Tuesday, February 21, 2012

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 √3Eph where Eph  is induced e.m.f. per phase.
       While if it is delta connected line value of e.m.f. is same as Eph .
       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 = 250 r.p.m. ,    f = 50 Hz
           N = 120f/P
...        250 = (120 x 50)/P
...         P = 24
...         n = Slots/Pole = 216/24 = 9
...         m = n/3 = 3
           β = 180o/9 = 20o
              = 0.9597
           K= 1 as full pitch coils.
       Total no. of conductors  Z = 216 x 5 = 1080
...          Zph  = Z/3 = 1080/3
                     = 360
              Tph   = Zph/2                       ..... 2 conductors 1 turn
                            = 360/2 = 180
...           Eph   = 4.44 KKd f Φ  Tph.
                      = 4.44 x 1 x 0.9597 x 30 x 10-3 x 50 x 180
                      = 1150.48 V
              Eline   = √3 Eph                               ........... 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 = 375 r.p.m.
       Slots = 144,    Conductors / slots = 10
       Eline  = 2.657 kV
       Ns  = 120f/P
...     375 = (120 x f)/16
...     f = 50 Hz
       Assuming full pitch winding , Kc = 1
...     n = Slots/pole = 144/16
           = 9
...     m = n/3
            = 3
...     β = 180o/9
          = 20o
          = 0.9597
       Total conductors = Slots x condutors/Slot
       i.e.              Z = 144 x 10 = 1440
...        Zph  = Z/3 = 1440/3
              = 480
            Tph    = Zph /2 = 480/2
                             = 240
            Eph  = Eline/√3 = 2.657/√3
                                = 1.534 kV
Now      Eph  = 4.44 Kc K f Φ Tph 
...           1.534 x 10-3 = 4.44 x 1 x 0.9597 x Φ x 50 x 240
...           Φ = 0.03 Wb

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