Lecture Notes Aerospace Propulsion

Bioengineering and Aerospace Engineering Department

Universidad Carlos III de Madrid

These lecture notes were originally based on a course that has been taught several years at MIT by Manuel Mart´ınez-S´anchez. Many of the figures have been taken or adapted from the lecture notes from that course and from 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 Mart´ınez-S´anchez, Manuel Garc´ıa Villalba, Pablo Fajardo, Mario Merino, and Andrea Ianiro. If you discover any errata, or have any suggestions, please contact us at:

Contents

1 Introduction to aerospace propulsion 6

1.1 Thrust generation and jet propulsion . . . 6

1.2 Effect of external expansion on thrust . . . 7

1.3 Global performance parameters . . . 8

1.3.1 Range of aircraft . . . 9

1.3.2 Efficiencies . . . 11

2 Aircraft Engine Modeling: the Turbojet 13 2.1 Thrust equation . . . 13

2.2 Shaft balance for the turbojet . . . 17

2.3 Fuel consumption . . . 17

2.4 Design parameters. Effect of mass flow on thrust. . . 17

2.4.1 Note on Ramjets . . . 19

2.5 Propulsive efficiency . . . 20

2.6 Thermal and overall efficiencies . . . 20

3 Introduction to Component Matching and Off-Design Operation 22 3.1 Discussion on nozzle choking . . . 22

3.2 Component matching . . . 23

3.3 Effects of Mach number . . . 27

3.4 Examples . . . 28

3.5 Compressor-turbine matching. Gas generators. . . 29

4 Turbofan Engines 31 4.1 Ideal turbofan model . . . 31

4.2 Shaft balance . . . 33

4.3 Velocity matching condition . . . 33

4.4 Optimal compression ratio . . . 34

5 Inlets and Nozzles 36 5.1 Inlets or Diffusers . . . 36

5.2 Subsonic Inlets . . . 36

5.3 Supersonic Inlets . . . 37

5.4 Exhaust nozzles . . . 42

6 Principles of Compressors and Fans 47 6.1 Euler’s equation . . . 48

6.2 Velocity triangles . . . 50

6.3 Isentropic efficiency and compressor map . . . 51

7 Compressor Blading, design and multi-staging 53 7.1 Diffusion factor. Stall and surge . . . 53

7.2 Compressor blading and radial variations. . . 54

7.3 Multi-staging and flow area variation . . . 56

7.4 Mach Number Effects . . . 57

7.5 The Polytropic Efficiency . . . 57

8 Turbines. Stage characteristics. Degree of reaction 60

8.1 Euler’s Equation . . . 60

8.2 Degree of Reaction . . . 60

8.3 Radial variations . . . 64

8.4 Rotating blade temperature . . . 64

9 Turbine solidity. Mass flow limits. Internal cooling. 66 9.1 Solidity and aerodynamic loading . . . 66

9.2 Mass flow per unit of annulus area and blade stress . . . 67

9.3 Turbine cooling. General trends and systems. Internal cooling. . . 68

10 Film cooling. Thermal stresses. Impingement. 72 10.1 Film cooling . . . 72

10.2 Impingement cooling . . . 74

10.3 Thermal stresses . . . 74

10.4 How to design cooled blades . . . 75

11 Combustion: Combustors and Pollutants 77 11.1 Combustion process . . . 77

11.2 Combustor chambers . . . 79

11.3 Combustor sizing . . . 80

11.4 Afterburners . . . 81

11.5 Pollutants: regulations . . . 82

11.6 Mechanisms for pollutant formation . . . 84

11.7 Upper-Atmospheric Emissions . . . 86

12 Introduction to engine noise and aeroacoustics 89 12.1 Noise propagation . . . 89

12.2 Acoustic energy density and power flux . . . 90

12.3 Noise sources and noise modeling . . . 91

12.4 Jet Noise . . . 94

12.5 Turbomachinery noise . . . 95

13 Engine rotating structures 97 13.1 Blade loads . . . 97

13.2 Centrifugal stresses and disc design . . . 97

14 Fundamentals of rotordynamics 100 14.1 Bearings and engine arrangements . . . 100

14.2 Lumped mass model . . . 101

14.3 Critical speed . . . 103

14.4 Forces on bearings . . . 103

Main nomenclature

General engine and flight magnitudes

Isp Specific impulse [m/s] (NB: also used in [s] in older texts)

F Thrust [N]

D Drag [N]

L Lift [N]

W Weight [N]

˙

mf Fuel mass flow [kg/s]

˙

m0; ˙m Mass flow (of air) [kg/s]

¯

mi 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 = ˙mf/ ˙m0 [-]

h Specific heat value of the fuel [J/kg]

η, ηp, ηt Overall, propulsive and thermal efficiency; η = ηtηp [-]

M0 Flight Mach number [-]

Mi Flow Mach number at stage i [-]

u0 Flight velocity [m/s]

u9 Exhaust velocity [m/s]

R Aircraft range [m]

Component subindices

c 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, Tti Stagnation (total) pressure and temperature at station i [Pa, K]

pi, Ti Static pressure and temperature at station i [Pa, K]

πc, πt, etc Stagnation pressure ratios of compressor, turbine, etc. [-]

τc, τt, etc Stagnation temperature ratios of compressor, turbine, etc. [-]

θi Non-dimensional stagnation temperature at station i (normalized with T0) [-]

δi Non-dimensional stagnation pressure at station i (normalized with p0) [-]

θt Non-dimensional stagnation temperature at turbine inlet, θt= Tt4/T0 [-]

Θ turbine inlet to free-stream Stagnation temperature ratio, Θ = Tt4/Tt0 [-]

ai Sound velocity at station i, [m/s]. For an ideal gas, a =pγRgT

Mi Mach number at station i, [-]

