Lecture Notes Aerospace Propulsion

8/16/2019 Lecture Notes Aerospace Propulsion

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Lecture Notes on Aerospace Propulsion

Bioengineering and Aerospace Engineering Department

Universidad Carlos III de Madrid

2015-2016

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Aerospace Propulsion Lecture notes

These lecture notes were originally based on a course that has been taught several years at MIT by Manuel

Mart́ınez-Sánchez. Many of the figures have been taken or adapted from the lecture notes from that course andfrom the book Aircraft Engines and Gas Turbines by J.L. Kerrebrock published by MIT press.The lecture notes have been gradually improved upon by Manuel Mart́ınez-Sánchez, Manuel Garćıa Villalba,

Pablo Fajardo, Mario Merino, and Andrea Ianiro. If you discover any errata, or have any suggestions, pleasecontact us at:

• [email protected]

2 Bioengineering and Aerospace Engineering Dept. v. 2015-2016


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Contents

1 Introduction to aerospace propulsion 6

1.1 Thrust generation and jet propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 Effect of external expansion on thrust . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Global performance parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Range of aircraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3.2 Efficiencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Aircraft Engine Modeling: the Turbojet 13

2.1 Thrust equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Shaft balance for the turbojet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Fuel consumption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.4 Design parameters. Effect of mass flow on thrust. . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.4.1 Note on Ramjets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Propulsive efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.6 Thermal and overall efficiencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Introduction to Component Matching and Off-Design Operation 22

3.1 Discussion on nozzle choking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 Component matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.3 Effects of Mach number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.5 Compressor-turbine matching. Gas generators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Turbofan Engines 31

4.1 Ideal turbofan model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2 Shaft balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.3 Velocity matching condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.4 Optimal compression ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Inlets and Nozzles 36

5.1 Inlets or Diffusers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Subsonic Inlets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.3 Supersonic Inlets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.4 Exhaust nozzles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6 Principles of Compressors and Fans 47

6.1 Euler’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486.2 Velocity triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3 Isentropic efficiency and compressor map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7 Compressor Blading, design and multi-staging 53

7.1 Diffusion factor. Stall and surge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537.2 Compressor blading and radial variations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547.3 Multi-staging and flow area variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567.4 Mach Number Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577.5 The Polytropic Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577.6 Starting and Low-Speed Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59



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8 Turbines. Stage characteristics. Degree of reaction 60

8.1 Euler’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608.2 Degree of Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608.3 Radial variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 648.4 Rotating blade temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

9 Turbine solidity. Mass flow limits. Internal cooling. 66

9.1 Solidity and aerodynamic loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669.2 Mass flow per unit of annulus area and blade stress . . . . . . . . . . . . . . . . . . . . . . . . . . 679.3 Turbine cooling. General trends and systems. Internal cooling. . . . . . . . . . . . . . . . . . . . 68

10 Film cooling. Thermal stresses. Impingement. 72

1 0 . 1 F i l m c o o l i n g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 721 0 . 2 I m p i n g e m e n t c o o l i n g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

10.3 T hermal stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7410.4 How to design cooled blades . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

11 Combustion: Combustors and Pollutants 77

11.1 C ombustion process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771 1 . 2 C o m b u s to r c h a m b e r s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791 1 . 3 C o m b u s to r s i z i n g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8011.4 A fterburners . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8111.5 Pollutants: regulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8211.6 Mechanisms for pollutant formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8411.7 Upper-Atmospheric Emissions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

12 Introduction to engine noise and aeroacoustics 89

12.1 N oise propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8912.2 Acoustic energy density and power flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9012.3 Noise sources and noise modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9112.4 Jet Noise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9412.5 Turbomachinery noise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

13 Engine rotating structures 97

13.1 B lade loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9713.2 Centrifugal stresses and disc design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

14 Fundamentals of rotordynamics 100

14.1 Bearings and engine arrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10014.2 L umped mass model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10114.3 C ritical speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10314.4 Forces on bearings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10314.5 Comments on blade vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104



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Main nomenclature

General engine and flight magnitudesI sp Specific impulse [m/s] (NB: also used in [s] in older texts)F Thrust [N]D Drag [N]L Lift [N]W Weight [N]ṁf Fuel mass flow [kg/s]ṁ0; ṁ Mass flow (of air) [kg/s]m̄i Nondimensional mass-flow (with respect to the critical mass flow rate) [-]SFC Specific fuel comsumption [s/m]f Fuel mass ratio to air mass flow, f = ṁf / ṁ0 [-]

h Specific heat value of the fuel [J/kg]η, η p, ηt Overall, propulsive and thermal efficiency; η = ηtη p [-]M 0 Flight Mach number [-]M i Flow Mach number at stage i [-]u0 Flight velocity [m/s]u9 Exhaust velocity [m/s]R Aircraft range [m]

Component subindicesc Subindex for ‘compressor’t Subindex for ‘turbine’b Subindex for ‘burner’ (combustor)a Subindex for ‘afterburner’d Subindex for ‘diffuser’n Subindex for ‘nozzle’f Subindex for ‘fan’

Thermodynamic properties: stagnation states and ratios pti, T ti Stagnation (total) pressure and temperature at station i [Pa, K] pi, T i Static pressure and temperature at station i [Pa, K]πc, πt, etc Stagnation pressure ratios of compressor, turbine, etc. [-]τ c, τ t, etc Stagnation temperature ratios of compressor, turbine, etc. [-]θi Non-dimensional stagnation temperature at station i (normalized with T 0) [-]δ i Non-dimensional stagnation pressure at station i (normalized with p0) [-]θt Non-dimensional stagnation temperature at turbine inlet, θt = T t4/T 0 [-]Θ turbine inlet to free-stream Stagnation temperature ratio, Θ = T t4/T t0 [-]ai Sound velocity at station i, [m/s]. For an ideal gas, a = γRgT M i Mach number at station i, [-]

Γ(γ ) Choked nondimensional mass flow parameter, Γ(γ ) =√

γ

2γ +1

γ+12(γ−1)

[-]

Turbomachineryu,v,w Radial, Azimuthal (tangential), and axial velocity components [m/s]V Velocity modulus [m/s]D Diffusion factor [-]ψz Zweifel coefficient [-]R Degree of reaction [-]φ, ψ Flow and power coefficients [-]



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1 Introduction to aerospace propulsion

1.1 Thrust generation and jet propulsion

Propulsion is the action of exerting a forward force F (termed thrust ) on a vehicle. According to Newton’ssecond law,

d (mv)

dt = F + D (v) , (1.1)

this thrust is necessary to accelerate the vehicle (i.e., change its momentum) or to compensate any resistiveforces that oppose to the vehicle’s movement in order to maintain a uniform velocity (in the case of aircraft,this force is the aerodynamic drag ).

Newton’s third law establishes that the existance of this force is necessarily linked to the existance of areaction force equal in magitude but opposite in direction, acting somewhere else, e.g. the ground for a groundvehicle. For an aircraft in flight, the reaction is exerted on the air that has been admitted by the engine for the

purpose, and that is therefore accelerated downstream, together with some fuel products. On a rocket in space,where there is no air to be admitted, the reaction force can only be exerted on some material that is carriedon board and expelled from the vehicle at high speed. As we will see, it is common to evaluate thrust in termsof the momentum gained by the medium (i.e., the reaction force), but it must be kept in mind that this is anindirect procedure, and the thrust force is, by definition, the force exerted by the medium on the vehicle (incomplicated ways, to be sure).

Conceptually, the simplest aerospace propulsion device is the Rocket. In a rocket combustion chamber,some propellants carried on board are reacted together to generate heated gas at high pressure, and this gas isallowed to flow through an open duct (a nozzle ), where it pushes forward on the walls and is pushed backwardsby the walls. The speed u9 of the gas as it leaves the nozzle is a measure of the momentum per unit mass ithas gained, and so, if the mass flow rate is ṁ9, the thrust is

F

≈ ṁ9u9. (1.2)

As it can be easily shown by applying the momentum equation to an appropriate control volume traveling withthe rocket, this equation is only formally correct if the nozzle is working under pressure-matching conditions.Otherwise, the gas may continue to expand and accelerate outside the engine, and a correction is necessary, aswill be shown in section 1.2.

