analysis of transient heat conduction in different geometries - ethesis ...

Differentiating the above equation with respect to x we get
Applying first boundary condition we have
∂ θ
= a1+ 2a2x
∂ x
a1+ 2a2x= 0
(3.112)
Thus
a 1 = 0
(3.113)
Applying second boundary condition we have
Thus
a + 2a
=−Bθ
1 2
a
2
Bθ
=−
2
We can also write the second boundary condition as
Using the above expression we have
Average temperature for long slab can be written as
37
(3.114)
(3.115)
∂ θ
=− B( a0 + a1+ a2)
∂ x
(3.116)
⎛ B ⎞
a0 = θ ⎜1+ ⎟
⎝ 2 ⎠ (3.117)
1
0 dx
θ θ = ∫
Substituting the value of θ and integrating we have
Bθ
θ = θ +
3
Integrating non-dimensional governing equation we have
(3.118)