SESA2005LN2006 Propulsion

School of Engineering Sciences.University of Southampton.

SESA2005 Propulsion lecturenotes.

Lecturer: Prof. S. I. Chernyshenko

2006/2007

Copyright c© 2006 University of Southampton

Contents

Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Lecture 1. Schemes of air-breathing turbo-engines. . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Lecture 2. Thrust equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Lecture 3. Engine performance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Lecture 4. Ramjet engine. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14

Lecture 5. Ideal ramjet calculation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Lecture 6. Ideal turbojet. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Lecture 7. Ideal turbofan. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Lecture 8. The effect of losses. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Note on turboprop and turboshaft engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Lecture 9. Typical parameters of real engines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Lecture 10. Rocket propulsion I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Lecture 11. Rocket propulsion II. Composition of combustion products. . . . . 40

Introduction.

This course has only two major goals. The first goal is to achieve an in-tuitive understanding of the basic relations between the flight regime, designparameters, and the engine performance. For example, at the end of the coursethe student should be able to figure out, intuitively, how the maximum (allowedby material properties, say) temperature in the engine affects the optimal com-pressor pressure ratio, or how the flight Mach number affects the propulsionefficiency etc. The variety of possible questions like that is endless. The an-swers to them cannot be memorised, however, once understanding is achievedone can answer most of these questions relatively easily. In the following ta-ble the upper row lists the flight regime and design parameters that determinethe performance characteristics listed in the first column. If, for example, youunderstand intuitively how, say, β affects TSFC, tic the empty cell at the in-tersection of the β column and TSFC row. Once the table is filled you havecompletely achieved the first goal.

M T0b prc prf β ηd ηc Ta ηf ηfn rb QR ηt ηn ηb

F/ma

TSFCηp

ηth

η0

The second goal is to teach the student how to calculate the engine perfor-mance, say, the fuel consumption rate, for given performance characteristicsof the engine components (diffuser, compressor, burner, turbine etc.) and theflight regime. In other words, given the numerical values of the parameters inthe upper row of the table and the air properties, you will be able to calculatethe numerical values of the parameters in the left column. If you forget any ofthe necessary formulae, you will be able to derive them.

In addition, you will have an understanding of how the engine characteris-tics affect the aircraft range. In the rocket propulsion part, you will learn theprinciples of calculating the rocket engine performance taking into account thechemical processes in the engine. However, the rocket part is relatively shortand provides only the basic ideas and illustrations.

These lecture notes consist of 11 rather terse lectures, which are difficult tounderstand without further explanations. More material will be given in thelectures, and having these notes cannot substitute attending the lectures.

At the time of writing the course website was at

http://www.afm.ses.soton.ac.uk/∼sergei/SESA2005/SESA2005.html,

where more material will be given as the course progresses, including the formerexam papers with solutions.

The course materials are also available through Blackboard.

The present course gives only the first introduction to propusion. Furtherpropulsion courses are available as an option for SES students.

Recommended reading: P. Hill and C.Peterson, Mechanics and Thermody-namics of Propulsion, second edition. Addison-Wesley, 1995.

The course lecturer, Prof. Sergei I. Chernyshenko, can be found in Room5069 of the Tizard building, or contacted by e-mail at [email protected].

The main tool for obtaining feedback on your progress is the tur-bofan applet. At the time of writing it was located at

http://www.soton.ac.uk/∼ge102/Jet.htmlPlease visit it and learn to use it as early in the course as possible.

6 SESA2005 Propulsion Lecture Notes 2006/7.

Lecture 1. Schemes of air-breathing turbo-engines.

CompressorAir inlet

Burner

fuel

Tur

bineShaft

Hot

high−pressure

gas

Basic gas generator

CompressorAir inlet

Burner

fuelT

urbi

ne ExhaustnozzleShaft Hot

high−speedjet

Turbojet

Compressor

Burner

fuel

Tur

bine

Tur

bine

Air inlet

Hot

moderate−speedjet

moderate−speedjet

Cold

Fan

nozzleExhaust

Shaft

Turbofan

Compressor

Tur

bine

Tur

bine

Burner

fuel

Gea

r

jetlow−speedCold

Air inlet

Pro

pelle

r

Shaft

Exhaustnozzle

Hotlow−speedexhaust

Turboprop

Lecture 1. Schemes of air-breathing turbo-engines. 7

Gas turbine engines. From Hill and Peteresen’s book.


Lecture 2. Thrust equation.

u

u m

u

p

−F

A

B C

D

m e eaa. .pe ,

(thrust reaction)

pa

m.

side

u

Figure 1: Thrust-producing device with a single jet.

By the second Newton law the rate of change of momentum of a body equalsthe force acting on the body:

dmv

dt= F

It is convenient to consider force F as a momentum flux. Then this secondlaw asserts that momentum is conserved: the change in momentum dmv equalsthe amount Fdt of momentum added to the body. This way of treating forces iscommon in modern physics. For a fixed volume in space momentum can flow inor out of it not only due to the action of force but also due to mass moving in orout of it. A unit mass leaving the volume carries away an amount of momentumequal to its velocity.

We will limit ourselves to momentum in the direction of the thrust forcewhich, in turn, will be assumed to be parallel to the velocity of the incomingflow. Then, the amount of momentum entering the control volume ABCD inFig. 1 through AB per unit time is

paAAB + mABu

where AAB is the area of the face AB of the control volume, mAB is themass flow rate through it, and u is the flight velocity. The pressure pa and thevelocity u are assumed constant, which is a good approximation if AB is farupstream from the engine.

The side of the control volume (BC and AD) is parallel to the direction ofthe motion and, therefore, the pressure force does not produce any momentum

Lecture 2. Thrust equation. 9

flux through it. If BC and AD are far from the engine we can assume, also asan approximation, that the velocity component in thrust direction is u. Thenthe amount of momentum leaving the control volume through its side per unittime is

msideu,

where mside is the flow rate from the control volume through its side.The reaction force, equal and of opposite direction to the thrust F , creates a

momentum flux into ABCD through BC. This flux flows through the structuralsupport of the engine shown in Fig. 1.

The amount of momentum leaving the control volume through DC per unittime is

pa(ADC − Ae) + peAe + (mDC − me)u + meue

Here, the pressure and velocity on DC outside the jet are assumed to coincidewith their values far upstream. This neglects the changes to the velocity profiledue to the nacelle1 drag. Ae is the nozzle outlet area. In supersonic jet theoutlet pressure pe may be not equal to pa.

The flow is assumed to be steady. Therefore, the amount of momentumentering the control volume per unit time should be equal to the amount ofmomentum leaving it per unit time:

paAAB+mABu+F = pa(ADC−Ae)+peAe+(mDC−me)u+meue+msideu (1)

Now, we introduce the mass flow rate into the engine intake ma and the fuelflow rate mf , so that me = ma+mf , mDC = mAB−mside+mf , and the fuel-airratio f = mf/ma. Then after simple transformation (note that ADC = AAB)(1) becomes

F = ma[(1 + f)ue − u] + (pe − pa)Ae (2)

Turbofan and turboprop engines have two exhaust streams, with differentexhaust velocities, as sketched in Fig. 2. In this case the thrust equation has tobe modified. While the amount of momentum entering the control volume perunit time through AB is the same as before, the amount of momentum leaving itthrough DC is now (mDC −me−mc)u+mcuc +meue. Usually for turbofan andturboprop engines the exhaust pressure is approximately equal to the outsidepressure, pe = pa. Assuming this, the thrust equation then becomes

F = ma[(1 + f)ue − u] + mc(uc − u) (3)

It is recommended to perform the full detailed derivation of the equations(2) and (3) during the revision since it is the simplest way to memorise them.

1nacelle is a streamlined enclosure for an aircraft engine.


u

u m

u

p

−F

A

B C

D

e eaa. .pe ,

(thrust reaction)

pa

ucc

m

m.

Figure 2: Thrust-producing device with two exhaust jets.

