Unformatted text preview: PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity 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 Fourier’s law, q cond = q x = q ′′ ⋅ A = -k x T −T dT ⋅ A = kA 1 2 . dx L Solving for T2 gives T2 = T1 − q cond L . kA Substituting numerical values, find T2 = 415 C - 3000W × 0.025m 0.2W / m ⋅ K × 10m2 T2 = 415 C - 37.5 C T2 = 378 C. COMMENTS: Note direction of heat flow and fact that T2 must be less than T1. < PROBLEM 1.2 KNOWN: 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 38°C. 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 Fourier’s law, it is evident that the gradient, dT dx = − q′′ k , is a constant, and x hence the temperature distribution is linear, if q′′ and k are each constant. The heat flux must be x constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C are ) ( 25 C − −15 C dT T1 − T2 q′′ = − k =k = 1W m ⋅ K = 133.3W m 2 . x dx L 0.30 m q 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 ≤ 38°C, with different wall thermal conductivities, k. 3500 Heat loss, qx (W) 2500 1500 500 -500 -1500 -20 -10 0 10 20 30 40 Ambient air temperature, T2 (C) Wall thermal conductivity, k = 1.25 W/m.K k = 1 W/m.K, concrete wall k = 0.75 W/m.K For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearily from +2667 W to -867 W and is zero when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with increasing thermal conductivity. COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane wall would not be linear. PROBLEM 1.3 KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency of 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 is T −T 7°C q = k ( LW ) 1 2 = 1.4 W / m ⋅ K (11m × 8 m ) = 4312 W t 0.20 m < The daily cost of natural gas that must be combusted to compensate for the heat loss is Cd = q Cg ηf ( ∆t ) = 4312 W × $0.01/ MJ 0.9 ×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 insulation between it and the concrete. PROBLEM 1.4 KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed thickness. FIND: Thermal conductivity, k, of the wood. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be determined from Fourier’s law, Eq. 1.2. Rearranging, k=q′′ x L W = 40 T − T2 m2 1 k = 0.10 W / m ⋅ K. 0.05m ( 40-20 ) C < COMMENTS: Note that the °C or K temperature units may be used interchangeably when evaluating a temperature difference. PROBLEM 1.5 KNOWN: 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-state conditions, (3) Constant properties. ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from Fourier’s law, Eq. 1.2. T −T q′′ = k 1 2 x L W (15-5 ) C q′′ = 1.4 x m ⋅ K 0.005m q′′ = 2800 W/m 2 . x Since the heat flux is uniform over the surface, the heat loss (rate) is q = q ′′ × A x q = 2800 W / m2 × 3m2 q = 8400 W. COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions. < PROBLEM 1.6 KNOWN: Width, height, thickness and thermal conductivity of a single pane window and the air space of a double pane window. Representative winter surface temperatures of single pane and air space. FIND: Heat loss through single and double pane windows. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-state conditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy induced motion). ANALYSIS: From Fourier’s law, the heat losses are Single Pane: $ T1 − T2 2 35 C = 19, 600 W qg = k g A = 1.4 W/m ⋅ K 2m L 0.005m () () T −T 25 $C Double Pane: qa = k a A 1 2 = 0.024 2m2 = 120 W L 0.010 m COMMENTS: Losses associated with a single pane are unacceptable and would remain excessive, even if the thickness of the glass were doubled to match that of the air space. The principal advantage of the double pane construction resides with the low thermal conductivity of air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, use of the double pane construction would also increase the surface temperature of the glass exposed to the room (inside) air. PROBLEM 1.7 KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures. FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribed value. SCHEMATIC: ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 5 2 walls of area A = 4m , (3) Steady-state conditions, (4) Constant properties. ANALYSIS: Using Fourier’s law, Eq. 1.2, the heat rate is q = q ′′ ⋅ A = k ∆T A total L 2 Solving for L and recognizing that Atotal = 5×W , find 5 k ∆ T W2 L= q L= () 5 × 0.03 W/m ⋅ K 35 - (-10 ) C 4m 2 500 W L = 0.054m = 54mm. < COMMENTS: The corners will cause local departures from one-dimensional conduction and a slightly larger heat loss. PROBLEM 1.8 KNOWN: Dimensions and thermal conductivity of food/beverage container. Inner and outer surface temperatures. FIND: Heat flux through container wall and total heat load. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through bottom wall, (3) Uniform surface temperatures and one-dimensional conduction through remaining walls. ANALYSIS: From Fourier’s 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 2 L 0.025 m < Since the flux is uniform over each of the five walls through which heat is transferred, the heat load is q = 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, the effect is negligible. PROBLEM 1.9 KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that through 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-dimensional wall follows from Fourier’s law, Eq. 1.2, q ′′ = k ∆T L where ∆T represents the difference in surface temperatures. Since ∆T is the same for both walls, it follows that L1 = L2 k1 q ′′ ⋅ 2. k2 q1 ′′ With the heat fluxes related as q1′ = 0.8 q ′′ ′ 2 L1 = 100mm 0.75 W / m ⋅ K 1 × = 375mm. 0.25 W / m ⋅ K 0.8 < COMMENTS: Not knowing the temperature difference across the walls, we cannot find the value of the heat rate. PROBLEM 1.10 KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil water. 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 Fourier’s law, the rate of heat transfer by conduction through the bottom of the pan is T −T q = kA 1 2 L Hence, T1 = T2 + qL kA 2 where A = π D2 / 4 = π (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 2 600W (0.005 m ) ( 390 W/m ⋅ K 0.0314 m2 ) ) = 110.40 $C = 110.25 $C COMMENTS: Although the temperature drop across the bottom is slightly larger for aluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible for both 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 1.11 KNOWN: 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 heat dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction in chip. ANALYSIS: All of the electrical power dissipated at the back surface of the chip is transferred by conduction through the chip. Hence, from Fourier’s law, P = q = kA ∆T t or ∆T = t ⋅P kW 2 = ∆T = 1.1 C. 0.001 m × 4 W 2 150 W/m ⋅ K ( 0.005 m ) < COMMENTS: For fixed P, the temperature drop across the chip decreases with increasing k and W, as well as with decreasing t. PROBLEM 1.12 KNOWN: Heat flux gage with thin-film thermocouples on upper and lower surfaces; output voltage, 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) Fourier’s law applied to the gage can be written as q ′′ = k ∆T ∆x and the gradient can be expressed as ∆T ∆E / N = ∆x SABt where N is the number of differentially connected thermocouple junctions, SAB is the Seebeck coefficient for type K thermocouples (A-chromel and B-alumel), and ∆x = t is the gage thickness. Hence, q ′′ = k∆E NSABt q ′′ = 1.4 W / m ⋅ K × 350 × 10-6 V = 9800 W / m2 . -6 V / $ C × 0.25 × 10-3 m 5 × 40 × 10 < (b) The major precaution to be taken with this type of gage is to match its thermal conductivity with that of the material on which it is installed. If the gage is bonded between laminates (see sketch above) and its thermal conductivity is significantly different from that of the laminates, one dimensional heat flow will be disturbed and the gage will read 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 1.13 KNOWN: 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 under normal room conditions. SCHEMATIC: ASSUMPTIONS: (1) Temperature is uniform over the hand’s surface, (2) Convection coefficient is uniform over the hand, and (3) Negligible radiation exchange between hand and surroundings in the case of air flow. ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss. The heat loss can be determined from Newton’s law of cooling, Eq. 1.3a, written as q′′ = h ( Ts − T∞ ) For the air stream: q′′ = 40 W m 2 ⋅ K 30 − ( −5) K = 1, 400 W m 2 air < For the water stream: 2 2 q′′ water = 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 when in the air stream for the given temperature and convection coefficient conditions. In contrast, the heat loss in a normal room environment is only 30 W/m2 which is a factor of 400 times less than the loss in the 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 is excessively high. PROBLEM 1.14 KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinder with an imbedded electrical heater for different air velocities. FIND: (a) Determine the convection coefficient for each of the air velocity conditions and display the results graphically, and (b) Assuming that the convection coefficient depends upon air velocity as h = CVn, determine the parameters C and n. SCHEMATIC: V(m/s) Pe′ (W/m) h (W/m2⋅K) 1 450 22.0 2 658 32.2 4 983 48.1 8 1507 73.8 12 1963 96.1 ASSUMPTIONS: (1) Temperature is uniform over the cylinder surface, (2) Negligible radiation exchange 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 the electrical heater is transferred by convection to the air stream. Using Newtons law of cooling on a per unit length basis, ′ Pe = h (π D )(Ts − T∞ ) ′ where Pe is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/s condition, using the data from the table above, find h = 450 W m π × 0.025 m 300 − 40 C = 22.0 W m 2⋅K ( ) < Repeating the calculations, find the convection coefficients for the remaining conditions which are tabulated 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/m2⋅K(s/m)n, assuring a match at V = 1, we can readily find the exponent n from the slope of the 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 < 100 80 60 40 20 0 2 4 6 8 10 12 Air velocity, V (m/s) Data, smooth curve, 5-points Coefficient, h (W/m^2.K) Coefficient, h (W/m^2.K) choice. Hence, C = 22.12 and n = 0.6. 100 80 60 40 20 10 1 2 4 6 Air velocity, V (m/s) Data , smooth curve, 5 points h = C * V^n, C = 22.1, n = 0.5 n = 0.6 n = 0.8 8 10 PROBLEM 1.15 KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required to 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 direction normal to flow. ANALYSIS: The convection heat rate from the cylinder per unit length of the cylinder has the form q′ = h (π D ) ( Ts − T∞ ) and solving for the heat transfer convection coefficient, find h= q′ . π D (Ts − T∞ ) Substituting numerical values for the water and air situations: Water hw = Air ha = 28 × 103 W/m π × 0.030m (90-25 ) C 400 W/m π × 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, yet hw ≈ 70 × ha. These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases. See Table 1.1. PROBLEM 1.16 KNOWN: Dimensions of a cartridge heater. Heater power. Convection coefficients in air and 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 transferred to the fluid by convection, (3) Negligible heat transfer from ends. ANALYSIS: With P = qconv, Newton’s law of cooling yields P=hA (Ts − T∞ ) = hπ DL ( Ts − T∞ ) P Ts = T∞ + . hπ DL In water, Ts = 20 C + 2000 W 5000 W / m ⋅ K × π × 0.02 m × 0.200 m 2 Ts = 20 C + 31.8 C = 51.8 C. < In air, Ts = 20 C + 2000 W 50 W / m ⋅ K × π × 0.02 m × 0.200 m 2 Ts = 20 C + 3183 C = 3203 C. < COMMENTS: (1) Air is much less effective than water as a heat transfer fluid. Hence, the cartridge 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 1.17 KNOWN: Length, diameter and calibration of a hot wire anemometer. Temperature of air stream. Current, voltage drop and surface temperature of wire for a particular application. FIND: Air velocity SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from the wire by natural convection or radiation. ANALYSIS: If all of the electric energy is transferred by convection to the air, the following equality must be satisfied Pelec = EI = hA (Ts − T∞ ) where A = π DL = π ( 0.0005m × 0.02m ) = 3.14 × 10−5 m 2 . Hence, h= EI 5V × 0.1A = = 318 W/m 2 ⋅ K A (Ts − T∞ ) 3.14 × 10−5m 2 50 $C ( ( ) V = 6.25 × 10−5 h 2 = 6.25 ×10−5 318 W/m 2 ⋅ K ) 2 = 6.3 m/s < COMMENTS: The convection coefficient is sufficiently large to render buoyancy (natural convection) and radiation effects negligible. PROBLEM 1.18 KNOWN: Chip width and maximum allowable temperature. Coolant conditions. FIND: Maximum allowable chip power for air and liquid coolants. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and bottom, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer by radiation in air. ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection to the coolant. Hence, P=q and from Newton’s law of cooling, 2 P = hA(T - T∞) = h W (T - T∞). In air, 2 2 Pmax = 200 W/m ⋅K(0.005 m) (85 - 15) ° C = 0.35 W. < In the dielectric liquid 2 2 Pmax = 3000 W/m ⋅K(0.005 m) (85-15) ° C = 5.25 W. < COMMENTS: Relative to liquids, air is a poor heat transfer fluid. Hence, in air the chip can dissipate far less energy than in the dielectric liquid. PROBLEM 1.19 KNOWN: Length, diameter and maximum allowable surface temperature of a power transistor. Temperature and convection coefficient for air cooling. FIND: Maximum allowable power dissipation. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through base of transistor, (3) Negligible heat transfer by radiation from surface of transistor. ANALYSIS: Subject to the foregoing assumptions, the power dissipated by the transistor is equivalent to the rate at which heat is transferred by convection to the air. Hence, Pelec = q conv = hA (Ts − T∞ ) ( ) 2 where A = π DL + D2 / 4 = π 0.012m × 0.01m + ( 0.012m ) / 4 = 4.90 ×10−4 m 2 . For a maximum allowable surface temperature of 85°C, the power is ( Pelec = 100 W/m2 ⋅ K 4.90 × 10−4 m 2 ) (85 − 25)$ 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 base could be significant, thereby permitting operation at a larger power. (2) The local convection coefficient varies over the surface, and hot spots could exist if there are locations at which the local value of h is substantially smaller than the prescribed average value. PROBLEM 1.20 KNOWN: 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 the measurements. In this case, the electric power dissipated in the chip would be transferred from the chip by radiation and conduction (to the substrate), as well as by convection to the jet. An energy balance for the chip yields q elec = q conv + q cond + q rad . Hence, with q conv = hA ( Ts − T∞ ) , where A = 100 mm2 is the surface area of the chip, q − q cond − q rad h = elec A (Ts − T∞ ) (1) While the electric power ( q elec ) and the jet ( T∞ ) and surface ( Ts ) temperatures may be measured, losses from 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 associated with determining the conduction and radiation losses. A second approach, Case (b), could involve fabrication of a heater assembly for which the conduction and radiation losses are controlled and minimized. A 10 mm × 10 mm copper block (k ~ 400 W/m⋅K) could be inserted in a poorly conducting substrate (k < 0.1 W/m⋅K) and a patch heater could be applied to the back of the block and insulated from below. If conduction to both the substrate and insulation could thereby be rendered negligible, heat would be transferred almost exclusively through the block. If radiation were rendered negligible by applying a low emissivity coating (ε < 0.1) to the surface of the copper block, virtually all of the heat would be transferred by convection to the jet. Hence, q cond and q rad may be neglected in equation (1), and the expression may be used to accurately determine h from the known (A) and measured ( q elec , Ts , T∞ ) quantities. COMMENTS: Since convection coefficients associated with gas flows are generally small, concurrent heat transfer by radiation and/or conduction must often be considered. However, jet impingement is one of the more effective means of transferring heat by convection and convection coefficients well in excess of 100 W/m2⋅K may be achieved. PROBLEM 1.21 KNOWN: Upper temperature set point, Tset, of a bimetallic switch and convection heat transfer coefficient between clothes dryer air and exposed surface of switch. FIND: Electrical power for heater to maintain Tset when air temperature is T∞ = 50°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Electrical heater is perfectly insulated from dryer wall, (3) Heater and switch are isothermal at Tset, (4) Negligible heat transfer from sides of heater or switch, (5) Switch surface, As, loses heat only by convection. ANALYSIS: Define a control volume around the bimetallic switch which experiences heat input from the heater and convection heat transfer to the dryer air. That is, Ein - Eout = 0 qelec - 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 temperatures with a single set-point, inexpensive, bimetallic-thermostatic switch by adjusting power levels to the heater. (2) Will the heater power requirement increase or decrease if the insulation pad is other than perfect? PROBLEM 1.22 KNOWN: Hot vertical plate suspended in cool, still air. Change in plate temperature with time at the instant when the plate temperature is 225°C. FIND: Convection heat transfer coefficient for this condition. SCHEMATIC: ASSUMPTIONS: (1) Plate is isothermal and of uniform temperature, (2) Negligible radiation exchange with surroundings, (3) Negligible heat lost through suspension wires. ANALYSIS: As shown in the cooling curve above, the plate temperature decreases with time. The condition of interest is for time to. For a control surface about the plate, the conservation of energy requirement is E in - E out = E st dT − 2hA s ( Ts − T∞ ) = M c p dt where As is the surface area of one side of the plate. Solving for h, find h= h= Mcp dT 2As (Ts − T∞ ) dt 3.75 kg × 2770 J/kg ⋅ K 2 × ( 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, determine whether radiation exchange with the surroundings at 25°C is negligible compared to convection. (2) We will later consider the criterion for determining whether the isothermal plate assumption is reasonable. If the thermal conductivity of the present plate were high (such as aluminum or copper), the criterion would be satisfied. PROBLEM 1.23 KNOWN: Width, input power and efficiency of a transmission. Temperature and convection coefficient 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 Newton’s law of cooling, q = hAs ( Ts − T∞ ) = 6 hW 2 ( Ts − T∞ ) where the output power is η Pi and the heat rate is q = Pi − Po = Pi (1 − η ) = 150 hp × 746 W / hp × 0.07 = 7833 W Hence, Ts = T∞ + q 6 hW 2 = 30°C + 7833 W 6 × 200 W / m 2 ⋅ K × (0.3m ) 2 = 102.5°C COMMENTS: There will, in fact, be considerable variability of the local convection coefficient over the transmission case and the prescribed value represents an average over the surface. < PROBLEM 1.24 KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficient and 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 chilled feeling is associated with excessive heat loss. Because the temperature of the room air is fixed, the different summer and winter comfort levels can not be attributed to convection heat transfer from the body. In both cases, the heat flux is 2 $ 2 Summer and Winter: q′′ conv = h ( Ts − T∞ ) = 2 W/m ⋅ K × 12 C = 24 W/m However, the heat flux due to radiation will differ, with values of Summer: ) ( ) ( −8 4 4 24 4 4 4 2 q ′′ = εσ Ts − Tsur = 0.9 × 5.67 × 10 W/m ⋅ K 305 − 300 K = 28.3 W/m rad ( ) ) ( 4 Winter: q ′′ = εσ Ts4 − Tsur = 0.9 × 5.67 × 10−8 W/m 2 ⋅ K 4 3054 − 287 4 K 4 = 95.4 W/m 2 rad There is a significant difference between winter and summer radiation fluxes, and the chilled condition is attributable to the effect of the colder walls on radiation. 2 COMMENTS: 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 is significant and if maintained over a 24 h period would amount to 2,950 kcal. PROBLEM 1.25 KNOWN: Diameter and emissivity of spherical interplanetary probe. Power dissipation within 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 the probe and radiation emission from the probe surface. Hence, at any instant -E out + E g = 0 εA sσTs4 = E g Eg Ts = επ D2σ 1/ 4 1/ 4 150W Ts = 0.8π 0.5 m 2 5.67 × 10−8 W/m2 ⋅ K 4 ( ) Ts = 254.7 K. < COMMENTS: Incident radiation, as, for example, from the sun, would increase the surface temperature. PROBLEM 1.26 KNOWN: Spherical shaped instrumentation package with prescribed surface emissivity within a large space-simulation chamber having walls at 77 K. FIND: Acceptable power dissipation for operating the package surface temperature in the range Ts = 40 to 85°C. 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 the spherical package, and (3) Steady-state conditions. ANALYSIS: From an overall energy balance on the package, the internal power dissipation Pe will be transferred by radiation exchange between the package and the chamber walls. From Eq. 1.7, 4 4 q rad = Pe = ε Asσ Ts − Tsur ) ( For the condition when Ts = 40°C, with As = πD2 the power dissipation will be 4 Pe = 0.25 (π × 0.10 m ) × 5.67 ×10−8 W m 2 ⋅ K 4 × ( 40 + 273) − 77 4 K 4 = 4.3 W < Repeating this calculation for the range 40 ≤ Ts ≤ 85°C, we can obtain the power dissipation as a function of surface temperature for the ε = 0.25 condition. Similarly, with 0.2 or 0.3, the family of curves shown below has been obtained. Power dissipation, Pe (W) 10 8 6 4 2 40 50 60 70 80 90 Surface temperature, Ts (C) Surface emissivity, eps = 0.3 eps = 0.25 eps = 0.2 COMMENTS: (1) As expected, the internal power dissipation increases with increasing emissivity and surface temperature. Because the radiation rate equation is non-linear with respect to temperature, 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 exceed 85°C? What kind of a coating should be applied to the instrument package in order to approach this limiting condition? PROBLEM 1.27 KNOWN: Area, emissivity and temperature of a surface placed in a large, evacuated chamber of prescribed temperature. FIND: (a) Rate of surface radiation emission, (b) Net rate of radiation exchange between surface 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 is 4 q emit = E ⋅ A = ε A σ 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 chamber walls is ( 44 q = ε A σ Ts − Tsur ( ) ) 4 4 q = 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 at the instant the surface is placed in the chamber. The surface would, of course, cool due to this heat 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 surface reached that of the surroundings. PROBLEM 1.28 KNOWN: Length, diameter, surface temperature and emissivity of steam line. Temperature and convection coefficient associated with ambient air. Efficiency and fuel cost for gas fired furnace. FIND: (a) Rate of heat loss, (b) Annual cost of heat loss. SCHEMATIC: ASSUMPTIONS: (1) Steam line operates continuously throughout year, (2) Net radiation transfer is between small surface (steam line) and large enclosure (plant walls). ANALYSIS: (a) From Eqs. (1.3a) and (1.7), the heat loss is ( ) 4 4 q = qconv + q rad = A h ( Ts − T∞ ) + εσ Ts − Tsur where A = π DL = π ( 0.1m × 25m ) = 7.85m 2 . Hence, ( ) q = 7.85m2 10 W/m2 ⋅ K (150 − 25) K + 0.8 × 5.67 × 10−8 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 is E = qt = 18, 405 W × 3600 s/h × 24h/d × 365 d/y = 5.80 ×1011 J With a furnace energy consumption of Ef = E/ηf = 6.45 ×1011 J, the annual cost of the loss is C = Cg Ef = 0.01 $/MJ × 6.45 ×105MJ = $6450 < COMMENTS: The heat loss and related costs are unacceptable and should be reduced by insulating the steam line. PROBLEM 1.29 KNOWN: Exact and approximate expressions for the linearized radiation coefficient, hr and hra, respectively. FIND: (a) Comparison of the coefficients with ε = 0.05 and 0.9 and surface temperatures which may exceed that of the surroundings (Tsur = 25°C) by 10 to 100°C; also comparison with a free convection coefficient correlation, (b) Plot of the relative error (hr - rra)/hr as a function of the furnace temperature associated with a workpiece at Ts = 25°C having ε = 0.05, 0.2 or 0.9. ASSUMPTIONS: (1) Furnace walls are large compared to the workpiece and (2) Steady-state conditions. ANALYSIS: (a) The linearized radiation coefficient, Eq. 1.9, follows from the radiation exchange rate equation, 2 2 h r = εσ (Ts + Tsur ) Ts + Tsur ) ( If Ts ≈ Tsur, the coefficient may be approximated by the simpler expression h r,a = 4εσ T3 T = ( Ts + Tsur ) 2 For the condition of ε = 0.05, Ts = Tsur + 10 = 35°C = 308 K and Tsur = 25°C = 298 K, find that h r = 0.05 × 5.67 × 10−8 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 ×10−8 W m 2 ⋅ K 4 ((308 + 298 ) 2 ) K3 = 0.32 W m 2 ⋅ K 3 < < The free convection coefficient with Ts = 35°C and T∞ = Tsur = 25°C, find that h = 0.98∆T1/ 3 = 0.98 (Ts − T∞ ) 1/ 3 = 0.98 (308 − 298 ) 1/ 3 < = 2.1W m 2 ⋅ K For the range Ts - Tsur = 10 to 100°C with ε = 0.05 and 0.9, the results for the coefficients are tabulated below. For this range of surface and surroundings temperatures, the radiation and free convection coefficients are of comparable magnitude for moderate values of the emissivity, say ε > 0.2. The approximate expression for the linearized radiation coefficient is valid within 2% for these conditions. (b) The above expressions for the radiation coefficients, hr and hr,a, are used for the workpiece at Ts = 25°C placed inside a furnace with walls which may vary from 100 to 1000°C. The relative error, (hr hra)/hr, will be independent of the surface emissivity and is plotted as a function of Tsur. For Tsur > 150°C, the approximate expression provides estimates which are in error more than 5%. The approximate expression should be used with caution, and only for surface and surrounding temperature differences of 50 to 100°C. Ts (°C) 35 135 ε 0.05 0.9 0.05 0.9 Coefficients (W/m ⋅K) hr,a h hr 0.32 0.32 2.1 5.7 5.7 0.51 0.50 4.7 9.2 9.0 Relative error, (hr-hra)/hr*100 (%) 30 2 20 10 0 100 300 500 700 Surroundings temperature, Tsur (C) 900 PROBLEM 1.30 KNOWN: Chip width, temperature, and heat loss by convection in air. Chip emissivity and temperature of large surroundings. FIND: Increase in chip power due to radiation. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between small surface and large enclosure. ANALYSIS: Heat transfer from the chip due to net radiation exchange with the surroundings is ( 4 q rad = ε W 2σ T 4 - Tsur ) ( ) q rad = 0.9 ( 0.005 m ) 5.67 × 10−8 W/m 2 ⋅ K 4 3584 - 2884 K 4 2 q rad = 0.0122 W. The percent increase in chip power is therefore q ∆P 0.0122 W × 100 = rad × 100 = × 100 = 35%. . P q conv 0.350 W COMMENTS: For the prescribed conditions, radiation effects are small. Relative to convection, the effect of radiation would increase with increasing chip temperature and decreasing convection coefficient. < PROBLEM 1.31 KNOWN: Width, surface emissivity and maximum allowable temperature of an electronic chip. Temperature of air and surroundings. Convection coefficient. 2 1/4 FIND: (a) Maximum power dissipation for free convection with h(W/m ⋅K) = 4.2(T - T∞) , (b) 2 Maximum 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 a large enclosure, (3) Negligible heat transfer from sides of chip or from back of chip by conduction through the substrate. ANALYSIS: Subject to the foregoing assumptions, electric power dissipation by the chip must be balanced by convection and radiation heat transfer from the chip. Hence, from Eq. (1.10), ( 4 4 Pelec = q conv + q rad = hA (Ts − T∞ ) + ε Aσ Ts − Tsur ) 2 where A = L2 = (0.015m ) = 2.25 × 10−4 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 × 10−4 m2 ( 60K )5 / 4 = 0.158 W ( ) ( ) q rad = 0.60 2.25 × 10−4 m2 5.67 × 10−8 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 × 10−4 m2 ( 60K ) = 3.375 W Pelec = 3.375 W + 0.065 W = 3.44 W < COMMENTS: Clearly, radiation and natural convection are inefficient mechanisms for transferring 2 heat from the chip. For Ts = 85°C and T∞ = 25°C, the natural convection coefficient is 11.7 W/m ⋅K. 2 Even for forced convection with h = 250 W/m ⋅K, the power dissipation is well below that associated with many of today’s processors. To provide acceptable cooling, it is often necessary to attach the chip to a highly conducting substrate and to thereby provide an additional heat transfer mechanism due to conduction from the back surface. PROBLEM 1.32 KNOWN: Vacuum enclosure maintained at 77 K by liquid nitrogen shroud while baseplate is maintained 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 (εp = 0.09) is bonded to baseplate surface. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) No heat losses from backside of heater or sides of plate, (3) Vacuum enclosure large compared to baseplate, (4) Enclosure is evacuated with negligible convection, (5) Liquid nitrogen (LN2) is heated only by heat transfer to the shroud, and (6) Foil is intimately 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 = 0 q elec - q rad = 0 and using Eq. 1.7 for radiative exchange between the baseplate and shroud, ( ) 4 qelec = ε p A pσ Tp - T 4 . sh ( ) Substituting numerical values, with A p = π D 2 / 4 , find p ( ) 2 qelec = 0.25 π ( 0.3 m ) / 4 5.67 × 10−8 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 stream causing evaporation, E in - E out = 0 q rad - m LN2 h fg = 0 where 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 (εp = 0.09) were bonded to the upper surface of the baseplate, ( ) q rad,foil = q rad ε f / ε p = 8.1 W ( 0.09/0.25 ) = 2.9 W and the liquid nitrogen consumption rate would be reduced by (0.25 - 0.09)/0.25 = 64% to 0.083 kg/h. < PROBLEM 1.33 KNOWN: Width, input power and efficiency of a transmission. Temperature and convection coefficient 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, which may be expressed as q = Pi – Po = Pi (1 - η) = 150 hp × 746 W/hp × 0.07 = 7833 W. Heat transfer from the case is by convection and radiation, in which case ( ) 4 4 q = As h ( Ts − T∞ ) + εσ Ts − Tsur 2 where As = 6 W . Hence, ( ) 2 7833 W = 6 ( 0.30 m ) 200 W / m 2 ⋅ K ( Ts − 303K ) + 0.8 × 5.67 × 10−8 W / m 2 ⋅ K 4 Ts4 − 3034 K 4 A trial-and-error solution yields Ts ≈ 373K = 100°C < COMMENTS: (1) For Ts ≈ 373 K, qconv ≈ 7,560 W and qrad ≈ 270 W, in which case heat transfer is dominated by convection, (2) If radiation is neglected, the corresponding surface temperature is Ts = 102.5°C. PROBLEM 1.34 KNOWN: 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 within the values for E , E , E g in out st 3 the resistor, q (W/m ), (c) Neglecting radiation from the resistor, determine the convection coefficient. SCHEMATIC: ASSUMPTIONS: (1) Electrical power is dissipated uniformly within the resistor, (2) Temperature of the resistor is uniform, (3) Negligible electrical power dissipated in the lead wires, (4) Negligible radiation exchange between the resistor and the surroundings, (5) No heat transfer occurs from the battery, (5) Steady-state conditions. ANALYSIS: (a) Referring to Section 1.3.1, the conservation of energy requirement for a control volume at an instant of time, Eq 1.11a, is Ein + Eg − Eout = Est where Ein , E out correspond to surface inflow and outflow processes, respectively. The energy generation 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 with changes 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 = 24V×6A = 144 W. Control volume: Resistor. E in = 0 < Eout = 144 W Eg = 144 W Est = 0 The E g term is due to conversion of electrical energy to thermal energy. The term E out is due to convection from the resistor surface to the air. Continued... PROBLEM 1.34 (Cont.) Control volume: Battery-Resistor System. E in = 0 E =0 g Eout = 144 W E = −144 W < st The Est term represents the decrease in the chemical energy within the battery. The conversion of chemical energy to electrical energy and its subsequent conversion to thermal energy are processes internal to the system which are not associated with Est or E g . The E out term is due to convection from the resistor surface to the air. (b) From the energy balance on the resistor with volume, ∀ = (πD2/4)L, E g = q∀ ( ) 144 W = q π (0.06 m ) / 4 × 0.25 m 2 q = 2.04 ×105 W m3 < (c) From the energy balance on the resistor and Newton's law of cooling with As = πDL + 2(πD2/4), Eout = qcv = hAs ( Ts − T∞ ) ) ( 144 W = h π × 0.06 m × 0.25 m + 2 π × 0.062 m 2 4 (95 − 25 ) C 144 W = h [0.0471 + 0.0057 ] m 2 (95 − 25 ) C h = 39.0 W m 2⋅K COMMENTS: (1) In using the conservation of energy requirement, Eq. 1.11a, it is important to recognize that Ein and E out will always represent surface processes and E g and Est , volumetric processes. The generation term E g is associated with a conversion process from some form of energy 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), we conclude that the resistor is experiencing forced, rather than free, convection. < PROBLEM 1.35 KNOWN: Thickness and initial temperature of an aluminum plate whose thermal environment is changed. FIND: (a) Initial rate of temperature change, (b) Steady-state temperature of plate, (c) Effect of emissivity 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) No internal heat generation. ANALYSIS: (a) Applying an energy balance, Eq. 1.11a, at an instant of time to a control volume about the plate, E in − E out = E st , it follows for a unit surface area. ()() () ( ) 2 2 αSGS 1m 2 − E 1m 2 − q′′ conv 1m = ( d dt )( McT ) = ρ 1m × L c ( dT dt ) . Rearranging and substituting from Eqs. 1.3 and 1.5, we obtain dT dt = (1 ρ Lc ) αSGS − εσ Ti4 − h ( Ti − T∞ ) . ( dT dt = 2700 kg m3 × 0.004 m × 900 J kg ⋅ K ) −1 × 0.8 × 900 W m 2 − 0.25 × 5.67 × 10−8 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 to αSGS = εσ T 4 + h ( T − T∞ ) < (2) 0.8 × 900 W m 2 = 0.25 × 5.67 × 10−8 W m 2 ⋅ K 4 × T 4 + 20 W m 2 ⋅ K ( T − 293 K ) The solution yields T = 321.4 K = 48.4°C. (c) Using the IHT First Law Model for an Isothermal Plane Wall, parametric calculations yield the following results. Plate temperature, T (C) 70 60 50 40 30 20 0 0.2 0.4 0.6 0.8 1 Coating emissivity, eps Solar absorptivity, alphaS = 1 alphaS = 0.8 alphaS = 0.5 COMMENTS: The surface radiative properties have a significant effect on the plate temperature, which decreases with increasing ε and decreasing αS. If a low temperature is desired, the plate coating should be characterized by a large value of ε/αS. The temperature also decreases with increasing h. < PROBLEM 1.36 KNOWN: 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, at any instant E in - E out + E g = 0. It follows that, with the solar input, ′′ αSAsqS − As E + P=0 4 ′′ αSAsqS − Asεσ Ts + P=0 1/ 4 α A q′′ + P Ts = S s S Asεσ . ′′ In the shade ( qS = 0 ) , 1/ 4 1000 W Ts = 1 m 2 × 1× 5.67 × 10−8 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 ×10−8 W/m 2 ⋅ K 4 = 380 K. < COMMENTS: In orbit, the space station would be continuously cycling between shade and sunshine, and a steady-state condition would not exist. PROBLEM 1.37 KNOWN: Daily hot water consumption for a family of four and temperatures associated with ground water 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 a heat 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): ρ = vf 1 = 993 kg/m3, cp,f = 4.178 kJ/kg⋅K. ANALYSIS: From Eq. 1.11c, the daily heating requirement is Qdaily = ∆U t = Mc∆T = ρ 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 kJ The 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, is C = 6380 kWh ($0.08/kWh ) = $510 < If a heat pump is used, Qannual = COP ( Welec ). Hence, Welec = Qannual /( COP ) = 6380kWh/(3) = 2130 kWh The corresponding cost is C = 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 factors such as geography and seasonal variations in COP, as well as the time value of money. PROBLEM 1.38 KNOWN: Initial temperature of water and tank volume. Power dissipation, emissivity, length and diameter of submerged heaters. Expressions for convection coefficient associated with natural convection in water and air. FIND: (a) Time to raise temperature of water to prescribed value, (b) Heater temperature shortly 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 in thermal energy storage for heaters, (4) Constant properties, (5) Surroundings afforded by tank wall are large relative to heaters. ANALYSIS: (a) Application of conservation of energy to a closed system (the water) at an instant, Eq. (1.11d), yields dU dT dT = Mc = ρ∀c = q = 3q1 dt dt dt Tf t dt = ( ρ∀c/3q1 ) ∫ dT ∫0 Ti Hence, t= ( ) 990 kg/m3 ×10gal 3.79 ×10−3m3 / gal 4180J/kg ⋅ K 3 × 500 W (335 − 295) K = 4180 s < (b) From Eq. (1.3a), the heat rate by convection from each heater is 4/3 ′′ q1 = Aq1 = Ah ( Ts − T ) = (π DL ) 370 ( Ts − T ) Hence, 3/ 4 3/ 4 500 W q1 Ts = T + = T+ = ( T + 24 ) K 370π DL 370 W/m 2 ⋅ K 4/3 × π × 0.025 m × 0.250 m With water temperatures of Ti ≈ 295 K and Tf = 335 K shortly after the start of heating and at the end of heating, respectively, Ts,i = 319 K < Ts,f = 359 K Continued ….. PROBLEM 1.38 (Continued) (c) From Eq. (1.10), the heat rate in air is ( ) 4 4 q1 = π DL 0.70 ( Ts − T∞ )4 / 3 + εσ Ts − Tsur Substituting the prescribed values of q1, D, L, T∞ = Tsur and ε, an iterative solution yields Ts = 830 K < COMMENTS: In part (c) it is presumed that the heater can be operated at Ts = 830 K without experiencing burnout. The much larger value of Ts for air is due to the smaller convection 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 1.39 KNOWN: Power consumption, diameter, and inlet and discharge temperatures of a hair dryer. 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 and kinetic energy changes of air flow, (4) Negligible work done by fan, (5) Negligible heat transfer from casing of dryer to ambient air (Part (a)), (6) Radiation exchange between a small surface 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 to m ( u + pv ) − m ( u + pv ) + q = 0 i o where u + pv = i and q = Pelec. Hence, with m (ii − io ) = mcp (Ti − To ) , mcp (To − Ti ) = Pelec m= Pelec 500 W = = 0.0199 kg/s cp ( To − Ti ) 1007 J/kg ⋅ K 25$C () m 0.0199 kg/s = 0.0181 m3 / s ∀= = ρ 1.10 kg/m3 Vo = < 4∀ 4 × 0.0181 m3 / s ∀ = = = 4.7 m/s 2 Ac π D2 π (0.07 m ) < (b) Heat transfer from the casing is by convection and radiation, and from Eq. (1.10) ( 4 4 q = hAs ( Ts − T∞ ) + ε Asσ Ts − Tsur ) Continued ….. PROBLEM 1.39 (Continued) where As = π DL = π (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 × 10−8 W/m2 ⋅ K 4 3134 − 2934 K 4 q = 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 loss is justified. COMMENTS: Although the mass flow rate is invariant, the volumetric flow rate increases as 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 ρ, and hence the effect on ∀, is small. PROBLEM 1.40 KNOWN: Speed, width, thickness and initial and final temperatures of 304 stainless steel in an annealing process. Dimensions of annealing oven and temperature, emissivity and convection coefficient 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 and potential energy. ( ) PROPERTIES: Table A.1, St.St.304 T = (Ti + To )/2 = 775 K : ρ = 7900 kg/m3, c p = 578 J/kg⋅K; Table A.3, Concrete, T = 300 K: k c = 1.4 W/m⋅K. ANALYSIS: The rate of energy addition to the oven must balance the rate of energy transfer to the steel sheet and the rate of heat loss from the oven. With Ein − Eout = 0, it follows that Pelec + m ( u i − u o ) − q = 0 where heat is transferred from the oven. With m = ρ Vs ( Ws t s ) , ( u i − u o ) = cp ( Ti − To ) , and ( ) 4 4 q = ( 2Ho Lo + 2Ho Wo + Wo Lo ) × h ( Ts − T∞ ) + ε sσ Ts − Tsur + k c ( Wo Lo )(Ts − Tb )/t c , it follows that Pelec = ρ Vs ( Ws t s ) cp ( To − Ti ) + ( 2Ho Lo + 2H o Wo + Wo Lo ) × ) ( h ( T T ) + ε σ T 4 − T 4 + k ( W L )(T − T )/t s o s s sur bc c oo s Pelec = 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 × 10−8 W/m 2 ⋅ K 4 3504 − 3004 K 4 ] + 1.4W/m ⋅ K ( 2.4m × 25m )(350 − 300 ) K/0.5m Continued.…. PROBLEM 1.40 (Cont.) Pelec = 694, 000W + 169.6m 2 (500 + 313 ) W/m2 + 8400W = ( 694, 000 + 84,800 + 53,100 + 8400 ) W = 840kW COMMENTS: 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 and radiation losses can both be reduced by adding insulation to the side and top surfaces, which would reduce the corresponding value of Ts . < PROBLEM 1.41 KNOWN: Hot plate-type wafer thermal processing tool based upon heat transfer modes by conduction 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 wafer temperatures 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 hot plate temperatures 300 < Th < 1300°C. Comment on the relative importance of the modes and the influence of the gap distance. Under what conditions could a wafer be heated to 900°C in less than 10 seconds? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions for flux calculations, (2) Diameter of hot plate and wafer much larger than gap spacing, approximating plane, infinite planes, (3) One-dimensional conduction through gas, (4) Hot plate and wafer are blackbodies, (5) Negligible heat losses from wafer backside, and (6) Wafer temperature is uniform at the onset of heating. 3 PROPERTIES: Wafer: ρ = 2700 kg/m , c = 875 J/kg⋅K; Gas in gap: k = 0.0436 W/m⋅K. ANALYSIS: (a) The radiative heat flux between the hot plate and wafer for Th = 600°C and Tw = 20° C follows from the rate equation, ( ) 4 4 q′′ = σ Th − Tw = 5.67 × 10−8 W / m 2 ⋅ K 4 rad ((600 + 273) 4 − ( 20 + 273) 4 )K 4 = 32.5 kW / m2 < The conduction heat flux through the gas in the gap with L = 0.2 mm follows from Fourier’s law, (600 − 20) K = 126 kW / m2 Th − Tw q′′ = 0.0436 W / m ⋅ K cond = k L 0.0002 m < The initial time rate of change of the wafer can be determined from an energy balance on the wafer at the instant of time the heating process begins, ′′ Ein − E′′ = E′′ out st dT E′′ = ρ c d w st dt i where E ′′ ut = 0 and E ′′n = q ′′ or q ′′ . Substituting numerical values, find o i rad cond dTw dt q′′ 32.5 × 103 W / m 2 = rad = = 17.6 K / s i,rad ρ cd 2700 kg / m3 × 875 J / kg ⋅ K × 0.00078 m < dTw dt q′′ = cond = 68.4 K / s ρ cd i,cond < Continued ….. PROBLEM 1.41 (Cont.) (b) Using the foregoing equations, the heat fluxes and initial rate of temperature change for each mode can be calculated for selected gap separations L and range of hot plate temperatures Th with Tw = 20°C. 200 Initial rate of change, dTw/dt (K.s^-1) 400 Heat flux (kW/m^2) 300 200 100 150 100 50 0 0 3 00 500 700 900 1100 1300 300 500 700 900 1100 Hot plate temperature, Th (C) Hot plate temperature, Th (C) q''rad q''cond, L = 1.0 mm q''cond, L = 0.5 mm q''cond, L = 0.2 mm q''rad q''cond, L = 1.0 m m q''cond, L = 0.5 m m q''cond, L = 0.2 m m In the left-hand graph, the conduction heat flux increases linearly with Th and inversely with L as expected. The radiative heat flux is independent of L and highly non-linear with Th, but does not approach that for the highest conduction heat rate until Th approaches 1200°C. The general trends for the initial temperature-time change, (dTw/dt)i, follow those for the heat fluxes. To reach 900°C 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 the smallest L. 1300 PROBLEM 1.42 KNOWN: Silicon wafer, radiantly heated by lamps, experiencing an annealing process with known backside temperature. FIND: Whether temperature difference across the wafer thickness is less than 2°C in order to avoid damaging 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 a large enclosure, and (4) Vacuum condition in chamber, no convection. PROPERTIES: Wafer: k = 30 W/m⋅K, ε = α = 0.65. ANALYSIS: Perform a surface energy balance on the upper surface of the wafer to determine Tw ,u . The processes include the absorbed radiant flux from the lamps, radiation exchange with the chamber walls, and conduction through the wafer. E′′n − E′′ = 0 i out α q′′ − q′′ − q′′ = 0 s rad cd ( ) 4 4 α q′′ − εσ Tw,u − Tsur − k s Tw,u − Tw, L =0 ( 4 0.65 × 3.0 × 105 W / m 2 − 0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 Tw,u − ( 27 + 273) 4 )K 4 −30W / m ⋅ K Tw,u − (997 + 273) K / 0.00078 m = 0 Tw ,u = 1273K = 1000°C < COMMENTS: (1) The temperature difference for this steady-state operating condition, Tw ,u − Tw,l , is larger than 2°C. Warping of the wafer and inducing slip planes in the crystal structure could 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 to show the control surfaces, and then identify the rate processes associated with the surfaces. Make sure the directions (in or out) of the process are consistent with the energy balance equation. PROBLEM 1.43 KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces exposed to hot and cool zones, respectively. FIND: (a) Initial rate of change of the wafer temperature corresponding to the wafer temperature Tw ,i = 300 K, and (b) Steady-state temperature reached if the wafer remains in this position. How significant is convection for this situation? Sketch how you’d expect the wafer temperature to vary as a function of vertical distance. SCHEMATIC: ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Transient conditions when wafer is initially positioned, (3) Hot and cool zones have uniform temperatures, (3) Radiation exchange is between small surface (wafer) and large enclosure (chamber, hot or cold zone), and (4) Negligible heat losses from wafer to mounting pin holder. ANALYSIS: The energy balance on the wafer illustrated in the schematic above includes convection from the upper (u) and lower (l) surfaces with the ambient gas, radiation exchange with the hot- and cool-zone (chamber) surroundings, and the rate of energy storage term for the transient condition. E′′n − E′′ = E′′ i out st q′′ rad,h + q′′ rad,c − q′′ cv,u − q′′ = ρ cd cv,l ( )( d Tw dt ) 4 4 4 4 εσ Tsur,h − Tw + εσ Tsur,c − Tw − h u (Tw − T∞ ) − h l (Tw − T∞ ) = ρ cd d Tw dt (a) For the initial condition, the time rate of temperature change of the wafer is determined using the energy balance above with Tw = Tw,i = 300 K, ( ) ( ) 0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 15004 − 3004 K 4 + 0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 3304 − 3004 K 4 −8 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 be solved for the steady-state wafer temperature, Tw = Tw,ss . Continued ….. PROBLEM 1.43 (Cont.) ) ( ) ( 4 4 0.65 σ 15004 − Tw,ss K 4 + 0.65 σ 3304 − Tw,ss K 4 ( ) ( ) 8 W / m 2 ⋅ K Tw,ss − 700 K − 4 W / m 2 ⋅ K Tw,ss − 700 K = 0 Tw ,ss = 1251 K < To determine the relative importance of the convection processes, re-solve the energy balance above ignoring those processes to find ( d Tw / dt )i = 101 K / s and Tw,ss = 1262 K. We conclude that the radiation exchange processes control the initial time rate of temperature change and the steady-state temperature. If the wafer were elevated above the present operating position, its temperature would increase, since the lower surface would begin to experience radiant exchange with progressively more of the hot zone chamber. Conversely, by lowering the wafer, the upper surface would experience less radiant exchange with the hot zone chamber, and its temperature would decrease. The temperature-distance trend might appear as shown in the sketch. PROBLEM 1.44 KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive wastes. Surface convection conditions. FIND: Total energy generation rate and surface temperature. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature drop across thin container wall. ANALYSIS: The rate of energy generation is r 2 Eg = ∫ qdV=qo ∫ o 1- ( r/ro ) 2π rLdr 0 = 2π Lq r 2 / 2 − r 2 / 4 o o Eg o ( ) or per unit length, πq r E′ = o o . g 2 2 < Performing an energy balance for a control surface about the container yields, at an instant, E′ − E′ = 0 g out and substituting for the convection heat rate per unit length, 2 π qo ro = h ( 2π ro )( Ts − T∞ ) 2 Ts = T∞ + q o ro . 4h < COMMENTS: The temperature within the radioactive wastes increases with decreasing r from Ts at ro to a maximum value at the centerline. PROBLEM 1.45 KNOWN: Rod of prescribed diameter experiencing electrical dissipation from passage of electrical current 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 convection 2 coefficients of 50, 100 and 250 W/m ⋅K. Will variations in the surface emissivity have a significant effect on the rod temperature? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform rod temperature, (3) Radiation exchange between the outer surface of the rod and the surroundings is between a small surface and large enclosure. ANALYSIS: The energy balance on the rod for steady-state conditions has the form, q′ conv + q′ = E′ rad gen ) ( 4 π Dh (T − T∞ ) + π Dεσ T 4 − Tsur = I 2R ′ e Using this equation in the Workspace of IHT, the rod temperature is calculated and plotted as a function of current for selected convection coefficients. 150 R o d te m p e ra tu re , T (C ) 125 100 75 50 25 0 0 2 4 6 8 10 C u rre n t, I (a m p e re s ) h = 5 0 W /m ^2 .K h = 1 0 0 W /m ^2 .K h = 2 5 0 W /m ^2 .K COMMENTS: (1) For forced convection over the cylinder, the convection heat transfer coefficient is 0.6 dependent upon air velocity approximately as h ~ V . Hence, to achieve a 5-fold change in the 2 convection coefficient (from 50 to 250 W/m ⋅K), the air velocity must be changed by a factor of nearly 15. Continued ….. PROBLEM 1.45 (Cont.) 2 (2) For the condition of I = 4 A with h = 50 W/m ⋅K with T = 63.5°C, the convection and radiation exchange rates per unit length are, respectively, q ′ v = 5.7 W / m and q ′ = 0.67 W / m. We conclude c rad that convection is the dominate heat transfer mode and that changes in surface emissivity could have 2 only 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 air flow would cease? (4) The Workspace for the IHT program to calculate the heat losses and perform the parametric analysis to generate the graph is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. It is also good practice to show plots in customary units, that is, the units used to prescribe the problem. As such the graph of the rod temperature is shown above with Celsius units, even though the calculations require temperatures in kelvins. // Energy balance; from Ex. 1.3, Comment 1 -q'cv - q'rad + Edot'g = 0 q'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 form Edot'g = I^2*R'e qdot = I^2*R'e / (pi*D^2/4) // Input parameters D = 0.001 Tsur = 300 T_C = T – 273 eps = 0.8 Tinf = 300 h = 100 //h = 50 //h = 250 I = 5.2 //I = 4 R'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'g T T_C q'cv q'rad qdot D I R'e Tinf Tsur eps h sigma 10.82 332.6 59.55 10.23 0.5886 1.377E7 0.001 5.2 0.4 300 300 0.8 100 5.67E-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/m Edot'g T T_C q'cv q'rad qdot D I R'e Tinf Tsur eps h sigma 6.4 336.5 63.47 5.728 0.6721 8.149E6 0.001 4 0.4 300 300 0.8 50 5.67E-8 */ PROBLEM 1.46 KNOWN: 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 65°C; compare relative importance of convection and radiation exchange heat rates, and (b) Show graphically 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 to the total heat transfer for these conditions. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Bus bar and conduit are very long in direction normal to page, (3) Uniform bus-bar temperature, (4) Radiation exchange between the outer surface of the bus bar and the conduit is between a small surface and a large enclosure. PROPERTIES: Bus-bar material, ρe = ρe,o [1 + α ( T − To )] , ρ e,o = 0.0171µΩ ⋅ m, To = 25°C, α = 0.00396 K −1. ANALYSIS: An energy balance on the bus-bar for a unit length as shown in the schematic above has the form E′ − E′ + E′ = 0 in out gen 2 −q ′ − q ′ rad conv + I R ′ = 0 e ) ( 4 −επ Dσ T 4 − Tsur − hπ D ( T − T∞ ) + I2 ρe / Ac = 0 where R ′ = ρ e / A c and A c = π D 2 / 4. Using the relations for ρ e ( T ) and h ( T, D ) , and substituting e numerical values with T = 65°C, find ( q′ = 0.85 π ( 0.020m ) × 5.67 × 10−8 W / m 2 ⋅ K 4 [65 + 273] − [30 + 273] rad 4 4 )K 4 = 223 W / m 2 q′ conv = 7.83W / m ⋅ K π ( 0.020m )( 65 − 30 ) K = 17.2 W / m < < −0.25 where 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 × 10−6 Ω ⋅ m / π (0.020 ) m 2 / 4 = 6.31× 10−5 I2 W / m e where 2 ρe = 0.0171× 10−6 Ω ⋅ m 1 + 0.00396 K −1 (65 − 25) K = 198.2 µΩ ⋅ m The maximum allowable current capacity and the ratio of the convection to total heat transfer rate are I = 1950 A q′ / ( q′ + q′ ) = q′ / q′tot = 0.072 cv cv rad cv < 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 calculated and plotted as a function of the current for the range 100 ≤ I ≤ 5000 A for diameters of 10, 20 and 40 mm. Also shown below is the corresponding graph of the ratio (expressed in percentage units) of the heat transfer by convection to the total heat transfer, q ′ v / q ′tot . c Continued ….. PROBLEM 1.46 (Cont.) 13 11 Ratio q'cv / q'tot, (%) Ba r te m p e ra tu re , Ts (C ) 100 80 60 40 9 7 5 3 20 0 1000 2000 3000 4000 1 5000 20 40 C u rre n t, I (A) 60 80 Bus bar temperature, T (C) D = 10 m m D = 20 m m D = 40 m m D = 10 mm D = 20 mm D = 40 mm COMMENTS: (1) The trade-off between current-carrying capacity, operating temperature and bar diameter is shown in the first graph. If the surface temperature is not to exceed 65°C, the maximum current 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 a c small fraction of the total heat transfer. That is, radiation is the dominant mode of heat transfer. Note also that the convection contribution increases with increasing diameter. (3) The Workspace for the IHT program to perform the parametric analysis and generate the graphs is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. /* Results: base-case conditions, Part (a) I R'e cvovertot hbar q'cv Tsur_C eps 1950 6.309E-5 7.171 7.826 17.21 30 0.85 */ q'rad rhoe D 222.8 1.982E-8 0.02 Tinf_C Ts_C 30 65 // Energy balance, on a per unit length basis; steady-state conditions // Edot'in - Edot'out + Edot'gen = 0 -q'cv - q'rad + Edot'gen = 0 q'cv = hbar * P * (Ts - Tinf) P = pi * D q'rad = eps * sigma * (Ts^4 - Tsur^4) sigma = 5.67e-8 Edot'gen = I^2 * R'e R'e = rhoe / Ac rhoe = rhoeo * (1 + alpha * (Ts - To) ) To = 25 + 273 Ac = pi * D^2 / 4 // Convection coefficient hbar = 1.21 * (D^-0.25) * (Ts - Tinf)^0.25 // Convection vs. total heat rates cvovertot = q'cv / (q'cv + q'rad) * 100 // Input parameters D = 0.020 // D = 0.010 // D = 0.040 // I = 1950 rhoeo = 0.01711e-6 alpha = 0.00396 Tinf_C = 30 Tinf = Tinf_C + 273 Ts_C = 65 Ts = Ts_C + 273 Tsur_C = 30 Tsur = Tsur_C + 273 eps = 0.85 // Compact convection coeff. correlation // Values of diameter for parameter study // Base case condition unknown // Base case condition to determine current 100 PROBLEM 1.47 KNOWN: Elapsed times corresponding to a temperature change from 15 to 14°C for a reference sphere 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 change from 15 to 14°C, (3) Spheres experience same convection heat transfer rate when the time rates of surface 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, after a sphere has been immersed in the ice-water mixture at T∞. Ein − E out = Est −q conv = Mc dT dt where q conv = h A s ( T − T∞ ). Since the temperatures of the spheres are uniform, the change in energy storage term can be represented with the time rate of temperature change, dT/dt. The convection heat rates are equal at this instant of time, and hence the change in energy storage terms for the reference (r) and test (t) spheres must be equal. M r cr dT dT = M t ct dt r dt t Approximating the instantaneous differential change, dT/dt, by the difference change over a short period of time, ∆T/∆t, the specific heat of the test sphere can be calculated. 0.515 kg × 447 J / kg ⋅ K c t = 132 J / kg ⋅ K (15 − 14 ) K 6.35s = 1.263kg × c t × (15 − 14 ) K 4.59s < COMMENTS: Why was it important to perform the experiments with the reference and test spheres over the same temperature range (from 15 to 14°C)? Why does the analysis require that the spheres have uniform temperatures at all times? PROBLEM 1.48 KNOWN: Inner surface heating and new environmental conditions associated with a spherical shell of prescribed dimensions and material. FIND: (a) Governing equation for variation of wall temperature with time. Initial rate of temperature change, (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: ρ = 8055 kg/m3, c p = 510 J/kg⋅K. 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, 4 dT 2 3 q′′ 4π ri2 − h 4π ro ( T − T∞ ) = ρ π ro − ri3 cp i 3 dt ( )( ) ( ) dT 3 q′′ r 2 − hr 2 ( T − T ) . = o ∞ 3 − r3 i i dt ρ c r po i ( ) Substituting numerical values for the initial condition, find dT dt i W W 3 105 (0.5m )2 − 500 2 (0.6m )2 (500 − 300 ) K m2 m ⋅K = kg J 8055 510 (0.6 )3 − (0.5 )3 m3 3 kg ⋅ K m dT = −0.089 K/s . dt i < (b) Under steady-state conditions with Est = 0, it follows that 2 q′′ 4π ri2 = h 4π ro ( T − T∞ ) i ( )( ) Continued ….. PROBLEM 1.48 (Cont.) 2 2 q′′ ri 105 W/m2 0.5m i T = T∞ + = 300K + = 439K h ro 500W/m 2 ⋅ K 0.6m < (c) Parametric calculations were performed using the IHT First Law Model for an Isothermal Hollow Sphere. As shown below, there is a sharp increase in temperature with decreasing values of h < 1000 W/m2⋅K. For T > 380 K, boiling will occur at the canister surface, and for T > 410 K a condition known as film boiling (Chapter 10) will occur. The condition corresponds to a precipitous reduction in h and increase in T. 1000 Temperature, T(K) 900 800 700 600 500 400 300 100 400 800 2000 6000 10000 Convection coefficient, h(W/m^2.K) Although the canister remains well below the melting point of stainless steel for h = 100 W/m2⋅K, boiling should be avoided, in which case the convection coefficient should be maintained at h > 1000 W/m2⋅K. COMMENTS: The governing equation of part (a) is a first order, nonhomogenous differential equation with constant coefficients. Its solution is θ = (S/R ) 1 − e−Rt + θ i e−Rt , where θ ≡ T − T∞ , ) ( 3 2 3 ′′ S ≡ 3qi ri2 / ρ cp ( ro − ri3 ) , R = 3hro / ρ cp ( ro − ri3 ) . Note results for t → ∞ and for S = 0. PROBLEM 1.49 KNOWN: 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 equals boiling point of oxygen. ANALYSIS: (a) Applying an energy balance to a control surface about the container, it follows that, at any instant, Ein − Eout = 0 or qconv + q rad − q evap = 0 . The evaporative heat loss is equal to the product of the mass rate of vapor production and the heat of vaporization. Hence, h ( T − T ) + εσ T 4 − T 4 A − m (1) s sur s s evap h fg = 0 ∞ ( ( ) ) h ( T − T ) + εσ T 4 − T 4 π D2 ∞ s sur s mevap = h fg ( ) 10 W m 2 ⋅ K ( 298 − 263) K + 0.2 × 5.67 × 10−8 W m 2 ⋅ K 4 2984 − 2634 K 4 π (0.5 m )2 mevap = 214 kJ kg mevap = (350 + 35.2 ) W / m2 (0.785 m2 ) = 1.41×10−3 kg s . < 214 kJ kg (b) Using the energy balance, Eq. (1), the mass rate of vapor production can be determined for the range of emissivity 0.2 to 0.94. The effect of increasing emissivity is to increase the heat rate into the container and, hence, increase the vapor production rate. Evaporation rate, mdot*1000 (kg/s) 1.9 1.8 1.7 1.6 1.5 1.4 0.2 0.4 0.6 0.8 1 Surface emissivity, eps COMMENTS: To reduce the loss of oxygen due to vapor production, insulation should be applied to the outer surface of the container, in order to reduce qconv and qrad. Note from the calculations in part (a), that heat transfer by convection is greater than by radiation exchange. PROBLEM 1.50 KNOWN: Frost formation of 2-mm thickness on a freezer compartment. Surface exposed to convection 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 away from the exposed surface, (3) Negligible radiation exchange at the exposed surface, and (4) Backside surface of frost formation is adiabatic. PROPERTIES: Frost, ρ 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 applying an energy balance, Eq 1.11b, over the differential time interval dt and to a differential control volume extending inward from the surface of the layer dx. From the schematic above, the energy in is the convection heat flux over the time period dt and the change in energy storage is the latent energy change within the control volume, As⋅dx. Ein − E out = Est q′′ onv As dt = dUat c h As ( T∞ − Tf ) dt = − ρf As h sf dx Integrating both sides of the equation and defining appropriate limits, find t 0 h ( T∞ − Tf ) m dt = − ρf hsf dx 0 xo ∫ tm = tm = ∫ ρf h sf x o h ( T∞ − Tf ) 700 kg / m3 × 334 × 103 J / kg × 0.002m 2 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 total heat in by convection during the time interval t m ( q′′ ⋅ t m ) must be equal to the total latent energy for cv melting the frost layer ( ρ x o h sf ). This equality is directly comparable to the derived expression above for tm. (2) Explain why the energy storage term in the analysis has a negative sign, and the limits of integration are as shown. Hint: Recall from the formulation of Eq. 1.11b, that the storage term represents the change between the final and initial states. PROBLEM 1.51 KNOWN: 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 . 2 FIND: Instantaneous rate of melting per unit area, m′′ (kg/s⋅m ), and the material removed in a m period of 2 s, (a) Neglecting heat transfer from the irradiated surface by convection and radiation exchange, and (b) Allowing for convection and radiation exchange. SCHEMATIC: ASSUMPTIONS: (1) Woods metal slab is isothermal at the fusion temperature, Tf, and (2) The melt runs off the irradiated surface. ANALYSIS: (a) The instantaneous rate of melting per unit area may be determined by applying an energy balance, Eq 1.11a, on the metal slab at an instant of time neglecting convection and radiation exchange from the irradiated surface. E′′n − E′′ = E′′ i out st αG = d dM′′ ( −M′′h sf ) = −hsf dt dt where dM′′ / dt = m′′ is the time rate of change of mass in the control volume. Substituting values, m m m 0.4 × 5000 W / m 2 = −33, 000 J / kg × m′′ m′′ = −60.6 × 10−3 kg / s ⋅ m 2 The material removed in a 2s period per unit area is M′′ = m′′ ⋅ ∆t = 121 g / m 2 2s m (b) The energy balance considering convection and radiation exchange with the surroundings yields < < m α G − q′′ − q′′ = −h sf m′′ cv rad q′′ = h ( Tf − T∞ ) = 15 W / m 2 ⋅ K ( 72 − 20 ) K = 780 W / m 2 cv ( ) ( 4 q′′ = εσ Tf4 − T∞ = 0.4 × 5.67 × 10−8 W / m 2 ⋅ K [72 + 273] − [20 + 273] rad m m′′ = −32.3 × 10−3 kg / s ⋅ m 2 4 M 2s = 64 g / m 2 4 )K 4 = 154 W / m 2 < COMMENTS: (1) The effects of heat transfer by convection and radiation reduce the estimate for the material removal rate by a factor of two. The heat transfer by convection is nearly 5 times larger than by radiation exchange. (2) Suppose the work piece were horizontal, rather than vertical, and the melt puddled on the surface rather than ran off. How would this affect the analysis? (3) Lasers are common heating sources for metals processing, including the present application of melting (heat transfer with phase change), as well as for heating work pieces during milling and turning (laser-assisted machining). PROBLEM 1.52 KNOWN: Hot formed paper egg carton of prescribed mass, surface area and water content exposed to infrared heater providing known radiant flux. FIND: Whether water content can be reduced from 75% to 65% by weight during the 18s period 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 occurs from bottom side. PROPERTIES: Water (given): hfg = 2400 kJ/kg. ANALYSIS: Define a control surface about the carton, and write the conservation of energy requirement for an interval of time, ∆t, E in − E o ut = ∆ E st = 0 where Ein is due to the absorbed radiant flux, q ′′ , from the h heater and Eout is the energy leaving due to evaporation of water from the carton. Hence. q ′′ ⋅ A s ⋅ ∆t = ∆M ⋅ h fg . h For the prescribed radiant flux q ′′ , h q ′′A s∆t 5000 W / m2 × 0.0625 m2 × 18s ∆M = h = = 0.00234 kg. h fg 2400 kJ / kg The chief engineer’s requirement was to remove 10% of the water content, or ∆M req = M × 0.10 = 0.220 kg × 0.10 = 0.022 kg which is nearly an order of magnitude larger than the evaporative loss. Considering heat losses by convection and radiation, the actual water removal from the carton will be less than ∆M. Hence, the purchase should not be recommended, since the desired water removal cannot be achieved. < PROBLEM 1.53 KNOWN: Average heat sink temperature when total dissipation is 20 W with prescribed air and surroundings 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 transferred to the sink, (3) Sink is isothermal, (4) Surroundings and air temperature remain the same for both power levels, (5) Convection coefficient is the same for both power levels, (6) Heat sink is a small surface within a large enclosure, the surroundings. ANALYSIS: Define a control volume around the heat sink. Power dissipated within the devices is transferred into the sink, while the sink loses heat to the ambient air and surroundings by convection and radiation exchange, respectively. Ein − Eout = 0 (1) 44 Pe − hAs ( Ts − T∞ ) − Asεσ Ts − Tsur = 0. ) ( Consider the situation when Pe = 20 W for which Ts = 42°C; find the value of h. ( ) 44 h= Pe / As − εσ Ts − Tsur / ( Ts − T∞ ) h= 20 W/0.045 m 2 − 0.8 × 5.67 × 10−8 W/m 2 ⋅ K 4 3154 − 3004 K 4 / (315 − 300 ) K 2 h = 24.4 W / m ⋅ K. ) ( For the situation when Pe = 30 W, using this value for h with Eq. (1), obtain 30 W - 24.4 W/m2 ⋅ K × 0.045 m 2 (Ts − 300 ) K ( ) 4 −0.045 m 2 × 0.8 × 5.67 × 10−8 W/m 2 ⋅ K 4 Ts − 3004 K 4 = 0 ( ) 4 30 = 1.098 (Ts − 300 ) + 2.041× 10−9 Ts − 3004 . By trial-and-error, find Ts ≈ 322 K = 49 C. < COMMENTS: (1) It is good practice to express all temperatures in kelvin units when using energy balances involving radiation exchange. (2) Note that we have assumed As is the same for the convection and radiation processes. Since not all portions 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 1.54 KNOWN: Number and power dissipation of PCBs in a computer console. Convection coefficient associated with heat transfer from individual components in a board. Inlet temperature of cooling air and fan power requirement. Maximum allowable temperature rise of air. Heat flux from component most susceptible to thermal failure. FIND: (a) Minimum allowable volumetric flow rate of air, (b) Preferred location and corresponding surface temperature of most thermally sensitive component. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and kinetic energy changes of air flow, (4) Negligible heat transfer from console to ambient air, (5) Uniform convection coefficient for all components. ANALYSIS: (a) For a control surface about the air space in the console, conservation of energy for an open system, Eq. (1.11e), reduces to m ( u + pv ) − m ( u + pv ) + q − W = 0 i o where u + pv = i, q = 5Pb , and W = − Pf . Hence, with m (ii − io ) = mcp (Ti − To ) , mcp (To − Ti ) = 5 Pb + Pf For a maximum allowable temperature rise of 15°C, the required mass flow rate is m= 5 Pb + Pf 5 × 20 W + 25 W = = 8.28 ×10−3 kg/s cp ( To − Ti ) 1007 J/kg ⋅ K 15 $C ( ) The corresponding volumetric flow rate is ∀= m 8.28 × 10−3 kg/s = = 7.13 × 10−3 m3 / s 3 ρ 1.161 kg/m < (b) The component which is most susceptible to thermal failure should be mounted at the bottom of one of the PCBs, where the air is coolest. From the corresponding form of Newton’s law of cooling, q′′ = h ( Ts − Ti ) , the surface temperature is 4 2 q′′ $ C + 1× 10 W/m = 70$ C Ts = Ti + = 20 h 200 W/m2 ⋅ K < COMMENTS: (1) Although the mass flow rate is invariant, the volumetric flow rate increases as the air is heated in its passage through the console, causing a reduction in the density. However, for the prescribed temperature rise, the change in ρ, and hence the effect on ∀, is small. (2) If the thermally sensitive component were located at the top of a PCB, it would be exposed to warmer air (To = 35°C) and the surface temperature would be Ts = 85°C. PROBLEM 1.55 ′′ KNOWN: Top surface of car roof absorbs solar flux, qS,abs , and experiences for case (a): convection with 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 an instant of time, Ein − E out = 0. Neglecting radiation emission, the relevant processes are convection ′′ between the plate and the air, q′′ conv , and the absorbed solar flux, qS,abs . Considering the roof to have an area As , ′′ qS,abs ⋅ As − hAs ( Ts − T∞ ) = 0 ′′ Ts = T∞ + qS,abs /h Ts = 20 C + 800W/m 2 12W/m 2 ⋅ K = 20 C + 66.7 C = 86.7 C < (b) With radiation emission from the surface, the energy balance has the form ′′ qS,abs ⋅ As − q conv − E ⋅ As = 0 ′′ qS,abs As − hAs ( Ts − T∞ ) − ε Asσ Ts4 = 0 . Substituting numerical values, with temperature in absolute units (K), 800 W m2 − 12 W m2 ⋅ K (Ts − 293K ) − 0.8 × 5.67 ×10−8 W T4 = 0 2 ⋅ K4 s m 4 12Ts + 4.536 × 10−8 Ts = 4316 It follows that Ts = 320 K = 47°C. < 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 ambient air. For a convection coefficient of h = 40 W/m2⋅K, which would be typical for a velocity of 55 mph, the roof temperature would exceed the ambient temperature by less than 10°C. 360 Temperature, Ts(K) 350 340 330 320 310 300 290 0 20 40 60 80 100 120 140 160 180 200 Convection coefficient, h(W/m^2.K) COMMENTS: By considering radiation emission, Ts decreases, as expected. Note the manner in which q ′′ is formulated using Newton’s law of cooling; since q ′′ is shown leaving the control surface, the conv conv rate equation must be h ( Ts − T∞ ) and not h ( T∞ − Ts ) . PROBLEM 1.56 KNOWN: Detector and heater attached to cold finger immersed in liquid nitrogen. Detector surface of ε = 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) required to 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 to enclosure surface. PROPERTIES: Cold finger (given): k = 10 W/m⋅K. ANALYSIS: Define a control volume about detector and heater and apply conservation of energy requirement on a rate basis, Eq. 1.11a, Ein − Eout = 0 where (1) Ein = q rad + q elec ; E out = q cond (2,3) Combining Eqs. (2,3) with (1), and using the appropriate rate equations, ) ( 4 4 ε Asσ Tsur − Ts + q elec = kAs (Ts − TL )/L . (4) (a) Where q elec = 0, substituting numerical values ( ) 4 0.9 × 5.67 ×10−8 W/m 2 ⋅ K 4 3004 − Ts K 4 = 10W/m ⋅ K (Ts − 77 ) K/0.050 m ( ) 5.103 × 10−8 3004 − Ts4 = 200 (Ts − 77 ) Ts = 79.1K < Continued.…. PROBLEM 1.56 (Cont.) (b) When Ts = 195 K, Eq. (4) yields ) ( 0.9 × [π ( 0.005 m ) / 4] × 5.67 × 10−8 W/m 2 ⋅ K 4 3004 − 1954 K 4 + q elec 2 = 10W/m ⋅ K × [π (0.005 m ) /4] × (195 − 77 ) K / 0.050 m 2 < qelec = 0.457 W = 457 mW (c) Calculations were performed using the First Law Model for a Nonisothermal Plane Wall. With net radiative transfer to the detector fixed by the prescribed values of Ts and Tsur , Eq. (4) indicates that q elec increases linearly with increasing k. Heater power, qelec(W) 19 17 15 13 11 9 7 5 3 1 -1 0 100 200 300 400 Thermal conductivity, k(W/m.K) Heat transfer by conduction through the finger material increases with its thermal conductivity. Note that, for k = 0.1 W/m⋅K, q elec = -2 mW, where the minus sign implies the need for a heat sink, rather than a heat source, to maintain the detector at 195 K. In this case q rad exceeds q cond , and a heat sink would be needed to dispose of the difference. A conductivity of k = 0.114 W/m⋅K yields a precise balance between q rad and q cond . Hence to circumvent heaving to use a heat sink, while minimizing the heater power requirement, k should exceed, but remain as close as possible to the value of 0.114 W/m⋅K. 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/m⋅K at an average finger temperature of T = 136 K. For this value, q elec = 18 mW COMMENTS: The heater power requirement could be further reduced by decreasing ε. PROBLEM 1.57 KNOWN: Conditions at opposite sides of a furnace wall of prescribed thickness, thermal conductivity 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) Negligible radiation exchange at surface 1, (4) Surface 2 is exposed to large surroundings. ANALYSIS: The unknown temperatures may be obtained by simultaneously solving energy balance equations for the two surface. At surface 1, q′′ onv,1 = q′′ c cond ( ) h1 T∞,1 − T1 = k (T1 − T2 )/L (1) At surface 2, q′′ cond = q′′ conv + q′′ rad ( ) ( 4 4 k ( T1 − T2 )/L = h 2 T2 − T∞,2 + εσ T2 − Tsur ) (2) Surface temperature, T(K) Using the IHT First Law Model for a Nonisothermal Plane Wall, we obtain 1700 1500 1300 1100 900 700 500 300 0 0.1 0.2 0.3 0.4 0.5 Wall thickness, L(m) Inner surface temperature, T1(K) Outer surface temperature, T2(K) Continued ….. PROBLEM 1.57 (Cont.) Both q′′ cond and T2 decrease with increasing wall thickness, and for the prescribed value of h 2 = 10 2 W/m ⋅K, a value of L ≥ 0.275 m is needed to maintain T2 ≤ 373 K = 100 °C. Note that inner surface temperature 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 obtain 1700 1500 1300 1100 900 700 500 300 0 10 20 30 40 50 Convection 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/m2⋅K is needed to maintain T2 ≤ 373 K. The variation has a negligible effect on T1 , causing it to decrease slightly with increasing h 2 , but does have a strong influence on T2 . COMMENTS: If one wishes to avoid use of active (forced convection) cooling on side 2, reliance will have to be placed on free convection, for which h 2 ≈ 5 W/m2⋅K. The minimum wall thickness would then be L = 0.40 m. PROBLEM 1.58 KNOWN: Furnace wall with inner surface temperature T1 = 352°C and prescribed thermal conductivity 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 or increasing the convection coefficient h by a factor of two; benefit of applying a low emissivity coating (ε < 0.8); comment on the effectiveness of these strategies to reduce risk of burn injury when 2 T2 ≤ 65°C; and (b) Calculate and plot T2 as a function of h for the range 20 ≤ h ≤ 100 W/m ⋅K for three materials with k = 0.3, 0.6, and 1.2 W/m⋅K; what conditions will provide for safe outer surface temperatures. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Radiation exchange is between small surface and large enclosure, (4) Inner surface temperature remains constant for all conditions. ANALYSIS: (a) The surface (x = L) energy balance is ( T −T 4 4 k 1 2 = h ( T2 − T∞ ) + εσ T2 − Tsur L ) With T1 = 352°C, the effects of parameters h, k and ε on the outer surface temperature are calculated and tabulated below. Conditions Example 1.5 Decrease k by ½ Increase h by 2 Change k and h Decrease ε k (W / m ⋅ K ) 1.2 0.6 1.2 0.6 1.2 ( h W / m2 ⋅ K 20 20 40 40 20 ) ε T2 (°C ) 0.8 0.8 0.8 0.8 0.1 100 69 73 51 115 (b) Using the energy balance relation in the Workspace of IHT, the outer surface temperature can be calculated and plotted as a function of the convection coefficient for selected values of the wall thermal conductivity. Continued ….. O u te r su rfa ce te m p e ra tu re , T2 (C ) PROBLEM 1.58 (Cont.) 100 80 60 40 20 20 40 60 80 100 C o n ve ctio n co e fficie n t, h (W /m ^2 .K ) k = 1 .2 W /m .K k = 0 .6 W /m .K k = 0 .3 W /m .K COMMENTS: (1) From the parameter study of part (a), note that decreasing the thermal conductivity is more effective in reducing T2 than is increasing the convection coefficient. Only if both 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 that before you did the analysis? Give a physical explanation for this result. (3) From the parameter study graph we conclude that safe wall conditions (T2 ≤ 65°C) can be 2 maintained for these conditions: with k = 1.2 W/m⋅K when h > 55 W/m ⋅K; with k = 0.6 W/m⋅K 2 when h > 25 W/m ⋅K; and with k = 0.3 W/m⋅K when h > 20 W/m⋅K. PROBLEM 1.59 KNOWN: Inner surface temperature, thickness and thermal conductivity of insulation exposed 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 the insulation, (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 ) = h L (T2 − T∞ ). Solving for T2, find ( ) () 0.1 W/m ⋅ K W k 400 C + 500 35 C T1 + h T∞ 2 ⋅K 0.025m m T2 = L = k W 0.1 W/m ⋅ K h+ 500 + L 0.025m m2 ⋅ K T2 = 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 T2 is small. In this case convection makes the dominant contribution to heat transfer from the outer surface, and assumption (3) is excellent. PROBLEM 1.60 KNOWN: Thickness and thermal conductivity, k, of an oven wall. Temperature and emissivity, ε, of front 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 ε. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Radiation exchange with large surroundings. ANALYSIS: (a) Applying an energy balance, Eq. 1.13, at an instant of time to the front surface and substituting the appropriate rate equations, Eqs. 1.2, 1.3a and 1.7, find ) ( T −T 4 4 k 1 2 = h ( T2 − T∞ ) + εσ T2 − Tsur . L Substituting numerical values, find T1 − T2 = W W −8 ( 400 K )4 − (300 K )4 = 200 K . 20 2 100 K + 0.8 × 5.67 × 10 0.7 W/m ⋅ K m ⋅ K m2 ⋅ 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 Plane Wall. Changes in k strongly influence conditions for k < 20 W/m⋅K, but have a negligible effect for larger values, as T2 approaches T1 and the heat fluxes approach the corresponding limiting values 10000 Heat flux, q''(W/m^2) Temperature, T2(K) 600 500 8000 6000 4000 2000 400 0 0 100 200 300 Thermal conductivity, k(W/m.K) 300 0 100 200 300 Thermal conductivity, k(W/m.K) 400 Conduction heat flux, q''cond(W/m^2) Convection heat flux, q''conv(W/m^2) Radiation heat flux, q''rad(W/m^2) 400 PROBLEM 1.60 (Cont.) The implication is that, for k > 20 W/m⋅K, heat transfer by conduction in the wall is extremely efficient relative to heat transfer by convection and radiation, which become the limiting heat transfer processes. Larger fluxes could be obtained by increasing ε and h and/or by decreasing T∞ and Tsur . With increasing h, the front surface is cooled more effectively ( T2 decreases), and although q′′ rad decreases, the reduction is exceeded by the increase in q′′ conv . With a reduction in T2 and fixed values of k and L, q′′ cond must also increase. 30000 Heat flux, q''(W/m^2) Temperature, T2(K) 600 500 20000 10000 0 0 100 200 Convection coefficient, h(W/m^2.K) 400 0 100 Conduction heat flux, q''cond(W/m^2) Convection heat flux, q''conv(W/m^2) Radiation heat flux, q''rad(W/m^2) 200 Convection coefficient, h(W/m^2.K) The surface temperature also decreases with increasing ε, and the increase in q′′ exceeds the reduction rad in q′′ , allowing q′′ to increase with ε. conv cond 10000 Heat flux, q''(W/m^2) 575 Temperature, T2(K) 570 565 560 8000 6000 4000 2000 0 555 0 0.2 0.4 0.6 0.8 Emissivity 550 0 0.2 0.4 0.6 Emissivity 0.8 1 Conduction 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 prescribed conditions, q′′ cond = q′′ conv + q′′ . rad 1 PROBLEM 1.61 KNOWN: Temperatures at 10 mm and 20 mm from the surface and in the adjoining airflow for a thick steel casting. FIND: Surface convection coefficient, h. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in the x-direction, (3) Constant properties, (4) Negligible generation. ANALYSIS: From a surface energy balance, it follows that q ′′ ′′ cond = q conv where the convection rate equation has the form q′′ conv = h ( T∞ − T0 ) , and q ′′ cond can be evaluated from the temperatures prescribed at surfaces 1 and 2. That is, from Fourier’s law, T1 − T2 q′′ cond = k x 2 − x1 (50 − 40 ) C = 15, 000 W/m2 . W q′′ cond = 15 m ⋅ K ( 20-10 ) ×10−3 m Since the temperature gradient in the solid must be linear for the prescribed conditions, it follows that T0 = 60°C. Hence, the convection coefficient is h= h= q ′′ cond T∞ − T0 15,000 W / m2 40 C = 375 W / m2 ⋅ K. < COMMENTS: The accuracy of this procedure for measuring h depends strongly on the validity of the assumed conditions. PROBLEM 1.62 KNOWN: Duct wall of prescribed thickness and thermal conductivity experiences prescribed heat flux q′′ at outer surface and convection at inner surface with known heat transfer coefficient. o FIND: (a) Heat flux at outer surface required to maintain inner surface of duct at Ti = 85°C, (b) Temperature of outer surface, To , (c) Effect of h on To and q′′ . o SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Constant properties, (4) Backside of heater perfectly insulated, (5) Negligible radiation. ANALYSIS: (a) By performing an energy balance on the wall, recognize that balance on the top surface, it follows that q′′ = q′′ o cond . From an energy q′′ ond = q′′ c conv = q′′ . Hence, using the convection rate equation, o 2 2 q′′ = q′′ o conv = h ( Ti − T∞ ) = 100 W / m ⋅ K (85 − 30 ) C = 5500W /m . (b) Considering the duct wall and applying Fourier’s Law, q′′ = k o < T −T ∆T =k o i L ∆X q′′ L 5500 W/m 2 × 0.010 m = (85 + 2.8 ) C = 87.8 C . To = Ti + o = 85 C + k 20 W/m ⋅ K < (c) For Ti = 85°C, the desired results may be obtained by simultaneously solving the energy balance equations T − Ti T − Ti q′′ = k o ko = h ( Ti − T∞ ) and o L L Using the IHT First Law Model for a Nonisothermal Plane Wall, the following results are obtained. 91 Surface temperature, To(C) Heat flux, q''o(W/m^2) 12000 10000 8000 6000 4000 2000 90 89 88 87 86 0 85 0 40 80 120 160 Convection coefficient, h(W/m^2.K) 200 0 40 80 120 160 200 Convection coefficient, h(W/m^2.K) Since q′′ conv increases linearly with increasing h, the applied heat flux q′′ must be balanced by an o increase in q′′ , which, with fixed k, Ti and L, necessitates an increase in To . cond COMMENTS: The temperature difference across the wall is small, amounting to a maximum value of (To − Ti ) = 5.5°C for h = 200 W/m2⋅K. If the wall were thinner (L < 10 mm) or made from a material with higher conductivity (k > 20 W/m⋅K), this difference would be reduced. PROBLEM 1.63 KNOWN: Dimensions, average surface temperature and emissivity of heating duct. Duct air inlet temperature and velocity. Temperature of ambient air and surroundings. Convection coefficient. FIND: (a) Heat loss from duct, (b) Air outlet temperature. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and kinetic energy changes of air flow, (4) Radiation exchange between a small surface and a large enclosure. ANALYSIS: (a) Heat transfer from the surface of the duct to the ambient air and the surroundings is given by Eq. (1.10) ( 4 4 q = hAs ( Ts − T∞ ) + ε Asσ Ts − Tsur ) 2 where 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 × 10−8 W/m2 ⋅ K 4 3234 − 2784 K 4 q = 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 ) = q where the sign on q has been reversed to reflect the fact that heat transfer is from the system. With m = ρ VAc = 1.10 kg/m3 × 4 m/s (0.35m × 0.20m ) = 0.308 kg/s, the outlet temperature is q 5268 W To = Ti − = 58$ C − = 41$ C mcp 0.308 kg/s ×1008 J/kg ⋅ K < COMMENTS: The temperature drop of the air is large and unacceptable, unless the intent is to use the duct to heat the basement. If not, the duct should be insulated to insure maximum delivery of thermal energy to the intended space(s). PROBLEM 1.64 KNOWN: Uninsulated pipe of prescribed diameter, emissivity, and surface temperature in a room with 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 two 2 the convection coefficient (from 15 to 7.5 W/m ⋅K) or the emissivity (from 0.8 to 0.4) and (b) Show 2 graphically the heat loss as a function of the convection coefficient for the range 5 ≤ h ≤ 20 W/m ⋅K for emissivities of 0.2, 0.4 and 0.8. Comment on the relative efficacy of reducing heat losses associated with the convection and radiation processes. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between pipe and the room is between a small surface in a much larger enclosure, (3) The surface emissivity and absorptivity are equal, and (4) Restriction of the air flow does not alter the radiation exchange process between the pipe and the room. ANALYSIS: (a) The heat rate from the pipe to the room per unit length is ( 4 4 q′ = q′ / L = q′ conv + q′ = h (π D )( Ts − T∞ ) + ε (π D )σ Ts − Tsur rad ) Substituting numerical values for the two options, the resulting heat rates are calculated and compared with those for the conditions of Example 1.2. We conclude that both options are comparably effective. ( h W / m2 ⋅ K Conditions Base case, Example 1.2 Reducing h by factor of 2 Reducing ε by factor of 2 ) ε 15 7.5 15 0.8 0.8 0.4 q′ ( W / m ) 998 788 709 (b) Using IHT, the heat loss can be calculated as a function of the convection coefficient for selected values of the surface emissivity. Heat loss, q' (/m) 1200 800 400 0 5 10 15 20 Convection coefficient, h (W/m^2.K) eps = 0.8, bare pipe eps = 0.4, coated pipe eps = 0.2, coated pipe Continued ….. PROBLEM 1.64 (Cont.) COMMENTS: (1) In Example 1.2, Comment 3, we read that the heat rates by convection and radiation exchange were comparable for the base case conditions (577 vs. 421 W/m). It follows that reducing the key transport parameter (h or ε) by a factor of two yields comparable reductions in the heat loss. Coating the pipe to reduce the emissivity might to be the more practical option as it may be difficult to control air movement. (2) For this pipe size and thermal conditions (Ts and T∞), the minimum possible convection coefficient 2 is 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 heat loss as a function of the convection coefficient for selected emissivities is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. // Heat loss per unit pipe length; rate equation from Ex. 1.2 q' = q'cv + q'rad q'cv = pi*D*h*(Ts - Tinf) q'rad = pi*D*eps*sigma*(Ts^4 - Tsur^4) sigma = 5.67e-8 // Input parameters D = 0.07 Ts_C = 200 // Representing temperatures in Celsius units using _C subscripting Ts = Ts_C +273 Tinf_C = 25 Tinf = Tinf_C + 273 h = 15 // For graph, sweep over range from 5 to 20 Tsur_C = 25 Tsur = Tsur_C + 273 eps = 0.8 //eps = 0.4 // Values of emissivity for parameter study //eps = 0.2 /* Base case results Tinf Ts Tsur eps h 298 473 298 0.8 15 q' q'cv sigma 997.9 577.3 5.67E-8 */ q'rad D Tinf_C Ts_C Tsur_C 420.6 0.07 25 200 25 PROBLEM 1.65 KNOWN: Conditions associated with surface cooling of plate glass which is initially at 600°C. Maximum allowable temperature gradient in the glass. FIND: Lowest allowable air temperature, T∞ SCHEMATIC: 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 the instant 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 ) ( dT 4 4 − h ( Ts − T∞ ) − εσ Ts − Tsur = 0 dx or, with (dT/dx)max = -15°C/mm = -15,000°C/m and Tsur = T∞, −1.4 C W W −15, 000 = 5 (873 − T∞ ) K m⋅K m m2 ⋅ K +0.8 × 5.67 × 10−8 W 8734 − 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, 21,000 W W W ≈ 1275 2 + 19,730 2 . m m m 2 Hence the lowest allowable air temperature is T∞ ≈ 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 the cooling process. Accordingly, T∞ could be decreasing with increasing time and still keep within the maximum allowable temperature gradient. PROBLEM 1.66 KNOWN: Hot-wall oven, in lieu of infrared lamps, with temperature Tsur = 200°C for heating a coated 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 ≤ 250°C for ambient air temperatures of 20, 40 and 60°C; identify conditions for which acceptable curing temperatures between 100 and 110°C 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. E′′n − E′′ = 0 i out ( or ) q′′ − q′′ rad conv = 0 4 εσ Tsur − Ts4 − h (Ts − T∞ ) = 0 0.5 × 5.67 × 10−8 W / m 2 ⋅ K 4 ([200 + 273] − T ) K 4 4 s 4 − 15 W / m 2 ⋅ K (Ts − [20 + 273]) K = 0 < Ts = 357 K = 84°C (b) Using the energy balance relation in the Workspace of IHT, the plate temperature can be calculated and plotted as a function of oven wall temperature for selected ambient air temperatures. Plate temperature, Ts (C) 150 100 50 150 175 200 225 250 Oven wall temperature, Tsur (C) Tinf = 60 C Tinf = 40 C Tinf = 20 C COMMENTS: From the graph, acceptable cure temperatures between 100 and 110°C can be maintained for these conditions: with T∞ = 20°C when 225 ≤ Tsur ≤ 240°C; with T∞ = 40°C when 205 ≤ Tsur ≤ 220°C; and with T∞ = 60°C when 175 ≤ Tsur ≤ 195°C. PROBLEM 1.67 KNOWN: Operating conditions for an electrical-substitution radiometer having the same receiver temperature, Ts, in electrical and optical modes. FIND: Optical power of a laser beam and corresponding receiver temperature when the indicated electrical power is 20.64 mW. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Conduction losses from backside of receiver negligible in optical mode, (3) Chamber walls form large isothermal surroundings; negligible effects due to aperture, (4) Radiation exchange between the receiver surface and the chamber walls is between small surface and large enclosure, (5) Negligible convection effects. PROPERTIES: Receiver surface: ε = 0.95, αopt = 0.98. ANALYSIS: The schematic represents the operating conditions for the electrical mode with the optical beam blocked. The temperature of the receiver surface can be found from an energy balance on the receiver, considering the electrical power input, conduction loss from the backside of the receiver, and the radiation exchange between the receiver and the chamber. Ein − Eout = 0 Pelec − q loss − q rad = 0 ( ) 4 4 Pelec − 0.05 Pelec − ε Asσ Ts − Tsur = 0 20.64 × 10 −3 ( ) ( ) 2 2 4 4 4 4 2 −8 W (1 − 0.05 ) − 0.95 π 0.015 / 4 m × 5.67 × 10 W / m ⋅ K Ts − 77 K = 0 Ts = 213.9 K < For the optical mode of operation, the optical beam is incident on the receiver surface, there is no electrical power input, and the receiver temperature is the same as for the electrical mode. The optical power of the beam can be found from an energy balance on the receiver considering the absorbed beam power and radiation exchange between the receiver and the chamber. Ein − Eout = 0 α opt Popt − q rad = 0.98 Popt − 19.60 mW = 0 Popt = 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 the same for both modes, an energy balance could be directly written in terms of the absorbed optical power and equivalent electrical power, αopt Popt = Pelec - qloss. PROBLEM 1.68 KNOWN: Surface temperature, diameter and emissivity of a hot plate. Temperature of surroundings and ambient air. Expression for convection coefficient. FIND: (a) Operating power for prescribed surface temperature, (b) Effect of surface temperature on power requirement and on the relative contributions of radiation and convection to heat transfer from the surface. SCHEMATIC: ASSUMPTIONS: (1) Plate is of uniform surface temperature, (2) Walls of room are large relative to plate, (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 ′′ ). rad 1/3 Substituting for the area of the plate and from Eqs. (1.3a) and (1.7), with h = 0.70 (Ts - T∞) , it follows that ( 2 Pelec = π D / 4 ) ( 4 4 4/3 0.70 ( Ts − T∞ ) + εσ Ts − Tsur ) ( 4 4 −8 2 4/3 Pelec = π ( 0.3m ) / 4 0.70 (175 ) 473 − 298 + 0.8 × 5.67 × 10 Pelec = 0.0707 m ) W/m 2 2 2 685 W/m + 1913 W/m = 48.4 W + 135.2 W = 190.6 W 2 < (b) As shown graphically, both the radiation and convection heat rates, and hence the requisite electric power, 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 s H e a t ra te (W ) 500 400 300 200 100 0 100 150 200 250 300 S u rfa c e te m p e ra tu re (C ) P e le c q ra d q co n v However, because of its dependence on the fourth power of the surface temperature, the increase in radiation is more pronounced. The significant relative effect of radiation is due to the small 2 convection coefficients characteristic of natural convection, with 3.37 ≤ h ≤ 5.2 W/m ⋅K for 100 ≤ Ts < 300°C. COMMENTS: Radiation losses could be reduced by applying a low emissivity coating to the surface, which would have to maintain its integrity over the range of operating temperatures. PROBLEM 1.69 KNOWN: 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 operating temperature of the bus bar as a function of the convection coefficient for the range 10 ≤ h ≤ 100 2 W/m ⋅K. Minimum convection coefficient required to maintain a safe-operating temperature below 120°C. 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 a small surface and a large enclosure. PROPERTIES: Bus-bar material, ρ e = ρ e,o [1 + α ( T − To )], ρ e,o = 0.0828 µΩ ⋅ m, To = 25°C, α = 0.0040 K −1 . ANALYSIS: (a) An energy balance on the bus-bar for a unit length as shown in the schematic above has the form ′ Ein − E′ + E′ = 0 out gen 2 −q ′ − q ′ rad conv + I R ′ = 0 e ) ( 4 −ε Pσ T 4 − Tsur − h P (T − T∞ ) + I 2 ρe / Ac = 0 where P = 2 ( H + W ) , R ′ = ρ e / A c and A c = H × W. Substituting numerical values, e 4 −0.8 × 2 0.600 + 0.200 m × 5.67 × 10−8 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 ×10 −6 ) } Ω ⋅ m 1 + 0.0040 K −1 ( T − [25 + 273]) K / ( 0.600 × 0.200 ) m 2 = 0 Solving for the bus-bar temperature, find < T = 426 K = 153°C. (b) Using the energy balance relation in the Workspace of IHT, the bus-bar operating temperature is 2 calculated as a function of the convection coefficient for the range 10 ≤ h ≤ 100 W/m ⋅K. From this graph we can determine that to maintain a safe operating temperature below 120°C, the minimum convection coefficient required is h 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 radiation as a function of the bus-bar temperature. 