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Unit 4 · Lesson 4-1
Component Losses and Efficiencies
2h 5m1300 XP
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Component Losses and Efficiencies

Isentropic and polytropic efficiency, pressure recovery, burner and mechanical losses, and the efficiency threshold of a gas turbine

Kristiania, 1903
The Norwegian engineer Ægidius Elling had patented a gas turbine in 1884. In 1903 he finally built one that did something no gas turbine had done before: it produced more power than its own compressor consumed — 11 horsepower, about 8 kW, net (Ægidius Elling). Eleven horsepower is nothing. What it proved was everything. A gas turbine's turbine must first pay for its compressor, and only the difference is useful. With the compressors and turbines of the time, that difference was negative: the machine could spin, but not drive itself. Unit 3 assumed perfect components and found engines with 60 % thermal efficiency. This lesson puts the losses back in — and shows why, for most of history, they ate the whole engine.
Every component of a real engine falls short of its ideal in a measurable way:
  • The inlet loses stagnation pressure to friction and shocks: pressure recovery πd=pt2/pt0\pi_d = p_{t2}/p_{t0}.
  • The fan, compressor and turbines are not isentropic: they have efficiencies.
  • The burner loses pressure to friction, mixing and heat addition (Lesson 2-3), πb=pt4/pt3≈0.95\pi_b = p_{t4}/p_{t3} \approx 0.95–0.970.97, and does not quite burn all its fuel: combustion efficiency ηb≈0.995\eta_b \approx 0.995.
  • Shafts and bearings lose a little work: mechanical efficiency ηm≈0.99\eta_m \approx 0.99.
  • Nozzles lose stagnation pressure: πn≈0.98\pi_n \approx 0.98–0.990.99.
  • The gas in the hot section is not the gas in the cold section: different cpc_p and γ\gamma.
ηc=Tt3s−Tt2Tt3−Tt2=πc(γ−1)/γ−1τc−1,ηt=Tt4−Tt5Tt4−Tt5s=1−τt1−πt(γ−1)/γ\eta_c = \frac{T_{t3s} - T_{t2}}{T_{t3} - T_{t2}} = \frac{\pi_c^{(\gamma - 1)/\gamma} - 1}{\tau_c - 1}, \qquad \eta_t = \frac{T_{t4} - T_{t5}}{T_{t4} - T_{t5s}} = \frac{1 - \tau_t}{1 - \pi_t^{(\gamma - 1)/\gamma}}

Isentropic efficiencies of a compressor and a turbine: ideal over actual temperature change for the compressor, actual over ideal for the turbine, at the same pressure ratio.

