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```Board Exam Problems
on
AC MACHINES
(Alternators)
41
1.
The number of cycles generated in a 10 pole alternator in one revolution is
A. 2
B. 5
C. 10
Solution:
ππ. ππ
ππ¦ππππ
πππ£
ο·
2.
1 ππ¦πππ
=(
ππ. ππ
2 ππππ
ππππππ
πππ
)(
10 πππππ
πππ£
.
D. 20
)
= π ππππππ/πππ
What is the number of electrical degrees made per revolution for a special purpose 40 pole
alternator.
A. 28, 800
B. 14, 400
C. 7, 200
D. 1, 600
Solution:
ππ. ππ πππππ‘πππππ πππ/πππ£ = (
180°
ππππ
ο·
40 πππππ
)(
πππ£
)
ππ. ππ ππππππππππ πππ/πππ = π, πππ°/πππ
3.
Alternators are usually designed to generate
A. variable frequency
B. definite currents
C. definite frequencies
D. definite p.f
4.
A 4-pole, 3-phase, star-connected alternator armature has 12 slots with 24 conductors per slot and
the flux per pole is 0.1 Wb sinusoidally distributed. Calculate the line emf generated at 50 Hz.
A. 1, 066 V
B. 3, 198 V
C.1, 846 V
D. 5, 538 V
Solution:
24 πππππ’ππ‘πππ
πβ = (
ο·
π πππ‘
)(12 π πππ‘π )(
1 π‘π’ππ
2 πππππ’ππ‘πππ
)(
1
3 πβππ π
)
πβ = 48 π‘π’πππ /πβππ π
1 turn
1 conductor
1 conductor
πΈβ = 4.44ππβ β
πΈβ = 4.44(50 π»π§)(48 π‘π’πππ /β)(0.1
ο·
ππ
ππππ
)
πΈβ = 1, 065.6 π
πΈπΏ = β3πΈβ
πΈπΏ = β3(1, 065.6)
ο· π¬π³ = π, πππ. ππ π½
REE β May 2008
5. A three-phase wye-connected 50 Hz, 2-pole synchronous machine has a stator with 2, 000 turns of
wire per phase. What rotor flux would be required to produce a terminal (line to line) voltage of 5
kV?
A. 8.4 mWb
B. 6.5 mWb
C. 5.2 mWb
D. 7.8 mWb
Solution:
πΈβ = 4.44ππβ β
πΈβ =
ο·
πΈπΏ
β3
=
5,000
β3
πΈβ = 2, 886.75 ππππ‘π
42
πΈβ
β=
4.44ππβ
2,886.75
β=
4.44(50)(2,000)
ο·
β = π. π × ππβπ πΎπ
REE β April 2006
6. A 3-phase, 8-pole generator is delta-connected. The terminal voltage is 2, 400 Volts while the line
current is 500 Amperes. If the machine is converted to wye, what will be the terminal voltage in
volts?
A. 4, 800 V
B. 4, 157 V
C. 4, 600 V
D. 3, 800 V
Solution:
β π€βππ πππππππ‘ππ ππ ππππ‘π
ο· πΈβ = πΈπΏ = 2, 400 ππππ‘π
β π€βππ πππππππ‘ππ ππ π€π¦π
πΈπΏ = β3πΈβ
πΈπΏ = β3(2, 400)
ο· π¬π³ = π, πππ. ππ π½ππππ
REE β September 2005
7. A three-phase, 8-pole, 2,400 Volts delta connected generator has a line current of 500 Amperes. If
converted in wye, what is the tolerable current?
A. 371.5 A
B. 288.7 A
C. 245.1 A
D. 315.2 A
Solution:
β π€βππ πππππππ‘ππ ππ ππππ‘π
πΌβ =
πΌπΏ
β3
=
500
β3
ο· πΌβ = 288.68 π΄ππππππ
β π€βππ πππππππ‘ππ ππ π€π¦π
ο· π°π³ = π°β = πππ. ππ π¨ππππππ
8.
If the alternator winding has fractional pitch of 4/5, the pitch factor ππ is
A. 0.833
Solution:
B. 0.966
4
π° = (180°)
5
C. 0.972
.
D. 0.951
full-pitch coil span or 1 pole span
ο· π° = 144
π°
ππ = π ππ( )
ππ = π ππ(
2
144
2
)
ο· ππ = π. πππ
9. A six pole, three-phase alternator has 72 slots and a coil span of 1 to 10, what is the
pitch factor?
A. 0.924
B. 0.939
C. 0.966
D. 0.985
Solution:
43
10-1= 9 slots
1
π° =
ο·
10
(10β1)π πππ‘π
72 π πππ‘π
6 ππππ
(180°)
π° = 135
π°
ππ = π ππ( )
ππ = π ππ(
ο·
2
135
2
)
ππ = π. πππ
10. The power factor of an alternator is 75%. The operator is ordered to increase the power factor to
80%. What shall he do?
A. increase the voltage
B. operate the governor
C. increase the excitation
D. decrease the excitation
11. Calculate the distribution factor for a 36-slot, 4-pole three phase winding.
A. 0.924
B. 0.951
C. 0.960
Solution:
π=
D. 0.975
36 π πππ‘π
(4 ππππ)(3 πβππ ππ )
π πππ‘π
ο·
π° =
π=3
ππππ
/πβππ π
180°
(
)(4 πππππ )
ππππ
ο·
36 π ππ‘π
π° = 20°
π°
ππ° =
ππ° =
ο·
π ππ π( 2 )
π°
2
20°
π ππ 3( 2 )
20°
3 π ππ( )
2
π π ππ( )
ππ° = π. πππ
12. A 144-slot stator has a whole-coiled 12 pole three-phase winding. What is the number of coils per
phase and per group, respectively?
A. 12 and 4
B. 24 and 4
C. 36 and 4
D. 48 and 4
Solution:
ππ. ππ πππππ  πππ πβππ π =
1 ππππ
)(144 π πππ‘)
π πππ‘
(
3 πβππ π
44
ο·
ππ. ππ πππππ πππ πππππ = ππ πππππ πππ πππππ
ππ. ππ πππππ  πππ ππππ’π =
ο·
48 πππππ /πβππ π
12 ππππ’ππ /πβππ π
ππ. ππ πππππ πππ πππππ = π πππππ πππ πππππ
13. A part of an alternator winding consists of six coils in series, each coil having an e.m.f of 10 Volts
(r.m.s) induced in it. The coils are placed in successive slots and between each slot and the next, there
is an electrical phasor displacement of 30°. Find the e.m.f of the six coils in series.
A. 77.28 V
B. 9.66 V
C. 38.64 V
D. 19.32 V
Solution:
πΈπ = πΈπ1 + πΈπ2 + πΈπ3 + πΈπ4 + πΈπ5 + πΈπ6
πΈπ = 10β 0° + 10β 30° + 10β 60° + 10β 90° + 10β 120° + 10β 150°
ο· π¬πΉ = ππ. ππβ ππ° π½ππππ
14. The following information is given in connection with an alternator: π πππ‘π  = 144; πππππ  = 8; πππ =
900; π‘π’πππ  πππ ππππ = 6; β = 1.8 × 106 ; ππππ π πππ = π πππ‘π  1 π‘π 16; π€ππππππ πππππππ‘ππππ  = π π‘ππ.
What is the voltage generated between terminals?
