40 Fundamentals of Propulsion Systems

Introduction

In the equations of motion developed in the previous chapter, thrust was introduced as the primary propulsive force. This chapter now explains how that thrust is generated. All flight vehicles require a propulsion system to sustain flight, the principal exception being a glider or sailplane. For an aircraft, propulsion typically involves an air-breathing engine comprising a mechanical power source and a propulsor that performs work on the air passing through it, thereby increasing its momentum and converting the engine’s power into thrust. This increase in momentum may be produced either by a jet efflux in the form of a jet velocity or by the action of a propeller on the air to give an increase in its slipstream velocity, as shown in Figure 1.

An air-breathing engine must produce enough thrust to overcome the aircraft’s aerodynamic drag and propel it forward, which is achieved by increasing the momentum of the airflow passing through the propulsion system.

In a rocket engine, which is a non-air-breathing propulsion system, thrust is generated by expelling high-speed gases produced by the combustion of a fuel and an oxidizer through a nozzle, as shown in Figure 2. For a launch vehicle, this thrust must be sufficient to initially overcome the vehicle’s weight and ultimately accelerate it to an orbital velocity. Rocket engines are also used to change satellite orbits and to provide spacecraft with the velocity needed to escape Earth’s gravitational field for deep-space missions. In all cases, thrust production is governed by Newton’s laws and the conservation of mass, momentum, and energy.

A rocket engine needs sufficient excess propulsive thrust to overcome the weight of a launch vehicle and accelerate the payload into space.

Learning Objectives

  • Be able to distinguish between the basic engine types used to power flight vehicles.
  • Learn about the fundamental differences between air-breathing engines, such as piston engines, turbojets, turbofans, and turboprops.
  • Understand the basic physical principles of thrust production from air-breathing and rocket engines.
  • Know how to determine the efficiency of a propulsive system in terms of specific fuel consumption and specific impulse.

Types of Propulsion Systems

Not all flight vehicles are created equally, and they require different propulsion systems depending on their flight conditions and mission requirements. For an aircraft in steady flight, the propulsion system must produce sufficient thrust to balance the aerodynamic drag and thereby sustain the required airspeed; this drag force is typically an order of magnitude less than the aircraft’s weight. To accelerate or climb, however, thrust must exceed drag, so the propulsion system must provide excess thrust (and power), particularly during takeoff and climb. Propeller-driven aircraft are fundamentally power-limited, whereas jet-powered aircraft are fundamentally thrust-limited. For spacecraft, such as launch vehicles, the rocket engine(s) must initially overcome the vehicle’s entire weight, requiring a thrust-to-weight ratio greater than unity. As fuel is consumed and the vehicle mass decreases, the thrust-to-weight ratio increases, allowing continued acceleration to an orbital velocity.

The needed propulsion on a flight vehicle can be achieved using at least one of the following systems, as shown in Figure 3:

  • A propeller and engine combination, such as a reciprocating (piston) engine.
  • A turbojet, which is a basic jet engine that produces pure jet thrust.
  • A turbofan, which is a jet engine with a bypass fan that directs a significant mass flow rate around the engine’s core.
  • A turboshaft, in which all power goes to a reduction gearbox and transmission system, such as the one used on helicopters.
  • A turboprop, which is a turboshaft engine driving a propeller from a power turbine with little jet thrust.
  • A rocket engine, which, unlike the preceding types, is not air-breathing.

Examples of different propulsion systems used for flight vehicles.

Each propulsion system differs in its functional design, but in all cases, the objective is to convert stored energy into propulsive power and thrust. In combustion-based systems, the stored energy is released by burning fuel; in battery-electric systems, the stored chemical energy is converted electrochemically into electrical power, which then drives a motor and propulsor. The performance of any flight vehicle depends strongly on the thrust produced, the power required, and the rate at which fuel or stored energy must be consumed. As shown in Figure 4, the appropriate propulsion system for a given flight vehicle depends primarily on its intended flight speed, typically expressed as a Mach number.

Propeller-turboshaft combinations, or turboprops, typically power low-speed transport aircraft, whereas smaller general aviation airplanes often employ piston engine–propeller systems. In the limiting case, as the airspeed and Mach number approach zero, corresponding to hover and vertical takeoff and landing (VTOL) flight, thrust must be generated without any forward motion of the vehicle. Under these conditions, the propulsive efficiency becomes very low, and the required power increases substantially because it must be supplied entirely through the vertical acceleration of the airflow to produce lift. Helicopters are the most efficient VTOL aircraft and are generally powered by turboshaft engines, although smaller designs may use piston engines.

At higher subsonic speeds, turbofan engines are generally preferred, especially for airliners, which operate predominantly under cruise conditions at nearly constant airspeed and thrust. Their high propulsive efficiency and relatively low fuel consumption make them well-suited to this role. At transonic and supersonic Mach numbers, however, turbojets and low-bypass turbofans become more suitable. Military aircraft, in particular, require a substantial excess of thrust to achieve rapid acceleration and overcome the much higher drag encountered in these flight regimes. Accordingly, they are designed with large thrust margins and often employ afterburners (reheat) to provide additional thrust for short periods. In contrast, rocket engines operate independently of the atmosphere and can produce very high thrust at high pressures and temperatures, although usually only for relatively short durations. Beyond about Mach 4, the use of conventional air-breathing engines becomes increasingly difficult.

Piston Engines

A reciprocating internal-combustion engine driving a propeller is commonly used to power low- to moderate-performance airplanes. The propeller is connected directly to the engine’s crankshaft, converting shaft power into thrust, as illustrated in Figure 5. The engine produces shaft power, which the rotating propeller converts into a change in the airflow’s momentum, thereby generating thrust. Lower-performance airplanes generally use fixed-pitch propellers, whereas higher-performance designs employ variable-pitch (constant-speed) propellers to maintain their efficient operation over a broad range of flight conditions. Such aeroengines are fundamentally different from automotive engines in that they are air-cooled and rotate at relatively low shaft speeds, keeping the propeller tips well below the speed of sound. This type of propulsion system for an airplane is mechanically simple, relatively inexpensive, and capable of good propulsive efficiency at low flight speeds where propellers operate most effectively.

A piston engine driving a propeller is a relatively simple and robust design and is widely used on general aviation aircraft.

