Air-Standard Cycles: Otto, Diesel and Brayton

This chapter completes Section XE4.7 of XE4 Thermodynamics, one of the optional sections of the Engineering Sciences (XE) paper, and answers the last line of the heading “Thermodynamic Cycles”: “Air-standard cycles - Otto, Diesel, and Brayton cycles”. It teaches the air-standard assumptions, the Otto cycle (compression ratio, η = 1 − r^(1−γ), mean effective pressure), the Diesel cycle (cut-off ratio and its efficiency), the Brayton cycle (pressure ratio, η = 1 − r_p^((1−γ)/γ), work ratio, the effect of compressor and turbine irreversibility, and the pressure ratio for maximum net work) and the comparison of Otto and Diesel. The numericals are efficiencies, mean effective pressure and Brayton net work, each with γ, R, c_p and every temperature stated in the stem. Sign convention: heat added to the air and work done by the air are positive.

1. Air-standard assumptions

A real engine burns fuel, changes composition and exchanges mass with the atmosphere. To analyse the cycle, replace it by an air-standard cycle:

  • A fixed mass of air, an ideal gas, circulates in a closed loop.
  • Combustion is replaced by heat addition from an external source; exhaust and intake by heat rejection to the surroundings.
  • All processes are internally reversible.
  • The cold-air-standard version also takes constant specific heats at room temperature: c_p = 1.005 kJ/kg·K, c_v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, γ = 1.4.

Two geometric terms recur for piston engines. The compression ratio is r = V_max/V_min = V₁/V₂. The mean effective pressure is the constant pressure which, acting over the swept volume for one stroke, would give the same net work: mep = w_net/(v₁ − v₂). It lets engines of different size be compared. All three cycles below are the ideal cycles for the real devices they model, and their efficiencies overestimate real ones.

2. The Otto cycle: constant-volume heat addition

Spark-ignition engine. 1→2 isentropic compression, 2→3 constant-volume heat addition, 3→4 isentropic expansion (power stroke), 4→1 constant-volume heat rejection. Derivation: q_in = c_v(T₃ − T₂), q_out = c_v(T₄ − T₁). Because 1→2 and 3→4 are isentropes between the same two volumes, T₂/T₁ = T₃/T₄ = r^(γ−1), so (T₄ − T₁)/(T₃ − T₂) = T₁/T₂ and

η_Otto = 1 − T₁/T₂ = 1 − 1/r^(γ−1) = 1 − r^(1−γ).

So η depends only on r and γ. It does not depend on the heat added, on T₁ or on T₃. It rises with r, but with diminishing returns, and a real petrol engine cannot raise r freely because the compressed mixture would auto-ignite (knock). For r = 8 and γ = 1.4: 8^0.4 = 2.297, η = 1 − 0.4353 = 56.5 %.

Worked example (mean effective pressure): 1 kg of air, p₁ = 100 kPa, T₁ = 300 K, r = 8, q_in = 1800 kJ/kg, R = 0.287, γ = 1.4. Then v₁ = RT₁/p₁ = 0.287 × 300/100 = 0.861 m³/kg, w_net = η q_in = 0.5647 × 1800 = 1016.5 kJ/kg, swept volume v₁ − v₂ = v₁(1 − 1/8) = 0.7534 m³/kg, and mep = 1016.5/0.7534 = 1349 kPa. Always divide by the swept volume v₁ − v₂, not by v₁.

3. The Diesel cycle: constant-pressure heat addition

Compression-ignition engine. Only air is compressed, then fuel is injected and burns while the piston moves out, so heat addition 2→3 is at constant pressure. Steps 1→2 (isentropic compression), 3→4 (isentropic expansion) and 4→1 (constant-volume rejection) are as in the Otto cycle. The new parameter is the cut-off ratio ρ = V₃/V₂ = T₃/T₂, the fraction of the stroke over which heat is added.

With q_in = c_p(T₃ − T₂), q_out = c_v(T₄ − T₁) and the isentrope relations, the result is

η_Diesel = 1 − (1/r^(γ−1)) × (ρ^γ − 1)/[γ(ρ − 1)].

