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Board Exam Problems on AC MACHINES (Transformers)
1. The high-voltage coil of a transformer is wound with 700 turns of wire, and the lowvoltage coil is wound with 292 turns. When used as a step-up transformer (the lowvoltage coil is used as the primary), the load current is 10.5 A. Find the load component of the primary current. A. 43.5 A B. 4.38 A C. 25.18 A D. 2.518 A Solution: ππ ππ
=
πΌπ =
πΌπ πΌπ πΌπ ππ ππ
=
(700)(10.5) 292
ο· π°π = ππ. ππ π¨ππππππ REE β May 2008 2. A transformer has a primary winding of 2, 000 turns and of 2, 400 Volts and current of 8.66 β π5 Ampere with an impedance π2 connected across the secondary winding. If the secondary winding has 500 turns, what is the value of the secondary current? A. 20 β π34.64 π΄ B. ππ. ππ β πππ π¨ C. 34.64 + π20 π΄ D. 20 + π34.64 π΄ Solution: ππ ππ
=
πΌπ =
πΌπ πΌπ πΌπ ππ ππ
ο·
=
(8.66βπ5)(2,000) 500
π°π = ππ. ππ β πππ π¨ππππππ
3. A 120 V to 27.5 V, 400 Hz step-down transformer is to be operated at 60 Hz. What is the highest safe input voltage? A. 200 V B. 400 V C. 120 V D. 18 V Solution: πΈ1 πΈ2
=
πΈ2 =
π1 π2 πΈ1 π2 π1
=
(120)(60) 400
ο· π¬π = ππ π½ππππ REE β September 2011 4. When a welding transformer is used in a resistance welding, it will A. step up voltage B. step down voltage C. step up current D. step down current β πππ‘π: πΌπ π π€ππππππ π‘ππππ ππππππ, π βππβ π£πππ’π ππ ππ’πππππ‘ ππ ππππππ π‘π ππππ‘ πππ πππ¦ π‘βπ ππππ.
5. A 4, 600/230 V, 60 Hz step-down transformer has core dimension of 76.2 mm by 111.8 mm. A maximum flux density of 0.93 ππ/π2 is to be used. Assuming 9 percent loss of area due to stacking factor of laminations, calculate the primary and secondary turns required. A. 2, 395 and 120 B. 120 and 2, 395 C. 2, 180 and 109 D. 109 and 2, 180 Solution:
π΄πππ = (1 β 0.09)(76.2 ππ Γ 111.8 ππ) π΄πππ = 7, 752.436ππ2 (
1π 1,000 ππ β3 2
ο· π΄πππ = 7.752 Γ 10 β
π = π½π π΄πππ β
π = (0.93
ππ π2
)
2
π
) (7.752 Γ 10β3 π2 )
ο· β
π = 7.209 πππ πΈπππ = 4.44πβ
π ππ =
πΈπππ 4.44πβ
=
4,600 (4.44)(60)(7.209Γ10β3 )
ο· π΅π = π, πππ πππππ ππ =
πΈπππ 4.44πβ
=
230 (4.44)(60)(7.209Γ10β3 )
ο· π΅π = πππ πππππ REE β October 1997 6. A small single-phase transformer has 10.2 watts no-load loss. The core has a volume of 750 cubic cm. The maximum flux density is 10, 000 gauss and the hysteresis constant of the core is 5 Γ 10β4 , using the Steinmetz law to find the hysteresis, determine the eddy current loss. A. 4.55 Watts B. 5.55 Watts C. 3.55 Watts D. 2.55 Watts Solution: β ππ‘πππππππ‘π§ πππ€
πβ πΌ π½π 1.6 β πβ = (5 Γ 10β4 )(750) ο· πβ = 0.375 πβ = (0.375)(60)(10, 000)1.6 ο· πβ = 5.652 πππ‘π‘π ππ = πππ β πβ ππ = 10.2 β 5.652 ο· π·π = π. πππ πΎππππ
ππ¦ππβππ π ππ
(
1π 105 ππ¦ππ
)(
1π 100 ππ
)
