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.
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 }.
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
= 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
= 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
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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