sm3-081sm3-098sm3-105sm3-117sm3-146sm3-154PROBLEM 3.81KNOWN: Plane wall with internal heat generation which is insulated at the inner surface andsubjected to a convection process at the outer surface.FIND: Maximum temperature in the wall.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with uniformvolumetric heat generation, (3) Inner surface is adiabatic.ANALYSIS: The temperature at the inner surface is given by Eq. 3.48 and is the maximumtemperature within the wall,2o sT qL / 2k+T .The outer surface temperature follows from Eq. 3.51,s6 2s3T T qL/hWT 92 C+0.3 10 0.1m/500W/m K=92 C+60 C=152 C.m It follows that 26 3oT 0.3 10 W/m 0.1m / 2 25W/m K+152 C oT 60 C+152 C=212 C. <COMMENTS: The heat flux leaving the wall can be determined from knowledge of h, Tsand Tusing Newton’s law of cooling. 2 2conv sq h T 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 theformg outE E 0 whereg out convE qAL and E q A. Hence,6 3 2convq qL=0.3 10 W/m 0.1m=30kW/m . PROBLEM 3.98 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. 3.54 and 3.28, the heat equations for the fuel (f) and cladding (c) are () ()cf112dTdT1d q 1dr0rrr0rrrr dr dr r dr drfk=− ≤ ≤ = ≤ ≤⎛⎞⎛⎞⎜⎟⎜⎟⎝⎠⎝⎠& Hence, integrating both equations twice, 2f1 1f2ff ffdT qr C qr CTlnrCdr 2k k r 4k k=− + =− + +&& (1,2) c3 3c4ccdT C CTlnrCdr k r k==+ (3,4) The corresponding boundary conditions are: )()()ff1c1r0dT dr 0 T r T r=== (5,6) ()[]112ccfc c c2rrrr rrdT dTdTfkk khTrTdr dr dr∞===−=− − =−⎞⎞⎞⎟⎟⎟⎠⎠⎠ (7,8) 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, 2f2fqrTC4k=− +& (9) From Eq. (6), it follows that 231124fcClnrqrCC4k k−+= +& (10) Continued...PROBLEM 3.98 (Cont.) Also, from Eq. (7), 231131Cqr qror C2r 2=− =−&& (11) Finally, from Eq. (8), 33242cCChlnrCTrk∞−= +−⎡⎤⎢⎥⎣⎦ or, substituting for C3 and solving for C4 2211422cqr qrClnrT2r h 2k∞=+ +&& (12) Substituting Eqs. (11) and (12) into (10), it follows that 22 2 21111 122fc2cqr qr ln r qr qrClnrT4k 2k 2r h 2k∞=− + + +&& & & 22 211212fc12qr qr r qrClnT4k 2k r 2r h∞=+ + +&& & (13) Substituting Eq. (13) into (9), ()2222121f1fc12qqrrqrTrrln T4k 2k r 2r h∞=−+ ++&&& (14)< Substituting Eqs. (11) and (12) into (4), 22121cc2qr r qrTln T2k r 2r h∞=++&&. (15)< (b) Applying Eq. (14) at r = 0, the maximum fuel temperature for h = 2000 W/m2⋅K is ()() ()2283 83f2 10 W m 0.006m 2 10 W m 0.006m0.009mT0 ln4 2WmK 2 25WmK 0.006m×× ××=+×⋅ × ⋅ ()()28322 10 W m 0.006m300K2 0.009m 2000 W m K×++×⋅ () ( )fT 0 900 58.4 200 300 K 1458K=+++ = . < (c) Temperature distributions for the prescribed values of h are as follows: 0 0.001 0.002 0.004 0.005 0.006Radius in fuel element, r(m)300500700900110013001500Temperature, Tf(K)h = 2000 W/m^2.Kh = 5000 W/m^2.Kh = 10000 W/m^2.K 0.006 0.007 0.008 0.009Radius in cladding, r(m)300400500600Temperature, Tc(K)h = 2000 W/m^2.Kh = 5000 W/m^2.Kh = 10000 W/m^2.K Continued...PROBLEM 3.98 (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. COMMENTS: For the prescribed conditions, Eq. (14) yields, Tf(0) - Tf(r1) = 21fqr 4k& = (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 3.105 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 T∞ = 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 22os2oqrrT(r) 1 T6kr=−+⎛⎞⎜⎟⎜⎟⎝⎠& (1) To determine Ts, perform an energy balance on the apple as shown in the sketch above, with volume V = 433πro, in out g cvEE E0 qq0−+= −+∀=&& && ()()()23os oh4r T T q4r 3 0ππ∞−−+=& (2) ()()()222 333s7.5W m K 4 0.040 m T 5 C 38.9 W m 4 0.040 m 3 0ππ−⋅× −+ × =o where the volumetric generation rate is q 4000J kg day=⋅& ()()3q 4000J kg day 840kg m 1day 24hr 1hr 3600s=⋅× × ×& 3q38.9Wm=& and solving for Ts, find sT5.14C=o < From Eq. (1), at r = 0, with Ts, find 32238.9W m 0.040 mT(0) 5.14 C 0.12 C 5.14 C 5.26 C60.5WmK×= +=+=×⋅oooo < Continued...PROBLEM 3.105 (Cont.) (b) With the convection coefficient depending upon velocity, 0.4251hCV= 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. 0 0.2 0.4 0.6 0.8 1Ventilation air velocity, V (m/s)5.25.35.4Center temperature, T(0) (C) COMMENTS: (1) While the temperature within the apple is nearly isothermal, the
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