table of contents - calvin collegepribeiro/courses/power systems interim... · draft and incomplete...

DRAFT and INCOMPLETE

Table of Contents

from

A. P. Sakis Meliopoulos Power System Modeling, Analysis and Control

Chapter 10 ____________________________________________________________ 3

Automatic Generation Control ____________________________________________ 3 10.1 Introduction ___________________________________________________________ 3 10.2 The Primary Generation Control System ___________________________________ 9

10.2.1 Governor/Hydraulic Actuator Model ___________________________________________ 10 10.2.2 Turbine Models____________________________________________________________ 13 10.2.3 Generator Model___________________________________________________________ 14 10.2.4 Electric Load Model ________________________________________________________ 16 10.2.5 Summary Of Models________________________________________________________ 22

10.3 Steady State Response of the Primary Generation Control System _____________ 24 10.4 Transient Response of the Primary Generation Control System _______________ 28

10.4.1 Transient Response Neglecting Governor/Turbine dynamics ________________________ 28 10.4.2 Transient Response of Full System_____________________________________________ 29 10.4.3 Load Allocation Among Generating Units _______________________________________ 32

10.5 The Secondary Generation Control System ________________________________ 34 10.5.1 The Need for Integral (Reset) Control __________________________________________ 35 10.5.2 Time Domain Model of the Secondary Generation Control System ___________________ 37 10.5.3 Steady State Response of the Secondary System __________________________________ 38 10.5.4 Transient Response of the Secondary System ____________________________________ 38

10.6 Multi-Machine Systems ________________________________________________ 42 10.6.1 Model of a Multi-Machine System _____________________________________________ 42 10.6.2 Generation Control of a Multi-Machine System___________________________________ 49 10.6.3. Steady State Response of a Two Machine System ________________________________ 51 10.6.4 Transient Response of a Two Machine System ___________________________________ 57

10.7 Automatic Generation Control in Modern Energy Management Systems________ 59 10.7.1 Allocation Rules ___________________________________________________________ 61 10.7.2 Past Practices and Regulations ________________________________________________ 63 10.7.3 New Practices and Regulations________________________________________________ 65

10.8 Summary and Discussion _______________________________________________ 67

Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos

10.9 Problems _____________________________________________________________ 68

Copyright © A. P. Sakis Meliopoulos – 1990-2006


Chapter 10

Automatic Generation Control

10.1 Introduction The Automatic Generation Control (AGC) system is a set of equipment and computer programs that applies closed loop feedback control to achieve the following objectives:

• To regulate frequency to a scheduled value. • To maintain all scheduled power transactions to the contract value as well as the

net power interchange at the value required by the interchange contracts. • To maintain each unit's operation at the most economic value (Economic

Dispatch). Modern automatic generation control functions are performed at a central location, the Energy Management System (EMS). The functional diagram of the automatic generation control is illustrated in Figure 10.1. Note that the AGC is driven by the scheduled frequency and the scheduled net interchange of the controlled entity. The data acquisition system collects the actual operating conditions of the system and the deviations (errors) are computed. The errors are used to determine the control signals to the individual generators in such a way that the frequency returns to the scheduled value and the power transactions are regulated to the scheduled value. Note that the generation of the signals is controlled by the economic dispatch (or an optimal power flow) thus integrating economic scheduling into the automatic generation control loop. The details of this process are described in this chapter. The success or failure of the automatic generation control is measured by the ability to maintain the frequency of the system near the synchronous speed and the net power interchange close to the scheduled values. Modern power system AGC is very successful in this respect. Even during a large disturbance, such as the one which led to the New York blackout on July 13, 1977, the frequency of the system did not change more than 1.0%. Figure 10.2 illustrates the variation of frequency during that event. Note that prior to the event, the system frequency varies in the range 60±0.02 Hz. This variation is normal. Also, not shown in the figure, the net interchange at utility interfaces is controlled near a scheduled value. At a certain time the NY utility lost a large generating unit at Indian Point. Following this disturbance, the system frequency, recorded in Birmingham, AL, decreased to 59.94 Hz. At the same time, not shown in the figure, the power flow on the lines to the NY utility was increased. This increase caused the tripping of the tie lines after approximately 15 minutes leading to the separation of the New York system. Later attempts to close the NY tie lines were unsuccessful leading to the NY blackout.



Measurements of . Generator output Pgi . Frequency . Tie Line Flows . Electric Load

ScheduledNetInterchange

Economic DispatchorOptimal Power Flow

ScheduledFrequency Computation of Error

Ptie =

Allocation Rules . Regulating . EconomicUnit Desired Output P gi

des

Ptie

s c h- Ptie

Pgi = Pgi

s c h- Pgi

f = f - fs c h

ACE

Control Commands

P gides

Data Acquisition andProcessing Subsystem

Figure 10.1 Functional Diagram of the Automatic Generation Control Loop



N.Y

. Los

t Gen

erat

orat

Indi

an P

oint

N.Y

. Sep

arat

ed

60.0

060

.02

60.0

459

.94

59.9

259

.96

Time5 min

Atte

mpt

s to

Pic

k U

pN

.Y. L

oado

n Ti

e Li

nes

New York BlackoutWednesday, July 13, 1977

Manned Downtown SubstationBirmingham, AL

59.9

8

Frequency(Hz)

Figure 10.2 Frequency Variation During the 1977 New York Blackout, Recorded in a Birmingham, AL Substation

The described disturbance is rather severe. Its effect, a blackout in this case, is rare. In Chapter 11 we shall discuss how to deal with these disturbances and how to avoid system collapse. In this chapter we shall discuss the AGC function which deals with disturbances generated by the continuous variation of the electric load. A medium size utility may experience load changes at a rate as high as 30 to 50 MW per minute. The automatic generation control loop responds to these load changes and continuously adjusts generation to match the load. It should be recognized that different generating units contribute differently to the control problem for a number of technical and economic reasons. A brief discussion is presented: Baseload units include nuclear and large fossil fired units. These units are normally run fully loaded on a 24 hour basis. The reasons for this type of operation are economic and technical. Typically, these units are characterized with low operating cost and therefore are loaded before any other type of unit. Also, reactor cores and huge boilers do not



tolerate fast power changes once thermal balance has been reached. As a result, these units do not participate in the regulation of frequency and net interchange (AGC). Controllable units comprise hydrogenerators and smaller fossil units. All such units have limitations on the rate by which they can change their output. The rate limit of a unit, expressed in MW/sec, depends on the size of the units and the type of the plant. Examples are illustrated in Figure 10.3. A typical drum type steam turbine is illustrated in Figure 10.4. Peak loading units, which can pick up load relatively fast. Common in this category are combustion turbine-driven generators (CT) and hydro units. Peak loading units also include generators driven from short-time energy storage facilities such as pump hydro, compressed gas, or thermal storage. A better perspective of the generation control problem can be obtained by considering all the control loops which exist in a typical generating unit. Figure 10.5 illustrates the control loops of a typical unit. In general, there are four distinct control loops:

• The Voltage Regulation Loop • The Power System Stabilizer Loop • The Primary Load-Frequency Control Loop • The Secondary Load-Frequency Control loop

.05

0 30sec

Oil, Gas

Coal

.05

0 30sec

.05

0 30sec

.5

0 20min

.5

0 20min

.5

0 20minDrum Type Steam-Turbine Leading

Steam Unit- Boiler Leading

Hydro



Figure 10.3 Typical Responses of the Three Types of Controllable Generating Units

LUMPEDSTORAGE

SUPERHEATER

BOILER

REHEAT

COOLINGTOWER

GeneratorTurbineTurbine

Figure 10.4 Schematic Representation of a Drum-Type Steam Turbine Power Plant

PRIMEMOVER

G

TRANSMISSIONSYSTEMAND LOAD

Pg f, V, TIELINE

Governor E(s)

PSS

-+

+

Vref

Pg

-

+

f

L(s)

U.C.E.Pdes

K(s)

+

+

BIASBf

f sched+

+

P sched

InterchangeError

∆ Pint

∆ f

ACE



Figure 10.5 Schematic Representation of Existing Control Loops in a Power Plant

The first and second control loops are exercised on the unit exciter, while the third and fourth controls are exercised on the unit prime mover which rotates the generator. The first and second controls are applied to the unit exciter and thus are classified as excitation control, while the third and forth controls are exercised on the prime mover and are classified as frequency control. While there is some dynamic interaction among all four control loops, the two excitation control loops can be considered independent from the two frequency control loops for the following reasons:

• Frequency control under normal operating conditions comprises smooth transition from one operating point to another. The design of the control loop is such that the time constants involved are relatively long.