!"#$%&

()

Figure 1.1: Mechanism of thrust generation in a rocket

In an airbreathing engine, such as a Turbojet, air is admitted from the atmosphere at a rate ṁ0 with arelative speed u0 (equal to the flight speed), and fuel is added so that the expelled mass flow rate is ṁ9. Then,the new expresssion for thrust is

F = ṁ9u9 − ṁ0u0. (1.3)



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In this equation, we are again assuming that the nozzle is operating at pressure matching conditions. The besic

parts of a Turbojet are the air intake, a compressor, a combustion chamber, a turbine and a nozzle.

Figure 1.2: Turbojet schematic

Clearly, if we move a larger airflow ṁ0, we need to accelerate it less (i.e., the exhaust velocity will be lower)to provide the same thrust: this is intuitively more advantageous, as the amount of jet momentum left behinduselessly is decreased (in the limit, one would like to react against the whole planet, as in a ground vehicle).This idea will be understood more clearly in section 1.3.2, where the propulsive efficiency is introduced, andis the main argument for replacing the turbojet by the Turbofan. In the turbofan, the admitted air is split

into core air, which reacts with the fuel and generates the torque required to drive the fan, and bypassed air,moved by the fan. The core may be basically a turbojet with an added low-pressure turbine, and the bypassactually handles most of the flow. The turbofans that are nowadays in operation reach up to a 9:1 mass flowratio between the bypass and the core streams.

The thrust is now (again, aside from external expansion corrections) the sum of the momenta added to bothstreams,

F = ṁcoreucore + ṁbypassubypass − ṁ0u0 (1.4)In terms of propulsive efficiency, the next logical step would be to eliminate (to the extent possible) the

core and obtain a simple Propeller, or in a modified form, a Helicopter Rotor. This propeller has to be drivenby an engine, generally the core of a turbojet (optimized to transmit most power to the propeller instead of increasing the energy of the exhaust gases of the turbojet) or a reciprocating piston engine. The historical orderhas been, of course, the reverse (propellers first, followed by turbojets and later turbofans), but this had to dowith details of technological feasibility that you will appreciate better as we progress in this course.

1.2 Effect of external expansion on thrust

We will only discuss in detail the case of a rocket, and the others can be easily understood by extension. Weneed to calculate the total force exerted by the gas applying the momentum equation to a control volumecomprising the system. If the pressure p9 at the nozzle exit is not equal to p0, the undisturbed outside pressure,the internal gas is pushed forward by the gas outside by a force ( p9 − p0)A9, where A9 is the exit plane area.This adds a second contribution to the thrust force:

F = ṁu9 + ( p9 − p0)A9. (1.5)



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This is the rational definition, but historically, specific impulse has been defined as thrust per unit weight flow

rate of fuelĨ s =

F

g ṁf (in s). (1.7)

In this notes we are going to try to use most of the times the first definition, although students should alwayscheck units carefully to be sure about which definition is being used. For a rocket, the only flow rate is that of the fuel, but for an airbreathing engine, care must be taken to use the fuel flow rate in the calculation of I s,not the air or the total flow rate.

The inverse of I s is also in common use, especially for airbreathing engines, and it is called the Specific FuelConsumption (or Thrust Specific Fuel Consumption), often expressed in picturesque units such as “pounds of fuel mass per hour-per pound of thrust force”, or the metric equivalent,

SFC or TSFC = 1

I s=

ṁf F

(in s/m). (1.8)

1.3.1 Range of aircraft

For an aircraft, the simplest measure of performance is the cruise fuel consumption and the resulting range. Atone time this was also the critical performance measure for transports, bombers and fighters. This is less truenowadays for transports because the ranges accessible with modern engines and airframes are in the order of 13,000 km. For bombers, aerial refueling extends the range to the extent that again range is no longer sucha challenge (although in-flight refueling is quite expensive). Range is still important for fighters because therequirements for high speed and maneuverability conflict with those for long range.

Figure 1.4: Equilibrium of forces in straight level flight.

Consider an aircraft in straight, level flight as shown in Fig. 1.4. The thrust has to compensate the drag,F = D, and the lift has to compensate the weight, L = W . So that we may write

F = D = DL

L =

L

L/D =

W

L/D. (1.9)

During cruise the rate of change of aircraft weight W is equal to g times the mass flow rate at which fuel isburned, that is

dW

dt = −g ṁf = −g F

I s= −g W

I sL/D, (1.10)



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where the first definition of I s, eq. (1.6), has been used.

If we assume that I s and L/D are constant (here we are implying models for both the propulsion systemand the aircraft), then

dW

W = −g dt

I sL/D, (1.11)

log

W (t)

W 0

= log

m(t)

m0

= −g t

I sL/D. (1.12)

Now we can define the range of an aircraft as R = u0t, where u0 is the flight cruise speed, and introducingt = R/u0 into eq. (1.12) we obtain

R = u0I sg

L

D log

m0m(R)

(1.13)

where m(R) is the aircraft mass at the range R. This is the Breguet range equation, named after one of the aviation pioneers, the French aircraft designer and builder Louis Charles Breguet. This equation is alsoexpressed in terms of the SFC instead of using the specific impulse as

R = u01

g · SFCL

D log

m0m(R)

. (1.14)

It is useful to divide the initial mass of the aircraft m0 into several parts: ‘empty’ mass (all the mass requiredfor the aircraft to operate but the fuel, that is: structure + engines + crew), ‘payload’ and ‘fuel’:

m0 = mempty + m pay + mfuel, (1.15)

or, in terms of mass fractions

1 =

mempty

m0 +

m pay

m0 +

mfuel

m0 . (1.16)

If the fuel is all expended at R,

m(R)

m0=

memptym0

+ m pay

m0= 1− mfuel

m0. (1.17)

For a fixed structure and engines, we can trade off between payload and fuel, hence between range and payload.We can now re-write eq.(1.13), as

R = u0I sg

L

D log

m0

mempty + m pay

(1.18)

and solving for m pay, we obtain

m paym0

= exp

− Rg

u0I sL/D

− mempty

m0. (1.19)

From this we can construct a range vs. payload chart. As an example suppose

mempty = 0.7m0,

u0 = 300m/s,

I s = 39200m/s,

L/D = 15,



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so that

u0I sg

L

D = 1.8 · 107

m = 1.8 · 104

km, (1.20)

and we obtain the chart displayed in Fig. 1.5. Note that this is not a straight line, although it is close. Thisgraph presents the maximum payload capacity of the aircraft as a function of the range required, showing thetheoretic interchangeability between payload and range. This behavior is obviously limited by the maximumcapacity of the fuel deposits in the considered aircraft.

mpaym0

0 1000 2000 3000 4000 5000 6000 70000

0.05

0.1

0.15

0.2

0.25

0.3

R [km]

Figure 1.5: Range vs. payload chart.

1.3.2 Efficiencies

For airbreathing engines, three kinds of efficiency are in use, that will be discussed more fully later

• Overall efficiency, defined as the propulsive power (thrust times flight velocity) over the input thermalpower, which is the product of the fuel flow rate and the “fuel heat value” (heat of combustion) h, i.e.,the amount of heat released in combustion per kg of fuel

η = F · u0

h ṁf =

u0h

I s, (1.21)

where the last equality follows from the definition of specific impulse.

In the case of the turbojet or the turbofan, the exhaust mass flow can be approximated as

ṁ9 = ṁ0 + ṁf ṁ0. (1.22)

• Propulsive efficiency, defined as the ratio of propulsive power (useful propulsion power) to the rate of addition of jet kinetic energy to the jet

η p = F u0

ṁ0 (u29 − u20) /2, (1.23)

using here F = ṁ0 (u9 − u0), equation (1.23) reduces to

η p = 2u0u9 + u0

. (1.24)



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Notice here that η p tends to 1 when u9 tends to u0. For this to mean a non-zero thrust, the mass flow

rate must be large; this is again the argument for the turbofan or the propeller, as opposed to the pureturbojet: ideally, we would like to exert the reaction force to our thrust in as large an air flow as possible,to minimize the kinetic energy deposited in the air and maximize η p. Observe that for a rocket thepropulsive efficiency is undefined and does not make sense (flight velocity becomes irrelevant, and thereis no external airflow flowing through the engine).

