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The gas generator & Brayton cycle

At the heart of every turbine engine sits the same three-part machine — compressor, combustor, turbine — called the gas generator. Its thermodynamic cycle is the Brayton cycle: the jet-engine counterpart of the piston course’s Otto cycle, with one big difference — everything happens continuously, in separate places, at constant pressure rather than constant volume.

The station numbers

Propulsion people label positions along the engine with standard numbers, and the demos’ readouts use them constantly, so learn them once:

Fig. 1. Standard stations: 0 freestream, 2 compressor face, 3 burner entry (compressor exit), 4 turbine entry — the hottest, most precious gas in the machine — 5 turbine exit, 9 nozzle exit.

Temperatures come in two flavours. Static temperature $T$ is what a thermometer riding with the flow reads. Total (stagnation) temperature $T_t$ is what the gas would reach if brought smoothly to rest — static plus the kinetic energy converted back to heat:

T_t = T\left(1 + \frac{\gamma-1}{2}M^2\right)

(1)

Cycle analysis lives almost entirely in total quantities (that’s what the subscript in $T_{t4}$ means) because total temperature only changes when you add heat or do work on the flow — exactly the two things an engine does.

The Brayton cycle

0 → 3: compression. The inlet slows the flow (ram compression, free) and the compressor squeezes it further, spending shaft work. Ideally isentropic.
3 → 4: heat addition at constant pressure. Fuel burns in the combustor. Nothing pushes a piston here — the flow simply gets hot at (ideally) unchanged total pressure.
4 → 9: expansion. The turbine extracts exactly enough work to drive the compressor; the nozzle turns every remaining degree of temperature into jet velocity. Ideally isentropic.
9 → 0: heat rejection. The exhaust cools back to ambient — outside the engine. The sky is the radiator.

Fig. 2. The Brayton cycle on temperature–entropy axes. Vertical legs are the ideal (isentropic) compression and expansion; the curved legs are the two constant-pressure processes. Enclosed area = net work per kg of air.

The efficiency — same trick as Otto

Both heat exchanges are at constant pressure, so $q = c_p \Delta T_t$ , and:

\eta_{th} = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{T_{t9} - T_{t0}}{T_{t4} - T_{t3}}

(2)

The two isentropic legs span the same pressure ratio $\pi$ , and isentropic compression ties temperature to pressure by $T_t \propto p_t^{(\gamma-1)/\gamma}$ . Exactly as in the Otto derivation, the temperature differences cancel, leaving:

\eta_{th} = 1 - \frac{1}{\pi^{(\gamma-1)/\gamma}} \;=\; 1 - \frac{1}{\tau_r\,\tau_c}

(3)

$\pi$: overall pressure ratio, inlet ram × compressor (πr·πc) [–]
$\tau_r$: ram temperature ratio 1 + (γ−1)/2·M₀² [–]
$\tau_c$: compressor temperature ratio = πc^((γ−1)/γ) [–]

Overall pressure ratio

\pi

\eta_{th} = 1 - 30^{-(\gamma-1)/\gamma}

\eta_{th} = 62.2\,\%

Modern military and single-aisle territory. Every extra bar of squeeze pays off directly in efficiency.

Where the turbine-inlet temperature comes in

Equation (3) says efficiency needs pressure ratio — it never mentions $T_{t4}$ . What $T_{t4}$ buys is specific work: the hotter the turbine entry, the more net work each kilogram of air yields, so the smaller and lighter the engine for a given thrust. The ceiling is brutal: modern $T_{t4}$ runs 1700–1900 K, hundreds of kelvin above the melting point of the turbine blades, survivable only through internal cooling channels, film-cooling holes and ceramic coatings. Every readout in the six demos treats $T_{t4}$ as the throttle, because in a real engine fuel flow is exactly what sets it.