3000 H e a t ra te s, q 'cv o r q 'ra d (W /m ) 175 B a r te m p e ra tu re , T (C ) 150 125 100 75 2000 1000 0 50 25 50 75 100 125 150 175 B u s b a r te m p e ra tu re , T (C ) 25 0 20 40 60 80 100 C o n ve c tio n h e a t flu x, q 'cv R a d ia tio n e xc h a n g e , q 'ra d , e p s = 0 .8 C 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 (120°C), convection and radiation exchange heat transfer rates become comparable. Notice that the relative importance of the radiation exchange rate increases with increasing bus-bar temperature. COMMENTS: (1) It follows from the second graph that increasing the surface emissivity will be only 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 is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. /* Results for base case conditions: Ts_C q'cv q'rad rhoe H eps h 153.3 1973 1786 1.253E-7 0.6 0.8 10 */ I Tinf_C Tsur_C W alpha 6E4 30 30 0.2 0.004 // Surface energy balance on a per unit length basis -q'cv - q'rad + Edot'gen = 0 q'cv = h * P * (Ts - Tinf) P = 2 * (W + H) // perimeter of the bar experiencing surface heat transfer q'rad = eps * sigma * (Ts^4 - Tsur^4) * P sigma = 5.67e-8 Edot'gen = I^2 * Re' Re' = rhoe / Ac rhoe = rhoeo * ( 1 + alpha * (Ts - Teo)) Ac = W * H // Input parameters I = 60000 alpha = 0.0040 rhoeo = 0.0828e-6 Teo = 25 + 273 W = 0.200 H = 0.600 Tinf_C = 30 Tinf = Tinf_C + 273 h = 10 eps = 0.8 Tsur_C = 30 Tsur = Tsur_C + 273 Ts_C = Ts - 273 // temperature coefficient, K^-1; typical value for cast aluminum // electrical resistivity at the reference temperature, Teo; microohm-m // reference temperature, K PROBLEM 1.70 KNOWN: Solar collector designed to heat water operating under prescribed solar irradiation and loss conditions. FIND: (a) Useful heat collected per unit area of the collector, q ′′ , (b) Temperature rise of the water u flow, 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/kg⋅K. ANALYSIS: (a) Defining the collector as the control volume and writing the conservation of energy requirement on a per unit area basis, find that E in − E out + E gen = E st . Identifying processes as per above right sketch, q ′′ − q ′′ − q ′′ − q ′′ = 0 solar rad conv u where q ′′ ′′ solar = 0.9 q s ; that is, 90% of the solar flux is absorbed in the collector (Eq. 1.6). Using the appropriate rate equations, the useful heat rate per unit area is ( ) 4 4 q′′ = 0.9 q′′ − εσ Tcp − Tsky − h ( Ts − T∞ ) u s W W W q′′ = 0.9 × 700 3034 − 2634 K 4 − 10 − 0.94 × 5.67 × 10−8 (30 − 25) C u 2 2 K4 2K m m⋅ m⋅ ) ( 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, the u useful heat causes an enthalpy change of the flowing water. That is, q′′ ⋅ A=mcp ( Ti − To ) u or (Ti − To ) = 386 W/m2 × 3m2 / 0.01kg/s × 4179J/kg ⋅ K=27.7C. ( )( ) (c) The efficiency is η = 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 temperature Tsky. PROBLEM 1.71 KNOWN: Surface-mount transistor with prescribed dissipation and convection cooling conditions. FIND: (a) Case temperature for mounting arrangement with air-gap and conductive paste between case and circuit board, (b) Consider options for increasing E g , subject to the constraint that Tc = 40°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Transistor case is isothermal, (3) Upper surface experiences convection; negligible losses from edges, (4) Leads provide conduction path between case and 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/m⋅K; Paste, k g,p = 0.12 W/m⋅K; Metal leads, k = 25 W/m⋅K. ANALYSIS: (a) Define the transistor as the system and identify modes of heat transfer. Ein − E out + Eg = ∆Est = 0 −q conv − q cond,gap − 3q lead + E g = 0 T − Tb T − Tb − hAs ( Tc − T∞ ) − k g As c − 3k Ac c + Eg = 0 t L where 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/m⋅K) { ( ) Tc = 50W/m 2 ⋅ K × 32 × 10−6 m 2 × 20 C + 0.0263W/m ⋅ K × 32 × 10−6 m 2 /0.2 × 10−3 m −8 2 −3 −3 −3 −3 +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/m⋅K), Tc = 39.9°C. As expected, the effect of the conductive paste 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 the desired calculations. For values of k = 200 and 400 W/m⋅K and convection coefficients in the range from 50 to 250 W/m2⋅K, the energy balance equation may be used to compute the power dissipation for a maximum allowable case temperature of 40°C. 0.7 0.6 0.5 0.4 0.3 50 100 150 200 250 Convection coefficient, h(W/m^2.K) kl = 400 W/m.K kl = 200 W/m.K As indicated by the energy balance, the power dissipation increases linearly with increasing h, as well as with increasing k . For h = 250 W/m2⋅K (enhanced air cooling) and k = 400 W/m⋅K (copper leads), the transistor may dissipate up to 0.63 W. COMMENTS: Additional benefits may be derived by increasing heat transfer across the gap separating the 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: qS ′′ Incident solar radiation, a large portion of which q S,abs , is absorbed by the asphalt ′′ surface, q ′′ rad Radiation emitted by the surface to the air, q conv Convection heat transfer from the surface to the air, and ′′ q 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 . ′′ ′′ ′′ rad COMMENTS: (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 the underlying pavement and the air, in which case heat transfer by conduction and convection are from the surface. (3) For simplicity, radiation incident on the pavement due to atmospheric emission has been ignored (see Section 12.8 for a discussion). Eq. 1.6 may then be used for the absorbed solar irradiation and Eq. 1.5 may be used to obtain the emitted radiation q ′′ . rad (4) With the rate equations, the energy balance becomes 4 ′′ qS,abs − ε σ Ts − h ( Ts − T∞ ) = − k dT . dx s PROBLEM 1.72(b) KNOWN: Physical mechanism for microwave heating. FIND: Comparison of (a) cooking in a microwave oven with a conventional radiant or convection oven and (b) a microwave clothes dryer with a conventional dryer. (a) Microwave cooking occurs as a result of volumetric thermal energy generation throughout the food, without heating of the food container or the oven wall. Conventional cooking relies on radiant heat transfer from the oven walls and/or convection heat transfer from the air space to 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 heating the water, energy would go directly into evaporation, unlike a conventional dryer where the walls and air are first heated electrically or by a gas heater, and thermal energy is subsequently transferred to the wet clothes. The microwave dryer would still require a rotating drum and air flow to remove the water vapor, but is able to operate more efficiently and at lower temperatures. For a more detailed description of microwave drying, see Mechanical Engineering, 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 processes are: q conv Convection heat transfer from the skin to the room air, and q rad Net 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∞ = 15°C), the inner surface temperature of the outside walls, Tsur, will decrease with decreasing outside air temperature. Upon exposure to these walls, body heat loss will be larger due to increased qrad. COMMENTS: The foregoing reasoning assumes that the thermostat measures the true room air 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,f Radiation emitted by the tungsten filament, a portion of which is transmitted through the glass, q conv,f Free convection from filament to air of temperature Ta,i < Tf , q rad,g,i Radiation emitted by inner surface of glass, a small portion of which is intercepted by the filament, q conv,g,i Free convection from air to inner glass surface of temperature Tg,i < Ta,i , q cond,g Conduction through glass wall, q conv,g,o Free convection from outer glass surface to room air of temperature Ta,o < Tg,o , and q rad,g-sur Net radiation heat transfer between outer glass surface and surroundings, such as the walls of a room, of temperature Tsur < Tg,o . COMMENTS: If the glass bulb is evacuated, no convection is present within the bulb; that is, 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 surface slabs. 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 is transferred through the composite by q cond,i Conduction through the inner solid slab, q conv,hc Free convection through the cellular airspace, q cond,hc Conduction through the honeycomb wall, q rad,hc Radiation between the honeycomb surfaces, and q cond,o Conduction through the outer solid slab. Heat may then be transferred from the outer surface by convection and radiation. Note that for 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 thermal conductivity, k, and emissivity, ε. Evacuating the airspace would enhance performance by eliminating heat transfer due to free convection. PROBLEM 1.72(f) KNOWN: A thermocouple junction is used, with or without a radiation shield, to measure the temperature of a gas flowing through a channel. The wall of the channel is at a temperature much less than that of the gas. FIND: (a) Relevant heat transfer processes, (b) Temperature of junction relative to that of gas, (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 rad Net radiation transfer from the junction to the walls, and q conv Convection transfer from the gas to the junction. (b) From a surface energy balance on the junction, q conv = q rad or from Eqs. 1.3a and 1.7, ( ) ( ) 4 4 h A Tj − Tg = ε A σ Tj − Ts . To satisfy this equality, it follows that Ts < 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 the schematic. The junction now exchanges radiation with the shield, whose temperature must exceed that of the channel wall. The radiation loss from the junction is therefore reduced, and its 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,1 Radiation from flames and cavity wall, portions of which are absorbed and transmitted by the two panes, q rad,2 Emission from inner surface of inner pane to cavity, q rad,3 Net radiation exchange between outer surface of inner pane and inner surface of outer pane, q rad,4 Net radiation exchange between outer surface of outer pane and walls of room, q conv,1 Convection between cavity gases and inner pane, q conv2 Convection across air space between panes, q conv,3 Convection from outer surface to room air, q cond,1 Conduction across inner pane, and q cond,2 Conduction across outer pane. COMMENTS: (1) Much of the luminous portion of the flame radiation is transmitted to the room 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 right schematic) glass panes include. q conv,1 Convection from room air to inner surface of first pane, q rad,1 Net radiation exchange between room walls and inner surface of first pane, q cond,1 Conduction through first pane, q conv,s Convection across airspace between panes, q rad,s Net radiation exchange between outer surface of first pane and inner surface of second pane (across airspace), q cond,2 Conduction through a second pane, q conv,2 Convection from outer surface of single (or second) pane to ambient air, q rad,2 Net radiation exchange between outer surface of single (or second) pane and surroundings such as the ground, and qS Incident solar radiation during day; fraction transmitted to room is smaller for double pane. COMMENTS: Heat loss from the room is significantly reduced by the double pane construction. 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: qS Incident solar radiation, a large portion of which is absorbed by the absorber plate. Reduced with use of cover plate (primarily due to reflection off cover plate). q rad,∞ Net radiation exchange between absorber plate or cover plate and surroundings, q conv,∞ Convection from absorber plate or cover plate to ambient air, q rad,a-c Net radiation exchange between absorber and cover plates, q conv,a-c Convection heat transfer across airspace between absorber and cover plates, q cond Conduction through insulation, and q conv Convection to working fluid. COMMENTS: The cover plate acts to significantly reduce heat losses by convection and radiation 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 the cover plates, the relevant processes are: q conv,a-i Convection from inside air to inner surface, q rad,p-i Net radiation transfer from absorber plates to inner surface, q conv,i-o Convection across airspace between covers, q rad,i-o Net radiation transfer from inner to outer cover, q conv,o-∞ Convection from outer cover to ambient air, q rad,o Net radiation transfer from outer cover to surroundings, and qS Incident solar radiation. Additional processes relevant to the absorber plates and airspace are: q S,t Solar radiation transmitted by cover plates, q conv,p-a Convection from absorber plates to inside air, and q cond Conduction 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: qS Incident solar radiation including contribution due to reflection off panel (most is transmitted), q conv,o Convection heat transfer from outer surface to ambient air, q rad,o-sur Net rate of radiation heat exchange between outer surface of outer tube and the surroundings, including the panel, q S,t Solar radiation transmitted through outer tube and incident on inner tube (most is absorbed), q rad,i-o Net rate of radiation heat exchange between outer surface of inner tube and inner surface of outer tube, and q conv,i Convection heat transfer to working fluid. There is also conduction heat transfer through the inner and outer tube walls. If the walls are thin, the temperature drop across the walls will be small. PROBLEM 2.1 KNOWN: 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) No internal heat generation. ANALYSIS: Performing an energy balance on the object according to Eq. 1.11a, E in − E out = 0, it follows that E in − E out = q x $ and that q x ≠ q x x . That is, the heat rate within the object is everywhere constant. From Fourier’s law, q x = − kA x dT , dx and since qx and k are both constants, it follows that Ax dT = Constant. dx That is, the product of the cross-sectional area normal to the heat rate and temperature gradient remains a constant and independent of distance x. It follows that since Ax increases with x, then dT/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? x PROBLEM 2.2 KNOWN: 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) No internal heat generation, (4) Insulation has uniform properties independent of temperature and position. ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional (cylindrical) radial system has the form q r = − kA r 16 dT dT = − k 2πr dr dr where A r = 2πr 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 requires E in = E out since E g = E st = 0. Hence qr = Constant. That is, qr is independent of radius (r). Since the thermal conductivity is also constant, it follows that r 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 must appear as shown in the sketch. COMMENTS: (1) Note that, while qr is a constant and independent of r, q ′′ is not a constant. How r 16 does q ′′ r vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with r increasing radius. PROBLEM 2.3 KNOWN: 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 (spherical coordinates) direction, (3) No internal generation, (4) Constant properties. ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional, radial (spherical coordinate) system has the form qr = −k Ar ( ) dT dT = − k 4π r 2 dr dr where Ar is the surface area of a sphere. For steady-state conditions, an energy balance on the system yields 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 is constant, it follows that dT r 2 = Constant. dr This relation requires that the product of the radial temperature gradient, dT/dr, and the radius 2 squared, r , remains constant throughout the shell. Hence, the temperature distribution appears as shown 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? r PROBLEM 2.4 KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature distribution 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, since E in = E out , q x = Constant ≠ f ( x ). Using Fourier’s law, Eq. 2.1, with appropriate expressions for Ax and T, yields dT dx d K 6000W=-k ⋅ (1-x ) m2 ⋅ 300 1 − 2x-x3 . m dx q x = −k A x ) ( Solving for k and recognizing its units are W/m⋅K, k= -6000 ( ) (1-x ) 300 −2 − 3x 2 = 20 (1 − x ) ( 2 + 3x 2 ) . < COMMENTS: (1) At x = 0, k = 10W/m⋅K and k → ∞ as x → 1. (2) Recognize that the 1-D assumption is an approximation which becomes more inappropriate as the area change with x, and hence two-dimensional effects, become more pronounced. PROBLEM 2.5 KNOWN: 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. x SCHEMATIC: 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, that for a differential control volume, E in = E out or q x = q x+dx . Hence qx is independent of x. $ Since A(x) increases with increasing x, it follows that q ′′ = q x / A x decreases with increasing x. x Since T decreases with increasing x, k increases with increasing x. Hence, from Fourier’s law, Eq. 2.2, q ′′ = − k x dT , dx it follows that | dT/dx | decreases with increasing x. PROBLEM 2.6 KNOWN: Temperature dependence of the thermal conductivity, k(T), for heat transfer through a plane wall. FIND: Effect of k(T) on temperature distribution, T(x). ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) No internal heat generation. ANALYSIS: From Fourier’s law and the form of k(T), q ′′ = − k x 1 6 dT dT = − k o + aT . dx dx (1) 2 2 The 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, x 1 ! 6 "#$ 2 d 2 T dT " −1 k o + aT6 2 − a ! dx #$ = 0. dx dq ′′ dT x =- d k o + aT =0 dx dx dx Hence, d 2T "#2 !$ -a dT = 2 k o + aT dx dx %k o + aT = k > 0 K where & dT " 2 K! dx #$ > 0 ' from which it follows that for a > 0: d 2 T / dx 2 < 0 a = 0: d 2 T / dx 2 = 0 a < 0: d 2 T / dx 2 > 0. COMMENTS: The shape of the distribution could also be inferred from Eq. (1). Since T decreases with increasing x, a > 0: k decreases with increasing x = > | dT/dx | increases with increasing x a = 0: k = ko = > dT/dx is constant a < 0: k increases with increasing x = > | dT/dx | decreases with increasing x. PROBLEM 2.7 KNOWN: Thermal conductivity and thickness of a one-dimensional system with no internal heat generation 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-state conditions, (4) Constant properties. ANALYSIS: The rate equation and temperature gradient for this system are dT dT T1 − T2 = q′′ = −k and . x dx dx L Using Eqs. (1) and (2), the unknown quantities can be determined. dT ( 400 − 300 ) K (a) = = 200 K/m dx 0.5m q′′ = −25 x (b) W m⋅K × 200 K m < = −5000 W/m 2 . K = 6250 W/m 2 m⋅K m K dT T2 = T1 − L = 1000 C-0.5m -250 m dx q′′ = −25 x W × −250 T2 = 225 C. (c) q′′ = −25 x W m⋅K × 200 < K m = −5000 W/m 2 T2 = 80 C-0.5m 200 (d) K = −20 C. m < q′′ 4000 W/m 2 K =− x =− = −160 dx k 25 W/m ⋅ K m dT T1 = L () dT + T = 0.5m -160 K + −5 C dx 2 m T1 = −85 C. (e) (1,2) dT =− q′′ x (−3000 W/m2 ) = 120 K =− 25 W/m ⋅ K K T2 = 30 C-0.5m 120 = −30 C. m dx k m 2 a +b 2 < PROBLEM 2.8 KNOWN: 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 heat generation, (4) Constant properties. ANALYSIS: The rate equation and temperature gradient for this system are dT dT T2 − T1 q′′ = −k and . = x dx dx L Using Eqs. (1) and (2), the unknown quantities for each case can be determined. (a) (b) (c) dT ( −20 − 50 ) K = −280 K/m 0.25m W K q′′ = −50 × −280 = 14.0 kW/m 2 . x m⋅K m dx dT = q′′ = −50 x T2 = L ⋅ m⋅K W dT dx × 160 K m < = −8.0 kW/m 2 + T1 = 0.25m × 160 K + 70 C. m < T2 = 110 C. (d) q′′ = −50 x m⋅K W T1 = T2 − L ⋅ × −80 dT dx K m = 4.0 kW/m 2 = 40 C − 0.25m −80 K m .∈ T1 = 60 C. q′′ = −50 x < m⋅K W × 200 K 2 = −10.0 kW/m m dT K = 30 C − 0.25m 200 = −20 C. T1 = T2 − L ⋅ dx m (e) < ( −10 − ( −30 )) K = 80 K/m 0.25m W K × 80 = −4.0 kW/m 2 . q′′ = −50 x m⋅K m dx = (1,2) < PROBLEM 2.9 KNOWN: Plane wall with prescribed thermal conductivity, thickness, and surface temperatures. FIND: Heat flux, q ′′ , and temperature gradient, dT/dx, for the three different coordinate systems x shown. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional heat flow, (2) Steady-state conditions, (3) No internal generation, (4) Constant properties. ANALYSIS: The rate equation for conduction heat transfer is q ′′ = − k x dT , dx (1) where the temperature gradient is constant throughout the wall and of the form $ $ dT T L − T 0 = . dx L (2) Substituting numerical values, find the temperature gradients, $ < 400 − 600 K dT T1 − T2 = = = −2000 K / m dx L 0.100m $ < 600 − 400 K dT T2 − T1 = = = 2000 K / m. dx L 0.100m $ < (a) 600 − 400 K dT T2 − T1 = = = 2000 K / m dx L 0.100m (b) (c) The heat rates, using Eq. (1) with k = 100 W/m⋅K, are (a) q ′′ = −100 x W × 2000 K / m = -200 kW / m2 m⋅ K < (b) q ′′ = −100 x W ( −2000 K / m) = +200 kW / m2 m⋅ K < (c) q ′′ = −100 x W × 2000 K / m = -200 kW / m2 m⋅ K < PROBLEM 2.10 KNOWN: 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) Constant properties. ANALYSIS: The heat rate from Fourier’s law for the radial (cylindrical) system has the form q r = − kA r dT . dr 2 Substituting for the temperature distribution, T(r) = a + br , 16 q r = − k 2πrL 2br = - 4πkbLr 2 . At the outer surface ( r = ro), the conduction heat rate is 2 q r=ro = −4πkbLro . < From a surface energy balance at r = ro, 1 6 16 q r=ro = q conv = h 2πro L T ro − T∞ , Substituting for q r=ro and solving for T∞, 16 T∞ = T ro + 2 kbro h 2 T∞ = a + bro + ! 2 kbro h T∞ = a + bro ro + "# $ 2k . h < PROBLEM 2.11 KNOWN: Two-dimensional body with specified thermal conductivity and two isothermal surfaces of 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 heat generation, (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 an isothermal surface. The heat rate at the surface A is given by Fourier’s law written as q′ = −k ⋅ wA y,A W K ∂T = −10 × 2 m × 30 = −600W / m. m⋅ K m ∂y A On the surface B, it follows that ∂T / ∂y$B = 0 < in order to satisfy the requirement that the heat flux vector be normal to the isothermal surface B. Using the conservation of energy requirement, Eq. 1.11a, on the body, find q′ − q′ = 0 y,A x,B Note that, q′ = −k ⋅ w B x,B or q′ = q′ . x,B y,A ∂T ∂x B and hence ′ −− y,A ∂T / ∂x$B = −qw B = 10 W600 ⋅W /×m$ = 60 K / m. k⋅ / m K 1m < COMMENTS: Note that, in using the conservation requirement, q ′ = + q ′ in y,A and q ′ = + q ′ . out x,B PROBLEM 2.12 KNOWN: 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) Constant properties. ANALYSIS: Temperatures at the top and bottom of the shaft are, respectively, T(0) = 100°C < T(L) = -40°C. Applying Fourier’s law, Eq. 2.1, " ' $ dT = −25 W / m ⋅ K 0.005 m2 −150 + 20x C / m dx q x = 0125 150 - 20x W. . q x = − kA $ Hence, qx(0) = 18.75 W qx(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 shaft cross-section. Hence, specification of T as a function of only x is an approximation. PROBLEM 2.13 KNOWN: A rod of constant thermal conductivity k and variable cross-sectional area Ax(x) = Aoeax where Ao and a are constants. FIND: (a) Expression for the conduction heat rate, qx(x); use this expression to determine the temperature distribution, T(x); and sketch of the temperature distribution, (b) Considering the presence of volumetric heat generation rate, q = q o exp ( −ax ) , obtain an expression for qx(x) when the 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 = 0 q x − q x + dx + q ⋅ A ( x ) ⋅ dx = 0 The conduction heat rate terms can be expressed as a Taylor series and substituting expressions for q and A(x), − d (q x ) + qo exp ( −ax ) ⋅ Ao exp (ax ) = 0 dx q x = −k ⋅ A ( x ) (1) dT dx (2) (a) With no internal generation, q o = 0, and from Eq. (1) find − d (q x ) = 0 dx < indicating that the heat rate is constant with x. By combining Eqs. (1) and (2) − d dT −k ⋅ A ( x ) = 0 dx dx or A (x )⋅ dT = C1 dx (3) < Continued... PROBLEM 2.13 (Cont.) That is, the product of the cross-sectional area and the temperature gradient is a constant, independent of x. Hence, with T(0) > T(L), the temperature distribution is exponential, and as shown in the sketch above. Separating variables and integrating Eq. (3), the general form for the temperature distribution can be determined, Ao exp ( ax ) ⋅ dT = C1 dx − dT = C1Ao 1 exp ( − ax ) dx T ( x ) = −C1Ao a exp ( −ax ) + C2 < We could use the two temperature boundary conditions, To = T(0) and TL = T(L), to evaluate C1 and C2 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 = 0 dx or q x = qo Ao x < That is, the heat rate increases linearly with x. COMMENTS: In part (b), you could determine the temperature distribution using Fourier’s law and knowledge of the heat rate dependence upon the x-coordinate. Give it a try! PROBLEM 2.14 KNOWN: 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 of 50°C = 323K and are summarized below. ANALYSIS: The heat transfer rate may be obtained from Fourier’s law. Since the axial temperature gradient is linear, this expression reduces to $ 100 − 0$° C = 0.491m⋅° C$ ⋅ k π 0.025m T −T q = kA 1 2 = k L 4 2 0.1m Cu Al St.St. SiN Oak Magnesia Pyrex (pure) (2024) (302) (85%) _______________________________________________________________ k(W/m⋅K) 401 177 16.3 14.9 0.19 0.052 1.4 q(W) 197 87 8.0 7.3 0.093 0.026 0.69 < 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 obtained by linear interpolation; and the values for oak and pyrex are for 300 K. PROBLEM 2.15 KNOWN: One-dimensional system with prescribed surface temperatures and thickness. FIND: Heat flux through system constructed of these materials: (a) pure aluminum, (b) plain carbon steel, (c) AISI 316, stainless steel, (d) pyroceram, (e) teflon and (f) concrete. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No heat generation, (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, Fourier’s law can be written as q ′′ = − k x dT T −T = −k 2 1 . dx L Substituting numerical values, the heat flux is q ′′ = − k x 275- 325$K = +2500 K ⋅ k 20 × 10-3 m m 2 where q ′′ will have units W/m if k has units W/m⋅K. The heat fluxes for each system follow. x Thermal conductivity Material (a) (b) (c) (d) (e) (f) Pure Aluminum Plain carbon steel AISI 316, S.S. Pyroceram Teflon Concrete Table A-1 A-1 A-1 A-2 A-3 A-3 k(W/m⋅K) 237 60.5 13.4 3.98 0.35 1.4 Heat flux " q ′′ kW / m2 x 593 151 33.5 9.95 0.88 3.5 ' < COMMENTS: Recognize that the thermal conductivity of these solid materials varies by more than two orders of magnitude. PROBLEM 2.16 KNOWN: Different thicknesses of three materials: rock, 18 ft; wood, 15 in; and fiberglass insulation, 6 in. FIND: The insulating quality of the materials as measured by the R-value. PROPERTIES: Table A-3 (300K): Material Thermal conductivity, W/m⋅K Limestone Softwood Blanket (glass, fiber 10 kg/m3) 2.15 0.12 0.048 ANALYSIS: The R-value, a quantity commonly used in the construction industry and building technology, is defined as R≡ $ L in " ' k Btu ⋅ in / h ⋅ ft 2 ⋅ F . 2 The R-value can be interpreted as the thermal resistance of a 1 ft cross section of the material. Using the conversion factor for thermal conductivity between the SI and English systems, the R-values are: Rock, Limestone, 18 ft: in −1 ft R= = 14.5 Btu/h ⋅ ft 2 ⋅ F W Btu/h ⋅ ft ⋅ F in 2.15 × 0.5778 ×12 m⋅K W/m ⋅ K ft 18 ft × 12 ( ) " ' Wood, Softwood, 15 in: 15 in R= 0.12 W Btu / h ⋅ ft ⋅ F in × 0.5778 × 12 m⋅ K W / m⋅ K ft = 18 Btu / h ⋅ ft 2 ⋅ F −1 Insulation, Blanket, 6 in: 6 in R= 0.048 W Btu / h ⋅ ft ⋅ F in × 0.5778 × 12 m⋅ K W / m⋅ K ft " ' = 18 Btu / h ⋅ ft 2 ⋅ F COMMENTS: The R-value of 19 given in the advertisement is reasonable. −1 PROBLEM 2.17 KNOWN: Electrical heater sandwiched between two identical cylindrical (30 mm dia. × 60 mm length) samples whose opposite ends contact plates maintained at To. FIND: (a) Thermal conductivity of SS316 samples for the prescribed conditions (A) and their average 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 for which ∆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 the samples which are presumed identical. Apply Fourier’s law to a sample q = kA c k= ∆T ∆x $ 0.5 100V × 0.353A × 0.015 m q∆x = = 15.0 W / m ⋅ K. 2 A c ∆T π 0.030 m / 4 × 25.0 C $ < The total temperature drop across the length of the sample is ∆T1(L/∆x) = 25°C (60 mm/15 mm) = 100°C. Hence, the heater temperature is Th = 177°C. 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 good agreement. (b) For Case B, we assume that the thermal conductivity of the SS316 sample is the same as that found in Part (a). The heat rate through the Armco iron sample is Continued ….. PROBLEM 2.17 (CONT.) q iron q iron $ π 0.030 m 15.0° C = q heater − q ss = 100V × 0.601A − 15.0 W / m ⋅ K × × 4 0.015 m = 601 − 10.6 W = 49.5 W . 2 $ where q ss = k ssA c ∆T2 / ∆x2 . Applying Fourier’s law to the iron sample, k iron = 49.5 W × 0.015 m q iron ∆x2 = = 70.0 W / m ⋅ K. 2 A c ∆T2 π 0.030 m / 4 × 15.0° C $ < The total drop across the iron sample is 15°C(60/15) = 60°C; the heater temperature is (77 + 60)°C = 137°C. 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 good agreement. (c) The principal advantage of having two identical samples is the assurance that all the electrical power dissipated in the heater will appear as equivalent heat flows through the samples. With only one 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 when the sample thermal conductivity is comparable to that of the insulating material. Hence, the method is suitable 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, if the insulation were improperly applied along the lateral surfaces, it is possible that heat leakage will occur, causing ∆T1 ≠ ∆T2. PROBLEM 2.18 KNOWN: Comparative method for measuring thermal conductivity involving two identical samples stacked with a reference material. FIND: (a) Thermal conductivity of test material and associated temperature, (b) Conditions for which ∆Tt,1 ≠ ∆Tt,2 SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer through samples and 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 the same diameter, is the same, it follows from Fourier’s law that kt ∆Tt,1 ∆x kt = kr = kr ∆Tt,2 ∆Tr = kt ∆x ∆x ∆Tr 2.49° C = 69.2 W / m ⋅ K = 519 W / m ⋅ K. . ∆Tt 3.32° C We 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 highly dependent upon temperature. Also, if there is heat leakage out the lateral surface, we can expect ∆Tt,2 < ∆Tt,1. Leakage could be influential, if the thermal conductivity of the test material were less than an order of magnitude larger than that of the insulating material. PROBLEM 2.19 KNOWN: Identical samples of prescribed diameter, length and density initially at a uniform temperature Ti, sandwich an electric heater which provides a uniform heat flux q ′′ for a period of o time ∆to. Conditions shortly after energizing and a long time after de-energizing heater are prescribed. 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 samples and heater, and apply conservation of energy over the time interval from t = 0 to ∞ E in − E out = ∆E = E f − E i 16 P∆t o − 0 = Mc p T ∞ − Ti where energy inflow is prescribed by the Case A power condition and the final temperature Tf by Case B. Solving for cp, cp = P∆t o 15 W × 120 s = 3 M T ∞ − Ti 2 × 3965 kg / m π × 0.0602 / 4 m2 × 0.010 m 33.50 - 23.00 ° C 16 2 7 < c p = 765 J / kg ⋅ K 2 where M = ρV = 2ρ(πD /4)L is the mass of both samples. For Case A, the transient thermal response of the heater is given by Continued ….. PROBLEM 2.19 (Cont.) t "#1/ 2 To 1 t 6 − Ti = 2q ′′ o !πρcp k #$ 2 t 2q o " ′′ k= # πρc p ! To 1 t 6 − Ti $ 30 s k= π × 3965 kg / m3 × 765 J / kg ⋅ K 2 × 2653 W / m "# !124.57 - 23.006° C $ 2 2 = 36.0 W / m ⋅ K < where q ′′ = o P P 15 W = = = 2653 W / m2 . 2 2 2 2A s 2 πD / 4 2 π × 0.060 / 4 m 4 94 9 With the following properties now known, ρ = 3965 kg/m 3 cp = 765 J/kg⋅K k = 36 W/m⋅K entries in Table A.1 are scanned to determine whether these values are typical of a metallic material. Consider the following, • metallics with low ρ 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 have much higher specific heats, • more than likely, the material is nonmetallic. From Table A.2, the second entry, polycrystalline aluminum oxide, has properties at 300 K corresponding to those found for the samples. < PROBLEM 2.20 KNOWN: Temperature distribution, T(x,y,z), within an infinite, homogeneous body at a given instant 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, must satisfy the heat equation. For the three-dimensional cartesian coordinate system, with constant properties and no internal heat generation, the heat equation, Eq. 2.15, has the form ∂ 2T ∂x 2 + ∂ 2T ∂y 2 + ∂ 2T ∂z 2 = 1 ∂T . α ∂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. 1 6 1 6 1 6 1 ∂T ∂ ∂ ∂ 2x - y + 2 z + 2y = −4 y - x + 2z + . ∂x ∂y ∂z α ∂t Performing the differentiations, 2−4+2 = 1 ∂T . α ∂t Hence, ∂T =0 ∂t which 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 the temperature distribution, T(x,y,z), at any future time. We can only determine that, for this special instant of time, the temperature will not change. PROBLEM 2.21 KNOWN: Diameter D, thickness L and initial temperature Ti of pan. Heat rate from stove to bottom of 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 is uniformly distributed over surface of pan in contact with the stove, (3) Constant properties. ANALYSIS: Stage 1 ∂ 2T Heat Equation: ∂x 2 = 1 ∂T α ∂t Initial Condition: −k qo ∂T = q′′ = o ∂x x = 0 π D2 / 4 −k Boundary Conditions: ∂T = h T ( L, t ) − T∞ ( t ) ∂x x = L ( ) T ( x, 0 ) = Ti Stage 2 Heat Equation: Boundary Conditions: d 2T dx 2 −k =0 dT = q′′ o dx x = 0 T ( L ) = TL COMMENTS: Stage 1 is a transient process for which T∞(t) must be determined separately. As a first approximation, it could be estimated by neglecting changes in thermal energy storage by the pan bottom and assuming that all of the heat transferred from the stove acted to increase thermal energy storage within the water. Hence, with q ≈ Mcp d T∞/dt, where M and cp are the mass and specific heat of the water in the pan, T∞(t) ≈ (q/Mcp) t. PROBLEM 2.22 KNOWN: Steady-state temperature distribution in a cylindrical rod having uniform heat generation of q1 = 5 × 107 W / m3 . FIND: (a) Steady-state centerline and surface heat transfer rates per unit length, q ′ . (b) Initial time r rate of change of the centerline and surface temperatures in response to a change in the generation rate from 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 ′′ = − k r ∂T ∂r q = -kA r ∂T . ∂r Hence, 1 6 ∂∂ T r q r = − k 2πrL or q ′ = −2πkr r ∂T ∂r where ∂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 is 16 < q ′ 0 = 0. r At r = ro, the temperature gradient is "# $ "# $ ! "#1 6 $ 4 91 6 ∂T K = −2 4.167 × 105 2 ro = −2 4.167 × 105 0.025m ∂ r r=r m o ∂T = −0.208 × 105 K / m. ∂ r r=r o Continued ….. PROBLEM 2.22(Cont.) Hence, the heat rate at the outer surface (r = ro) per unit length is 16 1 q ′ 1 ro 6 = 0.980 × 105 W / m. r 6 q ′ ro = −2π 30 W / m ⋅ K 0.025m −0.208 × 105 K / m r < (b) Transient (time-dependent) conditions will exist when the generation is changed, and for the prescribed assumptions, the temperature is determined by the following form of the heat equation, Eq. 2.20 ! "# $ ∂T ∂T 1∂ kr + q 2 = ρc p ∂r ∂t r ∂r Hence ! ! "# $ "# $ ∂T ∂T 1 1∂ kr = + q2 . ∂ t ρc p r ∂ r ∂r However, initially (at t = 0), the temperature distribution is given by the prescribed form, T(r) = 800 52 4.167×10 r , and ! "# $ 4 ∂T 1∂ k∂ kr r -8.334 × 105 ⋅ r = ∂r r ∂r r ∂r = 4 k −16.668 × 105 ⋅ r r 9 9 = 30 W / m ⋅ K -16.668 × 105 K / m2 1 6 = −5 × 107 W / m3 the original q = q1 . Hence, everywhere in the wall, ∂T 1 = −5 × 107 + 108 W / m3 3 ∂ t 1100 kg / m × 800 J / kg ⋅ K or ∂T = 56.82 K / s. ∂t < COMMENTS: (1) The value of (∂T/∂t) will decrease with increasing time, until a new steady-state condition 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 the 16 1 6 49 2 in gen steady-state condition, E ′ − E ′ + E ′ = 0. Hence q ′ 0 − q ′ ro = − q1 πro . out r r PROBLEM 2.23 KNOWN: Temperature distribution in a one-dimensional wall with prescribed thickness and thermal conductivity. FIND: (a) The heat generation rate, q, in the wall, (b) Heat fluxes at the wall faces and relation to q. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) Constant properties. ANALYSIS: (a) The appropriate form of the heat equation for steady-state, one-dimensional conditions with constant properties is Eq. 2.15 re-written as "# !$ Substituting the prescribed temperature distribution, d d d q = -k 4a + bx2 9"#$ = − k dx 2bx = −2bk dx ! dx q = -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 Fourier’s law, dT " q ′′ 1 x6 = − k x dx # x $. q = -k d dT dx dx < Using the temperature distribution T(x) to evaluate the gradient, find 16 q ′′ x = − k x d a + bx 2 = −2 kbx. dx The fluxes at x = 0 and x = L are then 16 q ′′ 1 L6 = −2 kbL = -2 × 50W / m ⋅ K4-2000 C / m2 9 × 0.050m x q ′′ 1 L6 = 10,000 W / m2 . x q ′′ 0 = 0 x < COMMENTS: From an overall energy balance on the wall, it follows that, for a unit area, E in − E out + E g = 0 16 16 16 1 6 q ′′ 0 − q ′′ L + qL = 0 x x q ′′ L − q ′′ 0 10,000 W / m2 − 0 x q= x = = 2.0 × 105 W / m3. L 0.050m < PROBLEM 2.24 KNOWN: 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 Fourier’s law, q ′′ = − k x $ ∂T = 200 − 60x ⋅ k ∂x q ′′ = q ′′ = 200 in x=0 °C W ×1 = 200 W / m2 m m⋅ K $ q ′′ = q ′′ = 200 − 60 × 0.3 ° C / m × 1 W / m ⋅ K = 182 W / m2 . out x=L < < Applying an energy balance to a control volume about the wall, Eq. 1.11a, ′′ E ′′ − E ′′ = E st in out E ′′ = q ′′ − q ′′ = 18 W / m2 . st in out < (b) Applying a surface energy balance at x = L, $ q ′′ = h T L − T∞ out q ′′ 182 W / m2 out h= = T L − T∞ 142.7 -100 ° C $ $ < h = 4.3 W / m2 ⋅ K. COMMENTS: (1) From the heat equation, 2 2 (∂T/∂t) = (k/ρcp) ∂ T/∂x = 60(k/ρcp), 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 2.25 KNOWN: Analytical expression for the steady-state temperature distribution of a plane wall experiencing 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 fluxes x x related 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 the x distribution; (f) If the source of heat generation is suddenly deactivated ( q = 0), what is the rate of change of energy stored at this instant; (g) Determine the temperature that the wall will reach eventually with q = 0; determine the energy that must be removed by the fluid per unit area of the wall to reach this state. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform volumetric heat generation, (3) Constant properties. ANALYSIS: (a) Using the analytical expression in the Workspace of IHT, the temperature distribution appears as shown below. The significant features include (1) parabolic shape, (2) maximum does not occur at the mid-plane, T(-5.25 mm) = 83.3°C, (3) the gradient at the x = +L surface is greater than at x = -L. Find also that T(-L) = 78.2°C and T(+L) = 69.8°C for use in part (d). Temperature distribution 90 Temperature, T(x) (C) 85 80 75 70 -20 -10 0 10 20 x-coordinate, x (mm) (b) Substituting the temperature distribution expression into the appropriate form of the heat diffusion equation, Eq. 2.15, the rate of volumetric heat generation can be determined. d dT q + =0 dx dx k where T ( x ) = a + bx + cx 2 d q q (0 + b + 2cx ) + = (0 + 2c ) + = 0 dx k k Continued ….. PROBLEM 2.25 (Cont.) ) ( q = −2ck = −2 −2 ×104°C / m 2 5 W / m ⋅ K = 2 ×105 W / m3 < (c) The heat fluxes at the two boundaries can be determined using Fourier’s law and the temperature distribution expression. q′′ ( x ) = − k x dT dx T ( x ) = a + bx + cx 2 where q′′ ( − L ) = − k [0 + b + 2cx ]x x =− L = − [b − 2cL] k ( ) q′′ ( − L ) = − −210°C / m − 2 −2 × 104°C / m 2 0.020m × 5 W / m ⋅ K = −2950 W / m 2 x < q′′ ( + L ) = − ( b + 2cL ) k = +5050 W / m 2 x < 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 = 0 x x or − 2950 W / m 2 − 5050 W / m 2 + 8000 W / m 2 = 0 where 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 conduction fluxes at the boundaries. See the surface energy balances in the sketch above. See also part (a) result for T(-L) and T(+L). q′′ = q′′ ( − L ) cv, x h l T∞ − T ( − L ) = h l [20 − 78.2] K = −2950 W / m 2 h l = 51W / m 2 ⋅ K < q′′ = q′′ ( + L ) cv,r x h r T ( + L ) − T∞ = h r [69.8 − 20] K = +5050 W / m 2 h r = 101W / m 2 ⋅ K < (e) The expression for the heat flux distribution can be obtained from Fourier’s law with the temperature distribution q′′ ( x ) = − k x dT = − k [0 + b + 2cx ] dx ( ) q′′ ( x ) = −5 W / m ⋅ K −210°C / m + 2 −2 × 104°C / m 2 x = 1050 + 2 × 105 x x Continued ….. < PROBLEM 2.25 (Cont.) The distribution is linear with the x-coordinate. The maximum temperature will occur at the location where q′′ ( x max ) = 0, x x max = − 1050 W / m 2 2 ×105 W / m3 = −5.25 × 10−3 m = −5.25 mm < (f) If the source of the heat generation is suddenly deactivated so that q = 0, the appropriate form of the heat diffusion equation for the ensuing transient conduction is k ∂ ∂T ∂T = ρ cp ∂x ∂x ∂t 2 At the instant this occurs, the temperature distribution is still T(x) = a + bx + cx . The right-hand term represents the rate of energy storage per unit volume, E′′ = k st ∂ ∂x [0 + b + 2cx ] = k [0 + 2c] = 5 W / m ⋅ K × 2 (−2 ×104°C / m2 ) = −2 ×105 W / m3 < (g) With no heat generation, the wall will eventually (t → ∞) come to equilibrium with the fluid, T(x,∞) = T∞ = 20°C. 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 is that corresponding to the steady-state temperature distribution, Ti, and the “final” state has Tf = 20°C. We’ve used T∞ as the reference condition for the energy terms. E′′ − E′′ = ∆E′′ = E′′ − E′′ in out st f i − E′′ = ρ cp 2L ( Tf − T∞ ) − ρ cp ∫ out E′′ = ρ cp ∫ out with E′′ = 0. in +L (T − T∞ ) dx −L i +L +L a + bx + cx 2 − T∞ dx = ρ cp ax + bx 2 / 2 + cx 3 / 3 − T∞ x −L −L E′′ = ρ cp 2aL + 0 + 2cx 3 / 3 − 2T∞ L out ( E′′ = 2600 kg / m3 × 800 J / kg ⋅ K 2 × 82°C × 0.020m + 2 −2 ×104°C / m2 out ) (0.020m )3 / 3 − 2 (20°C ) 0.020m E′′ = 4.94 ×106 J / m 2 out < 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 are evaluated. Continued ….. PROBLEM 2.25 (Cont.) (2) In evaluating the conduction heat fluxes, q′′ ( x ) , it is important to recognize that this flux x is in the positive x-direction. See how this convention is used in formulating the energy balance in part (c). (3) It is good practice to represent energy balances with a schematic, clearly defining the system 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) as − d dT −k +q =0 dx dx recognize that the term in parenthesis is the heat flux. From the differential equation, note that if the differential of this term is a constant ( q / k ) , then the term must be a linear function of 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 the volumetric heat generation was deactivated. Did you notice that Est = −2 × 105 W / m3 is the same value of the deactivated q ? How do you explain this? PROBLEM 2.26 KNOWN: 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 is insulated. FIND: (a) Calculate the internal energy generation rate, q , by applying an overall energy balance to the wall, (b) Determine the coefficients a, b, and c, by applying the boundary conditions to the prescribed 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, and the 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 the 2 convection coefficient remains unchanged (h = 500 W/m ⋅K); plot the temperature distribution and label as Case 3. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with constant properties 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 balance on the wall as shown in the schematic below. E′′ − E′′ + E′′ = 0 in out gen ′′ Ein = q′′ conv where h ( 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 applying the boundary conditions at x = 0 and x = L. See Table 2.1 for representation of the boundary conditions, and the schematic above for the relevant surface energy balances. Boundary condition at x = 0, convection surface condition E′′ − E′′ = q ′′ in out conv − q′′ ( 0 ) = 0 x where q′′ (0 ) = − k x dT dx x = 0 h ( T∞ − To ) − − k (0 + b + 2cx )x = 0 = 0 Continued ….. 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 surface Ein − E out = −q′′ ( L ) = 0 x q′′ ( L ) = − k x where dT dx x = L k [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 = 120°C, find T ( 0 ) = 120°C = a + b ⋅ 0 + c ⋅ 0 or a = 120°C (4) < Using the foregoing coefficients with the expression for T(x) in the Workspace of IHT, the temperature 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 × 106 3 W/m and other parameters remain unchanged except that To ≠ 120°C. We can determine a, b, and c for 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 + 20°C = 220°C < 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, 6 3 2 q 3 = 2q = 2 × 10 W / m , h = 500 W/m ⋅K, and other parameters remain unchanged except that To ≠ 120°C. Following the same analysis as part (c), the coefficients for the new temperature distribution, T (x), are a = 220°C b = 2 × 104 K / m c = −2 × 105 K / m2 < and the distribution is plotted as Case 3 below. Continued ….. PROBLEM 2.26 (Cont.) 800 700 Te m p e ra tu re , T (C ) 600 500 400 300 200 100 0 5 10 15 20 25 30 35 40 45 50 W 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 ^3 2 . h = 2 5 0 W /m ^2 .K , q d o t = 1 e 6 W /m ^3 3 . h = 5 0 0 W /m ^2 .K , q d o t = 2 e 6 W /m ^3 COMMENTS: 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 insulated boundary. By halving the convection coefficient for Case 2, we expect the surface temperature To to increase relative to the Case 1 value, since the same heat flux is removed from the wall ( qL ) but the convection resistance has increased. By doubling the generation rate for Case 3, we expect the surface temperature To to increase relative to 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 are quite different? Can you explain the relative magnitudes of T(L) for the three cases? PROBLEM 2.27 KNOWN: Temperature distribution and distribution of heat generation in central layer of a solar pond. FIND: (a) Heat fluxes at lower and upper surfaces of the central layer, (b) Whether conditions are steady 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) Constant properties ANALYSIS: (a) The desired fluxes correspond to conduction fluxes in the central layer at the lower and upper surfaces. A general form for the conduction flux is "# ! $ Hence, A e-aL + B"# q ′′ = q cond 1 x=L6 = − k ′′ l ! ka $ q ′′ cond = − k ∂T A -ax e +B . = −k ∂x ka 16 (b) Conditions are steady if ∂T/∂t = 0. Applying the heat equation, ∂ 2T ∂x 2 + q 1 ∂T = k α ∂t - ! "# $ A q ′′ = q ′′ +B . u cond x=0 = − k ka < A -ax A -ax 1 ∂ T e+e= α ∂t k k Hence conditions are steady since ∂T/∂t = 0 < (for all 0 ≤ × ≤ L). (c) For the central layer, the energy generation is I I L L E ′′ = q dx = A e-ax dx g 0 A E g = − e -ax a 0 L 0 =− 4 9 4 9 A -aL A e −1 = 1 − e -aL . a a Alternatively, 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 $a q ′′ − q1 + E g = 0 ′′ ′′ 2 < 1 69 g E ′′ = q1 − q ′′ = − q ′′ ′′ 2 cond x=0 − − q ′′ cond x=L COMMENTS: Conduction is in the negative x-direction, necessitating use of minus signs in the above energy balance. PROBLEM 2.28 KNOWN: Temperature distribution in a semi-transparent medium subjected to radiative flux. 16 FIND: (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 internal volumetric heat generation term q x . 16 ANALYSIS: (a) Knowing the temperature distribution, the surface heat fluxes are found using Fourier’s law, dT "# = − k - A 1−a6e-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 ′′ = − k x < < (b) The heat diffusion equation for the medium is d dT or q = -k dx dx d A -ax " q1 x6 = − k + e + B# = Ae-ax . dx ! ka $ d dT q + =0 dx dx k < (c) Performing an energy balance on the medium, E in − E out + E g = 0 recognize that E g represents the absorbed irradiation. On a unit area basis 16 1 6 4 9 A g ′′ ′′ E ′′ = − E in + E out = − q ′′ 0 + q ′′ L = + 1 − e-aL . x x a g Alternatively, evaluate E ′′ by integration over the volume of the medium, I 16 I 4 9 LA L L A E ′′ = q x dx = Ae -ax dx = - e-ax = 1 − e-aL . g 0 0 0 a a < PROBLEM 2.29 KNOWN: Steady-state temperature distribution in a one-dimensional wall of thermal 3 2 conductivity, T(x) = Ax + Bx + Cx + D. FIND: Expressions for the heat generation rate in the wall and the heat fluxes at the two wall faces (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 is d 2T q + =0 dx 2 k or q = -k d 2T dx 2 . Hence, the generation rate is q = -k "# !$ d dT d = −k 3Ax2 + 2 Bx + C + 0 dx dx dx < q = -k 6Ax + 2B which is linear with the coordinate x. The heat fluxes at the wall faces can be evaluated from Fourier’s law, q ′′ = − k x dT = − k 3Ax 2 + 2Bx + C dx using the expression for the temperature gradient derived above. Hence, the heat fluxes are: Surface x=0: 16 < q ′′ 0 = − kC x Surface x=L: 16 < q ′′ L = − k 3AL2 + 2BL + C . x COMMENTS: (1) From an overall energy balance on the wall, find E ′′ − E ′′ + E ′′ = 0 in out g 16 1 6 1 61 6 ′′ ′′ q ′′ 0 − q ′′ L + E g = − kC − − k 3AL2 + 2 BL + C + E g = 0 x x ′′ E g = −3AkL2 − 2 BkL. From integration of the volumetric heat rate, we can also find E ′′ as g I 16 I L L L E ′′ = q x dx = -k 6Ax + 2B dx = -k 3Ax 2 + 2 Bx g 0 0 E ′′ = −3AkL − 2 BkL. g 2 0 PROBLEM 2.30 KNOWN: Plane wall with no internal energy generation. FIND: Determine whether the prescribed temperature distribution is possible; explain your reasoning. With the temperatures T(0) = 0°C and T∞ = 20°C fixed, compute and plot the temperature T(L) as a function of the convection coefficient for the range 10 ≤ h ≤ 100 W/m2⋅K. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal energy generation, (3) Constant properties, (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 the surface x = L as shown above in the Schematic, must be satisfied. E in − E out ? = ? 0 q′′ ( L ) − q′′ ? = ? 0 (1,2) x cv where 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 2 x dx x = L L q′′ = h [T ( L ) − T∞ ] = 30 W m 2 ⋅ K × (120 − 20 ) C = 3000 W m2 cv Substituting the heat flux values into Eq. (2), find (-3000) - (3000) ≠ 0 and therefore, the temperature distribution is not possible. (b) With T(0) = 0°C and T∞ = 20°C, the temperature at the surface x = L, T(L), can be determined from an overall energy balance on the wall as shown above in the Schematic, T ( L ) − T (0 ) −k − h [T ( L ) − T∞ ] = 0 E in − E out = 0 q′′ (0) − q′′ = 0 x cv L −4.5 W m ⋅ K T ( L ) − 0 C 0.18 m − 30 W m 2 ⋅ K T ( L ) − 20 C = 0 < T(L) = 10.9°C 20 Surface temperature, T(L) (C) Using this same analysis, T(L) as a function of the convection coefficient can be determined and plotted. We don’t expect T(L) to be linearly dependent upon h. Note that as h increases to larger values, T(L) approaches T∞ . To what value will T(L) approach as h decreases? 16 12 8 4 0 0 20 40 60 Convection cofficient, h (W/m^2.K) 80 100 PROBLEM 2.31 KNOWN: Coal pile of prescribed depth experiencing uniform volumetric generation with convection, absorbed irradiation and emission on its upper surface. FIND: (a) The appropriate form of the heat diffusion equation (HDE) and whether the prescribed temperature distribution satisfies this HDE; conditions at the bottom of the pile, x = 0; sketch of the temperature distribution with labeling of key features; (b) Expression for the conduction heat rate at the 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 GS and h, compute and plot Ts and T(0) for (1) h = 5 W/m2⋅K with 50 ≤ GS ≤ 500 W/m2, (2) GS = 400 W/m2 with 5 ≤ h ≤ 50 W/m2⋅K. 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.K ANALYSIS: (a) For one-dimensional, steady-state conduction with uniform volumetric heat generation and constant properties the heat diffusion equation (HDE) follows from Eq. 2.16, d dT q + =0 dx dx k (1) Substituting the temperature distribution into the HDE, Eq. (1), qL2 x 2 d qL2 2x q 1 − T ( x ) = Ts + 0 + 0 − 2 + ? = ?0 2 2k L dx 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 is q′′ ( 0 ) = − k x qL2 dT 2x = −k 0 + =0 0 − 2 dx x = 0 2k L x =0 so that the gradient at x = 0 is zero. Hence, the bottom is insulated. (b) From an overall energy balance on the pile, the conduction heat flux at the surface must be q′′ ( L ) = E′′ = qL x g < Continued... PROBLEM 2.31 (Cont.) From a surface energy balance per unit area shown in the Schematic above, q′′ ( L ) − q′′ + GS,abs − E = 0 x cv E in − E out + E g = 0 qL − h ( Ts − T∞ ) + 0.95GS − εσ Ts4 = 0 (4) 20 W m ×1m − 5 W m ⋅K ( Ts − 298 K ) + 0.95 × 400 W m − 0.95 × 5.67 × 10 3 2 2 −8 2 44 W m ⋅K Ts = 0 < Ts = 295.7 K =22.7°C From Eq. (2) with x = 0, find 30 W m 2 × (1m ) qL2 T ( 0 ) = Ts + = 22.7 C + = 61.1 C 2k 2 × 0.26 W m ⋅ K 2 (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/m2⋅K for 50 ≤ GS ≤ 500 W/m2 and (2) with GS = 400 W/m2 for 5 ≤ h ≤ 50 W/m2⋅K. Solar irradiation, GS = 400 W/m^2 80 Convection coefficient, h = 5 W/m^2.K Temperature, Ts or T(0) (C) Temperature, Ts or T(0) (C) 80 60 40 20 60 40 20 0 0 10 20 30 40 50 Convection coefficient, h (W/m^2.K) -20 0 100 200 300 400 500 T0_C Ts_C Solar irradiation, GS (W/m^2) T0_C Ts_C From the T vs. h plot with GS = 400 W/m2, note that the convection coefficient does not have a major influence on the surface or bottom coal pile temperatures. From the T vs. GS plot with h = 5 W/m2⋅K, note that the solar irradiation has a very significant effect on the temperatures. The fact that Ts is less than the ambient air temperature, T∞ , and, in the case of very low values of GS, below freezing, is a consequence of the large magnitude of the emissive power E. COMMENTS: In our analysis we ignored irradiation from the sky, an environmental radiation effect 4 you’ll consider in Chapter 12. Treated as large isothermal surroundings, Gsky = σ Tsky where Tsky = 30°C for very clear conditions and nearly air temperature for cloudy conditions. For low GS conditions we should consider Gsky, the effect of which will be to predict higher values for Ts and T(0). PROBLEM 2.32 KNOWN: Cylindrical system with negligible temperature variation in the r,z directions. FIND: (a) Heat equation beginning with a properly defined control volume, (b) Temperature distribution T(φ) 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 = ridφ⋅∆r⋅L where L is length normal to page. Apply the conservation of energy requirement, Eq. 1.11a, E in − E out + E g = E st where q φ − q φ +dφ + qV = ρVc 1 6 r∂∂Tφ i q φ = − k ∆r ⋅ L ∂T ∂t q φ +dφ = q φ + (1,2) 38 ∂ q dφ . ∂φ φ (3,4) Eqs. (3) and (4) follow from Fourier’s law, Eq. 2.1, and from Eq. 2.7, respectively. Combining Eqs. (3) and (4) with Eq. (2) and canceling like terms, find 1∂ ∂T ∂T + q = ρc . k ∂t ri2 ∂ φ ∂ φ (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 "# !$ d dT = 0. k dφ dφ (6) With constant properties, it follows that dT/dφ is constant which implies T(φ) is linear in φ. That is, 1 dT T2 − T1 1 = = + T2 − T1 π dφ φ 2 − φ 1 6 or 16 T φ = T1 + 1 6 1 T − T φ. π21 (7,8) < (c) The heat rate for the conditions of Part (b) follows from Fourier’s law, Eq. (3), using the temperature gradient of Eq. (7). That is, 1 1 6 r1i !+ π 1T2 − T16"#$ = − k ! roπ−i ri "#$L1T2 − T16. r q φ = − k ∆r ⋅ L (9) < COMMENTS: Note the expression for the temperature gradient in Fourier’s law, Eq. (3), is ∂T/ri∂φ not ∂T/∂φ. For the conditions of Parts (b) and (c), note that qφ is independent of φ; this is first indicated by Eq. (6) and confirmed by Eq. (9). PROBLEM 2.33 KNOWN: 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 = 2πr ⋅ dr ⋅ 1, with unit thickness normal to page. Using the conservation of energy requirement, Eq. 1.11a, E in − E out + E gen = E st q r − q r +dr + qV = ρVc p ∂T . ∂t Fourier’s law, Eq. 2.1, for this one-dimensional coordinate system is q r = − kA r ∂T ∂T = − k × 2πr ⋅ 1 × . ∂r ∂r At the outer surface, r+dr, the conduction rate is q r+dr = q r + ∂ ∂ ∂T (q r ) dr=q r + −k ⋅ 2π r ⋅ dr. ∂r ∂ r ∂ r Hence, the energy balance becomes ∂ ∂ T ∂T q r − q r + − k2π r dr + q ⋅ 2π rdr=ρ ⋅ 2π rdr ⋅ cp ∂ r ∂ r ∂t Dividing by the factor 2πr dr, we obtain ∂T 1 ∂ ∂ T kr ∂ r + q=ρ cp ∂ t . r∂ r < COMMENTS: (1) Note how the result compares with Eq. 2.20 when the terms for the φ,z coordinates are eliminated. (2) Recognize that we did not require q and k to be independent of r. PROBLEM 2.34 KNOWN: Heat diffusion with internal heat generation for one-dimensional spherical, radial coordinate system. FIND: Heat diffusion equation. SCHEMATIC: ASSUMPTIONS: (1) Homogeneous medium. 2 ANALYSIS: Control volume has the volume, V = Ar ⋅ dr = 4πr dr. Using the conservation of energy requirement, Eq. 1.11a, E in − E out + E gen = E st q r − q r +dr + qV = ρVc p ∂T . ∂t Fourier’s law, Eq. 2.1, for this coordinate system has the form q r = − kA r ∂T ∂T . = − k ⋅ 4πr 2 ⋅ ∂r ∂r At the outer surface, r+dr, the conduction rate is q r+dr = q r + ∂ ∂ ∂T (q r ) dr = q r + −k ⋅ 4π r 2 ⋅ dr. ∂r ∂ r ∂ r Hence, the energy balance becomes ∂ ∂T 2 ∂ T 2 2 q r − q r + − k ⋅ 4π r ⋅ ∂ r dr + q ⋅ 4π r dr=ρ ⋅ 4π r dr ⋅ cp ∂ t . ∂ r Dividing by the factor 4πr 2 dr, we obtain 1 ∂ 2 ∂ T ∂T kr . + q=ρ cp ∂ r ∂t r2 ∂ r < COMMENTS: (1) Note how the result compares with Eq. 2.23 when the terms for the θ,φ directions are eliminated. (2) Recognize that we did not require q and k to be independent of the coordinate r. PROBLEM 2.35 KNOWN: Three-dimensional system – described by cylindrical coordinates (r,φ,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 volume given as V = dr⋅rdφ⋅dz. From the conservation of energy requirement, (1) q r − q r +dr + q φ − q φ +dφ + q z − q z+dz + E g = E st . The generation and storage terms, both representing volumetric phenomena, are 1 E g = qV = q dr ⋅ rdφ ⋅ dz 6 1 6 E g = ρVc∂ T / ∂ t = ρ dr ⋅ rdφ ⋅ dz c ∂ T / ∂ t. Using a Taylor series expansion, we can write ∂ ∂ q r +dr = q r + q r dr, q φ +dφ = q φ + q φ dφ , ∂r ∂φ 16 38 q z+dz = q z + (2,3) 16 ∂ q z dz. ∂z (4,5,6) Using Fourier’s law, the expressions for the conduction heat rates are 1 6 q φ = − kA φ ∂ T / r∂φ = − k 1dr ⋅ dz6∂ T / r∂φ q z = − kA z∂ T / ∂ z = − k1dr ⋅ rdφ 6∂ T / ∂ z. q r = − kA r ∂ T / ∂ r = − k rdφ ⋅ dz ∂ T / ∂ r (7) (8) (9) Note from the above, right schematic that the gradient in the φ-direction is ∂T/r∂φ and not ∂T/∂φ. Substituting Eqs. (2), (3) and (4), (5), (6) into Eq. (1), − 16 16 38 1 6 ∂ ∂ ∂ ∂T . q r dr − q φ dφ − q z dz + q dr ⋅ rdφ ⋅ dz = ρ dr ⋅ rdφ ⋅ dz c ∂r ∂φ ∂z ∂t Substituting Eqs. (7), (8) and (9) for the conduction rates, find − 1 6 "# 1 ! ! $ ∂T + q dr ⋅ rdφ ⋅ dz = ρ1dr ⋅ rdφ ⋅ dz6c . ∂t 1 ! 6 "#$ (10) 6 "#$ ∂ ∂T ∂ ∂T ∂ ∂T − k rdφ ⋅ dz dr − − k drdz dφ − − k dr ⋅ rdφ dz ∂r ∂r ∂φ ∂z ∂z r∂φ (11) Dividing Eq. (11) by the volume of the CV, Eq. 2.20 is obtained. ! "# $ ! "# $ ! "# $ ∂T ∂T ∂ ∂T ∂T 1∂ 1∂ kr +2 k + k + q = ρc ∂r ∂φ ∂z ∂z ∂t r∂r r ∂φ < PROBLEM 2.36 KNOWN: Three-dimensional system – described by cylindrical coordinates (r,φ,θ) – experiences transient 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 = dr⋅rsinθdφ⋅rdθ, and the conduction terms are identified in Figure 2.10. Conservation of energy requires q r − q r +dr + q φ − q φ +dφ + qθ − qθ +dθ + E g = E st . (1) The generation and storage terms, both representing volumetric phenomena, are ∂T ∂T E st = ρVc = ρ dr ⋅ r sinθdφ ⋅ rdθ c . ∂t ∂t E g = qV = q dr ⋅ r sinθdφ ⋅ rdθ (2,3) Using a Taylor series expansion, we can write q r +dr = q r + 16 ∂ q r dr, ∂r qφ +dφ = qφ + 38 ∂ qφ dφ , ∂φ qθ +dθ = qθ + 16 ∂ qθ dθ . ∂θ (4,5,6) From Fourier’s law, the conduction heat rates have the following forms. q r = − kA r ∂ T / ∂ r = − k r sinθdφ ⋅ rdθ ∂ T / ∂ r (7) q φ = − kA φ ∂ T / r sinθ∂φ = − k dr ⋅ rdθ ∂ T / r sinθ∂φ (8) qθ = − kAθ ∂ T / r∂θ = − k dr ⋅ r sinθdφ ∂ T / r∂θ . (9) Substituting Eqs. (2), (3) and (4), (5), (6) into Eq. (1), the energy balance becomes − 16 16 38 ∂T ∂ ∂ ∂ q r dr − q φ dφ − qθ dθ + q dr ⋅ r sinθdφ ⋅ rdθ = ρ dr ⋅ r sinθdφ ⋅ rdθ c ∂t ∂r ∂φ ∂θ Substituting Eqs. (7), (8) and (9) for the conduction rates, find − − ! ! ! "# $ (10) "# $ ∂T ∂ ∂T ∂ dr − dφ − k r sinθdφ ⋅ rdθ − k dr ⋅ rdθ r sinθ∂φ ∂r ∂φ ∂θ "# $ ∂ ∂T ∂T dθ + q dr ⋅ r sinθdφ ⋅ rdθ = ρ dr ⋅ r sinθdφ ⋅ rdθ c − k dr ⋅ r sinθdφ ∂θ ∂t r∂θ (11) Dividing Eq. (11) by the volume of the control volume, V, Eq. 2.23 is obtained. ! "# $ ! "# $ ! "# $ ∂T ∂ ∂T ∂ ∂T ∂T 1∂ 1 1 kr 2 k k sinθ . +2 2 +2 + q = ρc 2 ∂r ∂r ∂θ ∂t r r sin θ ∂φ ∂ φ r sinθ ∂θ < COMMENTS: Note how the temperature gradients in Eqs. (7) - (9) are formulated. The numerator is always ∂T while the denominator is the dimension of the control volume in the specified coordinate direction. PROBLEM 2.37 KNOWN: Temperature distribution in steam pipe insulation. FIND: Whether conditions are steady-state or transient. Manner in which heat flux and heat rate vary with radius. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties. ANALYSIS: From Equation 2.20, the heat equation reduces to 1∂ ∂T 1 ∂T r = . r∂r ∂r α ∂t Substituting for T(r), 1 ∂T 1 ∂ C r 1 = 0. = α ∂t r∂r r < Hence, steady-state conditions exist. From Equation 2.19, the radial component of the heat flux is q ′′ = − k r ∂T C = −k 1 . ∂r r 1 6 Hence, q ′′ decreases with increasing r q ′′α 1/ r . r r < At any radial location, the heat rate is q r = 2πrLq ′′ = −2πkC1L r Hence, qr is independent of r. < COMMENTS: The requirement that qr is invariant with r is consistent with the energy conservation requirement. If qr is constant, the flux must vary inversely with the area perpendicular to the direction of heat flow. Hence, q ′′ varies inversely with r. r PROBLEM 2.38 KNOWN: Inner and outer radii and surface temperatures of a long circular tube with internal energy generation. 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 to k d dT r +q =0 r 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 (call it C1). It follows that kd ( r C1 ) + q = 0 r dr Ck q=− 1 r < where C1 = (T2 - T1)/(r2 - r1). Hence, if the generation rate varies inversely with radial location, the radial temperature distribution is linear. COMMENTS: Conditions for which q ∝ (1/r) would be unusual. PROBLEM 2.39 KNOWN: Radii and thermal conductivity of conducting rod and cladding material. Volumetric rate of 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) Constant properties. ANALYSIS: From Equation 2.20, the appropriate forms of the heat equation are Conducting Rod: kr d dT r r +q = 0 r dr dt Cladding: < d dT r c = 0. dr dr < Appropriate boundary conditions are: < (a) dTr / dr|r =0 = 0 (b) Tr ri = Tc ri (c) kr dTr dT |ri = k c c |ri dr dr < (d) kc dTc |r = h Tc ro − T∞ dr o 16 < 16 16 < COMMENTS: Condition (a) corresponds to symmetry at the centerline, while the interface conditions at r = ri (b,c) correspond to requirements of thermal equilibrium and conservation of energy. Condition (d) results from conservation of energy at the outer surface. PROBLEM 2.40 KNOWN: Steady-state temperature distribution for hollow cylindrical solid with volumetric heat generation. FIND: (a) Determine the inner radius of the cylinder, ri, (b) Obtain an expression for the volumetric rate 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) r Determine the radial distribution of the heat flux at the end faces of the cylinder, q′′ ( r, + z o ) and z q′′ ( r, −z o ) , and the corresponding heat rates; are the heat rates in or out of the cylinder; (e) z Determine the relationship of the surface heat rates to the heat generation rate; is an overall energy balance satisfied? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction with constant properties and volumetric heat generation. ANALYSIS: (a) Since the inner boundary, r = ri, is adiabatic, then q′′ ( ri , z ) = 0. Hence the r temperature gradient in the r-direction must be zero. ∂T = 0 + 2bri + c / ri + 0 = 0 ∂r r i 1/ 2 c ri = + − 2b 1/ 2 −12°C = − 2 × 150°C / 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 conduction 1 ∂ ∂T ∂ ∂T q r ∂ + ∂ ∂ + = 0 r ∂r r z z k 1∂ q (r [0 + 2br + c / r + 0]) + ∂∂z (0 + 0 + 0 + 2dz ) + k = 0 r ∂r 1 q [4br + 0] + 2d + = 0 r k q = − k [4b − 2d ] = −16 W / m ⋅ K 4 × 150°C / m 2 − 2 −300°C / m2 ( ) q = 0 W / m3 < (c) The heat flux and the heat rate at the outer surface, r = ro, may be calculated using Fourier’s law. Note that the sign of the heat flux in the positive r-direction is negative, and hence the heat flow is into the cylinder. () q ′′ ro, z = − k r ∂T = − k [0 + 2bro + c / ro + 0 ] ∂r r o Continued ….. PROBLEM 2.40 (Cont.) 2 2 q ′′ ( ro, z ) = −16 W / m ⋅ K 2 × 150°C / m × 1 m − 12°C / 1 m = −4608 W / m r () q r ( ro ) = A r q′′ ro, z r < A r = 2π ro ( 2z o ) where q r ( ro ) = −4π × 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 Fourier’s law. The direction of the heat rate in or out of the end face is determined by the sign of the heat flux in the positive z-direction. < At the upper end face, z = + zo: heat rate is out of the cylinder q ′′ ( r, + z o ) = − k z ∂T ∂z z o = − k [0 + 0 + 0 + 2dz o ] ( q ′′ ( r, + z o ) = −16 W / m ⋅ K × 2 −300°C / m z q z ( + z o ) = A z q ′′ ( r, + z o ) z ( q z ( + z o ) = π 1 − 0.2 2 2 )m ) 2.5 m = +24, 000 W / m where A = π (r 2 z 2 < 2 2 2 o − ri ) < < 2 × 24, 000 W / m = +72, 382 W At the lower end face, z = - zo: heat rate is out of the cylinder q ′′ ( r, − z o ) = − k z ∂T ∂z − z o = − k [0 + 0 + 0 + 2dz o ] q ′′ ( r, − z o ) = −16 W / m ⋅ K × 2 ( −300°C / m )( −2.5 m ) = −24, 000 W / m z 2 2 q z ( − z o ) = −72, 382 W < < (e) The heat rates from the surfaces and the volumetric heat generation can be related through an overall energy balance on the cylinder as shown in the sketch. E in − E out + E gen = 0 where E gen = q∀ = 0 E in = −q r ( ro ) = − ( −144, 765 W ) = +144, 765 W E 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 Fourier’s law, the heat flux q′′ denotes the heat flux in the positive zz direction. At a boundary, the sign of the numerical value will determine whether heat is flowing into or out of the boundary. PROBLEM 2.41 KNOWN: An electric cable with an insulating sleeve experiences convection with adjoining air and radiation exchange with large surroundings. FIND: (a) Verify that prescribed temperature distributions for the cable and insulating sleeve satisfy their 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,1 r and Ts,2; apply a surface energy balance to the cable to obtain an alternative expression for q′ in r terms of q and r1; (c) Apply surface energy balance around the outer surface of the sleeve to obtain an expression for which Ts,2 can be evaluated; (d) Determine Ts,1, Ts,2, and To for the specified geometry and operating conditions; and (e) Plot Ts,1, Ts,2, and To as a function of the outer radius for the range 15.5 ≤ r2 ≤ 20 mm. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Uniform volumetric heat generation in cable, (3) Negligible thermal contact resistance between the cable and sleeve, (4) Constant properties in cable and sleeve, (5) Surroundings large compared to the sleeve, and (6) Steady-state conditions. ANALYSIS: (a) The appropriate forms of the heat diffusion equation (HDE) for the insulation and cable 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,2 ln r r ) ln ((r r2 )) (1) 12 and the appropriate HDE (radial coordinates, SS, q = 0), Eq. 2.20, d dT r =0 dr 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 as qr 2 r 2 T ( r ) = Ts,1 + 1 1 − 4k c r 2 1 and the appropriate HDE (radial coordinates, SS, q uniform), Eq. 2.20, (2) Continued... PROBLEM 2.41 (Cont.) 1 d dT q =0 r + r dr dr k c 1d qr 2 r 0 + 1 r dr 4k c 2r q 0 − + ? = ? 0 2 k r1 c 1d qr 2 2r 2 q − 1 + ? = ?0 r dr 4k c r 2 k c 1 2 1 qr1 4r q − + ? = ?0 r 4k c r 2 k c 1 < 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 are as follows: (1) Zero gradient, symmetry condition, (2) Increasing gradient with increasing radius, r, because of q , (3) Discontinuous T(r) across cable-sleeve interface because of different thermal conductivities, (4) Decreasing gradient with increasing radius, r, since heat rate is constant. (b) Using Fourier’s law for the radial-cylindrical coordinate, the heat rate through the insulation (sleeve) per unit length is q′ = − kA′ r r dT dr = − k2π r dT < dr and substituting for the temperature distribution, Eq. (1), ( q′ = − k s 2π r 0 + Ts,1 − Ts,2 r ) ln (1 r r ) = 2π ks r1 2 (Ts,1 − Ts,2 ) ln ( r2 r1 ) (3) < Applying an energy balance to a control surface placed around the cable, Ein − Eout = 0 c q∀′ − q′ = 0 r where q∀c represents the dissipated electrical power in the cable Continued... PROBLEM 2.41 (Cont.) () 2 q π r1 − q′ = 0 r (4) < (5) 2 q′ = π qr1 r or < (c) Applying an energy balance to a control surface placed around the outer surface of the sleeve, E in − E out = 0 q′ − q′ − q′ = 0 r cv rad ( ( ) ) 2 4 4 π qr1 − h ( 2π r2 ) Ts,2 − T∞ − ε ( 2π r2 )σ Ts,2 − Tsur = 0 This relation can be used to determine Ts,2 in terms of the variables q , r1, r2, h, T∞, ε and Tsur. (d) Consider a cable-sleeve system with the following prescribed conditions: kc = 200 W/m⋅K ks = 0.15 W/m⋅K r1 = 15 mm r2 = 15.5 mm h = 25 W/m2⋅K T∞ = 25°C ε = 0.9 Tsur = 35°C For 250 A with R ′ = 0.005 Ω/m, the volumetric heat generation rate is e (π r12 ) 2 q = ( 250 A ) × 0.005 Ω / m (π × 0.0152 m 2 ) = 4.42 × 105 W m3 q = I 2 R ′ ∀′ = I 2 R ′ ec e Substituting numerical values in appropriate equations, we can evaluate Ts,1, Ts,2 and To. Sleeve outer surface temperature, Ts,2: Using Eq. (5), ( π × 4.42 × 105 W m3 × ( 0.015m ) − 25 W m 2 ⋅ K × ( 2π × 0.0155m ) Ts,2 − 298K 2 ( ) ) 4 −0.9 × ( 2π × 0.0155m ) × 5.67 × 10−8 W m 2 ⋅ K 4 Ts,2 − 3084 K 4 = 0 < Ts,2 = 395 K = 122 C Sleeve-cable interface temperature, Ts,1: Using Eqs. (3) and (4), with Ts,2 = 395 K, 2 π qr1 = 2π k s (Ts,1 − Ts,2 ) ln ( r2 r1 ) π × 4.42 × 105 W m3 × ( 0.015 m ) = 2π × 0.15 W m ⋅ K 2 Ts,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 = 133°C, To = T(0) = Ts,1 + 2 qr1 4k c To = 133 C + 4.42 × 105 W m3 × ( 0.015 m ) 2 ( 4 × 200 W m ⋅ 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) 200 180 160 140 120 100 15 16 17 18 19 20 Sleeve outer radius, r2 (mm) Inner sleeve, r1 Outer sleeve, r2 On the plot above To would show the same behavior as Ts,1 since the temperature rise between cable center and its surface is 0.12°C. With increasing r2, we expect Ts,2 to decrease since the heat flux decreases with increasing r2. We expect Ts,1 to increase with increasing r2 since the thermal resistance of the sleeve increases. PROBLEM 2.42 KNOWN: Temperature distribution in a spherical shell. FIND: Whether conditions are steady-state or transient. Manner in which heat flux and heat rate vary with radius. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties. ANALYSIS: From Equation 2.23, the heat equation reduces to 1 ∂ 2∂T 1 ∂T r = . ∂r α ∂t r2 ∂ r Substituting for T(r), 1 ∂T 1 ∂ 2 C1 =− 2 r 2 = 0. α ∂t r ∂r r < Hence, steady-state conditions exist. From Equation 2.22, the radial component of the heat flux is q ′′ = − k r ∂T C = −k 1 . ∂r r2 4 9 Hence, q ′′ decreases with increasing r 2 q ′′α 1/ r 2 . r r < At any radial location, the heat rate is q r = 4πr 2q ′′ = 4πkC1. r Hence, qr is independent of r. < COMMENTS: The fact that qr is independent of r is consistent with the energy conservation requirement. If qr is constant, the flux must vary inversely with the area perpendicular to the direction 2 of heat flow. Hence, q ′′ varies inversely with r . r PROBLEM 2.43 KNOWN: Spherical container with an exothermic reaction enclosed by an insulating material whose outer surface experiences convection with adjoining air and radiation exchange with large surroundings. FIND: (a) Verify that the prescribed temperature distribution for the insulation satisfies the appropriate form of the heat diffusion equation; sketch the temperature distribution and label key features; (b) Applying Fourier's law, verify the conduction heat rate expression for the insulation layer, qr, in terms of Ts,1 and Ts,2; apply a surface energy balance to the container and obtain an alternative expression for qr in terms of q and r1; (c) Apply a surface energy balance around the outer surface of the insulation to obtain an expression to evaluate Ts,2; (d) Determine Ts,2 for the specified geometry and operating conditions; (e) Compute and plot the variation of Ts,2 as a function of the outer radius for the range 201 ≤ r2 ≤ 210 mm; explore approaches for reducing Ts,2 ≤ 45°C to eliminate potential risk for burn injuries to personnel. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, radial spherical conduction, (2) Isothermal reaction in container so that To = Ts,1, (2) Negligible thermal contact resistance between the container and insulation, (3) Constant properties in the insulation, (4) Surroundings large compared to the insulated vessel, and (5) Steady-state conditions. ANALYSIS: The appropriate form of the heat diffusion equation (HDE) for the insulation follows from Eq. 2.23, 1 d 2 dT r =0 r 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: ( ) ? = ? 0 0 + r1 r 2 1 d 2 r 0 − Ts,1 − Ts,2 1 − ( r r ) r 2 dr 12 ( ) 1 d r1 + ( Ts,1 − Ts,2 ) ? = ?0 1 − ( r1 r2 ) r 2 dr < and since the expression in parenthesis is independent of r, T(r) does indeed satisfy the HDE. The temperature 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 through the insulation. (b) Using Fourier’s law for the radial-spherical coordinate, the heat rate through the insulation is dT q r = − kA r dr ( ) dT dr < = − k 4π r 2 and substituting for the temperature distribution, Eq. (2), q r = − kπ r qr = 2 ( 0 − Ts,1 − Ts,2 ( 4π k Ts,1 − Ts,2 ( ) 0 + r1 r 2 ) 1 − (r 1 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 = 0 q∀ − q r = 0 where q∀ represents the generated heat in the container, 3 q r = ( 4 3 )π r1 q (c) Applying an energy balance to a control surface placed around the outer surface of the insulation, E in − E out = 0 q r − q cv − q rad = 0 ( ) ( ) 4 4 q r − hAs Ts,2 − T∞ − ε Asσ Ts,2 − Tsur = 0 (5) < Continued... PROBLEM 2.43 (Cont.) where 2 As = 4π r2 (6) These relations can be used to determine Ts,2 in terms of the variables q , r1, r2, h, T∞ , ε and Tsur. (d) Consider the reactor system operating under the following conditions: h = 5 W/m2⋅K T∞ = 25°C r1 = 200 mm r2 = 208 mm k = 0.05 W/m⋅K ε = 0.9 Tsur = 35°C The heat generated by the exothermic reaction provides for a volumetric heat generation rate, q = q o exp ( − A To ) q o = 5000 W m3 A = 75 K (7) where the temperature of the reaction is that of the inner surface of the insulation, To = Ts,1. The following system of equations will determine the operating conditions for the reactor. Conduction rate equation, insulation, Eq. (3), qr = ( 4π × 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 3π ( 0.200 m ) q 3 ( (9) q = 5000 W m3 exp − 75 K Ts,1 ) (10) Surface energy balance, insulation, Eqs. (5) and (6), ( ) ( 4 q r − 5 W m 2 ⋅ K As Ts,2 − 298 K − 0.9As 5.67 × 10−8 W m 2⋅K 4 Ts,2 − (308 K ) 4 )=0 As = 4π (0.208 m ) 2 (11) (12) Solving these equations simultaneously, find that Ts,1 = 94.3 C < Ts,2 = 52.5 C That is, the reactor will be operating at To = Ts,1 = 94.3°C, very close to the desired 95°C operating condition. (e) From the above analysis, we found the outer surface temperature Ts,2 = 52.5°C represents a potential burn risk to plant personnel. Using the above system of equations, Eqs. (8)-(12), we have explored the effects of changes in the convection coefficient, h, and the insulation thermal conductivity, k, as a function of insulation thickness, t = r2 - r1. Continued... PROBLEM 2.43 (Cont.) 120 100 50 Reaction temperature, To (C) Outer surface temperature, Ts2 (C) 55 45 40 35 0 2 4 6 8 Insulation thickness, (r2 - r1) (mm) k = 0.05 W/m.K, h = 5 W/m^2.K k = 0.01 W/m.K, h = 5 W/m^2.K k = 0.05 W/m.K, h = 15 W/m^2.K 10 80 60 40 20 0 2 4 6 8 10 Insulation thickness, (r2-r1) (mm) k = 0.05 W/m.K, h = 5 W/m^2.K k = 0.01 W/m.K, h = 5 W/m^2.K k = 0.05 W/m.K, h = 15 W/m^2.K In the Ts,2 vs. (r2 - r1) plot, note that decreasing the thermal conductivity from 0.05 to 0.01 W/m⋅K slightly increases Ts,2 while increasing the convection coefficient from 5 to 15 W/m2⋅K markedly decreases 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 reaction temperature. With k = 0.01 W/m⋅K, the reaction temperature increases beyond 95°C with less than 2 mm insulation. For the case with h = 15 W/m2⋅K, the reaction temperature begins to approach 95°C with insulation thickness around 10 mm. We conclude that by selecting the proper insulation thickness and controlling the convection coefficient, the reaction could be operated around 95°C such that the outer surface temperature would not exceed 45°C. PROBLEM 2.44 KNOWN: One-dimensional system, initially at a uniform temperature Ti, is suddenly exposed 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) Temperature distributions for following conditions: initial condition (t ≤ 0), and several times after heater is energized; will a steady-state condition be reached; (c) Heat flux at x = 0, L/2, L as a function of time; (d) Expression for uniform temperature, Tf, reached after heater has been switched 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, the appropriate boundary and initial conditions are: 16 Initial condition: ∂ 2T ∂x 2 = Boundary conditions: x=0 q o = − k∂ T / ∂ x) 0 ′′ x=L 1 ∂T α ∂t T x,0 = Ti ∂ T / ∂ x) L = 0 Uniform 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 eventually reach a uniform temperature, Tf. Perform an energy balance on the system, Eq. 1.11b, for an interval of time ∆t = te, E in = E st It follows that E in = Q in = I te 0 q o A sdt = q ′′A s t e ′′ o 1 q ′′A s t e = Mc Tf − Ti o 6 or 1 E st = Mc Tf − Ti q ′′A t Tf = Ti + o s e . Mc 6 < PROBLEM 2.45 KNOWN: Plate of thickness 2L, initially at a uniform temperature of Ti = 200°C, is suddenly 2 quenched in a liquid bath of T∞ = 20°C 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), the steady-state condition (t → ∞), and two intermediate times; (b) On q ′′ − t coordinates, sketch the x variation with time of the heat flux at x = L, (c) Determine the heat flux at x = L and for t = 0; what is the temperature gradient for this condition; (d) By performing an energy balance on the plate, 2 determine the amount of energy per unit surface area of the plate (J/m ) that is transferred to the bath over the time required to reach steady-state conditions; and (e) Determine the energy transferred to the bath 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 heat generation. 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 with x increasing 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 with x the surface temperature as T(L,0) = Ti. 2 2 q′′ ( L, 0 ) = q′′ x conv ( t = 0 ) = h ( Ti − T∞ ) = 100 W / m ⋅ K ( 200 − 20 ) °C = 18.0 kW / m From a surface energy balance as shown in the sketch considering the conduction and convection fluxes at the surface, the temperature gradient can be calculated. Continued ….. < PROBLEM 2.45 (Cont.) Ein − Eout = 0 q′′ ( L, 0 ) − q′′ x conv ( t = 0 ) = 0 with q′′ ( L, 0 ) = − k x ∂T ∂x x = L ∂T 3 2 = −q′′ conv ( t = 0 ) / k = −18 × 10 W / m / 50 W / m ⋅ K = −360 K / m ∂x L,0 < (d) The energy transferred from the plate to the bath over the time required to reach steady-state conditions can be determined from an energy balance on a time interval basis, Eq. 1.11b. For the initial state, the plate has a uniform temperature Ti; for the final state, the plate is at the temperature of the bath, T∞. E′′ − E′′ = ∆E′′ = E′′ − E′′ in out st f i with E′′ = 0, in − E′′ = ρ cp ( 2L )[T∞ − Ti ] out E′′ = −2770 kg / m3 × 875 J / kg ⋅ K ( 2 × 0.010 m )[20 − 200] K = +8.73 ×106 J / m 2 out < (e) The energy transfer from the plate to the bath during the quenching process can be evaluated from knowledge of the surface heat flux as a function of time. The area under the curve in the q ′′ ( L, t ) vs. x time plot (see schematic above) represents the energy transferred during the quench process. E′′ = 2 ∫ out ∞ ∞ q′′ ( L, t ) dt = 2 Ae − Bt dt t =0 x t =0 ∫ ∞ 1 1 E′′ = 2A − e− Bt = 2A − (0 − 1) = 2A / B out B 0 B E′′ = 2 × 1.80 ×104 W / m 2 / 4.126 × 10−3 s −1 = 8.73 × 106 J / m 2 out < COMMENTS: (1) Can you identify and explain the important features in the temperature distributions of part (a)? (2) The maximum heat flux from the plate occurs at the instant the quench process begins and is equal to the convection heat flux. At this instant, the gradient in the plate at the surface is a maximum. If the 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 the quenching process. We cannot determine the rate at which cooling of the plate occurs without solving the heat diffusion equation. PROBLEM 2.46 KNOWN: 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 the temperature 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) 3 Expression for total energy transferred to wall per unit volume (J/m ). SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal heat generation. ANALYSIS: (a) For one-dimensional conduction with constant properties, the heat equation has the form, ∂ 2T ∂x 2 = and the conditions are: 1 ∂T α ∂t %Initial, t ≤ 0: T1x,06 = Ti KBoundaries: x = 0 ∂ T / ∂ x) = 0 & 0 K x = L − k∂ T / ∂ x) L = h T1L, t6 − T∞ K ' uniform adiabatic convection (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 the gradient 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 x x = L. Continued ….. PROBLEM 2.46 (Cont.) 16 16 For the surface at x = 0, q ′′ 0, t = 0 since it is adiabatic. At x = L and t = 0, q ′′ L,0 is a x x maximum 16 16 q ′′ L,0 = h T L,0 − T∞ x where T(L,0) = Ti. The gradient, and hence the flux, decrease with time. (d) The total energy transferred to the wall may be expressed as E in = I ∞ 0 q conv A sdt ′′ E in = hA s I2 ∞ 0 1 67 T∞ − T L, t dt Dividing both sides by AsL, the energy transferred per unit volume is I 16 E in h ∞ T∞ − T L, t dt = V L0 J / m3 COMMENTS: Note that the heat flux at x = L is into the wall and is hence in the negative x direction. PROBLEM 2.47 KNOWN: Plane wall, initially at a uniform temperature Ti, is suddenly exposed to convection with a fluid at T∞ at one surface, while the other surface is exposed to a constant heat flux q ′′ . o FIND: (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 a x function of time, (d) Expression for the steady-state temperature of the heater, T(0,∞), in terms of q ′′ , T∞ , k, h and L. o SCHEMATIC: 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, x evident from Part (a), and Fourier’s 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, Fourier’s law gives q ′′ = − k o $$ T 0, ∞ − T L, ∞ dT =k . dx L Combine Eqs. (1), (2), (3) to find: $ T 0, ∞ = T∞ + qo ′′ . 1/ h + L / k (3) PROBLEM 2.48 KNOWN: Plane wall, initially at a uniform temperature To, has one surface (x = L) suddenly exposed 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 temperature will 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. x 1 6 SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Uniform volumetric generation, (4) To < T∞ and q large enough that T(x,∞) > T∞. ANALYSIS: (a) The initial and boundary conditions for the wall can be written as Initial (t ≤ 0): T(x,0) = To Uniform temperature Boundary: x = 0 T(0,t) = To Constant temperature x=L −k 16 ∂T = h T L, t − T∞ ∂ x x=L Convection process. The temperature distributions are shown on the T-x coordinates below. Note the special condition when the heat flux at (x = L) is zero. 16 16 (b) The heat flux as a function of time at the boundaries, q ′′ 0, t and q ′′ L, t , can be inferred from x x the temperature distributions using Fourier’s law. COMMENTS: Since T ( x,∞ ) > T∞ and T∞ > To , heat transfer at both boundaries must be out of the wall. Hence, it follows from an overall energy balance on the wall that + q′′ ( 0, ∞ ) − q′′ ( L,∞ ) + qL = 0. x x PROBLEM 2.49 KNOWN: Plane wall, initially at a uniform temperature To, has one surface (x = L) suddenly exposed 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 temperature will exceed 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 heat flux q ′′ vs. t at the boundaries x = 0 and L; identify key features of the distributions. x 1 6 SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Uniform volumetric generation, (4) T∞ < To and q large enough that T(x,∞) > To. ANALYSIS: (a) The initial and boundary conditions for the wall can be written as Initial (t ≤ 0): T(x,0) = To Boundary: x=L Uniform temperature Constant temperature x = 0 T(0,t) = To ∂T −k = h T L, t − T∞ ∂ x x=L 16 Convection process. The temperature distributions are shown on the T-x coordinates below. Note that the maximum temperature occurs under steady-state conditions not at the midplane, but to the right toward the surface experiencing convection. The temperature gradients at x = L increase for t > 0 since the convection heat rate from the surface increases as the surface temperature increases. 16 16 (b) The heat flux as a function of time at the boundaries, q ′′ 0, t and q ′′ L, t , can be inferred from x x the temperature distributions using Fourier’s 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, the x convection 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 from an overall energy balance on the wall that + q′′ ( 0, ∞ ) − q′′ ( L, ∞ ) + qL = 0. x x PROBLEM 2.50 KNOWN: 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 the interface, consideration of the temperature distribution may be restricted to 0 ≤ x ≤ L. During early stages of the process, heat transfer is into the material from the outer surface, as well as from the interface. During later stages and the eventual steady state, heat is transferred from the material at the outer surface. At steady-state, dT/dx = − ( q′′ 2 ) k = const . and o T(0,t) = To + ( q′′ 2 ) L k . o (b) At the outer surface, the heat flux is initially negative, but increases with time, approaching q′′ /2. It is zero when dT dx x = L = 0 . o PROBLEM 2.51 KNOWN: Temperature distribution in a plane wall of thickness L experiencing uniform volumetric heating q having one surface (x = 0) insulated and the other exposed to a convection process characterized by T∞ and h. Suddenly the volumetric heat generation is deactivated while convection continues to occur. FIND: (a) Determine the magnitude of the volumetric energy generation rate associated with the initial 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, x sketch 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 of x 2 energy removed from the wall per unit area (J/m ) by the fluid stream as the wall cools from its initial to steady-state condition. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, and (3) Uniform internal volumetric heat generation for t < 0. ANALYSIS: (a) The volumetric heating rate can be determined by substituting the temperature distribution for the initial condition into the appropriate form of the heat diffusion equation. d dT q + =0 dx dx k where d q q (0 + 2bx ) + = 0 + 2b + = 0 dx k k ( T ( x, 0 ) = a + bx 2 ) q = −2kb = −2 × 90 W / m ⋅ K −1.0 ×104°C / 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 decreases x with increasing time as shown in the sketch above. The heat flux at t = 0 is equal to the convection heat flux with the surface temperature T(L,0). See the surface energy balance represented in the schematic. 2 5 2 q′′ ( L, 0 ) = q′′ x conv ( t = 0 ) = h ( T ( L, 0 ) − T∞ ) = 1000 W / m ⋅ K ( 200 − 20 ) °C = 1.80 × 10 W / m < 2 where T ( L, 0 ) = a + bL2 = 300°C − 1.0 × 104°C / m 2 ( 0.1m ) = 200°C. (d) The energy removed from the wall to the fluid as it cools from its initial to steady-state condition can be determined from an energy balance on a time interval basis, Eq. 1.11b. For the initial state, the 2 wall has the temperature distribution T(x,0) = a + bx ; for the final state, the wall is at the temperature of the fluid, Tf = T∞. We have used T∞ as the reference condition for the energy terms. ′′ Ein − E′′ = ∆E′′ = E′′ − E′′ out st f i with E′′ = 0 in x =L T ( x, 0 ) − T∞ dx − E′′ = ρ cp L [Tf − T∞ ] − ρ cp ∫ out x =0 E′′ = ρ cp ∫ out L x =L a + bx 2 − T∞ dx = ρ cp ax + bx 3 / 3 − T∞ x 0 x =0 3 E′′ = 7000 kg / m3 × 450 J / kg ⋅ K 300 × 0.1 − 1.0 × 104 (0.1) / 3 − 20 × 0.1 K ⋅ m out E′′ = 7.77 ×107 J / m 2 out < COMMENTS: (1) In the temperature distributions of part (a), note these features: initial condition has quadratic form with zero gradient at the adiabatic boundary; for the steady-state condition, the wall has reached the temperature of the fluid; for all distributions, the gradient at the adiabatic boundary is zero; 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 the cooling process. However, we cannot determine the rate at which cooling of the wall occurs without solving the heat diffusion equation. PROBLEM 2.52 KNOWN: Temperature as a function of position and time in a plane wall suddenly subjected to a change 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 of temperature distribution at selected times and surface heat flux variation with time, (d) Effect of thermal 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 the heat equation and boundary conditions. Substituting the distribution into Equation 2.15, it follows that ∂ 2T ∂x 2 = 1 ∂T α ∂t π 2 αt π 2 cos π x 1 6 4 L2 2L 2 L π 2 α exp − π 2 αt cos π x . C = − 1 1Ti − Ts 6 α 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 that 1 6 1 6 < < ∂T π 2 αt πx Cπ |x=0 = − 1 Ti − Ts exp − sin |x=0 = 0 ∂x 2L 4 L2 2L and 16 T L, t = Ts + C1 Ti − Ts exp − π 2 αt πx cos |x=L = Ts . 2 4L 2L Hence, the boundary conditions are also satisfied. (b) The heat flux has the form 1 6 ∂T π 2 αt πx kC1π q ′′ = − k =+ Ti − Ts exp − sin . x 2 ∂x 2L 4L 2L Continued ….. PROBLEM 2.52 (Cont.) 16 < q ′′ 0 = 0, x Hence, 16 q ′′ L = + x 6 1 π 2 αt kC1π Ti − Ts exp − . 2L 4 L2 < (c) The temperature distribution and surface heat flux variations are: (d) For materials A and B of different α, 16 16 ! 6 "## $ 2 T x, t − Ts A = exp − π αA −α B t T x, t − Ts 4 L2 B 16 1 16 Hence, if α A > α B , T x, t → Ts more rapidly for Material A. If α A < α B , T x, t → Ts more rapidly for Material B. < COMMENTS: Note that the prescribed function for T(x,t) does not reduce to Ti for t → 0. For times at or close to zero, the function is not a valid solution of the problem. At such times, the solution for T(x,t) must include additional terms. The solution is consideed in Section 5.5.1 of the text. PROBLEM 2.53 2 KNOWN: Thin electrical heater dissipating 4000 W/m sandwiched between two 25-mm thick plates whose 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 and explain key features; (b) Case 2: sudden loss of coolant causing existence of adiabatic condition on the x = +L surface; sketch temperature distribution on same T-x coordinates as part (a) and calculate values for x = 0, ± L; explain key features; (c) Case 3: further loss of coolant and existence of adiabatic condition on the x = - L surface; situation goes undetected for 15 minutes at which time power to the heater is deactivated; determine the eventual (t → ∞) uniform, steady-state temperature distribution; sketch temperature distribution on same T-x coordinates as parts (a,b); and (d) On T-t coordinates, sketch the temperature-time history at the plate locations x = 0, ± L during the transient period between the steady-state distributions for Case 2 and Case 3; at what location and when will the temperature in the system achieve a maximum value? SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal volumetric generation in plates, and (3) Negligible thermal resistance between the heater surfaces and the plates. ANALYSIS: (a) Since the system is symmetrical, the heater power results in equal conduction fluxes through the plates. By applying a surface energy balance on the surface x = +L as shown in the schematic, determine the temperatures at the mid-point, x = 0, and the exposed surface, x + L. Ein − Eout = 0 q′′ ( + L ) − q′′ x conv = 0 q′′ ( + L ) = q′′ / 2 x o where q′′ / 2 − h T ( + L ) − T∞ = 0 o ( ) T1 ( + L ) = q′′ / 2h + T∞ = 4000 W / m 2 / 2 × 400 W / m 2 ⋅ K + 20°C = 25°C o < From Fourier’s law for the conduction flux through the plate, find T(0). q′′ = q′′ / 2 = k T ( 0 ) − T ( + L ) / L x o T1 ( 0 ) = T1 ( + L ) + q′′ L / 2k = 25°C + 4000 W / m 2 ⋅ K × 0.025m / ( 2 × 5 W / m ⋅ K ) = 35°C o The temperature distribution is shown on the T-x coordinates below and labeled Case 1. The key features of the distribution are its symmetry about the heater plane and its linear dependence with distance. Continued ….. < PROBLEM 2.53 (Cont.) (b) Case 2: sudden loss of coolant with the existence of an adiabatic condition on surface x = +L. For this situation, all the heater power will be conducted to the coolant through the left-hand plate. From a surface energy balance and application of Fourier’s law as done for part (a), find T2 ( − L ) = q′′ / h + T∞ = 4000 W / m 2 / 400 W / m 2 ⋅ K + 20°C = 30°C o T2 ( 0 ) = T2 ( − L ) + q′′ L / k = 30°C + 4000 W / m 2 × 0.025 m / 5 W / m ⋅ K = 50°C o The temperature distribution is shown on the T-x coordinates above and labeled Case 2. The distribution is linear in the left-hand plate, with the maximum value at the mid-point. Since no heat flows through the right-hand plate, the gradient must zero and this plate is at the maximum temperature as well. The maximum temperature is higher than for Case 1 because the heat flux through 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 heat 2 transfer out of either plate, so that for a 15-minute period, ∆to, the heater dissipates 4000 W/m and then is deactivated. To determine the eventual, uniform steady-state temperature distribution, apply the conservation of energy requirement on a time-interval basis, Eq. 1.11b. The initial condition corresponds 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 the energy terms. E′′ − E′′ + E′′ = ∆E′′ = E′′ − E′′ (1) in out gen st f i Note that E ′′n − E ′′ = 0 , and the dissipated electrical energy is i out E′′ = q′′ ∆t o = 4000 W / m 2 (15 × 60 ) s = 3.600 × 106 J / m 2 gen o For the final condition, E′′ = ρ c ( 2L )[Tf − T∞ ] = 2500 kg / m3 × 700 J / kg ⋅ K ( 2 × 0.025m ) [Tf − 20 ]°C f E′′ = 8.75 × 104 [Tf − 20] J / m2 f where Tf = T3, the final uniform temperature, Case 3. For the initial condition, E′′ = ρ c ∫ i +L T2 ( x ) − T∞ dx = ρ c −L [ {∫ 0 +L T2 ( x ) − T∞ dx + T2 (0 ) − T∞ dx 0 −L [ ∫ [ 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 ≤ 0 T2 ( x ) = 50°C + [50 − 30]°Cx / 0.025m T2 ( x ) = 50°C + 800x (5) Substituting for T2 ( x ) , Eq. (5), into Eq. (4) Continued ….. PROBLEM 2.53 (Cont.) 0 E′′ = ρ c ∫ [50 + 800x − T∞ ] dx + T2 ( 0 ) − T∞ L i −L 0 E′′ = ρ c 50x + 400x 2 − T∞ x + T2 (0 ) − T∞ L i −L { } ′′ Ei = ρ c − −50L + 400L2 + T∞ L + T2 ( 0 ) − T∞ L E′′ = ρ cL {+50 − 400L − T∞ + T2 (0 ) − T∞ } i E′′ = 2500 kg / m3 × 700 J / kg ⋅ K × 0.025 m {+50 − 400 × 0.025 − 20 + 50 − 20}K i E′′ = 2.188 × 106 J / m 2 i (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 2 T3 = ( 66.1 + 20 ) °C = 86.1°C < The temperature distribution is shown on the T-x coordinates above and labeled Case 3. The distribution 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 between the 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 surface x = - L becomes adiabatic. These temperatures correspond to the distribution for Case 2. The heater remains 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 2.54 KNOWN: Radius and length of coiled wire in hair dryer. Electric power dissipation in the wire, and temperature and convection coefficient associated with air flow over the wire. FIND: (a) Form of heat equation and conditions governing transient, thermal behavior of wire during start-up, (b) Volumetric rate of thermal energy generation in the wire, (c) Sketch of temperature distribution 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) Uniform volumetric 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 = ρ c p ∂T = 1 ∂T r ∂r ∂r k ∂t α ∂t 1∂ The initial condition is T ( r, 0 ) = Ti The boundary conditions are: < ∂T / ∂r r = 0 = 0 −k ∂T ∂r r = r o < < < = h [T ( ro , t ) − T∞ ] (b) The volumetric rate of thermal energy generation is q= Eg P 500 W 8 3 = elec = = 3.18 × 10 W / m 2 2 ∀ π ro L π ( 0.001m ) ( 0.5m ) < Under steady-state conditions, all of the thermal energy generated within the wire is transferred to the air by convection. Performing an energy balance for a control surface about the wire, − E out + E g = 0, it follows that −2π ro L q ′′ ( ro , t → ∞ ) + Pelec = 0. Hence, q ′′ ( ro , t → ∞ ) = Pelec 2π ro L = 500 W 2π ( 0.001m ) 0.5m 5 = 1.59 × 10 W / m 2 < COMMENTS: The symmetry condition at r = 0 imposes the requirement that ∂T / ∂r r = 0 = 0, and hence 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 by convection from the surface at all times, ∂T / ∂r r = r also increases during the start-up. o PROBLEM 3.1 KNOWN: 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 , k x and L. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties, (4) Negligible radiation, (5) No generation. ANALYSIS: For the foregoing conditions, the general solution to the heat diffusion equation is of the form, Equation 3.2, T ( x ) = C1x + C2 . (1) The constants of integration, C1 and C2, are determined by using surface energy balance conditions 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=0 dx x=L For 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 substitute C1 into Eq. (4) to obtain C2. The results are T∞,1 − T∞,2 T∞,1 − T∞,2 C1 = − C2 = − + T∞,1 1 1 1 L 1 L k + h1 + + + h1 h 2 k h1 h 2 k T∞,1 − T∞,2 x 1 T (x ) = − < + + T∞,1. 1 1 L k h1 h + h + k 2 1 ( ) ( ( ) ) From Fourier’s law, the heat flux is a constant and of the form T∞,1 − T∞,2 dT q′′ = k . = − k C1 = + x dx 1 1 L h + h + k 2 1 ( ) < PROBLEM 3.2 KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfaces of 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 of the 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 radiation effects, (4) Constant properties. PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) The heat flux may be obtained from Eqs. 3.11 and 3.12, ) ( 40 C − −10 C T∞,i − T∞,o q′′ = = 1L1 1 0.004 m 1 ++ + + 2 ho k hi 65 W m ⋅ K 1.4 W m ⋅ K 30 W m2 ⋅ K q′′ = 50 C (0.0154 + 0.0029 + 0.0333) m ( 2 ⋅K W = 968 W m 2 . ) Hence, with q′′ = h i T∞,i − T∞,o , the inner surface temperature is Ts,i = T∞ ,i − q′′ hi = 40 C − 968 W m 2 2 30 W m ⋅ K < = 7.7 C ( ) Similarly for the outer surface temperature with q′′ = h o Ts,o − T∞,o find Ts,o = T∞,o − q′′ = −10 C − 968 W m 2 2 = 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 air temperature, T∞,o, for outer convection coefficients of ho = 2, 65, and 100 W/m2⋅K. As expected, Ts,i and Ts,o are linear with changes in the outside air temperature. The difference between Ts,i and Ts,o increases with increasing convection coefficient, since the heat flux through the window likewise increases. This difference is larger at lower outside air temperatures for the same reason. Note that with ho = 2 W/m2⋅K, Ts,i - Ts,o, is too small to show on the plot. ho Continued ….. Surface temperatures, Tsi or Tso (C) PROBLEM 3.2 (Cont.) 40 30 20 10 0 -10 -20 -30 -30 -25 -20 -15 -10 -5 0 Outside air temperature, Tinfo (C) Tsi; ho = 100 W/m^2.K Tso; ho = 100 W/m^2.K Tsi; ho = 65 W/m^2.K Tso; ho = 65 W/m^2.K Tsi or Tso; ho = 2 W/m^.K COMMENTS: (1) The largest resistance is that associated with convection at the inner surface. The values 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 generate the above plot. The Workspace is shown below. // Thermal Resistance Network Model: // The Network: // Heat rates into node j,qij, through thermal resistance Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 + q43 = 0 q4 - q43 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tinfo // Outside air temperature, C //q1 = // Heat rate, W T2 = Tso // Outer surface temperature, C q2 = 0 // Heat rate, W; node 2, no external heat source T3 = Tsi // Inner surface temperature, C q3 = 0 // Heat rate, W; node 2, no external heat source T4 = 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, C ho = 65 // Convection coefficient, W/m^2.K; outer surface L = 0.004 // Thickness, m; glass k = 1.4 // Thermal conductivity, W/m.K; glass Tinfi = 40 // Inside air temperature, C hi = 30 // Convection coefficient, W/m^2.K; inner surface As = 1 // Cross-sectional area, m^2; unit area PROBLEM 3.3 KNOWN: Desired inner surface temperature of rear window with prescribed inside and outside air conditions. FIND: (a) Heater power per unit area required to maintain the desired temperature, and (b) Compute and plot the electrical power requirement as a function of T∞,o for the range -30 ≤ T∞,o ≤ 0°C with ho of 2, 20, 65 and 100 W/m2⋅K. Comment on heater operation needs for low ho. If h ~ Vn, where V is the vehicle speed and n is a positive exponent, how does the vehicle speed affect the need for heater operation? