SIMPLE algorithm-Please help
hi,
i tried to write a code for SIMPLE algorithm in Fortran but found it difficult. can anyone give a sample code on SIMPLE.my mail id is rajandhayalan@gmail.com thank you regards Thiyaga. |
are you writing it with colocated grids and staggered grids
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Simple
hi,
thanks for the reply.iam using staggered grid. reg Thiyaga. |
have u tried veersteeg's book, coz the algorithm in it is almost the code. I have implimented it in mathematica.
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hi,
thx , i will try... |
Hi Ammar,
I tried with verteeg's book only. But the solution is blowing out (simple algorithm). Please check whether my following boundary conditions are correct: firstof all i am using staggered grid (backward) and using Explicit scheme. Assume it is a duct flow --> two walls on the upper and lower side (v velocity = 0), leftside --> inlet (u velocity) and rightside -->outlet (pressure outlet) but the u velocity grid would be the last on the right side I have just created a dummy nodes outside the main domain (as mentioned in the book). my domain size is 0.09m by 0.09m, height 0.08m. Grid size is 20 by 20 1. On the inlet i am giving u velocity = 0.1 m/s 2. On the outlet i am extrapolating this u velocity as given in the book 3. On the top and bottom wall my v velocity = 0 4. on the right side (just before the u node -- outlet) i am fixing the pressure as 0 I have doubt in the pressure correction (pprime) bc. 1. On all the sides i am setting the value of pprime = 0. Withall this my solution is blowing out after few iterations (5 iterations) Also for the simplicity i am keeping the value of meu and row as constant in all the equations (common row(density) taking out from the equations and canceling). I am not doing any averaging on this scalars. If you wish i can send my code also (its fully commented) Your suggestion is highly appreciable. thanks jyo |
Dear friend. you should tell more details about your problem. your geometry.boundary conditions and...
maybe I can help you Quote:
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what are the values of under-relaxation for velocity and pressure?
you should set the pprime value zero at the beginning of each iteration. for B.cs usually the outlet boundary is difficult to set. Quote:
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pprime initialisation
Hi Artmiss,
I did the initialization of pprime. My urf for u and v, pressure is 0.3. for the outlet boundary since i am giving the pressure outlet (0 value -- gauge) my pprime would be zero at the outlet as there are no corrections my u and v velocities are going very high values and finally diverges. any suggestions thanks jyo |
Dear jyo,
Put The relaxation factor for U and V around 0.3-0.4 and for Pressure around 0.1. You should notice that relaxation factor for pressure should be less than for U and V ones. The other thing is before you wanna call your pressure boundary and solver you should put pprime=0.0 in the main Do of your program. hope it will work. Quote:
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Dear friend. perform the suggestions of Mr. babak.The SIMPLE algorithm is very sensitive to urf's especially when Re increased.Don't use the values suggested in commercial soft wares. I doubt the reality of convergence with this urf's!
for final check tell me the method of calculation of u and v at solid boundaries.it is important Quote:
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Thanks and i will do that
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Hi,
Again i have restructured my grid pattern in the following manner. I am now giving pressure inlet and pressure outlet boundary. Even on the wall (top and bottom) also i have my pressure and u velocity grid to coincide. This way i am sure about the pprime values at the boundary. Now my pprime value on the left and right side (inlet and exit) boundary will be zero. On the wall i am doing pwall = p(i,1) = 2*p(i,2) - p(i,3), similarly i am doing for pprime also. One more thing that i want to clarify is that i am fixing the value of row and meu everywhere (in the scalar grid points). Is that fine what i am doing. (still on the process of writing the code) Any suggestions ? thanks jyo |
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