However, several limitations arise with such piston aeroengines. The power-to-weight ratio declines rapidly as power requirements increase, making them less suitable for larger or higher-performance aircraft. In addition, propeller performance degrades at higher flight speeds because of compressibility effects and increasing helical-tip Mach numbers, thereby reducing propulsive efficiency and increasing noise. At higher altitudes, the reduced air density further limits both engine power and propeller thrust. For these reasons, piston engine-propeller combinations are generally confined to lower-speed, lower-altitude applications. Supercharging or turbocharging may be used to increase the mass flow rate of air through the engine, thereby increasing power and/or maintaining power output as the air density decreases with altitude. When higher airspeeds and improved high-altitude performance are required, a turboprop configuration will usually be preferred.

Turbojet, Turbofans & Turboshaft Engines

The turbojet, turbofan, turboprop, and turboshaft are closely related gas-turbine engine types, but they differ in how the energy extracted from the flow is used to produce thrust or shaft power. In a turbojet engine, as shown in Figure 6, essentially all of the thrust is produced by expelling high-speed exhaust gases through a nozzle at the rear of the engine. The compressor raises the pressure and temperature of the incoming air to support combustion, and the turbine extracts just enough energy from the flow to drive the compressor. The remaining energy appears as a high exhaust jet velocity, so thrust is produced primarily by accelerating the mass flow through the engine to a high speed as it exits the nozzle, often reaching supersonic conditions under appropriate operating states.

In a turbojet engine, thrust is produced by expelling hot gases at high “jet” velocity through a nozzle.

Turbofan

In a turbofan engine, one or more large fans at the front increase the total mass flow through the engine, as shown in Figure 7. A significant fraction of this flow bypasses the core, producing thrust at a lower exhaust velocity. Because the propulsive efficiency increases when a larger mass flow is accelerated by a smaller velocity increment, turbofans are more efficient than turbojets at subsonic speeds. The relative importance of the bypass flow can be quantified by the bypass ratio (BPR), i.e.,

(1)   \begin{equation*} {\text{\small BPR}} = \frac{\text{\small Mass~flow~rate~of~bypass~stream}}{\text{\small Mass~flow~rate~of~flow~through~core}} \end{equation*}

which represents the ratio of the mass flow in the bypass stream to that through the core. Modern high-bypass turbofans typically have BPR values between about 8 and 11. In such engines, most of the thrust is produced by the fan, with the core contributing a smaller fraction. For this reason, turbofan engines are more commonly used today than turbojets, for example, on commercial airliners, because they are more efficient and consume less fuel, and are also much quieter.

In a turbofan engine, a large fraction of the thrust is produced by the fan, improving propulsive efficiency.

Turboprops

Turboprops operate on the same basic thermodynamic cycle as turbofans, but most of the available energy is extracted by the turbine and delivered as shaft power to drive a propeller (Figure 8). The propeller then produces nearly all of the thrust, with only a small residual jet contribution through the exhaust nozzle. Because the propeller accelerates a large mass of air at relatively low velocity, turboprops achieve high propulsive efficiency at low flight speeds. Turboprop airplanes also have relatively high climb rates, making them suitable for smaller airports with shorter runways. However, like piston engines driving propellers, their performance is limited by compressibility effects on the propeller, particularly as the helical tip Mach number approaches unity. Consequently, turboprop-powered aircraft are generally confined to flight Mach numbers below about 0.4–0.5. Advanced propeller shapes and airfoils have extended this limitation to Mach numbers of about 0.6.

In a turboprop, most of the energy is converted to shaft power to drive a propeller, which produces nearly all of the thrust.

Turboshafts

Turboshaft engines (Figure 9) are closely related in operational principle to turboprops but are designed to deliver power to a shaft to drive a gearbox rather than to produce propulsive thrust directly from the shaft using a propeller. In most designs, a free-power turbine is used, allowing the gas generator and the output shaft to rotate independently at speeds appropriate to their respective functions. This configuration is particularly well-suited to helicopter applications, where the engine must supply power to a rotor system operating under widely varying load conditions. As with turboprops, only a small fraction of the total energy appears as jet thrust at the exhaust.

A turboshaft engine delivers power to a shaft, typically to drive a helicopter rotor.

Propulsion Fundamentals

Despite their different configurations, all propulsion systems operate on a common set of physical principles. Thrust is generated by processing a working fluid (normally air or an oxidizer such as oxygen) through a sequence of compression, heat addition via combustion of a suitable fuel, and subsequent expansion to produce useful work. Hydrocarbon fuels are relatively light and contain an exceptional amount of energy per unit mass. As previously mentioned, thrust is fundamentally produced by a change in momentum of the working fluid as it passes through the propulsive device. Therefore, it is helpful to establish the fundamental roles of compression and combustion, and to introduce a simplified flow model that captures the essential features of a generic propulsion system. Together, these elements provide the conceptual and analytical framework needed to interpret the operation, performance, and efficiency of practical propulsion systems.

Compression & Combustion

While combustion can still occur without prior compression, practical propulsion and power systems require combustion to proceed rapidly and with sufficient intensity to produce useful work. This requirement is met only by significantly compressing the working fluid before combustion, thereby increasing its temperature, density, and pressure. Compression raises the temperature of the working fluid from state 1 to state 2 according to

(2)   \begin{equation*} T_2 = T_1 \left( \frac{p_2}{p_1} \right)^{(\gamma - 1)/\gamma} \end{equation*}

where \gamma is the ratio of specific heats of the working fluid. This temperature rise is essential because chemical reaction rates depend critically on temperature. Chemical reactions require molecules to have enough energy to overcome an activation barrier during collisions. As the temperature increases, more molecules have sufficient energy, so the reaction rate increases exponentially with temperature. Without sufficient compression, the fluid-fuel mixture may not ignite or will burn too slowly to be useful. This effect is fundamental to compression-ignition engines. Compression also increases the density of the mixture from state 1 to state 2 according to the equation of state, i.e.,

(3)   \begin{equation*} \varrho_2 = \varrho_1 \left( \frac{p_2}{p_1} \right)^{1/\gamma} \end{equation*}

so that more fuel and oxidizer molecules are present per unit volume. This increase in density raises the frequency of molecular collisions, thereby increasing the rate of energy release during combustion. Therefore, compression establishes high initial pressure, temperature, and density so that, when fuel is added and burned (combustion), the resulting heat release produces a substantial rise in the flow’s pressure and thermodynamic enthalpy. The resulting pressure and enthalpy can then be converted into useful work through a turbine, piston, or nozzle.

