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Thermodynamics
Ideal Diesel cycle efficiency
Thermal efficiency of the air-standard Diesel cycle, with heat added at constant pressure.
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- THE-19
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- v1
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- Thermodynamics
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- ηD
The ideal Diesel cycle adds heat at constant pressure rather than constant volume. Its efficiency is η = 1 − (1/rγ−1) (rcγ − 1)/[γ(rc − 1)], with r the compression ratio, rc the cutoff ratio (the volume after heat addition divided by the volume before) and γ the ratio of specific heats. The cutoff ratio lies between 1 and r. As rc approaches 1 the result tends to the Otto efficiency at the same r.
Thermal efficiency of the air-standard Diesel cycle, with heat added at constant pressure. It rises with compression ratio and falls as the cutoff ratio, the ratio of volumes during heat addition, increases.
Calculator
- Otto efficiency at the same compression ratio0.6853
- Efficiency loss relative to the Otto cycle5.373 %
- Expansion ratio, r/rc9
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A Diesel cycle with r = 18, rc = 2 and γ = 1.4 has an ideal efficiency of 63.2 %, below the Otto value 68.5 % at the same r.
- The Diesel cycle’s practical advantage is that it can use a much higher compression ratio: a petrol (Otto) engine limited to r = 9 by knock would give an ideal 58.5 %. Real engines reach only a part of these ideal values.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| γ Ratio of specific heats | 1.1 to 1.67 | 0.2214 | 0.8116 | 0.5902 | +10% | +17.1 % |
| r Compression ratio | 8 to 25 | 0.4904 | 0.6769 | 0.1865 | +10% | +2.18 % |
| rc Cutoff ratio (between 1 and r) | 1.1 to 5 | 0.5213 | 0.6791 | 0.1578 | +10% | −1.45 % |
Sweep of Cutoff ratio (between 1 and r)
| Cutoff ratio (between 1 and r) | Diesel cycle efficiency |
|---|---|
| 1.1 | 0.6791 |
| 1.588 | 0.6519 |
| 2.075 | 0.6281 |
| 2.563 | 0.6067 |
| 3.05 | 0.5872 |
| 3.538 | 0.5691 |
| 4.025 | 0.5523 |
| 4.513 | 0.5364 |
| 5 | 0.5213 |
Sweep one variable
Variables and how each one contributes
Diesel cycle efficiency
What the equation gives you. Everything else feeds it.
Compression ratio
The compression ratio, the volume at the start of compression divided by the volume at the end. Raises the efficiency, with diminishing returns: a higher compression ratio means a higher mean temperature of heat addition. Raising it by 1 (to 19) moves the result by +1.25 %, from 0.6316 to 0.6395.
Cutoff ratio (between 1 and r)
The cutoff ratio: how far the piston has moved when heat addition ends, as a volume ratio (between 1 and r). A larger value means more fuel per cycle. Lowers the efficiency: more heat is added later in the stroke, at a lower effective compression ratio. Raising it by 0.25 (to 2.25) moves the result by −1.8 %, from 0.6316 to 0.6202.
Ratio of specific heats
The ratio of specific heats of the working gas; 1.4 for air at ordinary temperatures. Raises the efficiency: a larger γ gives a larger temperature rise on compression. Raising it by 0.02 (to 1.42) moves the result by +2.83 %, from 0.6316 to 0.6495.
Worked example with the values above
- Compression term: 1/rγ−1 = 1/180.4 = 0.3147.
- Cutoff term: (rcγ − 1)/[γ(rc − 1)] = (2.639 − 1)/(1.4 × 1) = 1.171.
- η = 1 − 0.3147 × 1.171 = 0.6316.
Assumptions and limits
- An air-standard cycle: constant specific heats and an ideal gas as the working fluid.
- Reversible compression and expansion, heat added at constant pressure and rejected at constant volume.
Common mistakes
- Entering a cutoff ratio of 1 or below, or above the compression ratio.
- Treating the ideal value as the actual engine efficiency, which is much lower.
Statistics and uncertainty
The ideal efficiency is sensitive to γ: the value of 1.4 applies to cold air, while hot combustion gas has a lower γ (about 1.3), which lowers the efficiency by several percentage points. Use a representative γ for the mixture temperature.
Where it is used
- Comparing compression-ignition and spark-ignition engine cycles at the same compression ratio.
- Seeing how fuel quantity (the cutoff ratio) affects efficiency at part load.
Maths you need
- Compression ratio: maximum volume divided by minimum volume
- Ratio of specific heats: γ = cp/cv