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assignment8 - PROBLEM 16.3 KNOWN: One-dimensional system...

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Unformatted text preview: PROBLEM 16.3 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: l‘*—’l— L =2.0mm 7;=325/< 7;=Z75/< Mafer'ia/ of known k l_"x 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 = 300 K. Property values and table identification are shown below. ANALYSIS: For this system, Fourier’s law can be written as dT T2 — T1 ” = —k — = —k . qx dx L Substituting numerical values, the heat flux is 275-325 K q; = —k(——T)— = +25005-k 20 x 10- m m where q; will have units W/m2 if k has units W/m-K. The heat fluxes for each system follow. Thermal conductivity Heat flux Material Table k(W/m-K) q; (kW/ m2) —————_______—_ (a) Pure Aluminum HT-l 237 593 < (b) Plain carbon steel HT-l 60.5 151 (c) AISI 316, SS. HT—1 13.4 33.5 (d) Pyroceram HT-l 3.98 9.95 (e) Teflon HT-2 0.35 0.88 (1) Concrete HT-2 1.4 3.5 COMMENTS: Recognize that the thermal conductivity of these solid materials varies by more than two orders of magnitude. PROBLEM 16.6 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 volumetric 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. SCHEMATIC: T(x) = a + bx + cx2 k = 5 W/m-K, q' T(0) = To = 120°C Tm: 20°C . If x , ’ = 2- ® ' h 500 W/m K is L_> x L=50mm Insulated boundary ASSUMPTIONS: (l) 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. Ein — E[out + Egen = 0 Where Ein = qlconv h(T°°—TO)+qL=O (1) q=—h(r°° —TO)/L=—500W/m2 -K(20—120)°C/0.050 m=1.0><106 W/m3 < To 1’ V q., a can.) Z conv —> a. ""7?! d I i.‘ 4 1 % Egan | —L> x = L L—> x L (3) Overall energy balance (b) Surface energy balances (b) The coefficients of the temperature distribution, T(x) = a + bx + cxz, can be evaluated by applying the boundary conditions at x = 0 and x = L. See Table 16.1 for representation of the boundary conditions, and the schematic above for the relevant surface energy balances. Boundary condition at x = 0, convection surface condition Ein _ E[out = Ciiionv — (lit (0) = 0 Where (1x (0) = ‘k— h(T°° —To)—[—k(0+b+20x)x=0] = 0 b=—h(T°°—TO)/k=—500W/m2.K(20—120)°C/5W/m-K=1.0><104K/m < Continued . . . .. PROBLEM 16.6 (Cont.) Boundary condition at x = L, adiabatic or insulated surface . . , , dT Ein —E0ut = —qx (L) = 0 where qx (L) =—k— dX =L k[0+b+2cx]x=L =0 (3) c=—b/2L=—1.0><104K/m/(2><0.050m)=—1.0><105K/m2 < Since the surface temperature at x = 0 is known, T(0) = T0 = 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 in the graph below. 400 6 ’_ 300 9 J E (T: Q g 200 ‘ F. 100 0 1O 20 30 4O 50 Wall position, x (mm) COMMENTS: The temperature distribution has a quadratic form as expected for steady-state conduction with constant properties and uniform volumetric energy generation (see Eq. 16.43). The maximum temperature is at the insulated (adiabatic) boundary, x = L, where the temperature gradient must be zero. The temperature gradient at x = 0 is, of course, finite since the conduction heat flux must be equal to the convective heat flux to the ambient fluid. PROBLEM 16.8 KNOWN: Dimensions of oven window materials as described in Example 16.1. The outside convection coefficient is increased to 35 W/m2 -K , while all other conditions remain the same. FIND: (a) Temperature of the outer surface, Ts,o , and (b) Temperature at the interface between the two plastics (A and B). SCHEMATIC: LA = 41.5 mm L3 = 20.7 mm Tsi = 335°C 5'" 5 Ts o “ Ts,i TAB Ts,o Too ' v ' q ———> WW—uvw. fl 6311’} R’bd.A Rae "cw, ifiggii/mz-K kA = 0.15 W/m-K k3 = 0.08 W/m-K ASSUMPTIONS: (1) Steady-state conditions, (2) One dimensional conduction through windows, (3) Negligible contact resistance between plastic materials, (4) Negligible radiation transfer through windows, and (5) Constant properties. ANALYSIS: (a) From the thermal circuit above representing the composite window with its boundary conditions, find the outer surface temperature, Ts,o , from an energy balance q, _ Ts,i "Too _ Ts,o ‘ Too R'cd,A + R'c':d,B + Rcv,o R'cvp The thermal resistances associated with conduction through the windows and convection from the surface are Rad’A =L_A=_M=o_277 m2 .K/w kA 0.15 W/m-K RgdB=£fi=m=0259 m2.K/w ’ kB 0.08 W/m-K Rgv,0=-1— 1 =0.0286 m2 -K/W ho ’M Substituting numerical values in the energy balance expression, find (385—25)°C _ T“, —25°C (0.277+0.259+0.0286)m2 -K/W _ 0.0286 m2 -K/ w T“, =43°C < (b) Referring to the thermal circuit, write the energy balance q" _ Ts,i "Too _ Ts,i ‘TAB REd,A +R2d,B +R2v,o REd,A Continued ..... PROBLEM 16.8 (Cont) and substituting numerical values, find TAB, (385-25)°C _ 385°C—TAB 0.5646 m2 -K/W 0.277 m2 -K/W TAB = 208°C COMNIENTS: Note that Rgv’o is nearly 6 times smaller than the thermal resistances of the windows. As such, changes in the convection coefficient will have a slight effect on the window surface temperature, as well as on the heat rate through the window. PROBLEM 16.14 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 T0 = 40°C. SCHEMATIC: Insulation, k = 0.05 W/m-K To = 40°C Too i Ti To Too o Qi Rafi Rad vao = 25°C .. h0 = 10 W/m2-K qrad = 100 Wm2 M Ti Tm = 300°C hi=30W/m2-K \ L IE ,I ASSUNIPTIONS: (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 0- nodes finding Too,i -Ti + T0 —Ti . ,, +(1" d = 0 (1) Rcv,i cd ra Ti —T0 +Too,o-T0 =0 (2) cd R'c’:v,o where the thermal resistances are 11ng =1/hi = 0.0333 m2 .K/w (3) R301 =L/k=L/0.05 mZ-K/W (4) Ram =1/h0 =0.0100 m2 -K/W (5) Substituting numerical values, and solving Eqs. (1) and (2) simultaneously, find L = 86 m < COMNIENTS: (l) The temperature at the inner surface can be found from an energy balance on the i-node using the value found for L. T -——T- T —T- «391 l_+_____w’o 1 = 0 = II n n . cv,o Rcd+ cv,1 It follows that Ti is close to Tm since the wall represents the dominant resistance of the system. (2) Verify that qi' = 50 W / m2 and q"O = 150 W / m2 . Is the overall energy balance on the system satisfied? PROBLEM 16.19 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: QED = 1000 mm2 Tco = 25°C A“ h=1000Wlm2-K ———p L=2mm Al cover Chip {c = 0.5x10'4 m2-KNV pm TO M = 85cc ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Negligible heat loss from sides and bottom, (4) Chip is isothermal. PROPERTIES: Table HT -1 , Aluminum (T z 325 K): k = 238 W/m-K. ANALYSIS: For a control surface about the chip, conservation of energy yields Pelec = (lo where the heat transfer rate from the chip, qc, can be expressed in terms of the conduction, contact and thermal resistances, P _ (Tc — Too ) As 616° _ [(L/k) + Rgc +(1/h)] Substituting numerical values with T0 = Tmax, the maximum chip power is (85—25)°c(10‘4m2) Pelec=——‘—“—T_——2_—— [(0.002/238)+0.5x10 +(1/1000)]m -K/W 60x10‘4 OC-m2 Pelee: "" _6 _" ———_3 2 "’— (8.4x10 +0.5x10 +10 )m -K/W Palec = 5.7 W. < COMMENTS: The dominant resistance is that due to convection (Rconv > Rt,c >> Roond ...
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assignment8 - PROBLEM 16.3 KNOWN: One-dimensional system...

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