Page 237 - Computational Fluid Dynamics for Engineers

226 7. Boundary-Layer Equations

n
This test is done in MAIN. If this test is satisfied, we set r/ e (x n+1 ) = rj e(x ).
Otherwise, we call GROWTH and set J new = J Q\d + £, where t is a number of
r
(
points, say t = 1. In this case we also specify values of /f > ^?>^?>&?) f° the
n
1
new rij points. We take the values of u 3; = 1, v^ = 0, j = (r/j — r]^)u j + j , and
/
/
1
1 1
1
Vf = V}. This is also done for the values of /J " , v?" , and 6?" .
7.4.5 Subroutine COEF3
This is one of the most important subroutines of BLP. It defines the coefficients
of the linearized momentum equation given by Eqs. (7.3.23) and (7.3.26).
7.4.6 Subroutine SOLV3

The solution of Eq. (4.4.29) by the block-elimination method discussed in sub-
section 4.4.3 can be obtained by using the recursion formulas given by Eqs.
(4.4.32) and (4.4.34), and determining the expressions such as Aj, Fj, Wj and
6j. To describe the procedure let us first consider Eq. (4.4.32). Noting that the
7
i~j matrix has the same structure as Bj and denoting the elements of j, by 7 ^
(i,k = 1, 2, 3), we can write Tj as
(711) (7l2)j (7is)j
r,= (721) (722 )j (723)j (7.4.11a)
0 0 0

Similarly, if the elements of Aj are denoted by o ^ we can write Aj as [note
that the third row of Aj follows from the third row of Aj according to Eq.
(4.4.32c)]

(an)j (an)j («i3)j
^ (a 2i)j (o 2 2 )j (a 23)j 0 < j < J - 1 (7.4.11b)
0 - 1 -hj+i/2

and for j = J, the first two rows are the same as the first two rows in Eq.
(4.4.32b), but the elements of the third row, which correspond to the boundary
conditions at j = J, are (0,1,0).
For j — 0, AQ = AQ; therefore the values of (0^)0 are
(an)o = 1 (012)0 = 0 (013)0 = 0
(7.4.12a)
(021)0 = 0 (0^2)0 = 1 (023)0 = 0

and the values of (7^)1 are

(711)1 = - 1 (712)1 = --hi (713)1 = 0
(7.4.12b)
(721)1 = (54)1 (723)1 = - 2 —— (722)l = (56)l + (723)l