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September 11, 2019, 00:32 |
About compressible and incomprehensible flow
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#1 |
Senior Member
Mandeep Shetty
Join Date: Apr 2016
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I have recently started studying solution methods for compressible and incompressible flows. From what I understand compressible flows are high Mach number flows and incompressible flows are low Mach number flow. Compressible and incompressible are the property of the flow and not of the fluid. For compressible flow, we have a relationship between density pressure via an equation of state.
a) What about flow through a centrifugal gas compressor or during combustion constant pressure)? Both these cases density is changing but the Mach number is low? Should I use compressible flow solver even though the Mach number is very low? According to https://knowledge.autodesk.com/suppo...5B008-htm.html it is not compressible flow. b) Does the equation of state hold only for high mach number flows or does it hold even for low Mach number flows but with changes in density (like the examples above) ? |
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September 11, 2019, 06:39 |
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#2 | |
Senior Member
Filippo Maria Denaro
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Quote:
Compressible effects due to the combustion can be taken into account in the low-Mach formulations, you can search in the literature. The state equation is valid also for subsonic low Mach flows. The problem is only of numerical issues for M->0 as the equations are stiff. |
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September 11, 2019, 07:14 |
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#3 |
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Alex
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Attributing "compressible" to high Mach numbers and vice versa is just a rule of thumb for many engineering applications. For example, gas flow through micro-channels can have pronounced compressibility effects despite very low Mach numbers.
Same problem with gases vs liquids. For many engineering applications, approximating liquids as incompressible is perfectly valid. Despite liquids being compressible given high enough pressure differences. |
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September 11, 2019, 07:50 |
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#4 |
Senior Member
Filippo Maria Denaro
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A similar question was asked here
https://www.researchgate.net/post/wh...ressible_fluid |
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September 11, 2019, 10:20 |
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#5 |
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The term "compressibility" is most often used for describing a coupling between pressure and density, such as for high Mach-number flows, but sometimes people also use it when they actually mean "constant density", which creates some confusion. If, for example, you have a free convection flow of course you have to take a varying density into account even though your Mach number is practically zero, otherwise there would be no weight difference.
You can therefore have an incompressibe flow with constant density, e.g. water in a pipe, and incompressibe flow with variable density, e.g. combustion at low Mach number. So remember: Incompressible does not mean constant density! |
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September 12, 2019, 05:48 |
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#6 |
Senior Member
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Textbook versions of the incompressible flow methods assume a constant density but, in general, it can still vary, within the same method, as function of temperature and species concentration.
In the same way, textbook versions of compressible flow methods use naive approaches that, then, fail at low Mach numbers. However, practical implementations actually use preconditioning, which makes them suitable for all the Mach regimes. If suitable, one tends to use incompressible methods because they have a much lighter memory footprint and may be, in certain cases, faster. Using a method instead of the other essentially revolves around the stiffness of the equations at low Mach numbers. One can show, by a low Mach expansion of the equations (e.g., https://nalu.readthedocs.io/en/lates...erivation.html), that the pressure affecting the motion of the fluid trough its gradient is of higher order with respect to the thermodynamic pressure entering the equation of state. Thus, while physically the equation of state, of course, doesn't break down, it becomes less useful from the computational point of view. In a certain sense, we can say that a certain method makes the problem computationally well behaved by actually modifying the used equation of state, in a form or another, and exploiting the resulting simplifications, if any. However, industry level compressible solver should have no problem with any Mach regime. |
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September 13, 2019, 00:15 |
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#7 |
Senior Member
Mandeep Shetty
Join Date: Apr 2016
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a) Why does state equations become less relevant as we move from high Mach number to low Mach number? Many pieces of literature just say there is no state equation for low Mach number flow (i.e. wit the incompressibility approximation.)
b) When it is said that, in the incompressibility approximation, the pressure doesn't remain a thermodynamic variable anymore and just takes a value to satisfy the zero divergences of the velocity field, does it mean that pressure doesn't remain a function nof density and temperature (ie vary with temperature and density)? c)I understand that incompressible flow doesn't mean that density doesn't vary with time it mean density doesn't vary with pressure variations, but when deriving the equation for incompressible flow we do use the assumption that d(rho)/dt = 0. |
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September 13, 2019, 04:33 |
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#8 |
Senior Member
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Watch out, I wrote "less useful from the computational point of view". If you look at the link I posted above, you see that for low Mach numbers one can split the pressure in a thermodynamic part and a dynamic part. The thermodynamic part enters the energy equation with its time derivative which, in turn, is determined by the change of volume of the domain or, more precisely, by the net mass flux in it and some thermal effects as well (yet, I'm not an expert in this method and there might be some caveat). So, you still have an equation of state, but that doesn't include the same pressure you have in the momentum equations.
Constant density is, in itself, an equation of state, but of such a special kind that you actually exploit it in formulating your final computational approach and never invoke it anymore. Of course, the pressure that enters your momentum equations at low Mach, and for which you solve a Poisson equation, that isn't thermodynamic and doesn't directly depend from the temperature (e.g., at rest). What I mean by less relevant is simply that you cannot use it as it is at low Mach. You either have to split the pressure in two parts or assume that the thermodynamic pressure is just constant. Using your assumptions directly on the density equation is not practically useful: either you have a constant density or not. But I am aware that this is how the matter is typically presented. In order for you to understand the matter, I suggest you to read papers on the low mach approach. Some references are cited in the link I posted. Another classical reference is https://link.springer.com/article/10...:1025669715376 You can also google "low mach pressure expansion" |
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September 13, 2019, 04:52 |
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#9 | |
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Filippo Maria Denaro
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Quote:
Let me clarify a concept. When one assumes an incompressible model, that says the density is homogeneous, that is a constant in space and time. Let's look to the density equation: d rho /dt + v.grad rho = -rho div v so that each term in the LHS is zero identically and the flow must be divergence-free. Comversely, if you say only that the flow is divergence-free, the density equation says D rho /Dt =0 that does not mean that the density is constant in space or time! Density remains constant along the trajectory dx/dt =v ... Then, if you look at the momentum equation inserting the state equation in to the pressure gradient, while assuming homogeneous density, you get Grad p = rho0*R* grad T that is the only way you can transmit a pressure gradient is by the temperature gradient. Of course, if the flow is assumed also at homogenoeus temperature you would get grad p =0. As a conclusion, what we retain while still writing grad p (but you can write grad f or grad phi, highlighting a general function) is just the presence of an auxiliary function (not of thermodynamics nature) that has the role of a potential function. Its role is implied by the Hodge-Helmholtz decomposition theorem. Low-Mach formulations relax the previous constraint and consider a linear expansion for the thermodynamics variables so that you can use a state equation. |
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Tags |
compressible flow, incompressible flow |
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