Microsoft Word - Synhronous Machines Units 5-8.doc

Synchronous Machine Dr. VishwanathHegde

Synchronous Machine Dr. VishwanathHegde

Synchronous Machine

Numerical Problems:

1.A 3, 50 Hz, star connected salient pole alternator has 216 slots with 5 conductors per slot. All the conductors of each phase are connected in series; the winding is distributed and full pitched. The flux per pole is 30 mwb and the alternator runs at 250 rpm. Determine the phase and line voltages of emf induced.

Slon: Ns = 250 rpm, f = 50 Hz,

P = 120 x f/Ns = 120 x 50/250 = 24 poles

m = number of slots/pole/phase = 216/(24 x 3) = 3

= 1800 / number of slots/pole = 1800 / (216/24) = 200

Hence distribution factor Kd = ( sin m/2) / (m sin /2)

= ( sin 3 x 20 / 2) / (3 sin 20/2)

= 0.9597

Pitch factor Kp= 1 for full pitched winding.

We have emf induced per conductor

Tph= Zph/2 ;Zph= Z/3

Z = conductor/ slot x number of slots

Tph= Z/6 = 216 x 5 /6 = 180

Therefore Eph = 4.44 KpKd f Tphvlolts

= 4.44 x 1 x 0.9597 x 50 x 30 x 10-3 x 180

= 1150.488 volts

Hence the line Voltage EL = 3 x phase voltage = 3 Eph

= 3 x1150.488

= 1992.65 volts

2.A 3, 16 pole, star connected salient pole alternator has 144 slots with 10 conductors per slot. The alternator is run at 375 rpm. The terminal voltage of the generator found to be 2.657 kV. Dteremine the frequency of the induced emf and the flux per pole.

Soln: Ns = 375 rpm, p =16, slots = 144, Total no. of conductors = 144 x 10 = 1440

EL = 2.657 kV,

e = P Ns/120 = 16 x 375/120 = 50 Hz

Assuming full pitched winding kp = 1

Number of slots per pole per phase = 144/(16 x 3) = 3

Slot angle = 1800 / number of slots/pole = 1800 /9 = 200

Hence distribution factor Kd = ( sin m/2) / (m sin /2)

= ( sin 3 x 20 / 2) / (3 sin 20/2)

= 0.9597

Turns per phase Tph = 144 x 10/ 6 = 240

Eph = EL/3 = 2.657/3 = 1.534 kV

Eph = 4.44 KpKd f Tphvlolts

1534.0 = 4.44 x 1 x 0.9597 x 50 x x 240

= 0.03 wb = 30 mwb

3. A 4 pole, 3 phase, 50 Hz, star connected alternator has 60 slots with 4 conductors per slot. The coils are short pitched by 3 slots. If the phase spread is 600, find the line voltage induced for a flux per pole of 0.943 wb.

Slon: p = 4, f = 50 Hz, Slots = 60, cond/slot = 4 , short pitched by 3 slots, phase spread = 600, = 0.943 wb

Number of slots/pole/phase m = 60/(4 x 3) = 5

Slot angle = phase spread/ number of slots per pole/phase

= 60/5 = 12

Distribution factor kd = (sin m/2) / (m sin/2)

= sin ( 5 x 12/2) / 5 sin(12/2)

= 0.957

Pitch factor = cos/2

Coils are short chorded by 3 slots

Slot angle = 180/number of slots/pole

= 180/15 = 12

Therefore coil is short pitched by = 3 x slot angle = 3 x 12 = 360

Hence pitch factor kp = cos/2 = cos 36/2 = 0.95

Number of turns per phase Tph= Zph/2 = (Z/3)/2 = Z /6 = 60 x 4 /6 = 40

EMF induced per phase Eph = 4.44 kpkdf Tph volts

= 4.44 x 0.95 x 0.957 x 50 x 0.943 x 40

= 7613 volts

Line voltage EL = 3 x Eph

= 3 x 7613 = 13185 volts

4.In a 3 phase star connected alternator, there are 2 coil sides per slot and 16 turns per coil. The stator has 288 slots. When run at 250 rpm the line voltage is 6600 volts at 50 Hz. The coils are shot pitched by 2 slots. Calculate the flux per pole.

Slon: Ns = 250 rpm, f = 50 Hz, slots = 288, EL= 6600 volts, 2 coilsides/slot, 16 turns /coil Short pitched by 2 slots

Number of poles = 120f/ Ns = 120 x 50/250 = 24

Number of slots /pole/phase m = 288 / ( 24 x 3) = 4

Number of slots /pole = 288 / 24 = 12

Slot angle = 180/ number of slots per pole

= 180 / 12 = 150

Distribution factor kd = (sin m/2) / (m sin/2)

= sin ( 4 x 15/2) / 4 sin(15/2)

= 0.9576

Coils are short chorded by 2 slots

Slot angle = 15

Therefore coil is short pitched by = 2 x slot angle = 2 x 15 = 300

Hence pitch factor kp = cos/2 = cos 30/2 = 0.9659

Two coil sides per slot and 16 turns per coil

Total number of conductors per slot = 2 x 16 = 32 turns

Total conductors = 32 x 288

Turns per phase = 32 x 288 / 6

= 1536

Eph= 6600 / 3 = 3810.51 volts,

We have EMF induced per phase Eph = 4.44 kpkdf Tph volts

3810.51 = 4.44 x 0.9659 x 0.9576 x 50 x x 1536

= 0.02 wb

5.A 10 pole, 600 rpm, 50Hz, alternator has the following sinusoidal flux density distribution. B = sin + 0.4 sin 3 + 0.2 sin 5wb/m2. The alternator has 180 slots with 2 layer 3 turn coils with a coil span of 15 slots. The coils are connected in 600 groups. If the armature diameter is 1.2 m and core length is 0.4 m, calculate (a) the expression for instantaneous emf/conductor (b) the expression for instantaneous emf/coil (c) the phase and line voltages if the machine is star connected.