Γ(γ) Choked nondimensional mass flow parameter, Γ(γ) = √γ 2 γ+1

2(γ−1)γ+1

[-] Turbomachinery

u, 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 [-]

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’s second 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 resistive forces 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 a reaction forceequal in magitude but opposite in direction, acting somewhere else, e.g. the ground for a ground vehicle. 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 carried on board and expelled from the vehicle at high speed. As we will see, it is common to evaluate thrust in terms of the momentum gained by the medium (i.e., the reaction force), but it must be kept in mind that this is an indirect procedure, and the thrust force is, by definition, the force exerted by the medium on the vehicle (in complicated 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 is allowed to flow through an open duct (a nozzle), where it pushes forward on the walls and is pushed backwards by the walls. The speed u9 of the gas as it leaves the nozzle is a measure of the momentum per unit mass it

has gained, and so, if the mass flow rate is ˙m9, the thrust is

F _{≈ ˙}m9u9. (1.2)

As it can be easily shown by applying the momentum equation to an appropriate control volume traveling with the 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, as will be shown in section 1.2.

Session 1.-Introductory Lecture

Propulsion Overview.

Propulsion is the action of exerting a forward force on a vehicle. This force must react on

something else, like the ground for a ground vehicle. For an aircraft in flight, the reaction is

exerted on air that has been admitted for the purpose, and that is therefore accelerated

backwards, together with some fuel products. On a rocket in space, where no air can be

admitted, the reaction can only be exerted on some material that is carried on board and

ex!"##"$%&'%()*(%+!""$,%-+%."%.)##%+""/%)'%)+%012213%'1%"4&#5&'"%'("%25'5&#%6170"%89'(75+':;%

in terms of the momentum gained by the medium, but it must be kept in mind that this is an

indirect procedure, and the thrust force is truly exerted by the medium on the vehicle (in

complicated ways, to be sure).

Conceptually, the simplest propulsion device is the Rocket. Some propellants carried on

board are reacted together to generate heated gas at high pressure, and this gas is allowed to

flow through an open duct 8&%931<<#":;/%.("7"%)'%!5+("+%617.&7$%13%'("%.&##+%&3$%)+%!5+("$%

backwards by the walls. The speed u

of the gas as it leaves the nozzle is a measure of the

momentum per unit mass it has gained, and so, if the mass flow rate is !"#$the thrust is

=>$!"%

' More precisely, the gas may continue to expand and accelerate outside the engine,

and a correction is necessary, as we will see later.

Mechanism of thrust generation in a rocket

In an airbreathing engine, such as a Turbojet, air is admitted at a rate !

" with a relative

speed u

(equal to the flight speed), and fuel is added so that the expelled mass flow rate is

!

" . Then, aside from the correction for external acceleration, the new expression for thrust

is

$$$$$) * !"

%

+ !"

%

!"#$%&'

'

(

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 ˙m0 with a

relative speed u0 (equal to the flight speed), and fuel is added so that the expelled mass flow rate is ˙m9. Then,

the new expresssion for thrust is

F = ˙m9u9− ˙m0u0. (1.3)

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.

Turbojet schematic

It is fairly intuitive that it would be advantageous to react against a larger mass of air, to

minimize the amount of jet momentum left behind uselessly (in the limit, one would like to

react against the whole planet, as in a ground vehicle). This is the main argument for

replacing the turbojet by the Turbofan. Here, the admitted air is split into core air, which

reacts with the fuel and generates the torque required for the fan, and bypassed air, moved

by the fan. The core may be basically a turbojet with an added low-pressure turbine, and the

bypass actually handles most of the flow:

Schematic of a Turbofan engine

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

momenta added to both streams:

! " #$

)

+#$

)

/ 0#$

1 #$

2)

#$

₍

)

₍

!

#$

₃

)

₃

!

F = ˙

m

u

+ ˙

m

u

− ( ˙m

+ ˙

m

)u

p

p

(p

−p

)A

A

˙

mu

F = ˙

mu

+ (p

− p

)A

.

F = ˙

mc

c

c

c

u

p

= p

p

p

= p

Figure 1.2: Turbojet schematic

Clearly, if we move a larger airflow ˙m0, 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 behind uselessly 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, and is 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 bypass actually handles most of the flow. The turbofans that are nowadays in operation reach up to a 9:1 mass flow ratio between the bypass and the core streams.

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

F = ˙mcoreucore+ ˙mbypassubypass− ˙m0u0 (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 driven by 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 order has been, of course, the reverse (propellers first, followed by turbojets and later turbofans), but this had to do with 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. We need to calculate the total force exerted by the gas applying the momentum equation to a control volume comprising the system. If the pressure p9at 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 = ˙mu9+ (p9− p0)A9. (1.5)

Aerospace Propulsion Lecture notes

Turbojet schematic

It is fairly intuitive that it would be advantageous to react against a larger mass of air, to

minimize the amount of jet momentum left behind uselessly (in the limit, one would like to

react against the whole planet, as in a ground vehicle). This is the main argument for

replacing the turbojet by the Turbofan. Here, the admitted air is split into core air, which

reacts with the fuel and generates the torque required for the fan, and bypassed air, moved

by the fan. The core may be basically a turbojet with an added low-pressure turbine, and the

bypass actually handles most of the flow:

Schematic of a Turbofan engine

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

momenta added to both streams:

! " #$

)

+#$

)

/ 0#$

1 #$

2)

#$

₍

)

₍

!

#$

₃

)

₃

!

Figure 1.3: Schematic of a Turbofan engine

This is sometimes written as F = ˙mc, where c is the “effective jet velocity”. In reflection, one can see that c really means the speed the gases will achieve after full equilibration of the jet and external pressures, some distance behind the vehicle. The difference between c and u9is not very large in general (a few per cent), and it

is literally zero at the pressure matching condition, when p9= p0; this can be achieved by controling the engine

internal pressure, by variation of the nozzle degree of expansion, or by flying at the appropriate altitude. For a given jet engine, if the outer (ambient) pressure is fixed and we can vary the nozzle exit area to change the pressure p9, maximum thrust is obtained when p9 = p0. The reason is that if p9 < p0, the region of the

nozzle interior next to the exit will produce suction (negative thrust), whereas if p9> p0, one could add positive

thrust by adding some extra diverging area to the nozzle. This fact will become clear in the study of nozzle flows. These comments can be extended easily to other engine types.

