Chapter 14, Problem 25P

(a)

To determine

The total power generated by the reactor.

Answer to Problem 25P

The total power generated by the reactor is $3333\text{ MW}$.

Explanation of Solution

Write the equation for the efficiency of the reactor.

  $e=\frac{P_{\text{delivered}}}{P_{\text{out}}}$

Here, $e$ is the efficiency, $P_{\text{delivered}}$ is the electrical power output of the plant and $P_{\text{out}}$ is the total power generated by the reactor.

Rewrite the above equation for $P_{\text{out}}$.

  $P_{\text{out}}=\frac{P_{\text{delivered}}}{e}$        (I)

Conclusion:

Substitute $1\text{ GW}$ for $P_{\text{delivered}}$ and $0.30$ for $e$ in equation (I) to find $P_{\text{out}}$.

  $\begin{array}{c}{P}_{\text{out}}=\frac{1\text{ GW}}{0.30}\\ =3.333\text{ GW}\frac{1000\text{ MW}}{1\text{ GW}}\\ =3333\text{ MW}\end{array}$

Therefore, the total power generated by the reactor is $3333\text{ MW}$.

(b)

To determine

The amount of power discharged to the environment as waste heat.

Answer to Problem 25P

The amount of power discharged to the environment as waste heat is $2333\text{ MW}$ .

Explanation of Solution

Write the equation for the amount of power discharged to the environment as waste heat.

  $P_{\text{heat}}=\left(1-e\right)P_{\text{out}}$        (II)

Here, $P_{\text{heat}}$ is the amount of power discharged to the environment as waste heat.

Conclusion:

Substitute $0.30$ for $e$ and $3333\text{ MW}$ for $P_{\text{out}}$ in equation (II) to find $P_{\text{heat}}$ .

  $\begin{array}{c}{P}_{\text{heat}}=\left(1-0.30\right)\left(3333\text{ MW}\right)\\ =2333\text{ MW}\end{array}$

Therefore, the amount of power discharged to the environment as waste heat is $2333\text{ MW}$.

(c)

To determine

The rate of fission events in the reactor core.

Answer to Problem 25P

The rate of fission events in the reactor core is $1.04\times {10}^{20}\text{ events/s}$ .

Explanation of Solution

The value of $P_{\text{out}}$ gives the total energy generated by the plant in one second. Thus, $\text{3333 MJ}$ of energy is produced per second in the core.

Write the equation for the rate of fission events in the reactor core

  $R=\frac{E}{Q}$        (III)

Here, $R$ is the rate of fission events in the reactor core, $E$ is the total energy generated by the plant in one second and $Q$ is the disintegration energy of a single fission reaction of ${}^{235}\text{U}$ .

Conclusion:

The value of $Q$ for ${}^{235}\text{U}$ is $208\text{ MeV}$ .

Substitute $\text{3333 MJ}$ for $E$ and $208\text{ MeV}$ for $Q$ in equation (III) to find $R$ .

  $\begin{array}{c}R=\frac{\left(\text{3333 MJ}\frac{{10}^6\text{ J}}{1\text{ MJ}}\frac{1\text{ eV}}{1.60\times {10}^{-19}\text{ J}}\frac{1\text{ MeV}}{{10}^6\text{ eV}}\right)}{208\text{ MeV}}\\ =1.04\times {10}^{20}\text{ events/s}\end{array}$

Therefore, the rate of fission events in the reactor core is $1.04\times {10}^{20}\text{ events/s}$ .

(d)

To determine

The mass of ${}^{235}\text{U}$ used up in one year.

Answer to Problem 25P

The mass of ${}^{235}\text{U}$ used up in one year is $1.17\text{ kg}$ .

Explanation of Solution

Write the equation for the mass of ${}^{235}\text{U}$ used up in one second.

  $\Delta m=\frac{E}{c^2}$        (IV)

Here, $\Delta m$ is the mass of ${}^{235}\text{U}$ used up in one second and $c$ is the speed of light in vacuum.

Write the equation for the mass of ${}^{235}\text{U}$ used up in one year.

  $\Delta M=\Delta m\Delta t$        (V)

Here, $\Delta M$ is the mass of ${}^{235}\text{U}$ used up in one year and $\Delta t$ is the time.

Conclusion:

The value of $c$ is $3.0\times {10}^8\text{ m/s}$ .

Substitute $3333\text{ MJ}$ for $E$ and $3.0\times {10}^8\text{ m/s}$ for $c$ in equation (IV) to find $\Delta m$ .

  $\begin{array}{c}\Delta m=\frac{3333\text{ MJ}\frac{{10}^6\text{ J}}{1\text{ MJ}}}{{\left(3.0\times {10}^8\text{ m/s}\right)}^2}\\ =3.70\times {10}^{-8}\text{ kg/s}\end{array}$

Substitute $3.70\times {10}^{-8}\text{ kg/s}$ for $\Delta m$ and $1\text{ yr}$  for $\Delta t$ in equation (V) to find $\Delta M$ .

  $\begin{array}{c}\Delta M=\left(3.70\times {10}^{-8}\text{ kg/s}\right)\left(1\text{ yr}\frac{365.25\text{ days}}{1\text{ yr}}\frac{24\text{ h}}{1\text{ day}}\frac{60\text{ min}}{1\text{ h}}\frac{60\text{ s}}{1\text{ min}}\right)\\ =1.17\text{kg}\end{array}$

Therefore, the mass of ${}^{235}\text{U}$ used up in one year is $1.17\text{ kg}$ .

(e)

To determine

The rate at which the fuel is converted to energy in the reactor core and to compare it with result from (d).

Answer to Problem 25P

The rate at which the fuel is converted to energy in the reactor core is $3.70\times {10}^{-8}\text{ kg/s}$ and it is in agreement with result from (d).

Explanation of Solution

Write the equation for the rate at which the fuel is converted to energy.

  $\text{rate}=\frac{\Delta m}{\Delta t}$

Put equation (IV) in the above equation.

  $\begin{array}{c}\text{rate}=\frac{\frac{E}{c^2}}{\Delta t}\\ =\frac{E}{c^2\Delta t}\end{array}$        (VI)

Conclusion:

Substitute $3333\text{ MJ}$ for $E$ , $3.0\times {10}^8\text{ m/s}$ for $c$ and $1\text{ s}$ for $\Delta t$ in equation (VI) to find the rate .

  $\begin{array}{c}\text{rate}=\frac{3333\text{ MJ}\frac{{10}^6\text{ J}}{1\text{ MJ}}}{{\left(3.0\times {10}^8\text{ m/s}\right)}^2\left(1\text{ s}\right)}\\ =3.70\times {10}^{-8}\text{ kg/s}\end{array}$

The result is in agreement with part (d).

Therefore, the rate at which the fuel is converted to energy in the reactor core is $3.70\times {10}^{-8}\text{ kg/s}$ and it is in agreement with result from (d).