From: Axil Axil * Alumina pulls in electrons rather than pushing them out, Is that correct?
Technically alumina is a good electrical insulator, as seen in the white ceramic part of a spark plug. Beta-alumina however is different, and can be produced in such a way as to conduct only positive ions. As such those ions would “pull in” electrons after the ions moved through the alumina. This, in fact, is the way that some sodium batteries operate, using beta-alumina as a solid electrolyte (oversimplified). I do not think Rossi is doing this. The simplest way to get direct current, if one had a glow tube reactor, operating inside a larger metal vacuum tube, which functions as an anode, as you suggest – is to wind the tube with heater wire which also is a good thermionic emitter like tungsten, then that heater coil itself could also function as a cathode with a small change in the circuitry. In this case, it would be wise to use thoriated tungsten as the heater wire, which is known as a good emitter but needs to be in a vacuum as it is easily oxidized in air. In operation, electrons emitted from the heater coil would decrease the heat given to the fuel (the Edison effect is a cooling effect). Also, they would require emf to overcome the space charge inside the gap (like any vacuum tube). However, if the LENR fuel (by this time) has reached strong self-sustain mode, with its own ability to produce heat without electrical input, then this device could be made to function almost like a self-powered amplifier tube of old. It could possibly function without a grid accelerator, if enough light was being produced inside the glow tube (to provide emf and overcome space charge). This essentially means that a glow tube which has gone into self-powered mode (infinite OU) could indeed be arranged to produce electrical current flow as a side effect, when properly designed inside a vacuum, if the incandescent photons provide the emf. This is more likely what Rossi is doing. In fact, when net electrical current is being produced, that would be a STRONG indication that the tube has gone into self-powered mode. As such, this might even be a better approach then the Parkhomov type of replication if one has a good vacuum system and a proper Dewar. The $64 question is can he provide electrical output with no input for longer than a few minutes. The thermal inertia of a very hot system could allow tens of seconds, but not minutes. Jones
