On Tue, Nov 29, 2011 at 12:18 PM, Joshua Cude <joshua.c...@gmail.com> wrote:
> I could have made a mistake, but by my calculations, if the power > increases at a constant Pdot = 160 W/s, then 13.5 L of water (per > unit) will begin to boil in t = sqrt(2Q/Pdot) = 40 minutes, not 90. > And the power at 90 minutes will be 160*90*60 = 864 kW, and at > 13:22, it will be 160*144*60 = 1.4 MW. > On the other hand, if boiling begins at 11:00, then the power > increase should be Pdot = 2Q/t^2 = 31 W/s, giving a power of > 31*144*60 = 270 kW at 13:22. Indeed, I got that wrong, 160 W/s doesn't make sense, but 45 W/s may fit. Here are parameters that seem to fit: 0) Let time t0 be the start of the warmup period. I have selected the earliest time at which the output temperature exceeds its 5th percentile. This is the 717th record in the XLS file and the timestamp is 11:00:04. 1) At t0, total reactor power (electrical + alleged nuclear) is P0 = 0. Note: Actually, I think P0 > 0 because the output temperature is 30 degrees. P0 is probably small, maybe 12 kW, given that the energy consumption in self-sustaining mode was 66 kWh over 5.5 h. 2) I still think the pump was turned on at 13:22:50 or t0 + 10000 seconds, because we have a nice linear increase in input temperature starting right at that moment, that also coincides with the sharp input temperature drop. This also fits with the water clog theory. I assume that at this point the power is P1 = 450 kW approximately. 3) So power went from ~0 to 450 kW in 10 000 seconds or an increase rate of beta = 45 W/s. 4) Assume all this power goes into heating a thermal mass C expressed in J/K. The formula T_out = 27.34 + 2.124e-6 (t - t0)^2 is a good fit. Call alpha = 2.124e-6 [K/s^2] the rate of increase of temperature increase. Then E = C T_out and thus P = dE/dT = C dT_out/ dt = 2 alpha C. Thus we have : beta = 2 alpha C 5) With beta = 45 W/s and alpha given above, we deduce : C = beta / (2 * alpha) = 10.6 MJ/K. 6) The quantity of water required for this thermal mass is 2523 kg. The reactor is, of course, not pure water, so that's an upper limit. 7) Given that we have 321 submodules, this is 7.9 l per submodule, a reasonable amount. I still find that quantity a bit high. Do we know the water-containing volume of the e-Cats? > Since the ecats are not full, water will not flow out of them when > the pumps are turned on and turning them on only adds a volume flow > rate of 675L/h = .19 L/s, or 1.8 mL/s per module, which is a tiny > fraction of a per cent of the total volume flow rate. In fact, since > adding cool water will remove heat otherwise going into steam, the > total volume flow rate of steam stays the same (at 490 kW). > Therefore this increase in the temperature of the input reservoir > cannot correlate with the pump turning on as you describe. Water did not necessarily flow out directly. I'd like to invoke the "water clog" theory (see my other post). To resume: (a) You have 2523 liters of water boiling in the e-cats. (b) The heaters are not fully immersed. (c) Therefore each submodule has a portion of its heating surfaces that is exposed and heating steam. (d) Steam has about 1/25th the thermal conductivity of water. (e) Therefore the exposed portion is significantly hotter than the unexposed portion. (f) When pumps are turned on, water is added and the level rises until evaporation. (g) Thus the contact area between water and the heating elements increases momentarily. (h) In addition, the formerly exposed and hotter area of the heating elements is now in contact with water. (i) The cold water picks up some energy, but that's only about 15% of the vaporization energy. The rate of steam production still increases significantly, because heat transfer is proportional to area of contact and increases dramatically with the temperature difference. (By a couple of orders of magnitude over 30 degrees above the boiling point, see the graph I posted.) Of course these are all transient effects. The extra water also takes up some volume. (j) Because there is a water clog somewhere (I guess in the condenser), the pressure increases. (k) The increased pressure finally overcomes the lukewarm water clog, which goes back into the reservoir. I concede that point (i) should be analyzed in further detail. > There are many other problems, but these two are significant enough > to leave it at that. I'm listening. > Finally, Rossi and his engineer claim a constant input flow rate for > the full 5.5 hours, and this is not an assumption about which they > could be reasonably be mistaken (like the output flow rate, or the > effectiveness of their trap). They don't explicitly claim that. They manifestly computed the flow rate of 675.6 l/h by dividing 3716 l by 5.5 h, because they state that the regulated flow rate is 700 l (2 x 350 l/h). If the pumps did turn on at 13:22:50 (presumably automatically, maybe based on the water level), then the flow time would have been 4.69 h for a rate of 792 l/h, not too different from 675.6 l, and the average power would be close to 570 kW. Note that the installed power is 1070 kW. > What's the point of analyzing the measurements, if you're going to > ignore some of them and make up your own? The report does not go into too much detail, is often ambiguous and it contains small errors. We don't have the plans for the 1 MW demo, so minor adjustments and reasonable hypotheses are needed to make sense of it. So far I haven't found anything significantly wrong with the 1 MW demo. Also I still don't understand your "instantaneous power transfer discontinuity" argument. -- Berke Durak
