What's the way to wind an 8-pole induction motor?

First of all think of a DC Motor. If it is an 8-pole motor is will have 8 field coils evenly spaced all the way around it. Now, in an AC Motor with 3-phases you will have the 8-poles 3 times. That means you would have 24-groups of coils. If the motor has 48 slots and 48 coils you would wind the stator with 24 groups of 2 coils. If you were to go round the 24 groups of coils and number them 1, 2, 3 repeat all the way around you would have 8 number 1's, 8 number 2's and 8 number 3's. All the number 1's are phase number 1. All the number 2's are phase number 2. All the like numbers are connected together in series leaving only the start of the phase and the end of the phase. To complete the circuit the phases are connected in series or parallel depending on the winding configuration. Induction motor
In a basic 2-coils per slot configuration the number of groups of coils in a phase will match the number of poles. Multiply that by the number of phases and you have the number of groups of coils. E.G. 4 pole X 3-phases - 12 groups of coils.

Due to the increase induction motor pole number and thus windings as stated above, the size of the motor increases and so does the frame size. This means that you can't necessarily directly replace a 4 pole motor with an 8 pole motor and expect it to fit into the same base holes or have the same shaft center line. The greater the number of poles, the greater the torque. Torque = Mechanical power x 9550 / speed. So the slower it goes the more torque you have for the same power. Because of the large available torque, the shaft is thicker (Larger diameter).

30kW, 525v. 2 pole = 97Nm. 4 pole = 194Nm. 6 pole = 290Nm. 8 pole = 387Nm.

The speed is related to the pole pairs and the frequency with a bit of slip. N = (Frequency x 60)/pole pairs. 60 converts from cycles per second to cycles per minute and then the speed becomes revolutions per minute. For 50Hz, (50 x 60) / 4 = 750 rpm. This is the Nominal speed. The induction motor will run slightly slower due to slip. On the same 30kW, 525v, 8 pole motor the speed will be 740 rpm. These would be used on fans, mixers, shredder, conveyors and axial flow pumps.

If you need to run a motor at say 300 rpm to 1500 rpm and want to use a variable frequency drive (VFD), and you select a standard 4 pole motor to run at 1500 nominal your frequency will range from 10Hz to 50Hz. At this slow speed, cooling becomes a problem as the fan on the shaft is only going at 1/5th the speed, 1/25th of the air and the motor can fail due to overheating. If you select an 8 pole motor, nominal speed 750 rpm, the same 300 to 1500 rpm relates to 20Hz to 100Hz. The cooling is not a problem and the motor can quit happily run at 2 x nominal speed without a bearing change. When a motor running on a VFD is operating at less than 50% of its rated speed I always recommend a forced air blower. They are readily available off the shelf nowadays.

I find that 6 pole motors are still stock but there are very few 8 pole motors stock, especially the larger kilowatts. So the answer may vary from increased torque, don't require a gearbox, or VFD application.

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8/26/2019 2:39 PM
You couldn't explain winding logic straightforward. if we have 48 slots 3 phase 8 pole motor ;

48 poles /3 phases = 16 slots that will be filled per phase

Since this motor has 8 poles
in order to fill 16 slots with 8 poles
we will fill 2 slots for each pole
As as result 2 winding group x 8 poles = 16 slots

Do the Same thing to the other phases and in total 48 slots complete.