Fluid Mechanics: Incompressible flow and pressure

[Niles]

Most liquids can be assumed to be incompressible, since the Mach-number is much smaller than 1. That means that the density variations are negligible and from the relation between pressure p and density ρ,
<br /> p=c_s^2 \rho<br />
we see that the pressure in constant as well. Now, say that I look at a pipe with the following geometry:

From Bernoulli's equation we get that the pressure and velocity will be different between the large-radius part of the pipe and the small-radius part. How does this varying pressure conform with the constant pressure/density obtained from the equation of state?

 

[AlephZero]

The relation $p = c_s^2\rho$ is only meaningful for compressible fluids. In an incompressible fluid (which doesn't exist, but it is a useful approximation), the speed of sound $c_s$ would be "infinitely fast".

If you rewrite your formula as $c_s^2 = \frac {\partial p}{\partial \rho}$, you can't differentiate with respect to $\rho$ if $\rho$ is constant.

 

[Entanglement]

Liquids are not incompressible btw

 

[Niles]

My book (Frank White, Fluid Mechanics) says that most liquids can be assumed to be incompressible since their speed-of-sound is much larger than usual flow velocities. So I agree that the speed of sound can be assumed to be "infinite"/very large, but I believe my question is still valid

 

[AlephZero]

My book (Frank White, Fluid Mechanics) says that most liquids can be assumed to be incompressible since their speed-of-sound is much larger than usual flow velocities.

The important word in that quote is assumed. A better word would be approximated. If you try to make inconsistent approximations, you tie your self in knots. If you want to assume the density is constant, then you can't talk about the speed of sound.

So I agree that the speed of sound can be assumed to be "infinite"/very large, but I believe my question is still valid

Believe what you like, but remember Bertrand Russell's definition: "belief is that for which there is no evidence". It's not a useful concept for doing science.

 

[Niles]

The important word in that quote is assumed. A better word would be approximated.
Believe what you like, but remember Bertrand Russell's definition: "belief is that for which there is no evidence". It's not a useful concept for doing science.

I have seen several papers on arXiv that use that relation for incompressible flows.

Thanks for your input

 

[AlephZero]

Assuming the fluid is incompressible is often very useful, but you need to be careful what your quote actually means.

A longer version: If the flow velocity in a compressible fluid is much smaller than the speed of sound in the compressible fluid, then we can assume the fluid is incompressible.

Note that the above says nothing about "the speed of sound in an incompressible fluid" which is how you were wrongly interpreting it in your OP.

 

[Niles]

Assuming the fluid is incompressible is often very useful, but you need to be careful what your quote actually means.

A longer version: If the flow velocity in a compressible fluid is much smaller than the speed of sound in the compressible fluid, then we can assume the fluid is incompressible.

Note that the above says nothing about "the speed of sound in an incompressible fluid" which is how you were wrongly interpreting it in your OP.

I see, I was being not very accurate in my OP, I admit that. OK, so starting all over again:

We can approximate many fluids as being incompressible, i.e. their density ρ is constant throughout the fluid. For our fluid we have the relation p = c_s²ρ, where c_s is the speed of sound in the fluid. Knowing this constant relation between ρ and p, how should I interpret the pressure changes that occur in Bernoulli's equation?

Now that has to be a valid question.

 

[SteamKing]

It's not only pressure which is changing in the flow. The velocities will be different at different points in the flow if the pressures are different.

All the Bernoulli relation is saying is that under certain assumptions, the total energy in the flow is constant from point to point. If pressure drops, then velocity must increase to maintain constant energy in the flow stream.

 

[Niles]

It's not only pressure which is changing in the flow. The velocities will be different at different points in the flow if the pressures are different.

All the Bernoulli relation is saying is that under certain assumptions, the total energy in the flow is constant from point to point. If pressure drops, then velocity must increase to maintain constant energy in the flow stream.

Thanks. But in my book (Frank White, Fluid Mechanics page 231) he says that the pressure changes Δp are roughly related to density changes Δρ as Δp=Δρ*c_s², where c_s² is the speed of sound squared. Knowing that density changes are negligible, this means pressure changes should be negligible too.

But this contradicts the Bernoulli equation and your post, basically. Where is my reasoning wrong?

 

[AlephZero]

The real fluid is compressible. Therefore it has a finite speed of sound.

A mathematical model that says the fluid is incompressible is approximate. If you have an approximate model, some of its predictions will be wrong.

The math of the incompressible model can not tell you the speed of sound in the fluid. Your attempt to get the speed of sound from the math is like asking "if $y = 4x^2$, what is $\frac{\partial y}{\partial 4}$"? If your model assumes that $\rho$ is a constant, then $\frac{\partial p}{\partial\rho}$ is meaningless.