In case of a rocket engine ma = 0, and fuel and propellant2 are effectivelythe same. The thrust then is

F = mfue + (pe − pa)Ae

Exercise: derive this equation using the control volume approach similaras it was done above for air-breathing engines.

2propellant is the substance which is accelerated by the engine in order to create thrust.Thus, for a car the tarmac of the road is the propellant.

Lecture 3. Engine performance. 11

Lecture 3. Engine performance.

Propulsion efficiency ηp is the ratio of the work done by the thrust perunit time, or thrust power, to the rate of production of the propellant kineticenergy. For a single propellant stream (jet)

ηp =Fu

ma[(1 + f)u2e/2 − u2/2]

Usually f ≪ 1 and pe ≈ pa so that F ≈ ma(ue − u). Therefore (deriveyourself!)

ηp ≈ 2u/ue

1 + u/ue(4)

Maximizing ηp by taking u/ue close to 1 means that large ma is neededfor creating finite thrust. This is approached in turboprop engines but is notpractical for turbojets.

Thermal efficiency ηth is the rate of addition of kinetic energy to thepropellant divided by the energy consumption rate, mfQR, where QR is theheat of reaction of fuel

ηth =ma[(1 + f)u2

e/2 − u2/2]

mfQR(5)

For turboprop and turboshaft engines the output is largely shaft power Ps,and

ηth =Ps

mfQR

Propeller efficiency ηpr is the ratio of thrust power to shaft power,

ηpr =Fpru

Ps

Turboprop engines can produce a fraction of thrust from the hot exhaust,which may need to be taken into account.

Overall efficiency ηo is the ratio of the thrust power to the rate of energyconsumption:

ηo =Fu

mfQR(6)

It is equal to ηpηth or ηprηth, as appropriate.For turbojet (=single exhaust) from (4)

ηo ≈ 2ηthu/ue

1 + u/ue


The overall efficiency is less than unity because part of the energy of the fuelis lost as kinetic energy of the exhaust and another part as the thermal energyof the exhaust (however, we neglect here the kinetic energy of the fuel, see theend of Lecture 5). Decreasing ue toward u reduces losses of kinetic energy butcan often increase losses of thermal energy, so that an optimum can exist.

Takeoff thrust. At take-off u = 0. Neglecting f ≪ 1 and pe −pa and usingthe thrust equation (F = ma[(1+f)ue−u]+(pe−pa)Ae, see the correspondinglecture) and (5) gives

Ftakeoff =2ηthQRmf

ue(7)

Therefore, for a given fuel consumption rate take-off thrust is inversely pro-portional to the exhaust velocity. In the engine fuel is mixed with air and thenburns. The temperature of the products of combustion is, in rough approxima-tion, proportional to the fuel-to-air ratio, and for optimized engines this peaktemperature is, again roughly, equal to the highest temperature at which theturbine is not damaged yet. So, for optimized engine fuel flow rate is approx-imately proportional to the hot-air flow rate, which, in turn, is approximatelyproportional to engine size. Therefore, for given take-off thrust smaller ue meanssmaller engine size.

Another important characteristic, specific thrust, is defined as F/ma. Ifcertain engine design ensures higher F/ma then for the same thrust smaller ma

that is smaller engine size (or fewer engines per aircraft) is required.

Aircraft range: Breguet’s range formula. In level flight at constantspeed F = D = L(D/L) = mg/(L/D), where D is drag, L is lift, m is aircraftmass, and g is gravity acceleration. Therefore, the thrust power is

Fu =mgu

L/D

If m is the aircraft mass including the mass of the fuel then

dm

dt= u

dm

dl= −mf

where l is the distance along the flight path. Using (6) and integrating givesthe Breguet’s range formula

l = ηoL

D

QR

gln

m1

m2(8)

where m1 and m2 are initial and final masses of the aircraft.Look at the picture and think why long-range passenger aircraft fly at high

subsonic M . Very high M causes an increase in the thermal and mechanicalloads on the aircraft structure which, hence, has to be heavier.

Thrust specific fuel consumption, TSFC, is defined as

TSFC = mf/F

Lecture 3. Engine performance. 13

0

5

10

15

20

0.5 1 1.5 2 2.5 30.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

L/D

L/D

η0

0η

M

Figure 3: Typical variation of the overall efficiency, lift-to-drag ratio and theirproduct as a function of Mach number for existing engine materials and the bestdesign configurations.

Using (6) reduces (8) to

s =L

D

u

g

1

TSFCln

m1

m2

Therefore, for a given flight speed the range of an aircraft is inversely pro-portional to TSFC. This formula is only approximate. Indeed, derivation of (8)assumed that L/D remains constant during the flight. This is not exactly truefor a flight with constant u.

Typical values of TSFC are: for ramjet (M = 2) 0.17-0.26, for turbojet(static) 0.075-0.11, for turbofan (static) 0.03-0.05 kg/(N·hr).

For turboprop engines a brake specific fuel consumption is defined byBSFC= mf/Ps, where Ps is the shaft power, and, if the exhaust thrust issubstantial, an equivalent brake specific fuel consumption is defined byEBSFC= mf/(Ps+Feu), where Fe is the thrust produced by the engine exhaust.Typically, EBSFC=0.27-0.36 kg/(kWh).


Lecture 4. Ramjet engine.

Diffuser NozzleCombustion chamber

Fuel injection

Exhaust jet

(burner)

a d b e

Figure 4: The ramjet.

To ram means to force in. In ramjet air is forced into the engine air intakeby the sheer drive of the speed of flight. Ramjet, in principle, can work atsubsonic speed but it can be practical only at supersonic speed. In a ramjet,air undergoes compression in the diffuser, then fuel is added and burnt in theburner, and then the combustion products expand through the nozzle. It ishelpful to consider first a simplified model of an ideal ramjet. For ideal ramjetit is assumed that compression and expansion processes are reversible and adi-abatic, that combustion occurs at constant pressure, that the air/combustionproducts properties (specific heat ratio γ and the gas constant R) are constantthroughout the engine, and, although this is not necessary, that the outlet pres-sure is equal to the ambient pressure, in other words, that the nozzle is in thedesign regime. The usual tool for analysis of the processes in engines is theso-called enthalpy-entropy diagram.

The thermodynamic state of air is determined by two independent parame-ters. If a point in h−s diagram is given then all other parameters, like pressure,temperature, density, internal energy etc can be calculated. When a unit massof air moves through the engine the properties are changing and the point thatindicates the state is moving accordingly. The use of the enthalpy h and en-tropy s is especially convenient for the following reasons. In adiabatic reversibleprocess s remains constant, and, therefore, the path of such a process is a ver-tical line in h − s diagram. Since irreversibility usually lead to deterioration ofperformance, engines are designed so as to be as close to reversible processes aspossible. If the process is irreversible then entropy at the end of it is greaterthan entropy at the end of the corresponding reversible process. Therefore, inh− s diagram it is easy to anticipate the effect of irreversibility on the shape ofthe diagram. The advantage of using enthalpy as the other parameter followsfrom the form of the energy conservation law for open steady-state system:

Lecture 4. Ramjet engine. 15

p=p =p =consta e

p=p =p =const

d b

a

d

b

e

s

h

ambient temperature

kinetic energyadded

heat releasedby the jet whenit cools to the

kinetic energyof incomingair returned

air

kinetic energyof incoming

heat added

Figure 5: Enthalpy-entropy diagram for an ideal ramjet

W

m=

Q

m− h0out + h0in =

Q

m− (hout +

u2out

2) + (hin +

u2in

2)

where Wm and Q

m are respectively the work done by and the heat addedto a unit mass of air passing from station “in” to station “out”. Therefore,changes in h can readily be interpreted as work done, heat added, or kineticenergy variation. For a perfect gas h = cpT , therefore, one can consider haxis as approximate temperature axis. Therefore, a restriction of the maximumtemperature in the engine can be interpreted as a restriction on maximum h.

