Chapter 13. Automatic Generation Control in a Restructured Power System

340



Electrical Power Systems



generation on the same footing as the IPPs. Figure 13.2 shows the deregulated utility structure.

In Fig. 13.2, GENCOs which will compete in a free market to sell electricity they produce. It can

be assumed that the retail customer will continue for some time to buy from the local distribution

company. Such distribution companies have been designated as DISCOs. The entities that will

wheel this power between GENCOs and DISCOs have been designated as TRANSCOs. Although

it is conceptually clean to have separate functionalities for the GENCOs, TRANSCOs and

DISCOs, in reality there will exist companies with combined or partial responsibilities.



Fig. 13.2: Deregulated utility structure.



With the emergence of the distinct identities of GENCOs, TRANSCOs, DISCOs and the

ISO, many of the ancillary services of a VIU will have a different role to play and hence have

to be modeled differently. Among these ancillary services is the Automatic Generation Control

(AGC). In the new scenario, a DISCO can contract individually with a GENCO for power and

these transactions will be made under the supervision of ISO.



13.2 TRADITIONAL VS RESTRUCTURED SCENARIO

In the previous section vertically integrated utility (VIU) structure and deregulated utility

structure have been discussed. As there are several GENCOs and DISCOs in the deregulated

structure, a DISCO has the freedom to have a contract with any GENCO for transaction of

power. A DISCO may have a contract with a GENCO in another control area. Such transactions

are called “bilateral transactions.” All the transactions have to be cleared through an impartial

entity called an independent system operator (ISO). The ISO has to control a number of socalled “ancillary services”, one of which is AGC.



13.3 DISCO PARTICIPATION MATRIX (DPM)

In the restructured environment, GENCOs sell power to various DISCOs at competitive prices.

Thus, DISCOs have the liberty to choose the GENCOs for contracts. They may or may not have

contracts with the GENCOs in their own area. This makes various combinations of GENCODISCO contracts possible in practice.

We will describe here DISCO participation matrix (DPM) to make the visualization of

contracts easier. DPM is a matrix with the number of rows equal to the number of GENCOs and

the number of columns equal to the number of DISCOs in the system. For the purpose of



Automatic Generation Control in a Restructured Power System 341



explanation, consider a two-area system in which each area has two GENCOs and two DISCOs

in it. Let GENCO1, GENCO2, DISCO1 and DISCO2 are in area-1, and GENCO3, GENCO4,

DISCO3 and DISCO4 are in area-2 as shown in Fig. 13.3.



Fig. 13.3: Schematic of a two-area system in a restructured power system.



The DPM of Fig. 13.3 can be given as:

DISCO1



DISCO3



DISCO4



GENCO1

DPM =



DISCO2



cpf 11



cpf 12



cpf 13



cpf 14



GENCO2



cpf 21



cpf 22



cpf 23



cpf 24



GENCO3



cpf 31



cpf 32



cpf 33



cpf 34



GENCO4



cpf 41



cpf 42



cpf 43



...(13.1)



cpf 44



Each entry in eqn. (13.1) can be thought of as a fraction of a total load contracted by a

DISCO (column) toward a GENCO (row). Thus, the ij-th entry corresponds to the fraction of the

total load power contracted by DISCO j from a GENCO i. The sum of all the entries in a column

in this matrix is unity. DPM shows the participation of a DISCO in a contract with a GENCO,

and hence the “DISCO participation matrix”. In eqn. (13.1) , cpfij refers to “contract participation

factor”. For the purpose of explanation, suppose that DISCO2 demands 0.1 pu MW power, out

of which 0.02 pu MW is demanded from GENCO1, 0.035 pu MW demanded from GENCO2,

0.025 pu MW demanded from GENCO3 and 0.02 pu MW demanded from GENCO4.

