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Unit 3 · Lesson 3-2
The Ideal Turbofan
2h 5m1300 XP
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The Ideal Turbofan

Bypass ratio and fan pressure ratio, two-stream thrust, the optimum fan ratio, and the trade between specific thrust and TSFC

4 February 1971
Rolls-Royce, the company that built the Merlin, is placed into receivership. The cause is a single engine: the RB211, a three-shaft, high-bypass turbofan for Lockheed's L-1011 TriStar. Its fan was to be made of Hyfil, a new carbon-fibre composite — much lighter than titanium. In May 1970 a Hyfil fan stage shattered when a chicken carcass was fired into it in the bird-ingestion test. The titanium blades held in reserve meant extra weight and extra cost, and by January 1971 the company was insolvent (the RB211). Because of its strategic importance the government nationalised the company, and the RB211 entered service in 1972. Its descendants, the Trents, power much of the world's long-haul fleet today. Why would a company stake everything on a bigger fan? Because of what this lesson computes: with the same core, a fan cuts the fuel burned for each newton of thrust almost in half.
A turbofan splits the incoming air. A fraction goes through the core — compressor, burner, turbine — exactly as in a turbojet. The rest goes only through the fan and leaves through its own nozzle. The bypass ratio α=m˙F/m˙C\alpha = \dot m_F/\dot m_C measures the split; the fan pressure ratio πf\pi_f sets how hard the fan pushes the bypass air.
The core's turbine now drives two things: the compressor and the fan. Every joule the turbine sends to the fan is taken from the core jet and given to a much larger, slower stream — the propulsive-efficiency argument of Lesson 1-2, built into the machine.
Fm˙0=a01+α[(V9a0−M0)+α(V19a0−M0)],TSFC=f(1+α) F/m˙0\frac{F}{\dot m_0} = \frac{a_0}{1 + \alpha}\left[\left(\frac{V_9}{a_0} - M_0\right) + \alpha\left(\frac{V_{19}}{a_0} - M_0\right)\right], \qquad \mathrm{TSFC} = \frac{f}{(1 + \alpha)\,F/\dot m_0}

Ideal separate-flow turbofan: specific thrust per unit of total air flow ṁ0 = ṁC + ṁF, and TSFC. The fuel–air ratio f refers to the core flow.

