HeLP with COP of motor

[gyulasun]

Well,  we learn sooner or later, as usual....    If he does not come up with news, videos etc,  then it is also an answer, isn't it.

You also agree with COP of 2 if  his data are correct?

rgds,  Gyula

[Honk]

He asked what the COP of a magnet motor was.

Motor Setup: 
the rotor is of one arm and at both ends there are two permanent magnets fixed.   
The stator is one electromagnet.
                     
His measurements: 
output torque is 0.31 Nm at 193 RPM

input to the electromagnet in case of one of the magnets: 254V @2A   0.4msec pulse
input to the electromagnet in case of the other magnet: 262V @ 9A   1.2msec pulse

Maybe I am mistaken but I figured a COP of about 2.

Hi, I believe I can answer that question.
Output @ 193RPM = (0,22847*2*3,14*193)/33000 = 0,00839135693*746 = 6,26 watts
Input = (254*2*0,0004)+(262*9*0,0012) = 3,0328 watts
COP = 6,26/3,0328 = 2,064

But I don't really understand the input to one magnet versus the other magnet???
Is the rotor magnets not equally sized or what? Or is there dual electromagnets?
The COP is dependent on the total average ON time of input power vs the output.
There is no info on how the coils is fired or the rise time to reach 2 vs 9 amps.
This is crucial to precisely calculate the input power.
Coils are funny, they don't respond like a resistor. A charge takes time to build-up to the
desired level of current flow and is accelerated by voltage. The charge time is not liniear.
And there is no info whether he is recycling the inductive kickback to save energy
or just release the coil energy to obtain a fast turn-off.

[gyulasun]

Hi Honk,

Thank you for the check of the COP, I figured it the same way you showed but did not bother much with the decimals.

Yes, the only explanation for huge difference in current in case of one of the magnets I could think of ala his deleted short text was the magnet strength was very different.  Or he realised something else in the meantime was wrong so deleted all.

Would like to ask you how much of this 3 Watts input power could be regained in practice? Maybe 65-70% of it is a pretty good catch already in practice?

I ask this because in the Pulse Generator thread user Slayer007 just reported he seems to find a way to increase the flyback pulse energy without any increase in the input energy...
He regained about twice the current and the voltage of the 'normal'  flyback pulse by using a bifilar coil and quote: The BEMF I'm running back thru my second winding in the positive and back out thru negative. unquote.  Very interesting.
Here is the link to his posts (continues on the next page too):
http://www.overunity.com/index.php?topic=5954.msg143053#msg143053  Maybe we could figure out how he does it...

Many thanks for dropping in!

rgds,  Gyula

[Honk]

Would like to ask you how much of this 3 Watts input power could be regained in practice? Maybe 65-70% of it is a pretty good catch already in practice?

The recoverable energy level is totaly dependent on winding resistance, hysteresis, eddy currents loss and switching frequency.
My own personal experience with MPP cores from Arnold Technology show it's possible to regain 99% or better.
I haven't tried to measure or calculate the exact performance on a single core but my best switched power supply design so far
had a total efficiency of 99,5% at peak level. But that was using the best and very expensive components and it dropped to 98%
in production just to lower the cost. But 98% is not bad when compared to the competitors that had 74% efficiency in a similar design.

I have not observed any gain in efficiency when using bifilar winding.
I think Slayer007 have made measurement errors. It takes "know how" and the right equipment to take good measurements.
It seems like he's not having access to a digital oscilloscope to perform the right metering.
Inductive kickback is very well understood and it's the basis for all switched power supplies and it have been studied for to long
and by to many (even using bifilar winding) to miss any sign of overunity from the flyback pulse. I'm sure Slayer007 is wrong here.

[terry1094]

Hi Ergo,

Well I try and hope Metalspider agrees here....

He asked what the COP of a magnet motor was.

Motor Setup: 

the rotor is of one arm and at both ends there are two permanent magnets fixed.   
The stator is one electromagnet.
                     
His measurements: 

output torque is 0.31 Nm at 193 RPM

input to the electromagnet in case of one of the magnets: 254V @2A   0.4msec pulse
input to the electromagnet in case of the other magnet: 262V @ 9A   1.2msec pulse

Maybe I am mistaken but I figured a COP of about 2.

rgds,  Gyula
           

Ein= (254x2x0.0004)+(262x9x0.0012) = 3.03 Joules (per cycle)

Eout= 0.31*2*pi=1.95 Joules (per cycle)

COP = Eout/Ein = 1.95/3.03 = 0.64