Diesel Cycle vs Otto Cycle
Complete Comparison — Efficiency Formulas, PV Diagrams, Compression Ratios & Applications
Last Updated: March 2026
📌 Key Takeaways
- Otto cycle: constant-volume heat addition (spark ignition). Diesel cycle: constant-pressure heat addition (compression ignition).
- Otto efficiency: η = 1 − 1/r(γ−1). Diesel efficiency: η = 1 − [1/r(γ−1)] × [(ργ − 1) / γ(ρ − 1)].
- For the same compression ratio, Otto is more efficient. But diesel engines use higher compression ratios (14–22 vs 8–12), so real diesel engines are more efficient overall.
- The cut-off ratio (ρ) is unique to the Diesel cycle — it measures how much the gas expands during constant-pressure heat addition.
- The Dual cycle (mixed cycle) combines both constant-volume and constant-pressure heat addition and is the most realistic model for modern engines.
1. The Diesel Cycle Explained
The Diesel cycle is the ideal thermodynamic cycle for compression-ignition (CI) engines — engines that ignite fuel by compressing air to a temperature high enough to cause spontaneous combustion, without needing a spark plug. Diesel engines are used in trucks, buses, ships, locomotives, generators, and heavy machinery.
Named after Rudolf Diesel, who patented the compression-ignition engine in 1893, the Diesel cycle differs from the Otto cycle in one critical way: heat addition occurs at constant pressure rather than constant volume. This models the real combustion process in diesel engines, where fuel is injected gradually into hot compressed air and burns as it enters, maintaining approximately constant pressure during combustion.
2. Diesel Cycle — Four Processes
| Process | Type | What Happens | Heat |
|---|---|---|---|
| 1→2 | Isentropic compression | Air is compressed from BDC to TDC. Temperature and pressure rise sharply. Only air is compressed (no fuel yet). | Q = 0 |
| 2→3 | Constant-pressure heat addition | Fuel is injected into hot compressed air and burns at approximately constant pressure. Piston moves down as volume increases from V&sub2; to V&sub3;. | QH added |
| 3→4 | Isentropic expansion | Combustion gases expand further, pushing the piston to BDC. Power stroke continues. | Q = 0 |
| 4→1 | Constant-volume heat rejection | Exhaust valve opens, pressure drops at constant volume. Gases expelled. | QL rejected |
The key difference from Otto: process 2→3 is constant-pressure (Diesel) instead of constant-volume (Otto). This means in the Diesel cycle, the piston continues to move during combustion, and the volume increases from V&sub2; to V&sub3;.
3. Diesel Cycle Efficiency Formula
For the Diesel cycle with air-standard assumptions:
QH = mCp(T&sub3; − T&sub2;) (constant pressure, using Cp)
QL = mCv(T&sub4; − T&sub1;) (constant volume, using Cv)
Diesel Cycle Efficiency
ηDiesel = 1 − [1/r(γ−1)] × [(ργ − 1) / γ(ρ − 1)]
Where:
- r = compression ratio = V&sub1;/V&sub2; (same definition as Otto)
- ρ = cut-off ratio = V&sub3;/V&sub2; (unique to Diesel cycle)
- γ = ratio of specific heats = Cp/Cv (1.4 for air)
Notice that the Diesel efficiency formula has an extra factor [(ργ − 1) / γ(ρ − 1)] compared to the Otto formula. Since this factor is always greater than 1 for ρ > 1, the Diesel cycle is less efficient than the Otto cycle at the same compression ratio.
When ρ → 1 (cut-off ratio approaches 1, meaning negligible heat addition at constant pressure), the Diesel formula reduces to the Otto formula — confirming that Otto is the limiting case of Diesel with zero constant-pressure heat addition.
4. Cut-Off Ratio — What It Is
Cut-Off Ratio
ρ = V&sub3; / V&sub2;
The ratio of cylinder volume at the end of constant-pressure combustion to the volume at the start (TDC volume).
The cut-off ratio indicates how long fuel injection lasts:
- Low ρ (close to 1): Short fuel injection, less heat added, higher efficiency but less power.
