# Outflow boundary condition

 Register Blogs Members List Search Today's Posts Mark Forums Read

 May 21, 2023, 23:50 Outflow boundary condition #1 Member   BM Join Date: Sep 2021 Posts: 35 Rep Power: 4 I have a simple setup with 2 boundary surfaces: one is a velocity inlet, and the other is outflow. The flow is single phase and incompressible, with no energy equation. My question is: if pressure is not specified anywhere in the domain, then how is pressure calculated? I understand that since the flow is incompressible, only the change in pressure is relevant, but doesn't Fluent still need to fix pressure at some location to solve for the pressure field? A similar follow up question would be if you specify one surface as a pressure inlet and the other as a pressure outlet, then how is the velocity field calculated since the velocity (and hence mass flow) is not specified anywhere?

May 22, 2023, 02:33
#2
Senior Member

Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,713
Rep Power: 66
Fluent fixes the operating pressure at the reference location. The default is the cell closest to the origin.

Quote:
 Originally Posted by blyatman A similar follow up question would be if you specify one surface as a pressure inlet and the other as a pressure outlet, then how is the velocity field calculated since the velocity (and hence mass flow) is not specified anywhere?
Have you ever tried turning on a water faucet? There is a fixed pressure upstream and downstream and yet somehow... the flow does indeed flow out of the faucet at a fixed rate. Hmmms... Nature doesn't use algorithms to determine the flowrate. Fixed pressures constraints are very practical and real boundary conditions.

May 22, 2023, 07:39
#3
Member

BM
Join Date: Sep 2021
Posts: 35
Rep Power: 4
Quote:
 Originally Posted by LuckyTran Have you ever tried turning on a water faucet? There is a fixed pressure upstream and downstream and yet somehow... the flow does indeed flow out of the faucet at a fixed rate. Hmmms... Nature doesn't use algorithms to determine the flowrate. Fixed pressures constraints are very practical and real boundary conditions.
I meant how is it solved numerically? Like suppose I have a pipe of uniform cross-section, and I specify the pressures at both ends. There is an infinite number of solutions right? So how does Fluent "pick" the velocity?

 May 22, 2023, 23:41 #4 Senior Member   Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,713 Rep Power: 66 Within reason, there's only one solution that satisfies conservation of mass and momentum. Nature doesn't pick velocity and neither does Fluent. Fluent solves transport equations. If you turn on a faucet an infinite number of flows doesn't come out. Of course due to non-linearity you need to supply initial guesses and there is a procedure for how the transport equations actually do get solved, but I don't think that was your question. But if it was, you discretize the navier-stokes onto the computational grid via the FVM method and apply the Gauss divergence theorem. And then discretize all the terms and this gives you a system of equations that need to be solved over the entire mesh that you then solve using a linear solver. These are just details in how the equation get solved, it is akin to asking how does one numerically solve x+1=2 and x + 2 = 3. Regardless of how Fluent might attempt to solve this system, the answer is obviously 1. Now substitute instead of x+1 and x+2 the conservation of mass and momentum over a computational cell.

May 30, 2023, 04:41
#5
Member

BM
Join Date: Sep 2021
Posts: 35
Rep Power: 4
Quote:
 Originally Posted by LuckyTran Within reason, there's only one solution that satisfies conservation of mass and momentum.
This was the part I was confused about. If there was just a pipe with uniform cross-section, then any velocity would satisfy conservation of mass. Likewise, wouldn't momentum also be conserved for any velocity since the only thing that matters is the pressure difference? So if there was a 10 Pa pressure difference, how would we uniquely determine velocity?

Sorry if my questions seem super basic.

 May 31, 2023, 15:45 #6 Senior Member   Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,713 Rep Power: 66 You have no slip walls as a boundary condition on the pipe walls. A pressure drop of 10 Pa (or any other number) must be exactly equal to the integrated wall shear stress on the pipe walls or you will not satisfy the momentum balance. So, you have to solve for the velocity field that has this wall shear stress that is also continuous. Hence, when you turn on a faucet, it has a supply pressure and atmospheric pressure. This driving pressure difference is exactly matched with the friction in the system and water flows at a fixed rate out of your faucet.

June 1, 2023, 07:30
#7
Member

BM
Join Date: Sep 2021
Posts: 35
Rep Power: 4
Quote:
 Originally Posted by LuckyTran You have no slip walls as a boundary condition on the pipe walls. A pressure drop of 10 Pa (or any other number) must be exactly equal to the integrated wall shear stress on the pipe walls or you will not satisfy the momentum balance. So, you have to solve for the velocity field that has this wall shear stress that is also continuous. Hence, when you turn on a faucet, it has a supply pressure and atmospheric pressure. This driving pressure difference is exactly matched with the friction in the system and water flows at a fixed rate out of your faucet.
Ah right, I forgot about the pressure loss due to friction. Thanks for clearing that up!

 Tags outflow