As a final point, you may be wondering how it is possible to have p9 different from p0 at the exit from the

nozzle. Indeed, this would not be possible if the flow were subsonic there, because there cannot be a significant pressure difference across a low-speed nearly parallel flow. But if the flow is supersonic, as it almost always is in propulsive nozzles, the pressure equalization takes place through more complex flow features, such as oblique shocks anchored at the nozzle lip, and the internal pressure at the exit plane can be different from the outside pressure.

1.3

Global performance parameters

For all the devices discussed, thrust is proportional to fuel flow rate, and it is advantageous to minimize this flow rate, or to maximize the ratio known as specific impulse

Is=

Thrust

Fuel mass flow rate = F ˙ mf

This is the rational definition, but historically, specific impulse has been defined as thrust per unit weight flow rate of fuel ˜ Is= F g ˙mf (in s). (1.7)

In this notes we are going to try to use most of the times the first definition, although students should always check 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 Is,

not the air or the total flow rate.

The inverse of Isis also in common use, especially for airbreathing engines, and it is called the Specific Fuel

Consumption (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 Is

= m˙f

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. At one time this was also the critical performance measure for transports, bombers and fighters. This is less true nowadays 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 such a challenge (although in-flight refueling is quite expensive). Range is still important for fighters because the requirements for high speed and maneuverability conflict with those for long range.

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,27('+%6J63('+23+6+8J913<+617KJ63'

LMN

!

H

E !

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 is burned, that is dW dt =−g ˙mf =−g F Is =_−g W IsL/D , (1.10)

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

If we assume that Is and L/D are constant (here we are implying models for both the propulsion system

and the aircraft), then

dW W =−g dt IsL/D , (1.11) log W (t) W0  = log m(t) m0  =−g_I t sL/D. (1.12)

Now we can define the range of an aircraft as R = u0t, where u0 is the flight cruise speed, and introducing

t = R/u0into eq. (1.12) we obtain

R = u0 Is g L Dlog  _m 0 m(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 also expressed in terms of the SFC instead of using the specific impulse as

R = u0 1 g_{· SFC} L Dlog  _m 0 m(R)  . (1.14)

It is useful to divide the initial mass of the aircraft m0into several parts: ‘empty’ mass (all the mass required

for the aircraft to operate but the fuel, that is: structure + engines + crew), ‘payload’ and ‘fuel’:

m0= mempty+ mpay+ mf uel, (1.15)

or, in terms of mass fractions

1 = mempty m0 +mpay m0 +mf uel m0 . (1.16)

If the fuel is all expended at R, m(R) m0 = mempty m0 +mpay m0 = 1−m_mf uel 0 . (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 = u0Is g L Dlog  m0 mempty+ mpay  (1.18) and solving for mpay, we obtain

mpay m0 = exp  −_u Rg 0IsL/D  −mempty_m 0 . (1.19)

From this we can construct a range vs. payload chart. As an example suppose mempty= 0.7m0,

u0= 300m/s,

Is= 39200m/s,

so that u0 Is g L D = 1.8· 10 7_{m = 1.8} · 104km, (1.20)

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

mpay m0 0 1000 2000 3000 4000 5000 6000 7000 0 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 thermal power, 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 ˙mf

=u0

hIs, (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 ˙

m9= ˙m0+ ˙mf ' ˙m0. (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

ηp=

F u0

˙

m0(u29− u20) /2

, (1.23)

using here F = ˙m0(u9− u0), equation (1.23) reduces to

ηp=

2u0

u9+ u0

Notice here that ηp 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 pure turbojet: 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 ηp. Observe that for a rocket the

propulsive efficiency is undefined and does not make sense (flight velocity becomes irrelevant, and there is no external airflow flowing through the engine).

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

2

Aircraft Engine Modeling: the Turbojet

All aircraft engines are Heat Engines They use the thermal energy derived from combustion of fossil fuels to produce 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, the thermodynamic efficiency of the device cannot exceed Carnot’s efficiency limit,

ηt≤ 1 − Tlow/Thigh, (2.1)

here being Thigh the temperature after combustion and Tlow 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 the approaches of Thermodynamics. Here the change in thermodynamic state of the air as it passes through the 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 derive in this chapter.

• Fluid mechanical, in which we relate the changes in pressure, temperature and velocity of the air, to the physical characteristics of the engine, e.g. the flow field over the blades of the compressor, in the combustion 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 = ˙m9u9− ˙m0u0+ (p9− p0)A9, (2.2)

where ˙m9 = (1 + f ) ˙m0 includes the fuel mass flow added to the airflow, and f = ˙mf/ ˙m0. Subscripts indicate

the standard flow station as shown in Fig. 2.1. We write this more conveniently in dimensionless form as F ˙ m0u0 = (1 + f )u9 u0 − 1 + p0A9 ˙ m0u0  p9 p0 − 1  . (2.3)

In our modeling of the aircraft engine we will often assume that the nozzle is operating at pressure matching conditions (p9 = p0), and usually take f  1, so, using the flight Mach number M0 = u0/a0, this expression

becomes simply F ˙ m0a0 = M0  u9 u0 − 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 supersonic underexpansion, 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 ˙m0, which will depend on the engine front area, flight velocity, etc., and

2. To estimate u9

u0. 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.

