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sharifi November 5, 2013 10:11

Gravity/Buoyancy/Buoyancy reference density in two phase flow
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im trying to modell a gravity driven two phase flow(air - water) in a pipe in CFX. (pic1.)

Velocity Inlet
Pressure outlet up
pressure outlet down

inhomogene Euler/Euler

Accordint to cfx, flows in which gravity is important can be modeled by CFX by the inclusion of buoyancy source terms.

For buoyancy calculations, a source term is added to the momentum equations as follows:
S= (rho - rhoRef) * g

For multiphase flows, it is important to correctly set the buoyancy reference density(rhoRef). For a flow containing a continuous phase and a dilute dispersed phase, you should set the buoyancy reference density to that of the continuous phase. so the reference density of the continuous phase cancels out buoyancy( eq. 1) and pressure gradients in the momentum equation of continous phase.

For non-dilute cases (which include all free surface cases), all terms can be equally important for each fluid. if there is a significant difference in density you should choose the density of the lighter fluid (that cancels out buoyancy of lighter phase) because this gives an intuitive interpretation of pressure (that is, constant in the light fluid and hydrostatic in the heavier fluid).

I've tried using the two limiting cases. if (rhoRef = rhoAir), no mass flows through the upper outlet. if (rhoRef= rhoWater) no mass through the lower outlet, also the hole tank is full of water and bubles come out.

If (rhoAir <rhoRef < rhoWater) it works, but i think it is not phiscally correct.
In my opinion it shoulde be define as a function. When I try to define rhoRef as a function of Volume fraction,
(rhoRef = Air.Vf * rho.Water + (1- Air.Vf) * rho.Air)
ther is this error message:

The parameter 'Buoyancy Reference Density' is defined to be "Single Valued" but it depends on the following field valued variables: , Air.density, Air.vf, Water.density.

have any one any idea, how can I solve this problem?

thx for ur time

ghorrocks November 5, 2013 18:19

I think you misunderstand the purpose of the reference value. The exact value of this parameter should not matter and it certainly does not vary over the domain. It is a single value used to reduce numerical round-off. You should be able to double or halve it and results are unaffected.

If changing the reference value changes your results then you have a very sensitive simulation and you should be using double precision numerics.

sharifi November 6, 2013 02:26

thx for your antwort.

Correct me if I wrong.

I dont think so.
Simply do the tut 9 and change the ref.Density and see the diffrence.
However When you see the NS.eq, you find out, that ref.Den has enormous influence.

Viele Gre,

ghorrocks November 6, 2013 06:58

The tutorials are only to show you how to get the models running. They unfortunately do not show you good CFD practise. It the tutorial result changes with different reference densities then it needs double precision numerics as well.

You cannot have a result which relies on precise definition of a reference condition. You cannot do accurate CFD unless this is the case.

sharifi November 17, 2013 06:50

hat any one any other idea to simulate such a separator???

evcelica November 20, 2013 00:14

What values are you using for your pressure outlets? These two values are determining your flow, and also the height of your liquid head along with the buoyancy forces. Sounds like you are setting them at the same pressure? Which is not true, and would explain why you are getting your results.

sharifi November 20, 2013 10:45

thx for the antwort.
I use the same pressure for both pressure outlet, p(atm)
physically this should be so, since both leave out in the environment.
can you explain me plz more, I should put pressure at out differently??!!!

sharifi November 20, 2013 10:59

Also, I do not understand it.
the water is at the bottom when Rho(ref)=Rho(air). This should physically lead to lower outlet is clogged and thus the air goes out from the other outlet, but it does not, it even comes out of the lower but not omit from the upper. I mean the transport eq for air hat no extra force-term that it force to go out from upper outlet. I mean, the only reson why air shoud goes out from the upper outlet is that the lower outlet is clogged with water or am I wrong???

ghorrocks November 20, 2013 17:15

A picture showing what you are describing would help.

evcelica November 21, 2013 14:52

The botom and top should not be at the same pressure. The bottom would be at a higher pressures
due to the liquid head.

atit.sut June 9, 2014 10:23

Adding the momentum source should not provide the correct answer. After specifying the Buoyancy reference density and the gravitational acceleration components, CFX will manage the buoyant effect for you. Try to disable this momentum source. It might work. Good luck.

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