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440 0 V 470 12 = 331 156 A 022 + j 30
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The real power consumed by this machine is
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P = 3V I A cos = 3 ( 440 V )( 331 A ) cos (156 ) = 421 kW
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The reactive power supplied by this machine is
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Q = 3V I A sin = 3 ( 440 V )( 331 A ) sin (156 ) = +117 kVAR
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6: Parallel Operation of Synchronous Generators
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6-1 A 480-V 400-kVA 085-PF-lagging 50-Hz four-pole -connected generator is driven by a 500-hp diesel engine and is used as a standby or emergency generator This machine can also be paralleled with the normal power supply (a very large power system) if desired (a) What are the conditions required for paralleling the emergency generator with the existing power system What is the generator s rate of shaft rotation after paralleling occurs (b) If the generator is connected to the power system and is initially floating on the line, sketch the resulting magnetic fields and phasor diagram (c) The governor setting on the diesel is now increased Show both by means of house diagrams and by means of phasor diagrams what happens to the generator How much reactive power does the generator supply now (d) With the diesel generator now supplying real power to the power system, what happens to the generator as its field current is increased and decreased Show this behavior both with phasor diagrams and with house diagrams SOLUTION (a) To parallel this generator to the large power system, the required conditions are: 1 The generator must have the same voltage as the power system 2 The phase sequence of the oncoming generator must be the same as the phase sequence of the power system 3 The frequency of the oncoming generator should be slightly higher than the frequency of the running system 4 The circuit breaker connecting the two systems together should be shut when the above conditions are met and the generator is in phase with the power system After paralleling, the generator s shaft will be rotating at
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nm =
120 f e 120(50 Hz ) = = 1500 r/min P 4
BR B net I
The magnetic field and phasor diagrams immediately after paralleling are shown below:
EA V jX I
(c) When the governor setpoints on the generator are increased, the emergency generator begins to supply more power to the loads, as shown below:
EA jX I PG P2 P1 Psys I
A S A
Note that as the load increased with E A constant, the generator began to consume a small amount of reactive power (d) With the generator now supplying power to the system, an increase in field current increases the reactive power supplied to the loads, and a decrease in field current decreases the reactive power supplied to the loads
E A1 E A2 I A1 Q
E A3
I A2 I A3
jX I
E A2 E A1 I A2 Q Q Q
jX I V
I A1
6-2
A 135-kV 20-MVA 08-PF-lagging 60-Hz two-pole Y-connected steam-turbine generator has a synchronous reactance of 50 per phase and an armature resistance of 05 per phase This generator is operating in parallel with a large power system (infinite bus) (a) What is the magnitude of E A at rated conditions (b) What is the torque angle of the generator at rated conditions (c) If the field current is constant, what is the maximum power possible out of this generator How much reserve power or torque does this generator have at full load (d) At the absolute maximum power possible, how much reactive power will this generator be supplying or consuming Sketch the corresponding phasor diagram (Assume I F is still unchanged) 104
SOLUTION (a) The phase voltage of this generator at rated conditions is
V =
VT 135 kV = = 7794 V 3 3 S 20,000,000 VA = = 855 A 3 VT 3 (13,500 V )
The armature current per phase at rated conditions is
IA =
Therefore, the internal generated voltage at rated conditions is
E A = 7794 0 + ( 05 )( 855 3687 A ) + j ( 50 )( 855 3687 A ) E A = 11,160 165 V
The magnitude of E A is 11,160 V (b) (c) The torque angle of the generator at rated conditions is = 165 Ignoring R A , the maximum output power of the generator is given by
E A = V + R A I A + jX S I A
PMAX =
3V E A XS
3 ( 7794 V )(11,160 V ) = 522 MW 5
The phasor diagram at these conditions is shown below: EA
jX I
R I V
Under these conditions, the armature current is
IA =
E A V
R A + jX S
11,650 90 V - 7794 0 V = 2790 395 A 05 + j50
The reactive power produced by the generator is
Q = 3 V I A sin = 3 ( 7794 V )( 2790 A ) sin ( 0 -395 ) = 415 MVAR
The generator is actually consuming reactive power at this time 6-3 A 480-V 200-kW two-pole three-phase 50-Hz synchronous generator s prime mover has a no-load speed of 3040 r/min and a full-load speed of 2975 r/min It is operating in parallel with a 480-V 150-kW four105
pole 50-Hz synchronous generator whose prime mover has a no-load speed of 1500 r/min and a full-load speed of 1485 r/min The loads supplied by the two generators consist of 200 kW at 085 PF lagging (a) Calculate the speed droops of generator 1 and generator 2 (b) Find the operating frequency of the power system (c) Find the power being supplied by each of the generators in this system (d) If VT is 460 V, what must the generator s operators do to correct for the low terminal voltage SOLUTION The no-load frequency of generator 1 corresponds to a frequency of
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