Page 261 - Computational Fluid Dynamics for Engineers

8.2 Solution of the Orr-Sommerfeld Equation 251

To summarize one part of the iteration process for a fixed value of R and
for assumed values of a r and CJ, we solve Eq. (4.4.29). If the initial estimates
of a r and u satisfy Eq. (8.2.11) and that /o — 0, then of course there is no
need to compute new values of a r and u. On the other hand, if fo ^ 0, then
new estimates of a r and u are obtained by Newton's method. With the right-
hand side of Eqs. (8.2.15a,b) given by Eqs. (8.2.16) and (8.2.17), we solve Eqs.
(8.2.15a,b) to compute df/da r and df/dou. Then we compute 6a r and 8u from
Eqs. (8.2.14) and insert them in Eqs. (8.2.12) so that we can solve Eq. (4.4.29)
with new estimates of a r and UJ and satisfy Eq. (8.2.11). This process is repeated
until the increments |&J| and |<5a r| are less than a specified tolerance parameter.
Once a solution to Eq. (8.2.11) is obtained at a specified Reynolds number
i?o> we can determine (da r/dR)o and (duj/dR)o at R = Ro and decide whether
we are going to solve the eigenvalue problem in which we compute (a r,uj) for a
fixed i?, as we did above, or compute (i?, u) for a fixed a r. For this purpose we
take the total derivative of Eq. (8.2.11) with respect to R and after separating
the real and imaginary parts of the resulting expression, we get:
(Ofr\
(8.2.18a)
\dRJo
da r) 0\dRj 0 \dco) 0\dR) 0
dfi\
(8.2.18b)
o v ^ i o dRJo
da r \doj) 0\dRj 0
It follows from Eqs. (8.2.18) that
(da r\ 1
(8.2.19a)
\dRj 0 dRj 0\duj) 0 \dRj 0 dojJo.
/dcv\ _ l_
(8.2.19b)
\dRj 0~ Ao dR da r/0 dRJo da r
where
'dfr\ (dfi\ _(9fr\ fdfi.
A 0 = (8.2.19c)
To evaluate the derivatives of f r and /j with respect to R, this time we differ-
entiate Eq. (4.4.29) to obtain the variational equations with respect to R,

dAy V
6 = r (8.2.20)
dRJ

The right hand side vectors of the above equation with (ri)j = (r2)j 0 for
0 < j < J, and with {r%)j and {r±)j for 1 < j < J are:

(r 3 )i-i = 0 (8.2.21a)
dc<2
, o f
( ^ ) i - i = 2 ( ^ (8.2.21b)