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Problems in understanding BuoyantBoussinesqSimpleFoam |
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August 24, 2011, 12:41 |
Problems in understanding BuoyantBoussinesqSimpleFoam
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#1 |
Senior Member
Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 168
Rep Power: 16 |
Dear Foamers,
I still habe problems in understanding the source code of BuoyantBoussinesqSimpleFoam. In the attachment you see the momentum equation which has to be solved where GT= rho g= (1-beta(T-T_0))*g In pEqn.H the main code is settled. First the velocity is solved without taking into account pressure or density. HTML Code:
U = rAU*UEqn().H(); HTML Code:
phi = fvc::interpolate(U) & mesh.Sf(); HTML Code:
surfaceScalarField buoyancyPhi(rAUf*ghf*fvc::snGrad(rhok)*mesh.magSf()); The flux is corrected with the buoyancy flux HTML Code:
phi -= buoyancyPhi; HTML Code:
fvm::laplacian(rAUf, p_rgh) == fvc::div(phi) HTML Code:
phi -= p_rghEqn.flux(); HTML Code:
U -= rAU*fvc::reconstruct((buoyancyPhi + p_rghEqn.flux())/rAUf); HTML Code:
p_rgh = p - rhok*gh; HTML Code:
phi = fvc::interpolate(U) & mesh.Sf() - (rAUf*ghf*fvc::snGrad(rhok)*mesh.magSf()) - p_rghEqn.flux(); Regarding the link http://openfoamwiki.net/index.php/Bu...sinesqPisoFoam in the solver BuoyantBoussinesqPisoFoam the flux looks like this HTML Code:
phi =fvc::interpolate(U) & mesh.Sf()) + rUAf*fvc::interpolate(rhok)*(g & mesh.Sf())-pEqn.flux(); I can not find any explanation in the forum. Who can help me to understand the code? |
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September 5, 2011, 09:58 |
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#2 |
Senior Member
Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 168
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No one for an answer???
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September 5, 2011, 10:39 |
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#3 |
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Roman Thiele
Join Date: Aug 2009
Location: Eindhoven, NL
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I am not sure, but you might have to look up in a numerical methods book, or somewhere else how the SIMPLE algorithm works, maybe that can give you some hints.
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~roman |
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September 8, 2011, 10:20 |
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#4 |
Senior Member
Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 168
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Thank you for answering!
Now I understand the implementation. If someone is interested I could post that, too. |
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September 8, 2011, 12:24 |
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#5 |
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Bernhard
Join Date: Sep 2009
Location: Delft
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If it is not on this forum yet, then it can be very interesting for future reference if you post your findings here!
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September 12, 2011, 10:02 |
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#6 |
Senior Member
Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 168
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In the momentum equation we have in z-direction (in direction of buoyancy) the terms
- dp / dz + rho*g In OpenFOAM, g is a vector (0 0 -9,81) which ensures that the buoyancy is only valid for the right coordinate direction. In order to guarantee, that in the pressure correction of the simple algorithm also the buoyancy term rho*g*z is taken into account, the pressure and the buoyancy are melted together in one term p_rgh = p - rho*g*z Instead of the normal gradient of p, the normal gradient of p_rgh is on the RHS. - d/dn (p_rgh) which equals in OF to HTML Code:
- fvc :: snGrad (p_rgh) - d/dz [ p - rho * g *z] = - dp/dz + rho*g + g*z* d rho/dz So the third term g*z* d rho/dz is "too much" and is therefore substracted via HTML Code:
- ghf *fvc::snGrad(rhok) |
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September 14, 2011, 03:08 |
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#7 |
Senior Member
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Hallo Anne!
Would you please clarify following points: 1. What do you mean by term "two much" (is it phisical meaning or regards difference in some mathematical formulation). 2. Did you use some reference to understand SIMPLE/PISO treatment of the buoyancy? Thank you in advance! Mfg, Alexander
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Best regards, Dr. Alexander VAKHRUSHEV Christian Doppler Laboratory for "Metallurgical Applications of Magnetohydrodynamics" Simulation and Modelling of Metallurgical Processes Department of Metallurgy University of Leoben http://smmp.unileoben.ac.at |
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September 14, 2011, 04:47 |
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#8 |
Senior Member
Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 168
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Hallo Alexander,
I will try to answer your questions: 1. With "too much" I mean the term g*z* d rho/dz which is the result of taking the derivative of - [ p - rho * g *z] in z-direction. It is an additional term due to the product rule. Basically we have the terms - dp / dz + rho*g So by taking the derivative of both, p and rho*g*z we obtain together with the correction term - g*z* d rho/ dz: - d/dz [ p - rho * g *z] - g*z* d rho/ dz= - dp /dz + rho*g + g*z* d rho/ dz - g*z* d rho/ dz = - dp/ dz + rho*g and this is the orignial term in the buoyancy driven momentum equation. In OpenFOAM- code this equals to the right-hand side of the UEqn HTML Code:
- fvc::snGrad(p_rgh) - ghf*fvc::snGrad(rhok) |
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September 14, 2011, 05:16 |
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#9 |
Senior Member
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Thank you Anne for the detailed explanation! I will search for the algorithm description in Peric's book.
Got it, thanx)
__________________
Best regards, Dr. Alexander VAKHRUSHEV Christian Doppler Laboratory for "Metallurgical Applications of Magnetohydrodynamics" Simulation and Modelling of Metallurgical Processes Department of Metallurgy University of Leoben http://smmp.unileoben.ac.at |
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June 8, 2019, 06:27 |
rhok in T and U equation
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#10 |
New Member
siamak rafizadeh
Join Date: May 2018
Posts: 8
Rep Power: 8 |
Hi
As anyone knows in buossinesq approximation for density is written as : rho=rhoRef*(1-beta*(T-Tref)) and in oF rhok=1-beta*(T-Tref) and in Chalmer university file: Cite as: Venkatesh, B V.: Tutorial of convective heat transfer in a vertical slot. In Proceedings of CFD with OpenSource Software, 2016, Edited by Nilsson. H., http://www.tfd.chalmers.se/~hani/kurser/OS_CFD_2016 CFD with OpenSource software A course at Chalmers University of Technology Taught by Håkan Nilsson we have: The governing equation for temperature is solved, given as ∇.(ρuT ) − ∇.αef f ∇T = Sradiation + ST (2.4) Where, αef f = νt /prt + ν/pr , Sradiation and ST are source terms due to radiation and user defined source term respectively. The density is updated as ρ = 1 − β(T − Tref ) and phi is defined as phi=rhok*U now question is how rhoRef is deleted in above equstion while alpha have it: α=k/(ρ*Cp) please help in making it clear. thanks in advance |
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