Why do we take outlet/exhaust pressure (station 6) as the same as in inlet (station 1) ?

In T-S diagram, we see that outlet temperature is much higher than inlet temperature since flow in outlet has higher kinetic energy. It is hot and has more velocity than inlet.

So I know that static pressure of inlet and outlet is the same, but we use stagnation pressure/temperature in station 1 and 6, so I think that stagnation pressure of station 6 is higher than 1 because it has more velocity and temperature.

What do you think?

enter image description here

  • $\begingroup$ What do you think the pressures at #1 and #6 are? Why would they be different from ambient? $\endgroup$
    – jwh20
    Commented Mar 22, 2023 at 15:30
  • $\begingroup$ EPR, or Engine Pressure Ratio, is an interesting subject. It seems, in reality, EPR > 1 as more throttle is applied. In theory, if P6 > P1, V6 may be less than optimal. But jets seem to like EPR more around 1.2 - 1.4. May have something to do with less flow turbulence around the nozzle exit. Learning more about it at present. $\endgroup$ Commented Mar 24, 2023 at 8:55

2 Answers 2


My memory on this is a little foggy -- but I think your observation is correct.

When doing cycle analysis, I think we apply a static pressure condition to the nozzle exit (not stagnation).

I think it is drawn that way in the cycle diagram because eventually the air returns to ambient conditions (after lots of mixing and heat transfer). The dotted line is basically a part of the overall cycle that we don't model in detail. I wouldn't assume it follows a strict constant stagnation pressure line.


Well, the trick to keeping that thing running is to keep everything moving.

Piston engines are batch thrust producers, and jets are continuous thrust producers.

Piston engines produce thrust?

What would you call a linear force in a given direction? How about this: $\vec{F}$. On the pistons. Which turn the crankshaft.

Piston thrust over time produces an average force.

Look at a radial engine. Look at 4 of them bolted together.

The mass of the moving engine parts and the prop also help smooth things out by acting as a momentum conserving flywheel.

because piston engines are used to create circular mechanical motion, their Force output is measured in torque, or pound-feet.

Or ... Horsepower as pound-feet/second.

all by expanding the volume of gas by heating it.

Pressure × Volume = Gas Density x Temperature

As you heat gas either the pressure or the volume will increase. After spark and combustion, increased pressure in the cylinder creates the FORCE that moves on the piston head until cylinder pressure is closer to incoming pressure.

the result is higher VOLUME of hotter gas

Then pushed out into the exhaust, it's work done.

Now on to jets. They, as everyone knows create thrust ... and their thrust is also unaffected by velocity ... and they also create mechanical force by expanding gasses ... which turn a turbine ... which can (for turboprops) be measured as pound-feet or torque per second!

so what is it about jets?

The flow in and out must be continous or the jet will not run properly. While piston engines make thrust in little batches$^2$ within a closed cylinder with variable pressure, jets produce a continous "push" on the turbine with constant pressure.

This is a very delicate balancing act, and a real challenge for early jet designers. If incoming and exiting pressures are out of balance, the compresser can "surge" or stall and the continous thrust is lost.

This is why computers, such as FADEC, have greatly reduced the frequency of these events occurring.

Theoretically, the faster exit gasses should be of lower pressure, and indeed, pressure reduces passing through the energy extracting turbines, but the jet engine operator is most concerned with keeping flow rate through engine, and pressure in the engine, steady by adjusting intake and exhaust volume and temperature.

Pressure = Density × Temperature/Volume

As in piston engines, the exhaust pressure may be slightly higher than inlet. And like a piston near the end of its power stroke, the pressure difference is so little that:

it isn't worth the weight of an extra turbine to try to extract any more mechanical energy

So it goes out the back as ... thrust!$^1$

$^1$ design depends on the airspeed. At lower airspeeds, turboprops (and helicopter rotors) are far more efficient than turbojets, so instead of just making thrust as (m$_{\text{2}}$V$_{\text{e}}$) - (m$_{\text{1}}$V$_{\text{o}}$), the mechanical energy from the turbine is used to turn a rotating airfoil.

$^2$ these "little batches" of thrust produce a "pulse" in exhaust pressure, which can be "tuned" to increase efficiency of 2 or 4 cycle engines.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .