Electrical Power Systems

Example 16.6: Consider Ex-16.5, with generators limits
125 < P g1 < 300
175 < P g2 < 350
Optimal System Operation 427
100 < P <
g3 300
Determine the optimal dispatch by iterative technique using gradient method.
Solution:
We have seen in Ex-16.5 that equality constraint was satisfied in two iterations and there
was no generator limit. In this case constraint is imposed on generator output.
After 2nd iteration (Ex-16.5), we have,
Pg1 = 258.669 MW
Pg2 = 373.0035 MW
Pg3 = 218.335 MW.
But Pg2 exceeds its upper limit. Thus, this unit is pegged at its upper limit. Hence Pg2 = 350 MW
and is kept constant at this value. Thus, the new imbalance in power is
\ DP g (2) = 23 MW.
rom eqn. (16.52), we have,
DP g (2) = 850 – (258.669 + 350.0 + 218.335)
Dl (2) =
23
1 1
+
2 ´ 015 . 2 ´ 018 .
= 3.763
\ l (3) = l (2) + Dl (2) = 118.6007 + 3.763
\ l (3) = 122.3637.
or the 3rd iteration, we have
and
122. 3637 - 41
Pg1 = = 271.21 MW
2 ´ 015 .
Pg2 = 350 MW
P g3 =
122. 3637 - 40
= 228.79 MW
2 ´ 018 .
DP g (3) = 850 – (271.21 + 350 + 228.79) = 0.0
DP g (3) = 0.0, and the equality comstraint is met and Pg1 and P g3 are within their limits. Thus, the
optimal dispatch is
Pg1 = 271.21 MW
Pg2 = 350.0 MW