# 32.5 Fusion  (Page 5/12)

 Page 5 / 12

## Calculating energy and power from fusion

(a) Calculate the energy released by the fusion of a 1.00-kg mixture of deuterium and tritium, which produces helium. There are equal numbers of deuterium and tritium nuclei in the mixture.

(b) If this takes place continuously over a period of a year, what is the average power output?

Strategy

According to ${}^{2}\text{H}+{}^{3}\text{H}\to {}^{4}\text{He}+n$ , the energy per reaction is 17.59 MeV. To find the total energy released, we must find the number of deuterium and tritium atoms in a kilogram. Deuterium has an atomic mass of about 2 and tritium has an atomic mass of about 3, for a total of about 5 g per mole of reactants or about 200 mol in 1.00 kg. To get a more precise figure, we will use the atomic masses from Appendix A. The power output is best expressed in watts, and so the energy output needs to be calculated in joules and then divided by the number of seconds in a year.

Solution for (a)

The atomic mass of deuterium ( ${}^{2}\text{H}$ ) is 2.014102 u, while that of tritium ( ${}^{3}\text{H}$ ) is 3.016049 u, for a total of 5.032151 u per reaction. So a mole of reactants has a mass of 5.03 g, and in 1.00 kg there are $\left(\text{1000 g}\right)/\left(\text{5.03 g/mol}\right)\text{=198}\text{.}8 mol of reactants$ . The number of reactions that take place is therefore

$\left(\text{198.8 mol}\right)\left(6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{-1}\right)=1.20×{\text{10}}^{\text{26}}\phantom{\rule{0.25em}{0ex}}\text{reactions}.$

The total energy output is the number of reactions times the energy per reaction:

$\begin{array}{}E=\left(1.20×{\text{10}}^{\text{26}}\phantom{\rule{0.25em}{0ex}}\text{reactions}\right)\left(\text{17.59 MeV/reaction}\right)\left(1.602×{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J/MeV}\right)\\ \text{}=3\text{.}\text{37}×{\text{10}}^{\text{14}}\phantom{\rule{0.25em}{0ex}}\text{J}\text{.}\end{array}$

Solution for (b)

Power is energy per unit time. One year has $3\text{.}\text{16}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{s}$ , so

$\begin{array}{lll}P& =& \frac{E}{t}=\frac{3\text{.}\text{37}×{\text{10}}^{\text{14}}\phantom{\rule{0.25em}{0ex}}\text{J}}{3\text{.}\text{16}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{s}}\\ & =& \text{}1\text{.}\text{07}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{W}=\text{10}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{MW}\text{.}\end{array}$

Discussion

By now we expect nuclear processes to yield large amounts of energy, and we are not disappointed here. The energy output of $3\text{.}\text{37}×{\text{10}}^{\text{14}}\phantom{\rule{0.25em}{0ex}}\text{J}$ from fusing 1.00 kg of deuterium and tritium is equivalent to 2.6 million gallons of gasoline and about eight times the energy output of the bomb that destroyed Hiroshima. Yet the average backyard swimming pool has about 6 kg of deuterium in it, so that fuel is plentiful if it can be utilized in a controlled manner. The average power output over a year is more than 10 MW, impressive but a bit small for a commercial power plant. About 32 times this power output would allow generation of 100 MW of electricity, assuming an efficiency of one-third in converting the fusion energy to electrical energy.

## Section summary

• Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus. It releases energy when light nuclei are fused to form medium-mass nuclei.
• Fusion is the source of energy in stars, with the proton-proton cycle,

being the principal sequence of energy-producing reactions in our Sun.

• The overall effect of the proton-proton cycle is

where the 26.7 MeV includes the energy of the positrons emitted and annihilated.

• Attempts to utilize controlled fusion as an energy source on Earth are related to deuterium and tritium, and the reactions play important roles.
• Ignition is the condition under which controlled fusion is self-sustaining; it has not yet been achieved. Break-even, in which the fusion energy output is as great as the external energy input, has nearly been achieved.
• Magnetic confinement and inertial confinement are the two methods being developed for heating fuel to sufficiently high temperatures, at sufficient density, and for sufficiently long times to achieve ignition. The first method uses magnetic fields and the second method uses the momentum of impinging laser beams for confinement.

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