# How much energy does it take to fuse 1m$^3$ of seawater, and how much energy is released by that fusion?

So, assume that you have 1m$^3$ of normal Terran seawater you can adiabatically raise to an arbitrarily high pressure/temperature in situ (say, by compressing it with nuclear shockwaves from all directions simultaneously).

Just how much energy would those shockwaves need to impart to said 1m$^3$, and how much energy would be released by the resulting fusion reactions, for these four cases:

• Only D-T fusion takes place, no proton-proton or deuteron-deuteron reactions, and no reactions involving heavy nuclei
• D-T and D-D fusion takes place, but no proton-proton or heavy nucleus fusion takes place
• Proton-proton (at incidence levels comparable to that found within Sol), D-D, and D-D fusion all take place, but no heavy nuclei fuse
• Proton-proton, D-D, and D-T fusion take place, sufficient to convert the entire hydrogen contents of our 1m$^3$ cube into helium, but no heavy nuclei fuse

Furthermore, if at least one of the above is net exergonic to the point where a chain reaction is plausible under the given priors, what would prevent such a fusion chain reaction from spreading throughout a water-bearing, rocky planet's oceans?

• Quick check: by hard-science rules, we have to recognize that we've only recently gotten a fusion reaction to put out as much energy as we put into making it fuse, and that was with carefully prepared materials. Are you prepared for an answer where the amount of energy in is far far more than the energy out? Dec 11, 2016 at 22:19
• @CortAmmon -- yes, although one difference is that I'm talking about uncontrolled conditions here (think H-bomb, not fusion reactor). Dec 11, 2016 at 22:59
• I was doing some research to see if I could answer this question, and it is looking like it is ridiculously unlikely. I can't find any solid numbers on fusion of non-enriched sea water, but given that CANDU enriches its water to 99.75% and nuclear weapons tend to enrich to similar levels, I'd be surprised if it was plausible to cause fusion in sea water, with its 0.0156% heavy-water content. I'd be far more frightened by the massive nuclear arsenal which it'd take to fuse it, if it is even possible. Did you intend to permit enriching that 1m^3 of water first? Dec 11, 2016 at 23:28
• @CortAmmon -- we can cover the enriched case as a Mythbusters-style bonus :) Dec 11, 2016 at 23:36
• Similarly to what @CortAmmon says in their first comment - it depends on how efficiently you do it. This may turn out to be opinion-based if there are multiple hard-science ideas behind how efficiently we can fuse. Dec 12, 2016 at 0:39

First, references: Bosch and Hale, 1992 has cross section and S-function values for the D and T reactions, as does Nuclear Cross Sections for Technology from the US Department of Commerce, 1979.

# D-T fusion

The natural deuterium abundance is 0.0156%. 1 m$^3$ of seawater has a mass of 1000 kg, or 5.5e4 mols of H$_2$O. Since there are two hydrogens per water molecule, that gives us about 17 mol of deuterium per cubic meter of seawater. The atomic density of deuterium is then 1.1e25 atoms / m$^3$.

Present day tritium concentrations range from 2 TU (1 TU = 1e-18 tritium per hydrogen) in the Arctic to 0.15 TU in the Southern Ocean. Lets establish a 1 TU concentration baseline, so using the same calculations we have 1.1e-13 mols of tritium per cubic meter and 6.9e10 atoms / m$^3$.

A nuclear reaction rate can be approximated by $$\text{Reaction Rate} = \Phi N\sigma,$$ where $\Phi$ is particle flux of the faster reactant, $N$ is the density of the slower reactant, and $\sigma$ is the cross section for interaction.

To calculate both flux and cross section for interaction we need the temperature of the plasma that we must turn the seawater into to initiate fusion. This is the relevant graph of cross sections for fusion of the various reactions based on center-of-mass kinetic energy of the plasma.

We can see for there that the reaction rate of D-T, the optimum temperature for cross section of fusion is 100 keV (about 1.1 billion kelvin) and at that rate the cross-section for fusion is 5 barns (1 barn is equal to 1e-28 m$^2$).

