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Unformatted text preview: Homework 2 due Wednesday, September 1 1. A plane wall of thickness 2L = 40 mm and thermal conductivity k = 5 W/m‐K experiences uniform volumetric heat generation q(dot), while convection heat transfer occurs at both of its surfaces, each of which is exposed to a fluid of temperature T∞ = 20o C. Under steady‐state conditions, the temperature distribution in the wall is T(x) = 80 ‐ 200x ‐ 1.5 X 104x2 (x is in meters). The origin of the x‐coordinate is at the midplane of the wall. The density and specific heat of the wall are 2600 kg/m3 and 800 J/kg‐K respectively. a) Graph the temperature distribution in the wall. Identify significant physical features. b) What is the volumetric rate of heat generation? c) What are the heat fluxes at the two surfaces? How are they related to the heat generation rate? d) What are convection coefficients at the surfaces? e) Obtain an expression for the heat flux distribution. Explain any significant features of the distribution. f) If the source of the heat generation is suddenly deactivated (q(dot) = 0), what is the rate of change of energy stored in the wall at this instant? g) What temperature will the wall eventually reach with q(dot) = 0? How much energy must be removed by the fluid per unit area (J/m2) to reach this state? a) Data: x T ‐0.02 78 82 ‐0.015 79.625 80 ‐0.01 80.5 78 ‐0.005 80.625 76 0 80 74 T 72 0.005 78.625 70 0.01 76.5 68 0.015 73.625 ‐0.03 ‐0.02 ‐0.01 0 0.01 0.02 0.03 0.02 70 T b. k d 2T d 2T W & & + q = 0 q = − k 2 = −k * 2c = −5 * 2 * −1.5 X 10 4 = 150,000 3 2 dx dx m
dT = − k (b + 2cx) = −5(−200 + 2 * −15000 x) dx c. q x " = − k at x = L, q” = ‐2000 W/m2 at x = ‐L, q” = 4000/m2 & Ein Back to the energy balance: ‐2000‐4000 + q(dot)*2L = 0 does it? ‐2000‐4000+(150000*2*.02) = 0 d. h(T∞ − TL ) = − q" ( − L ) at x =‐ L T = 78, h = ‐4000/20‐78 = 69 W/m2‐K; & & − Eout + q = 0 at x = L, h = 2000/78‐70 = 250 W/m2‐K e. the flux is linear with x. The max would be ‐3.33 mm, beyond the wall. f. Now the form of the heat equation is: q x " = −k dT = − k (b + 2cx) = −5(−200 + 2 * −15000 x) dx k d 2T dT = ρc p 2 dx dt = k * 2c = 5 * 2 * −1.5 X 10 4 = −150,000 g. With no heat generation, the temperature of the wall becomes the temperature of the outside fluid, 20o C. So: Ein” – Eout” = ΔEst = Ef – Ei Ein = 0 Ef occurs when T = T∞, Ei corresponds to the steady‐state temperature distribution, which is a function of x. So we integrate from –L to L: W m3 ⎡ ⎤ ⎡ ⎤ bx 2 cx 3 2cL3 E"out = ρc p ∫ a + bx + cx − T∞ dx = ρc p ⎢ax + + − T∞ x ⎥ = ρc p ⎢2aL + − 2T∞ L ⎥ = 2 3 3 ⎣ ⎦ −L ⎣ ⎦ −L
L [ 2 L 5.16 X 106 J/m2 2. A composite wall of an oven consists of three materials. kA = 20 W/m‐K, and kC = 50 W/m‐K. LA = 0.30 m, LB = 0.15 m, and LC = 0.15 m. The outside surface temperature of the oven is 20o C and the inside surface temperature is 613o C. The oven air temperature is 800 o C, with a heat transfer coefficient h = 25 W/m2‐K. a) Draw the circuit for this system. b) What is kB? c) What material do you think B is made of? a) b) q= Ts ,i − Ts ,o 613 − 20 593 = = 0.15 .3m 0.15m 0.15m L A LB LC + + 0.015 + + .003 + + kB kB 20W / m − K 50W / m − K k A k B kC q = h(T∞ − Ts ,i ) = 25W / m 2 − K (800 − 613) = 4675 W / m 2
kB =1.38 Wm‐K c) Silicon dioxide 3. Consider the problem we did in class with the composite spherical shell with heat generation. What would the generation have to be to make this proposal flawed? For the proposal to be flawed, the T1 must equal the melting point of lead which equals 601. q = T1 ‐ T∞/Rtot = 601 – 10o C/0.00372 = 158871 W ...
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This note was uploaded on 11/12/2010 for the course CHEN 3320 at Colorado.
 '10
 FALCONER,J

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