Fuel of a fission power plant

Problem

A fission power plant produces $1\ \text{GW}$ of electrical power with efficiency $\eta = 0{,}33$. Each fission of $^{235}\text{U}$ releases about $200\ \text{MeV}$. Estimate the mass of fissile fuel consumed in one year (assuming complete burn-up of only the $^{235}\text{U}$) and compare it with the mass of coal for an equivalent power plant ($\sim 30\ \text{MJ/kg}$).

Induced fission of $^{235}\text{U}$: a neutron triggers the split into two fragments, releasing further neutrons and about $200\ \text{MeV}$.

Solution

Thermal power. The electrical power is only a fraction $\eta$ of the thermal power generated by the fissions:

$P_{\text{term}} = \frac{P_e}{\eta} = \frac{1\ \text{GW}}{0{,}33} \approx 3{,}0\ \text{GW} = 3{,}0\cdot 10^{9}\ \text{W}$

Energy per fission. With $1\ \text{MeV} = 1{,}6\cdot 10^{-13}\ \text{J}$:

$E_{\text{fiss}} = 200\ \text{MeV} = 200\cdot 1{,}6\cdot 10^{-13}\ \text{J} = 3{,}2\cdot 10^{-11}\ \text{J}$

Fissions per second and per year.

$n = \frac{P_{\text{term}}}{E_{\text{fiss}}} = \frac{3{,}0\cdot 10^{9}}{3{,}2\cdot 10^{-11}} \approx 9{,}5\cdot 10^{19}\ \text{fissions/s}$

In one year ($3{,}156\cdot 10^{7}\ \text{s}$):

$N_{\text{fiss}} = 9{,}5\cdot 10^{19}\cdot 3{,}156\cdot 10^{7} \approx 3{,}0\cdot 10^{27}\ \text{fissions}$

Mass of $^{235}\text{U}$. Each fission consumes one nucleus; with $N_A = 6{,}022\cdot 10^{23}$ and molar mass $235\ \text{g/mol}$:

$m = \frac{N_{\text{fiss}}}{N_A}\cdot 235\ \text{g} = \frac{3{,}0\cdot 10^{27}}{6{,}022\cdot 10^{23}}\cdot 235\ \text{g} \approx 1{,}17\cdot 10^{6}\ \text{g} \approx \ev{m \approx 1{,}2\cdot 10^{3}\ \text{kg/year}}$

About $1{,}2$ tonnes of uranium-235 per year.

Comparison with coal. The annual thermal energy is the same:

$E_{\text{year}} = P_{\text{term}}\cdot t = 3{,}0\cdot 10^{9}\cdot 3{,}156\cdot 10^{7} \approx 9{,}5\cdot 10^{16}\ \text{J}$

$m_{\text{coal}} = \frac{9{,}5\cdot 10^{16}}{3\cdot 10^{7}\ \text{J/kg}} \approx 3{,}2\cdot 10^{9}\ \text{kg} \approx 3\ \text{million tonnes}$

The ratio is about $2{,}7\cdot 10^{6}$ in favour of nuclear: the same energy requires millions of times less fuel.

$\ev{m \approx 1{,}2\cdot 10^{3}\ \text{kg/year};\quad \sim 10^{6}\ \text{times less than coal}}$

Typo in the textbook answer key $\sim 10^{-3}\ \text{kg/year}$: this is a plain typo for $\sim 10^{3}\ \text{kg/year}$, since the calculation gives $\sim 1200\ \text{kg}$. (A consumption of one milligram per year for a $1\ \text{GW}$ plant would be physically absurd.)

The textbook’s answer key writes

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Topics: Nuclear physics Concepts: Nuclear fission Skills: Fermi estimation