Chapter 7, Problem 1P

To determine

Find the amount of utilization of lithium per year to produce tritium.

Answer to Problem 1P

The mass of Lithium is $\underset{¯}{2.64\times {10}^7\text{kg}}$.

Explanation of Solution

Take the world’s total primary energy use as $5.7\times {10}^{20}\text{J}$ per year from the preface of the textbook.

The energy per d-t fusion is $17.6\text{MeV}$from Equation (7.6) in the textbook.

The additional energy released from breeding reaction is $4.8\text{MeV}$ from Equation (7.10) in the textbook.

Calculation:

Find the total energy per d-t fusion as follows:

$\begin{array}{c}\text{Total}=\text{Energy}\text{per d-t fusion}+\text{Additional}\text{energy}\\ \text{=17}\text{.6}\text{MeV+4}\text{.8}\text{MeV}\\ =\text{22}\text{.4}\text{MeV}\text{per}\text{fusion}\end{array}$

Convert the unit from MeV to J.

$\begin{array}{c}\text{Total energy}=\text{22}\text{.4}\text{MeV}\times \left(1.6\times {10}^{-13}\text{J/MeV}\right)\\ =3.58\times {10}^{-12}\text{J}\text{per fusion}\end{array}$

Find the required World’s primary energy as follows:

$\begin{array}{c}\text{World's primary energy}=\text{World's}\text{total}\text{energy}\text{use}\times \text{ total energy per fusion}\\ =\left(5.7\times {10}^{20}\text{J}\right)\times \left(3.58\times {10}^{-12}\text{J}\text{per fusion}\right)\\ =1.59\times {10}^{32}\text{fusions}\end{array}$

Every d-t fusion needs 1 tritium molecules, which is generates from one $\text{L}\text{i}$ conferring to Equation (7.10).

$\text{L}\text{i}+\text{n}\to \text{H}\text{+}\text{H}\text{e}$

The world energy needs require $1.59\times {10}^{32}\text{L}\text{i}$ atoms.

Consider $\text{L}\text{i}$ is 7% naturally abundant. It is calculated as follows:

$\frac{1.59\times {10}^{32}}{\left(\frac{7}{100}\right)}=2.27\times {10}^{33}$

Consider that the mass of one Li atom is $7u\times 1.66\times {10}^{-27}\text{kg/u}=1.16\times {10}^{-26}\text{kg/atom}$.

Find the amount of utilization of lithium per year to produce tritium as follows:

$2.27\times {10}^{33}\text{atoms}\times 1.16\times {10}^{-26}\text{kg/atom}=2.64\times {10}^7\text{kg}$

The amount of utilization of lithium per year to produce tritium is $2.64\times {10}^7\text{kg}$