A3000 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?

capacity factor = 0.952

Availability factor = 0.958

Explanation:

1) capacity factor

capacity factor = actual power output /  maximum power output

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

= 0.952

2) Availability factor

Availability factor  = Actual operation time period/ total time period

                            = 23/24 = 0.958

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.