r 2qr 2 C1+ C2 .6krThe boundary conditions are:kd 2 dT qr 2r= dr dr kdT =0dr r=0henceC1 = 0, anddT = h T ( ro ) T .dr roSubstituting into the second boundary condition (r = ro), find qr 2o2qroqro qro= h + C2 T ++ T .C2 =33h 6k 6kThe temperature distribution has the formq 2 2 qroT (r ) =ro r ++ T .6k3hCOMMENTS: To verify the above result, obtain T(ro) = Ts,qrTs = o + T3hApplying energy balance to the control volume about the sphere,()4 32q ro = h4 ro ( Ts T )3findTs =qro+ T .3h<PROBLEM 3.94KNOWN: Radial distribution of heat dissipation of a spherical container of radioactivewastes. Surface convection conditions.FIND: Radial temperature distribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties, (4) Negligible temperature drop across container wall.ANALYSIS: The appropriate form of the heat equation is1r2qod 2 dT qr= =dr dr kk 2 1 r . ro q r3 r5 dT + C1= o 2drk 3 5ro q r2r 4 C1+ C2 .T= o 2k 6 20ro rFrom the boundary conditions,r2HencedT/dr |r=0 = 0and kdT/dr |r=ro = h T ( ro ) T it follows that C1 = 0 and q r2 r2 rrqo o o = h o o o + C2 T k 6 20 3 5C2 =22roq o 7q o ro++ T .15h60k24 22ro q o qro 7 1 r 1r HenceT ( r ) = T ++ + .15hk 60 6 ro 20 ro COMMENTS: Applying the above result at ro yields<Ts = T ( ro ) = T + ( 2ro q o /15h ).The same result may be obtained by applying an energy balance to a control surface about thecontainer, where Eg = qconv . The maximum temperature exists at r = 0.PROBLEM 3.95KNOWN: Dimensions and thermal conductivity of a spherical container. Thermal conductivity andvolumetric energy generation within the container. Outer convection conditions.FIND: (a) Outer surface temperature, (b) Container inner surface temperature, (c) Temperaturedistribution within and center temperature of the wastes, (d) Feasibility of operating at twice the energygeneration rate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional radialconduction.ANALYSIS: (a) For a control volume which includes the container, conservation of energy yieldsE g E out = 0 , or qV q conv = 0 . Hence()(2q ( 4 3 ) ri3 = h4 ro Ts,o T5)3and with q = 10 W/m ,Ts,o = T +qri323hro= 25 C +105 W m 2 ( 0.5 m )33000 W m K ( 0.6 m )22<= 36.6 C .(b) Performing a surface energy balance at the outer surface, E in E out = 0 or q cond q conv = 0 .Hence4 k ss Ts,i Ts,o2= h4 ro Ts,o T(1 ri ) (1 ro )()()()21000 W m K ro Ts,i = Ts,o + r 1 ro ( Ts,o T ) = 36.6 C + 15 W m K ( 0.2 ) 0.6 m 11.6 C = 129.4 C .k ss ih<(c) The heat equation in spherical coordinates isd dT 2k rw r 2 + qr = 0 .dr dr Solving,r2dT=qr 3+ C1dr3k rwApplying the boundary conditions,dTand=0dr r = 0C1 = 0andandT (r ) = qr 2C 1 + C26k rwrT ( ri ) = Ts,iC2 = Ts,i + qri2 6k rw .Continued...PROBLEM 3.95 (Cont.)Hence(ri2 r2 )6k rwqT ( r ) = Ts,i +<At r = 0,T ( 0 ) = Ts,i +qri26k rw= 129.4 C +105 W m3 ( 0.5 m )26 ( 20 W m K )<= 337.7 C(d) The feasibility assessment may be performed by using the IHT model for one-dimensional, steadystate conduction in a solid sphere, with the surface boundary condition prescribed in terms of the totalthermal resistance()ri22R tot,i = 4 ri R tot = R cnd,i + R cnv,i =[(1 ri ) (1 ro )] + 1 ri 2k ssh ro where, for ro = 0.6 m and h = 1000 W/m2K, R nd,i = 5.56 10-3 m2K/W, R nv,i = 6.94 10-4 m2K/W,ccCenter temperature, T(0) (C)and R ot,i = 6.25 10-3 m2K/W. Results for the center temperature are shown below.t6756255755254750200040006000800010000Convection coefficient, h(W/m^2.K)ro = 0.54 mro = 0.60 mClearly, even with ro = 0.54 m = ro,min and h = 10,000 W/m2K (a practical upper limit), T(0) > 475C andthe desired condition can not be met. The corresponding resistances are R nd,i = 2.47 10-3 m2K/W,c-52-32R cnv,i = 8.57 10 m K/W, and R ot,i = 2.56 10 m K/W. The conduction resistance remainstdominant, and the effect of reducing R nv,i by increasing h is small. The proposed extension is notcfeasible.COMMENTS: A value of q = 1.79 105 W/m3 would allow for operation at T(0) = 475C with ro =20.54 m and h = 10,000 W/m K.PROBLEM 3.96KNOWN: Carton of apples, modeled as 80-mm diameter spheres, ventilated with air at 5C andexperiencing internal volumetric heat generation at a rate of 4000 J/kgday.FIND: (a) The apple center and surface temperatures when the convection coefficient is 7.5 W/m2K,and (b) Compute and plot the apple temperatures as a function of air velocity, V, for the range 0.1 V 1 m/s, when the convection coefficient has the form h = C1V0.425, where C1 = 10.1 W/m2K(m/s)0.425.SCHEMATIC:ASSUMPTIONS: (1) Apples can be modeled as spheres, (2) Each apple experiences flow ofventilation air at TW = 5C, (3) One-dimensional radial conduction, (4) Constant properties and (5)Uniform heat generation.ANALYSIS: (a) From Eq. C.24, the temperature distribution in a solid sphere (apple) with uniformgeneration isT(r) =2qro r2 1 + Ts6k r 2 o(1)To determine Ts, perform an energy balance on the apple as shown in the sketch above, with volume V =34 3 5ro , q cv + q = 0E in E out + E g = 0( )(Ts T ) + q (4 3 ro3 ) = 07.5 W m 2 K ( 4 0.0402 m 2 )(Ts 5 C ) + 38.9 W m3 ( 4 3 0.0403 m3 ) = 02 h 4 ro(2)where the volumetric generation rate isq = 4000 J kg dayq = 4000 J kg day 840 kg m3 (1day 24 hr ) (1hr 3600 s )q = 38.9 W m3and solving for Ts, find<Ts = 5.14 CFrom Eq. (1), at r = 0, with Ts, findT(0) =38.9 W m3 0.0402 m 26 0.5 W m K+ 5.14 C = 0.12 C + 5.14 C = 5.26 C<Continued...PROBLEM 3.96 (Cont.)(b) With the convection coefficient depending upon velocity,h = C1V 0.425with C1 = 10.1 W/m2K(m/s)0.425, and using the energy balance of Eq. (2), calculate and plot Ts as afunction of ventilation air velocity V. With very low velocities, the center temperature is nearly 0.5Chigher than the air. From our earlier calculation we know that T(0) - Ts = 0.12C and is independent ofV.Center temperature, T(0) (C)5.45.35.200.20.40.60.81Ventilation air velocity, V (m/s)COMMENTS: (1) While the temperature within the apple is nearly isothermal, the center temperaturewill track the ventilation air temperature which will increase as it passes through stacks of cartons.(2) The IHT Workspace used to determine Ts for the base condition and generate the above plot is shownbelow.// The temperature distribution, Eq (1),T_r = qdot * ro^2 / (4 * k) * ( 1- r^2/ro^2 ) + Ts// Energy balance on the apple, Eq (2)- qcv + qdot * Vol = 0Vol = 4 / 3 * pi * ro ^3// Convection rate equation:qcv = h* As * ( Ts - Tinf )As = 4 * pi * ro^2// Generation rate:qdot = qdotm * (1/24) * (1/3600) * rho// Assigned variables:ro = 0.080k = 0.5qdotm = 4000rho = 840r=0h = 7.5//h = C1 * V^0.425//C1 = 10.1//V = 0.5Tinf = 5// Generation rate, W/m^3; Conversions: days/h and h/sec// Radius of apple, m// Thermal conductivity, W/m.K// Generation rate, J/kg.K// Specific heat, J/kg.K// Center, m; location for T(0)// Convection coefficient, W/m^2.K; base case, V = 0.5 m/s// Correlation// Air velocity, m/s; range 0.1 to 1 m/s// Air temperature, CPROBLEM 3.97KNOWN: Plane wall, long cylinder and sphere, each with characteristic length a, thermalconductivity k and uniform volumetric energy generation rate q.FIND: (a) On the same graph, plot the dimensionless temperature, [ T ( x or r ) T ( a ) ]/[ q a /2k], vs.the dimensionless characteristic length, x/a or r/a, for each shape; (b) Which shape has the smallesttemperature difference between the center and the surface? Explain this behavior by comparing theratio of the volume-to-surface area; and (c) Which shape would be preferred for use as a nuclear fuelelement? Explain why?2SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties and (4) Uniform volumetric generation.ANALYSIS: (a) For each of the shapes, with T(a) = Ts, the dimensionless temperature distributionscan be written by inspection from results in Appendix C.3.2T ( x ) TsxPlane wall, Eq. C.22= 1 aqa 2 / 2kLong cylinder, Eq. C.23Sphere, Eq. C.2421 r 1 qa 2 / 2k 2 a T ( r ) Ts 1 r 2 = 1 qa 2 / 2k 3 a T ( r ) Ts=The dimensionless temperature distributions using the foregoing expressions are shown in the graphbelow.Dimensionless temperature distribution(T_x,r-Ts) / (qdot*a^2/2*k)10.80.60.40.2000.20.40.60.81Dimensionless length, x/a or r/aPlane wall, 2aLong cylinder, aSphere, aContinued ..PROBLEM 3.97 (Cont.)(b) The sphere shape has the smallest temperature difference between the center and surface, T(0) T(a). The ratio of volume-to-surface-area, /As, for each of the shapes isPlane wall a (1 1)==aAs(11)Long cylinder a 2 1 a==As 2 a 1 2Sphere 4 a 3 / 3 a==As34 a 2The smaller the /As ratio, the smaller the temperature difference, T(0) T(a).(c) The sphere would be the preferred element shape since, for a given /As ratio, which controls thegeneration and transfer rates, the sphere will operate at the lowest temperature.PROBLEM 3.98KNOWN: Radius, thickness, and incident flux for a radiation heat gauge.FIND: Expression relating incident flux to temperature difference between center and edge ofgauge.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligibletemperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5)Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contactresistance between foil and heat sink.ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr,q r + q ( 2 rdr ) = q r+dr ,iq r = k ( 2 rt )dT,drq r+dr = q r +dq rdr.drRearranging, find thatq ( 2 rdr ) =iddT ( k2 rt ) dr drdr d dT qi r dr = kt r.dr Integrating,rdTqr 2= i + C1dr2ktandqr 2T ( r ) = i + C1lnr+C2 .4ktWith dT/dr|r=0 =0, C1 = 0 and with T(r = R) = T(R),qR 2T (R ) = i+ C24ktorqR 2C2 = T ( R ) + i.4ktHence, the temperature distribution isT (r ) =()qiR 2 r 2 + T ( R ).4ktApplying this result at r = 0, it follows thatq =i4kt4kt T ( 0 ) T ( R ) =T.R2R2COMMENTS: This technique allows for determination of a radiation flux frommeasurement of a temperature difference. It becomes inaccurate if emission from the foilbecomes significant.<PROBLEM 3.98KNOWN: Radius, thickness, and incident flux for a radiation heat gauge.FIND: Expression relating incident flux to temperature difference between center and edge ofgauge.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligibletemperature drop across foil thickness), (3) Constant properties, (4) Uniform incident flux, (5)Negligible heat loss from foil due to radiation exchange with enclosure wall, (6) Negligible contactresistance between foil and heat sink.ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr,q r + q ( 2 rdr ) = q r+dr ,iq r = k ( 2 rt )dT,drq r+dr = q r +dq rdr.drRearranging, find thatq ( 2 rdr ) =iddT ( k2 rt ) dr drdr d dT qi r dr = kt r.dr Integrating,rdTqr 2= i + C1dr2ktandqr 2T ( r ) = i + C1lnr+C2 .4ktWith dT/dr|r=0 =0, C1 = 0 and with T(r = R) = T(R),qR 2T (R ) = i+ C24ktorqR 2C2 = T ( R ) + i.4ktHence, the temperature distribution isT (r ) =()qiR 2 r 2 + T ( R ).4ktApplying this result at r = 0, it follows thatq =i4kt4kt T ( 0 ) T ( R ) =T.R2R2COMMENTS: This technique allows for determination of a radiation flux frommeasurement of a temperature difference. It becomes inaccurate if emission from the foilbecomes significant.<PROBLEM 3.99KNOWN: Net radiative flux to absorber plate.FIND: (a) Maximum absorber plate temperature, (b) Rate of energy collected per tube.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (x) conduction alongabsorber plate, (3) Uniform radiation absorption at plate surface, (4) Negligible losses byconduction through insulation, (5) Negligible losses by convection at absorber plate surface,(6) Temperature of absorber plate at x = 0 is approximately that of the water.PROPERTIES: Table A-1, Aluminum alloy (2024-T6): k 180 W/mK.ANALYSIS: The absorber plate acts as an extended surface (a conduction-radiation system),and a differential equation which governs its temperature distribution may be obtained byapplying Eq.1.11a to a differential control volume. For a unit length of tubeqx + q ( dx ) qx+dx = 0.radWithqx+dx = qx +andqx = ktdqxdxdxdTdxit follows that,q radd 2Tdx 2ddT kt dx = 0dx q+ rad = 0ktIntegrating twice it follows that, the general solution for the temperature distribution has theform,qT ( x ) = rad x 2 + C1x+C2 .2ktContinued ..PROBLEM 3.99 (Cont.)The boundary conditions are:T ( 0 ) = TwdT =0dx x=L/2C2 = Twq LC1 = rad2ktHence,qT ( x ) = rad x ( L x ) + Tw .2ktThe maximum absorber plate temperature, which is at x = L/2, is thereforeq L2Tmax = T ( L/2 ) = rad + Tw .8ktThe rate of energy collection per tube may be obtained by applying Fouriers law at x = 0.That is, energy is transferred to the tubes via conduction through the absorber plate. Hence,dT q=2 k tdx x=0 where the factor of two arises due to heat transfer from both sides of the tube. Hence,q= Lq .rad800W(0.2m )22m+ 60 CW8 180(0.006m )mKHenceTmax =orTmax = 63.7 Candq = 0.2m 800 W/m 2orq = 160 W/m.<<COMMENTS: Convection losses in the typical flat plate collector, which is not evacuated,would reduce the value of q .PROBLEM 3.100KNOWN: Surface conditions and thickness of a solar collector absorber plate. Temperature ofworking fluid.FIND: (a) Differential equation which governs plate temperature distribution, (b) Form of thetemperature distribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Adiabaticbottom surface, (4) Uniform radiation flux and convection coefficient at top, (5) Temperature ofabsorber plate at x = 0 corresponds to that of working fluid.ANALYSIS: (a) Performing an energy balance on the differential control volume,qx + dq = qx+dx + dqradconvwhereqx+dx = qx + ( dqx / dx ) dxdq = q dxradraddq= h ( T T ) dxconvHence,q dx= ( dq / dx ) dx+h ( T T ) dx.radxFrom Fouriers law, the conduction heat rate per unit width isqd 2T hqx = k t dT/dx(T T ) + rad = 0.2dxkT<kt(b) Defining = T T , d 2T/dx 2 = d 2 / dx 2 and the differential equation becomes,d 2dx 2qh + rad = 0.ktktIt is a second-order, differential equation with constant coefficients and a source term, and its generalsolution is of the form = C1e+ x + C2e- x + S/ 21/ 2 = ( h/kt ) ,S=q / kt.whereradAppropriate boundary conditions are: (0 ) = To T 0 ,d /dx) x=L = 0. o = C1 + C2 + S/ 2Hence,d /dx) x=L = C1 e+ L C2 e- L = 0)((C1 = 0 S/ 2 / 1 + e2 LHence,( = 0 S/ 2))C2 = C1 e2 L()(C2 = 0 S/ 2 / 1 + e-2 L e xe - x + + S/ 2 .2 L 1+e-2 L 1+e)<PROBLEM 3.101KNOWN: Dimensions of a plate insulated on its bottom and thermally joined to heat sinks at itsends. Net heat flux at top surface.FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperaturedistribution and heat loss to heat sinks.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in x (W,L>>t), (3) Constantproperties, (4) Uniform surface heat flux, (5) Adiabatic bottom, (6) Negligible contact resistance.ANALYSIS: (a) Applying conservation of energy to the differential control volume, qx + dq= qx +dx, where qx+dx = qx + (dqx/dx) dx and dq=q ( W dx ). Hence, ( dq x / dx ) q W=0.ooFrom Fouriers law, q x = k ( t W ) dT/dx. Hence, the differential equation for thetemperature distribution isq+ o = 0.dx 2 ktd 2TddT o ktW dx q W=0dx <(b) Integrating twice, the general solution is,qT ( x ) = o x 2 + C1 x +C22ktand appropriate boundary conditions are T(0) = To, and T(L) = To. Hence, To = C2, andqTo = o L2 + C1L+C22ktandq LC1 = o .2ktHence, the temperature distribution is()q LT ( x ) = o x 2 Lx + To .2kt<Applying Fouriers law at x = 0, and at x = L,q WLL q q ( 0 ) = k ( Wt ) dT/dx) x=0 = kWt o x = o2 x=02 kt q WLL q q ( L ) = k ( Wt ) dT/dx) x=L = kWt o x =+ o2 x=L2 kt Hence the heat loss from the plates is q=2 ( q WL/2 ) = q WL.oo<COMMENTS: (1) Note signs associated with q(0) and q(L). (2) Note symmetry about x =L/2. Alternative boundary conditions are T(0) = To and dT/dx)x=L/2=0.PROBLEM 3.102KNOWN: Dimensions and surface conditions of a plate thermally joined at its ends to heat sinks atdifferent temperatures.FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperaturedistribution and an expression for the heat rate from the plate to the sinks, and (c) Compute and plottemperature distribution and heat rates corresponding to changes in different parameters.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x (W,L >> t), (3)Constant properties, (4) Uniform surface heat flux and convection coefficient, (5) Negligible contactresistance.ANALYSIS: (a) Applying conservation of energy to the differential control volumeq x + dq o = q x + dx + dq convwhereq x + dx = q x + ( dq x dx ) dxdq conv = h ( T T )( W dx )Hence,dq xq x + q ( W dx ) = q x + ( dq x dx ) dx + h ( T T )( W dx )odxUsing Fouriers law, q x = k ( t W ) dT dx , ktWd2T+ hW ( T T ) = qod 2Thdx 2dx 2 kt(b) Introducing T T , the differential equation becomes(T T ) +qokt+ hW ( T T ) = q W .o<= 0.q + o =0.ktdx 2 ktThis differential equation is of second order with constant coefficients and a source term. Withd 2h 2 h kt and S q kt , it follows that the general solution is of the formo = C1e + x + C2 e x + S 2 .Appropriate boundary conditions are: (0) = To T o(1) (L) = TL T L(2,3)Substituting the boundary conditions, Eqs. (2,3) into the general solution, Eq. (1), o = C1e0 + C2e0 + S 2 L = C1e + L + C2e L + S 2To solve for C2, multiply Eq. (4) by -e+L and add the result to Eq. (5),()()C2 = ( L o e + L ) S 2 ( e + L + 1) ( e+ L + e L )(4,5) o e + L + L = C2 e + L + e L + S 2 e + L + 1(6)Continued...PROBLEM 3.102 (Cont.)Substituting for C2 from Eq. (6) into Eq. (4), find{()) (e+L + eL )} S 2(C1 = o L o e+ L S 2 e + L + 1 (7)Using C1 and C2 from Eqs. (6,7) and Eq. (1), the temperature distribution can be expressed as (x) = e+ x()(sinh ( x ) + L sinh ( x )+ L sinh ( x )+ Le+ 1 e o + sinh ( L ) L + 1 esinh ( L )sinh ( L )) S (8)2<The heat rate from the plate is q p = q x ( 0 ) + q x ( L ) and using Fouriers law, the conduction heat rates,with Ac = Wt, areq x ( 0 ) = kAcd e L o +L= kA c e0 dx x = 0sinh ( L ) sinh ( L ) 1 e+ L+ sinh ( L )q x ( L ) = kAcS 2 < cosh ( L )d e L cosh ( L ) o +L= kAc e L dx x = Lsinh ( L )sinh ( L ) 1 e+ L+ sinh ( L )S 2 cosh ( L ) e + L <(c) For the prescribed base-case conditions listed below, the temperature distribution (solid line) is shownin the accompanying plot. As expected, the maximum temperature does not occur at the midpoint, butslightly toward the x-origin. The sink heat rates areq ( 0 ) = 17.22 Wxq ( L ) = 23.62 Wx<Temperature, T(x) (C)3002001000020406080100Distance, x (mm)q''o = 20,000 W/m^2; h = 50 W/m^2.Kq''o = 30,000 W/m^2; h = 50 W/m^2.Kq''o = 20,000 W/m^2; h = 200 W/m^2.Kq''o = 4927 W/m^2 with q''x(0) = 0; h = 200 W/m^2.KThe additional temperature distributions on the plot correspond to changes in the following parameters,with all the remaining parameters unchanged: (i) q = 30,000 W/m2, (ii) h = 200 W/m2K, (iii) the valueoof q for which q (0) = 0 with h = 200 W/m2K. The condition for the last curve is q = 4927 W/m2xoofor which the temperature gradient at x = 0 is zero.Base case conditions are: q = 20,000 W/m2, To = 100C, TL = 35C, T = 25C, k = 25 W/mK, h = 50oW/m2K, L = 100 mm, t = 5 mm, W = 30 mm.PROBLEM 3.103KNOWN: Thin plastic film being bonded to a metal strip by laser heating method; strip dimensions andthermophysical properties are prescribed as are laser heating flux and convection conditions.FIND: (a) Expression for temperature distribution for the region with the plastic strip, -w1/2 x w1/2,(b) Temperature at the center (x = 0) and the edge of the plastic strip (x = w1/2) when the laser flux is10,000 W/m2; (c) Plot the temperature distribution for the strip and point out special features.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction only, (3)Plastic film has negligible thermal resistance, (4) Upper and lower surfaces have uniform convectioncoefficients, (5) Edges of metal strip are at air temperature (T), that is, strip behaves as infinite fin sothat w2 , (6) All the incident laser heating flux q is absorbed by the film.oPROPERTIES: Metal strip (given): = 7850 kg/m3, cp = 435 J/kgm3, k = 60 W/mK.ANALYSIS: (a) The strip-plastic film arrangement can be modeled as an infinite fin of uniform crosssection a portion of which is exposed to the laser heat flux on the upper surface. The general solutionsfor the two regions of the strip, in terms of T ( x ) T , are1 ( x ) = C1e0 x w1 2+ mx+ C 2e mx+M m 2 ( x ) = C3 e+ mx+ C4 e(1)m = ( 2h kd )M = q P 2kA c = q kdoow1 2 x 21/ 2 mx.Four boundary conditions can be identified to evaluate the constants:d1At x = 0:(0 ) = 0 = C1me0 C 2 me0 + 0dx ( w1 2 ) = 2 ( w1 2 )At x = w1/2:C1e+ m w1 2+ C 2e m w1 22+ mw1 2d1 ( w1 2 ) / dx = d 2 ( w1 2 ) / dxAt x = w1/2:mC1e+ m w1 2 mC 2 e m w1 2+ 0 = mC3e(4)C1 = C 2+ M m = C 3e+ m w1 2+ C 4eMm2e(5) m w1 2 mC 4 eAt x : 2 ( ) = 0 = C3 e + C 4 e C3 = 0With C3 = 0 and C1 = C2, combine Eqs. (6 and 7) to eliminate C4 to findC1 = C 2 = (2,3) mw1 2(6)(7)(8)2m w1 2.(9)and using Eq. (6) with Eq. (9) findC 4 = M m sinh ( mw1 2 ) e2 mx1 / 2(10)Continued...PROBLEM 3.103 (Cont.)Hence, the temperature distribution in the region (1) under the plastic film, 0 x w1/2, is2Mm1 ( x ) = 2em w1 w(e+ mx+e mxMM) + m = m (1 e2 m w1 22cosh mx)(11)<and for the region (2), x w1/2,2 (x ) =Mm2sinh ( mw1 2 ) e mx(12)(b) Substituting numerical values into the temperature distribution expression above, 1(0) and 1(w1/2)can be determined. First evaluate the following parameters:M = 10, 000 W m(260 W m K 0.00125 m = 133, 333 K m2m = 2 10 W m K 60 W m K 0.00125 m)1/ 22= 16.33 m1Hence, for the midpoint x = 0,1 ( 0 ) =133, 333 K m(16.33 m)21 2()1 exp 16.33 m 1 0.020 m cosh ( 0 ) = 139.3 K<T1 ( 0 ) = 1 ( 0 ) + T = 139.3 K + 25 C = 164.3 C .For the position x = w1/2 = 0.020 m,(1 ( w1 2 ) = 500.0 1 0.721cosh 16.33 m1 0.020 m) = 120.1K<T1 ( w1 2 ) = 120.1K + 25 C = 145.1 C .(c) The temperature distributions, 1(x) and 2(x), are shown in the plot below. Using IHT, Eqs. (11) and(12) were entered into the workspace and a graph created. The special features are noted:(1) No gradient at midpoint, x = 0; symmetricaldistribution.(3) Temperature excess and gradient approachzero with increasing value of x.Strip temperature, T (C)(2) No discontinuity of gradient at w1/2(20 mm).1801401006020050100150200250300x-coordinate, x (mm)Region 1 - constant heat flux, q''oRegion 2 - x >= w1/2COMMENTS: How wide must the strip be in order to satisfy the infinite fin approximation such that 2(x ) = 0? For x = 200 mm, find 2(200 mm) = 6.3C; this would be a poor approximation. When x= 300 mm, 2(300 mm) = 1.2C; hence when w2/2 = 300 mm, the strip is a reasonable approximation toan infinite fin.PROBLEM 3.104KNOWN: Thermal conductivity, diameter and length of a wire which is annealed by passing an electricalcurrent through the wire.FIND: Steady-state temperature distribution along wire.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction along the wire, (3) Constantproperties, (4) Negligible radiation, (5) Uniform convection coefficient h.ANALYSIS: Applying conservation of energy to a differential control volume,q x + Eg dq conv q x+dx = 0q x+dx = q x +dq xdxdx()Eg = q ( D2 / 4 ) dx.q x = k D2 / 4 dT/dxdq conv = h ( D dx ) ( T T )Hence,(k D2 / 4) dx 2 dx+q ( D2 / 4)dx h ( Ddx ) (T T ) = 0d 2Td 2or, with T T ,dx 24hq + =0kDkThe solution (general and particular) to this nonhomogeneous equation is of the formq = C1 emx + C2 e-mx +km 22where m = (4h/kD). The boundary conditions are:d = 0 = m C1 e0 mC2 e0 dx x=0()q ( L ) = 0 = C1 emL + e-mL +km 2C1 = C2C1 =q/km 2emL + e-mL= C2The temperature distribution has the formT = T emx + e-mx q cosh mx 1 = T 1 .km 2 emL +e-mL km 2 cosh mL q<COMMENTS: This process is commonly used to anneal wire and spring products. To checkthe result, note that T(L) = T(-L) = T.PROBLEM 3.105KNOWN: Electric power input and mechanical power output of a motor. Dimensions of housing, mounting padand connecting shaft needed for heat transfer calculations. Temperature of ambient air, tip of shaft, and base ofpad.FIND: Housing temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in pad and shaft, (3) Constantproperties, (4) Negligible radiation.ANALYSIS: Conservation of energy yieldsPelec Pmech q h q p q s = 0q h = h h A h ( Th T ) , L = 0,qs =( 2 / 4 D3h kss(tcosh mL L / bsinh mL1/ 21/ 2mL = 4h s L2 / k s D,Henceqs = M2 3M= D hsks 4)((Th T ) ,qp = k p W 2)1/ 2(Th T ).(Th T )tanh 4h s L2 / k s D)1/ 2Substituting, and solving for (Th - T),Th T =(()()1/ 2(4hsL2 / ksD) = 3.87,h h A h + k p W 2 / t+ 2 / 4 D3h s k s(( / 4) D h k )2Pelec Pmech31/ 2ss= 6.08 W/K,1/ 2/ tanh 4h s L2 / k s D)1/ 2tanhmL=0.999104 W( 25 15 ) 103 WTh T ==10 2+0.5 ( 0.7 )2 / 0.05 + 6.08 / 0.999 W/K ( 20+4.90+6.15 ) W/KTh T = 322.1KTh = 347.1 C<COMMENTS: (1) Th is large enough to provide significant heat loss by radiation from the()44housing. Assuming an emissivity of 0.8 and surroundings at 25C, q rad = A h Th Tsur = 4347W, which compares with q conv = hA h ( Th T ) = 5390 W. Radiation has the effect ofdecreasing Th. (2) The infinite fin approximation, qs = M, is excellent.PROBLEM 3.106KNOWN: Dimensions and thermal conductivity of pipe and flange. Inner surface temperature ofpipe. Ambient temperature and convection coefficient.FIND: Heat loss through flange.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction in pipe and flange, (3)Constant thermal conductivity, (4) Negligible radiation exchange with surroundings.ANALYSIS: From the thermal circuit, the heat loss through the flanges isq=Ts,i TR t,w + R t,f=Ts,i Tn ( Do / Di ) / 4 tk + (1/ hAf f )Since convection heat transfer only occurs from one surface of a flange, the connected flanges may bemodeled as a single annular fin of thickness t = 2t = 30 mm. Hence, r2c = ( Df / 2 ) + t / 2 = 0.140 m,()()()222A f = 2 r2c r1 = 2 r2c Do / 2 = 2 0.1402 0.062 m 2 = 0.101m 2 , Lc = L + t / 2 =( Df()1/ 22/22h / kA p= 0.188. With r2c/r1 = Do ) / 2 + t = 0.065 m, A p = Lc t = 0.00195 m , Lcr2c/(Do/2) = 1.87, Fig. 3.19 yields f = 0.94. Hence,q=q=300C 20C(n (1.25 ) / 4 0.03m 40 W / m K + 1/10 W / m 2 K 0.101m 2 0.94280C= 262 W(0.0148 + 1.053) K / W)<COMMENTS: Without the flange, heat transfer from a section of pipe of width t = 2t is1q = Ts,i T / R t,w + R t,cnv , where R t,cnv = ( h Do t ) = 7.07 K / W. Hence,q = 39.5 W, and there is significant heat transfer enhancement associated with the extended surfacesafforded by the flanges.()()PROBLEM 3.107KNOWN: TC wire leads attached to the upper and lower surfaces of a cylindrically shaped solderbead. Base of bead attached to cylinder head operating at 350C. Constriction resistance at base andTC wire convection conditions specified.FIND: (a) Thermal circuit that can be used to determine the temperature difference between the twointermediate metal TC junctions, (T1 T2); label temperatures, thermal resistances and heat rates; and(b) Evaluate (T1 T2) for the prescribed conditions. Comment on assumptions made in building themodel.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in solder bead; nolosses from lateral and top surfaces; (3) TC wires behave as infinite fins, (4) Negligible thermalcontact resistance between TC wire terminals and bead.ANALYSIS: (a) The thermal circuit is shown above. Note labels for the temperatures, thermalresistances and the relevant heat fluxes. The thermal resistances are as follows:Constriction (con) resistance, see Table 4.1, case 10R con = 1/ ( 2k bead Dsol ) = 1/ ( 2 40 W / m K 0.006 m ) = 2.08 K / WTC (tc) wires, infinitely long fins; Eq. 3.800.5R tc,1 = R tc,2 = R fin = ( hPk w Ac )(P = D w , A c = D2 / 4wR tc = 100 W / m 2 K 2 ( 0.003 m ) 70 W / m K / 43)0.5= 46.31 K / WSolder bead (sol), cylinder Dsol and LsolR sol = Lsol / ( k sol Asol )(2Asol = Dsol / 4)R sol = 0.010 m / 10 W / m K (0.006 m ) / 4 = 35.37 K / W2(b) Perform energy balances on the 1- and 2-nodes, solve the equations simultaneously to find T1 andT2, from which (T1 T2) can be determined.Continued ..PROBLEM 3.107 (Cont.)Node 1T2 T1 Thead T1 T T1++=0R solR conR tc,1Node 2T T2 T1 T2+=0R tc,2R solSubstituting numerical values with the equations in the IHT Workspace, findT1 = 359CT2 = 199.2CT1 T2 = 160CCOMMENTS: (1) With this arrangement, the TC indicates a systematically low reading of thecylinder head. The size of the solder bead (Lsol) needs to be reduced substantially.(2) The model neglects heat losses from the top and lateral sides of the solder bead, the effect ofwhich would be to increase our estimate for (T1 T2). Constriction resistance is important; note thatThead T1 = 26C.PROBLEM 3.108KNOWN: Rod (D, k, 2L) that is perfectly insulated over the portion of its length L x 0 andexperiences convection (T, h) over the portion 0 x + L. One end is maintained at T1 and theother is separated from a heat sink at T3 with an interfacial thermal contact resistance R c .tFIND: (a) Sketch the temperature distribution T vs. x and identify key features; assume T1 > T3 >T2; (b) Derive an expression for the mid-point temperature T2 in terms of thermal and geometricparameters of the system, (c) Using, numerical values, calculate T2 and plot the temperaturedistribution. Describe key features and compare to your sketch of part (a).SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in rod for L x 0,(3) Rod behaves as one-dimensional extended surface for 0 x +L, (4) Constant properties.ANALYSIS: (a) The sketch for the temperature distribution is shown below. Over the insulatedportion of the rod, the temperature distribution is linear. A temperature drop occurs across thethermal contact resistance at x = +L. The distribution over the exposed portion of the rod is nonlinear. The minimum temperature of the system could occur in this portion of the rod.(b) To derive an expression for T2, begin with the general solution from the conduction analysis for afin of uniform cross-sectional area, Eq. 3.66. ( x ) = C1emx + C2e mx0 x +L(1)where m = (hP/kAc)conditions.1/2and = T(x) - T. The arbitrary constants are determined from the boundaryAt x = 0, thermal resistance of rodq x ( 0 ) = kAc (0 )d = kAc 1dx x = 0Lm C1e0 m C 2e0 =(1 = T1 T)1 C1e0 + C2e0 1L(2)Continued ..PROBLEM 3.108 (Cont.)At x=L, thermal contact resistanceq x ( + L ) = kAc (L ) 3d =dx x = L R / A ctc 3 = T3 T1 mL k m C1emL m C2e mL =+ C 2e mL 3 R C1etc (3)Eqs. (2) and (3) cannot be rearranged easily to find explicit forms for C1 and C2. The constraints willbe evaluated numerically in part (c). Knowing C1 and C2, Eq. (1) gives 2 = (0 ) = T2 T = C1 e0 + C2e0(4)(c) With Eqs. (1-4) in the IHT Workspace using numerical values shown in the schematic, find T2 =62.1C. The temperature distribution is shown in the graph below.Temperature distribution in rodTemperature, T(x) (C)200150100500-50-30-10103050x-coordinate, x (mm)COMMENTS: (1) The purpose of asking you to sketch the temperature distribution in part (a) wasto give you the opportunity to identify the relevant thermal processes and come to an understanding ofthe system behavior.(2) Sketch the temperature distributions for the following conditions and explain their key features:(a) R c = 0, (b) R c , and (c) the exposed portion of the rod behaves as an infinitely long fin;ttthat is, k is very large.PROBLEM 3.109KNOWN: Long rod in oven with air temperature at 400C has one end firmly pressedagainst surface of a billet; thermocouples imbedded in rod at locations 25 and 120 mm fromthe billet indicate 325 and 375C, respectively.FIND: The temperature of the billet, Tb.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Rod is infinitely long with uniform crosssectional area, (3) Uniform convection coefficient along rod.ANALYSIS: For an infinitely long rod of uniform cross-sectional area, the temperaturedistribution is ( x ) = be-mx(1)where ( x ) = T ( x ) T b = T (0 ) T = Tb T .Substituting values for T1 and T2 at their respective distances, x1 and x2, into Eq. (1), it ispossible to evaluate m, ( x1 ) be-mx1-m x x== e ( 1 2)-mx 2 (x2 ) eb(325 400 ) C = e-m(0.0250.120 )m(375-400 ) Cm=11.56.Using the value for m with Eq. (1) at location x1, it is now possible to determine the rod baseor billet temperature, ( x1 ) = T1 T = (Tb T ) e-mx(325 400 ) C= (Tb 400 ) C e11.560.025Tb = 300 C.COMMENTS: Using the criteria mL 2.65 (see Example 3.8) for the infinite finapproximation, the insertion length should be 229 mm to justify the approximation,<PROBLEM 3.110KNOWN: Temperature sensing probe of thermal conductivity k, length L and diameter D is mountedon a duct wall; portion of probe Li is exposed to water stream at T,i while other end is exposed toambient air at T,o ; convection coefficients hi and ho are prescribed.FIND: (a) Expression for the measurement error, Terr = Ttip T,i , (b) For prescribed T,i andT,o , calculate Terr for immersion to total length ratios of 0.225, 0.425, and 0.625, (c) Computeand plot the effects of probe thermal conductivity and water velocity (hi) on Terr .SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in probe, (3) Probe isthermally isolated from the duct, (4) Convection coefficients are uniform over their respectiveregions.PROPERTIES: Probe material (given): k = 177 W/mK.ANALYSIS: (a) To derive an expression forTerr = Ttip - T,i , we need to determine thetemperature distribution in the immersedlength of the probe Ti(x). Consider the probeto consist of two regions: 0 xi Li, theimmersed portion, and 0 xo (L - Li), theambient-air portion where the origincorresponds to the location of the duct wall.Use the results for the temperature distributionand fin heat rate of Case A, Table 3.4:Temperature distribution in region i:Ti ( x i ) T,i cosh ( mi ( Li xi )) + ( hi mi k ) sinh ( Li xi )i== b,iTo T,icosh ( mi Li ) + ( h i mi k ) sinh ( mi Li )(1)and the tip temperature, Ttip = Ti(Li) at xi = Li, isTtip T,iTo T,i=A=cosh (0 ) + ( hi mi k ) sinh ( 0 )(2)cosh ( mi Li ) + ( h i mi k ) sinh ( mi Li )and hence(Terr = Ttip T,i = A To T,i)(3)<where To is the temperature at xi = xo = 0 which at present is unknown, but can be found by setting thefin heat rates equal, that is,q f ,o q f ,i(4)Continued...PROBLEM 3.110 (Cont.)( h o PkAc )1/ 2 b,o B = ( hi PkAc )1/ 2 b,i CSolving for To, find b,o b,i=To T,oTo T,i= ( h i PkAc )1/ 21/ 2 hi CTo = T,o + T,i B ho b,i C1/ 2 hi C1 + B ho (5)where the constants B and C are,B=sinh ( mo Lo ) + ( h o mo k ) cosh ( mo Lo )cosh ( mo Lo ) + ( h o mo k ) sinh ( mo Lo )(6)C=sinh ( mi Li ) + ( h i mi k ) cosh ( mi Li )cosh ( mi Li ) + ( h i mi k ) sinh ( mi Li )(7)(b) To calculate the immersion error for prescribed immersion lengths, Li/L = 0.225, 0.425 and 0.625,we use Eq. (3) as well as Eqs. (2, 6, 7 and 5) for A, B, C, and To, respectively. Results of thesecalculations are summarized below.Li/L0.225Lo (mm)155Li (mm)45A0.2328B0.5865C0.9731To (C)76.7Terr (C)-0.760.425115850.03960.46390.99277.5-0.100.625751250.00670.32050.999978.2-0.012.5Temperature error, Tinfo - Ttip (C)(c) The probe behaves as a fin havingends exposed to the cool ambient airand the hot ambient water whosetemperature is to be measured. If thethermal conductivity is decreased, heattransfer along the probe length islikewise decreased, the tip temperaturewill be closer to the water temperature.If the velocity of the water decreases,the convection coefficient willdecrease, and the difference between thetip and water temperatures will increase.<<<21.510.500.20.30.40.50.6Immersion length ratio, Li/LBase case: k = 177 W/m.K; ho = 1100 W/m^2.KLow velocity flow: k = 177 W/m.K; ho = 500 W/m^2.KLow conductivity probe: k = 50 W/m.K; ho = 1100 W/m^2.K0.7PROBLEM 3.111KNOWN: Rod protruding normally from a furnace wall covered with insulation of thickness Linswith the length Lo exposed to convection with ambient air.FIND: (a) An expression for the exposed surface temperature To as a function of the prescribedthermal and geometrical parameters. (b) Will a rod of Lo = 100 mm meet the specified operatinglimit, T0 100C? If not, what design parameters would you change?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in rod, (3) Negligiblethermal contact resistance between the rod and hot furnace wall, (4) Insulated section of rod, Lins,experiences no lateral heat losses, (5) Convection coefficient uniform over the exposed portion of therod, Lo, (6) Adiabatic tip condition for the rod and (7) Negligible radiation exchange between rod andits surroundings.ANALYSIS: (a) The rod can be modeled as a thermal network comprised of two resistances inseries: the portion of the rod, Lins, covered by insulation, Rins, and the portion of the rod, Lo,experiencing convection, and behaving as a fin with an adiabatic tip condition, Rfin. For the insulatedsection:R ins = Lins kA c(1)For the fin, Table 3.4, Case B, Eq. 3.76,1R fin = b q f =( hPkAc )1/ 2 tanh ( mLo )(2)m = ( hP kA c )Ac = D2 4P = DFrom the thermal network, by inspection,To TT TR finTo = T +=w(Tw T )R finR ins + R finR ins + R fin(b) Substituting numerical values into Eqs. (1) - (6) with Lo = 200 mm,6.298To = 25 C +( 200 25 ) C = 109 C6.790 + 6.2981/ 2R ins =0.200 m60 W m K 4.909 10R fin = 1(( hPkA c ) =4m20.0347 W 2 K 2(15 W m2= 6.790 K W)1/ 2A c = ( 0.025 m )2(3,4,5)(6)<<4 = 4.909 104m2tanh ( 6.324 0.200 ) = 6.298 K W K ( 0.025 m ) 60 W m K 4.909 104m2) = 0.0347 W2K2Continued...PROBLEM 3.111 (Cont.)m = ( hP kA c )1/ 2(= 15 W m K ( 0.025 m ) 60 W m K 4.909 1024m2)1/ 2= 6.324 m1Consider the following design changes aimed at reducing To 100C. (1) Increasing length of the finportions: with Lo = 200 mm, the fin already behaves as an infinitely long fin. Hence, increasing Lowill not result in reducing To. (2) Decreasing the thermal conductivity: backsolving the aboveequation set with T0 = 100C, find the required thermal conductivity is k = 14 W/mK. Hence, wecould select a stainless steel alloy; see Table A.1. (3) Increasing the insulation thickness: find thatfor To = 100C, the required insulation thickness would be Lins = 211 mm. This design solution mightbe physically and economically unattractive. (4) A very practical solution would be to introducethermal contact resistance between the rod base and the furnace wall by tack welding (rather than acontinuous bead around the rod circumference) the rod in two or three places. (5) A less practicalsolution would be to increase the convection coefficient, since to do so, would require an air handlingunit.COMMENTS: (1) Would replacing the rod by a thick-walled tube provide a practical solution?(2) The IHT Thermal Resistance Network Model and the Thermal Resistance Tool for a fin with anadiabatic tip were used to create a model of the rod. The Workspace is shown below.// Thermal Resistance Network Model:// The Network:// Heat rates into node j,qij, through thermal resistance Rijq21 = (T2 - T1) / R21q32 = (T3 - T2) / R32// Nodal energy balancesq1 + q21 = 0q2 - q21 + q32 = 0q3 - q32 = 0/* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodalpoints at which there is no external source of heat. */T1 = Tw// Furnace wall temperature, C//q1 =// Heat rate, WT2 = To// To, beginning of rod exposed lengthq2 = 0// Heat rate, W; node 2; no external heat sourceT3 = Tinf// Ambient air temperature, C//q3 =// Heat rate, W// Thermal Resistances:// Rod - conduction resistanceR21 = Lins / (k * Ac)// Conduction resistance, K/WAc = pi * D^2 / 4// Cross sectional area of rod, m^2// Thermal Resistance Tools - Fin with Adiabatic Tip:R32 = Rfin// Resistance of fin, K/W/* Thermal resistance of a fin of uniform cross sectional area Ac, perimeter P, length L, and thermalconductivity k with an adiabatic tip condition experiencing convection with a fluid at Tinf and coefficient h, */Rfin = 1/ ( tanh (m*Lo) * (h * P * k * Ac ) ^ (1/2) )// Case B, Table 3.4m = sqrt(h*P / (k*Ac))P = pi * D// Perimeter, m// Other Assigned Variables:Tw = 200// Furnace wall temperature, Ck = 60// Rod thermal conductivity, W/m.KLins = 0.200// Insulated length, mD = 0.025// Rod diameter, mh = 15// Convection coefficient, W/m^2.KTinf = 25// Ambient air temperature,CLo = 0.200// Exposed length, mPROBLEM 3.112KNOWN: Rod (D, k, 2L) inserted into a perfectly insulating wall, exposing one-half of its length toan airstream (T, h). An electromagnetic field induces a uniform volumetric energy generation ( q )in the imbedded portion.FIND: (a) Derive an expression for Tb at the base of the exposed half of the rod; the exposed regionmay be approximated as a very long fin; (b) Derive an expression for To at the end of the imbeddedhalf of the rod, and (c) Using numerical values, plot the temperature distribution in the rod anddescribe its key features. Does the rod behave as a very long fin?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in imbedded portionof rod, (3) Imbedded portion of rod is perfectly insulated, (4) Exposed portion of rod behaves as aninfinitely long fin, and (5) Constant properties.ANALYSIS: (a) Since the exposed portion of the rod (0 x + L) behaves as an infinite fin, the finheat rate using Eq. 3.80 isq x ( 0 ) = q f = M = ( hPkAc )1/ 2( Tb T )(1)From an energy balance on the imbedded portion of the rod,qf = q Ac L(2)2Combining Eqs. (1) and (2), with P = D and Ac = D /4, find1/ 21/ 2Tb = T + q f ( hPkAc )(3)= T + qA1/ 2 L ( hPk )c(b) The imbedded portion of the rod (-L x 0) experiences one-dimensional heat transfer withuniform q . From Eq. 3.43,To =qL2+ Tb2k(c) The temperature distribution T(x) for the rod is piecewise parabolic and exponential,T ( x ) Tb =T ( x ) TTb TqL2 x 22k L = exp ( mx )L x 00 x +LContinued ..<<PROBLEM 3.112 (Cont.)The gradient at x = 0 will be continuous since we used this condition in evaluating Tb. Thedistribution is shown below with To = 105.4C and Tb = 55.4C.T(x) over embedded and exposed portions of rod120Temperature, T(x)10080604020-50 -40 -30 -20 -10010Radial position, x20304050PROBLEM 3.113KNOWN: Very long rod (D, k) subjected to induction heating experiences uniform volumetricgeneration ( q ) over the center, 30-mm long portion. The unheated portions experience convection(T, h).FIND: Calculate the temperature of the rod at the mid-point of the heated portion within the coil, To,and at the edge of the heated portion, Tb.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with uniform q inportion of rod within the coil; no convection from lateral surface of rod, (3) Exposed portions of rodbehave as infinitely long fins, and (4) Constant properties.ANALYSIS: The portion of the rod within the coil, 0 x + L, experiences one-dimensionalconduction with uniform generation. From Eq. 3.43,qL2To =+ Tb2k(1)The portion of the rod beyond the coil, L x , behaves as an infinitely long fin for which the heatrate from Eq. 3.80 isq f = q x ( L ) = ( hPkAc )1/ 2(Tb T )(2)2where P = D and Ac = D /4. From an overall energy balance on the imbedded portion of the rod asillustrated in the schematic above, find the heat rate asEin Eout + Egen = 0q f + qA c L = 0q f = qAc L(3)Combining Eqs. (1-3),1/ 2cTb = T + qA1/ 2 L ( hPk )(4)qL21/ 2cTo = T ++ qA1/ 2 L ( hPk )2k(5)and substituting numerical values findTo = 305CTb = 272C<PROBLEM 3.114KNOWN: Dimensions, end temperatures and volumetric heating of wire leads. Convection coefficientand ambient temperature.FIND: (a) Equation governing temperature distribution in the leads, (b) Form of the temperaturedistribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in x, (3) Uniform volumetricheating, (4) Uniform h (both sides), (5) Negligible radiation.ANALYSIS: (a) Performing an energy balance for the differential control volume,Ein E out + Eg = 0dT dT d dT kAc kAc dx hPdx (T T ) + qAc dx = 0dx dx dx dx kAcd 2Tdx 2q x q x + dx dq conv + qdV = 0hPq(T T ) + = 0kAck<(b) With a reduced temperature defined as T T ( qA c /hP ) and m 2 hP/kA c , thedifferential equation may be rendered homogeneous, with a general solution and boundary conditionsasshownd 2 ( x ) = C1emx + C2emx m 2 = 02dxb = C1 + C2c = C1emL + C2emLit follows thatC1 =b emL ce mL emLC2 =c b emLe mL emL(bemL c ) emx + (c bemL )emx (x ) =e mL emLCOMMENTS: If q is large and h is small, temperatures within the lead may readily exceed theprescribed boundary temperatures.<PROBLEM 3.115KNOWN: Disk-shaped electronic device (D, Ld, kd) dissipates electrical power (Pe) at one of itssurfaces. Device is bonded to a cooled base (To) using a thermal pad (Lp, kA). Long fin (D, kf) isbonded to the heat-generating surface using an identical thermal pad. Fin is cooled by convection (T,h).FIND: (a) Construct a thermal circuit of the system, (b) Derive an expression for the temperature of theheat-generating device, Td, in terms of circuit thermal resistance, To and T; write expressions for thethermal resistances; and (c) Calculate Td for the prescribed conditions.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through thermal padsand device; no losses from lateral surfaces; (3) Fin is infinitely long, (4) Negligible contact resistancebetween components of the system, and (5) Constant properties.ANALYSIS: (a) The thermal circuit is shown below with thermal resistances associated withconduction (pads, Rp; device, Rd) and for the long fin, Rf.(b) To obtain an expression for Td, perform an energy balance about the d-nodeEin Eout = q a + q b + Pe = 0(1)Using the conduction rate equation with the circuitqa =To TdRf + RdT Tqb = dR p + RfCombine with Eq. (1), and solve for Td,Td =()((2,3)Pe + To / R p + R d + T / R p + R f()(1/ R p + R d + 1/ R p + R f))(4)2where the thermal resistances with P = D and Ac = D /4 are1/ 2R f = ( hPk f Ac )R d = Ld / k d A c(c) Substituting numerical values with the foregoing relations, findR p = Lp / k p AcR p = 1.061 K / WR d = 4.244 K / W(5,6,7)R f = 5.712 K / Wand the device temperature asTd = 62.4C<COMMENTS: What fraction of the power dissipated in the device is removed by the fin? Answer:qb/Pe = 47%.PROBLEM 3.116KNOWN: Dimensions and thermal conductivity of a gas turbine blade. Temperature and convectioncoefficient of gas stream. Temperature of blade base and maximum allowable blade temperature.FIND: (a) Whether blade operating conditions are acceptable, (b) Heat transfer to blade coolant.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3) Adiabaticblade tip, (4) Negligible radiation.ANALYSIS: Conditions in the blade are determined by Case B of Table 3.4.(a) With the maximum temperature existing at x = L, Eq. 3.75 yieldsT ( L ) TTb T=1cosh mLm = ( hP/kA c )1/ 2(= 250W/m 2 K 0.11m/20W/m K 6 104 m 2m = 47.87 m-1 and)1/ 2mL = 47.87 m-1 0.05 m = 2.39From Table B.1, cosh mL = 5.51. Hence,T ( L ) = 1200 C + (300 1200) C/5.51 = 1037 C<and the operating conditions are acceptable.((b) With M = ( hPkA c )1/ 2 b = 250W/m 2 K 0.11m 20W/m K 6 10 4 m 2) (900 C ) = 517W ,1/ 2Eq. 3.76 and Table B.1 yieldqf = M tanh mL = 517W ( 0.983) = 508WHence, q b = q f = 508W<COMMENTS: Radiation losses from the blade surface and convection from the tip will contribute toreducing the blade temperatures.PROBLEM 3.117KNOWN: Dimensions of disc/shaft assembly. Applied angular velocity, force, and torque. Thermalconductivity and inner temperature of disc.FIND: (a) Expression for the friction coefficient , (b) Radial temperature distribution in disc, (c) Valueof for prescribed conditions.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant k,(4) Uniform disc contact pressure p, (5) All frictional heat dissipation is transferred to shaft from base ofdisc.ANALYSIS: (a) The normal force acting on a differential ring extending from r to r+dr on the contactsurface of the disc may be expressed as dFn = p2 rdr . Hence, the tangential force is dFt = p2 rdr ,in which case the torque may be expressed asd = 2 pr 2drFor the entire disc, it follows thatr23 pr2 = 2 p 2 r 2dr =o32where p = F r2 . Hence,=32 Fr2<(b) Performing an energy balance on a differential control volume in the disc, it follows thatqcond,r + dq fric q cond,r + dr = 0()()2With dq fric = d = 2 F r 2 r2 dr , q cond,r + dr = q cond,r + dq cond,r dr dr , andqcond,r = k ( 2 rt ) dT dr , it follows thatd ( rdT dr )22 F r 2 r2 dr + 2 ktdr = 0dr(ord ( rdT dr )drIntegrating twice,)= F 2r2 ktr2Continued...PROBLEM 3.117 (Cont.)dT F 2 C1r+=2drr3 ktr2T= F29 ktr2r3 + C1nr + C2Since the disc is well insulated at r = r2 , dT drC1 =r2= 0 and F r23 ktWith T ( r1 ) = T1 , it also follows thatC2 = T1 + Fr3 C1nr1219 ktr2Hence,T ( r ) = T1 F29 ktr2(r3 r13 ) + 3Fktr2 n rr1<(c) For the prescribed conditions,=38N m= 0.3332 200N ( 0.18m )<Since the maximum temperature occurs at r = r2, F r2Tmax = T ( r2 ) = T1 9 kt 3 1 r1 + F r2 n r2 r2 3 kt r1 With ( F r2 3 kt ) = ( 0.333 200N 40rad/s 0.18m 3 15W/m K 0.012m ) = 282.7 C ,3282.7 C 0.02 0.18 1 Tmax = 80 C + 282.7 Cn 3 0.18 0.02 Tmax = 80 C 94.1 C + 621.1 C = 607 CCOMMENTS: The maximum temperature is excessive, and the disks should be actively cooled (byconvection) at their outer surfaces.<PROBLEM 3.118KNOWN: Extended surface of rectangular cross-section with heat flow in the longitudinal direction.FIND: Determine the conditions for which the transverse (y-direction) temperature gradient isnegligible compared to the longitudinal gradient, such that the 1-D analysis of Section 3.6.1 is validby finding: (a) An expression for the conduction heat flux at the surface, q ( t ) , in terms of Ts andyTo, assuming the transverse temperature distribution is parabolic, (b) An expression for theconvection heat flux at the surface for the x-location; equate the two expressions, and identify theparameter that determines the ratio (To Ts)/(Ts - T); and (c) Developing a criterion for establishingthe validity of the 1-D assumption used to model an extended surface.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform convection coefficient and (3) Constantproperties.ANALYSIS: (a) Referring to the schematics above, the conduction heat flux at the surface y = t atany x-location follows from Fouriers law using the parabolic transverse temperature distribution.q ( t ) = ky2y 2kT Ts ( x ) To ( x )= k Ts ( x ) To ( x ) =2ty y = tt y=t(1)(b) The convection heat flux at the surface of any x-location follows from the rate equationq = h Ts ( x ) T cv(2)Performing a surface energy balance as represented schematically above, equating Eqs. (1) and (2)providesq ( t ) = qycv2kTs ( x ) To ( x ) = h Ts ( x ) T tTs ( x ) To ( x )ht= 0.5 = 0.5 BiTs ( x ) T ( x )k(3)where Bi = ht/k, the Biot number, represents the ratio of the convection to the conduction thermalresistances,Bi =R cd = t / kR cv 1/ h(4)(c) The transverse gradient (heat flow) will be negligible compared to the longitudinal gradient whenBi << 1, say, 0.1, an order of magnitude smaller. This is the criterion to validate the one-dimensionalassumption used to model extended surfaces.COMMENTS: The coefficient 0.5 in Eq. (3) is a consequence of the parabolic distributionassumption. This distribution represents the simplest polynomial expression that could approximatethe real distribution.PROBLEM 3.119KNOWN: Long, aluminum cylinder acts as an extended surface.FIND: (a) Increase in heat transfer if diameter is tripled and (b) Increase in heat transfer ifcopper is used in place of aluminum.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties, (4) Uniform convection coefficient, (5) Rod is infinitely long.PROPERTIES: Table A-1, Aluminum (pure): k = 240 W/mK; Table A-1, Copper (pure): k= 400 W/mK.ANALYSIS: (a) For an infinitely long fin, the fin heat rate from Table 3.4 isqf = M = ( hPkAc )1/ 2(bqf = h D k D2 / 4)1/ 2b =( hk )1/ 2 D3/2 b .223/2where P = D and Ac = D /4 for the circular cross-section. Note that qf Dthe diameter is tripled,q f (3D )qf (D)= 33/ 2 = 5.2<and there is a 420% increase in heat transfer.(b) In changing from aluminum to copper, since qf kq f ( Cu ). Hence, if1/ 2k = Cu q f ( A1) k A1 1/ 2 400 = 240 1/2, it follows that= 1.29and there is a 29% increase in the heat transfer rate.<COMMENTS: (1) Because fin effectiveness is enhanced by maximizing P/Ac = 4/D, the useof a larger number of small diameter fins is preferred to a single large diameter fin.(2) From the standpoint of cost and weight, aluminum is preferred over copper.PROBLEM 3.120KNOWN: Length, diameter, base temperature and environmental conditions associated with a brass rod.FIND: Temperature at specified distances along the rod.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4)Negligible radiation, (5) Uniform convection coefficient h.)(PROPERTIES: Table A-1, Brass T = 110 C : k = 133 W/m K.ANALYSIS: Evaluate first the fin parameter1/ 2 hP m= kA c 1/ 2 h D = k D 2 / 4 1/ 2 4h = kD 1/ 2 4 30 W/m 2 K =133 W/m K 0.005m m = 13.43 m-1.Hence, m L = (13.43)0.1 = 1.34 and from the results of Example 3.8, it is advisable not to make theinfinite rod approximation. Thus from Table 3.4, the temperature distribution has the form=cosh m ( L x ) + ( h/mk ) sinh m ( L x )cosh mL + ( h/mk ) sinh mLbEvaluating the hyperbolic functions, cosh mL = 2.04 and sinh mL = 1.78, and the parameterh30 W/m 2 K== 0.0168,mk 13.43m-1 (133 W/m K )with b = 180C the temperature distribution has the formcosh m ( L x ) + 0.0168 sinh m ( L x ))(180 C .2.07The temperatures at the prescribed location are tabulated below.=x(m)cosh m(L-x)sinh m(L-x)T(C)x1 = 0.0251.551.19136.5156.5x2 = 0.051.240.725108.9128.9L = 0.101.000.0087.0107.0<<<COMMENTS: If the rod were approximated as infinitely long: T(x1) = 148.7C, T(x2) =112.0C, and T(L) = 67.0C. The assumption would therefore result in significantunderestimates of the rod temperature.PROBLEM 3.121KNOWN: Thickness, length, thermal conductivity, and base temperature of a rectangular fin. Fluidtemperature and convection coefficient.FIND: (a) Heat rate per unit width, efficiency, effectiveness, thermal resistance, and tip temperaturefor different tip conditions, (b) Effect of convection coefficient and thermal conductivity on the heatrate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along fin, (3) Constantproperties, (4) Negligible radiation, (5) Uniform convection coefficient, (6) Fin width is much longerthan thickness (w >> t).ANALYSIS: (a) The fin heat transfer rate for Cases A, B and D are given by Eqs. (3.72), (3.76) and21/221/2(3.80), where M (2 hw tk) (Tb - T) = (2 100 W/m K 0.001m 180 W/mK) (75C) w =1/221/2-1-1450 w W, m (2h/kt) = (200 W/m K/180 W/mK 0.001m) = 33.3m , mL 33.3m 0.010m2-1= 0.333, and (h/mk) (100 W/m K/33.3m 180 W/mK) = 0.0167. From Table B-1, it followsthat sinh mL 0.340, cosh mL 1.057, and tanh mL 0.321. From knowledge of qf, Eqs. (3.86),(3.81) and (3.83) yieldf qfh ( 2L + t ) b, fqf, R t,f = bht bqfCase A: From Eq. (3.72), (3.86), (3.81), (3.83) and (3.70),q =ff =f =M sinh mL + ( h / mk ) cosh mLw cosh mL + ( h / mk ) sinh mL151 W / m100 W / m K ( 0.021m ) 75C2151 W / m100 W / m K ( 0.001m ) 75C2T ( L ) = T += 450 W / m0.340 + 0.0167 1.0571.057 + 0.0167 0.340= 151 W / m<= 0.96= 20.1, R t,f =bcosh mL + ( h / mk ) sinh mL<75C151 W / m= 25C +<= 0.50 m K / W75C1.057 + ( 0.0167 ) 0.340= 95.6C<Case B: From Eqs. (3.76), (3.86), (3.81), (3.83) and (3.75)q =fMwtanh mL = 450 W / m ( 0.321) = 144 W / m<<<f = 0.92, f = 19.2, R t,f = 0.52 m K / WT ( L ) = T +bcosh mL= 25C +75C1.057= 96.0CContinued ..PROBLEM 3.121 (Cont.)Case D (L ): From Eqs. (3.80), (3.86), (3.81), (3.83) and (3.79)q =fMw<= 450 W / mf = 0, f = 60.0, R t,f = 0.167 m K / W, T ( L ) = T = 25C<(b) The effect of h on the heat rate is shown below for the aluminum and stainless steel fins.Va ria tio n o f q f' w ith h (k= 1 8 0 W /m .K )H e a t ra te , q f'(W /m )15001000500002004006008001000C o n ve ctio n co e fficie n t, h (W /m ^2 .K )q fA'q fB 'q fD 'Va ria tio n o f q f' w ith h (k = 1 5 W /m .K )H e a t ra te , q f'(W /m )400300200100002004006008001000C o n ve c tio n c o e ffic ie n t, h (W /m ^2 .K )q fA'q fB 'q fD 'For both materials, there is little difference between the Case A and B results over the entire range ofh. The difference (percentage) increases with decreasing h and increasing k, but even for the worst2case condition (h = 10 W/m K, k = 180 W/mK), the heat rate for Case A (15.7 W/m) is only slightlylarger than that for Case B (14.9 W/m). For aluminum, the heat rate is significantly over-predicted bythe infinite fin approximation over the entire range of h. For stainless steel, it is over-predicted for2small values of h, but results for all three cases are within 1% for h > 500 W/m K.COMMENTS: From the results of Part (a), we see there is a slight reduction in performance(smaller values of q , f and f , as well as a larger value of R t ,f ) associated with insulating the tip.fAlthough f = 0 for the infinite fin, q and f are substantially larger than results for L = 10 mm,findicating that performance may be significantly improved by increasing L.PROBLEM 3.122KNOWN: Thickness, length, thermal conductivity, and base temperature of a rectangular fin. Fluidtemperature and convection coefficient.FIND: (a) Heat rate per unit width, efficiency, effectiveness, thermal resistance, and tip temperaturefor different tip conditions, (b) Effect of fin length and thermal conductivity on the heat rate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along fin, (3) Constantproperties, (4) Negligible radiation, (5) Uniform convection coefficient, (6) Fin width is much longerthan thickness (w >> t).ANALYSIS: (a) The fin heat transfer rate for Cases A, B and D are given by Eqs. (3.72), (3.76) and21/221/2(3.80), where M (2 hw tk) (Tb - T) = (2 100 W/m K 0.001m 180 W/mK) (75C) w =1/221/2-1-1450 w W, m (2h/kt) = (200 W/m K/180 W/mK 0.001m) = 33.3m , mL 33.3m 0.010m2-1= 0.333, and (h/mk) (100 W/m K/33.3m 180 W/mK) = 0.0167. From Table B-1, it followsthat sinh mL 0.340, cosh mL 1.057, and tanh mL 0.321. From knowledge of qf, Eqs. (3.86),(3.81) and (3.83) yieldf qfh ( 2L + t ) b, f qf, R t,f = bht bqfCase A: From Eq. (3.72), (3.86), (3.81), (3.83) and (3.70),q =ff =f =M sinh mL + ( h / mk ) cosh mLw cosh mL + ( h / mk ) sinh mL151 W / m100 W / m K ( 0.021m ) 75C2151 W / m100 W / m K ( 0.001m ) 75C2T ( L ) = T += 450 W / m0.340 + 0.0167 1.0571.057 + 0.0167 0.340= 151 W / m<= 0.96= 20.1, R t,f =bcosh mL + ( h / mk ) sinh mL<75C151 W / m= 25C +<= 0.50 m K / W75C1.057 + ( 0.0167 ) 0.340= 95.6C<Case B: From Eqs. (3.76), (3.86), (3.81), (3.83) and (3.75)q =fMwtanh mL = 450 W / m ( 0.321) = 144 W / m<<<f = 0.92, f = 19.2, R t,f = 0.52 m K / WT ( L ) = T +bcosh mL= 25C +75C1.057= 96.0CContinued ..PROBLEM 3.122 (Cont.)Case D (L ): From Eqs. (3.80), (3.86), (3.81), (3.83) and (3.79)q =fMw<= 450 W / mf = 0, f = 60.0, R t,f = 0.167 m K / W, T ( L ) = T = 25C<(b) The effect of L on the heat rate is shown below for the aluminum and stainless steel fins.Va ria tio n o f q f' w ith L (k= 1 8 0 W /m .K )H e a t ra te , q f'(W /m )500400300200100000 .0 10 .0 20 .0 30 .0 40 .0 5Fin le n g th , L (m )q fA'q fB 'q fD 'Va ria tio n o f q f' w ith L (k= 1 5 W /m .K )H e a t ra te , q f'(W /m )150120906030000 .0 10 .0 20 .0 30 .0 40 .0 5Fin le n g th , L (m )q fA'q fB 'q fD 'For both materials, differences between the Case A and B results diminish with increasing L and arewithin 1% of each other at L 27 mm and L 13 mm for the aluminum and steel, respectively. At L= 3 mm, results differ by 14% and 13% for the aluminum and steel, respectively. The Case A and Bresults approach those of the infinite fin approximation more quickly for stainless steel due to thelarger temperature gradients, |dT/dx|, for the smaller value of k.COMMENTS: From the results of Part (a), we see there is a slight reduction in performance(smaller values of q , f and f , as well as a larger value of R t ,f ) associated with insulating the tip.fAlthough f = 0 for the infinite fin, q and f are substantially larger than results for L = 10 mm,findicating that performance may be significantly improved by increasing L.PROBLEM 3.123KNOWN: Length, thickness and temperature of straight fins of rectangular, triangular and parabolicprofiles. Ambient air temperature and convection coefficient.FIND: Heat rate per unit width, efficiency and volume of each fin.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties, (4)Negligible radiation, (5) Uniform convection coefficient.ANALYSIS: For each fin,q = qfmax = f hA b ,fV = A p1/221/2-1where f depends on the value of m = (2h/kt) = (100 W/m K/185 W/mK 0.003m) = 13.4m-1and the product mL = 13.4m 0.015m = 0.201 or mLc = 0.222. Expressions for f, A and Ap arefobtained from Table 3-5.Rectangular Fin:f =tanh mLc 0.218== 0.982, A = 2 Lc = 0.033mfmLc0.222<)(q = 0.982 50 W / m 2 K 0.033m (80C ) = 129.6 W / m, V = tL = 4.5 105 m 2<Triangular Fin:f =1 I1 ( 2mL )mL I0 ( 2 mL )(=0.205( 0.201)1.0421/ 22= 0.978, A = 2 L + ( t / 2 )f2)q = 0.978 50 W / m K 0.030m (80C ) = 117.3 W / m, V = ( t / 2 ) L = 2.25 102<= 0.030m5m2<Parabolic Fin:f =1/ 2 4 ( mL )2 + 1(()= 0.963, A = C1L + L2 / t ln ( t / L + C1 ) = 0.030mf2+1)q = 0.963 50 W / m 2 K 0.030m (80C ) = 115.6 W / m, V = ( t / 3) L = 1.5 105 m 2f<<COMMENTS: Although the heat rate is slightly larger (~10%) for the rectangular fin than for thetriangular or parabolic fins, the heat rate per unit volume (or mass) is larger and largest for thetriangular and parabolic fins, respectively.PROBLEM 3.124KNOWN: Melting point of solder used to join two long copper rods.FIND: Minimum power needed to solder the rods.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction along therods, (3) Constant properties, (4) No internal heat generation, (5) Negligible radiationexchange with surroundings, (6) Uniform h, and (7) Infinitely long rods.PROPERTIES: Table A-1: Copper T = ( 650 + 25 ) C 600K: k = 379 W/m K.ANALYSIS: The junction must be maintained at 650C while energy is transferred byconduction from the junction (along both rods). The minimum power is twice the fin heat ratefor an infinitely long fin,q min = 2q f = 2 ( hPkAc )1/ 2(Tb T ).Substituting numerical values,Wq min = 2 10( 0.01m ) m2 K1/ 2W2379 m K 4 ( 0.01m ) (650 25) C.Therefore,q min = 120.9 W.COMMENTS: Radiation losses from the rods may be significant, particularly near thejunction, thereby requiring a larger power input to maintain the junction at 650C.<PROBLEM 3.125KNOWN: Dimensions and end temperatures of pin fins.FIND: (a) Heat transfer by convection from a single fin and (b) Total heat transfer from a 12m surface with fins mounted on 4mm centers.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along rod, (3) Constantproperties, (4) No internal heat generation, (5) Negligible radiation.PROPERTIES: Table A-1, Copper, pure (323K): k 400 W/mK.ANALYSIS: (a) By applying conservation of energy to the fin, it follows thatq conv = qcond,i q cond,owhere the conduction rates may be evaluated from knowledge of the temperature distribution.The general solution for the temperature distribution is ( x ) = C1 emx + C2 e-mx T T .The boundary conditions are (0) o = 100C and (L) = 0. Hence o = C1 + C20 = C1 emL + C2 e-mLC2 = C1 e2mLTherefore,C1 = o e2mLo,C2 = 1 e 2mL1 e 2mLand the temperature distribution has the form=oemx e2mL-mx .1 e2mL The conduction heat rate can be evaluated by Fouriers law,qcond = kAckAc odm emx + e 2mL-mx =dx1 e 2mL or, with m = ( hP/kA c )1/ 2, o ( hPkAc )1/ 2q cond = 1 e2mL emx + e2mL-mx .Continued ..PROBLEM 3.125 (Cont.)Hence at x = 0, o ( hPkAc )1/ 2q cond,i = 1 e2mL(1 + e2mL )at x = L ( hPkA c )q cond,o = o1 e2mLEvaluating the fin parameters:1/ 21/ 2 hP m= kA c 1/ 2 4h = kD 1/ 2( hPkAc )1/ 2 2 3 D hk =4(2emL )1/ 2 4 100 W/m 2 K = 400 W/m K 0.001m 1/ 2 2WW3= ( 0.001m ) 100 400mK4m2 KmL = 31.62 m-1 0.025m = 0.791,The conduction heat rates areq cond,i =q cond,o =(100K 9.93 10-3 W/K(= 31.62 m-13.865emL = 2.204,WKe2mL = 4.865) 5.865 = 1.507 W100K 9.93 10-3 W/K-3.865and from the conservation relation,= 9.93 103) 4.408 = 1.133 W<q conv = 1.507 W 1.133 W = 0.374 W.(b) The total heat transfer rate is the heat transfer from N = 250250 = 62,500 rods and the2heat transfer from the remaining (bare) surface (A = 1m - NAc). Hence,()q = N q cond,i + hA o = 62,500 (1.507 W ) + 100W/m 2 K 0.951 m 2 100Kq = 9.42 104 W+0.95 104 W=1.037 105 W.COMMENTS: (1) The fins, which cover only 5% of the surface area, provide for more than90% of the heat transfer from the surface.(2) The fin effectiveness, q cond,i / hA c o , is = 192, and the fin efficiency, ( q conv / h DL o ) , is = 0.48.(3) The temperature distribution, (x)/o, and the conduction term, qcond,i, could have beenobtained directly from Eqs. 3.77and 3.78, respectively.(4) Heat transfer by convection from a single fin could also have been obtained from Eq. 3.73.PROBLEM 3.126KNOWN: Pin fin of thermal conductivity k, length L and diameter D connecting two devices (Lg,kg)experiencing volumetric generation of thermal energy (q ). Convection conditions are prescribed (T,h).FIND: Expression for the device surface temperature Tb in terms of device, convection and finparameters.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Pin fin is of uniform cross-section with constant h,(3) Exposed surface of device is at a uniform temperature Tb, (4) Backside of device is insulated, (5)Device experiences 1-D heat conduction with uniform volumetric generation, (6) Constant properties,and (7) No contact resistance between fin and devices.ANALYSIS: Recognizing symmetry, the pin fin is modeled as a fin of length L/2 with insulated tip.Perform a surface energy balance,E in E out = 0q d qs q f = 0(1)The heat rate qd can be found from anenergy balance on the entire device to findE in E out + E g = 0q d + qV = 0q d = qA g Lg(2)The fin heat rate, qf, follows from Case B,q f = M tanh mL/2 = ( hPkAc )1/ 2)(Table 3.4(Tb T ) tanh ( mL/2 ) ,P/A c = D/ D 2 / 4 = 4 / Dandm = ( hP/kAc )1/ 2PA c = 2D3 / 4.(3,4)(5,6)Hence, the heat rate expression can be written as((()qAg Lg = h Ag Ac (Tb T ) + hk 2 D3 / 4))1/ 2 4h 1/ 2 L tanh ( Tb T ) kD 2(7)Solve now for Tb, 4h 1/ 2 L 1/ 2 h Ag Ac + hk 2D3 / 4Tb = T + qAg Lg /tanh kD 2 ()(())(8) <PROBLEM 3.127KNOWN: Positions of equal temperature on two long rods of the same diameter, butdifferent thermal conductivity, which are exposed to the same base temperature and ambientair conditions.FIND: Thermal conductivity of rod B, kB.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Rods are infinitely long fins of uniformcross-sectional area, (3) Uniform heat transfer coefficient, (4) Constant properties.ANALYSIS: The temperature distribution for the infinite fin has the formT ( x ) T== e-mxTo Tb1/ 2 hP m= kAc .(1,2)For the two positions prescribed, xA and xB, it was observed thatTA ( x A ) = TB ( x B )or A ( x A ) = B ( x B ).(3)Since b is identical for both rods, Eq. (1) with the equality of Eq. (3) requires thatm A x A = m Bx BSubstituting for m from Eq. (2) gives1/ 2 hP k A Ac 1/ 2 hP xA = k BAc x B.Recognizing that h, P and Ac are identical for each rod and rearranging,2x kB = B kA xA 2 0.075m kB = 70 W/m K = 17.5 W/m K. 0.15m COMMENTS: This approach has been used as a method for determining the thermalconductivity. It has the attractive feature of not requiring power or temperaturemeasurements, assuming of course, a reference material of known thermal conductivity isavailable.<PROBLEM 3.128KNOWN: Slender rod of length L with ends maintained at To while exposed to convectioncooling (T < To, h).FIND: Temperature distribution for three cases, when rod has thermal conductivity (a) kA,(b) kB < kA, and (c) kA for 0 x L/2 and kB for L/2 x L.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constantproperties, and (4) Negligible thermal resistance between the two materials (A, B) at the midspan for case (c).ANALYSIS: (a, b) The effect of thermal conductivity on the temperature distribution whenall other conditions (To, h, L) remain the same is to reduce the minimum temperature withdecreasing thermal conductivity. Hence, as shown in the sketch, the mid-span temperaturesare TB (0.5L) < TA (0.5L) for kB < kA. The temperature distribution is, of course,symmetrical about the mid-span.(c) For the composite rod, the temperature distribution can be reasoned by considering theboundary condition at the mid-span.q ( 0.5L ) = q ( 0.5L )x,Ax,Bk AdT dT = k Bdx A,x=0.5Ldx B,x=0.5LSince kA > kB, it follows that dT dT <. dx A,x=0.5L dx B,x=0.5LIt follows that the minimum temperature in the rod must be in the kB region, x > 0.5L, and thetemperature distribution is not symmetrical about the mid-span.COMMENTS: (1) Recognize that the area under the curve on the T-x coordinates isproportional to the fin heat rate. What conclusions can you draw regarding the relativemagnitudes of qfin for cases (a), (b) and (c)?(2) If L is increased substantially, how would the temperature distribution be affected?PROBLEM 3.129KNOWN: Base temperature, ambient fluid conditions, and temperatures at a prescribeddistance from the base for two long rods, with one of known thermal conductivity.FIND: Thermal conductivity of other rod.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along rods, (3) Constantproperties, (4) Negligible radiation, (5) Negligible contact resistance at base, (6) Infinitelylong rods, (7) Rods are identical except for their thermal conductivity.ANALYSIS: With the assumption of infinitely long rods, the temperature distribution isT T== e-mx b Tb Tor1/ 2T T hP ln= mx = Tb T kA xHence, for the two rods,T T ln A 1/ 2 Tb T = k B TB T k A ln Tb T T T ln Aln Tb T = 200 1/ 21/2 = k1/2kB()AT T lnln B Tb T k B = 56.6 W/m K.75 25100 25 = 7.52460 25100 25<COMMENTS: Providing conditions for the two rods may be maintained nearly identical, theabove method provides a convenient means of measuring the thermal conductivity of solids.PROBLEM 3.130KNOWN: Arrangement of fins between parallel plates. Temperature and convection coefficient ofair flow in finned passages. Maximum allowable plate temperatures.FIND: (a) Expressions relating fin heat transfer rates to end temperatures, (b) Maximum powerdissipation for each plate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fins, (3) Constantproperties, (4) Negligible radiation, (5) All of the heat is dissipated to the air, (6) Uniform h, (7)Negligible variation in T, (8) Negligible contact resistance.PROPERTIES: Table A.1, Aluminum (pure), 375 K: k = 240 W/mK.ANALYSIS: (a) The general solution for the temperature distribution in fin is ( x ) T ( x ) T = C1emx + C 2e-mx (0 ) = o = To T ,Boundary conditions:Hence o = C1 + C2 ( L ) = L = TL T . L = C1emL + C2e-mL L = C1e mL + ( o C1 ) e-mLC1 =Hence L oe-mLemL e-mL (x ) =C2 = o L o e-mLemL e-mL= o emL LemL e-mL.m x-Lm L-x Lemx oe ( ) + oe ( ) Le-mxe mL e-mL(m L-x-m L-x o e ( ) e ( ) + L emx e-mx (x ) = mL e-mLe) sinh m (L-x ) + Lsinh mx. (x ) = osinh mLThe fin heat transfer rate is thenq f = kAcHencemdT mcosh m ( L x ) + Lcosh mx .= kDt odxsinh mL sinh mLm o mL tanh mL sinh mL <m omL. sinh mL tanh mL <q f,o = kDt q f,L = kDt Continued ..PROBLEM 3.130 (Cont.)1/ 2(b) hP m= kAc 1/ 2 50 W/m 2 K ( 2 0.1 m+2 0.001 m ) = 240 W/m K 0.1 m 0.001 m = 35.5 m-1mL = 35.5 m-1 0.012 m = 0.43sinh mL = 0.439tanh mL = 0.401 o = 100 K L = 50 K 100 K 35.5 m-1 50 K 35.5 m-1 q f,o = 240 W/m K 0.1 m 0.001 m 0.4010.439q f,o = 115.4 W(from the top plate) 100 K 35.5 m-1 50 K 35.5 m-1 = 240 W/m K 0.1 m 0.001 m q f,L0.4390.401q f,L = 87.8 W.(into the bottom plate)Maximum power dissipations are thereforeq o,max = Nf q f,o + ( W Nf t ) Dh oq o,max = 50 115.4 W+ ( 0.200 50 0.001) m 0.1 m 150 W/m 2 K 100 K<q o,max = 5770 W+225 W = 5995 Wq L,max = N f q f,L + ( W Nf t ) Dh oq L,max = 50 87.8W + ( 0.200 50 0.001) m 0.1 m 150 W/m 2 K 50 K<q L,max = 4390 W+112W = 4278 W.COMMENTS: (1) It is of interest to determine the air velocity needed to prevent excessive heating of the air asit passes between the plates. If the air temperature change is restricted tomair =T = 5 K, its flowrate must beq tot1717 W== 0.34 kg/s.cp T 1007 J/kg K 5 KIts mean velocity is thenmair0.34 kg/sVair === 163 m/s.3 0.012 m 0.2 50 0.001 mair Ac 1.16 kg/m()Such a velocity would be impossible to maintain. To reduce it to a reasonable value, e.g. 10m/s, Ac would have to be increased substantially by increasing W (and hence the spacebetween fins) and by increasing L. The present configuration is impractical from thestandpoint that 1717 W could not be transferred to air in such a small volume.(2) A negative value of qL,max implies that heat must be transferred from the bottom plate tothe air to maintain the plate at 350 K.PROBLEM 3.131KNOWN: Conditions associated with an array of straight rectangular fins.FIND: Thermal resistance of the array.SCHEMATIC:ASSUMPTIONS: (1) Constant properties, (2) Uniform convection coefficient, (3) Symmetry aboutmidplane.ANALYSIS: (a) Considering a one-half section of the array, the corresponding resistance is1R t,o = (o hA t )where A t = NA f + A b . With S = 4 mm and t = 1 mm, it follows that N = W1 /S = 250,2(L/2)W2 = 0.008 m2, Ab = W2(W1 - Nt) = 0.75 m2, and At = 2.75 m2.The overall surface efficiency iso = 1 NA fAtAf =(1 f )where the fin efficiency isf =tanh m ( L 2 )m (L 2)1/ 2and hP kA c m= h ( 2t + 2W2 ) ktW21/ 2=1/ 2 2h kt = 38.7m 1With m(L/2) = 0.155, it follows that f = 0.992 and o = 0.994. Hence(R t,o = 0.994 150W/m 2 K 2.75m 2)1= 2.44 10 3 K/W<(b) The requirements that t 0.5 m and (S - t) > 2 mm are based on manufacturing and flow passagerestriction constraints. Repeating the foregoing calculations for representative values of t and (S - t),we obtainS (mm)2.5334455N400333333250250200200t (mm)0.50.510.520.53Rt,o (K/W)0.001690.001930.002020.002340.002680.002640.00334COMMENTS: Clearly, the thermal performance of the fin array improves (Rt,o decreases) withincreasing N. Because f 1 for the entire range of conditions, there is a slight degradation inperformance (Rt,o increases) with increasing t and fixed N. The reduced performance is associatedwith the reduction in surface area of the exposed base. Note that the overall thermal resistance for the-2entire fin array (top and bottom) is Rt,o/2 = 1.22 10 K/W.PROBLEM 3.132KNOWN: Width and maximum allowable temperature of an electronic chip. Thermal contactresistance between chip and heat sink. Dimensions and thermal conductivity of heat sink.Temperature and convection coefficient associated with air flow through the heat sink.FIND: (a) Maximum allowable chip power for heat sink with prescribed number of fins, finthickness, and fin pitch, and (b) Effect of fin thickness/number and convection coefficient onperformance.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal chip, (4)Negligible heat transfer from top surface of chip, (5) Negligible temperature rise for air flow, (6)Uniform convection coefficient associated with air flow through channels and over outer surfaces ofheat sink, (7) Negligible radiation.ANALYSIS: (a) From the thermal circuit,T TTc Tqc = c =R totR t,c + R t,b + R t,o()where R t,c = R / W 2 = 2 10 6 m 2 K / W / ( 0.02m )2 = 0.005 K / W and R t,b = L b / k W 2t,c= 0.003m /180 W / m K ( 0.02m ) = 0.042 K / W. From Eqs. (3.103), (3.102), and (3.99)2R t,o =1,o h A to = 1 N Af(1 f ) ,At-42A t = N Af + A b22where Af = 2WLf = 2 0.02m 0.015m = 6 10 m and Ab = W N(tW) = (0.02m) 11(0.182-3-4 21/22 10 m 0.02m) = 3.6 10 m . With mLf = (2h/kt) Lf = (200 W/m K/180 W/mK 0.182 -3 1/210 m) (0.015m) = 1.17, tanh mLf = 0.824 and Eq. (3.87) yieldsf =tanh mLf 0.824== 0.704mLf1.17-32It follows that At = 6.96 10 m , o = 0.719, Rt,o = 2.00 K/W, andqc =(85 20 ) C(0.005 + 0.042 + 2.00 ) K / W<= 31.8 W(b) The following results are obtained from parametric calculations performed to explore the effect ofdecreasing the number of fins and increasing the fin thickness.Continued ..PROBLEM 3.132 (Cont.)Nt(mm)f678910111.8331.3140.9250.6220.3800.1820.9570.9410.9190.8850.8260.704Rt,o (K/W)2.762.402.151.971.892.002qc (W)At (m )23.226.629.732.233.531.80.003780.004420.005050.005690.006320.00696Although f (and o) increases with decreasing N (increasing t), there is a reduction in At whichyields a minimum in Rt,o, and hence a maximum value of qc, for N = 10. For N = 11, the effect of hon the performance of the heat sink is shown below.Heat rate as a function of convection coefficient (N=11)Heat rate, qc(W)150100500100 200 300 400 500 600 700 800 900 1000Convection coefficient, h(W/m2.K)2With increasing h from 100 to 1000 W/m K, Rt,o decreases from 2.00 to 0.47 K/W, despite adecrease in f (and o) from 0.704 (0.719) to 0.269 (0.309). The corresponding increase in qc issignificant.COMMENTS: The heat sink significantly increases the allowable heat dissipation. If it were notused and heat was simply transferred by convection from the surface of the chip with h = 10022W/m K, Rtot = 2.05 K/W from Part (a) would be replaced by Rcnv = 1/hW = 25 K/W, yielding qc =2.60 W.PROBLEM 3.133KNOWN: Number and maximum allowable temperature of power transistors. Contact resistancebetween transistors and heat sink. Dimensions and thermal conductivity of heat sink. Temperatureand convection coefficient associated with air flow through and along the sides of the heat sink.FIND: (a) Maximum allowable power dissipation per transistor, (b) Effect of the convectioncoefficient and fin length on the transistor power.