Fuels & Combustion

Combustion is fundamentally a process of adding chemical energy to a compressed working fluid by oxidizing a fuel. The combustion of aviation fuels such as Jet-A or 100LL (Avgas) releases a large amount of energy because the reactants are in a relatively high chemical-energy state compared to the products formed when they combine with atmospheric oxygen. During the reaction, the weaker C–C and C–H bonds in the fuel are broken and replaced by much stronger C=O bonds in CO2 and O–H bonds in H2O, resulting in a substantial decrease in chemical potential energy. This difference in bond energies is liberated as heat, giving hydrocarbon fuels their high energy content. Because these fuels are composed primarily of light elements, this energy release also corresponds to a high energy per unit mass, which is why liquid hydrocarbon fuels remain so effective for propulsion.

Rocket propellants such as RP-1 and liquid oxygen (LOX) release large amounts of energy for essentially the same thermodynamic reason as air-breathing fuels, i.e., the reactants are in a relatively high chemical energy state compared to the products of combustion. In this case, RP-1 (a refined kerosene) serves as the fuel, and liquid oxygen (LOX) provides the oxidizer, so the system carries both components needed for combustion rather than relying on atmospheric oxygen. When the mixture reacts, the C–C and C–H bonds in the fuel and the O=O bonds in the oxygen are replaced by much stronger C=O and O–H bonds in CO2 and H2O, producing a large decrease in chemical potential energy that is released as heat. Because the oxidizer is supplied in pure form and at high density, the reaction can proceed rapidly and at very high temperatures, giving rocket propellants their high energy release rate and making them effective for propulsion in the absence of air.

Flow Model

Consider a general propulsive device, as represented by the control volume in Figure 10, moving through the air with freestream velocity {V_{\infty}}. In an air-breathing propulsion system, air is entrained at the inlet, compressed, and directed to the combustion chamber. The energy released by fuel combustion increases the momentum and kinetic energy of the flow, which is then discharged into the slipstream at a higher jet velocity V_j. In a rocket engine, there is no freestream inlet velocity, and the fuel and oxidizer are introduced directly to the combustion chamber; however, the same fundamental principles apply.

Control volume analysis of a general propulsive device, which works on the air to increase its streamwise momentum and produce a reaction force (thrust).

Assume for the following exposition that there are no external pressure forces and that thrust is produced only by the time rate of change of momentum of the flow. If the mass flow of air into the device is denoted by \overbigdot{m}_{\rm air} and the mass flow rate of fuel is {\overbigdot{m}_{\rm fuel}}, then conservation of momentum applied to the flow gives the thrust produced as

(4)   \begin{equation*} T = \left( \overbigdot{m}_{\rm air} + \overbigdot{m}_{\rm fuel} \right) V_j - \overbigdot{m}_{\rm air} V_{\infty} \end{equation*}

Now, if it assumed that \overbigdot{m}_{\rm air} \gg \overbigdot{m}_{\rm fuel}, which is readily justified,[1] then

(5)   \begin{equation*} T \approx \overbigdot{m}_{\rm air} \left( V_j - V_{\infty} \right) = \overbigdot{m} \left( V_j - V_{\infty} \right) \end{equation*}

i.e., the thrust T is equal to the time rate of increase of momentum of the flow through the device and is proportional to the velocity increment V_j - V_{\infty}. The increased velocity of the flow also appears as a gain in kinetic energy in the slipstream, which is irrecoverable and so represents a loss, i.e.,

(6)   \begin{equation*} P_{\rm loss} = \overbigdot{KE}_{\rm loss} = \frac{1}{2} \overbigdot{m} \left( V_j - V_{\infty} \right)^2 \end{equation*}

It is clear from Eq. 5 that thrust depends on both the mass flow rate through the device and the increase in flow velocity. Therefore, the same thrust can be produced either by accelerating a large mass flow by a small velocity increment or a smaller mass flow by a larger velocity increment. As will now be shown, this fundamental trade-off governs the device’s propulsive efficiency.

Quantifying Propulsive Efficiency

Generally, a propulsion system’s overall efficiency is measured by the thrust it produces per unit of fuel consumed. However, the propulsive efficiency of the device is also important, i.e., the aerodynamic efficiency with which it creates a useful propulsive force by doing work on the air. Efficiency in producing thrust is directly related to the kinetic energy of the exit flow from the propulsion system, which represents the energy per unit time remaining in the slipstream or wake. This definition of efficiency does not account for the engine’s thermodynamic efficiency or internal mechanical losses within the propulsion system.

Relative Efficiency

In terms of quantifying the efficiency of thrust production, consider the useful power supplied by the device to propel the aircraft forward, which will be the product of the force produced by the propulsive device and the flight velocity {V_{\infty}} (true airspeed) of the aircraft, i.e.,

(7)   \begin{equation*} P_{\rm useful} = T \, V_{\infty} \end{equation*}

This expression applies in a ground-fixed reference frame; in a frame moving with the aircraft, the thrust does no work. The relative propulsive efficiency can now be defined as

(8)   \begin{equation*} \eta_p = \frac{\mbox{\small Useful power produced}}{\mbox{\small Total power expended }} = \frac{T \, V_{\infty}}{ T \, V_{\infty} + \frac{1}{2} \overbigdot{m} \left( V_j - V_{\infty} \right)^2} \end{equation*}

which recognizes the losses that appear as a gain in kinetic energy in the downstream jet flow.

Recall that based on the conservation of momentum, the thrust is given by Eq. 5, so substituting in the previous equation gives

(9)   \begin{equation*} \eta_p = \frac{T \, V_{\infty}}{T \, V_{\infty} + \frac{1}{2} \overbigdot{m} \left( V_j - V_{\infty} \right)^2} = \frac{\overbigdot{m} \left( V_j - V_{\infty} \right) V_{\infty}}{ \overbigdot{m} \left( V_j - V_{\infty} \right) V_{\infty} + \frac{1}{2} \overbigdot{m} \left( V_j - V_{\infty} \right)^2} \end{equation*}

which, after some simplification, gives

(10)   \begin{equation*} \eta_p = \frac{2}{1 + \left( \displaystyle{\frac{V_j}{V_{\infty} }}\right) } \end{equation*}

where to be meaningful V_j > V_{\infty} for positive thrust. Interestingly, if the exhaust velocity equals the flight speed, i.e., V_j = V_{\infty}, the propulsive efficiency becomes 100% in the limit. However, Eq. 5 shows that no thrust is produced in this case.