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Uniform heater flux, q′′ , (4) Constant properties, (5) Negligible radiation effects, (6) Negligible film resistance. h PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) From an energy balance at the inner surface and the thermal circuit, it follows that for a unit surface area, T∞ ,i − Ts,i 1 hi q ′′ = h Ts,i − T∞ ,o + q ′′ = h L k + 1 ho Ts,i − T∞ ,o L k + 1 ho − T∞ ,i − Ts,i 1 hi ( = 0.004 m 1.4 W m ⋅ K q ′′ = (1370 − 100 ) W m = 1270 W m h 2 15 C − −10 C + ) − 1 2 65 W m ⋅ K 25 C − 15 C 1 2 10 W m ⋅ K < 2 (b) The heater electrical power requirement as a function of the exterior air temperature for different exterior convection coefficients is shown in the plot. When ho = 2 W/m2⋅K, the heater is unecessary, since the glass is maintained at 15°C by the interior air. If h ~ Vn, we conclude that, with higher vehicle speeds, the exterior convection will increase, requiring increased heat power to maintain the 15°C condition. Heater power (W/m^2) 3500 3000 2500 2000 1500 1000 500 0 -30 -20 -10 0 Exterior air temperature, Tinfo (C) h = 20 W/m^2.K h = 65 W/m^2.K h = 100 W/m^2.K COMMENTS: With q′′ = 0, the inner surface temperature with T∞, o = -10°C would be given by h T∞ ,i − Ts,i T∞ ,i − T∞ ,o = 1 hi 1 hi + L k + 1 ho = 0.10 0.118 = 0.846, or () Ts,i = 25 C − 0.846 35 C = − 4.6 C . PROBLEM 3.4 KNOWN: Curing of a transparent film by radiant heating with substrate and film surface subjected to known thermal conditions. FIND: (a) Thermal circuit for this situation, (b) Radiant heat flux, q′′ (W/m2), to maintain bond at o curing temperature, To, (c) Compute and plot q′′ as a function of the film thickness for 0 ≤ Lf ≤ 1 mm, o and (d) If the film is not transparent, determine q′′ required to achieve bonding; plot results as a function o of Lf. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) All the radiant heat flux q′′ is absorbed at the bond, (4) Negligible contact resistance. o ANALYSIS: (a) The thermal circuit for this situation is shown at the right. Note that terms are written on a per unit area basis. (b) Using this circuit and performing an energy balance on the film-substrate interface, ′′ q′′ = q1 + q′′ o 2 q′′ = o To − T∞ T −T +o 1 ′′ R ′′ + R ′′ Rs cv f where the thermal resistances are R ′′ = 1 h = 1 50 W m 2 ⋅ K = 0.020 m 2 ⋅ K W cv R ′′ = L f k f = 0.00025 m 0.025 W m ⋅ K = 0.010 m 2 ⋅ K W f R ′′ = Ls k s = 0.001m 0.05 W m ⋅ K = 0.020 m 2 ⋅ K W s q′′ = o (60 − 20 ) C [0.020 + 0.010] m2 ⋅ K + W (60 − 30 ) C 2 0.020 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 Lf is 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 is o necessary 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) 7000 6000 5000 4000 3000 2000 0 0.2 0.4 0.6 0.8 1 Film thickness, Lf (mm) Opaque film Transparent film COMMENTS: (1) When the film is transparent, the radiant flux is absorbed on the bond. The flux required decreases with increasing film thickness. Physically, how do you explain this? Why is the relationship not linear? (2) When the film is opaque, the radiant flux is absorbed on the surface, and the flux required increases with increasing thickness of the film. Physically, how do you explain this? Why is the relationship linear? (3) The IHT Thermal Resistance Network Model was used to create a model of the film-substrate system and generate the above plot. The Workspace is shown below. // Thermal Resistance Network Model: // The Network: // Heat rates into node j,qij, through thermal resistance Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 + q43 = 0 q4 - q43 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tinf // Ambient air temperature, C //q1 = // Heat rate, W; film side T2 = Ts // Film surface temperature, C q2 = 0 // Radiant flux, W/m^2; zero for part (a) T3 = To // Bond temperature, C q3 = 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, C h = 50 // Convection coefficient, W/m^2.K Lf = 0.00025 // Thickness, m; film kf = 0.025 // Thermal conductivity, W/m.K; film To = 60 // Cure temperature, C Ls = 0.001 // Thickness, m; substrate ks = 0.05 // Thermal conductivity, W/m.K; substrate Tsub = 30 // Substrate temperature, C As = 1 // Cross-sectional area, m^2; unit area PROBLEM 3.5 KNOWN: Thicknesses and thermal conductivities of refrigerator wall materials. Inner and outer air temperatures and convection coefficients. FIND: Heat gain per surface area. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Negligible contact resistance, (4) Negligible radiation, (5) Constant properties. ANALYSIS: From the thermal circuit, the heat gain per unit surface area is q′′ = q′′ = q′′ = T∞,o − T∞,i (1/ h i ) + (Lp / k p ) + ( Li / k i ) + (Lp / k p ) + (1/ h o ) ( ) ( 25 − 4 ) °C 2 1/ 5 W / m 2 ⋅ K + 2 (0.003m / 60 W / m ⋅ K ) + (0.050m / 0.046 W / m ⋅ K ) 21°C (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 the insulation. PROBLEM 3.6 KNOWN: Design and operating conditions of a heat flux gage. FIND: (a) Convection coefficient for water flow (Ts = 27°C) and error associated with neglecting conduction in the insulation, (b) Convection coefficient for air flow (Ts = 125°C) and error associated with neglecting conduction and radiation, (c) Effect of convection coefficient on error associated with neglecting conduction for Ts = 27°C. 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 conduction through the insulation. An energy balance applied to a control surface about the foil therefore yields ′′ Pelec = q ′′ conv + q ′′ cond = h ( Ts − T∞ ) + k (Ts − Tb ) L Hence, h= h= ′′ Pelec − k ( Ts − Tb ) L Ts − T∞ ( 2000 − 8) W 2K m2 = 2000 W m 2 − 0.04 W m ⋅ K ( 2 K ) 0.01m 2K < = 996 W m 2 ⋅ K If conduction is neglected, a value of h = 1000 W/m2⋅K 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 balance yields ( ) 4 4 ′′ Pelec = q ′′ conv + q ′′ + q′′ rad cond = h (Ts − T∞ ) + εσ Ts − Tsur + k ( Ts − Tb ) L Hence, h= ( ) 4 ′′ Pelec − εσ Ts4 − Tsur − k ( Ts − T∞ ) L Ts − T∞ 2 = = 2000 W m − 0.15 × 5.67 × 10 −8 2 W m ⋅K 4 (398 4 − 298 4 )K 4 − 0.04 W m ⋅ K (100 K) / 0.01m 100 K ( 2000 − 146 − 400 ) W 100 K m2 = 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 the percentage errors are 18.5 W/m2⋅K (27.6%), 16 W/m2⋅K (10.3%), and 20 W/m2⋅K (37.9%). 2 (c) For a fixed value of Ts = 27°C, the conduction loss remains at q′′ cond = 8 W/m , which is also the ′′ fixed difference between Pelec and q′′ onv . Although this difference is not clearly shown in the plot for c 10 ≤ h ≤ 1000 W/m2⋅K, it is revealed in the subplot for 10 ≤ 100 W/m2⋅K. 200 Power dissipation, P''elec(W/m^2) Power dissipation, P''elec(W/m^2) 2000 1600 1200 800 400 0 0 200 400 600 800 Convection coefficient, h(W/m^2.K) No conduction With conduction 1000 160 120 80 40 0 0 20 40 60 80 100 Convection coefficient, h(W/m^2.K) No conduction With conduction Errors associated with neglecting conduction decrease with increasing h from values which are significant for small h (h < 100 W/m2⋅K) to values which are negligible for large h. COMMENTS: In liquids (large h), it is an excellent approximation to neglect conduction and assume that all of the dissipated power is transferred to the fluid. PROBLEM 3.7 KNOWN: A layer of fatty tissue with fixed inside temperature can experience different outside convection conditions. FIND: (a) Ratio of heat loss for different convection conditions, (b) Outer surface temperature for different convection conditions, and (c) Temperature of still air which achieves same cooling as moving air (wind chill effect). SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-state conditions, (3) Homogeneous medium with constant properties, (4) No internal heat generation (metabolic effects are negligible), (5) Negligible radiation effects. PROPERTIES: Table A-3, Tissue, fat layer: k = 0.2 W/m⋅K. ANALYSIS: The thermal circuit for this situation is Hence, the heat rate is q= Ts,1 − T∞ R tot = Ts,1 − T∞ L/kA + 1/ hA . Therefore, L 1 k + h windy q′′ calm = . q′′ L 1 windy k + h calm Applying a surface energy balance to the outer surface, it also follows that q′′ cond = q′′ conv . Continued ….. PROBLEM 3.7 (Cont.) Hence, ( )( k Ts,1 − Ts,2 = h Ts,2 − T∞ L k T∞ + Ts,1 hL Ts,2 = . k 1+ hL ) To determine the wind chill effect, we must determine the heat loss for the windy day and use it to evaluate the hypothetical ambient air temperature, T∞ , which would provide the same ′ heat loss on a calm day, Hence, q′′ = ′ Ts,1 − T∞ Ts,1 − T∞ = L 1 L 1 + k + h windy k h calm From these relations, we can now find the results sought: (a) 0.003 m 1 + q′′ calm = 0.2 W/m ⋅ K 65 W/m 2 ⋅ K = 0.015 + 0.0154 0.003 m 1 q′′ 0.015 + 0.04 windy + 0.2 W/m ⋅ K 25 W/m 2 ⋅ K q′′ calm = 0.553 q′′ windy < −15 C + (b) Ts,2 calm = 1+ ( 0.2 W/m ⋅ K (c) windy = 36 C = 22.1 C 0.2 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 ) C 0.2 W/m ⋅ K 36 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 11.3°C and -1 increase in the heat loss by a factor of (0.553) = 1.81. < PROBLEM 3.8 KNOWN: Dimensions of a thermopane window. Room and ambient air conditions. FIND: (a) Heat loss through window, (b) Effect of variation in outside convection coefficient for double and triple pane construction. SCHEMATIC (Double Pane): ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Negligible radiation effects, (5) Air between glass is stagnant. PROPERTIES: Table A-3, Glass (300 K): kg = 1.4 W/m⋅K; Table A-4, Air (T = 278 K): ka = 0.0245 W/m⋅K. ANALYSIS: (a) From the thermal circuit, the heat loss is T∞,i − T∞,o q= 1 1 L L L 1 + + + + A hi kg ka kg ho q= ( 20 C − −10 C ) 1 1 0.007 m 0.007 m 0.007 m 1 + + + + 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 ) K W = 30 C 1.021K W = 29.4 W < (b) For the triple pane window, the additional pane and airspace increase the total resistance from 1.021 K/W to 1.749 K/W, thereby reducing the heat loss from 29.4 to 17.2 W. The effect of ho on the heat loss is plotted as follows. 30 Heat loss, q(W) 27 24 21 18 15 10 28 46 64 82 100 Outside convection coefficient, ho(W/m^2.K) Double pane Triple pane Continued... PROBLEM 3.8 (Cont.) Changes in ho influence the heat loss at small values of ho, for which the outside convection resistance is not negligible relative to the total resistance. However, the resistance becomes negligible with increasing ho, particularly for the triple pane window, and changes in ho have little effect on the heat loss. COMMENTS: The largest contribution to the thermal resistance is due to conduction across the enclosed air. Note that this air could be in motion due to free convection currents. If the corresponding convection coefficient exceeded 3.5 W/m2⋅K, the thermal resistance would be less than that predicted by assuming conduction across stagnant air. PROBLEM 3.9 KNOWN: Thicknesses of three materials which form a composite wall and thermal conductivities 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) Constant properties, (4) Negligible contact resistance, (5) Negligible radiation effects. ANALYSIS: Referring to the thermal circuit, the heat flux may be expressed as Ts,i − Ts,o (600 − 20 ) C q′′ = = 0.3 m 0.15 m 0.15 m L A L B LC + + + + kB 50 W/m ⋅ K k A k B k C 20 W/m ⋅ K q′′= 580 W/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), find 0.15 580 580 = − 0.018 = − 0.018 = 0.098 kB q′′ 5000 k B = 1.53 W/m ⋅ K. COMMENTS: Radiation effects are likely to have a significant influence on the net heat flux at the inner surface of the oven. < PROBLEM 3.10 KNOWN: Properties and dimensions of a composite oven window providing an outer surface safe2 to-touch temperature Ts,o = 43°C with outer convection coefficient ho = 30 W/m ⋅K and ε = 0.9 when the oven wall air temperatures are Tw = Ta = 400°C. See Example 3.1. FIND: Values of the outer convection coefficient ho required to maintain the safe-to-touch condition when the oven wall-air temperature is raised to 500°C or 600°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in window with no contact 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 inner and outer surfaces are used to determine the required value of ho when Ts,o = 43°C and Tw,i = Ta = 500 or 600°C. ) ( 4 4 εσ 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 ) ) ( 4 4 = εσ Ts,o − Tw,o + h o Ts,o − T∞ ) Using these relations in IHT, the following results were calculated: Tw,i, Ts(°C) 400 500 600 Ts,i(°C) 392 493 594 2 ho(W/m ⋅K) 30 40.4 50.7 COMMENTS: Note that the window inner surface temperature is closer to the oven air-wall temperature as the outer convection coefficient increases. Why is this so? PROBLEM 3.11 KNOWN: Drying oven wall having material with known thermal conductivity sandwiched between thin metal sheets. Radiation and convection conditions prescribed on inner surface; convection conditions on outer surface. FIND: (a) Thermal circuit representing wall and processes and (b) Insulation thickness required to maintain outer wall surface at To = 40°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Thermal resistance of metal sheets negligible. ANALYSIS: (a) The thermal circuit is shown above. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. (b) Perform energy balances on the i- and o- nodes finding T∞,i − Ti R ′′ cv,i T −T + o i + q′′ = 0 rad R ′′ cd (1) Ti − To T∞,o − To + =0 R ′′ R ′′ cd cv,o (2) where the thermal resistances are R ′′ = 1/ h i = 0.0333 m 2 ⋅ K / W cv,i (3) R ′′ = L / k = L / 0.05 m 2 ⋅ K / W cd (4) 2 R ′′ cv,o = 1/ h o = 0.0100 m ⋅ K / W (5) Substituting numerical values, and solving Eqs. (1) and (2) simultaneously, find < L = 86 mm COMMENTS: (1) The temperature at the inner surface can be found from an energy balance on the i-node using the value found for L. T∞,i − Ti R ′′ cv,o + T∞,o − Ti R ′′ + R ′′ cd cv,i + q′′ = 0 rad Ti = 298.3°C It 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 system o i satisfied? PROBLEM 3.12 KNOWN: 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/m⋅K; urethane, kf = 0.026 W/m⋅K; wood, kw = 0.12 W/m⋅K; glass, kg = 1.4 W/m⋅K. Table A.4: air, ka = 0.0263 W/m⋅K. ANALYSIS: (a) The heat loss may be obtained by dividing the overall temperature difference by the total 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 2 q= 35 C 2.33 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 ) A q= 35 C ( 0.2 + 0.002 + 0.067 ) m 2 ⋅ K W 1m 2 = 35 C 0.269 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 ) A q= 35 C ( 0.2 + 0.004 + 0.190 + 0.067 ) m 2 ⋅ K W 1m 2 = 35 C 0.461K W = 75.9 W < COMMENTS: The composite wall is clearly superior from the standpoint of reducing heat loss, and the dominant contribution to its total thermal resistance (82%) is associated with the foam insulation. Even with double pane construction, heat loss through the window is significantly larger than that for the composite wall. PROBLEM 3.13 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer 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 : Fiberglass 3 blanket, 28 kg/m , kb = 0.038 W/m⋅K; Plywood siding, ks = 0.12 W/m⋅K; Plasterboard, kp = 0.17 W/m⋅K. ANALYSIS: (a) The expression for the total thermal resistance of the house wall follows from Eq. 3.18. Lp L L 1 1 R tot = + + b+ s+ . < hiA k pA k bA ksA hoA (b) The total heat loss through the house wall is q = ∆T/R tot = ( Ti − To ) / R tot . Substituting numerical values, find 1 R tot = 0.01m + + 0.10m 30W/m 2 ⋅ K × 350m 2 0.17W/m ⋅ K × 350m 2 0.038W/m ⋅ K × 350m 2 0.02m 1 + + 2 60W/m 2 ⋅ K × 350m 2 0.12W/m ⋅ K × 350m R tot = [9.52 + 16.8 + 752 + 47.6 + 4.76]× 10−5 C/W = 831× 10−5 C/W The heat loss is then, q= 20- (-15 ) C/831×10-5 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 a 0.5% increase in q. (d) From the expression for Rtot in part (b), note that the insulation resistance, Lb/kbA, is 752/830 ≈ 90% of the total resistance. Hence, this material layer controls the resistance of the wall. From part (c) note that a 5-fold decrease in the outer convection resistance due to an increase in the wind velocity has a negligible effect on the heat loss. PROBLEM 3.14 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer surfaces. FIND: Daily heat loss for prescribed diurnal variation in ambient air temperature. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wall thermal energy storage over 24h period), (2) Negligible contact resistance. 3 PROPERTIES: Table A-3, T ≈ 300 K: Fiberglass blanket (28 kg/m ), kb = 0.038 W/m⋅K; Plywood, ks = 0.12 W/m⋅K; Plasterboard, kp = 0.17 W/m⋅K. ANALYSIS: The heat loss may be approximated as Q = 24h T − T ∞,i ∞,o ∫ 0 R tot dt where L p L b Ls 1 + + + + A hi k p k b ks h o 1 1 0.01m 0.1m 0.02m 1 R tot = + + + + 2 2 2 200m 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 = 11 R tot = 0.01454 K/W. Hence the heat rate is 12h 1 2π Q= ∫ 293 − 273 + 5 sin R tot 24 0 Q = 68.8 t dt + 24h ∫ 12 2π 293 − 273 + 11 sin 24 t dt W 2π t 12 2π t 24 24 24 20t+5 cos 0 + 20t+11 2π cos 24 12 K ⋅ h K 24 2π 60 132 Q = 68.8 240 + ( −1 − 1) + 480 − 240 + (1 + 1) W ⋅ h π π Q = 68.8 {480-38.2+84.03} W ⋅ h Q=36.18 kW ⋅ h=1.302 × 108J. COMMENTS: From knowledge of the fuel cost, the total daily heating bill could be determined. For example, at a cost of 0.10$/kW⋅h, the heating bill would be $3.62/day. < PROBLEM 3.15 KNOWN: Dimensions and materials associated with a composite wall (2.5m × 6.5m, 10 studs each 2.5m 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/m⋅K; Hardwood, 3 kB = 0.16 W/m⋅K; Gypsum, kC = 0.17 W/m⋅K; Insulation (glass fiber paper faced, 28 kg/m ), kD = 0.038 W/m⋅K. ANALYSIS: Using the isothermal surface assumption, the thermal circuit associated with a single unit (enclosed by dashed lines) of the wall is ( LA / k A A A ) = 0.008m = 0.0524 K/W 0.094 W/m ⋅ K (0.65m × 2.5m ) ( LB / k BA B ) = 0.13m = 8.125 K/W 0.16 W/m ⋅ K ( 0.04m × 2.5m ) ( LD /k D A D ) = 0.13m = 2.243 K/W 0.038 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 is −1 R eq = (1/ R B + 1/ R D ) −1 = (1/ 8.125 + 1/ 2.243) = 1.758 K/W and the total unit resistance is R 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 thermal circuit and the value of Rtot will differ. < PROBLEM 3.16 KNOWN: Conditions associated with maintaining heated and cooled conditions within a refrigerator compartment. FIND: Coefficient of performance (COP). SCHEMATIC: ASSUMPTIONS: (1) Steady-state operating conditions, (2) Negligible radiation, (3) Compartment completely sealed from ambient air. ANALYSIS: The Case (a) experiment is performed to determine the overall thermal resistance to heat transfer between the interior of the refrigerator and the ambient air. Applying an energy balance to a control surface about the refrigerator, it follows from Eq. 1.11a that, at any instant, E g − E out = 0 Hence, q elec − q out = 0 ( where q out = T∞,i − T∞,o ) R t . It follows that T∞,i − T∞,o (90 − 25 ) C = = 3.25 C/W q elec 20 W For Case (b), heat transfer from the ambient air to the compartment (the heat load) is balanced by heat transfer to the refrigerant (qin = qout). Hence, the thermal energy transferred from the refrigerator over the 12 hour period is T∞,i − T∞,o Qout = q out ∆t = qin ∆t = ∆t Rt Rt = ( 25 − 5 ) C (12 h × 3600 s h ) = 266, 000 J 3.25 C W The coefficient of performance (COP) is therefore Q 266, 000 COP = out = = 2.13 Win 125, 000 COMMENTS: The ideal (Carnot) COP is Tc 278 K COP )ideal = = = 13.9 Th − Tc ( 298 − 278 ) K Qout = and the system is operating well below its peak possible performance. < PROBLEM 3.17 KNOWN: 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 floors which minimize heat loss for a prescribed floor space and distance between floors. Corresponding heat loss, 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. From Fig. (a) As = W 2 + 4WH = W 2 + 4WNf Hf where Nf = Af W 2 Hence, As = W 2 + 4WAf Hf W 2 = W 2 + 4A f Hf W The optimum value of W corresponds to dAs 4Af Hf = 2W − =0 dW W2 or Wop = ( 2Af Hf ) 1/ 3 < The competing effects of W on the areas of the roof and sidewalls, and hence the basis for an optimum, is shown 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 = Af W2 = 32, 768 m 2 (64 m )2 =8 < and 2 2 4 × 32, 768 m × 4 m q = UAs ∆T = 1W m 2 ⋅ K ( 64 m ) + 25 C = 307, 200 W 64 m < For Nf = 2, W = (Af/Nf)1/2 = (32,768 m2/2)1/2 = 128 m 2 2 4 × 32, 768 m × 4 m q = 1W m 2 ⋅ K (128 m ) + 25 C = 512,000 W 128 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 3.18 KNOWN: Concrete wall of 150 mm thickness experiences a flash-over fire with prescribed radiant flux and hot-gas convection on the fire-side of the wall. Exterior surface condition is 300°C, typical ignition temperature for most household and office materials. FIND: (a) Thermal circuit representing wall and processes and (b) Temperature at the fire-side of the wall; 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) Constant properties. PROPERTIES: Table A-3, Concrete (stone mix, 300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) The thermal cirucit is shown above. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. (b) To determine the fire-side wall surface temperatures, perform an energy balance on the o-node. T∞ − To T −T + q′′ = L o rad R ′′ R ′′ cv cd where the thermal resistances are R ′′ = 1/ h i = 1/ 200 W / m 2 ⋅ K = 0.00500 m 2 ⋅ K / W cv R ′′ = L / k = 0.150 m /1.4 W / m ⋅ K = 0.107 m 2 ⋅ K / W cd Substituting numerical values, ( 400 − To ) K 0.005 m 2 ⋅ K / W To = 515°C + 25, 000 W / m 2 (300 − To ) K 0.107 m 2 ⋅ K / W =0 < COMMENTS: (1) The fire-side wall surface temperature is within the 350 to 600°C range for which explosive spalling could occur. It is likely the wall will experience structural collapse for these conditions. (2) This steady-state condition is an extreme condition, as the wall may fail before near steady-state conditions can be met. PROBLEM 3.19 KNOWN: Representative dimensions and thermal conductivities for the layers of fire-fighter’s protective clothing, a turnout coat. FIND: (a) Thermal circuit representing the turnout coat; tabulate thermal resistances of the layers and processes; and (b) For a prescribed radiant heat flux on the fire-side surface and temperature of Ti =.60°C 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) Constant properties. PROPERTIES: Table A-4, Air (470 K, 1 atm): kab = kcd = 0.0387 W/m⋅K. ANALYSIS: (a) The thermal circuit is shown with labels for the temperatures and thermal resistances. The conduction thermal resistances have the form R ′′ = L / k while the radiation thermal cd resistances across the air gaps have the form R ′′ = rad 1 h rad = 1 3 4σ Tavg The linearized radiation coefficient follows from Eqs. 1.8 and 1.9 with ε = 1 where Tavg represents the average temperature of the surfaces comprising the gap ( ) 2 2 3 h rad = σ ( T1 + T2 ) T1 + T2 ≈ 4σ Tavg For 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 / W cd 0.0259 0.04583 0.0259 0.00921 R ′′ rad 2 0.04264 -- 0.04264 -- -- R ′′ gap 2 0.01611 -- 0.01611 -- -- -- 0.1043 ( ) 0.01702 (m ⋅ K / W ) -(m ⋅ K / W ) -2 R ′′ total -- -- -- -- Total (tot) -- From the thermal circuit, the resistance across the gap for the conduction and radiation processes is 1 1 1 = + R ′′ gap R ′′ cd R ′′ rad and the total thermal resistance of the turn coat is R ′′ = R ′′ + R ′′ tot cd,s gap,a − b + R ′′ cd,mb + R ′′ gap,c − d + R ′′ cd,tl 2 (b) If the heat flux through the coat is 0.25 W/cm , the fire-side surface temperature To can be calculated from the rate equation written in terms of the overall thermal resistance. q′′ = ( To − Ti ) / R ′′ tot ( To = 66°C + 0.25 W / cm 2 × 102 cm / m ) × 0.1043 m2 ⋅ K / W 2 To = 327°C COMMENTS: (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 the thermal liner layer. (2) The air gap conduction and radiation resistances were calculated based upon the average temperature of 470 K. This value was determined by setting Tavg = (To + Ti)/2 and solving the equation set using IHT with kair = kair (Tavg). PROBLEM 3.20 KNOWN: Materials and dimensions of a composite wall separating a combustion gas from a liquid 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/m⋅K; Table A-2, Beryllium Oxide (T ≈ 1500K): k = 21.5 W/m⋅K. ANALYSIS: (a) The desired heat flux may be expressed as q′′= T∞,1 − T∞,2 ( 2600 − 100 ) C = 1 LA L 1 0.02 1 m 2 .K 1 0.01 + + R t,c + B + + + 0.05 + + h1 k A k B h 2 50 21.5 25.4 1000 W q′′=34,600 W/m 2 . < (b) The composite surface temperatures may be obtained by applying appropriate rate equations. From the fact that q′′=h1 T∞,1 − Ts,1 , it follows that ( Ts,1 = T∞,1 − ( ) q′′ 34, 600 W/m 2 1908 C. = 2600 C − 2 ⋅K h1 50 W/m ) With q′′= ( k A / LA ) Ts,1 − Tc,1 , it also follows that L q′′ 0.01m × 34,600 W/m 2 Tc,1 = Ts,1 − A = 1908 C − = 1892 C. kA 21.5 W/m ⋅ K ( ) Similarly, with q′′= Tc,1 − Tc,2 / R t,c Tc,2 = Tc,1 − R t,c q′′=1892C − 0.05 m2 ⋅ K W × 34, 600 = 162 C 2 W m Continued ….. PROBLEM 3.20 (Cont.) ( ) and with q′′= ( k B / LB ) Tc,2 − Ts,2 , L q′′ 0.02m × 34,600 W/m 2 Ts,2 = Tc,2 − B = 162 C − = 134.6 C. kB 25.4 W/m ⋅ K The 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 the stainless steel. For improved accuracy the calculations should be repeated using k values corresponding to T ≈ 1900°C for the oxide and T ≈ 115°C for the steel. (3) The major contributions to the total resistance are made by the combustion gas boundary layer and the contact, where the temperature drops are largest. PROBLEM 3.21 KNOWN: Thickness, overall temperature difference, and pressure for two stainless steel plates. 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/m⋅K. ANALYSIS: (a) With R ′′ ≈ 15 × 10−4 m 2 ⋅ K/W from Table 3.1 and t,c L 0.01m = = 6.02 × 10−4 m 2 ⋅ K/W, k 16.6 W/m ⋅ K it follows that R ′′ = 2 ( L/k ) + R ′′ ≈ 27 × 10−4 m 2 ⋅ K/W; tot t,c hence q′′= 100 C ∆T = = 3.70 × 104 W/m 2 . -4 m 2 ⋅ K/W R ′′ tot 27 × 10 < (b) From the thermal circuit, −4 2 R ′′ ∆Tc t,c 15 × 10 m ⋅ K/W = = = 0.556. Ts,1 − Ts,2 R ′′ 27 × 10-4 m 2 ⋅ K/W tot Hence, ( ) ( ) ∆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. t PROBLEM 3.22 KNOWN: Temperatures and convection coefficients associated with fluids at inner and outer surfaces of a composite wall. Contact resistance, dimensions, and thermal conductivities associated 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, 1 L L 1 + A + R t,c + B + h1A k A A k BA h 2 A 0.01 0.3 0.02 1 K 1 R tot = + + + + 10 × 5 0.1× 5 5 0.04 × 5 20 × 5 W K K R tot = [0.02 + 0.02 + 0.06 + 0.10 + 0.01] = 0.21 W W R tot = q= T∞,1 − T∞,2 R tot ( 200 − 40 ) C = 762 W. = 0.21 K/W (b) It follows that Ts,1 = T∞,1 − TA = Ts,1 − q 762 W = 200 C − = 184.8 C h1A 50 W/K qL A = 184.8 C − kAA 762W × 0.01m 0.1 W m⋅K × 5m = 169.6 C 2 K TB = TA − qR t,c = 169.6 C − 762W × 0.06 Ts,2 = TB − qL B k BA T∞,2 = Ts,2 − = 123.8 C − q h 2A W 762W × 0.02m 0.04 = 47.6 C − W m⋅K × 5m 762W 100W/K = 123.8 C = 47.6 C 2 = 40 C < PROBLEM 3.23 KNOWN: Outer and inner surface convection conditions associated with zirconia-coated, Inconel turbine 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, with and without the TBC. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constant properties, (3) Negligible radiation. ANALYSIS: For a unit area, the total thermal resistance with the TBC is −1 −1 R ′′ tot,w = h o + ( L k )Zr + R ′′ + ( L k )In + h i t,c ( ) −3 −4 −4 −4 −3 R ′′ m 2 ⋅ K W = 3.69 × 10−3 m 2 ⋅ K W tot,w = 10 + 3.85 × 10 + 10 + 2 × 10 + 2 × 10 With a heat flux of T∞,o − T∞,i 1300 K q′′ = = = 3.52 × 105 W m 2 w −3 2 R ′′ 3.69 × 10 m ⋅ K W tot,w the 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 K w ( Ts,o(w) = T∞ ,i + (1 h i ) + ( L k )In q ′′ = 400 K + 2 × 10 w −3 + 2 × 10 −4 )m 2 ( 5 ⋅ K W 3.52 × 10 W m 2 ) = 1174 K −1 −1 −3 2 ′′ ′′ Without 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)/3.20×10-3 m2⋅K/W = 4.06×105 W/m2. The inner and outer surface temperatures of the Inconel are then ( ) Ts,i(wo) = T∞ ,i + ( q′′ h i ) = 400 K + 4.06 × 105 W m2 500 W m 2 ⋅ K = 1212 K wo Ts,o(wo) = T∞ ,i + [(1 h i ) + ( L k )In ] q′′wo = 400 K + ( 2 × 10−3 + 2 × 10−4 ) m 2 ⋅ K ( 5 W 4.06 × 10 W m 2 ) = 1293 K Continued... PROBLEM 3.23 (Cont.) Temperature, T(K) 1300 1260 1220 1180 1140 1100 0 0.001 0.002 0.003 0.004 0.005 Inconel location, x(m) With TBC Without TBC Use of the TBC facilitates operation of the Inconel below Tmax = 1250 K. COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increases with increasing thickness, limits to the thickness are associated with reliability considerations. PROBLEM 3.24 KNOWN: Size and surface temperatures of a cubical freezer. Materials, thicknesses and interface resistances 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/m⋅K. Table A-1, Carbon steel AISI 1010 (~295K): kst = 64 W/m⋅K. Table A-3 (~300K): kins = 0.039 W/m⋅K. ANALYSIS: For a unit wall surface area, the total thermal resistance of the composite wall is L L L R ′′ = al + R ′′ + ins + R ′′ + st tot t,c t,c k al k ins kst R ′′ = tot 0.00635m m2 ⋅ K 0.100m m2 ⋅ K 0.00635m + 2.5 × 10−4 + + 2.5 ×10−4 + 173 W / m ⋅ K W 0.039 W / m ⋅ K W 64 W / m ⋅ K ) ( R ′′ = 3.7 × 10−5 + 2.5 ×10−4 + 2.56 + 2.5 ×10−4 + 9.9 × 10−5 m 2 ⋅ K / W tot Hence, the heat flux is q′′ = Ts,o − Ts,i R ′′ tot 22 − ( −6 ) °C W = = 10.9 2.56 m 2 ⋅ K / W m2 and the cooling load is q = As q′′ = 6 W 2 q′′ = 54m 2 ×10.9 W / m 2 = 590 W COMMENTS: Thermal resistances associated with the cladding and the adhesive joints are negligible compared to that of the insulation. < PROBLEM 3.25 KNOWN: 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 prescribed conditions, (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 is q′′ − q′′ with × 100% = 1 − q′′ with R q = wo q′′ q′′ wo wo R ′′ tot,wo × 100% = 1 − × 100% R ′′tot, with where the total thermal resistances without and with the insulation, respectively, are R ′′ tot,wo = R ′′ cnv,o + R ′′ cnd,w + R ′′ cnv,i = 1 Lw 1 + + ho k w hi 2 2 R ′′ tot,wo = ( 0.050 + 0.004 + 0.200 ) m ⋅ K / W = 0.254 m ⋅ K / W R ′′ tot,with = R ′′ cnv,o + R ′′ cnd,w + R ′′ + R ′′ t,c cnd,ins + R ′′ cnv,i = L 1 Lw 1 + + R ′′ + ins + t,c ho k w kins h i 2 2 R ′′ tot,with = (0.050 + 0.004 + 0.002 + 0.926 + 0.500 ) m ⋅ K / W = 1.482 m ⋅ K / W R q = (1 − 0.254 /1.482 ) × 100% = 82.9% < 2 (b) With As = 12 m , the heat losses without and with the insulation are 2 2 q wo = As T∞,i − T∞,o / R ′′ tot,wo = 12 m × 32°C / 0.254 m ⋅ K / W = 1512 W ( ) ( ) 2 2 q with = As T∞,i − T∞,o / R ′′ tot,with = 12 m × 32°C /1.482 m ⋅ K / W = 259 W < < (c) With the windows covered for 12 hours per day, the daily savings are S= ( q wo − q with ) ηf ∆t C g × 10 −6 MJ / J = (1512 − 259 ) W 0.8 12h × 3600 s / h × $0.01 / MJ × 10 −6 MJ / J = $0.677 COMMENTS: (1) The savings may be insufficient to justify the cost of the insulation, as well as the daily tedium of applying and removing the insulation. However, the losses are significant and unacceptable. The owner of the building should install double pane windows. (2) The dominant contributions to the total thermal resistance are made by the insulation and convection at the inner surface. PROBLEM 3.26 KNOWN: Surface area and maximum temperature of a chip. Thickness of aluminum cover and 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/m⋅K. ANALYSIS: For a control surface about the chip, conservation of energy yields Eg − Eout = 0 or Pc − (Tc − T∞ ) A =0 ( L/k ) + R ′′ + (1/ h ) t,c (85 − 25 ) C Pc,max = Pc,max = (10-4m2 ) ( 0.002 / 238 ) + 0.5 × 10−4 + (1/1000 ) m 2 ⋅ K/W −4 C ⋅ m 2 60 × 10 (8.4 ×10-6 + 0.5 ×10−4 + 10−3 ) 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 3.27 KNOWN: Operating conditions for a board mounted chip. FIND: (a) Equivalent thermal circuit, (b) Chip temperature, (c) Maximum allowable heat dissipation for dielectric liquid (ho = 1000 W/m2⋅K) and air (ho = 100 W/m2⋅K). Effect of changes in circuit board temperature and contact resistance. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible chip thermal resistance, (4) Negligible radiation, (5) Constant properties. PROPERTIES: Table A-3, Aluminum oxide (polycrystalline, 358 K): kb = 32.4 W/m⋅K. ANALYSIS: (a) (b) Applying conservation of energy to a control surface about the chip ( E in − E out = 0 ) , ′′ o q′′ − qi − q′′ = 0 c q′′ = c Tc − T∞,i 1 h i + ( L k )b + R ′′ t,c + Tc − T∞ ,o 1 ho With q TT 3 ` 10 4 W m 2 , ho = 1000 W/m2⋅K, kb = 1 W/m⋅K and R ′′ = 10 −4 m 2 ⋅ K W , c t,c 3 × 104 W m 2 = Tc − 20 C (1 40 + 0.005 1 + 10−4 ) m2 ⋅ K W + Tc − 20 C (1 1000 ) m 2 ⋅ K W 3 × 104 W m 2 = (33.2Tc − 664 + 1000Tc − 20, 000 ) W m 2 ⋅ K 1003Tc = 50,664 < Tc = 49°C. (c) For Tc = 85°C and ho = 1000 W/m2⋅K, the foregoing energy balance yields < q′′ = 67,160 W m 2 c with q′′ = 65,000 W/m and q′′ = 2160 W/m . Replacing the dielectric with air (ho = 100 W/m ⋅K), the o i following results are obtained for different combinations of kb and R ′′,c . t 2 2 2 Continued... PROBLEM 3.27 (Cont.) kb (W/m⋅K) R ′′ t,c q′′ (W/m ) i q ′′ (W/m ) o q ′′ (W/m ) c 2159 2574 2166 2583 6500 6500 6500 6500 8659 9074 8666 9083 2 2 2 (m2⋅K/W) 1 32.4 1 32.4 10-4 10-4 10-5 10-5 < COMMENTS: 1. For the conditions of part (b), the total internal resistance is 0.0301 m2⋅K/W, while the outer resistance is 0.001 m2⋅K/W. Hence ( ( T − T∞,o q′′ o= c q′′ Tc − T∞,i i ) R ′′ = 0.0301 = 30 . o ) R ′′ 0.001 i and only approximately 3% of the heat is dissipated through the board. ′′ o 2. With ho = 100 W/m2⋅K, the outer resistance increases to 0.01 m2⋅K/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, although ′′ measures to reduce R i would have a negligible effect on q′′ for the liquid coolant, some improvement c may be gained for air-cooled conditions. As shown in the table of part (b), use of an aluminum oxide ′′ board increase q′′ by 19% (from 2159 to 2574 W/m2) by reducing R i from 0.0301 to 0.0253 m2⋅K/W. i ′′ Because the initial contact resistance ( R ′′ = 10 −4 m 2 ⋅ K W ) is already much less than R i , any reduction t,c in its value would have a negligible effect on q′′ . The largest gain would be realized by increasing hi, i since the inside convection resistance makes the dominant contribution to the total internal resistance. PROBLEM 3.28 KNOWN: Dimensions, thermal conductivity and emissivity of base plate. Temperature and convection coefficient of adjoining air. Temperature of surroundings. Maximum allowable temperature of transistor case. Case-plate interface conditions. FIND: (a) Maximum allowable power dissipation for an air-filled interface, (b) Effect of convection coefficient on maximum allowable power dissipation. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Negligible heat transfer from the enclosure, to the surroundings. (3) One-dimensional conduction in the base plate, (4) Radiation exchange at surface of base plate is with large surroundings, (5) Constant thermal conductivity. 5 2 PROPERTIES: Aluminum-aluminum interface, air-filled, 10 µm roughness, 10 N/m contact pressure (Table 3.1): R ′′ = 2.75 × 10−4 m 2 ⋅ K / W. t,c ANALYSIS: (a) With all of the heat dissipation transferred through the base plate, Pelec = q = Ts,c − T∞ (1) R tot where R tot = R t,c + R cnd + (1/ R cnv ) + (1/ R rad ) R tot = and R ′′ t,c Ac + L kW 2 ( h r = εσ Ts,p + Tsur + −1 1 1 W2 h + hr (2) 2 2 ) (Ts,p + Tsur ) (3) To obtain Ts,p, the following energy balance must be performed on the plate surface, q= Ts,c − Ts,p R t,c + R cnd -4 2 ( ) ( = qcnv + q rad = hW 2 Ts,p − T∞ + h r W 2 Ts,p − Tsur -4 ) 2 (4) -4 2 With Rt,c = 2.75 × 10 m ⋅K/W/2×10 m = 1.375 K/W, Rcnd = 0.006 m/(240 W/m⋅K × 4 × 10 m ) = 0.0625 K/W, and the prescribed values of h, W, T∞ = Tsur and ε, Eq. (4) yields a surface temperature of Ts,p = 357.6 K = 84.6°C and a power dissipation of Continued ….. PROBLEM 3.28 (Cont.) < Pelec = q = 0.268 W The convection and radiation resistances are Rcnv = 625 m⋅K/W and Rrad = 345 m⋅K/W, where hr = 2 7.25 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 be derived by increasing the value of h. 4 .5 4 3 .5 3 2 .5 2 1 .5 1 0 .5 0 0 20 40 60 80 100 120 140 160 180 200 C o n ve ctio n c o e ffic ie n t, h (W /m ^2 .K ) 2 For h = 200 W/m ⋅K, Rcnv = 12.5 m⋅K/W and Ts,p = 351.6 K, yielding Rrad = 355 m⋅K/W. The effect of radiation is then negligible. 2 COMMENTS: (1) The plate conduction resistance is negligible, and even for h = 200 W/m ⋅K, the contact resistance is small relative to the convection resistance. However, Rt,c could be rendered negligible by using indium foil, instead of an air gap, at the interface. From Table 3.1, R ′′ = 0.07 × 10−4 m 2 ⋅ K / W, in which case Rt,c = 0.035 m⋅K/W. t,c 2 (2) Because Ac < W , heat transfer by conduction in the plate is actually two-dimensional, rendering the conduction resistance even smaller. PROBLEM 3.29 KNOWN: Conduction in a conical section with prescribed diameter, D, as a function of x in 1/2 the 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/m⋅K. ANALYSIS: (a) Based upon the assumptions, and following the same methodology of Example 3.3, qx is a constant independent of x. Accordingly, q x = − kA 2 using A = πD /4 where D = ax 4q x x dx π a 2k ) ( 2 dT dT = − k π ax1/2 / 4 dx dx ∫x1 x 1/2 T T1 = −∫ (1) . Separating variables and identifying limits, (2) dT. Integrating and solving for T(x) and then for T2, T ( x ) = T1 − 4q x π a 2k ln x x1 T2 = T1 − 4q x x ln 2 . π a 2k x1 (3,4) Solving Eq. (4) for qx and then substituting into Eq. (3) gives the results, qx = − π2 a k ( T1 − T2 ) /1n ( x1 / x 2 ) 4 T ( 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 shown above. (b) The heat rate follows from Eq. (5), qx = π W 25 × 0.52 m × 236 (600 − 400 ) K/ln = 5.76kW. 4 m⋅K 125 < PROBLEM 3.30 KNOWN: 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/m⋅K. ( ) ANALYSIS: (a) From Fourier’s law, Eq. (2.1), with A=π D 2 / 4 = π a 2 / 4 x 3 , it follows that 4q x dx π a 2 x3 = − kdT. Hence, since qx is independent of x, T 4q x x dx = − k ∫ dT 2 ∫x1 x 3 T1 πa or x 4q x 1 − = − k ( T − T1 ). π a 2 2x 2 x1 Hence 1 − 1 . 2 π a 2 k x 2 x1 (b) From the foregoing expression, it also follows that T = T1 + 2q x < T2 − T1 π a 2k 2 1/x 2 − 1/ x 2 1 2 -1 238 W/m K π 1m ⋅ ( 20 − 100 ) C qx = × 2 ( 0.225 )−2 − ( 0.075)−2 m-2 qx = () q x = 189 W. < COMMENTS: The foregoing results are approximate due to use of a one-dimensional model in treating what is inherently a two-dimensional problem. PROBLEM 3.31 KNOWN: 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-state conditions, (3) No internal heat generation. ANALYSIS: For the assumed conditions, qx and A(x) are constant and Eq. 3.21 gives L T q′′ ∫ dx = − ∫ 1 ( k o + aT )dT x 0 q′′ = x To ( ) 1 a2 2 k o ( To − T1 ) + 2 To − T1 . L From Fourier’s law, q′′ = − ( k o + aT ) dT/dx. x Hence, since the product of (ko+aT) and dT/dx) is constant, decreasing T with increasing x implies, a > 0: decreasing (ko+aT) and increasing |dT/dx| with increasing x a = 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 3.32 KNOWN: 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 Fourier’s law is dT dT = − k ( 2π rL ) dr dr dT q′ = −2π kr r dr dT q′ = −2π rk o (1 + aT ) . r dr q r = − kA r Separating variables, q′ dr −r = k o (1 + aT ) dT 2π r and integrating across the wall, find q′ −r 2π q′ −r 2π q′ −r 2π ro dr ∫ri r ln o ri r ln o ri r T = k o ∫ o (1+aT ) dT Ti aT 2 To = ko T + 2 Ti a2 = k o ( To − Ti ) + To − Ti2 2 ( ) a (T − T ) q′ = −2π k o 1 + ( To + Ti ) o i . r 2 ln ( ro / ri ) It follows that the overall thermal resistance per unit length is ln ( ro / ri ) ∆T R ′t = . = q′ a r 2π k o 1 + ( To + Ti ) 2 COMMENTS: Note the necessity of the stated assumptions to treating q′ as independent of r. r < < PROBLEM 3.33 KNOWN: Steady-state temperature distribution of convex shape for material with k = ko(1 + αT) where α is a constant and the mid-point temperature is ∆To higher than expected for a linear temperature distribution. FIND: Relationship to evaluate α in terms of ∆To and T1, T2 (the temperatures at the boundaries). SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat generation, (4) α is positive and constant. ANALYSIS: At any location in the wall, Fourier’s law has the form dT q′′ = − k o (1 + α T ) . (1) x dx Since q′′ is a constant, we can separate Eq. (1), identify appropriate integration limits, and x integrate to obtain L T2 x ∫0 q′′ dx = − ∫T1 k o (1 + α T )dT 2 α T2 α T2 k o T2 + − T1 + 1 . q′′ = − x L 2 2 We could perform the same integration, but with the upper limits at x = L/2, to obtain 2 α T2 α T1 2k o TL/2 + L/2 − T1 + q′′ = − x L 2 2 where T +T TL/2 = T ( L/2 ) = 1 2 + ∆To . 2 Setting Eq. (3) equal to Eq. (4), substituting from Eq. (5) for TL/2, and solving for α, it follows that 2∆To α= . 2 2 2 T2 + T1 / 2 − ( T1 + T2 ) / 2 + ∆To ( ) (2) (3) (4) (5) < PROBLEM 3.34 KNOWN: 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+dr for steady-state, one-dimensional conditions with no heat generation. With Fourier’s law, q r = − kA dT dT = − k ( 2π rL ) dr dr (1) where A = 2πrL is the area normal to the direction of heat transfer. Since qr is constant, Eq. (1) may be separated and expressed in integral form, To q r ro dr ∫ri r = −∫Ti k (T ) dT. 2π L Assuming k is constant, the heat rate is qr = 2π Lk ( Ti − To ) ln ( ro / ri ) . Remembering that the thermal resistance is defined as R t ≡ ∆T/q it follows that for the hollow cylinder, Rt = ln ( ro / ri ) 2π LK . COMMENTS: Compare the alternative method used in this analysis with the standard method employed in Section 3.3.1 to obtain the same result. < PROBLEM 3.35 KNOWN: 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 surface temperature. (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/m⋅K. ANALYSIS: (a) From Eq. 3.27 with Ts,2 = 490 K, the heat rate per unit length is ( 2π k Ts,1 − Ts,2 q′ = q r L = q′ = ln ( r2 r1 ) ) 2π ( 0.089 W m ⋅ K )(800 − 490 ) K ln ( 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 that q′ cond = q′ conv + q′ rad Ts,1 − Ts,2 ln ( r2 r1 ) 2π k ( = Ts,2 − T∞ + Ts,2 − Tsur 1 ( 2π r2 h ) 1 ( 2π r2 h r ) where h r = εσ Ts,2 + Tsur 2 2 ) (Ts,2 + Tsur ) . Solving this equation for Ts,2, the heat rate may be determined from ( ) ( ) q′ = 2π r2 h Ts,2 − T∞ + h r Ts,2 − Tsur Continued... PROBLEM 3.35 (Cont.) and from Eq. 3.26 the temperature distribution is T(r) = Ts,1 − Ts,2 ln ( r1 r2 ) r + Ts,2 r2 ln As shown below, the outer surface temperature of the insulation Ts,2 and the heat loss q′ decay precipitously with increasing insulation thickness from values of Ts,2 = Ts,1 = 800 K and q′ = 11,600 W/m, respectively, at r2 = r1 (no insulation). 800 10000 Heat loss, qprime(W/m) Temperature, Ts2(K) 700 600 500 400 1000 300 100 0 0.04 0.08 0.12 0 0.04 Insulation thickness, (r2-r1) (m) 0.08 0.12 Insulation thickness, (r2-r1) (m) Outer surface temperature Heat loss, qprime When plotted as a function of a dimensionless radius, (r - r1)/(r2 - r1), the temperature decay becomes more pronounced with increasing r2. Temperature, T(r) (K) 800 700 600 500 400 300 0 0.2 0.4 0.6 0.8 1 Dimensionless radius, (r-r1)/(r2-r1) r2 = 0.20m r2 = 0.14m r2= 0.10m Note that T(r2) = Ts,2 increases with decreasing r2 and a linear temperature distribution is approached as r2 approaches r1. COMMENTS: An insulation layer thickness of 20 mm is sufficient to maintain the outer surface temperature and heat rate below 350 K and 1000 W/m, respectively. PROBLEM 3.36 KNOWN: Temperature and volume of hot water heater. Nature of heater insulating material. Ambient air 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 the water (Ts,1 = 55°C), (4) Constant properties, (5) Negligible radiation. PROPERTIES: Table A.3, Urethane Foam (T = 300 K): k = 0.026 W/m⋅K. ANALYSIS: To minimize heat loss, tank dimensions which minimize the total surface area, As,t, should ( ) be selected. With L = 4∀/πD2, As,t = π DL + 2 π D 2 4 = 4 ∀ D + π D 2 2 , and the tank diameter for which As,t is an extremum is determined from the requirement dAs,t dD = − 4∀ D2 + π D = 0 It follows that D = ( 4∀ π ) 1/ 3 L = ( 4∀ π ) 1/ 3 and With d 2 As,t dD 2 = 8∀ D3 + π > 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 m The total heat loss through the side and end walls is q= Ts,1 − T∞ ln ( r2 r1 ) 2π kLop + 1 h2π r2 Lop ( 2 Ts,1 − T∞ + ( δ 2 k π Dop 4 + ) 1 2 ) h (π 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 + δ = 0.417 m, it follows that Continued... PROBLEM 3.36 (Cont.) q= ln ( 0.417 0.392 ) (55 − 20 ) C 2π ( 0.026 W m ⋅ K ) 0.784 m (2 W m ⋅ K ) 2π (0.417 m ) 0.784 m 2 2 (55 − 20 ) C + 0.025 m (0.026 W q= 1 + (2 W m2 ⋅ K )π 4 (0.784 m )2 2 (35 C ) = ( 48.2 + 23.1) W = 71.3 W m ⋅ K )π 4 ( 0.784 m ) 2 35 C (0.483 + 0.243) K 1 + W + (1.992 + 1.036 ) K W The annual energy loss is therefore ( ) Qannual = 71.3 W (365 days ) ( 24 h day ) 10−3 kW W = 625 kWh With a unit electric power cost of $0.08/kWh, the annual cost of the heat loss is C = ($0.08/kWh)625 kWh = $50.00 Hence, an insulation thickness of δ = 25 mm < will satisfy the prescribed cost requirement. COMMENTS: Cylindrical containers of aspect ratio L/D = 1 are seldom used because of floor space constraints. Choosing L/D = 2, ∀ = πD3/2 and D = (2∀/π)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 annual cost of the heat loss is small, providing little justification for using the optimal heater dimensions. PROBLEM 3.37 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface and is exposed to a fluid of prescribed h and T∞. Thermal contact resistance between heater and tube wall and wall inner surface temperature. FIND: Heater power per unit length required to maintain a heater temperature of 25°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across heater. ANALYSIS: The thermal circuit has the form Applying an energy balance to a control surface about the heater, q′ = q′ + q′ a b To − Ti T −T q′ = +o ∞ ln ( ro / ri ) (1/hπ Do ) + R ′t,c 2π k 25 − ( −10 ) C ( 25-5 ) C q′= + ln (75mm/25mm ) m ⋅ K 1/ 100 W/m 2 ⋅ K × π × 0.15m + 0.01 2π × 10 W/m ⋅ K W ( ) q′ = ( 728 + 1649 ) W/m q′=2377 W/m. COMMENTS: The conduction, contact and convection resistances are 0.0175, 0.01 and 0.021 m ⋅K/W, respectively, < PROBLEM 3.38 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface. Inner and outer wall temperatures. Temperature of fluid adjoining outer wall. FIND: Effect of wall thermal conductivity, thermal contact resistance, and convection coefficient on total 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′ + q′ i o q′ = To − Ti T − T∞ +o ln ( ro ri ) (1 2π ro h ) + R′ t,c 2π k Selecting nominal values of k = 10 W/m⋅K, R ′ ,c = 0.01 m⋅K/W and h = 100 W/m2⋅K, the following t parametric variations are obtained 3500 3000 3000 2500 Heat rate(W/m) Heat rate (W/m) 2500 2000 1500 1000 2000 1500 1000 500 500 0 0 0 50 100 150 200 0 0.02 Thermal conductivity, k(W/m.K) qi q qo 0.04 0.06 0.08 0.1 Contact resistance, Rtc(m.K/W) qi q qo Continued... PROBLEM 3.38 (Cont.) 20000 Heat rate(W/m) 16000 12000 8000 4000 0 0 200 400 600 800 1000 Convection coefficient, h(W/m^2.K) qi q qo ′ For a prescribed value of h, q′ is fixed, while qi , and hence q′ , increase and decrease, respectively, o with increasing k and R ′ ,c . These trends are attributable to the effects of k and R ′ ,c on the total t t (conduction plus contact) resistance separating the heater from the inner surface. For fixed k and R ′ ,c , t ′ qi is fixed, while q′ , and hence q′ , increase with increasing h due to a reduction in the convection o resistance. COMMENTS: For the prescribed nominal values of k, R ′ ,c and h, the electric power requirement is t q′ = 2377 W/m. To maintain the prescribed heater temperature, q′ would increase with any changes which reduce the conduction, contact and/or convection resistances. PROBLEM 3.39 KNOWN: Wall thickness and diameter of stainless steel tube. Inner and outer fluid temperatures and convection coefficients. FIND: (a) Heat gain per unit length of tube, (b) Effect of adding a 10 mm thick layer of insulation to outer surface of tube. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties, (4) Negligible contact resistance between tube and insulation, (5) Negligible effect of radiation. PROPERTIES: Table A-1, St. St. 304 (~280K): kst = 14.4 W/m⋅K. ANALYSIS: (a) Without the insulation, the total thermal resistance per unit length is R ′tot = R ′ conv,i + R ′ cond,st + R ′ conv,o = R ′tot = 1 2π (0.018m ) 400 W / m 2 ⋅ K + ln ( r2 / ri ) 1 1 + + 2π ri h i 2π k st 2π r2 h o ln ( 20 /18 ) 2π (14.4 W / m ⋅ K ) + 1 2π (0.020m ) 6 W / m 2 ⋅ K ) ( R ′tot = 0.0221 + 1.16 × 10−3 + 1.33 m ⋅ K / W = 1.35 m ⋅ K / W The heat gain per unit length is then q′ = T∞,o − T∞,i R′ tot = ( 23 − 6 ) °C 1.35 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 ′ ′ +R′ conv,o , where R conv,i and R cond,st remain the same. The thermal resistance of the insulation is R′ cond,ins = ln ( r3 / r2 ) 2π k ins = ln (30 / 20 ) 2π (0.05 W / m ⋅ K ) = 1.29 m ⋅ K / W and the outer convection resistance is now R′ conv,o = 1 1 = = 0.88 m ⋅ K / W 2π r3h o 2π ( 0.03m ) 6 W / m 2 ⋅ K The total resistance is now ( ) R ′tot = 0.0221 + 1.16 × 10−3 + 1.29 + 0.88 m ⋅ K / W = 2.20 m ⋅ K / W Continued ….. PROBLEM 3.39 (Cont.) and the heat gain per unit length is q′ = T∞,o − T∞,i R ′tot = 17°C = 7.7 W / m 2.20 m ⋅ K / W COMMENTS: (1) The validity of assuming negligible radiation may be assessed for the worst case condition 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 ε = 0.7, the heat gain due to net ( ) 4 radiation exchange with the surroundings is q ′ = εσ ( 2π r2 ) Tsur − Ts4 = 7.7 W / m. Hence, the net rad rate of heat transfer by radiation to the tube surface is comparable to that by convection, and the assumption of negligible radiation is inappropriate. (2) If heat transfer from the air is by natural convection, the value of ho with the insulation would actually be less than the value for the bare tube, thereby further reducing the heat gain. Use of the insulation would also increase the outer surface temperature, thereby reducing net radiation transfer from the surroundings. (3) The critical radius is rcr = kins/h ≈ 8 mm < r2. Hence, as indicated by the calculations, heat transfer is reduced by the insulation. PROBLEM 3.40 KNOWN: Diameter, wall thickness and thermal conductivity of steel tubes. Temperature of steam flowing through the tubes. Thermal conductivity of insulation and emissivity of aluminum sheath. Temperature of ambient air and surroundings. Convection coefficient at outer surface and maximum allowable surface temperature. FIND: (a) Minimum required insulation thickness (r3 – r2) and corresponding heat loss per unit length, (b) Effect of insulation thickness on outer surface temperature and heat loss. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction, (3) Negligible contact resistances 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) Large surroundings. ANALYSIS: (a) To determine the insulation thickness, an energy balance must be performed at the outer surface, where q ′ = q ′ onv,o + q ′rad . With q ′ c conv,o = 2π r3h o ( Ts,o − T∞ ,o ) , q ′ ad = 2π r3 εσ r (T ) ( )( ) 4 4 ′ ′ ′ ′ s,o − Tsur , q = Ts,i − Ts,o / R cond,st + R cond,ins , R cond,st = n ( r2 / r1 ) / 2π k st , and R ′ ond,ins c = n ( r3 / r2 ) / 2π k ins , it follows that ( 2π Ts,i − Ts,o ) ( ) ( ) 4 4 = 2π r3 h o Ts,o − T∞,o + εσ Ts,o − Tsur n ( r2 / r1 ) n ( r3 / r2 ) + k st k ins 2π (848 − 323 ) K ( n 0.18 / 0.15 ) 35 W / m ⋅ K + ( n r3 / 0.18 ) = 2π r3 6 W / m 2 ⋅ K ( 323 − 300 ) K + 0.20 × 5.67 × 10−8 W / m 2 ⋅ K 4 ( 4 323 − 300 4 ) K 4 0.10 W / m ⋅ K A trial-and-error solution yields r3 = 0.394 m = 394 mm, in which case the insulation thickness is t ins = r3 − r2 = 214 mm < The heat rate is then q′ = 2π (848 − 323) K = 420 W / m n ( 0.18 / 0.15 ) n (0.394 / 0.18 ) + 35 W / m ⋅ K 0.10 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 , C 240 200 160 120 80 40 0 .2 0 .2 6 0 .3 2 0 .3 8 0 .4 4 0 .5 O u te r ra d iu s o f in s u la tio n , m Ts ,o H e a t ra te s , W /m 2500 2000 1500 1000 500 0 0 .2 0 .2 6 0 .3 2 0 .3 8 0 .4 4 0 .5 O u te r ra d iu s o f in s u la tio n , m To ta l h e a t ra te C o n ve ctio n h e a t ra te R a d ia tio n h e a t ra te Beyond r3 ≈ 0.40m, there are rapidly diminishing benefits associated with increasing the insulation thickness. COMMENTS: Note that the thermal resistance of the insulation is much larger than that for the tube wall. For the conditions of Part (a), the radiation coefficient is hr = 1.37 W/m, and the heat loss by radiation is less than 25% of that due to natural convection ( q ′ = 78 W / m, q ′ rad conv,o = 342 W / m ) . PROBLEM 3.41 KNOWN: Thin electrical heater fitted between two concentric cylinders, the outer surface of which experiences convection. FIND: (a) Electrical power required to maintain outer surface at a specified temperature, (b) Temperature at the center SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Steady-state conditions, (3) Heater element has negligible thickness, (4) Negligible contact resistance between cylinders and heater, (5) Constant properties, (6) No generation. ANALYSIS: (a) Perform an energy balance on the composite system to determine the power required to maintain T(r2) = Ts = 5°C. E′ − E′ + E gen = Est in out + q′ − q′ elec conv = 0. Using Newton’s law of cooling, q′ elec = q′ conv = h ⋅ 2π r2 ( Ts − T∞ ) q′ elec = 50 W m2 ⋅ K × 2π ( 0.040m ) 5 − ( −15 ) C=251 W/m. < (b) From a control volume about Cylinder A, we recognize that the cylinder must be isothermal, that is, T(0) = T(r1). Represent Cylinder B by a thermal circuit: q′= T ( r1 ) − Ts R′ B For the cylinder, from Eq. 3.28, R ′ = ln r2 / r1 / 2π k B B giving T ( r1 ) = Ts + q′R ′ = 5 C+253.1 B W ln 40/20 = 23.5 C π × 1.5 W/m ⋅ K m2 Hence, T(0) = T(r1) = 23.5°C. Note that kA has no influence on the temperature T(0). < PROBLEM 3.42 KNOWN: 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) Negligible radial temperature gradients in wire, (6) Large surroundings. ANALYSIS: (a) The rates of energy generation per unit length and volume are, respectively, 2 E′ = I 2 R ′ g elec = ( 20 A ) (0.01Ω / m ) = 4 W / m < 2 g q = E′ / Ac = 4 E′ / π D2 = 16 W / m / π ( 0.002m ) = 1.27 × 106 W / m3 g < (b) Without the insulation, an energy balance at the surface of the wire yields 4 4 E′ = q′ = q ′ g conv + q′ = π D h ( T − T∞ ) + π D ε wσ T − Tsur rad ) ( where h = 1.25 [( T − T∞ ) / D ]1/ 4 . Substituting, 4 W / m = 1.25π ( 0.002m ) 3/ 4 (T − 293 )5 / 4 + π (0.002m ) 0.3 × 5.67 × 10−8 W / m 2 ⋅ K 4 (T 4 ) − 2934 K 4 and a trial-and-error solution yields < T = 331K = 58°C (c) Performing an energy balance at the outer surface, ( ) ( 4 4 E′ = q′ = q ′ g conv + q′ = π D h Ts,2 − T∞ + π D ε iσ Ts,2 − Tsur rad 4 W / m = 1.25π ( 0.006m ) 3/ 4 ) 4 (Ts,2 − 293)5 / 4 + π (0.006m ) 0.9 × 5.67 × 10−8 W / m 2 ⋅ K 4 (Ts,2 − 2934 ) K 4 and an iterative solution yields the following value of the surface temperature < Ts,2 = 307.8 K = 34.8°C The inner surface temperature may then be obtained from the following expression for heat transfer by conduction in the insulation. Continued ….. PROBLEM 3.42 (Cont.) q′ = Ts,i − T2 R′ cond 4W = = Ts,i − Ts,2 n ( r2 / r1 ) / 2π k i ( 2π ( 0.25 W / m ⋅ K ) Ts,i − 307.8 K ) n 3 < Ts,i = 310.6 K = 37.6°C As shown below, the effect of increasing the insulation thickness is to reduce, not increase, the surface temperatures. S u rfa ce te m p e ra tu re s , C 50 45 40 35 30 0 1 2 3 4 In s u la tio n th ickn e s s , m m In n e r s u rfa ce te m p e ra tu re , C O u te r s u rfa ce te m p e ra tu re , C This behavior is due to a reduction in the total resistance to heat transfer with increasing r2. Although ( ) 2 2 the convection, h, and radiation, h r = εσ (Ts,2 + Tsur ) Ts,2 + Tsur , coefficients decrease with increasing r2, the corresponding increase in the surface area is more than sufficient to provide for a reduction in the total resistance. Even for an insulation thickness of t = 4 mm, h = h + hr = (7.1 + 5.4) 2 2 2 W/m ⋅K = 12.5 W/m ⋅K, and rcr = k/h = 0.25 W/m⋅K/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 3.43 KNOWN: 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 is independent of the contact resistance. From the thermal circuit, we may write E′ = q′ = g Tin,i − T∞ R′ cond + R ′ conv = Tin,i − T∞ ( ) ( n rin,o / rin,i / 2π k + 1/ 2π rin,o h ) where rin,i = D / 2 = 0.001m, rin,o = rin,i + t = 0.003m, and Tin,i = Tmax = 50°C yields the maximum allowable power dissipation. Hence, (50 − 20 ) °C E′ g,max = n 3 2π × 0.13 W / m ⋅ K + = 1 30°C (1.35 + 5.31) m ⋅ K / W = 4.51 W / m < 2π ( 0.003m )10 W / m ⋅ K 2 The critical insulation radius is also unaffected by the contact resistance and is given by rcr = k 0.13 W / m ⋅ K = = 0.013m = 13mm h 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 = 12 g mm). COMMENTS: The contact resistance affects the temperature of the wire, and for q ′ = E ′ ,max g = 4.51 W / m, the outer surface temperature of the wire is Tw,o = Tin,i + q ′ R ′ = 50°C + ( 4.51 W / m ) t,c (3 × 10 −4 ) m ⋅ K / W / π ( 0.002m ) = 50.2°C. Hence, the temperature change across the contact 2 resistance is negligible. PROBLEM 3.44 KNOWN: Long rod experiencing uniform volumetric generation of thermal energy, q, concentric with a hollow ceramic cylinder creating an enclosure filled with air. Thermal resistance per unit length due to radiation exchange between enclosure surfaces is R ′ . The free convection rad 2 coefficient 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 the rod, 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 the hollow cylinder, (3) The enclosure surfaces experience free convection and radiation exchange. ANALYSIS: (a) The thermal circuit is shown below. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. Enclosure, radiation exchange (given): R ′ = 0.30 m ⋅ K / W rad Enclosure, free convection: 1 1 = = 0.80 m ⋅ K / W 2 ⋅ K × π × 0.020m hπ Dr 20 W / m 1 1 R′ = = 0.40 m ⋅ K / W cv,cer = 2 ⋅ K × π × 0.040m hπ Di 20 W / m R′ cv,rod = Ceramic cylinder, conduction: R′ = cd n ( Do / Di ) n (0.120 / 0.040 ) = = 0.10 m ⋅ K / W 2π k 2π ×1.75 W / m ⋅ K The thermal resistance between the enclosure surfaces (r-i) due to convection and radiation exchange is 1 1 1 = + R′ enc R ′ rad R ′ cv,rod + R ′ cv,cer 1 1 R′ = + enc 0.30 0.80 + 0.40 −1 m ⋅ K / W = 0.24 m ⋅ K / W The total resistance between the rod surface (r) and the outer surface of the cylinder (o) is R ′tot = R ′ + R ′ = ( 0.24 + 0.1) m ⋅ K / W = 0.34 m ⋅ K / W enc cd Continued ….. PROBLEM 3.44 (Cont.) (b) From an energy balance on the rod (see schematic) find Tr. ′ Ein − E′ + E′ = 0 out gen −q + q∀ = 0 ( ) − (Tr − Ti ) / R ′tot + q π D 2 / 4 = 0 r ( ) − (Tr − 25 ) K / 0.34 m ⋅ K / W + 2 × 106 W / m3 π × 0.020m 2 / 4 = 0 Tr = 239°C < COMMENTS: In evaluating the convection resistance of the air space, it was necessary to define an average air temperature (T∞) and consider the convection coefficients for each of the space surfaces. As you’ll learn later in Chapter 9, correlations are available for directly estimating the convection coefficient (henc) for the enclosure so that qcv = henc (Tr – T1). PROBLEM 3.45 KNOWN: 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/m2⋅K and TW = 20°C. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible convection resistance for refrigerant flow T∞,i = Ts,1 , (3) Negligible tube wall conduction resistance, (4) Negligible radiation exchange at outer surface. ( ) ANALYSIS: (a) The cooling capacity in the defrosted condition (δ = 0) corresponds to the rate of heat extraction from the airflow. Hence, ( ) q′ = h2π r1 T∞,o − Ts,1 = 100 W m 2 ⋅ K ( 2π × 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 extraction rate is T∞,o − Ts,1 T∞,o − Ts,1 q′ = = R′ conv + R ′ cond 1 ( h2π r2 ) + ln ( r2 r1 ) 2π k For 5 ≤ r2 ≤ 9 mm and k = 0.4 W/m⋅K, this expression yields Thermal resistance, Rt(m.K/W) Heat extraction, qprime(W/m) 50 45 40 0.4 0.3 0.2 0.1 0 35 0 0.001 0.002 0.003 Frost layer thickness, delta(m) Heat extraction, qprime(W/m) 0.004 0 0.001 0.002 0.003 0.004 Frost 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 frost layer thickness due to an increase in the total resistance to heat transfer. Although the convection resistance decreases with increasing δ, the reduction is exceeded by the increase in the conduction resistance. (c) The time tm required to melt a 2 mm thick frost layer may be determined by applying an energy balance, Eq. 1.11b, over the differential time interval dt and to a differential control volume extending inward from the surface of the layer. E in dt = dEst = dU lat ( ) h ( 2π rL ) T∞,o − Tf dt = −h sf ρ d∀ = −h sf ρ ( 2π rL ) dr ( h T∞,o − Tf tm = 1 ) ∫0t m dt = − ρ hsf ∫rr2 dr ρ 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/m⋅K/100 W/m2⋅K = 0.004 m, in which case any frost formation will reduce the performance of the coil. PROBLEM 3.46 KNOWN: 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 inner heat flows and conditions for which ratio is minimized. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant properties, (3) Isothermal heater, (4) Negligible contact resistance(s). ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are as shown in the schematic. (b) Performing an energy balance for the heater, Ein = E out , it follows that q′′ ( 2π r2 ) = q′ + q′ = h i o Th − T∞,i −1 ( h i 2π r1 ) ln ( r2 r1 ) + 2π k B + Th − T∞,o −1 (h o 2π r3 ) ln ( r3 r2 ) + 2π k A < (c) From the circuit, ln ( r2 r1 ) q′ 2π k B o= × ln ( r r ) q′ Th − T∞,i i ( h o 2π r3 )−1 + 3 2 2π k A ( ( Th − T∞,o ) ) ( h i 2π r1 )−1 + To reduce q′ q′ , one could increase kB, hi, and r3/r2, while reducing kA, ho and r2/r1. oi COMMENTS: Contact resistances between the heater and materials A and B could be important. < PROBLEM 3.47 KNOWN: Electric current flow, resistance, diameter and environmental conditions associated with a cable. FIND: (a) Surface temperature of bare cable, (b) Cable surface and insulation temperatures for a thin coating of insulation, (c) Insulation thickness which provides the lowest value of the maximum 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 rate of heat generation in the cable. Performing an energy balance for a control surface about the cable, it follows that E = q or, for the bare cable, I2 R ′ L=h (π D L )(T − T ). With g q′=I2 R ′ = ( 700A ) e 2 Ts = T∞ + e (6 ×10−4 Ω / m) = 294 W/m, it follows that i s ∞ q′ 294 W/m = 30 C+ hπ D i 25 W/m 2 ⋅ K π (0.005m ) ( ) Ts = 778.7 C. < (b) With a thin coating of insulation, there exist contact and convection resistances to heat transfer from the cable. The heat transfer rate is determined by heating within the cable, however, and therefore remains the same. Ts − T∞ Ts − T∞ q= = 1 R ′′ 1 t,c R t,c + + hπ Di L π Di L hπ Di L π Di (Ts − T∞ ) q′= R ′′ + 1/ h t,c and solving for the surface temperature, find q′ Ts = π Di 1 294 W/m m2 ⋅ K m2 ⋅ K ′′ R t,c + h + T∞ = π ( 0.005m ) 0.02 W + 0.04 W + 30 C Ts = 1153 C. < Continued ….. PROBLEM 3.47 (Cont.) The insulation temperature is then obtained from T −T q= s i R t,c or Ti = Ts − qR t,c = 1153 C − q R ′′ t,c π Di L = 1153 C − 294 W m2 ⋅ K × 0.02 m W π (0.005m ) Ti = 778.7 C. < (c) The maximum insulation temperature could be reduced by reducing the resistance to heat transfer from 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 ⋅ K Hence, Dcr = 0.04m > Di = 0.005m. To minimize the maximum temperature, which exists at the inner surface of the insulation, add insulation in the amount D − Di Dcr − Di (0.04 − 0.005 ) m t= o = = 2 2 2 < t = 0.0175m. The cable surface temperature may then be obtained from q′= R ′′ t,c π Di + Ts − T∞ ln ( Dcr / Di ) 2π k + 1 hπ Dcr = Ts − 30 C ln ( 0.04/0.005 ) 0.02 m 2 ⋅ K/W + + 2π ( 0.5 W/m ⋅ K ) π ( 0.005m ) 1 25 W 2 m ⋅K π ( 0.04m ) Hence, Ts − 30 C Ts − 30 C W 294 = = m (1.27+0.66+0.32 ) m ⋅ K/W 2.25 m ⋅ K/W Ts = 692.5 C Recognizing that q = (Ts - Ti)/Rt,c, find Ti = Ts − qR t,c = Ts − q Ti = 318.2 C. R ′′ t,c π Di L = 692.5 C − 294 W m2 ⋅ K × 0.02 m W π (0.005m ) < COMMENTS: Use of the critical insulation thickness in lieu of a thin coating has the effect of reducing the maximum insulation temperature from 778.7°C to 318.2°C. Use of the critical insulation thickness also reduces the cable surface temperature to 692.5°C from 778.7°C with no insulation or from 1153°C with a thin coating. PROBLEM 3.48 KNOWN: 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 back period for insulation. SCHEMATIC: Steam Costs: 9 $4 for 10 J Insulation Cost: $100 per meter Operation time: 7500 h/yr ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Negligible pipe wall resistance, (5) Negligible steam side convection resistance (pipe inner surface temperature is equal to steam temperature), (6) Negligible contact 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/m⋅K. ANALYSIS: (a) Without the insulation, the heat loss may be expressed in terms of radiation and convection rates, ( ) 4 4 q′=επ Dσ Ts − Tsur + h (π D )( Ts − T∞ ) W q′=0.8π ( 0.2m ) 5.67 × 10−8 4864 − 2984 K 4 2 K4 m⋅ W +20 (π × 0.2m ) ( 486-298) K m2 ⋅ K ( ) q′= (1365+2362 ) W/m=3727 W/m. < With the insulation, the thermal circuit is of the form Continued ….. PROBLEM 3.48 (Cont.) From an energy balance at the outer surface of the insulation, q′ cond = q′ conv + q′ rad Ts,i − Ts,o 4 4 = hπ Do Ts,o − T∞ + εσπ Do Ts,o − Tsur ln ( Do / Di ) / 2π k 486 − Ts,o K W π (0.3m ) Ts,o − 298K = 20 2 ⋅K ln ( 0.3m/0.2m ) m 2π ( 0.058 W/m ⋅ K ) W 4 π (0.3m ) Ts,o − 2984 K 4 . +0.8 × 5.67 ×10-8 2 ⋅ K4 m ( ( ) ) ( ) ( ) ( ) By trial and error, we obtain Ts,o ≈ 305K in which case q′= ( 486-305) K = 163 W/m. ln ( 0.3m/0.2m ) 2π (0.055 W/m ⋅ K ) < (b) The yearly energy savings per unit length of pipe due to use of the insulation is Savings Energy Savings Cost = × Yr ⋅ m Yr. Energy Savings J s h $4 = (3727 − 163) × 3600 × 7500 × Yr ⋅ m s⋅m h Yr 109 J Savings = $385 / Yr ⋅ m. Yr ⋅ m The pay back period is then Pay Back Period = Insulation Costs $100 / m = Savings/Yr. ⋅ m $385/Yr ⋅ m Pay Back Period = 0.26 Yr = 3.1 mo. < COMMENTS: Such a low pay back period is more than sufficient to justify investing in the insulation. PROBLEM 3.49 KNOWN: Temperature and convection coefficient associated with steam flow through a pipe of 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/m⋅K. ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outer surface that T∞,i − Ts,o Ts,o − T∞,o Ts,o − Tsur = + R conv,i + R cond R conv,o R rad or from Eqs. 3.9, 3.28 and 1.7, (1/π T∞,i − Ts,o Di hi ) + ln ( Do / Di ) / 2π k 523K − Ts,o = Ts,o − T∞,o (1/π Do h o ) ( 4 4 + επ Doσ Ts,o − Tsur = ) Ts,o − 293K −1 ln (75/60 ) π × 0.075m × 25 W/m 2 ⋅ K 2π × 56.5 W/m ⋅ K 4 +0.8π × ( 0.075m ) × 5.67 × 10−8 W/m 2 ⋅ K 4 Ts,o − 2934 K 4 523 − Ts,o Ts,o − 293 −8 T 4 − 2934 . = + 1.07 × 10 s,o 0.0106+0.0006 0.170 ( π × 0.6m × 500 W/m 2 ⋅ K ) −1 + ( ) From a trial-and-error solution, Ts,o ≈ 502K. Hence the heat loss is ( ) ( 4 4 q′=π Do h o Ts,o − T∞,o + επ Doσ Ts,o − Tsur q′=π ( 0.075m ) 25 W/m 2 ⋅ K (502-293) + 0.8 π ( 0.075m ) 5.67 × 10−8 ) 5024 − 2434 K 4 m ⋅K W 2 4 q′=1231 W/m+600 W/m=1831 W/m. COMMENTS: The thermal resistance between the outer surface and the surroundings is much larger than that between the outer surface and the steam. < PROBLEM 3.50 KNOWN: Temperature and convection coefficient associated with steam flow through a pipe of prescribed inner and outer radii. Emissivity of outer surface magnesia insulation, and convection coefficient. Temperature of ambient air and surroundings. FIND: Heat loss per unit length q ′ and outer surface temperature Ts,o as a function of insulation thickness. Recommended insulation thickness. Corresponding annual savings and temperature distribution. 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): ks = 56.5 W/m⋅K. Table A-3, Magnesia, 85% (T ≈ 365 K): km = 0.055 W/m⋅K. ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outer surface that T∞,i − Ts,o Ts,o − T∞,o Ts,o − Tsur = + R′ R′ R′ conv,i + R ′ cond,s + R ′ cond,m conv,o rad or from Eqs. 3.9, 3.28 and 1.7, (1 2π r1h i ) + ln ( r2 T∞ ,i − Ts,o r1 ) 2π k s + ln ( r3 r2 ) 2π k m = Ts,o − T∞ ,o Ts,o − Tsur + (1 2π r3h o ) ( 2π r3 )εσ (Ts,o + Tsur ) (T 2 2 s,o + Tsur ) −1 This expression may be solved for Ts,o as a function of r3, and the heat loss may then be determined by evaluating either the left-or right-hand side of the energy balance equation. The results are plotted as follows. Continued... PROBLEM 3.50 (Cont.) 2000 Thermal resistance, Rprime(K/m.W) 2 Heat loss, qprime(W/m) 1600 1200 800 400 1.5 1 0.5 0 0.035 0 0.035 0.045 0.055 0.065 0.075 0.045 0.055 0.065 0.075 Outer radius of insulation, r3(m) Outer radius of insulation, r3(m) Insulation conduction resistance, Rcond,m Outer convection resistance, Rconv,o Radiation resistance, Rrad q1 The rapid decay in q′ with increasing r3 is attributable to the dominant contribution which the insulation begins to make to the total thermal resistance. The inside convection and tube wall conduction resistances are fixed at 0.0106 m⋅K/W and 6.29×10-4 m⋅K/W, respectively, while the resistance of the insulation increases to approximately 2 m⋅K/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 insulation thickness is increased to r3 = 0.0775 m. Hence, an insulation thickness of (r3 - r2) = 0.020 m is recommended, for which q′ = 172 W/m. The corresponding annual savings (AS) in energy costs is therefore $4 h s × 7000 × 3600 = $167 / m AS = [(1830 − 172 ) W m ] y h 109 J < The corresponding temperature distribution is Local temperature, T(K) 500 460 420 380 340 300 0.038 0.042 0.046 0.05 0.054 0.058 Radial location in insulation, r(m) Tr The 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 are substantial, as is the reduction in the outer surface temperature (from Ts,o ≈ 502 K for r3 = r2, to 309 K for r3 = 0.0575 m). 2. The increase in R ′ to a maximum value of 0.63 m⋅K/W at r3 = 0.0455 m and the subsequent decay rad is due to the competing effects of hrad and A′ = (1 2π r3 ) . Because the initial decay in T3 = Ts,o with 3 increasing r3, and hence, the reduction in hrad, is more pronounced than the increase in A′ , R ′ 3 rad increases with r3. However, as the decay in Ts,o, and hence hrad, becomes less pronounced, the increase in A′ becomes more pronounced and R ′ decreases with increasing r3. rad 3 PROBLEM 3.51 KNOWN: Pipe wall temperature and convection conditions associated with water flow through the pipe and ice layer formation on the inner surface. FIND: Ice layer thickness δ. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Negligible pipe wall thermal resistance, (3) negligible ice/wall contact resistance, (4) Constant k. PROPERTIES: Table A.3, Ice (T = 265 K): k ≈ 1.94 W/m⋅K. ANALYSIS: Performing an energy balance for a control surface about the ice/water interface, it follows that, for a unit length of pipe, q′ conv = q′ cond ( ) h i ( 2π r1 ) T∞,i − Ts,i = Ts,i − Ts,o ln ( r2 r1 ) 2π k Dividing both sides of the equation by r2, ln ( r2 r1 ) ( r2 r1 ) = Ts,i − Ts,o k 1.94 W m ⋅ K 15 C × = × = 0.097 2 ⋅ K 0.05 m h i r2 T∞,i − Ts,i 3 C 2000 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 layer thickness is δ = r2 − r1 = 0.005 m = 5 mm < COMMENTS: With no flow, hi → 0, in which case r1 → 0 and complete blockage could occur. The pipe should be insulated. PROBLEM 3.52 KNOWN: Inner surface temperature of insulation blanket comprised of two semi-cylindrical shells of different materials. 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 contact resistance between materials, (4) Constant properties. ANALYSIS: (a) The thermal circuit is, R′ conv,A = R ′ conv,B = 1 / π r2 h R′ cond ( A ) = R′ cond ( B ) = ln ( r2 / r1 ) < π kA ln ( r2 / ri ) π kB The conduction resistances follow from Section 3.3.1 and Eq. 3.28. Each resistance is larger by a factor of 2 than the result of Eq. 3.28 due to the reduced area. (b) Evaluating the thermal resistances and the heat rate ( q′=q′ + q′ ) , A B ( 2 R′ conv = π × 0.1m × 25 W/m ⋅ K R′ cond ( A ) = q′= q′= ln ( 0.1m/0.05m ) π × 2 W/m ⋅ K Ts,1 − T∞ R′ conv cond ( A ) + R ′ + ) −1 = 0.1273 m ⋅ K/W = 0.1103 m ⋅ K/W R′ cond ( B ) = 8 R ′ cond ( A ) = 0.8825 m ⋅ K/W Ts,1 − T∞ R′ conv cond ( 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 are W m⋅K × 0.1103 = 407K m W W m⋅K Ts,2( B) = Ts,1 − q′ R ′ = 325K. B cond ( B ) = 500K − 198 × 0.8825 m W Ts,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 , −1 −1 −1 ′ ′ ′ ′ 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 3.53 KNOWN: Surface temperature of a circular rod coated with bakelite and adjoining fluid conditions. FIND: (a) Critical insulation radius, (b) Heat transfer per unit length for bare rod and for insulation 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/m⋅K. ANALYSIS: (a) From Example 3.4, the critical radius is k 1.4 W/m ⋅ K rcr = = = 0.01m. h 140 W/m 2 ⋅ K < (b) For the bare rod, q′=h (π Di ) ( Ti − T∞ ) q′=140 W m2 ⋅ K (π × 0.01m ) ( 200 − 25) C=770 W/m < For the critical insulation thickness, ( 200 − 25) C Ti − T∞ q′= = ln ( rcr / ri ) ln (0.01m/0.005m ) 1 1 + + 2π rcr h 2π k 2π × 1.4 W/m ⋅ K 2π × (0.01m ) × 140 W/m 2 ⋅ K q′= 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) C Ti − T∞ W q′= = = 577 ln ( r/ri ) ln ( r/0.005m ) m 1 1 + + 2 ⋅ K 2π × 1.4 W/m ⋅ K 2π rh 2π k 2π ( r )140 W/m From a trial-and-error solution, find r ≈ 0.06 m. The desired insulation thickness is then δ = ( r − ri ) ≈ (0.06 − 0.005 ) m=55 mm. < PROBLEM 3.54 KNOWN: Geometry of an oil storage tank. Temperature of stored oil and environmental conditions. FIND: Heater power required to maintain a prescribed inner surface temperature. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial direction, (3) Constant properties, (4) Negligible radiation. PROPERTIES: Table A-3, Pyrex (300K): k = 1.4 W/m⋅K. ANALYSIS: The rate at which heat must be supplied is equal to the loss through the cylindrical and hemispherical sections. Hence, q=qcyl + 2q hemi = qcyl + qspher or, from Eqs. 3.28 and 3.36, q= q= Ts,i − T∞ ln ( Do / Di ) 1 + 2π Lk π Do Lh + Ts,i − T∞ 1 1 1 1 D − D + 2 2π k i o π Do h ( 400 − 300 ) K ln 1.04 1 + 2π ( 2m )1.4 W/m ⋅ K π (1.04m ) 2m 10 W/m 2 ⋅ K ( + ( 400 − 300 ) K ) 1 1 (1 − 0.962 ) m-1 + 2 2π (1.4 W/m ⋅ K ) π (1.04m ) 10 W/m 2 ⋅ K 100K 100K q= + -3 K/W + 15.30 × 10-3 K/W 4.32 ×10-3 K/W + 29.43 × 10-3 2.23 ×10 q = 5705W + 2963W = 8668W. < PROBLEM 3.55 KNOWN: Diameter of a spherical container used to store liquid oxygen and properties of insulating material. 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 of container wall and contact resistance between wall and insulation, (3) Container wall at boiling point of liquid oxygen. ANALYSIS: (a) Applying an energy balance to a control surface about the insulation, E in − E out = 0, it follows that q conv + q rad = q cond = q . Hence, T∞ − Ts,2 Tsur − Ts,2 Ts,2 − Ts,1 + = =q R t,conv R t,rad R t,cond ( 2 where R t,conv = 4π r2 h ) −1 ( 2 , R t,rad = 4π r2 h r ( (1) ) −1 1.9, the radiation coefficient is h r = εσ Ts,2 + Tsur , R t,cond = (1 4π k )[(1 r1 ) − (1 r2 )] , and, from Eq. 2 2 ) (Ts,2 + Tsur ) . With t = 10 mm (r2 = 260 mm), ε = 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 that the heat gain is qw ≈ 2.72 W Without the insulation, the heat gain is q wo = T∞ − Ts,1 Tsur − Ts,1 + R t,conv R t,rad where, with r2 = r1, Ts,1 = 90 K, Rt,conv = 0.127 K/W and Rt,rad = 3.14 K/W. Hence, qwo = 1702 W With 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 w m wo q − qw × 100% = wo × 100% q wo (1702 − 2.7 ) W 1702 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. 10000 0.01 Evaporation rate, mdot(kg/s) Heat gain, q(W) 1000 100 10 1 0.1 0.001 0.0001 1E-5 1E-6 0.25 0.26 0.27 0.28 0.29 Outer radius of insulation, r2(m) 0.3 0.25 0.26 0.27 0.28 0.29 0.3 Outer radius of insulation, r2(m) Because of its extremely low thermal conductivity, significant benefits are associated with using even a thin layer of insulation. Nearly three-order magnitude reductions in q and m are achieved with r2 = 0.26 -3 m. With increasing r2, q and m decrease from values of 1702 W and 8×10 kg/s at r2 = 0.25 m to 0.627 W and 2.9×10-6 kg/s at r2 = 0.30 m. COMMENTS: Laminated metallic-foil/glass-mat insulations are extremely effective and corresponding conduction resistances are typically much larger than those normally associated with surface convection and radiation. PROBLEM 3.56 KNOWN: Sphere of radius ri, covered with insulation whose outer surface is exposed to a convection 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 tot where R tot = R t,conv + R t,cond and R t,conv = 1 1 = hAs 4π hr 2 (3.9) R t,cond = 1 1 1 − 4π k ri r (3.36) If q is a maximum or minimum, we need to find the condition for which d R tot = 0. dr It follows that d 1 1 1 1 11 1 1 − + − =0 = + dr 4π k ri r 4π hr 2 4π k r 2 2π h r3 giving k h The second derivative, evaluated at r = rcr, is d dR tot 1 1 3 1 =− + dr dr 2π k r3 2π h r 4 r=r rcr = 2 cr = 1 − 1 ( 2k/h )3 2π k + 3 1 1 1 3 −1 + > 0 = 3 2π k 2π h 2k/h ( 2k/h ) 2 Hence, it follows no optimum Rtot exists. We refer to this condition as the critical insulation radius. See Example 3.4 which considers this situation for a cylindrical system. PROBLEM 3.57 KNOWN: Thickness of hollow aluminum sphere and insulation layer. Heat rate and inner surface 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 at outer surface. PROPERTIES: Table A-1, Aluminum (523K): k ≈ 230 W/m⋅K. ANALYSIS: From the thermal circuit, T −T T1 − T∞ q= 1 ∞ = 1/r1 − 1/ r2 1/ r2 − 1/ r3 1 R tot + + 2 4π k A1 4π k I h4π r3 q= ( 250 − 20 ) C 1/0.15 − 1/ 0.18 1/ 0.18 − 1/ 0.30 K 1 + + 2 4π k I 4π ( 230 ) 30 ( 4π )( 0.3) W = 80 W or 3.84 × 10−4 + 0.177 230 + 0.029 = = 2.875. kI 80 Solving for the unknown thermal conductivity, find kI = 0.062 W/m⋅K. COMMENTS: The dominant contribution to the total thermal resistance is made by the insulation. Hence uncertainties in knowledge of h or kA1 have a negligible effect on the accuracy of the kI measurement. < PROBLEM 3.58 KNOWN: Dimensions of spherical, stainless steel liquid oxygen (LOX) storage container. Boiling point 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 resistances associated with internal and external convection, conduction in the container wall, and contact between wall and insulation, (3) Negligible radiation at exterior surface, (4) Constant insulation thermal conductivity. PROPERTIES: Table A.1, 304 Stainless steel (T = 100 K): ks = 9.2 W/m⋅K; Table A.3, Reflective, aluminum foil-glass paper insulation (T = 150 K): ki = 0.000017 W/m⋅K. ANALYSIS: The heat gain associated with a loss of 1 kg/day is q = mh fg = 1kg day 86, 400 s day (2.13 ×105 J kg ) = 2.47 W ( ) With an overall temperature difference of T∞ − Tbp = 150 K, the corresponding total thermal resistance is ∆T 150 K R tot = = = 60.7 K W q 2.47 W Since the conduction resistance of the steel wall is R t,cond,s = 1 1 1 1 1 −3 − = 0.35 m − 0.40 m = 2.4 × 10 K W 4π k s r1 r2 4π (9.2 W m ⋅ K ) 1 it is clear that exclusive reliance must be placed on the insulation and that a special insulation of very low thermal conductivity should be selected. The best choice is a highly reflective foil/glass matted insulation which was developed for cryogenic applications. It follows that R t,cond,i = 60.7 K W = 1 1 1 1 1 1 − − = 4π k i r2 r3 4π (0.000017 W m ⋅ K ) 0.40 m r3 which yields r3 = 0.4021 m. The minimum insulation thickness is therefore δ = (r3 - r2) = 2.1 mm. COMMENTS: The heat loss could be reduced well below the maximum allowable by adding more insulation. Also, in view of weight restrictions associated with launching space vehicles, consideration should be given to fabricating the LOX container from a lighter material. PROBLEM 3.59 KNOWN: Diameter and surface temperature of a spherical cryoprobe. Temperature of surrounding tissue 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 resistance between probe and frozen tissue, (3) Constant properties. ANALYSIS: Performing an energy balance for a control surface about the phase front, it follows that q conv − q cond = 0 Hence, ( )(T∞ − Ts,2 ) = [(1 r1Ts,2(1−rTs,1] 4π k )− 2 ) 2 h 4π r2 2 r2 [(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 ) = 4.56 It follows that r2 = 6.84 mm and the thickness of the frozen tissue is δ = r2 − r1 = 5.34 mm < PROBLEM 3.60 KNOWN: Inner diameter, wall thickness and thermal conductivity of spherical vessel containing heat generating medium. Inner surface temperature without insulation. Thickness and thermal conductivity of insulation. Ambient air temperature and convection coefficient. FIND: (a) Thermal energy generated within vessel, (b) Inner surface temperature of vessel with insulation. 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 the pharmaceuticals, E g = q, in which case, without the insulation Eg = q = Ts,1 − T∞ 1 1 1 r − r + 4π k w 1 2 4π r 2 h 2 1 Eg = q = ( = (50 − 25 ) °C 1 1 − 1 + 4π (17 W / m ⋅ K ) 0.50m 0.51m 4π ( 0.51m )2 6 W / m 2 ⋅ K 25°C 1 ) 1.84 × 10−4 + 5.10 × 10−2 K / W = 489 W < (b) With the insulation, 1 1 1 1 1 1 1 Ts,1 = T∞ + q − + − + 4π k w r1 r2 4π k i r2 r3 4π r 2 h 3 K 1 1 1 1 1.84 × 10−4 + Ts,1 = 25°C + 489 W − + 4π ( 0.04 ) 0.51 0.53 4π ( 0.53)2 6 W K Ts,1 = 25°C + 489 W 1.84 × 10−4 + 0.147 + 0.047 W = 120°C < COMMENTS: The thermal resistance associated with the vessel wall is negligible, and without the insulation the dominant resistance is due to convection. The thermal resistance of the insulation is approximately three times that due to convection. PROBLEM 3.61 KNOWN: Spherical tank of 1-m diameter containing an exothermic reaction and is at 200°C when 2 the ambient air is at 25°C. Convection coefficient on outer surface is 20 W/m ⋅K. FIND: Determine the thickness of urethane foam required to reduce the exterior temperature to 40°C. Determine the percentage reduction in the heat rate achieved using the insulation. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial (spherical) conduction through 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 ambient surroundings. PROPERTIES: Table A-3, urethane, rigid foam (300 K): k = 0.026 W/m⋅K. ANALYSIS: (a) The heat transfer situation for the heat rate from the tank can be represented by the thermal circuit shown above. The heat rate from the tank is T − T∞ q= t R cd + R cv where the thermal resistances associated with conduction within the insulation (Eq. 3.35) and convection for the exterior surface, respectively, are R cd = (1/ rt − 1/ ro ) = (1/ 0.5 − 1/ ro ) 4π k = (1/ 0.5 − 1/ ro ) K / W 4π × 0.026 W / m ⋅ K 0.3267 1 1 1 − R cv = = = = 3.979 × 10−3 ro 2 K / W 2 4π × 20 W / m 2 ⋅ K × r 2 hAs 4π hro o To determine the required insulation thickness so that To = 40°C, perform an energy balance on the onode. Tt − To T∞ − To + =0 R cd R cv ( 200 − 40 ) K ro = 0.5135 m ( 25 − 40 ) K =0 2 3.979 ×10−3 ro K / W t = 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 kW Tt − T∞ = 1/ 4π hrt2 0.01592 K / W qins = ( 200 − 25 ) Tt − T∞ = = 0.994 kW R cd + R cv (0.161 + 0.01592 ) K / W Hence, the percentage reduction in heat loss achieved with the insulation is, qins − qo 0.994 − 10.99 ×100 = − ×100 = 91% qo 10.99 < PROBLEM 3.62 KNOWN: Dimensions and materials used for composite spherical shell. Heat generation associated with stored material. FIND: Inner surface temperature, T1, of lead (proposal is flawed if this temperature exceeds the melting point). SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties at 300K, (4) Negligible contact resistance. PROPERTIES: Table A-1, Lead: k = 35.3 W/m⋅K, MP = 601K; St.St.: 15.1 W/m⋅K. ANALYSIS: From the thermal circuit, it follows that T −T 4 3 q= 1 ∞ = q π r1 R tot 3 Evaluate the thermal resistances, 1 1 R Pb = 1/ ( 4π × 35.3 W/m ⋅ K ) 0.25m − 0.30m = 0.00150 K/W 1 1 R St.St. = 1/ ( 4π ×15.1 W/m ⋅ K ) 0.30m − 0.31m = 0.000567 K/W ) ( R conv = 1/ 4π × 0.312 m 2 × 500 W/m 2 ⋅ K = 0.00166 K/W R tot = 0.00372 K/W. The heat rate is q=5 × 105 W/m3 ( 4π / 3)( 0.25m ) = 32, 725 W. The inner surface 3 temperature is T1 = 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 thermal contact. A protective outer coating should be applied to prevent long term corrosion of the stainless steel. PROBLEM 3.63 KNOWN: Dimensions and materials of composite (lead and stainless steel) spherical shell used to store radioactive wastes with constant heat generation. Range of convection coefficients h available for cooling. FIND: (a) Variation of maximum lead temperature with h. Minimum allowable value of h to maintain maximum lead temperature at or below 500 K. (b) Effect of outer radius of stainless steel shell on maximum lead temperature for h = 300, 500 and 1000 W/m2⋅K. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties at 300 K, (4) Negligible contact resistance. PROPERTIES: Table A-1, Lead: k = 35.3 W/m⋅K, St. St.: 15.1 W/m⋅K. ANALYSIS: (a) From the schematic, the maximum lead temperature T1 corresponds to r = r1, and from the thermal circuit, it may be expressed as T1 = T∞ + R tot q 3 where q = q ( 4 3 )π r1 = 5 × 105 W m3 ( 4π 3)( 0.25 m ) = 32, 725 W . The total thermal resistance is 3 R tot = R cond,Pb + R cond,St.St + R conv where expressions for the component resistances are provided in the schematic. Using the Resistance Network model and Thermal Resistance tool pad of IHT, the following result is obtained for the variation of T1 with h. Maximum Pb Temperature, T(r1) (K) 700 600 500 400 300 100 200 300 400 500 600 700 800 900 1000 Convection coefficient, h(W/m^2.K) T_1 Continued... PROBLEM 3.63 (Cont.) To maintain T1 below 500 K, the convection coefficient must be maintained at < h ≥ 181 W/m2⋅K Maximum 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. 600 550 500 450 400 350 0.3 0.35 0.4 0.45 0.5 Outer radius of steel shell, r3(m) h = 300 W/m^2.K h = 500 W/m^2.K h = 1000 W/m^2.k For h = 300, 500 and 1000 W/m2⋅K, 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 the total thermal resistance is Rtot = (T1 - T∞)/q, or Rtot = (500 - 283)K/32,725 W = 0.00663 K/W. Hence, any increase in Rcond,St.St due to increasing r3 must be accompanied by an equivalent reduction in Rconv. PROBLEM 3.64 KNOWN: Representation of the eye with a contact lens as a composite spherical system subjected to convection processes at the boundaries. FIND: (a) Thermal circuits with and without contact lens in place, (b) Heat loss from anterior chamber for both cases, and (c) Implications of the heat loss calculations. SCHEMATIC: r1=10.2mm k1=0.35 W/m⋅K r2=12.7mm k2=0.80 W/m⋅K r3=16.5mm 2 T∞,i=37°C hi=12 W/m ⋅K T∞,o=21°C ho=6 W/m ⋅K 2 ASSUMPTIONS: (1) Steady-state conditions, (2) Eye is represented as 1/3 sphere, (3) Convection coefficient, 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 the thermal resistance from the above circuits. Without lens: R t,wo = 3 ( 12W/m 2 ⋅ K4π 10.2 × 10-3m 3 + 2 ( -3 6 W/m ⋅ K4π 12.7 × 10 m ) 2 R t,w = 191.2 K/W+13.2 K/W+ + 3 ( -3 6W/m ⋅ K4π 16.5 × 10 m ) 2 2 1 1 1 − m 4π × 0.35 W/m ⋅ K 10.2 12.7 10−3 3 = 191.2 K/W+13.2 K/W+246.7 K/W=451.1 K/W With lens: 2 ) + 1 1 1 12.7 − 16.5 −3 m 4π × 0.80 W/m ⋅ K 10 3 = 191.2 K/W+13.2 K/W+5.41 K/W+146.2 K/W=356.0 K/W Hence the heat loss rates from the anterior chamber are Without lens: q wo = (37 − 21) C/451.1 K/W=35.5mW With 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 contact lens is in place, implying that the outer radius, r3, is less than the critical radius. PROBLEM 3.65 KNOWN: Thermal conductivity and inner and outer radii of a hollow sphere subjected to a uniform heat flux at its outer surface and maintained at a uniform temperature on the inner surface. FIND: (a) Expression for radial temperature distribution, (b) Heat flux required to maintain prescribed 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 by integrating Fourier’s law, Eq. 3.33. That is, T q r r dr q1r dT or − r = −k ∫ = −k T − Ts,1 . Ts,1 4π ∫r1 r 2 4π r r1 Hence, q 1 1 T ( r ) = Ts,1 + r − 4π k r r1 ( ) 2 or, with q′′ ≡ q r / 4π r2 , 2 2 q′′ r2 1 1 2 T ( r ) = Ts,1 + − k r r1 < (b) Applying the above result at r2, q′′ = 2 ( k Ts,2 − Ts,1 2 1 1 r2 − 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 solving the appropriate form of the heat equation, d 2 dT r =0 dr dr and applying the boundary conditions T ( r1 ) = Ts,1 and − k dT = q′′ . 2 dr r 2 (2) The negative sign on q ′′ implies heat transfer in the negative r direction. 2 PROBLEM 3.66 KNOWN: Volumetric heat generation occurring within the cavity of a spherical shell of prescribed 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 is d 2 dT r =0 dr dr Integrate twice to obtain, dT r2 and = C1 dr C T = − 1 + C2 . r (1,2) The boundary conditions may be obtained from energy balances at the inner and outer surfaces. At the inner surface (ri), E = q 4/3π r 3 = q dT/dr) = −qr / 3k. = −k 4π r 2 dT/dr) g ( i ) cond,i ( i ) ri ri i (3) At the outer surface (ro), 2 2 q cond,o = − k4π ro dT/dr)ro = q conv = h4π 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 k 3kro 3 3 qr qr C2 = i − i + T∞ . 2 3hro 3ro k Hence, the temperature distribution is qri3 qri3 1 1 qri3 T= − + + T∞ . 2 3k r ro 3hro COMMENTS: Note that E g = q cond,i = q cond,o = q conv . < PROBLEM 3.67 KNOWN: Spherical tank of 3-m diameter containing LP gas at -60°C with 250 mm thickness of insulation having thermal conductivity of 0.06 W/m⋅K. Ambient air temperature and convection 2 coefficient on the outer surface are 20°C and 6 W/m ⋅K, respectively. FIND: (a) Determine the radial position in the insulation at which the temperature is 0°C and (b) If the insulation is pervious to moisture, what conclusions can be reached about ice formation? What effect 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) conduction through the insulation, and (3) Negligible radiation exchange between the insulation outer surface and the ambient surroundings. ANALYSIS: (a) The heat transfer situation can be represented by the thermal circuit shown above. The heat gain to the tank is q= 20 − ( −60 ) K T∞ − Tt = = 612.4 W R ins + R cv 0.1263 + 4.33 × 10−3 K / W ( ) where the thermal resistances for the insulation (see Table 3.3) and the convection process on the outer surface are, respectively, −1 1/ r − 1/ ro (1/1.50 − 1/1.75 ) m R ins = i = = 0.1263 K / W 4π k 4π × 0.06 W / m ⋅ K 1 1 1 R cv = = = = 4.33 ×10−3 K / W 2 hAs h4π ro 6 W / m 2 ⋅ K × 4π (1.75 m )2 To determine the location within the insulation where Too (roo) = 0°C, use the conduction rate equation, Eq. 3.35, −1 1 4π k (Too − Tt ) 4π k (Too − Tt ) q= roo = − q (1/ ri − 1/ roo ) ri and substituting numerical values, find −1 1 4π × 0.06 W / m ⋅ K (0 − ( −60 )) K roo = − 612.4 W 1.5 m = 1.687 m < (b) With roo = 1.687 m, we’d expect the region of the insulation ri ≤ r ≤ roo to be filled with ice formations if the insulation is pervious to water vapor. The effect of the ice formation is to substantially increase the heat gain since kice is nearly twice that of kins, and the ice region is of thickness (1.687 – 1.50)m = 187 mm. To avoid ice formation, a vapor barrier should be installed at a radius larger than roo. PROBLEM 3.68 KNOWN: Radius and heat dissipation of a hemispherical source embedded in a substrate of prescribed 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-infinite medium is equivalent to spherical source in infinite medium (with q = 8 W) and heat transfer is one-dimensional in the radial direction, (2) Steady-state conditions, (3) Constant properties, (4) No generation. ANALYSIS: Heat equation reduces to 1 d 2 dT r r 2dT/dr=C1 =0 2 dr dr r T ( r ) = −C1 / r+C2 . Boundary conditions: T ( ∞ ) = T∞ T ( ro ) = Ts Hence, C2 = T∞ and Ts = −C1 / ro + T∞ and C1 = ro ( T∞ − Ts ) . The temperature distribution has the form T ( r ) = T∞ + ( Ts − T∞ ) ro / r < and the heat rate is q=-kAdT/dr = − k2π r 2 − ( Ts − T∞ ) ro / r 2 = k2π ro ( Ts − T∞ ) It follows that Ts − T∞ = q 4W = = 50.9 C -4 m k2π ro 125 W/m ⋅ K 2π 10 ( ) Ts = 77.9 C. COMMENTS: For the semi-infinite (or infinite) medium approximation to be valid, the substrate dimensions must be much larger than those of the transistor. < PROBLEM 3.69 KNOWN: Critical and normal tissue temperatures. Radius of spherical heat source and radius of tissue to be maintained above the critical temperature. Tissue thermal conductivity. FIND: General expression for radial temperature distribution in tissue. Heat rate required to maintain prescribed thermal conditions. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant k. ANALYSIS: The appropriate form of the heat equation is 1 d dT r =0 r 2 dr dr Integrating twice, dT C1 = dr r 2 C T ( r ) = − 1 + C2 r ) ( 2 2 2 Since T → Tb as r → ∞, C2 = Tb. At r = ro, q = − k 4π ro dT dr = −4π kro C1 ro = -4πkC1. ro Hence, C1 = -q/4πk and the temperature distribution is T (r ) = q + Tb 4π kr < It follows that q = 4π kr T ( r ) − Tb Applying this result at r = rc, q = 4π ( 0.5 W m ⋅ K )( 0.005 m )( 42 − 37 ) C = 0.157 W < COMMENTS: At ro = 0.0005 m, T(ro) = q ( 4π kro ) + Tb = 92°C. Proximity of this temperature to the boiling point of water suggests the need to operate at a lower power dissipation level. PROBLEM 3.70 KNOWN: 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 is 1 d dT q r + = 0 r dr dr k The general solution is T (r ) = − q2 r + C1 ln r + C2 4k Applying the boundary conditions, it follows that T ( r1 ) = Ts,1 = − q2 r1 + C1 ln r1 + C2 4k T ( r2 ) = Ts,2 = − q2 r2 + C1 ln r2 + C2 4k which may be solved for ( ) ( ) 2 2 C1 = ( q/4k ) r2 − r1 + Ts,2 − Ts,1 ln ( r2 /r1 ) 2 C2 = Ts,2 + ( q 4k ) r2 − C1 ln r2 Hence, ln ( r/r2 ) 2 2 2 T ( 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 is q′′ ( r ) = q 2 ( ) ( ) 2 2 k ( 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′′ (2πrL), the heat rate distribution is q ( r ) = π Lqr − 2 ) ( ( ) 2 2 2π Lk ( q 4k ) r2 − r1 + Ts,2 − Ts,1 < ln ( r2 /r1 ) (b) For the spherical shell, the heat equation and general solution are 1 d 2 dT q r + = 0 2 dr dr k r T(r) = − ( q 6k ) r 2 − C1/r + C2 Applying the boundary conditions, it follows that 2 T ( r1 ) = Ts,1 = − ( q 6k ) r1 − C1/r1 + C2 2 T ( r2 ) = Ts,2 = − ( q 6k ) r2 − C1/r2 + C2 Hence, ( ) ( ) [(1 r1 ) − (1 r2 )] 2 2 C1 = ( q 6k ) r2 − r1 + Ts,2 − Ts,1 2 C2 = Ts,2 + ( q 6k ) r2 + C1/r2 and (1 r ) − (1 r2 ) 2 2 2 T ( r ) = Ts,2 + ( q 6k ) r2 − r 2 − ( q 6k ) r2 − r1 + Ts,2 − Ts,1 (1 r ) − (1 r ) 1 2 ) ( ( ) ( ) < With q′′ (r) = - k dT/dr, the heat flux distribution is q′′ ( r ) = q 3 ( ( ) ( q 6 ) r 2 − r 2 + k ( T − T ) 21 s,2 s,1 1 r− (1 r1 ) − (1 r2 ) < r2 ) and, with q = q′′ 4π r 2 , the heat rate distribution is 4π q 3 q (r ) = r− 3 ( ) ( ) 2 2 4π (q 6 ) r2 − r1 + k Ts,2 − Ts,1 (1 r1 ) − (1 r2 ) < PROBLEM 3.71 KNOWN: Temperature distribution in a composite wall. FIND: (a) Relative magnitudes of interfacial heat fluxes, (b) Relative magnitudes of thermal conductivities, and (c) Heat flux as a function of distance x. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties. 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′′ > q′′ 3 4 However, the linear temperature distributions in A and B indicate no generation, in which case ′′ q′′ = q3 2 (b) Since conservation of energy requires that q′′ ,B = q′′ and dT/dx)B < dT/dx)C , it follows 3 3,C from Fourier’s law that k B > kC. Similarly, since q′′ ,A = q′′ and dT/dx)A > dT/dx) B , it follows that 2 2,B k 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 3.72 KNOWN: Plane wall with internal heat generation which is insulated at the inner surface and subjected 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 uniform volumetric 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 is the maximum temperature within the wall, To = qL2 / 2k+Ts . The outer surface temperature follows from Eq. 3.46, Ts = T∞ + qL/h W Ts = 92 C+0.3 × 106 × 0.1m/500W/m2 ⋅ K=92C+60C=152C. 3 m It follows that To = 0.3 × 106 W/m3 × ( 0.1m ) / 2 × 25W/m ⋅ K+152C 2 To = 60 C+152C=212C. < COMMENTS: The heat flux leaving the wall can be determined from knowledge of h, Ts and T∞ using Newton’s law of cooling. 2 2 q′′ conv = 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 the form Eg − Eout = 0 where Eg = qAL and Eout = q′′ conv ⋅ A. Hence, 6 3 2 q′′ conv = qL=0.3 × 10 W/m × 0.1m=30kW/m . PROBLEM 3.73 KNOWN: Composite wall with outer surfaces exposed to convection process. FIND: (a) Volumetric heat generation and thermal conductivity for material B required for special conditions, (b) Plot of temperature distribution, (c) T1 and T2, as well as temperature distributions corresponding to loss of coolant condition where h = 0 on surface A. SCHEMATIC: LA = 30 mm LB = 30 mm LC = 20 mm kA = 25 W/m⋅K kC = 50 W/m⋅K ASSUMPTIONS: (1) Steady-state, one-dimensional heat transfer, (2) Negligible contact resistance at interfaces, (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 = 0 2 ′′ q B = ( q1 + q ′′ ) 2L B . 