In the following, we employ the defining relations for the stagnation properties1 Tt = T  1 +γ− 1 2 M 2  , (2.5) pt = p  1 +γ− 1 2 M 2 γ−1γ . (2.6)

Remember that a process that does not add or remove heat or work to the fluid always has Tt = const, and

that if Ttis constant and the process is isentropic, then ptis 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 letter that will identify the component):

• Diffuser (d) or inlet: Brings airflow from the flight Mach number M0, to the axial Mach number M2,

required by the compressor. Ideally ∆S = 0 and ∆Tt= 0, so ∆pt= 0. In reality, especially for M0 > 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 to

it, as isentropically as possible. If ideal, then pt3 pt2

= Tt3 Tt2

_γ−1γ

, with Tt3 > Tt2 (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), ideally at near-constant stagnation pressure (Brayton cycle). In practice, some stagnation pressure losses always occur. The maximum temperature that can be obtained with the combustion is fixed by the maximum temperature that the turbine can withstand.

1_{Check the companion notes, “Basic Relations of Gas Dynamics”, for a brief review of stagnation properties and other central}

concepts.

• Turbine (t): Extracts mechanical energy from the airflow, dropping its temperature and pressure, as nearly isentropically as possible. Tt5 < Tt4 (work extracted) with an ideal pressure drop pt5

pt4

= Tt5 Tt4

γ−1γ

. 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 limitations due to the thermal stresses in the turbine.

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

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

u9 u0 = M9 M0 r T9 T0 . (2.7)

It is more efficient (and easier) to find the exit Mach number and temperature by keeping track of the stagnation temperatures and pressures through the several components. The following procedure works for all aircraft 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 and distinguish them from the static or thermodynamic properties of the gas, because in general it is the stagnation properties that most conveniently represent the effect of the components on the fluid as it flows through the engine. 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 Tt’s will be denoted by the symbol τ .

• A ratio of a stagnation temperature (at one station) to the ambient static temperature T0 will be denoted

by θ with the corresponding station subscript.

• Similarly, a ratio of a stagnation pressure to the ambient static pressure p0will be denoted by δ.

So, for the flow upstream of the engine, Tt0 T0 =  1 +γ− 1 2 M 2 0  = θ0, (2.8) pt0 p0 =  1 +γ− 1 2 M 2 0 γ−1γ = δ0. (2.9)

As a particular case, due to its importance in the operation of the turbojet, the turbine-inlet temperature is represented by Tt4 T0 = θt (2.10) or, alternatively, by Θ = θt θ0 = Tt4 Tt0 ≡ Tt4 Tt2 (2.11)

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

(i.e., constant total temperature), since neither work or heat is added to the flow. For the compressor (ideal) pt3 pt2 = πc, (2.12) Tt3 Tt2 = τc, (2.13) πc = τ γ γ−1 c , (2.14)

and for the turbine (also ideal by assumption), pt5 pt4 = πt, (2.15) Tt5 Tt4 = τt, (2.16) πt = τ γ γ−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 the engine. Temperature accounting:

Tt9= T9  1 + γ− 1 2 M 2 9  = T0θ0τcτbτt= T0θtτt. (2.18) Pressure accounting: pt9= p9  1 +γ− 1 2 M 2 9 γ−1γ = p0δ0πcπbπt (2.19)

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

 1 +γ− 1 2 M 2 9  = (δ0πcπt) γ−1 γ _{= θ} 0τcτt (2.20)

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

M92=

2

γ_{− 1}(θ0τcτt− 1) . (2.21)

It is very important to realize that although this expression for the exit Mach number is written in terms of 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 stagnation pressure to the ambient pressure, not at all on the temperature ratio T9/T0. Now, from eq. (2.18)

T9 T0 = θtτt 1 + γ−1₂ M2 9
= θt θ0τc = Θ τc = τb. (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

u9 u0 = 1 M0 s 2 γ− 1(θ0τcτt− 1) θt θ0τc . (2.23)

Finally, the thrust per unit of mass flow (times the speed of sound to make it dimensionless) is F ˙ m0a0 = s 2 γ− 1(θ0τcτt− 1) θt θ0τc − M0. (2.24)

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 relationship between the compressor and turbine. The fact that distinguishes the turbojet engine from other engines we may consider later is that the turbine power, ˙m9(ht4− ht5) equals the compressor power, ˙m0(ht3− ht2), so:

ht3− ht2' ht4− ht5, (2.25)

cpTt0(τc− 1) = cpTt4(1− τt). (2.26)

This equation is known as the power balance in the shaft which can be written as τt= 1− θ0 θt (τc− 1) = 1 − 1 Θ(τc− 1). (2.27)

So finally for the Turbojet Engine F ˙ m0a0 = s 2 γ_{− 1}  θt− θ0(τc− 1) − θt θ0τc  − M0. (2.28)

2.3

Fuel consumption

We are also interested in the fuel consumption. We get ˙mf from a combustor heat balance (i.e., the heat power

released by burning ˙mf fuel per unit time must equal the increase in total enthalpy of the flow across the

combustor):

˙

mfh = ˙m9cpTt4− ˙m0cpTt3 ' ˙m0cp(Tt4− Tt3), (2.29)

where h is the heat released by burning a unit mass of fuel. In other words, ˙

mf = ˙m0

cpT0

h (θt− θ0τc) (2.30)

so that the fuel-specific impulse, Is= F/ ˙mf is

Is=  ha0 cpT0  ₁ (θt− θ0τc)  _F ˙ m0a0  (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 specified performance at the design conditions, and how the performance varies if these parameters are changed, still at 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/( ˙m0a0) on the main parameters, τc, M0 and θt. We can view them this way

• τc is a design choice (compressor pressure ratio)

• M0is the flight speed (flight conditions)

• θtor Θ is the combustor outlet temperature, an operating variable (we can choose how much fuel to burn),

Assuming again that the nozzle is matched, we can re-write equation (2.28) as F ˙ m0a0 = s 2 γ_{− 1}  θt  1−_θ1 0τc  − θ0(τc− 1)  − M0. (2.32)

we can see that since θ0τc > 1, F/( ˙m0a0) always increases with θt. This relationship is displayed in Fig. 2.2 for

a subsonic case. This reflects the fact that, the more energy we add to the flow, the larger the thrust we obtain. As already indicated, θtcannot be increased without limit (turbine survivability sets a maximum value).