The propulsive efficiency can also be written in a perhaps more intuitive form as

(11)   \begin{equation*} \eta_p = \frac{2 \left( \displaystyle{ \frac{V_{\infty}}{V_j}}\right)} {\left( \displaystyle{ \frac{V_{\infty}}{V_j}}\right) + 1} \quad \mbox{for} ~ \frac{V_{\infty}}{V_j}\le 1 \end{equation*}

by rearranging Eq. 10. This form indicates that propulsive efficiency increases with decreasing jet velocity at a given airspeed, as illustrated in Figure 11. This outcome is obtained because the kinetic energy losses become a smaller fraction of the total propulsive power.

The propulsive efficiency of an air-breathing engine increases with airspeed and as V_j approaches the flight speed.

Maximizing Propulsive Efficiency

In summary, taking the thrust and efficiency equations together, i.e.,

(12)   \begin{equation*} T = \overbigdot{m} \left( V_j - V_{\infty} \right) \quad\quad \mbox{and} \quad\quad \eta_p = \frac{2}{1 + \left( \displaystyle{\frac{V_j}{V_{\infty} }} \right)} \end{equation*}

reveals an important outcome regarding the efficiency of thrust production. Notice that a low value of V_j for a given thrust and a higher propulsive efficiency can only be achieved by having a significant mass flow rate, \overbigdot{m}, through the engine.

Notice also that the thrust equation can be written in terms of specific thrust, T/\overbigdot{m}, as

(13)   \begin{equation*} \frac{T}{\overbigdot{m} \, V_{\infty} } = \left( \frac{V_j}{V_{\infty} } \right) - 1 \end{equation*}

The specific thrust, T/\overbigdot{m}, is the thrust produced per unit air mass flow rate. It is not an efficiency, but it is useful for comparing propulsion systems because it indicates how much velocity increment is being given to the flow. Different engines will have different specific thrusts; however, higher specific thrust generally corresponds to higher jet velocities and therefore lower propulsive efficiency. Conversely, lower specific thrust requires a higher mass flow rate but yields higher propulsive efficiency. Therefore, yet another form of the efficiency equation is

(14)   \begin{equation*} \eta_p = \frac{2}{2 + \left( \displaystyle{ \frac{T}{\overbigdot{m} \, V_{\infty}}} \right) } \end{equation*}

noting that the grouping T/\overbigdot{m}V_{\infty} is dimensionless. This result is plotted in Figure 12 as a function of \overbigdot{m} V_{\infty}/T to emphasize the direct effect of mass flow rate. It will be apparent then that for a given airspeed, V_{\infty}, higher propulsive efficiency is obtained by increasing the mass flow rate and reducing the specific thrust.

The efficiency of thrust production increases with increasing mass flow rate and decreasing specific thrust.

Therefore, it is generally more efficient to create thrust by accelerating a large mass flow at a lower jet velocity than a smaller mass flow at a higher jet velocity. This is why turbofan engines and propellers, which are “high bypass” devices, achieve the highest propulsive efficiency. Nevertheless, efficiency is always less than 100% because of the increased kinetic energy of the exit (jet) flow, an inevitable loss associated with thrust production.

Specific Fuel Consumption

An aircraft’s overall performance is strongly influenced by the engine’s fuel consumption, quantified as specific fuel consumption (SFC). The BSFC (brake-specific fuel consumption) is the fuel weight used per brake unit of power generated, which is a measure of efficiency, i.e.,

(15)   \begin{equation*} {\rm BSFC}\equiv c_b = \frac{\mbox{\small Weight of fuel consumed}}{\mbox{\small (Unit power output)} \,\mbox{\small (Unit time)}} \end{equation*}

For turboshaft or piston engines, the shaft power produced must be converted into aerodynamic power by a device, such as a propeller or fan; therefore, the propeller or fan efficiency is also considered when determining the overall efficiency of the engine-propeller system. Notice that the term “brake” in BSFC refers to the shaft power delivered by the engine, historically measured using a mechanical brake or dynamometer. Accordingly, BSFC is defined by the usable output power at the engine shaft rather than by the internal power developed within the engine.

In the case of a thrust-producing engine, the fuel consumption is quantified using thrust-specific fuel consumption (TSFC), which is measured in terms of the weight of fuel used per unit thrust per unit time, again a measure of efficiency, i.e.,

(16)   \begin{equation*} {\rm TSFC} \equiv c_t = \frac{ \mbox{\small Weight of fuel consumed}}{\mbox{\small (Unit thrust output) (Unit time)}} \end{equation*}

Usually, the period for which TSFC and BSFC values are quoted is 1 hour.

To better understand the connection between thrust-specific fuel consumption and efficiency, consider that the overall efficiency of a jet engine is the product of thermal efficiency, \eta_{\text{th}}, and propulsive efficiency, \eta_p, with the latter defined as

(17)   \begin{equation*} \eta_p = \frac{2}{1 + \left( \displaystyle\frac{V_j}{V_{\infty}} \right)} \end{equation*}

The engine’s useful power output is the thrust power, given by T V_{\infty}, and the fuel energy input per unit time is \overbigdot{m}_f \, h_f, where h_f is the heating value of the fuel. Then, the overall efficiency is

(18)   \begin{equation*} \eta = \frac{T V_{\infty}}{\overbigdot{m}_f \, h_f} \end{equation*}

Solving for thrust-specific fuel consumption (TSFC) gives

(19)   \begin{equation*} \text{TSFC} = \frac{\overbigdot{m}_f}{T} = \frac{V_{\infty}}{\eta \, h_f} = \frac{V_{\infty}}{\eta_{\text{th}} \, \eta_p \, h_f} \end{equation*}

Over decades of continuous development, the SFC of aircraft engines has markedly improved (reduced), as summarized in Figure 13. Remember that these values measure the engine’s thrust-producing fuel efficiency; therefore, the reductions shown are significant and commensurate with advances in engineering technology more generally.

Improvements in subsonic engine performance in terms of thrust-producing specific fuel consumption (TSFC). The values are normalized to a cruise Mach number of 0.8 at ISA standard conditions.