REE β September 2006 7. The primary of transformer has 200 turns and is excited by a 240 V, 60 Hz source. What is the maximum value of the core flux? A. 4.04 mWb B. 4.40 mWb C. 4.13 mWb D. 4.32 mWb Solution:
β
π =
πΈ
240
4.44ππ
= (4.44)(60)(200)
ο· β
π = π. ππ ππΎπ REE β September 2008 8. A transformer is rated 1 kVA, 220/110 V, 60 Hz. Because of an emergency this transformer has to be used on a 50 Hz system. If the flux density in the transformer core is to be kept the same as at 60 Hz and 220 V, what is the kilovolt-ampere rating at 50 Hz. A. 0.890 kVA B. 0.833 kVA C. 0.909 kVA D. 0.871 kVA Solution:
ππΌπ π1 π2
π1
=
π2 =
π2 π1 π2 π1
=
(1)(50) 60
ο· πΊπ = π. πππ ππ½π¨ 9. A single-phase transformer has a no-load power input of 250 Watts, when supplied at 250 V, 50 Hz has a p.f of 0.25. What is the magnetizing component of the no-load current? A. 4.00 A B. 3.87 A C. 1.00 A D. none of these Solution:
π=
π π.π
=
250 0.25
ο· π = 1, 000 ππ΄ ππ = πππ β1 0.25 ο· ππ = 75.52Β° π = ππΌ πΌ=
1,000β 75.52 250
= 4β 75.52Β° π΄ππππππ
ο· πΌ = 1 + π 3.873 π΄ππππππ ο· π°π = π. πππ π¨ππππππ
REE β September 2011 10. A 4, 400 V, 60 Hz transformer has a core loss of 840 Watts, of which one-third is eddy current loss. What is the core loss when the xβformer is connected to a 4, 600 V, 50 Hz source? A. 977 Watts B. 907 Watts C. 927 Watts D. 944 Watts Solution: 1 3
πππππ = πππππ¦ 1
ππ1 = (840) 3
ο· ππ1 = 280 πππ‘π‘π πππππ = πβ + ππ πβ1 = 840 β 280 ο· πβ1 = 560 πππ‘π‘π ππ2
=
ππ1
π2 (πΈ2 )2 π1 (πΈ1 )2 4,6002
ππ2 = 280 (
4,4402
)
ο· ππ2 = 306.033 πππ‘π‘π 1.6
πβ2 πβ1
=
πΈ π2 ( 20.6 ) π 2 πΈ1 1.6 π1 ( 0.6 ) π1
πβ2 = 560 [
4,6001.6 ) 500.6 4,4001.6 ( ) 600.6
(
]
ο· πβ2 = 670.788 πππ‘π‘π πππππ2 = πβ2 + ππ2 = 670.788 + 306.033 ο· π·πππππ = πππ. πππ πΎππππ REE β September 2004 11. In an ideal transformer, what is the efficiency? A. 100% B. 90% C. 80%
D. 70%
12. A 100 kVA distribution transformer has a full-load copper loss of 1, 180 Watts. For what kilowatt load, at a power factor of 0.71, will the copper losses in the transformer be 1, 500 Watts? A. 90.25 B. 71 C. 112.75 D. 80.05 Solution: πππ’βπππ¦ 2 2
πππ’βπΉπΏ π
π=
1,500 1,180
(
=(
ππππ¦ ππππ‘ππ
)
2
π.π
=(
πΎπππππ /0.71 2 100
πΎπππππ 2 0.71
)
) = 1002 (
1,500 1,180
)
πΎπππππ = β(0.71)2 (100)2 (
1,500 1,180
)
ο· π²πΎππππ
= ππ. πππ ππΎ 13. Given a 10-kVA transformer with full-load losses amounting to 70 Watts in the iron and 140 Watts in the copper. Calculate the efficiency at half-load unity power factor. A. 98.62% B. 97.97% C. 97.28% D. 97.94% Solution: ππΆπ’ π»πΏ ππΆπ’ πΉπΏ
=(
ππΆπ’ π»πΏ =
ππ»πΏ 2 ππΉπΏ
)
140(52 ) 102
ο· ππΆπ’ π»πΏ = 35 πππ‘π‘π Ι³ π»πΏ =
ππ π»πΏ ππ π»πΏ+πππππ +ππΆπ’ π»πΏ
Γ 100% =
5,000 5,000+35
Γ 100%