In a diffuser no heat is added to air and no work is done, but the velocitydecreases. Therefore, h increases, as shown in Fig. 5 by a straight line a-d,with hd − ha equal to the kinetic energy of a unit mass in the incoming flow.Then, heat is added in the combustion process d-b. Heat addition leads to anincrease in s in accordance with the formula dQ = Tds for reversible processes.Therefore, d-b goes in the direction of increasing s along the curve p = const.Along this curve hb − hd represents the amount of heat added per unit mass ofair. In the nozzle no heat is added and no work is done, therefore, the totalenthalpy is constant, but kinetic energy increases and h decreases, with hb − he

giving the kinetic energy of a unit mass in the exhaust jet.For a perfect gas in h − s diagram the p = const curves have the form

h = Const exp(s/cp), and, therefore, the distance b-e is greater than d-a. Thisis a very important feature. It is also true for real gases even though the curvesmay be not exactly exponential. Therefore, the exhaust velocity is greater thanthe flight velocity, and thrust is created.

Naturally, exhaust gases do not go back into the air intake. However, theydo cool down to the ambient temperature. This process occurs at constant


pressure p = pa and is shown as curve e-a in the figure. The enthalpy differencehe − ha is the amount of heat per unit mass released by the jet when it coolsdown. We can see that only part of the heat added in the combustion processturns into the useful kinetic energy.

It is possible now to analyse the effects of various parameters on the engineperformance.

Decreasing the flight speed, that is decreasing the Mach number, decreasesthe kinetic energy hd − ha (see Fig. 5). Imagine that it became very small.Then d-b will almost coincide with a-e. As a result, the added kinetic energywill become very small. At the same time the heat released by the jet will bealmost equal to the heat added in the burner and will remain finite. That isburning a finite amount of fuel will produce almost no thrust, that is the thrustspecific fuel consumption, TSFC, will become quite high. Also, since the addedkinetic energy is small, specific thrust, that is thrust-to-air-mass-flow rate ratiowill also be small. Then, for producing finite thrust the air mass flow rate hasto be large, and, accordingly, the engine size has to be quite large. This is whyramjet is not practical at small Mach numbers.

If, for a fixed Mach number, that is fixed hd − ha, the added heat hb − hd

is increased the kinetic energy added will also be increased (see again Fig. 5).Therefore, the specific thrust will be increased thus allowing smaller engine forthe same thrust. The thermal efficiency will not be affected strongly as onecan also see from the figure. However, an increase in hb means increase in themaximum temperature, and this is limited by the material properties of theengine walls.

Suppose that the maximum temperature, and, hence, hb is fixed. Considerthe variation of the specific thrust and TSFC with the Mach number M . Atsmall M , point d is close to point a, and, as discussed, TSFC is high andspecific thrust is small. An increase in M reduces TSFC. As for the specificthrust, one has to take into account that with hb fixed increase in hd results in adecrease in the amount of heat added. For small M this effect is, clearly (see thefigure again) is less important and specific thrust increases with M . However,when hd approaches hb, the kinetic energy added, being only a fraction of theadded heat, tends to zero. Accordingly, the specific thrust also tends to zero.Therefore, specific thrust has a maximum at some value of M between zeroand that value of M at which isentropic compression in the diffuser leads totemperature attaining the maximum allowed value.

It is possible to calculate all characteristics of an ideal ramjet. However, theuse of h − s diagram makes it possible to achieve an intuitive understandingof the engine performance, as this was illustrated above. One can now, forexample, anticipate the rationale of the turbojet. Try it.

Lecture 5. Ideal ramjet calculation. 17

Lecture 5. Ideal ramjet calculation.

This is a terse version. Listen to the lecture :-) or consult the lecture notesof your friend.

The following basic relationships/ideas are assumed to be knownfrom courses studied earlier.

Speed of sound c =√

γRT , Mach number M = u/c.Enthalpy h = cpT for perfect gas.Entropy s = cp lnT − R ln p (for perfect gas) or s = s(p, T ) for real gas

(the functional form depends on the gas) remains constant in adiabatic (that iswithout heat addition) and reversible process.

Stagnation temperature

T0 = (1 +γ − 1

2M2)T (9)

For s1 = s2,p1/p2 = (T1/T2)

γ/(γ−1) (10)

Stagnation enthalpy h0 = h + u2/2.Mass conservation law for an open system with a steady flow

∑

outlets

mout =∑

inlets

min

where m-s are mass flow rates.Energy conservation law for an open system with a steady flow

W +∑

outlets

mouth0out = Q +∑

inlets

minh0in (11)

where W is the work per unit time done by the system (positive for turbine,negative for compressor) and Q is heat added per unit time (positive for aburner).

Notation convention. All symbols have their usual meaning. The sub-scripts denote:

a is the incoming stream, apart from the incoming stream velocity u andMach number M which have no subscripts.

d is the diffuser outletb is the burner outlete is the nozzle outlet (exhaust).

Calculating the ideal ramjet performance.

At least the following parameters should be known before the calculationstarts:

flight Mach number M , ambient temperature (at the diffuser entrance) Ta,temperature at the burner outlet T0b ( prescribing it makes sense since it is the


0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 80

0.05

0.1

0.15

0.2

0.25

0.3

Spe

cific

thru

st

TS

FC

M

Specific thrust , kN*s/kg, Tmax=2500KTSFC, kg/(kN*s), Tmax=2500

Specific thrust, kN*s/kg, Tmax=2000KTSFC, kg/(kN*s), Tmax=2000

Figure 6: Ideal ramjet thrust and fuel consumption, QR = 45000 kJ/kg, γ = 1.4,Ta = 225K. Note that higher Tmax allows smaller engine. Note that as M increasesabove 3, the engine size starts to increase.

maximum temperature in the engine), fuel heating value QR, specific heat ratioγ, gas constant R or specific heat at constant pressure cp = γR/(γ − 1).

The assumption are isentropic adiabatic flow in the diffuser and in the noz-zle, zero velocity and constant pressure in the combustor (hence whatever0 =whatever), the nozzle in the design regime (hence pe = pa).

Thread 1, leading to fuel-to-air ratio f :

By (9): T0d = Ta(1 + γ−12 M2)

By (11): (1+f)h0b = h0d +fQR with h = cpT ⇒ f = (T0b−T0d)/(QR/cp−T0b).

Thread 2, leading to ue:

By (10) p0d/pa = (T0d/Ta)γ/(γ−1), then p0b = p0d, then by inverse of (10)Te = T0b(pe/p0b)

(γ−1)/γ , then by (11) cpT0b = cpTe + u2e/2 ⇒ ue = . . . (derive

yourself!).

Thread 3, leading to specific thrust, TSFC, and efficiencies:

u = M√

γRTa, then specific thrust F/ma = (1 + f)ue − u by the thrustequation, then TSFC = mf/F = f/(F/ma), then the efficiencies as defined inLecture 3.

Lecture 5. Ideal ramjet calculation. 19

0

0.2

0.4

0.6

0.8

1

1.2

1 2 3 4 5 6 7 8

M

Overall efficiencyPropulsion efficiency

Thermal efficiency

Figure 7: Ideal ramjet efficiencies, QR = 45000 kJ/kg, γ = 1.4, Tmax = 2500K,Ta = 225K. Note that efficiencies continue to grow up to the highest M .

So, for very large M the compression in the diffuser may be so strong and,hence, the temperature of the compressed air so high that is even higher thanTmax allowed by the properties of the materials. The way to overcome thisdifficulty is not to bring the air velocity almost to zero in the diffuser, thusensuring less compression. This idea leads to development of scramjet, thatis supersonic combustion ramjet, in which the air speed remains supersonicthroughout the engine. At the moment this idea has not been implemented inpractice.

In our definition of the efficiencies in Lecture 3 the kinetic energy of the fuelwas ignored. However, the typical fuel has QR = 45000kJ/kg, and for M = 7 thekinetic energy of the fuel is u2/2 ≈ 3000kJ/kg (check yourself!), that is about6% of the chemical energy. This can result in the efficiencies being greater than1 (look at the plots)! Note that both specific thrust and TSFC are not affected,after all, so we will use the traditional definition. However, for hypersonic flight(scramjets!) the kinetic energy of the fuel is not negligible.