The efficiency of several propulsive devices is shown in Figure 14 as specific fuel consumption, i.e., the fuel used per unit thrust per unit time, which is inversely proportional to the overall efficiency of the propulsion system (and so influenced by both thermal and propulsive efficiency). Therefore, the higher the efficiency, the lower the specific fuel consumption. Note that “bypass” increases mass flow through the engine, thereby reducing the exit (jet) velocity. The bypass ratio (BPR) is the ratio of the mass flow through the fan relative to the mass flow through the engine’s core.

The thrust-producing specific fuel consumption (TSFC) of various types of jet engines and propulsion systems in terms of bypass ratio (BPR).

As an example of what has just been concluded in the preceding equations, a turbofan engine (which has a high BPR) has much better fuel efficiency (i.e., lower fuel burn rate) than a turbojet because the fan stage helps to increase mass flow through the engine, but without increasing the net jet velocity (V_j) substantially. A higher BPR also helps to decrease jet noise because of the lower values of V_j, where the noise increases quickly with increasing jet velocity. This characteristic is precisely why high-bypass turbofans are used on most modern airliners rather than turbojets: they can achieve efficiencies exceeding 60% and produce relatively low noise.

By similar arguments, a propeller-engine combination has a higher propulsive efficiency (about 70%) because it produces a high mass flow rate through the propeller, and the downstream V_j is relatively small. On a turbojet, V_j is always relatively high, and the propulsive efficiency of this type of engine may be only of the order of 20%.

Nevertheless, it is not always possible to have a propulsion system that is well-suited to the flight conditions in which it operates and highly efficient. This issue means that engineers designing new aircraft often must conduct studies to determine the relative benefits of one propulsion system over another, depending on the aircraft’s cruise speed and Mach number, as well as other design factors. Engineers often call these trade studies, in which the relative merits of one concept are traded off against those of another.

Units of BSFC and TSFC

The units of BSFC are typically expressed in lb hp^{-1}~hr^{-1} in the U.S. customary (USC) system, or in kg kW^{-1}~hr^{-1} or grams per kilowatt-hour (g/kWh) in the SI system. Notice that in the SI units, BSFC is defined using a unit of mass (kilograms or grams), not weight, which is a well-known anomaly in the SI system. However, in both systems, the time unit is consistently set to an hour because BSFC values are commonly quoted as fuel usage per hour of operation.

The thrust-specific fuel consumption (TSFC), by contrast, is expressed in lb lb^{-1} hr^{-1} in the USC system or kg kN^{-1} hr^{-1} in the SI system. Once again, the SI convention uses units of mass (kilograms) rather than units of weight (Newtons or kiloNewtons), so care must be taken when interpreting or converting these values. This discrepancy arises from the SI system treating the kilogram as a unit of mass, whereas thrust is inherently a force and should ideally be expressed in units of force. Further caution is warranted because some publications and engine manufacturers may report TSFC using alternative or legacy units, such as values per second or per Newton.

Published BSFC and TSFC values for different engines are typically referenced at the engine’s maximum-rated output (either power or thrust) under standard atmospheric conditions at sea level (ISA). Such values enable meaningful comparisons between engine types but must be adjusted or interpreted carefully under off-design conditions, such as high-altitude or partial-power operation.

Typical values of brake-specific fuel consumption (BSFC) and thrust-specific fuel consumption (TSFC) for various engine types under nominal operating conditions.
Engine Type BSFC (g/kWh) TSFC (lb/lb hr) Conditions
Piston (reciprocating) engine 250–300 — Sea level, cruise
Turboprop 200–250 — Sea level, cruise
Turbofan (high bypass, modern) — 0.3–0.6 Sea level, cruise
Turbofan (low bypass) — 0.5–0.9 Sea level, military thrust
Turbojet (dry) — 0.8–1.2 Sea level, maximum thrust
Turbojet (with afterburner) — 2.0–3.5 Sea level, full reheat
Ramjet — 1.5–2.5 Supersonic flight, Mach 2+

For example, a turbojet engine may have a quoted thrust-specific fuel consumption of

(20)   \begin{equation*} \text{TSFC} = 1.0 \ \frac{\text{lb}}{\text{lb} \, \text{hr}} \end{equation*}

If fuel weight and thrust are expressed in the same force units, this is dimensionally equivalent to inverse time, i.e.,

(21)   \begin{equation*} \text{TSFC} = 1.0 \ \frac{\text{force}}{\text{force} \, \text{hr}} = 1.0~\text{hr}^{-1} = \frac{1.0}{3600}~\text{s}^{-1} = 2.78 \times 10^{-4}~\text{s}^{-1} \end{equation*}

This form shows that TSFC is dimensionally equivalent to inverse time when fuel consumption is expressed as fuel weight flow per unit thrust. In practice, however, TSFC is retained as a fuel-flow-per-thrust quantity because that interpretation is more useful for performance calculations.

Likewise, a typical turboprop engine may have a quoted brake-specific fuel consumption of

    \[ \text{BSFC} = 0.50 \ \frac{\text{lb}}{\text{hp} \, \text{hr}} = 0.50 \ \text{lb} \, \text{hp}^{-1} \, \text{hr}^{-1} \]

Because 1~\text{hp} = 550~\text{ft} \,\text{lb} \, \text{s}^{-1} and 1~\text{hr} = 3{,}600~\text{s}, then

    \[ \text{BSFC} = \frac{0.50}{550 \times 3{,}600} \ \frac{\text{lb}}{\text{ft} \, \text{lb} \, \text{s}} = 2.53 \times 10^{-7} \ \text{lb} \, \text{ft}^{-1} \, \text{lb}^{-1} \, \text{s}^{-1} \equiv 2.53 \times 10^{-7} \ \text{ft}^{-1} \, \text{s}^{-1} \]

which shows that BSFC is dimensionally equivalent to s^{-1} in USC force-based units, although it is physically interpreted as the fuel weight flow required per unit shaft power.

Unit Warning: In U.S. customary engineering units, the term “pound-force” is properly written as lb. The unit notation lbf or lbf is deprecated and should be avoided. This text uses “lb” for force and “slug” or “sl” for mass consistently.

