---
title: "Ideal diesel engine CR 20: efficiency and MEP"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:diesel-cr20-max-2200k"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/diesel-cr20-max-2200k
site: BCT Engineering
---

# Ideal diesel engine CR 20: efficiency and MEP

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

- Asked in: 2073 Bhadra, 8 marks
- Video: [Numerical on Diesel Cycle](https://www.youtube.com/watch?v=gZ9CC-mckxE) — BHUSHAN KUMAR
- 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-cr20-max-2200k
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is 95kPa and 20°C. If the maximum temperature in the cycle is not to exceed 2200K, determine a) the thermal efficiency and b) the mean effective pressure. [Take Cp = 1005J/kgK, and γ = 1.4]

## Short answer

Given r=20, T_max=2200 K, cp=1005 J/kgK, gamma=1.4. T2=971.6 K, cutoff ratio rc=T3/T2=2.264. Thermal efficiency = 63.5%. MEP = 932 kPa.

## Answer

**Given:** r = 20, P₁ = 95 kPa, T₁ = 20 + 273.15 = 293.15 K, T₃ = T_max = 2200 K, cp = 1.005 kJ/kg·K, γ = 1.4.
So cv = cp/γ = 0.7179 kJ/kg·K and R = cp − cv = 0.2871 kJ/kg·K.

#### State temperatures
- T₂ = T₁ r^(γ−1) = 293.15 × 20^0.4 = 293.15 × 3.3145 = **971.6 K**
- T₃ = 2200 K
- Cut-off ratio: r_c = T₃/T₂ = 2200/971.6 = **2.264**
- T₄: v₄ = v₁, so T₄ = T₃ (r_c/r)^(γ−1) = 2200 × (2.264/20)^0.4 = **920.4 K**

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

r_c^γ = 2.264^1.4 = 3.1397

η = 1 − (1/3.3145) × (3.1397 − 1)/(1.4 × 1.264) = **63.5 %**

Check by heat: q_in = cp (T₃ − T₂) = 1.005 × (2200 − 971.6) = 1234.5 kJ/kg; q_out = cv (T₄ − T₁) = 0.7179 × 627.3 = 450.3 kJ/kg; w_net = 784.2 kJ/kg; η = 784.2/1234.5 = 63.5 %.

#### (b) Mean effective pressure
v₁ = RT₁/P₁ = 0.2871 × 293.15/95 = 0.8861 m³/kg
v₂ = v₁/20 = 0.0443 m³/kg

MEP = w_net/(v₁ − v₂) = 784.2/(0.8861 − 0.0443) = 932 kPa

**Answer: η ≈ 63.5 %, MEP ≈ 932 kPa (about 9.3 bar).**

## Check yourself

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

1. How is the cut-off ratio found when T_max is given?
   A. r_c = T3/T1
   B. r_c = T4/T1
   C. r_c = T3/T2
   D. r_c = r/T3
   Answer: C. 2-3 is at constant pressure, so r_c = v3/v2 = T3/T2 = 2200/971.6 = 2.264. (Key point: Cut-off ratio)

2. T2 is calculated from:
   A. T1 r
   B. T3/r
   C. T1 r^g
   D. T1 r^(g-1)
   Answer: D. Isentropic compression: T2 = 293.15 x 20^0.4 = 971.6 K. (Key point: T2)

3. The thermal efficiency is about:
   A. 63.5%
   B. 48%
   C. 55%
   D. 72%
   Answer: A. From q_in = 1234.5 and w_net = 784.2 kJ/kg, eta = 63.5%. (Key point: Efficiency)

4. MEP is obtained by:
   A. w_net/v2
   B. w_net/(v1 - v2)
   C. q_out/(v1 - v2)
   D. q_in/v1
   Answer: B. w_net/(0.8861 - 0.0443) = 932 kPa. (Key point: MEP)

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