questions regarding rewinding and stator insulation
questions regarding rewinding and stator insulation
Ok, bear with me a bit here...
Im rewinding a 415v 4kW 3 phase asyncronous AC motor down to 100v (96v, but the math works better this way)
Ive began cutting the old copper out of the stator, it currently has 66 turns per coil of #22 wire (0.75mm diameter) with the coils wound 2 in hand, so 33 turns at 415v, 8.4A at full load (4.2A per coil winding)
I figure that if I divide 415 by 33 I get 12.575, which should equate closely to volts per turn.
If I then calculate 100(Volts) divide by 12.575(V per turn) I get 7.95, so, 8 turns of wire at 12.575v on each coil should give me a nominal operating voltage of 100.6v
Essentially a rough ratio of turns/volts.
The area of #22 wire (0.75mm) is 0.41mm2
0.41 x 66 (wires total per slot) = 27.06mm2  This gives me the available space in the slots for the new conductors.
From there I figure that I need to put the thickest conductors I can in that area while still making 8 turns in my coil.
27.06 divide by 8 = 3.38mm2 per new conductor, #8 wire is 3.24mm2
3.24mm2 x 8 = 25.92mm2 total conductor area (compared to 27.06mm2 previously)
As voltage decreases current increases, so if I apply the same multiplication ratio to the current figures it should get me very close to where I want to be.
The motor takes 8.4A at 415v, 1380rpm. If I divide 415v by 100v i get 4.15. If I then mulitply 8.4A by 4.15 I get 37.8A which should be what the motor draws at 100v and 1400rpm (ignoring rotor slip)
From there I can roughly map out the current draw at different rpms
2800rpm  75.6A
4200rpm  113A
5600rpm  151.2A
7000rpm  189A
8400rpm  226.8
So, If my ramblings are correct, I need to rewind 8 turns with #8 wire and that'll bring my operating voltage down to 100v while still allowing sufficient current capacity.
Next up is the question about insulation, specifically the plastic/paper inserts in the slots of the stator.
I understand why they are there, but my question is, why does it have to be paper/plastic inserts? Could the same insulation be acheived with a paint on coating before inserting the coils?
Could I varnish the stator slots and then insert the coils? Or careful application of Mylar tape?
If I can reduce the thickness of the insulation strips I beleive I will be able to increase to the next conductor size, allowing a higher amperage capacity of the motor.
I dont know anything about anything, this is all self taught from reading, watching and playing, If you can see where ive gone wrong or gone completely off track, please explain it to me and share your knowledge, its how we all learn.
Im rewinding a 415v 4kW 3 phase asyncronous AC motor down to 100v (96v, but the math works better this way)
Ive began cutting the old copper out of the stator, it currently has 66 turns per coil of #22 wire (0.75mm diameter) with the coils wound 2 in hand, so 33 turns at 415v, 8.4A at full load (4.2A per coil winding)
I figure that if I divide 415 by 33 I get 12.575, which should equate closely to volts per turn.
If I then calculate 100(Volts) divide by 12.575(V per turn) I get 7.95, so, 8 turns of wire at 12.575v on each coil should give me a nominal operating voltage of 100.6v
Essentially a rough ratio of turns/volts.
The area of #22 wire (0.75mm) is 0.41mm2
0.41 x 66 (wires total per slot) = 27.06mm2  This gives me the available space in the slots for the new conductors.
From there I figure that I need to put the thickest conductors I can in that area while still making 8 turns in my coil.
27.06 divide by 8 = 3.38mm2 per new conductor, #8 wire is 3.24mm2
3.24mm2 x 8 = 25.92mm2 total conductor area (compared to 27.06mm2 previously)
As voltage decreases current increases, so if I apply the same multiplication ratio to the current figures it should get me very close to where I want to be.
The motor takes 8.4A at 415v, 1380rpm. If I divide 415v by 100v i get 4.15. If I then mulitply 8.4A by 4.15 I get 37.8A which should be what the motor draws at 100v and 1400rpm (ignoring rotor slip)
From there I can roughly map out the current draw at different rpms
2800rpm  75.6A
4200rpm  113A
5600rpm  151.2A
7000rpm  189A
8400rpm  226.8
So, If my ramblings are correct, I need to rewind 8 turns with #8 wire and that'll bring my operating voltage down to 100v while still allowing sufficient current capacity.
Next up is the question about insulation, specifically the plastic/paper inserts in the slots of the stator.
