43 Thermodynamics of Propulsion Systems

Introduction

Propulsion systems convert stored energy into useful work, shaft power, or thrust. In aerospace applications, this energy may be stored chemically in a fuel, propellant, or a battery. In combustion-based systems, energy conversion occurs through the release of heat, compression, expansion, and acceleration of a working fluid. In electrically powered systems, electrochemical energy from a battery is converted into electrical energy, which is then converted into mechanical shaft power by an electric motor. Thermodynamics provides the framework for analyzing these processes. The First Law of Thermodynamics establishes the accounting of energy through heat transfer, electrical and shaft work, enthalpy changes, and kinetic-energy changes. The Second Law establishes the limits on these conversions through irreversibility, entropy generation, and exergy destruction. Together, these laws explain why real propulsion systems cannot convert all of the available energy into useful thrust or shaft power.

Building on the thermodynamic foundations developed in Chapter 12, this chapter examines the principal components used in aerospace propulsion. Nozzles and diffusers convert between enthalpy and kinetic energy, compressors and turbines require or produce shaft work, and combustors add chemical energy to the working fluid. The chapter also introduces propulsion-specific performance measures, including isentropic efficiency, combustion efficiency, fuel-air ratio, fuel heating value, exergy destruction, and specific impulse. These component models and performance measures are used here to develop the thermodynamic basis of piston-engine, gas-turbine, and rocket propulsion cycles. Their detailed design, operation, and performance are treated in the following chapters.

Learning Objectives

  • Apply the steady-flow energy equation to nozzles, diffusers, compressors, turbines, and combustors.
  • Explain how stagnation enthalpy and stagnation temperature are used in propulsion-system analysis.
  • Define and use isentropic efficiencies for nozzles, compressors, and turbines.
  • Explain how entropy generation and exergy destruction limit the performance of real propulsion devices.
  • Describe the Otto, Diesel, Brayton, and rocket propulsion cycles and identify their main thermodynamic processes.
  • Relate compression ratio, pressure ratio, turbine inlet temperature, and cut-off ratio to ideal cycle efficiency.
  • Explain how supercharging, turbocharging, and afterburning affect power, thrust, and thermal efficiency.
  • Use fuel heating values, fuel-air ratio, equivalence ratio, and combustion efficiency in propulsion calculations.
  • Explain how rocket chamber conditions, nozzle expansion, characteristic velocity, thrust coefficient, and specific impulse are related.

Roadmap of Propulsion Thermodynamics

The thermodynamic analysis of propulsion systems begins with individual components and then builds toward complete engine cycles. The first step is to apply the First Law of Thermodynamics to steady-flow devices. For nozzles and diffusers, the dominant process is the conversion between enthalpy and kinetic energy, accompanied by changes in pressure. For compressors and turbines, the dominant exchange is between shaft work and changes in the flow’s enthalpy. Finally, for combustors, it is the conversion of fuel chemical energy into an increase in the flow’s enthalpy.

The Second Law of Thermodynamics then establishes the limits on these processes. Flows through nozzles, compressors, turbines, and combustors are irreversible because of friction, turbulence, shocks, mixing, heat transfer across finite temperature differences, and finite-rate chemical reactions, which generate entropy and destroy exergy. Isentropic and exergy efficiencies provide practical measures of how closely each component approaches the ideal reversible limit.

With these component models in place, complete propulsion cycles can then be analyzed and understood. Piston engines are represented by the Otto and Diesel cycles, gas-turbine engines by the Brayton cycle, and rocket engines by open-flow propulsion cycles in which stored propellants are converted into a high-speed jet exhaust. Combustion thermodynamics then connects the fuel chemistry, heating value, mixture ratio, and adiabatic flame temperature to the engine performance. In this way, the roadmap proceeds from component energy balances, to irreversibility and efficiency, to complete propulsion cycles and their practical performance measures.

Component Thermodynamics

Before complete propulsion cycles can be analyzed, the behavior of the individual components must be understood. Each component changes the state of the working fluid in a particular way. Nozzles and diffusers convert between static enthalpy and kinetic energy, compressors and turbines exchange shaft work with the flow, and combustors release chemical energy that increases the stagnation enthalpy of the working fluid. These devices are usually analyzed using steady-flow control volumes, so the steady-flow energy equation (SFEE) developed in Chapter 12 serves as the common starting point.

The First Law of Thermodynamics simplifies in different ways when applied to the principal components of an aerospace propulsion system. These devices are commonly modeled as steady-flow control volumes with negligible changes in potential energy. Additional assumptions, such as adiabatic operation or negligible changes in kinetic energy, are introduced as appropriate for each component. The resulting energy equations are developed in the following sections.

Nozzles & Diffusers

In a nozzle, as shown in Figure 1 for a subsonic flow, there is no shaft work, so the process is primarily a conversion between static enthalpy and kinetic energy. For a steady, adiabatic nozzle with negligible change in potential energy, the SFEE becomes

(1)   \begin{equation*} h_1+\frac{V_1^2}{2} = h_2+\frac{V_2^2}{2} \end{equation*}

or, in terms of stagnation enthalpy,

(2)   \begin{equation*} h_0=h+\frac{V^2}{2}=\text{constant} \end{equation*}

Therefore, an increase in flow velocity is accompanied by a decrease in static enthalpy. If the inlet velocity is negligible, the nozzle exit velocity can be obtained directly from

(3)   \begin{equation*} V_2 \simeq \sqrt{2\left(h_0-h_2\right)} \end{equation*}

The behavior of pressure and velocity is opposite in subsonic and supersonic flows. For subsonic flow (M<1) through a converging nozzle, the pressure decreases (p_2<p_1), the velocity increases (V_2>V_1), and the static temperature decreases (T_2<T_1).

Nozzle and diffuser characteristics for a subsonic flow.

For supersonic flow (M>1), the area rule is reversed. A diverging passage acts as a nozzle, so the pressure decreases (p_2<p_1), the velocity increases (V_2>V_1), and the static temperature decreases (T_2<T_1). A converging passage acts as a diffuser, so the pressure rises (p_2>p_1), the velocity decreases (V_2<V_1), and the static temperature increases (T_2>T_1). For an adiabatic nozzle or diffuser with no shaft work, the stagnation temperature remains constant, whereas the stagnation pressure decreases in real flows because of irreversibility.

Stagnation Temperature

In compressible flow, it is important to distinguish between the static temperature T of the moving fluid and the stagnation temperature T_0. The static temperature characterizes the local thermodynamic state and the distribution of molecular energy, as would be measured by a thermometer moving with the fluid. The stagnation temperature is the temperature the fluid would reach if it were decelerated isentropically to zero velocity.

For a calorically ideal gas,

(4)   \begin{equation*} T_0 = T\left(1+\frac{\gamma-1}{2}M^2\right) \end{equation*}

where M is the Mach number and \gamma is the ratio of specific heats. For low Mach numbers, the static and stagnation temperatures are approximately equal.

For a calorically ideal gas, stagnation enthalpy and stagnation temperature are related by

(5)   \begin{equation*} h_0=c_pT_0 \end{equation*}

Therefore, the stagnation temperature provides a measure of the flow’s total energy content per unit mass. The static temperature may change as kinetic energy is converted into static enthalpy or static enthalpy is converted into kinetic energy, while the stagnation temperature remains constant in adiabatic flow with no shaft work.

For example, when a flow is compressed adiabatically through a diffuser, its static temperature increases as kinetic energy is converted into static enthalpy. During an adiabatic expansion through a nozzle, the static temperature decreases as static enthalpy is converted into kinetic energy. In both cases, the stagnation temperature remains unchanged.

Heat transfer or shaft work changes the stagnation enthalpy, thereby changing the stagnation temperature. Compressors increase stagnation temperature by adding shaft work, turbines decrease it by extracting shaft work, and combustors increase it by releasing chemical energy into the flow.

Choked Flow

An important consequence of compressible-flow behavior is the phenomenon known as choking. The term can be misleading because a choked flow is not blocked or stopped. Instead, choking means that the mass flow rate through a passage has reached its maximum possible value for the specified upstream stagnation pressure, stagnation temperature, gas properties, and controlling flow area. The behavior in a converging-diverging nozzle is shown in Figure 2.

Pressure distributions and flow characteristics in a converging-diverging nozzle for different back-pressure ratios.

As the back pressure p_b is reduced, the gas accelerates through the converging portion of the nozzle and the mass flow rate increases. For sufficiently high back pressure, represented by Curves A and B in Figure 2, the flow remains subsonic throughout the nozzle. At the critical condition represented by Curve C, the Mach number reaches unity at the minimum-area section, or throat, while the flow remains subsonic elsewhere. The nozzle is then said to be choked.

Further reduction of the back pressure cannot increase the mass flow rate unless the upstream stagnation conditions or the throat area are changed. Instead, the downstream flow pattern changes. Curves D and E represent cases in which the flow accelerates to supersonic speed downstream of the throat and then passes through a normal shock in the diverging section, producing a transition back to subsonic flow. Curve F represents flow that remains supersonic downstream of the throat.

The physical reason for choking is associated with the propagation of pressure disturbances. In a subsonic flow, pressure disturbances can travel both upstream and downstream relative to the nozzle. At the sonic throat, however, an upstream-traveling pressure disturbance propagates through the gas at the same speed as the gas moves downstream. Its velocity relative to the nozzle is zero, so changes imposed downstream cannot propagate upstream through the throat or alter the mass flow rate.

For a calorically perfect ideal gas undergoing one-dimensional isentropic flow, the mass flow rate through an area A can be written as

(6)   \begin{equation*} \overbigdot{m} = A \, \frac{p_0}{\sqrt{T_0}} \sqrt{\frac{\gamma}{R}} \, M \left( 1+\frac{\gamma-1}{2}M^2 \right)^{-\tfrac{\gamma+1}{2(\gamma-1)}} \end{equation*}

where p_0 and T_0 are the upstream stagnation pressure and stagnation temperature. For fixed values of A, p_0, and T_0, this expression reaches its maximum value when M=1. The corresponding choked mass flow rate is

(7)   \begin{equation*} \overbigdot{m}_{\max} = A^* \, \frac{p_0}{\sqrt{T_0}} \sqrt{\frac{\gamma}{R}} \left( \frac{2}{\gamma+1} \right)^{\tfrac{\gamma+1}{2(\gamma-1)}} \end{equation*}

where A^* is the sonic throat area.

The critical static-to-stagnation pressure ratio at the sonic throat is

(8)   \begin{equation*} \frac{p^*}{p_0} = \left( \frac{2}{\gamma+1} \right)^{\tfrac{\gamma}{\gamma-1}} \end{equation*}

where the superscript * denotes the sonic condition. For air with \gamma = 1.4, then

(9)   \begin{equation*} \frac{p^*}{p_0} \approx 0.528 \end{equation*}

Therefore, the sonic pressure at the throat is approximately 52.8% of the upstream stagnation pressure. Choking begins when the back pressure is reduced sufficiently for the throat pressure to reach this critical value. Further reduction of p_b changes the downstream supersonic expansion and shock-wave structure, but it does not increase the mass flow rate.

