Handbook of Electrical Engineering

48 HANDBOOK OF ELECTRICAL ENGINEERING

Similarly for generators Nos. 2 and 3,

fz 2 = 61 .6Hzandfz 3 = 61 .5Hz

Step 2. The common system frequency after the load increases is found from (2.66), (2.67) and (2.68).

a=

1

60. 0

(

60. 9 × 20. 0

0. 03

+

61. 6 × 15. 0

0. 04

+

61. 5 × 10. 0

0. 05

)

= 1266. 67

b=

1

60

(

20. 0

0. 03

+

15. 0

0. 04

+

10. 0

0. 05

)

= 20. 6945

f=

a−P

b

=

1266. 67 − 40. 5

20. 6945

=59.25101 Hz

Step 3. Find the new load on each generator

P 1 =(fz 1 −f)

G 1

D 1 fo

=( 60. 9 − 59. 25101 )

20. 0

0. 03 × 60. 0

=18.3221 MW (91.61%)

Similarly for generators Nos. 2 and 3,

P 2 =14.6819 MW (97.88%) andP 3 =7.4966 MW (74.97%)

Note,

Pnew=P 1 +P 2 +P 3 = 18. 3221 + 14. 6819 + 7. 4966

=40.5 MW as required.

Step 4. Find the new set points that will recover the frequency to 60 Hz.

If a changePiinPiis added to the (2.69) then the change in the set point will be,

fzi=

DiPifo

GI

(or 60. 0 −f)

For generator No. 1,

fz 1 =

0. 03 ×( 18. 3221 − 10. 0 ) 60. 0

20

= 0. 74899

And so the new set-point isfz 1 +fz 1 = 61 .6489 Hz

Similarly for generators Nos. 2 and 3

fz 2 +fz 2 =62.3491 Hz, andfz 3 +fz 3 =62.2489 Hz

Step 5. Find the set points that will enable the generators to be equally loaded.