Lecture 6. Ideal turbojet.

Diffuser

a d e

chamber (burner)

Fuel injection

Combustion

b

Nozzle

tc

Com

pres

sor

Tur

bine

Exhaust jet

Figure 8: The turbojet.

Subscripts will correspond to Fig. 8.One would like to be able to take off but ramjet does not work at zero

flight velocity, and one would like to be able to fly at about M = 0.8 − 0.9(due to the drop in L/D at M ≈ 1, see Lecture 3). However, at small Mramjet performance is poor because of insufficient compression. The idea ofa turbojet is to remedy this by introducing a compressor between the diffuserand the burner as shown in Fig. 8. Since the compressor has to be driven, aturbine is added after the burner. An ideal turbojet is a turbojet in which allthe processes are reversible, and the combustion is assumed to occur at constantpressure. The enthalpy-entropy diagram is shown in Fig. 9.

In an ideal turbojet the work done by compressor is equal to the work doneon the turbine, that is d − c is equal to b − t on the diagram. The compressorpressure ratio prc = pb/pd is an important design parameter. By comparingFig. 9 with the h − s diagram of the ideal ramjet (Lecture 4) it can be seenthat the distance a− c in Fig. 9 plays the role of the distance a− d in the idealramjet diagram. Therefore, the dependence of the specific thrust and TSFC ofa turbojet on prc is similar to their dependence on M for a ramjet. (Do notgive this argument during the exams! Just repeat the reasoning for the ramjethere.) We should expect that TSFC decrease with increasing prc, and there isprc such that the specific thrust is at its maximum there. Notice now that thisoptimum (from specific thrust viewpoint) prc corresponds to a certain distancea− c. If for fixed c M increases then d goes up, and d− c is reduced. Therefore,prc is reduced. That is at higher M the prc ensuring maximum specific thrustis smaller. (As calculations show, at about M = 3 the optimum prc = 1 so thatcompressor is not needed at all, and turbojet becomes a ramjet.)

Calculation of ideal turbojet engine is similar to calculation of an idealramjet. In addition to the parameters which should be prescribed for an ideal

Lecture 6. Ideal turbojet. 21

p=p =p =consta e

p=p =p =const

cb

heat added

s

a

airof incomingkinetic energy

e

ambient temperatureit cools to the by the jet when

air returnedof incomingkinetic energy

d

bh

of airon unit massby compressorwork done

work done bya unit mass ofair on theturbine

kinetic energy

added

c

t

work

maxTc p

heat released

rejected heat, that is

Figure 9: The ideal turbojet h − s diagram

ramjet, for turbojet the compressor pressure ratio should be given. Refer toLecture 5 and check that you understand why each specific formula is usedbelow.

Thread 1, leading to fuel-to-air ratio f :T0d = Ta(1 + γ−1

2 M2)

p0d/pa = (T0d/Ta)γ/(γ−1), thenp0c = prcp0d

T0c = (prc)(γ−1)/γT0d

(1 + f)h0b = h0c + fQR with h = cpT ⇒ f = (T0b − T0c)/(QR/cp − T0b).

Thread 2, leading to ue:p0b = p0c since combustion is assumed to occur at constant pressureNeglecting the difference in flow rates through compressor and turbine, com-

pressor work = turbine work ⇒ h0c − h0d = h0b − h0t, or, with h = cpT ,T0t = T0b − (T0c − T0d)p0t = p0b(T0t/T0b)

γ/(γ−1)

Te = T0t(pe/p0t)(γ−1)/γ

cpT0t = cpTe + u2e/2 ⇒ ue = . . .

Thread 3, leading to specific thrust, TSFC, and efficiencies, is exactly thesame as for the ramjet.

Reheat. Note that fuel-to-air ratio f , being limited by the requirementon turbine inlet temperature, is well below the stoichiometric ratio. Therefore,


s

burn

er

afte

rbur

ner

com

pres

sor

diffu

ser

turbine

nozzle

Figure 10: The ideal turbojet with reheat h − s diagram.

after the burner the air still contains a lot of oxygen. In the turbine the temper-ature drops down. In order to increase the thrust, it is possible to add to thedesign another burner between the turbine and the nozzle, and reheat the airin this afterburner. The corresponding process is shown in Fig. 10. Since theafterburner isobar (constant pressure curve) is closer to the ambient pressureisobar, burning fuel in the afterburner is less efficient from the thermodynamicsviewpoint (consult again Lecture 4). However, the temperature at the outletof the afterburner may be much higher than the temperature at the turbineinlet, since the nozzle is less sensitive to it. Therefore, an increase in thrust canbe substantial. For example, General Electric J85-21 turbojet, which poweredthe Northrop F5E, gives 22.2 kN thrust with afterburning but only 15.6 kNwithout afterburning. However, the TSFC is 60.3 mg/(N s) and 28.3 mg/(N s)respectively. Military aircraft can cruise without afterburning and accelerate byengaging reheat when necessary.

Lecture 7. Ideal turbofan. 23

Lecture 7. Ideal turbofan.

This is terse again, see again Lecture 5. High time to get used to it :-).

Subscripts will correspond to Fig. 11.

e

Exhaust jet

Com

pres

sor

d (burner)c b

Tur

bine

Nozzle

Diffuser

Fannozzle

Fannozzle

t

f

f

fn

fn

a

Fan exhaust jet

Fan exhaust jet

Fan

Fan

Figure 11: The turbojet.

As it follows from Brequet’s range formula, an aircraft range is propor-tional to the overall efficiency, which in turn is proportional to the productof propulsion and thermal efficiencies (see Lecture 3). The exhaust jet, whilecreating thrust, carries away thermal and kinetic energy. Increasing the ther-mal efficiency reduces energy losses in the form of thermal energy in the ex-haust. Overall efficiency can also be improved by reducing losses with kineticenergy of the jet, that is by increasing the propulsion efficiency. To create acertain amount of momentum, an aircraft can accelerate a mass of air, say M ,to the velocity, say, U , thus getting, due to the action of the reaction force, theamount of momentum ∆I = MU . This mass of air will, then, have a kineticenergy MU2/2, which is the extra cost we have to pay for getting ∆I. Since,MU2/2 = U∆I/2 = (∆I)2/(2M), this extra cost decreases when U is decreasedand M is correspondingly increased. In other words, it is better to use morepropellant with less kinetic energy added to a unit mass of propellant.

For a turbojet, neglecting f in the equations for thrust and for kinetic energyand assuming pe = pa gives

ηp =2u

u + ue

(Oops. Think better, but if you do not understand this even after that, goback and reread the previous lectures). So, ηp can be increased by reducing thespecific thrust (Oops again . . .), but that would increase the size of the engine.Turbofan is designed to get larger mass of propellant to be accelerated thus


p=p =p =consta e

p=p =p =const

cb

heat added

s

a

airof incomingkinetic energy

e

ambient temperatureit cools to the by the jet whenheat released

air returnedof incomingkinetic energy

d

b

of airon unit massby compressorwork done

work done bya unit mass ofair on theturbine

c

work to compressor

addedkinetic energy

workto fan

t

max

h

Tcp

rejected heat, that is

Figure 12: The ideal turbofan h − s diagram.

obtaining higher propulsion efficiency without increasing the size of the basicgas generator.

Turbofan has also the advantage of relative simplicity of thrust reversal.During landing the fan jet (but not the hot exhaust jet) can be deflected tocreate negative thrust, thus eliminating the need for expensive and unreliablehydraulic braking systems in the wheels.

Another advantage of turbofan engines is that, due to smaller exhaust ve-locity, they produce less noise.

Thermodynamicly, the idea of turbofan is to take more power from theturbine then is needed to drive the compressor, thus reducing the kinetic energyof the exhaust, and use this additional power to drive the fan, thus increasingthe total mass flow rate of the propellant, see Fig. 12.

