Body
Ole,
I have had caps charge much faster in an LC tank than otherwise, but it
could also have to do with noise in the core material as I was no using
an air inductor.
On 11/16/2018 1:19 AM, [email protected] [EVGRAY] wrote:
>
> Hi Mick,
>
> "If a reactive LC tank capacitor is then disconnected from the
> reactive source then discharged to a resistive load then reconnected
> energized back to the standing wave modality recharged disconnected
> discharged etc. If the reactive energy is harvested in this manner
> are you claiming the same losses will occur as power factor correcting
> the reactive back into real power?"
>
> Yes that's correct.
>
> Common power factor correction in AC is for eliminating the reactive
> power that doesn't do any work but just heats the wires between the
> power plant and the reactive load. This is expensive for the owners of
> the power transmission lines and thus should be avoided. First storing
> the energy in a capacitor (or inductor or battery) and then dumping it
> to an active load will still make the power go in just one direction
> even though it happens in pulses. The energy is then moving from the
> source to the load. Reactive power has the energy moving to and from
> the load and power source. The power moving between the capacitor and
> inductor in an LC-tank is pure reactive except for the loss because of
> the parasitic resistance in the components and wires.
>
> Regards
> Ole
>
>
> ---In [email protected], <mkjekyll@...> wrote :
>
> Ole,
>
> Disregarding the issue at hand with the patent terminology and not
> concerning ourselves with the patent just a very high Q tank.
>
> If a reactive LC tank capacitor is then disconnected from the reactive
> source then discharged to a resistive load then reconnected energized
> back to the standing wave modality recharged disconnected discharged
> etc. If the reactive energy is harvested in this manner are you
> claiming the same losses will occur as power factor correcting the
> reactive back into real power?
>
>
>
> On 11/15/2018 5:36 PM, onielsen@... <mailto:onielsen@...> [EVGRAY] wrote
>
>> I have only read to the following paragraph cited below. I wonder
>> if the applicant knows about what he is writing.
>>
>> Citing paragraph 106 of the patent application
>> (https://patentimages.storage.googleapis.com/53/7d/29/f7b4bd0194992d/US20070296373A1.pdf):
>> "[0106] There is voltage across the inductor connected in series
>> to the capacitor and there is voltage across the inductor
>> connected in parallel to the capacitor. Since
>> Power=voltage×current, a single power input produces two branches
>> of electromagnetic power output, increasing the power output. In
>> the alternative, the present disclosure may have one
>> electromagnetic power output with less energy input."
>>
>>
>>
>> It isn't as simple as this. If using an oscilloscope for
>> measuring the current and voltage it will show some phase
>> displacement between the current and voltage which means reactive
>> power. Look at the following small videos to see how one can be
>> misled if not knowing how the phase between the current and
>> voltage is: Only the the resistive part consumes (transforms)
>> energy like R in figure 8.
>>
>> 0:42
>> <https://www.youtube.com/watch?v=yJEZMCCBV3U>
>>
>>
>> Overunity Device Cop 2,5 ? explanation part 3.
>> <https://www.youtube.com/watch?v=yJEZMCCBV3U>
>>
>> 9.5K views5 years ago
>> 1:01
>> <https://www.youtube.com/watch?v=Jamj3-2w0eE>
>>
>>
>> Overunity Device Cop 2,5 ? explanation part 2.
>> <https://www.youtube.com/watch?v=Jamj3-2w0eE>
>>
>> 10K views5 years ago
>> 1:00
>> <https://www.youtube.com/watch?v=P_E0ck3v40Y>
>> <https://www.youtube.com/watch?v=P_E0ck3v40Y>
>>
>>
>> Overunity Device Cop 2,5 ? part 1. but,
>> <https://www.youtube.com/watch?v=P_E0ck3v40Y>
>>
>> 8K views5 years ago
>>
>> The numbers without the phase shift looks like overunity which
>> isn't the case when taking into account the phase shift. Only
>> real power is useful in doing work. Reactive power doesn't do
>> work except for the small active part heating the the wires and
>> other components because of their parasitic resistance. This part
>> caused by current doing work against the resistance is active
>> power and not actually reactive power. This is why the power
>> distributing companies may charge costumers for having high
>> reactive power going through their cables which heats the power
>> lines without transferring active power to their costumers. It's
>> just pure loss because the wires have resistance. Resistance
>> converts electric power into heat.
>>
>> Just looked at the figures and found the description to figure 46:
>> "[0149]FIG. 46 illustrates an embodiment of the present
>> disclosure that conserves light bulb energy. In a test conducted,
>> on Nov. 16, 1998, a light bulb was installed according to the
>> schematic disclosed in FIG. 46 and is still running seven years
>> later.
>>
>> [0150]Since Apparent Power (VA)=Current (A)×Voltage (V), the
>> calculations below illustrate energy conservation for a 60 W
>> light bulb. The voltage applied at the power input is 115 V. The
>> apparent power for the Power Input Line (Line PIN) is measured at
>> 126.5 VA. The apparent power measured for the shed motor fan is
>> 126.5 VA and the apparent power measured for the light bulb is 65
>> VA. Therefore, 126 VA is measured at the input and a total of
>> 191.5 VA is measured at the output, indicating an approximately
>> 1.5 gain. Calculations are also shown for the energy savings for
>> a 75W Bulb."
>>
>> It is the apparent power that is increased. The apparent power is
>> the vector sum of the real power and the reactive power which
>> makes it always greater than or equal to the real (or active or
>> true) power.
>>
>>
>> Source with description at:
>> https://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/true-reactive-and-apparent-power/
>>
>> Regards
>> Ole
>
>