Sesa2005ln2006 Propulsion

School of Engineering Sciences. University of Southampton.

SESA2005 Propulsion lecture notes.

Lecturer: Prof. S. I. Chernyshenko

2006/2007

c 2006 University of Southampton Copyright

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 intuitive understanding of the basic relations between the flight regime, design parameters, and the engine performance. For example, at the end of the course the student should be able to figure out, intuitively, how the maximum (allowed by material properties, say) temperature in the engine affects the optimal compressor pressure ratio, or how the flight Mach number affects the propulsion efficiency etc. The variety of possible questions like that is endless. The answers to them cannot be memorised, however, once understanding is achieved one can answer most of these questions relatively easily. In the following table the upper row lists the flight regime and design parameters that determine the performance characteristics listed in the first column. If, for example, you understand intuitively how, say, β affects TSFC, tic the empty cell at the intersection of the β column and TSFC row. Once the table is filled you have completely achieved the first goal. M

T0b prc prf

β ηd ηc Ta ηf

ηf n rb QR ηt ηn ηb

F/m ˙a TSFC ηp ηth η0 The second goal is to teach the student how to calculate the engine performance, say, the fuel consumption rate, for given performance characteristics of the engine components (diffuser, compressor, burner, turbine etc.) and the flight regime. In other words, given the numerical values of the parameters in the upper row of the table and the air properties, you will be able to calculate the numerical values of the parameters in the left column. If you forget any of the necessary formulae, you will be able to derive them. In addition, you will have an understanding of how the engine characteristics affect the aircraft range. In the rocket propulsion part, you will learn the principles of calculating the rocket engine performance taking into account the chemical processes in the engine. However, the rocket part is relatively short and provides only the basic ideas and illustrations. These lecture notes consist of 11 rather terse lectures, which are difficult to understand without further explanations. More material will be given in the lectures, 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 former exam papers with solutions. The course materials are also available through Blackboard.

The present course gives only the first introduction to propusion. Further propulsion courses are available as an option for SES students. Recommended reading: P. Hill and C. Peterson, Mechanics and Thermodynamics of Propulsion, second edition. Addison-Wesley, 1995. The course lecturer, Prof. Sergei I. Chernyshenko, can be found in Room 5069 of the Tizard building, or contacted by e-mail at [email protected]. The main tool for obtaining feedback on your progress is the turbofan applet. At the time of writing it was located at http://www.soton.ac.uk/∼ge102/Jet.html Please visit it and learn to use it as early in the course as possible.

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SESA2005 Propulsion Lecture Notes 2006/7.

Turbine

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

Shaft

Compressor Air inlet

Hot high−pressure

Burner

gas

fuel

Basic gas generator

Air inlet

Exhaust nozzle

Turbine

Shaft

Compressor

Hot high−speed jet

Burner

fuel

Compressor

Shaft

Air inlet

Turbine

Cold moderate−speed jet

Turbine

Fan

Turbojet

Exhaust Hot nozzle moderate−speed jet

Burner

fuel

Turbofan Cold low−speed jet

Shaft

Propeller

Burner

fuel

Turboprop

Turbine

Compressor

Turbine

Gear

Air inlet

Hot Exhaust low−speed nozzle exhaust

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

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

7

8

SESA2005 Propulsion Lecture Notes 2006/7.

Lecture 2. Thrust equation. . m side

−F

(thrust reaction)

u

B

C pa u

u pe

. ma

pa

A

. ue , m e

D

Figure 1: Thrust-producing device with a single jet. By the second Newton law the rate of change of momentum of a body equals the force acting on the body: dmv =F dt It is convenient to consider force F as a momentum flux. Then this second law asserts that momentum is conserved: the change in momentum dmv equals the amount F dt of momentum added to the body. This way of treating forces is common in modern physics. For a fixed volume in space momentum can flow in or out of it not only due to the action of force but also due to mass moving in or out of it. A unit mass leaving the volume carries away an amount of momentum equal to its velocity. We will limit ourselves to momentum in the direction of the thrust force which, in turn, will be assumed to be parallel to the velocity of the incoming flow. Then, the amount of momentum entering the control volume ABCD in Fig. 1 through AB per unit time is pa AAB + m ˙ AB u where AAB is the area of the face AB of the control volume, m ˙ AB is the mass flow rate through it, and u is the flight velocity. The pressure pa and the velocity u are assumed constant, which is a good approximation if AB is far upstream from the engine. The side of the control volume (BC and AD) is parallel to the direction of the 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 as an approximation, that the velocity component in thrust direction is u. Then the amount of momentum leaving the control volume through its side per unit time is m ˙ side u, where m ˙ side 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 structural support of the engine shown in Fig. 1. The amount of momentum leaving the control volume through DC per unit time is pa (ADC − Ae ) + pe Ae + (m ˙ DC − m ˙ e )u + m ˙ e ue Here, the pressure and velocity on DC outside the jet are assumed to coincide with their values far upstream. This neglects the changes to the velocity profile due to the nacelle1 drag. Ae is the nozzle outlet area. In supersonic jet the outlet pressure pe may be not equal to pa . The flow is assumed to be steady. Therefore, the amount of momentum entering the control volume per unit time should be equal to the amount of momentum leaving it per unit time:

pa AAB + m ˙ AB u+F = pa (ADC −Ae )+pe Ae +(m ˙ DC − m ˙ e )u+ m ˙ e ue + m ˙ side u (1) Now, we introduce the mass flow rate into the engine intake m ˙ a and the fuel flow rate m ˙ f , so that m ˙e=m ˙ a +m ˙ f, m ˙ DC = m ˙ AB − m ˙ side + m ˙ f , and the fuel-air ratio f = m ˙ f /m ˙ a . Then after simple transformation (note that ADC = AAB ) (1) becomes F =m ˙ a [(1 + f )ue − u] + (pe − pa )Ae

(2)

Turbofan and turboprop engines have two exhaust streams, with different exhaust velocities, as sketched in Fig. 2. In this case the thrust equation has to be modified. While the amount of momentum entering the control volume per unit time through AB is the same as before, the amount of momentum leaving it through DC is now (m ˙ DC − m ˙ e −m ˙ c )u + m ˙ c uc + m ˙ e ue . Usually for turbofan and turboprop engines the exhaust pressure is approximately equal to the outside pressure, pe = pa . Assuming this, the thrust equation then becomes F =m ˙ a [(1 + f )ue − u] + m ˙ c (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. 1 nacelle

is a streamlined enclosure for an aircraft engine.

10

SESA2005 Propulsion Lecture Notes 2006/7. −F

(thrust reaction)

C

B

pa u u

. mc

pa

. ma

A

uc pe

. ue , m e

D

Figure 2: Thrust-producing device with two exhaust jets. In case of a rocket engine m ˙ a = 0, and fuel and propellant2 are effectively the same. The thrust then is F =m ˙ f ue + (pe − pa )Ae Exercise: derive this equation using the control volume approach similar as it was done above for air-breathing engines.

