Specifications

Double layer windings

Lap winding

Problem 1 : Develop the layout of a lap winding for a 3 phase ac machine having 4 pole and 24

slots. There are 2 coil sides per slot.

Solution : Coil groups per phase = 3 x 4 = 12

Slots per pole per phase, m = = 2

Angle between adjacent slots, â =

180

= 30

ε.

For full pitch coils, the coil span, á = 0, ie., angle between the two sides of the same coil is

180

corresponds to = = 6 slots.

There are 3 phase groups per pole, each comprising of 2 slots. The distribution of slots, of

phase sequence RYB or ABC, is shown in Table 4.4.

Table 4.4 Polar Groups

Poles N1 S1 N2 S2

Phase

R 1,2 7,8 13,14 19,20

B 3,4 9,10 15,16 21,22

Y 5,6 11,12 17,18 23,24

The start of the phases must be displaced by 120

and so must be finishes. If the start of

R or A phase lies in slot no. 1, the start of Y or B phase must be in slot no. (1 +

120

=) 5 and

that of B or C phase in slot no. (5 +

120

= ) 9. This makes the phase sequence RYB or

ABC.

The top coil side in slot no. 1 is to be connected to bottom coil side in slot no. (1 + 6 =) 7 or

back pitch, Y

= 13, in terms of coil sides, ie., if slot no. 1 is at the beginning of the first North

Pole, N1, the slot no. 7 will be at the beginning of the first South Pole, S1.

The winding pitch, Y = +2 (progressive winding).

The front pitch, Y

= Y

– Y = 13 – 2 = 11.

Now, coil sides 1 and 14 form a coil. Coil side 14 is connected to coil side (14 – Y

= ) 3, and

coil side 3 is connected to coil side (3 + Y

= ) 16. So coil sides 3 and 16 form the second and

the last coils of this pole phase groups.

3x4

180