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momentum equation--pressure correction with SIMPLE

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Old   October 24, 2009, 22:31
Default momentum equation--pressure correction with SIMPLE
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Hi,

I am very new to CFD, I have been unable to find a resource that shows how the discretised momentum equation is derived. Every reference starts with the differential momentum equation, then the next line is what seems to be a random equation with a_i_J multiplying the first term, with no explanation of the meaning of this coefficient.

The derivations of the pressure correction equation is similar. There is a d in it, which is related to this a_i_J term in some way that I cannot understand.

Can anyone point me to a web based reference, or provide an explaination.

Thanks
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Old   October 25, 2009, 05:40
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Hi fredp
I suggest you to study chapters 4, 5 and 6 of “An Introduction to Computational Fluid Dynamics - Versteeg”.
It is very good.
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Old   October 25, 2009, 10:52
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Quote:
Originally Posted by mhm View Post
Hi fredp
I suggest you to study chapters 4, 5 and 6 of “An Introduction to Computational Fluid Dynamics - Versteeg”.
It is very good.
I have read this--it wasn't helpful. It has a bunch of undefined coefficients. I would love to find an explanation of the meaning of each of the terms in the momentum equation in words.
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Old   October 30, 2009, 10:14
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refer to introduction to computational fluid methods by schafer ...its good....neway i found limited info on the staggered grid....try reading anil dates papers on pressure velocity corrections....here has well enuf explained the terms and their relevance.i used the collocated method ...i find that more convinient...all the best
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Old   October 30, 2009, 14:29
Default Simple algorithm
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Dear friends,

I have written my SIMPLE algorithm on staggered grid. But encountering the following problem : before that my boundary conditions are as follows:

Staggered arrangment is as follows:

left and right side (basically a duct) ---> pressure inlet and outlet
top and bottom ---> pressure and u velocity

Left side ---> pressure inlet (32 pa)
right side ---> pressure outlet (0 pa)
Top and bottom side ---> dp/dy = 0, dpprime/dy = 0, u=0. URF for pprime (or pressure) = 0.1 and for u and v equal to 0.3 and 0.4 respectively.

Problems:

(i) pprime values are very very small (so pressure correction also would be very small)
(ii) only at the exit my u velocities are reasonable

I have run my outer iterations = 3000 and my inner iteration (pprime) = 300. Still no change in the pressure value from the inlet to exit.

Please help me in this regard.

thanks

jyo
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Old   October 31, 2009, 05:53
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check with your urf ....it might be a bit low....
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Old   October 31, 2009, 09:46
Default initialisation problem
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Hi zobekenobe,

Thanks for your mail. I tried with a simple couette problem which is given in the CFD book by Anderson. But the formulation is from Malalasekara. I am getting a linear profile as given in the book. But when i change the initial value (it is given as zero in the book especially pressure) even for a small change my solution is blowing. Really don't know what is happening. My urf's are 0.1 for pressure and 0.3,0.4 for u and v velocity. ( I am using staggered grid and explicit method). I tried with both upwind and central difference. There is no improvement among the two.

thanks

jyothish
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Old   October 31, 2009, 09:58
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Hi,

Previous to my last message, I want to convey that effect of dp (between inlet and outlet) is reflected only at the outlet and not in the interior of the domain.

thanks

jyo
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Old   November 4, 2009, 09:22
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Hi,

Still my SIMPLE formulation is not giving the correct solution. i am using staggered grid arrangment. I am giving pressure inlet and pressure outlet boundary conditions. Here the pressure inlet is the total pressure. Top and bottom i am giving wall boundary conditions. From Po i am calculating Ps (static pressure) using Po = Ps + 0.5*row*u^2 and this Ps will go into the momentum equations (pressure gradient). Is this right ?. My solution is blowing after some thounsand iterations.

I am taking the dimension from anderson book (for couette flow). It is as follows :

domain size ----> 0.003m (height) and 0.15 m (length)
grid size ----> 21 x 11 for pressure
----> 22 x 11 for u velocity
------> 23 x 12 for v velocity
I am using explicit scheme (finite volume) with upwind discretization.

Please help me in this regard.

Thanks

jyothishkumar
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Old   November 12, 2009, 08:29
Default SIMPLE algorithm in colocated grids
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Hi,

Can you please check the following steps to follow to implement SIMPLE algorithm in colocated grids. Also i have added my questions (if there is any) immediately below the steps. please help me in this regard.


1. First initializing values at the nodes (p, control volume center) Up*, Vp* Pp* etc
2. Finding Up and Vp from the general discretized momentum equation from above guessed value.
3. Find Ue at the cell faces (rhie and chow’s intrapolation ) using the adjacent nodal U values
For eg. At face ‘e’ à Ue = f(Up, UE, PE, Pp, PEE, dp, dE, etc)
4. For the pressure correction equation we need Ue ,Up, de(?), dp
We can get the value of dp from the discretized momentum equation (CV center)
How to get this de (should also be through interpolation like de = (dP + dE) / 2 ?)
5. From the above pressure correction formulation we can get the value of Pprime.
6. Then PP = PP* + (alpha * Pprime)
7. Velocity correction
Up = Up* + dP(Pprime(i-1,j)-Pprime(I,j)) -------à without correction
Up = (alpha * Up (without correction)) + (1- alpha)*Up old (from previous iteration)
8. Above process is repeated for VP.

thanks

jyo
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Old   November 16, 2009, 10:51
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Hi,

Can you pls check my following steps for SIMPLE Algorithm in colocated grids.

1. Find Up's at all points from discretized momentum equation (without pressure term)
2. Apply rho and chi's interpolation for finding the face velocity Ue(interpolated from UP and UE (nodal points or cell centered)
3. substitute this face velocities in the correction equation.

Pls help me in this regard.

thanks

jyothish
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