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 December 8, 1999, 16:21 Need help. #1 Yangang Bao Guest   Posts: n/a Hi, I am just a beginner for CFD. I am trying to write code to solve some simple problems. I am using steady incompressible N-S equations. Here is my problem: Given the cavity, 2 opposite sides are wall (u=v=0), on the other 2 side, pressure p0, and p1(say 0) are given. Newtonian fluid is assumed and convective term is ommited. I tried many different standard ways, FDM, FVM and FEM, only found that all the final linear systems are singular. I found the cause of singular is due to continuity equation. Anybody can comment on this?

 December 8, 1999, 17:55 Re: Need help. #2 John C. Chien Guest   Posts: n/a (1).Follow existing methods, (2). Define your problem as if it actually exists in real world. That is, can you test your problem in real life based on the boundary conditions you have specified? (3). The incompressible flow is not as easy to solve as the compressible flow. So, try to follow the existing methods and test cases first.

 December 8, 1999, 18:17 Re: Need help. #3 Yangang Bao Guest   Posts: n/a This simple example has analytical solution: p is linear changed. u is a quadratic function of y v is 0 so dp/dx = d(du/dx)/dx+d(du/dy)/dy dp/dy = d(dv/dx)/dx+d(dv/dy)/dy du/dx+dv/dy=0 is satisfied. But I just cannot solve it numerically, especially don't know how to get rid of the singularity of the continuity. If I introduce the bubble function, then some errors appear which can never be eliminated. I hope somebody can give me a straw. Thanks a lot. yangang bao

 December 9, 1999, 02:08 Re: Need help. #4 John C. Chien Guest   Posts: n/a (1). If you set the velocity component, v=0 everywhere, and look for the special solution, then from the continuity equation, you get du/dx=0. (v=0 everywhere, so dv/dy=0) (2). du/dx=0 means that the velocity component, u is not a function of x. So, the u profile becomes a function of y only, and this function does not change in x direction. For duct flow, it is fully-developed profile. And now u=u(y), the problem is 1-D. So much from the continuity equation. (3). Now with v=0, and du/dx=0, the x-momentum equation becomes: dp/dx=mu*(d(du/dx)/dx + d(du/dy)/dy), the contribution from the convection terms drop out. Since, v=0, and du/dx=0 is for the whole flow field, d(du/dx)/dx also is zero. (4). Now the x-momentum equation becomes dp/dx=mu*d(du/dy)/dy. At this point, we have used the continuity equation and the assumption of v=0. So, the continuity equation can no longer exist as a separate equation. (5). For the y-momentum equation, v=0 causes the convection and the diffusion term to vanish. And the resultant equation is dp/dy=0, which means p is not a function of y. Thus p is a function of x only, that is p=p(x). This is a consequence of the assumption of v=0 and the continuity equation. (6). So, with v=0, and the continuity equation, the resultant governing equation becomes: p=p(x), dp/dx=mu*d(du/dy)/dy. (7). So, the y-momentum equation becomes d(du/dy)/dy=(dp/dx)/mu. This is the second-order equation for the velocity u. (8). The finite difference form of the equation using central difference approximation is u(j+1)-2.0*u(j)+u(j-1)=(deltx*deltx)*(dp/dx)/mu. This can be used to solve for u distribution in y, as long as (dp/dx) is known. (9). If p(i+1)=p(i-1), and dp/dx=0, that is no pressure gradient, then u(j)=0.5*(u(j+1)+u(j-1)). For two walls boundary condition, u(jtop)=0, u(jbottom)=0, u(j) will become zero. (using numerical solution) (10). If dp/dx is constant, that is dp/dx=CONST, then the governing equation becomes, u(j)=0.5*(u(j+1)+u(j-1)) -CONST*(deltx*deltx)/mu. The solution will be a second-order parabolic function of y. (using numerical solution)

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