0195136047.pdf

13.3 SYNCHRONOUS MACHINES 591

maintained on either side of the synchronizing switch. Thus,synchronizingrequires the following
conditions of the incoming machine:

• Correct phase sequence


• Phase voltages in phase with those of the system


• Frequency almost exactly equal to that of the system


• Machine terminal voltage approximately equal to the system voltage
  Asynchroscopeis used for indicating the appropriate moment for synchronization. After the
  machine has been synchronized and is part of the system, it can be made to take its share of the
  active and reactive power by appropriate adjustments of its prime-mover throttle and field rheostat.
  The system frequency and the division of active power among the generators are controlled
  by means of prime-mover throttles regulated by governors and automatic frequency regulators,
  whereas the terminal voltage and the reactive volt-ampere division among the generators are
  controlled by voltage regulators acting on the generator-field circuits and by transformers with
  automatic tap-changing devices.

EXAMPLE 13.3.3

Two three-phase, 6.6-kV, wye-connected synchronous generators, operating in parallel, supply a
load of 3000 kW at 0.8 power factor lagging. The synchronous impedance per phase of machineA
is 0. 5 +j 10 , and of machineBit is 0. 4 +j 12 . The excitation of machineAis adjusted so that
it delivers 150 A at a lagging power factor, and the governors are set such that the load is shared
equally between the two machines. Determine the armature current, power factor, excitation
voltage, and power angle of each machine.

Solution

One phase of each generator and one phase of the equivalent wye of the load are shown in Figure
E13.3.3(a). The load currentI ̄Lis calculated as

I ̄L=√^3 ,^000

3 × 6. 6 × 0. 8

 −cos−^10. 8 = 328 ( 0. 8 −j 0. 6 )= 262. 4 −j 196 .8A

For machineA,

cosφA=

1500

√

3 × 6. 6 × 150

= 0 .875 lagging;φA=29°;sinφA= 0. 485

I ̄A= 150 ( 0. 874 −j 0. 485 )= 131. 1 −j 72 .75 A

For machineB,

I ̄B=I ̄L−I ̄A= 131. 3 −j 124 = 180. 6  −cos

(

131. 3

180. 6

)

cosφB=

131. 3

180. 6

= 0 .726 lagging

With the terminal voltageV ̄ras reference, we have

E ̄fA=V ̄t+I ̄AZ ̄A=( 6. 6 /

√

3 )+( 131. 1 −j 72. 75 )( 0. 5 +j 10 )× 10 −^3

= 4. 6 +j 1 .27 kV per phase

Power angleδA=tan−^1 ( 1. 27 / 4. 6 )= 15 .4°