Air standard Otto, Diesel and dual cycles

Air-standard assumptions, the Otto, Diesel and dual (limited-pressure) cycles with their efficiency formulas, state temperatures, mean effective pressure and cycle comparisons.

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Why it matters

Piston engines still power trainer aircraft, light aircraft, UAVs and ground equipment, and the Otto and Diesel cycles are the models used to compare them. Air-standard analysis shows at a glance why compression ratio dominates efficiency, why diesels run at higher compression, and how the shape of heat addition changes performance. GATE uses these cycles for quick, formula-driven numericals.

Key ideas

Air-standard assumptions. The working fluid is air, an ideal gas, in a closed cycle. Combustion is replaced by heat addition from an external source and exhaust/intake by heat rejection at constant volume. All processes are internally reversible. Cold-air-standard adds constant specific heats at room temperature (γ = 1.4). These cycles overestimate real efficiency but show the right trends.

Terminology. Compression ratio r = V_max/V_min = V₁/V₂. Swept (displacement) volume = V₁ − V₂. Mean effective pressure (MEP) is the constant pressure that, acting over the full stroke, would give the same net work: MEP = w_net/(v₁ − v₂). It lets engines of different sizes be compared.

Otto cycle (spark ignition, idealised):

  • 1→2 isentropic compression
  • 2→3 constant-volume heat addition
  • 3→4 isentropic expansion
  • 4→1 constant-volume heat rejection

Efficiency depends only on r and γ: η = 1 − 1/r^(γ−1). r is limited (about 8–12) by knock — auto-ignition of the fuel–air mixture.

Diesel cycle (compression ignition): as Otto, but heat is added at constant pressure (2→3). The cut-off ratio rc = V₃/V₂ measures how long fuel injection continues. Because only air is compressed, knock is not a limit, so r = 14–22 is common. For the same r, Diesel is less efficient than Otto because the bracketed factor is greater than 1; diesels win in practice because they run at much higher r. Efficiency falls as rc increases (more load).

Dual (limited-pressure) cycle: heat added partly at constant volume (pressure ratio rp = p₃/p₂) and partly at constant pressure (cut-off ratio rc). It has five processes — isentropic compression, constant-volume addition, constant-pressure addition, isentropic expansion and constant-volume rejection — and fits real high-speed CI engines better. With rp = 1 it becomes Diesel; with rc = 1 it becomes Otto.

Comparisons.

  • Same r and same heat input: η_Otto > η_Dual > η_Diesel.
  • Same maximum pressure and temperature (same heat rejected): η_Diesel > η_Dual > η_Otto. This comparison is the more realistic one, because peak pressure is the structural limit.

Formulas

η_Otto = 1 − 1/r^(γ−1) η_Diesel = 1 − [1/r^(γ−1)]·(rc^γ − 1)/[γ·(rc − 1)] η_Dual = 1 − [1/r^(γ−1)]·(rp·rc^γ − 1)/[(rp − 1) + γ·rp·(rc − 1)] T₂ = T₁·r^(γ−1); p₂ = p₁·r^γ q_in = cv·(T₃ − T₂) (Otto); q_in = cp·(T₃ − T₂) (Diesel) q_out = cv·(T₄ − T₁) (all three) MEP = w_net / (v₁ − v₂) = w_net / [v₁·(1 − 1/r)]

  • r: compression ratio; rc: cut-off ratio V₃/V₂; rp: pressure ratio p₃/p₂ in the dual cycle
  • γ = cp/cv; cp = 1.005, cv = 0.718, R = 0.287 kJ/(kg·K) for air
  • T: K; p: kPa; v: m³/kg; q, w: kJ/kg; MEP: kPa

Worked examples

Example 1 (standard). An air-standard Otto cycle has r = 8. At the start of compression air is at 100 kPa and 300 K; 1800 kJ/kg of heat is added. Find T₂, T₃, T₄, the efficiency, the net work and the MEP.

