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If a diesel engine has a compression ratio of 24 and the cut-off takes place at 5% of stroke, the air standard efficiency will be ___________

a) 60.66%
b) 64.50%
c) 66.60%
d) 65.40%

Answer: c
Explanation: Here, r = V1/V2 = 24
V1 = 24 V2
V3 = V2 + 0.05 (V1-V2)
V3 = V2 + 0.05(24 V2-V2)
V3/V2 = 2.15
Efficiency: 1 – (1/r)γ-1 * [rcγ – 1] / r [rc – 1]
= 1 – (1/24)1.4-1 [2.151.4 – 1] / 1.4 [2.15-1]
= 66.60%.

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