---
title: "Diesel cycle: cut off 8% of stroke, CR 16 to 20"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 6
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:diesel-cr16-cutoff-8-percent"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/diesel-cr16-cutoff-8-percent
site: BCT Engineering
---

# Diesel cycle: cut off 8% of stroke, CR 16 to 20

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

- Asked in: 2072 Asoj, 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/diesel-cr16-cutoff-8-percent
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6

## Question

An engine working on a diesel cycle has a compression ratio of 16 and the cut off takes place at 8% of the stroke. Determine its air standard efficiency. What will be new efficiency if compression ratio is increased to 20? [Take γ = 1.4].

## Short answer

Cutoff ratio rc = 1 + 0.08(r-1). With r=16, rc=2.2, air-standard efficiency = 60.4%. Raising compression ratio to 20 yields rc=2.52 and efficiency = 62.5%.

## Answer

**Given:** r = 16 (later 20), γ = 1.4. Cut-off takes place at 8 % of the stroke.

#### Cut-off ratio
Stroke volume V_s = V₁ − V₂ = V₂ (r − 1). Cut-off at 8 % of the stroke means V₃ − V₂ = 0.08 V_s.

r_c = V₃/V₂ = 1 + 0.08 (r − 1)

Diesel efficiency: η = 1 − (1/r^(γ−1)) × (r_c^γ − 1)/(γ (r_c − 1))

#### Case 1: r = 16
r_c = 1 + 0.08 × 15 = 2.2
r^(γ−1) = 16^0.4 = 3.0314
r_c^γ = 2.2^1.4 = 3.0157

η = 1 − (1/3.0314) × (3.0157 − 1)/(1.4 × 1.2)
η = 1 − 0.3299 × 1.1998 = **60.4 %**

#### Case 2: r = 20
r_c = 1 + 0.08 × 19 = 2.52
r^(γ−1) = 20^0.4 = 3.3145
r_c^γ = 2.52^1.4 = 3.6472

η = 1 − (1/3.3145) × (3.6472 − 1)/(1.4 × 1.52)
η = 1 − 0.3017 × 1.2440 = **62.5 %**

**Answer: air-standard efficiency = 60.4 % at r = 16, rising to 62.5 % at r = 20.** (The compression ratio is raised, with the cut-off fixed at 8 % of stroke, so the efficiency improves by about 2.0 percentage points.)

## Check yourself

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

1. If cut-off is at 8% of the stroke, r_c equals:
   A. 1 + 0.08 (r - 1)
   B. 0.08 r
   C. r/0.08
   D. 1 + 0.08 r
   Answer: A. V3 - V2 = 0.08 V2 (r - 1), so r_c = 1 + 0.08(r - 1). (Key point: Cut-off ratio formula)

2. For r = 16, r_c is:
   A. 1.08
   B. 1.2
   C. 2.2
   D. 2.28
   Answer: C. r_c = 1 + 0.08 x 15 = 2.2. (Key point: Case 1)

3. Efficiency at r = 16 is about:
   A. 62.5%
   B. 60.4%
   C. 70%
   D. 56.5%
   Answer: B. eta = 1 - 0.3299 x 1.1998 = 60.4%. (Key point: Result r=16)

4. Raising r to 20 with the cut-off fixed at 8% of stroke:
   A. Leaves it unchanged
   B. Gives 100%
   C. Raises efficiency to 62.5%, even though r_c rises to 2.52
   D. Lowers efficiency
   Answer: C. Higher r raises efficiency, outweighing the increase in r_c. (Key point: Effect of raising r)

## Related in Fundamental of Thermodynamics & Heat Transfer

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