What is the energy content of Earth’s Water for DD (DeuteriumDeuterium) fusion?
A guest article by Joseph Friedlander for Brian Wang’s Next Big Future
What is the energy content of Earth’s water for DD (DeuteriumDeuterium) fusion?
 Each cubic meter of seawater contains about 1.028 x 10e25 atoms of deuterium with a mass of around 34.4 grams is equivalent to the heat combustion of 269 metric tons of coal, or of 1,360 barrels of crude oil.
 The world resource of seawater’s deuterium is around a billion times the fossil fuel reserve.
 Each cubic meter of seawater has a deuterium energy content equivalent to a 1.8=2.8 kiloton bomb.
 269 metric tons of coal per cubic meter of seawater’s deuterium (34.4 grams by Hubbert) gives a weight equivalency of 7819767 times the deuterium’s weight in coal. Or 269 times the water’s weight in coal.
 If we could use dilute deuterium (we can’t) in DD fusion reactors (there aren’t any) at $80 coal equivalent fuel value for a cubic meter of seawater (269 metric tons coal equivalent) would be $21520. But no one would pay coal rates for cheap fusion—prices would go down and demand explode. (I hope not literally:>))

Plainly energy would be much, much cheaper, I have calculated about 1900 times cheaper here–

(Quick summary: at $10 million a 1 megaton device, the energy generated is as expensive as coal.
IF the device gets cheaper or larger the price goes down accordingly)
If you like calculating for nonexplosive fusion reactors, a gigawatt year thermal is about equivalent to 7.5 megatons TNT. So that’s equivalent to $75 million for a one year lifespan gigawatt thermal reactor—as expensive as coal. Make it bigger or cheaper and the price goes down. Smaller or more expensive and it can’t compete…And if the lifespan of the reactor increases, the price goes down, but only to a limit because of the time value of money (discounting).
=============Sources and derivations=================
In a book carried on Google Books
Nuclear. Power and The Public. A. Symposium, Minneapolis, Oct. 1969 Harry Foreman et al
And in echo in the
Congressional report “GROWTH AND ITS IMPLICATIONS
FOR THE FUTURE”
by the United States Congress Subcommittee on Fisheries and Wildlife, Conservation and the Environment of the Committee on Merchant Marine and Fisheries, Ninetythird Congress
In both cases the statistics appear to be based on calculations generated by M. King Hubbert
(Yes, THAT Hubbert, who predicted the US peak oil production domestically—in 1956—for about the time these documents were published in 196970)
which basically pointed out that deuterium reserves of Earth were around a billion times then known fossil fuel reserves. (Because of the scarcity of lithium6, lithium deuterium fusion basically was only equal to fossil fuel reserves, this explains why the billion times more abundant deuterium is such a big deal, and why I write frequently on DD fusion. In that era,
Hubbert (and Caltech geochemist Harrison Brown, who wrote The Challenge of Man’s Future (Viking, New York, 1954) warned that on a closed and finite Earth, the graph of oil consumption over time (considered on a scale of thousands of years must appear like a single sharp vertical spike on a nearflat graph). Brown considered the Earth a finite and closed system and considered nuclear power (and thermonuclear) a necessity if industrial civilization was not to collapse over millennial timescales.
On page 205 of the Harry Foreman book, M. King Hubbert writes
“…whereas the deuterium in the oceans 1.5 x 10e 48 atoms, or about a billion times the lithium6 supply.
The energy obtainable from the lithiumdeuterium reaction based on the total supply of lithium6 would be about 2.4 x 10e23 joules approximately equal to that of the (1970 estimate—JF) world’s supply of the fossil fuels. The energy obtainable from a deuteriumdeuterium fusion would be of the order of a billion times larger.
For a more tangible future, 1 cubic meter of seawater contains about (~1.028—JF) 1.03 x 10e25 atoms of deuterium having a mass of 34.4 grams, and a potential fusion energy of 8.16 x 10e12 joules.
Ths is equivalent to the heat combustion of 269 metric tons of coal, or of 1,360
barrels of crude oil. The deuterium in 28 cubic kilometers of seawater would have an energy equivalent to the worlds coal supply.”
Friedlander here– In this book Hubbert estimates that fossil fuels will be only able to supply the major part of the world’s power for 3 centuries. He dismisses water power because of silting of reservoirs, geothermal and tidal power as being vast but of small tappable magnitude. Hubbert regards nuclear power as the sole source of sufficient magnitude to supply Man for the long haul.
It is interesting that those who quote Hubbert on the need to conserve oil never quote his nuclear conclusions—and also interesting that at the exact time he was writing this Gerard K. O’Neill was calculating that space solar power alone could supply man with no need for nuclear power. So this is an illustration that powerful intellects can come to utterly different conclusions based upon their starting assumptions.
