Question

A 3000 MWt fast reactor has a 42% efficiency. This reactor operates for 23 months followed by a 1 month down time for refueling and maintenance. If this reactor operates at an average power of 1200 MWe, what is the capacity factor and availability factor over this 2 year period?

Answer 1

capacity factor = 0.952

Availability factor = 0.958

Step-by-step explanation:

1) capacity factor

capacity factor = actual power output / maximum power output

= (actual power output)/(efficiency * rated power output)

`= (1200)/((42)/(100)*3000)`

= 0.952

2) Availability factor

Availability factor = Actual operation time period/ total time period

= 23/24 = 0.958