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Example 62
The left side of a large wall is held at 300°F while the right
side faces an evacuated chamber, as shown. Heat is
transferred by conduction through the wall and by
radiation from the right side of the wall, which has an
emissivity of 0.89.
The wall has a thermal conductivity of
0.041 Btu/h·ft·°F. The far side of the evacuated chamber
is black at a temperature of 70°F. Calculate the heat
transfer through a 1 ft
2
section of the wall by two methods:
a) Compute the radiative heat transfer coefficient and add
resistances in series. To find
h
rad
, assume the right wall
is at the average of
T
w
and
T
surr
.
b) Calculate the exact temperature of the right side of the
wall by equating conduction and radiation heat transfer
rates. Use this temperature to compute the heat
transferred. Compare to part a)
Solution:
a)
Assume the right wall is at the average of the other two
temperatures:
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o
300
70
185 F
22
ws
u
r
r
s
TT
T
+
+
==
=
185
460
645R
s
T
=+=
322
3
rad
s
s
surr
s
surr
surr
hT
T
T
T
T
T
εσ
=+++
70
460
530R
surr
T
=+ =
()()
8
24
3
33
Btu
0.89
0.1714
10
hf
t R
645
645
530
645
530
530
R
rad
h
−
⎛⎞
=×
⎜⎟
⋅⋅
⎝⎠
×+
+
+
2
Btu
1.25
rad
h
=
Calculate the resistances for radiation and conduction
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 Spring '08
 BORCATASCIUC

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