• Thermodynamic efficiency, ηt , defined as the ratio of jet kinetic power to input thermal power. Clearly,

η = ηt · η p (1.25)



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2 Aircraft Engine Modeling: the Turbojet

All aircraft engines are Heat Engines They use the thermal energy derived from combustion of fossil fuels toproduce heat and then transform it into mechanical energy in the form of kinetic energy of an exhaust jet.The excess of momentum of the exhaust jet over that of the incoming airflow produces thrust. As such, thethermodynamic efficiency of the device cannot exceed Carnot’s efficiency limit,

ηt ≤ 1− T low/T high, (2.1)

here being T high the temperature after combustion and T low the ambient temperature.In studying these devices we thus employ two types of models:

• Thermodynamic, in which the conversion of thermal energy in mechanical energy from is studied by theapproaches of Thermodynamics. Here the change in thermodynamic state of the air as it passes throughthe engine is studied. The detail of the fluid movement within each component is not identified. Rather,the processes are specified by pressure and temperature ratios. This is the type of model that we derivein this chapter.

• Fluid mechanical, in which we relate the changes in pressure, temperature and velocity of the air, tothe physical characteristics of the engine, e.g. the flow field over the blades of the compressor, in thecombustion chamber and in the turbine.

2.1 Thrust equation

With these ideas in mind, let us first outline a general approach to the modeling of aircraft propulsion systems.Our general expression for thrust, in which we have a main interest, is

F = ṁ9u9−

ṁ0u0 + ( p9−

p0)A9, (2.2)

where ṁ9 = (1 + f ) ṁ0 includes the fuel mass flow added to the airflow, and f = ṁf / ṁ0. Subscripts indicatethe standard flow station as shown in Fig. 2.1. We write this more conveniently in dimensionless form as

F

ṁ0u0= (1 + f )

u9u0− 1 + p0A9

ṁ0u0

p9 p0

− 1

. (2.3)

In our modeling of the aircraft engine we will often assume that the nozzle is operating at pressure matchingconditions ( p9 = p0), and usually take f 1, so, using the flight Mach number M 0 = u0/a0, this expressionbecomes simply

F

ṁ0a0= M 0

u9u0− 1

, (2.4)

but it should be recalled that the behavior of the nozzle is somewhat more complex. In practice the devi-ation from ideal expansion becomes important for supersonic flight. In particular, there can be supersonicunderexpansion, with an exhaust pressure p9 > p0. This topic will be analyzed in detail in future lectures.

Our tasks in estimating F are then

1. To estimate ṁ0, which will depend on the engine front area, flight velocity, etc., and

2. To estimate u9u0

. We will start by discussing this first.

Many of the engines we deal with (Turbofans) will have 2 exhaust streams. In this case we apply eq. (2.4)separately to each stream.



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In the following, we employ the defining relations for the stagnation properties1

T t = T

1 +

γ − 12

M 2

, (2.5)

pt = p

1 +

γ − 12

M 2 γ

γ−1

. (2.6)

Remember that a process that does not add or remove heat or work to the fluid always has T t = const, andthat if T t is constant and the process is isentropic, then pt is constant too.

Let us begin with a general Turbojet Engine, shown in Fig. 2.1 with the standard station numbering2.Station 0 refers to the unperturbed airflow, far upstream.

Figure 2.1: Schematic diagram of turbojet engine. This is the standard station numbering developed by SAE.

And, let us split the engine into a set of Components with functions as follows (in parenthesis, the letterthat will identify the component):

• Diffuser (d) or inlet: Brings airflow from the flight Mach number M 0, to the axial Mach number M 2,required by the compressor. Ideally ∆S = 0 and ∆T t = 0, so ∆ pt = 0. In reality, especially for M 0 > 1,the process is not isentropic. The diffuser is the region located between flow stations 1 and 2 in Fig. 2.1.

• Compressor (c): Raises the pressure (and the temperature) of the airflow by adding mechanical energy toit, as isentropically as possible. If ideal, then

pt3 pt2

= T t3T t2

γγ−1

, with T t3 > T t2 (work added). Losses and

real effects reduce the achievable pt3 pt2

. The compression process in a large engine might be split in several

steps, e.g. low pressure and high pressure compressor.

• Combustor (b): Raises temperature by adding heat (from the chemical energy of the burnt fuel), ideallyat near-constant stagnation pressure (Brayton cycle). In practice, some stagnation pressure losses alwaysoccur. The maximum temperature that can be obtained with the combustion is fixed by the maximumtemperature that the turbine can withstand.

1Check the companion notes, “Basic Relations of Gas Dynamics”, for a brief review of stagnation properties and other centralconcepts.

2Students should be aware of slight differences between this station numbering and that used in the book from Kerrebrock.



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• Turbine (t): Extracts mechanical energy from the airflow, dropping its temperature and pressure, as nearlyisentropically as possible. T t5 < T t4 (work extracted) with an ideal pressure drop pt5

pt4=

T t5T t4

γγ−1. The

work extracted in the turbine is used to drive the compressor and sometimes also some auxiliary systems.

• Afterburner (a): Only present in military turbojets, the afterburner heats air again by burning extra fuel,at nearly constant stagnation pressure. Here the air temperature can be increased without the limitationsdue to the thermal stresses in the turbine.

• Nozzle (n): Expands hot gases to produce a high-velocity jet. Station 8 denotes the nozzle throat (sectionof minimal area in the nozzle).

In order to calculate the velocity ratio u9/u0, we first note that

u9u0 =

M 9M 0 T 9

T 0 . (2.7)

It is more efficient (and easier) to find the exit Mach number and temperature by keeping track of thestagnation temperatures and pressures through the several components. The following procedure works for allaircraft engines, so it’s worth your paying some attention to the procedure itself, as well as the result.

It is very helpful to define a set of symbols that represent explicitly ratios of the stagnation properties anddistinguish them from the static or thermodynamic properties of the gas, because in general it is the stagnationproperties that most conveniently represent the effect of the components on the fluid as it flows through theengine. Thus,

• A ratio of pt’s through a component will be denoted by the symbol π and the subindex of that component.

• Similarly, a ratio of T t’s will be denoted by the symbol τ .

• A ratio of a stagnation temperature (at one station) to the ambient static temperature T 0 will be denotedby θ with the corresponding station subscript.

• Similarly, a ratio of a stagnation pressure to the ambient static pressure p0 will be denoted by δ .So, for the flow upstream of the engine,

T t0T 0

=

1 +

γ − 12

M 20

= θ0, (2.8)

pt0 p0

=

1 +

γ − 12

M 20

γγ−1

= δ 0. (2.9)

As a particular case, due to its importance in the operation of the turbojet, the turbine-inlet temperature isrepresented byT t4T 0

= θt (2.10)

or, alternatively, by

Θ = θtθ0

= T t4T t0

≡ T t4T t2

(2.11)

which is more convenient for scaling purposes, since it relates two engine total temperatures, a ratio that isoften independent of ambient conditions. Note that the ideal process in the diffuser is at constant total enthalpy



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(i.e., constant total temperature), since neither work or heat is added to the flow. For the compressor (ideal)

pt3 pt2

= πc, (2.12)

T t3T t2

= τ c, (2.13)

πc = τ γ

γ−1c , (2.14)

and for the turbine (also ideal by assumption),

pt5 pt4

= πt, (2.15)

T t5T t4

= τ t, (2.16)

πt = τ

γγ−1

t . (2.17)Now let us use this notation system to develop expressions for the Thrust and Specific Impulse of the

Turbojet Engine. We begin by tracking the changes of stagnation temperature and pressure through theengine. Temperature accounting:

T t9 = T 9

1 +

γ − 12

M 29

= T 0θ0τ cτ bτ t = T 0θtτ t. (2.18)

Pressure accounting:

pt9 = p9

1 +

γ − 12

M 29

γγ−1

= p0δ 0πcπbπt (2.19)

From equation (2.19), if p9 = p0 (ideally expanded nozzle) and if πb ≈ 1, the equation becomes

1 + γ − 12

M 29

= (δ 0πcπt)γ−1γ = θ0τ cτ t (2.20)

where the second equality assumes that the compression and expansion processes are reversible adiabatics. Fromthis we find an expression for the exit Mach number,

M 29 = 2

γ − 1 (θ0τ cτ t − 1) . (2.21)

It is very important to realize that although this expression for the exit Mach number is written in termsof temperature ratios, it comes from the pressure changes in the engine (i.e., it was calculated from Eq. (2.19)alone). This is a general result, namely that the exit Mach number depends on the ratio of jet stagnationpressure to the ambient pressure, not at all on the temperature ratio T 9/T 0. Now, from eq. (2.18)

