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Hi metalspider,
just wanted to write you seem to have a COP of 2, if I am not mistaken.
I wonder why you deleted your text?
rgds, Gyula
Hi.
What was his words before deleting his text....
Can you remember and recreate it, Gyula?
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
Interesting, a new magnet motor or just the old trusty one!
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
Quote from: gyulasun on December 10, 2008, 05:18:21 PM
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.
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
Quote from: gyulasun on December 10, 2008, 06:22:01 PM
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.
Quote from: gyulasun on December 10, 2008, 05:18:21 PM
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
Quote from: terry1094 on December 17, 2008, 08:56:22 PM
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
I agree with your input energy calculations.
You may wish to rethink your output energy calculations.
Once the RPM is given, you have to consider it.
See this link, out of many : http://www.engineeringtoolbox.com/electrical-motors-torques-d_651.html
rgds,
Gyula
Okay, I'll rethink it. Done. Nope, it doesn't matter how fast the motor is spinning as long as your energy calculations are done per unit. In this case the per unit calc is one revolution.
Think of it this way, work is force times distance. With the newton-meter, you are applying one newton of force to a wheel with a radius of one meter. The distance over which the force acts in a single cycle is the circumference of the wheel, pi x dia, or 2 x pi x the radius.
Hence, the mechanical energy expended in rotating the wheel one complete revolution is Newton x 2 x pi x radius. It makes no difference in the ENERGY calculation whether the wheel turns its cycle in one second or one year. The same amount of energy is expended.
The RPM does come into play when using POWER to calculate COP (a far more difficult calculation). Then you would take the Joules (which is a Watt-second) times the revolutions per second to get average power. But the Ein calculation becomes a bear since you have to average the pulses over the time for a cycle to get input power.
Nope, stick with the energy calculation. Far simplier.
Terry
I understand want you mean.
But if the input pulses becomes shorter at higher RPM and average power stays the same while the output
increases as the RPM increases at the same accessible torque it should theoretically increase COP.
But then again, the average input pulse power will most likely increase as the RPM increases due to the
affect of the electromagnet inductance. It takes more power to run the EM faster. Right or wrong?
Try a couple of COP calculations using power. You'll find that RPM is in both the numerator and denominator and thusly cancels.
As far as what happens to available torque in reality as you increase RPM, it must surely decrease as you reach the mechanical limits of the bearings and windage.
I have never built one; so, I do not know. But, yes, it should take more input power to run faster.
Terry