Page 69 - Handbook of Electrical Engineering

48 HANDBOOK OF ELECTRICAL ENGINEERING

Similarly for generators Nos. 2 and 3,

f z2 = 61.6Hzand f z3 = 61.5Hz

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

1 60.9 × 20.0 61.6 × 15.0 61.5 × 10.0
a = + + = 1266.67
60.0 0.03 0.04 0.05
1 20.0 15.0 10.0
b = + + = 20.6945
60 0.03 0.04 0.05
a − P 1266.67 − 40.5
f = =
b 20.6945
= 59.25101 Hz

Step 3. Find the new load on each generator

G 1 20.0
P 1 = (f z1 − f) = (60.9 − 59.25101)
D 1 f o 0.03 × 60.0
= 18.3221 MW (91.61%)

Similarly for generators Nos. 2 and 3,
P 2 = 14.6819 MW (97.88%) and P 3 = 7.4966 MW (74.97%)

Note,
P new = 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 change P i in P i is added to the (2.69) then the change in the set point will be,

D i P i f o
f zi = (or 60.0 − f)
G I
For generator No. 1,
0.03 × (18.3221 − 10.0)60.0
f z1 = = 0.74899
20
And so the new set-point is f z1 + f z1 = 61.6489 Hz

Similarly for generators Nos. 2 and 3

f z2 + f z2 = 62.3491 Hz, and f z3 + f z3 = 62.2489 Hz

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