---
title: "Air standard diesel: efficiency and mean effective pressure"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:diesel-cycle-cutoff-2-efficiency-mep"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/diesel-cycle-cutoff-2-efficiency-mep
site: BCT Engineering
---

# Air standard diesel: efficiency and mean effective pressure

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

- Asked in: 2074 Bhadra, 8 marks
- Video: [Solved: Diesel Cycle Problem | to Find Compression Ratio 16, Cutoff Ratio 2 | Cengel 9th Edition](https://www.youtube.com/watch?v=hFfCeCMp_Ss) — Education Shop
- 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-cutoff-2-efficiency-mep
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

At the beginning of a compression stroke of an air standard diesel cycle having a compression ratio of 16, the temperature is 300 K and the pressure is 100 kPa. If the cut off ratio for the cycle is 2, determine (a) the thermal efficiency (b) the mean effective pressure. [Take γ = 1.4, R = 287 J/kg.K].

## Short answer

Using air-standard diesel cycle with r=16, rc=2, gamma=1.4. Heat added and rejected give net work. Thermal efficiency = 61.4%. Mean effective pressure MEP = w_net/(v1-v2) = 695 kPa.

## Answer

**Given:** r = 16, r_c = 2, T₁ = 300 K, P₁ = 100 kPa, γ = 1.4, R = 0.287 kJ/kg·K.
So cv = R/(γ−1) = 0.7175 and cp = γ cv = 1.0045 kJ/kg·K.

#### (a) Thermal efficiency
η = 1 − (1/r^(γ−1)) × (r_c^γ − 1)/(γ (r_c − 1))

r^(γ−1) = 16^0.4 = 3.0314
r_c^γ = 2^1.4 = 2.6390

η = 1 − (1/3.0314) × (2.6390 − 1)/(1.4 × 1)
η = 1 − 0.3299 × 1.1707 = 0.6138

**η = 61.4 %**

#### (b) Mean effective pressure
State temperatures:
- T₂ = T₁ r^0.4 = 300 × 3.0314 = 909.4 K
- T₃ = T₂ r_c = 1818.9 K
- T₄ = T₁ r_c^γ = 300 × 2.6390 = 791.7 K

Heat added (2-3): q_in = cp (T₃ − T₂) = 1.0045 × 909.4 = 913.5 kJ/kg
Heat rejected (4-1): q_out = cv (T₄ − T₁) = 0.7175 × 491.7 = 352.8 kJ/kg
Net work: w_net = 913.5 − 352.8 = 560.7 kJ/kg (check: η q_in = 560.7)

Volumes:
v₁ = RT₁/P₁ = 0.287 × 300/100 = 0.861 m³/kg
v₂ = v₁/16 = 0.0538 m³/kg

MEP = w_net/(v₁ − v₂) = 560.7/(0.861 − 0.0538) = 695 kPa

**Answer: η ≈ 61.4 %, MEP ≈ 695 kPa (about 6.9 bar).**

## Check yourself

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

1. Which formula gives diesel efficiency?
   A. 1 - r_c/r
   B. 1 - (r_c - 1)/r
   C. 1 - 1/r^(g-1)
   D. 1 - (1/r^(g-1)) (r_c^g - 1)/(g(r_c - 1))
   Answer: D. This is the air-standard diesel efficiency with cut-off ratio r_c. (Key point: Efficiency formula)

2. For r = 16, r_c = 2, g = 1.4, the efficiency is about:
   A. 45%
   B. 75%
   C. 61.4%
   D. 56.5%
   Answer: C. eta = 1 - 0.3299 x 1.1707 = 0.614. (Key point: Result)

3. How is T3 found?
   A. T3 = T2 / r_c
   B. T3 = T2 x r_c
   C. T3 = T4 x r
   D. T3 = T1 r_c
   Answer: B. Heat addition at constant pressure: T3/T2 = v3/v2 = r_c, so T3 = 1818.9 K. (Key point: T3)

4. How is MEP defined here?
   A. w_net/v1
   B. w_net x (v1 - v2)
   C. q_in/(v1 - v2)
   D. w_net/(v1 - v2)
   Answer: D. MEP is net work per unit swept volume: 560.7/(0.861 - 0.0538) = 695 kPa. (Key point: MEP)

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