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Unit 4 · Lesson 4-2
Design-Point Analysis of a Real Turbofan
1h 50m1300 XP
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Design-Point Analysis of a Real Turbofan

Station-by-station analysis of a two-spool separate-flow turbofan: fuel–air ratio, turbine work balances, nozzles, thrust, TSFC and sensitivities

Two weeks from closing
The CFM56 — built by CFM International, a partnership of GE and the French Snecma, now Safran — first ran on 20 June 1974. Its core came from GE's F101. By April 1979 the joint venture had not received a single order in five years and was two weeks away from being dissolved. The programme was saved when Delta Air Lines, United Airlines and Flying Tigers chose the CFM56 to re-engine their Douglas DC-8s. It went on to power the Boeing 737 Classic and Next Generation and the Airbus A320 and A340, and became the most used turbofan in the world: 32,645 built by June 2018 (the CFM56). Before an engine like that exists as metal, it exists as a design point — one flight condition at which every temperature, pressure and flow in the engine is chosen. This lesson computes one.
On-design (parametric) analysis fixes a flight condition — here Mach 0.8 at 10.7 km — and the engine's design choices: bypass ratio α\alpha, fan pressure ratio πf\pi_f, overall pressure ratio πc\pi_c and turbine entry temperature Tt4T_{t4}. With the component losses of Lesson 4-1 it then follows the air station by station and returns specific thrust, TSFC and the efficiencies.
It answers "what engine should we build?" Lesson 8-2's off-design analysis answers the next question: "how does that engine behave everywhere else?"
💡The procedure
For a two-spool separate-flow turbofan — a high-pressure spool (compressor and HP turbine) inside a low-pressure spool (fan and LP turbine):
  1. Free stream and inlet: Tt0T_{t0}, pt0p_{t0} from the flight Mach number; pt2=πd pt0p_{t2} = \pi_d\,p_{t0}.
  2. Fan (station 13) and compressor (station 3): temperatures from the polytropic efficiencies.
  3. Burner: the fuel–air ratio from an energy balance; pt4=πb pt3p_{t4} = \pi_b\,p_{t3}.
  4. HP turbine (station 45) gives the compressor its work; LP turbine (station 5) gives the fan its work.
  5. Nozzles (9, 19): jet velocities from the remaining stagnation pressures.
  6. Thrust, TSFC and efficiencies from the jet velocities and the fuel flow.
f=cptTt4−cpcTt3ηbhPR−cptTt4,Tt45=Tt4−cpc(Tt3−Tt2)(1+f) cpt ηm,Tt5=Tt45−α cpc(Tt13−Tt2)(1+f) cpt ηmf = \frac{c_{pt}T_{t4} - c_{pc}T_{t3}}{\eta_b h_{PR} - c_{pt}T_{t4}}, \qquad T_{t45} = T_{t4} - \frac{c_{pc}(T_{t3} - T_{t2})}{(1 + f)\,c_{pt}\,\eta_m}, \qquad T_{t5} = T_{t45} - \frac{\alpha\,c_{pc}(T_{t13} - T_{t2})}{(1 + f)\,c_{pt}\,\eta_m}

The burner energy balance and the two turbine work balances, per kilogram of core air. Turbine pressure ratios then follow from the polytropic efficiencies.