Afterburning

A supersonic aircraft generally requires a propulsion system capable of producing substantial excess thrust, often achieved with a turbojet or low-bypass turbofan equipped with an afterburner, to overcome the sharp rise in aerodynamic drag in the transonic regime and sustain flight at high supersonic speeds. As the aircraft approaches the speed of sound, it encounters compressibility effects and shock waves, leading to a rise in transonic drag, commonly referred to as the “sound barrier.” To pass through this regime, the afterburner injects additional fuel into the jet pipe downstream of the turbine, where it burns in the hot exhaust, as shown in Figure 15.

Control volume analysis of a propulsion system with an afterburner.

This combustion significantly increases the exhaust velocity from V_j to V_j + \Delta V_j, which raises the net thrust from T to T + \Delta T, as described by the momentum thrust relation, i.e.,

(22)   \begin{equation*} T + \Delta T = \left( \overbigdot{m}_{\rm air} + \overbigdot{m}_{\rm fuel} + \Delta \overbigdot{m}_{\rm fuel} \right) \left( V_j + \Delta V_j \right) - \overbigdot{m}_{\rm air} V_\infty \end{equation*}

where \overbigdot{m} denotes the mass flow rate and {V_{\infty}} is the freestream velocity. If the additional fuel mass flow rate, i.e., \Delta \overbigdot{m}_{\rm fuel}, is small compared to the air mass flow rate, so that \overbigdot{m} \approx \overbigdot{m}_{\rm air}, then the thrust equation simplifies to

(23)   \begin{equation*} T + \Delta T = \overbigdot{m} \left( V_j + \Delta V_j - V_\infty \right) = \overbigdot{m} \left( V_j - V_\infty \right) + \overbigdot{m} \, \Delta V_j \end{equation*}

This shows that the additional thrust produced by the afterburner is approximately equal to \overbigdot{m} \, \Delta V_j, representing the extra momentum gained from the increase in jet exhaust velocity.

The principal advantage of afterburning lies in its simplicity and effectiveness. Afterburning allows for a rapid and substantial increase in thrust without requiring larger compressor stages or a redesign of the engine core. This makes it especially beneficial during combat maneuvers, takeoff from short runways, or when accelerating through transonic drag rise. Afterburners provide thrust augmentation on demand, enabling aircraft to exceed the performance limits of dry (non-afterburning) engines. However, this increase in exhaust speed comes at the cost of lower propulsive efficiency. The efficiency with which the engine converts fuel energy into useful thrust is given by

(24)   \begin{equation*} { \eta_p = \frac{2}{1 + \left( \displaystyle\frac{V_j + \Delta V_j}{V_{\infty}} \right)} = \frac{2}{1 + \left( \displaystyle\frac{V_j}{V_{\infty}} + \frac{\Delta V_j}{V_{\infty}} \right)} } \end{equation*}

This expression shows that as \Delta V_j is added to the jet velocity by the afterburner, then the efficiency \eta_p decreases significantly.

The implications of this relationship are significant when considering the afterburning operation. An afterburner increases thrust by injecting additional fuel downstream of the turbine to raise the exhaust velocity from V_j to V_j + \Delta V_j. While this increases the net thrust T, it does so by disproportionately increasing the jet velocity relative to the freestream speed, thereby reducing the propulsive efficiency \eta_p. According to the relationship

(25)   \begin{equation*} \eta_p = \frac{2}{1 + \left( \displaystyle\frac{V_j + \Delta V_j}{V_{\infty}} \right)} \end{equation*}

a large increase in V_j lowers \eta_p significantly, which, according to the inverse relationship in the TSFC expression, then

    \[ \text{TSFC} = \frac{V_{\infty}}{\eta_{\text{th}} \, \eta_p \, h_f} \]

leading to a sharp increase in fuel consumption per unit thrust. In practice, this means that the thrust gained from afterburning comes at the cost of much higher fuel burn. For example, while a turbojet engine might exhibit a TSFC of around 1.0 lb/(lb hr) \equiv 1.0 kg/(kg hr) in dry mode, this value can rise to 2.5 to 3.5 lb/(lb hr) \equiv 2.5 to 3.5 kg/(kg hr) under full afterburner, depending on flight conditions. This is why afterburning is used sparingly, typically for takeoff, transonic acceleration, or combat scenarios, where maximum thrust is prioritized over fuel efficiency. The photograph in Figure 16 shows the SR-71 Blackbird with both engines operating in full afterburner, producing bright exhaust plumes. This configuration enabled the aircraft to pass through transonic drag rise and maintain cruise speeds well above Mach 3.

SR-71 Blackbird with the engines on full afterburner. The high-speed exhaust contains numerous shock waves, which appear as “diamonds.”

Check Your Understanding # 1 – Efficiency of propulsion

An airplane’s engine consumes 500 liters of fuel per hour and produces 2,000 N of thrust. The fuel’s energy content is 35 MJ per liter. Calculate the engine’s propulsive efficiency, assuming the airplane flies at a true airspeed of 250 kts.

Show solution/hide solution.

First, calculate the energy input from the fuel consumption using

    \[ \text{\small Energy input} = \text{\small Rate of fuel consumption} \times \text{\small Energy content of fuel} \]

Convert liters of fuel to energy content in MJ using

    \[ \text{\small Energy input} = 500 \times 35 = 17,500 \text{ MJ/hour} \]

The power output of the engine can be found using the thrust produced and the aircraft’s speed, i.e.,

    \[ \text{\small Power output} = \text{\small Thrust} \times \text{\small Aircraft speed} \]

An airspeed of 250 kts is approximately 129.12 m/s (1 knot {\approx} 0.514 m/s). The useful power output is

    \[ \text{\small Power output} = 2,000 \times 129.12 = 258,240 \text{W} = 258.24 \text{kW} \]

The propulsive efficiency, \eta, is defined as the ratio of the useful power output (propulsive power) to the energy input from the fuel per unit time, i.e.,

    \[ \eta = \frac{\text{\small Power output}}{\text{\small Energy input per unit time}} \]

Converting all of the numerical values into consistent base units, then

    \[ \eta = \frac{258.24 \times 1,000 \times 3,600}{17,500 \times 10^6} = 53.18\% \]

Rocket Propulsion Fundamentals

Rocket engines, which are non-airbreathing, are used for spacecraft and launch vehicles; they must carry fuel and an oxidizer during flight. The photograph in Figure 17 shows a rocket engine under test that uses hydrogen and oxygen as propellants, producing a very clean exhaust and a lot of steam. Rocket engines may also power some types of high-speed aircraft. The first aircraft to break the speed of sound, the Bell X-1, was powered by a rocket engine, as was the first hypersonic aircraft, the X-15. Rocket engines are extremely powerful and operate at high pressures and temperatures, but only for relatively short periods.

