Prev
Unit 8 · Lesson 8-1
Component Matching
1h 50m1300 XP
Next

Component Matching

Mass, work and speed compatibility, corrected flow and speed, the choked turbine throat, the running line, and what moves it

The F-111's intakes
The F-111 of the 1960s pioneered variable-sweep wings and afterburning turbofans together (the F-111). Its Pratt & Whitney TF30s had problems with inlet compatibility, and many blamed the intakes' position behind the disturbed air of the wing (the TF30). NASA, the US Air Force and General Dynamics redesigned the inlet twice. The "Triple Plow II" intakes of later models stood four inches (100 mm) further from the fuselage, so the engines no longer swallowed the sluggish boundary-layer air known to stall them. An engine is a set of components that must agree — on mass flow, on work, on shaft speed — and every disagreement shows up as lost margin somewhere. This lesson finds the one line on the compressor map where they all agree.
Put a compressor, a burner, a turbine and a nozzle on one shaft and three conditions tie them together at every operating point:
  • Mass: the turbine passes what the compressor delivers, plus fuel, minus bleed; the nozzle passes what the turbine delivers.
  • Work: the turbine drives the compressor, cpc(Tt3−Tt2)=ηm(1+f) cpt(Tt4−Tt5)c_{pc}(T_{t3} - T_{t2}) = \eta_m (1 + f)\,c_{pt}(T_{t4} - T_{t5}) per kilogram of air.
  • Speed: compressor and turbine turn together.
Each component has its own map. The engine runs only where all three conditions hold at once.
m˙c=m˙θδ,Nc=Nθ,θ=Tt2288.15 K,δ=pt2101 325 Pa\dot m_c = \frac{\dot m\sqrt{\theta}}{\delta}, \qquad N_c = \frac{N}{\sqrt{\theta}}, \qquad \theta = \frac{T_{t2}}{288.15\ \mathrm{K}}, \qquad \delta = \frac{p_{t2}}{101\,325\ \mathrm{Pa}}

Corrected mass flow and corrected speed: a compressor at the same corrected point behaves the same on any day and at any altitude.