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal transistors, (4)Negligible heat transfer from top surface of heat sink (all heat transfer is through the heat sink), (5)Negligible temperature rise for the air flow, (6) Uniform convection coefficient, (7) Negligibleradiation.ANALYSIS: (a) From the thermal circuit,Nt qt =Tt T(R t,c )equiv + R t,b + R t,oFor the array of transistors, the corresponding contact resistance is the equivalent resistanceassociated with the component resistances, in which case,(R t,c )equiv = N t (1/ R t,c )11= (9 / 0.045 K / W )= 5 103 K / WThe thermal resistance associated with the base of the heat sink isR t,b =Lbk (W )2=0.006m180 W / m K ( 0.150m )2= 1.48 103 K / WFrom Eqs. (3.103), (3.102) and (3.99), the thermal resistance associated with the fin array and thecorresponding overall efficiency and total surface area areR t,o =1,o h A to = 1 Nf Af(1 f ) ,AtA t = Nf Af + Ab-32Each fin has a surface area of Af 2 W Lf = 2 0.15m 0.03m = 9 10 m , and the area of the22-2 2exposed base is Ab = W Nf (tW) = (0.15m) 25 (0.003m 0.15m) = 1.13 10 m . With mLf =1/221/2(2h/kt) Lf = (200 W/m K/180 W/mK 0.003m) (0.03m) = 0.577, tanh mLf = 0.520 and Eq.(3.87) yieldsf =tanh mLf 0.520== 0.902mLf0.577-3-222Hence, with At = [25 (9 10 ) + 1.13 10 ]m = 0.236m ,Continued ..o = 1 PROBLEM 3.133 (Cont.)(25 0.009m 20.236m 2) (1 0.901) = 0.907(R t,o = 0.907 100 W / m 2 K 0.236m 2)1= 0.0467 K / WThe heat rate per transistor is thenqt =(100 27 ) C1= 152 W9 (0.0050 + 0.0015 + 0.0467 ) K / W<(b) As shown below, the transistor power dissipation may be enhanced by increasing h and/or Lf.Tra n s is to r p o w e r (W )6005004003002001000100200300400500600700800900 1000C o n ve c tio n c o e ffic ie n t, h (W /m ^2 .K )250240H e a t ra te , q t(W )2302202102001901801701601500 .0 30 .0 40 .0 50 .0 60 .0 70 .0 80 .0 90 .1Fin le n g th , L f(m )However, in each case, the effect of the increase diminishes due to an attendant reduction in f. For2example, as h increases from 100 to 1000 W/m K for Lf = 30 mm, f decreases from 0.902 to 0.498.COMMENTS: The heat sink significantly increases the allowable transistor power. If it were not222used and heat was simply transferred from a surface of area W = 0.0225 m with h = 100 W/m K,2 -1the corresponding thermal resistance would be Rt,cnv = (hW ) K/W = 0.44 and the transistor powerwould be qt = (Tt - T)/Nt Rt,cnv = 18.4 W.PROBLEM 3.134KNOWN: Geometry and cooling arrangement for a chip-circuit board arrangement.Maximum chip temperature.FIND: (a) Equivalent thermal circuit, (b) Maximum chip heat rate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in chipboard assembly, (3) Negligible pin-chip contact resistance, (4) Constant properties, (5)Negligible chip thermal resistance, (6) Uniform chip temperature.PROPERTIES: Table A.1, Copper (300 K): k 400 W/mK.ANALYSIS: (a) The thermal circuit isRf =cosh mL+ ( h o / mk ) sinh mLb=16q f 16 h PkA 1/ 2 sinh mL+ h / mk cosh mL (o)( o c,f ) (b) The maximum chip heat rate isq c = 16q f + q b + qi .Evaluate these parametershPm= o kAc,f(1/ 2 4h= o kDp1/ 2)1/ 2 4 1000 W/m 2 K = 400 W/m K 0.0015 m mL = 81.7 m-1 0.015 m = 1.23,( h/mk ) =1000 W/m 2 K81.7 m-1 400 W/m K(2M = h o D p k Dp / 4()1/ 2sinh mL = 1.57,= 81.7 m-1cosh mL = 1.86= 0.0306b)1/ 23M = 1000 W/m 2 K 2 / 4 ( 0.0015 m ) 400 W/m K (55 C) = 3.17 W.Continued ..PROBLEM 3.134 (Cont.)The fin heat rate isqf = Msinh mL+ ( h/mk ) cosh mLcosh mL+ ( h/mk ) sinh mL= 3.17 W1.57+0.0306 1.861.86+0.0306 1.57q f = 2.703 W.The heat rate from the board by convection is22q b = h o A b b = 1000 W/m 2 K ( 0.0127 m ) (16 / 4 )( 0.0015 m ) 55 Cq b = 7.32 W.The convection heat rate is(55 C)qi ==(1/hi + R + Lb / k b ) (1/ Ac ) (1/40+10-4 + 0.005 /1) m2 K/Wt,c(0.0127 m )2Tc T,iqi = 0.29 W.Hence, the maximum chip heat rate isq c = 16 ( 2.703) + 7.32 + 0.29 W = [43.25 + 7.32 + 0.29] W<q c = 50.9 W.COMMENTS: (1) The fins are extremely effective in enhancing heat transfer from the chip2(assuming negligible contact resistance). Their effectiveness is = q f / Dp / 4 h o b = 2.703()W/0.097 W = 27.822(2) Without the fins, qc = 1000 W/m K(0.0127 m) 55C + 0.29 W = 9.16 W. Hence the finsprovide for a (50.9 W/9.16 W) 100% = 555% enhancement of heat transfer.2(3) With the fins, the chip heat flux is 50.9 W/(0.0127 m) or q = 3.16 105 W/m 2 = 31.6c2W/cm .(4) If the infinite fin approximation is made, qf = M = 3.17 W, and the actual fin heat transferis overestimated by 17%.PROBLEM 3.135KNOWN: Geometry of pin fin array used as heat sink for a computer chip. Array convection and chipsubstrate conditions.FIND: Effect of pin diameter, spacing and length on maximum allowable chip power dissipation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in chip-boardassembly, (3) Negligible pin-chip contact resistance, (4) Constant properties, (5) Negligible chip thermalresistance, (6) Uniform chip temperature.ANALYSIS: The total power dissipation is q c = qi + q t , whereTc T,iqi == 0.3W1 h i + R + L b k b A ct,c()andqt =Tc T,oR t,oThe resistance of the pin array is1R t,o = (o h o A t )whereo = 1 NAfAt(1 f )A t = NA f + A b(A f = D p L c = D p L p + D p /4)Subject to the constraint that N D p U 9 mm, the foregoing expressions may be used to compute qt as afunction of D p for L p = 15 mm and values of N = 16, 25 and 36. Using the IHT PerformanceCalculation, Extended Surface Model for the Pin Fin Array, we obtainContinued...1/ 2PROBLEM 3.135 (CONT.)35Heat rate, qt(W)3025201510500.50.91.31.72.12.5Pin diameter, Dp(mm)N = 36N = 25N = 16Clearly, it is desirable to maximize the number of pins and the pin diameter, so long as flow passages arenot constricted to the point of requiring an excessive pressure drop to maintain the prescribed convectioncoefficient. The maximum heat rate for the fin array ( q t = 33.1 W) corresponds to N = 36 and D p = 1.5mm. Further improvement could be obtained by using N = 49 pins of diameter D p = 1.286 mm, whichyield q t = 37.7 W.Exploring the effect of L p for N = 36 and D p = 1.5 mm, we obtainHeat rate, qt(W)605040301020304050Pin length, Lp(mm)N = 36, Dp = 1.5 mmClearly, there are benefits to increasing L p , although the effect diminishes due to an attendant reductionin f (from f = 0.887 for L p = 15 mm to f = 0.471 for L p = 50 mm). Although a heat dissipationrate of q t = 56.7 W is obtained for L p = 50 mm, package volume constraints could preclude such alarge fin length.COMMENTS: By increasing N, D p and/or L p , the total surface area of the array, A t , is increased,thereby reducing the array thermal resistance, R t ,o . The effects of D p and N are shown for L p = 15mm.Resistance, Rt,o(K/W)864200.511.5Pin diameter, Dp(mm)N = 16N = 25N = 3622.5PROBLEM 3.136KNOWN: Copper heat sink dimensions and convection conditions.FIND: (a) Maximum allowable heat dissipation for a prescribed chip temperature and interfacialchip/heat-sink contact resistance, (b) Effect of fin length and width on heat dissipation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in chip-heat sinkassembly, (3) Constant k, (4) Negligible chip thermal resistance, (5) Negligible heat transfer fromback of chip, (6) Uniform chip temperature.ANALYSIS: (a) For the prescribed system, the chip power dissipation may be expressed asTc Tqc =R t,c + R cond,b + R t,owhere R t,c =R t,c2WcR cond,b ==5 10 6 m 2 K W(0.016m )2Lb=2kWc= 0.0195 K W0.003m400 W m K ( 0.016m )2= 0.0293 K WThe thermal resistance of the fin array is1R t,o = (o hA t )where o = 1 andNAfAt(1 f )(2A t = NA f + A b = N ( 4wL c ) + Wc Nw 2)Continued...PROBLEM 3.136 (Cont.)With w = 0.25 mm, S = 0.50 mm, Lf = 6 mm, N = 1024, and Lc Lf + w 4 = 6.063 103 m, itfollows that A f = 6.06 106 m 2 and A t = 6.40 103 m 2 . The fin efficiency isf =tanh mLcmLc1/ 21/ 2where m = ( hP kA c )= 245 m-1 and mLc = 1.49. It follows that f = 0.608 and= ( 4h kw )o = 0.619, in which case)(R t,o = 0.619 1500 W m 2 K 6.40 103 m 2 = 0.168 K Wand the maximum allowable heat dissipation isqc =(85 25) C(0.0195 + 0.0293 + 0.168 ) KW<= 276W(b) The IHT Performance Calculation, Extended Surface Model for the Pin Fin Array has been usedto determine q c as a function of Lf for four different cases, each of which is characterized by theclosest allowable fin spacing of (S - w) = 0.25 mm.Maximum heat rate, qc(W)CaseABCDw (mm)0.250.350.450.55S (mm)0.500.600.700.80N1024711522400340330320310300290280270678910Fin length, Lf(mm)w = 0.25 mm, S = 0.50 mm, N =1024w = 0.35 mm, S = 0.60 mm, N = 711w = 0.45 mm, S = 0.70 mm, N = 522w = 0.55 mm, S = 0.80 mm, N = 400With increasing w and hence decreasing N, there is a reduction in the total area A t associated withheat transfer from the fin array. However, for Cases A through C, the reduction in A t is more thanbalanced by an increase in f (and o ), causing a reduction in R t ,o and hence an increase in q c .As the fin efficiency approaches its limiting value of f = 1, reductions in A t due to increasing ware no longer balanced by increases in f , and q c begins to decrease. Hence there is an optimumvalue of w, which depends on Lf . For the conditions of this problem, Lf = 10 mm and w = 0.55mm provide the largest heat dissipation.Problem 3.137KNOWN: Two finned heat sinks, Designs A and B, prescribed by the number of fins in the array, N,fin dimensions of square cross-section, w, and length, L, with different convection coefficients, h.FIND: Determine which fin arrangement is superior. Calculate the heat rate, qf, efficiency, f, andeffectiveness, f, of a single fin, as well as, the total heat rate, qt, and overall efficiency, o, of thearray. Also, compare the total heat rates per unit volume.SCHEMATIC:Fin dimensionsCross sectionLengthw x w (mm)L (mm)1x1303x37DesignABNumber offins6x914 x 17Convectioncoefficient(W/m2K)125375ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fins, (3)Convection coefficient is uniform over fin and prime surfaces, (4) Fin tips experience convection,and (5) Constant properties.ANALYSIS: Following the treatment of Section 3.6.5, the overall efficiency of the array, Eq. (3.98),iso =qtq max=qthA t b(1)where At is the total surface area, the sum of the exposed portion of the base (prime area) plus the finsurfaces, Eq. 3.99,A t = N Af + A b(2)where the surface area of a single fin and the prime area areAf = 4 ( L W ) + w 2(3)A b = b1 b2 N A c(4)Combining Eqs. (1) and (2), the total heat rate for the array isq t = Nf hAf b + hA b b(5)where f is the efficiency of a single fin. From Table 4.3, Case A, for the tip condition withconvection, the single fin efficiency based upon Eq. 3.86,f =qfhAf b(6)Continued...PROBLEM 3.137 (Cont.)whereqf = Msinh(mL) + ( h mk ) cosh(mL)(7)cosh(mL) + ( h mk ) sinh(mL)M = ( hPkA c )1/ 2bm = ( hP kAc )1/ 2P = 4wAc = w 2(8,9,10)The single fin effectiveness, from Eq. 3.81,f =qfhA c b(11)Additionally, we want to compare the performance of the designs with respect to the array volume,volq = qf = qff( b1 b2 L )(12)The above analysis was organized for easy treatment with equation-solving software. Solving Eqs.(1) through (11) simultaneously with appropriate numerical values, the results are tabulated below.DesignABqt(W)113165qf(W)1.800.475off0.8040.9090.7790.87331.925.3q TTTf(W/m3)1.251067.81106COMMENTS: (1) Both designs have good efficiencies and effectiveness. Clearly, Design B issuperior because the heat rate is nearly 50% larger than Design A for the same board footprint.Further, the space requirement for Design B is four times less ( = 2.1210-5 vs. 9.0610-5 m3) andthe heat rate per unit volume is 6 times greater.(2) Design A features 54 fins compared to 238 fins for Design B. Also very significant to theperformance comparison is the magnitude of the convection coefficient which is 3 times larger forDesign B. Estimating convection coefficients for fin arrays (and tube banks) is discussed in Chapter7.6. Of concern is how the fins alter the flow past the fins and whether the convection coefficient isuniform over the array.(3) The IHT Extended Surfaces Model, for a Rectangular Pin Fin Array could have been used to solvethis problem.PROBLEM 3.138KNOWN: Geometrical characteristics of a plate with pin fin array on both surfaces. Inner and outerconvection conditions.FIND: (a) Heat transfer rate with and without pin fin arrays, (b) Effect of using silver solder to join thepins and the plate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant k, (3) Negligible radiation.PROPERTIES: Table A-1: Copper, T 315 K, k = 400 W/mK.ANALYSIS: (a) The heat rate may be expressed asT,i T,oq=R t,o(c),i + R w + R t,o(c),owhere(R t,o(c) = o(c) hA to(c) = 1 )1 ,NA f f 1 ,A t C1 A t = NA f + A b ,A f = D p Lc D p ( L + D 4 ) ,()A b = W 2 NAc,b = W 2 N D 2 4 ,pf =tanh mLcmLc,(m = 4h kD p)1/ 2 ,Continued...PROBLEM 3.138 (Cont.)()C1 = 1 + f hA f R A c,b ,t,candRw =LwW2k.Calculations may be expedited by using the IHT Performance Calculation, Extended Surface Model forthe Pin Fin Array. For R ,c = 0, C1 = 1, and with W = 0.160 m, Rw = 0.005 m/(0.160 m)2 400 W/mK =t4.88 10-4 K/W. For the prescribed array geometry, we also obtain A c,b = 1.26 10-5 m2, A f = 2.64 10-4 m2, A b = 2.06 10-2 m2, and At = 0.126 m2.On the outer surface, where h o = 100 W/m2K, m = 15.8 m-1, f = 0.965, o = 0.970 and R t ,o = 0.0817K/W. On the inner surface, where h i = 5 W/m2K, m = 3.54 m-1, f = 0.998, o = 0.999 and R t ,o =1.588 K/W.Hence, the heat rate isq=(65 20 ) C(1.588 + 4.88 104)+ 0.0817 K W<= 26.94WWithout the fins,q=T,i T,o(65 20 ) C== 5.49W(1 h i A w ) + R w + (1 h o A w ) 7.81 + 4.88 104 + 0.39)(<Hence, the fin arrays provide nearly a five-fold increase in heat rate.(b) With use of the silver solder, o(c),o = 0.962 and R t,o(c),o = 0.0824 K/W. Also, o(c),i = 0.998and R t,o(c),i = 1.589 K/W. Henceq=((65 20 ) C)1.589 + 4.88 104 + 0.0824 K W= 26.92WHence, the effect of the contact resistance is negligible.COMMENTS: The dominant contribution to the total thermal resistance is associated with internalconditions. If the heat rate must be increased, it should be done by increasing hi.<PROBLEM 3.139KNOWN: Long rod with internal volumetric generation covered by an electrically insulating sleeve andsupported with a ribbed spider.FIND: Combination of convection coefficient, spider design, and sleeve thermal conductivity whichenhances volumetric heating subject to a maximum centerline temperature of 100C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial heat transfer in rod, sleeveand hub, (3) Negligible interfacial contact resistances, (4) Constant properties, (5) Adiabatic outersurface.ANALYSIS: The system heat rate per unit length may be expressed asT1 T2q = q ro =Rsleeve + R hub + R t,owhere()Rsleeve =ln ( r1 ro ), Rhub =2 k sNAf (1 ) ,o = 1 fAtf =tanh m ( r3 r2 )m ( r3 r2 ),ln ( r2 r1 )= 3.168 104 m K W , R t,o =2 k rA = 2 ( r3 r2 ) ,fm = ( 2h k r t )1/ 21,o hAtA = NA + ( 2 r3 Nt ) ,tf.The rod centerline temperature is related to T1 throughqr 2To = T ( 0 ) = T1 + o4kCalculations may be expedited by using the IHT Performance Calculation, Extended Surface Model forthe Straight Fin Array. For base case conditions of ks = 0.5 W/mK, h = 20 W/m2K, t = 4 mm and N =612, R sleeve = 0.0580 mK/W, R ,o = 0.0826 mK/W, f = 0.990, q = 387 W/m, and q = 1.23 10tW/m3. As shown below, q may be increased by increasing h, where h = 250 W/m2K represents areasonable upper limit for airflow. However, a more than 10-fold increase in h yields only a 63%increase in q .Continued...Heat generation, qdot(W/m^3)PROBLEM 3.139 (Cont.)2E61.8E61.6E61.4E61.2E61E6050100150200250Convection coefficient, h(W/m^2.K)t = 4 mm, N = 12, ks = 0.5 W/m.KThe difficulty is that, by significantly increasing h, the thermal resistance of the fin array is reduced to0.00727 mK/W, rendering the sleeve the dominant contributor to the total resistance.Heat generation, qdotx1E-6(W/mSimilar results are obtained when N and t are varied. For values of t = 2, 3 and 4 mm, variations of N inthe respective ranges 12 N 26, 12 N 21 and 12 N 17 were considered. The upper limit on Nwas fixed by requiring that (S - t) 2 mm to avoid an excessive resistance to airflow between the ribs.As shown below, the effect of increasing N is small, and there is little difference between results for thethree values of t.2.12.082.062.042.0221214161820222426Number of ribs, Nt = 2 mm, N: 12 - 26, h = 250 W/m^2.Kt = 3 mm, N: 12 - 21, h = 250 W/m^2.Kt = 4 mm, N: 12 -17, h = 250 W/m^2.KHeat generation, qdot(W/m^3)In contrast, significant improvement is associated with changing the sleeve material, and it is onlynecessary to have ks 25 W/mK (e.g. a boron sleeve) to approach an upper limit to the influence of ks.4E63.6E63.2E62.8E62.4E62E6020406080100Sleeve conductivity, ks(W/m.K)t = 4 mm, N = 12, h = 250 W/m^2.KFor h = 250 W/m2K and ks = 25 W/mK, only a slight improvement is obtained by increasing N. Hence,the recommended conditions are:h = 250 W m 2 K,k s = 25 W m K,N = 12,t = 4mm<COMMENTS: The upper limit to q is reached as the total thermal resistance approaches zero, in which2case T1 T. Hence q max = 4k ( To T ) ro = 4.5 106 W m3 .PROBLEM 3.140KNOWN: Geometrical and convection conditions of internally finned, concentric tube air heater.FIND: (a) Thermal circuit, (b) Heat rate per unit tube length, (c) Effect of changes in fin array.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in radial direction, (3)Constant k, (4) Adiabatic outer surface.ANALYSIS: (a) For the thermal circuit shown schematically,1Rconv,i = ( h i 2 r1 )whereo = 1 NAf 1 ,( f)At(b)q =,Rcond = ln ( r2 r1 ) 2 k , andA = 2L = 2 ( r3 r2 ) ,fAt = NA + ( 2 r2 Nt ) , andf(T,i T,o )Rconv,i + R cond + R t,oSubstituting the known conditions, it follows that(2Rconv,i = 5000 W m K 2 0.013m)11R = (o h o A )t,otf =,tanh mLmL.= 2.45 103 m K W3Rcond = ln (0.016m 0.013m ) 2 ( 20 W m K ) = 1.65 10 m K W(R = 0.575 200 W m 2 K 0.461mt,o)1= 18.86 10 3 m K Wwhere f = 0.490. Hence,(90 25 ) Cq =( 2.45 + 1.65 + 18.86 ) 103 m K= 2831W m<W(c) The small value of f suggests that some benefit may be gained by increasing t, as well as byincreasing N. With the requirement that Nt 50 mm, we use the IHT Performance Calculation,Extended Surface Model for the Straight Fin Array to consider the following range of conditions: t = 2mm, 12 N 25; t = 3 mm, 8 N 16; t = 4 mm, 6 N 12; t = 5 mm, 5 N 10. Calculations basedon the foregoing model are plotted as follows.Continued...PROBLEM 3.140 (Cont.)Heat rate, q'(w/m)5000400030002000510152025Number of fins, Nt = 2 mmt = 3 mmt = 4 mmt = 5 mmBy increasing t from 2 to 5 mm, f increases from 0.410 to 0.598. Hence, for fixed N, q increaseswith increasing t. However, from the standpoint of maximizing q , it is clearly preferable to use thetlarger number of thinner fins. Hence, subject to the prescribed constraint, we would choose t = 2 mmand N = 25, for which q = 4880 W/m.COMMENTS: (1) The air side resistance makes the dominant contribution to the total resistance, andefforts to increase q by reducing R ,o are well directed. (2) A fin thickness any smaller than 2 mmtwould be difficult to manufacture.PROBLEM 3.141KNOWN: Dimensions and number of rectangular aluminum fins. Convection coefficient with and without fins.FIND: Percentage increase in heat transfer resulting from use of fins.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4)Negligible radiation, (5) Negligible fin contact resistance, (6) Uniform convection coefficient.PROPERTIES: Table A-1, Aluminum, pure: k 240 W/mK.ANALYSIS: Evaluate the fin parametersLc = L+t/2 = 0.05025mA p = Lc t = 0.05025m 0.5 10-3m=25.13 10-6 m 21/ 2()= ( 0.05025m )()30 W/m 2 K 240 W/m K 25.13 10-6m 2 = 0.794L3/2 h w / kA pcL3/2 h w / kA pc1/ 21/ 23/ 2It follows from Fig. 3.18 that f 0.72. Hence,q f = f q max = 0.72 h w 2wL bqf = 0.72 30 W/m 2 K 2 0.05m ( w b ) = 2.16 W/m K ( w b )With the fins, the heat transfer from the walls isq w = N qf + (1 Nt ) w h w b)(W( w b ) + 1m 250 5 104 m 30 W/m2 K ( w b )mKWq w = (540 + 26.3)( w b ) = 566 w b .mKq w = 250 2.16Without the fins, qwo = hwo 1m w b = 40 w b. Hence the percentage increase in heattransfer isq w q wo (566 40 ) w b<== 13.15 = 1315%q wo40 w b1/2COMMENTS: If the infinite fin approximation is made, it follows that qf = (hPkAc)1/2-4 1/2=[hw2wkwt] b = (30 2 240 510 )overestimated.w b = 2.68 w b. Hence, qf isbPROBLEM 3.142KNOWN: Dimensions, base temperature and environmental conditions associated with rectangular andtriangular stainless steel fins.FIND: Efficiency, heat loss per unit width and effectiveness associated with each fin.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,(4) Negligible radiation, (5) Uniform convection coefficient.PROPERTIES: Table A-1, Stainless Steel 304 (T = 333 K): k = 15.3 W/mK.ANALYSIS: For the rectangular fin, with Lc = L + t/2, evaluate the parameter1/ 21/ 23/ 2 L3/ 2 h kA p= ( 0.023m )c(75 W m 2 K15.3 W m K ( 0.023m )( 0.006 m ) )= 0.66 .Hence, from Fig. 3.18, the fin efficiency is<f 0.79From Eq. 3.86, the fin heat rate is q f = f hAf b = f hPLc b = f h2wLc b or, per unit width,()qq = f = 0.79 75 W m 2 K 2 ( 0.023m ) 80 C = 218 W m .fw<From Eq. 3.81, the fin effectiveness isf =qfq w218 W mf=== 6.06 .2 K 0.006 m 80 ChA c,b b h ( t w ) b 75 W m()<For the triangular fin with1/ 21/ 23/ 2 L3/ 2 h kA p= ( 0.02 m )c()75 W m 2 K (15.3 W m K )( 0.020 m )( 0.003m ) = 0.81 ,find from Figure 3.18,<f 0.78 ,From Eq. 3.86 and Table 3.5 find1/ 22q = f hA b = f h2 L2 + ( t 2 ) ffb1/ 222q = 0.78 75 W m 2 K 2 ( 0.02 ) + ( 0.006 2 ) f()m 80 C = 187 W m .<and from Eq. 3.81, the fin effectiveness isf =q w187 W mf== 5.19h ( t w ) b 75 W m 2 K ( 0.006 m ) 80 CCOMMENTS: Although it is 14% less effective, the triangular fin offers a 50% weight savings.<PROBLEM 3.143KNOWN: Dimensions, base temperature and environmental conditions associated with a triangular,aluminum fin.FIND: (a) Fin efficiency and effectiveness, (b) Heat dissipation per unit width.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,(4) Negligible radiation and base contact resistance, (5) Uniform convection coefficient.PROPERTIES: Table A-1, Aluminum, pure (T 400 K): k = 240 W/mK.ANALYSIS: (a) With Lc = L = 0.006 m, findA p = Lt 2 = ( 0.006 m )( 0.002 m ) 2 = 6 106 m 2 ,1/ 23/ 2L3 / 2 h kA p= ( 0.006 m )c(1/ 2240 W m K 6 106 m 2 )40 W m2 K= 0.077and from Fig. 3.18, the fin efficiency is<f 0.99 .From Eq. 3.86 and Table 3.5, the fin heat rate is1/ 22q f = f q max = f hAf (tri) b = 2f hw L2 + ( t 2 ) b .From Eq. 3.81, the fin effectiveness is1/ 2f =qfhAc,b b=22f hw L2 + ( t 2 ) g ( w t ) bb1/ 2=22f L2 + ( t 2 ) t1/ 2f =222 0.99 ( 0.006 ) + ( 0.002 2 ) m0.002 m<= 6.02(b) The heat dissipation per unit width is1/ 22q = ( q f w ) = 2f h L2 + ( t 2 ) fb1/ 222q = 2 0.99 40 W m 2 K ( 0.006 ) + ( 0.002 2 ) fm ( 250 20 ) C = 110.8 W m .<COMMENTS: The triangular profile is known to provide the maximum heat dissipation per unit finmass.PROBLEM 3.144KNOWN: Dimensions and base temperature of an annular, aluminum fin of rectangular profile.Ambient air conditions.FIND: (a) Fin heat loss, (b) Heat loss per unit length of tube with 200 fins spaced at 5 mm increments.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,(4) Negligible radiation and contact resistance, (5) Uniform convection coefficient.PROPERTIES: Table A-1, Aluminum, pure (T 400 K): k = 240 W/mK.ANALYSIS: (a) The fin parameters for use with Figure 3.19 arer2c = r2 + t 2 = (12.5 mm + 10 mm ) + 0.5 mm = 23mm = 0.023mr2c r1 = 1.84Lc = L + t 2 = 10.5 mm = 0.0105 mA p = Lc t = 0.0105 m 0.001m = 1.05 105 m 21/ 23/ 2 L3/ 2 h kA p= (0.0105 m )c()1/ 25 2 240 W m K 1.05 10 m 25 W m 2 K= 0.15 .Hence, the fin effectiveness is f 0.97, and from Eq. 3.86 and Fig. 3.5, the fin heat rate is()22q f = f q max = f hAf (ann) b = 2f h r2,c r1 b22q f = 2 0.97 25 W m 2 K ( 0.023m ) ( 0.0125 m ) 225 C = 12.8 W .<(b) Recognizing that there are N = 200 fins per meter length of the tube, the total heat rate consideringcontributions due to the fin and base (unfinned surfaces isq = Nq f + h (1 Nt ) 2 r1 b()q = 200 m 1 12.8 W + 25 W m 2 K 1 200 m 1 0.001m 2 ( 0.0125 m ) 225 Cq = ( 2560 W + 353 W ) m = 2.91kW m .<COMMENTS: Note that, while covering only 20% of the tube surface area, the tubes account for morethan 85% of the total heat dissipation.PROBLEM 3.145KNOWN: Dimensions and base temperature of aluminum fins of rectangular profile. Ambient airconditions.FIND: (a) Fin efficiency and effectiveness, (b) Rate of heat transfer per unit length of tube.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction in fins, (3)Constant properties, (4) Negligible radiation, (5) Negligible base contact resistance, (6) Uniformconvection coefficient.PROPERTIES: Table A-1, Aluminum, pure (T 400 K): k = 240 W/mK.ANALYSIS: (a) The fin parameters for use with Figure 3.19 arer2c = r2 + t 2 = 40 mm + 2 mm = 0.042 mLc = L + t 2 = 15 mm + 2 mm = 0.017 mr2c r1 = 0.042 m 0.025 m = 1.68A p = Lc t = 0.017 m 0.004 m = 6.8 10 5 m 2(L3 / 2 h kA pc)1/ 2 = (0.017 m )3 / 2 40 W1/ 2m 2 K 240 W m K 6.8 105 m 2 = 0.11The fin efficiency is f 0.97. From Eq. 3.86 and Fig. 3.5,22q f = f q max = f hAf (ann) b = 2f h r2c r1 b22q f = 2 0.97 40 W m 2 K ( 0.042 ) ( 0.025 ) m 2 180 C = 50 W<From Eq. 3.81, the fin effectiveness isf =qfhAc,b b=50 W40 W m 2 K 2 ( 0.025 m )( 0.004 m )180 C= 11.05<(b) The rate of heat transfer per unit length isq = Nq f + h (1 Nt )( 2 r1 ) bq = 125 50 W m + 40 W m 2 K (1 125 0.004 )( 2 0.025 m ) 180 Cq = ( 6250 + 565 ) W m = 6.82 kW mCOMMENTS: Note the dominant contribution made by the fins to the total heat transfer.<PROBLEM 3.146KNOWN: Dimensions, base temperature, and contact resistance for an annular, aluminum fin. Ambientfluid conditions.FIND: Fin heat transfer with and without base contact resistance.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,(4) Negligible radiation, (5) Uniform convection coefficient.PROPERTIES: Table A-1, Aluminum, pure (T 350 K): k 240 W/mK.ANALYSIS: With the contact resistance, the fin heat loss is q f =Tw TR t,c + R fwhereR t,c = R A b = 2 104 m 2 K W 2 ( 0.015 m )( 0.002 m ) = 1.06 K W .t,cFrom Eqs. 3.83 and 3.86, the fin resistance isbb1.Rf = b ===q f f q max f hA f b 2 h r 2 r 2f 2,c 1(Evaluating parameters,r2,c = r2 + t 2 = 30 mm + 1mm = 0.031m(Lc = L + t 2 = 0.016 mA p = Lc t = 3.2 105 m 2r2c r1 = 0.031 0.015 = 2.07 ZL3 / 2 h kA pc))1/ 2 = (0.016 m )3 / 2 75 W1/ 2m 2 K 240 W m K 3.2 105 m 2 = 0.20find the fin efficiency from Figure 3.19 as f = 0.94. Hence,Rf =qf =()1222 75 W m K 0.94 ( 0.031m ) ( 0.015 m ) 2= 3.07 K W(100 25 ) C= 18.2 W .(1.06 + 3.07 ) K W<Without the contact resistance, Tw = Tb andqf =bRf=75 C3.07 K W= 24.4 W .COMMENTS: To maximize fin performance, every effort should be made to minimize contactresistance.<PROBLEM 3.147KNOWN: Dimensions and materials of a finned (annular) cylinder wall. Heat flux andambient air conditions. Contact resistance.FIND: Surface and interface temperatures (a) without and (b) with an interface contactresistance.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3)Uniform h over surfaces, (4) Negligible radiation.ANALYSIS: The analysis may be performed per unit length of cylinder or for a 4 mm longsection. The following calculations are based on a unit length. The inner surface temperaturemay be obtained fromT Tq = i = q ( 2 ri ) = 105 W/m 2 2 0.06 m = 37, 700 W/miR totwhere R tot = R + R t,c + R w + R cequiv ;1Requiv = (1/ R + 1/ R )fb.R , Conduction resistance of cylinder wall:cR =cln ( r1 / ri )ln ( 66/60 )=2 k2 (50 W/m K )R , Contact resistance:t,c= 3.034 104 m K/WR t,c = R / 2 r1 = 104 m 2 K/W/2 0.066 m = 2.411104 m K/Wt,cw , Conduction resistance of aluminum base:RR w =ln ( rb / r1 )2 k=ln ( 70/66 )2 240 W/m K= 3.902 105 m K/WR , Resistance of prime or unfinned surface:bR =b11== 454.7 104 m K/W2 K 0.5 2 0.07 mhA 100 W/mb()R , Resistance of fins: The fin resistance may be determined fromfT T1R = b =fqf hAffThe fin efficiency may be obtained from Fig. 3.19,r2c = ro + t/2 = 0.096 mLc = L+t/2 = 0.026 mContinued ..PROBLEM 3.147 (Cont.)A p = Lc t = 5.2 105 m 2r2c / r1 = 1.45(L3/2 h/kA pc)1/ 2= 0.375Fig. 3.19 f 0.88.The total fin surface area per meter length22A = 250 ro rb 2 = 250 m -1 2 0.0962 0.07 2 m 2 = 6.78 m.f)(Hence)(R = 0.88 100 W/m 2 K 6.78 m f1(= 16.8 104 m K/W)44 W/m K = 617.2 W/m K1/ R equiv = 1/16.8 10 + 1/ 454.7 104Requiv = 16.2 10 m K/W.Neglecting the contact resistance,R tot = (3.034 + 0.390 + 16.2 )104 m K/W = 19.6 104 m K/WTi = qR tot + T = 37, 700 W/m 19.6 10-4 m K/W+320 K = 393.9 K<T1 = Ti qR w = 393.9 K 37, 700 W/m 3.034 10-4 m K/W = 382.5 K<Tb = T1 qR = 382.5 K 37, 700 W/m 3.902 10-5 m K/W = 381.0 K. <bIncluding the contact resistance,()R tot = 19.6 104 + 2.411 104 m K/W = 22.0 104 m K/WTi = 37, 700 W/m 22.0 10-4 m K/W+320 K = 402.9 K<T1,i = 402.9 K 37, 700 W/m 3.034 10-4 m K/W = 391.5 K<T1,o = 391.5 K 37, 700 W/m 2.411 10-4 m K/W = 382.4 K<Tb = 382.4 K 37, 700 W/m 3.902 10-5 m K/W = 380.9 K.<COMMENTS: (1) The effect of the contact resistance is small.(2) The effect of including the aluminum fins may be determined by computing Ti without thefins. In this case R tot = R + R cconv , where11== 241.1 104 m K/W.Rconv =2 K 2 0.066 mh2 r1 100 W/m()Hence, R tot = 244.1104 m K/W, andTi = qR tot + T = 37, 700 W/m 244.110-4 m K/W+320 K = 1240 K.Hence, the fins have a significant effect on reducing the cylinder temperature.(3) The overall surface efficiency iso = 1 ( A / A )(1 f ) = 1 6.78 m/7.00 m (1 0.88 ) = 0.884.ftIt follows that q=o h o At b = 37, 700 W/m, which agrees with the prescribed value.PROBLEM 3.148KNOWN: Dimensions and materials of a finned (annular) cylinder wall. Combustion gas and ambientair conditions. Contact resistance.FIND: (a) Heat rate per unit length and surface and interface temperatures, (b) Effect of increasing thefin thickness.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Uniform hover surfaces, (4) Negligible radiation.ANALYSIS: (a) The heat rate per unit length isTg Tq =R totwhere R ot = R + R + R + R + R , andtgwt,cbt,o(R = h g 2 rigR =w)1 = (150 Wln ( r1 ri )ln (66 60 )=2 k wm 2 K 2 0.06m2 ( 50 W m K )())1= 0.0177m K W ,= 3.03 104 m K W ,R = R 2 r1 = 104 m 4 K W 2 0.066m = 2.41 104 m K Wt,ct,cR =bln ( rb r1 )=2 ko = 1 = 3.90 105 m K W ,2 240 W m K1R t,o = (o hA )tln ( 70 66 ),NA f(1 f ) ,At(22A f = 2 roc rb)At = NA f + (1 Nt ) 2 rbf =( 2rbm ) K1 ( mrb ) I1 ( mroc ) I1 ( mrb ) K1 ( mroc )2(roc rb2 ) I0 (mr1 ) K1 (mroc ) + K0 (mrb ) I1 (mroc )roc = ro + ( t 2 ) ,m = ( 2h kt )1/ 2Continued...PROBLEM 3.148 (Cont.)Once the heat rate is determined from the foregoing expressions, the desired interface temperatures maybe obtained fromTi = Tg qR g()T1,o = Tg q ( R + R + R t,c )gwTb = Tg q ( R + R + R + R )gwt,cbT1,i = Tg q R + R gwFor the specified conditions we obtain A = 7.00 m, f = 0.902, o = 0.906 and R ,o = 0.00158ttmK/W. It follows thatq = 39, 300 W mTi = 405K,T1,i = 393K,T1,o = 384K,Tb = 382K<<(b) The Performance Calculation, Extended Surface Model for the Circular Fin Array may be used toassess the effects of fin thickness and spacing. Increasing the fin thickness to t = 3 mm, with = 2 mm,reduces the number of fins per unit length to 200. Hence, although the fin efficiency increases ( Kf =0.930), the reduction in the total surface area ( A = 5.72 m) yields an increase in the resistance of the fintarray ( R ,o = 0.00188 mK/W), and hence a reduction in the heat rate ( q = 38,700 W/m) and an increasetin the interface temperatures ( Ti = 415 K, T1,i = 404 K, T1,o = 394 K, and Tb = 393 K).COMMENTS: Because the gas convection resistance exceeds all other resistances by at least an orderof magnitude, incremental changes in R t ,o will not have a significant effect on q or the interfacetemperatures.PROBLEM 3.149KNOWN: Dimensions of finned aluminum sleeve inserted over transistor. Contact resistance andconvection conditions.FIND: Measures for increasing heat dissipation.SCHEMATIC: See Example 3.10.ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from top and bottom oftransistor, (3) One-dimensional radial heat transfer, (4) Constant properties, (5) Negligible radiation.ANALYSIS: With 2 r2 = 0.0188 m and Nt = 0.0084 m, the existing gap between fins is extremely small(0.87 mm). Hence, by increasing N and/or t, it would become even more difficult to maintainsatisfactory airflow between the fins, and this option is not particularly attractive.Because the fin efficiency for the prescribed conditions is close to unity ( f = 0.998), there is littleadvantage to replacing the aluminum with a material of higher thermal conductivity (e.g. Cu with k ~ 400W/mK). However, the large value of f suggests that significant benefit could be gained by increasingthe fin length, L = r3 r2 .It is also evident that the thermal contact resistance is large, and from Table 3.2, its clear that asignificant reduction could be effected by using indium foil or a conducting grease in the contact zone.Specifically, a reduction of R ,c from 10-3 to 10-4 or even 10-5 m2K/W is certainly feasible.tTable 1.1 suggests that, by increasing the velocity of air flowing over the fins, a larger convectioncoefficient may be achieved. A value of h = 100 W/m2K would not be unreasonable.As options for enhancing heat transfer, we therefore use the IHT Performance Calculation, ExtendedSurface Model for the Straight Fin Array to explore the effect of parameter variations over the ranges 10 L 20 mm, 10-5 R ,c 10-3 m2K/W and 25 h 100 W/m2K. As shown below, there is atsignificant enhancement in heat transfer associated with reducing R ,c from 10-3 to 10-4 m2K/W, fortwhich R t ,c decreases from 13.26 to 1.326 K/W. At this value of R ,c , the reduction in R t ,o fromt23.45 to 12.57 K/W which accompanies an increase in L from 10 to 20 mm becomes significant, yieldinga heat rate of q t = 4.30 W for R ,c = 10-4 m2K/W and L = 20 mm. However, since R t ,o >> R t,c , littletbenefit is gained by further reducing R ,c to 10-5 m2K/W.tHeat rate, qt(W)5432100.010.0120.0140.0160.0180.02Fin length, L(m)h = 25 W/m^2.K, R''t,c = E-3 m^2.K/Wh = 25 W/m^2.K, R''t,c = E-4 m^2.K/Wh = 25 W/m^2.K, R''t,c = E-5m^2.K/WContinued...PROBLEM 3.149 (Cont.)To derive benefit from a reduction in R ,c to 10-5 m2K/W, an additional reduction in R t ,o must betmade. This can be achieved by increasing h, and for L = 20 mm and h = 100 W/m2K, R t ,o = 3.56 K/W.With R ,c = 10-5 m2K/W, a value of q t = 16.04 W may be achieved.tHeat rate, qt(W)2016128400.010.0120.0140.0160.0180.02Fin length, L(m)h = 25 W/m^2.K, R''t,c = E-5 m^2.K/Wh = 50 W/m^2.K, R''t,c = E-5 m^2.K/Wh = 100 W/m^2.K, R''t,c = E-5 m^2.K/WCOMMENTS: In assessing options for enhancing heat transfer, the limiting (largest) resistance(s)should be identified and efforts directed at their reduction.PROBLEM 3.150KNOWN: Diameter and internal fin configuration of copper tubes submerged in water. Tube walltemperature and temperature and convection coefficient of gas flow through the tube.FIND: Rate of heat transfer per tube length.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) One-dimensional fin conduction, (3) Constant properties, (4)Negligible radiation, (5) Uniform convection coefficient, (6) Tube wall may be unfolded and representedas a plane wall with four straight, rectangular fins, each with an adiabatic tip.ANALYSIS: The rate of heat transfer per unit tube length is:q = o hA Tg Tstt(o = 1 NAf 1 (ftA))NA = 4 2L = 8 ( 0.025m ) = 0.20mfA = NA + A = 0.20m + ( D 4t ) = 0.20m + ( 0.05m 4 0.005m ) = 0.337mtfbFor an adiabatic fin tip,qM tanh mLf = f =q max h ( 2L 1) Tg Ts(M = [h2 (1m + t ) k (1m t )]1/ 2mL = {[h2 (1m + t )])(Tg Ts ) 30 W(m K ( 2m ) 400 W m K 0.005m230 W m 2 K ( 2m )1/ 2[k (1m t )]} L 2 400 W m K 0.005m(2)1/ 2( 400K ) = 4382W1/ 2)0.025m = 0.137Hence, tanh mL = 0.136, and4382W ( 0.136 )595Wf === 0.99222600W30 W m K 0.05m ( 400K )(o = 1 0.200.337()(1 0.992 ) = 0.995)q = 0.995 30 W m 2 K 0.337m ( 400K ) = 4025 W mt()COMMENTS: Alternatively, q = 4q + h ( A A ) Tg Ts . Hence, q = 4(595 W/m) + 30tftf2W/m K (0.137 m)(400 K) = (2380 + 1644) W/m = 4024 W/m.PROBLEM 3.151KNOWN: Internal and external convection conditions for an internally finned tube. Fin/tubedimensions and contact resistance.FIND: Heat rate per unit tube length and corresponding effects of the contact resistance, number of fins,and fin/tube material.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constantproperties, (4) Negligible radiation, (5) Uniform convection coefficient on finned surfaces, (6) Tube wallmay be unfolded and approximated as a plane surface with N straight rectangular fins.PROPERTIES: Copper: k = 400 W/mK; St.St.: k = 20 W/mK.ANALYSIS: The heat rate per unit length may be expressed asTg Twq =Rt,o(c) + R cond + R conv,owhere()R t,o(c) = o(c) h g A ,tA = NA + ( 2 r1 Nt ) ,tfRcond =ln ( r2 r1 )2 k, ando(c) = 1 NAfAtA = 2r1 ,f f 1 , C1 ((f = tanh mr1 mr1 , m = 2h g kt1Rconv,o = ( 2 r2 h w ))C1 = 1 + f h g A R Aft,cc,b ,)1/ 2A ,b = t ,c.Using the IHT Performance Calculation, Extended Surface Model for the Straight Fin Array, thefollowing results were obtained. For the base case, q = 3857 W/m, where R ,o(c) = 0.101 mK/W,t-5Rcond = 7.25 10 mK/W and R onv,o = 0.00265 mK/W. If the contact resistance is eliminatedc( R = 0), q T = 3922 W/m, where R ,o = 0.0993 mK/W. If the number of fins is increased to N = 8,t,ctq = 5799 W/m, with R ,o(c) = 0.063 mK/W. If the material is changed to stainless steel, q = 3591tW/m, with R ,o(c) = 0.107 mK/W and R ond = 0.00145 mK/W.tcCOMMENTS: The small reduction in q T associated with use of stainless steel is perhaps surprising, inview of the large reduction in k. However, because h g is small, the reduction in k does not significantlyreduce the fin efficiency ( f changes from 0.994 to 0.891). Hence, the heat rate remains large. Theinfluence of k would become more pronounced with increasing h g .PROBLEM 3.152KNOWN: Design and operating conditions of a tubular, air/water heater.FIND: (a) Expressions for heat rate per unit length at inner and outer surfaces, (b) Expressions for innerand outer surface temperatures, (c) Surface heat rates and temperatures as a function of volumetricheating q for prescribed conditions. Upper limit to q .SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Constant properties, (3) One-dimensional heat transfer.PROPERTIES: Table A-1: Aluminum, T = 300 K, k a = 237 W/mK.ANALYSIS: (a) Applying Equation C.8 to the inner and outer surfaces, it follows that qr 2 r 2 o 1 i + ( Ts,o Ts,i )ln ( ro ri ) 4k s r 2 o22 k s qro ri2 2 ( ro ) = q ro q 1 + ( Ts,o Ts,i )ln ( ro ri ) 4k s r 2 o2 k sq ( ri ) = q ri2 <<(b) From Equations C.16 and C.17, energy balances at the inner and outer surfaces are of the form()h i T,i Ts,i =(qri)2Uo Ts,o T,o =qro2 qr 2 r 2 k s o 1 i + ( Ts,o Ts,i )2 4k s ro <ri ln ( ro ri ) qr 2 r 2 k s o 1 i + Ts,o Ts,i 4k s r 2 o(ro ln ( ro ri ))<Accounting for the fin array and the contact resistance, Equation 3.104 may be used to cast the overallheat transfer coefficient U o in the formUo =(q ( ro )A Ts,o T,ow)=1A= t o(c) h oA R w t,o(c) Awwhere o(c) is determined from Equations 3.105a,b and A = 2 ro .wContinued...PROBLEM 3.152 (Cont.)Surface temperatures, Ts(K)(c) For the prescribed conditions and a representative range of 107 q 108 W/m3, use of the relationsof part (b) with the capabilities of the IHT Performance Calculation Extended Surface Model for aCircular Fin Array yields the following graphical results.5004604203803403001E72E73E74E75E76E77E78E79E71E8Heat generation, qdot(W/m^3)Inner surface temperature, Ts,iOuter surface temperature, Ts,oIt is in this range that the upper limit of Ts,i = 373 K is exceeded for q = 4.9 107 W/m3, while thecorresponding value of Ts,o = 379 K is well below the prescribed upper limit. The expressions of partSurface heat rates, q'(W/m)(a) yield the following results for the surface heat rates, where heat transfer in the negative r directioncorresponds to q ( ri ) < 0.500003000010000-10000-30000-500001E72E73E74E75E76E77E78E79E71E8Heat generation, qdot(W/m^3)q'(ri)q'(ro)For q = 4.9 107 W/m3, q ( ri ) = -2.30 104 W/m and q ( ro ) = 1.93 104 W/m.COMMENTS: The foregoing design provides for comparable heat transfer to the air and water streams.This result is a consequence of the nearly equivalent thermal resistances associated with heat transfer1from the inner and outer surfaces. Specifically, R conv,i = ( h i 2 ri ) = 0.00318 mK/W is slightlysmaller than R t,o(c) = 0.00411 mK/W, in which case q ( ri ) is slightly larger than q ( ro ) , while Ts,iis slightly smaller than Ts,o . Note that the solution must satisfy the energy conservation requirement,2 ro ri2 q = q ( ri ) + q ( ro ) .()PROBLEM 4.1KNOWN: Method of separation of variables (Section 4.2) for two-dimensional, steady-state conduction.2FIND: Show that negative or zero values of , the separation constant, result in solutions whichcannot satisfy the boundary conditions.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.2ANALYSIS: From Section 4.2, identification of the separation constant leads to the two ordinarydifferential equations, 4.6 and 4.7, having the formsd 2Xd 2Y+ 2X = 0 2Y = 0(1,2)22dxdyand the temperature distribution is ( x,y ) = X ( x ) Y (y ).(3)2Consider now the situation when = 0. From Eqs. (1), (2), and (3), find thatX = C1 + C 2x,Y = C3 + C 4y and ( x,y ) = ( C1 + C 2x ) ( C3 + C 4y ) .(4)Evaluate the constants - C1, C2, C3 and C4 - by substitution of the boundary conditions:x = 0: ( 0,y) = ( C1 + C 2 0 ) ( C3 + C4 y ) = 0C1 = 0y =0: ( x,0) = ( 0 + C2 X)( C3 + C4 0 ) = 0C3 = 0x = L: ( L,0) = ( 0 + C2 L)( 0 + C4 y ) = 0C2 = 0y = W: ( x,W ) = ( 0 + 0 x )( 0 + C4 W ) = 1012The last boundary condition leads to an impossibility (0 1). We therefore conclude that a valueof zero will not result in a form of the temperature distribution which will satisfy the boundary2conditions. Consider now the situation when < 0. The solutions to Eqs. (1) and (2) will beandX = C5e- x + C6e + x ,Y = C7cos y + C8sin y ( x,y ) = C5 e- x + C6 e + x [ C7 cos y + C8 sin y ] .Evaluate the constants for the boundary conditions.y = 0 : ( x,0) = C5 e- x + C6 e- x [ C7 cos 0 + C8 sin 0 ] = 0 C e0 + C e 0 [ 0 + C sin y] = 0x = 0 : ( 0,y) = 568(5,6)(7)C7 = 0C8 = 0If C8 = 0, a trivial solution results or C5 = -C6.x = L: ( L,y ) = C5 e-xL e+xL C8 sin y = 0.From the last boundary condition, we require C5 or C8 is zero; either case leads to a trivial solutionwith either no x or y dependence.PROBLEM 4.2KNOWN: Two-dimensional rectangular plate subjected to prescribed uniform temperature boundaryconditions.FIND: Temperature at the mid-point using the exact solution considering the first five non-zero terms;assess error resulting from using only first three terms. Plot the temperature distributions T(x,0.5) andT(1,y).SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.ANALYSIS: From Section 4.2, the temperature distribution isn +1+ 1 n x sinh ( n y L )T T1 2 ( 1) ( x, y ) sin .(1,4.19)=T2 T1 n L sinh ( n W L )n =1Considering now the point (x,y) = (1.0,0.5) and recognizing x/L = 1/2, y/L = 1/4 and W/L = 1/2,n +1+ 1 n sinh ( n 4 )T T1 2 ( 1) (1, 0.5) sin .=T2 T1 n 2 sinh ( n 2 )n =1When n is even (2, 4, 6 ...), the corresponding term is zero; hence we need only consider n = 1, 3, 5, 7and 9 as the first five non-zero terms. (1, 0.5) =2 sinh ( 4 ) 2 3+ sin 2sin 2 sinh ( 2 ) 3 22 5sin 5 2 (1, 0.5) = sinh (5 4 ) 2 7+ sin sinh (5 2 ) 72 sinh (3 4 )+ sinh (3 2 ) sinh (7 4 ) 2 9+ sin sinh (7 2 ) 92 sinh (9 4 ) sinh (9 2 ) 2[0.755 0.063 + 0.008 0.001 + 0.000] = 0.445(2)T (1, 0.5) = (1, 0.5)( T2 T1 ) + T1 = 0.445 (150 50 ) + 50 = 94.5 C .<Using Eq. (1), and writing out the first fiveterms of the series, expressions for (x,0.5) orT(x,0.5) and (1,y) or T(1,y) were keyboardedinto the IHT workspace and evaluated forsweeps over the x or y variable. Note that forT(1,y), that as y 1, the upper boundary,T(1,1) is greater than 150C. Upon examinationof the magnitudes of terms, it becomes evidentthat more than 5 terms are required to providean accurate solution.T(x,0.5) or T(1,y), CIf only the first three terms of the series, Eq. (2), are considered, the result will be (1,0.5) = 0.46; that is,there is less than a 0.2% effect.15013011090705000.20.40.60.8x or y coordinate (m)T(1,y)T(x,0.5)1PROBLEM 4.3KNOWN: Temperature distribution in the two-dimensional rectangular plate of Problem 4.2.FIND: Expression for the heat rate per unit thickness from the lower surface (0 x 2, 0) and resultbased on first five non-zero terms of the infinite series.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.ANALYSIS: The heat rate per unit thickness from the plate along the lower surface isx =2x =2x =2Tq = dqy ( x, 0 ) = kdx = k ( T2 T1 ) dxout y y =0yx =0x =0x =0y =0(1)where from the solution to Problem 4.2,n +1+ 1 n x sinh (n y L )T T1 2 ( 1)sin .=T2 T1 n L sinh ( n W L )(2)n =1Evaluate the gradient of from Eq. (2) and substitute into Eq. (1) to obtainq = k ( T2 T1 )outx=2x =0n +1+ 1 n x ( n L ) cosh ( n y L )2 ( 1)sin nsinh ( n W L )Ln =1dxy =0n +12+12 ( 1)1 cos n x q = k ( T2 T1 ) outnsinh ( n W L ) L x =0 n =1n +1+12 ( 1)11 cos ( n )q = k ( T2 T1 ) outnsinh ( n L ) <n =1To evaluate the first five, non-zero terms, recognize that since cos(n) = 1 for n = 2, 4, 6 ..., only the nodd terms will be non-zero. Hence,Continued ..PROBLEM 4.3 (Cont.)q = 50 W m K (150 50 ) Cout+( 1)6 + 151sinh (5 2 )( 1) + 1 2 ( 1) + 111 (2)(2) +1sinh ( 2 )3sinh (3 2 )(2 ) +2( 1)8 + 1741sinh ( 7 2 )(2 ) +( 1)10 + 191sinh ( 9 2 )q = 3.183kW m [1.738 + 0.024 + 0.00062 + (...)] = 5.611kW mout( 2 )<COMMENTS: If the foregoing procedure were used to evaluate the heat rate into the upper surface,qin = x =2dqy ( x, W ) , it would follow thatx =0q = k ( T2 T1 )inn +1+12 ( 1)coth ( n 2 ) 1 cos ( n )nn =1However, with coth(n/2) 1, irrespective of the value of n, and with (1)n +1 + 1n =1n being adivergent series, the complete series does not converge and q . This physically untenableincondition results from the temperature discontinuities imposed at the upper left and right corners.PROBLEM 4.4KNOWN: Rectangular plate subjected to prescribed boundary conditions.FIND: Steady-state temperature distribution.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.ANALYSIS: The solution follows the method of Section 4.2. The product solution is(T ( x,y ) = X ( x ) Y ( y ) = ( C1cos x + C2 sin x ) C3e - y + C4e + y)and the boundary conditions are: T(0,y) = 0, T(a,y) = 0, T(x,0) = 0, T(x.b) = Ax. ApplyingBC#1, T(0,y) = 0, find C1 = 0. Applying BC#2, T(a,y) = 0, find that = n /a with n = 1,2,.Applying BC#3, T(x,0) = 0, find that C3 = -C4. Hence, the product solution isnT ( x,y ) = X ( x ) Y ( y ) = C2 C4 sin x e + y e- y .a Combining constants and using superposition, find(T ( x,y ) =) n x n y Cn sin sinh a .an =1To evaluate Cn, use orthogonal functions with Eq. 4.16 to findan x n b a2 n x Cn = Ax sin a dx/sinh a 0 sin a dx,0noting that y = b. The numerator, denominator and Cn, respectively, are:a22 a 2 n x ax cos n x = Aa cos n = Aa 1 n+1 ,A x sin dx = A sin[ ( ) ] n ( ) a n a 0an n 0an xa1a n b a 2 n x n b 1 2n x n b sinh 0 sin a dx = sinh a 2 x 4n sin a = 2 sinh a ,a 0Aa 2an bn b ( 1) n+1 / sinh = 2Aa ( 1) n+1 / n sinh a a .n2Hence, the temperature distribution is n y sinh 1 n+12 Aa()n x a .T ( x,y ) = n sin a n =1 sinh n b aCn =<PROBLEM 4.5KNOWN: Long furnace of refractory brick with prescribed surface temperatures and materialthermal conductivity.FIND: Shape factor and heat transfer rate per unit length using the flux plot methodSCHEMATIC:ASSUMPTIONS: (1) Furnace length normal to page, l, >> cross-sectional dimensions, (2) Twodimensional, steady-state conduction, (3) Constant properties.ANALYSIS: Considering the cross-section, the cross-hatched area represents a symmetricalelement. Hence, the heat rte for the entire furnace per unit length isq =qS= 4 k ( T1 T2 )ll(1)where S is the shape factor for the symmetrical section. Selecting three temperature increments ( N =3), construct the flux plot shown below.From Eq. 4.26,and from Eq. (1),S=MlNorS M 8.5=== 2.83lN3q = 4 2.83 1.2W( 600 60 )o C = 7.34 kW/m.m K<<COMMENTS: The shape factor can also be estimated from the relations of Table 4.1. Thesymmetrical section consists of two plane walls (horizontal and vertical) with an adjoining edge. Usingthe appropriate relations, the numerical values are, in the same order,S=0.75m0.5ml + 0.54l +l = 3.04l0.5m0.5mNote that this result compares favorably with the flux plot result of 2.83l.PROBLEM 4.6KNOWN: Hot pipe embedded eccentrically in a circular system having a prescribed thermalconductivity.FIND: The shape factor and heat transfer per unit length for the prescribed surface temperatures.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Length l >>diametrical dimensions.ANALYSIS: Considering the cross-sectional view of the pipe system, the symmetrical sectionshown above is readily identified. Selecting four temperature increments (N = 4), construct the fluxplot shown below.For the pipe system, the heat rate per unit length isq =qW= kS ( T1 T2 ) = 0.5 4.26 (150 35 )o C = 245 W/m.lm K<COMMENTS: Note that in the lower, right-hand quadrant of the flux plot, the curvilinear squaresare irregular. Further work is required to obtain an improved plot and, hence, obtain a moreaccurate estimate of the shape factor.PROBLEM 4.7KNOWN: Structural member with known thermal conductivity subjected to a temperature difference.FIND: (a) Temperature at a prescribed point P, (b) Heat transfer per unit length of the strut, (c) Sketchthe 25, 50 and 75C isotherms, and (d) Same analysis on the shape but with adiabatic-isothermalboundary conditions reversed.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constantproperties.ANALYSIS: (a) Using the methodology of Section 4.3.1, construct a flux plot. Note the line ofsymmetry which passes through the point P is an isotherm as shown above. It follows thatT ( P ) = ( T1 + T2 ) 2 = (100 + 0 ) C 2 = 50 C .<(b) The flux plot on the symmetrical section is now constructed to obtain the shape factor from which theheat rate is determined. That is, from Eq. 4.25 and 4.26,q = kS ( T1 T2 ) and S = M N .(1,2)From the plot of the symmetrical section,So = 4.2 4 = 1.05 .For the full section of the strut,M = M o = 4.2but N = 2No = 8. Hence,S = So 2 = 0.53and with q = q , givingq = 75 W m K 0.53 (100 0 ) C = 3975 W m .<(c) The isotherms for T = 50, 75 and 100C are shown on the flux plot. The T = 25C isotherm issymmetric with the T = 75C isotherm.(d) By reversing the adiabatic and isothermal boundary conditions, the two-dimensional shape appears asshown in the sketch below. The symmetrical element to be flux plotted is the same as for the strut,except the symmetry line is now an adiabat.Continued...PROBLEM 4.7 (Cont.)From the flux plot, find Mo = 3.4 and No = 4, and from Eq. (2)So = M o N o = 3.4 4 = 0.85S = 2So = 1.70and the heat rate per unit length from Eq. (1) isq = 75 W m K 1.70 (100 0 ) C = 12, 750 W m<From the flux plot, estimate thatT(P) 40C.<COMMENTS: (1) By inspection of the shapes for parts (a) and (b), it is obvious that the heat rate forthe latter will be greater. The calculations show the heat rate is greater by more than a factor of three.(2) By comparing the flux plots for the two configurations, and corresponding roles of the adiabats andisotherms, would you expect the shape factor for parts (a) to be the reciprocal of part (b)?PROBLEM 4.8KNOWN: Relative dimensions and surface thermal conditions of a V-grooved channel.FIND: Flux plot and shape factor.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constantproperties.ANALYSIS: With symmetry about the midplane, only one-half of the object need be considered asshown below.Choosing 6 temperature increments (N = 6), it follows from the plot that M 7. Hence from Eq.4.26, the shape factor for the half section isM7S = l = l = 1.17l.N6For the complete system, the shape factor is thenS = 2.34l.<PROBLEM 4.9KNOWN: Long conduit of inner circular cross section and outer surfaces of square cross section.FIND: Shape factor and heat rate for the two applications when outer surfaces are insulated ormaintained at a uniform temperature.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties and (3)Conduit is very long.ANALYSIS: The adiabatic symmetry lines for each of the applications is shown above. Using the fluxplot methodology and selecting four temperature increments (N = 4), the flux plots are as shown below.For the symmetrical sections, S = 2So, where So = M /N and the heat rate for each application is q =2(So/ )k(T1 - T2).ApplicationABM10.36.2N44So/ 2.581.55q (W/m)11,5886,975<<COMMENTS: (1) For application A, most of the heat lanes leave the inner surface (T1) on the upperportion.(2) For application B, most of the heat flow lanes leave the inner surface on the upper portion (that is,lanes 1-4). Because the lower, right-hand corner is insulated, the entire section experiences small heatflows (lane 6 + 0.2). Note the shapes of the isotherms near the right-hand, insulated boundary and thatthey intersect the boundary normally.PROBLEM 4.10KNOWN: Shape and surface conditions of a support column.FIND: (a) Heat transfer rate per unit length. (b) Height of a rectangular bar of equivalent thermalresistance.SCHEMATIC:ASSUMPTIONS: (1)Steady-state conditions, (2) Negligible three-dimensional conduction effects,(3) Constant properties, (4) Adiabatic sides.PROPERTIES: Table A-1, Steel, AISI 1010 (323K): k = 62.7 W/mK.ANALYSIS: (a) From the flux plot for thehalf section, M 5 and N 8. Hence for thefull sectionMl 1.25lNq = Sk ( T1 T2 )S= 2q 1.25 62.7W(100 0)o Cm K<q 7.8 kW/m.(b) The rectangular bar provides for one-dimensional heat transfer. Hence,q=kAHence,( T1 T2 )H(T T )= k ( 0.3l ) 1 2H()o0.3k ( T1 T2 ) 0.3m ( 62.7 W/m K ) 100 CH=== 0.24m.q7800 W/m<COMMENTS: The fact that H < 0.3m is consistent with the requirement that the thermalresistance of the trapezoidal column must be less than that of a rectangular bar of the same height andtop width (because the width of the trapezoidal column increases with increasing distance, x, fromthe top). Hence, if the rectangular bar is to be of equivalent resistance, it must be of smaller height.PROBLEM 4.11KNOWN: Hollow prismatic bars fabricated from plain carbon steel, 1m in length with prescribedtemperature difference.FIND: Shape factors and heat rate per unit length.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties.PROPERTIES: Table A-1, Steel, Plain Carbon (400K), k = 57 W/mK.ANALYSIS: Construct a flux plot on the symmetrical sections (shaded-regions) of each of thebars.The shape factors for the symmetrical sections are,So,A =Ml 4= l = 1lN4So,B =Ml 3.5=l = 0.88l.N4Since each of these sections is of the bar cross-section, it follows thatSA = 4 1l = 4lSB = 4 0.88l = 3.5l.<The heat rate per unit length is q = q/l = k ( S/l ) (T1 T2 ) ,qA = 57W 4 (500 300 ) K = 45.6 kW/mmK<qB = 57W 3.5 ( 500 300 ) K = 39.9 kW/m.m K<PROBLEM 4.12KNOWN: Two-dimensional, square shapes, 1 m to a side, maintained at uniform temperatures asprescribed, perfectly insulated elsewhere.FIND: Using the flux plot method, estimate the heat rate per unit length normal to the page if thethermal conductivity is 50 W/mKASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: Use the methodology of Section 4.3.1 to construct the flux plots to obtain the shape factorsfrom which the heat rates can be calculated. With Figure (a), begin at the lower-left side making theisotherms almost equally spaced, since the heat flow will only slightly spread toward the right. Startsketching the adiabats in the vicinity of the T2 surface. The dashed line represents the adiabat whichseparates the shape into two segments. Having recognized this feature, it was convenient to identifypartial heat lanes. Figure (b) is less difficult to analyze since the isotherm intervals are nearly regular inthe lower left-hand corner.The shape factors are calculated from Eq. 4.26 and the heat rate from Eq. 4.25.S =M 0.5 + 3 + 0.5 + 0.5 + 0.2=N6S =M 4.5== 0.90N5S = 0.70q = kS ( T1 T2 )q = kS ( T1 T2 )q = 10 W m K 0.70 (100 0 ) K = 700 W m q = 10 W m K 0.90 (100 0 ) K = 900 W m<COMMENTS: Using a finite-element package with a fine mesh, we determined heat rates of 956 and915 W/m, respectively, for Figures (a) and (b). The estimate for the less difficult Figure (b) is within2% of the numerical method result. For Figure (a), our flux plot result was 27% low.PROBLEM 4.13KNOWN: Uniform media of prescribed geometry.FIND: (a) Shape factor expressions from thermal resistance relations for the plane wall, cylindricalshell and spherical shell, (b) Shape factor expression for the isothermal sphere of diameter D buried inan infinite medium.ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform properties.ANALYSIS: (a) The relationship between the shape factor and thermal resistance of a shape followsfrom their definitions in terms of heat rates and overall temperature differences.q = kST( 4.25 ),q=TRt(3.19 ) ,S = 1/kR t(4.27)Using the thermal resistance relations developed in Chapter 3, their corresponding shape factors are:Rt =Plane wall:Rt =Cylindrical shell:ln ( r2 / r1 )S=2 LkLkA2 Llnr2 / r .1S=A.L<<(L into the page)Rt =Spherical shell:1 1 1 4 k r1 r2 S=4.l/r1 l/r2<(b) The shape factor for the sphere of diameter D in aninfinite medium can be derived using the alternativeconduction analysis of Section 3.1. For this situation, qr isa constant and Fouriers law has the formdTq r = k 4 r 2.() drSeparate variables, identify limits and integrate.T2qdr r D / 2= T dT14 kr2Dq r = 4 k ( T1 T 2 )2qr 1 q2= r 0 = ( T2 T1 ) r4 k D/24 k DorS = 2 D.<COMMENTS: Note that the result for the buried sphere, S = 2 D, can be obtained from theexpression for the spherical shell with r2 = . Also, the shape factor expression for the isothermalsphere buried in a semi-infinite medium presented in Table 4.1 provides the same result with z .PROBLEM 4.14KNOWN: Heat generation in a buried spherical container.FIND: (a) Outer surface temperature of the container, (b) Representative isotherms and heat flowlines.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Soil is a homogeneous medium with constantproperties.PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/mK.&ANALYSIS: (a) From an energy balance on the container, q = Eg and from the first entry in Table4.1,q=2 Dk ( T1 T 2 ).l D/4zHence,T1 = T2 +q 1 D/4z500W 1 2m/40m= 20o C+= 92.7 o CWk 2 D2 ( 2m )0.52m K(b) The isotherms may be viewed as spherical surfaces whose center moves downward withincreasing radius. The surface of the soil is an isotherm for which the center is at z = .<PROBLEM 4.15KNOWN: Temperature, diameter and burial depth of an insulated pipe.FIND: Heat loss per unit length of pipe.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through insulation,two-dimensional through soil, (3) Constant properties, (4) Negligible oil convection and pipe wallconduction resistances.PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/mK; Table A-3, Cellular glass (365K): k= 0.069 W/mK.ANALYSIS: The heat rate can be expressed asT Tq= 1 2R totwhere the thermal resistance is Rtot = Rins + Rsoil. From Eq. 3.28,R ins =ln ( D2 / D1 )=ln ( 0.7m/0.5m )2 Lk ins2 L 0.069 W/m KFrom Eq. 4.27 and Table 4.1,R soil ==0.776m K/W.Lcosh -1 ( 2z/D2 )cosh -1 ( 3/0.7 )10.653===m K/W.SKsoil2 Lksoil2 ( 0.52 W/m K ) LLHence,q=(120 0)o C1m K( 0.776 + 0.653)LWq = q/L = 84 W/m.= 84WLm<COMMENTS: (1) Contributions of the soil and insulation to the total resistance are approximatelythe same. The heat loss may be reduced by burying the pipe deeper or adding more insulation.(2) The convection resistance associated with the oil flow through the pipe may be significant, inwhich case the foregoing result would overestimate the heat loss. A calculation of this resistance maybe based on results presented in Chapter 8.(3) Since z > 3D/2, the shape factor for the soil can also be evaluated from S = 2 L/ ln (4z/D) ofTable 4.1, and an equivalent result is obtained.PROBLEM 4.16KNOWN: Operating conditions of a buried superconducting cable.FIND: Required cooling load.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Two-dimensionalconduction in soil, (4) One-dimensional conduction in insulation.ANALYSIS: The heat rate per unit length isq =q =Tg TnR + RIgTg Tn kg ( 2 /ln ( 4z/Do ) ) 1+ ln ( Do / Di ) / 2 kiwhere Tables 3.3 and 4.1 have been used to evaluate the insulation and ground resistances,respectively. Hence,q =( 300 77 ) K(1.2 W/m K ) ( 2 /ln ( 8/0.2 ) ) 223 Kq =( 0.489+22.064 ) m K/Wq = 9.9 W/m.1+ ln ( 2 ) / 2 0.005 W/m K<COMMENTS: The heat gain is small and the dominant contribution to the thermal resistance ismade by the insulation.PROBLEM 4.17KNOWN: Electrical heater of cylindrical shape inserted into a hole drilled normal to the surface ofa large block of material with prescribed thermal conductivity.FIND: Temperature reached when heater dissipates 50 W with the block at 25 C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Block approximates semi-infinite medium withconstant properties, (3) Negligible heat loss to surroundings above block surface, (4) Heater can beapproximated as isothermal at T1.ANALYSIS: The temperature of the heater surface follows from the rate equation written asT1 = T2 + q/kSwhere S can be estimated from the conduction shape factor given in Table 4.1 for a vertical cylinderin a semi-infinite medium,S = 2 L/l n ( 4L/D ) .Substituting numerical values, find4 0.1m S = 2 0.1m/l n 0.005m = 0.143m.The temperature of the heater is thenT1 = 25 C + 50 W/(5 W/mK 0.143m) = 94.9 C.<COMMENTS: (1) Note that the heater has L >> D, which is a requirement of the shape factorexpression.(2) Our calculation presumes there is negligible thermal contact resistance between the heater and themedium. In practice, this would not be the case unless a conducting paste were used.(3) Since L >> D, assumption (3) is reasonable.(4) This configuration has been used to determine the thermal conductivity of materials frommeasurement of q and T1.PROBLEM 4.18KNOWN: Surface temperatures of two parallel pipe lines buried in soil.FIND: Heat transfer per unit length between the pipe lines.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties, (4) Pipe lines are buried very deeply, approximating burial in an infinite medium, (5) Pipelength >> D1 or D2 and w > D1 or D2.ANALYSIS: The heat transfer rate per unit length from the hot pipe to the cool pipe isq =qS= k ( T1 T 2 ) .LLThe shape factor S for this configuration is given in Table 4.1 asS=2 L222-1 4w D1 D2 cosh2D1D2.Substituting numerical values,222S1 4 ( 0.5m ) ( 0.1m ) ( 0.075m ) = 2 /cosh -1(65.63)= 2 /coshL2 0.1m 0.075mS= 2 /4.88 = 1.29.LHence, the heat rate per unit length isoq = 1.29 0.5W/m K (175 5) C = 110 W/m.<COMMENTS: The heat gain to the cooler pipe line will be larger than 110 W/m if the soiltemperature is greater than 5 C. How would you estimate the heat gain if the soil were at 25 C?PROBLEM 4.19KNOWN: Tube embedded in the center plane of a concrete slab.FIND: (a) The shape factor and heat transfer rate per unit length using the appropriate tabulatedrelation, (b) Shape factor using flux plot method.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constantproperties, (4) Concrete slab infinitely long in horizontal plane, L >> z.PROPERTIES: Table A-3, Concrete, stone mix (300K): k = 1.4 W/mK.ANALYSIS: (a) If we relax the restriction that z >> D/2, the embedded tube-slab systemcorresponds to the fifth case of Table 4.1. Hence,S=2 Lln (8z/ D )where L is the length of the system normal to the page, z is the half-thickness of the slab and D is thediameter of the tube. Substituting numerical values, findS = 2 L/l n( 8 50mm/ 50mm ) = 6.72L.Hence, the heat rate per unit length isq =qSW= k ( T1 T2 ) = 6.72 1.4( 85 20 )o C = 612 W.LLm K(b) To find the shape factor using the flux plot method, first identify the symmetrical section boundedby the symmetry adiabats formed by the horizontal and vertical center lines. Selecting fourtemperature increments (N = 4), the flux plot can then be constructed.From Eq. 4.26, the shape factor of the symmetrical section isSo = ML/N = 6L/4 = 1.5L.For the tube-slab system, it follows that S = 4So = 6.0L, which compares favorably with the resultobtained from the shape factor relation.PROBLEM 4.20KNOWN: Dimensions and boundary temperatures of a steam pipe embedded in a concrete casing.FIND: Heat loss per unit length.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible steam side convection resistance,pipe wall resistance and contact resistance (T1 = 450K), (3) Constant properties.PROPERTIES: Table A-3, Concrete (300K): k = 1.4 W/mK.ANALYSIS: The heat rate can be expressed asq = SkT1-2 = Sk ( T1 T2 )From Table 4.1, the shape factor isS=2 L.1.08 w ln DHence, q 2 k ( T1 T2 )q = = L ln 1.08 w Dq =2 1.4W/m K ( 450 300 ) K= 1122 W/m.1.08 1.5m ln 0.5m <COMMENTS: Having neglected the steam side convection resistance, the pipe wall resistance,and the contact resistance, the foregoing result overestimates the actual heat loss.PROBLEM 4.21KNOWN: Thin-walled copper tube enclosed by an eccentric cylindrical shell; intervening space filledwith insulation.FIND: Heat loss per unit length of tube; compare result with that of a concentric tube-shellarrangement.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Thermal resistances ofcopper tube wall and outer shell wall are negligible, (4) Two-dimensional conduction in insulation.ANALYSIS: The heat loss per unit length written in terms of the shape factor S isq = k ( S/ l ) (T1 T2 ) and from Table 4.1 for this geometry, D2 + d 2 4z 2 S= 2 /cosh-1 .l2DdSubstituting numerical values, all dimensions in mm,222S-1 120 + 30 4 ( 20 ) = 2 /cosh-1 1.903 = 4.991.= 2 /cosh()l2 120 30Hence, the heat loss isoq = 0.05W/m K 4.991( 85 35 ) C = 12.5 W/m.<If the copper tube were concentric withthe shell, but all other conditions werethe same, the heat loss would beqc =2 k ( T1 T2 )ln ( D2 / D1 )using Eq. 3.27. Substituting numericalvalues,(W( 85 35 )o C/ln 120 / 30m Kqc = 11.3 W/m.qc = 2 0.05)COMMENTS: As expected, the heat loss with the eccentric arrangement is larger than that for theconcentric arrangement. The effect of the eccentricity is to increase the heat loss by (12.5 - 11.3)/11.3 11%.PROBLEM 4.22KNOWN: Cubical furnace, 350 mm external dimensions, with 50 mm thick walls.FIND: The heat loss, q(W).SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties.PROPERTIES: Table A-3, Fireclay brick ( T = (T1 + T2 ) / 2 = 610K ) : k 1.1 W/m K.ANALYSIS: Using relations for the shape factor from Table 4.1,A 0.25 0.25m 2== 1.25mL0.05mPlane Walls (6)SW =Edges (12)SE = 0.54D = 0.54 0.25m = 0.14mCorners (8)SC = 0.15L = 0.15 0.05m = 0.008m.The heat rate in terms of the shape factors isq = kS ( T1 T2 ) = k ( 6S W + 12SE + 8SC ) ( T1 T2 )Wq = 1.1( 6 1.25m+12 0.14m+0.15 0.008m ) ( 600 75 )o CmKq = 5.30 kW.COMMENTS: Note that the restrictions for SE and SC have been met.<PROBLEM 4.23KNOWN: Dimensions, thermal conductivity and inner surface temperature of furnace wall. Ambientconditions.FIND: Heat loss.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Uniform convection coefficient over entire outer surface ofcontainer.ANALYSIS: From the thermal circuit, the heat loss isq=Ts,i TR cond(2D) + R convwhere Rconv = 1/hAs,o = 1/6(hW2) = 1/6[5 W/m2K(5 m)2] = 0.00133 K/W. From Eq. (4.27), the twodimensional conduction resistance isR cond(2D) =1Skwhere the shape factor S must include the effects of conduction through the 8 corners, 12 edges and 6plane walls. Hence, using the relations for Cases 8 and 9 of Table 4.1,S = 8 (0.15 L ) + 12 0.54 ( W 2L ) + 6 As,i Lwhere As,i = (W - 2L)2. Hence,S = 8 (0.15 0.35 ) + 12 0.54 ( 4.30 ) + 6 (52.83) mS = ( 0.42 + 27.86 + 316.98 ) m = 345.26mand Rcond(2D) = 1/(345.26 m 1.4 W/mK) = 0.00207 K/W. Hence(1100 25) Cq=(0.00207 + 0.00133) KW= 316 kW<COMMENTS: The heat loss is extremely large and measures should be taken to insulate the furnace.PROBLEM 4.24KNOWN: Platen heated by passage of hot fluid in poor thermal contact with cover plates exposedto cooler ambient air.FIND: (a) Heat rate per unit thickness from each channel, q , (b) Surface temperature of coveriplate, Ts , (c) qi and Ts if lower surface is perfectly insulated, (d) Effect of changing centerlinespacing on qi and TsSCHEMATIC:D=15 mmLA=30 mmTi=150 CLo=60 mmLB=7.5 mm2hi=1000 W/m K2T=25 Cho=200 W/m KkA=20 W/mK kB=75 W/mKR = 2.0 10t,c42m K/WASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in platen, butone-dimensional in coverplate, (3) Temperature of interfaces between A and B is uniform, (4)Constant properties.ANALYSIS: (a) The heat rate per unit thickness from each channel can be determined from thefollowing thermal circuit representing the quarter section shown.The value for the shape factor is S = 1.06 as determined from the flux plot shown on the next page.Hence, the heat rate isqi = 4 ( Ti T ) / Rtot(1)R tot = [1/1000 W/m 2 K ( 0.015m/4 ) + 1/20 W/m K 1.06+ 2.0 104 m2 K/W ( 0.060m/2 ) + 0.0075m/75 W/m K ( 0.060m/2)+ 1/200 W/m2 K ( 0.060m/2 )]R tot = [0.085 + 0.047 + 0.0067 + 0.0033 + 0.1667 ] m K/WR tot = 0.309 m K/Wqi = 4 (150 25) K/0.309 m K/W = 1.62 kW/m.<(b) The surface temperature of the cover plate also follows from the thermal circuit asTs Tqi / 4 =1/ho ( Lo / 2 )Continued ..(2)PROBLEM 4.24 (Cont.)q11.62 kWTs = T + i= 25o C + 0.167 m K/W4 ho ( Lo / 2 )4Ts = 25o C + 67.6o C 93oC.<(c,d) The effect of the centerline spacing on qi and Ts can be understood by examining the relativemagnitudes of the thermal resistances. The dominant resistance is that due to the ambient airconvection process which is inversely related to the spacing Lo. Hence, from Eq. (1), the heat ratewill increase nearly linearly with an increase in Lo,qi ~11~ L o.Rtot 1 / h o ( Lo / 2 )From Eq. (2), findq1oT = T T = i~ qi L-1 ~ Lo L-1 1.so4 h o ( Lo / 2 )Hence we conclude that T will not increase with a change in Lo. Does this seem reasonable?What effect does Lo have on Assumptions (2) and (3)?If the lower surface were insulated, the heat rate would be decreased nearly by half. This followsagain from the fact that the overall resistance is dominated by the surface convection process. Thetemperature difference, Ts - T, would only increase slightly.PROBLEM 4.25KNOWN: Long constantan wire butt-welded to a large copper block forming a thermocouple junctionon the surface of the block.FIND: (a) The measurement error (Tj - To) for the thermocouple for prescribed conditions, and (b)Compute and plot (Tj - To) for h = 5, 10 and 25 W/m2K for block thermal conductivity 15 k 400W/mK. When is it advantageous to use smaller diameter wire?SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Thermocouple wire behaves as a fin with constantheat transfer coefficient, (3) Copper block has uniform temperature, except in the vicinity of the junction.PROPERTIES: Table A-1, Copper (pure, 400 K), kb = 393 W/mK; Constantan (350 K), kt 25W/mK.ANALYSIS: The thermocouple wire behaves as a long fin permitting heat to flow from the surfacethereby depressing the sensing junction temperature below that of the block To. In the block, heat flowsinto the circular region of the wire-block interface; the thermal resistance to heat flow within the block isapproximated as a disk of diameter D on a semi-infinite medium (kb, To). The thermocouple-blockcombination can be represented by a thermal circuit as shown above. The thermal resistance of the finfollows from the heat rate expression for an infinite fin, Rfin = (hPktAc)-1/2.From Table 4.1, the shape factor for the disk-on-a-semi-infinite medium is given as S = 2D and henceRblock = 1/kbS = 1/2kbD. From the thermal circuit,R block1.27To Tj =(To T ) =(125 25 ) C 0.001(125 25 ) C = 0.1 C .R fin + R block1273 + 1.27<with P = D and Ac = D2/4 and the thermal resistances as(R fin = 10 W m 2 K ( 4 ) 25 W m K 1 103 m) 31/ 2= 1273 K WR block = (1 2 ) 393 W m K 103 m = 1.27 K W .(b) We keyed the above equations into the IHT workspace, performed a sweep on kb for selected valuesof h and created the plot shown. When the block thermal conductivity is low, the error (To - Tj) is larger,increasing with increasing convection coefficient. A smaller diameter wire will be advantageous for lowvalues of kb and higher values of h.5Error, To-Tj (C)432100100200300Block thermal conductivity, kb (W/m.K)h = 25 W/m^2.K; D = 1 mmh = 10 W/m^2.K; D = 1mmh = 5 W/m^2.K; D = 1mmh = 25 W/m^2.K; D = 5 mm400PROBLEM 4.26KNOWN: Dimensions, shape factor, and thermal conductivity of square rod with drilled interior hole.Interior and exterior convection conditions.FIND: Heat rate and surface temperatures.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Uniformconvection coefficients at inner and outer surfaces.ANALYSIS: The heat loss can be expressed asq=T,1 T,2R conv,1 + R cond(2D) + R conv,2where1R conv,1 = ( h1 D1L )1R cond(2D) = (Sk )(= 50 W m 2 K 0.25 m 2 m1= (8.59 m 150 W m K )1R conv,2 = ( h 2 4wL )()1= 0.01273K W= 0.00078 K W= 4 W m 2 K 4 m 1m)1= 0.0625 K WHence,(300 25 ) C = 3.62 kWq=0.076 K WT1 = T,1 qR conv,1 = 300 C 46 C = 254 CT2 = T,2 + qR conv,2 = 25 C + 226 C = 251 C<<<COMMENTS: The largest resistance is associated with convection at the outer surface, and theconduction resistance is much smaller than both convection resistances. Hence, (T2 - T,2) > (T,1 - T1)>> (T1 - T2).PROBLEM 4.27KNOWN: Long fin of aluminum alloy with prescribed convection coefficient attached to different basematerials (aluminum alloy or stainless steel) with and without thermal contact resistance R j at thet,junction.FIND: (a) Heat rate qf and junction temperature Tj for base materials of aluminum and stainless steel,(b) Repeat calculations considering thermal contact resistance, R j , and (c) Plot as a function of h fort,the range 10 h 1000 W/m2K for each base material.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Infinite fin.PROPERTIES: (Given) Aluminum alloy, k = 240 W/mK, Stainless steel, k = 15 W/mK.ANALYSIS: (a,b) From the thermal circuits, the heat rate and junction temperature areT TTb Tqf = b =R totR b + R t, j + R f(1)Tj = T + qf R f(2)and, with P = D and Ac = D2/4, from Tables 4.1 and 3.4 find1R b = 1 Sk b = 1 ( 2Dk b ) = ( 2 0.005 m k b )R t, j = R j A c = 3 105 m 2 K W ( 0.005 m )t,21/ 2R f = ( hPkA c )4 = 1.528 K W2= 50 W m 2 K 2 (0.005 m ) 240 W m K 4 1/ 2Without R jt,BaseAl alloySt. steelRb (K/W)0.4176.667qf (W)4.463.26Tj (C)98.278.4= 16.4 K WWith R jt,qf (W)4.093.05Tj (C)92.175.1(c) We used the IHT Model for Extended Surfaces, Performance Calculations, Rectangular Pin Fin tocalculate qf for 10 h 100 W/m2K by replacing R (thermal resistance at fin base) by the sum of thetccontact and spreading resistances, R j + R .t,bContinued...PROBLEM 4.27 (Cont.)Fin heat rate, qf (W)654321020406080100Convection coefficient, h (W/m^2.K)Base material - aluminum alloyBase material - stainless steelCOMMENTS: (1) From part (a), the aluminum alloy base material has negligible effect on the fin heatrate and depresses the base temperature by only 2C. The effect of the stainless steel base material issubstantial, reducing the heat rate by 27% and depressing the junction temperature by 25C.(2) The contact resistance reduces the heat rate and increases the temperature depression relatively morewith the aluminum alloy base.(3) From the plot of qf vs. h, note that at low values of h, the heat rates are nearly the same for bothmaterials since the fin is the dominant resistance. As h increases, the effect of R becomes morebimportant.PROBLEM 4.28KNOWN: Igloo constructed in hemispheric shape sits on ice cap; igloo wall thickness andinside/outside convection coefficients (hi, ho) are prescribed.FIND: (a) Inside air temperature T,i when outside air temperature is T,o = -40C assuming occupantsprovide 320 W within igloo, (b) Perform parameter sensitivity analysis to determine which variables havesignificant effect on Ti.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Convection coefficient is the same on floor andceiling of igloo, (3) Floor and ceiling are at uniform temperature, (4) Floor-ice cap resembles disk onsemi-infinite medium, (5) One-dimensional conduction through igloo walls.PROPERTIES: Ice and compacted snow (given): k = 0.15 W/mK.ANALYSIS: (a) The thermal circuit representing the heat loss from the igloo to the outside air andthrough the floor to the ice cap is shown above. The heat loss isT,i T,oT,i Ticq=.+R cv,c + R wall + R cv,o R cv,f + R cap2Convection, ceiling:R cv,c =Convection, outside:R cv,o =Convection, floor:R cv,f =Conduction, wall:Conduction, ice cap:R cap ==2R wall = 2 ()h i 4 ri22()2h o 4 ro1()h i ri2==2215 W m 2 K 4 ( 2.3 m )216 W m 2 K (1.8 m )2= 0.00201K W= 0.01637 K W11 1 1 1 1 r r = 2 4 0.15 W m K 1.8 2.3 m = 0.1281K W 4 k i o 1=11=kS 4kri 4 0.15 W m K 1.8 mwhere S was determined from the shape factor of Table 4.1. Hence,T,i ( 40 ) Cq = 320 W == 0.00819 K W6 W m K 4 (1.8 m )2( 0.00818 + 0.1281 + 0.0020 ) K= 0.9259 K WT,i ( 20 ) CW320 W = 7.232( T,i + 40) + 1.06( T,i + 20)+( 0.01637 + 0.9259 ) KT,i = 1.1C.W<Continued...PROBLEM 4.28 (Cont.)