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Thermodynamics · Chapter 9 of 10 · Advanced

Gas Power Cycles

Engines that burn fuel with air, from a car engine to a jet, are modelled as ideal gas cycles. A single compression ratio or pressure ratio sets the efficiency of each.

Readiness check

This chapter assembles cycles from earlier tools. Tick only what you can do closed-notes.

Use the isentropic relation T₂/T₁ = (P₂/P₁)^(k−1)/k.
Apply Q = mcΔT with c_v and c_p.
Compute thermal efficiency as net work over heat in.
Recall the Carnot limit for comparison.
Raise a ratio to a fractional power.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit isentropic relations in Chapter 7.

3 or more weak itemsReview specific heats in Chapter 4.

The core idea

A real combustion engine is idealised as a closed air cycle with two isentropic strokes and two heat-exchange strokes. For the Otto cycle the thermal efficiency depends only on the compression ratio.

η_Otto = 1 − 1/r^k−1η_Brayton = 1 − 1/r_p^(k−1)/k

The air-standard model treats the working fluid as air behaving as an ideal gas, replaces combustion with heat addition, and assumes the compression and expansion strokes are isentropic. The Otto cycle (spark ignition) adds heat at constant volume; the Diesel cycle (compression ignition) adds it at constant pressure; the Brayton cycle (gas turbine) adds and rejects heat at constant pressure in steady flow. In every case, a higher compression or pressure ratio means a higher efficiency.

The skill works when: you track the four states, use the isentropic relations, and identify the single ratio that sets efficiency.

The skill breaks down when: the wrong specific heat is used for a stroke, or the result is allowed to exceed the Carnot limit.

The concept. The air-standard Otto cycle on a P-V diagram: isentropic compression (1 to 2), constant-volume heat addition (2 to 3), isentropic expansion (3 to 4), and constant-volume heat rejection (4 to 1). The enclosed area is the net work.

The skills, taught in order

Five skills set the air-standard assumptions and then build the Otto, Diesel, and Brayton cycles with their efficiencies.

9.1 The air-standard assumptions

To make a combustion engine tractable, we assume the working fluid is air as an ideal gas, the cycle is closed, combustion is replaced by heat addition from outside, and all processes are internally reversible. The cold-air-standard version further takes constant specific heats at room temperature. These idealisations capture the trends while keeping the algebra simple.

9.2 The Otto cycle

The Otto cycle models the spark-ignition engine: isentropic compression, constant-volume heat addition, isentropic expansion, and constant-volume heat rejection. Its efficiency is η = 1 − 1/r^k−1, where r = V₁/V₂ is the compression ratio. Higher compression raises efficiency, until knock limits it in practice.

9.3 The Diesel cycle

The Diesel cycle adds heat at constant pressure instead, modelling compression ignition. Its efficiency, η = 1 − (1/r^k−1) [(r_c^k − 1)/(k(r_c − 1))], carries an extra cutoff-ratio factor r_c that is always greater than one, so for the same compression ratio a Diesel cycle is slightly less efficient than an Otto. Diesels win in practice by running much higher compression ratios.

Cycle	Heat addition	Thermal efficiency
Otto	constant volume	1 − 1/r^k−1
Diesel	constant pressure	1 − (1/r^k−1)[(r_c^k − 1)/(k(r_c − 1))]
Brayton	constant pressure (steady flow)	1 − 1/r_p^(k−1)/k

9.4 The Brayton cycle

The Brayton cycle is the gas turbine: a compressor, a constant-pressure combustor, a turbine, and constant-pressure heat rejection, all in steady flow. Its efficiency is η = 1 − 1/r_p^(k−1)/k, set by the pressure ratio r_p across the compressor. It powers jet engines and many power stations.

9.5 The back-work ratio

In a gas turbine the compressor consumes a large share of the turbine's output, often 40 to 50 percent. This back-work ratio, w_compressor/w_turbine, is far higher than in a steam plant, where pumping a liquid is cheap, and it makes the gas turbine sensitive to component efficiencies.

Engineering connection: the Brayton cycle pairs with the Rankine steam cycle of Chapter 10 in combined-cycle power plants, the most efficient fossil plants built.