F ˙ m0a0 1 2 3 4 5 6 7 8 0 0.5 1 1.5 2 2.5 3 3.5 τc

Figure 2.2: Non-dimensional thrust, F ˙ m0a0

, vs. compressor temperature ratio τc for an ideal turbojet with

matched exit at M0= 0.85, and a range of turbine inlet temperatures θt= 3, 4, 5, 6, 7, 8. The arrow indicates

increasing values of θt. γ = 1.4.

For a given θt, what is the variation with τc? By inspection we see that there is a maximum at the maximum

of the bracketed quantity in Eq. (2.32), so at the value of τc that satisfies

∂ ∂τc  θt  1₋ 1 θ0τc  − θ0(τc− 1)  = θt θ0 1 τ2 c − θ 0= 0 (2.33) This value is τc|max(F )= √ θt θ0 (2.34) This result can be seen to be equivalent to Tt3=√T0Tt4, namely, the compressor exhaust should be at the

geometrical 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 little work 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, is immaterial. Also, as far as this argument goes, the compression Tt3/T0 can be arbitrarily divided between ram

S T T 0 T t4 T t3 S T T 0 T t4 T t3 S T T 0 T t4 T t3

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 and Is for engines optimized for thrust per unit of airflow:

F ˙ m0a0     max(F ) = s 2 γ_{− 1} p θt− 1 2 + M2 0 − M0, (2.35) Is|max(F )=  _ha 0 CpT0  ₁ (θt−√θt)  _F ˙ m0a0  . (2.36)

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

ha0 gCpT0 = (4.3· 10 7_{J/kg)(283m/s)} (1004J/kgK)(200K) = 60547m/s (2.37) leading to F ˙ m0a0    _{max(F )} = q 11.25 + M2 0 − M0, (2.38) Is|_{max(F )}= 16157.5m/s F ˙ m0a0    _{max(F )} . (2.39) 2.4.1 Note on Ramjets

Since θ0 increases with Mach number, there is an upper limit on M0 reached when the compressor ratio for

maximum 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 M0> 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 τc = 1. No turbine is now needed, so τt= 1 as well. The thrust

follows from (2.28) F ˙ m0a0 = s 2 γ_{− 1}(θ0− 1) θt θ0− M 0= M0 r θt θ0 − 1 ! . (2.40)

where eq. (2.8) has been used to simplify the expression. Notice that in our ideal model, both p9= p0(matched

T9> T0. In fact u9 u0 =r T9 T0 = v u u u u u t Tt9 Tt0 1 +γ− 1 2 M 2 0 1 +γ− 1 2 M 2 9 =r Tt9 Tt0 =r θt θ0 (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, we consider the Propulsive Efficiency, defined as done in eq. (1.23) (again, for adapted nozzle):

ηp= power to airplane power in jet = F u0 1 2m˙0(u29− u20) = 2 ˙m0(u9− u0) u0 ˙ m0(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 ˙ m0a0 = M0  u9 u0 − 1  , (2.43) u9 u0 = 1 M0  _F ˙ m0a0  + 1 = F ˙ m0u0 + 1, (2.44)

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

ηp= 2 F ˙ m0u0 + 2 = 2 1 M0 F ˙ m0a0 ! + 2 (2.45)

Since F/( ˙m0a0)∼ 2 to 3 for low M0, ηp is not good for the turbojet at low Mach numbers. We will see later

how 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 isentropic compression, (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 exhaust thermalizes with the ambient, closing the cycle). The Thermal Efficiency is defined for the Turbojet Engine as

ηt=

power in jet power in fuel flow =

1 2m˙0 u 2 9− u20
˙ mfh = ˙ m0cp(Tt9− T9− Tt0+ T0) ˙ m0cp(Tt4− Tt3) . (2.46)

Let us write this result in a more convenient way for the considered ideal case. Observe that, according to the shaft balance, Tt4− Tt9= Tt3− Tt0, so we can substitute Tt9− Tt0= Tt4− Tt3 in the numerator. Hence,

ηt= 1−

T9− T0

For the denominator, on the one hand we are considering ideal process in compressor and turbine flow, thus Tt4 = T9(pt4/p9)(γ−1)/γ and Tt3 = T0(pt3/p0)(γ−1)/γ, and on the other hand, pt3 = pt4 (no pressure drop in

the combustion chamber) and p0= p9(pressure matching conditions in the nozzle). Therefore,

ηt= 1−  p0 pt3 γ−1γ = 1₋ T0 Tt3 = 1₋ T9 Tt4 . (2.48)

Finally we can define an Overall Efficiency as

η = power to airplane power in fuel flow=

F u0 ˙ mfh (2.49) We see that η = ηtηp. (2.50)

It is also important that the overall efficiency is directly related to the specific impulse η = F u0 ˙ mfh = F ˙ mf u0 h = Is u0 h. (2.51)

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 operating condition. 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 M2, τc,

τt, 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 flow continuity 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 forward to this point.

