My Physics is a bit rusty, but bertil's calcuations seemed a bit wrong to me:
> The book gives the energy gained from fusion of one atom of deuterium
> (including all byproducts etc) to 13E-13 Joules.
O.K. I don't know any different here.
> The occurance of deuterium in hydrogene it states as 15 to 100,000.
> (Deuterium is a hydrogene isotope).
Fine
> One gram (1E-3 kilos, 1E-6 tons) of hydrogene contains 6.023E23 hydrogene
> atoms, and will thus yield 11E9 Jouls.
Here is where I have problems:
Via Avogradro's number one mole contains 6.02217E23 atoms (minor correction)
One mole of hydrogen weighs 1.008 grams (again minor correction)
1 gram contains 5.9742E23 hydrogen atoms
at 15:100,000 this is 8.9613E19 deuterium atoms
at 13E-13 joules each this is 1.1656E8 Joules
> One ton of Tech15 plant uses 0.009 tons of H2 per hour.
> This translates to about 99E12 MWatts per per kl of plant per hour or, in
> other words, 99,000,000 MW per kl of plant...
0.009 tonnes is 9000 grams, at 1.1656E8 Joules each...
now we have 1.049E12 Joules per kl over 1 hour, one Watt is 1 Joule/second
divide by 3600 seconds in an hour, gives 2.9140E8 Watts
which is 291.4 MW of output per kl of plant... (assuming 100% efficiency,and
non-enriched fuel).
> (I hope I havn't done any errors too big to explain away:-)
> -bertil-
I hope I have not corrected you in error...
-Dan
Dan Corrin, System Manager, Mechanical Engineering, UWO, London, Ontario
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