Question

The natural fission reactor discussed in Module 43-3 is estimated to have generated 15 gigawatt-years of energy during its lifetime.

(a) If the reactor lasted for 200,000 y, at what average power level did it operate?

(b) How many kilograms of ${}^{\mathbf{235}}\mathbf{U}$did it consume during its lifetime?

Short answer

a) The average power of the reactor is75 kW.

b)$5.8\times {10}^3\mathrm{kg}$

Step-by-step solution

Introduction

Nuclear reactors are the heart of a nuclear power plant. They contain and control nuclear chain reactions that produce heat through a physical process called fission. That heat is used to make steam that spins a turbine to create electricity.

Given data

The energy generated by the natural fission reactor is,

$E=15\mathrm{GW}.\mathrm{y}\left(\frac{{10}^9\mathrm{W}.\mathrm{y}}{1\mathrm{GW}.\mathrm{y}}\right)=15\times {10}^9\mathrm{W}.\mathrm{y}$

The time t = 200.000 y.

(a) Determine the average power of the reactor

The average power of the reactor is,

$$\begin{gathered} p_{avg}=\frac{E}{t} \\ =\frac{\left(15\times {10}^9\mathrm{W}.\mathrm{y}\right)}{\left(200,000\mathrm{y}\right)} \\ =75\times {10}^3\mathrm{W} \\ =\left(75\times {10}^3\mathrm{W}\right)\left(\frac{1\mathrm{k}\mathrm{W}}{{10}^3\mathrm{W}}\right) \\ =\boxed{75\mathrm{kW}} \end{gathered}$$

Hence, the average power of the reactor is 75 kW.

(b) Determine how much it consumes

The mass of the uranium nucleus is,

$$\begin{gathered} m_{\mathrm{U}}=235\mathrm{u} \\ =\left(235\mathrm{u}\right)\left(1.67\times {10}^{-27}\mathrm{kg}/\mathrm{u}\right) \\ =392.45\times {10}^{-27}\mathrm{kg} \\ \approx 392.45\times {10}^{-27}\mathrm{kg} \end{gathered}$$

The energy released in each fission reaction is,

$$\begin{gathered} \mathrm{Q}=200\mathrm{MeV} \\ =\left(200\mathrm{MeV}\right)\left(\frac{1.6\times {10}^{-13}\mathrm{J}}{1\mathrm{MeV}}\right) \\ =320\times {10}^{-13}\mathrm{J} \end{gathered}$$

The energy generated by the natural fission reactor is,

$$\begin{gathered} E=15\times {10}^{9}W.y \\ =\left(15\times {10}^9\mathrm{W}.\mathrm{y}\right)\left(\frac{3.156\times {10}^7\mathrm{s}}{1}\right) \\ =47.34\times {10}^7\mathrm{s} \end{gathered}$$

Hence, the amount of mass of ${}^{235}\mathrm{U}$did it consume during its lifetime is,

$$\begin{gathered} M=\frac{m_{U}E}{Q} \\ =\frac{\left(392.45\times {10}^{-27}\mathrm{kg}\right)\left(47.34\times {10}^{16}\mathrm{s}\right)}{\left(320\times {10}^{-13}\mathrm{J}\right)} \\ =\boxed{5.8\times {10}^3\mathrm{kg}} \end{gathered}$$

Therefore, $5.8\times {10}^3\mathrm{kg}$it consumes during its lifetime.