Choked flow does not mean stopped flow.

In compressible-flow terminology, the word “choked” does not mean that the flow is obstructed or has stopped. It means that a sonic condition has formed at a controlling section and that the mass flow rate has reached its maximum value for the specified upstream stagnation state, gas properties, and flow area. Lowering the downstream pressure further may change the expansion and shock-wave structure downstream of the throat, but it cannot change the sonic throat condition or increase the mass flow rate through the nozzle. Choking is not limited to converging nozzles. A sonic controlling condition can also arise in converging-diverging nozzles, orifices, valves, ducts with friction, and flows subjected to heat addition. The detailed mechanism differs among isentropic nozzle flow, Fanno flow, and Rayleigh flow, but the defining feature remains the attainment of a limiting mass flow rate associated with a sonic condition.

Compressors & Turbines

Compressors and turbines, as illustrated in Figure 3, exchange shaft work with the working fluid. Under the control-volume sign convention used here, shaft work done by the control volume is positive, while shaft work done on the control volume is negative. In propulsion calculations, however, compressor work input and turbine work output are usually quoted as positive magnitudes: Compressor or pump: shaft work is supplied to the fluid. Turbine: shaft work is extracted from the fluid.

Compressors do work on a flow, and turbines extract work.

For a steady, adiabatic compressor with negligible changes in kinetic and potential energy, the required shaft work input per unit mass is

(10)   \begin{equation*} w_{\text{shaft,in}}=h_2-h_1 \end{equation*}

For a steady, adiabatic turbine with negligible changes in kinetic and potential energy, the shaft work output per unit mass is

(11)   \begin{equation*} w_{\text{shaft,out}}=h_1-h_2 \end{equation*}

In a compressor, the addition of shaft work increases the fluid pressure and stagnation temperature, so generally p_2>p_1 and T_{0,2}>T_{0,1}. The change in flow velocity is usually modest because the compressor is designed primarily to raise the stagnation pressure rather than the flow velocity. In a turbine, shaft work is extracted, so the pressure and stagnation temperature decrease. The flow velocity may increase or decrease depending on the blade design and the distribution of expansion between the stator and rotor.

Mach number strongly affects losses, choking, shock formation, and component efficiency. Excessive relative Mach numbers in compressor blade passages can produce shock waves, entropy generation, flow separation, and reduced pressure rise. For this reason, compressor designers carefully manage the relative Mach number through the rotor and stator passages. Many compressor stages operate with predominantly subsonic relative flow, although modern high-pressure-ratio axial compressors may contain transonic regions, especially near the blade tips.

Turbines may also contain both subsonic and supersonic regions, particularly in nozzle guide vanes where the hot gases are accelerated before entering the rotor. Controlled acceleration and expansion are acceptable, but uncontrolled shocks, flow separation, leakage, and highly nonuniform inflow reduce efficiency and blade durability. Therefore, both compressors and turbines require careful management of the local and relative Mach numbers, although the aerodynamic constraints differ between the two devices.

Combustors

In a combustor, there is no shaft work, and chemical energy released by the fuel increases the stagnation enthalpy of the working fluid. If the combustor is treated as adiabatic and changes in kinetic and potential energy are neglected, a commonly used approximate energy balance per unit mass of air is

(12)   \begin{equation*} h_{0,\rm out}-h_{0,\rm in} \approx \eta_b\,f\,\mathrm{LHV} \end{equation*}

or equivalently,

(13)   \begin{equation*} \eta_b \approx \frac{h_{0,\rm out}-h_{0,\rm in}} {f\,\mathrm{LHV}} \end{equation*}

where \scriptstyle f is the fuel-air ratio, \eta_b is the combustion efficiency, and LHV is the lower heating value of the fuel.

The lower heating value excludes the recovery of latent heat associated with water vapor in the combustion products. This convention is appropriate for aerospace propulsion systems because the exhaust water normally remains in the vapor phase and its latent heat is not recovered.

A more complete combustor balance accounts for the added fuel mass and the enthalpy of the incoming fuel, i.e.,

(14)   \begin{equation*} h_{0,\rm in} + f\,h_{f,\rm in} + \eta_b\,f\,\mathrm{LHV} = (1+f)\,h_{0,\rm out} \end{equation*}

on a per-unit-mass-of-air basis. When \scriptstyle f is small, and the fuel sensible enthalpy is negligible relative to its heating value, the simpler approximate relation is often adequate.

Check Your Understanding #1 – Flow through a nozzle

Air enters a nozzle with negligible velocity at a stagnation temperature of T_0 = 900 K. Assuming ideal-gas behavior with c_p = 1,005 J kg^{-1} K^{-1}, determine the exit velocity if the exit static temperature is T = 600 K.

Show solution/hide solution.

For a nozzle (no shaft work) and negligible heat transfer, the total enthalpy is conserved, i.e.,

    \[ h_0 = h + \frac{V^2}{2} \]

With h = c_p T, the flow velocity is

    \[ V = \sqrt{2c_p (T_0 - T)} \]

Substituting the numerical values gives

    \[ V = \sqrt{2 \times 1{,}005 \times (900-600)} = \sqrt{603{,}000} = 776.5~\mathrm{m/s} \]

Check Your Understanding #2 – Work input to a compressor

Air enters a compressor at T_1 = 300 K and exits at T_2 = 450 K. Assuming constant specific heat of c_p = 1,005 J kg^{-1} K^{-1}, calculate the shaft work input per unit mass of air.
Show solution/hide solution.

The shaft work for an adiabatic compressor (neglecting potential and kinetic energy changes) is the enthalpy change, i.e.,

    \[ w_{\text{shaft}} = h_2 - h_1 = c_p (T_2 - T_1) \]

Substituting the values gives

    \[ w_{\text{shaft}} = 1,005 \times (450 - 300) = 150{,}750~\mathrm{J\,kg^{-1}} \]

Therefore, the required shaft work is 150.8 kJ/kg of air.

Second Law Efficiencies

The performance of real aerospace devices is limited by irreversibilities, including friction, turbulence, and losses associated with shock-wave formation. To quantify these effects, each device can be compared with an idealized reversible process in which entropy is constant (isentropic). The ratio of ideal to actual performance defines an isentropic efficiency. The concept of isentropic efficiency provides a convenient way to quantify the losses in practical components relative to their idealized, reversible counterparts.

For a compressor, a given pressure ratio requires a larger actual enthalpy rise than in the ideal isentropic case. This efficiency is defined as

(15)   \begin{equation*} \eta_c = \frac{h_{2s} - h_1}{h_2 - h_1} \end{equation*}

where h_{2s} denotes the enthalpy at the exit state if the compression were isentropic.

In a turbine, the actual enthalpy drop is smaller than the ideal isentropic enthalpy drop for the same pressure ratio. The efficiency is expressed in terms of the positive work output as

(16)   \begin{equation*} \eta_t = \frac{h_1 - h_2}{h_1 - h_{2s}} \end{equation*}

where h_{2s} is the enthalpy at the turbine exit in the ideal isentropic expansion.

For a nozzle, the actual exhaust velocity is less than the ideal isentropic value obtained from the same inlet state and exit static pressure. The nozzle efficiency is written as

(17)   \begin{equation*} \eta_n = \frac{V_{\text{actual}}^{2}-V_1^2} {V_{\text{ideal}}^{2}-V_1^2} = \frac{h_1-h_2} {h_1-h_{2s}} \end{equation*}

where h_{2s} is the exit enthalpy for an isentropic expansion from the same inlet state to the same exit pressure. When the inlet velocity is negligible, this expression reduces to \eta_n=V_{\text{actual}}^2/V_{\text{ideal}}^2.

Therefore, these efficiencies measure how closely real devices approach the reversible limit. A value of unity, although practically unattainable, corresponds to an ideal isentropic process, while lower values reflect greater entropy generation. The Second Law sets the ultimate bounds on propulsion performance. Compressors and turbines require multiple stages to reduce losses; supersonic nozzles must be carefully contoured to minimize shock-induced irreversibility; and even advanced rocket engines cannot convert all chemical energy into thrust. Minimizing entropy generation remains a guiding principle of aerospace thermodynamics.

Entropy, Availability, & Exergy Analysis

The Second Law of Thermodynamics establishes that entropy generation is nonnegative in real processes and also introduces the idea that not all energy can be converted into useful work. In aerospace propulsion, this is a crucial concept because the chemical energy of fuel or the enthalpy of a hot gas stream must be transformed into thrust or shaft power. However, irreversibilities limit the amount of that energy available for practical purposes. The framework for quantifying this available portion is called exergy analysis.

Entropy Generation & Irreversibility

When entropy is generated in a process, part of the available energy is degraded into a form that cannot be recovered as useful work. The entropy balance for a control volume is

(18)   \begin{equation*} \frac{dS_{\text{cv}}}{dt} = \sum_{\text{in}} \overbigdot{m} \, s - \sum_{\text{out}} \overbigdot{m} \, s + \sum_j \frac{\overbigdot{Q}_j}{T_{b,j}} + \overbigdot{S}_{\text{gen}} \end{equation*}

where S_{\text{cv}} is the total entropy in the control volume, s is the specific entropy carried by the mass flow, \overbigdot{Q}_j is the heat-transfer rate across boundary portion j, T_{b,j} is the corresponding boundary temperature, and \overbigdot{S}_{\text{gen}} \geq 0 represents entropy generation. This term quantifies irreversibilities resulting from friction, turbulence, mixing, chemical reactions, and shock waves. In aerospace systems, minimizing \overbigdot{S}_{\text{gen}} is directly linked to maximizing an engine’s efficiency.

Availability (Exergy)

Availability or exergy is the maximum theoretical work that can be obtained from a system when it comes into equilibrium with its surroundings. It measures the useful portion of the energy relative to a reference environment. In aerospace propulsion, the reference environment is typically defined by the ambient temperature T_{\infty} and ambient pressure p_{\infty}.

The specific physical flow exergy, excluding chemical exergy, is defined as

(19)   \begin{equation*} \psi = (h - h_{\infty}) - T_{\infty} (s - s_{\infty}) + \frac{V^2}{2} + g \, (z-z_{\infty}) \end{equation*}

where h, s, and V are the local flow properties, and the subscript {\infty} denotes the reference environment. The combined quantity (h-h_{\infty})-T_{\infty}(s-s_{\infty}) represents the physical exergy associated with the thermodynamic state of the flow relative to the environment. The last two terms, V^2/2 and g\,(z-z_{\infty}), account for the kinetic energy and the potential energy relative to the reference environment, respectively.

This formulation highlights the direct connection between thermodynamics and propulsion by quantifying the portion of the flow energy that can, in principle, be converted into thrust or shaft power under ideal conditions.

Exergy Destruction

In real processes, exergy is destroyed by irreversibility. The rate of exergy destruction is proportional to entropy generation, i.e.,

(20)   \begin{equation*} \overbigdot{X}_{\text{destroyed}} = T_{\infty} \, \overbigdot{S}_{\text{gen}} \end{equation*}

which shows that each increment in entropy generation reduces the maximum useful work that can be extracted.

