Page 152 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

144
The Galerkin formulation and the element equations are similar to those for two-
dimensional heat transfer problems, but are different owing to the ring nature of the
elements. STEADY STATE HEAT CONDUCTION IN MULTI-DIMENSIONS
The differential equation in a cylindrical coordinate system (r, z) for steady state is
2
2
2
∂ T k r ∂T k θ ∂ T ∂ T
k r 2 + + 2 2 + k z 2 + G = 0 (5.57)
∂r r ∂r r ∂θ ∂z
An axisymmetric problem is independent of the angle θ and hence Equation 5.57
reduces to
2
∂T 2 k r ∂T ∂ T
k r 2 + + k z 2 + G = 0 (5.58)
∂r r ∂r ∂z
This can be rewritten, if the thermal conductivity in the radial direction, k r is constant, as
2
1 ∂ ∂T ∂ T
k r r + k z + G = 0 (5.59)
r ∂r ∂r ∂z 2
The boundary conditions are

T = T b on 1
∂T ∂T
k r l + k z n + h(T − T a ) + q = 0 on 2 (5.60)
∂r ∂z
The temperature distribution is described as follows:

T = N i T i + N j T j + N k T k (5.61)
which is similar in form to that of a linear triangular plane element, where

1
N i = (a i + b i r + c i z)
2A
1
N j = (a j + b j r + c j z)
2A
1
N k = (a k + b k r + c k z) (5.62)
2A
The area, A, is calculated from

& &
& 1 r i z i &
& &
2A = det 1 r j z j & (5.63)
&
& &
1 r k z k
& &
Other constants in Equation 5.62 are defined as
a i = r j z k − r k z j ; b i = z j − z k ; c i = r k − r j
a j = r k z i − r i z k ; b j = z k − z i ; c j = r i − r k
a k = r i z j − r j z i ; b k = z i − z j ; c k = r j − r i (5.64)