Calculation of ideal turbofan engine is similar to calculation of an idealturbojet. In addition to the parameters which should be prescribed for anideal turbojet, for turbofan the fan pressure ratio prf and the bypass ratioβ = mfan/ma (ratio of the bypass airflow rate to the core engine airflow rate)should be given.

Thread 1, leading to fuel-to-air ratio f , is the same as for the turbojet, seeLecture 6.

Lecture 7. Ideal turbofan. 25

Thread “Fan”, leading to the fan jet velocity uef :

p0d/pa = (T0d/Ta)γ/(γ−1) (I will not say “oops” anymore. You know whatto do), then

p0f = prfp0d

T0f = (prf )(γ−1)/γT0d

we assume pef = pa,Tef = T0f (pef/p0f )(γ−1)/γ

cpT0f = cpTef + u2ef/2 ⇒ uef = . . .

Thread 2, leading to ue:

p0b = p0c as for turbojet.Then, similar to turbojet, neglecting the difference in flow rates through

compressor and turbine, compressor work+fan work = turbine work ⇒h0c−h0d +β(h0f −h0d) = h0b−h0t. Here we assumed that both compressor

and fan use the same diffuser. For ideal engine there is no other way, really,since h0d = h0a for any ideal diffuser.

With h = cpT ,T0t = T0b − (T0c − T0d) − β(T0f − T0d)p0t = p0b(T0t/T0b)

γ/(γ−1)

Te = T0t(pe/p0t)(γ−1)/γ

cpT0t = cpTe + u2e/2 ⇒ ue = . . .

Thread 3, leading to specific thrust, TSFC, and efficiencies.The specific thrust now isF/ma = (1 + f)ue + βuef − (1 + β)u,and TSFC= fma/F .The kinetic energy added to the propellant, calculated per unit mass of the

core airflow, now is((1 + f)u2

e + βu2ef − (1 + β)u2)/2. Therefore

ηp =2u((1 + f)ue + βuef − (1 + β)u)

((1 + f)u2e + βu2

ef − (1 + β)u2)

ηth =((1 + f)u2

e + βu2ef − (1 + β)u2)

2fQR

ηo = ηpηth

Optimization of ideal turbofan, that is the choice of the optimum valuesof prf and β is not really possible. The power available for fan is limited, so, noneof these two parameters can be increased indefinitely while the other parameteris held constant. However, it is possible to increase β and decrease prf at thesame time so that the power ( = the position of the point t in Fig. 12) remainsconstant. Then, increase in β will increase the mass flow rate of the propellant,hence, increase the propulsion efficiency. So, the optimum β is infinitely large.In practice, of course, losses, nacelle drag, allowable stresses limit β.


Lecture 8. The effect of losses.

In real engines the processes are not reversible. However, the heat transferfrom the air to the engine can usually be neglected. Accordingly, the h − sdiagram changes so that, at the end of each process, the entropy is somewhatgreaterthan it would be in an ideal case. Compare the diagrams in Fig. with thediagram for the ideal ramjet (Lecture 4) and observe how, in each case, theirreversibility leads to the decrease in the kinetic energy of the exhaust jet andthe increase in its thermal energy.

In real engine, due to combustion and temperature changes, the chemicalcomposition of the gas varies. It can be approximately taken into accountby using different values of γ and R for different components. Apart fromthis, gas is a two-parametric medium, that is its state is determined by twoparameters, say, pressure and temperature. Fortunately, since the heat transferbetween the engine and the air can be neglected, only one empirical constant isrequired in order to describe the deviation of the real performance from the idealperformance. For engine components in which the stagnation pressure does notchange in an ideal case, that is in diffusers, burners, and nozzles, the performancemay be characterized by stagnation pressure ratios rd = p0d/p0a, rb = p0b/p0d,and rn = p0e/p0t respectively. In general, the most commonly used parametersfor describing component performance are the so-called adiabatic efficiencies3:

for diffuser:

ηd =< ideal enthalpy change needed to ensure the same p0d/pa >

< actual enthalpy change >=

h0ds − ha

h0d − ha

for compressor (and similar for fan adiabatic efficiency ηf ): ηc =

< work required in isentropic process to ensure the same p0c/p0d >

< actual work >=

h0cs − h0d

h0c − h0d

for turbine: ηt =

< actual work obtained from the turbine >

< work that would be obtained in isentropic process to the same p0t/p0b >=

h0b − h0t

h0b − h0ts

for nozzle (and similar for fan nozzle ηfn):

ηn =< actual kinetic energy of the jet >

< jet kinetic energy in isentropic expansion to the same pe >=

h0t − he

h0t − hes

Also, the burner efficiency ηb is just the fraction of the chemical energy of thefuel that is released in the burner.

3Typically 0.7 < ηd < 0.9 (depending strongly on M), 0.85 < ηc < 0.9, 0.9 < ηt < 0.95,0.95 < ηn < 0.98, 0.97 < ηb < 0.99

Lecture 8. The effect of losses. 27

kinetic energyof the exhaustjet

kinetic energyof incomingair

a

e

s


ambient temperature

b

d

h

heat added

Ramjet h-s diagram for a real diffuser



a

e

s


ambient temperature

b

d

h

heat added

Ramjet h-s diagram for a real burner



a

e

s


ambient temperature

b

d

h

heat added

Ramjet h-s diagram for a real nozzle

Figure 13: Ramjet h-s diagrams


Now, we can get the full set of formulae for calculating a real turbofanengine.

The following quantities should be given: M , T0b, prc, prf , β, γ, R, ηd, γc, ηc,Ta, γf , ηf , ηfn, Rf , rb, QR, cp, γt, ηt, γn, ηn, Rn, ηb. The core exhaust pressureis assumed to be equal to pa, which implies a subsonic or design supersonicnozzle.

Diffuser: T0d =

(

1 +γ − 1

2M2

)

Ta, T0ds = Ta + ηd(T0d − Ta),

p0d

pa=

(

T0ds

Ta

)γ/(γ−1)

orp0d

pa=

(

1 + ηd

(

T0d

Ta− 1

))γ/(γ−1)

Compressor:p0c

pa=

p0d

paprc

T0cs

T0d= p(γc−1)/γc

rc , T0c = T0d +1

ηc(T0cs − T0d),

or T0c =

(

1 +1

ηc

(

p(γc−1)/γcrc − 1

)

)

T0d

Fan, similarly: T0f =

(

1 +1

ηf

(

p(γf−1)/γf

rf − 1)

)

T0d,p0f

pa=

p0d

paprf

f =T0b − T0c

(ηbQR/cp) − T0b

Turbine:p0b

pa=

p0c

parb, T0t = T0b − (T0c − T0d) − β(T0f − T0d)

T0ts = T0b −1

ηt(T0b − T0t),

p0t

p0b=

(

T0ts

T0b

)γt/(γt−1)

orp0t

pa=

p0b

pa

(

1 − 1

ηt

(

1 − T0t

T0b

))γt/(γt−1))

Core engine nozzle and fan nozzle. Thrust:

cpf =γf

γf − 1Rf , h0f − hefs = cpfT0f

(

1 −(

p0f

pa

)(1−γf )/γf

)

,

uef =√

2ηfn(h0f − hefs) or uef =

√

√

√

√2ηfnγf

γf − 1RfT0f

[

1 −(

p0f

pa

)(1−γf )/γf

]

.

