Sharing ideas on how to make a more efficent motor using Flyback (MODERATED)

[woopy]

Hi Luc

amazing work

If i understand well.

1- the motor is idling and spins the generator under no load and the input power is around 9 watts.

2- When you load the generator, the input power to the motor is always around 9 watts, but now the generator outputs 5 watts
     So it seems that 5 watts are created at the expense of only a decrease in RPM but not due to an increase in input power???

3- The flyback energy is not used in this test, i mean that you only redirect the flyback energy to the charge battery but not to the     génerator or the motor ?

This is a "scratching head" and  very interesting test

Thank's for sharing

Laurent

[gyulasun]

Hi Luc,

I think that regardless of the unchanging input power we could calculate efficiency in a 'normal way'.  That is, the total input power is 3 x 9.4 W = 28.2 W (assuming the other two phases consume similarly 383 mA current from 24.3 V battery voltages which is also 9.4W for each).

Then the output power consists of the 5 W AC output from the alternator but we do not know the self efficiency of this alternator so I assume it has say 60%, this means the shaft load on your motor may have been equivalent to 5 W times 1.4= 7W output instead of the 5W dissipated in the 5 Ohm.

And the flyback recovery output which presently charges the other batteries is also an output, and to be more precise here, the charging batteries could be replaced also with real power resistors that would represent equivalent load for the flyback output the batteries do now.  So summing up these  V² /R_load  outputs with the roughly 7W load on the shaft would give total output. If this sum is equal or higher than the total input power, then you surely have a very decent achievent. 

Of course I do not mean to lessen the significance of the unchanging input power under the load of the alternator but this would be really significant if the total ouput would be at least comparable to the total input.

Hi Laurent,

the input power to the motor is 3 times the 9 Watts because the other two phases also consume the same 9 W from the other two battery banks (there are 3 separate battery banks for the input).

Gyula

[gotoluc]

Hi Luc

amazing work

If i understand well.

1- the motor is idling and spins the generator under no load and the input power is around 9 watts.

2- When you load the generator, the input power to the motor is always around 9 watts, but now the generator outputs 5 watts
     So it seems that 5 watts are created at the expense of only a decrease in RPM but not due to an increase in input power???

3- The flyback energy is not used in this test, i mean that you only redirect the flyback energy to the charge battery but not to the     génerator or the motor ?

This is a "scratching head" and  very interesting test

Thank's for sharing

Laurent

Bonjour Laurent,

I wish it was only 9 Watts total input but it is 9 Watts per phase. So 9w x 3 = 27 Watts total input.

I checked the Flyback Recovery and it is 4 Watts per Phase. So 4w x 3 = 12 Watts total Recovery.

So the motor uses 15 Watts to turn the Alternator under no load.

When the alternator is on load, the motor rpm decreases which lowers the frequency of the Flyback so now the Recovery drops to 3 Watts per phase. So 3w x 3 = 9 Watts total Recovery, so we lost 3 Watts with the frequency drop, however, we are delivering 5 Watts on our Alternator load.

Summary:

Input is 27 Watts in all conditions
Recovery is 12 Watts (no load)

With Alternator on load delivers 5 Watts
Recovery is 9 Watts (on load)

So the motor with NO load is 27w - 12w recovery = 15 Watts
and with motor ON load it is  27w - 14w recovery =  13 Watts


Hope this helps better explain the results?
This is just a first basic test and I will get some more precise measurement data in the next several days.

Luc

[Over Goat]

So the motor with NO load is 27w - 12w recovery = 15 Watts
and with motor ON load it is  27w - 14w recovery =  13 Watts


Luc

so it is looking to be 50% more efficient at least , at least 50% recovery?
and over 50% recovery in the second instance. 
are there tweaks which you could to get this even better, not that this is not amazing enough,
great work Luc

[Magluvin]

Bonjour Laurent,

I wish it was only 9 Watts total input but it is 9 Watts per phase. So 9w x 3 = 27 Watts total input.

I checked the Flyback Recovery and it is 4 Watts per Phase. So 4w x 3 = 12 Watts total Recovery.

So the motor uses 15 Watts to turn the Alternator under no load.

When the alternator is on load, the motor rpm decreases which lowers the frequency of the Flyback so now the Recovery drops to 3 Watts per phase. So 3w x 3 = 9 Watts total Recovery, so we lost 3 Watts with the frequency drop, however, we are delivering 5 Watts on our Alternator load.

Summary:

Input is 27 Watts in all conditions
Recovery is 12 Watts (no load)

With Alternator on load delivers 5 Watts
Recovery is 9 Watts (on load)

So the motor with NO load is 27w - 12w recovery = 15 Watts
and with motor ON load it is  27w - 14w recovery =  13 Watts


Hope this helps better explain the results?
This is just a first basic test and I will get some more precise measurement data in the next several days.

Luc

Hey Luc

Seems odd that recovery from the motor, with the other motor added, is less, because the switching on times should be longer if the motor is slowed down by the other loaded motor.  Not sure. Maybe Im not fully understanding.  Just thinking longer on time for the coils shouldnt end with less bemf than free running short pulses. Possibly the number of pulses per second being less as the rpm is lower is the culprit as in total bemf output. Like if you end up with the same out per pulse, whether it is high or low rpm, the high will have more pulses equaling more out over time.
But neat stuff to say the least.

Mags