Page 90 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

82
From the above equation, we can write

2
d θ 2 THE FINITE ELEMENT METHOD
− µ θ = 0 (3.208)
dζ 2
ζ i = 1
2
Substituting Equation 3.207 into 3.208, with ζ = 1/2, we have
2 !
1
2
2B − µ 1 − 1 − B = 0 (3.209)
2
which gives
2
µ
2
B = (3.210)
3 2
1 + µ
8
Substituting into Equation 3.177, the solution is obtained as
2
µ
θ(ζ) 2 2
= 1 − (1 − ζ ) (3.211)
θ b 3 2
1 + µ
8
2
For a problem with µ = 3, then
θ(ζ) 12 2
= 1 − (1 − ζ ) (3.212)
θ b 17
Sub-domain method
The weighting function w i = 1 that results in the sub-domain formulation being

2
1 d θ
2
(1) 2 − µ θ dζ = 0 (3.213)
0 dζ
Substituting Equation 3.177 and integrating, we get
µ 2
2
B = 2 (3.214)
µ
1 +
3
The solution becomes
2
µ
θ(ζ) 2 2
= 1 − (1 − ζ ) (3.215)
θ b µ 2
1 +
3
2
For the particular case of µ = 3
θ(ζ) 3 2
= 1 − (1 − ζ ) (3.216)
θ b 4