I understand why they are there, but my question is, why does it have to be paper/plastic inserts? Could the same insulation be acheived with a paint on coating before inserting the coils?
Could I varnish the stator slots and then insert the coils? Or careful application of Mylar tape?
If I can reduce the thickness of the insulation strips I beleive I will be able to increase to the next conductor size, allowing a higher amperage capacity of the motor.
I dont know anything about anything, this is all self taught from reading, watching and playing, If you can see where ive gone wrong or gone completely off track, please explain it to me and share your knowledge, its how we all learn.
 coulomb
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Re: questions regarding rewinding and stator insulation
I'm no motor expert, so I'll let others comment on the rest of your questions.
But 96 V sounds like a nominal battery voltage. I hope you realise that a 100V battery will produce less than 70 VAC phase to phase. Alternatively, for 100 VAC pp you'll need at least a 150 V battery. That's assuming you want to generate sine waves, and I think you do.
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Re: questions regarding rewinding and stator insulation
The current draw is not what gives you rpm it's voltage.
Current gives you torque.
I'm a bit rusty on motors but...
If you rewind the motor from 1400rpm@415V like that it will end up as a 1400rpm@100V. Meaning if you want to spin the motor at 8400rpm you will need more volts.
What you want to calculate is the rpm per volt.
The original motor is 415v@1400rpm
1400/415= 3.37rpm per Volt applied
To make it spin at 8400rpm it's going to need 2500V
If it's rewound down to be 100V at the 1400rpm it ends up being
1400/100=14rpm per Volt
To get up to 8400rpm
8400/14=600V
Current gives you torque.
I'm a bit rusty on motors but...
If you rewind the motor from 1400rpm@415V like that it will end up as a 1400rpm@100V. Meaning if you want to spin the motor at 8400rpm you will need more volts.
What you want to calculate is the rpm per volt.
The original motor is 415v@1400rpm
1400/415= 3.37rpm per Volt applied
To make it spin at 8400rpm it's going to need 2500V
If it's rewound down to be 100V at the 1400rpm it ends up being
1400/100=14rpm per Volt
To get up to 8400rpm
8400/14=600V

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Re: questions regarding rewinding and stator insulation
each winding wire also needs to be insulated relative to all the other turns of the same winding and that of other windings.
given that you will likely be increasing the voltage differential between turns due to a higher operating rpm, you will likely need more insulation on the wire
given that you will likely be increasing the voltage differential between turns due to a higher operating rpm, you will likely need more insulation on the wire
Matt
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Re: questions regarding rewinding and stator insulation
Coloumb, yea, 96v is my 'nominal' battery voltage, based roughly on what voltage the eventual controller will take. I figured it was somewhat easier to set a nominal voltage that was common across many controllers. I can wire my battery up pretty much any way I want, but currently looking at 7S4P Liion 18650 cell modules in parallel to give 25.9v nominal over 4 batteries in series. (Still investigating BMS)
Is the phase to phase voltage going to be a controlled function output of the controller? I know EV's are a complicated learning curve but from what ive seen its as 'simple' (lol) as matching a motor,controller and battery to the same voltage range, then the equipment does its magic as designed.
im not looking to build my own motor controller. While I could learn, I really dont want to, not at this stage while trying to learn everything else.
Maybe EV version 2, lol.
Perhaps the sharp corners of the stator cause an issue with the insulation breaking down? maybe a layer of mylar tape would also help?
eg, I'll strip the stator down and clean it all, then varnish it and cook it off, then i'll add a strip of mylar tape inside each slot, to give added protection at the corners where the wire bends.
Then i'll add the copper coil of enamelled wire, tie it up, make connections ect, then varnish and cook the whole stator assembled.
Thoughts? would this be a suitable alternative to plastic/paper? I suppose it would also transfer heat from the copper wire into the stator mass better too, so thats probably a small bonus.
I could possibly also machine all the sharp edges off, but thats probably a lot of complicated tool time.
But yes, Current gives torque. However at a higher RPM there would be a higher current draw to make that extra power. If the voltage stays the same and the frequency increases, the current should increase too
I'll have to think about my calculations for a bit
No, I didnt realise this, i'll have to research a bit into this to learn about it and how it'll affect things.I hope you realise that a 100V battery will produce less than 70 VAC phase to phase. Alternatively, for 100 VAC pp you'll need at least a 150 V battery. That's assuming you want to generate sine waves, and I think you do.
