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This disclosure is concerning a successful approach to my original theory as explained in post #20. That is, with an application of RLE to a two winding transformer with equal inductances, one would be able to reach a theoretical maximum COP=2. For an explanation of what constitutes RLE, see my paper "Reducing the Lenz Effect in a Transformer" in post #6.
The simulation below utilizes the RLE concept in a unique circuit configuration that allows a COP>1. The transformer model for this sim is taken from an actual bench device with two equal windings L1 and L2 of 140 turns each, spaced apart by ~.040" to produce the K=.8 coupling factor. The core is a 1/4" P-42510 ferrite by Magnetics and has a .010 gap in all legs and the core remains in a linear mode throughout all current excursions.
Prior to the sim start, secondary L2 is charged to a -100ma, the constant current source L3 is charged to +100ma, and the primary L1 has no charge. Conventional current flow is through the coils toward the dot.
L1 is now connected to the 20vdc supply Vs via S1. Due to the current bias in the secondary L2, the current in L1 uniquely begins to ramp in a negative direction due to standard transformer action. After 13.941us, the current in L1 is equal to the current in L2 and of -97.78ma and at this point in time the simulation is stopped. The reason for the lack of equivalent current drop in the secondary is due to the RLE effect or constant current sourcing by inductor L3 in this case.
What we now have is a drop in current in L3 and L2 to [97.78ma] and the primary L1 and secondary L2 have equal currents and polarities. Therefore, the transformer has now converted to an equivalent aiding inductance of 10.46mH if these winding were connected in series.
For the calculation of the energies involved, we'll start with the input. From V(vs)*I(V1) we see the input is actually -13.644uJ. IOW, energy is supplied back to the power supply from the circuit.
Energy produced by the circuit action is calculated from the identical L1 and L2 currents of -97.78ma in the resulting 10.46mH series inductance which is .09778^2*.01046/2 = 50uJ. Therefore, the total energy produced by the circuit is 13.644uJ+50uJ = 63.644uJ.
For the energy consumption we'll first examine L2. L2 was initially charged to [100ma] which required an energy of .1^2*.0029/2 = 14.5uJ. We will consider all this energy as consumed because the ending current in L2 is calculated in the ending series calculation with the primary.
Next, L3 was initially charged to 100ma but ended with 97.8 ma for a loss of (.1^2-.0978^2)*.1/2 = 21.7uJ. So, the total energy consumed by the circuit is 14.5uJ+21.7uJ = 36.2uJ.
Therefore, the total apparent COP=63.644/36.2=1.76. This is short of the theoretical COP=2 primarily due to the voltage drop across L3 thus rendering it as being less than a perfect current source.
Although not shown, those skilled in the art would be able to conceive how to connect the L1 and L2 winding in series and discharge them into the power supply plus, restore the current in L3 back to 100ma for periodic operation. This scheme should allow a reasonably easy to build OU device that would be able to self charge a battery system for example, but certainly not limited to this application by any means.
Regards,
Pm
« Last Edit: 2019-12-18, 01:08:15 by partzman »
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