Turbofan, turboprop, propfan

Welcome to part two, where the component models of part one become complete engines. We start with the separate-exhaust turbofan and its two-spool power balance, then solve exercise 1 A, the C F M fifty-six at cruise, and 1 B, the same engine at take-off, step by step. Then we review propeller momentum theory and the turboprop, and finish with the two conversions of the exercises: a propfan at cruise and an open rotor on the ground.

A separate-exhaust turbofan splits the inlet flow into a core stream and a bypass, or secondary, stream, and the bypass ratio alpha is their ratio. So the core flow is m dot zero over one plus alpha, and the bypass flow is alpha times m dot zero over one plus alpha. Fuel is referred to the core, so the fuel flow is f times the core flow, and the T S F C carries a factor one plus alpha in the denominator: forgetting it is, according to the manual, the most frequent error in turbofan calculations. The turbofan has three design variables where the turbojet had two: besides the compressor pressure ratio and T t four, the fan pressure ratio and the bypass ratio, which are coupled through the turbine. In exercise 1 A, with alpha equal to 6 and a total flow of 170 kilograms per second, the core receives 24.29 and the bypass 145.71 kilograms per second.

Real turbofans have two spools turning at different speeds, and the exercises are posed that way. The high pressure spool is simple: the high pressure turbine drives only the high pressure compressor, so the compressor power equals eta m H times the turbine power, which gives T t four point five. The low pressure turbine drives both the fan and the low pressure compressor, working on gas already cooled by the high pressure turbine, which gives T t five. Careful with the mass flows: the fan compresses the whole inlet flow, while the low pressure compressor only compresses the core; mixing the two conventions changes the fan power by a factor one plus alpha. The overall pressure ratio is the product of the three compression ratios, so in exercise 1 A the high pressure compressor must provide 30.6 divided by 1.32 times 5, which is 4.636, and each compressor starts from the temperature the previous one left behind.

Thrust now comes from two streams, each with its own momentum, ram drag and pressure term. The thrust ratio F R compares how hard each stream works per unit of its own mass flow: it is the ratio of specific thrusts, not of thrusts, and values around one minimise fuel consumption in civil engines. The kinetic energy increment adds both streams, and the thermal and propulsive efficiencies follow as before. In the manual's worked example, using the closed-form propulsive efficiency with under-expanded nozzles loses eight percentage points, so always compute it from the full thrust. In exercise 1 A, F R comes out at about 10: the core stream works ten times harder per kilogram than the fan stream, which says there is room for a larger bypass ratio, exactly the trend of modern engines.

Here is exercise 1 A, the template for the whole partial. The C F M fifty-six has two shafts: the high pressure turbine drives the high pressure compressor, and the low pressure turbine drives the low pressure compressor and the fan, with pressure ratios 5 and 1.32, a bypass ratio of 6, an overall pressure ratio of 30.6 and T t four of 1 600 kelvin. Both nozzles are convergent, and the design point is cruise at 11 km, Mach 0.8, with a total mass flow of 170 kilograms per second. We must draw the block diagram, compute the spool powers, the thrust, T S F C and efficiencies, and sketch the T s diagram. All the component data are isentropic efficiencies, so we use the isentropic forms throughout, with 0.98 for the inlet and 0.9 for the burner pressure ratios.

Start at the inlet: the free-stream total temperature is 244.55 kelvin, and the total pressure drops by 2 % in the diffuser. The fan raises the total temperature by its isentropic relation to 268.30 kelvin, matching the official table. The low pressure compressor, with a pressure ratio of 5 and an efficiency of 0.83, reaches 457.02 kelvin. The high pressure compressor, with 4.636 and 0.80, reaches 771.2 kelvin, and the total pressure is now 10.39 bar. The powers follow: the fan works on all 170 kilograms per second for 4 056 kilowatts, and the low pressure compressor on the 24.29 kilograms per second of core for 4 604 kilowatts. The high pressure compressor absorbs 7 665 kilowatts, so the low pressure spool must supply 8 660 kilowatts to its compressors and the high pressure spool 7 665, matching the official solution to within rounding.

Now the burner: plugging T t three of 771.2 and T t four of 1 600 into the fuel ratio formula gives f equal to 0.0301. The hot flow is 25.02 kilograms per second, and the high pressure spool balance gives the turbine exit temperature: 1 600 minus 7 665 kilowatts divided by 0.95 times 25.02 times 1.239, which is 1 339.7 kelvin. Inverting the isentropic efficiency of 0.80 gives a pressure ratio of 0.373, so the total pressure after the high pressure turbine is 3.49 bar. The low pressure spool must deliver the fan plus the low pressure compressor powers, so the gas cools further to 1 045.6 kelvin. With an efficiency of 0.81, the low pressure turbine pressure ratio is 0.254, leaving 0.887 bar at the core nozzle entry; notice that these two turbine ratios are exactly those of the manual's table on polytropic efficiencies.

