Air-standard Otto, Diesel and dual cycles

The air-standard Otto, Diesel and dual cycles: assumptions, process sequences, efficiency and MEP formulas, how the cycles compare, with Otto and dual-cycle worked examples.

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Why it matters

Air-standard cycles are the yardstick for every reciprocating engine: they give the highest efficiency a petrol or diesel engine could reach at a given compression ratio, and they explain why diesels run at high compression and why raising the compression ratio pays. Almost every engine numerical in university exams and GATE starts from one of these three cycles.

Key ideas

Air-standard assumptions. The working fluid is a fixed mass of air, treated as an ideal gas with constant specific heats (cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, γ = 1.4, R = 0.287 kJ/kg·K unless stated). Combustion is replaced by heat addition from an external source, and exhaust plus intake are replaced by heat rejection at constant volume. Compression and expansion are reversible adiabatic, i.e. isentropic. Friction, heat transfer to the walls, valve timing and pumping work are ignored. The cycle is closed, so the same air is reused.

Geometry. Swept volume Vs = V1 − V2, clearance volume Vc = V2, compression ratio r = V1/V2 = (Vs + Vc)/Vc.

Otto cycle (constant-volume cycle, ideal SI engine). 1–2 isentropic compression, 2–3 heat addition at constant volume, 3–4 isentropic expansion, 4–1 heat rejection at constant volume. The efficiency depends only on r and γ. Petrol engines are limited to r ≈ 8–12 because a higher ratio causes knock (covered in the SI combustion topic).

Diesel cycle (constant-pressure cycle, ideal slow-speed CI engine). 1–2 isentropic compression, 2–3 heat addition at constant pressure, 3–4 isentropic expansion, 4–1 heat rejection at constant volume. The cut-off ratio ρ = V3/V2 (> 1) measures how long fuel is "burnt" at constant pressure. Efficiency falls as ρ rises, so a diesel loses efficiency at high load.

Dual cycle (mixed or limited-pressure cycle, ideal high-speed CI engine). 1–2 isentropic compression, 2–3 heat addition at constant volume, 3–4 heat addition at constant pressure, 4–5 isentropic expansion, 5–1 heat rejection at constant volume. The pressure ratio α = p3/p2 describes the constant-volume part; ρ = V4/V3 the constant-pressure part. With ρ = 1 it becomes the Otto cycle; with α = 1 it becomes the Diesel cycle. It is closest to a real modern diesel, where part of the fuel burns rapidly (premixed) and the rest burns while the piston moves.

Comparisons (standard results).

  • Same compression ratio and same heat input: η_Otto > η_Dual > η_Diesel.
  • Same maximum pressure and temperature (and same heat rejection): η_Diesel > η_Dual > η_Otto. This is the fair comparison for real engines, because the diesel is allowed a much higher r, which is why diesels are more efficient in practice.
  • Efficiency rises with r and with γ; it does not depend on the amount of heat added for the Otto cycle, but does for Diesel and dual (through ρ).

Mean effective pressure (MEP). The constant pressure that, acting over one full stroke, would give the net work of the cycle. It is the measure of how much work a given cylinder size delivers, so it lets you compare engines of different size.

Limits. Real engines reach roughly 50–60 % of the air-standard efficiency because specific heats rise with temperature, gases dissociate, combustion takes time and heat is lost to the walls. These are treated in the next topic, fuel-air and actual cycles.

Formulas

r = V1 / V2 = (Vs + Vc) / Vc r compression ratio (–); Vs swept volume (m³); Vc clearance volume (m³).

η_Otto = 1 − 1 / r^(γ−1) γ = cp/cv (–). Ideal Otto cycle, constant specific heats.

η_Diesel = 1 − [1 / r^(γ−1)] · (ρ^γ − 1) / [γ·(ρ − 1)] ρ = V3/V2 cut-off ratio (–). Ideal Diesel cycle. The bracketed factor is always > 1, so η_Diesel < η_Otto at the same r.

