Electrical Power Systems

Automatic Generation Control in a Restructured Power System 355
= Load of DISCO 1 + load of DISCO 2
= DPL1 + DPuc, 1 + DPL2 = 0.04 + 0.04 + 0.04
= 0.12 pu MW.
Similarly, the total load in area-2
= Load of DISCO3 + load of DISCO4 = DPL3 + DPL4 = 0.04 + 0.04 = 0.08 pu MW.
igure 13.7 shows the dynamic responses under this condition. rom ig. 13.7, it is seen
that frequency deviations vanish in the steady state. As DISCO participation matrix (DPM) is
same as in the case-2 and the excess load demand is taken up by GENCOs in the same area
(area-1), the tie-line power is the same as in case-2 in the steady-state. In the steady-state, the
generation of GENCO3 and GENCO4 are not affected by the excess load of DISCO1, i.e.,
DPg3, steady-state = 0.078 pu MW
DPg4, steady-state = 0.022 pu MW.
The uncontracted load of DISCO1 is reflected in the generations of GENCO1 and GENCO2. ACE participation factors of area-1 decide the distribution of uncontracted load in the steadystate.
Therefore in this case,
and
DPg1, steady-state = DP1 + a11.D ¢ Puc1
= 0.042 + 0.75 ´ 0.04 = 0.072 pu MW
DPg2, steady-state = DP2 + a12 ¢ P .D uc1
= 0.018 + 0.25 ´ 0.04
= 0.028 pu MW.
This is also shown in ig. 13.7
EXERCISE–13
13.1 Consider three area power system (i) radial interconnected power system (ii) loop interconnected
power system. Area-1 has two GENCOs and two DISCOs, Area-2 has three GENCOs and two
DISCOs and Area-3 has two GENCOs and three DISCOs. Draw the block diagram for both the
cases, and construct DPM. Also obtain the expressions of scheduled tie-line power flows.
13.2 Consider a two area interconnected power system. Area-1 consists of two GENCOs and two
DISCOs and Area-2 also consists of two GENCOs and two DISCOs. Contract participation factors
are cpf11 = 0.40, cpf21 = 0.30, cpf31 = 0.10, cpf41 = 0.20, cpf12 = cpf22 = cpf32 = cpf42 = 0.25, cpf13 = cpf23 = 0.0, cpf33 = cpf43 = 0.50, cpf14 = 0.30, cpf24 = 0.0, cpf34 = 0.70, cpf44 = 0.0. Assume that each DISCO
demands 0.10 pu MW power from GENCOs as defined by contract participation factors and each
GENCO participates in AGC as defined by following ACE participation factors. a¢ 11 = 0.75,
a¢ 12 = 0.25 and a¢ 21 = a¢ 22 = 0.50. Determine the output of each GENCO in the steady-state. Also
determine the scheduled power flow on the tie-line and its direction. If DISCO1 in area-1 and
DISCO4 in area-2 demands 0.10 pu MW (each) of excess power, calculate the output of each
GENCO in the steady-state. Note that GENCO1 , GENCO2 , DISCO1 and DISCO2 are in area-1 and
GENCO3 , GENCO4 , DISCO3 and DISCO4 are in area-2.