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

T −T
θ =
T −T
0
∞
∞
,
hR
B =
k
40
,
αt
τ = 2
R
,
r
x =
R
For simplicity, Eq. (3.126) and boundary conditions can be rewritten in dimensionless form
(3.130)
∂θ 1 ∂ ⎛ ∂θ⎞
= ⎜x⎟ ∂τ x ∂x⎝ ∂x
⎠ (3.131)
∂ θ
= 0
∂ x at x = 0
(3.132)
∂ θ
= −Bθ
∂ x at x = 1
(3.133)
θ = 1 at τ = 0
(3.134)
For a long cylinder with the same Biot number in both sides, temperature distribution is the same
for each half, and so just one half can be considered
3.7.2 PROFILE 1
The guess temperature profile is assumed as
( ) ( ) ( ) 2
θ τ τ τ
p = a0 + a1 x+ a2 x
Differentiating the above equation with respect to x we get
Applying first boundary condition we have
(3.135)
∂ θ
= a1+ 2a2x
∂ x
(3.136)
a1+ 2a2x= 0
(3.137)
Thus
a 1 = 0
(3.138)
Applying second boundary condition we have