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(^{238}U) = 238.050784 u, m(^{95}Sr) ...
fissions **produced** by neutrons by having a large **amount** of fissionable **material**.

For more information, see **Fission** | **Physics**

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

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 ...

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