Alan Smith... "ultrasound/sonic triggering to reduce the input energy cost"
Maybe Suhas , Mizuno and Egely can get together
Japanese Saying : Sannin yoreba Monju no Chie: Three men together and there is Monju's wisdom.
Robbie Burns: The well-laid plans of mice and men gang oft aglay
Just to add an extra check. Enthalpy of combustion of H2 is 280kJ/mol or 0.14MJ/g
So if (let us say) atmospheric O2 could get in to the reactor we have available chemical energy at approximately this level - I'm ignoring the fact that we have H here as Hydride, which would maybe increase the enthalpy a bit?
So for this output you'd need about 17g of H2 stored (in this case 34g of D2). At a loading of 1 that (whether H2 or D2) turns into roughly 900g of Ni - so we cannot possibly store anything like enough H or D in the 20g Ni mesh.
However the reactor is 20kg stainless steel. The average loading, given 17g of H (or 34g D) is about 5%.
This does not look impossible to me - but I'm no expert on the diffusion of hydrogen into stainless steel so cannot say.
The conditions here look vaguely plausible for this mechanism. High pressure for the loading. near vacuum for the unloading. You would need a slow O2 leak and some way to take the H2O out of the system (I guess a vacuum pump would do that).
There are a lot of unknowns to check here which could knock this down, but worth checking.
EDIT - not high pressure for loading, if 500Pa!
but worth checking.
probably checked already by Mizuno last year.
There are 3.4 mg of D2 in the reactor.
Enthalpy of combustion 0.07 Mj/g of D2
If Xs energy due to combustion of D2
3.4 mg x 0.07 MJ/g = 0.000238 MJ.
Xs energies observed are much larger
Socceroos versus Samurai Blues 0-0 at halftime, Saitama
Mizuno report
"
A test with this reactor lasts almost 30 days. Typical excess heat during the time is estimated as 300 W. Total energy is thus ~2.6×108 J. The amount of D2 used was 20 cm3 STP. Assuming that the reaction is D+D fusion, and assuming that all gases react, the amount of gas required to generate this much energy is approximately 12 cm3 STP. Although this is a very rough calculation, this value coincides with the amount of gas consumed. "
Display Morebut worth checking.
probably checked already by Mizuno last year.
There are 3.4 mg of D2 in the reactor.
Enthalpy of combustion 0.07 Mj/g of D2
If Xs energy due to combustion of D2
3.4 mg x 0.07 MJ/g = 0.000238 MJ.
Xs energies observed are much larger
Socceroos versus Samurai Blues 0-0 at halftime, Saitama
How is this 3.4mg of D2 computed? During the prep phase it does not seem clear to me how much D2 was added? In the report it refers to this quantity as being used during the test, but does not say whether that is just the heat production part, though this is implicit because it is comparing this usage with possible fusion enthalpy. In which case it leaves open how much was used in the prep part.
1 mole of H2 is 2.0158 grams.
Density = mass/volume.
In g/L it would be (2.0158)/(22.4) =0.089 g/L 0.089 mg/ml
D2 =twice = 0.178 mg/ml
20 ml = ~3.5 mg at STP give or take a few nanograms
at 65 minutes Samurai blues are 1 Socceroos 0 in Saitama
It is academic, because the IH replication failed. Still, I'd like to debug this. But not sure I have the motivation to spend long amts of time on it given the IH work.
Display More1 mole of H2 is 2.0158 grams.
Density = mass/volume.
In g/L it would be (2.0158)/(22.4) =0.089 g/L 0.089 mg/ml
D2 =twice = 0.178 mg/ml
20 ml = ~3.5 mg at STP give or take a few nanograms
at 65 minutes Samurai blues are 1 Socceroos 0 in Saitama
bocjin - I did the same calculations up above, you will note (though a different way round).
The issue at hand here is whether the amt of D2 during the prep phase was limited to 20ml.
You could take into account the pressure.
ballpark
760 pascals or so.. compared to atmospheric 100.000
then factor in the volume of the reactor chamber say 2.5 litres
760/100,000 x 2500 mls = 20???
JPn 2 Australia 0 out of 2018 world cup run
I didn't realize that tritium wants out more than H2.
Sorry if I was unclear, but I was speaking generically about hydrogen. There may be an isotope effect on leak rates but I don't know that for sure.
TH huxley ". But not sure I have the motivation to spend long amts of time"
Yosh. Ganbatte ne?
THHuxley."Still, I'd like to debug this"
Maybe you could start here?
"100 W × 71.46 ks = 71.46 MJ, "
As for me..time to sleep.
"100 W × 71.46 ks = 71.46 MJ, " === Error * Big time interval = great big error
Kirk.
There's another one. Found them in my first reading a while back
Preprint is being corrected.
You should see the booboos I made in my submission to a journal.
Its very hard to proofread your own stuff in another language.
Not sure what David Nagel was doing.
Now I really need to get some kip.
So if (let us say) atmospheric O2 could get in to the reactor we have available chemical energy at approximately this level
This is ruled out. The pressure is monitored and recorded in the spreadsheet. If enough O2 got in to cause significant heat, the pressure would rise.
... "he tried developing a 'gas cat' with ultrasound/sonic triggering ..."
AR in disguise:
https://patentscope.wipo.int/s…566/PDOC/WO2016026720.pdf
(Check "Ad Maiora LLC" at http://search.sunbiz.org/Inquiry/CorporationSearch/ByName and the (fake) inventor's address.)
This is ruled out. The pressure is monitored and recorded in the spreadsheet. If enough O2 got in to cause significant heat, the pressure would rise.
The real world is a bitch, and you learn not to make assumptions.
In this case we have:
2D2 [+ O2] -> 2D2O
so adding oxygen to make water actually keeps pressure the same, in theory.
Beyond that, since for significant heat we need D out from metal, the issue is what happens to the D2O? If it stays as a gas you are correct, but in a complex environment like this I would not want to bet on what it does chemically.
correct me please if i am wrong here but you have 3 mol of gas on the left side of your equation and 2 mol of gas/liquid on the other side. From my understanding the pressure should drop. If you have a Knallgas reaction occur of course thats another thing...
correct me please if i am wrong here but you have 3 mol of gas on the left side of your equation and 2 mol of gas/liquid on the other side. From my understanding the pressure should drop. If you have a Knallgas reaction occur of course thats another thing...
You are wrong. The suggestion is ingress of O2 from outside, so one of the mol on the LHS does not count. That is why I put it in brackets...
But only if your water is evaporated...
