SHEET 5
SOLVED EXAMPLES IN SYNCHRONOUS MACHINES
Example 1 :
Calculate the percent voltage regulation for a three-phase wye-connected 2500 kVA
6600-V turboalternator operating at full-load and 0.8 power factor lagging. The per
phase synchronous reactance and the armature resistance are 10.4 Ω and 0.071 Ω,
respectively?
Solution:
Clearly, we have XS >> R a, The phasor diagram for the lagging power factor
neglecting the effect of R a is shown in Fig. (a). The numerical values are as follows:
Vt = 6600 / √3 = 3810 V
Ia = ( 2500 * 1000 ) / ( √3 * 6600) = 218.7 A
Eo = 3810 + 218.7( 0.8 – j 0.6) j10.4 = 5485∟19.3º
Percentage regulation = ( 5485 – 3810 )/ 3810 *100 = 44%
Example 2 :
Repeat Ex.1 calculations with 0.8 power factor leading as shown in Fig. (b) ?
Solution:
Eo = 3810 + 218.7( 0.8 + j 0.6) j10.4 = 3048∟36.6º
Percentage regulation = ( 3048 – 3810 )/ 3810 *100 = -20%
Fff
H.W. : Repeat Ex.1 calculations with unity power factor , and draw the phasor
diagram for this case ?
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Example 3 :
A 20-KVA, 220 V, 60 Hz, way-connected three phase salient-pole synchronous
generated supplies rated load at 0.707 lagging power factor.The phase constants of
the machine are Ra = 0.5 Ω and Xd =2 Xq = 4 Ω. Calculate the voltage regulation at
the specified load. ?
Solution:
Vt = 220 / √3 = 127 V
Ia = 20000 / (√3 * 220) = 52.5 A
φ = cos-1 0.707 = 45º
tan δ = (Ia Xq cos φ) / (Vt + Ia Xq sin φ)
= 52.5 * 2 * 0.707 / ( 127 * 52.5 * 2 * 0.707)
= 0.37
δ = 20.6º
Id = 52.5 sin ( 20.6 + 45 ) = 47.5 A
Id Xd = 47.5 * 4 = 190 V
Eo = Vt cos δ + Id Xd
= 127 cos 20.6 + 190 = 308 V
Percent regulation = (Eo – Vt ) / Vt *100%
= ( 308 – 127 ) / 127 *100%
= 142 %
Example 4:
The stator core of a 4-pole, 3-phase a.c. machine has 36 slots. It carries a short pitch
3-phase winding with coil span equal to 8 slots. Determine the distribution and coil
pitch factors?
Solution:
Number of slots per pole = 36 / 4 = 9
Angular displacement between slots = β = 180º / 9 = 20º
Coil span = 20º * 8 =160º
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Therefore, pitch factor = k p = cos (θ/2)
k p = cos 10º = 0.985
,
θ = 180º – 160º = 20º
Now, number of slots per pole per phase,
m = 9 /3 = 3
Distribution factor,
kd = ( Length of long chord ) / (Sum of lengths of short chords )
= [sin (mβ/2)] / [m sin (β/2)]
= sin 30º / (3 sin 10º) = 0.96
Example 5:
3- phase, 50 Hz generator has 120 turns per phase. The flux per pole is 0.07
weber, assume sinusoidally distributed. Find (a) the e.m f. generated per phase. (b)
e.m.f. between the line terminals with star connection. Assume full pitch winding
and distribution factor equal to 0.96 ?
Solution:
(a) E.m.f. generated per phase = 4 k f k p k d f T Φp
volts
= 4 * 1.11 * 0.96 * 1 * 0.07 * 50 * 120
=1792 volts
(b) E.m.f. between line terminals= √ 3 E.m.f. generated per phase =√ 3 * 1792
= 3100 volts
Example 6:
A three phase, 16-pole, star-connected alternator, has 192 stator slots with eight
conductors per slot and the conductors of each phase are connected in series. The
coil span is 150 electrical degrees. Determine the phase and line voltages if the
machine runs at 375 r.p m. and the flux per pole is 64 mWb distributed sinusoidally
over the pole?
Solution:
The flux is sinusoidally distributed over the pole, hence from factor, k f = 1.11
Pitch factor , k p = cos [(180º - 150º) / 2 ] = cos 15º = 0.966
Distribution factor, kd = [sin (mβ/2)] / [m sin (β/2)]
m = Number of slots per pole per phase = 192 / (16 * 3) = 4
β = 180º / (No. of slots/pole) = 180º / (192/16) = 15º
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kd = [ sin (4 * 15º / 2)] / [4 * sin (15º/2)] = sin 30º / (4 * sin 7.5º) = 0.958
Now total number of conductors per phase
=Total number of slots per phase * number of conductors per slot.
