Reactive power - Reactive Generator research from GotoLuc - discussion thread

[d3x0r]

A way to try and explain this reactive power situation using water and a paddle wheel
In the video,we see water flowing out of a chanel(the waterfall),and onto a paddle wheel.
No matter how much we load the paddle wheel,no reflection or restriction of the water flow will be detected at the output of the water channel(top of the waterfall).But what would happen if we reduce or stop the water flow at the top of the waterfall?. We also know that the paddle wheel would not have enough energy to return all the water driving it,back up to the top of the waterfall.

http://www.youtube.com/watch?v=gKx-Jms624A

A simple test for you all(one i carried out) in reguards to the main's/MOT setup.First,get it to read 0 watt's,and measure voltage across load resistor on the secondary. Next test-place a 1k ohm non inductive resistor on the active of the mains line,befor the watt meter.Now run again,and measure voltage across load resistor on MOT secondary. If we were indeed drawing 0 watts of power,then no drop in voltage across the load resistor should be detected. I could tell you my results,but this is something you must see for your self.

okay but now turn the paddle wheel on its side, and make a vortex, which has a combined resonant motion greater than the input...

backyard science version

industrial scale version

Edit: generator in india Had to add this one too...

Would be best if the impeller could just react to the surface flow ,which is the fastest portion...

viktor schauburger aparatus for generating power at his cabin... top is a splined pipe falling on an impeller

[wings]

the max allowable reactive power should be within 90% of active power

INPUT
1260 Watts and 3000 VAR -
3300 total input power V*A (200 Volts 5-6 amperes)



cost (3.23*1260 + 4.21*3000)/3600= 4.6 centEuro (KV*A*hours)


OUTPUT
9080 Watts and 165 VAR
9100 total output power V*A (220Volts 13.5 amperes)

cost (3.23*9080 + 4.21*165)/3600= 8.4 centEuro (KV*A*hours)


amplifications of 10 times active power 3 times full power - not bad

I was not able to download
http://x-faq.ru/index.php?PHPSESSID=2ff6a3659cbfc6464117764804a0dbaa&action=dlattach;topic=1316.0;attach=6677;image


http://x-faq.ru/index.php?topic=1316.240


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here more:
http://nshb.at.ua/load/svobodnaja_ehnergija/rezonansnyj_usilitel_moshhnosti_stepanova/21-1-0-589
http://www.docstoc.com/docs/104616649/Stepanov-electric-power-amplifier

[d3x0r]

Or maybe the load is reacting like a brown dipper (image)

American dipper under water
more of the same

Dipper at Tyresta Nyfors




Was reflecting; if you have a twisted pipe and can guarantee its negative resistance flow rate, could put the water back uphill easier... even if it wasn't negative and it was just lower resistance would be good....
and that's all doable using hydrodynamics alone

[Farmhand]

Maybe the Russian Guys device takes the 50 Hz grid supply and chops it up to high frequency which causes the output meter to read incorrectly. We have to assume the light bulbs are 500 Watts each, I would have thought 10 kW of 500 Watt light bubs would be difficult to look at and film properly if fully powered. Then there is the reactance from the globes, if the output of the device is high frequency and the bulbs have inductance there are more issues for the meter which is designed for 50/60 Hz Sine wave power. If the output is not 50/60 Hz Sine wave power the meter will not be reading correctly.

Then there is the problem of it being connected to the grid, none of these systems that require connection to the grid will have any future if the power company finds out about it, just like the guys in Brazil, the power company will not allow other people to make use of their grid to do work for free. For there to be any uses for this stuff the device needs to be able to make use of a motor driven AC generator/alternator as the power supply or some other way of obtaining the AC supply on site, connecting to the grid takes out the cost of producing the AC power to begin with, it's simply cheating.

If it won't work when used with say a 5 kVA petrol/gas powered generator then it probably isn't "making" any energy.

It requires energy to keep the grid supply there and waiting.

It takes energy to power a load.

Someone beg borrow or otherwise obtain a small AC petrol generator and use that for the power supply, then factor in the cost of fuel and maintenance for that power supply for the duration of the tests to get the real efficiency. Notice the way generators are in rated kVA not Watts output.

..

[wings]

Maybe the Russian Guys device takes the 50 Hz grid supply and chops it up to high frequency which causes the output meter to read incorrectly. We have to assume the light bulbs are 500 Watts each, I would have thought 10 kW of 500 Watt light bubs would be difficult to look at and film properly if fully powered. Then there is the reactance from the globes, if the output of the device is high frequency and the bulbs have inductance there are more issues for the meter which is designed for 50/60 Hz Sine wave power. If the output is not 50/60 Hz Sine wave power the meter will not be reading correctly.

Then there is the problem of it being connected to the grid, none of these systems that require connection to the grid will have any future if the power company finds out about it, just like the guys in Brazil, the power company will not allow other people to make use of their grid to do work for free. For there to be any uses for this stuff the device needs to be able to make use of a motor driven AC generator/alternator as the power supply or some other way of obtaining the AC supply on site, connecting to the grid takes out the cost of producing the AC power to begin with, it's simply cheating.

If it won't work when used with say a 5 kVA petrol/gas powered generator then it probably isn't "making" any energy.

It requires energy to keep the grid supply there and waiting.

It takes energy to power a load.

Someone beg borrow or otherwise obtain a small AC petrol generator and use that for the power supply, then factor in the cost of fuel and maintenance for that power supply for the duration of the tests to get the real efficiency. Notice the way generators are in rated kVA not Watts output.

..

i like this diagram
http://fotki.yandex.ru/users/nshb/view/420216/
the use of variable reactor (or magnetic amplifier) that sense the load
http://fotki.yandex.ru/users/nshb/view/420217/
http://fotki.yandex.ru/users/nshb/view/420214/

all it seem well done
http://nshb.at.ua/load/svobodnaja_ehnergija/rezonansnyj_usilitel_moshhnosti_stepanova/21-1-0-589