The Brayton Cycle and the Ideal Turbojet

The Brayton cycle, its thermal efficiency, temperature-ratio analysis of the ideal turbojet and the best compressor ratio

RAF Cranwell — 15 May 1941

Gloster's chief test pilot, Gerry Sayer, takes the small Gloster E.28/39 off from the long runway at RAF Cranwell in Lincolnshire and flies it for 17 minutes on the power of Frank Whittle's jet engine (the E.28/39). It is Britain's first jet flight. In Germany, Hans von Ohain's engine had flown in the Heinkel He 178 on 27 August 1939, for six minutes (the He 178). Whittle had patented the idea in January 1930. In January 1935 he could not afford the £5 renewal fee, the Air Ministry refused to pay it, and the patent lapsed (Frank Whittle). Six years later the engine was in the air. His W.1 squeezed air about 3.8 times, burned kerosene in it, and spent some of the hot gas's energy turning the compressor. This lesson asks what such an engine could do if every one of those steps were perfect.

Every gas turbine runs the same cycle, named after the American engineer George Brayton: compress the air, heat it at constant pressure, expand it. In a turbojet the expansion happens in two parts. The turbine takes out just enough work to drive the compressor; the nozzle turns the rest into jet velocity.

Ideal cycle analysis keeps the cycle and removes every loss, so that what remains is the limit set by thermodynamics alone.

💡What 'ideal' means here

Air is a perfect gas with constant $c_p$ and $\gamma$ throughout (1004.5 J/(kg·K) and 1.4).
Compression and expansion are isentropic; the burner has no pressure loss.
The fuel mass is small compared with the air ( $f \ll 1$ ), and its heating value is $h_{PR} = 43$ MJ/kg.
The nozzle expands the jet exactly to ambient pressure, $p_9 = p_0$ .

Results are written with temperature ratios:

\tau_r = T_{t0}/T_0 = 1 + \frac{\gamma - 1}{2}M_0^2

(ram),

\tau_c = T_{t3}/T_{t2} = \pi_c^{(\gamma - 1)/\gamma}

(compressor),

\tau_t = T_{t5}/T_{t4}

(turbine) and

\tau_\lambda = T_{t4}/T_0

(the burner's temperature limit, relative to ambient).

\eta_{th} = 1 - \frac{1}{\tau_r\,\tau_c}

The thermal efficiency of the ideal Brayton cycle. It depends only on the total compression temperature ratio, ram times compressor — not on Tt4.

Derivation

The Brayton efficiency

Heat goes in at constant pressure in the burner and comes out at constant pressure in the atmosphere, where the exhaust cools from

T_9

back to

T_0

\eta_{th} = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{c_p(T_9 - T_0)}{c_p(T_{t4} - T_{t3})}.

Compression from

p_0

p_{t3}

and expansion from

p_{t4} = p_{t3}

back to

p_0

span the same pressure ratio, so for isentropic processes they span the same temperature ratio:

\frac{T_{t3}}{T_0} = \frac{T_{t4}}{T_9} = \tau_r\tau_c \quad\Longrightarrow\quad T_9 - T_0 = \frac{T_{t4} - T_{t3}}{\tau_r\tau_c}.

Substituting gives

\eta_{th} = 1 - 1/(\tau_r\tau_c)

. At rest (

\tau_r = 1

) with

\pi_c = 30

1 - 30^{-0.2857} = 62.2\,\%

📐From the W.1 to the GE9X

The ideal Brayton efficiency at rest, against the overall pressure ratios of real engines (Lab 3 adds real-air properties from CoolProp, which take about 3 points off each value):

Whittle's W.1, about 3.8: 31.7 %
$\pi_c = 10$ : 48.2 %
$\pi_c = 30$ : 62.2 %
The GE9X, 60: 69.0 %

Sixty years of compressor development are, in this one equation, sixty years of thermal efficiency. The W.1 burned 38.5 g of fuel per kilonewton-second at full thrust (the W.1); the big turbofan of Lesson 1-1 needs 7.6.

The ideal turbojet, station by station. Follow a kilogram of air. The inlet brings it to rest relative to the engine (

\tau_r

), the compressor raises its stagnation temperature by

\tau_c

, the burner heats it to

T_{t4}

. The turbine must give back exactly the work the compressor absorbed, which fixes how far it expands the gas:

\tau_t = 1 - \frac{\tau_r}{\tau_\lambda}(\tau_c - 1).

Whatever stagnation pressure is left drives the nozzle, and the jet velocity follows.

\frac{V_9}{a_0} = \sqrt{\frac{2}{\gamma - 1}\,\frac{\tau_\lambda}{\tau_r\tau_c}\left(\tau_r\tau_c\tau_t - 1\right)}, \qquad \frac{F}{\dot m_0} = a_0\left(\frac{V_9}{a_0} - M_0\right), \qquad f = \frac{c_p T_0(\tau_\lambda - \tau_r\tau_c)}{h_{PR}}

Ideal turbojet: jet velocity, specific thrust and fuel–air ratio. TSFC = f/(F/ṁ0).