💡Polytropic efficiency
The isentropic efficiency depends on the pressure ratio, so it cannot compare a 1.5:1 fan with a 40:1 compressor. Engineers use the polytropic (small-stage) efficiency ee instead: the efficiency of each infinitesimal step of compression or expansion, which reflects how well the blades are designed and is nearly the same for machines of the same technology. τc=πc(γ−1)/(γec),τt=πt(γ−1)et/γ.\tau_c = \pi_c^{(\gamma - 1)/(\gamma e_c)}, \qquad \tau_t = \pi_t^{(\gamma - 1)e_t/\gamma}. Today's values are about ef=0.92e_f = 0.92 for fans, ec=0.90e_c = 0.90 for compressors, et=0.89e_t = 0.89–0.910.91 for turbines.
Derivation
From polytropic to isentropic efficiency
For one small compression step, the isentropic temperature rise is dTs=T γ−1γdppdT_s = T\,\frac{\gamma - 1}{\gamma}\frac{dp}{p}, and the actual rise is dT=dTs/ecdT = dT_s/e_c:
dTT=γ−1γec dpp⟹Tt3Tt2=πc(γ−1)/(γec).\frac{dT}{T} = \frac{\gamma - 1}{\gamma e_c}\,\frac{dp}{p} \quad\Longrightarrow\quad \frac{T_{t3}}{T_{t2}} = \pi_c^{(\gamma - 1)/(\gamma e_c)} .
Substituting into the isentropic efficiency:
ηc=πc(γ−1)/γ−1πc(γ−1)/(γec)−1.\eta_c = \frac{\pi_c^{(\gamma - 1)/\gamma} - 1}{\pi_c^{(\gamma - 1)/(\gamma e_c)} - 1}.
Because each stage receives air already heated by the inefficiency of the stages before it, and compressing hotter air takes more work, ηc<ec\eta_c < e_c — and the gap grows with pressure ratio. In a turbine the reheat helps the later stages, so ηt>et\eta_t > e_t.
📐One technology, many efficiencies
With ec=et=0.90e_c = e_t = 0.90 (Lab 1):
  • Compressor isentropic efficiency: 0.890 at a pressure ratio of 2, 0.840 at 40, 0.832 at 60.
  • Turbine: 0.909 at 2, 0.941 at 40.
  • A 40:1 compressor from 288 K: isentropic exit 827 K, real exit 929 K — 103 K of extra heating that the burner can no longer use.
💡The work ratio and the threshold
A gas turbine's net work is the small difference of two large numbers: what the turbine gives minus what the compressor takes. The work ratio — net work over turbine work — measures how small. When it is small, a few points of component efficiency decide whether there is any net work at all. For a simple cycle with compressor and turbine of equal efficiency η\eta (Lab 2):
  • Early-1900s conditions (πc=4\pi_c = 4, 800 K): no net work at all until η=0.732\eta = 0.732; at 0.85, a work ratio of only 0.26.
  • A modern core (πc=30\pi_c = 30, 1700 K): the threshold falls to 0.669, and at 0.85 the work ratio is 0.38.
Derivation
Net work of a real simple cycle
Compressor work, from T1T_1 through pressure ratio πc\pi_c with τ=πc(γ−1)/γ\tau = \pi_c^{(\gamma - 1)/\gamma}:
wc=cpT1(τ−1)ηc.w_c = \frac{c_p T_1(\tau - 1)}{\eta_c}.
Turbine work, expanding from Tt4T_{t4} back through the same pressure ratio (ignoring the burner's pressure loss):
wt=cpTt4 ηt(1−1τ).w_t = c_p T_{t4}\,\eta_t\left(1 - \frac{1}{\tau}\right).
Net work w=wt−wcw = w_t - w_c vanishes when Tt4ηtηc(1−1/τ)=T1(τ−1)T_{t4}\eta_t\eta_c(1 - 1/\tau) = T_1(\tau - 1), i.e. ηcηt=τT1/Tt4\eta_c\eta_t = \tau T_1/T_{t4} — a low turbine entry temperature demands excellent components. Elling's problem in one line.
Cold gas and hot gas. A real calculation uses γc=1.4\gamma_c = 1.4, cpc=1004.5c_{pc} = 1004.5 J/(kg·K) before the burner and γt=1.33\gamma_t = 1.33, cpt=1156c_{pt} = 1156 J/(kg·K) after it. Even that is an approximation: air's cpc_p rises from 1006 to 1119 J/(kg·K) through a 40:1 compressor (Lab 3, with CoolProp). Assuming a constant cpc_p gets the compressor work within 0.5 % but puts the exit temperature 31 K too high. Lesson 4-3 deals with it properly.
The production of heat alone is not sufficient to give birth to the impelling power: it is necessary that there should also be cold; without it, the heat would be useless.
– Sadi Carnot, Reflections on the Motive Power of Fire (1824) (Wikiquote)
🕰Elling's second machine
By 1912 Elling had a gas turbine with the compressor and turbine as separate units in series, the arrangement now universal (Ægidius Elling). His work remained almost unknown outside Norway in his lifetime.
💡Where you will use this
Lesson 4-2 puts every one of these losses into a design-point turbofan. Unit 6 explains where compressor and turbine losses come from — blade profiles, tip clearance, shocks. Lesson 7-3 designs the burner around its pressure loss.
📚Further reading
  • H. I. H. Saravanamuttoo, G. F. C. Rogers, H. Cohen, P. V. Straznicky and A. C. Nix, Gas Turbine Theory (7th ed., Pearson, 2017), chapter 2, on component losses and polytropic efficiency.
  • S. Larry Dixon and Cesare A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed., Butterworth-Heinemann, 2014), chapter 2.
  • Jack D. Mattingly and Keith M. Boyer, Elements of Propulsion: Gas Turbines and Rockets (2nd ed., AIAA, 2016), on component performance.
↓But there is a problem
We now have a number for every loss. Put them all into one engine and design it — station by station, fuel flow by fuel flow. On 20 June 1974 GE and Snecma ran such an engine for the first time. Five years later it had not won a single order, and the partnership was two weeks from being dissolved.