- High ρ (2–3+): Longer injection, more heat and power, but lower efficiency.
The cut-off ratio is determined by the amount of fuel injected per cycle and the engine operating conditions. At full load (maximum fuel injection), ρ is higher; at part load, ρ is lower. This means diesel engine efficiency varies with load — part-load efficiency is higher than full-load efficiency, which is the opposite of what many students expect.
5. Otto vs Diesel — Side-by-Side Comparison
| Feature | Otto Cycle | Diesel Cycle |
|---|---|---|
| Heat addition | Constant volume | Constant pressure |
| Heat rejection | Constant volume | Constant volume |
| Ignition method | Spark plug (spark ignition) | Self-ignition by compression (compression ignition) |
| Fuel | Petrol / gasoline | Diesel fuel |
| Compression ratio | 8:1 to 12:1 | 14:1 to 22:1 |
| What is compressed | Air-fuel mixture | Air only (fuel injected later) |
| Efficiency formula | η = 1 − 1/r(γ−1) | η = 1 − [1/r(γ−1)] × [(ργ−1)/γ(ρ−1)] |
| Peak pressure | Higher (constant-V combustion causes pressure spike) | Lower (constant-P combustion limits peak pressure) |
| Knock problem | Yes — limits compression ratio | No — fuel injected after compression |
| Typical applications | Cars, motorcycles, small generators | Trucks, buses, ships, generators, heavy machinery |
| Real-world efficiency | 25–30% | 35–45% |
| Weight and cost | Lighter and cheaper | Heavier and more expensive (higher pressures require stronger components) |
6. Efficiency Comparison — Same r vs Different r
Same Compression Ratio
At the same compression ratio, Otto efficiency > Diesel efficiency. The constant-pressure heat addition in the Diesel cycle adds an extra loss factor [(ργ−1)/γ(ρ−1)] > 1, making it inherently less efficient at the same r.
Same Peak Pressure
At the same maximum pressure, Diesel efficiency > Otto efficiency. The Diesel engine reaches peak pressure at the end of compression (state 2), while the Otto engine reaches peak pressure after combustion (state 3). So for the same peak pressure, the Diesel engine has a higher compression ratio and extracts more work.
Same Heat Input
At the same heat input and compression ratio, Otto efficiency > Diesel efficiency (same reasoning as the first case).
Actual Operating Conditions
In practice, diesel engines operate at r = 14–22 while petrol engines operate at r = 8–12. At these respective compression ratios, diesel engines achieve higher real-world efficiency (35–45% vs 25–30%) despite the theoretical disadvantage at equal r.
| Comparison Basis | More Efficient Cycle |
|---|---|
| Same compression ratio | Otto > Diesel |
| Same peak pressure | Diesel > Otto |
| Same heat input & same r | Otto > Diesel |
| Real operating conditions | Diesel > Otto (due to higher r) |
GATE tip: Questions on this topic almost always specify the comparison basis. Read carefully whether it says “same compression ratio,” “same peak pressure,” or “same heat rejection.” The answer changes depending on the basis.
7. The Dual Cycle (Mixed Cycle)
In reality, combustion in both petrol and diesel engines occurs partially at constant volume and partially at constant pressure. The Dual cycle (also called the mixed or limited-pressure cycle) combines both:
- Process 1→2: Isentropic compression
- Process 2→2′: Constant-volume heat addition (partial combustion)
- Process 2′→3: Constant-pressure heat addition (remaining combustion)
- Process 3→4: Isentropic expansion
- Process 4→1: Constant-volume heat rejection
Dual Cycle Efficiency
ηDual = 1 − [1/r(γ−1)] × [(rp × ργ − 1) / (rp − 1 + γ × rp(ρ − 1))]
Where: rp = pressure ratio during constant-volume heat addition = P&sub2;′/P&sub2;
The Dual cycle reduces to the Otto cycle when ρ = 1 (no constant-pressure heat addition) and to the Diesel cycle when rp = 1 (no constant-volume heat addition). It is the most realistic air-standard model for modern engines.
8. Worked Numerical Examples
Example 1: Diesel Cycle Efficiency
Problem: A diesel engine has a compression ratio of 18 and a cut-off ratio of 2.5. Using γ = 1.4, calculate the air-standard efficiency.
Solution
η = 1 − [1/r(γ−1)] × [(ργ − 1) / γ(ρ − 1)]
r0.4 = 180.4 = e0.4 × ln(18) = e0.4 × 2.89 = e1.156 = 3.177
ργ = 2.51.4 = e1.4 × ln(2.5) = e1.4 × 0.916 = e1.283 = 3.607
Factor = (3.607 − 1) / [1.4 × (2.5 − 1)] = 2.607 / 2.1 = 1.241
η = 1 − (1/3.177) × 1.241 = 1 − 0.3148 × 1.241 = 1 − 0.3907
η = 0.609 = 60.9%
Example 2: Otto vs Diesel at Same r
Problem: Compare the efficiency of an Otto cycle and a Diesel cycle, both with r = 18 and γ = 1.4. The Diesel cycle has ρ = 2.5.
Solution
ηOtto = 1 − 1/180.4 = 1 − 1/3.177 = 1 − 0.3148 = 68.5%
ηDiesel = 60.9% (from Example 1)
Difference = 68.5% − 60.9% = 7.6 percentage points
At the same compression ratio, the Otto cycle is 7.6 percentage points more efficient. However, a real petrol engine cannot operate at r = 18 due to knock — it would be limited to r ≈ 10, giving ηOtto = 60.2%, which is actually lower than the Diesel at r = 18.
9. Common Mistakes Students Make
- Stating “Diesel is always more efficient than Otto” without qualification: This is only true in practice because diesel engines operate at higher compression ratios. At the same r, Otto is more efficient. Always state the comparison basis.
- Forgetting the cut-off ratio in Diesel efficiency: The Diesel formula requires both r and ρ. Leaving out the cut-off ratio factor gives you the Otto formula instead — a completely different answer.
- Confusing cut-off ratio with compression ratio: The compression ratio r = V&sub1;/V&sub2; (total to clearance volume). The cut-off ratio ρ = V&sub3;/V&sub2; (end of combustion to start of combustion volume). They measure different things.
- Not recognising the Dual cycle as the most realistic model: Both Otto and Diesel are simplifications. Real engines have combustion that occurs partially at constant volume and partially at constant pressure. The Dual cycle captures this better.
- Using Cv for constant-pressure heat addition: In the Diesel cycle, heat is added at constant pressure, so use QH = mCpΔT, not mCvΔT. Using Cv is correct only for constant-volume processes (Otto heat addition, and heat rejection in both cycles).
10. Frequently Asked Questions
What is the main difference between the Diesel cycle and the Otto cycle?
The fundamental difference is the mode of heat addition. In the Otto cycle, heat is added at constant volume (modelling rapid spark-ignited combustion in petrol engines). In the Diesel cycle, heat is added at constant pressure (modelling the gradual combustion that occurs as fuel is injected into hot compressed air). This difference leads to different efficiency formulas, different compression ratio limits, and different engine characteristics.
Which is more efficient — Diesel or Otto cycle?
It depends on the basis of comparison. For the same compression ratio, Otto is more efficient. For the same peak pressure, Diesel is more efficient. In real-world operating conditions, diesel engines achieve higher efficiency (35–45%) than petrol engines (25–30%) because they operate at much higher compression ratios (14–22 vs 8–12), more than compensating for the theoretical disadvantage at equal r.
What is the cut-off ratio in the Diesel cycle?
The cut-off ratio ρ = V&sub3;/V&sub2; is the ratio of the volume at the end of constant-pressure heat addition to the volume at the beginning (clearance volume). It represents the fraction of the expansion stroke during which fuel injection and combustion occur. Higher ρ means more fuel is injected, producing more power but reducing efficiency. Typical values range from 1.5 to 3.0 depending on engine load.