CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
-   Main CFD Forum (http://www.cfd-online.com/Forums/main/)
-   -   A problem on the pressure (http://www.cfd-online.com/Forums/main/2799-problem-pressure.html)

 John November 15, 2000 17:45

A problem on the pressure

Hi there,

I have to re-post the following message, and hope somebody could help me.

I have a problem in understanding the role that the pressure (p) plays in the incompressible NS equations. It is well known that only the pressure gradients contribute to the flow motion. You can force the pressure at any point in the flow domain to zero as a reference datum. But, for outgoing (external) flows, one can also set the traction (Fx, Fy) at the downstream exit face to vanish:

Fx = (-p + 2/Re du/dx) nx + (du/dy + dv/dx) ny = 0 Fy = (dv/dx + du/dy) nx + (-p + 2/Re dv/dy) ny = 0

in which (nx, ny) is the direction vector normal to the exit. The problem is that the pressure at the exit can be any value if different reference points are set. What would you say?

Regards,

 Abhijit Tilak November 16, 2000 00:47

Re: A problem on the pressure

hi,Jhon

read the chapter on determining flow field which discusses relative nature of pressure (from Numerical Heat Transfer and Fluid Flow by S.V.Patankar) what is important is pressure gradient not the absolute value of pressure. (dp/dx) will be same between two points even if u choose differnet reference pressure values. Pref will cancel when u calculate dp/dx for a particular Pref that u choose. u can hence choose any value of Pref for ur computations. I hope this helps.

Abhijit

 John November 16, 2000 03:03

Re: A problem on the pressure

Hi Abhijit,

I understand that only the pressure gradient matters in incompressible flow. But how about the traction on the exit open boundaries ? If it vanishes, pressure must be p = 2ux/Re (say, the exit pointing to x-direction). I do not think you have answered my question. Thx anyway.

John

 andy November 16, 2000 05:38

Re: A problem on the pressure

If you choose to impose a value of pressure at an exit (or set of exits) then you have specified your reference pressure here. You cannot specify a second unrelated one somewhere else (although you probably could impose a physically related one). Also, having introduced a velocity/pressure exit condition you must:

(a) Impose a compatible inlet condition.

(b) Ensure your pressure equation still has a valid set of boundary conditions (still satisfies continuity and, usually, a low order approximation to the governing pressure equation).

Inlet and exit boundary conditions for the velocity and pressure are coupled together and cannot be chosen arbitrarily. They come in valid sets.

 John November 17, 2000 13:39

Re: A problem on the pressure

Hi, Andy,

Thx for ur reply. It is quite helpful. Still, one more question:

If the traction-free BC is used to the exit, the pressure must be ballanced with the viscous force, say p = 2du/dx/Re in x-direction. So, the pressure reference is indeed given, but not arbitrarily !! Is this contrary to the statement that only pressure gradient matters?

Regards,

 Jim Park November 17, 2000 19:12

Re: A problem on the pressure

Am I right that your equation comes from integrating a reduced momentum equation; pressure gradient is balanced by the gradient of the viscous normal stress?

If so, you need to set an integration constant, which will involve a pressure and normal stress at some other point.

So the exit pressure depends on where you choose that point and is thus still arbitrary.

 andy November 18, 2000 07:41

Re: A problem on the pressure

Only the pressure gradient matters within the solution region for the isothermal, constant density, incompressible formulation of the Navier-Stokes equations (qualifications because the formulation of the density/energy treatment for incompressible flows may introduce an absolute pressure dependence e.g. through a gas law).

However, various boundary condition treatments (inlet/exit pairs particularly) can involve reference to "external" absolute pressures (or other quantities) to determine the values on the boundary of the solution variables. This can couple the boundary pressure and velocity components. Note that this arises through a choice of boundary conditions and not governing equations. You can still add the first number you thought of to the pressure field and it will satisfy the incompressible N-S equations (plus qualifications) within the solution region. It simply may not satisfy the chosen boundary conditions.

The choice of valid sets of boundary conditions for the full and reduced forms of the Navier-Stokes equations is a large topic.

 All times are GMT -4. The time now is 04:18.