ο· Ι³ π―π³ = ππ. ππ% REE β April 2004 14. Instrument transformers are used in indicating and metering and with protective devices, they are used for . A. measuring B. detecting C. relaying D. sensing REE β September 2003 15. What type of transformer bank is used to convert 2-phase to 3-phase power? A. open-delta B. scott-T C. wye-delta D. delta-wye
16. A 100-kVA 2, 400/240-Volt 60 cycle transformer has the following constants: ππ = 0.42 β¦, ππ = 0.72 β¦; ππ = 0.0038 β¦, ππ = 0.0068 β¦. What is the equivalent impedance in primary terms? A. 0.016 β¦ B. 1.612 β¦ C. 0.161 β¦ D. 16.12 β¦ Solution:
ππ π = (ππ + π2 ππ ) + π(ππ + π2 ππ ) π=
ππ ππ
=
2,400 240
ο· π = 10 ππ π = (0.42) + (102 )(0.0038) + π(0.72) + (10)2 (0.0068) ππ π = 0.8 + π1.4 = 1.61β 60.26Β° ο· |ππ π | = π. ππ ππππ 17. Calculate the all-day efficiency of a 100-kVA transformer operating under the following conditions: 6 hours on a load of 50 kW at 0.73 power factor; 3 hours on a load of 90 kW at 0.82 power factor; 15 hours with no load on secondary. The iron loss is 1, 000 Watts and the full-load copper loss is 1, 060 Watts. A. 96.31% B. 94.87% C. 95.33% D. 95.29% Solution: πππ’π‘ = (50, 000)(6) + (90, 000)(3) + (0)(15) ο· πππ’π‘ = 570 ππβπ πππππ = (1, 000)(24) ο· πππππ = 24 ππβπ ππΆπ’ 1 ππΆπ’ πΉπΏ
πππ΄
2
= (πππ΄ 1 ) πΉπΏ
ππΆπ’ 1 = 1, 060 (
50/0.73 2 100
)
ο· ππΆπ’ 1 = 497.279 πππ‘π‘π ππΆπ’ 2 = 1, 060 (
90/0.82 2 100
)
ο· ππΆπ’ 2 = 1, 276.919 πππ‘π‘π ο· ππΆπ’ 3 = 0 πππ‘π‘π ππΆπ’ π‘ππ‘ππ = ππΆπ’ 1 (6) + ππΆπ’ 2 (3) + ππΆπ’ 3 (15) ππΆπ’ π‘ππ‘ππ = (487.279)(6) + (1, 276.919)(3) + (0)(15) ο· ππΆπ’ π‘ππ‘ππ = 6.814 ππβπ Ι³πππ πππ¦ =
πππ’π‘ πππ’π‘ +πππππ+ππΆπ’
Γ 100% =
570 570+24+6.814
Γ 100%
ο· Ι³πππ π
ππ = ππ. ππ% REE β September 2005 18. A 50-kVA, single-phase transformer has 96% efficiency when it operates at full-load unity power factor for 8 hours per day. What is the all-day efficiency of the transformer if the copper loss is 60% of full-load losses? A. 92% B. 90% C. 89.5% D. 93% Solution:
0.96 =
50
50+ππππ π ππ πΉπΏ 50 ππππ π ππ πΉπΏ = β 50 0.96
ο· ππππ π ππ πΉπΏ = 2.083 ππ ππΆπ’ πΉπΏ = (0.6)(2, 083) ο· ππΆπ’ πΉπΏ = 1, 249.8 π πππππ = ππππ π + ππΆπ’ πΉπΏ = 2, 083 β 1, 249.8 ο· πππππ = 833.2 πππ‘π‘π πππ’π‘ π‘ππ‘ππ = (50 πππ΄)(1.0)(8) ο· πππ’π‘ π‘ππ‘ππ = 400ππβπ πππππ π‘ππ‘ππ = (833.2)(24) ο· πππ’π‘ π‘ππ‘ππ = 19.997 ππβπ ππΆπ’ π‘ππ‘ππ = (1, 249.8)(8) ο· ππΆπ’ π‘ππ‘ππ = 9.998 ππβπ Ι³πππ πππ¦ =
400 400+19.997+9.998
Γ 100%
ο· Ι³πππ π
ππ = ππ. ππ% Asst. EE β October 1991 19. A 10 kVA, 2, 400/240 V, single-phase transformer has the following resistances and leakage reactances; ππ = 3 β¦ ππ = 0.03 β¦ ππ = 15 β¦ ππ = 0.15 β¦ Find the primary voltage required to produce 240 V at the secondary terminals at full-load, when the load power factor is 0.8 lagging. A. 2, 400 V B. 2, 496.5 V C. 2, 348 V D. 2, 445.5 V Solution:
π=
2,400 240
ο· π = 10
πππ = 10(240) ο· πππ = 2, 400 πΌπ‘ =
10,000 β πππ β1 (0.8) 2,400
