Determine the efficiency of a bus engine consuming 63 kg of fuel q = 4.33 * 10 ^ 4

Determine the efficiency of a bus engine consuming 63 kg of fuel q = 4.33 * 10 ^ 4 railroad / kg for 2.5 hours of operation with an average power of 70 kW.

m = 63 kg.

q = 4.33 * 10 ^ 4 J / kg.

t = 2.5 h = 9000 s.

N = 70 kW = 70,000 W.

Efficiency -?

Let’s write down the definition for the efficiency of the bus engine: efficiency = Ap * 100% / Az.

We will find the useful work of the engine Ap by the formula: Ap = N * t, where N is the engine power, t is the engine operating time.

Ap = 70,000 W * 9000 s = 630,000,000 J.

The expended work Az is expressed by the formula: Az = m * q, where m is the mass of the fuel that has burned in the engine, q is the specific heat of combustion of the fuel.

Az = 63 kg * 4.33 * 10 ^ 4J / kg = 2727900 J.

Efficiency = 630,000,000 J * 100% / 2727900 J = 23094%.

The efficiency cannot be more than 100%.

Answer: a mistake was made in the problem statement.