The Brayton Cycle — Thermodynamics of a Jet Engine
The Brayton Cycle — Thermodynamics of a Jet Engine
Every commercial jet, military fighter, and gas-turbine power plant runs on the same thermodynamic cycle, discovered by engineer George Brayton in the 1870s. Four simple processes — compress, heat, expand, exhaust — turn fuel into thrust, and every step is governed by equations you can write on a napkin.
The Four Processes
- 1 → 2: Isentropic Compression. The compressor squeezes air through a pressure ratio $r_p = P_2/P_1$. No heat is added — the temperature rises purely from work done on the gas.
- 2 → 3: Constant-Pressure Heat Addition. Fuel burns continuously in the combustor, raising temperature at constant pressure up to the turbine inlet temperature $T_3$ — the single most important design limit in any gas turbine.
- 3 → 4: Isentropic Expansion. Hot gas expands through the turbine (and then the jet nozzle), doing work and accelerating the flow.
- 4 → 1: Heat Rejection. Exhaust leaves the engine; fresh air enters. The atmosphere acts as the heat sink.
The Key Equation — Isentropic Temperature Ratio
For air (ideal gas, $\gamma = 1.4$), isentropic compression through pressure ratio $r_p$ raises temperature by:
$$T_2 = T_1 \cdot r_p^{(\gamma-1)/\gamma} = T_1 \cdot r_p^{0.286}$$
At $r_p = 40$: $T_2 = 288 \times 40^{0.286} \approx 288 \times 4.18 \approx 1{,}204$ K — the air reaches over 930°C before any fuel is burned! That's why modern compressors are made of exotic titanium alloys and cooled with bleed air.
Thermal Efficiency — Why Higher Pressure Ratio Wins
The ideal Brayton cycle thermal efficiency depends only on pressure ratio:
$$\eta_{th} = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}} = 1 - \frac{T_1}{T_2}$$
This is the fraction of heat input converted to useful work. Real engines add compressor and turbine isentropic efficiencies ($\eta_c$, $\eta_t$) to account for friction and flow losses:
$$T_{2,\text{actual}} = T_1 + \frac{T_2 - T_1}{\eta_c}, \qquad T_{4,\text{actual}} = T_3 - \eta_t (T_3 - T_4)$$
Inlet: $T_1 = 288$ K, $r_p = 32$, $T_3 = 1{,}550$ K, $\eta_c = 88\%$, $\eta_t = 90\%$. Find thermal efficiency and net specific work.
- Ideal: $T_2 = 288 \times 32^{0.286} = 288 \times 3.84 = 1{,}106$ K
- Actual:
- $T_{2a} = 288 + (1{,}106 - 288)/0.88 = 288 + 930 = 1{,}218$ K
- Ideal: $T_4 = T_3/r_p^{0.286} = 1{,}550/3.84 = 403$ K
- Actual: $T_{4a} = 1{,}550 - 0.90 \times (1{,}550 - 403) = 1{,}550 - 1{,}032 = 518$ K
- $W_{net} = c_p[(T_3 - T_{4a}) - (T_{2a} - T_1)] = 1.005[(1{,}550-518) - (1{,}218-288)] = 1.005 \times 102 = 102$ kJ/kg
- $\eta_{th} = W_{net}/Q_{in} = 102/[1.005 \times (1{,}550-1{,}218)] = 102/334 = 30.5\%$
The GE9X engine (Boeing 777X) achieves an overall pressure ratio above 60:1 and turbine inlet temperatures above 1,700°C — pushing materials to their absolute limits. Every percentage point of efficiency improvement saves airlines millions of dollars in fuel annually. The math is the same; the engineering is extraordinary.