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

˙ m = ˙m4= Γ(γ) pt4A4 √ RTt4 , (3.1)

where A4 is the effective flow area of these nozzles and

Γ(γ) =√γ  2 γ + 1 _2(γ−1)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 introduction to the operation of nozzles in chocked conditions is presented. If we want to have a supersonic adapted exhaust (M9≥ 1), from eq. (2.21) we must have,

2

γ_{− 1}(θ0τcτt− 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 _p

9 ≥ 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

θ0τc  1₋τc− 1 Θ  ≥γ + 1₂ . (3.4)

which, in case of equality makes M9 = 1 while still p9 = p0. This limit can be rearranged into a quadratic

equation for τc τc2− (Θ + 1)τc+  γ + 1 2  Θ θ0 = 0 (3.5)

with the two solutions

τc+,−= Θ + 1 2 ± s  Θ + 1 2 2 − γ + 1₂  Θ_θ 0 . (3.6)

It can be verified that τc must be in the range between these two roots to ensure M9> 1. The τc+ is normally

very high, so the relevant condition is τ−

c . Values of τc− are tabulated in Table 3.1 as a function of the flight

Mach number and of the parameter Θ.

3_{See the additional materials on basic gas dynamics relations for a review of these equations}

4_{In the following lectures we will see that the only possible situation for having a supersonic jet at the exhaust of a nozzle is to}

Θ = 4 Θ = 6 Θ = 8

M0= 0 1.296 1.253 1.234

M0= 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 (M0 = 0) engine conditions, so the

assumption of a choked nozzle is a good one in general. Whether or not the nozzle is also matched is a different question, 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 series imposes very strong constraints on the flow conditions. The flow can be expressed at the throat as

˙ m = ˙m8= Γ(γ) pt8A8 √ RTt8 , (3.7)

and equating equations (3.1) and (3.7)

pt8 pt4 r Tt4 Tt8 =A4 A8 . (3.8)

For a non-afterburning turbojet, pt8' pt5 and Tt8 = Tt5, therefore

πt √_τ t = A4 A8 , (3.9)

and if the turbine is ideal, πt= τ

γ γ−1 t , and we obtain τt=  A4 A8 2(γ−1)γ+1 (3.10) and then πt=  A4 A8 γ+12γ . (3.11)

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

1. Compressor ratios. In terms of Θ = Tt4/Tt0, equation (2.27)

τc = 1 + Θ (1− τt) , (3.12)

πc = τ

γ γ−1

c . (3.13)

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

2. Dimensionless air flow. The flow at compressor inlet is generally subsonic, so we express the flow rate there as ˙ m = ˙m2= Γ(γ) pt2A2 √ RTt2 m2(M2), (3.14)

with m2(M2) = M2     γ + 1 2 1 +γ− 1 2 M 2 2     γ+1 2(γ−1) . (3.15)

The dimensionless flow function m2(M2) (mass flow rate as a fraction of the critical mass flow, ˙m∗2)

increases to a maximum of 1 when M2= 1, then decreases again, as shown (for γ = 1.4) in Fig. 3.1.

m2 0 1 2 3 4 5 0 0.2 0.4 0.6 0.8 1 M2

Figure 3.1: Dimensionless function m2 as a function of M2for γ = 1.4.

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

m2= pt4 pt2 r Tt2 Tt4 A4 A2 . (3.16)

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

γ γ−1 c , Tt2= Tt0, m2= τ γ γ−1 c √ Θ A4 A2 . (3.17)

Since τc = τc(Θ), we see now that m2= m2(Θ) as well. This is very useful for scaling from one operating

condition to another.

3. Mach number at compressor inlet (M2). Returning to equation (3.14), we see that a certain dimensionless

mass flow, for a given gas, corresponds to two possible Mach numbers (the subonic and the supersonic solution). Since the non-dimensional mass flow is only function of Θ, M2= M2(Θ) (the supersonic solution

for M2 can be disregarded).

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

and using Tt2= Tt0 and f = ˙mf/ ˙m0,

f h CpTt0

= Θ− τc(Θ) (3.19)

so this quantity is another function of Θ alone. But notice that, for a given fuel (h)and gas (Cp) and at

a fixed T0, f itself does depend on M0.

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 _p 8 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 !γ−1γ =  ₂ γ + 1 γ−1γ (3.22) pt3 pt2 = πc= τ γ γ−1 c (3.23) pt5 pt4 = πt= τ γ γ−1 t (3.24) (3.25) we obtain p8 pt0 =  ₂ γ + 1τtτc(Θ) γ−1γ (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 thrust by 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 = ˙ m(u9− u0) pt0A2 = m2Γ pt0A2 √ RTt0 (u9− u0) pt0A2 . (3.27)

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

u9= a0 s 2 γ_{− 1}(θ0τcτt− 1) Θ τc . (3.28) and using a0 √ RTt0 =r γRT0 RTt0 =r γ θ0 (3.29) we obtain ϕ2= m2Γ r γ θ0 "s 2Θ γ_{− 1} (θ0τcτt− 1) τc − M0 # . (3.30)

Here the quantities m2 and τc depend on Θ only, but we can see that the Mach number M0 appears

explicitly (as M0and as θ0= 1 + γ− 1

2 M

2

7. Thrust (convergent-only, underexpanded nozzle). We now have M9= M8= 1, but p9= p8> p0 so ϕ2= F pt0A2 = m(u˙ 9− u0) + (p9− p0)A9 pt0A2 = m2Γ (u9− u0) √ RTt0 + p9 pt0 − p0 pt0  A9 A2 , (3.31)

and this time M9= 1, leading to

u9=pγRT9= r γR 2 γ + 1Tt5= r 2γ γ + 1RTt0Θτt (3.32) so that √u9 RTt0

depends on Θ alone. Since we also know that m2and 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 A9 A2  −  m2Γ u0 √_RT t0 + p0 pt0 A9 A2  , (3.33) ϕ2= " m2Γ r 2γ γ + 1Θτt+  ₂ γ + 1τtτc γ−1γ _A 9 A2 # − " m2ΓM0 r γ θ0 + 1 θ γ γ−1 0 A9 A2 # . (3.34)

In equation (3.34) the first bracket is a function of Θ only, ϕ∗2(Θ). Once again, the normalized thrust

depends on both, Θ and M0, 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 can be 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 the

compressor (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 rather of the rest of the engine. We can calculate this line with the information we have now, before deciding what particular compressor to use. From eq. (3.17),

m2= A4 A2 τ γ γ−1 c √ Θ, (3.35)

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

Θ = τc− 1 1_{− τ}t

, (3.36)

where we recall that τtis fixed for a fixed geometry. Eliminating Θ,

m2=A4 A2 τ γ γ−1 c r 1 − τt τc− 1 , (3.37) or, in terms of πc m2= A4 A2 πc s 1_{− τ}t π γ−1 γ 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 to

place this line in a ”good” place on the map, i.e., below the stall line and through the best efficiency points. Since m2depends on Θ = Tt4/Tt0, varying Tt4moves 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 selected Θ, the engine settles to a πc, a M2, a (normalized) rotation rate, etc.

