Corrected Data on Fission verses Fusion Energy Yield
The following information on energy yield of fission and fusion reactions is taken from
The Effect of Nuclear Weapons, 1977 Edition, edited by Samuel Glasstone et al., and can be located here:
http://www.abomb1.org/nukeffct/index.html
It is:
This is the definitive, unclassified text on nuclear weapons effects.
I am more familiar with it as DA Pam 50-3.
Complete fission of 0.057 kg (57 grams or 2 ounces) fissionable material
Fission of 1.45 x 10(exp25) nuclei
10(exp12) calories
2.6 x 10(exp25) million electron volts
4.48 x 10(exp19) ergs (4.18 x 10(exp12) joules)
1.16 x 10(exp6) kilowatt-hours
3.97 x 10(exp6) British thermal units
From this, the complete fissioning of 2 ounces of U-3=235 (57 grams) yields 1.16 Million Kilowatt hours of electricity.
In the following quote, D stands for Deuterium, T stands for Tritium, He is Helium, either 3 or 4, and H is straight Hydrogen.
1.69 Four thermonuclear fusion reactions appear to be of interest for the production of energy because they are expected to occur sufficiently rapidly at realizable temperatures; these are:
D + D = He-3 + n + 3.2 MeV
D + D = T + H + 4.0 MeV
T + D = He-4 + n + 17.6 MeV
T + T = He-4 + 2n + 11.3 MeV,
where He is the symbol for helium and n (mass = 1) represents a neutron. The energy liberated in each case is given in million electron volt (MeV) units. The first two of these reactions occur with almost equal probability at the temperatures associated with nuclear explosions (several tens of million degrees Kelvin), whereas the third reaction has a much higher probability and the fourth a much lower probability. Thus, a valid comparison of the energy released in fusion reactions with that produced in fission can be made by noting that, as a result of the first three reactions given above, five deuterium nuclei, with a total mass of 10 units, will liberate 24.8 MeV upon fusion. On the other hand, in the fission process, e.g., of uranium-235, a mass of 235 units will produce a total of about 200 MeV of energy (¤ 1.43). Weight for weight, therefore, the fusion of deuterium nuclei would produce nearly three times as much energy as the fission of uranium or plutonium.
In my earlier post, I did not include the D + T reaction, but because it is highly likely to occur, I should. On a weight for weight basis, 235 units of Deuterium will yield 582.8 MeV of energy, compared to either my earlier figure of 215 MeV for U-235 fission, or Glasstone's 200 MeV for U-235 fission. Using Glasstone's figure, as the more conservative, that would give an equal weight of Deuterium 2.914 times the energy of U-235. This would mean that the complete fusion of 2 ounces or 57 grams of Deuterium would 3.38 Million Kilowatt-hours of electricity.
Therefore, the complete fusion of One pound or 453.6 grams of Deuterium would yeild 27.04 Million Kilowatt-hours of electricity. That would give 3,086.75 Kilowatts of power per hour of a 365 day year with 24 hours per day. The complete fusion of 2.2046 pounds of Deuterium or 1 Kilogram, will yield 6,805 kilowatts of electricity per hour. Remember that is the fusion yield for a year of 2.2046 pounds/1 kilogram of Deuterium. I suspect that one kilogram of Deuterium per year would supply the basic ship power for a Scout Ship, or a much larger ship for that matter.
Based on all of this, I would say that providing 1 Traveller dTon for Liquid Deuterium would be more than adequate for supplying the base power of a small star ship for one year. As a Traveller dTon of Liquid Deuterium would weight 2192.4 kilograms, more than twice the weight of Liquid Hydrogen, that would give an ample fuel reserve. For my small-ship universe, I think I will go with 1 dTon of Liquid D times hull size divided by 1,000 rounded up, with a minimum of 1 dTon. That does not include fuel for Maneuvering however, just the typical ship power load. I will have to work on the fuel for Maneuvering next.