(b) Begin the parameter sensitivity analysis to determine important variables which have a significantinfluence on the inside air temperature by examining the thermal resistances associated with theprocesses present in the system and represented by the network.ProcessConvection, outsideConduction, wallConvection, ceilingConvection, floorConduction, ice capSymbolsRcv,oRwallRcv,cRcv,fRcapValue (K/W)0.00200.12810.00820.01640.9259R21R32R43R54R65It follows that the convection resistances are negligible relative to the conduction resistance across theigloo wall. As such, only changes to the wall thickness will have an appreciable effect on the inside airtemperature relative to the outside ambient air conditions. We dont want to make the igloo walls thinnerand thereby allow the air temperature to dip below freezing for the prescribed environmental conditions.Using the IHT Thermal Resistance Network Model, we used the circuit builder to construct the networkand perform the energy balances to obtain the inside air temperature as a function of the outsideconvection coefficient for selected increased thicknesses of the wall.Air temperature, Tinfi (C)2520151050020406080100Outside coefficient, ho (W/m^2.K)Wall thickness, (ro-ri) = 0.5 m(ro-ri) = 0.75 m(ro-ri) = 1.0 mCOMMENTS: (1) From the plot, we can see that the influence of the outside air velocity whichcontrols the outside convection coefficient ho is negligible.(2) The thickness of the igloo wall is the dominant thermal resistance controlling the inside airtemperature.PROBLEM 4.29KNOWN: Diameter and maximum allowable temperature of an electronic component. Contactresistance between component and large aluminum heat sink. Temperature of heat sink andconvection conditions at exposed component surface.FIND: (a) Thermal circuit, (b) Maximum operating power of component.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat lossfrom sides of chip.ANALYSIS: (a) The thermal circuit is:where R2D,cond is evaluated from the shape factor S = 2D of Table 4.1.(b) Performing an energy balance for a control surface about the component,Tc TbP = q air + q sink = h D2 / 4 ( Tc T ) +R / D2 / 4 + 1/2Dkt,c()P = 25 W/m 2 K ( /4 )( 0.01 m ) 2 75o C +P = 0.15 W +()75o C 0.5 10 -4 / ( / 4 )( 0.01)2 + ( 0.02 237 )1 K/W{75o C= 0.15 W + 88.49 W=88.6 W.( 0.64+0.21) K/W}<COMMENTS: The convection resistance is much larger than the cumulative contact andconduction resistance. Hence, virtually all of the heat dissipated in the component is transferredthrough the block. The two-dimensional conduction resistance is significantly underestimated by useof the shape factor S = 2D. Hence, the maximum allowable power is less than 88.6 W.PROBLEM 4.30KNOWN: Disc-shaped electronic devices dissipating 100 W mounted to aluminum alloy block withprescribed contact resistance.FIND: (a) Temperature device will reach when block is at 27C assuming all the power generated by thedevice is transferred by conduction to the block and (b) For the operating temperature found in part (a),the permissible operating power with a 30-pin fin heat sink.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Device is at uniform temperature,T1, (3) Block behaves as semi-infinite medium.PROPERTIES: Table A.1, Aluminum alloy 2024 (300 K): k = 177 W/mK.ANALYSIS: (a) The thermal circuit for the conduction heat flow between the device and the blockshown in the above Schematic where Re is the thermal contact resistance due to the epoxy-filledinterface,R e = R A c = R t,ct,c( D2 4)(R e = 5 105 K m 2 W ( 0.020 m )2) 4 = 0.159 K WThe thermal resistance between the device and the block is given in terms of the conduction shape factor,Table 4.1, asR b = 1 Sk = 1 ( 2Dk )R b = 1 ( 2 0.020 m 177 W m K ) = 0.141K WFrom the thermal circuit,T1 = T2 + qd ( R b + R e )T1 = 27 C + 100 W ( 0.141 + 0.159 ) K WT1 = 27 C + 30 C = 57 C<(b) The schematic below shows the device with the 30-pin fin heat sink with fins and base material ofcopper (k = 400 W/mK). The airstream temperature is 27C and the convection coefficient is 1000W/m2K.Continued...PROBLEM 4.30 (Cont.)The thermal circuit for this system has two paths for the device power: to the block by conduction, qcd,and to the ambient air by conduction to the fin array, qcv,qd =T1 T2T1 T+R b + R e R e + R c + R fin(3)where the thermal resistance of the fin base material isRc =Lc0.005 m== 0.03979 K Wk c Ac 400 W m K 0.022 4 m 2)((4)and Rfin represents the thermal resistance of the fin array (see Section 3.6.5),1o hA t(5, 3.103)NAf(1 f )At(6, 3.102)R fin = R t,o =o = 1 where the fin and prime surface area isA t = NAf + A b(3.99)()22A t = N ( Df L ) + Dd 4 N Df 4 where Af is the fin surface area, Dd is the device diameter and Df is the fin diameter.22A t = 30 ( 0.0015 m 0.015 m ) + (0.020 m ) 4 30 (0.0015 m ) 4 ()At = 0.06362 m2 + 0.0002611 m2 = 0.06388 m2Using the IHT Model, Extended Surfaces, Performance Calculations, Rectangular Pin Fin, find the finefficiency asf = 0.8546(7)Continued...PROBLEM 4.30 (Cont.)Substituting numerical values into Eq. (6), findo = 1 30 0.0015 m 0.015 m0.06388 m 2(1 0.8546 )o = 0.8552and the fin array thermal resistance isR fin =10.8552 1000 W m 2 K 0.06388 m 2= 0.01831K WReturning to Eq. (3), with T1 = 57C from part (a), the permissible heat rate isqd =(57 27 ) C(0.141 + 0.159 ) KW+(57 27 ) C(0.159 + 0.03979 + 0.01831) KWqd = 100 W + 138.2 W = 238 W<COMMENTS: (1) Recognize in the part (b) analysis, that thermal resistances of the fin base and arrayare much smaller than the resistance due to the epoxy contact interfaces.(2) In calculating the fin efficiency, f, using the IHT Model it is not necessary to know the basetemperature as f depends only upon geometric parameters, thermal conductivity and the convectioncoefficient.PROBLEM 4.31KNOWN: Dimensions and surface temperatures of a square channel. Number of chips mounted onouter surface and chip thermal contact resistance.FIND: Heat dissipation per chip and chip temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady state, (2) Approximately uniform channel inner and outer surfacetemperatures, (3) Two-dimensional conduction through channel wall (negligible end-wall effects), (4)Constant thermal conductivity.ANALYSIS: The total heat rate is determined by the two-dimensional conduction resistance of thechannel wall, q = (T2 T1)/Rt,cond(2D), with the resistance determined by using Eq. 4.27 with Case 11of Table 4.1. For W/w = 1.6 > 1.4R t,cond(2D) =0.930 ln ( W / w ) 0.0502 L k=0.3872 ( 0.160m ) 240 W / m K= 0.00160 K / WThe heat rate per chip is thenqc =(50 20 ) C = 156.3 WT2 T1=N R t,cond ( 2D ) 120 ( 0.0016 K / W )<and, with qc = (Tc T2)/Rt,c, the chip temperature isTc = T2 + R t,c qc = 50C + ( 0.2 K / W )156.3 W = 81.3C<COMMENTS: (1) By acting to spread heat flow lines away from a chip, the channel wall providesan excellent heat sink for dissipating heat generated by the chip. However, recognize that, in practice,there will be temperature variations on the inner and outer surfaces of the channel, and if theprescribed values of T1 and T2 represent minimum and maximum inner and outer surfacetemperatures, respectively, the rate is overestimated by the foregoing analysis. (2) The shape factormay also be determined by combining the expression for a plane wall with the result of Case 8 (Table4.1). With S = [4(wL)/(W-w)/2] + 4(0.54 L) = 2.479 m, Rt,cond(2D) = 1/(Sk) = 0.00168 K/W.PROBLEM 4.32KNOWN: Dimensions and thermal conductivity of concrete duct. Convection conditions of ambientair. Inlet temperature of water flow through the duct.FIND: (a) Heat loss per duct length near inlet, (b) Minimum allowable flow rate corresponding tomaximum allowable temperature rise of water.SCHEMATIC:ASSUMPTIONS: (1) Steady state, (2) Negligible water-side convection resistance, pipe wallconduction resistance, and pipe/concrete contact resistance (temperature at inner surface of concretecorresponds to that of water), (3) Constant properties, (4) Negligible flow work and kinetic andpotential energy changes.ANALYSIS: (a) From the thermal circuit, the heat loss per unit length near the entrance isq =Ti TTi T=R1conv ln (1.08 w / D ) +cond ( 2D ) + R 2 kh ( 4w )where R cond ( 2D ) is obtained by using the shape factor of Case 6 from Table 4.1 with Eq. (4.27).Hence,q =90C(90 0 ) C=ln (1.08 0.3m / 0.15m )1(0.0876 + 0.0333 ) K m / W+22 (1.4 W / m K )25 W / m K (1.2m )= 745 W / m<(b) From Eq. (1.11e), with q = qL and ( Ti To ) = 5C,m=745 W / m (100m )qLqL=== 3.54 kg / su i u o c (Ti To ) 4207 J / kg K (5C )<COMMENTS: The small reduction in the temperature of the water as it flows from inlet to outletinduces a slight departure from two-dimensional conditions and a small reduction in the heat rate perunit length. A slightly conservative value (upper estimate) of m is therefore obtained in part (b).PROBLEM 4.33KNOWN: Dimensions and thermal conductivities of a heater and a finned sleeve. Convectionconditions on the sleeve surface.FIND: (a) Heat rate per unit length, (b) Generation rate and centerline temperature of heater, (c)Effect of fin parameters on heat rate.SCHEMATIC:ASSUMPTIONS: (1) Steady state, (2) Constant properties, (3) Negligible contact resistancebetween heater and sleeve, (4) Uniform convection coefficient at outer surfaces of sleeve, (5) Uniformheat generation, (6) Negligible radiation.ANALYSIS: (a) From the thermal circuit, the desired heat rate isq =Ts TT T=s RR tott,ocond ( 2D ) + R The two-dimensional conduction resistance, may be estimated from Eq. (4.27) and Case 6 of Table4.2Rcond ( 2D ) =ln (1.08w / D )ln ( 2.16 )1=== 5.11 104 m K / WSk s2 k s2 ( 240 W / m K )The thermal resistance of the fin array is given by Eq. (3.103), where o and At are given by Eqs.1/2(3.102) and (3.99) and f is given by Eq. (3.89). With Lc = L + t/2 = 0.022 m, m = (2h/kst) = 32.3-1m and mLc = 0.710,f =tanh mLc 0.61== 0.86mLc0.71A = NA + A = N ( 2L + t ) + ( 4w Nt ) = 0.704m + 0.096m = 0.800mtfbo = 1 NAf 1 = 1 0.704m 0.14 = 0.88( f)()At0.800m1R t,o = (o h At )q =((= 0.88 500 W / m 2 K 0.80m(300 50 ) C)5.11 104 + 2.84 103 m K / W)1= 2.84 103 m K / W= 74, 600 W / m(b) Eq. (3.55) may be used to determine q, if h is replaced by an overall coefficient based on thesurface area of the heater. From Eq. (3.32),<PROBLEM 4.33 (Cont.)1Us A = Us D = ( R )stot1= (3.35 m K / W )= 298 W / m KUs = 298 W / m K / ( 0.02m ) = 4750 W / m 2 K)(q = 4 Us ( Ts T ) / D = 4 4750 W / m 2 K ( 250C ) / 0.02m = 2.38 108 W / m3<From Eq. (3.53) the centerline temperature isT (0 ) =2q (D / 2 )4 kh2.38 108 W / m3 (0.01m )2+ Ts =4 ( 400 W / m K )+ 300C = 315C<(c) Subject to the prescribed constraints, the following results have been obtained for parametervariations corresponding to 16 N 40, 2 t 8 mm and 20 L 40 mm.N161628324040t(mm)484322L(mm)fq ( W / m )2020202020400.860.910.860.830.780.5174,40077,000107,900115,200127,800151,300Clearly there is little benefit to simply increasing t, since there is no change in A and only atmarginal increase in f . However, due to an attendant increase in A , there is significant benefit totincreasing N for fixed t (no change in f ) and additional benefit in concurrently increasing N whiledecreasing t. In this case the effect of increasing A exceeds that of decreasing f . The same isttrue for increasing L, although there is an upper limit at which diminishing returns would be reached.The upper limit to L could also be influenced by manufacturing constraints.COMMENTS: Without the sleeve, the heat rate would be q = Dh ( Ts T ) = 7850 W / m,which is well below that achieved by using the increased surface area afforded by the sleeve.PROBLEM 4.34KNOWN: Dimensions of chip array. Conductivity of substrate. Convection conditions. Contactresistance. Expression for resistance of spreader plate. Maximum chip temperature.FIND: Maximum chip heat rate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Constant thermal conductivity, (3) Negligible radiation, (4)All heat transfer is by convection from the chip and the substrate surface (negligible heat transferfrom bottom or sides of substrate).ANALYSIS: From the thermal circuit,T TTh Tq = q h + qsp = h+R h,cnv R t (sp ) + R t,c + R sp,cnv(R h,cnv = h As,nR t (sp ) =1)()= hL2h12= 100 W / m 2 K ( 0.005m ) 1 1.410 A r + 0.344 A3 + 0.043 A5 + 0.034 A7rrr4 k sub LhR t,c =R t,cL2h=(0.5 104 m 2 K / W(0.005m )2)R sp,cnv = h Asub As,h q=1=1= 400 K / W1 0.353 + 0.005 + 0 + 04 (80 W / m K )( 0.005m )= 2.000 K / W()= 100 W / m 2 K 0.010m 2 0.005m 2 = 0.408 K / W170C70C+= 0.18 W + 0.52 W = 0.70 W400 K / W ( 0.408 + 2 + 133.3) K / W= 133.3 K / W<COMMENTS: (1) The thermal resistances of the substrate and the chip/substrate interface are muchless than the substrate convection resistance. Hence, the heat rate is increased almost in proportion tothe additional surface area afforded by the substrate. An increase in the spacing between chips (Sh)would increase q correspondingly.(2) In the limit A r 0, R t (sp ) reduces to 2 1/ 2 k sub D h for a circular heat source and 4k sub L hfor a square source.PROBLEM 4.35KNOWN: Internal corner of a two-dimensional system with prescribed convection boundaryconditions.FIND: Finite-difference equations for these situations: (a) Horizontal boundary is perfectly insulatedand vertical boundary is subjected to a convection process (T,h), (b) Both boundaries are perfectlyinsulated; compare result with Eq. 4.45.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties, (4) No internal generation.ANALYSIS: Consider the nodal network shown above and also as Case 2, Table 4.2. Havingdefined the control volume the shaded area of unit thickness normal to the page next identify theheat transfer processes. Finally, perform an energy balance wherein the processes are expressedusing appropriate rate equations.(a) With the horizontal boundary insulated and the vertical boundary subjected to a convection process,the energy balance results in the following finite-difference equation:&&Ein Eout = 0q1 + q2 + q3 + q4 + q5 + q6 = 0T Tm,nT x T y k ( y 1) m-1,n+ k 1 m,n-1 m,n + h 1 T Tm,nxy2 2 () Tm,nTT y T+ 0 + k 1 m+1,n+ k ( x 1) m,n+1 m,n = 0.xy2 Letting x = y, and regrouping, find()()2 Tm-1,n + Tm,n+1 + Tm+1,n + Tm,n-1 +hxhx T 6 +Tm,n = 0.kk<(b) With both boundaries insulated, the energy balance would have q3 = q4 = 0. The same result wouldbe obtained by letting h = 0 in the previous result. Hence,()()2 Tm-1,n + Tm,n+1 + Tm+1,n + Tm,n-1 6 Tm,n = 0.<Note that this expression compares exactly with Eq. 4.45 when h = 0, which corresponds to insulatedboundaries.PROBLEM 4.36KNOWN: Plane surface of two-dimensional system.FIND: The finite-difference equation for nodal point on this boundary when (a) insulated; compareresult with Eq. 4.46, and when (b) subjected to a constant heat flux.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional, steady-state conduction with no generation, (2) Constantproperties, (3) Boundary is adiabatic.ANALYSIS: (a) Performing an energy balance on the control volume, (x/2) y, and using theconduction rate equation, it follows that&&E in E out = 0q1 + q2 + q3 = 0(1,2)T Tm,nTT x T x Tk ( y 1) m-1,n+ k 1 m,n-1 m,n + k 1 m,n+1 m,n = 0. (3)xyy2 2 Note that there is no heat rate across the control volume surface at the insulated boundary.Recognizing that x =y, the above expression reduces to the form2Tm-1,n + Tm,n-1 + Tm,n+1 4Tm,n = 0.(4) <The Eq. 4.46 of Table 4.3 considers the same configuration but with the boundary subjected to aconvection process. That is,( 2Tm-1,n + Tm,n-1 + Tm,n+1 ) + 2hkx T 2 hx + 2 Tm,n = 0.k(5)Note that, if the boundary is insulated, h = 0 and Eq. 4.46 reduces to Eq. (4).(b) If the surface is exposed to a constant heat flux, q , the energy balance has the formoq1 + q2 + q3 + q y = 0 and the finite difference equation becomesoq x2Tm-1,n + Tm,n-1 + Tm,n+1 4Tm,n = o .k<COMMENTS: Equation (4) can be obtained by using the interior node finite-difference equation,Eq. 4.33, where the insulated boundary is treated as a symmetry plane as shown below.PROBLEM 4.37KNOWN: External corner of a two-dimensional system whose boundaries are subjected toprescribed conditions.FIND: Finite-difference equations for these situations: (a) Upper boundary is perfectly insulated andside boundary is subjected to a convection process, (b) Both boundaries are perfectly insulated;compare result with Eq. 4.47.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties, (4) No internal generation.ANALYSIS: Consider the nodal point configuration shown in the schematic and also as Case 4, Table4.2. The control volume about the node shaded area above of unit thickness normal to the page has dimensions, (x/2)(y/2) 1. The heat transfer processes at the surface of the CV are identified asq1, q2 . Perform an energy balance wherein the processes are expressed using the appropriate rateequations.(a) With the upper boundary insulated and the side boundary subjected to a convection process, theenergy balance has the form&&E in E out = 0q1 + q2 + q3 + q4 = 0 Tm,n y T xk 1 m-1,n+k x2 2TT y1 m,n-1 m,n + h y2(1,2)1 T Tm,n + 0 = 0.()Letting x = y, and regrouping, findTm,n-1 + Tm-1,n +hx1 hx T 2 2 k + 1 Tm,n = 0.k(3) <(b) With both boundaries insulated, the energy balance of Eq. (2) would have q3 = q4 = 0. The sameresult would be obtained by letting h = 0 in the finite-difference equation, Eq. (3). The result isTm,n-1 + Tm-1,n 2Tm,n = 0.<Note that this expression is identical to Eq. 4.47 when h = 0, in which case both boundaries areinsulated.COMMENTS: Note the convenience resulting from formulating the energy balance by assumingthat all the heat flow is into the node.PROBLEM 4.38KNOWN: Conduction in a one-dimensional (radial) cylindrical coordinate system with volumetricgeneration.FIND: Finite-difference equation for (a) Interior node, m, and (b) Surface node, n, with convection.SCHEMATIC:(a) Interior node, m(b) Surface node with convection, nASSUMPTIONS: (1) Steady-state, one-dimensional (radial) conduction in cylindrical coordinates,(2) Constant properties.ANALYSIS: (a) The network has nodes spaced at equal r increments with m = 0 at the center;hence, r = mr (or nr). The control volume is V = 2 r r l = 2 ( m r ) r l. The energy&&&balance is Ein +E g = q a +qb +qV = 0k 2 r Tm-1 Tm+ k 2r 2 l r2 r Tm+1 Tm &+ q 2 ( m r ) rl = 0.r+ 2 l rRecognizing that r = mr, canceling like terms, and regrouping find&11qmr 2m Tm-1 + m+ Tm+1 2mTm += 0.2k<(b) The control volume for the surface node is V = 2 r ( r/2 ) l. The energy balance is&&&E in +E g = q d + q conv + qV=0. Use Fouriers law to express qd and Newtons law of cooling forqconv to obtaink 2r r T Tnn-1&+ h [ 2 rl ] ( T Tn ) + q 2 ( nr ) l = 0.r 2 l r2Let r = nr, cancel like terms and regroup to find& 1 hn r qn r 2 hn r 1n Tn-1 n + Tn + 2k + k T = 0.22k<COMMENTS: (1) Note that when m or n becomes very large compared to , the finite-differenceequation becomes independent of m or n. Then the cylindrical system approximates a rectangular one.(2) The finite-difference equation for the center node (m = 0) needs to be treated as a special case.The control volume isV = ( r / 2 ) l and the energy balance is22 r T T & r &&&E in +E g = q a + qV = k 2 l 1 0 + q l = 0. 2 r 2 Regrouping, the finite-difference equation is To + T1 +&2qr4k= 0.PROBLEM 4.39KNOWN: Two-dimensional cylindrical configuration with prescribed radial (r) and angular ( )spacings of nodes.FIND: Finite-difference equations for nodes 2, 3 and 1.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in cylindricalcoordinates (r, ), (3) Constant properties.ANALYSIS: The method of solution is to define the appropriate control volume for each node, toidentify relevant processes and then to perform an energy balance.(a) Node 2. This is an interior node with control volume as shown above. The energy balance is&E = q + q + q + q = 0. Using Fouriers law for each process, findinabcd3 (T T )(T T )k ri + r 5 2 + k ( r ) 3 2 +2r( ri + r ) ( Ti T2 )(T T )1+ k ri + r + k ( r ) 1 2 = 0.2 r( ri + r ) Canceling terms and regrouping yields,( r ) 2 1 T r 3 r T( r )21 2+ i+ 5+2 ( ri + r ) +( T3 + T1) + ri + r Ti = 0.22( ) 2 ( ri + r ) ( ri + r ) ( ) 2(b) Node 3. The adiabatic surface behaves as a symmetry surface. We can utilize the result of Part(a) to write the finite-difference equation by inspection as2( r )2 1 32 ( r )1( ri + r ) + T3 + ri + r T6 +2 T + r + r Ti = 0.2 ( r + r ) 2 2 i 22( ) i( ri + r ) ( )(c) Node 1. The energy balance is q + qb + q + q = 0. Substituting,acd3 ( T4 T1 )( T2 T1) +k ri + r + k ( r )2 2r( ri + r ) 1 ( Ti T1 )+ k ri + r + h ( r )( T T ) = 012 2 rThis expression could now be rearranged.<PROBLEM 4.40KNOWN: Heat generation and thermal boundary conditions of bus bar. Finite-difference grid.FIND: Finite-difference equations for selected nodes.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties.ANALYSIS: (a) Performing an energy balance on the control volume, (x/2)(y/2) 1, find the FDEfor node 1,k ( y/2 1)To T1 x + hu 1 ( T T1) +( T2 T1 )R / ( y/2) 1x2 t,ck ( x/2 1)+( T6 T ) + & ( x/2 )( y/2) 1 = 01 qyx/kR To + ( h u x/k ) T +T 2 + T6t,c()()2&+ q ( x ) / 2 k x/kR + ( hu x/k ) + 2 T = 0.t,c1<(b) Performing an energy balance on the control volume, (x)(y/2) 1, find the FDE for node 13,h l ( x 1) ( T T13 ) + ( k/ x ) ( y/2 1) ( T12 T13 )&+ ( k/ y )( x 1)( T8 T ) + ( k/ x )( y/2 1)( T T ) + q ( x y/2 1) = 0131413&( h l x/k ) T +1 / 2 (T12 + 2T8 + T14 ) + q ( x )2 /2k ( h l x/k + 2 ) T13 = 0.<COMMENTS: For fixed To and T, the relative amounts of heat transfer to the air and heat sinkare determined by the values of h and R ,c.tPROBLEM 4.41KNOWN: Nodal point configurations corresponding to a diagonal surface boundary subjected to aconvection process and to the tip of a machine tool subjected to constant heat flux and convectioncooling.FIND: Finite-difference equations for the node m,n in the two situations shown.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.ANALYSIS: (a) The control volume about node m,n has triangular shape with sides x and y whilethe diagonal (surface) length is2 x. The heat rates associated with the control volume are due toconduction, q1 and q2, and to convection, qc . Performing an energy balance, find&&E in E out = 0q 1 + q 2 + qc = 0Tm,n-1 Tm,nT Tm,nk ( x 1)+ k ( y 1) m+1,n+hyx()()2 x 1 T Tm,n = 0.Note that we have considered the tool to have unit depth normal to the page. Recognizing that x =y, dividing each term by k and regrouping, findTm,n-1 + Tm+1,n + 2 hxhx T 2 + 2 Tm,n = 0.kk<(b) The control volume about node m,n has triangular shape with sides x/2 and y/2 while the lowerdiagonal surface length is2 ( x/2 ). The heat rates associated with the control volume are due tothe constant heat flux, qa, to conduction, qb, and to the convection process, qc . Perform an energybalance,&&E in E out = 0q a + q b + qc = 0x x y Tm+1,n Tm,nq 1 + k 1+h 2 T Tm,n = 0.ox22 2 ()Recognizing that x = y, dividing each term by k/2 and regrouping, findTm+1,n + 2 hxx hx T + q 1+ 2 Tm,n = 0.okkkCOMMENTS: Note the appearance of the term hx/k in both results, which is a dimensionlessparameter (the Biot number) characterizing the relative effects of convection and conduction.<PROBLEM 4.42KNOWN: Nodal point on boundary between two materials.FIND: Finite-difference equation for steady-state conditions.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties, (4) No internal heat generation, (5) Negligible thermal contact resistance at interface.ANALYSIS: The control volume is defined about nodal point 0 as shown above. The conservation ofenergy requirement has the form6 qi = q1 + q2 + q3 + q4 + q5 + q6 = 0i =1since all heat rates are shown as into the CV. Each heat rate can be written using Fouriers law,y T1 T0T Ty T3 T0+ k A x 2 0 + k A 2xy2xy T3 T0T4 T0y T1 T0+ kB + kB x + kB = 0.2xy2xkA Recognizing that x = y and regrouping gives the relation,1kA1kBT0 + T1 +T2 + T3 +T = 0.42 (k A + k B )42 (k A + kB ) 4<COMMENTS: Note that when kA = kB, the result agrees with Eq. 4.33 which is appropriate for aninterior node in a medium of fixed thermal conductivity.PROBLEM 4.43KNOWN: Two-dimensional grid for a system with no internal volumetric generation.FIND: Expression for heat rate per unit length normal to page crossing the isothermal boundary.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, (3) Constantproperties.ANALYSIS: Identify the surface nodes (Ts ) and draw control volumes about these nodes. Sincethere is no heat transfer in the direction parallel to the isothermal surfaces, the heat rate out of theconstant temperature surface boundary isq = qa + qb + qc + qd + qe + qfFor each q , use Fouriers law and pay particular attention to the manner in which the crossisectional area and gradients are specified.T TT TT Tq = k ( y/2 ) 1 s + k ( y ) 2 s + k ( y ) 3 sxxxT5 TsT6 TsT7 Ts+ k ( x )+ k ( x )+ k ( x/2)yyyRegrouping with x = y, findq = k [ 0.5T1 + T2 + T3 + T5 + T6 + 0.5T7 5Ts ].<COMMENTS: Looking at the corner node, it is important to recognize the areas associated withqc and qd (y and x, respectively).PROBLEM 4.44KNOWN: One-dimensional fin of uniform cross section insulated at one end with prescribed basetemperature, convection process on surface, and thermal conductivity.FIND: Finite-difference equation for these nodes: (a) Interior node, m and (b) Node at end of fin, n,where x = L.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction.ANALYSIS: (a) The control volume about node m is shown in the schematic; the node spacing andcontrol volume length in the x direction are both x. The uniform cross-sectional area and finperimeter are Ac and P, respectively. The heat transfer process on the control surfaces, q1 and q2,represent conduction while qc is the convection heat transfer rate between the fin and ambient fluid.Performing an energy balance, find&&E in E out = 0q 1 + q 2 + qc = 0Tm-1 TmTTkAc+ kAc m+1 m + hPx ( T Tm ) = 0.xxMultiply the expression by x/kAc and regroup to obtainTm-1 + Tm+1 +hPhP x 2T 2 +x 2 Tm = 0kAckAc1<m<n<Considering now the special node m = 1, then the m-1 node is Tb, the base temperature. The finitedifference equation would beTb + T2 +hPhPx 2 T 2 +x 2 T1 = 0kAc kAcm=1<(b) The control volume of length x/2 about node n is shown in the schematic. Performing an energybalance,&&E in E out = 0q 3 + q 4 + qc = 0T TxkAc n-1 n + 0 + hP( T T ) = 0.x2nNote that q4 = 0 since the end (x = L) is insulated. Multiplying by x/kAc and regrouping, hP x 2 hP x 2Tn-1 +T +1 Tn = 0.kAc 2 kAc 2<COMMENTS: The value of x will be determined by the selection of n; that is, x = L/n. Note thatthe grouping, hP/kAc , appears in the finite-difference and differential forms of the energy balance.PROBLEM 4.45KNOWN: Two-dimensional network with prescribed nodal temperatures and thermal conductivity ofthe material.FIND: Heat rate per unit length normal to page, q.SCHEMATIC:Node1234567Ti(C)120.55120.64121.29123.89134.57150.49147.14ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, (3) No internalvolumetric generation, (4) Constant properties.ANALYSIS: Construct control volumes around the nodes on the surface maintained at the uniformtemperature Ts and indicate the heat rates. The heat rate per unit length is q = qa + qb + qc + qd + qeor in terms of conduction terms between nodes,q = q + q2 + q3 + q4 + q + q 7.15Each of these rates can be written in terms of nodal temperatures and control volume dimensions usingFouriers law,x T1 TsT TT TT T+ k x 2 s + k x 3 s + k x 4 s2yyyyT5 Tsy T7 Ts+ k x+ k.y2xq = k and since x =y,q = k[ (1/2 ) ( T1 Ts ) + ( T2 Ts ) + ( T3 Ts )+ ( T4 Ts ) + ( T5 Ts ) + (1 / 2)( T7 Ts )].Substituting numerical values, findq = 50 W/m K[ (1/2 ) (120.55 100 ) + (120.64 100 ) + (121.29 100 )+ (123.89 100) + (134.57 100) + (1 / 2)(147.14 100 )]q = 6711 W/m.COMMENTS: For nodes a through d, there is no heat transfer into the control volumes in the xdirection. Look carefully at the energy balance for node e, q e = q5 + q7 , and how q and q 7 are5evaluated.<PROBLEM 4.46KNOWN: Nodal temperatures from a steady-state, finite-difference analysis for a one-eighthsymmetrical section of a square channel.FIND: (a) Beginning with properly defined control volumes, derive the finite-difference equations fornodes 2, 4 and 7, and determine T2, T4 and T7, and (b) Heat transfer loss per unit length from the channel,q .SCHEMATIC:Node1368,9T(C)430394492600ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) No internalvolumetric generation, (4) Constant properties.ANALYSIS: (a) Define control volumes about the nodes 2, 4, and 7, taking advantage of symmetrywhere appropriate and performing energy balances, Ein E out = 0 , with x = y,Node 2: q + q + q + q = 0abcdT TT TT Thx ( T T2 ) + k ( y 2 ) 3 2 + kx 6 2 + k ( y 2 ) 1 2 = 0xyxT2 = 0.5T1 + 0.5T3 + T6 + ( hx k ) T 2 + ( hx k )T2 = 0.5 430 + 0.5 394 + 492 + 50 W m 2 K 0.01m 1W m K 300 K(T2 = 422 K)[2 + 0.50]<Node 4: q + q + q = 0abcT Th ( x 2 )( T T4 ) + 0 + k ( y 2 ) 3 4 = 0xT4 = T3 + ( hx k ) T 1 + ( hx k )T4 = [394 + 0.5 300] K [1 + 0.5] = 363K<Continued...PROBLEM 4.46 (Cont.)Node 7: From the first schematic, recognizing that the diagonal is a symmetry adiabat, we can treat node7 as an interior node, henceT7 = 0.25 ( T3 + T3 + T6 + T6 ) = 0.25 (394 + 394 + 492 + 492 ) K = 443K<(b) The heat transfer loss from the upper surface can be expressed as the sum of the convection ratesfrom each node as illustrated in the first schematic,q = q1 + q + q + qcv234q = h ( x 2 )( T1 T ) + hx ( T2 T ) + hx ( T3 T ) + h ( x 2 )( T4 T )cvq = 50 W m 2 K 0.1m ( 430 300 ) 2 + ( 422 300 ) + (394 300 ) + (363 300 ) 2 Kcvq = 156 W mcv<COMMENTS: (1) Always look for symmetry conditions which can greatly simplify the writing of thenodal equation as was the case for Node 7.(2) Consider using the IHT Tool, Finite-Difference Equations, for Steady-State, Two-Dimensional heattransfer to determine the nodal temperatures T1 - T7 when only the boundary conditions T8, T9 and ( T ,h)are specified.PROBLEM 4.47KNOWN: Steady-state temperatures (K) at three nodes of a long rectangular bar.FIND: (a) Temperatures at remaining nodes and (b) heat transfer per unit length from the bar using&nodal temperatures; compare with result calculated using knowledge of q.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.ANALYSIS: (a) The finite-difference equations for the nodes (1,2,3,A,B,C) can be written byinspection using Eq. 4.39 and recognizing that the adiabatic boundary can be represented by asymmetry plane.273&22 / k = 0 and qx = 5 10 W/m ( 0.005m ) = 62.5K.&Tneighbors 4Ti + q xkNode A (to find T2):Node 3 (to find T3):Node 1 (to find T1):20 W/m K&2T2 + 2TB 4TA + q x2 / k = 01T2 = ( 2 374.6 + 4 398.0 62.5) K = 390.2K2&Tc + T2 + TB + 300K 4T3 + q x2 / k = 01T3 = ( 348.5 + 390.2 + 374.6 + 300 + 62.5 ) K = 369.0K4&300 + 2TC + T2 4T1 + q x2 / k = 01T1 = ( 300 + 2 348.5 + 390.2 + 62.5 ) = 362.4K4<<<(b) The heat rate out of the bar is determined by calculating the heat rate out of each control volumearound the 300K nodes. Consider the node in the upper left-hand corner; from an energy balance&&&E in E out + E g = 0Hence, for the entire bar&&&qa = qa,in + E g where E g = qV.qbar = qa + qb + q + q + qe + qf , orcdorT 300x y T1 300 & x y & x y +&q bar = k+q + k y C+q y + q xx 2 2 a 2b 2 2 c2TC 300x y + kx T3 300 + q x y + k x TB 300 + q x y .&&&kx y + q 2 2 2 d y2 e 2y fSubstituting numerical values, find qbar = 7,502.5 W/m. From an overall energy balance on the bar,&&&qbar = Eg = qV/ l = q ( 3x 2y ) = 5 107 W/m 3 6 ( 0.005m ) 2 = 7,500 W/m.As expected, the results of the two methods agree. Why must that be?<PROBLEM 4.48KNOWN: Steady-state temperatures at selected nodal points of the symmetrical section of a flowchannel with uniform internal volumetric generation of heat. Inner and outer surfaces of channelexperience convection.FIND: (a) Temperatures at nodes 1, 4, 7, and 9, (b) Heat rate per unit length (W/m) from the outersurface A to the adjacent fluid, (c) Heat rate per unit length (W/m) from the inner fluid to surface B,and (d) Verify that results are consistent with an overall energy balance.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: (a) The nodal finite-difference equations are obtained from energy balances on controlvolumes about the nodes shown in the schematics below.Node 1q + q + q + q + E = 0abcdgT TT T0 + k ( y / 2 ) 2 1 + k ( x / 2 ) 3 1 + 0 + q ( x y / 4 ) = 0xyT1 = ( T2 + T3 ) / 2 + qx 2 / 2kT1 = (95.47 + 117.3) C / 2 + 106 W / m3 ( 25 25 ) 106 m 2 / ( 2 10 W / m K ) = 122.0CNode 4q + q + q + q + q + q + E = 0abcdefgT Tk ( x / 2 ) 2 4 + h i ( y / 2 ) T,i T4 + h i ( x / 2 )(T T4 ) +y()Continued ..PROBLEM 4.48 (Cont.)T TT TT Tk ( y / 2 ) 5 4 + k ( x ) 8 4 + k ( y ) 3 4 + q (3x y / 4 ) = 0xyx)(T4 = T2 + 2T3 + T5 + 2T8 + 2 ( hi x / k ) T,i + 3qx 2 / 2k 6 + 2 ( h i x / k )<T4 = 94.50CNode 7q + q + q + q + E = 0abcdgT TT Tk ( x / 2 ) 3 7 + k ( y / 2 ) 8 7 + h o ( x / 2 ) T,o T7 + 0 + q ( x y / 4 ) = 0yx(T7 = T3 + T8 + ( h o x / k ) T,o + qx 2 / 2k )( 2 + h o x / k )<T7 = 95.80CNode 9q + q + q + q + E = 0abcdgT TT Tk ( x ) 5 9 + k ( y / 2 ) 10 9 + h o ( x ) T,o T9yyT T+ k ( y / 2 ) 8 9 + q ( x y / 2 ) = 0xT9 = T5 + 0.5T8 + 0.5T10 + ( h o x / k ) T,o + qx 2 / 2k / ( 2 + h o x / k )()<T9 = 79.67C(b) The heat rate per unit length from the outer surface A to the adjacent fluid, q , is the sum of theAconvection heat rates from the outer surfaces of nodes 7, 8, 9 and 10.q = h o ( x / 2 ) T7 T ,o + x T8 T ,o + x T9 T,o + ( x / 2 ) T10 T ,o A()()()(q = 250 W / m 2 K ( 25 / 2 )(95.80 25 ) + 25 (87.28 25 )A+25 (79.67 25 ) + ( 25 / 2 )( 77.65 25 ) 103 m KContinued ..)PROBLEM 4.48 (Cont.)q = 1117 W / mA<(c) The heat rate per unit length from the inner fluid to the surface B, qB , is the sum of theconvection heat rates from the inner surfaces of nodes 2, 4, 5 and 6.()()()()q = h i ( y / 2 ) T ,i T2 + ( y / 2 + x / 2 ) T ,i T4 + x T,i T5 + ( x / 2 ) T ,i T6 Bq = 500 W / m 2 K ( 25 / 2 )(50 95.47 ) + ( 25 / 2 + 25 / 2 )(50 94.50 )B+25 (50 79.79 ) + ( 25 / 2 )(50 77.29 ) 103 m Kq = 1383 W / mB<(d) From an overall energy balance on the section, we see that our results are consistent since theconservation of energy requirement is satisfied.E E + E = q + q + E = (1117 1383 + 2500)W / m = 0inoutgenABgen6 263where Egen = q = 10 W / m [25 50 + 25 50 ]10 m = 2500 W / mCOMMENTS: The nodal finite-difference equations for the four nodes can be obtained by usingIHT Tool Finite-Difference Equations | Two-Dimensional | Steady-state. Options are provided tobuild the FDEs for interior, corner and surface nodal arrangements including convection and internalgeneration. The IHT code lines for the FDEs are shown below./* Node 1: interior node; e, w, n, s labeled 2, 2, 3, 3. */0.0 = fd_2d_int(T1,T2,T2,T3,T3,k,qdot,deltax,deltay)/* Node 4: internal corner node, e-n orientation; e, w, n, s labeled 5, 3, 2, 8. */0.0 = fd_2d_ic_en(T4,T5,T3,T2,T8,k,qdot,deltax,deltay,Tinfi,hi,q a4q a4 = 0// Applied heat flux, W/m^2; zero flux shown/* Node 7: plane surface node, s-orientation; e, w, n labeled 8, 8, 3. */0.0 = fd_2d_psur_s(T7,T8,T8,T3,k,qdot,deltax,deltay,Tinfo,ho,q a7q a7=0// Applied heat flux, W/m^2; zero flux shown/* Node 9: plane surface node, s-orientation; e, w, n labeled 10, 8, 5. */0.0 = fd_2d_psur_s(T9, T10, T8, T5,k,qdot,deltax,deltay,Tinfo,ho,q a9q a9 = 0// Applied heat flux, W/m^2; zero flux shownPROBLEM 4.49KNOWN: Outer surface temperature, inner convection conditions, dimensions and thermalconductivity of a heat sink.FIND: Nodal temperatures and heat rate per unit length.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, (2) Two-dimensional conduction, (3) Uniform outer surfacetemperature, (4) Constant thermal conductivity.ANALYSIS: (a) To determine the heat rate, the nodal temperatures must first be computed from thecorresponding finite-difference equations. From an energy balance for node 1,T TT Th ( x / 2 1)( T T1 ) + k ( y / 2 1) 2 1 + k ( x 1) 5 1 = 0xyhx hxT = 03+ T1 + T2 + 2T5 +kk(1)With nodes 2 and 3 corresponding to Case 3 of Table 4.2,2hx hxT1 2 T = 0+ 2 T2 + T3 + 2T6 +kk(2)hx hxT2 T = 0+ 2 T3 + T7 +kk(3)where the symmetry condition is invoked for node 3. Applying an energy balance to node 4, weobtain2T4 + T5 + Ts = 0(4)The interior nodes 5, 6 and 7 correspond to Case 1 of Table 4.2. Hence,T1 + T4 4T5 + T6 + Ts = 0T2 + T5 4T6 + T7 + Ts = 0T3 + 2T6 4T7 + Ts = 0(5)(6)(7)where the symmetry condition is invoked for node 7. With Ts = 50C, T = 20C, andhx / k = 5000 W / m K ( 0.005m ) / 240 W / m K = 0.1042, the solution to Eqs. (1) (7) yields2T1 = 46.61C, T2 = 45.67C, T3 = 45.44C, T4 = 49.23C<T5 = 48.46C, T6 = 48.00C, T7 = 47.86CContinued ..PROBLEM 4.49 (Cont.)The heat rate per unit length of channel may be evaluated by computing convection heat transfer fromthe inner surface. That is,q = 8h x / 2 ( T1 T ) + x ( T2 T ) + x / 2 ( T3 T )q = 8 5000 W / m 2 K 0.0025m ( 46.61 20 ) C + 0.005m ( 45.67 20 ) C+0.0025m ( 45.44 20 ) C] = 10,340 W / m<(b) Since h = 5000 W / m 2 K is at the high end of what can be achieved through forced convection,we consider the effect of reducing h. Representative results are as follows()h W / m 2 K T1 (C )T2 (C )T3 (C )20010002000500049.8049.0248.1145.6749.7948.9748.0045.4449.8449.2448.5346.61T4 (C ) T5 (C )49.9649.8349.6649.2349.9349.6549.3348.46T6 (C )T7 (C ) q ( W / m )49.9149.5549.1348.0049.9049.5249.0647.864772325451010,340There are two resistances to heat transfer between the outer surface of the heat sink and the fluid, thatdue to conduction in the heat sink, R cond ( 2D ), and that due to convection from its inner surface to thefluid, R conv . With decreasing h, the corresponding increase in R conv reduces heat flow andincreases the uniformity of the temperature field in the heat sink. The nearly 5-fold reduction in q corresponding to the 5-fold reduction in h from 1000 to 200 W / m 2 K indicates that the convection()resistance is dominant R conv >> R cond ( 2D ) .COMMENTS: To check our finite-difference solution, we could assess its consistency withconservation of energy requirements. For example, an energy balance performed at the inner surfacerequires a balance between convection from the surface and conduction to the surface, which may beexpressed asq = k ( x 1)(T5 T1 ) + ky( x 1)T6 T2T T+ k ( x / 2 1) 7 3yySubstituting the temperatures corresponding to h = 5000 W / m 2 K, the expression yieldsq = 10, 340 W / m, and, as it must be, conservation of energy is precisely satisfied. Results of the()analysis may also be checked by using the expression q = ( Ts T ) / R ( 2D ) + R condconv , where, for2h = 5000 W / m K, R = 1/ 4hw = 2.5 103 m K / W, and from Eq. (4.27) and Case 11 ofconv()4Table 4.1, R cond = [0.930 ln ( W / w ) 0.05] / 2 k = 3.94 10 m K / W. Hence,(q = (50 20 ) C / 2.5 103+ 3.94 104) m K / W = 10, 370 W / m, and the agreement with thefinite-difference solution is excellent. Note that, even for h = 5000 W / m 2 K, R conv >> R cond ( 2D ).PROBLEM 4.50KNOWN: Steady-state temperatures (C) associated with selected nodal points in a two-dimensionalsystem.FIND: (a) Temperatures at nodes 1, 2 and 3, (b) Heat transfer rate per unit thickness from thesystem surface to the fluid.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: (a) Using the finite-difference equations for Nodes 1, 2 and 3:Node 1, Interior node, Eq. 4.33: T1 =T1 =1 Tneighbors41(172.9 + 137.0 + 132.8 + 200.0 )o C = 160.7oC4<Node 2, Insulated boundary, Eq. 4.46 with h = 0, Tm,n = T2()1Tm-1,n + Tm+1,n + 2Tm,n-141T2 = (129.4 + 45.8 + 2 103.5 )o C = 95.6oC4T2 =<Node 3, Plane surface with convection, Eq. 4.46, Tm,n = T32h x h x + 2 T = 2T2T3m-1,n + Tm,n+1 +Tm,n-1 +kk()h x/k = 50W/m 2 K 0.1m/1.5W/m K = 3.332 ( 3.33 + 2 ) T3 = ( 2 103.5 + 45.8 + 67.0 ) C + 2 3.33 30 o C1( 319.80 + 199.80) C=48.7C10.66(b) The heat rate per unit thickness from the surface to the fluid is determined from the sum of theconvection rates from each control volume surface.qconv = qa + qb + qc + q dT3 =q i = hy i ( T T )iqconv = 50 0.1 m 45.8 30.0 C +()m K 2W20.1m ( 48.7 30.0 ) C +0.1m ( 67.0 30.0 ) C +0.1m+( 200.0 30.0) C2qconv = ( 39.5 + 93.5 + 185.0 + 425 ) W/m = 743 W/m.<<PROBLEM 4.51KNOWN: Nodal temperatures from a steady-state finite-difference analysis for a cylindrical fin ofprescribed diameter, thermal conductivity and convection conditions ( T , h).FIND: (a) The fin heat rate, qf, and (b) Temperature at node 3, T3.SCHEMATIC:T0 = 100.0CT1 = 93.4CT2 = 89.5CASSUMPTIONS: (a) The fin heat rate, qf, is that of conduction at the base plane, x = 0, and can befound from an energy balance on the control volume about node 0, Ein E out = 0 ,qf + q1 + qconv = 0orqf = q1 qconv .Writing the appropriate rate equation for q1 and qconv, with Ac = D2/4 and P = D,T1 T0 kD2q f = kAc hP ( x 2 )( T T0 ) = (T1 T0 ) ( 2 ) Dhx (T T0 )x4xSubstituting numerical values, with x = 0.010 m, find 15 W m K (0.012 m )2qf = (93.4 100 ) C4 0.010 m 0.012 m 25 W m 2 K 0.010 m ( 25 100 ) C2qf = (1.120 + 0.353) W = 1.473 W .<(b) To determine T3, derive the finite-difference equation for node 2, perform an energy balance on thecontrol volume shown above, Ein E out = 0 ,qcv + q3 + q1 = 0T TT ThPx ( T T2 ) + kA c 3 2 + kAc 1 2 = 0xxT3 = T1 + 2T2 hPx 2 2x [T T2 ]kAcSubstituting numerical values, findT2 = 89.2 CCOMMENTS: Note that in part (a), the convection heat rate from the outer surface of the controlvolume is significant (25%). It would have been poor approximation to ignore this term.<PROBLEM 4.52KNOWN: Long rectangular bar having one boundary exposed to a convection process (T, h) while theother boundaries are maintained at a constant temperature (Ts).FIND: (a) Using a grid spacing of 30 mm and the Gauss-Seidel method, determine the nodaltemperatures and the heat rate per unit length into the bar from the fluid, (b) Effect of grid spacing andconvection coefficient on the temperature field.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: (a) With the grid spacing x = y = 30 mm, three nodes are created. Using the finitedifference equations as shown in Table 4.2, but written in the form required of the Gauss-Seidel method(see Section 4.5.2), and with Bi = hx/k = 100 W/m2K 0.030 m/1 W/mK = 3, we obtain:Node 1:T1 =111(T2 + Ts + BiT ) = (T2 + 50 + 3 100 ) = (T2 + 350)55( Bi + 2 )(1)Node 2:T2 =111(T1 + 2Ts + T3 ) = (T1 + T3 + 2 50 ) = (T1 + T3 + 100 )444(2)Node 3:T3 =111(T2 + 3Ts ) = (T2 + 3 50 ) = ( T2 + 150 )444(3)kDenoting each nodal temperature with a superscript to indicate iteration step, e.g. T1 , calculate valuesas shown below.k0T185T260T3 (C)55123482.0081.8581.7181.6959.2558.5458.4658.4552.3152.1452.1252.11 initialguessBy the 4th iteration, changes are of order 0.02C, suggesting that further calculations may not benecessary.Continued...PROBLEM 4.52 (Cont.)In finite-difference form, the heat rate from the fluid to the bar isqconv = h ( x 2 )( T Ts ) + hx ( T T1 ) + h ( x 2 )( T Ts )qconv = hx ( T Ts ) + hx ( T T1 ) = hx ( T Ts ) + ( T T1 )2qconv = 100 W m K 0.030 m (100 50 ) + (100 81.7 ) C = 205 W m .<(b) Using the Finite-Difference Equations option from the Tools portion of the IHT menu, the followingtwo-dimensional temperature field was computed for the grid shown in schematic (b), where x and y arein mm and the temperatures are in C.y\x01530456075900505050505050501580.3363.5856.2752.9151.3250.51503085.1667.7358.5854.0751.8650.72504580.3363.5856.2752.9151.3250.51506050505050505050The improved prediction of the temperature field has a significant influence on the heat rate, where,accounting for the symmetrical conditions,q = 2h ( x 2 )( T Ts ) + 2h ( x )( T T1 ) + h ( x )( T T2 )q = h ( x ) ( T Ts ) + 2 ( T T1 ) + ( T T2 )q = 100 W m 2 K ( 0.015 m ) 50 + 2 (19.67 ) + 14.84 C = 156.3 W m<Additional improvements in accuracy could be obtained by reducing the grid spacing to 5 mm, althoughthe requisite number of finite-difference equations would increase from 12 to 108, significantlyincreasing problem set-up time.An increase in h would increase temperatures everywhere within the bar, particularly at theheated surface, as well as the rate of heat transfer by convection to the surface.COMMENTS: (1) Using the matrix-inversion method, the exact solution to the system of equations (1,2, 3) of part (a) is T1 = 81.70C, T2 = 58.44C, and T3 = 52.12C. The fact that only 4 iterations wererequired to obtain agreement within 0.01C is due to the close initial guesses.(2) Note that the rate of heat transfer by convection to the top surface of the rod must balance the rate ofheat transfer by conduction to the sides and bottom of the rod.NOTE TO INSTRUCTOR: Although the problem statement calls for calculations with x = y = 5mm and for plotting associated isotherms, the instructional value and benefit-to-effort ratio are small.Hence, it is recommended that this portion of the problem not be assigned.PROBLEM 4.53KNOWN: Square shape subjected to uniform surface temperature conditions.FIND: (a) Temperature at the four specified nodes; estimate the midpoint temperature To, (b) Reducingthe mesh size by a factor of 2, determine the corresponding nodal temperatures and compare results, and(c) For the finer grid, plot the 75, 150, and 250C isotherms.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: (a) The finite-difference equation for each node follows from Eq. 4.33 for an interior pointwritten in the form, Ti = 1/4Tneighbors. Using the Gauss-Seidel iteration method, Section 4.5.2, the finitedifference equations for the four nodes are:)(kkkkkT2 = 0.25 (100 + 200 + T4 1 + T1 1 ) = 0.25T1 1 + 0.25T4 1 + 75.0kkkkkT3 = 0.25 ( T1 1 + T4 1 + 300 + 50 ) = 0.25T1 1 + 0.25T4 1 + 87.5kkkkkT4 = 0.25 ( T2 1 + 200 + 300 + T3 1 ) = 0.25T2 1 + 0.25T3 1 + 125.0kkkkkT1 = 0.25 100 + T2 1 + T3 1 + 50 = 0.25T2 1 + 0.25T3 1 + 37.5The iteration procedure using a hand calculator is implemented in the table below. Initial estimates areentered on the k = 0 row.k01234567T1(C)100112.50123.44119.93119.05118.83118.77118.76T2(C)150165.63158.60156.40156.40156.29156.26156.25T3(C)150178.13171.10169.34168.90168.79168.76168.76T4(C)250210.94207.43206.55206.33206.27206.26206.25<Continued...PROBLEM 4.53 (Cont.)By the seventh iteration, the convergence is approximately 0.01C. The midpoint temperature can beestimated asTo = ( T1 + T2 + T3 + T4 ) 2 = (118.76 + 156.25 + 168.76 + 206.25) C 4 = 162.5 C(b) Because all the nodes are interior ones, the nodal equations can be written by inspection directly intothe IHT workspace and the set of equations solved for the nodal temperatures (C).MeshCoarseFineT1118.76117.4To162.5162.5T2156.25156.1T3168.76168.9T4206.25207.6The maximum difference for the interior points is 1.5C (node 4), but the estimate at the center, To, is thesame, independently of the mesh size. In terms of the boundary surface temperatures,To = (50 + 100 + 200 + 300 ) C 4 = 162.5 CWhy must this be so?(c) To generate the isotherms, it would be necessary to employ a contour-drawing routine using thetabulated temperature distribution (C) obtained from the finite-difference solution. Using these valuesas a guide, try sketching a few isotherms.5050505050-10086.088.299.6123.0173.4300100105.6117.4137.1168.9220.7300100119138.7162.5194.9240.6300100131.7156.1179.2207.6246.8300100151.6174.6190.8209.4239.0300200200200200200-COMMENTS: Recognize that this finite-difference solution is only an approximation to thetemperature distribution, since the heat conduction equation has been solved for only four (or 25)discrete points rather than for all points if an analytical solution had been obtained.PROBLEM 4.54KNOWN: Long bar of square cross section, three sides of which are maintained at a constanttemperature while the fourth side is subjected to a convection process.FIND: (a) The mid-point temperature and heat transfer rate between the bar and fluid; a numericaltechnique with grid spacing of 0.2 m is suggested, and (b) Reducing the grid spacing by a factor of 2, findthe midpoint temperature and the heat transfer rate. Also, plot temperature distribution across the surfaceexposed to the fluid.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: (a) Considering symmetry, the nodal network is shown above. The matrix inversionmethod of solution will be employed. The finite-difference equations are:Nodes 1, 3, 5 Nodes 2, 4, 6 Nodes 7, 8 -Interior nodes, Eq. 4.33; written by inspection.Also can be treated as interior points, considering symmetry.On a plane with convection, Eq. 4.46; noting that hx/k =10 W/m2K 0.2 m/2W/mK = 1, findNode 7: (2T5 + 300 + T8) + 21100 - 2(1+2)T7 = 0Node 8: (2T6 + T7 + T7) + 21100 - 2(1+2)T8 = 0The solution matrix [T] can be found using a stock matrix program using the [A] and [C] matrices shownbelow to obtain the solution matrix [T] (Eq. 4.52). Alternatively, the set of equations could be enteredinto the IHT workspace and solved for the nodal temperatures. 4 1 1 0 0 0 0 0 2 4 0 1 0 0 0 0 1 0 4 1 1 0 0 0 2 4 0 1 0 0 A=0 10 0 1 0 4 1 1 0 0 0 0 12 4 0 1 0 0 0 0 2 0 6 1 0 0 0 0 0 2 2 6 600 300 300 C= 0 3000 500 200 292.2 289.2 279.7 T = 272.2 254.5 240.1198.1 179.4 From the solution matrix, [T], find the mid-point temperature asT4 = 272.2C<Continued...PROBLEM 4.54 (Cont.)The heat rate by convection between the bar and fluid is given as,qconv = 2 ( q + q + q )abcqconv = 2x h ( x 2 )( T8 T ) + h ( x )( T7 T ) + h ( x 2 )(300 T )2qconv = 2x 10 W m K ( 0.2 m 2 ) (179.4 100 ) + 2 (198.1 100 ) + (300 100 ) K <qconv = 952 W m .(b) Reducing the grid spacing by a factor of 2, the nodal arrangement will appear as shown. The finitedifference equation for the interior and centerline nodes were written by inspection and entered into theIHT workspace. The IHT Finite-Difference Equations Tool for 2-D, SS conditions, was used to obtainthe FDE for the nodes on the exposed surface.The midpoint temperature T13 and heat rate for the finer mesh areT13 = 271.0Cq = 834 W/m<COMMENTS: The midpoint temperatures for the coarse and finer meshes agree closely, T4 = 272C vs.T13 = 271.0C, respectively. However, the estimate for the heat rate is substantially influenced by themesh size; q = 952 vs. 834 W/m for the coarse and finer meshes, respectively.PROBLEM 4.55KNOWN: Volumetric heat generation in a rectangular rod of uniform surface temperature.FIND: (a) Temperature distribution in the rod, and (b) With boundary conditions unchanged, heatgeneration rate causing the midpoint temperature to reach 600 K.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Uniformvolumetric heat generation.ANALYSIS: (a) From symmetry it follows that six unknown temperatures must be determined. Sinceall nodes are interior ones, the finite-difference equations may be obtained from Eq. 4.39 written in theformTi = 1 2 Tneighbors + 1 4 ( q ( xy1) k ) .With q ( xy ) 4k = 62.5 K, the system of finite-difference equations isT1 = 0.25 ( Ts + T2 + T4 + Ts ) + 15.625(1)T2 = 0.25 ( Ts + T3 + T5 + T1 ) + 15.625(2)T3 = 0.25 ( Ts + T2 + T6 + T2 ) + 15.625(3)T4 = 0.25 ( T1 + T5 + T1 + Ts ) + 15.625(4)T5 = 0.25 ( T2 + T6 + T2 + T4 ) + 15.625(5)T6 = 0.25 ( T3 + T5 + T3 + T5 ) + 15.625(6)With Ts = 300 K, the set of equations was written directly into the IHT workspace and solved for thenodal temperatures,T1T2T3T4T5T6 (K)348.6368.9374.6362.4390.2398.0<(b) With the boundary conditions unchanged, the q required for T6 = 600 K can be found using the sameset of equations in the IHT workspace, but with these changes: (1) replace the last term on the RHS(15.625) of Eqs. (1-6) by q (xy)/4k = (0.005 m)2 q /420 W/mK = 3.125 10-7 q and (2) set T6 =600 K. The set of equations has 6 unknown, five nodal temperatures plus q . Solving findq = 1.53 108 W m3<PROBLEM 4.56KNOWN: Flue of square cross section with prescribed geometry, thermal conductivity and innerand outer surface temperatures.FIND: Heat loss per unit length from the flue, q.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Nointernal generation.ANALYSIS: Taking advantage of symmetry, the nodal network using the suggested 75mm gridspacing is shown above. To obtain the heat rate, we first need to determine the unknowntemperatures T1, T2, T3 and T4. Recognizing that these nodes may be treated as interior nodes, thenodal equations from Eq. 4.33 are(T2 + 25 + T2 + 350) - 4T1 = 0(T1 + 25 + T3 + 350) - 4T2 = 0(T2 + 25 + T4 + 350) - 4T3 = 0(T3 + 25 + 25 + T3) - 4T4 = 0.The Gauss-Seidel iteration method is convenient for this system of equations and following theprocedures of Section 4.5.2, they are rewritten as,kk-1T1 = 0.50 T2 + 93.75kkk-1T2 = 0.25 T1 + 0.25 T3 + 93.75kkk-1T3 = 0.25 T2 + 0.25 T4 + 93.75kkT4 = 0.50 T3 + 12.5.The iteration procedure is implemented in the table on the following page, one row for each iterationk. The initial estimates, for k = 0, are all chosen as (350 + 25)/2 185 C. Iteration is continueduntil the maximum temperature difference is less than 0.2 C, i.e., < 0.2 C.Note that if the system of equations were organized in matrix form, Eq. 4.52, diagonal dominancewould exist. Hence there is no need to reorder the equations since the magnitude of the diagonalelement is greater than that of other elements in the same row.Continued ..PROBLEM 4.56 (Cont.)kT1( C)T2( C)T3( C)T4( C)01234567185186.3187.1187.4184.9184.2184.0183.9185186.6187.2182.3180.8180.4180.3180.3185186.6167.0163.3162.5162.3162.3162.2185105.896.094.293.893.793.693.6 initial estimate <0.2 CFrom knowledge of the temperature distribution, the heat rate may be obtained by summing the heatrates across the nodal control volume surfaces, as shown in the sketch.The heat rate leaving the outer surface of this flue section is,q = qa + qb + qc + qd + qex 1q = k 2 ( T1 25 ) + (T 2 25) + ( T3 25 ) + ( T4 25 ) + 0 y W 1q = 0.85 2 (183.9 25 ) + (180.3 25 ) + (162.2 26 ) + ( 93.6 25) m Kq = 374.5 W/m.Since this flue section is 1/8 the total cross section, the total heat loss from the flue isq = 8 374.5 W/m = 3.00 kW/m.<COMMENTS: The heat rate could have been calculated at the inner surface, and from the abovesketch has the formq = kx 1 2 ( 350 T1 ) + (350 T2 ) + (350 T3 ) = 374.5 W/m.y This result should compare very closely with that found for the outer surface since the conservationof energy requirement must be satisfied in obtaining the nodal temperatures.PROBLEM 4.57KNOWN: Flue of square cross section with prescribed geometry, thermal conductivity and inner andouter surface convective conditions.FIND: (a) Heat loss per unit length, q , by convection to the air, (b) Effect of grid spacing andconvection coefficients on temperature field; show isotherms.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: (a) Taking advantage of symmetry, the nodal network for a 75 mm grid spacing is shownin schematic (a). To obtain the heat rate, we need first to determine the temperatures Ti. Recognize thatthere are four types of nodes: interior (4-7), plane surface with convection (1, 2, 8-11), internal cornerwith convection (3), and external corner with convection (12). Using the appropriate relations fromTable 4.2, the finite-difference equations areNode1234567891011122h i x h xT,i i + 2 T1 = 0kk2h xh x( 2T5 + T3 + T1 ) + i T,i 2 i + 2 T2 = 0kk( 2T4 + T2 + T2 ) +2 ( T6 + T6 ) + ( T2 + T2 ) +2h i xk(T8 + T5 + T1 + T5 ) 4T4 = 0(T9 + T6 + T2 + T4 ) 4T5 = 0(T10 + T7 + T3 + T5 ) 4T6 = 0(T11 + T11 + T6 + T6 ) 4T7 = 0( 2T4 + T9 + T9 ) +2h o xk h x T,i 2 3 + i T3 = 0k h xT,o 2 o + 2 T8 = 0k4.464.454.334.334.334.334.46 h xT,o 2 o + 2 T9 = 0kk2h xh x( 2T6 + T11 + T9 ) + o T,o 2 o + 2 T10 = 0kk4.46 h x + 2 T = 0T ,o 2 o 11kk2h o xh o xT ,o 2 + 1 T12 = 0(T11 + T11 ) +kk4.46( 2T5 + T10 + T8 ) +( 2T7 + T12 + T10 ) +2h o xEquation4.462h o x4.464.47Continued...PROBLEM 4.57 (Cont.)The Gauss-Seidel iteration is convenient for this system of equations. Following procedures of Section4.5.2, the system of equations is rewritten in the proper form. Note that diagonal dominance is present;hence, no re-ordering is necessary.k 1kT1 = 0.09239T2k 1+ 0.09239T4+ 285.3kkkkT2 = 0.04620T1 + 0.04620T3 1 + 0.09239T5 1 + 285.3kkkT3 = 0.08457T2 + 0.1692T6 1 + 261.2kkkkT4 = 0.25T1 + 0.50T5 1 + 0.25T8 1kkkkkT5 = 0.25T2 + 0.25T4 + 0.25T6 1 + 0.25T9 1kkkkkT6 = 0.25T3 + 0.25T5 + 0.25T7 1 + 0.25T9 1kkkT7 = 0.50T6 + 0.50T111kkkT8 = 0.4096T4 + 0.4096T9 1 + 4.52kkkkT9 = 0.4096T5 + 0.2048T8 + 0.2048T101 + 4.52kkkkT10 = 0.4096T6 + 0.2048T9 + 0.2048T111 + 4.52kkkkT11 = 0.4096T7 + 0.2048T10 + 0.2048T121 + 4.52kkT12 = 0.6939T11 + 7.65The initial estimates (k = 0) are carefully chosen to minimize calculation labor; let < 1.0.k01234567T1340338.9338.3338.8339.4339.8340.1340.3T2330336.3337.4338.4338.8339.2339.4339.5T3315324.3328.0328.2328.9329.3329.7329.9T4250237.2241.4247.7251.6254.0255.4256.4T5225232.1241.5245.7248.7250.5251.7252.5T6205225.4226.6230.6232.9234.5235.7236.4T7195175.2178.6180.5182.3183.7184.7185.5T8160163.1169.6175.6178.7180.6181.8182.7T9150161.7170.0173.7176.0177.5178.5179.1T10140155.6158.9161.2162.9164.1164.7165.6T11125130.7130.4131.6132.8133.8134.5135.1T1211098.398.198.999.8100.5101.0101.4The heat loss to the outside air for the upper surface (Nodes 8 through 12) is of the form11q = h o x T8 T ,o + T9 T,o + T10 T ,o + T11 T ,o + T12 T ,o 22()()()()() 1 182.7 25 + 179.1 25 + 165.6 25 + 135.1 25 + 1 101.4 25 C = 195 W m)()()()()2 (2Hence, for the entire flue cross-section, considering symmetry,q = 5 W m K 0.075 m2<qtot = 8 q = 8 195 W m = 1.57 kW mThe convection heat rate at the inner surface is11q = 8 h i x T,i T1 + T,i T2 + T,i T3 = 8 190.5 W m = 1.52 kW mtot22which is within 2.5% of the foregoing result. The calculation would be identical if = 0.()()()Continued...PROBLEM 4.57 (Cont.)(b) Using the Finite-Difference Equations option from the Tools portion of the IHT menu, the followingtwo-dimensional temperature field was computed for the grid shown in the schematic below, where x andy are in mm and the temperatures are in C.y\x02550751001251500180.7204.2228.9255.0282.4310.9340.025180.2203.6228.3254.4281.8310.5340.050178.4201.6226.2252.4280.1309.3339.675175.4198.2222.6248.7276.9307.1339.1100171.1193.3217.2243.1271.6303.2337.9125165.3186.7209.7235.0263.3296.0335.3150158.1178.3200.1223.9250.5282.2324.7175149.6168.4188.4209.8232.8257.5200140.1157.4175.4194.1213.5225129.9145.6161.6177.8250119.4133.4147.5275108.7121.030098.0Agreement between the temperature fields for the (a) and (b) grids is good, with the largest differencesoccurring at the interior and exterior corners. Ten isotherms generated using FEHT are shown on thesymmetric section below. Note how the heat flow is nearly normal to the flue wall around the midsection. In the corner regions, the isotherms are curved and wed expect that grid size might influencethe accuracy of the results. Convection heat transfer to the inner surface is()()()()()q = 8h i x T,i T1 2 + T,i T2 + T,i T3 + T,i T4+ T,i T5 + T,i T6 + T,i T7 2 = 1.52 kW m()()and the agreement with results of the coarse grid is excellent.The heat rate increases with increasing hi and ho, while temperatures in the wall increase anddecrease, respectively, with increasing hi and ho.PROBLEM 4.58KNOWN: Rectangular air ducts having surfaces at 80C in a concrete slab with an insulated bottomand upper surface maintained at 30C.FIND: Heat rate from each duct per unit length of duct, q.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) No internalvolumetric generation, (4) Constant properties.PROPERTIES: Concrete (given): k = 1.4 W/mK.ANALYSIS: Taking advantage of symmetry, thenodal network, using the suggested grid spacingx = 2y = 37.50 mmy = 0.125L = 18.75 mmwhere L = 150 mm, is shown in the sketch. Toevaluate the heat rate, we need the temperatures T1,T2, T3, T4, and T5. All the nodes may be treated asinterior nodes ( considering symmetry for those nodes oninsulated boundaries), Eq. 4.33. Use matrix notation, Eq. 4.52,[A][T] = [C], and perform the inversion.The heat rate per unit length from the prescribed section ofthe duct follows from an energy balance on the nodes at the top isothermal surface.q = q + q2 + q3 + q4 + q51T1 TsT TT TT TT Tq = k ( x/2 )+ k x 2 s + k x 3 s + k x 4 s + k ( x/2 ) 5 syyyyy = k ( T1 Ts ) + 2 ( T2 Ts ) + 2 ( T3 Ts ) + 2 ( T4 Ts ) + ( T5 Ts )qq = 1.4 W/m K ( 41.70 30 ) + 2 ( 44.26 30 ) + 2 ( 53.92 30 ) + 2 ( 54.89 30 ) + ( 54.98 30 ) q = 228 W/m.Since the section analyzed represents one-half of the region about an air duct, the heat loss per unitlength for each duct is,qduct = 2xq = 456 W/m.<Continued ..PROBLEM 4.58 (Cont.)Coefficient matrix [A]PROBLEM 4.59KNOWN: Dimensions and operating conditions for a gas turbine blade with embedded channels.FIND: Effect of applying a zirconia, thermal barrier coating.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Negligibleradiation.ANALYSIS: Preserving the nodal network of Example 4.4 and adding surface nodes for the TBC,finite-difference equations previously developed for nodes 7 through 21 are still appropriate, while newequations must be developed for nodes 1c-6c, 1o-6o, and 1i-6i. Considering node 3c as an example, anenergy balance yieldsk ( yc 2 )k ( y 2 )k xh o x T,o T3c + c(T2c T3c ) + c c (T4c T3c ) + c (T3o T3c ) = 0xxyc()or, with x = 1 mm and yc = 0.5 mm,h x h x0.25 ( T2c + T4c ) + 2T3o 2.5 + o T3c = o T ,okc kcSimilar expressions may be obtained for the other 5 nodal points on the outer surface of the TBC.Applying an energy balance to node 3o at the inner surface of the TBC, we obtaink c xyc(T3c T3o ) +k c ( yc 2 )x( T2o T3o ) +k c ( yc 2 )x(T4o T3o ) +x(T3i T3o ) = 0R t,cor,2T3c + 0.25 ( T2o + T4o ) +xT3ik c R t,c 2.5 +xk c R t,c T3o = 0Similar expressions may be obtained for the remaining nodal points on the inner surface of the TBC(outer region of the contact resistance).Continued...PROBLEM 4.59 (Cont.)Applying an energy balance to node 3i at the outer surface of the turbine blade, we obtaink ( y 2 )k ( y 2 )xkx(T3o T3i ) +(T2i T3i ) +(T4i T3i ) +(T9 T3i ) = 0xxyR t,cor,xxT3o + 0.5 T2,i + T4,i + T9 2 +kR kR t,ct,c() T3i = 0Similar expressions may be obtained for the remaining nodal points on the inner region of the contactresistance.The 33 finite-difference equations were entered into the workspace of IHT from the keyboard(model equations are appended), and for ho = 1000 W/m2K, T,o = 1700 K, hi = 200 W/m2K and T,i =400 K, the following temperature field was obtained, where coordinate (x,y) locations are in mm andtemperatures are in C.y\x00.50.51.52.53.5015361473145614501446144511535147214561450144514432153414711454144714411438315331469145214461438041533146814511444143705153214681451144414360Note the significant reduction in the turbine blade temperature, as, for example, from a surfacetemperature of T1 = 1526 K without the TBC to T1i = 1456 K with the coating. Hence, the coating isserving its intended purpose.COMMENTS: (1) Significant additional benefits may still be realized by increasing hi. (2) Theforegoing solution may be used to determine the temperature field without the TBC by setting kc and R ,c 0.tPROBLEM 4.60KNOWN: Bar of rectangular cross-section subjected to prescribed boundary conditions.FIND: Using a numerical technique with a grid spacing of 0.1m, determine the temperaturedistribution and the heat transfer rate from the bar to the fluid.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties.ANALYSIS: The nodal network has x = y = 0.1m. Note the adiabat corresponding to systemsymmetry. The finite-difference equations for each node can be written using either Eq. 4.33, forinterior nodes, or Eq. 4.46, for a plane surface with convection. In the case of adiabatic surfaces,Eq. 4.46 is used with h = 0. Note thathx 50W/m2 K 0.1m== 3.333.k1.5 W/m KNodeFinite-Difference Equations123456789101112131415-4T1 + 2T2 + 2T4 = 0-4T2 + T1 + T3 + 2T5 = 0-4T3 + 200 + 2T6 + T2 = 0-4T4 + T1 + 2T5 + T7 = 0-4T5 + T2 + T6 + T8 + T4 = 0-4T6 + T5 + T3 + 200 + T9 = 0-4T7 + T4 + 2T8 + T10 = 0-4T8 + T7 +T5 + T9 + T11 = 0-4T9 + T8 + T6 + 200 + T12 = 0-4T10 + T7 + 2T11 + T13 = 0-4T11 + T10 + T8 + T12 + T14 = 0-4T12 + T11 + T9 +200 + T15 = 02T10 + T14 + 6.66630-10.666 T13 = 02T11 + T13 + T15 + 6.66630-2(3.333+2)T14 = 02T12 +T14 + 200 + 6.66630-2(3.333+2) T15 = 0Using the matrix inversion method, Section 4.5.2, the above equations can be written in the form [A][T] = [C] where [A] and [C] are shown on the next page. Using a stock matrix inversion routine,the temperatures [T] are determined.Continued .. -4101000[A] = 0000000020-4 11 -40010010000000000000000000 0 -2000 0 -2000 [C] = 0 -2000 0 -200 -200 -200 -400200-4101000000000202-41010000000PROBLEM 4.60 (Cont.)00000000000000000000-2 0 0 0 0 0 000001000000001010000000-4 0 0 1 0 0 00000 -4 2 0 1 0 00000 1 -4 1 0 1 00001 0 1 -4 0 0 10000 1 0 0 -4 2 0 1000 0 1 0 -1 -4 10100 0 0 1 0 1 -40010 0 0 0 2 0 0 -10.66 200 0 0 0 0 2 0 1 -10.66 1 000000201 -10.66T1 153.9 T2 159.7 T3 176.4 T4 148.0 T5 154.4 T6 172.9 T 129.4 7[T] = T8 = 137.0 o CT 160.7 T9 95.6 10 103.5 T11 T12 132.8 T13 45.8 T14 48.7 T15 67.0 ()Considering symmetry, the heat transfer rate to the fluid is twice the convection rate from the surfaces of the control volumesexposed to the fluid. Using Newtons law of cooling, considering a unit thickness of the bar, findyyq conv = 2 h 2 ( T13 T ) + h y ( T14 T ) + h y ( T15 T ) + h 2 ( 200 T ) 11q conv = 2h y (T13 T ) + ( T14 T ) + ( T15 T ) + ( 200 T ) 22W11q conv = 2 50 0.1m ( 45.8 30 ) + ( 48.7 30 ) + ( 67.0 30 ) + ( 200 30 ) 22m2 Kq conv = 1487 W/m.<PROBLEM 4.61KNOWN: Upper surface and grooves of a plate are maintained at a uniform temperature T1, while thelower surface is maintained at T2 or is exposed to a fluid at T.FIND: (a) Heat rate per width of groove spacing (w) for isothermal top and bottom surfaces using afinite-difference method with x = 40 mm, (b) Effect of grid spacing and convection at bottom surface.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: (a) Using a space increment of x = 40 mm, the symmetrical section shown in schematic(a) corresponds to one-half the groove spacing. There exist only two interior nodes for which finitedifference equations must be written.Node a:Node b:4Ta (T1 + Tb + T2 + T1 ) = 04Ta ( 200 + Tb + 20 + 200 ) = 0or4Ta Tb = 420(1)or2Ta + 4Tb = 220(2)4Tb (T1 + Ta + T2 + Ta ) = 04Tb ( 200 + 2Ta + 20 ) = 0Multiply Eq. (2) by 2 and subtract from Eq. (1) to obtain7Tb = 860orTb = 122.9CFrom Eq. (1),4Ta - 122.9 = 420orTa = (420 + 122.9)/4 = 135.7C.The heat transfer through the symmetrical section is equal to the sum of heat flows through controlvolumes adjacent to the lower surface. From the schematic,T T2 x T1 T2 x Tb T2q = q1 + q + q = k .+ k ( x ) a+ k23y 2 y 2 yContinued...PROBLEM 4.61 (Cont.)Noting that x = y, regrouping and substituting numerical values, find11q = k ( T1 T2 ) + ( Ta T2 ) + ( Tb T2 )2211q = 15 W m K ( 200 20 ) + (135.7 20 ) + (122.9 20 ) = 3.86 kW m .22<For the full groove spacing, qtotal = 2 3.86 kW/m = 7.72 kW/m.(b) Using the Finite-Difference Equations option from the Tools portion of the IHT menu, the followingtwo-dimensional temperature field was computed for the grid shown in schematic (b), where x and y arein mm and the nodal temperatures are in C. Nodes 2-54 are interior nodes, with those along thesymmetry adiabats characterized by Tm-1,n = Tm+1,n, while nodes 55-63 lie on a plane surface.y\x010203040506070800200141.497.0957.6920102030200175.4134.394.6256.8320200182.4160.3125.790.2755.0120200186.7169.5148.9118.085.7352.952040200191177.2160.1140.1111.681.7351.042050200186.6171.2153.4133.5106.778.5149.462060200184.3167.5149.0128.7103.176.1748.312070200183.1165.5146.4125.7100.974.7347.602080200182.8164.8145.5124.4100.174.2447.3620The foregoing results were computed for h = 107 W/m2K (h ) and T = 20C, which is tantamountto prescribing an isothermal bottom surface at 20C. Agreement between corresponding results for thecoarse and fine grids is surprisingly good (Ta = 135.7C T23 = 140.1C; Tb = 122.9C T27 =124.4C). The heat rate isq = 2 k ( T46 T55 ) 2 + ( T47 T56 ) + ( T48 T57 ) + ( T49 T58 ) + ( T50 T59 )+ ( T51 T60 ) + (T52 T61 ) + ( T53 T62 ) + ( T54 T63 ) 2 q = 2 15 W m K [18.84 + 36.82 + 35.00 + 32.95 + 31.04 + 29.46<+28.31 + 27.6 + 13.68] C = 7.61kW mThe agreement with q T = 7.72 kW/m from the coarse grid of part (a) is excellent and a fortuitousconsequence of compensating errors. With reductions in the convection coefficient from h to h =1000, 200 and 5 W/m2K, the corresponding increase in the thermal resistance reduces the heat rate tovalues of 6.03, 3.28 and 0.14 kW/m, respectively. With decreasing h, there is an overall increase innodal temperatures, as, for example, from 191C to 199.8C for T2 and from 20C to 196.9C for T55.NOTE TO INSTRUCTOR: To reduce computational effort, while achieving the same educationalobjectives, the problem statement has been changed to allow for convection at the bottom, rather than thetop, surface.PROBLEM 4.62KNOWN: Rectangular plate subjected to uniform temperature boundaries.FIND: Temperature at the midpoint using a finite-difference method with space increment of 0.25mSCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties.ANALYSIS: For the nodal network above, 12 finite-difference equations must be written. It followsthat node 8 represents the midpoint of the rectangle. Since all nodes are interior nodes, Eq. 4.33 isappropriate and is written in the form4Tm Tneighbors = 0.For nodes on the symmetry adiabat, the neighboring nodes include two symmetrical nodes. Hence, forNode 4, the neighbors are Tb, T8 and 2T3. Because of the simplicity of the finite-difference equations,we may proceed directly to the matrices [A] and [C] see Eq. 4.52 and matrix inversion can beused to find the nodal temperatures Tm. 4 1 0 0 11 4 1 0 00 1 4 1 000 2 4 010 0 0 40A = 0 1 0 0 10010000 1 0000 0 1000 0 0000 0 0000 0 00000010000010000010010010410011 41000 2 400000 4 110 0 1 4010010010000000010014200000001001-4 200 150 150 150 50 C = 0 00 100 50 50 50 96.5 112.9 118.9 120.4 73.2T = 86.2 92.3 94.0 59.9 65.5 69.9 71.0The temperature at the midpoint (Node 8) isT (1,0.5) = T8 = 94.0o C.<COMMENTS: Using the exact analytical, solution see Eq. 4.19 and Problem 4.2 the midpointtemperature is found to be 94.5C. To improve the accuracy of the finite-difference method, it wouldbe necessary to decrease the nodal mesh size.PROBLEM 4.63KNOWN: Long bar with trapezoidal shape, uniform temperatures on two surfaces, and two insulatedsurfaces.FIND: Heat transfer rate per unit length using finite-difference method with space increment of10mm.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties.ANALYSIS: The heat rate can be found after the temperature distribution has been determined.Using the nodal network shown above with x = 10mm, nine finite-difference equations must bewritten. Nodes 1-4 and 6-8 are interior nodes and their finite-difference equations can be writtendirectly from Eq. 4.33. For these nodesTm,n+1 + Tm,n-1 + Tm+1,n + Tm-1,n 4Tm,n = 0m = 1 4, 6 8.(1)For nodes 5 and 9 located on the diagonal, insulated boundary, the appropriate finite-difference equationfollows from an energy balance on the control volume shown above (upper-right corner of schematic),&&E in E out = q a + qb = 0T Tm,nTTk ( y 1) m-1,n+ k ( x 1) m,n-1 m,n = 0.xySince x = y, the finite-difference equation for nodes 5 and 9 is of the formTm-1,n + Tm,n-1 2Tm,n = 0m = 5,9.(2)The system of 9 finite-difference equations is first written in the form of Eqs. (1) or (2) and thenwritten in explicit form for use with the Gauss-Seidel iteration method of solution; see Section 4.5.2.Node123456789Finite-difference equationT2+T2+T6+100-4T1 = 0T3+T1+T7+100-4T2 = 0T4+T2+T8+100-4T3 = 0T5+T3+T9+100-4T4 = 0100+T4-2T5 = 0T7+T7+25+T1-4T6 = 0T8+T6+25+T2-4T7 = 0T9+T7+25+T3-4T8 = 0T4+T8-2T9 = 0Gauss-Seidel formT1 = 0.5T2+0.25T6+25T2 = 0.25(T1+T3+T7)+25T3 = 0.25(T2+T4+T8)+25T4 = 0.25(T3+T5+T9)+25T5 = 0.5T4+50T6 = 0.25T1+0.5T7+6.25T7 = 0.25(T2+T6+T8)+6.25T8 = 0.25(T3+T7+T9)+6.25T9 = 0.5(T4+T8)Continued ..PROBLEM 4.63 (Cont.)The iteration process begins after an initial guess (k = 0) is made. The calculations are shown in thetable below.kT1T2T3T4T5T6T7T8T9( C)01234567575.075.776.376.376.676.67576.376.977.077.377.377.58080.080.080.280.280.380.38586.386.386.386.386.386.49092.593.293.293.293.293.25050.051.351.351.751.751.95052.552.252.752.752.952.96057.557.557.357.557.457.57572.571.971.971.871.971.9Note that by the sixth iteration the change is less than 0.3C; hence, we assume the temperaturedistribution is approximated by the last row of the table.The heat rate per unit length can be determined by evaluating the heat rates in the x-direction for thecontrol volumes about nodes 6, 7, and 8. From the schematic, find thatq = q1 + q2 + q3T 25T 25y T6 25q = ky 8+ ky 7+kxx2xRecognizing that x = y and substituting numerical values, findq = 20W1( 57.5 25) + ( 52.9 25 ) + 2 ( 51.9 25) Km K q = 1477 W/m.<COMMENTS: (1) Recognize that, while the temperature distribution may have been determined to areasonable approximation, the uncertainty in the heat rate could be substantial. This follows since theheat rate is based upon a gradient and hence on temperature differences.(2) Note that the initial guesses (k = 0) for the iteration are within 5C of the final distribution. Thegeometry is simple enough that the guess can be very close. In some instances, a flux plot may behelpful and save labor in the calculation.(3) In writing the FDEs, the iteration index (superscript k) was not included to simplify expression ofthe equations. However, the most recent value of Tm,n is always used in the computations. Note thatthis system of FDEs is diagonally dominant and no rearrangement is required.PROBLEM 4.64KNOWN: Edge of adjoining walls (k = 1 W/mK) represented by symmetrical element bounded by thediagonal symmetry adiabat and a section of the wall thickness over which the temperature distribution isassumed to be linear.FIND: (a) Temperature distribution, heat rate and shape factor for the edge using the nodal network with= x = y = 10 mm; compare shape factor result with that from Table 4.1; (b) Assess the validity ofassuming linear temperature distributions across sections at various distances from the edge.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties, and (3) Lineartemperature distribution at specified locations across the section.ANALYSIS: (a) Taking advantage of symmetry along the adiabat diagonal, all the nodes may be treatedas interior nodes. Across the left-hand boundary, the temperature distribution is specified as linear. Thefinite-difference equations required to determine the temperature distribution, and hence the heat rate,can be written by inspection.T3 = 0.25 ( T2 + T4 + T6 + Tc )T4 = 0.25 ( T2 + T5 + T7 + T3 )T5 = 0.25 ( T2 + T2 + T4 + T4 )T6 = 0.25 ( T3 + T7 + T8 + Tb )T7 = 0.25 ( T4 + T4 + T6 + T6 )T8 = 0.25 ( T6 + T6 + Ta + Ta )The heat rate for both surfaces of the edge isqtot = 2 [q + q + q + q ]abcdq = 2 [k ( x 2 )( Tc T2 ) y + kx ( T3 T2 ) y + kx ( T4 T2 ) y + kx ( T5 T2 ) x ]totThe shape factor for the full edge is defined asqtot = kS ( T1 T2 )Solving the above equation set in IHT, the temperature (C) distribution isContinued...PROBLEM 4.64 (Cont.)0000025 18.75 12.5 6.2550 37.5 25.075 56.2500<and the heat rate and shape factor areqtot = 100 W m<S =1From Table 4.1, the edge shape factor is 0.54, considerably below our estimate from this coarse gridanalysis.(b) The effect of the linear temperature distribution on the shape factor estimate can be explored using amore extensive grid as shown below. The FDE analysis was performed with the linear distributionimposed as the different sections a, b, c, d, e. Following the same approach as above, findLocation of linear distributionShape factor, S(a)0.797(b)0.799(c)0.809(d)0.857(e)1.00The shape factor estimate decreases as the imposed linear temperature distribution section is locatedfurther from the edge. We conclude that assuming the temperature distribution across the section directlyat the edge is a poor-one.COMMENTS: The grid spacing for this analysis is quite coarse making the estimates in poor agreementwith the Table 4.1 result. However, the analysis does show the effect of positioning the lineartemperature distribution condition.PROBLEM 4.65KNOWN: Long triangular bar insulated on the diagonal while sides are maintained at uniformtemperatures Ta and Tb.FIND: (a) Using a nodal network with five nodes to the side, and beginning with properly definedcontrol volumes, derive the finite-difference equations for the interior and diagonal nodes and obtain thetemperature distribution; sketch the 25, 50 and 75C isotherms and (b) Recognizing that the insulateddiagonal surface can be treated as a symmetry line, show that the diagonal nodes can be treated asinterior nodes, and write the finite-difference equations by inspection.