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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.

Wyatt
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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.)

Rob McDonald
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  • 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? – Wyatt Feb 02 '24 at 18:53
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    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. – Rob McDonald Feb 02 '24 at 19:37
  • ah, I understand. Do you know where I could see more about entropy? I don't quite understand how it works. – Wyatt Feb 02 '24 at 19:54
  • 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. – Rob McDonald Feb 02 '24 at 20:24
  • I see. Last thing I'll ask before I read more about that subject : Increases in temperature always comes with an increase in pressure, right? (From what I've read so far, entropy is basically how random the molecules are, or that's what I've interpreted it as) If that's the case, then after the lambda shock, pressure shouldn't be equal on both sides of the slip line even though it is? (because higher temperature comes with higher pressure, and both sides of the slip line experience different temperatures) – Wyatt Feb 02 '24 at 20:53
  • "Increases in temperature always comes with an increase in pressure, right?" Wrong. Look at the ideal gas law. There are three other variables in the equation besides P and T. – Organic Marble Feb 02 '24 at 20:55
  • Both density and temperature are different across the slip line. Pressure is the only thing equal. The product of temperature and density must be equal before/after. You are confused because these ideas are often introduced with Boyle's law and Charles' law and Gay-Lussac's law. These are simplified versions of the ideal gas law -- but they only apply in special conditions such as constant temperature, constant pressure, or constant volume (say a sealed bottle of gas). The general solution is the ideal gas law $P=\rho,R,T$ or $P,v=n,R,T$. – Rob McDonald Feb 02 '24 at 21:10
  • Confusion can arise with $P=\rho,R,T$ vs. $P,v=n,R,T$. Aerospace engineers tend to use the former version, with R being a specific gas constant for air and density being the inverse of specific volume. Scientists tend to use the latter form with volume and the universal gas constant R, but then they must know how many particles they are dealing with (which is never a real thing). You can also use a version in terms of specific volume and molecular weight. These are all the same idea, get comfortable with the one(s) used in your field. – Rob McDonald Feb 02 '24 at 21:15
  • I see, thanks for the modified version of the equation. So in P=pRT, the pressure is P, the density is p, R is the gas constant, and T is temperature. Considering this, that's all pressure relies on. So behind a shockwave, the density is higher, the gas constant stays the same (I think), and the temperature is higher. What I'm getting at is this : What in that equation differs and changes to make the pressure the same above and below the slip line? What values have to be what, if that makes sense. (also I'm probably wrong about some of those things) – Wyatt Feb 02 '24 at 23:25
  • The flow went through different shocks. Different shocks with different strengths cause different jumps in properties. Not all shocks are the same. – Rob McDonald Feb 03 '24 at 02:20
  • Ah I see, makes sense. This is the last little thing I was curious about; I don't want to bore you with questions. : R doesn't stay the same as said in your answer. What is the fundamental thing that makes it change? (definitely my last follow up question, I scrolled up and I've asked a lot of them) – Wyatt Feb 03 '24 at 02:49
  • You should read the Wikipedia article for Gas Constant. R for a specific gas is equal to the universal R divided by the molar mass of the gas. Of course air is not a single gas, but a mixture of gases. The average molar mass of air is about 28.964917. Air is 78% Nitrogen, Nitrogen's molar mass is about 28. Air is 21% oxygen. Oxygen's molar mass is about 32. R will be different when you consider the mixture of combusted gases exiting an engine. R is different when you consider a helicopter flying in the mostly C02 atmosphere of Mars. – Rob McDonald Feb 03 '24 at 04:18
  • R will also be different in the extreme environment around a spacecraft re-entering the earth. In particular, in a hypersonic flow, the bow shock around the spacecraft can be so strong that the gas will get so hot that Oxygen $O_2$ and Nitrogen $N_2$ will disassociate into $2O$ and $2N$. When this happens, the average molar mass changes because there are now twice the number of particles for the same mass. – Rob McDonald Feb 03 '24 at 04:23
  • Oh okay, cool. (well re-entry isn’t very cool). That Wikipedia article definitely answered my questions, thanks for that. I find it intriguing that so much heat will change the molecular makeup of the air. Thanks for all of your intuition, I greatly appreciate it! – Wyatt Feb 03 '24 at 05:02
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    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. – Rob McDonald Feb 03 '24 at 05:57