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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:
Field
Current
(A)
10 20 30 40
Terminal 1200 2100 2830 3460
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.
Volts (OC)SC
Current
13.2 26.0
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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
Current (A)
90 135 180 225
a) Calculate the regulation at a pf of 0.8 lagging.
b) Calculate the regulation for a load of unity pf.
TerminalVolts (OC)
9800 13000 14900 15800
SC Current 195 291
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Example 3: A 3-phase, 800 kVA, 3000 V, 50 Hz alternator gave thefollowing results:
ExcitingCurrent (A)
30 35 40 50 60 65 70 75 77.5 80 85 90 100 110
O.C. volt(line)
_ _ _ 2560 3000
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 loadcurrent on short circuit of the alternator. The armature resistance perphase is 0.27225 . 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.
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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:
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.
Field
Current
(A)
2 4 7 9 12 15 20 22 24 25
Terminal
Volts
155 287 395 440 475 530 555 560 610 650
. . . , .
also the vector diagram. Solution:IL = 30 kVA /(3) (650) = 26.6469355 AZS = [ 650 / (3) ] / 26.6469355 = 14.08333333 A
Ra=0; XS = ZSIXS = 375.2776749 V
Eph = Vph + IL ( Ra + j XL ) ; Vph + IXS < 53.13010235Eph = 622.7425899 < 28.8224976 VVR% = 622.7425899 (555/ 3)
(555/ 3)= 94.34627131 %
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Solution:IL = 30 kVA /(3) (550) = 31.49183286 AVph = 550 / 3 = 317.5426481 V
IRa= 4.72377493 V/phase
Eph = Vph + IL ( Ra + j XL ) ; Vph + IRa < 36.86989765Eph = 321.3341678 < 0.50537273 VELL = 556.5671049 V
If = 20
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:Solution
( )191188
144toslots
P
S ==
=
2
10x3cospk
== 1018
180
1. The following information is given in
connection with an alternator: slots = 144;
poles = 8; rpm = 900; conductors/slot = 6; fluxper 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
( ) conductorsZT 8641446 ==
864==
965925826.0=pk
6
3
8
144
==
phase
poles
slots
m
=
2
10sin6
2
10x6sin
dk 95614277.0=dk
3p
phaseturnsT /1442
288==
( )( )( )( )( )( )86 10x16014410x8.144.4 = pdg kkE VEg 7283753.637=
( )
V
E LLg
577947.1104
7283753.6373
=
=
1 pt
1 pt
1 pt
1 pt
8 pts
8 pts
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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. Whendriven 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
VELL 6600=
VV
Eg 511777.38103
6600==
( )1311224
288toslots
P
S==
180 424
288
==poles
slots
m( )
polesN
fP 24
120==
==12
=
2
15x2cospk
965925826.0=pk 1 pt
=
2
15sin4
2
15x4sin
dk 957662196.0=dk
1 pt
slot
cond
slots
turns
coil
turnsx
slot
coils .6432162==
( ) conductorsZT 1843228864==
61443
18432==phZ
phaseturnsT /30722
6144
==
6 pts
6 pts
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:2. Solutionnoofoncontinuati
( ) ( )( )( )50307244.4511777.3810 pd kk=
mWb040223388.6= 6 pts
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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
:Solution
( ) Hzf 50120
60010==
( ) conductorsZT 10801290 ==
3603
1080==phZ 1 pt
phaseturnsT /1802 ==
Wye:
( )( )( )( )( )180501.0144.4 dg kE =
VEg 16.3836= 6 pts
1 pt
( )
V
E LLg
424026.6644
16.38363
=
=
6 pts
Delta:
( )( )( )( )( )180501.0144.4 dg kE =
LLg EVE == 16.3836
6 pts
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1. Three non-inductive resistances, each of100 , are connected in star to 3-phase, 440
< 0O V supply. Three equal choking coilseach of reactance 100 are also connectedin 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 Wye load:
:Solution
100
303
440
1
== IIL
AIIL 30540341184.21 ==
For Delta load:
a)
100
0440
jI =
AI 904.4 =
AIL 120621023553.72 =
b) total line current
AIII LLT 5650512.101033264177.821 =+=
( ) ( ) WIP LT 3333331.64510021 ==
( ) ( ) VarsIQ LT 19361002
2 ==
VAUT 723183.2040=
)(316227765.0 laggingU
Ppf
T
T ==
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2. A symmetrical 3-phase, 3-wire supply with aline voltage of 173 < 0OV supplies two balanced
3-phase loads; one Y-connected with eachbranch impedance equal to (6 + j8) ohm andthe other -connected with each branchimpedance 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 circuitd) total real power and apparent power
For Wye load:
86
303
173
1j
IIL+
==
AIIL 309881.91 ==
a)
:Solution
For Delta load:
2418 jI +=
AI 13010235.53766666667.5 =
AIL 13010235.83988159757.92 =
b) total line current
AIII LLT 13010235.8397631931.1921 =+=
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Solution continuation No.2:
( ) ( ) WIP LL 58.5986211 ==
VAUL 6333333.9971 =
( ) ( ) WIP LL 5800001.59862
22 ==
VAUL 6333334.9972 =
WPPP LLT 16.119721 =+=
VAUUU LLT 266667.199521 =+=
)(6.0 laggingU
Ppf
T
T ==
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pfIVP LLT 3=
= 78833062.31
( )( )( )85.0554003=TP
WPT 28511.35628=
3. A 440-V, 50-Hz induction motor takes a linecurrent of 55 A at a power factor of 0.85
(lagging). Three -connected capacitors areinstalled to improve the power factor to 0.9(lagging). Calculate the kVA of the capacitorbank and the capacitance of each capacitor.
power factor of 0.85 (lagging).
TNEW PQ )tan(=
= 8419327.25
VARSQNEW 56604.17255=
TOLD PQ )tan(=
VARSQOLD 42799.22080=
power factor of 0.9 (lagging).
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NEWOLDCAP QQQ =
Continuation of No. 3 solution:
VARSQCAP 861954.4824=
VARS
VARS
QCAP 287318.16083
861954.4824
==
VARS65519745.3
287318.1608==
== 376501.120440
C
CI
VX
( )( )FC
44285915.26
376501.120502
1==