Negative Static Pressure

sdarmayuda · August 30, 2018, 12:39

Hello everyone,

I know this topic has been discussed a few times in this forum. But I don't really understand. I am trying to do some CFD Simulation for a small hydrodynamic coupling/fluid coupling.

My Modell
1. The fluid is oil (incompressible) and has constant Rho and viscousity (no heat transfer)
2. I am using k-epsilon, realizable, scalable wall function
3. For every simulation, I am giving a different rotational velocity to my Pump-cell zone and my Turbine-cellzone. So I can meassure the Torque for every slip condition (0-100%). The rotational velocity is about 5000-8000 rpm.
4. The diameter of the coupling is about 66 mm. And the total width is 20 mm.
5. The reference pressure: 0 Pa
6. No gravitational force.

Problem:
At certain point (for example Pump: 8000 rpm, Turbine: 0 rpm, it means Slip 100%), the static preassure (volume average) is around -1,8 bar. This doesn't make any sense, right? Although, the total pressure=static pressure+dynamic pressure is fulfilled.

1. After reading a bit in this forum. I have found, that it can happen sometimes in Fluent for incompressible fluid. The negative static pressure only show the difference of the pressure. But I don't really understand why?
Can anyone explain to me a bit?

2. Another posibility is a cavitation is expected at this condition to happen. How should I prove this phenomenon?

Another feedback would be much appreciated. If you need any information, please don't hesitate to let me know. Thank you in advance.

Best regards,
Darma

blackmask · August 30, 2018, 21:41

For incompressible flow, only the pressure gradient rather than the pressure itself appears in the governing equation, which means that you can add any constant to the pressure field without affecting the governing equation. The arbitrariness of such constant is removed by any Dirichlet-type boundary conditions for pressure or by the operating pressure if no such boundary exists.

LuckyTran · August 30, 2018, 22:59

Cavitation (a phase change phenomenon) can't occur for a constant density fluid so you will never get this phenomenon. You need to change your model so that it is at least possible first.

sdarmayuda · August 31, 2018, 01:14

Hello blackmask,

Thank you very much for your response. Ah I see. So that means, my initial pressure in this model shouldn‘t be 0 bar (gauge)? And if I add some higher pressure for the operating pressure, I would get the same delta pressure. Am I correct?

Best regards,
Darma

sdarmayuda · August 31, 2018, 01:22

Hello LuckyTran,

Thank you very much for your response. Do we need multiphase for the cavitation? At first I thought, that a negative static pressure is a sign of a cavitation and if we want to look furthermore, we need to activate the multiphase. Also, we can not identify a cavitation just by looking at the pressure itself without activating the multiphase?

Best regards
Darma

blackmask · August 31, 2018, 02:19

By initial pressure do you mean operating pressure? It does not matter if you set the operating pressure to zero. In fact such a value is not used unless there is no Dirichlet-type pressure boundary condition in your case setup, which is rare in common practice. You only need to interpret pressure as "pressure difference". For example, say you have a pressure outlet where the static pressure is zero pa, and there is a point where pressure is -10000 Pa, then it should be interpreted as the pressure difference from such a point to the outlet is -10000Pa.

sdarmayuda · September 4, 2018, 13:06

Hello blackmask,

Thank you very much for your reply. Yes I mean operating pressure. Sorry, my mistake. So I've got several question.
1. I've tried to set the operating pressure (the one in boundary condition) and the reference pressure (the one in reference value) to 1 bar. And the value of the pressure I got is the same like the one with 0 bar. I thought if I change the ref pressure to 1 bar, the fluent will show us the pressure in absolut instead of a gauge pressure. Is it because my model just doesn't have Dirichlet type boundary pressure? If that's the case, what should I do, if I want Fluent to show absolute pressure instead of gauge pressure?

2. Regarding your example, static pressure I got on the result is about -1,8 bar. And by pressure difference here you mean the pressure difference before the simulation and after the simulation, right? And how do I know, which point is my reference point here?

Best regards,
Darma

blackmask · September 4, 2018, 21:10

Quote:

Originally Posted by sdarmayuda

Hello blackmask,

Thank you very much for your reply. Yes I mean operating pressure. Sorry, my mistake. So I've got several question.
1. I've tried to set the operating pressure (the one in boundary condition) and the reference pressure (the one in reference value) to 1 bar. And the value of the pressure I got is the same like the one with 0 bar. I thought if I change the ref pressure to 1 bar, the fluent will show us the pressure in absolut instead of a gauge pressure. Is it because my model just doesn't have Dirichlet type boundary pressure? If that's the case, what should I do, if I want Fluent to show absolute pressure instead of gauge pressure?

If you want the gauge pressure to be the same of absolute pressure, then you have to set operating pressure to zero. In fact, absolute pressure = gauge pressure + operating pressure.

Quote:

2. Regarding your example, static pressure I got on the result is about -1,8 bar. And by pressure difference here you mean the pressure difference before the simulation and after the simulation, right? And how do I know, which point is my reference point here?

Best regards,
Darma

By pressure difference I mean the spatial difference, i.e., pressure difference between two locations.

sdarmayuda · September 7, 2018, 07:26

Hello blackmask,
Thank you very much for your reply. Then I would like to ask further.

Quote:

Originally Posted by blackmask

By pressure difference I mean the spatial difference, i.e., pressure difference between two locations.

I see. I thought it was pressure drop. If it is spatial difference, how do I know which one is my reference point. I have measured the pressure at multiple points of the coupling. Is the reference point for each meassured pressure the same?

