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I've read that Mach 0.3 is pretty much the upper limit for treating air as an incompressible fluid. The sources I've read seem to treat this as a given, without proof or justification.

Why is this the limit? Is there a mathematical justification for this? Also, does this limit only apply to air? If not, then what does the limit depend on?

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Wikipedia gives the reason for Mach 0.3 as due to the fact that this achieves >5% change in density.

I found a NASA page that describes (analytically!) the relationship. I cited the source, but I'll reproduce the work here for posterity, in the event their links change.

Start with conservation of momentum:

$$ (\rho V) dV = -dp \\ $$

where $\rho$ is the fluid density, $V$ is the velocity, and $p$ is the pressure. for isentropic flow:

$$ \frac{dp}{p} = \gamma \frac{d\rho}{\rho} \\ dp = \left( \frac{\gamma p}{\rho} \right) d\rho \\ $$

where $\gamma$ is the specific heat ratio. The ideal gas law gives:

$$ p = \rho R T \\ $$

where $R$ is the specific gas constant and $T$ is the absolute temperature. So, substituting:

$$ dp = \gamma R T d\rho $$

The speed of sound can be calculated by:

$$ \gamma R T = a^2 \\ $$

where $a$ is the speed of sound, so:

$$ dp = a^2 d\rho \\ $$

Substituting the expression above into the conservation of momentum equation gives:

$$ (\rho V)dV = -a^2 d\rho \\ -\left(\frac{V^2}{a^2}\right)dV/V = d\rho/\rho \\ -M^2 dV/V = d\rho/\rho \\ $$

where $M$ is the Mach number. This gives a Mach number of 0.3 to be approximately a 10% change in density (vs. Wikipedia's 5% - I'd go with NASA's calculations).

As a note, this is based on the Mach number, which in turn is dependent on the speed of sound in the gas, so it's automatically adjusted on a per-gas basis.

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@Paul this is derived from the conservation of momentum. it is not so much a "rule" as a suggestion. if you don't care about 10% (or higher) changes in density (or other quantities), go ahead and use the incompressible relations for high Mach numbers. if you do care about small changes in density, then use the compressible relations even for low Mach numbers – costrom Feb 9 at 17:02
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It's not just density. When we non-dimensionalize equations, we get dimensionless groups out. The rule of thumb is that if a dimensionless group is less than 0.1 we can ignore the relevant terms. In the case of the Mach number it shows up squared. So we want the (Mach number)^2<0.1. This approximately gives 0.3. It's not just the density - basically all things that change at higher speed are going to be affected by roughly 10% once the Mach number reaches 0.3. – Joel Feb 9 at 21:03
    
@Joel - For context, OP was asking about compressibility specifically, which is why this answer only covers density. – Chuck Feb 9 at 21:22
2  
I would make it clearer that it's not a sharp dividing line. If you have a lower tolerance for errors, start using the compressible solution at lower mach numbers. If you don't care as much, keep assuming incompressibility at higher mach numbers. 10% is just an arbitrary choice of how much error "really matters", and 0.3 falls out of that mathematically, but no less arbitrarily. – hobbs Feb 9 at 22:53
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@chuck - nitpicking here, but treating something as an incompressible fluid means I get to say that the divergence of the velocity field is 0. That affects a lot more than just density - to the point that when I go to a talk and someone says he's assuming its an incompressible fluid it's usually not a statement about density. – Joel Feb 10 at 4:11

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