Suppose a plant that has installed capacity of 20 MW produces annual output of 7.35 × 10⁶ kWh and remains in operation for 2190 hours in a year. Find the plant use factor.

Question

Question

This question was previously asked in

SSC JE EE Previous Paper 9 (Held on: 29 Oct 2020 Evening)

Answer 1

Concept:

Plant use factor: It is the ratio of kWh generated to the product of plant capacity and the number of hours for which the plant was in operation.

\(Plant\;use\;factor = \frac{{Station\;output\;}}{{Plant\;capacity \times hours\;of\;use}}\)

Calculation:

Given that, station output = 7.35 × 10⁶ kWh = 7.35 × 10³ MWh

Plant capacity = 20 MW

Hours of use = 2190

Plant use factor \( = \frac{{7.35 \times {{10}^3}}}{{20 \times 2190}} \times 100 = 16.78\% \)