Page 125 - Solutions Manual to accompany Electric Machinery Fundamentals

P  s  f  f 
1 P 1 nl1 sys
200 MW  85 MW/Hz  f nl1  60 Hz

f  62.35 Hz
nl1
The no-load frequency of the fourth generator is

P  4 s P 4  f nl4  f sys 

 50 MW   f nl1  60 Hz  85 MW/Hz

f nl1  60.59 Hz

(b) The setpoints of generators 1, 2, 3, and 4 do not change, so the new system frequency will be

P LOAD  s P 1  f nl1  f sys  s P 2  f nl2  f sys  s P 3  f nl3  f sys  s P 4  f nl4  f sys 

725 MW   85  62.35  f sys   62.35 85  f sys   62.35 85  f sys    85  60.59  f sys 



725 21049.40 340 f
sys
 20324.40   340 f
sys
f sys  59.78 Hz

(c) The governor setpoints of the swing generator must be increased until the system frequency rises
back to 60 Hz. At 60 Hz, the other three generators will be supplying 200 MW each, so the swing
generator must supply 725 MW – 3(200 MW) = 125 MW at 60 Hz. Therefore, the swing generator’s
setpoints must be set to
P  4 s P 4  f nl4  f sys 

125 MW  85 MW/Hz  f nl1  60 Hz 
f nl1  61.47 Hz

(d) If the swing generator trips off the line, the other three generators would have to supply all 725 MW
of the load. Therefore, the system frequency will become
P LOAD  s P 1  f nl1  f sys  s P 2  f nl2  f sys  s P 3  f nl3  f sys 

725 MW   85  62.35  f sys   62.35 85  f sys   62.35 85  f sys 



725 15899.25 255 f
sys
-15174.25  255 f
sys
f sys  59.51 Hz

Each generator will supply 241.7 MW to the loads.
4-29. A 100-MVA, 14.4-kV 0.8-PF-lagging, Y-connected synchronous generator has a negligible armature
resistance and a synchronous reactance of 1.0 per-unit. The generator is connected in parallel with a 60-
Hz, 14.4-kV infinite bus that is capable of supplying or consuming any amount of real or reactive power
with no change in frequency or terminal voltage.

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