# A probelm on the pressure

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 November 7, 2000, 16:43 A probelm on the pressure #1 John Guest   Posts: n/a Hi, there, I have a problem in understanding the role that the pressure (p) plays in 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 flows, one can set the traction (Fx, Fy) at the 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,

 November 8, 2000, 10:24 Re: A probelm on the pressure #2 Chidu Guest   Posts: n/a Hi John, The "system" pressure drops out of the NS equations only for incompressible flows since density no longer depends on pressure. So your first point is valid only under this constraint. As far as the traction condition, is this intended for internal flows or external flows? By that I mean is your boundary near to a wall or far away from a wall? Somehow, I need something more concrete to proceed ahead. regards, chidu...

 November 8, 2000, 12:49 Re: A probelm on the pressure #3 John Guest   Posts: n/a Hi, Chidu, Thx for your reply. You are right that I was talking about the incompressible viscous flow. And, the traction BC will be applied to the external flow that the boundary is far away from a wall. regards,

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