HW6 solution

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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).
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online