2 propellant

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 per unit time, or thrust power, to the rate of production of the propellant kinetic energy. For a single propellant stream (jet) ηp =

Fu m ˙ a [(1 + f )u2e /2 − u2 /2]

Usually f ≪ 1 and pe ≈ pa so that F ≈ m ˙ a (ue − u). Therefore (derive yourself!) ηp ≈

2u/ue 1 + u/ue

(4)

Maximizing ηp by taking u/ue close to 1 means that large m ˙ a is needed for creating finite thrust. This is approached in turboprop engines but is not practical for turbojets. Thermal efficiency ηth is the rate of addition of kinetic energy to the propellant divided by the energy consumption rate, m ˙ f QR , where QR is the heat of reaction of fuel ηth

m ˙ a [(1 + f )u2e /2 − u2 /2] = m ˙ f QR

(5)

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

Ps m ˙ f QR

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

Fpr u 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 energy consumption: ηo =

Fu m ˙ f QR

It is equal to ηp ηth or ηpr ηth , as appropriate. For turbojet (=single exhaust) from (4) ηo ≈ 2ηth

u/ue 1 + u/ue

(6)

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SESA2005 Propulsion Lecture Notes 2006/7.

The overall efficiency is less than unity because part of the energy of the fuel is lost as kinetic energy of the exhaust and another part as the thermal energy of the exhaust (however, we neglect here the kinetic energy of the fuel, see the end of Lecture 5). Decreasing ue toward u reduces losses of kinetic energy but can 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 using the thrust equation (F = m ˙ a [(1 + f )ue − u] + (pe − pa )Ae , see the corresponding lecture) and (5) gives 2ηth QR m ˙f (7) ue Therefore, for a given fuel consumption rate take-off thrust is inversely proportional to the exhaust velocity. In the engine fuel is mixed with air and then burns. The temperature of the products of combustion is, in rough approximation, proportional to the fuel-to-air ratio, and for optimized engines this peak temperature is, again roughly, equal to the highest temperature at which the turbine is not damaged yet. So, for optimized engine fuel flow rate is approximately proportional to the hot-air flow rate, which, in turn, is approximately proportional to engine size. Therefore, for given take-off thrust smaller ue means smaller engine size. Another important characteristic, specific thrust, is defined as F/m ˙ a . If certain engine design ensures higher F/m ˙ a then for the same thrust smaller m ˙a that is smaller engine size (or fewer engines per aircraft) is required. Ftakeoff =

Aircraft range: Breguet’s range formula. In level flight at constant speed F = D = L(D/L) = mg/(L/D), where D is drag, L is lift, m is aircraft mass, and g is gravity acceleration. Therefore, the thrust power is mgu L/D If m is the aircraft mass including the mass of the fuel then Fu =

dm dm =u = −m ˙f dt dl where l is the distance along the flight path. Using (6) and integrating gives the Breguet’s range formula L QR m1 ln (8) D g m2 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 mechanical loads on the aircraft structure which, hence, has to be heavier. l = ηo

Thrust specific fuel consumption, TSFC, is defined as TSFC = m ˙ f /F

Lecture 3. Engine performance.

13

20

0.5

0.45

η

L/D

15

0

0.4

0.35

10

0.3

0.25

η L/D 0

5

0.2

0.15

0 0.5

0.1 1

1.5

2

2.5

3

M

Figure 3: Typical variation of the overall efficiency, lift-to-drag ratio and their product as a function of Mach number for existing engine materials and the best design configurations. Using (6) reduces (8) to s=

m1 Lu 1 ln D g TSFC m2

Therefore, for a given flight speed the range of an aircraft is inversely proportional to TSFC. This formula is only approximate. Indeed, derivation of (8) assumed that L/D remains constant during the flight. This is not exactly true for 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 by BSFC= m ˙ f /Ps , where Ps is the shaft power, and, if the exhaust thrust is substantial, an equivalent brake specific fuel consumption is defined by EBSFC= m ˙ f /(Ps +Fe u), where Fe is the thrust produced by the engine exhaust. Typically, EBSFC=0.27-0.36 kg/(kWh).

14

SESA2005 Propulsion Lecture Notes 2006/7.

Lecture 4. Ramjet engine.

Diffuser

Combustion chamber (burner)

Nozzle

Fuel injection

Exhaust jet

a

d

b

e

Figure 4: The ramjet. To ram means to force in. In ramjet air is forced into the engine air intake by the sheer drive of the speed of flight. Ramjet, in principle, can work at subsonic 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 the burner, and then the combustion products expand through the nozzle. It is helpful to consider first a simplified model of an ideal ramjet. For ideal ramjet it is assumed that compression and expansion processes are reversible and adiabatic, that combustion occurs at constant pressure, that the air/combustion products properties (specific heat ratio γ and the gas constant R) are constant throughout the engine, and, although this is not necessary, that the outlet pressure is equal to the ambient pressure, in other words, that the nozzle is in the design regime. The usual tool for analysis of the processes in engines is the so-called enthalpy-entropy diagram. The thermodynamic state of air is determined by two independent parameters. 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 mass of air moves through the engine the properties are changing and the point that indicates the state is moving accordingly. The use of the enthalpy h and entropy s is especially convenient for the following reasons. In adiabatic reversible process s remains constant, and, therefore, the path of such a process is a vertical line in h − s diagram. Since irreversibility usually lead to deterioration of performance, engines are designed so as to be as close to reversible processes as possible. If the process is irreversible then entropy at the end of it is greater than entropy at the end of the corresponding reversible process. Therefore, in h − s diagram it is easy to anticipate the effect of irreversibility on the shape of the diagram. The advantage of using enthalpy as the other parameter follows from the form of the energy conservation law for open steady-state system:

Lecture 4. Ramjet engine.