A. 1, 276 V
B. 2, 210 V
C. 635 V
D. 1, 100 V
Solution:
(16β1)π πππ‘π
π° =
144 π πππ‘π
8 πππππ
ο·
(
180°
ππππ
)
π° = 150°
π°
150°
2
2
ππ° = π ππ ( ) = π ππ(
ο·
π=
ππ° = 0.966
144 π πππ‘π
(8 πππππ )(3 β)
π πππ‘π
ο·
π° =
π=6
ππππ
/πβππ π
180°
(
)(8 πππππ )
ππππ
144 π πππ‘π
ο·
π° = 10°/π πππ‘
π°
ππ° =
π ππ π( 2 )
π°
π π ππ( )
2
ο·
π=
)
10°
=
π ππ 6( 2 )
10°
6 π ππ( )
2
ππ° = 0.956
π×πππ
120
ο·
=
2(900)
120
π = 60 π»π§
6 π‘π’πππ
πβ = (
ππππ
)(
1 ππππ
π πππ‘
)(144 π πππ‘π )(
1
3 πβππ π)
ο· πβ = 288 π‘π’πππ /πβππ π
πΈβ = 4.44(ππ°)(ππ°)ππβ β × 10β8
πΈβ = 4.44(0.966)(0.956)(60)(288)(1.8 × 106 )(10β8 )
ο· πΈβ = 1, 275.36 ππππ‘π
πΈπΏ = β3πΈβ = β3(1, 275.36)
ο· π¬π³ = π, πππ π½ππππ
15. The disadvantage of a short-pitch coil is that
.
A. harmonics are introduced
B. waveform becomes non-sinusoidal
C. voltage around the coil is reduced D. both a and b
45
16. A 4-pole alternator, on open circuit, generates 200 Volts at 50 Hz when the field current is 4
Amperes. Determine the generated e.m.f at a speed of 1, 200 rpm and a field current of 3 Amperes,
neglecting saturation in the iron parts.
A. 40 V
B. 240 V
C. 60 V
D. 120 V
Solution:
πΈ = πππΌπ
πΈ1 = ππ1 πΌπ
1
πΈ1
π=
π1 πΌπ
(50)(4)
1
ο·
π2 =
200
=
π=1
π×πππ
120
=
4(1,200)
120
ο· π2 = 40 π»π§
πΈ2 = ππ2 πΌπ
2
πΈ2 = 1(40)(3)
ο· π¬π = πππ π½ππππ
17. A 3-phase. Wye-connected alternator having a sinusoidal line potential with a 5% 5 th harmonic
content, 3% 7th harmonic content, 1.5% 11th harmonic content and 0.85% 13 th harmonic content.
What is the total harmonic distortion (THD) of the system line voltage?
A. 11%
B. 9%
C. 6%
D. 3%
Solution:
ππ»π· = β
ππ»π· = β
ο·
πΈ5 2 +πΈ7 2 +πΈ11 2 +πΈ13 2
πΈ1
(0.05πΈ1 )2 +(0.03πΈ1 )2 +(0.015πΈ1 )2 +(0.0085πΈ1 )2
πΈ1
π»π―π« = π. ππ%
18. A 100 MVA, 13.8 kV, three-phase, wye-connected alternator will have a per phase nominal
impedance of
.
β¦
A. 3.5 ββ
B. 2.9 β¦ββ
C. 1.9 β¦ββ
D. 7.5 β¦ββ
Solution:
πβ =
πβ =
ππ 2
πππ΄
13.82
100
πβ = π. π β¦ββ
ο·
REE β March 1998
19. A generator is rated 100 MW, 13.8 kV and 90% power factor. The effective resistance to ohmic
resistance is 1.5. The ohmic resistance is obtained by connecting two terminals to a d.c source. The
current and voltage are 87.6 amperes and 6 Volts, respectively. What is the resistance per phase?
A. 0.0617 β¦
C. 0.0685 β¦
D. 0.342 β¦
Solution:
ππ‘ =
πππ
πΌππ
ο·
ππ ππ =
=
6
87.6
ππ‘ = 0.0685 β¦
ππ‘
2
=
β dc ohmic resistance between terminals
0.0685
2
ππ ππ = 0.03425 β¦
46
ο·
π΄πΆ ππππ‘ππ =
ππ
ππ ππ
ππ = 1.5(0.03425)
ο· πΉπ = π. ππππ β¦/β
REE β April 2006
20. A single-phase alternator gives 250 Amperes at 1, 200 Volts. What is the output of the machine in
KVA?
