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For a two cylinder, four stroke cycle engine the bore is 108 mm and the stroke is 135mm. During a test with the engine running at a speed of 16.5 revs/s the fuel consumption inan hour is 2.5 kg. The calorific value of the fuel is 45.5 MJ/kg. The indicated mean effectivepressure is 558 kN/m2 and the mechanical efficiency is 80 %. Determine;(i) The indicated power and the brake power. (8 marks)(ii) The brake specific fuel consumption. (4 marks)(iii) The thermal efficiency on a brake and on an indicated power basis. (6 marks)(iv) If 40 % of the total energy supplied in the fuel is accounted for by friction, exhaustgases and losses to the surroundings, estimate the rise in temperature of the coolingwater. The cooling water flows at a rate of 215 kg/hour. Take the specific heatcapacity of water as 4200 J/kgK

Question

For a two cylinder, four stroke cycle engine the bore is 108 mm and the stroke is 135mm. During a test with the engine running at a speed of 16.5 revs/s the fuel consumption inan hour is 2.5 kg. The calorific value of the fuel is 45.5 MJ/kg. The indicated mean effectivepressure is 558 kN/m2 and the mechanical efficiency is 80 %. Determine;(i) The indicated power and the brake power. (8 marks)(ii) The brake specific fuel consumption. (4 marks)(iii) The thermal efficiency on a brake and on an indicated power basis. (6 marks)(iv) If 40 % of the total energy supplied in the fuel is accounted for by friction, exhaustgases and losses to the surroundings, estimate the rise in temperature of the coolingwater. The cooling water flows at a rate of 215 kg/hour. Take the specific heatcapacity of water as 4200 J/kgK

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Solution

(i) The indicated power (IP) can be calculated using the formula:

IP = IMEP * L * A * N * n / 2

where: IMEP = Indicated Mean Effective Pressure = 558 kN/m^2 = 558000 N/m^2 L = Stroke Length = 135 mm = 0.135 m A = Area of the cylinder = π * (bore/2)^2 = π * (0.108/2)^2 = 0.00915 m^2 N = Engine Speed = 16.5 rev/s n = Number of cylinders = 2

Substituting these values into the formula gives:

IP = 558000 * 0.135 * 0.00915 * 16.5 * 2 / 2 = 47.2 kW

The brake power (BP) can be calculated using the formula:

BP = IP * Mechanical Efficiency

Substituting the given mechanical efficiency of 80% gives:

BP = 47.2 * 0.8 = 37.76 kW

(ii) The brake specific fuel consumption (BSFC) can be calculated using the formula:

BSFC = Fuel Consumption / BP

Substituting the given fuel consumption of 2.5 kg/hour and the calculated BP gives:

BSFC = 2.5 / 37.76 = 0.066 kg/kWh

(iii) The thermal efficiency on a brake power basis (ηb) can be calculated using the formula:

ηb = BP / (Fuel Consumption * Calorific Value)

Substituting the calculated BP, the given fuel consumption and calorific value gives:

ηb = 37.76 / (2.5 * 45.5) = 0.33 or 33%

The thermal efficiency on an indicated power basis (ηi) can be calculated using the formula:

ηi = IP / (Fuel Consumption * Calorific Value)

Substituting the calculated IP, the given fuel consumption and calorific value gives:

ηi = 47.2 / (2.5 * 45.5) = 0.41 or 41%

(iv) If 40% of the total energy supplied in the fuel is accounted for by friction, exhaust gases and losses to the surroundings, then the energy absorbed by the cooling water is 60%. The rise in temperature of the cooling water can be calculated using the formula:

ΔT = Energy Absorbed by Water / (Mass Flow Rate of Water * Specific Heat Capacity of Water)

Substituting the given values gives:

ΔT = 0.6 * 2.5 * 45.5 * 10^6 / (215 * 4200) = 80.4 K

This problem has been solved

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