T 9

T 0 =

θtτ t1 + γ −12 M 29

=

θt

θ0τ c =

Θ

τ c = τ b. (2.22)

So far these are quite general expressions applicable to any gas stream engine. Substituting eqs. (2.22) and(2.21) in our expressions for the velocity ratio, eq. (2.7), we have

u9u0

= 1

M 0

2

γ − 1 (θ0τ cτ t − 1) θtθ0τ c

. (2.23)

Finally, the thrust per unit of mass flow (times the speed of sound to make it dimensionless) is

F

ṁ0a0=

2

γ − 1 (θ0τ cτ t − 1) θtθ0τ c

−M 0. (2.24)



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2.2 Shaft balance for the turbojet

So far we have not made this particular to the turbojet engine, because we have not included the relationshipbetween the compressor and turbine. The fact that distinguishes the turbojet engine from other engines wemay consider later is that the turbine power, ṁ9(ht4 − ht5) equals the compressor power, ṁ0(ht3 − ht2), so:

ht3 − ht2 ht4 − ht5, (2.25)

c pT t0(τ c − 1) = c pT t4(1− τ t). (2.26)This equation is known as the power balance in the shaft which can be written as

τ t = 1− θ0θt

(τ c − 1) = 1 − 1Θ

(τ c − 1). (2.27)

So finally for the Turbojet Engine

F

ṁ0a0=

2

γ − 1

θt − θ0(τ c − 1)− θtθ0τ c

−M 0. (2.28)

2.3 Fuel consumption

We are also interested in the fuel consumption. We get ṁf from a combustor heat balance (i.e., the heat powerreleased by burning ṁf fuel per unit time must equal the increase in total enthalpy of the flow across thecombustor):

ṁf h = ṁ9c pT t4 − ṁ0c pT t3 ṁ0c p(T t4 − T t3), (2.29)where h is the heat released by burning a unit mass of fuel. In other words,

ṁf = ṁ0c pT 0

h (θt − θ0τ c) (2.30)

so that the fuel-specific impulse, I s = F / ṁf is

I s =

ha0c pT 0

1

(θt − θ0τ c)

F

ṁ0a0

(2.31)

2.4 Design parameters. Effect of mass flow on thrust.

In this section we examine the question of how to choose the key parameters of the engine to obtain some specifiedperformance at the design conditions, and how the performance varies if these parameters are changed, stillat the design conditions. Later we will look at a complementary question, namely, how the performance of a

particular design changes when conditions are different from design conditions.With the results that we worked out in the previous section for the Turbojet engine, let us look at the

dependence of F /( ṁ0a0) on the main parameters, τ c, M 0 and θt. We can view them this way

• τ c is a design choice (compressor pressure ratio)• M 0 is the flight speed (flight conditions)• θt or Θ is the combustor outlet temperature, an operating variable (we can choose how much fuel to burn),

limited by turbine materials to some maximum value.



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Assuming again that the nozzle is matched, we can re-write equation (2.28) as

F

ṁ0a0=

2

γ − 1

θt

1− 1

θ0τ c

− θ0(τ c − 1)

−M 0. (2.32)

we can see that since θ0τ c > 1, F /( ṁ0a0) always increases with θt. This relationship is displayed in Fig. 2.2 fora subsonic case. This reflects the fact that, the more energy we add to the flow, the larger the thrust we obtain.As already indicated, θt cannot be increased without limit (turbine survivability sets a maximum value).

F ṁ0a0

1 2 3 4 5 6 7 8

0

0.5

1

1.5

2

2.5

3

3.5

τ c

Figure 2.2: Non-dimensional thrust, F

ṁ0a0, vs. compressor temperature ratio τ c for an ideal turbojet with

matched exit at M 0 = 0.85, and a range of turbine inlet temperatures θt = 3, 4, 5, 6, 7, 8. The arrow indicatesincreasing values of θt. γ = 1.4.

For a given θt, what is the variation with τ c? By inspection we see that there is a maximum at the maximumof the bracketed quantity in Eq. (2.32), so at the value of τ c that satisfies

∂

∂τ c

θt

1− 1

θ0τ c

− θ0(τ c − 1)

=

θtθ0

1

τ 2c− θ0 = 0 (2.33)

This value is

τ c|max(F ) =√

θtθ0

(2.34)

This result can be seen to be equivalent to T t3 =√

T 0T t4, namely, the compressor exhaust should be at thegeometrical mean of the ambient and combustor exhaust temperatures. If it was much lower or much higher,the T-S diagram of the equivalent Brayton cycle would be too “skinny”, and enclose too little area (too littlework per unit mass) as shown in Fig. 2.3.

Whether this power is utilized as jet kinetic energy, as in the turbojet, or as shaft power in a turboprop, isimmaterial. Also, as far as this argument goes, the compression T t3/T 0 can be arbitrarily divided between ramcompression (θ0) and mechanical compression (τ c). What is the meaning of this for the turbojet? Putting this



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S

T

T0

Tt4

Tt3

S

T

T0

Tt4

Tt3

S

T

T0

Tt4

Tt3

Figure 2.3: Left, too much compression. Middle, optimal compression. Right, too little compression.

value into Eq. (2.32) and the corresponding expression for the specific impulse (2.31), we have the thrust andI s for engines optimized for thrust per unit of airflow:

F

ṁ0a0

max(F )

=

2

γ − 1

θt − 12

+ M 20 − M 0, (2.35)

I s|max(F ) =

ha0C pT 0

1

(θt −√

θt)

F

ṁ0a0

. (2.36)

As an example take: θt = 6.25, γ = 1.4, and

ha0gC

pT 0

= (4.3 · 107J/kg)(283m/s)

(1004J/kgK)(200K) = 60547m/s (2.37)

leading toF

ṁ0a0

max(F )

=

11.25 + M 20 − M 0, (2.38)

I s|max(F ) = 16157.5m/s F

ṁ0a0

max(F )

. (2.39)

2.4.1 Note on Ramjets

Since θ0 increases with Mach number, there is an upper limit on M 0 reached when the compressor ratio formaximum thrust is equal to 1. This theoretical limit is reached for the flight Mach number at which θ0 =

√ θt.

Considering typical value of θt = 9, if θ0 > 3 or M 0 > 3.9 the preferred engine is a ramjet.A ramjet is essentially a turbojet without compressor or turbine. In a Ramjet, all compression is due to the

ram (dynamic) effect (θ0 > 1) so we can put τ c = 1. No turbine is now needed, so τ t = 1 as well. The thrustfollows from (2.28)

F

ṁ0a0=

2

γ − 1 (θ0 − 1)θtθ0−M 0 = M 0

θtθ0− 1

. (2.40)

where eq. (2.8) has been used to simplify the expression. Notice that in our ideal model, both p9 = p0 (matchednozzle) and pt9 = pt0 (no losses). Taken together, this implies M 9 = M 0. But of course, u9 > u0, because



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T 9 > T 0. In fact

u9u0

=

T 9T 0

=

T t9T t01 + γ − 1

2 M 20

1 + γ − 1

2 M 29

=

T t9T t0

=

θtθ0

(2.41)

which is consistent with eqs. (2.4) and (2.40).

2.5 Propulsive efficiency

The Turbojet engine is attractive for its simplicity and its good thrust behavior at high Mach numbers. Un-fortunately it is not very efficient at low Mach numbers, because its jet velocity is too high. To see this, weconsider the Propulsive Efficiency, defined as done in eq. (1.23) (again, for adapted nozzle):

η p = power to airplanepower in jet

= F u012

ṁ0 (u29 − u20) = 2 ṁ0 (u9 − u0) u0

ṁ0 (u29 − u20) = 2u0

u9 + u0(2.42)

From this we see that there is a direct conflict between the desire for high jet velocity to give high thrust, and jet velocity near the flight velocity, to maximize the propulsive efficiency.

In terms of our expression for thrust, since

F

ṁ0a0= M 0

u9u0− 1

, (2.43)

u9u0

= 1

M 0

F

ṁ0a0

+ 1 =

F

ṁ0u0+ 1, (2.44)

and we can write the expression for the propulsive efficiency in terms of our expression for thrust

η p = 2

F

ṁ0u0+ 2

= 2

1

M 0

F

ṁ0a0

+ 2

(2.45)

Since F/( ṁ0a0) ∼ 2 to 3 for low M 0, η p is not good for the turbojet at low Mach numbers. We will see laterhow this deficiency is remedied by adding a fan to the engine to produce a Turbofan.

