[akhenathon]

hi all again me Akhenathon

first off all i very pleased to previous helps of you and i thank you.


my new problem is;

im working on a 2D invisdic, incompressible flow. i want to determine inlet and outlet pressures and velocities. But i didnt find with these together. there is a velocity inlet boundary condition and pressure outlet condition. if i want to give an edge two boundary condition( velocity and pressure values together), what i must do?

[jpo]

I am unsure if one can use velocity inlet and pressure outlet in the case setup.

The easiest thing to do is: use "velocity inlet" and "outflow" boundary conditions. Usually these converge very easily. Once converged one can check all observables on any boundary.

Report->Result reports... opens a menu

Choose Area-weighted-average on the left
choose static pressure on the right
select your outlet from the list below

You will have the value of pressure

Similarly, one can calculate many other properties at any boundary

[akhenathon]

i mean that

inlet boundary condition: Velecity 10m/s and pressure 10 Pa
outlet boundary condition: Velocity 10 m/s and presssure 10 Pa

is it possible?? İf it is, how can i do that?

[jpo]

if you use density-based solver, you can input a pressure reading in the velocity inlet window. But it will be used only in a case of backflow

[ComputerGuy]

Originally Posted by akhenathon

i mean that

inlet boundary condition: Velecity 10m/s and pressure 10 Pa
outlet boundary condition: Velocity 10 m/s and presssure 10 Pa

is it possible?? İf it is, how can i do that?

akhenathon,

If you think about flow in a pipe, I believe you're trying to over-specify the problem.

You can specify the mass flow and one pressure (inlet or outlet) and the solver will solve the other.
You can specify both pressures and the solver will calculate a mass flow.
You can specify both mass flow and the pressure distribution will be calculated.

You are trying to specify the mass flow and the pressure at both boundaries. If I'm correct, there is nothing left to solve.

Regards,
ComputerGuy