Flux is the mean speed of particles divided by density and can be calculated from thermal kinetic energy in three dimensions. We want the mean magnitude of velocity $v_{th}$, which we can get from $$v_{th} = \sqrt{\frac{8k_BT}{m\pi}}$$

where $m$ is the mass of the particle. $T$ is the temperature, and can be converted to Kelvin from electron-volts as ratio of the Boltzmann constant ($k_B$) as $$T_K = \frac{1.6\times10^{-19} \text{J/eV}}{k_B}T_{eV}$$ giving us $$v_{th} = \sqrt{\frac{8\cdot1.6\times10^{-19} \text{ J/eV} \cdot 100000 \text{ eV}}{2\mu\pi}} = 3.5\times10^6 \text{ m/s}$$ for deuterium where $\mu$ is one one atomic mass unit (1.66e-27 kg).

Deuterium flux is then $$3.5\times10^6 \text{ m/s}\cdot1.1\times10^{25} \text{atoms/m}^3 = 3.8\times10^{31} \text{ atoms / s}\cdot\text{m}^2.$$ We now have all the parts to calculate overall reaction rate: $$\text{Reaction Rate} = \left(3.8\times10^{31} \text{ 1 / s}\cdot\text{m}^2\right)\left(5\times10^{-28}\text{ m}^2\right)\left(6.9\times10^{10} \text{ 1 / m}^3\right)$$ which equals $1.3\times10^{15} \text{fusions / m}^3\cdot\text{s}$.

Now that we have got that tricky bit of math done, lets figure out the answer. The energy required to heat 5.5e4 mols of water to 100 keV is about 5.4e14 J. The energy produced by 1.3e15 fusions is 14.1 MeV per fusion or 2.9 kJ total, per second.

D-T fusion costs you 5e14 J to start and produces 3e3 J per second. So that is not worth it, even remotely. There just isn't enough tritium in the water

# D-D fusion

First, lets point something out. At the temperature for the above reaction, the cross section for reaction is much lower for D-D than D-T, but given that you have much, much more deuterium, you will get much more energy.

Using the above calculations, lets let cross section be 0.02 barns (from the chart). Deuterium flux is the same, while tritium density is replaced by deuterium density for a fusion rate of 8.4e26 fusions per second per cubic meter. D-D does two different reactions, into both tritium and helium-3. Since these each have a 50% chance of occurrence, we can average the two energies for an expected 2.73 MeV per fusion. This gives us 4e14 W of fusion output on 5e14 J of input. Almost return on investment! Of course, your plasma is so hot there is no way you would get even 1 second of output before that plasma expanded like...well...a thermonuclear bomb.

To figure out the best possible results from fusing a bunch of seawater we can plot the center-of-mass kinetic energy against output fusion power. Fusion output power can be calculated by combining the D-D fusion and D-T fusion, as calculated above.

The red line is total output, the blue line is D-D output, and the green line is D-T output. As you can see from the lack of a blue line, D-D ouptut dominates due to low availability of tritium. Generally you get more out the more you put in, up to about 3 MeV. However, it takes a lot more heat to get your plasma to 3 MeV in the first place. Let us instead plot energy input in J against energy output in J.

Here the black line is break even: 1 J out per second per 1 J in. Again, in reality, you would never be able to hold this plasma together for even a second. Even so, you don't really get back what you put in. The closest you come is right about a 100 keV, like we calculated at the start of this section.

# P-P fusion

The rates for most stellar nuclear reactions are inferred by extrapolating measurements at higher energies to stellar reaction energies. However, the rate for the fundamental p + p $\rightarrow$ $^2$D+e$^{+}$+$\nu_e$ reaction is too small to be measured in the laboratory. Instead, the cross section for the p-p reaction must be calculated from standard weak-interaction theory.

This explains why I haven't found any good graphs of cross-section versus temperature as I was able to find for D-T and D-D reactions.

Given that most of the sun's fusion is of the pp variety, and that the sun produces less energy per volume than the human body, we can assume that this interaction will produce a negligible amount of energy in return for the terajoules necessary to start it.

• So at the densities we see in seawater, hitting the broad side of a barn is actually difficult? =) Nice work, by the way! Dec 12, 2016 at 4:57
• Nice work. Do you know if quantum tunneling would have any effect, as it does in stars, to overcome the Coulomb barrier, or is that not applicable here? Dec 12, 2016 at 15:00
• @HDE226868 It does! Both references describe how the cross section must be calculated in two separate effects, the coulomb cross section, which varies by about a factor of 1e14 from 1eV to 1MeV, and the 'astrophysical S-function' which varies by about a factor of 20 from 1eV to 1MeV. My understanding of the reading is that the Coulomb barrier is more important to determine when fusion can or cannot happen, while the S-function dominates the 'rate' of the reaction at near-optimal fusion temperatures. Dec 12, 2016 at 15:05