Fluid Mechanics: Incompressible flow and pressure

[Malverin]

But it is not total pressure in that formula, it is static. I know that because this is the pressure that eventually becomes the ∇p-term in the NS-equations

Show us some reference about that then.
I don't pretend i understand that better than you.
I just can not understand the reason for your assumptions.

If you can show us some reference it will be easier to understand where the problem is.

 

[Niles]

Show us some reference about that then.
I don't pretend i understand that better than you.
I just can not understand the reason for your assumptions.

If you can show us some reference it will be easier to understand where the problem is.

From my post #19: "Please take a look at this paper, on the second page equation (5). There they state that $p=c_s^2\rho$".

In eq. (25) they use this very same p in the NS-equations.

 

[Malverin]

From my post #19: "Please take a look at this paper, on the second page equation (5). There they state that $p=c_s^2\rho$".

In eq. (25) they use this very same p in the NS-equations.

LB models
can be used in a low Mach-number regime to simulate
incompressible flows. In that case, the pressure p of the
fluid is related to the particle density through the equation of state for an ideal gas:

So it is written for ideal gas
Nothing said, about gravity

When there is no gravity => there is no hydrostatic pressure

it is written this is valid for 'low Mach number'

That means they consider dynamic pressure negligible , or they mean something else?
It is not clear to me.

 

[Niles]

So it is written for ideal gas
Nothing said, about gravity

When there is no gravity => there is no hydrostatic pressure

it is written this is valid for 'low Mach number'

That means they consider dynamic pressure negligible , or they mean something else?
It is not clear to me.

Low Mach number just means low velocity, so basically fluid can be approximated as incompressible. But Bernoulli's eqs. is still valid!

 

[SteamKing]

I think you have to be careful picking formulas from a random paper and automatically thinking they are necessarily applicable to all circumstances. The particular paper referenced does not include a table explaining the definition of the variables used, and it takes one picking through the discussion to find out what the symbols mean.

In Section II.A., ρ is defined as something called a 'particle density'. ρ itself is also defined as the sum of several state variables 'f', which are not defined explicitly, but it is implied that the state variables f are related to the manner, or lattice, of how the fluid is discretized for the analysis described later in the paper.

Now, I will admit I'm no great shakes at understanding all of the fluid dynamics which go into the author's paper, but it seems that the 'ρ' used in this particular work is not the same quantity as 'ρ' which is generally used in other works to denote density, that is, ρ = mass per unit volume.

 

[Malverin]

Low Mach number just means low velocity, so basically fluid can be approximated as incompressible. But Bernoulli's eqs. is still valid!

Any gas is compressible even at low speed.
So I don't think this is sufficient condition for incompressibility.

 

[cjl]

Any gas is compressible even at low speed.
So I don't think this is sufficient condition for incompressibility.

Low mach number gas flows are generally assumed to be incompressible, since the density changes within the flow are negligible (because the pressure changes are relatively small). This is a standard simplifying assumption that is commonly made in fluid mechanics.

 

[boneh3ad]

Yikes, so much nonsense all in one thread...

So you agree with me, there really is an inconsistency somewhere, right?

No I am not saying that. I do not have that particular book from White and I don't have any other texts containing that relationship that I can recall (or find by searching briefly through my library). What I will say is that generally,
$$c_s = \left(\dfrac{\partial p}{\partial \rho}\right)_s$$
where the subscript $s$ denotes the derivative is taken at constant entropy. The only way you arrive at your equation is by assuming that this relation can be expressed here as an ordinary derivative, separating the two sides and then assuming the sound speed is constant so that it doesn't get affected by the integral, then assuming the constant of integration is zero. I do not know Frank White's assumptions made when deriving the relationship you cited and without seeing what he did, I can't really comment on why it doesn't work for your situation.

What I can say is that based on the latest paper you cited, they are using this equation as an equation of state relating the thermodynamic variables $p$ and $\rho$. The speed of sound is determined elsewhere and is treated as a known quantity.

atmospheric and hydrostatic pressure are not the same thing

Based on the typical definition of atmospheric pressure, yes, it is a form of hydrostatic pressure. It simply varies with height, and since you have two fluids stacked on top of one another, you have to include the weight of one in the hydrostatic pressure of the other. That doesn't change the fact that atmospheric pressure is the hydrostatic pressure from the air.

