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

Thus,
a
2
Bθ
=−
2
We can also write the second boundary condition as
Using equation (3.55) , (3.58)and (3.60-3.61) we have
Average temperature equation used in this problem is
Substituting the value of θ we have
Integrating equation (3.64) we have
Substituting the value of 0 a , 1 a , 2 a we have
29
(3.60)
∂ θ
=− B( a0 + a1+ a2)
∂ x
(3.61)
⎛ B ⎞
a0 = θ ⎜1+ ⎟
⎝ 2 ⎠ (3.62)
1
0 dx
θ θ = ∫ (3.63)
2
( 0 1 2 )
1
∫ a a x a x dx
0 (3.64)
θ = + +
a1 a2
θ = a0
+ +
2 3
Integrating non-dimensional governing equation we have
(3.65)
⎛ B ⎞
θ = θ⎜1+ ⎟
⎝ 3 ⎠ (3.66)
∂θ ∂ θ
dx dx Gdx
∂τ∂x 2
1 1 1
= +
0 0 2 0
∫ ∫ ∫