Вот, к примеру, описание одного из экспериментов. Там они подсчитывают количество энергии, необходимое для нагрева воды до точки кипения и энергию, необходимую для испарения данного количества воды. Далее сравнивают полученную мощность за время эксперимента с мощностью, которая отбиралась от сети. Походу упоминается, что от сети, которая не является силовой, даже невозможно отобрать мощность, полученную на выходе.
http://lenr-canr.org/wordpress/?p=643Цитата:
RESULTS
The test run on January 14 lasted for 1 hour. After the first 30 minutes the outlet flow became dry steam. The outlet temperature reached 101°C. The enthalpy during the last 30 minutes can be computed very simply, based on the heat capacity of water (4.2 kJ/kgK) and heat of vaporization of water (2260 kJ/kg):
Mass of water 8.8 kg
Temperature change 87°C
Energy to bring water to 100°C: 87°C*4.2*8.8 kg = 3,216 kJ
Energy to vaporize 8.8 kg of water: 2260*8.8 = 19,888 kJ
Total: 23,107 kJ
Duration 30 minutes = 1800 seconds
Power 12,837 W, minus auxiliary power ~12 kW
There were two potential ways in which input power might have been measured incorrectly: heater power, and the hydrogen, which might have burned if air had been present in the cell.
The heater power was measured at 400 W. It could not have been much higher than this, because it is plugged into an ordinary wall socket, which cannot supply 12 kW. Even if a wall socket could supply 12 kW, the heater electric wire would burn.
During the test runs less than 0.1 g of hydrogen was consumed. 0.1 g of hydrogen is 0.1 mole, which makes 0.05 mole of water. The heat of formation of water is 286 kJ/mole, so if the hydrogen had been burned it would have produced less than 14.3 kJ.