ο· πΌπ‘ = 4.17 β β 36.87Β° π΄ππππππ ππ π = [3 + (10)2 (0.03)] + π[15 + (10)2 (0.15)] ο· ππ π = 6 + π30 πβππ ππ = πΌπ‘ ππ π + πππ ππ = (4.17 β β 36.87)(6 + π30 ) + 2, 400 ππ = 2, 496.526 β 1.95Β° ππππ‘π ο· |π½π | = π, πππ. πππ π½ππππ 20. A 500 kVA, single-phase, 13, 200/2, 400 Volts transformer has 4% reactance and 1% resistance. The leakage reactance and resistance of the high voltage (primary) winding are 6.34 β¦ and 1.83 β¦, respectively. The core loss under rated condition is 1, 800 Watts. Calculate the leakage reactance and resistance of the low voltage (secondary) winding. A. 7.56 β¦ and 1.66 β¦ B. 13.69 β¦ and 3.42 β¦ C. 0.25 β¦ and 0.055 β¦ D. 13.9 β¦ and 3.48 β¦ Solution:
ππ =
(ππ )2 ππ
=
(13,200)2 500,000
ο· ππ = 348.48 πβππ π
π π = π
ππ’ ππππ π = (0.01)(348.48) ο· π
π π = 3.4848 πβππ ππ π = πππ’ ππππ π = (0.04)(348.48) ο· ππ π = 13.9392 πβππ ππππππππ π‘π πππππππ¦ π
π π = ππ + π2 ππ π=
13,200 2,400
ο· π = 5.5 ππ =
π
π π βππ π2
=
3.4848β1.83 5.52
ο· ππ = π. πππ ππ π = ππ + π2 ππ
ππ =
ππ πβππ π2
=
13.9392β6.34 5.52
ο· πΏπ = π. πππ ππππ 21. In Problem No.20, calculate the %V.R and efficiency of the transformer at full-load, 0.85 p.f. lagging and 2, 400 Volts. A. 4% and 97.8% B. 6% and 95.4% C. 5% and 96.8% D. 3% and 98.4% Solution: ππ π2
1.83
=
5.52 ππ
ο· ππ π2
= ο·
π2 6.34 5.52 ππ
= 0.0605 πβππ
= 0.21 πβππ
π2 β1
π = πππ (0.85) ο· π = 31.79Β° πΌπ =
500,000 β β31.79Β° 2,400
ο· πΌπ = 208.33 β β 31.79Β° ππ π = (
ππ π2
+ ππ ) + π (
ππ π2
+ ππ ) = (0.0605 + 0.055) + π(0.21 + 0.25)
ο· ππ π = 0.1155 + π0.46 β¦ ππ = πΌπ ππ π + ππ ππ = (208.33 β β 31.79Β°)(0.1155 + π0.46) + 2, 400 ο· ππ = 2, 471.89 β 1.6Β° ππππ‘π %ππ
=
2,471.89β2,400 2,400
Γ 100
ο· %π½πΉ = π% πππ’π‘ = (500 πππ΄)(0.85) ο· πππ’π‘ = 425 ππ ο· πππππ = 1, 800 π πππ’ = πΌπ 2 π
π = (208.33)2 (0.1155) ο· πππ’ = 5, 012.86 π Ι³=
425,000 425,000+1,800+5,012.86
ο· Ι³ = ππ %
Γ 100
22. An 11, 000/230 V, 150 kVA, single-phase, 50 Hz transformer has a core loss of 1.4 kW and a full-load copper loss of 1.6 kW. What is the value of maximum efficiency at unity p.f? A. 98.17% B. 98.04% C. 97.22% D. 97.64% Solution:
πππ’ πππ₯ = πππππ ο· πππ’ πππ₯ = 1, 400 π πππ’ πππ₯ πππ’ πΉπΏ
=(
πππ΄πππ₯ 2 πππ΄πΉπΏ
)
πππ΄πππ₯ = β(1502 ) (
1,400 1,600
)
ο· πππ΄πππ₯ = 140.31 πππ΄ πππππ₯ = πππ΄πππ₯ (π. π) = (140.31)(1.0) ο· πππππ₯ = 140.13 ππ Ι³=
140.13 140.13+2(1.4)
Γ 100
ο· Ι³ = ππ. ππ % 23. A 300-kVA, single-phase transformer is designed to have a resistance of 1.5% and maximum efficiency occurs at a load of 173.2 kVA. Find its efficiency when supplying full-load at 0.8 p.f. lagging at normal voltage and frequency. A. 97.56% B. 96.38% C. 98.76% D. 95.89% Solution:
ππ.π’ =
πππ’ πΉπΏ ππΉπΏ
πππ’ πΉπΏ = ππ.π’ ππΉπΏ = (0.015)(300, 000) ο· πππ’ πΉπΏ = 4, 500 πππ‘π‘π πππ’ πππ₯ 4,500
=(
173.2 2 300
)
