PROBLEM 3.1
KNOWN:
Onedimensional, plane wall separating hot and cold fluids at T
and T
,1
,2
∞∞
,
respectively.
FIND:
Temperature distribution, T(x), and heat flux,
′′
q
x
, in terms of T
T
h
,1
,2
1
,,
,
h
2
, k
and L.
SCHEMATIC:
ASSUMPTIONS:
(1) Onedimensional conduction, (2) Steadystate conditions, (3) Constant
properties, (4) Negligible radiation, (5) No generation.
ANALYSIS:
For the foregoing conditions, the general solution to the heat diffusion equation
is of the form, Equation 3.2,
()
12
Tx Cx C.
=+
(1)
The constants of integration, C
1
and C
2
, are determined by using surface energy balance
conditions at x = 0 and x = L, Equation 2.23, and as illustrated above,
1,
1
2
,
2
x=0
x=L
dT
dT
k
h
T
T 0
k
h
T L
T
.
dt
dx
−=
−
−
=−
(2,3)
For the BC at x = 0, Equation (2), use Equation (1) to find
(
)
11
,
1
1
2
kC 0 h T
C 0 C
∞
−+
=
−
⋅
+
(4)
and for the BC at x = L to find
()(
)
1
2
,
2
kC 0 h
CL C
T
.
∞
=
+
−
(5)
Multiply Eq. (4) by h
2
and Eq. (5) by h
1
, and add the equations to obtain C
1
.
Then substitute
C
1
into Eq. (4) to obtain C
2
.
The results are
,1
,2
,1
,2
,
1
1
TT
C
C
T
L
L
kh
hh
k
k
∞
−−
+
++
,1
,2
,1
1
x1
Tx
T .
L
k
∞
−
+
+
<
From Fourier’s law, the heat flux is a constant and of the form
,1
,2
dT
qk
k
C
.
dx
L
k
−
′′ =−
<
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View Full DocumentPROBLEM 3.2
KNOWN:
Temperatures and convection coefficients associated with air at the inner and outer surfaces
of a rear window.
FIND:
(a) Inner and outer window surface temperatures, T
s,i
and T
s,o
, and (b) T
s,i
and T
s,o
as a function of
the outside air temperature T
∞
,o
and for selected values of outer convection coefficient, h
o
.
SCHEMATIC:
ASSUMPTIONS:
(1) Steadystate conditions, (2) Onedimensional conduction, (3) Negligible radiation
effects, (4) Constant properties.
PROPERTIES:
Table A3
, Glass (300 K):
k = 1.4 W/m
⋅
K.
ANALYSIS:
(a) The heat flux may be obtained from Eqs. 3.11 and 3.12,
( )
,i
,o
22
oi
40 C
10 C
TT
q
1
L
1
1
0.004m
1
h
k
h
1.4 W m K
65W m
K
30 W m
K
∞∞
−−
−
′′
==
++
+
+
⋅
⋅⋅
()
2
2
50 C
q9
6
8
W
m
0.0154
0.0029
0.0333 m
K W
⋅
.
Hence, with
i,
i
,
o
qh
T
T
=−
, the inner surface temperature is
2
s,i
,i
2
i
6
8
W
m
T
T
40 C
7.7 C
h
30 W m
K
∞
=
−
=
⋅
<
Similarly for the outer surface temperature with
o
s,o
,o
T
T
∞
find
2
2
o
6
8
W
m
T
T
10 C
4.9 C
h
K
∞
=
=
⋅
<
(b) Using the same analysis, T
s,i
and T
s,o
have been computed and plotted as a function of the outside air
temperature, T
∞
,o
, for outer convection coefficients of h
o
= 2, 65, and 100 W/m
2
⋅
K.
As expected, T
s,i
and
T
s,o
are linear with changes in the outside air temperature.
The difference between T
s,i
and T
s,o
increases
with increasing convection coefficient, since the heat flux through the window likewise increases.
This
difference is larger at lower outside air temperatures for the same reason.
Note that with h
o
= 2 W/m
2
⋅
K,
T
s,i
 T
s,o
, is too small to show on the plot.
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 Fall '07
 SAVASYAVUZKURT
 Heat, Heat Transfer, tc, heat loss

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