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Va b = VA VC = V P 0 V P 120 = 3V P 30
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Note that the line voltage on the secondary side lags the line voltage on the primary side by 30 3-21 A single-phase 10-kVA 480/120-V transformer is to be used as an autotransformer tying a 600-V distribution line to a 480-V load When it is tested as a conventional transformer, the following values are measured on the primary (480-V) side of the transformer:
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Open-circuit test VOC = 480 V IOC = 041 A VOC = 38 W
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Short-circuit test VSC = 100 V ISC = 106 A PSC = 26 W
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(a) Find the per-unit equivalent circuit of this transformer when it is connected in the conventional manner What is the efficiency of the transformer at rated conditions and unity power factor What is the voltage regulation at those conditions (b) Sketch the transformer connections when it is used as a 600/480-V step-down autotransformer (c) What is the kilovoltampere rating of this transformer when it is used in the autotransformer connection (d) Answer the questions in (a) for the autotransformer connection SOLUTION (a) The base impedance of this transformer referred to the primary side is
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(VP ) 2 = ( 480 V ) 2 = 2304 =
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S 10 kVA
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The open circuit test yields the values for the excitation branch (referred to the primary side):
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YEX = = 041 A = 0000854 S 480 V
= cos 1
POC 38 W 1 = cos ( 480 V ) ( 041 A ) = 7887 VOC I OC
YEX = GC jBM = 0000854 7887 = 0000165 j 0000838 RC = 1/ GC = 6063
X M = 1/ BM = 1193
The excitation branch elements can be expressed in per-unit as
RC = 6063 = 263 pu 2304 VSC 100 V = = 0943 I SC 106 A PSC 26 W 1 = cos (100 V ) (106 A ) = 758 VSC I SC XM = 1193 = 518 pu 2304
The short circuit test yields the values for the series impedances (referred to the primary side):
Z EQ =
= cos 1
Z EQ = REQ + jX EQ = 0943 758 = 0231 + j 0915
The resulting per-unit impedances are
REQ = 0231 = 0010 pu 2304 X EQ = 0915 = 00397 pu 2304
The per-unit equivalent circuit is
REQ 0010
jXEQ j00397
RC 263
jXM j518
The core losses (in resistor RC ) would be
Pcore V 2 (10) = = = 000380 pu 263 RC
At rated conditions and unity power factor, the input power to this transformer would be PIN = 10 pu
The copper losses (in resistor REQ ) would be
PCU = I 2 REQ = (10) ( 0010) = 0010 pu
The output power of the transformer would be
POUT = POUT PCU Pcore = 10 0010 00038 = 0986
and the transformer efficiency would be
POUT 0986 100% = 100% = 986% PIN 10
The output voltage of this transformer is
VOUT = VIN IZ EQ = 10 (10 0 )( 001 + j 00397 ) = 0991 23
The voltage regulation of the transformer is
VR = 10-0991 100% = 09% 0991
The autotransformer connection for 600/480 V stepdown operation is
NSE 600 V
+ VSE + +
480 V
When used as an autotransformer, the kVA rating of this transformer becomes: 61
SIO =
N C + N SE 4 +1 SW = (10 kVA ) = 50 kVA N SE 1
As an autotransformer, the per-unit series impedance Z EQ is decreased by the reciprocal of the
0010 = 0002 pu 5 00397 = = 000794 pu 5
power advantage, so the series impedance becomes
REQ = X EQ
while the magnetization branch elements are basically unchanged At rated conditions and unity power factor, the input power to this transformer would be PIN = 10 pu The core losses (in resistor RC ) would be
Pcore V 2 (10) = = = 000380 pu 263 RC
The copper losses (in resistor REQ ) would be
PCU = I 2 REQ = (10) ( 0002) = 0002 pu
The output power of the transformer would be
POUT = POUT PCU Pcore = 10 0002 00038 = 0994
and the transformer efficiency would be
POUT 0994 100% = 100% = 994% PIN 10
The output voltage of this transformer is
VOUT = VIN IZ EQ = 10 (10 0 )( 0002 + j 000794) = 0998 05
The voltage regulation of the transformer is
VR = 10-0998 100% = 02% 0998
3-22
Figure P3-4 shows a power system consisting of a three-phase 480-V 60-Hz generator supplying two loads through a transmission line with a pair of transformers at either end (a) Sketch the per-phase equivalent circuit of this power system (b) With the switch opened, find the real power P, reactive power Q, and apparent power S supplied by the generator What is the power factor of the generator (c) With the switch closed, find the real power P, reactive power Q, and apparent power S supplied by the generator What is the power factor of the generator (d) What are the transmission losses (transformer plus transmission line losses) in this system with the switch open With the switch closed What is the effect of adding Load 2 to the system
Region 1 S base1 = 1000 kVA
Region 2 S base2 = 1000 kVA
Region 3 S base3 = 1000 kVA
VL ,base2 = 480 V
VL ,base2 = 13,800 V
VL ,base3 = 480 V
SOLUTION This problem can best be solved using the per-unit system of measurements The power system can be divided into three regions by the two transformers If the per-unit base quantities in Region 1 are chosen to be S base1 = 1000 kVA and VL ,base1 = 480 V, then the base quantities in Regions 2 and 3 will be as shown above The base impedances of each region will be: 2 2 3V 3 ( 277 V ) Z base1 = 1 = = 0238 Sbase1 1000 kVA
Z base2 = Z base3 = 3V 2 3V 3
Sbase2
3 ( 7967 V ) = = 1904 1000 kVA
Sbase3
3 ( 277 V ) = 0238 1000 kVA
(a) To get the per-unit, per-phase equivalent circuit, we must convert each impedance in the system to per-unit on the base of the region in which it is located The impedance of transformer T1 is already in per-unit to the proper base, so we don t have to do anything to it: R1,pu = 0010
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