Is the phase to phase voltage going to be a controlled function output of the controller? I know EV's are a complicated learning curve but from what ive seen its as 'simple' (lol) as matching a motor,controller and battery to the same voltage range, then the equipment does its magic as designed.
im not looking to build my own motor controller. While I could learn, I really dont want to, not at this stage while trying to learn everything else.
Maybe EV version 2, lol.
Yes, I would be using enamelled wire, same as what is used in all motors, I guess im wondering why if the varnish coating on the wire is acceptable to insulate between shorting of the coils, it should also be acceptable to insulate the varnished insulated wires from the stator, the insulation layer would be thinner than plastic or paper allowing more copper to occupy the limited space in the slots.each winding wire also needs to be insulated relative to all the other turns of the same winding and that of other windings.
given that you will likely be increasing the voltage differential between turns due to a higher operating rpm, you will likely need more insulation on the wire
Perhaps the sharp corners of the stator cause an issue with the insulation breaking down? maybe a layer of mylar tape would also help?
eg, I'll strip the stator down and clean it all, then varnish it and cook it off, then i'll add a strip of mylar tape inside each slot, to give added protection at the corners where the wire bends.
Then i'll add the copper coil of enamelled wire, tie it up, make connections ect, then varnish and cook the whole stator assembled.
Thoughts? would this be a suitable alternative to plastic/paper? I suppose it would also transfer heat from the copper wire into the stator mass better too, so thats probably a small bonus.
I could possibly also machine all the sharp edges off, but thats probably a lot of complicated tool time.
I was under the impression is was frequency that gave you RPM?The current draw is not what gives you rpm it's voltage.
Current gives you torque.
But yes, Current gives torque. However at a higher RPM there would be a higher current draw to make that extra power. If the voltage stays the same and the frequency increases, the current should increase too
Now i see it written like that, i understand where i went wrong.I'm a bit rusty on motors but...
If you rewind the motor from 1400rpm@415V like that it will end up as a 1400rpm@100V. Meaning if you want to spin the motor at 8400rpm you will need more volts.
What you want to calculate is the rpm per volt
I'll have to think about my calculations for a bit
Re: questions regarding rewinding and stator insulation
random question too.
Why dont hairpin windings need 30+ turns to make a coil in a single stator slot?
from pics I see on google they only seem to have 4 or 5 winds to a coil and the coil seems to be spread over multiple stator slots?
Im kind of wondering if maybe I could hand make hairpin windings for this motor...
Why dont hairpin windings need 30+ turns to make a coil in a single stator slot?
from pics I see on google they only seem to have 4 or 5 winds to a coil and the coil seems to be spread over multiple stator slots?
Im kind of wondering if maybe I could hand make hairpin windings for this motor...
 coulomb
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Re: questions regarding rewinding and stator insulation
Yes. But there is a limit. The peak of an AC sine wave is √2 times the RMS value. It's a little more complex for threephase, but it turns out that the answer is still the same; you still need √2 times more DC voltage than AC voltage. There is also a voltage drop across the MOSFETs or IGBTs; budget about 5 V extra on the DC side.
Edit:
True. But you need to vary the volts pretty strictly with the rpm as well. At a certain rpm, there will be a certain backEMF (voltage "generated" by the motor), and you need to be applying just a little more than that voltage, or else the motor will draw ridiculously large current.I was under the impression is was frequency that gave you RPM?
Edit 2: The controller can use "field weakening" to reduce the backemf. The reduction of field will reduce the torque, but the extra speed will more or less cancel that effect, so you end up with very roughly constant power from the motor. Not all controllers handle field weakening properly or well. So IF your controller is good at field weakening, then you don't need nearly as much voltage for higher speed. A purpose designed invertermotor set can run at nearly constant power out to quite high speed, say 13,000 rpm. An industrial AC motor probably won't be safe beyond 6000 rpm, probably 5000 rpm or less, unless it is properly balanced.
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Re: questions regarding rewinding and stator insulation
High rpm bearings and get the stator precision balanced, it shall be done!
That should help out the top RPM.
The stator appears to be solid laminated steel with some longditudinal holes through the stack. Instinct tells me it should survive centripital forces from high speeds pretty well.
only one way to find out...
Re: questions regarding rewinding and stator insulation
That depends on torque
Higher frequency can mean less amps
Volts x amps = power = Torque x rpm/9.5
So if you increase the rpm (with frequency) and the power output is the same then torque needs to go down.
