Ive a bit concern about the total radiation. But Im not expert in calorimetry, neither in heat transfer.
Radiation is emission of IR (infrared) from a warm surface. The Boltzmanns law give the energy radiated from a surface at a given temperature and kind of material. Inside the Celanis cell, there are equipments (included wires) which emit also IR. The total surface inside the cell is not negligible and warmer than the borosilicate glass. What happens to the IR from the inside of the cell when they arrive to the borosilicate glass? Does the IR heat the glass or did they pass through it? Or is it like a greenhouse effects? The IR wavelength emitted inside the cell is around 5 ~8 µm. What is the transmission of those IR through the glass used by Celani? From Duran In the spectral range from about 310 to 2200 nm the absorption of DURAN® is negligibly low. What about above the 2200 nm? <http://www.duran-group.com/en/about-duran/duran-properties/optical-properti es-of-duran.html> http://www.duran-group.com/en/about-duran/duran-properties/optical-propertie s-of-duran.html My thinking could be here completely false. _____ From: Robert Lynn [mailto:robert.gulliver.l...@gmail.com] Sent: samedi 18 août 2012 05:08 To: vortex-l@eskimo.com Subject: Re: [Vo]:Some doubts expressed about Celani demonstration >From those numbers (30°C room, 120°C at 48W and 140°C when LENR active) I calculate 16W excess if you assume all radiative heat transfer. But it will actually be slightly less than that because the hotter tube surface will convect heat away at a rate that is roughly proportional to the air to tube temperature difference. The next level of complication is that the natural convection air flow will also be slightly faster due to the increased buoyancy, so the heat transfer coefficient will increase as temperature increases too, typically at a rate proportional to the temperature differential to the power of 0.25. I'll do the calculation assuming constant heat transfer coefficient and then with variable heat transfer coefficient caused by increased temperature, shouldn't be much difference due to relatively small relative temperature increase. >From his paper he says that the tube dimensions are Ø40mm OD and 280mm long, I will use the full length assume that the temperature is the same everywhere due to internal convection of that most magical of heat transfer fluids hydrogen. Borosilicate glass has emissivity of about 0.9 so the tube is radiating about 27.4W at 120°C and 36.7W at 140°C in a 30°C environment. So 48-27.4=20.6W convected at 120°C and 20.6x(140-30)/(120-30)=25.2W at 140°C. Add that 25.2 to the 36.7 and subtract 48 input and you get 14W excess. Assuming that the heat transfer coefficient increases in proportion to the temperature differential to the power of 0.25 then the convected and therefore excess heat rises by about 1.2W to 15.2W All the same calculations repeated for a 25°C ambient temperature instead of 30°C drop the excess heat from 15.2W to 14.6W, again not much difference There might be a little more complication with the end caps etc, but I think you can pretty confidently state that it is over 10W. Also perhaps someone did a check on the temperature at the top and bottom of the outside of the tube to see if there was a significant temperature difference? I think it is pretty unlikely but you never know. On 18 August 2012 01:53, Jed Rothwell <jedrothw...@gmail.com> wrote: Several experts in calorimetry expressed doubts about the Celani demonstration at ICCF17. Mike McKubre in particular feels that it is impossible to judge whether it really produced heat or not, because the method is poor. He does not say he is sure there was no heat; he simply does not know. Others feel that he exaggerates the problem. There were concerns because Celani has programmed in the Stephan-Boltzmann law which multiplies things to the a 4-th power. Srinivasan worried that he makes a mountain out of a molehill. The temperature is measured at one point on the surface of the tube. I asked Brian of NI to give me the actual temperature readings. With 48 W of input power only, before excess heat or with the Ar calibration, in a room with 30 deg C ambient temperature, the temperature rose to 120 deg C. When the excess heat appeared it rose to 140 deg C. Celani says that equals 14 W excess, and that is what was displayed by the instrument. McKubre and others worry this may be caused by decreased pressure in the cell. However, the pressure fell only gradually, and stabilized in the last 2 days. They also worried about changes in conduction within the tube, and uneven heat on the surface. I do not think that such effects can account for a 20 deg C temperature rise, especially given the smooth line produced when there is no heat, with H or Ar. The temperature returned to the same level with 48 W, in Italy, Texas and Korea, after the gas had been changed out twice. Anyway, I would like to note that these people have doubts. Others agree with me that the method is crude but unlikely to produce such a large error. Celani hopes to run it in self-sustaining mode with better insulation. That will put to rest all questions about calorimetry. He hopes to do this as quickly as 2 weeks from now! More power to him. He has run it for as long as a month, so a 1 or 2 week self-sustaining run should not be a problem. Given the mass of wire, even 10 minutes would be convincing. - Jed
