The diesel engine consumes 1 ton of diesel fuel at a distance of 1200 km. Engine efficiency 35%.

| January 15, 2021 | education

The diesel engine consumes 1 ton of diesel fuel at a distance of 1200 km. Engine efficiency 35%. Find the average driving speed if the engine power is 90 kW. (q = 42 • 10 ^ 6 J / kg)

m = 1 t = 1000 kg.
S = 1200 km = 12 * 10 ^ 5 m.
N = 90 kW = 90 * 10 ^ 3 W.
q = 42 * 10 ^ 6 J / kg.
Efficiency = 35%.
V -?
Let’s write down the definition for the efficiency of a diesel engine: efficiency = Ap * 100% / Az.
Ap = efficiency * Az / 100%.
The expended work Az is expressed by the formula: Az = q * m.
Ap = efficiency * q * m / 100%.
Ap = 35% * 42 * 10 ^ 6 J / kg * 1000 kg / 100% = 147 * 10 ^ 8 J.
The useful work of the engine An is expressed by the formula: An = N * t.
t = An / N.
t = 147 * 10 ^ 8 J / 90 * 10 ^ 3 W = 1.63 * 105 s.
We express the average speed of movement V by the formula: V = S / t.
V = 12 * 10 ^ 5 m / 1.63 * 10 ^ 5 s = 7.36 m / s.
Answer: the average speed of movement is V = 12 * 10 ^ 5 m / 1.63 * 10 ^ 5 s = 7.36 m / s.



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