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April 29, 2011, 08:05 
pisoFoam laplace equation

#1 
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fisch
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Hello,
i'm currently reading jasaks thesis and have a look inside the pisofoam solver. Here is one thing i didnt understand in the following part of the solver: U = rUA*UEqn.H(); phi = (fvc::interpolate(U) & mesh.Sf())+(maybe something) ... fvScalarMatrix pEqn ( fvm::laplacian(rUA, p) == fvc::div(phi) ); Is the equation for phi representing the hole (2nd) r.h.s. of equation 3.141 in the thesis (page 146)? I thought so, because we are evaluating here the UEqn.H/a_p on the Surface multiplied by the Area. Why are we then taking the divergence of phi in the laplace equation? This would not match the formula 3.141 or 3.143... Can anybody help me and/or tell me my missinterpretations? This would be great. Thanks for any advice, rupert 

April 30, 2011, 02:40 

#2 
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Alberto Passalacqua
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Hi, think to discretizing the momentum equation as
A*U = H  grad(p), then U = H/A  1/A * grad(p) The face flux is then phi = U_f . S = (H/A)_f . S  (1/A)_f S snGrad(p) OpenFOAM first computes phi = (H/A)_f . S, then solves the pressure equation obtained imposing div(phi) = 0, which leads to laplacian (1/A*p) = div(phi) Once this equation is solved, the flux is corrected using the second part of the flux definition.
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April 30, 2011, 08:37 

#3  
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fisch
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Quote:
Hello Alberto, thank you for the reply. Maybe i have problems understanding the commands... I understand that taking the divergence of equation phi = U_f . S = (H/A)_f . S  (1/A)_f S snGrad(p) removes the left hand side and the right hand side = 0 is building then the poisson equation... right? But I thought that (1/A)_f S snGrad(p) == div[(1/A) * grad(p)] == laplace (1/A * p) Herein snGrad(p) is the gradient of p at the surfaces. is this not right? Having a look on the equation for pressure on http://openfoamwiki.net/index.php/Th...hm_in_OpenFOAM at chapter 2 (the last equation) this would lead (for me) to a pressure equation like: laplacian (1/A*p) = (H/A)_f . S = phi (and not div(phi)) Do you understand my problem? where is my misinterpretation??? Thank you a lot, rupert 

April 30, 2011, 16:57 

#4  
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Alberto Passalacqua
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Quote:
Quote:
phi = (H/A)_f . S  (1/A)_f S snGrad(p) you have div(phi) = 0 implies div((H/A)_f . S) = laplacian(1/A*p), which is exactly div(phi) = laplacian(1/A*p) Best,
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Alberto Passalacqua GeekoCFD  A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541) OpenQBMM  An opensource implementation of quadraturebased moment methods. To obtain more accurate answers, please specify the version of OpenFOAM you are using. Last edited by alberto; May 2, 2011 at 03:12. Reason: Added missing S 

May 1, 2011, 06:47 

#5 
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fisch
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Good morning,
i hope that i'm not bugging you, but: you confirmed that a) (1/A)_f S snGrad(p) == div[(1/A) * grad(p)] == laplace (1/A * p) right? If a) works it will make the phi equation from phi = (H/A)_f  (1/A)_f S snGrad(p) to b) phi = (H/A)_f  laplace (1/A*p) div(b) leads to 0 = div((H/A)_f)  div(laplace (1/A*p)) or not? I think that my problem lies in a) If (1/A)_f S snGrad(p) would be equal to (1/A) * grad(p) this would solve my problems... Is this the case? This would mean that snGrad(p) == grad(p) (defined at the cellcenter) * normalvector (at the surfaces) Hope you can finally help me Thanks, rupert 

May 1, 2011, 18:19 

#6  
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Alberto Passalacqua
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div(1/A grad(p)) be equal to (1/A)_f S snGrad(p) ? You have to think to the velocity predictor U = H/A  1/A grad(p), interpolate it on cell faces, and obtain U_f. At that point you dot it with the surface vector, and, at that point, you have the snGrad(p). Best,
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Alberto Passalacqua GeekoCFD  A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541) OpenQBMM  An opensource implementation of quadraturebased moment methods. To obtain more accurate answers, please specify the version of OpenFOAM you are using. 

May 2, 2011, 02:58 

#7  
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fisch
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Quote:
using your definition for phi: "phi = (H/A)_f  (1/A)_f S snGrad(p)" Div of this equation leads to " 0 = div((H/A)_f)  div( (1/A)_f) S snGrad(p))" How can you say here that div( (1/A)_f) S snGrad(p)) is equal to laplacian(1/A*p)??? 

May 2, 2011, 03:22 

#8  
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Alberto Passalacqua
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The only element that probably can cause confusion is the norm of the surface, which is not in the code, since operators take care of it. Best,
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Alberto Passalacqua GeekoCFD  A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541) OpenQBMM  An opensource implementation of quadraturebased moment methods. To obtain more accurate answers, please specify the version of OpenFOAM you are using. 

May 2, 2011, 05:04 

#9 
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fisch
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thank you very much


October 29, 2011, 08:14 
learn a lot!

#10 
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charlse
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I learn a lot! Thank you!
For me, the relationship of div(phi)=0 is doubted. Now I know, phi is a scalar, and its div is zero! Last edited by star shower; October 29, 2011 at 09:11. 

July 16, 2013, 22:13 

#11  
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Dongyue Li
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