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

a + 2a
=−Bθ
1 2
Subtracting the equation (3.30) from (3.29) we get
a
2
Bθ − Q
=
2 1
( ε − )
Substituting the value at equation (3.30) we have
Using second boundary conditions we have
( 1)
( )
25
(3.33)
(3.34)
−Bθ ε − − Bθ −Q
a1
=
ε − 1
(3.35)
⎛a1+ 2a2
⎞
a0 = −⎜ ⎟−a1−a2
⎝ B ⎠ (3.36)
Thus substituting the value of and at yhe expression of we get the following value
a
0
( − ) + B ( − ) + ( B −Q)
2( ε −1)
2θ ε 1 2 θ ε 1 θ
=
We may write the average temperature equation as
Where m=1 for cylinderical co-ordinate
Thus the above equation may be written as
1
m
x dx
mε θ = ∫ θ
Substituting the value of and integrating equation (3.35) we get
1
(3.37)
θ = ∫ θxdx
ε
(3.38)
2 2 3 4
⎛a0 a1 a2 ⎞ ⎛a0εa1εa1ε a2ε
⎞
θ = ⎜ + + ⎟−
⎜ + + + ⎟
⎝ 2 3 4 ⎠ ⎝ 2 2 3 4 ⎠ (3.39)
is the ratio of inside diameter and outside diameter of the cylinder
2 2 3 4
⎛a0εa1ε a1ε a2ε
⎞
⎜ + + + ⎟=
0
⎝ 2 2 3 4 ⎠ (3.40)
Thus considering ε = 0 and substituting the value of
a0, a1, a 2 at equation (3.40) we get the
value of θ as