A. 300
B. 450
C. 400
D. 350
Solution:
πΎππ΄ππ’π‘ππ’π‘ =
πΎππ΄ππ’π‘ππ’π‘ =
ο·
ππΏ πΌ πΏ
1,000
(1,200)(250)
1,000
π²π½π¨ππππππ = πππ
21. High speed alternators have a rotor construction
A. similar to d.c. machines
B. of non-salient type
C. of salient type
D. none of these
22. The imaginary or fictitious part of synchronous reactance on alternator takes care of
A. inductive reactance
B. leakage reactance
C. armature reaction
D. copper losses
REE β April 2006
23. A single-phase generator delivers 80 A at 240 V and 75% lagging p.f. What kind of load is the
generator supplying?
A. capacitive
B. resistive
C. inductive
D. reactive
REE β September 2004
24. A 3,600 Volts, 500 kVA, 60 Hz, three-phase, Y-connected generator on test yields the following
results: Mechanical and iron losses are 10 kW; field current at full load 100% p.f is 50 Amperes; field
current at full load 80% p.f is 70 Amperes; resistance per phase of armature winding is 0.4 ohm. The
exciter voltage is constant at 120 Volts and voltage control is done by means of rheostat. Determine
the full load armature current at 80% lagging p.f
A. 96.5 A
B. 99.8 A
C. 64.2 A
D. 80.2 A
Solution:
πΌπ = πΌπΏ =
πΌπ =
πΎππ΄ππ’π‘ππ’π‘ (1,000)
β3
(500)(1,000)
ο·
β3
π°π = ππ. ππ π¨ππππππ
REE β April 2005
25. A single-phase 2, 400 Volts synchronous generator delivers 450 Amperes at unity power factor. The
synchronous impedance of the generator is 0.08 + π2.8 β¦. What is the regulation in percent?
A. 14.2
B. 16.3
C. 20.1
D. 18.4
Solution:
πΈπ = ππ‘ + πΌπ ππ
πΈπ = 2, 400β 0° + (450β 0°)(0.08 + π2.8)
47
ο·
πΈπ = 2, 742.57β 27.35° ππππ‘π
%π. π =
%π. π =
ο·
πΈπ βππ‘
× 100
ππ‘
2,742.57β2,400
2,400
× 100
%π½. πΉ = ππ. ππ%
26. A 25 kVa alternator has a total loss of 2, 000 watts when it delivers rated kVA to a load at a power
factor of 0.76. Calculate its percent efficiency.
A. 90.48%
B. 92.32%
C. 91.58%
D. 89.12%
Solution:
πππ’π‘ = πππ΄(π. π) = 25(0.76)
ο· πππ’π‘ = 19 ππ
Ι³=
Ι³=
πππ’π‘
× 100
πππ’π‘ +π‘ππ‘ππ πππ π
19,000
19,000+2,000
× 100
ο· Ι³ = ππ. ππ%
REE β September 2010
27. A three-phase, wye-connected wound rotor synchronous generator rated at 10 kVA, 230 V has a
synchronous reactance of 1.2 ohms/phase and an armature resistance of 0.5 ohm/phase. What is the
percent voltage regulation at full-load with 85% lagging power factor?
A. 18.3%
B. 24.7%
C. 20.8%
D. 22.5%
Solution:
πβ =
ππΏ
β3
ο·
β 0° =
230
β3
β 0°
πβ = 132.79β 0° ππππ‘π
πΌπ = πΌπΏ =
πππ‘ππ πππ ×1,000
β3 ππΏπππ‘ππ
=
10,000
β3 (230)
πΌπ = 25.1β  β 31.79° π΄ππππππ
πΈπ = πβ + πΌπ ππ
πΈπ = 132.79β 0° + (25.1β  β 31.79° )(0.5 + π1.2)
ο·
πΈπ = 160.45β 6.8° ππππ‘π
%π. π =
%π. π =
ο·
πΈπ βπβ
× 100
πβ
160.45β132.79
132.79
100
%π½. πΉ = ππ. π%
REE β September 2010
28. A three-phase, wye-connected wound rotor synchronous generator rated at 10 kVA, 230 V has a
synchronous reactance of 1.2 ohms/phase and armature resistance of 0.5 ohm/phase. What is the
A. β 0.92%
B. β 5.14%
C. β 3.08%
D. β 3.91%
Solution:
πβ =
ο·
ππΏ
β3
β 0° =
230
β3
β 0°
πβ = 132.79β 0° ππππ‘π
πΌπ = πΌπΏ =
10,000
β3 (230)
β πππ  β1 0.80
ο· πΌπ = 25.1β 36.87° π΄ππππππ
πΈπ = πβ + πΌπ ππ  = 132.79β 0° + (25.1β 36.87° )(0.5 + π1.2)
48
ο·
πΈπ = 128.69β 14.22° ππππ‘π
%π. π =
ο·
πΈπ βπβ
πβ
× 100
%π½. πΉ = βπ. ππ%
REE β May 2010
29. A three-phase wye-connected wound rotor synchronous generator rated at 10 kVA, 230 V has
synchronous reactance of 1.2 ohms per phase and armature resistance of 0.5 ohm per phase. What is
the power factor such that the voltage regulation at full load is zero?
