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pressure initialisation in a closed system

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Old   August 22, 2018, 10:32
Default pressure initialisation in a closed system
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M. Fenster
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Dear community,


I have a question about the static-pressure initialisation in CFX.
The graphic below this post should clarify the situation.
I want to run a CFD simulation of a mixer in a big cylindrical tank with a tube inside.


In the following, a short summery of the setup in CFX-pre:


- steady state simulation
- incompressible fluid
- ref. pressure 1 atm
- buoyancy Model activ
- isothermal
- turbulence model SST


The model is divided in two parts which are separated by the interfaces.
The impeller domain is rotating with 5000 rpm and the tank-tube domain is stationary.
On the top of the tank a "Free-Slip-Wall" boundary was choosen.


1) The first question I have is, how can I initilize a static-pressure field in this closed system?


I started a CFD simulation with the setting above and I found the following text in the out-file:


"Pressure has not been set at any boundary conditions.
The pressure will be set to 0.00000E+00 at the following location:
Domain : Mixer_SinglePhase
Node : 1 (equation 1)
Coordinates : (-6.20517E-05, 5.46441E-04,-2.61476E+00)."


As you can see, on node 1 the pressure is set to 0. Is this value representative for the whole domain?



After the simulation was finished, I made a contour-plot of the static-pressure and I was wondering why the pressure was negative in the tank domain, because actually I estimated that the pressure should be 0.



2) I have a problem to imagine the behaviour of the static-pressure field in a closed system with a roatating impeller inside. Maybe you have any idea?


best regards M3tt





Konzept.jpg

Last edited by M3tt; August 23, 2018 at 02:39.
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Old   August 22, 2018, 20:18
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Glenn Horrocks
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As this is a compressible simulation you will need to set the pressure level correctly. It can be ignored in incompressible simulations, but in compressible simulations the density is a function of pressure so you have to get it right.

If the top free slip wall represents a free surface which is basically flat, then you could but a small section of that face as an opening with the local atmospheric pressure.
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Old   August 23, 2018, 02:38
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Hi ghorrocks and thanks for your reply,

Sorry that was my fault ... the fluid is incompreesible not compressible. I will edit that post.
You are right the free slip surface is representing an opening.
In the end I am interested in the shaft-power, so I thought that the pressure level should be set correctly, in order to get the correct value for the torque.

Best regards

M3tt
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Old   August 23, 2018, 02:44
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If the flow is incompressible the pressure level does not matter, provided it converges. If your simulations are converging then it might be OK.
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Old   August 23, 2018, 02:49
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Thanks for the fast reply and your efforts .
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