Line Value of Induced E.M.F.

E.M.F. Equation of an Alternator : Part5 
       If the armature winding of three phase alternator is start connected, then the value of induced e.m.f. across the terminals is √3E_ph where E_ph  is induced e.m.f. per phase.

       While if it is delta connected line value of e.m.f. is same as E_ph .

       This is shown in the Fig. 1(a) and (b).

Fig. 1

       Practically most of the alternators are star connected due to following reasons :

1. Neutral point can be earthed from safety point of view.

2. For the same phase voltage, available across the terminal is more than delta connection.

3. For the same terminal voltage, the phase voltage in star is 1/√3 times line value.

       This reduce stains on the insulation of the armature winding.

Example 1 : 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

              = 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.

Example 2 : 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

          = 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

Related Articles 

- E.M.F. Equation of an Alternator 
- Pitch Factor or Coil Span Factor (Kc) 
- Distribution Factor (Kd)
- Generalized Expression for E.M.F. Equation of an Alterntor

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