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

m0pT2/298.15K/(p2/101325Pa) and corrected rotational velocity NpT2/298.15K are customarily used in these

diagrams 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 the same non-dimensional conditions throughout, which, as we have seen, can be done by maintaining for example a constant compressor inlet Mach number M2. This, in turn guarantees a constant Θ = Tt4/Tt0,

but since now we have a varying Mach number, so that Tt0 increases with M0, we may find that the

turbine inlet temperature Tt4 needs to become too high at the higher Mach numbers. For example, Tt4

would have to be 1.8 times higher at M0= 2 than at static conditions, and 2.25 times at M0= 2.5.

A more reasonable assumption is that the ratio θt = Tt4/T0 can be maintained the same at all Mach

numbers, since at least in the tropopause (between 11 km and 20 km above mean sea level), T0is almost

invariant. The compressor temperature ratio now follows from τc= 1 +

θt(1− τt)

θ0

, where the numerator is a constant; thus, τc will be lowered as the Mach number increases, but less strongly than as would

be 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 Mach number from Eq. (3.15). Once these parameters are known, we can use Eq. (3.27) to calculate the normalized thrust; since we are interested in the effect of Mach number, it makes sense to re-normalize

thrust by p0A2, or F p0A2 = ϕ2θ γ γ−1 0 . (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 an aircraft engine are:

(a) Take-off conditions (M0' 0.25, T0' 290K) , and

(b) End-of-climb conditions (M0' 0.85, T0' 220K).

Using γ = 1.4, the total temperatures are Tt0 = 290 1 + 0.2× 0.252
= 294K (take-off) and Tt0 =

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 Tt4, which will have to be maintained for many hours of cruise,

is selected at a conservative Tt4 = 1600K. We then have Θ = Tt4/Tt0= 6.35 at this condition. If we now

decided to maintain Θ = 6.35 also for take-off, we would need then Tt4= 6.35× 294 = 1868K. While this

is too high for long-term operation (creep, corrosion), it may be acceptable for the few minutes per cycle that the engine will be at take-off maximum power. If this actually done, the engine operates at a fixed nondimensional 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

2

0AwcL. Now, the lift coefficient will be kept close to that for optimum

aerodynamic efficiency, L/D_|max, and the Mach number M0is unlikely to change much, as it will stay just

below the transonic drag peak, and so u2

0 will be proportional to T0due to the speed of sound variation.

Together with the density part of lift, we can see that the ambient pressure p0 must be decreasing in

proportion to the airplane’s weight, i.e., the plane must be climbing gradually. Turning now to the forward force balance, given a constant L/D, the drag, and hence the engine thrust, must also be decreasing in time in the same proportion as the ambient pressure. Therefore, from Eq. (3.27), the nondimensional thrust ϕ2(Θ, M0) will remain constant, and since M0 does too, the peak temperature ratio Θ will also

remain constant, and with it all the important ratios like τc, M2, etc.

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

A numerical example

We want to analyze the parameters during the engine operation at different regimes (different values of M0), but keeping a constant θt= 7, or Tt4 = 1540K (T0 = 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 τt such as to obtain maximum thrust at M0 = 1.

From the shaft balance equation,

τt= 1−

θ0

θt

(τc− 1) (3.40)

and we put now θ0= 1.2 and τc =√7/1.2 = 2.2048 (at M0= 1). This fixes τt = 0.7935. Similarly, the

area ratio A4/A2must have been fixed, and we select it here so as to obtain at M0= 1 a compressor-face

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 3.5r θ₀ 1.2 (3.41)

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

M0 0 1 2 2.5 θ0 1 1.2 1.8 2.25 τc 2.4458 2.2048 1.8032 1.6426 m2 0.9796 0.7464 0.4523 0.3648 M2 0.8486 0.5 0.2737 0.2172 ϕ2 2.9117 1.5531 0.5795 0.3503 F/(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 flow in the engine inlet. The previous model has been computed assuming constant altitude. However, we should note that an aircraft normally flies at increasing altitude as the Mach Number increases, so that dynamic pressure γp0M02 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 heart of any gas turbine engine. Once we understand its behavior we can appreciate most of the real characteristics of 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”, which is 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 flow area ratio A4/A8 is set.

2. One additional single parameter is sufficient to specify all the other gas generator parameters. This can be the temperature ratio Θ = Tt4/Tt2, or the compressor temperature ratio τc, or the engine-face Mach

number M2, 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 be calculated once the ratios of all the flow areas are known. For a given choice of one of the parameters listed 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 additional information. This parameter is therefore to be added to our list of possible parameters (as is done in the figure 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 normalized fuel factor, or the peak temperature ratio Tt4/Tt2, since these closely relate to the engine throttle control.

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 detailed behavior 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 a fraction of its design value). The quantity in the denominator is the non-dimensional compressor-face tem-perature θ2 = T2/T0, because the blade speed ωr is made non-dimensional with the speed of sound at station

2.