• Excitation control becomes active when large and abrupt deviations in the operating point occur (stability and voltage control). The response of the excitation control loop is relatively fast resulting in relatively short time constants.

Because the excitation control action is fast compared to the frequency control action, the frequency control problem is traditionally examined independently from the excitation control problem. The frequency control function was the first one to be automated in power systems. Advances in computer technology made it possible to integrate the frequency control problem with economic functions, such as economic dispatch, interchange control, scheduling functions, etc. A generic term to describe all these functions is energy management. Chapters 9 and 10 are devoted in examining these functions. Specifically, in Chapter 9 we have examined the economic optimization methods for an electric power system. In this Chapter 10 we examine the basic frequency control mechanism and the integration of the economic functions in the generation control loop.



To Turbine

Steam

A B

f

Pc

HighPressureOil

PilotValve

MainPiston

Servomotor

c

PilotValve

Figure 10.6 Simplified Diagram of a Speed Governor Mechanism

10.2 The Primary Generation Control System Each generating unit is equipped with a speed governor mechanism. The speed governor mechanism is a hydraulic system which controls the opening of the steam valves and, therefore, the mechanical power input of the unit. A simplified system is illustrated in Figure 10.6. Inputs to the system are:

• The position of the power changer (indicated as input ∆Pc to the servomotor).



• The speed of the generator which is proportional to the frequency, f, of the generated voltages (indicated as input ∆f to the servomotor).

These inputs are identified in Figure 10.6 as ∆Pc and∆f , respectively, going into the governor. These inputs determine the position of the steam valve. The system is dynamic and therefore any change in the inputs will generate some transients before the system acquires the new steady state position. In subsequent paragraphs, simple mathematical models of the system components will be developed. 10.2.1 Governor/Hydraulic Actuator Model In this section, we develop the model of the governor with the hydraulic system that controls the position of the steam valve. Consider the simplified diagram of Figure 10.6. The inputs to the governor are: ∆Pc and∆f . These inputs determine the position of point A. A change in point A position will cause a change in the pilot valve position which allows high pressure oil to run into the main chamber and move the piston that is directly connected to the steam valve. Specifically, point A is controlled by the servomotor which responds to the inputs ∆ and∆ . Thus, the position of the pilot valve is determined by the quantities ∆ and∆ . With this observation and utilizing the geometry illustrated in Figure 10.6, the mathematical model of the governor/hydraulic actuator can be constructed.

f Pcf Pc

Assume a specific operating condition (f P , etc.). Assume that at this operating condition the position of point A is XA. In case of a frequency change,∆ or a set point change,∆ or combination of the two, the position of point A will change by the action of the servomotor. The servomotor is designed in such a way that the change in the position of point A is:

go , o

fPc

∆ ∆ ∆X a f a PA C= −1 2 where a1 and a2 are constants. Note also that the construction of the hydraulic amplifier in Figure 10.6 is such that the following relationships are valid: ∆ ∆ ∆X a X a XB A C= +3 4

'

∆ ∆ X a X dC B

' ( )= −∫5 t ∆XC

' denotes the deviation of the point C position (valve position) and ∆ denotes the deviation of the point B position. Note that the first equation expresses the fact that the three points A, B and C are on a straight line (rigid lever). The second equation expresses the fact that the position of the main piston location is determined by the amount of

XB



hydraulic fluid into the chamber and assumes that the rate of hydraulic fluid flow is proportional to the opening expressed with ∆XB. Upon elimination of the variables ∆XA and ∆ from above equations: XB dtXaPaafaaaX CCC )( '

432315' ∆−∆+∆−=∆ ∫

Differentiation of the last equation yields:

)( '

4

32

4

3154

'

CCC XP

aaaf

aaaaa

dtXd

∆−∆+∆−=∆

Define

Ta aG =

1

4 5

G a aaG = 2 3

4

R aa

= 2

1

Then

)1(1 ''

CG

GC

G

C PfRT

GXTdt

Xd∆+∆−+∆−=

∆

Without loss of generality, we can assume that GG equals 1.0. This is equivalent of scaling the parameters R and . Then the equation reads: CP∆

CGG

CG

C PT

fRT

XTdt

Xd∆+∆−∆−=

∆ 111 ''

Taking the Laplace transform of above equation and solving for ∆ (s) XC

'

∆ ∆X s H sR

f s P sC G C' ( ) ( )( ( ) ( ))= − +

1∆ (10.1)

where G

G sTH

+=

11 is the transfer function of the speed governor system.



Equation (10.1) is a very simple mathematical model of the speed governor system and it is represented with the block diagram of Figure 10.7a. The parameters appearing in Figure 10.7a are defined as follows:

GT time constant of the speed governing system R speed regulation due to governor action

f

XcPc +

+-1/sTg

-11/R

f

XcPc +

+-1/sTg

-11/R

(a)

(b)

∆

∆

∆

∆

∆

∆

Figure 10.7 Signal Flow Diagram of the Speed Governor System a) Without Rate Limiting Function

b) With Rate Limiting Function In practice, the speed governor system is not as simple as discussed. In modern large plants, more than one hydraulic amplification stages are needed. In addition, signals going into the servomotor may be processed by appropriate circuits (lead-lag compensators, etc.). A very important function of the signal conditioning system is the limitation on the rate of increase/decrease of the steam supply. This function is illustrated with the signal limiter in Figure 10.7b. The speed regulation R is selected in such a way that even a large deviation in power results in a small frequency deviation. Traditionally, the practice has been such that R is selected in the order of 0.05 to 0.06 pu (0.05 in Europe and 0.06 in the U.S.). This means that for a change of power equal to the nominal power of the unit, the frequency will change by only 5 or 6%.



10.2.2 Turbine Models Prime movers for generators are almost exclusively steam or hydro turbines. In general, there are two types of steam turbines (non-reheat and reheat), and two types of hydro turbines (Pelt and Francis). Reheat type steam turbines provide improved plant efficiency and always preferred in large plants. Pelt hydro turbines provide high efficiency for high head hydro plants while Francis turbines provide high efficiency for low head hydro plants. The various types of turbines are illustrated in Figure 10.10. A turbine is a complex system. When the valve (steam or hydro) opens wider, more steam or water will flow through the turbine and the mechanical output at the shaft of the turbine will increase. Because the system is dynamic, the variation of the mechanical power output to changes in valve opening is governed by a dynamic equation. Let ∆PG be the change in mechanical power output due to a change in valve opening . In the simplest form, the turbine can be assumed to be a first order dynamic system expressed with the following simple equation:

∆XC'

'1C

T

TG

T

G XTGP

TdtPd

∆+∆−=∆

where is the gain of the system in MW per valve displacement GT T is the time constant of the turbine system. T Time constants for single steam turbines are in the subsecond region. Without loss of generality we can assume that the gain GT equals 1.0. This corresponds to a scaling of the variable ∆ , or equivalently, we can introduce the variable XC

' ∆XC =G . Note that the units of the variable ∆ is power. The equation then becomes.

T∆XC'

XC

'11C

TG

T

G XT

PTdt

Pd∆+∆−=

∆

For reheat units with multiple turbines, the differential equations describing the response of the turbines are more complex.

TurbineSteamChest

To Condenser

∆PT

∆Pv



(a)

TurbineSteamChest

To Condenser

LPHPTurbine

Reheater

∆Pv

∆PT

(b)

Penstock

'Head'

∆Pv

∆PT

(c)

Figure 10.8 Three Types of Turbines Used in Power Plants

Non-Reheat Steam (a), Reheat Steam (b), Hyrdo (Pelt) (c) 10.2.3 Generator Model The synchronous generator is a very complex device. However, for purposes of studying the load/frequency response of the system, a simple model will suffice. For this purpose, the generator is modeled as a rotating mass of certain inertia constant J. The applied torques are the mechanical torque and the electrical torque. Thus

emm TT

dttdJ −=2

2 )(θ

where m(t): the position of the generator rotor θ Tm : mechanical torque applied by the turbine Te : electromagnetic torque developed by the generator. Let )()( ttt om δωθ +=



dt

tddt

tdt om

m)()()( δωθω +==

where ω : the synchronous angular speed (mechanical) o

ωm(t): the actual angular speed of the generator rotor. Then

emm TTdt

tdJ −=)(ω

(10.2)

Consider the per unit inertia constant, H, of a generating unit.

