[GDS]

I cannot figure out why I keep getting this error. From previous posts, this usually happens because rho is not defined in the controlDict file, but I did define it there.


Screen output:


// * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * //
Create time

Create mesh for time = 0


SIMPLE: convergence criteria
field p tolerance 1e-06
field U tolerance 1e-05
field "(k|omega|e)" tolerance 1e-05

Reading field p

Reading field U

Reading/calculating face flux field phi

Selecting incompressible transport model Newtonian
Selecting turbulence model type RAS
Selecting RAS turbulence model SpalartAllmaras
Selecting patchDistMethod meshWave
SpalartAllmarasCoeffs
{
Cb2 0.622;
Cb1 0.1355;
kappa 0.41;
sigmaNut 0.66666;
Cw3 2;
Cv2 5;
Cw2 0.3;
Cv1 7.1;
Cs 0.3;
}

No MRF models present

No finite volume options present

Starting time loop

forces forces1:
Not including porosity effects

forceCoeffs forceCoeffs:
p: p
U: U
rho: rhoInf
Freestream density (rhoInf) set to 1
Not including porosity effects

Time = 1

smoothSolver: Solving for Ux, Initial residual = 1, Final residual = 0.04419339, No Iterations 4
smoothSolver: Solving for Uy, Initial residual = 1, Final residual = 0.074409196, No Iterations 2
smoothSolver: Solving for Uz, Initial residual = 1, Final residual = 0.043509528, No Iterations 4
GAMG: Solving for p, Initial residual = 1, Final residual = 0.051297006, No Iterations 4
time step continuity errors : sum local = 0.00010679296, global = 7.7896282e-06, cumulative = 7.7896282e-06
smoothSolver: Solving for nuTilda, Initial residual = 0.99999991, Final residual = 0.077435556, No Iterations 3
ExecutionTime = 179.44 s ClockTime = 181 s



--> FOAM FATAL ERROR:
Could not find rho:rho

From void Foam::functionObjects::forces::initialise()
in file forces/forces.C at line 235.

FOAM exiting

[Yann]

Hi,

You have defined rho in the forceCoeffs function object, but not in forces1 function. You need to define it there too.

Yann

[GDS]

Thanks, that worked. I don't understand, though. I was using the case files from an example that solved with rhoSimpleFoam, and that ran with no problem.

[Yann]

Since simpleFoam is an incompressible solver, it doesn't compute rho. But you need to know rho is order to compute the forces, this is why you need to define it in the forces function objects.

rhoSimpleFoam is a compressible solver, rho is computed by the solver and retrieved by the forces function object to compute the forces.

This is why your function object definition works with rhoSimpleFoam and not with simpleFoam.

Cheers,
Yann

[pranay10]

Originally Posted by Yann

Since simpleFoam is an incompressible solver, it doesn't compute rho. But you need to know rho is order to compute the forces, this is why you need to define it in the forces function objects.

rhoSimpleFoam is a compressible solver, rho is computed by the solver and retrieved by the forces function object to compute the forces.

This is why your function object definition works with rhoSimpleFoam and not with simpleFoam.

Cheers,
Yann

Hey Yann,
I am using simplefoam for power law fluid. Pressure output in openfoam is kinematic. So what is the rho in this case? and how do I change that rho?

[Yann]

Hello pranay10,

The power law model is only used to compute viscosity.
OpenFOAM does not always use kinematic pressure (only for incompressible solvers)
simpleFoam indeed uses kinematic pressure so it doesn't need rho and this is why it is not defined anywhere. (and this is also why you need to define it explicitly in the forces function object, as it requires a rho value to compute the forces)

So to answer your question: rho is whatever value you want it to be and you don't need to define it to run simpleFoam. Just remember to multiply the pressure p calculated by simpleFoam by rho in order to get the static pressure. (or use the pressure function object to automatically compute fields like static and total pressure. You will have to define rho in the function object definition)

Cheers,
Yann

[pranay10]

Yann,
Thank You for the response.
I have a doubt since simpleFoam uses kinematic pressure (p/rho), then what is the default value of rho in this case? As without a rho being defined, it's not possible to do calculations right? I read through the documentation and code as well but coulnd't find it.

[pranay10]

Originally Posted by Yann

Hello pranay10,

The power law model is only used to compute viscosity.
OpenFOAM does not always use kinematic pressure (only for incompressible solvers)
simpleFoam indeed uses kinematic pressure so it doesn't need rho and this is why it is not defined anywhere. (and this is also why you need to define it explicitly in the forces function object, as it requires a rho value to compute the forces)

So to answer your question: rho is whatever value you want it to be and you don't need to define it to run simpleFoam. Just remember to multiply the pressure p calculated by simpleFoam by rho in order to get the static pressure. (or use the pressure function object to automatically compute fields like static and total pressure. You will have to define rho in the function object definition)

Cheers,
Yann

Yann,
Thank You for the response.
I have a doubt since simpleFoam uses kinematic pressure (p/rho), then what is the default value of rho in this case? As without a rho being defined, it's not possible to do calculations right? I read through the documentation and code as well but coulnd't find it.

[Yann]

Originally Posted by pranay10

I have a doubt since simpleFoam uses kinematic pressure (p/rho), then what is the default value of rho in this case? As without a rho being defined, it's not possible to do calculations right?

This is the other way around: since simpleFoam is an incompressible solver, rho is constant and you can totally remove it from the equations you solve. This is what is done in simpleFoam, and as a consequence the pressure field ends up being normalized by rho. simpleFoam uses the kinematic pressure in order to totally get rid of rho in the equations it solves. (so there is no default value, there is no rho at all)

The value of p by itself does not matter in simpleFoam, only the pressure difference matters.