# UDF for source term in momentum equation

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 November 4, 2004, 05:50 UDF for source term in momentum equation #1 Enrico Guest   Posts: n/a 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; }

 November 4, 2004, 06:50 Re: UDF for source term in momentum equation #2 Andrew Garrard Guest   Posts: n/a you haven't defined A?

 November 5, 2004, 04:26 Re: UDF for source term in momentum equation #3 Sun Guest   Posts: n/a 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

 November 6, 2004, 05:26 Re: UDF for source term in momentum equation #4 enrico Guest   Posts: n/a A is a constant

 November 9, 2004, 05:27 Re: UDF for source term in momentum equation #5 Bowling Guest   Posts: n/a 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

 November 10, 2004, 07:06 Re: UDF for source term in momentum equation #6 enrico Guest   Posts: n/a Hi ,Bowling , but why dS[eqn]=0 ? dS[eqn]=dS/dU

 November 12, 2004, 01:58 Re: UDF for source term in momentum equation #7 Bowling Guest   Posts: n/a 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

 November 13, 2004, 04:33 Re: UDF for source term in momentum equation #8 enrico Guest   Posts: n/a Hi Bowling , should i use UDS for calculating dS[eqn] ?

 November 15, 2004, 01:20 Re: UDF for source term in momentum equation #9 Bowling Guest   Posts: n/a 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 Rashed likes this.

 May 30, 2014, 12:34 #10 Senior Member   Join Date: Jan 2012 Posts: 193 Rep Power: 11 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