Chapter 14, Problem 37P

(a)

To determine

The net power output of a fusion reaction.

Answer to Problem 37P

The net power output of a fusion reaction is $\underset{\_}{3000\text{MW}}$.

Explanation of Solution

Write the expression for the net power output,

    $P_{\text{net}}=P_{\text{output}}-P_{\text{input}}$        (I)

Here, $P_{\text{net}}$ is the net power output, $P_{\text{output}}$ is the power output and $P_{\text{input}}$ the power input.

Write the expression for the number of ${}_1^2\text{H}$ pairs,

    $N_{\text{pair}}=\left(\frac{m}{M}\right)N_a$        (II)

Here, $N_{\text{pair}}$ is the number of ${}_1^2\text{H}$ pairs, $m$ is the mass of ${}_1^2\text{H}$ pairs, $M$ is the molar mass and $N_a$ is the Avogadro number.

Write the expression for the power input,

    $P_{\text{input}}=\frac{W}{t}$        (III)

Here, $W$ is the work done and $t$ is the time taken.

Write the expression for the output,

    $P_{\text{output}}=\frac{N_{pair}nQ}{t}$        (IV)

Here, $Q$ the released energy in fusion reaction.

Conclusion:

Substitute $3\times {10}^{-3}\text{g}$ $m$, $6.02\times {10}^{23}\text{pairs/mole}$ for $N_a$ and $5.0\text{g/mole}$ (II),

    $\begin{array}{c}{N}_{\text{pair}}=\frac{\left(3\times {10}^{-3}\text{g}\right)\left(6.02\times {10}^{23}\text{pairs/mole}\right)}{5.0\text{g/mole}}\\ =3.61\times {10}^{23}\text{pair}\end{array}$

Substitute $3.61\times {10}^{23}\text{pair}$ $N_{\text{pair}}$ , $17.6\text{MeV/fusion}$ for $Q$ , $\left(10\right)\left(0.3\right)$ for $n$ and $1\text{s}$ for $t$ in (IV),

    $\begin{array}{c}{P}_{\text{output}}=\frac{\left(10\right)\left(0.3\right)\left(3.61\times {10}^{23}\text{pair}\right)\left(17.6\text{MeV/fusion}\right)\left(1.60\times {10}^{-13}\text{J/MeV}\right)}{1\text{s}}\\ =3.1\times {10}^9\text{W}\end{array}$

Substitute $5\times {10}^{14}\text{J/s}$ for $W$ and ${10}^{-8}\text{s}$ for $t$ in (III),

    $\begin{array}{c}{P}_{\text{input}}=\frac{\left(10\right)\left(5\times {10}^{14}\text{J/s}\right)}{{10}^{-8}\text{s}}\\ =5\times {10}^7\text{W}\end{array}$

Substitute $5\times {10}^7\text{W}$ for $P_{\text{input}}$ and $3.1\times {10}^9\text{W}$ for $P_{\text{output}}$ in (I),

    $\begin{array}{c}{P}_{\text{net}}=3.1\times {10}^9\text{W}-5\times {10}^7\text{W}\\ \simeq 3.0\times {10}^9\text{W}\\ =3.\times {10}^9\text{W}\left(\frac{1\text{MW}}{{10}^6\text{W}}\right)\\ =3000\text{MW}\end{array}$

Therefore, the net power output of a fusion reaction is $\underset{\_}{3000\text{MW}}$.

(b)

To determine

The equivalent of oil operates in a dayin liters.

Answer to Problem 37P

The equivalent of oil operates in a day in liters is $\underset{\_}{5.2\times {10}^6\text{liters}}$.

Explanation of Solution

Write the expression for the fusion energy in a day,

    $Q=P_{\text{net}}t$        (V)

Here, $Q$ is the fusion energy in a day

In this case, $2\text{L}$ of oil produce $100\text{MJ}$ , so the energy per liter,

    $k=\frac{E}{L}$        (VI)

Here, $k$ is the energy per liter, $E$ is the released energy and $L$ is the oil in liter.

Write the expression for the equivalent value of burned oil in liters,

    $eq=\frac{Q}{k}$        (VII)

Here, $eq$ is the equivalent value of burned oil in liters.

Conclusion:

Substitute $3000\text{MW}$ for $P_{\text{net}}$ and $1\text{day}$ for $t$ in (V),

    $\begin{array}{c}Q=\left(3000\text{MW}\right)\left(1\text{day}\right)\left(\frac{24\text{hr}}{1\text{day}}\right)\left(\frac{60\text{min}}{1\text{hr}}\right)\left(\frac{60\text{s}}{1\min }\right)\\ =2.6\times {10}^{14}\text{J}\end{array}$

Substitute $2\text{L}$ for $L$ and $100\times {10}^6\text{J}$ for $E$ in (VI),

    $\begin{array}{c}k=\frac{100\times {10}^6\text{J}}{2\text{L}}\\ =50\times {10}^{16}\text{J/liter}\end{array}$

Substitute $50\times {10}^{16}\text{J/liter}$ for $k$ and $2.6\times {10}^{14}\text{J}$ for $Q$ in (VII),

    $\begin{array}{c}eq=\frac{2.6\times {10}^{14}\text{J}}{50\times {10}^{16}\text{J/liter}}\\ =5.2\times {10}^6\text{liter}\end{array}$

Therefore, theequivalent of oil operates in a day in liters is $\underset{\_}{5.2\times {10}^6\text{liters}}$.