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, and (3) Constantproperties.ANALYSIS: (a) For the nodal network shown above, nodes 2, 4, 5, 7, 8 and 9 are interior nodes and,since x = y, the corresponding finite-difference equations are of the form, Eq. 4.33,Tj = 1 4 Tneighbors(1)For a node on the adiabatic, diagonal surface, an energy balance, E in E out = 0 , yieldsq + q + q = 0abcT TT T3=00 + kx 5 3 + ky 2yxT3 = 1 2 ( T2 + T5 )(2)That is, for the diagonal nodes, m,Tm = 1 2 Tneighbors(3)To obtain the temperature distributions, enter Eqs. (1, 2, 3) into the IHT workspace and solve for thenodal temperatures (C), tabulated according to the nodal arrangement:Continued...PROBLEM 4.65 (Cont.)00 85.7100 71.43 50.0000 50.00 28.57 14.29000The 25, 50 and 75C isotherms are sketched below, using an interpolation scheme to scale the isothermson the triangular bar.(b) If we consider the insulated surface as a symmetry plane, the nodal network appears as shown. Assuch, the diagonal nodes can be treated as interior nodes, as Eq. (1) above applies. Recognize the form isthe same as that of Eq. (2) or (3).COMMENTS: Always look for symmetry conditions which can greatly simplify the writing of nodalequations. In this situation, the adiabatic surface can be treated as a symmetry plane such that the nodescan be treated as interior nodes, and the finite-difference equations can be written by inspection.PROBLEM 4.66KNOWN: Straight fin of uniform cross section with prescribed thermal conditions and geometry; tipcondition allows for convection.FIND: (a) Calculate the fin heat rate, q , and tip temperature, TL , assuming one-dimensional heatftransfer in the fin; calculate the Biot number to determine whether the one-dimensional assumption isvalid, (b) Using the finite-element software FEHT, perform a two-dimensional analysis to determinethe fin heat rate and the tip temperature; display the isotherms; describe the temperature field and theheat flow pattern inferred from the display, and (c) Validate your FEHT code against the 1-Danalytical solution for a fin using a thermal conductivity of 50 and 500 W/mK.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conduction with constant properties, (2) Negligible radiationexchange, (3) Uniform convection coefficient.ANALYSIS: (a) Assuming one-dimensional conduction, q and TL can be determined using Eqs.L3.72 and 3.70, respectively, from Table 3.4, Case A. Alternatively, use the IHT Model | ExtendedSurfaces | Temperature Distribution and Heat Rate | Straight Fin | Rectangular. These results aretabulated below and labeled as 1-D. The Biot number for the fin ish t/2500 W / m 2 K 0.020 m / 2Bi =(k)=(5 W / mK) =1(b, c) The fin can be drawn as a two-dimensional outline in FEHT with convection boundaryconditions on the exposed surfaces, and with a uniform temperature on the base. Using a fine mesh (atleast 1280 elements), solve for the temperature distribution and use the View | Temperature Contourscommand to view the isotherms and the Heat Flow command to determine the heat rate into the finbase. The results of the analysis are summarized in the table below.k(W/mK)550500Bi10.10.01(Tip temperature, TL (C)1-D2-D100100100.3100123.8124Fin heat rate, q (W/m)f1-D1010319498122-D80529909563Difference*(%)206.42.5)* Difference = q ,1D q ,2D 100 / q ,1DfffCOMMENTS: (1) From part (a), since Bi = 1 > 0.1, the internal conduction resistance is notnegligible. Therefore significant transverse temperature gradients exist, and the one-dimensionalconduction assumption in the fin is a poor one.Continued ..PROBLEM 4.66 (Cont.)(2) From the table, with k = 5 W/mK (Bi = 1), the 2-D fin heat rate obtained from the FEA analysis is20% lower than that for the 1-D analytical analysis. This is as expected since the 2-D model accountsfor transverse thermal resistance to heat flow. Note, however, that analyses predict the same tiptemperature, a consequence of the fin approximating an infinitely long fin (mL = 20.2 >> 2.56; see Ex.3.8 Comments).(3) For the k = 5 W/mK case, the FEHT isotherms show considerable curvature in the region near thefin base. For example, at x = 10 and 20 mm, the difference between the centerline and surfacetemperatures are 15 and 7C.(4) From the table, with increasing thermal conductivity, note that Bi decreases, and the onedimensional heat transfer assumption becomes more appropriate. The difference for the case when k =500 W/mK is mostly due to the approximate manner in which the heat rate is calculated in the FEAsoftware.PROBLEM 4.67KNOWN: Long rectangular bar having one boundary exposed to a convection process (T, h) whilethe other boundaries are maintained at constant temperature Ts.FIND: Using the finite-element method of FEHT, (a) Determine the temperature distribution, plotthe isotherms, and identify significant features of the distribution, (b) Calculate the heat rate per unitlength (W/m) into the bar from the air stream, and (c) Explore the effect on the heat rate of increasingthe convection coefficient by factors of two and three; explain why the change in the heat rate is notproportional to the change in the convection coefficient.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two dimensional conduction, (2) Constant properties.ANALYSIS: (a) The symmetrical section shown in the schematic is drawn in FEHT with thespecified boundary conditions and material property. The View | Temperature Contours command isused to represent ten isotherms (isopotentials) that have minimum and maximum values of 53.9C and85.9C, respectively.Because of the symmetry boundary condition, the isotherms are normal to the center-plane indicatingan adiabatic surface. Note that the temperature change along the upper surface of the bar issubstantial ( 40C), whereas the lower half of the bar has less than a 3C change. That is, the lowerhalf of the bar is largely unaffected by the heat transfer conditions at the upper surface.(b, c) Using the View | Heat Flows command considering the upper surface boundary with selectedconvection coefficients, the heat rates into the bar from the air stream were calculated.h W / m2 K100200 300(q ( W / m ))128175206Increasing the convection coefficient by factors of 2 and 3, increases the heat rate by 37% and 61%,respectively. The heat rate from the bar to the air stream is controlled by the thermal resistances ofthe bar (conduction) and the convection process. Since the conduction resistance is significant, weshould not expect the heat rate to change proportionally to the change in convection resistance.PROBLEM 4.68KNOWN: Log rod of rectangular cross-section of Problem 4.55 that experiences uniform heatgeneration while its surfaces are maintained at a fixed temperature. Use the finite-element softwareFEHT as your analysis tool.FIND: (a) Represent the temperature distribution with representative isotherms; identify significantfeatures; and (b) Determine what heat generation rate will cause the midpoint to reach 600 K withunchanged boundary conditions.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, and (2) Two-dimensional conduction with constantproperties.ANALYSIS: (a) Using FEHT, do the following: in Setup, enter an appropriate scale; Draw theoutline of the symmetrical section shown in the above schematic; Specify the Boundary Conditions(zero heat flux or adiabatic along the symmetrical lines, and isothermal on the edges). Also Specify theMaterial Properties and Generation rate. Draw three Element Lines as shown on the annotatedversion of the FEHT screen below. To reduce the mesh, hit Draw/Reduce Mesh until the desiredfineness is achieved (256 elements is a good choice).Continued PROBLEM 4.68 (Cont.)After hitting Run, Check and then Calculate, use the View/Temperature Contours and select the 10isopotential option to display the isotherms as shown in an annotated copy of the FEHT screen below.The isotherms are normal to the symmetrical lines as expected since those surfaces are adiabatic. Theisotherms, especially near the center, have an elliptical shape. Along the x = 0 axis and the y = 10mm axis, the temperature gradient is nearly linear. The hottest point is of course the center for whichthe temperature is(T(0, 10 mm) = 401.3 K.<The temperature of this point can be read using the View/Temperatures or View|Tabular Outputcommand.(b) To determine the required generation rate so that T(0, 10 mm) = 600 K, it is necessary to re-run themodel with several guessed values of q . After a few trials, findq = 1.48 108 W / m3<PROBLEM 4.69KNOWN: Symmetrical section of a flow channel with prescribed values of q and k, as well as thesurface convection conditions. See Problem 4.5(S).FIND: Using the finite-element method of FEHT, (a) Determine the temperature distribution and plotthe isotherms; identify the coolest and hottest regions, and the region with steepest gradients; describethe heat flow field, (b) Calculate the heat rate per unit length (W/m) from the outer surface A to theadjacent fluid, (c) Calculate the heat rate per unit length (W/m) to surface B from the inner fluid, and(d) Verify that the results are consistent with an overall energy balance on the section.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.ANALYSIS: (a) The symmetrical section shown in the schematic is drawn in FEHT with thespecified boundary conditions, material property and generation. The View | Temperature Contourscommand is used to represent ten isotherms (isopotentials) that have minimum and maximum valuesof 82.1C and 125.2C.The hottest region of the section is the upper vertical leg (left-hand corner). The coolest region is inthe lower horizontal leg at the far right-hand boundary. The maximum and minimum sectiontemperatures (125C and 77C), respectively, are higher than either adjoining fluid. Rememberingthat heat flow lines are normal to the isotherms, heat flows from the hottest corner directly to the innerfluid and downward into the lower leg and then flows out surface A and the lower portion of surfaceB.Continued ..PROBLEM 4.69 (Cont.)(b, c) Using the View | Heat Flows command considering the boundaries for surfaces A and B, theheat rates are:q = 1135 W / msq = 1365 W / m.B<From an energy balance on the section, we note that the results are consistent since conservation ofenergy is satisfied.E E + E g = 0inoutq + q + q = 0AB1135 W / m + ( 1365 W / m ) + 2500 W / m = 0<where q = 1 106 W / m3 [25 50 + 25 50] 106 m 2 = 2500 W / m.COMMENTS: (1) For background on setting up this problem in FEHT, see the tutorial example ofthe Users Manual. While the boundary conditions are different, and the internal generation term is tobe included, the procedure for performing the analysis is the same.(2) The heat flow distribution can be visualized using the View | Temperature Gradients command.The direction and magnitude of the heat flow is represented by the directions and lengths of arrows.Compare the heat flow distribution to the isotherms shown above.PROBLEM 4.70KNOWN: Hot-film flux gage for determining the convection coefficient of an adjoining fluid streamby measuring the dissipated electric power, Pe , and the average surface temperature, Ts,f.FIND: Using the finite-element method of FEHT, determine the fraction of the power dissipation thatis conducted into the quartz substrate considering three cases corresponding to convection coefficients2of 500, 1000 and 2000 W/m K.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant substrate properties,(3) Uniform convection coefficient over the hot-film and substrate surfaces, (4) Uniform powerdissipation over hot film.ANALYSIS: The symmetrical section shown in the schematic above (right) is drawn into FEHTspecifying the substrate material property. On the upper surface, a convection boundary condition((T,h) is specified over the full width W/2. Additionally, an applied uniform flux Pe , W / m 2)boundary condition is specified for the hot-film region (w/2). The remaining surfaces of the twodimensional system are specified as adiabatic. In the schematic below, the electrical power dissipationPe (W/m) in the hot film is transferred by convection from the film surface, q v,f , and from thecadjacent substrate surface, q v,s .cThe analysis evaluates the fraction, F, of the dissipated electrical power that is conducted into thesubstrate and convected to the fluid stream,F = q / Pe = 1 qcv,s cv,f / Pewhere Pe = Pe ( w / 2 ) = 5000 W / m 2 ( 0.002 m ) = 10 W / m.After solving for the temperature distribution, the View|Heat Flow command is used to evaluate q v,fcfor the three values of the convection coefficient.Continued ..PROBLEM 4.70 (Cont.)2h(W/m K)Case123qcv,f ( W / m )500100020005.646.747.70F(%)Ts,f (C)43.632.623.330.928.627.0COMMENTS: (1) For the ideal hot-film flux gage, there is negligible heat transfer to the substrate,and the convection coefficient of the air stream is calculated from the measured electrical power, Pe ,the average film temperature (by a thin-film thermocouple), Ts,f, and the fluid stream temperature, T,as h = Pe / Ts,f T . The purpose in performing the present analysis is to estimate a correction()factor to account for heat transfer to the substrate.(2) As anticipated, the fraction of the dissipated electrical power conducted into the substrate, F,decreases with increasing convection coefficient. For the case of the largest convection coefficient, Famounts to 25%, making it necessary to develop a reliable, accurate heat transfer model to estimate theapplied correction. Further, this condition limits the usefulness of this gage design to flows with highconvection coefficients.(3) A reduction in F, and hence the effect of an applied correction, could be achieved with a substratematerial having a lower thermal conductivity than quartz. However, quartz is a common substratematerial for fabrication of thin-film heat-flux gages and thermocouples. By what other means couldyou reduce F?(4) In addition to the tutorial example in the FEHT Users Manual, the solved models for Examples4.3 and 4.4 are useful for developing skills helpful in solving this problem.PROBLEM 4.71KNOWN: Hot-plate tool for micro-lithography processing of 300-mm silicon wafer consisting of analuminum alloy equalizing block (EB) heated by ring-shaped main and trim electrical heaters (MHand TH) providing two-zone control.FIND: The assignment is to size the heaters, MH and TH, by specifying their applied heat fluxes,q and q , and their radial extents, rmh and rth , to maintain an operating temperature ofmhth140C with a uniformity of 0.1C. Consider these steps in the analysis: (a) Perform an energybalance on the EB to obtain an initial estimate for the heater fluxes with q h = q extending overmththe full radial limits; using FEHT, determine the upper surface temperature distribution and commenton whether the desired uniformity has been achieved; (b) Re-run your FEHT code with differentvalues of the heater fluxes to obtain the best uniformity possible and plot the surface temperaturedistribution; (c) Re-run your FEHT code for the best arrangement found in part (b) using therepresentative distribution of the convection coefficient (see schematic for h(r) for downward flowinggas across the upper surface of the EB; adjust the heat flux of TH to obtain improved uniformity; and(d) Suggest changes to the design for improving temperature uniformity.SCHEMATIC:<ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction with uniform andconstant properties in EB, (3) Lower surface of EB perfectly insulated, (4) Uniform convectioncoefficient over upper EB surface, unless otherwise specified and (5) negligible radiation exchangebetween the EB surfaces and the surroundings.ANALYSIS: (a) To obtain initial estimates for the MH and TH fluxes, perform an overall energybalance on the EB as illustrated in the schematic below.Ein Eout = 0()()22222q r2 r1 + q r4 r3 h ro + 2 ro w ( Ts T ) = 0mhthContinued ..PROBLEM 4.71 (Cont.)Substituting numerical values and letting q h = q , findmthq = q = 2939 W / m 2mhth<Using FEHT, the analysis is performed on an axisymmetric section of the EB with the nodalarrangement as shown below.The Temperature Contour view command is used to create the temperature distribution shown below.The temperatures at the center (T1) and the outer edge of the wafer (r = 150 mm, T14) are read fromthe Tabular Output page. The Temperature Gradients view command is used to obtain the heat flowdistribution when the line length is proportional to the magnitude of the heat rate.From the analysis results, for this base case design ( q = q ) , the temperature difference acrossmhththe radius of the wafer is 1.7C, much larger than the design goal of 0.1C. The upper surfacetemperature distribution is shown in the graph below.Continued ..PROBLEM 4.71 (Cont.)EB surface temperature distribution141.5141T(r,z), (C)140.5140139.5139138.5020406080100120140160Radial position, r (mm)(b) From examination of the results above, we conclude that if q h is reduced and qh increased, themtEB surface temperature uniformity could improve. The results of three trials compared to the basecase are tabulated below.Trial(qmhW/m2)(qthW / m2)T1T14(C )(C )T1 T14( C )Base29392939141.1139.31.812880(-2%)2997(+2%)141.1139.41.722880(-2%)3027(+3%)141.7140.01.732910(-1%)2997(+2%)141.7139.91.829392939141.7139.12.6Part (d) 2939k=150 W/mK2939140.4139.50.9Part (d) 2939k=300 W/mK2939140.0139.60.4Part (c)The strategy of changing the heater fluxes (trials 1-3) has not resulted in significant improvements inthe EB surface temperature uniformity.Continued ..PROBLEM 4.71 (Cont.)(c) Using the same FEHT code as with part (b), base case, the boundary conditions on the uppersurface of the EB were specified by the function h(r) shown in the schematic. The value of h(r)2ranged from 5.4 to 13.5 W/m K between the centerline and EB edge. The result of the analysis istabulated above, labeled as part (c). Note that the temperature uniformity has become significantlypoorer.(d) There are at least two options that should be considered in the re-design to improve temperatureuniformity. Higher thermal conductivity material for the EB. Aluminum alloy is the material mostwidely used in practice for reasons of low cost, ease of machining, and durability of the heatedsurface. The results of analyses for thermal conductivity values of 150 and 300 W/mK are tabulatedabove, labeled as part (d). Using pure or oxygen-free copper could improve the temperatureuniformity to better than 0.5C.Distributed heater elements. The initial option might be to determine whether temperature uniformitycould be improved using two elements, but located differently. Another option is a single elementheater spirally embedded in the lower portion of the EB. By appropriately positioning the element asa function of the EB radius, improved uniformity can be achieved. This practice is widely used whereprecise and uniform temperature control is needed.PROBLEM 4.72KNOWN: Straight fin of uniform cross section with insulated end.FIND: (a) Temperature distribution using finite-difference method and validity of assuming onedimensional heat transfer, (b) Fin heat transfer rate and comparison with analytical solution, Eq. 3.76, (c)Effect of convection coefficient on fin temperature distribution and heat rate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fin, (3) Constantproperties, (4) Uniform film coefficient.ANALYSIS: (a) From the analysis of Problem 4.45, the finite-difference equations for the nodalarrangement can be directly written. For the nodal spacing x = 4 mm, there will be 12 nodes. With >> w representing the distance normal to the page,)(hPh 2h 2 2500 W m 2 K 24 103 mm = 0.0533 x 2 x 2 =x =3 mkA ck wkw50 W m K 6 10Node 1:Node n:Node 12:100 + T2 + 0.0533 30 ( 2 + 0.0533) T1 = 0 or-2.053T1 + T2 = -101.6Tn +1 + Tn 1 + 1.60 2.0533Tn = 0orT11 + ( 0.0533 2 ) 30 ( 0.0533 2 + 1) T12 = 0 orTn 1 2.053Tn + Tn 1 = 1.60T11 1.0267T12 = 0.800Using matrix notation, Eq. 4.52, where [A] [T] = [C], the A-matrix is tridiagonal and only the non-zeroterms are shown below. A matrix inversion routine was used to obtain [T].Tridiagonal Matrix ANonzero Termsa1,1a1,2a2,1a2,2a2,3a3,2a3,3a3,4a4,3a4,4a4,5a5,4a5,5a5,6a6,5a6,6a6,7a7,6a7,7a7,8a8,7a8,8a8,9a9,8a9,9a9,10a10,9a10,10a10,11a11,10a11,11a11,12a12,11a12,12a12,1311111111111Column MatricesValues-2.053-2.053-2.053-2.053-2.053-2.053-2.053-2.053-2.053-2.053-2.053-1.027111111111111Node123456789101112C-101.6-1.6-1.6-1.6-1.6-1.6-1.6-1.6-1.6-1.6-1.6-0.8T85.874.565.658.653.148.845.543.041.239.939.238.9The assumption of one-dimensional heat conduction is justified when Bi h(w/2)/k < 0.1. Hence, withBi = 500 W/m2K(3 10-3 m)/50 W/mK = 0.03, the assumption is reasonable.Continued...PROBLEM 4.72 (Cont.)(b) The fin heat rate can be most easily found from an energy balance on the control volume about Node0,T0 T1 x + h2q = q1 + q ( T0 T )fconv = k wx 2(q = 50 W m K 6 103 mf)(100 85.8 ) C4 103+ 500 W m2 K 2 m4 103 m 2 (100 30 ) Cq = (1065 + 140 ) W m = 1205 W m .fFrom Eq. 3.76, the fin heat rate isq = ( hPkA c )1/ 2< b tanh mL .Substituting numerical values with P = 2(w + ) 2 and Ac = w , m = (hP/kAc)1/2 = 57.74 m-1 and M= (hPkAc)1/2 = 17.32 W/K. Hence, with b = 70C,q = 17.32 W K 70 K tanh (57.44 0.048 ) = 1203 W mand the finite-difference result agrees very well with the exact (analytical) solution.(c) Using the IHT Finite-Difference Equations Tool Pad for 1D, SS conditions, the fin temperaturedistribution and heat rate were computed for h = 10, 100, 500 and 1000 W/m2K. Results are plotted asfollows.100180015008070Heat rate, q'(W/m)Temperature, T(C)90605040300816243240481200900600300Fin location, x(mm)h = 10 W/m^2.Kh = 100 W/m^2.Kh = 500 W/m^2.Kh = 1000 W/m^2.K002004006008001000Convection coefficient, h(W/m^2.K)The temperature distributions were obtained by first creating a Lookup Table consisting of 4 rows ofnodal temperatures corresponding to the 4 values of h and then using the LOOKUPVAL2 interpolatingfunction with the Explore feature of the IHT menu. Specifically, the function T_EVAL =LOOKUPVAL2(t0467, h, x) was entered into the workspace, where t0467 is the file name given to theLookup Table. For each value of h, Explore was used to compute T(x), thereby generating 4 data setswhich were placed in the Browser and used to generate the plots. The variation of q with h was simplygenerated by using the Explore feature to solve the finite-difference model equations for values of hincremented by 10 from 10 to 1000 W/m2K.Although q increases with increasing h, the effect of changes in h becomes less pronounced. Thisftrend is a consequence of the reduction in fin temperatures, and hence the fin efficiency, with increasingh. For 10 h 1000 W/m2K, 0.95 f 0.24. Note the nearly isothermal fin for h = 10 W/m2K andthe pronounced temperature decay for h = 1000 W/m2K.PROBLEM 4.73KNOWN: Pin fin of 10 mm diameter and length 250 mm with base temperature of 100C experiencingradiation exchange with the surroundings and free convection with ambient air.FIND: Temperature distribution using finite-difference method with five nodes. Fin heat rate andrelative contributions by convection and radiation.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fin, (3) Constantproperties, (4) Fin approximates small object in large enclosure, (5) Fin tip experiences convection andradiation, (6) hfc = 2.89[0.6 + 0.624(T - T)1/6]2.ANALYSIS: To apply the finite-difference method, define the 5-node system shown above where x =L/5. Perform energy balances on the nodes to obtain the finite-difference equations for the nodaltemperatures.Interior node, n = 1, 2, 3 or 4Ein Eout = 0qa + q b + qc + qd = 0(1)TTTTh r,n Px (Tsur Tn ) + kA c n +1 n + h fc,n Px (T Tn ) + kA c n 1 n = 0xx(2)where the free convection coefficient is1/ 6 2h fc,n = 2.89 0.6 + 0.624 ( Tn T )and the linearized radiation coefficient is22h r,n = (Tn + Tsur ) Tn + Tsurwith P = D and Ac = D2/4.()(3)(4)(5,6)Tip node, n = 5Ein Eout = 0qa + q b + qc + qd + qe = 0h r,5 ( Px 2 )(Tsur T5 ) + h r,5A c (Tsur T5 ) + h fc,5Ac (T T5 )T T+ h fc,5 ( Px 2 )( T T5 ) + kAc 4 5 = 0x(7)Continued...PROBLEM 4.73 (Cont.)Knowing the nodal temperatures, the heat rates are evaluated as:Fin Heat Rate: Perform an energy balancearound Node b.Ein Eout = 0qa + q b + q c + q fin = 0h r,b ( Px 2 )(Tsur Tb ) + h fc,b ( Px 2 )(T Tb ) + kA c(T1 Tb ) + qxfin = 0(8)where hr,b and hfc,b are evaluated at Tb.Convection Heat Rate: To determine the portion of the heat rate by convection from the fin surface, weneed to sum contributions from each node. Using the convection heat rate terms from the foregoingenergy balances, for, respectively, node b, nodes 1, 2, 3, 4 and node 5.qcv = q b )b qc )1 4 ( qc + q d )5(9)Radiation Heat Rate: In the same manner,q rad = qa )b q b )1 4 ( qa + q b )5The above equations were entered into the IHT workspace and the set of equations solved for the nodaltemperatures and the heat rates. Summary of key results including the temperature distribution and heatrates is shown below.Nodeb12345FinTj (C)qcv (W)qfin (W)qrad (W)hcv (W/m2K)hrod (W/m2K)1000.60310.11.558.50.4518.61.440.90.1837.31.333.10.0816.41.329.80.0435.71.228.80.0155.51.21.3751.6040.229-<COMMENTS: From the tabulated results, it is evident that free convection is the dominant node. Notethat the free convection coefficient varies almost by a factor of two over the length of the fin.PROBLEM 4.74KNOWN: Thin metallic foil of thickness, t, whose edges are thermally coupled to a sink at temperatureTsink is exposed on the top surface to an ion beam heat flux, q TT , and experiences radiation exchange withsthe vacuum enclosure walls at Tsur.FIND: Temperature distribution across the foil.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, steady-state conduction in the foil, (2) Constant properties, (3)Upper and lower surfaces of foil experience radiation exchange, (4) Foil is of unit length normal to thepage.ANALYSIS: The 10-node network representing the foil is shown below.From an energy balance on node n, E in E out = 0 , for a unit depth,q + q + q + q + q = 0abcdeq x + h r,n x ( Tsur Tn ) + k ( t )( Tn +1 Tn ) x + h r,n x ( Tsur Tn ) + k ( t )( Tn 1 Tn ) x = 0 (1)swhere the linearized radiation coefficient for node n is(22h r,n = ( Tsur + Tn ) Tsur + TnSolving Eq. (1) for Tn find,)((2))()()Tn = ( Tn +1 + Tn 1 ) + 2h r,n x 2 kt Tsur + x 2 kt qs h r,n x 2 kt + 2 (3)which, considering symmetry, applies also to node 1. Using IHT for Eqs. (3) and (2), the set of finitedifference equations was solved for the temperature distribution (K):T1374.1T2374.0T3373.5T4372.5T5370.9T6368.2T7363.7T8356.6T9345.3T10327.4Continued...PROBLEM 4.74 (Cont.)COMMENTS: (1) If the temperature gradients were excessive across the foil, it would wrinkle; mostlikely since its edges are constrained, the foil will bow.(2) The IHT workspace for the finite-difference analysis follows:// The nodal equations:T1 = ( (T2 + T2) + A1 * Tsur + B *q''s ) / ( A1 + 2)A1= 2 * hr1 * deltax^2 / (k * t)hr1 = eps * sigma * (Tsur + T1) * (Tsur^2 + T1^2)sigma = 5.67e-8B = deltax^2 / (k * t)T2 = ( (T1 + T3) + A2 * Tsur + B *q''s ) / ( A2 + 2)A2= 2 * hr2 * deltax^2 / (k * t)hr2 = eps * sigma * (Tsur + T2) * (Tsur^2 + T2^2)T3 = ( (T2 + T4) + A3 * Tsur + B *q''s ) / ( A3 + 2)A3= 2 * hr3 * deltax^2 / (k * t)hr3 = eps * sigma * (Tsur + T3) * (Tsur^2 + T3^2)T4 = ( (T3 + T5) + A4 * Tsur + B *q''s ) / ( A4 + 2)A4= 2 * hr4 * deltax^2 / (k * t)hr4 = eps * sigma * (Tsur + T4) * (Tsur^2 + T4^2)T5 = ( (T4 + T6) + A5 * Tsur + B *q''s ) / ( A5 + 2)A5= 2 * hr5 * deltax^2 / (k * t)hr5 = eps * sigma * (Tsur + T5) * (Tsur^2 + T5^2)T6 = ( (T5 + T7) + A6 * Tsur + B *q''s ) / ( A6 + 2)A6= 2 * hr6 * deltax^2 / (k * t)hr6 = eps * sigma * (Tsur + T6) * (Tsur^2 + T6^2)T7 = ( (T6 + T8) + A7 * Tsur + B *q''s ) / ( A7 + 2)A7= 2 * hr7 * deltax^2 / (k * t)hr7 = eps * sigma * (Tsur + T7) * (Tsur^2 + T7^2)T8 = ( (T7 + T9) + A8 * Tsur + B *q''s ) / ( A8 + 2)A8= 2 * hr8 * deltax^2 / (k * t)hr8 = eps * sigma * (Tsur + T8) * (Tsur^2 + T8^2)T9 = ( (T8 + T10) + A9 * Tsur + B *q''s ) / ( A9 + 2)A9= 2 * hr9 * deltax^2 / (k * t)hr9 = eps * sigma * (Tsur + T9) * (Tsur^2 + T9^2)T10 = ( (T9 + Tsink) + A10 * Tsur + B *q''s ) / ( A10 + 2)A10= 2 * hr10 * deltax^2 / (k * t)hr10 = eps * sigma * (Tsur + T10) * (Tsur^2 + T10^2)// Assigned variablesdeltax = L / 10L = 0.150t = 0.00025eps = 0.45Tsur = 300k = 40Tsink = 300q''s = 600// Spatial increment, m// Foil length, m// Foil thickness, m// Emissivity// Surroundings temperature, K// Foil thermal conductivity, W/m.K// Sink temperature, K// Ion beam heat flux, W/m^2/* Data Browser results: Temperature distribution (K) and linearized radiation cofficients(W/m^2.K):T1T2374.1 374T3373.5T4372.5T5370.9T6368.2T7363.7T8356.6T9345.3T10327.4hr13.956hr33.943hr43.926hr53.895hr63.845hr73.765hr83.639hr93.444hr103.157 */hr23.953PROBLEM 4.75KNOWN: Electrical heating elements with known dissipation rate embedded in a ceramic plate ofknown thermal conductivity; lower surface is insulated, while upper surface is exposed to a convectionprocess.FIND: (a) Temperature distribution within the plate using prescribed grid spacing, (b) Sketch isothermsto illustrate temperature distribution, (c) Heat loss by convection from exposed surface (compare withelement dissipation rate), (d) Advantage, if any, in not setting x = y, (e) Effect of grid size andconvection coefficient on the temperature field.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction in ceramic plate, (2) Constantproperties, (3) No internal generation, except for Node 7 (or Node 15 for part (e)), (4) Heating elementapproximates a line source of negligible wire diameter.ANALYSIS: (a) The prescribed grid for the symmetry element shown above consists of 12 nodal points.Nodes 1-3 are points on a surface experiencing convection; nodes 4-6 and 8-12 are interior nodes. Node7 is a special case of the interior node having a generation term; because of symmetry, q = 25 W/m.htThe finite-difference equations are derived as follows:Continued...PROBLEM 4.75 (Cont.)Surface Node 2. From an energy balance on the prescribed control volume with x/y = 3,Ein E out = q + q + q + q = 0;abcdkT Ty T1 T2y T3 T2+ hx (T T2 ) + k+ kx 5 2 = 0 .2 x2xyRegrouping, find221 + 2N x + 1 + 2 x = T1 + T3 + 2 x T5 + 2N x TT2yy y y where N = hx/k = 100 W/m2K 0.006 m/2 W/mK = 0.30 K. Hence, with T = 30C,T2 = 0.04587T1 + 0.04587T3 + 0.82569T5 + 2.4771(1)From this FDE, the forms for nodes 1 and 3 can also be deduced.Interior Node 7. From an energy balance on the prescribed control volume, with x/y = 3,E E = 0 , where E = 2 q and En represents the conduction terms. Hence,inggihtq + q + q + q + 2q = 0 , orabcdhtT8 T7T T7T T7T T7ky+ kx 4+ ky 8+ kx 10+ 2q = 0htxyxyRegrouping,222 x 22q x x x x 1 + + 1 + = T8 + T4 + T8 + T10 + ht T7k y y y y y Recognizing that x/y = 3, q t = 25 W/m and k = 2 W/mK, the FDE ishT7 = 0.0500T8 + 0.4500T4 + 0.0500T8 + 0.4500T10 + 3.7500(2)The FDEs for the remaining nodes may be deduced from this form. Following the procedure describedin Section 4.5.2 for the Gauss-Seidel method, the system of FDEs has the form:kk 1kkkkT1 = 0.09174T2k 1+ 0.8257T4k 1T2 = 0.04587T1 + 0.04587T3k 1T3 = 0.09174T2 + 0.8257T6+ 2.4771k 1+ 0.8257T5+ 2.4771+ 2.4771kkk 1kkkk 1kkkk 1kkk 1kkkk 1kkkk 1kkk 1kkk 1kkkT4 = 0.4500T1 + 0.1000T5k 1+ 0.4500T7T5 = 0.4500T2 + 0.0500T4 + 0.0500T6k 1+ 0.4500T8T6 = 0.4500T3 + 0.1000T5 + 0.4500T9T7 = 0.4500T4 + 0.1000T8k 1+ 0.4500T10T8 = 0.4500T5 + 0.0500T7 + 0.0500T9+ 3.7500k 1+ 0.4500T11T9 = 0.4500T6 + 0.1000T8 + 0.4500T12T10 = 0.9000T7 + 0.1000T11T11 = 0.9000T8 + 0.0500T10k 1+ 0.0500T12T12 = 0.9000T9 + 0.1000T11Continued ..PROBLEM 4.75 (Cont.)Note the use of the superscript k to denote the level of iteration. Begin the iteration procedure withrational estimates for Ti (k = 0) and prescribe the convergence criterion as 0.1.k/Ti12345678910111201255.057.457.150.051.751.645.046.046.961.060.459.754.053.853.247.048.148.765.063.564.356.054.654.349.049.649.960.062.763.455.054.854.550.050.150.455.8049.9347.6759.0351.7249.1963.8952.9850.1462.8453.3550.46The last row with k = corresponds to the solution obtained by matrix inversion. It appears that at least20 iterations would be required to satisfy the convergence criterion using the Gauss-Seidel method.(b) Selected isotherms are shown in the sketch of the nodal network.Note that the isotherms are normal to the adiabatic surfaces.(c) The heat loss by convection can be expressed as11qconv = h x ( T T ) + x ( T2 T ) + x ( T3 T )1222qconv = 100 W m K 0.006 m 1 (55.80 30 ) + ( 49.93 30 ) + 1 ( 47.67 30 ) = 25.00 W m .22<As expected, the heat loss by convection is equal to the heater element dissipation. This follows from theconservation of energy requirement.(d) For this situation, choosing x = 3y was advantageous from the standpoint of precision and effort.If we had chosen x = y = 2 mm, there would have been 28 nodes, doubling the amount of work, butwith improved precision.(e) Examining the effect of grid size by using the Finite-Difference Equations option from the Toolsportion of the IHT Menu, the following temperature field was computed for x = y = 2 mm, where xand y are in mm and the temperatures are in C.y\x0246055.0458.7166.5663.14253.8856.6159.7059.71452.0354.1755.9056.33650.3252.1453.3953.80849.0250.6751.7352.091048.2449.8050.7751.111247.9749.5150.4650.78Continued ..PROBLEM 4.75 (Cont.)Agreement with the results of part (a) is excellent, except in proximity to the heating element, where T15= 66.6C for the fine grid exceeds T7 = 63.9C for the coarse grid by 2.7C.For h = 10 W/m2K, the maximum temperature in the ceramic corresponds to T15 = 254C, and the heatercould still be operated at the prescribed power. With h = 10 W/m2K, the critical temperature of T15 =400C would be reached with a heater power of approximately 82 W/m.COMMENTS: (1) The method used to obtain the rational estimates for Ti (k = 0) in part (a) is asfollows. Assume 25 W/m is transferred by convection uniformly over the surface; find Tsurf 50C.Set T2 = 50C and recognize that T1 and T3 will be higher and lower, respectively. Assume 25 W/m isconducted uniformly to the outer nodes; find T5 - T2 4C. For the remaining nodes, use intuition toguess reasonable values. (2) In selecting grid size (and whether x = y), one should consider the regionof largest temperature gradients.NOTE TO INSTRUCTOR: Although the problem statement calls for calculations with x = y = 1mm, the instructional value and benefit-to-effort ratio are small. Hence, consideration of this grid size isnot recommended.PROBLEM 4.76KNOWN: Silicon chip mounted in a dielectric substrate. One surface of system is convectivelycooled while the remaining surfaces are well insulated.FIND: Whether maximum temperature in chip will exceed 85C.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Negligiblecontact resistance between chip and substrate, (4) Upper surface experiences uniform convectioncoefficient, (5) Other surfaces are perfectly insulated.ANALYSIS: Performing an energy balance on the chip assuming it is perfectly insulated from thesubstrate, the maximum temperature occurring at the interface with the dielectric substrate will be,according to Eqs. 3.43 and 3.46,Tmax =2&q ( H/4 )2k c+&q ( H/4 )h7+ T =10 W/m3( 0.003 m ) 22 50 W/m K7+10 W/m3( 0.003 m)2500 W/m Koo+ 20 C = 80.9 C.Since Tmax < 85C for the assumed situation, for the actual two-dimensional situation with theconducting dielectric substrate, the maximum temperature should be less than 80C.Using the suggested grid spacing of 3 mm, construct thenodal network and write the finite-difference equation foreach of the nodes taking advantage of symmetry of thesystem. Note that we have chosen to n ot locate nodes onthe system surfaces for two reasons: (1) fewer totalnumber of nodes, 20 vs. 25, and (2) Node 5 correspondsto center of chip which is likely the point of maximumtemperature. Using these numerical values,hx 500 W/m2 K 0.003 m== 0.30ks5 W/m Khx 500 W/m2 K 0.003 m== 0.030kc5 W/m K&qxy= 1.800kc=22== 1.818( k s / k c ) +1 5/50 +122== 0.182( k c / k s ) +1 50/5 +11== 0.0910kc / ks + 1=find the nodal equations:Node 1T TT Tk sx 6 1 + k sy 2 1 + hx ( T T ) = 01yxContinued ..PROBLEM 4.76 (Cont.) 2 +h x T1 + T2 + T6 = k Tks sh x 2.30T + T2 + T6 = 6.001(1)T1 3.3T2 + T3 + T7 = 6.00Node 2(2)Node 3T T3k s y 2+xT4 T3T T3+ k s x 8+ h x ( T T ) = 03( x/2 ) / kc y + ( x/2 ) / ks yyT2 ( 2 + + ( h x/k s ) T3 ) + T4 + T8 = ( h x/k ) TT2 4.12T3 + 1.82T4 + T8 = 6.00(3)Node 4T3 T4T T4T9 T4+ k c y 5+( x/2 ) / ks y + ( x/2 ) / kc yx( y/2 ) / ks x + ( y/2 ) k cx&+ q ( xy ) + h x ( T T4 ) = 0& T3 (1 + 2 + [ h x/k c ]) T4 + T5 + T9 = ( h x/kc ) T q x y/kc0.182T3 1.39T4 + T5 + 0.182T9 = 2.40(4)Node 5T T5T10 T5&k c y 4++ h ( x/2 )( T T ) + q y ( x/2 ) = 05x( y/2) / ks ( x/2 ) + ( y/2 ) / k c ( x/2 )2T4 2.21T5 + 0.182T10 = 2.40(5)Nodes 6 and 11k sx ( T1 T 6 ) / y + ksy ( T7 T 6 ) / x + k sx ( T11 T6 ) / y = 0T1 3T6 + T7 + T11 = 0T6 3T11 + T + T = 01216(6,11)Nodes 7, 8, 12, 13, 14 Treat as interior points,T2 + T6 4T7 + T8 + T12 = 0T3 + T7 4T8 + T9 + T13 = 0T7 + T11 4T12 + T + T = 01317(7,8)(12,13)(14)T8 + T 4T + T + T = 012131418T9 + T13 4T14 + T + T = 01519Node 9T T9T4 T9T T9T T9k s y 8++ k sy 10+ k sx 14=0x( y/2 ) / k c x + ( y/2 ) / k sxxy1.82T4 + T8 4.82T9 + T10 + T = 014(9)Node 10 Using the result of Node 9 and considering symmetry,1.82T5 + 2T9 4.82T10 + T = 015Node 15 Interior point considering symmetry T10 + 2T14 4T15 + T20 = 0Node 16 By inspection,(10)(15)T11 2T16 + T = 017(16)Continued ..PROBLEM 4.76 (Cont.)Nodes 17, 18, 19, 20T12 + T16 3T17 + T18 = 0T14 + T18 3T19 + T20 = 0T13 + T 3T18 + T = 01719T15 + 2T19 3T20 = 0(17,18)(19,20)Using the matrix inversion method, the above system of finite-difference equations is written in matrixnotation, Eq. 4.52, [A][T] = [C] where-2.3 10001 -3.3 1000 1 -4.12 1.82 00 0 .182 -1.39 10002 -2.211000001000001000 0 0 1.82 00000 1.82[A] = 0 0 00000000000000000000000000000000000000000000000010000-310001000000000010001-41000100000000000 0000000000000 0000000000100 00000000000 .182 0 0 0 0 0 0 0 0 0 0 000 .182 0 0 0 0 0 0 0 0 0 0000 1000000000100 0100000000-410 00100000001 -4.82 1 0 0 0 1 0 0 0 0 0 00 2 -4.82 0 0 0 0 1 0 0 0 0 0000 -3 1 0 0 0 1 0 0 0 0000 1 -4 1 0 0 0 1 0 0 0100 0 1 -4 1 0 0 0 1 0 0010 0 0 1 -4 1 0 0 0 1 0001 0 0 0 2 -4 0 0 0 0 1000 1 0 0 0 0 -2 1 0 0 0000 0 1 0 0 0 1 -3 1 0 0000 0 0 1 0 0 0 1 -3 1 0000 0 0 0 1 0 0 0 1 -3 1000 0 0 0 0 1 0 0 0 2 -3[C] =-6-6-6-2.4-2.4000000000000000and the temperature distribution (C), in geometrical representation, is34.4637.1338.5639.1636.1338.3739.3839.7740.4140.8540.8140.7645.8843.8042.7241.7046.2344.5142.7842.06The maximum temperature is T5 = 46.23C which is indeed less than 85C.<COMMENTS: (1) The convection process for the energybalances of Nodes 1 through 5 were simplified by assumingthe node temperature is also that of the surface. ConsideringNode 2, the energy balance processes for qa, qb and qc areidentical (see Eq. (2)); however,q conv =T T2 h ( T T2 )1/h + y/2k2-4where hy/2k = 5 W/m K0.003 m/250 W/mK = 1.510simplification is justified.<< 1. Hence, for this situation, thePROBLEM 4.77KNOWN: Electronic device cooled by conduction to a heat sink.FIND: (a) Beginning with a symmetrical element, find the thermal resistance per unit depth between thedevice and lower surface of the sink, R ,d s (mK/W) using the flux plot method; compare result withtthermal resistance based upon assumption of one-dimensional conduction in rectangular domains of (i)width wd and length L and (ii) width ws and length L; (b) Using a coarse (5x5) nodal network, determineR s ; (c) Using nodal networks with finer grid spacings, determine the effect of grid size on thet,dprecision of the thermal resistance calculation; (d) Using a fine nodal network, determine the effect ofdevice width on R ,d s with wd/ws = 0.175, 0.275, 0.375 and 0.475 keeping ws and L fixed.tSCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, and (3) Nointernal generation, (4) Top surface not covered by device is insulated.ANALYSIS: (a) Begin the flux plot for the symmetricalelement noting that the temperature drop along the lefthand symmetry line will be almost linear. Choosing tosketch five isotherms and drawing the adiabats, findN=5M = 2.75so that the shape factor for the device to the sinkconsidering two symmetrical elements per unit depth isS = 2S = 2oMN= 1.10and the thermal resistance per unit depth isR s,fp = 1 kS = 1 300 W m K 1.10 = 3.03 103 m K Wt,d<The thermal resistances for the two rectangular domains are represented schematically below.Continued...PROBLEM 4.77 (Cont.)R s,i =t,dR s,ii =t,dLkw d=Lkw s0.024 m300 W m K 0.018 m=0.024 m300 W m K 0.048 m= 4.44 103 m K W<= 1.67 103 m K W<We expect the flux plot result to be bounded by the results for the rectangular domains. The spreadingeffect can be seen by comparing R ,d s,fp with R ,d s,i .ft(b) The coarse 5x5 nodal network isshown in the sketch including thenodes adjacent to the symmetry linesand the adiabatic surface. As such,all the finite-difference equations areinterior nodes and can be written byinspection directly onto the IHTworkspace. Alternatively, one coulduse the IHT Finite-DifferenceEquations Tool. The temperaturedistribution (C) is tabulated in thesame arrangement as the nodalnetwork.85.0085.0065.7663.8550.3249.1737.1836.7025.0025.0062.3155.4945.8035.4725.0053.2650.0043.0634.3725.0050.7348.2042.0733.9525.00The thermal resistance between the device and sink per unit depth isT TsR d = dt,s2qtotPerforming an energy balance on the device nodes, findq = q + q + qtotabcT T1T T5T T4q = k ( y 2 ) d+ kx d+ k ( x 2 ) dtotxyyq = 300 W m K [(85 62.31) 2 + (85 63.85 ) + (85 65.76 ) 2] K = 1.263 104 W mtotR d =t,s(85 25 ) K4= 2.38 10 3 m K W<2 1.263 10 W m(c) The effect of grid size on the precision of the thermal resistance estimate should be tested bysystematically reducing the nodal spacing x and y. This is a considerable amount of work even withIHT since the equations need to be individually entered. A more generalized, powerful code would beContinued...PROBLEM 4.77 (Cont.)required which allows for automatically selecting the grid size. Using FEHT, a finite-element package,with eight elements across the device, representing a much finer mesh, we foundR d = 3.64 103 m K Wt,s(d) Using the same tool, with the finest mesh, the thermal resistance was found as a function of wd/wswith fixed ws and L.As expected, as wd increases, R ,d s decreases, and eventually will approach the value for thetrectangular domain (ii). The spreading effect is shown for the base case, wd/ws = 0.375, where thethermal resistance of the sink is less than that for the rectangular domain (i).COMMENTS: It is useful to compare the results for estimating the thermal resistance in terms ofprecision requirements and level of effort,R s 103 (mK/W)t,dRectangular domain (i)Flux plotRectangular domain (ii)FDE, 5x5 networkFEA, fine mesh4.443.031.672.383.64PROBLEM 4.78KNOWN: Nodal network and boundary conditions for a water-cooled cold plate.FIND: (a) Steady-state temperature distribution for prescribed conditions, (b) Means by which operationmay be extended to larger heat fluxes.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constantproperties.ANALYSIS: Finite-difference equations must be obtained for each of the 28 nodes. Applying theenergy balance method to regions 1 and 5, which are similar, it follows thatNode 1:Node 5:( y( yx ) T2 + ( x y ) T6 ( y x ) + ( x y ) T1 = 0x ) T4 + ( x y ) T10 ( y x ) + ( x y ) T5 = 0Nodal regions 2, 3 and 4 are similar, and the energy balance method yields a finite-difference equation ofthe formNodes 2,3,4:( y()x ) Tm 1,n + Tm +1,n + 2 ( x y ) Tm,n 1 2 ( y x ) + ( x y ) Tm,n = 0Energy balances applied to the remaining combinations of similar nodes yield the following finitedifference equations.Continued...PROBLEM 4.78 (Cont.)( x( xNodes 7, 15:y ) T1 + ( y x ) T7 [( x y ) + ( y x ) + ( hx k )] T6 = ( hx k ) T( yNodes 6, 14:x )( T6 + T8 ) + 2 ( x y ) T2 2 [( y x ) + ( x y ) + ( hx k )] T7 = ( 2hx k ) T( y( yNodes 8, 16:y ) T19 + ( y x ) T15 [( x y ) + ( y x ) + ( hx k )] T14 = ( hx k ) Tx )( T14 + T16 ) + 2 ( x y ) T20 2 [( y x ) + ( x y ) + ( hx k )] T15 = ( 2hx k ) Tx ) T7 + 2 ( y x ) T9 + ( x y ) T11 + 2 ( x y ) T3 [3 ( y x ) + 3 ( x y )+ ( h k )( x + y )] T8 = ( h k )( x + y ) Tx ) T15 + 2 ( y x ) T17 + ( x y ) T11 + 2 ( x y ) T21 [3 ( y x ) + 3 ( x y )( y+ ( h k )( x + y )] T16 = ( h k )( x + y ) TNode 11: ( x y ) T8 + ( x y ) T16 + 2 ( y x ) T12 2 [( x y ) + ( y x ) + ( hy k )] T11 = ( 2hy k ) TNodes 9, 12, 17, 20, 21, 22:( yx ) Tm 1,n + ( y x ) Tm +1,n + ( x y ) Tm,n +1 + ( x y ) Tm,n 1 2 [( x y ) + ( y x )] Tm,n = 0Nodes 10, 13, 18, 23:( xNode 19:Nodes 24, 28:( x( x( xy ) Tn +1,m + ( x y ) Tn 1,m + 2 ( y x ) Tm 1,n 2 [( x y ) + ( y x )] Tm,n = 0y ) T14 + ( x y ) T24 + 2 ( y x ) T20 2 [( x y ) + ( y x )] T19 = 0y ) T19 + ( y x ) T25 [( x y ) + ( y x )] T24 = ( q x k )oy ) T23 + ( y x ) T27 [( x y ) + ( y x )] T28 = ( q x k )oNodes 25, 26, 27:( yx ) Tm 1,n + ( y x ) Tm +1,n + 2 ( x y ) Tm,n +1 2 [( x y ) + ( y x )] Tm,n = ( 2q x k )oEvaluating the coefficients and solving the equations simultaneously, the steady-state temperaturedistribution (C), tabulated according to the node locations, is:23.7723.4123.9123.6228.9030.7232.7728.7630.6732.7424.2724.3125.7028.2630.5732.6924.6124.8926.1828.3230.5332.6624.7425.0726.3328.3530.5232.65Alternatively, the foregoing results may readily be obtained by accessing the IHT Tools pat and using the2-D, SS, Finite-Difference Equations options (model equations are appended). Maximum and minimumcold plate temperatures are at the bottom (T24) and top center (T1) locations respectively.(b) For the prescribed conditions, the maximum allowable temperature (T24 = 40C) is reached when qo= 1.407 105 W/m2 (14.07 W/cm2). Options for extending this limit could include use of a copper coldplate (k 400 W/mK) and/or increasing the convection coefficient associated with the coolant. With k =400 W/mK, a value of q = 17.37 W/cm2 may be maintained. With k = 400 W/mK and h = 10,000oW/m2K (a practical upper limit), q = 28.65 W/cm2. Additional, albeit small, improvements may beorealized by relocating the coolant channels closer to the base of the cold plate.COMMENTS: The accuracy of the solution may be confirmed by verifying that the results satisfy theoverall energy balanceq ( 4x ) = h [( x 2 ) ( T6 T ) + x ( T7 T ) + ( x + y )( T8 T ) 2o+y ( T11 T ) + ( x + y )( T16 T ) 2 + x (T15 T ) + ( x 2 )( T14 T )] .PROBLEM 4.79KNOWN: Heat sink for cooling computer chips fabricated from copper with microchannels passingfluid with prescribed temperature and convection coefficient.FIND: (a) Using a square nodal network with 100 m spatial increment, determine the temperaturedistribution and the heat rate to the coolant per unit channel length for maximum allowable chiptemperature Tc,max = 75C; estimate the thermal resistance betweeen the chip surface and the fluid,R f (mK/W); maximum allowable heat dissipation for a chip that measures 10 x 10 mm on a side;t,c(b) The effect of grid spacing by considering spatial increments of 50 and 25 m; and (c) Consistent withthe requirement that a + b = 400 m, explore altering the sink dimensions to decrease the thermalresistance.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, and (3)Convection coefficient is uniform over the microchannel surface and independent of the channeldimensions and shape.ANALYSIS: (a) The square nodal network with x = y = 100 m is shown below. Consideringsymmetry, the nodes 1, 2, 3, 4, 7, and 9 can be treated as interior nodes and their finite-differenceequations representing nodal energy balances can be written by inspection. Using the, IHT FiniteDifference Equations Tool, appropriate FDEs for the nodes experiencing surface convection can beobtained. The IHT code including results is included in the Comments. Having the temperaturedistribution, the heat rate to the coolant per unit channel length for two symmetrical elements can beobtained by applying Newtons law of cooling to the surface nodes,q = 2 [h ( y 2 + x 2 )( T5 T ) + h ( x 2 )( T6 T ) + h ( y )( T8 T ) h ( y 2 )( T10 T )]cvq = 2 30, 000 W m K 100 10cv26m [( 74.02 25 ) + ( 74.09 25 ) 2 + ( 73.60 25 ) + ( 73.37 25 ) 2 ] K<q = 878 W mcvThe thermal resistance between the chip and fluid per unit length for each microchannel isT T ( 75 25 ) CR f = c== 5.69 102 m K Wt,cq878 W mcv<The maximum allowable heat dissipation for a 10 mm 10 mm chip isPchip,max = q A chip = 2.20 106 W m 2 (0.01 0.01) m 2 = 220 Wcwhere Achip = 10 mm 10 mm and the heat flux on the chip surface (wf + ws) isq = qccv(wf+ w s ) = 878 W m( 200 + 200 ) 106 m = 2.20 106 W<m2Continued...PROBLEM 4.79 (Cont.)(b) To investigate the effect of grid spacing, the analysis was repreated with a spatial increment of 50 m(32 nodes as shown above) with the following resultsR f = 5.67 10 2 m K Wt,cq = 881W mcv<Using a finite-element package with a mesh around 25 m, we found R f = 5.70 102 m K Wt,cwhich suggests the grid spacing effect is not very significant.(c) Requring that the overall dimensions of the symmetrical element remain unchanged, we exploredwhat effect changes in the microchannel cross-section would have on the overall thermal resistance,R f . It is important to recognize that the sink conduction path represents the dominant resistance,t,csince for the convection process()262Rt,cv = 1 A = 1 30, 000 W m K 600 10 m = 5.55 10 m K Wswhere A = (wf + 2b) = 600 m.sUsing a finite-element package, the thermal resistances per unit length for three additional channel crosssections were determined and results summarized below.Microchannel (m)CaseABCDHeight200133300250Half-width100150100150R s 10t,c2(mK/W)5.706.124.294.25Continued...PROBLEM 4.79 (Cont.)COMMENTS: (1) The IHT Workspace for the 5x5 coarse node analysis with results follows.// Finite-difference equations - energy balances// First row - treating as interior nodes considering symmetryT1 = 0.25 * ( Tc + T2 + T4 + T2 )T2 = 0.25 * ( Tc + T3 + T5 + T1 )T3 = 0.25 * ( Tc + T2 + T6 + T2 )/* Second row - Node 4 treat as interior node; for others, use Tools: Finite-Difference Equations,Two-Dimensional, Steady-State; be sure to delimit replicated q''a = 0 equations. */T4 = 0.25 * ( T1 + T5+ T7 + T5 )/* Node 5: internal corner node, e-s orientation; e, w, n, s labeled 6, 4, 2, 8. */0.0 = fd_2d_ic_es(T5,T6,T4,T2,T8,k,qdot,deltax,deltay,Tinf,h,q''a)q''a = 0// Applied heat flux, W/m^2; zero flux shown/* Node 6: plane surface node, s-orientation; e, w, n labeled 5, 5, 3 . */0.0 = fd_2d_psur_s(T6,T5,T5,T3,k,qdot,deltax,deltay,Tinf,h,q''a)//q''a = 0// Applied heat flux, W/m^2; zero flux shown/* Third row - Node 7 treat as interior node; for others, use Tools: Finite-Difference Equations,Two-Dimensional, Steady-State; be sure to delimit replicated q''a = 0 equations. */T7 = 0.25 * (T4 + T8 + T9 + T8)/* Node 8: plane surface node, e-orientation; w, n, s labeled 7, 5, 10. */0.0 = fd_2d_psur_e(T8,T7,T5,T10,k,qdot,deltax,deltay,Tinf,h,q''a)//q''a = 0// Applied heat flux, W/m^2; zero flux shown/* Fourth row - Node 9 treat as interior node; for others, use Tools: Finite-Difference Equations,Two-Dimensional, Steady-State; be sure to delimit replicated q''a = 0 equations. */T9 = 0.25 * (T7 + T10 +T7 + T10)/* Node 10: plane surface node, e-orientation; w, n, s labeled 9, 8, 8. */0.0 = fd_2d_psur_e(T10,T9,T8,T8,k,qdot,deltax,deltay,Tinf,h,q''a)//q''a = 0// Applied heat flux, W/m^2; zero flux shown// Assigned variables// For the FDE functions,qdot = 0deltax = deltaydeltay = 100e-6Tinf = 25h = 30000// Sink and chip parametersk = 400Tc = 75wf = 200e-6ws = 200e-6// Volumetric generation, W/m^3// Spatial increments// Spatial increment, m// Microchannel fluid temperature, C// Convection coefficient, W/m^2.K// Sink thermal conductivity, W/m.K// Maximum chip operating temperature, C// Channel width, m// Sink width, m/* Heat rate per unit length, for two symmetrical elements about one microchannel, */q'cv= 2 * (q'5 + q'6 + q'8 + q'10)q'5 = h* (deltax / 2 + deltay / 2) * (T5 - Tinf)q'6 = h * deltax / 2 * (T6 - Tinf)q'8 = h * deltax * (T8 - Tinf)q'10 = h * deltax / 2 * (T10 - Tinf)/* Thermal resistance between chip and fluid, per unit channel length, */R'tcf = (Tc - Tinf) / q'cv// Thermal resistance, m.K/W// Total power for a chip of 10mm x 10mm, Pchip (W),q''c = q'cv / (wf + ws)// Heat flux on chip surface, W/m^2Pchip = Achip * q''c// Power, WAchip = 0.01 * 0.01// Chip area, m^2/* Data Browser results: chip power, thermal resistance, heat rates and temperature distributionPchipR'tcfq''cq'cv219.50.05694 2.195E6 878.1T174.53T274.52T374.53T474.07T574.02T674.09T773.7T873.6T973.53T1073.37 */PROBLEM 4.80KNOWN: Longitudinal rib (k = 10 W/mK) with rectangular cross-section with length L= 8 mm andwidth w = 4 mm. Base temperature Tb and convection conditions, T and h, are prescribed.FIND: (a) Temperature distribution and fin base heat rate using a finite-difference method with x = y= 2 mm for a total of 5 3 = 15 nodal points and regions; compare results with those obtained assumingone-dimensional heat transfer in rib; and (b) The effect of grid spacing by reducing nodal spacing to x =y = 1 mm for a total of 9 3 = 27 nodal points and regions considering symmetry of the centerline; and(c) A criterion for which the one-dimensional approximation is reasonable; compare the heat rate for therange 1.5 L/w 10, keeping L constant, as predicted by the two-dimensional, finite-difference methodand the one-dimensional fin analysis.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, and (3) Convection coefficientuniform over rib surfaces, including tip.ANALYSIS: (a) The rib is represented by a 5 3 nodal grid as shown above where the symmetry planeis an adiabatic surface. The IHT Tool, Finite-Difference Equations, for Two-Dimensional, Steady-Stateconditions is used to formulate the nodal equations (see Comment 2 below) which yields the followingnodal temperatures ( C)45454539.340.039.335.736.435.733.534.033.532.232.632.2Note that the fin tip temperature isTtip = T12 = 32.6 C<The fin heat rate per unit width normal to the page, q , can be determined from energy balances on thefinthree base nodes as shown in the schematic below.q = q + q + q + q + qfinabcde = h ( x 2 )( Tb T )qaq = k ( y 2 )( Tb T1 ) xbq = k ( y )( Tb T5 ) xcq = k ( y 2 )( Tb T9 ) xd = h ( x 2 )( Tb T )q3Continued...PROBLEM 4.80 (Cont.)Substituting numerical values, findq = (12.0 + 28.4 + 50.0 + 28.4 + 12.0 ) W m = 130.8 W mfinUsing the IHT Model, Extended Surfaces, Heat Rate and Temperature Distributions for Rectangular,Straight Fins, with convection tip condition, the one-dimensional fin analysis yields<q = 131W mTtip = 32.2 Cf(b) With x = L/8 = 1 mm and x = 1 mm, for a total of 9 3 = 27 nodal points and regions, the gridappears as shown below. Note the rib centerline is a plane of symmetry.<Using the same IHT FDE Tool as above with an appropriate expression for the fin heat rate, Eq. (1), thefin heat rate and tip temperature were determined.1-D analysisTtip (C)32.2q (W/m)fin2-D analysis (nodes)( 5 3)(9 3)32.632.6131131129<<(c) To determine when the one-dimensional approximation is reasonable, consider a rib of constantlength, L = 8 mm, and vary the thickness w for the range 1.5 L/w 10. Using the above IHT model forthe 27 node grid, the fin heat rates for 1-D, q1d , and 2-D, q d , analysis were determined as a function of2w with the error in the approximation evaluated asError ( % ) = ( q q1d ) 100 q1d2d4Error (%)20-2-40246810Length / width, L/wNote that for small L/w, a thick rib, the 1-D approximation is poor. For large L/w, a thin rib whichapproximates a fin, we would expect the 1-D approximation to become increasingly more satisfactory.The discrepancy at large L/w must be due to discretization error; that is, the grid is too coarse toaccurately represent the slender rib.PROBLEM 4.81KNOWN: Bottom half of an I-beam exposed to hot furnace gases.FIND: (a) The heat transfer rate per unit length into the beam using a coarse nodal network (5 4)considering the temperature distribution across the web is uniform and (b) Assess the reasonableness ofthe uniform web-flange interface temperature assumption.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, and (2) Constant properties.ANALYSIS: (a) The symmetrical section of the I-beam is shown in the Schematic above indicating theweb-flange interface temperature is uniform, Tw = 100C. The nodal arrangement to represent thissystem is shown below. The nodes on the line of symmetry have been shown for convenience in derivingthe nodal finite-difference equations.Using the IHT Finite-Difference Equations Tool, the set of nodal equations can be readily formulated.The temperature distribution (C) is tabulated in the same arrangement as the nodal network.100.00166.6211.7241.4100.00177.1219.5247.2215.8222.4241.9262.9262.9255.0262.7279.3284.8272.0274.4292.9The heat rate to the beam can be determined from energy balances about the web-flange interface nodesas shown in the sketch below.Continued...PROBLEM 4.81 (Cont.)q = q + q + qwabcT TT TwT Tw+ k ( x 2 ) 4q = k ( y 2 ) 1 w + k ( x ) 5wxyyq = 10 W m K [( 215.8 100 ) 2 + (177.1 100 ) + (166.6 100 ) 2] K = 1683 W mw<(b) The schematic below poses the question concerning the reasonableness of the uniform temperatureassumption at the web-flange interface. From the analysis above, note that T1 = 215.8C vs. Tw = 100Cindicating that this assumption is a poor one. This L-shaped section has strong two-dimensionalbehavior. To illustrate the effect, we performed an analysis with Tw = 100C located nearly 2 timesfurther up the web than it is wide. For this situation, the temperature difference at the web-flangeinterface across the width of the web was nearly 40C. The steel beam with its low thermal conductivityhas substantial internal thermal resistance and given the L-shape, the uniform temperature assumption(Tw) across the web-flange interface is inappropriate.PROBLEM 4.82KNOWN: Plane composite wall with exposed surfaces maintained at fixed temperatures. Material Ahas temperature-dependent thermal conductivity.FIND: Heat flux through the wall (a) assuming a uniform thermal conductivity in material Aevaluated at the average temperature of the section, and considering the temperature-dependentthermal conductivity of material A using (b) a finite-difference method of solution in IHT with aspace increment of 1 mm and (c) the finite-element method of FEHT.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) No thermal contact resistancebetween the materials, and (3) No internal generation.ANALYSIS: (a) From the thermal circuit in the above schematic, the heat flux isq =xT1 T2T T= AB 2R + R R ABB(1, 2)and the thermal resistances of the two sections areR = LA / k AAR = LB / k BB(3, 4)The thermal conductivity of material A is evaluated at the average temperature of the section{}k A = k o 1 + ( T1 + TAB ) / 2 To (5)Substituting numerical values and solving the system of equations simultaneously in IHT, findTAB = 563.2 Kq = 52.64 kW / m 2x<(b) The nodal arrangement for the finite-difference method of solution is shown in the schematicbelow. FDEs must be written for the internal nodes (02 10, 12 15) and the A-B interface node(11) considering in section A, the temperature-dependent thermal conductivity.Interior Nodes, Section A (m = 02, 03 10)Referring to the schematic below, the energy balance on node m is written in terms of the heat fluxesat the control surfaces using Fouriers law with the thermal conductivity based upon the averagetemperature of adjacent nodes. The heat fluxes into node m areContinued ..PROBLEM 4.82 (Cont.)TTq = k a ( m, m + 1) m +1 mcx(1)TTq = k a ( m 1, m ) m 1 mdx(2)and the FDEs are obtained from the energy balance written asq + q = 0cd(3)TTTTk a ( m, m + 1) m +1 m + k a ( m 1, m ) m 1 m = 0xx(4)where the thermal conductivities averaged over the path between the nodes are expressed as{}k a ( m, m + 1) = k o {1 + ( Tm + Tm +1 ) / 2 To }k a ( m 1, m ) = k o 1 + ( Tm 1 + Tm ) / 2 To (5)(6)A-B Interface Node 11Referring to the above schematic, the energy balance on the interface node, q + q = 0, has the formcdT TT Tk b 12 11 + k a (10,11) 10 11 = 0xx(7)where the thermal conductivity in the section A path is{}k (10,11) = k o 1 + (T10 + T11 ) / 2 To (8)Interior Nodes, Section B (n = 12 15)Since the thermal conductivity in Section B is uniform, the FDEs have the formTn = ( Tn 1 + Tn +1 ) / 2(9)And the heat flux in the x-direction isT Tn +1q = k b nxx(10)Finite-Difference Method of SolutionThe foregoing FDE equations for section A nodes (m = 02 to 10), the AB interface node and theirrespective expressions for the thermal conductivity, k (m, m +1), and for section B nodes are enteredinto the IHT workspace and solved for the temperature distribution. The heat flux can be evaluatedusing Eq. (2) or (10). A portion of the IHT code is contained in the Comments, and the results of theanalysis are tabulated below.T11 = TAB = 563.2 Kq = 52.64 kW / m 2xContinued ..<PROBLEM 4.82 (Cont.)(c) The finite-element method of FEHT can be used readily to obtain the heat flux considering thetemperature-dependent thermal conductivity of section A. Draw the composite wall outline withproperly scaled section thicknesses in the x-direction with an arbitrary y-direction dimension. In theSpecify | Materials Properties box for the thermal conductivity, specify ka as 4.4*[1 + 0.008*(T 300)] having earlier selected Set | Temperatures in K. The results of the analysis areTAB = 563 Kq = 5.26 kW / m 2x<COMMENTS: (1) The results from the three methods of analysis compare very well. Because thethermal conductivity in section A is linear, and moderately dependent on temperature, the simplestmethod of using an overall section average, part (a), is recommended. This same method isrecommended when using tabular data for temperature-dependent properties.(2) For the finite-difference method of solution, part (b), the heat flux was evaluated at several nodeswithin section A and in section B with identical results. This is a consequence of the technique foraveraging ka over the path between nodes in computing the heat flux into a node.(3) To illustrate the use of IHT in solving the finite-difference method of solution, lines of code forrepresentative nodes are shown below.// FDEs Section Ak01_02 * (T01-T02)/deltax + k02_03 * (T03-T02)/deltax = 0k01_02 = ko * (1+ alpha * ((T01 + T02)/2 To))k02_03 = ko * (1 + alpha * ((T02 + T03)/2 To))k02_03 * (T02 T03)/deltax + k03_04 * (T04 T03)/deltax = 0k03_04 = ko * (1 + alpha * ((T03 + T04)/2 To))// Interface, node 11k11 * (T10 T11)/deltax + kb * (T12 T11)/deltax =0k11 = ko * (1 + alpha * ((T10 + T11)/2 To))// Section B (using Tools/FDE/One-dimensional/Steady-state)/* Node 12: interior node; */0.0 = fd_1d_int(T12, T13, T11, kb, qdot, deltax)(4) The solved models for Text Examples 4.3 and 4.4, plus the tutorial of the Users Manual, providebackground for developing skills in using FEHT.PROBLEM 4.83KNOWN: Upper surface of a platen heated by hot fluid through the flow channels is used to heat aprocess fluid.FIND: (a) The maximum allowable spacing, W, between channel centerlines that will provide auniform temperature requirement of 5C on the upper surface of the platen, and (b) Heat rate per unitlength from the flow channel for this condition.SCHEMATIC:ASSUMPTIONS: (1) Steady-state, two-dimensional conduction with constant properties, and (2)Lower surface of platen is adiabatic.ANALYSIS: As shown in the schematic above for a symmetrical section of the platen-flow channelarrangement, the temperature uniformity requirement will be met when T1 T2 = 5C. The maximumtemperature, T1, will occur directly over the flow channel centerline, while the minimum surfacetemperature, T2, will occur at the mid-span between channel centerlines.We chose to use FEHT to obtain the temperature distribution and heat rate for guessed values of thechannel centerline spacing, W. The following method of solution was used: (1) Make an initial guessvalue for W; try W = 100 mm, (2) Draw an outline of the symmetrical section, and assign propertiesand boundary conditions, (3) Make a copy of this file so that in your second trial, you can use theDraw | Move Node option to modify the section width, W/2, larger or smaller, (4) Draw element lineswithin the outline to create triangular elements, (5) Use the Draw | Reduce Mesh command togenerate a suitably fine mesh, then solve for the temperature distribution, (6) Use the View |Temperatures command to determine the temperatures T1 and T2, (7) If, T1 T2 5C, use the View |Heat Flows command to find the heat rate, otherwise, change the width of the section outline andrepeat the analysis. The results of our three trials are tabulated below.Trial123W (mm)1006080T1 (C)T2 (C)T1 T2 (C)108119113981181081015q (W/m)--1706COMMENTS: (1) In addition to the tutorial example in the FEHT Users Manual, the solved modelsfor Examples 4.3 and 4.4 of the Text are useful for developing skills in using this problem-solvingtool.(2) An alternative numerical method of solution would be to create a nodal network, generate thefinite-difference equations and solve for the temperature distribution and the heat rate. The FDEsshould allow for a non-square grid, x y, so that different values for W/2 can be accommodated bychanging the value of x. Even using the IHT tool for building FDEs (Tools | Finite-DifferenceEquations | Steady-State) this method of solution is very labor intensive because of the large numberof nodes required for obtaining good estimates.PROBLEM 4.84KNOWN: Silicon chip mounted in a dielectric substrate. One surface of system is convectivelycooled, while the remaining surfaces are well insulated. See Problem 4.77. Use the finite-elementsoftware FEHT as your analysis tool.FIND: (a) The temperature distribution in the substrate-chip system; does the maximum temperatureexceed 85C?; (b) Volumetric heating rate that will result in a maximum temperature of 85C; and (c)Effect of reducing thickness of substrate from 12 to 6 mm, keeping all other dimensions unchanged73with q = 110 W/m ; maximum temperature in the system for these conditions, and fraction of thepower generated within the chip removed by convection directly from the chip surface.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in system, and (3)Uniform convection coefficient over upper surface.ANALYSIS: Using FEHT, the symmetrical section is represented in the workspace as two connectedregions, chip and substrate. Draw first the chip outline; Specify the material and generationparameters. Now, Draw the outline of the substrate, connecting the nodes of the interfacing surfaces;Specify the material parameters for this region. Finally, Assign the Boundary Conditions: zero heatflux for the symmetry and insulated surfaces, and convection for the upper surface. Draw ElementLines, making the triangular elements near the chip and surface smaller than near the lower insulatedboundary as shown in a copy of the FEHT screen on the next page. Use the Draw|Reduce Meshcommand and Run the model.(a) Use the View|Temperature command to see the nodal temperatures through out the system. Asexpected, the hottest location is on the centerline of the chip at the bottom surface. At this location, thetemperature is<T(0, 9 mm) = 46.7C(b) Run the model again, with different values of the generation rate until the temperature at thislocation is T(0, 9 mm) = 85C, findingq = 2.43 107 W / m3<Continued ..PROBLEM 4.84 (Cont.)(c) Returning to the model code with the conditions of part (a), reposition the nodes on the lowerboundary, as well as the intermediate ones, to represent a substrate that is of 6-mm, rather than 12-mmthickness. Find the maximum temperature asT ( 0,3 mm ) = 47.5C<Using the View|Heat Flow command, click on the adjacent line segments forming the chip surfaceexposed to the convection process. The heat rate per unit width (normal to the page) isqchip,cv = 60.26 W / mThe total heat generated within the chip isqtot = q ( L / 6 H / 4 ) = 1 107 W / m3 ( 0.0045 0.003) m 2 = 135 W / mso that the fraction of the power dissipated by the chip that is convected directly to the coolant streamisF = qchip,cv / q = 60.26 /135 = 45%tot<COMMENTS: (1) Comparing the maximum temperatures for the system with the 12-mm and 6-mmthickness substrates, note that the effect of halving the substrate thickness is to raise the maximumtemperature by less than 1C. The thicker substrate does not provide significantly improved heatremoval capability.(2) Without running the code for part (b), estimate the magnitude of q that would make T(0, 9 mm) =7385C. Did you get q = 2.4310 W/m ? Why?PROBLEM 5.1KNOWN: Electrical heater attached to backside of plate while front surface is exposed toconvection process (T,h); initially plate is at a uniform temperature of the ambient air and suddenlyheater power is switched on providing a constant q .oFIND: (a) Sketch temperature distribution, T(x,t), (b) Sketch the heat flux at the outer surface,q ( L,t ) as a function of time.xSCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible heatloss from heater through insulation.ANALYSIS: (a) The temperature distributions for four time conditions including the initialdistribution, T(x,0), and the steady-state distribution, T(x, ), are as shown above.Note that the temperature gradient at x = 0, -dT/dx)x=0, for t > 0 will be a constant since the flux,q ( 0 ), is a constant. Noting that To = T(0,), the steady-state temperature distribution will bexlinear such thatT T ( L, )q = k o= h T ( L, ) T .oL(b) The heat flux at the front surface, x = L, is given by q ( L,t ) = k ( dT/dx ) x=L . From thextemperature distribution, we can construct the heat flux-time plot.COMMENTS: At early times, the temperature and heat flux at x = L will not change from theirinitial values. Hence, we show a zero slope for q ( L,t ) at early times. Eventually, the value ofxq ( L,t ) will reach the steady-state value which is q .xoPROBLEM 5.2KNOWN: Plane wall whose inner surface is insulated and outer surface is exposed to anairstream at T. Initially, the wall is at a uniform temperature equal to that of the airstream.Suddenly, a radiant source is switched on applying a uniform flux, q , to the outer surface.oFIND: (a) Sketch temperature distribution on T-x coordinates for initial, steady-state, andtwo intermediate times, (b) Sketch heat flux at the outer surface, q ( L,t ) , as a function ofxtime.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internalgeneration, Eg = 0, (4) Surface at x = 0 is perfectly insulated, (5) All incident radiant poweris absorbed and negligible radiation exchange with surroundings.ANALYSIS: (a) The temperature distributions are shown on the T-x coordinates and labeledaccordingly. Note these special features: (1) Gradient at x = 0 is always zero, (2) gradient ismore steep at early times and (3) for steady-state conditions, the radiant flux is equal to theconvective heat flux (this follows from an energy balance on the CS at x = L),q = q oconv = h [T ( L, ) T ].(b) The heat flux at the outer surface, q ( L,t ) , as a function of time appears as shown above.xCOMMENTS: The sketches must reflect the initial and boundary conditions:T(x,0) = TTkx=0 = 0xTkx=L = h T ( L,t ) T qoxuniform initial temperature.insulated at x = 0.surface energy balance at x = L.PROBLEM 5.3KNOWN: Microwave and radiant heating conditions for a slab of beef.FIND: Sketch temperature distributions at specific times during heating and cooling.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Uniform internal heat generation formicrowave, (3) Uniform surface heating for radiant oven, (4) Heat loss from surface of meat tosurroundings is negligible during the heating process, (5) Symmetry about midplane.ANALYSIS:COMMENTS: (1) With uniform generation and negligible surface heat loss, the temperaturedistribution remains nearly uniform during microwave heating. During the subsequent surfacecooling, the maximum temperature is at the midplane.(2) The interior of the meat is heated by conduction from the hotter surfaces during radiant heating,and the lowest temperature is at the midplane. The situation is reversed shortly after cooling begins,and the maximum temperature is at the midplane.PROBLEM 5.4KNOWN: Plate initially at a uniform temperature Ti is suddenly subjected to convection process(T,h) on both surfaces. After elapsed time to, plate is insulated on both surfaces.FIND: (a) Assuming Bi >> 1, sketch on T - x coordinates: initial and steady-state (t )temperature distributions, T(x,to) and distributions for two intermediate times to < t < , (b) Sketchon T - t coordinates midplane and surface temperature histories, (c) Repeat parts (a) and (b)assuming Bi << 1, and (d) Obtain expression for T(x, ) = Tf in terms of plate parameters (M,cp),thermal conditions (Ti, T, h), surface temperature T(L,t) and heating time to.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internalgeneration, (4) Plate is perfectly insulated for t > to, (5) T(0, t < to) < T.ANALYSIS: (a,b) With Bi >> 1, appreciable temperature gradients exist in the plate followingexposure to the heating process.On T-x coordinates: (1) initial, uniform temperature, (2) steady-state conditions when t , (3)distribution at to just before plate is covered with insulation, (4) gradients are always zero(symmetry), and (5) when t > to (dashed lines) gradients approach zero everywhere.(c) If Bi << 1, plate is space-wise isothermal (no gradients). On T-x coordinates, the temperaturedistributions are flat; on T-t coordinates, T(L,t) = T(0,t).(d) The conservation of energy requirement for the interval of time t = to isEin Eout = E = Efinal Einitialt2 o hAs T T ( L,t ) dt 0 = Mc p ( T Ti )f0where Ein is due to convection heating over the period of time t = 0 to. With knowledge ofT(L,t), this expression can be integrated and a value for Tf determined.PROBLEM 5.5KNOWN: Diameter and initial temperature of steel balls cooling in air.FIND: Time required to cool to a prescribed temperature.SCHEMATIC:ASSUMPTIONS: (1) Negligible radiation effects, (2) Constant properties.ANALYSIS: Applying Eq. 5.10 to a sphere (Lc = ro/3),2hLc h ( ro / 3) 20 W/m K (0.002m )Bi ==== 0.001.kk40 W/m KHence, the temperature of the steel remains approximately uniform during the cooling process, andthe lumped capacitance method may be used. From Eqs. 5.4 and 5.5,()3Ti T D / 6 c p Ti Tt=ln=lnhAsT TT Th D2 Vcpt=7800kg/m 3 ( 0.012m ) 600J/kg K6 20 W/m 2 Kt = 1122 s = 0.312hln1150 325400 325<COMMENTS: Due to the large value of Ti, radiation effects are likely to be significant during theearly portion of the transient. The effect is to shorten the cooling time.PROBLEM 5.6KNOWN: The temperature-time history of a pure copper sphere in an air stream.FIND: The heat transfer coefficient between the sphere and the air stream.SCHEMATIC:ASSUMPTIONS: (1) Temperature of sphere is spatially uniform, (2) Negligible radiationexchange, (3) Constant properties.3PROPERTIES: Table A-1, Pure copper (333K): = 8933 kg/m , cp = 389 J/kgK, k = 398W/mK.ANALYSIS: The time-temperature history is given by Eq. 5.6 with Eq. 5.7. (t)t= exp i R tC t whereRt =1hAsCt = Vcp = T T.A s = D2 D3V=6Recognize that when t = 69s,o ( t ) ( 55 27 ) Ct== 0.718 = exp i t( 66 27 )o Cand noting that t = R tC t find t = 208s.Hence,h= VcpAs t=( 69s = exp t )8933 kg/m 3 0.0127 3 m3 / 6 389J/kg K 0.0127 2m 2 208sh = 35.3 W/m 2 K.COMMENTS: Note that with Lc = Do/6,Bi =hLc0.0127= 35.3 W/m 2 K m/398 W/m K = 1.88 10 -4.k6Hence, Bi < 0.1 and the spatially isothermal assumption is reasonable.