There are many much more interesting questions related to the pole number of induction motors, e.g.:

1. Does the induction motor supplied by the main grid (say, 50 Hz) increases its torque capability in "p" times with growing pole number "p" since its speed decreases in "p" time (like in a gearbox)?

2. Let's we have an induction motor with p=2 and feed it from 50 Hz grid. Then we re-connect the winding coils to arrange p=4 and feed if grow 100 Hz grid. Are performances of these two motors different or the same? Please note excluding frequency and inter-coil connections all remained the same.

It depends on the required speed. *n* (rpm) = (60 x *f*) / *N* where:- *f* = frequency and *N* = number of pole pairs. The 60 is there to convert from revolutions per second to revolutions per minute as the frequency is in cycles per second. Pole pairs is there because that any pole must be constructed in a pairs top and bottom / left right, so with one cycle it will move half the distance.

If you are using 50Hz and have a two pole motor 60 x 50 / 1 = 3000 rpm. The induction motor will run at a slight less speed due to "slip" which is what gives the motor its torque. For example 5.5kW, 400v, 2 pole motor will run at approximately 2880 rpm.

For a four pole machine, 60 x 50 / 2 = 1500 rpm so the same size motor at 5.5kW, 400v but 4 poles will have a nominal speed of 1500rpm but will run near 1455 rpm.

When selecting a three phase motor, the number of poles is chosen to achieve the speed of rotation that you require. Here are two tables, one for a 50 Hz power supply and one for a 60 Hz power supply:

The formula is *n* = 60 x *f */*p* where *n* = synchronous speed; *f* = supply frequency & *p* = pairs of poles per phase. The actual running speed is the synchronous speed minus the slip speed.

For a 50 Hz three phase supply:

2 poles or 1 pair of poles = 3,000 RPM (minus the slip speed = about 2,750 RPM or 6 -7% *n*)

4 poles or 2 pairs of poles = 1,500 RPM

6 poles or 3 pairs of poles = 1,000 RPM

8 poles or 4 pairs of poles = 750 RPM

10 poles or 5 pairs of poles = 600 RPM

12 poles or 6 pairs of poles = 500 RPM

16 poles or 8 pairs of poles = 375 RPM

For a 60 Hz three phase supply:

2 poles or 1 pair of poles = 3,600 RPM (minus the slip speed = about 2,750 RPM or 6 -7% *n*)

4 poles or 2 pairs of poles = 1,800 RPM

6 poles or 3 pairs of poles = 1,200 RPM

8 poles or 4 pairs of poles = 900 RPM

10 poles or 5 pairs of poles = 720 RPM

12 poles or 6 pairs of poles = 600 RPM

16 poles or 8 pairs of poles = 450 RPM

To determine the number of poles, you can read the data plate directly or calculate it from the RPM stated on the data plate or you can count the coils and divide by 3 (poles per phase) or by 6 (pairs of poles per phase). Where the power of the induction motor is constant, the torque increases at the rate that the speed decreases.

With the advent of variable frequency drive (VFD), you can have any frequency / rated volts you desire. I often see name plates with things like 575VAC, 42.5 Hz etc. When these "specials" are made I usually see 6 pole machines - but that may be just a manufacturer's preference.

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No of coil set in 3phse AC Motor

=no of poles×no of phase.

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How much pole i need

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Number of pole ?

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My point is we should surely have a formula relating the following:-

No of turns / coils per pole to No to No of poles.

In fact I have a 1750 and 420 rmp 19.3 Hp . 48 slots Two speed ac motor. The 420 rmp winding has been stripped off with no trace it.

Does your Easy formula apply in this case?

No of coil set in 3phase Ac motor=no of poles X NO of Phase. Quite confusing "coil set" and " no of turns per pole"

Anyway thanks for to consider my case. Regards Farook.

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If we got a 400v ,50Hz 4 pole ...?

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As well as how to slip percent ?....

awating kind cooperation ..

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B. Melon

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Only speed at no load and at full load.

How can I determine number of pole in induction motor?

Eg. A three-phase 60 hz induction motor runs at 715 r/min and at 670 r/min at full load. How many poles does this motor have?

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# of poles x 3 phase = # of groups

# of slots / # of groups = # of coils in each group

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Thanks for the page its' GREAT

Happy New Year to you all.

Ha!Ha! No I'm not a robot.

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NO.OF TURNS 266/126

NOW,CALCULTE TOTAL IMPEDANCE OF I.MOTOR ?

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