  1. T₂ = T₁·r^(γ−1) = 300 × 8⁰·⁴ = 689.2 K
  2. T₃ = T₂ + q_in/cv = 689.2 + 1800/0.718 = 3196.2 K
  3. T₄ = T₃/r^(γ−1) = 3196.2/2.2974 = 1391.2 K
  4. q_out = cv·(T₄ − T₁) = 0.718 × 1091.2 = 783.5 kJ/kg
  5. w_net = q_in − q_out = 1016.5 kJ/kg; η = 1016.5/1800 = 0.565 (= 1 − 1/8⁰·⁴ ✓)
  6. v₁ = R·T₁/p₁ = 0.287 × 300/100 = 0.861 m³/kg; MEP = 1016.5/(0.861 × 0.875) = 1349 kPa

Answer: T₂ = 689 K, T₃ = 3196 K, T₄ = 1391 K, η = 56.5%, w_net = 1016.5 kJ/kg, MEP ≈ 1.35 MPa

Example 2 (GATE level). An air-standard Diesel cycle has r = 18 and rc = 2, starting at 100 kPa and 300 K. Find the temperatures, the efficiency and the MEP.

  1. T₂ = 300 × 18⁰·⁴ = 953.3 K
  2. Constant pressure: T₃ = T₂·rc = 1906.6 K
  3. Expansion from V₃ to V₄ = V₁: T₄ = T₃·(rc/r)^(γ−1) = 1906.6 × (2/18)⁰·⁴ = 791.7 K
  4. q_in = cp·(T₃ − T₂) = 1.005 × 953.3 = 958.1 kJ/kg
  5. q_out = cv·(T₄ − T₁) = 0.718 × 491.7 = 353.0 kJ/kg
  6. w_net = 605.0 kJ/kg; η = 605.0/958.1 = 0.631
  7. Check with the formula: 1 − (1/18⁰·⁴) × (2¹·⁴ − 1)/(1.4 × 1) = 1 − 0.3147 × 1.1707 = 0.632 ✓
  8. MEP = 605.0/(0.861 × (1 − 1/18)) = 744 kPa

Answer: T₂ = 953 K, T₃ = 1907 K, T₄ = 792 K, η ≈ 63.2%, MEP ≈ 744 kPa

Common mistakes

  • Using cp for Otto heat addition or cv for Diesel heat addition.
  • Writing T₄ = T₃/r^(γ−1) for the Diesel cycle; the expansion ratio is r/rc, not r.
  • Stating that the Diesel cycle is more efficient than Otto "at the same compression ratio" — it is the other way round.
  • Forgetting that the dual cycle has two constant-volume processes (one heat addition, one rejection).
  • Computing MEP with the total volume instead of the swept volume.
  • Using 1/r^γ instead of 1/r^(γ−1).

For GATE AE

Expect direct efficiency calculations for Otto and Diesel cycles, temperature at the end of compression or expansion, MEP, and comparisons of the three cycles under "same r" or "same peak pressure" conditions. Conceptual questions ask why diesels use higher compression ratios and how efficiency varies with cut-off ratio. Practise reaching every state temperature from T₁ in a few lines.

Quick check

  1. Find the Otto efficiency for r = 10, γ = 1.4.
  2. In a Diesel cycle, what happens to efficiency as the cut-off ratio increases at fixed r?
  3. For the same r and heat input, which is more efficient: Otto or Diesel?
  4. What does the dual cycle reduce to when the cut-off ratio is 1?
  5. Find the Diesel efficiency for r = 16, rc = 2, γ = 1.4.

Answers: 1. 60.2%. 2. It decreases. 3. Otto. 4. The Otto cycle. 5. 61.4%.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is an air standard Otto cycle?Concept

    The air standard Otto cycle is an idealized thermodynamic cycle that describes the functioning of a typical spark ignition piston engine. It consists of two adiabatic processes and two isochoric processes. The cycle is used to model the performance of gasoline engines.