When fluid does not move the total pressure is static pressure (atmospheric + hydrostatic)

This depends on how you define total pressure. Sometimes total pressure includes the $\rho g z$ term factoring in potential energy, so be careful. Of course, the important thing is that either way, total pressure is not useful as a thermodynamic property, as it is frame-dependent. For that reason, most equations, such as the Navier-Stokes equatiuons and the equation for speed of sound discussed here, use static pressure, which is frame-independent, confirming what Niles has been saying.

So it is written for ideal gas
Nothing said, about gravity

When there is no gravity => there is no hydrostatic pressure

Be careful here again. There may not be gravity used in a given problem, but the static pressure of a stagnant fluid, for example, is still likely a result of hydrostatic pressure. Gravity is often simply not included because its contribution to the change in pressure in a given situation is negligible. Either way, that is irrelevant to the discussion since the speed of sound relationship depends on static pressure regardless of the "source" of that static pressure, or more correctly, it depends on the rate of change of the pressure with respect to the density at constant entropy.

it is written this is valid for 'low Mach number'

That means they consider dynamic pressure negligible , or they mean something else?
It is not clear to me.

Any gas is compressible even at low speed.
So I don't think this is sufficient condition for incompressibility.

That is absolutely not what low Mach number implies. At sea level in air, you can be moving at a "low Mach number" (typically meaning M < 0.3) and still be moving at over 100 m/s (224 mph)! There will certainly be some non-negligible dynamic pressure there. What it implies is that for a low enough Mach number, the flow can be considered incompressible. If you need convincing of this, I worked out the math in this thread not too long ago showing why the Mach number is a good indicator of compressibility.

Also, ladies and gents, keep in mind that "incompressible" does not mean constant density. It means that the material derivative of density is zero, or alternatively that the divergence of the velocity field is zero (from the continuity equation). You can certainly still have variable density in an incompressible flow provided that this condition is met.

Also remember that this incompressible assumption is, as AlephZero has stressed, an approximation. No fluid is truly incompressible.

 

[Malverin]

Yikes, so much nonsense all in one thread...



No I am not saying that. I do not have that particular book from White and I don't have any other texts containing that relationship that I can recall (or find by searching briefly through my library). What I will say is that generally,
$$c_s = \left(\dfrac{\partial p}{\partial \rho}\right)_s$$
where the subscript $s$ denotes the derivative is taken at constant entropy. The only way you arrive at your equation is by assuming that this relation can be expressed here as an ordinary derivative, separating the two sides and then assuming the sound speed is constant so that it doesn't get affected by the integral, then assuming the constant of integration is zero. I do not know Frank White's assumptions made when deriving the relationship you cited and without seeing what he did, I can't really comment on why it doesn't work for your situation.

What I can say is that based on the latest paper you cited, they are using this equation as an equation of state relating the thermodynamic variables $p$ and $\rho$. The speed of sound is determined elsewhere and is treated as a known quantity.





Be careful here again. There may not be gravity used in a given problem, but the static pressure of a stagnant fluid, for example, is still likely a result of hydrostatic pressure. Gravity is often simply not included because its contribution to the change in pressure in a given situation is negligible. Either way, that is irrelevant to the discussion since the speed of sound relationship depends on static pressure regardless of the "source" of that static pressure, or more correctly, it depends on the rate of change of the pressure with respect to the density at constant entropy.




That is absolutely not what low Mach number implies. At sea level in air, you can be moving at a "low Mach number" (typically meaning M < 0.3) and still be moving at over 100 m/s (224 mph)! There will certainly be some non-negligible dynamic pressure there. What it implies is that for a low enough Mach number, the flow can be considered incompressible. If you need convincing of this, I worked out the math in this thread not too long ago showing why the Mach number is a good indicator of compressibility.



Also remember that this incompressible assumption is, as AlephZero has stressed, an approximation. No fluid is truly incompressible.

Also, ladies and gents, keep in mind that "incompressible" does not mean constant density. It means that the material derivative of density is zero, or alternatively that the divergence of the velocity field is zero (from the continuity equation). You can certainly still have variable density in an incompressible flow provided that this condition is met.