η_Dual = 1 − [1 / r^(γ−1)] · (α·ρ^γ − 1) / [(α − 1) + γ·α·(ρ − 1)] α = p3/p2 pressure ratio (–); ρ cut-off ratio (–). Ideal dual cycle.

T2 = T1·r^(γ−1), p2 = p1·r^γ Isentropic compression; T in K, p in kPa or Pa (same unit on both sides).

q_in = cv·(T3 − T2) (constant volume), q_in = cp·(T3 − T2) (constant pressure) q in kJ/kg; cp, cv in kJ/kg·K; T in K.

η = w_net / q_in = 1 − q_out / q_in w_net net work per kg (kJ/kg).

MEP = w_net / (v1 − v2) = w_net / [v1·(1 − 1/r)] MEP in kPa when w in kJ/kg and v in m³/kg; v1 = R·T1/p1.

Worked examples

Example 1 (standard): Otto cycle. An air-standard Otto cycle has r = 8. At the start of compression p1 = 100 kPa, T1 = 300 K. Heat added is 1800 kJ/kg. Take γ = 1.4, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K. Find efficiency, peak temperature, net work and MEP.

  1. η = 1 − 1/r^(γ−1) = 1 − 1/8^0.4 = 1 − 1/2.2974 = 1 − 0.4353 = 0.5647, i.e. 56.5 %.
  2. T2 = T1·r^(γ−1) = 300 × 2.2974 = 689.2 K; p2 = p1·r^γ = 100 × 18.379 = 1838 kPa.
  3. T3 = T2 + q_in/cv = 689.2 + 1800/0.718 = 689.2 + 2507.0 = 3196 K.
  4. w_net = η·q_in = 0.5647 × 1800 = 1016.5 kJ/kg.
  5. v1 = R·T1/p1 = 0.287 × 300/100 = 0.861 m³/kg; swept volume per kg = v1(1 − 1/8) = 0.7534 m³/kg.
  6. MEP = w_net/(v1 − v2) = 1016.5/0.7534 = 1349 kPa ≈ 13.5 bar.

Example 2 (GATE level): dual cycle. An air-standard dual cycle has r = 15, pressure ratio α = 1.5 and cut-off ratio ρ = 1.6. p1 = 100 kPa, T1 = 300 K; cp = 1.005, cv = 0.718 kJ/kg·K, γ = 1.4, R = 0.287 kJ/kg·K. Find the maximum pressure, efficiency and MEP.

  1. Compression: T2 = T1·r^(γ−1) = 300 × 15^0.4 = 300 × 2.9542 = 886.3 K; p2 = p1·r^γ = 100 × 44.313 = 4431 kPa.
  2. Constant-volume heating: T3 = α·T2 = 1.5 × 886.3 = 1329.4 K; p3 = α·p2 = 6647 kPa ≈ 66.5 bar (maximum pressure).
  3. Constant-pressure heating: T4 = ρ·T3 = 1.6 × 1329.4 = 2127.0 K.
  4. Expansion from V4 = ρ·V2 to V5 = V1: T5 = T4·(ρ/r)^(γ−1) = 2127.0 × (1.6/15)^0.4 = 2127.0 × 0.40852 = 868.9 K.
  5. Heat added: q_in = cv(T3 − T2) + cp(T4 − T3) = 0.718 × 443.1 + 1.005 × 797.6 = 318.2 + 801.6 = 1119.8 kJ/kg.
  6. Heat rejected: q_out = cv(T5 − T1) = 0.718 × 568.9 = 408.5 kJ/kg.
  7. η = 1 − q_out/q_in = 1 − 408.5/1119.8 = 0.635 (63.5 %). Check with the formula: (α·ρ^γ − 1)/[(α − 1) + γα(ρ − 1)] = (1.5 × 1.9309 − 1)/(0.5 + 1.26) = 1.8963/1.76 = 1.0775; η = 1 − 1.0775/2.9542 = 0.635.
  8. w_net = 1119.8 − 408.5 = 711.3 kJ/kg; v1 = 0.861 m³/kg; v1(1 − 1/15) = 0.8036 m³/kg; MEP = 711.3/0.8036 = 885 kPa ≈ 8.85 bar.