Similarly ue =

√

√

√

√2ηnγn

γn − 1RnT0t

[

1 −(

p0t

pa

)(1−γn)/γn

]

.

u = M√

γRTa, F/ma = ((1 + f)ue + βuef − (1 + β)u), TSFC = fma/F

ηp = 2u((1 + f)ue + βuef − (1 + β)u)/((1 + f)u2e + βu2

ef − (1 + β)u2)

ηth = ((1 + f)u2e + βu2

ef − (1 + β)u2)/(2fQR), ηo = ηpηth

Lecture 9. Typical parameters of real engines. 29

Note on turboprop and turboshaft engines

The approach used to calculate the performance of the turbofan engine canbe used for calculating turboprop and turboshaft engines with minimal correc-tions. For turbofan, the only point were the existence of the fan affects thecore engine calculation is the energy equation for the turbine (see Lecture 8):T0t = T0b − (T0c − T0d) − β(T0f − T0d)

Here, the power supplied to the fan is maβcp(T0f − T0d). In turboshaft en-gine, this should be replaced with the shaft power, and the energy equation canthen be used to calculate this shaft power if T0t is known. Now, using the for-mulae from Lecture 8, T0t can be found by prescribing the pressure immediatelyafter turbine p0t to be just slightly above the ambient pressure, since no thrustis expected to be produced from the exhaust jet.

In a turboprop engine the hot exhaust jet can produce a significant part ofthe total thrust. For a given gas generator and flight speed there is an optimumhot-gas exhaust velocity that gives maximum thrust. Let us rewrite the sameenergy equation in the form

P/ma = h0b − (h0c − h0d) − h0t

where P is the power passed through to the propeller (for turboshaft engineP is the shaft power, by the way). Given the gear ηg and the propeller ηpr

efficiencies, the propeller thrust then isFpr = ηprηgP/u.Then the expression for the specific thrust isF/ma = (1 + f)ue − u + Fpr/ma

In these formulae P or h0t can be considered as the design parameter. If h0t

increases then the propeller thrust decreases but the exhaust velocity increases.By varying h0t over the possible range one can see that the total thrust indeedhas a maximum inside that range.


Lecture 9. Typical parameters of real engines.

The following is just a set of citations and figures from the correspondentchapter of the recommended textbook, (P. Hill and C. Peterson, Mechanics andThermodynamics of Propulsion, second edition. Addison-Wesley, 1995.)

We now consider examples of turbojet thrust and fuel consumption varia-tion with compressor pressure ratio, turbine inlet temperature, and flight Machnumber. For sample calculations we make the assumptions about componentefficiency, fluid properties, and engine conditions shown in Table 5.1. Eachflight Mach number is assumed to correspond to a different altitude and thus adifferent pair of ambient pressure and temperature values pa and Ta. In thesecalculations we assume, as before, that the product mcp is the same in theturbine as in the compressor.

TABLE 5.1 Turbojet calculation parameters (Fuel heating value 45,000 kJ/kg)

Component Adiabatic efficiency Average specific heat ratioDiffuser ηd= 0.97 1.40Compressor ηc= 0.85 1.37Burner ηb= 1.00 1.35Turbine ηt= 0.90 1.33Nozzle ηn= 0.98 1.36Flight altitude Ambient pressure Ambient temperature(cruise Mach no.) (kPa) (K)Sea level (0) 101.30 288.240,000 ft (12,200 m) (0.85) 18.75 216.760,000 ft (18,300 m) (2.0) 7.170 216.780,000 ft (24,400 m) (3.0) 2.097 216.7

Figures 5.19 to 5.22, inclusive, show the thrust and specific fuel consumptioncalculated under these conditions for three values of turbine inlet temperatureand four values of flight Mach number. Both of these parameters, as well asthe compressor ratio, strongly influence the engine performance. We see inparticular from these graphs:


1. At a given flight Mach number and a given turbine inlet temperature, thepressure ratio that maximizes specific thrust does not provide minimumfuel consumption. Since the mass of the engine depends strongly on air-flow rate and substantially affects aircraft carrying capacity, both specificthrust and specific fuel consumption have to be considered in selecting thebest compressor pressure ratio. The choice of pressure ratio for best cruis-ing range will require a compromise that takes into account both engineand aircraft characteristics.

2. Raising turbine inlet temperature can substantially improve specific thrust.The maximum temperature shown, 1700 K, is far below the maximum sto-ichiometric combustion of hydrocarbon fuels but requires not only high-temperature alloys for the turbine blades but also quite intensive bladecooling as well.

3. With a given compressor pressure ratio, raising turbine inlet temperaturemay or may not raise fuel consumption per unit thrust. For pressureratios associated with minimum fuel consumption, increasing turbine inlettemperature can reduce specific fuel consumption somewhat. (It may,though, require an increase in the rate of cooling air extracted from thecompressor, and the present calculations have not taken this into account.)

4. The compressor pressure ratio required to minimize specific fuel consump-tion is much less for supersonic than for subsonic flight. At Mach 3, peakspecific thrust occurs with a compressor pressure ratio of 1; in this caseno compressor or turbine is required, and the turbojet becomes a ramjet.The ramjet could, of course, tolerate much higher maximum temperature,not having exposure of the hot gas to highly stressed turbine blades.

Figures 5.19 to 5.22 do not display the sensitivity of engine performance toengine component efficiencies. The designer has great incentive to make thecompression, combustion, and expansion processes as efficient as possible.

In Chapters 6-9 we discuss the physical factors that explain engine perfor-mance limitations. Since these factors strongly influence the configuration ofactual machines, it is appropriate to point them out here before making com-ponent analyses. Basically they are of two kinds: material limits, as expressed


by allowable stress and temperature levels, and aerodynamic limits, imposedmainly by the behavior of boundary layers in the presence of rising pressure.

As it does in all flight applications, the desire for minimum engine massleads to highly stressed engine components as well as high peak temperatures.The largest stresses on turbine blades are due to centrifugal force, and theallowable stress in these blades is directly related to the temperature at whichthey must operate. Thus compromise is needed between the turbine blade stressand combustion temperature. With currently available material, the maximumstagnation temperature of the turbine inlet gas is limited to about 1200 K foruncooled blades. If the blades are cooled internally by passing air through smallinterior passages, it is possible to use stagnation temperatures of 1700 K ormore. Operation at such high temperatures, however, may require a substantialfraction of the compressor air flow (as much as 10% or more) to be diverted toturbine cooling rather than to entering the combustor.

Owing primarily to the necessity of avoiding boundary layer separation, and(to a lesser degree) to the high losses associated with compressibility, the pres-sure ratio that can be accommodated in a single state of a turbomachine islimited. (A stage may be defined as a single circumferential row of rotatingblades and an associated set of stationary guide vanes, stators, or nozzles.)This limitation is most severe in compressors where the fluid necessarily flowsagainst a rising pressure gradient (see Chapter 4). Primarily for this reason,axial-compressor-stage pressure ratios are much lower than turbine-stage pres-sure ratios. Hence axial compressors of 10 to 20 stages can be driven by turbinesof 2 or 3 stages.

To illustrate the performance improvement possible with bypass engines, weuse the same calculation assumptions as shown in Table 5.1 and the bypass fan


assumptions shown in Table 5.2 to obtain the results shown in Figs. 5.29 to5.34.

We can see the effect of bypass ratio on takeoff thrust by comparing Figs.5.19 and 5.29. Other conditions remaining the same, the takeoff thrust per unitcore engine flow rate ma has nearly doubled, as the bypass ratio has changedfrom 0 to 5. At the same time the thrust specific fuel consumption has decreasedsubstantially.

Comparison of Figs. 5.20 and 5.30 shows the same kinds of benefits for highsubsonic (M = 0.85) cruise thrust, though the relative gains are considerablyless. For supersonic flows the bypass idea would have little potential benefit, andthe problem of shock losses associated with a large nacelle would be formidable,if not prohibitive.

Figure 5.31 shows the efficiencies of the turbojet engine for M = 0.85 cruiseconditions. In contrast, Fig. 5.32 shows the corresponding efficiencies for theturbofan engine with β = 5. The overall efficiency is significantly greater thanfor the turbojet (β = 0).

TABLE 5.2 Bypass fan characteristicsComponent Efficiency Specific heat ratioDiffuser ηd = 0.97 γd = 1.4Fan ηf= 0.85 γf = 1.4Fan nozzle ηfn= 0.97 γn = 1.4Fan pressure ratio pfr = 1.50

Figures 5.33 and 5.34 show large values of the overall efficiency potentiallyavailable with supersonic flight. The compressor pressure ratio that maximizesη0 drops rapidly as the design flight Mach number increases above, say, 2. Foran aircraft that is required to cruise at both subsonic and supersonic speeds,the choice of engine compression ratio implies an important compromise.

