[mcc007]

Hello everybody,

I'd like to use fluent to solve Poisson equation. The Poisson equation is attached. But the simulation result is much higher than expected. I am not sure if I need to add some other terms (e.g., velocity term, transient term). in the equation. Or my code is wrong. Here is part of my code:

DEFINE_SOURCE(potential,c,t,dS,eqn)
{
real source;
source=F*(-C_UDSI(c,t,2)+C_UDSI(c,t,3));
dS[eqn]=0.0;
return source;
}

Thanks a lot

[vinerm]

Why do you have more than one UDS? And what are c, z, and F? Source should only have the right hand part of the equation.

[mcc007]

Originally Posted by vinerm

Why do you have more than one UDS? And what are c, z, and F? Source should only have the right hand part of the equation.

I have defined 6 UDS. In the equation, c = charge of ions, z = ion concentrations, and F = Faraday constant. The reseason why I have two UDS in the RHS is because I need to sum all ions in the solution. And there are two different ions in the solution. That's why I have two UDS in the RHS.

Another question is that the LHS is the second derivative of phi. I checked the UDF manual. And I found all of the equation on the LHS is the first derivative. I am not sure if I need to ingrate the LHS to make it first derivative of phi.

Can you help me with this?

Thanks a lot

[vinerm]

Do you have six different scalar equations to be solved? If yes, then six UDS are required. If you only have one Poisson equation with one variable, then you only need one UDS. Poisson equation always has Laplace operator, i.e., second derivative. It corresponds to diffusion term. So, no convection term needs to be included. is the diffusion coefficient for UDS. And the right hand side is the source term.

[mcc007]

Originally Posted by vinerm

Do you have six different scalar equations to be solved? If yes, then six UDS are required. If you only have one Poisson equation with one variable, then you only need one UDS. Poisson equation always has Laplace operator, i.e., second derivative. It corresponds to diffusion term. So, no convection term needs to be included. is the diffusion coefficient for UDS. And the right hand side is the source term.

Yes I have six UDS. But they are used for different purposes. In the Poisson equation, only two UDSs are being used. In the code that I posted, I am not sure if I need to change something or not. What do you think?

[vinerm]

DEFINE_SOURCE is used to provide source term. The RHS of the equation, i.e., , is the source and the UDF should return this value. But the UDF you have is not returning this.

[mcc007]

Originally Posted by vinerm

DEFINE_SOURCE is used to provide source term. The RHS of the equation, i.e., , is the source and the UDF should return this value. But the UDF you have is not returning this.

Thanks for your reply. In my code, the source term is Source = F*(-C_UDSI(c,t,2)+C_UDSI(c,t,3)). When you say the UDF doesn't return the value, what do you mean? Can you be more specific? Because I followed the UDF manual, the summation of the RHS is F*(-C_UDSI(c,t,2)+C_UDSI(c,t,3)).

Thanks a lot

[bestucan]

Is't your source should be something like this ?

for(j = 0; j < i; j++) {
     source += f * c_j * z_j
 }

[mcc007]

Originally Posted by bestucan

Is't your source should be something like this ?

for(j = 0; j < i; j++) {
     source += f * c_j * z_j
 }

No, I don't think so.

[vinerm]

Originally Posted by mcc007

Thanks for your reply. In my code, the source term is Source = F*(-C_UDSI(c,t,2)+C_UDSI(c,t,3)). When you say the UDF doesn't return the value, what do you mean? Can you be more specific? Because I followed the UDF manual, the summation of the RHS is F*(-C_UDSI(c,t,2)+C_UDSI(c,t,3)).

Thanks a lot

The value returned by your UDF is a product of F with difference of two scalars. Assuming these two scalars represent c and z in the RHS of the Poisson equation, then both scalars must be multiplied and not subtracted. The source term also depends on the summation parameter, i. If i implies discrete locations, then summation is not required in the UDF, however, if i represents some other parameter, such as ions, then you have to sum over all ions in each cell. Furthermore, UDF manual talks about logic and implementation; it does not mention anything about specific equations. And in case c and z are functions of potential, then dS[eqn] cannot be 0.