15

h b kinetic energy added

heat added

kinetic energy of incoming air returned

kinetic energy of incoming air

st =con p = b =pd

d

p

e heat released by the jet when it cools to the ambient temperature

=const p=pa=p e a s

Figure 5: Enthalpy-entropy diagram for an ideal ramjet

˙ W u2 u2 Q˙ Q˙ = − h0out + h0in = − (hout + out ) + (hin + in ) m ˙ m ˙ m ˙ 2 2 ˙

˙

Q where W m ˙ and m ˙ are respectively the work done by and the heat added to 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 kinetic energy variation. For a perfect gas h = cp T , therefore, one can consider h axis as approximate temperature axis. Therefore, a restriction of the maximum temperature 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 velocity decreases. 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 an increase in s in accordance with the formula dQ = T ds 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 of air. In the nozzle no heat is added and no work is done, therefore, the total enthalpy 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. This is a very important feature. It is also true for real gases even though the curves may be not exactly exponential. Therefore, the exhaust velocity is greater than the flight velocity, and thrust is created. Naturally, exhaust gases do not go back into the air intake. However, they do cool down to the ambient temperature. This process occurs at constant

16

SESA2005 Propulsion Lecture Notes 2006/7.

pressure p = pa and is shown as curve e-a in the figure. The enthalpy difference he − ha is the amount of heat per unit mass released by the jet when it cools down. We can see that only part of the heat added in the combustion process turns into the useful kinetic energy. It is possible now to analyse the effects of various parameters on the engine performance. Decreasing the flight speed, that is decreasing the Mach number, decreases the 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 energy will become very small. At the same time the heat released by the jet will be almost equal to the heat added in the burner and will remain finite. That is burning a finite amount of fuel will produce almost no thrust, that is the thrust specific fuel consumption, TSFC, will become quite high. Also, since the added kinetic energy is small, specific thrust, that is thrust-to-air-mass-flow rate ratio will also be small. Then, for producing finite thrust the air mass flow rate has to be large, and, accordingly, the engine size has to be quite large. This is why ramjet 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 for the same thrust. The thermal efficiency will not be affected strongly as one can also see from the figure. However, an increase in hb means increase in the maximum temperature, and this is limited by the material properties of the engine walls. Suppose that the maximum temperature, and, hence, hb is fixed. Consider the variation of the specific thrust and TSFC with the Mach number M . At small M , point d is close to point a, and, as discussed, TSFC is high and specific thrust is small. An increase in M reduces TSFC. As for the specific thrust, one has to take into account that with hb fixed increase in hd results in a decrease in the amount of heat added. For small M this effect is, clearly (see the figure again) is less important and specific thrust increases with M . However, when hd approaches hb , the kinetic energy added, being only a fraction of the added heat, tends to zero. Accordingly, the specific thrust also tends to zero. Therefore, specific thrust has a maximum at some value of M between zero and that value of M at which isentropic compression in the diffuser leads to temperature attaining the maximum allowed value. It is possible to calculate all characteristics of an ideal ramjet. However, the use of h − s diagram makes it possible to achieve an intuitive understanding of the engine performance, as this was illustrated above. One can now, for example, 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 notes of your friend. The following basic relationships/ideas are assumed to be known from courses studied earlier. √ Speed of sound c = γRT , Mach number M = u/c. Enthalpy h = cp T for perfect gas. Entropy s = cp ln T − 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 is without heat addition) and reversible process. Stagnation temperature T0 = (1 + For s1 = s2 ,

γ−1 2 M )T 2

p1 /p2 = (T1 /T2 )γ/(γ−1)

(9)

(10)

Stagnation enthalpy h0 = h + u2 /2. Mass conservation law for an open system with a steady flow X X m ˙ out = m ˙ in outlets

inlets

where m-s ˙ are mass flow rates. Energy conservation law for an open system with a steady flow X X ˙ + W m ˙ out h0out = Q˙ + m ˙ in h0in outlets

(11)

inlets

˙ is the work per unit time done by the system (positive for turbine, where W negative for compressor) and Q˙ is heat added per unit time (positive for a burner). Notation convention. All symbols have their usual meaning. The subscripts denote: a is the incoming stream, apart from the incoming stream velocity u and Mach number M which have no subscripts. d is the diffuser outlet b is the burner outlet e is the nozzle outlet (exhaust). Calculating the ideal ramjet performance. At least the following parameters should be known before the calculation starts: 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

18

SESA2005 Propulsion Lecture Notes 2006/7.

1

0.3 Specific thrust , kN*s/kg, Tmax=2500K TSFC, kg/(kN*s), Tmax=2500 Specific thrust, kN*s/kg, Tmax=2000K TSFC, kg/(kN*s), Tmax=2000 0.25

0.8

0.6 0.15

TSFC

Specific thrust

0.2

0.4 0.1

0.2 0.05

0

0 1

2

3

4

5

6

7

8

M

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 increases above 3, the engine size starts to increase. maximum temperature in the engine), fuel heating value QR , specific heat ratio γ, 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 nozzle, 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 : 2 By (9): T0d = Ta (1 + γ−1 2 M ) By (11): (1 + f )h0b = h0d + f QR with h = cp T ⇒ 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) cp T0b = cp Te + u2e /2 ⇒ ue = . . . (derive yourself!). Thread 3, leading to specific thrust, TSFC, and efficiencies: √ u = M γRTa , then specific thrust F/m ˙ a = (1 + f )ue − u by the thrust equation, then T SF C = m ˙ f /F = f /(F/m ˙ a ), then the efficiencies as defined in Lecture 3.

Lecture 5. Ideal ramjet calculation.

19

1.2 Overall efficiency Propulsion efficiency Thermal efficiency 1

0.8

0.6

0.4

0.2

0 1

2

3

4

5

6

7

8

M

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 than Tmax allowed by the properties of the materials. The way to overcome this difficulty is not to bring the air velocity almost to zero in the diffuser, thus ensuring less compression. This idea leads to development of scramjet, that is supersonic combustion ramjet, in which the air speed remains supersonic throughout the engine. At the moment this idea has not been implemented in practice. In our definition of the efficiencies in Lecture 3 the kinetic energy of the fuel was ignored. However, the typical fuel has QR = 45000kJ/kg, and for M = 7 the kinetic energy of the fuel is u2 /2 ≈ 3000kJ/kg (check yourself!), that is about 6% of the chemical energy. This can result in the efficiencies being greater than 1 (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.

20

SESA2005 Propulsion Lecture Notes 2006/7.

Diffuser

Combustion chamber (burner)

Nozzle

Turbine

Compressor

Lecture 6. Ideal turbojet.

