Propulsion: fundamentals

Welcome to the first of three lectures for the aircraft propulsion partial exam. Part one builds the toolbox: thermodynamic fundamentals, the station numbering, the model of every engine component, the thrust equation and the efficiencies, and the complete turbojet procedure with a worked example. Part two applies it to turbofans, turboprops and open rotors, solving exercises 1 A and 1 B in full, and part three covers compressor and turbine stages with exercises 2 A, 2 B, 3 A and 3 B. Every number shown has been recomputed with a cycle solver that reproduces the official solutions.

First, the rules, from the Exams page that the professors updated on the twenty-second of October. There is a conflict on the date: the timetable viewer says Monday the second of November, but both the Atenea Exams page and the course overview say Thursday the fifth, from 15:00 to 18:00 in room 2.6 of TR5; two course sources agree on the fifth, but confirm it by email with both professors. The exam lasts two and a half hours with no breaks, and leaving the room means handing in the exam. You may bring water, your ID, pens of any colour except red, and a non-programmable calculator. And you may bring a cheat sheet of two pages, front and back, handwritten or printed, which is collected with the exam, so keep a photocopy. The partial is worth 30 % of the course and covers the first half of the course plan: on-design analysis and turboprops, compressors and turbines, exactly the topics of exercises 1 A to 3 B.

The whole course rests on a few hypotheses. The gas is calorically perfect, with one set of properties for the cold section, air with gamma 1.4, and another for the hot section, combustion products with gamma about 1.3; the flow is one-dimensional, steady and adiabatic in ducts, and isentropic only where explicitly stated. The equation of state is P equals rho R T, with R equal to c p times gamma minus one over gamma, and mass conservation is rho v A constant. The first law balances heat and work against the change of total enthalpy, with the sign convention that work is positive when extracted, as in a turbine. Entropy changes as c p log of the temperature ratio minus R log of the pressure ratio, and this is the relation to remember, because losses will always appear as total pressure drops at constant total temperature. Finally, the isentropic relation links temperature and pressure ratios with the exponent gamma minus one over gamma, and the speed of sound is the square root of gamma R T.

Total, or stagnation, quantities are the backbone of every engine calculation, and there is one asymmetry you must understand. Total temperature only requires the deceleration to be adiabatic and without work: T t equals T plus v squared over two c p. Total pressure requires, in addition, that the deceleration be reversible. In terms of Mach number, T t over T is one plus gamma minus one over two M squared, and P t over P is that same bracket raised to gamma over gamma minus one. The consequence is that in a duct with no work and no heat the total temperature is constant whatever the losses, while the total pressure drops with the entropy generated, P t 2 over P t 1 equals e to the minus delta s over R, as across a normal shock. That is why every component is described by a total temperature ratio tau and a total pressure ratio pi, and why an adiabatic inlet has tau d equal to one but pi d below one.

Whenever continuity meets the total quantities, the same group appears: the mass flow parameter, m dot times the square root of R T t, divided by the area normal to the flow and by P t. It depends only on gamma and the Mach number: the square root of gamma times M, times one plus gamma minus one over two M squared, raised to minus gamma plus one over two times gamma minus one, the formula in the title of the plot, which you must copy onto your cheat sheet. It has a maximum at Mach one: for given total conditions, the largest mass flow per unit area is sonic, which is the statement that the flow is choked; as an example, for gamma 1.3 and Mach 1.11, the turbine stator exit of exercise 3 A, it is 0.661. We use it to size every station: the area equals m dot times root R T t over P t times M F P; in turbomachinery the normal area is the annulus area times the cosine of the flow angle. One warning from the manual: Mattingly's tables use a dimensional M F P that absorbs the square root of R, so if an area comes out wrong by a factor of about 17, the root of 287, you have mixed the two conventions.

