UDF for source term in momentum equation
 

Hi all I write a UDF for source term in momentum equation this source term is : A.( du/dx) , where u is the x-component of the velocity ...

please can somebody tell me what is the mistacke I have do in this UDF

DEFINE_SOURCE(xmom_source, c, t, dS, eqn) {

real source;

source = A*C_DUDX(c,t);

return source; }

Re: UDF for source term in momentum equation
 

you haven't defined A?

Re: UDF for source term in momentum equation
 

Not only that you have not defined A, but also you have to add one more statement namely:

dS[eqn]=0;

Good Luck to you. Sun

Re: UDF for source term in momentum equation
 

A is a constant

Re: UDF for source term in momentum equation
 

Yes, A is a constant but you must define it. for example; DEFINE_SOURCE(xmom_source, c, t, dS, eqn) { real source; int A; A=22; source = A*C_DUDX(c,t); dS[eqn]=0;

return source; }

"A" can be real or integer. It is up to you.

Hope I can help you. Bowling

Re: UDF for source term in momentum equation
 

Hi ,Bowling , but why dS[eqn]=0 ? dS[eqn]=dS/dU

Re: UDF for source term in momentum equation
 

You are misunderstand. dS[eqn] in here means d(source)/d(your parameter) eg. x-momentum, your parameter is U your source is A*(dU/dX), so dS[eqn] is second derivative of dU/dX equal d^2U/dX^2. Diff your source by your parameter.

dS[eqn] is the way to define your source solving. I don't know how to explain. You can read more infomation in fluent document.

Bowling

Re: UDF for source term in momentum equation
 

Hi Bowling , should i use UDS for calculating dS[eqn] ?

Re: UDF for source term in momentum equation
 

I don't think so. It is the condition in DEFINE_SOURCE. It is the way to solve in the finite volume method not a scalar.

Bowling

how do we know this formula: source = A*C_DUDX(c,t);

I have a equation for velocity gradient: dudz=const

However, I have no idea how to transfer my equation into this kind of form: source = A*C_DUDX(c,t);

Thanks