SSE JE Electrical Engineering 10 Dec 2020 Papers

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