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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 15cm 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 A15:
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 732 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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View Full DocumentFor 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)
1.875E05
ν
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 Spring '08
 Staff

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