Problem

An alternator runs at 250 r.p.m. and generates an e.m.f. at 50 Hz. There are 216 slots each containing 5 conductors. The winding is distributed and full pitch. All the conductors of each phase are in series and flux per pole is 30 mWb which is sinusoidally distributed. If the winding is star connected, determine the value of induced e.m.f. available across the terminals.

Solution :

           N_s  = 250 r.p.m. ,    f = 50 Hz

           N_s  = 120f/P

.^..        250 = (120 x 50)/P

.^..         P = 24

.^..         n = Slots/Pole = 216/24 = 9

.^..         m = n/3 = 3

           β = 180^o/9 = 20^o

             Kd = 0.9597

           K_c  = 1 as full pitch coils.

       Total no. of conductors  Z = 216 x 5 = 1080

.^..          Z_ph  = Z/3 = 1080/3

                     = 360

              T_ph   = Z_ph/2                       ..... 2 conductors → 1 turn

                            = 360/2 = 180

.^..           E_ph   = 4.44 K_c  K_d f Φ  T_ph.

                      = 4.44 x 1 x 0.9597 x 30 x 10^-3 x 50 x 180

                      = 1150.48 V

              E_line   = √3 E_ph                               ........... star connection

                        = √3 x 1150.48

                        = 1992.70 V.