Incompresible means constant volume.
So if the volume is constant and the mass is constant too, there can not be variable density.

If the speed of sound is determined only by static pressure, according to Bernoulli principal, there will be a huge difference in sound speed for stationary and moving (with 100 m/s as you say) fluid, because when fluid moves, static pressure will drop significantly, especially in water.

This depends on how you define total pressure. Sometimes total pressure includes the $\rho g z$ term factoring in potential energy, so be careful.

$\rho g z$ is nothing more than... hydrostatic pressure

Based on the typical definition of atmospheric pressure, yes, it is a form of hydrostatic pressure. It simply varies with height, and since you have two fluids stacked on top of one another, you have to include the weight of one in the hydrostatic pressure of the other. That doesn't change the fact that atmospheric pressure is the hydrostatic pressure from the air.

They have very different density .
And much more different hydrostatic heights.
So you have to calculate them separately.

Of course, the important thing is that either way, total pressure is not useful as a thermodynamic property, as it is frame-dependent. For that reason, most equations, such as the Navier-Stokes equatiuons and the equation for speed of sound discussed here, use static pressure, which is frame-independent, confirming what Niles has been saying.

Speed is frame dependent too.
So 'low Mach number' is a very relative term too

 

[boneh3ad]

Incompresible means constant volume.
So if the volume is constant and the mass is constant too, there can not be variable density.

In continuum mechanics that is only true if you specifically mean that a given fluid element remains at constant volume. To illustrate, the continuity equation is

$$\dfrac{D\rho}{Dt} + \rho(\nabla \cdot \vec{V}) = 0.$$

For an incompressible flow, we say that $\frac{D\rho}{Dt} = 0$, which is therefore the same as requiring that $\nabla \cdot \vec{V}=0$, which says that there is no dilatation of a fluid element. That completely allows for variable density under certain conditions. For example, if I place a probe on the hood of my car that somehow measures density and then drove around indoors at a constant speed, the probe very well could (and would) measure density fluctuations despite the flow being "incompressible".

Think of it this way: the speed of sound is a measure of how fast pressure, density, or temperature pulses and perturbations are transmitted through a given medium. Using density as an example, if you have, say, a sphere moving much more slowly than the speed of sound in a fluid, then the effect the sphere has on the density around it is going to propagate away from it at the speed of sound. These perturbations are going to be generated simply from moving through the fluid, but since the speed of sound is so much faster than the source of these perturbations (the sphere), they can basically get out of the way of one another and simply push molecules aside (i.e. stretch but not compress a fluid element).

Once this sphere starts moving at a rate closer to the speed of sound, these small perturbations are generated from a moving source that is moving at a speed that is comparable to the speed of sound, and these perturbations that are generated don't outrun each other quite as easily. The peaks and valleys of successive waves are closer together and the resulting pressure disturbance is large enough that it starts having to squeeze the fluid elements in order to conserve mass. That is when a flow becomes compressible.

If the speed of sound is determined only by static pressure, according to Bernoulli principal, there will be a huge difference in sound speed for stationary and moving (with 100 m/s as you say) fluid, because when fluid moves, static pressure will drop significantly, especially in water.

That's not what I said. The sound speed depends on the rate of change of static pressure with respect to density. For many fluids, that is nearly constant over fairly wide ranges of static pressure provided you hold other quantities constant. In an idea gas, for example, the speed of sound varies with the square root of temperature, but remains constant provided whatever process is isothermal regardless of how high or low the pressure is. So, unless the temperature changes appreciably, there would be nearly zero change in sound speed between stationary and moving 100 m/s.

They have very different density .
And much more different hydrostatic heights.
So you have to calculate them separately.

Be that as it may, atmospheric pressure is still a hydrostatic pressure and can be calculated in the same way as any other hydrostatic pressure minus any relatively minute changes due to weather. I never said you calculate them all in one go. My point was that you are making distinctions where there are none.

