---
title: "Diesel cycle with compression ratio 15:1"
subject: "Power Plant Equipment"
chapter: 5
key: "semester_7/power_plant_equipment:diesel-cycle-compression-15"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/diesel-cycle-compression-15
site: BCT Engineering
---

# Diesel cycle with compression ratio 15:1

Power Plant Equipment · Chapter 5: Diesel Power Plant

- Asked in: 2078 Bhadra, 8 marks
- Video: [Thermodynamics DIESEL CYCLE in 10 Minutes!](https://www.youtube.com/watch?v=TMAJMGCl-tI) — Less Boring Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/diesel-cycle-compression-15
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-5

## Question

In a diesel power plant engine is coupled with an alternator. The engine has compression ratio 15:1. The compression begins at 0.1MPa and 40°C. The heat added at the end of compression is 1675MJ/kg. Consider air constant, R = 287 kJ/kgK and Find: (i) The maximum temperature and pressure in the cycle (ii) The cut off ratio of the engine (iii) Net work done per kg of air (iv) Thermal efficiency of the engine (v) Mean effective pressure of the cycle.

## Short answer

Assuming 1675 kJ/kg heat added, T2=924.66 K and p2=4.431 MPa. Maximum temperature is T3=2591.3 K and maximum pressure is 4.431 MPa; cut-off ratio is 2.802. Net work is 948.6 kJ/kg, thermal efficiency is 56.64%, and mean effective pressure is 1131.5 kPa (11.31 bar).

## Answer

**Assumptions:** the engine works on the air-standard **Diesel cycle** (isentropic compression 1–2, constant-pressure heat addition 2–3, isentropic expansion 3–4, constant-volume heat rejection 4–1). The heat added is taken as **1675 kJ/kg** (1675 MJ/kg is clearly a misprint), $R = 0.287$ kJ/kg K, $c_p = 1.005$, $c_v = 0.718$ kJ/kg K, $\gamma = 1.4$.

Data: $r = 15$, $p_1 = 0.1$ MPa, $T_1 = 40 + 273 = 313$ K.

#### (i) Maximum temperature and pressure
Process 1–2 (isentropic compression):
$$
\begin{aligned}
T_2 &= T_1 r^{\gamma-1} = 313 \times 15^{0.4} = 924.66\ \text{K} \\
p_2 &= p_1 r^{\gamma} = 0.1 \times 15^{1.4} = 4.431\ \text{MPa}
\end{aligned}
$$
Process 2–3 (heat added at constant pressure):
$$
T_3 = T_2 + \frac{q_s}{c_p} = 924.66 + \frac{1675}{1.005} = 2591.32\ \text{K}
$$
Maximum temperature $T_3 = 2591.3$ K; maximum pressure $p_3 = p_2 = 4.431$ MPa.

#### (ii) Cut-off ratio
$$
\rho = \frac{v_3}{v_2} = \frac{T_3}{T_2} = \frac{2591.32}{924.66} = 2.802
$$

#### (iii) Net work per kg of air
$$
\begin{aligned}
T_4 &= T_3\left(\frac{\rho}{r}\right)^{\gamma-1} = 2591.32\left(\frac{2.802}{15}\right)^{0.4} = 1324.65\ \text{K} \\
q_r &= c_v (T_4 - T_1) = 0.718(1324.65 - 313) = 726.37\ \text{kJ/kg} \\
W_{net} &= q_s - q_r = 1675 - 726.37 = 948.63\ \text{kJ/kg}
\end{aligned}
$$

#### (iv) Thermal efficiency
$$
\eta = \frac{W_{net}}{q_s} = \frac{948.63}{1675} = 0.5664 = 56.64\%
$$
Check with the formula $\eta = 1 - \dfrac{1}{r^{\gamma-1}}\cdot\dfrac{\rho^{\gamma}-1}{\gamma(\rho-1)} = 56.64\%$.

#### (v) Mean effective pressure
$$
\begin{aligned}
v_1 &= \frac{R T_1}{p_1} = \frac{0.287 \times 313}{100} = 0.8983\ \text{m}^3/\text{kg} \\
v_2 &= \frac{v_1}{r} = 0.05989\ \text{m}^3/\text{kg} \\
p_m &= \frac{W_{net}}{v_1 - v_2} = \frac{948.63}{0.8983 - 0.05989} = 1131.5\ \text{kPa}
\end{aligned}
$$

**Answer:** $T_{max}$ = 2591.3 K, $p_{max}$ = 4.431 MPa, cut-off ratio = 2.802, $W_{net}$ = 948.6 kJ/kg, $\eta$ = 56.64 %, MEP = 1131.5 kPa ≈ 11.31 bar.

## Check yourself

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

1. Which formula gives T2 after isentropic compression?
   A. T2 = T1 x r
   B. T2 = T1 x r^(gamma-1)
   C. T2 = T1 / r^(gamma-1)
   D. T2 = T1 + q/cv
   Answer: B. For isentropic compression T2 = T1 r^(gamma-1) = 313 x 15^0.4 = 924.66 K. (Key point: Compression temperature)

2. How is T3 found in the diesel cycle?
   A. T3 = T2 + q_s/cv
   B. T3 = T2 + q_s/cp, since heat is added at constant pressure
   C. T3 = T2 x r
   D. T3 = T1 + q_s/cp
   Answer: B. Constant pressure heating uses cp: T3 = 924.66 + 1675/1.005 = 2591.3 K. (Key point: Maximum temperature)

3. What is the maximum pressure of the cycle?
   A. p1 = 0.1 MPa
   B. p2 = p3 = 4.431 MPa
   C. p4 = 1.0 MPa
   D. p3 = 44.31 MPa
   Answer: B. Heat addition is at constant pressure, so p3 = p2 = p1 r^gamma = 4.431 MPa. (Key point: Maximum pressure)

4. How is the cut-off ratio computed?
   A. rho = T3/T2 (= v3/v2)
   B. rho = T4/T1
   C. rho = p3/p2
   D. rho = r x T3
   Answer: A. At constant pressure v3/v2 = T3/T2 = 2591.32/924.66 = 2.802. (Key point: Cut-off ratio)

5. How is heat rejected q_r calculated?
   A. cp(T4 - T1)
   B. cv(T4 - T1), since 4-1 is constant volume
   C. cv(T3 - T2)
   D. cp(T3 - T4)
   Answer: B. Rejection at constant volume uses cv: 0.718(1324.65-313) = 726.37 kJ/kg, so W_net = 1675 - 726.37. (Key point: Heat rejection)

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