From Carnot's theorem, we know that efficiency is

$$ \eta = 1-\dfrac{T_c}{T_h}$$

where $T_c$ is the temperature of the cold reservoir and $T_h$ the temperature of the hot reservoir. But what do we have to consider for both?

Is $T_c$ the temperature of the air coming into the compressor from the fan or the temperature of air coming into the combustion chamber after being compressed and heated by the compressor?

Is $T_h$ the temperature of the gases in the combustion chamber (1500C°) or at the exhaust nozzle?


1 Answer 1


The Carnot efficiency is the maximum attainable efficiency in terms of extracting work from a temperature differential, not the actual efficiency in a turbine engine. For the actual efficiency, you should look into the Brayton cycle, which approximates a turbine engine much better.

With that out of the way, let's look at the Carnot cycle. This cycle assumes a heat source that can provide arbitrary amounts of energy at a certain temperature $T_h$, and a heat sink that can sink arbitrary amounts of heat at a temperature $T_c$. These are the isothermal parts of the cycle. To get the working fluid to either temperature, adiabatic compression is used.

These heat sources and sinks do not exist in reality, but by approximation, the atmosphere is an excellent heat sink, and burning fuel is an excellent heat source. The low temperature is thus taken as the atmospheric temperature which can be as low as -50°C (~220 K, remember that Carnot's efficiency is defined in terms of absolute temperature) and the high temperature is the temperature in the combustion chamber.

Sometimes you hear that $T_h$ is the hottest temperature in the cycle, and $T_c$ the coldest temperature. While this is often true in practice, this does not need to be the case for theoretical considerations. The only thing that matters for the Carnot efficiency is the temperature of the heat sinks/sources. You can't get more efficient than the Carnot cycle by hooking your engine up to a fridge. Indeed, the Carnot cycle is just a way to express the Second Law of Thermodynamics, which states that entropy in a closed system can never increase.

  • $\begingroup$ I see. But if we, for example, consider the outside air temperature to be -50C° and the temperature in the combustion chamber to be 1500C° for the Carnot theorem we would obtain an efficiency over the value of 1, which is clearly not correct. Looking at the Brayton Cicle, I see effciency depends on pressure ratios. But what pressures does it consider? Outside air pressure? Pressure in the compressor and combustion chamber, or at the exhaust nozzle? $\endgroup$ Jun 6, 2018 at 9:18
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    $\begingroup$ @franycio Carnot efficiency is expressed in kelvin (or really, any absolute temperature scale), not in degrees Celsius. Try again with 223K and 1773K. $\endgroup$
    – Sanchises
    Jun 6, 2018 at 9:23
  • $\begingroup$ right, forgot that very detail. Thank you. $\endgroup$ Jun 6, 2018 at 9:26
  • $\begingroup$ @franycio Regarding the pressure ratios, this is also based in theoretical considerations, namely, a heat sink at a low pressure and temperature, and a heat source at high pressure and temperature. So here too take the combustion chamber and the atmosphere pressures. Take a look at thermodynamic p-V diagrams and really try to understand what's happening and how that applies to reality (for example, the combustion chamber in a turbine does not tend to have a large pressure differential - it's just a chamber - so whatever process happens there must be approximated as isobaric). $\endgroup$
    – Sanchises
    Jun 6, 2018 at 9:32

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