ο· πππ’ πππ₯ = 1, 499.91 πππ‘π‘π πππ’ πππ₯ = πππππ Ι³πΉπΏ =
πππ’π‘ πππ’π‘ +πππππ +πππ’ πΉπΏ
300(0.8)
Γ 100 = (300)(0.8)+1.5+4.5 Γ 100
ο· Ι³ππ³ = ππ. ππ %
REE β September 2002 24. A 20 kV/7.87 kV autotransformer has 200 A current in the common winding. What is the secondary line current? A. 143.52 B. 200 C. 56.48 D. 329 Solution: πΌπ πΌπ
=πβ1
π=
20,000 7,870
ο· π = 2.54 200 πΌπ
= 2.54 β 1
ο· πΌπ = 129.76 π΄ππππππ πΌπ = πΌπ + πΌπ = 129.76 + 200 ο· π°π = πππ. ππ π¨ππππππ 25. An autotransformer is adjusted for an output voltage of 85.3 Volts when operated from a 117 Volts line. The variable power load draws 3.63 kW at unity power factor at this setting. Determine the transformed power and the connected power from the source to the load. A. 980 Watts and 2, 650 Watts B. 1, 343 Watts and 2, 287 Watts C. 1, 815 Watts and 1, 815 Watts D. 1, 210 Watts and 2, 420 Watts Solution:
π=
117 85.3
ο· π = 1.37 1
1
π
1.37
ππ‘ππππ = πππ (1 β ) = (3, 630) (1 β
)
ο· π·πππππ = πππ. ππ πΎππππ ππππ =
πππ π
=
3,630 1.37
ο· π·πππ = π, πππ. ππ πΎππππ REE β April 2006 26. What would happen if you connect a transformer to a dc circuit with a voltage of 20% of nameplate ratings after a steady state condition is reached? A. No voltage is registered at the secondary B. Rated no-load current flows to the secondary C. primary current is equal to voltage over equivalent primary impedance D. Voltage is established at secondary
27. A short-circuit test was performed upon a 10 kVA, 2, 300/230-Volt transformer with the following results: πΈπ π = 137 ππππ‘π ; ππ π = 192 πππ‘π‘π ; πΌπ π = 4.34 π΄ππππππ . Calculate in secondary terms the transformer equivalent. A. 29.88 β¦ B. 2.988 β¦ C. 0.2988 β¦ D. 298.8 β¦ Solution:
π
=
ππ π
ο·
π=
=
πΌπ π 2
192 4.34 2
π
= 10.19 β¦ πΈπ π πΌπ π
ο·
=
137 4.34
π = 31.57 β¦
π = βπ 2 β π
2 = β31.572 β 10.192
ο· πΏ = ππ. ππ β¦ REE β April 2007 28. A transformer is rated 500 kVA, 4, 800/480 V, 60 Hz when it is operated as a conventional two winding transformer. This transformer is to be used as a 5280/4800 V stepdown autotransformer in a power distribution system. In the autotransformer, what is the transformer rating when used in this manner? A. 5 MVA B. 6 MVA C. 5.5 MVA D. 6.5 MVA Solution: π
500,000
πΌπ = π =
4,800
π
ο·
πΌπ = 104.17 π΄ππππππ π
500,000
π
4,80
πΌπ = π =
ο· πΌπ = 1, 041.7 π΄ππππππ πΌπ β² = πΌπ ο· πΌπ β² = 1, 041.7 π΄ππππππ π = ππ β²πΌπ β² = (5, 280)(1, 041.7)
ο· πΊ = π. π π΄π½π¨ REE β September 2008 29. Two identical transformers bank on open delta serve a balanced three-phase load of 26 kVA at 240 V, 60 Hz. What is the minimum size of each in kVA needed to serve this load? A. 25 B. 10 C. 30 D. 15 Solution: πβ
πππ‘ππ =
πβ
ππππ (β3)
=
26,000 β3
ο· πΊβ
πππππ
= ππ ππ½π¨
