Field 10 20 30 Current (A) Terminal 1200 2100 2830 Volts (OC) SC Current
40
3460
13.2 26.0
Calculate the synchronous impedance and the synchronous reactance per phase for this machine, using the highest point given on the saturation or open circuit voltage curve to obtain the values.
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Example 1 . A three-phase, Y-connected generator is rated at 100 kVa, 60 cycles, 2300 volts. The effective resistance of the armature is 1.5 ohms per leg. The test data are given below:
Example 2 . A three-phase, slow speed, Y-connected alternator is rated at 5000 kVA and 13,200 volts. The resistance of the armature between terminals is 0.192 ohm at 75 C. The effective resistance is 1.6 times the dc-value at 75 C.. The test data on this machine is given below. Field 90 Current (A)
135
Terminal Volts (OC)
9800 13000
SC Current
195
180
225
14900
15800
291
a) Calculate the regulation at a pf of 0.8 lagging. b) Calculate the regulation for a load of unity pf.
Example 3: A 3-phase, 800 kVA, 3000 V, 50 Hz alternator gave the following results: Exciting Current (A)
30
35
40
50
60
65
70
75
77.5
80
85
90
100
110
O.C. volt (line)
_
_
_
2560
300 0
3250
3300
3450
3500
3600
3700
3800
3960
4050
S.C. current
140
150
170
190
_
_
_
_
_
_
_
_
_
_
a) A field current of 110 A is found necessary to circulate a full load current on short circuit of the alternator. The armature resistance per phase is 0.27225 Ω. Calculate the voltage regulation at 0.8 p.f. lagging and 0.9 p.f leading, using synchronous impedance method. Show also the vector diagram.
Example 4 . A 30-kVA, Y-connected alternator rated at 555 volts at 50 Hz has the open circuit characteristics given by the following data: Field Current (A) Terminal Volts
2
4
7
9
12
15
20
22
24
25
155
287
395
440
475
530
555
560
610
650
A field current of 25 A is found necessary to circulate a full load current on short circuit of the alternator. Calculate the voltage regulation at 0.8 p.f. lagging and 0.8 p.f. leading, using synchronous impedance method. Show Solution: also the vector diagram. IL = 30 kVA /(√3) (650) = 26.6469355 A ZS = [ 650 / (√3) ] / 26.6469355 = 14.08333333 A
Ra=0; XS = ZS IXS = 375.2776749 V Eph = Vph + IL ( Ra + j XL ) ; Vph + IXS < 53.13010235 Eph = 622.7425899 < 28.8224976 V VR% = 622.7425899 – (555/ √3) (555/ √3) = 94.34627131 %
Solution: IL = 30 kVA /(√3) (550) = 31.49183286 A Vph = 550 / √3 = 317.5426481 V IRa= 4.72377493 V/phase Eph = Vph + IL ( Ra + j XL ) ; Vph + IRa < 36.86989765 Eph = 321.3341678 < 0.50537273 V ELL = 556.5671049 V If = 20<90.50537273 + 7<216.8698976 A = 16.82207726 < 110.0830856 A
Solution :
S 144 = = 18 (slots 1to19 ) P 8 180 = 10 ° = β 18
1. The following information is given in connection with an alternator: slots = 144; poles = 8; rpm = 900; conductors/slot = 6; flux per pole = 1.8 x 106 maxwells; coil span = slots 1 to 16; winding connection = star. Calculate: (a) the voltage per phase; (b) the voltage between terminals. 20 PTS
Z T = 6(144 ) = 864 conductors
3x10 ° k p = cos 2
864 = 288 1 pt 3 288 T= = 144 turns / phase 2
Z ph =
k p = 0.965925826 1 pt
144 slots 8 poles m= =6 3 phase 6 x10 ° sin 2 kd = 10 6 sin 2
(
1 pt
)
(
E g φ = 4.44 (k d )(k p ) 1.8 x10 6 (144 )(60 ) 1x10 −8 E g φ = 637 .7283753 V
8 pts
E g LL = 3 (637 .7283753 )
k d = 0.95614277 1 pt
= 1104 .577947 V
8 pts
)
2. In a 3 phase, start connected alternator, there are 2 coil per slot and 16 turns per coil. Armature has 288 slots on its periphery. When driven at 250 rpm it produces 6600 V between the lines at 50 Hz. The pitch of the coil is 2 slots less than the full pitch. Calculate the flux per pole, total number of conductors and turns per phase. 20 PTS