NASA image
In this case, a rocket engine under test uses hydrogen and oxygen as propellants, hence the almost invisible exhaust as superheated steam.

The principle of thrust generation for a rocket engine is the reaction force associated with accelerating a mass of gas at high velocity, the gas being a byproduct of the combustion of the fuel and oxidizer, thereby increasing the momentum of the ejected gas; see Figure 18. The thrust force on the rocket engine is then opposite to the direction of the gas exit velocity, which is a result of Newton’s third law. Notice that there is no external ambient mass flow into the engine, such as there would be with an air-breathing engine.

Flow model used to calculate the thrust produced by a rocket motor.

Using the principles of conservation of momentum, the thrust T produced by the rocket engine will be

(26)   \begin{equation*} T = \left( \overbigdot{m}_{\rm ox} + \overbigdot{m}_{\rm fuel} \right) V_e = \overbigdot{m} V_e \end{equation*}

where V_e is the equivalent average exit velocity and \overbigdot{m} is the propellant net mass flow rate; there is no incoming momentum.

Therefore, the higher the mass flow of propellant into the motor and the higher the exit velocity from the nozzle, the higher the thrust will be. Rocket engines typically have much higher mass flow rates and exit velocities than air-breathing engines. In this case, all of the energy and power liberated to create thrust is manifest as lost kinetic energy, i.e.,

(27)   \begin{equation*} \frac{d (KE_{\rm loss})}{dt} = \frac{1}{2} \overbigdot{m} \, V_j ^2 \end{equation*}

Relative Efficiency

The efficiency of thrust production by a rocket engine at a flight velocity {V_{\infty}} can be defined as

(28)   \begin{equation*} \eta_p = \frac{\mbox{\small Useful power produced}}{\mbox{\small Total power expended }} = \frac{T \, V_{\infty}}{ T \, V_{\infty} + \frac{1}{2} \overbigdot{m} \, V_j^2 } \end{equation*}

where {V_{\infty}} is the flight speed.

Using the conservation of momentum, then

(29)   \begin{equation*} \eta_p = \frac{\overbigdot{m} V_j \, V_{\infty}}{ \overbigdot{m} V_j V_{\infty} + \frac{1}{2} \overbigdot{m} \, V_j^2 } \end{equation*}

which, after simplification, gives

(30)   \begin{equation*} \eta_p = \frac{2 }{2 + \left( \displaystyle{\frac{V_j}{V_{\infty} }}\right)} \end{equation*}

noting that this is a different result for the efficiency compared to an air-breathing engine, i.e., Eq. 10.

An alternative form of the previous equation is

(31)   \begin{equation*} \eta_p = \frac{2 \left( \displaystyle{\frac{V_{\infty}}{V_j }}\right) }{2 \left( \displaystyle{\frac{V_{\infty}}{V_j }} \right) + 1 } \end{equation*}

as shown in Figure 19. Notice that, unlike air-breathing engines, the flight velocity of rocket engines can exceed their exhaust velocity.

The propulsive efficiency of a rocket continues to increase with increasing flight speed.

Specific Impulse

In the case of rocket engines, it is customary to define their efficiency in terms of specific impulse, I_{\rm sp}, which is the thrust produced divided by the flow rate of the weight of propellant, i.e.,

(32)   \begin{equation*} I_{\rm sp} = \frac{T}{\overbigdot{m} \, g_0} \end{equation*}

where {g_0} is the acceleration under gravity at MSL. Notice that I_{\rm sp} has units of seconds, and the specific impulse is inversely proportional to the specific fuel consumption. For rockets, values of I_{\rm sp} typically range from 280 to 465 seconds, depending on the fuel and oxidizer used and the rocket engine design.

Check Your Understanding #2 – Units of specific impulse

Show that the specific impulse has units of seconds.

Show solution/hide solution.

The equation for specific impulse is

    \[ \small I_{\rm sp} = \frac{T}{\overbigdot{m} \, g_0} \]

In terms of dimensions, then

    \[ \small \left[ I_{\rm sp} \right] = \frac{\rm M L T^{-2}}{(\rm M T^{-1}) L T^{-2} } = \frac{1}{\rm T^{-1}} = \rm T \]

Therefore, the units of specific impulse are seconds. Of course, an advantage of this definition is that the numerical value is the same in both SI and USC units.

As shown in Figure 20, air-breathing engines have a specific impulse that is typically one order of magnitude higher than that of a rocket. This latter characteristic is because an air-breathing engine uses ambient air as the oxidizer for combustion, which does not have to be carried. The upshot is lower kinetic energy losses. An air-breathing engine uses far less energy to generate a given thrust than a rocket. As previously discussed, turbofans and turboprops are the most efficient types of air-breathing engines because they accelerate a significant air mass flow rate to a relatively low “jet” velocity.

The specific impulse is the thrust generated per unit mass flow rate of fuel and can be viewed as a measure of propulsive efficiency.

Electric Propulsion

The engine that drives an airplane’s propeller is usually a piston or turboprop engine. However, engineers have been experimenting with electric propulsion systems, including hybrid systems, which use another engine type to augment the electric system. Electric propulsion shifts the limitation from fuel flow to energy storage, i.e., battery energy density.

Compared to a piston or gas turbine engine, an electric motor has certain advantages, including its small, compact size and relatively few moving parts. It is also relatively inexpensive and quiet compared to an internal combustion engine, and it produces no direct emissions during operation, although the overall environmental impact depends on how the electrical energy is generated. As a point of reference, electric motors can convert a large fraction of electrical energy into usable shaft power (often exceeding 90% at the motor level). In comparison, conventional fossil-fuel engines (e.g., Avgas or Jet A-1) have significantly lower thermal efficiencies.

