Inductive heating may disturb the thermocouple if its in the magnetic field.
Its may be good to test before. On Fri, 20 Mar 2015 15:07:05 -0400, Alberto De Souza wrote: To eliminate doubts about the current used on each of the two reactors, one just needs to put the two heater coils in series; the current will be exactly the same (at DC or, to a very good approximation, 60Hz). If the coils are made with the same type of wire and the wires have the same size, we have the same resistance. If the reactors are in the same room and close together, about the same heat dissipation. I would make one completely empty and the other with LENR fuel; so, we will have only one variable under test (the heat or excess heat - the dependent variable), only one independent variable (the fuel), and all remaining variables (control variables) under control. With COP 3 the difference in temperature will be huge and the experiment a great success (proof of excess heat). Alberto. On Fri, Mar 20, 2015 at 2:46 PM, Axil Axil wrote: Induction Heater Circuit ~ FULL explanation & schematic https://www.youtube.com/watch?v=pVYMLnXW9uo [2] The problem with inductive heating is the lack of control over temperature that this method of heating gives. It will also produce overkill if the design is not well designed. The Russian has shown that it takes 12 hours of gradually increasing heat to control the LENR effect. It will take an inductive heater with a very small output to heat a gram of fuel over 12 hours; maybe just a few milliwatts of power. To calculate the COP, we must convert or calibrate the RF power delivered to the fuel into heat output. On Fri, Mar 20, 2015 at 2:10 PM, David Roberson wrote: Jones, eventually you can adjust the shape and type of the fuel until it becomes conductive enough and has sufficient area to capture the time changing field and absorb power. The Russian team that uses an inductive heating technique described by MFMP pressed their fuel into pellets that have the right area and resistivity to work with their RF generator. In that case the normal heating resistor coil is not needed. I have not studied the standard cooking drivers but would be surprised to find that they would work efficiently into a object with a small surface area. You would be wise to construct a drive coil that has an inner area that comes closer to matching the fuel pellet. That way much of the magnetic flux inside the main coil is linked to the pellet. A tighter coupling would allow the reflected resistive component due to the fuel losses to appear larger in the main drive loop. RF current flowing within the main loop would induce power into the reflected resistance from the pellet and if the unloaded 'Q' of the main loop inductor is large enough, most of the input power ends up in the pellet and not as losses within the drive system. You can use resonating capacitors to cancel the input inductive component if you are skilled in the RF field. With careful matching of this type, you can come up with an overall system that efficiently converts the DC input power into pellet heating. But, it takes very careful and skillful design to make it happen. Dave -----Original Message----- From: Jones Beene To: vortex-l Sent: Fri, Mar 20, 2015 12:20 pm Subject: RE: [Vo]:Am I the only one.. Dave Ø Jones, even at 40 kHz it is going to be extremely difficult to get enough current to flow inside a coil of wire. Remember, they normally drive the expansive sheet of resistive metal that has an effective resistance that is much less than an ohm. The coils that we are using is in the vicinity of 10 ohms. Yes, that is true but don't forget that the tube fill mix can be made conductive as well. This is the reason I suggested to Jack to use Fe3O4 instead of Fe2O3 as the bulk fill (or support material) with an inductor setup. The former is 6 orders of magnitude more electrically conductive than the later. So, you have a magnetic field that enters a much larger area of resistive metal when a pan is placed upon the unit than with the small coil. Then, the length of wire used in the coil has a large series resistance whereas the pan is more of a parallel resistance and much less in total value. Both of these effects are working against you. I agree but Fe3O4 is highly conductive - although we do not know what happens at elevated temperature in the presence of reducing compounds, but as long as it is not further oxidized, Fe3O4 should be in the few Ohm range, no? Not to mention acting as a transformer coil, to an extent. Jones Links: ------ [1] mailto:janap...@gmail.com [2] https://www.youtube.com/watch?v=pVYMLnXW9uo [3] mailto:dlrober...@aol.com [4] mailto:jone...@pacbell.net [5] mailto:vortex-l@eskimo.com
