If all of the world energy needs were provided by d-t fusion, how much lithium per year would be utilized to produce tritium? Consider the total energy associated with fusion as being equivalent to the world's current primary energy use.

Question

Quizplus · Accepted Answer

The Answer is From the Preface the world's total primary energy use is

5.7 imes 10 ^ { 20 } \mathrm {~J} ext { per vear. The }

energy per d-t fusion is given by equation (7.6) as 17.6 MeV. An additional 4.8 MeV is released in the breeding reaction given in equation (7.10) for a total of (17.6+4.8) MeV =
22.4 MeV. This is converted into joules to yield

( 22.4 \mathrm { MeV } ) imes \left( 1.6 imes 10 ^ { - 13 } \mathrm {~J} / \mathrm { MeV } ight) = 3.58 imes 10 ^ { - 12 } \mathrm {~J}

thus the worlds primary energy needs require

\left( 5.7 imes 10 ^ { 20 } \mathrm {~J} ight) \left( 3.58 imes 10 ^ { - 12 } \mathrm {~J} / ext { fusion } ight) = 1.59 imes 10 ^ { 32 } ext { fusions }

Each d-t fusion requires 1 tritium nucleus, which is produced from one

{ } ^ { 6 } \mathrm { Li }

according to equation (7.10)

{ } ^ { 6 } \mathrm { Li } + \mathrm { n } ightarrow { } ^ { 3 } \mathrm { H } + { } ^ { 4 } \mathrm { He }

Thus the world energy needs require

1.59 imes 10 ^ { 32 } { } ^ { 6 } \mathrm { Li }

atoms. Since

{ } ^ { 6 } \mathrm { Li }

is

7 \%

naturally abundant then

\left( 1.59 imes 10 ^ { 32 } / 0.07 ight) = 2.27 imes 10 ^ { 33 }

total

\mathrm { Li }

atoms are needed. Since the mass of one

\mathrm { Li }

atom is

\approx 7 \mathrm { u } imes 1.66 imes 10 ^ { - 27 } \mathrm {~kg} / \mathrm { u } = 1.16 imes 10 ^ { - 26 } \mathrm {~kg}

then the mass of

\mathrm { Li }

is

\left( 2.27 imes 10 ^ { 33 } ext { atoms } ight) imes \left( 1.16 imes 10 ^ { - 26 } \mathrm {~kg} / ext { atom } ight) = 2.64 imes 10 ^ { 7 } \mathrm {~kg}

If All of the World Energy Needs Were Provided by D-T