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For isentropic $P = k\rho^\gamma$ steady irrotational $\vec{u} = \nabla\phi$ flow, the momentum equation implies the Bernoulli relation $$\tfrac{1}{2}|\nabla\phi|^2+\frac{k\gamma}{\gamma-1}\rho^{\gamma-1} = C$$ where $C$ is constant throughout the (connected) fluid. So far so good, I think, at least for smooth solutions. Now it is also true that the speed of sound is given by $c^2 = P'(\rho) = k\gamma\rho^{\gamma-1}$, which ought to vary throughout the flow. But then wherever the speed $|\nabla\phi|$ is equal to the sound speed you could solve the Bernoulli relation to find $$ c^2 = \frac{\gamma-1}{\gamma+1}\frac{C}{3}.$$ But this is constant. What am I doing wrong?

Bob Terrell
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  • At the moment, this is a guess (I plan to dig deeper), but suspect that you are dealing with a mathematical inconsistency in the underlying derivations - sound (and sound speed) are typically developed under the implied assumption that the underlying fluid is at rest. – John Polcari Jul 19 '18 at 13:39
  • This is certainly not intended to replace the answer below, nor intended to sow any confusion, but your question intrigued me sufficiently to work through how the acoustic wave equation changes if one assumes a (simple) underlying flow field. If you are interested in seeing how sound advection pops out of that, see here. – John Polcari Jul 20 '18 at 14:27

1 Answers1

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Aside from an algebraic error, you have solved for the critical value of sound speed where the fluid attains Mach 1 -- hence, you have done nothing wrong. The sound speed will assume different values throughout the flow as a function of local temperature. Nevertheless the fluid will attain a condition where the velocity equals a specific sound speed (Mach 1) that depends on the density where the fluid is at rest. There are other constraints to isentropic flow, for example, a maximum attainable velocity which corresponds to the pressure falling to $0$.

At Mach 1 the fluid velocity equals the critical speed of sound $c_*$at local conditions, so $|\nabla\phi|^2 = c_*^2$ and substituting in the Bernoulli equation we have

$$\frac{c^2}{2} + \frac{k\gamma \rho^{\gamma-1}}{\gamma-1} = \frac{c_*^2}{2} + \frac{c_*^2}{\gamma-1} = C \\ \implies c_*^2 \frac{\gamma +1 }{2(\gamma -1)}= C \\ \implies c_*^2 = 2C \frac{\gamma-1}{\gamma+1} \quad (*)$$.

The constant C can be related to the speed of sound $c_0$ when the fluid is at rest. Under such stagnation conditions the fluid velocity is $0$ and, again, using the Bernoulli equation we have

$$ \frac{k\gamma \rho_0^{\gamma-1}}{\gamma-1} = \frac{c_0^2}{\gamma-1} = C $$

Substituting for C in (*) we obtain

$$c_*^2 = 2\frac{c_0^2}{\gamma-1} \frac{\gamma-1}{\gamma+1} = \frac{2c_0^2}{\gamma+1}, $$

which yields

$$\frac{c_*}{c_0} = \sqrt{\frac{2}{\gamma+1}}$$

Typical values for air are $\gamma = 1.4$ and $c_* \approx 0.913 c_0$.

RRL
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  • you helped me understand that my error was one of logic (aside from the algebra error); If I have it straight now, the correct conclusion was not that the sound speed is constant everywhere, which I knew wasn't right, but rather that every point where the local Mach number is 1 has the same sound speed. This kind of flow is very special! Thanks. – Bob Terrell Jul 20 '18 at 00:25
  • @BobTerrell: You're welcome. Compressible flow, especially transonic and supersonic, behaves quite differently than incompressible flow. The exchange between mechanical and thermal energy accounts for this type of behavior. – RRL Jul 20 '18 at 03:43