Engineering Thermodynamics

Chapter 9
Gas Power and Propulsion Cycles

Bachelors Program in Mechnical Engineering
Engineering Thermodynamics   •   Bachelor

Gas Power and Propulsion Cycles

Introduction

Gas power cycles are thermodynamic models used to analyze internal-combustion engines, gas-turbine plants, aircraft engines, and related propulsion systems. Unlike vapor-power cycles, which generally involve phase change and a condenser, gas-power cycles operate primarily with gases and usually model combustion as heat addition. The working fluid may be air, combustion products, or an idealized gas with constant or temperature-dependent properties.

Ideal-cycle analysis does not reproduce every detail of a real engine. Instead, it separates the principal thermodynamic effects of compression, heat addition, expansion, and exhaust. This makes it possible to compare engine architectures, estimate performance, identify losses, and understand why practical improvements such as intercooling, reheating, and regeneration are used.

The most important cycles in this chapter are the Otto cycle for spark-ignition engines, the Diesel cycle for compression-ignition engines, the dual cycle for a more realistic representation of diesel combustion, and the Brayton cycle for gas turbines. The chapter also introduces component efficiencies, regenerative gas turbines, multistage compression with intercooling, multistage expansion with reheating, and turbojet performance.

Learning objectives

After studying this chapter, the student should be able to:

  1. Explain the ideal assumptions used in gas-power-cycle analysis.
  2. Analyze Otto, Diesel, dual, and Brayton cycles using property relations.
  3. Determine thermal efficiency, work output, back-work ratio, and net power.
  4. Explain the effects of pressure ratio, cutoff ratio, maximum temperature, and component losses.
  5. Analyze regeneration, intercooling, and reheating in gas-turbine plants.
  6. Calculate compressor and turbine exit states using isentropic efficiencies.
  7. Interpret the thrust and propulsive efficiency of a turbojet.
  8. Recognize practical, environmental, safety, and design limitations of ideal-cycle models.

1. Modeling gas-power cycles

1.1 Air-standard assumptions

The air-standard model commonly uses the following assumptions:

  • The working fluid is a fixed mass of air circulating in a closed cycle.
  • Air behaves as an ideal gas.
  • Combustion is replaced by external heat addition.
  • Exhaust is replaced by heat rejection that returns the air to its initial state.
  • Compression and expansion are treated as internally reversible unless a component efficiency is specified.
  • For the cold-air-standard model, specific heats are assumed constant, usually cp ≈ 1.005\ kJ/(kg·K) and cv ≈ 0.718\ kJ/(kg·K) near ordinary temperatures.

The ratio of specific heats is

γ = cp/cv

For air with constant properties, γ ≈ 1.4. At high temperatures, specific heats increase and γ decreases. Consequently, constant-property calculations often overestimate the efficiency of high-temperature cycles.

For an ideal gas,

Pv = RT

where P is absolute pressure, v is specific volume, R is the gas constant, and T is absolute temperature. For air, R = 0.287\ kJ/(kg·K) when pressure is in kPa and specific volume is in m³/kg.

For an internally reversible adiabatic process with constant γ,

T2/T1 = (v1/v2)^(γ−1) = (P2/P1)^((γ−1)/γ)

The corresponding pressure-volume relation is

Pv^γ = constant

These equations apply only to an ideal gas with constant specific heats undergoing an isentropic process. They must not be used directly for a real compressor or turbine without accounting for component inefficiency.

1.2 Performance measures

The thermal efficiency of a heat engine is

ηth = Wnet,out/Qin = 1 − Qout/Qin

where Wnet,out is net work output per unit mass and Qin and Qout are heat supplied and rejected per unit mass. For a steady-flow gas-turbine plant, the net work is turbine work minus compressor work:

wnet = wt − wc

The back-work ratio is

BWR = wc/wt

A high back-work ratio means that a large fraction of turbine work is consumed by the compressor. This is a central issue in gas turbines because the compressor can require a substantial portion of the turbine output.

For an open gas-turbine plant, the thermal efficiency is commonly written as

ηth = wnet/qin

All work and heat quantities in these expressions are specific quantities, usually in kJ/kg, so their ratios are dimensionless.