The station numbers follow the SAE standard A S 755, and you must use them without hesitation. In a turbojet: 0 is the free stream, 2 the compressor face, 3 the burner entry, 4 the burner exit and turbine entry, 5 the turbine exit and 9 the nozzle exit. A turbofan adds 13 at the fan exit and 19 at the fan nozzle exit, where the bypass stream leaves through its own nozzle, and two spools drawn as shafts: the high pressure turbine drives the high pressure compressor, and the low pressure turbine drives the fan and the booster; 2.5 sits between the compressors and 4.5 between the turbines. In a turboprop, the low pressure, or free, turbine drives the propeller through a gearbox, and the propeller stream is an external stream with no nozzle, handled with momentum theory. An open rotor has the same block diagram, with the rotor, often a pusher behind the gas generator, driven by the low pressure turbine; these block diagrams, with their shafts and streams, appear in the exercises, so practise drawing them quickly.

Every component ratio is written pi for total pressure and tau for total temperature, with a subscript for the component. Two members of the family are special: pi r and tau r describe the free-stream deceleration, so they belong to the flight condition and not to any hardware. The other is tau lambda, c p t times T t four divided by c p c times T zero: it compares the hottest total enthalpy in the engine with the coldest static enthalpy, and almost every performance trend is best read as a function of it. In the exercises the cold gas has gamma 1.4 and c p 1 004.5, and the hot gas gamma 1.3 and c p 1 239 joules per kilogram kelvin. And the fuel-to-air ratio f is always referred to the core mass flow entering the burner, which is the source of the most common error in turbofan fuel consumption.

Now the components, one by one, starting with the flight condition and the inlet. The free-stream total temperature and pressure follow from the flight Mach number, assuming a reversible deceleration, so all real inlet losses go into pi d. Pi d is the friction and diffusion loss, pi d max, times the shock recovery of the military standard: one for subsonic flight, and one minus 0.075 times M zero minus one to the 1.35 for supersonic flight, 0.925 at Mach 2. The inlet is adiabatic, so tau d equals one, and if an isentropic efficiency is used instead, it compares the ideal and actual pressure recovery. The inlet delivers the air to the fan or compressor at a Mach number of about 0.4 to 0.6, and in exercise 1 A, at 11 km and Mach 0.8, the total temperature is 244.55 kelvin and pi d is 0.98.

The compressor raises the total pressure by pi c, a design choice, and the temperature rise follows from its figure of merit. With a polytropic efficiency e c, tau c equals pi c to the power gamma minus one over gamma e c. With an isentropic efficiency, which is how the exercises give the data, tau c equals one plus pi c to the gamma minus one over gamma, minus one, all divided by eta c. The power absorbed is m dot c p times the total temperature rise, and the kinetic energy change is already inside the total enthalpy. The fan is treated exactly like a compressor, but remember that it compresses the whole inlet flow, and its pressure ratio is limited to about 1.4 to 1.7 in civil engines by the tip Mach number. Applying these formulas to exercise 1 A gives 268.3 kelvin after the fan, 457.0 after the low pressure compressor, and 771.2 at the burner entry.

Two efficiencies exist for every turbomachine, and the exam expects you to know the difference. The polytropic efficiency is the efficiency of an infinitesimal stage, applied uniformly along the machine; the isentropic efficiency compares the whole machine with one reversible process between the same pressures. Because constant pressure lines diverge on the T s diagram, a compressor's isentropic efficiency is lower than its polytropic one, and the gap grows with the pressure ratio; in a turbine it is the reverse, the reheat effect. This table, from the manual, uses exactly the components of exercise 1 A: the fan's 0.85 is a polytropic 0.856, the low pressure compressor's 0.83 is 0.863, while the turbines' 0.80 and 0.81 are polytropic 0.78. Use the polytropic efficiency to compare designs with different pressure ratios, and the isentropic one when analysing a specific engine whose data are given that way, as in the exercises.

