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Old   October 28, 1998, 13:03
Default grid generation
  #1
Klass Nilsson
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I am a biginner of CFD. Currently, I am taking a course of CFD. I had a problem with the grid generation. The Thompson method (Differential Equation Method) is used. However, the two elliptic equations in the computational domain are very complicated. My question is how to numerically solve these two equations. I am looking for the answer from the expext. I hope I can get into this wonderful area.
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Old   October 28, 1998, 13:49
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John C. Chien
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Joe F. Thompson's old book on "Numerical Grid Generation" has Fortran listings of various methods for grid generation in the Appendix section ( working condition ). The book is now online. Look for the home page address in the Resources Section, cfd-online, this site. (so, you can read the book from Internet directly)
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Old   October 29, 1998, 08:10
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Mickeal Wang
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I am also a beginner. What I do not understand in this issue is how to derive the following interative equation: XTEMP=.5*(G22*(P(I,J)*XXI+XXIXI)+G11*(Q(I,J)*XETA+ XETA2) & -2.*G12*XXIETA)/(G11+G22) Could somebody tell me how to get this? Thanks a lot.
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Old   October 29, 1998, 18:40
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John C. Chien
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(1). equation-18a,18b,18c define coordinate transformation factors between (X,Y) and (XI,ETA) coordinate systems.(2). equation-20 is a vector equation for vector-r, which need to be written in two separate equations, one for X and one for Y. (3). for X-equation, it now takes the form: g22*(X,xi,xi + P * X,xi ) + g11 * ( X,eta,eta + Q * X,eta ) - 2.0* g12 * X,xi,eta =0. (4). the first order derivatives ( or the transformation factors ) can be evaluated during the iteration as, say, X,xi = ( X(i+1,j)-X(i-1,j) )/ ( xi(i+1,j)- xi(i-1,j) ). if we assume the transformed mesh ( XI, YI ) is divided into one unit square, then (xi(i+1,j)-xi(i-1,j))=2 units. and the first order derivative X,xi = 0.5 * ( X(i+1,j) - X(i-1,j) ). (5). for the second order derivatives, such as X,xi,xi= ( X(i+1,j) - 2.0 * X(i,j) + X(i-1,j) ), it is a second order central difference. (6). the mixed derivative is done is two steps, first find the first order derivatives at j+1 and j-1 locations, then use these two to find the second order mixed derivatives. (7). when you substitute these finite-difference form of the transformation factors into the equation in step-3 above, you are going to get an equation with a lot of X(i,j), X(i+1,j), X(i-1,j),......So, it is very important to group terms with X(i,j) into one group, and keep the rest of the terms in another group. (8). rearrange the equation into the form: X(i,j) = ( the rest of the terms which include X(i+1,j), X(i-1,j),...etc...), (9). the S.O.R. method says that you can update the field point ( X(i,j) ) one-by-one in a loop ( loop 750) by using "old neighboring point values ( X(i+1,j),.. just calculated )". (10). the newly calculated value of X(i,j) is stored at XTEMP and the O.R. (over-relaxation) part is carried out using the orve-relaxation factor w (set equal to 1.8).(11). there are a couple of typo in the code such as YTENP, XTENP. (12). time's up , I have to go now.
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