Otto Cycle
The Thermodynamic Cycle Behind Petrol Engines — Efficiency Formula, PV Diagram & Solved Problems
Last Updated: March 2026
📌 Key Takeaways
- The Otto cycle models spark-ignition (SI) petrol/gasoline engines — the most common type of car engine worldwide.
- Four processes: Isentropic compression → Constant-volume heat addition → Isentropic expansion → Constant-volume heat rejection.
- Efficiency formula: η = 1 − 1/r(γ−1), where r = compression ratio and γ = Cp/Cv.
- Efficiency depends only on the compression ratio — not on the heat added or the peak temperature.
- Typical compression ratios for petrol engines: 8:1 to 12:1. Higher r is limited by engine knock.
- Named after Nikolaus Otto, who built the first successful four-stroke engine in 1876.
1. What is the Otto Cycle?
The Otto cycle is the ideal thermodynamic cycle that models the operation of spark-ignition (SI) internal combustion engines — the type of engine used in most cars, motorcycles, and small aircraft running on petrol (gasoline). It describes the sequence of compression, combustion, expansion, and exhaust that occurs inside the engine cylinder.
In reality, an SI engine operates on a four-stroke cycle (intake, compression, power, exhaust) involving a complex mixture of air and fuel with chemical combustion. The Otto cycle is the air-standard idealisation — it replaces combustion with constant-volume heat addition and replaces the exhaust-intake process with constant-volume heat rejection, using air as the working fluid throughout. This simplification makes thermodynamic analysis tractable while preserving the essential physics.
The key feature that distinguishes the Otto cycle from the Diesel cycle is that heat addition occurs at constant volume (modelling rapid combustion triggered by a spark plug), whereas in the Diesel cycle, heat addition occurs at constant pressure (modelling slower combustion as fuel is injected into hot compressed air).
2. The Four Processes
Process 1→2: Isentropic Compression
The piston moves from Bottom Dead Centre (BDC) to Top Dead Centre (TDC), compressing the air-fuel mixture. This process is adiabatic and reversible (ideally), so entropy remains constant. Both temperature and pressure increase significantly.
Process 2→3: Constant-Volume Heat Addition
This models the combustion of the air-fuel mixture triggered by the spark plug. In the ideal cycle, heat QH is added instantaneously at constant volume. Temperature and pressure rise sharply while volume remains at its minimum (TDC). This is the point of maximum pressure and temperature in the cycle.
Process 3→4: Isentropic Expansion (Power Stroke)
The high-pressure, high-temperature gas expands and pushes the piston from TDC back to BDC, producing useful work. This is the power stroke — the only process that generates net work output. Entropy remains constant.
Process 4→1: Constant-Volume Heat Rejection
This models the exhaust process. In the ideal cycle, heat QL is rejected at constant volume (at BDC), bringing the gas back to its initial state. In a real engine, this corresponds to opening the exhaust valve and expelling combustion gases.
| Process | Type | Volume | Heat | Real Engine Equivalent |
|---|---|---|---|---|
| 1→2 | Isentropic compression | Decreases (BDC → TDC) | Q = 0 | Compression stroke |
| 2→3 | Constant-volume heat addition | Constant (at TDC) | QH added | Spark ignition / combustion |
| 3→4 | Isentropic expansion | Increases (TDC → BDC) | Q = 0 | Power stroke |
| 4→1 | Constant-volume heat rejection | Constant (at BDC) | QL rejected | Exhaust blowdown |
3. PV and TS Diagrams
PV Diagram
On a PV diagram, the Otto cycle has a distinctive shape:
- 1→2: Steep curve rising to the left — adiabatic compression reduces volume and increases pressure.
- 2→3: Vertical line upward — pressure increases at constant volume (heat addition at TDC).
- 3→4: Steep curve falling to the right — adiabatic expansion increases volume and decreases pressure.
- 4→1: Vertical line downward — pressure decreases at constant volume (heat rejection at BDC).
The enclosed area represents net work output. Point 3 has the highest pressure and temperature; point 1 has the lowest.
TS Diagram
On a TS diagram:
- 1→2: Vertical line upward — entropy constant (isentropic), temperature increases.
- 2→3: Curve moving up and to the right — temperature and entropy increase (heat added at constant volume).
- 3→4: Vertical line downward — entropy constant, temperature decreases.