Fuel injection

Exhaust jet

a

d

c

b

t

e

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 M ramjet performance is poor because of insufficient compression. The idea of a turbojet is to remedy this by introducing a compressor between the diffuser and the burner as shown in Fig. 8. Since the compressor has to be driven, a turbine is added after the burner. An ideal turbojet is a turbojet in which all the processes are reversible, and the combustion is assumed to occur at constant pressure. The enthalpy-entropy diagram is shown in Fig. 9. In an ideal turbojet the work done by compressor is equal to the work done on the turbine, that is d − c is equal to b − t on the diagram. The compressor pressure ratio prc = pb /pd is an important design parameter. By comparing Fig. 9 with the h − s diagram of the ideal ramjet (Lecture 4) it can be seen that the distance a − c in Fig. 9 plays the role of the distance a − d in the ideal ramjet diagram. Therefore, the dependence of the specific thrust and TSFC of a turbojet on prc is similar to their dependence on M for a ramjet. (Do not give this argument during the exams! Just repeat the reasoning for the ramjet here.) We should expect that TSFC decrease with increasing prc , and there is prc such that the specific thrust is at its maximum there. Notice now that this optimum (from specific thrust viewpoint) prc corresponds to a certain distance a − 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 thrust is smaller. (As calculations show, at about M = 3 the optimum prc = 1 so that compressor is not needed at all, and turbojet becomes a ramjet.) Calculation of ideal turbojet engine is similar to calculation of an ideal ramjet. In addition to the parameters which should be prescribed for an ideal

Lecture 6. Ideal turbojet.

c p Tmax

21

h

b work done by a unit mass of air on the turbine t

heat added kinetic energy added rk wo

work done by compressor on unit mass of air

st con = pb pc= = p

kinetic energy of incoming air returned

c e

heat released by the jet when it cools to the ambient temperature

d

kinetic energy of incoming air

rejected heat, that is

const a

= p=pa=p e

s

Figure 9: The ideal turbojet h − s diagram ramjet, for turbojet the compressor pressure ratio should be given. Refer to Lecture 5 and check that you understand why each specific formula is used below. Thread 1, leading to fuel-to-air ratio f : 2 T0d = Ta (1 + γ−1 2 M ) p0d /pa = (T0d /Ta )γ/(γ−1) , then p0c = prc p0d T0c = (prc )(γ−1)/γ T0d (1 + f )h0b = h0c + f QR with h = cp T ⇒ f = (T0b − T0c )/(QR /cp − T0b ). Thread 2, leading to ue : p0b = p0c since combustion is assumed to occur at constant pressure Neglecting the difference in flow rates through compressor and turbine, compressor work = turbine work ⇒ h0c − h0d = h0b − h0t , or, with h = cp T , T0t = T0b − (T0c − T0d ) p0t = p0b (T0t /T0b )γ/(γ−1) Te = T0t (pe /p0t )(γ−1)/γ cp T0t = cp Te + u2e /2 ⇒ ue = . . . Thread 3, leading to specific thrust, TSFC, and efficiencies, is exactly the same as for the ramjet. Reheat. Note that fuel-to-air ratio f , being limited by the requirement on turbine inlet temperature, is well below the stoichiometric ratio. Therefore,

22

SESA2005 Propulsion Lecture Notes 2006/7.

n ur

diffuser compressor

b

turbine

er

nozzle

r ne r bu tf er a

s 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 temperature drops down. In order to increase the thrust, it is possible to add to the design another burner between the turbine and the nozzle, and reheat the air in this afterburner. The corresponding process is shown in Fig. 10. Since the afterburner isobar (constant pressure curve) is closer to the ambient pressure isobar, burning fuel in the afterburner is less efficient from the thermodynamics viewpoint (consult again Lecture 4). However, the temperature at the outlet of the afterburner may be much higher than the temperature at the turbine inlet, since the nozzle is less sensitive to it. Therefore, an increase in thrust can be substantial. For example, General Electric J85-21 turbojet, which powered the Northrop F5E, gives 22.2 kN thrust with afterburning but only 15.6 kN without 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 by engaging 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 :-).

Fan

Subscripts will correspond to Fig. 11.

f

Fan nozzle

Fan exhaust jet

d

a Fan

f

Turbine

Diffuser

Compressor

fn

c (burner) b

Fan nozzle

Nozzle

Exhaust jet

t Fan exhaust jet

e

fn

Figure 11: The turbojet. As it follows from Brequet’s range formula, an aircraft range is proportional to the overall efficiency, which in turn is proportional to the product of propulsion and thermal efficiencies (see Lecture 3). The exhaust jet, while creating thrust, carries away thermal and kinetic energy. Increasing the thermal efficiency reduces energy losses in the form of thermal energy in the exhaust. Overall efficiency can also be improved by reducing losses with kinetic energy of the jet, that is by increasing the propulsion efficiency. To create a certain 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, the amount of momentum ∆I = M U . This mass of air will, then, have a kinetic energy M U 2 /2, which is the extra cost we have to pay for getting ∆I. Since, M U 2 /2 = U ∆I/2 = (∆I)2 /(2M ), this extra cost decreases when U is decreased and M is correspondingly increased. In other words, it is better to use more propellant with less kinetic energy added to a unit mass of propellant. For a turbojet, neglecting f in the equations for thrust and for kinetic energy and assuming pe = pa gives 2u ηp = u + ue (Oops. Think better, but if you do not understand this even after that, go back and reread the previous lectures). So, ηp can be increased by reducing the specific 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

24

SESA2005 Propulsion Lecture Notes 2006/7. h b

cp T max

work done by a unit mass of air on the turbine

work to fan

heat added

or

w

work done by compressor on unit mass of air

s con = pb pc= = p

t

o kt

co

r sso e r p

m

t kinetic energy added kinetic energy of incoming air returned

c e

heat released by the jet when it cools to the ambient temperature

d

kinetic energy of incoming air

rejected heat, that is

=const a p=pa=p e s

Figure 12: The ideal turbofan h − s diagram. obtaining higher propulsion efficiency without increasing the size of the basic gas 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 to create negative thrust, thus eliminating the need for expensive and unreliable hydraulic braking systems in the wheels. Another advantage of turbofan engines is that, due to smaller exhaust velocity, they produce less noise. Thermodynamicly, the idea of turbofan is to take more power from the turbine then is needed to drive the compressor, thus reducing the kinetic energy of the exhaust, and use this additional power to drive the fan, thus increasing the total mass flow rate of the propellant, see Fig. 12. Calculation of ideal turbofan engine is similar to calculation of an ideal turbojet. In addition to the parameters which should be prescribed for an ideal turbojet, for turbofan the fan pressure ratio prf and the bypass ratio β=m ˙ f an /m ˙ a (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, see Lecture 6.

Lecture 7. Ideal turbofan.

25

Thread “Fan”, leading to the fan jet velocity ue f : p0d /pa = (T0d /Ta )γ/(γ−1) (I will not say “oops” anymore. You know what to do), then p0f = prf p0d T0f = (prf )(γ−1)/γ T0d we assume pef = pa , Tef = T0f (pef /p0f )(γ−1)/γ cp T0f = cp Tef + 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 = cp T , T0t = T0b − (T0c − T0d ) − β(T0f − T0d ) p0t = p0b (T0t /T0b )γ/(γ−1) Te = T0t (pe /p0t )(γ−1)/γ cp T0t = cp Te + u2e /2 ⇒ ue = . . . Thread 3, leading to specific thrust, TSFC, and efficiencies. The specific thrust now is F/m ˙ a = (1 + f )ue + βuef − (1 + β)u, and TSFC= f m ˙ a /F . The kinetic energy added to the propellant, calculated per unit mass of the core airflow, now is ((1 + f )u2e + βu2ef − (1 + β)u2 )/2. Therefore ηp = ηth

2u((1 + f )ue + βuef − (1 + β)u) ((1 + f )u2e + βu2ef − (1 + β)u2 )

((1 + f )u2e + βu2ef − (1 + β)u2 ) = 2f QR ηo = ηp ηth

Optimization of ideal turbofan, that is the choice of the optimum values of prf and β is not really possible. The power available for fan is limited, so, none of these two parameters can be increased indefinitely while the other parameter is held constant. However, it is possible to increase β and decrease prf at the same time so that the power ( = the position of the point t in Fig. 12) remains constant. 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.