The column 2 entries in eqn. (13.1) can easily be defined as:

cpf12 =



0.02

= 0.20;

0.1



cpf32 =



0.025

0.02

= 0.25; cpf42 =

= 0.20;

0.1

0.1



cpf22 =



0.035

= 0.35;

0.1



Note that cpf12 + cpf22 + cpf32 + cpf42 = 1.0

Other cpfs are defined easily to obtain the entire DPM. In general



åi cpf ij



= 1.0



...(13.2)



13.4 BLOCK DIAGRAM REPRESENTATION

In chapter-12, Block diagram representation of two area system has been presented for

conventional AGC. Here, we will formulate the block diagram for a two area AGC system in the



342



Electrical Power Systems



deregulated scenario. Whenever a load demanded by a DISCO changes, it is reflected as a local

load in the area to which this DISCO belongs. This corresponds to the local loads DPL1 and DPL2

and should be reflected in the deregulated AGC system block diagram at the point of input to

the power system block. As there are many GENCOs in each area, ACE signal has to be

distributed among them in proportion to their participation in AGC. Coefficients that distribute

ACE to several GENCOs are termed as “ACE participation factors”.

Note that

NGENCO j



¢

å a ji



= 1.0



...(13.3)



i=1



Where



¢

a ji = participation factor of i-th GENCO in j-th area



NGENCOj = Number of GENCOs in j-th area.

Unlike the traditional AGC system, a DISCO asks/demands a particular GENCO or GENCOs

for load power. These demands must be reflected in the dynamics of the system. Turbine and

governor units must respond to this power demand. Thus, as a particular set of GENCOs are

supposed to follow the load demanded by a DISCO, information signals must flow from a

DISCO to a particular GENCO specifying corresponding demands. The demands are specified

by cpfs (elements of DPM) and the pu MW load of a DISCO. These signals carry information as

to which GENCO has to follow a load demanded by which DISCO.

The scheduled steady state power flow on the tie-line is given as:

scheduled =

DPtie 12



\



(Demand of DISCOs in area-2 from GENCOs in area-1) –

(Demand of DISCOs in area-1 from GENCOs in area-2)



2

4

S cpf ij

scheduled = S

DPtie 12

i =1 j=3



4



DPL j – S



2



S cpfij DPL j



i =3 j=1



...(13.4)



error

At any given time, the tie-line power error DPtie,12 is defined as:



error

actual

scheduled

DPtie 12 = DPtie 12 – DPtie 12



...(13.5)



error

DPtie 12 vanishes in the steady-state as the actual tie-line power flow reaches the scheduled



power flow. This error signal is used to generate the respective ACE signals as in the traditional

scenario:

error

ACE1 = B1 D F1 + DPtie 12



...(13.6)



error

ACE2 = B2 D F2 + a12 DPtie 12



...(13.7)



For two area system as shown in Fig. 13.3, contracted power supplied by i-th GENCO is

given as:

NDISCO=4



DP i =



å cpfijDPLj

j=1



...(13.8)



Automatic Generation Control in a Restructured Power System 343



The block diagram of two area AGC system in a deregulated environment is shown in Fig.

13.4(a). Simplified version of Fig. 13.4(a) is shown in Fig. 13.4(b). In fig. 13.4(b), for i = 1,

...(13.9)

D P1 = cpf11 D PL1 + cpf12 D PL2 + cpf13 D PL3 + cpf14 D PL4



Fig. 13.4(=): Block diagram of two-area deregulated power system.



344

Electrical Power Systems



Fig. 13.4(>): Simplified representation of fig. 13.4(a).



Automatic Generation Control in a Restructured Power System 345



Similarly, DP2, DP3 and DP4 can easily be obtained from eqn. (13.8). In Fig. 13.4(b), DPuc1 and

DPuc2 are uncontracted power demand (if any).

Also note that DPL1,LOC = DPL1 + DPL2 and DPL2,LOC = DPL3 + DPL4. In the proposed AGC

implementation, contracted load is fed forward through the DPM matrix to GENCO setpoints.

This is shown in Fig. 13.4(b) i.e., DP1, DP2, DP3 and DP4. The actual loads affect system dynamics

via the inputs DPL,LOC to the power system blocks. Any mismatch between actual and contracted

demands will result in a frequency deviation that will result in a frequency deviation that will

¢

¢

¢

drive AGC to redispatch GENCOs according to ACE participation factors, i.e., a11 , a12 , a21 ,

and a22 . The AGC scheme does not require measurement of actual loads. The inputs DPL1,LOC

¢

and DPL2,LOC in the block diagram of Fig. 13.4 (a) & (b) are part of the power system model, not

part of AGC.



