Threshhold compressible flow Ma = 0.3

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The discussion focuses on the differing approaches to determining the threshold for considering flow compressible at Mach number (M) = 0.3. One method, based on Anderson's equation, suggests a 5% error in density relative to total density is acceptable for incompressibility, while other textbooks propose a 10% error based on percent change in density. The participants note that both methods yield different error thresholds yet arrive at the same conclusion regarding M = 0.3 as the cutoff. The conversation highlights the importance of understanding how each method defines and measures error, as well as the implications for fluid dynamics. Ultimately, both approaches aim to establish when density changes are negligible enough to treat the flow as incompressible.
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Hi guys,

I'm trying to delve into compressible flow and in the various textbooks I'm reading I've found two school of thoughts when it comes to how to calculate the threshold (or error compared to incompressible flow) of 'M' when to consider a flow compressible.

In Fundamentals of Aerdnyamics (Anderson) this equation is used to calculate the 'error' between considering the fluid compressible and not
$$ \frac{\rho_0} {\rho}= (1+ \frac {\gamma -1}{2} M^2)^{\frac{1} {\gamma-1}} $$
For air (gamma = 1.4), this would give a deviation of (nearly) 5% between ##\rho## and ##\rho_0## at M = 0.3. Anderson uses this to show with an error of merely 5% you might as well consider the flow incompressible. Whereas in other textbooks (White; Kundu, Cohen) this equation is used to determine said deviation
$$ -M^2 \frac {dV}{V} = \frac {d\rho}{\rho}$$
This would give us a 9% error at M = 0.3 (for some reason literally every textbook rounds this up to 10%. I haven't found a single textbook that said otherwise). In these textbooks the authors derive at the conclusion that with an error of 10% (or less) the flow can be considered incompressible and ignore all the extra kerfuffle that compressible flow brings to the table.

My question is: What differentiates the two approaches to calculate the error? I mean obviously in the first equation the fluid is important (gamma value), but other than that? I think the two approaches describe two different "errors", but I can't seem to pinpoint what exactly the difference is.

If there is a link or a textbook that describes or answers my question, there is no need to spoonfeed me, just point me in the right direction.

Thanks!
 
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You are looking at the error in two different variables. In the first it's a density change relative to the total density and in the second it's the percent change in density.
 
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Oh, right! Thank you! I have a follow up question tho: Why is it that for the density change relative to the total densitiy an error of (or under) 5% is good enough to be considered incompressible, whereas for the perecent change in density an error of 10% is considered OK? I know that its just a guide value, but its kinda nagging me, because at the end of the day both methods justify the same thing ("when is it OK to consider a flow incompressible").

Like to me it looks like one of the two methods was 'reverse-engineered' (even though I know that is not the case), to say: Both methods arrive at the conclusion that M=0.3 is the cutoff value, while both methods use different variables AND have different results (errors). Its kinda as if M=0.3 was set as the cutoff value using one method, and then the error for the other method was calculated using M=0.3.
 
Personally, I don't like the Anderson definition. Ultimately, the assumption of incompressibility is tantamount to assuming the density isn't changing, so you have to pick a criteria that you feel makes that true enough. Most of the time, it amounts to assuming that the density fluctuations (relative to density) are much smaller than 1, i.e.
\dfrac{d\rho}{\rho}\ll 1.
The implication is that variations in other quantities like ##du/u## are going to be order 1, so ##d\rho/\rho \ll 1## makes those fluctuations irrelevant by comparison. usually, that means it would need to be at least an order of magnitude less than 1, or
d\rho/\rho < 0.1.
Since, as you've already mentioned, the Mach number squared is proportional to ##d\rho/\rho##,
M^2\propto \dfrac{d\rho}{\rho},
then
M^2\ll 1,
or
M^2< 0.1.
The square root of that is
M<0.3.
 
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