26

SESA2005 Propulsion Lecture Notes 2006/7.

Lecture 8. The effect of losses. In real engines the processes are not reversible. However, the heat transfer from the air to the engine can usually be neglected. Accordingly, the h − s diagram changes so that, at the end of each process, the entropy is somewhat greater than it would be in an ideal case. Compare the diagrams in Fig. with the diagram for the ideal ramjet (Lecture 4) and observe how, in each case, the irreversibility leads to the decrease in the kinetic energy of the exhaust jet and the increase in its thermal energy. In real engine, due to combustion and temperature changes, the chemical composition of the gas varies. It can be approximately taken into account by using different values of γ and R for different components. Apart from this, gas is a two-parametric medium, that is its state is determined by two parameters, say, pressure and temperature. Fortunately, since the heat transfer between the engine and the air can be neglected, only one empirical constant is required in order to describe the deviation of the real performance from the ideal performance. For engine components in which the stagnation pressure does not change in an ideal case, that is in diffusers, burners, and nozzles, the performance may be characterized by stagnation pressure ratios rd = p0d /p0a , rb = p0b /p0d , and rn = p0e /p0t respectively. In general, the most commonly used parameters for describing component performance are the so-called adiabatic efficiencies 3 : for diffuser: ηd =

h0ds − ha < ideal enthalpy change needed to ensure the same p0d /pa > = < actual enthalpy change > h0d − ha

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

< work required in isentropic process to ensure the same p0c /p0d > h0cs − h0d = < actual work > 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 ηf n ): ηn =

h0t − he < actual kinetic energy of the jet > = < jet kinetic energy in isentropic expansion to the same pe > h0t − hes

Also, the burner efficiency ηb is just the fraction of the chemical energy of the fuel that is released in the burner. 3 Typically

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

h b kinetic energy of the exhaust jet heat added

d

e

kinetic energy of incoming air

heat released by the jet when it cools to the ambient temperature

a

s

Ramjet h-s diagram for a real diffuser h b kinetic energy of the exhaust jet heat added

d

e

kinetic energy of incoming air

heat released by the jet when it cools to the ambient temperature

a

s

Ramjet h-s diagram for a real burner h b kinetic energy of the exhaust jet heat added

d

e

kinetic energy of incoming air

heat released by the jet when it cools to the ambient temperature

a

s

Ramjet h-s diagram for a real nozzle Figure 13: Ramjet h-s diagrams

28

SESA2005 Propulsion Lecture Notes 2006/7.

Now, we can get the full set of formulae for calculating a real turbofan engine. The following quantities should be given: M , T0b , prc , prf , β, γ, R, ηd , γc , ηc , Ta , γf , ηf , ηf n , Rf , rb , QR , cp , γt , ηt , γn , ηn , Rn , ηb . The core exhaust pressure is assumed to be equal to pa , which implies a subsonic or design supersonic nozzle.   γ−1 2 Diffuser: T0d = 1 + M Ta , T0ds = Ta + ηd (T0d − Ta ), 2  γ/(γ−1)  γ/(γ−1)  T0ds p0d T0d p0d = or = 1 + ηd −1 pa Ta pa Ta p0c p0d Compressor: = prc pa pa T0cs (γc −1)/γc = prc , T0d  or T0c = 1 +  1 Fan, similarly: T0f = 1 + ηf

Turbine:

p0c p0b = rb , pa pa

1 (T0cs − T0d ), ηc  1  (γc −1)/γc p −1 T0d ηc rc   p0d p0f (γf −1)/γf prf − 1 T0d , = prf pa pa T0b − T0c f= (ηb QR /cp ) − T0b T0c = T0d +

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

 γ /(γ −1) 1 T0ts t t p0t T0ts = T0b − (T0b − T0t ), = ηt p0b T0b   γt /(γt −1)) p0b 1 T0t p0t = 1− 1− or pa pa ηt T0b Core engine nozzle and fan nozzle. Thrust: (1−γf )/γf !  γf p0f cpf = , Rf , h0f − hef s = cpf T0f 1 − γf − 1 pa v " u (1−γf )/γf #  q u γ p f 0f Rf T0f 1 − . uef = 2ηf n (h0f − hef s ) or uef = t2ηf n γf − 1 pa v " u  (1−γn )/γn # u p0t γ n . Rn T0t 1 − Similarly ue = t2ηn γn − 1 pa p u = M γRTa , F/m ˙ a = ((1 + f )ue + βuef − (1 + β)u), T SF C = f m ˙ a /F ηp = 2u((1 + f )ue + βuef − (1 + β)u)/((1 + f )u2e + βu2ef − (1 + β)u2 ) ηth = ((1 + f )u2e + βu2ef − (1 + β)u2 )/(2f QR ),

ηo = ηp ηth

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 can be used for calculating turboprop and turboshaft engines with minimal corrections. For turbofan, the only point were the existence of the fan affects the core 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 m ˙ a βcp (T0f − T0d ). In turboshaft engine, this should be replaced with the shaft power, and the energy equation can then be used to calculate this shaft power if T0t is known. Now, using the formulae from Lecture 8, T0t can be found by prescribing the pressure immediately after turbine p0t to be just slightly above the ambient pressure, since no thrust is expected to be produced from the exhaust jet. In a turboprop engine the hot exhaust jet can produce a significant part of the total thrust. For a given gas generator and flight speed there is an optimum hot-gas exhaust velocity that gives maximum thrust. Let us rewrite the same energy equation in the form P/m ˙ a = h0b − (h0c − h0d ) − h0t where P is the power passed through to the propeller (for turboshaft engine P is the shaft power, by the way). Given the gear ηg and the propeller ηpr efficiencies, the propeller thrust then is Fpr = ηpr ηg P/u. Then the expression for the specific thrust is F/m ˙ a = (1 + f )ue − u + Fpr /m ˙a 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 indeed has a maximum inside that range.

30

SESA2005 Propulsion Lecture Notes 2006/7.

