Reactive power - Reactive Generator research from GotoLuc - discussion thread

[TinselKoala]

I'm not sure I fully understand all of your question.

A DC coupled trace will produce a zero MEAN if the area of excursions above and below the zero ref line are equal.

An AC coupled trace will always produce a zero MEAN I think because the zero ref line always settles to the average value of the wave form.

It is irrelevant if the voltage or current trace have a MEAN of zero, because the MEAN of the trace doesn't enter into the multiplication. We're only taking the MEAN of the product. Make sense?

Yes, makes sense. Yes, that is exactly what I was asking. Yes, it is in line with my prior understanding and with my empirical experience with scopes.
The issue isn't irrelevant though when the change from AC to DC coupling is considered. It's the change in the trace mean that carries over to the math trace and causes the differences there, even if the trace mean itself isn't explicitly computed. I think.

[poynt99]

The issue isn't irrelevant though when the change from AC to DC coupling is considered. It's the change in the trace mean that carries over to the math trace and causes the differences there, even if the trace mean itself isn't explicitly computed. I think.

Partially true. In this case though I don't believe we are seeing a change in the MEAN value before multiplication.

You don't have to have a shift in the MEAN value of one of the multiplicands in order to see the resulting product exhibit a different MEAN value. A simple phase shift can produce the same result.

[nilrehob]

Hi Luc,

I just watched your video reactive battery charging. Why the resistor? Since the current through the resistor is not constant the voltage is not either and when the moment the voltage across the resistor is lower than the batteries the batteries are drained and not charged.You don't need the resistor. The high voltage is not seen by the battery, it only sees the current.

To match a cap to a particular battery as to not get too high current do this calculation:

C = Ah / (20 V 2 pi f)

The voltage of the battery is not important as long as it is relative low to V. As an example in my case:

Ah = 24, V = 230, f = 50
C = 24/(20 * 230 * 2 * pi * 50) = 17 uF

http://www.youtube.com/watch?v=hSssnbLzTIw

This can also be done with a coil, but the calculation needs to be changed:

L = 20 V / (Ah 2 pi f)

As an example in my case:

Ah = 24, V = 230, f = 50
L = 20 * 230 / (24 * 2 * pi * 50) = 610 mH

I tried this with one of my generator-coils but only my MOT test is on video:

http://www.youtube.com/watch?v=9LSTBMwC_lc

And of cource if You put the cap and the coil in the usual resonance formula You get:

1/(2 pi sqrt(L C)) = 1/(2 pi sqrt(610e-3 * 17e-6)) = 50 Hz (close to)

Back to the battery; it functions as a resistor with external resistance:

RB = VB/I

where RB = battery external resistance and VB = battery voltage

This means that the resistance decrease with higher current, and when the current goes to zero the resistance goes to infinity. So when the current and the voltage are close to 90 deg apart, the battery allows the current to pass but "blocks" the voltage. Kind of.

/Hob

[Hoppy]

On the question of whether Luc used AC or DC coupling, he shows in this video (from 03:18) that all channels are set to AC.

[tinman]

This post is for TinMan,

I was thinking over the the battery charging of my last video and thought you are well setup to take advantage of this interesting effect since your prime mover is DC.

After I made the video for you of the meters on the exciter side of my Gen which was tuned to output the most AC on a 15 ohm load, I decided to test with a Series FWBR instead of the 15 ohm load and connect the DC side to a battery. Nothing special other than I could reduce my series cap bank to 15uf instead of 25uf. Then I decided to add a second battery in series so now I would have 24.6vdc on the 15 ohm load, now that was special since I didn't change the 15uf cap and the batteries maintained the voltage and we now have 40 watts on the load. I didn't add a third battery to see if it could keep going up since I decided to test it on the 240vac grid to see the effects and that is the last video I shared.

I would suggest you test it and see how far you can take it and who knows, maybe there will eventually be enough power to feed the prime mover. This is the goal of this research and you are well set up for it now.

I'm also starting to think the smaller batteries not maintaining there charge maybe due to the load being above the batteries charge rate capability.

All the best with your tests

Luc

Hi Luc

It just so happens that i had the same idea,and started testing this very circuit you are talking about yesterday. I must admit that at first i was interested in this setup of your's,but then realised that the power could have been coming from the exciter circuit-and we now know it was.But when i tried the above mentioned circuit (with batteries),one thing led to another,and im now seeing something in this setup that is leading toward you being correct Luc.

So here is the reason i say this. The generator draws 51 watt's at speed 34 on my setup,and this speed is very close to constant regardless of load,until we exceed the rated power of the motors 400 watts.Once the exciter circuit fires up,the motor draws 135-140 watts. So lets look at what real,reactive and apparent power are.
Quote: Practical loads have resistance, inductance, and capacitance, so both real and reactive power will flow to real loads. Power engineers measure apparent power as the magnitude of the vector sum of real and reactive power. Apparent power is the product of the root-mean-square of voltage and current.

Engineers care about apparent power, because even though the current associated with reactive power does no work at the load, it heats the wires, wasting energy. Conductors, transformers and generators must be sized to carry the total current, not just the current that does useful work.

The above statement makes no sense to me,and seems to be an oxymoron.
Quote:Engineers care about apparent power, because even though the current associated with reactive power does no work at the load, it heats the wires, wasting energy.

Is not heating wires doing work,in the form of creating heat?.This is reactive power doing useful work as far as im concerned.

Anyway,once the exciter circuit is loaded,we can see a decent amount of power being disipated within the circuit,and my 5 watt resistor gets red hot very quickly.Now by useing the right size cap on the generators 240 volt output,i can remove all the power being disipated in the exciter circuit,and bring the prime mover power draw back down to 51 watt's. So lets say that the power we see in the exciter circuit,is the power being delivered to our home's via the grid, and if this is the case,then i see it as being very possable to draw power from the grid,without there being any load placed on the generator at the power station-as im doing it in my workshop. But there is a trick to this,so as no real or active power draw is taken from the system as a whole.

My generator is now basically two reactive circuit's that are 180* out of phase with each other. What i mean by this is one cap(on the gen output) is on the high side of the wave,while the cap on the exciter circuit is on the low side of the wave-this is in reference to how each reactive circuit is hooked to the FWBR. Now the system is unballanced,as the 240 volt windings in the generator are heaver that the exciter circuit winding's-but the results are still good. One AC leg of the FWBR is conected to the high side of one of the reactive circuit's,while the other AC leg of the FWBR is conected to the low side of the other reactive circuit. It seems that because we have only one wire hooked to each reactive circuit,there is no load place on the generator when we draw power from the FWBR.

As time permit's,i will tune this setup so as it's as good as i can get it ,with the parts i have-i dont have a lot of high voltage AC cap's,most of my stuff is DC. Once i have it as best i can get it,i will make a video showing my result's.