A power station’s maximum demand is 50 MW, capacity factor is 0.6 and utilization factor is 0.85. Calculate the following:

a. Load factor

b. Annual energy producedThis question was previously asked in

MPSC Assistant Engineer EE Mains 2018 - Paper 1

Option 4 : ^{6} MWh

Load factor = 0.7058

Annual energy: 0.3 × 10CT 1: Engineering Mathematics

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**Concept:**

Load factor = average load/maximum demand

= Units generated in T hours/ (T × Maximum demand)

Utilization factor: It is the ratio of maximum demand on the power station to the rated capacity of the power station.

Utilization factor = maximum demand / rated capacity

Plant capacity factor: It is the ratio of actual energy produced to the maximum possible energy that could have been produced during a given period.

__Calculation:__

Maximum demand = 50 MW

Capacity factor = 0.6

Utilization factor = 0.85

Utilization factor = 0.85 = 50/rated capacity

⇒ Rated capacity = 58.82 MW

Capacity factor = 0.6 = Actual energy/58.82

⇒ Actual energy = 35.29 MW

Number of hours in year = 8760

Energy generated per annum = 35.29 × 8760 = 0.3 × 10^{6} MWh

Load factor = 35.29/50 = 0.7058