Lecture 9. Typical parameters of real engines. The following is just a set of citations and figures from the correspondent chapter of the recommended textbook, (P. Hill and C. Peterson, Mechanics and Thermodynamics of Propulsion, second edition. Addison-Wesley, 1995.) We now consider examples of turbojet thrust and fuel consumption variation with compressor pressure ratio, turbine inlet temperature, and flight Mach number. For sample calculations we make the assumptions about component efficiency, fluid properties, and engine conditions shown in Table 5.1. Each flight Mach number is assumed to correspond to a different altitude and thus a different pair of ambient pressure and temperature values pa and Ta . In these calculations we assume, as before, that the product mc ˙ p is the same in the turbine as in the compressor. TABLE 5.1 Turbojet calculation parameters (Fuel heating value 45,000 k J/kg) Component Adiabatic efficiency Average specific heat ratio Diffuser ηd = 0.97 1.40 Compressor ηc = 0.85 1.37 Burner ηb = 1.00 1.35 Turbine ηt = 0.90 1.33 Nozzle ηn = 0.98 1.36 Flight altitude Ambient pressure Ambient temperature (cruise Mach no.) (kPa) (K) Sea level (0) 101.30 288.2 40,000 ft (12,200 m) (0.85) 18.75 216.7 60,000 ft (18,300 m) (2.0) 7.170 216.7 80,000 ft (24,400 m) (3.0) 2.097 216.7 Figures 5.19 to 5.22, inclusive, show the thrust and specific fuel consumption calculated under these conditions for three values of turbine inlet temperature and four values of flight Mach number. Both of these parameters, as well as the compressor ratio, strongly influence the engine performance. We see in particular from these graphs:

Lecture 9. Typical parameters of real engines.

31

1. At a given flight Mach number and a given turbine inlet temperature, the pressure ratio that maximizes specific thrust does not provide minimum fuel consumption. Since the mass of the engine depends strongly on airflow rate and substantially affects aircraft carrying capacity, both specific thrust and specific fuel consumption have to be considered in selecting the best compressor pressure ratio. The choice of pressure ratio for best cruising range will require a compromise that takes into account both engine and aircraft characteristics. 2. Raising turbine inlet temperature can substantially improve specific thrust. The maximum temperature shown, 1700 K, is far below the maximum stoichiometric combustion of hydrocarbon fuels but requires not only hightemperature alloys for the turbine blades but also quite intensive blade cooling as well. 3. With a given compressor pressure ratio, raising turbine inlet temperature may or may not raise fuel consumption per unit thrust. For pressure ratios associated with minimum fuel consumption, increasing turbine inlet temperature can reduce specific fuel consumption somewhat. (It may, though, require an increase in the rate of cooling air extracted from the compressor, and the present calculations have not taken this into account.) 4. The compressor pressure ratio required to minimize specific fuel consumption is much less for supersonic than for subsonic flight. At Mach 3, peak specific thrust occurs with a compressor pressure ratio of 1; in this case no 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 to engine component efficiencies. The designer has great incentive to make the compression, combustion, and expansion processes as efficient as possible. In Chapters 6-9 we discuss the physical factors that explain engine performance limitations. Since these factors strongly influence the configuration of actual machines, it is appropriate to point them out here before making component analyses. Basically they are of two kinds: material limits, as expressed

32

SESA2005 Propulsion Lecture Notes 2006/7.

by allowable stress and temperature levels, and aerodynamic limits, imposed mainly by the behavior of boundary layers in the presence of rising pressure. As it does in all flight applications, the desire for minimum engine mass leads to highly stressed engine components as well as high peak temperatures. The largest stresses on turbine blades are due to centrifugal force, and the allowable stress in these blades is directly related to the temperature at which they must operate. Thus compromise is needed between the turbine blade stress and combustion temperature. With currently available material, the maximum stagnation temperature of the turbine inlet gas is limited to about 1200 K for uncooled blades. If the blades are cooled internally by passing air through small interior passages, it is possible to use stagnation temperatures of 1700 K or more. Operation at such high temperatures, however, may require a substantial fraction of the compressor air flow (as much as 10% or more) to be diverted to turbine 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 pressure ratio that can be accommodated in a single state of a turbomachine is limited. (A stage may be defined as a single circumferential row of rotating blades and an associated set of stationary guide vanes, stators, or nozzles.) This limitation is most severe in compressors where the fluid necessarily flows against a rising pressure gradient (see Chapter 4). Primarily for this reason, axial-compressor-stage pressure ratios are much lower than turbine-stage pressure ratios. Hence axial compressors of 10 to 20 stages can be driven by turbines of 2 or 3 stages.

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

Lecture 9. Typical parameters of real engines.

33

assumptions shown in Table 5.2 to obtain the results shown in Figs. 5.29 to 5.34. 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 unit core engine flow rate m ˙ a has nearly doubled, as the bypass ratio has changed from 0 to 5. At the same time the thrust specific fuel consumption has decreased substantially. Comparison of Figs. 5.20 and 5.30 shows the same kinds of benefits for high subsonic (M = 0.85) cruise thrust, though the relative gains are considerably less. For supersonic flows the bypass idea would have little potential benefit, and the 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 cruise conditions. In contrast, Fig. 5.32 shows the corresponding efficiencies for the turbofan engine with β = 5. The overall efficiency is significantly greater than for the turbojet (β = 0). TABLE 5.2 Component Diffuser Fan Fan nozzle Fan pressure ratio

Bypass fan characteristics Efficiency Specific heat ratio ηd = 0.97 γd = 1.4 ηf = 0.85 γf = 1.4 ηf n = 0.97 γn = 1.4 pf r = 1.50

Figures 5.33 and 5.34 show large values of the overall efficiency potentially available with supersonic flight. The compressor pressure ratio that maximizes η0 drops rapidly as the design flight Mach number increases above, say, 2. For an 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 a bypass rather than a turbojet engine. The higher the turbine inlet temperature (and thus the higher the exhaust velocity of the corresponding turbojet), the greater the benefit. The questions of optimum bypass ratio and optimum fan pressure ratio deserve serious exploration and may lead to further refinements of turbofan designs. In the absence of geared power transmission to the fan, the problem of speed mismatch between the fan and its driving turbine tends to limit the bypass ratio of turbofan engines. Other considerations affecting the choice of bypass ratio include the structural weight and the aerodynamic drag of the engine nacelle. Designing a gearbox to transmit 30,000 kW would mean confronting serious questions concerning gearbox weight and reliability. Nevertheless, the geared fan is a future possibility.

34

SESA2005 Propulsion Lecture Notes 2006/7.