Solution :
E LL = 6600V E gφ =
6600V = 3810 .511777 V 3
S 288 = = 12 (slots 1to13 ) P 24 180 = 15 ° = β 12
120 ( f ) P= = 24 poles N
2coils 16turns 32turns 64 cond . x = = slot coil slots slot
2 x15 ° k p = cos 2
Z T = 64 (288 ) = 18432 conductors
k p = 0.965925826 1 pt
4 x15 ° sin 2 kd = 15 4 sin 2
288 slots 24 poles m= =4 3 phase
18432 = 6144 6 pts 3 6144 T= = 3072 turns / phase 2 6 pts
Z ph = k d = 0.957662196 1 pt
continuati on of no.2 Solution : 3810 .511777 = 4.44 (k d )(k p )(φ )(3072 )(50 )
φ = 6.040223388 mWb
6 pts
Solution :
f =
10 (600 ) = 50 Hz 120
Z T = 90 (12 ) = 1080 conductors Z ph = T=
1080 = 360 3
3. A 3-phase, 10-pole alternator has 90 slots, each containing 12 conductors. If the speed is 600 r.p.m. and the flux per pole is 0.1 Wb, calculate the line e.m.f. and voltage per phase when the phases are (i) star connected (ii) delta connected. Assume the winding factor to be 0.96 and the flux sinusoidally distributed. 20 PTS
1 pt
360 = 180 turns / phase 2
1 pt
Wye:
Delta:
E g φ = 4.44 (k d )(1)(0.1)(50 )(180 )
E g φ = 4.44 (k d )(1)(0.1)(50 )(180 )
E g φ = 3836 .16V
E g φ = 3836 .16V = E LL
6 pts
E g LL = 3 (3836 .16 ) = 6644 .424026 V
6 pts
6 pts
Solution : a) For Wye load:
I L1
440 ∠ − 30 = Iφ = 3 100
I L1 = I φ = 2.540341184 ∠ − 30 A
1. Three non-inductive resistances, each of 100 Ω, are connected in star to 3-phase, 440 < 0O V supply. Three equal choking coils each of reactance 100 Ω are also connected in delta to the same supply. Calculate: a) line current of each 3-phase load (in polar form) b) the total line current (in polar form) c) power factor of the system
For Delta load: c) pf of the system
440 ∠ 0 Iφ = j100 I φ = 4.4∠ − 90 A
PT = (I L1 ) (100 ) = 645 .3333331W 2
QT = (I L 2 ) (100 ) = 1936Vars 2
U T = 2040 .723183 VA
I L 2 = 7.621023553 ∠ − 120 A b) total line current
I T = I L1 + I L 2 = 8.033264177 ∠ − 101 .5650512 A
PT pf = = 0.316227765 (lagging ) UT
Solution : a) For Wye load:
I L1
173 ∠ − 30 = Iφ = 3 6 + j8 I L1 = I φ = 9.9881∠ − 30 A
For Delta load:
2. A symmetrical 3-phase, 3-wire supply with a line voltage of 173 < 0O V supplies two balanced 3-phase loads; one Y-connected with each branch impedance equal to (6 + j8) ohm and the other Δ-connected with each branch impedance equal to (18 + j24) ohm. Calculate: a) line current taken by each 3-phase load (in polar form) b) the total line current (in polar form) c) power factor of the entire load circuit d) total real power and apparent power
I φ = 18 + j 24 I φ = 5.766666667 ∠ − 53 .13010235 A I L 2 = 9.988159757 ∠ − 83 .13010235 A b) total line current
I T = I L1 + I L 2 = 19 .97631931 ∠ − 83 .13010235 A
Solution continuation No.2:
PL1 = (I L1 ) (6 ) = 598 .58W
PL 2 = (I L 2 ) (6 ) = 598 .5800001W
U L1 = 997 .6333333 VA
U L 2 = 997 .6333334 VA
2
2
PT = PL1 + PL 2 = 1197 .16W U T = U L1 + U L 2 = 1995 .266667 VA
PT pf = = 0.6(lagging ) UT
PT = 3VL I L pf PT = 3 (400 )(55 )(0.85 ) PT = 35628 .28511W
3. A 440-V, 50-Hz induction motor takes a line current of 55 A at a power factor of 0.85 (lagging). Three Δ-connected capacitors are installed to improve the power factor to 0.9 (lagging). Calculate the kVA of the capacitor bank and the capacitance of each capacitor.
power factor of 0.85 (lagging).
θ = 31 .78833062 ° QOLD = tan(θ ) PT QOLD = 22080 .42799 VARS power factor of 0.9 (lagging).
β = 25 .8419327 ° Q NEW = tan( β ) PT Q NEW = 17255 .56604 VARS
Continuation of No. 3 solution:
QCAP = QOLD − Q NEW QCAP = 4824 .861954 VARS
QCAP φ =
I Cφ =
4824 .861954 VARS = 1608 .287318 VARS 3
1608 .287318 VARS = 3.65519745 A 400V X Cφ =
C=
440V = 120 .376501 Ω I Cφ
1 = 26 .44285915 µF 2π (50 )(120 .376501 )