Worked example 1: the Otto cycle efficiency

An air-standard Otto cycle has a compression ratio of 8. Air enters compression at 300 K, and the peak temperature after heat addition is 1800 K. With k = 1.4 and c_v = 0.718 kJ/kg·K, find the temperature after compression, the thermal efficiency, and the net work per kilogram.

Figure 1. The compression ratio alone fixes the Otto efficiency. The peak temperature then sets how much heat is added and therefore the net work per kilogram.

ProblemFind T₂, η, and the net work for the Otto cycle in Figure 1.
Given / findr = 8, T₁ = 300 K, T₃ = 1800 K, k = 1.4, c_v = 0.718 kJ/kg·K. Find T₂, η, w_net.
AssumptionsCold-air-standard cycle: ideal gas, constant specific heats, isentropic compression and expansion.
ModelThe isentropic relation gives T₂; the efficiency follows from r; heat added is c_v(T₃ − T₂) and net work is η times that.
EquationsT₂ = T₁ r^k−1η = 1 − 1/r^k−1w_net = η q_in = η c_v(T₃ − T₂)
SolveT₂ = 300 × 8^0.4 = 300 × 2.297 = 689 K. η = 1 − 1/2.297 = 0.565. q_in = 0.718 × (1800 − 689) = 797.5 kJ/kg, so w_net = 0.565 × 797.5 = 450 kJ/kg.
CheckThe Carnot limit between 300 K and 1800 K is 1 − 300/1800 = 0.833, comfortably above 0.565, as it must be. Real spark engines reach only about 0.30 because of friction, heat loss, and incomplete combustion.
ConclusionThe compression ratio is the master dial of the Otto engine. Raising it raises efficiency, which is why engineers push it to the edge of knock.

Result. T₂ = 689 K, η = 0.565, and w_net = 450 kJ/kg.

Worked example 2: the Brayton gas-turbine cycle

An air-standard Brayton cycle has a pressure ratio of 10. Air enters the compressor at 300 K and the turbine at 1300 K. With k = 1.4 and c_p = 1.005 kJ/kg·K, find the compressor-exit temperature, the thermal efficiency, the net work, and the back-work ratio.

Figure 2. In a gas turbine, air is compressed, heated in the burner, and expanded through the turbine. The compressor reclaims a large share of the turbine work, the back-work ratio.

ProblemFind T₂, η, w_net, and the back-work ratio for the Brayton cycle in Figure 2.
Given / findr_p = 10, T₁ = 300 K, T₃ = 1300 K, k = 1.4, c_p = 1.005 kJ/kg·K. Find T₂, η, w_net, bwr.
AssumptionsCold-air-standard, steady flow, isentropic compressor and turbine, constant-pressure heat exchange.
ModelIsentropic relations give T₂ and T₄; the work terms are enthalpy differences; efficiency is net work over heat in; the back-work ratio is compressor over turbine work.
EquationsT₂ = T₁ r_p^(k−1)/k, T₄ = T₃ / r_p^(k−1)/kw_t = c_p(T₃ − T₄), w_c = c_p(T₂ − T₁)η = 1 − 1/r_p^(k−1)/k
SolveT₂ = 300 × 10^0.2857 = 579 K, T₄ = 1300/1.931 = 673 K. w_t = 1.005(1300 − 673) = 629.8 kJ/kg, w_c = 1.005(579 − 300) = 280.6 kJ/kg, so w_net = 349 kJ/kg. η = 1 − 1/1.931 = 0.482, and bwr = 280.6/629.8 = 0.45.
CheckThe efficiency from the work balance (349.2/724.4 = 0.482) matches the closed-form value exactly. The back-work ratio near 0.45 is typical of gas turbines and far above a steam plant's.
ConclusionThe pressure ratio sets Brayton efficiency, but the high back-work ratio means small drops in compressor or turbine efficiency cost a lot of net output.