30. Two single-phase, 100 kVA transformers are connected in V (open delta) bank supplying a balanced three-phase load. If the balanced three-phase load is 135 kW at 0.82 p.f lagging and 0.823 efficiency, determine the overload kVA on each transformer. A. 10.5 B. 5.5 C. 15.5 D. 20.5 Solution:
ππΏ =
π π.π
=
135,000 0.82
ο· ππΏ = 164.634 πππ΄ Ι³=
πππ’π‘ πππ 135
πππ =
0.823
ο· πππ = 164.034 ππ πππ =
πππ π.π
=
164.034 ππ 0.82
ο· πππ = 200.041 πππ΄ πβ
=
πππ β3
=
200.041 β3
ο· πβ
= 115.494 πππ΄ ο· πβ
πππ‘ππ = 100 πππ΄ ππ.πΏ = ππΏβ
β πβ
πππ‘ππ = 115.494 β 100 ο· πΊπΆ.π³ = ππ. πππ ππ½π¨ 31. In problem No. 30, determine the p.f of each transformer secondary. A. 0.820 lagging and 0.820 lagging B. 0.996 lagging and 0.424 leading C. 0.996 lagging and 0.424 lagging D. 0.410 lagging and 0.410 lagging Solution:
π = πππ β1 (0.82) ο· π = 34.92Β° π. π1 = ππ π (30 + π) = ππ π (30 + 34.92) ο· π. ππ = π. πππ πππππππ π. π2 = ππ π (30 β π) = ππ π (30 β 34.92) ο· π. ππ = π. πππ πππππππ REE β April 2005 32. What is the normal secondary circuit current of a current transformer? A. 15 A B. 20 A C. 5 A D. 10 A
33. In Problem No.30, what is the minimum size in kVAR of a capacitor bank to be connected across the load so that each transformer is loaded 96% of its rated capacity? A. 87 kVAR B. 114 kVAR C. 27 kVAR D. 66 kVAR Solution:
πβ
= (0.96)(πβ
πππ‘ππ ) = (0.96)(100) ο· πβ
= 96 πππ΄ ο· ππ = 164.034 ππ ο· π = 200.041 πππ΄ π = βπ 2 β ππ 2 = β(200.041)2 β (164.034)2 ο· π = 114.5 πππ΄π
ππππ€ = (96 πππ΄)(β3) ο· ππππ€ = 166.277 πππ΄ π = π2 + π2 2
π = βππππ€ 2 β ππ 2 = β(166.277)2 β (164.034)2 ο· πΈ = ππ. ππ ππ½π¨πΉ 34. A polarity test is performed upon a 1, 150/115 V transformer. If the input voltage is 116, calculate the voltmeter reading if the polarity is subtractive. A. 127.6 V B. 106 V C. 126 V D. 104.4 V Solution: π π’ππ‘ππππ‘ππ£π π=
1,150
ο·
115
π = 10
ππππππππ = 116 β ο·
116 10
π½ππππ
πππ = πππ. π π½ππππ
35. A 20:1 potential transformer is used with a 150-V voltmeter. If the instrument deflection is 118 Volts, calculate the line voltage. A. 3, 000 V B. 2, 850 V C. 2, 360 V D. 2, 242 V Solution:
π= 20 1
=
ππΏ ππππππππ ππΏ 118
ο· π½π³ = π, πππ π½ππππ
REE βSeptember 2010 36. A three-phase wye-delta connected, 50 MVA, 345/34.5 kV transformer is protected by differential protection. The current transformer on the high side for differential protection is 150:5. What is the current on the secondary side of CTβs? A. 3.83 A B. 2.53 A C. 4.50 A D. 4.83 A Solution: π
πΌβ
π =
=
3πβ
π
50,000,000 3(
345,000 ) β3
ο· πΌβ
π = 83.67 π΄ππππππ π ππππ π€π¦π β πππππππ‘πππ πΌβ
π = πΌπΏ π ο· πΌπΆπ π = πΌπΆπ π = 83.67 π΄ππππππ πΏ
π= 150 5
πΌπ πΌπ
=
=
β
ππ
ππ 83.67 πΌπΆπ π
β
πΌπΆπ π =
(83.67)(5) 150
β
ο· πΌπΆπ π = 2.789 π΄ππππππ β