<PROBLEM 5.7KNOWN: Solid steel sphere (AISI 1010), coated with dielectric layer of prescribed thickness andthermal conductivity. Coated sphere, initially at uniform temperature, is suddenly quenched in an oilbath.FIND: Time required for sphere to reach 140C.SCHEMATIC:()oPROPERTIES: Table A-1, AISI 1010 Steel T = [500 +140 ] C/2 = 320oC 600K : = 7832 kg/m 3 , c = 559 J/kg K, k = 48.8 W/m K.ASSUMPTIONS: (1) Steel sphere is space-wise isothermal, (2) Dielectric layer has negligiblethermal capacitance compared to steel sphere, (3) Layer is thin compared to radius of sphere, (4)Constant properties.ANALYSIS: The thermal resistance to heat transfer from the sphere is due to the dielectric layer andthe convection coefficient. That is,l10.002m1m2 K= ( 0.050 + 0.0003 ) = 0.0503,W3300 W/m 2 Kor in terms of an overall coefficient, U = 1/R = 19.88 W/m2 K. The effective Biot number isR =k+h=0.04 W/m K+2ULc U ( ro /3 ) 19.88 W/m K ( 0.300/6 ) mBi e ==== 0.0204kk48.8 W/m Kwhere the characteristic length is Lc = ro/3 for the sphere. Since Bie < 0.1, the lumped capacitanceapproach is applicable. Hence, Eq. 5.5 is appropriate with h replaced by U,t= c V i c V T ( 0 ) Tln=ln.U As o U As T ( t ) TSubstituting numerical values with (V/As ) = ro/3 = D/6,o7832 kg/m 3 559 J/kg K 0.300m ( 500 100 ) Ct=ln6o19.88 W/m 2 K(140 100 )t = 25,358s = 7.04h.C<COMMENTS: (1) Note from calculation of R that the resistance of the dielectric layer dominatesand therefore nearly all the temperature drop occurs across the layer.PROBLEM 5.8KNOWN: Thickness, surface area, and properties of iron base plate. Heat flux at inner surface.Temperature of surroundings. Temperature and convection coefficient of air at outer surface.FIND: Time required for plate to reach a temperature of 135C. Operating efficiency of iron.SCHEMATIC:ASSUMPTIONS: (1) Radiation exchange is between a small surface and large surroundings, (2)Convection coefficient is independent of time, (3) Constant properties, (4) Iron is initially at roomtemperature (Ti = T).ANALYSIS: Biot numbers may be based on convection heat transfer and/or the maximum heattransfer by radiation, which would occur when the plate reaches the desired temperature (T = 135C).()2From Eq. (1.9) the corresponding radiation transfer coefficient is hr = (T +Tsur) T 2 + Tsur = 0.8 -82422225.67 10 W/m K (408 + 298) K (408 + 298 ) K = 8.2 W/m K. Hence,Bi =2hL 10 W / m K ( 0.007m )== 3.9 104k180 W / m K2h r L 8.2 W / m K (0.007m )Bi r === 3.2 104k180 W / m KWith convection and radiation considered independently or collectively, Bi, Bir, Bi + Bir << 1 and thelumped capacitance analysis may be used.The energy balance, Eq. (5.15), associated with Figure 5.5 may be applied to this problem. WithEg = 0, the integral form of the equation isT Ti =(As t 440 qh h (T T ) T Tsur Vc )dtIntegrating numerically, we obtain, for T = 135C,t = 168s<COMMENTS: Note that, if heat transfer is by natural convection, h, like hr, will vary during theprocess from a value of 0 at t = 0 to a maximum at t = 168s.PROBLEM 5.9KNOWN: Diameter and radial temperature of AISI 1010 carbon steel shaft. Convectioncoefficient and temperature of furnace gases.FIND: Time required for shaft centerline to reach a prescribed temperature.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties.PROPERTIES: AISI 1010 carbon steel, Table A.1 ( T = 550 K ) : = 7832 kg / m 3 , k = 51.2-5W/mK, c = 541 J/kgK, = 1.21102m /s.ANALYSIS: The Biot number isBi =2hro / 2 100 W/m K ( 0.05 m/2)== 0.0488.k51.2 W/m KHence, the lumped capacitance method can be applied. From Equation 5.6, hAs T T 4h = exp t = exp cD t Ti T Vc 4 100 W/m 2 K 800 1200 ln t = 0.811 = 300 1200 7832 kg/m 3 ( 541 J/kg K ) 0.1 m<t = 859 s.COMMENTS: To check the validity of the foregoing result, use the one-term approximation to theseries solution. From Equation 5.49c,(To T 4002== 0.444 = C1 exp 1 FoTi T 900)For Bi = hro/k = 0.0976, Table 5.1 yields 1 = 0.436 and C1 = 1.024. Hence ( 0.436)2(1.2105 m2 / s) t = ln (0.434) = 0.835( 0.05 m ) 2t = 915 s.The results agree to within 6%. The lumped capacitance method underestimates the actual time,since the response at the centerline lags that at any other location in the shaft.PROBLEM 5.10KNOWN: Configuration, initial temperature and charging conditions of a thermal energy storage unit.FIND: Time required to achieve 75% of maximum possible energy storage. Temperature of storagemedium at this time.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible radiationexchange with surroundings.()PROPERTIES: Table A-1, Aluminum, pure T 600K = 327o C : k = 231 W/mK, c = 10333J/kgK, = 2702 kg/m .ANALYSIS: Recognizing the characteristic length is the half thickness, findhL 100 W/m 2 K 0.025mBi =k== 0.011.231 W/m KHence, the lumped capacitance method may be used. From Eq. 5.8,Q = ( Vc ) i 1 exp ( t/ t ) = Est(1)E st,max = ( Vc ) i.(2)Dividing Eq. (1) by (2),Est / Est,max = 1 exp ( t/ th ) = 0.75.Solving for th = Vc Lc 2702 kg/m 3 0.025m 1033 J/kg K=== 698s.hAsh100 W/m2 KHence, the required time isexp ( t/698s ) = 0.25or<t = 968s.From Eq. 5.6,T T= exp ( t/ th )Ti T()T = T + ( Ti T ) exp ( t/ th ) = 600o C 575o C exp ( 968/698 )T = 456oC.<COMMENTS: For the prescribed temperatures, the property temperature dependence is significantand some error is incurred by assuming constant properties. However, selecting properties at 600Kwas reasonable for this estimate.PROBLEM 5.11KNOWN: Diameter, density, specific heat and thermal conductivity of aluminum spheres used inpacked bed thermal energy storage system. Convection coefficient and inlet gas temperature.FIND: Time required for sphere to acquire 90% of maximum possible thermal energy and thecorresponding center temperature. Potential advantage of using copper in lieu of aluminum.SCHEMATIC:ASSUMPTIONS: (1) Negligible heat transfer to or from a sphere by radiation or conduction due tocontact with other spheres, (2) Constant properties.ANALYSIS: To determine whether a lumped capacitance analysis can be used, first compute Bi =2h(ro/3)/k = 75 W/m K (0.025m)/150 W/mK = 0.013 < 0.1. Hence, the lumped capacitanceapproximation may be made, and a uniform temperature may be assumed to exist in the sphere at anytime. From Eq. 5.8a, achievement of 90% of the maximum possible thermal energy storagecorresponds toQ= 0.90 = 1 exp ( t / t ) cViwhere t = Vc / hA s = Dc / 6h = 2700 kg / m3 0.075m 950 J / kg K / 6 75 W / m 2 K = 427s. Hence,t = t ln ( 0.1) = 427s 2.30 = 984s<From Eq. (5.6), the corresponding temperature at any location in the sphere is()T (984s ) = Tg,i + Ti Tg,i exp ( 6ht / Dc )T (984s ) = 300C 275C exp 6 75 W / m K 984s / 2700 kg / m 0.075m 950 J / kg K()T (984 ) s = 272.5C<233Obtaining the density and specific heat of copper from Table A-1, we see that (c)Cu 8900 kg/m 6363400 J/kgK = 3.56 10 J/m K > (c)Al = 2.57 10 J/m K. Hence, for an equivalent spherediameter, the copper can store approximately 38% more thermal energy than the aluminum.COMMENTS: Before the packed bed becomes fully charged, the temperature of the gas decreasesas it passes through the bed. Hence, the time required for a sphere to reach a prescribed state ofthermal energy storage increases with increasing distance from the bed inlet.PROBLEM 5.12KNOWN: Wafer, initially at 100C, is suddenly placed on a chuck with uniform and constanttemperature, 23C. Wafer temperature after 15 seconds is observed as 33C.FIND: (a) Contact resistance, R , between interface of wafer and chuck through which helium slowlytcflows, and (b) Whether R will change if air, rather than helium, is the purge gas.tcSCHEMATIC:PROPERTIES: Wafer (silicon, typical values): = 2700 kg/m3, c = 875 J/kgK, k = 177 W/mK.ASSUMPTIONS: (1) Wafer behaves as a space-wise isothermal object, (2) Negligible heat transferfrom wafer top surface, (3) Chuck remains at uniform temperature, (4) Thermal resistance across theinterface is due to conduction effects, not convective, (5) Constant properties.ANALYSIS: (a) Perform an energy balance on the wafer as shown in the Schematic. Ein E + Eg = Estout(1)qcond = Est(2)T ( t ) TcdTw= wc wR dttc(3)Separate and integrate Eq. (3)tTdTwdt=w0 wcR Twi Tw TctcTw ( t ) Tct= exp Twi Tctc wcR (4)(5)Substituting numerical values for Tw(15s) = 33C,(33 23) C exp = 2700 kg(100 23) Cm3 0.758 103 m 875 J kg K R tc 15sR = 0.0041m 2 K Wtc(b) R will increase since kair < khelium. See Table A.4.tcCOMMENTS: Note that Bi = Rint/Rext = (w/k)/ R = 0.001. Hence the spacewise isothermaltcassumption is reasonable.(6)<PROBLEM 5.13KNOWN: Inner diameter and wall thickness of a spherical, stainless steel vessel. Initialtemperature, density, specific heat and heat generation rate of reactants in vessel. Convectionconditions at outer surface of vessel.FIND: (a) Temperature of reactants after one hour of reaction time, (b) Effect of convectioncoefficient on thermal response of reactants.SCHEMATIC:ASSUMPTIONS: (1) Temperature of well stirred reactants is uniform at any time and is equal toinner surface temperature of vessel (T = Ts,i), (2) Thermal capacitance of vessel may be neglected, (3)Negligible radiation exchange with surroundings, (4) Constant properties.ANALYSIS: (a) Transient thermal conditions within the reactor may be determined from Eq. (5.25),which reduces to the following form for Ti - T = 0.T = T + ( b / a ) 1 exp ( at )where a = UA/Vc and b = E g / Vc = q / c. From Eq. (3.19) the product of the overall heat transfer-1coefficient and the surface area is UA = (Rcond + Rconv) , where from Eqs. (3.36) and (3.9),R t,cond =1 111=2 k Di Do 2 (17 W / m K )R t,conv =11== 0.0438K / W2hAo6 W / m 2 K (1.1m )114 = 8.51 10 K / W1.0m 1.1m )(Hence, UA = 24.4 W/K. It follows that, with V = D3 / 6,ia=6 ( 22.4 W / K )UA== 1.620 105 s 13 1m 3 2400 J / kg K Vc 1100 kg / m()q104 W / m3b=== 3.788 103 K / s3 2400 J / kg K c 1100 kg / mWith (b/a) = 233.8C and t = 18,000s,()T = 25C + 233.8C 1 exp 1.62 105 s1 18, 000s = 84.1C<Neglecting the thermal capacitance of the vessel wall, the heat rate by conduction through the wall isequal to the heat transfer by convection from the outer surface, and from the thermal circuit, we knowthatContinued ..PROBLEM 5.13 (Cont.)T Ts,oTs,o TTs,o ==R t,condR t,conv=8.51104 K / W= 0.01940.0438 K / WT + 0.0194T 84.1C + 0.0194 ( 25C )== 83.0C1.01941.0194<22(b) Representative low and high values of h could correspond to 2 W/m K and 100 W/m K for freeand forced convection, respectively. Calculations based on Eq. (5.25) yield the following temperaturehistories.R e a cto r te m p e ra tu re (C )10080604020036007200108001440018000P ro ce s s tim e (s )h =2 W /m ^2 .Kh =6 W /m ^2 .Kh =1 0 0 W /m ^2 .KForced convection is clearly an effective means of reducing the temperature of the reactants andaccelerating the approach to steady-state conditions.COMMENTS: The validity of neglecting thermal energy storage effects for the vessel may beassessed by contrasting its thermal capacitance with that of the reactants. Selecting values of =38000 kg/m and c = 475 J/kgK for stainless steel from Table A-1, the thermal capacitance of the()3vessel is Ct,v = (Vc)st = 6.57 10 J/K, where V = ( / 6 ) D3 Di . With Ct,r = (Vc)r = 2.64 o5610 J/K for the reactants, Ct,r/Ct,v 4. Hence, the capacitance of the vessel is not negligible andshould be considered in a more refined analysis of the problem.PROBLEM 5.14KNOWN: Volume, density and specific heat of chemical in a stirred reactor. Temperature andconvection coefficient associated with saturated steam flowing through submerged coil. Tubediameter and outer convection coefficient of coil. Initial and final temperatures of chemical and timespan of heating process.FIND: Required length of submerged tubing. Minimum allowable steam flowrate.SCHEMATIC:ASSUMPTIONS: (1) Constant properties, (2) Negligible heat loss from vessel to surroundings, (3)Chemical is isothermal, (4) Negligible work due to stirring, (5) Negligible thermal energy generation(or absorption) due to chemical reactions associated with the batch process, (6) Negligible tube wallconduction resistance, (7) Negligible kinetic energy, potential energy, and flow work changes forsteam.ANALYSIS: Heating of the chemical can be treated as a transient, lumped capacitance problem,wherein heat transfer from the coil is balanced by the increase in thermal energy of the chemical.Hence, conservation of energy yieldsdUdT= Vc= UAs (Th T )dtdtT dTUAs tIntegrating,Ti Th T = Vc o dtT T UAs t ln h= VcTh TiAs = ( Vc Th TlnUtTh TiU = hi 1 + h o 1)1= (1/10, 000 ) + (1/ 2000 )(1)1W / m2 KU = 1670 W / m 2 K(1200 kg / m3 )(2.25m3 )(2200 J / kg K ) ln 500 450 = 1.37m2As = 500 300(1670 W / m2 K )(3600s)L=As1.37m 2== 21.8m D (0.02m )<COMMENTS: Eq. (1) could also have been obtained by adapting Eq. (5.5) to the conditions of thisproblem, with T and h replaced by Th and U, respectively.PROBLEM 5.15KNOWN: Thickness and properties of furnace wall. Thermal resistance of film on surface of wallexposed to furnace gases. Initial wall temperature.FIND: (a) Time required for surface of wall to reach a prescribed temperature, (b) Correspondingvalue of film surface temperature.SCHEMATIC:ASSUMPTIONS: (1) Constant properties, (2) Negligible film thermal capacitance, (3) Negligibleradiation.3PROPERTIES: Carbon steel (given): = 7850 kg/m , c = 430 J/kgK, k = 60 W/mK.ANALYSIS: The overall coefficient for heat transfer from the surface of the steel to the gas is1U = ( R )tot1= + R fh11=+ 102 m2 K/W 25 W/m 2 K1= 20 W/m 2 K.Hence,UL 20 W/m 2 K 0.01 m== 0.0033k60 W/m Kand the lumped capacitance method can be used.Bi =(a) It follows thatT T= exp ( t/ t ) = exp ( t/RC ) = exp ( Ut/ Lc )Ti T7850 kg/m3 ( 0.01 m ) 430 J/kg K 1200 1300 Lc T Tt=ln=lnUTi T300 130020 W/m 2 K<t = 3886s = 1.08h.(b) Performing an energy balance at the outer surface (s,o),h T Ts,o = Ts,o Ts,i / R f(Ts,o =)()hT + Ts,i / R 25 W/m 2 K 1300 K + 1200 K/10-2 m2 K/Wf=h + (1 / R )f( 25 +100 ) W/m2 KTs,o = 1220 K.COMMENTS: The film increases t by increasing Rt but not Ct.<PROBLEM 5.16KNOWN: Thickness and properties of strip steel heated in an annealing process. Furnace operatingconditions.FIND: (a) Time required to heat the strip from 300 to 600C. Required furnace length for prescribedstrip velocity (V = 0.5 m/s), (b) Effect of wall temperature on strip speed, temperature history, andradiation coefficient.SCHEMATIC:ASSUMPTIONS: (1) Constant properties, (2) Negligible temperature gradients in transverse directionacross strip, (c) Negligible effect of strip conduction in longitudinal direction.PROPERTIES: Steel: = 7900 kg/m3, cp = 640 J/kgK, k = 30 W/mK, = 0.7.ANALYSIS: (a) Considering a fixed (control) mass of the moving strip, its temperature variation withtime may be obtained from an energy balance which equates the change in energy storage to heat transferby convection and radiation. If the surface area associated with one side of the control mass isdesignated as As, As,c = As,r = 2As and V = As in Equation 5.15, which reduces to c()dT4= 2 h ( T T ) + T 4 Tsur dtor, introducing the radiation coefficient from Equations 1.8 and 1.9 and integrating,Tf Ti = tf1o h (T T ) + h r (T Tsur )dt c ( 2 )Using the IHT Lumped Capacitance Model to integrate numerically with Ti = 573 K, we find that Tf =873 K corresponds totf 209s<in which case, the required furnace length isL = Vt f 0.5 m s 209s 105 m<(b) For Tw = 1123 K and 1273 K, the numerical integration yields tf 102s and 62s respectively. Hence,for L = 105 m , V = L/tf yieldsV ( Tw = 1123K ) = 1.03m sV ( Tw = 1273K ) = 1.69 m s<Continued...PROBLEM 5.16 (Cont.)which correspond to increased process rates of 106% and 238%, respectively. Clearly, productivity canbe enhanced by increasing the furnace environmental temperature, albeit at the expense of increasingenergy utilization and operating costs.If the annealing process extends from 25C (298 K) to 600C (873 K), numerical integrationyields the following results for the prescribed furnace temperatures.600200Radiation coefficient, hr(W/m^2.K)Temperature, T(C)5004003002001000050100150200Annealing time, t(s)Tsur = Tinf = 1000 CTsur = Tinf = 850 CTsur = Tinf = 700 C25030015010050050100150200250300Annealing time, t(s)Tsur = Tinf = 1000 CTsur = Tinf = 850 CTsur = Tinf = 700 CAs expected, the heating rate and time, respectively, increase and decrease significantly with increasingTw. Although the radiation heat transfer rate decreases with increasing time, the coefficient hr increaseswith t as the strip temperature approaches Tw.COMMENTS: To check the validity of the lumped capacitance approach, we calculate the Biot numberbased on a maximum cumulative coefficient of (h + hr) 300 W/m2K. It follows that Bi = (h + hr)(/2)/k= 0.06 and the assumption is valid.PROBLEM 5.17KNOWN: Diameter, resistance and current flow for a wire. Convection coefficient and temperatureof surrounding oil.FIND: Steady-state temperature of the wire. Time for the wire temperature to come within 1C ofits steady-state value.SCHEMATIC:ASSUMPTIONS: (1) Constant properties, (2) Wire temperature is independent of x.3PROPERTIES: Wire (given): = 8000 kg/m , cp = 500 J/kgK, k = 20 W/mK, R = 0.01 /m.eANALYSIS: SinceBi =h ( ro / 2)k=(500 W/m 2 K 2.5 10-4m20 W/m K) = 0.006 < 0.1the lumped capacitance method can be used. The problem has been analyzed in Example 1.3, andwithout radiation the steady-state temperature is given by Dh ( T T ) = I 2R .eHenceI 2R (100A )2 0.01 / me = 25o C +T = T += 88.7 o C.2 K Dh 0.001 m 500 W/m()<With no radiation, the transient thermal response of the wire is governed by the expression (Example1.3)dTI 2R 4he=(T T ) .dt c D2 / 4 c pDp()With T = Ti = 25C at t = 0, the solution isT T I 2R e / Dh() = exp 4h t . cp D Ti T ( I 2 Re / Dh )Substituting numerical values, find87.7 25 63.74 500 W/m 2 K= expt 8000 kg/m3 500 J/kg K 0.001 m 25 25 63.7t = 8.31s.COMMENTS: The time to reach steady state increases with increasing , cp and D and withdecreasing h.<PROBLEM 5.18KNOWN: Electrical heater attached to backside of plate while front is exposed to a convection process(T, h); initially plate is at uniform temperature T before heater power is switched on.FIND: (a) Expression for temperature of plate as a function of time assuming plate is spacewiseisothermal, (b) Approximate time to reach steady-state and T() for prescribed T, h and q when wallomaterial is pure copper, (c) Effect of h on thermal response.SCHEMATIC:ASSUMPTIONS: (1) Plate behaves as lumped capacitance, (2) Negligible loss out backside of heater,(3) Negligible radiation, (4) Constant properties.PROPERTIES: Table A-1, Copper, pure (350 K): k = 397 W/mK, cp = 385 J/kgK, = 8933 kg/m3.ANALYSIS: (a) Following the analysis of Section 5.3, the energy conservation requirement for thesystem is Ein E out = Est or q h ( T T ) = Lcp dT dt . Rearranging, and with R = 1/h andtoC = Lcp,tT T q h = R C dT dtott(1)Defining ( t ) T T q h with d = dT, the differential equation iso = R Cttd.dt(2)Separating variables and integrating,t dt d=0 R Ci ttit follows thatt= exp itt R C (3)where i = ( 0 ) = Ti T ( q h )o<(4)(b) For h = 50 W/m2 K, the steady-state temperature can be determined from Eq. (3) with t ; that is, ( ) = 0 = T ( ) T q hT ( ) = T + q h ,oroogiving T() = 27C + 5000 W/m2 /50 W/m2K = 127C. To estimate the time to reach steady-state, firstdetermine the thermal time constant of the system,)11 t = R C = cp L = 8933kg m3 385 J kg K 12 103 m = 825stt 50 W m 2 K h()(Continued...PROBLEM 5.18 (Cont.)When t = 3t = 3825s = 2475s, Eqs. (3) and (4) yield (3 t ) = T (3 t ) 27 C 5000 W m 2 = e3 27 C 27 C 50 W m 2 K50 W m 2 K 5000 W m 2<T(3t) = 122C(c) As shown by the following graphical results, which were generated using the IHT LumpedCapacitance Model, the steady-state temperature and the time to reach steady-state both decrease withincreasing h.125Temperature, T(C)1058565452505001000150020002500Time, t(s)h = 50 W/m^2.Kh = 100 W/m^2.Kh = 200 W/m^2.KCOMMENTS: Note that, even for h = 200 W/m2K, Bi = hL/k << 0.1 and assumption (1) is reasonable.PROBLEM 5.19KNOWN: Electronic device on aluminum, finned heat sink modeled as spatially isothermal object withinternal generation and convection from its surface.FIND: (a) Temperature response after device is energized, (b) Temperature rise for prescribedconditions after 5 min.SCHEMATIC:ASSUMPTIONS: (1) Spatially isothermal object, (2) Object is primarily aluminum, (3) Initially, objectis in equilibrium with surroundings at T.()oPROPERTIES: Table A-1, Aluminum, pure T = ( 20 + 100 ) C/2 333K : c = 918 J/kgK.ANALYSIS: (a) Following the general analysis of Section 5.3, apply the conservation of energyrequirement to the object,dT&Eg hAs (T T ) = Mcdt&&&&Ein + Eg -Eout = Est(1)where T = T(t). Consider now steady-state conditions, in which case the storage term of Eq. (1) iszero. The temperature of the object will be T() such that&E g = hAs ( T ( ) T ) .&Substituting for Eg using Eq. (2) into Eq. (1), the differential equation is(2)Mc d(3,4)hAs dtwith T - T() and noting that d = dT. Identifying R t = 1/hA s and C t = Mc, the differentialMc dTT ( ) T [ T T ] =hAs dtor =equation is integrated with proper limits,1t d0 dt = i R t Ctort= exp i R t Ct (5)<where i = (0) = Ti - T() and Ti is the initial temperature of the object.(b) Using the information about steady-state conditions and Eq. (2), find first the thermal resistance andcapacitance of the system,T ( ) T (100 20 )o C1Rt ==== 1.33 K/W&hAsEg60 WC t = Mc = 0.31 kg 918 J/kg K = 285 J/K.Using Eq. (5), the temperature of the system after 5 minutes is 5minT 5min T T 5min 100o C(i)= ()( )= ()Ti T ( )( 20 100 )o C5 60s= exp = 0.453 1.33 K/W 285 J/K oT ( 5min ) = 100o C + ( 20 100 ) C 0.453 = 63.8oC&COMMENTS: Eq. 5.24 may be used directly for Part (b) with a = hAs/Mc and b = E g /Mc.<PROBLEM 5.20KNOWN: Spherical coal pellet at 25C is heated by radiation while flowing through a furnacemaintained at 1000C.FIND: Length of tube required to heat pellet to 600C.SCHEMATIC:ASSUMPTIONS: (1) Pellet is suspended in air flow and subjected to only radiative exchange withfurnace, (2) Pellet is small compared to furnace surface area, (3) Coal pellet has emissivity, = 1.PROPERTIES: Table A-3, Coal()T = ( 600 + 25 ) C/2 = 585K, however, only 300K data available : = 13503kg/m ,cp = 1260 J/kgK, k = 0.26 W/mK.ANALYSIS: Considering the pellet as spatially isothermal, use the lumped capacitance method ofSection 5.3 to find the time required to heat the pellet from To = 25C to TL = 600C. From anenergy balance on the pellet Ein = Est wheredT44Ein = q rad = As Tsur TsEst = cp(())44As Tsur Ts = cpgivingdtdT.dtSeparating variables and integrating with limits shown, thetemperature-time relation becomesTLAs tdTdt = .04cpT o Tsur T 4The integrals are evaluated in Eq. 5.18 givingt=ln34As Tsur cpTsur + T lnTsur TT T Tsur + Ti+ 2 tan -1 tan -1 i .Tsur Ti Tsur Tsur 23Recognizing that As = D and = D /6 or As/ = 6/D and substituting values,1350 kg/m3 0.001 m 1260 J/kg Kt=()1273 + 298 1273 + 873 lnln31273 29824 5.67 10-8 W/m 2 K 4 (1273 K ) 1273 873 873 -1 298 +2 tan -1 tan 1273 = 1.18s. 1273 Hence,L = Vt = 3m/s1.18s = 3.54m.<The validity of the lumped capacitance method requires Bi = h( /As)k < 0.1. Using Eq. (1.9) for h =hr and /As = D/6, find that when T = 600C, Bi = 0.19; but when T = 25C, Bi = 0.10. At earlytimes, when the pellet is cooler, the assumption is reasonable but becomes less appropriate as thepellet heats.PROBLEM 5.21KNOWN: Metal sphere, initially at a uniform temperature Ti, is suddenly removed from a furnace andsuspended in a large room and subjected to a convection process (T, h) and to radiation exchange withsurroundings, Tsur.FIND: (a) Time it takes for sphere to cool to some temperature T, neglecting radiation exchange, (b)Time it takes for sphere to cool to some temperature t, neglecting convection, (c) Procedure to obtaintime required if both convection and radiation are considered, (d) Time to cool an anodized aluminumsphere to 400 K using results of Parts (a), (b) and (c).SCHEMATIC:ASSUMPTIONS: (1) Sphere is spacewise isothermal, (2) Constant properties, (3) Constant heattransfer convection coefficient, (4) Sphere is small compared to surroundings.PROPERTIES: Table A-1, Aluminum, pure ( T = [800 + 400] K/2 = 600 K): = 2702 kg/m3, c = 1033J/kgK, k = 231 W/mK, = k/c = 8.276 10-5 m2/s; Aluminum, anodized finish: = 0.75, polishedsurface: = 0.1.ANALYSIS: (a) Neglecting radiation, the time to cool is predicted by Eq. 5.5,t= Vc i Dc Ti Tln =lnhAs 6hT T(1)<where V/As = (D3/6)/(D2) = D/6 for the sphere.(b) Neglecting convection, the time to cool is predicted by Eq. 5.18,t=TT + Ti Tsur + T ln sur+ 2 tan 1 ln3Tsur Ti24 Tsur Tsur T Tsur Dc1 Ti tan Tsur (2)where V/As,r = D/6 for the sphere.(c) If convection and radiation exchange are considered, the energy balance requirement results in Eq. 5.15 (with qs = E g = 0). Hence()dT64=h ( T T ) + T 4 Tsur dt Dc (3)<where As(c,r) = As = D2 and V/As(c,r) = D/6. This relation must be solved numerically in order to evaluatethe time-to-cool.(d) For the aluminum (pure) sphere with an anodized finish and the prescribed conditions, the times tocool from Ti = 800 K to T = 400 K are:Continued...PROBLEM 5.21 (Cont.)Convection only, Eq. (1)t=2702 kg m3 0.050 m 1033J kg K6 10 W m 2 Kln800 300= 3743s = 1.04h400 300<Radiation only, Eq. (2)t=2702 kg m3 0.050 m 1033J kg K 400 + 300800 + 300 ln ln+3800 300 24 0.75 5.67 108 W m 2 K 4 (300 K ) 400 300400800 2 tan 1 tan 1300300 t = 5.065 103 {1.946 0.789 + 2 ( 0.927 1.212 )} = 2973s = 0.826h<Radiation and convection, Eq. (3)Using the IHT Lumped Capacitance Model, numerical integration yieldst 1600s = 0.444hIn this case, heat loss by radiation exerts the stronger influence, although the effects of convection are byno means negligible. However, if the surface is polished ( = 0.1), convection clearly dominates. Foreach surface finish and the three cases, the temperature histories are as follows.800800700Temperature, T(K)Temperature, T(K)7006005004006005004000400800 1200 1600 2000 2400 2800 3200 3600 400000.511.5Time, t(s)h = 10 W/m^2.K, eps = 0.75h = 0,eps = 0.75h = 10 W/m^2.K, eps = 022.5Time, t x E-4 (s)h = 10 W/m^2.K, eps = 0.1h = 10 W/ m^2.K, eps = 0h = 0,eps = 0.1COMMENTS: 1. A summary of the analyses shows the relative importance of the various modes ofheat loss:Active ModesConvection onlyRadiation onlyBoth modesTime required to cool to 400 K (h) = 0.75 = 0.11.0400.8270.4441.0406.1940.8892. Note that the spacewise isothermal assumption is justified since Be << 0.1. For the convection-onlyprocess,2-4Bi = h(ro/3)/k = 10 W/m K (0.025 m/3)/231 W/mK = 3.6 10PROBLEM 5.22KNOWN: Droplet properties, diameter, velocity and initial and final temperatures.FIND: Travel distance and rejected thermal energy.SCHEMATIC:ASSUMPTIONS: (1) Constant properties, (2) Negligible radiation from space.3PROPERTIES: Droplet (given): = 885 kg/m , c = 1900 J/kgK, k = 0.145 W/mK, = 0.95.ANALYSIS: To assess the suitability of applying the lumped capacitance method, use Equation 1.9to obtain the maximum radiation coefficient, which corresponds to T = Ti.h r = Ti3 = 0.95 5.67 10 8 W/m 2 K 4 ( 500 K )3 = 6.73 W/m 2 K.HenceBi r =h r ( ro / 3 )( 6.73 W/m2 K ) (0.25 103 m/3) = 0.0039=0.145 W/m Kkand the lumped capacitance method can be used. From Equation 5.19,((3L c D / 6t= =V 3 D2 L=)) 1 1 T3fTi3 ( 0.1 m/s ) 885 kg/m3 ( 1900 J/kg K ) 0.5 103 m 11 1 3003 5003 K318 0.95 5.67 10-8 W/m2 K4<L = 2.52 m.The amount of energy rejected by each droplet is equal to the change in its internal energy.5 104 m )3 (E i E f = Vc ( Ti Tf ) = 885 kg/m1900 J/kg K ( 200 K )36E i E f = 0.022 J.<COMMENTS: Because some of the radiation emitted by a droplet will be intercepted by otherdroplets in the stream, the foregoing analysis overestimates the amount of heat dissipated by radiationto space.PROBLEM 5.23KNOWN: Initial and final temperatures of a niobium sphere. Diameter and properties of the sphere.Temperature of surroundings and/or gas flow, and convection coefficient associated with the flow.FIND: (a) Time required to cool the sphere exclusively by radiation, (b) Time required to cool thesphere exclusively by convection, (c) Combined effects of radiation and convection.SCHEMATIC:ASSUMPTIONS: (1) Uniform temperature at any time, (2) Negligible effect of holding mechanismon heat transfer, (3) Constant properties, (4) Radiation exchange is between a small surface and largesurroundings.ANALYSIS: (a) If cooling is exclusively by radiation, the required time is determined from Eq.32(5.18). With V = D /6, As,r = D , and = 0.1,t=8600 kg / m3 ( 290 J / kg K ) 0.01m 298 + 573298 + 1173 lnln3298 117324 ( 0.1) 5.67 108 W / m 2 K 4 ( 298K ) 298 573 573 1 1173 +2 tan 1 tan 298 298 t = 6926s {1.153 0.519 + 2 (1.091 1.322 )} = 1190s( = 0.1)<( = 0.6 )<If = 0.6, cooling is six times faster, in which case,t = 199s(b) If cooling is exclusively by convection, Eq. (5.5) yields3 cD Ti T 8600 kg / m ( 290 J / kg K ) 0.010m 875 t=ln ln =6h 275 1200 W / m 2 K Tf T <t = 24.1s(c) With both radiation and convection, the temperature history may be obtained from Eq. (5.15).() D3 / 6 c()dT4= D2 h ( T T ) + T 4 Tsur dtIntegrating numerically from Ti = 1173 K at t = 0 to T = 573K, we obtain<t = 21.0sContinued ..PROBLEM 5.23 (Cont.)Cooling times corresponding to representative changes in and h are tabulated as follows2h(W/m K) ||t(s)|2000.621.02001.019.4200.6102.85000.69.1For values of h representative of forced convection, the influence of radiation is secondary, even for amaximum possible emissivity of 1.0. Hence, to accelerate cooling, it is necessary to increase h.However, if cooling is by natural convection, radiation is significant. For a representative natural2convection coefficient of h = 20 W/m K, the radiation flux exceeds the convection flux at the surfaceof the sphere during early to intermediate stages of the transient.H e a t flu xe s (W /m ^2 .K )700006000050000400003000020000100000020406080100C o o lin g tim e (s )C o n ve ctio n flu x (h = 2 0 W /m ^2 .K )R a d ia tio n flu x (e p s = 0 .6 )22COMMENTS: (1) Even for h as large as 500 W/m K, Bi = h (D/6)/k = 500 W/m K (0.01m/6)/63W/mK = 0.013 < 0.1 and the lumped capacitance model is appropriate. (2) The largest value of hr-824corresponds to Ti =1173 K, and for = 0.6 Eq. (1.9) yields hf = 0.6 5.67 10 W/m K (1173 +2222298)K (1173 + 298 )K = 73.3 W/m K.PROBLEM 5.24KNOWN: Diameter and thermophysical properties of alumina particles. Convection conditionsassociated with a two-step heating process.FIND: (a) Time-in-flight (ti-f) required for complete melting, (b) Validity of assuming negligibleradiation.SCHEMATIC:ASSUMPTIONS: (1) Particle behaves as a lumped capacitance, (2) Negligible radiation, (3) Constantproperties.ANALYSIS: (a) The two-step process involves (i) the time t1 to heat the particle to its melting point and(ii) the time t2 required to achieve complete melting. Hence, ti-f = t1 + t2, where from Eq. (5.5),t1 =t1 = p Vc p p DpcpT Tln i =ln i hAs6hTmp T)(3970 kg m3 50 106 m 1560 J kg K(6 30, 000 W m 2 K)ln(300 10, 000 ) = 4 104 s( 2318 10, 000 )Performing an energy balance for the second step, we obtaint1 + t 2t1q conv dt = Estwhere qconv = hAs(T - Tmp) and Est = pVhsf. Hence,t2 =pDp6hh sf(T Tmp )=(3970 kg m3 50 106 m(6 30, 000 W m 2 KHence t i f = 9 104 s 1ms)) 3.577 106 J kg(10, 000 2318 ) K= 5 104 s<82(b) Contrasting the smallest value of the convection heat flux, q conv,min = h ( T Tmp ) = 2.3 10 W m()4452to the largest radiation flux, qrad,max = Tmp Tsur = 6.5 10 W/m , we conclude that radiationis, in fact, negligible.COMMENTS: (1) Since Bi = (hrp/3)/k 0.05, the lumped capacitance assumption is good. (2) In anactual application, the droplet should impact the substrate in a superheated condition (T > Tmp), whichwould require a slightly larger ti-f.PROBLEM 5.25KNOWN: Diameters, initial temperature and thermophysical properties of WC and Co in compositeparticle. Convection coefficient and freestream temperature of plasma gas. Melting point and latentheat of fusion of Co.FIND: Times required to reach melting and to achieve complete melting of Co.SCHEMATIC:ASSUMPTIONS: (1) Particle is isothermal at any instant, (2) Radiation exchange with surroundingsis negligible, (3) Negligible contact resistance at interface between WC and Co, (4) Constantproperties.ANALYSIS: From Eq. (5.5), the time required to reach the melting point ist1 =( Vc )tot2h DolnTi TTmp Twhere the total heat capacity of the composite particle is( Vc )tot = ( Vc )c + ( Vc )s = 16, 000 kg / m3 (1.6 105 m)((3+8900 kg / m3 / 6 2.0 105 m 1.6 105 m)() / 6 300 J / kg K3) 750 J / kg K3 = 1.03 108 + 1.36 108 J / K = 2.39 108 J / Kt1 =(2.39 108 J / K)()220, 000 W / m 2 K 2.0 105 mln(300 10, 000 ) K = 1.56 104 s(1770 10, 000 ) K<The time required to melt the Co may be obtained by applying the first law, Eq. (1.11b) to a controlsurface about the particle. It follows that233Ein = h Do T Tmp t 2 = Est = s ( / 6 ) Do Di h sf((2 10 m ) (1.6 10 m ) 2.59 10(20, 000 W / m K ) (2 10 m ) (10, 000 1770 ) K8900 kg / mt2 =)()35( / 6 ) 2355235J / kg= 2.28 105s<COMMENTS: (1) The largest value of the radiation coefficient corresponds to hr = (Tmp + Tsur)(T)22mp + Tsur .22For the maximum possible value of = 1 and Tsur = 300K, hr = 378 W/m K << h =20,000 W/m K. Hence, the assumption of negligible radiation exchange is excellent. (2) Despite thelarge value of h, the small values of Do and Di and the large thermal conductivities (~ 40 W/mK and70 W/mK for WC and Co, respectively) render the lumped capacitance approximation a good one.(3) A detailed treatment of plasma heating of a composite powder particle is provided by Demetriou,Lavine and Ghoniem (Proc. 5th ASME/JSME Joint Thermal Engineering Conf., March, 1999).PROBLEM 5.26KNOWN: Dimensions and operating conditions of an integrated circuit.FIND: Steady-state temperature and time to come within 1 C of steady-state.SCHEMATIC:ASSUMPTIONS: (1) Constant properties, (2) Negligible heat transfer from chip to substrate.3PROPERTIES: Chip material (given): = 2000 kg/m , c = 700 J/kgK.ANALYSIS: At steady-state, conservation of energy yields&& Eout + Eg = 0()()& h L2 ( Tf T ) + q L2 t = 0&qtTf = T +hTf = 20o C +9 106 W/m3 0.001 m150 W/m 2 K= 80o C.<From the general lumped capacitance analysis, Equation 5.15 reduces todT& L2 t c= q L2 t h ( T T ) L2 .dtWith()ab()h150 W/m 2 K== 0.107 s -1 tc 2000 kg/m3 ( 0.001 m )( 700 J/kg K )()9 106 W/m33&q=c2000 kg/m()( 700 J/kg K )= 6.429 K/s.From Equation 5.24,exp ( at ) =t=T T b/a ( 79 20 60 ) K== 0.01667Ti T b/a ( 20 20 60 ) Kln ( 0.01667 )0.107 s -1= 38.3 s.<COMMENTS: Due to additional heat transfer from the chip to the substrate, the actual values ofTf and t are less than those which have been computed.PROBLEM 5.27KNOWN: Dimensions and operating conditions of an integrated circuit.FIND: Steady-state temperature and time to come within 1C of steady-state.SCHEMATIC:ASSUMPTIONS: (1) Constant properties.3PROPERTIES: Chip material (given): = 2000 kg/m , cp = 700 J/kgK.ANALYSIS: The direct and indirect paths for heat transfer from the chip to the coolant are inparallel, and the equivalent resistance is1)(R equiv = hL2 + R -1 = 3.75 103 + 5 103 W/K tThe corresponding overall heat transfer coefficient is1= 114.3 K/W.1(R equiv )U=L2=0.00875 W/K(0.005 m )2= 350 W/m 2 K.To obtain the steady-state temperature, apply conservation of energy to a control surface aboutthe chip.E + E = 0 UL2 ( T T ) + q L2 t = 0outgf()qt9 106 W/m3 0.001 m= 20 C += 45.7 C.2 KU350 W/mFrom the general lumped capacitance analysis, Equation 5.15 yieldsdT L2 t c= q L2 t U (T T ) L2 .dtWithTf = T +()<()U350 W/m 2 Ka== 0.250 s-1 tc2000 kg/m3 (0.001 m )(700 J/kg K )()()q9 106 W/m3b=== 6.429 K/s3 700 J/kg Kc2000 kg/m ()Equation 5.24 yieldsT T b/a ( 44.7 20 25.7 ) Kexp ( at ) === 0.0389Ti T b/a(20 20 25.7 ) Kt = ln ( 0.0389 ) / 0.250 s-1 = 13.0 s.COMMENTS: Heat transfer through the substrate is comparable to that associated withdirect convection to the coolant.<PROBLEM 5.28KNOWN: Dimensions, initial temperature and thermophysical properties of chip, solder andsubstrate. Temperature and convection coefficient of heating agent.FIND: (a) Time constants and temperature histories of chip, solder and substrate when heated by anair stream. Time corresponding to maximum stress on a solder ball. (b) Reduction in time associatedwith using a dielectric liquid to heat the components.SCHEMATIC:ASSUMPTIONS: (1) Lumped capacitance analysis is valid for each component, (2) Negligible heattransfer between components, (3) Negligible reduction in surface area due to contact betweencomponents, (4) Negligible radiation for heating by air stream, (5) Uniform convection coefficientamong components, (6) Constant properties.ANALYSIS: (a) From Eq. (5.7), t = ( Vc ) / hAChip:()V = L ch t ch = ( 0.015m )22(0.002m ) = 4.50 10 7 m3 , As == 2 ( 0.015m ) + 4 ( 0.015m ) 0.002m = 5.70 102t =4m2300 kg / m3 4.50 107 m3 710 J / kg K50 W / m 2 K 5.70 104 m 2V = D / 6 = ( 0.002m ) / 6 = 4.19 103Solder:t =(39(2L2ch + 4L ch t ch)2<= 25.8sm , A s = D = ( 0.002m ) = 1.26 103211, 000 kg / m3 4.19 109 m3 130 J / kg K50 W / m 2 K 1.26 105 m 225m2<= 9.5s)Substrate: V = L2 t sb = ( 0.025m )2 ( 0.01m ) = 6.25 10 6 m3 , As = L2 = ( 0.025m )2 = 6.25 10 4 m 2sbsbt =4000 kg / m3 6.25 106 m3 770 J / kg K50 W / m 2 K 6.25 104 m 2<= 616.0sSubstituting Eq. (5.7) into (5.5) and recognizing that (T Ti)/(T - Ti) = 1 (/i), in which case (T Ti)/(T -Ti) = 0.99 yields /i = 0.01, it follows that the time required for a component to experience99% of its maximum possible temperature rise ist 0.99 = ln ( i / ) = ln (100 ) = 4.61Hence,Chip: t = 118.9s,Solder: t = 43.8s,<Substrate: t = 2840Continued ..PROBLEM 5.28 (Cont.)Histories of the three components and temperature differences between a solder ball and its adjoiningcomponents are shown below.Te m p e ratu re (C )80655035200100200300400500Tim e (s )Te m p e ra tu re d iffe re n ce (C )Ts dTchTs b6050403020100020406080100Tim e (s )Ts d -TchTs d -Ts bCommensurate with their time constants, the fastest and slowest responses to heating are associatedwith the solder and substrate, respectively. Accordingly, the largest temperature difference is betweenthese two components, and it achieves a maximum value of 55C att ( maximum stress ) 40s<(b) With the 4-fold increase in h associated with use of a dielectric liquid to heat the components, thetime constants are each reduced by a factor of 4, and the times required to achieve 99% of themaximum temperature rise areChip: t = 29.5s,Solder: t = 11.0s,Substrate: t = 708s<The time savings is approximately 75%.COMMENTS: The foregoing analysis provides only a first, albeit useful, approximation to theheating problem. Several of the assumptions are highly approximate, particularly that of a uniformconvection coefficient. The coefficient will vary between components, as well as on the surfaces ofthe components. Also, because the solder balls are flattened, there will be a reduction in surface areaexposed to the fluid for each component, as well as heat transfer between components, which reducesdifferences between time constants for the components.PROBLEM 5.29KNOWN: Electrical transformer of approximate cubical shape, 32 mm to a side, dissipates 4.0 W2when operating in ambient air at 20C with a convection coefficient of 10 W/m K.FIND: (a) Develop a model for estimating the steady-state temperature of the transformer, T(), andevaluate T(), for the operating conditions, and (b) Develop a model for estimating the temperaturetime history of the transformer if initially the temperature is Ti = T and suddenly power is applied.Determine the time required to reach within 5C of its steady-state operating temperature.SCHEMATIC:ASSUMPTIONS: (1) Transformer is spatially isothermal object, (2) Initially object is in equilibriumwith its surroundings, (3) Bottom surface is adiabatic.ANALYSIS: (a) Under steady-state conditions, for the control volume shown in the schematic above,the energy balance isEin Eout + Egen = 00 q cv + Pe = h As T ( ) T + Pe = 02-3(1)2where As = 5 L = 5 0.032m 0.032m = 5.12 10 m , find()T ( ) = T + Pe / h As = 20C + 4 W / 10 W / m 2 K 5.12 103 m 2 = 98.1C<(b) Under transient conditions, for the control volume shown above, the energy balance isEin E out + E gen = Est0 q cv + Pe = McdTdt(2)Substitute from Eq. (1) for Pe, separate variables, and define the limits of integration. h T ( t ) T + h T ( ) T = Mc h T ( t ) T ( ) = McdTdt o dtoh 0 dt = i Mcd( T T ( ))dtwhere = T(t) T(); i = Ti T() = T - T(); and o = T(to) T() with to as the time when o =- 5C. Integrating and rearranging find (see Eq. 5.5),to =to =Mcn ih As o0.28 kg 400 J / kg K10 W / m 2 K 5.12 103 m 2n( 20 98.1) C5C= 1.67 hour<COMMENTS: The spacewise isothermal assumption may not be a gross over simplification sincemost of the material is copper and iron, and the external resistance by free convection is high.However, by ignoring internal resistance, our estimate for to is optimistic.PROBLEM 5.30KNOWN: Series solution, Eq. 5.39, for transient conduction in a plane wall with convection.FIND: Midplane (x*=0) and surface (x*=1) temperatures * for Fo=0.1 and 1, using Bi=0.1, 1 and 10with only the first four eigenvalues. Based upon these results, discuss the validity of the approximatesolutions, Eqs. 5.40 and 5.41.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties.ANALYSIS: The series solution, Eq. 5.39a, is of the form,* = Cnn =1)((exp - 2 Fo cos n x*n)where the eigenvalues, n , and the constants, Cn, are from Eqs. 5.39b and 5.39c.Cn = 4sin n / ( 2 n + sin ( 2 n ) ) .The eigenvalues are tabulated in Appendix B.3; note, however, that 1 and C1 are available from Table 5.1.The values of n and Cn used to evaluate * are as follows: n tan n = BiBi1C10.11100.31110.86031.42892C23C31.01601.11911.26203.17313.42564.3058-0.0197-0.1517-0.39346.29916.43737.228140.00500.04660.2104C49.43549.529310.2003-0.0022-0.0217-0.1309****Using n and Cn values, the terms of * , designated as 1 , 2 , 3 and 4 , are as follows:Fo=0.1Bi=1.0Bi=0.1x**1*2*3*4*010Bi=101011.00620.95791.03930.67781.02890.1455-0.00720.0072-0.04690.0450-0.06160.02440.00010.00010.00070.00070.00110.0006-2.9910-73.0010-72.4710-62.4610-7-3.9610-62.8310-60.99910.96520.99310.72350.96840.1705Continued ..PROBLEM 5.30(Cont.)Fo=1Bi=1.0Bi=0.1x**1*2*3*4*010Bi=101010.92230.87800.53390.34820.16380.02328.3510-78.3510-7-1.2210-51.1710-63.4910-91.3810-97.0410-20-4.7010-20-4.3010-24-4.7710-42-7.9310-42-8.5210-47-0.92230.87800.5339(0.34820.16380.0232)The tabulated results for * = * x* , Bi, Fo demonstrate that for Fo=1, the first eigenvalue is sufficient toaccurately represent the series. However, for Fo=0.1, three eigenvalues are required for accuraterepresentation.A more detailed analysis would show that a practical criterion for representation of the series solution by oneeigenvalue is Fo>0.2. For these situations the approximate solutions, Eqs. 5.40 and 5.41, are appropriate.For the midplane, x*=0, the first two eigenvalues for Fo=0.2 are:Fo=0.2x*=0Bi0.11.010*1*20.99650.96510.8389-0.00226-0.0145-0.0096*0.99390.95060.8293+0.26+1.53+1.16Error,%The percentage error shown in the last row of the above table is due to the effect of the second term. ForBi=0.1, neglecting the second term provides an error of 0.26%. For Bi=1, the error is 1.53%.Hence we conclude that the approximate series solutions (with only one eigenvalue) provides systematicallyhigh results, but by less than 1.5%, for the Biot number range from 0.1 to 10.PROBLEM 5.31KNOWN: One-dimensional wall, initially at a uniform temperature, Ti, is suddenly exposed to aconvection process (T, h). For wall #1, the time (t1 = 100s) required to reach a specifiedtemperature at x = L is prescribed, T(L1, t1) = 315 C.FIND: For wall #2 of different thickness and thermal conditions, the time, t2, required for T(L2, t2)= 28 C.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.ANALYSIS: The properties, thickness and thermal conditions for the two walls are:2WallL(m) (m /s)120.100.401510-625102k(W/mK)Ti( C)T( C)h(W/m K)501003003040020200100-6The dimensionless functional dependence for the one-dimensional, transient temperature distribution,Eq. 5.38, is =T ( x,t ) TTi T(= f x , Bi, Fo)wherex = x/LBi = hL/kFo = t/L2 .If the parameters x*, Bi, and Fo are the same for both walls, then 1 = 2 . Evaluate theseparameters:Wall12where1 =x*11315 400= 0.85300 400Bi0.400.402 =Fo0.150-41.56310 t2*0.850.8528.5 20= 0.85.30 20It follows thatFo 2 = Fo1t 2 = 960s.1.563 10 -4t 2 = 0.150<PROBLEM 5.32KNOWN: The chuck of a semiconductor processing tool, initially at a uniform temperature of Ti =100C, is cooled on its top surface by supply air at 20C with a convection coefficient of 50 W/m2K.FIND: (a) Time required for the lower surface to reach 25C, and (b) Compute and plot the time-to-coolas a function of the convection coefficient for the range 10 h 2000 W/m2K; comment on theeffectiveness of the head design as a method for cooling the chuck.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction in the chuck, (2) Lower surface isperfectly insulated, (3) Uniform convection coefficient and air temperature over the upper surface of thechuck, and (4) Constant properties.PROPERTIES: Table A.1, Aluminum alloy 2024 ( (25 + 100)C / 2 = 335 K): = 2770 kg/m3, cp =880 J/kg K, k = 179 W/mK.ANALYSIS: (a) The Biot number for the chuck with h = 50 W/m2K isBi =hL 50 W m 2 K 0.025 m== 0.007 0.1k179 W m K(1)so that the lumped capacitance method is appropriate. Using Eq. 5.5, with V/As = L,t= Vc ilnhAs = T T i = Ti T(100 20 ) Ct = 2770 kg m3 0.025 m 880 J kg K 50 W m 2 K ln(t = 3379s = 56.3min)( 25 20 ) C<Continued...PROBLEM 5.32 (Cont.)(b) When h = 2000 W/m2K, using Eq. (1), find Bi = 0.28 > 0.1 so that the series solution, Section 5.51,for the plane wall with convection must be used. Using the IHT Transient Conduction, Plane WallModel, the time-to-cool was calculated as a function of the convection coefficient. Free convectioncooling conduction corresponds to h 10 W/m2K and the time-to-cool is 282 minutes. With the coolinghead design, the time-to-cool can be substantially decreased if the convection coefficient can beincreased as shown below.Time-to-cool, t (min)604020001000Convection coefficient, h (W/m^2.K)2000PROBLEM 5.33KNOWN: Configuration, initial temperature and charging conditions of a thermal energy storage unit.FIND: Time required to achieve 75% of maximum possible energy storage and correspondingminimum and maximum temperatures.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible radiationexchange with surroundings.ANALYSIS: For the system, find firsthL 100 W/m 2 K 0.025mBi =k=0.7 W/m K= 3.57indicating that the lumped capacitance method cannot be used.Groeber chart, Fig. D.3:=Q/Qo = 0.75k0.7 W/m K== 4.605 107 m 2 / s c 1900 kg/m3 800 J/kg K(2100 W/m 2 K2Fo = h t =Bik2) ( 4.605107 m2 / s) t (s) = 9.4 103t2( 0.7 W/m K ) 22Find Bi Fo 11, and substituting numerical valuest = 11/9.4 10-3 = 1170s.<Heisler chart, Fig. D.1: Tmin is at x = 0 and Tmax at x = L, with t 4.605 10 7 m 2 / s 1170 sFo =2L=( 0.025m )From Fig. D.1, o 0.33. Hence,2Bi -1 = 0.28.= 0.86()To T + 0.33 ( Ti T ) = 600o C + 0.33 575o C = 410o C = Tmin .From Fig. D.2, /o 0.33 at x = L, for whichoTx =L T + 0.33 ( To T ) = 600o C + 0.33 ( 190 ) C = 537 oC = Tmax .<<COMMENTS: Comparing masonry (m) with aluminum (Al), see Problem 5.10, ( c)Al > ( c)m andkAl > km. Hence, the aluminum can store more energy and can be charged (or discharged) morequickly.PROBLEM 5.34KNOWN: Thickness, properties and initial temperature of steel slab. Convection conditions.FIND: Heating time required to achieve a minimum temperature of 550C in the slab.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible radiation effects, (3) Constantproperties.2ANALYSIS: With a Biot number of hL/k = (250 W/m K 0.05m)/48 W/mK = 0.260, a lumpedcapacitance analysis should not be performed. At any time during heating, the lowest temperature inthe slab is at the midplane, and from the one-term approximation to the transient thermal response of aplane wall, Eq. (5.41), we obtain(550 800 ) C 0.417 C exp 2 Fo T To = o ===11Ti T ( 200 800 ) C()With 1 0.488 rad and C1 1.0396 from Table 5.1 and = k / c = 1.115 10 5 m 2 / s,)(21 t / L2 = ln (0.401) = 0.914t=0.841(0.05m )20.914 L2=21 (0.488 ) 1.115 105 m 2 / s2<= 861sCOMMENTS: The surface temperature at t = 861s may be obtained from Eq. (5.40b), where( ) = 0.417 cos (0.488 rad ) = 0.368. Hence, T (L, 792s ) T = T = o cos 1xs + 0.368 ( Ti T )= 800C 221C = 579C. Assuming a surface emissivity of = 1 and surroundings that are atTsur = T = 800C, the radiation heat transfer coefficient corresponding to this surface temperature is()h r = ( Ts + Tsur ) Ts + Tsur = 205 W / m K. Since this value is comparable to the convection222coefficient, radiation is not negligible and the desired heating will occur well before t = 861s.PROBLEM 5.35KNOWN: Pipe wall subjected to sudden change in convective surface condition. See Example 5.4.FIND: (a) Temperature of the inner and outer surface of the pipe, heat flux at the inner surface, andenergy transferred to the wall after 8 min; compare results to the hand calculations performed for theText Example; (b) Time at which the outer surface temperature of the pipe, T(0,t), will reach 25C; (c)Calculate and plot on a single graph the temperature distributions, T(x,t) vs. x, for the initial condition,the final condition and the intermediate times of 4 and 8 min; explain key features; (d) Calculate andplot the temperature-time history, T(x,t) vs. t, for the locations at the inner and outer pipe surfaces, x =0 and L, and for the range 0 t 16 min. Use the IHT | Models | Transient Conduction | Plane Wallmodel as the solution tool.SCHEMATIC:ASSUMPTIONS: (1) Pipe wall can be approximated as a plane wall, (2) Constantproperties, (3) Outer surface of pipe is adiabatic.ANALYSIS: The IHT model represents the series solution for the plane wall providingtemperatures and heat fluxes evaluated at (x,t) and the total energy transferred at the innerwall at (t). Selected portions of the IHT code used to obtain the results tabulated below areshown in the Comments.(a) The code is used to evaluate the tabulated parameters at t = 8 min for locations x = 0 and L.The agreement is very good between the one-term approximation of the Example and the multipleterm series solution provided by the IHT model.Text Ex 5.445.242.9q ( L, 8 min ) , W / m 2x-2.72-7400Q (8 min ) 107 , J / m45.443.1-2.73T(L, 8min), CT(0, 8 min), CIHT Model-7305(b) To determine the time to for which T(0,t) = 25C, the IHT model is solved for to after setting x = 0and T_xt = 25C. Find, to = 4.4 min.<T e m p e ra tu re , T(x ,t) (C )(c) The temperature distributions, T(x,t) vs x, for the initial condition (t = 0), final condition ( t )and intermediate times of 4 and 8 min. are shown on the graph below.T e m p e ra tu re d is trib u tio n s , T(x,t) vs . x6040200-2 0010203040W a ll lo c a tio n , x (m m )In itia l co n d itio n , t = 0t = 4 m int = 8 m inS te a d y-s ta te c o n d itio n , t > 3 0 m inContinued ..PROBLEM 5.35 (Cont.)The final condition corresponds to the steady-state temperature, T (x,) = T. For the intermediatetimes, the gradient is zero at the insulated boundary (x = 0, the pipe exterior). As expected, thetemperature at x = 0 will be less than at the boundary experiencing the convection process with the hotoil, x = L. Note, however, that the difference is not very significant. The gradient at the inner wall, x= L, decreases with increasing time.(d) The temperature history T(x,t) for the locations at the inner and outer pipe surfaces areshown in the graph below. Note that the temperature difference between the two locations isgreatest at the start of the transient process and decreases with increasing time. After a 16min. duration, the pipe temperature is almost uniform, but yet 3 or 4C from the steady-statecondition.T e m p e ra tu re -tim e h is to ry, T (x,t) vs . tT e m p e ra tu re , T (x ,t) (C )6040200-2 00246810121416T im e , t (m in )O u te r s u rfa c e , x = 0In n e r s u rfa c e , x = LCOMMENTS: (1) Selected portions of the IHT code for the plane wall model are shown below.Note the relation for the pipe volume, vol, used in calculating the total heat transferred per unit lengthover the time interval t.// Models | Transient Conduction | Plane Wall// The temperature distribution isT_xt = T_xt_trans("Plane Wall",xstar,Fo,Bi,Ti,Tinf) // Eq 5.39//T_xt = 25// Part (b) surface temperature, x = 0// The heat flux in the x direction isq''_xt = qdprime_xt_trans("Plane Wall",x,L,Fo,Bi,k,Ti,Tinf) // Eq 2.6// The total heat transfer from the wall over the time interval t isQoverQo = Q_over_Qo_trans("Plane Wall",Fo,Bi) // Eq 5.45Qo = rho * cp * vol * (Ti - Tinf) // Eq 5.44//vol = 2 * As * L// Appropriate for wall of 2L thicknessvol = pi * D * L// Pipe wall of diameter D, thickness L and unit lengthQ = QoverQo * Qo// Total energy transfered per unit length(2) Can you give an explanation for why the inner and outer surface temperatures are not verydifferent? What parameter provides a measure of the temperature non-uniformity in a system duringa transient conduction process?PROBLEM 5.36KNOWN: Thickness, initial temperature and properties of furnace wall. Convection conditions atinner surface.FIND: Time required for outer surface to reach a prescribed temperature. Correspondingtemperature distribution in wall and at intermediate times.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3)Adiabatic outer surface, (4) Fo > 0.2, (5) Negligible radiation from combustion gases.ANALYSIS: The wall is equivalent to one-half of a wall of thickness 2L with symmetric convection2conditions at its two surfaces. With Bi = hL/k = 100 W/m K 0.15m/1.5 W/mK = 10 and Fo > 0.2,the one-term approximation, Eq. 5.41 may be used to compute the desired time, where o = (To T ) / (Ti T ) = 0.215. From Table 5.1, C1 = 1.262 and 1 = 1.4289. Hence,ln o / C1ln ( 0.215 /1.262 )Fo = (21)== 0.867(1.4289 )20.867 ( 0.15m )Fo L2== 33,800s3 1000 J / kg K1.5 W / m K / 2600 kg / m2t=)(<The corresponding temperature distribution, as well as distributions at t = 0, 10,000, and 20,000 s areplotted below1000Te m p e ra tu re , C800600400200000 .20 .40 .60 .81D im e n s io n le s s lo ca tio n , x/Lt= 0 st= 1 0 ,0 0 0 st= 2 0 ,0 0 0 st= 3 3 ,8 0 0 sCOMMENTS: Because Bi >>1, the temperature at the inner surface of the wall increases muchmore rapidly than at locations within the wall, where temperature gradients are large. Thetemperature gradients decrease as the wall approaches a steady-state for which there is a uniformtemperature of 950C.PROBLEM 5.37KNOWN: Thickness, initial temperature and properties of steel plate. Convection conditions at bothsurfaces.FIND: Time required to achieve a minimum temperature.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in plate, (2) Symmetric heating on both sides, (3)Constant properties, (4) Negligible radiation from gases, (5) Fo > 0.2.2ANALYSIS: The smallest temperature exists at the midplane and, with Bi = hL/k = 500 W/m K 0.050m/45 W/mK = 0.556 and Fo > 0.2, may be determined from the one-term approximation of Eq.5.41. From Table 5.1, C1 = 1.076 and 1 = 0.682. Hence, with o = (To - T)/(Ti - T) = 0.375,Fo = (ln o / C121) = ln (0.375 /1.076) = 2.266(0.682 )22.266 ( 0.05m )Fo L2== 491s3 500 J / kg K45 W / m K / 7800 kg / m2t=()COMMENTS: From Eq. 5.40b, the corresponding surface temperature isTs = T + ( Ti T ) o cos (1 ) = 700C 400C 0.375 0.776 = 584CBecause Bi is not much larger than 0.1, temperature gradients in the steel are moderate.<PROBLEM 5.38KNOWN: Plate of thickness 2L = 25 mm at a uniform temperature of 600C is removed from a hotpressing operation. Case 1, cooled on both sides; case 2, cooled on one side only.FIND: (a) Calculate and plot on one graph the temperature histories for cases 1 and 2 for a 500second cooling period; use the IHT software; Compare times required for the maximum temperature inthe plate to reach 100C; and (b) For both cases, calculate and plot on one graph, the variation withtime of the maximum temperature difference in the plate; Comment on the relative magnitudes of thetemperature gradients within the plate as a function of time.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the plate, (2) Constant properties, and (3) Forcase 2, with cooling on one side only, the other side is adiabatic.3PROPERTIES: Plate (given): = 3000 kg/m , c = 750 J/kgK, k = 15 W/mK.ANALYSIS: (a) From IHT, call up Plane Wall, Transient Conduction from the Models menu. Forcase 1, the plate thickness is 25 mm; for case 2, the plate thickness is 50 mm. The plate center (x = 0)temperature histories are shown in the graph below. The times required for the center temperatures toreach 100C aret1 = 164 s<t2 = 367 s(b) The plot of T(0, t) T(1, t), which represents the maximum temperature difference in the plateduring the cooling process, is shown below.Plate center temperature historiesTemperature difference history600150T(0,t) - T(L,t) (C)T(0,t) (C)50040030020010050100000100200300Time, t (s)Cooling - both sidesCooling - one side only4005000100200300400500Time (s)Cooling - both sidesCooling - one side onlyCOMMENTS: (1) From the plate center-temperature history graph, note that it takes more than twiceas long for the maximum temperature to reach 100C with cooling on only one side.(2) From the maximum temperature-difference graph, as expected, cooling from one side creates alarger maximum temperature difference during the cooling process. The effect could causemicrostructure differences, which could adversely affect the mechanical properties within the plate.PROBLEM 5.39KNOWN: Properties and thickness L of ceramic coating on rocket nozzle wall. Convection conditions.Initial temperature and maximum allowable wall temperature.FIND: (a) Maximum allowable engine operating time, tmax, for L = 10 mm, (b) Coating inner and outersurface temperature histories for L = 10 and 40 mm.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3)Negligible thermal capacitance of metal wall and heat loss through back surface, (4) Negligible contactresistance at wall/ceramic interface, (5) Negligible radiation.ANALYSIS: (a) Subject to assumptions (3) and (4), the maximum wall temperature corresponds to theceramic temperature at x = 0. Hence, for the ceramic, we wish to determine the time tmax at which T(0,t)= To(t) = 1500 K. With Bi = hL/k = 5000 W/m2K(0.01 m)/10 W/mK = 5, the lumped capacitancemethod cannot be used. Assuming Fo > 0.2, obtaining 1 = 1.3138 and C1 = 1.2402 from Table 5.1, and*evaluating o = ( To T ) ( Ti T ) = 0.4, Equation 5.41 yieldsFo = (*ln o C121) = ln (0.4 1.2402) = 0.656(1.3138 )2confirming the assumption of Fo > 0.2. Hence,t max =( ) = 0.656 (0.01m )2 = 10.9sFo L2<6 106 m 2 s(b) Using the IHT Lumped Capacitance Model for a Plane Wall, the inner and outer surface temperaturehistories were computed and are as follows:Temperature, T(K)23001900150011007003000306090120150Time, t(s)L = 0.01, x = LL = 0.01, x = 0L = 0.04, x = LL = 0.04, x = 0Continued...PROBLEM 5.39 (Cont.)The increase in the inner (x = 0) surface temperature lags that of the outer surface, but within t 45s bothtemperatures are within a few degrees of the gas temperature for L = 0.01 m. For L = 0.04 m, theincreased thermal capacitance of the ceramic slows the approach to steady-state conditions. The thermalresponse of the inner surface significantly lags that of the outer surface, and it is not until t 137s thatthe inner surface reaches 1500 K. At this time there is still a significant temperature difference acrossthe ceramic, with T(L,tmax) = 2240 K.COMMENTS: The allowable engine operating time increases with increasing thermal capacitance ofthe ceramic and hence with increasing L.PROBLEM 5.40KNOWN: Initial temperature, thickness and thermal diffusivity of glass plate. Prescribed surfacetemperature.FIND: (a) Time to achieve 50% reduction in midplane temperature, (b) Maximum temperaturegradient at that time.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.ANALYSIS: Prescribed surface temperature is analogous to h and T = Ts . Hence, Bi = .Assume validity of one-term approximation to series solution for T (x,t).(a) At the midplane,( T T2 o = o s = 0.50 = C1exp 1 FoTi Ts)1tan 1 = Bi = 1 = /2.HenceC1 =4sin14= = 1.27321 + sin ( 21 ) Fo = (ln o / C121) = 0.3792FoL2 0.379 ( 0.01 m )t=== 63 s.6 107 m 2 / s(<)2(b) With = C1exp 1 Fo cos 1x(T T ) T ( Ti Ts ) 2== i s 1C1exp 1 Fo sin 1x xLLx()300o C T/ x max = T/ x = 0.5 = 2.36 10 4 oC/m.x =10.01 m 2COMMENTS: Validity of one-term approximation is confirmed by Fo > 0.2.<PROBLEM 5.41KNOWN: Thickness and properties of rubber tire. Convection heating conditions. Initial and finalmidplane temperature.FIND: (a) Time to reach final midplane temperature. (b) Effect of accelerated heating.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in a plane wall, (2) Constant properties, (3)Negligible radiation.ANALYSIS: (a) With Bi = hL/k = 200 W/m2K(0.01 m)/0.14 W/mK = 14.3, the lumped capacitancemethod is clearly inappropriate. Assuming Fo > 0.2, Eq. (5.41) may be used with C1 = 1.265 and 1 1.458 rad from Table 5.1 to obtain)(* T T2 o = o = C1 exp 1 Fo = 1.265exp ( 2.126 Fo )Ti T*With o = ( To T ) ( Ti T ) = (-50)/(-175) = 0.286, Fo = ln ( 0.286 1.265 ) 2.126 = 0.70 = t f L20.7 ( 0.01m )2tf =6.35 108 m 2 s<= 1100s(b) The desired temperature histories were generated using the IHT Transient Conduction Model for aPlane Wall, with h = 5 104 W/m2K used to approximate imposition of a surface temperature of 200C.Temperature, T(C)200150100500020040060080010001200Time, t(s)x = 0, h = 200 W/m^2.Kx = L, h = 200 W/m^2.Kx = 0, h = 5E4 W/m^2.Kx = L, h = 5E4W/m^2.KThe fact that imposition of a constant surface temperature (h ) does not significantly accelerate theheating process should not be surprising. For h = 200 W/m2K, the Biot number is already quite large (Bi= 14.3), and limits to the heating rate are principally due to conduction in the rubber and not toconvection at the surface. Any increase in h only serves to reduce what is already a small component ofthe total thermal resistance.COMMENTS: The heating rate could be accelerated by increasing the steam temperature, but an upperlimit would be associated with avoiding thermal damage to the rubber.PROBLEM 5.42KNOWN: Stack or book comprised of 11 metal plates (p) and 10 boards (b) each of 2.36 mmthickness and prescribed thermophysical properties.FIND: Effective thermal conductivity, k, and effective thermal capacitance, (cp).SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible contact resistancebetween plates and boards.3PROPERTIES: Metal plate (p, given): p = 8000 kg/m , cp,p = 480 J/kgK, kp = 123W/mK; Circuit boards (b, given): b = 1000 kg/m , cp,b = 1500 J/kgK, kb = 0.30 W/mK.ANALYSIS: The thermal resistance of the book is determined as the sum of the resistance ofthe boards and plates,R = NR + MR totbpwhere M,N are the number of plates and boards in the book, respectively, and R = Li / kiiwhere Li and ki are the thickness and thermal conductivities, respectively.()R = M L p / k p + N ( L b / k b )totR = 11 ( 0.00236 m/12 W/m K ) + 10 ( 0.00236 m/0.30 W/m K )totR = 2.163 10tot3+ 7.867 102= 8.083 102K/W.The effective thermal conductivity of the book of thickness (10 + 11) 2.36 mm is0.04956 mk = L/R =tot8.083 10-2 K/WThe thermal capacitance of the stack is()C = M p L p c p + N ( b L b c b )tot(<= 0.613 W/m K.)(C = 11 8000 kg/m 0.00236 m 480 J/kg K + 10 1000 kg/m 0.00236 m 1500 J/kg Ktot34453)2C = 9.969 10 + 3.540 10 = 1.35 10 J/m K.totThe effective thermal capacitance of the book is( cp ) = Ctot / L = 1.351105 J/m2 K/0.04956 m = 2.726 106 J/m3 K. <COMMENTS: The results of the analysis allow for representing the stack as a homogeneous-7 2medium with effective properties: k = 0.613 W/mK and = (k/cp) = 2.24910 m /s. Seefor example, Problem 5.38.PROBLEM 5.43KNOWN: Stack of circuit board-pressing plates, initially at a uniform temperature, is subjected byupper/lower platens to a higher temperature.FIND: (a) Elapsed time, te, required for the mid-plane to reach cure temperature when platens aresuddenly changed to Ts = 190C, (b) Energy removal from the stack needed to return its temperatureto Ti.SCHEMATIC:63-7PROPERTIES: Stack (given): k = 0.613 W/mK, cp = 2.7310 J/m K; = k/cp = 2.245102m /s.ANALYSIS: (a) Recognize that sudden application of surface temperature corresponds to h , or-1Bi = 0 (Heisler chart) or Bi (100, Table 5.1). With Ts = T, T (0,t ) Ts = (170 190 ) C = 0.114.o =Ti Ts(15 190 ) CUsing Eq. 5.41 with values of 1 = 1.552 and C1 = 1.2731 at Bi = 100 (Table 5.1), find Fo2 o = C1exp 1 Fo(Fo = 12where Fo = t/L ,t=()ln o / C1 = 21FoL2)=(1(1.552 )21.002 25 103 m)ln ( 0.114/1.2731) = 1.00222.245 107 m 2 / s= 2.789 103 s = 46.5 min.<The Heisler chart, Figure D.1, could also be used to find Fo from values of o and Bi -1 = 0.(b) The energy removal is equivalent to the energy gained by the stack per unit area for the timeinterval 0 te. With Q corresponding to the maximum amount of energy that could be transferred,o(Q = c ( 2L )( Ti T ) = 2.73 10 J/m K 2 25 10o63-3)m (15 190 ) K = 2.389 10 J/m72.Q may be determined from Eq. 5.46,sin (1.552rad )Qsin1 o = 1 = 1 0.114 = 0.795Q1.552rad1oWe conclude that the energy to be removed from the stack per unit area to return it to Ti isQ = 0.795Q = 0.795 2.389 107 J/m2 = 1.90 107 J/m 2 .o<PROBLEM 5.44KNOWN: Car windshield, initially at a uniform temperature of -20C, is suddenly exposed on itsinterior surface to the defrost system airstream at 30C. The ice layer on the exterior surface acts as aninsulating layer.FIND: What airstream convection coefficient would allow the exterior surface to reach 0C in 60 s?SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction in the windshield, (2) Constant properties,(3) Exterior surface is perfectly insulated.PROPERTIES: Windshield (Given): = 2200 kg/m3, cp = 830 J/kgK and k = 1.2 W/mK.ANALYSIS: For the prescribed conditions, from Equations 5.31 and 5.33, (0, 60s ) o T (0, 60s ) T(0 30 ) C = 0.6===iiTi T( 20 30 ) CFo =kt cL2=1.2 W m K 602200 kg m3 830 J kg K (0.005 m )2= 1.58The single-term series approximation, Eq. 5.41, along with Table 5.1, requires an iterative solution tofind an appropriate Biot number. Alternatively, the Heisler charts, Appendix D, Figure D.1, for themidplane temperature could be used to findBi 1 = k hL = 2.5h = 1.2 W m K 2.5 0.005 m = 96 W m 2 K<COMMENTS: Using the IHT, Transient Conduction, Plane Wall Model, the convection coefficientcan be determined by solving the model with an assumed h and then sweeping over a range of h until theT(0,60s) condition is satisfied. Since the model is based upon multiple terms of the series, the result of h= 99 W/m2K is more precise than that found using the chart.PROBLEM 5.45KNOWN: Thickness, initial temperature and properties of plastic coating. Safe-to-touchtemperature. Convection coefficient and air temperature.FIND: Time for surface to reach safe-to-touch temperature. Corresponding temperature atplastic/wood interface.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in coating, (2) Negligible radiation, (3) Constantproperties, (4) Negligible heat of reaction, (5) Negligible heat transfer across plastic/wood interface.2ANALYSIS: With Bi = hL/k = 200 W/m K 0.002m/0.25 W/mK = 1.6 > 0.1, the lumpedcapacitance method may not be used. Applying the approximate solution of Eq. 5.40a, with C1 =1.155 and 1 = 0.990 from Table 5.1,()( )( 42 25 ) C T T2== 0.0971 = C1 exp 1 Fo cos 1x = 1.155exp ( 0.980 Fo ) cos ( 0.99 )s = sTi T ( 200 25 ) CHence, for x = 1,0.09712Fo = ln / ( 0.99 ) = 1.914 1.155cos ( 0.99 ) Fo L2 1.914 ( 0.002m )== 63.8s7 m 2 / s1.20 102t=<From Eq. 5.41, the corresponding interface temperature is()2To = T + ( Ti T ) exp 1 Fo = 25C + 175C exp ( 0.98 1.914 ) = 51.8C<COMMENTS: By neglecting conduction into the wood and radiation from the surface, the coolingtime is overpredicted and is therefore a conservative estimate. However, if energy generation due tosolidification of polymer were significant, the cooling time would be longer.PROBLEM 5.46KNOWN: Inlet and outlet temperatures of steel rods heat treated by passage through an oven.FIND: Rod speed, V.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction (axial conduction is negligible),(2) Constant properties, (3) Negligible radiation.PROPERTIES: Table A-1, AISI 1010 Steel ( T 600K ) : k = 48.8 W/mK, = 7832 kg/m ,3-5 2cp = 559 J/kgK, = (k/cp) = 1.1110 m /s.ANALYSIS: The time needed to traverse the rod through the oven may be found from Fig.D.4.600 750 T To = o == 0.214Ti T50 750k48.8 W/m KBi-1 == 15.6.hro 125 W/m 2 K (0.025m )Hence,2Fo = t/ro 12.22t = 12.2 ( 0.025m ) /1.11 105 m 2 / s = 687 s.The rod velocity isV=L 5m== 0.0073 m/s.t 687sCOMMENTS: (1) Since (h ro/2)/k = 0.032, the lumped capacitance method could have beenused. From Eq. 5.5 it follows that t = 675 s.(2) Radiation effects decrease t and hence increase V, assuming there is net radiant transferfrom the oven walls to the rod.(3) Since Fo > 0.2, the approximate analytical solution may be used. With Bi = hro/k=0.0641, Table 5.1 yields 1 = 0.3549 rad and C1 = 1.0158. Hence from Eq. 5.49c ln o = 12.4, C1 which is in good agreement with the graphical result.()2Fo = 11PROBLEM 5.47KNOWN: Hot dog with prescribed thermophysical properties, initially at 6C, is immersed inboiling water.FIND: Time required to bring centerline temperature to 80C.SCHEMATIC:ASSUMPTIONS: (1) Hot dog can be treated as infinite cylinder, (2) Constant properties.ANALYSIS: The Biot number, based upon Eq. 5.10, is2-3h Lc h ro / 2 100 W/m K 10 10 m/2Bi === 0.96(k0.52 W/m Kk)Since Bi > 0.1, a lumped capacitance analysis is not appropriate. Using the Heisler chart, Figure D.4withhr100W/m 2 K 10 10-3mBi o == 1.92 or Bi-1 = 0.520.52 W/m KkT ( 0,t ) T (80 100 ) Co = o === 0.21Ti Ti(6-100 ) Cand(1)(10 10-3m )2findFo = t =t2ro2rot = Fo == 0.81.764 107 m 2 / s< 0.8 = 453.5s = 7.6 min = k/ c = 0.52 W/m K/880 kg/m3 3350 J/kg K = 1.764 107 m 2 / s.whereCOMMENTS: (1) Note that Lc = ro/2 when evaluating the Biot number for the lumped capacitanceanalysis; however, in the Heisler charts, Bi hro/k.(2) The surface temperature of the hot dog follows from use of Figure D.5 with r/ro = 1 and Bi-1=0.52; find (1,t)/o 0.45. From Eq. (1), note that o = 0.21 i giving (1, t ) = T ( ro , t ) T = 0.45 o = 0.45 (0.21[Ti T ]) = 0.45 0.21[6 100] C = 8.9CT ( ro , t ) = T 8.9 C = (100 8.9 ) C = 91.1C(3) Since Fo 0.2, the approximate solution for *, Eq. 5.49, is valid. From Table 5.1 with Bi = 1.92,find that 1 = 1.3245 rad and C1 = 1.2334. Rearranging Eq. 5.49 and substituting values,Fo = ()ln o / C1 =211 0.213 ln = 1.0021.2334 (1.3245 rad )1This result leads to a value of t = 9.5 min or 20% higher than that of the graphical method.PROBLEM 5.48KNOWN: Long rod with prescribed diameter and properties, initially at a uniform temperature, isheated in a forced convection furnace maintained at 750 K with a convection coefficient of h = 1000W/m2K.FIND: (a) The corresponding center temperature of the rod, T(0, to), when the surface temperature T(ro,to) is measured as 550 K, (b) Effect of h on centerline temperature history.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial conduction in rod, (2) Constant properties, (3) Rod,when initially placed in furnace, had a uniform (but unknown) temperature, (4) Fo 0.2.ANALYSIS: (a) Since the rod was initially at a uniform temperature and Fo 0.2, the approximatesolution for the infinite cylinder is appropriate. From Eq. 5.49b,()()* * r* , Fo = o (Fo ) J 0 1r*(1)where, for r* = 1, the dimensionless temperatures are, from Eq. 5.31, * (1, Fo ) =T ( ro , t o ) T* o ( Fo ) =Ti TT (0, t o ) TTi T(2,3)Combining Eqs. (2) and (3) with Eq. (1) and rearranging,T ( ro , t o ) TTi T=T ( 0, t o ) = T +T (0, t o ) TTi TJ 0 (1 1)1 T ( ro , t o ) T J 0 (1 ) (4)The eigenvalue, 1 = 1.0185 rad, follows from Table 5.1 for the Biot numberBi =2hro 1000 W m K (0.060 m 2 )== 0.60 .k50 W m KFrom Table B-4, with 1 = 1.0185 rad, J0(1.0185) = 0.7568. Hence, from Eq. (4)T ( 0, t o ) = 750 K +1[550 750] K = 486 K0.7568<(b) Using the IHT Transient Conduction Model for a Cylinder, the following temperature histories weregenerated.Continued...PROBLEM 5.48 (Cont.)Centerline temperature, To(K)5004003000100200300400Time, t(s)h = 100 W/m^2.Kh = 500 W/m^2.Kh = 1000 W/m^2.KThe times required to reach a centerline temperature of 500 K are 367, 85 and 51s, respectively, for h =100, 500 and 1000 W/m2K. The corresponding values of the Biot number are 0.06, 0.30 and 0.60.Hence, even for h = 1000 W/m2K, the convection resistance is not negligible relative to the conductionresistance and significant reductions in the heating time could still be effected by increasing h to valuesconsiderably in excess of 1000 W/m2K.COMMENTS: For Part (a), recognize why it is not necessary to know Ti or the time to. We requirethat Fo 0.2, which for this sphere corresponds to t 14s. For this situation, the time dependence of thesurface and center are the same.PROBLEM 5.49KNOWN: A long cylinder, initially at a uniform temperature, is suddenly quenched in a large oil bath.FIND: (a) Time required for the surface to reach 500 K, (b) Effect of convection coefficient on surfacetemperature history.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Fo > 0.2.ANALYSIS: (a) Check first whether lumped capacitance method is applicable. For h = 50 W/m2K,Bic =2hLc h ( ro 2 ) 50 W m K ( 0.015 m / 2 )=== 0.221 .kk1.7 W m KSince Bic > 0.1, method is not suited. Using the approximate series solution for the infinite cylinder,)(()()2 * r* , Fo = C1 exp 1 Fo J 0 1r*(1)Solving for Fo and setting r* = 1, find *ln 21 C1J 0 (1 ) T ( ro , t o ) T (500 350 ) Kwhere * = (1, Fo ) === 0.231 .Ti T(1000 350 ) KFo = 1From Table 5.1, with Bi = 0.441, find 1 = 0.8882 rad and C1 = 1.1019. From Table B.4, find J0(1) =0.8121. Substituting numerical values into Eq. (2),Fo = 1(0.8882 )2ln [0.231 1.1019 0.8121] = 1.72 .2From the definition of the Fourier number, Fo = t ro , and = k/c,r22 ct = Fo o = Fo rokt = 1.72 ( 0.015 m ) 400 kg m3 1600 J kg K 1.7 W m K = 145s .2<(b) Using the IHT Transient Conduction Model for a Cylinder, the following surface temperaturehistories were obtained.Continued...PROBLEM 5.49 (Cont.)Surface temperature, T(K)1000900800700600500400300050100150200250300Time, t(s)h = 250 W/m^2.Kh = 50 W/m^2.KIncreasing the convection coefficient by a factor of 5 has a significant effect on the surface temperature,greatly accelerating its approach to the oil temperature. However, even with h = 250 W/m2K, Bi = 1.1and the convection resistance remains significant. Hence, in the interest of accelerated cooling,additional benefit could be achieved by further increasing the value of h.COMMENTS: For Part (a), note that, since Fo = 1.72 > 0.2, the approximate series solution isappropriate.PROBLEM 5.50KNOWN: Long pyroceram rod, initially at a uniform temperature of 900 K, and clad with a thinmetallic tube giving rise to a thermal contact resistance, is suddenly cooled by convection.FIND: (a) Time required for rod centerline to reach 600 K, (b) Effect of convection coefficient oncooling rate.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Thermal resistance and capacitance ofmetal tube are negligible, (3) Constant properties, (4) Fo 0.2.PROPERTIES: Table A-2, Pyroceram ( T = (600 + 900)K/2 = 750 K): = 2600 kg/m3, c = 1100J/kgK, k = 3.13 W/mK.ANALYSIS: (a) The thermal contact and convection resistances can be combined to give an overall heattransfer coefficient. Note that R ,c [mK/W] is expressed per unit length for the outer surface. Hence,tfor h = 100 W/m2K,U=11== 57.0 W m 2 K .1 h + R ( D ) 1 100 W m 2 K + 0.12 m K W ( 0.020 m )t,cUsing the approximate series solution, Eq. 5.50c, the Fourier number can be expressed as2*Fo = 1 1 ln o C1 .( )()From Table 5.1, find 1 = 0.5884 rad and C1 = 1.0441 forBi = Uro k = 57.0 W m 2 K ( 0.020 m 2 ) 3.13 W m K = 0.182 .The dimensionless temperature is* o (0, Fo ) =T ( 0, t ) TTi T=(600 300 ) K = 0.5.(900 300 ) KSubstituting numerical values to find Fo and then the time t,Fo =1(0.5884 )2ln0.5= 2.1271.0441r22 ct = Fo o = Fo rokt = 2.127 ( 0.020 m 2 ) 2600 kg m3 1100 J kg K 3.13 W m K = 194s .2<(b) The following temperature histories were generated using the IHT Transient conduction Model for aCylinder.Continued...PROBLEM 5.50 (Cont.)900800800Centerline temperature, (K)Surface temperature, (K)900700600500400300700600500400300050100150Time, t(s)r = ro, h = 100 W/m^2.Kr = ro, h = 500 W/m^2.Kr = ro, h = 1000 W/m^2.K200250300050100150200250300Time, t(s)r = 0, h = 100 W/m^2.Kr = 0, h = 500 W/m^2.Kr = 0, h = 1000 W/m^2.KWhile enhanced cooling is achieved by increasing h from 100 to 500 W/m2K, there is little benefitassociated with increasing h from 500 to 1000 W/m2K. The reason is that for h much above 500W/m2K, the contact resistance becomes the dominant contribution to the total resistance between thefluid and the rod, rendering the effect of further reductions in the convection resistance negligible. Notethat, for h = 100, 500 and 1000 W/m2K, the corresponding values of U are 57.0, 104.8 and 117.1W/m2K, respectively.COMMENTS: For Part (a), note that, since Fo = 2.127 > 0.2, Assumption (4) is satisfied.PROBLEM 5.51KNOWN: Sapphire rod, initially at a uniform temperature of 800K is suddenly cooled by a convectionprocess; after 35s, the rod is wrapped in insulation.FIND: Temperature rod reaches after a long time following the insulation wrap.