Speed is frame dependent too.
So 'low Mach number' is a very relative term too

Yes, speed is definitely frame dependent, and you can find a variety of Mach numbers relative to different frames of reference. The one that determines the relative compressibility of of a flow, however, is pretty simple to understand, and is the velocity of the free stream relative to the body in question. That goes for aircraft (the velocity of the air relative to the moving aircraft), shock tubes (the velocity of the air relative to the stationary wall or any test model present), explosions (the velocity of the blast wave relative to the ground).

 

[Niles]

Also, ladies and gents, keep in mind that "incompressible" does not mean constant density. It means that the material derivative of density is zero, or alternatively that the divergence of the velocity field is zero (from the continuity equation). You can certainly still have variable density in an incompressible flow provided that this condition is met.

Also remember that this incompressible assumption is, as AlephZero has stressed, an approximation. No fluid is truly incompressible.

In the thread you are referring to, you showed that when Ma²<<1 then Δρ/ρ<<1, i.e. density can be approximated constant in the material in this limit. Hence the fluid itself, not only its flow, can be approximated as incompressible (https://en.wikipedia.org/wiki/Incom...ence_between_incompressible_flow_and_material).But I completely understand the difference between and incompressible flow and fluid. Based on your derivation I just believe you showed the latter (and thus the former), and not only the former.

 

[Malverin]

In continuum mechanics that is only true if you specifically mean that a given fluid element remains at constant volume. To illustrate, the continuity equation is

$$\dfrac{D\rho}{Dt} + \rho(\nabla \cdot \vec{V}) = 0.$$

For an incompressible flow, we say that $\frac{D\rho}{Dt} = 0$, which is therefore the same as requiring that $\nabla \cdot \vec{V}=0$, which says that there is no dilatation of a fluid element. That completely allows for variable density under certain conditions. For example, if I place a probe on the hood of my car that somehow measures density and then drove around indoors at a constant speed, the probe very well could (and would) measure density fluctuations despite the flow being "incompressible".

You are speaking about a constant mass flow.
Any incompressible fluid has constant mass flow, but constant mass flow doesn't ensure incompressibility.

Any compression is change in volume. So if you have a change in volume, there is a compression.

Liquids and gases cannot bear steady uniaxial or biaxial compression, they will deform promptly and permanently and will not offer any permanent reaction force. However they can bear isotropic compression, and may be compressed in other ways momentarily, for instance in a sound wave.


Tightening a corset applies biaxial compression to the waist.
Every ordinary material will contract in volume when put under isotropic compression, contract in cross-section area when put under uniform biaxial compression, and contract in length when put into uniaxial compression.

http://en.wikipedia.org/wiki/Compression_(physical)

 

[boneh3ad]

You are speaking about a constant mass flow.
Any incompressible fluid has constant mass flow, but constant mass flow doesn't ensure incompressibility.

That's not true. You can very certainly have unsteady, incompressible flows. It's true that constant mass flow does not necessarily imply incompressibility, but incompressibility does not imply constant mass flow either. The flow field around a swinging pendulum, for example, is time-varying, including the mass flow around the pendulum, and it is most certainly incompressible.

Any compression is change in volume. So if you have a change in volume, there is a compression.

You simply have to be careful about what you are saying is changing. In continuum mechanics, the concept of incompressibility says that a given fluid element does not change volume. This is, of course, assuming that the given fluid element is not gaining mass for any reason. A stronger statement is that a given element's density does not change. That still allows for a given fluid element to be highly deformed under the action of the stress tensor but remain at constant volume (or density) and it allows for for many fluid elements of varying volume (or density) to flow by a fixed measurement point.

 

[Malverin]

That's not true. You can very certainly have unsteady, incompressible flows. It's true that constant mass flow does not necessarily imply incompressibility, but incompressibility does not imply constant mass flow either. The flow field around a swinging pendulum, for example, is time-varying, including the mass flow around the pendulum, and it is most certainly incompressible.



You simply have to be careful about what you are saying is changing. In continuum mechanics, the concept of incompressibility says that a given fluid element does not change volume. This is, of course, assuming that the given fluid element is not gaining mass for any reason. A stronger statement is that a given element's density does not change. That still allows for a given fluid element to be highly deformed under the action of the stress tensor but remain at constant volume (or density) and it allows for for many fluid elements of varying volume (or density) to flow by a fixed measurement point.

In this case, incompresible fluid will have constant mass flow for sure.