|September 17, 2012, 12:19||
comparing values p_rgh and p
Join Date: May 2012
Posts: 42Rep Power: 6
I have a problem at the moment that I unluckily can't solve using the search function of this board.
I need to compare values of total pressure (sum of dynamic, hydraulic and static pressure as in the Bernoulli equation) between two different cases, that use the same geometry but different solver and (now it becomes tricky) also a different OpenFoam version.
My case is new (OF 2.1.0, SIMPLE) and pressure is defined in p-file as [m^2/s^2].
To obtain total pressure from this, I think I have to multiply p*rho (=static pressure) and add rho/2*mag(U)^2 (=dynamic pressure). The hydraulic pressure should be neglectable (gas with a height difference of 10cm........).
The old case that I have to compare with, was set up in OF 2.0 (I believe, interFoam), but pressure is defined as p_rgh [kg/ (m*s^2)].
I found out that p_rgh=p-rho*g*h, but
1) is p in the equation above the same as static pressure in the Bernoulli equation? Would (static pressure in the Bernoulli equation) than be p.stat=p_rgh+rho*g*h ?
2) what steps should i perform to obtain a comparable total pressure from this?
Pressure differs from 1800 to 270, so I tried all kinds of conversions but end up with wrong numbers.
I am not sure if the flaw is within my case setup or caused by my misunderstanding of p_rgh.
Thanks for everyone taking time to answer my stupid questions.
|August 15, 2014, 10:32||
Join Date: May 2014
Location: Paris, France
Posts: 108Rep Power: 4
I hope you have found an answer since your post but i prefer to post this answer to help you if you are waiting for this response facing your computer, with a very long hair.
p_rgh is a dynamic pressure, the exact definition is :
p_rgh = p - rho*g*h.
So you are right when you say :
p (the static one) = p_rgh + rho*g*h.
Have a good day.