13.5 STATE SPACE REPRESENTATION OF THE TWO-AREA SYSTEM

IN DEREGULATED ENVIRONMENT

The closed loop system shown in Fig. 13.4 (b) is characterized in state space form as

•



X = AX + BU + GP + g p

...(13.10)

For this case, A is 11 × 11 matrix, B is 11 × 2 matrix, G is 11 ´ 4 matrix and g is 11 × 2 matrix.

Details of this matrices are given below:



LM –1

MM T

MM 0

MM 2pT

MM 0

MM –1

MM R T

0

A = M

MM –1

MM R T

MM 0

MM 0

MM

MM 0

MM 0

N



p1



12



1 g1



2 g2



0

–1

Tp2

–2 pT12



– K p1



K p1



Tp1

– K p2 a12



Tp1



Tp2

0



0



K p1

Tp1



0



0

K p2



0

0



0

K p2



0



0



0



0



0

1

Tt1

–1

T g1



0



0



Tp2

0



0



Tp2

0



0



0



0



0



0



0



0



0



0



0



1

Tt2

–1

Tg2



0



0



0



0



0



0



1

Tt3

–1

Tg3



0



0



0

–1

Tt1



0



0



0



0



0



0



0



–1

Tt2



0



0



0



0



0



0



0



0



0



0



0



–1

R3 Tg3



–1

Tt3



0



0



0



0



0



0



0



0



0



0



0



0



0



0



–1

R4 Tg4



–1

Tt4



0



0



0



0



0



0



0



0



0

0



OP

PP

0 P

P

0 P

P

0 P

PP

0 P

PP

0 P

PP

0

PP

0 P

PP

0 P

P

1 P

P

T P

–1 P

P

T Q

0



t4



g4



346



Electrical Power Systems



LM 0

MM 0

MM 0

0

a

MMT¢

MM 0

a

MMT¢

MM 0

MM 0

MM 0

MNM 0



11

g1



B=



12

g2



LM -K

0 O

P MM T

0 P

0

0 P

PP MMM 0

0

0 P

0

PP MMM cpf

0 P

MT

, G= M

0 P

0

PP MM cpf

0

T

a¢ P

PP MMM cpf

0

T

0 P

PP MMM T0

a¢

T Q

P MM cpf

MN T



- K p1



p1



Tp1



p1



0

0

0

cpf12

Tg1

0

cpf22

Tg2

0

cpf32

Tg3

0

cpf42

Tg4



11



g1

21



21



g3



22



g4



g2

31



g3

41



g4



LMu OP

U = M P,

NMu QP

1



2



0



0



- K p2



- K p2



Tp2

0

0

cpf13

Tg1

0

cpf23

Tg2

0

cpf33

Tg3

0

cpf43

Tg4



Tp2

0

0

cpf14

Tg1

0

cpf24

Tg2

0

cpf34

Tg3

0

cpf44

Tg4



LMDPL OP

DPL P

P= M

MMDPL PP

NDPL Q



OP LM -K

PP MM T

PP MM 0

PP MM 0

PP MM 0

PP g = MM 0

PP MM 0

PP MM 0

PP MM 0

PP MM 0

PP MM 0

PQ MN 0



1

2

3



p1



p1



LMDP

MNMDP



uc1



and p =



uc2



4



0

- K p2

Tp2

0

0

0

0

0

0

0

0

0



OP

PP

PP

PP

PP

PP

PP

PP

PP

PP

PP

PQ



OP

PQP



Integral control law for area-1 and area-2 are given as:



z

z



U1 = – K I1 ACE1dt



...(13.11)



U2 = – K I2 ACE2 dt



...(13.12)



KI1 and KI2 are the integral gain settings of area-1 and area-2 respectively,



Case-1

Consider a case where the GENCOs in each area participate equally in AGC, i.e., ACE

participation factors are a11 = 0.5, a12 = 1– a11 = 0.5; a21 = 0.5, a22 = 1– a21 = 0.50. Assuming

¢

¢

¢

¢

¢

¢

that the load change occurs only in area-1. Thus, the load is demanded only by DISCO1 and