In a compressor, irreversibilities such as viscous losses and boundary layers degrade the flow’s exergy. In turbines, additional sources of destruction include leakage through tip clearances and shock interactions in transonic or supersonic stages. In a nozzle, shock waves and non-isentropic expansions generate entropy, thereby reducing the thrust that could otherwise be achieved. In a combustor, the dominant causes are chemical-reaction irreversibility, mixing, heat transfer across finite temperature differences, and stagnation-pressure loss. Incomplete combustion may introduce an additional loss when combustion efficiency is less than unity. Taken together, these examples illustrate how an exergy analysis provides a unifying framework for quantifying losses across all propulsion components.

Exergy Efficiency

While the isentropic efficiencies can be used to compare real processes to reversible baselines, exergy efficiency measures how effectively available energy is utilized, i.e.,

(21)   \begin{equation*} \eta_{\text{exergy}} = \frac{\text{\small Useful power output}}{\text{\small Exergy input rate}} \end{equation*}

For example, in a gas turbine, the exergy input rate is primarily associated with the chemical exergy rate of the fuel, and the useful output may be shaft or propulsive power. In rocket engines, an exergy analysis quantifies the fraction of the propellant’s chemical potential lost to irreversibilities during mixing, combustion, and nozzle expansion.

Implications for Aerospace Propulsion

An exergy analysis identifies where improvements can yield the most significant gains in propulsion efficiency. In compressors and turbines, for example, pressure ratios are distributed across multiple stages, allowing each stage to operate much closer to an ideal reversible process, thereby reducing overall irreversibility. In nozzles operating at supersonic speeds, careful contouring is necessary to prevent shock waves and minimize entropy generation.

Combustors are designed to produce nearly complete, uniform combustion, as incomplete combustion and poor mixing result in substantial exergy destruction. Ultimately, the success of high-bypass-ratio turbofans hinges on accelerating a large mass flow with only a small change in velocity, thereby minimizing the kinetic energy and exergy carried away in the exhaust.

Ultimately, the Second Law and an exergy analysis provide not just a statement of thermodynamic limits, but also a diagnostic framework. By identifying where entropy is generated and exergy is destroyed, engineers can focus design efforts on minimizing irreversibility and maximizing the useful work obtained from each component of a propulsion system.

Thermodynamics of Internal Combustion Engines

An internal combustion (IC) engine is a device in which the combustion of a fuel with an oxidizer, most often ambient air, occurs within the engine itself, as illustrated in the schematic in Figure 4. The fundamental idea is that the chemical energy in the fuel is released through combustion, producing hot, high-pressure gases. These gases expand within the engine and perform work on moving components, such as pistons in reciprocating engines or turbine blades in gas turbines, or they are accelerated through a nozzle to generate a propulsive jet.

An IC engine is a device in which the combustion of a fuel with an oxidizer inside the engine creates useful work.

The overall energy transformation may be summarized as

(22)   \begin{equation*} \text{\small Fuel + Air (input)} \;\;\xrightarrow{\;\;\text{\normalsize Combustion}\;\;}\;\; \text{\small Power or thrust (output) + Exhaust gases + Heat} \end{equation*}

This description is entirely general and applies to all IC engine types. Reciprocating piston engines, rotary engines, gas turbines, and jet engines are all specific realizations of the same principle, i.e., the direct conversion of fuel chemical energy into useful mechanical power or thrust through combustion within the engine. In every case, the outputs also include exhaust products and waste heat, which are necessary byproducts because no engine can achieve 100% efficiency.

Gas Power Thermodynamic Cycles

The First and Second Laws of thermodynamics describe how energy and entropy change in individual processes. In practice, aerospace propulsion and power systems are analyzed using thermodynamic cycles or corresponding open-flow process sequences. In an ideal closed cycle, the working fluid undergoes a series of processes and ultimately returns to its initial state, whereas an open propulsion system continuously admits and exhausts mass. The net effect of a cycle is to produce useful work from a flow (as in engines and turbines) or to absorb work otherwise, such as to drive a pump or compressor.

Thermodynamic cycles are central to aerospace engineering analyses because they set the limits of propulsive performance. The thermal efficiency of a piston engine, the specific thrust of a turbojet, and the exhaust velocity of a rocket all trace back to the principles of these idealized cycles, corrected for actual irreversibilities. By studying the classic cycles, engineers can identify what is achievable, what is lost to irreversibility, and where design improvements have the greatest impact. Although there are more, five cycles are important in aerospace applications:

  • Otto cycle: The model for spark-ignition piston engines, relevant to all types of piston-powered airplanes.
  • Diesel cycle: The model for compression-ignition engines, although not widely used in aviation.
  • Brayton cycle: The fundamental cycle for gas turbines, turbofans, turbojets, and turboprops, making it the workhorse of modern aviation.
  • Rankine cycle: A vapor power cycle, not typical for flight propulsion, but essential in auxiliary and space-based power systems.
  • Rocket propulsion cycles: Variants of open and closed cycles that rapidly convert chemical energy directly into high-velocity exhaust jets to produce thrust.

Otto Cycle (Spark-Ignition Engines)

The Otto cycle is the idealized thermodynamic cycle that represents the operation of a spark-ignition piston-powered IC engine. Named after Nikolaus Otto, who built the first successful four-stroke piston engine in 1876, the cycle represents the basis of modern gasoline-fueled engines. It consists of four internally reversible processes applied to a fixed mass of working gas. Under the air-standard approximation, the working fluid is modeled as an ideal gas consisting of air alone, and combustion and exhaust processes are replaced by equivalent heat-addition and heat-rejection processes.

Process Stages

The cycle begins at state 1, with the piston at bottom dead center (BDC) and the cylinder containing the fixed mass of air used in the air-standard model, as shown in Figure 5. The first process, state 1 to 2, is an isentropic compression in which the piston moves upward in the cylinder with the inlet and exhaust valves closed, reducing the cylinder volume from the maximum value \mathcal{V}_1 at bottom dead center (BDC) to the minimum value \mathcal{V}_2 at top dead center (TDC). The pressure and temperature rise significantly during this compression process, but no heat is exchanged with the surroundings. The ratio \mathcal{V}_1/\mathcal{V}_2 is called the geometric compression ratio of the engine, and is typically at least 8.5:1 in normally aspirated (non-supercharged) forms. Supercharging increases the intake manifold pressure and the mass of charge inducted into the cylinder, thereby increasing the intake charge density and the mass of air-fuel mixture inducted into the cylinder, and so the engine’s power output.

The process stages, as depicted by the p\mathcal{V} and Ts diagrams of the Otto cycle.

The second process, from state 2 to state 3, represents the rapid combustion initiated by the spark plug in the real engine. In the air-standard cycle, combustion is replaced by constant-volume heat addition to the fixed mass of air, so the volume remains at \mathcal{V}_2 while the pressure and temperature rise as an amount of heat per unit mass, q_{\text{in}}, is supplied to the working fluid. The third process, from state 3 to 4, is an isentropic expansion. The piston moves downward from TDC back to BDC, increasing the volume from \mathcal{V}_2 to \mathcal{V}_1. During this expansion, the hot gas does work on the piston, causing the pressure and temperature to fall accordingly. This is the engine’s power stroke. The fourth process, state 4 to 1, closes the ideal cycle. This is a constant-volume heat-rejection process in which heat is removed from the fixed mass of working gas. The pressure and temperature drop at fixed volume \mathcal{V}_1 by an amount corresponding to the rejected heat per unit mass, q_{\text{out}}. In a real four-stroke engine, this idealized process corresponds only approximately to the exhaust blowdown and subsequent exhaust stroke.

Process Diagrams

The ideal p\mathcal{V} diagram of the Otto cycle (shown in Figure 5 above) consists of two adiabatic (isentropic) curves, representing compression and expansion, joined by two vertical constant-volume lines, representing the parts of the cycle where heat is added and heat is rejected. On the p\mathcal{V} diagram, the compression and expansion legs are curves because isentropic processes follow p \,\mathcal{V}^{\,\gamma} = constant. Because the force on the piston is the instantaneous cylinder pressure times the piston area, i.e., F_p(t) = p(t)\, A_p, the incremental work during an infinitesimal piston displacement dx is

(23)   \begin{equation*} d W = F_p\, dx = p(t)\, A_p \, dx \end{equation*}

The corresponding change in cylinder volume is A_p\, dx = d\mathcal{V}, then, over one complete thermodynamic cycle, the net work is

(24)   \begin{equation*} W = \oint dW = \oint p \, d\mathcal{V} \end{equation*}

which is the (signed) area enclosed by the closed loop on the p\mathcal{V} diagram. A clockwise loop gives W > 0 (engine: net work output), whereas a counterclockwise loop gives W < 0 (compressor/pump: net work input).

The corresponding Ts diagram shows two vertical isentropic lines and two constant-volume curves. During the constant-volume process from state 2 to 3, heat is added to the gas, raising its temperature from T_2 to T_3. Conversely, during the process from state 4 to 1, heat is rejected, reducing the temperature from T_4 back to T_1. The heat-addition and heat-rejection legs are curved constant-pressure and constant-volume processes, respectively. For the constant-volume heat-addition process, then

(25)   \begin{equation*} s_3 - s_2 = c_v \ln\!\left(\frac{T_3}{T_2}\right) \end{equation*}

whereas for the constant-volume heat-rejection process, then

(26)   \begin{equation*} s_1 - s_4 = c_v \ln\!\left(\frac{T_1}{T_4}\right) \end{equation*}

For the internally reversible idealized processes shown on a Ts diagram, the heat transfer per unit mass along each reversible portion of the cycle is represented by the area under the process curve, i.e.,

(27)   \begin{equation*} q = \int T \, ds \end{equation*}

The area enclosed by the closed loop indicates the net heat exchanged per unit mass, which equals the net work per unit mass of the cycle according to the First Law of Thermodynamics. Therefore, the specific heat interactions are

(28)   \begin{equation*} q_{\text{in}} = c_v\,(T_3 - T_2) \qquad \text{and} \qquad q_{\text{out}} = c_v\,(T_4 - T_1) \end{equation*}

so the net specific work is w_{\text{net}} = q_{\text{in}} - q_{\text{out}}. From the isentropic compression and expansion relations, then

(29)   \begin{equation*} \frac{T_2}{T_1} = r^{\,\gamma - 1} \qquad \text{and} \qquad \frac{T_3}{T_4} = r^{\,\gamma - 1} \end{equation*}

where r=\mathcal{V}_1/\mathcal{V}_2 is the compression ratio. Substituting these relations gives the thermal efficiency of the Otto cycle as

(30)   \begin{equation*} \eta_{\text{Otto}} = 1 - \frac{1}{r^{\,\gamma - 1}} \end{equation*}

which shows that efficiency depends solely on the compression ratio and the specific heat ratio. In practice, r is limited by abnormal combustion, particularly end-gas autoignition or “knock,” which is strongly influenced by the octane rating of the fuel.

