On Feb 24, 2011, at 12:19 PM, mix...@bigpond.com wrote:
In reply to Horace Heffner's message of Thu, 24 Feb 2011 11:39:36
-0900:
Hi,
[snip]
..we also don't know how much of the H remained in the Ni after the
reaction was
finished.
Yes, very true. The 25.4 keV is a *minimum* energy per hy
In reply to Peter Gluck's message of Thu, 24 Feb 2011 22:48:52 +0200:
Hi,
[snip]
>Robin,
>I don't understand- excuse where is the pressure of hydrogen measured? It is
>adsorbed absorbed in the nanometric nickel, the temperature increases there
>up to say 400 C- I don't think the reactor has a mano
In reply to Horace Heffner's message of Thu, 24 Feb 2011 11:39:36 -0900:
Hi,
[snip]
>> ..we also don't know how much of the H remained in the Ni after the
>> reaction was
>> finished.
>
>Yes, very true. The 25.4 keV is a *minimum* energy per hydrogen
>atom. However, if 30% of the Ni was conv
Robin,
I don't understand- excuse where is the pressure of hydrogen measured? It is
adsorbed absorbed in the nanometric nickel, the temperature increases there
up to say 400 C- I don't think the reactor has a manometer on it.
Peter
On Thu, Feb 24, 2011 at 10:39 PM, Horace Heffner wrote:
>
> On Fe
On Feb 23, 2011, at 5:47 PM, mix...@bigpond.com wrote:
In reply to Horace Heffner's message of Tue, 22 Feb 2011 13:35:03
-0900:
Hi,
[snip]
This 270kWh per 0.4 g if hydrogen is obviously well beyond chemical
if the consumables actually are H and Ni. The energy E per H is:
E = (270kwh)
In reply to Horace Heffner's message of Tue, 22 Feb 2011 13:35:03 -0900:
Hi,
[snip]
>This 270kWh per 0.4 g if hydrogen is obviously well beyond chemical
>if the consumables actually are H and Ni. The energy E per H is:
>
>E = (270kwh) /(0.4 g * Na / (1.00797 gm/mol)) = 2.54x10^4 eV / H
>
On Feb 22, 2011, at 2:11 PM, Jed Rothwell wrote:
Horace Heffner wrote:
The above chart is merely a very approximate visual aid to show
feasible reaction product probabilities by a rule of thumb
estimate. Copper is visualized as a most likely product.
Izzatso? So you think the reports of
Horace Heffner wrote:
The above chart is merely a very approximate visual aid to show
feasible reaction product probabilities by a rule of thumb estimate.
Copper is visualized as a most likely product.
Izzatso? So you think the reports of copper can be explained by your theory?
- Jed
On Feb 22, 2011, at 11:34 AM, Jed Rothwell wrote:
Here is some additional info on the 18-hour test. I do not think I
will add this to the News section. It can wait for a paper from
Levi. This may have been reported here by Cousin Peter:
Approximately 0.4 g of hydrogen was consumed in 18 ho
Here is some additional info on the 18-hour test. I do not think I will add
this to the News section. It can wait for a paper from Levi. This may have
been reported here by Cousin Peter:
Approximately 0.4 g of hydrogen was consumed in 18 hours. This is based on
what sounds like a crude estimate to
More notes
I do not know if they used a pump, or simply let the water flow from the
tap. I have used both methods at various times, and so has Dennis Cravens,
although not for such a large flow rate.
They said they checked the flow rate "several times" which I assume means it
was measured manually
A source close to the recent 18-hour test of the Rossi device gave me the
following figures. These are approximations.
Flow rate: 3,000 L/h = 833 ml/s.
Input temperature: 15°C
Output temperature ~20°C
Input power from control electronics: variable, average 80 W, closer to 20 W
for 6 hours
Note
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