Air-standard Otto, Diesel and dual cycles

Cold air-standard Otto, Diesel and dual cycles: efficiency formulas, state temperatures, heat and work, mean effective pressure and how the cycles compare, with Otto and Diesel worked examples.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

Petrol engines, diesel engines, gensets and the engines in hybrid vehicles are all reciprocating internal-combustion engines. Their real cycles involve combustion, heat loss and changing gas composition, but air-standard cycles capture the main trends: why efficiency rises with compression ratio, why diesels can be more efficient than petrol engines, and how heat-addition style matters. These ideal cycles are the starting point for engine design and for the performance topic that follows.

Key ideas

Air-standard assumptions. The working fluid is air, an ideal gas, in a closed system; combustion is replaced by heat addition from an external source; the exhaust and intake strokes are replaced by heat rejection at constant volume; all processes are internally reversible. With constant specific heats at room temperature (c_p = 1.005, c_v = 0.718 kJ/kg·K, γ = 1.4) this is the cold air-standard analysis. Real engines reach lower efficiencies — roughly half to two-thirds of these values.

Geometric definitions. Compression ratio r = V₁/V₂ = (V_s + V_c)/V_c, where V_s is swept volume and V_c clearance volume. Cut-off ratio r_c = V₃/V₂ (Diesel). Pressure (explosion) ratio r_p = p₃/p₂ for constant-volume heat addition.

Otto cycle (spark-ignition model): 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. SI engines are limited to r ≈ 8–12 by knock (auto-ignition of the end gas).

Diesel cycle (slow-speed compression-ignition model): heat added at constant pressure (2→3) while the piston moves, then isentropic expansion and constant-volume rejection. For the same r, η_Diesel < η_Otto, because the cut-off term (r_c^γ − 1)/(γ(r_c − 1)) is always greater than 1. But diesels run at r ≈ 14–22 (no knock limit, since only air is compressed), so in practice they are more efficient. Efficiency falls as r_c (that is, load) increases.

Dual (limited-pressure, Sabathé) cycle (modern high-speed CI engines): part of the heat added at constant volume (2→x), the rest at constant pressure (x→3). It has two isentropic, two constant-volume and one constant-pressure process. It reduces to Otto when r_c = 1 and to Diesel when r_p = 1.

Comparisons (memorise the conditions).

  • 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 comparison that reflects real engines, where peak pressure is the structural limit.

Mean effective pressure (MEP). The constant pressure that, acting over the full stroke, would give the same net work per cycle. It compares engines of different sizes: a higher MEP means more work from the same displacement.

Formulas

r = V₁ / V₂, V_s = V₁ − V₂, r = 1 + V_s / V_c T₂ = T₁ r^(γ−1), p₂ = p₁ r^γ — isentropic compression (T in K). η_Otto = 1 − 1 / r^(γ−1) η_Diesel = 1 − [1 / r^(γ−1)] · (r_c^γ − 1) / (γ (r_c − 1)) η_Dual = 1 − [1 / r^(γ−1)] · (r_p r_c^γ − 1) / ((r_p − 1) + γ r_p (r_c − 1)) q_in = c_v (T₃ − T₂) (Otto), q_in = c_p (T₃ − T₂) (Diesel), q_in = c_v (T_x − T₂) + c_p (T₃ − T_x) (Dual) — kJ/kg. q_out = c_v (T₄ − T₁) — all three cycles. T₄ = T₃ (V₃ / V₄)^(γ−1) — isentropic expansion; for Diesel V₃/V₄ = r_c / r. MEP = w_net / (v₁ − v₂) = w_net / (v₁ (1 − 1/r)) — kPa when w is in kJ/kg and v in m³/kg. v₁ = R T₁ / p₁ — R = 0.287 kJ/kg·K for air.

Worked examples

Example 1 (standard, Otto cycle). An Otto cycle has r = 8. At the start of compression air is at 100 kPa, 300 K; 800 kJ/kg of heat is added. Cold air-standard (c_v = 0.718 kJ/kg·K, γ = 1.4, R = 0.287 kJ/kg·K). Find the peak temperature, efficiency, net work and MEP.

