---
title: "Otto cycle CR 8: state points and efficiency"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:otto-cycle-cr8-2000kj-heat"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/otto-cycle-cr8-2000kj-heat
site: BCT Engineering
---

# Otto cycle CR 8: state points and efficiency

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

- Asked in: 2070 Magh, 8 marks
- 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/otto-cycle-cr8-2000kj-heat
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

A compression ratio of an air standard otto cycle is 8. At the beginning of the compression process, the pressure and temperature of air are 100 kPa and 20°C respectively. The heat added per kg of air during the cycle is 2000 kJ/kg. Determine:
a) The pressure and temperature at the end of each process of the cycle.
b) The thermal efficiency. [Take Cv = 718 J/kg.K, γ = 1.4]

## Short answer

Isentropic compression with r=8 gives T2=673.5 K, P2=1838 kPa. Constant-volume heat addition of 2000 kJ/kg yields T3=3459 K, P3=9440 kPa. Expansion to T4=1506 K, P4=514 kPa. Efficiency η = 1 - 1/8^0.4 = 56.5%.

## Answer

**Given:** r = 8, P₁ = 100 kPa, T₁ = 20 + 273.15 = 293.15 K, heat added q_in = 2000 kJ/kg (in process 2-3), cv = 0.718 kJ/kg·K, γ = 1.4.

#### (a) Pressure and temperature at each state
r^(γ−1) = 8^0.4 = 2.2974; r^γ = 8^1.4 = 18.3792

**State 1:** P₁ = 100 kPa, T₁ = 293.15 K

**State 2 (after isentropic compression):**
T₂ = T₁ r^0.4 = 293.15 × 2.2974 = **673.5 K**
P₂ = P₁ r^1.4 = 100 × 18.3792 = **1838 kPa**

**State 3 (after constant-volume heat addition):**
q_in = cv (T₃ − T₂), so T₃ = T₂ + 2000/0.718 = 673.5 + 2785.5 = **3459.0 K**
P₃ = P₂ T₃/T₂ = 1838 × 3459.0/673.5 = **9440 kPa**

**State 4 (after isentropic expansion):**
T₄ = T₃/r^0.4 = 3459.0/2.2974 = **1505.6 K**
P₄ = P₃/r^1.4 = 9440/18.3792 = **513.6 kPa**

| State | P (kPa) | T (K) |
|---|---|---|
| 1 | 100 | 293.1 |
| 2 | 1838 | 673.5 |
| 3 | 9440 | 3459.0 |
| 4 | 514 | 1505.6 |

#### (b) Thermal efficiency
q_out = cv (T₄ − T₁) = 0.718 × (1505.6 − 293.1) = 870.6 kJ/kg
w_net = 2000 − 870.6 = 1129.4 kJ/kg

η = w_net/q_in = 1129.4/2000 = 0.5647

Check: η = 1 − 1/r^(γ−1) = 1 − 1/2.2974 = 0.5647.

**Answer: η = 56.5 %.**

## Check yourself

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

1. T2 is:
   A. T1 r^1.4
   B. T1/r
   C. T1 r^0.4 = 673.5 K
   D. T1 + 2000
   Answer: C. Isentropic compression. (Key point: T2)

2. T3 is found from:
   A. T2 + q_in/cv
   B. T2 x q_in
   C. T2 - q_in/cv
   D. T2 + q_in/cp
   Answer: A. Constant-volume addition: T3 = 673.5 + 2000/0.718 = 3459 K. (Key point: T3)

3. P3 follows from:
   A. P3 = P1 r
   B. P3 = P2 T2/T3
   C. P3 = P2 T3/T2 (constant volume)
   D. P3 = P2
   Answer: C. At constant volume pressure scales with temperature: 9440 kPa. (Key point: P3)

4. Efficiency is:
   A. 56.5%, from 1 - 1/r^0.4
   B. 70%
   C. 100%
   D. 43.5%
   Answer: A. Equals w_net/q_in = 1129.4/2000. (Key point: Efficiency)

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