The combustor burns fuel and delivers the gas to the turbine at a uniform temperature T t four, which is limited by the turbine materials and is the single most important design parameter. The energy balance says that the combustion efficiency times the fuel flow times the heating value equals the total enthalpy leaving, with the hot c p, minus the total enthalpy entering, with the cold c p. Solving for f gives c p t T t four minus c p c T t three, over eta b h P R minus c p t T t four, or the equivalent form with tau lambda and tau r tau c. The burner loses total pressure, roughly one minus epsilon gamma over two times the burner Mach number squared, about 0.95 for a Mach of 0.2, which is why burners are wide; the exercises use 0.90. For exercise 1 A, with T t three of 771 kelvin, T t four of 1 600 kelvin and a heating value of 42.5 megajoules per kilogram, f is 0.0301; careful with a claim you may read, the formula only blows up at about 34 000 kelvin, which is meaningless, while the real limit is chemistry: the stoichiometric fuel ratio of kerosene, about 0.068, corresponds to roughly 2 300 to 2 500 kelvin, and engines stay well below it.

The turbine extracts power from the hot gas to drive the compressor, the fan and the accessories. Its power is the hot mass flow, one plus f times the core flow, times c p t times the total temperature drop, and the shaft delivers a fraction eta m of it to the compressor. Equating the two gives tau t, and this power balance is the heart of the calculation, the only equation that links the hot section back to the cold one; for a turbojet it reads one minus tau r times tau c minus one over eta m, one plus f, tau lambda. Only then do you get the pressure ratio, either with the polytropic exponent or, as in the exercises, by inverting the isentropic efficiency: pi t equals one minus one minus tau t over eta t, to the gamma over gamma minus one. Notice that the polytropic efficiency divides the exponent for a compressor and multiplies it for a turbine, because losses increase the compressor work and decrease the turbine work; in exercise 1 A this gives 1 339.7 kelvin and 3.49 bar after the high pressure turbine and 1 045.6 kelvin after the low pressure one.

The nozzle accelerates the gas, and it is adiabatic, so T t nine equals T t five. For a convergent-divergent nozzle the exit pressure ratio is a design choice, and the exit Mach number follows from the total-to-static pressure ratio. For a purely convergent nozzle, which is the case in the exercises, you must use hypothesis and check: first assume the nozzle is choked, Mach one at the exit, where P t over P equals gamma plus one over two to the gamma over gamma minus one, 1.832 for hot gas and 1.893 for air. If that critical pressure is at least the ambient pressure, the nozzle is choked and the exit pressure is the critical one; otherwise the nozzle is matched, the exit pressure equals ambient, and you compute a subsonic exit Mach number. With a nozzle isentropic efficiency, as in the exercises, the critical static temperature is two T t over gamma plus one, and the critical pressure comes from the efficiency definition. In exercise 1 A the core nozzle's critical pressure, 0.447 bar, is above the ambient 0.227 bar, so it is choked with 581 metres per second, while the fan nozzle is matched at Mach 0.97 and 292.5 metres per second; it is normal for the two nozzles of one engine to be in different regimes.

The thrust equation comes from a momentum balance on a control volume whose upstream boundary is pushed into the undisturbed free stream, which eliminates the unknown pre-entry pressure force. The result is F equals m dot nine v nine minus m dot zero v zero plus P nine minus P zero times A nine: a momentum term, the ram drag of bringing the air on board, and a pressure term that exists only when the nozzle is not matched. For a turbofan you add the fan stream terms, and the exit areas come from continuity with the static exit state; never forget the pressure term of a choked nozzle. The installed thrust subtracts the nacelle and additive drags, but the engine efficiencies use the uninstalled thrust. A nice insight, due to Farokhi: writing the force on each component as the change of impulse function shows that the compressor and the burner carry large forward forces, while the turbine and the convergent nozzle pull rearwards; the net thrust is the small difference of large forces; the manual's text swaps these signs, so trust this version.

Every engine is judged with the same handful of numbers. Specific thrust is the thrust per unit of inlet mass flow, and it sets the size of the engine; the thrust specific fuel consumption is the fuel flow per unit thrust, usually in milligrams per second per newton, and enters the Breguet range equation. The thermal efficiency is the mechanical power produced, the increase of kinetic energy of all the streams, divided by the fuel power, m dot f times h P R. The propulsive efficiency is the useful power, F v zero, divided by that kinetic energy increase, and the manual stresses that the full thrust, including pressure terms, must be used. The overall efficiency is their product, F v zero over the fuel power, which equals v zero over h P R times T S F C: a perfect cross-check, since for exercise 1 A, 236 metres per second over 42.5 megajoules times 32.49 milligrams per newton second gives exactly the official 17.1 %.

