Question

1. Give an example design for a typical alternator, induction motor and synchronous motor in India.

( I'm not really sure if it is a diagram or what. But based on the question above it is a design.)

Answer 1

Design of Induction Motor

The Induction Motor consists of three stator windings. The windings are arranged in the same stator core. Stator winding designed for 415 V is connected across the supply till rated speed is attained. Based on the shaft load, other two coils are energized. Stator windings of three winding induction motor can be arranged with different shift angles. Three sets of windings are arranged with zero degree displacement between them to obtain optimum utilization.

Design of three phase three phase induction motor is affected by various constraints such as thermal limit, overload capacity and utility of stator slots. The energy conserving three stator winding induction motor is ideal to be used for low power operations due to the limitation in thermal insulation value. The proposed model uses semi-enclosed deeper slots.

The stator consists of three sets of three phase windings placed in the same stator slots and therefore slot utility is increased. Slot utility factor for the designed motor is 53% but for conventional induction motor, it is about 29%. The suggested design is suitable only for small capacity induction motors.

Design detail of stator

Capacity= 2.2kW

Rated voltage = 415V

Rated current= 4.5A

Speed = 1440 rpm

Stator slots, Ss =36 Rotor slots, Sr = 44 Diameter of bore = 150mm Length of the bore =120mm

Stator poles = 4

Air gap length = 1 mm

Standard wire used for winding 1 = 22 SWG

Standard wire used for winding 2 = 18 SWG

Standard wire used for winding 3 = 17 SWG

Design of first stator winding:

Operating voltage = 200 V

Rated current = 1.2A

Flux= Bav x L xt= 0.4 x 0.12x0.079 = 3.8 mWb

Eph= 4.44x 3.8x 10-3 x 50 x Tph x 0.955

115.5= 4.44 x 3.8 x 10-3 x 50 x Tphx 0.955

Tph= 144

Zss= 6 x Tph/ 36 = 24

Cross sectional area= 1.2/ 5.5 = 0.22 mm²Total conductor area = 0.22 x 24 = 5.28 mm²

Design of second stator winding:

Operating voltage = 350 V

Rated current = 3.5A

Flux= 0.4 x 0.079 x 0.12 76= 3.8 mWb

Tph= 250

Zss= 6 x Tph/ 36 = 42

Cross sectional area = 3.5/ 5.5 = 0.636 mm²

Total conductor area = 0.636 x 42 = 26.73 mm²

Design of third stator winding:

Operating voltage = 415V

Rated current = 4.5A

Flux= Bav x L x = 0.4 x 0.079 x 0.12 = 3.8 mWb

Eph= 4.44x 3.8x 10-3 x 50 x Tph x 0.955

Tph= 298 Zss= 6 x Tph/ 36 = 50

Cross sectional area= 4.5/ 5.5 = 0.818 mm²

Total conductor area = 0.818 x 50 = 40.91 mm²

Slot Dimensions Slot width = 7.9 mm

Slot depth = 17.4 mm

Slot area = Slot width x Slot depth = 7.9 x 17.4 = 137.46

Space factor = area of conductors / area of slot = 72.98/137.46 = 0.53 77

For conventional induction motor, the space factor will be around 0.29. For the designed three winding induction motor, the space factor is 0.53. Slot area is effectively utilized by the conductors.

The design is not as easy as you think. As the time is not enough I cant update more details here.