[I never got my first "Fusion rocket science" message back.]
Someone who is much better at arithmetic than I am should please compare Table 1 in this paper (http://gltrs.grc.nasa.gov/reports/1996/TM-107030.pdf) to the following calculations.
I assume "Cat-DD" means catalyzed DD fusion, and I assume the Energy release and Converted mass fraction would be the same for this as for cold fusion. On that basis I reckon:
Deuterium fusion yields 3.45E14 J/kg of fuel. In other words, pure deuterium gas yields 345 million MJ per kilogram. Gasoline has 45 MJ/kg (or 132 MJ/gallon), so a kilogram of deuterium has roughly as much energy as 7.6 million kilograms of gasoline (2.6 million gallons).
One mole of heavy water consists of 16 g of oxygen and 4 g of deuterium, so deuterium gas has five times more energy per kilogram than heavy water. One kilogram of heavy water produces 69 million MJ (523,000 gallons of gasoline).
One kilogram of ordinary water contains 0.015 at% deuterium, or 1 deuterium atom for 6700 hydrogen atoms. (Some sources say 1 in 5400.) When fused the deuterium in ordinary water yields 13,000 MJ (98 gallons of gasoline).
The entire world consumes 403 quads. A quad is a quadrillion Btu; 10E15 Btu = 1.055 Exajoules, or 1.1E12 MJ. 403 quads equals 4.3E14 MJ. If all the energy in the world came from cold fusion (or plasma fusion), it would consume 6,162 tons of heavy water.
. . .
The 6,162 tons of heavy water we would use for worldwide energy production would be converted into 4,930 tons of free oxygen, 1,227 tons of helium, and 5 tons of the mass would be annihilated, converting into energy, according to Einstein's special relativity formula E=mc^2.
- Jed
