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August 31, 2012, 05:47 
UDF to define the vector transport equation

#1 
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Vaze
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Hi
Through the UDS, scalar transport equation can be defined. I would like to define the vector transport equation. How can i define it? regards Mvee 

September 3, 2012, 00:03 

#3 
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Hi Amir
Can you describe how and which UDS to use? Mvee 

September 3, 2012, 03:38 

#4 
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Suppose you wanna model momentum equation via UDS. You have to define 3 UDSs; u, v and w; and provide appropriate transport terms such as convection, diffusion, unsteady and source terms for each of them. A UDS has a general form of transport equation which you can modify for your purpose.
Bests,
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Amir 

September 3, 2012, 04:07 

#5 
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Hi Amir
Problem: Welding arc simulation Description: 1) No. of zones: 3 2) Solid zone: electrode and workpiece (copper) 3) Fluid: argon (electrically conductive) Current is imposed on the electrode which passes from the fluid and workpiece (which is at zero potential). Due to the current electromagnetic field is generated which is governed by Maxwell’s equation (consists of current continuity and potential vector equation). These equations are having same form as that of our NS equations. I have defined one UDS for voltage to solve current continuity equation. In the similar form do you mean to define UDS for vector A. Can you describe how shall I follow for the physics mentioned above? Also the output of this UDS will go as source term in momentum equation. Thanks Mvee 

September 3, 2012, 05:29 

#6  
Senior Member

Quote:
Ok, you'll also need a vector transport equation. You can consider a vector transport equation as 3 scalar transport equation (one for each component) and model each of them via a UDS. Here you have a vector A, so define 3 UDS as Ax, Ay and Az as its components. So consider your vector equation and decompose it into 3 scalar equation. Finally with simple UDFs you can add each component of your added equations into each momentum components without any trouble. Bests,
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Amir 

September 3, 2012, 06:13 

#7 
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Hi Amir
Thank you very much for your help, still have some doubt. My algorithm will work in following manner  (1) Current continuit equation: =0 /* = applied voltage and = Laplacian operator*/ (2) current density, j =  (3) Potential vector equation: = First and third point can be applied through UDS. How to interface (required to solve point 2) between point 1 and 3? Is it required UDF? 

September 3, 2012, 07:34 

#8  
Senior Member

Quote:
It depends. It doesn't need to evaluate j explicitly if you just want to obtain A! you can substitute j from equation 2 into 3 directly and remove the j variable. But if you want to obtain j as well; yes, you'll need a UDF to store the phi gradient into a UDM. It should be noted that you would also need UDFs for defining source terms of equation 3. Bests,
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Amir 

September 10, 2012, 01:27 

#9 
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Hi Amir
Thank you very much for your kind cooperation. I have solved these equations. Thank you Mvee 

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