Derivation
Burner and turbine balances
Burner. Air enters with enthalpy cpcTt3c_{pc}T_{t3} per kilogram; fuel adds ηbhPR\eta_b h_{PR} per kilogram of fuel; (1+f)(1 + f) kilograms of gas leave with cptTt4c_{pt}T_{t4} each:
cpcTt3+f ηbhPR=(1+f) cptTt4⟹f=cptTt4−cpcTt3ηbhPR−cptTt4.c_{pc}T_{t3} + f\,\eta_b h_{PR} = (1 + f)\,c_{pt}T_{t4} \quad\Longrightarrow\quad f = \frac{c_{pt}T_{t4} - c_{pc}T_{t3}}{\eta_b h_{PR} - c_{pt}T_{t4}}.
HP spool. The (1+f)(1 + f) kilograms of gas through the HP turbine give, after mechanical losses, the work one kilogram of core air needs in the compressor: (1+f)cpt(Tt4−Tt45)ηm=cpc(Tt3−Tt2)(1 + f)c_{pt}(T_{t4} - T_{t45})\eta_m = c_{pc}(T_{t3} - T_{t2}).
LP spool. The LP turbine drives the fan, which compresses the core air's α\alpha kilograms of bypass air: (1+f)cpt(Tt45−Tt5)ηm=αcpc(Tt13−Tt2)(1 + f)c_{pt}(T_{t45} - T_{t5})\eta_m = \alpha c_{pc}(T_{t13} - T_{t2}).
Each turbine's pressure ratio follows from πt=τtγt/((γt−1)et)\pi_t = \tau_t^{\gamma_t/((\gamma_t - 1)e_t)}, and the nozzle exit velocity from V2=2cpTt[1−(p0/pt)(γ−1)/γ]V^2 = 2c_p T_t\left[1 - (p_0/p_t)^{(\gamma - 1)/\gamma}\right].
📐A design point
Cruise at Mach 0.8, 10.7 km; α=10\alpha = 10, πf=1.7\pi_f = 1.7, πc=40\pi_c = 40, Tt4=1700T_{t4} = 1700 K, today's component efficiencies (Lab 1):
  • Compressor exit: Tt3=796T_{t3} = 796 K, pt3=1446p_{t3} = 1446 kPa. HP turbine exit 1231 K, LP turbine exit 854 K.
  • Fuel–air ratio 0.0286; core jet 651 m/s, fan jet 371 m/s.
  • F/m˙0=160.6F/\dot m_0 = 160.6 N·s/kg and TSFC = 16.2 mg/(N·s) = 0.571 lb/(lbf·h).
  • ηth=48.5 %\eta_{th} = 48.5\,\%, ηp=70.4 %\eta_p = 70.4\,\%, η0=34.1 %\eta_0 = 34.1\,\%.
  • 60 kN of cruise thrust needs 374 kg/s of air, only 34 kg/s of it through the core.
The ideal engine of Unit 3 would have a thermal efficiency near 69 %; the losses take about 20 points of it.
💡Which component is worth most?
Raise one quantity by 0.01 at a time (Lab 2):
  • Inlet pressure recovery: TSFC −1.1 % — every stream passes through it.
  • Combustion efficiency: −1.0 % — trivially, unburned fuel is wasted fuel.
  • LP turbine and fan efficiency: −0.3 % each; HP turbine −0.28 %; burner pressure ratio −0.18 %.
  • Compressor efficiency: only −0.12 % TSFC, but +1.0 % thrust.
The last one surprises everybody. At fixed pressure ratio and Tt4T_{t4}, a better compressor delivers cooler air; more fuel is then burned to reach Tt4T_{t4}, so the engine gives more thrust for about the same TSFC. The gain shows up as a smaller engine for the same thrust.
📐Designing under limits
Sweep the overall pressure ratio and the turbine entry temperature with the fan fixed (Lab 3). Two limits shape the answer:
  • The compressor exit temperature, limited by the last discs' material — here 900 K in cruise. At πc=60\pi_c = 60 it is 905 K, so that whole row is out.
  • For a fixed fan, each pressure ratio has a best Tt4T_{t4}. Too hot, and the core jet is too fast. Too cold, and the core cannot drive the fan (πc=60\pi_c = 60 at 1500 K is not possible at all).
The best design within the limit is πc=50\pi_c = 50 with Tt4=1600T_{t4} = 1600 K: TSFC 15.4 mg/(N·s) at 143 N·s/kg.
Failure is only the opportunity to more intelligently begin again. There is no disgrace in honest failure; there is disgrace in fearing to fail.
– Henry Ford, My Life and Work (1922) (Wikiquote)
⚠Design point is not the whole story
An engine designed for cruise must also take off on a hot day, climb, idle and relight. Choices that look best at the design point — a very high pressure ratio, a very low specific thrust — can fail elsewhere: an overheated compressor exit at take-off, too little thrust at the top of the climb. Real design iterates between on- and off-design (Unit 8).
💡Where you will use this
Lesson 8-1 takes this design and matches its components off design. Lesson 10-1 uses the same procedure for adaptive-cycle engines with a third stream. In industry the procedure is the core of every cycle code — GasTurb, NPSS, pyCycle.
📚Further reading
  • Jack D. Mattingly and Keith M. Boyer, Elements of Propulsion: Gas Turbines and Rockets (2nd ed., AIAA, 2016), on the parametric cycle analysis of real engines.
  • Philip P. Walsh and Paul Fletcher, Gas Turbine Performance (2nd ed., Blackwell, 2004).
  • The CFM56 on Wikipedia, for how one design point became 30,000 engines.
↓But there is a problem
Every number in this lesson used two constant values of cpc_p — one for cold air, one for hot gas. At 1700 K, with water vapour and carbon dioxide in the gas, that is an approximation, and engine makers fight over fractions of a per cent. The newest big engine, the GE9X, runs a 60:1 pressure ratio and puts ceramics where nickel alloys used to be. To see why, we need the real properties of the gas.