  2. 2.Explain the Diesel cycle and its significance in engine design.Concept

    The air-standard Diesel cycle models a compression-ignition engine: isentropic compression, heat addition at constant pressure (fuel injection over a cut-off ratio rc = V₃/V₂), isentropic expansion and constant-volume heat rejection. Its efficiency is η = 1 − (1/r^(γ−1))·(rc^γ − 1)/[γ(rc − 1)], which for the same compression ratio is lower than Otto's and falls as rc (load) rises. Its significance is that only air is compressed, so knock does not limit r; diesels run at r ≈ 14–22 and therefore achieve higher real efficiencies than spark-ignition engines.

  3. 3.What is a dual cycle, and how does it differ from the Otto and Diesel cycles?Concept

    The dual (limited-pressure) cycle adds heat partly at constant volume and partly at constant pressure, so it has five processes: isentropic compression, constant-volume heat addition, constant-pressure heat addition, isentropic expansion and constant-volume heat rejection. It is described by r, the pressure ratio rp = p₃/p₂ and the cut-off ratio rc; with rc = 1 it becomes the Otto cycle and with rp = 1 the Diesel cycle. It represents high-speed compression-ignition engines better, because their combustion is partly rapid and partly controlled by injection.

  4. 4.Why is the compression ratio important in the Otto cycle?Application

    The compression ratio in the Otto cycle is crucial because it directly affects the thermal efficiency of the engine. A higher compression ratio leads to higher thermal efficiency, as it allows the engine to extract more work from the combustion process. However, it is limited by the onset of engine knocking.

  5. 5.What happens if the compression ratio is increased in a Diesel cycle?Application

    Increasing the compression ratio in a Diesel cycle generally improves the thermal efficiency of the engine. Diesel engines can operate at higher compression ratios than gasoline engines because they do not suffer from knocking. This is one reason why diesel engines are more fuel-efficient than gasoline engines.

  6. 6.How does the presence of an isobaric process in the Diesel cycle affect its efficiency?Application

    Adding heat at constant pressure means part of the heat goes in while the gas is already expanding, so on average it is added at a lower effective expansion ratio than in constant-volume addition. For the same compression ratio and heat input the Diesel cycle is therefore less efficient than the Otto cycle, and its efficiency falls as the cut-off ratio grows. The advantage of the Diesel cycle is indirect: constant-pressure addition with air-only compression permits much higher compression ratios and limits peak pressure, so at the same peak pressure the Diesel cycle is the more efficient.

  7. 7.Calculate the thermal efficiency of an Otto cycle with a compression ratio of 8:1. Assume γ (gamma) = 1.4.Numerical

    For the air-standard Otto cycle η = 1 − 1/r^(γ−1). With r = 8 and γ − 1 = 0.4: 8⁰·⁴ = 2.297, so 1/8⁰·⁴ = 0.4353 and η = 1 − 0.4353 = 0.565, i.e. about 56.5%. Real spark-ignition engines achieve far less because of combustion, heat-transfer and friction losses.

  8. 8.For a Diesel cycle, if the compression ratio is 16:1 and the cutoff ratio is 2, calculate the thermal efficiency. Assume γ = 1.4.Numerical

    η = 1 − (1/r^(γ−1))·(rc^γ − 1)/[γ(rc − 1)]. Here 1/16⁰·⁴ = 0.3299 and (2¹·⁴ − 1)/(1.4 × 1) = (2.639 − 1)/1.4 = 1.1707. So η = 1 − 0.3299 × 1.1707 = 1 − 0.3862 = 0.614, about 61.4%. An Otto cycle at the same r would give 1 − 0.3299 = 67.0%.

  9. 9.Explain how the dual cycle can be used to model real engine performance more accurately than the Otto or Diesel cycles alone.Application

    The dual cycle incorporates both constant volume and constant pressure heat addition processes, making it a more flexible model for real engine performance. It accounts for the fact that in actual engines, combustion does not occur entirely at constant volume or constant pressure. By combining elements of both the Otto and Diesel cycles, the dual cycle provides a more accurate representation of the thermodynamic processes occurring in real engines.

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