# Can artificial fusion have the same power density as in Sun's core?

I recently read about solar core, and found that the power density is only 276.5 watts per cubic meter. This seems rather low to me, but on the other hand the Sun is an enormous object, so the total power output can be large.

When trying to replicate fusion on planet Earth, the attempt is probably for much higher power densities. A fusion power plant having a density of 276.5 watts per cubic meter would not be of much use. For example, ITER plasma volume is 840 cubic meters, so ITER would have 232.26 kW heat output given this power density, about the same as a typical car engine (at 25% efficiency, 232.26 kW heat output corresponds to 58 kW of mechanical power).

Is it possible in theory to create a self-sustaining megawatt-scale fusion power plant that has the same power density as in the solar core? I'm not interested in the economical viability (it's not viable unless the power density is made much larger), but rather the physical viability. Can a megawatt-scale fusion power plant maintain the fuel in a high enough temperature for fusion to occur, if the power density is only 276.5 watts per cubic meter?

• As the linked article mentions, the solar core density is 150 times that of liquid water, at a temperature of 15 million kelvins. If you can replicate that, you'll get fusion of plain hydrogen. Good luck doing that without using far more energy than 276.5 watts per cubic metre. ;) See en.wikipedia.org/wiki/Fusor & fusor.net for fusion devices that don't waste quite as much energy. – PM 2Ring Dec 28 '19 at 9:31

To get the same fusion rate as in the Sun, you need not only the high temperature but also the high density, which requires high pressure.

According to the cited wiki article the pressure at the core is 265 billion bar. The deepest oceans are 1,000 bar and Diamond anvil cells Schematics of the core of a diamond anvil cell. The culets (tip) of the two diamond anvils are typically 100–250 microns across.

can achieve 7,700,000 bar in a tiny sample. The latter is 32,000 times less than the required core pressure.

Interesting question! Let's try to answer it.

Magnetic confinement fusion is at the moment the most successful attempt trying to build a fusion reactor on Earth. You mentioned ITER which belongs to this class. ITER will have a plasma volume of $$840\,\mathrm{m}^3$$ and a fusion power of $$500\,\mathrm{MW}$$, resulting in a power density of roughly $$600\,\mathrm{kW}/\mathrm{m}^{3}$$. You are asking if it is possible to have a self-sustained fusion power-plant in the megawatt-range with the same power density as the sun. I guess that you have a typo there and wanted to write gigawatt-range instead (I am assuming $$1\,\mathrm{GW}$$ here).

Let's first estimate the volume of your plasma: $$\begin{equation} V = \frac{1\,\mathrm{GW}}{276.5\,\mathrm{W}/\mathrm{m}^3} \approx 3.6\cdot10^6\,\mathrm{m}^3. \end{equation}$$ This corresponds to a sphere with a radius of $$r\approx95\,\mathrm{m}$$ or a torus with a large radius of $$R\approx118\,\mathrm{m}$$ (assuming a typical aspect ratio of $$A=3$$).

Let's now estimate the number of fusion reactions required to keep your plasma burning for a time of $$1\,\mathrm{s}$$. The energy released over that time period amounts to $$\begin{equation} W = P\cdot \Delta t = 1\,\mathrm{GW}\cdot1\mathrm{s} = 1\,\mathrm{GJ}. \end{equation}$$ The energy released in a single fusion reaction can be calculated from the mass difference of the fusion product compared to its initial ingredients. The mass difference is for D-T fusion $$\Delta m=0.018835\,\mathrm{u}$$, with $$\mathrm{u}$$ the atomic mass unit, which corresponds to an energy of $$\Delta E\approx2.81\,\mathrm{nJ}$$. We can now estimate the number of fusion reactions to $$\begin{equation} N = \frac{1\,\mathrm{GJ}}{\Delta E} \approx 3.6 \cdot 10^{20}. \end{equation}$$

Next, we will have a look at the fusion reaction rate coefficient $$\mathcal{R}$$, which is the number of reactions per volume per time. I am assuming D-T fusion here (and not p-p fusion which is mostly what happens in our sun). $$\mathcal{R}$$ is defined as $$\begin{equation} \mathcal{R} = n_\mathrm{deuterium} n_\mathrm{tritium} \left<\sigma v \right> = \left(\frac{n}{2}\right)^2\left<\sigma v \right>, \end{equation}$$ with $$\left<\sigma v \right>$$ the fusion reactivity (depending on the cross section $$\sigma$$ and the relative velocity $$v$$ of the two fusing particles). Since $$\mathcal{R}$$ corresponds to the number of reactions per volume per time, we can calculate it as $$\begin{equation} \mathcal{R} = \frac{N}{V\cdot\Delta t} \approx \frac{3.6\cdot10^{20}}{3.6\cdot10^6\,\mathrm{m}^3\cdot 1\,\mathrm{s}} = 10^{14}\,\frac{1}{\mathrm{m}^3\mathrm{s}} \end{equation}$$

Looking at the fusion reactivity will now allow us to estimate the plasma density. Using a wikipedia article as reference, we have a close look at the plot of the fusion reactivity (plotted as a function of the plasma temperature). We see that the DT reactivity has a maximum at a temperature of around $$50\,\mathrm{keV}$$. The maximum value is roughly $$\begin{equation} \left<\sigma v \right>_\mathrm{max} \approx 10^{-21}\,\mathrm{m}^3/\mathrm{s}. \end{equation}$$ Rearranging the equation for the reactivity, we can estimate the plasma density: $$\begin{equation} n = \sqrt{ 2\frac{\mathcal{R}}{\left<\sigma v \right>_\mathrm{max}} } \approx \sqrt{ 2 \frac{10^{14}\,\frac{1}{\mathrm{m}^3\mathrm{s}}}{10^{-21}\,\mathrm{m}^3/\mathrm{s}} } = \sqrt{ 2\cdot 10^{35}\,\mathrm{m}^{-6} } \approx 4.5\cdot10^{17}\,\mathrm{m}^{-3}. \end{equation}$$

Finally, we apply the Lawson criterion (or the triple product), again a link to the corresponding wikipedia article, which let's us estimate the required energy confinement time (since we have set plasma density $$n$$ and temperature $$T$$ already) to get a self-sustained plasma: $$\begin{equation} n \cdot T \cdot \tau_E > 3\cdot 10^{24}\,\frac{\mathrm{eV}\,\mathrm{s}}{\mathrm{m}^3}. \end{equation}$$ Rearranging and inserting the values for $$n$$ and $$T$$ yields $$\begin{equation} \tau_E > \frac{ 3\cdot 10^{24}\,\frac{\mathrm{eV}\,\mathrm{s}}{\mathrm{m}^3} } {4.5\cdot10^{17}\,\mathrm{m}^{-3} \cdot 50\,\mathrm{keV}} \approx 130\,\mathrm{s}. \end{equation}$$ Note that this value is two orders of magnitude above what is currently possible. Apart from that, you now know what should be required to operate a fusion power plant with the same power density as the sun.