Page 107 - Introduction to Computational Fluid Dynamics

P1: IWV
11:7
May 25, 2005
0 521 85326 5
CB908/Date
0521853265c04
86
2D BOUNDARY LAYERS
where p x is the pressure gradient and D j contains source terms arising from other
body forces. To solve this equation by TDMA, let the postulated equation be
u j = A j u j+1 + B j − R j p x , (4.79)
where R 1 = R JN = 0. Then, the recurrence relations will take the following form:
AN j AS j B j−1 + D j AS j R j−1 + V j
A j = , B j = , R j = , (4.80)
DE N DE N DE N
where DE N = AP j − AS j A j−1 . Note that A(2), B(2), and R(2) can be recovered
from Equation 4.78. Therefore, the coefficients in Equation 4.80 can be determined
for j = 3to JN − 1 by recurrence. Now, let u j be further postulated as
u j = F j − G j p x , (4.81)

where, again by recurrence, F j and G j can be determined for j = JN − 1to2by

F j = A j F j+1 + B j , G j = A j G j+1 + R j , (4.82)
where A(JN) = G(JN) = 0. Thus, it is possible to replace u j in Equation 4.77 by
Equation 4.81. The replacement yields a nonlinear equation in p x :

ω j

− C = 0. (4.83)
ρ j (F j − G j p x )
This equation can be solved by Newton–Raphson iterative procedure:
C − S 1
∗
p x = p + ,
x
S 2
ω j

S 1 = ,
ρ j (F j − G j p )
∗
x
ω j G j

S 2 = , (4.84)
∗ 2
ρ j (F j − G j p )
x
∗
where p is the guessed pressure gradient. Iterations are continued until |C − S 1 | <
x
10 −4 C. Usually, about five iterations suffice.
Finally, we note that in free-shear flows, the pressure gradient is zero.
4.7.2 Q and R k
The source terms in the energy and mass transfer equation depend on the problem
at hand. In general, however,
Dp
˙ ˙ ˙
Q = Q rad + Q cr + µ v + + Q md , (4.85)
Dt
˙
˙
where Q rad = ∂q rad,y /∂y represents the radiation contribution, Q cr repre-
sents the generation rate due to endothermic or exothermic chemical reac-
2
tions, µ v = µ(∂u/∂y) represents the viscous dissipation effect, Dp/Dt =