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 xcomponent 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. xmomentum, 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 
All times are GMT 4. The time now is 04:48. 