The bracket (ρ^γ − 1)/[γ(ρ − 1)] is always greater than 1 for ρ > 1 and tends to 1 as ρ → 1, when the Diesel cycle becomes the Otto cycle. Hence at the same r, η_Diesel < η_Otto, and at fixed r, a larger cut-off (heavier load, more heat added) lowers η.

Worked example: r = 18, ρ = 2, γ = 1.4. 18^(−0.4) = 0.3147; (2^1.4 − 1)/(1.4 × 1) = (2.639 − 1)/1.4 = 1.1708; η = 1 − 0.3147 × 1.1708 = 0.6316, i.e. 63.2 %. A Diesel engine nevertheless beats a petrol engine in practice because only air is compressed, so no knock limits r and it runs at r of 15 to 20 against about 8 to 10.

4. The Brayton cycle: the gas-turbine cycle

Steady-flow cycle: 1→2 isentropic compression in a compressor from p₁ to p₂ (pressure ratio r_p = p₂/p₁), 2→3 constant-pressure heat addition in a combustor, 3→4 isentropic expansion in a turbine, 4→1 constant-pressure heat rejection. Isentropic relations give T₂/T₁ = T₃/T₄ = r_p^((γ−1)/γ). Write x = r_p^((γ−1)/γ). Then q_in = c_p(T₃ − T₂), q_out = c_p(T₄ − T₁), and

η_Brayton = 1 − T₁/T₂ = 1 − 1/r_p^((γ−1)/γ) = 1 − r_p^((1−γ)/γ).

Efficiency depends only on r_p, but the net work depends on both r_p and the peak temperature T₃. Define the turbine work w_t = c_p(T₃ − T₄), the compressor work w_c = c_p(T₂ − T₁), the back work ratio w_c/w_t = T₁x/T₃ and the work ratio w_net/w_t = 1 − T₁x/T₃. The back work ratio of a gas turbine is large, often 40 to 60 % or more, since a gas is compressed, whereas for the Rankine cycle it is only about 1 % or less. That is why a small loss in compressor or turbine efficiency hurts a gas turbine so much.

⚠️ r_p and r are different
For the Brayton cycle the exponent is (γ − 1)/γ on the pressure ratio, not (γ − 1) on a volume ratio. Using r^(γ−1) with a pressure ratio, or r_p^(γ−1) in place of r_p^((γ−1)/γ), is the commonest slip. For γ = 1.4 the exponent is 0.2857.

5. Irreversibility and the pressure ratio for maximum net work

Let τ = T₃/T₁. For the ideal cycle w_net = c_pT₁[τ(1 − 1/x) − (x − 1)]. It is zero at r_p = 1 (x = 1) and zero again when T₂ = T₃ (x = τ); in between there is a maximum. Setting d(w_net)/dx = 0 gives τ/x² = 1, so x = √τ, i.e.

r_p,opt = (T₃/T₁)^(γ/[2(γ−1)]), at which T₂ = T₄ = √(T₁T₃) and w_net,max = c_pT₁(√τ − 1)².

For T₁ = 300 K, T₃ = 1200 K (τ = 4), γ = 1.4: r_p,opt = 4^1.75 = 11.31 and w_net,max = 1.005 × 300 × (2 − 1)² = 301.5 kJ/kg. The efficiency at that point is 1 − 1/√τ = 50 %. Note that the efficiency keeps rising with r_p beyond r_p,opt even though the net work falls: the highest-η and highest-work points are different.

With isentropic efficiencies η_c and η_t, w_t = η_t c_pT₃(1 − 1/x) and w_c = c_pT₁(x − 1)/η_c, and the actual compressor exit is T₂ = T₁[1 + (x − 1)/η_c]. The optimum becomes x² = η_cη_tτ, so irreversibility lowers the optimum pressure ratio as well as the work and efficiency.