The fact that a model is "wrong" doesn't make it useless. Every math model in every branch of physics is "wrong" to some extent. But you always have to remember that "the map is not the territory," and not try to use a model for things that it can't do.

 

[cjl]

I see a lot of nitpicking in this thread about whether or not liquids are truly incompressible (and thus whether the sound speed is infinite), and honestly, it's not really relevant in this case. I'm more curious where the equation P = c_s²ρ came from. That equation is not generally true for any fluid, since it implies the pressure is purely a function of the sound speed and the density of the fluid, which is clearly incorrect. Obviously, for a completely incompressible fluid, c_s is infinite, making the relation even more irrelevant, but that's not the real problem here. Where did you get that relation from?

 

[Niles]

It is derived on page 605 in Frank White's book on Fluid Mechanics, and in the following links it is derived as well

https://www.physicsforums.com/showthread.php?t=728695
http://www.sjsu.edu/faculty/watkins/sound.htm

 

[cjl]

That's not the same equation though - c_s² = dP/dρ is correct, but P = c_s²ρ is not correct. Yes, you can integrate the former to get the latter, but you have to remember the integration constant as well (so P = c_s²ρ + c is fine, but the + c means that the pressure can be pretty much anything, since that constant can be pretty much anything). As a result, the whole premise is incorrect - the pressure is not constant in an incompressible fluid.

 

[boneh3ad]

Also don't forget that's a partial derivative, so in general that integration constant is not actually a constant but an unknown function of other variables.

 

[AlephZero]

That's not the same equation though - c_s² = dP/dρ is correct, but P = c_s²ρ is not correct.

I don't have White's book, but this could be just a matter of notation. For small amplitude waves in a fluid (for example in acoustics) you typically write $P(x,y,z,t) = P_0(x,y,z) + p(x,y,z,t)$ and similarly for the density. $p$ and $\rho$ are small changes from the steady state conditions. In that case, assuming the fluid is incompressible is the same as assuming $\rho_0$ is constant, and $\rho = 0$.

I don't know how many times we need to repeat this till the OP "gets the point", but the derivation in http://www.sjsu.edu/faculty/watkins/sound.htm assumes that the fluid is compressible.

 

[Niles]

I don't know how many times we need to repeat this till the OP "gets the point", but the derivation in http://www.sjsu.edu/faculty/watkins/sound.htm assumes that the fluid is compressible.

I do get the point now. If the fluid itself is not compressible, the derivation doesn't make sense.

 

[AlephZero]

When you start learning physics in high school, it's easy to get the wrong idea that empirical "laws" like Hooke's law of elasticity or Coulomb's law of friction are in some sense "true". They are only approximately true, in some situations. But while you are learning how to use them to solve problems, that fact can disappear under the radar.

"Classical" Fluid mechanics (i.e. most models of fluid behavior which lead to equations that you actually solve using algebra and calculus) is rather different, in that almost all the models ignore things that are critical to the way fluids behave in the real world - such as compressibility and viscosity. Or they include those things as "fudges" that only work in particular situations, like boundary layer theory.

Getting insight into what simplified models work in what circumstances is a big part of the learning process in fluid mechanics (or at least it should be, IMO). The good news is, you just took one important step along that road!

 

[Niles]

Hi all

Please take a look at this paper, on the second page equation (5). There they state that p=c_s^2\rho.

I agree with everything said in this thread and that the above relation is clearly missing an integration shift, which depends on some other variable (e.g. spatial, such that Bernoulli's equation in not conflicted).

However, I wanted you to know that the relation is being used in various papers without this shift -- and that it seems there are some details we haven't touched upon here.

 

[Malverin]

Most liquids can be assumed to be incompressible, since the Mach-number is much smaller than 1. That means that the density variations are negligible and from the relation between pressure p and density ρ,
<br /> p=c_s^2 \rho<br />
we see that the pressure in constant as well. Now, say that I look at a pipe with the following geometry:

From Bernoulli's equation we get that the pressure and velocity will be different between the large-radius part of the pipe and the small-radius part. How does this varying pressure conform with the constant pressure/density obtained from the equation of state?

Bernoulli's principal says that the SUM of all types of pressure is constant

when the speed grows, static pressure drops.
But the SUM of pressures is always the same.