HJ

S

o

B=

12

2ω

Then:

BHSJE == 202

1 ω

Multiplication of Equation (10.2) by ωo and substitution of J in terms of E yields:

eGeomom PPTT

dtdE

−=−= ωωωω0

2 (10.3)

Note that fm πω 2= and fπω 20 = o. Linearization of above equation around an operating point, and substituting ωm and ω yields: o

eGo

PPdt

fdfE

∆−∆=∆2 (10.4)

The electric power output of the generator, ∆Pe , must be always equal to the electric load. Considering the electric load, it can be observed that: (1) the electric load has a component which varies as customers switch in and out, and (2) the electric load has certain sensitivity to system frequency. This sensitivity of the electric load will be discussed in the next paragraph. We postulate that the electric load variations will be: )(tgfDPe +∆=∆ D is the sensitivity of the electric load to frequency variations. The load model will be examined in the next paragraph.



A signal flow model for the generator can be derived by taking the Laplace transform of equation (10.4): (10.5) )()()( sPsPsFsT eGp ∆−∆=∆where:

s

p fET 02

=

Figure 10.9 illustrates the signal flow diagram for the generator model.

f(s)Pg(s) +1/sTp

Pe(s)

∆

∆

∆

Figure 10.9 Signal Flow Diagram of the Generator Model 10.2.4 Electric Load Model The electric load model comprises many apparatus with different characteristics. In general, these characteristics are voltage and frequency dependent. In this section, the dependence of the electric load on frequency will be discussed. Three types of loads are of importance: (a) Constant Resistance/ Inductance Load, (b) Constant Resistance/ Capacitance Load and (c) Induction Motors. Constant Resistance/Inductance Loads: Such a load is shown in Figure 10.10a. The absorbed power is:

22

2

)2(3

fLRRV

P phD π+= (10.6)



(a) (b)

Figure 10.10 Constant Impedance Electric Load (a) Constant R-L Electric Load, (b) Constant R-C Electric Load

Linearization of equation (10.6) around the operating frequency fo and voltage Vo

ph yields:

)()()( ophph

ph

Do

DoDD VV

VPff

fPfPP −+−+≅

∂∂

∂∂

where

)(

)(222

2

L

L

o

oDD

XRX

ffPD

fP

+−==

∂∂

)(

622

L

ph

ph

D

XRRV

VP

+=

∂∂

LfX oL π2= Constant Resistance/Capacitance Load: The electric power absorbed by such an electric load (Figure 10.10b) is:

22

2

)2

1(

3

fCR

RVP ph

D

π+

= (10.7)

Linearization of equation (10.7) around the operating frequency fo and voltage Vo

ph yields:

)()()( ophph

ph

Do

DoDD VV

VPff

fPfPP −+−+≅

∂∂

∂∂



where

)(

)(222

2

C

C

o

oDD

XRX

ffPD

fP

+==

∂∂

)(

622

C

ph

ph

D

XRRV

VP

+=

∂∂

Cf

Xo

C π21

=

Induction Motors: The induction motors have a torque-frequency characteristic as illustrated in Figure 10.11a. A typical mechanical load characteristic is superimposed on the same figure. The intersection determines the operating point. Whenever the network frequency changes, the induction motor curve will shift (to the left for a frequency decrease and to the right for a frequency increase). Thus a frequency decrease will result in less power absorbed by the induction motor and a frequency increase will increase the electric power intake. Mathematically, the induction motor is represented with the equivalent circuit of Figure

10.11b. The power absorbed at the resistor s

sr −12 is the power which is converted into

mechanical power. s is the slip defined with em s ωω )1( −= (for a two pole machine). For simplicity of analysis, the shunt term is neglected. The electric power absorbed is:

22

12

21

21

2

)()(

)(3

srrxx

srrV

Pph

e

+++

+=

The mechanical power output is

))()(

)(1(322

12

21

2

2

srrxx

Vs

srP phm

+++

−=

Let T = f (ωm) be the mechanical load speed-torque characteristic. The operation of the motor is determined from the solution of the equation obtained when the mechanical power is equated to the electrical:



Torque-Speed Characteristic

SpeedNetworkFrequency

Torq

ueMechamical Load Torque

(a)

1-ss

r1 r2x2

gmbm

x1

r2

(b)

Figure 10.11 Induction Motor Models, (a) Torque-Speed Characteristic,

(b) Equivalent Circuit

))()(

)(1(3)()1(22

12

21

2

2

srrxx

Vs

srfs phme

+++

−=− ωω

Solution of above expression provides the slip s. Let it be s1. A frequency decrease will cause a decrease in the electrical angular frequency from ωe to

. Equating the mechanical and electrical powers yields another equation whose solution will yield the new slip s'. At this slip s', let the mechanical or electrical power be Pm'. Then the linearized model of the electric load is

ω e'

fPPfPee

mm ∆−−

=∆ πωω

2)( '

'



The quantity P Pm m

e e

−−

'

'ω ωπ2 is the sensitivity of the load to the frequency and will be

designated with D. Thus fDfP ∆=∆ )( The outlined procedure for the computation of the linearized model is very complex. A simplified approach is presented next. In the simplified analysis, the parameters x1, r1, x2, gm, bm of the motor are neglected. Then the electric power absorbed is

e

mephphe r

Vs

rV

Pωωω −

⋅==2

2

2

2 33 (10.7)

The mechanical power output is

)(3

)1(3

22

2

2

2

mee

mphphm r

Vss

rV

P ωωωω

−=−=

Let T = f (ωm) be the torque speed characteristic of the mechanical load. Then

)(3

)( 22

2

mee

mphmm r

Vf ωω

ωωωω −=

or

2

22 3)(

)( rV

f phm

me

e =−

ωωω

ω

Solution of above expression provides the slip s. Now assume a drop in the frequency of the network, ω ω . Differentiation of above equation yields: ωe e d→ + e

2 2

2ω ω ω ω

ω ωω ωe e m e

e mm ef d( )

( )( )− −

−+[ ω

ω ωωe

e mmf

2

2( )( )

−+ ωω ω

∂ω

e

e m m

f2

( )−]d mω = 0

or

e

mmeeme

memem dff

fd ω

∂ω∂ωωωωω

ωωωωω ⋅−+

−=

)()(

)()2(22

2



This relationship suggests that a decrease in electrical frequency causes a decrease in rotating speed.

In general, the term ω ω ω∂ωe e m

m

f2 ( )− is very small compared to the term ω ω and

thus it can be omitted. Then

e mf2 ( )

ee

emm dd ω

ωωωω ⋅

−=

2

The incremental change of the electric load is now computed upon differentiation of Equation (10.7).

)1(3

22

2

me

ee

mphe dd

rV

dP ωω

ωωω

−=

Upon substitution of dωm

dffPdPd

rV

dPo

e

e

eee

e

mephe ==

−=

ωωω

ωωω2

2

23

Thus the sensitivity D of induction motor power to frequency changes is approximately

o

e

fPD =

In summary, the electric load of a system is a complex function of voltage and frequency. A linearized model of the load versus frequency and voltage has been derived. The load model is VDfDtgPP VDe ∆+∆+=∆=∆ )( In this chapter we shall be concerned with changes in system frequency only. For this reason we shall neglect the term D Vv∆ . Figure 10.12 illustrates the variation of the electric load versus frequency of the various electric load types discussed.



P

ffo

P

ffo

ffo

P

(a)

(b)

(c) Figure 10.12 Absorbed Power versus Frequency of Various Electric Loads

(a) Induction Motors, (b) Inductive Load, (c) Capacitive Load 10.2.5 Summary Of Models The mathematical model describing the governor/turbine/generator and load is:

)(222

tgEf

fDEf

PEf

dtfd oo

Go −∆−∆=

∆

CT

GT

G XT

PTdt

Pd∆+∆−=

∆ 11

)1(11C

GC

G

C PfRT

XTdt

Xd∆+∆−+∆−=

∆



In compact matrix notation:

BuAxdtdx

+=

where

, , ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆∆

=

C

G

XPf

x ⎥⎦

⎤⎢⎣

⎡∆

=CPtg

u)(

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−

−

=

GG

TT

oo

TRT

TT

Ef

EDf

A

101

110

022

,

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡−

=

G

o

T

Ef

B1000

02

The signal flow diagram of the system is illustrated in Figure 10.13. Note that the inputs to the system are changes to the plant set point, ∆PC , and changes to the electric load, g(t). The input g(t) is exogenous. The input ∆PC is an external input (i.e. the operator or an external control loop such as the secondary generation control loop). Note that there is a limiter at the input of the governor model. This represents the usual constraint that a unit cannot respond to commands that require a large change in unit output within a short time. The clipped values of the limiter represent the maximum rate by which a unit can change its power output.