2. Reciprocating-engine cycles

2.1 Otto cycle

The ideal Otto cycle represents a spark-ignition engine. It consists of:

  1. Isentropic compression from state 1 to state 2.
  2. Constant-volume heat addition from state 2 to state 3.
  3. Isentropic expansion from state 3 to state 4.
  4. Constant-volume heat rejection from state 4 to state 1.

The compression ratio is

r = v1/v2

For the cold-air-standard Otto cycle,

ηOtto = 1 − 1/r^(γ−1)

This result shows that ideal Otto efficiency increases with compression ratio. It also shows why increasing compression ratio is attractive. In a real spark-ignition engine, however, excessive compression can cause knock, an uncontrolled autoignition that produces high pressure oscillations and possible damage.

The temperature relations are

T2 = T1 r^(γ−1)

and

T4 = T3/r^(γ−1)

For constant-volume heat addition and rejection,

qin = cv(T3 − T2)
qout = cv(T4 − T1)

The ideal Otto cycle is useful for studying compression ratio and ignition timing, but it does not represent finite-duration combustion, heat transfer to the cylinder walls, friction, pumping losses, gas exchange, or variable specific heats.

2.2 Diesel cycle

The ideal Diesel cycle represents a compression-ignition engine. Its processes are:

  1. Isentropic compression.
  2. Constant-pressure heat addition.
  3. Isentropic expansion.
  4. Constant-volume heat rejection.

The cutoff ratio is

rc = v3/v2

For constant specific heats, the ideal Diesel efficiency is

ηDiesel = 1 − [1/r^(γ−1)] [(rc^γ − 1)/(γ(rc − 1))]

The temperature relations are

T2 = T1 r^(γ−1)

and, because heat addition occurs at constant pressure,

T3/T2 = v3/v2 = rc

During isentropic expansion from state 3 to state 4, the volume ratio is v4/v3 = r/rc, giving

T4 = T3 (rc/r)^(γ−1)

The Diesel cycle generally has a lower efficiency than an Otto cycle at the same compression ratio and heat input conditions because heat is added over a volume-increasing process. Actual diesel engines, however, operate at higher compression ratios than spark-ignition engines and can achieve high practical efficiency.

2.3 Dual cycle

The dual cycle, also called the mixed cycle, models heat addition partly at constant volume and partly at constant pressure. It is therefore more representative of combustion in many compression-ignition engines.

Its processes are:

  1. Isentropic compression.
  2. Constant-volume heat addition.
  3. Constant-pressure heat addition.
  4. Isentropic expansion.
  5. Constant-volume heat rejection.

Two useful ratios are the pressure ratio during constant-volume heat addition,

α = P3/P2

and the cutoff ratio during constant-pressure heat addition,

ρ = v4/v3

Using constant specific heats, the ideal dual-cycle efficiency is

ηdual = 1 − [1/(r^(γ−1))] [(α ρ^γ − 1)/( (α − 1) + γα(ρ − 1) )]

Here r = v1/v2. The exact form follows from the heat supplied during the constant-volume and constant-pressure portions and the heat rejected during constant-volume rejection. The dual cycle approaches the Otto cycle when the constant-pressure portion vanishes, and approaches the Diesel cycle when the constant-volume portion becomes negligible.

3. Brayton cycle and gas-turbine plants

3.1 Ideal Brayton cycle

The ideal Brayton, or Joule, cycle models a simple gas-turbine plant. In its closed-cycle representation:

  1. State 1–2: isentropic compression in the compressor.
  2. State 2–3: constant-pressure heat addition.
  3. State 3–4: isentropic expansion in the turbine.
  4. State 4–1: constant-pressure heat rejection.

The pressure ratio is

rp = P2/P1 = P3/P4

For an ideal gas with constant γ, the compressor and turbine temperature ratios are

T2/T1 = rp^((γ−1)/γ)
T3/T4 = rp^((γ−1)/γ)

The ideal Brayton thermal efficiency is

ηBrayton = 1 − 1/rp^((γ−1)/γ)

This expression resembles the Otto result, but the controlling ratio is pressure ratio rather than volume ratio. Increasing pressure ratio increases ideal efficiency, but it does not always increase net work. At a specified turbine inlet temperature, an excessively high pressure ratio can raise compressor work and reduce net specific work.

