Question

Compute the thermal utilization factor for a nuclear reactor core where there are 50 moles of heavy water per mole of 5% enriched uranium monocarbide (UC). What would be the value if light water replaced heavy water?

Answer 1

Ordinary water is composed of 2 atoms of ordinary hydrogen and one atom of oxygen.Each hydrogen atom has only one proton but neutron is absent and one electron is revolving around proton.But deuterium has both proton and neutron in its nucleus, these special hydrogen atoms weigh about twice as much as an ordinary hydrogen atom we call them as heavy hydrogen or deuterium.when water is made by such heavy hydrogen we call them as heavy water(D₂O).

In a nuclear reactor, after fission reaction heavy nucleus splits into lighter and heavy fragments along with some neutrons also liberates,if one of these neutron hits an atom of nuclear fuel, such as U-235,that atom can split and keep the chain reaction going .Each atom that splits sends out several neutrons ,but most of them are lost.some leak out of the reactor and some get absorbed into the U-238 atoms.There are several tricks used to conserve neutrons ,one of those trick is slow down the neuton speed and to slow neutrons down,moderator is needed,we may use hydrogen for this purpose but hydrogen gas doesn't work but water works well because it packs a lot of hydrogen atoms together in small space.

A neutron might bounce off and slow down but it also might get absorbed by the hydrogen atom present in normal light water ,making it an atom of deuterium so in order to avoid this heavy water is used consequently none of the neutrons are absorbed in the water , all the hydrogen atoms in heavy water already contain neutron and they don't want extra neutron. So the neutrons bounce and bounce a little more and now there are enough of them left over that they can keep a chain reaction going even in natural.