+

+-

1/R

HG(s)-

D

-HT(s)

Hp(s)

GP∆

)(tg

fD∆

cP∆

f∆

Figure 10.13 Signal Flow Diagram of the Governor, Turbine, Generator and

Load Model



g(t) Pd

G

+

-

fsch

f

Pc

Primemover

Governor

Figure 10.14 Block Diagram of The Primary AGC Loop System In subsequent paragraphs the response of the system will be examined. Specifically, first the steady state response will be analyzed and then the dynamic response.

10.3 Steady State Response of the Primary Generation Control System The primary AGC loop is illustrated in Figure 10.14. The inputs to the system are changes to the plant set point, ∆ , and changes to the electric load, g(t). Quantities of interest are the mechanical power output of the turbine,

PC∆PG , and the frequency, f, of the

system. Other quantities of interest are the valve opening ∆XC and the changes to electric load. The mathematical model of the system is:

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

∆

−

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆∆

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−

−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆∆

G

CC

G

GG

TTC

G

TP

Etgf

xPf

TRT

TT

Ef

EDf

xPf

dtd 0

2)(

101

110

022 0

00

(10.8)

For the computation of the steady state response, the time derivatives must be set equal to zero. In this case and solving for the state of the system

ββ

CPtgf∆

+−=∆)(

ββ

CG

PDtg

RP

∆+=∆

)(1



ββ

CC

PDtg

RX

∆+=∆

)(1

where

R

D 1+=β

The quantity β is characteristic of the generating unit speed regulation, R, and the electric load characteristic, D. It shall be called the area frequency response. Above expressions indicate that: (a) an increase (decrease) in load will cause a decrease (increase) in frequency, (b) an increase (decrease) in the unit set point will increase (decrease) the frequency, and (c) an increase (decrease) in either the load or the unit set point will increase (decrease) the power output of the turbine. Of interest is the case where the unit set point, PC, is kept constant. In this case, the frequency of the system and the turbine PG will vary as the electric load of the system varies. Specifically, in this case 0)(,0 ≠=∆ tgandPC

The following relationships are valid in this case

β

)(tgf −=∆ (10.9)

)(11 tgR

fR

PG β=∆−=∆ (10.10)

)(11 tgR

fR

X C β=∆−=∆ (10.11)

Another case of interest is the required adjustment to the unit set point in such a way that the frequency of the system remains constant equal to the rated frequency. In this case, ∆f = 0. Substituting into equations (10.9-10.11): )(tgPC =∆

CG PP ∆=∆

CC PX ∆=∆ These cases are illustrated in Figure 10.15.



slope = 1

Pg

Pc

f=0f1

slope = -1/R

f

(a)

(b)

∆

Pg∆

∆

∆

∆

∆∆

Pc1∆

Pc=0∆Pc2∆

Figure 10.15 Steady State Response of Governor/Turbine System (a) versus GP∆ CP∆ , parameter f∆ , (b) versus GP∆ f∆ , parameter CP∆

Example E10.1: Consider a single 800 MVA generator supplying a service area. At a given instant of the time, the system operates at 60 Hz. The electric load is PD = 600 MW, the per unit inertia constant is H = 3.5 sec, the speed regulation (droop characteristic) is R = 0.05 pu and the load frequency characteristic is D=15 MW/Hz. (a)Compute the steady state condition of the system after a 10 MW increase in the electric load. (b) Compute the required change in the set point to insure a 60 Hz operation. Solution: (a) The frequency deviation at steady state will be:

∆f a

DR

= −+

1



R = 0.05 60800

0 00375 = . /Hz MW

DR

MW Hz MWHz

MW HZ+ = + =1 15 1

0 05800

60281 66/

.( ) . /

∆f Hz= − = −10

281 660 0355

.. Hz

∆P a

R(DR

G =+

= =110 9 47

).MW

(0.00375)(281.66)

∆ ∆P D f MWD = = −0 53. Analysis of events: New Generation: PG = 600 + 9.47 = 609.47 MW New Frequency: f = 60 - 0.0355 = 59.9645 Hz New Load: PD = 600 - 0.53 + 10 = 609.47 MW (b) Now it is required that

Hz

RD

bf 0355.09645.590.601

=−=+

=∆

Upon solution we obtain: b = 10 MW. Note that:

MWb

RD

DPG 53.01

=+

=∆

MWfDPD 53.0=∆=∆ The new steady state operation will be at:

New Frequency: f = 59.9645 + 0.0355 = 60.0 New Generation: PG = 609.47 + 0.53 = 610 New Load: PD = 610



Above completes the solution.

10.4 Transient Response of the Primary Generation Control System The primary AGC loop has as inputs the unit power set point and the electric load variations. The system is dynamic. Thus, any changes in the inputs will cause some transients. The system will relax at a point defined in previous sections. In this section, the transients will be examined. A simplified case will be examined first and then typical responses will be given for the more detailed model. 10.4.1 Transient Response Neglecting Governor/Turbine dynamics

In this case, it is assumed that TT = TG = 0. The model in this case becomes:

Gooo PEftg

EffD

Ef

dtfd

∆+−∆−=∆

2)(

22

0 = − +∆ ∆P XG C

0 1

= − − +∆ ∆XR

f PC C∆

Elimination of the variables ∆ and XC ∆PG from above equations yields:

Cooo PEftg

Eff

RD

Ef

dtfd

∆+−∆+−=∆

2)(

2)1(

2

The response of the system to a step change of the electric load g(t) = a u-1(t) is

)1()( ts eatf α

β−−−=∆

where

R

D 1+=β is the area frequency response

E

fo

2βα =



The step response is illustrated in Figure 10.16. The response of the system to a general variation of the electric load g(t) will be

τβατ ατdetg

dttfdtgtf

ts −∫ −−=

∆=∆

0

)()(*)()(

f

Timea-β

Figure 10.16 Response of the Primary AGC Loop to a Step Load Change Neglecting Governor-Turbine Dynamics

10.4.2 Transient Response of Full System In this case, the governor dynamics are not neglected. The model is given with equation (10.8). Assuming the unit power set point is constant, the typical response of the system to a step change in the load or any general load change can be computed numerically. Typical responses of the primary AGC loop are illustrated with the following example. Example E10.2: Consider an electric power system comprising an 800 MVA generator supplying a load area. At a given instant of time the system operates at 60 Hz while supplying a 600 MW electric load. Other parameters of the system are: H = 3.5 sec R = 0.05 D = 10 MW/Hz TT = 0.6 sec TG = 0.15 sec Assume that no control is exercised in the plant set point, i.e., ∆PC = 0 0. .

(a) Compute the transient response to a 10 MW increase in the electric load neglecting governor and turbine dynamics.

(b) Compute the transient response of the system to a 10 MW increase in the electric load using a linear model (no limitations on unit rate of response).

(c) Compute the transient response of the system to a 10 MW increase in the electric load assuming that the unit has a limit of 50 MW/min.



Solution: The model of the system upon substitution of numerical values is:

)(0107143.00107143.0107143.0 tgPfdt

fdG −∆+∆−=

∆

CGG XP

dtPd

∆+∆−=∆

666.1666.1

CCC PXf

dtXd

∆+∆−∆−=∆

666.6666.677.1777

(a) Assuming TG→0 and TT→0 the last two equations yield: CG XP ∆+∆−=0 CC PXf ∆+∆−∆−= 67.2660 Upon solution of these equations: CG PfX ∆+∆−=∆ 67.266 CC PfX ∆+∆−=∆ 67.266 Upon substitution in the first equation

))((0107143.09643.2 tgPfdt

fdC −∆+∆−=

∆

The response of this system to a 10MW load change will be a simple exponential with a time constant of 0.3373 seconds (1/2.9643). (b) The response of this system to a 10MW load change is given in Figure E10.2a.



Program XfmCtr - Page 1 of 1

c:\books\md_psa\examples\psa-ch10-ex10-2b - Nov 27, 2001, 00:44:38.000000 - 10000.0 samples/sec - 150000 Samples

00:44:3800:44:3900:44:4000:44:4100:44:4200:44:4300:44:4400:44:4500:44:4600:44:4700:44:4800:44:4900:44:5000:44:5100:44:52Tuesday, November 27

-66.13 m

-57.87 m

-49.60 m

-41.33 m

-33.07 m

-24.80 m

-16.53 m

-8.267 m

0.000 System_Frequency (Hz)

0.000 p

1.730

3.461

5.191

6.921

8.652

10.38

12.11

13.84 Unit_Mechanical_Power (MW)

Figure E10.2a System Frequency Response of the Linear System -

Neglecting Rate Limiting (c) The rate limiting function can be represented with the following model:

)(0107143.00107143.0107143.0 tgPfdt

fdG −∆+∆−=

∆

CGG XP

dtPd

∆+∆−=∆

666.1666.1



)666.6666.677.1777( CCC PXfh

dtXd

∆+∆−∆−=∆

where

833.0

833.0833.0

,833.0,833.0

,)(

−<><

⎪⎩

⎪⎨

⎧

−=

xxxx

xh

The response of this system is illustrated in Figure E10.1b.

