This is in response to the following comment.
(Ships carry fuel only for the power plant and J-drive, not M-drive fuel. Power plant fuel is consumed at the same rate regardless of whether the M-drive is in use or not. Therefore, the M-drive does not consume fuel or any other visible reaction mass.)
The following material is taken from
The Effects of Nuclear Weapons, Dept. of Army Pamphlet 50-3, March 1977. Said document is an official US government publication and is therefore in the public domain and copyright free.
The complete fission of 0.057 kg. (57 grams or 2 ounces of fissionable material is equivalent to 1 kiloton of TNT. Equivalents to this are:
10 to the 12th Calories
2.6 X 10 to the 25th Million Electron Volts
4.18 X 10 to the 19th Ergs
4.18 X 10 to the 12th Joules
1.16 X 10 to the 6th Kilowatt-hours
3.97 X 10 to the 9th British Thermal Units
Weight for weight, therefore, the fusion of deuterium nuclei would produce nearly three times as much energy as the fission of uranium or plutonium.
I understand that fuel for power plants in Traveller is Liquid Hydrogen, not Liquid Deuterium, but all that is going to do is increase your energy yield per unit mass from the fusion of hydrogen into deuterium. However, for the purpose of the following discussion, I will use the deuterium energy value.
A standard 200 ton Free Trader uses 10 metric tons or 10,000 kilograms of liquid hydrogen for its power plant every 28 days. That means an average fuel consumption of 14.88 kilograms per hour, or 14,880 grams per hour. Now if you assume that 57 grams of liquid H will yield 3 X 10 to the 6th kilowatt-hours (that assumes your fusion reaction is equal to 0.86 that of fission, the actual figure is 0.9475, so this is quite conservative), or 3 X 10 to the 3rd Megawatt-hours, or 3 Gigawatt-hour of power, the the fusion yield of the power plant is going to be 783 Gigawatts of power per hour on the average. If you are not using that energy, which is quite a lot, for maneuvering your Free Trader, what is the ship using it for?
In
Thrust into Space, by Maxwell Hunter, he calculates that it takes 4 kilowatt-hours of power to place 1 pound of mass into Earth orbit. That equates to 1.764 Gigawatt-hours to put a 200 metric ton mass into Earth Orbit. Your power plant is producing, on the average, 443.9 times the power needed to place a 200 metric ton Free Trader into Earth Orbit.
Now, if you are getting ready to jump, then your power plant is going to use 20 metric tons, or 20,000 kilograms of fuel in a very short time. Twenty metric tons of liquid hydrogen will supply 1,052,631 Gigawatt-hours of power. To put this in context, the US in 2008 used 4,401,698 Gigawatt-hours of power. You also have 20 metric tons, essentially, of extremely hot Helium plasma to get rid of in some manner.
The MegaTraveller Imperial Encyclopedia, on 81, lists the power plant output for a 200 ton Free Trader at 990 Megawatt-hours. Based on the data from
Thrust into Space, mentioned earlier, that would be sufficient to place a 200 ton Free Trader into Earth orbit in 2 hours. However, based on the fuel consumption of 14.88 kilograms per hour, the power plant is using only about 0.126% of the potential energy of the fuel if complete fusion takes place. That does leaves you with 14.88 kilograms of extremely hot Helium plasma to dispose of in some manner. If you actually need Terawatts of power to make a jump, that less than a Gigawatt power plant is going to take a long time to generate that much power.
If you are sitting on a planet, with a D or E class starport, which I would assume that a Free Trader would do on a regular basis, for what exactly are you using those 990 Megawatts, going with the MegaTraveller reference? A 729 Gross Register Ton former US Coast Guard buoy tender, which would equate to 153 displacement tons in Traveller uses two 170 Kilowatt ship service generators for all of its power needs, to include A/C, full ship lighting (using incandescent lights for the most part), electronics and computers, galley, etc. It actually only needs one, but has to have two for a backup. A 1945 aircraft carrier was requiring somewhere between 5 and 10 Megawatts of power for full operation. That would be roughly a 7500 ton ship under Traveller's displacement ton rules.
Again, if the maneuver drive is not consuming power from the power plant, where is all of that power going?