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May 21, 2023, 23:50 
Outflow boundary condition

#1 
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BM
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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 
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Lucky
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Fluent fixes the operating pressure at the reference location. The default is the cell closest to the origin.
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 

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May 22, 2023, 23:41 

#4 
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Lucky
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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 nonlinearity 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 navierstokes 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 

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Sorry if my questions seem super basic. 

May 31, 2023, 15:45 

#6 
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Lucky
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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 

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