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Posted: 1 decade ago 8 mai 2013, 06:56 UTC−4

Hi, try with: velocity at the inlet, pressure+no viscous stress at the outlet. if you set a constant velocity at the inlet, for a "long enough" channel you'll see exactly the parabolic profile for the velocity at the outlet. You may find this setting a bit "unphysical", but the reasons stand behind the nature of the PDE you're solving with FEM. Mattia

Posted: 1 decade ago 8 mai 2013, 09:29 UTC−4

[QUOTE] Hi, try with: velocity at the inlet, pressure+no viscous stress at the outlet. if you set a constant velocity at the inlet, for a "long enough" channel you'll see exactly the parabolic profile for the velocity at the outlet. You may find this setting a bit "unphysical", but the reasons stand behind the nature of the PDE you're solving with FEM. Mattia [/QUOTE] Thanks very much for your kind reply. I have already tried your boundary condition, but it seems there is still problem with the b.c.. In fact, I have tried all kinds of combinations of inlet and outlet boundary conditions. First, for your boundary condition, let's simplify it. We can set the "velocity" boundary condition at the inlet-----u=s*(1-s),v=0; and the "pressure+no viscous stress" boundary condition at the outlet. Since the horizontal velocity u is already parabolic, we do not need a "long enough" channel. With a channel of any finite length, we hope we can get the same profile of u at the outlet as the input at the inlet, and we hope to get zero or near-zero v. But the result is not like this, because the v velocity at the outlet is finite instead of zero. This can be seen very easily in the contour plot of pressure near the outlet(i.e. the contours are not parallel vertical lines). So I can still not obtain a fully developed flow under this boundary condition. Second, from all the trials, I found the combination of "velocity" at the inlet and "pressure" at the outlet is OK(of course the velocity b.c. is parabolic as above). Under this b.c. I can get the a fully developed flow with v equals zero or near-zero everywhere. And in this situation the profile of u does not change with x and the pressure does not vary with y(i.e. the contours of pressure are parallel vertical lines). So, if we apply "velocity" b.c.at the inlet for Poiseuille flow, I think the problem is what kind of b.c. should be applied at the outlet. Though I got fully developed flow with the "pressure" b.c. at the outlet, but I think in FE theory the b.c. at the outlet in this situation should be the natural boundary condition(i.e. (-p*I+Mu*(grad(u)+(grad(u))T)*n=0 ). So I got confused again with the paradox. And what b.c. should be applied for Poiseuille flow on earth when the inlet b.c. is "velocity"???? Any help will be grateful. Pai Liu

Posted: 1 decade ago 8 mai 2013, 10:52 UTC−4

Hi, I'm not quite sure if I'm following you... what you're saying is that: parabolic profile @inlet, pressure&no viscous stress @outlet does not work but parabolic profile @inlet, pressure @outlet does work. Is that correct? Anyway, about "FEM theory", in Stoke's equation the gradient of -p*I+Mu*(grad(u)+(grad(u))T) leads to an integral for the pressure gradient and a boundary integral which vanishes with the natural boundary conditions of no viscous stress. So if you also add a boundary condition for the pressure you close the problem. However Comsol threats this term differently. It does not apply the continuity equation, and a third integral (over the domain) is used to enforced coupling between velocity components. If you derive equations by hand mind also that term... Analytic and numerical solution are not quite exactly the same thing.. Back to your problem... what I'd do first is: 1. ask myself what's the point of solve this problem with a parabolic inlet profile. You already insert the solution.. if you wanna check Comsol's capabilities try with a "physically wrong" boundary conditions: constant velocity at the inlet. 2. Try finer and finer meshes. 3. Look up for the 'cylinder_flow" into the model library. It solves Navies Stokes... If you load it, neglect inertial term, click on "incompressible fluid" you'll have a correct model :) Have fun, Mattia