Derivation
The two jet velocities
Fan stream. The bypass air is compressed by the ram and the fan only, and expands isentropically to p0p_0: Tt19/T19=τrτfT_{t19}/T_{19} = \tau_r\tau_f with Tt19=T0τrτfT_{t19} = T_0\tau_r\tau_f, so
(V19a0)2=2γ−1(τrτf−1).\left(\frac{V_{19}}{a_0}\right)^2 = \frac{2}{\gamma - 1}\left(\tau_r\tau_f - 1\right).
Work balance. The turbine drives the compressor (core flow) and the fan (all the flow): τt=1−τrτλ[(τc−1)+α(τf−1)]\tau_t = 1 - \frac{\tau_r}{\tau_\lambda}\left[(\tau_c - 1) + \alpha(\tau_f - 1)\right].
Core stream. As for the turbojet, T9=T0τλ/(τrτc)T_9 = T_0\tau_\lambda/(\tau_r\tau_c) and V92=2cpT9(τrτcτt−1)V_9^2 = 2c_p T_9(\tau_r\tau_c\tau_t - 1); substituting τt\tau_t:
(V9a0)2=2γ−1[τλ−τr(τc−1)−ατr(τf−1)−τλτrτc].\left(\frac{V_9}{a_0}\right)^2 = \frac{2}{\gamma - 1}\left[\tau_\lambda - \tau_r(\tau_c - 1) - \alpha\tau_r(\tau_f - 1) - \frac{\tau_\lambda}{\tau_r\tau_c}\right].
When the bracket reaches zero the core has no energy left for its own jet: that fan is too big for that core.
📐One core, with and without a fan
Cruise at Mach 0.85 and 11 km, πc=36\pi_c = 36, Tt4=1700T_{t4} = 1700 K (Lab 1):
  • Turbojet: F/m˙0=955F/\dot m_0 = 955 N·s/kg, TSFC = 24.7 mg/(N·s), V9=1206V_9 = 1206 m/s, ηp=0.34\eta_p = 0.34.
  • Turbofan, α=10\alpha = 10, πf=1.6\pi_f = 1.6: F/m˙0=161F/\dot m_0 = 161 N·s/kg, TSFC = 13.3 mg/(N·s), core jet 860 m/s, fan jet 367 m/s, ηp=0.64\eta_p = 0.64.
The thermal efficiency is the same 68.6 % — it is the same core. The fan cuts TSFC by 46 %, at the price of six times less thrust per kilogram of air: a much bigger engine for the same thrust.
💡The best fan ratio: equal jet speeds
For a given core and bypass ratio, which fan pressure ratio burns the least fuel per newton? A low πf\pi_f leaves most of the energy in a fast core jet; a high one starves the core jet. TSFC is lowest when both jets leave at the same speed, V19=V9V_{19} = V_9: τf∗=τλ−τr(τc−1)−τλτrτc+ατr+1τr(1+α).\tau_f^* = \frac{\tau_\lambda - \tau_r(\tau_c - 1) - \dfrac{\tau_\lambda}{\tau_r\tau_c} + \alpha\tau_r + 1}{\tau_r(1 + \alpha)}. For the core above at α=10\alpha = 10: πf∗=2.21\pi_f^* = 2.21, both jets at 435 m/s, TSFC = 11.6 mg/(N·s) (Lab 2 finds it with minimize_scalar and checks the closed form).
Derivation
Why equal jet velocities
With the core fixed, the fuel flow is fixed, and so is the kinetic energy the engine can give the two streams (per unit core flow):
E=12(V92−V02)+α2(V192−V02)=const.E = \tfrac12\left(V_9^2 - V_0^2\right) + \tfrac{\alpha}{2}\left(V_{19}^2 - V_0^2\right) = \text{const}.
Minimising TSFC means maximising thrust, F∝(V9−V0)+α(V19−V0)F \propto (V_9 - V_0) + \alpha(V_{19} - V_0), under that constraint. With a Lagrange multiplier λ\lambda:
∂∂V9:1=λV9,∂∂V19:α=λαV19⟹V9=V19.\frac{\partial}{\partial V_9}: 1 = \lambda V_9, \qquad \frac{\partial}{\partial V_{19}}: \alpha = \lambda\alpha V_{19} \quad\Longrightarrow\quad V_9 = V_{19}.
Any difference between the jets is kinetic energy that buys less thrust than it would in the slower stream. Setting the two velocity formulas equal and solving for τf\tau_f gives τf∗\tau_f^*.
📐The carpet
Specific thrust and TSFC over bypass ratios 2–12 and fan ratios 1.4–2.5 (Lab 3) make a carpet plot:
  • At α=2\alpha = 2, TSFC is 19–22 mg/(N·s) and specific thrust 360–415 N·s/kg: a compact, thirsty engine.
  • At α=12\alpha = 12, πf=2.0\pi_f = 2.0: 11.2 mg/(N·s) at 161 N·s/kg.
  • Two corners are impossible: at α=10\alpha = 10 and 12 a fan ratio of 2.5 asks the core for more power than it has.
Engine design is choosing a point on this carpet, and the aircraft decides how much engine size it can carry.
Separate or mixed? Airliner engines usually exhaust the two streams separately. Military low-bypass turbofans usually mix them in a common nozzle, which suits an afterburner and, if the two streams arrive at equal stagnation pressure, even gains a little thrust. Lesson 4-2 analyses both with real component losses.
Strive for perfection in everything we do. Take the best that exists and make it better. When it does not exist, design it. Accept nothing nearly right or good enough.
– Henry Royce; the slogan of the Rolls-Royce company, as quoted in Transforming the Organization (1996) (Wikiquote)
🕰The fan blade that came back
Composite fan blades returned in 1995 with the GE90, and the GE9X of Lesson 1-1 has 16 of them, each made of carbon-fibre composite. The bird-strike test that broke Hyfil is still part of every engine's certification.
💡Where you will use this
Lesson 4-2 repeats this analysis with real losses and finds the design point of a real turbofan. Lesson 8-1 shows how the fan, core and nozzles match off design. Lesson 10-2 follows the bypass ratio's next steps: geared fans and open rotors.
📚Further reading
  • Jack D. Mattingly and Keith M. Boyer, Elements of Propulsion: Gas Turbines and Rockets (2nd ed., AIAA, 2016), the ideal turbofan and its optimum.
  • Nicholas Cumpsty and Andrew Heyes, Jet Propulsion (3rd ed., Cambridge University Press, 2015), on turbofan design choices.
  • NASA Glenn Research Center, Turbofan thrust.
↓But there is a problem
The turbofan wins by slowing the jet down. Now push the other way and fly faster: ram compression starts doing the compressor's job — and heating the air before the compressor has done anything. By Mach 3 the SR-71's engines were sending a fifth of their compressor air around the core, straight into the afterburner.