Figure E10.2b Actual System Frequency Response (Rate Limiting is

Modeled)

10.4.3 Load Allocation Among Generating Units Consider two units connected in parallel. We are interested in determining how a specific load change will affect the output of each one of the two units. The mathematical model for this system is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆

−+=

11

11

111

/0

2/)()(

)(

GC

o

TP

EtgftxA

dttdx

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆

−+=

22

22

222

/0

2/)()(

)(

GC

o

TP

EtgftxA

dttdx

where

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆∆

=

1

1

1

1

)()(

C

G

XP

tftx

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∆∆∆

=

2

2

2

2

)()(

C

G

XP

tftx



Consider a net change of the electric load equal to: g1(t) + g2(t) = g(t). Assuming that there is no external control, i.e. ∆PC =0 for each of the units, the steady state operating point will be:

2

2

2

2

1

1

1

121

)()(ββββ

CC PtgPtgfff

∆+−=

∆+−=∆=∆=∆

212121 )()()()( cc PPftgtgtg ∆+∆+∆+−=+= ββ

fR

tgR

PG ∆−==∆11

1

11

1)(1β

fR

X C ∆−=∆1

11

fR

PG ∆−=∆2

21

fR

X C ∆−=∆2

21

The above results are illustrated in Figure 10.17.

f

slope=-1/R1

slope=-1/R2

f

∆Pg1

∆Pg2

∆Pg

Figure 10.17. Steady State Response of a Two Unit System to Electric Load

Change Under no Control

Note that the real power output change for each unit is inversely proportional to its regulation characteristic (droop characteristic). Since on a per unit basis each unit has the



same droop characteristic, the real power output change for each unit will be proportional to its rating. This conclusion can become obvious when the regulation characteristic be written as:

iui S

fRR 0=

where: Ru is the regulation characteristic in p.u., f0 is the scheduled frequency, and Si is the generator rating. Substitution into the equations for the real power output change, yields:

ffR

SP

uG ∆−=∆

0

11

ffR

SP

uG ∆−=∆

0

22

It should be apparent from above equations that if the per unit values of R are the same for each unit, then the real power output change will be proportional to each unit’s rating. In other words the two units will share the new load proportionally to their ratings.

10.5 The Secondary Generation Control System The primary automatic generation control loop usually yields frequency deviations that are not acceptable in modern power systems. This is where the secondary loop enters the picture. It performs slow "reset" adjustments of the frequency by changing the plant set points PC. This is accomplished with an integral feedback control loop. Specifically the plant set point is controlled with the integral of the frequency deviations. Following a sudden load increase, g(t), the turbine output, ∆PG , is increased to a new value as rapidly as the primary AGC loop will permit. Typically, the turbine response sets the pace. The turbine response is limited by the availability of steam. Thus, in addition to the turbine time constant, the boiler capability to increase steam production must be considered. This limitation is modeled by imposing a limit on the rate by which the steam valve position is changed. The end effect is that the frequency decreases after a load increase. Thus, a negative frequency error is generated. This negative error remains until the turbine increases the mechanical power output to compensate for the load increase. The frequency error causes an increasing integral of frequency error. The integral of the frequency error is used to control the unit set point. The control action will not stop until the integral of the frequency error is zero. This control action may take a long time depending on the magnitude of the load increase and the control loop parameters. For example, for a desired increase of 20 MW in the unit set point and assuming that the unit capability to increase its output is 1 MW/sec, the secondary control action will take approximately 20 seconds.



Frequency accuracy is not the only reason that integral control is required. Utilities are interconnected. Thus, they must operate in synchronism. This operation imposes the following basic requirements: • Constant frequency throughout the interconnected system. • Constant time (accurate time readings are taken from the National Bureau of

Standards). Note that time equals the integral of frequency. In a modern power system this is achieved centrally at the control center. The control center receives the exact time from the National Bureau of Standards (via satellite) and makes adjustments such that the integral of the frequency is in pace with the real time. This operation insures that the synchronous clocks are always showing accurate time. The US utilities have made it their practice not to exceed 3 seconds in time error. 10.5.1 The Need for Integral (Reset) Control Integral control is needed to insure accurate regulation of frequency, i.e. to assure that the frequency error is equal to zero. Without integral control this objective cannot be achieved no matter how well the control loop is designed. To illustrate the point, consider the first order approximation (neglecting governor and turbine dynamics) of the primary AGC loop

CPbtCgfadt

fd∆+−∆−=

∆ )(

where:

βHf

a o

2=

C = fo/2E b = fo/2E Assume a feedback control law for the minimization of the frequency deviation of the form (proportional feedback control): )(tfkPC ∆−=∆ In this case the system response will obey the equation

)()( tCgfbkadt

fd−∆+−=

∆ (10.12)

For any change g(t) in the electric load, the steady state frequency of the system will have an error equal to



bka

Cgf ss +−=∆

Obviously, no matter what the gain of the feedback law is (parameter k), the frequency error cannot be zero. Now assume that integral control is applied. Specifically, the unit power set point is adjusted based on the integral of the frequency error:

ττ dfkPt

C )(∆−=∆ ∫∞−

Above feedback control law can be easily implemented. Consider the integral. The frequency deviation is expressed as the difference of the actual frequency minus the scheduled frequency yielding:

∫∫∫∞−∞−∞−

−=∆t

schtt

dfdfdf τττττ )()(

The first integral can be easily measured with a frequency counter. The second integral represents the actual time. It can be obtained with a precision clock such as a Global Positioning System (GPS) receiver. Thus the feedback quantity can be easily measured high very high precision. Now let's compute the steady state frequency deviation due to an electric load change of g assuming the above control law. For this purpose, define

∫∞−

∆=t

dftx ττ )()(

Upon differentiation of above equation and substitution in equation (10.2), the model becomes:

)()( tfdt

tdx∆=

)()( tCgtbkxfadt

fd−−∆−=

∆

The steady state response is obtained by substituting zero for the derivatives yielding 0 = ∆f t( ) 0 = − − −a f bkx t Cg t∆ ( ) ( )



Thus in this case, the steady state frequency deviation is exactly zero and it is independent from the gain of the integral control law (parameter k). 10.5.2 Time Domain Model of the Secondary Generation Control System The equation that defines the integral feedback control law is defined with

ττ dfkPt

C )(∆−=∆ ∫∞−

Upon differentiation

fkdtPd C ∆−=

∆

Incorporating above equation into the rest of the model, one obtains

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆∆∆∆

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆∆∆∆

000

g(t) 2/

000/1/10/10/1/10002/2/ Ef

PXPf

KTTRT

TTEfEDf

PXPf

dtd

o

C

C

G

GGG

TT

oo

C

C

G

Above model totally describes the performance of the generation control loop. The gain, k, of the integral control must be selected according to technical and economic criteria. The pictorial view of this system is illustrated in Figure 10.17. Note that the only exogenous parameter to the system is the electric load that can change at any time.

g(t) Pd

G

+

-

fsch

f

SteamLine

Governor

-K



Figure 10.18 Block Diagram of Proportional and Integral Control for a

Generating Unit 10.5.3 Steady State Response of the Secondary System We examine the steady state response of the generation control loop under integral control. For this purpose we consider the model equations and we set the time derivatives equal to zero. The solution of the resulting equations provides the steady state response of the system. For example, consider the steady state response to a step load change, i.e., g(t) = a At the steady state all derivatives will vanish, yielding the following equations:

0222

=−∆−∆ aEf

fDEf

PEf oo

Go

011=∆+∆− C

TG

T

XT

PT

0)1(11=∆+∆−+∆− C

GC

G

PfRT

XT

0=∆− fk Solution of these equations yields: 0=∆f aPG =∆ aX C =∆ aPC =∆ Above result can be interpreted as follows: In the presence of integral control the frequency of the system at steady state will be always equal to the scheduled frequency. Any load change will result in an equal change of the generation. Thus the integral control results in precise balancing of generation and load while regulating the frequency of the system always to the scheduled frequency. 10.5.4 Transient Response of the Secondary System



The transient response of the frequency control loop of an electric power system can be computed with numerical simulation techniques. Specifically, the differential equations of the system can be integrated using anyone of a number of numerical integration methods [???] to yield the solution of the differential equations [???]. This analysis can illustrate the impact of the control loop gain on the system performance as well as the system stress. The transient response of the secondary generation control loop for different gains of the control law will be illustrated with an example. Example E10.3: Consider the electric power system of the Example 10.2. Assume the following two integral control laws:

(a) ∫∞−

∆−=∆t

C dfP ττβ )(1.0

(b) ∫∞−

∆−=∆t

C dfP ττβ )(2.0

For each one of the above control laws repeat parts b and c of Example E10.2. Solution: (a) The model for this case is the same as the one in problem E10.2 appended with the equation for the feedback.