Result. T₂ = 579 K, η = 0.482, w_net = 349 kJ/kg, back-work ratio 0.45.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Wrong specific heat per stroke	Heat added or work off by the factor k	"Is this stroke at constant volume or pressure?"	Use c_v for constant-volume heat, c_p for constant-pressure.
Efficiency above Carnot	η exceeds 1 − T_L/T_H	"Is my result below the Carnot limit?"	Any cycle must fall under the Carnot value for its temperatures.
Ignoring the back-work ratio	Net work overstated for a gas turbine	"Did I subtract the compressor work?"	Net work is turbine minus compressor work.
Confusing the ratios	Otto and Brayton formulas swapped	"Is the controlling ratio volume or pressure?"	Otto uses compression ratio r; Brayton uses pressure ratio r_p.

Practice ladder

Level 1 · Direct skill

Find the Otto-cycle efficiency for a compression ratio of 10 (k = 1.4).

Show answer

η = 1 − 1/10^0.4 = 1 − 1/2.512 = 0.602, or 60.2 percent. Raising r from 8 to 10 lifts efficiency by about 4 points.

Level 2 · Mixed concept

The Brayton cycle of Worked Example 2 keeps the same temperatures but the pressure ratio rises to 16. What is the new efficiency?

Show answer

η = 1 − 1/16^0.2857 = 1 − 1/2.297 = 0.565, up from 0.482. A higher pressure ratio raises efficiency, though net work per kilogram has its own optimum.

Level 3 · Independent problem

A Diesel cycle has compression ratio 18 and cutoff ratio 2 (k = 1.4). Find its thermal efficiency.

Show answer

η = 1 − (1/18^0.4)[(2^1.4 − 1)/(1.4(2 − 1))] = 1 − (1/3.178)(1.6390/1.4) = 1 − 0.3147 × 1.1707 = 1 − 0.368 = 0.632, about 63 percent.

Level 4 · Transfer to real engineering

Explain why a Diesel engine, though slightly less efficient than an Otto engine at the same compression ratio, is more efficient in practice.

What good work looks like

The cutoff factor makes Diesel less efficient at equal r, but Diesels are not knock-limited and run compression ratios of 15 to 22 versus 8 to 11 for petrol, and the much higher r more than overcomes the cutoff penalty.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my cycle efficiency stays below the Carnot limit for these temperatures."

"Give me three cycles; I will name the ratio that controls each efficiency."

"What is the Otto efficiency?" Applying 1 − 1/r^k−1 yourself is the skill.

"Solve this gas turbine." Tracking the four states and the back-work is the point.

Portfolio task

Pick a real engine (a car engine, a jet, a peaking gas turbine). Estimate its controlling ratio, compute the ideal air-standard efficiency, and compare it to the real value, naming the losses.

Must include: a controlling ratio, an ideal efficiency below Carnot, and a real value with the gap explained.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. List the air-standard assumptions.

Air as an ideal gas, a closed cycle, heat addition in place of combustion, and internally reversible processes (constant specific heats for cold-air-standard).

2. Write the Otto efficiency.

η = 1 − 1/r^k−1, with r the compression ratio.

3. Write the Brayton efficiency.

η = 1 − 1/r_p^(k−1)/k, with r_p the pressure ratio.

4. What is the back-work ratio?

The compressor work divided by the turbine work, often 40 to 50 percent in a gas turbine.

5. How do Otto and Diesel differ?

Otto adds heat at constant volume, Diesel at constant pressure; the Diesel efficiency carries an extra cutoff-ratio factor.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the Otto and Brayton results from a blank page.

+3 daysCompute a Diesel efficiency and compare cycles.

+7 daysCarry cycle analysis into vapor cycles, Chapter 10.

+30 daysCombine Brayton and Rankine into a combined cycle.

Textbook mapping

Item	Mapping
Primary source	Borgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 12 (Power and Refrigeration Systems, Gaseous Working Fluids)
Cross-reference	Çengel and Boles, Ch. 9 · Moran, Ch. 9
Core topics	9.1 Air-standard assumptions · 9.2 Otto · 9.3 Diesel · 9.4 Brayton · 9.5 Back-work ratio
Engineering connection	The Brayton cycle joins the Rankine cycle in combined-cycle power plants.
Read next	Chapter 10: Vapor and Refrigeration Cycles.