Worked example: τ = 4, c_p = 1.005, γ = 1.4, η_c = 0.80, η_t = 0.85, r_p = 8. x = 8^0.2857 = 1.8114. w_t = 0.85 × 1.005 × 1200 × (1 − 1/1.8114) = 459.2 kJ/kg; w_c = 1.005 × 300 × 0.8114/0.80 = 305.8 kJ/kg; w_net = 153.4 kJ/kg, against 1.005 × 300 × [4 × 0.4479 − 0.8114] = 295.6 for the ideal machines. The back work ratio is 305.8/459.2 = 66.6 %. T₂ = 300 + 300 × 0.8114/0.80 = 604.3 K, q_in = 1.005 × (1200 − 604.3) = 598.7, so η = 153.4/598.7 = 25.6 %, against 44.8 % ideal. The optimum with these losses is x = √(0.8 × 0.85 × 4) = 1.649, r_p = 5.76, well below the ideal 11.31.

6. Comparing Otto and Diesel

Which cycle is better depends on what is held equal
Basis of comparisonMore efficientReason
Same compression ratio and same heat inputOttoDiesel bracket > 1, so η_Diesel < η_Otto
Same compression ratio and same heat rejectionOttosame states 1, 2, 4; on T-s the constant-volume line is steeper, so Otto adds more heat
Same maximum pressure and same heat inputDieselthe Diesel cycle can use a higher compression ratio within the pressure limit, so more of its heat is added at high temperature

The first two rows are the familiar “Otto beats Diesel” statements and the third is the reverse. A question must always state the basis. Real practice follows the third row, since the Diesel engine is limited by peak pressure rather than by knock. In practice the comparison is also affected by the fuel, the engine speed and the mean effective pressure achieved.

7. The three cycles side by side

Summary of the air-standard cycles
CycleHeat addition / rejectionDesign variableIdeal efficiency
Ottoconstant volume / constant volumer1 − r^(1−γ)
Dieselconstant pressure / constant volumer and ρ1 − r^(1−γ)(ρ^γ − 1)/[γ(ρ − 1)]
Braytonconstant pressure / constant pressurer_p (and T₃)1 − r_p^((1−γ)/γ)

On the p-v diagram the Otto and Diesel cycles differ only in the heat-addition leg, a vertical line against a horizontal line; on T-s the Otto heat-addition leg is a constant-volume curve and the Diesel one a flatter constant-pressure curve. For the Otto and Brayton cycles the ideal efficiency can be written as η = 1 − T₁/T₂, with T₂ the temperature at the end of the isentropic compression (the Diesel formula carries the extra cut-off bracket). A larger T₂/T₁ gives a higher efficiency, so what matters is how much temperature rise the compression gives, not the numerical value of r or r_p alone.

Check: Otto with r = 8 has T₂/T₁ = 8^0.4 = 2.297 and η = 56.5 %. Brayton with r_p = 8 has T₂/T₁ = 8^0.2857 = 1.811 and η = 1 − 1/1.811 = 44.8 %. The Brayton figure is smaller because a pressure ratio of 8 compresses by a smaller temperature ratio than a volume ratio of 8 does.

Key takeaways

  • Air-standard analysis uses a fixed mass of ideal-gas air, heat addition and rejection in place of combustion and exhaust, and internally reversible processes; the cold-air-standard value is γ = 1.4.
  • Otto: η = 1 − r^(1−γ), independent of heat input; mep = w_net/(v₁ − v₂), with the swept volume in the denominator.
  • Diesel: η = 1 − r^(1−γ)(ρ^γ − 1)/[γ(ρ − 1)]; the bracket exceeds 1, so at the same r it is less efficient than Otto, and a larger cut-off lowers η.
  • Brayton: η = 1 − r_p^((1−γ)/γ) depends on r_p alone, but net work depends on r_p and T₃; the back work ratio is large.
  • Maximum net work: r_p,opt = (T₃/T₁)^(γ/[2(γ−1)]) for ideal machines (T₂ = T₄), and x² = η_cη_tτ with irreversibility, which lowers r_p,opt.
  • Otto beats Diesel at equal compression ratio; Diesel beats Otto at equal maximum pressure and heat input; always state the basis.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which of the following is NOT an assumption of the cold-air-standard analysis of an engine cycle?

    1. The working fluid is air behaving as an ideal gas with constant specific heats
    2. All the processes of the cycle are internally reversible
    3. Combustion is replaced by heat addition from an external source
    4. Fuel raises the fluid mass and changes its composition
    Show answer

    Answer: D — Fuel raises the fluid mass and changes its composition

    The air-standard model keeps a fixed mass of unchanging air in a closed loop. Fuel addition and composition change are exactly the real-engine features it removes; the other three statements are part of the model.