)(0107143.00107143.0107143.0 tgPfdt

fdG −∆+∆−=

∆

CGG XP

dtPd

∆+∆−=∆

666.1666.1

CCC PXf

dtXd

∆+∆−∆−=∆

666.6666.677.1777

fdtPd C ∆−=

∆6666.27

The last equation is the derivative of the feedback control law. Note that the variable beta is 276.666 MW/Hz. Numerical integration of above equations for g(t)=10 MW yields the results of Figure E10.3a.




c:\books\md_psa\examples\psa-ch10-ex10-3a - Nov 27, 2001, 00:52:57.000000 - 10000.0 samples/sec - 800000 Samples

00:53:00 00:53:10 00:53:20 00:53:30 00:53:40 00:53:50 00:54:00 00:54:10Tuesday, November 27

-65.90

-52.72

-39.54

-26.36

-13.18

-3.806 u System_Frequency__Hz_ (V)

4.297 u

2.932 k

5.864 k

8.796 k

11.73 k

14.66 k Unit_Mechanical_Power__MW_ (V)

-8.188 u

1.970 k

3.940 k

5.909 k

7.879 k

9.849 k Unit_Set_Point__MW_ (V)

Figure E10.3a. Transient Response of the System for k=0.1

(b) The model for this case is the same as the one in problem E10.2 appended with the equation for the feedback.

)(0107143.00107143.0107143.0 tgPfdt

fdG −∆+∆−=

∆

CGG XP

dtPd

∆+∆−=∆

666.1666.1



CCC PXf

dtXd

∆+∆−∆−=∆

666.6666.677.1777

fdtPd C ∆−=

∆3333.55

The last equation is the derivative of the feedback control law. Numerical integration of above equations for g(t)=10 MW yields the results of Figure E10.3b.


c:\books\md_psa\examples\psa-ch10-ex10-3b - Nov 27, 2001, 01:01:35.000000 - 10000.0 samples/sec - 800000 Samples

01:01:40 01:01:50 01:02:00 01:02:10 01:02:20 01:02:30 01:02:40 01:02:50Tuesday, November 27

-65.13

-52.10

-39.08

-26.05

-13.03

6.802 u System_Frequency__Hz_ (V)

-2.579 u

3.088 k

6.177 k

9.265 k

12.35 k

15.44 k Unit_Mechanical_Power__MW_ (V)

11.89 u

2.000 k

3.999 k

5.999 k

7.999 k

9.999 k Unit_Set_Point__MW_ (V)

Figure E10.3b. Transient Response of the System for k=0.2



10.6 Multi-Machine Systems Modern power systems are interconnected for a variety of economic and technical reasons. Figure 10.18 illustrates four interconnected power systems. Interconnections are accomplished with power lines which are called tie lines. In general, utilities operate and cooperate on the basis of agreement. This means that the net power flow in tie lines should be maintained equal to mutually agreed scheduled values. The agreed values are referred to as scheduled interchange power. Thus the tie lines (interconnections) serve the useful purpose of accommodating power transactions. At the same time, the tie lines transmit disturbances from one system to another when abnormal conditions occur. For example, the random tripping of a generating unit in one system can generate severe oscillations in a tie line power flow. It is apparent that in an interconnected system, the generation control system, in addition to the usual objective, it should also minimize the oscillations in the tie lines. The objectives of the generation control system in an interconnected power system are summarized below:

1. Maintain constant frequency close to scheduled value 2. Maintain accurate real time 3. Maintain net interchange power equal to scheduled values 4. Minimize equipment wear 5. Mutual assistance among interconnected systems in case of emergencies.

These objectives can be accomplished with the cooperation among the individual system. In subsequent paragraphs the model of a multimachine-multiarea system will be developed. The properties of the model will be studied. The control strategies of modern power systems will be introduced and their performance will be studied. 10.6.1 Model of a Multi-Machine System This section presents a simplified model of a multimachine-multiarea power system. The model of the generating unit is identical to the one developed in earlier sections. It should be recognized that the electric load for each unit will consist of the local load plus the electric power flow on the lines that connect the unit to the rest of the system. Let the quantity Pei represent the total electric load of unit i. Then: ∑+=

jijDiei PPP

where Pij is the power flow in the circuit from unit I to another point of the network j. As in previous sections, the electric load PDi is modeled with iiiDi fDtgP ∆+=∆ )( The power flow in the transmission line i,j, Pij can be represented with the following model:



)sin( jiij

jiij x

VVP δδ −= (10.13)

Linearization of Equation (10.13) around an operating point yields:

)(C= )()cos(

i jijjiij

jijiij x

VVP δδδδ

δδ∆−∆∆−∆

−=∆

where Cij is given by:

ij

jij x

VV=iC

Upon substitution, one obtains

)(2

)(222 ji

jij

i

oi

i

oii

i

oG

i

oi CEf

tgEf

fDEf

PEf

dtfd

iδδ ∆−∆−−∆−∆=

∆ ∑

In addition, the phase angle δ of the voltage Vi is dependent on the frequency of the system. Specifically

i

isii ttft δωπθ +== 2)( Upon differentiation

dt

df

dtd

f io

isi

δπ

δωπ +=+= 22

Upon rearrangement and substituting oii fff −=∆

ii f

dtd

∆=∆

πδ

2



Area 3Area 1

Area 4 Area 2

Tie Lines

Vi e Ptiei, v V ejδi

jδ33

V ejδ 2

2

Figure 10.19. Schematic Diagram of a Multi-Machine/ Multi-Area System.

The equations for a generating unit i in an interconnected system are summarized below:

)(2

)(222 ji

jij

i

oi

i

oii

i

oG

i

oi CEf

tgEf

fDEf

PEf

dtfd

iδδ ∆−∆−−∆−∆=

∆ ∑

ii f

dtd

∆=∆

πδ

2

CiTi

GiTi

Gi XT

PTdt

Pd∆+∆−=

∆ 11

)1(11Cii

iGiCi

Gi

Ci PfRT

XTdt

Xd∆+∆−+∆−=

∆

For a system with m units one set of above equations will be written for each unit. The resulting equations can be written in the following compact matrix notation.



)(tCgBuAxdtdx

++=

where

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆

∆∆

=

Cm

C

C

P

PP

u...

2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

)(...

)()(

)( 2

1

tg

tgtg

tg

m

The described model is suitable for representing a multimachine system. Usually a multimachine system results from the interconnection of several individual systems. For a two unit system the equations are:

δ∆−∆+−∆+−=∆

121

11

11

11

11

1

22)(

2)1(

2C

Ef

PEf

tgEf

fR

DEf

dtfd o

Cooo

δ∆−∆+−∆+−=∆

212

22

22

22

22

2

22)(

2)1(

2C

Ef

PEf

tgEf

fR

DEf

dtfd o

Cooo

The feedback control law for this system is given by the equations: 1112011 fKCKPC ∆+∆=∆ δ

2221022 fKCKPC ∆+∆=∆ δ Upon substitution of above expressions into the equations:

δβ ∆−+∆−−=∆

12011

1111

1 )1(2

)(2

CKEf

fKEf

dtfd oo

)(2

)1(2

)(2 2

21022

2222

2 tgEf

CKEf

fKEf

dtfd ooo −∆−+∆−−=

∆δβ

The modeling procedure is illustrated with an example of a two area system.



