[Overdue]

Hey Guys,

I have some problems to understand the behaviour of a simple flow simulation in a pipe with pressure boundaries (total or static). For example: water flows through a straight pipe with a length of 1m and a diameter of 50mm.
To get a velocity of 1m/s, I can calculate the necessary pressure difference:

delta_p= rho/2*v² = 997 kg/m³/2 * 1 m/s = 498,5 Pa, that’s easy.


For my first simulation I choose the following boundaries:

• Inlet: total pressure = 498,5 Pa
• Outlet: Average static pressure, relative pressure = 0 Pa
• The remaining surface as a free slip wall

The result is correct, the velocity is about 1 m/s



And here is my problem of understanding in a second simulation:
When I choose static pressure instead of total pressure as my Inlet boundary, the result is completely different. The velocity is approx. 5,29 m/s. The min. and max. total pressure are 13939,5 Pa and 14449,1 Pa.


Can anyone explain how velocity and total pressure are calculated in this case?

Thanks in advance!

Best regards,
Luke

[ghorrocks]

Your second simulation sounds badly posed to me. You specify a static pressure difference to drive the flow, but you have free slip walls and just a straight pipe so there is no flow resistance. I would suspect your answer is not converged, and it is not possible to converge it.

[Opaque]

As Glenn said, your second setup is badly (ill) posed.

When working with numerical methods, additional care must be taken when setting up boundary conditions. For example in this pipe simulation, the change in static pressure between inlet and outlet MUST exactly match the losses through the system (including the numerical ones). Such losses are a function of the mesh resolution until a "mesh independent" setup is reached.

It is nearly impossible to force the equations to match such requirements. The total pressure boundary condition is a more flexible boundary condition, and gives room for the numerical error to be reduced as the mesh resolution is improved. In addition, from the physics if you use free slip walls, the change in static pressure should be 0; therefore, any velocity should satisfy the problem, correct ?

You can setup similar setup for heat transfer: two walls at specified inlet and outlet, and specified matching heat flux on the rest.

Hope the above helps,

[Overdue]

My intention why I choose free slip wall, is that I want to compare the results with my own calculation with the frictionless Bernoulli equitation. I thought, I won’t have any friction losses with a free slip wall
The calculation converges, but with a lot of timesteps (see picture)

So when I understand that correctly, if I just choose static pressure for inlet and outlet, the total pressure and velocity in the pipe are calculated randomly.
I’ll think more about the points you mentioned. Thank you very much for your help so far!

[Jiricbeng]

Well, if there is a free slip, it might not mean there is no loss. Because pressure loss depends on velocity gradient in the liquid which can be presented even in case of free slip wall.

Bernoullie eq: p1/rho + 0.5*v1^2 = p2/rho + 0.5*v2^2
Continuity: v1 * S1 = v2 * S2, for straight pipe you have v1 = v2

so if v1 = v2, the Bernoullo eq. is reduced to:
p1/rho = p2/rho /where rho is constant
that is p1 = p2
And your boundary condition is that p1>p2. Well the Bernoulli equation assumes no losses, obviously the CFX solver tries to find such velocity magnitude and velocity profile, whose gradient will correspond to the static pressure difference (p1 - p2). Probably the velocity received from the solver need not be random.

[ghorrocks]

You show that p1 must equal p2, so how can boundary conditions where p1>p2 converge in steady state for a straight pipe with a constant flow across the cross section?

I appreciate that if the pipe cross section varies or if there is a non-constant velocity across the duct cross section then you will get pressure changes along the length. Also in transient flow a pressure difference causes the flow to accelerate.

[Jiricbeng]

Originally Posted by ghorrocks

You show that p1 must equal p2, so how can boundary conditions where p1>p2 converge in steady state for a straight pipe with a constant flow across the cross section?

This is what I wanted to explain, it converges to certain mass flow, because even if you set free slip on walls, it does not mean there is not any pressure loss. There still is, because the flow is viscid, it does not matter you have free slip on walls, because pressure loss is a function of velocity gradient, not function of friction along the wall.

[ghorrocks]

I do not understand your logic. In this case there is no velocity gradient and no friction and therefore no pressure gradient. Therefore this simulation will never converge for a Navier Stokes solver. Have I missed something?

[Overdue]

If it helps, here is the CL Code:

[ghorrocks]

The CCL just confirms what you are saying. I think Jiri is wrong here (but I am keen for him to explain his reasons if he disagrees), and as Opaque and myself have said this simulation is badly posed and will never converge.

[Jiricbeng]

Well, I assume the following: pressure loss (based on Navier Stokes eq.) is a function of velocity gradient of position (kinematic viscosity multiplied by second derivative of velocity vector V_i with respect to the coordinate X_j). I mean if you set no friction on the wall, it does not have to mean there is no loss, because in the liquid there might be areas with different velocity vectors (gradient) and that is why I think there can be non-zero pressure loss even in this frictionless wall.
For example, if you simulated flow through a valve, even if there was no friction on the walls, the pressure loss would be non-zero, as there would be pressure loss due to vorticity of the flow caused by the valve. I know we are talking about straight pipe in this case, but I try to consider this in general.

[ghorrocks]

Yes, I see your point Jiri.

Luke, can you say whether the inlet or initial condition has any velocity variation in it?

[Overdue]

Of course, here are pictures of the velocity profile at Inlet and Outlet:

[ghorrocks]

Thanks. So the velocity gradients are small, but they are not zero.