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,
()
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
−
′′ =−
<