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Why are pressures equal across the slip line of a lambda shock, but density is not? In this answer, it says this :

The surface Σ is a slip line between zones 3 and 4. Velocities will be parallel across that line. Pressures will be equal across the slip line. Temperature, density, entropy, etc. will be different across the slip line.

So why are the pressures equal across the slip line? I thought low density almost always came with low pressure, and vice versa. In that case, how could you have different density's with the same pressures? Thanks.

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As I mentioned in one of my many answers / replies -- think of the slip line like an invisible thin membrane.

Velocity must be parallel because otherwise it would pierce the membrane.

Pressures must be equal because otherwise they would push the membrane up/down.

Everything else (density, temperature, enthalpy, etc.) can be different.

This is compressible flow. When you go through a shock, you have an abrupt change in properties -- including a jump in entropy. Therefore Bernoulli's equation does not work across a shock.

When you're considering a lifting flow with shocks, the flow around the upper and lower surface will go through different shocks -- the amount of entropy change will be different for the top and bottom surface flows. Therefore, Bernoulli's equation can not be used to compare top/bottom flows.

$P=\rho\,R\,T$

If $P$ and $\rho$ are not going hand-in-hand, then it must be because Temperature is different (or you've had a change in gas composition, $R$, but if that has happened, you've also had changes in temperature.)

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  • $\begingroup$ I see, thanks. So put simply, the shocks change the gas composition/properties, that changes the entropy along with other things, allowing the sides of the slip line to have different density's? $\endgroup$
    – Wyatt
    Commented Feb 2 at 18:53
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    $\begingroup$ A high temperature shock will change the composition and properties more. A more mild shock won't change the composition -- but it certainly will add entropy and will cause the temperature to jump. The entropy change is enough to make Bernoulli not work. The temperature change shows how P and rho can vary differently. $\endgroup$
    – Rob McDonald
    Commented Feb 2 at 19:37
  • $\begingroup$ ah, I understand. Do you know where I could see more about entropy? I don't quite understand how it works. $\endgroup$
    – Wyatt
    Commented Feb 2 at 19:54
  • $\begingroup$ In the context of compressible flow, you'll want to look at a Gas Dynamics book. In general, there are three things in a fluid flow that are not isentropic. 1) friction - in the boundary layer, entropy changes. 2) work addition - flow that goes through an engine, propeller, or wind turbine. 3) shocks. Maybe there are some others, but those three will get you most of the way. $\endgroup$
    – Rob McDonald
    Commented Feb 2 at 20:24
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    $\begingroup$ The ideal gas law can actually be derived from theory based on a bunch of particles bouncing around in a box. The same path will also lead to a derivation of the speed of sound $a=\sqrt{\gamma\,R\,T}$. Knowing this helps to explain why changing the number of particles is such a big deal -- 2x the particles, many more collisions, very different behavior. $\endgroup$
    – Rob McDonald
    Commented Feb 3 at 5:57

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