= ( 192 / 3 ) * 8 = 512
Number of turns per phase = T = 512 / 2 = 256
Frequency of generated e.m.f. = f = P N / 120
=[ Number of poles * Synchronous Speed] / 120
=[ 16 * 375] / 120 = 50 cycle per second
Flux per pole = Φp=0.064 wb
Therefore, e.m.f. per phase = 4 k f k p k d f T Φp
volts
= 4 * 1.11 * 0.966 * 0.958 * 0.064 * 50 * 256
= 3367 volts
Hence for star connected alternator, Line voltage = √3 * 3367 = 5830 volts
Example 7:
A 2200 volt, star-connected, 50 Hz alternator has 12 poles. The stator has 108 slots
each with 5 conductors per slot. Calculate the necessary flux per pole to give 2200
volts on no-load. The winding is full pitch ?
Solution:
Slots per phase = 108 / 3 = 36
Therefore, conductors per phase = 36 * 5 = 180
and number of turns per phase = 180 / 2 = 90
Slots per pole per phase = 36 / 12 =12
Frequency,f =50 cycle per second
kf = 1.11
k p = 1.0 for full pitch winding
kd = [sin (mβ/2)] / [m sin (β/2)]
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Where β = 180º / (No. Of slots/pole) = 180º / (108/12) = 20º
m = Number of voltage vectors= Number of slots per pole per phase
= 108 / (12*3) = 3
kd = [sin (3*20º/2)] / [ 3 sin (20º/2)] = [sin 30º] / [3 sin 10º] = 0.96
Also e.m.f. per phase = Eph = 2200 /√3 volts
The e.m.f. equation for Alternator is
Eph = 4 k f k p k d f T Φp
volts
Φp = Flux per pole = Eph / [4 k f k p k d f T ] =( 2200 /√3) / [ 4*1.11*1*0.96*50*90]
= 0.066 wb
Example 8:
Find the number of armature conductors in series per phase required for the
armature of a 3 phase, 50 Hz, 10 pole alternator with 90 slots. The winding is to
be star connected to give a line voltage of 11,000 volts. The flux per pole is 0.16
webers?
Solution:
Number of slots per pole = 90 / 10 = 9
Therefore,
β = 180º / 9 = 20º
Slots per pole per phase, m = 9 / 3 = 3
Distribution factor,
kd = [sin (mβ/2)] / [m sin (β/2)]
kd = [sin (3*20º/2)] / [ 3 sin (20º/2)] = [sin 30º] / [3 sin 10º] = 0.96
Since there is no mention about the type of winding, the full pitched winding is assumed.
Hence , k p = 1.0
Also , Eph = 11000 / √3 = 6352 volts.
Now , Eph = 4 k f k p k d f T Φp
volts
T = [Eph] / [4 k f k p k d f Φp]
= [6352] / [4 * 1.11 * 0.96 * 1 * 0.16 * 50]
= 1863
Number of Armature conductors in serues per phase = 1863 * 2 = 3726
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Example 9 :
A six-pole, 3-phase, 60 cycle alternator has 12 slots per pole and 4 conductors
per slot. The winding is 5/6 th pitched. There is 0.025 wb flux entering the
armature from each north pole and the flux is sinusoidally distributed along
the air gap. The armature coils are all connected in series. The winding is starconnected. Determine the open circuit e.m.f of the alternatorper phase?
Solution:
Here, number of conductors connected in series per phase
=12 * 6 * 4 / 3 = 96
Therefore, number of turns per phase , T = 96 / 2 = 48
Flux per pole , Φp = 0.025 wb
Frequency,
f = 5 0 c yc l e p e r s e c o n d .
k f = 1.11
k p = cos [180º(1 – 5/6) / 2] = cos 15º = 0.966.
Number of slots per pole per phase, m = 12 / 3 = 4
β = 180º / 12 = 15º
Distribution factor ,
kd = [sin (mβ/2)] / [m sin (β/2)]
kd = [sin (4*15º/2)] / [ 4 sin (15º/2)] = [sin 30º] / [4 sin 7.5º] = 0.958
Hence induced e.m.f.,
Eph = 4 k f k p k d f T Φp
volts
= 4 * 1.11 * 0.966 * 0.958 * 0.025 * 50 * 48
= 295 volts.
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