For constant specific heats,

wc = cp(T2 − T1)
wt = cp(T3 − T4)
qin = cp(T3 − T2)

For the ideal cycle, the expansion ratio equals the compression pressure ratio. In an open-cycle plant, air enters the compressor from the atmosphere, fuel is burned in a combustor, and turbine exhaust is discharged to the surroundings.

3.2 Optimum pressure ratio for maximum specific work

For a simple ideal Brayton cycle with fixed minimum temperature T1 and maximum temperature T3, maximum net work occurs at a particular pressure ratio. Let

x = rp^((γ−1)/γ)

Then the net work is

wnet/cp = (T3 − T1 x) − (T3/x − T1)

Differentiation with respect to x gives the maximum-work condition

x = √(T3/T1)

Therefore,

rp,opt = (T3/T1)^[γ/(2(γ−1))]

This is a maximum-specific-work result for the ideal simple Brayton cycle, not a maximum-efficiency result. Real optimum pressure ratios differ because of component efficiencies, pressure losses, cooling flows, variable properties, and operating constraints.

Worked Example 1: Ideal Brayton cycle

An ideal Brayton cycle operates with T1 = 300\ K, T3 = 1400\ K, pressure ratio rp = 8, γ = 1.4, and cp = 1.005\ kJ/(kg·K). Determine the compressor exit temperature, turbine exit temperature, net work, and thermal efficiency.

First,

rp^((γ−1)/γ) = 8^(0.4/1.4) ≈ 1.811

Thus,

T2 = 300(1.811) ≈ 543.3 K
T4 = 1400/1.811 ≈ 773.0 K

Compressor work is

wc = 1.005(543.3 − 300) ≈ 244.5 kJ/kg

Turbine work is

wt = 1.005(1400 − 773.0) ≈ 630.1 kJ/kg

Therefore,

wnet ≈ 630.1 − 244.5 = 385.6 kJ/kg

Heat addition is

qin = 1.005(1400 − 543.3) ≈ 860.9 kJ/kg

and

ηth = 385.6/860.9 ≈ 0.448

The ideal thermal efficiency is therefore approximately 44.8%. Real simple-cycle gas turbines have lower efficiency because of non-isentropic compression and expansion, combustor pressure loss, heat transfer, mechanical losses, and nonconstant properties.

4. Real gas-turbine components

4.1 Compressor isentropic efficiency

For a compressor, the isentropic efficiency compares the ideal minimum work with the actual work:

ηc = (h2s − h1)/(h2a − h1)

For constant cp, this becomes

ηc = (T2s − T1)/(T2a − T1)

where subscript s denotes the isentropic exit state and a denotes the actual exit state. Since a real compressor requires more work, T2a > T2s and ηc < 1.

4.2 Turbine isentropic efficiency

For a turbine, the isentropic efficiency compares actual work with ideal maximum work:

ηt = (h3 − h4a)/(h3 − h4s)

For constant cp,

ηt = (T3 − T4a)/(T3 − T4s)

A real turbine has T4a > T4s, so its actual work is lower than the isentropic work.

4.3 Combustor and mechanical efficiency

Combustor efficiency may be defined as

ηcomb = actual\ heat\ release/theoretical\ heat\ release

Fuel–air ratio, incomplete combustion, and wall heat loss influence this quantity. Combustors also cause pressure loss, which reduces turbine expansion work.

Mechanical efficiency accounts for shaft and bearing losses. If Wt is turbine power available at the shaft and Wc is compressor power requirement, the net shaft power may be approximated by

Ẇshaft = ηm(Ẇt − Ẇc)

where ηm is mechanical efficiency and the overdot denotes power. Generator efficiency must also be included when electrical output is required.

5. Improvements to the Brayton cycle

5.1 Regeneration

A regenerator transfers heat from the hot turbine exhaust to the compressed air leaving the compressor. It can reduce fuel consumption when the turbine exhaust temperature exceeds the compressor exit temperature.

