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

Thus
a + 2a
=−Bθ
1 2
a
2
Bθ
=−
2
We can also write the second boundary condition as
From equation (3.117) we get
Average temperature for long cylinder can be written as
41
(3.139)
(3.140)
∂ θ
=− B( a0 + a1+ a2)
∂ x
(3.141)
⎛ B ⎞
a0 = θ ⎜1+ ⎟
⎝ 2 ⎠ (3.142)
( ) 1
1
θ = + ∫ θ
m
m x dx
0
Substituting the value of θ , m and integrating we get
Integrating non-dimensional governing equation we have
⎛ B ⎞
θ = θ⎜1+ ⎟
⎝ 4 ⎠ (3.143)
∂θ ∂ ⎛ ∂θ
⎞
x dx = ⎜x⎟dx ∂τ∂x⎝ ∂x
⎠
1
m
1
m
0 0
∫ ∫
(3.144)
Substituting the value of θ at equation (3.120) and from equation (3.114), (3.116), (118) we get
∫
1
0
∂θ
dx = −Bθ
∂τ
Considering the average temperature we may write
(3.145)
∂ θ
=−2Bθ
∂ τ
(3.146)