Common mistakes

  • Writing the Diesel factor as (ρ^(γ−1) − 1)/[γ(ρ − 1)]. The numerator is ρ^γ − 1. The wrong form gives a Diesel efficiency higher than Otto, which is impossible at the same r.
  • Calling all four Otto processes "adiabatic". Only compression and expansion are adiabatic (and reversible, so isentropic); heat addition and rejection are at constant volume.
  • Using cp for the constant-volume heat addition, or cv for the constant-pressure part of the dual cycle.
  • Stating "Diesel is more efficient than Otto" without the condition. At the same r it is less efficient; at the same peak pressure it is more efficient.
  • Using temperatures in °C in T2 = T1·r^(γ−1). Always use kelvin.
  • Dividing work by the total volume v1 instead of the swept volume v1 − v2 when finding MEP.
  • Using the cut-off ratio as a fraction of stroke (cut-off percentage) directly in the formula; convert first: ρ = 1 + (cut-off fraction)·(r − 1).

For GATE ME

Expect numericals asking for Otto or Diesel efficiency from r, ρ and γ; peak temperature or pressure after a given heat input; MEP and net work per kg or per cycle; and conversion of cut-off percentage of stroke to cut-off ratio. Conceptual questions test the order of efficiencies under "same r and heat input" versus "same peak pressure", and the process sequence of each cycle. Practise the T–s and p–v diagrams for all three cycles and do the arithmetic quickly with powers of r.

Quick check

  1. An Otto cycle has r = 10 and γ = 1.4. What is its efficiency?
  2. For the same compression ratio and heat input, rank Otto, Diesel and dual cycles by efficiency.
  3. In a Diesel cycle, what happens to efficiency when the cut-off ratio increases at fixed r?
  4. A dual cycle with α = 1 reduces to which cycle?
  5. Cut-off occurs at 5 % of the stroke in a Diesel engine with r = 15. What is the cut-off ratio?

Answers: 1. 1 − 1/10^0.4 = 60.2 %. 2. Otto > dual > Diesel. 3. It decreases. 4. The Diesel cycle. 5. ρ = 1 + 0.05 × 14 = 1.7.

Interview questions

All Automotive Engines, Emissions and Alternate Fuels interview questions

Try answering each one aloud before you open it.

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

    It is the ideal cycle for a spark-ignition engine, using a fixed mass of air as an ideal gas with constant specific heats. It has four processes: isentropic compression, heat addition at constant volume, isentropic expansion and heat rejection at constant volume. Its efficiency is η = 1 − 1/r^(γ−1), so it depends only on the compression ratio and γ, not on the amount of heat added. With r = 8 and γ = 1.4 it gives about 56.5 %.

  2. 2.Explain the air-standard Diesel cycle.Concept

    The Diesel cycle is the ideal cycle for a slow-speed compression-ignition engine: isentropic compression, heat addition at constant pressure, isentropic expansion and heat rejection at constant volume. The constant-pressure heat addition is described by the cut-off ratio ρ = V3/V2. Its efficiency is η = 1 − [1/r^(γ−1)]·(ρ^γ − 1)/[γ(ρ − 1)]; the bracketed factor exceeds 1, so efficiency falls as ρ (load) increases.

  3. 3.What is a dual cycle in thermodynamics?Concept

    The dual (mixed or limited-pressure) cycle adds heat partly at constant volume and partly at constant pressure: isentropic compression, constant-volume heating, constant-pressure heating, isentropic expansion and constant-volume heat rejection. It is described by the pressure ratio α = p3/p2 and cut-off ratio ρ. It models modern high-speed diesel engines, where part of the fuel burns almost instantly and the rest burns as the piston moves down. With ρ = 1 it becomes the Otto cycle and with α = 1 the Diesel cycle.

