The reactor in numbers

Problem

A $^{235}\text{U}$ fission power plant produces an electrical power $P_e = 800\ \text{MW}$ with an efficiency $\eta = 0{,}32$. Each fission releases on average $\sim 200\ \text{MeV}$.

• (a) How many fissions per second occur in the core?
• (b) How many grams of $^{235}\text{U}$ are fissioned per day?
• (c) Compare this with the coal burned by a coal-fired power plant of equal thermal power (calorific value $\approx 30\ \text{MJ/kg}$).

Diagram of the core: fuel rods (blue), water-steam circuit (green) and cadmium control rods (red).

Solution

Thermal power. The efficiency links electrical and thermal power: $P_t = P_e/\eta = 800/0{,}32 = 2500\ \text{MW} = 2{,}5\times 10^{9}\ \text{W}$.

Energy per fission in joules. $E_f = 200\ \text{MeV} = 200 \times 1{,}6\times 10^{-13}\ \text{J} = 3{,}2\times 10^{-11}\ \text{J}$.

(a) Fissions per second. The number of fissions per second is the thermal power divided by the energy of one fission:

$R = \frac{P_t}{E_f} = \frac{2{,}5\times 10^{9}}{3{,}2\times 10^{-11}} \approx 7{,}8\times 10^{19}\ \text{s}^{-1}$

(b) Mass fissioned per day. In one day ($86\,400\ \text{s}$) the number of fissions is $N = R \cdot 86\,400 \approx 6{,}74\times 10^{24}$. Each fission consumes one $^{235}\text{U}$ nucleus, so the mass is the number of moles times the molar mass:

$m_U = \frac{N}{N_A}\cdot 235 = \frac{6{,}74\times 10^{24}}{6{,}022\times 10^{23}}\cdot 235\ \text{g} \approx 2{,}63\times 10^{3}\ \text{g} \approx 2{,}65\ \text{kg/day}$

(c) Comparison with coal. For equal thermal power, the energy to be supplied in one day is $E = P_t \cdot 86\,400 = 2{,}5\times 10^{9}\cdot 86\,400 \approx 2{,}16\times 10^{14}\ \text{J}$. With a calorific value of $30\ \text{MJ/kg} = 3\times 10^{7}\ \text{J/kg}$:

$m_C = \frac{2{,}16\times 10^{14}}{3\times 10^{7}} \approx 7{,}2\times 10^{6}\ \text{kg} \approx 7000\ \text{t/day}$

The ratio between coal mass and uranium mass is $\approx 7{,}2\times 10^{6}/2{,}65 \approx 2{,}7\times 10^{6}$: roughly two and a half million times more fuel needed for coal. This is a manifestation of the enormous energy density of nuclear fuel.

$\ev{\text{(a) } \approx 7{,}8\times 10^{19}\ \text{s}^{-1};\quad \text{(b) } \approx 2{,}65\ \text{kg/day};\quad \text{(c) factor } \approx 2{,}5\times 10^{6}}$

Connections

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