The regenerator effectiveness is

εreg = (T3 − T2)/(T5 − T2)

for the arrangement in which state 5 is the turbine exhaust entering the regenerator and state 3 is the heated compressed air leaving it. More generally, effectiveness is the actual heat transfer divided by the maximum possible heat transfer:

εreg = qactual/qmax

The maximum possible heat transfer is based on the minimum heat-capacity-rate stream. For equal mass flow and constant specific heats, it is proportional to the smaller of the hot- and cold-stream temperature differences.

Regeneration improves thermal efficiency but usually does not increase net work. It is ineffective when the compressor exit temperature is close to or higher than the turbine exhaust temperature. Pressure losses, leakage, fouling, cost, and size are important practical limitations.

5.2 Intercooling

Intercooling divides compression into two or more stages and cools the air between stages. Since cooler air has lower specific volume, the work required for compression decreases. For two ideal compressor stages with perfect intercooling and equal stage pressure ratios, minimum work occurs when the intermediate pressure is approximately

Pintermediate = √(P1 P3)

where P1 is the inlet pressure and P3 is the final pressure. This result assumes equal gas properties, equal stage efficiencies, negligible pressure loss, and perfect intercooling to the initial inlet temperature.

Intercooling increases net work because compressor work decreases, but it also lowers the temperature entering the combustor. Without regeneration, this can increase the heat input required and may reduce thermal efficiency. Intercooling is therefore especially attractive when the objective is high specific power rather than maximum simple-cycle efficiency.

5.3 Reheating

Reheating divides turbine expansion into two or more stages and adds heat between stages. It increases the average temperature at which heat is supplied and generally increases turbine work. For ideal two-stage expansion with reheating and equal pressure ratios, the intermediate pressure is approximately the geometric mean of inlet and final pressures.

Reheating raises the turbine exhaust temperature. It can therefore work well with regeneration: reheating increases turbine work, while the hotter exhaust provides more energy for the regenerator. However, reheating requires additional combustor equipment and fuel and may reduce thermal efficiency if the added heat is not effectively converted into net work.

5.4 Combined arrangements

Intercooling, reheating, and regeneration may be combined. Their effects are not simply additive:

ModificationPrincipal benefitMain disadvantage
IntercoolingReduces compressor work and increases specific powerMay reduce efficiency without regeneration
ReheatingIncreases turbine work and specific powerRequires extra fuel and increases exhaust temperature
RegenerationReduces fuel required for a given turbine inlet temperatureAdds pressure loss, volume, cost, and maintenance

A sound design evaluates net power, thermal efficiency, emissions, capital cost, transient behavior, and maintenance rather than selecting modifications from efficiency alone.

6. Jet propulsion

6.1 Turbojet operation

A turbojet consists of an inlet, compressor, combustor, turbine, and nozzle. The turbine extracts only enough work to drive the compressor and accessories; the remaining gas energy is converted into a high exit velocity through the nozzle.

For a steady, adiabatic nozzle with negligible potential-energy change and negligible inlet velocity relative to the nozzle energy, the ideal exhaust velocity is approximated by

Ve = √[2 cp T5 (1 − (P0/P5)^((γ−1)/γ))]

where T5 and P5 are nozzle-entry temperature and pressure, and P0 is ambient pressure. This relation assumes an ideal gas, constant cp and γ, and an isentropic nozzle. If the nozzle is imperfect, a nozzle efficiency must be applied.

The thrust equation for a one-dimensional steady turbojet is

F = ṁa(Ve − V0) + (Pe − P0)Ae

where F is net thrust, ṁa is air mass-flow rate, V0 is flight speed, Ve is exit speed, Pe is exit pressure, P0 is ambient pressure, and Ae is exit area. The second term is pressure thrust. It vanishes for a perfectly expanded nozzle, where Pe = P0. Fuel mass flow can be included by replacing ṁa in the momentum term with total exhaust mass flow ṁa + ṁf.

Specific thrust is thrust per unit air mass flow:

F/ṁa = (1 + f)Ve − V0 + [(Pe − P0)Ae/ṁa]

where f = ṁf/ṁa is the fuel–air ratio.