Now the Congressional report “GROWTH AND ITS IMPLICATIONS
FOR THE FUTURE”
Has this paragraph (apparently starting with the Hubbert stats as a starting point) —“Since a cubic kilometer contains 10e9 cubic meters, if follows that the fuel equivalents of 1 cubic kilometer of sea water are 269 billion tons of coal, or 1,360 billion barrels of crude oil. The latter figure is approximately equal to the lower of the two estimates of ultimate world resources of crude oil.
Since the ultimate world coal resources as estimated by Averitt are about
7,600 X 10e9 metric tons, the volume of sea water required to be equivalent
to this would be about 28 cubic kilometers. The total volume of the oceans
is about 1.5 billion cubic kilometers. Should enough deuterium be withdrawn
to reduce the initial concentration by 1 percent, the energy released by
fusion would amount to about 500,000 times that of the world’s initial
supply of fossil fuels.
From Carey Sublette’s Nuclear Weapons Faq http://nuclearweaponarchive.org/Nwfaq/Nfaq0.html
Fusion of pure deuterium: 82.2 kt/kg
1 kt (equivalent –JF)
4.19x10e12 joules
(a joule is a watt second –JF)
so 344.418 x10e12 joules ( per kilogram of DD fusion—JF)
34.4 grams, (after Hubbert) of DD per cubic meter of sea water
1 in 6400 hydrogen atoms is deuterium
Ordinary water is 88.8% oxygen to about 11.2% hydrogen by mass
Heavy water is typically around 20% deuterium by mass (if pure)
World Water supply around 1.37 billion cubic kilometers –calculating for pure water
1.37 billion billion cubic meters (tons) 153,440 grams of ordinary hydrogen (at 11.2% weight ) per cubic meter water
34.4 grams deuterium so 1/4460 by weight. This is .7 of the 1/6400 value so there may be some disconnect between the figures.
Wikipedia says–
Between 3200 and 4460 is a ratio of 1.4375 or .695. I will note this and neglect it, moving on with the article. Possibly Hubbert’s figures are wrong, but his point is still valid.
Assuming Hubbert’s figures are right and applying Sublette’s calculation of: 82.2 kt/kg
82.2 tons TNT equivalent per gram Deuterium
So the equivalent in TNT of each cubic meter of seawater’s deuterium is 2827.68 tons TNT or a 2.8 kiloton bomb!
According however to the classic 7:1 rule (7 kilotons of TNT equals a kiloton of coal)
the equivalent yield would be 269 tons of coal x 7 or about 1.8 kilotons (1883 tons TNT)
The discrepancy may be due to errors, the fact that coal is a variable quality resource and so on. But the magnitude of energy is clear.
269 metric tons of coal per cubic meter of water’s deuterium (34.4 grams by Hubbert) gives a weight equivalency of 7819767 times the deuterium’s weight in coal. Or 269 times the seawater’s weight in coal.
======================
Convenient Energy Content Approximations
Fission of U233: 17.8 kt/kg
Fission of U235: 17.6 kt/kg
Fission of Pu239: 17.3 kt/kg
Fusion of pure deuterium: 82.2 kt/kg
Fusion of tritium and deuterium (50/50): 80.4 kt/kg
Fusion of lithium6 deuteride: 64.0 kt/kg
Fusion of lithium7 deuteride ?
Total conversion of matter to energy: 21.47 Mt/kg
Fission of 1.11 g U235: 1 megawattday (thermal)
1 kt (equivalent –JF)
10^12 calories
4.19×10^12 joules
4.19×10^19 ergs
2.62×10^31 eV
2.62×10^25 MeV
fission of 0.241 moles of material (1.45×10^23 nuclei)
fission of approx. 57 g of material
Deuterium produces energy through four reactions:
D + T > He4 + n + 17.588 MeV
D + D > He3 + n + 3.268 MeV
D + D > T + p + 4.03 MeV
He3 + D > He4 + p + 18.34 MeV
If only reactions 13 contribute significantly, corresponding to the combustion of 25% of the deuterium fuel or less, then the energy output is 57 kT/kg. If reaction 4 contributes to the maximum extent, the output is 82.4 kT/kg. The maximum temperature generated by an efficient burn reaches 350 million K.
The net effect of reactions 14 together is:
6 D > 2 He4 + 2p + 2n + 43.243 MeV
Of the two neutrons produced, one is high energy (14.07 MeV) and one is moderate energy (2.450). The ratio of high energy neutrons produced to deuterons consumed (or energy produced) is significant for driving fast fission reactions. If the He3 is converted to tritium instead of being burned directly, the net reaction is the same with the exception that two high energy neutrons are produced.
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