Find the efficiency of a tractor engine that develops a power of 110 kW and consumes 28 kg of diesel fuel per hour.

univerkov | November 25, 2020 | education

GIVEN:
g = 42 × 10 ^ 6J / kg
m = 28kg
N = 110kW
Decision:
N = A / t
A = Nt = 110000W × 3600s = 396 × 10 ^ 6J
Efficiency = Ap / Q × 100%
Q = gm = 42 × 10 ^ 6 × 28 = 1176 × 10 ^ 6J
Efficiency = 396 × 10 ^ 6/1176 × 10 ^ 6 × 100% = 33.67%
Answer: Efficiency = 33.67%

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