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In $P=ρRT$, ρ is density, R is the gas constant, and T is the temperature in Kelvin. In the boundary layer, the temperature and density varies, so why doesn't the pressure?

If you have a slip line in a shock system, the density might be different across it and the pressure might be the same. This is different than the pressure/density/temperature relation in the BL, because the air went through different shock systems in the slip line example, so the properties are bound to be different.

Basically, in this specific example, I'm not sure how increased temperature/density wouldn't increase the pressure.

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    $\begingroup$ downvoter, please explain why you downvoted. I can't improve anything unless I know what's wrong with this question... $\endgroup$
    – Wyatt
    Mar 25 at 23:22

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Because the pressure is constant across the boundary layer.

If the pressure at the wall was higher, it would do something strange -- such as push the streamlines away from the wall.

The pressure at the wall of a boundary layer is set by the pressure on the outside of the boundary layer. Period. Full Stop. You don't get to set it twice.

Increasing temperature does not increase pressure.

The fact that $P$ is on the left in $P=\rho R T$ gives it no special place in the world. We just prefer multiplication to division, we can equally well write $T=\frac{P}{\rho R}$ or $\rho=\frac{P}{T R}$.

You might be thinking about Boyle's law, Charles's law, or other such things. These are each special cases of the $P=\rho R T$ gas law. They are limited to constant volume, constant pressure, or constant something else.

Edits:

There are lots of situations we can think of where pressure and volume will be related. Here are four containers of gas....

enter image description here

The first is totally sealed and rigid.

It is a constant volume container. Since it is sealed, it is also a constant density container. If you heat it up, the pressure will change because the density is constrained to be constant.

The second has an idealized piston. We imagine that this piston creates a perfect seal, but has no friction and no mass.

This is a constant pressure container. The pressure inside the container will always be equal to the ambient pressure outside the container. The amount of gas inside the container (mass) is constant, but if you (for example) cool the gas, it will shrink and the piston will move down to maintain constant pressure.

The third has an idealized piston, but there are a stack of weights on top of the piston. It also forms a constant pressure volume, but this time the pressure is the sum of the ambient pressure and also the amount of weight divided by the area of the piston.

The fourth is a fun one. On top of our idealized piston, there is a spring that is rigidly fixed at the top. What do you think the pressure-volume relationship will be for this case?

These are simple examples of physical systems where it is easy to see and understand the relationship between pressure and volume (and hopefully thereby density and pressure).

In all of these cases, $P=\rho R T$.

In each of these cases, something else determines the relationship between pressure and volume (or whatever the process entails.) Here, that is determined by the structural rigidity of the box, by the behavior of the piston, by the stack of masses, or the deflection of the spring. I.e. additional physical principles and equations are needed to specify the PV relationship.

In fluid dynamics, the additional physical principles that determine the PV or P$\rho$ relationship are the conservation of mass, momentum, and energy.

BTW, The best (and easiest) way to render $\rho$ here is \$\rho\$.

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  • $\begingroup$ Oh okay, thanks for your answer. If you do an example equation for P = ρRT, you get a certain number for P. If you increase T, in my case, I got a higher number for P. Does that not apply for the BL? Or does it not apply for the BL because it's not a confined space? $\endgroup$
    – Wyatt
    Mar 26 at 1:33
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    $\begingroup$ @Wyatt As long as the flow is free to expand, all a temperature increase does is to lower density, so more volume is taken up by the same amount of air. P will only increase if volume is constrained (such as inside a pressure cooker). $\endgroup$ Mar 26 at 4:25
  • $\begingroup$ The boundary layer is not a confined space. Instead, it is a constant pressure space. $\endgroup$ Mar 26 at 5:04

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