SIMPLE algorithmPlease 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

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.

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:

what are the values of underrelaxation 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:

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 0.30.4 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:

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:

Thanks and i will do that

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 
All times are GMT 4. The time now is 23:56. 