2 To determine the heat fluxes, q1T and q TT , construct thermal circuits for A and C: T 2 ′′ q1 = ( T1 − T∞ ) (1 h + L A k A ) q ′′ = ( T2 − T∞ ) ( L C k C + 1 h ) 2 1 0.030 m + 1000 W m 2 ⋅ K 25 W m ⋅ K ′′ q1 = ( 261 − 25 ) C ′′ q1 = 236 C ( 0.001 + 0.0012 ) m ⋅ K W q ′′ = 186 C ( 0.0004 + 0.001) m ⋅ K W 2 2 ′′ q1 = 107, 273 W m 0.020 m 1 + 50 W m ⋅ K 1000 W m 2 ⋅ K q ′′ = ( 211 − 25 ) C 2 2 2 q ′′ = 132, 857 W m 2 2 ′′ Using the values for q1 and q′′ in Eq. (1), find 2 ( q B = 106, 818 + 132,143 W m 2 ) 2 × 0.030 m = 4.00 ×10 6 < 3 Wm. To determine kB, use the general form of the temperature and heat flux distributions in wall B, T(x) = − qB 2k B 2 x + C1x + C 2 q ′′ (x) = − k B − x q kB x + 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 = − qB 2k B T ( + L B ) = T2 = − qB 2k B ( − L B )2 − C1L B + C 2 where T1 = 261°C (3) ( + L B )2 + C1L B + C2 where T2 = 211°C (4) ′′ where q1 = 107,273 W/m2 (5) q ′′ q ′′ ( − L B ) = −q1 = − k B − B ( − L B ) + C1 x kB Using 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 temperature distribution is shown in the plot below. The important features are (1) Distribution is quadratic in B, but non-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 plot below when h = 0 on the surface of A. Since the left boundary is adiabatic, material A will be isothermal at T1. Find T1 = 835°C < T2 = 360°C Loss of coolant on surface A 400 Temperature, T (C) Temperature, T (C) 800 300 200 100 600 400 200 -60 -40 -20 0 20 Wall position, x-coordinate (mm) T_xA, kA = 25 W/m.K T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3 T_x, kC = 50 W/m.K 40 -60 -40 -20 0 20 Wall position, x-coordinate (mm) T_xA, kA = 25 W/m.K; adiabatic surface T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3 T_x, kC = 50 W/m.K 40 PROBLEM 3.74 KNOWN: Composite wall exposed to convection process; inside wall experiences a uniform heat generation. FIND: (a) Neglecting interfacial thermal resistances, determine T1 and T2, as well as the heat fluxes through walls A and C, and (b) Determine the same parameters, but consider the interfacial contact resistances. Plot temperature distributions. SCHEMATIC: k A = 25 W m K L A = 30 mm k B = 15 W m ⋅ K L B = 30 mm k C = 50 W m ⋅ K L C = 20 mm q B 4 ` 10 6 W m 3 ASSUMPTIONS: (1) One-dimensional, steady-state heat flow, (2) Negligible contact resistance between 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 x T − T2 . +1 T ( x ) = B B 1 − + 2 2k B 2 LB 2 LB 2 2 (1) The heat fluxes to the neighboring walls are found using Fourier’s law, q′′ = − k x dT dx . q T −T ′′ At x = −L B : q′′ ( − L B ) − k B + B ( L B ) + 2 1 = q1 (2) x 2L B kB q T −T At x = + L B : q′′ ( L B ) − k B − B ( L B ) + 2 1 = q′′ (3) x 2 2L B kB ′′ The heat fluxes, q1 and q′′ , can be evaluated by thermal circuits. 2 Substituting 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 ⋅ K 2 q ′′ = ( T2 − 25 ) C 2 ( 0.001 + 0.0004 ) K W = 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 two equations with two unknowns is obtained. T −T ′′ Eq. (2): −4 × 106 W m3 × 0.03 m + 15 W m ⋅ K 2 1 = q1 2 × 0.03 m −1.2 × 105 W m 2 − 2.5 × 102 ( T2 − T1 ) W m 2 = 454.6 ( 25 − T1 ) 704.6 T1 − 250 T2 = 131, 365 Eq. (3): (6) +4 × 106 W m3 × 0.03 m − 15 W m ⋅ K T2 − T1 2 × 0.03 m = q′′ 2 +1.2 × 105 W m 2 − 2.5 × 102 (T2 − T1 ) W m 2 = 714.3 ( T2 − 25 ) 250 T1 − 964 T2 = −137,857 Solving Eqs. (6) and (7) simultaneously, find (7) T2 = 210.0°C T1 = 260.9°C From 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 . 2 Note directions of the heat fluxes. (b) Considering interfacial contact resistances, we will use a different approach. The general solution for the temperature and heat flux distributions in each of the materials is TA ( x ) = C1x + C2 TB ( x ) = − qB 2k B q′′ = − k A C1 x x 2 + C3 x + C4 TC ( x ) = C5 x + C6 − (LA + LB ) ≤ x ≤ −LB (1,2) q q′′ = − B x + C3 x kB −LB ≤ x ≤ LB (3,4) + L B ≤ x ≤ ( L B + LC ) (5,6) q′′ = − k CC5 x To determine C1 ... C6 and the distributions, we need to identify boundary conditions using surface energy balances. At x = -(LA + LB): (7) −q′′ ( −L A − L B ) + q′′ = 0 x cv − ( − 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 the contact resistance. q′′ ( − L B ) = q′′ ( − L B ) x,A x,B (9) q ′′ ( − L B ) = [T1A ( − L B ) − T1B ( − L B )] R ′′ x,A tc,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,B x,C (11) q ′′ ( + L B ) = [T2B ( + L B ) − T2C ( + L B )] R ′′ x,B tc,BC (12) At x = +(LB + LC): − q x,C ( L B + LC ) − q ′′ = 0 cv (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 of equations above for the constants C1 ... C6 for conditions with negligible and prescribed values for the ′′ interfacial constant resistances. The results are tabulated and plotted below; q1 and q′′ represent heat 2 fluxes leaving surfaces A and C, respectively. T1B (°C) 260 T2B (°C) 210 T2C (°C) 210 ′′ q1 (kW/m ) q′′ (kW/m ) 2 R ′ = 0 tc T1A (°C) 260 106.8 132.0 R ′′ ≠ 0 tc 233 470 371 227 94.6 144.2 Conditions 2 500 Temperature, T (C) Temperature, T (C) 500 2 300 100 300 100 -60 -40 -20 0 20 40 -60 -40 Wall position, x-coordinate (mm) -20 0 20 40 Wall position, x-coordinate (mm) T_xA, kA = 25 W/m.K T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3 T_x, kC = 50 W/m.K T_xA, kA = 25 W/m.K T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3 T_x, kC = 50 W/m.K COMMENTS: (1) The results for part (a) can be checked using an energy balance on wall B, E in − E out = − E g ′′ q1 − q ′′ = −q B × 2L B 2 where 2 ′′ q1 − q ′′ = −107, 240 − 132,146 = 239, 386 W m 2 − q B L B = −4 × 10 W m × 2 ( 0.03 m ) = −240, 000 W m . 6 3 2 Hence, 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 3.75 KNOWN: Composite wall of materials A and B. Wall of material A has uniform generation, while wall B has no generation. The inner wall of material A is insulated, while the outer surface of material B experiences convection cooling. Thermal contact resistance between the materials is R ′′ = 10 t,c −4 2 m ⋅ 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 constant properties, 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 is parabolic with zero slope at the inner boundary, and that the distribution in material B is linear. At the interface between the two materials, x = LA, the temperature distribution will show a discontinuity. TA ( x ) = q L2 A x2 1 − +T 2 1A LA 0 ≤ x ≤ LA 2 kA TB ( x ) = T1B − ( T1B − T2 ) x − LA LA ≤ x ≤ LA + LB LB Considering the thermal circuit above (see also Ex. 3.6) including the thermal contact resistance, T − T∞ T1B − T∞ T − T∞ = =2 q′′ = q L A = 1A R ′′ R ′′ R ′′ tot cond,B + R ′′ conv conv find TA(0) = 147.5°C, T1A = 122.5°C, T1B = 115°C, and T2 = 105°C. Using the foregoing equations in IHT, the temperature distributions for each of the materials can be calculated and are plotted on the Effect of therm al contact res is tance on tem perature distribution graph below. 150 140 T (C ) 130 120 110 100 0 10 20 30 40 50 60 70 x (m m ) COMMENTS: (1) The effect of the thermal contact resistance between the materials is to increase the maximum temperature of the system. (2) Can you explain why the temperature distribution in the material B is not affected by the presence of the thermal contact resistance at the materials’ interface? PROBLEM 3.76 KNOWN: 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 = 50°C. For case 2, the boundary conditions are the same, but a thin dielectric strip with thermal resistance R ′′ = 0.0005 m 2 ⋅ K / W is inserted at the mid-plane. t FIND: (a) Sketch the temperature distribution for case 1 on T-x coordinates and describe key features; identify and calculate the maximum temperature in the wall, (b) Sketch the temperature distribution for case 2 on the same T-x coordinates and describe the key features; (c) What is the temperature difference between the two walls at x = 0 for case 2? And (d) What is the location of the maximum temperature of the composite wall in case 2; calculate this temperature. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the plane and composite walls, and (3) Constant properties. ANALYSIS: (a) For case 1, the temperature distribution, T1(x) vs. x, is parabolic as shown in the schematic below and the gradient is zero at the insulated boundary, x = -L. From Eq. 3.43, 2 2 q 2L 5 × 106 W / m3 2 × 0.020 m T1 ( − L ) − T1 ( + L ) = () 2k ( = 2 × 50 W / m ⋅ K ) = 80°C and since T1(+L) = To = 50°C, the maximum temperature occurs at x = -L, T1 ( − L ) = T1 ( + L ) + 80°C = 130°C (b) For case 2, the temperature distribution, T2(x) vs. x, is piece-wise parabolic, with zero gradient at x = -L and a drop across the dielectric strip, ∆TAB. The temperature gradients at either side of the dielectric strip are equal. (c) For case 2, the temperature drop across the thin dielectric strip follows from the surface energy balance shown above. q′′ ( 0 ) = ∆TAB / R ′′ x t q′′ ( 0 ) = qL x ∆TAB = R ′′ qL = 0.0005 m 2 ⋅ K / W × 5 × 106 W / m3 × 0.020 m = 50°C. t (d) For case 2, the maximum temperature in the composite wall occurs at x = -L, with the value, T2 ( − L ) = T1 ( − L ) + ∆TAB = 130°C + 50°C = 180°C < PROBLEM 3.77 KNOWN: Geometry and boundary conditions of a nuclear fuel element. FIND: (a) Expression for the temperature distribution in the fuel, (b) Form of temperature distribution 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 fuel and cladding. ANALYSIS: (a) The general solution to the heat equation, Eq. 3.39, d 2T dx 2 is + T=− q =0 kf ( −L ≤ x ≤ +L ) q2 x + C1x+C2 . 2k f The insulated wall at x = - (L+b) dictates that the heat flux at x = - L is zero (for an energy balance applied to a control volume about the wall, Ein = Eout = 0). Hence dT q = − ( − L ) + C1 = 0 dx x =− L kf T=− C1 = − or qL kf q 2 qL x− x+C2 . 2k f kf The value of Ts,1 may be determined from the energy conservation requirement that Eg = qcond = qconv , or on a unit area basis. ( )( ) k q ( 2L ) = s Ts,1 − Ts,2 = h Ts,2 − T∞ . b Hence, Ts,1 = Ts,1 = q ( 2 Lb ) ks q ( 2 Lb ) ks + Ts,2 + q ( 2L ) h where Ts,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 () 2 3 qL + T∞ = − + C2 2 kf which yields 2b 2 3 L C2 = T∞ + qL + + ks h 2 k f Hence, the temperature distribution for ( − L ≤ x ≤ +L ) is T=− q 2 qL x− x+qL 2k f kf 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 3.78 KNOWN: Thermal conductivity, heat generation and thickness of fuel element. Thickness and thermal conductivity of cladding. Surface convection conditions. FIND: (a) Temperature distribution in fuel element with one surface insulated and the other cooled by 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,2 1 − + + 2k f L2 2 L 2 (1) With an insulated surface at x = -L, Eq. C.10 yields Ts,1 − Ts,2 = 2 q L2 kf (2) and with convection at x = L + b, Eq. C.13 yields ( ) ( k U Ts,2 − T∞ = q L − f Ts,2 − Ts,1 2L Ts,1 − Ts,2 = ) 2 LU 2 q L2 Ts,2 − T∞ − kf kf ( ) (3) Substracting Eq. (2) from Eq. (3), 0= 2LU 4 q L2 Ts,2 − T∞ − kf kf ( Ts,2 = T∞ + ) 2 qL U (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 ) = − -1 2 3 L q 2 qL x− x + qL + 2 kf kf U 2 kf + T∞ -1 or, with U = h + b/ks, T (x ) = − 2b 2 3 L q 2 qL x− x + qL + + 2 kf kf ks h 2 k f + T∞ (6) < The maximum temperature occurs at x = - L and is b 1 L T ( − L ) = 2 qL + + ks h k f + T∞ 0.003m 1 0.015 m + + + 200°C = 530°C 15 W / m ⋅ K 2 10, 000 W / m ⋅ K 60 W / m ⋅ K T ( − L ) = 2 × 2 × 10 W / m × 0.015 m 7 3 < The lowest temperature is at x = + L and is T (+L ) = − 2b 2 3 L 3 qL2 + qL + + 2 kf ks h 2 k f + T∞ = 380°C < (b) If a convection condition is maintained at x = - L, Eq. C.12 reduces to ( ) ( k U T∞ − Ts,1 = −qL − f Ts,2 − Ts,1 2L ) 2 LU 2 qL2 Ts,1 − Ts,2 = Ts,1 − T∞ − kf kf ( ) (7) Subtracting Eq. (7) from Eq. (3), 0= ( 2 LU Ts,2 − T∞ − Ts,1 + T∞ kf ) or Ts,1 = Ts,2 Hence, from Eq. (7) Continued ….. PROBLEM 3.78 (Cont.) Ts,1 = Ts,2 = 1 b qL + T∞ = qL + + T∞ U h ks (8) Substituting into Eq. (1), the temperature distribution is T (x ) = 1 b qL2 x 2 1 − + qL + + T∞ 2 k f L2 h ks (9) < The maximum temperature is at x = 0 and is T (0 ) = 7 2 × 10 W / m 3 (0.015 m )2 2 × 60 W / m ⋅ K 7 3 1 0.003 m + + 200°C 10, 000 W / m 2 ⋅ K 15 W / m ⋅ K + 2 × 10 W / m × 0.015 m T ( 0 ) = 37.5°C + 90°C + 200°C = 327.5°C < The minimum temperature at x = ± L is 1 0.003m + 200°C = 290°C Ts,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. 550 Te m p e ra tu re , T(C ) 500 450 400 350 300 250 200 -0 .0 1 5 -0 .0 0 9 -0 .0 0 3 0 .0 0 3 0 .0 0 9 0 .0 1 5 Fu 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 e S ym m e trica l co n ve c tio n c o n d itio n s The amount of heat generation is the same for both cases, but the ability to transfer heat from both surfaces 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 and equivalent to Ts,1 = 530°C. < PROBLEM 3.79 KNOWN: Wall of thermal conductivity k and thickness L with uniform generation q ; strip heater with uniform heat flux q′′ ; prescribed inside and outside air conditions (hi, T∞,i, ho, T∞,o). o FIND: (a) Sketch temperature distribution in wall if none of the heat generated within the wall is lost to 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), if o ′′ corresponds to the value calculated in (c). q=0 but q o SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform volumetric generation, (4) Constant properties. ANALYSIS: (a) If none of the heat generated within the wall is lost to the outside of the chamber, the gradient at x = 0 must be zero. Since q is uniform, the temperature distribution is parabolic, with T(L) > T∞,i. (b) To find temperatures at the boundaries of wall, begin with the general solution to the appropriate form of the heat equation (Eq.3.40). T (x ) = − q2 x + C1x+C2 2k (1) From the first boundary condition, dT =0 dx x=o → C1 = 0. (2) Two approaches are possible using different forms for the second boundary condition. Approach No. 1: With boundary condition → T (0 ) = T1 T (x ) = − q2 x + T1 2k (3) To find T1, perform an overall energy balance on the wall Ein − E out + Eg = 0 − h T ( L ) − T∞,i + qL=0 T ( L ) = T2 = T∞,i + qL h (4) Continued ….. PROBLEM 3.79 (Cont.) and from Eq. (3) with x = L and T(L) = T2, T (L) = − q2 L + T1 2k or T1 = T2 + q2 qL qL2 L = T∞,i + + 2k h 2k (5,6) Substituting numerical values into Eqs. (4) and (6), find T2 = 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. < 2 Approach No. 2: Using the boundary condition −k dT = h T ( L ) − T∞,i dx x=L yields the following temperature distribution which can be evaluated at x = 0,L for the required temperatures, T (x ) = − ) ( q 2 2 qL x −L + + T∞,i . 2k h (c) The value of q′′ when T(0) = T1 = 65°C o follows from the circuit q′′ = o T1 − T∞,o 1/ h o q′′ = 5 W/m2 ⋅ K ( 65-25) C=200 W/m 2 . o < (d) With q=0, the situation is represented by the thermal circuit shown. Hence, q′′ = q′′ + q′′ o a b q′′ = o T1 − T∞ ,o 1/ h o + T1 − T∞,i L/k+1/h i which yields T1 = 55 C. < PROBLEM 3.80 KNOWN: Wall of thermal conductivity k and thickness L with uniform generation and strip heater with uniform heat flux q′′ ; prescribed inside and outside air conditions ( T∞,i , hi, T∞, o , ho). Strip heater o acts 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, 50 o and 70°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform volumetric generation, (4) Constant properties. ANALYSIS: If no heat generated within the wall 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 distribution must 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 , the o analysis begins by considering the temperature distribution in the wall and then surface energy balances at 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 the wall is, Eq. 3.40, T(x) = − q 2k x 2 + C1x + C2 and the guard condition at the outer wall, x = 0, requires that the conduction heat flux be zero. Using Fourier’s law, dT (1) q′′ (0) = − k = − kC1 = 0 (C1 = 0 ) x dx x = 0 At the outer wall, x = 0, T(0) = C2 (2) Surface energy balance, x = 0: E in − E out = 0 q ′′ − q ′′ o cv,o − q ′′ ( 0 ) = 0 x ( ) ′′ 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 = 0 q ′′ (L) − q ′′ = 0 x cv,i q ′′ (L) = − k x dT dx x = L (5) = + qL (6) q ′′ v,i = h T(L) − T∞ ,i c q L2 + T 0 − T ( ) ∞,i 2k q ′′ = h − cv,i (7) 400 300 200 100 0 0 500 1000 1500 2000 Volumetric generation rate, qdot (W/m^3) Tinfi = 30 C Tinfi = 50 C Tinfi = 70 C Wall temperature, T(0) (C) Heater flux, q''o (W/m^2) Solving Eqs. (1) through (7) simultaneously with appropriate numerical values and performing the parametric analysis, the results are plotted below. 120 100 80 60 40 20 0 500 1000 1500 2000 Volumetric generation rate, qdot (W/m^3) Tinfi = 30 C Tinfi = 50 C Tinfi = 70 C From the first plot, the heater flux q′′ is a linear function of the volumetric generation rate q . As o expected, 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 the x required q′′ . The outer wall temperature T(0) is also linearly dependent upon q . From our knowledge o of the temperature distribution, it follows that for any q condition, the outer wall temperature T(0) will track changes in T∞,i . PROBLEM 3.81 KNOWN: Plane wall with prescribed nonuniform volumetric generation having one boundary 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 is d dT q dx + k = 0. dx Noting that q = q ( x ) = qo (1 − x/L ) , substitute for q ( x ) into the above equation, separate variables and then integrate, x2 x − + C1. 2L Separate variables and integrate again to obtain the general form of the temperature distribution in the wall, q q x 2 x3 x2 dT = − o x − T (x ) = − o − dx+C1dx + C1x+C2 . k 2L k 2 6L q dT d = − o k dx x 1 − L dx q dT =− o dx k Identify the boundary conditions at x = 0 and x = L to evaluate C1 and C2. At x = 0, q T ( 0 ) = To = − o ( 0 − 0 ) + C1 ⋅ 0 + C2 hence, C2 = To k At x = L, q qL dT L2 = 0 = − o L − hence, C1 = o + C1 dx x=L k 2L 2k The temperature distribution is qo x 2 x 3 qo L T (x ) = − x+To . − + k 2 6L 2k < COMMENTS: It is good practice to test the final result for satisfying BCs. The heat flux at x = 0 can be found using Fourier’s law or from an overall energy balance L Eout = Eg = ∫ qdV 0 to obtain q′′ = q o L/2. out PROBLEM 3.82 KNOWN: Distribution of volumetric heating and surface conditions associated with a quartz window. 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 emission from outer surface, (4) Constant properties. ANALYSIS: The appropriate form of the heat equation for the quartz is obtained by substituting the prescribed form of q into Eq. 3.39. d 2T α (1 − β ) q′′ -α x o e + =0 2 k dx Integrating, (1 − β ) q′′ e-α x + C dT o =+ 1 dx k T=− (1 − β ) q′′ e-α x + C x+C o kα 1 2 − k dT/dx) x=o = β q′′ o − k dT/dx) x=L = h T ( L ) − T∞ Boundary Conditions: (1-β ) −k q′′ + C1 = β q′′ o o k C1 = −q′′ / k o Hence, at x = 0: At x = L: (1-β ) (1-β ) −k q′′ e-α L + C1 = h q′′ e-α L + C1L+C2 − T∞ o o k kα Substituting for C1 and solving for C2, q′′ C2 = o h Hence, q′′ o 1 − (1 − β ) e-α L + q′′ + o(1-β ) e-α L + T . ∞ k kα T (x ) = (1 − β ) q′′ e-α L − e-α x + q′′ o o kα k (L − x ) + q′′ o 1 − (1 − β ) e-α L + T . < ∞ h COMMENTS: The temperature distribution depends strongly on the radiative coefficients, α and β. For α → ∞ or β = 1, the heating occurs entirely at x = 0 (no volumetric heating). PROBLEM 3.83 KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive wastes. Surface convection conditions. FIND: Radial temperature distribution. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across container wall. ANALYSIS: The appropriate form of the heat equation is q r2 1 d dT q = − = − o 1 − r 2 r dr dr k k ro r q r 2 qr 4 dT =− o + + C1 2 dr 2k 4kro q r2 q r4 T = − o + o + C1 ln r+C2 . 4k 16kr 2 o From the boundary conditions, dT |r=0 = 0 → C1 = 0 dr −k dT |r=ro = h T ( ro ) − T∞ ) dr q r2 q r2 q o ro q o ro + − = h − o o + o o + C2 − T∞ 2 4 16k 4k qr 3q r 2 C2 = o o + o o + T∞ . 4h 16k Hence qr q r2 3 1 r T ( r ) = T∞ + o o + o o − 4h k 16 4 ro 2 4 1r + . 16 ro < COMMENTS: Applying the above result at ro yields Ts = T ( ro ) = T∞ + ( qo ro ) / 4h The same result may be obtained by applying an energy balance to a control surface about the container, where Eg = qconv . The maximum temperature exists at r = 0. PROBLEM 3.84 KNOWN: Cylindrical shell with uniform volumetric generation is insulated at inner surface and 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 conditions are T (r ) = − q2 r + C1 ln r+C2 4k dT q 1 = 0 = − ri + C1 + 0 dr r 2k ri at r = ri: C1 = i dT = h T ( ro ) − T∞ dr r −k at r = ro: q2 r 2k i surface energy balance o q q q 2 q 1 k − ro + ri2 ⋅ = h − ro + ri2 ln ro + C2 − T∞ ro 2k 2k 4k 2k qr C2 = − o 2h 2 2 1 + ri + qro ro 2k 2 1 − ri ln ro + T∞ 2 ro Hence, ( ) 2 r qro q 2 2 qri T (r ) = ro − r + ln − 4k 2k ro 2h 2 1 + ri + T∞ . ro < (b) From an overall energy balance on the shell, q′ ( r ) = E′ = qπ r 2 − r 2 . ro g (o i ) < Alternatively, the heat rate may be found using Fourier’s law and the temperature distribution, q′ ( r ) = − k ( 2π ro ) r q qr 2 dT 2 − ro + i 1 + 0 + 0 = qπ ro − r 2 = −2π kro 2k i dr r 2k ro o ( ) PROBLEM 3.85 KNOWN: The solid tube of Example 3.7 with inner and outer radii, 50 and 100 mm, and a thermal conductivity of 5 W/m⋅K. The inner surface is cooled by a fluid at 30°C with a convection coefficient 2 of 1000 W/m ⋅K. 5 FIND: Calculate and plot the temperature distributions for volumetric generation rates of 1 × 10 , 5 5 6 3 × 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) Constant properties 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 + ) ( q22 q 2 r2 r2 − r − r "n 4k 2k 2 r r1 ≤ r ≤ r2 The temperature at the inner boundary, Ts,1, follows from the surface energy balance, Eq. (10), 2 2 π q r2 − r1 = h2π 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 = r1 and T(r) = Ts,1, are solved simultaneously to find Ts,2 = 69.3°C. Eq. (1), with Ts,2 now a known parameter, can be used to determine the temperature distribution, T(r). The results for different values of the generation rate are shown in the graph. Effect of generation rate on temperature distributions Temperature, T(C) 500 400 300 200 100 0 50 60 70 80 90 100 Radial location, r (mm) qdot = 1e5 W/m^3 qdot = 5e5 W/m^3 qdot = 1e6 W/m^3 COMMENTS: (1) The temperature distributions are parabolic with a zero gradient at the insulated outer boundary, r = r2. The effect of increasing q is to increase the maximum temperature in the tube, 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 3.7 T_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. 7 Ts1 = Ts2 + qdot / (4*k) * (r2^2 – r1^2) – qdot / (2*k) * r2^2 * ln (r2/r1) // The energy balance on the surface, from Eq. 10 pi * qdot * (r2^2 – r1^2) = h * 2 * pi * r1 * (Ts1 – Tinf) PROBLEM 3.86 KNOWN: Diameter, resistivity, thermal conductivity, emissivity, voltage, and maximum temperature of 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 maximum volumetric heat generation may be obtained from Eq. (3.55), but with the total heat transfer coefficient, ht = h + hr, used in lieu of the convection coefficient h. With ( ) h r = εσ ( Ts + Tsur ) Ts + Tsur = 0.20 × 5.67 × 10 2 2 −8 2 W/m ⋅K 4 (1473 + 323 ) K ( 2 1473 + 323 ) 2 2 2 K = 46.3 W / m ⋅ K h t = ( 250 + 46.3) W / m 2 ⋅ K = 296.3 W / m 2 ⋅ K ) ( 2 296.3 W / m 2 ⋅ K 2 ht q max = (Ts − T∞ ) = (1150°C ) = 1.36 × 109 W / m3 ro 0.0005m Hence, with 2 I 2 R e I ( ρe L / A c ) I 2 ρ e I2 ρe q= = = = 2 2 ∀ LAc Ac π D2 / 4 ) ( 1/ 2 q Imax = max ρe 1/ 2 π D 2 1.36 × 109 W / m3 = 10−6 Ω ⋅ m 4 π (0.001m ) = 29.0 A 4 2 < Also, with ∆E = I Re = I (ρeL/Ac), ∆E ⋅ Ac L= = Imax ρe 2 110 V π ( 0.001m ) / 4 = 2.98m −6 Ω ⋅ m 29.0 A 10 ( ) < and the power rating is Pelec = ∆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 be used to compute the centerline temperature corresponding to q max and a surface temperature of 1200°C. It follows that To = 2 q ro 4k ( 0.0005m )2 4 ( 25 W / m ⋅ K ) 9 + Ts = 1.36 × 10 W / m 3 + 1200°C = 1203°C. With only a 3°C temperature difference between the centerline and surface of the wire, the assumption is excellent. PROBLEM 3.87 KNOWN: Energy generation in an aluminum-clad, thorium fuel rod under specified operating conditions. FIND: (a) Whether prescribed operating conditions are acceptable, (b) Effect of q and h on acceptable operating 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 between aluminum and thorium. PROPERTIES: Table A-1, Aluminum, pure: M.P. = 933 K; Table A-1, Thorium: M.P. = 2023 K, k ≈ 60 W/m⋅K. ANALYSIS: (a) System failure would occur if the melting point of either the thorium or the aluminum were exceeded. From Eq. 3.53, the maximum thorium temperature, which exists at r = 0, is T(0) = 2 qro 4k + Ts = TTh,max where, from the energy balance equation, Eq. 3.55, the surface temperature, which is also the aluminum temperature, is Ts = T∞ + qro 2h = TAl Hence, TAl = Ts = 95 C + 7 × 108 W m3 × 0.0125 m 2 14, 000 W m ⋅ K 7 × 108 W m3 ( 0.0125m ) = 720 C = 993 K 2 TTh,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 melt under the proposed operating conditions. The problem could be eliminated by decreasing q , increasing h 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, the following radial temperature distributions were obtained for parametric variations in q and h. Continued... PROBLEM 3.87 (Cont.) 1600 1200 Temperature, T(K) Temperature, T(K) 1500 1400 1300 1200 1100 1000 1000 800 600 900 400 800 0 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.002 0.004 0.006 0.008 0.01 0.012 0.014 Radius, r(m) Radius, r(m) qdot = 2E8, h = 2000 W/m^2.K qdot = 2E8, h = 3000 W/m^2.K qdot = 2E8, h = 5000 W/m^2.K qdot = 2E8, h = 10000 W/m^2.K h = 10000 W/m^2.K, qdot = 7E8 W/m^3 h = 10000 W/m^2.K, qdot = 8E8 W/m^3 h = 10000 W/m^2.K, qdot = 9E9 W/m^3 For h = 10,000 W/m2⋅K, which represents a reasonable upper limit with water cooling, the temperature of the aluminum would be well below its melting point for q = 7 × 108 W/m3, but would be close to the 8 3 melting point for q = 8 × 10 W/m and would exceed it for q = 9 × 108 W/m3. Hence, under the best of 8 3 conditions, q ≈ 7 × 10 W/m corresponds to the maximum allowable energy generation. However, if coolant flow conditions are constrained to provide values of h < 10,000 W/m2⋅K, volumetric heating would have to be reduced. Even for q as low as 2 × 108 W/m3, operation could not be sustained for h = 2 2000 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) 2000 2000 1600 1200 800 400 0 1600 1200 800 400 0 1E8 1E8 2.8E8 4.6E8 6.4E8 8.2E8 Energy generation, qdot (W/m^3) h = 2000 W/m^2.K h = 5000 W/m^2.K h = 10000 W/m^2.K 1E9 2.8E8 4.6E8 6.4E8 8.2E8 1E9 Energy generation, qdot (W/m^3) h = 2000 W/m^2.K h = 5000 W/m^2.K h = 10000 W/m^2.K For h = 2000 and 5000 W/m2⋅K, the melting point of thorium would be approached for q ≈ 4.4 × 108 and 8 3 2 8.5 × 10 W/m , respectively. For h = 2000, 5000 and 10,000 W/m ⋅K, the melting point of aluminum would be approached for q ≈ 1.6 × 108, 4.3 × 108 and 8.7 × 108 W/m3. Hence, the envelope of acceptable operating conditions must call for a reduction in q with decreasing h, from a maximum of q 8 3 2 ≈ 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 h would decrease drastically. PROBLEM 3.88 KNOWN: Radii and thermal conductivities of reactor fuel element and cladding. Fuel heat generation rate. Temperature and convection coefficient of coolant. FIND: (a) Expressions for temperature distributions in fuel and cladding, (b) Maximum fuel element temperature for prescribed conditions, (c) Effect of h on temperature distribution. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible contact resistance, (4) Constant properties. ANALYSIS: (a) From Eqs. 3.49 and 3.23, the heat equations for the fuel (f) and cladding (c) are 1 d dTf q =− kf r dr dr 1 d dTc r =0 r dr dr (0 ≤ r ≤ r1 ) ( r1 ≤ r ≤ r2 ) Hence, integrating both equations twice, dTf dr dTc dr =− = qr 2k f + C1 kf r C3 k cr Tf = − qr 2 C + 1 ln r + C 2 4k f k f C Tc = 3 ln r + C4 kc (1,2) (3,4) The corresponding boundary conditions are: dTf dr )r = 0 = 0 −k f dTf dr dT = −k c c dr r = r r = r1 1 Tf ( r1 ) = Tc ( r1 ) − kc dTc = h [Tc ( r2 ) − T∞ ] dr r = r (5,6) (7,8) 2 Note that Eqs. (7) and (8) are obtained from surface energy balances at r1 and r2, respectively. Applying Eq. (5) to Eq. (1), it follows that C1 = 0. Hence, Tf = − qr 2 + C2 4k f From Eq. (6), it follows that C ln r qr 2 − 1 + C2 = 3 1 + C4 4k f kc (9) (10) Continued... PROBLEM 3.88 (Cont.) Also, from Eq. (7), qr1 C =− 3 2 r1 qr 2 C3 = − 1 2 or (11) C C Finally, from Eq. (8), − 3 = h 3 ln r2 + C4 − T∞ or, substituting for C3 and solving for C4 r2 kc C4 = 2 qr1 2 qr1 ln r2 + T∞ + 2r2 h 2k c Substituting Eqs. (11) and (12) into (10), it follows that C2 = 2 qr1 4k f 2k c + 2 qr1 2r2 h + 2 qr1 2k c ln r2 + T∞ r qr 2 ln 2 + 1 T∞ 4k f 2k c r1 2r2 h Substituting Eq. (13) into (9), C2 = 2 qr1 − 2 qr1 ln r1 (12) + ( 2 qr1 ) (13) r qr 2 ln 2 + 1 + T∞ 4k f 2k c r1 2r2 h Substituting Eqs. (11) and (12) into (4), Tf = Tc = q 2 r1 − r 2 + 2 qr1 (14) (15) 2 qr1 r qr 2 ln 2 + 1 + T∞ . 2k c r 2r2 h < < (b) Applying Eq. (14) at r = 0, the maximum fuel temperature for h = 2000 W/m2⋅K is Tf ( 0 ) = 2 × 108 W m3 × ( 0.006 m ) 2 × 108 W m3 × (0.006 m ) 2 4× 2 W m⋅ K 2 + 2 × 25 W m ⋅ K 2 × 108 W m3 ( 0.006 m ) ln 0.009 m 0.006 m 2 + 2 × ( 0.09 m ) 2000 W m 2 ⋅ K + 300 K Tf ( 0 ) = (900 + 58.4 + 200 + 300 ) K = 1458 K . (c) Temperature distributions for the prescribed values of h are as follows: 600 1300 Temperature, Tc(K) Temperature, Tf(K) 1500 < 1100 900 700 500 300 0 0.001 0.002 0.004 0.005 Radius in fuel element, r(m) h = 2000 W/m^2.K h = 5000 W/m^2.K h = 10000 W/m^2.K 0.006 500 400 300 0.006 0.007 0.008 0.009 Radius in cladding, r(m) h = 2000 W/m^2.K h = 5000 W/m^2.K h = 10000 W/m^2.K Continued... 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 low thermal conductivity of the fuel material. 2 COMMENTS: For the prescribed conditions, Eq. (14) yields, Tf(0) - Tf(r1) = qr1 4k f = (2×108 W/m3)(0.006 m)3/8 W/m⋅K = 900 K, in which case, with no cladding and h → ∞, Tf(0) = 1200 K. To reduce Tf(0) below 1000 K for the prescribed material, it is necessary to reduce q . PROBLEM 3.89 KNOWN: Dimensions and properties of tubular heater and external insulation. Internal and external convection conditions. Maximum allowable tube temperature. FIND: (a) Maximum allowable heater current for adiabatic outer surface, (3) Effect of internal convection 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) Uniform heat 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 2 r2 ln − r2 − r1 2k r1 4k Ts,2 − Ts,1 = and Ts,1 = T∞ ,1 + ( 2 2 q r2 − r1 ) (1) ) (2) 2h1r1 Hence, eliminating Ts,1, we obtain Ts,2 − T∞ ,1 = ) ( 2 qr2 ) ( r2 1 k 22 2 2 ln r − 2 1 − r1 r2 + h r 1 − r1 r2 2k 1 11 Substituting the prescribed conditions (h1 = 100 W/m2⋅K), ( )( Ts,2 − T∞ ,1 = 1.237 × 10−4 m3 ⋅ K W q W m3 ) Hence, with Tmax corresponding to Ts,2, the maximum allowable value of q is 1400 − 400 q max = = 8.084 × 106 W m3 −4 1.237 × 10 with q= I 2 Re ∀ = I2 ρe L Ac ρe I2 = 2 LA c π r 2 − r 21 ( 2 2 I max = π r2 − r1 ) 1/ 2 q ρe ) ( ( 2 = π 0.035 − 0.025 2 ) 6 3 1/ 2 2 8.084 × 10 W m m = 6406 A −6 0.7 × 10 Ω ⋅ m < Continued ….. PROBLEM 3.89 (Cont.) (b) Using the one-dimensional, steady-state conduction model of IHT (hollow cylinder; convection at inner surface and adiabatic outer surface), the following temperature distributions were obtained. Temperature, T(K) 1500 1300 1100 900 700 500 300 0.025 0.027 0.029 0.031 0.033 0.035 Radius, r(m) h = 100 W/m^2.K h = 500 W/m^2.K h = 1000 W/m^2.K The results are consistent with key implications of Eqs. (1) and (2), namely that the value of h1 has no effect on the temperature drop across the tube (Ts,2 - Ts,1 = 30 K, irrespective of h1), while Ts,1 decreases with increasing h1. For h1 = 100, 500 and 1000 W/m2⋅K, respectively, the ratio of the temperature drop between 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, the heat rate to the airflow is fixed by the value of q and, irrespective of h1, ( ) 2 2 q′ ( r1 ) = π r2 − r1 q = −15, 240 W < (c) Heat loss from the outer surface of the tube to the surroundings depends on the total thermal resistance ln ( r3 r2 ) 1 R tot = + 2π Lk i 2π r3Lh 2 or, for a unit area on surface 2, r2 ln ( r3 r2 ) r R ′′ +2 tot,2 = ( 2π r2 L ) R tot = ki r3h 2 Again using the capabilities of IHT (hollow cylinder; convection at inner surface and heat transfer from outer surface through R ′′ot,2 ), the following temperature distributions were determined for the tube and t insulation. Insulation Temperature, T(K) Tube temperature, T(K) 1200 1160 1120 1080 1040 1000 0.025 1200 1100 1000 900 800 700 600 500 0.027 0.029 0.031 Radius, r(m) delta =0.025 m delta = 0.050 m 0.033 0.035 0 0.2 0.4 0.6 0.8 1 Dimensionless radius, (r-r2)/(r3-r2) r3 = 0.060 m r3 = 0.085 m Continued... PROBLEM 3.89 (Cont.) Heat losses through the insulation, q′ ( r2 ) , are 4250 and 3890 W/m for δ = 25 and 50 mm, respectively, with corresponding values of q′ ( r1 ) equal to -10,990 and -11,350 W/m. Comparing the tube temperature distributions with those predicted for an adiabatic outer surface, it is evident that the losses reduce tube wall 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 should be reduced by selecting an insulation material with a significantly smaller thermal conductivity. PROBLEM 3.90 KNOWN: Electric current I is passed through a pipe of resistance R ′ to melt ice under e steady-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 pipe is adiabatic, (6) Inner surface is at a constant temperature, Tm. 3 PROPERTIES: Table A-3, Ice (273K): ρ = 920 kg/m ; Handbook Chem. & Physics, Ice: 5 Latent heat of fusion, hsf = 3.34×10 J/kg. ANALYSIS: (a) The appropriate form of the heat equation is Eq. 3.49, and the general solution, Eq. 3.51 is T (r ) = − q2 r + C1lnr+C2 4k where q= ( I2R ′ e 22 π r2 − r1 ) . Applying the boundary condition ( dT/dr )r = 0, it follows that 2 0= qr2 C1 + 2k r2 2 qr2 Hence C1 = and qr q T ( r ) = − r 2 + 2 lnr+C2 . 4k 2k 2k 2 Continued ….. PROBLEM 3.90 (Cont.) Applying the second boundary condition, T ( r1 ) = Tm , it follows that 2 q 2 qr Tm = − r1 + 2 lnr1 + C2 . 4k 2k Solving for C2 and substituting into the expression for T(r), find T ( r ) = Tm + 2 qr2 2k ln ) ( rq22 − r − 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 the wall during the melt period. Hence from Eq. 1.11b ∆Est = Ein − Eout + Egen ∆Est = E m = t m ⋅ qcond,r1 or, for a unit length of pipe, dT 2 ρ π r1 h sf = t m − k ( 2π r1 ) dr r1 () qr 2 qr 2 ρ π r1 h sf = −2π r1kt m 2 − 1 2kr1 2k () () ) ( 2 22 ρ π r1 h sf = − t m qπ r2 − r1 . Dropping the minus sign, which simply results from the fact that conduction is in the negative r direction, it follows that tm = 2 ρ h sf r1 ( 2 2 q r2 − r1 ) = 2 ρ h sf π r1 I2R ′ e . With r1 = 0.05m, I = 100 A and R ′ = 0.30 Ω/m, it follows that e 920kg/m3 × 3.34 ×105 J/kg × π × (0.05m ) 2 tm = or (100A )2 × 0.30Ω / m t m = 804s. < COMMENTS: The foregoing expression for tm could also be obtained by recognizing that all of the energy which is generated by electrical heating in the pipe wall must be transferred to the ice. Hence, 2 I2 R ′ t m = ρ h sf π r1 . e PROBLEM 3.91 KNOWN: 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 for different 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 rate equation T − T∞ q′ = 2 R′ tot where R′ = tot ln ( r3 r2 ) 2π k g + 1 2π r3h = ln (14 11) 2π (3 W m ⋅ K ) + ( 1 2π ( 0.014 m ) 2000 W m ⋅ K 2 ) = 0.0185 m ⋅ K W and the heat rate per unit length may be determined by applying an energy balance to a control surface about the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that ) ( ( ) 2 2 q′ = qπ r2 − r1 = 108 W m3 × π 0.0112 − 0.0082 m 2 = 17, 907 W m Hence, T2 = q′R ′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 yields T1 = T2 + 2 qr2 r 2 qr 2 r 1 − 1 − 1 ln 2 4k t r 2 2k t r1 2 T1 = 931K + 2 2 108 W m3 ( 0.011m ) 0.008 2 108 W m3 ( 0.008 m ) 0.011 ln 1 − − 4 × 57 W m ⋅ K 2 × 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, steady state conduction in a hollow tube. For the fuel element ( q > 0), an adiabatic surface condition is prescribed at r1, while heat transfer from the outer surface at r2 to the coolant is governed by the thermal resistance R ′′ot,2 = 2π r2 R ′ = 2π(0.011 m)0.0185 m⋅K/W = 0.00128 m2⋅K/W. For the graphite ( q = t tot 0), 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, the following distributions are obtained. 2500 2100 2100 Temperature, T(K) Temperature, T(K) 2500 1700 1300 900 500 0.008 0.009 0.01 Radial location in fuel, r(m) qdot = 5E8 qdot = 3E8 qdot = 1E8 0.011 1700 1300 900 500 0.011 0.012 0.013 0.014 Radial location in graphite, r(m) qdot = 5E8 qdot = 3E8 qdot = 1E8 The 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. Operation at q = 5 × 108 W/m3 is clearly unacceptable, since the melting points of thorium and graphite are exceeded and approached, respectively. To prevent softening of the materials, which would occur below their 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 the fuel element, thereby reducing the maximum allowable value of q . PROBLEM 3.92 KNOWN: Long rod experiencing uniform volumetric generation encapsulated by a circular sleeve 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-state conditions, (3) Uniform volumetric generation in rod, (4) Negligible contact resistance between rod and sleeve. ANALYSIS: (a) Construct a thermal circuit for the sleeve, where 2 2 q′=E′ = qπ D1 / 4 = 24, 000 W/m3 × π × ( 0.20 m ) / 4 = 754.0 W/m gen ′ Rs = ln ( r2 / r1 ) 2π k s R conv = = ln ( 400/200 ) 2π × 4 W/m ⋅ K = 2.758 × 10−2 m ⋅ K/W 1 1 = = 3.183 × 10−2 m ⋅ K/W hπ D 2 25 W/m 2 ⋅ K × π × 0.400 m The rate equation can be written as q′= T1 − T∞ T −T =2 ∞ R′ + R′ R′ s conv conv ) ( -2 −2 T1 = T∞ + q′ ( R ′ + R ′ K/W ⋅ m=71.8C s conv ) = 27 C+754 W/m 2.758 × 10 + 3.183 × 10 -2 T2 = T∞ + q′R ′ conv = 27 C+754 W/m × 3.183 × 10 m ⋅ K/W=51.0 C. < < (b) The temperature at the center of the rod is T ( 0 ) = To = 2 qr1 4k r 24, 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 are comparable. Will increasing the sleeve outer diameter cause the surface temperature T2 to increase or decrease? PROBLEM 3.93 KNOWN: Radius, thermal conductivity, heat generation and convection conditions associated 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=0 r 2 dr 1 r2 dT qr3 =− + C1 dr 3k T (r ) = − or dT qr C =− + 1 dr 3k r 2 qr 2 C1 − + C2 . 6k r The boundary conditions are: −k d 2 dT qr 2 r =− dr dr k dT =0 dr r=0 hence C1 = 0, and dT = h T ( ro ) − T∞ . dr r o Substituting into the second boundary condition (r = ro), find qr 2 o 2 qro qro qro = h + C2 − T∞ + + T∞ . C2 = 3 3h 6k 6k The temperature distribution has the form q 2 2 qro T (r ) = ro − r + + T∞ . 6k 3h COMMENTS: To verify the above result, obtain T(ro) = Ts, qr Ts = o + T∞ 3h Applying energy balance to the control volume about the sphere, ( ) 4 3 2 q π ro = h4π ro ( Ts − T∞ ) 3 find Ts = qro + T∞ . 3h < PROBLEM 3.94 KNOWN: Radial distribution of heat dissipation of a spherical container of radioactive wastes. Surface convection conditions. FIND: Radial temperature distribution. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across container wall. ANALYSIS: The appropriate form of the heat equation is 1 r2 qo d 2 dT q r =− =− dr dr k k 2 1 − r . ro q r3 r5 dT + C1 =− o − 2 dr k 3 5ro q r2 r 4 C1 − + C2 . T=− o − 2 k 6 20ro r From the boundary conditions, r2 Hence dT/dr |r=0 = 0 and − kdT/dr |r=ro = h T ( ro ) − T∞ it follows that C1 = 0 and q r2 r2 r r qo o − o = h − o o − o + C2 − T∞ k 6 20 3 5 C2 = 2 2roq o 7q o ro + + T∞ . 15h 60k 2 4 2 2ro q o qro 7 1 r 1r Hence T ( r ) = T∞ + + − + . 15h k 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 the container, where Eg = qconv . The maximum temperature exists at r = 0. PROBLEM 3.95 KNOWN: Dimensions and thermal conductivity of a spherical container. Thermal conductivity and volumetric energy generation within the container. Outer convection conditions. FIND: (a) Outer surface temperature, (b) Container inner surface temperature, (c) Temperature distribution within and center temperature of the wastes, (d) Feasibility of operating at twice the energy generation rate. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional radial conduction. ANALYSIS: (a) For a control volume which includes the container, conservation of energy yields E g − E out = 0 , or qV − q conv = 0 . Hence () ( 2 q ( 4 3 ) π ri3 = h4π ro Ts,o − T∞ 5 ) 3 and with q = 10 W/m , Ts,o = T∞ + qri3 2 3hro = 25 C + 105 W m 2 ( 0.5 m ) 3 3000 W m ⋅ K ( 0.6 m ) 2 2 < = 36.6 C . (b) Performing a surface energy balance at the outer surface, E in − E out = 0 or q cond − q conv = 0 . Hence 4π k ss Ts,i − Ts,o 2 = h4π ro Ts,o − T∞ (1 ri ) − (1 ro ) ( ) ( ) ( ) 2 1000 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 i h < (c) The heat equation in spherical coordinates is d dT 2 k rw r 2 + qr = 0 . dr dr Solving, r2 dT =− qr 3 + C1 dr 3k rw Applying the boundary conditions, dT and =0 dr r = 0 C1 = 0 and and T (r ) = − qr 2 C − 1 + C2 6k rw r T ( ri ) = Ts,i C2 = Ts,i + qri2 6k rw . Continued... PROBLEM 3.95 (Cont.) Hence (ri2 − r2 ) 6k rw q T ( r ) = Ts,i + < At r = 0, T ( 0 ) = Ts,i + qri2 6k rw = 129.4 C + 105 W m3 ( 0.5 m ) 2 6 ( 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 total thermal resistance () ri2 2 R ′′ tot,i = 4π ri R tot = R ′′ cnd,i + R ′′ cnv,i = [(1 ri ) − (1 ro )] + 1 ri 2 k ss h ro where, for ro = 0.6 m and h = 1000 W/m2⋅K, R ′′ nd,i = 5.56 × 10-3 m2⋅K/W, R ′′ nv,i = 6.94 × 10-4 m2⋅K/W, c c Center temperature, T(0) (C) and R ′′ot,i = 6.25 × 10-3 m2⋅K/W. Results for the center temperature are shown below. t 675 625 575 525 475 0 2000 4000 6000 8000 10000 Convection coefficient, h(W/m^2.K) ro = 0.54 m ro = 0.60 m Clearly, even with ro = 0.54 m = ro,min and h = 10,000 W/m2⋅K (a practical upper limit), T(0) > 475°C and the desired condition can not be met. The corresponding resistances are R ′′ nd,i = 2.47 × 10-3 m2⋅K/W, c -5 2 -3 2 R ′′ cnv,i = 8.57 × 10 m ⋅K/W, and R ′′ot,i = 2.56 × 10 m ⋅K/W. The conduction resistance remains t dominant, and the effect of reducing R ′′ nv,i by increasing h is small. The proposed extension is not c feasible. COMMENTS: A value of q = 1.79 × 105 W/m3 would allow for operation at T(0) = 475°C with ro = 2 0.54 m and h = 10,000 W/m ⋅K. PROBLEM 3.96 KNOWN: Carton of apples, modeled as 80-mm diameter spheres, ventilated with air at 5°C and experiencing internal volumetric heat generation at a rate of 4000 J/kg⋅day. FIND: (a) The apple center and surface temperatures when the convection coefficient is 7.5 W/m2⋅K, 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/m2⋅K⋅(m/s)0.425. SCHEMATIC: ASSUMPTIONS: (1) Apples can be modeled as spheres, (2) Each apple experiences flow of ventilation air at TW = 5°C, (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 uniform generation is T(r) = 2 qro r2 1 − + Ts 6k r 2 o (1) To determine Ts, perform an energy balance on the apple as shown in the sketch above, with volume V = 3 4 3 5ro , − q cv + q∀ = 0 E in − E out + E g = 0 ( )(Ts − T∞ ) + q (4 3π ro3 ) = 0 −7.5 W m 2 ⋅ K ( 4π × 0.0402 m 2 )(Ts − 5 C ) + 38.9 W m3 ( 4 3 π × 0.0403 m3 ) = 0 2 − h 4π ro (2) where the volumetric generation rate is q = 4000 J kg ⋅ day q = 4000 J kg ⋅ day × 840 kg m3 × (1day 24 hr ) × (1hr 3600 s ) q = 38.9 W m3 and solving for Ts, find < Ts = 5.14 C From Eq. (1), at r = 0, with Ts, find T(0) = 38.9 W m3 × 0.0402 m 2 6 × 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 0.425 with C1 = 10.1 W/m2⋅K⋅(m/s)0.425, and using the energy balance of Eq. (2), calculate and plot Ts as a function of ventilation air velocity V. With very low velocities, the center temperature is nearly 0.5°C higher than the air. From our earlier calculation we know that T(0) - Ts = 0.12°C and is independent of V. Center temperature, T(0) (C) 5.4 5.3 5.2 0 0.2 0.4 0.6 0.8 1 Ventilation air velocity, V (m/s) COMMENTS: (1) While the temperature within the apple is nearly isothermal, the center temperature will 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 shown below. // 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 = 0 Vol = 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 = 0.080 k = 0.5 qdotm = 4000 rho = 840 r=0 h = 7.5 //h = C1 * V^0.425 //C1 = 10.1 //V = 0.5 Tinf = 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, C PROBLEM 3.97 KNOWN: Plane wall, long cylinder and sphere, each with characteristic length a, thermal conductivity 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 smallest temperature difference between the center and the surface? Explain this behavior by comparing the ratio of the volume-to-surface area; and (c) Which shape would be preferred for use as a nuclear fuel element? Explain why? 2 SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties and (4) Uniform volumetric generation. ANALYSIS: (a) For each of the shapes, with T(a) = Ts, the dimensionless temperature distributions can be written by inspection from results in Appendix C.3. 2 T ( x ) − Ts x Plane wall, Eq. C.22 = 1− a qa 2 / 2k Long cylinder, Eq. C.23 Sphere, Eq. C.24 2 1 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 graph below. Dimensionless temperature distribution (T_x,r-Ts) / (qdot*a^2/2*k) 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 Dimensionless length, x/a or r/a Plane wall, 2a Long cylinder, a Sphere, a Continued ….. 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 is Plane wall ∀ a (1× 1) = =a As (1×1) Long cylinder ∀ π a 2 ×1 a = = As 2π a ×1 2 Sphere ∀ 4π a 3 / 3 a = = As 3 4π a 2 The 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 the generation and transfer rates, the sphere will operate at the lowest temperature. PROBLEM 3.98 KNOWN: Radius, thickness, and incident flux for a radiation heat gauge. FIND: Expression relating incident flux to temperature difference between center and edge of gauge. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible temperature 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 contact resistance between foil and heat sink. ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr, q r + q′′ ( 2π rdr ) = q r+dr , i q r = − k ( 2π rt ) dT , dr q r+dr = q r + dq r dr. dr Rearranging, find that q′′ ( 2π rdr ) = i d dT ( −k2π rt ) dr dr dr d dT q′′ i r dr = − kt r. dr Integrating, r dT q′′r 2 = − i + C1 dr 2kt and q′′r 2 T ( r ) = − i + C1lnr+C2 . 4kt With dT/dr|r=0 =0, C1 = 0 and with T(r = R) = T(R), q′′R 2 T (R ) = − i + C2 4kt or q′′R 2 C2 = T ( R ) + i . 4kt Hence, the temperature distribution is T (r ) = ( ) ′′ qi R 2 − r 2 + T ( R ). 4kt Applying this result at r = 0, it follows that q′′ = i 4kt 4kt T ( 0 ) − T ( R ) = ∆T. R2 R2 COMMENTS: This technique allows for determination of a radiation flux from measurement of a temperature difference. It becomes inaccurate if emission from the foil becomes significant. < PROBLEM 3.98 KNOWN: Radius, thickness, and incident flux for a radiation heat gauge. FIND: Expression relating incident flux to temperature difference between center and edge of gauge. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r (negligible temperature 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 contact resistance between foil and heat sink. ANALYSIS: Applying energy conservation to a circular ring extending from r to r + dr, q r + q′′ ( 2π rdr ) = q r+dr , i q r = − k ( 2π rt ) dT , dr q r+dr = q r + dq r dr. dr Rearranging, find that q′′ ( 2π rdr ) = i d dT ( −k2π rt ) dr dr dr d dT q′′ i r dr = − kt r. dr Integrating, r dT q′′r 2 = − i + C1 dr 2kt and q′′r 2 T ( r ) = − i + C1lnr+C2 . 4kt With dT/dr|r=0 =0, C1 = 0 and with T(r = R) = T(R), q′′R 2 T (R ) = − i + C2 4kt or q′′R 2 C2 = T ( R ) + i . 4kt Hence, the temperature distribution is T (r ) = ( ) ′′ qi R 2 − r 2 + T ( R ). 4kt Applying this result at r = 0, it follows that q′′ = i 4kt 4kt T ( 0 ) − T ( R ) = ∆T. R2 R2 COMMENTS: This technique allows for determination of a radiation flux from measurement of a temperature difference. It becomes inaccurate if emission from the foil becomes significant. < PROBLEM 3.99 KNOWN: 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 along absorber plate, (3) Uniform radiation absorption at plate surface, (4) Negligible losses by conduction 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/m⋅K. 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 by applying Eq.1.11a to a differential control volume. For a unit length of tube q′x + q′′ ( dx ) − q′x+dx = 0. rad With q′x+dx = q′x + and q′x = − kt dq′x dx dx dT dx it follows that, q′′ − rad d 2T dx 2 d dT − kt dx = 0 dx q′′ + rad = 0 kt Integrating twice it follows that, the general solution for the temperature distribution has the form, q′′ T ( x ) = − rad x 2 + C1x+C2 . 2kt Continued ….. PROBLEM 3.99 (Cont.) The boundary conditions are: T ( 0 ) = Tw dT =0 dx x=L/2 C2 = Tw q′′ L C1 = rad 2kt Hence, q′′ T ( x ) = rad x ( L − x ) + Tw . 2kt The maximum absorber plate temperature, which is at x = L/2, is therefore q′′ L2 Tmax = T ( L/2 ) = rad + Tw . 8kt The rate of energy collection per tube may be obtained by applying Fourier’s law at x = 0. That is, energy is transferred to the tubes via conduction through the absorber plate. Hence, dT q′=2 − k t dx x=0 where the factor of two arises due to heat transfer from both sides of the tube. Hence, q′= − Lq′′ . rad 800 W (0.2m )2 2 m + 60 C W 8 180 (0.006m ) m⋅K Hence Tmax = or Tmax = 63.7 C and q′ = −0.2m × 800 W/m 2 or q′ = −160 W/m. < < COMMENTS: Convection losses in the typical flat plate collector, which is not evacuated, would reduce the value of q ′. PROBLEM 3.100 KNOWN: Surface conditions and thickness of a solar collector absorber plate. Temperature of working fluid. FIND: (a) Differential equation which governs plate temperature distribution, (b) Form of the temperature distribution. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Adiabatic bottom surface, (4) Uniform radiation flux and convection coefficient at top, (5) Temperature of absorber plate at x = 0 corresponds to that of working fluid. ANALYSIS: (a) Performing an energy balance on the differential control volume, q′x + dq′ = q′x+dx + dq′ rad conv where q′x+dx = q′x + ( dq′x / dx ) dx dq′ = q′′ ⋅ dx rad rad dq′ = h ( T − T∞ ) ⋅ dx conv Hence, q′′ dx= ( dq′ / dx ) dx+h ( T − T∞ ) dx. rad x From Fourier’s law, the conduction heat rate per unit width is q′′ d 2T h q′x = − k t dT/dx − (T − T∞ ) + rad = 0. 2 dx kT < kt (b) Defining θ = T − T∞ , d 2T/dx 2 = d 2θ / dx 2 and the differential equation becomes, d 2θ dx 2 − q′′ h θ + rad = 0. kt kt It is a second-order, differential equation with constant coefficients and a source term, and its general solution is of the form θ = C1e+λ x + C2e-λ x + S/λ 2 1/ 2 λ = ( h/kt ) , S=q′′ / kt. where rad Appropriate boundary conditions are: θ (0 ) = To − T∞ ≡ θ 0 , dθ /dx) x=L = 0. θ o = C1 + C2 + S/λ 2 Hence, dθ /dx) x=L = C1 λ e+λ L − C2 λ e-λ L = 0 )( ( C1 = θ 0 − S/λ 2 / 1 + e2λ L Hence, ( θ = θ 0 − S/λ 2 ) ) C2 = C1 e2λ L ( )( C2 = θ 0 − S/λ 2 / 1 + e-2λ L eλ x e -λ x + + S/λ 2 . 2λ L 1+e-2λ L 1+e ) < PROBLEM 3.101 KNOWN: Dimensions of a plate insulated on its bottom and thermally joined to heat sinks at its ends. Net heat flux at top surface. FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperature distribution and heat loss to heat sinks. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in x (W,L>>t), (3) Constant properties, (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. o o From Fourier’s law, q x = − k ( t ⋅ W ) dT/dx. Hence, the differential equation for the temperature distribution is − q′′ + o = 0. dx 2 kt d 2T d dT o ktW dx − q′′ W=0 dx < (b) Integrating twice, the general solution is, q′′ T ( x ) = − o x 2 + C1 x +C2 2kt and appropriate boundary conditions are T(0) = To, and T(L) = To. Hence, To = C2, and q′′ To = − o L2 + C1L+C2 2kt and q′′ L C1 = o . 2kt Hence, the temperature distribution is ( ) q′′ L T ( x ) = − o x 2 − Lx + To . 2kt < Applying Fourier’s law at x = 0, and at x = L, q′′ WL L q′′ q ( 0 ) = − k ( Wt ) dT/dx) x=0 = −kWt − o x − =− o 2 x=0 2 kt q′′ WL L q′′ q ( L ) = − k ( Wt ) dT/dx) x=L = − kWt − o x − =+ o 2 x=L 2 kt Hence the heat loss from the plates is q=2 ( q′′ WL/2 ) = q′′ WL. o o < 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 3.102 KNOWN: Dimensions and surface conditions of a plate thermally joined at its ends to heat sinks at different temperatures. FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperature distribution and an expression for the heat rate from the plate to the sinks, and (c) Compute and plot temperature 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 contact resistance. ANALYSIS: (a) Applying conservation of energy to the differential control volume q x + dq o = q x + dx + dq conv where q x + dx = q x + ( dq x dx ) dx dq conv = h ( T − T∞ )( W ⋅ dx ) Hence, dq x q x + q ′′ ( W ⋅ dx ) = q x + ( dq x dx ) dx + h ( T − T∞ )( W ⋅ dx ) o dx Using Fourier’s law, q x = − k ( t ⋅ W ) dT dx , − ktW d2T + hW ( T − T∞ ) = q′′ o d 2T − h dx 2 dx 2 kt (b) Introducing θ ≡ T − T∞ , the differential equation becomes (T − T∞ ) + q′′ o kt + hW ( T − T∞ ) = q ′′ W . o < = 0. q′′ θ + o =0. kt dx 2 kt This differential equation is of second order with constant coefficients and a source term. With d 2θ − h λ 2 ≡ h kt and S ≡ q′′ kt , it follows that the general solution is of the form o θ = C1e + λ x + C2 e−λ x + S λ 2 . Appropriate boundary conditions are: θ (0) = To − T∞ ≡ θ o (1) θ (L) = TL − T∞ ≡ θ L (2,3) Substituting the boundary conditions, Eqs. (2,3) into the general solution, Eq. (1), θ o = C1e0 + C2e0 + S λ 2 θ L = C1e + λ L + C2e − λ L + S λ 2 To solve for C2, multiply Eq. (4) by -e+λL and add the result to Eq. (5), ( )( ) C2 = (θ L − θ o e + λ L ) − S λ 2 ( −e + λ L + 1) ( −e+ λ L + e −λ L ) (4,5) −θ o e + λ L + θ L = C2 −e + λ L + e − λ L + S λ 2 −e + λ L + 1 (6) Continued... PROBLEM 3.102 (Cont.) Substituting for C2 from Eq. (6) into Eq. (4), find {( ) ) (−e+λL + e−λL )} − S λ 2 ( C1 = θ o − θ L − θ o e+ λ L − S λ 2 −e + λ L + 1 (7) Using C1 and C2 from Eqs. (6,7) and Eq. (1), the temperature distribution can be expressed as θ (x) = e +λ x − ( ) ( sinh ( λ x ) + λ L sinh ( λ x ) + λ L sinh ( λ x ) +λ L e + 1− e θ o + sinh ( λ L ) θ L + − 1 − e sinh ( λ L ) sinh ( λ L ) ) λS (8) 2 < The heat rate from the plate is q p = −q x ( 0 ) + q x ( L ) and using Fourier’s law, the conduction heat rates, with Ac = W⋅t, are q x ( 0 ) = − kAc dθ eλ L λ λ θo + θL = − kA c λ e0 − dx x = 0 sinh ( λ L ) sinh (λ L ) 1 − e+ λ L + − sinh (λ L ) q x ( L ) = − kAc S λ2 λ − λ < λ cosh ( λ L ) dθ eλ L λ cosh ( λ L ) θ o + θL = − kAc λ eλ L − dx x = L sinh ( λ L ) sinh ( λ L ) 1 − e+ λ L + − sinh (λ L ) S λ2 λ cosh ( λ L ) − λ e + λ L < (c) For the prescribed base-case conditions listed below, the temperature distribution (solid line) is shown in the accompanying plot. As expected, the maximum temperature does not occur at the midpoint, but slightly toward the x-origin. The sink heat rates are q′′ ( 0 ) = −17.22 W x q′′ ( L ) = 23.62 W x < Temperature, T(x) (C) 300 200 100 0 0 20 40 60 80 100 Distance, x (mm) q''o = 20,000 W/m^2; h = 50 W/m^2.K q''o = 30,000 W/m^2; h = 50 W/m^2.K q''o = 20,000 W/m^2; h = 200 W/m^2.K q''o = 4927 W/m^2 with q''x(0) = 0; h = 200 W/m^2.K The 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/m2⋅K, (iii) the value o of q′′ for which q′′ (0) = 0 with h = 200 W/m2⋅K. The condition for the last curve is q′′ = 4927 W/m2 x o o for which the temperature gradient at x = 0 is zero. Base case conditions are: q′′ = 20,000 W/m2, To = 100°C, TL = 35°C, T∞ = 25°C, k = 25 W/m⋅K, h = 50 o W/m2⋅K, L = 100 mm, t = 5 mm, W = 30 mm. PROBLEM 3.103 KNOWN: Thin plastic film being bonded to a metal strip by laser heating method; strip dimensions and thermophysical 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 is 10,000 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 convection coefficients, (5) Edges of metal strip are at air temperature (T∞), that is, strip behaves as infinite fin so that w2 → ∞, (6) All the incident laser heating flux q′′ is absorbed by the film. o PROPERTIES: Metal strip (given): ρ = 7850 kg/m3, cp = 435 J/kg⋅m3, k = 60 W/m⋅K. ANALYSIS: (a) The strip-plastic film arrangement can be modeled as an infinite fin of uniform cross section a portion of which is exposed to the laser heat flux on the upper surface. The general solutions for the two regions of the strip, in terms of θ ≡ T ( x ) − T∞ , are θ1 ( x ) = C1e 0 ≤ x ≤ w1 2 + mx + C 2e − mx +M m θ 2 ( x ) = C3 e + mx + C4 e (1) m = ( 2h kd ) M = q ′′ P 2kA c = q ′′ kd o o w1 2 ≤ x ≤ ∞ 2 1/ 2 − mx . Four boundary conditions can be identified to evaluate the constants: dθ1 At x = 0: (0 ) = 0 = C1me0 − C 2 me−0 + 0 dx θ ( w1 2 ) = θ 2 ( w1 2 ) At x = w1/2: C1e + m w1 2 + C 2e − m w1 2 → 2 + mw1 2 dθ1 ( w1 2 ) / dx = dθ 2 ( w1 2 ) / dx At 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 4e Mm 2e (5) − m w1 2 − mC 4 e ∞ −∞ At x → ∞: θ 2 ( ∞ ) = 0 = C3 e + C 4 e → C3 = 0 With C3 = 0 and C1 = C2, combine Eqs. (6 and 7) to eliminate C4 to find C1 = C 2 = − (2,3) − mw1 2 (6) (7) (8) 2 m w1 2 . (9) and using Eq. (6) with Eq. (9) find C 4 = M m sinh ( mw1 2 ) e 2 − mx1 / 2 (10) Continued... PROBLEM 3.103 (Cont.) Hence, the temperature distribution in the region (1) under the plastic film, 0 ≤ x ≤ w1/2, is 2 Mm θ1 ( x ) = − 2e m w1 w (e + mx +e − mx M M ) + m = m (1 − e 2 − m w1 2 2 cosh mx ) (11) < and for the region (2), x ≥ w1/2, θ2 (x ) = M m 2 sinh ( mw1 2 ) e − mx (12) (b) Substituting numerical values into the temperature distribution expression above, θ1(0) and θ1(w1/2) can be determined. First evaluate the following parameters: M = 10, 000 W m ( 2 60 W m ⋅ K × 0.00125 m = 133, 333 K m 2 m = 2 × 10 W m ⋅ K 60 W m ⋅ K × 0.00125 m ) 1/ 2 2 = 16.33 m −1 Hence, for the midpoint x = 0, θ1 ( 0 ) = 133, 333 K m ( 16.33 m ) 2 −1 2 ( ) 1 − exp −16.33 m −1 × 0.020 m × cosh ( 0 ) = 139.3 K < T1 ( 0 ) = θ1 ( 0 ) + T∞ = 139.3 K + 25 C = 164.3 C . For the position x = w1/2 = 0.020 m, ( θ1 ( w1 2 ) = 500.0 1 − 0.721cosh 16.33 m −1 × 0.020 m ) = 120.1K < T1 ( w1 2 ) = 120.1K + 25 C = 145.1 C . (c) The temperature distributions, θ1(x) and θ2(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; symmetrical distribution. (3) Temperature excess and gradient approach zero with increasing value of x. Strip temperature, T (C) (2) No discontinuity of gradient at w1/2 (20 mm). 180 140 100 60 20 0 50 100 150 200 250 300 x-coordinate, x (mm) Region 1 - constant heat flux, q''o Region 2 - x >= w1/2 COMMENTS: How wide must the strip be in order to satisfy the infinite fin approximation such that θ2 (x → ∞) = 0? For x = 200 mm, find θ2(200 mm) = 6.3°C; this would be a poor approximation. When x = 300 mm, θ2(300 mm) = 1.2°C; hence when w2/2 = 300 mm, the strip is a reasonable approximation to an infinite fin. PROBLEM 3.104 KNOWN: Thermal conductivity, diameter and length of a wire which is annealed by passing an electrical current through the wire. FIND: Steady-state temperature distribution along wire. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction along the wire, (3) Constant properties, (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 = 0 q x+dx = q x + dq x dx dx ( ) Eg = q (π D2 / 4 ) dx. q x = − k π D2 / 4 dT/dx dq conv = h (π D dx ) ( T − T∞ ) Hence, ( k π D2 / 4 ) dx 2 dx+q (π D2 / 4)dx − h (π Ddx ) (T − T∞ ) = 0 d 2T d 2θ or, with θ ≡ T − T∞ , dx 2 − 4h q θ + =0 kD k The solution (general and particular) to this nonhomogeneous equation is of the form q θ = C1 emx + C2 e-mx + km 2 2 where m = (4h/kD). The boundary conditions are: dθ = 0 = m C1 e0 − mC2 e0 → dx x=0 ( ) q θ ( L ) = 0 = C1 emL + e-mL + → km 2 C1 = C2 C1 = −q/km 2 emL + e-mL = C2 The temperature distribution has the form T = 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 check the result, note that T(L) = T(-L) = T∞. PROBLEM 3.105 KNOWN: Electric power input and mechanical power output of a motor. Dimensions of housing, mounting pad and connecting shaft needed for heat transfer calculations. Temperature of ambient air, tip of shaft, and base of pad. FIND: Housing temperature. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in pad and shaft, (3) Constant properties, (4) Negligible radiation. ANALYSIS: Conservation of energy yields Pelec − Pmech − q h − q p − q s = 0 q h = h h A h ( Th − T∞ ) , θ L = 0, qs = ( π 2 / 4 D3h k ss ( t cosh mL − θ L / θ b sinh mL 1/ 2 1/ 2 mL = 4h s L2 / k s D , Hence qs = M π2 3 M= D hsks 4 ) ( (Th − T∞ ) , qp = k p W 2 ) 1/ 2 (Th − T∞ ). (Th − T∞ ) tanh 4h s L2 / k s D ) 1/ 2 Substituting, and solving for (Th - T∞), Th − T∞ = (( ) ( ) 1/ 2 (4hsL2 / ksD) = 3.87, h h A h + k p W 2 / t+ π 2 / 4 D3h s k s ((π / 4) D h k ) 2 Pelec − Pmech 3 1/ 2 ss = 6.08 W/K, 1/ 2 / tanh 4h s L2 / k s D ) 1/ 2 tanhmL=0.999 104 W ( 25 − 15 ) ×103 W Th − T∞ = = 10 × 2+0.5 ( 0.7 )2 / 0.05 + 6.08 / 0.999 W/K ( 20+4.90+6.15 ) W/K Th − T∞ = 322.1K Th = 347.1 C < COMMENTS: (1) Th is large enough to provide significant heat loss by radiation from the ( ) 4 4 housing. Assuming an emissivity of 0.8 and surroundings at 25°C, q rad = ε A h Th − Tsur = 4347 W, which compares with q conv = hA h ( Th − T∞ ) = 5390 W. Radiation has the effect of decreasing Th. (2) The infinite fin approximation, qs = M, is excellent. PROBLEM 3.106 KNOWN: Dimensions and thermal conductivity of pipe and flange. Inner surface temperature of pipe. 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 is q= Ts,i − T∞ R t,w + R t,f = Ts,i − T∞ n ( Do / Di ) / 4π tk + (1/ hAf ηf ) Since convection heat transfer only occurs from one surface of a flange, the connected flanges may be modeled as a single annular fin of thickness t ′ = 2t = 30 mm. Hence, r2c = ( Df / 2 ) + t ′ / 2 = 0.140 m, ( )( ) ( ) 2 2 2 A f = 2π r2c − r1 = 2π r2c − Do / 2 = 2 0.1402 − 0.062 m 2 = 0.101m 2 , Lc = L + t ′ / 2 = ( Df ( ) 1/ 2 2/2 2 h / kA p = 0.188. With r2c/r1 = − Do ) / 2 + t = 0.065 m, A p = Lc t ′ = 0.00195 m , Lc r2c/(Do/2) = 1.87, Fig. 3.19 yields ηf = 0.94. Hence, q= q= 300°C − 20°C ( n (1.25 ) / 4π × 0.03m × 40 W / m ⋅ K + 1/10 W / m 2 ⋅ K × 0.101m 2 × 0.94 280°C = 262 W (0.0148 + 1.053) K / W ) < COMMENTS: Without the flange, heat transfer from a section of pipe of width t ′ = 2t is −1 q = Ts,i − T∞ / R t,w + R t,cnv , where R t,cnv = ( h × π Do t ′ ) = 7.07 K / W. Hence, q = 39.5 W, and there is significant heat transfer enhancement associated with the extended surfaces afforded by the flanges. ( )( ) PROBLEM 3.107 KNOWN: TC wire leads attached to the upper and lower surfaces of a cylindrically shaped solder bead. Base of bead attached to cylinder head operating at 350°C. Constriction resistance at base and TC wire convection conditions specified. FIND: (a) Thermal circuit that can be used to determine the temperature difference between the two intermediate 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 the model. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in solder bead; no losses from lateral and top surfaces; (3) TC wires behave as infinite fins, (4) Negligible thermal contact resistance between TC wire terminals and bead. ANALYSIS: (a) The thermal circuit is shown above. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. The thermal resistances are as follows: Constriction (con) resistance, see Table 4.1, case 10 R con = 1/ ( 2k bead Dsol ) = 1/ ( 2 × 40 W / m ⋅ K × 0.006 m ) = 2.08 K / W TC (tc) wires, infinitely long fins; Eq. 3.80 −0.5 R tc,1 = R tc,2 = R fin = ( hPk w Ac ) ( P = π D w , A c = π D2 / 4 w R tc = 100 W / m 2 ⋅ K × π 2 × ( 0.003 m ) × 70 W / m ⋅ K / 4 3 ) −0.5 = 46.31 K / W Solder bead (sol), cylinder Dsol and Lsol R sol = Lsol / ( k sol Asol ) ( 2 Asol = π Dsol / 4 ) R sol = 0.010 m / 10 W / m ⋅ K × π (0.006 m ) / 4 = 35.37 K / W 2 (b) Perform energy balances on the 1- and 2-nodes, solve the equations simultaneously to find T1 and T2, from which (T1 – T2) can be determined. Continued ….. PROBLEM 3.107 (Cont.) Node 1 T2 − T1 Thead − T1 T∞ − T1 + + =0 R sol R con R tc,1 Node 2 T∞ − T2 T1 − T2 + =0 R tc,2 R sol Substituting numerical values with the equations in the IHT Workspace, find T1 = 359°C T2 = 199.2°C T1 − T2 = 160°C COMMENTS: (1) With this arrangement, the TC indicates a systematically low reading of the cylinder 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 of which would be to increase our estimate for (T1 – T2). Constriction resistance is important; note that Thead – T1 = 26°C. PROBLEM 3.108 KNOWN: Rod (D, k, 2L) that is perfectly insulated over the portion of its length –L ≤ x ≤ 0 and experiences convection (T∞, h) over the portion 0 ≤ x ≤ + L. One end is maintained at T1 and the other is separated from a heat sink at T3 with an interfacial thermal contact resistance R ′′c . t FIND: (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 geometric parameters of the system, (c) Using, numerical values, calculate T2 and plot the temperature distribution. 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 insulated portion of the rod, the temperature distribution is linear. A temperature drop occurs across the thermal 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 a fin of uniform cross-sectional area, Eq. 3.66. θ ( x ) = C1emx + C2e− mx 0 ≤ x ≤ +L (1) where m = (hP/kAc) conditions. 1/2 and θ = T(x) - T∞. The arbitrary constants are determined from the boundary At x = 0, thermal resistance of rod q x ( 0 ) = − kAc θ − θ (0 ) dθ = kAc 1 dx x = 0 L m C1e0 − m C 2e0 = ( θ1 = T1 − T∞ ) 1 θ − C1e0 + C2e0 1 L (2) Continued ….. PROBLEM 3.108 (Cont.) At x=L, thermal contact resistance q x ( + L ) = − kAc θ (L ) − θ3 dθ = dx x = L R ′′ / A c tc θ 3 = T3 − T∞ 1 mL − k m C1emL − m C2e− mL = + C 2e− mL − θ 3 R ′′ C1e tc (3) Eqs. (2) and (3) cannot be rearranged easily to find explicit forms for C1 and C2. The constraints will be evaluated numerically in part (c). Knowing C1 and C2, Eq. (1) gives θ 2 = θ (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 = 62.1°C. The temperature distribution is shown in the graph below. Temperature distribution in rod Temperature, T(x) (C) 200 150 100 50 0 -50 -30 -10 10 30 50 x-coordinate, x (mm) COMMENTS: (1) The purpose of asking you to sketch the temperature distribution in part (a) was to give you the opportunity to identify the relevant thermal processes and come to an understanding of the 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; t t that is, k is very large. PROBLEM 3.109 KNOWN: Long rod in oven with air temperature at 400°C has one end firmly pressed against surface of a billet; thermocouples imbedded in rod at locations 25 and 120 mm from the billet indicate 325 and 375°C, 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 temperature distribution is θ ( x ) = θ be-mx (1) where θ ( x ) = T ( x ) − T∞ θ b = T (0 ) − T∞ = Tb − T∞ . Substituting values for T1 and T2 at their respective distances, x1 and x2, into Eq. (1), it is possible to evaluate m, θ ( x1 ) θ be-mx1 -m x − x = = e ( 1 2) -mx 2 θ (x2 ) θ e b (325 − 400 ) C = e-m(0.025−0.120 )m (375-400 ) C m=11.56. Using the value for m with Eq. (1) at location x1, it is now possible to determine the rod base or billet temperature, θ ( x1 ) = T1 − T∞ = (Tb − T∞ ) e-mx (325 − 400 ) C= (Tb − 400 ) C e−11.56×0.025 Tb = 300 C. COMMENTS: Using the criteria mL ≥ 2.65 (see Example 3.8) for the infinite fin approximation, the insertion length should be ≥ 229 mm to justify the approximation, < PROBLEM 3.110 KNOWN: Temperature sensing probe of thermal conductivity k, length L and diameter D is mounted on a duct wall; portion of probe Li is exposed to water stream at T∞,i while other end is exposed to ambient 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 and T∞,o , calculate ∆Terr for immersion to total length ratios of 0.225, 0.425, and 0.625, (c) Compute and 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 is thermally isolated from the duct, (4) Convection coefficients are uniform over their respective regions. PROPERTIES: Probe material (given): k = 177 W/m⋅K. ANALYSIS: (a) To derive an expression for ∆Terr = Ttip - T∞,i , we need to determine the temperature distribution in the immersed length of the probe Ti(x). Consider the probe to consist of two regions: 0 ≤ xi ≤ Li, the immersed portion, and 0 ≤ xo ≤ (L - Li), the ambient-air portion where the origin corresponds to the location of the duct wall. Use the results for the temperature distribution and 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 ) θi = = θ b,i To − T∞,i cosh ( mi Li ) + ( h i mi k ) sinh ( mi Li ) (1) and the tip temperature, Ttip = Ti(Li) at xi = Li, is Ttip − T∞,i To − 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 the fin heat rates equal, that is, q f ,o q f ,i (4) Continued... PROBLEM 3.110 (Cont.) ( h o PkAc )1/ 2 θ b,o ⋅ B = − ( hi PkAc )1/ 2 θ b,i ⋅ C Solving for To, find θ b,o θ b,i = To − T∞,o To − T∞,i = − ( h i PkAc ) 1/ 2 1/ 2 hi C To = T∞,o + T∞,i B ho θ b,i ⋅ C 1/ 2 hi C 1 + 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 these calculations are summarized below. Li/L 0.225 Lo (mm) 155 Li (mm) 45 A 0.2328 B 0.5865 C 0.9731 To (°C) 76.7 ∆Terr (°C) -0.76 0.425 115 85 0.0396 0.4639 0.992 77.5 -0.10 0.625 75 125 0.0067 0.3205 0.9999 78.2 -0.01 2.5 Temperature error, Tinfo - Ttip (C) (c) The probe behaves as a fin having ends exposed to the cool ambient air and the hot ambient water whose temperature is to be measured. If the thermal conductivity is decreased, heat transfer along the probe length is likewise decreased, the tip temperature will be closer to the water temperature. If the velocity of the water decreases, the convection coefficient will decrease, and the difference between the tip and water temperatures will increase. < < < 2 1.5 1 0.5 0 0.2 0.3 0.4 0.5 0.6 Immersion length ratio, Li/L Base case: k = 177 W/m.K; ho = 1100 W/m^2.K Low velocity flow: k = 177 W/m.K; ho = 500 W/m^2.K Low conductivity probe: k = 50 W/m.K; ho = 1100 W/m^2.K 0.7 PROBLEM 3.111 KNOWN: Rod protruding normally from a furnace wall covered with insulation of thickness Lins with the length Lo exposed to convection with ambient air. FIND: (a) An expression for the exposed surface temperature To as a function of the prescribed thermal and geometrical parameters. (b) Will a rod of Lo = 100 mm meet the specified operating limit, T0 ≤ 100°C? If not, what design parameters would you change? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in rod, (3) Negligible thermal 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 the rod, Lo, (6) Adiabatic tip condition for the rod and (7) Negligible radiation exchange between rod and its surroundings. ANALYSIS: (a) The rod can be modeled as a thermal network comprised of two resistances in series: 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 insulated section: R ins = Lins kA c (1) For the fin, Table 3.4, Case B, Eq. 3.76, 1 R fin = θ b q f = ( hPkAc )1/ 2 tanh ( mLo ) (2) m = ( hP kA c ) Ac = π D2 4 P = πD From the thermal network, by inspection, To − T∞ T − T∞ R fin To = T∞ + =w (Tw − T∞ ) R fin R ins + R fin R ins + R fin (b) Substituting numerical values into Eqs. (1) - (6) with Lo = 200 mm, 6.298 To = 25 C + ( 200 − 25 ) C = 109 C 6.790 + 6.298 1/ 2 R ins = 0.200 m 60 W m ⋅ K × 4.909 × 10 R fin = 1 ( ( hPkA c ) = −4 m 2 0.0347 W 2 K 2 (15 W m 2 = 6.790 K W ) 1/ 2 A c = π ( 0.025 m ) 2 (3,4,5) (6) < < 4 = 4.909 × 10 −4 m 2 tanh ( 6.324 × 0.200 ) = 6.298 K W ⋅ K × π ( 0.025 m ) × 60 W m ⋅ K × 4.909 × 10 −4 m 2 ) = 0.0347 W 2 K 2 Continued... PROBLEM 3.111 (Cont.) m = ( hP kA c ) 1/ 2 ( = 15 W m ⋅ K × π ( 0.025 m ) 60 W m ⋅ K × 4.909 × 10 2 −4 m 2 ) 1/ 2 = 6.324 m −1 Consider the following design changes aimed at reducing To ≤ 100°C. (1) Increasing length of the fin portions: with Lo = 200 mm, the fin already behaves as an infinitely long fin. Hence, increasing Lo will not result in reducing To. (2) Decreasing the thermal conductivity: backsolving the above equation set with T0 = 100°C, find the required thermal conductivity is k = 14 W/m⋅K. Hence, we could select a stainless steel alloy; see Table A.1. (3) Increasing the insulation thickness: find that for To = 100°C, the required insulation thickness would be Lins = 211 mm. This design solution might be physically and economically unattractive. (4) A very practical solution would be to introduce thermal contact resistance between the rod base and the furnace wall by “tack welding” (rather than a continuous bead around the rod circumference) the rod in two or three places. (5) A less practical solution would be to increase the convection coefficient, since to do so, would require an air handling unit. 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 an adiabatic 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 Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 // Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tw // Furnace wall temperature, C //q1 = // Heat rate, W T2 = To // To, beginning of rod exposed length q2 = 0 // Heat rate, W; node 2; no external heat source T3 = Tinf // Ambient air temperature, C //q3 = // Heat rate, W // Thermal Resistances: // Rod - conduction resistance R21 = Lins / (k * Ac) // Conduction resistance, K/W Ac = 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 thermal conductivity 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 3.4 m = sqrt(h*P / (k*Ac)) P = pi * D // Perimeter, m // Other Assigned Variables: Tw = 200 // Furnace wall temperature, C k = 60 // Rod thermal conductivity, W/m.K Lins = 0.200 // Insulated length, m D = 0.025 // Rod diameter, m h = 15 // Convection coefficient, W/m^2.K Tinf = 25 // Ambient air temperature,C Lo = 0.200 // Exposed length, m PROBLEM 3.112 KNOWN: Rod (D, k, 2L) inserted into a perfectly insulating wall, exposing one-half of its length to an 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 region may be approximated as a very long fin; (b) Derive an expression for To at the end of the imbedded half of the rod, and (c) Using numerical values, plot the temperature distribution in the rod and describe its key features. Does the rod behave as a very long fin? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in imbedded portion of rod, (3) Imbedded portion of rod is perfectly insulated, (4) Exposed portion of rod behaves as an infinitely long fin, and (5) Constant properties. ANALYSIS: (a) Since the exposed portion of the rod (0 ≤ x ≤ + L) behaves as an infinite fin, the fin heat rate using Eq. 3.80 is q 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) 2 Combining Eqs. (1) and (2), with P = πD and Ac = πD /4, find −1/ 2 −1/ 2 Tb = 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 with uniform q . From Eq. 3.43, To = qL2 + Tb 2k (c) The temperature distribution T(x) for the rod is piecewise parabolic and exponential, T ( x ) − Tb = T ( x ) − T∞ Tb − T∞ qL2 x 2 2k L = exp ( −mx ) −L ≤ x ≤ 0 0 ≤ x ≤ +L Continued ….. < < PROBLEM 3.112 (Cont.) The gradient at x = 0 will be continuous since we used this condition in evaluating Tb. The distribution is shown below with To = 105.4°C and Tb = 55.4°C. T(x) over embedded and exposed portions of rod 120 Temperature, T(x) 100 80 60 40 20 -50 -40 -30 -20 -10 0 10 Radial position, x 20 30 40 50 PROBLEM 3.113 KNOWN: Very long rod (D, k) subjected to induction heating experiences uniform volumetric generation ( 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 in portion of rod within the coil; no convection from lateral surface of rod, (3) Exposed portions of rod behave as infinitely long fins, and (4) Constant properties. ANALYSIS: The portion of the rod within the coil, 0 ≤ x ≤ + L, experiences one-dimensional conduction with uniform generation. From Eq. 3.43, qL2 To = + Tb 2k (1) The portion of the rod beyond the coil, L ≤ x ≤ ∞, behaves as an infinitely long fin for which the heat rate from Eq. 3.80 is q f = q x ( L ) = ( hPkAc ) 1/ 2 (Tb − T∞ ) (2) 2 where P = πD and Ac = πD /4. From an overall energy balance on the imbedded portion of the rod as illustrated in the schematic above, find the heat rate as Ein − Eout + Egen = 0 −q f + qA c L = 0 q f = qAc L (3) Combining Eqs. (1-3), −1/ 2 c Tb = T∞ + qA1/ 2 L ( hPk ) (4) qL2 −1/ 2 c To = T∞ + + qA1/ 2 L ( hPk ) 2k (5) and substituting numerical values find To = 305°C Tb = 272°C < PROBLEM 3.114 KNOWN: Dimensions, end temperatures and volumetric heating of wire leads. Convection coefficient and ambient temperature. FIND: (a) Equation governing temperature distribution in the leads, (b) Form of the temperature distribution. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in x, (3) Uniform volumetric heating, (4) Uniform h (both sides), (5) Negligible radiation. ANALYSIS: (a) Performing an energy balance for the differential control volume, Ein − E out + Eg = 0 dT dT d dT − − kAc − kAc dx − hPdx (T − T∞ ) + qAc dx = 0 dx dx dx dx − kAc d 2T dx 2 q x − q x + dx − dq conv + qdV = 0 − hP q (T − T∞ ) + = 0 kAc k < (b) With a reduced temperature defined as Θ ≡ T − T∞ − ( qA c /hP ) and m 2 ≡ hP/kA c , the differential equation may be rendered homogeneous, with a general solution and boundary conditionsas shown d 2Θ Θ ( x ) = C1emx + C2e−mx − m 2Θ = 0 2 dx Θb = C1 + C2 Θc = C1emL + C2e−mL it follows that C1 = Θb e−mL − Θc e− mL − emL C2 = Θc − Θb emL e− mL − emL (Θbe−mL − Θc ) emx + (Θc − ΘbemL )e−mx Θ (x ) = e− mL − emL COMMENTS: If q is large and h is small, temperatures within the lead may readily exceed the prescribed boundary temperatures. < PROBLEM 3.115 KNOWN: Disk-shaped electronic device (D, Ld, kd) dissipates electrical power (Pe) at one of its surfaces. Device is bonded to a cooled base (To) using a thermal pad (Lp, kA). Long fin (D, kf) is bonded 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 the heat-generating device, Td, in terms of circuit thermal resistance, To and T∞; write expressions for the thermal resistances; and (c) Calculate Td for the prescribed conditions. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through thermal pads and device; no losses from lateral surfaces; (3) Fin is infinitely long, (4) Negligible contact resistance between components of the system, and (5) Constant properties. ANALYSIS: (a) The thermal circuit is shown below with thermal resistances associated with conduction (pads, Rp; device, Rd) and for the long fin, Rf. (b) To obtain an expression for Td, perform an energy balance about the d-node Ein − Eout = q a + q b + Pe = 0 (1) Using the conduction rate equation with the circuit qa = To − Td Rf + Rd T −T qb = ∞ d R p + Rf Combine 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) 2 where the thermal resistances with P = πD and Ac = πD /4 are −1/ 2 R f = ( hPk f Ac ) R d = Ld / k d A c (c) Substituting numerical values with the foregoing relations, find R p = Lp / k p Ac R p = 1.061 K / W R d = 4.244 K / W (5,6,7) R f = 5.712 K / W and the device temperature as Td = 62.4°C < COMMENTS: What fraction of the power dissipated in the device is removed by the fin? Answer: qb/Pe = 47%. PROBLEM 3.116 KNOWN: Dimensions and thermal conductivity of a gas turbine blade. Temperature and convection coefficient 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) Adiabatic blade 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 yields T ( L ) − T∞ Tb − T∞ = 1 cosh mL m = ( hP/kA c ) 1/ 2 ( = 250W/m 2 ⋅ K × 0.11m/20W/m ⋅ K × 6 ×10−4 m 2 m = 47.87 m-1 and ) 1/ 2 mL = 47.87 m-1 × 0.05 m = 2.39 From 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 Θ b = 250W/m 2 ⋅ K × 0.11m × 20W/m ⋅ K × 6 × 10 −4 m 2 ) (−900 C ) = −517W , 1/ 2 Eq. 3.76 and Table B.1 yield qf = M tanh mL = −517W ( 0.983) = −508W Hence, q b = −q f = 508W < COMMENTS: Radiation losses from the blade surface and convection from the tip will contribute to reducing the blade temperatures. PROBLEM 3.117 KNOWN: Dimensions of disc/shaft assembly. Applied angular velocity, force, and torque. Thermal conductivity and inner temperature of disc. FIND: (a) Expression for the friction coefficient µ, (b) Radial temperature distribution in disc, (c) Value of µ 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 of disc. ANALYSIS: (a) The normal force acting on a differential ring extending from r to r+dr on the contact surface of the disc may be expressed as dFn = p2π rdr . Hence, the tangential force is dFt = µ p2π rdr , in which case the torque may be expressed as dτ = 2πµ pr 2dr For the entire disc, it follows that r 2π 3 µ pr2 τ = 2πµ p 2 r 2dr = o 3 ∫ 2 where p = F π r2 . Hence, µ= 3τ 2 Fr2 < (b) Performing an energy balance on a differential control volume in the disc, it follows that qcond,r + dq fric − q cond,r + dr = 0 ( ) ( ) 2 With dq fric = ω dτ = 2 µ Fω r 2 r2 dr , q cond,r + dr = q cond,r + dq cond,r dr dr , and qcond,r = − k ( 2π rt ) dT dr , it follows that d ( rdT dr ) 2 2 µ Fω r 2 r2 dr + 2π kt dr = 0 dr ( or d ( rdT dr ) dr Integrating twice, ) =− µ Fω 2 r 2 π ktr2 Continued... PROBLEM 3.117 (Cont.) dT µ Fω 2 C1 r+ =− 2 dr r 3π ktr2 T=− µ Fω 2 9π ktr2 r3 + C1nr + C2 Since the disc is well insulated at r = r2 , dT dr C1 = r2 = 0 and µ Fω r2 3π kt With T ( r1 ) = T1 , it also follows that C2 = T1 + µ Fω r3 − C1nr1 21 9π ktr2 Hence, T ( r ) = T1 − µ Fω 2 9π ktr2 (r3 − r13 ) + µ3Fπωktr2 n rr1 < (c) For the prescribed conditions, µ= 3 8N ⋅ m = 0.333 2 200N ( 0.18m ) < Since the maximum temperature occurs at r = r2, µ Fω r2 Tmax = T ( r2 ) = T1 − 9π kt 3 1 − r1 + µ Fω r2 n r2 r2 3π kt r1 With ( µ Fω r2 3π kt ) = ( 0.333 × 200N × 40rad/s × 0.18m 3π × 15W/m ⋅ K × 0.012m ) = 282.7 C , 3 282.7 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 C COMMENTS: The maximum temperature is excessive, and the disks should be actively cooled (by convection) at their outer surfaces. < PROBLEM 3.118 KNOWN: 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 is negligible compared to the longitudinal gradient, such that the 1-D analysis of Section 3.6.1 is valid by finding: (a) An expression for the conduction heat flux at the surface, q′′ ( t ) , in terms of Ts and y To, assuming the transverse temperature distribution is parabolic, (b) An expression for the convection heat flux at the surface for the x-location; equate the two expressions, and identify the parameter that determines the ratio (To – Ts)/(Ts - T∞); and (c) Developing a criterion for establishing the validity of the 1-D assumption used to model an extended surface. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform convection coefficient and (3) Constant properties. ANALYSIS: (a) Referring to the schematics above, the conduction heat flux at the surface y = t at any x-location follows from Fourier’s law using the parabolic transverse temperature distribution. q′′ ( t ) = − k y 2y 2k ∂T Ts ( x ) − To ( x ) = − k Ts ( x ) − To ( x ) =− 2 t ∂y y = t t y=t (1) (b) The convection heat flux at the surface of any x-location follows from the rate equation q′′ = h Ts ( x ) − T∞ cv (2) Performing a surface energy balance as represented schematically above, equating Eqs. (1) and (2) provides q′′ ( t ) = q′′ y cv 2k Ts ( x ) − To ( x ) = h Ts ( x ) − T∞ t Ts ( x ) − To ( x ) ht = −0.5 = −0.5 Bi Ts ( x ) − T∞ ( x ) k − (3) where Bi = ht/k, the Biot number, represents the ratio of the convection to the conduction thermal resistances, Bi = R ′′ cd = t / k R ′′ cv 1/ h (4) (c) The transverse gradient (heat flow) will be negligible compared to the longitudinal gradient when Bi << 1, say, 0.1, an order of magnitude smaller. This is the criterion to validate the one-dimensional assumption used to model extended surfaces. COMMENTS: The coefficient 0.5 in Eq. (3) is a consequence of the parabolic distribution assumption. This distribution represents the simplest polynomial expression that could approximate the real distribution. PROBLEM 3.119 KNOWN: Long, aluminum cylinder acts as an extended surface. FIND: (a) Increase in heat transfer if diameter is tripled and (b) Increase in heat transfer if copper is used in place of aluminum. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Uniform convection coefficient, (5) Rod is infinitely long. PROPERTIES: Table A-1, Aluminum (pure): k = 240 W/m⋅K; Table A-1, Copper (pure): k = 400 W/m⋅K. ANALYSIS: (a) For an infinitely long fin, the fin heat rate from Table 3.4 is qf = M = ( hPkAc ) 1/ 2 ( θb qf = h π D k π D2 / 4 ) 1/ 2 θb = π ( hk )1/ 2 D3/2 θ b . 2 2 3/2 where P = πD and Ac = πD /4 for the circular cross-section. Note that qf α D the 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 α k q f ( Cu ) . Hence, if 1/ 2 k = Cu q f ( A1) k A1 1/ 2 400 = 240 1/2 , it follows that = 1.29 and there is a 29% increase in the heat transfer rate. < COMMENTS: (1) Because fin effectiveness is enhanced by maximizing P/Ac = 4/D, the use of 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 3.120 KNOWN: 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 parameter 1/ 2 hP m= kA c 1/ 2 hπ D = kπ 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 the infinite rod approximation. Thus from Table 3.4, the temperature distribution has the form θ= cosh m ( L − x ) + ( h/mk ) sinh m ( L − x ) cosh mL + ( h/mk ) sinh mL θb Evaluating the hyperbolic functions, cosh mL = 2.04 and sinh mL = 1.78, and the parameter h 30 W/m 2 ⋅ K = = 0.0168, mk 13.43m-1 (133 W/m ⋅ K ) with θb = 180°C the temperature distribution has the form cosh m ( L − x ) + 0.0168 sinh m ( L − x ) ) ( 180 C . 2.07 The temperatures at the prescribed location are tabulated below. θ= x(m) cosh m(L-x) sinh m(L-x) θ T(°C) x1 = 0.025 1.55 1.19 136.5 156.5 x2 = 0.05 1.24 0.725 108.9 128.9 L = 0.10 1.00 0.00 87.0 107.0 < < < COMMENTS: If the rod were approximated as infinitely long: T(x1) = 148.7°C, T(x2) = 112.0°C, and T(L) = 67.0°C. The assumption would therefore result in significant underestimates of the rod temperature. PROBLEM 3.121 KNOWN: Thickness, length, thermal conductivity, and base temperature of a rectangular fin. Fluid temperature and convection coefficient. FIND: (a) Heat rate per unit width, efficiency, effectiveness, thermal resistance, and tip temperature for different tip conditions, (b) Effect of convection coefficient and thermal conductivity on the heat rate. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along fin, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient, (6) Fin width is much longer than thickness (w >> t). ANALYSIS: (a) The fin heat transfer rate for Cases A, B and D are given by Eqs. (3.72), (3.76) and 2 1/2 2 1/2 (3.80), where M ≈ (2 hw tk) (Tb - T∞) = (2 × 100 W/m ⋅K × 0.001m × 180 W/m⋅K) (75°C) w = 1/2 2 1/2 -1 -1 450 w W, m≈ (2h/kt) = (200 W/m ⋅K/180 W/m⋅K ×0.001m) = 33.3m , mL ≈ 33.3m × 0.010m 2 -1 = 0.333, and (h/mk) ≈ (100 W/m ⋅K/33.3m × 180 W/m⋅K) = 0.0167. From Table B-1, it follows that 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) yield ηf ≈ q′ f h ( 2L + t )θ b , εf ≈ q′ f θ , R ′t,f = b ht θ b q′ f Case A: From Eq. (3.72), (3.86), (3.81), (3.83) and (3.70), q′ = f ηf = εf = M sinh mL + ( h / mk ) cosh mL w cosh mL + ( h / mk ) sinh mL 151 W / m 100 W / m ⋅ K ( 0.021m ) 75°C 2 151 W / m 100 W / m ⋅ K ( 0.001m ) 75°C 2 T ( L ) = T∞ + = 450 W / m 0.340 + 0.0167 × 1.057 1.057 + 0.0167 × 0.340 = 151 W / m < = 0.96 = 20.1, R ′t,f = θb cosh mL + ( h / mk ) sinh mL < 75°C 151 W / m = 25°C + < = 0.50 m ⋅ K / W 75°C 1.057 + ( 0.0167 ) 0.340 = 95.6°C < Case B: From Eqs. (3.76), (3.86), (3.81), (3.83) and (3.75) q′ = f M w tanh mL = 450 W / m ( 0.321) = 144 W / m < < < ηf = 0.92, ε f = 19.2, R ′t,f = 0.52 m ⋅ K / W T ( L ) = T∞ + θb cosh mL = 25°C + 75°C 1.057 = 96.0°C Continued ….. PROBLEM 3.121 (Cont.) Case D (L → ∞): From Eqs. (3.80), (3.86), (3.81), (3.83) and (3.79) q′ = f M w < = 450 W / m ηf = 0, ε f = 60.0, R ′t,f = 0.167 m ⋅ K / W, T ( L ) = T∞ = 25°C < (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 ) 1500 1000 500 0 0 200 400 600 800 1000 C 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 ) 400 300 200 100 0 0 200 400 600 800 1000 C 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 of h. The difference (percentage) increases with decreasing h and increasing k, but even for the worst 2 case condition (h = 10 W/m ⋅K, k = 180 W/m⋅K), the heat rate for Case A (15.7 W/m) is only slightly larger than that for Case B (14.9 W/m). For aluminum, the heat rate is significantly over-predicted by the infinite fin approximation over the entire range of h. For stainless steel, it is over-predicted for 2 small 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 ′ , ηf and ε f , as well as a larger value of R ′t ,f ) associated with insulating the tip. f Although ηf = 0 for the infinite fin, q′ and εf are substantially larger than results for L = 10 mm, f indicating that performance may be significantly improved by increasing L. PROBLEM 3.122 KNOWN: Thickness, length, thermal conductivity, and base temperature of a rectangular fin. Fluid temperature and convection coefficient. FIND: (a) Heat rate per unit width, efficiency, effectiveness, thermal resistance, and tip temperature for 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) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient, (6) Fin width is much longer than thickness (w >> t). ANALYSIS: (a) The fin heat transfer rate for Cases A, B and D are given by Eqs. (3.72), (3.76) and 2 1/2 2 1/2 (3.80), where M ≈ (2 hw tk) (Tb - T∞) = (2 × 100 W/m ⋅K × 0.001m × 180 W/m⋅K) (75°C) w = 1/2 2 1/2 -1 -1 450 w W, m≈ (2h/kt) = (200 W/m ⋅K/180 W/m⋅K ×0.001m) = 33.3m , mL ≈ 33.3m × 0.010m 2 -1 = 0.333, and (h/mk) ≈ (100 W/m ⋅K/33.3m × 180 W/m⋅K) = 0.0167. From Table B-1, it follows that 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) yield ηf ≈ q′ f h ( 2L + t )θ b , εf ≈ q′ f θ , R ′t,f = b ht θ b q′ f Case A: From Eq. (3.72), (3.86), (3.81), (3.83) and (3.70), q′ = f ηf = εf = M sinh mL + ( h / mk ) cosh mL w cosh mL + ( h / mk ) sinh mL 151 W / m 100 W / m ⋅ K ( 0.021m ) 75°C 2 151 W / m 100 W / m ⋅ K ( 0.001m ) 75°C 2 T ( L ) = T∞ + = 450 W / m 0.340 + 0.0167 × 1.057 1.057 + 0.0167 × 0.340 = 151 W / m < = 0.96 = 20.1, R ′t,f = θb cosh mL + ( h / mk ) sinh mL < 75°C 151 W / m = 25°C + < = 0.50 m ⋅ K / W 75°C 1.057 + ( 0.0167 ) 0.340 = 95.6°C < Case B: From Eqs. (3.76), (3.86), (3.81), (3.83) and (3.75) q′ = f M w tanh mL = 450 W / m ( 0.321) = 144 W / m < < < ηf = 0.92, ε f = 19.2, R ′t,f = 0.52 m ⋅ K / W T ( L ) = T∞ + θb cosh mL = 25°C + 75°C 1.057 = 96.0°C Continued ….. PROBLEM 3.122 (Cont.) Case D (L → ∞): From Eqs. (3.80), (3.86), (3.81), (3.83) and (3.79) q′ = f M w < = 450 W / m ηf = 0, ε f = 60.0, R ′t,f = 0.167 m ⋅ K / W, T ( L ) = T∞ = 25°C < (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 ) 500 400 300 200 100 0 0 0 .0 1 0 .0 2 0 .0 3 0 .0 4 0 .0 5 Fin 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 ) 150 120 90 60 30 0 0 0 .0 1 0 .0 2 0 .0 3 0 .0 4 0 .0 5 Fin 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 are within 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 B results approach those of the infinite fin approximation more quickly for stainless steel due to the larger 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 ′ , ηf and ε f , as well as a larger value of R ′t ,f ) associated with insulating the tip. f Although ηf = 0 for the infinite fin, q′ and εf are substantially larger than results for L = 10 mm, f indicating that performance may be significantly improved by increasing L. PROBLEM 3.123 KNOWN: Length, thickness and temperature of straight fins of rectangular, triangular and parabolic profiles. 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′ = q′ f max = ηf hA′ θ b , f V′ = A p 1/2 2 1/2 -1 where ηf depends on the value of m = (2h/kt) = (100 W/m ⋅K/185 W/m⋅K × 0.003m) = 13.4m -1 and the product mL = 13.4m × 0.015m = 0.201 or mLc = 0.222. Expressions for ηf, A′ and Ap are f obtained from Table 3-5. Rectangular Fin: ηf = tanh mLc 0.218 = = 0.982, A′ = 2 Lc = 0.033m f mLc 0.222 < ) ( q′ = 0.982 50 W / m 2 ⋅ K 0.033m (80°C ) = 129.6 W / m, V′ = tL = 4.5 × 10−5 m 2 < Triangular Fin: ηf = 1 I1 ( 2mL ) mL I0 ( 2 mL ) ( = 0.205 ( 0.201)1.042 1/ 2 2 = 0.978, A ′ = 2 L + ( t / 2 ) f 2 ) q ′ = 0.978 50 W / m ⋅ K 0.030m (80°C ) = 117.3 W / m, V ′ = ( t / 2 ) L = 2.25 × 10 2 < = 0.030m −5 m 2 < Parabolic Fin: ηf = 1/ 2 4 ( mL )2 + 1 ( () = 0.963, A′ = C1L + L2 / t ln ( t / L + C1 ) = 0.030m f 2 +1 ) q′ = 0.963 50 W / m 2 ⋅ K 0.030m (80°C ) = 115.6 W / m, V′ = ( t / 3) L = 1.5 × 10−5 m 2 f < < COMMENTS: Although the heat rate is slightly larger (~10%) for the rectangular fin than for the triangular or parabolic fins, the heat rate per unit volume (or mass) is larger and largest for the triangular and parabolic fins, respectively. PROBLEM 3.124 KNOWN: 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 the rods, (3) Constant properties, (4) No internal heat generation, (5) Negligible radiation exchange 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 650°C while energy is transferred by conduction from the junction (along both rods). The minimum power is twice the fin heat rate for an infinitely long fin, q min = 2q f = 2 ( hPkAc ) 1/ 2 (Tb − T∞ ). Substituting numerical values, W q min = 2 10 (π × 0.01m ) m2 ⋅ K 1/ 2 Wπ 2 379 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 the junction, thereby requiring a larger power input to maintain the junction at 650°C. < PROBLEM 3.125 KNOWN: Dimensions and end temperatures of pin fins. FIND: (a) Heat transfer by convection from a single fin and (b) Total heat transfer from a 1 2 m surface with fins mounted on 4mm centers. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction along rod, (3) Constant properties, (4) No internal heat generation, (5) Negligible radiation. PROPERTIES: Table A-1, Copper, pure (323K): k ≈ 400 W/m⋅K. ANALYSIS: (a) By applying conservation of energy to the fin, it follows that q conv = qcond,i − q cond,o where the conduction rates may be evaluated from knowledge of the temperature distribution. The general solution for the temperature distribution is θ ( x ) = C1 emx + C2 e-mx θ ≡ T − T∞ . The boundary conditions are θ(0) ≡ θo = 100°C and θ(L) = 0. Hence θ o = C1 + C2 0 = C1 emL + C2 e-mL C2 = C1 e2mL Therefore, C1 = θ o e2mL θo , C2 = − 1 − e 2mL 1 − e 2mL and the temperature distribution has the form θ= θo emx − e2mL-mx . 1 − e2mL The conduction heat rate can be evaluated by Fourier’s law, qcond = − kAc kAcθ o dθ m emx + e 2mL-mx =− dx 1 − e 2mL or, with m = ( hP/kA c ) 1/ 2 , θ o ( hPkAc ) 1/ 2 q cond = − 1 − e2mL emx + e2mL-mx . Continued ….. PROBLEM 3.125 (Cont.) Hence at x = 0, θ o ( hPkAc ) 1/ 2 q cond,i = − 1 − e2mL (1 + e2mL ) at x = L θ ( hPkA c ) q cond,o = − o 1 − e2mL Evaluating the fin parameters: 1/ 2 1/ 2 hP m= kA c 1/ 2 4h = kD 1/ 2 ( hPkAc ) 1/ 2 π 2 3 D hk = 4 (2emL ) 1/ 2 4 × 100 W/m 2 ⋅ K = 400 W/m ⋅ K × 0.001m 1/ 2 π 2 W W 3 = × ( 0.001m ) × 100 × 400 m⋅K 4 m2 ⋅ K mL = 31.62 m-1 × 0.025m = 0.791, The conduction heat rates are q cond,i = q cond,o = ( −100K 9.93 × 10-3 W/K ( = 31.62 m-1 −3.865 emL = 2.204, W K e2mL = 4.865 ) × 5.865 = 1.507 W −100K 9.93 × 10-3 W/K -3.865 and from the conservation relation, = 9.93 × 10−3 ) × 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 = 250×250 = 62,500 rods and the 2 heat transfer from the remaining (bare) surface (A = 1m - NAc). Hence, ( ) q = N q cond,i + hAθ o = 62,500 (1.507 W ) + 100W/m 2 ⋅ K 0.951 m 2 100K q = 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 than 90% of the heat transfer from the surface. (2) The fin effectiveness, ε ≡ q cond,i / hA cθ o , is ε = 192, and the fin efficiency, η ≡ ( q conv / hπ DLθ o ) , is η = 0.48. (3) The temperature distribution, θ(x)/θo, and the conduction term, qcond,i, could have been obtained 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 3.126 KNOWN: 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 fin parameters. 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 = 0 q d − qs − q f = 0 (1) The heat rate qd can be found from an energy balance on the entire device to find E in − E out + E g = 0 −q d + qV = 0 q 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 = π D/ π D 2 / 4 = 4 / D and m = ( hP/kAc ) 1/ 2 PA c = π 2D3 / 4. (3,4) (5,6) Hence, the heat rate expression can be written as ( (( ) qAg Lg = h Ag − Ac (Tb − T∞ ) + hk π 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 π 2D3 / 4 Tb = T∞ + qAg Lg / tanh ⋅ kD 2 ( ) (( )) (8) < PROBLEM 3.127 KNOWN: Positions of equal temperature on two long rods of the same diameter, but different thermal conductivity, which are exposed to the same base temperature and ambient air conditions. FIND: Thermal conductivity of rod B, kB. