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Marat Hoshim September 20, 2000 03:47

negative static pressure !?!
 
Hi all,

I have some difficulties interpreting a negative absolute pressure. Imagine a channel in which some kind of rotor is rotating. It is an incompressible calculation so the reference pressure is set to 100000 Pa at the inlet of the channel. What am I'm wondering about is the result, that the pressure at some point at the rotor falls below -100000 Pa, which means the the absolute pressure is negative ! Could there be some physical reason for that or did I make something wrong in setting up the run ?

Any enlightment on meaning of pressure in incompressible flow will be appreciated !

Marat

Maurizio Barbato September 20, 2000 04:04

Re: negative static pressure !?!
 
Hi Marat,

is your flow really incompressible?

Cheers

Maurizio

Sebastien Perron September 20, 2000 06:47

Re: negative static pressure !?!
 
In incompressible flow, the pressure you'll find is relative to a pressure reference. Actually, you are not interested in knowing the pressure but the pressure gradient. If the scheme you use is correct, the pressure field is correct. But, you have to had or substract a pressure reference everywhere in order to get the real pressure. In fact, in most applications, you're not interested by the real pressure (but it depends on the problem). As far as I can remember, the pressure you'll find is the dynamic pressure induce by the fluid mouvement. The hydrostatic pressure is lost.

About BCs, are you sure about the inlet boundary condition. We often use imposed normal velocity, but it depends on the problem. In general, the pressure is inposed at the outlet.

By the way, what kind of projection scheme do you use? (Chorin, Simple family, usawa....)

Marat Hoshim September 20, 2000 08:17

Re: negative static pressure !?!
 
Hi Sebastien,

in my run at the inlet there's also the normal velocity and the turbulence quantities given. The reference pressure can be set at any cell to a specific value. I have choosen to set it at a cell next to the inlet boundary to the value of 100000 Pa. So that pressure had not anything to do with the inlet boundary condition. My difficulty is now to explain the the pressure values of -120000 Pa near the rotor, which means an absolute pressure of (100000 Pa +(-120000 Pa))=-20000 Pa !

Marat


Sebastien Perron September 20, 2000 08:36

Re: negative static pressure !?!
 
1) The pressure can't be set anywhere. When you use projection schemes (chorin and simple), you don't specify the pressure when the velocity is given. Because when you will correct the pressure field to get the zero divergence velocity you will be able to correct the velocity. If you impose the pressure field at an inlet, the velocity won't be corrected.

2) If you have an inlet and an outlet (it must be if you have an inlet), may I suggest (again) that you impose the pressure at the outlet.

2) I 'dont think you get the absolute pressure. -120000 Pa can be interpreted as a difference off 220000Pa between the inlet pressure and the pressure near the rotor.

I hope it helps you.

Marat Hoshim September 20, 2000 09:42

Re: negative static pressure !?!
 
Hi Sebastien,

thanks for your hints concerning the pressure in incompressible flows. So I only get pressure differences out of my calcs. But how can I interpret a pressure difference between the inlet a rotor point of -120000 Pa, if in reality the pressure at the inlet is the atmospheric pressure with 100000 Pa ?

Thanks for your patience,

Marat

Maurizio Barbato September 20, 2000 09:45

Re: negative static pressure !?!
 
Hi Marat,

which velocities have you got near to the rotor??

Maurizio

Marat Hoshim September 20, 2000 09:57

Re: negative static pressure !?!
 
The velocity is about 15 m/s at some points. The inlet velocity is about 0.1 m/s. So the low pressure of at the rotor of -120000 Pa corresponds to the value of the dynamic pressure 0.5*rho*v^2, with rho=1000 kg/m^3.

But what would happen in the real flow, if the inlet pressure would be the atmospheric pressure with 100000 Pa ?

Marat

Maurizio Barbato September 20, 2000 10:11

Re: negative static pressure !?!
 
Dear Marat,

from the density you cited, you are dealing with water. Therefore I think what you could have in a real device working as your model, is cavitation due to excessive acceleration of the liquid.

Just to tell you, negative pressures are dinamically possible in liquids and the reaction of the liquid to this situation is to form cavities which expand and riequilibrate the negative pressure.

Cheers

Maurizio


Marat Hoshim September 20, 2000 10:31

Re: negative static pressure !?!
 
Dear Maurizio,

so if I understood you right, every pressure difference in my calculation that lowers the absolute pressure below the vapour pressure of water would launch cavitation in reality. And as the numerical method does not know anything about cavitation the pressure lowers to irreal values, which would have caused cavitation in reality much earlier.

