Air-standard cycles: Otto, Diesel and Brayton

Air-standard assumptions; Otto, Diesel, dual and Brayton cycles; efficiency, mean effective pressure, back-work ratio and the optimum pressure ratio.

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Why it matters

Petrol engines, diesel engines and gas turbines power vehicles, generator sets, aircraft and many factory utilities. Real combustion engines are messy, but air-standard cycles strip them down to a few ideal processes so you can see what controls efficiency and work output — compression ratio, pressure ratio and peak temperature — and compare designs quickly.

Key ideas

Air-standard assumptions.

  1. The working fluid is air, an ideal gas with constant specific heats (c_p = 1.005, c_v = 0.718 kJ/kg·K, γ = 1.4) — the "cold air-standard" model.
  2. All processes are internally reversible.
  3. Combustion is replaced by heat addition from an external source; exhaust is replaced by heat rejection that returns the air to its initial state, so the cycle is closed.
  4. Intake and exhaust strokes are ignored. Air-standard efficiencies are always higher than real engine efficiencies, but the trends they show are correct.

Engine terms. Top dead centre (TDC) and bottom dead centre (BDC); clearance volume V_c; swept volume V_s = (π/4)·D²·L; compression ratio r = (V_c + V_s)/V_c = V_max/V_min. Mean effective pressure (MEP) is the constant pressure that, acting over the full stroke, would give the same net work: MEP = W_net/V_s. It allows comparison of engines of different sizes.

Otto cycle (spark ignition). 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 and rises with r. In a petrol engine the fuel–air mixture is compressed, so r is limited to about 8–12 by knock (auto-ignition of the end gas).

Diesel cycle (compression ignition). Same as Otto except heat is added at constant pressure (2→3). The cut-off ratio ρ = V₃/V₂ describes how long fuel is injected. Because only air is compressed, there is no knock limit, and r is typically 14–22. For the same compression ratio, Otto is more efficient than Diesel (the bracket term in the Diesel formula is greater than 1); in practice diesels are more efficient because they run at much higher r. Diesel efficiency falls as the cut-off ratio (load) increases.

Dual (limited-pressure) cycle. Heat is added partly at constant volume and partly at constant pressure; it models modern high-speed diesels better. For the same r and heat input: η_Otto > η_Dual > η_Diesel. For the same maximum pressure and temperature: η_Diesel > η_Dual > η_Otto.

Brayton (Joule) cycle — gas turbines. 1→2 isentropic compression in a compressor; 2→3 constant-pressure heat addition in the combustor; 3→4 isentropic expansion in a turbine; 4→1 constant-pressure heat rejection. It is a steady-flow cycle, so work is calculated with enthalpy changes. Efficiency depends only on the pressure ratio r_p. The compressor absorbs a large fraction of the turbine work (the back-work ratio, often 40–60%), so component efficiencies matter a great deal. For given T₁ and T₃ (limited by turbine-blade materials), net work per kg is maximum at r_p,opt = (T₃/T₁)^(γ/(2(γ−1))), where T₂ = T₄ = √(T₁T₃).

Improvements to Brayton. Regeneration (heating compressed air with turbine exhaust) raises efficiency when T₄ > T₂, i.e. at low pressure ratios. Intercooling between compressor stages and reheating between turbine stages raise net work.

Formulas

η_Otto = 1 − 1/r^(γ−1)

  • r: compression ratio; γ = c_p/c_v.

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

  • ρ: cut-off ratio V₃/V₂.

η_Brayton = 1 − 1/r_p^((γ−1)/γ)

  • r_p: pressure ratio p₂/p₁.

T₂/T₁ = r^(γ−1) (piston engines), T₂/T₁ = r_p^((γ−1)/γ) (gas turbines)

  • Isentropic relations; T in K.

q_in = c_v·(T₃ − T₂) (Otto), q_in = c_p·(T₃ − T₂) (Diesel and Brayton)

  • Heat added per kg (kJ/kg).