DISCO2. Let the value of this load demand be 0.04 pu MW for each of them, i.e., DPL1 = 0.04

pu MW, DPL2 = 0.04 pu MW, DPL3 = DPL4 = 0.0. DISCO participation matrix (DPM), referring

to eqn. (13.1) is considered as



LM0.50

0.50

DPM = M

MM 0

N0



0.50

0.50

0

0



0

0

0

0



0

0

0

0



OP

PP

PQ



Automatic Generation Control in a Restructured Power System 347



Note that DISCO3 and DISCO4 do not demand power from any GENCOs and hence the

corresponding contract participation factors (columns 3 and 4) are zero . DISCO1 and DISCO2

demand identically from their local GENCOs, viz., GENCO1 and GENCO2. Therefore, cpf11 =

cpf12 = 0.50 and cpf21 = cpf22 = 0.50.



Fig. 13.5(=): D.1 (Hz) vs. time(sec).



Fig. 13.5(>): D.2 (Hz) vs. time (sec).



Fig. 13.5(?): D2tiel–2act (pu Mw) vs. time (sec).



348



Electrical Power Systems



Fig. 13.5(@): D2tiel–2error (pu Mw) vs. time (sec).



Fig. 13.5(A): D2g1 (pu Mw) vs. time (sec).



Fig. 13.5(B): D2g2 (pu Mw) vs. time (sec).



Fig. 13.5(C): D2g3 (pu Mw) vs. time (sec).



Automatic Generation Control in a Restructured Power System 349



Fig. 13.5(D): D2g4 (pu Mw) vs. time (sec).

Fig. 13.5: Dynamic responses for Case-1.



Figure 13.5 shows the results of this load change: area frequency deviations, actual power

flow on the tie-line ( in a direction from area-1 to area-2), and the generated powers of various

GENCOs, following a step change in the load demands of DISCO1 and DISCO2.

The frequency deviation in each area goes to zero in the steady state. Since, there are no

contracts of power between a GENCO in one area and a DISCO in another area, the scheduled

steady state power flow over the tie-line is zero.

In the steady state, generation of a GENCO must match the demand of the DISCOs in

contract with it. Expanding eqn. (13.8), we have,

DPi = cpf i1 DPL1 + cpf i2 DPL2 + cpf i3 DPL3 + cpf i4 DPL4



...(13.13)



For the case under consideration, we have,

DPg1,steady-state = DP1 = 0.5 ´ 0.04 + 0.50 ´ 0.04 = 0.04 pu MW.



Similarly

DPg2,steady-state = DP2 = 0.04 pu MW

DPg3,steady-state = DP3 = 0.0 pu MW

DPg4,steady-state = DP4 = 0.0 pu MW



Figure 13.5 also shows the actual generated powers of the GENCOs reach the desired

values in the steady state GENCO3 and GENCO4 are not contracted by any DISCO for a

transaction of power, hence, their change in generated power is zero in the steady-state, i.e.,

DPg3,steady-state = 0.0 pu MW and DPg4,steady-state = 0.0 pu MW.



Case-2

Here we will consider that all the DISCOs have contract with the GENCOs for power as per the

following DISCO participation Matrix (DPM):



LM0.50

0.20

DPM = M

MM 0.0

N0.30



0.25

0.25

0.25

0.25



0.0

0.0

10

.

0.0



0.30

0.0

0.70

0.0



OP

PP

PQ



350



Electrical Power Systems



In this case, it is also assumed that each DISCO demands 0.04 pu MW power from GENCOs

as defined by cpfs in DPM matrix and each GENCO participates in AGC as defined by following:

apfs: a11 = 0.75, a12 = 0.25; a21 = a22 = 0.50

¢

¢

¢

¢

Note that ACE participation factors (apfs) affect only the transient behaviour of the

system and not the steady-state behaviour when uncontracted loads are absent, i.e.,

DPuc1 = DPuc2 = 0.0.

The scheduled power flow on the tie-line in the direction from area-1 to area-2 is (eqn. 13.4):

scheduled



DPtie,1-2



2



4



4



2



i=1



=



j= 3



i =3



j=1



å å cpfijDPL j – å å cpfijDPLj



Fig. 13.6(=): D.1(Hz) vs. time (sec).



Fig. 13.6(>): D.2(Hz) vs. time (sec).



Fig. 13.6(?): D2tiel2, actual (pu Mw) vs. time (sec).