Hi THH, JR wrote to you:
There is no input power for most of the boil off. That is what the graphs show, and what common sense tells you must be the case. Got that? No Input Power. NO INPUT POWER, for crying out loud. Do you understand what that means? Ascoli does not, but you probably have some knowledge of everyday physics.
I strongly recommend you to follow his hint, and look at the graphs.
IMO, the most important graph produced by F&P, and one of the most revealing of the whole CF history, is the one shown in Figure 8 of the F&P's "simplicity paper": https://www.lenr-canr.org/acrobat/Fleischmancalorimetra.pdf .
Its caption says: "Expansion of the temperature-time portion of Fig 6B during the final period of rapid boiling and evaporation."
The temperature-time portion starts at about 1,597,000 s, when temperature of the electrolyte is about 86 °C. Then it slowly reaches a maximum temperature of about 101°C, and suddenly drops at 1,657,000 s (at the right end of the horizontal arrow). So the rising part lasts 60,000 s, that is 16h40'.
You should locate the fig.8 portion into the whole temperature trend of fig.6B. This last shows the voltage too. Before the beginning of fig.8 portion, both temperature and voltage in fig.6B are rising (on average, of course), and both are accelerating their increasing rates. The second derivative is positive for both curves.
Voltage increases due to increasing of resistance of circuit, but, as known, in electrolyte the resistivity decreases on increasing temperature because viscosity of electrolyte decreases. So the resistance can only increase for the phenomenon highlighted by Lonchampt, that is build up of deposits on electrodes. This slow phenomenon explain the needing for F&P to run very long experiments.
On the other hand, electrolyte temperature increases because cell must dissipate more extra heat by means of radiative losses, which vary with the 4th power of temperature.
F&P calculated the radiative losses at boiling point. On page 16, they wrote that, at 101°C, the cell loses 6,700 J in 600 s, i.e. 11.1 W, toward an ambient at 20°C. Therefore, at 86 °C, at the beginning of fig.8, the radiative dissipation is about 8.5 W. So, at a constant current of 0.5 A, the radiative losses, in the portion shown on fig.8, accounts for a voltage difference ranging from 17 to 22.2 V, that is 18.5 to 23.7 V after having included the 1.54 V absorbed by the electrolysis. But fig.6B shows that, in this final boil-off period, voltage skyrockets well above 25 V. Where does it go all this extra power?
Only a tiny fraction of it can be stored as sensible heat in the electrolyte, because its temperature is already close to the boiling point, and it increases more and more slowly, by levelling its trend. As a consequence, as clearly shown by fig.8, the second derivative of temperature is negative, while the second derivative of voltage continues to be positive for a while, until voltage reaches the maximum allowed value of 100 V, that is 50 W. Nearly 40 W more than the radiative losses in that period!
Again, where does it go all this extra power?
Well, the answer is simple: it generates steam!
Evaporation is the only way in which the F&P cell can dissipate the extra heat produced by the electrolyte current during "the final period of rapid boiling and evaporation". But this period doesn't last 600 s, as F&P assumed in their calculation at page 16, it actually lasted many hours as shown by the levelling trend of the curve in their fig.8. Probably yet at the beginning of the curve, at 1,597,000 s, but surely at 1,620,000 s (that is 10 hours before dry out(*)), water is evaporating at the cathode surface. It means that the cathode temperature is above boiling point, hence it is higher than the electrolyte's. And this over temperature surely increases with the increasing of the extra power.
In conclusion, if you look carefully at fig.8 of the F&P "simplicity paper", and interpret it correctly, you can desume that:
- there have been input power for ALL of the boil off (not only for most of it);
- this input power is largely sufficient to explain the dry out of the cell (no excess heat is required);
- the temperature of the cathode rises well above the electrolyte's.
(*) Please, pay attention, the dry out doesn't happen in the instant indicated by F&P on fig.8, in fact the vertical arrows are misplaced, the dry out happens when temperature drops.
