PROBLEM 3.101
KNOWN:
Dimensions of a plate insulated on its bottom and thermally joined to heat sinks at its
ends.
Net heat flux at top surface.
FIND:
(a) Differential equation which determines temperature distribution in plate, (b) Temperature
distribution and heat loss to heat sinks.
SCHEMATIC:
ASSUMPTIONS:
(1) Steadystate, (2) Onedimensional conduction in x (W,L>>t), (3) Constant
properties, (4) Uniform surface heat flux, (5) Adiabatic bottom, (6) Negligible contact resistance.
ANALYSIS:
(a) Applying conservation of energy to the differential control volume, q
x
+ dq
= q
x +dx
, where q
x+dx
= q
x
+ (dq
x
/dx) dx and
()
o
dq=q W dx .
′′
⋅
Hence,
xo
dq / dx
q W=0.
−
From Fourier’s law,
x
q
k t W dT/dx.
=−
⋅
Hence, the differential equation for the
temperature distribution is
2
o
o
2
q
dd
T
d
T
ktW
q W=0
0.
dx
dx
kt
dx
−−
+
=
<
(b) Integrating twice, the general solution is,
2
o
12
q
T x
x
C x +C
2kt
+
and appropriate boundary conditions are T(0) = T
o
, and T(L) = T
o
.
Hence, T
o
= C
2
, and
2
oo
o1
2
1
qq
L
T
L
C L+C
and
C
.
2kt
2kt
+
=
Hence, the temperature distribution is
( )
2
o
o
qL
Tx
x
L
x T.
2kt
−
+
<
Applying Fourier’s law at x = 0, and at x = L,
() ( )
x=0
x=0
W
L
L
q 0
k Wt dT/dx)
kWt
x
kt
2
2
−
−
x=L
x=L
W
L
L
q L
k Wt dT/dx)
kWt
x
kt
2
2
−
−
=+
Hence the heat loss from the plates is
q=2 q WL/2
q WL.
=
<
COMMENTS:
(1) Note signs associated with q(0) and q(L).
(2) Note symmetry about x =
L/2.
Alternative boundary conditions are T(0) = T
o
and dT/dx)
x=L/2
=0.
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View Full DocumentPROBLEM 3.102
KNOWN:
Dimensions and surface conditions of a plate thermally joined at its ends to heat sinks at
different temperatures.
FIND:
(a) Differential equation which determines temperature distribution in plate, (b) Temperature
distribution and an expression for the heat rate from the plate to the sinks, and (c) Compute and plot
temperature distribution and heat rates corresponding to changes in different parameters.
SCHEMATIC:
ASSUMPTIONS:
(1) Steadystate conditions, (2) Onedimensional conduction in x (W,L >> t), (3)
Constant properties, (4) Uniform surface heat flux and convection coefficient, (5) Negligible contact
resistance.
ANALYSIS:
(a) Applying conservation of energy to the differential control volume
xo
x
d
xc
o
n
v
qd
qq
d
q
+
+=
+
where
()
xd
x
x
x
q
q
dq
dx dx
+
=+
(
)
conv
dq
h T
T
W dx
∞
=−
⋅
Hence,
(
)
(
)
x
x
q
q
W dx
q
dq
dx dx
h T
T
W dx
∞
′′
+⋅
=
+
+
−
⋅
x
o
dq
hW T
T
q W
dx
∞
+−
=
.
Using Fourier’s law,
x
qk
t
W
d
T
d
x
⋅
,
2
o
2
dT
ktW
hW T
T
q
dx
∞
−+
−
=
2
o
2
q
dT h
TT
0
kt
kt
dx
∞
−−
+
=
.
<
(b) Introducing
T
T
θ
∞
≡−
, the differential equation becomes
2
o
2
q
dh
0
kt
kt
dx
=
.
This differential equation is of second order with constant coefficients and a source term.
With
2
hk
t
λ
≡
and
o
Sqk
t
≡
, it follows that the general solution is of the form
xx
2
12
Ce
C e
S
λλ
θλ
=++
.(
1
)
Appropriate boundary conditions are:
oo
LL
(0)
T
T
(L)
T
T
θθ
∞∞
≡
≡
(2,3)
Substituting the boundary conditions, Eqs. (2,3) into the general solution, Eq. (1),
002
o1
2
S
2
L1
2
S
(4,5)
To solve for C
2
, multiply Eq. (4) by e
+
λ
L
and add the result to Eq. (5),
L
2
L
oL
2
eC
e
e
S
e
1
++
−
+
=
−
+
+
−
+
(
)
L2
L
L
L
2L
o
S
e
1
e
e
+
−
−
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 Fall '07
 SAVASYAVUZKURT
 Heat, Heat Transfer, Eqs., temperature distribution

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