Slon: Area under one pole pitch = DL/p = x 1.2 x 0.4/10 = 0.1508 m2

Fundamental flux/pole, 1 = average flux density x area

= 2/ x 1 x 0.1508

= 0.096wb

(a) rms value of emf induced/conductor = 2.22f 1 = 2.22 x 50 x 0.096 = 10.656 volts

maximum value of emf/conductor = 2 x 10.656 = 15.07 volts 3rd harmonic voltage = 0.4 x 15.07 = 6.02 volts

5th harmonic voltage = 0.2 x 15.07 = 3.01 volts

the expression for instantaneous emf/conductor e = 15.07 sin + 6.02 sin 3 + 3.01 sin 5 volts

(b) conductors/slot = 6 = conductors/coil, slots = 180, coil span = 15 slots slots/pole = 18 slot angle = 180/number of slots/ pole = 180/ 18 = 100

coil is short chorded by 3 slots hence = 300

Pitch factor kpn = cos n /2 kp1 = cos /2 = cos 30/2 = 0.9659 kp3 = cos 3 x 30/2 = 0.707

kp5 = cos 5 x 30/2 = 0.2588

Fundamental rms value of emf induced/coil = 2.22 kpf 1 Z

= 2.22 x 0.9659 x 50 x 0.096 x 6

= 61.76 volts

Maximum value of emf induced/coil = 2 x 61.76 = 87.34 volts

Similarly 3rd harmonic voltage = 25.53 volts

5th harmonic voltage = 4.677 volts

expression for instantaneous emf/coil e = 87.34 sin + 25.53 sin 3 + 4.677 sin 5 volts

slot angle = 180/number of slots/ pole = 180/ 18 = 100 number of slots/pole/phase = 180/(10 x 3) = 6

Distribution factor kdn = (sin m n/2) / (m sin n/2)

kd1 = sin ( 6 x 10/2) / 6 sin(10/2)

= 0.956

kd3 = sin ( 6 x 3 x 10/2) / 6 sin (3 x 10/2)

= 0.644

kd5 = sin ( 6 x 5 x 10/2) / 6 sin (5 x 10/2)

= 0.197

Turns/phase Tph = 180 x 6/6 = 180

rms value of emf induced = 4.44 kpnkdn (nf ) nTph for any nth harmonic

fundamental voltage Eph1 = 4.44 kp1 kd1 f 1Tph

= 4.44 x 0.9659 x 0.956 x 50 x 0.096 x 180

= 3542.68 volts

Similarly 3rd harmonic voltage Eph3 = 697.65 volts

5th harmonic voltage Eph5= 39.09 volts

Phase voltage = (E2ph1 + E2ph3 + E2ph5)

= (3542.682 + 697.652 + 39.092)

= 3610.93 volts

Line voltage = 3 x (E2ph1+ E2ph5)

= 3 x (3542.682 + 39.092)

= 6136.48 volts

6.A 3 phase 10 pole 600 rpm star connected alternator has 12 slots/pole with 8 conductors per slot. The windings are short chorded by 2 slots. The flux per pole contains a fundamental of 100 mwb, the third harmonic having an amplitude of 33% and fifth harmonic of 20% of the fundamental. Determine the rms value of the phase and line voltages.

Soln: P = 10, Ns= 600 rpm, 12 slots/pole, 8 cond/slot star connected

Slots/ploe/phase m = 4, slot angle = 180/number of slots/ pole = 180/ 12 = 150 chording angle = 2 x slot angle = 2 x 15 =300

Air gap fluxes

1= 100 mwb;

3 = 33% of 1 = 0.33 x 100 = 33 mwb

Pitch factors

5 = 20% of 1 = 0.2 x 100 = 20 mwb

kp1 = cos /2 = cos 30/2 = 0.9659 kp3 = cos 3 x 30/2 = 0.707

kp5 = cos 5 x 30/2 = 0.2588

Distribution factors

kd1 = sin ( 4 x 15/2) / 4 sin(15/2)

= 0.9576

kd3 = sin ( 4 x 3 x 15/2) / 4 sin (3 x 15/2)

= 0.6532

kd5 = sin ( 4 x 5 x 15/2) / 4 sin (5 x 15/2)

= 0.2053

Total number of conductors = cond/slot x slot/pole x no. of poles

= 8 x 12 x 10

= 960

Turns/phase = Z/6 = 960 /6 = 160

emf induced for any nth harmonic En ph = 4.44 kpnkdn (nf ) nTph

fundamentalvoltage Eph1 = 4.44 kp1 kd1 f 1Tph

= 4.44 x 0.9659 x 0.9576 x 50 x 0.1 x 160

= 3285.4volts

Similarly 3rd harmonic voltage Eph3 = 541.39 volts

5th harmonic voltage Eph5= 37.74 volts

Phase voltage = (E2ph1 + E2ph3 + E2ph5)

= (3285.42 + 541.392 + 37.742)

= 3329.92 volts

Line voltage = 3 x (E2ph1+ E2ph5)

= 3 x (3285.42 + 37.742)

= 5690.85 volts

7.A three phase 600 kVA, 400 volts, delta connected alternator is reconnected in star. Calc