2.6 Thermal and overall efficiencies

We have described the evolution of the working fluid in the turbojet as an ideal Bryton cycle: (1) an isentropiccompression, (2) an isobaric (at constant stagnation pressure) heating, (3) an isentropic expansion, and (4)

isobaric (at constant static pressure) heat release (which takes place outside of the engine, as the exhaustthermalizes with the ambient, closing the cycle). The Thermal Efficiency is defined for the Turbojet Engine as

ηt = power in jet

power in fuel flow =

12 ṁ0

u29 − u20

ṁf h

= ṁ0c p (T t9 − T 9 − T t0 + T 0)

ṁ0c p (T t4 − T t3) . (2.46)

Let us write this result in a more convenient way for the considered ideal case. Observe that, according to theshaft balance, T t4 − T t9 = T t3 − T t0, so we can substitute T t9 − T t0 = T t4 − T t3 in the numerator. Hence,

ηt = 1− T 9 − T 0T t4 − T t3 . (2.47)



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For the denominator, on the one hand we are considering ideal process in compressor and turbine flow, thus

T t4 = T 9 ( pt4/p9)(γ −1)/γ

and T t3 = T 0 ( pt3/p0)(γ −1)/γ

, and on the other hand, pt3 = pt4 (no pressure drop inthe combustion chamber) and p0 = p9 (pressure matching conditions in the nozzle). Therefore,

ηt = 1−

p0 pt3

γ−1γ

= 1− T 0T t3

= 1− T 9T t4

. (2.48)

Finally we can define an Overall Efficiency as

η = power to airplane

power in fuel flow =

F u0ṁf h

(2.49)

We see thatη = ηtη p. (2.50)

It is also important that the overall efficiency is directly related to the specific impulse

η = F u0ṁf h

= F

ṁf

u0h

= I su0h

. (2.51)



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3 Introduction to Component Matching and Off-Design Operation

In last lecture we derived a thermodynamic model that can be used to design a turbojet for a given operatingcondition. In this chapter we consider how the engine, already built, behaves in the different operating points.

At this point it is adequate to reflect on which of the many parameters we have introduced (like M 2, τ c,τ t, θt, f , etc.) can be controlled by the pilot, and what are the inter-relationships that determine the others.This connectivity is in part mechanical, like the shaft power balance (equation 2.27), but it also comes via flowcontinuity among components. This topic is usually relegated to the very end of the study of engine components,where it is introduced under the section of ”Component Matching”. We find it advantageous to move it forwardto this point.

The price to pay for the insight to be gained is the need to introduce one assumption at this point (tobe justified later). This is the assumption that the stators leading to the turbine (the “turbine nozzles” arechoked). This means the mass flow rate can be written as

ṁ = ṁ4 = Γ(γ ) pt4A4√ RT t4 , (3.1)

where A4 is the effective flow area of these nozzles and

Γ(γ ) =√

γ

2

γ + 1

γ+12(γ−1)

(3.2)

This condition is obtained from the chocked 1D channel flow equation3.

3.1 Discussion on nozzle choking

The analysis of the flow in nozzles will be done in detail in future lectures. In this subsection an introductionto the operation of nozzles in chocked conditions is presented. If we want to have a supersonic adapted exhaust(M 9

≥1), from eq. (2.21) we must have,

2

γ − 1 (θ0τ cτ t − 1) ≥ 1, (3.3)

which may not be satisfied at low power and/or low Mach number.Equation (3.3) is the condition for the exhaust to be supersonic, with4 p9 ≥ p0. It involves both the

compressor and the turbine temperature ratios, but we can eliminate the turbine ratio using the shaft balance(Eq. 2.27), so that the condition is now

θ0τ c

1− τ c − 1

Θ

≥ γ + 1

2 . (3.4)

which, in case of equality makes M 9 = 1 while still p9 = p0. This limit can be rearranged into a quadraticequation for τ c

τ 2c − (Θ + 1)τ c + γ + 12 Θθ0 = 0 (3.5)with the two solutions

τ +,−c = Θ + 1

2 ±

Θ + 1

2

2−

γ + 1

2

Θ

θ0. (3.6)

It can be verified that τ c must be in the range between these two roots to ensure M 9 > 1. The τ +c is normally

very high, so the relevant condition is τ −c . Values of τ −

c are tabulated in Table 3.1 as a function of the flightMach number and of the parameter Θ.

3See the additional materials on basic gas dynamics relations for a review of these equations4In the following lectures we will see that the only possible situation for having a supersonic jet at the exhaust of a nozzle is to

have p9 > p0. The use of a convergent-divergent nozzle is required.



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Θ = 4 Θ = 6 Θ = 8

M 0 = 0 1.296 1.253 1.234M 0 = 0.85 1.066 1.059 1.056

Table 3.1: Values of τ −c

These values are fairly low compressor ratios, even for stationary (M 0 = 0) engine conditions, so theassumption of a choked nozzle is a good one in general. Whether or not the nozzle is also matched is a differentquestion, as noted before.

3.2 Component matching

Passing the same flow through two choked apertures (the turbine inlet, 4, and the nozzle throat, 8) in seriesimposes very strong constraints on the flow conditions. The flow can be expressed at the throat as

ṁ = ṁ8 = Γ(γ ) pt8A8√

RT t8, (3.7)

and equating equations (3.1) and (3.7)

pt8 pt4

T t4T t8

= A4A8

. (3.8)

For a non-afterburning turbojet, pt8 pt5 and T t8 = T t5, thereforeπt√

τ t=

A4A8

, (3.9)

and if the turbine is ideal, πt = τ γ

γ−1

t , and we obtain

τ t =

A4A8

2(γ−1)γ+1

(3.10)

and then

πt =

A4A8

2γγ+1

. (3.11)

This is a strong result: as long as both the turbine nozzles and the exhaust throat remain choked, the turbinemaintains the same pressure and temperature ratios (same operating point), regardless of fuel flow, Machnumber, altitude, etc. We can now trace the variability of other quantities:

1. Compressor ratios. In terms of Θ = T t4/T t0, equation (2.27)

τ c = 1 + Θ ( 1− τ t) , (3.12)πc = τ

γγ−1c . (3.13)

Thus τ c and πc do vary, but only as a function of the single quantity Θ, τ c = τ c(Θ) for a given engine.

2. Dimensionless air flow. The flow at compressor inlet is generally subsonic, so we express the flow ratethere as

ṁ = ṁ2 = Γ(γ ) pt2A2√

RT t2m2(M 2), (3.14)



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with

m2(M 2) = M 2

γ + 121 +

γ − 12

M 22

γ+1

2(γ−1)

. (3.15)

The dimensionless flow function m2(M 2) (mass flow rate as a fraction of the critical mass flow, ṁ∗

2)increases to a maximum of 1 when M 2 = 1, then decreases again, as shown (for γ = 1.4) in Fig. 3.1.

m2

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

M 2

Figure 3.1: Dimensionless function m2 as a function of M 2 for γ = 1.4.

Equating (3.14) to (3.1), we see that

m2 = pt4 pt2

T t2T t4

A4A2

. (3.16)

For an ideal combustor, pt4 = pt3, and so, using πc = τ γ

γ−1c , T t2 = T t0,

m2 = τ

γγ−1c√

Θ

A4A2

. (3.17)

Since τ c = τ c(Θ), we see now that m2 = m2(Θ) as well. This is very useful for scaling from one operatingcondition to another.

3. Mach number at compressor inlet (M 2). Returning to equation (3.14), we see that a certain dimensionlessmass flow, for a given gas, corresponds to two possible Mach numbers (the subonic and the supersonicsolution). Since the non-dimensional mass flow is only function of Θ, M 2 = M 2(Θ) (the supersonic solutionfor M 2 can be disregarded).

4. Fuel/air ratio. The combustor heat balance is

ṁf h = ṁ0C p (T t4 − T t3) = ṁ0C pT t2 (Θ − τ c) (3.18)



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and using T t2 = T t0 and f = ṁf / ṁ0,

f hC pT t0= Θ− τ c(Θ) (3.19)

so this quantity is another function of Θ alone. But notice that, for a given fuel (h)and gas (C p) and ata fixed T 0, f itself does depend on M 0.