Solution:
πΌπ = πΌπΏ =
10,000
β3 (230)
β 0°
ο· πΌπ = 25.1β 0° π΄ππππππ
πΌπ ππ  = (25.1β 0° )((0.5 + π1.2)
ο· πΌπ ππ  = 32.63β 67.38° ππππ‘π
πβ =
ο·
ππΏ
β3
β 0° =
230
β3
β 0° and
%π. π =
πΈπ βπβ
πβ
× 100 = 0
πΈβ = πβ = 132.79β 0° ππππ‘π
πΌπ ππ
πΈβ
πΌπ ππ
πΌπ = πΌπΏ
67.38°
πΌ
π
πΌπ ππ
πβ
β ππ¦ πππ πππ πππ€:
πΈβ 2 = πβ 2 + (πΌπ ππ  )2 β 2(πβ )(πΌπ ππ  )πππ πΌ
(132.73)2 = (132.73)2 + (32.63)2 β 2(132.73)(32.63)πππ πΌ
πΌ = πππ  β1 [
32.632
]
2(132.73)(32.63)
ο· πΌ = 82.94°
π = 180° β 82.94° β 63.78°
ο· π = 29.68°
π. π = πππ  π = cos 29.68°
ο· π. π = π. πππ πππππππ
30. A 1, 000 kVA, 3kV, 50 Hz, thee-phase star-connected alternator has an armature effective resistance
of 0.2 β¦. A field current of 40 A produces a short-circuit current of 200 A and an open-circuit e.m.f of
1, 040 V (line value). Calculate the full load percentage regulation at a power factor of 0.8 lagging.
49
A. β21.43%
Solution:
B. 21.43%
C. β24.31%
D. 24.31
3,000
πβ =
β3
ο·
πβ = 1, 732.1β 0° ππππ‘π
πΌπ = πΌπΏ =
ο·
β3 (3,000)
β  β πππ  β1 0.80
πΌπ = 192.45β  β 36.87° π΄ππππππ
πΈππΆ
ππ  =
1,000(1,000)
=
πΌππΆ
ο·
1,040/β3
200
ππ  = 3 πβππ
ππ  = βππ  2 β ππ 2 = β32 β 0.22
ο· ππ  = 2.99 πβππ
πΈπ = πβ + πΌπ ππ  = 1, 732.1β 0° + (192.45β  β 36.87° )(3 + π2.99)
ο·
πΈπ = 2, 152.92β 11.72° ππππ‘π
%π. π =
ο·
πΈπ βπβ
πβ
× 100 =
2,152.92β1,732.1
1,732.1
× 100
%π½. πΉ = ππ. π%
31. Adjustment of field excitation on one of the two alternators operating in parallel will
B. change its frequency
D. change its power factor
REE β September 2001
32. A generator is being synchronized to a large system. The actual system voltage and frequency are
13.7 kV and 60 Hz, respectively. The generator voltage and frequency are 13.6 kV and 60 Hz,
respectively. When the generator is switched to the system, choose which one happens,
A. generator delivers MVAR
B. generator takes MVAR
C. generator delivers MW
D. generator delivers both MW and MVAR
Solution:
π ππππ π‘βπ πππππππ‘ππ π£πππ‘πππ ππ  πππ π  π‘βππ π‘βπ π π¦π π‘ππ π£πππ‘πππ
π‘βπ πππππππππ πππππ π΄π½π¨πΉ ππππ π‘βπ π π¦π π‘ππ
33. A 25 kVA, 220 Volts, three-phase alternator delivers rated kVA at a power factor of 0.84. The effective
ac resistance between armature winding terminals is 0.18 β¦. The field takes 9.3 Amperes at 115
Volts. If friction and windage loss is 460 watts and the core loss is 610 Watts. Calculate the percent
efficiency.
