During this time, he conducted research at the National Spherical Torus Experiment (NSTX), the DIIID tokamak, and the LArge Plasma Device (LAPD) facilities. In January 2003 he joined the Electrical and Computer Engineering and Physics and Astronomy Departments at the University of New Mexico as an Assistant Professor. He currently has research projects in basic plasma physics (turbulence and transport in laboratory plasmas, as well as active feedback control of turbulent transport, laboratory astrophysics, and plasma sources), plasma diagnostics, magnetoinertial fusion (MIF), pulsed power technology, and space weather
http://www.ece.unm.edu/~gilmore/gilmore_bio_cv.html
Chapter 6
Seems Davis's advisor Mark Gilmore has a good background in this sort of thing:
I like the pB11 resonance peak at 50 KV acceleration. In2 years we'll know.
Fuel mass for 100 MW gross reactor
I wanted to double check the massperhour needed for a 100 MW reactor. Here's what I did:
Instead of calculating from the energy per reaction, I found the energytomass conversion ratio per reaction. With that ratio and E=mc^2, I found an answer.
If you don't want to follow through my calculations, the conclusion is that you'll need 5.16 grams of boronhydrogen fuel per hour for a reactor of 100 MW gross power output.
Actual power output would, of course, be less; or, actual fuel use would be more, depending on how you want to spin the numbers.
This number seems to be closer to D.Tibbit's calculations than they are to MSimon's....
Very likely, I'm missing something from MSimon's conclusions, which seem to be 4 times my numbers. Perhaps he's calculating for 100MW of *net* power?
MSimon, would you be willing to explain?
~~~~~~~~~~~
My Calculations:
The amount of mass released as energy per reaction can be found by subtracting the mass of the resulting ash from that of the original fuel :
(one boron11 atom + one light hydrogen atom)  (three helium atoms) = total mass released as energy
(11.0093054 amu + 1.0078250 amu )(3 * 4.0026032 amu) = 0.009320669 amu [Masses found at wikipedia's isotope entries for each element.]
Next, find the ratio of mass released in one reaction to the mass of the fuel for that reaction:
0.009320669 amu / (11.0093054 amu + 1.0078250 amu) = 0.0007756152 [This is a ratio and is a unitless number, equivalent to 0.0775% or 775 parts per million.]
This is the masstoenergy conversion ratio for this reaction.
In other words, if you put 1kg of mass into your reactor, you'll get a gross total energy output equivalent to 0.7756152 grams.
Since E=mc^2 and c=299,792,458 m/s, converting to joules is easy:
0.0007756152 kg * (299792458 m/s)^2 = 6.970882e13 Joule
And since 1 joule = 2.777778e7 kWh (*):
6.970882e13 joule * 2.777778e7 kWh/joule = 19,363,561 kWh
So, you should get about 19.4 GWh (gross) out of 1kg of boronhydrogen fuel.
Conversion efficiencies and energy overhead would then be applied to get actual output of the plant.
If 1kg of fuel puts out 19.363 GWh, then 1g of fuel puts out 19.363 MWh
100 MWh / 19.363 MWh/g = 5.16 grams
So, you'll need 5.16 grams of boronhydrogen fuel per hour for a 100 MW (gross) reactor.
You can also run the calculation backwards:
You want a 100MW reactor; IE, 100e6 joules per second. That's 360e9 Joules per hour.
Convert this to mass equivalents then divide by our reaction's mass ratio:
(360e9 / 299792458^2) / 0.0007756152 = 5.164339e03 kg/h = 5.16 g/h
Can anyone find a mistake?
~~~~~~~~~
(*)
1 kilowatt = 1000 joule/sec
...therefore:
1 kWh = 1000 joules/sec * 3600 sec/hour = 3,600,000 joules/hour
...therefore:
1 joule = 1 / 3,600,000 kWh = 2.777778e7 kWh
~~~~~~~~~
Instead of calculating from the energy per reaction, I found the energytomass conversion ratio per reaction. With that ratio and E=mc^2, I found an answer.