  1. r^(γ−1) = 8^0.4 = 2.2974; T₂ = 300 × 2.2974 = 689.2 K.
  2. T₃ = T₂ + q_in / c_v = 689.2 + 800/0.718 = 689.2 + 1114.2 = 1803.4 K.
  3. T₄ = T₃ / 2.2974 = 785.0 K; q_out = 0.718 × (785.0 − 300) = 348.2 kJ/kg.
  4. w_net = 800 − 348.2 = 451.8 kJ/kg; η = 451.8/800 = 0.565. Check: 1 − 1/2.2974 = 0.565 ✓.
  5. v₁ = 0.287 × 300/100 = 0.861 m³/kg; v₁ − v₂ = 0.861 × (1 − 1/8) = 0.7534 m³/kg; MEP = 451.8/0.7534 = 599.7 kPa.

T_max ≈ 1803 K, η ≈ 56.5 %, w_net ≈ 452 kJ/kg, MEP ≈ 600 kPa

Example 2 (GATE level, Diesel cycle). A Diesel cycle has r = 18 and r_c = 2; air at 100 kPa, 300 K at the start of compression. Take c_p = 1.005, c_v = 0.718 kJ/kg·K, γ = 1.4. Find η, net work and MEP.

  1. 18^0.4 = 3.1777; T₂ = 300 × 3.1777 = 953.3 K; p₂ = 100 × 18^1.4 = 5720 kPa.
  2. Constant-pressure heating: T₃ = r_c T₂ = 1906.6 K; q_in = 1.005 × (1906.6 − 953.3) = 958.1 kJ/kg.
  3. Expansion: T₄ = T₃ (r_c/r)^0.4 = 1906.6 × (2/18)^0.4 = 1906.6 × 0.41525 = 791.7 K.
  4. q_out = 0.718 × (791.7 − 300) = 353.0 kJ/kg; w_net = 958.1 − 353.0 = 605.1 kJ/kg.
  5. η = 605.1/958.1 = 0.632. Formula check: 1 − (1/3.1777) × (2^1.4 − 1)/(1.4 × 1) = 1 − 0.31469 × 1.6390/1.4 = 1 − 0.3684 = 0.632 ✓.
  6. MEP = 605.1 / (0.861 × (1 − 1/18)) = 605.1/0.8132 = 744 kPa.

η ≈ 63.2 %, w_net ≈ 605 kJ/kg, MEP ≈ 744 kPa

Example 3 (quick comparison). For r = 16 and γ = 1.4: η_Otto = 1 − 1/16^0.4 = 1 − 1/3.0314 = 67.0 %; η_Diesel with r_c = 2 is 61.4 %; a dual cycle with r_p = 1.5 and r_c = 1.6 gives 64.5 %. Same r, so Otto > Dual > Diesel, as expected.

Common mistakes

  • Writing r_c^(γ−1) instead of r_c^γ in the Diesel efficiency formula; this gives impossibly high efficiencies.
  • Using c_v for the constant-pressure heat addition of the Diesel cycle, or c_p for the heat rejection (which is at constant volume, so c_v).
  • Treating 8^0.4 as about 2.5 or forgetting to use kelvin; check powers on a calculator: 8^0.4 = 2.297, 16^0.4 = 3.031, 18^0.4 = 3.178.
  • Stating "Diesel is more efficient than Otto" without the condition. For the same r it is less efficient.
  • Taking the expansion ratio of a Diesel cycle as r. It is r/r_c.
  • Computing MEP with total volume v₁ instead of swept volume v₁ − v₂.

For GATE ME

This is a regular source of numericals: efficiency from r (and r_c, r_p), temperatures and pressures at each state, heat added and net work, MEP, and finding r from a given efficiency. Conceptual questions test the comparison of the three cycles under the two conditions and the effect of cut-off ratio. Practise the state-by-state chain with powers of r computed carefully.

Quick check

  1. Otto cycle with r = 10, γ = 1.4: efficiency?
  2. An Otto engine has a swept volume of 450 cm³ and a clearance volume of 50 cm³. What is r?
  3. For the same r and heat input, rank Otto, Diesel and Dual efficiencies.
  4. What happens to Diesel-cycle efficiency as the cut-off ratio increases?
  5. What compression ratio gives an Otto efficiency of 50 % with γ = 1.4?

