Kundel Motor

[gn0stik]

Fans are very easy to turn,
you don?t need 20 Watts for this, you only have the low wind friction !
Until no prony brake measurements are made, these
above calculations are just speculations,
for instance if the normal 30 Watts fan motor
would only have a 20 to 30  % efficiency the
calculation would be quite different, also it is a question,
if the 30 Watt fan motor would also run at 30 Watts
input all the time or if the input is lower, but the motor
is just rated at maximum 30 Watts input....
questions over questions..

NO THEY ARE NOT.. The fan motor calcs are NOT speculation. IT'S HARD MATH. Just because a blade is easy to spin doesn't mean the motor is not capable of more work than that. Torque curves are based on RPM, PERIOD. More RPM less torque, and vice versa.. Efficiency losses are based on LOAD. Both motors calcs were figured with NO LOAD. And I even allowed for losses, and load in the effective output.

[hartiberlin]

Yes, but you don?t know, at what RPM the Kundel motor really is rotating
and what the windfriction loss power is at that RPM, probably less than 2 Watts...

[gn0stik]

Yes, but you don?t know, at what RPM the Kundel motor really is rotating
and what the windfriction loss power is at that RPM, probably less than 2 Watts...

I got the figures for RPM from his forums, and yes, you are correct, I don't know what the wind friction, and rpm figures are under load, so that's why I did my calculations based on no load. Remember his motor consumes less power under load, due to the half duty cycle, and lower rpms, and hence slower switching speed/power consumption. This improves efficiency under load, as compared to a fan motor, rather than decrease it.

Again, lack of data forced me to do calculations with no load, and use known values of existing conversion methods estimate the effective output. I was very conservative in my calculations, and never gave his motor the benefit of any doubt. My effective output took 40-60% losses into consideration. Even if I halved that conversion efficiency and took values of 70-80% losses, it would still be 200-375% effective efficiency. Let's be extremely conservative, and say the conversion to electricity is only 10-15% efficient. The output would be 100 to 188% effective efficiency. That's after 85-90% losses during conversion.

The only thing that would make the calculations before conversion to electricity invalid would be his estimated torque.

Understand I'm approaching this very skeptically, but the math doesn't lie.

The only way to know for sure would be to build it, and close the loop. I have already said that. These calculations were simply to get an idea of whether or not it would be worth the effort. And the figures were very promising.