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) No heat lossesfrom the rod when insulation is applied.3PROPERTIES: Table A-2, Aluminum oxide, sapphire (550K): = 3970 kg/m , c = 1068 J/kgK, k =-522.3 W/mK, = 5.259102m /s.ANALYSIS: First calculate the Biot number with Lc = ro/2,2h Lc h ( ro / 2 ) 1600 W/m K ( 0.020 m/2)Bi =k==k22.3 W/m K= 0.72.Since Bi > 0.1, the rod cannot be approximated as a lumped capacitance system. The temperaturedistribution during the cooling process, 0 t 35s, and for the time following the application ofinsulation, t > 35s, will appear asEventually (t ), the temperature of the rod will be uniform at T ( ) . To find T ( ) , write theconservation of energy requirement for the rod on a time interval basis,E in E out = E E final E initial .Using the nomenclature of Section 5.5.3 and basing energy relative to T, the energy balance becomesQ = cV ( T ( ) T ) Qowhere Qo = cV(Ti - T). Dividing through by Qo and solving for T ( ) , findT ( ) = T + ( Ti T ) (1 Q/Qo ) .From the Groeber chart, Figure D.6, withhr1600 W/m 2 K 0.020mBi = o == 1.43k(22.3 W/m K)()2Bi 2Fo = Bi 2 t/ro = (1.43 ) 2 5.259 10-6 m 2 /s 35s/ ( 0.020m ) 2 = 0.95.find Q/Qo 0.57. Hence,T ( ) = 300K + ( 800 300 ) K (1-0.57 ) = 515 K.<COMMENTS: From use of Figures D.4 and D.5, find T(0,35s) = 525K and T(ro,35s) = 423K.PROBLEM 5.52KNOWN: Long bar of 70 mm diameter, initially at 90 C, is suddenly immersed in a water bath2(T = 40 C, h = 20 W/m K).FIND: (a) Time, tf, that bar should remain in bath in order that, when removed and allowed toequilibrate while isolated from surroundings, it will have a uniform temperature T(r, ) = 55 C.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties.3PROPERTIES: Bar (given): = 2600 kg/m , c = 1030 J/kgK, k = 3.50 W/mK, = k/c =-6 21.3110 m /s.ANALYSIS: Determine first whether conditions are space-wise isothermal2hLc h ( ro / 2) 20 W/m K ( 0.035 m/2)Bi ==== 0.10kk3.50 W/m Kand since Bi 0.1, a Heisler solution is appropriate.(a) Consider an overall energy balance on the bar during the time interval t = tf (the time the bar isin the bath).Ein Eout = E0 Q = E final Einitial = Mc ( Tf T ) Mc ( Ti T )Q = Mc ( Tf T ) Q oQT T( 55 40 )o C = 0.70= 1 f= 1QoTi T( 90 40 )o Cwhere Qo is the initial energy in the bar (relative to T; Eq. 5.44). With Bi = hro/k = 0.20 and22Q/Qo = 0.70, use Figure D.6 to find Bi Fo = 0.15; hence Fo = 0.15/Bi = 3.75 and2t f = Fo r2 / = 3.75 ( 0.035 m ) /1.3110 6 m 2 / s = 3507 s.o<-1(b) To determine T(ro, tf), use Figures D.4 and D.5 for (ro,t)/i (Fo = 3.75, Bi = 5.0) and o/i-1(Bi = 5.0, r/ro = 1, respectively, to findT (ro , t f ) = T + ( ro , t ) o i = 40o C + 0.25 0.90 ( 90 50 )o C = 49oC.oi<PROBLEM 5.53KNOWN: Long plastic rod of diameter D heated uniformly in an oven to Ti and then allowed toconvectively cool in ambient air (T, h) for a 3 minute period. Minimum temperature of rod shouldnot be less than 200 C and the maximum-minimum temperature within the rod should not exceed10 C.FIND: Initial uniform temperature Ti to which rod should be heated. Whether the 10 C internaltemperature difference is exceeded.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Uniformand constant convection coefficients.3PROPERTIES: Plastic rod (given): k = 0.3 W/mK, cp = 1040 kJ/m K.ANALYSIS: For the worst case condition, the rod cools for 3 minutes and its outer surface is atleast 200 C in order that the subsequent pressing operation will be satisfactory. Hence,hro 8 W/m 2 K 0.015 mBi === 0.40k0.3 W/m Ktkt0.3 W/m K3 60sFo ==== 0.2308.22 cp ro 1040 103 J/m3 K ( 0.015 m )2roUsing Eq. 5.49a and 1 = 0.8516 rad and C1 = 1.0932 from Table 5.1,T (ro , t ) T2 == C1J 0 1 ro exp 1 Fo .Ti T()()With ro = 1, from Table B.4, J 0 (1 1) = J o ( 0.8516 ) = 0.8263, giving200 25= 1.0932 0.8263exp 0.85162 0.2308Ti = 254o C.<Ti 25At this time (3 minutes) what is the difference between the center and surface temperatures of therod? From Eq. 5.49b, T (ro , t ) T200 25=== J 0 1ro = 0.8263oT ( 0,t ) TT ( 0,t ) 25)(()which gives T(0,t) = 237 C. Hence,oT = T ( 0,180s ) T ( ro ,180s ) = ( 237 200 ) C = 37oC.<Hence, the desired max-min temperature difference sought (10 C) is not achieved.COMMENTS: T could be reduced by decreasing the cooling rate; however, h can not be mademuch smaller. Two solutions are (a) increase ambient air temperature and (b) non-uniformly heatrod in oven by controlling its residence time.PROBLEM 5.54KNOWN: Diameter and initial temperature of roller bearings. Temperature of oil bath andconvection coefficient. Final centerline temperature. Number of bearings processed per hour.FIND: Time required to reach centerline temperature. Cooling load.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial conduction in rod, (2) Constant properties.()3PROPERTIES: Table A.1, St. St. 304 T = 548 K : =7900 kg/m , k = 19.0 W/mK, cp = 546-62J/kgK, = 4.40 10 m /s.ANALYSIS: With Bi = h (ro/2)/k = 0.658, the lumped capacitance method can not be used. Fromthe one-term approximation of Eq. 5.49 c for the centerline temperature,()50 3022 T To = o == 0.0426 = C1 exp 1 Fo = 1.1382 exp ( 0.9287 ) Fo Ti T 500 30where, for Bi = hro/k = 1.316, C1 = 1.1382 and 1 = 0.9287 from Table 5.1.Fo = n ( 0.0374 ) / 0.863 = 3.812t f = Fo ro / = 3.81( 0.05 m ) / 4.40 106 = 2162s = 36 min2<From Eqs. 5.44 and 5.51, the energy extracted from a single rod is 2 Q = cV ( Ti T ) 1 o J1 (1 )1With J1 (0.9287) = 0.416 from Table B.4, 0.0852 0.416 27Q = 7900 kg / m3 546 J / kg K ( 0.05m ) 1m 470 K 1 = 1.53 10 J0.9287The nominal cooling load isq=N Q 10 1.53 107 J== 70,800 W = 7.08 kWtf2162sCOMMENTS: For a centerline temperature of 50C, Eq. 5.49b yields a surface temperature ofT ( ro , t ) = T + ( Ti T ) o Jo (1 ) = 30C + 470C 0.0426 0.795 = 45.9C<PROBLEM 5.55KNOWN: Long rods of 40 mm- and 80-mm diameter at a uniform temperature of 400C in acuring oven, are removed and cooled by forced convection with air at 25C. The 40-mmdiameter rod takes 280 s to reach a safe-to-handle temperature of 60C.FIND: Time it takes for a 80-mm diameter rod to cool to the same safe-to-handle temperature.Comment on the result? Did you anticipate this outcome?SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial (cylindrical) conduction in the rods, (2) Constantproperties, and (3) Convection coefficient same value for both rods.3PROPERTIES: Rod (given): = 2500 kg/m , c = 900 J/kgK, k = 15 W/mK.ANALYSIS: Not knowing the convection coefficient, the Biot number cannot be calculated todetermine whether the rods behave as spacewise isothermal objects. Using the relations fromSection 5.6, Radial Systems with Convection, for the infinite cylinder, Eq. 5.50, evaluate2Fo = t / ro , and knowing T(ro, to), a trial-and-error solution is required to find Bi = h ro/k andhence, h. Using the IHT Transient Conduction model for the Cylinder, the following results arereadily calculated for the 40-mm rod. With to = 280 s,Fo = 4.667Bi = 0.264h = 197.7 W / m 2 KFor the 80-mm rod, with the foregoing value for h, with T(ro, to) = 60C, findBi = 0.528Fo = 2.413t o = 579 s<COMMENTS: (1) The time-to-cool, to, for the 80-mm rod is slightly more than twice that forthe 40-mm rod. Did you anticipate this result? Did you believe the times would be proportionalto the diameter squared?(2) The simplest approach to explaining the relationship between to and the diameter followsfrom the lumped capacitance analysis, Eq. 5.13, where for the same /i, we expect BiFoo to be aconstant. That is,h ro t o=C2kro2yielding to ~ ro (not ro ).PROBLEM 5.56KNOWN: Initial temperature, density, specific heat and diameter of cylindrical rod. Convectioncoefficient and temperature of air flow. Time for centerline to reach a prescribed temperature.Dependence of convection coefficient on flow velocity.FIND: (a) Thermal conductivity of material, (b) Effect of velocity and centerline temperature andtemperature histories for selected velocities.SCHEMATIC:ASSUMPTIONS: (1) Lumped capacitance analysis can not be used but one-term approximation foran infinite cylinder is appropriate, (2) One-dimensional conduction in r, (3) Constant properties, (4)Negligible radiation, (5) Negligible effect of thermocouple hole on conduction.ANALYSIS: (a) With o =[To(0,1136s) - T]/(Ti - T) = (40 25)/(100 25) = 0.20, Eq. 5.49cyieldsFo =t2ro=kt2 c p ro=k (1136 s )1200 kg / m 1250 J / kg K ( 0.02 m )232= ln ( 0.2 / C1 ) / 1(1)Because C1 and 1 depend on Bi = hro/k, a trial-and-error procedure must be used. For example, avalue of k may be assumed and used to calculate Bi, which may then be used to obtain C1 and 1from Table 5.1. Substituting C1 and 1 into Eq. (1), k may be computed and compared with theassumed value. Iteration continues until satisfactory convergence is obtained, with<k 0.30 W / m Kand, hence, Bi = 3.67, C1 = 1.45, 1 = 1.87 and Fo = 0.568. For the above value of k,2 ln (0.2 / C1 ) / 1 = 0.567, which equals the Fourier number, as prescribed by Eq. (1).20.6180.6182.618yields a value of C = 16.8 Ws/m(b) With h = 55 W/m K for V = 6.8 m/s, h = CVK.The desired variations of the centerline temperature with velocity (for t = 1136 s) and time (for V = 3,10 and 20 m/s) are as follows:Continued ..PROBLEM 5.56 (Cont.)100C e n te rline te m p e ra tu re , To (C )C e n te rlin e te m p e ra tu re , To (C )50454035300510157550250205001000Tim e , t(s )Air ve lo city, V(m /s )V=3 m /sV=1 0 m /sV=2 0 m /s2With increasing V from 3 to 20 m/s, h increases from 33 to 107 W/m K, and the enhanced coolingreduces the centerline temperature at the prescribed time. The accelerated cooling associated withincreasing V is also revealed by the temperature histories, and the time required to achieve thermalequilibrium between the air and the cylinder decreases with increasing V.2COMMENTS: (1) For the smallest value of h = 33 W/m K, Bi h (ro/2)/k = 1.1 >> 0.1, and use ofthe lumped capacitance method is clearly inappropriate.(2) The IHT Transient Conduction Model for a cylinder was used to perform the calculations of Part(b). Because the model is based on the exact solution, Eq. 5.47a, it is accurate for values of Fo < 0.2,as well as Fo > 0.2. Although in principle, the model may be used to calculate the thermalconductivity for the conditions of Part (a), convergence is elusive and may only be achieved if theinitial guesses are close to the correct results.1500PROBLEM 5.57KNOWN: Diameter, initial temperature and properties of stainless steel rod. Temperature andconvection coefficient of coolant.FIND: Temperature distributions for prescribed convection coefficients and times.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties.ANALYSIS: The IHT model is based on the exact solution to the heat equation, Eq. 5.47. Theresults are plotted as followsh =1 0 0 0 W /m ^2 -Kh =1 0 0 W /m ^2 -K325325275Te m p e re a tu re , CTe m p e ra tu re , C2752251751252251751257575252500 .20 .40 .60 .8010 .20 .810 .81t=0 st=1 0 st=5 0 st= 0t= 1 0 0 st= 5 0 0 sh =5 0 0 0 W /m ^2 -K2larger than [T (ro,t) - T].0 .6D im e n s io n le s s ra d iu s , r*D im e n s io n le s s ra d iu s , r*325275Te m p e ra tu re , CFor h = 100 W/m K, Bi = hro/k = 0.1, and asexpected, the temperature distribution is nearlyuniform throughout the rod. For h = 10002W/m K (Bi = 1), temperature variationswithin the rod are not negligible. In this casethe centerline-to-surface temperaturedifference is comparable to the surface-to-fluid2temperature difference. For h = 5000 W/m K(Bi = 5), temperature variations within the rodare large and [T (0,t) T (ro,t)] is substantially0 .4225175125752500 .20 .40 .6D im e n s io n le s s ra d iu s , r*t=0 st=1 st=5 st=2 5 sCOMMENTS: With increasing Bi, conduction within the rod, and not convection from the surface,becomes the limiting process for heat loss.PROBLEM 5.58KNOWN: A ball bearing is suddenly immersed in a molten salt bath; heat treatment to harden occursat locations with T > 1000K.FIND: Time required to harden outer layer of 1mm.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Fo 0.2.ANALYSIS: Since any location within the ball whose temperature exceeds 1000K will be hardened,the problem is to find the time when the location r = 9mm reaches 1000K. Then a 1mm outer layerwill be hardened. Begin by finding the Biot number.5000 W/m2 K ( 0.020m/2 )hrBi = o == 1.00.50 W/m KkUsing the one-term approximate solution for a sphere, findFo = ()1ln / C1sin 1r211r1.From Table 5.1 with Bi = 1.00, for the sphere find 1 = 15708 rad and C1 = 1.2732. With r*.= r/ro = (9mm/10mm) = 0.9, substitute numerical values. (1000 1300 ) K11Fo =ln /1.2732sin (1.5708 0.9 rad ) = 0.441.1.5708 0.9(1.5708)2 (300 1300 ) KFrom the definition of the Fourier number with = k/c,22ro2 c = 0.441 0.020m 7800 kg 500t = Fo = Fo ro3k2mJ/ 50 W/m K = 3.4s.kg K<COMMENTS: (1) Note the very short time required to harden the ball. At this time it can be easilyshown the center temperature is T(0,3.4s) = 871 K.(2) The Heisler charts can also be used. From Fig. D.8, with Bi0.69(0.03). Since = T T = 1000 1300 = 300K o=,i o i o / i = 0.30 / 0.69 = 0.43 ( 0.02 ).andSince= 1.0 and r/ro = 0.9, read /o = i = Ti T = 1000Kit follows that= 0.30.i-1then= 0.69 oii-1From Fig. D.7 at o/i=0.43, Bi =1.0, read Fo = 0.45 (0.03) and t = 3.5 (0.2)s. Note the use oftolerances associated with reading the charts to 5%.PROBLEM 5.59KNOWN: An 80mm sphere, initially at a uniform elevated temperature, is quenched in an oil bathwith prescribed T, h.FIND: The center temperature of the sphere, T(0,t) at a certain time when the surface temperatureis T(ro,t) = 150 C.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Initial uniform temperature withinsphere, (3) Constant properties, (4) Fo 0.2.ANALYSIS: Check first to see if the sphere is spacewise isothermal.h ( ro / 3 ) 1000 W/m 2 K 0.040m/3hLBi c = c === 0.26.kk50 W/m KSince Bic > 0.1, lumped capacitance method is not appropriate. Recognize that when Fo 0.2, thetime dependence of the temperature at any point within the sphere will be the same as the center.Using the Heisler chart method, Fig. D.8 provides the relation between T(ro,t) and T(0,t). Find firstthe Biot number,hro 1000 W/m 2 K 0.040m== 0.80.k50 W/m K-1With Bi = 1/0.80 = 1.25 and r/ro =1, read from Fig. D.8,Bi =T ( ro , t ) T== 0.67. o T ( 0,t ) TIt follows thatT ( 0,t ) = T +11 T (ro , t ) T = 50o C +[150 50 ]o C = 199 oC.0.670.67<COMMENTS: (1) There is sufficient information to evaluate Fo; hence, we require that the timebe sufficiently long after the start of quenching for this solution to be appropriate.(2)The approximate series solution could also be used to obtain T(0,t). For Bi = 0.80 from Table5.1, 1 = 1.5044 rad. Substituting numerical values, r* = 1,o=T (ro , t ) TT ( 0,t ) T=It follows that T(0,t) = 201 C.()1sin 1r =sin (1.5044 rad ) = 0.663.1.50441r1PROBLEM 5.60KNOWN: Steel ball bearings at an initial, uniform temperature are to be cooled byconvection while passing through a refrigerated chamber; bearings are to be cooled to atemperature such that 70% of the thermal energy is removed.FIND: Residence time of the balls in the 5m-long chamber and recommended drive velocityfor the conveyor.SCHEMATIC:ASSUMPTIONS: (1) Negligible conduction between ball and conveyor surface, (2)Negligible radiation exchange with surroundings, (3) Constant properties, (4) Uniformconvection coefficient over balls surface.ANALYSIS: The Biot number for the lumped capacitance analysis is2hLc h ( ro / 3) 1000 W/m K (0.1m/3)Bi === 0.67.kk50 W/m KSince Bi > 0.1, lumped capacitance analysis is not appropriate. In Figure D.9, the internalenergy change is shown as a function of Bi and Fo. ForQ= 0.70Qohro 1000 W/m 2 K 0.1mBi === 2.0,k50 W/m Kand2find Bi Fo 1.2. The Fourier number isFo =t2ro=2 105 m 2 / s t(0.1 m )2= 2.0 103 tgivingt=Fo2.0 10-3=1.2 / Bi 22.0 10-31.2 / ( 2.0 )2=2.0 103= 150s.The velocity of the conveyor is expressed in terms of the length L and residence time t. HenceV=L 5m== 0.033m/s = 33mm/s.t 150sCOMMENTS: Referring to Eq. 5.10, note that for a sphere, the characteristic length is432rLc = V/As = ro / 4 ro = o .33However, when using the Heisler charts, note that Bi h ro/k.<PROBLEM 5.61KNOWN: Diameter and initial temperature of ball bearings to be quenched in an oil bath.FIND: (a) Time required for surface to cool to 100 C and the corresponding center temperature,(b) Oil bath cooling requirements.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction in ball bearings, (2) Constant properties.PROPERTIES: Table A-1, St. St., AISI 304, (T 500 C): k = 22.2 W/mK, cp = 579 J/kgK,3-6 2 = 7900 kg/m , = 4.8510 m /s.ANALYSIS: (a) To determine whether use of the lumped capacitance method is suitable, firstcomputeh ( ro / 3) 1000 W/m 2 K ( 0.010m/3)Bi === 0.15.k22.2 W/m KWe conclude that, although the lumped capacitance method could be used as a first approximation,the Heisler charts should be used in the interest of improving accuracy. Hence, withBi -1 =k22.2 W/m K== 2.22hro 1000 W/m 2 K ( 0.01m )andr= 1,roFig. D.8 gives ( ro , t ) 0.80.o (t )Hence, with ( ro , t )i=T (ro , t ) TTi T=100 40= 0.074,850 40Continued ..PROBLEM 5.61 (Cont.)it follows that o ( ro , t ) / i 0.074=== 0.093. i ( ro , t ) / o 0.80From Fig. D.7, with o / i = 0.093 and Bi -1 = k/hro = 2.22, findt = Fo 2.02( 0.01m )2 ( 2.0 )ro Fot=== 41s.4.85 106 m 2 / s<Also, o = To T = 0.093( Ti T ) = 0.093 ( 850 40 ) = 75o CTo = 115o C2<2(b) With Bi Fo = (1/2.2) 2.0 = 0.41, where Bi (hro/k) = 0.45, it follows from Fig. D.9 that fora single ballQ 0.93.QoHence, from Eq. 5.44,Q = 0.93 c pV ( Ti T )Q = 0.93 7900 kg/m 3 579 J/kg K Q = 1.44 104J( 0.02m )3 810o C6is the amount of energy transferred from a single ball during the cooling process. Hence, the oil bathcooling rate must beq = 104 Q/3600sq = 4 10 4 W = 40 kW.<COMMENTS: If the lumped capacitance method is used, the cooling time, obtained from Eq. 5.5,would be t = 39.7s, where the ball is assumed to be uniformly cooled to 100 C. This result, and thefact that To - T(ro) = 15 C at the conclusion, suggests that use of the lumped capacitance methodwould have been reasonable. Note that, when using the Heisler charts, accuracy to better than 5%is seldom possible.PROBLEM 5.62KNOWN: Diameter and initial temperature of hailstone falling through warm air.FIND: (a) Time, tm, required for outer surface to reach melting point, T(ro,tm) = Tm = 0 C, (b)Centerpoint temperature at that time, (c) Energy transferred to the stone.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties.3PROPERTIES: Table A-3, Ice (253K): = 920 kg/m , k = 2.03 W/mK, cp = 1945 J/kgK; -6 2= k/cp = 1.13 10 m /s.ANALYSIS: (a) Calculate the lumped capacitance Biot number,h ( ro / 3) 250 W/m 2 K ( 0.0025m/3)== 0.103.k2.03 W/m KSince Bi > 0.1, use the Heisler charts for which ( ro , t m ) T ( ro , tm ) T0 5=== 0.143iTi T30 5k2.03 W/m KBi -1 === 3.25.hro 250 W/m 2 K 0.0025mBi =From Fig. D.8, find ( ro , t m ) 0.86.o ( t m )It follows thato ( tm ) ( ro , t m ) / i0.143= 0.17.i ( ro , t m ) / o ( t m ) 0.86From Fig. D.7 find Fo 2.1. Hence,22.1 ( 0.0025) 2Fo rotm == 12s.1.13 106 m2 / s<(b) Since (o/i) 0.17, findTo T 0.17 ( Ti T ) 0.17 ( 30 5 ) 6.0o CTo ( tm ) 1.0o C.2<2(c) With Bi Fo = (1/3.25) 2.1 = 0.2, from Fig. D.9, find Q/Qo 0.82. From Eq. 5.44,(Qo = Vc p i = 920 kg/m 3) ( /6)(0.005m )31945 (J/kg K )( 35K ) = 4.10 JQ = 0.82 Qo = 0.82 ( 4.10 J ) = 3.4 J.<PROBLEM 5.63KNOWN: Sphere quenching in a constant temperature bath.FIND: (a) Plot T(0,t) and T(ro,t) as function of time, (b) Time required for surface to reach 415 K, t ,(c) Heat flux when T(ro, t ) = 415 K, (d) Energy lost by sphere in cooling to T(ro, t ) = 415 K, (e)Steady-state temperature reached after sphere is insulated at t = t , (f) Effect of h on center and surfacetemperature histories.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Uniform initialtemperature.ANALYSIS: (a) Calculate Biot number to determine if sphere behaves as spatially isothermal object,2hLc h ( ro 3) 75 W m K ( 0.015m 3)Bi ==== 0.22 .kk1.7 W m KHence, temperature gradients exist in the sphere and T(r,t) vs. t appears as shown above.(b) The Heisler charts may be used to find t when T(ro, t ) = 415 K. Using Fig. D.8 with r/ro = 1 and Bi-1= k/hro = 1.7 W/mK/(75 W/m2K 0.015 m) = 1.51, (1, t ) o 0.72 . In order to enter Fig. D.7, weneed to determine o/i, which is o (1, t ) (1, t ) ( 415 320 ) K0.72 = 0.275=ioi(800 320 ) K2Hence, for Bi-1 = 1.51, Fo t ro 0.87 and2 cp 2ro400 kg m3 1600 J kg K2t = Fo = Fo ro 0.87 ( 0.015m ) = 74sk1.7 W m K(c) The heat flux at the outer surface at time t T is given by Newtons law of coolingq = h T ( ro , t ) T = 75 W m 2 K [415 320] K = 7125 W / m2 . .<<The manner in which q is calculated indicates that energy is leaving the sphere.(d) The energy lost by the sphere during the cooling process from t = 0 to t can be determined from theGroeber chart, Fig. D.9. With Bi = 1/1.51 = 0.67 and Bi2Fo = (1/1.51)2 0.87 0.4, the chart yieldsQ Qo 0.75 . The energy loss by the sphere with V = (D3)/6 is therefore()Q 0.85Qo = 0.85 D3 6 cp ( Ti T )()Q 0.85 400 kg m3 [0.030 m ] 6 1600 J kg K (800 320 ) K = 3691J3<Continued...PROBLEM 5.63 (Cont.)(e) If at time t the surface of the sphere is perfectly insulated, eventually the temperature of the spherewill be uniform at T(). Applying conservation of energy to the sphere over a time interval, Ein - Eout =E Efinal - Einitial. Hence, -Q = cV[T() - T] - Qo, where Qo cV[Ti - T]. Dividing by Qo andregrouping, we obtainT ( ) = T + (1 Q Qo )( Ti T ) 320 K + (1 0.75)(800 320 ) K = 440 K<(f) Using the IHT Transient Conduction Model for a Sphere, the following graphical results weregenerated.80090000Heat flux, q''(ro,t) (W/m^2.K)Temperature, T(K)70060050040030005010060000300001500Time, t (s)h = 75 W/m^2.K, r = roh = 75 W/m^2.K, r = 0h = 200 W/m^2.K, r = roh = 200 W/m^2.K, r = 0050100150Time, t(s)h = 75 W/m^2.Kh = 200 W/m^2.KThe quenching process is clearly accelerated by increasing h from 75 to 200 W/m2K and is virtuallycompleted by t 100s for the larger value of h. Note that, for both values of h, the temperaturedifference [T(0,t) - T(ro,t)] decreases with increasing t. Although the surface heat flux for h = 200W/m2K is initially larger than that for h = 75 W/m2K, the more rapid decline in T(ro,t) causes it tobecome smaller at t 30s.COMMENTS: 1. There is considerable uncertainty associated with reading Q/Qo from the Groeberchart, Fig. D.9, and it would be better to use the one-term approximation solutions of Section 5.6.2. WithBi = 0.662, from Table 5.1, find 1 = 1.319 rad and C1 = 1.188. Using Eq. 5.50, find Fo = 0.852 and t =72.2 s. Using Eq. 5.52, find Q/Qo = 0.775 and T() = 428 K.2. Using the Transient Conduction/Sphere model in IHT based upon multiple-term series solution, thefollowing results were obtained: t = 72.1 s; Q/Qo = 0.7745, and T() = 428 K.PROBLEM 5.64KNOWN: Two spheres, A and B, initially at uniform temperatures of 800K and simultaneouslyquenched in large, constant temperature baths each maintained at 320K; properties of the spheres andconvection coefficients.FIND: (a) Show in a qualitative manner, on T-t coordinates, temperatures at the center and the outersurface for each sphere; explain features of the curves; (b) Time required for the outer surface of eachsphere to reach 415K, (c) Energy gained by each bath during process of cooling spheres to a surfacetemperature of 415K.SCHEMATIC:Sphere A150ro (mm)3 (kg/m )c (J/kgK)k (W/mK)2h (W/m K)16004001705Sphere B1540016001.750ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Uniform properties, (3) Constantconvection coefficient.ANALYSIS: (a) From knowledge of the Biot number and the thermal time constant, it is possible toqualitatively represent the temperature distributions. From Eq. 5.10, with Lc = ro/3, find5 W/m K ( 0.150m/3 )2Bi A =Bi =h ( ro / 3 )k170 W/m K50 W/m K ( 0.015m/3 )= 1.47 103(1)2Bi B =1.7 W/m K= 0.147(2)The thermal time constant for a lumped capacitance system from Eq. 5.7 is1 = ( Vc ) hAs = ro c3hB =A =1600 kg/m3 ( 0.150m ) 400 J/kg K3 5 W/m 2 K400 kg/m3 ( 0.015m )1600 J/kg K3 50 W/m 2 K= 64s= 6400s(3)(4)When Bi << 0.1, the sphere will cool in aspacewise isothermal manner (Sphere A).For sphere B, Bi > 0.1, hence gradients willbe important. Note that the thermal timeconstant of A is much larger than for B;hence, A will cool much slower. See sketchfor these features.(b) Recognizing that BiA < 0.1, Sphere A can be treated as spacewise isothermal and analyzedusing the lumped capacitance method. From Eq. 5.6 and 5.7, with T = 415 KT T== exp ( t/ )(5)i Ti TContinued ..PROBLEM 5.64 (Cont.) T T 415 320 t A = A ln= 10,367s = 2.88h. = 6400s ln 800 320 Ti T <Note that since the sphere is nearly isothermal, the surface and inner temperatures areapproximately the same.Since BiB > 0.1, Sphere B must be treated by the Heisler chart method of solution beginningwith Figure D.8. Using2hro 50 W/m K (0.015m )Bi B == 0.44k1.7 W/m KBi-1 = 2.27,Borfind that for r/ro = 1, (1, t ) T ( ro , t ) T ( 415 320 )=== 0.8.ooo(6)Using Eq. (6) and Figure D.7, find the Fourier number, o (T ( ro , t ) T ) / 0.8 ( 415 320 ) K/0.8=== 0.25Ti Ti(800 320 ) KFo =t2ro= 1.3.21.3 ( 0.015m )Fo rotB === 110s = 1.8 min2.656 106 m 2 / s3-6 2where = k/c = 1.7 W/mK/400 kg/m 1600 J/kgK = 2.65610 m /s.2<(c) To determine the energy change by the spheres during the cooling process, apply theconservation of energy requirement on a time interval basis.Sphere A:Ein Eout = E Q A = E = E ( t ) E (0 ).QA = cV T ( t ) Ti = 1600kg/m3 400J/kg K ( 4/3) (0.150m ) [415 800] K3QA = 3.483 106 J.Note that this simple expression is a consequence of the spacewise isothermal behavior.Sphere B:< QB = E ( t ) E (0 ).Ein E out = EFor the nonisothermal sphere, the Groeber chart, Figure D.9, can be used to evaluate QB.22With Bi = 0.44 and Bi Fo = (0.44) 1.3 = 2.52, find Q/Qo = 0.74. The energy transfer fromthe sphere during the cooling process, using Eq. 5.44, isQB = 0.74 Qo = 0.74 cV ( Ti T )QB = 0.75 400kg/m3 1600J/kg K ( 4/3) (0.015m ) (800 320 ) K = 3257 J.3COMMENTS: (1) In summary:SphereABBi = hro/k-34.41100.44s$t(s)Q(J)64006410,3701103.481032576<PROBLEM 5.65KNOWN: Spheres of 40-mm diameter heated to a uniform temperature of 400C are suddenlyremoved from an oven and placed in a forced-air bath operating at 25C with a convection coefficient2of 300 W/m K.FIND: (a) Time the spheres must remain in the bath for 80% of the thermal energy to be removed,and (b) Uniform temperature the spheres will reach when removed from the bath at this condition andplaced in a carton that prevents further heat loss.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction in the spheres, (2) Constant properties, and(3) No heat loss from sphere after removed from the bath and placed into the packing carton.3PROPERTIES: Sphere (given): = 3000 kg/m , c = 850 J/kgK, k = 15 W/mK.ANALYSIS: (a) From Eq. 5.52, the fraction of thermal energy removed during the time interval t =to isQ3= 1 3 o / 1 sin (1 ) 1 cos (1 )Qo(1)where Q/Qo = 0.8. The Biot number isBi = hro / k = 300 W / m 2 K 0.020 m /15 W / m K = 0.40and for the one-term series approximation, from Table 5.1,1 = 1.0528 radC1 = 1.1164(2)The dimensionless temperature o , Eq. 5.31, follows from Eq. 5.50.(2 o = C1 exp 1 Fo)(3)2where Fo = t o / ro . Substituting Eq. (3) into Eq. (1), solve for Fo and to.)(Q23= 1 3 C1 exp 1 Fo / 1 sin (1 ) 1 cos (1 )Qo(4)Fo = 1.45<t o = 98.6 s(b) Performing an overall energy balance on the sphere during the interval of time to t ,Ein E out = E = E f Ei = 0(5)where Ei represents the thermal energy in the sphere at to,Ei = (1 0.8 ) Qo = (1 0.8 ) cV (Ti T )(6)and Ef represents the thermal energy in the sphere at t = ,(Ef = cV Tavg T)(7)Combining the relations, find the average temperature() cV Tavg T (1 0.8 )(Ti T ) = 0Tavg = 100C<PROBLEM 5.66KNOWN: Diameter, density, specific heat and thermal conductivity of Pyrex spheres in packed bedthermal energy storage system. Convection coefficient and inlet gas temperature.FIND: Time required for sphere to acquire 90% of maximum possible thermal energy and thecorresponding center temperature.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional radial conduction in sphere, (2) Negligible heat transfer to orfrom a sphere by radiation or conduction due to contact with adjoining spheres, (3) Constantproperties.2ANALYSIS: With Bi h(ro/3)/k = 75 W/m K (0.0125m)/1.4 W/mK = 0.67, the approximatesolution for one-dimensional transient conduction in a sphere is used to obtain the desired results. Wefirst use Eq. (5.52) to obtain o .o =31Q1 3 sin (1 ) 1 cos (1 ) Qo With Bi hro/k = 2.01, 1 2.03 and C1 1.48 from Table 5.1. Hence,30.1( 2.03)0.837=o== 0.1553 0.896 2.03 ( 0.443)5.386The center temperature is therefore()To = Tg,i + 0.155 Ti Tg,i = 300C 42.7C = 257.3CFrom Eq. (5.50c), the corresponding time is r2t = o ln o 21 C1 (<)where = k / c = 1.4 W / m K / 2225 kg / m 3 835 J / kg K = 7.54 10 7 m 2 / s.t=(0.0375m )2 ln (0.155 /1.48 ) 1, 020s=7 m 2 / s 2.03 27.54 10()<COMMENTS: The surface temperature at the time of interest may be obtained from Eq. (5.50b).With r = 1, o sin (1 ) 0.155 0.896 Ts = Tg,i + Ti Tg,i= 300C 275C = 280.9C12.03()<PROBLEM 5.67KNOWN: Initial temperature and properties of a solid sphere. Surface temperature after immersion in afluid of prescribed temperature and convection coefficient.FIND: (a) Time to reach surface temperature, (b) Effect of thermal diffusivity and conductivity onthermal response.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties.ANALYSIS: (a) For k = 15 W/mK, the Biot number isBi =h ( ro 3)k=300 W m 2 K ( 0.05 m 3)15 W m K= 0.333 .Hence, the lumped capacitance method cannot be used. From Equation 5.50a,))(sin 1r*T T2 Fo.= C1 exp 1Ti T1r*(At the surface, r* = 1. From Table 5.1, for Bi = 1.0, 1 = 1.5708 rad and C1 = 1.2732. Hence,)(60 75sin 90= 0.30 = 1.2732 exp 1.57082 Fo25 751.5708exp(-2.467Fo) = 0.370Fo =t2ro= 0.4032ro(0.05 m ) = 100st = 0.403 = 0.403105 m 2 s2<(b) Using the IHT Transient Conduction Model for a Sphere to perform the parametric calculations, theeffect of is plotted for k = 15 W/mK.Continued...PROBLEM 5.67 (Cont.)756565Center temperature, T(C)Surface temperature, T(C)755545352555453525050100150200250300050100Time, t(s)150200250300Time, t(s)k = 15 W/m.K, alpha = 1E-4 m^2/sk = 15 W/m.K, alpha = 1E-5 m^2/sk = 15 W/m.K, alpha = 1E-6m^2/sk = 15 W/m.K, alpha = 1E-4 m^2/sk = 15 W/m.K, alpha = 1E-5 m^2/sk = 15 W/m.K, alpha = 1E-6 m^2/sFor fixed k and increasing , there is a reduction in the thermal capacity (cp) of the material, and hencethe amount of thermal energy which must be added to increase the temperature. With increasing , thematerial therefore responds more quickly to a change in the thermal environment, with the response atthe center lagging that of the surface.The effect of k is plotted for = 10-5 m2/s.756565Center temperature, T(C)Surface temperature, T(C)755545352555453525050100150200Time, t(s)k = 1.5 W/m.K, alpha = 1E-5 m^2/sk = 15 W/m.K, alpha = 1E-5 m^2/sk = 150W/m.K, alpha = 1E-5 m^2/s250300050100150200250300Time, t(s)k = 1.5 W/m.K, alpha = 1E-5 m^2/sk = 15 W/m.K, alpha = 1E-5 m^2/sk =150 W/m.K, alpha = 1E-5m^2/sWith increasing k for fixed alpha, there is a corresponding increase in cp, and the material thereforeresponds more slowly to a thermal change in its surroundings. The thermal response of the center lagsthat of the surface, with temperature differences, T(ro,t) - T(0,t), during early stages of solidificationincreasing with decreasing k.COMMENTS: Use of this technique to determine h from measurement of T(ro) at a prescribed trequires an interative solution of the governing equations.PROBLEM 5.68KNOWN: Properties, initial temperature, and convection conditions associated with cooling of glassbeads.FIND: (a) Time required to achieve a prescribed center temperature, (b) Effect of convection coefficienton center and surface temperature histories.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties, (3) Negligibleradiation, (4) Fo 0.2.ANALYSIS: (a) With h = 400 W/m2K, Bi h(ro/3)/k = 400 W/m2K(0.0005 m)/1.4 W/mK = 0.143 andthe lumped capacitance method should not be used. From the one-term approximation for the centertemperature, Eq. 5.50c,80 15* T T2o o == 0.141 = C1 exp 1 FoTi T 477 15For Bi hro/k = 0.429, Table 5.1 yields 1 = 1.101 rad and C1 = 1.128. Hence,)(Fo = * 1 0.141 ln o = ln = 1.71522 1.128 1 C1 1.101)(12t = 1.715ro cpk32 2200 kg m 800 J kg K= 1.715 ( 0.0015 m )1.4 W m K<= 4.85sFrom Eq. 5.50b, the corresponding surface (r* = 1) temperature is* sin 1 = 15 C + 462 C 0.141 0.892 = 67.8 CT ( ro , t ) = T + ( Ti T ) o11.101(b) The effect of h on the surface and center temperatures was determined using the IHT TransientConduction Model for a Sphere.)(500400400Surface temperature, T(C)Center temperature, T(C)500<3002001000300200100004812Time, t(s)h = 100 W/m^2.K, r = 0h = 400 W/m^2.K, r = 0h = 1000 W/m^2.K, r = 01620048121620Time, t(s)h = 100 W/m^2.K, r = roh = 400 W/m^2.K, r = roh = 1000 W/m^2.K, r = roContinued...PROBLEM 5.68 (Cont.)The cooling rate increases with increasing h, particularly from 100 to 400 W/m2K. The temperaturedifference between the center and surface decreases with increasing t and, during the early stages ofsolidification, with decreasing h.COMMENTS: Temperature gradients in the glass are largest during the early stages of solidificationand increase with increasing h. Since thermal stresses increase with increasing temperature gradients, thepropensity to induce defects due to crack formation in the glass increases with increasing h. Hence, thereis a value of h above which product quality would suffer and the process should not be operated.PROBLEM 5.69KNOWN: Temperature requirements for cooling the spherical material of Ex. 5.4 in air and in awater bath.FIND: (a) For step 1, the time required for the center temperature to reach T(0,t) = 335C while2cooling in air at 20C with h = 10 W/m K; find the Biot number; do you expect radial gradients to beappreciable?; compare results with hand calculations in Ex. 5.4; (b) For step 2, time required for the2center temperature to reach T(0,t) = 50C while cooling in water bath at 20C with h = 6000 W/m K;and (c) For step 2, calculate and plot the temperature history, T(x,t) vs. t, for the center and surface ofthe sphere; explain features; when do you expect the temperature gradients in the sphere to the largest?Use the IHT Models | Transient Conduction | Sphere model as your solution tool.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the radial direction, (2) Constant properties.ANALYSIS: The IHT model represents the series solution for the sphere providing the temperaturesevaluated at (r,t). A selected portion of the IHT code used to obtain results is shown in the Comments.(a) Using the IHT model with step 1 conditions, the time required for T(0,ta) = T_xt = 335C with r =0 and the Biot number are:t a = 94.2 s<Bi = 0.0025Radial temperature gradients will not be appreciable since Bi = 0.0025 << 0.1. The sphere behaves asspace-wise isothermal object for the air-cooling process. The result is identical to the lumpedcapacitance analysis result of the Text example.(b) Using the IHT model with step 2 conditions, the time required for T(0,tw) = T_xt = 50C with r = 0and Ti = 335C is<t w = 3.0 sRadial temperature gradients will be appreciable, since Bi = 1.5 >> 0.1. The sphere does not behaveas a space-wise isothermal object for the water-cooling process.(c) For the step 2 cooling process, the temperature histories for the center and surface of the sphere arecalculated using the IHT model.Continued ..PROBLEM 5.69 (Cont.)Te m p e ra tu re -tim e h is to ry, S te p 2Te m p e ra tu re , T(r,t) (C )40030020010000123456Tim e , t (s )S u rfa c e , r = roC e n te r, r = 0At early times, the difference between the center and surface temperature is appreciable. It is in thistime region that thermal stresses will be a maximum, and if large enough, can cause fracture. Within 6seconds, the sphere has a uniform temperature equal to that of the water bath.COMMENTS: Selected portions of the IHT sphere model codes for steps 1 and 2 are shown below./* Results, for part (a), step 1, air cooling; clearly negligible gradientBiFotT_xtTirro0.002525.1394.2233540000.005 */// Models | Transient Conduction | Sphere - Step 1, Air cooling// The temperature distribution T(r,t) isT_xt = T_xt_trans("Sphere",rstar,Fo,Bi,Ti,Tinf) // Eq 5.47T_xt = 335// Surface temperature/* Results, for part (b), step 2, water cooling; Ti = 335 CBiFotT_xtTirro1.50.7936 2.9765033500.005 */// Models | Transient Conduction | Sphere - Step 2, Water cooling// The temperature distribution T(r,t) isT_xt = T_xt_trans("Sphere",rstar,Fo,Bi,Ti,Tinf) // Eq 5.47//T_xt = 335// Surface temperature from Step 1; initial temperature for Step 2T_xt = 50// Center temperature, end of Step 2PROBLEM 5.70KNOWN: Two large blocks of different materials like copper and concrete at roomtemperature, 23C.FIND: Which block will feel cooler to the touch?SCHEMATIC:ASSUMPTIONS: (1) Blocks can be treated as semi-infinite solid, (2) Hand or fingertemperature is 37C.3PROPERTIES: Table A-1, Copper (300K): = 8933 kg/m , c = 385 J/kgK, k = 4013W/mK; Table A-3, Concrete, stone mix (300K): = 2300 kg/m , c = 880 J/kgK, k = 1.4W/mK.ANALYSIS: Considering the block as a semi-infinite solid, the heat transfer situationcorresponds to a sudden change in surface temperature, Case 1, Figure 5.7. The sensation ofcoolness is related to the heat flow from the hand or finger to the block. From Eq. 5.58, thesurface heat flux isq ( t ) = k ( Ts Ti ) / ( t )s(1)q ( t ) ~ ( k c )s(2)1/ 2or1/ 2since = k/ c.Hence for the same temperature difference, Ts Ti , and elapsed time, it follows that the heatfluxes for the two materials are related as1/ 2WkgJ1/ 2( k c )copper 401 m K 8933 m3 385 kg K qs,copper=== 22.11/21/ 2qs,concrete ( k c )WkgJconcrete1.4 m K 2300 3 880 kg K mHence, the heat flux to the copper block is more than 20 times larger than to the concreteblock. The copper block will therefore feel noticeably cooler than the concrete one.PROBLEM 5.71KNOWN: Asphalt pavement, initially at 50 C, is suddenly exposed to a rainstorm reducing thesurface temperature to 20 C.2FIND: Total amount of energy removed (J/m ) from the pavement for a 30 minute period.SCHEMATIC:ASSUMPTIONS: (1) Asphalt pavement can be treated as a semi-infinite solid, (2) Effect ofrainstorm is to suddenly reduce the surface temperature to 20 C and is maintained at that level forthe period of interest.3PROPERTIES: Table A-3, Asphalt (300K): = 2115 kg/m , c = 920 J/kgK, k = 0.062W/mK.ANALYSIS: This solution corresponds to Case 1, Figure 5.7, and the surface heat flux is given byEq. 5.58 asq ( t ) = k ( Ts Ti ) / ( t )s1/2(1)The energy into the pavement over a period of time is the integral of the surface heat flux expressedasQ = q ( t ) dt.s0t(2)Note that q ( t ) is into the solid and, hence, Q represents energy into the solid. Substituting Eq. (1)sfor q ( t ) into Eq. (2) and integrating findstQ = k ( Ts Ti ) / ( )1/2 t -1/2dt =0k ( Ts Ti )( )1/2 2 t1/2 .(3)Substituting numerical values into Eq. (3) withk0.062 W/m K=== 3.18 10 8 m 2 / s c 2115 kg/m 3 920 J/kg Kfind that for the 30 minute period,0.062 W/m K ( 20 50 ) KQ = 2 ( 30 60s )1/2 = 4.99 105 J/m 2 .1/2 3.18 10-8m2 / s()COMMENTS: Note that the sign for Q is negative implying that energy is removed from thesolid.<PROBLEM 5.72KNOWN: Thermophysical properties and initial temperature of thick steel plate. Temperature ofwater jets used for convection cooling at one surface.FIND: Time required to cool prescribed interior location to a prescribed temperature.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in slab, (2) Validity of semi-infinite mediumapproximation, (3) Negligible thermal resistance between water jets and slab surface (Ts = T), (4)Constant properties.ANALYSIS: The desired cooling time may be obtained from Eq. (5.57). With T(0.025m, t) = 50C,T ( x, t ) TsTi Ts=(50 25) C = 0.0909 = erf x (300 25) C 2 t x= 0.08072 tt=x2(0.0807 ) 42=(0.025m )2(0.0261 1.34 105 m 2 / s3)<= 1793s-52where = k/c = 50 W/mK/(7800 kg/m 480 J/kgK) = 1.34 10 m /s.42COMMENTS: (1) Large values of the convection coefficient (h ~ 10 W/m K) are associated withwater jet impingement, and it is reasonable to assume that the surface is immediately quenched to thetemperature of the water. (2) The surface heat flux may be determined from Eq. (5.58). In principle,1/2the flux is infinite at t = 0 and decays as t .PROBLEM 5.73KNOWN: Temperature imposed at the surface of soil initially at 20C. See Example 5.5.FIND: (a) Calculate and plot the temperature history at the burial depth of 0.68 m for selected soil72thermal diffusivity values, 10 = 1.0, 1.38, and 3.0 m /s, (b) Plot the temperature distribution over-7 2the depth 0 x 1.0 m for times of 1, 5, 10, 30, and 60 days with = 1.38 10 m /s, (c) Plot thesurface heat flux, q ( 0, t ) , and the heat flux at the depth of the buried main, q ( 0.68m, t ) , as axx-72function of time for a 60 day period with = 1.38 10 m /s. Compare your results with those in theComments section of the example. Use the IHT Models | Transient Conduction | Semi-infiniteMedium model as the solution tool.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Soil is a semi-infinitemedium, and (3) Constant properties.ANALYSIS: The IHT model corresponds to the case 1, constant surface temperature suddenboundary condition, Eqs. 5.57 and 5.58. Selected portions of the IHT code used to obtain thegraphical results below are shown in the Comments.(a) The temperature history T(x,t) for x = 0.68 m with selected soil thermal diffusivities is shownbelow. The results are directly comparable to the graph shown in the Ex. 5.5 comments.x = 0.68 m, T(0,t) = Ts = -15C, T(x,0) = 20CT(0.68 m, t) (C)20100-10015304560Time, t (days)alpha = 1.00e-7 m^2/salpha = 1.38e-7 m^2/salpha = 3.00e-7 m^2/sContinued ..PROBLEM 5.73 (Cont.)(b) The temperature distribution T(x,t) for selected times is shown below. The results are directlycomparable to the graph shown in the Ex. 5.5 comments.alpha = 1.38e-7 m^2/s , T(0,t) = -15C, T(x,0) = 20 C2015T(x,t) (C)1050-5-10-1500.250.50.751Depth, x (m)1 day5 days10 days30 days60 days(c) The heat flux from the soil, q ( 0, t ) , and the heat flux at the depth of the buried main,xq ( 0.68m, t ) , are calculated and plotted for the time period 0 t 60 days.xH e a t flu x , q ''(x ,t) (W /m ^ 2 )0a lp h a = 1 .3 8 e -7 m ^2 /s , k = 0 .5 2 W /m -K , T(0 ,t) = -1 5 C-5 0-1 0 0-1 5 0-2 0 0015304560Tim e , t (d a ys )S u rfa c e h e a t flu x, x = 0B u rie d -m a in d e p th , x = 0 .6 8 mBoth the surface and buried-main heat fluxes have a negative sign since heat is flowing in the negativex-direction. The surface heat flux is initially very large and, in the limit, approaches that of the buriedmain heat flux. The latter is initially zero, and since the effect of the sudden change in surfacetemperature is delayed for a time period, the heat flux begins to slowly increase.Continued ..PROBLEM 5.73 (Cont.)COMMENTS: (1) Can you explain why the surface and buried-main heat fluxes are nearly the sameat t = 60 days? Are these results consistent with the temperature distributions? What happens to theheat flux values for times much greater than 60 days? Use your IHT model to confirm yourexplanation.(2) Selected portions of the IHT code for the semi-infinite medium model are shown below.// Models | Transient Conduction | Semi-infinite Solid | Constant temperature Ts/* Model: Semi-infinite solid, initially with a uniform temperature T(x,0) = Ti, suddenly subjected toprescribed surface boundary conditions. */// The temperature distribution (Tx,t) isT_xt = T_xt_semi_CST(x,alpha,t,Ts,Ti) // Eq 5.55// The heat flux in the x direction isq''_xt = qdprime_xt_semi_CST(x,alpha,t,Ts,Ti,k) //Eq 5.56// Input parameters/* The independent variables for this system and their assigned numerical values are */Ti = 20// initial temperature, Ck = 0.52// thermal conductivity, W/m.K; base case conditionalpha = 1.38e-7// thermal diffusivity, m^2/s; base case//alpha = 1.0e-7//alpha = 3.0e-7// Calculating at x-location and time t,x=0// m, surface// x = 0.68// m, burial deptht = t_day * 24 * 3600// seconds to days time covnersion//t_day = 60//t_day = 1//t_day = 5//t_day = 10//t_day = 30t_day = 20// Surface condition: constant surface temperatureTs = -15// surface temperature, KPROBLEM 5.74KNOWN: Tile-iron, 254 mm to a side, at 150C is suddenly brought into contact with tile over asubflooring material initially at Ti = 25C with prescribed thermophysical properties. Tile adhesivesoftens in 2 minutes at 50C, but deteriorates above 120C.FIND: (a) Time required to lift a tile after being heated by the tile-iron and whether adhesivetemperature exceeds 120C, (2) How much energy has been removed from the tile-iron during the time ithas taken to lift the tile.SCHEMATIC:ASSUMPTIONS: (1) Tile and subflooring have same thermophysical properties, (2) Thickness ofadhesive is negligible compared to that of tile, (3) Tile-subflooring behaves as semi-infinite solidexperiencing one-dimensional transient conduction.PROPERTIES: Tile-subflooring (given): k = 0.15 W/mK, cp = 1.5 106 J/m3K, = k/cp = 1.00 10-7 m2/s.ANALYSIS: (a) The tile-subflooring can be approximated as a semi-infinite solid, initially at a uniformtemperature Ti = 25C, experiencing a sudden change in surface temperature Ts = T(0,t) = 150C. Thiscorresponds to Case 1, Figure 5.7. The time required to heat the adhesive (xo = 4 mm) to 50C followsfrom Eq. 5.57T ( x o , t o ) TsTi Tsxo= erf 2 ( t )1/ 2 o50 1500.004 m= erf 25 150 2 1.00 107 m 2 s t 1/ 2 o((0.80 = erf 6.325t o 1/ 2))to = 48.7s = 0.81 minusing error function values from Table B.2. Since the softening time, ts, for the adhesive is 2 minutes,the time to lift the tile ist = t o + t s = ( 0.81 + 2.0 ) min = 2.81min .<To determine whether the adhesive temperature has exceeded 120C, calculate its temperature at t =2.81 min; that is, find T(xo, t )T ( x o , t ) 1500.004 m= erf 25 150 2 1.0 107 m 2 s 2.81 60s 1/ 2 ()Continued...PROBLEM 5.74 (Cont.)T ( x o , t ) 150 = 125erf ( 0.4880 ) = 125 0.5098T ( x o , t ) = 86 C<Since T(xo, t ) < 120C, the adhesive will not deteriorate.(b) The energy required to heat a tile to the lift-off condition istq ( 0, t ) As dt .0xQ=Using Eq. 5.58 for the surface heat flux q (t) = q (0,t), findsx2k ( Ts Ti )t k ( Ts Ti )dtAsAs t1/ 2=1/ 21/ 21/ 20tQ=Q=( )( )2 0.15 W m K (150 25 ) C( 1.00 107 m 2 s)1/ 2 ( 0.254 m ) ( 2.81 60s )21/ 2= 56 kJ<COMMENTS: (1) Increasing the tile-iron temperature would decrease the time required to soften theadhesive, but the risk of burning the adhesive increases.(2) From the energy calculation of part (b) we can estimate the size of an electrical heater, if operatingcontinuously during the 2.81 min period, to maintain the tile-iron at a near constant temperature. Thepower required isP = Q t = 56 kJ 2.81 60s = 330 W .Of course a much larger electrical heater would be required to initially heat the tile-iron up to theoperating temperature in a reasonable period of time.PROBLEM 5.75KNOWN: Heat flux gage of prescribed thickness and thermophysical properties (, cp, k)initially at a uniform temperature, Ti, is exposed to a sudden change in surface temperatureT(0,t) = Ts.FIND: Relationships for time constant of gage when (a) backside of gage is insulated and (b)gage is imbedded in semi-infinite solid having the same thermophysical properties. Compare()with equation given by manufacturer, = 4d 2 cp / 2k.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.ANALYSIS: The time constant is defined as the time required for the gage to indicate,following a sudden step change, a signal which is 63.2% that of the steady-state value. Themanufacturers relationship for the time constant)( = 4d 2 cp / 2kcan be written in terms of the Fourier number ask4Fo ==== 0.4053.d 2 cp d 2 2The Fourier number can be determined for the two different installations.(a) For the gage having its backside insulated, thesurface and backside temperatures are Ts andT(0,t), respectively. From the sketch it followsthato =T (0, ) TsTi TsFrom Eq. 5.41,= 0.368.()2 o = 0.368 = C1exp 1 FoUsing Table 5.1 with Bi = 100 (as the best approximation for Bi = hd/k , correspondingto sudden surface temperature change with h ), 1 = 1.5552 rad and C1 = 1.2731. Hence,0.368 = 1.2731exp(1.55522 Foa )<Foa = 0.513.Continued ..PROBLEM 5.75 (Cont.)(b) For the gage imbedded in a semi-infinitemedium having the same thermophysicalproperties, Table 5.7 (case 1) and Eq. 5.57 yieldT ( x, ) Ts1/ 2= 0.368 = erf d/2 ( ) Ti Ts1/ 2d/2 ( )= 0.3972Fob =d2=1( 2 0.3972 )2= 1.585<COMMENTS: Both models predict higher values of Fo than that suggested by themanufacturer. It is understandable why Fob > Foa since for (b) the gage is thermallyconnected to an infinite medium, while for (a) it is isolated. From this analysis we concludethat the gages transient response will depend upon the manner in which it is applied to thesurface or object.PROBLEM 5.76KNOWN: Procedure for measuring convection heat transfer coefficient, which involvesmelting of a surface coating.FIND: Melting point of coating for prescribed conditions.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in solid rod (negligible losses toinsulation), (2) Rod approximated as semi-infinite medium, (3) Negligible surface radiation,(4) Constant properties, (5) Negligible thermal resistance of coating.-42PROPERTIES: Copper rod (Given): k = 400 W/mK, = 10 m /s.ANALYSIS: Problem corresponds to transient conduction in a semi-infinite solid. Themalresponse is given byT ( x,t ) TiT Ti1/ 2 h ( t )t x . erfc+ 2 ( t )1/ 2k 2x exp hx + h = erfc1/ 2 kk2 2 ( t )For x = 0, erfc(0) = 1 and T(x,t) = T(0,t) = Ts. Hence h 2Ts Ti= 1 exp k2T Ti h ( t )1/ 2 t erfc kwithh ( t m )k1/ 2=(200 W/m 2 K 10-4 m 2 / s 400 s400 W/m K)1/ 2= 0.1Ts = Tm = Ti + ( T Ti ) 1 exp ( 0.01) erfc ( 0.1)Ts = 25 C + 275 C [1-1.01 0.888] = 53.5 C.COMMENTS: Use of the procedure to evaluate h from measurement of tm necessitatesiterative calculations.<PROBLEM 5.77KNOWN: Irreversible thermal injury (cell damage) occurs in living tissue maintained at T 48C for aduration t 10s.FIND: (a) Extent of damage for 10 seconds of contact with machinery in the temperature range 50 to100C, (b) Temperature histories at selected locations in tissue (x = 0.5, 1, 5 mm) for a machinerytemperature of 100C.SCHEMATIC:ASSUMPTIONS: (1) Portion of workers body modeled as semi-infinite medium, initially at a uniformtemperature, 37C, (2) Tissue properties are constant and equivalent to those of water at 37C, (3)Negligible contact resistance.PROPERTIES: Table A-6, Water, liquid (T = 37C = 310 K): = 1/vf = 993.1 kg/m3, c = 4178 J/kgK,k = 0.628 W/mK, = k/c = 1.513 10-7 m2/s.ANALYSIS: (a) For a given surface temperature -- suddenly applied -- the analysis is directed towardfinding the skin depth xb for which the tissue will be at Tb 48C for more than 10s? From Eq. 5.57,T ( x b , t ) TsTi Ts= erf x b 2 ( t )1/ 2 = erf [ w ] .For the two values of Ts, the left-hand side of the equation isTs = 100 C :( 48 100 ) C 0.825=(37 100 ) CTs = 50 C :( 48 50 ) C 0.154=(37 50 ) CThe burn depth isx b = [ w ] 2 ( t )1/ 2(= [ w ] 2 1.513 107 m 2 s t)1/ 2= 7.779 104 [ w ] t1/ 2 .Continued...PROBLEM 5.77 (Cont.)Using Table B.2 to evaluate the error function and letting t = 10s, find xb asTs = 100C:xb = 7.779 10-4 [0.96](10s)1/2 = 2.362 103 m = 2.36 mmTs = 50C:xb = 7.779 10-4 [0.137](10s)1/2 = 3.37 103 m = 0.34 mm<<Recognize that tissue at this depth, xb, has not been damaged, but will become so if Ts is maintained forthe next 10s. We conclude that, for Ts = 50C, only superficial damage will occur for a contact period of20s.(b) Temperature histories at the prescribed locations are as follows.97Temperature, T(C)87776757473701530Time, t(s)x = 0.5 mmx = 1.0 mmx = 2.0 mmThe critical temperature of 48C is reached within approximately 1s at x = 0.5 mm and within 7s at x = 2mm.COMMENTS: Note that the burn depth xb increases as t1/2.PROBLEM 5.78KNOWN: Thermocouple location in thick slab. Initial temperature. Thermocouple measurementtwo minutes after one surface is brought to temperature of boiling water.FIND: Thermal conductivity of slab material.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Slab is semi-infinite medium, (3)Constant properties.3PROPERTIES: Slab material (given): = 2200 kg/m , c = 700 J/kgK.ANALYSIS: For the semi-infinite medium from Eq. 5.57,x= erfTi Ts 2 ( t )1/2 65 1000.01m= erf 30 100 2 ( 120s )1/2 0.01m = 0.5.erf 2 ( 120s )1/2 T ( x,t ) TsFrom Appendix B, find for erf w = 0.5 that w = 0.477; hence,0.01m2 ( 120s )1/2= 0.477( 120)1/2 = 0.0105 = 9.156 107 m2 /s.It follows that since = k/c,k = ck = 9.156 10-7 m 2 / s 2200 kg/m 3 700 J/kg Kk = 1.41 W/mK.<PROBLEM 5.79KNOWN: Initial temperature, density and specific heat of a material. Convection coefficient andtemperature of air flow. Time for embedded thermocouple to reach a prescribed temperature.FIND: Thermal conductivity of material.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Sample behaves as a semi-infinitemodium, (3) Constant properties.ANALYSIS: The thermal response of the sample is given by Case 3, Eq. 5.60,T ( x, t ) TiT Ti hx h 2 t xh t xexp erfc = erfc ++ 2 tkk 2 t k 2 where, for x = 0.01m at t = 300 s, [T(x,t) Ti]/(T - Ti) = 0.533. The foregoing equation must besolved iteratively for k, with = k/cp. The result isk = 0.45 W / m K-7<2with = 4.30 10 m /s.COMMENTS: The solution may be effected by inserting the Transient Conduction/Semi-infiniteSolid/Surface Conduction Model of IHT into the work space and applying the IHT Solver. However,the ability to obtain a converged solution depends strongly on the initial guesses for k and .PROBLEM 5.80KNOWN: Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to a surfaceconvection cooling process (T,h).FIND: (a) Temperatures at the surface and 45 mm depth after 3 minutes, (b) Effect of thermaldiffusivity and conductivity on temperature histories at x = 0, 0.045 m.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Plate approximates semi-infinite medium, (3)Constant properties, (4) Negligible radiation.ANALYSIS: (a) The temperature distribution for a semi-infinite solid with surface convection is givenby Eq. 5.60.T ( x, t ) TiT Ti1/ 2 hx h 2 t h ( t )xx exp + .= erfc + erfc 2 ( t )1/ 2 k 2 ( t )1/ 2kk 2 At the surface, x = 0, and for t = 3 min = 180s,T ( 0,180s ) 325 C(15 325 ) C 1002 W 2 m 4 K 2 5.6 106 m 2 s 180s = erfc ( 0 ) exp 0 +2( 20 W m K )1/ 2 26 2 100 W m K 5.6 10 m s 180s erfc 0 +20 W m K()= 1 [exp ( 0.02520 )] [erfc ( 0.159 )] = 1 1.02552 (1 0.178 )T ( 0,180s ) = 325 C (15 325 ) C (1 1.0255 0.822 )T ( 0,180s ) = 325 C 49.3 C = 276 C .At the depth x = 0.045 m, with t = 180s,T ( 0.045m,180s ) 325 C(15 325 ) C< 100 W m 2 K 0.045 m0.045 m= erfc + 0.02520 exp 1/ 220 W m K 2 5.6 106 m 2 s 180s 0.045 m erfc + 0.159 1/ 2 2 5.6 106 m 2 s 180s())(= erfc ( 0.7087 ) + [exp ( 0.225 + 0.0252 )] [erfc ( 0.7087 + 0.159 )] .T ( 0.045m,180s ) = 325 C + (15 325 ) C [(1 0.684 ) 1.284 (1 0.780 )] = 315 C<Continued...PROBLEM 5.80 (Cont.)(b) The IHT Transient Conduction Model for a Semi-Infinite Solid was used to generate temperaturehistories, and for the two locations the effects of varying and k are as follows.275Temperature, T(C)325300Temperature, T(C)32527525022522517512520075175050100150200250030050100200250300250300k = 2 W/m.K, alpha = 5.6E-6m^2/s, x = 0k = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 0k = 200 W/m.K, alpha = 5.6E-6m^2/s, x = 0k = 20 W/m.K, alpha = 5.6E-5 m^2/s, x = 0k = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 0k = 20 W/m.K, alpha = 5.6E-7m^2/s, x = 0325325300305Temperature, T(C)Temperature, T(C)150Time, t(s)Time, t(s)275250225285265245200225050100150200250Time, t(s)k = 20 W/m.K, alpha = 5.6E-5 m^2.K, x = 45 mmk = 20 W/m.K, alpha = 5.6E-6m^2.K, x = 45 mmk = 20 W/m.K, alpha = 5.6E-7m^2.K, x = 45mm300050100150200Time, t(s)k = 2 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mmk = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mmk = 200 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mmmFor fixed k, increasing alpha corresponds to a reduction in the thermal capacitance per unit volume (cp)of the material and hence to a more pronounced reduction in temperature at both surface and interiorlocations. Similarly, for fixed , decreasing k corresponds to a reduction in cp and hence to a morepronounced decay in temperature.COMMENTS: In part (a) recognize that Fig. 5.8 could also be used to determine the requiredtemperatures.PROBLEM 5.81KNOWN: Thick oak wall, initially at a uniform temperature of 25C, is suddenly exposed tocombustion products at 800C with a convection coefficient of 20 W/m2K.FIND: (a) Time of exposure required for the surface to reach an ignition temperature of 400C, (b)Temperature distribution at time t = 325s.SCHEMATIC:ASSUMPTIONS: (1) Oak wall can be treated as semi-infinite solid, (2) One-dimensional conduction,(3) Constant properties, (4) Negligible radiation.PROPERTIES: Table A-3, Oak, cross grain (300 K): = 545 kg/m3, c = 2385 J/kgK, k = 0.17 W/mK, = k/c = 0.17 W/mK/545 kg/m3 2385 J/kgK = 1.31 10-7 m2/s.ANALYSIS: (a) This situation corresponds to Case 3 of Figure 5.7. The temperature distribution isgiven by Eq. 5.60 or by Figure 5.8. Using the figure withT ( 0, t ) TiT Ti=400 25= 0.48800 25xand2 ( t )1/ 2=0we obtain h(t)1/2/k 0.75, in which case t (0.75k/h1/2)2. Hence,22 K 1.31 107 m 2 s 1/ 2 = 310st 0.75 0.17 W m K 20 W m)(<(b) Using the IHT Transient Conduction Model for a Semi-infinite Solid, the following temperaturedistribution was generated for t = 325s.400Temperature, T(C)3252501751002500.0050.010.0150.020.0250.03Distance from the surface, x(m)The temperature decay would become more pronounced with decreasing (decreasing k, increasing cp)and in this case the penetration depth of the heating process corresponds to x 0.025 m at 325s.COMMENTS: The result of part (a) indicates that, after approximately 5 minutes, the surface of thewall will ignite and combustion will ensue. Once combustion has started, the present model is no longerappropriate.PROBLEM 5.82KNOWN: Thickness, initial temperature and thermophysical properties of concrete firewall.Incident radiant flux and duration of radiant heating. Maximum allowable surface temperatures at theend of heating.FIND: If maximum allowable temperatures are exceeded.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in wall, (2) Validity of semi-infinite mediumapproximation, (3) Negligible convection and radiative exchange with the surroundings at theirradiated surface, (4) Negligible heat transfer from the back surface, (5) Constant properties.ANALYSIS: The thermal response of the wall is described by Eq. (5.60)T ( x, t ) = Ti +2 q ( t / )o1/ 2k x 2 q xx o erfc exp 4 t k 2 t where, = k / c p = 6.92 10 7 m 2 / s and for t = 30 min = 1800s, 2q ( t / )1/ 2 / k = 284.5 K. Hence,oat x = 0,T ( 0,30 min ) = 25C + 284.5C = 309.5C < 325C(<)At x = 0.25m, x 2 / 4 t = 12.54, q x / k = 1, 786K, and x / 2 ( t )1/ 2 = 3.54. Hence,o()T ( 0.25m, 30 min ) = 25C + 284.5C 3.58 106 1786C ( ~ 0 ) 25C<Both requirements are met.COMMENTS: The foregoing analysis is conservative since heat transfer at the irradiated surfacedue to convection and net radiation exchange with the environment have been neglected. If theemissivity of the surface and the temperature of the surroundings are assumed to be = 1 and Tsur =()4298K, radiation exchange at Ts = 309.5C would be q = Ts4 Tsur = 6, 080 W / m 2 K,radwhich is significant (~ 60% of the prescribed radiation).PROBLEM 5.83KNOWN: Initial temperature of copper and glass plates. Initial temperature and properties offinger.FIND: Whether copper or glass feels cooler to touch.SCHEMATIC:ASSUMPTIONS: (1) The finger and the plate behave as semi-infinite solids, (2) Constantproperties, (3) Negligible contact resistance.3PROPERTIES: Skin (given): = 1000 kg/m , c = 4180 J/kgK, k = 0.625 W/mK; Table A-13(T = 300K), Copper: = 8933 kg/m , c = 385 J/kgK, k = 401 W/mK; Table A-3 (T = 300K),3Glass: = 2500 kg/m , c = 750 J/kgK, k = 1.4 W/mK.ANALYSIS: Which material feels cooler depends upon the contact temperature Ts given byEquation 5.63. For the three materials of interest,( k c )1/2 = (0.625 1000 4180 )1 / 2 = 1,616 J/m 2 K s1/2skin1/2( k c ) cu = ( 401 8933 385 )1/2 = 37,137 J/m2 K s1/2( k c )1/2 = (1.4 2500 750 )1/2 = 1,620 J/m2 K s1/2.glassSince ( k c )1/2 >> ( k c )1/2 , the copper will feel much cooler to the touch. From Equationcuglass5.63,( k c )1/2 TA,i + ( k c )1/2 TB,iABTs =1/21/2( k c ) A + ( k c ) BTs( cu ) =1,616 ( 310 ) + 37,137 ( 300 )Ts( glass ) =1,616 + 37,137= 300.4 K1,616 ( 310 ) + 1,620 ( 300 )1,616 + 1,620= 305.0 K.<<COMMENTS: The extent to which a materials temperature is affected by a change in its thermal1/2environment is inversely proportional to (kc) . Large k implies an ability to spread the effect byconduction; large c implies a large capacity for thermal energy storage.PROBLEM 5.84KNOWN: Initial temperatures, properties, and thickness of two plates, each insulated on onesurface.FIND: Temperature on insulated surface of one plate at a prescribed time after they are pressedtogether.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligiblecontact resistance.3PROPERTIES: Stainless steel (given): = 8000 kg/m , c = 500 J/kgK, k = 15 W/mK.ANALYSIS: At the instant that contact is made, the plates behave as semi-infinite slabs and, sincethe (kc) product is the same for the two plates, Equation 5.63 yields a surface temperature ofTs = 350K.The interface will remain at this temperature, even after thermal effects penetrate to the insulatedsurfaces. The transient response of the hot wall may therefore be calculated from Equations 5.40and 5.41. At the insulated surface (x* = 0), Equation 5.41 yields(To Ts2= C1 exp 1 FoTi Ts)where, in principle, h and T Ts . From Equation 5.39c, Bi yields 1 = 1.5707, andfrom Equation 5.39bC1 =4sin 1= 1.27321 + sin ( 21 )tAlso,Fo =Hence,To 350= 1.273exp 1.5707 2 0.563 = 0.318400 350L2=3.75 10 6 m 2 / s ( 60s )( 0.02 m )2(= 0.563.)To = 365.9 K.COMMENTS: Since Fo > 0.2, the one-term approximation is appropriate.<PROBLEM 5.85KNOWN: Thickness and properties of liquid coating deposited on a metal substrate. Initial temperatureand properties of substrate.FIND: (a) Expression for time required to completely solidify the liquid, (b) Time required to solidifyan alumina coating.SCHEMATIC:ASSUMPTIONS: (1) Substrate may be approximated as a semi-infinite medium in which there is onedimensional conduction, (2) Solid and liquid alumina layers remain at fusion temperature throughoutsolidification (negligible resistance to heat transfer by conduction through solid), (3) Negligible contactresistance at the coating/substrate interface, (4) Negligible solidification contraction, (5) Constantproperties.ANALYSIS: (a) Performing an energy balance on the solid layer, whose thickness S increases with t,the latent heat released at the solid/liquid interface must be balanced by the rate of heat conduction intothe solid. Hence, per unit surface area, h sfdS= qconddt1/ 2where, from Eq. 5.58, qcond = k ( Tf Ti ) ( t ) . It follows thatdS ks (Tf Ti )=dt(s t )1/ 2k s ( Tf Ti ) t dto dS = h 1/ 2 o t1/ 2sf ( s ) h sf= Tf Ti 1/ 2t(s )1/ 2 hsf 2k s hsf t= s24k s Tf Ti 2<(b) For the prescribed conditions, 4 105 m 2 st=() 0.002 m 3970 kg m3 3.577 106 J kg 2 = 0.43s24 (120 W m K ) 2018 K<COMMENTS: Such solidification processes occur over short time spans and are typically termed rapidsolidification.PROBLEM 5.86KNOWN: Properties of mold wall and a solidifying metal.FIND: (a) Temperature distribution in mold wall at selected times, (b) Expression for variation of solidlayer thickness.SCHEMATIC:ASSUMPTIONS: (1) Mold wall may be approximated as a semi-infinite medium in which there is onedimensional conduction, (2) Solid and liquid metal layers remain at fusion temperature throughoutsolidification (negligible resistance to heat transfer by conduction through solid), (3) Negligible contactresistance at mold/metal interface, (4) Constant properties.ANALYSIS: (a) As shown in schematic (b), the temperature remains nearly uniform in the metal (at Tf)throughout the process, while both the temperature and temperature penetration increase with time in themold wall.(b) Performing an energy balance for a control surface about the solid layer, the latent energy releaseddue to solidification at the solid/liquid interface is balanced by heat conduction into the solid, q =latqcond is given by Eq. 5.58. Hence,lcond , where qat = h sf dS dt and q h sfSdS k w ( Tf Ti )=dt( w t )1/ 2o dS = S=k w (Tf Ti )t dt1/ 2 o t1/ 2h sf ( w )2k w ( Tf Ti ) 1/ 2t1/ 2 h sf ( w )<COMMENTS: The analysis of part (b) would only apply until the temperature field penetrates to theexterior surface of the mold wall, at which point, it may no longer be approximated as a semi-infinitemedium.PROBLEM 5.87KNOWN: Steel (plain carbon) billet of square cross-section initially at a uniformtemperature of 30C is placed in a soaking oven and subjected to a convection heating processwith prescribed temperature and convection coefficient.FIND: Time required for billet center temperature to reach 600C.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction in x1 and x2 directions, (2) Constantproperties, (3) Heat transfer to billet is by convection only.PROPERTIES: Table A-1, Steel, plain carbon (T = (30+600)C/2 = 588K = 600K): =3-5 27854 kg/m , cp = 559 J/kgK, k = 48.0 W/mK, =k/cp = 1.093 10 m /s.ANALYSIS: The billet corresponds to Case (e), Figure 5.11 (infinite rectangular bar).Hence, the temperature distribution is of the form ( x1, x 2 , t ) = P ( x1, t ) P ( x 2 , t )where P(x,t) denotes the distribution corresponding to the plane wall. Because of symmetry inthe x1 and x2 directions, the P functions are identical. Hence, (0, 0, t ) o (0, t ) =i i 2Plane wall = T Twhere i = Ti T o = T (0,t ) Tand L = 0.15m.Substituting numerical values, find o (0, t ) T (0,0,t ) T =i Ti T 1/ 21/ 2 ( 600 750 ) C = (30 750 ) C = 0.46.Consider now the Heisler chart for the plane wall, Figure D.1. For the values o = o 0.46ifindt = Fo =tL2Bi-1 =k48.0 W/m K== 3.2hL 100 W/m 2 K 0.15m 3.2.Hence,3.2 (0.15m )3.2 L2== 6587s = 1.83h.1.093 105 m 2 / s2t=<PROBLEM 5.88KNOWN: Initial temperature of fire clay brick which is cooled by convection.FIND: Center and corner temperatures after 50 minutes of cooling.SCHEMATIC:ASSUMPTIONS: (1) Homogeneous medium with constant properties, (2) Negligibleradiation effects.3PROPERTIES: Table A-3, Fire clay brick (900K): = 2050 kg/m , k = 1.0 W/mK, cp =-6 2960 J/kgK. = 0.51 10 m /s.ANALYSIS: From Fig. 5.11(h), the center temperature is given byT ( 0,0,0,t ) TTi T= P1 (0, t ) P2 ( 0, t ) P3 ( 0, t )where P1, P2 and P3 must be obtained from Fig. D.1.L1 = 0.03m:Bi1 =h L1= 1.50kFo1 =L2 = 0.045m:Bi 2 =h L2= 2.25kFo 2 =L3 = 0.10m:Bi3 =h L3= 5.0kFo3 =tL21tL22tL23= 1.70= 0.756= 0.153Hence from Fig. D.1,P1 (0, t ) 0.22Hence,P2 ( 0, t ) 0.50T ( 0,0,0,t ) TTi TP3 (0, t ) 0.85. 0.22 0.50 0.85 = 0.094and the center temperature isT ( 0,0,0,t ) 0.094 (1600 313) K + 313K = 434K.<Continued ..PROBLEM 5.88 (Cont.)The corner temperature is given byT ( L1, L 2 , L3 , t ) TTi T= P ( L1, t ) P ( L2 , t ) P ( L3 , t )whereP ( L1, t ) = ( L1, t ) P1 (0, t ) , etc.oand similar forms can be written for L2 and L3. From Fig. D.2, ( L1, t ) 0.55o ( L2 , t ) 0.43o ( L3 , t ) 0.25.oHence,P ( L1, t ) 0.55 0.22 = 0.12P ( L 2 , t ) 0.43 0.50 = 0.22P ( L3 , t ) 0.85 0.25 = 0.21andT ( L1, L 2 , L3 , t ) TTi T 0.12 0.22 0.21 = 0.0056orT ( L1, L2 , L3 , t ) 0.0056 (1600 313) K + 313K.The corner temperature is thenT ( L1, L2 , L3 , t ) 320K.<COMMENTS: (1) The foregoing temperatures are overpredicted by ignoring radiation,which is significant during the early portion of the transient.(2) Note that, if the time required to reach a certain temperature were to be determined, aniterative approach would have to be used. The foregoing procedure would be used to computethe temperature for an assumed value of the time, and the calculation would be repeated untilthe specified temperature were obtained.PROBLEM 5.89KNOWN: Cylindrical copper pin, 100mm long 50mm diameter, initially at 20C; end faces aresubjected to intense heating, suddenly raising them to 500C; at the same time, the cylindrical surfaceis subjected to a convective heating process (T,h).FIND: (a) Temperature at center point of cylinder after a time of 8 seconds from sudden applicationof heat, (b) Consider parameters governing transient diffusion and justify simplifying assumptionsthat could be applied to this problem.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction, (2) Constant properties and convection heattransfer coefficient.3PROPERTIES: Table A-1, Copper, pure T (500 + 20 ) C/2 500K : = 8933 kg/m , c = 407)(3-4 2J/kgK, k = 386 W/mK, = k/c = 386 W/mK/8933 kg/m 407 J/kgK = 1.064 10 m /s.ANALYSIS: (1) The pin can be treated as a two-dimensional system comprised of an infinitecylinder whose surface is exposed to a convection process (T,h) and of a plane wall whose surfacesare maintained at a constant temperature (Te). This configuration corresponds to the short cylinder,Case (i) of Fig. 5.11, ( r,x,t )= C ( r,t ) P ( x,t ) .i(1)For the infinite cylinder, using Fig. D.4, withBi =hrok=(100 W/m 2 K 25 10-3m385 W/m KC ( 0,8s ) =find) = 6.47 103Fo =andt2ro1.064 104m=2s 8s(25 10 m )-3 ( 0,8s ) 2= 1.36, 1. cyli(2)For the infinite plane wall, using Fig. D.1, withBi =findhLkorP ( 0,8s ) =Bi-10andFo =tL2=1.064 104 m 2 / s 8s(50 10 m )-3 ( 0,8s ) i 0.5. wallCombining Eqs. (2) and (3) with Eq. (1), find2= 0.34,(3) ( 0, 0,8s )i=T ( 0,0,8s ) TTi T 1 0.5 = 0.5T ( 0,0,8s ) = T + 0.5 ( Ti T ) = 500 + 0.5 ( 20 500 ) = 260 C.(b) The parameters controlling transient conduction with convective boundary conditions are the Biotand Fourier numbers. Since Bi << 0.1 for the cylindrical shape, we can assume radial gradients arenegligible. That is, we need only consider conduction in the x-direction.<PROBLEM 5.90KNOWN: Cylindrical-shaped meat roast weighing 2.25 kg, initially at 6C, is placed in anoven and subjected to convection heating with prescribed (T,h).FIND: Time required for the center to reach a done temperature of 80C.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction in x and r directions, (2) Uniform andconstant properties, (3) Properties approximated as those of water.()PROPERTIES: Table A-6, Water, liquid T = (80 + 6 ) C/2 315K : = 1/vf = 1/1.009 -3 33-7 210 m /kg = 991.1 kg/m , cp,f = 4179 J/kgK, k = 0.634 W/mK, = k/c = 1.531 10 m /s.ANALYSIS: The dimensions of the roast are determined from the requirement ro = L andknowledge of its weight and density,1/ 31/ 3 2.25 kg2 or r = L = M M = V = 2L ro(1)== 0.0712m.o 2 2 991.1 kg/m3 The roast corresponds to Case (i), Figure 5.11, and the temperature distribution may beT ( x,r,t ) Texpressed as the product of one-dimensional solutions,= P ( x,t ) C ( r,t ) , whereTi TP(x,t) and C(r,t) are defined by Eqs. 5.65 and 5.66, respectively. For the center of thecylinder,T ( 0,0,t ) TTi T(80 175 ) C = 0.56.=(6 175) CIn terms of the product solutions,T ( 0,0,t ) TT ( 0,t ) T = 0.56 =Ti TTi T (2)T ( 0,t ) T Ti T wallcylinder(3)For each of these shapes, we need to find values of o/i such that their product satisfies Eq.(3). For both shapes,Bi =h ro hL 15 W/m 2 K 0.0712m=== 1.68kk0.634 W/m KorBi-1 0.62Fo = t/ro = t/L2 = 1.53 107 m 2 / s t/ (0.0712m ) = 3.020 105 t.2Continued ..PROBLEM 5.90 (Cont.)A trial-and-error solution is necessary. Begin by assuming a value of Fo; obtain the respectiveo/i values from Figs. D.1 and D.4; test whether their product satisfies Eq. (3). Two trialsare shown as follows:o o Trial Fot(hrs) o / i )wall o / i )cyl i w i cyl120.40.33.682.750.720.780.500.680.360.53For Trial 2, the product of 0.53 agrees closely with the value of 0.56 from Eq. (2). Hence, itwill take approximately 2 hours to roast the meat.PROBLEM 5.91KNOWN: A long alumina rod, initially at a uniform temperature of 850K, is suddenlyexposed to a cooler fluid.FIND: Temperature of the rod after 30s, at an exposed end, T(0,0,t), and at an axial distance6mm from the end, T(0, 6mm, t).SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction in (r,x) directions, (2) Constantproperties, (3) Convection coefficient is same on end and cylindrical surfaces.PROPERTIES: Table A-2, Alumina, polycrystalline aluminum oxide (assume3T (850 + 600 ) K/2 = 725K): = 3970 kg/m , c = 1154 J/kgK, k = 12.4 W/mK.ANALYSIS: First, check if system behaves as a lumped capacitance. FindBi =hLc h ( ro / 2 ) 500 W/m K (0.010m/2 )=== 0.202.kk12.4 W/m KSince Bi > 0.1, rod does not behave as spacewise isothermal object. Hence, treat rod as asemi-infinite cylinder, the multi-dimensional system Case (f), Fig. 5.11.The product solution can be written as ( r,x,t ) = ( r,x,t ) ( r,t ) ( x,t )== C r , t S x , tiii)(()Infinite cylinder, C(r*,t*). Using the Heisler charts with r* = r = 0 and h ro Bi-1 =k1 500 W/m 2 K 0.01m =12.4 W/m K1= 2.48.2Evaluate = k/c = 2.71 10 m /s, find Fo = t/ro = 2.71 106 m 2 / s 30s/(0.01m) =-6 22-10.812. From the Heisler chart, Fig. D.4, with Bi = 2.48 and Fo = 0.812, read C(0,t*) =(0,t)/i = 0.61.Continued ..PROBLEM 5.91 (Cont.)Semi-infinite medium, S(x*,t*). Recognize this as Case (3), Fig. 5.7. From Eq. 5.60, note thatthe LHS needs to be transformed as follows,T TiT TT T= 1S ( x,t ) =.T TiTi TTi TThus,1/ 2 2 h ( t ) xx exp hx + h t erfc .S ( x,t ) = 1 erfc +2 k 2 ( t )1/ 2 2 ( t )1/ 2kk Evaluating this expression at the surface (x = 0) and 6mm from the exposed end, find2500 W/m 2 K 2.71 106 m 2 / s 30s S (0,30s ) = 1 erfc (0 ) exp 0 +2(12.4 W/m K ))()1/ 2 500 W/m 2 K 2.71 10-6 m 2 / s 30serfc 0 +12.4 W/m K({}S (0,30s ) = 1 1 exp (0.1322 ) erfc ( 0.3636 ) = 0.693.Note that Table B.2 was used to evaluate the complementary error function, erfc(w).0.006mS (6mm,30s ) = 1 erfc 2 2.7110-6m 2 / s 30s 1/ 2 )( 500 W/m 2 K 0.006mexp erfc ( 0.3327 + 0.3636 ) = 0.835.+ 0.1322 12.4 W/m KThe product solution can now be evaluated for each location. At (0,0),T ( 0,0,30s ) T (0, 0, t ) == C 0,t S 0,t = 0.61 0.693 = 0.423.Ti T()()T ( 0,0,30s ) = T + 0.423 (Ti T ) = 350K + 0.423 (850 350 ) K = 561K.Hence,At (0,6mm),()(<) (0, 6mm,t ) = C 0,t S 6mm,t = 0.61 0.835 = 0.509T ( 0,6mm,30s ) = 604K.<COMMENTS: Note that the temperature at which the properties were evaluated was a goodestimate.PROBLEM 5.92KNOWN: Stainless steel cylinder of Ex. 5.7, 80-mm diameter by 60-mm length, initially at 600 K,2suddenly quenched in an oil bath at 300 K with h = 500 W/m K. Use the Transient Conduction,Plane Wall and Cylinder models of IHT to obtain the following solutions.FIND: (a) Calculate the temperatures T(r,x,t) after 3 min: at the cylinder center, T(0, 0, 3 min), at thecenter of a circular face, T(0, L, 3 min), and at the midheight of the side, T(ro, 0, 3 min); compare yourresults with those in the example; (b) Calculate and plot temperature histories at the cylinder center,T(0, 0, t), the mid-height of the side, T(ro, 0, t), for 0 t 10 min; comment on the gradients and whateffect they might have on phase transformations and thermal stresses; and (c) For 0 t 10 min,calculate and plot the temperature histories at the cylinder center, T(0, 0, t), for convection coefficients2of 500 and 1000 W/m K.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction in r- and x-coordinates, (2) Constant properties.3PROPERTIES: Stainless steel (Example 5.7): = 7900 kg/m , c = 526 J/kgK, k = 17.4 W/mK.ANALYSIS: The following results were obtained using the Transient Conduction models for thePlane Wall and Cylinder of IHT. Salient portions of the code are provided in the Comments.(a) Following the methodology for a product solution outlined in Example 5.7, the following resultswere obtained at t = to = 3 min(r, x, t)P(x, t)C(r, t)0, 0, to0, L, toro, 0, to0.63570.43650.63570.53880.53880.3273T(r, x, t)-IHT(K)402.7370.5362.4T(r, x, t)-Ex(K)405372365Continued ..PROBLEM 5.92 (Cont.)The temperatures from the one-term series calculations of the Example 5.7 are systematically higherthan those resulting from the IHT multiple-term series model, which is the more accurate method.(b) The temperature histories for the center and mid-height of the side locations are shown in the graphbelow. Note that at early times, the temperature difference between these locations, and hence thegradient, is large. Large differences could cause variations in microstructure and hence, mechanicalproperties, as well as induce residual thermal stresses.(c) Effect of doubling the convection coefficient is to increase the quenching rate, but much less thanby a factor of two as can be seen in the graph below.Effect of increased conv. coeff. on quenching rateQuenching with h = 500 W/m^2.K500500T(0, 0, t) (C)T(x, r, t) (C)60060040040030030000246810246810Time, t (min)Time, t (min)h = 500 W/m^2.Kh = 1000 W/m^2.KMid-height of side (0,ro)Center (0, 0)COMMENTS: From IHT menu for Transient Conduction | Plane Wall and Cylinder, the modelswere combined to solve the product solution. Key portions of the code, less the input variables, arecopied below.