5. Throat pressure (normalized). With the hypothesis considered in this section (ideal turbojet)

p8 pt0

= p8 pt8

pt8 pt5

pt5 pt4

pt4 pt3

pt3 pt2

pt2 pt0

, (3.20)

is reduced to p8 pt0

= p8 pt8

pt5 pt4

pt3 pt2

. (3.21)

Introducing the chocked flow condition at the nozzle throat

p8 pt8

= 1

1 + γ − 1

2 × 1

γγ−1

=

2

γ + 1

γγ−1

(3.22)

pt3 pt2

= πc = τ γ

γ−1c (3.23)

pt5 pt4

= πt = τ γ

γ−1

t (3.24)

(3.25)

we obtain p8

pt0 = 2

γ + 1 τ tτ c(Θ)

γγ−1

(3.26)

which is yet another function of Θ alone.

6. Thrust (matched nozzle). We already have equation (2.28), but it is sometimes better to normalize thrustby the total free-stream pressure on the compressor inlet, pt0A2, which is known from flight conditions.If p9 = p0 (e.g. variable area nozzle, or just design point for a fixed nozzle),

ϕ2 ≡ F pt0A2

= ṁ(u9 − u0)

pt0A2= m2Γ

pt0A2√ RT t0

(u9 − u0) pt0A2

. (3.27)

We already had an expression for u9, from eq. (2.23)

u9 = a0 2γ − 1

(θ0τ cτ t−

1) Θ

τ c. (3.28)

and usinga0√ RT t0

=

γRT 0RT t0

=

γ

θ0(3.29)

we obtain

ϕ2 = m2Γ

γ

θ0

2Θ

γ − 1(θ0τ cτ t − 1)

τ c−M 0

. (3.30)

Here the quantities m2 and τ c depend on Θ only, but we can see that the Mach number M 0 appears

explicitly (as M 0 and as θ0 = 1 + γ − 1

2 M 20 ), so the normalized thrust ϕ2 depends on both Θ and M 0.



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7. Thrust (convergent-only, underexpanded nozzle). We now have M 9 = M 8 = 1, but p9 = p8 > p0 so

ϕ2 = F pt0A2

= ṁ(u9 − u0) + ( p9 − p0)A9 pt0A2

= m2Γ (u9 − u0)√

RT t0+

p9 pt0

− p0 pt0

A9A2

, (3.31)

and this time M 9 = 1, leading to

u9 =

γRT 9 =

γR

2

γ + 1T t5 =

2γ

γ + 1RT t0Θτ t (3.32)

so that u9√

RT t0depends on Θ alone. Since we also know that m2 and p9/pt0 are functions of Θ alone, it

makes sense to separate out equation (3.31) in the form

ϕ2 =

m2Γ

u9√ RT t0

+ p9 pt0

A9A2

−

m2Γ u0√

RT t0+

p0 pt0

A9A2

, (3.33)

ϕ2 =

m2Γ

2γ γ + 1

Θτ t +

2γ + 1

τ tτ c γ

γ−

1 A9A2

− m2ΓM 0 γ θ0

+ 1

θγ

γ−1

0

A9A2

. (3.34)

In equation (3.34) the first bracket is a function of Θ only, ϕ∗2(Θ). Once again, the normalized thrustdepends on both, Θ and M 0, but the structure is fairly simple, and in particular, the portion ϕ

∗

2 of ϕ2(neglecting the incoming momentum and the external pressure) is a function of Θ alone. This portion canbe very easily scaled between conditions, and the rest can be subtracted separately.

8. The Operating Line in the compressor map. Compressor performance is typically presented as a map of πc vs. m2, with lines of constant normalized rotational speed ω and ηc, the isentropic efficiency of thecompressor (to be defined in future lectures), superimposed. The details are the subject of later Lectures,but the general shape is as shown in Fig. 3.2, where the flow and speed variables are renormalized by the“Design” values.

Actually the “nominal operating line” shown in the figure is not a property of the compressor, but ratherof the rest of the engine. We can calculate this line with the information we have now, before decidingwhat particular compressor to use. From eq. (3.17),

m2 = A4A2

τ γ

γ−1c√

Θ, (3.35)

and from the shaft power balance (Eq. 3.12),

Θ = τ c − 11− τ t , (3.36)

where we recall that τ t is fixed for a fixed geometry. Eliminating Θ,

m2

= A4

A2τ

γγ−1c 1− τ t

τ c − 1, (3.37)

or, in terms of πc

m2 = A4A2

πc

1− τ tπ

γ−1γ

c − 1(3.38)

which is the equation for the operating line (written in reverse).

If the compressor is already available, we see from eq. (3.38) that we can adjust the nozzle area A4 toplace this line in a ”good” place on the map, i.e., below the stall line and through the best efficiencypoints. Since m2 depends on Θ = T t4/T t0, varying T t4 moves the operating point along the operating line,and this is what the pilot does with the throttle stick to power the engine up or down. At each selectedΘ, the engine settles to a πc, a M 2, a (normalized) rotation rate, etc.



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Figure 3.2: Performance map for a typical high-pressure-ratio compressor. The corrected airflowṁ0

T 2/298.15K/( p2/101325Pa) and corrected rotational velocity N

T 2/298.15K are customarily used in thesediagrams to account for ambient condition variations at the entrance of the compressor.

3.3 Effects of Mach number

If we look at operation of a given engine at different flight Mach numbers, we may try to maintain thesame non-dimensional conditions throughout, which, as we have seen, can be done by maintaining forexample a constant compressor inlet Mach number M 2. This, in turn guarantees a constant Θ = T t4/T t0,but since now we have a varying Mach number, so that T t0 increases with M 0, we may find that the

turbine inlet temperature T t4 needs to become too high at the higher Mach numbers. For example, T t4would have to be 1.8 times higher at M 0 = 2 than at static conditions, and 2.25 times at M 0 = 2.5.

A more reasonable assumption is that the ratio θt = T t4/T 0 can be maintained the same at all Machnumbers, since at least in the tropopause (between 11 km and 20 km above mean sea level), T 0 is almost

invariant. The compressor temperature ratio now follows from τ c = 1 + θt(1− τ t)

θ0, where the numerator

is a constant; thus, τ c will be lowered as the Mach number increases, but less strongly than as wouldbe required in order to maintain maximum thrust per unit flow

√ θt/θ0

. The flow parameter is now

determined by Eq. (3.17), i.e. compressor-turbine flow matching, and then the compressor-inlet Machnumber from Eq. (3.15). Once these parameters are known, we can use Eq. (3.27) to calculate thenormalized thrust; since we are interested in the effect of Mach number, it makes sense to re-normalize



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thrust by p0A2, orF

p0A2 = ϕ2θ

γ

γ−

10 . (3.39)

3.4 Examples

A note on Θ: the near-constancy of the engine operating point

In this section different examples of aircraft operation are given in order to show the possibility of main-taining a near-constant value of the parameter Θ. Two important points in the flight envelope of anaircraft engine are:

(a) Take-off conditions (M 0 0.25, T 0 290K) , and(b) End-of-climb conditions (M 0 0.85, T 0 220K).Using γ = 1.4, the total temperatures are T

t0 = 290 1 + 0.2× 0.252 = 294K (take-off) and T t0 =220 1 + 0.2× 0.852 = 252K (end of climb). Suppose the engine is dimensioned for end-of-climb, which is

common, and that the peak temperature T t4, which will have to be maintained for many hours of cruise,is selected at a conservative T t4 = 1600K. We then have Θ = T t4/T t0 = 6.35 at this condition. If we nowdecided to maintain Θ = 6.35 also for take-off, we would need then T t4 = 6.35× 294 = 1868K. While thisis too high for long-term operation (creep, corrosion), it may be acceptable for the few minutes per cyclethat the engine will be at take-off maximum power. If this actually done, the engine operates at a fixednondimensional point all the way from take-off through start of cruise.

Consider now a commercial jet in a long cruise. As the fuel is consumed and the weight decreases,

the lift must decrease L = 1

2ρ0u

20AwcL. Now, the lift coefficient will be kept close to that for optimum

aerodynamic efficiency, L/D|max, and the Mach number M 0 is unlikely to change much, as it will stay justbelow the transonic drag peak, and so u20 will be proportional to T 0 due to the speed of sound variation.