Both nozzles are convergent, so we apply the hypothesis and check with the nozzle efficiencies. For the core, the critical static temperature is two T t over gamma plus one, 909.2 kelvin, and the efficiency definition gives a critical static pressure of 0.447 bar. That is above the ambient 0.227 bar, so the core nozzle is choked: Mach one, and a velocity of 581.3 metres per second from the enthalpy drop. The fan nozzle, with only 0.45 bar of total pressure, gives a critical pressure below ambient, so it is matched and its exit pressure equals ambient. Then its exit temperature, with the 0.90 efficiency, is 225.7 kelvin, giving 292.5 metres per second and Mach 0.97, all matching the official table. For the choked core nozzle we still need the exit area, 0.25 square metres from continuity, because its pressure term contributes thrust.

Now the thrust, term by term: the core jet gives 14.54 kilonewtons of momentum, the fan jet 42.62, the ram drag of the 170 kilograms per second subtracts 40.14, and the choked core nozzle's pressure term adds 5.49, for a total of 22.52 kilonewtons. The fuel flow is 0.0301 times the core flow, 0.7315 kilograms per second, so the T S F C is 32.49 milligrams per second per newton, exactly the official value. The kinetic energy increment of the two streams is 5.72 megawatts. So the thermal efficiency is 18.4 %, low because of the modest component efficiencies, and the propulsive efficiency is 92.9 %, high because both jets are slow compared with the flight speed. The overall efficiency is 17.1 %, and the quick check, v zero over h P R times T S F C, confirms it.

Part d asks for a sketch of the thermodynamic cycle, which you should be able to draw from your station table. Watch the core stream being built station by station from the computed totals: first the free ram compression from the static free-stream to the total state. Then the fan and the two compressors climb to 771 kelvin, each process drifting to the right because losses generate entropy. The burner adds heat up to 1 600 kelvin at slightly falling pressure, and the two turbines expand the gas down to 1 046 kelvin. Finally the nozzle accelerates the flow to its choked exit at 909 kelvin static, and the vertical extent of the diagram is bounded by T t four. Two cautions for the sketch: this is an open trace of total states with losses, not a closed reversible cycle, so the area under or inside it is not the net work; and the bypass stream is a separate short branch, from 2 to 13 through the fan and then expanding in its own nozzle to 19, which you should add as a second line.

Exercise 1 B takes the same engine to sea level static conditions, at 450 kilograms per second, so the core gets 64.29. Without ram compression but with warmer air, the compressor exit is hotter, 909 kelvin, and the overall pressure is 30.4 bar. All spool powers scale up with the mass flow, about three times, and the fuel ratio is slightly lower, 0.0267, because the burner entry is hotter. At sea level the ambient pressure is high, so both nozzles turn out matched, the core at Mach 0.71 and the fan at Mach 0.58, a good reminder to always check. The take-off thrust is 104.3 kilonewtons with a T S F C of 16.44 milligrams per second per newton, both matching the official solution. And since the aircraft is not moving, no useful power is produced: the propulsive and overall efficiencies are zero by definition, even though the thrust is large; only the thermal efficiency, 18.1 %, is meaningful.

Two architectures from the manual deserve a short mention. In a mixed-exhaust turbofan, the core and bypass streams join in a mixer and leave through one nozzle, the standard layout of military engines with afterburners. Because the mixer requires matching total pressures, the turbine pressure ratio is fixed first, and the power balance then delivers the bypass ratio: alpha becomes an output, the single structural difference from the separate-exhaust engine. Mixing recovers thrust by removing the velocity mismatch between the streams, a few per cent, but on a large civil engine the full-length duct costs more in weight and drag than it returns. The turboshaft splits the turbine: the high pressure turbine drives the compressor, and a free turbine drives the output shaft, judged by its power specific fuel consumption; its propulsive efficiency only makes sense together with its rotor or propeller.

A propeller produces thrust by giving a small velocity increase to a large mass of air, the opposite of a turbojet, and momentum theory captures it with an actuator disc. Bernoulli on each side of the disc and the momentum balance give the central result: the velocity at the disc is the average of the upstream and far-downstream velocities. With the axial induction factor a, the disc velocity is v zero times one plus a, the wake velocity v zero times one plus two a, and the thrust two m dot a v zero. The disc power, thrust times disc velocity, equals the kinetic energy given to the slipstream, and the ideal, Froude, propulsive efficiency is one over one plus a. Solving the quadratic for the disc velocity shows that the loss is governed by the disc loading, thrust over disc area: a big diameter means high efficiency, but also high tip speeds, ground clearance problems and root stresses.

A real propeller absorbs more shaft power than the ideal disc for the same thrust, because of profile drag, swirl and tip losses, collected in a loss efficiency eta L, so the propeller efficiency is eta L times eta P. For cycle work the course uses the algebraic model plotted here: ten M zero eta max at very low speed, constant at eta max up to Mach 0.7, and falling beyond, when the blade tips approach drag divergence. The procedure is: thrust from the shaft power, disc velocity from the disc area, then the induction factor, the disc power and eta L, which must not exceed one; otherwise the disc is too small for the assumed efficiency. At zero speed the efficiency is meaningless, so the static thrust uses the figure of merit, F equals two rho A to the one third times F M times the power to the two thirds, with F M around 0.7 to 0.8, while the model's own limit, ten eta max times the power over the speed of sound, can violate momentum theory, as exercise 1 B will show.