// Plane wall temperature distribution// The temperature distribution isT_xtP = T_xt_trans("Plane Wall",xstar,FoP,BiP,Ti,Tinf)// The dimensionless parameters arexstar = x / LBiP = h * L / k// Eq 5.9FoP= alpha * t / L^2// Eq 5.33alpha = k/ (rho * cp)// Dimensionless representation, P(x,t)P_xt = (T_xtP - Tinf ) / (Ti - Tinf)// Cylinder temperature distribution// The temperature distribution T(r,t) isT_rtC = T_xt_trans("Cylinder",rstar,FoC,BiC,Ti,Tinf)// The dimensionless parameters arerstar = r / roBiC = h * ro / kFoC= alpha * t / ro^2// Dimensionless representation, C(r,t)C_rt= (T_rtC - Tinf ) / (Ti - Tinf)// Product solution temperature distribution(T_xrt - Tinf) / (Ti - Tinf) = P_xt * C_rt// Eq 5.39// Eq 5.47PROBLEM 5.93pKNOWN: Stability criterion for the explicit method requires that the coefficient of the Tmterm of the one-dimensional, finite-difference equation be zero or positive.pFIND: For Fo > 1/2, the finite-difference equation will predict values of Tm+1 which violatethe Second law of thermodynamics. Consider the prescribed numerical values.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant properties, (3) Nointernal heat generation.ANALYSIS: The explicit form of the finite-difference equation, Eq. 5.73, for an interiornode is()p+1pppTm = Fo Tm+1 + Tm-1 + (1 2 Fo ) Tm .pThe stability criterion requires that the coefficient of Tm be zero or greater. That is,1orFo .(1 2 Fo ) 02For the prescribed temperatures, consider situations for which Fo = 1, and and calculatep+1Tm .Fo = 1Fo = 1/2Fo = 1/4p+1Tm = 1(100 + 100 ) C + (1 2 1) 50 C = 250Cp+1Tm = 1/ 2 (100 + 100 ) C + (1 2 1/ 2 ) 50 C = 100Cp+1Tm = 1/ 4 (100 + 100 ) C + (1 2 1/ 4 ) 50 C = 75C.pPlotting these distributions above, note that when Fo = 1, Tm+1 is greater than 100C, whilepfor Fo = and , Tm+1 100C. The distribution for Fo = 1 is thermodynamicallyimpossible: heat is flowing into the node during the time period t, causing its temperature torise; yet heat is flowing in the direction of increasing temperature. This is a violation of theSecond law. When Fo = or , the node temperature increases during t, but thetemperature gradients for heat flow are proper. This will be the case when Fo , verifyingthe stability criterion.PROBLEM 5.94KNOWN: Thin rod of diameter D, initially in equilibrium with its surroundings, Tsur,suddenly passes a current I; rod is in vacuum enclosure and has prescribed electricalresistivity, e, and other thermophysical properties.FIND: Transient, finite-difference equation for node m.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings aremuch larger than rod, (3) Properties are constant and evaluated at an average temperature, (4)No convection within vacuum enclosure.ANALYSIS: The finite-difference equation is derived fromthe energy conservation requirement on the control volume,Acx, where Ac = D 2 / 4The energy balance has the formP = D.andpT p+1 Tmq a + q b q rad + I2 R e = cV m.tEin E out + Eg = Estwhere Eg = I 2R e and R e = e x/A c . Using Fouriers law to express the conduction terms,qa and qb, and Eq. 1.7 for the radiation exchange term, qrad, find)(pppT p TmT p Tm xT p+1 Tm4,p4+ kA c m+1 Px Tm Tsur + I2 e = cAc x mkAc m-1.xxtAcpDivide each term by cAc x/t, solve for Tm+1 and regroup to obtainp+1Tm =) P t 4,pI2 e t4Tm Tsur ) +.2Ac c (Ac c(pttkkpp 1 TmTm-1 + Tm+1 2 c x 2 c x 2 2Recognizing that Fo = t/x , regroup to obtain()p+1pppTm = Fo Tm-1 + Tm+1 + (1 2 Fo ) Tm () Px 2I2 e x 24,p4 Fo Tm Tsur + Fo.2kAckAcpThe stability criterion is based upon the coefficient of the Tm term written asFo .<COMMENTS: Note that we have used the forward-difference representation for the time derivative;see Section 5.9.1. This permits convenient treatment of the non-linear radiation exchange term.PROBLEM 5.95KNOWN: One-dimensional wall suddenly subjected to uniform volumetric heating andconvective surface conditions.FIND: Finite-difference equation for node at the surface, x = -L.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3)Uniform q.ANALYSIS: There are two types of finite-difference equations for the explicit and implicitmethods of solution. Using the energy balance approach, both types will be derived.Explicit Method. Perform an energy balance on the surface node shown above,pT p+1 Toq conv + q cond + qV = cV otEin E out + Eg = Est(1))(ppp+1pT1 Tox x To Toph (1 1) T To + k (1 1).+ q 1 1 = c 1 1 22xt(2)For the explicit method, the temperatures on the LHS are evaluated at the previous time (p).The RHS provides a forward-difference approximation to the time derivative. Divide Eq. (2)pby cx/2t and solve for To +1.htkttp+1ppppTo = 2T To + 2T1 To + q+ To . cxc cx 2Introducing the Fourier and Biot numbers,(Fo ( k/ c ) t/x 2)()Bi hx/kpqx 2 p+1pTo = 2 Fo T1 + Bi T + + (1 2 Fo 2 Fo Bi ) To .2k(3)pThe stability criterion requires that the coefficient of To be positive. That is,(1 2 Fo 2 Fo Bi ) 0orFo 1/2 (1 + Bi ) .(4) <Implicit Method. Begin as above with an energy balance. In Eq. (2), however, thetemperatures on the LHS are evaluated at the new (p+1) time. The RHS provides a backwarddifference approximation to the time derivative.()p+1p+1p+1pT1 To Tox x Tp+1 h T To+k+q = c oxt22p+1p(1 + 2 Fo (Bi + 1)) To 2 Fo T1p+1 = To + 2Bi Fo T + Fo qkx(5)2.(6) <COMMENTS: Compare these results (Eqs. 3, 4 and 6) with the appropriate expression inTable 5.2.PROBLEM 5.96KNOWN: Plane wall, initially having a linear, steady-state temperature distribution with boundariesmaintained at T(0,t) = T1 and T(L,t) = T2, suddenly experiences a uniform volumetric heat generation dueto the electrical current. Boundary conditions T1 and T2 remain fixed with time.FIND: (a) On T-x coordinates, sketch the temperature distributions for the following cases: initialconditions (t 0), steady-state conditions (t ) assuming the maximum temperature exceeds T2, andtwo intermediate times; label important features; (b) For the three-nodal network shown, derive thefinite-difference equation using either the implicit or explicit method; (c) With a time increment of t =5 s, obtain values of Tm for the first 45s of elapsed time; determine the corresponding heat fluxes at theboundaries; and (d) Determine the effect of mesh size by repeating the foregoing analysis using grids of 5and 11 nodal points.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional, transient conduction, (2) Uniform volumetric heat generationfor t 0, (3) Constant properties.PROPERTIES: Wall (Given): = 4000 kg/m3, c = 500 J/kgK, k = 10 W/mK.ANALYSIS: (a) The temperature distributionon T-x coordinates for the requested cases areshown below. Note the following key features:(1) linear initial temperature distribution, (2)non-symmetrical parabolic steady-statetemperature distribution, (3) gradient at x = L isfirst positive, then zero and becomes negative,and (4) gradient at x = 0 is always positive.(b) Performing an energy balance on the control volume about node m above, for unit area, findEin E out + Eg = EstpT2 TmT1 TmT p +1 Tm (1) x = (1) cx mk (1)+ k (1)+qxxt tqppFo [T1 + T2 2Tm ] += Tm+1 Tm cpFor the Tm term in brackets, use p for explicit and p+1 for implicit form,ppppTm+1 = Fo T1 + T2 + (1 2Fo ) Tm + qt cpExplicit:(Implicit:())ppppTm+1 = Fo T1 +1 + T2 +1 + q t cp + Tm (1 + 2Fo )(1)<(2)<Continued...PROBLEM 5.96 (Cont.)(c) With a time increment t = 5s, the FDEs, Eqs. (1) and (2) becomeExplicit:ppTm+1 = 0.5Tm + 75(3)Implicit:ppTm+1 = Tm + 75 1.5(4))(whereFo =qtkt cx 2=10 W m K 5s4000 kg m3 500 J kg K ( 0.010 m )2= 0.252 107 W m3 5s= 50 K4000 kg m3 500 J kg KPerforming the calculations, the results are tabulated as a function of time,c=pt (s)T1 (C)0123456789051015202530354045Tm (C)0000000000Explicit50100.00125.00137.50143.75146.88148.44149.22149.61149.80T2 (C)Implicit5083.33105.55120.37130.25136.83141.22144.15146.10147.40100100100100100100100100100100<The heat flux at the boundaries at t = 45s follows from the energy balances on control volumes about thepboundary nodes, using the explicit results for Tm ,Node 1:E in E out + E g = EstTp Tq ( 0, t ) + k m 1 + q ( x 2 ) = 0xx(pq ( 0, t ) = k Tm T1x) x qx 2(5)q ( 0, 45s ) = 10 W m K (149.8 0 ) K 0.010 m 2 107 W m3 0.010 m 2xq ( 0, 45s ) = 149,800 W m 2 100, 000 W m2 = 249,800 W m2xNode 2:<T p T2km q ( L, t ) + q ( x 2 ) = 0xx(pq ( L, t ) = k Tm T2x) x + qx 2 = 0(6)Continued...PROBLEM 5.96 (Cont.)q ( L, t ) = 10 W m K (149.80 100 ) C 0.010 m + 2 107 W m3 0.010 m 2xq ( L, t ) = 49,800 W m 2 + 100, 000 W m 2 = +149,800 W m2x<(d) To determine the effect of mesh size, the above analysis was repeated using grids of 5 and 11 nodalpoints, x = 5 and 2 mm, respectively. Using the IHT Finite-Difference Equation Tool, the finitedifference equations were obtained and solved for the temperature-time history. Eqs. (5) and (6) werepused for the heat flux calculations. The results are tabulated below for t = 45s, where Tm (45s) is thecenter node,Mesh Sizex(mm)1052pTm (45s)(C)149.8149.3149.4q (0,45s)xq (L,45s)xkW/m2-249.8-249.0-249.1kW/m2+149.8+149.0+149.0COMMENTS: (1) The center temperature and boundary heat fluxes are quite insensitive to mesh sizefor the condition.(2) The copy of the IHT workspace for the 5 node grid is shown below.// Mesh size - 5 nodes, deltax = 5 mm// Nodes a, b(m), and c are interior nodes// Finite-Difference Equations Tool - nodalequations/* Node a: interior node; e and w labeled b and1. */rho*cp*der(Ta,t) =fd_1d_int(Ta,Tb,T1,k,qdot,deltax)/* Node b: interior node; e and w labeled c anda. */rho*cp*der(Tb,t) =fd_1d_int(Tb,Tc,Ta,k,qdot,deltax)/* Node c: interior node; e and w labeled 2 andb. */rho*cp*der(Tc,t) =fd_1d_int(Tc,T2,Tb,k,qdot,deltax)// Assigned Variables:deltax = 0.005k = 10rho = 4000cp = 500qdot = 2e7T1 = 0T2 = 100/* Initial Conditions:Tai = 25Tbi = 50Tci = 75 *//* Data Browser Results - Nodaltemperatures at 45sTaTbTct99.5149.3149.5 45 */// Boundary Heat Fluxes - at t = 45sq''x0 = - k * (Taa - T1 ) / deltax - qdot* deltax / 2q''xL = k * (Tcc - T2 ) / deltax + qdot *deltax /2//where Taa = Ta (45s), Tcc =Tc(45s)Taa = 99.5Tcc = 149.5/* Data Browser resultsq''x0q''xL-2.49E5 1.49E5 */PROBLEM 5.97KNOWN: Solid cylinder of plastic material ( = 6 10-7 m2/s), initially at uniform temperature of Ti =20C, insulated at one end (T4), while other end experiences heating causing its temperature T0 toincrease linearly with time at a rate of a = 1C/s.FIND: (a) Finite-difference equations for the 4 nodes using the explicit method with Fo = 1/2 and (b)Surface temperature T0 when T4 = 35C.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction in cylinder, (2) Constant properties, and(3) Lateral and end surfaces perfectly insulated.ANALYSIS: (a) The finite-difference equations using the explicit method for the interior nodes (m = 1,2, 3) follow from Eq. 5.73 with Fo = 1/2,)((ppppppTm+1 = Fo Tm +1 + Tm 1 + (1 2Fo ) Tm = 0.5 Tm +1 + Tm 1)(1)From an energy balance on the control volume node 4 as shown above yields with Fo = 1/2(ppqa + q b + 0 = cV T4 +1 T4Ein E out + Eg = Est(pp0 + k T3 T4) t) x = c (x 2)(T4p+1 T4p ) tppppT4 +1 = 2FoT3 + (1 2Fo ) T4 = T3(2)(b) Performing the calculations, the temperature-time history is tabulated below, where T0 = Ti +atwhere a = 1C/s and t = pt with,Fo = t x 2 = 0.5p01234567t(s)0306090120150180210t = 0.5 (0.006 m )2T0(C)205080110140170200230T1(C)2020355068.7587.5108.1-When T4(210s, p = 7) = 35C, find T0(210s) = 230C.6 107 m 2 s = 30sT2(C)20202027.53546.2557.5-T3(C)2020202023.7527.535-T4(C)202020202023.7527.535<PROBLEM 5.98KNOWN: A 0.12 m thick wall, with thermal diffusivity 1.5 10-6 m2/s, initially at a uniformtemperature of 85C, has one face suddenly lowered to 20C while the other face is perfectly insulated.FIND: (a) Using the explicit finite-difference method with space and time increments of x = 30 mmand t = 300s, determine the temperature distribution within the wall 45 min after the change in surfacetemperature; (b) Effect of t on temperature histories of the surfaces and midplane.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties.ANALYSIS: (a) The finite-difference equations for the interior points, nodes 0, 1, 2, and 3, can bedetermined from Eq. 5.73,)(ppppTm+1 = Fo Tm 1 + Tm +1 + (1 2 Fo ) Tm(1)withFo = t x 2 = 1.5 106 m 2 s 300s (0.03m ) = 1/ 2 .2(2)Note that the stability criterion, Eq. 5.74, Fo 1/2, is satisfied. Hence, combining Eqs. (1) and (2),pppTm+1 = 1/ 2 Tm 1 + Tm +1 for m = 0, 1, 2, 3. Since the adiabatic plane at x = 0 can be treated as a)(symmetry plane, Tm-1 = Tm+1 for node 0 (m = 0). The finite-difference solution is generated in the tablebelow using t = pt = 300 p (s) = 5 p (min).p0123456789t(min)01020304045T08585858576.976.968.868.861.761.7T185858576.976.968.868.861.761.755.6T2858568.868.860.760.754.654.649.549.5T38552.552.544.444.440.440.437.337.334.8TL(C)20202020202020202020<The temperature distribution can also be determined from the Heisler charts. For the wall,Fo =tL2=1.5 106 m 2 s ( 45 60 ) s(0.12 m )2= 0.28andBi 1 =k= 0.hLContinued...PROBLEM 5.98 (Cont.)From Figure D.1, for Bi-1 = 0 and Fo = 0.28, find o i 0.55. Hence, for x = 0=oTi T iTo TTo = T ( 0, t ) = T + o ( Ti T ) = 20 C + 0.55 (85 20 ) C = 55.8 C .iorThis value is to be compared with 61.7C for the finite-difference method.(b) Using the IHT Finite-Difference Equation Tool Pad for One-Dimensional Transient Conduction,temperature histories were computed and results are shown for the insulated surface (T0) and themidplane, as well as for the chilled surface (TL).Temperature, T(C)8575655545352515020004000600080001000012000140001600018000Time, t(s)T0, deltat = 75 sT2, deltat = 75 sTLT0, deltat = 300 sT2, deltat = 300 sApart from small differences during early stages of the transient, there is excellent agreement betweenresults obtained for the two time steps. The temperature decay at the insulated surface must, of course,lag that of the midplane.PROBLEM 5.99KNOWN: Thickness, initial temperature and thermophysical properties of molded plastic part.Convection conditions at one surface. Other surface insulated.FIND: Surface temperatures after one hour of cooling.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in product, (2) Negligible radiation, at cooledsurface, (3) Negligible heat transfer at insulated surface, (4) Constant properties.ANALYSIS: Adopting the implicit scheme, the finite-difference equation for the cooled surfacenode is given by Eq. (5.88), from which it follows thatppp(1 + 2 Fo + 2 FoBi ) T10+1 2Fo T9 +1 = 2 FoBi T + T10The general form of the finite-difference equation for any interior node (1 to 9) is given by Eq. (5.89),pppp(1 + 2 Fo ) Tm+1 Fo Tm+1 + Tm+1 = Tm1+1)(The finite-difference equation for the insulated surface node may be obtained by applying theppsymmetry requirement to Eq. (5.89); that is, Tm +1 = Tm 1. Hence,pp(1 + 2 Fo ) To +1 2 Fo T1p +1 = To2For the prescribed conditions, Bi = hx/k = 100 W/m K (0.006m)/0.30 W/mK = 2. If the explicitmethod were used, the most restrictive stability requirement would be given by Eq. (5.79). Hence, for2-7 2Fo (1+Bi) 0.5, Fo 0.167. With Fo = t/x and = k/c = 1.67 10 m /s, the correspondingrestriction on the time increment would be t 36s. Although no such restriction applies for theimplicit method, a value of t = 30s is chosen, and the set of 11 finite-difference equations is solvedusing the Tools option designated as Finite-Difference Equations, One-Dimensional, and Transientfrom the IHT Toolpad. At t = 3600s, the solution yields:T10 (3600s ) = 24.1CT0 (3600s ) = 71.5C<COMMENTS: (1) More accurate results may be obtained from the one-term approximation to theexact solution for one-dimensional, transient conduction in a plane wall. With Bi = hL/k = 20, Table25.1 yields 1 = 1.496 rad and C1 = 1.2699. With Fo = t/L = 0.167, Eq. (5.41) then yields To = T +()2(Ti - T) C1 exp 1 Fo = 72.4C, and from Eq. (5.40b), Ts = T + (Ti - T) cos (1 ) = 24.5C.Since the finite-difference results do not change with a reduction in the time step (t < 30s), thedifference between the numerical and analytical results is attributed to the use of a coarse grid. Toimprove the accuracy of the numerical results, a smaller value of x should be used.Continued ..PROBLEM 5.99 (Cont.)(2) Temperature histories for the front and back surface nodes are as shown.80Te m p e ra tu re (C )706050403020060012001800240030003600Tim e (s )In s u la te d s u rfa ceC o o le d s u rfa ceAlthough the surface temperatures rapidly approaches that of the coolant, there is a significant lag inthe thermal response of the back surface. The different responses are attributable to the small value of and the large value of Bi.PROBLEM 5.100KNOWN: Plane wall, initially at a uniform temperature Ti = 25C, is suddenly exposed to convectionwith a fluid at T = 50C with a convection coefficient h = 75 W/m2K at one surface, while the other isexposed to a constant heat flux q = 2000 W/m2. See also Problem 2.43.oFIND: (a) Using spatial and time increments of x = 5 mm and t = 20s, compute and plot thetemperature distributions in the wall for the initial condition, the steady-state condition, and twointermediate times, (b) On q -x coordinates, plot the heat flux distributions corresponding to the fourxtemperature distributions represented in part (a), and (c) On q -t coordinates, plot the heat flux at x = 0xand x = L.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction and (2) Constant properties.ANALYSIS: (a) Using the IHT Finite-Difference Equations, One-Dimensional, Transient Tool, theequations for determining the temperature distribution were obtained and solved with a spatial incrementof x = 5 mm. Using the Lookup Table functions, the temperature distributions were plotted as shownbelow.(b) The heat flux, q (x,t), at each node can be evaluated considering the control volume shown with thexschematic above(q ( m, p ) = q + qxx,ax,b)ppTTp T Tppp k (1) m 1 m + k (1) m m +1 2 = k Tm 1 Tm +12=xx() 2xFrom knowledge of the temperature distribution, the heat flux at each node for the selected times iscomputed and plotted below.Heat flux, q''x(x,t) (W/m^2)Temperature, T(x,t) (C)16014012010080604020010203040200015001000500050Wall coordinate, x (mm)Initial condition, t<=0sTime = 150sTime = 300sSteady-state conditions, t>1200s01020304050Wall coordinate, x (mm)Initial condition, t<=0sTime = 150sTime = 300sSteady-state conditions, t>1200s(c) The heat fluxes for the locations x = 0 and x = L, are plotted as a function of time. At the x = 0surface, the heat flux is constant, q TT = 2000 W/m2. At the x = L surface, the heat flux is given byoNewtons law of cooling, q (L,t) = h[T(L,t) - T ]; at t = 0, q (L,0) = -1875 W/m2. For steady-statexxconditions, the heat flux q (x,) is everywhere constant at q TT .oxContinued...PROBLEM 5.100 (Cont.)2000Heat flux (W/m^2)10000-1000-2000020040060080010001200Elapsed time, t (s)q''x(0,t) - Heater fluxq''x(L,t) - Convective fluxComments: The IHT workspace using the Finite-Difference Equations Tool to determine thetemperature distributions and heat fluxes is shown below. Some lines of code were omitted to save spaceon the page.// Finite-Difference Equations, One-Dimensional, Transient Tool:// Node 0 - Applied heater flux/* Node 0: surface node (w-orientation); transient conditions; e labeled 1. */rho * cp * der(T0,t) = fd_1d_sur_w(T0,T1,k,qdot,deltax,Tinf0,h0,q''a0)q''a0 = 2000// Applied heat flux, W/m^2;Tinf0 = 25// Fluid temperature, C; arbitrary value since h0 is zero; no convection processh0 = 1e-20// Convection coefficient, W/m^2.K; made zero since no convection process// Interior Nodes 1 - 9:/* Node 1: interior node; e and w labeled 2 and 0. */rho*cp*der(T1,t) = fd_1d_int(T1,T2,T0,k,qdot,deltax)/* Node 2: interior node; e and w labeled 3 and 1. */rho*cp*der(T2,t) = fd_1d_int(T2,T3,T1,k,qdot,deltax)............/* Node 9: interior node; e and w labeled 10 and 8. */rho*cp*der(T9,t) = fd_1d_int(T9,T10,T8,k,qdot,deltax)// Node 10 - Convection process:/* Node 10: surface node (e-orientation); transient conditions; w labeled 9. */rho * cp * der(T10,t) = fd_1d_sur_e(T10,T9,k,qdot,deltax,Tinf,h,q''a)q''a = 0// Applied heat flux, W/m^2; zero flux shown// Heat Flux Distribution at Interior Nodes, q''m:q''1 = k / deltax * (T0 - T2) / 2q''2 = k / deltax * (T1 - T3) / 2............q''9 = k / deltax * (T8 - T10) / 2// Heat flux at boundary x= L, q''10q''xL = h * (T10 - Tinf)// Assigned Variables:deltax = 0.005k = 1.5alpha = 7.5e-6cp = 1000alpha = k / (rho * cp)qdot = 0Ti = 25Tinf = 50h = 75// Spatial increment, m// thermal conductivity, W/m.K// Thermal diffusivity, m^2/s// Specific heat, J/kg.K; arbitrary value// Defintion from which rho is calculated// Volumetric heat generation rate, W/m^3// Initial temperature, C; used also for plotting initial distribution// Fluid temperature, K// Convection coefficient, W/m^2.K// Solver Conditions: integrated t from 0 to 1200 with 1 s step, log every 2nd valuePROBLEM 5.101KNOWN: Plane wall, initially at a uniform temperature To = 25C, has one surface (x = L) suddenlyexposed to a convection process with T = 50C and h = 1000 W/m2K, while the other surface (x = 0) ismaintained at To. Also, the wall suddenly experiences uniform volumetric heating with q = 1 1073W/m . See also Problem 2.44.FIND: (a) Using spatial and time increments of x = 4 mm and t = 1s, compute and plot thetemperature distributions in the wall for the initial condition, the steady-state condition, and twointermediate times, and (b) On q -t coordinates, plot the heat flux at x = 0 and x = L. At what elapsedxtime is there zero heat flux at x = L?SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction and (2) Constant properties.ANALYSIS: (a) Using the IHT Finite-Difference Equations, One-Dimensional, Transient Tool, thetemperature distributions were obtained and plotted below.(b) The heat flux, q TT (L,t), can be expressed in terms of Newtons law of cooling,x()q ( L, t ) = h T10 T .xpFrom the energy balance on the control volume about node 0 shown above,(q ( 0, t ) + E g + q = 0xaq ( 0, t ) = q ( x 2 ) k T1 Toxp) xFrom knowledge of the temperature distribution, the heat fluxes are computed and plotted.120100080Heat flux (W/m^2)Temperature, T(x,t) (C)1000006040-1E5-2E52001020Wall coordinate, x (mm)Initial condition, t<=0sTime = 60sTime = 120sSteady-state conditions, t>600s3040-3E50100200300400500600Elapsed time, t(s)q''x(0,t)q''x(L,t)COMMENTS: The steady-state analytical solution has the form of Eq. 3.40 where C1 = 6500 m-1/Cand C2 = 25C. Find q ( 0, ) = 3.25 105 W / m 2 and q ( L ) = +7.5 104 W / m 2 . Comparing withxxthe graphical results above, we conclude that steady-state conditions are not reached in 600 x.PROBLEM 5.102KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250C withno internal generation; suddenly a uniform generation, q = 108 W/m3 , occurs when theelement is inserted into the core while the surfaces experience convection (T,h).FIND: Temperature distribution 1.5s after element is inserted into the core.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3)q = 0, initially; at t > 0, q is uniform.ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used.Using the nodal network of Example 5.8, the same finite-difference equations may be used.Interior nodes, m = 1, 2, 3, 42q ( x ) p+1ppp + (1 2 Fo ) Tm .Tm = Fo Tm-1 + Tm+1 +2Midplane node, m = 0(1)ppSame as Eq. (1), but with Tm-1 = Tm+1.Surface node, m = 52q ( x ) p+1p + (1 2Fo 2Bi Fo ) T p .(2)T5 = 2 Fo T4 + Bi T +52k The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) 1/2. Consider thefollowing parameters:2hx 1100W/m K ( 0.002m )Bi === 0.0733k30W/m K1/2Fo = 0.466(1 + Bi )Fo ( x )2t = 0.466(0.002m )25 106 m 2 / s= 0.373s.Continued ..PROBLEM 5.102 (Cont.)To be well within the stability limit, select t = 0.3s, which corresponds toFo =t=5 106 m 2 / s 0.3sx 2(0.002m )2t = pt = 0.3p (s ).= 0.375Substituting numerical values with q = 108 W/m3 , the nodal equations become2p+1ppT0 = 0.375 2T1 + 108 W/m3 ( 0.002m ) / 30W/m K + (1 2 0.375) T0p+1ppT0 = 0.375 2T1 + 13.33 + 0.25 T0p+1pppT1 = 0.375 T0 + T2 + 13.33 + 0.25 T1p+1pppT2 = 0.375 T1 + T3 + 13.33 + 0.25 T2(3)(4)(5)p+1pppT3 = 0.375 T2 + T4 + 13.33 + 0.25 T3p+1pppT4 = 0.375 T3 + T5 + 13.33 + 0.25 T4(6)(7)13.33 pp+1pT5 = 2 0.375 T4 + 0.0733 250 ++ (1 2 0.375 2 0.0733 0.375 ) T52p+1ppT5 = 0.750 T4 + 24.99 + 0.195 T5 .(8)The initial temperature distribution is Ti = 250C at all nodes. The marching solution,following the procedure of Example 5.8, is represented in the table below.p01234t(s)00.30.60.91.2T0250255.00260.00265.00270.00T1250255.00260.00265.00270.00T2250255.00260.00265.00270.00T3250255.00260.00265.00269.96T4250255.00260.00264.89269.74T5(C)250254.99259.72264.39268.9751.5275.00275.00274.98274.89274.53273.50The desired temperature distribution T(x, 1.5s), corresponds to p = 5.COMMENTS: Note that the nodes near the midplane (0,1) do not feel any effect of thecoolant during the first 1.5s time period.<PROBLEM 5.103KNOWN: Conditions associated with heat generation in a rectangular fuel element with surfacecooling. See Example 5.8.FIND: (a) The temperature distribution 1.5 s after the change in operating power; compare yourresults with those tabulated in the example, (b) Calculate and plot temperature histories at the midplane (00) and surface (05) nodes for 0 t 400 s; determine the new steady-state temperatures, andapproximately how long it will take to reach the new steady-state condition after the step change inoperating power. Use the IHT Tools | Finite-Difference Equations | One-Dimensional | Transientconduction model builder as your solution tool.SCHEMATIC:ASSUMPTIONS: (1) One dimensional conduction in the x-direction, (2) Uniform generation, and (3)Constant properties.ANALYIS: The IHT model builder provides the transient finite-difference equations for the implicitmethod of solution. Selected portions of the IHT code used to obtain the results tabulated below areshown in the Comments.(a) Using the IHT code, the temperature distribution (C) as a function of time (s) up to 1.5 s after thestep power change is obtained from the summarized results copied into the workspace123456t00.30.60.91.21.5T00357.6358.1358.6359.1359.6360.1T01356.9357.4357.9358.4358.9359.4T02354.9355.4355.9356.4356.9357.4T03351.6352.1352.6353.1353.6354.1T04346.9347.4347.9348.4348.9349.3T05340.9341.4341.9342.3342.8343.2(b) Using the code, the mid-plane (00) and surface (05) node temperatures are plotted as a function oftime.Te m p e ra tu re h is to ry a fte r s te p ch a n g e in p o w e rTe m p e ra tu re , T(x,t) (C )4804404003603200100200300400Tim e , t (s )T0 0 , Mid -p la n e , x = 0T0 5 , S u rfa c e , x = LContinued ..PROBLEM 5.103 (Cont.)Note that at t 240 s, the wall has nearly reached the new steady-state condition for which the nodaltemperatures (C) were found as:T00465T01463.7T02459.7T03453T04443.7T05431.7COMMENTS: (1) Can you validate the new steady-state nodal temperatures from part (b) bycomparison against an analytical solution?(2) Will using a smaller time increment improve the accuracy of the results? Use your code with t =0.15 s to justify your explanation.(3) Selected portions of the IHT code to obtain the nodal temperature distribution using spatial andtime increments of x = 2 mm and t = 0.3 s, respectively, are shown below. For the solveintegration step, the initial condition for each of the nodes corresponds to the steady-state temperaturedistribution with q1.// Tools | Finite-Difference Equations | One-Dimensional | Transient/* Node 00: surface node (w-orientation); transient conditions; e labeled 01. */rho * cp * der(T00,t) = fd_1d_sur_w(T00,T01,k,qdot,deltax,Tinf01,h01,q''a00)q''a00 = 0// Applied heat flux, W/m^2; zero flux shownTinf01 = 20// Arbitrary valueh01 = 1e-8// Causes boundary to behave as adiabatic/* Node 01: interior node; e and w labeled 02 and 00. */rho*cp*der(T01,t) = fd_1d_int(T01,T02,T00,k,qdot,deltax)/* Node 02: interior node; e and w labeled 03 and 01. */rho*cp*der(T02,t) = fd_1d_int(T02,T03,T01,k,qdot,deltax)/* Node 03: interior node; e and w labeled 04 and 02. */rho*cp*der(T03,t) = fd_1d_int(T03,T04,T02,k,qdot,deltax)/* Node 04: interior node; e and w labeled 05 and 03. */rho*cp*der(T04,t) = fd_1d_int(T04,T05,T03,k,qdot,deltax)/* Node 05: surface node (e-orientation); transient conditions; w labeled 04. */rho * cp * der(T05,t) = fd_1d_sur_e(T05,T04,k,qdot,deltax,Tinf05,h05,q''a05)q''a05 = 0// Applied heat flux, W/m^2; zero flux shownTinf05 = 250// Coolant temperature, Ch05 = 1100// Convection coefficient, W/m^2.K// Input parametersqdot = 2e7// Volumetric rate, W/m^3, step changedeltax = 0.002// Space incrementk = 30// Thermophysical propertiesalpha = 5e-6rho = 1000alpha = k / (rho * cp)/* Steady-state conditions, with qdot1 = 1e7 W/m^3; initial conditions for step changeT_x = 16.67 * (1 - x^2/L^2) + 340.91// See textSeek T_x for x = 0, 2, 4, 6, 8, 10 mm; results used for Ti areNode T_x00357.601356.902354.903351.604346.905340.9*/PROBLEM 5.104KNOWN: Conditions associated with heat generation in a rectangular fuel element with surfacecooling. See Example 5.8.FIND: (a) The temperature distribution 1.5 s after the change in the operating power; compare resultswith those tabulated in the Example, and (b) Plot the temperature histories at the midplane, x = 0, andthe surface, x = L, for 0 t 400 s; determine the new steady-state temperatures, and approximatelyhow long it takes to reach this condition. Use the finite-element software FEHT as your solution tool.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Uniform generation, (3)Constant properties.ANALYSIS: Using FEHT, an outline of the fuel element is drawn of thickness 10 mm in the xdirection and arbitrary length in the y-direction. The boundary conditions are specified as follows: onthe y-planes and the x = 0 plane, treat as adiabatic; on the x = 10 mm plane, specify the convectionoption. Specify the material properties and the internal generation with q1 . In the Setup menu, clickon Steady-state, and then Run to obtain the temperature distribution corresponding to the initialtemperature distribution, Ti ( x, 0 ) = T ( x, q1 ) , before the change in operating power to q 2 .Next, in the Setup menu, click on Transient; in the Specify | Internal Generation box, change the valueto q 2 ; and in the Run command, click on Continue (not Calculate).(a) The temperature distribution 1.5 s after the change in operating power from the FEHT analysis andfrom the FDE analysis in the Example are tabulated below.x/LT(x/L, 1.5 s)FEHT (C)FDE (C)00.20.4360.1360.08359.4359.41357.4357.410.60.81.0354.1 349.3354.07 349.37343.2343.27The mesh spacing for the FEHT analysis was 0.5 mm and the time increment was 0.005 s. For theFDE analyses, the spatial and time increments were 2 mm and 0.3 s. The agreement between theresults from the two numerical methods is within 0.1C.(b) Using the FEHT code, the temperature histories at the mid-plane (x = 0) and the surface (x = L) areplotted as a function of time.Continued ..PROBLEM 5.104 (Cont.)From the distribution, the steady-state condition (based upon 98% change) is approached in 215 s.The steady-state temperature distributions after the step change in power from the FEHT and FDEanalysis in the Example are tabulated below. The agreement between the results from the twonumerical methods is within 0.1Cx/LT(x/L, )FEHT (C)FDE (C)00.20.40.60.8465.0465.15463.7463.82459.6459.82453.0453.15443.6443.821.0431.7431.82COMMENTS: (1) For background information on the Continue option, see the Run menu in theFEHT Help section. Using the Run/Calculate command, the steady-state temperature distribution wasdetermined for the q1 operating power. Using the Run|Continue command (after re-setting thegeneration to q 2 and clicking on Setup | Transient), this steady-state distribution automaticallybecomes the initial temperature distribution for the q 2 operating power. This feature allows forconveniently prescribing a non-uniform initial temperature distribution for a transient analysis (ratherthan specifying values on a node-by-node basis).(2) Use the View | Tabular Output command to obtain nodal temperatures to the maximum number ofsignificant figures resulting from the analysis.(3) Can you validate the new steady-state nodal temperatures from part (b) (with q 2 , t ) bycomparison against an analytical solution?PROBLEM 5.105KNOWN: Thickness, initial temperature, speed and thermophysical properties of steel in a thin-slabcontinuous casting process. Surface convection conditions.FIND: Time required to cool the outer surface to a prescribed temperature. Corresponding value ofthe midplane temperature and length of cooling section.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible radiation at quenched surfaces,(3) Symmetry about the midplane, (4) Constant properties.ANALYSIS: Adopting the implicit scheme, the finite-difference equaiton for the cooled surfacenode is given by Eq. (5.88), from which it follows thatppp(1 + 2 Fo + 2 FoBi ) T10+1 2 Fo T9 +1 = 2 FoBi T + T10The general form of the finite-difference equation for any interior node (1 to 9) is given by Eq. (5.89),p(1 + 2 Fo ) Tm+1 Foppp(Tm+11 + Tm++11 ) = TmThe finite-difference equation for the midplane node may be obtained by applying the symmetrypprequirement to Eq. (5.89); that is, Tm +1 = Tm 1. Hence,pp(1 + 2 Fo ) T0 +1 2 Fo T1p +1 = T02For the prescribed conditions, Bi = hx/k = 5000 W/m K (0.010m)/30 W/mK = 1.67. If the explicitmethod were used, the stability requirement would be given by Eq. (5.79). Hence, for Fo(1 + Bi) 2-6 20.5, Fo 0.187. With Fo = t/x and = k/c = 5.49 10 m /s, the corresponding restriction onthe time increment would be t 3.40s. Although no such restriction applies for the implicit method,a value of t = 1s is chosen, and the set of 11 finite-difference equations is solved using the Toolsoption designated as Finite-Difference Equations, One-Dimensional and Transient from the IHTToolpad. For T10 (t) = 300C, the solution yields<t = 161sContinued ..PROBLEM 5.105 (Cont.)T0 ( t ) = 1364C<With a casting speed of V = 15 mm/s, the length of the cooling section isLcs = Vt = 0.015 m / s (161s ) = 2.42m<2COMMENTS: (1) With Fo = t/L = 0.088 < 0.2, the one-term approximation to the exact solutionfor one-dimensional conduction in a plane wall cannot be used to confirm the foregoing results.However, using the exact solution from the Models, Transient Conduction, Plane Wall Option of IHT,values of T0 = 1366C and Ts = 200.7C are obtained and are in good agreement with the finitedifference predictions. The accuracy of these predictions could still be improved by reducing thevalue of x.(2) Temperature histories for the surface and midplane nodes are plotted for 0 < t < 600s.1500T e m p e ra tu re (C )130011009007005003001000100200300400500600T im e (s )Mid p la n eC o o le d s u rfa c eWhile T10 (600s) = 124C, To (600s) has only dropped to 879C. The much slower thermalresponse at the midplane is attributable to the small value of and the large value of Bi =16.67.PROBLEM 5.106KNOWN: Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to aconvection cooling process (T,h).FIND: Temperatures at the surface and a 45mm depth after 3 minutes using finite-differencemethod with space and time increments of 15mm and 18s.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Plate approximates semi-infinitemedium, (3) Constant properties.ANALYSIS: The grid network representing the plate is shown above. The finite-differenceequation for node 0 is given by Eq. 5.82 for one-dimensional conditions or Eq. 5.77,()p+1ppT0 = 2 Fo T1 + Bi T + (1 2 Fo 2 Bi Fo ) T0 .(1)The numerical values of Fo and Bi areFo =tx 2=5.6 106 m2 /s 18s( 0.015m )2= 0.448()2-3hx 100 W/m K 15 10 mBi === 0.075.k20 W/m KRecognizing that T = 15 C, Eq. (1) has the formp+1ppT0 = 0.0359 T0 + 0.897 T1 + 1.01.(2)It is important to satisfy the stability criterion, Fo (1+Bi) 1/2. Substituting values,0.448 (1+0.075) = 0.482 1/2, and the criterion is satisfied.The finite-difference equation for the interior nodes, m = 1, 2, follows from Eq. 5.73,()p+1pppTm = Fo Tm+1 + Tm-1 + (1 2Fo ) Tm .(3)Recognizing that the stability criterion, Fo 1/2, is satisfied with Fo = 0.448,()p+1pppTm = 0.448 Tm+1 + Tm-1 + 0.104Tm.(4)Continued ..PROBLEM 5.106 (Cont.)The time scale is related to p, the number of steps in the calculation procedure, and t, the timeincrement,t = pt.(5)The finite-difference calculations can now be performed using Eqs. (2) and (4). The results aretabulated below.p012345678910t(s)01836547290108126144162180T0325304.2303.2294.7293.0287.6285.6281.8279.8276.7274.8T1325324.7315.3313.7307.8305.8301.6299.5296.2294.1291.3T2325325324.5320.3318.9315.2313.5310.5308.6306.0304.1T3325325325324.5322.5321.5319.3317.9315.8314.3312.4T4325325325325324.5323.5322.7321.4320.4319.0T5325325325325325324.5324.0323.3322.5T6T7(K)325325325325325325325325325325325325324.5 325324.2Hence, find10T ( 0, 180s ) = T0 = 275o C10T ( 45mm, 180s ) = T3 = 312o C.<COMMENTS: (1) The above results can be readily checked against the analytical solutionrepresented in Fig. 5.8 (see also Eq. 5.60). For x = 0 and t = 180s, findx=01/22 ( t )(1/2 100 W/m 2 K 5.60 10-6m 2 / s 180sh ( t )=k20 W/m Kfor which the figure givesT Ti= 0.15T Tiso that,)1/2= 0.16T ( 0, 180s ) = 0.15 ( T T i ) +T i = 0.15 (15 325)o C + 325oCT ( 0, 180s ) = 278o C.For x = 45mm, the procedure yields T(45mm, 180s) = 316 C. The agreement with the numericalsolution is nearly within 1%.PROBLEM 5.107KNOWN: Sudden exposure of the surface of a thick slab, initially at a uniform temperature,to convection and to surroundings at a high temperature.FIND: (a) Explicit, finite-difference equation for the surface node in terms of Fo, Bi, Bir, (b)Stability criterion; whether it is more restrictive than that for an interior node and does itchange with time, and (c) Temperature at the surface and at 30mm depth for prescribedconditions after 1 minute exposure.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Thick slab may beapproximated as semi-infinite medium, (3) Constant properties, (4) Radiation exchange isbetween small surface and large surroundings.ANALYSIS: (a) The explicit form of the FDE forthe surface node may be obtained by applying anenergy balance to a control volume about the node.E E = qinoutconv + q + qradcond = Est)(()ppT1 Topph T To + h r Tsur To + k 1xp+1p To x T(1)= c 1 ot2 where the radiation process has been linearized, Eq. 1.8. (See also Comment 4, Example 5.9), p2pp2h r = h p To , Tsur = To + Tsur T0 + Tsur .(2)r Divide Eq. (1) by cx/2t and regroup using these definitions to obtain the FDE:)((Fo ( k/ c ) t/x 2()Bi hx/k)Bi r h r x/k(3,4,5)p+1ppTo = 2Fo Bi T + Bir Tsur + T1 + (1 2 Bi Fo 2Bir Fo 2Fo ) To .(6) <p(b) The stability criterion for Eq. (6) requires that the coefficient of To be positive.1 2Fo ( Bi + Bi r + 1) 0orFo 1/2 ( Bi + Bi r + 1).(7) <The stability criterion for an interior node, Eq. 5.74, is Fo 1/2. Since Bi + Bir > 0, thestability criterion of the surface node is more restrictive. Note that Bir is not constant butpdepends upon hr which increases with increasing To (time). Hence, the restriction on Fopincreases with increasing To (time).Continued ..PROBLEM 5.107 (Cont.)(c) Consider the prescribed conditions with negligible convection (Bi = 0). The FDEs for thethick slab are:Surface (0)()p+1pTo = 2Fo Bi Fo + Bi r Tsur + T1 + (1 2Bi Fo 2Bi r Fo 2Fo ) Top)((8)p+1pppTm = Fo Tm+1 + Tm-1 + (1 2Fo ) TmInterior (m1)(9,5,7,3)The stability criterion from Eq. (7) with Bi = 0 is,Fo 1/2 (1 + Bi r )(10)To proceed with the explicit, marching solution, we need to select a value of t (Fo) that willsatisfy the stability criterion. A few trial calculations are helpful. A value of t = 15s2provides Fo = 0.105, and using Eqs. (2) and (5), hr(300K, 1000K) = 72.3 W/m K and Bir =p0.482. From the stability criterion, Eq. (10), find Fo 0.337. With increasing To , hr and Bir2increase: hr(800K, 1000K) = 150.6 W/m K, Bir = 1.004 and Fo 0.249. Hence, ifpTo < 800K, t = 15s or Fo = 0.105 satisfies the stability criterion.Using t = 15s or Fo = 0.105 with the FDEs, Eqs. (8) and (9), the results of the solution aretabulated below. Note how h p and Bi p are evaluated at each time increment. Note that t =rrpt, where t = 15s.pt(s)To / h p / Bi rr0030072.30.482300115370.86779.5770.530523034T3T4300300300300300300300426.07985.9840.5733307.44130030030045470.25691.6190.6108319.117300.78130030060502.289333.061302.624300.082300After 60s(p = 4),T1(K)T2To(0, 1 min) = 502.3K and T3(30mm, 1 min) = 300.1K..<COMMENTS: (1) The form of the FDE representing the surface node agrees with Eq. 5.82if this equation is reduced to one-dimension.(2) We should recognize that the t = 15s time increment represents a coarse step. Toimprove the accuracy of the solution, a smaller t should be chosen.PROBLEM 5.108KNOWN: Thick slab of copper, initially at a uniform temperature, is suddenly exposed to a constantnet radiant flux at one surface. See Example 5.9.FIND: (a) The nodal temperatures at nodes 00 and 04 at t = 120 s; that is, T00(0, 120 s) and T04(0.15m, 120 s); compare results with those given by the exact solution in Comment 1; will a time incrementof 0.12 s provide more accurate results?; and, (b) Plot the temperature histories for x = 0, 150 and 600mm, and explain key features of your results. Use the IHT Tools | Finite-Difference Equations | OneDimensional | Transient conduction model builder to obtain the implicit form of the FDEs for theinterior nodes. Use space and time increments of 37.5 mm and 1.2 s, respectively, for a 17-nodenetwork. For the surface node 00, use the FDE derived in Section 2 of the Example.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Slab of thickness 600 mmapproximates a semi-infinite medium, and (3) Constant properties.ANALYSIS: The IHT model builder provides the implicit-method FDEs for the interior nodes, 01 15. The +x boundary condition for the node-16 control volume is assumed adiabatic. The FDE for thesurface node 00 exposed to the net radiant flux was derived in the Example analysis. Selected portionsof the IHT code used to obtain the following results are shown in the Comments.(a) The 00 and 04 nodal temperatures for t = 120 s are tabulated below using a time increment of t =1.2 s and 0.12 s, and compared with the results given from the exact analytical solution, Eq. 5.59.Node0004FDE results (C)t = 1.2 s119.345.09t = 0.12 s119.445.10Analytical result (C)Eq. 5.59120.045.4The numerical FDE-based results with the different time increments agree quite closely with oneanother. At the surface, the numerical results are nearly 1 C less than the result from the exactanalytical solution. This difference represents an error of -1% ( -1 C / (120 20 ) C x 100). At thex = 150 mm location, the difference is about -0.4 C, representing an error of 1.5%. For thissituation, the smaller time increment (0.12 s) did not provide improved accuracy. To improve theaccuracy of the numerical model, it would be necessary to reduce the space increment, in addition tousing the smaller time increment.(b) The temperature histories for x = 0, 150 and 600 mm (nodes 00, 04, and 16) for the range 0 t 150 s are as follows.Continued ..PROBLEM 5.108 (Cont.)Te m p e ra tu re h is to rie s fo r N o d e s 0 0 , 0 4 , a n d 1 6Te m p e ra tu re , T(x,t)12080400050100150Tim e , t (s )T0 0 = T(0 , t)T0 4 = T(1 5 0 m m , t)T0 0 = T(6 0 0 m m , t)As expected, the surface temperature, T00 = T(0,t), increases markedly at early times. As thermalpenetration increases with increasing time, the temperature at the location x = 150 mm, T04 = T(150mm, t), begins to increase after about 20 s. Note, however, the temperature at the location x = 600mm, T16 = T(600 mm, t), does not change significantly within the 150 s duration of the appliedsurface heat flux. Our assumption of treating the +x boundary of the node 16 control volume asadiabatic is justified. A copper plate of 600-mm thickness is a good approximation to a semi-infinitemedium at times less than 150 s.COMMENTS: Selected portions of the IHT code with the nodal equations to obtain the temperaturedistribution are shown below. Note how the FDE for node 00 is written in terms of an energy balanceusing the der (T,t) function. The FDE for node 16 assumes that the east boundary is adiabatic.// Finite-difference equation, node 00; from Examples solution derivation; implicit methodq''o + k * (T01 - T00) / deltax = rho * (deltax / 2) *cp * der (T00,t)// Finite-difference equations, interior nodes 01-15; from Tools/* Node 01: interior node; e and w labeled 02 and 00. */rho*cp*der(T01,t) = fd_1d_int(T01,T02,T00,k,qdot,deltax)rho*cp*der(T02,t) = fd_1d_int(T02,T03,T01,k,qdot,deltax)....rho*cp*der(T14,t) = fd_1d_int(T14,T15,T13,k,qdot,deltax)rho*cp*der(T15,t) = fd_1d_int(T15,T16,T14,k,qdot,deltax)// Finite-difference equation node 16; from Tools, adiabatic surface/* Node 16: surface node (e-orientation); transient conditions; w labeled 15. */rho * cp * der(T16,t) = fd_1d_sur_e(T16,T15,k,qdot,deltax,Tinf16,h16,q''a16)q''a16 = 0// Applied heat flux, W/m^2; zero flux shownTinf16 = 20// Arbitrary valueh16 = 1e-8// Causes boundary to behave as adiabaticPROBLEM 5.109KNOWN: Thick slab of copper as treated in Example 5.9, initially at a uniform temperature, issuddenly exposed to large surroundings at 1000C (instead of a net radiant flux).FIND: (a) The temperatures T(0, 120 s) and T(0.15 m, 120s) using the finite-element software FEHTfor a surface emissivity of 0.94 and (b) Plot the temperature histories for x = 0, 150 and 600 mm, andexplain key features of your results.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Slab of thickness 600 mmapproximates a semi-infinite medium, (3) Slab is small object in large, isothermal surroundings.ANALYSIS: (a) Using FEHT, an outline of the slab is drawn of thickness 600 mm in the x-directionand arbitrary length in the y-direction. Click on Setup | Temperatures in K, to enter all temperatures inkelvins. The boundary conditions are specified as follows: on the y-planes and the x = 600 mm plane,treat as adiabatic; on the surface (0,y), select the convection coefficient option, enter the linearizedradiation coefficient after Eq. 1.9 written as0.94 * 5.67e-8 * (T + 1273) * (T^2 + 1273^2)and enter the surroundings temperature, 1273 K, in the fluid temperature box. See the Comments for aview of the input screen. From View|Temperatures, find the results:T(0, 120 s) = 339 K = 66C<T(150 mm, 120 s) = 305K = 32C(b) Using the View | Temperatures command, the temperature histories for x = 0, 150 and 600 mm (10mm mesh, Nodes 18, 23 and 15, respectively) are plotted. As expected, the surface temperatureincreases markedly at early times. As thermal penetration increases with increasing time, thetemperature at the location x = 150 mm begins to increase after about 30 s. Note, however, that thetemperature at the location x = 600 mm does not change significantly within the 150 s exposure to thehot surroundings. Our assumption of treating the boundary at the x = 600 mm plane as adiabatic isjustified. A copper plate of 600 mm is a good approximation to a semi-infinite medium at times lessthan 150 s.Continued ..PROBLEM 5.109 (Cont.)COMMENTS: The annotated Input screen shows the outline of the slab, the boundary conditions,and the triangular mesh before using the Reduce-mesh option.PPROBLEM 5.110KNOWN: Electric heater sandwiched between two thick plates whose surfaces experienceconvection. Case 2 corresponds to steady-state operation with a loss of coolant on the x = -L surface.Suddenly, a second loss of coolant condition occurs on the x = +L surface, but the heater remainsenergized for the next 15 minutes. Case 3 corresponds to the eventual steady-state condition followingthe second loss of coolant event. See Problem 2.53.FIND: Calculate and plot the temperature time histories at the plate locations x = 0, L during thetransient period between steady-state distributions for Case 2 and Case 3 using the finite-elementapproach with FEHT and the finite-difference method of solution with IHT (x = 5 mm and t = 1 s).SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Heater has negligiblethickness, and (4) Negligible thermal resistance between the heater surfaces and the plates.PROPERTIES: Plate material (given); = 2500 kg/m3, c = 700 J/kgK, k = 5 W/mK.ANALYSIS: The temperature distribution for Case 2 shown in the above graph represents the initialcondition for the period of time following the second loss of coolant event. The boundary conditions2at x = L are adiabatic, and the heater flux is maintained at q = 4000 W/m for 0 t 15 min.oUsing FEHT, the heater is represented as a plate of thickness Lh = 0.5 mm with very low thermalcapacitance ( = 1 kg/m and c = 1 J/kgK), very high thermal conductivity (k= 10,000 W/mK), and a3uniform volumetric generation rate of q = q / L h = 4000 W / m 2 / 0.0005 m = 8.0 106 W/m for 0 t o900 s. In the Specify | Generation box, the generation was prescribed by the lookup file (see FEHTHelp): hfvst,1,2,Time. This Notepad file is comprised of four lines, with the values on each lineseparated by a single tab space:090090150008e68e600The temperature-time histories are shown in the graph below for the surfaces x = - L (lowest curve,13) and x = +L (19) and the center point x = 0 (highest curve, 14). The center point experiences themaximum temperature of 89C at the time the heater is deactivated, t = 900 s.Continued ..PROBLEM 5.110For the finite-difference method of solution, the nodal arrangement for the system is shown below.The IHT model builder Tools | Finite-Difference Equations | One Dimensional can be used to obtainthe FDEs for the internal nodes (02-04, 07-10) and the adiabatic boundary nodes (01, 11).For the heater-plate interface node 06, the FDE for the implicit method is derived from an energybalance on the control volume shown in the schematic above.E E + E = Einoutgenstq + q + q = Eabostp +1p +1p+p+p+pT05 T06T07 1 T06 1T06 1 T06k+k+ q = cxoxxtThe IHT code representing selected nodes is shown below for the adiabatic boundary node 01, interiornode 02, and the heater-plates interface node 06. Note how the foregoing derived finite-differenceequation in implicit form is written in the IHT Workspace. Note also the use of a Lookup Table forrepresenting the heater flux vs. time.Continued ..PROBLEM 5.110 (Cont.)// Finite-difference equations from Tools, Nodes 01, 02/* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */rho * cp * der(T01,t) = fd_1d_sur_w(T01,T02,k,qdot,deltax,Tinf01,h01,q''a01)q''a01 = 0// Applied heat flux, W/m^2; zero flux shownqdot = 0// No internal generationTinf01 = 20// Arbitrary valueh01 = 1e-6// Causes boundary to behave as adiabatic/* Node 02: interior node; e and w labeled 03 and 01. */rho*cp*der(T02,t) = fd_1d_int(T02,T03,T01,k,qdot,deltax)// Finite-difference equation from energy balance on CV, Node 06k * (T05 - T06) / deltax + k * (T07 - T06)/ deltax + q''h = rho * cp * deltax * der(T06,t)q''h = LOOKUPVAL(qhvst,1,t,2)// Heater flux, W/m^2; specified by Lookup Table/* See HELP (Solver, Lookup Tables). The Look-up table file name "qhvst" contains040009004000900.5 05000 0*/The temperature-time histories using the IHT code for the plate locations x = 0, L are shown in thegraphs below. We chose to show expanded presentations of the histories at early times, just after thesecond loss of coolant event, t = 0, and around the time the heater is deactivated, t = 900 s.6090Temperature, T (C)Temperature, T (C)85504080753070050100150200800900Time, t (s)Surface x = -LCenter point, x = 0Surface x = +L100011001200Time, t (s)Surface x = -LCenter point, x = 0Surface x = +LCOMMENTS: (1) The maximum temperature during the transient period is at the center point andoccurs at the instant the heater is deactivated, T(0, 900s) = 89C. After 300 s, note that the two surfacetemperatures are nearly the same, and never rise above the final steady-state temperature.(2) Both the FEHT and IHT methods of solution give identical results. Their steady-state solutionsagree with the result of an energy balance on a time interval basis yielding Tss = 86.1C.PROBLEM 5.111KNOWN: Plane wall of thickness 2L, initially at a uniform temperature, is suddenly subjected toconvection heat transfer.FIND: The mid-plane, T(0,t), and surface, T(L,t), temperatures at t = 50, 100, 200 and 500 s, usingthe following methods: (a) the one-term series solution; determine also the Biot number; (b) thelumped capacitance solution; and (c) the two- and 5-node finite-difference numerical solutions.Prepare a table summarizing the results and comment on the relative differences of the predictedtemperatures.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, and (2) Constant properties.ANALYSIS: (a) The results are tabulated below for the mid-plane and surface temperatures using theone-term approximation to the series solution, Eq. 5.40 and 5.41. The Biot number for the heattransfer process isBi = h L / k = 500 W / m 2 K 0.020 m / 15 W / m K = 0.67Since Bi >> 0.1, we expect an appreciable temperature difference between the mid-plane and surfaceas the tabulated results indicate (Eq. 5.10).(b) The results are tabulated below for the wall temperatures using the lumped capacitance method(LCM) of solution, Eq. 5.6. The LCM neglects the internal conduction resistance and since Bi = 0.67>> 0.1, we expect this method to predict systematically lower temperatures (faster cooling) at themidplane compared to the one-term approximation.Solution method/Time(s)50100200500Mid-plane, T(0,t) (C)One-term, Eqs. 5.40, 5.41Lumped capacitance2-node FDE5-node FDE207.1181.7210.6207.5160.5133.9163.5160.999.9777.69100.5100.237.7030.9737.1737.77160.1181.7163.7160.2125.4133.9125.2125.680.5677.6979.4080.6734.4130.9733.7734.45Surface, T(L,t) (C)One-term, Eqs. 5.40, 5.41Lumped capacitance2-node FDE5-node FDE(c) The 2- and 5-node nodal networks representing the wall are shown in the schematic above. Theimplicit form of the finite-difference equations for the mid-plane, interior (if present) and surfacenodes can be derived from energy balances on the nodal control volumes. The time-rate of change ofthe temperature is expressed in terms of the IHT integral intrinsic function, der(T,t).Continued ..PROBLEM 5.111 (Cont.)Mid-plane nodek (T 2 T 1) / x = c ( x / 2 ) der (T 1, t )Interior node (5-node network)k (T 1 T 2 ) / x + k (T 3 T 2 ) / x = c x der (T 2, t )Surface node (shown for 5-node network)k (T 4 T 5 ) / x + h (T inf T 5) = c ( x / 2 ) der (T 5, t )With appropriate values for x, the foregoing FDEs were entered into the IHT workspace and solvedfor the temperature distributions as a function of time over the range 0 t 500 s using an integrationtime step of 1 s. Selected portions of the IHT codes for each of the models are shown in theComments. The results of the analysis are summarized in the foregoing table.COMMENTS: (1) Referring to the table above, we can make the following observations about therelative differences and similarities of the estimated temperatures: (a) The one-term series modelestimates are the most reliable, and can serve as the benchmark for the other model results; (b) TheLCM model over estimates the rate of cooling, and poorly predicts temperatures since the modelneglects the effect of internal resistance and Bi = 0.67 >> 0.1; (c) The 5-node model results are inexcellent agreement with those from the one-term series solution; we can infer that the chosen spaceand time increments are sufficiently small to provide accurate results; and (d) The 2-node model underestimates the rate of cooling for early times when the time-rate of change is high; but for late times,the agreement is improved.(2) See the Solver | Intrinsic Functions section of IHT|Help or the IHT Examples menu (Example 5.3)for guidance on using the der(T,t) function.(3) Selected portions of the IHT code for the 2-node network model are shown below.// Writing the finite-difference equations 2-node model// Node 1k * (T2 - T1)/ deltax = rho * cp * (deltax / 2) * der(T1,t)// Node 2k * (T1 - T2)/ deltax + h * (Tinf - T2) = rho * cp * (deltax / 2) * der(T2,t)// Input parametersL = 0.020deltax = Lrho = 7800// density, kg/m^3cp = 440// specific heat, J/kgKk = 15// thermal conductivity, W/m.Kh = 500// convection coefficient, W/m^2KTinf = 25// fluid temperature, K(4) Selected portions of the IHT code for the 5-node network model are shown below.// Writing the finite-difference equations 5-node model// Node 1 - midplanek * (T2 - T1)/ deltax = rho * cp * (deltax / 2) * der(T1,t)// Interior nodesk * (T1 - T2)/ deltax + k * (T3 - T2 )/ deltax = rho * cp * deltax * der(T2,t)k * (T2 - T3)/ deltax + k * (T4 - T3 )/ deltax = rho * cp * deltax * der(T3,t)k * (T3 - T4)/ deltax + k * (T5 - T4 )/ deltax = rho * cp * deltax * der(T4,t)// Node5 - surfacek * (T4 - T5)/ deltax + h * (Tinf - T5) = rho * cp * (deltax / 2) * der(T5,t)// Input parametersL = 0.020deltax = L / 4....PROBLEM 5.1122KNOWN: Plastic film on metal strip initially at 25C is heated by a laser (85,000 W/m forton = 10 s), to cure adhesive; convection conditions for ambient air at 25C with coefficient2of 100 W/m K.FIND: Temperature histories at center and film edge, T(0,t) and T(x1,t), for 0 t 30 s,using an implicit, finite-difference method with x = 4mm and t = 1 s; determine whetheradhesive is cured (Tc 90C for tc = 10s) and whether the degradation temperature of 200Cis exceeded.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Uniformconvection coefficient on upper and lower surfaces, (4) Thermal resistance and mass of plasticfilm are negligible, (5) All incident laser flux is absorbed.3PROPERTIES: Metal strip (given): = 7850 kg/m , cp = 435 J/kgK, k = 60 W/mK, =-5 2k/cp = 1.757 10 m /s.ANALYSIS: (a) Using a space increment of x = 4mm, set up the nodal network shownbelow. Note that the film half-length is 22mm (rather than 20mm as in Problem 3.97) tosimplify the finite-difference equation derivation.Consider the general control volume and use the conservation of energy requirement to obtainthe finite-difference equation.Ein E out = EstpT p+1 Tmqa + q b + q laser + qconv = Mcp mtContinued ..PROBLEM 5.112 (Cont.)k ( d 1)p+1p+1Tm-1 Tmx+ k (d 1)p+1p+1Tm+1 Tm(x)pT p+1 Tmp+1 ( x 1) + 2h ( x 1) T Tm = ( x d 1) cp m+ qotpp+1Tm = (1 + 2Fo + 2Fo Bi ) Tm(1))(p+1p+1 Fo Tm+1 + Tm+1 2Fo Bi T Fo QwhereFo =Bi =Q=tx 2=(1.757 105 m 2 / s 1sh x 2 / dk(= 1.098(2)) = 100 W/m2 K (0.0042 / 0.00125) m = 0.021360 W/m K(3)) = 85, 000 W/m2 (0.0042 / 0.0015) m = 18.133.60 W/m K(4)(0.004 m )2q x 2 / dokThe results of the matrix inversion numerical method of solution (x = 4mm, t = 1s) areshown below. The temperature histories for the center (m = 1) and film edge (m = 5) nodes,T(0,t) and T(x1,t), respectively, permit determining whether the adhesive has cured (T 90Cfor 10 s).Certainly the center region, T(0,t), is fully cured and furthermore, the degradation temperature(200C) has not been exceeded. From the T(x1,t) distribution, note that tc 8 sec, which is20% less than the 10 s interval sought. Hence, the laser exposure (now 10 s) should beslightly increased and quite likely, the maximum temperature will not exceed 200C.PROBLEM 5.113KNOWN: Insulated rod of prescribed length and diameter, with one end in a fixture at 200C, reaches auniform temperature. Suddenly the insulating sleeve is removed and the rod is subjected to a convectionprocess.FIND: (a) Time required for the mid-length of the rod to reach 100C, (b) Temperature history T(x,t t1), where t1 is time at which the midlength reaches 50C. Temperature distribution at 0, 200s, 400s andt1.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional transient conduction in rod, (2) Uniform h along rod and at end,(3) Negligible radiation exchange between rod and surroundings, (4) Constant properties.ANALYSIS: (a) Choosing x = 0.016 m, the finite-difference equations for the interior and end nodesare obtained.ppTm+1 Tmq a + q b + q c = A c x c p Interior Point, m:tk AcppTm 1 Tmx+ kA cppTm +1 TmxppTm+1 Tmp+ hPx T Tm = A c xc p)(tRegrouping,ppppTm+1 = Tm (1 2Fo Bi Fo ) + Fo Tm 1 + Tm +1 + Bi FoT(whereFo =tx 2(2))Bi = h x 2 ( Ac P ) k .(1)(3)From Eq. (1), recognize that the stability of the numerical solution will be assured when the first term onthe RHS is positive; that isContinued...PROBLEM 5.113 (Cont.)(1 2Fo Bi Fo ) 0Fo 1 ( 2 + Bi ) .or(4)pNodal Point 1: Consider Eq. (1) for the special case that Tm 1 = To, which is independent of time.Hence,pppT1 +1 = T1 (1 2Fo Bi Fo ) + Fo To + T2 + Bi FoT .(5))(End Nodal Point 10:ppT10+1 T10xqa + q b + qc = Ac cp2tp +1pxx T10 T10ppk Ac+ hA c T T10 + hPT T10 = A ccpx22tp +1ppRegrouping, T10 = T10 (1 2Fo 2N Fo Bi Fo ) + 2FoT9 + T ( 2N Fo + Bi Fo )where N = hx/k.)(ppT9 T10)((6)(7)The stability criterion is Fo 1/2(1 + N + Bi/2).(8)With the finite-difference equations established, we can now proceed with the numerical solution.Having already specified x = 0.016 m, Bi can now be evaluated. Noting that Ac = D2/4 and P = D,giving Ac/P = D/4, Eq. (3) yields2 0.010 m Bi = 30 W m 2 K ( 0.016 m )(9) 14.8 W m K = 0.2084From the stability criteria, Eqs. (4) and (8), for the finite-difference equations, it is recognized that Eq.(8) requires the greater value of Fo. Hence10.208 Fo = 1 + 0.0324 + = 0.44022where from Eq. (7), N =(10)30 W m 2 K 0.016 m= 0.0324 .14.8 W m K(11)From the definition of Fo, Eq. (2), we obtain the time increment2Fo ( x )2t == 0.440 0.016 m3.63 106 m 2 s = 31.1s(12)and the time relation is t = pt = 31.1t.(13)()Using the numerical values for Fo, Bi and N, the finite-difference equations can now be written (C).Nodal Point m (2 m 9):ppppTm+1 = Tm (1 2 0.440 0.208 0.440 ) + 0.440 Tm 1 + Tm +1 + 0.208 0.440 25)(()ppppTm+1 = 0.029Tm + 0.440 Tm 1 + Tm +1 + 2.3Nodal Point 1:p+ppppT1 1 = 0.029T1 + 0.440 200 + T2 + 2.3 = 0.029T1 + 0.440T2 + 90.3()(14)(15)Nodal Point 10:ppppT10+1 = 0 T10 + 2 0.440T9 + 25 ( 2 0.0324 0.440 + 0.208 0.440 ) = 0.880T9 + 3.0 (16)Continued...PROBLEM 5.113 (Cont.)Using finite-difference equations (14-16) with Eq. (13), the calculations may be performed to obtainp012345678t(s)031.162.293.3124.4155.5186.6217.7248.8T1200184.1175.6168.6163.3158.8155.2152.1145.1T2200181.8166.3154.8145.0137.1130.2124.5119.5T3200181.8165.3150.7138.8128.1119.2111.3104.5T4200181.8165.3150.7137.0125.3114.8105.797.6T5200181.8165.3150.7137.0124.5113.4103.594.8T6200181.8165.3150.7137.0124.3113.0102.9T7200181.8165.3150.7136.5124.2112.6102.4T8200181.8165.3149.7136.3123.4112.3T9200181.8164.0149.2135.0123.0111.5T10(C)200179.0163.0147.3134.3121.8111.2<Using linear interpolation between rows 7 and 8, we obtain T(L/2, 230s) = T5 100C.(b) Using the option concerning Finite-Difference Equations for One-Dimensional TransientConduction in Extended Surfaces from the IHT Toolpad, the desired temperature histories werecomputed for 0 t t1 = 930s. A Lookup Table involving data for T(x) at t = 0, 200, 400 and 930s wascreated.t(s)/x(mm)0200400930020020020020016200157.8146.2138.132200136.7114.999.2348200127.097.3274.9864200122.787.759.9480200121.082.5750.6796200120.279.844.99112200119.678.1441.53128200118.676.8739.44144200117.175.638.2160200114.774.1337.55and the LOOKUPVAL2 interpolating function was used with the Explore and Graph feature of IHT tocreate the desired plot.Temperature, T(C)225200175150125100755025020406080100120140160Fin location, x(mm)t= 0t = 200 st = 400 st = 930 sTemperatures decrease with increasing x and t, and except for early times (t < 200s) and locations inproximity to the fin tip, the magnitude of the temperature gradient, |dT/dx|, decreases with increasing x.The slight increase in |dT/dx| observed for t = 200s and x 160 mm is attributable to significant heatloss from the fin tip.COMMENTS: The steady-state condition may be obtained by extending the finite-differencecalculations in time to t 2650s or from Eq. 3.70.PROBLEM 5.114KNOWN: Tantalum rod initially at a uniform temperature, 300K, is suddenly subjected to acurrent flow of 80A; surroundings (vacuum enclosure) and electrodes maintained at 300K.FIND: (a) Estimate time required for mid-length to reach 1000K, (b) Determine the steadystate temperature distribution and estimate how long it will take to reach steady-state. Use afinite-difference method with a space increment of 10mm.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings aremuch larger than rod, (3) Properties are constant and evaluated at an average temperature.PROPERTIES: Table A-1, Tantalum ( T = (300+1000 ) K/2 = 650K ) : = 16,600 kg/m , c33= 147 J/kgK, k = 58.8 W/mK, and = k/c = 58.8 W/mK/16,600 kg/m 147 J/kgK =-5 22.410 10 m /s.ANALYSIS: From the derivation of the previous problem, the finite-difference equation wasfound to be)(p+1pppTm = Fo Tm-1 + Tm+1 + (1 2Fo ) Tm Fo = t/x 2where()I2 e x 2 Px 24,p4Fo Tm Tsur + Fo2kAckA cA c = D2 / 4P = D.(1)(2,3,4)From the stability criterion, let Fo = 1/2 and numerically evaluate terms of Eq. (1).20.1 5.67 108 W/m 2 K 4 ( 0.01m ) 4 1 4,p4p+1 1 ppTm = Tm-1 + Tm+1 Tm [300K ] +258.8 W/m K ( 0.003m )2)((80A )2 95 108 m (0.01m )2 1+(58.8 W/m K [0.003m ] / 4p+1Tm =(2))2()21pp4,pTm-1 + Tm+1 6.4285 1012 Tm + 103.53.2(5)Note that this form applies to nodes 0 through 5. For node 0, Tm-1 = Tm+1 = T1. Since Fo =1/2, using Eq. (2), find thatt = x 2 Fo/ = (0.01m ) 1/ 2 / 2.410 105 m 2 / s = 2.07s.2(6)Hence, t = pt = 2.07p.(7)Continued ..PROBLEM 5.114 (Cont.)(a) To estimate the time required for the mid-length to reach 1000K, that is To = 1000K,perform the forward-marching solution beginning with Ti = 300K at p = 0. The solution, astabulated below, utilizes Eq. (5) for successive values of p. Elapsed time is determined by Eq.(7).Pt(s)T0T1T2T3T401234567891011121314150300403.5506.9610.0712.6814.5915.21010.91104.71190.91274.11348.21419.71479.81542.61605.3300403.5506.9610.0712.6814.5911.91007.91096.81183.51261.61336.71402.41465.51538.21569.3300403.5506.9610.0712.6808.0902.4988.91073.81150.41224.91290.61353.91408.41460.91514.0300403.5506.9610.0699.7788.8867.4945.01014.01081.71141.51199.81250.51299.81341.21381.6300403.5506.9584.1661.1724.7787.9842.3896.1943.2989.41029.91069.41103.61136.91164.810.420.731.1T5T6(C)300403.5455.1506.7545.2583.5615.1646.6673.6700.3723.6746.5766.5786.0802.9819.3Note that, at p 6.9 or t = 6.9 2.07 = 14.3s, the mid-point temperature is To 1000K.300300300300300300300300300300300300300300300300<(b) The steady-state temperature distribution can be obtained by continuing the marchingsolution until only small changes in Tm are noted. From the table above, note that at p = 15 ort = 31s, the temperature distribution is still changing with time. It is likely that at least 15more calculation sets are required to see whether steady-state is being approached.COMMENTS: (1) This problem should be solved with a computer rather than a handcalculator. For such a situation, it would be appropriate to decrease the spatial increment inorder to obtain better estimates of the temperature distribution.(2) If the rod were very long, the steady-state temperaturedistribution would be very flat at the mid-length x = 0.Performing an energy balance on the small control volumeshown to the right, findEg Eout = 0 x44I2 e Px To Tsur = 0.Ac()Substituting numerical values, find To = 2003K. It is unlikely that the present rod would everreach this steady-state, maximum temperature. That is, the effect of conduction along the rodwill cause the center temperature to be less than this value.PROBLEM 5.115KNOWN: Support rod spanning a channel whose walls are maintained at Tb = 300 K. Suddenly the rodis exposed to cross flow of hot gases with T = 600 K and h = 75 W/m2K. After the rod reachessteady-state conditions, the hot gas flow is terminated and the rod cools by free convection and radiationexchange with surroundings.FIND: (a) Compute and plot the midspan temperature as a function of elapsed heating time; compare thesteady-state temperature distribution with results from an analytical model of the rod and (b) Computethe midspan temperature as a function of elapsed cooling time and determine the time required for therod to reach the safe-to-touch temperature of 315 K.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Constant properties, (3) Duringheating process, uniform convection coefficient over rod, (4) During cooling process, free convectioncoefficient is of the form h = CTn where C = 4.4 W/m2K1.188 and n = 0.188, and (5) During coolingprocess, surroundings are large with respect to the rod.ANALYSIS: (a) The finite-difference equations for the 10-node mesh shown above can be obtainedusing the IHT Finite-Difference Equation, One-Dimensional, Transient Extended Surfaces Tool. Thetemperature-time history for the midspan position T10 is shown in the plot below. The steady-statetemperature distribution for the rod can be determined from Eq. 3.75, Case B, Table 3.4. This case istreated in the IHT Extended Surfaces Model, Temperature Distribution and Heat Rate, Rectangular PinFin, for the adiabatic tip condition. The following table compares the steady-state temperaturedistributions for the numerical and analytical methods.MethodAnalyticalNumericalTemperatures (K) vs. Position x (mm)10203040386.1443.4479.5499.4386.0443.2479.3499.2030030050505.8505.6The comparison is excellent indicating that the nodal mesh is sufficiently fine to obtain precise results.Midspan temperature, T10 (K)6005004003000100200300400500600Elapsed heating time, t (s)Continued...PROBLEM 5.115 (Cont.)(b) The same finite-difference approach can be used to model the cooling process. In using the IHT tool,the following procedure was used: (1) Set up the FDEs with the convection coefficient expressed as hm =hfc,m + hr,m, the sum of the free convection and linearized radiation coefficients based upon nodaltemperature Tm.(ph fc,m = C Tm T))( )( p2p2h r,m = Tm + Tsur Tm + Tsur (2) For the initial solve, set hfc,m = hr,m = 5 W/m2K and solve, (3) Using the solved results as the InitialGuesses for the next solve, allow hfc,m and hr,m to be unknowns. The temperature-time history for themidspan during the cooling process is shown in the plot below. The time to reach the safe-to-touchptemperature, T10 = 315 K , is<t = 550 sMidspan temperature, T10 (K)6005004003000200400600Elapsed cooling time, t (s)8001000PROBLEM 5.116KNOWN: Thin metallic foil of thickness, w, whose edges are thermally coupled to a sink attemperature, Tsink, initially at a uniform temperature Ti = Tsink, is suddenly exposed on the top surface toan ion beam heat flux, qs , and experiences radiation exchange with the vacuum enclosure walls at Tsur.Consider also the situation when the foil is operating under steady-state conditions when suddenly theion beam is deactivated.FIND: (a) Compute and plot the midspan temperature-time history during the heating process;determine the elapsed time that this point on the foil reaches a temperature within 1 K of the steady-statevalue, and (b) Compute and plot the midspan temperature-time history during the cooling process fromsteady-state operation; determine the elapsed time that this point on the foil reaches the safe-to-touchtemperature of 315 K.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional, transient conduction in the foil, (2) Constant properties, (3)Upper and lower surfaces of foil experience radiation exchange with the large surroundings, (4) Ion beamincident on upper surface only, (4) Foil is of unit width normal to the page.ANALYSIS: (a) The finite-difference equations for the 10-node mesh shown above can be obtainedusing the IHT Finite-Difference Equation, One-Dimensional, Transient, Extended Surfaces Tool. Informulating the energy-balance functions, the following steps were taken: (1) the FDE functioncoefficient h must be identified for each node, e.g., h1 and (2) coefficient can be represented by the()22linearized radiation coefficient, e.g., h1 = (T1 + Tsur ) T1 + Tsur , (3) set q = q 2 since the ionaobeam is incident on only the top surface of the foil, and (4) when solving, the initial conditioncorresponds to Ti = 300 K for each node. The temperature-time history of the midspan position is shownbelow. The time to reach within 1 K of the steady-state temperature (374.1 K) isT10 ( t h ) = 373K<t h = 136s(b) The same IHT workspace may be used to obtain the temperature-time history for the cooling processby taking these steps: (1) set q = 0, (2) specify the initial conditions as the steady-state temperature (K)sdistribution tabulated below,T1374.1T2374.0T3373.5T4372.5T5370.9T6368.2T7363.7T8356.6T9345.3T10327.4(3) when performing the integration of the independent time variable, set the start value as 200 s and (4)save the results for the heating process in Data Set A. The temperature-time history for the heating andcooling processes can be made using Data Browser results from the Working and A Data Sets. The timerequired for the midspan to reach the safe-to-touch temperature isT10 ( t c ) = 315 Kt c = 73s<Continued...PROBLEM 5.116 (Cont.)Midpsan temperature, T1 (K)4003803603403203000100200300400500Heating or cooling time, t (s)Heating processCooling processCOMMENTS: The IHT workspace using the Finite-Difference Equations Tool to determine thetemperature-time distributions is shown below. Some of the lines of code were omitted to save space onthe page.// Finite Difference Equations Tool: One-Dimensional, Transient, Extended Surface/* Node 1: extended surface interior node; transient conditions; e and w labeled 2 and 2. */rho * cp * der(T1,t) = fd_1d_xsur_i(T1,T2,T2,k,qdot,Ac,P,deltax,Tinf, h1,q''a)q''a1 = q''s / 2// Applied heat flux, W/m^2; on the upper surface onlyh1 = eps * sigma * (T1 + Tsur) * (T1^2 + Tsur^2)sigma = 5.67e-8// Boltzmann constant, W/m^2.K^4/* Node 2: extended surface interior node; transient conditions; e and w labeled 3 and 1. */rho * cp * der(T2,t) = fd_1d_xsur_i(T2,T3,T1,k,qdot,Ac,P,deltax,Tinf, h2,q''a2)q''a2 = 0// Applied heat flux, W/m^2; zero flux shownh2 = eps * sigma * (T2+ Tsur) * (T2^2 + Tsur^2)............../* Node 10: extended surface interior node; transient conditions; e and w labeled sk and 9. */rho * cp * der(T10,t) = fd_1d_xsur_i(T10,Tsk,T9,k,qdot,Ac,P,deltax,Tinf, h10,q''a)q''a10 = 0// Applied heat flux, W/m^2; zero flux shownh10 = eps * sigma * (T10 + Tsur) * (T10^2 + Tsur^2)// Assigned variablesdeltax = L / 10Ac = w * 1P=2*1L = 0.150w = 0.00025eps = 0.45Tinf = TsurTsur = 300k = 40Tsk = 300q''s = 600q''s = 0qdot = 0alpha = 3e-5rho = 1000alpha = k / (rho * cp)// Spatial increment, m// Cross-sectional area, m^2// Perimeter, m// Overall length, m// Foil thickness, m// Foil emissivity// Fluid temperature, K// Surroundings temperature, K// Foil thermal conductivity// Sink temperature, K// Ion beam heat flux, W/m^2; for heating process// Ion beam heat flux, W/m^2; for cooling process// Foil volumetric generation rate, W/m^3// Thermal diffusivity, m^2/s// Density, kg.m^3; arbitrary value// DefinitionPROBLEM 5.117KNOWN: Stack or book of steel plates (sp) and circuit boards (b) subjected to a prescribedplaten heating schedule Tp(t). See Problem 5.42 for other details of the book.FIND: (a) Using the implicit numerical method with x = 2.36mm and t = 60s, find the mid-planetemperature T(0,t) of the book and determine whether curing will occur (> 170 C for 5 minutes), (b)Determine how long it will take T(0,t) to reach 37 C following reduction of the platen temperature to15 C (at t = 50 minutes), (c) Validate code by using a sudden change of platen temperature from 15to 190 C and compare with the solution of Problem 5.38.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible contact resistance betweenplates, boards and platens.3PROPERTIES: Steel plates (sp, given): sp = 8000 kg/m , cp,sp = 480 J/kgK, ksp = 123W/mK; Circuit boards (b, given): b = 1000 kg/m , cp,b = 1500 J/kgK, kb = 0.30 W/mK.ANALYSIS: (a) Using the suggested space increment x = 2.36mm, the model grid spacingtreating the steel plates (sp) and circuit boards (b) as discrete elements, we need to derive the nodalequations for the interior nodes (2-11) and the node next to the platen (1). Begin by definingappropriate control volumes and apply the conservation of energy requirement.Effective thermal conductivity, ke: Consider an adjacent steel plate-board arrangement. Thethermal resistance between the nodes i and j isx x/2 x/2=+kekbksp22==W/m K1 / k b+ + 1 / k sp 1/0.3 + 1/12= 0.585 W/m K.R =ijkekeOdd-numbered nodes, 3 m 11 - steel plates (sp): Treat as interior nodes using Eq. 5.89 withke0.585 W/m K== 1.523 10 7 m 2 / ssp csp 8000 kg/m3 480 J/kg Kspt 1.523 10 7 m 2 / s 60sFo m === 1.6412x 2( 0.00236 m )sp Continued ..PROBLEM 5.117 (Cont.)to obtain, with m as odd-numbered,p+1(1 + 2Fom ) Tm Fomp+1p+1p( Tm-1 + Tm+1 ) = Tm(1)Even-numbered nodes, 2 n 10 - circuit boards (b): Using Eq. 5.89 and evaluating b andFonk b = e = 3.900 107 m 2 /sFon = 4.201 bc bp+1(1 + 2Fon ) Tn Fonp+1p+1(Tn-1 + Tn+1 ) = Tnp(2)Plate next to platen, n = 1 - steel plate (sp): The finite-difference equation for the plate node (n =1) next to the platen follows from a control volume analysis.