Timerover51
SOC-14 5K
Can you give a rationalle for that assumption? At this point we are just beginning to see a break-even point for fusion power. How would we know how much fuel a power plant would need 'in reality?'
The following comments are based on information from this website at the University of Pennsylvania:
http://www.sas.upenn.edu/~kennethp/nuclearenergy.pdf
The complete fission of one pound of U-235 results in an energy equivalent to one million gallons of gasoline. One pound of gasoline is rated at about 18,000 BTUs of energy, and one kilowatt is equal to 3412 BTU. So, figure a pound of gasoline is good for 5 kilowatts of power. A gallon of gasoline weighs 6 pounds, so a gallon of gasoline will yield 30 kilowatts of power if burned with 100% efficiency. Therefore, the fission of one pound of U-235, at 100% efficiency, will yield 30 million kilowatts of power, or 3424 kilowatts per hour, assuming a 24 hour and 365 day years. Currently, assuming a high-efficiency reactor, you might get about 25% or so efficiency at energy conversion, so figure that you would need to burn 4 pounds or so of U-235 per year to get that power.
The following website gives some fusion reaction yields and compares fusion and fission.
http://hyperphysics.phy-astr.gsu.edu/Hbase/nucene/fusion.html
The Deuterium-Tritium fusion reaction is rated at an energy yield of 17.6 MeV per fusion, while U-235 is rated at 215 MeV per fission. However, on an equal weight basis, as there are a lot more fusion reactions per weight, the fusion reaction produces 3.84 times the power by weight of reactants. So one pound of equal parts Deuterium and Tritium completely reacted will yield, at a little more than 25% efficiency, the same amount of power as 4 pounds of U-235.
However, the Tritium in the above reaction is not really found naturally, and would have to be produced in some manner. The Deuterium-Deuterium fusion reaction has two possible outcomes, one yielding 3.27 MeV and one 4.03 MeV per fusion. They average out to 7.3 MeV for every 2 fusion reactions of a pair of Deuterium atoms. That is quite a bit less energy from about 8 Atomic Mass Units, compared to 5 AMU of the Deuterium-Tritium reaction, but Deuterium is a naturally occurring stable isotope of Hydrogen, and simply has to be separated out. Dividing U-235 mass by 8, you get 29.375. Multiplying by the average energy yield of the fusion of 4 Deuterium atoms of 7.3 MeV, you get 214.4 MeV for the same weight of Deuterium as one atom of U-235. Therefore, for every pound of Deuterium completely reacted in fusion, you get just about the same energy as a pound of U-235.
Now, to avoid one horrendous heat exchange problem, the fusion plants in Traveller have to be fantastically efficient at converting fusion energy to electricity, so one pound of Deuterium per year would give about 3400 kilowatts of power for every hour of a year. One kilogram of Deuterium will give, at 2.0246 pounds per kilogram, 7,495 kilowatts of power per hour for an entire year. I suspect that much power would be more than adequate for the basic power needs of a starship in Traveller, unless it is a REALLY, REALLY BIG ONE. For your average Scout Ship, or say a cargo/passenger ship of under 1000 tons or so, that should be far more than needed.
Now, you could require 6.5 tons of Liquid Hydrogen per year for a ship, as Deuterium exists in the ratio of 1 Deuterium atom per 6,420 hydrogen atoms, so 6,420 kilograms of Hydrogen will contain 1 kilogram of Deuterium for my Deuterium to Deuterium fusion reactor. That is if you want to extract the Deuterium onboard. Otherwise, a ton of Liquid Dee should prove more than enough for a year or so for my power plant.
Mental note to Poster: Based on the above fission yields, give serious thought to small fission plants for long-endurance vehicles. Maybe HBP direct conversion of nuclear to electricity units? Need to ponder.
Edit Note 2: The Effects of Nuclear Weapons, 1977 edition, gives the yield of 2 ounces of completely fissioned U-235 as 1.16 Million Kilowatt hours, so I will need to adjust the above figures by a factor of 3. Therefore a pound of U-235 will yield 9.28 Million Kilowatt hours, rather than 30 Million Kilowatt hours. Will correct on Jan. 31, 2015. This does not significantly change the result.
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