If you don't want to follow through my calculations, the conclusion is that you'll need 5.16 grams of boronhydrogen fuel per hour for a reactor of 100 MW gross power output.
Actual power output would, of course, be less; or, actual fuel use would be more, depending on how you want to spin the numbers.
This number seems to be closer to D.Tibbit's calculations than they are to MSimon's....
Very likely, I'm missing something from MSimon's conclusions, which seem to be 4 times my numbers. Perhaps he's calculating for 100MW of *net* power?
MSimon, would you be willing to explain?
~~~~~~~~~~~
My Calculations:
The amount of mass released as energy per reaction can be found by subtracting the mass of the resulting ash from that of the original fuel :
(one boron11 atom + one light hydrogen atom)  (three helium atoms) = total mass released as energy
(11.0093054 amu + 1.0078250 amu )(3 * 4.0026032 amu) = 0.009320669 amu [Masses found at wikipedia's isotope entries for each element.]
Next, find the ratio of mass released in one reaction to the mass of the fuel for that reaction:
0.009320669 amu / (11.0093054 amu + 1.0078250 amu) = 0.0007756152 [This is a ratio and is a unitless number, equivalent to 0.0775% or 775 parts per million.]
This is the masstoenergy conversion ratio for this reaction.
In other words, if you put 1kg of mass into your reactor, you'll get a gross total energy output equivalent to 0.7756152 grams.
Since E=mc^2 and c=299,792,458 m/s, converting to joules is easy:
0.0007756152 kg * (299792458 m/s)^2 = 6.970882e13 Joule
And since 1 joule = 2.777778e7 kWh (*):
6.970882e13 joule * 2.777778e7 kWh/joule = 19,363,561 kWh
So, you should get about 19.4 GWh (gross) out of 1kg of boronhydrogen fuel.
Conversion efficiencies and energy overhead would then be applied to get actual output of the plant.
If 1kg of fuel puts out 19.363 GWh, then 1g of fuel puts out 19.363 MWh
100 MWh / 19.363 MWh/g = 5.16 grams
So, you'll need 5.16 grams of boronhydrogen fuel per hour for a 100 MW (gross) reactor.
You can also run the calculation backwards:
You want a 100MW reactor; IE, 100e6 joules per second. That's 360e9 Joules per hour.
Convert this to mass equivalents then divide by our reaction's mass ratio:
(360e9 / 299792458^2) / 0.0007756152 = 5.164339e03 kg/h = 5.16 g/h
Can anyone find a mistake?
~~~~~~~~~
(*)
1 kilowatt = 1000 joule/sec
...therefore:
1 kWh = 1000 joules/sec * 3600 sec/hour = 3,600,000 joules/hour
...therefore:
1 joule = 1 / 3,600,000 kWh = 2.777778e7 kWh
~~~~~~~~~
zDarby,
Nice calculation.
you may have missed this, it was earlier in the thread
Nice calculation.
you may have missed this, it was earlier in the thread
MSimon wrote:I believe the .5 kg / day was net/net (I also had a .230Kg per day number floating in my head so who knows?). In any case we are now in the correct ball park so I have no further nits to pick. On that topic.
Near term, cheap, dark horse fusion hits the air waves, GF  TED, LM  Announcement. The race is on.

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Better store the borax in several different places, just in case an incoming shell hits your fuel bunker. It'd be bad if the ship was stranded for the lack of a cup of borax. Maybe hide several tonnes of it around, just in case.paperburn1 wrote:We still have not solved the problem of where to store the borax.....zDarby wrote:You're quite right: I did miss that.
Thank you, mvanwink5.
Aero
You need at least three nines B11. Maybe five nines.dweigert wrote:Wouldn't it just be easier to pull it out of the seawater itself? Kinda sucks for a fresh water navy, but you should be able to find room on board a naval vessel to put a boron recovery plant and materials needed in the steps to refine it.
Engineering is the art of making what you want from what you can get at a profit.