PROBLEM 3.1
KNOWN:
One-dimensional, 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) One-dimensional conduction, (2) Steady-state 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,
( )
1
2
T x
C x
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
(
)
(
)
1
1
,1
1
2
k C
0
h
T
C
0
C
∞
−
+
=
−
⋅
+
(4)
and for the BC at x = L to find
(
)
(
)
1
2
1
2
,2
k C
0
h
C L
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
2
,1
1
1
2
1
2
T
T
T
T
C
C
T
1
1
L
1
1
L
k
h
h
h
k
h
h
k
∞
∞
∞
∞
∞
−
−
= −
= −
+
+
+
+
+
( )
(
)
,1
,2
,1
1
1
2
T
T
x
1
T x
T
.
k
h
1
1
L
h
h
k
∞
∞
∞
−
= −
+
+
+
+
<
From Fourier’s law, the heat flux is a constant and of the form
(
)
,1
,2
x
1
1
2
T
T
dT
q
k
k C
.
dx
1
1
L
h
h
k
∞
∞
−
′′ = −
= −
= +
+
+
<