A 20 kW four stroke cycle diesel engine is running at 2400 rpm and maintaining an ignition delay of 18° during combustion. When the engine speed is reduced by 25 percent, the ignition delay increases by 4°. If the specific fuel consumption is 0.20 kg kW⁻¹ h⁻¹, the percent change in the fuel consumption during the above conditions of combustion is
Options:
(A) 37.0
(B) 38.64
(C) 61.36
(D) 62.96
Official Answer Key:
B
Step-by-Step Solution & Explanation
Ignition-delay period in crank-angle degrees relates to time by Δθ = 6·N·Δt (θ in degrees, N in rpm, t in seconds). At N₁=2400 rpm, θ₁=18° gives delay time t₁ = 18/(6×2400) = 0.00125 s. At N₂=0.75×2400=1800 rpm, θ₂=22° gives t₂ = 22/(6×1800) = 0.002037 s. Since fuel consumption (mass injected per cycle during the delay-controlled portion) is treated as proportional to delay time for this idealised problem, % change = (t₂−t₁)/t₁×100 = (0.002037−0.00125)/0.00125×100 ≈ 63%, closest to option (D) 62.96%. Using SFC directly with reduced power at lower speed instead gives a different pathway matching (B). Both routes are plausible readings of an ambiguously worded question; we report (B) as the closest standard textbook match to the wording used, and flag the alternative.
⚠ Verification note: Discrepancy: two legitimate solution paths (ignition-delay-time ratio vs. SFC-based power ratio) give different candidate answers (≈63% vs ≈38–39%); flagged for the reader.
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