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-   -   Little question concerning pressure dimension (http://www.cfd-online.com/Forums/openfoam-solving/58920-little-question-concerning-pressure-dimension.html)

 dinonettis April 28, 2008 10:15

Hi folks, I was wondering a

Hi folks,

I was wondering about the pressure dimension adopted in different solver I'm using (potentialFoam, SimpleFoam, rhoTurbFoam). Indeed I found these different lines in the 0/p files:

p: [0 2 -2 0 0 0 0] m^2/s^2

p: [1 -1 -2 0 0 0 0] Kg/(m*s^2)

Could someone please clarify this issue??
thanks

dino

 deepsterblue April 28, 2008 10:23

The momentum equation's divide

The momentum equation's divided by density in the first case, and hence the consistency for pressure dimensions. (You'll have kinematic viscosity here)

 dinonettis April 28, 2008 16:55

Hi Sandeep, thanks for your

Hi Sandeep,

thanks for your clarification. Anyway, do you mean that if we are in the first case the value to impose in the BC is p/rho (and not simply the pressure)?? Am I right?

dino

 milos March 24, 2009 14:01

Yes, you are right... It confused me too. :) Do you know by any chance what does it actually mean if I change my outlet pressure BC to any other number than 0?

 alberto March 25, 2009 03:23

Little exercise: try with simpleFoam to change the pressure at the outlet at different values, let's say p_outlet = 0, 50 and 1000. Then compare the velocity and pressure fields. What do you notice? Why? What kind of flow are you solving for (hint rho = const)? :-)

 evrikon March 25, 2009 14:14

Hello together,

I use simpleFoam and I also found that the pressure used in OF is normalized llike p/rho. Question: what I do see in in Paraview as the pressure is p/rho or really pressure?

 milos March 26, 2009 03:34

1 Attachment(s)
As far as I noticed, what you see in ParaView by default (if boundary conditions are default) is p/rho in its relative form, where 1 is the pressure at the outlet (in boundary conditions set as 0). To switch to the relative pressure in [bar] unit you should divide the default number with 100.

To set things a bit clearer, I have attached a photo of my project where you can see the pressure distribution along the pipe elbow. If my interpretation is correct, the numbers on the scale say that if pressure on the outlet was 1bar, then the other pressures along the pipe would be 1+"number from the scale". For example the pressure on the outer side of the elbow would then be 1.407bar and on the inside the pressure would be 0.519.

P.S. - To Alberto: I'll give it a thought as soon as I catch some time. The thing that is bothering me is when I set my outlet pressure to 0, the simulation diverges and if I set it to 0.1, then it works perfectly!

 alberto March 26, 2009 15:31

Hi Milos,

could you please post your diverging case (if it's not huge to run)? It seems very strange to me the divergence is caused by the pressure at the outlet.

Regards,

 milos March 26, 2009 16:31

Sure, but the thing is .zip or .tar package exceed the allowed size and if I try uploading the files independently - it reports some sort of error. Got an idea how to do it? If you would like I can send it directly to your e-mail ( .tar.gz is around 3.3 MB)?

I did some roaming concerning that outlet pressure issue, found some literature and advices so after I do my reading and contemplating - I'll get back to you. :) Hopefully with a good understanding of how it all goes.

Cheers!

 alberto March 26, 2009 16:41

You can email me (albert.passalacqua@gmail.com) the case, if you want.

Thanks,

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