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

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%.

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].

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.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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