A. 87.82%
B. 84.27%
C. 86.41%
D. 88.33%
Solution:
πππ’π‘ = 25(0.84)
ο· πππ’π‘ = 21 ππ
πΌπ =
25,000
β3 (220)
ο· πΌπ = 65.61 π΄ππππππ
β π β πππππππ‘ππ
ππ =
ο·
ππ‘
2
=
0.18
2
ππ = 0.09 πβππ
π³πππππ:
1. πππππ‘π’ππ π€ππππππ
50
ππ = 3πΌπ 2 ππ = 3(65.612 )(0.09)
ππ = 1, 162.26 πππ‘π‘π
2. πππππ π€πππππππ
ππ = ππ πΌπ = (115)(9.3)
ππ = 1. 069.5 πππ‘π‘π
3. πππππ‘πππ πππ π€ππππππ πππ π
πππ€ = 460 πππ‘π‘π
4. ππππ πππ π
ππ = 610 πππ‘π‘π
π‘ππ‘ππ πππ π ππ  = ππ + ππ + πππ€ + ππ = 1, 162.26 + 1, 069.5 + 460 + 610
ο·
Ι³=
π‘ππ‘ππ πππ π ππ  = 3, 301.76 πππ‘π‘π
πππ’π‘
πππ’π‘ +ππππ π ππ
ο·
=
21
21+3.302
Ι³ = ππ. ππ%
34. When the speed of an alternator is changed from 3, 600 rpm to 1, 800 rpm, the generated emf/phase
will become
A. one half
B. twice
C. four times
D. one fourth
REE β September 2009
35. Two alternators operating in parallel supply 2, 500 kW to a load at a power factor of 80% lagging. If
one machine delivers 1, 200 kW at a power factor of 95% lagging, what is the power supplied by the
other machine?
A. 1, 400 kW
B. 1, 600 kW
C. 1, 300 kW
D. 1, 500 kW
Solution:
ππ‘ = π1 + π2
π2 = ππ‘ β π1
π2 = 2, 500 β 1, 200
ο· π·π = π, πππ ππΎ
36. Two exactly similar turbo-alternators are rated 20 MW each. They are running in parallel. The speed
load characteristics of the driving turbines are such that the frequency of alternator one drops
uniformly from 50 Hz on no-load to 48 Hz on full-load, that of the alternator two from 50 Hz to 48.5
Hz. How will the two machines share a load of 30, 000 kW?
A. 14.56 MW, 15.44 MW
B. 12.80 MW, 17.20 MW
C. 17.47 MW, 12.53 MW
D. 16.92 MW, 13.08 MW
Solution:
β πππ ππππππππππ π
50βπ
50β48
=
ππ1
20
ο· ππ1 = 500 β 10π
β πππ ππππππππππ π
50βπ
50β48.5
=
β ππππππππ π
ππ2
20
ππ2 = 666.67 β 13.33π
30 β ππ1 = 666.67 β 13.33π
ο· ππ1 = 13.33π β 636.67 β ππππππππ π
β ππππππππ π = ππππππππ π
500 β 10π = 13.33π β 636.67
51
π=
500+636.67
13.33+10
ο·
π = 48.72 π»π§
ππ1 = 500 β 10π
ππ1 = 500 β 10(48.72)
ο· π΄πΎπ = ππ. π π΄πΎ
ππ2 = 666.67 β 13.33π
ππ2 = 666.67 β 13.33(48.72)
ο· π΄πΎπ = ππ. π π΄πΎ
37. Each of two single-phase alternators has an armature winding whose resistance and synchronous
reactance are respectively, 0.025 and 0.06 ohm. If the machines are operating without load, calculate
the circulating current in the windings if the alternators are paralleled, when the emfs are equal at