Best regards,
Darma

blackmask · September 7, 2018, 22:21

It is like
$\int^x \nabla p dr = p(x)+p_0 \int_{x_1}^{x_2} \nabla p dr = p_2 - p_1$
When you know $\nabla p$ only, the pressure at a specific location can vary be an arbitrary constant. However, the pressure difference between any two locations can be uniquely determined.

The reference point is nothing but a way to determine the otherwise arbitrary constant.

August 30, 2018, 12:39	Negative Static Pressure	#1
sdarmayuda New Member Darma Yuda Join Date: Sep 2017 Location: Germany Posts: 18 Rep Power: 9	Hello everyone, I know this topic has been discussed a few times in this forum. But I don't really understand. I am trying to do some CFD Simulation for a small hydrodynamic coupling/fluid coupling. My Modell 1. The fluid is oil (incompressible) and has constant Rho and viscousity (no heat transfer) 2. I am using k-epsilon, realizable, scalable wall function 3. For every simulation, I am giving a different rotational velocity to my Pump-cell zone and my Turbine-cellzone. So I can meassure the Torque for every slip condition (0-100%). The rotational velocity is about 5000-8000 rpm. 4. The diameter of the coupling is about 66 mm. And the total width is 20 mm. 5. The reference pressure: 0 Pa 6. No gravitational force. Problem: At certain point (for example Pump: 8000 rpm, Turbine: 0 rpm, it means Slip 100%), the static preassure (volume average) is around -1,8 bar. This doesn't make any sense, right? Although, the total pressure=static pressure+dynamic pressure is fulfilled. 1. After reading a bit in this forum. I have found, that it can happen sometimes in Fluent for incompressible fluid. The negative static pressure only show the difference of the pressure. But I don't really understand why? Can anyone explain to me a bit? 2. Another posibility is a cavitation is expected at this condition to happen. How should I prove this phenomenon? Another feedback would be much appreciated. If you need any information, please don't hesitate to let me know. Thank you in advance. Best regards, Darma

August 30, 2018, 22:59		#3
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,761 Rep Power: 66	Cavitation (a phase change phenomenon) can't occur for a constant density fluid so you will never get this phenomenon. You need to change your model so that it is at least possible first. sdarmayuda likes this.

August 31, 2018, 02:19		#6
blackmask Senior Member Join Date: Aug 2011 Posts: 421 Blog Entries: 1 Rep Power: 22	By initial pressure do you mean operating pressure? It does not matter if you set the operating pressure to zero. In fact such a value is not used unless there is no Dirichlet-type pressure boundary condition in your case setup, which is rare in common practice. You only need to interpret pressure as "pressure difference". For example, say you have a pressure outlet where the static pressure is zero pa, and there is a point where pressure is -10000 Pa, then it should be interpreted as the pressure difference from such a point to the outlet is -10000Pa. sdarmayuda likes this.

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August 30, 2018, 21:41		#2
blackmask Senior Member Join Date: Aug 2011 Posts: 421 Blog Entries: 1 Rep Power: 22	For incompressible flow, only the pressure gradient rather than the pressure itself appears in the governing equation, which means that you can add any constant to the pressure field without affecting the governing equation. The arbitrariness of such constant is removed by any Dirichlet-type boundary conditions for pressure or by the operating pressure if no such boundary exists.

August 31, 2018, 01:14		#4
sdarmayuda New Member Darma Yuda Join Date: Sep 2017 Location: Germany Posts: 18 Rep Power: 9	Hello blackmask, Thank you very much for your response. Ah I see. So that means, my initial pressure in this model shouldn‘t be 0 bar (gauge)? And if I add some higher pressure for the operating pressure, I would get the same delta pressure. Am I correct? Best regards, Darma

August 31, 2018, 01:22		#5
sdarmayuda New Member Darma Yuda Join Date: Sep 2017 Location: Germany Posts: 18 Rep Power: 9	Hello LuckyTran, Thank you very much for your response. Do we need multiphase for the cavitation? At first I thought, that a negative static pressure is a sign of a cavitation and if we want to look furthermore, we need to activate the multiphase. Also, we can not identify a cavitation just by looking at the pressure itself without activating the multiphase? Best regards Darma

September 4, 2018, 13:06		#7
sdarmayuda New Member Darma Yuda Join Date: Sep 2017 Location: Germany Posts: 18 Rep Power: 9	Hello blackmask, Thank you very much for your reply. Yes I mean operating pressure. Sorry, my mistake. So I've got several question. 1. I've tried to set the operating pressure (the one in boundary condition) and the reference pressure (the one in reference value) to 1 bar. And the value of the pressure I got is the same like the one with 0 bar. I thought if I change the ref pressure to 1 bar, the fluent will show us the pressure in absolut instead of a gauge pressure. Is it because my model just doesn't have Dirichlet type boundary pressure? If that's the case, what should I do, if I want Fluent to show absolute pressure instead of gauge pressure? 2. Regarding your example, static pressure I got on the result is about -1,8 bar. And by pressure difference here you mean the pressure difference before the simulation and after the simulation, right? And how do I know, which point is my reference point here? Best regards, Darma

September 7, 2018, 22:21		#10
blackmask Senior Member Join Date: Aug 2011 Posts: 421 Blog Entries: 1 Rep Power: 22	It is like $\int^x \nabla p dr = p(x)+p_0 \int_{x_1}^{x_2} \nabla p dr = p_2 - p_1$ When you know $\nabla p$ only, the pressure at a specific location can vary be an arbitrary constant. However, the pressure difference between any two locations can be uniquely determined. The reference point is nothing but a way to determine the otherwise arbitrary constant.