(a) How much energy can be extracted by deuterium fusion from a gallon of sea water? Assume that an average D-D fusion yield is about per atom.

(b) A modem super tanker can hold$\text{9}\times {\text{10}}^{\text{7}}$ gallons. How many "water tankers" would be needed to supply the energy need of greater Los Angeles, consuming electricity at a rate of about $\text{20\hspace{0.17em}GW}$, for 1 year? Assume that only$\text{20\%}$ of the available energy actually becomes electrical energy.

Given:

• The energy produced per atom in a fusion of D-D is given as $Q_{D-D}=2MeV$.
• The volume of a water tank is $V_t=9\times {10}^7$gallons.
• Power consumption of the city is $20GW=20\times {10}^{9}W$.
• The efficiency of converting nuclear energy to electrical energy is $\text{20\%}$.
• The abundance of deuterium atoms among hydrogen molecules is $\text{0.0115\%}$.

The energy released in this decay depends upon the mass difference between the parent and daughter nucleus.

We can calculate the kinetic energy Q of the emitted particle using the formula: $Q=(m_p-m_D)c^2$

In this formula ${\text{m}}_P$is the mass of the parent nucleus. And ${\text{m}}_D$is the mass of the daughter nucleus.

The mass of one gallon of water is ${\text{M}}_{\text{g}}$, it can be calculated as:

$\begin{array}{c}{M}_g=\rho \times V\\ =1000\frac{kg}{m^3}\times 0.00454m^3\\ =4.54kg\end{array}$

The number of water molecules in a gallon of water can be calculated as:

$\begin{array}{c}{N}_g=\frac{M_g}{m_{H_{2}O}}\\ =\frac{4.54kg}{18u\times 1.66\times {10}^{-27}\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$u$}\right.}\\ =1.5194\times {10}^{26}\end{array}$

The number of hydrogen in water molecules can be calculated as:

$\begin{array}{c}{N}_H=2\times N_g\\ =3.03\times {10}^{26}\end{array}$

The number of deuterium can be calculated as:

$\begin{array}{c}{N}_D=\frac{0.0115}{100}\times 3.03\times {10}^{26}\\ =3.484\times {10}^{22}\end{array}$

The number of deuterium atoms in a gallon of water is $N_D=3.484\times {10}^{22}$.

Released energy can be calculated as:

$\begin{array}{c}{Q}_g=Q_{D-D}\times N_D\\ =2MeV\times 3.484\times {10}^{22}\\ =6.968\times {10}^{22}MeV\\ =11.14\times {10}^{9}J\end{array}$

The kinetic energy released from the fusion of deuterium from a gallon of water is $Q_g=11.14\times {10}^{9}J$.

The electrical power consumption of the city is $P=20\times {10}^{9}W$.

The electrical energy used by the city in the entire year is

$\begin{array}{c}E=20\times {10}^{9}W\times 3.16\times {10}^{7}s\\ =6.32\times {10}^{17}J\end{array}$

The total nuclear energy required for the production of electrical$\text{E}$energy is:

$\begin{array}{c}Q=\frac{E}{20\%}\\ =6.32\times {10}^{17}\times \frac{100}{20}\\ =3.16\times {10}^{18}J\end{array}$

K.E energy that can be produced from the fusion of D atoms from a truck of water is:

$\begin{array}{c}{Q}_t=V_t\times Q_g\\ =9\times {10}^7\times 1.114\times {10}^{10}J\\ =10.26\times {10}^{17}J\end{array}$

The number of trucks required to meet the electric needs of the city for a year is:

$\begin{array}{c}{T}_n=\frac{Q}{Q_t}\\ =\frac{3.16\times {10}^{18}J}{1.026\times {10}^{18}J}\\ T_n=3.08\end{array}$

Therefore, the number of trucks required to meet the electric needs of the city for a year is 3.