I made a very rough estimate of the heat from the pump by comparing the cooling rates on the night of Oct. 22 with the pump working, to Oct. 21 when pump failed. I think the pump adds ~1 watt to the water.
Both example are 7.7 hours long, with ~3 deg C temperature difference between the water and ambient. I looked for segments when conditions were similar and ambient changes were small, plus I wanted a temperature difference similar to the peak in the Oct. 21 data, when the water was just over 2 deg C warmer than ambient. The coefficient for Newton's law of cooling is cribbed from: http://www.endmemo.com/physics/coollaw.php (To work this calculator, you fill in all but one field, click calculate, and it solves for whatever you left blank.) Results: On Oct. 22 with the pump working, the reactor + Dewar lose heat at 1.99 W. On Oct. 21, with the pump not working: Dewar (3000 g water) loses heat at 0.36 W, Reactor + 1 kg water (equivalent to 6920 g of water) loses 2.57 W; average weighed to thermal masses, 2.92 W. Here are the spreadsheet cells. The format will be gutted by Vortex; use your imagination. Oct 22 night, pump working Ambient Water Water-Ambient Hour 10, cell 1470 19.96 23.29 3.33 Hour 17.7, cell 2603 18.30 21.94 3.64 Change: 1.66 1.35 Avg: 19.13 Newton Coefficient: 0.051 Duration: 27724 s 7.7 hours Total thermal mass: 9920 g water equiv. 55175 J loss 1.99 W loss Oct 21 night, pump fails Ambient Water Reactor R Reactor L Water-Ambient Hour 9.5, cell 1397 22.28 25.33 24.08 24.11 3.04 Hour 17.2, cell 2530 20.31 24.52 21.53 21.67 4.21 Change: 1.97 0.80 2.55 2.44 Avg: 21.30 2.50 Newton Coefficient: 0.029 0.324 0.263 Duration s, h: 27724 7.7 2.50 Average Thermal mass Dewar 3000 g water Thermal mass reactor 6920 g water 9937 71133 81071 Joules 0.36 2.57 2.92 Watts