  2. An air-standard Otto cycle with γ = 1.4 has its compression ratio doubled from 8 to 16. The ideal thermal efficiency changes from 56.5 % to about

    1. 113 %, since efficiency doubles with compression ratio
    2. 67.0 %, an increase of about 10.5 percentage points
    3. 79.0 %, because the efficiency follows 1 − 1/r
    4. 56.5 %, because it depends only on the heat input
    Show answer

    Answer: B — 67.0 %, an increase of about 10.5 percentage points

    η = 1 − r^(−0.4): for r = 16, 16^0.4 = 3.031 and η = 1 − 0.330 = 0.670. The gain is real but diminishing. Using 1 − 1/r (the exponent γ − 1 omitted) would give 93.8 %, not 79 %, and a value above 100 % is impossible; the efficiency depends on r and γ, not on the heat input.

  3. For the same compression ratio and the same heat input, the air-standard Otto cycle compared with the Diesel cycle has

    1. a lower thermal efficiency, because heat is added at constant volume
    2. the same thermal efficiency, because both have the same compression ratio
    3. a higher thermal efficiency, because the Diesel cut-off bracket exceeds 1
    4. a higher thermal efficiency only when the cut-off ratio exceeds 2
    Show answer

    Answer: C — a higher thermal efficiency, because the Diesel cut-off bracket exceeds 1

    η_Diesel = 1 − r^(1−γ) × (ρ^γ − 1)/[γ(ρ − 1)] and the bracket is above 1 for every ρ > 1, so the Otto cycle (bracket = 1) is always better at equal r. Equal compression ratio alone does not give equal efficiency, and there is no threshold on ρ.

  4. In an air-standard Diesel cycle the compression ratio is held fixed and the engine load is increased, so more heat is added and the cut-off ratio rises. The thermal efficiency

    1. decreases, since the Diesel bracket rises with ρ
    2. increases, since the mean temperature of heat addition rises
    3. is unchanged with load, exactly as it is in the Otto cycle
    4. first increases and then decreases, with a maximum near ρ = γ
    Show answer

    Answer: A — decreases, since the Diesel bracket rises with ρ

    The bracket equals 1 at ρ = 1 and grows as ρ grows, so at fixed r the efficiency falls with load. It is not constant as in the Otto cycle, where η is independent of heat input, and there is no maximum at ρ = γ.

  5. For an ideal air-standard Brayton cycle with fixed compressor inlet temperature T₁ and fixed turbine inlet temperature T₃, as the pressure ratio is raised from 1 upwards the efficiency

    1. rises steadily, while the net work first rises and then falls
    2. rises, and the net work rises with it up to the limit r_p = τ^(γ/(γ−1))
    3. peaks at the pressure ratio of maximum net work and then falls
    4. is constant, since it depends only on T₃/T₁
    Show answer

    Answer: A — rises steadily, while the net work first rises and then falls

    η = 1 − r_p^(−(γ−1)/γ) increases monotonically with r_p, while the net work is zero at r_p = 1, rises to a maximum at r_p,opt and returns to zero when T₂ reaches T₃. So the pressure ratio for maximum efficiency and for maximum work are different. Efficiency does not depend on T₃/T₁.

  6. A gas turbine has compressor and turbine isentropic efficiencies below 100 % at a given pressure ratio. Compared with the ideal cycle at the same pressure ratio and temperatures T₁ and T₃, the actual cycle has

    1. a lower compressor exit temperature and a higher net work, because less heat is needed
    2. a higher thermal efficiency, since less heat is rejected in the exhaust
    3. an unchanged net work, since the work ratio absorbs the losses
    4. higher compressor work, lower turbine and net work, higher back work ratio
    Show answer

    Answer: D — higher compressor work, lower turbine and net work, higher back work ratio

    The compressor needs w_c = c_pT₁(x − 1)/η_c, more than ideal, and the turbine delivers only η_t times the ideal drop, so w_net falls and w_c/w_t rises. The hotter compressor exit slightly reduces the heat needed, but efficiency still falls because w_net falls faster. The exhaust is hotter, not cooler.