Answers: 1. 1 − 1/10^0.4 = 1 − 0.398 = 60.2 %. 2. (450 + 50)/50 = 10. 3. Otto > Dual > Diesel. 4. It decreases. 5. r = 2^(1/0.4) = 5.66.

Interview questions

All Thermal Engineering 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 assumes that the working fluid is air, which behaves as an ideal gas, and that the combustion process is replaced by a heat addition process at constant volume.

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

    The air-standard Diesel cycle is an idealized thermodynamic cycle that models the operation of a compression-ignition engine. It consists of two adiabatic processes, one isochoric process, and one isobaric process. In this cycle, the heat is added at constant pressure, which differentiates it from the Otto cycle where heat is added at constant volume.

  3. 3.What is the dual cycle in thermal engineering?Concept

    The dual (limited-pressure or Sabathé) cycle adds heat partly at constant volume and partly at constant pressure. It has five processes: isentropic compression, constant-volume heat addition, constant-pressure heat addition, isentropic expansion and constant-volume heat rejection. It models modern high-speed compression-ignition engines better than the pure Diesel cycle, and it reduces to the Otto cycle when the cut-off ratio is 1 and to the Diesel cycle when the pressure ratio is 1.

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

    In a spark-ignition engine a premixed fuel–air charge burns very quickly after the spark, while the piston is near top dead centre, so heat release is close to constant volume — exactly the Otto cycle's assumption. Its efficiency 1 − 1/r^(γ−1) rises with compression ratio, but in petrol engines r is limited to about 8–12 by knock, the auto-ignition of the unburnt end gas.

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

    For a fixed cut-off ratio the efficiency rises, since η = 1 − (1/r^(γ−1))·(r_c^γ − 1)/(γ(r_c − 1)). In a real CI engine a higher r also gives hotter compressed air and a shorter ignition delay, which reduces diesel knock and aids cold starting. The limits are higher peak pressures, heavier engine structure, more friction and higher NOx, so practical ratios are about 14–22.

  6. 6.How does the efficiency of the dual cycle compare with the Otto and Diesel cycles?Application

    It depends on what is held fixed. For the same compression ratio and heat input, the order is Otto > Dual > Diesel, because constant-volume heat addition happens at the highest temperatures. For the same peak pressure and temperature, which is closer to real design limits, the order reverses: Diesel > Dual > Otto. In both cases the dual cycle lies between the other two.

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

    η = 1 − 1/r^(γ−1) = 1 − 1/8^0.4. Since 8^0.4 = 2.297, η = 1 − 0.435 = 0.565, or about 56.5 %. A common slip is to take 8^0.4 as about 2.5, which wrongly gives about 60 %.

  8. 8.For a Diesel cycle with compression ratio 16 and cut-off ratio 2, calculate the thermal efficiency. Assume γ = 1.4.Numerical

    η = 1 − (1/r^(γ−1))·(r_c^γ − 1)/(γ(r_c − 1)). Here 16^0.4 = 3.031, so 1/r^(γ−1) = 0.330, and 2^1.4 = 2.639, so the bracket is (2.639 − 1)/1.4 = 1.171. Then η = 1 − 0.330 × 1.171 = 1 − 0.386 = 0.614, or about 61.4 %, lower than the Otto value of 67 % at the same compression ratio.

  9. 9.Explain why the air-standard assumptions are used in analyzing these cycles.Concept

    The air-standard assumptions simplify the analysis of thermodynamic cycles by assuming that the working fluid is air, which behaves as an ideal gas. These assumptions eliminate the complexities of combustion and variable specific heats, allowing for a more straightforward mathematical treatment. While not entirely accurate, they provide a useful approximation for understanding the fundamental behavior of engine cycles.

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

    The Otto cycle adds heat at constant volume; the Diesel cycle adds it at constant pressure up to the cut-off point; both reject heat at constant volume. For the same compression ratio the Otto cycle is more efficient, because the Diesel efficiency carries an extra factor (r_c^γ − 1)/(γ(r_c − 1)) greater than 1. Real diesel engines are still more efficient than petrol engines because they compress only air and can use compression ratios of 14–22 without knock, against about 8–12 for SI engines.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousBrayton cycle and gas turbinesNext IC engine performance and combustion basics