On Feb 21, 2011, at 6:27 PM, Jed Rothwell wrote:
Horace Heffner <hheff...@mtaonline.net> wrote:
As a double check on concepts, if you plug x=0.02856 into x/((x+(1-
x)*0.0006)) then you get 0.98. That is to say, 98% of the mass of
the volume expelled is water, and 2% steam - your starting
assumptions.
As a double check on this discussion, you should note that they
have now run the cell with hot water only, no phase change, and
they found it recovered even more heat than with the phase change.
So this speculation about wet steam and greatly reduced enthapy is
incorrect.
Evidently Dr. Galantini was correct, and the steam was dry. Either
that or these estimates of the enthalpy of wet steam are incorrect.
I do not know which true, and it does not matter. A different
method has now been used to confirm the original conclusion.
- Jed
I look forward to the report. This 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 gm/mol)) = 2.52x10^4 eV / H = 25
kEv per atom of H.
On Feb 21, 2011, at 8:47 PM, Peter Gluck wrote:
This morning I have received this from Giuseppe Levi re this test
:
Average flux in that test was 1 liter per second (measured by me
many times during the test). No steam. MINIMUM power measured was
15 kW for 18h. 0.4g H2 consumed.
This means that a 270 kWh = 972 MJ where at least produced. This is
an under estimation.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
