This is a resend test to see if this shows up in the archives this time.

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/




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