Ill round numbers here for easy maths
100V x 10A = 1000w = 19Nm x 500rpm/9.5
But if you increase frequency to make it spin at 1000rpm
100V x 10A = 1000w = 9.5Nm x 1000rpm/9.5
However if you increase the frequency you will also increase the impedance of the circuit so if you have a motor and you give it 100V @50hz which draws 10A and spins at 2000rpm. (1000W and 4.75Nm)
and you then apply 100V @ 100hz to get 4000rpm because the impedance increases the current will go down (I=V/Z)
And if the current goes down then power goes down (P=VI)
So you will be spinning faster but making less power.
Or if you want the same power you need to crank up the voltage.
I was trying to do the maths but I couldn't get it to work. I feel I'm missing something...
I'll keep working at it
Re: questions regarding rewinding and stator insulation
Edit 4
Ignore the numbers , Im not thinking straight apparently, thanks @coulomb for helping me see clearly what was staring me in the face.
Here's what I've got
415V x 10A = 4150W
50Hz 415V / 10A = 41.5 Ohms (total impedance and resistance)
Impedance @ 50Hz = 31.42
Impedance @ 100Hz = 62.83
(Edit 3, I've made an assumption here and that is what is stuffing me up)
At 50Hz
41.5^2 = R^2 + 31.42^2
R = 27 Ohms (actual wire resistance)
So at 100Hz
R(total)^2 = 27^2 + 62.83^2
R(total) = 68.4 Ohms @ 100hz
415V / 68.4 Ohms = 6A draw at 100hz
415V x 6A = 2490W
You have doubled the rpm there is actually only 2/3 of the power available so the torque available has taken a dive.
To get the power back up you will need to increase the voltage as well.
Edit
I've got to be missing something. (or doing it wrong)
If I bring it down to 100V 2a it's 200W @50Hz and 127W @100Hz.
Seems too low
Edit 2
I think my problem is the assumption about the impedance.
Edit 3
Yeah, 100V with a total resistance of 41.5 Ohms
Makes for 2.4A. That's only 240W. But you are rewinding it so the resistance and impedance will be all over the show. So the numbers I made don't work but the theory still holds true, your going to need more Volts as well as frequency
Edit 4
As above, don't listen to the calculations...
Ignore the numbers , Im not thinking straight apparently, thanks @coulomb for helping me see clearly what was staring me in the face.
Here's what I've got
415V x 10A = 4150W
50Hz 415V / 10A = 41.5 Ohms (total impedance and resistance)
Impedance @ 50Hz = 31.42
Impedance @ 100Hz = 62.83
(Edit 3, I've made an assumption here and that is what is stuffing me up)
At 50Hz
41.5^2 = R^2 + 31.42^2
R = 27 Ohms (actual wire resistance)
So at 100Hz
R(total)^2 = 27^2 + 62.83^2
R(total) = 68.4 Ohms @ 100hz
415V / 68.4 Ohms = 6A draw at 100hz
415V x 6A = 2490W
You have doubled the rpm there is actually only 2/3 of the power available so the torque available has taken a dive.
To get the power back up you will need to increase the voltage as well.
Edit
I've got to be missing something. (or doing it wrong)
If I bring it down to 100V 2a it's 200W @50Hz and 127W @100Hz.
Seems too low
Edit 2
I think my problem is the assumption about the impedance.
Edit 3
Yeah, 100V with a total resistance of 41.5 Ohms
Makes for 2.4A. That's only 240W. But you are rewinding it so the resistance and impedance will be all over the show. So the numbers I made don't work but the theory still holds true, your going to need more Volts as well as frequency
Edit 4
As above, don't listen to the calculations...
Last edited by brendon_m on Sun, 09 Feb 2020, 14:53, edited 1 time in total.
 coulomb
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Re: questions regarding rewinding and stator insulation
Err, that's like an equivalent impedance for the mechanical load (plus a small winding resistance and reactance).