💡Why corrected quantities
A compressor's behaviour is set by the Mach numbers in its blade passages: the axial Mach number (from the flow) and the blade Mach number (from the speed). Both depend on the inlet temperature, and the flow also on the inlet pressure. Dividing them out leaves the corrected flow and speed above, and dimensional analysis says the pressure ratio and efficiency depend on nothing else. So one map serves every day and every altitude, and every test-cell measurement is corrected to standard conditions before anyone looks at it.
📐One map point, three conditions
This unit's turbojet at its design point — 50 kg/s at 12,000 rpm and Tt4=1500T_{t4} = 1500 K at sea level, 15 °C — and the same corrected point elsewhere (Lab 1):
  • A Dubai afternoon at 40 °C: the rotor must turn 4.2 % faster, 12,510 rpm, and the turbine entry must reach 1630 K. Held to 12,000 rpm, the engine runs at a corrected speed of only 95.9 %.
  • 11 km, Mach 0.85: 11,132 rpm, only 19.3 kg/s of air, and Tt4=1291T_{t4} = 1291 K.
Derivation
The choked turbine throat
The turbine's first nozzle guide vanes run choked over most of the operating range, so they pass m˙=A4 Γ pt4/Tt4\dot m = A_4\,\Gamma\,p_{t4}/\sqrt{T_{t4}}, with Γ\Gamma fixed by the gas (Lesson 2-1). With pt4=πb πc pt2p_{t4} = \pi_b\,\pi_c\,p_{t2}, divide both sides by pt2/Tt2p_{t2}/\sqrt{T_{t2}}:
m˙Tt2pt2=A4 Γ πb πc Tt2Tt4.\frac{\dot m\sqrt{T_{t2}}}{p_{t2}} = A_4\,\Gamma\,\pi_b\,\pi_c\,\sqrt{\frac{T_{t2}}{T_{t4}}}.
At a fixed Tt4/Tt2T_{t4}/T_{t2} the corrected flow is proportional to the pressure ratio: a straight line through the origin of the map — the line Lesson 6-2 called the operating line.
Derivation
The running line
If the exhaust nozzle is choked as well, the turbine runs between two choked throats, and its pressure ratio and temperature ratio τt\tau_t are fixed by the area ratio A4/A8A_4/A_8 (Lab 3 solves for them). The work balance then fixes Tt4/Tt2T_{t4}/T_{t2} for every compressor temperature ratio:
Tt4Tt2=cpccpt τc−11−τt.\frac{T_{t4}}{T_{t2}} = \frac{c_{pc}}{c_{pt}}\,\frac{\tau_c - 1}{1 - \tau_t}.
Put this into the throat condition (Lab 2 does it with SymPy) and the constants collapse:
m˙c∝πcτc−1,τc−1=πc(γ−1)/γ−1ηc.\dot m_c \propto \frac{\pi_c}{\sqrt{\tau_c - 1}}, \qquad \tau_c - 1 = \frac{\pi_c^{(\gamma - 1)/\gamma} - 1}{\eta_c}.
That is one line on the map: the running line. Each throttle setting is a point on it, and the speed line through that point is the speed the engine runs at.
📐The running line on the map
On the compressor map of Lesson 6-2, through the design point (corrected flow 1.039, pressure ratio 15.58, Tt4=1500T_{t4} = 1500 K), Lab 2 finds:
  • N=1.00N = 1.00: pressure ratio 15.58, surge margin 18.7 %.
  • N=0.80N = 0.80: pressure ratio 10.20, Tt4=1185T_{t4} = 1185 K, surge margin 12.5 %.
  • N=0.70N = 0.70: pressure ratio 7.88, Tt4=1012T_{t4} = 1012 K, surge margin 10.5 %.
At N=0.80N = 0.80 with the design Tt4/Tt2T_{t4}/T_{t2} — Lesson 6-2's straight line — the margin would be only 5.0 %. That is where an engine runs while it accelerates (Lesson 8-3).
💡What moves the running line
  • Exhaust nozzle area: close it and the turbine expands less, so a hotter Tt4T_{t4} is needed for the same work and the running line climbs towards surge; open it and the line drops. Afterburning engines open their nozzles as the afterburner lights for exactly this reason (Lesson 3-3).
  • Bleed: letting air off the compressor at low speed lowers the running line where the margin is smallest.
  • Inlet distortion leaves the running line where it is but lowers the surge line — the F-111's problem.
📐Nozzle area and distortion
With the nozzle throat and the turbine throat both choked (Lab 3):
  • Close the nozzle by 15 %: the full-speed surge margin falls from 18.7 % to 12.0 %, and at N=0.80N = 0.80 from 12.5 % to 6.8 %.
  • Open it by 10 %: 22.6 % at full speed.
  • Inlet distortion that takes 8 % off the surge pressure ratio leaves only 3.5 % at N=0.80N = 0.80.
When we try to pick out anything by itself, we find it hitched to everything else in the universe.
– John Muir, My First Summer in the Sierra (1869) (Wikiquote)
💡Why part speed is the dangerous part
The running line's margin shrinks as the speed falls — from 18.7 % at full speed to 10.5 % at N=0.70N = 0.70 in Lab 2. Real multistage compressors are worse: at low speed the front stages are pushed towards stall while the rear stages choke. Variable stators and handling bleeds (Lesson 6-2) exist to rescue exactly this corner of the map.
💡Where you will use this
Lesson 8-2 solves the same balances for a two-spool turbofan at any flight condition. Lesson 8-3 leaves the running line while the engine accelerates. Every engine test cell corrects its measurements with the same θ\theta and δ\delta.
📚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), on off-design performance and component matching.
  • Philip P. Walsh and Paul Fletcher, Gas Turbine Performance (2nd ed., Blackwell, 2004).
  • Jack D. Mattingly and Keith M. Boyer, Elements of Propulsion: Gas Turbines and Rockets (2nd ed., AIAA, 2016), on engine off-design performance.
↓But there is a problem
One running line and one design point. But an airliner takes off on a 40 °C afternoon in Dubai and cruises in air at −56 °C eleven kilometres up. Same engine, same throttle — and very different thrust. Why?