Question

A 200 MW fission reactor consumes half its fuel in 3.00 y . How much ${}^{\mathbf{235}}\mathbf{U}$did it contain initially? Assume that all the energy generated arises from the fission of ${}^{\mathbf{235}}\mathbf{U}$ and that this nuclide is consumed only by the fission process.

Short answer

The mass of the Uranium is 462 kg.

Step-by-step solution

Write the given data from the question:

The power of the reactor, $P=200\mathrm{MW}$

The time, $∆t=3\mathrm{years}$

The molar number of the uranium, $M=235\mathrm{g}/\mathrm{mol}$

Determine the formulas to calculate the initially contain mass of the U235 :

The expression to calculate the released energy is given as follows.

$\mathit{E}\mathbf{=}\mathit{P}\mathbf{∆}\mathit{t}$ …… (1)

The expression to calculate the number of fission is given as follows.

$\mathit{n}\mathbf{=}\frac{\mathbf{E}}{\mathbf{Q}}$ …… (2)

Here,Qis the energy per fission.

The expression to calculate the number of nuclei is given as follows.

$\mathit{N}\mathbf{=}\mathbf{2}\mathit{n}$ …… (3)

Calculate the initially contain mass of the U235 :

The typical energy released per fission is $Q=200\mathrm{MeV}$

Calculate the released energy.

Substitute 200 MW for P and 3 years into equation (1).

$$\begin{gathered} E=200\times {10}^6\times \left(3\times 365\times 24\times 3600\right) \\ =200\times {10}^6\times 9.46\times {10}^7 \\ =1.894\times {10}^{16}\mathrm{J} \\ =1.183\times {10}^{29}\mathrm{MeV} \end{gathered}$$

Calculate the number of the fission.

Substitute $1.183\times {10}^{29}\mathrm{MeV}$ for E and 200 MeV for Q into equation (2).

$$\begin{gathered} n=\frac{1.183\times {10}^{29}}{200} \\ =0.005913\times {10}^{29} \\ =5.913\times {10}^{26} \end{gathered}$$

calculate the number of the nuclei.

Substitute $5.913\times {10}^{26}$ for n into equation (3).

$$\begin{gathered} N=2\times 5.913\times {10}^{26} \\ =11.8375\times {10}^{26} \end{gathered}$$

Calculate the mass of the uranium.

$$\begin{gathered} M_U=11.8375\times {10}^{26}\times 235\times u \\ =2.15\times {10}^{26}\times 235\times 1.661\times {10}^{-27} \\ =4620.59\times {10}^{-1} \\ =462\mathrm{kg} \end{gathered}$$

Hence, the mass of the Uranium is $462\mathrm{kg}$.