Together with the density part of lift, we can see that the ambient pressure p0 must be decreasing inproportion to the airplane’s weight, i.e., the plane must be climbing gradually. Turning now to the forwardforce balance, given a constant L/D, the drag, and hence the engine thrust, must also be decreasing intime in the same proportion as the ambient pressure. Therefore, from Eq. (3.27), the nondimensionalthrust ϕ2(Θ, M 0) will remain constant, and since M 0 does too, the peak temperature ratio Θ will alsoremain constant, and with it all the important ratios like τ c, M 2, etc.

In other words, Θ may not vary much among (important) flight conditions, and the engine will be operatingat a fixed nondimensional condition (constant compression ratio, nondimensional flow, compressor inletMach number, etc.). But of course, the dimensional quantities (flow rate, peak pressure, etc.) will bedifferent, depending on p0, etc.

A numerical example

We want to analyze the parameters during the engine operation at different regimes (different values of M 0), but keeping a constant θt = 7, or T t4 = 1540K (T 0 = 220K) in the stratosphere. The geometry of the engine must have been specified in advance. This means that the turbine temperature ratio (Eq. 3.10)is a known fixed number. For the example, we select τ t such as to obtain maximum thrust at M 0 = 1.From the shaft balance equation,

τ t = 1− θ0θt

(τ c − 1) (3.40)

and we put now θ0 = 1.2 and τ c =√

7/1.2 = 2.2048 (at M 0 = 1). This fixes τ t = 0.7935. Similarly, thearea ratio A4/A2 must have been fixed, and we select it here so as to obtain at M 0 = 1 a compressor-faceMach number M 2 = 0.5, which, from Eq. (3.15) implies m2 = 0.7464.



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In order to analyze how m2 is scaled at different conditions, using eq. (3.17) we can write,

m2 = 0.7464 τ c

2.2048

3.5 θ01.2

(3.41)

and the rest of the steps are as described above. The table 3.2 summarizes the results.

M 0 0 1 2 2.5θ0 1 1.2 1.8 2.25τ c 2.4458 2.2048 1.8032 1.6426m2 0.9796 0.7464 0.4523 0.3648M 2 0.8486 0.5 0.2737 0.2172ϕ2 2.9117 1.5531 0.5795 0.3503F/( p0A2) 2.9117 2.9399 4.534 5.985

Table 3.2: Summary of results of the numerical example

We find that at a fixed altitude the thrust is nearly constant up to Mach 1, then it increases rapidly.Actually the increase is less rapid than this simple model predicts, because of losses in the supersonic flowin the engine inlet. The previous model has been computed assuming constant altitude. However, weshould note that an aircraft normally flies at increasing altitude as the Mach Number increases, so thatdynamic pressure γ p0M

20 is roughly constant. In this case the change in F between Mach 1 and Mach 2

is actually a thrust reduction.

3.5 Compressor-turbine matching. Gas generators.

Assembling the compressor and the turbine with a combustor between gives us a gas generator, which is the heartof any gas turbine engine. Once we understand its behavior we can appreciate most of the real characteristicsof aircraft engines, and graduate from thinking of them as a lot of abstract equations.

Actually, we have already done in the previous section most of the work needed to understand the perfor-mance of an ideal Gas Generator, including the important concept of the “Compressor operating line”, whichis set by the flow passing characteristics of the rest of the engine. We summarize here the main findings

1. If the nozzle and the turbine stators are both choked, the turbine temperature ratio is fixed once the flowarea ratio A4/A8 is set.

2. One additional single parameter is sufficient to specify all the other gas generator parameters. This canbe the temperature ratio Θ = T t4/T t2, or the compressor temperature ratio τ c, or the engine-face Machnumber M 2, or the normalized air mass flow, or the fuel flow ratio.

3. A “Compressor Operating Line” in the plane of compressor pressure ratio vs. normalized flow can becalculated once the ratios of all the flow areas are known. For a given choice of one of the parameterslisted above, a point is selected along this operating line.

4. All the above is independent of the compressor specifics. After the compressor has been selected, its per-formance map (pressure ratio vs. normalized flow) contains normalized rotational speed lines as additionalinformation. This parameter is therefore to be added to our list of possible parameters (as is done in thefigure at the end of this lecture), each of which uniquely specifies the state of the gas generator.

It follows from the above that a single degree of freedom is left to the pilot (or to the engine controller),unless geometry can be varied. It is probably most intuitive to think of this unique freedom as the normalizedfuel factor, or the peak temperature ratio T t4/T t2, since these closely relate to the engine throttle control.



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It is to be noted, however, that many idealizations have been made to obtain these simple results. If the

turbine or the nozzle un-choke, or if the engine inefficiencies are rigorously accounted for, the overall detailedbehavior is more complex, but its main qualitative features are not too different.Using these ideas, one can generate and plot a set of Gas Generator Characteristics, such as those below.

With this Gas Generator, we can analyze several engines

1. Turbojet

2. Turbofan

3. Turboprop

4. Unducted fan

5. Helicopter-Turboshaft

Notice that the single free variable chosen for this particular plot is the normalized rotational speed (as afraction of its design value). The quantity in the denominator is the non-dimensional compressor-face tem-perature θ2 = T 2/T 0, because the blade speed ωr is made non-dimensional with the speed of sound at station2.

Figure 3.3: Pumping characteristics for a gas generator with T t4/T t2 = 6



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4 Turbofan Engines

In §2.4 we saw that for low M 0, η p is not very high for turbojets. In essence, there is too much kinetic energy inthe exhaust jet (per unit mass). This is the main reason for using the Turbofan engine. The turbofan separatesthe inflow into a core airstream and a secondary or bypass airstream. The core stream goes into a gas generatorwhich is essentially a turbojet. The bypass stream is simply moved by a fan, which is essentially a big ductedpropeller that increases the total pressure and temperature of the air. This fan is driven by the turbine insidethe core part of the engine, which usually has two shafts: the inner shaft rigidly connects the low-pressureturbine with the fan, and the outer shaft (or spool) connects the high-pressure turbine with the high-pressurecompressor. The spool and the shaft rotate freely using a bearing system. The jets that result from the core andbypass streams can be merged at the exit or before a possible afterburner stage, or be released independentlyinto the ambient.

If it is designed for subsonic cruise flight it looks like the sketch of Fig. 4.1.

Figure 4.1: Schematic diagram of turbofan engine

If designed for both subsonic cruise and for supersonic flight with afterburning it looks more like Fig. 4.2.In both of these diagrams the inlet has been greatly simplified, of course. The numbering of the stations

follows the standard SAE AS755: numbers in the bypass stream are prepended with a “1”, to express that this

is a different air stream from the core one (could be regarded as prepended by index “0”).

4.1 Ideal turbofan model

Let us now see how we can model these engines thermodynamically, following the same process as with theturbojet. The total thrust of the device is given by Eq. 1.4. Notice first that we now have two air streams: thecore stream through the gas generator, with a mass flow ṁ, and the bypass stream, with a mass flow that canbe written as a fraction of the former, α ṁ. We call α the bypass ratio (BPR). To obtain a simplified model of the device sketched in Fig. 4.1, we will take the following assumptions, which will be familiar from the turbojetmodel:

• isentropic processes in the fan, compressor, turbine



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Figure 4.2: Schematic diagram of turbofan engine with afterburner

• constant total pressure through combustor• fuel mass flow neglected with respect to air flow• the nozzle of the core jet and the bypass stream are both adapted to the ambient pressure ( p9 = p19 = p0).

Of course, this may be not the case in many circumstances: as in the turbojet, a fixed convergent nozzleis more likely to be sonic and under-expanded, namely, to have M 8 = M 18 = 1 and p9, p19 > p0. Thederivation of the model with non-matched conditions is left to the student (like in the turbojet case).