πΌπΆπ π ππ = β3 (2.789) ο· π°πͺπ» πππ = π. πππ π¨ππππππ REE β October 2000 37. The CT ratio and PT ratio used to protect a line are 240 and 2,000, respectively. If the impedance of each line is 10 β¦, what is the relay impedance to protect the line from fault? A. 83.33 ohms B. 1.2 ohms C. 48, 000 ohms D. 12 ohms Solution:
ππ =
πππ πΆππ
=
2,000 240
ο· ππ = 8.33 πβππ ππ =
πππππ
ππππππ¦ 10 ππππππ¦ = 8.33
ο· ππππππ = π. π ππππ
38. Two transformers 1 and 2 are connected in parallel supplying a common load of 120 kVA. Transformer 1 is rated 50 kVA, 7, 620/240-V single-phase and has an equivalent impedance of 8.5 β¦ while transformer 2 is rated 75 kVA, 7, 620/240-V single-phase and has an equivalent impedance of 5.1 β¦. The two transformers operate with the same power factors. What is the kVA load of each transformer? A. 48 & 72 B. 45 & 75 C. 42 & 78 D. 40 & 80 Solution: ππ 1 ππ 2 8.5 5.1
= =
π2 π1 π2 π1
ο· π2 = 1.67π1 β ππβ²π 1 ο· ππΏ = π1 + π2 β ππβ² π 2 ππ’ππ π‘ππ‘π’π‘π ππβπ 1 ππ 2 120 = π1 + 1.67π1 ο· πΊπ = ππ. ππ ππ½π¨ π2 = 1.67(44.94) ο· πΊπ = ππ. ππ ππ½π¨ 39. Two single-phase transformers are connected in parallel at no-load. One has a turns ratio of 5, 000/440 and rating of 200 kVA, the other has a ratio of 5, 000/480 and rating of 350 kVA the leakage reactance of each is 3.5%. The no-load circulating current is . A. 207 A B. 702 A C. 720 A D. 270 A Solution:
ππππ π1 =
(ππππ π1 )
2
=
ππππ π1
(440)2 200,000
ο· ππππ π1 = 0.968 β¦ ππππ π2 =
(ππππ π2 ) ππππ π2
2
=
(480)2 350,000
ο· ππππ π2 = 0.658 β¦ ππβπ 1 = ππ ππππ π1 = (0.035)(0.968) ο· πΏπβππ = π. ππππ β¦ ππβπ 2 = ππ ππππ π2 = (0.035)(0.658) ο· πΏπβππ = π. πππ β¦
REE β October 1997 40. A power transformer rated 50, 000 kVA, 34.5/13.8 kV is connected Y-grounded primary and delta on the secondary. Determine the full load phase current at the secondary side. A. 2, 092 A B. 1, 725 A C. 1, 449 A D. 1, 208 A Solution: π = β3 ππΏ πΌπΏ 50,000,000
πΌπΏ π =
(34,500)(β3)
ο· πΌπΏ π = 836.74 π΄ππππππ πΌπΏ π = πΌβ
π ππ ππ
πΌ
= πΌπ
π
πΌβ
π =
(34,500)(836.74) 13,800 (β3)
ο·
π°π β
= π, πππ. ππ π¨ππππππ
ππππ‘βππ π πππ’π‘πππ: πΌπΏ π = ( ο· πΌβ
π = ο·
50,000,000 β3) (13,800)
πΌπΏ π = 2, 091.849 πΌπΏ π 2,091.849 β3
=
π΄ππππππ
β3
π°β
π = π, πππ. ππ π¨ππππππ
REE β April 2006 41. A 2, 000 kW, 2, 400-V, 75% p.f load is to be supplied from a 34, 5000-V, 3-phase line through a single bank of transformers. Give the primary and secondary line currents in amperes for the wye-wye connections. A. 50/700 B. 48/650 C. 60/800 D. 45/642 Solution: π β β πΆππππππ‘πππ
πΌπΏ π = [
β 2 ππ β πππ β1 (0.75) 0.75 β3 (34,500)
]
πΌπΏ π = 44.626β 44.41Β° ο·
π°π³ π = ππ. πππ π¨ππππππ β 2 ππ β πππ β1 (0.75) 0.75
πΌπΏ π = [
β3 (2,400)
]
πΌπΏ π = 641.5β 44.41Β° ο·
π°π³ π = πππ. π π¨ππππππ