Diesel Cycle (Compression-Ignition Engines)

The Diesel cycle is the idealized thermodynamic model for compression-ignition internal combustion engines. First demonstrated by Rudolf Diesel in 1897, it utilizes self-ignition (auto-ignition), in which fuel is injected into highly compressed, hot air, causing it to burn spontaneously. Diesel engines are valued for their high efficiency, durability, and ability to operate on heavier fuels than gasoline engines. They are used extensively for automobiles, trucks, buses, ships, locomotives, and stationary power generation. In aerospace, their application has been more limited, although lightweight compression-ignition engines have been adapted for certain UAVs and light aircraft.

The efficiency advantage of diesel engines stems primarily from their higher compression ratios, which allow a larger fraction of the fuel energy to be converted into work, as well as from lean operation and relatively low throttling losses. Their robust construction also provides them with a long service life and the ability to withstand continuous high-load operation. However, disadvantages include greater weight, higher initial cost, and increased particulate and nitrogen oxide emissions, all of which necessitate sophisticated after-treatment systems to meet environmental standards.

Process Stages

The Diesel cycle consists of four internally reversible processes applied to a fixed mass of an ideal gas, as shown in Figure 6. The first process, from state 1 to state 2, is an isentropic compression from the maximum cylinder volume \mathcal{V}_1 at BDC to the clearance volume \mathcal{V}_2 at TDC, which raises pressure and temperature without heat transfer. The second process, from state 2 to 3, involves adding heat to the fixed mass of working gas at constant pressure. During this idealized process, the piston moves downward in the cylinder, increasing the temperature and volume from \mathcal{V}_2 to \mathcal{V}_3 at essentially constant pressure. In a real Diesel engine, this heat-addition process corresponds approximately to fuel injection and auto-ignition in the hot compressed air. The third process, from state 3 to 4, is an isentropic expansion back to \mathcal{V}_1, during which the gas does work on the piston, and its pressure and temperature decrease. The fourth process, from state 4 to 1, is a constant-volume heat-rejection process that removes heat from the fixed mass of the working gas, thereby reducing pressure and temperature while holding volume constant to complete the ideal cycle. In a real engine, this idealized process corresponds only approximately to exhaust blowdown and the subsequent exhaust stroke.

The process stages with p\mathcal{V} and Ts diagrams of the Diesel cycle.

Process Diagrams

The ideal p\mathcal{V} diagram of the Diesel cycle consists of two adiabatic (isentropic) curves, representing compression and expansion, joined by a horizontal constant-pressure line for heat addition and a vertical constant-volume line for heat rejection. The heat-addition and heat-rejection legs are curved constant-volume processes. For the constant-pressure heat-addition process,

(31)   \begin{equation*} s_3 - s_2 = c_p \ln\!\left(\frac{T_3}{T_2}\right) \end{equation*}

whereas for the constant-volume heat-rejection process,

(32)   \begin{equation*} s_1 - s_4 = c_v \ln\!\left(\frac{T_1}{T_4}\right) \end{equation*}

The corresponding Ts diagram has two vertical isentropic lines, a curve representing constant-pressure heat addition, and another curve representing constant-volume heat rejection. Heat is input during states 2 to 3 at constant pressure, while heat is rejected during states 4 to 1 at constant volume.

The specific heat interactions are

(33)   \begin{equation*} q_{\text{in}} = c_p\,(T_3 - T_2) \qquad \text{and} \qquad q_{\text{out}} = c_v\,(T_4 - T_1) \end{equation*}

with the net specific work w_{\text{net}} = q_{\text{in}} - q_{\text{out}}. Using the temperature relations for isentropic compression and expansion, then

(34)   \begin{equation*} \frac{T_2}{T_1} = r^{\,\gamma - 1} \qquad \text{and} \qquad \frac{T_4}{T_3} = \left(\frac{\beta}{r}\right)^{\,\gamma - 1} \end{equation*}

and noting that T_3/T_2=\beta, where r=\mathcal{V}_1/\mathcal{V}_2 is the compression ratio and \beta =\mathcal{V}_3/\mathcal{V}_2 is the cutoff ratio. Therefore, the efficiency becomes

(35)   \begin{equation*} \eta_{\text{Diesel}} = 1 - \frac{1}{r^{\,\gamma - 1}} \left(\frac{\beta^{\,\gamma} - 1}{\gamma(\beta - 1)}\right) \end{equation*}

This expression indicates that Diesel efficiency depends on both the attainable compression ratio and the cutoff ratio. Higher compression ratios yield improved performance, and the cutoff ratio moderates the effect of heat addition. In practice, Diesel engines often achieve higher indicated and brake thermal efficiencies than comparable spark-ignition engines, particularly at reduced loads, because they tolerate higher compression ratios and operate with overall lean mixtures.

Cycle Work and Power

For any reciprocating engine, the cylinder volume varies between the clearance volume at TDC, denoted \mathcal{V}_{\rm TDC}, and the maximum volume at BDC, denoted \mathcal{V}_{\rm BDC}. The displacement (swept) volume is \mathcal{V}_d = \mathcal{V}_{\rm BDC} - \mathcal{V}_{\rm TDC}. It is convenient to normalize this to the displacement volume by defining the indicated mean effective pressure (IMEP), i.e.,

(36)   \begin{equation*} \mathrm{IMEP} = \frac{\displaystyle{\oint p \, d\mathcal{V}}} {\mathcal{V}_d} = \frac{W}{\mathcal{V}_d} \end{equation*}

This value is the hypothetical constant-pressure equivalent which, if applied during the expansion over \mathcal{V}_d, would yield the same net cycle work. If the crankshaft rotates at N revolutions per minute and the engine has n_{\rm cyl} cylinders, the cycle frequency per cylinder is f_{\rm cyc} = N/120 for a four-stroke cycle and f_{\rm cyc} = N/60 for a two-stroke cycle, expressed in cycles per second. The indicated power of the engine is then

(37)   \begin{equation*} P_{\rm ind} = \overbigdot{W}_{\rm ind} = n_{\rm cyl}\, f_{\rm cyc}\, W = n_{\rm cyl}\, f_{\rm cyc} \, \mathrm{IMEP} \,\mathcal{V}_d \end{equation*}

Therefore, the power produced from the Otto or Diesel cycles is directly proportional to the indicated mean effective pressure, the total displacement volume, and the number of thermodynamic cycles per second.

Supercharging and Turbocharging

Aircraft piston engines operating on the Otto or Diesel cycles experience a progressive loss of power as altitude increases. The underlying reason is the decrease in ambient air pressure and density at higher altitudes, which reduces the mass of air inducted into the engine per cycle. Because engine power depends directly on the mass of the air-fuel mixture that can be compressed and burned, altitude severely limits the performance of a normally aspirated intake system. The solution is to raise the intake pressure by compressing the air before it enters the cylinder, a process known as forced induction, which increases the IMEP value.

As shown in Figure 7, supercharging uses a mechanically driven compressor, powered directly by the crankshaft, to increase manifold pressure. Turbocharging utilizes a turbine positioned in the exhaust stream to harness energy from the exhaust gases, which then drives a compressor. Both methods achieve the same basic goal: raising the pressure and density of the intake charge to restore power otherwise lost at altitude. In either case, the thermodynamic process involves compressing the working fluid before the cycle begins.

Superchargers and turbochargers can maintain the power output from aeroengines at higher flight altitudes.

For an ideal gas undergoing an isentropic compression process, the relation between pressure and temperature is

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

but in all compressors, the actual temperature rise exceeds this ideal value. An intercooler or charge-air cooler can be installed downstream of a compressor or between compression stages to lower the air temperature. Intercooling between stages reduces the work required for a specified overall pressure ratio, while charge cooling also increases the density of the air supplied to the engine. The isentropic efficiency can be introduced to quantify the irreversibilities, i.e.,

(39)   \begin{equation*} \eta_c = \frac{T_{2s} - T_1}{T_2 - T_1} \end{equation*}

where T_{2s} is the ideal (isentropic) exit temperature for a given pressure ratio and T_2 is the actual exit temperature. This framework applies equally to mechanically driven superchargers and turbine-driven turbochargers.

The effect of turbo-boosting can be visualized directly on the p\mathcal{V} and Ts diagrams of the Otto and Diesel cycles. In the p\mathcal{V} diagram, compression begins from a higher intake pressure, so the entire cycle is elevated relative to the unboosted case. This increases both the indicated mean effective pressure and the net work per cycle. On the Ts diagram, boosting changes the intake state and modifies the subsequent compression and expansion paths. The practical benefit is that the enclosed cycle area, proportional to the work output, increases with turbocharging, even though the fundamental thermodynamic cycle remains the same.

In spark-ignition (Otto) engines, the degree of supercharging is limited by abnormal combustion, particularly end-gas autoignition or knock, which constrains the allowable combination of compression ratio, boost pressure, mixture strength, and ignition timing. In compression-ignition (Diesel) engines, supercharging is widely used to improve both power and efficiency because knock is not a limiting factor. In modern practice, exhaust-gas-driven turbochargers are the most common method for boosting power in Diesel engines.

Check Your Understanding #3 – Power from a single cylinder piston engine

A single-cylinder four-stroke engine has a displacement volume \mathcal{V}_d = 500 cm^{3} = 5.0\times 10^{-4} m^{3} and operates at N = 3,000 rpm. The indicated mean effective pressure is \mathrm{IMEP} = 900 kPa.  Determine the net indicated work per cycle and the indicated power output for this engine.

Show solution/hide solution.

The net indicated work per cycle is found by multiplying the IMEP by the displacement volume. With \mathrm{IMEP} = 9.0\times 10^{5}~\mathrm{Pa} and \mathcal{V}_d = 5.0\times 10^{-4}~\mathrm{m}^3, the result is

    \[ W = \mathrm{IMEP}\,\mathcal{V}_d = (9.0\times 10^{5})(5.0\times 10^{-4}) = 450~\mathrm{J/cycle} \]

In a four-stroke engine, one thermodynamic cycle is completed every two crankshaft revolutions. At 3,000 rpm, the cycle frequency is

    \[ f_{\rm cyc} = \frac{N}{120} = \frac{3000}{120} = 25~\mathrm{cycles/s} \]

Multiplying the work per cycle by the cycle frequency gives the indicated power for this single cylinder, i.e.,

    \[ P_{\rm ind} = f_{\rm cyc}\,W = 25 \times 450 = 1.125\times 10^{4}~\mathrm{W} = 11.25~\mathrm{kW} \]

Therefore, the engine develops about 11.3 kW of indicated power under these conditions. Notice that for a multi-cylinder engine, the total indicated power is obtained by multiplying by the number of cylinders.

Brayton Cycle (Gas Turbine Engines)

The Brayton cycle is the fundamental thermodynamic model for gas turbine engines, which power most of modern aviation in the form of turbojets, turbofans, and turboprops. First proposed in the 1870s by George Brayton for an early gas engine, the cycle became practical only with the development of axial compressors and turbines in the 20th century. It remains central to aerospace propulsion because it shows how chemical energy in fuel is converted into a high-velocity exhaust jet, as in a turbojet, or into mechanical shaft work, as in turbofans and turboprops, thereby producing thrust.

