# Fluid Mechanics: Incompressible flow and pressure

1. Dec 30, 2013

### 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,
$$p=c_s^2 \rho$$
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?

2. Dec 30, 2013

### 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.

Last edited: Dec 30, 2013
3. Dec 30, 2013

### Entanglement

Liquids are not incompressible btw

4. Dec 30, 2013

### 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

5. Dec 30, 2013

### AlephZero

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.

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.

6. Dec 30, 2013

### Niles

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

7. Dec 30, 2013

### 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.

8. Dec 30, 2013

### Niles

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 = cs2ρ, where cs 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.

9. Dec 30, 2013

### SteamKing

Staff Emeritus
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.

10. Dec 30, 2013

### Niles

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*cs2, where cs2 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?

Last edited: Dec 30, 2013
11. Dec 30, 2013

### 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.

12. Dec 30, 2013

### 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 = cs2ρ 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, cs is infinite, making the relation even more irrelevant, but that's not the real problem here. Where did you get that relation from?

13. Dec 30, 2013

### Niles

14. Dec 30, 2013

### cjl

That's not the same equation though - cs2 = dP/dρ is correct, but P = cs2ρ 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 = cs2ρ + 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.

15. Dec 30, 2013

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.

16. Dec 30, 2013

### AlephZero

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.

17. Dec 30, 2013

### Niles

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

18. Dec 30, 2013

### 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!

19. Jan 2, 2014

### 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.

20. Jan 2, 2014

### Malverin

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.

Last edited: Jan 2, 2014