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As per Bryton Cycles, most of TurboJet engines' Thermodynamic efficiency should be around 40%. Rest of energy gets rejected as waste heat in exit gases of after the turbine.

But if these hot gases from the turbine are passed through Nozzle of 95% efficiency (either Convergent or CD), then overall efficiency should be 95% Right? Because Nozzle is scavenging all the waste heat and pressure, converting it into Kinetic energy of gases.

As per my knowledge nozzles are extremely efficient devices with efficiencies above 95% (Please do comment if that's not the case). And for the sake of discussion please assume Combustion efficiency as 100% and no mechanical loss. And here I'm considering whole purpose of turbojet engine it propell the gases as fast as possible. So overall efficiency here is kinetic energy of exit gases vs chemical energy released by fuel.

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    $\begingroup$ No, it isn't scavenging anything. If you place a 95% efficient nozzle over a candle flame, what does it do? $\endgroup$
    – Phil Sweet
    Jun 22, 2019 at 2:50

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Thruth to be told, you cannot have an engine 95 % efficient. Why? You have a combustion process and combustion is irreversible (entropy rises).

And no, you cannot calculate overall efficiency like that. In a turbojet you calculate it this way:

https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node85.html

Read this too: How efficient is a turbojet engine

How do you intend to recover the waste heat? It's WASTE, its exergy is too low!

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The complete Brayton cycle of an aircraft gas turbine engine has a couple of sources of energy loss - loss from the propulsive thrust power that is. The nozzle you mention is only part of the whole cycle, and yes can be very efficient if shaped properly.

from an old uni book, English terms are my own translations

Above image quantifies the energy streams for a bypass engine with a high bypass ratio. Every stage in energy transformation comes with a loss, quantified in the stage efficiency. The stages in the energy transformation are:

  1. Chemical energy => heat. Has a very high efficiency, heat radiation to the environment is minimal, as is CO and soot formation.
  2. Heat energy => gas power. We're converting heat into useful power, which can never be fully accomplished: after full expansion, gas temperature has increased.
  3. Gas power => propulsive power. Either into shaft power for turboprops/turboshafts, or into kinetic power for jet engines. Cannot occur loss free (isentropically) due to heat transfer, friction losses etc.
  4. Propulsive power (of the medium = air, combustion gasses) into thrust power (of the aircraft). The medium is propelled with a higher velocity than that of the aircraft, which is an efficiency loss. A necessary one of course.

End efficiency of the whole cycle is about 34%, probably a bit higher nowadays since the depiction is several decades old. Notice that the jet nozzle (exhaust) mentioned in the OP is only one aspect, with a high efficiency indeed.

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