For subsonic flight there is a strong propulsion efficiency advantage in using abypass rather than a turbojet engine. The higher the turbine inlet temperature(and thus the higher the exhaust velocity of the corresponding turbojet), thegreater the benefit. The questions of optimum bypass ratio and optimum fanpressure ratio deserve serious exploration and may lead to further refinementsof turbofan designs. In the absence of geared power transmission to the fan,the problem of speed mismatch between the fan and its driving turbine tendsto limit the bypass ratio of turbofan engines. Other considerations affectingthe choice of bypass ratio include the structural weight and the aerodynamicdrag of the engine nacelle. Designing a gearbox to transmit 30,000 kW wouldmean confronting serious questions concerning gearbox weight and reliability.Nevertheless, the geared fan is a future possibility.


Lecture 10. Rocket propulsion I 35

Lecture 10. Rocket propulsion I

The general overview of the rocket propulsion, including the types of rocketengines and their performance characteristics, is given in the parallel (Astro-nautics Part 2) course. As it follows from the Tsiolkovsky (or rocket) equationderived in that course, the major performance characteristic of a rocket engineis the specific impulse, defined as

Isp = F/(mg)

where F is the thrust, m is the propellant flow rate, and g is the gravityacceleration at the earth’s surface. The presence of g in the definition is dueto historical tradition only. The higher the specific impulse, the greater thefraction of the payload in the total weight of the spacecraft launched to orbit.

In the present course we will limit the scope to chemical rockets only, andwill consider only one very simple example of a rocket calculation.

The expression for the thrust of the rocket engine can be obtained from thegeneral equation of the thrust for a turbojet engine (see Eq. (2 in Lecture 2)

F = ma[(1 + f)ue − u] + (pe − pa)Ae

in which, first, using f = mf/ma, we eliminate f , then put ma = 0, andmf = m, because the rocket engine does not swallow any air, and the onlypropellant it uses is its own fuel and oxidizer. Therefore, one has

F = mue + (pe − pa)Ae (12)

Here, pe is the pressure in the exhaust jet, pa is the ambient pressure, ue isthe exhaust velocity, and Ae is the cross-section area of the exhaust jet at thenozzle outlet. As it is known from nozzle analysis (see parallel AA208 (Aerother-modynamics) course, current lecturer Dr. G. N. Coleman), the maximum thrustcorresponds to pe = pa. However, when a rocket ascends to space, pa changesfrom atmospheric pressure at the launch pad level to zero in the orbit, and thishas to be taken into account in accurate calculations.

For pe = pa the specific impulse depends only on the exhaust velocity: Isp =ue/g, that is the greater the exhaust velocity the better.

The simplest scheme of a rocket engine consists of only two components:combustion chamber and a nozzle. As a united system they are also calleda thrust chamber. In case of a rocket with a liquid propellant the propellantis pumped into the combustion chamber. Solid propellant rockets have theirpropellant stored in the combustion chamber itself.

Nozzle calculation. In fact, nozzle calculation is the same as for air-breathing engines. However, for the analysis of a rocket it is useful to derive aspecific form of the expression for the exhaust velocity. For simplicity, an idealnozzle is considered, that is we assume that inside the nozzle the exhaust gascan be approximated by a perfect gas of constant composition and with constant


specific heat, and that the expansion is steady, one-dimensional, and isentropic.We neglect the gas velocity in the combustion chamber, and denote T0 and p0

the stagnation temperature and pressure there.Then the energy equation for an adiabatic process givesu2

e/2 = h0 − he = cp(T0 − Te)Then, using formulae for isentropic processes (see Lecture 5), after simple

transformations one obtains

ue =

√

√

√

√2cpT0

[

1 −(

pe

p0

)(γ−1)/γ]

Expressing cp in terms of γ, the universal gas constant R, and the propellantmolecular weight M gives (compare with the formula for ue in Lecture 8)

ue =

√

√

√

√

2γR

(γ − 1)MT0

[

1 −(

pe

p0

)(γ−1)/γ]

(13)

The flow rate through the nozzle can be expressed via p0, T0, and the nozzlethroat area A∗ as the product of density at the nozzle throat, A∗, and thevelocity at the nozzle throat (again see AA208 course). The result is

m =A∗p0√RT0

√

γ

(

2

γ + 1

)(γ+1)/(γ−1)

(14)

where R = R/M . Using (12) one now obtains

F

A∗p0=

√

√

√

√

2γ2

γ − 1

(

2

γ + 1

)(γ+1)/(γ−1)[

1 −(

pe

p0

)(γ−1)/γ]

+

(

pe

p0− pa

p0

)

Ae

A∗

(15)There is no need to memorize this but notice that, for given γ the thrust

depends only on the pressures and nozzle geometry. Since pe/p0 also dependson the geometry only, the designer is left with p0 and the geometry as the designparameters. It seems attractive to have high p0 since that would mean that thesame thrust can be achieved with smaller A∗, that is smaller nozzle. However,increase in p0 leads to higher stresses in the walls of the combustion chamber, so,optimization of the chamber pressure requires an analysis on the entire-rocket-and-mission level. In what follows we will assume that p0 is prescribed.

The amount of propellant needed for a specific mission depends mostly onue (recollect again the Tsiolkovsky equation). From (13) it follows that ue canbe increased by increasing T0 and/or by decreasing M . This implies an im-portant trade-off between the desire to have the fuel-to-oxidizer ratio closer tostoichiometric for increasing T0 and having the fuel-to-oxidizer ratio giving thecombustion products of smaller molecular weight. In case of hydrogen-oxygen


combination, for example, the fuel-reach mixture is more advantageous. Calcu-lating T0 and the combustion product composition requires certain background.

Mixtures of gases. The properties of the mixture may be determinedfrom the properties of the constituents by using the Gibbs-Dalton Law. For amixture of n constituents:

Temperature T = T1 = T2 = · · · = Tn

Pressure p = p1 + p2 + · · · + pn

Density ρ = ρ1 + ρ2 + · · · + ρn

Internal energy per unit mass e = c1e1 + c2e2 + · · · + cnen

Entropy per unit mass s = c1s1 + c2s2 + · · · + cnsn

Enthalpy per unit mass h = c1h1 + c2h2 + · · · + cnhn

where ρi is the density of the i-th constituent, that is the mass of thatconstituent in a unit volume of the mixture, ci = ρi/ρ is the mass concentrationof the constituent, and pi is the partial pressure. The partial pressure is relatedto the density and temperature by the same formula as for pure substance:

pi = ρiRiTi = ρiRiT = ρiR

MiT

Here, R is the universal gas constant and Mi is the molecular weight of thei-th constituent.

A mole of a substance is the amount of substance the mass of which isthe molecular weight times the mass unit. For example, the molecular weightof water (H2O) is 2+16=18. Therefore, in the system of units where mass ismeasured in grams, one mole of water is 18 grams of water. If the mass unitis, say, a pound, then a mole of water is 18 pounds of water, etc. To avoidconfusion, notations like gm-mole and kg-mole are often used.

It is quite easy to obtain that

pi

p=

ni

n1 + n2 + · · · + nn(16)

where ni is the number of moles of the i-th constituent in the mixture.The molecular weight of the mixture can be calculated from the equation

(derive yourself assuming that p = ρRT/M should hold true for the mixture)

1

M=

c1

M1+

c2

M2+ · · · + cn

Mn(17)

For fixed chemical composition cv and cp of the mixture can be found usingthe above formulae and the fact that cv = de/dT , cp = dh/dT .