If the impedance doubled at double the frequency, then it must be a pure inductance, which means zero load. The above is simply invalid.Impedance @ 100Hz = 62.83
The actual resistance can't be nearly that high. At 10 A, you'd be dropping V = I·R = 10·27 = 270 V across the resistance. That's pure loss. That figure doesn't pass the sanity check.R = 27 Ohms (actual wire resistance)
I think the essential thing you're missing is the effect of back EMF. The same thing happens in a DC motor. Imagine 10 A into a 100 V DC motor. Is the resistance of the motor windings therefore 10 Ω? No! Because then all the 1000 W would be lost in the resistance heating. There's something like 98V of back EMF, so only 2 V appears across the internal resistance of the motor. R = V / I = 2.0 / 10 = 0.2 Ω = 200 mΩ. So the resistance loss would in this case be P = I²R = 10·10·0.2 = 20 W, or 2% lost to resistance heating.I've got to be missing something. (or doing it wrong)
Edit: if you like, the resistance equivalent to the mechanical load is 98 Ω, in series with the internal resistance of the motor (2 Ω), causing the 100 / (98 + 2) = 10 A of current to flow. But it's 98 Ω due to the particulars of this example; the same mechanical load could look like a different resistance in another example. So resistance (and impedance for AC motors) isn't helpful in describing the load.
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Patching PIP4048/5048 inverterchargers.
Re: questions regarding rewinding and stator insulation
coulomb wrote: ↑Sun, 09 Feb 2020, 14:36Err, that's like an equivalent impedance for the mechanical load (plus a small winding resistance and reactance).
If the impedance doubled at double the frequency, then it must be a pure inductance, which means zero load. The above is simply invalid.Impedance @ 100Hz = 62.83
The actual resistance can't be nearly that high. At 10 A, you'd be dropping V = I·R = 10·27 = 270 V across the resistance. That's pure loss. That figure doesn't pass the sanity check.R = 27 Ohms (actual wire resistance)
I think the essential thing you're missing is the effect of back EMF. The same thing happens in a DC motor. Imagine 10 A into a 100 V DC motor. Is the resistance of the motor windings therefore 10 Ω? No! Because then all the 1000 W would be lost in the resistance heating. There's something like 98V of back EMF, so only 2 V appears across the internal resistance of the motor. R = V / I = 2.0 / 10 = 0.2 Ω = 200 mΩ. So the resistance loss would in this case be P = I²R = 10·10·0.2 = 20 W, or 2% lost to resistance heating.I've got to be missing something. (or doing it wrong)
Edit: if you like, the resistance equivalent to the mechanical load is 98 Ω, in series with the internal resistance of the motor (2 Ω), causing the 100 / (98 + 2) = 10 A of current to flow. But at twice the speed, that equivalent resistance doesn't double.
Yep, back emf is what I wasn't taking into account (and the energy that goes to the wheels, apparently actual mechanical work is being done, it's not just a crappy heater)
Thanks for setting me straight. I just couldn't get the numbers to work. And it's because I was trying to force them into a line without taking a step back and thinking about why it wasnt working

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Re: questions regarding rewinding and stator insulation
Likely Becuase the voltage between windings is 5 to 10v, but the voltage between coils is 400v.Alan82 wrote: ↑Sun, 09 Feb 2020, 11:39Yes, I would be using enamelled wire, same as what is used in all motors, I guess im wondering why if the varnish coating on the wire is acceptable to insulate between shorting of the coils, it should also be acceptable to insulate the varnished insulated wires from the stator, the insulation layer would be thinner than plastic or paper allowing more copper to occupy the limited space in the slots.each winding wire also needs to be insulated relative to all the other turns of the same winding and that of other windings.
given that you will likely be increasing the voltage differential between turns due to a higher operating rpm, you will likely need more insulation on the wire
Matt
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1998 prius  needs Batt
1999 Prius  needs batt
2000 prius  has 200 x headway 38120 cells
Re: questions regarding rewinding and stator insulation
But the coils of varnished wire sit directly next to each other in the heads, tied together with string, all 3 phases side by side. so that proposition doesnt hold.antiscab wrote: ↑Sun, 09 Feb 2020, 17:41Likely Becuase the voltage between windings is 5 to 10v, but the voltage between coils is 400v.Alan82 wrote: ↑Sun, 09 Feb 2020, 11:39Yes, I would be using enamelled wire, same as what is used in all motors, I guess im wondering why if the varnish coating on the wire is acceptable to insulate between shorting of the coils, it should also be acceptable to insulate the varnished insulated wires from the stator, the insulation layer would be thinner than plastic or paper allowing more copper to occupy the limited space in the slots.each winding wire also needs to be insulated relative to all the other turns of the same winding and that of other windings.
given that you will likely be increasing the voltage differential between turns due to a higher operating rpm, you will likely need more insulation on the wire
Besides, im talking about the insulation between the coils and the stator housing, not each coil. The voltage difference would be the operating voltage of the motor attempting to ground itself to chassis. That voltage difference would be the same as between one of 2 phases.
Im not sure you understand what it is that im querying.