6.2 Propulsive efficiency

When pressure thrust is negligible, propulsive efficiency is approximately

ηp = useful\ propulsive\ power/jet\ power = 2V0/(Ve + V0)

This result shows that propulsive efficiency improves when exhaust velocity is closer to aircraft speed. Very high exhaust velocity produces strong thrust but wastes more kinetic energy in the wake. This is why turbofans, which accelerate a larger mass of air by a smaller velocity increase, are generally more efficient than pure turbojets for subsonic transport aircraft.

The thermal efficiency of propulsion measures how effectively fuel chemical energy becomes jet kinetic-energy increase. Overall efficiency is commonly expressed as

ηo = ηth ηp

The product form is useful conceptually, although exact definitions must be applied consistently to the selected control volume and fuel heating value.

7. Applied case: choosing a gas-turbine configuration

Background

Consider an illustrative industrial facility requiring approximately 35 MW of electrical power. The plant operates continuously, has access to natural gas, and faces strict limits on fuel consumption and nitrogen-oxide emissions. Ambient temperature varies significantly between winter and summer.

Problem

A simple-cycle gas turbine can meet peak demand, but its efficiency falls during hot weather. Management is considering inlet-air cooling, regeneration, or a combined-cycle upgrade. The engineering team must distinguish measures that increase power from those that primarily improve efficiency.

Analysis

High ambient temperature increases compressor inlet temperature. For a fixed pressure ratio, compressor exit temperature and compressor work increase. Air density decreases, so the mass flow through a fixed-volume inlet generally falls. Both effects reduce net power.

Regeneration can reduce fuel consumption if turbine exhaust remains hotter than compressor discharge. However, a large industrial regenerator introduces pressure losses and substantial capital cost. Its value is greatest when the plant operates many hours at part load or when fuel cost is high.

Inlet-air cooling increases air density and may restore mass flow and power during hot conditions. It does not directly recover exhaust heat and can require water or refrigeration energy.

A combined-cycle plant directs gas-turbine exhaust to a heat-recovery steam generator and a steam turbine. It usually provides substantially higher overall efficiency than a simple-cycle plant, but it has greater capital cost, startup time, water requirements, and operational complexity.

Approach and solution

The engineering team should establish a common basis: net electrical power, ambient conditions, fuel lower heating value, emissions, annual operating hours, water availability, and maintenance schedule. It should then compare:

  • simple-cycle operation;
  • simple cycle with inlet cooling;
  • regenerative operation;
  • combined-cycle operation.

Each option should be evaluated using component efficiencies and pressure losses rather than ideal-cycle efficiency alone. A life-cycle cost model should include fuel, carbon cost, maintenance, availability, and replacement expenditure.

Outcome

For a plant operating primarily at peak demand, inlet cooling may provide valuable additional capacity. For high annual operating hours, combined cycle generally offers the strongest fuel-efficiency case. Regeneration may be attractive in smaller or distributed applications where a steam bottoming cycle is impractical. The final decision depends on local fuel prices, electricity prices, environmental requirements, and required flexibility.

Learning points

  1. Maximum thermal efficiency is not the only design objective.
  2. Intercooling and inlet cooling primarily address power loss associated with compression and air density.
  3. Regeneration is useful only when an adequate temperature difference exists.
  4. Real pressure losses and component efficiencies can reverse conclusions drawn from ideal models.
  5. Environmental and economic constraints must be included in thermodynamic design.

8. Limitations, safety, and contemporary practice

Ideal cycles are diagnostic models, not complete engine simulations. Important omitted effects include variable specific heats, combustion chemistry, dissociation, finite-rate combustion, heat transfer, friction, leakage, pressure losses, blade cooling, turbine clearance, compressor surge, choking, and transient operation.

Gas turbines operate at high pressure and temperature. Professional practice requires pressure-vessel compliance, fuel-system safeguards, overspeed protection, fire detection, emergency shutdown systems, combustion monitoring, and systematic hazard analysis. Engineers must also address carbon dioxide, nitrogen oxides, carbon monoxide, unburned hydrocarbons, noise, and water consumption.