Combustion product temperature. The most important feature of amole is that a mole of any substance contains the same number (Avogadronumber) of molecules. Because of that, chemical reaction written down as, say,

2H2 + O2 → 2H2O

can be interpreted as “two moles of H2 react with one mole of O2 to producetwo moles of H2O”, and, hence, H2O here is interpreted as a mole of H2O. Moregenerally, one can have, say,


αH2 + βO2 → nH2OH2O + nH2H2 + nO2

O2

This can happen if hydrogen and oxygen are not in stoichiometric ratio unlikethe previous formula, or due to the dissociation of the combustion products. Inthe latter case the right hand side can also include H, O, and HO, but forsimplicity we will neglect their presence (not a good assumption in practice,though).

Assume now that the numbers of moles in this formula, that is α, β, nH2O,nH2

, nO2are known, and also that the reaction occurs at constant pressure. If

the initial temperature T1 of the reactants is given, what is the temperature ofthe combustion products?

By definition, the enthalpy of formation of a substance is the amount ofheat needed to be added at a constant temperature and pressure for this sub-stance to be formed in a reaction from its components as they occur naturallyat this temperature. Since this amount depends on the temperature, its valueis important, and this temperature is called reference temperature. Enthalpyof formation can be measured experimentally and is available from the tablesfor many substances. If the formation reaction releases heat, enthalpy of for-mation is negative. Usually, enthalpy of formation is tabulated per mole of thesubstance.

Now, we can take α moles of H2 and β moles of O2 at the initial temperatureT1 and bring them at constant pressure to the reference temperature Tf . Forthis we will have to add the amount of heat equal to

cp(H2)(Tf − T1)αMH2+ cp(O2)(Tf − T1)βMO2

.

Quite often the specific heat per mole cp = Mcp is used in such calculations,and then the amount of heat required can be rewritten as

(αcp(H2) + βcp(O2))(Tf − T1).

Then we can decompose these reactants into their natural components, andfor that we will have to subtract the amount of heat equal to their formationethalpies at Tf times number of moles:

α∆Hof (H2) + β∆Ho

f (O2).

(For usual values of Tf formation enthalpies of hydrogen and oxygen will bezero because they will naturally occur as they are :-), but for other fuels andoxidizers this step can be necessary.)

Then we can compose the products, and for this we will have to add theamount of heat equal to

nH2O∆Hof (H2O) + nH2

∆Hof (H2) + nO2

∆Hof (O2)

Finally, we will bring them to the final temperature T2, and for that weshould add the heat in the amount equal to

(nH2O cp(H2O) + nH2cp(H2) + nO2

cp(O2))(T2 − Tf )

Therefore, in total, in order to change the temperature of the mixture fromT1 to T2 and its chemical composition according to the reaction, we have to addthe amount of heat equal to


Qtotal = (αcp(H2) + βcp(O2))(Tf − T1) − (α∆Hof (H2) + β∆Ho

f (O2))+

nH2O∆Hof (H2O) + nH2

∆Hof (H2) + nO2

∆Hof (O2)+

(nH2O cp(H2O) + nH2cp(H2) + nO2

cp(O2))(T2 − Tf )(18)

If this transformation occurs in the combustion chamber adiabaticly, thenQtotal = 0 and (18) can be considered as an equation for determining T2, thatis the temperature of the combustion products.

Note, that these formulae imply that h = cpT , that is that cp is constant.At high temperatures characteristic for rocket engines this assumption is notvalid and, hence, in the above formulae cp denotes some averaged values. Inmore accurate calculations h = cpT has to be replaced with h = h(T ), withh(T ) given by more complicated formulae or tables.

Note that since we neglect velocity in the combustion chamber, T2 in theabove formulae is the stagnation temperature T02, and that in formulae for thenozzle we omitted this second subscript, so that in fact T2 = T0.


Lecture 11. Rocket propulsion II. Composition of com-bustion products.

Consider again the reaction

αH2 + βO2 → nH2OH2O + nH2H2 + nO2

O2 (19)

How can one find the composition of the combustion products, that is nH2O,nH2

, and nO2?

Intuitively it may seem that hydrogen and oxygen will always react, andthat the reaction stops only when there is no more hydrogen or no more oxygenleft, that is either nH2

or nO2is equal to zero. However, at high temperature

the water molecules dissociate into various parts like H, O, HO, H2, and O2.In other words, reactions

2H2 + O2 → 2H2O2H2O → H2 + O2

(20)

occur simultaneously. As a result, in mixture the concentrations of all possibleconstituents are never exactly zero and tend to a certain equilibrium. So, whatis the mixture composition at that equilibrium?

Atom balance. One obvious condition is that the number of atoms of eachelement does not change. Since a mole of any substance contains the samenumber of atoms, we can count the number of atoms of each element on theright and left sides of (19). This gives:

2α = 2nH2O + 2nH2(21)

2β = nH2O + 2nO2(22)

Here, α and β may be considered as given design parameters. Let as denote

x = nH2(23)

Then from (21)

nH2O = α − x (24)

and from (22)

nO2= β − (α − x)/2 (25)

Therefore, atom balance allows to express all concentrations via only one un-known x. If, however, we would take into account other dissociation products onthe right-hand side of (19), as we certainly should do in real applications, thenwe would have even more unknowns yet, and some other means of determiningthe remaining unknowns are necessary.

Equilibrium condition. Remarkably, at equilibrium the partial pressuresare at a certain relation to each other, namely, if we have a reaction like

Lecture 11. Rocket propulsion II. Composition of combustion products. 41

n1A1 + n2A2 ↔ n3A3, where A1, A2, and A3 represent chemical substances(not necessarily elements) then

pn1(A1)pn2(A2)

pn3(A3)= Kp(T )

is, for mixtures of perfect gases, a function of temperature only. This can beproved by thermodynamic analysis, see the recommended text. The term Kp

is called the equilibrium constant for this reaction. For many reactions theequilibrium constants are tabulated or given in plots, as in Fig. 2-12 from therecommended text.

In particular, for our reaction (19), or, more exactly, for ’pure’ reactions (20)we have

p2(H2)p(O2)

p2(H2O)= K3(T ) (26)

In order to use it one should substitute partial pressures with their expres-sions via the mole fractions, Eq. (16) from Lecture 10. The total number of moles


in the right-hand side of (19) is nH2O +nH2+nO2

= α−x+x+β− (α−x)/2 =β + (α + x)/2, and (26) becomes

x2[β − (α − x)/2]

(α − x)2[β + (α + x)/2]=

K3(T )

p(27)

Now, within our simple model of a rocket engine we neglect velocity in thecombustion chamber. Therefore, T = T0 and p = p0.

From this equation one can find x if T0 is known. Usually, however, it is notknown. Instead, one has to solve a system of simultaneous equations: (23), (24),(25), (27), and Eq. (18) from Lecture 10, assuming Qtotal = 0, T = T2 = T0.

When more terms are taken into account in the right-hand side of (19), moreequations are needed. For example, if we would not neglect the atomic hydrogenH, that would add the unknown mole number nH . Then the reaction

H + H ↔ H2

would be considered and additional equation involving K5 in Fig. 2-12 wouldbe used.

An exercise.

Assume α = 2, β = 1, T1 = 300◦K, p0 = 100 atm, cp(H2) = 28 kJ/(kg ·mol ·◦C) (note the dimensions!), cp(O2) = 32 kJ/(kg ·mol ·◦ C), cp(H2O) = 44 kJ/(kg ·mol ·◦ C), ∆Ho

f (H2) = 0, ∆Hof (O2) = 0, ∆Ho

f (H2O) = −230000 kJ/(kg · mole).Plot combustion chamber temperature as a function of x using (27) and

Fig. 2.12 from recommended text (given above) together with combustion cham-ber temperature as a function of x (figure out how!) from (18) of the previouslecture (remember that Qtotal = 0). You should obtain the following picture:

2000

2500

3000

3500

4000

4500

5000

5500

6000

0 0.2 0.4 0.6 0.8 1

T(x) from (18) and Fig.2-12T(x) from (18)

Looking at this plot, tell what would be the combustion chamber temper-ature if there were no dissociation at all. What happened to the temperaturedue to the dissociation?

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