HEAT TRANSFER 787

dharm

\M-therm\Th15-1.pm5

15.2.5. Heat conduction through plane and composite walls

15.2.5.1. Heat conduction through a plane wall
Refer Fig. 15.3 (a). Consider a plane wall of homogeneous material through which heat is
flowing only in x-direction.
Let, L = Thickness of the plane wall,
A = Cross-sectional area of the wall,
k = Thermal conductivity of the wall material, and
t 1 , t 2 = Temperatures maintained at the two faces 1 and 2 of the wall, respectively.
The general heat conduction equation in cartesian coordinates is given by :

∂

∂

+ ∂

∂

+∂

∂

+=∂

∂τ

2

2

2

2

2

2

t 1

x

t

y

t

z

q

k

g t

α

. ...[Eqn. 15.13]

If the heat conduction takes place under the conditions,

steady state ∂
∂τ

F =

HG

I

KJ

t 0 , one-dimensional ∂

∂

=∂

∂

=

L

N

M

M

O

Q

P

P

2

2

2

2 0

t

y

t

z and

with no internal heat generation

q

k

F g=

HG

I

KJ

(^0) then the above
equation is reduced to :
∂
∂

2
2 0
t
x
,ordt
dx
2
2 =^0 ...(15.21)
By integrating the above differential twice, we have
∂
∂
t=
x
C 1 and t = C 1 x + C 2 ...(15.22)
where C 1 and C 2 are the arbitrary constants. The values of
these constants may be calculated from the known boundary
conditions as follows :
At x = 0 t = t 1
At x = Lt = t 2
Substituting the values in the eqn. (15.22), we get
t 1 = O + C 2 and t 2 = C 1 L + C 2
After simplification, we have, C 2 = t 1 and C 1 =
tt
L
21 −
Thus, the eqn. (15.22) reduces to :
t =
tt
L
F 21 − xt 1
HG
I
KJ + ...(15.23)
The eqn. (15.23) indicates that temperature distribution across a wall is linear and is
independent of thermal conductivity. Now heat through the plane wall can be found by using
Fourier’s equation as follows :
Q = – kA dt
dx
dt
dx
, whereFHG =temperature gradientIKJ ...[Eqn. (1.1)]
Q t^1 t^2 Q
(b)
(R )th cond.=kAL
x
Q dt Q
dx
K
t 2
t 1
t
1 2
L
(a)
Plane wall
Fig. 15.3. Heat conduction
through a plane wall.