MEP = W_net / (V₁ − V₂)

  • kPa if W in kJ and V in m³ (or per kg with specific volumes).

r_p,opt = (T₃/T₁)^(γ/(2(γ−1)))

  • Brayton pressure ratio for maximum net work.

Worked examples

Example 1 (standard): Otto cycle. Given: air-standard Otto cycle, r = 8, start of compression 100 kPa and 300 K, heat added 1500 kJ/kg. c_v = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, γ = 1.4. Find T₂, T₃, η, net work and MEP.

  1. T₂ = T₁·r^(γ−1) = 300 × 8^0.4 = 689.2 K
  2. T₃ = T₂ + q_in/c_v = 689.2 + 1500/0.718 = 2778 K
  3. η = 1 − 1/8^0.4 = 0.5647
  4. w_net = η·q_in = 0.5647 × 1500 = 847.1 kJ/kg
  5. v₁ = RT₁/p₁ = 0.287 × 300/100 = 0.861 m³/kg; v₂ = v₁/8 = 0.1076 m³/kg
  6. MEP = w_net/(v₁ − v₂) = 847.1/0.7534 = 1124 kPa Answer: T₂ ≈ 689 K, T₃ ≈ 2778 K, η ≈ 56.5%, w_net ≈ 847 kJ/kg, MEP ≈ 1.12 MPa

Example 2 (GATE level): Brayton cycle. Given: ideal Brayton cycle, r_p = 10, T₁ = 300 K, turbine inlet T₃ = 1300 K, c_p = 1.005 kJ/kg·K, γ = 1.4. Find the net work, efficiency and back-work ratio, and the pressure ratio for maximum work.

  1. (γ−1)/γ = 0.2857; 10^0.2857 = 1.9307
  2. T₂ = 300 × 1.9307 = 579.2 K; T₄ = 1300/1.9307 = 673.3 K
  3. w_c = c_p·(T₂ − T₁) = 1.005 × 279.2 = 280.6 kJ/kg
  4. w_t = c_p·(T₃ − T₄) = 1.005 × 626.7 = 629.8 kJ/kg
  5. w_net = 629.8 − 280.6 = 349.2 kJ/kg
  6. q_in = c_p·(T₃ − T₂) = 1.005 × 720.8 = 724.4 kJ/kg; η = 349.2/724.4 = 0.482 (check: 1 − 1/1.9307 = 0.482 ✓)
  7. Back-work ratio = 280.6/629.8 = 0.446
  8. r_p,opt = (1300/300)^(1.4/0.8) = 4.333^1.75 = 13.0 Answer: w_net ≈ 349 kJ/kg, η ≈ 48.2%, back-work ratio ≈ 0.45, r_p for maximum work ≈ 13

Example 3 (quick): Diesel cycle. With r = 16 and ρ = 2: η = 1 − (1/16^0.4) × (2^1.4 − 1)/(1.4 × 1) = 1 − 0.3299 × 1.1707 = 0.614, i.e. about 61.4%.

Common mistakes

  • Using (γ−1)/γ as the exponent with a volume (compression) ratio, or γ−1 with a pressure ratio.
  • Using c_p for the constant-volume heat addition in the Otto cycle.
  • Comparing Otto and Diesel without stating what is held equal (compression ratio or peak pressure).
  • Dividing net work by total volume V₁ instead of swept volume V₁ − V₂ for MEP.
  • Forgetting that regeneration helps only when the turbine exhaust is hotter than the compressor delivery.
  • Entering temperatures in °C.

For GATE PI

Expect direct-formula efficiency questions for Otto, Diesel and Brayton cycles, temperature calculations at cycle points, MEP, and Brayton net work and the optimum pressure ratio. Conceptual questions compare Otto, Diesel and dual cycles under different constraints and ask why diesels use higher compression ratios. Practise the isentropic relations with both volume and pressure ratios until the exponents are automatic.

Quick check

  1. What is the efficiency of an air-standard Otto cycle with r = 10 (γ = 1.4)?
  2. For the same compression ratio, which is more efficient: Otto or Diesel?
  3. Brayton cycle with r_p = 8: what is the efficiency?
  4. Why can a diesel engine use a higher compression ratio than a petrol engine?