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Rods are infinitely long fins of uniform cross-sectional area, (3) Uniform heat transfer coefficient, (4) Constant properties. ANALYSIS: The temperature distribution for the infinite fin has the form T ( x ) − T∞ θ = = e-mx To − T∞ θb 1/ 2 hP m= kAc . (1,2) For the two positions prescribed, xA and xB, it was observed that TA ( x A ) = TB ( x B ) or θ A ( x A ) = θ B ( x B ). (3) Since θb is identical for both rods, Eq. (1) with the equality of Eq. (3) requires that m A x A = m Bx B Substituting for m from Eq. (2) gives 1/ 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, 2 x 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 thermal conductivity. It has the attractive feature of not requiring power or temperature measurements, assuming of course, a reference material of known thermal conductivity is available. < PROBLEM 3.128 KNOWN: Slender rod of length L with ends maintained at To while exposed to convection cooling (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) Constant properties, 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 when all other conditions (To, h, L) remain the same is to reduce the minimum temperature with decreasing thermal conductivity. Hence, as shown in the sketch, the mid-span temperatures are 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 the boundary condition at the mid-span. q′′ ( 0.5L ) = q′′ ( 0.5L ) x,A x,B −k A dT dT = −k B dx A,x=0.5L dx B,x=0.5L Since kA > kB, it follows that dT dT < . dx A,x=0.5L dx B,x=0.5L It follows that the minimum temperature in the rod must be in the kB region, x > 0.5L, and the temperature distribution is not symmetrical about the mid-span. COMMENTS: (1) Recognize that the area under the curve on the T-x coordinates is proportional to the fin heat rate. What conclusions can you draw regarding the relative magnitudes of qfin for cases (a), (b) and (c)? (2) If L is increased substantially, how would the temperature distribution be affected? PROBLEM 3.129 KNOWN: Base temperature, ambient fluid conditions, and temperatures at a prescribed distance 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) Constant properties, (4) Negligible radiation, (5) Negligible contact resistance at base, (6) Infinitely long rods, (7) Rods are identical except for their thermal conductivity. ANALYSIS: With the assumption of infinitely long rods, the temperature distribution is T − T∞ θ = = e-mx θ b Tb − T∞ or 1/ 2 T − T∞ hP ln = − mx = Tb − T∞ kA x Hence, for the two rods, T − T ln A ∞ 1/ 2 Tb − T∞ = k B TB − T∞ k A ln Tb − T∞ T − T∞ ln A ln Tb − T∞ = 200 1/ 2 1/2 = k1/2 kB () A T − T ln ln B ∞ Tb − T∞ k B = 56.6 W/m ⋅ K. 75 − 25 100 − 25 = 7.524 60 − 25 100 − 25 < COMMENTS: Providing conditions for the two rods may be maintained nearly identical, the above method provides a convenient means of measuring the thermal conductivity of solids. PROBLEM 3.130 KNOWN: Arrangement of fins between parallel plates. Temperature and convection coefficient of air flow in finned passages. Maximum allowable plate temperatures. FIND: (a) Expressions relating fin heat transfer rates to end temperatures, (b) Maximum power dissipation for each plate. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fins, (3) Constant properties, (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/m⋅K. ANALYSIS: (a) The general solution for the temperature distribution in fin is θ ( x ) ≡ T ( x ) − T∞ = C1emx + C 2e-mx θ (0 ) = θ o = To − T∞ , Boundary conditions: Hence θ o = C1 + C2 θ ( L ) = θ L = TL − T∞ . θ L = C1emL + C2e-mL θ L = C1e mL + (θ o − C1 ) e-mL C1 = Hence θ L − θ oe-mL emL − e-mL θ (x ) = C2 = θ o − θ L − θ o e-mL emL − e-mL = θ o emL − θ L emL − e-mL . m x-L m L-x θ Lemx − θ oe ( ) + θ oe ( ) − θ Le-mx e mL − e-mL ( m L-x -m L-x θ o e ( ) − e ( ) + θ L emx − e-mx θ (x ) = mL − e-mL e ) θ sinh m (L-x ) + θ Lsinh mx . θ (x ) = o sinh mL The fin heat transfer rate is then q f = − kAc Hence θm dT θm cosh m ( L − x ) + L cosh mx . = − kDt − o dx sinh mL sinh mL θm θo m −L tanh mL sinh mL < θm θom −L . 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-1 mL = 35.5 m-1 × 0.012 m = 0.43 sinh mL = 0.439 tanh mL = 0.401 θ o = 100 K θ 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 − 0.401 0.439 q 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,L 0.439 0.401 q f,L = 87.8 W. (into the bottom plate) Maximum power dissipations are therefore q o,max = Nf q f,o + ( W − Nf t ) Dhθ o q 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 W q L,max = − N f q f,L + ( W − Nf t ) Dhθ o q 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 as it passes between the plates. If the air temperature change is restricted to mair = ∆T∞ = 5 K, its flowrate must be q tot 1717 W = = 0.34 kg/s. cp ∆T∞ 1007 J/kg ⋅ K × 5 K Its mean velocity is then mair 0.34 kg/s Vair = = = 163 m/s. 3 × 0.012 m 0.2 − 50 × 0.001 m ρair Ac 1.16 kg/m ( ) Such a velocity would be impossible to maintain. To reduce it to a reasonable value, e.g. 10 m/s, Ac would have to be increased substantially by increasing W (and hence the space between fins) and by increasing L. The present configuration is impractical from the standpoint 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 to the air to maintain the plate at 350 K. PROBLEM 3.131 KNOWN: 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 about midplane. ANALYSIS: (a) Considering a one-half section of the array, the corresponding resistance is −1 R t,o = (ηo 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 is ηo = 1 − NA f At Af = (1 − ηf ) where the fin efficiency is ηf = tanh m ( L 2 ) m (L 2) 1/ 2 and hP kA c m= h ( 2t + 2W2 ) ktW2 1/ 2 = 1/ 2 2h kt ≈ = 38.7m −1 With m(L/2) = 0.155, it follows that ηf = 0.992 and ηo = 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 passage restriction constraints. Repeating the foregoing calculations for representative values of t and (S - t), we obtain S (mm) 2.5 3 3 4 4 5 5 N 400 333 333 250 250 200 200 t (mm) 0.5 0.5 1 0.5 2 0.5 3 Rt,o (K/W) 0.00169 0.00193 0.00202 0.00234 0.00268 0.00264 0.00334 COMMENTS: Clearly, the thermal performance of the fin array improves (Rt,o decreases) with increasing N. Because ηf ≈ 1 for the entire range of conditions, there is a slight degradation in performance (Rt,o increases) with increasing t and fixed N. The reduced performance is associated with the reduction in surface area of the exposed base. Note that the overall thermal resistance for the -2 entire fin array (top and bottom) is Rt,o/2 = 1.22 × 10 K/W. PROBLEM 3.132 KNOWN: Width and maximum allowable temperature of an electronic chip. Thermal contact resistance 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, fin thickness, and fin pitch, and (b) Effect of fin thickness/number and convection coefficient on performance. 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 of heat sink, (7) Negligible radiation. ANALYSIS: (a) From the thermal circuit, T −T Tc − T∞ qc = c ∞ = R tot R 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 2 t,c = 0.003m /180 W / m ⋅ K ( 0.02m ) = 0.042 K / W. From Eqs. (3.103), (3.102), and (3.99) 2 R t,o = 1 , ηo h A t ηo = 1 − N Af (1 − ηf ) , At -4 2 A t = N Af + A b 2 2 where Af = 2WLf = 2 × 0.02m × 0.015m = 6 × 10 m and Ab = W – N(tW) = (0.02m) – 11(0.182 -3 -4 2 1/2 2 × 10 m × 0.02m) = 3.6 × 10 m . With mLf = (2h/kt) Lf = (200 W/m ⋅K/180 W/m⋅K × 0.182 × -3 1/2 10 m) (0.015m) = 1.17, tanh mLf = 0.824 and Eq. (3.87) yields ηf = tanh mLf 0.824 = = 0.704 mLf 1.17 -3 2 It follows that At = 6.96 × 10 m , ηo = 0.719, Rt,o = 2.00 K/W, and qc = (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 of decreasing the number of fins and increasing the fin thickness. Continued ….. PROBLEM 3.132 (Cont.) N t(mm) ηf 6 7 8 9 10 11 1.833 1.314 0.925 0.622 0.380 0.182 0.957 0.941 0.919 0.885 0.826 0.704 Rt,o (K/W) 2.76 2.40 2.15 1.97 1.89 2.00 2 qc (W) At (m ) 23.2 26.6 29.7 32.2 33.5 31.8 0.00378 0.00442 0.00505 0.00569 0.00632 0.00696 Although ηf (and ηo) increases with decreasing N (increasing t), there is a reduction in At which yields a minimum in Rt,o, and hence a maximum value of qc, for N = 10. For N = 11, the effect of h on the performance of the heat sink is shown below. Heat rate as a function of convection coefficient (N=11) Heat rate, qc(W) 150 100 50 0 100 200 300 400 500 600 700 800 900 1000 Convection coefficient, h(W/m2.K) 2 With increasing h from 100 to 1000 W/m ⋅K, Rt,o decreases from 2.00 to 0.47 K/W, despite a decrease in ηf (and ηo) from 0.704 (0.719) to 0.269 (0.309). The corresponding increase in qc is significant. COMMENTS: The heat sink significantly increases the allowable heat dissipation. If it were not used and heat was simply transferred by convection from the surface of the chip with h = 100 2 2 W/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 3.133 KNOWN: Number and maximum allowable temperature of power transistors. Contact resistance between transistors and heat sink. Dimensions and thermal conductivity of heat sink. Temperature and 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 convection coefficient 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) Negligible radiation. ANALYSIS: (a) From the thermal circuit, Nt qt = Tt − T∞ (R t,c )equiv + R t,b + R t,o For the array of transistors, the corresponding contact resistance is the equivalent resistance associated with the component resistances, in which case, (R t,c )equiv = N t (1/ R t,c ) −1 −1 = (9 / 0.045 K / W ) = 5 × 10−3 K / W The thermal resistance associated with the base of the heat sink is R t,b = Lb k (W ) 2 = 0.006m 180 W / m ⋅ K ( 0.150m ) 2 = 1.48 ×10−3 K / W From Eqs. (3.103), (3.102) and (3.99), the thermal resistance associated with the fin array and the corresponding overall efficiency and total surface area are R t,o = 1 , ηo h A t ηo = 1 − Nf Af (1 − ηf ) , At A t = Nf Af + Ab -3 2 Each fin has a surface area of Af ≈ 2 W Lf = 2 × 0.15m × 0.03m = 9 × 10 m , and the area of the 2 2 -2 2 exposed base is Ab = W – Nf (tW) = (0.15m) – 25 (0.003m × 0.15m) = 1.13 × 10 m . With mLf = 1/2 2 1/2 (2h/kt) Lf = (200 W/m ⋅K/180 W/m⋅K × 0.003m) (0.03m) = 0.577, tanh mLf = 0.520 and Eq. (3.87) yields ηf = tanh mLf 0.520 = = 0.902 mLf 0.577 -3 -2 2 2 Hence, with At = [25 (9 × 10 ) + 1.13 × 10 ]m = 0.236m , Continued ….. ηo = 1 − PROBLEM 3.133 (Cont.) ( 25 0.009m 2 0.236m 2 ) (1 − 0.901) = 0.907 ( R t,o = 0.907 ×100 W / m 2 ⋅ K × 0.236m 2 ) −1 = 0.0467 K / W The heat rate per transistor is then qt = (100 − 27 ) °C 1 = 152 W 9 (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 ) 600 500 400 300 200 100 0 100 200 300 400 500 600 700 800 900 1000 C o n ve c tio n c o e ffic ie n t, h (W /m ^2 .K ) 250 240 H e a t ra te , q t(W ) 230 220 210 200 190 180 170 160 150 0 .0 3 0 .0 4 0 .0 5 0 .0 6 0 .0 7 0 .0 8 0 .0 9 0 .1 Fin le n g th , L f(m ) However, in each case, the effect of the increase diminishes due to an attendant reduction in ηf. For 2 example, as h increases from 100 to 1000 W/m ⋅K for Lf = 30 mm, ηf decreases from 0.902 to 0.498. COMMENTS: The heat sink significantly increases the allowable transistor power. If it were not 2 2 2 used and heat was simply transferred from a surface of area W = 0.0225 m with h = 100 W/m ⋅K, 2 -1 the corresponding thermal resistance would be Rt,cnv = (hW ) K/W = 0.44 and the transistor power would be qt = (Tt - T∞)/Nt Rt,cnv = 18.4 W. PROBLEM 3.134 KNOWN: 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/m⋅K. ANALYSIS: (a) The thermal circuit is Rf = cosh mL+ ( h o / mk ) sinh mL θb = 16q f 16 h PkA 1/ 2 sinh mL+ h / mk cosh mL (o ) ( o c,f ) (b) The maximum chip heat rate is q c = 16q f + q b + qi . Evaluate these parameters hP m= o kAc,f ( 1/ 2 4h = o kDp 1/ 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 ⋅ K 81.7 m-1 × 400 W/m ⋅ K ( 2 M = h oπ D p kπ Dp / 4 ( ) 1/ 2 sinh mL = 1.57, = 81.7 m-1 cosh mL = 1.86 = 0.0306 θb ) 1/ 2 3 M = 1000 W/m 2 ⋅ K π 2 / 4 ( 0.0015 m ) 400 W/m ⋅ K (55 C) = 3.17 W. Continued ….. PROBLEM 3.134 (Cont.) The fin heat rate is qf = M sinh mL+ ( h/mk ) cosh mL cosh mL+ ( h/mk ) sinh mL = 3.17 W 1.57+0.0306 × 1.86 1.86+0.0306 × 1.57 q f = 2.703 W. The heat rate from the board by convection is 2 2 q b = h o A bθ b = 1000 W/m 2 ⋅ K ( 0.0127 m ) − (16π / 4 )( 0.0015 m ) 55 C q 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/W t,c (0.0127 m )2 Tc − T∞,i qi = 0.29 W. Hence, the maximum chip heat rate is q 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 chip 2 (assuming negligible contact resistance). Their effectiveness is ε = q f / π Dp / 4 h oθ b = 2.703 ( ) W/0.097 W = 27.8 2 2 (2) Without the fins, qc = 1000 W/m ⋅K(0.0127 m) 55°C + 0.29 W = 9.16 W. Hence the fins provide 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 = 31.6 c 2 W/cm . (4) If the infinite fin approximation is made, qf = M = 3.17 W, and the actual fin heat transfer is overestimated by 17%. PROBLEM 3.135 KNOWN: Geometry of pin fin array used as heat sink for a computer chip. Array convection and chip substrate 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-board assembly, (3) Negligible pin-chip contact resistance, (4) Constant properties, (5) Negligible chip thermal resistance, (6) Uniform chip temperature. ANALYSIS: The total power dissipation is q c = qi + q t , where Tc − T∞,i qi = = 0.3W 1 h i + R ′′ + L b k b A c t,c ( ) and qt = Tc − T∞,o R t,o The resistance of the pin array is −1 R t,o = (ηo h o A t ) where ηo = 1 − NAf At (1 − ηf ) A t = NA f + A b ( A f = π D p L c = π 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 a function of D p for L p = 15 mm and values of N = 16, 25 and 36. Using the IHT Performance Calculation, Extended Surface Model for the Pin Fin Array, we obtain Continued... 1/ 2 PROBLEM 3.135 (CONT.) 35 Heat rate, qt(W) 30 25 20 15 10 5 0 0.5 0.9 1.3 1.7 2.1 2.5 Pin diameter, Dp(mm) N = 36 N = 25 N = 16 Clearly, it is desirable to maximize the number of pins and the pin diameter, so long as flow passages are not constricted to the point of requiring an excessive pressure drop to maintain the prescribed convection coefficient. The maximum heat rate for the fin array ( q t = 33.1 W) corresponds to N = 36 and D p = 1.5 mm. Further improvement could be obtained by using N = 49 pins of diameter D p = 1.286 mm, which yield q t = 37.7 W. Exploring the effect of L p for N = 36 and D p = 1.5 mm, we obtain Heat rate, qt(W) 60 50 40 30 10 20 30 40 50 Pin length, Lp(mm) N = 36, Dp = 1.5 mm Clearly, there are benefits to increasing L p , although the effect diminishes due to an attendant reduction in ηf (from ηf = 0.887 for L p = 15 mm to ηf = 0.471 for L p = 50 mm). Although a heat dissipation rate of q t = 56.7 W is obtained for L p = 50 mm, package volume constraints could preclude such a large 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 = 15 mm. Resistance, Rt,o(K/W) 8 6 4 2 0 0.5 1 1.5 Pin diameter, Dp(mm) N = 16 N = 25 N = 36 2 2.5 PROBLEM 3.136 KNOWN: Copper heat sink dimensions and convection conditions. FIND: (a) Maximum allowable heat dissipation for a prescribed chip temperature and interfacial chip/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 sink assembly, (3) Constant k, (4) Negligible chip thermal resistance, (5) Negligible heat transfer from back of chip, (6) Uniform chip temperature. ANALYSIS: (a) For the prescribed system, the chip power dissipation may be expressed as Tc − T∞ qc = R t,c + R cond,b + R t,o where R t,c = R ′′ t,c 2 Wc R cond,b = = 5 × 10 −6 m 2 ⋅ K W (0.016m ) 2 Lb = 2 kWc = 0.0195 K W 0.003m 400 W m ⋅ K ( 0.016m ) 2 = 0.0293 K W The thermal resistance of the fin array is −1 R t,o = (ηo hA t ) where ηo = 1 − and NAf At (1 − ηf ) ( 2 A 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 × 10−3 m, it follows that A f = 6.06 × 10−6 m 2 and A t = 6.40 ×10−3 m 2 . The fin efficiency is ηf = tanh mLc mLc 1/ 2 1/ 2 where m = ( hP kA c ) = 245 m-1 and mLc = 1.49. It follows that ηf = 0.608 and = ( 4h kw ) ηo = 0.619, in which case ) ( R t,o = 0.619 × 1500 W m 2 ⋅ K × 6.40 × 10−3 m 2 = 0.168 K W and the maximum allowable heat dissipation is qc = (85 − 25) C (0.0195 + 0.0293 + 0.168 ) K W < = 276W (b) The IHT Performance Calculation, Extended Surface Model for the Pin Fin Array has been used to determine q c as a function of Lf for four different cases, each of which is characterized by the closest allowable fin spacing of (S - w) = 0.25 mm. Maximum heat rate, qc(W) Case A B C D w (mm) 0.25 0.35 0.45 0.55 S (mm) 0.50 0.60 0.70 0.80 N 1024 711 522 400 340 330 320 310 300 290 280 270 6 7 8 9 10 Fin length, Lf(mm) w = 0.25 mm, S = 0.50 mm, N =1024 w = 0.35 mm, S = 0.60 mm, N = 711 w = 0.45 mm, S = 0.70 mm, N = 522 w = 0.55 mm, S = 0.80 mm, N = 400 With increasing w and hence decreasing N, there is a reduction in the total area A t associated with heat transfer from the fin array. However, for Cases A through C, the reduction in A t is more than balanced by an increase in ηf (and ηo ), causing a reduction in R t ,o and hence an increase in q c . As the fin efficiency approaches its limiting value of ηf = 1, reductions in A t due to increasing w are no longer balanced by increases in ηf , and q c begins to decrease. Hence there is an optimum value of w, which depends on Lf . For the conditions of this problem, Lf = 10 mm and w = 0.55 mm provide the largest heat dissipation. Problem 3.137 KNOWN: 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, ηf, and effectiveness, εf, of a single fin, as well as, the total heat rate, qt, and overall efficiency, ηo, of the array. Also, compare the total heat rates per unit volume. SCHEMATIC: Fin dimensions Cross section Length w x w (mm) L (mm) 1x1 30 3x3 7 Design A B Number of fins 6x9 14 x 17 Convection coefficient (W/m2⋅K) 125 375 ASSUMPTIONS: (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), is ηo = qt q max = qt hA tθ b (1) where At is the total surface area, the sum of the exposed portion of the base (prime area) plus the fin surfaces, Eq. 3.99, A t = N ⋅ Af + A b (2) where the surface area of a single fin and the prime area are Af = 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 is q t = Nηf hAf θ b + hA bθ b (5) where ηf is the efficiency of a single fin. From Table 4.3, Case A, for the tip condition with convection, the single fin efficiency based upon Eq. 3.86, ηf = qf hAf θ b (6) Continued... PROBLEM 3.137 (Cont.) where qf = M sinh(mL) + ( h mk ) cosh(mL) (7) cosh(mL) + ( h mk ) sinh(mL) M = ( hPkA c ) 1/ 2 θb m = ( hP kAc ) 1/ 2 P = 4w Ac = w 2 (8,9,10) The single fin effectiveness, from Eq. 3.81, εf = qf hA cθ b (11) Additionally, we want to compare the performance of the designs with respect to the array volume, vol q′′′ = qf ∀ = qf f ( 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. Design A B qt (W) 113 165 qf (W) 1.80 0.475 ηo ηf εf 0.804 0.909 0.779 0.873 31.9 25.3 q TTT f (W/m3) 1.25×106 7.81×106 COMMENTS: (1) Both designs have good efficiencies and effectiveness. Clearly, Design B is superior 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 (∀ = 2.12×10-5 vs. 9.06×10-5 m3) and the 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 the performance comparison is the magnitude of the convection coefficient which is 3 times larger for Design B. Estimating convection coefficients for fin arrays (and tube banks) is discussed in Chapter 7.6. Of concern is how the fins alter the flow past the fins and whether the convection coefficient is uniform over the array. (3) The IHT Extended Surfaces Model, for a Rectangular Pin Fin Array could have been used to solve this problem. PROBLEM 3.138 KNOWN: Geometrical characteristics of a plate with pin fin array on both surfaces. Inner and outer convection conditions. FIND: (a) Heat transfer rate with and without pin fin arrays, (b) Effect of using silver solder to join the pins 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/m⋅K. ANALYSIS: (a) The heat rate may be expressed as T∞,i − T∞,o q= R t,o(c),i + R w + R t,o(c),o where ( R t,o(c) = ηo(c) hA t ηo(c) = 1 − )−1 , NA f ηf 1 − , A t C1 A t = NA f + A b , A f = π D p Lc ≈ π D p ( L + D 4 ) , ( ) A b = W 2 − NAc,b = W 2 − N π D 2 4 , p ηf = tanh mLc mLc , ( m = 4h kD p )1/ 2 , Continued... PROBLEM 3.138 (Cont.) ( ) C1 = 1 + ηf hA f R ′′ A c,b , t,c and Rw = Lw W2k . Calculations may be expedited by using the IHT Performance Calculation, Extended Surface Model for the 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/m⋅K = t 4.88 × 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/m2⋅K, m = 15.8 m-1, ηf = 0.965, ηo = 0.970 and R t ,o = 0.0817 K/W. On the inner surface, where h i = 5 W/m2⋅K, m = 3.54 m-1, ηf = 0.998, ηo = 0.999 and R t ,o = 1.588 K/W. Hence, the heat rate is q= (65 − 20 ) C (1.588 + 4.88 ×10 −4 ) + 0.0817 K W < = 26.94W Without 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 ×10−4 + 0.39 ) ( < Hence, the fin arrays provide nearly a five-fold increase in heat rate. (b) With use of the silver solder, ηo(c),o = 0.962 and R t,o(c),o = 0.0824 K/W. Also, ηo(c),i = 0.998 and R t,o(c),i = 1.589 K/W. Hence q= ( (65 − 20 ) C ) 1.589 + 4.88 × 10−4 + 0.0824 K W = 26.92W Hence, the effect of the contact resistance is negligible. COMMENTS: The dominant contribution to the total thermal resistance is associated with internal conditions. If the heat rate must be increased, it should be done by increasing hi. < PROBLEM 3.139 KNOWN: Long rod with internal volumetric generation covered by an electrically insulating sleeve and supported with a ribbed spider. FIND: Combination of convection coefficient, spider design, and sleeve thermal conductivity which enhances volumetric heating subject to a maximum centerline temperature of 100°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial heat transfer in rod, sleeve and hub, (3) Negligible interfacial contact resistances, (4) Constant properties, (5) Adiabatic outer surface. ANALYSIS: The system heat rate per unit length may be expressed as T1 − T∞ 2 q′ = q π ro = R′ sleeve + R ′ hub + R ′ t,o where () R′ sleeve = ln ( r1 ro ) , R′ hub = 2π k s NA′ f (1 − η ) , ηo = 1 − f A′t ηf = tanh m ( r3 − r2 ) m ( r3 − r2 ) , ln ( r2 r1 ) = 3.168 × 10−4 m ⋅ K W , R ′t,o = 2π k r A′ = 2 ( r3 − r2 ) , f m = ( 2h k r t ) 1/ 2 1 , ηo hA′t A′ = NA′ + ( 2π r3 − Nt ) , t f . The rod centerline temperature is related to T1 through qr 2 To = T ( 0 ) = T1 + o 4k Calculations may be expedited by using the IHT Performance Calculation, Extended Surface Model for the Straight Fin Array. For base case conditions of ks = 0.5 W/m⋅K, h = 20 W/m2⋅K, t = 4 mm and N = 6 12, R ′ sleeve = 0.0580 m⋅K/W, R ′ ,o = 0.0826 m⋅K/W, ηf = 0.990, q′ = 387 W/m, and q = 1.23 × 10 t W/m3. As shown below, q may be increased by increasing h, where h = 250 W/m2⋅K represents a reasonable 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.) 2E6 1.8E6 1.6E6 1.4E6 1.2E6 1E6 0 50 100 150 200 250 Convection coefficient, h(W/m^2.K) t = 4 mm, N = 12, ks = 0.5 W/m.K The difficulty is that, by significantly increasing h, the thermal resistance of the fin array is reduced to 0.00727 m⋅K/W, rendering the sleeve the dominant contributor to the total resistance. Heat generation, qdotx1E-6(W/m Similar results are obtained when N and t are varied. For values of t = 2, 3 and 4 mm, variations of N in the respective ranges 12 ≤ N ≤ 26, 12 ≤ N ≤ 21 and 12 ≤ N ≤ 17 were considered. The upper limit on N was 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 the three values of t. 2.1 2.08 2.06 2.04 2.02 2 12 14 16 18 20 22 24 26 Number of ribs, N t = 2 mm, N: 12 - 26, h = 250 W/m^2.K t = 3 mm, N: 12 - 21, h = 250 W/m^2.K t = 4 mm, N: 12 -17, h = 250 W/m^2.K Heat generation, qdot(W/m^3) In contrast, significant improvement is associated with changing the sleeve material, and it is only necessary to have ks ≈ 25 W/m⋅K (e.g. a boron sleeve) to approach an upper limit to the influence of ks. 4E6 3.6E6 3.2E6 2.8E6 2.4E6 2E6 0 20 40 60 80 100 Sleeve conductivity, ks(W/m.K) t = 4 mm, N = 12, h = 250 W/m^2.K For h = 250 W/m2⋅K and ks = 25 W/m⋅K, 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 which 2 case T1 → T∞. Hence q max = 4k ( To − T∞ ) ro = 4.5 × 106 W m3 . PROBLEM 3.140 KNOWN: 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, −1 R′ conv,i = ( h i 2π r1 ) where ηo = 1 − NA′ f 1 −η , ( f) A′t (b) q′ = , R′ cond = ln ( r2 r1 ) 2π k , and A′ = 2L = 2 ( r3 − r2 ) , f A′t = NA′ + ( 2π r2 − Nt ) , and f (T∞,i − T∞,o ) R′ conv,i + R ′ cond + R ′ t,o Substituting the known conditions, it follows that ( 2 R′ conv,i = 5000 W m ⋅ K × 2π × 0.013m ) −1 −1 R ′ = (ηo h o A′ ) t,o t ηf = , tanh mL mL . = 2.45 × 10−3 m ⋅ K W −3 R′ cond = ln (0.016m 0.013m ) 2π ( 20 W m ⋅ K ) = 1.65 × 10 m ⋅ K W ( R ′ = 0.575 × 200 W m 2 ⋅ K × 0.461m t,o ) −1 = 18.86 × 10 −3 m ⋅ K W where ηf = 0.490. Hence, (90 − 25 ) C q′ = ( 2.45 + 1.65 + 18.86 ) ×10−3 m ⋅ K = 2831W m < W (c) The small value of ηf suggests that some benefit may be gained by increasing t, as well as by increasing 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 = 2 mm, 12 ≤ N ≤ 25; t = 3 mm, 8 ≤ N ≤ 16; t = 4 mm, 6 ≤ N ≤ 12; t = 5 mm, 5 ≤ N ≤ 10. Calculations based on the foregoing model are plotted as follows. Continued... PROBLEM 3.140 (Cont.) Heat rate, q'(w/m) 5000 4000 3000 2000 5 10 15 20 25 Number of fins, N t = 2 mm t = 3 mm t = 4 mm t = 5 mm By increasing t from 2 to 5 mm, ηf increases from 0.410 to 0.598. Hence, for fixed N, q′ increases with increasing t. However, from the standpoint of maximizing q′ , it is clearly preferable to use the t larger number of thinner fins. Hence, subject to the prescribed constraint, we would choose t = 2 mm and N = 25, for which q′ = 4880 W/m. COMMENTS: (1) The air side resistance makes the dominant contribution to the total resistance, and efforts to increase q′ by reducing R ′ ,o are well directed. (2) A fin thickness any smaller than 2 mm t would be difficult to manufacture. PROBLEM 3.141 KNOWN: 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/m⋅K. ANALYSIS: Evaluate the fin parameters Lc = L+t/2 = 0.05025m A p = Lc t = 0.05025m × 0.5 × 10-3m=25.13 × 10-6 m 2 1/ 2 ( ) = ( 0.05025m ) ( ) 30 W/m 2 ⋅ K 240 W/m ⋅ K × 25.13 × 10-6m 2 = 0.794 L3/2 h w / kA p c L3/2 h w / kA p c 1/ 2 1/ 2 3/ 2 It follows from Fig. 3.18 that ηf ≈ 0.72. Hence, q f = ηf q max = 0.72 h w 2wL θ b qf = 0.72 × 30 W/m 2 ⋅ K × 2 × 0.05m × ( w θ b ) = 2.16 W/m ⋅ K ( w θ b ) With the fins, the heat transfer from the walls is q w = N qf + (1 − Nt ) w h w θ b ) ( W ( w θ b ) + 1m − 250 × 5 ×10−4 m × 30 W/m2 ⋅ K ( w θ b ) m⋅K W q w = (540 + 26.3) ( w θ b ) = 566 w θ b . m⋅K q w = 250 × 2.16 Without the fins, qwo = hwo 1m × w θb = 40 w θb. Hence the percentage increase in heat transfer is q w − q wo (566 − 40 ) w θ b < = = 13.15 = 1315% q wo 40 w θ b 1/2 COMMENTS: If the infinite fin approximation is made, it follows that qf = (hPkAc) 1/2 -4 1/2 =[hw2wkwt] θb = (30 × 2 × 240 × 5×10 ) overestimated. w θb = 2.68 w θb. Hence, qf is θb PROBLEM 3.142 KNOWN: Dimensions, base temperature and environmental conditions associated with rectangular and triangular 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/m⋅K. ANALYSIS: For the rectangular fin, with Lc = L + t/2, evaluate the parameter 1/ 2 1/ 2 3/ 2 L3/ 2 h kA p = ( 0.023m ) c ( 75 W m 2 ⋅ K 15.3 W m ⋅ K ( 0.023m )( 0.006 m ) ) = 0.66 . Hence, from Fig. 3.18, the fin efficiency is < ηf ≈ 0.79 From Eq. 3.86, the fin heat rate is q f = ηf hAf θ b = ηf hPLcθ b = ηf h2wLcθ b or, per unit width, ( ) q q′ = f = 0.79 75 W m 2 ⋅ K 2 ( 0.023m ) 80 C = 218 W m . f w < From Eq. 3.81, the fin effectiveness is εf = qf q′ × w 218 W m f = = = 6.06 . 2 ⋅ K 0.006 m 80 C hA c,bθ b h ( t × w )θ b 75 W m ( ) < For the triangular fin with 1/ 2 1/ 2 3/ 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, < ηf ≈ 0.78 , From Eq. 3.86 and Table 3.5 find 1/ 2 2 q′ = ηf hA′ θ b = ηf h2 L2 + ( t 2 ) f f θb 1/ 2 2 2 q′ = 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 is εf = q′ × w 187 W m f = = 5.19 h ( t × w )θ b 75 W m 2 ⋅ K ( 0.006 m ) 80 C COMMENTS: Although it is 14% less effective, the triangular fin offers a 50% weight savings. < PROBLEM 3.143 KNOWN: 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/m⋅K. ANALYSIS: (a) With Lc = L = 0.006 m, find A p = Lt 2 = ( 0.006 m )( 0.002 m ) 2 = 6 × 10−6 m 2 , 1/ 2 3/ 2 L3 / 2 h kA p = ( 0.006 m ) c ( 1/ 2 240 W m ⋅ K × 6 × 10−6 m 2 ) 40 W m2 ⋅ K = 0.077 and from Fig. 3.18, the fin efficiency is < ηf ≈ 0.99 . From Eq. 3.86 and Table 3.5, the fin heat rate is 1/ 2 2 q f = ηf q max = ηf hAf (tri)θ b = 2ηf hw L2 + ( t 2 ) θb . From Eq. 3.81, the fin effectiveness is 1/ 2 εf = qf hAc,bθ b = 2 2ηf hw L2 + ( t 2 ) g ( w ⋅ t )θ b θb 1/ 2 = 2 2ηf L2 + ( t 2 ) t 1/ 2 εf = 2 2 2 × 0.99 ( 0.006 ) + ( 0.002 2 ) m 0.002 m < = 6.02 (b) The heat dissipation per unit width is 1/ 2 2 q′ = ( q f w ) = 2ηf h L2 + ( t 2 ) f θb 1/ 2 2 2 q′ = 2 × 0.99 × 40 W m 2 ⋅ K ( 0.006 ) + ( 0.002 2 ) f m × ( 250 − 20 ) C = 110.8 W m . < COMMENTS: The triangular profile is known to provide the maximum heat dissipation per unit fin mass. PROBLEM 3.144 KNOWN: 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/m⋅K. ANALYSIS: (a) The fin parameters for use with Figure 3.19 are r2c = r2 + t 2 = (12.5 mm + 10 mm ) + 0.5 mm = 23mm = 0.023m r2c r1 = 1.84 Lc = L + t 2 = 10.5 mm = 0.0105 m A p = Lc t = 0.0105 m × 0.001m = 1.05 ×10−5 m 2 1/ 2 3/ 2 L3/ 2 h kA p = (0.0105 m ) c ( ) 1/ 2 −5 2 240 W m ⋅ K × 1.05 ×10 m 25 W m 2 ⋅ K = 0.15 . Hence, the fin effectiveness is ηf ≈ 0.97, and from Eq. 3.86 and Fig. 3.5, the fin heat rate is ( ) 2 2 q f = ηf q max = ηf hAf (ann)θ b = 2πηf h r2,c − r1 θ b 2 2 q f = 2π × 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 considering contributions due to the fin and base (unfinned surfaces is q′ = N′q f + h (1 − N′t ) 2π r1θ b ( ) q′ = 200 m −1 × 12.8 W + 25 W m 2 ⋅ K 1 − 200 m −1 × 0.001m × 2π × ( 0.0125 m ) 225 C q′ = ( 2560 W + 353 W ) m = 2.91kW m . < COMMENTS: Note that, while covering only 20% of the tube surface area, the tubes account for more than 85% of the total heat dissipation. PROBLEM 3.145 KNOWN: Dimensions and base temperature of aluminum fins of rectangular profile. Ambient air conditions. 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) Uniform convection coefficient. PROPERTIES: Table A-1, Aluminum, pure (T ≈ 400 K): k = 240 W/m⋅K. ANALYSIS: (a) The fin parameters for use with Figure 3.19 are r2c = r2 + t 2 = 40 mm + 2 mm = 0.042 m Lc = L + t 2 = 15 mm + 2 mm = 0.017 m r2c r1 = 0.042 m 0.025 m = 1.68 A p = Lc t = 0.017 m × 0.004 m = 6.8 × 10 −5 m 2 ( L3 / 2 h kA p c )1/ 2 = (0.017 m )3 / 2 40 W 1/ 2 m 2 ⋅ K 240 W m ⋅ K × 6.8 × 10−5 m 2 = 0.11 The fin efficiency is ηf ≈ 0.97. From Eq. 3.86 and Fig. 3.5, 2 2 q f = ηf q max = ηf hAf (ann)θ b = 2πηf h r2c − r1 θ b 2 2 q f = 2π × 0.97 × 40 W m 2 ⋅ K ( 0.042 ) − ( 0.025 ) m 2 × 180 C = 50 W < From Eq. 3.81, the fin effectiveness is εf = qf hAc,bθ b = 50 W 40 W m 2 ⋅ K 2π ( 0.025 m )( 0.004 m )180 C = 11.05 < (b) The rate of heat transfer per unit length is q′ = N′q f + h (1 − N′t )( 2π r1 )θ b q′ = 125 × 50 W m + 40 W m 2 ⋅ K (1 − 125 × 0.004 )( 2π × 0.025 m ) × 180 C q′ = ( 6250 + 565 ) W m = 6.82 kW m COMMENTS: Note the dominant contribution made by the fins to the total heat transfer. < PROBLEM 3.146 KNOWN: Dimensions, base temperature, and contact resistance for an annular, aluminum fin. Ambient fluid 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/m⋅K. ANALYSIS: With the contact resistance, the fin heat loss is q f = Tw − T∞ R t,c + R f where R t,c = R ′′ A b = 2 × 10−4 m 2 ⋅ K W 2π ( 0.015 m )( 0.002 m ) = 1.06 K W . t,c From Eqs. 3.83 and 3.86, the fin resistance is θ θb θb 1 . Rf = b = = = q f ηf q max ηf hA f θ b 2π hη r 2 − r 2 f 2,c 1 ( Evaluating parameters, r2,c = r2 + t 2 = 30 mm + 1mm = 0.031m ( Lc = L + t 2 = 0.016 m A p = Lc t = 3.2 × 10−5 m 2 r2c r1 = 0.031 0.015 = 2.07 Z L3 / 2 h kA p c ) )1/ 2 = (0.016 m )3 / 2 75 W 1/ 2 m 2 ⋅ K 240 W m ⋅ K × 3.2 × 10−5 m 2 = 0.20 find the fin efficiency from Figure 3.19 as ηf = 0.94. Hence, Rf = qf = ( ) 1 2 2 2π 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 and qf = θb Rf = 75 C 3.07 K W = 24.4 W . COMMENTS: To maximize fin performance, every effort should be made to minimize contact resistance. < PROBLEM 3.147 KNOWN: Dimensions and materials of a finned (annular) cylinder wall. Heat flux and ambient air conditions. Contact resistance. FIND: Surface and interface temperatures (a) without and (b) with an interface contact resistance. 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 long section. The following calculations are based on a unit length. The inner surface temperature may be obtained from T −T q′ = i ∞ = q′′ ( 2π ri ) = 105 W/m 2 × 2π × 0.06 m = 37, 700 W/m i R ′tot where R ′tot = R ′ + R ′t,c + R ′w + R ′ c equiv ; −1 R′ equiv = (1/ R ′ + 1/ R ′ ) f b . R ′ , Conduction resistance of cylinder wall: c R′ = c ln ( r1 / ri ) ln ( 66/60 ) = 2π k 2π (50 W/m ⋅ K ) R ′ , Contact resistance: t,c = 3.034 ×10−4 m ⋅ K/W R ′t,c = R ′′ / 2π r1 = 10−4 m 2 ⋅ K/W/2π × 0.066 m = 2.411×10−4 m ⋅ K/W t,c ′w , Conduction resistance of aluminum base: R R ′w = ln ( rb / r1 ) 2π k = ln ( 70/66 ) 2π × 240 W/m ⋅ K = 3.902 × 10−5 m ⋅ K/W R ′ , Resistance of prime or unfinned surface: b R′ = b 1 1 = = 454.7 × 10−4 m ⋅ K/W 2 ⋅ K × 0.5 × 2π 0.07 m hA′ 100 W/m b ( ) R ′ , Resistance of fins: The fin resistance may be determined from f T −T 1 R′ = b ∞ = f q′ ηf hA′ f f The fin efficiency may be obtained from Fig. 3.19, r2c = ro + t/2 = 0.096 m Lc = L+t/2 = 0.026 m Continued ….. PROBLEM 3.147 (Cont.) A p = Lc t = 5.2 × 10−5 m 2 r2c / r1 = 1.45 ( L3/2 h/kA p c ) 1/ 2 = 0.375 Fig. 3.19 → ηf ≈ 0.88. The total fin surface area per meter length 22 A′ = 250 π ro − rb × 2 = 250 m -1 2π 0.0962 − 0.07 2 m 2 = 6.78 m. f ) ( Hence ) ( R ′ = 0.88 ×100 W/m 2 ⋅ K × 6.78 m f −1 ( = 16.8 ×10−4 m ⋅ K/W ) −4 −4 W/m ⋅ K = 617.2 W/m ⋅ K 1/ R ′ equiv = 1/16.8 ×10 + 1/ 454.7 × 10 −4 R′ equiv = 16.2 × 10 m ⋅ K/W. Neglecting the contact resistance, R ′tot = (3.034 + 0.390 + 16.2 )10−4 m ⋅ K/W = 19.6 × 10−4 m ⋅ K/W Ti = q′R ′tot + T∞ = 37, 700 W/m × 19.6 × 10-4 m ⋅ K/W+320 K = 393.9 K < T1 = Ti − q′R ′w = 393.9 K − 37, 700 W/m × 3.034 × 10-4 m ⋅ K/W = 382.5 K < Tb = T1 − q′R ′ = 382.5 K − 37, 700 W/m × 3.902 × 10-5 m ⋅ K/W = 381.0 K. < b Including the contact resistance, ( ) R ′tot = 19.6 ×10−4 + 2.411× 10−4 m ⋅ K/W = 22.0 ×10−4 m ⋅ K/W Ti = 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 the fins. In this case R ′tot = R ′ + R ′ c conv , where 1 1 = = 241.1× 10−4 m ⋅ K/W. R′ conv = 2 ⋅ K 2π 0.066 m h2π r1 100 W/m ( ) Hence, R tot = 244.1×10−4 m ⋅ K/W, and Ti = q′R ′tot + T∞ = 37, 700 W/m × 244.1×10-4 m ⋅ K/W+320 K = 1240 K. Hence, the fins have a significant effect on reducing the cylinder temperature. (3) The overall surface efficiency is ηo = 1 − ( A′ / A′ )(1 − ηf ) = 1 − 6.78 m/7.00 m (1 − 0.88 ) = 0.884. f t It follows that q′=ηo h o A′tθ b = 37, 700 W/m, which agrees with the prescribed value. PROBLEM 3.148 KNOWN: Dimensions and materials of a finned (annular) cylinder wall. Combustion gas and ambient air conditions. Contact resistance. FIND: (a) Heat rate per unit length and surface and interface temperatures, (b) Effect of increasing the fin thickness. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Uniform h over surfaces, (4) Negligible radiation. ANALYSIS: (a) The heat rate per unit length is Tg − T∞ q′ = R ′tot where R ′ ot = R ′ + R ′ + R ′ + R ′ + R ′ , and t g w t,c b t,o ( R ′ = h g 2π ri g R′ = w )−1 = (150 W ln ( r1 ri ) ln (66 60 ) = 2π k w m 2 ⋅ K × 2π × 0.06m 2π ( 50 W m ⋅ K ) ( ) ) −1 = 0.0177m ⋅ K W , = 3.03 × 10−4 m ⋅ K W , R ′ = R ′′ 2π r1 = 10−4 m 4 ⋅ K W 2π × 0.066m = 2.41 × 10−4 m ⋅ K W t,c t,c R′ = b ln ( rb r1 ) = 2π k ηo = 1 − = 3.90 × 10−5 m ⋅ K W , 2π × 240 W m ⋅ K −1 R t,o = (ηo hA′ ) t ln ( 70 66 ) , N′A f (1 − ηf ) , A′t ( 2 2 A f = 2π roc − rb ) A′t = N′A f + (1 − N′t ) 2π rb ηf = ( 2rb m ) 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/ 2 Continued... PROBLEM 3.148 (Cont.) Once the heat rate is determined from the foregoing expressions, the desired interface temperatures may be obtained from Ti = Tg − q′R ′ g ( ) T1,o = Tg − q′ ( R ′ + R ′ + R ′t,c ) g w Tb = Tg − q′ ( R ′ + R ′ + R ′ + R ′ ) g w t,c b T1,i = Tg − q′ R ′ + R ′ g w For the specified conditions we obtain A′ = 7.00 m, ηf = 0.902, ηo = 0.906 and R ′ ,o = 0.00158 t t m⋅K/W. It follows that q′ = 39, 300 W m Ti = 405K, T1,i = 393K, T1,o = 384K, Tb = 382K < < (b) The Performance Calculation, Extended Surface Model for the Circular Fin Array may be used to assess the effects of fin thickness and spacing. Increasing the fin thickness to t = 3 mm, with δ = 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 fin t array ( R ,o = 0.00188 m⋅K/W), and hence a reduction in the heat rate ( q′ = 38,700 W/m) and an increase t in 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 order of magnitude, incremental changes in R t ,o will not have a significant effect on q or the interface temperatures. PROBLEM 3.149 KNOWN: Dimensions of finned aluminum sleeve inserted over transistor. Contact resistance and convection 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 of transistor, (3) One-dimensional radial heat transfer, (4) Constant properties, (5) Negligible radiation. ANALYSIS: With 2π 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 maintain satisfactory airflow between the fins, and this option is not particularly attractive. Because the fin efficiency for the prescribed conditions is close to unity ( ηf = 0.998), there is little advantage to replacing the aluminum with a material of higher thermal conductivity (e.g. Cu with k ~ 400 W/m⋅K). However, the large value of ηf suggests that significant benefit could be gained by increasing the fin length, L = r3 r2 . It is also evident that the thermal contact resistance is large, and from Table 3.2, it’s clear that a significant 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 m2⋅K/W is certainly feasible. t Table 1.1 suggests that, by increasing the velocity of air flowing over the fins, a larger convection coefficient may be achieved. A value of h = 100 W/m2⋅K would not be unreasonable. As options for enhancing heat transfer, we therefore use the IHT Performance Calculation, Extended Surface 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 m2⋅K/W and 25 ≤ h ≤ 100 W/m2⋅K. As shown below, there is a t significant enhancement in heat transfer associated with reducing R ′′,c from 10-3 to 10-4 m2⋅K/W, for t which R t ,c decreases from 13.26 to 1.326 K/W. At this value of R ′′,c , the reduction in R t ,o from t 23.45 to 12.57 K/W which accompanies an increase in L from 10 to 20 mm becomes significant, yielding a heat rate of q t = 4.30 W for R ′′,c = 10-4 m2⋅K/W and L = 20 mm. However, since R t ,o >> R t,c , little t benefit is gained by further reducing R ′′,c to 10-5 m2⋅K/W. t Heat rate, qt(W) 5 4 3 2 1 0 0.01 0.012 0.014 0.016 0.018 0.02 Fin length, L(m) h = 25 W/m^2.K, R''t,c = E-3 m^2.K/W h = 25 W/m^2.K, R''t,c = E-4 m^2.K/W h = 25 W/m^2.K, R''t,c = E-5m^2.K/W Continued... PROBLEM 3.149 (Cont.) To derive benefit from a reduction in R ′′,c to 10-5 m2⋅K/W, an additional reduction in R t ,o must be t made. This can be achieved by increasing h, and for L = 20 mm and h = 100 W/m2⋅K, R t ,o = 3.56 K/W. With R ′′,c = 10-5 m2⋅K/W, a value of q t = 16.04 W may be achieved. t Heat rate, qt(W) 20 16 12 8 4 0 0.01 0.012 0.014 0.016 0.018 0.02 Fin length, L(m) h = 25 W/m^2.K, R''t,c = E-5 m^2.K/W h = 50 W/m^2.K, R''t,c = E-5 m^2.K/W h = 100 W/m^2.K, R''t,c = E-5 m^2.K/W COMMENTS: In assessing options for enhancing heat transfer, the limiting (largest) resistance(s) should be identified and efforts directed at their reduction. PROBLEM 3.150 KNOWN: Diameter and internal fin configuration of copper tubes submerged in water. Tube wall temperature 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 represented as 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′ = ηo hA′ Tg − Ts t t ( ηo = 1 − NA′ f 1 −η (f ′t A ) ) NA′ = 4 × 2L = 8 ( 0.025m ) = 0.20m f A′ = NA′ + A′ = 0.20m + (π D − 4t ) = 0.20m + (π × 0.05m − 4 × 0.005m ) = 0.337m t f b For an adiabatic fin tip, q M tanh mL ηf = f = q max h ( 2L ⋅ 1) Tg − Ts ( M = [h2 (1m + t ) k (1m × t )] 1/ 2 mL = {[h2 (1m + t )] ) (Tg − Ts ) ≈ 30 W ( m ⋅ K ( 2m ) 400 W m ⋅ K 0.005m 2 30 W m 2 ⋅ K ( 2m ) 1/ 2 [k (1m × t )]} L ≈ 2 400 W m ⋅ K 0.005m ( 2 ) 1/ 2 ( 400K ) = 4382W 1/ 2 ) 0.025m = 0.137 Hence, tanh mL = 0.136, and 4382W ( 0.136 ) 595W ηf = = = 0.992 2 2 600W 30 W m ⋅ K 0.05m ( 400K ) ( ηo = 1 − 0.20 0.337 ( ) (1 − 0.992 ) = 0.995 ) q′ = 0.995 30 W m 2 ⋅ K 0.337m ( 400K ) = 4025 W m t ( ) COMMENTS: Alternatively, q′ = 4q′ + h ( A′ − A′ ) Tg − Ts . Hence, q′ = 4(595 W/m) + 30 t f t f 2 W/m ⋅K (0.137 m)(400 K) = (2380 + 1644) W/m = 4024 W/m. PROBLEM 3.151 KNOWN: Internal and external convection conditions for an internally finned tube. Fin/tube dimensions 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) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient on finned surfaces, (6) Tube wall may be unfolded and approximated as a plane surface with N straight rectangular fins. PROPERTIES: Copper: k = 400 W/m⋅K; St.St.: k = 20 W/m⋅K. ANALYSIS: The heat rate per unit length may be expressed as Tg − Tw q′ = R′ t,o(c) + R ′ cond + R ′ conv,o where ( ) R t,o(c) = ηo(c) h g A′ , t A′ = NA′ + ( 2π r1 − Nt ) , t f R′ cond = ln ( r2 r1 ) 2π k , and ηo(c) = 1 − NA′ f A′ t A′ = 2r1 , f ηf 1 − , C1 ( ( ηf = tanh mr1 mr1 , m = 2h g kt −1 R′ conv,o = ( 2π r2 h w ) ) C1 = 1 + ηf h g A′ R ′′ A′ f t,c c,b , )1/ 2 A′ ,b = t , c . Using the IHT Performance Calculation, Extended Surface Model for the Straight Fin Array, the following results were obtained. For the base case, q′ = 3857 W/m, where R ′ ,o(c) = 0.101 m⋅K/W, t -5 R′ cond = 7.25 × 10 m⋅K/W and R ′ onv,o = 0.00265 m⋅K/W. If the contact resistance is eliminated c ( R ′′ = 0), q T = 3922 W/m, where R ′ ,o = 0.0993 m⋅K/W. If the number of fins is increased to N = 8, t,c t q′ = 5799 W/m, with R ′ ,o(c) = 0.063 m⋅K/W. If the material is changed to stainless steel, q′ = 3591 t W/m, with R ′ ,o(c) = 0.107 m⋅K/W and R ′ ond = 0.00145 m⋅K/W. t c COMMENTS: The small reduction in q T associated with use of stainless steel is perhaps surprising, in view of the large reduction in k. However, because h g is small, the reduction in k does not significantly reduce the fin efficiency ( ηf changes from 0.994 to 0.891). Hence, the heat rate remains large. The influence of k would become more pronounced with increasing h g . PROBLEM 3.152 KNOWN: 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 inner and outer surface temperatures, (c) Surface heat rates and temperatures as a function of volumetric heating 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/m⋅K. 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 o 2 2π k s qro ri2 2 ′ ( ro ) = qπ ro − q 1 − + ( Ts,o − Ts,i ) ln ( ro ri ) 4k s r 2 o 2π k s q′ ( ri ) = qπ 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 ) 2 Uo Ts,o − T∞,o = − qro 2 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 overall heat transfer coefficient U o in the form Uo = ( q′ ( ro ) A′ Ts,o − T∞,o w ) = 1 A′ = t ηo(c) h o A′ R ′ w t,o(c) A′ w where ηo(c) is determined from Equations 3.105a,b and A′ = 2π ro . w Continued... 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 relations of part (b) with the capabilities of the IHT Performance Calculation Extended Surface Model for a Circular Fin Array yields the following graphical results. 500 460 420 380 340 300 1E7 2E7 3E7 4E7 5E7 6E7 7E7 8E7 9E7 1E8 Heat generation, qdot(W/m^3) Inner surface temperature, Ts,i Outer surface temperature, Ts,o It is in this range that the upper limit of Ts,i = 373 K is exceeded for q = 4.9 × 107 W/m3, while the corresponding value of Ts,o = 379 K is well below the prescribed upper limit. The expressions of part Surface heat rates, q'(W/m) (a) yield the following results for the surface heat rates, where heat transfer in the negative r direction corresponds to q′ ( ri ) < 0. 50000 30000 10000 -10000 -30000 -50000 1E7 2E7 3E7 4E7 5E7 6E7 7E7 8E7 9E7 1E8 Heat 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 transfer −1 from the inner and outer surfaces. Specifically, R ′ conv,i = ( h i 2π ri ) = 0.00318 m⋅K/W is slightly smaller than R ′ t,o(c) = 0.00411 m⋅K/W, in which case q′ ( ri ) is slightly larger than q′ ( ro ) , while Ts,i is slightly smaller than Ts,o . Note that the solution must satisfy the energy conservation requirement, 2 π ro − ri2 q = q′ ( ri ) + q′ ( ro ) . ( ) PROBLEM 4.1 KNOWN: Method of separation of variables (Section 4.2) for two-dimensional, steady-state conduction. 2 FIND: Show that negative or zero values of λ , the separation constant, result in solutions which cannot satisfy the boundary conditions. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. 2 ANALYSIS: From Section 4.2, identification of the separation constant λ leads to the two ordinary differential equations, 4.6 and 4.7, having the forms d 2X d 2Y + λ 2X = 0 − λ 2Y = 0 (1,2) 2 2 dx dy and the temperature distribution is θ ( x,y ) = X ( x ) ⋅ Y (y ). (3) 2 Consider now the situation when λ = 0. From Eqs. (1), (2), and (3), find that X = C1 + C 2x, Y = C3 + C 4y and θ ( 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: θ ( 0,y) = ( C1 + C 2 ⋅ 0 ) ( C3 + C4 y ) = 0 C1 = 0 y =0: θ ( x,0) = ( 0 + C2 X)( C3 + C4 ⋅ 0 ) = 0 C3 = 0 x = L: θ ( L,0) = ( 0 + C2 L)( 0 + C4 y ) = 0 C2 = 0 y = W: θ ( x,W ) = ( 0 + 0 ⋅ x )( 0 + C4 W ) = 1 0≠1 2 The last boundary condition leads to an impossibility (0 ≠ 1). We therefore conclude that a λ value of zero will not result in a form of the temperature distribution which will satisfy the boundary 2 conditions. Consider now the situation when λ < 0. The solutions to Eqs. (1) and (2) will be and X = C5e-λ x + C6e +λ x , Y = C7cos λ y + C8sin λy θ ( x,y ) = C5 e-λ x + C6 e +λ x [ C7 cos λ y + C8 sin λ y ] . Evaluate the constants for the boundary conditions. y = 0 : θ ( x,0) = C5 e-λ x + C6 e-λ x [ C7 cos 0 + C8 sin 0 ] = 0 C e0 + C e 0 [ 0 + C sin λ y] = 0 x = 0 : θ ( 0,y) = 5 6 8 (5,6) (7) C7 = 0 C8 = 0 If C8 = 0, a trivial solution results or C5 = -C6. x = L: θ ( L,y ) = C5 e-xL − e+xL C8 sin λ y = 0. From the last boundary condition, we require C5 or C8 is zero; either case leads to a trivial solution with either no x or y dependence. PROBLEM 4.2 KNOWN: Two-dimensional rectangular plate subjected to prescribed uniform temperature boundary conditions. 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) and T(1,y). SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS: From Section 4.2, the temperature distribution is n +1 + 1 nπ x sinh ( nπ y L ) T − T1 2 θ ( −1) θ ( x, y ) ≡ sin . (1,4.19) = ⋅ T2 − T1 π n L sinh ( nπ W L ) n =1 Considering 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 nπ sinh ( nπ 4 ) T − T1 2 θ ( −1) θ (1, 0.5) ≡ sin . = ⋅ T2 − T1 π n 2 sinh ( nπ 2 ) n =1 When n is even (2, 4, 6 ...), the corresponding term is zero; hence we need only consider n = 1, 3, 5, 7 and 9 as the first five non-zero terms. ∑ ∑ θ (1, 0.5) = 2 π sinh (π 4 ) 2 3π + sin 2sin π 2 sinh (π 2 ) 3 2 2 5π sin 5 2 θ (1, 0.5) = sinh (5π 4 ) 2 7π + sin sinh (5π 2 ) 7 2 sinh (3π 4 ) + sinh (3π 2 ) sinh (7π 4 ) 2 9π + sin sinh (7π 2 ) 9 2 sinh (9π 4 ) sinh (9π 2 ) 2 [0.755 − 0.063 + 0.008 − 0.001 + 0.000] = 0.445 π (2) T (1, 0.5) = θ (1, 0.5)( T2 − T1 ) + T1 = 0.445 (150 − 50 ) + 50 = 94.5 C . < Using Eq. (1), and writing out the first five terms of the series, expressions for θ(x,0.5) or T(x,0.5) and θ(1,y) or T(1,y) were keyboarded into the IHT workspace and evaluated for sweeps over the x or y variable. Note that for T(1,y), that as y → 1, the upper boundary, T(1,1) is greater than 150°C. Upon examination of the magnitudes of terms, it becomes evident that more than 5 terms are required to provide an accurate solution. T(x,0.5) or T(1,y), C If only the first three terms of the series, Eq. (2), are considered, the result will be θ(1,0.5) = 0.46; that is, there is less than a 0.2% effect. 150 130 110 90 70 50 0 0.2 0.4 0.6 0.8 x or y coordinate (m) T(1,y) T(x,0.5) 1 PROBLEM 4.3 KNOWN: 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 result based 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 is x =2 x =2 x =2 ∂T ∂θ q′ = − ∫ dq′y ( x, 0 ) = − ∫ − k dx = k ( T2 − T1 ) ∫ dx out ∂ y y =0 ∂y x =0 x =0 x =0 y =0 (1) where from the solution to Problem 4.2, n +1 + 1 nπ x sinh (nπ y L ) T − T1 2 ∞ ( −1) θ≡ sin . =∑ T2 − T1 π n L sinh ( nπ W L ) (2) n =1 Evaluate the gradient of θ from Eq. (2) and substitute into Eq. (1) to obtain q′ = k ( T2 − T1 ) out x=2 ∫ x =0 n +1 + 1 nπ x ( nπ L ) cosh ( nπ y L ) 2 ∞ ( −1) sin ∑ π n sinh ( nπ W L ) L n =1 dx y =0 n +1 2 +1 2 ∞ ( −1) 1 − cos nπ x q′ = k ( T2 − T1 ) ∑ out π n sinh ( nπ W L ) L x =0 n =1 n +1 +1 2 ∞ ( −1) 1 1 − cos ( nπ ) q′ = k ( T2 − T1 ) ∑ out π n sinh ( nπ L ) < n =1 To evaluate the first five, non-zero terms, recognize that since cos(nπ) = 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 ) C out + ( −1)6 + 1 5 ⋅ 1 sinh (5π 2 ) ( −1) + 1 ⋅ 2 ( −1) + 1 1 1 ⋅ ⋅ (2) (2) + π 1 sinh (π 2 ) 3 sinh (3π 2 ) (2 ) + 2 ( −1)8 + 1 7 4 ⋅ 1 sinh ( 7π 2 ) (2 ) + ( −1)10 + 1 9 ⋅ 1 sinh ( 9π 2 ) q′ = 3.183kW m [1.738 + 0.024 + 0.00062 + (...)] = 5.611kW m out ( 2 ) < COMMENTS: If the foregoing procedure were used to evaluate the heat rate into the upper surface, ′ qin = − x =2 ∫ dq′y ( x, W ) , it would follow that x =0 q′ = k ( T2 − T1 ) in n +1 +1 2 ∞ ( −1) coth ( nπ 2 ) 1 − cos ( nπ ) ∑ π n n =1 However, with coth(nπ/2) ≥ 1, irrespective of the value of n, and with ∞ ∑ (−1)n +1 + 1 n =1 n being a divergent series, the complete series does not converge and q′ → ∞ . This physically untenable in condition results from the temperature discontinuities imposed at the upper left and right corners. PROBLEM 4.4 KNOWN: 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 ) = ( C1cosλ x + C2 sinλ x ) C3e -λ y + C4e +λ y ) and the boundary conditions are: T(0,y) = 0, T(a,y) = 0, T(x,0) = 0, T(x.b) = Ax. Applying BC#1, T(0,y) = 0, find C1 = 0. Applying BC#2, T(a,y) = 0, find that λ = nπ /a with n = 1,2,…. Applying BC#3, T(x,0) = 0, find that C3 = -C4. Hence, the product solution is nπ T ( x,y ) = X ( x ) ⋅ Y ( y ) = C2 C4 sin x e + λ y − e- λ y . a Combining constants and using superposition, find ( T ( x,y ) = ) ∞ nπ x nπ y Cn sin sinh a . a n =1 ∑ To evaluate Cn, use orthogonal functions with Eq. 4.16 to find a nπ x nπ b a 2 nπ x Cn = ∫ Ax ⋅ sin a ⋅ dx/sinh a ∫0 sin a dx, 0 noting that y = b. The numerator, denominator and Cn, respectively, are: a 2 2 a 2 nπ x − ax cos nπ x = Aa − cos n π = Aa −1 n+1 , A ∫ x ⋅ sin ⋅ dx = A sin [ ( ) ] nπ ( ) a nπ a 0 a nπ nπ 0 a nπ x a 1 a nπ b a 2 nπ x nπ b 1 2nπ x nπ b sinh ∫0 sin a ⋅ dx = sinh a 2 x − 4n π sin a = 2 ⋅ sinh a , a 0 Aa 2 a nπ b nπ b ( −1) n+1 / sinh = 2Aa ( −1) n+1 / nπ sinh a a . nπ 2 Hence, the temperature distribution is nπ y sinh ∞ −1 n+1 2 Aa () nπ x a . T ( x,y ) = ∑ n ⋅ sin a π n =1 sinh nπ b a Cn = < PROBLEM 4.5 KNOWN: Long furnace of refractory brick with prescribed surface temperatures and material thermal conductivity. FIND: Shape factor and heat transfer rate per unit length using the flux plot method SCHEMATIC: 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 symmetrical element. Hence, the heat rte for the entire furnace per unit length is q′ = q S = 4 k ( T1 − T2 ) l l (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= Ml N or S M 8.5 = = = 2.83 lN 3 q′ = 4 × 2.83 ×1.2 W ( 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. The symmetrical section consists of two plane walls (horizontal and vertical) with an adjoining edge. Using the appropriate relations, the numerical values are, in the same order, S= 0.75m 0.5m l + 0.54l + l = 3.04l 0.5m 0.5m Note that this result compares favorably with the flux plot result of 2.83l. PROBLEM 4.6 KNOWN: Hot pipe embedded eccentrically in a circular system having a prescribed thermal conductivity. 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 section shown above is readily identified. Selecting four temperature increments (N = 4), construct the flux plot shown below. For the pipe system, the heat rate per unit length is q′ = q W = kS ( T1 − T2 ) = 0.5 × 4.26 (150 35 )o C = 245 W/m. l m ⋅K < COMMENTS: Note that in the lower, right-hand quadrant of the flux plot, the curvilinear squares are irregular. Further work is required to obtain an improved plot and, hence, obtain a more accurate estimate of the shape factor. PROBLEM 4.7 KNOWN: 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) Sketch the 25, 50 and 75°C isotherms, and (d) Same analysis on the shape but with adiabatic-isothermal boundary conditions reversed. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: (a) Using the methodology of Section 4.3.1, construct a flux plot. Note the line of symmetry which passes through the point P is an isotherm as shown above. It follows that T ( 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 the heat 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 = 4.2 but N = 2No = 8. Hence, S = So 2 = 0.53 and with q′ = q , giving q′ = 75 W m ⋅ K × 0.53 (100 − 0 ) C = 3975 W m . < (c) The isotherms for T = 50, 75 and 100°C are shown on the flux plot. The T = 25°C isotherm is symmetric with the T = 75°C isotherm. (d) By reversing the adiabatic and isothermal boundary conditions, the two-dimensional shape appears as shown 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 = 0.85 S = 2So = 1.70 and the heat rate per unit length from Eq. (1) is q′ = 75 W m ⋅ K × 1.70 (100 − 0 ) C = 12, 750 W m < From the flux plot, estimate that T(P) ≈ 40°C. < COMMENTS: (1) By inspection of the shapes for parts (a) and (b), it is obvious that the heat rate for the 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 and isotherms, would you expect the shape factor for parts (a) to be the reciprocal of part (b)? PROBLEM 4.8 KNOWN: 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) Constant properties. ANALYSIS: With symmetry about the midplane, only one-half of the object need be considered as shown 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 is M 7 S = l = l = 1.17l. N 6 For the complete system, the shape factor is then S = 2.34l. < PROBLEM 4.9 KNOWN: 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 or maintained 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 flux plot 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). Application A B M 10.3 6.2 N 4 4 So/ 2.58 1.55 q′ (W/m) 11,588 6,975 < < COMMENTS: (1) For application A, most of the heat lanes leave the inner surface (T1) on the upper portion. (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 heat flows (lane 6 + 0.2). Note the shapes of the isotherms near the right-hand, insulated boundary and that they intersect the boundary normally. PROBLEM 4.10 KNOWN: Shape and surface conditions of a support column. FIND: (a) Heat transfer rate per unit length. (b) Height of a rectangular bar of equivalent thermal resistance. 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/m⋅K. ANALYSIS: (a) From the flux plot for the half section, M ≈ 5 and N ≈ 8. Hence for the full section Ml ≈ 1.25l N q = Sk ( T1 − T2 ) S= 2 q′ ≈ 1.25 × 62.7 W (100 − 0)o C m ⋅K < q′ ≈ 7.8 kW/m. (b) The rectangular bar provides for one-dimensional heat transfer. Hence, q=kA Hence, ( T1 − T2 ) H (T − T ) = k ( 0.3l ) 1 2 H ( ) o 0.3k ( T1 − T2 ) 0.3m ( 62.7 W/m ⋅ K ) 100 C H= = = 0.24m. q′ 7800 W/m < COMMENTS: The fact that H < 0.3m is consistent with the requirement that the thermal resistance of the trapezoidal column must be less than that of a rectangular bar of the same height and top width (because the width of the trapezoidal column increases with increasing distance, x, from the top). Hence, if the rectangular bar is to be of equivalent resistance, it must be of smaller height. PROBLEM 4.11 KNOWN: Hollow prismatic bars fabricated from plain carbon steel, 1m in length with prescribed temperature difference. FIND: Shape factors and heat rate per unit length. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. PROPERTIES: Table A-1, Steel, Plain Carbon (400K), k = 57 W/m⋅K. ANALYSIS: Construct a flux plot on the symmetrical sections (shaded-regions) of each of the bars. The shape factors for the symmetrical sections are, So,A = Ml 4 = l = 1l N4 So,B = Ml 3.5 = l = 0.88l. N 4 Since each of these sections is ¼ of the bar cross-section, it follows that SA = 4 ×1l = 4l SB = 4 × 0.88l = 3.5l. < The heat rate per unit length is q′ = q/l = k ( S/l ) (T1 − T2 ) , q′A = 57 W × 4 (500 − 300 ) K = 45.6 kW/m m⋅K < q′B = 57 W × 3.5 ( 500 − 300 ) K = 39.9 kW/m. m ⋅K < PROBLEM 4.12 KNOWN: Two-dimensional, square shapes, 1 m to a side, maintained at uniform temperatures as prescribed, perfectly insulated elsewhere. FIND: Using the flux plot method, estimate the heat rate per unit length normal to the page if the thermal conductivity is 50 W/m⋅K ASSUMPTIONS: (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 factors from which the heat rates can be calculated. With Figure (a), begin at the lower-left side making the isotherms almost equally spaced, since the heat flow will only slightly spread toward the right. Start sketching the adiabats in the vicinity of the T2 surface. The dashed line represents the adiabat which separates the shape into two segments. Having recognized this feature, it was convenient to identify partial heat lanes. Figure (b) is less difficult to analyze since the isotherm intervals are nearly regular in the 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 = N 6 S′ = M 4.5 = = 0.90 N 5 S′ = 0.70 q′ = 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 and 915 W/m, respectively, for Figures (a) and (b). The estimate for the less difficult Figure (b) is within 2% of the numerical method result. For Figure (a), our flux plot result was 27% low. PROBLEM 4.13 KNOWN: Uniform media of prescribed geometry. FIND: (a) Shape factor expressions from thermal resistance relations for the plane wall, cylindrical shell and spherical shell, (b) Shape factor expression for the isothermal sphere of diameter D buried in an infinite medium. ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform properties. ANALYSIS: (a) The relationship between the shape factor and thermal resistance of a shape follows from their definitions in terms of heat rates and overall temperature differences. q = kS∆T ( 4.25 ), q= ∆T Rt (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= 2π Lk L kA 2π L lnr2 / r . 1 S= A . L < < (L into the page) Rt = Spherical shell: 1 1 1 − 4π k r1 r2 S= 4π . l/r1 − l/r2 < (b) The shape factor for the sphere of diameter D in an infinite medium can be derived using the alternative conduction analysis of Section 3.1. For this situation, qr is a constant and Fourier’s law has the form dT q r = −k 4π r 2 . ( ) dr Separate variables, identify limits and integrate. ∞ T2 q dr − r ∫D / 2 = ∫ T dT 1 4π k r2 D q r = 4π k ( T1 − T 2 ) 2 qr 1 ∞ q 2 − = − r 0 − = ( T2 − T1 ) − r 4π k D/2 4π k D or S = 2π D. < COMMENTS: Note that the result for the buried sphere, S = 2π D, can be obtained from the expression for the spherical shell with r2 = ∞. Also, the shape factor expression for the “isothermal sphere buried in a semi-infinite medium” presented in Table 4.1 provides the same result with z → ∞. PROBLEM 4.14 KNOWN: Heat generation in a buried spherical container. FIND: (a) Outer surface temperature of the container, (b) Representative isotherms and heat flow lines. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Soil is a homogeneous medium with constant properties. PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/m⋅K. & ANALYSIS: (a) From an energy balance on the container, q = Eg and from the first entry in Table 4.1, q= 2π D k ( T1 −T 2 ). l − D/4z Hence, T1 = T2 + q 1− D/4z 500W 1− 2m/40m = 20o C+ = 92.7 o C W k 2π D 2π ( 2m ) 0.52 m ⋅K (b) The isotherms may be viewed as spherical surfaces whose center moves downward with increasing radius. The surface of the soil is an isotherm for which the center is at z = ∞. < PROBLEM 4.15 KNOWN: 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 wall conduction resistances. PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/m⋅K; Table A-3, Cellular glass (365K): k = 0.069 W/m⋅K. ANALYSIS: The heat rate can be expressed as T −T q= 1 2 R tot where the thermal resistance is Rtot = Rins + Rsoil. From Eq. 3.28, R ins = ln ( D2 / D1 ) = ln ( 0.7m/0.5m ) 2π Lk ins 2π L × 0.069 W/m ⋅ K From Eq. 4.27 and Table 4.1, R soil = = 0.776m ⋅ K/W . L cosh -1 ( 2z/D2 ) cosh -1 ( 3/0.7 ) 1 0.653 = = = m ⋅ K/W. SKsoil 2π Lksoil 2π × ( 0.52 W/m ⋅ K ) L L Hence, q= (120 − 0)o C 1 m ⋅K ( 0.776 + 0.653) L W q′ = q/L = 84 W/m. = 84 W ×L m < COMMENTS: (1) Contributions of the soil and insulation to the total resistance are approximately the 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, in which case the foregoing result would overestimate the heat loss. A calculation of this resistance may be based on results presented in Chapter 8. (3) Since z > 3D/2, the shape factor for the soil can also be evaluated from S = 2π L/ ln (4z/D) of Table 4.1, and an equivalent result is obtained. PROBLEM 4.16 KNOWN: Operating conditions of a buried superconducting cable. FIND: Required cooling load. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Two-dimensional conduction in soil, (4) One-dimensional conduction in insulation. ANALYSIS: The heat rate per unit length is q′ = q′ = Tg − Tn R′ + R′I g Tg − Tn kg ( 2π /ln ( 4z/Do ) ) −1 + ln ( Do / Di ) / 2π ki where 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 ) ( 2π /ln ( 8/0.2 ) ) 223 K q′ = ( 0.489+22.064 ) m ⋅ K/W q′ = 9.9 W/m. −1 + ln ( 2 ) / 2π × 0.005 W/m ⋅ K < COMMENTS: The heat gain is small and the dominant contribution to the thermal resistance is made by the insulation. PROBLEM 4.17 KNOWN: Electrical heater of cylindrical shape inserted into a hole drilled normal to the surface of a 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 with constant properties, (3) Negligible heat loss to surroundings above block surface, (4) Heater can be approximated as isothermal at T1. ANALYSIS: The temperature of the heater surface follows from the rate equation written as T1 = T2 + q/kS where S can be estimated from the conduction shape factor given in Table 4.1 for a “vertical cylinder in a semi-infinite medium,” S = 2π L/l n ( 4L/D ) . Substituting numerical values, find 4 × 0.1m S = 2π × 0.1m/l n 0.005m = 0.143m. The temperature of the heater is then T1 = 25° C + 50 W/(5 W/m⋅K × 0.143m) = 94.9° C. < COMMENTS: (1) Note that the heater has L >> D, which is a requirement of the shape factor expression. (2) Our calculation presumes there is negligible thermal contact resistance between the heater and the medium. 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 from measurement of q and T1. PROBLEM 4.18 KNOWN: 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) Constant properties, (4) Pipe lines are buried very deeply, approximating burial in an infinite medium, (5) Pipe length >> D1 or D2 and w > D1 or D2. ANALYSIS: The heat transfer rate per unit length from the hot pipe to the cool pipe is q′ = qS = k ( T1 −T 2 ) . LL The shape factor S for this configuration is given in Table 4.1 as S= 2π L 2 2 2 -1 4w − D1 − D2 cosh 2D1D2 . Substituting numerical values, 2 2 2 S −1 4 × ( 0.5m ) − ( 0.1m ) − ( 0.075m ) = 2π /cosh -1(65.63) = 2π /cosh L 2 × 0.1m × 0.075m S = 2π /4.88 = 1.29. L Hence, the heat rate per unit length is o q′ = 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 soil temperature is greater than 5° C. How would you estimate the heat gain if the soil were at 25° C? PROBLEM 4.19 KNOWN: 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 tabulated relation, (b) Shape factor using flux plot method. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constant properties, (4) Concrete slab infinitely long in horizontal plane, L >> z. PROPERTIES: Table A-3, Concrete, stone mix (300K): k = 1.4 W/m⋅K. ANALYSIS: (a) If we relax the restriction that z >> D/2, the embedded tube-slab system corresponds to the fifth case of Table 4.1. Hence, S= 2π L ln (8z/ π D ) where L is the length of the system normal to the page, z is the half-thickness of the slab and D is the diameter of the tube. Substituting numerical values, find S = 2π L/l n( 8× 50mm/π 50mm ) = 6.72L. Hence, the heat rate per unit length is q′ = qS W = k ( T1 − T2 ) = 6.72 ×1.4 ( 85 − 20 )o C = 612 W. LL m ⋅K (b) To find the shape factor using the flux plot method, first identify the symmetrical section bounded by the symmetry adiabats formed by the horizontal and vertical center lines. Selecting four temperature increments (N = 4), the flux plot can then be constructed. From Eq. 4.26, the shape factor of the symmetrical section is So = ML/N = 6L/4 = 1.5L. For the tube-slab system, it follows that S = 4So = 6.0L, which compares favorably with the result obtained from the shape factor relation. PROBLEM 4.20 KNOWN: 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/m⋅K. ANALYSIS: The heat rate can be expressed as q = Sk∆T1-2 = Sk ( T1 − T2 ) From Table 4.1, the shape factor is S= 2π L . 1.08 w ln D Hence, q 2π k ( T1 − T2 ) q′ = = L ln 1.08 w D q′ = 2π × 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 4.21 KNOWN: Thin-walled copper tube enclosed by an eccentric cylindrical shell; intervening space filled with insulation. FIND: Heat loss per unit length of tube; compare result with that of a concentric tube-shell arrangement. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Thermal resistances of copper 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 is q′ = k ( S/ l ) (T1 − T2 ) and from Table 4.1 for this geometry, D2 + d 2 − 4z 2 S = 2π /cosh-1 . l 2Dd Substituting numerical values, all dimensions in mm, 2 2 2 S -1 120 + 30 − 4 ( 20 ) = 2π /cosh-1 1.903 = 4.991. = 2π /cosh ( ) l 2 ×120 × 30 Hence, the heat loss is o q′ = 0.05W/m ⋅ K × 4.991( 85 − 35 ) C = 12.5 W/m. < If the copper tube were concentric with the shell, but all other conditions were the same, the heat loss would be q′c = 2π k ( T1 − T2 ) ln ( D2 / D1 ) using Eq. 3.27. Substituting numerical values, ( W ( 85 − 35 )o C/ln 120 / 30 m⋅ K q′c = 11.3 W/m. q′c = 2 π × 0.05 ) COMMENTS: As expected, the heat loss with the eccentric arrangement is larger than that for the concentric arrangement. The effect of the eccentricity is to increase the heat loss by (12.5 - 11.3)/11.3 ≈ 11%. PROBLEM 4.22 KNOWN: 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) Constant properties. 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 = = 1.25m L 0.05m Plane Walls (6) SW = Edges (12) SE = 0.54D = 0.54 × 0.25m = 0.14m Corners (8) SC = 0.15L = 0.15 × 0.05m = 0.008m. The heat rate in terms of the shape factors is q = kS ( T1 − T2 ) = k ( 6S W + 12SE + 8SC ) ( T1 − T2 ) W q = 1.1 ( 6 ×1.25m+12 × 0.14m+0.15 × 0.008m ) ( 600 − 75 )o C m⋅K q = 5.30 kW. COMMENTS: Note that the restrictions for SE and SC have been met. < PROBLEM 4.23 KNOWN: Dimensions, thermal conductivity and inner surface temperature of furnace wall. Ambient conditions. FIND: Heat loss. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Uniform convection coefficient over entire outer surface of container. ANALYSIS: From the thermal circuit, the heat loss is q= Ts,i − T∞ R cond(2D) + R conv where Rconv = 1/hAs,o = 1/6(hW2) = 1/6[5 W/m2⋅K(5 m)2] = 0.00133 K/W. From Eq. (4.27), the twodimensional conduction resistance is R cond(2D) = 1 Sk where the shape factor S must include the effects of conduction through the 8 corners, 12 edges and 6 plane 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 L where As,i = (W - 2L)2. Hence, S = 8 (0.15 × 0.35 ) + 12 × 0.54 ( 4.30 ) + 6 (52.83) m S = ( 0.42 + 27.86 + 316.98 ) m = 345.26m and Rcond(2D) = 1/(345.26 m × 1.4 W/m⋅K) = 0.00207 K/W. Hence (1100 − 25) C q= (0.00207 + 0.00133) K W = 316 kW < COMMENTS: The heat loss is extremely large and measures should be taken to insulate the furnace. PROBLEM 4.24 KNOWN: Platen heated by passage of hot fluid in poor thermal contact with cover plates exposed to cooler ambient air. FIND: (a) Heat rate per unit thickness from each channel, q ′ , (b) Surface temperature of cover i plate, Ts , (c) q′i and Ts if lower surface is perfectly insulated, (d) Effect of changing centerline spacing on q′i and Ts SCHEMATIC: D=15 mm LA=30 mm Ti=150° C Lo=60 mm LB=7.5 mm 2 hi=1000 W/m ⋅K 2 T∞=25° C ho=200 W/m ⋅K kA=20 W/m⋅K kB=75 W/m⋅K R ′′ = 2.0 ×10 t,c −4 2 m ⋅ K/W ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in platen, but one-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 the following 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 is q′i = 4 ( Ti − T∞ ) / R′tot (1) R ′tot = [1/1000 W/m 2 ⋅ K ( π 0.015m/4 ) + 1/20 W/m ⋅ K ×1.06 + 2.0 × 10−4 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/W R ′tot = 0.309 m ⋅ K/W q′i = 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 as Ts − T∞ q′i / 4 = 1/ho ( Lo / 2 ) Continued ….. (2) PROBLEM 4.24 (Cont.) q′ 1 1.62 kW Ts = T∞ + i = 25o C + × 0.167 m ⋅ K/W 4 ho ( Lo / 2 ) 4 Ts = 25o C + 67.6o C ≈ 93oC. < (c,d) The effect of the centerline spacing on q′i and Ts can be understood by examining the relative magnitudes of the thermal resistances. The dominant resistance is that due to the ambient air convection process which is inversely related to the spacing Lo. Hence, from Eq. (1), the heat rate will increase nearly linearly with an increase in Lo, q′i ~ 1 1 ≈ ~ L o. R′ tot 1 / h o ( Lo / 2 ) From Eq. (2), find q′ 1 ′o ∆T = T − T∞ = i ~ qi ⋅ L-1 ~ Lo ⋅ L-1 ≈ 1. s o 4 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 follows again from the fact that the overall resistance is dominated by the surface convection process. The temperature difference, Ts - T∞, would only increase slightly. PROBLEM 4.25 KNOWN: Long constantan wire butt-welded to a large copper block forming a thermocouple junction on 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/m2⋅K for block thermal conductivity 15 ≤ k ≤ 400 W/m⋅K. When is it advantageous to use smaller diameter wire? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Thermocouple wire behaves as a fin with constant heat 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/m⋅K; Constantan (350 K), kt ≈ 25 W/m⋅K. ANALYSIS: The thermocouple wire behaves as a long fin permitting heat to flow from the surface thereby depressing the sensing junction temperature below that of the block To. In the block, heat flows into the circular region of the wire-block interface; the thermal resistance to heat flow within the block is approximated as a disk of diameter D on a semi-infinite medium (kb, To). The thermocouple-block combination can be represented by a thermal circuit as shown above. The thermal resistance of the fin follows 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 hence Rblock = 1/kbS = 1/2kbD. From the thermal circuit, R block 1.27 To − Tj = (To − T∞ ) = (125 − 25 ) C ≈ 0.001(125 − 25 ) C = 0.1 C . R fin + R block 1273 + 1.27 < with P = πD and Ac = πD2/4 and the thermal resistances as ( R fin = 10 W m 2 ⋅ K (π 4 ) 25 W m ⋅ K × 1 × 10−3 m ) 3 −1/ 2 = 1273 K W R block = (1 2 ) × 393 W m ⋅ K × 10−3 m = 1.27 K W . (b) We keyed the above equations into the IHT workspace, performed a sweep on kb for selected values of 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 low values of kb and higher values of h. 5 Error, To-Tj (C) 4 3 2 1 0 0 100 200 300 Block thermal conductivity, kb (W/m.K) h = 25 W/m^2.K; D = 1 mm h = 10 W/m^2.K; D = 1mm h = 5 W/m^2.K; D = 1mm h = 25 W/m^2.K; D = 5 mm 400 PROBLEM 4.26 KNOWN: 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) Uniform convection coefficients at inner and outer surfaces. ANALYSIS: The heat loss can be expressed as q= T∞,1 − T∞,2 R conv,1 + R cond(2D) + R conv,2 where −1 R conv,1 = ( h1π D1L ) −1 R cond(2D) = (Sk ) ( = 50 W m 2 ⋅ K × π × 0.25 m × 2 m −1 = (8.59 m × 150 W m ⋅ K ) −1 R 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 W Hence, (300 − 25 ) C = 3.62 kW q= 0.076 K W T1 = T∞,1 − qR conv,1 = 300 C − 46 C = 254 C T2 = T∞,2 + qR conv,2 = 25 C + 226 C = 251 C < < < COMMENTS: The largest resistance is associated with convection at the outer surface, and the conduction resistance is much smaller than both convection resistances. Hence, (T2 - T∞,2) > (T∞,1 - T1) >> (T1 - T2). PROBLEM 4.27 KNOWN: Long fin of aluminum alloy with prescribed convection coefficient attached to different base materials (aluminum alloy or stainless steel) with and without thermal contact resistance R ′′ j at the t, 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 for t, the range 10 ≤ h ≤ 1000 W/m2⋅K for each base material. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Infinite fin. PROPERTIES: (Given) Aluminum alloy, k = 240 W/m⋅K, Stainless steel, k = 15 W/m⋅K. ANALYSIS: (a,b) From the thermal circuits, the heat rate and junction temperature are T −T Tb − T∞ qf = b ∞ = R tot R b + R t, j + R f (1) Tj = T∞ + qf R f (2) and, with P = πD and Ac = πD2/4, from Tables 4.1 and 3.4 find −1 R b = 1 Sk b = 1 ( 2Dk b ) = ( 2 × 0.005 m × k b ) R t, j = R ′′ j A c = 3 ×10−5 m 2 ⋅ K W π ( 0.005 m ) t, 2 −1/ 2 R f = ( hPkA c ) 4 = 1.528 K W 2 = 50 W m 2 ⋅ K π 2 (0.005 m ) 240 W m ⋅ K 4 −1/ 2 Without R ′′ j t, Base Al alloy St. steel Rb (K/W) 0.417 6.667 qf (W) 4.46 3.26 Tj (°C) 98.2 78.4 = 16.4 K W With R ′′ j t, qf (W) 4.09 3.05 Tj (°C) 92.1 75.1 (c) We used the IHT Model for Extended Surfaces, Performance Calculations, Rectangular Pin Fin to calculate qf for 10 ≤ h ≤ 100 W/m2⋅K by replacing R ′′ (thermal resistance at fin base) by the sum of the tc contact and spreading resistances, R ′′ j + R ′′ . t, b Continued... PROBLEM 4.27 (Cont.) Fin heat rate, qf (W) 6 5 4 3 2 1 0 20 40 60 80 100 Convection coefficient, h (W/m^2.K) Base material - aluminum alloy Base material - stainless steel COMMENTS: (1) From part (a), the aluminum alloy base material has negligible effect on the fin heat rate and depresses the base temperature by only 2°C. The effect of the stainless steel base material is substantial, reducing the heat rate by 27% and depressing the junction temperature by 25°C. (2) The contact resistance reduces the heat rate and increases the temperature depression relatively more with 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 both materials since the fin is the dominant resistance. As h increases, the effect of R ′′ becomes more b important. PROBLEM 4.28 KNOWN: Igloo constructed in hemispheric shape sits on ice cap; igloo wall thickness and inside/outside convection coefficients (hi, ho) are prescribed. FIND: (a) Inside air temperature T∞,i when outside air temperature is T∞,o = -40°C assuming occupants provide 320 W within igloo, (b) Perform parameter sensitivity analysis to determine which variables have significant effect on Ti. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Convection coefficient is the same on floor and ceiling of igloo, (3) Floor and ceiling are at uniform temperature, (4) Floor-ice cap resembles disk on semi-infinite medium, (5) One-dimensional conduction through igloo walls. PROPERTIES: Ice and compacted snow (given): k = 0.15 W/m⋅K. ANALYSIS: (a) The thermal circuit representing the heat loss from the igloo to the outside air and through the floor to the ice cap is shown above. The heat loss is T∞,i − T∞,o T∞,i − Tic q= . + R cv,c + R wall + R cv,o R cv,f + R cap 2 Convection, ceiling: R cv,c = Convection, outside: R cv,o = Convection, floor: R cv,f = Conduction, wall: Conduction, ice cap: R cap = = 2 R wall = 2 () h i 4π ri2 2 () 2 h o 4π ro 1 () h i π ri2 = = 2 2 15 W m 2 ⋅ K × 4π ( 2.3 m ) 2 1 6 W m 2 ⋅ K × π (1.8 m ) 2 = 0.00201K W = 0.01637 K W 1 1 1 1 1 1 r − r = 2 4π × 0.15 W m ⋅ K 1.8 − 2.3 m = 0.1281K W 4π k i o 1 = 1 1 = kS 4kri 4 × 0.15 W m ⋅ K × 1.8 m where S was determined from the shape factor of Table 4.1. Hence, T∞,i − ( −40 ) C q = 320 W = = 0.00819 K W 6 W m ⋅ K × 4π (1.8 m ) 2 ( 0.00818 + 0.1281 + 0.0020 ) K = 0.9259 K W T∞,i − ( −20 ) C W 320 W = 7.232( T∞,i + 40) + 1.06( T∞,i + 20) + ( 0.01637 + 0.9259 ) K T∞,i = 1.1°C. W < Continued... PROBLEM 4.28 (Cont.) (b) Begin the parameter sensitivity analysis to determine important variables which have a significant influence on the inside air temperature by examining the thermal resistances associated with the processes present in the system and represented by the network. Process Convection, outside Conduction, wall Convection, ceiling Convection, floor Conduction, ice cap Symbols Rcv,o Rwall Rcv,c Rcv,f Rcap Value (K/W) 0.0020 0.1281 0.0082 0.0164 0.9259 R21 R32 R43 R54 R65 It follows that the convection resistances are negligible relative to the conduction resistance across the igloo wall. As such, only changes to the wall thickness will have an appreciable effect on the inside air temperature relative to the outside ambient air conditions. We don’t want to make the igloo walls thinner and 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 network and perform the energy balances to obtain the inside air temperature as a function of the outside convection coefficient for selected increased thicknesses of the wall. Air temperature, Tinfi (C) 25 20 15 10 5 0 0 20 40 60 80 100 Outside coefficient, ho (W/m^2.K) Wall thickness, (ro-ri) = 0.5 m (ro-ri) = 0.75 m (ro-ri) = 1.0 m COMMENTS: (1) From the plot, we can see that the influence of the outside air velocity which controls the outside convection coefficient ho is negligible. (2) The thickness of the igloo wall is the dominant thermal resistance controlling the inside air temperature. PROBLEM 4.29 KNOWN: Diameter and maximum allowable temperature of an electronic component. Contact resistance between component and large aluminum heat sink. Temperature of heat sink and convection 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 loss from 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 − Tb P = q air + q sink = h π D2 / 4 ( Tc − T∞ ) + R ′′ / π D2 / 4 + 1/2Dk t,c ( ) P = 25 W/m 2 ⋅ K (π /4 )( 0.01 m ) 2 75o C + P = 0.15 W + ( ) 75o C 0.5 ×10 -4 / (π / 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 and conduction resistance. Hence, virtually all of the heat dissipated in the component is transferred through the block. The two-dimensional conduction resistance is significantly underestimated by use of the shape factor S = 2D. Hence, the maximum allowable power is less than 88.6 W. PROBLEM 4.30 KNOWN: Disc-shaped electronic devices dissipating 100 W mounted to aluminum alloy block with prescribed contact resistance. FIND: (a) Temperature device will reach when block is at 27°C assuming all the power generated by the device 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/m⋅K. ANALYSIS: (a) The thermal circuit for the conduction heat flow between the device and the block shown in the above Schematic where Re is the thermal contact resistance due to the epoxy-filled interface, R e = R ′′ A c = R ′′ t,c t,c (π D2 4) ( R e = 5 × 10−5 K ⋅ m 2 W π ( 0.020 m ) 2 ) 4 = 0.159 K W The thermal resistance between the device and the block is given in terms of the conduction shape factor, Table 4.1, as R b = 1 Sk = 1 ( 2Dk ) R b = 1 ( 2 × 0.020 m × 177 W m ⋅ K ) = 0.141K W From the thermal circuit, T1 = T2 + qd ( R b + R e ) T1 = 27 C + 100 W ( 0.141 + 0.159 ) K W T1 = 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 of copper (k = 400 W/m⋅K). The airstream temperature is 27°C and the convection coefficient is 1000 W/m2⋅K. 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 − T2 T1 − T∞ + R b + R e R e + R c + R fin (3) where the thermal resistance of the fin base material is Rc = Lc 0.005 m = = 0.03979 K W k c Ac 400 W m ⋅ K π 0.022 4 m 2 ) ( (4) and Rfin represents the thermal resistance of the fin array (see Section 3.6.5), 1 ηo hA t (5, 3.103) NAf (1 − ηf ) At (6, 3.102) R fin = R t,o = ηo = 1 − where the fin and prime surface area is A t = NAf + A b (3.99) ( ) 2 2 A t = N (π Df L ) + π Dd 4 − N π Df 4 where Af is the fin surface area, Dd is the device diameter and Df is the fin diameter. 2 2 A t = 30 (π × 0.0015 m × 0.015 m ) + π (0.020 m ) 4 − 30 π (0.0015 m ) 4 ( ) At = 0.06362 m2 + 0.0002611 m2 = 0.06388 m2 Using the IHT Model, Extended Surfaces, Performance Calculations, Rectangular Pin Fin, find the fin efficiency as ηf = 0.8546 (7) Continued... PROBLEM 4.30 (Cont.) Substituting numerical values into Eq. (6), find ηo = 1 − 30 × π × 0.0015 m × 0.015 m 0.06388 m 2 (1 − 0.8546 ) ηo = 0.8552 and the fin array thermal resistance is R fin = 1 0.8552 ×1000 W m 2 ⋅ K × 0.06388 m 2 = 0.01831K W Returning to Eq. (3), with T1 = 57°C from part (a), the permissible heat rate is qd = (57 − 27 ) C (0.141 + 0.159 ) K W + (57 − 27 ) C (0.159 + 0.03979 + 0.01831) K W qd = 100 W + 138.2 W = 238 W < COMMENTS: (1) Recognize in the part (b) analysis, that thermal resistances of the fin base and array are much smaller than the resistance due to the epoxy contact interfaces. (2) In calculating the fin efficiency, ηf, using the IHT Model it is not necessary to know the base temperature as ηf depends only upon geometric parameters, thermal conductivity and the convection coefficient. PROBLEM 4.31 KNOWN: Dimensions and surface temperatures of a square channel. Number of chips mounted on outer 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 surface temperatures, (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 the channel wall, q = (T2 – T1)/Rt,cond(2D), with the resistance determined by using Eq. 4.27 with Case 11 of Table 4.1. For W/w = 1.6 > 1.4 R t,cond(2D) = 0.930 ln ( W / w ) − 0.050 2π L k = 0.387 2π ( 0.160m ) 240 W / m ⋅ K = 0.00160 K / W The heat rate per chip is then qc = (50 − 20 ) °C = 156.3 W T2 − T1 = N R t,cond ( 2D ) 120 ( 0.0016 K / W ) < and, with qc = (Tc – T2)/Rt,c, the chip temperature is Tc = T2 + R t,c qc = 50°C + ( 0.2 K / W )156.3 W = 81.3°C < COMMENTS: (1) By acting to spread heat flow lines away from a chip, the channel wall provides an 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 the prescribed values of T1 and T2 represent minimum and maximum inner and outer surface temperatures, respectively, the rate is overestimated by the foregoing analysis. (2) The shape factor may also be determined by combining the expression for a plane wall with the result of Case 8 (Table 4.1). With S = [4(wL)/(W-w)/2] + 4(0.54 L) = 2.479 m, Rt,cond(2D) = 1/(Sk) = 0.00168 K/W. PROBLEM 4.32 KNOWN: Dimensions and thermal conductivity of concrete duct. Convection conditions of ambient air. Inlet temperature of water flow through the duct. FIND: (a) Heat loss per duct length near inlet, (b) Minimum allowable flow rate corresponding to maximum allowable temperature rise of water. SCHEMATIC: ASSUMPTIONS: (1) Steady state, (2) Negligible water-side convection resistance, pipe wall conduction resistance, and pipe/concrete contact resistance (temperature at inner surface of concrete corresponds to that of water), (3) Constant properties, (4) Negligible flow work and kinetic and potential energy changes. ANALYSIS: (a) From the thermal circuit, the heat loss per unit length near the entrance is q′ = Ti − T∞ Ti − T∞ = R′ 1 conv ln (1.08 w / D ) + cond ( 2D ) + R ′ 2π k h ( 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′ = 90°C (90 − 0 ) °C = ln (1.08 × 0.3m / 0.15m ) 1 (0.0876 + 0.0333 ) K ⋅ m / W + 2 2π (1.4 W / m ⋅ K ) 25 W / m ⋅ K (1.2m ) = 745 W / m < (b) From Eq. (1.11e), with q = q′L and ( Ti − To ) = 5°C, m= 745 W / m (100m ) q′L q′L = = = 3.54 kg / s u i − u o c (Ti − To ) 4207 J / kg ⋅ K (5°C ) < COMMENTS: The small reduction in the temperature of the water as it flows from inlet to outlet induces a slight departure from two-dimensional conditions and a small reduction in the heat rate per unit length. A slightly conservative value (upper estimate) of m is therefore obtained in part (b). PROBLEM 4.33 KNOWN: Dimensions and thermal conductivities of a heater and a finned sleeve. Convection conditions 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 resistance between heater and sleeve, (4) Uniform convection coefficient at outer surfaces of sleeve, (5) Uniform heat generation, (6) Negligible radiation. ANALYSIS: (a) From the thermal circuit, the desired heat rate is q′ = Ts − T∞ T −T =s ∞ R′ R ′tot t,o cond ( 2D ) + R ′ The two-dimensional conduction resistance, may be estimated from Eq. (4.27) and Case 6 of Table 4.2 R′ cond ( 2D ) = ln (1.08w / D ) ln ( 2.16 ) 1 = = = 5.11× 10−4 m ⋅ K / W S′k s 2π k s 2π ( 240 W / m ⋅ K ) The thermal resistance of the fin array is given by Eq. (3.103), where ηo and At are given by Eqs. 1/2 (3.102) and (3.99) and ηf is given by Eq. (3.89). With Lc = L + t/2 = 0.022 m, m = (2h/kst) = 32.3 -1 m and mLc = 0.710, ηf = tanh mLc 0.61 = = 0.86 mLc 0.71 A′ = NA′ + A′ = N ( 2L + t ) + ( 4w − Nt ) = 0.704m + 0.096m = 0.800m t f b ηo = 1 − NA′ f 1 − η = 1 − 0.704m 0.14 = 0.88 ( f) () A′t 0.800m −1 R ′t,o = (ηo h A′t ) q′ = ( ( = 0.88 × 500 W / m 2 ⋅ K × 0.80m (300 − 50 ) °C ) 5.11× 10−4 + 2.84 × 10−3 m ⋅ K / W ) −1 = 2.84 ×10−3 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 the surface area of the heater. From Eq. (3.32), < PROBLEM 4.33 (Cont.) −1 Us A′ = Usπ D = ( R ′ ) s tot −1 = (3.35 m ⋅ K / W ) = 298 W / m ⋅ K Us = 298 W / m ⋅ K / (π × 0.02m ) = 4750 W / m 2 ⋅ K ) ( q = 4 Us ( Ts − T∞ ) / D = 4 4750 W / m 2 ⋅ K ( 250°C ) / 0.02m = 2.38 × 108 W / m3 < From Eq. (3.53) the centerline temperature is T (0 ) = 2 q (D / 2 ) 4 kh 2.38 × 108 W / m3 (0.01m ) 2 + Ts = 4 ( 400 W / m ⋅ K ) + 300°C = 315°C < (c) Subject to the prescribed constraints, the following results have been obtained for parameter variations corresponding to 16 ≤ N ≤ 40, 2 ≤ t ≤ 8 mm and 20 ≤ L ≤ 40 mm. N 16 16 28 32 40 40 t(mm) 4 8 4 3 2 2 L(mm) ηf q′ ( W / m ) 20 20 20 20 20 40 0.86 0.91 0.86 0.83 0.78 0.51 74,400 77,000 107,900 115,200 127,800 151,300 Clearly there is little benefit to simply increasing t, since there is no change in A′ and only a t marginal increase in ηf . However, due to an attendant increase in A′ , there is significant benefit to t increasing N for fixed t (no change in ηf ) and additional benefit in concurrently increasing N while decreasing t. In this case the effect of increasing A′ exceeds that of decreasing ηf . The same is t true 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′ = π Dh ( Ts − T∞ ) = 7850 W / m, which is well below that achieved by using the increased surface area afforded by the sleeve. PROBLEM 4.34 KNOWN: Dimensions of chip array. Conductivity of substrate. Convection conditions. Contact resistance. 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 transfer from bottom or sides of substrate). ANALYSIS: From the thermal circuit, T − T∞ Th − T∞ q = q h + qsp = h + R h,cnv R t (sp ) + R t,c + R sp,cnv ( R h,cnv = h As,n R t (sp ) = −1 ) () = hL2 h −1 2 = 100 W / m 2 ⋅ K ( 0.005m ) 1 − 1.410 A r + 0.344 A3 + 0.043 A5 + 0.034 A7 r r r 4 k sub Lh R t,c = R ′′ t,c L2 h = ( 0.5 × 10−4 m 2 ⋅ K / W (0.005m )2 ) R sp,cnv = h Asub − As,h q= −1 = −1 = 400 K / W 1 − 0.353 + 0.005 + 0 + 0 4 (80 W / m ⋅ K )( 0.005m ) = 2.000 K / W ( ) = 100 W / m 2 ⋅ K 0.010m 2 − 0.005m 2 = 0.408 K / W −1 70°C 70°C + = 0.18 W + 0.52 W = 0.70 W 400 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 much less than the substrate convection resistance. Hence, the heat rate is increased almost in proportion to the 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 2π 1/ 2 k sub D h for a circular heat source and 4k sub L h for a square source. PROBLEM 4.35 KNOWN: Internal corner of a two-dimensional system with prescribed convection boundary conditions. FIND: Finite-difference equations for these situations: (a) Horizontal boundary is perfectly insulated and vertical boundary is subjected to a convection process (T∞,h), (b) Both boundaries are perfectly insulated; compare result with Eq. 4.45. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties, (4) No internal generation. ANALYSIS: Consider the nodal network shown above and also as Case 2, Table 4.2. Having defined the control volume – the shaded area of unit thickness normal to the page – next identify the heat transfer processes. Finally, perform an energy balance wherein the processes are expressed using 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 = 0 q1 + q2 + q3 + q4 + q5 + q6 = 0 T − Tm,n −T ∆x T ∆y k ( ∆y ⋅1) m-1,n + k ⋅1 m,n-1 m,n + h ⋅1 T∞ − Tm,n ∆x ∆y 2 2 ( ) − Tm,n T −T ∆y T + 0 + k ⋅1 m+1,n + k ( ∆x ⋅ 1) m,n+1 m,n = 0. ∆x ∆y 2 Letting ∆x = ∆y, and regrouping, find ( )( ) 2 Tm-1,n + Tm,n+1 + Tm+1,n + Tm,n-1 + h∆x h∆x T∞ − 6 + Tm,n = 0. k k < (b) With both boundaries insulated, the energy balance would have q3 = q4 = 0. The same result would be 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 insulated boundaries. PROBLEM 4.36 KNOWN: Plane surface of two-dimensional system. FIND: The finite-difference equation for nodal point on this boundary when (a) insulated; compare result 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) Constant properties, (3) Boundary is adiabatic. ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2) ⋅∆y, and using the conduction rate equation, it follows that & & E in − E out = 0 q1 + q2 + q3 = 0 (1,2) T − Tm,n −T −T ∆x T ∆x T k ( ∆y ⋅1) m-1,n + k ⋅1 m,n-1 m,n + k ⋅1 m,n+1 m,n = 0. (3) ∆x ∆y ∆y 2 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 form 2Tm-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 a convection process. That is, ( 2Tm-1,n + Tm,n-1 + Tm,n+1 ) + 2hk∆x T∞ − 2 h∆x + 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 form o q1 + q2 + q3 + q ′′ ⋅∆y = 0 and the finite difference equation becomes o q ′′ ∆x 2Tm-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 4.37 KNOWN: External corner of a two-dimensional system whose boundaries are subjected to prescribed conditions. FIND: Finite-difference equations for these situations: (a) Upper boundary is perfectly insulated and side 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) Constant properties, (4) No internal generation. ANALYSIS: Consider the nodal point configuration shown in the schematic and also as Case 4, Table 4.2. 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 as q1, q2 ⋅⋅⋅. Perform an energy balance wherein the processes are expressed using the appropriate rate equations. (a) With the upper boundary insulated and the side boundary subjected to a convection process, the energy balance has the form & & E in − E out = 0 q1 + q2 + q3 + q4 = 0 − Tm,n ∆y T ∆x k ⋅1 m-1,n +k ∆x 2 2 −T T ∆y ⋅1 m,n-1 m,n + h ∆y 2 (1,2) ⋅1 T∞ − Tm,n + 0 = 0. ( ) Letting ∆x = ∆y, and regrouping, find Tm,n-1 + Tm-1,n + h∆x 1 h∆x 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 same result would be obtained by letting h = 0 in the finite-difference equation, Eq. (3). The result is Tm,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 are insulated. COMMENTS: Note the convenience resulting from formulating the energy balance by assuming that all the heat flow is into the node. PROBLEM 4.38 KNOWN: Conduction in a one-dimensional (radial) cylindrical coordinate system with volumetric generation. 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, n ASSUMPTIONS: (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 = m∆r (or n∆r). The control volume is V = 2π r ⋅ ∆r ⋅ l = 2π ( m ∆r ) ∆r ⋅ l. The energy & & & balance is Ein +E g = q a +qb +qV = 0 k 2π ∆r Tm-1 − Tm + k 2π r − 2 l ∆r 2 ∆r Tm+1 − Tm & + q 2π ( m ∆r ) ∆rl = 0. r+ 2 l ∆r Recognizing that r = m∆r, canceling like terms, and regrouping find & 1 1 qm∆r 2 m − Tm-1 + m+ Tm+1 − 2mTm + = 0. 2 k < (b) The control volume for the surface node is V = 2 π r ⋅ ( ∆r/2 ) ⋅ l. The energy balance is & & & E in +E g = q d + q conv + qV=0. Use Fourier’s law to express qd and Newton’s law of cooling for qconv to obtain k 2π ∆r ∆r T − Tn n-1 & + h [ 2π rl ] ( T∞ − Tn ) + q 2π ( n∆r ) l = 0. r − 2 l ∆r 2 Let r = n∆r, cancel like terms and regroup to find & 1 hn ∆r qn ∆r 2 hn ∆r 1 n − Tn-1 − n − + Tn + 2k + k T∞ = 0. 2 2 k < COMMENTS: (1) Note that when m or n becomes very large compared to ½, the finite-difference equation 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 is V = π ( ∆r / 2 ) l and the energy balance is 2 2 ∆r T − T & ∆r & & & E in +E g = q a + qV = k 2π l 1 0 + q π l = 0. 2 ∆r 2 Regrouping, the finite-difference equation is − To + T1 + &2 q∆r 4k = 0. PROBLEM 4.39 KNOWN: Two-dimensional cylindrical configuration with prescribed radial (∆r) and angular (∆φ ) spacings of nodes. FIND: Finite-difference equations for nodes 2, 3 and 1. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction in cylindrical coordinates (r,φ ), (3) Constant properties. ANALYSIS: The method of solution is to define the appropriate control volume for each node, to identify 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 Fourier’s law for each process, find in a b c d 3 (T − T ) (T − T ) k ri + ∆r ∆φ 5 2 + k ( ∆r ) 3 2 + 2 ∆r ( ri + ∆ r ) ∆φ ( Ti − T2 ) (T − T ) 1 + k ri + ∆r ∆φ + k ( ∆r ) 1 2 = 0. 2 ∆r ( ri + ∆r ) ∆φ Canceling terms and regrouping yields, ( ∆r ) 2 1 T r 3 r T ( ∆r )2 1 2+ i+ ∆ 5+ −2 ( ri + ∆r ) + ( T3 + T1) + ri + ∆r Ti = 0. 2 2 ( ∆φ ) 2 ( ri + ∆r ) ( ri + ∆ r ) ( ∆φ ) 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 as 2 ( ∆r )2 1 3 2 ( ∆r ) 1 ( ri + ∆r ) + T3 + ri + ∆r T6 + −2 ⋅ T + r + ∆r Ti = 0. 2 ( r + ∆r ) 2 2 i 2 2 ( ∆φ ) i ( ri + ∆ r ) ( ∆φ ) ′ (c) Node 1. The energy balance is q ′ + qb + q′ + q′ = 0. Substituting, a c d 3 ∆φ ( T4 − T1 ) ( T2 − T1) + k ri + ∆r + k ( ∆r ) 2 2 ∆r ( ri + ∆r ) ∆φ 1 ∆φ ( Ti − T1 ) + k ri + ∆r + h ( ∆r )( T∞ − T ) = 0 1 2 2 ∆r This expression could now be rearranged. < PROBLEM 4.40 KNOWN: 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) Constant properties. ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)(∆y/2) ⋅1, find the FDE for node 1, k ( ∆y/2 ⋅1) To − T1 ∆x + hu ⋅ 1 ( T∞ − T1) + ( T2 − T1 ) R′′ / ( ∆y/2) 1 ∆x 2 t,c k ( ∆x/2 ⋅1) + ( T6 − T ) + & ( ∆x/2 )( ∆y/2) 1 = 0 1 q ∆y ∆x/kR′′ To + ( h u ∆x/k ) T∞ +T 2 + T6 t,c ( ) ( ) 2 & + q ( ∆x ) / 2 k − ∆x/kR′′ + ( hu ∆x/k ) + 2 T = 0. t,c 1 < (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) = 0 13 14 13 & ( 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 sink are determined by the values of h and R ′′,c. t PROBLEM 4.41 KNOWN: Nodal point configurations corresponding to a diagonal surface boundary subjected to a convection process and to the tip of a machine tool subjected to constant heat flux and convection cooling. 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 while the diagonal (surface) length is 2 ∆x. The heat rates associated with the control volume are due to conduction, q1 and q2, and to convection, qc . Performing an energy balance, find & & E in − E out = 0 q 1 + q 2 + qc = 0 Tm,n-1 − Tm,n T − Tm,n k ( ∆x ⋅1) + k ( ∆y ⋅1) m+1,n +h ∆y ∆x ( )( ) 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, find Tm,n-1 + Tm+1,n + 2 ⋅ h∆x h∆x T∞ − 2 + 2 ⋅ Tm,n = 0. k k < (b) The control volume about node m,n has triangular shape with sides ∆x/2 and ∆y/2 while the lower diagonal surface length is 2 ( ∆x/2 ). The heat rates associated with the control volume are due to the constant heat flux, qa, to conduction, qb, and to the convection process, qc . Perform an energy balance, & & E in − E out = 0 q a + q b + qc = 0 ∆x ∆x ∆y Tm+1,n − Tm,n q′′ ⋅ ⋅1 + k ⋅ ⋅1 +h ⋅ 2 ⋅ T∞ − Tm,n = 0. o ∆x 2 2 2 ( ) Recognizing that ∆x = ∆y, dividing each term by k/2 and regrouping, find Tm+1,n + 2 ⋅ h∆x ∆x h∆x ⋅ T∞ + q′′ ⋅ − 1+ 2 ⋅ Tm,n = 0. o k k k COMMENTS: Note the appearance of the term h∆x/k in both results, which is a dimensionless parameter (the Biot number) characterizing the relative effects of convection and conduction. < PROBLEM 4.42 KNOWN: 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) Constant properties, (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 of energy requirement has the form 6 ∑ qi = q1 + q2 + q3 + q4 + q5 + q6 = 0 i =1 since all heat rates are shown as into the CV. Each heat rate can be written using Fourier’s law, ∆y T1 − T0 T −T ∆y T3 − T0 ⋅ + k A ⋅ ∆x ⋅ 2 0 + k A ⋅ ⋅ 2 ∆x ∆y 2 ∆x ∆y T3 − T0 T4 − T0 ∆y T1 − T0 + kB ⋅ ⋅ + kB ⋅ ∆x ⋅ + kB ⋅ ⋅ = 0. 2 ∆x ∆y 2 ∆x kA ⋅ Recognizing that ∆x = ∆y and regrouping gives the relation, 1 kA 1 kB −T0 + T1 + T2 + T3 + T = 0. 