Rossi said: The volume free for the water is about 30 liters, so that to fill up it are > necessary about 2 hours ( the pump of the primary circuit pumps about 15 > liters per hour) . . . >
This is confusing yet strangely helpful. Didn't he say there are four cells in each reactor box? There are 4 pumps on the table there. It is multiplexed. (Why not just make 4 boxes?) Anyway, there are multiple cells and only one cell was in use during this test. I assume each of the 4 cells has its own reservoir, so 30 L / 4 = 7.4 L. Let us assume the primary loop flow rate was 0.9 mL/s, and it took about 2 hours to fill up the reservoir. The graph shows no water or steam going into the heat exchanger for 2 hours, and Rossi said 2 hours above. 2 hours * 0.9 mL/s = ~6.5 L. Close enough! The temperature of the water is around 120°C, evidently under some pressure. Ambient is around 20°C so . . . 120°C - 20°C * 6500 g = 650,000 calories = 2.7 MJ stored heat in the water. Lewan's minimum estimate is 2 kW. 2.7 MJ / 2,000 J/s means all of the stored heat in the water must be extracted by the cooling water flow in 23 minutes, less time than it takes to replace all the water. It would have to magically surrender all 2.7 MJ and be room temperature 23 minutes after the power goes off. Yet 4 hours after it went off, the water was still boiling, the reactor surface was still hot, and the hose was still hot enough to burn someone. Krivit thinks the water "stored" 33 MJ, which magically came out twice as shown by the calorimetry: once before heat after death, and again during it. I guess Maxwell's Demon turns those megajoules around and pushes them right back in to the water. Anyway, assuming I am correct about the volume of the reservoir the temperature of that water would be 1230°C. - Jed