 

[Niles]

Yes, but p=c_s^2\rho is a local expression, it doesn't refer to a sum of pressures. That's what makes this strange

 

[Malverin]

Yes, but p=c_s^2\rho is a local expression, it doesn't refer to a sum of pressures. That's what makes this strange

Change in speed, depends on change in pressure/density ratio
The pressure is constant(it is not mentioned what kind of pressure because it is the total pressure)
so there is no change in speed, or pressure or density.

c = (dp / dρ)^1/2 (1)

where

c = sound velocity (m/s, ft/s)

dp = change in pressure (Pa, psi)

dρ = change in density (kg/m3, lb/ft3)

http://www.engineeringtoolbox.com/speed-sound-d_82.html

 

[Niles]

First, thanks for helping, I really appreciate having someone to talk to about this.

Change in speed, depends on change in pressure/density ratio
The pressure is constant(it is not mentioned what kind of pressure because it is the total pressure)
so there is no change in speed, or pressure or density.

c = (dp / dρ)^1/2 (1)

where

c = sound velocity (m/s, ft/s)

dp = change in pressure (Pa, psi)

dρ = change in density (kg/m3, lb/ft3)

http://www.engineeringtoolbox.com/speed-sound-d_82.html

The pressure p is the same that goes into the Navier-Stokes equations as a gradient, ∇p. This is in equation (24) of the paper I attached.

To my knowledge the pressure p in the Navier-Stokes equations does not refer to the total pressure.

 

[Malverin]

http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations

First, thanks for helping, I really appreciate having someone to talk to about this.





The pressure p is the same that goes into the Navier-Stokes equations as a gradient, ∇p. This is in equation (24) of the paper I attached.

To my knowledge the pressure p in the Navier-Stokes equations does not refer to the total pressure.

where v is the flow velocity, ρ is the fluid density, p is the pressure, T is the (deviatoric) component of the total stress tensor, which has the order two, and f represents body forces (per unit volume) acting on the fluid and ∇ is the del operator.

http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations

So why do you think it is not the total pressure?

 

[Niles]

http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations

http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations

So why do you think it is not the total pressure?

The Bernoulli equation refers to the total pressure energy across the cross section @ a specific point on a streamline. We know that is constant.

In the Navier-Stokes equation the p only refers to the "local" pressure, i.e. the first term p in the Bernoulli equation. Otherwise, if it really did refer to the total pressure, we would never be able to have a nonzero gradient ∇p, since the total pressure is constant.

 

[boneh3ad]

So why do you think it is not the total pressure?

Because he is right. The pressure in the N-S equations is static pressure.

 

[Niles]

Because he is right. The pressure in the N-S equations is static pressure.

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

 

[Malverin]

The Bernoulli equation refers to the total pressure energy across the cross section @ a specific point on a streamline. We know that is constant.

In the Navier-Stokes equation the p only refers to the "local" pressure, i.e. the first term p in the Bernoulli equation. Otherwise, if it really did refer to the total pressure, we would never be able to have a nonzero gradient ∇p, since the total pressure is constant.

Yes. You are right.
Here p is the static pressure

The left side of the equation describes acceleration, and may be composed of time dependent or convective effects (also the effects of non-inertial coordinates if present). The right side of the equation is in effect a summation of body forces (such as gravity) and divergence of stress (pressure and shear stress).

But that does not explain why do you think that is the same pressure as here

 

[Niles]

But that does not explain why do you think that is the same pressure as here

I believe the p in the definition of the speed of sound is the same pressure the one in Bernoulli's law (first pressure term).

 

[Malverin]

I believe the p in the definition of the speed of sound is the same pressure the one in Bernoulli's law (first pressure term).

Underwater acoustic propagation depends on many factors. The direction of sound propagation is determined by the sound speed gradients in the water. In the sea the vertical gradients are generally much larger than the horizontal ones. Combining this with a tendency towards increasing sound speed at increasing depth, due to the increasing pressure in the deep sea

Hydrostatic pressure increases with depth.
Static pressure is atmospheric pressure at the water surface and dynamic pressure is zero if water does not move
So when you go deeper total pressure increases, but static(the first one) remains the same

http://en.wikipedia.org/wiki/Underwater_acoustics

 

[boneh3ad]

Hydrostatic pressure is static pressure.

 

[Malverin]

Hydrostatic pressure is static pressure.

Yes it is.
In the equation it is considered separately (and it should be, because atmospheric and hydrostatic pressure are not the same thing)
so to be correct we have to consider it separately.

And to generalize all this

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

 

[Niles]

Yes it is.
In the equation it is considered separately (and it should be, because atmospheric and hydrostatic pressure are not the same thing)
so to be correct we have to consider it separately.

And to generalize all this

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

I am confused, does this mean you understand what the solution to the problem is?

 

[Malverin]

I am confused, does this mean you understand what the solution to the problem is?

There is no problem if you use total pressure in the 'speed of sound' formula

 

[Niles]

There is no problem if you use total pressure in the 'speed of sound' formula

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

 

[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.