simple algorithm method with colocated mesh and steady flow

sajad6 · May 23, 2016, 09:42

Hi
I've written a code for solving flow field by applying simple colocated algorithm. I have a problem with coefficient matrix of pressure correction equations. You know in steady flow for pressure correction equation we have:

Ap=-Aw-Ae-An-As

So we will have a matrix with zero determinant.In this situation, pressure correction equations is impossible to be solved:

[A][PP]=[B] => DET[A]=0 => PP EQUATIONS IS UNSOLVABLE

is there anyone who can help me with this dilemma?

hadian · May 23, 2016, 11:19

note that for internal nodes $A_{p}=\sum A_{nb}$ . at the boundaries you will have nodes that $A_{p}>\sum A_{nb}$ . so there is no problem for solving the equation.

mprinkey · May 23, 2016, 11:24

I assume that you are working with incompressible flow, and are building the pressure equation to enforce div(U) = 0.

The system of equations that you get is indeterminate ONLY if you are not enforcing a fixed pressure boundary condition. The pressure is this case is a gauge variable because it can get zero-shifted and the physics will not change. The typical solution is to pick a cell and set the pressure in that cell to a fixed value (like zero). That make the system determinate. Note that this is exactly what OpenFOAM does with a line like:

pEqn.setReference(pRefCell, pRefValue);

Good luck.

FMDenaro · May 23, 2016, 11:57

I agree with mprinkey that fixing a value in an arbitrary node makes resolvable the problem, however that is not necessary as a fixed value will be implied in the iterative solution algorithm provided that the compatibility condition is fulfilled. The compatibility relation is the integral of the Poisson equation over the computational domain, hence if your BC.s are well posed, that relation will be satisfied

sajad6 · May 23, 2016, 16:33

points of mprinkey and FMdinaro are correct but it will not change system of pressure correction eequations to make it solvable.because determinant of coefficient matrix is zero.

but Hadian's point can help my problem but I have a question about it.According to Hadian,I should add neighbour coefficient to Ap. consider I want to calculate Ap and B(source term) of a cell at inlet boundary:

Ap=-Aw(neighbour)-Ae-As-An

after that, should I add below term to cell's source term?

B=B+Aw(neighbour)*pp

If yes how can I apply it to my code while I haven't pp(pressure correction) before solving?

FMDenaro · May 23, 2016, 16:46

Quote:

Originally Posted by sajad6

points of mprinkey and FMdinaro are correct but it will not change system of pressure correction eequations to make it solvable.because determinant of coefficient matrix is zero.

but Hadian's point can help my problem but I have a question about it.According to Hadian,I should add neighbour coefficient to Ap. consider I want to calculate Ap and B(source term) of a cell at inlet boundary:

Ap=-Aw(neighbour)-Ae-As-An

after that, should I add below term to cell's source term?

B=B+Aw(neighbour)*pp

If yes how can I apply it to my code while I haven't pp(pressure correction) before solving?

no, you are wrong, if the compatibilty relation is satisfied, even if det(A)=0 you got an indeterminate solution. In other words, you have a solution determined apart an arbitrary constant.

hadian · May 23, 2016, 21:48

at inlet (that usually the zero gradient is used for velocity) and walls again $A_{p}=\sum {A_{nb}}$ . But point you have reference pressure, the pressure correction is zero and then $A_{p}>\sum {A_{nb}}$ . for example if the point is at east boundary, you will have

so after calculating $A_{p}$ using $A_{p}=\sum {A_{nb}}$ , you implement the boundary condition by setting $A_{E}=0$ at this point. and this point solve the problem!!!

May 23, 2016, 09:42	simple algorithm method with colocated mesh and steady flow	#1
sajad6 Member sajad Join Date: Apr 2014 Location: Iran Posts: 46 Rep Power: 11	Hi I've written a code for solving flow field by applying simple colocated algorithm. I have a problem with coefficient matrix of pressure correction equations. You know in steady flow for pressure correction equation we have: Ap=-Aw-Ae-An-As So we will have a matrix with zero determinant.In this situation, pressure correction equations is impossible to be solved: [A][PP]=[B] => DET[A]=0 => PP EQUATIONS IS UNSOLVABLE is there anyone who can help me with this dilemma?

May 23, 2016, 16:33	I got all of these points but ...	#5
sajad6 Member sajad Join Date: Apr 2014 Location: Iran Posts: 46 Rep Power: 11	points of mprinkey and FMdinaro are correct but it will not change system of pressure correction eequations to make it solvable.because determinant of coefficient matrix is zero. but Hadian's point can help my problem but I have a question about it.According to Hadian,I should add neighbour coefficient to Ap. consider I want to calculate Ap and B(source term) of a cell at inlet boundary: Ap=-Aw(neighbour)-Ae-As-An after that, should I add below term to cell's source term? B=B+Aw(neighbour)*pp If yes how can I apply it to my code while I haven't pp(pressure correction) before solving?

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May 23, 2016, 11:19		#2
hadian Member Mohammad Reza Hadian Join Date: Mar 2009 Location: Yazd, Iran Posts: 52 Rep Power: 17	note that for internal nodes $A_{p}=\sum A_{nb}$ . at the boundaries you will have nodes that $A_{p}>\sum A_{nb}$ . so there is no problem for solving the equation.

May 23, 2016, 11:24		#3
mprinkey Senior Member Michael Prinkey Join Date: Mar 2009 Location: Pittsburgh PA Posts: 363 Rep Power: 25	I assume that you are working with incompressible flow, and are building the pressure equation to enforce div(U) = 0. The system of equations that you get is indeterminate ONLY if you are not enforcing a fixed pressure boundary condition. The pressure is this case is a gauge variable because it can get zero-shifted and the physics will not change. The typical solution is to pick a cell and set the pressure in that cell to a fixed value (like zero). That make the system determinate. Note that this is exactly what OpenFOAM does with a line like: pEqn.setReference(pRefCell, pRefValue); Good luck.

May 23, 2016, 11:57		#4
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	I agree with mprinkey that fixing a value in an arbitrary node makes resolvable the problem, however that is not necessary as a fixed value will be implied in the iterative solution algorithm provided that the compatibility condition is fulfilled. The compatibility relation is the integral of the Poisson equation over the computational domain, hence if your BC.s are well posed, that relation will be satisfied

May 23, 2016, 21:48		#7
hadian Member Mohammad Reza Hadian Join Date: Mar 2009 Location: Yazd, Iran Posts: 52 Rep Power: 17	at inlet (that usually the zero gradient is used for velocity) and walls again $A_{p}=\sum {A_{nb}}$ . But point you have reference pressure, the pressure correction is zero and then $A_{p}>\sum {A_{nb}}$ . for example if the point is at east boundary, you will have so after calculating $A_{p}$ using $A_{p}=\sum {A_{nb}}$ , you implement the boundary condition by setting $A_{E}=0$ at this point. and this point solve the problem!!!