Quote:
Originally Posted by ale0502
If i calculate:
380V / 8,3A = 45,7ohms (resistance of coil)
so:
(120V / sqrt(2)) / 45,7ohms = 1,85A (new current with 120V)
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The impedance (resistance plus reactance from the inductor in the motor coil) calculates to 45,7 ohms at 50 Hz. The resistance by itself should be a minor part of that. The impedance rises as the frequency rises.
Quote:
My motor is a 2 poles, if I re-connect the series coils all in parallel, need only the middle of voltage. So the power is:
P = (2*120V) * sqrt(2) * sqrt(3/2) * 1,85A * 0,88 * 0,83
P = 562w
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If the coils will draw 1,85 amps (with some assumptions, at 50 Hz) when in series, R + R, then each coil should draw 3,70 amps with connected in parallel as R // R, 7,40 amps total, so you should see much more power than you show.
P = 120V * sqrt(2) * sqrt(3/2) * ((2 *1,85A) + (2 * 1,85A)) * 0,88 * 0,83
P = 1123w
Quote:
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If run it up to 200Hz (with Paul´s controller), can give me approximately 2,2Kw.
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I don't think you will be able to get 7,4 amps into the motor with only 120V at 200 Hz, since the impedance will be at least 3 times as high.
A 2-pole motor at about 2950 rpm, 50 Hz would be spinning at over
11,000
rpm at 200 Hz. This may be dangerous! I would not personally go over about 5000 rpm .. and I'd test it the first time, on a bench instead of inside the car, with a sturdy barrier between me and the motor, and no one near the motor.
For lower speed use, you can get *MUCH* more than 7,4 amps for the 2 coils in parallel. If the motor is rated for 8,3 amps, you should be able to run twice that, 16,6 amps, with the coils in parallel. And it should run all day without overheating. From experience, induction motors willl take at least 3.5 times rated current for short periods of acceleration.
Using this calculator
Online calculator: Power into Torque and vise versa with 2950 rpm, 4 kw
I get the rated torque out of your motor to be 12.9 N m
With about 3.5 times the rated current (from experience) you can get about 3 times the rated torque (this is a rule of thumb
). So 58 amps should get you about 39 N m of torque until you run out of battery voltage. But with a first gear of 12:1 or so, you get over 400 N m of torque .. that should accelerate your vehicle quite nicely.
When Paul's controller can no longer push 58 amps since the battery is only 120V, your torque will drop and you will accelerate more slowly as you speed up. As long as you are below the rated current of 16.66 on your re-wound motor you should not overheat your motor.
If you are using a transmission on your vehicle, changing gears should let you get up to reasonable speed.
I hope to do some testing on a bit larger motor - 5 HP (3.7 KW) 2 pole, 460V motor - to verify the rule of thumb .. 3.5X current gives 3X torque. But that won't be until September or October (or maybe later!)
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