4 2 (k A + k B ) 4 2 (k A + kB ) 4 < COMMENTS: Note that when kA = kB, the result agrees with Eq. 4.33 which is appropriate for an interior node in a medium of fixed thermal conductivity. PROBLEM 4.43 KNOWN: 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) Constant properties. ANALYSIS: Identify the surface nodes (Ts ) and draw control volumes about these nodes. Since there is no heat transfer in the direction parallel to the isothermal surfaces, the heat rate out of the constant temperature surface boundary is ′ ′ ′ ′ ′ ′ q′ = qa + qb + qc + qd + qe + qf For each q ′ , use Fourier’s law and pay particular attention to the manner in which the crossi sectional area and gradients are specified. T −T T −T T −T q′ = k ( ∆y/2 ) 1 s + k ( ∆y ) 2 s + k ( ∆y ) 3 s ∆x ∆x ∆x T5 − Ts T6 − Ts T7 − Ts + k ( ∆x ) + k ( ∆x ) + k ( ∆x/2) ∆y ∆y ∆y Regrouping with ∆x = ∆y, find q′ = k [ 0.5T1 + T2 + T3 + T5 + T6 + 0.5T7 − 5Ts ]. < COMMENTS: Looking at the corner node, it is important to recognize the areas associated with q′c and q′d (∆y and ∆x, respectively). PROBLEM 4.44 KNOWN: One-dimensional fin of uniform cross section insulated at one end with prescribed base temperature, 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 and control volume length in the x direction are both ∆x. The uniform cross-sectional area and fin perimeter 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 = 0 q 1 + q 2 + qc = 0 Tm-1 − Tm T −T kAc + kAc m+1 m + hP∆x ( T∞ − Tm ) = 0. ∆x ∆x Multiply the expression by ∆x/kAc and regroup to obtain Tm-1 + Tm+1 + hP hP ⋅ ∆x 2T∞ − 2 + ∆x 2 Tm = 0 kAc kAc 1<m<n < Considering now the special node m = 1, then the m-1 node is Tb, the base temperature. The finitedifference equation would be Tb + T2 + hP hP ∆x 2 T∞ − 2 + ∆x 2 T1 = 0 kAc kAc m=1 < (b) The control volume of length ∆x/2 about node n is shown in the schematic. Performing an energy balance, & & E in − E out = 0 q 3 + q 4 + qc = 0 T −T ∆x kAc n-1 n + 0 + hP ( T − T ) = 0. ∆x 2∞n Note that q4 = 0 since the end (x = L) is insulated. Multiplying by ∆x/kAc and regrouping, hP ∆x 2 hP ∆x 2 Tn-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 that the grouping, hP/kAc , appears in the finite-difference and differential forms of the energy balance. PROBLEM 4.45 KNOWN: Two-dimensional network with prescribed nodal temperatures and thermal conductivity of the material. FIND: Heat rate per unit length normal to page, q′. SCHEMATIC: Node 1 2 3 4 5 6 7 Ti(°C) 120.55 120.64 121.29 123.89 134.57 150.49 147.14 ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, (3) No internal volumetric generation, (4) Constant properties. ANALYSIS: Construct control volumes around the nodes on the surface maintained at the uniform ′ ′ ′ ′ temperature Ts and indicate the heat rates. The heat rate per unit length is q ′ = qa′ + qb + qc + qd + qe or in terms of conduction terms between nodes, ′ ′ ′ q′ = q′ + q2 + q3 + q4 + q′ + q ′7. 1 5 Each of these rates can be written in terms of nodal temperatures and control volume dimensions using Fourier’s law, ∆x T1 − Ts T −T T −T T −T ⋅ + k ⋅ ∆x ⋅ 2 s + k ⋅ ∆x 3 s + k ⋅ ∆x 4 s 2 ∆y ∆y ∆y ∆y T5 − Ts ∆y T7 − Ts + k ⋅ ∆x + k⋅ ⋅ . ∆y 2 ∆x q′ = 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, find q′ = 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 x′ direction. Look carefully at the energy balance for node e, q ′e = q′5 + q′7 , and how q ′ and q 7 are 5 evaluated. < PROBLEM 4.46 KNOWN: Nodal temperatures from a steady-state, finite-difference analysis for a one-eighth symmetrical section of a square channel. FIND: (a) Beginning with properly defined control volumes, derive the finite-difference equations for nodes 2, 4 and 7, and determine T2, T4 and T7, and (b) Heat transfer loss per unit length from the channel, q′ . SCHEMATIC: Node 1 3 6 8,9 T(°C) 430 394 492 600 ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) No internal volumetric generation, (4) Constant properties. ANALYSIS: (a) Define control volumes about the nodes 2, 4, and 7, taking advantage of symmetry where appropriate and performing energy balances, Ein − E out = 0 , with ∆x = ∆y, Node 2: q′ + q′ + q′ + q′ = 0 a b c d T −T T −T T −T h∆x ( T∞ − T2 ) + k ( ∆y 2 ) 3 2 + k∆x 6 2 + k ( ∆y 2 ) 1 2 = 0 ∆x ∆y ∆x T2 = 0.5T1 + 0.5T3 + T6 + ( h∆x k ) T∞ 2 + ( h∆x 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′ = 0 a b c T −T h ( ∆x 2 )( T∞ − T4 ) + 0 + k ( ∆y 2 ) 3 4 = 0 ∆x T4 = T3 + ( h∆x k ) T∞ 1 + ( h∆x 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 node 7 as an interior node, hence T7 = 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 rates from each node as illustrated in the first schematic, ′ q′ = q1 + q′ + q′ + q′ cv 2 3 4 q′ = h ( ∆x 2 )( T1 − T∞ ) + h∆x ( T2 − T∞ ) + h∆x ( T3 − T∞ ) + h ( ∆x 2 )( T4 − T∞ ) cv q′ = 50 W m 2 ⋅ K × 0.1m ( 430 − 300 ) 2 + ( 422 − 300 ) + (394 − 300 ) + (363 − 300 ) 2 K cv q′ = 156 W m cv < COMMENTS: (1) Always look for symmetry conditions which can greatly simplify the writing of the nodal equation as was the case for Node 7. (2) Consider using the IHT Tool, Finite-Difference Equations, for Steady-State, Two-Dimensional heat transfer to determine the nodal temperatures T1 - T7 when only the boundary conditions T8, T9 and ( T∞ ,h) are specified. PROBLEM 4.47 KNOWN: 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 by inspection using Eq. 4.39 and recognizing that the adiabatic boundary can be represented by a symmetry plane. 2 7 3 &2 2 / k = 0 and q∆x = 5 × 10 W/m ( 0.005m ) = 62.5K. & Tneighbors − 4Ti + q ∆x ∑ k Node A (to find T2): Node 3 (to find T3): Node 1 (to find T1): 20 W/m ⋅ K & 2T2 + 2TB − 4TA + q∆ x2 / k = 0 1 T2 = ( −2 × 374.6 + 4 × 398.0 − 62.5) K = 390.2K 2 & Tc + T2 + TB + 300K − 4T3 + q∆ x2 / k = 0 1 T3 = ( 348.5 + 390.2 + 374.6 + 300 + 62.5 ) K = 369.0K 4 & 300 + 2TC + T2 − 4T1 + q∆ x2 / k = 0 1 T1 = ( 300 + 2 × 348.5 + 390.2 + 62.5 ) = 362.4K 4 < < < (b) The heat rate out of the bar is determined by calculating the heat rate out of each control volume around the 300K nodes. Consider the node in the upper left-hand corner; from an energy balance & & & E in − E out + E g = 0 Hence, for the entire bar & & & q′a = q′a,in + E g where E g = qV. ′ ′ q′bar = qa + qb + q′ + q′ + q′e + q′f , or c d or T − 300 ∆x ∆y T1 − 300 & ∆x ∆y & ∆x ∆y + & q ′bar = k +q ⋅ + k ∆y C +q ⋅ ∆y + q ⋅ ∆x ∆x 2 2 a 2 b 2 2 c 2 TC − 300 ∆x ⋅ ∆y + k∆x T3 − 300 + q ∆x ⋅ ∆y + k ∆x TB − 300 + q ∆x ⋅ ∆y . & & & k∆x ∆y + q 2 2 2 d ∆y 2 e 2 ∆y f Substituting numerical values, find q′bar = 7,502.5 W/m. From an overall energy balance on the bar, & & & q′bar = E′g = qV/ l = q ( 3∆x ⋅ 2∆y ) = 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 4.48 KNOWN: Steady-state temperatures at selected nodal points of the symmetrical section of a flow channel with uniform internal volumetric generation of heat. Inner and outer surfaces of channel experience convection. FIND: (a) Temperatures at nodes 1, 4, 7, and 9, (b) Heat rate per unit length (W/m) from the outer surface 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 control volumes about the nodes shown in the schematics below. Node 1 q′ + q′ + q′ + q′ + E′ = 0 a b c d g T −T T −T 0 + k ( ∆y / 2 ) 2 1 + k ( ∆x / 2 ) 3 1 + 0 + q ( ∆x ⋅ ∆y / 4 ) = 0 ∆x ∆y T1 = ( T2 + T3 ) / 2 + q∆x 2 / 2k T1 = (95.47 + 117.3) °C / 2 + 106 W / m3 ( 25 × 25 ) × 10−6 m 2 / ( 2 × 10 W / m ⋅ K ) = 122.0°C Node 4 q′ + q′ + q′ + q′ + q′ + q′ + E′ = 0 a b c d e f g T −T k ( ∆x / 2 ) 2 4 + h i ( ∆y / 2 ) T∞,i − T4 + h i ( ∆x / 2 )(T∞ − T4 ) + ∆y ( ) Continued ….. PROBLEM 4.48 (Cont.) T −T T −T T −T k ( ∆y / 2 ) 5 4 + k ( ∆x ) 8 4 + k ( ∆y ) 3 4 + q (3∆x ⋅ ∆y / 4 ) = 0 ∆x ∆y ∆x ) ( T4 = T2 + 2T3 + T5 + 2T8 + 2 ( hi ∆x / k ) T∞,i + 3q∆x 2 / 2k 6 + 2 ( h i ∆x / k ) < T4 = 94.50°C Node 7 q′ + q′ + q′ + q′ + E′ = 0 a b c d g T −T T −T k ( ∆x / 2 ) 3 7 + k ( ∆y / 2 ) 8 7 + h o ( ∆x / 2 ) T∞,o − T7 + 0 + q ( ∆x ⋅ ∆y / 4 ) = 0 ∆y ∆x ( T7 = T3 + T8 + ( h o ∆x / k ) T∞,o + q∆x 2 / 2k ) ( 2 + h o ∆x / k ) < T7 = 95.80°C Node 9 q′ + q′ + q′ + q′ + E′ = 0 a b c d g T −T T −T k ( ∆x ) 5 9 + k ( ∆y / 2 ) 10 9 + h o ( ∆x ) T∞,o − T9 ∆y ∆y T −T + k ( ∆y / 2 ) 8 9 + q ( ∆x ⋅ ∆y / 2 ) = 0 ∆x T9 = T5 + 0.5T8 + 0.5T10 + ( h o ∆x / k ) T∞,o + q∆x 2 / 2k / ( 2 + h o ∆x / k ) ( ) < T9 = 79.67°C (b) The heat rate per unit length from the outer surface A to the adjacent fluid, q ′ , is the sum of the A convection 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 ) × 10−3 m ⋅ K Continued ….. ) PROBLEM 4.48 (Cont.) q′ = 1117 W / m A < (c) The heat rate per unit length from the inner fluid to the surface B, q′B , is the sum of the convection 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 B q′ = 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 ) × 10−3 m ⋅ K q′ = −1383 W / m B < (d) From an overall energy balance on the section, we see that our results are consistent since the conservation of energy requirement is satisfied. E′ − E′ + E′ = −q′ + q ′ + E′ = (−1117 − 1383 + 2500)W / m = 0 in out gen A B gen −6 2 6 3 where E′ gen = q∀′ = 10 W / m [25 × 50 + 25 × 50 ]×10 m = 2500 W / m COMMENTS: The nodal finite-difference equations for the four nodes can be obtained by using IHT Tool Finite-Difference Equations | Two-Dimensional | Steady-state. Options are provided to build the FDEs for interior, corner and surface nodal arrangements including convection and internal generation. 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• a4 q• 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• a7 q• 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• a9 q• a9 = 0 // Applied heat flux, W/m^2; zero flux shown PROBLEM 4.49 KNOWN: Outer surface temperature, inner convection conditions, dimensions and thermal conductivity 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 surface temperature, (4) Constant thermal conductivity. ANALYSIS: (a) To determine the heat rate, the nodal temperatures must first be computed from the corresponding finite-difference equations. From an energy balance for node 1, T −T T −T h ( ∆x / 2 ⋅1)( T∞ − T1 ) + k ( ∆y / 2 ⋅1) 2 1 + k ( ∆x ⋅1) 5 1 = 0 ∆x ∆y h∆x h∆x T∞ = 0 −3+ T1 + T2 + 2T5 + k k (1) With nodes 2 and 3 corresponding to Case 3 of Table 4.2, 2h∆x h∆x T1 − 2 T∞ = 0 + 2 T2 + T3 + 2T6 + k k (2) h∆x h∆x T2 − T∞ = 0 + 2 T3 + T7 + k k (3) where the symmetry condition is invoked for node 3. Applying an energy balance to node 4, we obtain −2T4 + 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 = 0 T2 + T5 − 4T6 + T7 + Ts = 0 T3 + 2T6 − 4T7 + Ts = 0 (5) (6) (7) where the symmetry condition is invoked for node 7. With Ts = 50°C, T∞ = 20°C, and h∆x / k = 5000 W / m ⋅ K ( 0.005m ) / 240 W / m ⋅ K = 0.1042, the solution to Eqs. (1) – (7) yields 2 T1 = 46.61°C, T2 = 45.67°C, T3 = 45.44°C, T4 = 49.23°C < T5 = 48.46°C, T6 = 48.00°C, T7 = 47.86°C Continued ….. PROBLEM 4.49 (Cont.) The heat rate per unit length of channel may be evaluated by computing convection heat transfer from the 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 ) 200 1000 2000 5000 49.80 49.02 48.11 45.67 49.79 48.97 48.00 45.44 49.84 49.24 48.53 46.61 T4 (°C ) T5 (°C ) 49.96 49.83 49.66 49.23 49.93 49.65 49.33 48.46 T6 (°C ) T7 (°C ) q ′ ( W / m ) 49.91 49.55 49.13 48.00 49.90 49.52 49.06 47.86 477 2325 4510 10,340 There are two resistances to heat transfer between the outer surface of the heat sink and the fluid, that due to conduction in the heat sink, R cond ( 2D ), and that due to convection from its inner surface to the fluid, R conv . With decreasing h, the corresponding increase in R conv reduces heat flow and increases 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 with conservation of energy requirements. For example, an energy balance performed at the inner surface requires a balance between convection from the surface and conduction to the surface, which may be expressed as q′ = k ( ∆x ⋅1) (T5 − T1 ) + k ∆y ( ∆x ⋅1) T6 − T2 T −T + k ( ∆x / 2 ⋅ 1) 7 3 ∆y ∆y Substituting the temperatures corresponding to h = 5000 W / m 2 ⋅ K, the expression yields q ′ = 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 ′ cond conv , where, for 2 h = 5000 W / m ⋅ K, R ′ = 1/ 4hw = 2.5 × 10−3 m ⋅ K / W, and from Eq. (4.27) and Case 11 of conv ( ) −4 Table 4.1, R ′ cond = [0.930 ln ( W / w ) − 0.05] / 2π k = 3.94 × 10 m ⋅ K / W. Hence, ( q ′ = (50 − 20 ) °C / 2.5 × 10 −3 + 3.94 × 10 −4 ) m ⋅ K / W = 10, 370 W / m, and the agreement with the finite-difference solution is excellent. Note that, even for h = 5000 W / m 2 ⋅ K, R ′ conv >> R ′ cond ( 2D ). PROBLEM 4.50 KNOWN: Steady-state temperatures (°C) associated with selected nodal points in a two-dimensional system. FIND: (a) Temperatures at nodes 1, 2 and 3, (b) Heat transfer rate per unit thickness from the system 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 ⋅ ∑ Tneighbors 4 1 (172.9 + 137.0 + 132.8 + 200.0 )o C = 160.7oC 4 < Node 2, Insulated boundary, Eq. 4.46 with h = 0, Tm,n = T2 ( ) 1 Tm-1,n + Tm+1,n + 2Tm,n-1 4 1 T2 = (129.4 + 45.8 + 2 ×103.5 )o C = 95.6oC 4 T2 = < Node 3, Plane surface with convection, Eq. 4.46, Tm,n = T3 2h ∆x h ∆x + 2 T = 2T 2 T∞ 3 m-1,n + Tm,n+1 +Tm,n-1 + k k ( ) h ∆x/k = 50W/m 2 ⋅ K × 0.1m/1.5W/m ⋅ K = 3.33 2 ( 3.33 + 2 ) T3 = ( 2 × 103.5 + 45.8 + 67.0 ) °C + 2 × 3.33 × 30 o C 1 ( 319.80 + 199.80) °C=48.7°C 10.66 (b) The heat rate per unit thickness from the surface to the fluid is determined from the sum of the convection rates from each control volume surface. ′ ′ ′ q′ conv = qa + qb + qc + q ′ d T3 = q i = h∆y i ( T − T ) i ∞ q′ conv = 50 0.1 m 45.8 − 30.0 °C + ( ) m ⋅K 2 W 2 0.1m ( 48.7 − 30.0 ) °C + 0.1m ( 67.0 − 30.0 ) °C + 0.1m + ( 200.0 − 30.0) °C 2 q′ conv = ( 39.5 + 93.5 + 185.0 + 425 ) W/m = 743 W/m. < < PROBLEM 4.51 KNOWN: Nodal temperatures from a steady-state finite-difference analysis for a cylindrical fin of prescribed diameter, thermal conductivity and convection conditions ( T∞ , h). FIND: (a) The fin heat rate, qf, and (b) Temperature at node 3, T3. SCHEMATIC: T0 = 100.0°C T1 = 93.4°C T2 = 89.5°C ASSUMPTIONS: (a) The fin heat rate, qf, is that of conduction at the base plane, x = 0, and can be found from an energy balance on the control volume about node 0, Ein − E out = 0 , qf + q1 + qconv = 0 or qf = −q1 − qconv . Writing the appropriate rate equation for q1 and qconv, with Ac = πD2/4 and P = πD, T1 − T0 π kD2 q f = − kAc − hP ( ∆x 2 )( T∞ − T0 ) = − (T1 − T0 ) − (π 2 ) Dh∆x (T∞ − T0 ) ∆x 4∆x Substituting numerical values, with ∆x = 0.010 m, find π × 15 W m ⋅ K (0.012 m )2 qf = − (93.4 − 100 ) C 4 × 0.010 m π − × 0.012 m × 25 W m 2 ⋅ K × 0.010 m ( 25 − 100 ) C 2 qf = (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 the control volume shown above, Ein − E out = 0 , qcv + q3 + q1 = 0 T −T T −T hP∆x ( T∞ − T2 ) + kA c 3 2 + kAc 1 2 = 0 ∆x ∆x T3 = −T1 + 2T2 − hP∆x 2 2 ∆x [T∞ − T2 ] kAc Substituting numerical values, find T2 = 89.2 C COMMENTS: Note that in part (a), the convection heat rate from the outer surface of the control volume is significant (25%). It would have been poor approximation to ignore this term. < PROBLEM 4.52 KNOWN: Long rectangular bar having one boundary exposed to a convection process (T∞, h) while the other boundaries are maintained at a constant temperature (Ts). FIND: (a) Using a grid spacing of 30 mm and the Gauss-Seidel method, determine the nodal temperatures and the heat rate per unit length into the bar from the fluid, (b) Effect of grid spacing and convection 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 = h∆x/k = 100 W/m2⋅K × 0.030 m/1 W/m⋅K = 3, we obtain: Node 1: T1 = 1 1 1 (T2 + Ts + BiT∞ ) = (T2 + 50 + 3 ×100 ) = (T2 + 350) 5 5 ( Bi + 2 ) (1) Node 2: T2 = 1 1 1 (T1 + 2Ts + T3 ) = (T1 + T3 + 2 × 50 ) = (T1 + T3 + 100 ) 4 4 4 (2) Node 3: T3 = 1 1 1 (T2 + 3Ts ) = (T2 + 3 × 50 ) = ( T2 + 150 ) 4 4 4 (3) k Denoting each nodal temperature with a superscript to indicate iteration step, e.g. T1 , calculate values as shown below. k 0 T1 85 T2 60 T3 (°C) 55 1 2 3 4 82.00 81.85 81.71 81.69 59.25 58.54 58.46 58.45 52.31 52.14 52.12 52.11 ← initial guess By the 4th iteration, changes are of order 0.02°C, suggesting that further calculations may not be necessary. Continued... PROBLEM 4.52 (Cont.) In finite-difference form, the heat rate from the fluid to the bar is q′ conv = h ( ∆x 2 )( T∞ − Ts ) + h∆x ( T∞ − T1 ) + h ( ∆x 2 )( T∞ − Ts ) q′ conv = h∆x ( T∞ − Ts ) + h∆x ( T∞ − T1 ) = h∆x ( T∞ − Ts ) + ( T∞ − T1 ) 2 q′ conv = 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 following two-dimensional temperature field was computed for the grid shown in schematic (b), where x and y are in mm and the temperatures are in °C. y\x 0 15 30 45 60 75 90 0 50 50 50 50 50 50 50 15 80.33 63.58 56.27 52.91 51.32 50.51 50 30 85.16 67.73 58.58 54.07 51.86 50.72 50 45 80.33 63.58 56.27 52.91 51.32 50.51 50 60 50 50 50 50 50 50 50 The 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, although the requisite number of finite-difference equations would increase from 12 to 108, significantly increasing problem set-up time. An increase in h would increase temperatures everywhere within the bar, particularly at the heated 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 = 81.70°C, T2 = 58.44°C, and T3 = 52.12°C. The fact that only 4 iterations were required to obtain agreement within 0.01°C 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 of heat transfer by conduction to the sides and bottom of the rod. NOTE TO INSTRUCTOR: Although the problem statement calls for calculations with ∆x = ∆y = 5 mm 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 4.53 KNOWN: Square shape subjected to uniform surface temperature conditions. FIND: (a) Temperature at the four specified nodes; estimate the midpoint temperature To, (b) Reducing the 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 250°C 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 point written in the form, Ti = 1/4∑Tneighbors. Using the Gauss-Seidel iteration method, Section 4.5.2, the finitedifference equations for the four nodes are: ) ( k k k k k T2 = 0.25 (100 + 200 + T4 −1 + T1 −1 ) = 0.25T1 −1 + 0.25T4 −1 + 75.0 k k k k k T3 = 0.25 ( T1 −1 + T4 −1 + 300 + 50 ) = 0.25T1 −1 + 0.25T4 −1 + 87.5 k k k k k T4 = 0.25 ( T2 −1 + 200 + 300 + T3 −1 ) = 0.25T2 −1 + 0.25T3 −1 + 125.0 k k k k k T1 = 0.25 100 + T2 −1 + T3 −1 + 50 = 0.25T2 −1 + 0.25T3 −1 + 37.5 The iteration procedure using a hand calculator is implemented in the table below. Initial estimates are entered on the k = 0 row. k 0 1 2 3 4 5 6 7 T1 (°C) 100 112.50 123.44 119.93 119.05 118.83 118.77 118.76 T2 (°C) 150 165.63 158.60 156.40 156.40 156.29 156.26 156.25 T3 (°C) 150 178.13 171.10 169.34 168.90 168.79 168.76 168.76 T4 (°C) 250 210.94 207.43 206.55 206.33 206.27 206.26 206.25 < Continued... PROBLEM 4.53 (Cont.) By the seventh iteration, the convergence is approximately 0.01°C. The midpoint temperature can be estimated as To = ( 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 into the IHT workspace and the set of equations solved for the nodal temperatures (°C). Mesh Coarse Fine T1 118.76 117.4 To 162.5 162.5 T2 156.25 156.1 T3 168.76 168.9 T4 206.25 207.6 The maximum difference for the interior points is 1.5°C (node 4), but the estimate at the center, To, is the same, independently of the mesh size. In terms of the boundary surface temperatures, To = (50 + 100 + 200 + 300 ) C 4 = 162.5 C Why must this be so? (c) To generate the isotherms, it would be necessary to employ a contour-drawing routine using the tabulated temperature distribution (°C) obtained from the finite-difference solution. Using these values as a guide, try sketching a few isotherms. 50 50 50 50 50 - 100 86.0 88.2 99.6 123.0 173.4 300 100 105.6 117.4 137.1 168.9 220.7 300 100 119 138.7 162.5 194.9 240.6 300 100 131.7 156.1 179.2 207.6 246.8 300 100 151.6 174.6 190.8 209.4 239.0 300 200 200 200 200 200 - COMMENTS: Recognize that this finite-difference solution is only an approximation to the temperature 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 4.54 KNOWN: Long bar of square cross section, three sides of which are maintained at a constant temperature 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 numerical technique with grid spacing of 0.2 m is suggested, and (b) Reducing the grid spacing by a factor of 2, find the midpoint temperature and the heat transfer rate. Also, plot temperature distribution across the surface exposed 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 inversion method 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 h∆x/k = 10 W/m2⋅K × 0.2 m/2W/m⋅K = 1, find Node 7: (2T5 + 300 + T8) + 2×1⋅100 - 2(1+2)T7 = 0 Node 8: (2T6 + T7 + T7) + 2×1⋅100 - 2(1+2)T8 = 0 The solution matrix [T] can be found using a stock matrix program using the [A] and [C] matrices shown below to obtain the solution matrix [T] (Eq. 4.52). Alternatively, the set of equations could be entered into 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 1 0 0 1 0 −4 1 1 0 0 0 0 1 2 −4 0 1 0 0 0 0 2 0 −6 1 0 0 0 0 0 2 2 −6 −600 −300 −300 C= 0 −300 0 −500 −200 292.2 289.2 279.7 T = 272.2 254.5 240.1 198.1 179.4 From the solution matrix, [T], find the mid-point temperature as T4 = 272.2°C < Continued... PROBLEM 4.54 (Cont.) The heat rate by convection between the bar and fluid is given as, q′ conv = 2 ( q′ + q′ + q′ ) a b c q′ conv = 2x h ( ∆x 2 )( T8 − T∞ ) + h ( ∆x )( T7 − T∞ ) + h ( ∆x 2 )(300 − T∞ ) 2 q′ conv = 2x 10 W m ⋅ K × ( 0.2 m 2 ) (179.4 − 100 ) + 2 (198.1 − 100 ) + (300 − 100 ) K < q′ conv = 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 the IHT workspace. The IHT Finite-Difference Equations Tool for 2-D, SS conditions, was used to obtain the FDE for the nodes on the exposed surface. The midpoint temperature T13 and heat rate for the finer mesh are T13 = 271.0°C q′ = 834 W/m < COMMENTS: The midpoint temperatures for the coarse and finer meshes agree closely, T4 = 272°C vs. T13 = 271.0°C, respectively. However, the estimate for the heat rate is substantially influenced by the mesh size; q′ = 952 vs. 834 W/m for the coarse and finer meshes, respectively. PROBLEM 4.55 KNOWN: Volumetric heat generation in a rectangular rod of uniform surface temperature. FIND: (a) Temperature distribution in the rod, and (b) With boundary conditions unchanged, heat generation rate causing the midpoint temperature to reach 600 K. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Uniform volumetric heat generation. ANALYSIS: (a) From symmetry it follows that six unknown temperatures must be determined. Since all nodes are interior ones, the finite-difference equations may be obtained from Eq. 4.39 written in the form Ti = 1 2 ∑ Tneighbors + 1 4 ( q ( ∆x∆y1) k ) . With q ( ∆x∆y ) 4k = 62.5 K, the system of finite-difference equations is T1 = 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 the nodal temperatures, T1 T2 T3 T4 T5 T6 (K) 348.6 368.9 374.6 362.4 390.2 398.0 < (b) With the boundary conditions unchanged, the q required for T6 = 600 K can be found using the same set 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 (∆x∆y)/4k = (0.005 m)2 q /4×20 W/m⋅K = 3.125 × 10-7 q and (2) set T6 = 600 K. The set of equations has 6 unknown, five nodal temperatures plus q . Solving find q = 1.53 × 108 W m3 < PROBLEM 4.56 KNOWN: Flue of square cross section with prescribed geometry, thermal conductivity and inner and 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) No internal generation. ANALYSIS: Taking advantage of symmetry, the nodal network using the suggested 75mm grid spacing is shown above. To obtain the heat rate, we first need to determine the unknown temperatures T1, T2, T3 and T4. Recognizing that these nodes may be treated as interior nodes, the nodal 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 the procedures of Section 4.5.2, they are rewritten as, k k-1 T1 = 0.50 T2 + 93.75 k k k-1 T2 = 0.25 T1 + 0.25 T3 + 93.75 k k k-1 T3 = 0.25 T2 + 0.25 T4 + 93.75 k k T4 = 0.50 T3 + 12.5. The iteration procedure is implemented in the table on the following page, one row for each iteration k. The initial estimates, for k = 0, are all chosen as (350 + 25)/2 ≈ 185° C. Iteration is continued until the maximum temperature difference is less than 0.2° C, i.e., ε < 0.2° C. Note that if the system of equations were organized in matrix form, Eq. 4.52, diagonal dominance would exist. Hence there is no need to reorder the equations since the magnitude of the diagonal element is greater than that of other elements in the same row. Continued ….. PROBLEM 4.56 (Cont.) k T1(° C) T2(° C) T3(° C) T4(° C) 0 1 2 3 4 5 6 7 185 186.3 187.1 187.4 184.9 184.2 184.0 183.9 185 186.6 187.2 182.3 180.8 180.4 180.3 180.3 185 186.6 167.0 163.3 162.5 162.3 162.3 162.2 185 105.8 96.0 94.2 93.8 93.7 93.6 93.6 ← initial estimate ← ε <0.2° C From knowledge of the temperature distribution, the heat rate may be obtained by summing the heat rates 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 + qe ∆x 1 q′ = k 2 ( T1 − 25 ) + (T 2 − 25) + ( T3 − 25 ) + ( T4 − 25 ) + 0 ∆y W 1 q′ = 0.85 2 (183.9 − 25 ) + (180.3 − 25 ) + (162.2 − 26 ) + ( 93.6 − 25) m ⋅ K q′ = 374.5 W/m. Since this flue section is 1/8 the total cross section, the total heat loss from the flue is q′ = 8 × 374.5 W/m = 3.00 kW/m. < COMMENTS: The heat rate could have been calculated at the inner surface, and from the above sketch has the form q′ = k ∆x 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 conservation of energy requirement must be satisfied in obtaining the nodal temperatures. PROBLEM 4.57 KNOWN: Flue of square cross section with prescribed geometry, thermal conductivity and inner and outer surface convective conditions. FIND: (a) Heat loss per unit length, q′ , by convection to the air, (b) Effect of grid spacing and convection 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 shown in schematic (a). To obtain the heat rate, we need first to determine the temperatures Ti. Recognize that there are four types of nodes: interior (4-7), plane surface with convection (1, 2, 8-11), internal corner with convection (3), and external corner with convection (12). Using the appropriate relations from Table 4.2, the finite-difference equations are Node 1 2 3 4 5 6 7 8 9 10 11 12 2h i ∆x h ∆x T∞,i − i + 2 T1 = 0 k k 2h ∆x h ∆x ( 2T5 + T3 + T1 ) + i T∞,i − 2 i + 2 T2 = 0 k k ( 2T4 + T2 + T2 ) + 2 ( T6 + T6 ) + ( T2 + T2 ) + 2h i ∆x k (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 ∆x k h ∆x T∞,i − 2 3 + i T3 = 0 k h ∆x T∞,o − 2 o + 2 T8 = 0 k 4.46 4.45 4.33 4.33 4.33 4.33 4.46 h ∆x T∞,o − 2 o + 2 T9 = 0 k k 2h ∆x h ∆x ( 2T6 + T11 + T9 ) + o T∞,o − 2 o + 2 T10 = 0 k k 4.46 h ∆x + 2 T = 0 T∞ ,o − 2 o 11 k k 2h o ∆x h o ∆x T∞ ,o − 2 + 1 T12 = 0 (T11 + T11 ) + k k 4.46 ( 2T5 + T10 + T8 ) + ( 2T7 + T12 + T10 ) + 2h o ∆x Equation 4.46 2h o ∆x 4.46 4.47 Continued... PROBLEM 4.57 (Cont.) The Gauss-Seidel iteration is convenient for this system of equations. Following procedures of Section 4.5.2, the system of equations is rewritten in the proper form. Note that diagonal dominance is present; hence, no re-ordering is necessary. k −1 k T1 = 0.09239T2 k −1 + 0.09239T4 + 285.3 k k k k T2 = 0.04620T1 + 0.04620T3 −1 + 0.09239T5 −1 + 285.3 k k k T3 = 0.08457T2 + 0.1692T6 −1 + 261.2 k k k k T4 = 0.25T1 + 0.50T5 −1 + 0.25T8 −1 k k k k k T5 = 0.25T2 + 0.25T4 + 0.25T6 −1 + 0.25T9 −1 k k k k k T6 = 0.25T3 + 0.25T5 + 0.25T7 −1 + 0.25T9 −1 k k k T7 = 0.50T6 + 0.50T11−1 k k k T8 = 0.4096T4 + 0.4096T9 −1 + 4.52 k k k k T9 = 0.4096T5 + 0.2048T8 + 0.2048T10−1 + 4.52 k k k k T10 = 0.4096T6 + 0.2048T9 + 0.2048T11−1 + 4.52 k k k k T11 = 0.4096T7 + 0.2048T10 + 0.2048T12−1 + 4.52 k k T12 = 0.6939T11 + 7.65 The initial estimates (k = 0) are carefully chosen to minimize calculation labor; let ε < 1.0. k 0 1 2 3 4 5 6 7 T1 340 338.9 338.3 338.8 339.4 339.8 340.1 340.3 T2 330 336.3 337.4 338.4 338.8 339.2 339.4 339.5 T3 315 324.3 328.0 328.2 328.9 329.3 329.7 329.9 T4 250 237.2 241.4 247.7 251.6 254.0 255.4 256.4 T5 225 232.1 241.5 245.7 248.7 250.5 251.7 252.5 T6 205 225.4 226.6 230.6 232.9 234.5 235.7 236.4 T7 195 175.2 178.6 180.5 182.3 183.7 184.7 185.5 T8 160 163.1 169.6 175.6 178.7 180.6 181.8 182.7 T9 150 161.7 170.0 173.7 176.0 177.5 178.5 179.1 T10 140 155.6 158.9 161.2 162.9 164.1 164.7 165.6 T11 125 130.7 130.4 131.6 132.8 133.8 134.5 135.1 T12 110 98.3 98.1 98.9 99.8 100.5 101.0 101.4 The heat loss to the outside air for the upper surface (Nodes 8 through 12) is of the form 1 1 q′ = h o ∆x T8 − T∞ ,o + T9 − T∞,o + T10 − T∞ ,o + T11 − T∞ ,o + T12 − T∞ ,o 2 2 ( )( )( )( ) ( ) 1 182.7 − 25 + 179.1 − 25 + 165.6 − 25 + 135.1 − 25 + 1 101.4 − 25 C = 195 W m )( )( )( )( ) 2 ( 2 Hence, for the entire flue cross-section, considering symmetry, q ′ = 5 W m ⋅ K × 0.075 m 2 < q′tot = 8 × q′ = 8 × 195 W m = 1.57 kW m The convection heat rate at the inner surface is 1 1 q′ = 8 × h i ∆x T∞,i − T1 + T∞,i − T2 + T∞,i − T3 = 8 × 190.5 W m = 1.52 kW m tot 2 2 which is within 2.5% of the foregoing result. The calculation would be identical if ε = 0. ( )( ) ( ) Continued... PROBLEM 4.57 (Cont.) (b) Using the Finite-Difference Equations option from the Tools portion of the IHT menu, the following two-dimensional temperature field was computed for the grid shown in the schematic below, where x and y are in mm and the temperatures are in °C. y\x 0 25 50 75 100 125 150 0 180.7 204.2 228.9 255.0 282.4 310.9 340.0 25 180.2 203.6 228.3 254.4 281.8 310.5 340.0 50 178.4 201.6 226.2 252.4 280.1 309.3 339.6 75 175.4 198.2 222.6 248.7 276.9 307.1 339.1 100 171.1 193.3 217.2 243.1 271.6 303.2 337.9 125 165.3 186.7 209.7 235.0 263.3 296.0 335.3 150 158.1 178.3 200.1 223.9 250.5 282.2 324.7 175 149.6 168.4 188.4 209.8 232.8 257.5 200 140.1 157.4 175.4 194.1 213.5 225 129.9 145.6 161.6 177.8 250 119.4 133.4 147.5 275 108.7 121.0 300 98.0 Agreement between the temperature fields for the (a) and (b) grids is good, with the largest differences occurring at the interior and exterior corners. Ten isotherms generated using FEHT are shown on the symmetric 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 we’d expect that grid size might influence the 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 and decrease, respectively, with increasing hi and ho. PROBLEM 4.58 KNOWN: Rectangular air ducts having surfaces at 80°C in a concrete slab with an insulated bottom and upper surface maintained at 30°C. FIND: Heat rate from each duct per unit length of duct, q′. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) No internal volumetric generation, (4) Constant properties. PROPERTIES: Concrete (given): k = 1.4 W/m⋅K. ANALYSIS: Taking advantage of symmetry, the nodal network, using the suggested grid spacing ∆x = 2∆y = 37.50 mm ∆y = 0.125L = 18.75 mm where L = 150 mm, is shown in the sketch. To evaluate the heat rate, we need the temperatures T1, T2, T3, T4, and T5. All the nodes may be treated as interior nodes ( considering symmetry for those nodes on insulated 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 of the duct follows from an energy balance on the nodes at the top isothermal surface. ′ ′ ′ ′ q′ = q′ + q2 + q3 + q4 + q5 1 T1 − Ts T −T T −T T −T T −T q′ = k ( ∆x/2 ) + k ⋅ ∆x 2 s + k ⋅ ∆x 3 s + k ⋅ ∆x 4 s + k ( ∆x/2 ) 5 s ∆y ∆y ∆y ∆y ∆y ′ = k ( T1 − Ts ) + 2 ( T2 − Ts ) + 2 ( T3 − Ts ) + 2 ( T4 − Ts ) + ( T5 − Ts ) q q′ = 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 unit length for each duct is, q′duct = 2xq′ = 456 W/m. < Continued ….. PROBLEM 4.58 (Cont.) Coefficient matrix [A] PROBLEM 4.59 KNOWN: 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) Negligible radiation. 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 new equations must be developed for nodes 1c-6c, 1o-6o, and 1i-6i. Considering node 3c as an example, an energy balance yields k ( ∆yc 2 ) k ( ∆y 2 ) k ∆x h o ∆x T∞,o − T3c + c (T2c − T3c ) + c c (T4c − T3c ) + c (T3o − T3c ) = 0 ∆x ∆x ∆yc ( ) or, with ∆x = 1 mm and ∆yc = 0.5 mm, h ∆x h ∆x 0.25 ( T2c + T4c ) + 2T3o − 2.5 + o T3c = − o T∞ ,o kc kc Similar 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 obtain k c ∆x ∆yc (T3c − T3o ) + k c ( ∆yc 2 ) ∆x ( T2o − T3o ) + k c ( ∆yc 2 ) ∆x (T4o − T3o ) + ∆x (T3i − T3o ) = 0 R ′′ t,c or, 2T3c + 0.25 ( T2o + T4o ) + ∆x T3i k c R ′′ t,c − 2.5 + ∆x k c R ′′ t,c T3o = 0 Similar 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 obtain k ( ∆y 2 ) k ( ∆y 2 ) ∆x k∆x (T3o − T3i ) + (T2i − T3i ) + (T4i − T3i ) + (T9 − T3i ) = 0 ∆x ∆x ∆y R ′′ t,c or, ∆x ∆x T3o + 0.5 T2,i + T4,i + T9 − 2 + kR ′′ kR ′′ t,c t,c ( ) T3i = 0 Similar expressions may be obtained for the remaining nodal points on the inner region of the contact resistance. The 33 finite-difference equations were entered into the workspace of IHT from the keyboard (model equations are appended), and for ho = 1000 W/m2⋅K, T∞,o = 1700 K, hi = 200 W/m2⋅K and T∞,i = 400 K, the following temperature field was obtained, where coordinate (x,y) locations are in mm and temperatures are in °C. y\x 0 0.5 0.5 1.5 2.5 3.5 0 1536 1473 1456 1450 1446 1445 1 1535 1472 1456 1450 1445 1443 2 1534 1471 1454 1447 1441 1438 3 1533 1469 1452 1446 1438 0 4 1533 1468 1451 1444 1437 0 5 1532 1468 1451 1444 1436 0 Note the significant reduction in the turbine blade temperature, as, for example, from a surface temperature of T1 = 1526 K without the TBC to T1i = 1456 K with the coating. Hence, the coating is serving its intended purpose. COMMENTS: (1) Significant additional benefits may still be realized by increasing hi. (2) The foregoing solution may be used to determine the temperature field without the TBC by setting kc → ∞ and R ′′,c → 0. t PROBLEM 4.60 KNOWN: Bar of rectangular cross-section subjected to prescribed boundary conditions. FIND: Using a numerical technique with a grid spacing of 0.1m, determine the temperature distribution and the heat transfer rate from the bar to the fluid. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. ANALYSIS: The nodal network has ∆x = ∆y = 0.1m. Note the adiabat corresponding to system symmetry. The finite-difference equations for each node can be written using either Eq. 4.33, for interior 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 that h∆x 50W/m2 ⋅K × 0.1m = = 3.333. k 1.5 W/m ⋅ K Node Finite-Difference Equations 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -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 = 0 2T10 + T14 + 6.666×30-10.666 T13 = 0 2T11 + T13 + T15 + 6.666×30-2(3.333+2)T14 = 0 2T12 +T14 + 200 + 6.666×30-2(3.333+2) T15 = 0 Using 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 ….. -4 1 0 1 0 0 0 [A] = 0 0 0 0 0 0 0 0 20 -4 1 1 -4 00 10 01 00 00 00 00 00 00 00 00 00 0 0 -200 0 0 -200 0 [C] = 0 -200 0 0 -200 -200 -200 -400 2 0 0 -4 1 0 1 0 0 0 0 0 0 0 0 0 2 0 2 -4 1 0 1 0 0 0 0 0 0 0 PROBLEM 4.60 (Cont.) 0000000 0 0 0 0000000 0 0 0 -2 0 0 0 0 0 0 0 0 0 0100000 0 0 0 1010000 0 0 0 -4 0 0 1 0 0 0 0 0 0 0 -4 2 0 1 0 0 0 0 0 0 1 -4 1 0 1 0 0 0 0 1 0 1 -4 0 0 1 0 0 0 0 1 0 0 -4 2 0 1 0 0 0 0 1 0 -1 -4 1 0 1 0 0 0 0 1 0 1 -4 0 0 1 0 0 0 0 2 0 0 -10.66 2 0 0 0 0 0 0 2 0 1 -10.66 1 00000020 1 -10.66 T1 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 C T 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 volumes exposed to the fluid. Using Newton’s law of cooling, considering a unit thickness of the bar, find ∆y ∆y q conv = 2 h ⋅ ∞ 2 ⋅ ( T13 − T∞ ) + h ⋅ ∆y ⋅ ( T14 − T ) + h ⋅ ∆y ( T15 − T∞ ) + h ⋅ 2 ( 200 − T∞ ) 1 1 q conv = 2h ⋅ ∆y (T13 − T∞ ) + ( T14 − T∞ ) + ( T15 − T∞ ) + ( 200 − T∞ ) 2 2 W 1 1 q conv = 2 × 50 × 0.1m ( 45.8 − 30 ) + ( 48.7 − 30 ) + ( 67.0 − 30 ) + ( 200 − 30 ) 2 2 m2 ⋅ K q conv = 1487 W/m. < PROBLEM 4.61 KNOWN: Upper surface and grooves of a plate are maintained at a uniform temperature T1, while the lower 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 a finite-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 ) = 0 4Ta − ( 200 + Tb + 20 + 200 ) = 0 or 4Ta − Tb = 420 (1) or −2Ta + 4Tb = 220 (2) 4Tb − (T1 + Ta + T2 + Ta ) = 0 4Tb − ( 200 + 2Ta + 20 ) = 0 Multiply Eq. (2) by 2 and subtract from Eq. (1) to obtain 7Tb = 860 or Tb = 122.9°C From Eq. (1), 4Ta - 122.9 = 420 or Ta = (420 + 122.9)/4 = 135.7°C. The heat transfer through the symmetrical section is equal to the sum of heat flows through control volumes adjacent to the lower surface. From the schematic, T − T2 ∆x T1 − T2 ∆x Tb − T2 ′ q′ = q1 + q′ + q′ = k . + k ( ∆x ) a + k 2 3 ∆y 2 ∆y 2 ∆y Continued... PROBLEM 4.61 (Cont.) Noting that ∆x = ∆y, regrouping and substituting numerical values, find 1 1 q′ = k ( T1 − T2 ) + ( Ta − T2 ) + ( Tb − T2 ) 2 2 1 1 q′ = 15 W m ⋅ K ( 200 − 20 ) + (135.7 − 20 ) + (122.9 − 20 ) = 3.86 kW m . 2 2 < For the full groove spacing, q′ total = 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 following two-dimensional temperature field was computed for the grid shown in schematic (b), where x and y are in mm and the nodal temperatures are in °C. Nodes 2-54 are interior nodes, with those along the symmetry adiabats characterized by Tm-1,n = Tm+1,n, while nodes 55-63 lie on a plane surface. y\x 0 10 20 30 40 50 60 70 80 0 200 141.4 97.09 57.69 20 10 20 30 200 175.4 134.3 94.62 56.83 20 200 182.4 160.3 125.7 90.27 55.01 20 200 186.7 169.5 148.9 118.0 85.73 52.95 20 40 200 191 177.2 160.1 140.1 111.6 81.73 51.04 20 50 200 186.6 171.2 153.4 133.5 106.7 78.51 49.46 20 60 200 184.3 167.5 149.0 128.7 103.1 76.17 48.31 20 70 200 183.1 165.5 146.4 125.7 100.9 74.73 47.60 20 80 200 182.8 164.8 145.5 124.4 100.1 74.24 47.36 20 The foregoing results were computed for h = 107 W/m2⋅K (h → ∞) and T∞ = 20°C, which is tantamount to prescribing an isothermal bottom surface at 20°C. Agreement between corresponding results for the coarse and fine grids is surprisingly good (Ta = 135.7°C ↔ T23 = 140.1°C; Tb = 122.9°C ↔ T27 = 124.4°C). The heat rate is q′ = 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 m The agreement with q T = 7.72 kW/m from the coarse grid of part (a) is excellent and a fortuitous consequence of compensating errors. With reductions in the convection coefficient from h → ∞ to h = 1000, 200 and 5 W/m2⋅K, the corresponding increase in the thermal resistance reduces the heat rate to values of 6.03, 3.28 and 0.14 kW/m, respectively. With decreasing h, there is an overall increase in nodal temperatures, as, for example, from 191°C to 199.8°C for T2 and from 20°C to 196.9°C for T55. NOTE TO INSTRUCTOR: To reduce computational effort, while achieving the same educational objectives, the problem statement has been changed to allow for convection at the bottom, rather than the top, surface. PROBLEM 4.62 KNOWN: Rectangular plate subjected to uniform temperature boundaries. FIND: Temperature at the midpoint using a finite-difference method with space increment of 0.25m SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. ANALYSIS: For the nodal network above, 12 finite-difference equations must be written. It follows that node 8 represents the midpoint of the rectangle. Since all nodes are interior nodes, Eq. 4.33 is appropriate and is written in the form 4Tm − ∑ Tneighbors = 0. For nodes on the symmetry adiabat, the neighboring nodes include two symmetrical nodes. Hence, for Node 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 be used to find the nodal temperatures Tm. −4 1 0 0 1 1 − 4 1 0 0 0 1 − 4 1 0 0 0 2 −4 0 1 0 0 0 −4 0 A = 0 1 0 0 1 0 01 0 0 00 1 0 0 00 0 1 0 00 0 0 0 00 0 0 0 00 0 0 0 0 000 1 0 000 0 1 000 0 0100 1 0010 −4 1001 1 −4 100 0 2 −4 00 00 0 −4 1 1 0 0 1 −4 01 001 0 010 0 0 0 0 0 0 0 1 0 0 1 −4 2 0 0 0 0 0 0 0 1 0 0 1 -4 −200 −150 −150 −150 −50 C = 0 0 0 −100 −50 −50 −50 96.5 112.9 118.9 120.4 73.2 T = 86.2 92.3 94.0 59.9 65.5 69.9 71.0 The temperature at the midpoint (Node 8) is T (1,0.5) = T8 = 94.0o C. < COMMENTS: Using the exact analytical, solution – see Eq. 4.19 and Problem 4.2 – the midpoint temperature is found to be 94.5°C. To improve the accuracy of the finite-difference method, it would be necessary to decrease the nodal mesh size. PROBLEM 4.63 KNOWN: Long bar with trapezoidal shape, uniform temperatures on two surfaces, and two insulated surfaces. FIND: Heat transfer rate per unit length using finite-difference method with space increment of 10mm. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. 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 be written. Nodes 1-4 and 6-8 are interior nodes and their finite-difference equations can be written directly from Eq. 4.33. For these nodes Tm,n+1 + Tm,n-1 + Tm+1,n + Tm-1,n − 4Tm,n = 0 m = 1 − 4, 6 − 8. (1) For nodes 5 and 9 located on the diagonal, insulated boundary, the appropriate finite-difference equation follows from an energy balance on the control volume shown above (upper-right corner of schematic), & & E in − E out = q a + qb = 0 T − Tm,n T −T k ( ∆y ⋅1) m-1,n + k ( ∆x ⋅1) m,n-1 m,n = 0. ∆x ∆y Since ∆x = ∆y, the finite-difference equation for nodes 5 and 9 is of the form Tm-1,n + Tm,n-1 − 2Tm,n = 0 m = 5,9. (2) The system of 9 finite-difference equations is first written in the form of Eqs. (1) or (2) and then written in explicit form for use with the Gauss-Seidel iteration method of solution; see Section 4.5.2. Node 1 2 3 4 5 6 7 8 9 Finite-difference equation T2+T2+T6+100-4T1 = 0 T3+T1+T7+100-4T2 = 0 T4+T2+T8+100-4T3 = 0 T5+T3+T9+100-4T4 = 0 100+T4-2T5 = 0 T7+T7+25+T1-4T6 = 0 T8+T6+25+T2-4T7 = 0 T9+T7+25+T3-4T8 = 0 T4+T8-2T9 = 0 Gauss-Seidel form T1 = 0.5T2+0.25T6+25 T2 = 0.25(T1+T3+T7)+25 T3 = 0.25(T2+T4+T8)+25 T4 = 0.25(T3+T5+T9)+25 T5 = 0.5T4+50 T6 = 0.25T1+0.5T7+6.25 T7 = 0.25(T2+T6+T8)+6.25 T8 = 0.25(T3+T7+T9)+6.25 T9 = 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 the table below. k T1 T2 T3 T4 T5 T6 T7 T8 T9(° C) 0 1 2 3 4 5 6 75 75.0 75.7 76.3 76.3 76.6 76.6 75 76.3 76.9 77.0 77.3 77.3 77.5 80 80.0 80.0 80.2 80.2 80.3 80.3 85 86.3 86.3 86.3 86.3 86.3 86.4 90 92.5 93.2 93.2 93.2 93.2 93.2 50 50.0 51.3 51.3 51.7 51.7 51.9 50 52.5 52.2 52.7 52.7 52.9 52.9 60 57.5 57.5 57.3 57.5 57.4 57.5 75 72.5 71.9 71.9 71.8 71.9 71.9 Note that by the sixth iteration the change is less than 0.3°C; hence, we assume the temperature distribution 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 the control volumes about nodes 6, 7, and 8. From the schematic, find that ′ ′ ′ q′ = q1 + q2 + q3 T − 25 T − 25 ∆y T6 − 25 q′ = k∆y 8 + k∆y 7 +k ∆x ∆x 2 ∆x Recognizing that ∆x = ∆y and substituting numerical values, find q′ = 20 W 1 ( 57.5 − 25) + ( 52.9 − 25 ) + 2 ( 51.9 − 25) K m ⋅K q′ = 1477 W/m. < COMMENTS: (1) Recognize that, while the temperature distribution may have been determined to a reasonable approximation, the uncertainty in the heat rate could be substantial. This follows since the heat rate is based upon a gradient and hence on temperature differences. (2) Note that the initial guesses (k = 0) for the iteration are within 5°C of the final distribution. The geometry is simple enough that the guess can be very close. In some instances, a flux plot may be helpful and save labor in the calculation. (3) In writing the FDEs, the iteration index (superscript k) was not included to simplify expression of the equations. However, the most recent value of Tm,n is always used in the computations. Note that this system of FDEs is diagonally dominant and no rearrangement is required. PROBLEM 4.64 KNOWN: Edge of adjoining walls (k = 1 W/m⋅K) represented by symmetrical element bounded by the diagonal symmetry adiabat and a section of the wall thickness over which the temperature distribution is assumed 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 of assuming linear temperature distributions across sections at various distances from the edge. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties, and (3) Linear temperature distribution at specified locations across the section. ANALYSIS: (a) Taking advantage of symmetry along the adiabat diagonal, all the nodes may be treated as interior nodes. Across the left-hand boundary, the temperature distribution is specified as linear. The finite-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 is q′tot = 2 [q′ + q′ + q′ + q′ ] a b c d q′ = 2 [k ( ∆x 2 )( Tc − T2 ) ∆y + k∆x ( T3 − T2 ) ∆y + k∆x ( T4 − T2 ) ∆y + k∆x ( T5 − T2 ) ∆x ] tot The shape factor for the full edge is defined as q′tot = kS′ ( T1 − T2 ) Solving the above equation set in IHT, the temperature (°C) distribution is Continued... PROBLEM 4.64 (Cont.) 0 0 0 0 0 25 18.75 12.5 6.25 50 37.5 25.0 75 56.25 00 < and the heat rate and shape factor are q′tot = 100 W m < S =1 From Table 4.1, the edge shape factor is 0.54, considerably below our estimate from this coarse grid analysis. (b) The effect of the linear temperature distribution on the shape factor estimate can be explored using a more extensive grid as shown below. The FDE analysis was performed with the linear distribution imposed as the different sections a, b, c, d, e. Following the same approach as above, find Location of linear distribution Shape factor, S (a) 0.797 (b) 0.799 (c) 0.809 (d) 0.857 (e) 1.00 The shape factor estimate decreases as the imposed linear temperature distribution section is located further from the edge. We conclude that assuming the temperature distribution across the section directly at the edge is a poor-one. COMMENTS: The grid spacing for this analysis is quite coarse making the estimates in poor agreement with the Table 4.1 result. However, the analysis does show the effect of positioning the linear temperature distribution condition. PROBLEM 4.65 KNOWN: Long triangular bar insulated on the diagonal while sides are maintained at uniform temperatures Ta and Tb. FIND: (a) Using a nodal network with five nodes to the side, and beginning with properly defined control volumes, derive the finite-difference equations for the interior and diagonal nodes and obtain the temperature distribution; sketch the 25, 50 and 75°C isotherms and (b) Recognizing that the insulated diagonal surface can be treated as a symmetry line, show that the diagonal nodes can be treated as interior nodes, and write the finite-difference equations by inspection. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, and (3) Constant properties. 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 , yields q′ + q′ + q′ = 0 a b c T −T T − T3 =0 0 + k∆x 5 3 + k∆y 2 ∆y ∆x T3 = 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 the nodal temperatures (°C), tabulated according to the nodal arrangement: Continued... PROBLEM 4.65 (Cont.) 00 85.71 00 71.43 50.00 00 50.00 28.57 14.29 0 0 0 The 25, 50 and 75°C isotherms are sketched below, using an interpolation scheme to scale the isotherms on the triangular bar. (b) If we consider the insulated surface as a symmetry plane, the nodal network appears as shown. As such, the diagonal nodes can be treated as interior nodes, as Eq. (1) above applies. Recognize the form is the same as that of Eq. (2) or (3). COMMENTS: Always look for symmetry conditions which can greatly simplify the writing of nodal equations. In this situation, the adiabatic surface can be treated as a symmetry plane such that the nodes can be treated as interior nodes, and the finite-difference equations can be written by inspection. PROBLEM 4.66 KNOWN: Straight fin of uniform cross section with prescribed thermal conditions and geometry; tip condition allows for convection. FIND: (a) Calculate the fin heat rate, q′ , and tip temperature, TL , assuming one-dimensional heat f transfer in the fin; calculate the Biot number to determine whether the one-dimensional assumption is valid, (b) Using the finite-element software FEHT, perform a two-dimensional analysis to determine the fin heat rate and the tip temperature; display the isotherms; describe the temperature field and the heat flow pattern inferred from the display, and (c) Validate your FEHT code against the 1-D analytical solution for a fin using a thermal conductivity of 50 and 500 W/m⋅K. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conduction with constant properties, (2) Negligible radiation exchange, (3) Uniform convection coefficient. ANALYSIS: (a) Assuming one-dimensional conduction, q′ and TL can be determined using Eqs. L 3.72 and 3.70, respectively, from Table 3.4, Case A. Alternatively, use the IHT Model | Extended Surfaces | Temperature Distribution and Heat Rate | Straight Fin | Rectangular. These results are tabulated below and labeled as “1-D.” The Biot number for the fin is h t/2 500 W / m 2 ⋅ K 0.020 m / 2 Bi = ( k )= ( 5 W / m⋅K ) =1 (b, c) The fin can be drawn as a two-dimensional outline in FEHT with convection boundary conditions on the exposed surfaces, and with a uniform temperature on the base. Using a fine mesh (at least 1280 elements), solve for the temperature distribution and use the View | Temperature Contours command to view the isotherms and the Heat Flow command to determine the heat rate into the fin base. The results of the analysis are summarized in the table below. k (W/m⋅K) 5 50 500 Bi 1 0.1 0.01 ( Tip temperature, TL (°C) 1-D 2-D 100 100 100.3 100 123.8 124 Fin heat rate, q′ (W/m) f 1-D 1010 3194 9812 2-D 805 2990 9563 Difference* (%) 20 6.4 2.5 ) * Difference = q′ ,1D − q′ ,2D × 100 / q′ ,1D f f f COMMENTS: (1) From part (a), since Bi = 1 > 0.1, the internal conduction resistance is not negligible. Therefore significant transverse temperature gradients exist, and the one-dimensional conduction assumption in the fin is a poor one. Continued ….. PROBLEM 4.66 (Cont.) (2) From the table, with k = 5 W/m⋅K (Bi = 1), the 2-D fin heat rate obtained from the FEA analysis is 20% lower than that for the 1-D analytical analysis. This is as expected since the 2-D model accounts for transverse thermal resistance to heat flow. Note, however, that analyses predict the same tip temperature, 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/m⋅K case, the FEHT isotherms show considerable curvature in the region near the fin base. For example, at x = 10 and 20 mm, the difference between the centerline and surface temperatures are 15 and 7°C. (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/m⋅K is mostly due to the approximate manner in which the heat rate is calculated in the FEA software. PROBLEM 4.67 KNOWN: Long rectangular bar having one boundary exposed to a convection process (T∞, h) while the other boundaries are maintained at constant temperature Ts. FIND: Using the finite-element method of FEHT, (a) Determine the temperature distribution, plot the isotherms, and identify significant features of the distribution, (b) Calculate the heat rate per unit length (W/m) into the bar from the air stream, and (c) Explore the effect on the heat rate of increasing the convection coefficient by factors of two and three; explain why the change in the heat rate is not proportional 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 the specified boundary conditions and material property. The View | Temperature Contours command is used to represent ten isotherms (isopotentials) that have minimum and maximum values of 53.9°C and 85.9°C, respectively. Because of the symmetry boundary condition, the isotherms are normal to the center-plane indicating an adiabatic surface. Note that the temperature change along the upper surface of the bar is substantial (≈ 40°C), whereas the lower half of the bar has less than a 3°C change. That is, the lower half 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 selected convection coefficients, the heat rates into the bar from the air stream were calculated. h W / m2 ⋅ K 100 200 300 ( q′ ( W / m ) ) 128 175 206 Increasing 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 of the bar (conduction) and the convection process. Since the conduction resistance is significant, we should not expect the heat rate to change proportionally to the change in convection resistance. PROBLEM 4.68 KNOWN: Log rod of rectangular cross-section of Problem 4.55 that experiences uniform heat generation while its surfaces are maintained at a fixed temperature. Use the finite-element software FEHT as your analysis tool. FIND: (a) Represent the temperature distribution with representative isotherms; identify significant features; and (b) Determine what heat generation rate will cause the midpoint to reach 600 K with unchanged boundary conditions. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, and (2) Two-dimensional conduction with constant properties. ANALYSIS: (a) Using FEHT, do the following: in Setup, enter an appropriate scale; Draw the outline 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 the Material Properties and Generation rate. Draw three Element Lines as shown on the annotated version of the FEHT screen below. To reduce the mesh, hit Draw/Reduce Mesh until the desired fineness 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. The isotherms, especially near the center, have an elliptical shape. Along the x = 0 axis and the y = 10 mm axis, the temperature gradient is nearly linear. The hottest point is of course the center for which the temperature is (T(0, 10 mm) = 401.3 K. < The temperature of this point can be read using the View/Temperatures or View|Tabular Output command. (b) To determine the required generation rate so that T(0, 10 mm) = 600 K, it is necessary to re-run the model with several guessed values of q . After a few trials, find q = 1.48 × 108 W / m3 < PROBLEM 4.69 KNOWN: Symmetrical section of a flow channel with prescribed values of q and k, as well as the surface convection conditions. See Problem 4.5(S). FIND: Using the finite-element method of FEHT, (a) Determine the temperature distribution and plot the isotherms; identify the coolest and hottest regions, and the region with steepest gradients; describe the heat flow field, (b) Calculate the heat rate per unit length (W/m) from the outer surface A to the adjacent 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 the specified boundary conditions, material property and generation. The View | Temperature Contours command is used to represent ten isotherms (isopotentials) that have minimum and maximum values of 82.1°C and 125.2°C. The hottest region of the section is the upper vertical leg (left-hand corner). The coolest region is in the lower horizontal leg at the far right-hand boundary. The maximum and minimum section temperatures (125°C and 77°C), respectively, are higher than either adjoining fluid. Remembering that heat flow lines are normal to the isotherms, heat flows from the hottest corner directly to the inner fluid and downward into the lower leg and then flows out surface A and the lower portion of surface B. Continued ….. PROBLEM 4.69 (Cont.) (b, c) Using the View | Heat Flows command considering the boundaries for surfaces A and B, the heat rates are: q′ = 1135 W / m s q′ = −1365 W / m. B < From an energy balance on the section, we note that the results are consistent since conservation of energy is satisfied. E′ − E′ + E g = 0 in out −q′ + q′ + q∀′ = 0 A B −1135 W / m + ( −1365 W / m ) + 2500 W / m = 0 < where q∀′ = 1× 106 W / m3 × [25 × 50 + 25 × 50]× 10−6 m 2 = 2500 W / m. COMMENTS: (1) For background on setting up this problem in FEHT, see the tutorial example of the User’s Manual. While the boundary conditions are different, and the internal generation term is to be 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 4.70 KNOWN: Hot-film flux gage for determining the convection coefficient of an adjoining fluid stream by 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 that is conducted into the quartz substrate considering three cases corresponding to convection coefficients 2 of 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 power dissipation over hot film. ANALYSIS: The symmetrical section shown in the schematic above (right) is drawn into FEHT specifying 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 dissipation ′ Pe (W/m) in the hot film is transferred by convection from the film surface, q′ v,f , and from the c adjacent substrate surface, q′ v,s . c The analysis evaluates the fraction, F, of the dissipated electrical power that is conducted into the substrate and convected to the fluid stream, ′ F = q′ / Pe = 1 − q′ cv,s ′ cv,f / Pe ′ ′′ where 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,f c for the three values of the convection coefficient. Continued ….. PROBLEM 4.70 (Cont.) 2 h(W/m ⋅K) Case 1 2 3 q′ cv,f ( W / m ) 500 1000 2000 5.64 6.74 7.70 F(%) Ts,f (°C) 43.6 32.6 23.3 30.9 28.6 27.0 COMMENTS: (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, F amounts to 25%, making it necessary to develop a reliable, accurate heat transfer model to estimate the applied correction. Further, this condition limits the usefulness of this gage design to flows with high convection coefficients. (3) A reduction in F, and hence the effect of an applied correction, could be achieved with a substrate material having a lower thermal conductivity than quartz. However, quartz is a common substrate material for fabrication of thin-film heat-flux gages and thermocouples. By what other means could you reduce F? (4) In addition to the tutorial example in the FEHT User’s Manual, the solved models for Examples 4.3 and 4.4 are useful for developing skills helpful in solving this problem. PROBLEM 4.71 KNOWN: Hot-plate tool for micro-lithography processing of 300-mm silicon wafer consisting of an aluminum alloy equalizing block (EB) heated by ring-shaped main and trim electrical heaters (MH and 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 of mh th 140°C with a uniformity of 0.1°C. Consider these steps in the analysis: (a) Perform an energy balance on the EB to obtain an initial estimate for the heater fluxes with q′′ h = q′ extending over m th the full radial limits; using FEHT, determine the upper surface temperature distribution and comment on whether the desired uniformity has been achieved; (b) Re-run your FEHT code with different values of the heater fluxes to obtain the best uniformity possible and plot the surface temperature distribution; (c) Re-run your FEHT code for the best arrangement found in part (b) using the representative distribution of the convection coefficient (see schematic for h(r) for downward flowing gas 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 and constant properties in EB, (3) Lower surface of EB perfectly insulated, (4) Uniform convection coefficient over upper EB surface, unless otherwise specified and (5) negligible radiation exchange between the EB surfaces and the surroundings. ANALYSIS: (a) To obtain initial estimates for the MH and TH fluxes, perform an overall energy balance on the EB as illustrated in the schematic below. Ein − Eout = 0 ( ) ( ) 22 22 2 q′′ π r2 − r1 + q′′ π r4 − r3 − h π ro + 2π ro w ( Ts − T∞ ) = 0 mh th Continued ….. PROBLEM 4.71 (Cont.) Substituting numerical values and letting q′′ h = q′′ , find m th q′′ = q′′ = 2939 W / m 2 mh th < Using FEHT, the analysis is performed on an axisymmetric section of the EB with the nodal arrangement 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 from the Tabular Output page. The Temperature Gradients view command is used to obtain the heat flow distribution 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 across mh th the radius of the wafer is 1.7°C, much larger than the design goal of 0.1°C. The upper surface temperature distribution is shown in the graph below. Continued ….. PROBLEM 4.71 (Cont.) EB surface temperature distribution 141.5 141 T(r,z), (C) 140.5 140 139.5 139 138.5 0 20 40 60 80 100 120 140 160 Radial position, r (mm) (b) From examination of the results above, we conclude that if q′′ h is reduced and q′′h increased, the m t EB surface temperature uniformity could improve. The results of three trials compared to the base case are tabulated below. Trial ( q′′ mh W/m 2 ) ( q′′ th W / m2 ) T1 T14 (°C ) (°C ) T1 − T14 ( °C ) Base 2939 2939 141.1 139.3 1.8 1 2880 (-2%) 2997 (+2%) 141.1 139.4 1.7 2 2880 (-2%) 3027 (+3%) 141.7 140.0 1.7 3 2910 (-1%) 2997 (+2%) 141.7 139.9 1.8 2939 2939 141.7 139.1 2.6 Part (d) 2939 k=150 W/m⋅K 2939 140.4 139.5 0.9 Part (d) 2939 k=300 W/m⋅K 2939 140.0 139.6 0.4 Part (c) The strategy of changing the heater fluxes (trials 1-3) has not resulted in significant improvements in the 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 upper surface of the EB were specified by the function h(r) shown in the schematic. The value of h(r) 2 ranged from 5.4 to 13.5 W/m ⋅K between the centerline and EB edge. The result of the analysis is tabulated above, labeled as part (c). Note that the temperature uniformity has become significantly poorer. (d) There are at least two options that should be considered in the re-design to improve temperature uniformity. Higher thermal conductivity material for the EB. Aluminum alloy is the material most widely used in practice for reasons of low cost, ease of machining, and durability of the heated surface. The results of analyses for thermal conductivity values of 150 and 300 W/m⋅K are tabulated above, labeled as part (d). Using pure or oxygen-free copper could improve the temperature uniformity to better than 0.5°C. Distributed heater elements. The initial option might be to determine whether temperature uniformity could be improved using two elements, but located differently. Another option is a single element heater spirally embedded in the lower portion of the EB. By appropriately positioning the element as a function of the EB radius, improved uniformity can be achieved. This practice is widely used where precise and uniform temperature control is needed. PROBLEM 4.72 KNOWN: 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) Constant properties, (4) Uniform film coefficient. ANALYSIS: (a) From the analysis of Problem 4.45, the finite-difference equations for the nodal arrangement 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, ) ( hP h ⋅ 2 h ⋅2 2 500 W m 2 ⋅ K × 2 4 × 10−3 mm = 0.0533 ⋅ ∆x 2 ≈ ∆x 2 = ∆x = −3 m kA c k ⋅⋅w kw 50 W m ⋅ K × 6 ×10 Node 1: Node n: Node 12: 100 + T2 + 0.0533 × 30 − ( 2 + 0.0533) T1 = 0 or -2.053T1 + T2 = -101.6 Tn +1 + Tn −1 + 1.60 − 2.0533Tn = 0 or T11 + ( 0.0533 2 ) 30 − ( 0.0533 2 + 1) T12 = 0 or Tn −1 − 2.053Tn + Tn −1 = −1.60 T11 − 1.0267T12 = −0.800 Using matrix notation, Eq. 4.52, where [A] [T] = [C], the A-matrix is tridiagonal and only the non-zero terms are shown below. A matrix inversion routine was used to obtain [T]. Tridiagonal Matrix A Nonzero Terms a1,1 a1,2 a2,1 a2,2 a2,3 a3,2 a3,3 a3,4 a4,3 a4,4 a4,5 a5,4 a5,5 a5,6 a6,5 a6,6 a6,7 a7,6 a7,7 a7,8 a8,7 a8,8 a8,9 a9,8 a9,9 a9,10 a10,9 a10,10 a10,11 a11,10 a11,11 a11,12 a12,11 a12,12 a12,13 1 1 1 1 1 1 1 1 1 1 1 Column Matrices Values -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -1.027 1 1 1 1 1 1 1 1 1 1 1 1 Node 1 2 3 4 5 6 7 8 9 10 11 12 C -101.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -0.8 T 85.8 74.5 65.6 58.6 53.1 48.8 45.5 43.0 41.2 39.9 39.2 38.9 The assumption of one-dimensional heat conduction is justified when Bi ≡ h(w/2)/k < 0.1. Hence, with Bi = 500 W/m2⋅K(3 × 10-3 m)/50 W/m⋅K = 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 Node 0, T0 − T1 ∆x ′ + h2 q′ = q1 + q′ ( T0 − T∞ ) f conv = k ⋅ w ∆x 2 ( q′ = 50 W m ⋅ K 6 × 10−3 m f ) (100 − 85.8 ) C 4 × 10 −3 + 500 W m2 ⋅ K 2 ⋅ m 4 × 10−3 m 2 (100 − 30 ) C q′ = (1065 + 140 ) W m = 1205 W m . f From Eq. 3.76, the fin heat rate is q = ( hPkA c ) 1/ 2 < ⋅ θ 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 θb = 70°C, q′ = 17.32 W K × 70 K × tanh (57.44 × 0.048 ) = 1203 W m and 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 temperature distribution and heat rate were computed for h = 10, 100, 500 and 1000 W/m2⋅K. Results are plotted as follows. 100 1800 1500 80 70 Heat rate, q'(W/m) Temperature, T(C) 90 60 50 40 30 0 8 16 24 32 40 48 1200 900 600 300 Fin location, x(mm) h = 10 W/m^2.K h = 100 W/m^2.K h = 500 W/m^2.K h = 1000 W/m^2.K 0 0 200 400 600 800 1000 Convection coefficient, h(W/m^2.K) The temperature distributions were obtained by first creating a Lookup Table consisting of 4 rows of nodal temperatures corresponding to the 4 values of h and then using the LOOKUPVAL2 interpolating function 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 the Lookup Table. For each value of h, Explore was used to compute T(x), thereby generating 4 data sets which were placed in the Browser and used to generate the plots. The variation of q′ with h was simply generated by using the Explore feature to solve the finite-difference model equations for values of h incremented by 10 from 10 to 1000 W/m2⋅K. Although q′ increases with increasing h, the effect of changes in h becomes less pronounced. This f trend is a consequence of the reduction in fin temperatures, and hence the fin efficiency, with increasing h. For 10 ≤ h ≤ 1000 W/m2⋅K, 0.95 ≥ ηf ≥ 0.24. Note the nearly isothermal fin for h = 10 W/m2⋅K and the pronounced temperature decay for h = 1000 W/m2⋅K. PROBLEM 4.73 KNOWN: Pin fin of 10 mm diameter and length 250 mm with base temperature of 100°C experiencing radiation exchange with the surroundings and free convection with ambient air. FIND: Temperature distribution using finite-difference method with five nodes. Fin heat rate and relative contributions by convection and radiation. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in fin, (3) Constant properties, (4) Fin approximates small object in large enclosure, (5) Fin tip experiences convection and radiation, (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 nodal temperatures. Interior node, n = 1, 2, 3 or 4 Ein − Eout = 0 qa + q b + qc + qd = 0 (1) T T −T −T h r,n P∆x (Tsur − Tn ) + kA c n +1 n + h fc,n P∆x (T∞ − Tn ) + kA c n −1 n = 0 ∆x ∆x (2) where the free convection coefficient is 1/ 6 2 h fc,n = 2.89 0.6 + 0.624 ( Tn − T∞ ) and the linearized radiation coefficient is 2 2 h r,n = εσ (Tn + Tsur ) Tn + Tsur with P = πD and Ac = πD2/4. ( ) (3) (4) (5,6) Tip node, n = 5 Ein − Eout = 0 qa + q b + qc + qd + qe = 0 h r,5 ( P∆x 2 )(Tsur − T5 ) + h r,5A c (Tsur − T5 ) + h fc,5Ac (T∞ − T5 ) T −T + h fc,5 ( P∆x 2 )( T∞ − T5 ) + kAc 4 5 = 0 ∆x (7) Continued... PROBLEM 4.73 (Cont.) Knowing the nodal temperatures, the heat rates are evaluated as: Fin Heat Rate: Perform an energy balance around Node b. Ein − Eout = 0 qa + q b + q c + q fin = 0 h r,b ( P∆x 2 )(Tsur − Tb ) + h fc,b ( P∆x 2 )(T∞ − Tb ) + kA c (T1 − Tb ) + q ∆x fin = 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, we need to sum contributions from each node. Using the convection heat rate terms from the foregoing energy 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 )5 The above equations were entered into the IHT workspace and the set of equations solved for the nodal temperatures and the heat rates. Summary of key results including the temperature distribution and heat rates is shown below. Node b 1 2 3 4 5 Fin Tj (°C) qcv (W) qfin (W) qrad (W) hcv (W/m2⋅K) hrod (W/m2⋅K) 100 0.603 10.1 1.5 58.5 0.451 8.6 1.4 40.9 0.183 7.3 1.3 33.1 0.081 6.4 1.3 29.8 0.043 5.7 1.2 28.8 0.015 5.5 1.2 1.375 1.604 0.229 - < COMMENTS: From the tabulated results, it is evident that free convection is the dominant node. Note that the free convection coefficient varies almost by a factor of two over the length of the fin. PROBLEM 4.74 KNOWN: Thin metallic foil of thickness, t, whose edges are thermally coupled to a sink at temperature Tsink is exposed on the top surface to an ion beam heat flux, q TT , and experiences radiation exchange with s the 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 the page. 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′ = 0 a b c d e q′′ ∆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) s where the linearized radiation coefficient for node n is ( 2 2 h r,n = εσ ( Tsur + Tn ) Tsur + Tn Solving 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): T1 374.1 T2 374.0 T3 373.5 T4 372.5 T5 370.9 T6 368.2 T7 363.7 T8 356.6 T9 345.3 T10 327.4 Continued... PROBLEM 4.74 (Cont.) COMMENTS: (1) If the temperature gradients were excessive across the foil, it would wrinkle; most likely 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-8 B = 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 variables deltax = L / 10 L = 0.150 t = 0.00025 eps = 0.45 Tsur = 300 k = 40 Tsink = 300 q''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): T1 T2 374.1 374 T3 373.5 T4 372.5 T5 370.9 T6 368.2 T7 363.7 T8 356.6 T9 345.3 T10 327.4 hr1 3.956 hr3 3.943 hr4 3.926 hr5 3.895 hr6 3.845 hr7 3.765 hr8 3.639 hr9 3.444 hr10 3.157 */ hr2 3.953 PROBLEM 4.75 KNOWN: Electrical heating elements with known dissipation rate embedded in a ceramic plate of known thermal conductivity; lower surface is insulated, while upper surface is exposed to a convection process. FIND: (a) Temperature distribution within the plate using prescribed grid spacing, (b) Sketch isotherms to illustrate temperature distribution, (c) Heat loss by convection from exposed surface (compare with element dissipation rate), (d) Advantage, if any, in not setting ∆x = ∆y, (e) Effect of grid size and convection coefficient on the temperature field. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, two-dimensional conduction in ceramic plate, (2) Constant properties, (3) No internal generation, except for Node 7 (or Node 15 for part (e)), (4) Heating element approximates 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. Node 7 is a special case of the interior node having a generation term; because of symmetry, q′ = 25 W/m. ht The 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; a b c d k T −T ∆y T1 − T2 ∆y T3 − T2 + h∆x (T∞ − T2 ) + k + k∆x 5 2 = 0 . 2 ∆x 2 ∆x ∆y Regrouping, find 2 2 1 + 2N ∆x + 1 + 2 ∆x = T1 + T3 + 2 ∆x T5 + 2N ∆x T∞ T2 ∆y ∆y ∆y ∆y where N = h∆x/k = 100 W/m2⋅K × 0.006 m/2 W/m⋅K = 0.30 K. Hence, with T∞ = 30°C, 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 E′n represents the conduction terms. Hence, in g g i ht q′ + q′ + q′ + q′ + 2q′ = 0 , or a b c d ht T8 − T7 T − T7 T − T7 T − T7 k∆y + k∆x 4 + k∆y 8 + k∆x 10 + 2q′ = 0 ht ∆x ∆y ∆x ∆y Regrouping, 2 2 2 ∆x 2 2q′ ∆x ∆x ∆x ∆x 1 + + 1 + = T8 + T4 + T8 + T10 + ht T7 k ∆y ∆y ∆y ∆y ∆y Recognizing that ∆x/∆y = 3, q′ t = 25 W/m and k = 2 W/m⋅K, the FDE is h T7 = 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 described in Section 4.5.2 for the Gauss-Seidel method, the system of FDEs has the form: k k −1 k k k k T1 = 0.09174T2 k −1 + 0.8257T4 k −1 T2 = 0.04587T1 + 0.04587T3 k −1 T3 = 0.09174T2 + 0.8257T6 + 2.4771 k −1 + 0.8257T5 + 2.4771 + 2.4771 k k k −1 k k k k −1 k k k k −1 k k k −1 k k k k −1 k k k k −1 k k k −1 k k k −1 k k k T4 = 0.4500T1 + 0.1000T5 k −1 + 0.4500T7 T5 = 0.4500T2 + 0.0500T4 + 0.0500T6 k −1 + 0.4500T8 T6 = 0.4500T3 + 0.1000T5 + 0.4500T9 T7 = 0.4500T4 + 0.1000T8 k −1 + 0.4500T10 T8 = 0.4500T5 + 0.0500T7 + 0.0500T9 + 3.7500 k −1 + 0.4500T11 T9 = 0.4500T6 + 0.1000T8 + 0.4500T12 T10 = 0.9000T7 + 0.1000T11 T11 = 0.9000T8 + 0.0500T10 k −1 + 0.0500T12 T12 = 0.9000T9 + 0.1000T11 Continued ….. PROBLEM 4.75 (Cont.) Note the use of the superscript k to denote the level of iteration. Begin the iteration procedure with rational estimates for Ti (k = 0) and prescribe the convergence criterion as ε ≤ 0.1. k/Ti 1 2 3 4 5 6 7 8 9 10 11 12 0 1 2 55.0 57.4 57.1 50.0 51.7 51.6 45.0 46.0 46.9 61.0 60.4 59.7 54.0 53.8 53.2 47.0 48.1 48.7 65.0 63.5 64.3 56.0 54.6 54.3 49.0 49.6 49.9 60.0 62.7 63.4 55.0 54.8 54.5 50.0 50.1 50.4 ∞ 55.80 49.93 47.67 59.03 51.72 49.19 63.89 52.98 50.14 62.84 53.35 50.46 The last row with k = ∞ corresponds to the solution obtained by matrix inversion. It appears that at least 20 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 as 1 1 q′ conv = h ∆x ( T − T∞ ) + ∆x ( T2 − T∞ ) + ∆x ( T3 − T∞ ) 1 2 2 2 q′ conv = 100 W m ⋅ K × 0.006 m 1 (55.80 − 30 ) + ( 49.93 − 30 ) + 1 ( 47.67 − 30 ) = 25.00 W m . 2 2 < As expected, the heat loss by convection is equal to the heater element dissipation. This follows from the conservation of energy requirement. (d) For this situation, choosing ∆x = 3∆y 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, but with improved precision. (e) Examining the effect of grid size by using the Finite-Difference Equations option from the Tools portion of the IHT Menu, the following temperature field was computed for ∆x = ∆y = 2 mm, where x and y are in mm and the temperatures are in °C. y\x 0 2 4 6 0 55.04 58.71 66.56 63.14 2 53.88 56.61 59.70 59.71 4 52.03 54.17 55.90 56.33 6 50.32 52.14 53.39 53.80 8 49.02 50.67 51.73 52.09 10 48.24 49.80 50.77 51.11 12 47.97 49.51 50.46 50.78 Continued ….. PROBLEM 4.75 (Cont.) Agreement with the results of part (a) is excellent, except in proximity to the heating element, where T15 = 66.6°C for the fine grid exceeds T7 = 63.9°C for the coarse grid by 2.7°C. For h = 10 W/m2⋅K, the maximum temperature in the ceramic corresponds to T15 = 254°C, and the heater could still be operated at the prescribed power. With h = 10 W/m2⋅K, the critical temperature of T15 = 400°C 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 as follows. Assume 25 W/m is transferred by convection uniformly over the surface; find Tsurf ≈ 50°C. Set T2 = 50°C and recognize that T1 and T3 will be higher and lower, respectively. Assume 25 W/m is conducted uniformly to the outer nodes; find T5 - T2 ≈ 4°C. For the remaining nodes, use intuition to guess reasonable values. (2) In selecting grid size (and whether ∆x = ∆y), one should consider the region of largest temperature gradients. NOTE TO INSTRUCTOR: Although the problem statement calls for calculations with ∆x = ∆y = 1 mm, the instructional value and benefit-to-effort ratio are small. Hence, consideration of this grid size is not recommended. PROBLEM 4.76 KNOWN: Silicon chip mounted in a dielectric substrate. One surface of system is convectively cooled while the remaining surfaces are well insulated. FIND: Whether maximum temperature in chip will exceed 85°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Negligible contact resistance between chip and substrate, (4) Upper surface experiences uniform convection coefficient, (5) Other surfaces are perfectly insulated. ANALYSIS: Performing an energy balance on the chip assuming it is perfectly insulated from the substrate, 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 ) h 7 + T∞ = 10 W/m 3 ( 0.003 m ) 2 2 × 50 W/m ⋅ K 7 + 10 W/m 3 ( 0.003 m) 2 500 W/m ⋅ K o o + 20 C = 80.9 C. Since Tmax < 85°C for the assumed situation, for the actual two-dimensional situation with the conducting dielectric substrate, the maximum temperature should be less than 80°C. Using the suggested grid spacing of 3 mm, construct the nodal network and write the finite-difference equation for each of the nodes taking advantage of symmetry of the system. Note that we have chosen to n ot locate nodes on the system surfaces for two reasons: (1) fewer total number of nodes, 20 vs. 25, and (2) Node 5 corresponds to center of chip which is likely the point of maximum temperature. Using these numerical values, h∆x 500 W/m2 ⋅K × 0.003 m = = 0.30 ks 5 W/m ⋅ K h∆x 500 W/m2 ⋅K × 0.003 m = = 0.030 kc 5 W/m ⋅ K & q∆x∆y = 1.800 kc α= 2 2 = = 1.818 ( k s / k c ) +1 5/50 +1 2 2 = = 0.182 ( k c / k s ) +1 50/5 +1 1 γ= = 0.0910 kc / ks + 1 β= find the nodal equations: Node 1 T −T T −T k s∆x 6 1 + k s∆y 2 1 + h∆x ( T∞ − T ) = 0 1 ∆y ∆x Continued ….. PROBLEM 4.76 (Cont.) − 2 + h ∆x T1 + T2 + T6 = − k T∞ ks s h ∆x − 2.30T + T2 + T6 = − 6.00 1 (1) T1 − 3.3T2 + T3 + T7 = −6.00 Node 2 (2) Node 3 T − T3 k s ∆y 2 + ∆x T4 − T3 T − T3 + k s∆ x 8 + h∆ x ( T∞ − T ) = 0 3 ( ∆x/2 ) / kc ∆y + ( ∆x/2 ) / ks ∆y ∆y T2 − ( 2 + α + ( h∆ x/k s ) T3 ) + α T4 + T8 = − ( h ∆x/k ) T∞ T2 − 4.12T3 + 1.82T4 + T8 = −6.00 (3) Node 4 T3 − T4 T − T4 T9 − T4 + k c ∆y 5 + ( ∆x/2 ) / ks ∆y + ( ∆x/2 ) / kc ∆y ∆x ( ∆y/2 ) / ks ∆x + ( ∆y/2 ) k c∆x & + q ( ∆ x∆y ) + h ∆ x ( T∞ − T4 ) = 0 & β T3 − (1 + 2 β + [ h∆ x/k c ]) T4 + T5 + β T9 = − ( h∆ x/kc ) T∞ − q∆ x∆ y/kc 0.182T3 − 1.39T4 + T5 + 0.182T9 = −2.40 (4) Node 5 T − T5 T10 − T5 & k c ∆y 4 + + h ( ∆x/2 )( T∞ − T ) + q ∆y ( ∆x/2 ) = 0 5 ∆x ( ∆y/2) / ks ( ∆x/2 ) + ( ∆y/2 ) / k c ( ∆x/2 ) 2T4 − 2.21T5 + 0.182T10 = −2.40 (5) Nodes 6 and 11 k s∆x ( T1 − T 6 ) / ∆ y + ks∆y ( T7 − T 6 ) / ∆x + k s∆x ( T11 − T6 ) / ∆y = 0 T1 − 3T6 + T7 + T11 = 0 T6 − 3T11 + T + T = 0 12 16 (6,11) Nodes 7, 8, 12, 13, 14 Treat as interior points, T2 + T6 − 4T7 + T8 + T12 = 0 T3 + T7 − 4T8 + T9 + T13 = 0 T7 + T11 − 4T12 + T + T = 0 13 17 (7,8) (12,13) (14) T8 + T − 4T + T + T = 0 12 13 14 18 T9 + T13 − 4T14 + T + T = 0 15 19 Node 9 T − T9 T4 − T9 T − T9 T − T9 k s ∆y 8 + + k s∆y 10 + k s∆x 14 =0 ∆x ( ∆y/2 ) / k c ∆x + ( ∆y/2 ) / k s∆x ∆x ∆y 1.82T4 + T8 − 4.82T9 + T10 + T = 0 14 (9) Node 10 Using the result of Node 9 and considering symmetry, 1.82T5 + 2T9 − 4.82T10 + T = 0 15 Node 15 Interior point considering symmetry T10 + 2T14 − 4T15 + T20 = 0 Node 16 By inspection, (10) (15) T11 − 2T16 + T = 0 17 (16) Continued ….. PROBLEM 4.76 (Cont.) Nodes 17, 18, 19, 20 T12 + T16 − 3T17 + T18 = 0 T14 + T18 − 3T19 + T20 = 0 T13 + T − 3T18 + T = 0 17 19 T15 + 2T19 − 3T20 = 0 (17,18) (19,20) Using the matrix inversion method, the above system of finite-difference equations is written in matrix notation, Eq. 4.52, [A][T] = [C] where -2.3 1 0 00 1 -3.3 1 00 0 1 -4.12 1.82 0 0 0 .182 -1.39 1 000 2 -2.21 100 0 0 010 0 0 001 0 0 0 0 0 1.82 0 000 0 1.82 [A] = 0 0 0 0 0 000 0 0 000 0 0 000 0 0 000 0 0 000 0 0 000 0 0 000 0 0 000 0 0 000 0 0 1 0 0 0 0 -3 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 -4 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0000000000 0 0 0 0000000000 1 0 0 0000000000 0 .182 0 0 0 0 0 0 0 0 0 0 0 0 0 .182 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000000 1 0 0 0100000000 -4 1 0 0010000000 1 -4.82 1 0 0 0 1 0 0 0 0 0 0 0 2 -4.82 0 0 0 0 1 0 0 0 0 0 00 0 -3 1 0 0 0 1 0 0 0 0 00 0 1 -4 1 0 0 0 1 0 0 0 10 0 0 1 -4 1 0 0 0 1 0 0 01 0 0 0 1 -4 1 0 0 0 1 0 00 1 0 0 0 2 -4 0 0 0 0 1 00 0 1 0 0 0 0 -2 1 0 0 0 00 0 0 1 0 0 0 1 -3 1 0 0 00 0 0 0 1 0 0 0 1 -3 1 0 00 0 0 0 0 1 0 0 0 1 -3 1 00 0 0 0 0 0 1 0 0 0 2 -3 [C] = -6 -6 -6 -2.4 -2.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 and the temperature distribution (°C), in geometrical representation, is 34.46 37.13 38.56 39.16 36.13 38.37 39.38 39.77 40.41 40.85 40.81 40.76 45.88 43.80 42.72 41.70 46.23 44.51 42.78 42.06 The maximum temperature is T5 = 46.23°C which is indeed less than 85°C. < COMMENTS: (1) The convection process for the energy balances of Nodes 1 through 5 were simplified by assuming the node temperature is also that of the surface. Considering Node 2, the energy balance processes for qa, qb and qc are identical (see Eq. (2)); however, q conv = T∞ − T2 ≈ h ( T∞ − T2 ) 1/h + ∆y/2k 2 -4 where h∆y/2k = 5 W/m ⋅K×0.003 m/2×50 W/m⋅K = 1.5×10 simplification is justified. << 1. Hence, for this situation, the PROBLEM 4.77 KNOWN: Electronic device cooled by conduction to a heat sink. FIND: (a) Beginning with a symmetrical element, find the thermal resistance per unit depth between the device and lower surface of the sink, R ′ ,d s (m⋅K/W) using the flux plot method; compare result with t thermal 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, determine R ′ − s ; (c) Using nodal networks with finer grid spacings, determine the effect of grid size on the t,d precision of the thermal resistance calculation; (d) Using a fine nodal network, determine the effect of device width on R ′ ,d − s with wd/ws = 0.175, 0.275, 0.375 and 0.475 keeping ws and L fixed. t SCHEMATIC: ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, and (3) No internal generation, (4) Top surface not covered by device is insulated. ANALYSIS: (a) Begin the flux plot for the symmetrical element noting that the temperature drop along the lefthand symmetry line will be almost linear. Choosing to sketch five isotherms and drawing the adiabats, find N=5 M = 2.75 so that the shape factor for the device to the sink considering two symmetrical elements per unit depth is S′ = 2S′ = 2 o M N = 1.10 and the thermal resistance per unit depth is R ′ −s,fp = 1 kS′ = 1 300 W m ⋅ K × 1.10 = 3.03 × 10−3 m ⋅ K W t,d < The thermal resistances for the two rectangular domains are represented schematically below. Continued... PROBLEM 4.77 (Cont.) R ′ −s,i = t,d R ′ −s,ii = t,d L kw d = L kw s 0.024 m 300 W m ⋅ K × 0.018 m = 0.024 m 300 W m ⋅ K × 0.048 m = 4.44 × 10−3 m ⋅ K W < = 1.67 × 10−3 m ⋅ K W < We expect the flux plot result to be bounded by the results for the rectangular domains. The spreading effect can be seen by comparing R ′ ,d − s,fp with R ′ ,d − s,i . f t (b) The coarse 5x5 nodal network is shown in the sketch including the nodes adjacent to the symmetry lines and the adiabatic surface. As such, all the finite-difference equations are interior nodes and can be written by inspection directly onto the IHT workspace. Alternatively, one could use the IHT Finite-Difference Equations Tool. The temperature distribution (°C) is tabulated in the same arrangement as the nodal network. 85.00 85.00 65.76 63.85 50.32 49.17 37.18 36.70 25.00 25.00 62.31 55.49 45.80 35.47 25.00 53.26 50.00 43.06 34.37 25.00 50.73 48.20 42.07 33.95 25.00 The thermal resistance between the device and sink per unit depth is T − Ts R′ −d = d t,s 2q′ tot Performing an energy balance on the device nodes, find q′ = q′ + q′ + q′ tot a b c T − T1 T − T5 T − T4 q ′ = k ( ∆y 2 ) d + k∆x d + k ( ∆x 2 ) d tot ∆x ∆y ∆y q′ = 300 W m ⋅ K [(85 − 62.31) 2 + (85 − 63.85 ) + (85 − 65.76 ) 2] K = 1.263 × 104 W m tot R ′ −d = t,s (85 − 25 ) K 4 = 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 by systematically reducing the nodal spacing ∆x and ∆y. This is a considerable amount of work even with IHT since the equations need to be individually entered. A more generalized, powerful code would be Continued... 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 found R ′ − d = 3.64 × 10−3 m ⋅ K W t,s (d) Using the same tool, with the finest mesh, the thermal resistance was found as a function of wd/ws with fixed ws and L. As expected, as wd increases, R ′ ,d − s decreases, and eventually will approach the value for the t rectangular domain (ii). The spreading effect is shown for the base case, wd/ws = 0.375, where the thermal 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 of precision requirements and level of effort, R ′ − s × 103 (m⋅K/W) t,d Rectangular domain (i) Flux plot Rectangular domain (ii) FDE, 5x5 network FEA, fine mesh 4.44 3.03 1.67 2.38 3.64 PROBLEM 4.78 KNOWN: Nodal network and boundary conditions for a water-cooled cold plate. FIND: (a) Steady-state temperature distribution for prescribed conditions, (b) Means by which operation may be extended to larger heat fluxes. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties. ANALYSIS: Finite-difference equations must be obtained for each of the 28 nodes. Applying the energy balance method to regions 1 and 5, which are similar, it follows that Node 1: Node 5: ( ∆y ( ∆y ∆x ) T2 + ( ∆x ∆y ) T6 − ( ∆y ∆x ) + ( ∆x ∆y ) T1 = 0 ∆x ) T4 + ( ∆x ∆y ) T10 − ( ∆y ∆x ) + ( ∆x ∆y ) T5 = 0 Nodal regions 2, 3 and 4 are similar, and the energy balance method yields a finite-difference equation of the form Nodes 2,3,4: ( ∆y ( ) ∆x ) Tm −1,n + Tm +1,n + 2 ( ∆x ∆y ) Tm,n −1 − 2 ( ∆y ∆x ) + ( ∆x ∆y ) Tm,n = 0 Energy balances applied to the remaining combinations of similar nodes yield the following finitedifference equations. Continued... PROBLEM 4.78 (Cont.) ( ∆x ( ∆x Nodes 7, 15: ∆y ) T1 + ( ∆y ∆x ) T7 − [( ∆x ∆y ) + ( ∆y ∆x ) + ( h∆x k )] T6 = − ( h∆x k ) T∞ ( ∆y Nodes 6, 14: ∆x )( T6 + T8 ) + 2 ( ∆x ∆y ) T2 − 2 [( ∆y ∆x ) + ( ∆x ∆y ) + ( h∆x k )] T7 = − ( 2h∆x k ) T∞ ( ∆y ( ∆y Nodes 8, 16: ∆y ) T19 + ( ∆y ∆x ) T15 − [( ∆x ∆y ) + ( ∆y ∆x ) + ( h∆x k )] T14 = − ( h∆x k ) T∞ ∆x )( T14 + T16 ) + 2 ( ∆x ∆y ) T20 − 2 [( ∆y ∆x ) + ( ∆x ∆y ) + ( h∆x k )] T15 = − ( 2h∆x k ) T∞ ∆x ) 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 ) T∞ ∆x ) 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 ) T∞ Node 11: ( ∆x ∆y ) T8 + ( ∆x ∆y ) T16 + 2 ( ∆y ∆x ) T12 − 2 [( ∆x ∆y ) + ( ∆y ∆x ) + ( h∆y k )] T11 = − ( 2h∆y k ) T∞ Nodes 9, 12, 17, 20, 21, 22: ( ∆y ∆x ) 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 = 0 Nodes 10, 13, 18, 23: ( ∆x Node 19: Nodes 24, 28: ( ∆x ( ∆x ( ∆x ∆y ) Tn +1,m + ( ∆x ∆y ) Tn −1,m + 2 ( ∆y ∆x ) Tm −1,n − 2 [( ∆x ∆y ) + ( ∆y ∆x )] Tm,n = 0 ∆y ) T14 + ( ∆x ∆y ) T24 + 2 ( ∆y ∆x ) T20 − 2 [( ∆x ∆y ) + ( ∆y ∆x )] T19 = 0 ∆y ) T19 + ( ∆y ∆x ) T25 − [( ∆x ∆y ) + ( ∆y ∆x )] T24 = − ( q′′ ∆x k ) o ∆y ) T23 + ( ∆y ∆x ) T27 − [( ∆x ∆y ) + ( ∆y ∆x )] T28 = − ( q′′ ∆x k ) o Nodes 25, 26, 27: ( ∆y ∆x ) Tm −1,n + ( ∆y ∆x ) Tm +1,n + 2 ( ∆x ∆y ) Tm,n +1 − 2 [( ∆x ∆y ) + ( ∆y ∆x )] Tm,n = − ( 2q ′′ ∆x k ) o Evaluating the coefficients and solving the equations simultaneously, the steady-state temperature distribution (°C), tabulated according to the node locations, is: 23.77 23.41 23.91 23.62 28.90 30.72 32.77 28.76 30.67 32.74 24.27 24.31 25.70 28.26 30.57 32.69 24.61 24.89 26.18 28.32 30.53 32.66 24.74 25.07 26.33 28.35 30.52 32.65 Alternatively, the foregoing results may readily be obtained by accessing the IHT Tools pat and using the 2-D, SS, Finite-Difference Equations options (model equations are appended). Maximum and minimum cold plate temperatures are at the bottom (T24) and top center (T1) locations respectively. (b) For the prescribed conditions, the maximum allowable temperature (T24 = 40°C) is reached when q′′ o = 1.407 × 105 W/m2 (14.07 W/cm2). Options for extending this limit could include use of a copper cold plate (k ≈ 400 W/m⋅K) and/or increasing the convection coefficient associated with the coolant. With k = 400 W/m⋅K, a value of q′′ = 17.37 W/cm2 may be maintained. With k = 400 W/m⋅K and h = 10,000 o W/m2⋅K (a practical upper limit), q′′ = 28.65 W/cm2. Additional, albeit small, improvements may be o realized 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 the overall energy balance q′′ ( 4∆x ) = h [( ∆x 2 ) ( T6 − T∞ ) + ∆x ( T7 − T∞ ) + ( ∆x + ∆y )( T8 − T∞ ) 2 o +∆y ( T11 − T∞ ) + ( ∆x + ∆y )( T16 − T∞ ) 2 + ∆x (T15 − T∞ ) + ( ∆x 2 )( T14 − T∞ )] . PROBLEM 4.79 KNOWN: Heat sink for cooling computer chips fabricated from copper with microchannels passing fluid with prescribed temperature and convection coefficient. FIND: (a) Using a square nodal network with 100 µm spatial increment, determine the temperature distribution and the heat rate to the coolant per unit channel length for maximum allowable chip temperature Tc,max = 75°C; estimate the thermal resistance betweeen the chip surface and the fluid, R ′ − f (m⋅K/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 µm; and (c) Consistent with the requirement that a + b = 400 µm, explore altering the sink dimensions to decrease the thermal resistance. 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 channel dimensions and shape. ANALYSIS: (a) The square nodal network with ∆x = ∆y = 100 µm is shown below. Considering symmetry, the nodes 1, 2, 3, 4, 7, and 9 can be treated as interior nodes and their finite-difference equations representing nodal energy balances can be written by inspection. Using the, IHT FiniteDifference Equations Tool, appropriate FDEs for the nodes experiencing surface convection can be obtained. The IHT code including results is included in the Comments. Having the temperature distribution, the heat rate to the coolant per unit channel length for two symmetrical elements can be obtained by applying Newton’s 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∞ )] cv q ′ = 2 × 30, 000 W m ⋅ K × 100 × 10 cv 2 −6 m [( 74.02 − 25 ) + ( 74.09 − 25 ) 2 + ( 73.60 − 25 ) + ( 73.37 − 25 ) 2 ] K < q′ = 878 W m cv The thermal resistance between the chip and fluid per unit length for each microchannel is T − T∞ ( 75 − 25 ) C R′ −f = c = = 5.69 × 10−2 m ⋅ K W t,c q′ 878 W m cv < The maximum allowable heat dissipation for a 10 mm × 10 mm chip is Pchip,max = q ′′ × A chip = 2.20 × 106 W m 2 × (0.01 × 0.01) m 2 = 220 W c where Achip = 10 mm × 10 mm and the heat flux on the chip surface (wf + ws) is q′′ = q′ c cv (wf + w s ) = 878 W m ( 200 + 200 ) × 10−6 m = 2.20 × 106 W < m2 Continued... PROBLEM 4.79 (Cont.) (b) To investigate the effect of grid spacing, the analysis was repreated with a spatial increment of 50 µm (32 nodes as shown above) with the following results R ′ − f = 5.67 × 10 −2 m ⋅ K W t,c q′ = 881W m cv < Using a finite-element package with a mesh around 25 µm, we found R ′ − f = 5.70 × 10−2 m ⋅ K W t,c which suggests the grid spacing effect is not very significant. (c) Requring that the overall dimensions of the symmetrical element remain unchanged, we explored what 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,c since for the convection process ( ) 2 −6 −2 R′ t,cv = 1 A′ = 1 30, 000 W m ⋅ K × 600 × 10 m = 5.55 × 10 m ⋅ K W s where A′ = (wf + 2b) = 600 µm. s Using a finite-element package, the thermal resistances per unit length for three additional channel crosssections were determined and results summarized below. Microchannel (µm) Case A B C D Height 200 133 300 250 Half-width 100 150 100 150 R ′ − s × 10 t,c 2 (m⋅K/W) 5.70 6.12 4.29 4.25 Continued... 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 symmetry T1 = 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 = 0 deltax = deltay deltay = 100e-6 Tinf = 25 h = 30000 // Sink and chip parameters k = 400 Tc = 75 wf = 200e-6 ws = 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^2 Pchip = Achip * q''c // Power, W Achip = 0.01 * 0.01 // Chip area, m^2 /* Data Browser results: chip power, thermal resistance, heat rates and temperature distribution Pchip R'tcf q''c q'cv 219.5 0.05694 2.195E6 878.1 T1 74.53 T2 74.52 T3 74.53 T4 74.07 T5 74.02 T6 74.09 T7 73.7 T8 73.6 T9 73.53 T10 73.37 */ PROBLEM 4.80 KNOWN: Longitudinal rib (k = 10 W/m⋅K) with rectangular cross-section with length L= 8 mm and width 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 assuming one-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 the range 1.5 ≤ L/w ≤ 10, keeping L constant, as predicted by the two-dimensional, finite-difference method and the one-dimensional fin analysis. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, and (3) Convection coefficient uniform over rib surfaces, including tip. ANALYSIS: (a) The rib is represented by a 5 × 3 nodal grid as shown above where the symmetry plane is an adiabatic surface. The IHT Tool, Finite-Difference Equations, for Two-Dimensional, Steady-State conditions is used to formulate the nodal equations (see Comment 2 below) which yields the following nodal temperatures (° C) 45 45 45 39.3 40.0 39.3 35.7 36.4 35.7 33.5 34.0 33.5 32.2 32.6 32.2 Note that the fin tip temperature is Ttip = T12 = 32.6 C < The fin heat rate per unit width normal to the page, q ′ , can be determined from energy balances on the fin three base nodes as shown in the schematic below. q′ = q′ + q′ + q′ + q′ + q′ fin a b c d e ′ = h ( ∆x 2 )( Tb − T∞ ) qa q′ = k ( ∆y 2 )( Tb − T1 ) ∆x b q′ = k ( ∆y )( Tb − T5 ) ∆x c q′ = k ( ∆y 2 )( Tb − T9 ) ∆x d ′ = h ( ∆x 2 )( Tb − T∞ ) q3 Continued... PROBLEM 4.80 (Cont.) Substituting numerical values, find q′ = (12.0 + 28.4 + 50.0 + 28.4 + 12.0 ) W m = 130.8 W m fin Using 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 m Ttip = 32.2 C f (b) With ∆x = L/8 = 1 mm and ∆x = 1 mm, for a total of 9 × 3 = 27 nodal points and regions, the grid appears 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), the fin heat rate and tip temperature were determined. 1-D analysis Ttip (°C) 32.2 q′ (W/m) fin 2-D analysis (nodes) ( 5 × 3) (9 × 3) 32.6 32.6 131 131 129 < < (c) To determine when the one-dimensional approximation is reasonable, consider a rib of constant length, L = 8 mm, and vary the thickness w for the range 1.5 ≤ L/w ≤ 10. Using the above IHT model for ′ the 27 node grid, the fin heat rates for 1-D, q1d , and 2-D, q′ d , analysis were determined as a function of 2 w with the error in the approximation evaluated as ′ ′ Error ( % ) = ( q′ − q1d ) × 100 q1d 2d 4 Error (%) 2 0 -2 -4 0 2 4 6 8 10 Length / width, L/w Note that for small L/w, a thick rib, the 1-D approximation is poor. For large L/w, a thin rib which approximates 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 to accurately represent the slender rib. PROBLEM 4.81 KNOWN: 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 of the 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 the web-flange interface temperature is uniform, Tw = 100°C. The nodal arrangement to represent this system is shown below. The nodes on the line of symmetry have been shown for convenience in deriving the 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. 100.00 166.6 211.7 241.4 100.00 177.1 219.5 247.2 215.8 222.4 241.9 262.9 262.9 255.0 262.7 279.3 284.8 272.0 274.4 292.9 The heat rate to the beam can be determined from energy balances about the web-flange interface nodes as shown in the sketch below. Continued... PROBLEM 4.81 (Cont.) q′ = q′ + q′ + q′ w a b c T −T T − Tw T − Tw + k ( ∆x 2 ) 4 q′ = k ( ∆y 2 ) 1 w + k ( ∆x ) 5 w ∆x ∆y ∆y q′ = 10 W m ⋅ K [( 215.8 − 100 ) 2 + (177.1 100 ) + (166.6 − 100 ) 2] K = 1683 W m w < (b) The schematic below poses the question concerning the reasonableness of the uniform temperature assumption at the web-flange interface. From the analysis above, note that T1 = 215.8°C vs. Tw = 100°C indicating that this assumption is a poor one. This L-shaped section has strong two-dimensional behavior. To illustrate the effect, we performed an analysis with Tw = 100°C located nearly 2 × times further up the web than it is wide. For this situation, the temperature difference at the web-flange interface across the width of the web was nearly 40°C. The steel beam with its low thermal conductivity has substantial internal thermal resistance and given the L-shape, the uniform temperature assumption (Tw) across the web-flange interface is inappropriate. PROBLEM 4.82 KNOWN: Plane composite wall with exposed surfaces maintained at fixed temperatures. Material A has temperature-dependent thermal conductivity. FIND: Heat flux through the wall (a) assuming a uniform thermal conductivity in material A evaluated at the average temperature of the section, and considering the temperature-dependent thermal conductivity of material A using (b) a finite-difference method of solution in IHT with a space increment of 1 mm and (c) the finite-element method of FEHT. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) No thermal contact resistance between the materials, and (3) No internal generation. ANALYSIS: (a) From the thermal circuit in the above schematic, the heat flux is q′′ = x T1 − T2 T −T = AB 2 R ′′ + R ′′ R ′′ A B B (1, 2) and the thermal resistances of the two sections are R ′′ = LA / k A A R ′′ = LB / k B B (3, 4) The thermal conductivity of material A is evaluated at the average temperature of the section { } k A = k o 1 + α ( T1 + TAB ) / 2 − To (5) Substituting numerical values and solving the system of equations simultaneously in IHT, find TAB = 563.2 K q′′ = 52.64 kW / m 2 x < (b) The nodal arrangement for the finite-difference method of solution is shown in the schematic below. 