- 4→1: Curve moving down and to the left — temperature and entropy decrease (heat rejected at constant volume).
4. Efficiency Formula — Derivation
Using the First Law for each process with air as an ideal gas:
Heat added (2→3, constant volume): QH = mCv(T&sub3; − T&sub2;)
Heat rejected (4→1, constant volume): QL = mCv(T&sub4; − T&sub1;)
η = 1 − QL/QH = 1 − (T&sub4; − T&sub1;)/(T&sub3; − T&sub2;)
For the isentropic processes (using TVγ−1 = constant):
T&sub1;V&sub1;γ−1 = T&sub2;V&sub2;γ−1 → T&sub2;/T&sub1; = (V&sub1;/V&sub2;)γ−1 = rγ−1
T&sub4;V&sub4;γ−1 = T&sub3;V&sub3;γ−1 → T&sub3;/T&sub4; = (V&sub4;/V&sub3;)γ−1 = rγ−1
Since T&sub2;/T&sub1; = T&sub3;/T&sub4; = rγ−1, we get T&sub4;/T&sub1; = T&sub3;/T&sub2;, which gives:
(T&sub4; − T&sub1;)/(T&sub3; − T&sub2;) = T&sub1;/T&sub2; = 1/rγ−1
Otto Cycle Efficiency
ηOtto = 1 − 1/r(γ−1)
Where:
- r = compression ratio = Vmax/Vmin = V&sub1;/V&sub2; (dimensionless)
- γ = ratio of specific heats = Cp/Cv (1.4 for air, 1.3 for exhaust gases)
Critical observation: Otto cycle efficiency depends only on the compression ratio r and the gas property γ. It does not depend on the amount of heat added, the peak temperature, the peak pressure, or the engine size. This is a powerful result — to increase efficiency, increase the compression ratio.
5. Compression Ratio — Why It Matters
The compression ratio r is the ratio of the maximum cylinder volume (at BDC) to the minimum cylinder volume (at TDC):
r = VBDC / VTDC = (Vclearance + Vswept) / Vclearance
Higher compression ratios give higher efficiency, but there is a practical limit:
| Compression Ratio (r) | Otto Efficiency (γ = 1.4) |
|---|---|
| 6 | 51.2% |
| 8 | 56.5% |
| 10 | 60.2% |
| 12 | 63.0% |
| 14 | 65.2% |
| 20 | 69.8% |
Notice that each additional unit of compression ratio gives diminishing returns — going from r = 6 to r = 8 gains 5.3 percentage points, while going from r = 12 to r = 14 gains only 2.2 points.
Engine knock limits compression ratio in petrol engines: When the air-fuel mixture is compressed too much, it reaches temperatures where it auto-ignites before the spark plug fires. This uncontrolled combustion (detonation/knock) creates extreme pressure waves that can damage pistons, cylinder walls, and bearings. Modern petrol engines typically use r = 9:1 to 12:1. Diesel engines, which compress only air (no fuel during compression), can safely use r = 14:1 to 22:1.
6. Real Engine vs Ideal Otto Cycle
| Aspect | Ideal Otto Cycle | Real SI Engine |
|---|---|---|
| Working fluid | Air (ideal gas, constant properties) | Air-fuel mixture, then combustion gases |
| Combustion | Instantaneous heat addition at constant V | Finite-duration combustion, not perfectly constant V |
| Compression/Expansion | Perfectly isentropic (no friction, no heat loss) | Friction and heat losses present |
| Exhaust | Constant-volume heat rejection | Complex exhaust and intake strokes |
| Cycle type | Closed cycle (same air recirculated) | Open cycle (fresh charge each cycle) |
| Efficiency at r = 10 | ~60% | ~25–30% |
The large gap between ideal and real efficiency comes from: combustion irreversibility, heat loss through cylinder walls, friction in piston rings and bearings, incomplete combustion, pumping losses (intake and exhaust strokes), and variable gas properties at high temperatures.
7. Worked Numerical Examples
Example 1: Basic Otto Efficiency
Problem: A petrol engine has a compression ratio of 9:1. Assuming air-standard conditions with γ = 1.4, calculate the Otto cycle efficiency.