Example E10.4: Consider the two generator system of Figure E10.4. The two systems are connected with a 115 kV transmission line. The total series reactance of the line is 45 ohms. At the present time the tie line carries a total load of 30 MW. Develop the system model in state space representation. System data are: Unit 1 Unit 2 Rating` 250MVA 300MVA Per Unit Inertia Constant (H) 2.5 sec 2.8 sec Unit Output 180 MW 220 MW Turbine Time Constant 0.2 sec 0.1 sec Governor Time Constant 0.25 sec 0.2 sec Regulation 0.05 pu 0.05 pu Load Frequency Characteristic 5 MW/Hz 6 MW/Hz

G1 G2

Area 1 Area 2

Figure E10.4. A Simplified Two Area System Solution: The equations for area 1 are:

)(2

)(222 1

1

'8

'412

12

111

1

1 tgEf

xxCEf

xEf

xDEf

dtdx oooo −−−+−=

31

21

2 11 xT

xTdt

dx

TT

+−=

11

31

111

3 111 uT

xT

xTRdt

dx

GGG

+−−=

1

'4 2 x

dtdx

π=



The equations for area 2 are:

)(2

)(222 2

2

'4

'812

26

252

2

5 tgEf

xxCEf

xEf

xDEf

dtdx oooo −−−+−=

72

62

6 11 xT

xTdt

dx

TT

+−=

22

72

522

7 111 uT

xT

xTRdt

dx

GGG

+−−=

5

'8 2 x

dtdx

π=

where

22

11

2'8

27

26

25

1'4

13

12

11

C

C

E

G

E

G

PuPu

x

XxPxfx

x

XxPxfx

∆=∆=∆=

∆=∆=∆=∆=

∆=∆=∆=

δ

δ

Note that only the difference of the variables x4' and x8' is needed. Thus, these two variables can be represented with a new variable x4= x4' - x8'. For this purpose, subtract the 4th and 8th equation to obtain

)(2)(

51

'8

'4 xxdt

xxd−=

−π

or

)(2 514 xx

dtdx

−= π

Equations 4 and 8 must be substituted with above equation. The resulting seven equations represent the model of the two area system.



Note that the numerical values of the parameters are: E1 = 625 MW.sec PG1 = 180 MW TT1 = 0.2 seconds TG1 = 0.25 seconds R1 = 0.012 Hz/MW D1 = 5 MW/Hz E2 = 840 MW.sec PG2 = 220 MW TT2 = 0.1 seconds TG2 = 0.2 seconds R2 = 0.01 Hz/MW D2 = 6 MW/Hz The constant C12 is computed as follows. First observe that the line flow is 30 MW, i.e.

MWx

VV30

)sin(

12

2121 =− δδ

Above equation yields the solution 21 δδ − = 5.85890 or 0.10225 rads. The constant C12 is:

9235.235.292)cos(

,1212

212112 ==

−= puCorMW

xVV

Cδδ

Upon substitution of numerical values, the seven equations read:

)(048.0032.14048.024.0 14211 tgxxx

dtdx

−−+−=

322 55 xx

dtdx

+−=

1313 4433.333 uxx

dtdx

+−−=

)(2 514 xx

dtdx

−= π



)(035713.044.10035713.021426.0 24655 tgxxx

dtdx

−++−=

766 1010 xx

dtdx

+−=

2757 55500 uxx

dtdx

+−−=

Above equations complete the model. The performance of a multimachine-multiarea system can be studied by exercising the above model. The performance of the system is dependent upon the feedback control law that is used. The impact of controls will be discussed in the next section. 10.6.2 Generation Control of a Multi-Machine System The objectives of the automatic generation control loop of a control area have been listed in a previous section. The objectives are achieved with a feedback control loop. The design of the feedback control law is always a compromise because of conflicting technical factors. Historically, an acceptable solution to the generation control problem has been found in the following simple integral feedback control law:

(10.14) τττ dfBPkP ii

t

jiCi ))()(( ij ∆+∆−=∆ ∫ ∑

∞−

where ki is the control loop gain and Bi is the so-called frequency bias, is the deviation of the net power interchange from the scheduled value and is the frequency deviation from the scheduled frequency. The feedback control law tries to “drive” the quantity in the integrand to zero and therefore it tries to “drive” the net interchange and frequency to their scheduled values. It will be shown in the subsequent paragraphs that the steady state performance of this system is not affected by the selected values of the parameters k

)(tPij∆

)(tf i∆

i and Bi. However, the performance of the system during transients is affected by the selection of these parameters. The judicial selection of the parameters of the feedback control law, ki and Bi, can be achieved in two ways: (a) by exhaustive simulation and (b) by use of optimal control theory. These methods will be discussed later. The derivative of the above defined control signal is called the Area Control Error (ACE), i.e. )()()( ij tfBtPtACE ii

ji ∆+∆= ∑



Thus, the area control error (ACE) is defined to be equal to the sum of the net interchange flow deviation and the weighted frequency deviation (with the frequency bias parameter). The integral feedback control law of Equation (10.14) accomplishes the objectives of the automatic generation control loop in steady state. Consider, for example, the four control areas of Figure 10.18. Assuming a feedback control law as in Equation (10.14), it is required that at steady state the area control error (ACE) of each system must be equal to zero, i.e.: 4and32,1,=i ,0ij =∆+∆∑ ii

j

fBP

Summation of above four equations yields 044332211

3,2,1j4,

4,2,1j3,

4,3,1j2,

4,3,2j1, =∆+∆+∆+∆+∆+∆+∆+∆ ∑∑∑∑

====

fBfBfBfBPPPPjjjj

Note that (by definition) 0

3,2,1j4,

4,2,1j3,

4,3,1j2,

4,3,2j1, =∆+∆+∆+∆ ∑∑∑∑

==== jjjj

PPPP

and that at steady state the frequency of any system will be equal to the frequency of any other system: fffff ∆=∆=∆=∆=∆ 4321

thus 0=∆f and for every system i 0ij =∆∑

j

P

It is concluded, therefore, that at steady state the system operation will be characterized with frequency equal to the scheduled value and the net interchange among areas equal to the scheduled values, independently of the selection of the feedback control parameters, ki and Bi. The steady state response of the integral feedback control law meets the objectives of the automatic generation control. The transient performance of the control loop, however, depends on the gain ki and the frequency bias Bi. The performance criteria in transient conditions are: (a) minimization of equipment wear, and (b) mutual assistance among interconnected control areas. The generation control loop parameters ki and Bi are



selected as to optimize the performance of the control loop measured with above two criteria. In subsequent paragraphs, the generation control problem will be formulated, and the steady state and transient response will be studied. 10.6.3. Steady State Response of a Two Machine System The steady state performance of a two area system can be determined utilizing the previously developed equations. Two cases will be examined: (1) the system is unregulated, i.e. u1= u2 = 0 and (2) the system is regulated, i.e. u1, u2 ≠0. Unregulated System. In this case it is assumed that the set points of the units remain unchanged, i.e. u1= u2 = 0. Assume that there is a step change a in the load of area 1, g1(t) = a, g2(t) = 0. The steady state response is computed by setting the derivatives in the model equal to zero and solving the resulting equations. The result is:

2121 ββ +

−=∆=∆aff

2111

1ββ +

=∆a

RPG

2122

1ββ +

=∆a

RPG

2111 ββ +

−=∆aDPD

2122 ββ +

−=∆aDPD

aP21

212 ββ

β+

−=∆

111

1R

D +=β

222

1R

D +=β

Above results indicate how the two areas help each other to improve frequency control. They also demonstrate how increases in load demand are distributed among the controlling generators. The load "assumed" by a generator is proportional to the inverse of its droop characteristic (parameter R). This behavior has been observed earlier for two units operating in parallel, see Figure 10.17 which offers a geometric interpretation. Similarly, the response to a step change in the plant set point of area 1, ∆P bC1 = ,

, is computed to be: ∆PC2 0=



2121 ββ +=∆=∆

bff

bD

PG21

211 ββ

β++

=∆

bR

PG )(1

2122 ββ +

−=∆

bP21

212 ββ

β+

=∆

From above results it is obvious how an interconnection helps the areas to improve frequency control. Regulated System. In this case it is assumed that the set points of the units are regulated with a feedback control law. Specifically, the feedback control law is assumed to be:

τττ dfBPkPt

C ))()(( 111211 ∆+∆−=∆ ∫∞−

τττ dfBPkPt

C ))()(( 222122 ∆+∆−=∆ ∫∞−

In this case the steady state response to a step load change in area 1, 0)(,)( 21 == tgatg is

00

0

12

2

1

11

=∆=∆=∆

=∆=∆

PP

aPff

G

G

Note that in this case, the first generator assumes the full load change of its area. This is desirable when the two generators belong to different economic entities. Note also that, the net interchange remains constant (∆P12=0). At steady state, the two generators respond only to load changes of their own areas. Clearly integral control achieves this objective. The steady state response will be demonstrated with an example. Example E10.5: Consider the system of example E10.4. Assume that an electric load change of g1(t) = 50 MW occurs in system 1.