Process Stages

The ideal Brayton cycle comprises four internally reversible processes, as illustrated in the schematic of Figure 8. The first process is an isentropic compression, in which air is compressed adiabatically from state 1 to state 2. The second process is constant-pressure heat addition, in which combustion of the fuel in the combustor raises the temperature of the working fluid from state 2 to state 3 while maintaining essentially constant pressure. The third process is an isentropic expansion, during which the high-temperature gas expands adiabatically through the turbine, producing shaft work as it expands from state 3 to state 4. The final process is constant-pressure heat rejection, which, conceptually, returns the working fluid to its initial state in the ideal closed-cycle representation. In a gas turbine or turbojet, the exhaust is discharged into the atmosphere, and any remaining useful enthalpy may be expanded through a nozzle to produce thrust.

Schematic of the Brayton cycle with the compressor, combustor, and a power-generating turbine.

A practical aircraft gas-turbine engine operates as an open system. However, in the ideal Brayton cycle, the working fluid can be imagined to follow a closed loop, allowing consistent representation of thermodynamic processes using known elements of thermodynamic modeling, as shown in Figure 9. Heat rejection from state 4 back to state 1 conceptually closes the cycle, even though in a real gas turbine, the exhaust is discharged to the atmosphere and replaced by new fresh air at the inlet. When the cycle is depicted in this manner, two external heat exchangers are necessary: one to add heat and another to reject heat, thereby returning the fluid to its initial state.

Schematic of the idealized closed-Brayton cycle with compressor, two heat exchangers, and a power-generating turbine.

Process Diagrams

On the pressure-volume (p\mathcal{V}) diagram as shown in Figure 10, the idealized Brayton cycle is represented by two adiabatic compression and expansion curves connected by two constant-pressure processes. The compression process from state 1 to 2 is an isentropic rise in pressure with decreasing volume, the heat addition process from state 2 to 3 is a horizontal line of constant pressure with increasing volume, the expansion process from 3 to 4 is an isentropic pressure drop with increasing volume, and the heat rejection process from 4 to 1 is a horizontal line at constant pressure returning to the initial state.

The p\mathcal{V} and Ts process diagrams of the idealized Brayton cycle.

The corresponding Ts diagram shows two vertical isentropic lines and two curved constant-pressure segments. The area enclosed by the cycle represents the net work output per unit mass for the ideal cycle. Heat is added during states 2 to 3 at constant pressure, whereas heat is rejected during states 4 to 1 at constant pressure.

Notice that in most textbooks, the Brayton cycle is usually drawn as a closed loop on Ts and p\mathcal{V} diagrams, with a heat-rejection leg (shown here as a dashed line) returning to the initial state. In reality, a jet engine is an open cycle in that the working fluid enters once and is exhausted to the atmosphere. The state 4-to-1 process in the ideal Brayton cycle (or state 5-to-ambient in the afterburning case) is only a conceptual closure, representing cooling by the environment in accordance with the principle of energy conservation. This simplification enables cyclic analysis, whereas the actual engine operates continuously, with the atmosphere serving as the sink.

The specific heat interactions are

(40)   \begin{equation*} q_{\text{in}} = c_p\,(T_3 - T_2) \qquad \text{and} \qquad q_{\text{out}} = c_p\,(T_4 - T_1) \end{equation*}

and the turbine and compressor work values per unit mass are

(41)   \begin{equation*} w_t = c_p\,(T_3 - T_4) \qquad \text{and} \qquad w_c = c_p\,(T_2 - T_1) \end{equation*}

so that w_{\text{net}} = w_t - w_c = q_{\text{in}} - q_{\text{out}}. From the isentropic relations, then

(42)   \begin{equation*} \frac{T_2}{T_1} = r_p^{(\gamma-1)/\gamma} \qquad \text{and} \qquad \frac{T_4}{T_3} = \frac{1}{r_p^{(\gamma-1)/\gamma}} \end{equation*}

where r_p = p_2/p_1 is the compressor pressure ratio. Substituting gives the thermal efficiency as

(43)   \begin{equation*} \eta_{\text{Brayton}} = 1 - \frac{T_4 - T_1}{T_3 - T_2} = 1 - \frac{1}{r_p^{(\gamma - 1)/\gamma}} \end{equation*}

Therefore, for the ideal cold-air-standard Brayton cycle with constant specific heats, the thermal efficiency increases with compressor pressure ratio. Increasing the turbine inlet temperature does not directly change this ideal efficiency, but it increases the specific work output and significantly improves practical engine performance, subject to limits imposed by materials, cooling, compressor design, and aircraft weight.

Back-Work Ratio

In a gas turbine cycle, the turbine must deliver sufficient work to drive the compressor and, depending on the engine type, a fan, a propeller, a rotor, or a gearbox. In turbojets and turbofans, propulsion is then produced by expanding the remaining high-enthalpy flow through a suitably shaped nozzle; in turboprops and turboshafts, additional turbine work is extracted as useful shaft power. The turbine work output per unit mass is

(44)   \begin{equation*} w_{\text{turb}} = c_p\,(T_3 - T_4) \end{equation*}

and the compressor work input per unit mass is

(45)   \begin{equation*} w_{\text{comp}} = c_p\,(T_2 - T_1) \end{equation*}

so that the net specific cycle work is

(46)   \begin{equation*} w_{\text{net}} = w_{\text{turb}} - w_{\text{comp}} \end{equation*}

The fraction of the turbine work consumed by the compressor is called the back-work ratio (BWR). It is defined as

(47)   \begin{equation*} \text{BWR} = \frac{w_{\text{comp}}}{w_{\text{turb}}} = \frac{c_p\,(T_2 - T_1)}{c_p\,(T_3 - T_4)} = \frac{T_2 - T_1}{T_3 - T_4} \end{equation*}

A high BWR value indicates that a significant fraction of the turbine output is used internally to drive the compressor, leaving less net work available. On the Ts diagram shown above, this can be illustrated by splitting the turbine expansion into two steps, i.e., state 3 to 4′ supplies exactly the compressor work. In contrast, state 4′ to 4 represents the net turbine work output. In many simple gas-turbine cycles, the back-work ratio can be a large fraction of the turbine output, often on the order of 40–60%. This analysis illustrates why gas turbine performance is highly dependent on achieving exceptionally high turbine inlet temperatures; the hotter the working fluid entering the turbine, the more expansion work is available to cover the compressor demand and still provide useful net power.

The interpretation of the BWR depends on the gas-turbine configuration. In a simple shaft-power Brayton cycle, the turbine produces more work than is required by the compressor, and the difference is available as net shaft work. The BWR then directly indicates the fraction of the total turbine output consumed by the compressor.

In a turbojet, however, the turbine is normally designed to extract approximately the work required to drive the compressor and accessories. With the definition used here, the ratio of compressor work to turbine work is therefore close to unity. The substantial stagnation enthalpy remaining downstream of the turbine is not counted as turbine work but is expanded through the exhaust nozzle to produce thrust. In a turbofan, the turbine must drive both the compressor and the fan, so nearly all of the extracted turbine work is likewise consumed internally.

In a turboprop or turboshaft engine, additional turbine expansion produces useful shaft power for the propeller, rotor, or gearbox. If the total work from all turbine stages is used in the denominator, the compressor work represents a smaller fraction of that total. Therefore, the BWR is most directly useful for shaft-power Brayton cycles and must be interpreted carefully when applied to thrust-producing engines.

Afterburning (Reheat) Cycle

The thrust of a turbojet (and some turbofans) can be increased temporarily by injecting fuel into the exhaust stream downstream of the turbine. This arrangement is called an afterburner or reheat system. The significant advantage of afterburning is the substantial temporary increase in thrust, which may be used for takeoff, acceleration, combat maneuvers, or sustained supersonic flight in aircraft designed for such operation. A significant disadvantage is the very low thermal efficiency, as the afterburner consumes a large amount of fuel for a given thrust level. For this reason, afterburners are used only intermittently when absolutely needed.

In thermodynamic terms, an afterburner can be modeled as a Brayton cycle with an additional approximately constant-pressure heat-addition process between the turbine exit and the exhaust nozzle. The turbine still provides the work required to drive the compressor, while the afterburner raises the exhaust stream’s stagnation temperature before expansion through the nozzle. If T_4 is the turbine exit temperature, then afterburning increases the stagnation temperature to T_5, i.e.,

(48)   \begin{equation*} T_5 = T_4 + \Delta T_{\text{AB}} \end{equation*}

where the \Delta T_{\text{AB}} boost depends on the amount of fuel injected and burned.

The effect on the Ts diagram shown above is an additional constant-pressure heat-addition process between the turbine exit and the nozzle inlet, with both temperature and entropy increasing. On the corresponding p\mathcal{V} diagram, the cycle does not change shape significantly; however, the expansion through the nozzle begins from a higher total temperature, resulting in a higher jet velocity and, consequently, greater thrust.

Connection to Carnot Efficiency

While the Otto, Diesel, and Brayton cycles describe the operation of real IC engines, it is useful to recall the Carnot efficiency. Carnot established that no heat engine operating between a hot reservoir at temperature T_H and a cold reservoir at T_C can exceed the limit

(49)   \begin{equation*} \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} \end{equation*}

This limit follows from the reversible Carnot cycle, which consists of two isothermal and two adiabatic processes, as shown in Figure 11. On the p\mathcal{V} process diagram, the process from state 1 to 2 is an isothermal expansion at T_H. On a per-unit-mass basis, this process satisfies p\,\mathsf{v} = R \, T_H and has heat input q_H > 0, where R is the specific gas constant. The process from state 2 to 3 is a reversible adiabatic expansion to T_C with p\,\mathsf{v}^{\,\gamma} = constant.

The process stages with p\mathcal{V} and Ts diagrams of the Carnot cycle.

The process from state 3 to 4 is an isothermal compression at T_C. On a per-unit-mass basis, this process satisfies p\,\mathsf{v} = R \, T_C and rejects heat to the cold reservoir, denoted here by the positive quantity |q_C|. Finally, the process from state 4 to 1 is a reversible adiabatic compression back to T_H. The area enclosed equals the specific net work.

On the Ts process diagram, the process from state 1 to 2 is a horizontal line at T_H, process 2 to 3 is a vertical adiabat, process 3 to 4 is a horizontal line at T_C, and process 4-1 is a vertical adiabat; the rectangular area equals the specific net work. Using p = R \, T/\mathsf{v} on a per-unit-mass basis, the isothermal heat transfers can be written as

(50)   \begin{equation*} q_H = \int_{1}^{2} p\,d\mathsf{v} = R T_H \ln\!\left(\frac{\mathsf{v}_2}{\mathsf{v}_1}\right) \end{equation*}

and

(51)   \begin{equation*} |q_C| = -\int_{3}^{4} p\,d\mathsf{v} = R T_C \ln\!\left(\frac{\mathsf{v}_3}{\mathsf{v}_4}\right) \end{equation*}

The reversible adiabatic condition is

(52)   \begin{equation*} T\,\mathsf{v}^{\,\gamma-1} = \text{constant} \end{equation*}

Applying this result from states 2 to 3 and states 4 to 1 gives

(53)   \begin{equation*} T_H\,\mathsf{v}_2^{\,\gamma-1} = T_C\,\mathsf{v}_3^{\,\gamma-1} \qquad \text{and} \qquad T_H\,\mathsf{v}_1^{\,\gamma-1} = T_C\,\mathsf{v}_4^{\,\gamma-1} \end{equation*}

which implies

(54)   \begin{equation*} \frac{\mathsf{v}_2}{\mathsf{v}_1} = \frac{\mathsf{v}_3}{\mathsf{v}_4} \end{equation*}

The same logarithmic factor appears on both isotherms, so

(55)   \begin{equation*} \frac{|q_C|}{q_H} = \frac{T_C}{T_H} \end{equation*}

so that

(56)   \begin{equation*} \eta_{\text{Carnot}} = \frac{w_{\text{net}}}{q_H} = \frac{q_H - |q_C|}{q_H} = 1 - \frac{|q_C|}{q_H} = 1 - \frac{T_C}{T_H} \end{equation*}

The Carnot result itself is independent of the equation of state and depends only on the reservoir temperatures and the assumption of reversibility.

