Look closer at this one:
http://www.mtaonline.net/~hheffner/RossiT2Pout.png
Let me give you a scenario. There is some back pressure on the E-Cat, so boiling temperature rises as high as 124 degrees.
Note: This is in the believer's favor. If atmospheric pressure is lower, then the boiling point is lower, and even less power is required for 124 degree steam (because the specific heat of steam is lower).
In 6 hours of operation, 19.656 kg of water flows through the E-Cat. (.91 g/s x 60 sec/min x 60 min/hr x 6 hours)
To raise all of the water from 24 degrees to 124 degrees, would take 1,965.6 kcal (19.656 kg x 100C)
To vaporize all of the incoming water, 10,614.24 kcal (540 cal/g x 19.656 kg)
This is 12,579.84 kcal over 6 hours, or 2,096,640 cal/hr, which is 2,436 Watts
2,436 Watts would completely vaporize the input water, and over that would deplete the water collected in the E-Cat.
If we could actually rely on the E-Cat performance data, before this test was over, the E-Cat would have been bone-dry, and the steam should have been climbing to ever-higher temperatures.
Please, anyone, tell me where this logic is flawed.
I've set this calculation up for 1 hour :
http://lenr.qumbu.com/ecatcalc.php?plot=Plot&ever=c&efzx0=0&efzy0=0&efzx9=9&efzy9=9&esl=1&epbr=1&enm=Oct+6++--+Input+Power+only&edh=1&edm=0&eds=0&eif=3.27&eip=2.5&ecp=0.06&eop=2.5&eoxr=1&et0=20&ep0=1&et1=15&ep2=1&er2=2
For the input-power-only phase, 1 bar, with 0.9 g/sec and 2.5kW -- should get 170 C superheated steam !!!!!
(Doesn't make much difference if it's 1 bar or 2)
If you double the flow, at 2 bars then you get quality 0.5 120 C steam from input power only.
http://lenr.qumbu.com/ecatcalc.php?plot=Plot&ever=c&efzx0=0&efzy0=0&efzx9=9&efzy9=9&esl=1&epbr=1&enm=Oct+6++--+Input+Power+only&edh=1&edm=0&eds=0&eif=6.5&eip=2.5&ecp=0.06&eop=2.5&eoxr=1&et0=20&ep0=1&et1=15&et2=120&er2=1