Lecture 10. Rocket propulsion I

35

Lecture 10. Rocket propulsion I The general overview of the rocket propulsion, including the types of rocket engines and their performance characteristics, is given in the parallel (Astronautics Part 2) course. As it follows from the Tsiolkovsky (or rocket) equation derived in that course, the major performance characteristic of a rocket engine is the specific impulse, defined as Isp = F/(mg) ˙ where F is the thrust, m ˙ is the propellant flow rate, and g is the gravity acceleration at the earth’s surface. The presence of g in the definition is due to historical tradition only. The higher the specific impulse, the greater the fraction 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, and will consider only one very simple example of a rocket calculation. The expression for the thrust of the rocket engine can be obtained from the general equation of the thrust for a turbojet engine (see Eq. (2 in Lecture 2) F =m ˙ a [(1 + f )ue − u] + (pe − pa )Ae in which, first, using f = m ˙ f /m ˙ a , we eliminate f , then put m ˙ a = 0, and m ˙ f = m, ˙ because the rocket engine does not swallow any air, and the only propellant it uses is its own fuel and oxidizer. Therefore, one has F = mu ˙ e + (pe − pa )Ae

(12)

Here, pe is the pressure in the exhaust jet, pa is the ambient pressure, ue is the exhaust velocity, and Ae is the cross-section area of the exhaust jet at the nozzle outlet. As it is known from nozzle analysis (see parallel AA208 (Aerothermodynamics) course, current lecturer Dr. G. N. Coleman), the maximum thrust corresponds to pe = pa . However, when a rocket ascends to space, pa changes from atmospheric pressure at the launch pad level to zero in the orbit, and this has 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 called a thrust chamber. In case of a rocket with a liquid propellant the propellant is pumped into the combustion chamber. Solid propellant rockets have their propellant stored in the combustion chamber itself. Nozzle calculation. In fact, nozzle calculation is the same as for airbreathing engines. However, for the analysis of a rocket it is useful to derive a specific form of the expression for the exhaust velocity. For simplicity, an ideal nozzle is considered, that is we assume that inside the nozzle the exhaust gas can be approximated by a perfect gas of constant composition and with constant

36

SESA2005 Propulsion Lecture Notes 2006/7.

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 gives u2e /2 = h0 − he = cp (T0 − Te ) Then, using formulae for isentropic processes (see Lecture 5), after simple transformations one obtains v " u  (γ−1)/γ # u pe ue = t2cp T0 1 − p0 ¯ and the propellant Expressing cp in terms of γ, the universal gas constant R, ¯ molecular weight M gives (compare with the formula for ue in Lecture 8) v " u  (γ−1)/γ # u 2γ R ¯ pe T0 1 − (13) ue = t ¯ p0 (γ − 1)M

The flow rate through the nozzle can be expressed via p0 , T0 , and the nozzle throat area A∗ as the product of density at the nozzle throat, A∗ , and the velocity at the nozzle throat (again see AA208 course). The result is s (γ+1)/(γ−1)  A∗ p0 2 (14) m ˙ =√ γ γ+1 RT0 ¯ M ¯ . Using (12) one now obtains where R = R/ v " u  (γ−1)/γ #   u 2γ 2  2 (γ+1)/(γ−1) F pe pe pa Ae t + 1− = − A∗ p0 γ−1 γ+1 p0 p0 p0 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 depends on the geometry only, the designer is left with p0 and the geometry as the design parameters. It seems attractive to have high p0 since that would mean that the same 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-rocketand-mission level. In what follows we will assume that p0 is prescribed. The amount of propellant needed for a specific mission depends mostly on ue (recollect again the Tsiolkovsky equation). From (13) it follows that ue can ¯ . This implies an imbe increased by increasing T0 and/or by decreasing M portant trade-off between the desire to have the fuel-to-oxidizer ratio closer to stoichiometric for increasing T0 and having the fuel-to-oxidizer ratio giving the combustion products of smaller molecular weight. In case of hydrogen-oxygen

Lecture 10. Rocket propulsion I

37

combination, for example, the fuel-reach mixture is more advantageous. Calculating T0 and the combustion product composition requires certain background. Mixtures of gases. The properties of the mixture may be determined from the properties of the constituents by using the Gibbs-Dalton Law. For a mixture of n constituents: Temperature T = T1 = T2 = · · · = Tn Pressure p = p1 + p2 + · · · + pn Density ρ = ρ1 + ρ2 + · · · + ρn Internal energy per unit mass e = c1 e1 + c2 e2 + · · · + cn en Entropy per unit mass s = c1 s1 + c2 s2 + · · · + cn sn Enthalpy per unit mass h = c1 h1 + c2 h2 + · · · + cn hn where ρi is the density of the i-th constituent, that is the mass of that constituent in a unit volume of the mixture, ci = ρi /ρ is the mass concentration of the constituent, and pi is the partial pressure. The partial pressure is related to the density and temperature by the same formula as for pure substance: ¯ R pi = ρi Ri Ti = ρi Ri T = ρi M ¯i T ¯ is the universal gas constant and M ¯ i is the molecular weight of the Here, R i-th constituent. A mole of a substance is the amount of substance the mass of which is the molecular weight times the mass unit. For example, the molecular weight of water (H2 O) is 2+16=18. Therefore, in the system of units where mass is measured in grams, one mole of water is 18 grams of water. If the mass unit is, say, a pound, then a mole of water is 18 pounds of water, etc. To avoid confusion, notations like gm-mole and kg-mole are often used. It is quite easy to obtain that pi ni = p n1 + n2 + · · · + nn

(16)

c2 cn c1 1 + + · · · + = ¯ ¯1 ¯2 M M M M¯n

(17)

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

For fixed chemical composition cv and cp of the mixture can be found using the above formulae and the fact that cv = de/dT , cp = dh/dT . Combustion product temperature. The most important feature of a mole is that a mole of any substance contains the same number (Avogadro number) of molecules. Because of that, chemical reaction written down as, say, 2H2 + O2 → 2H2 O can be interpreted as “two moles of H2 react with one mole of O2 to produce two moles of H2 O”, and, hence, H2 O here is interpreted as a mole of H2 O. More generally, one can have, say,

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SESA2005 Propulsion Lecture Notes 2006/7.

