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Extracting solenoidal field from 2D velocity field |
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January 13, 2016, 10:19 |
Extracting solenoidal field from 2D velocity field
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#1 |
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Hello everyone, I'm more into computer science but I'm working on a CFD project for the moment, so please bare with me. Here is my question:
I know that I can create a solenidal vector field by taking the curl of another vector field because the divergence of the curl is zero. However 2D vector fields result in a scalar function when taking the curl. So is it possible to transform a known 2D vector field to it's "nearest" divergence-free field by any other way? Thanks in advance |
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January 13, 2016, 10:53 |
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#2 | |
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Filippo Maria Denaro
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Quote:
no, not at all.... |
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January 13, 2016, 11:04 |
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#3 |
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Thank you for your reply, apparently that can be done by solving Navier-Stockes etc ... but the thing is I only have my velocity field generated by another mean not at all passing by these equations, any other recommendations will be highly appreciated.
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January 13, 2016, 11:26 |
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#4 |
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The curl is a vector operator. It cannot return a scalar. What it does return, in the case of a 2D flowfield, is a vector normal to the 2D plane. What you are looking for is variously called the Stokes-Helmholtz or Helmholtz-Hodge or Stokes-Helmholtz-Hodge decomposition.
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January 13, 2016, 11:30 |
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#5 | |
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Filippo Maria Denaro
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I totally agree , this is a very old theory of the vector calculus. If you are interested in details you could read the original book of Hodge which is, however, quite complex. I worked on this topic and some useful details are in https://www.researchgate.net/publica...ary_conditions |
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January 14, 2016, 10:36 |
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#6 | |
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Jonas T. Holdeman, Jr.
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Quote:
https://www.researchgate.net/publica...ble_fluid_flow https://www.researchgate.net/publica...inite_elements Similarly, you could extract the irrotational part of a field using vector elements which are the gradient of a potential. |
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January 14, 2016, 10:56 |
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#7 | |
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Filippo Maria Denaro
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Hello Jonas, I have a question about the Finite Element approach you and other authors used. What about if your decomposition is not orthogonal? I mean for example what happen when n.v is not zero at the boundaries and you cannot fulfill the orthogonality constraint. |
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January 14, 2016, 15:04 |
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#8 | |
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Jonas T. Holdeman, Jr.
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So what happens when the field to be interpolated, given at nodes, is not "divergence-free"? The interpolation will be divergence-free at the expense of introducing (perhaps unwanted) in/outflow between the nodes. Notice here I am now talking about interpolations of fixed data, not projections which might give new (consistent) velocity values at the nodes. But you may be asking, what would happen if the domain is closed and values on the boundary are not consistent net zero flow? The nodal boundary values will be respected, but there will be leakage between the nodes to satisfy the mathematical constraint. Just recognize this as a case of garbage in - garbage out. |
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January 14, 2016, 15:18 |
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#9 | |
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Filippo Maria Denaro
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I was asking for the case of a divergence-free flow but with not-vanishing normal component of the velocity (of course with total surface integral vanishing). Such a case has no longer the mathematical property of being a decomposition both orthogonal and unique. |
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January 14, 2016, 16:00 |
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#10 | |
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Jonas T. Holdeman, Jr.
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January 14, 2016, 16:17 |
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#11 | |
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Filippo Maria Denaro
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yes, you can always perform the decomposition (in continuous or discrete sense), it would turn to be just one between infinite decompositions...but without the orthogonality that fixes the unicity, the discrete error in the velocity enters into the pressure field. So, I was wondering if in Finite Element such an aspect is considered. |
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January 14, 2016, 17:38 |
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#12 | |
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Jonas T. Holdeman, Jr.
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January 17, 2016, 13:33 |
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#13 | |
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I just read that using some transformations in fourrier space, one can separate a vector field into it's irrotational and solenoidal componenets, that should solve my problem right , any confirmation on this ? |
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January 18, 2016, 04:11 |
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#14 | |
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January 18, 2016, 04:49 |
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#15 |
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Filippo Maria Denaro
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but what are you looking for?
the mathematical basis of the Hodge decomposition has nothing to do with the practical resolving method |
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