For comparison, the idealized cycle efficiencies are

(57)   \begin{eqnarray*} \eta_{\text{Otto}} & = & 1 - \frac{1}{r^{\gamma-1}} = 1 - \frac{c_v\,(T_4 - T_1)}{c_v\,(T_3 - T_2)} = 1 - \frac{T_4 - T_1}{T_3 - T_2} \\[6pt] \eta_{\text{Diesel}} & = & 1 - \frac{1}{r^{\gamma-1}} \left( \frac{\beta^{\gamma} - 1}{\gamma (\beta - 1)} \right) = 1 - \frac{c_v\,(T_4 - T_1)}{c_p\,(T_3 - T_2)} = 1 - \frac{1}{\gamma}\,\frac{T_4 - T_1}{T_3 - T_2} \\[6pt] \eta_{\text{Brayton}} & = & 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}} = 1 - \frac{c_p\,(T_4 - T_1)}{c_p\,(T_3 - T_2)} = 1 - \frac{T_4 - T_1}{T_3 - T_2} \end{eqnarray*}

where r is the compression ratio, \beta is the cut-off ratio for the Diesel cycle, r_p is the  compressor pressure ratio, and \gamma = c_p/c_v is the ratio of specific heats.

The temperature forms of the efficiencies show that the Otto cycle adds and rejects heat at constant volume, while the Brayton cycle adds and rejects heat at constant pressure. Consequently, the same specific heat appears in the heat-addition and heat-rejection terms for each cycle and cancels from the corresponding efficiency expression. In contrast, the Diesel cycle combines constant-pressure and constant-volume processes, introducing the 1/\gamma factor. Using a representative peak combustion temperature of T_H \sim 2,000 to 2,500 K and an environmental temperature of T_C \sim 300 K gives a nominal Carnot efficiency of approximately 85–88%. This comparison is only a theoretical reference because combustion engines do not generally receive heat isothermally from a single reservoir at the peak gas temperature. Real engines fall well short of this ceiling because of heat-transfer losses, finite-rate combustion, pressure losses, mechanical losses, and other irreversibilities, as shown in Figure 12.

Efficiency comparison of the Otto, Diesel, Brayton, and Carnot thermodynamic cycles.

For Otto-cycle spark-ignition engines, the ideal air-standard efficiency ranges from approximately 55–65% for compression ratios between r=8 and 12. Practical efficiencies are much lower because of heat transfer, finite-rate combustion, pumping and mechanical losses, and other irreversibilities. The allowable compression ratio and boost pressure are also constrained by abnormal combustion, including knock and pre-ignition, which are distinct phenomena.

In the Diesel cycle (compression-ignition engines), ideal efficiencies can reach 60-70% for compression ratios of r = 14-20 with modest cut-off ratios. Practical heavy-duty diesels can achieve 40-46%, and advanced designs have been reported to approach 50%. Higher cut-off ratios reduce the ideal efficiency, and in real engines, additional losses arise from incomplete combustion and heat transfer.

The Brayton cycle (used in gas turbines) has an ideal efficiency of approximately 55-60% for pressure ratios r_p = 30-40 and \gamma \approx 1.33. In practice, simple-cycle gas turbines achieve lower efficiencies than the ideal Brayton limit, while combined-cycle industrial gas-turbine plants can exceed 60% overall thermal efficiency by recovering exhaust heat. Gas turbines for airplanes, however, are limited to approximately 35-40% efficiency at cruise conditions because of turbine inlet temperature restrictions, cooling requirements, compressor inefficiencies, and weight limitations.

Therefore, while real cycles remain below the Carnot limit, no universal ordering of Otto, Diesel, and Brayton efficiencies exists without specifying the compression ratio, pressure ratio, cutoff ratio, peak temperature, component efficiencies, and operating constraints. For comparable idealized assumptions, each cycle efficiency follows from its own limiting parameters, i.e.,

(58)   \begin{equation*} \eta_{\text{Otto}} = 1 - \frac{1}{r^{\gamma-1}}, \qquad \eta_{\text{Diesel}} = 1 - \frac{1}{r^{\gamma-1}} \left( \frac{\beta^{\gamma} - 1}{\gamma(\beta - 1)} \right), \qquad \text{and} \qquad \eta_{\text{Brayton}} = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}} \end{equation*}

with all three remaining strictly below \eta_{\text{Carnot}} for engines operating between the same limiting reservoir temperatures.

As previously discussed, supercharging or turbocharging increases the intake pressure and density before compression. In the Otto cycle, the ideal efficiency expression is

(59)   \begin{equation*} \eta_{\text{Otto}} = 1 - \frac{1}{r^{\gamma-1}} \end{equation*}

where r=\mathcal{V}_1/\mathcal{V}_2 is the compression ratio. Therefore, if r is unchanged, supercharging does not directly change the ideal Otto-cycle efficiency. However, the higher intake pressure increases the mass of fuel-air charge per cycle, thereby increasing the indicated mean effective pressure and the engine’s power output. In practice, the allowable compression ratio and boost pressure are limited by abnormal combustion, including knock and pre-ignition, so the main thermodynamic benefit of supercharging in a spark-ignition engine is usually increased power rather than a large increase in ideal cycle efficiency.

In the Diesel cycle, supercharging can also improve efficiency. The ideal Diesel-cycle efficiency is

(60)   \begin{equation*} \eta_{\text{Diesel}} = 1 - \frac{1}{r^{\gamma-1}} \left( \frac{\beta^{\gamma} - 1}{\gamma(\beta - 1)} \right) \end{equation*}

where r=\mathcal{V}_1/\mathcal{V}_2 is the compression ratio and \beta=\mathcal{V}_3/\mathcal{V}_2 is the cut-off ratio. For a given required power, increasing the intake pressure and air mass can allow the same work output to be obtained with a smaller cut-off ratio. Because reducing \beta increases the ideal Diesel-cycle efficiency, supercharging can improve both power and efficiency in this cycle model. In the Brayton cycle, afterburning has the opposite effect on efficiency, i.e., heat is added downstream of the compressor-driving turbine, so afterburning increases thrust but reduces thermal efficiency.

Check Your Understanding #4 – Brayton cycle efficiency

Air enters a turbojet engine at T_1 = 300 K. The compressor pressure ratio is r_p = 12, and the turbine inlet temperature is T_3 = 1,400 K. Assume constant properties c_p = 1,004 J kg^{-1} K^{-1}, and \gamma = 1.4. For the real Brayton cycle, the compressor and turbine have isentropic efficiencies of \eta_c = 0.85 and \eta_t = 0.88. Calculate the thermal efficiency and back-work ratio for both the ideal and real cycles.

Show solution/hide solution.

For the ideal Brayton cycle, the isentropic temperature rise through the compressor is

    \[ T_{2s} = T_1 \, r_p^{(\gamma-1)/\gamma} \]

which gives T_{2s} = 610~\mathrm{K}. The isentropic temperature drop through the turbine is

    \[ T_{4s} = T_3 \, r_p^{-(\gamma-1)/\gamma} \]

which gives T_{4s} = 688~\mathrm{K}. The compressor work per unit mass is

    \[ w_c = c_p (T_{2s} - T_1) = 311~\mathrm{kJ/kg} \]

The turbine work per unit mass is

    \[ w_t = c_p (T_3 - T_{4s}) = 715~\mathrm{kJ/kg} \]

and the heat input per unit mass is

    \[ q_{\text{in}} = c_p (T_3 - T_{2s}) = 793~\mathrm{kJ/kg} \]

The net specific work output is

    \[ w_{\text{net}} = w_t - w_c = 403~\mathrm{kJ/kg} \]

so the thermal efficiency is

    \[ \eta_{\text{ideal}} = \frac{w_{\text{net}}}{q_{\text{in}}} = 0.51 \]

The back-work ratio is

    \[ \mathrm{BWR} = \frac{w_c}{w_t} = 0.44 \]

For the real Brayton cycle, the actual compressor outlet temperature is obtained from

    \[ \eta_c = \frac{T_{2s} - T_1}{T_2 - T_1} \]

so that

    \[ T_2 = T_1 + \frac{T_{2s} - T_1}{\eta_c} = 665~\mathrm{K} \]

The compressor work per unit mass is then

    \[ w_c = c_p (T_2 - T_1) = 366~\mathrm{kJ/kg} \]

For the turbine, the isentropic efficiency gives

    \[ \eta_t = \frac{T_3 - T_4}{T_3 - T_{4s}} \]

so that

    \[ T_4 = T_3 - \eta_t (T_3 - T_{4s}) = 774~\mathrm{K} \]

The turbine work per unit mass is

    \[ w_t = c_p (T_3 - T_4) = 629~\mathrm{kJ/kg} \]

and the heat input per unit mass is

    \[ q_{\text{in}} = c_p (T_3 - T_2) = 738~\mathrm{kJ/kg} \]

The net specific work is

    \[ w_{\text{net}} = w_t - w_c = 262~\mathrm{kJ/kg} \]

and so the thermal efficiency is

    \[ \eta_{\text{real}} = \frac{w_{\text{net}}}{q_{\text{in}}} = 0.36 \]

The back-work ratio is

    \[ \mathrm{BWR} = \frac{w_c}{w_t} = 0.58 \]

This example shows that the ideal Brayton cycle efficiency is about 51%, but real component inefficiencies reduce it to 36%. The increase in the back-work ratio from 0.44 to 0.58 indicates that compressor and turbine irreversibilities consume a significant fraction of the turbine output, thereby reducing net work.

Rocket Propulsion Cycles

Rocket engines require both fuel and oxidizer, collectively called propellants, to be carried on board the vehicle. When suitably combined and ignited, they produce high-temperature combustion products that are expanded through a nozzle to generate thrust, as shown in Figure 13. Thermodynamically, rockets are open-flow propulsion systems in which stored chemical energy is converted into the enthalpy of high-pressure combustion products, and then into the kinetic energy of the exhaust through a nozzle. Unlike air-breathing Brayton-cycle engines, rockets carry both fuel and oxidizer onboard. Thrust-to-weight ratio and specific impulse provide complementary measures of rocket performance.

A rocket motor requires both a fuel and an oxidizer to generate thrust.

