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SOLUTION
1 ∂
∂θ
1 ∂θ
∂ 2 θ 1 ∂θ
1 ∂θ
r
=
or
+
=
r ∂r
∂r
α ∂t
∂r 2
r ∂r
α ∂t
Boundary conditions:
∂θ �
i. r = 0,
�
= 0
∂r r =0
∂θ �
ii. r = r o , −k
�
= hθ r o
∂r r =r o
Separation of variables:
θ = R(r )τ(t )
∂τ
2
−αλ 2 t
+ λ τ = 0; τ = C 3 e
∂t
∂ 2 R
1 ∂R
+
+ λ 2 R = 0
∂r 2
r ∂r
R(r ) = C 1 J 0 (λr ) + C 2 Y 0 (λr )
Applying BCs
∂R �
= −C 1 λJ 1 (0) + C 2 λY 1 (0) = 0, J 1 (0) = 0, ⇒ C 2 = 0
∂r r =0
∂R �
h
−k
= kC 1 λJ 1 (λr o ) = hθ R , λJ 1 (λr o ) = J 0 (λr o ), λ n = λr o
∂r
k
r =r o
λ n J 1 (λ n ) − BiJ 0 (λ n ) = 0
where Bi = (hr o /k ),
∞
�
�
2
r
−(α/r )λ 2
o n
r o
θ =
C n e
t J 0 λ n
n=1
at t = 0, θ = 1
∞
�
�
r
1 =
C n J 0 λ n r o
n=1

r o

∫ 0 rJ 0 (λ n (r /r o )) dr
(r 2 /λ n )J 1 (λ n )
o
C n =
=
r o
2
∫ 0 rJ 0
2 (λ n (r /r o )) dr
(r o /2)[J 0
2 (λ n ) + J 1
2 (λ n )]
∞ 2J 1 (λ n )J 0 (λ n (r /r o ))
λ 2
n
θ =
· e − F 0
J 2
n=1 λ n 0 (λ n ) + J 1
2 (λ n )
where F 0 = (αt /r 2 ).
o
94
Analytical Heat Transfer