HW6 solution

# HW6 solution - CHE/ME 109 Homework 6 15 cm air 500C 5 m/s...

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CHE/ME 109 Homework 6 Name__ Solution _________ 15 cm air 50 0 C 5 m/s 15 W Problem I. A 15 x 15-cm circuit board dissipating 15 W of power uniformly is cooled by air, which approaches the circuit board at 50 0 C with a velocity of 5 m/s. Flow across the circuit board is turbulent due to the presence of the components. Disregarding any heat transfer from the back surface of the board, determine the surface temperature of the electronic components a.) at a distance of 1 cm from the leading edge of the circuit board Since we don’t have the surface temperature, determine the properties of air at the air temperature: at T = 50 0 C = 323.15K from Table A-15: k = 0.02735 W/m 0 C; Pr = 0.7202 and ν = 1.798 x 10 -5 . ν δ = V Re =(5 m/s)*(0.01 m)/ 1.798 x 10 -5 m 2 /s = 2,781 this is a laminar Re but we are told that flow along the board is turbulent due to the presence of the components, so we will use equation 7-32 for the Nu assuming uniform heat flux from the board: 333 . 8 . 0 Pr Re 0308 . 0 x x Nu = = 15.72 h x = Nu x k/x = 15.72*(0.02735 W/m 2 0 C)/0.01 m = 42.99 ) ( = T T hA Q s & => 15W = 41.87 (0.15m) 2 *(T s - 50 0 C) T s = 65.51 0 C Notice we should recalculate the properties of air to calculate the temperature more accurately. Just for the exercise, the film temperature would become: T f = (65.51+50)/2 = 57.75 TdegC = k(W/mC)= 0.027919 Pr= 0.72086 ν (m2/s) 1.873607 Re=2,668 Nu x =15.21 h x = 42.48 and Ts = 65.69 0 C very close. .. b.) at the end of the circuit board (15 cm from the leading edge).

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For calculation of the film temp, I will use the surface temperature calculated at the 1 cm from the leading edge as calculated in part a. T f = (65.69+50)/2 = 57.85 TdegC = k(W/mC)= 0.02793 Pr= 0.7208 ν (m2/s) 1.875E-05 ν
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HW6 solution - CHE/ME 109 Homework 6 15 cm air 500C 5 m/s...

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