  7. For an ideal Brayton cycle with a fixed compressor inlet temperature, if the maximum cycle temperature T₃ is raised, the pressure ratio for maximum net work

    1. decreases, since the hotter turbine inlet allows less compression
    2. is unchanged, since it is fixed by γ and the isentropic exponent alone
    3. increases, since r_p,opt = (T₃/T₁)^(γ/[2(γ−1)])
    4. increases only until T₃ reaches twice the value of T₁
    Show answer

    Answer: C — increases, since r_p,opt = (T₃/T₁)^(γ/[2(γ−1)])

    x² = T₃/T₁ at the maximum, so a higher T₃ needs a larger pressure ratio, and the maximum work itself rises as (√τ − 1)². The optimum depends on T₃/T₁ as well as γ, and there is no cap at 2T₁.

  8. Practical compression-ignition engines run at much higher compression ratios than spark-ignition engines mainly because

    1. the Diesel cycle has the higher efficiency at equal compression ratio
    2. the heat addition at constant pressure needs no compression
    3. only air is compressed, so no mixture can knock during compression
    4. the cut-off ratio of the Diesel cycle can be made negative
    Show answer

    Answer: C — only air is compressed, so no mixture can knock during compression

    In a spark-ignition engine the compressed fuel-air mixture limits r through knock. In a compression-ignition engine fuel is injected near the end of compression, so r can be 15 to 20. At equal r the Diesel cycle is actually less efficient than Otto, so the claim of higher efficiency at equal r is false, and the cut-off ratio is always above 1.

  9. Otto and Diesel air-standard cycles are compared for the same maximum cycle pressure and the same heat input (the Diesel cycle therefore runs at a higher compression ratio). The more efficient cycle is

    1. the Otto cycle, because it adds its heat at constant volume
    2. the Diesel cycle, since more heat is added hot and less is rejected
    3. neither cycle, since the peak pressure and heat input are equal for both
    4. the Otto cycle, because it has the larger cut-off ratio
    Show answer

    Answer: B — the Diesel cycle, since more heat is added hot and less is rejected

    On the T-s diagram, with equal peak pressure and equal heat added, the Diesel cycle starts its heat addition from a higher temperature after its larger compression and ends its expansion at a lower entropy, so less heat is rejected. The comparison basis here differs from the equal-compression-ratio one, which is why the answer reverses; the Otto cycle has no cut-off ratio.

  10. For an ideal air-standard Brayton cycle with fixed T₁ and T₃ (with τ = T₃/T₁ and constant specific heats), which statements are correct?

    1. The net work is zero at r_p = 1
    2. The net work is zero when the compressor exit temperature equals T₃
    3. At the pressure ratio of maximum net work, T₂ = T₄
    4. The efficiency is greatest at the pressure ratio of maximum net work
    Show answer

    Answer: A — The net work is zero at r_p = 1; B — The net work is zero when the compressor exit temperature equals T₃; C — At the pressure ratio of maximum net work, T₂ = T₄

    At r_p = 1 nothing is compressed or expanded. When T₂ = T₃ the turbine work equals the compressor work. The maximum satisfies x² = τ, i.e. T₂ = T₁x = T₃/x = T₄. The efficiency keeps increasing with r_p past the work maximum, so it is not greatest there.

  11. For the air-standard Diesel cycle with constant specific heats, which statements are correct?

    1. At fixed cut-off ratio, the efficiency increases with the compression ratio
    2. As the cut-off ratio tends to 1, the efficiency tends to that of the Otto cycle of the same compression ratio
    3. The heat rejection is at constant volume
    4. At the same compression ratio the Diesel cycle is more efficient than the Otto cycle
    Show answer

    Answer: A — At fixed cut-off ratio, the efficiency increases with the compression ratio; B — As the cut-off ratio tends to 1, the efficiency tends to that of the Otto cycle of the same compression ratio; C — The heat rejection is at constant volume

    The factor r^(1−γ) falls as r rises, which raises η. The cut-off bracket tends to 1 as ρ → 1, which gives the Otto result. Exhaust heat is rejected 4→1 at constant volume, as in the Otto cycle. At equal r the Diesel efficiency is lower than Otto, so the claim of Diesel superiority at equal r is wrong.