Let us start with the core jet. With these hypotheses, the expression for the thrust contribution of the core

jet is the same as for the turbojet before applying the shaft balance. Thus from equation (2.24) we get for thecore jet5

F cṁa0

=

2

γ − 1 (θ0τ cτ t − 1) θtθ0τ c

−M 0. (4.1)

For the bypass stream, there is no turbine, so let us repeat the argument

T t19 = T 19

1 +

γ − 12

M 219

= T 0θ0τ f , (4.2)

pt19 = p19

1 +

γ − 12

M 219

γγ−1

= p0δ 0πf = p0(θ0τ f )γ

γ−1 . (4.3)

So, for p19 = p0

1 + γ − 12

M 219 = θ0τ f , (4.4)

M 19 =

2

γ − 1 (θ0τ f − 1). (4.5)

Using (4.4), from eq. (4.2) we also have that T 19 = T 0 so

F BP α ṁa0

= M 0

u19u0

− 1

=

2

γ − 1 (θ0τ f − 1)−M 0. (4.6)

5Quick observation: τ f = T t13/T t2. For congruency with the definitions of previous chapters, we keep τ c = T t3/T t2, includingin it the possible compression of the core stream due to the central part of the fan.



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Adding this thrust to the thrust of the core jet, we find the total thrust

F

ṁa0=

2

γ − 1 (θ0τ cτ t − 1) θtθ0τ c

−M 0

+ α

2

γ − 1 (θ0τ f − 1)−M 0

(4.7)

4.2 Shaft balance

Now we need τ t, which has to be calculated with the applicable shaft balance in this case. We will see laterthat most engines have two shafts, so each will set a separate balance equation. For now, however, we will lumptheir power together; at off-design conditions, this would have to be modified. Applying the energy equation tothe compressor, fan and turbine, and forcing the work extracted by the latter to be consumed by the former,

ṁC p(T t4 − T t5) = ṁC p(T t3 − T t2) + α ṁC p(T t13 − T t2), (4.8)θt(1

−τ t) = θ0(τ c

−1) + αθ0(τ f

−1), (4.9)

leading to

τ t = 1− θ0θt

[(τ c − 1) + α(τ f − 1)]. (4.10)

Substituting this gives us our result for the thrust of the turbofan. It doesn’t really help to carry out thesubstitution at this point. Instead, let us think about how to simplify the expressions to make them more easilyunderstandable.

4.3 Velocity matching condition

For this engine there are more parameters than for the turbojet

• θt, τ c - as before• α, τ f - characterizing the fan flow.We can relate some of the parameters to others by noting that the highest propulsive efficiency is realized

when u19 = u9: this maximizes the ratio of

momentum

energy

in the jets6. Setting the two velocities to be the

same is referred to as velocity matching, and is usually imposed as a design constrain for the nominal operationpoint. For this situation, the two

√ ’s in the thrust equation are equal, and this requires that:

(θ0τ cτ t − 1) θtθ0τ c

= θ0τ f − 1 (4.11)6A simple derivation: for two streams 1 and 2,

momentum = ṁ1u1 + ṁ2u2,

kinetic energy = 1

2 ṁ1u

2

1 + 1

2 ṁ2u

2

2.

Now we maximize the momentum at a constrained energy. Using a Lagrange multiplier λ, we form the auxiliary function

φ = ṁ1u1 + ṁ2u2 + λ

1

2 ṁ1u

2

1 + 1

2 ṁ2u

2

2

,

and equate to zero the derivatives w.r.t. u1 and u2, leading to

ṁ1 + λ ṁ1u1 = 0,

ṁ2 + λ ṁ2u2 = 0.

So that u1 = u2 = −1

λ .



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Solving for τ f as a function of α, using eq. (4.10), after some algebra we obtain

τ f =1 + θt + θ0(1 + α − τ c)−

θt

θ0τ cθ0(1 + α)

. (4.12)

When this equation is satisfied, the thrust equation becomes simply

F

ṁa0= (1 + α)

2

γ − 1 (θ0τ f − 1)−M 0

(4.13)

4.4 Optimal compression ratio

Now, as for the turbojet, there is still the choice of the compression ratio. Generally we want to choose it for

maximum p ower, which in this case for given turbine inlet temperature θt and bypass ratio α means maximumthrust. To maximize F in this velocity-matching condition, we choose τ c to maximize τ f , since F increasesmonotonically with τ f . From the expression for τ f , eq. (4.12),

∂τ f ∂τ c

= −θ0 + θtθ0τ 2c

= 0 → τ c|max(F ) =√

θtθ0

. (4.14)

Notice that this is precisely the same result as for the Turbojet. Substituting in the expression for τ f

τ f |max(F ) =√

θt − 12

θ0(1 + α) + 1 (4.15)

F

ṁ0a0max(F ) = (1 + α)

2γ − 1

√ θt − 1

2

1 + α + θ0 − 1−M 0 (4.16)Now we have 3 parameters,

• θt - which we set at the maximum feasible value• M 0 - flight speed• α - the prime variable distinguishing the turbofanAs before, for u9 = u19 we still have:

η propulsive = 2

u9

u0 + 1

= 2

F

ṁa0M 0(1 + α)

+ 2

(4.17)

The variation of F /( ṁa0) and η p with M 0 and α is shown in Figures 4.3 and 4.4 for θt = 6.25. Of course forthe higher bypass ratios we are really only interested in the range of M 0


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F

ṁa0

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

10

1

5

20

α

M 0

Figure 4.3: Variation of F /( ṁa0) with M 0 and α for θt = 6.25 and γ = 1.4.

η p

0 0.5 1 1.5 2 2.5 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1

5

20

α

M 0

Figure 4.4: Variation of η p with M 0 and α for θt = 6.25 and γ = 1.4.



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5 Inlets and Nozzles

5.1 Inlets or Diffusers

While the Gas Generator, composed of the compressor, combustor and turbine, is the heart of any gas turbineengine, the overall performance of the propulsion system is strongly influenced by the inlet and the nozzle. Thisis especially true for high M 0 flight, when a major portion of the overall temperature and pressure rise of thecycle are in the inlet, and a correspondingly large part of the expansion in the nozzle. So it is important tounderstand how these components function and how they limit the performance of the propulsion system.

Inlets or DiffusersThese two titles are used interchangeably for the component that captures the oncoming propulsive stream-

tube and conditions it for entrance to the engine. The function of the inlet is to adjust the flow from the ambientflight condition, to that required for entry into the fan or compressor of the engine. It must do this over thefull flight speed range, from static (takeoff) to the highest M 0 the vehicle can attain. Comparing the simple

diagrams of subsonic and supersonic inlets, we can appreciate that the subsonic inlet has the simpler task:

Figure 5.1: Schematic diagram of diffusers

For any inlet the requirements are two

• To bring the inlet flow to the engine with the highest possible stagnation pressure. This is measured bythe Inlet Pressure Recovery, πd =

pt2 pt0

.

• To provide the required engine mass flow. As we shall see the mass flow can be limited by choking of theinlet.

5.2 Subsonic Inlets

The subsonic inlet must satisfy two basic requirements:

• Diffusion of the free-stream flow to the compressor inlet condition at cruise.• Acceleration of static air to the compressor inlet condition at takeoff.

There is a compromise to be made because a relatively thin ”lip” aligned with the entering flow is best forthe cruise condition. This is to avoid accelerating the flow, already at a Mach number of the order of 0.8, tosupersonic speeds that will lead to shock losses. But a more rounded lip will better avoid separation for thetakeoff condition because the air must be captured from a wide range of angles:

Usually subsonic diffusers consist of a divergent duct. Therefore, the minimum area is located at the inlet.The inlet area, A1, is set to just avoid choking (M 1 < 1) when M 2 (compressor inlet) has its largest value (this



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Figure 5.2: Schematic diagram of a subsonic diffuser

is set by the maximum blade Mach number and the blade shape). Since the total temperature and pressure arecommon, we must have

m1(M 1)A1 = m2(M 2)A2. (5.1)

A well designed subsonic inlet will produce a stagnation pressure recovery πd in the order of 0.97 at itsdesign condition.

5.3 Supersonic Inlets

At supersonic flight speeds the pressure and temperature rise in the inlet can be quite large. For the bestthermodynamic efficiency it is important that this compression is as nearly reversible (isentropic) as possible.At a flight Mach number of 3 the ideal pressure ratio is

δ 0 = pt0 p0

=

1 +

γ − 12

M 20

γγ−1

= 36.7 (5.2)

while the temperature rise isθ0 =

T t0T 0

= 1 + γ − 1

2 M 20 = 2.8 (5.3)

For the turbojet cycle the compression is partly in the inlet and partly in the compressor:If the diffusion from point 0 to point 2 is not reversible, the entropy increases, and this results in a lower

value of pt2. This has two effects:

• The expansion ratio of the nozzle is decrea

Lecture Notes Aerospace Propulsion

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