REE β April 2005 42. A 3, 000 kVA, 2, 400 V, 75% power factor load is to be supplied from a 34, 500-V, three-phase line through a single bank of transformers. What is the voltage rating of each transformer if the connection is wye-wye? A. 20, 000/1, 380 B. 18, 500/1, 350 C. 18, 000/1, 850 D. 19, 000/1, 350 Solution:
πβ
π = ο·
34,500 β3
π½β
π = ππ, πππ. πππ π½ππππ
πβ
π = ο·
2,400 β3
π½β
π = π, πππ. πππ π½ππππ
REE β March 1998 43. A 13.8 kV/480 V, 10 MVA three-phase transformer has 5% impedance. What is the impedance in ohms referred to the primary? A. 0.952 ohm B. 0.03 ohm C. 5.125 ohm D. 9.01 ohm Solution:
πππ’ = ππ =
ππππ‘π’ππ
ππ (ππ )2 ππ
=
(13,800)2 10,000,000
ο·
ππ π
ππ = 19.044 πβππ = πππ’ ππ = (0.05)(19.044)
ο·
ππ π = π. πππ ππππ
REE β May 2009 44. A three-phase transformer is rated 15 MVA, 69/13.2 kV has a series impedance of 5%. What is the new per unit impedance if the system study requires a 100 MVA base and 67 kV base? A. 0.354 B. 0.347 C. 0.372 D. 0.333 Solution π
π
ππ
ππ
πππ’ πππ€ = πππ’ πππ ( π ) ( π ) πππ’ πππ€ = (0.05) ( ο· πππ
πππ
100 15
2
69 2
)( )
= π. πππ
67
REE β April 2004 45. A transformer rated 2, 000 kVA, 34, 500/240 volts has 5.75% impedance. What is the per unit impedance? A. 0.0635 B. 0.0656 C. 0.0575 D. 34.2 Solution:
πππ’ = 57.5% ο· β΄ πππ’ = 0.0575 46. Three 5:1 transformers are connected in delta-wye to step up the voltage at the beginning of a 13, 200-Volt three-phase transmission line. Calculate the line voltage on the high side of the transformers. A. 114, 300 V B. 66, 000 V C. 132, 000 V D. 198, 000 V Solution: 5 1
=
πβ
2 πβ
1
=
πβ
2 13,200 π
πβ
2 = πππ πβ
2 = (5)(13, 200) ο· πβ
2 = 66, 000 ππππ‘π ππΏ = β3 πβ
2 = β3 (66, 000) ο· π½π³ = πππ, πππ. πππ π½ππππ 47. A 150 kVA, 2, 400/480-V, three-phase transformer with an equivalent impedance of 4%is connected to an infinite bus and without load. If a three-phase fault occurs at the secondary terminals, the fault current in amperes is . A. 4, 512 A B. 3, 908 A C. 7, 815 A D. 1, 504 A Solution: 3 β
πππ’ππ‘ ππ π‘βπ π ππππππππ¦ π‘ππππππππ
πΌπΉ 3β
= πΌπΉ 3β
=
ππ΅ β3 ππ΅ πππ’ 150,000 β3 (480)(0.04)
ο· π°π πβ
= π, πππ. πππ π¨ππππππ
48. Transformer 1 is in parallel with Transformer 2 Transformer 1 Transformer 2 150 kVA, single-phase 300 kVA, single-phase 6, 600/240 V 6, 600/240 V ππβπ 1 = 0.02425β 62.9Β° β¦ ππβπ 2 = 0.01067β 62.9Β° β¦ Determine the maximum kVA load the bank can carry without overloading any of the two transformers, assuming that the two transformers operate at the same power factors. A. 450 kVA B. 432 kVA C. 420 kVA D. 412 kVA Solution:
π1 + π2 = πππππ πΆππ π πΌ: π1 @ 150 πππ΄ ππβπ 1 ππβπ 2 ππβπ 1 ππβπ 2
ο·
= =
π2 π1 0.02425β 62.9Β°
0.01067β 62.9Β° ππβπ 1 ππβπ 2
2.273 =
= 2.273
π2 π1
π2 = 2.273 (150, 000) ο· π2 = 340.95 πππ΄ ο· β΄ ππ£ππππππ ππ‘ π2 πΆππ π πΌπΌ: π2 @ 300 πππ΄
π1 =
300,000 2.273
ο· π1 = 131.984 πππ΄ πππππ = π1 + π2 = (131.984) + (300) ο· πΊππππ
= πππ. πππ ππ½π¨