Derivation

Turbine work and jet velocity

Turbine work. With

f \ll 1

the same mass flows through compressor and turbine, so

c_p(T_{t4} - T_{t5}) = c_p(T_{t3} - T_{t2})

. Divide by

c_p T_{t4}

and use

T_{t2} = T_0\tau_r

T_{t3} = T_0\tau_r\tau_c

T_{t4} = T_0\tau_\lambda

1 - \tau_t = \frac{\tau_r(\tau_c - 1)}{\tau_\lambda}.

Jet velocity. The nozzle entry has

T_{t9} = T_{t4}\tau_t

, and its stagnation pressure ratio to ambient is the product of the ram, compressor and turbine ratios, so isentropically

T_{t9}/T_9 = \tau_r\tau_c\tau_t

. Then

T_9 = \frac{T_0\tau_\lambda\tau_t}{\tau_r\tau_c\tau_t} = \frac{T_0\tau_\lambda}{\tau_r\tau_c}, \qquad V_9^2 = 2c_p(T_{t9} - T_9) = 2c_p T_9(\tau_r\tau_c\tau_t - 1).

Dividing by

a_0^2 = (\gamma - 1)c_p T_0

gives the formula. Fuel. The burner energy balance

\dot m_f h_{PR} = \dot m_0 c_p(T_{t4} - T_{t3})

gives

f

📐An ideal turbojet in airliner cruise

Mach 0.85 at 11 km (

T_0 = 216.65

K),

T_{t4} = 1600

K (Lab 1):

$\pi_c = 20$ : $F/\dot m_0 = 910$ N·s/kg, TSFC = 26.1 mg/(N·s), $\eta_{th} = 62.9\,\%$ , $\eta_p = 35.5\,\%$ .
$\pi_c = 40$ : TSFC falls to 23.3 mg/(N·s) as $\eta_{th}$ rises to 69.5 %, but the specific thrust falls to 891 N·s/kg.

The jet leaves at about 1160 m/s from an aircraft flying at 251 m/s. The thermal efficiency is excellent; the propulsive efficiency is terrible.

💡The best compressor ratio for thrust

Efficiency rises with

\pi_c

forever, but specific thrust does not. Too little compression and the cycle converts little heat into work; too much and the compressor exit is already so hot that little fuel can be burned before

T_{t4}

. The work — and the thrust — peaks at

\tau_c^* = \frac{\sqrt{\tau_\lambda}}{\tau_r}.

At that point

T_{t3} = T_9 = \sqrt{T_0 T_{t4}}

, and the net work is

c_p\left(\sqrt{T_{t4}} - \sqrt{T_0}\right)^2

. For

T_{t4} = 1600

\pi_c^* = 20.6

in cruise at Mach 0.85, 4.2 at Mach 2, and 1 — no compressor at all — at Mach 2.93 (Lab 2).

Derivation

The optimum compressor temperature ratio

Expand

(V_9/a_0)^2\cdot\frac{\gamma - 1}{2}

using

\tau_t

from the work balance:

X = \frac{\tau_\lambda}{\tau_r\tau_c}\left(\tau_r\tau_c - \frac{\tau_r^2\tau_c(\tau_c - 1)}{\tau_\lambda} - 1\right) = \tau_\lambda - \tau_r(\tau_c - 1) - \frac{\tau_\lambda}{\tau_r\tau_c}.

Set

dX/d\tau_c = -\tau_r + \tau_\lambda/(\tau_r\tau_c^2) = 0

\tau_c^{*2} = \tau_\lambda/\tau_r^2

. The second derivative is negative, so this is a maximum. Lab 2 lets SymPy do the algebra.

It seems that perfection is attained not when there is nothing more to add, but when there is nothing more to remove.

– Antoine de Saint-Exupéry, Wind, Sand and Stars (1939), in the chapter on the aeroplane (Wikiquote)

💡Where you will use this

Lesson 3-2 adds a fan to the same core; 3-3 removes the compressor altogether. Unit 4 puts real component efficiencies into these formulas, and Unit 8 runs them off-design. The station numbers and stagnation-property bookkeeping used here are the language of every cycle code, from NASA's NPSS to the open-source pyCycle.

📚Further reading

Jack D. Mattingly and Keith M. Boyer, Elements of Propulsion: Gas Turbines and Rockets (2nd ed., AIAA, 2016), chapter on the parametric cycle analysis of ideal engines.
H. I. H. Saravanamuttoo, G. F. C. Rogers, H. Cohen, P. V. Straznicky and A. C. Nix, Gas Turbine Theory (7th ed., Pearson, 2017), chapters 2 and 3.
NASA Glenn Research Center, The Brayton cycle.

↓But there is a problem

The ideal turbojet makes power well and spends it badly: a 1160 m/s jet pushing a 251 m/s aircraft. Lesson 1-2 already said what to do — move more air, more slowly. Doing it nearly destroyed the most famous engine company in the world. In February 1971 Rolls-Royce, the maker of the Merlin, went into receivership because of a turbofan.