  12. For an air-standard Otto cycle with fixed γ and fixed compression ratio, which statements are correct?

    1. The efficiency does not change when the heat added per kg is doubled
    2. The net work per kg is proportional to the heat added per kg
    3. The heat rejected per kg equals q_in(1 − η)
    4. The efficiency rises as the peak temperature T₃ rises
    Show answer

    Answer: A — The efficiency does not change when the heat added per kg is doubled; B — The net work per kg is proportional to the heat added per kg; C — The heat rejected per kg equals q_in(1 − η)

    η = 1 − r^(1−γ) contains neither q_in nor T₃, so w_net = ηq_in is proportional to q_in and q_out = q_in − w_net = q_in(1 − η). A hotter T₃ at fixed r only means more heat is added; the efficiency does not change.

  13. An air-standard Otto cycle has a compression ratio of 8. Take air with γ = 1.4 (constant specific heats). The ideal thermal efficiency is ____ % (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 56.5

    η = 1 − r^(1−γ) = 1 − 8^(−0.4). 8^0.4 = e^(0.4 ln 8) = e^0.8318 = 2.2974, so η = 1 − 0.4353 = 0.5647 = 56.5 %. Using 1 − 1/8 = 87.5 % omits the exponent.

  14. One kilogram of air (R = 0.287 kJ/kg·K, γ = 1.4) executes an air-standard Otto cycle with compression ratio 8. At the start of compression p₁ = 100 kPa and T₁ = 300 K, and the heat added per cycle is 1800 kJ/kg. The mean effective pressure is ____ kPa (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 1349

    η = 1 − 8^(−0.4) = 0.56472; w_net = 0.56472 × 1800 = 1016.5 kJ/kg. v₁ = RT₁/p₁ = 0.287 × 300/100 = 0.861 m³/kg; v₁ − v₂ = 0.861 × (1 − 1/8) = 0.75338. mep = 1016.5/0.75338 = 1349.3 ≈ 1349 kPa. Dividing by v₁ gives 1181, the usual slip.

  15. An air-standard Diesel cycle has a compression ratio of 18 and a cut-off ratio of 2. Take γ = 1.4 (constant specific heats). The ideal thermal efficiency is ____ % (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 63.2

    η = 1 − r^(1−γ)(ρ^γ − 1)/[γ(ρ − 1)]. 18^(−0.4) = 0.3147; 2^1.4 = 2.6390, so (2.6390 − 1)/(1.4 × 1) = 1.1707. η = 1 − 0.3147 × 1.1707 = 0.6316 = 63.2 %. The Otto value at r = 18 would be 68.5 %, which shows the bracket cost.

  16. An ideal air-standard Brayton cycle has a compressor inlet temperature of 300 K and a turbine inlet temperature of 1200 K; take γ = 1.4 and constant specific heats. The pressure ratio that gives the maximum net work is ____ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 11.31

    r_p,opt = (T₃/T₁)^(γ/[2(γ−1)]) = 4^(1.4/0.8) = 4^1.75 = 2^3.5 = 11.31. Check: x = r_p^0.2857 = 2 = √τ, so T₂ = 600 K = T₄ = √(300 × 1200). Using 4^3.5 gives 128, the slip of dropping the factor 2.

  17. A gas turbine works on the Brayton cycle with T₁ = 300 K, T₃ = 1200 K, pressure ratio 8, c_p = 1.005 kJ/kg·K and γ = 1.4. The compressor isentropic efficiency is 0.80 and the turbine isentropic efficiency is 0.85. The net work output is ____ kJ/kg (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 153.4

    x = 8^0.2857 = 1.8114. w_c = c_pT₁(x − 1)/η_c = 1.005 × 300 × 0.8114/0.80 = 305.8. w_t = η_t c_pT₃(1 − 1/x) = 0.85 × 1.005 × 1200 × 0.4479 = 459.2. w_net = 459.2 − 305.8 = 153.4 kJ/kg (ideal machines would give 295.6).