- Nuclear reactors are useful in producing electricity.
- A nuclear reactor is an arrangement to generate electricity whichmakes use of nuclear fission.
Requirements for controlled nuclear fission in reactor:
- Neutrons to be slowed down.
- Neutrons are slowed down by using moderators which are lighter nuclei which slow down fast moving neutrons by elastic collision.
- Commonly used moderators:-
- Heavy Water
- Consequence of use of moderators:-
- Multiplication factor of neutrons increases: - When a neutron hits a target nucleus along with daughter nucleus,it produces 3 neutrons.These 3 neutrons are highly energetic but they need to be slowed down, so they can hit the target nucleus.
- As a result high multiplication factor results in uncontrolled chain reaction.
- Excess neutrons to be absorbed.
- As uncontrolled chain reaction is wanted therefore to absorb excess neutrons Control Rods are used.
- These control rods are inserted in the core of the nuclear reactor.
- Control rods are capable of initiating(while taking out of the reactor) and stopping(inserting in the nuclear reactor) the nuclear reaction.
- As they absorb all the excess neutronsthere are no neutrons left to start the reaction.
- Control rods are made up of neutron absorbing materials.
- They decrease the multiplication factor of neutrons to a very small value.
- Commonly used material is Cadmium.
Construction of Nuclear Reactor
- The core of the nuclear reactor consists of uranium (235U) in the form of cylindrical rods.These rods are dipped inside a liquid which is the moderator.
- Whenever one neutron strikes this uranium rod nuclear fission reaction starts and 3 fast moving neutrons are produced.
- Because of the moderator these 3 neutrons undergo elastic collision as a result they slow down before they strike the second rod.
- Geometry of the core is such that only one out of 3 neutrons which are emitted strike the next rod making the reaction a controlled one.
- When the control rods are inserted inside they will absorb all the extra neutrons.Since there are no neutrons nuclear fission reaction will stop.
- Large amount of energy is also released in the core.
- In order to extract the energy from the core water at very high pressure is passed through it.
- As hot water passes through it produces steam in the steam generators.
- This steam is used to run the turbines which in turn produce electricity.
- This process will keep on continuing till the uranium on the rods does not get over.Then the rods have to be replaced in the nuclear reactor.
- Energy released is extremely large.
- Needs fuel in extremely small quantity.
- Spent fuel is highly radioactive and extremely hazardous to all life forms.
- Accumulation of radioactive waste.
Problem:- A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much (23592U) did it containinitially? Assume that the reactor operates 80% of the time that all the energy generatedarises from the fission of (23592U) and that this nuclide is consumed only by the fissionprocess.
Answer:- Half-life of the fuel of the fission reactor, t(1/2)= 5years
= 5 × 365 × 24 × 60 × 60 s
We know that in the fission of 1 g of (23592U) nucleus, the energy released is equal to 200MeV.
1 mole, i.e., 235 g of (23592U) contains (6.023 × 1023) atoms.
1 g (23592U) contains = (6.023x1023)/ (235) atoms.
The total energy generated per gram of (23592U) is calculated as:
E = ((6.023x1023)/ (235)) x200MeV/g
= (200x6.023x1023x1.6x10-19x106)/ (235) =8.20x1010 J/g
The reactor operates only 80% of the time.
Hence, the amount of (23592U) consumed in 5 years by the 1000 MW fission reactor iscalculated as:
= (5x80x60x60x365x24x1000x106) g / (100x8.20x1010)
Therefore, Initial amount of (23592U) = 2 × 1538 = 3076 kg
Problem:- The fission properties of (23994Pu) are very similar to those of (23592U).The average energy released per fission is 180 MeV. How much energy, in MeV, isreleased if all the atoms in 1 kg of pure (23994Pu) undergo fission?
Answer:- Average energy released per fission of (23994Pu), Eav=180MeV
Amount of pure (23994Pu), m = 1 kg = 1000 g
NA= Avogadro number = 6.023 × 1023
Mass number of (23994Pu) = 239 g
1 mole of (23994Pu) contains NA atoms.
Therefore mg of (23994Pu) contains ((NA)/ (Mass number) x m) atoms
= ((6.023 × 1023)/ (239) x1000) =2.52 x1024 atoms.
Therefore Total energy released during the fission of 1 kg of (23994Pu) is calculated as:
E=Eav x 2.52x1024
=180x2.52x1024 = 4.536x1026MeV
Hence, 4.536x1026MeV is released if all the atoms in 1 kg of pure (23994Pu) undergo fission.