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Please read this page before going to my question.

When we derive the relation between Mach no. to the left side of the shock wave (M1) and that to the right side of the shock wave (M2),we do that with the help of the pressure wave speed equation i.e. $c=\sqrt{kRT}$. But this relation was obtained by assuming the process to be isentropic and propagation of a normal shock is by no means reversibly adiabatic due to the abrupt discontinuity in properties on either sides. Then how come we are able to use this relation for our derivation please someone explain.

Its said that $c$ is a thermodynamic property and once its derived by any process it can be used for any other process which may or may not be same. But I don't quite understand that.

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    Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Also, it would be preferable if you could include relevant parts from external sources in your question so it stands on its own. Links can change or break over time. Thanks! – jng224 Nov 14 '21 at 11:21
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    Questions (and answers) should be self-contained. – Tobias Fünke Nov 14 '21 at 11:21
  • Thank You so much for the help. About the link I'm sure it would never change because its of a reputed university still i would try making the question self contained. – Apoorv Mishra Nov 14 '21 at 13:09
  • What is c here? Is it the sound speed? If so, there is nothing assumed about isentropic across the shock, one just assumes an equation of state locally for determining the relationship between pressure and density and temperature. – honeste_vivere Nov 15 '21 at 15:36
  • @honeste_vivere Thank you so much for taking time to reply this. Yes c is the speed of sound and k is nothing but gamma that is ratio of specific heats at constant pressure and temperature. – Apoorv Mishra Nov 16 '21 at 14:47
  • @honeste_vivere My question actually was that the equation for c was derived using the isentropic expansion of gas and here we are using it to derive relation of between properties across the shock. Is it possible that we use something derived using an isentropic process and use it for derivation of some process that's not isentropic and how? – Apoorv Mishra Nov 16 '21 at 14:56
  • @honeste_vivere I'm concerned of this,(from the linked page itself) We recall that the expression for sonic speed as $c=\sqrt{kRT}$ was derived for an ideal gas undergoing an isentropic process . However, the process by virtue of which shock occurs is by no means isentropic (as it includes a discontinuity). The reason we are still able to use the derived expression for sonic speed is because we are using sonic speed as a property. And while the expressions for thermodynamic properties can be obtained using a certain process, is not dependent on the process by which it is derived. – Apoorv Mishra Nov 16 '21 at 15:07
  • Well, formally the speed of sound is given by (to first order) $C_{s}^{2} = \partial P / \partial \rho$, where $P$ is pressure and $\rho$ is mass density. You can leave things in that form and not define any equation of state for the pressure, but then you'll have other issues with your expression for enthalpy, for instance. I agree, the assumption of the adiabatic equation of state seems odd but the basic idea is that it's assumed things happen so fast as to preclude heat transfer. – honeste_vivere Nov 16 '21 at 15:11
  • @honeste_vivere Yes, the relation you've mentioned for the speed of pressure wave is exactly what is used to derive the relation $c=\sqrt{kRT}$ assuming reversibly adiabatic process i.e. $constant=\frac{P}{ρ^k}$ But isn't this relation specified only if the process under consideration is quasi-static, i.e reversible? In case of rightwards propogating pressure waves the pressures on either sides can be P(on right), P+dP(on left) similarly density and other properties so that there's continuity in properties and the process is reversible and adiabatic due to assumed negligible heat transfer. – Apoorv Mishra Nov 16 '21 at 17:29
  • @honeste_vivere But isn't this relation just specified for isentropic processes and shouldn't the processes that posses some change in entropy be out of bounds for this relation? Like the shock in which the pressure and other properties are in no manner infinitesimally different i.e. we cant write pressure and density in above manner because of finite discontinuity in properties across the sides. In the article its mentioned that we use C as thermodynamic property i.e. independent of the path of derivation but how was this idea first thought of? – Apoorv Mishra Nov 16 '21 at 17:36
  • See https://physics.stackexchange.com/a/611938/59023 – honeste_vivere Nov 16 '21 at 17:56
  • @honeste_vivere I'm sorry for the late reply I think I need to read more and develop a deeper understanding on the shockwave propogation before I go ahead with my doubt (if it ain't resolved by then) please be there to help me out when I need. The link you provided have been somewhat helpful. – Apoorv Mishra Nov 18 '21 at 18:30

1 Answers1

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But this relation was obtained by assuming the process to be isentropic and propagation of a normal shock is by no means reversibly adiabatic due to the abrupt discontinuity in properties on either sides.

You are correct. The fluid on either side of a shock wave is not in a succession of equilibrium states, i.e., energy dissipation is occurring. Energy dissipation here is just an irreversible transformation of energy. In this case, the energy being converted is the kinetic energy of the fluid flow across the shock, which is not constant.

You are correct. We cannot use the normal adiabatic equation of state to describe the pressure and density change across a shock, given by: $$ \frac{ d }{ dt } \left( P \ \rho^{-\gamma} \right) = 0 \tag{0} $$ where $P$ is the scalar thermal pressure, $\rho$ is the mass density, and $\gamma$ is the polytropic index or ratio of specific heats. We must use the Rankine–Hugoniot relations (RHRs), which are just a series of conservation relations.

Note that in the RHRs the immediate fluids on either side of the shock are not in equilibrium however the asymptotic states far from the shock are assumed to be in equilibrium. The RHRs assume that the shock transition is very abrupt, i.e., that there is no heat flux across the shock ramp.

Now to your question, the derivation of the speed of sound from the adiabatic equation of state is different. For instance, we can use the adiabatic equation of state to derive the internal energy, $\xi = \tfrac{ P }{ \gamma - 1 }$, and the enthalpy, $\zeta = \tfrac{ \gamma P }{ \gamma - 1 }$, without loss of generality because these functions of the state of the system and don't care about how the system got into that state. That is, the details for how the system evolved to the state you are trying to describe using $\xi$ and/or $\zeta$ do not matter.

Similarly, the local speed of sound does not care about the impending shock wave, it's just the local speed of sound. In deriving that, we can assume that the longitudinal oscillations are sufficiently fast and small that we need not worry about heat transfer during one oscillation. Thus, we can use the adiabatic equation of state to show that: $$ C_{s} \equiv \frac{ \partial P }{ \partial \rho } = \frac{ \partial }{ \partial \rho } \left( P \ \rho^{-\gamma} \right) = \frac{ \gamma \ P }{ \rho } \tag{1} $$

A sound wave doesn't care about or even know about the shock or what's on the other side of the shock, so the irreversibility happening within the ramp is not relevant to our formulation of the sound speed here.

Its said that $c$ is a thermodynamic property and once its derived by any process it can be used for any other process which may or may not be same.

I think they are implying that consistency is key. That is, once you define a form for your speed of sound, as long as you consistently use that form thereafter, the form doesn't necessarily matter (within physically meaningful limits, of course). It's a roundabout way of saying what I said above, namely that the local speed of sound formula does not care that the shock is irreversible.