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June 25, 2009, 11:42 
problematic source term in UDS

#1 
Member
Akour
Join Date: May 2009
Posts: 79
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Hello everyone,
I have posted a similar question as this previously, but no one has gotten back, so Ill be more detailed, could somebody please provide some assistance... I have 2 PDEs which I am trying to solve using UDSs PDE of scalar 1 (V) : 0=electric_diff*d/dx_i(dV/dx_i)+(density*Q)/e PDE of scalar 2 (Q) : 0=spcharge_diff*d/dx_i(dQ/dx_i)+(density^2*Q^2*K)/e+density*k*(dV/Dx_i)*(dQ/DX_i) I am having a very big problem with the term in bold, which I have implemented as a source term..defined in my UDF shown below as DEFINE_SOURCE(spcharge_lin,c,t,dS,eqn): The problem is that when I turn this term on, the solution diverges in very few iterations, it doesnt seem possible to get convergence even with tiny relaxation factors (.0001)...does someone have an idea as to why this may be happening? The UDF is shown below: #include "udf.h" #include "sg.h" static real kappa = .00000003; static real epsilon = .00000000000855; /*source of scalar 1 (V)*/ DEFINE_SOURCE(electric_source,c,t,dS,eqn) { dS[eqn] = 0; return (C_R(c,t)*C_UDSI(c,t,1)/epsilon); } /*diffusivity of scalar 1*/ DEFINE_DIFFUSIVITY(electric_diff, c, t, i) { return 3.54; } /*problematic source of scalar 2 (Q)*/ DEFINE_SOURCE(spcharge_lin,c,t,dS,eqn) { real source,source_a,source_b; source_a = C_R(c,t)*kappa*C_UDSI_G(c,t,0)[0]*C_UDSI_G(c,t,1)[0]; source_b = C_R(c,t)*kappa*C_UDSI_G(c,t,0)[1]*C_UDSI_G(c,t,1)[1]; source=source_a+source_b; dS[eqn] = 0; return source; } /*OK source of scalar 2*/ DEFINE_SOURCE(spcharge_quad,c,t,dS,eqn) { real source; source = (pow(C_R(c,t),2)*pow(C_UDSI(c,t,1),2)*kappa)/epsilon; dS[eqn] = (2*pow(C_R(c,t),2)*C_UDSI(c,t,1)*kappa)/epsilon; return source; } /*diffusivity of scalar 2*/ DEFINE_DIFFUSIVITY(spcharge_diff, c, t, i) { return kappa / 40; } 

July 10, 2009, 11:38 

#2 
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Join Date: Jul 2009
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Hi,
dS[eqn] = 0 means that FLUENT treats your source term explicitly and it could cause divergence problems. If you specify the derivative of your source term, FLUENT may treat your source term with an explicit and implicit part and this could help convergence. good luck, 

July 12, 2009, 15:57 

#3 
Member
Akour
Join Date: May 2009
Posts: 79
Rep Power: 10 
hi,
thanks for your reply greg...however I am solving for scalar Q, and my source term has a dQ/dX term so i dont think I can specify the derivative of that (with respect to Q)...so I am not sure what the problem is...
__________________
akour 

July 15, 2009, 09:30 
UDF with C_H_G "access violation"

#4 
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hameurlaine
Join Date: Jul 2009
Posts: 27
Rep Power: 10 
Hello,
I am solving the energy equation with the enthalpy formulation. I would like to add a source term via UDF that includes the gradient of enthalpy but I get an Access Violation error when I try to use the C_H_G(c,t) macro. "access violation" I was able to work around the problem by storing enthalpy in a UDS and then calculating the gradient using a C_UDSI_G macro. but a have also message error in compilation. physics formule is: (dh/dx+dh/dr); deriverd enthalpy axial+ derived enthalpy radial. please help me Thank you! 

December 25, 2013, 00:36 

#5  
New Member
xingangzheng
Join Date: Jul 2009
Posts: 11
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Dear ak6g08,
Have you solved your problems? I have read the similar simulation published in Fuel Processing Technology titled Particulate removal via electrostatic precipitatorsCFD simulation. And write a letter to find some consulatings, but no reply. I wonder wether they have treat the gradients very well in their simulation... Quote:


January 4, 2014, 22:38 

#6  
New Member
CAO Liushuai
Join Date: Apr 2012
Posts: 29
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Quote:
The gradients should be treated well to help the convergence, I think it is better to be a negetive expression. 

February 10, 2015, 01:19 

#7 
Senior Member
Vaze
Join Date: Jun 2009
Posts: 160
Rep Power: 10 
Hi Shuai
I am also facing same problem. I am not solving UDS but using it to store the value which is solution variable depedent. This is done to have gradient of UDS (C_UDSI_G) in source term. I felt this is not the correct way to achieve it. And in this case linearization is not requried because gradient itselt suggest to be linearlized the function. Can you suggest better way to calculate gradient of UDS which I am not solving? ak6g08, you can also share your experience. Thank you 

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