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 fluxes at the control surfaces using Fourier’s law with the thermal conductivity based upon the average temperature of adjacent nodes. The heat fluxes into node m are Continued ….. PROBLEM 4.82 (Cont.) T −T q′′ = k a ( m, m + 1) m +1 m c ∆x (1) T −T q′′ = k a ( m − 1, m ) m −1 m d ∆x (2) and the FDEs are obtained from the energy balance written as q′′ + q′′ = 0 c d (3) T T −T −T k a ( m, m + 1) m +1 m + k a ( m − 1, m ) m −1 m = 0 ∆x ∆x (4) where the thermal conductivities averaged over the path between the nodes are expressed as { } k a ( m, m + 1) = k o {1 + α ( Tm + Tm +1 ) / 2 − To } k a ( m − 1, m ) = k o 1 + α ( Tm −1 + Tm ) / 2 − To (5) (6) A-B Interface Node 11 Referring to the above schematic, the energy balance on the interface node, q ′′ + q ′′ = 0, has the form c d T −T T −T k b 12 11 + k a (10,11) 10 11 = 0 ∆x ∆x (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 form Tn = ( Tn −1 + Tn +1 ) / 2 (9) And the heat flux in the x-direction is T − Tn +1 q′′ = k b n x ∆x (10) Finite-Difference Method of Solution The foregoing FDE equations for section A nodes (m = 02 to 10), the AB interface node and their respective expressions for the thermal conductivity, k (m, m +1), and for section B nodes are entered into the IHT workspace and solved for the temperature distribution. The heat flux can be evaluated using Eq. (2) or (10). A portion of the IHT code is contained in the Comments, and the results of the analysis are tabulated below. T11 = TAB = 563.2 K q′′ = 52.64 kW / m 2 x Continued ….. < PROBLEM 4.82 (Cont.) (c) The finite-element method of FEHT can be used readily to obtain the heat flux considering the temperature-dependent thermal conductivity of section A. Draw the composite wall outline with properly scaled section thicknesses in the x-direction with an arbitrary y-direction dimension. In the Specify | 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 are TAB = 563 K q′′ = 5.26 kW / m 2 x < COMMENTS: (1) The results from the three methods of analysis compare very well. Because the thermal conductivity in section A is linear, and moderately dependent on temperature, the simplest method of using an overall section average, part (a), is recommended. This same method is recommended 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 nodes within section A and in section B with identical results. This is a consequence of the technique for averaging 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 for representative nodes are shown below. // FDEs – Section A k01_02 * (T01-T02)/deltax + k02_03 * (T03-T02)/deltax = 0 k01_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 = 0 k03_04 = ko * (1 + alpha * ((T03 + T04)/2 – To)) // Interface, node 11 k11 * (T10 –T11)/deltax + kb * (T12 –T11)/deltax =0 k11 = 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 User’s Manual, provide background for developing skills in using FEHT. PROBLEM 4.83 KNOWN: Upper surface of a platen heated by hot fluid through the flow channels is used to heat a process fluid. FIND: (a) The maximum allowable spacing, W, between channel centerlines that will provide a uniform temperature requirement of 5°C on the upper surface of the platen, and (b) Heat rate per unit length 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 channel arrangement, the temperature uniformity requirement will be met when T1 – T2 = 5°C. The maximum temperature, T1, will occur directly over the flow channel centerline, while the minimum surface temperature, 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 the channel centerline spacing, W. The following method of solution was used: (1) Make an initial guess value for W; try W = 100 mm, (2) Draw an outline of the symmetrical section, and assign properties and boundary conditions, (3) Make a copy of this file so that in your second trial, you can use the Draw | Move Node option to modify the section width, W/2, larger or smaller, (4) Draw element lines within the outline to create triangular elements, (5) Use the Draw | Reduce Mesh command to generate 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 ≈ 5°C, use the View | Heat Flows command to find the heat rate, otherwise, change the width of the section outline and repeat the analysis. The results of our three trials are tabulated below. Trial 1 2 3 W (mm) 100 60 80 T1 (°C) T2 (°C) T1 – T2 (°C) 108 119 113 98 118 108 10 1 5 q’ (W/m) --1706 COMMENTS: (1) In addition to the tutorial example in the FEHT User’s Manual, the solved models for Examples 4.3 and 4.4 of the Text are useful for developing skills in using this problem-solving tool. (2) An alternative numerical method of solution would be to create a nodal network, generate the finite-difference equations and solve for the temperature distribution and the heat rate. The FDEs should allow for a non-square grid, ∆x ≠ ∆y, so that different values for W/2 can be accommodated by changing the value of ∆x. Even using the IHT tool for building FDEs (Tools | Finite-Difference Equations | Steady-State) this method of solution is very labor intensive because of the large number of nodes required for obtaining good estimates. PROBLEM 4.84 KNOWN: Silicon chip mounted in a dielectric substrate. One surface of system is convectively cooled, while the remaining surfaces are well insulated. See Problem 4.77. Use the finite-element software FEHT as your analysis tool. FIND: (a) The temperature distribution in the substrate-chip system; does the maximum temperature exceed 85°C?; (b) Volumetric heating rate that will result in a maximum temperature of 85°C; and (c) Effect of reducing thickness of substrate from 12 to 6 mm, keeping all other dimensions unchanged 7 3 with q = 1×10 W/m ; maximum temperature in the system for these conditions, and fraction of the power 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 connected regions, chip and substrate. Draw first the chip outline; Specify the material and generation parameters. 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 heat flux for the symmetry and insulated surfaces, and convection for the upper surface. Draw Element Lines, making the triangular elements near the chip and surface smaller than near the lower insulated boundary as shown in a copy of the FEHT screen on the next page. Use the Draw|Reduce Mesh command and Run the model. (a) Use the View|Temperature command to see the nodal temperatures through out the system. As expected, the hottest location is on the centerline of the chip at the bottom surface. At this location, the temperature is < T(0, 9 mm) = 46.7°C (b) Run the model again, with different values of the generation rate until the temperature at this location is T(0, 9 mm) = 85°C, finding q = 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 lower boundary, as well as the intermediate ones, to represent a substrate that is of 6-mm, rather than 12-mm thickness. Find the maximum temperature as T ( 0,3 mm ) = 47.5°C < Using the View|Heat Flow command, click on the adjacent line segments forming the chip surface exposed to the convection process. The heat rate per unit width (normal to the page) is q′ chip,cv = 60.26 W / m The total heat generated within the chip is q′tot = q ( L / 6 × H / 4 ) = 1× 107 W / m3 × ( 0.0045 × 0.003) m 2 = 135 W / m so that the fraction of the power dissipated by the chip that is convected directly to the coolant stream is F = q′ chip,cv / q′ = 60.26 /135 = 45% tot < COMMENTS: (1) Comparing the maximum temperatures for the system with the 12-mm and 6-mm thickness substrates, note that the effect of halving the substrate thickness is to raise the maximum temperature by less than 1°C. The thicker substrate does not provide significantly improved heat removal capability. (2) Without running the code for part (b), estimate the magnitude of q that would make T(0, 9 mm) = 7 3 85°C. Did you get q = 2.43×10 W/m ? Why? PROBLEM 5.1 KNOWN: Electrical heater attached to backside of plate while front surface is exposed to convection process (T∞,h); initially plate is at a uniform temperature of the ambient air and suddenly heater power is switched on providing a constant q′′ . o FIND: (a) Sketch temperature distribution, T(x,t), (b) Sketch the heat flux at the outer surface, q′′ ( L,t ) as a function of time. x SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible heat loss from heater through insulation. ANALYSIS: (a) The temperature distributions for four time conditions including the initial distribution, 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 be x linear such that T − T ( L,∞ ) q′′ = k o = h T ( L,∞ ) − T∞ . o L (b) The heat flux at the front surface, x = L, is given by q′′ ( L,t ) = −k ( dT/dx ) x=L . From the x temperature 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 their initial values. Hence, we show a zero slope for q′′ ( L,t ) at early times. Eventually, the value of x q′′ ( L,t ) will reach the steady-state value which is q′′ . x o PROBLEM 5.2 KNOWN: Plane wall whose inner surface is insulated and outer surface is exposed to an airstream 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. o FIND: (a) Sketch temperature distribution on T-x coordinates for initial, steady-state, and two intermediate times, (b) Sketch heat flux at the outer surface, q′′ ( L,t ) , as a function of x time. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal generation, Eg = 0, (4) Surface at x = 0 is perfectly insulated, (5) All incident radiant power is absorbed and negligible radiation exchange with surroundings. ANALYSIS: (a) The temperature distributions are shown on the T-x coordinates and labeled accordingly. Note these special features: (1) Gradient at x = 0 is always zero, (2) gradient is more steep at early times and (3) for steady-state conditions, the radiant flux is equal to the convective heat flux (this follows from an energy balance on the CS at x = L), q ′′ = q ′′ o conv = h [T ( L,∞ ) − T∞ ]. (b) The heat flux at the outer surface, q′′ ( L,t ) , as a function of time appears as shown above. x COMMENTS: The sketches must reflect the initial and boundary conditions: T(x,0) = T∞ ∂T −k x=0 = 0 ∂x ∂T −k x=L = h T ( L,t ) − T∞ − q′′ o ∂x uniform initial temperature. insulated at x = 0. surface energy balance at x = L. PROBLEM 5.3 KNOWN: 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 for microwave, (3) Uniform surface heating for radiant oven, (4) Heat loss from surface of meat to surroundings is negligible during the heating process, (5) Symmetry about midplane. ANALYSIS: COMMENTS: (1) With uniform generation and negligible surface heat loss, the temperature distribution remains nearly uniform during microwave heating. During the subsequent surface cooling, 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 5.4 KNOWN: 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) Sketch on 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 internal generation, (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 following exposure 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 temperature distributions 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 is Ein − Eout = ∆E = Efinal − Einitial t 2 ∫ o hAs T∞ − T ( L,t ) dt − 0 = Mc p ( T − Ti ) f 0 where Ein is due to convection heating over the period of time t = 0 → to. With knowledge of T(L,t), this expression can be integrated and a value for Tf determined. PROBLEM 5.5 KNOWN: 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), 2 hLc h ( ro / 3) 20 W/m ⋅ K (0.002m ) Bi = = = = 0.001. k k 40 W/m ⋅ K Hence, the temperature of the steel remains approximately uniform during the cooling process, and the lumped capacitance method may be used. From Eqs. 5.4 and 5.5, ( ) 3 Ti − T∞ ρ π D / 6 c p Ti − T∞ t= ln = ln hAs T − T∞ T − T∞ hπ D2 ρ Vcp t= 7800kg/m 3 ( 0.012m ) 600J/kg ⋅ K 6 × 20 W/m 2 ⋅ K t = 1122 s = 0.312h ln 1150 − 325 400 − 325 < COMMENTS: Due to the large value of Ti, radiation effects are likely to be significant during the early portion of the transient. The effect is to shorten the cooling time. PROBLEM 5.6 KNOWN: 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 radiation exchange, (3) Constant properties. 3 PROPERTIES: Table A-1, Pure copper (333K): ρ = 8933 kg/m , cp = 389 J/kg⋅K, k = 398 W/m⋅K. ANALYSIS: The time-temperature history is given by Eq. 5.6 with Eq. 5.7. θ (t) t = exp − θi R tC t where Rt = 1 hAs Ct = ρ Vcp θ = T − T∞. A s = π D2 π D3 V= 6 Recognize that when t = 69s, o θ ( t ) ( 55 − 27 ) C t = = 0.718 = exp − θi τt ( 66 − 27 )o C and noting that τ t = R tC t find τ t = 208s. Hence, h= ρ Vcp Asτ t = ( 69s = exp − τt ) 8933 kg/m 3 π 0.0127 3 m3 / 6 389J/kg ⋅ K π 0.0127 2m 2 × 208s h = 35.3 W/m 2 ⋅ K. COMMENTS: Note that with Lc = Do/6, Bi = hLc 0.0127 = 35.3 W/m 2 ⋅ K × m/398 W/m ⋅ K = 1.88 ×10 -4. k 6 Hence, Bi < 0.1 and the spatially isothermal assumption is reasonable. < PROBLEM 5.7 KNOWN: Solid steel sphere (AISI 1010), coated with dielectric layer of prescribed thickness and thermal conductivity. Coated sphere, initially at uniform temperature, is suddenly quenched in an oil bath. FIND: Time required for sphere to reach 140°C. SCHEMATIC: ( ) o PROPERTIES: Table A-1, AISI 1010 Steel T = [500 +140 ] C/2 = 320oC ≈ 600K : ρ = 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 negligible thermal 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 and the convection coefficient. That is, l1 0.002m 1 m2 ⋅ K = ( 0.050 + 0.0003 ) = 0.0503 , W 3300 W/m 2 ⋅ K or in terms of an overall coefficient, U = 1/R′′ = 19.88 W/m2 ⋅ K. The effective Biot number is R ′′ = k + h = 0.04 W/m ⋅ K + 2 ULc U ( ro /3 ) 19.88 W/m ⋅ K × ( 0.300/6 ) m Bi e = = = = 0.0204 k k 48.8 W/m ⋅ K where the characteristic length is Lc = ro/3 for the sphere. Since Bie < 0.1, the lumped capacitance approach is applicable. Hence, Eq. 5.5 is appropriate with h replaced by U, t= ρ c V θ i ρ c V T ( 0 ) − T∞ ln = ln . U As θ o U As T ( t ) − T∞ Substituting numerical values with (V/As ) = ro/3 = D/6, o 7832 kg/m 3 × 559 J/kg ⋅ K 0.300m ( 500 −100 ) C t= ln 6 o 19.88 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 dominates and therefore nearly all the temperature drop occurs across the layer. PROBLEM 5.8 KNOWN: 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 135°C. 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 room temperature (Ti = T∞). ANALYSIS: Biot numbers may be based on convection heat transfer and/or the maximum heat transfer by radiation, which would occur when the plate reaches the desired temperature (T = 135°C). ( ) 2 From Eq. (1.9) the corresponding radiation transfer coefficient is hr = εσ(T +Tsur) T 2 + Tsur = 0.8 × -8 2 4 2 2 2 2 5.67 × 10 W/m ⋅K (408 + 298) K (408 + 298 ) K = 8.2 W/m ⋅K. Hence, Bi = 2 hL 10 W / m ⋅ K ( 0.007m ) = = 3.9 × 10−4 k 180 W / m ⋅ K 2 h r L 8.2 W / m ⋅ K (0.007m ) Bi r = = = 3.2 × 10−4 k 180 W / m ⋅ K With convection and radiation considered independently or collectively, Bi, Bir, Bi + Bir << 1 and the lumped capacitance analysis may be used. The energy balance, Eq. (5.15), associated with Figure 5.5 may be applied to this problem. With Eg = 0, the integral form of the equation is T − Ti = ( As t 4 4 ∫0 q′′h − h (T − T∞ ) − εσ T − Tsur ρ Vc )dt Integrating numerically, we obtain, for T = 135°C, t = 168s < COMMENTS: Note that, if heat transfer is by natural convection, h, like hr, will vary during the process from a value of 0 at t = 0 to a maximum at t = 168s. PROBLEM 5.9 KNOWN: Diameter and radial temperature of AISI 1010 carbon steel shaft. Convection coefficient 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 ) : ρ = 7832 kg / m 3 , k = 51.2 -5 W/m⋅K, c = 541 J/kg⋅K, α = 1.21×10 2 m /s. ANALYSIS: The Biot number is Bi = 2 hro / 2 100 W/m ⋅ K ( 0.05 m/2) = = 0.0488. k 51.2 W/m ⋅ K Hence, the lumped capacitance method can be applied. From Equation 5.6, hAs T − T∞ 4h = exp − t = exp − ρ cD t Ti − T∞ ρ 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 the series solution. From Equation 5.49c, ( To − T∞ −400 2 = = 0.444 = C1 exp −ς1 Fo Ti − T∞ −900 ) For Bi = hro/k = 0.0976, Table 5.1 yields ς1 = 0.436 and C1 = 1.024. Hence − ( 0.436) 2 (1.2×10−5 m2 / s) t = ln (0.434) = −0.835 ( 0.05 m ) 2 t = 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 5.10 KNOWN: 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 storage medium at this time. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible radiation exchange with surroundings. ( ) PROPERTIES: Table A-1, Aluminum, pure T ≈ 600K = 327o C : k = 231 W/m⋅K, c = 1033 3 J/kg⋅K, ρ = 2702 kg/m . ANALYSIS: Recognizing the characteristic length is the half thickness, find hL 100 W/m 2 ⋅ K × 0.025m Bi = k = = 0.011. 231 W/m ⋅ K Hence, the lumped capacitance method may be used. From Eq. 5.8, Q = ( ρ Vc ) θ i 1 − exp ( −t/ τ t ) =−∆ Est (1) −∆E st,max = ( ρ Vc )θ i. (2) Dividing Eq. (1) by (2), ∆Est / ∆Est,max = 1 − exp ( − t/τ th ) = 0.75. Solving for τ th = ρ Vc ρ Lc 2702 kg/m 3 × 0.025m ×1033 J/kg ⋅ K = = = 698s. hAs h 100 W/m2 ⋅ K Hence, the required time is −exp ( − t/698s ) = −0.25 or < t = 968s. From Eq. 5.6, T − T∞ = exp ( − t/ τ th ) Ti − T∞ ( ) T = T∞ + ( Ti − T∞ ) exp ( − t/ τ th ) = 600o C − 575o C exp ( −968/698 ) T = 456oC. < COMMENTS: For the prescribed temperatures, the property temperature dependence is significant and some error is incurred by assuming constant properties. However, selecting properties at 600K was reasonable for this estimate. PROBLEM 5.11 KNOWN: Diameter, density, specific heat and thermal conductivity of aluminum spheres used in packed 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 the corresponding 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 to contact with other spheres, (2) Constant properties. ANALYSIS: To determine whether a lumped capacitance analysis can be used, first compute Bi = 2 h(ro/3)/k = 75 W/m ⋅K (0.025m)/150 W/m⋅K = 0.013 < 0.1. Hence, the lumped capacitance approximation may be made, and a uniform temperature may be assumed to exist in the sphere at any time. From Eq. 5.8a, achievement of 90% of the maximum possible thermal energy storage corresponds to Q = 0.90 = 1 − exp ( − t / τ t ) ρ cVθi where τ t = ρ Vc / hA s = ρ Dc / 6h = 2700 kg / m3 × 0.075m × 950 J / kg ⋅ K / 6 × 75 W / m 2 ⋅ K = 427s. Hence, t = −τ 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 / ρ Dc ) T (984s ) = 300°C − 275°C exp −6 × 75 W / m ⋅ K × 984s / 2700 kg / m × 0.075m × 950 J / kg ⋅ K ( ) T (984 ) s = 272.5°C < 2 3 3 Obtaining the density and specific heat of copper from Table A-1, we see that (ρc)Cu ≈ 8900 kg/m × 6 3 6 3 400 J/kg⋅K = 3.56 × 10 J/m ⋅K > (ρc)Al = 2.57 × 10 J/m ⋅K. Hence, for an equivalent sphere diameter, 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 decreases as it passes through the bed. Hence, the time required for a sphere to reach a prescribed state of thermal energy storage increases with increasing distance from the bed inlet. PROBLEM 5.12 KNOWN: Wafer, initially at 100°C, is suddenly placed on a chuck with uniform and constant temperature, 23°C. Wafer temperature after 15 seconds is observed as 33°C. FIND: (a) Contact resistance, R ′′ , between interface of wafer and chuck through which helium slowly tc flows, and (b) Whether R ′′ will change if air, rather than helium, is the purge gas. tc SCHEMATIC: PROPERTIES: Wafer (silicon, typical values): ρ = 2700 kg/m3, c = 875 J/kg⋅K, k = 177 W/m⋅K. ASSUMPTIONS: (1) Wafer behaves as a space-wise isothermal object, (2) Negligible heat transfer from wafer top surface, (3) Chuck remains at uniform temperature, (4) Thermal resistance across the interface 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 = Est out (1) −q′′ cond = E′′ st (2) T ( t ) − Tc dT −w = ρ wc w R ′′ dt tc (3) Separate and integrate Eq. (3) t T dTw dt =w 0 ρ wcR ′′ Twi Tw − Tc tc −∫ ∫ Tw ( t ) − Tc t = exp − Twi − Tc tc ρ wcR ′′ (4) (5) Substituting numerical values for Tw(15s) = 33°C, (33 − 23) C exp = 2700 kg (100 − 23) C m3 × 0.758 × 10−3 m × 875 J kg ⋅ K × R ′′ tc 15s R ′′ = 0.0041m 2 ⋅ K W tc (b) R ′′ will increase since kair < khelium. See Table A.4. tc COMMENTS: Note that Bi = Rint/Rext = (w/k)/ R ′′ = 0.001. Hence the spacewise isothermal tc assumption is reasonable. (6) < PROBLEM 5.13 KNOWN: Inner diameter and wall thickness of a spherical, stainless steel vessel. Initial temperature, density, specific heat and heat generation rate of reactants in vessel. Convection conditions at outer surface of vessel. FIND: (a) Temperature of reactants after one hour of reaction time, (b) Effect of convection coefficient on thermal response of reactants. SCHEMATIC: ASSUMPTIONS: (1) Temperature of well stirred reactants is uniform at any time and is equal to inner 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/ρVc and b = E g / ρ Vc = q / ρ c. From Eq. (3.19) the product of the overall heat transfer -1 coefficient and the surface area is UA = (Rcond + Rconv) , where from Eqs. (3.36) and (3.9), R t,cond = 1 1 1 1 − = 2π k Di Do 2π (17 W / m ⋅ K ) R t,conv = 1 1 = = 0.0438K / W 2 hAo 6 W / m 2 ⋅ K π (1.1m ) 1 1 −4 − = 8.51× 10 K / W 1.0m 1.1m ) ( Hence, UA = 24.4 W/K. It follows that, with V = π D3 / 6, i a= 6 ( 22.4 W / K ) UA = = 1.620 × 10−5 s −1 3 × π 1m 3 2400 J / kg ⋅ K ρ Vc 1100 kg / m () q 104 W / m3 b= = = 3.788 × 10−3 K / s 3 × 2400 J / kg ⋅ K ρ c 1100 kg / m With (b/a) = 233.8°C and t = 18,000s, ( ) T = 25°C + 233.8°C 1 − exp −1.62 × 10−5 s−1 × 18, 000s = 84.1°C < Neglecting the thermal capacitance of the vessel wall, the heat rate by conduction through the wall is equal to the heat transfer by convection from the outer surface, and from the thermal circuit, we know that Continued ….. PROBLEM 5.13 (Cont.) T − Ts,o Ts,o − T∞ Ts,o = = R t,cond R t,conv = 8.51×10−4 K / W = 0.0194 0.0438 K / W T + 0.0194T∞ 84.1°C + 0.0194 ( 25°C ) = = 83.0°C 1.0194 1.0194 < 2 2 (b) Representative low and high values of h could correspond to 2 W/m ⋅K and 100 W/m ⋅K for free and forced convection, respectively. Calculations based on Eq. (5.25) yield the following temperature histories. R e a cto r te m p e ra tu re (C ) 100 80 60 40 20 0 3600 7200 10800 14400 18000 P ro ce s s tim e (s ) h =2 W /m ^2 .K h =6 W /m ^2 .K h =1 0 0 W /m ^2 .K Forced convection is clearly an effective means of reducing the temperature of the reactants and accelerating the approach to steady-state conditions. COMMENTS: The validity of neglecting thermal energy storage effects for the vessel may be assessed by contrasting its thermal capacitance with that of the reactants. Selecting values of ρ = 3 8000 kg/m and c = 475 J/kg⋅K for stainless steel from Table A-1, the thermal capacitance of the ( ) 3 vessel is Ct,v = (ρVc)st = 6.57 × 10 J/K, where V = (π / 6 ) D3 − Di . With Ct,r = (ρVc)r = 2.64 × o 5 6 10 J/K for the reactants, Ct,r/Ct,v ≈ 4. Hence, the capacitance of the vessel is not negligible and should be considered in a more refined analysis of the problem. PROBLEM 5.14 KNOWN: Volume, density and specific heat of chemical in a stirred reactor. Temperature and convection coefficient associated with saturated steam flowing through submerged coil. Tube diameter and outer convection coefficient of coil. Initial and final temperatures of chemical and time span 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 wall conduction resistance, (7) Negligible kinetic energy, potential energy, and flow work changes for steam. 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 yields dU dT = ρ Vc = UAs (Th − T ) dt dt T dT UAs t Integrating, ∫Ti Th − T = ρ Vc ∫o dt T − T UAs t − ln h = ρ Vc Th − Ti As = − ( ρ Vc Th − T ln Ut Th − Ti − − U = hi 1 + h o 1 ) −1 = (1/10, 000 ) + (1/ 2000 ) (1) −1 W / m2 ⋅ K U = 1670 W / m 2 ⋅ K (1200 kg / m3 )(2.25m3 )(2200 J / kg ⋅ K ) ln 500 − 450 = 1.37m2 As = − 500 − 300 (1670 W / m2 ⋅ K )(3600s) L= As 1.37m 2 = = 21.8m π D π (0.02m ) < COMMENTS: Eq. (1) could also have been obtained by adapting Eq. (5.5) to the conditions of this problem, with T∞ and h replaced by Th and U, respectively. PROBLEM 5.15 KNOWN: Thickness and properties of furnace wall. Thermal resistance of film on surface of wall exposed to furnace gases. Initial wall temperature. FIND: (a) Time required for surface of wall to reach a prescribed temperature, (b) Corresponding value of film surface temperature. SCHEMATIC: ASSUMPTIONS: (1) Constant properties, (2) Negligible film thermal capacitance, (3) Negligible radiation. 3 PROPERTIES: Carbon steel (given): ρ = 7850 kg/m , c = 430 J/kg⋅K, k = 60 W/m⋅K. ANALYSIS: The overall coefficient for heat transfer from the surface of the steel to the gas is −1 U = ( R′′ ) tot 1 = + R′′ f h −1 1 = + 10−2 m2 ⋅ K/W 25 W/m 2 ⋅ K −1 = 20 W/m 2 ⋅ K. Hence, UL 20 W/m 2 ⋅ K × 0.01 m = = 0.0033 k 60 W/m ⋅ K and the lumped capacitance method can be used. Bi = (a) It follows that T − T∞ = exp ( − t/ τ t ) = exp ( −t/RC ) = exp ( −Ut/ ρ Lc ) Ti − T∞ 7850 kg/m3 ( 0.01 m ) 430 J/kg ⋅ K 1200 −1300 ρ Lc T − T∞ t=− ln =− ln U Ti − T∞ 300 − 1300 20 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/W f = h + (1 / R′′ ) f ( 25 +100 ) W/m2 ⋅ K Ts,o = 1220 K. COMMENTS: The film increases τ t by increasing Rt but not Ct. < PROBLEM 5.16 KNOWN: Thickness and properties of strip steel heated in an annealing process. Furnace operating conditions. FIND: (a) Time required to heat the strip from 300 to 600°C. Required furnace length for prescribed strip velocity (V = 0.5 m/s), (b) Effect of wall temperature on strip speed, temperature history, and radiation coefficient. SCHEMATIC: ASSUMPTIONS: (1) Constant properties, (2) Negligible temperature gradients in transverse direction across strip, (c) Negligible effect of strip conduction in longitudinal direction. PROPERTIES: Steel: ρ = 7900 kg/m3, cp = 640 J/kg⋅K, k = 30 W/m⋅K, ε= 0.7. ANALYSIS: (a) Considering a fixed (control) mass of the moving strip, its temperature variation with time may be obtained from an energy balance which equates the change in energy storage to heat transfer by convection and radiation. If the surface area associated with one side of the control mass is designated as As, As,c = As,r = 2As and V = δAs in Equation 5.15, which reduces to ρ cδ ( ) dT 4 = −2 h ( T − T∞ ) + εσ T 4 − Tsur dt or, introducing the radiation coefficient from Equations 1.8 and 1.9 and integrating, Tf − Ti = − tf 1 ∫o h (T − T∞ ) + h r (T − Tsur )dt ρ c (δ 2 ) Using the IHT Lumped Capacitance Model to integrate numerically with Ti = 573 K, we find that Tf = 873 K corresponds to tf ≈ 209s < in which case, the required furnace length is L = 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 yields V ( Tw = 1123K ) = 1.03m s V ( Tw = 1273K ) = 1.69 m s < Continued... PROBLEM 5.16 (Cont.) which correspond to increased process rates of 106% and 238%, respectively. Clearly, productivity can be enhanced by increasing the furnace environmental temperature, albeit at the expense of increasing energy utilization and operating costs. If the annealing process extends from 25°C (298 K) to 600°C (873 K), numerical integration yields the following results for the prescribed furnace temperatures. 600 200 Radiation coefficient, hr(W/m^2.K) Temperature, T(C) 500 400 300 200 100 0 0 50 100 150 200 Annealing time, t(s) Tsur = Tinf = 1000 C Tsur = Tinf = 850 C Tsur = Tinf = 700 C 250 300 150 100 50 0 50 100 150 200 250 300 Annealing time, t(s) Tsur = Tinf = 1000 C Tsur = Tinf = 850 C Tsur = Tinf = 700 C As expected, the heating rate and time, respectively, increase and decrease significantly with increasing Tw. Although the radiation heat transfer rate decreases with increasing time, the coefficient hr increases with t as the strip temperature approaches Tw. COMMENTS: To check the validity of the lumped capacitance approach, we calculate the Biot number based on a maximum cumulative coefficient of (h + hr) ≈ 300 W/m2⋅K. It follows that Bi = (h + hr)(δ/2)/k = 0.06 and the assumption is valid. PROBLEM 5.17 KNOWN: Diameter, resistance and current flow for a wire. Convection coefficient and temperature of surrounding oil. FIND: Steady-state temperature of the wire. Time for the wire temperature to come within 1°C of its steady-state value. SCHEMATIC: ASSUMPTIONS: (1) Constant properties, (2) Wire temperature is independent of x. 3 PROPERTIES: Wire (given): ρ = 8000 kg/m , cp = 500 J/kg⋅K, k = 20 W/m⋅K, R ′ = 0.01 Ω/m. e ANALYSIS: Since Bi = h ( ro / 2) k = ( 500 W/m 2 ⋅ K 2.5 × 10-4m 20 W/m ⋅ K ) = 0.006 < 0.1 the lumped capacitance method can be used. The problem has been analyzed in Example 1.3, and without radiation the steady-state temperature is given by π Dh ( T − T∞ ) = I 2R ′ . e Hence I 2R ′ (100A )2 0.01Ω / m e = 25o C + T = T∞ + = 88.7 o C. 2 ⋅K π Dh π 0.001 m 500 W/m ( ) < With no radiation, the transient thermal response of the wire is governed by the expression (Example 1.3) dT I 2R ′ 4h e = − (T − T ) . ∞ dt ρ c π D2 / 4 ρ c pD p ( ) With T = Ti = 25°C at t = 0, the solution is T − T∞ − I 2R e / π Dh ′ ( ) = exp − 4h t . ρ cp D Ti − T∞ − ( I 2 R′e /π Dh ) Substituting numerical values, find 87.7 − 25 − 63.7 4 × 500 W/m 2 ⋅ K − = exp t 8000 kg/m3 × 500 J/kg ⋅ K × 0.001 m 25 − 25 − 63.7 t = 8.31s. COMMENTS: The time to reach steady state increases with increasing ρ , cp and D and with decreasing h. < PROBLEM 5.18 KNOWN: 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 spacewise isothermal, (b) Approximate time to reach steady-state and T(∞) for prescribed T∞, h and q′′ when wall o material 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/m⋅K, cp = 385 J/kg⋅K, ρ = 8933 kg/m3. ANALYSIS: (a) Following the analysis of Section 5.3, the energy conservation requirement for the system is Ein − E out = Est or q′′ − h ( T − T∞ ) = ρ Lcp dT dt . Rearranging, and with R ′′ = 1/h and t o C′′ = ρLcp, t T − T∞ − q′′ h = − R ′′ ⋅ C′′ dT dt o tt (1) Defining θ ( t ) ≡ T − T∞ − q′′ h with dθ = dT, the differential equation is o θ = − R ′′C′′ tt dθ . dt (2) Separating variables and integrating, t dt θ dθ =− 0 R ′′C′′ θi θ tt ∫ ∫ it follows that θ t = exp − θi tt R ′′C′′ (3) where θ i = θ ( 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, θ (∞ ) = 0 = T (∞ ) − T∞ − q′′ h T ( ∞ ) = T∞ + q′′ h , or o o giving T(∞) = 27°C + 5000 W/m2 /50 W/m2⋅K = 127°C. To estimate the time to reach steady-state, first determine the thermal time constant of the system, ) 1 1 τ t = R ′′C′′ = ρ cp L = 8933kg m3 × 385 J kg ⋅ K ×12 × 10−3 m = 825s tt 50 W m 2 ⋅ K h ( ) ( Continued... PROBLEM 5.18 (Cont.) When t = 3τt = 3×825s = 2475s, Eqs. (3) and (4) yield θ (3τ t ) = T (3τ t ) − 27 C − 5000 W m 2 = e−3 27 C − 27 C − 50 W m 2 ⋅ K 50 W m 2 ⋅ K 5000 W m 2 < T(3τt) = 122°C (c) As shown by the following graphical results, which were generated using the IHT Lumped Capacitance Model, the steady-state temperature and the time to reach steady-state both decrease with increasing h. 125 Temperature, T(C) 105 85 65 45 25 0 500 1000 1500 2000 2500 Time, t(s) h = 50 W/m^2.K h = 100 W/m^2.K h = 200 W/m^2.K COMMENTS: Note that, even for h = 200 W/m2⋅K, Bi = hL/k << 0.1 and assumption (1) is reasonable. PROBLEM 5.19 KNOWN: Electronic device on aluminum, finned heat sink modeled as spatially isothermal object with internal generation and convection from its surface. FIND: (a) Temperature response after device is energized, (b) Temperature rise for prescribed conditions after 5 min. SCHEMATIC: ASSUMPTIONS: (1) Spatially isothermal object, (2) Object is primarily aluminum, (3) Initially, object is in equilibrium with surroundings at T∞. ( ) o PROPERTIES: Table A-1, Aluminum, pure T = ( 20 + 100 ) C/2 ≈ 333K : c = 918 J/kg⋅K. ANALYSIS: (a) Following the general analysis of Section 5.3, apply the conservation of energy requirement to the object, dT & Eg − hAs (T − T∞ ) = Mc dt & && & Ein + Eg -Eout = Est (1) where T = T(t). Consider now steady-state conditions, in which case the storage term of Eq. (1) is zero. 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 dθ (3,4) hAs dt with θ ≡ T - T(∞) and noting that dθ = dT. Identifying R t = 1/hA s and C t = Mc, the differential Mc dT T ( ∞ ) − T∞ − [ T − T∞ ] = hAs dt or θ =− equation is integrated with proper limits, 1 t θ dθ ∫0 dt = −∫θi θ R t Ct or θ t = exp − θi R t Ct (5) < where θi = θ(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 and capacitance of the system, T ( ∞ ) − T∞ (100 − 20 )o C 1 Rt = = = = 1.33 K/W & hAs Eg 60 W C t = Mc = 0.31 kg × 918 J/kg ⋅ K = 285 J/K. Using Eq. (5), the temperature of the system after 5 minutes is θ 5min T 5min − T ∞ T 5min − 100o C ( θi )= ( ) ( )= ( ) Ti − T ( ∞ ) ( 20 − 100 )o C 5 × 60s = exp − = 0.453 1.33 K/W × 285 J/K o T ( 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 5.20 KNOWN: Spherical coal pellet at 25°C is heated by radiation while flowing through a furnace maintained at 1000°C. FIND: Length of tube required to heat pellet to 600°C. SCHEMATIC: ASSUMPTIONS: (1) Pellet is suspended in air flow and subjected to only radiative exchange with furnace, (2) Pellet is small compared to furnace surface area, (3) Coal pellet has emissivity, ε = 1. PROPERTIES: Table A-3, Coal ( ) T = ( 600 + 25 ) C/2 = 585K, however, only 300K data available : ρ = 1350 3 kg/m ,cp = 1260 J/kg⋅K, k = 0.26 W/m⋅K. ANALYSIS: Considering the pellet as spatially isothermal, use the lumped capacitance method of Section 5.3 to find the time required to heat the pellet from To = 25°C to TL = 600°C. From an energy balance on the pellet Ein = Est where dT 4 4 Ein = q rad = σ As Tsur − Ts Est = ρ∀cp ( ( ) ) 4 4 Asσ Tsur − Ts = ρ∀cp giving dt dT . dt Separating variables and integrating with limits shown, the temperature-time relation becomes TL Asσ t dT dt = ∫ . ∫0 4 ρ∀cp T o Tsur − T 4 The integrals are evaluated in Eq. 5.18 giving t= ln 3 4Asσ Tsur ρ∀cp Tsur + T − ln Tsur − T T T Tsur + Ti + 2 tan -1 − tan -1 i . Tsur − Ti Tsur Tsur 2 3 Recognizing that As = πD and ∀ = πD /6 or As/ ∀ = 6/D and substituting values, 1350 kg/m3 0.001 m 1260 J/kg ⋅ K t= ( ) 1273 + 298 1273 + 873 − ln ln 3 1273 − 298 24 × 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 = V⋅t = 3m/s×1.18s = 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 = 600°C, Bi = 0.19; but when T = 25°C, Bi = 0.10. At early times, when the pellet is cooler, the assumption is reasonable but becomes less appropriate as the pellet heats. PROBLEM 5.21 KNOWN: Metal sphere, initially at a uniform temperature Ti, is suddenly removed from a furnace and suspended in a large room and subjected to a convection process (T∞, h) and to radiation exchange with surroundings, 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 obtain time required if both convection and radiation are considered, (d) Time to cool an anodized aluminum sphere to 400 K using results of Parts (a), (b) and (c). SCHEMATIC: ASSUMPTIONS: (1) Sphere is spacewise isothermal, (2) Constant properties, (3) Constant heat transfer convection coefficient, (4) Sphere is small compared to surroundings. PROPERTIES: Table A-1, Aluminum, pure ( T = [800 + 400] K/2 = 600 K): ρ = 2702 kg/m3, c = 1033 J/kg⋅K, k = 231 W/m⋅K, α = k/ρc = 8.276 × 10-5 m2/s; Aluminum, anodized finish: ε = 0.75, polished surface: ε = 0.1. ANALYSIS: (a) Neglecting radiation, the time to cool is predicted by Eq. 5.5, t= ρ Vc θi ρ Dc Ti − T∞ ln = ln hAs θ 6h T − T∞ (1) < where V/As = (πD3/6)/(πD2) = D/6 for the sphere. (b) Neglecting convection, the time to cool is predicted by Eq. 5.18, t= T T + Ti Tsur + T − ln sur + 2 tan −1 ln 3 Tsur − Ti 24εσ Tsur Tsur − T Tsur ρ Dc −1 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 ( ) dT 6 4 = h ( T − T∞ ) + εσ T 4 − Tsur dt ρ Dc (3) < where As(c,r) = As = πD2 and V/As(c,r) = D/6. This relation must be solved numerically in order to evaluate the time-to-cool. (d) For the aluminum (pure) sphere with an anodized finish and the prescribed conditions, the times to cool 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 ⋅ K 6 ×10 W m 2 ⋅ K ln 800 − 300 = 3743s = 1.04h 400 − 300 < Radiation only, Eq. (2) t= 2702 kg m3 × 0.050 m × 1033J kg ⋅ K 400 + 300 800 + 300 ⋅ ln − ln + 3 800 − 300 24 × 0.75 × 5.67 × 10−8 W m 2 ⋅ K 4 × (300 K ) 400 − 300 400 800 2 tan −1 − tan −1 300 300 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 yields t ≈ 1600s = 0.444h In this case, heat loss by radiation exerts the stronger influence, although the effects of convection are by no means negligible. However, if the surface is polished (ε = 0.1), convection clearly dominates. For each surface finish and the three cases, the temperature histories are as follows. 800 800 700 Temperature, T(K) Temperature, T(K) 700 600 500 400 600 500 400 0 400 800 1200 1600 2000 2400 2800 3200 3600 4000 0 0.5 1 1.5 Time, t(s) h = 10 W/m^2.K, eps = 0.75 h = 0, eps = 0.75 h = 10 W/m^2.K, eps = 0 2 2.5 Time, t x E-4 (s) h = 10 W/m^2.K, eps = 0.1 h = 10 W/ m^2.K, eps = 0 h = 0, eps = 0.1 COMMENTS: 1. A summary of the analyses shows the relative importance of the various modes of heat loss: Active Modes Convection only Radiation only Both modes Time required to cool to 400 K (h) ε = 0.75 ε = 0.1 1.040 0.827 0.444 1.040 6.194 0.889 2. Note that the spacewise isothermal assumption is justified since Be << 0.1. For the convection-only process, 2 -4 Bi = h(ro/3)/k = 10 W/m ⋅K (0.025 m/3)/231 W/m⋅K = 3.6 × 10 PROBLEM 5.22 KNOWN: 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. 3 PROPERTIES: Droplet (given): ρ = 885 kg/m , c = 1900 J/kg⋅K, k = 0.145 W/m⋅K, ε = 0.95. ANALYSIS: To assess the suitability of applying the lumped capacitance method, use Equation 1.9 to obtain the maximum radiation coefficient, which corresponds to T = Ti. h r = εσ Ti3 = 0.95 × 5.67 ×10 −8 W/m 2 ⋅ K 4 ( 500 K )3 = 6.73 W/m 2 ⋅ K. Hence Bi r = h r ( ro / 3 ) ( 6.73 W/m2 ⋅K ) (0.25 × 10−3 m/3) = 0.0039 = 0.145 W/m ⋅ K k and the lumped capacitance method can be used. From Equation 5.19, ( ( 3 L ρc π D / 6 t= = V 3ε π D2 σ L= ) ) 1 − 1 T3 f Ti3 ( 0.1 m/s ) 885 kg/m3 ( 1900 J/kg ⋅ K ) 0.5 × 10−3 m 1 1 1 − 3003 5003 K3 18 × 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 ×10−4 m ) 3π ( E i − E f = ρ Vc ( Ti − Tf ) = 885 kg/m 1900 J/kg ⋅ K ( 200 K ) 3 6 E i − E f = 0.022 J. < COMMENTS: Because some of the radiation emitted by a droplet will be intercepted by other droplets in the stream, the foregoing analysis overestimates the amount of heat dissipated by radiation to space. PROBLEM 5.23 KNOWN: 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 the sphere exclusively by convection, (c) Combined effects of radiation and convection. SCHEMATIC: ASSUMPTIONS: (1) Uniform temperature at any time, (2) Negligible effect of holding mechanism on heat transfer, (3) Constant properties, (4) Radiation exchange is between a small surface and large surroundings. ANALYSIS: (a) If cooling is exclusively by radiation, the required time is determined from Eq. 3 2 (5.18). With V = πD /6, As,r = πD , and ε = 0.1, t= 8600 kg / m3 ( 290 J / kg ⋅ K ) 0.01m 298 + 573 298 + 1173 − ln ln 3 298 − 1173 24 ( 0.1) 5.67 × 10−8 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 (ε = 0.1) < (ε = 0.6 ) < If ε = 0.6, cooling is six times faster, in which case, t = 199s (b) If cooling is exclusively by convection, Eq. (5.5) yields 3 ρ 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). ( ) ρ π D3 / 6 c ( ) dT 4 = −π D2 h ( T − T∞ ) + εσ T 4 − Tsur dt Integrating numerically from Ti = 1173 K at t = 0 to T = 573K, we obtain < t = 21.0s Continued ….. PROBLEM 5.23 (Cont.) Cooling times corresponding to representative changes in ε and h are tabulated as follows 2 h(W/m ⋅K) | | ε t(s) | 200 0.6 21.0 200 1.0 19.4 20 0.6 102.8 500 0.6 9.1 For values of h representative of forced convection, the influence of radiation is secondary, even for a maximum 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 natural 2 convection coefficient of h = 20 W/m ⋅K, the radiation flux exceeds the convection flux at the surface of the sphere during early to intermediate stages of the transient. H e a t flu xe s (W /m ^2 .K ) 70000 60000 50000 40000 30000 20000 10000 0 0 20 40 60 80 100 C 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 ) 2 2 COMMENTS: (1) Even for h as large as 500 W/m ⋅K, Bi = h (D/6)/k = 500 W/m ⋅K (0.01m/6)/63 W/m⋅K = 0.013 < 0.1 and the lumped capacitance model is appropriate. (2) The largest value of hr -8 24 corresponds to Ti =1173 K, and for ε = 0.6 Eq. (1.9) yields hf = 0.6 × 5.67 × 10 W/m ⋅K (1173 + 2 22 2 298)K (1173 + 298 )K = 73.3 W/m ⋅K. PROBLEM 5.24 KNOWN: Diameter and thermophysical properties of alumina particles. Convection conditions associated with a two-step heating process. FIND: (a) Time-in-flight (ti-f) required for complete melting, (b) Validity of assuming negligible radiation. SCHEMATIC: ASSUMPTIONS: (1) Particle behaves as a lumped capacitance, (2) Negligible radiation, (3) Constant properties. 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 = ρ p Vc p ρ p Dpcp T −T θ ln i = ln i ∞ θ hAs 6h Tmp − T∞ ) ( 3970 kg m3 50 × 10−6 m 1560 J kg ⋅ K ( 6 30, 000 W m 2 ⋅ K ) ln (300 − 10, 000 ) = 4 × 10−4 s ( 2318 − 10, 000 ) Performing an energy balance for the second step, we obtain t1 + t 2 ∫t1 q conv dt = ∆Est where qconv = hAs(T∞ - Tmp) and ∆Est = ρpVhsf. Hence, t2 = ρpDp 6h h sf (T∞ − Tmp ) = ( 3970 kg m3 50 × 10−6 m ( 6 30, 000 W m 2 ⋅ K Hence t i − f = 9 × 10−4 s ≈ 1ms ) ) × 3.577 ×106 J kg (10, 000 − 2318 ) K = 5 × 104 s < 8 2 (b) Contrasting the smallest value of the convection heat flux, q ′′ conv,min = h ( T∞ − Tmp ) = 2.3 × 10 W m ( ) 4 4 5 2 to the largest radiation flux, q′′ rad,max = εσ Tmp − Tsur = 6.5 × 10 W/m , we conclude that radiation is, in fact, negligible. COMMENTS: (1) Since Bi = (hrp/3)/k ≈ 0.05, the lumped capacitance assumption is good. (2) In an actual application, the droplet should impact the substrate in a superheated condition (T > Tmp), which would require a slightly larger ti-f. PROBLEM 5.25 KNOWN: Diameters, initial temperature and thermophysical properties of WC and Co in composite particle. Convection coefficient and freestream temperature of plasma gas. Melting point and latent heat 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 surroundings is negligible, (3) Negligible contact resistance at interface between WC and Co, (4) Constant properties. ANALYSIS: From Eq. (5.5), the time required to reach the melting point is t1 = ( ρ Vc )tot 2 h π Do ln Ti − T∞ Tmp − T∞ where the total heat capacity of the composite particle is ( ρ Vc )tot = ( ρ Vc )c + ( ρ Vc )s = 16, 000 kg / m3 π ( 1.6 × 10−5 m )( ( 3 +8900 kg / m3 π / 6 2.0 × 10−5 m − 1.6 × 10−5 m ) ( ) / 6 300 J / kg ⋅ K 3 ) 750 J / kg ⋅ K 3 = 1.03 × 10−8 + 1.36 ×10−8 J / K = 2.39 ×10−8 J / K t1 = ( 2.39 × 10−8 J / K )( ) 2 20, 000 W / m 2 ⋅ K π 2.0 × 10−5 m ln (300 − 10, 000 ) K = 1.56 × 10−4 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 control surface about the particle. It follows that 2 3 3 Ein = hπ Do T∞ − Tmp t 2 = ∆Est = ρs (π / 6 ) Do − Di h sf ( (2 ×10 m ) − (1.6 ×10 m ) 2.59 ×10 (20, 000 W / m ⋅ K )π (2 × 10 m ) (10, 000 − 1770 ) K 8900 kg / m t2 = ) ( ) 3 −5 (π / 6 ) 2 3 −5 −5 2 3 5 J / kg = 2.28 × 10 −5 s < COMMENTS: (1) The largest value of the radiation coefficient corresponds to hr = εσ (Tmp + Tsur) (T ) 2 2 mp + Tsur . 2 2 For the maximum possible value of ε = 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 the large value of h, the small values of Do and Di and the large thermal conductivities (~ 40 W/m⋅K and 70 W/m⋅K 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 5.26 KNOWN: 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. 3 PROPERTIES: Chip material (given): ρ = 2000 kg/m , c = 700 J/kg⋅K. ANALYSIS: At steady-state, conservation of energy yields & & − Eout + Eg = 0 () () & − h L2 ( Tf − T∞ ) + q L2 ⋅ t = 0 & qt Tf = T∞ + h Tf = 20o C + 9 × 106 W/m3 × 0.001 m 150 W/m 2 ⋅ K = 80o C. < From the general lumped capacitance analysis, Equation 5.15 reduces to dT & ρ L2 ⋅ t c = q L2 ⋅ t − h ( T − T ) L2 . ∞ dt With () a≡ b≡ () h 150 W/m 2 ⋅ K = = 0.107 s -1 ρ tc 2000 kg/m3 ( 0.001 m )( 700 J/kg ⋅ K ) ( ) 9 × 106 W/m3 3 & q = ρc 2000 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 = = 0.01667 Ti − T∞ − b/a ( 20 − 20 − 60 ) K ln ( 0.01667 ) 0.107 s -1 = 38.3 s. < COMMENTS: Due to additional heat transfer from the chip to the substrate, the actual values of Tf and t are less than those which have been computed. PROBLEM 5.27 KNOWN: 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. 3 PROPERTIES: Chip material (given): ρ = 2000 kg/m , cp = 700 J/kg⋅K. ANALYSIS: The direct and indirect paths for heat transfer from the chip to the coolant are in parallel, and the equivalent resistance is −1 ) ( R equiv = hL2 + R -1 = 3.75 × 10−3 + 5 × 10−3 W/K t The corresponding overall heat transfer coefficient is −1 = 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 about the chip. −E + E = 0 − UL2 ( T − T ) + q L2 ⋅ t = 0 out g f ∞ () qt 9 × 106 W/m3 × 0.001 m = 20 C + = 45.7 C. 2 ⋅K U 350 W/m From the general lumped capacitance analysis, Equation 5.15 yields dT ρ L2 t c = q L2 t − U (T − T∞ ) L2 . dt With Tf = T∞ + () < () U 350 W/m 2 ⋅ K a≡ = = 0.250 s-1 ρ tc 2000 kg/m3 (0.001 m )(700 J/kg ⋅ K ) ( ) ( ) q 9 × 106 W/m3 b= = = 6.429 K/s 3 700 J/kg ⋅ K ρc 2000 kg/m ( ) Equation 5.24 yields T − T∞ − b/a ( 44.7 − 20 − 25.7 ) K exp ( −at ) = = = 0.0389 Ti − T∞ − b/a (20 − 20 − 25.7 ) K t = −ln ( 0.0389 ) / 0.250 s-1 = 13.0 s. COMMENTS: Heat transfer through the substrate is comparable to that associated with direct convection to the coolant. < PROBLEM 5.28 KNOWN: Dimensions, initial temperature and thermophysical properties of chip, solder and substrate. Temperature and convection coefficient of heating agent. FIND: (a) Time constants and temperature histories of chip, solder and substrate when heated by an air stream. Time corresponding to maximum stress on a solder ball. (b) Reduction in time associated with using a dielectric liquid to heat the components. SCHEMATIC: ASSUMPTIONS: (1) Lumped capacitance analysis is valid for each component, (2) Negligible heat transfer between components, (3) Negligible reduction in surface area due to contact between components, (4) Negligible radiation for heating by air stream, (5) Uniform convection coefficient among components, (6) Constant properties. ANALYSIS: (a) From Eq. (5.7), τ t = ( ρ Vc ) / hA Chip: () V = L ch t ch = ( 0.015m ) 2 2 (0.002m ) = 4.50 × 10 −7 m3 , As = = 2 ( 0.015m ) + 4 ( 0.015m ) 0.002m = 5.70 × 10 2 τt = −4 m 2300 kg / m3 × 4.50 ×10−7 m3 × 710 J / kg ⋅ K 50 W / m 2 ⋅ K × 5.70 × 10−4 m 2 V = π D / 6 = π ( 0.002m ) / 6 = 4.19 × 10 3 Solder: τt = ( 3 −9 (2L 2 ch + 4L ch t ch ) 2 < = 25.8s m , A s = π D = π ( 0.002m ) = 1.26 × 10 3 2 11, 000 kg / m3 × 4.19 × 10−9 m3 × 130 J / kg ⋅ K 50 W / m 2 ⋅ K × 1.26 × 10−5 m 2 2 −5 m 2 < = 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 2 sb sb τt = 4000 kg / m3 × 6.25 ×10−6 m3 × 770 J / kg ⋅ K 50 W / m 2 ⋅ K × 6.25 × 10−4 m 2 < = 616.0s Substituting Eq. (5.7) into (5.5) and recognizing that (T – Ti)/(T∞ - Ti) = 1 – (θ/θi), in which case (T – Ti)/(T∞ -Ti) = 0.99 yields θ/θi = 0.01, it follows that the time required for a component to experience 99% of its maximum possible temperature rise is t 0.99 = τ ln (θ i / θ ) = τ ln (100 ) = 4.61τ Hence, Chip: t = 118.9s, Solder: t = 43.8s, < Substrate: t = 2840 Continued ….. PROBLEM 5.28 (Cont.) Histories of the three components and temperature differences between a solder ball and its adjoining components are shown below. Te m p e ratu re (C ) 80 65 50 35 20 0 100 200 300 400 500 Tim e (s ) Te m p e ra tu re d iffe re n ce (C ) Ts d Tch Ts b 60 50 40 30 20 10 0 0 20 40 60 80 100 Tim e (s ) Ts d -Tch Ts d -Ts b Commensurate with their time constants, the fastest and slowest responses to heating are associated with the solder and substrate, respectively. Accordingly, the largest temperature difference is between these two components, and it achieves a maximum value of 55°C at t ( maximum stress ) ≈ 40s < (b) With the 4-fold increase in h associated with use of a dielectric liquid to heat the components, the time constants are each reduced by a factor of 4, and the times required to achieve 99% of the maximum temperature rise are Chip: 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 the heating problem. Several of the assumptions are highly approximate, particularly that of a uniform convection coefficient. The coefficient will vary between components, as well as on the surfaces of the components. Also, because the solder balls are flattened, there will be a reduction in surface area exposed to the fluid for each component, as well as heat transfer between components, which reduces differences between time constants for the components. PROBLEM 5.29 KNOWN: Electrical transformer of approximate cubical shape, 32 mm to a side, dissipates 4.0 W 2 when operating in ambient air at 20°C with a convection coefficient of 10 W/m ⋅K. FIND: (a) Develop a model for estimating the steady-state temperature of the transformer, T(∞), and evaluate 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 5°C of its steady-state operating temperature. SCHEMATIC: ASSUMPTIONS: (1) Transformer is spatially isothermal object, (2) Initially object is in equilibrium with 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 is Ein − Eout + Egen = 0 0 − q cv + Pe = − h As T ( ∞ ) − T∞ + Pe = 0 2 -3 (1) 2 where As = 5 × L = 5 × 0.032m × 0.032m = 5.12 × 10 m , find ( ) T ( ∞ ) = T∞ + Pe / h As = 20°C + 4 W / 10 W / m 2 ⋅ K × 5.12 ×10−3 m 2 = 98.1°C < (b) Under transient conditions, for the control volume shown above, the energy balance is Ein − E out + E gen = Est 0 − q cv + Pe = Mc dT dt (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 ( ∞ ) = Mc dT dt θ o dθ to h ∫ 0 dt = −∫θi θ Mc d ( T − T ( ∞ )) dt where θ = T(t) – T(∞); θi = Ti – T(∞) = T∞ - T(∞); and θo = T(to) – T(∞) with to as the time when θo = - 5°C. Integrating and rearranging find (see Eq. 5.5), to = to = θ Mc n i h As θ o 0.28 kg × 400 J / kg ⋅ K 10 W / m 2 ⋅ K × 5.12 × 10−3 m 2 n ( 20 − 98.1) °C −5°C = 1.67 hour < COMMENTS: The spacewise isothermal assumption may not be a gross over simplification since most 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 5.30 KNOWN: Series solution, Eq. 5.39, for transient conduction in a plane wall with convection. FIND: Midplane (x*=0) and surface (x*=1) temperatures θ* for Fo=0.1 and 1, using Bi=0.1, 1 and 10 with only the first four eigenvalues. Based upon these results, discuss the validity of the approximate solutions, 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, θ* = ∞ ∑ Cn n =1 )( ( exp -ζ 2 Fo cos ζ n x* n ) where the eigenvalues, ζ n , and the constants, Cn, are from Eqs. 5.39b and 5.39c. Cn = 4sin ζ n / ( 2ζ n + sin ( 2ζ n ) ) . The eigenvalues are tabulated in Appendix B.3; note, however, that ζ 1 and C1 are available from Table 5.1. The values of ζ n and Cn used to evaluate θ* are as follows: ζ n tan ζ n = Bi Bi ζ1 C1 0.1 1 10 0.3111 0.8603 1.4289 ζ2 C2 ζ3 C3 1.0160 1.1191 1.2620 3.1731 3.4256 4.3058 -0.0197 -0.1517 -0.3934 6.2991 6.4373 7.2281 ζ4 0.0050 0.0466 0.2104 C4 9.4354 9.5293 10.2003 -0.0022 -0.0217 -0.1309 *** * Using ζ n and Cn values, the terms of θ * , designated as θ1 , θ2 , θ3 and θ4 , are as follows: Fo=0.1 Bi=1.0 Bi=0.1 x* * θ1 * θ2 * θ3 * θ4 θ* 0 1 0 Bi=10 1 0 1 1.0062 0.9579 1.0393 0.6778 1.0289 0.1455 -0.0072 0.0072 -0.0469 0.0450 -0.0616 0.0244 0.0001 0.0001 0.0007 0.0007 0.0011 0.0006 -2.99×10-7 3.00×10-7 2.47×10-6 2.46×10-7 -3.96×10-6 2.83×10-6 0.9991 0.9652 0.9931 0.7235 0.9684 0.1705 Continued ….. PROBLEM 5.30(Cont.) Fo=1 Bi=1.0 Bi=0.1 x* * θ1 * θ2 * θ3 * θ4 θ* 0 1 0 Bi=10 1 0 1 0.9223 0.8780 0.5339 0.3482 0.1638 0.0232 8.35×10-7 8.35×10-7 -1.22×10-5 1.17×10-6 3.49×10-9 1.38×10-9 7.04×10-20 - 4.70×10-20 - 4.30×10-24 - 4.77×10-42 - 7.93×10-42 - 8.52×10-47 - 0.9223 0.8780 0.5339 ( 0.3482 0.1638 0.0232 ) The tabulated results for θ * = θ * x* , Bi, Fo demonstrate that for Fo=1, the first eigenvalue is sufficient to accurately represent the series. However, for Fo=0.1, three eigenvalues are required for accurate representation. A more detailed analysis would show that a practical criterion for representation of the series solution by one eigenvalue 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=0.2 x*=0 Bi 0.1 1.0 10 * θ1 * θ2 0.9965 0.9651 0.8389 -0.00226 -0.0145 -0.0096 θ* 0.9939 0.9506 0.8293 +0.26 +1.53 +1.16 Error,% The percentage error shown in the last row of the above table is due to the effect of the second term. For Bi=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 systematically high results, but by less than 1.5%, for the Biot number range from 0.1 to 10. PROBLEM 5.31 KNOWN: One-dimensional wall, initially at a uniform temperature, Ti, is suddenly exposed to a convection process (T∞, h). For wall #1, the time (t1 = 100s) required to reach a specified temperature 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: 2 Wall L(m) α (m /s) 1 2 0.10 0.40 15×10 -6 25×10 2 k(W/m⋅K) Ti(° C) T∞(° C) h(W/m ⋅K) 50 100 300 30 400 20 200 100 -6 The dimensionless functional dependence for the one-dimensional, transient temperature distribution, Eq. 5.38, is θ∗ = T ( x,t ) − T∞ Ti − T∞ ( = f x ∗ , Bi, Fo ) where x∗ = x/L Bi = hL/k Fo = α t/L2 . ∗ ∗ If the parameters x*, Bi, and Fo are the same for both walls, then θ1 = θ2 . Evaluate these parameters: Wall 1 2 where ∗ θ1 = x* 1 1 315 − 400 = 0.85 300 − 400 Bi 0.40 0.40 ∗ θ2 = Fo 0.150 -4 1.563×10 t2 θ* 0.85 0.85 28.5 − 20 = 0.85. 30 − 20 It follows that Fo 2 = Fo1 t 2 = 960s. 1.563 ×10 -4t 2 = 0.150 < PROBLEM 5.32 KNOWN: The chuck of a semiconductor processing tool, initially at a uniform temperature of Ti = 100°C, is cooled on its top surface by supply air at 20°C with a convection coefficient of 50 W/m2⋅K. FIND: (a) Time required for the lower surface to reach 25°C, and (b) Compute and plot the time-to-cool as a function of the convection coefficient for the range 10 ≤ h ≤ 2000 W/m2⋅K; comment on the effectiveness of the head design as a method for cooling the chuck. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, transient conduction in the chuck, (2) Lower surface is perfectly insulated, (3) Uniform convection coefficient and air temperature over the upper surface of the chuck, and (4) Constant properties. PROPERTIES: Table A.1, Aluminum alloy 2024 ( (25 + 100)°C / 2 = 335 K): ρ = 2770 kg/m3, cp = 880 J/kg⋅ K, k = 179 W/m⋅K. ANALYSIS: (a) The Biot number for the chuck with h = 50 W/m2⋅K is Bi = hL 50 W m 2 ⋅ K × 0.025 m = = 0.007 ≤ 0.1 k 179 W m ⋅ K (1) so that the lumped capacitance method is appropriate. Using Eq. 5.5, with V/As = L, t= ρ Vc θi ln hAs θ θ = T − T∞ θ i = Ti − T∞ (100 − 20 ) C t = 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/m2⋅K, 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 Wall Model, the time-to-cool was calculated as a function of the convection coefficient. Free convection cooling conduction corresponds to h ≈ 10 W/m2⋅K and the time-to-cool is 282 minutes. With the cooling head design, the time-to-cool can be substantially decreased if the convection coefficient can be increased as shown below. Time-to-cool, t (min) 60 40 20 0 0 1000 Convection coefficient, h (W/m^2.K) 2000 PROBLEM 5.33 KNOWN: Configuration, initial temperature and charging conditions of a thermal energy storage unit. FIND: Time required to achieve 75% of maximum possible energy storage and corresponding minimum and maximum temperatures. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible radiation exchange with surroundings. ANALYSIS: For the system, find first hL 100 W/m 2 ⋅ K × 0.025m Bi = k = 0.7 W/m ⋅ K = 3.57 indicating that the lumped capacitance method cannot be used. Groeber chart, Fig. D.3: α= Q/Qo = 0.75 k 0.7 W/m ⋅ K = = 4.605 × 10−7 m 2 / s ρ c 1900 kg/m3 × 800 J/kg ⋅ K ( 2 100 W/m 2 K 2Fo = h α t = Bi k2 ) × ( 4.605×10−7 m2 / s) × t (s) = 9.4 ×10−3t 2 ( 0.7 W/m ⋅ K ) 2 2 Find Bi Fo ≈ 11, and substituting numerical values t = 11/9.4 × 10-3 = 1170s. < Heisler chart, Fig. D.1: Tmin is at x = 0 and Tmax at x = L, with α t 4.605 ×10 −7 m 2 / s ×1170 s Fo = 2 L = ( 0.025m ) ∗ From Fig. D.1, θ o ≈ 0.33. Hence, 2 Bi -1 = 0.28. = 0.86 ( ) To ≈ T∞ + 0.33 ( Ti − T∞ ) = 600o C + 0.33 −575o C = 410o C = Tmin . From Fig. D.2, θ/θo ≈ 0.33 at x = L, for which o Tx =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, (ρ c)Al > (ρ c)m and kAl > km. Hence, the aluminum can store more energy and can be charged (or discharged) more quickly. PROBLEM 5.34 KNOWN: Thickness, properties and initial temperature of steel slab. Convection conditions. FIND: Heating time required to achieve a minimum temperature of 550°C in the slab. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible radiation effects, (3) Constant properties. 2 ANALYSIS: With a Biot number of hL/k = (250 W/m ⋅K × 0.05m)/48 W/m⋅K = 0.260, a lumped capacitance analysis should not be performed. At any time during heating, the lowest temperature in the slab is at the midplane, and from the one-term approximation to the transient thermal response of a plane wall, Eq. (5.41), we obtain (550 − 800 ) °C 0.417 C exp ζ 2 Fo ∗ T −T θo = o ∞ = = =1 −1 Ti − T∞ ( 200 − 800 ) °C ( ) With ζ1 ≈ 0.488 rad and C1 ≈ 1.0396 from Table 5.1 and α = k / ρ c = 1.115 × 10 −5 m 2 / s, ) ( 2 −ζ1 α t / L2 = ln (0.401) = −0.914 t= 0.841(0.05m ) 2 0.914 L2 = 2 ζ1 α (0.488 ) 1.115 ×10−5 m 2 / s 2 < = 861s COMMENTS: 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 θ = θ o cos ζ 1x ∗ s ∞ + 0.368 ( Ti − T∞ ) = 800°C − 221°C = 579°C. Assuming a surface emissivity of ε = 1 and surroundings that are at Tsur = T∞ = 800°C, the radiation heat transfer coefficient corresponding to this surface temperature is ( ) h r = εσ ( Ts + Tsur ) Ts + Tsur = 205 W / m ⋅ K. Since this value is comparable to the convection 2 2 2 coefficient, radiation is not negligible and the desired heating will occur well before t = 861s. PROBLEM 5.35 KNOWN: 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, and energy transferred to the wall after 8 min; compare results to the hand calculations performed for the Text Example; (b) Time at which the outer surface temperature of the pipe, T(0,t), will reach 25°C; (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 and plot 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 Wall model as the solution tool. SCHEMATIC: ASSUMPTIONS: (1) Pipe wall can be approximated as a plane wall, (2) Constant properties, (3) Outer surface of pipe is adiabatic. ANALYSIS: The IHT model represents the series solution for the plane wall providing temperatures and heat fluxes evaluated at (x,t) and the total energy transferred at the inner wall at (t). Selected portions of the IHT code used to obtain the results tabulated below are shown 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 5.4 45.2 42.9 q′′ ( L, 8 min ) , W / m 2 x -2.72 -7400 Q′ (8 min ) × 10−7 , J / m 45.4 43.1 -2.73 T(L, 8min), °C T(0, 8 min), °C IHT Model -7305 (b) To determine the time to for which T(0,t) = 25°C, the IHT model is solved for to after setting x = 0 and T_xt = 25°C. 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 . x 60 40 20 0 -2 0 0 10 20 30 40 W a ll lo c a tio n , x (m m ) In itia l co n d itio n , t = 0 t = 4 m in t = 8 m in S te a d y-s ta te c o n d itio n , t > 3 0 m in Continued ….. PROBLEM 5.35 (Cont.) The final condition corresponds to the steady-state temperature, T (x,∞) = T∞. For the intermediate times, the gradient is zero at the insulated boundary (x = 0, the pipe exterior). As expected, the temperature at x = 0 will be less than at the boundary experiencing the convection process with the hot oil, 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 are shown in the graph below. Note that the temperature difference between the two locations is greatest at the start of the transient process and decreases with increasing time. After a 16 min. duration, the pipe temperature is almost uniform, but yet 3 or 4°C from the steady-state condition. T e m p e ra tu re -tim e h is to ry, T (x,t) vs . t T e m p e ra tu re , T (x ,t) (C ) 60 40 20 0 -2 0 0 2 4 6 8 10 12 14 16 T im e , t (m in ) O u te r s u rfa c e , x = 0 In n e r s u rfa c e , x = L COMMENTS: (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 length over the time interval t. // Models | Transient Conduction | Plane Wall // The temperature distribution is T_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 is q''_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 is QoverQo = Q_over_Qo_trans("Plane Wall",Fo,Bi) // Eq 5.45 Qo = rho * cp * vol * (Ti - Tinf) // Eq 5.44 //vol = 2 * As * L // Appropriate for wall of 2L thickness vol = pi * D * L // Pipe wall of diameter D, thickness L and unit length Q = QoverQo * Qo // Total energy transfered per unit length (2) Can you give an explanation for why the inner and outer surface temperatures are not very different? What parameter provides a measure of the temperature non-uniformity in a system during a transient conduction process? PROBLEM 5.36 KNOWN: Thickness, initial temperature and properties of furnace wall. Convection conditions at inner surface. FIND: Time required for outer surface to reach a prescribed temperature. Corresponding temperature 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 convection 2 conditions at its two surfaces. With Bi = hL/k = 100 W/m ⋅K × 0.15m/1.5 W/m⋅K = 10 and Fo > 0.2, the one-term approximation, Eq. 5.41 may be used to compute the desired time, where ∗ θ o = (To − T∞ ) / (Ti − T∞ ) = 0.215. From Table 5.1, C1 = 1.262 and ζ1 = 1.4289. Hence, ∗ ln θ o / C1 ln ( 0.215 /1.262 ) Fo = − ( 2 ζ1 )=− = 0.867 (1.4289 )2 0.867 ( 0.15m ) Fo L2 = = 33,800s 3 × 1000 J / kg ⋅ K α 1.5 W / m ⋅ K / 2600 kg / m 2 t= ) ( < The corresponding temperature distribution, as well as distributions at t = 0, 10,000, and 20,000 s are plotted below 1000 Te m p e ra tu re , C 800 600 400 200 0 0 0 .2 0 .4 0 .6 0 .8 1 D im e n s io n le s s lo ca tio n , x/L t= 0 s t= 1 0 ,0 0 0 s t= 2 0 ,0 0 0 s t= 3 3 ,8 0 0 s COMMENTS: Because Bi >>1, the temperature at the inner surface of the wall increases much more rapidly than at locations within the wall, where temperature gradients are large. The temperature gradients decrease as the wall approaches a steady-state for which there is a uniform temperature of 950°C. PROBLEM 5.37 KNOWN: Thickness, initial temperature and properties of steel plate. Convection conditions at both surfaces. 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. 2 ANALYSIS: The smallest temperature exists at the midplane and, with Bi = hL/k = 500 W/m ⋅K × 0.050m/45 W/m⋅K = 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 ζ1 = 0.682. Hence, with θ o = (To - T∞)/(Ti - T∞) = 0.375, Fo = − ( ∗ ln θ o / C1 2 ζ1 ) = − ln (0.375 /1.076) = 2.266 (0.682 )2 2.266 ( 0.05m ) Fo L2 = = 491s 3 × 500 J / kg ⋅ K α 45 W / m ⋅ K / 7800 kg / m 2 t= ( ) COMMENTS: From Eq. 5.40b, the corresponding surface temperature is ∗ Ts = T∞ + ( Ti − T∞ )θ o cos (ζ1 ) = 700°C − 400°C × 0.375 × 0.776 = 584°C Because Bi is not much larger than 0.1, temperature gradients in the steel are moderate. < PROBLEM 5.38 KNOWN: Plate of thickness 2L = 25 mm at a uniform temperature of 600°C is removed from a hot pressing 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 in the plate to reach 100°C; and (b) For both cases, calculate and plot on one graph, the variation with time of the maximum temperature difference in the plate; Comment on the relative magnitudes of the temperature gradients within the plate as a function of time. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the plate, (2) Constant properties, and (3) For case 2, with cooling on one side only, the other side is adiabatic. 3 PROPERTIES: Plate (given): ρ = 3000 kg/m , c = 750 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: (a) From IHT, call up Plane Wall, Transient Conduction from the Models menu. For case 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 to reach 100°C are t1 = 164 s < t2 = 367 s (b) The plot of T(0, t) – T(1, t), which represents the maximum temperature difference in the plate during the cooling process, is shown below. Plate center temperature histories Temperature difference history 600 150 T(0,t) - T(L,t) (C) T(0,t) (C) 500 400 300 200 100 50 100 0 0 0 100 200 300 Time, t (s) Cooling - both sides Cooling - one side only 400 500 0 100 200 300 400 500 Time (s) Cooling - both sides Cooling - one side only COMMENTS: (1) From the plate center-temperature history graph, note that it takes more than twice as long for the maximum temperature to reach 100°C with cooling on only one side. (2) From the maximum temperature-difference graph, as expected, cooling from one side creates a larger maximum temperature difference during the cooling process. The effect could cause microstructure differences, which could adversely affect the mechanical properties within the plate. PROBLEM 5.39 KNOWN: 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 outer surface 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 contact resistance at wall/ceramic interface, (5) Negligible radiation. ANALYSIS: (a) Subject to assumptions (3) and (4), the maximum wall temperature corresponds to the ceramic 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/m2⋅K(0.01 m)/10 W/m⋅K = 5, the lumped capacitance method cannot be used. Assuming Fo > 0.2, obtaining ζ1 = 1.3138 and C1 = 1.2402 from Table 5.1, and * evaluating θ o = ( To − T∞ ) ( Ti − T∞ ) = 0.4, Equation 5.41 yields Fo = − ( * ln θ o C1 2 ζ1 ) = − ln (0.4 1.2402) = 0.656 (1.3138 )2 confirming the assumption of Fo > 0.2. Hence, t max = ( ) = 0.656 (0.01m )2 = 10.9s Fo L2 < 6 × 10−6 m 2 s α (b) Using the IHT Lumped Capacitance Model for a Plane Wall, the inner and outer surface temperature histories were computed and are as follows: Temperature, T(K) 2300 1900 1500 1100 700 300 0 30 60 90 120 150 Time, t(s) L = 0.01, x = L L = 0.01, x = 0 L = 0.04, x = L L = 0.04, x = 0 Continued... PROBLEM 5.39 (Cont.) The increase in the inner (x = 0) surface temperature lags that of the outer surface, but within t ≈ 45s both temperatures are within a few degrees of the gas temperature for L = 0.01 m. For L = 0.04 m, the increased thermal capacitance of the ceramic slows the approach to steady-state conditions. The thermal response of the inner surface significantly lags that of the outer surface, and it is not until t ≈ 137s that the inner surface reaches 1500 K. At this time there is still a significant temperature difference across the ceramic, with T(L,tmax) = 2240 K. COMMENTS: The allowable engine operating time increases with increasing thermal capacitance of the ceramic and hence with increasing L. PROBLEM 5.40 KNOWN: Initial temperature, thickness and thermal diffusivity of glass plate. Prescribed surface temperature. FIND: (a) Time to achieve 50% reduction in midplane temperature, (b) Maximum temperature gradient 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 −T 2 θ o = o s = 0.50 = C1exp −ζ1 Fo Ti − Ts ) ζ1tan ζ1 = Bi = ∞ → ζ1 = π /2. Hence C1 = 4sinζ1 4 = = 1.273 2ζ1 + sin ( 2ζ1 ) π Fo = − ( ∗ ln θ o / C1 2 ζ1 ) = 0.379 2 FoL2 0.379 ( 0.01 m ) t= = = 63 s. α 6 × 10−7 m 2 / s ( < ) 2 (b) With θ ∗ = C1exp −ζ1 Fo cos ζ1x∗ (T − T ) ∂ T ( Ti − Ts ) ∂θ ∗ 2 = = − i s ζ 1C1exp −ζ 1 Fo sinζ 1x ∗ ∗ ∂x L L ∂x ( ) 300o C π ∂ T/ ∂ x max = ∂ T/ ∂ x ∗ = − 0.5 = −2.36 ×10 4 oC/m. x =1 0.01 m 2 COMMENTS: Validity of one-term approximation is confirmed by Fo > 0.2. < PROBLEM 5.41 KNOWN: Thickness and properties of rubber tire. Convection heating conditions. Initial and final midplane 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/m2⋅K(0.01 m)/0.14 W/m⋅K = 14.3, the lumped capacitance method is clearly inappropriate. Assuming Fo > 0.2, Eq. (5.41) may be used with C1 = 1.265 and ζ1 ≈ 1.458 rad from Table 5.1 to obtain ) ( * T −T 2 θ o = o ∞ = C1 exp −ζ1 Fo = 1.265exp ( −2.126 Fo ) Ti − T∞ * With θ o = ( To − T∞ ) ( Ti − T∞ ) = (-50)/(-175) = 0.286, Fo = − ln ( 0.286 1.265 ) 2.126 = 0.70 = α t f L2 0.7 ( 0.01m ) 2 tf = 6.35 × 10−8 m 2 s < = 1100s (b) The desired temperature histories were generated using the IHT Transient Conduction Model for a Plane Wall, with h = 5 × 104 W/m2⋅K used to approximate imposition of a surface temperature of 200°C. Temperature, T(C) 200 150 100 50 0 0 200 400 600 800 1000 1200 Time, t(s) x = 0, h = 200 W/m^2.K x = L, h = 200 W/m^2.K x = 0, h = 5E4 W/m^2.K x = L, h = 5E4W/m^2.K The fact that imposition of a constant surface temperature (h → ∞) does not significantly accelerate the heating process should not be surprising. For h = 200 W/m2⋅K, 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 to convection at the surface. Any increase in h only serves to reduce what is already a small component of the total thermal resistance. COMMENTS: The heating rate could be accelerated by increasing the steam temperature, but an upper limit would be associated with avoiding thermal damage to the rubber. PROBLEM 5.42 KNOWN: Stack or book comprised of 11 metal plates (p) and 10 boards (b) each of 2.36 mm thickness and prescribed thermophysical properties. FIND: Effective thermal conductivity, k, and effective thermal capacitance, (ρcp). SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Negligible contact resistance between plates and boards. 3 PROPERTIES: Metal plate (p, given): ρp = 8000 kg/m , cp,p = 480 J/kg⋅K, kp = 12 3 W/m⋅K; Circuit boards (b, given): ρb = 1000 kg/m , cp,b = 1500 J/kg⋅K, kb = 0.30 W/m⋅K. ANALYSIS: The thermal resistance of the book is determined as the sum of the resistance of the boards and plates, R ′′ = NR ′′ + MR ′′ tot b p where M,N are the number of plates and boards in the book, respectively, and R ′′ = Li / ki i where Li and ki are the thickness and thermal conductivities, respectively. ( ) R ′′ = M L p / k p + N ( L b / k b ) tot R ′′ = 11 ( 0.00236 m/12 W/m ⋅ K ) + 10 ( 0.00236 m/0.30 W/m ⋅ K ) tot R ′′ = 2.163 × 10 tot −3 + 7.867 × 10 −2 = 8.083 × 10 −2 K/W. The effective thermal conductivity of the book of thickness (10 + 11) 2.36 mm is 0.04956 m k = L/R ′′ = tot 8.083 × 10-2 K/W The thermal capacitance of the stack is ( ) C′′ = M ρ p L p c p + N ( ρ 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 ⋅ K tot 3 4 4 5 3 ) 2 C′′ = 9.969 × 10 + 3.540 × 10 = 1.35 × 10 J/m ⋅ K. tot The effective thermal capacitance of the book is ( ρ cp ) = C′′tot / L = 1.351×105 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 2 medium with effective properties: k = 0.613 W/m⋅K and α = (k/ρcp) = 2.249×10 m /s. See for example, Problem 5.38. PROBLEM 5.43 KNOWN: Stack of circuit board-pressing plates, initially at a uniform temperature, is subjected by upper/lower platens to a higher temperature. FIND: (a) Elapsed time, te, required for the mid-plane to reach cure temperature when platens are suddenly changed to Ts = 190°C, (b) Energy removal from the stack needed to return its temperature to Ti. SCHEMATIC: 6 3 -7 PROPERTIES: Stack (given): k = 0.613 W/m⋅K, ρcp = 2.73×10 J/m ⋅K; α = k/ρcp = 2.245×10 2 m /s. ANALYSIS: (a) Recognize that sudden application of surface temperature corresponds to h → ∞, or -1 Bi = 0 (Heisler chart) or Bi → ∞ (100, Table 5.1). With Ts = T∞, ∗ T (0,t ) − Ts = (170 − 190 ) C = 0.114. θo = Ti − Ts (15 − 190 ) C Using Eq. 5.41 with values of ζ1 = 1.552 and C1 = 1.2731 at Bi = 100 (Table 5.1), find Fo 2 ∗ θ o = C1exp −ζ1 Fo ( Fo = − ζ1 2 where Fo = αt/L , t= α ( ) ∗ ln θ o / C1 = − 2 1 FoL2 ) = ( 1 (1.552 )2 1.002 25 ×10−3 m ) ln ( 0.114/1.2731) = 1.002 2 2.245 × 10−7 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 θ o and Bi -1 = 0. (b) The energy removal is equivalent to the energy gained by the stack per unit area for the time interval 0 → te. With Q′′ corresponding to the maximum amount of energy that could be transferred, o ( Q′′ = ρ c ( 2L )( Ti − T∞ ) = 2.73 × 10 J/m ⋅ K 2 × 25 × 10 o 6 3 -3 ) m (15 − 190 ) K = −2.389 × 10 J/m 7 2. Q′′ may be determined from Eq. 5.46, sin (1.552rad ) Q′′ sinζ1 ∗ θo = 1 − = 1− × 0.114 = 0.795 Q′′ 1.552rad ζ1 o We conclude that the energy to be removed from the stack per unit area to return it to Ti is Q′′ = 0.795Q′′ = 0.795 × 2.389 × 107 J/m2 = 1.90 × 107 J/m 2 . o < PROBLEM 5.44 KNOWN: Car windshield, initially at a uniform temperature of -20°C, is suddenly exposed on its interior surface to the defrost system airstream at 30°C. The ice layer on the exterior surface acts as an insulating layer. FIND: What airstream convection coefficient would allow the exterior surface to reach 0°C 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): ρ = 2200 kg/m3, cp = 830 J/kg⋅K and k = 1.2 W/m⋅K. ANALYSIS: For the prescribed conditions, from Equations 5.31 and 5.33, θ (0, 60s ) θ o T (0, 60s ) − T∞ (0 − 30 ) C = 0.6 = = = θi θi Ti − T∞ ( −20 − 30 ) C Fo = kt ρ cL2 = 1.2 W m ⋅ K × 60 2200 kg m3 × 830 J kg ⋅ K × (0.005 m ) 2 = 1.58 The single-term series approximation, Eq. 5.41, along with Table 5.1, requires an iterative solution to find an appropriate Biot number. Alternatively, the Heisler charts, Appendix D, Figure D.1, for the midplane temperature could be used to find Bi −1 = k hL = 2.5 h = 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 coefficient can be determined by solving the model with an assumed h and then sweeping over a range of h until the T(0,60s) condition is satisfied. Since the model is based upon multiple terms of the series, the result of h = 99 W/m2⋅K is more precise than that found using the chart. PROBLEM 5.45 KNOWN: Thickness, initial temperature and properties of plastic coating. Safe-to-touch temperature. Convection coefficient and air temperature. FIND: Time for surface to reach safe-to-touch temperature. Corresponding temperature at plastic/wood interface. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in coating, (2) Negligible radiation, (3) Constant properties, (4) Negligible heat of reaction, (5) Negligible heat transfer across plastic/wood interface. 2 ANALYSIS: With Bi = hL/k = 200 W/m ⋅K × 0.002m/0.25 W/m⋅K = 1.6 > 0.1, the lumped capacitance method may not be used. Applying the approximate solution of Eq. 5.40a, with C1 = 1.155 and ζ1 = 0.990 from Table 5.1, ( )( ) ( 42 − 25 ) °C ∗ T − T∞ ∗ 2 = = 0.0971 = C1 exp −ζ 1 Fo cos ζ 1x = 1.155exp ( −0.980 Fo ) cos ( 0.99 ) θs = s Ti − T∞ ( 200 − 25 ) °C Hence, for x∗ = 1, 0.0971 2 Fo = − ln / ( 0.99 ) = 1.914 1.155cos ( 0.99 ) Fo L2 1.914 ( 0.002m ) = = 63.8s −7 m 2 / s α 1.20 × 10 2 t= < From Eq. 5.41, the corresponding interface temperature is ( ) 2 To = T∞ + ( Ti − T∞ ) exp −ζ1 Fo = 25°C + 175°C exp ( −0.98 × 1.914 ) = 51.8°C < COMMENTS: By neglecting conduction into the wood and radiation from the surface, the cooling time is overpredicted and is therefore a conservative estimate. However, if energy generation due to solidification of polymer were significant, the cooling time would be longer. PROBLEM 5.46 KNOWN: 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/m⋅K, ρ = 7832 kg/m , 3 -5 2 cp = 559 J/kg⋅K, α = (k/ρcp) = 1.11×10 m /s. ANALYSIS: The time needed to traverse the rod through the oven may be found from Fig. D.4. 600 − 750 ∗ T −T θo = o ∞ = = 0.214 Ti − T∞ 50 − 750 k 48.8 W/m ⋅ K Bi-1 ≡ = = 15.6. hro 125 W/m 2 ⋅ K (0.025m ) Hence, 2 Fo = α t/ro ≈ 12.2 2 t = 12.2 ( 0.025m ) /1.11× 10−5 m 2 / s = 687 s. The rod velocity is V= L 5m = = 0.0073 m/s. t 687s COMMENTS: (1) Since (h ro/2)/k = 0.032, the lumped capacitance method could have been used. From Eq. 5.5 it follows that t = 675 s. (2) Radiation effects decrease t and hence increase V, assuming there is net radiant transfer from 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 ζ1 = 0.3549 rad and C1 = 1.0158. Hence from Eq. 5.49c θ ∗ ln o = 12.4, C1 which is in good agreement with the graphical result. () 2 Fo = − ζ1 −1 PROBLEM 5.47 KNOWN: Hot dog with prescribed thermophysical properties, initially at 6°C, is immersed in boiling water. FIND: Time required to bring centerline temperature to 80°C. SCHEMATIC: ASSUMPTIONS: (1) Hot dog can be treated as infinite cylinder, (2) Constant properties. ANALYSIS: The Biot number, based upon Eq. 5.10, is 2 -3 h Lc h ro / 2 100 W/m ⋅ K 10 × 10 m/2 Bi ≡ = = = 0.96 ( k 0.52 W/m ⋅ K k ) Since Bi > 0.1, a lumped capacitance analysis is not appropriate. Using the Heisler chart, Figure D.4 with hr 100W/m 2 ⋅ K × 10 × 10-3m Bi ≡ o = = 1.92 or Bi-1 = 0.52 0.52 W/m ⋅ K k T ( 0,t ) − T∞ (80 − 100 ) C ∗θ θo = o = = = 0.21 Ti − T∞ θi (6-100 ) C and (1) (10 ×10-3m ) 2 find Fo = t∗ = αt 2 ro 2 ro t = ⋅ Fo = = 0.8 α 1.764 × 10−7 m 2 / s < × 0.8 = 453.5s = 7.6 min α = k/ρ c = 0.52 W/m ⋅ K/880 kg/m3 × 3350 J/kg ⋅ K = 1.764 × 10−7 m 2 / s. where COMMENTS: (1) Note that Lc = ro/2 when evaluating the Biot number for the lumped capacitance analysis; 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 θ(1,t)/θo ≈ 0.45. From Eq. (1), note that θo = 0.21 θi giving θ (1, t ) = T ( ro , t ) − T∞ = 0.45θ o = 0.45 (0.21[Ti − T∞ ]) = 0.45 × 0.21[6 − 100] C = −8.9C T ( ro , t ) = T∞ − 8.9 C = (100 − 8.9 ) C = 91.1C (3) Since Fo ≥ 0.2, the approximate solution for θ*, Eq. 5.49, is valid. From Table 5.1 with Bi = 1.92, find that ζ1 = 1.3245 rad and C1 = 1.2334. Rearranging Eq. 5.49 and substituting values, Fo = − ( ) ∗ ln θ o / C1 = 2 1 ζ1 0.213 ln = 1.00 2 1.2334 (1.3245 rad ) 1 This result leads to a value of t = 9.5 min or 20% higher than that of the graphical method. PROBLEM 5.48 KNOWN: Long rod with prescribed diameter and properties, initially at a uniform temperature, is heated in a forced convection furnace maintained at 750 K with a convection coefficient of h = 1000 W/m2⋅K. 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 approximate solution for the infinite cylinder is appropriate. From Eq. 5.49b, ( ) () * θ * r* , Fo = θ o (Fo ) J 0 ζ1r* (1) where, for r* = 1, the dimensionless temperatures are, from Eq. 5.31, θ * (1, Fo ) = T ( ro , t o ) − T∞ * θ o ( Fo ) = Ti − T∞ T (0, t o ) − T∞ Ti − T∞ (2,3) Combining Eqs. (2) and (3) with Eq. (1) and rearranging, T ( ro , t o ) − T∞ Ti − T∞ = T ( 0, t o ) = T∞ + T (0, t o ) − T∞ Ti − T∞ J 0 (ζ1 ⋅1) 1 T ( ro , t o ) − T∞ J 0 (ζ1 ) (4) The eigenvalue, ζ1 = 1.0185 rad, follows from Table 5.1 for the Biot number Bi = 2 hro 1000 W m ⋅ K (0.060 m 2 ) = = 0.60 . k 50 W m ⋅ K From Table B-4, with ζ1 = 1.0185 rad, J0(1.0185) = 0.7568. Hence, from Eq. (4) T ( 0, t o ) = 750 K + 1 [550 − 750] K = 486 K 0.7568 < (b) Using the IHT Transient Conduction Model for a Cylinder, the following temperature histories were generated. Continued... PROBLEM 5.48 (Cont.) Centerline temperature, To(K) 500 400 300 0 100 200 300 400 Time, t(s) h = 100 W/m^2.K h = 500 W/m^2.K h = 1000 W/m^2.K The times required to reach a centerline temperature of 500 K are 367, 85 and 51s, respectively, for h = 100, 500 and 1000 W/m2⋅K. The corresponding values of the Biot number are 0.06, 0.30 and 0.60. Hence, even for h = 1000 W/m2⋅K, the convection resistance is not negligible relative to the conduction resistance and significant reductions in the heating time could still be effected by increasing h to values considerably in excess of 1000 W/m2⋅K. COMMENTS: For Part (a), recognize why it is not necessary to know Ti or the time to. We require that Fo ≥ 0.2, which for this sphere corresponds to t ≥ 14s. For this situation, the time dependence of the surface and center are the same. PROBLEM 5.49 KNOWN: 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 surface temperature 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/m2⋅K, Bic = 2 hLc h ( ro 2 ) 50 W m ⋅ K ( 0.015 m / 2 ) = = = 0.221 . k k 1.7 W m ⋅ K Since Bic > 0.1, method is not suited. Using the approximate series solution for the infinite cylinder, ) ( ( )() 2 θ * r* , Fo = C1 exp −ζ1 Fo × J 0 ζ1r* (1) Solving for Fo and setting r* = 1, find θ* ln 2 ζ1 C1J 0 (ζ1 ) T ( ro , t o ) − T∞ (500 − 350 ) K where θ * = (1, Fo ) = = = 0.231 . Ti − T∞ (1000 − 350 ) K Fo = − 1 From Table 5.1, with Bi = 0.441, find ζ1 = 0.8882 rad and C1 = 1.1019. From Table B.4, find J0(ζ1) = 0.8121. Substituting numerical values into Eq. (2), Fo = − 1 (0.8882 )2 ln [0.231 1.1019 × 0.8121] = 1.72 . 2 From the definition of the Fourier number, Fo = α t ro , and α = k/ρc, r2 2 ρc t = Fo o = Fo ⋅ ro α k t = 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 temperature histories were obtained. Continued... PROBLEM 5.49 (Cont.) Surface temperature, T(K) 1000 900 800 700 600 500 400 300 0 50 100 150 200 250 300 Time, t(s) h = 250 W/m^2.K h = 50 W/m^2.K Increasing 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/m2⋅K, Bi = 1.1 and 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 is appropriate. PROBLEM 5.50 KNOWN: Long pyroceram rod, initially at a uniform temperature of 900 K, and clad with a thin metallic 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 on cooling rate. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Thermal resistance and capacitance of metal tube are negligible, (3) Constant properties, (4) Fo ≥ 0.2. PROPERTIES: Table A-2, Pyroceram ( T = (600 + 900)K/2 = 750 K): ρ = 2600 kg/m3, c = 1100 J/kg⋅K, k = 3.13 W/m⋅K. ANALYSIS: (a) The thermal contact and convection resistances can be combined to give an overall heat transfer coefficient. Note that R ′ ,c [m⋅K/W] is expressed per unit length for the outer surface. Hence, t for h = 100 W/m2⋅K, U= 1 1 = = 57.0 W m 2 ⋅ K . 1 h + R ′ (π D ) 1 100 W m 2 ⋅ K + 0.12 m ⋅ K W (π × 0.020 m ) t,c Using the approximate series solution, Eq. 5.50c, the Fourier number can be expressed as 2 * Fo = − 1 ζ1 ln θ o C1 . ( )( ) From Table 5.1, find ζ1 = 0.5884 rad and C1 = 1.0441 for Bi = Uro k = 57.0 W m 2 ⋅ K ( 0.020 m 2 ) 3.13 W m ⋅ K = 0.182 . The dimensionless temperature is * θ o (0, Fo ) = T ( 0, t ) − T∞ Ti − T∞ = (600 − 300 ) K = 0.5. (900 − 300 ) K Substituting numerical values to find Fo and then the time t, Fo = −1 (0.5884 ) 2 ln 0.5 = 2.127 1.0441 r2 2 ρc t = Fo o = Fo ⋅ ro α k t = 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 a Cylinder. Continued... PROBLEM 5.50 (Cont.) 900 800 800 Centerline temperature, (K) Surface temperature, (K) 900 700 600 500 400 300 700 600 500 400 300 0 50 100 150 Time, t(s) r = ro, h = 100 W/m^2.K r = ro, h = 500 W/m^2.K r = ro, h = 1000 W/m^2.K 200 250 300 0 50 100 150 200 250 300 Time, t(s) r = 0, h = 100 W/m^2.K r = 0, h = 500 W/m^2.K r = 0, h = 1000 W/m^2.K While enhanced cooling is achieved by increasing h from 100 to 500 W/m2⋅K, there is little benefit associated with increasing h from 500 to 1000 W/m2⋅K. The reason is that for h much above 500 W/m2⋅K, the contact resistance becomes the dominant contribution to the total resistance between the fluid and the rod, rendering the effect of further reductions in the convection resistance negligible. Note that, for h = 100, 500 and 1000 W/m2⋅K, the corresponding values of U are 57.0, 104.8 and 117.1 W/m2⋅K, respectively. COMMENTS: For Part (a), note that, since Fo = 2.127 > 0.2, Assumption (4) is satisfied. PROBLEM 5.51 KNOWN: Sapphire rod, initially at a uniform temperature of 800K is suddenly cooled by a convection process; 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 losses from the rod when insulation is applied. 3 PROPERTIES: Table A-2, Aluminum oxide, sapphire (550K): ρ = 3970 kg/m , c = 1068 J/kg⋅K, k = -5 22.3 W/m⋅K, α = 5.259×10 2 m /s. ANALYSIS: First calculate the Biot number with Lc = ro/2, 2 h Lc h ( ro / 2 ) 1600 W/m ⋅ K ( 0.020 m/2) Bi = k = = k 22.3 W/m ⋅ K = 0.72. Since Bi > 0.1, the rod cannot be approximated as a lumped capacitance system. The temperature distribution during the cooling process, 0 ≤ t ≤ 35s, and for the time following the application of insulation, t > 35s, will appear as Eventually (t → ∞), the temperature of the rod will be uniform at T ( ∞ ) . To find T (∞ ) , write the conservation 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 becomes −Q = ρ cV ( T ( ∞ ) − T∞ ) − Qo where Qo = ρ cV(Ti - T∞). Dividing through by Qo and solving for T (∞ ) , find T ( ∞ ) = T∞ + ( Ti − T∞ ) (1 − Q/Qo ) . From the Groeber chart, Figure D.6, with hr 1600 W/m 2 ⋅ K × 0.020m Bi = o = = 1.43 k ( 22.3 W/m ⋅ K ) ( ) 2 Bi 2Fo = Bi 2 α 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 5.52 KNOWN: Long bar of 70 mm diameter, initially at 90° C, is suddenly immersed in a water bath 2 (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 to equilibrate while isolated from surroundings, it will have a uniform temperature T(r, ∞) = 55° C. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties. 3 PROPERTIES: Bar (given): ρ = 2600 kg/m , c = 1030 J/kg⋅K, k = 3.50 W/m⋅K, α = k/ρc = -6 2 1.31×10 m /s. ANALYSIS: Determine first whether conditions are space-wise isothermal 2 hLc h ( ro / 2) 20 W/m ⋅ K ( 0.035 m/2) Bi = = = = 0.10 k k 3.50 W/m ⋅ K and 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 is in the bath). Ein − Eout = ∆E 0 − Q = E final − Einitial = Mc ( Tf − T∞ ) − Mc ( Ti − T∞ ) −Q = Mc ( Tf − T∞ ) − Q o Q T − T∞ ( 55 − 40 )o C = 0.70 = 1− f = 1− Qo Ti − T∞ ( 90 − 40 )o C where Qo is the initial energy in the bar (relative to T∞; Eq. 5.44). With Bi = hro/k = 0.20 and 2 2 Q/Qo = 0.70, use Figure D.6 to find Bi Fo = 0.15; hence Fo = 0.15/Bi = 3.75 and 2 t f = Fo ⋅ r2 / α = 3.75 ( 0.035 m ) /1.31×10− 6 m 2 / s = 3507 s. o < -1 (b) To determine T(ro, tf), use Figures D.4 and D.5 for θ(ro,t)/θi (Fo = 3.75, Bi = 5.0) and θo/θi -1 (Bi = 5.0, r/ro = 1, respectively, to find T (ro , t f ) = T∞ + θ ( ro , t ) θ o ⋅ ⋅ θ i = 40o C + 0.25 × 0.90 ( 90 − 50 )o C = 49oC. θo θi < PROBLEM 5.53 KNOWN: Long plastic rod of diameter D heated uniformly in an oven to Ti and then allowed to convectively cool in ambient air (T∞, h) for a 3 minute period. Minimum temperature of rod should not be less than 200° C and the maximum-minimum temperature within the rod should not exceed 10° C. FIND: Initial uniform temperature Ti to which rod should be heated. Whether the 10° C internal temperature difference is exceeded. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Uniform and constant convection coefficients. 3 PROPERTIES: Plastic rod (given): k = 0.3 W/m⋅K, ρcp = 1040 kJ/m ⋅K. ANALYSIS: For the worst case condition, the rod cools for 3 minutes and its outer surface is at least 200° C in order that the subsequent pressing operation will be satisfactory. Hence, hro 8 W/m 2 ⋅K × 0.015 m Bi = = = 0.40 k 0.3 W/m ⋅ K αt k t 0.3 W/m ⋅ K 3 × 60s Fo = = ⋅= × = 0.2308. 2 2 ρ cp ro 1040 × 103 J/m3 ⋅ K ( 0.015 m )2 ro Using Eq. 5.49a and ζ1 = 0.8516 rad and C1 = 1.0932 from Table 5.1, T (ro , t ) − T∞ ∗ 2 θ∗ = = C1J 0 ζ1 ro exp −ζ1 Fo . Ti − T ()( ) ∗ With ro = 1, from Table B.4, J 0 (ζ1 ×1) = J o ( 0.8516 ) = 0.8263, giving 200 − 25 = 1.0932 × 0.8263exp −0.85162 × 0.2308 Ti = 254o C. < Ti − 25 At this time (3 minutes) what is the difference between the center and surface temperatures of the rod? From Eq. 5.49b, θ ∗ T (ro , t ) − T∞ 200 − 25 ∗ = = = J 0 ζ1ro = 0.8263 θo T ( 0,t ) − T∞ T ( 0,t ) − 25 ) ( () which gives T(0,t) = 237° C. Hence, o ∆T = 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 made much smaller. Two solutions are (a) increase ambient air temperature and (b) non-uniformly heat rod in oven by controlling its residence time. PROBLEM 5.54 KNOWN: Diameter and initial temperature of roller bearings. Temperature of oil bath and convection 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. ( ) 3 PROPERTIES: Table A.1, St. St. 304 T = 548 K : ρ=7900 kg/m , k = 19.0 W/m⋅K, cp = 546 -6 2 J/kg⋅K, α = 4.40 × 10 m /s. ANALYSIS: With Bi = h (ro/2)/k = 0.658, the lumped capacitance method can not be used. From the one-term approximation of Eq. 5.49 c for the centerline temperature, ( ) 50 − 30 2 2 ∗ T −T θo = o ∞ = = 0.0426 = C1 exp −ζ1 Fo = 1.1382 exp − ( 0.9287 ) Fo Ti − T∞ 500 − 30 where, for Bi = hro/k = 1.316, C1 = 1.1382 and ζ1 = 0.9287 from Table 5.1. Fo = −n ( 0.0374 ) / 0.863 = 3.81 2 t f = Fo ro / α = 3.81( 0.05 m ) / 4.40 × 10−6 = 2162s = 36 min 2 < From Eqs. 5.44 and 5.51, the energy extracted from a single rod is 2θ ∗ Q = ρ cV ( Ti − T∞ ) 1 − o J1 (ζ1 ) ζ1 With J1 (0.9287) = 0.416 from Table B.4, 0.0852 × 0.416 2 7 Q = 7900 kg / m3 × 546 J / kg ⋅ K π ( 0.05m ) 1m 470 K 1 − = 1.53 × 10 J 0.9287 The nominal cooling load is q= N Q 10 × 1.53 × 107 J = = 70,800 W = 7.08 kW tf 2162s COMMENTS: For a centerline temperature of 50°C, Eq. 5.49b yields a surface temperature of ∗ T ( ro , t ) = T∞ + ( Ti − T∞ )θ o Jo (ζ1 ) = 30°C + 470°C × 0.0426 × 0.795 = 45.9°C < PROBLEM 5.55 KNOWN: Long rods of 40 mm- and 80-mm diameter at a uniform temperature of 400°C in a curing oven, are removed and cooled by forced convection with air at 25°C. The 40-mm diameter rod takes 280 s to reach a safe-to-handle temperature of 60°C. 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) Constant properties, and (3) Convection coefficient same value for both rods. 3 PROPERTIES: Rod (given): ρ = 2500 kg/m , c = 900 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: Not knowing the convection coefficient, the Biot number cannot be calculated to determine whether the rods behave as spacewise isothermal objects. Using the relations from Section 5.6, Radial Systems with Convection, for the infinite cylinder, Eq. 5.50, evaluate 2 Fo = α t / ro , and knowing T(ro, to), a trial-and-error solution is required to find Bi = h ro/k and hence, h. Using the IHT Transient Conduction model for the Cylinder, the following results are readily calculated for the 40-mm rod. With to = 280 s, Fo = 4.667 Bi = 0.264 h = 197.7 W / m 2 ⋅ K For the 80-mm rod, with the foregoing value for h, with T(ro, to) = 60°C, find Bi = 0.528 Fo = 2.413 t o = 579 s < COMMENTS: (1) The time-to-cool, to, for the 80-mm rod is slightly more than twice that for the 40-mm rod. Did you anticipate this result? Did you believe the times would be proportional to the diameter squared? (2) The simplest approach to explaining the relationship between to and the diameter follows from the lumped capacitance analysis, Eq. 5.13, where for the same θ/θi, we expect Bi⋅Foo to be a constant. That is, h ⋅ ro α t o × =C 2 k ro 2 yielding to ~ ro (not ro ). PROBLEM 5.56 KNOWN: Initial temperature, density, specific heat and diameter of cylindrical rod. Convection coefficient 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 and temperature histories for selected velocities. SCHEMATIC: ASSUMPTIONS: (1) Lumped capacitance analysis can not be used but one-term approximation for an 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 θ o =[To(0,1136s) - T∞]/(Ti - T∞) = (40 – 25)/(100 – 25) = 0.20, Eq. 5.49c yields Fo = αt 2 ro = kt 2 ρ c p ro = k (1136 s ) 1200 kg / m × 1250 J / kg ⋅ K × ( 0.02 m ) 2 3 2 = − ln ( 0.2 / C1 ) / ζ1 (1) Because C1 and ζ1 depend on Bi = hro/k, a trial-and-error procedure must be used. For example, a value of k may be assumed and used to calculate Bi, which may then be used to obtain C1 and ζ1 from Table 5.1. Substituting C1 and ζ1 into Eq. (1), k may be computed and compared with the assumed value. Iteration continues until satisfactory convergence is obtained, with < k ≈ 0.30 W / m ⋅ K and, hence, Bi = 3.67, C1 = 1.45, ζ1 = 1.87 and Fo = 0.568. For the above value of k, 2 − ln (0.2 / C1 ) / ζ1 = 0.567, which equals the Fourier number, as prescribed by Eq. (1). 2 0.618 0.618 2.618 yields a value of C = 16.8 W⋅s /m (b) With h = 55 W/m ⋅K for V = 6.8 m/s, h = CV ⋅K. 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.) 100 C 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 ) 50 45 40 35 30 0 5 10 15 75 50 25 0 20 500 1000 Tim e , t(s ) Air ve lo city, V(m /s ) V=3 m /s V=1 0 m /s V=2 0 m /s 2 With increasing V from 3 to 20 m/s, h increases from 33 to 107 W/m ⋅K, and the enhanced cooling reduces the centerline temperature at the prescribed time. The accelerated cooling associated with increasing V is also revealed by the temperature histories, and the time required to achieve thermal equilibrium between the air and the cylinder decreases with increasing V. 2 COMMENTS: (1) For the smallest value of h = 33 W/m ⋅K, Bi ≡ h (ro/2)/k = 1.1 >> 0.1, and use of the 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 thermal conductivity for the conditions of Part (a), convergence is elusive and may only be achieved if the initial guesses are close to the correct results. 1500 PROBLEM 5.57 KNOWN: Diameter, initial temperature and properties of stainless steel rod. Temperature and convection 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. The results are plotted as follows h =1 0 0 0 W /m ^2 -K h =1 0 0 W /m ^2 -K 325 325 275 Te m p e re a tu re , C Te m p e ra tu re , C 275 225 175 125 225 175 125 75 75 25 25 0 0 .2 0 .4 0 .6 0 .8 0 1 0 .2 0 .8 1 0 .8 1 t=0 s t=1 0 s t=5 0 s t= 0 t= 1 0 0 s t= 5 0 0 s h =5 0 0 0 W /m ^2 -K 2 larger than [T (ro,t) - T∞]. 0 .6 D 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* 325 275 Te m p e ra tu re , C For h = 100 W/m ⋅K, Bi = hro/k = 0.1, and as expected, the temperature distribution is nearly uniform throughout the rod. For h = 1000 2 W/m ⋅K (Bi = 1), temperature variations within the rod are not negligible. In this case the centerline-to-surface temperature difference is comparable to the surface-to-fluid 2 temperature difference. For h = 5000 W/m ⋅K (Bi = 5), temperature variations within the rod are large and [T (0,t) – T (ro,t)] is substantially 0 .4 225 175 125 75 25 0 0 .2 0 .4 0 .6 D im e n s io n le s s ra d iu s , r* t=0 s t=1 s t=5 s t=2 5 s COMMENTS: With increasing Bi, conduction within the rod, and not convection from the surface, becomes the limiting process for heat loss. PROBLEM 5.58 KNOWN: A ball bearing is suddenly immersed in a molten salt bath; heat treatment to harden occurs at 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 layer will be hardened. Begin by finding the Biot number. 5000 W/m2 ⋅ K ( 0.020m/2 ) hr Bi = o = = 1.00. 50 W/m ⋅ K k Using the one-term approximate solution for a sphere, find Fo = − () 1 ln θ ∗ / C1 sin ζ1r∗ 2 ∗ ζ1 ζ1r 1 . From Table 5.1 with Bi = 1.00, for the sphere find ζ 1 = 15708 rad and C1 = 1.2732. With r* . = r/ro = (9mm/10mm) = 0.9, substitute numerical values. (1000 − 1300 ) K 1 −1 Fo = ln /1.2732 sin (1.5708 × 0.9 rad ) = 0.441. 1.5708 × 0.9 (1.5708)2 (300 − 1300 ) K From the definition of the Fourier number with α = k/ρc, 2 2 ro 2 ρ c = 0.441× 0.020m 7800 kg × 500 t = Fo = Fo ⋅ ro 3 α k 2 m J / 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 easily shown the center temperature is T(0,3.4s) = 871 K. (2) The Heisler charts can also be used. From Fig. D.8, with Bi 0.69(±0.03). Since θ = T − T∞ = 1000 − 1300 = −300K θ θ θo = ⋅, θi θo θi θ o / θ i = 0.30 / 0.69 = 0.43 ( ±0.02 ). and Since = 1.0 and r/ro = 0.9, read θ/θo = θ i = Ti − T∞ = −1000K it follows that θ = 0.30. θi -1 then θ θ = 0.69 o θi θi -1 From Fig. D.7 at θo/θi=0.43, Bi =1.0, read Fo = 0.45 (±0.03) and t = 3.5 (±0.2)s. Note the use of tolerances associated with reading the charts to ±5%. PROBLEM 5.59 KNOWN: An 80mm sphere, initially at a uniform elevated temperature, is quenched in an oil bath with prescribed T∞, h. FIND: The center temperature of the sphere, T(0,t) at a certain time when the surface temperature is T(ro,t) = 150° C. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Initial uniform temperature within sphere, (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/3 hL Bi c = c = = = 0.26. k k 50 W/m ⋅ K Since Bic > 0.1, lumped capacitance method is not appropriate. Recognize that when Fo ≥ 0.2, the time 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 first the Biot number, hro 1000 W/m 2 ⋅ K × 0.040m = = 0.80. k 50 W/m ⋅ K -1 With Bi = 1/0.80 = 1.25 and r/ro =1, read from Fig. D.8, Bi = T ( ro , t ) − T∞ θ = = 0.67. θ o T ( 0,t ) − T∞ It follows that T ( 0,t ) = T∞ + 1 1 T (ro , t ) − T∞ = 50o C + [150 − 50 ]o C = 199 oC. 0.67 0.67 < COMMENTS: (1) There is sufficient information to evaluate Fo; hence, we require that the time be 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 Table 5.1, ζ1 = 1.5044 rad. Substituting numerical values, r* = 1, θ∗ ∗ θo = T (ro , t ) − T∞ T ( 0,t ) − T∞ = It follows that T(0,t) = 201° C. () 1 sin ζ1r ∗ = sin (1.5044 rad ) = 0.663. 1.5044 ζ1r∗ 1 PROBLEM 5.60 KNOWN: Steel ball bearings at an initial, uniform temperature are to be cooled by convection while passing through a refrigerated chamber; bearings are to be cooled to a temperature such that 70% of the thermal energy is removed. FIND: Residence time of the balls in the 5m-long chamber and recommended drive velocity for the conveyor. SCHEMATIC: ASSUMPTIONS: (1) Negligible conduction between ball and conveyor surface, (2) Negligible radiation exchange with surroundings, (3) Constant properties, (4) Uniform convection coefficient over ball’s surface. ANALYSIS: The Biot number for the lumped capacitance analysis is 2 hLc h ( ro / 3) 1000 W/m ⋅ K (0.1m/3) Bi ≡ = = = 0.67. k k 50 W/m ⋅ K Since Bi > 0.1, lumped capacitance analysis is not appropriate. In Figure D.9, the internal energy change is shown as a function of Bi and Fo. For Q = 0.70 Qo hro 1000 W/m 2 ⋅ K × 0.1m Bi = = = 2.0, k 50 W/m ⋅ K and 2 find Bi Fo ≈ 1.2. The Fourier number is Fo = αt 2 ro = 2 × 10−5 m 2 / s × t (0.1 m ) 2 = 2.0 × 10−3 t giving t= Fo 2.0 × 10-3 = 1.2 / Bi 2 2.0 × 10-3 1.2 / ( 2.0 ) 2 = 2.0 × 10−3 = 150s. The velocity of the conveyor is expressed in terms of the length L and residence time t. Hence V= L 5m = = 0.033m/s = 33mm/s. t 150s COMMENTS: Referring to Eq. 5.10, note that for a sphere, the characteristic length is 43 2r Lc = V/As = π ro / 4π ro = o . 3 3 However, when using the Heisler charts, note that Bi ≡ h ro/k. < PROBLEM 5.61 KNOWN: 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/m⋅K, cp = 579 J/kg⋅K, 3 -6 2 ρ = 7900 kg/m , α = 4.85×10 m /s. ANALYSIS: (a) To determine whether use of the lumped capacitance method is suitable, first compute h ( ro / 3) 1000 W/m 2 ⋅ K ( 0.010m/3) Bi = = = 0.15. k 22.2 W/m ⋅ K We 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, with Bi -1 = k 22.2 W/m ⋅ K = = 2.22 hro 1000 W/m 2 ⋅ K ( 0.01m ) and r = 1, ro Fig. D.8 gives θ ( ro , t ) ≈ 0.80. θo (t ) Hence, with θ ( ro , t ) θi = T (ro , t ) − T∞ Ti − T∞ = 100 − 40 = 0.074, 850 − 40 Continued ….. PROBLEM 5.61 (Cont.) it follows that θ o θ ( ro , t ) / θ i 0.074 = = = 0.093. θ i θ ( ro , t ) / θ o 0.80 From Fig. D.7, with θ o / θ i = 0.093 and Bi -1 = k/hro = 2.22, find t ∗ = Fo ≈ 2.0 2 ( 0.01m )2 ( 2.0 ) ro Fo t= = = 41s. α 4.85 × 10−6 m 2 / s < Also, θ o = To − T∞ = 0.093( Ti − T∞ ) = 0.093 ( 850 − 40 ) = 75o C To = 115o C 2 < 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 for a single ball Q ≈ 0.93. Qo Hence, from Eq. 5.44, Q = 0.93 ρc pV ( Ti − T∞ ) Q = 0.93× 7900 kg/m 3 × 579 J/kg ⋅ K × Q = 1.44 × 104J π ( 0.02m )3 × 810o C 6 is the amount of energy transferred from a single ball during the cooling process. Hence, the oil bath cooling rate must be q = 104 Q/3600s q = 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 the fact that To - T(ro) = 15° C at the conclusion, suggests that use of the lumped capacitance method would have been reasonable. Note that, when using the Heisler charts, accuracy to better than 5% is seldom possible. PROBLEM 5.62 KNOWN: 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. 