Is this explanation senseful ? Can I interpret the result in this way, that to prevent cavitation the pressure difference must not increase this high that the atmospheric pressure plus pressure difference falls below the vapour pressure of water ?

Thanks,

Marat


John C. Chien September 20, 2000 11:28

Re: negative static pressure !?!
 
(1). If the flow is incompressible, then the pressure is decoupled from the flow field formally. (2). This means that the absolute level of the pressure has no effect on the flow field solution. and you can add or substract a constant pressure to the pressure field, without affecting the velocity field. (3). This can be seen easily by solving the stream function-vorticity equations without knowing the absolute pressure level. So, the pressure field in the incompressible flow is floating by a constant value. (4). But this does not mean that your cfd calculation is absolutely right. You still have to check the results to see if the solution is mesh independent or not. You also need to check the total pressure distribution to see whether there is something unusual related to the physics of the flow. (5). For water , there will be cavitation problem as the local pressure is dropped below the vapor pressure, or something like that. For other kind of liquid, this may not be a problem. (6). I would suggest that you check the total pressure distribution first to make sure that everything is all right, including the mesh independent solution. After that, all you need to know is that the absolute pressure in the incompressible flow is "floating" , that is, the pressure distribution can be moved up or down by a constant value. (depends on the actual operating condition, in that case, you will have to set a pressure level at a point somewhere in the flow field)

Maurizio Barbato September 20, 2000 11:38

Re: negative static pressure !?!
 
Dear Marat,

your explaination is sensful. I will just say that you should add a treshold for the pressure in your code which gives you a warning when the cavitation conditions are achieved. In particular when the static pressure of your fluid (imagine you start with water at atmospheric pressure i.e. 101325 Pa) reaches the water vapour pressure (about 1750 Pa) then you, putting yourself on the safe side, can assume that cavitation occurs.

I hope this helps

Ciao

Maurizio

Jas September 21, 2000 04:08

Re: negative static pressure !?!
 
I have had similar problems in the past on this type of problem. I found that if modelled in two dimensions pressure values near the rotor went negative absolute. If modelled in 3D pressure values near the rotor were close to experimental values. If you are modelling in 2D you are neglecting vital clearances on the front and back of the rotor which lead to a lowering of the velocities near the rotor since the area there is greater.

John C. Chien September 21, 2000 09:46

Re: negative static pressure !?!
 
(1). It is interesting to know that there are some problems with the incompressible flow calculations. I think, it must be related to the particular code or algorithm used. Could you give us some more details about the methods used? (2). I have used my own codes and commercial codes to solve 2-D and 3-D problems, but the negative pressure was never an issue,except in the case of using density based code for low Mach number flow calculations. (3). In the low Mach number flow (much lower than 0.1), the density based code will give oscillating pressure solution. So, in this case, it is possible to get negative density, pressure,etc. But, then the code will stop by itself, because in compressible flow (even though it is almost incompressible), the pressure must be positive to satisfy the equation of state.(p=rho*R*T). (4). Under the same condition, it is likely that the local area distribution for 2-D flow will be different from that of the 3-D flow for the same problem, thus the pressure distribution will be different also. (5). This is an interesting point, if the pressure is given at some point, the it is a good idea to simulate the real flow as close as possible to avoid non-realistic pressure distribution. (6). Otherwise, like the example I mentioned, the flow field solution using the stream-function-vorticity approach does not need any information about the pressure at all to give a velocity distribution. The pressure can be obtained from the velocity field by integration. (accuracy in mesh and velocity can affect the pressure integration) (7). At the mathematical solution level, the pressure field has an unknown constant, so, the absolute pressure level is not useful at all. (the pressure does not affect the density or anything else in incompressible flow) But, when comparing with the test data or the real world operating conditions, the geometry (corners), 2-D or 3-D can affect the pressure distributions of the cfd simulation. Thus make the comparison looks strange, I guess.

Marat Hoshim September 22, 2000 02:34

Re: negative static pressure !?!
 
As I have learned in this thread, it's not really a problem. Because I'm only dealing with pressure differences instead of absolute pressure. The only problem I faced was the interpretation of a negative pressure difference of 120000 Pa at the rotor relative to the inflow. Assuming approx. atmospheric pressure at the inlet in reality of 100000 Pa the absolute pressure would become negative. But as I have learned due to the fact that the solver doesn't know anything about cavitation the pressure field get unrealistic when pressure falls below the vapour pressure of water. BTW the code I used is a pressure based SIMPLE algorithm.

Marat

Jas October 26, 2000 11:48

Re: negative static pressure !?!
 
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