4

Turbofan Engines

In_{§2.4 we saw that for low M}0, ηp is not very high for turbojets. In essence, there is too much kinetic energy in

the exhaust jet (per unit mass). This is the main reason for using the Turbofan engine. The turbofan separates the inflow into a core airstream and a secondary or bypass airstream. The core stream goes into a gas generator which is essentially a turbojet. The bypass stream is simply moved by a fan, which is essentially a big ducted propeller that increases the total pressure and temperature of the air. This fan is driven by the turbine inside the core part of the engine, which usually has two shafts: the inner shaft rigidly connects the low-pressure turbine with the fan, and the outer shaft (or spool) connects the high-pressure turbine with the high-pressure compressor. The spool and the shaft rotate freely using a bearing system. The jets that result from the core and bypass streams can be merged at the exit or before a possible afterburner stage, or be released independently into the ambient.

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

19 9

13

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 the turbojet. The total thrust of the device is given by Eq. 1.4. Notice first that we now have two air streams: the core stream through the gas generator, with a mass flow ˙m, and the bypass stream, with a mass flow that can be written as a fraction of the former, α ˙m. 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 turbojet model:

7 9 13

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 nozzle is more likely to be sonic and under-expanded, namely, to have M8 = M18 = 1 and p9, p19 > p0. The

derivation 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 the core jet5 Fc ˙ ma0 = s 2 γ_{− 1}(θ0τcτt− 1) θt θ0τc − M0 . (4.1)

For the bypass stream, there is no turbine, so let us repeat the argument Tt19= T19  1 +γ− 1 2 M 2 19  = T0θ0τf, (4.2) pt19 = p19  1 +γ− 1 2 M 2 19 γ−1γ = p0δ0πf = p0(θ0τf) γ γ−1. (4.3) So, for p19= p0 1 +γ− 1 2 M 2 19= θ0τf, (4.4) M19= r 2 γ_{− 1}(θ0τf− 1). (4.5)

Using (4.4), from eq. (4.2) we also have that T19= T0 so

FBP α ˙ma0 = M0  u19 u0 − 1  = r 2 γ_{− 1}(θ0τf− 1) − M0. (4.6) 5_{Quick observation: τ}

f = Tt13/Tt2. For congruency with the definitions of previous chapters, we keep τc= Tt3/Tt2, including

Adding this thrust to the thrust of the core jet, we find the total thrust F ˙ ma0 = "s 2 γ− 1(θ0τcτt− 1) θt θ0τc − M0 # + α r ₂ γ− 1(θ0τf− 1) − M0  (4.7)

4.2

Shaft balance

Now we need τt, which has to be calculated with the applicable shaft balance in this case. We will see later

that most engines have two shafts, so each will set a separate balance equation. For now, however, we will lump their power together; at off-design conditions, this would have to be modified. Applying the energy equation to the compressor, fan and turbine, and forcing the work extracted by the latter to be consumed by the former,

˙ mCp(Tt4− Tt5) = mC˙ p(Tt3− Tt2) + α ˙mCp(Tt13− Tt2), (4.8) θt(1− τt) = θ0(τc− 1) + αθ0(τf− 1), (4.9) leading to τt= 1− θ0 θt[(τc− 1) + α(τf− 1)]. (4.10)

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

4.3

Velocity matching condition

For this engine there are more parameters than for the turbojet • θt, τc - as before

• α, τf - 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 operation point. For this situation, the two √ ’s in the thrust equation are equal, and this requires that:

(θ0τcτt− 1)

θt

θ0τc

= θ0τf− 1 (4.11)

6_{A simple derivation: for two streams 1 and 2,}

momentum = m˙1u1+ ˙m2u2, kinetic energy = 1 2m˙1u 2 1+ 1 2m˙2u 2 2.

Now we maximize the momentum at a constrained energy. Using a Lagrange multiplier λ, we form the auxiliary function φ = ˙m1u1+ ˙m2u2+ λ  1 2m˙1u 2 1+ 1 2m˙2u 2 2  , and equate to zero the derivatives w.r.t. u1 and u2, leading to

˙ m1+ λ ˙m1u1 = 0, ˙ m2+ λ ˙m2u2 = 0. So that u1= u2= −1 λ .

Solving for τf as a function of α, using eq. (4.10), after some algebra we obtain τf= 1 + θt+ θ0(1 + α− τc)− θt θ0τc θ0(1 + α) . (4.12)

When this equation is satisfied, the thrust equation becomes simply F ˙ ma0 = (1 + α) r ₂ γ− 1(θ0τf− 1) − M0  (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 power, which in this case for given turbine inlet temperature θtand bypass ratio α means maximum

thrust. To maximize F in this velocity-matching condition, we choose τc to maximize τf, since F increases

monotonically with τf. From the expression for τf, eq. (4.12),

∂τf ∂τc =−θ0+ θt θ0τc2 = 0 → τc|_{max(F )}= √ θt θ0 . (4.14)

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

τf|max(F ) = √ θt− 1
2 θ0(1 + α) + 1 (4.15) F ˙ m0a0    _{max(F )} = (1 + α)   v u u t 2 γ_{− 1} √ θt− 1
2 1 + α + θ0− 1 ! − M0   (4.16)

Now we have 3 parameters,

• θt- which we set at the maximum feasible value

• M0- flight speed

• α - the prime variable distinguishing the turbofan As before, for u9= u19 we still have:

ηpropulsive= 2 u9 u0 + 1 = 2 F ˙ ma0 M0(1 + α) + 2 (4.17)

The variation of F/( ˙ma0) and ηp with M0and α is shown in Figures 4.3 and 4.4 for θt= 6.25. Of course for

the higher bypass ratios we are really only interested in the range of M0< 1, but the lower bypass and higher

F ˙ ma0 0 0.5 1 1.5 2 2.5 3 0 1 2 3 4 5 6 7 8 9 10 1 5 20 α M0

Figure 4.3: Variation of F/( ˙ma0) with M0 and α for θt= 6.25 and γ = 1.4.

ηp 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 5 20 α M0