Page 264 - Introduction to Computational Fluid Dynamics

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8.4 SORENSON’S METHOD
Now, solving Equations 8.35 and 8.36 simultaneously, we can show that 16:28 243
−0.5

2 2

∂x 1 ds 0 ∂x 1 ∂x 2 ∂x 2 ∂x 1
| ξ 2 =0 =− cos θ 0 + sin θ 0 + ,
∂ξ 2 d ξ 2 ∂ξ 1 ∂ξ 1 ∂ξ 1 ∂ξ 1
ξ 2 =0
(8.37)
−0.5

2 2
∂x 2 ds 0 ∂x 1 ∂x 2 ∂x 2 ∂x 1
| ξ 2 =0 = sin θ 0 − cos θ 0 + .
∂ξ 2 d ξ 2 ∂ξ 1 ∂ξ 1 ∂ξ 1 ∂ξ 1
ξ 2 =0
(8.38)
Identical expressions can be developed for ∂x 1 /∂ξ 2 and ∂x 2 /∂ξ 2 at ξ 2 = ξ 2max .

8.4.2 Stretching Functions

Sorenson [71] defines P and Q functions as

P (ξ 1 ,ξ 2 ) = P (ξ 1 , 0) exp (−a ξ 2 ) + P (ξ 1 ,ξ 2max )exp {−c (ξ 2max − ξ 2 )} , (8.39)

Q (ξ 1 ,ξ 2 ) = Q (ξ 1 , 0) exp (−b ξ 2 ) + Q (ξ 1 ,ξ 2max )exp {−d (ξ 2max − ξ 2 )} , (8.40)

where a, b, c, and d are positive constants to be chosen by the analyst. Now, for
convenience, we introduce the following symbols:

L 1 =− (LHS of Equation 8.26)/J 2

and
2
L 2 =− (LHS of Equation 8.27)/J .
Thus,

∂x 1 ∂x 1
L 1 (ξ 2 = 0) = P (ξ 1 , 0) + Q (ξ 1 , 0) , (8.41)

∂ξ 1 ξ 2 =0 ∂ξ 2 ξ 2 =0

∂x 2 ∂x 2
L 2 (ξ 2 = 0) = P (ξ 1 , 0) + Q (ξ 1 , 0) . (8.42)

∂ξ 1 ξ 2 =0 ∂ξ 2 ξ 2 =0
Therefore, using the definition of J (see Equation 8.28), we get
1 ∂x 2 ∂x 1
P (ξ 1 , 0) = L 1 − L 2 , (8.43)
J
∂ξ 2
∂ξ 2
ξ 2 =0
1 ∂x 1 ∂x 2
Q (ξ 1 , 0) = L 2 − L 2 . (8.44)
J
∂ξ 1
∂ξ 1
ξ 2 =0