Efficiency is fixed by the pressure ratio
For constant specific heats and isentropic compression and expansion, the Brayton thermal efficiency is η = 1 − 1 / r_p^{(γ−1)/γ}. At r_p = 10 and γ = 1.4 that is about 48.2%. If you enter a compressor-inlet temperature T1, the isentropic compressor exit is T2 = T1 r_p^{(γ−1)/γ}; an optional turbine-inlet T3 then gives T4 = T3 / r_p^{(γ−1)/γ}, a heat input c_p(T3−T2), and a net work c_p((T3−T4) − (T2−T1)). The efficiency implied by those temperatures matches the pressure-ratio formula on the ideal cycle. This page uses c_p = 1,004.5 J/(kg·K) for the optional work numbers only.
A real engine has burner pressure loss, cooling bleed, and component efficiencies that all cut η. For a control-volume thrust from mass flow and velocity, see jet engine thrust; for a rocket, see specific impulse.
Sources
- Cengel and Boles, Thermodynamics: An Engineering Approach, air-standard Brayton cycle.
- Hill and Peterson, Mechanics and Thermodynamics of Propulsion.
- Mattingly, Elements of Propulsion.
Frequently asked questions
Why doesn’t T3 appear in the simple efficiency formula?
On the ideal cycle with constant cp, η depends only on the pressure ratio (or equivalently on T1/T2). T3 changes the work and the heat input in the same proportion.
What if T3 is lower than T2?
That is not a heat-engine cycle — you would be “adding” negative heat. The page rejects it.
Is this a turbofan or a ramjet?
It is the ideal Brayton (gas-turbine) cycle. A ramjet is a related Brayton process without the turbomachinery, and a turbofan adds a bypass stream this page does not model.