Digital control systems increasingly use real-time measurements and physics-informed models to detect compressor degradation, fouling, vibration, and abnormal combustion. Artificial intelligence can assist predictive maintenance and anomaly detection, but it should not replace independent safety systems, validated physical models, or qualified engineering judgment. Data quality, cybersecurity, explainability, and failure-mode testing are essential.

Contemporary development focuses on higher turbine inlet temperature, advanced cooling, ceramic-matrix composites, variable-geometry compressors, low-emission combustors, sustainable aviation fuels, hydrogen blending, hybrid-electric propulsion, and carbon capture for stationary plants. Each development introduces trade-offs involving materials, combustion stability, infrastructure, lifecycle emissions, and regulatory approval.

Best-practice analysis procedure

A reliable cycle analysis should follow this sequence:

  1. Define the control volume and whether the cycle is closed or open.
  2. State the working-fluid model and property assumptions.
  3. Establish inlet conditions, pressure ratios, maximum temperatures, and mass flow.
  4. Calculate ideal isentropic states first.
  5. Apply compressor, turbine, combustor, regenerator, and nozzle efficiencies.
  6. Include pressure losses at inlets, combustors, heat exchangers, and exhaust systems.
  7. Calculate net work, heat input, efficiency, power, thrust, and emissions indicators.
  8. Check signs, units, temperature scales, and physical plausibility.
  9. Compare results with realistic component performance and operating limits.
  10. Perform sensitivity analysis for ambient temperature, pressure ratio, turbine inlet temperature, and degradation.

Chapter Summary

Gas-power cycles provide idealized frameworks for analyzing engines and propulsion systems. The Otto cycle represents spark-ignition engines and shows the strong influence of compression ratio. The Diesel cycle models constant-pressure heat addition, while the dual cycle combines constant-volume and constant-pressure heat addition. The Brayton cycle represents gas turbines, with pressure ratio controlling ideal efficiency and compressor work strongly affecting net output.

Real gas turbines require compressor and turbine isentropic efficiencies, combustor efficiency, mechanical efficiency, and pressure-loss models. Regeneration reduces fuel input when hot exhaust can preheat compressed air. Intercooling reduces compressor work and generally increases specific power, while reheating increases turbine work but adds heat input. Their combined value must be assessed using net performance and economic conditions.

Turbojet thrust results from momentum change and, when present, pressure thrust. Propulsive efficiency improves when exhaust velocity is not excessively higher than flight velocity, which explains the importance of turbofans for many subsonic applications. Practical design must combine thermodynamics with materials, combustion, controls, safety, environmental regulation, economics, and maintainability.

Key Terms

Air-standard cycle: An idealized cycle in which air is treated as the working fluid and combustion is replaced by heat addition.

Back-work ratio: The ratio of compressor work to turbine work in a gas-turbine plant.

Brayton cycle: The ideal gas-turbine cycle consisting of isentropic compression and expansion with constant-pressure heat transfer.

Combustor efficiency: The fraction of theoretical fuel energy that is released effectively in the combustor.

Compression ratio: The ratio of maximum to minimum cylinder volume in a reciprocating engine.

Cutoff ratio: The volume ratio during constant-pressure heat addition in the Diesel cycle.

Dual cycle: A cycle with heat addition partly at constant volume and partly at constant pressure.

Isentropic efficiency: A measure comparing actual component performance with ideal reversible adiabatic performance.

Intercooling: Cooling between stages of compression.

Otto cycle: The ideal cycle associated with spark-ignition engines.

Pressure ratio: The ratio of compressor exit pressure to compressor inlet pressure in a gas turbine.

Propulsive efficiency: The fraction of jet power converted into useful propulsive power.

Regeneration: Transfer of turbine-exhaust heat to compressed air before combustion.

Reheating: Addition of heat between stages of turbine expansion.

Specific thrust: Thrust per unit air mass flow rate.

Thermal efficiency: Net work output divided by heat supplied.

Turbojet: A gas-turbine engine that produces thrust mainly through a high-speed exhaust jet.

Turbine inlet temperature: The temperature of gas entering the first turbine stage.

