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flow split boundary condition at outlets

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Old   April 17, 2023, 04:45
Default flow split boundary condition at outlets
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Hello all,

Sorry if I am asking a classic question. I have a domain with an inlet and two outlets. At the inlet I can impose a normal inflow BC but at outlets I have no idea about the pressures and I only have flow split between the two outlets.

I was wondering how can I safely impose this BC in ANSYS CFX? Any advice would be much appreciated. Thanks
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Old   April 17, 2023, 10:50
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You mention you have flow split between two outlets. Does that mean that you know how much is going through each? If so, then there are 2 reasonable options:
1) set normal inflow, 1 massflow on 1 outlet and 1 static pressure BC on the other outlet (P=0Pa).
2) set massflow on 1 outlet, set massflow on the other outlet as well, set total pressure on inlet (Ptot=0Pa).
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Old   April 17, 2023, 20:03
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As a general rule, you set the boundary conditions based on what you know about the flow at the boundary. So if you know the flow split at the outlets then you use this as your boundary condition. So you would define a mass flow boundary at the two outlets and a pressure boundary (total pressure probably) at the inlet. This is Gert-Jan's option 2.
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Old   April 18, 2023, 12:50
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Careful here. Though Gert-Jan's option (2) is desirable, my bet is on option (1).

Why? The mass flow split is not a user decision, but an observation in the experiments, or available data. However, the mass flow split is a function of the losses in the main section, and the two branches that create the split, i.e. it is pressure-based and not enforcement on the system. If lucky the split fits the mesh, and the discretized losses the model will converge; otherwise, it will refuse to converge. Think about it as the choke condition, a system cannot handle whatever flow we want, but whatever it can support based on physics.

Similarly, with the energy equation. We can impose whatever heat flux we want IN, but not OUT. For heat removal, we are limited by the system's maximum available energy and the timestep used. Even worse, we cannot remove more energy that it is available in a given control volume for a given iteration/timestep, i.e. timestep limitation.

With enough time, we can enumerate several examples to illustrate the point.

Please keep us posted. I am interested in the final solution approach.
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