Worked exercise — more thermonuclear than the Sun

Problem

In the experimental reactor ITER, the fusion reaction $\text{D} + \text{T} \to {}^{4}\text{He} + n + 17{,}6\ \text{MeV}$ is exploited, with a thermal power $P_{\text{out}} = 500\ \text{MW}$. How many grams of deuterium per second are needed, assuming the tritium is regenerated internally?

Solution

Number of reactions per second. Each reaction releases $E_{\text{reaction}} = 17{,}6\ \text{MeV}$. Converting to joules ($1\ \text{MeV} = 1{,}6\cdot 10^{-13}\ \text{J}$):

$E_{\text{reaction}} = 17{,}6\cdot 1{,}6\cdot 10^{-13}\ \text{J} \approx 2{,}82\cdot 10^{-12}\ \text{J}$

The number of reactions per second needed to deliver $P_{\text{out}} = 5\cdot 10^{8}\ \text{W}$ is:

$n = \frac{P_{\text{out}}}{E_{\text{reaction}}} = \frac{5\cdot 10^{8}}{2{,}82\cdot 10^{-12}} \approx 1{,}8\cdot 10^{20}\ \text{reactions/s}$

Mass of deuterium consumed. Each reaction consumes one deuterium nucleus, of mass $m_{\text{D}} \approx 3{,}34\cdot 10^{-27}\ \text{kg}$. The mass flow rate is therefore:

$\dot m = n\cdot m_{\text{D}} = 1{,}8\cdot 10^{20}\cdot 3{,}34\cdot 10^{-27}\ \text{kg} \approx \ev{\dot m \approx 0{,}6\ \text{mg/s}}$

Just over half a gram per minute: an extraordinarily small amount of fuel for a power-plant-scale output. This is a sign of the enormous energy density of nuclear fusion, the same process that powers the Sun.

Collegamenti

Argomenti: Fisica nucleare Concetti: Fusione nucleare · Equivalenza massa-energia