Electrical Power Systems

Automatic Generation Control: Conventional Scenario 333
Example 12.5: ig. 12.23 shows a two area interconnected system. The load in each area varies
1% for every 1% change in system frequency. f 0 = 50 Hz, R = 6% for all units. Area-1 is operating
with a spinning reserve of 1000 MW spread uniformly over a generation of 4000 MW capacity,
and area-2 is operating with a spinning reserve of 1000 MW spread uniformly over a generation
of 10,000 MW. D 1 = 380 MW/Hz and D 2 = 800 MW/Hz.
ig. 12.23
Determine the steady-state frequency, generation and load of each area, and tie-line power
for the following cases.
(a) Loss of 1000 MW load in area-1, assuming that no supplementary controls.
(b) Each of following contingencies, when the generation carrying spinning reserve in each
area is on supplementary control with B1 = 2500 MW/Hz and B2 = 5000 MW/Hz.
(i) Loss of 1000 MW load in area-1
(ii) Loss of 500 MW generation, carrying part of the spinning reserve, in area-1.
Solution
(a) 6 % regulation on 20000 MW generating capacity (induding spinning reserve of
1000 MW) in area-1 corresponds to 1
R
Similarly
\
R 2
\ D f ss =
1
1 20000
= ´ = 6666.67 MW/Hz
006 . 50
1
=
1 42000
´ = 14000 MW/Hz.
006 . 50
1
=
R
1 1
+
R 1 R 2
= 20666.67 MW/Hz
D = D1 + D2 = 380 + 800 = 1180 MW/Hz
I
HG KJ
\ D f ss = 0.04577 Hz
b g
-DPL
1
D +
R
=
--1000
b1180 + 20666. 67g
Load changes in the two areas due to increase in frequency are
DP d1 = D 1.Df ss = 380 ´ 0.04577 MW = 17.392 MW
DP d2 = D 2 .Df ss = 800 ´ 0.04577 MW = 36.616 MW
DPg1 = -Dfss
R1
DPg2 = -Dfss
R
2
= – 0.04577 ´ 6666.67 = – 305.13 MW
= – 0.04577 ´ 14000 = – 640.78 MW