Here is the central trade of the whole subject. For a single matched stream, the propulsive efficiency is approximately two over one plus v nine over v zero: the closer the jet speed to the flight speed, the higher the propulsive efficiency. Raising T t four raises the thermal efficiency but also the jet velocity, which costs propulsive efficiency; raising the pressure ratio also raises the thermal efficiency, but the jet velocity only grows up to the pressure ratio of maximum specific thrust and then falls, so beyond it the propulsive efficiency recovers, as the design-trend slide will show. The turbofan breaks the coupling: it takes the core power and spreads it over a much larger mass flow at a lower velocity, so the overall efficiency improves even if the thermal efficiency does not, and the propeller is the extreme case of the same idea. That is the reason bypass ratios have kept growing for four decades, and why open rotors are studied today.

All of this assembles into an explicit procedure with no iteration, the same for every engine. Step one, gas properties and flight condition; step two, the inlet. Step three, the compressor temperature ratio, and step four, the burner: tau lambda and the fuel ratio f. Step five, the turbine from the power balance, first tau t, then pi t; this is the step students most often forget. Step six, the nozzle: the total-to-static pressure ratio is the product of all the pi's times P zero over P nine, then the exit Mach number, with the choked or matched check for a convergent nozzle; step seven, the exit temperature and velocity, being careful with the different gamma R of the hot gas. Step eight, the performance: specific thrust, T S F C equal to f over the specific thrust, and the three efficiencies.

The manual's worked example, from Mattingly, lets you validate the procedure line by line: Mach 2 at 12 km, compressor pressure ratio 10, T t four 1 800 kelvin and a convergent-divergent nozzle with P zero over P nine of one half. The free-stream gives tau r 1.8 and pi r 7.82, and the shock recovery 0.925 brings pi d down to 0.879. The compressor gives tau c 2.08 with an isentropic efficiency of 0.864, below the polytropic 0.9 as expected, and the burner needs f equal to 0.0357. The power balance gives tau t 0.8155 and pi t 0.375, the nozzle expands to Mach 2.25, and the jet leaves at 4.25 times the free-stream speed of sound. The specific thrust is 807 newton seconds per kilogram and the T S F C 44.2 milligrams per second per newton, with efficiencies of 42, 74 and 31 %. The propulsive efficiency is high because at Mach 2 the jet is only about twice as fast as the aircraft, and the strong ram compression keeps the thermal efficiency high despite the inlet shock loss; that is why turbojets survive at high Mach numbers, where turbofans do not.

Two trends must be understood qualitatively. There is an optimum compressor pressure ratio: raising it raises the thermal efficiency, but the turbine must then expand more to drive it, leaving less pressure for the nozzle, so beyond a point the specific thrust falls. For the ideal cycle, maximum specific thrust occurs when tau c equals the square root of tau lambda divided by tau r, while the minimum fuel consumption occurs at a much higher pressure ratio, so real engines are designed between the two, as these ideal curves show. And as the flight Mach number grows, the ram compression does the job and the optimum pressure ratio falls, until above Mach 3 or so the compressor and turbine can be removed: the engine becomes a ramjet.

Let me summarise part one as the first page of your cheat sheet. Total temperature is conserved without work or heat, total pressure falls with losses, and each component is a tau and a pi, with tau lambda as the key cycle parameter. Write both forms of the compressor and turbine relations, polytropic and isentropic, the fuel ratio formula, and the reminder that the power balance gives tau t. Write the convergent nozzle check with its two critical ratios, and the mass flow parameter formula for sizing areas. And the thrust, T S F C and efficiency formulas, including the quick check that the overall efficiency is v zero over h P R times T S F C. In part two we apply all of it to the turbofan of exercise 1 A, to its static version 1 B, and to the propfan and open rotor conversions.