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February 8, 2003, 03:12 |
Has the pressure to be fixed at one node?
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#1 |
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Hello,
I am developing the mixed FEM (GLS) code for incompressible fluid flow calculation. I would like confirm whether a pressure value has to be fixed at a grid. Let's take a pipe flow as an example. BCs: velocity inlet and outflow condition at the outlet. I calculated two cases: first case, a pressure value was fixed to be zero at a node at the outlet, and there is no pressure fixed in the second case. However, the calculated fluid velocity and pressure fields are almost the same for the two cases. Can anyone explain the reasons? In theory, should a pressure value be fixed in the flow field? Thanks! |
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February 8, 2003, 04:19 |
Re: Has the pressure to be fixed at one node?
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#2 |
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In incompressible flow pressure is determined only upto a constant. The only thing that matters is the pressure gradient. Once you solve for the pressure you can add/subtract any constant from the pressure field. In compressible flow the equation of state has to be satisfied and hence pressure is absolutely fixed.
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February 8, 2003, 18:04 |
Re: Has the pressure to be fixed at one node?
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#3 |
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Thanks for your points, Praveen.
Do you mean it is not necessary to fix a pressure value at a node in the algorithm (code) or this value has to be fixed but it can be any number (say zero)? |
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February 9, 2003, 03:12 |
Re: Has the pressure to be fixed at one node?
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#4 |
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In your second case, the initial pressure (only one value in the whole field ?) is the constant.
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February 9, 2003, 13:07 |
Re: Has the pressure to be fixed at one node?
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#5 |
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In the second case, the initial pressure field was zero and no pressure was fixed in the field during the iterations.
In the first case, however, a pressure value at a node was fixed during the calculation. (Initial pressure field was also set to zero). My question: which treatment is correct in the two cases? Thanks! |
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February 9, 2003, 14:14 |
Re: Has the pressure to be fixed at one node?
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#6 |
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Choose some grid point P, compute the pressure and subtract the pressure at P from all the grid points. (So the pressure at P will be zero.) The correctness of the solution will not depend on the arbitrary constant since it is only pressure differences/gradients that are important. But you might want to remove the arbitrariness in the pressure value by doing the above for the pupose of visualization and comparison of results.
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