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Why does density decrease more above Mach 0.3, and not as much below Mach 0.3? From what I've heard, it isn't a linear relationship but why is that? Asked differently, what makes it a non linear relationship?

Thanks!

Link to chat room of this question.

Also, the last paragraph of this question has my guess of why the density doesn't change much below Mach ~0.3. It might be useful to see where I'm coming from.

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  • $\begingroup$ Some reading here. $\endgroup$ Jan 17 at 0:40
  • $\begingroup$ @RobertDiGiovanni Interesting concepts in there. Thanks $\endgroup$
    – Wyatt
    Jan 17 at 0:47
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    $\begingroup$ Q.1 asks what it does. Q.2 mentions the definition of nonlinearity, and then asks why. Q.3 asks the opposite of Q.1. The third paragraph asks yet another question. Help! Use fewer words, not more. $\endgroup$ Jan 18 at 20:14
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    $\begingroup$ Your edit just now improved the question, thanks! But if you can edit it down even farther so it asks just a single question, that would help a ton. Find the one best way to phrase the question, and leave it at that. Repeating with variations is useful for poetry and theology, but in the hard sciences it's only confusing. $\endgroup$ Jan 18 at 20:40
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    $\begingroup$ Ask one question per stack-exchange-question. How you do that is up to you! $\endgroup$ Jan 18 at 21:03

1 Answer 1

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When you combine the equations for isentropic flow you can plot "density compared to density at zero speed, M=0", i.e. $\rho / \rho_0$, as a function of M.

Anderson's Fundamentals of Aerodynamics presents that as $\rho_0 / \rho = (1+ \frac {\gamma -1}{2} M^2)^{\frac{1} {\gamma-1}}$ with $\gamma = 1.4$ for air (page 567, eq. 8.43).

Inverting, $\rho / \rho_0 = (1+ \frac {\gamma -1}{2} M^2)^{\frac{1} {1-\gamma}}$.

I've copied a plot of this below, but you can calculate it yourself if you like. The red curve shows that M = 0.3 isn't magical. (At M=0.3, the density ratio is 0.956.) It's just a convenient, practically useful threshold for everyday calculations.

Why isn't the red curve a straight line? Because these equations have proven to be a useful, accurate, predictive model of fluid flow in many fields, and these equations are nonlinear. You could replace them with linear ones if you liked, but the calculations you'd make with them would be less accurate.

enter image description here

Anderson's own plot, p. 573 fig. 8.6, agrees: enter image description here

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