The enthalpy calculations of Ahern do not appear to account for the change of the phase of water to steam at about 100 C. This is about 540 calories per gram and should add to the heating of the liquid phase over about 30 C.
This amounts to 540 /30 or about 1800% additional enthalpy—joules or calories whatever units you want-- IMHO. Bob Cook Sent from Mail for Windows 10 From: Jed Rothwell Sent: Wednesday, February 1, 2017 12:40 PM To: Vortex Subject: Re: [Vo]:I calculated his power output from his own data. It is veryexciting and he may have something real that he is blundering with. Seebelow. Brian Ahern <[email protected]> wrote: The water flow rate is 36000kG/day or 36,000kG x 1,000g/kG x 1 day/84,600 sec/day = 425.5 G/sec Note: 1. Rossi and Penon arbitrarily reduced the flow rate by 10%. That is what Rossi told Lewan in an interview. That is shown in this spreadsheet, in the "reduced flowed water (kg/d)" column. So, use 32,400 kg instead of 36,000 kg. 2. They used the wrong kind of flow meter, and it was installed in the gravity return pipe, which was only about half full of water. The manual for this flow meter says it does not work in a pipe that is half full, so the flow rates are far too high. It is difficult to say how far off they are, but they cannot be right. 3. The numbers are impossible in any case. No flow rate can be exactly the same, every day, for weeks. This meter measures to the nearest 1000 kg, which is ridiculous, but given that it does, it would record something like 35,000 kg one day, 34,000 the next, and 36,000 the next even if the flow was extremely consistent. The change in temperature is 69.1 C up to 103.9 = a temperature rise of34.8 degrees C. Heat capacity of water = 4.2 joules/gram/C The power needed for this temperature rise at that flow rate is: Flow rate (G/sec ) x Temp. rise (degrees C) x heat capacity of water (4.2 joules/G/degree C) 425.5g/sec x 34.8C x 4.2 Joules/gram/C leaves units of Joules/second = 62,191watts The authors claim that the water was vaporized, so they used the heat of vaporization. It could not have been vaporized, because there was some back pressure from the equipment. At these temperatures, even a little pressure will prevent vaporization. However, their calculations result in a COP of 82.3. Who knows where that came from? Probably the adjustments I just described account for it, but the data is fake and the instruments and configuration are preposterous, so it means nothing. - Jed
