Reactive power - Reactive Generator research from GotoLuc - discussion thread

[hartiberlin]

Hi Luc and all,
I checked the mechanical Watt meters over here in Germany and you are right,
they just register only Real active power and not
Apparent power, what I thought before.

So okay, you have these about 20 Watts for free with your circuit,
but the question really is, if the power company will like this.
Probably not, cause your house meter will not regirster it and
they can not charge you, but they have to pay the bigger transformers
to keep up with the reactive and apparent power needing higher
currents.

As they have to generate the current in their generators, the question
still is, if the cycling of your about 800 Watts getting in from the grid
and 800 Watts returning to the grid in a 120 Hz rhtyhm will need them to generate more power than
just idling the reactive power back and forth...Hmm...

You seem to have shown in your first experiment with the motor
driving the generator, that this additional reactive load
does not need any additional input.

The question then is why your motor really needs so many Watts to drive an
idle generator ? Are these just friction losses ?? There shouldn´t be so many losses
and you should have better used an at least 90 % efficient DC motor to drive
your generator to have a much easier measurement for the input power without any
power factor there as it is the case with your AC drive motor...

Anyway, I think this should be better tried at higher frequencies
with smaller ferrite type transformers and a sinewave opscillator that
drives these phase shifting ferrite transformers and then you can also use smaller
caps and the output power is then easier to scale up as you can transfer more energy
per timeframe this way with faster frequencies.

If you drive the sine wave oscillator from a DC battery you can then measure, if it will
draw more power from the battery, if you draw the reactive power from the output side.

Many thanks.

Regards, Stefan.

[hartiberlin]

Luc said:
Not quite. If you are setting the scope channel for current, you must have a current probe connected to it.

Luc just emailed me this:


Sorry but that changes nothing.

I just tried it. If I select Voltage instead of Current the number are exactly the same. The only difference is instead of VA at the end it displays VV

Nice try but you'll have to come up with something a little better than that to explain what's going on.

Don't forget, when the circuit is connected to a generator it has Zero effects to the prime mover. So something is going on so don't get caught up in scope shot power calculation details. I'm not trying to prove this works by just looking at a scope shot.  That's just for show and irrelevant!  what is relevant is the zero effect to the generator and delivering over 20 Watts to the load!... this is what needs to be understood.

Hope you can tell them what you thought was a mistake does not change anything.

Regards

Luc

[poynt99]

There is something not quite correct in those measurements.

Time will tell.

[hartiberlin]

So Luc and poynt99,

it is clear now that using a shunt this way works exactly the same as using a current probe.

But Luc did not yet answer  what  the exact real absolute value of the current was in his last test ?

Was it 320 mA or 3.2  Amps  ?

Many thanks.

Regards, Stefan.

[codeboundfuture]

poynt99,
The current sensing resistor is using ohm's law and an oscilloscope to detect the voltage drop.  This method is probably the most accurate method possible.  Luc's data is indeed valid.

"Multiply the current by 10 as CSR is 0.1 Ohm." - gotoluc
We should then have 3.2A.

Cheers,
matt