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Author Topic: ReSyMo  (Read 173 times)

Offline pemox1

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ReSyMo
« on: November 05, 2018, 04:16:49 PM »
I want to present this project, because I don'n know if I can manage the mechanical construction in that manner that it will work. And perhaps there're other people who are interested in such a project and who have ab better equipped mechanical workshop....
First a bit of theory:
Every e-motor has at least one coil for producing a magnetic field. Such a magnetic field is build with electric current and, other than in an ohmic resistance, current is not lost, but is given back when magnetic field becomes smaller. This is called idle current.

Now, the idea is: why not collecting this idle current and recycle it?

For this, we have to consider several points:
1: the inductivity  shouldn't change
2: we need a capacitor for collecting the current
3: coil and capacitor must be coordinated
4: as a consequence of this, we have precisely one frequency and one rotation speed
5: we need an electric control ho hold this frequency
6: because there's only one coil (though you can divide it into two) this coil should be wound around the stator and therefor the rotor must be fitted with permanent magnets
7: to make such a motor efficiency the ohmic resistance must be as small as possible (only a few mOhm). That's why wire should be really "fat" ones (larger than 6mm²)

For controlling, I think a Royer/Baxandall-converter would be perfect: https://www.mikrocontroller.net/articles/Royer_converter
Don't know, if there's an english translation...
This circuit is simple, robust and runs always with resonance frequency ( and it works!).
Because there's only one coil in this circuit, stator must look like this: picture1
Rotor must fitted with permanent magnets and must be mounted between the two poles. Magnets must be mounted alternately with N and S around the rotor.picture2
This results at least 6 magnets and always 4 magnets in addition. The more magnets the smaller becomes rotation speed or the higher can be frequency.

For the rotor I took an asynchronous motor, cut out the windings and the iron sheets. So I've got a rotor which will withstand even 1000 U/min.
Into the housing I cut two holes, one on the left and one on the right side, through which can be inserted the stator: picture3
The easiest way to construct a stator is probably to dismantle a big transformer. Cut out the bar in the middle and drill a hole in long bar for the rotor. Then there remains one long bar for winding the coil.
Now it's only the holder-system left. For this I've only vague ideas up to now...

Because such a motor  can not jump from zero to nominal frequency it needs a starting aid. The most elegant solution would be an arduino to start slowly and increase speed until nominal frequency. But for this, relais are necessary to cut off the power supply and this would pull large currents.
I think it's easier to take a starter motor which can drive the rotor to the nominal frequency and disconnect it soon afterwards - and after this switch on the power supply.
And because such a motor makes o sense without driving anything, there should be connected a generator to the shaft.

Now a few calculations:
One advantage of the Royer-converter is, that it runs with DC-current. That's why a normally car battery is enough for power supply. All following calculations are made for 12VDC.
Supposed rotor has 22 magnets, speed is exactly 1 U/sec with a frequency of 11 Hz ( 22 half-periods ). If speed should be 600 U/min nominal frequency is: ( 600 U/min / 60 sec ) * 11 Hz  =  110 Hz.
f = 110 Hz
Coil should have about 100 windings with isolated copper wire: diameter 3mm: A = r² * PI = 1,5² * PI   =  7mm². Wire length is about 16m:  Rc = ( 0,0174 * 16m ) / 7mm²  = 0,04 Ohm.
Rc  = 0,04 Ohm
With this, inductivity of the coil should be about 1 mH.
L  =  1 mH
With f = 110 Hz, inductive reactance becomes:  Xl = 2*PI*f*L  = 2 * PI * 110 Hz * 0,001 H  =  0,7 Ohm.
Xl =  0,7 Ohm
Consequently, capacitor C becomes:  C = 1 / ( 2 * PI * f * Xl )  =  1 / ( 2 * PI * 110 Hz * 0,7 Ohm )  = 0,002067 F
C  =  2067 uF  (AC)

There isn't a secondary load, so we can calculate with those values the idle current between coil and capacitor:
Mikrocontroller.net says,  Ires = Uo * PI * SQR( C / L )  = 12 V * PI * SQR( 0,002067 uF / 0,001 H ) =  54A.
I prefer to calculate:  Ires = Ures / Xl :

Ures = Ul(eff)  =  ( Uo * PI ) / ( SQR(2) )  =  ( 12V * PI ) / SQR( 2 )  =  26V
Ires =  Ures / Xl  = 26V / 0,7 Ohm  =  37A

This Ires is that current, that oscillates between coil and capacitor and this current is responsible for the magnetic field.Power of this motor is similar to a  conventional motor: U * I  = Ures * Ires    =  26V * 37A  =  962 VAr

Smotor =  962 VAr

Now let's see, how much power the battery has to deliver:
Because this is a parallel resonance circuit, we first have to calculate the parallel-resistance of this circuit.
This parallel-resistance is not the ohmic resistance of the coil, but is calculated to:
Rp = ( Rc² + Xl² ) / Rc   =  ( 0,04² Ohm + 0,7² Ohm ) / 0,04 Ohm  =  12,3 Ohm
Rp =  12, 3 Ohm
This resistance takes effect to the battery.
Battery current is now:  Ubat / Rp  =  12V / 12,3 Ohm  = 1 A
Ibat =  1 A
Therefor power, which battery must provide is:  Uo * Io =  12V * 1A  = 12 W
Pbat =  12W

I think, this is worth a try, isn't it?


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ReSyMo
« on: November 05, 2018, 04:16:49 PM »

Offline F6FLT

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Re: ReSyMo
« Reply #1 on: November 05, 2018, 05:31:59 PM »
...
Now, the idea is: why not collecting this idle current and recycle it?
...

Here's why. If we need mechanical energy from the motor, a countertorque must be applied to the motor shaft. This tends to create an opposite magnetic field and therefore a counter-EMF in the coil, so that the current would decrease if a generator were not there to compensate by providing more current. Current is lost during the production of work even if we had an ideal motor with zero resistance and no friction. Even if there was no payload, since friction is equivalent to a motor load, idling current cannot be recycled for the same reason.


Offline pemox1

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Re: ReSyMo
« Reply #2 on: November 08, 2018, 11:04:23 AM »
I try to explain it in other words.
Is the coil powered with AC-voltage, it creates a magnetic field and the rotor rotates with the frequency.

As long as phase shifting isn't 90° either the inductive or the capacitive part predominates and there's a huge amount of current required for powering the motor.
If phase shifting is exactly 90°, inductive and capacitive currents are balanced out and there remains only the (parallel) ohmic resistance for which current must provided by the source.

In both cases, the coil creates a magnetic field by what the rotor rotates.

Free Energy | searching for free energy and discussing free energy

Re: ReSyMo
« Reply #2 on: November 08, 2018, 11:04:23 AM »
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