3 PROPERTIES: Table A-3, Ice (253K): ρ = 920 kg/m , k = 2.03 W/m⋅K, cp = 1945 J/kg⋅K; α -6 2 = k/ρcp = 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. k 2.03 W/m ⋅ K Since Bi > 0.1, use the Heisler charts for which θ ( ro , t m ) T ( ro , tm ) − T∞ 0− 5 = = = 0.143 θi Ti − T∞ −30 − 5 k 2.03 W/m ⋅ K Bi -1 = = = 3.25. hro 250 W/m 2 ⋅K × 0.0025m Bi = From Fig. D.8, find θ ( ro , t m ) ≈ 0.86. θo ( t m ) It follows that θo ( tm ) θ ( ro , t m ) / θ i 0.143 = ≈ ≈ 0.17. θi θ ( ro , t m ) / θ o ( t m ) 0.86 From Fig. D.7 find Fo ≈ 2.1. Hence, 2 2.1 ( 0.0025) 2 Fo ro tm ≈ = = 12s. α 1.13 ×10−6 m2 / s < (b) Since (θo/θi) ≈ 0.17, find To − T∞ ≈ 0.17 ( Ti − T∞ ) ≈ 0.17 ( −30 − 5 ) ≈ −6.0o C To ( 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 = ρ Vc pθ i = 920 kg/m 3 ) (π /6)(0.005m )31945 (J/kg ⋅K )( −35K ) = −4.10 J Q = 0.82 Qo = 0.82 ( −4.10 J ) = −3.4 J. < PROBLEM 5.63 KNOWN: 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 surface temperature histories. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Uniform initial temperature. ANALYSIS: (a) Calculate Biot number to determine if sphere behaves as spatially isothermal object, 2 hLc h ( ro 3) 75 W m ⋅ K ( 0.015m 3) Bi = = = = 0.22 . k k 1.7 W m ⋅ K Hence, 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/m⋅K/(75 W/m2⋅K × 0.015 m) = 1.51, θ (1, t ′ ) θ o ≈ 0.72 . In order to enter Fig. D.7, we need to determine θo/θi, which is θ o θ (1, t ′ ) θ (1, t ′ ) ( 415 − 320 ) K 0.72 = 0.275 = ≈ θi θo θi (800 − 320 ) K 2 Hence, for Bi-1 = 1.51, Fo ≡ α t ′ ro ≈ 0.87 and 2 ρ cp 2 ro 400 kg m3 × 1600 J kg ⋅ K 2 t ′ = Fo = Fo ⋅ ⋅ ro ≈ 0.87 × ( 0.015m ) = 74s α k 1.7 W m ⋅ K (c) The heat flux at the outer surface at time t T is given by Newton’s law of cooling q′′ = 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 the Groeber chart, Fig. D.9. With Bi = 1/1.51 = 0.67 and Bi2Fo = (1/1.51)2 × 0.87 ≈ 0.4, the chart yields Q Qo ≈ 0.75 . The energy loss by the sphere with V = (πD3)/6 is therefore ( ) Q ≈ 0.85Qo = 0.85 ρ π D3 6 cp ( Ti − T∞ ) ( ) Q ≈ 0.85 × 400 kg m3 π [0.030 m ] 6 1600 J kg ⋅ K (800 − 320 ) K = 3691J 3 < Continued... PROBLEM 5.63 (Cont.) (e) If at time t ′ the surface of the sphere is perfectly insulated, eventually the temperature of the sphere will be uniform at T(∞). Applying conservation of energy to the sphere over a time interval, Ein - Eout = ∆E ≡ Efinal - Einitial. Hence, -Q = ρcV[T(∞) - T∞] - Qo, where Qo ≡ ρcV[Ti - T∞]. Dividing by Qo and regrouping, we obtain T ( ∞ ) = 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 were generated. 800 90000 Heat flux, q''(ro,t) (W/m^2.K) Temperature, T(K) 700 600 500 400 300 0 50 100 60000 30000 150 0 Time, t (s) h = 75 W/m^2.K, r = ro h = 75 W/m^2.K, r = 0 h = 200 W/m^2.K, r = ro h = 200 W/m^2.K, r = 0 0 50 100 150 Time, t(s) h = 75 W/m^2.K h = 200 W/m^2.K The quenching process is clearly accelerated by increasing h from 75 to 200 W/m2⋅K and is virtually completed by t ≈ 100s for the larger value of h. Note that, for both values of h, the temperature difference [T(0,t) - T(ro,t)] decreases with increasing t. Although the surface heat flux for h = 200 W/m2⋅K is initially larger than that for h = 75 W/m2⋅K, the more rapid decline in T(ro,t) causes it to become smaller at t ≈ 30s. COMMENTS: 1. There is considerable uncertainty associated with reading Q/Qo from the Groeber chart, Fig. D.9, and it would be better to use the one-term approximation solutions of Section 5.6.2. With Bi = 0.662, from Table 5.1, find ζ1 = 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, the following results were obtained: t′ = 72.1 s; Q/Qo = 0.7745, and T(∞) = 428 K. PROBLEM 5.64 KNOWN: Two spheres, A and B, initially at uniform temperatures of 800K and simultaneously quenched in large, constant temperature baths each maintained at 320K; properties of the spheres and convection coefficients. FIND: (a) Show in a qualitative manner, on T-t coordinates, temperatures at the center and the outer surface for each sphere; explain features of the curves; (b) Time required for the outer surface of each sphere to reach 415K, (c) Energy gained by each bath during process of cooling spheres to a surface temperature of 415K. SCHEMATIC: Sphere A 150 ro (mm) 3 ρ (kg/m ) c (J/kg⋅K) k (W/m⋅K) 2 h (W/m ⋅K) 1600 400 170 5 Sphere B 15 400 1600 1.7 50 ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Uniform properties, (3) Constant convection coefficient. ANALYSIS: (a) From knowledge of the Biot number and the thermal time constant, it is possible to qualitatively represent the temperature distributions. From Eq. 5.10, with Lc = ro/3, find 5 W/m ⋅ K ( 0.150m/3 ) 2 Bi A = Bi = h ( ro / 3 ) k 170 W/m ⋅ K 50 W/m ⋅ K ( 0.015m/3 ) = 1.47 × 10 −3 (1) 2 Bi B = 1.7 W/m ⋅ K = 0.147 (2) The thermal time constant for a lumped capacitance system from Eq. 5.7 is 1 τ = ( ρ Vc ) hAs τ= ρ ro c 3h τB = τA = 1600 kg/m3 × ( 0.150m ) 400 J/kg ⋅ K 3 × 5 W/m 2 ⋅ K 400 kg/m3 × ( 0.015m )1600 J/kg ⋅ K 3 × 50 W/m 2 ⋅ K = 64s = 6400s (3) (4) When Bi << 0.1, the sphere will cool in a spacewise isothermal manner (Sphere A). For sphere B, Bi > 0.1, hence gradients will be important. Note that the thermal time constant of A is much larger than for B; hence, A will cool much slower. See sketch for these features. (b) Recognizing that BiA < 0.1, Sphere A can be treated as spacewise isothermal and analyzed using the lumped capacitance method. From Eq. 5.6 and 5.7, with T = 415 K θ T − T∞ = = exp ( − t/τ ) (5) θi Ti − T∞ Continued ….. PROBLEM 5.64 (Cont.) T − T∞ 415 − 320 t A = −τ 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 are approximately the same. Since BiB > 0.1, Sphere B must be treated by the Heisler chart method of solution beginning with Figure D.8. Using 2 hro 50 W/m ⋅ K × (0.015m ) Bi B ≡ = = 0.44 k 1.7 W/m ⋅ K Bi-1 = 2.27, B or find that for r/ro = 1, θ (1, t ) T ( ro , t ) − T∞ ( 415 − 320 ) = = = 0.8. θo θo θo (6) Using Eq. (6) and Figure D.7, find the Fourier number, θ o (T ( ro , t ) − T∞ ) / 0.8 ( 415 − 320 ) K/0.8 = = = 0.25 Ti − T∞ θi (800 − 320 ) K Fo = αt 2 ro = 1.3. 2 1.3 ( 0.015m ) Fo ro tB = = = 110s = 1.8 min α 2.656 10−6 m 2 / s 3 -6 2 where α = k/ρc = 1.7 W/m⋅K/400 kg/m × 1600 J/kg⋅K = 2.656×10 m /s. 2 < (c) To determine the energy change by the spheres during the cooling process, apply the conservation of energy requirement on a time interval basis. Sphere A: Ein − Eout = ∆E − Q A = ∆E = E ( t ) − E (0 ). QA = ρ cV T ( t ) − Ti = 1600kg/m3 × 400J/kg ⋅ K × ( 4/3)π (0.150m ) [415 − 800] K 3 QA = 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 = ∆E For the nonisothermal sphere, the Groeber chart, Figure D.9, can be used to evaluate QB. 2 2 With Bi = 0.44 and Bi Fo = (0.44) ×1.3 = 2.52, find Q/Qo = 0.74. The energy transfer from the sphere during the cooling process, using Eq. 5.44, is QB = 0.74 Qo = 0.74 ρ cV ( Ti − T∞ ) QB = 0.75 × 400kg/m3 × 1600J/kg ⋅ K ( 4/3)π (0.015m ) (800 − 320 ) K = 3257 J. 3 COMMENTS: (1) In summary: Sphere A B Bi = hro/k -3 4.41×10 0.44 τs $ t(s) Q(J) 6400 64 10,370 110 3.48×10 3257 6 < PROBLEM 5.65 KNOWN: Spheres of 40-mm diameter heated to a uniform temperature of 400°C are suddenly removed from an oven and placed in a forced-air bath operating at 25°C with a convection coefficient 2 of 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 and placed 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. 3 PROPERTIES: Sphere (given): ρ = 3000 kg/m , c = 850 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: (a) From Eq. 5.52, the fraction of thermal energy removed during the time interval ∆t = to is Q ∗3 = 1 − 3θ o / ζ1 sin (ζ1 ) − ζ1 cos (ζ1 ) Qo (1) where Q/Qo = 0.8. The Biot number is Bi = hro / k = 300 W / m 2 ⋅ K × 0.020 m /15 W / m ⋅ K = 0.40 and for the one-term series approximation, from Table 5.1, ζ1 = 1.0528 rad C1 = 1.1164 (2) ∗ The dimensionless temperature θ o , Eq. 5.31, follows from Eq. 5.50. ( 2 ∗ θ o = C1 exp −ζ1 Fo ) (3) 2 where Fo = α t o / ro . Substituting Eq. (3) into Eq. (1), solve for Fo and to. ) ( Q 2 3 = 1 − 3 C1 exp −ζ1 Fo / ζ1 sin (ζ1 ) − ζ1 cos (ζ1 ) 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 ) ρ cV (Ti − T∞ ) (6) and Ef represents the thermal energy in the sphere at t = ∞, ( Ef = ρ cV Tavg − T∞ ) (7) Combining the relations, find the average temperature ( ) ρ cV Tavg − T∞ − (1 − 0.8 )(Ti − T∞ ) = 0 Tavg = 100°C < PROBLEM 5.66 KNOWN: Diameter, density, specific heat and thermal conductivity of Pyrex spheres in packed 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 the corresponding center temperature. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction in sphere, (2) Negligible heat transfer to or from a sphere by radiation or conduction due to contact with adjoining spheres, (3) Constant properties. 2 ANALYSIS: With Bi ≡ h(ro/3)/k = 75 W/m ⋅K (0.0125m)/1.4 W/m⋅K = 0.67, the approximate solution for one-dimensional transient conduction in a sphere is used to obtain the desired results. We ∗ first use Eq. (5.52) to obtain θ o . ∗ θo = 3 ζ1 Q 1 − 3 sin (ζ1 ) − ζ1 cos (ζ1 ) Qo With Bi ≡ hro/k = 2.01, ζ1 ≈ 2.03 and C1 ≈ 1.48 from Table 5.1. Hence, 3 0.1( 2.03) 0.837 ∗= θo = = 0.155 3 0.896 − 2.03 ( −0.443) 5.386 The center temperature is therefore ( ) To = Tg,i + 0.155 Ti − Tg,i = 300°C − 42.7°C = 257.3°C From Eq. (5.50c), the corresponding time is θ∗ r2 t = − o ln o 2 αζ1 C1 ( < ) where α = k / ρ 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 2 7.54 × 10 () < COMMENTS: The surface temperature at the time of interest may be obtained from Eq. (5.50b). With r ∗ = 1, ∗ θ o sin (ζ1 ) 0.155 × 0.896 Ts = Tg,i + Ti − Tg,i = 300°C − 275°C = 280.9°C ζ1 2.03 ( ) < PROBLEM 5.67 KNOWN: Initial temperature and properties of a solid sphere. Surface temperature after immersion in a fluid of prescribed temperature and convection coefficient. FIND: (a) Time to reach surface temperature, (b) Effect of thermal diffusivity and conductivity on thermal response. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Constant properties. ANALYSIS: (a) For k = 15 W/m⋅K, the Biot number is Bi = 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 ζ1r* T − T∞ 2 Fo . = C1 exp −ζ1 Ti − T∞ ζ1r* ( At the surface, r* = 1. From Table 5.1, for Bi = 1.0, ζ1 = 1.5708 rad and C1 = 1.2732. Hence, ) ( 60 − 75 sin 90 = 0.30 = 1.2732 exp −1.57082 Fo 25 − 75 1.5708 exp(-2.467Fo) = 0.370 Fo = αt 2 ro = 0.403 2 ro (0.05 m ) = 100s t = 0.403 = 0.403 α 10−5 m 2 s 2 < (b) Using the IHT Transient Conduction Model for a Sphere to perform the parametric calculations, the effect of α is plotted for k = 15 W/m⋅K. Continued... PROBLEM 5.67 (Cont.) 75 65 65 Center temperature, T(C) Surface temperature, T(C) 75 55 45 35 25 55 45 35 25 0 50 100 150 200 250 300 0 50 100 Time, t(s) 150 200 250 300 Time, t(s) k = 15 W/m.K, alpha = 1E-4 m^2/s k = 15 W/m.K, alpha = 1E-5 m^2/s k = 15 W/m.K, alpha = 1E-6m^2/s k = 15 W/m.K, alpha = 1E-4 m^2/s k = 15 W/m.K, alpha = 1E-5 m^2/s k = 15 W/m.K, alpha = 1E-6 m^2/s For fixed k and increasing α, there is a reduction in the thermal capacity (ρcp) of the material, and hence the amount of thermal energy which must be added to increase the temperature. With increasing α, the material therefore responds more quickly to a change in the thermal environment, with the response at the center lagging that of the surface. The effect of k is plotted for α = 10-5 m2/s. 75 65 65 Center temperature, T(C) Surface temperature, T(C) 75 55 45 35 25 55 45 35 25 0 50 100 150 200 Time, t(s) k = 1.5 W/m.K, alpha = 1E-5 m^2/s k = 15 W/m.K, alpha = 1E-5 m^2/s k = 150W/m.K, alpha = 1E-5 m^2/s 250 300 0 50 100 150 200 250 300 Time, t(s) k = 1.5 W/m.K, alpha = 1E-5 m^2/s k = 15 W/m.K, alpha = 1E-5 m^2/s k =150 W/m.K, alpha = 1E-5m^2/s With increasing k for fixed alpha, there is a corresponding increase in ρcp, and the material therefore responds more slowly to a thermal change in its surroundings. The thermal response of the center lags that of the surface, with temperature differences, T(ro,t) - T(0,t), during early stages of solidification increasing with decreasing k. COMMENTS: Use of this technique to determine h from measurement of T(ro) at a prescribed t requires an interative solution of the governing equations. PROBLEM 5.68 KNOWN: Properties, initial temperature, and convection conditions associated with cooling of glass beads. FIND: (a) Time required to achieve a prescribed center temperature, (b) Effect of convection coefficient on center and surface temperature histories. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in r, (2) Constant properties, (3) Negligible radiation, (4) Fo ≥ 0.2. ANALYSIS: (a) With h = 400 W/m2⋅K, Bi ≡ h(ro/3)/k = 400 W/m2⋅K(0.0005 m)/1.4 W/m⋅K = 0.143 and the lumped capacitance method should not be used. From the one-term approximation for the center temperature, Eq. 5.50c, 80 − 15 * T −T 2 θo ≡ o ∞ = = 0.141 = C1 exp −ζ1 Fo Ti − T∞ 477 − 15 For Bi ≡ hro/k = 0.429, Table 5.1 yields ζ1 = 1.101 rad and C1 = 1.128. Hence, ) ( Fo = − θ* 1 0.141 ln o = − ln = 1.715 2 2 1.128 ζ1 C1 1.101) ( 1 2 t = 1.715ro ρ cp k 3 2 2200 kg m × 800 J kg ⋅ K = 1.715 ( 0.0015 m ) 1.4 W m ⋅ K < = 4.85s From Eq. 5.50b, the corresponding surface (r* = 1) temperature is * sin ζ1 = 15 C + 462 C 0.141 0.892 = 67.8 C T ( ro , t ) = T∞ + ( Ti − T∞ )θ o ζ1 1.101 (b) The effect of h on the surface and center temperatures was determined using the IHT Transient Conduction Model for a Sphere. ) ( 500 400 400 Surface temperature, T(C) Center temperature, T(C) 500 < 300 200 100 0 300 200 100 0 0 4 8 12 Time, t(s) h = 100 W/m^2.K, r = 0 h = 400 W/m^2.K, r = 0 h = 1000 W/m^2.K, r = 0 16 20 0 4 8 12 16 20 Time, t(s) h = 100 W/m^2.K, r = ro h = 400 W/m^2.K, r = ro h = 1000 W/m^2.K, r = ro Continued... PROBLEM 5.68 (Cont.) The cooling rate increases with increasing h, particularly from 100 to 400 W/m2⋅K. The temperature difference between the center and surface decreases with increasing t and, during the early stages of solidification, with decreasing h. COMMENTS: Temperature gradients in the glass are largest during the early stages of solidification and increase with increasing h. Since thermal stresses increase with increasing temperature gradients, the propensity to induce defects due to crack formation in the glass increases with increasing h. Hence, there is a value of h above which product quality would suffer and the process should not be operated. PROBLEM 5.69 KNOWN: Temperature requirements for cooling the spherical material of Ex. 5.4 in air and in a water bath. FIND: (a) For step 1, the time required for the center temperature to reach T(0,t) = 335°C while 2 cooling in air at 20°C with h = 10 W/m ⋅K; find the Biot number; do you expect radial gradients to be appreciable?; compare results with hand calculations in Ex. 5.4; (b) For step 2, time required for the 2 center temperature to reach T(0,t) = 50°C while cooling in water bath at 20°C 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 of the 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 temperatures evaluated 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 = 335°C with r = 0 and the Biot number are: t a = 94.2 s < Bi = 0.0025 Radial temperature gradients will not be appreciable since Bi = 0.0025 << 0.1. The sphere behaves as space-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 = 50°C with r = 0 and Ti = 335°C is < t w = 3.0 s Radial temperature gradients will be appreciable, since Bi = 1.5 >> 0.1. The sphere does not behave as 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 are calculated 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 2 Te m p e ra tu re , T(r,t) (C ) 400 300 200 100 0 0 1 2 3 4 5 6 Tim e , t (s ) S u rfa c e , r = ro C e n te r, r = 0 At early times, the difference between the center and surface temperature is appreciable. It is in this time region that thermal stresses will be a maximum, and if large enough, can cause fracture. Within 6 seconds, 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 gradient Bi Fo t T_xt Ti r ro 0.0025 25.13 94.22 335 400 0 0.005 */ // Models | Transient Conduction | Sphere - Step 1, Air cooling // The temperature distribution T(r,t) is T_xt = T_xt_trans("Sphere",rstar,Fo,Bi,Ti,Tinf) // Eq 5.47 T_xt = 335 // Surface temperature /* Results, for part (b), step 2, water cooling; Ti = 335 C Bi Fo t T_xt Ti r ro 1.5 0.7936 2.976 50 335 0 0.005 */ // Models | Transient Conduction | Sphere - Step 2, Water cooling // The temperature distribution T(r,t) is T_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 2 T_xt = 50 // Center temperature, end of Step 2 PROBLEM 5.70 KNOWN: Two large blocks of different materials – like copper and concrete – at room temperature, 23°C. FIND: Which block will feel cooler to the touch? SCHEMATIC: ASSUMPTIONS: (1) Blocks can be treated as semi-infinite solid, (2) Hand or finger temperature is 37°C. 3 PROPERTIES: Table A-1, Copper (300K): ρ = 8933 kg/m , c = 385 J/kg⋅K, k = 401 3 W/m⋅K; Table A-3, Concrete, stone mix (300K): ρ = 2300 kg/m , c = 880 J/kg⋅K, k = 1.4 W/m⋅K. ANALYSIS: Considering the block as a semi-infinite solid, the heat transfer situation corresponds to a sudden change in surface temperature, Case 1, Figure 5.7. The sensation of coolness is related to the heat flow from the hand or finger to the block. From Eq. 5.58, the surface heat flux is q′′ ( t ) = k ( Ts − Ti ) / (πα t ) s (1) q′′ ( t ) ~ ( kρ c ) s (2) 1/ 2 or 1/ 2 since α = k/ρ c. Hence for the same temperature difference, Ts − Ti , and elapsed time, it follows that the heat fluxes for the two materials are related as 1/ 2 W kg J 1/ 2 ( kρ c )copper 401 m ⋅ K × 8933 m3 × 385 kg ⋅ K q′′ s,copper = = = 22.1 1/2 1/ 2 ′′ qs,concrete ( kρ c ) W kg J concrete 1.4 m ⋅ K × 2300 3 × 880 kg ⋅ K m Hence, the heat flux to the copper block is more than 20 times larger than to the concrete block. The copper block will therefore feel noticeably cooler than the concrete one. PROBLEM 5.71 KNOWN: Asphalt pavement, initially at 50° C, is suddenly exposed to a rainstorm reducing the surface temperature to 20° C. 2 FIND: 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 of rainstorm is to suddenly reduce the surface temperature to 20° C and is maintained at that level for the period of interest. 3 PROPERTIES: Table A-3, Asphalt (300K): ρ = 2115 kg/m , c = 920 J/kg⋅K, k = 0.062 W/m⋅K. ANALYSIS: This solution corresponds to Case 1, Figure 5.7, and the surface heat flux is given by Eq. 5.58 as q′′ ( t ) = k ( Ts − Ti ) / ( πα t ) s 1/2 (1) The energy into the pavement over a period of time is the integral of the surface heat flux expressed as Q′′ = ∫ q ′′ ( t ) dt. s 0 t (2) Note that q′′ ( t ) is into the solid and, hence, Q represents energy into the solid. Substituting Eq. (1) s for q′′ ( t ) into Eq. (2) and integrating find s t Q′′ = k ( Ts − Ti ) / ( πα )1/2 ∫ t -1/2dt = 0 k ( Ts − Ti ) (πα ) 1/2 × 2 t1/2 . (3) Substituting numerical values into Eq. (3) with k 0.062 W/m ⋅ K α= = = 3.18 ×10 −8 m 2 / s ρ c 2115 kg/m 3 × 920 J/kg ⋅ K find that for the 30 minute period, 0.062 W/m ⋅ K ( 20 − 50 ) K Q′′ = × 2 ( 30 × 60s )1/2 = −4.99 ×105 J/m 2 . 1/2 π × 3.18 ×10-8m2 / s ( ) COMMENTS: Note that the sign for Q′′ is negative implying that energy is removed from the solid. < PROBLEM 5.72 KNOWN: Thermophysical properties and initial temperature of thick steel plate. Temperature of water 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 medium approximation, (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) = 50°C, T ( x, t ) − Ts Ti − Ts = (50 − 25) °C = 0.0909 = erf x (300 − 25) °C 2 αt x = 0.0807 2 αt t= x2 (0.0807 ) 4α 2 = (0.025m )2 ( 0.0261 1.34 × 10−5 m 2 / s 3 ) < = 1793s -5 2 where α = k/ρc = 50 W/m⋅K/(7800 kg/m × 480 J/kg⋅K) = 1.34 × 10 m /s. 4 2 COMMENTS: (1) Large values of the convection coefficient (h ~ 10 W/m ⋅K) are associated with water jet impingement, and it is reasonable to assume that the surface is immediately quenched to the temperature of the water. (2) The surface heat flux may be determined from Eq. (5.58). In principle, 1/2 the flux is infinite at t = 0 and decays as t . PROBLEM 5.73 KNOWN: Temperature imposed at the surface of soil initially at 20°C. See Example 5.5. FIND: (a) Calculate and plot the temperature history at the burial depth of 0.68 m for selected soil 7 2 thermal diffusivity values, α × 10 = 1.0, 1.38, and 3.0 m /s, (b) Plot the temperature distribution over -7 2 the depth 0 ≤ x ≤ 1.0 m for times of 1, 5, 10, 30, and 60 days with α = 1.38 × 10 m /s, (c) Plot the surface heat flux, q′′ ( 0, t ) , and the heat flux at the depth of the buried main, q′′ ( 0.68m, t ) , as a x x -7 2 function of time for a 60 day period with α = 1.38 × 10 m /s. Compare your results with those in the Comments section of the example. Use the IHT Models | Transient Conduction | Semi-infinite Medium model as the solution tool. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Soil is a semi-infinite medium, and (3) Constant properties. ANALYSIS: The IHT model corresponds to the case 1, constant surface temperature sudden boundary condition, Eqs. 5.57 and 5.58. Selected portions of the IHT code used to obtain the graphical 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 shown below. 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) = 20C T(0.68 m, t) (C) 20 10 0 -10 0 15 30 45 60 Time, t (days) alpha = 1.00e-7 m^2/s alpha = 1.38e-7 m^2/s alpha = 3.00e-7 m^2/s Continued ….. PROBLEM 5.73 (Cont.) (b) The temperature distribution T(x,t) for selected times is shown below. The results are directly comparable 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 C 20 15 T(x,t) (C) 10 5 0 -5 -10 -15 0 0.25 0.5 0.75 1 Depth, x (m) 1 day 5 days 10 days 30 days 60 days (c) The heat flux from the soil, q′′ ( 0, t ) , and the heat flux at the depth of the buried main, x q′′ ( 0.68m, t ) , are calculated and plotted for the time period 0 ≤ t ≤ 60 days. x H e a t flu x , q ''(x ,t) (W /m ^ 2 ) 0 a 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 0 0 15 30 45 60 Tim e , t (d a ys ) S u rfa c e h e a t flu x, x = 0 B u rie d -m a in d e p th , x = 0 .6 8 m Both the surface and buried-main heat fluxes have a negative sign since heat is flowing in the negative x-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 surface temperature 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 same at t = 60 days? Are these results consistent with the temperature distributions? What happens to the heat flux values for times much greater than 60 days? Use your IHT model to confirm your explanation. (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 to prescribed surface boundary conditions. */ // The temperature distribution (Tx,t) is T_xt = T_xt_semi_CST(x,alpha,t,Ts,Ti) // Eq 5.55 // The heat flux in the x direction is q''_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, C k = 0.52 // thermal conductivity, W/m.K; base case condition alpha = 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 depth t = t_day * 24 * 3600 // seconds to days time covnersion //t_day = 60 //t_day = 1 //t_day = 5 //t_day = 10 //t_day = 30 t_day = 20 // Surface condition: constant surface temperature Ts = -15 // surface temperature, K PROBLEM 5.74 KNOWN: Tile-iron, 254 mm to a side, at 150°C is suddenly brought into contact with tile over a subflooring material initially at Ti = 25°C with prescribed thermophysical properties. Tile adhesive softens in 2 minutes at 50°C, but deteriorates above 120°C. FIND: (a) Time required to lift a tile after being heated by the tile-iron and whether adhesive temperature exceeds 120°C, (2) How much energy has been removed from the tile-iron during the time it has taken to lift the tile. SCHEMATIC: ASSUMPTIONS: (1) Tile and subflooring have same thermophysical properties, (2) Thickness of adhesive is negligible compared to that of tile, (3) Tile-subflooring behaves as semi-infinite solid experiencing one-dimensional transient conduction. PROPERTIES: Tile-subflooring (given): k = 0.15 W/m⋅K, ρcp = 1.5 × 106 J/m3⋅K, α = k/ρcp = 1.00 × 10-7 m2/s. ANALYSIS: (a) The tile-subflooring can be approximated as a semi-infinite solid, initially at a uniform temperature Ti = 25°C, experiencing a sudden change in surface temperature Ts = T(0,t) = 150°C. This corresponds to Case 1, Figure 5.7. The time required to heat the adhesive (xo = 4 mm) to 50°C follows from Eq. 5.57 T ( x o , t o ) − Ts Ti − Ts xo = erf 2 (α t )1/ 2 o 50 − 150 0.004 m = erf 25 − 150 2 1.00 × 10−7 m 2 s × t 1/ 2 o ( ( − 0.80 = erf 6.325t o 1/ 2 ) ) to = 48.7s = 0.81 min using error function values from Table B.2. Since the softening time, ∆ts, for the adhesive is 2 minutes, the time to lift the tile is t = t o + ∆t s = ( 0.81 + 2.0 ) min = 2.81min . < To determine whether the adhesive temperature has exceeded 120°C, calculate its temperature at t = 2.81 min; that is, find T(xo, t ) T ( x o , t ) − 150 0.004 m = erf 25 − 150 2 1.0 × 10−7 m 2 s × 2.81× 60s 1/ 2 ( ) Continued... PROBLEM 5.74 (Cont.) T ( x o , t ) − 150 = −125erf ( 0.4880 ) = 125 × 0.5098 T ( x o , t ) = 86 C < Since T(xo, t ) < 120°C, the adhesive will not deteriorate. (b) The energy required to heat a tile to the lift-off condition is t q′′ ( 0, t ) ⋅ As dt . 0x Q=∫ Using Eq. 5.58 for the surface heat flux q ′′ (t) = q ′′ (0,t), find s x 2k ( Ts − Ti ) t k ( Ts − Ti ) dt As As t1/ 2 = 1/ 2 1/ 2 1/ 2 0 t Q=∫ Q= (πα ) (πα ) 2 × 0.15 W m ⋅ K (150 − 25 ) C ( π ×1.00 ×10−7 m 2 s ) 1/ 2 × ( 0.254 m ) × ( 2.81× 60s ) 2 1/ 2 = 56 kJ < COMMENTS: (1) Increasing the tile-iron temperature would decrease the time required to soften the adhesive, 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 operating continuously during the 2.81 min period, to maintain the tile-iron at a near constant temperature. The power required is P = 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 the operating temperature in a reasonable period of time. PROBLEM 5.75 KNOWN: Heat flux gage of prescribed thickness and thermophysical properties (ρ, cp, k) initially at a uniform temperature, Ti, is exposed to a sudden change in surface temperature T(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, τ = 4d 2 ρ cp / π 2k. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties. ANALYSIS: The time constant τ 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. The manufacturer’s relationship for the time constant ) ( τ = 4d 2 ρ cp / π 2k can be written in terms of the Fourier number as k 4 ατ τ Fo = = ⋅ = = 0.4053. d 2 ρ cp d 2 π 2 The Fourier number can be determined for the two different installations. (a) For the gage having its backside insulated, the surface and backside temperatures are Ts and T(0,t), respectively. From the sketch it follows that ∗ θo = T (0,τ ) − Ts Ti − Ts From Eq. 5.41, = 0.368. ( ) 2 ∗ θ o = 0.368 = C1exp −ζ1 Fo Using Table 5.1 with Bi = 100 (as the best approximation for Bi = hd/k → ∞, corresponding to sudden surface temperature change with h → ∞), ζ1 = 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-infinite medium having the same thermophysical properties, Table 5.7 (case 1) and Eq. 5.57 yield T ( x,τ ) − Ts 1/ 2 = 0.368 = erf d/2 (ατ ) Ti − Ts 1/ 2 d/2 (ατ ) = 0.3972 Fob = ατ d2 = 1 ( 2 × 0.3972 )2 = 1.585 < COMMENTS: Both models predict higher values of Fo than that suggested by the manufacturer. It is understandable why Fob > Foa since for (b) the gage is thermally connected to an infinite medium, while for (a) it is isolated. From this analysis we conclude that the gage’s transient response will depend upon the manner in which it is applied to the surface or object. PROBLEM 5.76 KNOWN: Procedure for measuring convection heat transfer coefficient, which involves melting of a surface coating. FIND: Melting point of coating for prescribed conditions. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in solid rod (negligible losses to insulation), (2) Rod approximated as semi-infinite medium, (3) Negligible surface radiation, (4) Constant properties, (5) Negligible thermal resistance of coating. -4 2 PROPERTIES: Copper rod (Given): k = 400 W/m⋅K, α = 10 m /s. ANALYSIS: Problem corresponds to transient conduction in a semi-infinite solid. Themal response is given by T ( x,t ) − Ti T∞ − Ti 1/ 2 h (α t ) t x . erfc + 2 (α t )1/ 2 k 2 x − exp hx + h α = erfc 1/ 2 k k2 2 (α t ) For x = 0, erfc(0) = 1 and T(x,t) = T(0,t) = Ts. Hence h 2α Ts − Ti = 1 − exp k2 T∞ − Ti h (α t )1/ 2 t erfc k with h (α t m ) k 1/ 2 = ( 200 W/m 2 ⋅ K 10-4 m 2 / s × 400 s 400 W/m ⋅ K ) 1/ 2 = 0.1 Ts = 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 necessitates iterative calculations. < PROBLEM 5.77 KNOWN: Irreversible thermal injury (cell damage) occurs in living tissue maintained at T ≥ 48°C for a duration ∆t ≥ 10s. FIND: (a) Extent of damage for 10 seconds of contact with machinery in the temperature range 50 to 100°C, (b) Temperature histories at selected locations in tissue (x = 0.5, 1, 5 mm) for a machinery temperature of 100°C. SCHEMATIC: ASSUMPTIONS: (1) Portion of worker’s body modeled as semi-infinite medium, initially at a uniform temperature, 37°C, (2) Tissue properties are constant and equivalent to those of water at 37°C, (3) Negligible contact resistance. PROPERTIES: Table A-6, Water, liquid (T = 37°C = 310 K): ρ = 1/vf = 993.1 kg/m3, c = 4178 J/kg⋅K, k = 0.628 W/m⋅K, α = k/ρc = 1.513 × 10-7 m2/s. ANALYSIS: (a) For a given surface temperature -- suddenly applied -- the analysis is directed toward finding the skin depth xb for which the tissue will be at Tb ≥ 48°C for more than 10s? From Eq. 5.57, T ( x b , t ) − Ts Ti − Ts = erf x b 2 (α t ) 1/ 2 = erf [ w ] . For the two values of Ts, the left-hand side of the equation is Ts = 100 C : ( 48 − 100 ) C 0.825 = (37 − 100 ) C Ts = 50 C : ( 48 − 50 ) C 0.154 = (37 − 50 ) C The burn depth is x b = [ w ] 2 (α t ) 1/ 2 ( = [ w ] 2 1.513 × 10−7 m 2 s × t ) 1/ 2 = 7.779 × 10−4 [ w ] t1/ 2 . Continued... PROBLEM 5.77 (Cont.) Using Table B.2 to evaluate the error function and letting t = 10s, find xb as Ts = 100°C: xb = 7.779 × 10-4 [0.96](10s)1/2 = 2.362 × 103 m = 2.36 mm Ts = 50°C: 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 for the next 10s. We conclude that, for Ts = 50°C, only superficial damage will occur for a contact period of 20s. (b) Temperature histories at the prescribed locations are as follows. 97 Temperature, T(C) 87 77 67 57 47 37 0 15 30 Time, t(s) x = 0.5 mm x = 1.0 mm x = 2.0 mm The critical temperature of 48°C is reached within approximately 1s at x = 0.5 mm and within 7s at x = 2 mm. COMMENTS: Note that the burn depth xb increases as t1/2. PROBLEM 5.78 KNOWN: Thermocouple location in thick slab. Initial temperature. Thermocouple measurement two 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. 3 PROPERTIES: Slab material (given): ρ = 2200 kg/m , c = 700 J/kg⋅K. ANALYSIS: For the semi-infinite medium from Eq. 5.57, x = erf Ti − Ts 2 (α t )1/2 65 −100 0.01m = erf 30 −100 2 (α ×120s )1/2 0.01m = 0.5. erf 2 (α × 120s )1/2 T ( x,t ) − Ts From Appendix B, find for erf w = 0.5 that w = 0.477; hence, 0.01m 2 (α ×120s ) 1/2 = 0.477 (α × 120)1/2 = 0.0105 α = 9.156 ×10−7 m2 /s. It follows that since α = k/ρc, k = αρ c k = 9.156 ×10-7 m 2 / s × 2200 kg/m 3 × 700 J/kg ⋅ K k = 1.41 W/m⋅K. < PROBLEM 5.79 KNOWN: Initial temperature, density and specific heat of a material. Convection coefficient and temperature 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-infinite modium, (3) Constant properties. ANALYSIS: The thermal response of the sample is given by Case 3, Eq. 5.60, T ( x, t ) − Ti T∞ − Ti hx h 2α t x h α t x exp erfc = erfc − + + 2 αt k k 2 αt k 2 where, for x = 0.01m at t = 300 s, [T(x,t) – Ti]/(T∞ - Ti) = 0.533. The foregoing equation must be solved iteratively for k, with α = k/ρcp. The result is k = 0.45 W / m ⋅ K -7 < 2 with α = 4.30 × 10 m /s. COMMENTS: The solution may be effected by inserting the Transient Conduction/Semi-infinite Solid/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 α. PROBLEM 5.80 KNOWN: Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to a surface convection cooling process (T∞,h). FIND: (a) Temperatures at the surface and 45 mm depth after 3 minutes, (b) Effect of thermal diffusivity 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 given by Eq. 5.60. T ( x, t ) − Ti T∞ − Ti 1/ 2 hx h 2α t h (α t ) x x − exp + . = erfc + erfc 2 (α t )1/ 2 k 2 (α t )1/ 2 k k 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 × 10−6 m 2 s × 180s = erfc ( 0 ) − exp 0 + 2 ( 20 W m ⋅ K ) 1/ 2 2 −6 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 m 0.045 m = erfc + 0.02520 − exp 1/ 2 20 W m ⋅ K 2 5.6 × 10−6 m 2 s × 180s 0.045 m × erfc + 0.159 1/ 2 2 5.6 × 10−6 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 temperature histories, and for the two locations the effects of varying α and k are as follows. 275 Temperature, T(C) 325 300 Temperature, T(C) 325 275 250 225 225 175 125 200 75 175 0 50 100 150 200 250 0 300 50 100 200 250 300 250 300 k = 2 W/m.K, alpha = 5.6E-6m^2/s, x = 0 k = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 0 k = 200 W/m.K, alpha = 5.6E-6m^2/s, x = 0 k = 20 W/m.K, alpha = 5.6E-5 m^2/s, x = 0 k = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 0 k = 20 W/m.K, alpha = 5.6E-7m^2/s, x = 0 325 325 300 305 Temperature, T(C) Temperature, T(C) 150 Time, t(s) Time, t(s) 275 250 225 285 265 245 200 225 0 50 100 150 200 250 Time, t(s) k = 20 W/m.K, alpha = 5.6E-5 m^2.K, x = 45 mm k = 20 W/m.K, alpha = 5.6E-6m^2.K, x = 45 mm k = 20 W/m.K, alpha = 5.6E-7m^2.K, x = 45mm 300 0 50 100 150 200 Time, t(s) k = 2 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mm k = 20 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mm k = 200 W/m.K, alpha = 5.6E-6m^2/s, x = 45 mm m For fixed k, increasing alpha corresponds to a reduction in the thermal capacitance per unit volume (ρcp) of the material and hence to a more pronounced reduction in temperature at both surface and interior locations. Similarly, for fixed α, decreasing k corresponds to a reduction in ρcp and hence to a more pronounced decay in temperature. COMMENTS: In part (a) recognize that Fig. 5.8 could also be used to determine the required temperatures. PROBLEM 5.81 KNOWN: Thick oak wall, initially at a uniform temperature of 25°C, is suddenly exposed to combustion products at 800°C with a convection coefficient of 20 W/m2⋅K. FIND: (a) Time of exposure required for the surface to reach an ignition temperature of 400°C, (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): ρ = 545 kg/m3, c = 2385 J/kg⋅K, k = 0.17 W/m⋅K, α = k/ρc = 0.17 W/m⋅K/545 kg/m3 × 2385 J/kg⋅K = 1.31 × 10-7 m2/s. ANALYSIS: (a) This situation corresponds to Case 3 of Figure 5.7. The temperature distribution is given by Eq. 5.60 or by Figure 5.8. Using the figure with T ( 0, t ) − Ti T∞ − Ti = 400 − 25 = 0.48 800 − 25 x and 2 (α t ) 1/ 2 =0 we obtain h(αt)1/2/k ≈ 0.75, in which case t ≈ (0.75k/hα1/2)2. Hence, 2 2 ⋅ K 1.31× 10−7 m 2 s 1/ 2 = 310s t ≈ 0.75 × 0.17 W m ⋅ K 20 W m ) ( < (b) Using the IHT Transient Conduction Model for a Semi-infinite Solid, the following temperature distribution was generated for t = 325s. 400 Temperature, T(C) 325 250 175 100 25 0 0.005 0.01 0.015 0.02 0.025 0.03 Distance from the surface, x(m) The temperature decay would become more pronounced with decreasing α (decreasing k, increasing ρcp) 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 the wall will ignite and combustion will ensue. Once combustion has started, the present model is no longer appropriate. PROBLEM 5.82 KNOWN: Thickness, initial temperature and thermophysical properties of concrete firewall. Incident radiant flux and duration of radiant heating. Maximum allowable surface temperatures at the end of heating. FIND: If maximum allowable temperatures are exceeded. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in wall, (2) Validity of semi-infinite medium approximation, (3) Negligible convection and radiative exchange with the surroundings at the irradiated 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′′ (α t / π ) o 1/ 2 k − x 2 q′′ x x − o erfc exp 4α t k 2 αt where, α = k / ρ c p = 6.92 × 10 −7 m 2 / s and for t = 30 min = 1800s, 2q′′ (α t / π )1/ 2 / k = 284.5 K. Hence, o at x = 0, T ( 0,30 min ) = 25°C + 284.5°C = 309.5°C < 325°C ( < ) At x = 0.25m, − x 2 / 4α t = −12.54, q ′′ x / k = 1, 786K, and x / 2 (α t )1/ 2 = 3.54. Hence, o ( ) T ( 0.25m, 30 min ) = 25°C + 284.5°C 3.58 × 10−6 − 1786°C × ( ~ 0 ) ≈ 25°C < Both requirements are met. COMMENTS: The foregoing analysis is conservative since heat transfer at the irradiated surface due to convection and net radiation exchange with the environment have been neglected. If the emissivity of the surface and the temperature of the surroundings are assumed to be ε = 1 and Tsur = ( ) 4 298K, radiation exchange at Ts = 309.5°C would be q′′ = εσ Ts4 − Tsur = 6, 080 W / m 2 ⋅ K, rad which is significant (~ 60% of the prescribed radiation). PROBLEM 5.83 KNOWN: Initial temperature of copper and glass plates. Initial temperature and properties of finger. FIND: Whether copper or glass feels cooler to touch. SCHEMATIC: ASSUMPTIONS: (1) The finger and the plate behave as semi-infinite solids, (2) Constant properties, (3) Negligible contact resistance. 3 PROPERTIES: Skin (given): ρ = 1000 kg/m , c = 4180 J/kg⋅K, k = 0.625 W/m⋅K; Table A-1 3 (T = 300K), Copper: ρ = 8933 kg/m , c = 385 J/kg⋅K, k = 401 W/m⋅K; Table A-3 (T = 300K), 3 Glass: ρ = 2500 kg/m , c = 750 J/kg⋅K, k = 1.4 W/m⋅K. ANALYSIS: Which material feels cooler depends upon the contact temperature Ts given by Equation 5.63. For the three materials of interest, ( kρ c )1/2 = (0.625 ×1000 × 4180 )1 / 2 = 1,616 J/m 2 ⋅ K ⋅s1/2 skin 1/2 ( kρ c ) cu = ( 401 × 8933 × 385 )1/2 = 37,137 J/m2 ⋅ K ⋅ s1/2 ( kρ c )1/2 = (1.4 × 2500 × 750 )1/2 = 1,620 J/m2 ⋅ K ⋅ s1/2. glass Since ( kρ c )1/2 >> ( kρ c )1/2 , the copper will feel much cooler to the touch. From Equation cu glass 5.63, ( kρ c )1/2 TA,i + ( kρ c )1/2 TB,i A B Ts = 1/2 1/2 ( kρ c ) A + ( kρ c ) B Ts( cu ) = 1,616 ( 310 ) + 37,137 ( 300 ) Ts( glass ) = 1,616 + 37,137 = 300.4 K 1,616 ( 310 ) + 1,620 ( 300 ) 1,616 + 1,620 = 305.0 K. < < COMMENTS: The extent to which a material’s temperature is affected by a change in its thermal 1/2 environment is inversely proportional to (kρc) . Large k implies an ability to spread the effect by conduction; large ρc implies a large capacity for thermal energy storage. PROBLEM 5.84 KNOWN: Initial temperatures, properties, and thickness of two plates, each insulated on one surface. FIND: Temperature on insulated surface of one plate at a prescribed time after they are pressed together. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Negligible contact resistance. 3 PROPERTIES: Stainless steel (given): ρ = 8000 kg/m , c = 500 J/kg⋅K, k = 15 W/m⋅K. ANALYSIS: At the instant that contact is made, the plates behave as semi-infinite slabs and, since the (ρkc) product is the same for the two plates, Equation 5.63 yields a surface temperature of Ts = 350K. The interface will remain at this temperature, even after thermal effects penetrate to the insulated surfaces. The transient response of the hot wall may therefore be calculated from Equations 5.40 and 5.41. At the insulated surface (x* = 0), Equation 5.41 yields ( To − Ts 2 = C1 exp −ζ1 Fo Ti − Ts ) where, in principle, h → ∞ and T∞ → Ts . From Equation 5.39c, Bi → ∞ yields ζ 1 = 1.5707, and from Equation 5.39b C1 = 4sinζ 1 = 1.273 2ζ1 + sin ( 2ζ1 ) αt Also, Fo = Hence, To − 350 = 1.273exp −1.5707 2 × 0.563 = 0.318 400 − 350 L2 = 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 5.85 KNOWN: Thickness and properties of liquid coating deposited on a metal substrate. Initial temperature and properties of substrate. FIND: (a) Expression for time required to completely solidify the liquid, (b) Time required to solidify an 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 throughout solidification (negligible resistance to heat transfer by conduction through solid), (3) Negligible contact resistance at the coating/substrate interface, (4) Negligible solidification contraction, (5) Constant properties. 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 into the solid. Hence, per unit surface area, ρ h sf dS = q′′ cond dt 1/ 2 where, from Eq. 5.58, q′′ cond = k ( Tf − Ti ) (πα t ) . It follows that dS ks (Tf − Ti ) = dt (παs t )1/ 2 k s ( Tf − Ti ) t dt δ ∫o dS = ρ h πα 1/ 2 ∫o t1/ 2 sf ( s ) ρ h sf δ= Tf − Ti 1/ 2 t (παs )1/ 2 ρ hsf 2k s πα δρ hsf t= s 2 4k s Tf − Ti 2 < (b) For the prescribed conditions, π 4 10−5 m 2 s t= (× ) 0.002 m × 3970 kg m3 × 3.577 ×106 J kg 2 = 0.43s 2 4 (120 W m ⋅ K ) 2018 K < COMMENTS: Such solidification processes occur over short time spans and are typically termed rapid solidification. PROBLEM 5.86 KNOWN: Properties of mold wall and a solidifying metal. FIND: (a) Temperature distribution in mold wall at selected times, (b) Expression for variation of solid layer 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 throughout solidification (negligible resistance to heat transfer by conduction through solid), (3) Negligible contact resistance 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 the mold wall. (b) Performing an energy balance for a control surface about the solid layer, the latent energy released due to solidification at the solid/liquid interface is balanced by heat conduction into the solid, q′′ = lat q′′ cond is given by Eq. 5.58. Hence, l cond , where q′′at = ρ h sf dS dt and q′′ ρ h sf S dS k w ( Tf − Ti ) = dt (πα w t )1/ 2 ∫o dS = ρ S= k w (Tf − Ti ) t dt 1/ 2 ∫o t1/ 2 h sf (πα w ) 2k w ( Tf − Ti ) 1/ 2 t 1/ 2 ρ h sf (πα w ) < COMMENTS: The analysis of part (b) would only apply until the temperature field penetrates to the exterior surface of the mold wall, at which point, it may no longer be approximated as a semi-infinite medium. PROBLEM 5.87 KNOWN: Steel (plain carbon) billet of square cross-section initially at a uniform temperature of 30°C is placed in a soaking oven and subjected to a convection heating process with prescribed temperature and convection coefficient. FIND: Time required for billet center temperature to reach 600°C. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional conduction in x1 and x2 directions, (2) Constant properties, (3) Heat transfer to billet is by convection only. PROPERTIES: Table A-1, Steel, plain carbon (T = (30+600)°C/2 = 588K = ≈ 600K): ρ = 3 -5 2 7854 kg/m , cp = 559 J/kg⋅K, k = 48.0 W/m⋅K, α =k/ρcp = 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 θ ∗ ( 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 in the x1 and x2 directions, the P functions are identical. Hence, θ (0, 0, t ) θ o (0, t ) = θi θi 2 Plane wall θ = T − T∞ where θi = Ti − T∞ θ o = T (0,t ) − T∞ and L = 0.15m. Substituting numerical values, find θ o (0, t ) T (0,0,t ) − T∞ = θi Ti − T∞ 1/ 2 1/ 2 ( 600 − 750 ) C = (30 − 750 ) C = 0.46. Consider now the Heisler chart for the plane wall, Figure D.1. For the values ∗θ θ o = o ≈ 0.46 θi find t∗ = Fo = αt L2 Bi-1 = k 48.0 W/m ⋅ K = = 3.2 hL 100 W/m 2 ⋅ K × 0.15m ≈ 3.2. Hence, 3.2 (0.15m ) 3.2 L2 = = 6587s = 1.83h. α 1.093 × 10−5 m 2 / s 2 t= < PROBLEM 5.88 KNOWN: 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) Negligible radiation effects. 3 PROPERTIES: Table A-3, Fire clay brick (900K): ρ = 2050 kg/m , k = 1.0 W/m⋅K, cp = -6 2 960 J/kg⋅K. α = 0.51 × 10 m /s. ANALYSIS: From Fig. 5.11(h), the center temperature is given by T ( 0,0,0,t ) − T∞ Ti − 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 = 1.50 k Fo1 = L2 = 0.045m: Bi 2 = h L2 = 2.25 k Fo 2 = L3 = 0.10m: Bi3 = h L3 = 5.0 k Fo3 = αt L2 1 αt L2 2 αt L2 3 = 1.70 = 0.756 = 0.153 Hence from Fig. D.1, P1 (0, t ) ≈ 0.22 Hence, P2 ( 0, t ) ≈ 0.50 T ( 0,0,0,t ) − T∞ Ti − T∞ P3 (0, t ) ≈ 0.85. ≈ 0.22 × 0.50 × 0.85 = 0.094 and the center temperature is T ( 0,0,0,t ) ≈ 0.094 (1600 − 313) K + 313K = 434K. < Continued ….. PROBLEM 5.88 (Cont.) The corner temperature is given by T ( L1, L 2 , L3 , t ) − T∞ Ti − T∞ = P ( L1, t ) × P ( L2 , t ) × P ( L3 , t ) where P ( L1, t ) = θ ( L1, t ) ⋅ P1 (0, t ) , etc. θo and similar forms can be written for L2 and L3. From Fig. D.2, θ ( L1, t ) ≈ 0.55 θo θ ( L2 , t ) ≈ 0.43 θo θ ( L3 , t ) ≈ 0.25. θo Hence, P ( L1, t ) ≈ 0.55 × 0.22 = 0.12 P ( L 2 , t ) ≈ 0.43 × 0.50 = 0.22 P ( L3 , t ) ≈ 0.85 × 0.25 = 0.21 and T ( L1, L 2 , L3 , t ) − T∞ Ti − T∞ ≈ 0.12 × 0.22 × 0.21 = 0.0056 or T ( L1, L2 , L3 , t ) ≈ 0.0056 (1600 − 313) K + 313K. The corner temperature is then T ( 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, an iterative approach would have to be used. The foregoing procedure would be used to compute the temperature for an assumed value of the time, and the calculation would be repeated until the specified temperature were obtained. PROBLEM 5.89 KNOWN: Cylindrical copper pin, 100mm long × 50mm diameter, initially at 20°C; end faces are subjected to intense heating, suddenly raising them to 500°C; at the same time, the cylindrical surface is subjected to a convective heating process (T∞,h). FIND: (a) Temperature at center point of cylinder after a time of 8 seconds from sudden application of heat, (b) Consider parameters governing transient diffusion and justify simplifying assumptions that could be applied to this problem. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional conduction, (2) Constant properties and convection heat transfer coefficient. 3 PROPERTIES: Table A-1, Copper, pure T ≈ (500 + 20 ) C/2 ≈ 500K : ρ = 8933 kg/m , c = 407 ) ( 3 -4 2 J/kg⋅K, k = 386 W/m⋅K, α = k/ρc = 386 W/m⋅K/8933 kg/m × 407 J/kg⋅K = 1.064 × 10 m /s. ANALYSIS: (1) The pin can be treated as a two-dimensional system comprised of an infinite cylinder whose surface is exposed to a convection process (T∞,h) and of a plane wall whose surfaces are maintained at a constant temperature (Te). This configuration corresponds to the short cylinder, Case (i) of Fig. 5.11, θ ( r,x,t ) = C ( r,t ) × P ( x,t ) . θi (1) For the infinite cylinder, using Fig. D.4, with Bi = hro k = ( 100 W/m 2 ⋅ K 25 × 10-3m 385 W/m ⋅ K C ( 0,8s ) = find ) = 6.47 ×10−3 Fo = and αt 2 ro 1.064 × 10− 4m = 2 s × 8s (25 ×10 m ) -3 θ ( 0,8s ) 2 = 1.36, ≈ 1. cyl θi (2) For the infinite plane wall, using Fig. D.1, with Bi = find hL k →∞ or P ( 0,8s ) = Bi -1 →0 and Fo = αt L2 = 1.064 × 10−4 m 2 / s × 8s (50 ×10 m ) -3 θ ( 0,8s ) θi ≈ 0.5. wall Combining Eqs. (2) and (3) with Eq. (1), find 2 = 0.34, (3) θ ( 0, 0,8s ) θi = T ( 0,0,8s ) − T∞ Ti − T∞ ≈ 1 × 0.5 = 0.5 T ( 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 Biot and Fourier numbers. Since Bi << 0.1 for the cylindrical shape, we can assume radial gradients are negligible. That is, we need only consider conduction in the x-direction. < PROBLEM 5.90 KNOWN: Cylindrical-shaped meat roast weighing 2.25 kg, initially at 6°C, is placed in an oven and subjected to convection heating with prescribed (T∞,h). FIND: Time required for the center to reach a done temperature of 80°C. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional conduction in x and r directions, (2) Uniform and constant properties, (3) Properties approximated as those of water. ( ) PROPERTIES: Table A-6, Water, liquid T = (80 + 6 ) C/2 ≈ 315K : ρ = 1/vf = 1/1.009 × -3 3 3 -7 2 10 m /kg = 991.1 kg/m , cp,f = 4179 J/kg⋅K, k = 0.634 W/m⋅K, α = k/ρc = 1.531 × 10 m /s. ANALYSIS: The dimensions of the roast are determined from the requirement ro = L and knowledge of its weight and density, 1/ 3 1/ 3 2.25 kg 2 or r = L = M M = ρ V = ρ ⋅ 2L ⋅ π ro (1) = = 0.0712m. o 2π ρ 2π 991.1 kg/m3 The roast corresponds to Case (i), Figure 5.11, and the temperature distribution may be T ( x,r,t ) − T∞ expressed as the product of one-dimensional solutions, = P ( x,t ) × C ( r,t ) , where Ti − T∞ P(x,t) and C(r,t) are defined by Eqs. 5.65 and 5.66, respectively. For the center of the cylinder, T ( 0,0,t ) − T∞ Ti − T∞ (80 − 175 ) C = 0.56. = (6 − 175) C In terms of the product solutions, T ( 0,0,t ) − T∞ T ( 0,t ) − T∞ = 0.56 = Ti − T∞ Ti − T∞ (2) T ( 0,t ) − T∞ Ti − T∞ wall cylinder × (3) For each of these shapes, we need to find values of θo/θi such that their product satisfies Eq. (3). For both shapes, Bi = h ro hL 15 W/m 2 ⋅ K × 0.0712m = = = 1.68 k k 0.634 W/m ⋅ K or Bi-1 ≈ 0.6 2 Fo = α t/ro = α t/L2 = 1.53 ×10−7 m 2 / s × t/ (0.0712m ) = 3.020 × 10−5 t. 2 Continued ….. PROBLEM 5.90 (Cont.) A trial-and-error solution is necessary. Begin by assuming a value of Fo; obtain the respective θo/θi values from Figs. D.1 and D.4; test whether their product satisfies Eq. (3). Two trials are shown as follows: θo θo Trial Fo t(hrs) θ o / θi )wall θ o / θi )cyl × θ i w θ i cyl 1 2 0.4 0.3 3.68 2.75 0.72 0.78 0.50 0.68 0.36 0.53 For Trial 2, the product of 0.53 agrees closely with the value of 0.56 from Eq. (2). Hence, it will take approximately 2 ¾ hours to roast the meat. PROBLEM 5.91 KNOWN: A long alumina rod, initially at a uniform temperature of 850K, is suddenly exposed to a cooler fluid. FIND: Temperature of the rod after 30s, at an exposed end, T(0,0,t), and at an axial distance 6mm from the end, T(0, 6mm, t). SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional conduction in (r,x) directions, (2) Constant properties, (3) Convection coefficient is same on end and cylindrical surfaces. PROPERTIES: Table A-2, Alumina, polycrystalline aluminum oxide (assume 3 T ≈ (850 + 600 ) K/2 = 725K): ρ = 3970 kg/m , c = 1154 J/kg⋅K, k = 12.4 W/m⋅K. ANALYSIS: First, check if system behaves as a lumped capacitance. Find Bi = hLc h ( ro / 2 ) 500 W/m ⋅ K (0.010m/2 ) = = = 0.202. k k 12.4 W/m ⋅ K Since Bi > 0.1, rod does not behave as spacewise isothermal object. Hence, treat rod as a semi-infinite cylinder, the multi-dimensional system Case (f), Fig. 5.11. The product solution can be written as θ ∗ ( r,x,t ) = θ ( r,x,t ) θ ( r,t ) θ ( x,t ) = × = C r∗ , t ∗ × S x ∗ , t∗ θi θi θi )( ( ) Infinite cylinder, C(r*,t*). Using the Heisler charts with r* = r = 0 and h ro Bi-1 = k −1 500 W/m 2 ⋅ K × 0.01m = 12.4 W/m ⋅ K −1 = 2.48. 2 Evaluate α = k/ρc = 2.71 × 10 m /s, find Fo = α t/ro = 2.71× 10−6 m 2 / s × 30s/(0.01m) = -6 2 2 -1 0.812. From the Heisler chart, Fig. D.4, with Bi = 2.48 and Fo = 0.812, read C(0,t*) = θ(0,t)/θi = 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 that the LHS needs to be transformed as follows, T − Ti T − T∞ T − T∞ = 1− S ( x,t ) = . T∞ − Ti Ti − T∞ Ti − T∞ Thus, 1/ 2 2 h (α t ) x x − exp hx + h α t erfc . S ( x,t ) = 1 − erfc + 2 k 2 (α t )1/ 2 2 (α t )1/ 2 k k Evaluating this expression at the surface (x = 0) and 6mm from the exposed end, find 2 500 W/m 2 ⋅ K 2.71× 10−6 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 × 30s erfc 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). 0.006m S (6mm,30s ) = 1 − erfc − 2 2.71×10-6m 2 / s × 30s 1/ 2 ) ( 500 W/m 2 ⋅ K × 0.006m exp erfc ( 0.3327 + 0.3636 ) = 0.835. + 0.1322 12.4 W/m ⋅ K The product solution can now be evaluated for each location. At (0,0), T ( 0,0,30s ) − T∞ θ ∗ (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), ()( < ) θ ∗ (0, 6mm,t ) = C 0,t∗ × S 6mm,t ∗ = 0.61× 0.835 = 0.509 T ( 0,6mm,30s ) = 604K. < COMMENTS: Note that the temperature at which the properties were evaluated was a good estimate. PROBLEM 5.92 KNOWN: Stainless steel cylinder of Ex. 5.7, 80-mm diameter by 60-mm length, initially at 600 K, 2 suddenly 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 the center of a circular face, T(0, L, 3 min), and at the midheight of the side, T(ro, 0, 3 min); compare your results 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 what effect 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 coefficients 2 of 500 and 1000 W/m ⋅K. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional conduction in r- and x-coordinates, (2) Constant properties. 3 PROPERTIES: Stainless steel (Example 5.7): ρ = 7900 kg/m , c = 526 J/kg⋅K, k = 17.4 W/m⋅K. ANALYSIS: The following results were obtained using the Transient Conduction models for the Plane 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 results were obtained at t = to = 3 min (r, x, t) P(x, t) C(r, t) 0, 0, to 0, L, to ro, 0, to 0.6357 0.4365 0.6357 0.5388 0.5388 0.3273 T(r, x, t)-IHT (K) 402.7 370.5 362.4 T(r, x, t)-Ex (K) 405 372 365 Continued ….. PROBLEM 5.92 (Cont.) The temperatures from the one-term series calculations of the Example 5.7 are systematically higher than 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 graph below. Note that at early times, the temperature difference between these locations, and hence the gradient, is large. Large differences could cause variations in microstructure and hence, mechanical properties, as well as induce residual thermal stresses. (c) Effect of doubling the convection coefficient is to increase the quenching rate, but much less than by a factor of two as can be seen in the graph below. Effect of increased conv. coeff. on quenching rate Quenching with h = 500 W/m^2.K 500 500 T(0, 0, t) (C) T(x, r, t) (C) 600 600 400 400 300 300 0 0 2 4 6 8 10 2 4 6 8 10 Time, t (min) Time, t (min) h = 500 W/m^2.K h = 1000 W/m^2.K Mid-height of side (0,ro) Center (0, 0) COMMENTS: From IHT menu for Transient Conduction | Plane Wall and Cylinder, the models were combined to solve the product solution. Key portions of the code, less the input variables, are copied below. // Plane wall temperature distribution // The temperature distribution is T_xtP = T_xt_trans("Plane Wall",xstar,FoP,BiP,Ti,Tinf) // The dimensionless parameters are xstar = x / L BiP = h * L / k // Eq 5.9 FoP= alpha * t / L^2 // Eq 5.33 alpha = k/ (rho * cp) // Dimensionless representation, P(x,t) P_xt = (T_xtP - Tinf ) / (Ti - Tinf) // Cylinder temperature distribution // The temperature distribution T(r,t) is T_rtC = T_xt_trans("Cylinder",rstar,FoC,BiC,Ti,Tinf) // The dimensionless parameters are rstar = r / ro BiC = h * ro / k FoC= 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 5.47 PROBLEM 5.93 p KNOWN: Stability criterion for the explicit method requires that the coefficient of the Tm term of the one-dimensional, finite-difference equation be zero or positive. p FIND: For Fo > 1/2, the finite-difference equation will predict values of Tm+1 which violate the Second law of thermodynamics. Consider the prescribed numerical values. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant properties, (3) No internal heat generation. ANALYSIS: The explicit form of the finite-difference equation, Eq. 5.73, for an interior node is ( ) p+1 p p p Tm = Fo Tm+1 + Tm-1 + (1 − 2 Fo ) Tm . p The stability criterion requires that the coefficient of Tm be zero or greater. That is, 1 or Fo ≤ . (1 − 2 Fo ) ≥ 0 2 For the prescribed temperatures, consider situations for which Fo = 1, ½ and ¼ and calculate p+1 Tm . Fo = 1 Fo = 1/2 Fo = 1/4 p+1 Tm = 1(100 + 100 ) C + (1 − 2 ×1) 50 C = 250C p+1 Tm = 1/ 2 (100 + 100 ) C + (1 − 2 ×1/ 2 ) 50 C = 100C p+1 Tm = 1/ 4 (100 + 100 ) C + (1 − 2 ×1/ 4 ) 50 C = 75C. p Plotting these distributions above, note that when Fo = 1, Tm+1 is greater than 100°C, while p for Fo = ½ and ¼ , Tm+1 ≤ 100°C. The distribution for Fo = 1 is thermodynamically impossible: heat is flowing into the node during the time period ∆t, causing its temperature to rise; yet heat is flowing in the direction of increasing temperature. This is a violation of the Second law. When Fo = ½ or ¼, the node temperature increases during ∆t, but the temperature gradients for heat flow are proper. This will be the case when Fo ≤ ½, verifying the stability criterion. PROBLEM 5.94 KNOWN: 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 electrical resistivity, ρe, and other thermophysical properties. FIND: Transient, finite-difference equation for node m. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings are much 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 from the energy conservation requirement on the control volume, Ac∆x, where Ac = π D 2 / 4 The energy balance has the form P = π D. and p T p+1 − Tm q a + q b − q rad + I2 R e = ρ cV m . ∆t Ein − E out + Eg = Est where Eg = I 2R e and R e = ρe ∆x/A c . Using Fourier’s law to express the conduction terms, qa and qb, and Eq. 1.7 for the radiation exchange term, qrad, find ) ( p p p T p − Tm T p − Tm ρ ∆x T p+1 − Tm 4,p 4 + kA c m+1 − ε P∆xσ Tm − Tsur + I2 e = ρ cAc ∆x m kAc m-1 . ∆x ∆x ∆t Ac p Divide each term by ρcAc ∆x/∆t, solve for Tm+1 and regroup to obtain p+1 Tm = ) ε Pσ ∆t 4,p I2 ρ e ∆t 4 − ⋅ Tm − Tsur ) + ⋅ . 2 Ac ρ c ( Ac ρ c ( p ∆t ∆t k k p p ⋅ ⋅ − 1 Tm Tm-1 + Tm+1 − 2 ⋅ ρ c ∆x 2 ρ c ∆x 2 2 Recognizing that Fo = α ∆t/∆x , regroup to obtain ( ) p+1 p p p Tm = Fo Tm-1 + Tm+1 + (1 − 2 Fo ) Tm − ( ) ε Pσ∆x 2 I2 ρe ∆x 2 4,p 4 ⋅ Fo Tm − Tsur + ⋅ Fo. 2 kAc kAc p The stability criterion is based upon the coefficient of the Tm term written as Fo ≤ ½. < 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 5.95 KNOWN: One-dimensional wall suddenly subjected to uniform volumetric heating and convective 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 implicit methods of solution. Using the energy balance approach, both types will be derived. Explicit Method. Perform an energy balance on the surface node shown above, p T p+1 − To q conv + q cond + qV = ρ cV o ∆t Ein − E out + Eg = Est (1) ) ( p p p+1 p T1 − To ∆x ∆x To − To p h (1 ⋅1) T∞ − To + k (1 ⋅1) . + q 1 ⋅1 ⋅ = ρ c 1 ⋅1 ⋅ 2 2 ∆x ∆t (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) p by ρc∆x/2∆t and solve for To +1. h∆t k∆t ∆t p+1 p p p p To = 2 T∞ − To + 2 T1 − To + q + To . ρ c∆x ρc ρ c∆x 2 Introducing the Fourier and Biot numbers, ( Fo ≡ ( k/ρ c ) ∆t/∆x 2 ) ( ) Bi ≡ h∆x/k p q∆x 2 p+1 p To = 2 Fo T1 + Bi ⋅ T∞ + + (1 − 2 Fo − 2 Fo ⋅ Bi ) To . 2k (3) p The stability criterion requires that the coefficient of To be positive. That is, (1 − 2 Fo − 2 Fo ⋅ Bi ) ≥ 0 or Fo ≤ 1/2 (1 + Bi ) . (4) < Implicit Method. Begin as above with an energy balance. In Eq. (2), however, the temperatures on the LHS are evaluated at the new (p+1) time. The RHS provides a backwarddifference approximation to the time derivative. ( ) p+1 p+1 p+1 p T1 − To − To ∆x ∆x T p+1 h T∞ − To +k +q = ρ c o ∆x ∆t 2 2 p+1 p (1 + 2 Fo (Bi + 1)) To − 2 Fo ⋅ T1p+1 = To + 2Bi ⋅ Fo ⋅ T∞ + Fo q∆kx (5) 2 . (6) < COMMENTS: Compare these results (Eqs. 3, 4 and 6) with the appropriate expression in Table 5.2. PROBLEM 5.96 KNOWN: Plane wall, initially having a linear, steady-state temperature distribution with boundaries maintained at T(0,t) = T1 and T(L,t) = T2, suddenly experiences a uniform volumetric heat generation due to 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: initial conditions (t ≤ 0), steady-state conditions (t → ∞) assuming the maximum temperature exceeds T2, and two intermediate times; label important features; (b) For the three-nodal network shown, derive the finite-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 the boundaries; and (d) Determine the effect of mesh size by repeating the foregoing analysis using grids of 5 and 11 nodal points. SCHEMATIC: ASSUMPTIONS: (1) Two-dimensional, transient conduction, (2) Uniform volumetric heat generation for t ≥ 0, (3) Constant properties. PROPERTIES: Wall (Given): ρ = 4000 kg/m3, c = 500 J/kg⋅K, k = 10 W/m⋅K. ANALYSIS: (a) The temperature distribution on T-x coordinates for the requested cases are shown below. Note the following key features: (1) linear initial temperature distribution, (2) non-symmetrical parabolic steady-state temperature distribution, (3) gradient at x = L is first 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, find Ein − E out + Eg = Est p T2 − Tm T1 − Tm T p +1 − Tm (1) ∆x = ρ (1) c∆x m k (1) + k (1) +q ∆x ∆x ∆t ∆t q p p Fo [T1 + T2 − 2Tm ] + = Tm+1 − Tm ρ cp For the Tm term in brackets, use “p” for explicit and “p+1” for implicit form, p p p p Tm+1 = Fo T1 + T2 + (1 − 2Fo ) Tm + q∆t ρ cp Explicit: ( Implicit: ( ) ) p p p p Tm+1 = Fo T1 +1 + T2 +1 + q ∆t ρ cp + Tm (1 + 2Fo ) (1) < (2) < Continued... PROBLEM 5.96 (Cont.) (c) With a time increment ∆t = 5s, the FDEs, Eqs. (1) and (2) become Explicit: p p Tm+1 = 0.5Tm + 75 (3) Implicit: p p Tm+1 = Tm + 75 1.5 (4) ) ( where Fo = q∆t k∆t ρ c∆x 2 = 10 W m ⋅ K × 5s 4000 kg m3 × 500 J kg ⋅ K ( 0.010 m ) 2 = 0.25 2 × 107 W m3 × 5s = 50 K 4000 kg m3 × 500 J kg ⋅ K Performing the calculations, the results are tabulated as a function of time, ρc = p t (s) T1 (°C) 0 1 2 3 4 5 6 7 8 9 0 5 10 15 20 25 30 35 40 45 Tm (°C) 0 0 0 0 0 0 0 0 0 0 Explicit 50 100.00 125.00 137.50 143.75 146.88 148.44 149.22 149.61 149.80 T2 (°C) Implicit 50 83.33 105.55 120.37 130.25 136.83 141.22 144.15 146.10 147.40 100 100 100 100 100 100 100 100 100 100 < The heat flux at the boundaries at t = 45s follows from the energy balances on control volumes about the p boundary nodes, using the explicit results for Tm , Node 1: E in − E out + E g = Est Tp − T q′′ ( 0, t ) + k m 1 + q ( ∆x 2 ) = 0 x ∆x ( p q′′ ( 0, t ) = − k Tm − T1 x ) ∆x − q∆x 2 (5) q′′ ( 0, 45s ) = −10 W m ⋅ K (149.8 − 0 ) K 0.010 m − 2 × 107 W m3 × 0.010 m 2 x q′′ ( 0, 45s ) = −149,800 W m 2 − 100, 000 W m2 = −249,800 W m2 x Node 2: < T p − T2 km − q′′ ( L, t ) + q ( ∆x 2 ) = 0 x ∆x ( p q′′ ( L, t ) = k Tm − T2 x ) ∆x + q∆x 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 2 x q′′ ( L, t ) = 49,800 W m 2 + 100, 000 W m 2 = +149,800 W m2 x < (d) To determine the effect of mesh size, the above analysis was repeated using grids of 5 and 11 nodal points, ∆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) were p used for the heat flux calculations. The results are tabulated below for t = 45s, where Tm (45s) is the center node, Mesh Size ∆x (mm) 10 5 2 p Tm (45s) (°C) 149.8 149.3 149.4 q′′ (0,45s) x q′′ (L,45s) x kW/m2 -249.8 -249.0 -249.1 kW/m2 +149.8 +149.0 +149.0 COMMENTS: (1) The center temperature and boundary heat fluxes are quite insensitive to mesh size for 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 - nodal equations /* Node a: interior node; e and w labeled b and 1. */ rho*cp*der(Ta,t) = fd_1d_int(Ta,Tb,T1,k,qdot,deltax) /* Node b: interior node; e and w labeled c and a. */ rho*cp*der(Tb,t) = fd_1d_int(Tb,Tc,Ta,k,qdot,deltax) /* Node c: interior node; e and w labeled 2 and b. */ rho*cp*der(Tc,t) = fd_1d_int(Tc,T2,Tb,k,qdot,deltax) // Assigned Variables: deltax = 0.005 k = 10 rho = 4000 cp = 500 qdot = 2e7 T1 = 0 T2 = 100 /* Initial Conditions: Tai = 25 Tbi = 50 Tci = 75 */ /* Data Browser Results - Nodal temperatures at 45s Ta Tb Tc t 99.5 149.3 149.5 45 */ // Boundary Heat Fluxes - at t = 45s q''x0 = - k * (Taa - T1 ) / deltax - qdot * deltax / 2 q''xL = k * (Tcc - T2 ) / deltax + qdot * deltax /2 //where Taa = Ta (45s), Tcc = Tc(45s) Taa = 99.5 Tcc = 149.5 /* Data Browser results q''x0 q''xL -2.49E5 1.49E5 */ PROBLEM 5.97 KNOWN: Solid cylinder of plastic material (α = 6 × 10-7 m2/s), initially at uniform temperature of Ti = 20°C, insulated at one end (T4), while other end experiences heating causing its temperature T0 to increase linearly with time at a rate of a = 1°C/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 = 35°C. 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, ) ( ( p p p p p p Tm+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 ( p p qa + q b + 0 = ρ cV T4 +1 − T4 Ein − E out + Eg = Est ( p p 0 + k T3 − T4 ) ∆t ) ∆x = ρ c (∆x 2)(T4p+1 − T4p ) ∆t p p p p T4 +1 = 2FoT3 + (1 − 2Fo ) T4 = T3 (2) (b) Performing the calculations, the temperature-time history is tabulated below, where T0 = Ti +a⋅t where a = 1°C/s and t = p⋅∆t with, Fo = α∆t ∆x 2 = 0.5 p 0 1 2 3 4 5 6 7 t (s) 0 30 60 90 120 150 180 210 ∆t = 0.5 (0.006 m ) 2 T0 (°C) 20 50 80 110 140 170 200 230 T1 (°C) 20 20 35 50 68.75 87.5 108.1 - When T4(210s, p = 7) = 35°C, find T0(210s) = 230°C. 6 × 10−7 m 2 s = 30s T2 (°C) 20 20 20 27.5 35 46.25 57.5 - T3 (°C) 20 20 20 20 23.75 27.5 35 - T4 (°C) 20 20 20 20 20 23.75 27.5 35 < PROBLEM 5.98 KNOWN: A 0.12 m thick wall, with thermal diffusivity 1.5 × 10-6 m2/s, initially at a uniform temperature of 85°C, has one face suddenly lowered to 20°C while the other face is perfectly insulated. FIND: (a) Using the explicit finite-difference method with space and time increments of ∆x = 30 mm and ∆t = 300s, determine the temperature distribution within the wall 45 min after the change in surface temperature; (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 be determined from Eq. 5.73, ) ( p p p p Tm+1 = Fo Tm −1 + Tm +1 + (1 − 2 Fo ) Tm (1) with Fo = α∆t ∆x 2 = 1.5 × 10−6 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), p p p Tm+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 table below using t = p⋅∆t = 300 p (s) = 5 p (min). p 0 1 2 3 4 5 6 7 8 9 t(min) 0 10 20 30 40 45 T0 85 85 85 85 76.9 76.9 68.8 68.8 61.7 61.7 T1 85 85 85 76.9 76.9 68.8 68.8 61.7 61.7 55.6 T2 85 85 68.8 68.8 60.7 60.7 54.6 54.6 49.5 49.5 T3 85 52.5 52.5 44.4 44.4 40.4 40.4 37.3 37.3 34.8 TL(°C) 20 20 20 20 20 20 20 20 20 20 < The temperature distribution can also be determined from the Heisler charts. For the wall, Fo = αt L2 = 1.5 × 10−6 m 2 s × ( 45 × 60 ) s (0.12 m )2 = 0.28 and Bi −1 = k = 0. hL Continued... PROBLEM 5.98 (Cont.) From Figure D.1, for Bi-1 = 0 and Fo = 0.28, find θ o θ i ≈ 0.55. Hence, for x = 0 θ =o Ti − T∞ θ i To − T∞ θ To = T ( 0, t ) = T∞ + o ( Ti − T∞ ) = 20 C + 0.55 (85 − 20 ) C = 55.8 C . θi or This value is to be compared with 61.7°C 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 the midplane, as well as for the chilled surface (TL). Temperature, T(C) 85 75 65 55 45 35 25 15 0 2000 4000 6000 8000 10000 12000 14000 16000 18000 Time, t(s) T0, deltat = 75 s T2, deltat = 75 s TL T0, deltat = 300 s T2, deltat = 300 s Apart from small differences during early stages of the transient, there is excellent agreement between results obtained for the two time steps. The temperature decay at the insulated surface must, of course, lag that of the midplane. PROBLEM 5.99 KNOWN: 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 cooled surface, (3) Negligible heat transfer at insulated surface, (4) Constant properties. ANALYSIS: Adopting the implicit scheme, the finite-difference equation for the cooled surface node is given by Eq. (5.88), from which it follows that p p p (1 + 2 Fo + 2 FoBi ) T10+1 − 2Fo T9 +1 = 2 FoBi T∞ + T10 The general form of the finite-difference equation for any interior node (1 to 9) is given by Eq. (5.89), p p p p (1 + 2 Fo ) Tm+1 − Fo Tm+1 + Tm+1 = Tm −1 +1 ) ( The finite-difference equation for the insulated surface node may be obtained by applying the p p symmetry requirement to Eq. (5.89); that is, Tm +1 = Tm −1. Hence, p p (1 + 2 Fo ) To +1 − 2 Fo T1p +1 = To 2 For the prescribed conditions, Bi = h∆x/k = 100 W/m ⋅K (0.006m)/0.30 W/m⋅K = 2. If the explicit method were used, the most restrictive stability requirement would be given by Eq. (5.79). Hence, for 2 -7 2 Fo (1+Bi) ≤ 0.5, Fo ≤ 0.167. With Fo = α∆t/∆x and α = k/ρc = 1.67 ×10 m /s, the corresponding restriction on the time increment would be ∆t ≤ 36s. Although no such restriction applies for the implicit method, a value of ∆t = 30s is chosen, and the set of 11 finite-difference equations is solved using the Tools option designated as Finite-Difference Equations, One-Dimensional, and Transient from the IHT Toolpad. At t = 3600s, the solution yields: T10 (3600s ) = 24.1°C T0 (3600s ) = 71.5°C < COMMENTS: (1) More accurate results may be obtained from the one-term approximation to the exact solution for one-dimensional, transient conduction in a plane wall. With Bi = hL/k = 20, Table 2 5.1 yields ζ1 = 1.496 rad and C1 = 1.2699. With Fo = αt/L = 0.167, Eq. (5.41) then yields To = T∞ + ( ) 2 (Ti - T∞) C1 exp −ζ 1 Fo = 72.4°C, and from Eq. (5.40b), Ts = T∞ + (Ti - T∞) cos (ζ1 ) = 24.5°C. Since the finite-difference results do not change with a reduction in the time step (∆t < 30s), the difference between the numerical and analytical results is attributed to the use of a coarse grid. To improve 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. 80 Te m p e ra tu re (C ) 70 60 50 40 30 20 0 600 1200 1800 2400 3000 3600 Tim e (s ) In s u la te d s u rfa ce C o o le d s u rfa ce Although the surface temperatures rapidly approaches that of the coolant, there is a significant lag in the thermal response of the back surface. The different responses are attributable to the small value of α and the large value of Bi. PROBLEM 5.100 KNOWN: Plane wall, initially at a uniform temperature Ti = 25°C, is suddenly exposed to convection with a fluid at T∞ = 50°C with a convection coefficient h = 75 W/m2⋅K at one surface, while the other is exposed to a constant heat flux q′′ = 2000 W/m2. See also Problem 2.43. o FIND: (a) Using spatial and time increments of ∆x = 5 mm and ∆t = 20s, compute and plot the temperature distributions in the wall for the initial condition, the steady-state condition, and two intermediate times, (b) On q′′ -x coordinates, plot the heat flux distributions corresponding to the four x temperature distributions represented in part (a), and (c) On q′′ -t coordinates, plot the heat flux at x = 0 x and 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, the equations for determining the temperature distribution were obtained and solved with a spatial increment of ∆x = 5 mm. Using the Lookup Table functions, the temperature distributions were plotted as shown below. (b) The heat flux, q′′ (x,t), at each node can be evaluated considering the control volume shown with the x schematic above ( q′′ ( m, p ) = q′′ + q′′ x x,a x,b ) p p T Tp − T − Tp p p k (1) m −1 m + k (1) m m +1 2 = k Tm −1 − Tm +1 2= ∆x ∆x ( ) 2∆x From knowledge of the temperature distribution, the heat flux at each node for the selected times is computed and plotted below. Heat flux, q''x(x,t) (W/m^2) Temperature, T(x,t) (C) 160 140 120 100 80 60 40 20 0 10 20 30 40 2000 1500 1000 500 0 50 Wall coordinate, x (mm) Initial condition, t<=0s Time = 150s Time = 300s Steady-state conditions, t>1200s 0 10 20 30 40 50 Wall coordinate, x (mm) Initial condition, t<=0s Time = 150s Time = 300s Steady-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 = 0 surface, the heat flux is constant, q TT = 2000 W/m2. At the x = L surface, the heat flux is given by o Newton’s law of cooling, q′′ (L,t) = h[T(L,t) - T∞ ]; at t = 0, q′′ (L,0) = -1875 W/m2. For steady-state x x conditions, the heat flux q′′ (x,∞) is everywhere constant at q TT . o x Continued... PROBLEM 5.100 (Cont.) 2000 Heat flux (W/m^2) 1000 0 -1000 -2000 0 200 400 600 800 1000 1200 Elapsed time, t (s) q''x(0,t) - Heater flux q''x(L,t) - Convective flux Comments: The IHT workspace using the Finite-Difference Equations Tool to determine the temperature distributions and heat fluxes is shown below. Some lines of code were omitted to save space on 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 process h0 = 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) / 2 q''2 = k / deltax * (T1 - T3) / 2 ...... ...... q''9 = k / deltax * (T8 - T10) / 2 // Heat flux at boundary x= L, q''10 q''xL = h * (T10 - Tinf) // Assigned Variables: deltax = 0.005 k = 1.5 alpha = 7.5e-6 cp = 1000 alpha = k / (rho * cp) qdot = 0 Ti = 25 Tinf = 50 h = 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 value PROBLEM 5.101 KNOWN: Plane wall, initially at a uniform temperature To = 25°C, has one surface (x = L) suddenly exposed to a convection process with T∞ = 50°C and h = 1000 W/m2⋅K, while the other surface (x = 0) is maintained at To. Also, the wall suddenly experiences uniform volumetric heating with q = 1 × 107 3 W/m . See also Problem 2.44. FIND: (a) Using spatial and time increments of ∆x = 4 mm and ∆t = 1s, compute and plot the temperature distributions in the wall for the initial condition, the steady-state condition, and two intermediate times, and (b) On q′′ -t coordinates, plot the heat flux at x = 0 and x = L. At what elapsed x time 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, the temperature distributions were obtained and plotted below. (b) The heat flux, q TT (L,t), can be expressed in terms of Newton’s law of cooling, x ( ) q ′′ ( L, t ) = h T10 − T∞ . x p From the energy balance on the control volume about node 0 shown above, ( q′′ ( 0, t ) + E g + q′′ = 0 x a q′′ ( 0, t ) = −q ( ∆x 2 ) − k T1 − To x p ) ∆x From knowledge of the temperature distribution, the heat fluxes are computed and plotted. 120 100 0 80 Heat flux (W/m^2) Temperature, T(x,t) (C) 100000 60 40 -1E5 -2E5 20 0 10 20 Wall coordinate, x (mm) Initial condition, t<=0s Time = 60s Time = 120s Steady-state conditions, t>600s 30 40 -3E5 0 100 200 300 400 500 600 Elapsed 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/°C and C2 = 25°C. Find q ′′ ( 0, ∞ ) = −3.25 × 105 W / m 2 and q′′ ( L ) = +7.5 × 104 W / m 2 . Comparing with x x the graphical results above, we conclude that steady-state conditions are not reached in 600 x. PROBLEM 5.102 KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with no internal generation; suddenly a uniform generation, q = 108 W/m3 , occurs when the element 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, 4 2 q ( ∆x ) p+1 p p p + (1 − 2 Fo ) Tm . Tm = Fo Tm-1 + Tm+1 + 2 Midplane node, m = 0 (1) p p Same as Eq. (1), but with Tm-1 = Tm+1. Surface node, m = 5 2 q ( ∆x ) p+1 p + (1 − 2Fo − 2Bi ⋅ Fo ) T p . (2) T5 = 2 Fo T4 + Bi ⋅ T∞ + 5 2k The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the following parameters: 2 h∆x 1100W/m ⋅ K × ( 0.002m ) Bi = = = 0.0733 k 30W/m ⋅ K 1/2 Fo ≤ = 0.466 (1 + Bi ) Fo ( ∆x ) 2 ∆t ≤ α = 0.466 (0.002m )2 5 × 10−6 m 2 / s = 0.373s. Continued ….. PROBLEM 5.102 (Cont.) To be well within the stability limit, select ∆t = 0.3s, which corresponds to Fo = α∆t = 5 × 10−6 m 2 / s × 0.3s ∆x 2 (0.002m )2 t = p∆t = 0.3p (s ). = 0.375 Substituting numerical values with q = 108 W/m3 , the nodal equations become 2 p+1 p p T0 = 0.375 2T1 + 108 W/m3 ( 0.002m ) / 30W/m ⋅ K + (1 − 2 × 0.375) T0 p+1 p p T0 = 0.375 2T1 + 13.33 + 0.25 T0 p+1 p p p T1 = 0.375 T0 + T2 + 13.33 + 0.25 T1 p+1 p p p T2 = 0.375 T1 + T3 + 13.33 + 0.25 T2 (3) (4) (5) p+1 p p p T3 = 0.375 T2 + T4 + 13.33 + 0.25 T3 p+1 p p p T4 = 0.375 T3 + T5 + 13.33 + 0.25 T4 (6) (7) 13.33 p p+1 p T5 = 2 × 0.375 T4 + 0.0733 × 250 + + (1 − 2 × 0.375 − 2 × 0.0733 × 0.375 ) T5 2 p+1 p p T5 = 0.750 T4 + 24.99 + 0.195 T5 . (8) The initial temperature distribution is Ti = 250°C at all nodes. The marching solution, following the procedure of Example 5.8, is represented in the table below. p 0 1 2 3 4 t(s) 0 0.3 0.6 0.9 1.2 T0 250 255.00 260.00 265.00 270.00 T1 250 255.00 260.00 265.00 270.00 T2 250 255.00 260.00 265.00 270.00 T3 250 255.00 260.00 265.00 269.96 T4 250 255.00 260.00 264.89 269.74 T5(°C) 250 254.99 259.72 264.39 268.97 5 1.5 275.00 275.00 274.98 274.89 274.53 273.50 The 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 the coolant during the first 1.5s time period. < PROBLEM 5.103 KNOWN: Conditions associated with heat generation in a rectangular fuel element with surface cooling. See Example 5.8. FIND: (a) The temperature distribution 1.5 s after the change in operating power; compare your results 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, and approximately how long it will take to reach the new steady-state condition after the step change in operating power. Use the IHT Tools | Finite-Difference Equations | One-Dimensional | Transient conduction 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 implicit method of solution. Selected portions of the IHT code used to obtain the results tabulated below are shown in the Comments. (a) Using the IHT code, the temperature distribution (°C) as a function of time (s) up to 1.5 s after the step power change is obtained from the summarized results copied into the workspace 1 2 3 4 5 6 t 0 0.3 0.6 0.9 1.2 1.5 T00 357.6 358.1 358.6 359.1 359.6 360.1 T01 356.9 357.4 357.9 358.4 358.9 359.4 T02 354.9 355.4 355.9 356.4 356.9 357.4 T03 351.6 352.1 352.6 353.1 353.6 354.1 T04 346.9 347.4 347.9 348.4 348.9 349.3 T05 340.9 341.4 341.9 342.3 342.8 343.2 (b) Using the code, the mid-plane (00) and surface (05) node temperatures are plotted as a function of time. 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 r Te m p e ra tu re , T(x,t) (C ) 480 440 400 360 320 0 100 200 300 400 Tim e , t (s ) T0 0 , Mid -p la n e , x = 0 T0 5 , S u rfa c e , x = L Continued ….. PROBLEM 5.103 (Cont.) Note that at t ≈ 240 s, the wall has nearly reached the new steady-state condition for which the nodal temperatures (°C) were found as: T00 465 T01 463.7 T02 459.7 T03 453 T04 443.7 T05 431.7 COMMENTS: (1) Can you validate the new steady-state nodal temperatures from part (b) by comparison 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 and time 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 temperature distribution 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 shown Tinf01 = 20 // Arbitrary value h01 = 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 shown Tinf05 = 250 // Coolant temperature, C h05 = 1100 // Convection coefficient, W/m^2.K // Input parameters qdot = 2e7 // Volumetric rate, W/m^3, step change deltax = 0.002 // Space increment k = 30 // Thermophysical properties alpha = 5e-6 rho = 1000 alpha = k / (rho * cp) /* Steady-state conditions, with qdot1 = 1e7 W/m^3; initial conditions for step change T_x = 16.67 * (1 - x^2/L^2) + 340.91 // See text Seek T_x for x = 0, 2, 4, 6, 8, 10 mm; results used for Ti are Node T_x 00 357.6 01 356.9 02 354.9 03 351.6 04 346.9 05 340.9 */ PROBLEM 5.104 KNOWN: Conditions associated with heat generation in a rectangular fuel element with surface cooling. See Example 5.8. FIND: (a) The temperature distribution 1.5 s after the change in the operating power; compare results with those tabulated in the Example, and (b) Plot the temperature histories at the midplane, x = 0, and the surface, x = L, for 0 ≤ t ≤ 400 s; determine the new steady-state temperatures, and approximately how 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: on the y-planes and the x = 0 plane, treat as adiabatic; on the x = 10 mm plane, specify the convection option. Specify the material properties and the internal generation with q1 . In the Setup menu, click on Steady-state, and then Run to obtain the temperature distribution corresponding to the initial temperature 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 value to 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 and from the FDE analysis in the Example are tabulated below. x/L T(x/L, 1.5 s) FEHT (°C) FDE (°C) 0 0.2 0.4 360.1 360.08 359.4 359.41 357.4 357.41 0.6 0.8 1.0 354.1 349.3 354.07 349.37 343.2 343.27 The mesh spacing for the FEHT analysis was 0.5 mm and the time increment was 0.005 s. For the FDE analyses, the spatial and time increments were 2 mm and 0.3 s. The agreement between the results from the two numerical methods is within 0.1°C. (b) Using the FEHT code, the temperature histories at the mid-plane (x = 0) and the surface (x = L) are plotted 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 FDE analysis in the Example are tabulated below. The agreement between the results from the two numerical methods is within 0.1°C x/L T(x/L, ∞) FEHT (°C) FDE (°C) 0 0.2 0.4 0.6 0.8 465.0 465.15 463.7 463.82 459.6 459.82 453.0 453.15 443.6 443.82 1.0 431.7 431.82 COMMENTS: (1) 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 the q1 operating power. Using the Run|Continue command (after re-setting the generation to q 2 and clicking on Setup | Transient), this steady-state distribution automatically becomes the initial temperature distribution for the q 2 operating power. This feature allows for conveniently prescribing a non-uniform initial temperature distribution for a transient analysis (rather than specifying values on a node-by-node basis). (2) Use the View | Tabular Output command to obtain nodal temperatures to the maximum number of significant figures resulting from the analysis. (3) Can you validate the new steady-state nodal temperatures from part (b) (with q 2 , t → ∞) by comparison against an analytical solution? PROBLEM 5.105 KNOWN: Thickness, initial temperature, speed and thermophysical properties of steel in a thin-slab continuous casting process. Surface convection conditions. FIND: Time required to cool the outer surface to a prescribed temperature. Corresponding value of the 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 surface node is given by Eq. (5.88), from which it follows that p p p (1 + 2 Fo + 2 FoBi ) T10+1 − 2 Fo T9 +1 = 2 FoBi T∞ + T10 The 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 − Fo p p p (Tm+−11 + Tm++11 ) = Tm The finite-difference equation for the midplane node may be obtained by applying the symmetry p p requirement to Eq. (5.89); that is, Tm +1 = Tm −1. Hence, p p (1 + 2 Fo ) T0 +1 − 2 Fo T1p +1 = T0 2 For the prescribed conditions, Bi = h∆x/k = 5000 W/m ⋅K (0.010m)/30 W/m⋅K = 1.67. If the explicit method were used, the stability requirement would be given by Eq. (5.79). Hence, for Fo(1 + Bi) ≤ 2 -6 2 0.5, Fo ≤ 0.187. With Fo = α∆t/∆x and α = k/ρc = 5.49 × 10 m /s, the corresponding restriction on the 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 Tools option designated as Finite-Difference Equations, One-Dimensional and Transient from the IHT Toolpad. For T10 (t) = 300°C, the solution yields < t = 161s Continued ….. PROBLEM 5.105 (Cont.) T0 ( t ) = 1364°C < With a casting speed of V = 15 mm/s, the length of the cooling section is Lcs = Vt = 0.015 m / s (161s ) = 2.42m < 2 COMMENTS: (1) With Fo = αt/L = 0.088 < 0.2, the one-term approximation to the exact solution for 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 = 1366°C and Ts = 200.7°C are obtained and are in good agreement with the finitedifference predictions. The accuracy of these predictions could still be improved by reducing the value of ∆x. (2) Temperature histories for the surface and midplane nodes are plotted for 0 < t < 600s. 1500 T e m p e ra tu re (C ) 1300 1100 900 700 500 300 100 0 100 200 300 400 500 600 T im e (s ) Mid p la n e C o o le d s u rfa c e While T10 (600s) = 124°C, To (600s) has only dropped to 879°C. The much slower thermal response at the midplane is attributable to the small value of α and the large value of Bi = 16.67. PROBLEM 5.106 KNOWN: Very thick plate, initially at a uniform temperature, Ti, is suddenly exposed to a convection cooling process (T∞,h). FIND: Temperatures at the surface and a 45mm depth after 3 minutes using finite-difference method with space and time increments of 15mm and 18s. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Plate approximates semi-infinite medium, (3) Constant properties. ANALYSIS: The grid network representing the plate is shown above. The finite-difference equation for node 0 is given by Eq. 5.82 for one-dimensional conditions or Eq. 5.77, ( ) p+1 p p T0 = 2 Fo T1 + Bi ⋅ T∞ + (1 − 2 Fo − 2 Bi ⋅ Fo ) T0 . (1) The numerical values of Fo and Bi are Fo = α∆t ∆x 2 = 5.6 ×10−6 m2 /s × 18s ( 0.015m )2 = 0.448 ( ) 2 -3 h∆x 100 W/m ⋅ K × 15 ×10 m Bi = = = 0.075. k 20 W/m ⋅ K Recognizing that T∞ = 15° C, Eq. (1) has the form p+1 p p T0 = 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+1 p p p Tm = Fo Tm+1 + Tm-1 + (1 − 2Fo ) Tm . (3) Recognizing that the stability criterion, Fo ≤ 1/2, is satisfied with Fo = 0.448, ( ) p+1 p p p Tm = 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 time increment, t = p∆t. (5) The finite-difference calculations can now be performed using Eqs. (2) and (4). The results are tabulated below. p 0 1 2 3 4 5 6 7 8 9 10 t(s) 0 18 36 54 72 90 108 126 144 162 180 T0 325 304.2 303.2 294.7 293.0 287.6 285.6 281.8 279.8 276.7 274.8 T1 325 324.7 315.3 313.7 307.8 305.8 301.6 299.5 296.2 294.1 291.3 T2 325 325 324.5 320.3 318.9 315.2 313.5 310.5 308.6 306.0 304.1 T3 325 325 325 324.5 322.5 321.5 319.3 317.9 315.8 314.3 312.4 T4 325 325 325 325 324.5 323.5 322.7 321.4 320.4 319.0 T5 325 325 325 325 325 324.5 324.0 323.3 322.5 T6 T7(K) 325 325 325 325 325 325 325 325 325 325 325 325 324.5 325 324.2 Hence, find 10 T ( 0, 180s ) = T0 = 275o C 10 T ( 45mm, 180s ) = T3 = 312o C. < COMMENTS: (1) The above results can be readily checked against the analytical solution represented in Fig. 5.8 (see also Eq. 5.60). For x = 0 and t = 180s, find x =0 1/2 2 (α t ) ( 1/2 100 W/m 2 ⋅ K 5.60× 10-6m 2 / s × 180s h (α t ) = k 20 W/m ⋅ K for which the figure gives T − Ti = 0.15 T∞ − Ti so that, ) 1/2 = 0.16 T ( 0, 180s ) = 0.15 ( T∞ − T i ) +T i = 0.15 (15 − 325)o C + 325oC T ( 0, 180s ) = 278o C. For x = 45mm, the procedure yields T(45mm, 180s) = 316° C. The agreement with the numerical solution is nearly within 1%. PROBLEM 5.107 KNOWN: 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 it change with time, and (c) Temperature at the surface and at 30mm depth for prescribed conditions after 1 minute exposure. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Thick slab may be approximated as semi-infinite medium, (3) Constant properties, (4) Radiation exchange is between small surface and large surroundings. ANALYSIS: (a) The explicit form of the FDE for the surface node may be obtained by applying an energy balance to a control volume about the node. E′′ − E′′ = q′′ in out conv + q′′ + q′′ rad cond = E′′ st )( ( ) p p T1 − To p p h T∞ − To + h r Tsur − To + k ⋅1⋅ ∆x p+1 p − To ∆x T (1) = ρ c ⋅1 o ∆t 2 where the radiation process has been linearized, Eq. 1.8. (See also Comment 4, Example 5.9), p2 p p 2 h r = h p To , Tsur = εσ To + Tsur T0 + Tsur . (2) r Divide Eq. (1) by ρc∆x/2∆t and regroup using these definitions to obtain the FDE: )( ( Fo ≡ ( k/ρ c ) ∆t/∆x 2 ( ) Bi ≡ h∆x/k ) Bi r ≡ h r ∆x/k (3,4,5) p+1 p p To = 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) ≥ 0 or Fo ≤ 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, the stability criterion of the surface node is more restrictive. Note that Bir is not constant but p depends upon hr which increases with increasing To (time). Hence, the restriction on Fo p increases with increasing To (time). Continued ….. PROBLEM 5.107 (Cont.) (c) Consider the prescribed conditions with negligible convection (Bi = 0). The FDEs for the thick slab are: Surface (0) ( ) p+1 p To = 2Fo Bi ⋅ Fo + Bi r ⋅ Tsur + T1 + (1 − 2Bi ⋅ Fo − 2Bi r ⋅ Fo − 2Fo ) To p ) ( (8) p+1 p p p Tm = Fo Tm+1 + Tm-1 + (1 − 2Fo ) Tm Interior (m≥1) (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 will satisfy the stability criterion. A few trial calculations are helpful. A value of ∆t = 15s 2 provides Fo = 0.105, and using Eqs. (2) and (5), hr(300K, 1000K) = 72.3 W/m ⋅K and Bir = p 0.482. From the stability criterion, Eq. (10), find Fo ≤ 0.337. With increasing To , hr and Bir 2 increase: hr(800K, 1000K) = 150.6 W/m ⋅K, Bir = 1.004 and Fo ≤ 0.249. Hence, if p To < 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 are tabulated below. Note how h p and Bi p are evaluated at each time increment. Note that t = r r p⋅∆t, where ∆t = 15s. p t(s) To / h p / Bi r r 0 0 300 72.3 0.482 300 1 15 370.867 79.577 0.5305 2 30 3 4 T3 T4 300 300 300 300 300 300 300 426.079 85.984 0.5733 307.441 300 300 300 45 470.256 91.619 0.6108 319.117 300.781 300 300 60 502.289 333.061 302.624 300.082 300 After 60s(p = 4), T1(K) T2 To(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. 5.82 if this equation is reduced to one-dimension. (2) We should recognize that the ∆t = 15s time increment represents a coarse step. To improve the accuracy of the solution, a smaller ∆t should be chosen. PROBLEM 5.108 KNOWN: Thick slab of copper, initially at a uniform temperature, is suddenly exposed to a constant net 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(0.15 m, 120 s); compare results with those given by the exact solution in Comment 1; will a time increment of 0.12 s provide more accurate results?; and, (b) Plot the temperature histories for x = 0, 150 and 600 mm, 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 the interior nodes. Use space and time increments of 37.5 mm and 1.2 s, respectively, for a 17-node network. 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 mm approximates 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 the surface node 00 exposed to the net radiant flux was derived in the Example analysis. Selected portions of 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. Node 00 04 FDE results (°C) ∆t = 1.2 s 119.3 45.09 ∆t = 0.12 s 119.4 45.10 Analytical result (°C) Eq. 5.59 120.0 45.4 The numerical FDE-based results with the different time increments agree quite closely with one another. At the surface, the numerical results are nearly 1 °C less than the result from the exact analytical solution. This difference represents an error of -1% ( -1 °C / (120 – 20 ) °C x 100). At the x = 150 mm location, the difference is about -0.4 °C, representing an error of –1.5%. For this situation, the smaller time increment (0.12 s) did not provide improved accuracy. To improve the accuracy of the numerical model, it would be necessary to reduce the space increment, in addition to using 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 6 Te m p e ra tu re , T(x,t) 120 80 40 0 0 50 100 150 Tim 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 thermal penetration increases with increasing time, the temperature at the location x = 150 mm, T04 = T(150 mm, t), begins to increase after about 20 s. Note, however, the temperature at the location x = 600 mm, T16 = T(600 mm, t), does not change significantly within the 150 s duration of the applied surface heat flux. Our assumption of treating the +x boundary of the node 16 control volume as adiabatic is justified. A copper plate of 600-mm thickness is a good approximation to a semi-infinite medium at times less than 150 s. COMMENTS: Selected portions of the IHT code with the nodal equations to obtain the temperature distribution are shown below. Note how the FDE for node 00 is written in terms of an energy balance using 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 method q''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 shown Tinf16 = 20 // Arbitrary value h16 = 1e-8 // Causes boundary to behave as adiabatic PROBLEM 5.109 KNOWN: Thick slab of copper as treated in Example 5.9, initially at a uniform temperature, is suddenly exposed to large surroundings at 1000°C (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 FEHT for a surface emissivity of 0.94 and (b) Plot the temperature histories for x = 0, 150 and 600 mm, and explain key features of your results. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Slab of thickness 600 mm approximates 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-direction and arbitrary length in the y-direction. Click on Setup | Temperatures in K, to enter all temperatures in kelvins. 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 linearized radiation coefficient after Eq. 1.9 written as 0.94 * 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 a view of the input screen. From View|Temperatures, find the results: T(0, 120 s) = 339 K = 66°C < T(150 mm, 120 s) = 305K = 32°C (b) Using the View | Temperatures command, the temperature histories for x = 0, 150 and 600 mm (10 mm mesh, Nodes 18, 23 and 15, respectively) are plotted. As expected, the surface temperature increases markedly at early times. As thermal penetration increases with increasing time, the temperature at the location x = 150 mm begins to increase after about 30 s. Note, however, that the temperature at the location x = 600 mm does not change significantly within the 150 s exposure to the hot surroundings. Our assumption of treating the boundary at the x = 600 mm plane as adiabatic is justified. A copper plate of 600 mm is a good approximation to a semi-infinite medium at times less than 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 5.110 KNOWN: Electric heater sandwiched between two thick plates whose surfaces experience convection. 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 remains energized for the next 15 minutes. Case 3 corresponds to the eventual steady-state condition following the 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 the transient period between steady-state distributions for Case 2 and Case 3 using the finite-element approach 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 negligible thickness, and (4) Negligible thermal resistance between the heater surfaces and the plates. PROPERTIES: Plate material (given); ρ = 2500 kg/m3, c = 700 J/kg⋅K, k = 5 W/m⋅K. ANALYSIS: The temperature distribution for Case 2 shown in the above graph represents the initial condition for the period of time following the second loss of coolant event. The boundary conditions 2 at x = ±L are adiabatic, and the heater flux is maintained at q′′ = 4000 W/m for 0 ≤ t ≤ 15 min. o Using FEHT, the heater is represented as a plate of thickness Lh = 0.5 mm with very low thermal capacitance (ρ = 1 kg/m and c = 1 J/kg⋅K), very high thermal conductivity (k= 10,000 W/m⋅K), and a 3 uniform volumetric generation rate of q = q′′ / L h = 4000 W / m 2 / 0.0005 m = 8.0 × 106 W/m for 0 ≤ t ≤ o 900 s. In the Specify | Generation box, the generation was prescribed by the lookup file (see FEHT Help): ‘hfvst’,1,2,Time. This Notepad file is comprised of four lines, with the values on each line separated by a single tab space: 0 900 901 5000 8e6 8e6 0 0 The 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 the maximum temperature of 89°C at the time the heater is deactivated, t = 900 s. Continued ….. PROBLEM 5.110 For 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 obtain the 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 energy balance on the control volume shown in the schematic above. E′′ − E′′ + E′′ = E′′ in out gen st q′′ + q′′ + q′′ = E′′ a b o st p +1 p +1 p+ p+ p+ p T05 − T06 T07 1 − T06 1 T06 1 − T06 k +k + q′′ = ρ c∆x o ∆x ∆x ∆t The IHT code representing selected nodes is shown below for the adiabatic boundary node 01, interior node 02, and the heater-plates interface node 06. Note how the foregoing derived finite-difference equation in implicit form is written in the IHT Workspace. Note also the use of a Lookup Table for representing 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 shown qdot = 0 // No internal generation Tinf01 = 20 // Arbitrary value h01 = 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 06 k * (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" contains 0 4000 900 4000 900.5 0 5000 0 */ The temperature-time histories using the IHT code for the plate locations x = 0, ±L are shown in the graphs below. We chose to show expanded presentations of the histories at early times, just after the second loss of coolant event, t = 0, and around the time the heater is deactivated, t = 900 s. 60 90 Temperature, T (C) Temperature, T (C) 85 50 40 80 75 30 70 0 50 100 150 200 800 900 Time, t (s) Surface x = -L Center point, x = 0 Surface x = +L 1000 1100 1200 Time, t (s) Surface x = -L Center point, x = 0 Surface x = +L COMMENTS: (1) The maximum temperature during the transient period is at the center point and occurs at the instant the heater is deactivated, T(0, 900s) = 89°C. After 300 s, note that the two surface temperatures 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 solutions agree with the result of an energy balance on a time interval basis yielding Tss = 86.1°C. PROBLEM 5.111 KNOWN: Plane wall of thickness 2L, initially at a uniform temperature, is suddenly subjected to convection heat transfer. FIND: The mid-plane, T(0,t), and surface, T(L,t), temperatures at t = 50, 100, 200 and 500 s, using the following methods: (a) the one-term series solution; determine also the Biot number; (b) the lumped 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 predicted temperatures. 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 the one-term approximation to the series solution, Eq. 5.40 and 5.41. The Biot number for the heat transfer process is Bi = h L / k = 500 W / m 2 ⋅ K × 0.020 m / 15 W / m ⋅ K = 0.67 Since Bi >> 0.1, we expect an appreciable temperature difference between the mid-plane and surface as 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 the midplane compared to the one-term approximation. Solution method/Time(s) 50 100 200 500 Mid-plane, T(0,t) (°C) One-term, Eqs. 5.40, 5.41 Lumped capacita