A turboprop is a turboshaft whose free turbine drives a propeller through a gearbox. It handles two streams, the gas generator flow and the propeller flow, and their ratio beta behaves like a bypass ratio, but it is an output of momentum theory, not a design choice, because the propeller has no duct. The power reaching the propeller is the gearbox efficiency times the shaft's share of the low pressure turbine power minus the low pressure compressor power. The total thrust adds the jet and the propeller, and the kinetic energy increment uses the disc power, so propeller losses appear in the thermal efficiency; Mattingly uses the shaft power instead, which gives a higher thermal and a lower propulsive efficiency but the same overall efficiency. Finally, there is an optimum free-turbine expansion: more expansion sends power to the efficient propeller until the jet is starved, and the optimum shifts towards the jet as the flight speed grows.

Part 2 of exercise 1 A converts the engine into a propfan: the fan becomes a propeller of 3.93 metres, the nacelle opens, and we keep the core flow, the low pressure turbine pressure ratio and, as the official solution does, the high pressure compressor ratio of 4.636, so the overall pressure ratio drops from 30.6 to 5 times 4.636, 23.2, with 1.70 bar after the booster and 7.87 bar at the burner. Without the fan, the low pressure compressor starts from the inlet temperature, so it ends at 416.6 kelvin, the burner entry falls to 703 kelvin, and f rises to 0.0318. The compressor powers fall to 4 196 kilowatts and 6 987 kilowatts, while the low pressure turbine, with its pressure ratio kept, delivers 9 290 kilowatts. The propeller receives the gearbox efficiency times the shaft's share minus the low pressure compressor: 4 583 kilowatts; note that the gearbox efficiency multiplies both terms, as the official solution stresses. At Mach 0.8 the model is past 0.7, so the propeller efficiency is 0.9 times one minus 0.1 over 3, which is 0.87, and the propeller thrust is 16.89 kilonewtons. The choked core jet adds 13.38, for a total of 30.27 kilonewtons, with a T S F C of 25.54 milligrams per second per newton: 21 % better than the turbofan, exactly the official answer.

Now the propeller stream, from momentum theory with the full disc area, 12.13 square metres, and the cruise density of 0.365 kilograms per cubic metre. The disc velocity is 243.9 metres per second, so the induction factor is only 0.033. The ideal propulsive efficiency is therefore 0.97, and the loss efficiency eta L is 0.87 times 1.033, which is 0.90, below one as it must be: the official note says this factor collects the profile drag, swirl and tip losses that momentum theory ignores. The propeller moves 1 081 kilograms per second, so the total flow is 1 105 and the bypass ratio jumps from 6 to 44.5. The efficiencies are 23.6 % thermal, 92.2 % propulsive and 21.75 % overall, against 17.1 % for the turbofan: this is the promise of the propfan.

Exercise 1 B converts the take-off engine into an open rotor, with the rotor behind the gas generator, driven by the low pressure turbine. The same power bookkeeping gives 14 413 kilowatts for the rotor. At zero speed the efficiency model cannot be used directly, so we try its static limit, ten eta max times the power over the speed of sound, 381 kilonewtons, and compare it with an ideal disc of the same area, 183 kilonewtons. The implied figure of merit, 381 over 183 to the power three halves, is about 3, which is impossible, since no propeller beats the ideal disc: the algebraic model hides a disc loading, and here the disc is too small for it. So, as instructed, we use the figure of merit 0.75: the rotor gives 151.1 kilonewtons, and the matched core jet 20.4 more. The disc velocity is the square root of the propeller thrust over two rho A, 71.5 metres per second, so the propeller stream is rho A v p, 1 056 kilograms per second; the thermal efficiency is 17.8 %, while the propulsive and overall efficiencies are zero at zero flight speed. The total is 171.5 kilonewtons with a T S F C of 10.75 milligrams per second per newton, a total flow of 1 120 kilograms per second and a bypass ratio of 16.4, and here eta L equals the figure of merit, 0.75.

Let us put the four results side by side. With the same gas generator and the same turbine inlet temperature, replacing the ducted fan by an unducted propeller multiplies the bypass ratio several times and cuts the fuel consumption by about 20 % at cruise and 35 % on the ground. The overall efficiency rises from 17.1 to 21.75 % at cruise. The price is the tip Mach number limit on flight speed, about Mach 0.8 even with swept blades, plus noise, the gearbox and installation issues; that is the whole story of propfans and open rotors.

Here is the second page of your cheat sheet. The mass flows: core, bypass and fuel referred to the core, the fan on all the flow, and who drives whom on each spool. The procedure in one line, which is exactly the structure of exercise 1 A. Momentum theory and the propeller efficiency model, with the gearbox in the propeller power. And the static case: the figure of merit formula, the check that it cannot exceed one, and the reminder that propulsive and overall efficiencies are zero when the aircraft is not moving. In part three we go inside the compressor and the turbine, with velocity triangles and exercises 2 A to 3 B.