2
1-D Steady-State Heat Conduction
2.1 Conduction through Plane Walls
For 1-D steady-state heat conduction in plane wall shown in Figure 2.1, without heat generation, the heat conduction equation 1.17 can be simplified as
∂ 2 T = 0
(2.1)
∂x 2
dT = C 1
dx
Equation 2.1 has the general solution
T = c 1 x + c 2
(2.2)
with boundary conditions:
at x = 0, T = T s1 = c 1 · 0 + c 2 = c 2
at x = L, T = T s2 = c 1 L + c 2
Solve for c 1 and c 2 ,
T s2 − T s1
c 1 =
, c 2 = T s1 ,
L
Substituting c 1 and c 2 into Equation 2.2, the temperature distribution is
T s,1 − T s,2
T(x) = T s,1 −
x
(2.3)
L
Applying Fourier’s Conduction Law, one obtains the heat transfer rate
through the plane wall
∂T
T s,1 − T s,2
T s,1 − T s,2
q = −kA
= kA
=
(2.4)
∂x
L
(L/KA)
13
1-D Steady-State Heat Conduction
2.1 Conduction through Plane Walls
For 1-D steady-state heat conduction in plane wall shown in Figure 2.1, without heat generation, the heat conduction equation 1.17 can be simplified as
∂ 2 T = 0
(2.1)
∂x 2
dT = C 1
dx
Equation 2.1 has the general solution
T = c 1 x + c 2
(2.2)
with boundary conditions:
at x = 0, T = T s1 = c 1 · 0 + c 2 = c 2
at x = L, T = T s2 = c 1 L + c 2
Solve for c 1 and c 2 ,
T s2 − T s1
c 1 =
, c 2 = T s1 ,
L
Substituting c 1 and c 2 into Equation 2.2, the temperature distribution is
T s,1 − T s,2
T(x) = T s,1 −
x
(2.3)
L
Applying Fourier’s Conduction Law, one obtains the heat transfer rate
through the plane wall
∂T
T s,1 − T s,2
T s,1 − T s,2
q = −kA
= kA
=
(2.4)
∂x
L
(L/KA)
13