The thrust from a rocket motor follows directly from the conservation of momentum, i.e.,

(61)   \begin{equation*} F = \overbigdot m \, V_e + (p_e - p_a) \, A_e \end{equation*}

where \overbigdot m = \overbigdot m_f + \overbigdot m_o is the total mass flow rate of fuel and oxidizer, V_e is the exhaust or “jet” velocity, p_e is the nozzle exit pressure, p_a is the ambient pressure, and A_e is the exit area. It is generally considered convenient to define an effective exhaust velocity that absorbs the pressure term, i.e.,

(62)   \begin{equation*} V_{e,\mathrm{eff}} \equiv \frac{F}{\overbigdot m} = V_e + \frac{(p_e - p_a)A_e}{\overbigdot m} \end{equation*}

The specific impulse is then defined by

(63)   \begin{equation*} I_{\rm sp} = \frac{F}{\overbigdot m \, g_0} \qquad \text{so that} \qquad V_{e,\mathrm{eff}} = g_0 \, I_{\rm sp} \end{equation*}

Specific impulse, which serves as a standard measure of rocket propulsive performance, has units of seconds when defined as thrust divided by propellant weight-flow rate. This form makes it independent of the particular force and mass units used. Notice that for a rocket, g can vary as it reaches higher altitudes above the Earth’s surface, so the standard gravitational acceleration, {g_0}, is used as a reference value.

For an ideal isentropic nozzle or “bell,” the exhaust velocity is obtained from the chamber stagnation conditions and the nozzle expansion ratio, i.e.,

(64)   \begin{equation*} V_e = \sqrt{\frac{2\gamma}{\gamma-1}\,R\,T_c \left( 1-\left(\frac{p_e}{p_c}\right)^{(\gamma-1)/\gamma}\right)} \end{equation*}

where p_c and T_c are the chamber stagnation pressure and temperature, \gamma is the ratio of specific heats, and R = \overline{R}/M is the specific gas constant of the exhaust products. Lower values of molar mass M increase the value of R, leading to a higher speed of sound and greater exhaust velocity for a given T_c. This is why hydrogen-oxygen (LH_2-LOX) propellants perform better than heavier hydrocarbon-oxygen systems.

The nozzle exit pressure p_e is related to the area ratio \varepsilon = A_e/A_t through the isentropic area-Mach relations, i.e.,

(65)   \begin{equation*} \varepsilon = \frac{1}{M_e}\left[\frac{2}{\gamma+1} \left(1+\frac{\gamma-1}{2}M_e^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}} \qquad \text{and} \qquad \frac{p_e}{p_c} = \left(1+\frac{\gamma-1}{2}M_e^2\right)^{-\frac{\gamma}{\gamma-1}} \end{equation*}

The two aggregate performance parameters are the characteristic velocity and the thrust coefficient, which are given by

(66)   \begin{equation*} c^* = \frac{p_c \, A_t}{\overbigdot m} \qquad \text{and} \qquad C_F = \frac{F}{p_c \, A_t} \end{equation*}

This means that

(67)   \begin{equation*} F = \overbigdot m\, c^*\, C_F \qquad \text{and} \qquad I_{\rm sp} = \frac{c^*\,C_F}{g_0} \end{equation*}

where c^* measures the combustion performance in the chamber and is independent of nozzle design, while C_F measures the nozzle’s ability to convert the chamber’s pressure into thrust.

For an ideal equilibrium process, the value of c^* depends only on the propellant chemistry, T_c, and product properties. In practice, c^* is less than its theoretical value because of incomplete combustion, mixing limits, finite residence time, and chamber losses. The thrust coefficient C_F, by contrast, reflects the effects of nozzle geometry and ambient pressure conditions, with MSL and vacuum values given by

(68)   \begin{equation*} I_{sp,\mathrm{SL}} = \frac{c^*\,C_{F,\mathrm{MSL}}}{g_0} \qquad \text{and} \qquad I_{sp,\mathrm{vac}} = \frac{c^*\,C_{F,\mathrm{vac}}}{g_0} \end{equation*}

The difference between the sea-level and vacuum values is dominated by the pressure-thrust contribution in C_F and by the chosen expansion ratio \varepsilon.

The different rocket propulsion cycles are distinguished thermodynamically by the power source of the turbopumps and by whether the turbine exhaust is returned to the main chamber. In the gas-generator cycle, a small fraction of the propellant is burned in a separate chamber to drive the turbopumps, and the turbine exhaust is discharged separately rather than being returned to the main combustion chamber. The exhaust may be released through a separate outlet or introduced into a lower-pressure region of the nozzle. This arrangement is mechanically simpler than a staged-combustion cycle. However, the turbine-drive flow does not undergo the full main-chamber expansion, resulting in a lower specific impulse than an otherwise comparable closed cycle. To first order, a dumped propellant fraction \beta_{\rm dump} imposes an I_{\rm sp} penalty.

The operational principle of a rocket motor.

In the staged-combustion cycle, a preburner operates at either fuel-rich or oxidizer-rich conditions to generate turbine-drive gas. After the turbopumps are powered, the gas is fed into the main chamber, ensuring that all propellant contributes to thrust. This configuration avoids dump losses and permits high chamber pressure, giving higher overall performance and usually higher I_{\rm sp}, but at the cost of very demanding turbomachinery designs.

In the expander cycle, cryogenic fuel (usually liquid hydrogen) circulates through cooling channels around the chamber and nozzle, absorbing heat and vaporizing before driving the turbopumps and then entering the main chamber. This cycle is efficient and clean because no propellant is discarded. Still, it is thrust-limited by the available heat pickup, so the maximum \overbigdot{m} and chamber pressure are constrained by cooling power. A summary of these cycles is shown in the table below.

Cycle Pump drive/flow path Advantages Limitations Typical applications Indicative I_{\rm sp} (sea-level/vacuum)
Gas-generator A small propellant fraction burns in a gas generator → drives turbine → exhaust discharged separately or into a lower-pressure region of the nozzle; remaining flow enters the main chamber → nozzle. Relatively simple, robust, good throttle response, high thrust-to-weight. Turbine-drive flow does not undergo the full main-chamber expansion; lower I_{\rm sp} than comparable closed cycles. First stages & boosters where simplicity and T/W are prioritized. LOX/RP-1: 260–300 s / 300–330 s; LOX/LH_2: 360–380 s / ~430–445 s.
Staged-combustion Preburner (fuel-rich or oxidizer-rich) drives turbine; turbine exhaust routed to main chamber so all mass expands through nozzle. High I_{\rm sp}, high chamber pressure p_c, good propellant utilization. Complex, high stresses/temperatures, demanding materials & controls. Reusable boosters, high-performance cores, and upper stages. LOX/RP-1: 290–315 s / 330–360 s; LOX/LH_2: 375–395 s / 450–465 s.
Expander Cryogenic fuel heated in cooling channels → drives turbine → enters main chamber; oxidizer pumped to chamber → nozzle. Clean turbine drive, high efficiency, excellent reliability. Thrust limited by available heat pickup; typically LH_2 only. Cryogenic upper stages, long-duration restarts, precise orbital work. LOX/LH_2: 360–380 s / 440–465 s (thrust typically modest).

In practice, gas-generator cycles are widely used where mechanical simplicity, reliability, and high thrust-to-weight ratio are important. Staged-combustion cycles are used in many high-performance booster and upper-stage engines when high chamber pressure and propellant utilization justify the additional complexity. Expander cycles are primarily used in cryogenic upper-stage engines, where hydrogen is available, and precise orbital maneuvers are required. However, all cycles reflect the same thermodynamic principles but represent different trade-offs among engineering simplicity, efficiency, and thrust production capability.

Check Your Understanding #5 – Rocket nozzle performance

Given A_e/A_t = 3, chamber (stagnation) temperature T_0 = 3000 K, chamber pressure p_0 = 2.0 \times 10^{6} Pa, ambient pressure p_a = 1.0 \times 10^{5} Pa, specific heat ratio \gamma = 1.2, and molar mass M = 33.4 kg/kmol. Assume isentropic expansion to the exit. Find the thrust per unit throat area F/A_t and the specific impulse I_{\rm sp}.
Show solution/hide solution.

The specific gas constant is

    \[ R = \frac{\overline{R}}{M} = \frac{8.314 \times 10^{3}}{33.4} = 2.49 \times 10^{2}~\mathrm{J/(kg \, K)} \]

From the isentropic area-Mach number relation (supersonic branch), A_e/A_t = 3 with \gamma = 1.2 gives M_e \approx 2.40. The exit temperature is

    \[ \frac{T_e}{T_0} = \frac{1}{1 + \frac{\gamma - 1}{2} M_e^2} = \frac{1}{1 + 0.1 \times (2.40)^2} = 0.635 \]

so T_e \approx 1.90 \times 10^{3}~\mathrm{K}. The exit pressure is

    \[ \frac{p_e}{p_0} = \left( \frac{T_e}{T_0} \right)^{\gamma/(\gamma - 1)} = (0.635)^6 \approx 0.065 \]

hence p_e \approx 1.30 \times 10^{5}~\mathrm{Pa}. The exit velocity is obtained from

    \[ V_e = M_e \sqrt{\gamma \, R \, T_e} = 2.40 \, \sqrt{1.2 \, \times 2.49 \times 10^{2} \, \times 1.90 \times 10^{3}} \approx 1.81 \times 10^{3}~\mathrm{m/s} \]

The choked mass flux at the throat is

    \[ \frac{\overbigdot{m}}{A_t} = \frac{p_0}{\sqrt{T_0}} \sqrt{\frac{\gamma}{R}} \left( \frac{2}{\gamma + 1} \right)^{\frac{\gamma + 1}{2(\gamma - 1)}} \approx 1.50 \times 10^{3}~\mathrm{kg/(s \, m^2)} \]

The thrust per unit throat area is

    \[ \frac{F}{A_t} = \left( \frac{\overbigdot{m}}{A_t} \right) V_e + (p_e - p_a)\, \frac{A_e}{A_t} \]

so

    \[ \frac{F}{A_t} = (1.50 \times 10^{3})\times (1.81 \times 10^{3}) + (1.30 \times 10^{5} - 1.00 \times 10^{5}) \times 3 = 2.81 \times 10^{6}~\mathrm{N/m^2} \]

The specific impulse is

    \[ I_{\rm sp} = \frac{F/A_t}{(\overbigdot{m}/A_t)\, g_0} = \frac{2.81 \times 10^{6}}{(1.50 \times 10^{3})(9.81)} \approx 1.91 \times 10^{2}~\mathrm{s} \]

Thermodynamics of Combustion Processes

Combustion releases the chemical energy stored in a fuel and transfers much of that energy to a working fluid. The resulting increase in temperature and enthalpy allows the gas to perform work on a piston or turbine or to accelerate through a nozzle and produce thrust. Thermodynamics relates the fuel’s composition and heating value to the resulting state of the combustion products and establishes the limits on how much of the released energy can be converted into useful work.

Combustion must occur rapidly, reliably, and efficiently over a wide range of pressures, temperatures, and operating conditions. The principal quantities needed for an introductory analysis are the fuel heating value, the stoichiometric fuel-oxidizer ratio, the actual mixture ratio, the combustion efficiency, and the pressure loss through the combustor.

