This might just be a misunderstanding on my part, but in an ideal Brayton cycle, the flow through the compressor and turbine is isentropic, meaning $\Delta s = 0$. However, usually you are given the compressor pressure ratio to calculate the exit stagnation pressure, i.e. $$ p_{cr} \equiv \frac{p_{03}}{p_{02}}, $$ where station 3 is the exit and station 2 is the inlet. This makes completely no sense to me. In an isentropic process, the stagnation pressure is constant, so I don't understand how there can be a ratio. I do understand that the pressure increases in the compressor due to compressing the air and increasing the temperature, but these two points of view seem to be contradictory.

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By stagnation pressure I assume you mean total pressure. Stagnation is not a term commonly used in gas turbine engine performance, it’s more common with aerodynamics. The compressor increases the total pressure because the rotating blades are doing mechanical work on the fluid.

A couple of other comments, to explain in more detail;

First, the flow is not isentropic in a compressor. It only would be if a compressor was 100% efficient. The efficiency is actually calculated by the degree to which it is not isentropic. The efficiency is given by the isentropic temperature rise that would occur for the given total pressure rise, divided by the actual total temperature rise. An efficient compressor obtains the same pressure rise, for a smaller temperature rise. While you might think a higher exit temperature would be good, because it would mean we could burn less fuel, a greater temperature rise across the compressor would require more shaft power to compressor the air, which is taken from the turbine and means the turbine exit gas is now cooler and a lower pressure and now cannot produce the same thrust.

Secondly, the stagnation pressure is not constant (even if it was 100% isentropic). This maybe the case in a fluid flow where no energy exchange is occurring, but in a compressor, the rotating blades are doing work on the fluid. This increases its total pressure. The area at the entrance and exit determine its velocity (along with its density). Different areas, changes in density, and total pressure mean the static pressure isn’t the same at the entrance and exit, either.

I suspect you were trying to match the realities of an actual jet engine with the ideal conditions of a Brayton cycle. It’s kind of like pointing out the air is, in fact, not an ideal gas. Well true the behavior of each of those things is very close to the ideal models that we created for them

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