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X 1,pu = 0040
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The impedance of transformer T2 is already in per-unit, but it is per-unit to the base of transformer T2 , so it must be converted to the base of the power system (V ) 2 ( S ) (3-66) ( R, X , Z ) pu on base 2 = ( R, X , Z ) pu on base 1 base 1 2 base 2 (Vbase 2 ) ( Sbase 1 )
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R2,pu = 0020
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( 7967 V ) 2 (1000 kVA ) = 0040 ( 7967 V ) 2 (1000 kVA ) ( 7967 V ) 2 (1000 kVA ) = 0170 X 2,pu = 0085 ( 7967 V ) 2 (1000 kVA )
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The per-unit impedance of the transmission line is Z 15 + j10 Z line,pu = line = = 000788 + j 00525 Z base2 1904 The per-unit impedance of Load 1 is 63
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Z load1,pu =
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Z load1 05 3687 = = 1681 + j1261 Z base3 0238
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The per-unit impedance of Load 2 is j 0833 Z Z load2,pu = load2 = = j 35 Z base3 0238 The resulting per-unit, per-phase equivalent circuit is shown below:
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0010 j0040 000788 j00525 0040 j0170
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1 0
T2 1681 j1261 L1 L2 -j35
With the switch opened, the equivalent impedance of this circuit is
Z EQ = 0010 + j 0040 + 000788 + j 00525 + 0040 + j 0170 + 1681 + j1261 Z EQ = 17389 + j15235 = 2312 412
The resulting current is
1 0 = 04325 412 2312 412
The load voltage under these conditions would be
VLoad = VLoad,puVbase3 = ( 0909 )( 480 V ) = 436 V VLoad,pu = I Z Load = ( 04325 412 )(1681 + j1261) = 0909 43
The power supplied to the load is
PLoad,pu = I 2 RLoad = ( 04325) (1681) = 0314
PLoad = PLoad,pu S base = ( 0314 )(1000 kVA ) = 314 kW
The power supplied by the generator is
QG ,pu = VI sin = (1)( 04325) sin 412 = 0285 SG ,pu = VI = (1)( 04325) = 04325 PG ,pu = VI cos = (1)( 04325) cos 412 = 0325
QG = QG ,pu S base = ( 0285)(1000 kVA ) = 285 kVAR
PG = PG ,pu S base = ( 0325)(1000 kVA ) = 325 kW
SG = SG ,pu S base = ( 04325)(1000 kVA ) = 4325 kVA
The power factor of the generator is
PF = cos 412 = 0752 lagging
With the switch closed, the equivalent impedance of this circuit is
Z EQ = 0010 + j 0040 + 000788 + j 00525 + 0040 + j 0170 + Z EQ Z EQ = 2685 + j 0261 = 2698 56 j1261)( j 35) 1681 + j1261 j 35 = 0010 + j 0040 + 000788 + j 00525 + 0040 + j 0170 + (2627 j 00011)
(1681 +
The resulting current is
1 0 = 0371 56 2698 56
The load voltage under these conditions would be
VLoad = VLoad,puVbase3 = ( 0975)( 480 V ) = 468 V VLoad,pu = I Z Load = ( 0371 56 )( 2627 j 0011) = 0975 56
The power supplied to the two loads is
PLoad,pu = I 2 RLoad = ( 0371) ( 2627) = 0361
PLoad = PLoad,pu S base = ( 0361)(1000 kVA ) = 361 kW
The power supplied by the generator is
QG ,pu = VI sin = (1)( 0371) sin 56 = 0036 PG ,pu = VI cos = (1)( 0371) cos56 = 0369
SG ,pu = VI = (1)(0371) = 0371
QG = QG ,pu S base = ( 0036)(1000 kVA ) = 36 kVAR SG = SG ,pu S base = ( 0371)(1000 kVA ) = 371 kVA
PG = PG ,pu S base = ( 0369 )(1000 kVA ) = 369 kW
The power factor of the generator is
PF = cos 56 = 0995 lagging
The transmission losses with the switch open are: 2 Pline,pu = I 2 Rline = ( 04325) ( 000788) = 000147
Pline = Pline ,pu S base = ( 000147 )(1000 kVA ) = 147 kW
The transmission losses with the switch closed are: 2 Pline,pu = I 2 Rline = ( 0371) ( 000788) = 000108
Pline = Pline ,pu S base = ( 000108)(1000 kVA ) = 108 kW
Load 2 improved the power factor of the system, increasing the load voltage and the total power supplied to the loads, while simultaneously decreasing the current in the transmission line and the transmission line losses This problem is a good example of the advantages of power factor correction in power systems
4: AC Machinery Fundamentals
4-1 Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14 poles operating at frequencies of 50, 60, and 400 Hz SOLUTION The equation relating the speed of magnetic field rotation to the number of poles and electrical frequency is
nm =
120 f e P f e = 50 Hz
3000 r/min 1500 r/min 1000 r/min 750 r/min 600 r/min 500 r/min 4286 r/min
The resulting table is Number of Poles 2 4 6 8 10 12 14 4-2
f e = 60 Hz
3600 r/min 1800 r/min 1200 r/min 900 r/min 720 r/min 600 r/min 5143 r/min
f e = 400 Hz
24000 r/min 12000 r/min 8000 r/min 6000 r/min 4800 r/min 4000 r/min 3429 r/min
A three-phase two-pole winding is installed in six slots on a stator There are 80 turns of wire in each slot of the windings All coils in each phase are connected in series, and the three phases are connected in The flux per pole in the machine is 0060 Wb, and the speed of rotation of the magnetic field is 3600 r/min (a) What is the frequency of the voltage produced in this winding (b) What are the resulting phase and terminal voltages of this stator SOLUTION (a) The frequency of the voltage produced in this winding is
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