My professor usually asks us in our exams to plot the Thrust-Pressure ratio and TSFC - Pressure Ratio charts and justify the tendencies observed.

(TSFC = Thrust Specific Fuel Consumption).

We all know that, as the compressor pressure ratio ($\pi_c$) increases, the higher the Thrust ($F$) and the lower the $TSFC$. Here is a graph (which was taken from the book Fundamentals of Jet Propulsion with Applications, by Ronald D. Flack).


This chart was made after taking the final expressions for $F$ and the $TSFC$ (which are huge) and plotting them against $\pi_c$ using a computer. Obviously we don't have computers in the exam, nor we have enough time to derive such complex formulas. Here is an example for the (non-dimensionalized) $TSFC$:


Instead, I want to impress him with a convincing, non-mathematical, qualitative explanation of this behaviour.

Therefore my question reduces to: why, physically speaking, does the thrust increase with $\pi_c$? (The answer to "why does the TSFC decrease with $\pi_c$ can be answered saying that $F$ and $TSFC$ behave inversely to one another).

This is my attempt: As the compressor pressure ratio increases, the flow will be mixed and burned more efficiently in the combustion chamber, therefore increasing the energy available for the nozzle to accelerate the flow. Since the jet thrust increases with the exhaust speed, we can rest assured that a higher pressure ratio will translate into a higher thrust. The $(TSFC-\pi_c)$ behaviour will be inversely proportional to $(F-\pi_c)$, since $TSFC=\dot{m}_f/F$

NOTE: My explanation of the process, I think is a bit incomplete. For example, it doesn't explain why there is a $\pi_c$ for which $F$ will actually start decreasing (in the graph, this $\pi_c$ value is around 15).


1 Answer 1


Because the constant pressure lines are diverging with increasing enthalpy.

Have a look at the right most diagram. The two lines are constant pressure lines, with the higher line indicating the higher pressure, and as you can see, they are diverging.

1 is the beginning of the cycle, thus at ambient temperature

From 1 to 2, we increase the temperature, and go to a higher pressure in the compressor. This will cost some energy, which is delivered by the turbine.

From 2 to 3 we increase the Temperature at the the same pressure by adding energy.

From 3 to 4, we decrease the temperature and pressure in the turbine (and use some of the extracted energy to drive the turbine)

From 4 to 1, we use the energy in the flow to propel us, and we go back to ambient temperature.

In principle we could also skip the higher pressure, and simply go from 1 to 4 and back to 1 (which is just heating air, and propelling yourself by the heat.)

But by moving to a higher pressure, we can take advantage of the divergent pressure lines. This yields us a benefit because the distance 3-4 is larger than 2-1. This means that if we use the turbine to drive the compressor, we gain some 'free' energy.

enter image description here


Following OSUZorba's comment:

To illustrate the point of the 'free energy', look at the image below, here two extra stations are added the inlet, which isn't that important for now, and the nozzle. Note that the vertical distances of the compressor and the turbine are equal (they have to be, because one drives the other), thus $$ \begin{aligned}P_{comp} = P_{turb} &\Rightarrow \Delta T_{comp} = \Delta T_{turb} \\ &\Rightarrow (T_3 - T_2) = (T_4 - T_5) \end{aligned}$$

But because of the divergent nature of the lines, we have some extra energy left. We use the nozzle to optimally expand the high pressure, high temperature flow and extract the 'extra' energy.

enter image description here


  • $\begingroup$ So, the higher the pressure $p_2-p_1$, the higher $h_3-h_4$. Am I right here? This means that the turbine is working more efficiently, because it is able to extract more energy from the incoming flow. But I can't see a direct relationship with the thrust. Can you help here? Thanks for your awesome answer. $\endgroup$ Commented Jun 15, 2016 at 12:56
  • $\begingroup$ Oh, and by the way, would you be able to explain why there is a point at which further increasing the pressure won't increase the thrust, but actually decrease it? $\endgroup$ Commented Jun 15, 2016 at 12:58
  • $\begingroup$ I think, if you further increase the pressure, you might run into other issues. The general limiting factor is point 3, there is only so much temperature the metal in your combustion chamber (and the following turbine) can handle. Another issue might be the compressor itself, it might not be able to reach extremely high pressures with flow separation at the blades. $\endgroup$
    – ROIMaison
    Commented Jun 15, 2016 at 13:07
  • 1
    $\begingroup$ This is a very good response, but in a non-turboshaft engine there is another point between 3 and 4, which is the exit of the turbine and entrance of the nozzle. So as the pressure ratio increases less of the energy in the turbine is used to drive the compressor and more is available for thrust $\endgroup$
    – OSUZorba
    Commented Jun 15, 2016 at 14:15
  • 1
    $\begingroup$ Now I understand everything clearly!! Thank you ROIMaison. Also thanks to @OSUZorba for chiming in and adding extra value to the answer. $\endgroup$ Commented Jun 15, 2016 at 14:39

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