---
title: "Ideal diesel cycle: diagrams and efficiency"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:diesel-cycle-efficiency-derivation"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/diesel-cycle-efficiency-derivation
site: BCT Engineering
---

# Ideal diesel cycle: diagrams and efficiency

Fundamental of Thermodynamics & Heat Transfer · Chapter 6: Thermodynamic Cycles

- Asked in: 2072 Magh, 6 marks
- Video: [Diesel Cycle Efficiency Derivation, Thermodynamics, Thermal Engineering](https://www.youtube.com/watch?v=gVwIho5AV34) — Reference Book
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/diesel-cycle-efficiency-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

Sketch an ideal Diesel cycle on P-v and T-s diagram. Also derive an expression for its efficiency.

## Short answer

Efficiency = 1 - (1/r^(gamma-1)) * (r_c^gamma - 1) / (gamma (r_c - 1)). Derived from q_in = cp(T3-T2), q_out = cv(T4-T1), using isentropic relations: T2 = T1 r^(gamma-1), T3 = T1 r^(gamma-1) r_c, T4 = T1 r_c^gamma.

## Answer

The ideal diesel cycle is the air-standard cycle of a compression-ignition engine: two isentropic processes, one constant-pressure heat addition and one constant-volume heat rejection.

#### Diagrams
```text
 P-v diagram                      T-s diagram
 P                                T
 |   2 _______ 3                  |               3
 |   |          \                 |             .'|
 |   |           '.               |           .'  |
 |   |             '.             |         .'    |
 |    \              '-.          |    2 _.'      |
 |     \                '-. 4     |    |          4
 |      '.                  |     |    |        .'
 |        '-.               |     |    |      .'
 |           '-._           |     |    |    .'
 |               '--.._     |     |    1.-'
 |                     '--- 1     |
 +-------------------------> v    +--------------------> s
    v2                      v1
 1-2 isentropic compression       2-3 constant-P heat addition
 3-4 isentropic expansion         4-1 constant-v heat rejection
```

#### Derivation
Let r = v₁/v₂ (compression ratio) and r_c = v₃/v₂ (cut-off ratio). Air is an ideal gas with constant specific heats; γ = cp/cv.

Heat added (2-3, constant P): q_in = cp (T₃ − T₂)
Heat rejected (4-1, constant v): q_out = cv (T₄ − T₁)

η = 1 − q_out/q_in = 1 − (T₄ − T₁)/(γ (T₃ − T₂))

Express all temperatures in terms of T₁:
- 1-2 isentropic: T₂ = T₁ r^(γ−1)
- 2-3 constant pressure: T₃/T₂ = v₃/v₂ = r_c, so T₃ = T₁ r^(γ−1) r_c
- 3-4 isentropic, with v₄/v₃ = v₁/v₃ = r/r_c: T₄ = T₃ (r_c/r)^(γ−1) = T₁ r_c^γ

Substituting:

T₄ − T₁ = T₁ (r_c^γ − 1)
T₃ − T₂ = T₁ r^(γ−1) (r_c − 1)

η = 1 − T₁ (r_c^γ − 1)/(γ T₁ r^(γ−1) (r_c − 1))

**η_Diesel = 1 − (1/r^(γ−1)) × (r_c^γ − 1)/(γ (r_c − 1))**

Comparison with the Otto cycle (η = 1 − 1/r^(γ−1)): the extra bracket (r_c^γ − 1)/(γ(r_c − 1)) is always greater than 1, so for the same compression ratio the diesel cycle is less efficient. Diesel engines, however, work at higher compression ratios and so reach higher practical efficiency. As r_c → 1, the bracket → 1 and the diesel cycle becomes the Otto cycle.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Diesel heat added and rejected are:
   A. Both with cp
   B. Both with cv
   C. q_in = cp(T3 - T2), q_out = cv(T4 - T1)
   D. q_in = cv(T3 - T2), q_out = cp(T4 - T1)
   Answer: C. Addition is at constant pressure, rejection at constant volume. (Key point: Heat terms)

2. T4 in terms of T1 is:
   A. T1 r^g
   B. T1 r_c^g
   C. T1 r^(g-1)
   D. T1 r_c
   Answer: B. T4 = T3 (r_c/r)^(g-1) = T1 r_c^g. (Key point: T4)

3. Final efficiency expression:
   A. 1 - (1/r^(g-1)) (r_c^g - 1)/(g(r_c - 1))
   B. 1 - (r_c^g - 1)/r
   C. 1 - g(r_c - 1)/(r_c^g - 1)
   D. 1 - 1/r^(g-1)
   Answer: A. Substitute T4 - T1 and T3 - T2 into eta. (Key point: Final formula)

4. For the same compression ratio, diesel is:
   A. More efficient
   B. Not comparable
   C. Less efficient than Otto, since the bracket is greater than 1
   D. Equal
   Answer: C. The extra bracket > 1 reduces efficiency. (Key point: Comparison with Otto)

## Related in Fundamental of Thermodynamics & Heat Transfer

- [Ideal Brayton cycle: diagrams and efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-cycle-efficiency-derivation) — Fundamental of Thermodynamics & Heat Transfer, chapter 6, asked 2×
- [Ideal Otto cycle diagrams and efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aotto-cycle-efficiency-compression-ratio) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Ideal diesel cycle working principle](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Adiesel-cycle-working-principle) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Ideal diesel engine CR 20: efficiency and MEP](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Adiesel-cr20-max-2200k) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Diesel cycle: cut-off percentage, expansion temperature, efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Adiesel-1500c-cr16-cutoff-percentage) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Ideal Brayton cycle: net work per kg and efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-work-per-kg-efficiency) — Fundamental of Thermodynamics & Heat Transfer, chapter 6

## Related in other subjects

- [Diesel cycle efficiency compared with Otto](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Adiesel-cycle-versus-otto-numerical) — Internal Combustion Engines, chapter 1
- [Ideal Brayton cycle work and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aideal-brayton-numerical) — Turbo Machines, chapter 3
- [Reheat cycle at 14 bar: ideal efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Areheat-14-bar-efficiency) — Power Plant Equipment, chapter 7
- [Brayton cycle efficiency and optimum pressure ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Abrayton-cycle-efficiency-optimum-pressure-ratio) — Turbo Machines, chapter 3