Solution
η = 1 − 1/r(γ−1) = 1 − 1/9(1.4−1) = 1 − 1/90.4
90.4 = e0.4 × ln(9) = e0.4 × 2.197 = e0.879 = 2.408
η = 1 − 1/2.408 = 1 − 0.415
η = 0.585 = 58.5%
Example 2: Finding Peak Temperature and Pressure
Problem: An Otto cycle has r = 8, γ = 1.4. Initial conditions are T&sub1; = 300 K and P&sub1; = 100 kPa. Heat added is 800 kJ/kg. Cv = 0.718 kJ/kg·K. Find T&sub2;, T&sub3;, and the peak pressure P&sub3;.
Solution
Temperature after compression:
T&sub2; = T&sub1; × r(γ−1) = 300 × 80.4 = 300 × 2.297 = 689.1 K
Temperature after heat addition:
QH = Cv(T&sub3; − T&sub2;) → 800 = 0.718 × (T&sub3; − 689.1)
T&sub3; = 689.1 + 800/0.718 = 689.1 + 1114.2 = 1,803.3 K
Pressure after compression:
P&sub2; = P&sub1; × rγ = 100 × 81.4 = 100 × 18.38 = 1,838 kPa
Peak pressure (constant volume 2→3):
P&sub3;/P&sub2; = T&sub3;/T&sub2; (constant volume, ideal gas)
P&sub3; = 1838 × (1803.3/689.1) = 1838 × 2.617 = 4,810 kPa ≈ 48.1 bar
Example 3: GATE-Style — Comparing Two Compression Ratios
Problem: Engine A has r = 8 and Engine B has r = 10. Both use air with γ = 1.4. By what percentage does the efficiency increase from A to B?
Solution
ηA = 1 − 1/80.4 = 1 − 1/2.297 = 1 − 0.4353 = 0.5647 = 56.47%
ηB = 1 − 1/100.4 = 1 − 1/2.512 = 1 − 0.3981 = 0.6019 = 60.19%
Percentage increase = (60.19 − 56.47)/56.47 × 100 = 6.6%
Increasing the compression ratio from 8 to 10 improves efficiency by about 6.6%.
8. Common Mistakes Students Make
- Using (γ) instead of (γ−1) in the exponent: The formula is r(γ−1), NOT rγ. For air with γ = 1.4, the exponent is 0.4, not 1.4. This is the most frequent numerical error in Otto cycle problems.
- Confusing Otto (constant-V heat addition) with Diesel (constant-P heat addition): In Otto, combustion is modelled at constant volume (spark ignition). In Diesel, combustion is at constant pressure (fuel injection into hot air). The efficiency formulas are different.
- Forgetting that Otto efficiency is independent of heat input: η = 1 − 1/r(γ−1) has no Q term. Adding more fuel (more heat) does not change the ideal Otto efficiency — it only increases the peak temperature and work output.
- Not recognising that real engine efficiency is much lower: The air-standard Otto efficiency at r = 10 is about 60%, but a real petrol engine at the same compression ratio achieves about 25–30%. Do not quote air-standard values as real-world performance.
- Applying the Otto formula to diesel engines: Diesel engines use constant-pressure heat addition and have a different efficiency formula involving both the compression ratio and the cut-off ratio.
9. Frequently Asked Questions
What is the Otto cycle?
The Otto cycle is the ideal thermodynamic cycle for spark-ignition (petrol/gasoline) internal combustion engines. It consists of four processes: isentropic compression, constant-volume heat addition (modelling combustion), isentropic expansion (power stroke), and constant-volume heat rejection (modelling exhaust). It was named after Nikolaus Otto, who built the first practical four-stroke engine in 1876.
What is the efficiency formula for the Otto cycle?
η = 1 − 1/r(γ−1), where r is the compression ratio and γ is the ratio of specific heats (1.4 for air). The efficiency depends only on r and γ — not on the heat input, peak temperature, or engine size. Higher compression ratios give higher efficiency, with diminishing returns.
Why can’t petrol engines use very high compression ratios?
At high compression ratios, the air-fuel mixture reaches such high temperatures during compression that it auto-ignites before the spark plug fires. This phenomenon, called engine knock or detonation, creates violent pressure oscillations that can damage pistons, bearings, and cylinder walls. Typical petrol engines are limited to r = 9:1 to 12:1. Diesel engines avoid this problem by compressing only air and injecting fuel after compression, allowing compression ratios of 14:1 to 22:1.