&&&Ein Eout = Estq + q = sp xc spabTp+1 Tp11twhereq = k spap+1Tp ( t ) T1p+1T2 Tp+11q = k ebx/2xand Tp(t) = Tp(p) is the platen temperature which is changedwith time according to the heating schedule. Regrouping find, 2k sp p+1p+1 2k sppFomTp ( p) = T1 1 + Fom 1 + T1 Fom T2 ke kewhere 2ksp/ke = 2 12 W/mK/0.585 W/mK = 41.03.(3)Using the nodal Eqs. (1) -(3), an inversion method of solution was effected and the temperaturedistributions are shown on the following page.Temperature distributions - discussion: As expected, the temperatures of the nodes near thecenter of the book considerably lag those nearer the platen. The criterion for cure is T 170 C =443 K for tc = 5 min = 300 sec. From the temperature distributions, note that node 10 justreaches 443 K after 50 minutes and will not be cured. It appears that the region about node 5 willbe cured.(b) The time required for the book to reach 37 C = 310 K can likewise be seen from thetemperature distribution results. The plates/boards nearest the platen will cool to the safe handlingtemperature with 1000 s = 16 min, but those near the center of the stack will require in excess of2000 s = 32 min.Continued ..PROBLEM 5.117 (Cont.)(c) It is important when validating computer codes to have the program work a problem which hasan exact analytical solution. You should select the problem such that all features of the code aretested.PROBLEM 5.118KNOWN: Reaction and composite clutch plates, initially at a uniform temperature, Ti = 40C, aresubjected to the frictional-heat flux shown in the engagement energy curve, q vs. t .fFIND: (a) On T-t coordinates, sketch the temperature histories at the mid-plane of the reaction plate,at the interface between the clutch pair, and at the mid-plane of the composite plate; identify keyfeatures; (b) Perform an energy balance on the clutch pair over a time interval basis and calculate thesteady-state temperature resulting from a clutch engagement; (c) Obtain the temperature historiesusing the finite-element approach with FEHT and the finite-difference method of solution with IHT(x = 0.1 mm and t = 1 ms). Calculate and plot the frictional heat fluxes to the reaction andcomposite plates, q and q , respectively, as a function of time. Comment on the features of therpcptemperature and frictional-heat flux histories.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible heat transfer to the surroundings.3PROPERTIES: Steel, s = 7800 kg/m , cs = 500 J/kgK, ks = 40 W/mK; Friction material, fm =31150 kg/m , cfm = 1650 J/kgK, and kfm = 4 W/mK.ANALYSIS: (a) The temperature histories for specified locations in the system are sketched on T-tcoordinates below.Initially, the temperature at all locations is uniform at Ti. Since there is negligible heat transfer to thesurroundings, eventually the system will reach a uniform, steady-state temperature T(). During theengagement period, the interface temperature increases much more rapidly than at the mid-planes ofthe reaction (rp) and composite (cp) plates. The interface temperature should be the maximum withinthe system and could occur before lock-up, t = tlu.Continued ..PROBLEM 5.118 (Cont.)(b) To determine the steady-state temperature following the engagement period, apply theconservation of energy requirement on the clutch pair on a time-interval basis, Eq. 1.11b.The final and initial states correspond to uniform temperatures of T() and Ti, respectively. Theenergy input is determined from the engagement energy curve, q vs. t .fE E + E = Estnoutgent lu0Ein = E = 0out()q ( t ) dt = E E = scs L rp / 2 + Lcp / 2 + fm cfm Lfm (Tf Ti )ffiSubstituting numerical values, with Ti = 40C and Tf = T().()0.5 q t lu = s cs Lrp / 2 + Lcp / 2 + fm cfm Lfm ( T ( ) Ti )o0.5 1.6 107 W / m 2 0.100 s = 7800 kg / m3 500 J / kg K ( 0.001 + 0.0005 ) m+1150 kg / m3 1650 J / kg K 0.0005 m ( T ( ) 40 ) CT ( ) = 158C<(c) Finite-element method of solution, FEHT. The clutch pair is comprised of the reaction plate (1mm), an interface region (0.1 mm), and the composite plate (cp) as shown below.Continued (2)...PROBLEM 5.118 (Cont.)The external boundaries of the system are made adiabatic. The interface region provides the means torepresent the frictional heat flux, specified with negligible thermal resistance and capacitance. Thegeneration rate is prescribed asq = 1.6 1011 (1 Time / 0.1) W / m30 Time t luwhere the first coefficient is evaluated as q / 0.1 103 m and the 0.1 mm parameter is the thicknessoof the region. Using the Run command, the integration is performed from 0 to 0.1 s with a time step of-6110 s. Then, using the Specify|Generation command, the generation rate is set to zero and theRun|Continue command is executed. The temperature history is shown below.(c) Finite-difference method of solution, IHT. The nodal arrangement for the clutch pair is shownbelow with x = 0.1 mm and t = 1 ms. Nodes 02-10, 13-16 and 18-21 are interior nodes, and theirfinite-difference equations (FDE) can be called into the Workspace using Tools|Finite DifferenceEquations|One-Dimenisonal|Transient. Nodes 01 and 22 represent the mid-planes for the reaction andcomposite plates, respectively, with adiabatic boundaries. The FDE for node 17 is derived from anenergy balance on its control volume (CV) considering different properties in each half of the CV.The FDE for node 11 and 12 are likewise derived using energy balances on their CVs. At theinterface, the following conditions must be satisfiedT11 = T12q = q + qfrpcpThe frictional heat flux is represented by a Lookup Table, which along with the FDEs, are shown inthe IHT code listed in Comment 2.Continued (3)...PROBLEM 5.118 (Cont.)The temperature and heat flux histories are plotted below. The steady-state temperature was found as156.5 C, which is in reasonable agreement with the energy balance result from part (a).Tem perature his tory for clutch pair, 100 m s lock-up tim eHeat flux his tories for clutch pair during engagement2Heat flux, q'' (W/m^2 * 10^7)250Tem perature, T (C)2001501005001.510.50-0.502004006008001000020Engagem ent tim e, t (ms )Midplane, reaction plate, T01Interface, T11 or T12Midplane, com pos ite plate, T22406080Engagem ent tim e, t (ms )Frictional heat flux, q''fReaction plate, q''rpcom pos ite plate, q''cpCOMMENTS: (1) The temperature histories resulting from the FEHT and IHT based solutions are inagreement. The interface temperature peaks near 225C after 75 ms, and begins dropping toward thesteady-state condition. The mid-plane of the reaction plate peaks around 100 ms, nearly reaching200C. The temperature of the mid-plane of the composite plate increases slowly toward the steadystate condition.(2) The calculated temperature-time histories for the clutch pair display similar features as expectedfrom our initial sketches on T vs. t coordinates, part a. The maximum temperature for the composite isvery high, subjecting the bonded frictional material to high thermal stresses as well as acceleratingdeterioration. For the reaction steel plate, the temperatures are moderate, but there is a significantgradient that could give rise to thermal stresses and hence, warping. Note that for the composite plate,the steel section is nearly isothermal and is less likely to experience warping.(2) The IHT code representing the FDE for the 22 nodes and the frictional heat flux relation is shownbelow. Note use of the Lookup Table for representing the frictional heat flux vs. time boundarycondition for nodes 11 and 12.// Nodal equations, reaction plate (steel)/* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */rhos * cps * der(T01,t) = fd_1d_sur_w(T01,T02,ks,qdot,deltax,Tinf01,h01,q''a01)q''a01 = 0// Applied heat flux, W/m^2; zero flux shownTinf01 = 40// Arbitrary valueh01 = 1e-5// Causes boundary to behave as adiabaticqdot = 0/* Node 02: interior node; e and w labeled 03 and 01. */rhos*cps*der(T02,t) = fd_1d_int(T02,T03,T01,ks,qdot,deltax)./* Node 10: interior node; e and w labeled 11 and 09. */rhos*cps*der(T10,t) = fd_1d_int(T10,T11,T09,ks,qdot,deltax)/* Node 11: From an energy on the CV about node 11 */ks * (T10 - T11) / deltax + q''rp = rhos * cps * deltax / 2 * der(T11,t)Continued (4)...100PROBLEM 5.118 (Cont.)// Friction-surface interface conditionsT11 = T12q''f = LOOKUPVAL(HFVST16,1,t,2)// Applied heat flux, W/m^2; specified by Lookup Table/* See HELP (Solver, Lookup Tables). The look-up table, file name "HFVST16' contains016e60.101000*/q''rp + q''cp = q''f// Frictional heat flux// Nodal equations - composite plate// Frictional material, nodes 12-16/* Node 12: From an energy on the CV about node 12 */kfm * (T13 - T12) / deltax + q''cp = rhofm * cpfm * deltax / 2 * der(T12,t)/* Node 13: interior node; e and w labeled 08 and 06. */rhofm*cpfm*der(T13,t) = fd_1d_int(T13,T14,T12,kfm,qdot,deltax)/* Node 16: interior node; e and w labeled 11 and 09. */rhofm*cpfm*der(T16,t) = fd_1d_int(T16,T17,T15,kfm,qdot,deltax)// Interface between friction material and steel, node 17/* Node 17: From an energy on the CV about node 17 */kfm * (T16 - T17) / deltax + ks * (T18 - T17) / deltax = RHSRHS = ( (rhofm * cpfm * deltax /2) + (rhos * cps * deltax /2) ) * der(T17,t)// Steel, nodes 18-22/* Node 18: interior node; e and w labeled 03 and 01. */rhos*cps*der(T18,t) = fd_1d_int(T18,T19,T17,ks,qdot,deltax)./* Node 22: interior node; e and w labeled 21 and 21. Symmetry condition. */rhos*cps*der(T22,t) = fd_1d_int(T22,T21,T21,ks,qdot,deltax)// qdot = 0// Input variables// Ti = 40deltax = 0.0001rhos = 7800cps = 500ks = 40rhofm = 1150cpfm = 1650kfm = 4// Initial temperature; entered during Solve// Steel properties//Friction material properties// Conversions, to facilitate graphingt_ms = t * 1000qf_7 = q''f / 1e7qrp_7 = q''rp / 1e7qcp_7 = q''cp / 1e7PROBLEM 5.119KNOWN: Hamburger patties of thickness 2L = 10, 20 and 30 mm, initially at a uniform temperatureTi = 20C, are grilled on both sides by a convection process characterized by T = 100C and h =5000 W/m2K.FIND: (a) Determine the relationship between time-to-doneness, td, and patty thickness. Donenesscriteria is 60C at the center. Use FEHT and the IHT Models|Transient Conduction|Plane Wall. (b)Using the results from part (a), estimate the time-to-doneness if the initial temperature is 5 C ratherthan 20C. Calculate values using the IHT model, and determine the relationship between time-todoneness and initial temperature.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction, and (2) Constant properties are approximated asthose of water at 300 K.3PROPERTIES: Table A-6, Water (300K), = 1000 kg/m , c = 4179 J/kgK, k = 0.613 W/mK.ANALYSIS: (a) To determine T(0, td), the center point temperature at the time-to-doneness time, td,a one-dimensional shape as shown in the FEHT screen below is drawn, and the material properties,boundary conditions, and initial temperature are specified. With the Run|Calculate command, the earlyintegration steps are made very fine to accommodate the large temperature-time changes occurringnear x = L. Use the RunContinue command (see FEHT HELP) for the second and subsequent stepsof the integration. This sequence of Start-(Step)-Stop values was used: 0 (0.001) 0.1 (0.01 ) 1 (0.1)120 (1.0) 840 s.Continued ..PROBLEM 5.119 (Cont.)Using the View|Temperature vs. Time command, the temperature-time histories for the x/L = 0(center), 0.5, and 1.0 (grill side) are plotted and shown below for the 2L = 10 mm thick patty.Using the View|Temperatures command, the time slider can be adjusted to read td, when the centerpoint, x = 0, reaches 60C. See the summary table below.The IHT ready-to-solve model in Models|Transient Conduction|Plane Wall is based upon Eq. 5.40 andpermits direct calculation of td when T(0,td) = 60C for patty thickness 2L = 10, 20 and 30 mm andinitial temperatures of 20 and 5C. The IHT code is provided in Comment 3, and the results aretabulated below.Solution methodTime-to-doneness, t (s)Ti ( C)Patty thickness, 2L (mm)10FEHTIHTEq. 5.40 (seeComment 4)203066.267.780.2x264.5264.5312.2xx591590.4699.1x20205520Considering the IHT results for Ti = 20C, note that when the thickness is doubled from 10 to 20 mm,td is (264.5/67.7=) 3.9 times larger. When the thickness is trebled, from 10 to 30 mm, td is2(590.4/67.7=) 8.7 times larger. We conclude that, td is nearly proportional to L , rather than linearlyproportional to thickness.Continued ..PROBLEM 5.119 (Cont.)(b) The temperature span for the cooking process ranges from T = 100 to Ti = 20 or 5C. Thedifferences are (100-20 =) 80 or (100-5 =) 95C. If td is proportional to the overall temperature span,then we expect td for the cases with Ti = 5C to be a factor of (95/80 =) 1.19 higher (approximately20% ) than with Ti = 20C. From the tabulated results above, for the thickness 2L = 10, 20 and 30mm, the td with Ti = 5C are (80.2/67.7 = ) 1.18, (312 / 264.5 =) 1.18, and (699.1/590.4 =) 1.18,respectively, higher than with Ti = 20C. We conclude that td is nearly proportional to the temperaturespan (T - Ti).COMMENTS: (1) The results from the FEHT and IHT calculations are in excellent agreement. Forthis analysis, the FEHT model is more convenient to use as it provides direct calculations of the timeto-doneness. The FEHT tool allows the user to watch the cooking process. Use theViewTemperature Contours command, click on the from start-to-stop button, and observe how colorband changes represent the temperature distribution as a function of time.(2) It is good practice to check software tool analyses against hand calculations. Besides providingexperience with the basic equations, you can check whether the tool was used or functioned properly.Using the one-term series solution, Eq. 5.40: T ( 0, t d ) T= C1 exp 2 Foo =Ti T)(C1, = ( Bi ) , Table 5.1Fo = t d / L2Ti (C)2L (mm)oBiC11Fotd (s)20510300.50000.421124.4773.411.27071.27291.50681.54710.41080.462270.0709The results are slightly higher than those from the IHT model, which is based upon a multiple- ratherthan single-term series solution.(3) The IHT code used to obtain the tabulated results is shown below. Note that T_xt_trans is anintrinsic heat transfer function dropped into the Workspace from the Models window (see IHTHelp|Solver|Intrinsic Functions|Heat Transfer Functions).// Models | Transient Conduction | Plane Wall/* Model: Plane wall of thickness 2L, initially with a uniform temperature T(x,0) = Ti, suddenly subjectedto convection conditions (Tinf,h). */// The temperature distribution isT_xt = T_xt_trans("Plane Wall",xstar,Fo,Bi,Ti,Tinf)// Eq 5.39// The dimensionless parameters arexstar = x / LBi = h * L / k// Eq 5.9Fo= alpha * t / L^2// Eq 5.33alpha = k/ (rho * cp)// Input parametersx=0// Center point of meatL = 0.005// Meat half-thickness, m//L = 0.010//L = 0.015T_xt = 60// Doneness temperature requirement at center, x = 0; CTi = 20// Initial uniform temperature//Ti = 5rho = 1000// Water properties at 300 Kcp = 4179k = 0.613h = 5000// Convection boundary conditionsTinf = 100PROBLEM 5.120KNOWN: A process mixture at 200C flows at a rate of 207 kg/min onto a 1-m wide conveyor belttraveling with a velocity of 36 m/min. The underside of the belt is cooled by a water spray.FIND: The surface temperature of the mixture at the end of the conveyor belt, Te,s, using (a) IHT forwriting and solving the FDEs, and (b) FEHT. Validate your numerical codes against an appropriateanalytical method of solution.SCHEMATIC:ASSUMPTIONS: (1) One-dimensional conduction in the x-direction at any z-location, (2)Negligible heat transfer from mixture upper surface to ambient air, and (3) Constant properties.3PROPERTIES: Process mixture (m), m = 960 kg/m , cm = 1700 J/kgK, and km = 1.5 W/mK;3Conveyor belt (b), b = 8000 kg/ m , cb = 460 J/kgK, and kb = 15 W/mK.ANALYSIS: From the conservation of mass requirement, the thickness of the mixture on theconveyor belt can be determined.m = m Ac VwhereAc = W Lm207 kg / min 1 min/ 60s = 960 kg / m3 1m L m 36 m / min1min/ 60sLm = 0.0060 m = 6 mmThe time that the mixture is in contact with the steel conveyor belt, referred to as the residence time, ist res = Lc / V = 30 m / (36 m / min 1 min/ 60 s ) = 50 sThe composite system comprised of the belt, Lb = 3 mm, and mixture, Lm = 6 mm, as represented inthe schematic above, is initially at a uniform temperature T(x,0) = Ti = 200C while at location z = 0,and suddenly is exposed to convection cooling (T, h). We will calculate the mixture upper surfacetemperature after 50 s, T(0, tres) = Te,s .(a) The nodal arrangement for the composite system is shown in the schematic below. The IHT modelbuilder Tools|Finite-Difference Equations|Transient can be used to obtain the FDEs for nodes 01-12and 14-19.Continued ..PROBLEM 5.120 (Cont.)For the mixture-belt interface node 13, the FDE for the implicit method is derived from an energybalance on the control volume about the node as shown above.E E = Einoutstq + q = Eabst,m + Est,bkmppT12+1 T13+1x+ kbppT14+1 T13+1x= ( mc m + bc b )( x / 2 )ppT13+1 T13tIHT code representing selected FDEs, nodes 01, 02, 13 and 19, is shown in Comment 4 below (x =0.5 mm, t = 0.1 s). Note how the FDE for node 13 derived above is written in the Workspace. Fromthe analysis, find<Te,s = T(0, 50s) = 54.8C(b) Using FEHT, the composite system is drawn and the material properties, boundary conditions, andinitial temperature are specified. The screen representing the system is shown below in Comment 5with annotations on key features. From the analysis, find<Te,s = T(0, 50s) = 54.7CCOMMENTS: (1) Both numerical methods, IHT and FEHT, yielded the same result, 55C. For thesafety of plant personnel working in the area of the conveyor exit, the mixture exit temperature shouldbe lower, like 43C.(2) By giving both regions of the composite the same properties, the analytical solution for the planewall with convection, Section 5.5, Eq. 5.40, can be used to validate the IHT and FEHT codes. Usingthe IHT Models|Transient Conduction|Plane Wall for a 9-mm thickness wall with mixturethermophysical properties, we calculated the temperatures after 50 s for three locations: T(0, 50s) =91.4C; T(6 mm, 50s) = 63.6C; and T(3 mm, 50s) = 91.4C. The results from the IHT and FEHTcodes agreed exactly.(3) In view of the high heat removal rate on the belt lower surface, it is reasonable to assume thatnegligible heat loss is occurring by convection on the top surface of the mixture.Continued ..PROBLEM 5.120 (Cont.)(4) The IHT code representing selected FDEs, nodes 01, 02, 13 and 19, is shown below. The FDE fornode 13 was derived from an energy balance, while the others are written from the Tools pad.// Finite difference equations from Tools, Nodes 01 -12 (mixture) and 14-19 (belt)/* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */rhom * cm * der(T01,t) = fd_1d_sur_w(T01,T02,km,qdot,deltax,Tinf01,h01,q''a01)q''a01 = 0// Applied heat flux, W/m^2; zero flux shownqdot = 0Tinf01 = 20// Arbitrary valueh01 = 1e-6// Causes boundary to behave as adiabatic/* Node 02: interior node; e and w labeled 03 and 01. */rhom*cm*der(T02,t) = fd_1d_int(T02,T03,T01,km,qdot,deltax)/* Node 19: surface node (e-orientation); transient conditions; w labeled 18. */rhob * cb * der(T19,t) = fd_1d_sur_e(T19,T18,kb,qdot,deltax,Tinf19,h19,q''a19)q''a19 = 0// Applied heat flux, W/m^2; zero flux shownTinf19 = 30h19 = 3000// Finite-difference equation from energy balance on CV, Node 13km*(T12 - T13)/deltax + kb*(T14 - T13)/deltax = (rhom*cm + rhob*cb) *(deltax/2)*der(T13,t)(5) The screen from the FEHT analysis is shown below. It is important to use small time steps in theintegration at early times. Use the View|Temperatures command to find the temperature of themixture surface at tres = 50 s.PROBLEM 5.121KNOWN: Thin, circular-disc subjected to induction heating causing a uniform heatgeneration in a prescribed region; upper surface exposed to convection process.FIND: (a) Transient finite-difference equation for a node in the region subjected to inductionheating, (b) Sketch the steady-state temperature distribution on T-r coordinates; identifyimportant features.SCHEMATIC:ASSUMPTIONS: (1) Thickness w << ro, such that conduction is one-dimensional in rdirection, (2) In prescribed region, q is uniform, (3) Bottom surface of disc is insulated, (4)Constant properties.ANALYSIS: (a) Consider the nodal point arrangementfor the region subjected to induction heating. The size ofthe control volume is V = 2 rm r w. The energyconservation requirement for the node m has the formwithEin E out + Eg = Estq a + q b + qconv + qV = Est .Recognizing that qa and qb are conduction terms and qconv is the convection process,pppp Tm Tmr Tr Tk 2 rm w m-1+ k 2 rm + w m+12 2 rr+ h [2 rm r ]()p+1pTm Tmp [ 2 r r w ] = c [2 r r w ]T Tm + q.mpmtUpon regrouping, the finite-difference equation has the form,2 r p r p r T + qr + 1 2Fo Bi Fo r T pp+1Tm = Fo 1 Tm-1 + 1 +Tm+1 + Biw w mk 2rm 2rm whereFo = t/r 2Bi = hr/k.(b) The steady-state temperature distribution hasthese features:1. Zero gradient at r = 0, r02. No discontinuity at r1, r23. Tmax occurs in region r1 < r < r2Note also, distribution will not be linear anywhere;distribution is not parabolic in r1 < r < r2 region.<PROBLEM 5.122KNOWN: An electrical cable experiencing uniform volumetric generation; the lower half is wellinsulated while the upper half experiences convection.FIND: (a) Explicit, finite-difference equations for an interior node (m,n), the center node (0,0), andan outer surface node (M,n) for the convective and insulated boundaries, and (b) Stability criterion foreach FDE; identify the most restrictive criterion.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional (r,), transient conduction, (2) Constant properties, (3)Uniform q.ANALYSIS: The explicit, finite-difference equations may be obtained by applying energy balancesto appropriate control volumes about the node of interest. Note the coordinate system defined abovewhere (r,) (mr, n). The stability criterion is determined from the coefficient associated withthe node of interest.Interior Node (m,n). The control volumefor an interior node isV = rm r (with rm = mr, = 1 ) where is thelength normal to the page. Theconservation of energy requirement isEin E out + Eg = Estp+1pTT(q1 + q 2 )r + (q3 + q 4 ) + qV = cV m,n m,ntppppppTm-1,n Tm,nTm+1,n Tm,nTm,n+1 Tm,n11k m r + k m + r + k r 22rr( mr ) + k r ppTm,n-1 Tm,n( mr ) + q ( mr ) r = c ( mr ) r p+1pTm,n Tm,nt(1)Define the Fourier number ask t tFo =(2)= c r 2 r 2and then regroup the terms of Eq. (1) to obtain the FDE, m 1/2m + 1/2 p1q 2p+1pppTm,n = Fo Tm-1,n +Tm+1,n +T+ Tm,n-1 + r 2 m,n+1mkm( m )2p2 + + 1 Tm,n .(3) <+ Fo ( m )2 Continued ..()PROBLEM 5.122 (Cont.)The stability criterion requires that the last term on the right-hand side in braces be positive. That is,pthe coefficient of Tm,n must be positive and the stability criterion is2Fo 1/2 1 + 1/ ( m ) (4)2Note that, for m >> 1/2 and (m) >>1, the FDE takes the form of a 1-D cartesian system.Center Node (0,0). For the control volume,V = ( r/2 ) 1. The energy balance is2E E + E = E where E = q .inoutgstinnppT1,n Torr k 2 r + q 2 n =0N= c 2p2 p+1 r To To2t(5)where N = (2/) - 1, the total number of qn. Using the definition of Fo, findq1 N pp+1To = 4Fo T1,n + 4k r 2 + (1 4Fo ) Top . < N + 1 n =0By inspection, the stability criterion is Fo 1/4.(7)Surface Nodes (M,n). The control volumefor the surface node is V = (M - )rr/2.1.From the energy balance,E E + E = (q1 + q )r + ( q + q ) + qV = Einoutg234stk ( M 1/2 ) r pppTM-1,n TM,nr+ h ( Mr )()ppr TM,n+1 TM,npT TM,n +k 2p( Mr ) p+1pr TM,n-1 TM,nr r TM,n TM,n.+k + q ( M 1/4 ) r = c ( M 1/4 ) r 222( Mr ) t2Regrouping and using the definitions for Fo = t/r and Bi = hr/k, M 1/21qpppp+1Tm,n = Fo 2TM-1,n +TM,n+1 TM,n-1 + 2Bi T + r 2 2k( M-1/4 ) M ( ) M 1/4 M-1/2M1p TM,n .+ 1 2Fo + Bi +(8) <M 1/4 ( M 1/4 ) M ( )2 M-1/41 M 1/2M1 . (9)The stability criterion is+ Bi+Fo 2 M 1/4M 1/4 ( M 1/4 ) M ( )2 ()To determine which stability criterion is most restrictive, compare Eqs. (4), (7) and (9). The mostrestrictive (lowest Fo) has the largest denominator. For small values of m, it is not evident whetherEq. (7) is more restrictive than Eq. (4); Eq. (4) depends upon magnitude of . Likewise, it is notclear whether Eq. (9) will be more or less restrictive than Eq. (7). Numerical values must besubstituted.PROBLEM 5.123KNOWN: Initial temperature distribution in two bars that are to be soldered together; interfacecontact resistance.FIND: (a) Explicit FDE for T4,2 in terms of Fo and Bi = x/k R ,c ; stability criterion, (b) T4,2tone time step after contact is made if Fo = 0.01 and value of t; whether the stability criterion issatisfied.SCHEMATIC:PROPERTIES: Table A-1, Steel, AISI 1010 (1000K): k = 31.3 W/mK, c = 1168 J/kgK, = 78323kh/m .ASSUMPTIONS: (1) Two-dimensional transient conduction, (2) Constant properties, (3) Interfacialsolder layer has negligible thickness.ANALYSIS: (a) From an energy balance onthe control volume V = (x/2)y1.E in E out + E g = Estp+1pT4,2 T4,2q a + q b + q c + q d = cV.tNote that qa = T/R Ac while the remaining qi are conduction terms,t,c(1R t,c())(ppT3,2 T4,2 y + k ( x/2 )(+ k ( x/2 )ppT4,3 T4,2yppT4,1 T4,2y)+ k ( y()ppT5,2 T4,2)x) = c [(x / 2 ) y] Tp+1p4,2 T4,2t.Defining Fo ( k/ c ) t/x 2 and Bic y/R k, regroup to obtaint,c()T4,2 = Fo T4,3 + 2T5,2 + T4,1 + 2Bi T3,2 + (1 4Fo 2FoBi ) T4,2 .p+1ppppp<pThe stability criterion requires the coefficient of the T4,2 term be zero or positive,(1 4Fo 2FoBi ) 0Fo 1/ ( 4 + 2Bi )or()<(b) For Fo = 0.01 and Bi = 0.020m/ 2 10-5m 2 K/W 31.3W/m K = 31.95,T4,2 = 0.01 (1000 + 2 900 + 1000 + 2 31.95 700 ) K + (1 4 0.01 2 0.01 31.95 )1000Kp+1p+1T4,2 = 485.30K + 321.00K = 806.3K.With Fo = 0.01, the time step ist = Fo x 2 ( c/k ) = 0.01 ( 0.020m )2(7832kg/m3 1168J/kg K/31.3W/m K ) = 1.17s.With Bi = 31.95 and Fo = 0.01, the stability criterion, Fo 0.015, is satisfied.<<<PROBLEM 5.124KNOWN: Stainless steel cylinder of Ex. 5.7, 80-mm diameter by 60-mm length, initially at 600 K,2suddenly quenched in an oil bath at 300 K with h = 500 W/m K. Use the ready-to-solve model in theExamples menu of FEHT to obtain the following solutions.FIND: (a) Calculate the temperatures T(r, x ,t) after 3 min: at the cylinder center, T(0, 0, 3 mm), atthe center of a circular face, T(0, L, 3 min), and at the midheight of the side, T(ro, 0, 3 min); compareyour results with those in the example; (b) Calculate and plot temperature histories at the cylindercenter, T(0, 0, t), the mid-height of the side, T(ro, 0, t), for 0 t 10 min; use the View/Temperaturevs. Time command; comment on the gradients and what effect they might have on phasetransformations and thermal stresses; (c) Using the results for the total integration time of 10 min, usethe View/Temperature Contours command; describe the major features of the cooling process shownin this display; create and display a 10-isotherm temperature distribution for t = 3 min; and (d) For thelocations of part (a), calculate the temperatures after 3 min if the convection coefficient is doubled (h =21000 W/m K); for these two conditions, determine how long the cylinder needs to remain in the oilbath to achieve a safe-to touch surface temperature of 316 K. Tabulate and comment on the results ofyour analysis.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction in r- and x-coordinates, (2) Constant properties.3PROPERTIES: Stainless steel (Example 5.7): = 7900 kg/m , c = 256 J/kgK, k = 17.4 W/mK.ANALYSIS: (a) The FEHT ready-to-solve model for Example 5.7 is accessed through the Examplesmenu and the annotated Input page is shown below. The following steps were used to obtain thesolution: (1) Use the DrawReduce Mesh command three times to create the 512-element mesh; (2)In Run, click on Check, (3) In Run, press Calculate and hit OK to initiate the solver; and (4) Go to theView menu, select Tabular Output and read the nodal temperatures 4, 1, and 3 at t = to = 180 s. Thetabulated results below include those from the n-term series solution used in the IHT software.Continued ..PROBLEM 5.124 (Cont.)(r, x, to)0, 0, to0, L, toro, 0, toFEHT node413T(r, x, to) (K)FEHT402.7368.7362.5T(r, x, to) (K)1-term series405372365T(r, x, to) (K)n-term series402.7370.5362.4Note that the one-term series solution results of Example 5.7 are systematically lower than those fromthe 512-element, finite-difference FEHT analyses. The FEHT results are in excellent agreement withthe IHT n-term series solutions for the x = 0 plane nodes (4,3), except for the x = L plane node (1).(b) Using the View Temperature vs. Time command, the temperature histories for nodes 4, 1, and 3 areplotted in the graph shown below. There is very small temperature difference between the locationson the surface, (node 1; 0, L) and (node 3; ro, 0). But, the temperature difference between thesesurface locations and the cylinder center (node 4; 0, 0) is large at early times. Such differenceswherein locations cool at considerably different rates could cause variations in microstructure andhence, mechanical properties, as well as induce thermal stresses.Continued ..PROBLEM 5.124 (Cont.)(c) Use the View|Temperature Contours command with the shaded band option for the isothermcontours. Selecting the From Start to Stop time option, see the display of the contours as the cylindercools during the quench process. The movie shows that cooling initiates at the corner (ro,L,t) andthe isotherms quickly become circular and travel toward the center (0,0,t). The 10-isothermdistribution for t = 3 min is shown below.2(d) Using the FEHT model with convection coefficients of 500 and 1000 W/m K, the temperatures att = to = 180 s for the three locations of part (a) are tabulated below.22h = 500 W/m KT(0, 0, to), KT(0, L, to), KT(ro, 0, to), Kh = 1000 W/m K402.7368.7362.5352.8325.8322.1Note that the effect of doubling the convection coefficient is to reduce the temperature at theselocations by about 40C. The time the cylinder needs to remain in the oil bath to achieve the safe-totouch surface temperature of 316 K can be determined by examining the temperature history of thelocation (node1; 0, L). For the two convection conditions, the results are tabulated below. Doublingthe coefficient reduces the cooling process time by 40 %.T(0, L, to)3163162h (W/m K)5001000to (s)370219PROBLEM 5.125KNOWN: Flue of square cross-section, initially at a uniform temperature is suddenly exposed to hotflue gases. See Problem 4.57.FIND: Temperature distribution in the wall 5, 10, 50 and 100 hours after introduction of gases usingthe implicit finite-difference method.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional transient conduction, (2) Constant properties.-72PROPERTIES: Flue (given): k = 0.85 W/mK, = 5.5 10 m /s.ANALYSIS: The network representing the flue cross-sectional area is shown with x = y = 50mm.Initially all nodes are at Ti = 25C when suddenly the interior and exterior surfaces are exposed toconvection processes, (T,i, hi) and (T,o, ho), respectively Referring to the network above, note thatthere are four types of nodes: interior (02, 03, 06, 07, 10, 11, 14, 15, 17, 18, 20); plane surfaces withconvection (interior 01, 05, 09); interior corner with convection (13), plane surfaces with convection(exterior 04, 08, 12, 16, 19, 21); and, exterior corner with convection. The system of finitedifference equations representing the network is obtained using IHT|Tools|Finite-differenceequations|Two-dimensional|Transient. The IHT code is shown in Comment 2 and the results for t = 5,10, 50 and 100 hour are tabulated below.p+1p+1p+1p+1p(T18 + T14 + T18 + T14 ) = T17Node 17p+1(1 + 4Fo ) T17 FoNode 131 + 4Fo 1 + 1 Bi T p+1 2 Fo 2T p+1 + T + 2T p+1 + T p+1 = T p + 4 Bi Fo T9i ,i1414913 3 i 1333()p+1p+1p+1p(1 + 2Fo (2 + Bio )) T12 Fo (2T11 + T16 + T8p+1 ) = T12 + 2Bio Fo T,oNode 12p+1p+1p+1p(1 + 4Fo (1 + Bio )) T22 2Fo (T21 + T21 ) = T22 + 4Bio Fo T,oNode 22Numerical values for the relevant parameters are:t 5.5 106 m2 / s 3600sFo =x 2=(0.050m )2= 7.92000h x 5 W/m 2 K 0.050mBio = o== 0.29412k0.85 W/m Kh x 100 W/m 2 K 0.050mBii = i== 5.88235k0.85 W/m KThe system of FDEs can be represented in matrix notation, [A][T] = [C]. The coefficient matrix [A]and terms for the right-hand side matrix [C] are given on the following page.Continued ..PROBLEM 5.125 (Cont.)For this problem a stock computer program was used to obtain the solution matrix [T]. The0initial temperature distribution was Tm = 298K. The results are tabulated below.T(m,n) (C)Node/time(h)T0105105010025335.00338.90340.20340.20T0225248.00274.30282.90282.90T0325179.50217.40229.80229.80T0425135.80170.30181.60181.60T0525334.50338.50339.90339.90T0625245.30271.90280.80280.80T0725176.50214.60227.30227.30T0825133.40168.00179.50179.50T0925332.20336.60338.20338.20T1025235.40263.40273.20273.20T1125166.40205.40219.00219.00T1225125.40160.40172.70172.70T1325316.40324.30327.30327.30T1425211.00243.00254.90254.90T1525146.90187.60202.90202.90T1625110.90146.70160.20160.20T1725159.80200.50216.20216.20T1825117.40160.50177.50177.50T192590.97127.40141.80141.80T202590.62132.20149.00149.00T212572.43106.70120.60120.60T222559.4787.3798.8998.89COMMENTS: (1) Note that the steady-state condition is reached by t = 5 hours; this can be seen bycomparing the distributions for t = 50 and 100 hours. Within 10 hours, the flue is within a fewdegrees of the steady-state condition.Continued ..PROBLEM 5.125 (Cont.)(2) The IHT code for performing the numerical solution is shown in its entirety below. Use has beenmade of symmetry in writing the FDEs. The tabulated results above were obtained by copying fromthe IHT Browser and pasting the desired columns into EXCEL.// From Tools|Finite-difference equations|Two-dimensional|Transient// Interior surface nodes, 01, 05, 09, 13/* Node 01: plane surface node, s-orientation; e, w, n labeled 05, 05, 02 . */rho * cp * der(T01,t) = fd_2d_psur_s(T01,T05,T05,T02,k,qdot,deltax,deltay,Tinfi,hi,q''a)q''a = 0// Applied heat flux, W/m^2; zero flux shownqdot = 0rho * cp * der(T05,t) = fd_2d_psur_s(T05,T09,T01,T06,k,qdot,deltax,deltay,Tinfi,hi,q''a)rho * cp * der(T09,t) = fd_2d_psur_s(T09,T13,T05,T10,k,qdot,deltax,deltay,Tinfi,hi,q''a)/* Node 13: internal corner node, w-s orientation; e, w, n, s labeled 14, 09, 14, 09. */rho * cp * der(T13,t) = fd_2d_ic_ws(T13,T14,T09,T14,T09,k,qdot,deltax,deltay,Tinfi,hi,q''a)// Interior nodes, 02, 03, 06, 07, 10, 11, 14, 15, 18, 20/* Node 02: interior node; e, w, n, s labeled 06, 06, 03, 01. */rho * cp * der(T02,t) = fd_2d_int(T02,T06,T06,T03,T01,k,qdot,deltax,deltay)rho * cp * der(T03,t) = fd_2d_int(T03,T07,T07,T04,T02,k,qdot,deltax,deltay)rho * cp * der(T06,t) = fd_2d_int(T06,T10,T02,T07,T05,k,qdot,deltax,deltay)rho * cp * der(T07,t) = fd_2d_int(T07,T11,T03,T08,T06,k,qdot,deltax,deltay)rho * cp * der(T10,t) = fd_2d_int(T10,T14,T06,T11,T09,k,qdot,deltax,deltay)rho * cp * der(T11,t) = fd_2d_int(T11,T15,T07,T12,T10,k,qdot,deltax,deltay)rho * cp * der(T14,t) = fd_2d_int(T14,T17,T10,T15,T13,k,qdot,deltax,deltay)rho * cp * der(T15,t) = fd_2d_int(T15,T18,T11,T16,T14,k,qdot,deltax,deltay)rho * cp * der(T17,t) = fd_2d_int(T17,T18,T14,T18,T14,k,qdot,deltax,deltay)rho * cp * der(T18,t) = fd_2d_int(T18,T20,T15,T19,T17,k,qdot,deltax,deltay)rho * cp * der(T20,t) = fd_2d_int(T20,T21,T18,T21,T18,k,qdot,deltax,deltay)// Exterior surface nodes, 04, 08, 12, 16, 19, 21, 22/* Node 04: plane surface node, n-orientation; e, w, s labeled 08, 08, 03. */rho * cp * der(T04,t) = fd_2d_psur_n(T04,T08,T08,T03,k,qdot,deltax,deltay,Tinfo,ho,q''a)rho * cp * der(T08,t) = fd_2d_psur_n(T08,T12,T04,T07,k,qdot,deltax,deltay,Tinfo,ho,q''a)rho * cp * der(T12,t) = fd_2d_psur_n(T12,T16,T08,T11,k,qdot,deltax,deltay,Tinfo,ho,q''a)rho * cp * der(T16,t) = fd_2d_psur_n(T16,T19,T12,T15,k,qdot,deltax,deltay,Tinfo,ho,q''a)rho * cp * der(T19,t) = fd_2d_psur_n(T19,T21,T16,T18,k,qdot,deltax,deltay,Tinfo,ho,q''a)rho * cp * der(T21,t) = fd_2d_psur_n(T21,T22,T19,T20,k,qdot,deltax,deltay,Tinfo,ho,q''a)/* Node 22: external corner node, e-n orientation; w, s labeled 21, 21. */rho * cp * der(T22,t) = fd_2d_ec_en(T22,T21,T21,k,qdot,deltax,deltay,Tinfo,ho,q''a)// Input variablesdeltax = 0.050deltay = 0.050Tinfi = 350hi = 100Tinfo = 25ho = 5k = 0.85alpha = 5.55e-7alpha = k / (rho * cp)rho = 1000// arbitrary value(3) The results for t = 50 hour, representing the steady-state condition, are shown below, arrangedaccording to the coordinate system.Tmn (C)x/y (mm)0501001502002503000181.60179.50172.70160.20141.80120.6098.89149.0050229.80227.30219.00202.90177.50100282.90280.80273.20172.70216.20150340.20339.90338.20327.30In Problem 4.57, the temperature distribution was determined using the FDEs written for steady-stateconditions, but with a finer network, x = y = 25 mm. By comparison, the results for the coarsernetwork are slightly higher, within a fraction of 1C, along the mid-section of the flue, but notablyhigher in the vicinity of inner corner. (For example, node 13 is 2.6C higher with the coarser mesh.)PROBLEM 5.126KNOWN: Electrical heating elements embedded in a ceramic plate as described in Problem4.75; initially plate is at a uniform temperature and suddenly heaters are energized.FIND: Time required for the difference between the surface and initial temperatures to reach95% of the difference for steady-state conditions using the implicit, finite-difference method.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction, (2) Constant properties, (3) No internalgeneration except for Node 7, (4) Heating element approximates a line source; wire diameteris negligible.ANALYSIS: The grid for the symmetry element above consists of 12 nodes. Nodes 1-3 arepoints on a surface experiencing convection; nodes 4-12 are interior nodes; node 7 is a specialcase with internal generation and because of symmetry, q t = 25 W/m. Their finitehdifference equations are derived as followsSurface Node 2. From an energy balance on the prescribed control volume with x/y = 3,pT p+1 T2 = E = q + q + q + q = cV 2Einstabcdtp+1y T1k2p+1 T2xp+1y T3+k2(p+1+ hx T T2p+1 T2x+ kx)p+1p+1T5 T2yp+1p y T2 T2.= c x2tContinued ..PROBLEM 5.126 (Cont.)Divide by k, use the following definitions, and regroup to obtain the finite-differenceequations.N hx/k = 100 W/m 2 K 0.006m/2 W/m K = 0.3000Fo ( k/ c ) t/x y = t/x y =(1)1.5 106 m 2 / s 1s/ (0.006 0.002 ) m 2 = 0.1250()()(2)(1 y p+1p+1p+1 1 y p+1p+1T1 T2+ N T T2+ T3 T2 x 2 2 x x 1p+1p+1pT p+1 T2+ T5 T2=y 2Fo 2()())1 y p+1 x 1 p+1 1 x p+1 y T1 + N + ++ T3T2 x 2 y x 2Fo 2 y x p+11p+ T5 = NT T.2Fo 2 y (3)Substituting numerical values for Fo and N, and using T = 30C and x/y = 3, findp+1p+1p+1p+1p0.16667T1 7.63333T2 + 0.16667T3 + 3.00000T5 = 9.0000 4.0000T2 .(4)By inspection and use of Eq. (3), the FDEs for Nodes 1 and 3 can be inferred.Interior Node 7. From an energy balance on theprescribed control volume with x/y = 3,E + E = Estingwhere E = 2q and E represents theghtinconduction terms q + q + q + q ,abcdkyp+1p+1T8 T7x+ kxp+1p+1T4 T7yp+1p+1T10 T7+ kyp+1p+1T8 T7x+ 2q = c ( x y )htyUsing the definition of Fo, Eq. (2), and regrouping, find1 x p+1 x y 1 p+1 y T4 y + x + 2Fo T72 q1p y p+1 1 x p+1+ T8 + T10 = ht T2 y k2Fo 7 x + kxp+1pT7 T7t.(5)p+1p+1p+1p+1p1.50000T4 7.33333T7 + 0.33333T8 + 1.50000T10 = 12.5000 4.0000T7 .Continued ..(6)PROBLEM 5.126 (Cont.)Recognizing the form of Eq. (5), it is a simple matter to infer the FDE for the remaininginterior points for which q ht = 0. In matrix notation [A][T] = [C], the coefficient matrix [A]and RHS matrix [C] are:Recall that the problem asks for the time required to reach95% of the difference for steady-state conditions. Thisprovides information on approximately how long it takesfor the plate to come to a steady operating condition. Ifyou worked Problem 4.71, you know the steady-statetemperature distribution. Then you can proceed to find thepTm values with increasing time until the first node reachesthe required limit. We should not expect the nodes to reach their limit at the same time.Not knowing the steady-state temperature distribution, use the implicit FDE in matrix formabove to step through time to the steady-state solution; that is, proceed to p 10,20100 until the solution matrix [T] does not change. The results of the analysis aretabulated below. Column 1 labeled Tm() is the steady-state distribution. Column 2,Tm(95%), is the 95% limit being sought as per the graph directly above. The third column isthe temperature distribution at t = to = 248s, Tm(248s); at this elapsed time, Node 1 hasreached its limit. Can you explain why this node was the first to reach this limit? Whichnodes will be the last to reach their limits?Tm()55.8049.9347.6759.0351.7249.1963.8952.9850.1462.8453.3550.46Tm(95%)54.5148.9346.7857.5850.6348.2362.2051.8349.1361.2052.1849.43Tm(248s)54.5148.6446.3857.6450.3247.7962.4251.5248.6861.3551.8648.98<PROBLEM 5.127KNOWN: Nodal network and operating conditions for a water-cooled plate.FIND: Transient temperature response.SCHEMATIC:ASSUMPTIONS: (1) Steady-sate conditions, (2) Two-dimensional conduction.ANALYSIS: The energy balance method must be applied to each nodal region. Groupingsimilar regions, the following results are obtained.Nodes 1 and 5: 2t 2t p+1 2t p+1 2t p+1pTT+= T11 +T22 12226xy xy 2t 2t p+1 2t p+1 2t p+1pTT+= T51 +T22 5242 10xy xyNodes 2, 3, 4: 2t 2t p+1 t p+12t p+1t p+1pTTT+= Tm,n1 +T22 m,n x 2 m-1,n x 2 m+1,n y 2 m,n-1xy Nodes 6 and 14: 2t 2t 2ht p+1 2t p+1 2t p+1 2htpTTT +T6++=1 + T6 222127ky kyxyyx 2t 2t 2ht p+1 2t p+1 2t p+1 2htpTTT +T14++=1 + T14 222 152 19ky kyxyxyContinued ..PROBLEM 5.127 (Cont.)Nodes 7 and 15: 2t 2t 2ht p+1 2t p+1 t p+1 t p+1 2htp++=TTTT +T71 + T7 x 2 y2222628ky kyyxkx 2t 2t 2ht p+1 t p+1 t p+1 2t p+1 2htp++=TTTT +T151 + T15 x 2 y22 142 162 20ky kyxxyNodes 8 and 16: 2t 2t 2 ht 2 ht p+1 4 t p+1 2 t p+1TT+++1 + T8 3 y 2 33 x 2 7x 2 y 2 3 kx 3 ky 4 t p+1 2 t p+1 2 ht 11pT9 T11 =+ T + T83 x 23 y 23 k x y 2t 2t 2 ht 2 ht p+1 2 t p+1 2 t p+1TT++++1 + T16 3 y 2 113 x 2 15x 2 y 2 3 kx 3 ky 4 t p+1 4 t p+1 2 ht 11pT17 T21 =+ T + T16223 x3 y3 k x y Node 11: 2t 2t 2ht p+1 t p+1t p+1 t p+1 2htpT +T111 + 2 + 2 + T11 2 T8 2 2 T12 2 T16 =kx kxyyxy xNodes 9, 12, 17, 20, 21, 22: 2t 2t p+1 t p+1t p+1p+1p+1pTm,n+1 + Tm,n-1 Tm-1,n + Tm+1,n = Tm,n+1 + Tm,n x 2 y 2 y 2x 2Nodes 10, 13, 18, 23: 2t 2t p+1 t p+12t p+1p+1pTm,n+1 + Tm,n-1 Tm-1,n = Tm,n+1 + Tm,n x 2 y 2 y 2x 2Node 19: 2t 2t p+1 t p+1p+1 2t p+1pT14 + T24 T20 = T19+1 + T19 x 2 y 2 y 2x 2Nodes 24, 28: 2t 2t p+1 2t p+1 2t p+1 2q t pTT+T24+=o1 +T22 242 192 25kyxy yx 2t 2t p+1 2t p+1 2t p+1 2q t pTT+T28+=o1 +T22 282 232 27kyxy yx()()(())Continued ..PROBLEM 5.127 (Cont.)Nodes 25, 26, 27:)( 2t 2t p+1 2t p+12q t p+1t p+1p+1Tm,n+1 Tm-1,n + Tm+1,n = o+Tm,n+1 + Tm,n kyx 2 y 2 y 2x 2The convection heat rate isqconv = h [( x/2 )( T6 T ) + x ( T7 T ) + ( x + y )( T8 T ) / 2 + y ( T11 T ) + ( x+y )( T16 T ) / 2 + x ( T15 T ) + ( x/2 )( T14 T ) = q out.The heat input isq = q ( 4x )inoand, on a percentage basis, the ratio isn (qconv / q ) 100.inResults of the calculations (in C) are as follows:Time: 5.00 sec;n = 60.57%19.61219.44619.71219.59724.21725.65827.58124.07425.60827.55419.97420.10521.37023.55825.48527.49320.20620.49021.64723.49425.41727.446Time: 15.0 sec;23.36323.09628.29430.06332.09528.15530.01832.07223.71623.76125.14227.65229.90832.02120.29220.60921.73023.48325.39627.429n = 94.89%23.22822.896Time: 10.00 sec;Time: 23.00 sec;23.66323.31123.80223.51628.78230.59132.63628.64430.54632.61324.04224.31725.59427.69429.86731.98722.269 22.394 22.72321.981 22.167 22.79124.14327.216 27.075 26.56928.898 28.851 28.73830.901 30.877 30.823Time: 20.00 sec;24.16524.49125.73327.71929.85731.976n = 85.80%23.57423.22623.71223.43028.68230.48332.52528.54330.43832.50223.02523.30224.54826.58328.69030.78623.13723.46124.67326.59828.67730.773n = 98.16%24.07324.11025.50228.04230.33032.45224.40924.68225.97028.09430.29132.41924.53524.86126.11528.12230.28232.409n = 99.00%24.16524.20025.59528.14330.43832.56324.50324.77626.06728.19830.40032.53124.63024.95726.21428.22630.39232.520COMMENTS: Temperatures at t = 23 s are everywhere within 0.13C of the final steadystate values.PROBLEM 5.128KNOWN: Cubic-shaped furnace, with prescribed operating temperature and convection heat transferon the exterior surfaces.FIND: Time required for the furnace to cool to a safe working temperature corresponding to an innerwall temperature of 35C considering convection cooling on (a) the exterior surfaces and (b) on boththe exterior and interior surfaces.SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction through the furnace walls and (2) Constantproperties.ANALYSIS: Assuming two-dimensional conduction through the walls and taking advantage ofsymmetry for the cubical shape, the analysis considers the quarter section shown in the schematicabove. For part (a), with no cooling on the interior during the cool-down process, the inner surfaceboundary condition is adiabatic. For part (b), with cooling on both the exterior and interior, theboundary conditions are prescribed by the convection process. The boundaries through the centerlineof the wall and the diagonal through the corner are symmetry planes and considered as adiabatic. Wehave chosen to use the finite-element software FEHT as the solution tool.Using FEHT, an outline of the symmetrical wall section is drawn, and the material properties arespecified. To determine the initial conditions for the cool-down process, we will first find thetemperature distribution for steady-state operation. As such, specify the boundary condition for theinner surface as a constant temperature of 900C; the other boundaries are as earlier described. In theSetup menu, click on Steady-State, and then Run to obtain the steady-state temperature distribution.This distribution represents the initial temperature distribution, Ti (x, y, 0), for the wall at the onset ofthe cool-down process.Next, in the Setup menu, click on Transient; for the nodes on the inner surface, in the Specify |Boundary Conditions menu, deselect the Temperature box (900C) and set the Flux box to zero for theadiabatic condition (part (a)); and, in the Run command, click on Continue (not Calculate). Be sure tochange the integration time scale from seconds to hours.Because of the high ratio of wall section width (nearly 8.5 m) to the thickness (1 m), the conductionheat transfer through the section is nearly one-dimensional. We chose the x,y-section 1 m to the rightof the centerline (1 m, y) as the location for examining the temperature-time history, and determiningthe cool-down time for the inner surface to reach the safe working temperature of 35C.Continued ..PROBLEM 5.128 (Cont.)Time-to-cool, Part (a), Adiabatic inner surface. From the above temperature history, the cool-downtime, ta, corresponds to the condition when Ta (1 m, 0, ta) = 35C. As seen from the history, thislocation is the last to cool. From the View | Tabular Output, find that<t a = 1306 h = 54 daysContinued ..PROBLEM 5.128 (Cont.)Time-to-cool, Part (b), Cooled inner surface. From the above temperature history, note that the centerportion of the wall, and not the inner surface, is the last to cool. The inner surface cools to 35C inapproximately 175 h or 7 days. However, if the cooling process on the inner surface werediscontinued, its temperature would increase and eventually exceed the desired safe workingtemperature. To assure the safe condition will be met, estimate the cool down time as, tb,corresponding to the condition when Tb (1 m, 0.75 m, tb) = 35C. From the View | Tabular Output,find thatt b = 311 h = 13 days<COMMENTS: (1) Assuming the furnace can be approximated by a two-dimensional symmetricalsection greatly simplifies our analysis by not having to deal with three-dimensional corner effects. Wejustify this assumption on the basis that the corners represent a much shorter heat path than the straightwall section. Considering corner effects would reduce the cool-down time estimates; hence, ouranalysis provides a conservative estimate.(2) For background information on the Continue option, see the Run menu in the FEHT Help section.Using the Run | Calculate command, the steady-state temperature distribution was determined for thenormal operating condition of the furnace. Using the Run | Continue command (after clicking onSetup | Transient), this steady-state distribution automatically becomes the initial temperaturedistribution for the cool-down transient process. This feature allows for conveniently prescribing anon-uniform initial temperature distribution for a transient analysis (rather than specifying values on anode-by-node basis.PROBLEM 5.129KNOWN: Door panel with ribbed cross-section, initially at a uniform temperature of 275C, isejected from the hot extrusion press and experiences convection cooling with ambient air at 25C and2a convection coefficient of 10 W/m K.FIND: (a) Using the FEHT View|Temperature vs. Time command, create a graph with temperaturetime histories of selected locations on the panel surface, T(x,0,t). Comment on whether you seenoticeable differential cooling in the region above the rib that might explain the appearance defect; andUsing the View|Temperature Contours command with the shaded-band option for the isothermcontours, select the From start to stop time option, and view the temperature contours as the panelcools. Describe the major features of the cooling process you have seen. Use other options of thiscommand to create a 10-isotherm temperature distribution at some time that illustrates importantfeatures. How would you re-design the ribbed panel in order to reduce this thermally induced paintdefect situation, yet retain the stiffening function required of the ribs?SCHEMATIC:ASSUMPTIONS: (1) Two-dimensional conduction in the panel, (2) Uniform convection coefficientover the upper and lower surfaces of the panel, (3) Constant properties.3PROPERTIES: Door panel material (given): = 1050 kg/m , c = 800 J/kgK, k = 0.5 W/mK.ANALYSIS:(a) Using the Draw command, the shape of the symmetrical element of the panel (darkened region inschematic) was generated and elements formed as shown below. The symmetry lines representadiabatic surfaces, while the boundary conditions for the exposed web and rib surfaces arecharacterized by (T, h).Continued ..PROBLEM 5.129 (Cont.)After running the calculation for the time period 0 to 400 s with a 1-second time step, the temperaturetime histories for three locations were obtained and the graph is shown below.As expected, the region directly over the rib (0,0) cooled the slowest, while the extreme portion of theweb (0, 13 mm) cooled the fastest. The largest temperature differences between these two locationsoccur during the time period 50 to 150 s. The maximum difference does not exceed 25C.Continued ..PROBLEM 5.129 (Cont.)(b) It is possible that the temperature gradients within the web-rib regions rather than just the uppersurface temperature differentials might be important for understanding the panels response tocooling. Using the Temperature Contours command (with the From start to stop option), we saw thatthe center portion of the web and the end of the rib cooled quickly, but that the region on the ribcenterline (0, 3-5 mm), was the hottest region. The isotherms corresponding to t = 100 s are shownbelow. For this condition, the temperature differential is about 21C.From our analyses, we have identified two possibilities to consider. First, there is a significant surfacetemperature distribution across the panel during the cooling process. Second, the web and theextended portion of the rib cool at about the same rate, and with only a modest normal temperaturegradient. The last region to cool is at the location where the rib is thickest (0, 3-5 mm). The largetemperature gradient along the centerline toward the surface may be the cause of microstructurevariations, which could influence the adherence of paint. An obvious re-design consideration is toreduce the thickness of the rib at the web joint, thereby reducing the temperature gradients in thatregion. This fix comes at the expense of decreasing the spacing between the ribs.PROBLEM 6.1KNOWN: Variation of hx with x for laminar flow over a flat plate.FIND: Ratio of average coefficient, h x , to local coefficient, hx, at x.SCHEMATIC:ANALYSIS: The average value of hx between 0 and x is1xCx h x dx = x -1/2dxx0x0C 1/2= 2x = 2Cx -1/2x= 2h x .hx =hxhxHence,hx= 2.hx<COMMENTS: Both the local and average coefficients decrease with increasing distance xfrom the leading edge, as shown in the sketch below.PROBLEM 6.2KNOWN: Variation of local convection coefficient with x for free convection from a verticalheated plate.FIND: Ratio of average to local convection coefficient.SCHEMATIC:ANALYSIS: The average coefficient from 0 to x is1xC x -1/4h x = h x dx = xdxx0x04 C 3/4 44hx =x= C x -1/4 = h x .3x33Hence,hx 4=.hx 3<The variations with distance of the local and average convection coefficients are shown in thesketch.COMMENTS: Note that h x / h x = 4 / 3 is independent of x. Hence the average coefficient4for an entire plate of length L is h L = h L , where hL is the local coefficient at x = L. Note3also that the average exceeds the local. Why?PROBLEM 6.3KNOWN: Expression for the local heat transfer coefficient of a circular, hot gas jet at Tdirected normal to a circular plate at Ts of radius ro.FIND: Heat transfer rate to the plate by convection.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Flow is axisymmetric about the plate, (3)For h(r), a and b are constants and n -2.ANALYSIS: The convective heat transfer rate to the plate follows from Newtons law ofcoolingq conv = dq conv = h ( r ) dA ( T Ts ).AAThe local heat transfer coefficient is known to have the form,h ( r ) = a + br nand the differential area on the plate surface isdA = 2 r dr.Hence, the heat rate isq conv = ro0(a + brn ) 2 r dr (T Ts )rb n+2 oaq conv = 2 ( T Ts ) r 2 +rn+220b n+2 a 2q conv = 2 ro +ro ( T Ts ) .n+22<COMMENTS: Note the importance of the requirement, n -2. Typically, the radius of thejet is much smaller than that of the plate.PROBLEM 6.4KNOWN: Distribution of local convection coefficient for obstructed parallel flow over a flatplate.FIND: Average heat transfer coefficient and ratio of average to local at the trailing edge.SCHEMATIC:ANALYSIS: The average convection coefficient is()1L1Lh x dx = 0.7 + 13.6x 3.4x 2 dxL 0L01hL =0.7L + 6.8L2 1.13L3 = 0.7 + 6.8L 1.13L2LhL =()h L = 0.7 + 6.8 (3) 1.13 (9 ) = 10.9 W/m 2 K.<The local coefficient at x = 3m ish L = 0.7 + 13.6 (3) 3.4 (9 ) = 10.9 W/m 2 K.Hence,<h L / h L = 1.0.COMMENTS: The result h L / h L = 1.0 is unique to x = 3m and is a consequence of theexistence of a maximum for h x x . The maximum occurs at x = 2m, where(dh x / dx ) = 0 and$(d2h x / dx2 < 0.)PROBLEM 6.5KNOWN: Temperature distribution in boundary layer for air flow over a flat plate.FIND: Variation of local convection coefficient along the plate and value of average coefficient.SCHEMATIC:ANALYSIS: From Eq. 6.17,h=k T y y =0k ( 70 600x )=+(Ts T )(Ts T )where Ts = T(x,0) = 90C. Evaluating k at the arithmetic mean of the freestream and surfacetemperatures, T = (20 + 90)C/2 = 55C = 328 K, Table A.4 yields k = 0.0284 W/mK. Hence, withTs - TW = 70C = 70 K,h=0.0284 W m K ( 42, 000x ) K m70 K(= 17x W m 2 K)<and the convection coefficient increases linearly with x.The average coefficient over the range 0 x 5 m is51L17 517 x 2h = hdx = xdx == 42.5 W m 2 K00L5520<PROBLEM 6.6KNOWN: Variation of local convection coefficient with distance x from a heated plate with auniform temperature Ts.FIND: (a) An expression for the average coefficient h12 for the section of length (x2 - x1) in terms ofC, x1 and x2, and (b) An expression for h12 in terms of x1 and x2, and the average coefficients h1 andh 2 , corresponding to lengths x1 and x2, respectively.SCHEMATIC:ASSUMPTIONS: (1) Laminar flow over a plate with uniform surface temperature, Ts, and (2)Spatial variation of local coefficient is of the form h x = Cx 1/ 2 , where C is a constant.ANALYSIS: (a) The heat transfer rate per unit width from a longitudinal section, x2 - x1, can beexpressed asq12 = h12 ( x 2 x1 )( Ts T )(1)where h12 is the average coefficient for the section of length (x2 - x1). The heat rate can also bewritten in terms of the local coefficient, Eq. (6.3), asxxq12 = 2 h x dx ( Ts T ) = ( Ts T ) 2 h x dxx1x1Combining Eq. (1) and (2),x21h12 =h dx( x 2 x1 ) x1 xand substituting for the form of the local coefficient, h x = Cx 1/ 2 , find thatx2x1/ 2 x1/ 2x21C x1/ 2 1/ 2dx =1h12 =Cx = 2C 2x 2 x1 ) x1x 2 x1 1/ 2 x 2 x1( x1(b) The heat rate, given as Eq. (1), can also be expressed asq12 = h 2 x 2 ( Ts T ) h1x1 ( Ts T )(2)(3)(4)<(5)which is the difference between the heat rate for the plate over the section (0 - x2) and over the section(0 - x1). Combining Eqs. (1) and (5), find,h x h xh12 = 2 2 1 1x 2 x1(6)COMMENTS: (1) Note that, from Eq. 6.6,1x1 x 1/ 2hx =h x dx =Cxdx = 2Cx 1/ 220x0or h x = 2hx. Substituting Eq. (7) into Eq. (6), see that the result is the same as Eq. (4).<(7)PROBLEM 6.7KNOWN: Radial distribution of local convection coefficient for flow normal to a circulardisk.FIND: Expression for average Nusselt number.SCHEMATIC:ASSUMPTIONS: Constant propertiesANALYSIS: The average convection coefficient is1hdAsA s As1 ro knh=0 D Nuo 1 + a ( r/ro ) 2 rdr2 roh=rokNu o r 2ar n+2 +h=3nro 2 ( n + 2 ) ro 0where Nuo is the Nusselt number at the stagnation point (r = 0). Hence,ro r/r 2( o ) + a r n + 2 hDNu D == 2Nu o 2k( n+2 ) ro 0Nu D = Nu o 1 + 2a/ ( n + 2 )Nu D = 1+ 2a/ ( n + 2 ) 0.814Re1/2 Pr 0.36 .D<COMMENTS: The increase in h(r) with r may be explained in terms of the sharp turn whichthe boundary layer flow must make around the edge of the disk. The boundary layeraccelerates and its thickness decreases as it makes the turn, causing the local convectioncoefficient to increase.PROBLEM 6.8KNOWN: Convection correlation and temperature of an impinging air jet. Dimensions and initialtemperature of a heated copper disk. Properties of the air and copper.FIND: Effect of jet velocity on temperature decay of disk following jet impingement.SCHEMATIC:ASSUMPTIONS: (1) Validity of lumped capacitance analysis, (2) Negligible heat transfer from sidesand bottom of disk, (3) Constant properties.ANALYSIS: Performing an energy balance on the disk, it follows thatEst = Vc dT dt = As (qconv + q ) . Hence, with V = AsL,radh (T T ) + h r (T Tsur )dT= cLdt()2where, h r = (T + Tsur ) T 2 + Tsur and, from the solution to Problem 6.7,h=kk2a 1/ 2 0.36Nu D = 1 + 0.814 ReD PrDD n+2With a = 0.30 and n = 2, it follows thath = ( k D ) 0.936 Re1/ 2 Pr 0.36Dwhere ReD = VD/. Using the Lumped Capacitance Model of IHT, the following temperature historieswere determined.Continued ..PROBLEM 6.8 (Cont.)1000Temperature, T(K)900800700600500400300050010001500200025003000Time, t(s)V = 4 m/sV = 20 m/sV = 50 m/sThe temperature decay becomes more pronounced with increasing V, and a final temperature of 400 K isreached at t = 2760, 1455 and 976s for V = 4, 20 and 50 m/s, respectively.()COMMENTS: The maximum Biot number, Bi = h + h r L k Cu , is associated with V = 50 m/s(maximum h of 169 W/m K) and t = 0 (maximum hr of 64 W/m2K), in which case the maximum Biotnumber is Bi = (233 W/m2K)(0.025 m)/(386 W/mK) = 0.015 < 0.1. Hence, the lumped capacitanceapproximation is valid.2PROBLEM 6.9KNOWN: Local convection coefficient on rotating disk. Radius and surface temperature of disk.Temperature of stagnant air.FIND: Local heat flux and total heat rate. Nature of boundary layer.SCHEMATIC:ASSUMPTIONS: (1) Negligible heat transfer from back surface and edge of disk.ANALYSIS: If the local convection coefficient is independent of radius, the local heat flux at everypoint on the disk isq = h ( Ts T ) = 20 W / m 2 K (50 20 ) C = 600 W / m 2<Since h is independent of location, h = h = 20 W / m 2 K and the total power requirement is2Pelec = q = hAs (Ts T ) = h ro (Ts T )()Pelec = 20 W / m 2 K (0.1m )2(50 20 ) C = 18.9 W<If the convection coefficient is independent of radius, the boundary layer must be of uniformthickness . Within the boundary layer, air flow is principally in the circumferential direction. Thecircumferential velocity component u corresponds to the rotational velocity of the disk at the surface(y = 0) and increases with increasing r (u = r). The velocity decreases with increasing distance yfrom the surface, approaching zero at the outer edge of the boundary layer (y ).PROBLEM 6.10KNOWN: Form of the velocity and temperature profiles for flow over a surface.FIND: Expressions for the friction and convection coefficients.SCHEMATIC:ANALYSIS: The shear stress at the wall iss = u= A + 2By 3Cy 2 y=0 = A . y y=0Hence, the friction coefficient has the form,Cf =Cf =s2 u / 22A2u=2A2 u.<The convection coefficient is2 k f ( T/ y )y=0 k f E + 2Fy 3Gy y=0h==Ts TD Th=k f E.D T<COMMENTS: It is a simple matter to obtain the important surface parameters fromknowledge of the corresponding boundary layer profiles. However, it is rarely a simple matterto determine the form of the profile.PROBLEM 6.11KNOWN: Surface temperatures of a steel wall and temperature of water flowing over thewall.FIND: (a) Convection coefficient, (b) Temperature gradient in wall and in water at wallsurface.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in x, (3)Constant properties.PROPERTIES: Table A-1, Steel Type AISI 1010 (70C = 343K), ks = 61.7 W/mK; TableA-6, Water (32.5C = 305K), kf = 0.62 W/mK.ANALYSIS: (a) Applying an energy balance to the control surface at x = 0, it follows thatqx,cond qx,conv = 0and using the appropriate rate equations,Ts,2 Ts,1ks= h Ts,1 T .LHence,k s Ts,2 Ts,1 61.7 W/m K 60 Ch=== 705 W/m 2 K.L Ts,1 T0.35m15 C()<(b) The gradient in the wall at the surface is(dT/dx )s = Ts,2 Ts,1L=60 C= 171.4 C/m.0.35mIn the water at x = 0, the definition of h gives()(dT/dx )f,x=0 = hTs,1 Tkf(dT/dx )f,x=0 = 705 W/m 2 K 15 C = 17, 056 C/m.0.62 W/m K()COMMENTS: Note the relative magnitudes of the gradients. Why is there such a largedifference?<PROBLEM 6.12KNOWN: Boundary layer temperature distribution.FIND: Surface heat flux.SCHEMATIC:PROPERTIES: Table A-4, Air (Ts = 300K): k = 0.0263 W/mK.ANALYSIS: Applying Fouriers law at y = 0, the heat flux isu yTu= k ( T Ts ) Pr exp Pr y y=0 y=0 u= k ( T Ts ) Pr = 0.0263 W/m K (100K ) 0.7 5000 1/m.q = ksqsqsq = 9205 W/m 2 .sCOMMENTS: (1) Negative flux implies convection heat transfer to the surface.(2) Note use of k at Ts to evaluate q from Fouriers law.s<PROBLEM 6.13KNOWN: Air flow over a flat plate of length L = 1 m under conditions for which transition fromlaminar to turbulent flow occurs at xc = 0.5m based upon the critical Reynolds number, Re x ,c = 5 105.Forms for the local convection coefficients in the laminar and turbulent regions.FIND: (a) Velocity of the air flow using thermophysical properties evaluated at 350 K, (b) An expressionfor the average coefficient h lan ( x ) , as a function of distance from the leading edge, x, for the laminarregion, 0 x xc, (c) An expression for the average coefficient h turb ( x ) , as a function of distancefrom the leading edge, x, for the turbulent region, xc x L, and (d) Compute and plot the local andaverage convection coefficients, hx and h x , respectively, as a function of x for 0 x L.SCHEMATIC:ASSUMPTIONS: (1) Forms for the local coefficients in the laminar and turbulent regions, hlam =Clamx-0.5 and htirb = Cturbx-0.2 where Clam = 8.845 W/m3/2K, Cturb = 49.75 W/m2K0.8, and x has units (m).PROPERTIES: Table A.4, Air (T = 350 K): k = 0.030 W/mK, = 20.92 10-6 m2/s, Pr = 0.700.ANALYSIS: (a) Using air properties evaluated at 350 K with xc = 0.5 m,uxu = 5 105 x c = 5 105 20.92 10 6 m 2 s 0.5 m = 20.9 m sRe x,c = c = 5 105(b) From Eq. 6.5, the average coefficient in the laminar region, 0 x xc, isx1x11(1)hlam ( x ) = h lam ( x ) dx = Clam x 0.5dx = Clam x 0.5 = 2Clam x 0.5 = 2h lam ( x )ox0xx(c) The average coefficient in the turbulent region, xc x L, is<<xcxx1 xcx 0.5x 0.8h turb ( x ) =h lam ( x ) dx + h turb ( x ) dx = Clam+ C turbxcx 00.50.8h turb ( x ) =)(100.50.80.8 2Clam x c + 1.25C turb x x c xxc (2)<Convection coefficient (W/m^2.K)(d) The local and average coefficients, Eqs. (1) and (2) are plotted below as a function of x for the range0 x L.15010050000.5Distance from leading edge, x (m)Local - laminar, x <= xcLocal - turbulent, x => xcAverage - laminar, x <= xcAverage - turbulent, x => xc1PROBLEM 6.14KNOWN: Air speed and temperature in a wind tunnel.8FIND: (a) Minimum plate length to achieve a Reynolds number of 10 , (b) Distance fromleading edge at which transition would occur.SCHEMATIC:ASSUMPTIONS: (1) Isothermal conditions, Ts = T.-6 2PROPERTIES: Table A-4, Air (25C = 298K): = 15.71 10 m /s.ANALYSIS: (a) The Reynolds number isRe x = ux ux.=8To achieve a Reynolds number of 1 10 , the minimum plate length is then(6 28Re x 110 15.71 10 m / sLmin ==u50 m/s)<Lmin = 31.4 m.5(b) For a transition Reynolds number of 5 10xc =Re x,c u=(5 105 15.71 10-6 m 2 / sx c = 0.157 m.)50 m/s<COMMENTS: Note thatx c Re x,c=LReLThis expression may be used to quickly establish the location of transition from knowledge ofRe x,c and Re L .PROBLEM 6.15KNOWN: Transition Reynolds number. Velocity and temperature of atmospheric air, water,engine oil and mercury flow over a flat plate.FIND: Distance from leading edge at which transition occurs for each fluid.SCHEMATIC:ASSUMPTIONS: Transition Reynolds number is Re x,c = 5 105.PROPERTIES: For the fluids at T = 300K;FluidTable2v(m /s)A-415.89 10WaterA-60.858 10Engine OilA-5550 10MercuryA-5-6Air (1 atm)0.113 10-6-6-6ANALYSIS: The point of transition isx c = Re x,c5 105.=u 1 m/sSubstituting appropriate viscosities, findFluidAirWaterOilMercuryxc(m)7.950.432750.06<COMMENTS: Due to the effect which viscous forces have on attenuating the instabilitieswhich bring about transition, the distance required to achieve transition increases withincreasing .PROBLEM 6.16KNOWN: Two-dimensional flow conditions for which v = 0 and T = T(y).FIND: (a) Verify that u = u(y), (b) Derive the x-momentum equation, (c) Derive the energy equation.SCHEMATIC:Pressure & shear forcesEnergy fluxesASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3)Negligible body forces, (4) v = 0, (5) T = T(y) or T/x = 0, (6) Thermal energy generation occursonly by viscous dissipation.ANALYSIS: (a) From the mass continuity equation, it follows from the prescribed conditions thatu/x = 0. Hence u = u(y).(b) From Newtons second law of motion, Fx = (Rate of increase of fluid momentum)x,{} p p + p dx dy 1 + + + dy dx 1 = u u + u u dx dy 1 u u dy 1()()[( ) ] y x xHence, with = ( u/ y ) , it follows thatp 2u.=x y2p ( u ) u = 0+= x y x<(c) From the conservation of energy requirement and the prescribed conditions, it follows thatEin Eout = 0, or pu + u e + u 2 / 2 dy 1 + k T + u + ( u ) dy dx 1y y({ pu +or,x( pu ) dx + u)(e + u / 2) + x u (e + u / 2) dx} dy 1 u k T + y k T dy dx 1 = 0yy22 ( u ) T( pu ) u e + u 2 / 2 + k = 0 y yyx x()up 2T+uu+k= 0.yyx y2Noting that the second and third terms cancel from the momentum equation,2 2T u +k = 0. y2 y <PROBLEM 6.17KNOWN: Oil properties, journal and bearing temperatures, and journal speed for a lightlyloaded journal bearing.FIND: Maximum oil temperature.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constantproperties, (3) Clearance is much less than journal radius and flow is Couette.ANALYSIS: The temperature distribution corresponds to the result obtained in the textExample on Couette flow,T(y) = T0 +2 2 y y U .2kL L The position of maximum temperature is obtained fromdT 2 1 2y U =0=dy2k L L2 or,y = L/2.The temperature is a maximum at this point since d 2 T/dy2 < 0. Hence,Tmax = T ( L/2 ) = T0 +Tmax = 40 C + 2 1 1 U2U = T0 +2k8k2 410-2kg/s m (10m/s )Tmax = 40.83 C.28 0.15 W/m K<COMMENTS: Note that Tmax increases with increasing and U, decreases with increasingk, and is independent of L.PROBLEM 6.18KNOWN: Diameter, clearance, rotational speed and fluid properties of a lightly loaded journalbearing. Temperature of bearing.FIND: (a) Temperature distribution in the fluid, (b) Rate of heat transfer from bearing and operatingpower.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3)Couette flow.3-5 2-3PROPERTIES: Oil (Given): = 800 kg/m , = 10 m /s, k = 0.13 W/mK; = = 8 10kg/sm.ANALYSIS: (a) For Couette flow, the velocity distribution is linear, u(y) = U(y/L), and the energyequation and general form of the temperature distribution are222 du d 2TC UUkT = y 2 + 1 y + C2 .= = 2k L kL dy dy22Considering the boundary conditions dT/dy)y=L = 0 and T(0) = T0, find C2 = T0 and C1 = U /L.Hence,2T = T0 + U 2 / k ( y/L ) 1/ 2 ( y/L ) .()<(b) Applying Fouriers law at y = 0, the rate of heat transfer per unit length to the bearing isq = k ( D )dT dy y=0= ( D ) U2L(3= 75 10 m) 8 103kg/s m (14.14 m/s )20.25 10 3 m= 1507.5 W/mwhere the velocity is determined asU = ( D/2 ) = 0.0375m 3600 rev/min ( 2 rad/rev ) / ( 60 s/min ) = 14.14 m/s.The journal power requirement isP = F( y=L )U = s( y=L ) D U()P = 452.5kg/s 2 m 75 10-3m 14.14m/s = 1507.5kg m/s3 = 1507.5W/m<where the shear stress at y = L is 14.14 m/s U s( y=L ) = ( u/ y )y=L = = 8 103 kg/s m = 452.5 kg/s 2 m.-3m L 0.25 10COMMENTS: Note that q = P, which is consistent with the energy conservation requirement.PROBLEM 6.19KNOWN: Conditions associated with the Couette flow of air or water.FIND: (a) Force and power requirements per unit surface area, (b) Viscous dissipation, (c) Maximumfluid temperature.SCHEMATIC:ASSUMPTIONS: (1) Fully-developed Couette flow, (2) Incompressible fluid with constantproperties.-72-3PROPERTIES: Table A-4, Air (300K): = 184.6 10 Ns/m , k = 26.3 10 W/mK; Table A-6,-62Water (300K): = 855 10 Ns/m , k = 0.613 W/mK.ANALYSIS: (a) The force per unit area is associated with the shear stress. Hence, with the linearvelocity profile for Couette flow, = ( du/dy ) = ( U/L ) .Air: air = 184.6 107 N s/m 2 Air:( P/A )air =200 m/s= 0.738 N/m 20.005 m200 m/sWater: water = 855 106 N s/m 2 = 34.2 N/m 2 .0.005 mWith the required power given by P/A = U,Water:<(0.738 N/m2 ) 200 m/s = 147.6 W/m2( P/A )water = (34.2 N/m 2 ) 200 m/s = 6840 W/m 2 .<22(b) The viscous dissipation is = (du/dy ) = ( U/L ) . Hence,2N s 200 m/s = 2.95 10 4 W/m32 0.005 m mAir:( )air = 184.6 107Water:( )water = 855 1062N s 200 m/s = 1.37 106 W/m3.2 0.005 m m(c) From the solution to Part 4 of the text Example, the location of the maximum temperaturecorresponds to ymax = L/2. Hence, Tmax = T0 + U 2 / 8k and2-72Air:(Tmax )air = 27 C +Water:(Tmax )water = 27184.6 10N s/m( 200 m/s )= 30.5 C8 0.0263 W/m K855 10-6 N s/m 2 ( 200 m/s )2C+<8 0.613 W/m K<= 34.0 C.COMMENTS: (1) The viscous dissipation associated with the entire fluid layer, ( LA ) , must$$equal the power, P. (2) Although water >> air , k water >> k air . Hence,Tmax,water Tmax,air .PROBLEM 6.20KNOWN: Velocity and temperature difference of plates maintaining Couette flow. Meantemperature of air, water or oil between the plates.FIND: (a) PrEc product for each fluid, (b) PrEc product for air with plate at sonic velocity.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Couette flow, (3) Air is at 1 atm.PROPERTIES: Table A-4, Air (300K, 1atm), cp = 1007 J/kgK, Pr = 0.707, = 1.4, R=287.02 J/kgK; Table A-6, Water (300K): cp = 4179 J/kgK, Pr = 5.83; Table A-5, Engine oil(300K), cp = 1909 J/kgK, Pr = 6400.ANALYSIS: The product of the Prandtl and Eckert numbers is dimensionless,Pr Ec = PrU2m2 / s2m 2 / s2.cp T ( J/kg K ) Kkg m 2 / s 2 / kg)(Substituting numerical values, findAirPrEc 0.0028Water0.0056Oil13.41<(b) For an ideal gas, the speed of sound isc = ( R T )1/ 2where R, the gas constant for air, is Ru/ = 8.315 kJ/kmolK/(28.97 kg/kmol) = 287.02J/kgK. Hence, at 300K for air,U = c = (1.4 287.02 J/kg K 300K )1/ 2= 347.2 m/s.For sonic velocities, it follows thatPr Ec = 0.707(347.2 m/s )21007J / kg K 25K<= 3.38.COMMENTS: From the above results it follows that viscous dissipation effects must beconsidered in the high speed flow of gases and in oil flows at moderate speeds. For PrEc tobe less than 0.1 in air with T = 25C, U should be < 60 m/s.~PROBLEM 6.21KNOWN: Couette flow with moving plate isothermal and stationary plate insulated.FIND: Temperature of stationary plate and heat flux at the moving plate.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3)Couette flow.ANALYSIS: The energy equation is given by2 2T u 0= k+ y2 y Integrating twice find the general form of the temperature distribution,22 2T UT U= = y + C12yyk LT ( y) = k L U 2y + C1y + C2 .2k L 2Consider the boundary conditions to evaluate the constants, T/ y y=0 = 0 C1 = 0 and T ( L ) = TL C2 = TL +2U.2kHence, the temperature distribution is U2 T ( y ) = TL + 2k y 2 1 . L The temperature of the lower plate (y = 0) is U2 T ( 0 ) = TL + .<The heat flux to the upper plate (y = L) isT U2q ( L ) = k.=y=L< 2k yLCOMMENTS: The heat flux at the top surface may also be obtained by integrating the viscousdissipation over the fluid layer height. For a control volume about a unit area of the fluid layer,E = Egout2 u dy = q ( L )0 y L U2q ( L ) =.LPROBLEM 6.22KNOWN: Couette flow with heat transfer. Lower (insulated) plate moves with speed U and upper plateis stationary with prescribed thermal conductivity and thickness. Outer surface of upper plate maintainedat constant temperature, Tsp = 40C.FIND: (a) On T-y coordinates, sketch the temperature distribution in the oil and the stationary plate, and(b) An expression for the temperature at the lower surface of the oil film, T(0) = To, in terms of the platespeed U, the stationary plate parameters (Tsp, ksp, Lsp) and the oil parameters (, ko, Lo). Determine thistemperature for the prescribed conditions.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed Couette flow and (3) Incompressiblefluid with constant properties.ANALYSIS: (a) The temperature distribution is shown above with these key features: linear in plate,parabolic in oil film, discontinuity at plate-oil interface, and zero gradient at lower plate surface.(b) From Example 6.4, the general solution to the conservation equations for the temperature distributionin the oil film is2 U2 +C y+CTo ( y ) = AywhereA=342k o Lo and the boundary conditions are,At y = 0, insulated boundarydTo = 0;dy y = 0At y = Lo, heat fluxes in oil and plate are equal,C3 = 0q ( Lo ) = q ( Lo )ospContinued...PROBLEM 6.22 (Cont.)To ( Lo ) TspdT k o =dy y = LR spo dTo = 2ALody y = LR = L ksp sp spTo ( L ) = AL2 + C4ok Lsp C4 = Tsp + AL2 1 + 2 ooLo ksp Hence, the temperature distribution at the lower surface isTo ( 0 ) = A 0 + C4To ( 0 ) = Tsp +kU 2 1 + 2 o2k oLoLsp ksp <Substituting numerical values, findTo ( 0 ) = 40 C +0.799 N s m 20.145 3 = 116.9 C(5 m s )2 1 + 22 0.145 W m K51.5 <COMMENTS: (1) Give a physical explanation about why the maximum temperature occurs at thelower surface.(2) Sketch the temperature distribution if the upper plate moved with a speed U while the lower plate isstationary and all other conditions remain the same.PROBLEM 6.23KNOWN: Shaft of diameter 100 mm rotating at 9000 rpm in a journal bearing of 70 mm length.Uniform gap of 1 mm separates the shaft and bearing filled with lubricant. Outer surface of bearing iswater-cooled and maintained at Twc = 30C.FIND: (a) Viscous dissipation in the lubricant, (W/m3), (b) Heat transfer rate from the lubricant,assuming no heat lost through the shaft, and (c) Temperatures of the bearing and shaft, Tb and Ts.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed Couette flow, (3) Incompressiblefluid with constant properties, and (4) Negligible heat lost through the shaft.ANALYSIS: (a) The viscous dissipation, , Eq. 6.40, for Couette flow from Example 6.4, is222 47.1m s du U73 = = = 0.03 N s m 2 = 6.656 10 W mL dy 0.001m <where the velocity distribution is linear and the tangential velocity of the shaft isU = DN = ( 0.100 m ) 9000 rpm ( min 60s ) = 47.1m s .(b) The heat transfer rate from the lubricant volume through the bearing isq = = ( D L ) = 6.65 107 W m3 ( 0.100 m 0.001m 0.070 m ) = 1462 W<where = 70 mm is the length of the bearing normal to the page.Continued...PROBLEM 6.23 (Cont.)(c) From Fouriers law, the heat rate through the bearing material of inner and outer diameters, Di and Do,and thermal conductivity kb is, from Eq. (3.27),qr =2 k b (Tb Twc )ln ( Do Di )q ln ( Do Di )Tb = Twc + r2 k bTb = 30 C +1462 W ln ( 200 100