2-92 Heat is generated in a large plane wall whose one side is insulated while the other side is maintained at a specified
temperature. The mathematical formulation, the variation of temperature in the wall, and the temperature of the insulated
surface are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional
since the wall is large relative to its thickness, and there is thermal symmetry about the center plane. 3 Thermal conductivity
is constant. 4 Heat generation varies with location in the x direction.
Properties The thermal conductivity is given to be k = 30 W/m°C.
Analysis (a) Noting that heat transfer is steady and one-dimensional in x
direction, the mathematical formulation of this problem can be expressed as
0
)(
gen
2
2
=+ k
xe
dx
Td
where
and
= 8106 W/m3
and
(insulated surface at x = 0)
30C (specified surface temperature)
(b) Rearranging the differential equation and integrating,
1
/5.0
0
1
/5.0
0
/5.0
0
2
22
/5.0
Ce
k
Le
dx
dT
C
L
e
k
e
dx
dT
e
k
e
dx
Td Lx
Lx
Lx +=→+
−
−=→−= −
−
−
Integrating one more time,
4
)(
/5.0
2
)( 21
/5.0
2
0
21
/5.0
0CxCe
k
Le
xTCxC
L
e
k
Le
xT Lx
Lx
++−=→++
−
=−
−
(1)
Applying the boundary conditions:
B.C. at x = 0:
k
Le
CC
k
Le
Ce
k
Le
dx
dT L0
11
0
1
/05.0
02
2
0
2
)0( −=→+=→+= −
B. C. at x = L:
k
Le
e
k
Le
TCCLCe
k
Le
TLT LL
2
0
5.0
2
0
2221
/5.0
2
0
2
24
4
)(
++=→++−== −−
Substituting the C1 and C2 relations into Eq. (1) and rearranging give
)]/1(2)(4[)( /5.05.0
2
0
2Lxee
k
Le
TxT Lx −+−+= −−
which is the desired solution for the temperature distribution in the wall as a function of x.
(c) The temperature at the insulate surface (x = 0) is determined by substituting the known quantities to be
C314=
−+−
+=
−+−+=
−
−
)]02()1(4[
C) W/m30(
m) 05.0)( W/m10(8
C30
)]/02()(4[)0(
5.0
236
05.0
2
0
2
e
Lee
k
Le
TT
Therefore, there is a temperature difference of almost 300°C between the two sides of the plate.