αH2 + βO2 → nH2 O H2 O + nH2 H2 + nO2 O2 This can happen if hydrogen and oxygen are not in stoichiometric ratio unlike the previous formula, or due to the dissociation of the combustion products. In the latter case the right hand side can also include H, O, and HO, but for simplicity 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, nH2 O , nH2 , nO2 are known, and also that the reaction occurs at constant pressure. If the initial temperature T1 of the reactants is given, what is the temperature of the combustion products? By definition, the enthalpy of formation of a substance is the amount of heat needed to be added at a constant temperature and pressure for this substance to be formed in a reaction from its components as they occur naturally at this temperature. Since this amount depends on the temperature, its value is important, and this temperature is called reference temperature. Enthalpy of formation can be measured experimentally and is available from the tables for many substances. If the formation reaction releases heat, enthalpy of formation is negative. Usually, enthalpy of formation is tabulated per mole of the substance. Now, we can take α moles of H2 and β moles of O2 at the initial temperature T1 and bring them at constant pressure to the reference temperature Tf . For this we will have to add the amount of heat equal to ¯ H + cp (O2 )(Tf − T1 )β M ¯O . cp (H2 )(Tf − T1 )αM 2 2 ¯ cp is used in such calculations, Quite often the specific heat per mole c¯p = M and then the amount of heat required can be rewritten as (α¯ cp (H2 ) + β¯ cp (O2 ))(Tf − T1 ). Then we can decompose these reactants into their natural components, and for that we will have to subtract the amount of heat equal to their formation ethalpies at Tf times number of moles: α∆Hfo (H2 ) + β∆Hfo (O2 ). (For usual values of Tf formation enthalpies of hydrogen and oxygen will be zero because they will naturally occur as they are :-), but for other fuels and oxidizers this step can be necessary.) Then we can compose the products, and for this we will have to add the amount of heat equal to nH2 O ∆Hfo (H2 O) + nH2 ∆Hfo (H2 ) + nO2 ∆Hfo (O2 ) Finally, we will bring them to the final temperature T2 , and for that we should add the heat in the amount equal to (nH2 O c¯p (H2 O) + nH2 c¯p (H2 ) + nO2 c¯p (O2 ))(T2 − Tf ) Therefore, in total, in order to change the temperature of the mixture from T1 to T2 and its chemical composition according to the reaction, we have to add the amount of heat equal to

Lecture 10. Rocket propulsion I

Qtotal = (α¯ cp (H2 ) + β¯ cp (O2 ))(Tf − T1 ) − (α∆Hfo (H2 ) + β∆Hfo (O2 ))+ o nH2 O ∆Hf (H2 O) + nH2 ∆Hfo (H2 ) + nO2 ∆Hfo (O2 )+ (nH2 O c¯p (H2 O) + nH2 c¯p (H2 ) + nO2 c¯p (O2 ))(T2 − Tf )

39

(18) If this transformation occurs in the combustion chamber adiabaticly, then Qtotal = 0 and (18) can be considered as an equation for determining T2 , that is the temperature of the combustion products. Note, that these formulae imply that h = cp T , that is that cp is constant. At high temperatures characteristic for rocket engines this assumption is not valid and, hence, in the above formulae cp denotes some averaged values. In more accurate calculations h = cp T has to be replaced with h = h(T ), with h(T ) given by more complicated formulae or tables. Note that since we neglect velocity in the combustion chamber, T2 in the above formulae is the stagnation temperature T02 , and that in formulae for the nozzle we omitted this second subscript, so that in fact T2 = T0 .

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SESA2005 Propulsion Lecture Notes 2006/7.

Lecture 11. Rocket propulsion II. Composition of combustion products. Consider again the reaction αH2 + βO2 → nH2 O H2 O + nH2 H2 + nO2 O2

(19)

How can one find the composition of the combustion products, that is nH2 O , nH2 , and nO2 ? Intuitively it may seem that hydrogen and oxygen will always react, and that the reaction stops only when there is no more hydrogen or no more oxygen left, that is either nH2 or nO2 is 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 → 2H2 O 2H2 O → H2 + O2

(20)

occur simultaneously. As a result, in mixture the concentrations of all possible constituents are never exactly zero and tend to a certain equilibrium. So, what is the mixture composition at that equilibrium? Atom balance. One obvious condition is that the number of atoms of each element does not change. Since a mole of any substance contains the same number of atoms, we can count the number of atoms of each element on the right and left sides of (19). This gives: 2α = 2nH2 O + 2nH2

(21)

2β = nH2 O + 2nO2

(22)

Here, α and β may be considered as given design parameters. Let as denote x = nH2

(23)

nH2 O = α − x

(24)

nO2 = β − (α − x)/2

(25)

Then from (21) and from (22) Therefore, atom balance allows to express all concentrations via only one unknown x. If, however, we would take into account other dissociation products on the right-hand side of (19), as we certainly should do in real applications, then we would have even more unknowns yet, and some other means of determining the remaining unknowns are necessary. Equilibrium condition. Remarkably, at equilibrium the partial pressures are at a certain relation to each other, namely, if we have a reaction like

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

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n1 A1 + n2 A2 ↔ n3 A3 , where A1 , A2 , and A3 represent chemical substances (not necessarily elements) then pn1 (A1 )pn2 (A2 ) = Kp (T ) pn3 (A3 ) is, for mixtures of perfect gases, a function of temperature only. This can be proved by thermodynamic analysis, see the recommended text. The term Kp is called the equilibrium constant for this reaction. For many reactions the equilibrium constants are tabulated or given in plots, as in Fig. 2-12 from the recommended text.

In particular, for our reaction (19), or, more exactly, for ’pure’ reactions (20) we have p2 (H2 )p(O2 ) = K3 (T ) (26) p2 (H2 O) In order to use it one should substitute partial pressures with their expressions via the mole fractions, Eq. (16) from Lecture 10. The total number of moles

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SESA2005 Propulsion Lecture Notes 2006/7.

in the right-hand side of (19) is nH2 O + nH2 + nO2 = α − x + x + β − (α − x)/2 = β + (α + x)/2, and (26) becomes x2 [β − (α − x)/2] K3 (T ) = 2 (α − x) [β + (α + x)/2] p

(27)

Now, within our simple model of a rocket engine we neglect velocity in the combustion chamber. Therefore, T = T0 and p = p0 . From this equation one can find x if T0 is known. Usually, however, it is not known. 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), more equations are needed. For example, if we would not neglect the atomic hydrogen H, 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 would be used. An exercise. Assume α = 2, β = 1, T1 = 300◦ K, p0 = 100 atm, c¯p (H2 ) = 28 kJ/(kg · mol ·◦ C) (note the dimensions!), c¯p (O2 ) = 32 kJ/(kg · mol ·◦ C), c¯p (H2 O) = 44 kJ/(kg · mol ·◦ C), ∆Hfo (H2 ) = 0, ∆Hfo (O2 ) = 0, ∆Hfo (H2 O) = −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 chamber temperature as a function of x (figure out how!) from (18) of the previous lecture (remember that Qtotal = 0). You should obtain the following picture:

6000 T(x) from (18) and Fig.2-12 T(x) from (18)

5500 5000 4500 4000 3500 3000 2500 2000 0

0.2

0.4

0.6

0.8

1

Looking at this plot, tell what would be the combustion chamber temperature if there were no dissociation at all. What happened to the temperature due to the dissociation?