For an electrically powered propulsion system, the useful propulsive power is

(33)   \begin{equation*} P_{\rm useful} = T \, V_{\infty} \end{equation*}

and the corresponding electrical power required is

(34)   \begin{equation*} P_{\rm elec} = \frac{T \, V_{\infty}}{\eta_{\rm motor} \, \eta_{\rm prop}} \end{equation*}

where \eta_{\rm motor} is the motor efficiency and \eta_{\rm prop} is the propulsive efficiency. However, using electric engines for aircraft propulsion has many challenges. The main challenge is storing the energy needed for flight, which requires many batteries that are tightly packed inside the aircraft, an example being shown in Figure 21. The total energy required for a flight of duration t is

(35)   \begin{equation*} \mathcal{E}_{\rm req} = P_{\rm elec} \, t \end{equation*}

and the corresponding battery weight is

(36)   \begin{equation*} W_{\rm batt} = \frac{\mathcal{E}_{\rm req}}{e_{\rm batt}} \end{equation*}

where e_{\rm batt} is the specific energy of the batteries. However, batteries are heavy and bulky, and they must be connected to the motor with heavy cables. Fossil fuels have a much higher energy storage per unit weight, called energy density, than batteries, typically by an order of magnitude or more. Battery life is also reasonably short because of the chemical and thermal stresses inside the battery induced by the relatively fast discharge cycles required by the power draw for flight. Battery packs tend to be expensive to replace, with a replacement cycle significantly shorter than the time between overhauls of an internal combustion engine.

An example of an “all-electric” airplane powered by batteries and an electric motor driving a propeller.

A hybrid propulsion system can give an electric aircraft much greater flexibility. In this case, another engine and generator complement the electric system, providing additional power when needed. For example, such hybrid systems could be paired with a hydrogen fuel cell, gas turbine, or even a diesel engine coupled to a generator. The consequence is that a hybrid system can provide sustained power and extend flight range and/or endurance, while electric motors may be used where high efficiency and control are advantageous. For now, only relatively small electric-powered aircraft capable of short-duration missions are feasible.

Electric vertical takeoff and landing (eVTOL) aircraft are also being developed to transport goods and people in an urban environment, i.e., as an air taxi for increased urban mobility; see Figure 22. The AgustaWestland Project Zero was the world’s first eVTOL aircraft based on a tiltrotor concept. It was developed as a technology demonstrator and was used to investigate all-electric propulsion and other advanced technologies. The photograph below shows another eVTOL concept vehicle with rotors for vertical lift and a propeller for forward thrust.

An eVTOL “urban taxi” concept using rotors for vertical lift and a propeller for forward thrust. Lift is created by the wings when flying forward.

However, vertical flight imposes a fundamental energetic penalty. To hover, the propulsion system must generate a thrust equal to the vehicle’s weight, i.e., T = W, and the corresponding power required is obtained from the classic momentum theory as

(37)   \begin{equation*} P_{\rm hover} = N_r \, T_r \sqrt{\frac{T_r}{2 \, \varrho \, A_r}} = \frac{W^{3/2}}{\sqrt{2 \varrho \, N_r \, A_r}} = \frac{W^{3/2}}{\sqrt{2 \, \varrho \, A}} \end{equation*}

where T_r = W/N_r is the thrust carried by each rotor, \varrho is the air density, and A_r is the disk area of each rotor, so that N_r A_r = A is the total rotor disk area. This result shows that the power required for vertical flight scales with weight and decreases with increasing net disk area. Consequently, hover and vertical takeoff require significantly more power than forward flight for a given vehicle, leading to high energy consumption and reduced range. While many such eVTOL aircraft concepts are now under consideration, their technical and economic viability remains an open question. Another concern is how their flight operations, which could number hundreds at any given time, can be effectively and safely integrated into the national airspace system.

Summary & Closure

The production of thrust is essential for all powered flight vehicles. A propulsion system consists of an engine that produces power and work, and hence a force to propel the vehicle forward. Ultimately, the performance of any flight vehicle is determined by the balance between the thrust produced by the propulsion system and the aerodynamic forces acting on the airframe. Air-breathing engines include reciprocating piston engines driving a propeller, turboprops, turbojets, and turbofans. In each case, the purpose of the engine is to do work on the air, increasing its momentum and generating a reaction force, i.e., propulsive thrust. Rocket engines are not air-breathing and can operate outside of the atmosphere. However, the principle of thrust generation remains the same: thrust equals the time rate of change of the momentum of the gases exiting the engine.

Propulsive efficiency is also essential to understand, as it is directly related to the fuel required to produce thrust and to the aerodynamic effectiveness of the propulsion system. In general, creating thrust using higher mass flow rates and lower exit velocities is more efficient. To this end, turbofans and turboprops are attractive options for powering many airplanes. While electric propulsion concepts for aircraft are still in their infancy, they are likely to see increased use in the coming decades, particularly for smaller aircraft and urban mobility or VTOL concepts.

For Further Thought or Discussion

  • In designing a commuter passenger aircraft, consider some design trades in powering the aircraft using turboprops versus turbofans. Hint: Not all of these trades may have an engineering basis.
  • Why is the engine used to power a commercial transport aircraft more of what might be called a “point design” versus one used on a military fighter aircraft?
  • What are some of the engineering and other issues that might be associated with using an afterburner on a turbojet engine?
  • What engine type(s) are being proposed for supersonic business jets?
  • What is meant by a hybrid-electric propulsive system? List the relative merits of a hybrid system compared to a pure electric system.

5-Question Self-Assessment Quickquiz

Other Useful Online Resources

To gain a further understanding of propulsion systems, check out these online resources:

  • An excellent video with good graphics explaining the differences between engine types.
  • To learn more about air-breathing rockets, read How Air-breathing Rockets Will Work.
  • For more in-depth information on the Boeing 787 propulsion system, check out this article from AERO Magazine.
  • Aircraft engine types and propulsion systems – how do they work?
  • A great video on how jet engines work.
  • A video explaining future aircraft propulsion systems.
  • Electrified aircraft – a video presentation by NASA.
  • The reason why aircraft jet engines are so monstrously large today!
  • Backyard run of a Rolls-Royce Spey jet engine!
  • Great video on how an afterburner works.
  1. The mass flow rate of air through an engine is much greater than the mass flow rate of fuel because combustion requires a low fuel-air ratio by mass, typically on the order of 1:15 at stoichiometric conditions and usually even lower in practice. Therefore, only a small fuel mass flow is needed compared with the much larger air mass flow required to supply oxygen and produce thrust. ↵

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Introduction to Aerospace Flight Vehicles Copyright © 2022–2026 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n28

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