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Old   April 16, 2002, 07:27
Default negative pressure
  #1
Mark Render
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Hi all,

I simulate the flow around an hydrofoil under cavitating conditions. I use a simple VOF model, which converts water into vapour depending on the local pressure. Before the cavity starts at the leading edge there's a small area (0.02 chord) where the absolute pressure gets negative (like in calculations without considering cavitaion). Now my question: Is it physically possible or senseful to have negative absolute pressures in water ?

Regards,

Mark
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Old   April 16, 2002, 08:53
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Philipp Beierer
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In my opinion, in a given volume cannot be less than an absolute vacuum. Hence, you will never find a negative pressure in "real" life. Since the fluid is evaporating somwhat before that, your lowest absolute pressure should be the vapour pressure of the fluid. On the other hand I could imagine, too, that you can maybe find slightly lower pressures than that since cavitation must not necessarily start immediately as soon as the pressure is below p_vap (I'm thinking now e.g. at highly accelerated flow). Anyway, negative pressure is impossible!

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Old   April 16, 2002, 09:25
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bosko
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It depends what pressure you are specifying at the inlet. If you set an inlet pressure of 1Pa for example then the absolute pressure would be negative almost everywhere in the domain. Make sure you specify a reasonable value.
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Old   April 16, 2002, 10:01
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bosko
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It depends what pressure you are specifying at the inlet. If you set an inlet pressure of 1Pa for example then the absolute pressure would be negative almost everywhere in the domain. Make sure you specify a reasonable value.
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Old   April 16, 2002, 12:30
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dav
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it is possible in two-phase flow, i've a case. it is possible mathematically and physically. i.e. it depends in your formula, P = x + y + ....

cfd

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Old   April 19, 2002, 05:10
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Michael
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This is a matter how you define a negative pressure. Pressure can be given a direction in the alignment with the acting normal force. A negative pressure is a tension acting on a fluid volume normal to the surface, so a positive stress leads to a negative pressure. The ability of a liquid to withstand a tension is expressd in (negative) pressure units. See the book by Trevena "Cavitation and tension in liquids". Even Reynolds has measured a negative pressure of water of about -5bar. This "negative" pressure energy is used to build up a phase interface, before cavitation starts. Once the interface between vapour and liquid has been created, the negative pressure vanishes because the energy now goes into vaporization.

Current VOF cavitation models as e.g offered in Fluent and StarCD assume that cavitation starts at vapor pressure, so this "non-equilibrium" effect is not modelled ! In this case the negavite pressure is an artefact of the model due to an unphysical description of the pressure field and a wrong pressure - vaporization coupling. There a several ways to define a more realistic pressure value in a cavitating region, e.g. you can use a conditional ensemble averaging procedure.
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Old   April 19, 2002, 07:30
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Philipp Beierer
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Hi Michael,

I agree with you that a slight positive normal stress on a fluid particle is possible and that in accordance to that the pressure is negative.

On the other hand act impurities or dissolved gases in liquids as sites for cavitation. Hence, in practical situations, it is generally not possible to observe pressures falling below the vapourisation pressure of the liquid; not at least for longer than it takes for the cavitation process to occur. And, as you are mentioning that this "non-equilibrium" effect cannot be modeled by the commercial VOF cavitation models (as e.g. Fluent, etc), I believe, too, that in Mark's case the detected negative absolute pressure is a matter of the employed model rather than a copy of real phenomena.
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Old   April 19, 2002, 13:28
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Mark Render
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Hi Michael,

thanks for answer !

So did I understand you correct, that in reality it is possible that the pressure could get negative if the water is very pure ?

So why should a cavitation model that is based on bubble growth (Rayleigh-Plesset equation) not be able to predict that negative pressure just before the water vaporizes ?

Regards,

Mark

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Old   April 23, 2002, 07:02
Default Re: negative pressure
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Michael
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Hello Mark,

you can measure even higher negative pressures if the water is pure and the tube the water is contained in is also cleaned; like in the early experiments of Lauterborn (Acustica 22(1), 35-47 (1969/70)).

The Rayleigh equation in most models starts from a microbubble that is already there and is assumed to be at vapor pressure. So you already start with a (very small) vapor volume fraction. In this case no negative pressures should occur in the model because the bubble instantaneously starts to grow. You can assume a high surface tension so that p(vapor) - 2*sigma/r becomes negative. Unfortunately your Rayleigh bubble will dissolve if the surface tension is that high, so you have to fix a minimum radius. This minimum radius is a mere model constant. Equivallently you can assume a negative vapor pressure, which is as arbitrary as a minimum radius. Whatever you do, you have to know in advance what you actually intended to predict.

Despite tremendous research in nucleation, the whole details of cavitation inception are not yet fully understood or at least I never fully understood it. In water tunnels with a lot of free air bubbles you can have more confidence in your Rayleigh bubble analysis and negative pressures should not be of much concern. In oil hydraulic systems, especially when the dimensions get small, e.g. in small throttles, the viscous effects become more important. The relation of wall nuclii to free stream nuclii gets larger and the whole cavitation inception can be greatly influenced by molecular structures of the nearby surfaces. There is no general model available to describe cavitation inception or liquid tensions in this case.

It is much more difficult than it may seem to model a realistic pressure field in a cavitating liquid. It is not easy to measure it anyway. You need much more than a Rayleigh bubble equation to accomplish that. Even in more advanced two-fluid Euler-Euler models with whatever source terms in the continuity and momentum equations, it is still not possible to predict the onset of cavitation or possible tensions without additional model or "tuning" constants. In some commercial codes cavitation is calculated using an additional scalar equation, e.g. by using a mixed density equation. This means that you are using an artificial equation of state in an otherwise incompressible liquid to account for cavitation. But the pressure field is still calculated using the normal pressure correction method. So your pressure field can get negative not only before cavitation starts, but also in the cavitating region. But this is an unintended side effect from the modeling. It has nothing to do with tension in liquids or even the attempt to model real negative pressures.
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