230 Volts, but are displaced 30° from a position of phase opposition.
A. 915 A
B. 1, 830 A
C. 457.5 A
D. 119 A
Solution:
ππ  1 = ππ  πΏ = ππ + πππ
ππ  1 = 0.025 + π0.06
ο·
ππ  1 = 0.065β 67.38° β¦
π΅π¦ πΎππΏ:
πΈ1 β πΌπ (ππ  1 + ππ  2 ) β πΈ2 = 0
πΌπ =
πΌπ =
πΈ1 βπΈ2
ππ  1 +ππ  2
230β 0°β230β β30°
2(0.065β 67.38°)
ο·
π°π = πππ. ππβ π. ππ° π¨ππππππ
38. A 3-phase 11 kV, 10 MVA alternator has a sequence reactances of positive sequence, π1 = π0.15 π. π’ ;
negative sequence, π2 = π0.15 π. π’ and π0 = π0.15 π. π’, with negligible neutral wire reactance. If the
alternator is on no-load, calculate the line to ground and 3-phase fault currents.
A. 4, 499 A and 3, 499 A
B. 2, 814 A and 2, 187 A
C. 4, 499 A and 3, 030 A
D. 1, 842 A and 854 A
Solution:
β πππ ππππ π‘π ππππ’ππ πππ’ππ‘
πΌπΉ = 3πΌπΏ πππ‘ππ (
πΌπΉ = 3 [
1
π1 ππ’ +π2 ππ’ +π0 ππ’
10×106
](
β3 (11×103 )
)
1
)
0.15+0.15+0.15
ο· π°π­ = π, πππ π¨ππππππ
β πππ π‘βπππ β πβππ π πππ’ππ‘
πΌπΉ = 3πΌπΏ πππ‘ππ (
πΌπΉ = 3 [
ο·
1
π1 ππ’
10×106
)
](
β3 (11×103 )
1
0.15
)
π°π­ = π, πππ π¨ππππππ
39. In Problem No.37, when the voltages are in phase opposition, but one voltage is 230 and the other is
200 Volts.
A. 462 A
B. 231 A
C. 400 A
D. 800 A
52
Solution:
β π΅πππ: ππ %π2 ππ  πππ‘ πππ£ππ π‘βππ ππ π π’ππ π‘βππ‘ ππ‘ ππ  πππ’ππ π‘π %π1
πΌπ =
πΌπ =
πΌπ =
πΈ1 βπΈ2
ππ  1 +ππ  2
230β 0°β200β 0°
2(0.065β 67.38°)
230β 0°β200β 0°
2(0.065β 67.38°)
ο·
π°π = πππ. ππβ  β ππ. ππ° π¨ππππππ
40. A 600 kVA, 2.4kV, 3-phase alternator has a zero phase sequence reactance of 12%, and a positive and
negative phase sequence reactances of 8%. If the alternator resistance is negligible, determine the
fault currents that the alternator can sustain if a three-phase fault occurs at its terminals.
A. 1, 042 A
B. 1, 804 A
C. 3, 125 A
D. 602 A
Solution:
β πππ π‘βπππ β πβππ π πππ’ππ‘
πΌπΉ = πΌπΏ πππ‘ππ (
πΌπΉ = [
ο·
100
)
%π1
600(1,000) 100
β3 (2,400)
](
8
)
π°π­ = π, πππ. ππ π¨ππππππ
41. In Problem No. 40, determine the fault currents delivered if a phase to phase fault occurs at the
alternator terminals.
A. 1, 562 A
B. 902 A
C. 1, 804 A
D. 3, 125 A
Solution:
/πΌπ /=/πΌπ /= β3 [πΌβ πππ‘ππ (
100
%π1 +%π2
600(1,000) 100
/πΌπ /=/πΌπ /= β3 [
β3 (2,400)
](
8+8
)]
)
ο· /π°π /=/π°π /= π, πππ π¨ππππππ
a
πΌπ
πΌπ
fault
b
πΌπ
c
53
```
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