  4. 4.Why is the Otto cycle used to model petrol engines?Application

    In a spark-ignition engine the premixed charge burns very quickly after the spark, near TDC, while the piston barely moves, so heat release is close to constant volume. The Otto cycle represents this with constant-volume heat addition between isentropic compression and expansion, and replaces exhaust blowdown with constant-volume heat rejection. It gives the upper limit of efficiency for a given compression ratio, which is why raising r (up to the knock limit) is the main route to better efficiency.

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

    For a fixed cut-off ratio, the thermal efficiency rises because the 1/r^(γ−1) factor falls; the end-of-compression temperature and pressure also rise. In a real CI engine the higher compression temperature shortens ignition delay, which reduces diesel knock and helps cold starting. The penalties are higher peak pressures (heavier structure), more friction and higher NOx; beyond about 16–20 the friction and heat losses cancel most of the gain.

  6. 6.Why is the dual cycle considered a better model of a real diesel engine than the Otto or Diesel cycle?Application

    A real high-speed diesel burns fuel in two phases: the fuel injected during the delay period burns rapidly as a premixed charge (nearly constant volume), and the rest burns as it is injected while the piston moves (nearer constant pressure). The dual cycle captures both with α and ρ, so its peak pressure and efficiency are closer to measured values. It also shows that limiting peak pressure forces more of the heat into the constant-pressure part, at some cost in efficiency.

  7. 7.Calculate the thermal efficiency of an Otto cycle with a compression ratio of 8:1 and a specific heat ratio (γ) of 1.4.Numerical

    η = 1 − 1/r^(γ−1) = 1 − 1/8^0.4. Since 8^0.4 = 2.297, 1/2.297 = 0.435, so η = 1 − 0.435 = 0.565, i.e. about 56.5 %. This is the ideal upper limit; a real petrol engine at r = 8 reaches roughly half of it as brake thermal efficiency.

  8. 8.For a Diesel cycle, if the compression ratio is 16:1 and the cutoff ratio is 2, with a specific heat ratio (γ) of 1.4, calculate the thermal efficiency.Numerical

    η = 1 − [1/r^(γ−1)]·(ρ^γ − 1)/[γ(ρ − 1)]. Here 16^0.4 = 3.031, so 1/r^(γ−1) = 0.330; 2^1.4 = 2.639, so (2.639 − 1)/(1.4 × 1) = 1.171. Then η = 1 − 0.330 × 1.171 = 1 − 0.386 = 0.614, i.e. about 61.4 %. An Otto cycle at the same r = 16 would give 67.0 %.

  9. 9.Explain why the specific heat ratio (γ) is important in analyzing air-standard cycles.Concept

    γ = cp/cv sets the slope of the isentropes, through T2/T1 = r^(γ−1) and p2/p1 = r^γ, so it fixes how much the temperature rises on compression and falls on expansion. A higher γ gives a higher cycle efficiency at the same compression ratio. Air-standard analysis uses γ = 1.4, but real combustion products have γ nearer 1.25–1.3 at high temperature, which is one reason fuel-air cycle efficiencies are lower than air-standard values.

  10. 10.What are the main differences between the Otto and Diesel cycles in terms of process and application?Concept

    The Otto cycle adds heat at constant volume and models spark-ignition petrol engines; the Diesel cycle adds heat at constant pressure and models compression-ignition engines; both reject heat at constant volume. At the same compression ratio and heat input the Otto cycle is more efficient. In practice diesels are more efficient because they run at r ≈ 15–22 (no knock limit on the air-only compression), whereas petrol engines are limited to about 8–12; at the same peak pressure the Diesel cycle is the more efficient one.

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