Review Questions

Short Answer Questions

  1. State the four processes of the ideal Otto cycle.
  2. Define the pressure ratio of a Brayton cycle.
  3. Why does a real compressor have an exit temperature higher than the corresponding isentropic exit temperature?
  4. What is the purpose of a regenerator?
  5. Distinguish between thrust and propulsive efficiency.

Long Answer Questions

  1. Derive the cold-air-standard efficiency equation for the Otto cycle and explain the role of compression ratio.
  2. Explain the operation and efficiency characteristics of the Diesel and dual cycles.
  3. Describe the effects of compressor pressure ratio on ideal Brayton-cycle efficiency and net work.
  4. Compare intercooling, reheating, and regeneration in terms of thermodynamic benefit and practical limitation.
  5. Explain how compressor and turbine isentropic efficiencies influence a real gas-turbine plant.

Analytical / Application Questions

  1. An ideal Brayton cycle operates between T1 = 290\ K and T3 = 1350\ K with rp = 10, γ = 1.4, and cp = 1.005\ kJ/(kg·K). Determine compressor work, turbine work, net work, and thermal efficiency.
  2. A two-stage compressor has an overall pressure ratio of 16. Assuming perfect intercooling, equal stage pressure ratios, constant properties, and equal stage efficiencies, determine the intermediate pressure and explain why it minimizes ideal compression work.
  3. A turbojet flies at V0 = 250\ m/s and exhausts gas at Ve = 600\ m/s, with negligible pressure thrust. Estimate the propulsive efficiency using the ideal momentum-based expression and discuss one limitation of that estimate.

Multiple Choice Questions with Answers

Question 1

For the cold-air-standard Otto cycle, increasing the compression ratio generally causes the ideal thermal efficiency to:

A. Decrease

B. Increase

C. Remain constant

D. Become zero

Correct Answer: B. Increase

Question 2

The ideal Brayton-cycle efficiency is primarily a function of:

A. Cylinder compression ratio

B. Cutoff ratio only

C. Pressure ratio and specific-heat ratio

D. Fuel mass flow only

Correct Answer: C. Pressure ratio and specific-heat ratio

Question 3

The back-work ratio of a gas turbine is defined as:

A. Compressor work divided by turbine work

B. Turbine work divided by compressor work

C. Net work divided by heat input

D. Heat rejection divided by heat input

Correct Answer: A. Compressor work divided by turbine work

Question 4

A real compressor has an isentropic efficiency less than unity because it:

A. Requires less work than an ideal compressor

B. Requires more work than the corresponding isentropic compressor

C. Has no temperature rise

D. Rejects all supplied heat

Correct Answer: B. Requires more work than the corresponding isentropic compressor

Question 5

Intercooling in a multistage compressor primarily:

A. Increases compressor work

B. Reduces compressor work

C. Eliminates turbine work

D. Raises the compression inlet pressure automatically

Correct Answer: B. Reduces compressor work

Question 6

Regeneration is most useful when:

A. Turbine exhaust is cooler than compressor exit air

B. Turbine exhaust is hotter than compressor exit air

C. There is no temperature difference in the heat exchanger

D. The combustor has zero pressure loss

Correct Answer: B. Turbine exhaust is hotter than compressor exit air

Question 7

For a two-stage ideal compressor with perfect intercooling, minimum work is obtained approximately when the intermediate pressure is:

A. The arithmetic mean of inlet and final pressures

B. The difference between inlet and final pressures

C. The geometric mean of inlet and final pressures

D. Equal to atmospheric pressure in every case

Correct Answer: C. The geometric mean of inlet and final pressures

Question 8

For a perfectly expanded turbojet nozzle, the pressure-thrust term is:

A. Positive and maximum

B. Negative

C. Zero

D. Equal to compressor work

Correct Answer: C. Zero

Question 9

If a turbojet exhaust velocity becomes much larger than flight velocity, propulsive efficiency generally:

A. Improves toward 100%

B. Decreases

C. Becomes independent of velocity

D. Equals the combustor efficiency

Correct Answer: B. Decreases

Question 10

In the ideal Diesel cycle, heat addition occurs at:

A. Constant volume

B. Constant pressure

C. Constant entropy

D. Constant temperature

Correct Answer: B. Constant pressure

Engineering Thermodynamics
Chapter 9 of 15