second problem

A 3 phase, 16 pole, star connected alternators has 144 slots on the armature periphery. Each slot contains 10 conductors. It is driven at 375 r.p.m. The line value of e.m.f. available across the terminals is observed to be 2.657 kV. Find the frequency of the induced e.m.f. and flux per pole.

Solution :

                          P = 16,                      N_s  = 375 r.p.m.

       Slots = 144,    Conductors / slots = 10

       E_line  = 2.657 kV

       Ns  = 120f/P

.^..     375 = (120 x f)/16

.^..     f = 50 Hz

       Assuming full pitch winding , K_c = 1

.^..     n = Slots/pole = 144/16

           = 9

.^..     m = n/3

            = 3

.^..     β = 180^o/9

          = 20^o

          Kd= 0.9597

       Total conductors = Slots x condutors/Slot

       i.e.              Z = 144 x 10 = 1440

.^..        Z_ph  = Z/3 = 1440/3

              = 480

            T_ph    = Z_ph /2 = 480/2

                             = 240

            E_ph  = E_line/√3 = 2.657/√3

                                = 1.534 kV

Now      E_ph  = 4.44 K_c K_d  f Φ T_ph 

.^..           1.534 x 10^-3 = 4.44 x 1 x 0.9597 x Φ x 50 x 240

.^..           Φ = 0.03 Wb

                  = 30 mWb