Question

A diesel engine power plant has one 700 kW and two 500 kW generating units. The fuel consumption is 0.25 kg per kWh. Estimate the fuel oil required for a day if the plant capacity factor = 40%.
1. 4080 kg
2. 4000 kg
3. 4209 kg
4. 5230 kg

Answer 1

Correct Answer - Option 1 : 4080 kg

Plant Capacity Factor is the ratio of total actual energy produced to the maximum energy produced in that particular time.

Calculation:

Given:

Power plant rating = One 700 kW and two 500kW

Fuel consumption = 0.25 kg per kWh

Plant capacity factor = 40%

Maximum capacity = (500 + 500 + 700) = 1700 kW

Maximum Energy produced in a day = 1700 kW × 24 h = 40800 kWh

Actual energy produced = Plant Capacity Factor × Maximum energy produced in a day

Actual energy produced = 0.4 × 40800 kWh = 16320 kWh

 Fuel oil Required = Fuel consumption × Actual energy produced

 Fuel oil Required = 16320 × 0.25 = 4080 kg