Qalaxia Master Bot
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I found an answer from courses.lumenlearning.com

Fission | Physics


Fission is the opposite of fusion and releases energy only when heavy nuclei are split. ... The amount of energy per fission reaction can be large, even by nuclear standards. ... given the atomic masses to be m(238U) = 238.050784 u, m(95Sr) ... fissions produced by neutrons by having a large amount of fissionable material.


For more information, see Fission | Physics

Qalaxia Knowlege Bot
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I found an answer from large.stanford.edu

World Energy Resources | 2016


Oct 13, 2016 ... shale contains at least three times as much oil as conventional crude oil reserves , which ... Natural gas is the second largest energy source in power generation, ... 5. Global coal consumption increased by 64% from 2000 to 2014. ... 92% (10 936 MW) of all offshore wind installations are in European waters.


For more information, see World Energy Resources | 2016

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I found an answer from www.quora.com

If aircraft carriers and submarines can have nuclear power plants ...


I think Matthew Clifford's answer is on the right lines but I'm going to turn it on its head. ... I'll say that regular nuclear reactors aren't really that big. You cou... ... have 1 megawatt TRIGA research reactors like this one, made by General Atomics. ... You could probably fit a nuclear reactor core inside your bedroom at home.


For more information, see If aircraft carriers and submarines can have nuclear power plants ...

Krishna
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Given that

Fission reactor power P = 1000 MW = 10^3* 10^{6} W = 10^{9} W

Half life of the ^{235}_{92} U, T_{\frac{1}{2}} = 5 years  

Suppose 80% of the time the reactor operates = 80% of T_{\frac{1}{2}} = \frac{80}{100} * 5 = 4 years

\text{ 4 years } = 4*360*24 *60*60 . seconds

= 124416000 seconds


Step 1: Finding the number of atoms in the 1 kg fission matter

Avogadro number NA = 6.023 * 10^{23}

Atomic mass of   ^{235}_{92} U =. 235

Mass of a Uranium-235 m = 1kg = 10^{3} g   

Number of atoms in the 1 kg  _{92}^{235}U,\ \ N=\frac{NA}{A}*m

N = \frac{6.023 * 10^{23}}{235} * 10^{3}   


Step 2: Determine the total energy released by 1 kg mass of substance

The average amount of energy released per nuclear fusion = 200 MeV

E = \frac{6.023 * 10^{23}}{235} * 10^{3}* 200 MeV

E = 5.106 * 10^{26} MEV

E = 5.106 * 10^{26} * 1.6 * 10^{-13} Joules

E = 8.1696 * 10^{13} Joules


Step 3: Calculating the total energy consumed by substance in half life by fission reactor.

\text{ Energy consumed }= \frac{\text{ Energy of reactor released in 5 years }}{\text{ Energy release in 1 kg of substance }}

= \frac{124416000 * 10^{9}}{ 8.1696 * 10^{13} }

= 1522.9 kg  

Hence, energy consumed by Uranium-235 = 1522.9 kg  

Since the half life of Uranium-235 consumed = 1522.9 kg, the initially the amount will be doubled = 2 * 1522.9 = 3045.8   kg