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March 22, 2018, 10:07 |
Boundary Layer over a surface
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#1 |
New Member
Asif Shahriar Nafi
Join Date: Dec 2015
Location: Conway, SC, USA
Posts: 8
Rep Power: 11 |
Hi,
I am trying to solve this problem numerically (Boundary Layer over a flat plate, see attached). Matlab code: nx=2001;ny=51; xmax=1; ymax=0.25; vis = 1.4e-5; dx = xmax/(nx-1);dy = ymax / (ny - 1); %grid u = nan (nx,ny); v = nan (nx,ny); %Boundary condition u(1, : ) = 0; v(1,: ) = 0; u(nx,: ) = 10; %calc for p = 1:nx-1 for k = 2:ny-1 u(p+1,k)= u(p,k) + (vis*dx / u(p,k)*(dy^2)) * (u(p,k+1) - 2*u(p,k) ... + u(p,k-1)) - (v(p,k)*dx / 2*u(p,k)*dy) * (u(p,k+1) - u(p,k-1)); v(p+1,k) = v(p+1,k-1) - 0.5* dy * ((u(p+1,k) - u(p,k))/dx + ... (u(p+1,k-1) - u(p,k-1))/dx); end end But i dont get any value of u and v. I would very much appreciate your help Last edited by asifnafi; March 22, 2018 at 12:15. Reason: typo |
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March 26, 2018, 10:36 |
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#2 |
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Philipp Wiedemer
Join Date: Dec 2016
Location: Munich, Germany
Posts: 42
Rep Power: 9 |
Well you're dividing by u(p,k) in your formula for u(p+1,k) but your boundary condition is set to zero. As a result you get a NaN because the numerator is zero aswell (0/0 = NaN in Matlab). This NaN gets then propagated through your whole calculation because you always have NaN's from the row before (If you have a NaN in the formula the result will be a NaN aswell). Maybe you have the indices mixed up or something.
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March 26, 2018, 11:40 |
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#3 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,892
Rep Power: 73 |
You can solve this problem in a simplified equation using the von Mises transformation
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April 29, 2020, 23:27 |
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#4 | |
New Member
Colin
Join Date: Apr 2020
Posts: 7
Rep Power: 6 |
Quote:
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Tags |
boundary layer, fdm, flat plate, numerical simulation |
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