Low Mach Numbers: Incompressible flow or fluid

In summary, when a fluid is said to be incompressible at low Mach numbers, it means that the flow is incompressible (i.e. the material derivative is zero) and also that the fluid itself is behaving in an incompressible manner (with constant density). This is typically the case when the Mach number is less than 0.3, as the change in density is so small that it can effectively be ignored.
  • #1
Niles
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Hi

When we talk about a fluid moving at low Mach numbers, it is said to be incompressible. But does this mean that the flow is incompressible (i.e., material derivate is zero) or does it imply that the fluid itself is incompressible (constant density)?

If anybody has a reference (book, article, ...) explaining this matter, I would be happy to see it.
 
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  • #2
I don't understand this. If a material is incompressible then the velocity of sound will be infinite?
Is there some kind of shorthand involved in the original statement?
Or does it just mean that we assume the density and modulus can be considered to be constant over the range of amplitudes in the vibration cycle?
 
  • #3
It's the later, which implies the former. (Converse is not necessarily true.) Which also gives you zero divergence of the flow.
sophiecentaur said:
I don't understand this. If a material is incompressible then the velocity of sound will be infinite?
Yes, but you should be able to make fluid less compressible while keeping speed of sound constant. It doesn't really make sense to have compressibility be exactly zero, but I think all other properties work without contradiction in the limit.
 
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  • #4
There are no truly incompressible fluids. Even water, which is often treated as incompressible, can undergo compression; it just takes an extraordinarily large pressure to do it and for all practical purposes (other than sound speed) it is incompressible.

That said, the answer to the original question would typically be that the flow is incompressible, which, again, for all practical purposes (other than sound speed) means the fluid is incompressible in that situation as well.

The explanation for why the Mach number actually implies this is fairly simple if a little long-winded. Are you familiar with the Lagrangian and Eulerian frames of reference for fluid flows? The Lagrangian frame of reference is where the frame follows along with a moving fluid element. The Eulerian frame is stationary and sees fluid elements pass by. In the Lagrangian frame, compression can be defined as the change in volume of a fluid element versus a base state, and this quantity can be related back to pressure change using the bulk modulus, [itex]E[/itex]:
[tex]\Delta p = -E\dfrac{\Delta \mathcal{V}}{\mathcal{V}_0}.[/tex]
If this was a solid, [itex]E[/itex] is the equivalent of Young's modulus. In a fluid it is the negative reciprocal of the thermodynamic compressibility.

If you look from the Eulerian frame, the total volume doesn't change and instead you can view compressibility based on the change in density in the region in question and you can relate that to the volume using conservation of mass from some base state [itex](\rho_0,\mathcal{V}_0)[/itex] as
[tex](\mathcal{V}_0 + \Delta\mathcal{V})(\rho_0 + \Delta \rho) = \mathcal{V}_0\rho_0.[/tex]
If you expand this expression and take the limit as the [itex]\Delta[/itex] terms go to zero, you get
[tex]\dfrac{\Delta\mathcal{V}}{\mathcal{V}_0} = -\dfrac{\Delta\rho}{\rho_0}.[/tex]
Combining this with our bulk modulus definition results in
[tex]\dfrac{\Delta\rho}{\rho_0} = \dfrac{\Delta p}{E}.[/tex]

Generally speaking, a flow is incompressible when [itex]\Delta\rho/\rho_0 \ll 1[/itex]. From the integrated Euler equation [itex]\int dp/\rho + V^2/2 = \text{const.}[/itex] you can show that the change in pressure is always bounded by the dynamic pressure [itex]\rho V^2/2[/itex], so we can drop that into our previous equation
[tex]\dfrac{\Delta\rho}{\rho_0} \approx \dfrac{\rho V^2}{2E}.[/tex]
One of the definitions for the speed of sound in a continuum is [itex]a^2 = E/\rho[/itex], so
[tex]\dfrac{\Delta\rho}{\rho_0} \approx \dfrac{V^2}{2a^2} = \dfrac{1}{2}M^2.[/tex]
In reality, if you are familiar with order, it is probably more accurate to say
[tex]\dfrac{\Delta\rho}{\rho_0} = O\left(\dfrac{1}{2}M^2\right).[/tex]
So what that says is that the relative density change in the flow is directly related to the Mach number of that flow. If [itex]M^2 \ll 1[/itex], then [itex]\Delta\rho/\rho_0 \ll 1[/itex]. Generally speaking, this usually means that [itex]M^2[/itex] has to be an order of magnitude less than 1 or smaller, or [itex]M^2 < 0.1[/itex], or [itex]M < 0.316[/itex]. We usually just simplify that to [itex]M<0.3[/itex].

In other words, if the Mach number is less than 0.3, the change in density in a given flow is so minute that it can effectively be ignored for most practical purposes. That also implies that for that given situation, the fluid itself is also behaving in an incompressible manner.
 
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  • #5


Great question! The term "incompressible" in this context refers to the fluid itself being incompressible, meaning that its density remains constant regardless of changes in pressure or temperature. This is in contrast to compressible fluids, which experience changes in density due to changes in pressure or temperature.

At low Mach numbers, the fluid flow can be considered incompressible because the changes in density are negligible compared to the overall flow. However, it is important to note that even at low Mach numbers, there may still be small changes in density, but they are not significant enough to affect the overall flow behavior.

As for a reference, I would recommend the book "Fluid Mechanics" by Frank White, which discusses the concept of incompressible flow in detail. Additionally, the article "The Myth of Incompressibility" by John Anderson also provides a good explanation of the differences between incompressible and compressible fluids.

I hope this helps clarify the concept for you. Keep up the curiosity and keep asking great questions!
 

FAQ: Low Mach Numbers: Incompressible flow or fluid

What is the definition of low Mach numbers in fluid dynamics?

The Mach number is a dimensionless quantity that represents the ratio of an object's speed to the speed of sound in the surrounding medium. Low Mach numbers, typically less than 0.3, refer to fluid flows where the speed of the fluid is much smaller than the speed of sound. In this case, the fluid is considered to be incompressible.

How does the incompressibility assumption affect the governing equations for low Mach number flows?

The incompressibility assumption means that the density of the fluid remains constant and the governing equations for low Mach number flows can be simplified. The conservation of mass equation reduces to the continuity equation and the conservation of momentum equation reduces to the Navier-Stokes equation without the compressibility terms.

What are some examples of low Mach number flows?

Low Mach number flows are commonly found in everyday situations such as water flowing through pipes, air moving over a car, or blood flowing through arteries. They are also important in aerodynamics of aircraft at takeoff and landing speeds.

What are the limitations of using the low Mach number assumption in fluid dynamics?

The low Mach number assumption is only valid for flows where the Mach number is significantly less than 1. If the speed of the fluid approaches the speed of sound, the compressibility effects become significant and the incompressible flow assumption breaks down. Additionally, the low Mach number assumption cannot be applied to high-speed flows with shocks or supersonic flows.

How is the low Mach number assumption used in practical applications?

The incompressible flow assumption is commonly used in the design and analysis of engineering systems, such as aircraft wings, propellers, and pumps. It simplifies the mathematical equations and allows for easier and faster computations. However, in situations where compressibility effects are important, the low Mach number assumption must be used with caution or a more complex compressible flow model must be employed.

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