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.
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).
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| 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 :
Ns = 250 r.p.m. , f = 50 Hz
Ns = 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
Kc = 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 Kc Kd 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, Ns = 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 Kd 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
= 30 mWb
