Derivation of heat conduction through a plane wall. Reference: Heat and Mass Transfer,R.K.Rajput

1. HEAT CONDUCTION THROUGH A PLANE WALL
• Let us consider a plane wall of homogeneous material through which heat
is flowing in x-direction.
• Let Q +x
L = thickness of the wall T0
A = cross-sectional area of the wall k
k = thermal conductivity of wall material T1
T0 , T1 = temperature maintained at surfaces 1 and 2. 1 2
L
k

2. • General heat conduction equation is:
𝜕2 𝑇
𝜕𝑥2
+
𝜕2 𝑇
𝜕𝑦2
+
𝜕2 𝑇
𝜕𝑧2
+
𝑞
𝑘
=
1
𝛼
.
𝜕𝑇
𝜕𝑡
• For one dimensional steady state system (
𝜕𝑇
𝜕𝑡
= 0)
• With no heat generation (
𝑞
𝑘
= 0)
• One dimensional flow (
𝜕2 𝑇
𝜕𝑦2 =
𝜕2 𝑇
𝜕𝑧2 = 0)
• Then, heat equation will be
𝜕2 𝑇
𝜕𝑥2 = 0 or
𝑑2 𝑇
𝑑𝑥2 = 0

3. 𝑑2 𝑇
𝑑𝑥2
= 0
Integrating the above expression;
𝑑𝑇
𝑑𝑥
= 𝐶1
Integrating it again;
T = C1.x + C2 ……………….. (1)
Where C1 and C2 are arbitrary constants.

4. • At x = 0; T = T0
• At x = L; T = T1
• From the expression derived above T = C1.x + C2
• At x = 0;
T0 = C1 (0)+ C2
C2 = T0
• At x = L;
T1 = C1 . L + T0
C1 = (T1 - T0 )/L
Eqn. 1 can be re-written as
T = (
𝑻𝟏−𝑻𝟎
𝑳
) . 𝒙 + T0

5. • Inference:
1. Temperature distribution across the wall is linear.
2. Temperature distribution is independent of k.
From Fourier’s Law, we have
Q = - k. A.
𝑑𝑇
𝑑𝑥
𝑑
𝑑𝑥
(
𝑇1−𝑇0
𝐿
. 𝑥 + 𝑇0) =
𝑇1−𝑇0
𝐿
Fourier’s Law can be re-written as
Q = k. A.(
𝑇0−𝑇1
𝐿
)

6. • Another way of writing the Fourier’s Law is
Q =
𝑇0−𝑇1
𝐿/𝑘𝐴
Where L/kA = Thermal Resistance of heat conduction (Rth)cond.
(Rth)cond. = (L/kA)