Answers: 1. 1 − 10^(−0.4) ≈ 60.2% 2. Otto 3. 1 − 8^(−0.2857) ≈ 44.8% 4. It compresses only air, so there is no knock limit

Interview questions

All Thermal and Fluids Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is an air-standard cycle, and why is it used in thermodynamics?Concept

    An air-standard cycle is a simplified model of a thermodynamic cycle that assumes the working fluid is air, which behaves as an ideal gas. It is used to analyze the performance of internal combustion engines and gas turbines by simplifying the complex processes into idealized ones, making calculations more manageable.

  2. 2.Explain the Otto cycle and its significance in internal combustion engines.Concept

    The Otto cycle is a 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 significant because it models the operation of gasoline engines, allowing engineers to predict engine performance and efficiency.

  3. 3.Describe the Diesel cycle and how it differs from the Otto cycle.Concept

    The Diesel cycle models a compression-ignition engine: isentropic compression, constant-pressure heat addition, isentropic expansion and constant-volume heat rejection. The Otto cycle adds heat at constant volume instead. Because a diesel compresses only air and injects fuel near TDC, it has no knock limit and runs at compression ratios of about 14–22 against 8–12 for petrol engines. For the same compression ratio the Otto cycle is more efficient, but the higher ratio makes real diesels more efficient overall.

  4. 4.What is the Brayton cycle, and where is it commonly used?Concept

    The Brayton cycle is a thermodynamic cycle that describes the workings of a constant pressure heat engine, such as a gas turbine engine. It is commonly used in jet engines and power plants due to its ability to operate efficiently at high temperatures and pressures.

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

    The compression ratio in the Otto cycle is the ratio of the maximum to minimum volume in the cylinder. It is important because a higher compression ratio generally leads to higher thermal efficiency, as it allows the engine to extract more mechanical energy from a given amount of fuel.

  6. 6.What happens if the compression ratio is increased in a Diesel engine?Application

    Increasing the compression ratio in a Diesel engine typically improves its thermal efficiency and power output. However, it also increases the engine's weight and cost due to the need for stronger materials to withstand higher pressures.

  7. 7.Why is the Brayton cycle preferred for jet engines over the Otto or Diesel cycles?Application

    A gas turbine works on a continuous steady-flow Brayton cycle, so it can pass a very large mass flow of air through compact rotating machinery, giving a much higher power-to-weight ratio than a reciprocating engine. There are no reciprocating parts, so vibration is low and very high shaft speeds are possible. The hot, high-velocity exhaust leaving the turbine can be expanded in a nozzle to produce thrust directly. Its efficiency rises with pressure ratio and turbine inlet temperature, which modern materials and cooling allow.

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

    For the air-standard Otto cycle η = 1 − 1/r^(γ−1). With r = 8 and γ − 1 = 0.4, 8^0.4 = 2.297, so η = 1 − 1/2.297 = 1 − 0.435 = 0.565, i.e. about 56.5%. The efficiency depends only on compression ratio and γ, which is why raising r (up to the knock limit) is the main route to better efficiency.

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

    η = 1 − [1/r^(γ−1)]·(ρ^γ − 1)/(γ(ρ − 1)). Here 1/16^0.4 = 0.3299 and (2^1.4 − 1)/(1.4 × 1) = (2.639 − 1)/1.4 = 1.1707. So η = 1 − 0.3299 × 1.1707 = 1 − 0.386 = 0.614, about 61.4%. An Otto cycle at r = 16 would give 1 − 0.3299 = 67%, showing the penalty of constant-pressure heat addition.

  10. 10.Explain how regeneration improves the efficiency of the Brayton cycle.Application

    Regeneration in the Brayton cycle involves using a heat exchanger to transfer heat from the exhaust gases to the compressed air before it enters the combustion chamber. This process reduces the fuel required to reach the desired temperature, thereby improving the cycle's thermal efficiency.

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