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Negative values of concentration while solving temperature |
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February 9, 2009, 01:42 |
Hi Vishal,
I m not sure but j
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#1 |
Member
Sachin Kanetkar
Join Date: Mar 2009
Posts: 57
Rep Power: 17 |
Hi Vishal,
I m not sure but just a suggestion are your boundary conditions correct or check your domain if it is too complex.... try to simplify... if that works fine....reduce the grid size (inc no. of grid pts)and reduce time step size I think the equations would be making solution quite unstable Though I m not sure...just some suggestions Hope you find the error Regards Sachin |
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February 9, 2009, 03:13 |
Hi Vishal,
I guess these are
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#2 |
Member
Sachin Kanetkar
Join Date: Mar 2009
Posts: 57
Rep Power: 17 |
Hi Vishal,
I guess these are to stringent BC's may be they reach at some point where they dont satisfy them correctly My suggestion - for outlet just extrapolate the values by zerogradient...chk result...if this doesnt work ... ...i hope it does Dont modify the whole set at time....modify B.C's one by one ...try modify only for Phi then for C1 then for C2 ...see if u can get descent results no absurd values...if this also fails change one boundary everytime and run ...might find why it fails...I know this sounds wierd... Just one more question are you very sure of the eq you have discretised...i mean u have separated phi from conc ..ok but after they recombine do they give the equations back Sachin |
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February 9, 2009, 03:39 |
Hi Sachin,
Thanks for the r
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#3 |
Senior Member
Vishal Nandigana
Join Date: Mar 2009
Location: Champaign, Illinois, U.S.A
Posts: 208
Rep Power: 18 |
Hi Sachin,
Thanks for the response.. I would try them.. and would get back to you... The exact equations which I would like to solve are div {(D1*grad(C1) + D1*Z1*C1*grad(Phi))} = 0 -(1) div {(D2*grad(C2) + D2*Z2*C2*grad(Phi))} = 0 -(2) laplacian (phi) = - alpha * Z1*C1 - alpha*Z2 * C2 - (3) where C1 C2 and Phi are concentrations and potential respectively and D1,D2,Z1,Z2, alpha are constants Regards Vishal |
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