On Thu, Mar 14, 2013 at 4:21 PM, James Bowery <jabow...@gmail.com> wrote:
> On Thu, Mar 14, 2013 at 10:49 AM, Jones Beene <jone...@pacbell.net> wrote:
>
>> The chlorine-hydrogen photoactivated reaction is the only chemical
>> reaction which is known to produce nuclear reactions (when deuterium is
>> used in place of hydrogen). Neutrons are “stripped” from the deuterium in
>> that case.
>>
>
> Normal water is 0.02% D2O, so can't we expect:
>
> 2014101.77812 + 15994914.61957 => 16999131.75650 + 1007825.03223 +
> 2059.609uamu energy
> D + O16 => O17 + H + 2059.609uamu energy
>
> in appropriately dilute amounts?
>
Still awaiting the cite from Jones about H+Cl => HCl producing neutrons.
Meanwhile, here's a wrap-up of the arithmetic for this explanation for the
Papp engine's energy source
Starting with the energy of a molar reaction:
(6.0221415e+23*2059.609udalton*c^2)?J
(6.0221415E23 * [2059.609 * {micro*dalton}]) * (speed_of_light^2) ? joule
= 1.8510799E11 J
Now we take 100hp as the Papp engine output and as how moles per hour of
deuterium it would consume:
(6.0221415e+23*2059.609udalton*c^2)/mole;100hp?mole/hour
([{6.0221415E23 * (2059.609 * [micro*dalton])} * {speed_of_light^2}] /
mole)^-1* (100 * horsepower) ? mole / hour
= 0.0014502439 mole/hour
We take that and convert that to grams of deuterium:
0.0014502439 mole;2g/mole?g
(0.0014502439 * mole) * ([2 * gramm] / mole) ? gramm
= 0.0029004878 g
And since we know that deuterium is 0.0156% of hydrogen and hydrogen is 1/8
the mass of water we can ask how much water per hour is consumed per hour
by the Papp engine running at 100hp:
8*0.0029004878 g/0.000156;1kg/l?l
([8 * {0.0029004878 * gramm}] / 0.000156) * ([1 * {kilo*gramm}] / liter)^-1
? liter
= 0.14874296 l
or about a half cup of water per hour.
Of course, the power level would decrease as the deuterium is burned up and
is therefore more dilute.