Fuels

A combustion analysis begins with the fuel because its chemical composition determines the available energy, the amount of oxidizer required, and the composition of the products. Most aviation fuels are hydrocarbons chosen for their high specific energy, relatively high liquid density, and ease of storage and handling.

Aviation gasoline, such as 100LL, may be approximated by \mathrm{C}_8\mathrm{H}_{18} and has a lower heating value (LHV) of about 44 MJ/kg. Kerosene-based fuels such as Jet-A and the rocket fuel RP-1 are commonly represented approximately by \mathrm{C}_{12}\mathrm{H}_{23}. These fuels have similar specific energies but relatively high liquid densities, which provide good volumetric energy storage.

Hydrogen is important in high-performance rocket propulsion and experimental aeropropulsion. Its LHV is approximately 120 MJ/kg, which is much greater than that of hydrocarbon fuels. However, liquid hydrogen has a density of only about 71 kg/m^3, so it requires large cryogenic tanks and specialized handling systems.

The oxidizer is equally important. Air supplies oxygen to piston engines and gas turbines, whereas rockets must carry an oxidizer on board. Liquid oxygen (LOX) is widely used in rocket propulsion, while nitrous oxide and hydrogen peroxide are used in some specialized systems.

Chemical Equations – Conservation of Mass

A balanced chemical equation describes the proportions of fuel and oxidizer required for combustion and enforces conservation of each chemical element. For a hydrocarbon fuel \mathrm{C}_a\mathrm{H}_b burning in air modeled as oxygen and nitrogen in the ratio 1:3.76, the stoichiometric reaction is

(69)   \begin{equation*} \mathrm{C}_a\mathrm{H}_b + \nu_s\left(\mathrm{O_2}+3.76\,\mathrm{N_2}\right) \rightarrow a\,\mathrm{CO_2} + \frac{b}{2}\,\mathrm{H_2O} + 3.76\,\nu_s\,\mathrm{N_2} \end{equation*}

where the stoichiometric oxygen coefficient is

(70)   \begin{equation*} \nu_s=a+\frac{b}{4} \end{equation*}

This value supplies enough oxygen to convert all of the carbon to carbon dioxide and all of the hydrogen to water. The balanced equation determines the stoichiometric air-fuel ratio, which is the mass of air required to completely combust a unit mass of fuel.

The actual mixture may contain more or less fuel than the stoichiometric mixture. The equivalence ratio is defined as

(71)   \begin{equation*} \phi = \frac{(F/A)_{\rm actual}} {(F/A)_{\rm st}} \end{equation*}

where F/A is the fuel-air ratio. A mixture is lean when \phi<1, stoichiometric when \phi=1, and rich when \phi>1.

Energy Release and Combustion Efficiency

The lower heating value (LHV) represents the chemical energy released per unit mass of fuel when combustion is complete, and the water in the products remains as vapor. In an adiabatic combustor with no shaft work and negligible changes in kinetic and potential energy, the released energy increases the stagnation enthalpy of the working fluid.

On a per-unit-mass-of-air basis, a useful approximate balance is

(72)   \begin{equation*} h_{0,\rm out}-h_{0,\rm in} \approx \eta_b\,f\,\mathrm{LHV} \end{equation*}

where \scriptstyle f is the fuel-air ratio and \eta_b is the combustion efficiency. For a calorically ideal gas, the corresponding stagnation-temperature rise is

(73)   \begin{equation*} c_p\left(T_{0,\rm out}-T_{0,\rm in}\right) \approx \eta_b\,f\,\mathrm{LHV} \end{equation*}

The combustion efficiency accounts for the fact that a real combustor does not transfer all of the fuel’s theoretical chemical energy to the working fluid. It is defined as

(74)   \begin{equation*} \eta_b = \frac{\text{\small Actual heat released to the flow}} {\text{\small Theoretical heat release based on LHV}} \end{equation*}

Incomplete combustion, imperfect mixing, finite reaction rates, and heat transfer to surrounding structures reduce the combustion efficiency. A value of \eta_b=1 represents the ideal limit, while practical values remain below unity.

A combustor also incurs a loss of stagnation pressure from friction, turbulent mixing, heat addition, and other irreversible effects. This loss is commonly expressed as

(75)   \begin{equation*} \pi_b = \frac{p_{t3}}{p_{t2}} <1 \end{equation*}

where p_{t2} is the stagnation pressure at the combustor inlet and p_{t3} is the stagnation pressure at its outlet. An effective combustor must produce the required rise in stagnation temperature while maintaining high combustion efficiency and minimizing stagnation-pressure loss.

More advanced combustion analyses account for temperature-dependent specific heats, detailed product composition, finite-rate chemical reactions, and dissociation at very high temperatures. These effects are important in detailed gas-turbine and rocket calculations.

Aerospace Applications

In a gas turbine, combustion raises the stagnation temperature between the compressor exit and turbine inlet. For approximately constant specific heat, the required fuel-air ratio may be estimated from

(76)   \begin{equation*} f \approx \frac{c_{p,g}\left(T_{t3}-T_{t2}\right)} {\eta_b\,\mathrm{LHV}} \end{equation*}

where T_{t2} is the compressor-exit stagnation temperature and T_{t3} is the turbine-inlet stagnation temperature. Materials and cooling requirements limit the attainable turbine-inlet temperature, while the combustor pressure loss reduces the stagnation pressure available for expansion through the turbine and nozzle.

In a rocket engine, combustion establishes the chamber temperature, pressure, and product composition. These properties determine the characteristic and exhaust velocities achievable by the nozzle. High chamber temperature and low product molecular mass generally improve performance, but the mixture ratio must also provide efficient combustion, acceptable cooling, and suitable material temperatures. Therefore, combustion performance depends on more than fuel heating value alone. Stoichiometry, mixture ratio, combustion efficiency, pressure loss, chamber conditions, and product properties all influence the amount of useful work or thrust that can ultimately be obtained.

Check Your Understanding #6 – Jet-A combustion

Approximate Jet-A as \mathrm{C}_{12}\mathrm{H}_{23}. (a) Write the balanced stoichiometric reaction with air modeled as \mathrm{O_2} + 3.76\,\mathrm{N_2}. (b) Compute the stoichiometric air-fuel ratio (AFR) by mass. (c) If a gas-turbine combustor operates with \scriptstyle f = 0.022 kg of fuel per kg of air, compute the equivalence ratio \phi.

Show solution/hide solution.

(a) The general form of the combustion equation is fuel + oxygen → products. For Jet-A, then

    \[ \mathrm{C}_{12}\mathrm{H}_{23} + \nu_s(\mathrm{O_2}+3.76\,\mathrm{N_2}) \;\rightarrow\; 12\,\mathrm{CO_2} + \frac{23}{2}\,\mathrm{H_2O} + 3.76\,\nu_s\,\mathrm{N_2} \]

Balancing oxygen gives \nu_s = 12 + \dfrac{23}{4} = 17.75.

(b) The fuel molar mass is M_f = 12\times 12 + 23 = 167~\mathrm{g/mol}. Using M_{\mathrm{O_2}} = 32 and M_{\mathrm{N_2}} = 28 gives

    \[ \mathrm{AFR}_{\rm st} = \frac{17.75\,(32+3.76 \times 28)}{167} \approx 14.6 \]

Approximately 14.6 kg of air is required for every 1 kg of Jet-A fuel.

(c) The actual fuel-air ratio is f = 0.022. The equivalence ratio is

    \[ \phi = \frac{F/A}{(F/A)_{\rm st}} = f \, \mathrm{AFR}_{\rm st} = 0.022 \times 14.6 \approx 0.32 \]

This result indicates lean operation, typical of a gas turbine.

Electrochemical Energy Conversion

Not all aerospace propulsion systems release chemical energy through combustion. In a battery, electrochemical reactions convert chemical energy directly into electrical work, which an electric motor then converts into shaft power. Unlike a heat engine, a battery does not rely on combustion, compression, or expansion of a working fluid. Nevertheless, its performance remains governed by the First and Second Laws of Thermodynamics because electrical energy output, heat generation, internal losses, and efficiency must all satisfy energy and entropy balances. The detailed chemistry, construction, capacity, and performance of aerospace batteries are treated in Chapter 49 on electric propulsion.

Summary & Closure

This chapter has applied thermodynamic principles to aerospace propulsion systems. The steady-flow energy equation was used to interpret the behavior of nozzles, diffusers, compressors, turbines, and combustors. In each case, the same basic accounting applies: heat transfer, shaft work, enthalpy change, and kinetic-energy change determine how energy is transferred through the component.

The Second Law provides the corresponding limits. Real propulsion devices generate entropy through friction, turbulence, shocks, mixing, heat transfer, chemical reactions, and other irreversible processes. Isentropic efficiencies and exergy analysis provide practical measures of these losses. They show why real compressors require more work than ideal compressors, why real turbines deliver less work than ideal turbines, why nozzles do not achieve the ideal exhaust velocity, and why combustors inevitably destroy some of the available energy in the fuel.

The same ideas extend naturally to complete propulsion cycles. The Otto and Diesel cycles provide idealized models for piston engines. The Brayton cycle provides the thermodynamic basis for turbojets, turbofans, turboprops, and turboshafts. Rocket propulsion cycles convert stored fuel and oxidizer into high-pressure combustion products, which are then expanded into high-speed exhaust. In all cases, the achievable performance is governed by the combined effects of cycle temperature limits, pressure ratios, component efficiencies, fuel properties, and irreversibility.

Combustion thermodynamics connects fuel chemistry to propulsion performance. Heating value, fuel-air ratio, equivalence ratio, adiabatic flame temperature, combustion efficiency, and pressure loss all determine the amount of useful enthalpy added to the working fluid. For rockets, the same thermodynamic principles appear through chamber temperature, molecular weight, characteristic velocity, thrust coefficient, and specific impulse. Propulsion performance is not merely a matter of mechanical design because it is constrained at every stage by the First and Second Laws of Thermodynamics.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

  • How does the First Law of Thermodynamics lead directly to the nozzle relation between enthalpy drop and exit velocity?
  • Why does a real compressor require more shaft work than an ideal isentropic compressor operating over the same pressure ratio?
  • Why does a real turbine produce less shaft work than an ideal isentropic turbine expanding through the same pressure ratio?
  • How does entropy generation reduce stagnation pressure in nozzles, diffusers, inlets, and combustors?
  • Why is stagnation temperature conserved in an adiabatic nozzle or diffuser but changed by compressors, turbines, and combustors?
  • How do the Otto, Diesel, and Brayton cycles differ in the way heat is added and rejected?
  • Why does increasing compression ratio improve the ideal Otto-cycle efficiency, and why is this limited in real spark-ignition engines?
  • Why is the Brayton cycle so strongly affected by compressor pressure ratio and turbine inlet temperature?
  • Why does afterburning increase thrust while usually reducing thermal efficiency?
  • How do fuel heating value, mixture ratio, chamber temperature, molecular weight, and nozzle expansion affect rocket specific impulse?

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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.

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