(a) Compute the steady state response of the system for the unregulated system. Account for all power.

(b) Compute the steady state response of the system for the regulated system with integral control.

Solution: (a) Unregulated system. g1(t) = 0.5 pu Thus

257289.0

11

2.1106.005.0

5.0−=

+++−=∆f

2144.01 =∆ GP 2573.02 =∆ GP 0128.01 −=∆ DP 015437.02 −=∆ DP 2727.02,1 −=∆ tieP An accounting of the power is illustrated in Figure E10.5.

∆PG1 = +0.2144

∆Pd1 = -0.0128g(t) = +0.5

∆Ptie = -0.2727

∆Pd2 = -0.0154

∆PG2 = +0.2573

a) Figure E10.5 Accounting of Power Change (b) Regulated system.



000

0.05.0

0

2,1

2

1

2

1

=∆=∆=∆=∆=∆

=∆

tie

D

D

G

G

PPPPPf

Example E10.6: Consider a single unit electric power system connected to a very large power system through a 115 kV transmission line as illustrated in Figure E10.6. Assume that the voltage and frequency of the large system remain constant (infinite bus). The series reactance of the line is 30 ohms. The voltage at the terminals of the generator is maintained constant at 1.0 pu. Other system parameters are: SB = 350 MVA (generator rated power), H = 3 sec, D = 5 MW/Hz, TG = 0.15 sec, TT= 0.55 sec, R = 0.05 pu. The generator generates 150 MW of real power and the load is 200 MW. (a) Develop the dynamic equations of the system in the state space form

)()()()( tcgtbutAxdt

tdx++=

where u(t) = ∆ (t), the unit set point and g(t) is the electric load disturbance. PC (b) Compute the steady state frequency and generator power output for a 10 MW

increase of the electric load assuming ∆PC (t) = 0. (c) Compute the steady state frequency and generator power output for a 10 MW

increase of the electric load assuming the following feedback law: )()()( tftPtP tieC ∆−∆−=∆ β (d) Compute the steady state frequency and generator power output for a 10 MW increase

of the electric load assuming the following feedback law:

∫∞−

∆+∆−=∆t

tieC dfPtP ττβτ ))()((1.0)(



GTie Line

"Infinite System"

Figure E10.6. A Single Generator System Supplying an Area and Connected to a Large System

Solution: (a) The model of the single unit power system connected to the infinite bus is:

)(2222 1 tg

Ef

CEf

PEf

fDEf

dtfd oo

Goo −∆−∆+∆−=

∆∞ δ

CT

GT

G XT

PTdt

Pd∆+∆−=

∆ 11

)1(11C

GC

G

C PfRT

XTdt

Xd∆+∆−+∆−=

∆

fdt

d∆=

∆ πδ 2

The constant C is computed from the given data as follows: 1∞

MW50sin30

1152

1 −==∞ δP

Solution of above equations for δ yields: δ = -6.51o or –0.1137 rads. Thus

MW98.437cos30

1152

1 ==∞ δC

Next the parameters of this problem are converted into a consisted set of units:



sec.1050 MWHSE B == MWHzR /0086.0= HzMWD /5= ondsTT sec55.0= ondsTG sec15.0= HzMW /279.121=β Upon substitution of all numerical values:

)(

06.666

00

)(

000

0.0286-

+x(t)

0002832.60666.6019.7750818.1818.10514.1200286.01428.0

)( tPtgdt

tdxC∆

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−

=

where [ ]TCG ttXtPtftx )()()()()( δ∆∆∆∆= (b) In this case g(t) = 10 MW and ∆Pc(t) = 0.0. Substitution into the system model, and solving for the steady state (assuming all derivatives will vanish):

MW10-or pu 02857.002283.0

00

−=∆−=∆=∆=∆

tie

Gss

ss

P

Pf

δ

The last result implies that the new load g(t) = 10 MW will be served from the large system via the tie line. (c) In this case g(t )= 10 MW, and ∆PC(t) = -437.98∆δ(t) – 121.279∆f(t). Substitution into the system model, and solving for the steady state (assuming all derivatives will vanish): 0=∆ ssf 011415.0−=∆δ MW5.0 =∆ GP W0.5 MX C =∆ MW.05−=∆ tieP



These results indicate that the new load g(t) will be served partly by the unit and partly by the large system. (d) In this case, define as an additional variable )()(5 tPtx C∆= Upon differentiation of the feedback law equation

)(1279.12)(798.43)(5 tft

dttdx

∆−∆−= δ

Above equation is added to the previous system model. The resulting equations are:

)(

00000286.0

)(

0798.43001279.1200002832.6666.60666.6019.77500818.1818.100514.1200286.01428.0

)( tgtxdt

tdx

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡−

+

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−

−−

=

Upon solution of above equations under steady state conditions, i.e. all derivatives equal zero:

MW10=∆ GP MW10 =∆ CX MW 10=∆ CP

0=∆δ 0=∆ tieP These results illustrate that in this case the entire new electric load, g(t), will be served by the generating unit. The tie line flow will not change. 10.6.4 Transient Response of a Two Machine System It has been seen that in the presence of integral control, a change of electric load in an area, is totally absorbed by the units in that area. To ensure smooth operation, it is also important to minimize the transients following any load change. For example, it is desirable that the transients in one area due to load changes in another area be minimal. To insure this, the feedback gains must be so selected as to minimize these transients. In general, the parameter of the feedback control loop k and B should be selected in such a way that the system has: (a) a stable response to disturbances and (b) smooth response



(low control effort). To achieve these objectives, the feedback control parameters are selected with one of the following ways:

• Optimization through detailed simulation. • Modern optimal control theory

The two methods will be discussed next. Approach Based on Simulation. In this approach, the response of the system is computed for many different values of the design parameters involved. The optimal value of the design parameters is selected upon inspection of the results and evaluation of all conflicting factors. Usually, the performance of the system is measured with appropriate performance indices. The performance index provides a measure of the system oscillations and of the control effort. The general form, of this type of index is:

τττττ dRuuQxxJ TT ))()()()((21

0∫∞

+=

where x(t) is the system state and u(t) is the system control input. Next the system is simulated for specific disturbances (load changes) and the value of the performance index is plotted for various values of the feedback control variables k and B. A graph of the value of the performance index versus the control parameters is given in Figure 10.10. The conclusions from these simulations are the following. The best selection for the parameters k and B is:

• Select the parameter B to be equal to the system constant β • Select the parameter k to be in the range 0.1 to 0.3

B=Constant

Integral Gain , k

Obj

ectiv

e Fu

nctio

n



Figure 10.20 Variation of the Objective Function vs. the Integral Feedback

Gain k Approach Based on Optimal Control Theory. In this approach, the performance of the system is again measured with a performance index. The performance index penalizes system oscillations as well as control effort. Such a performance index has been introduced in the previous paragraph. Then we define an optimization problem with the objective to minimize the performance index subject to the system dynamic equations.

Minimize τττττ dRuuQxxJ TT ))()()()((21

0∫∞

+=

Subject to )()()( tButAxdt

tdx+=

0)0( xx = The solution of above optimization problem is obtained with the solution of the appropriate Riccatti Equation [???]. Specifically, the solution is given with: )()( tkxtu −=where:

PBRk T1−= and P is determined from the solution of the Riccatti Equation QPBPBRPAPA TT −=−+ −1

For more information on this method, consult the references.

10.7 Automatic Generation Control in Modern Energy Management Systems AGC was one of the first power system functions that were computerized. Experience gained in the field allowed the integration of many functions in the AGC loop, such as economic dispatch, security controls, etc. Today, AGC is performed with digital computers every 1 to 6 seconds. Figure 10.21 illustrates a typical modern configuration. The acceptance of digital systems is mainly due to the flexibility with which these systems can adapt to new control concepts and new operating policies. In addition, they present flexibility in integrating other functions, such as economic functions and security controls. Modern power plants implement these controls with digital computers.



Power System

Pdes1

Pdes2

Pdes3

L1(S) GovMover 1

L2(s)GovMover 2

L3(s)GovMover 3

+-+-+-

UCE1

UCE2

UCE3

G1

G2

G3

Tie Line

Tie Line

ESTIMATION

ACECalculation

K(s)AllocationRules

Security,Economic Dispatch,Etc.

EverySeveralMinutes

ACE

ACE = Ptie + B f

f Tie Flows

Every 1 to 6 sec.

PG1

PG2

PG3

Figure 10.21 Modern Automatic Generation Control Structure The AGC functions are typically performed every 1 to 6 seconds. At every operation of the AGC, the state of the system is measured or estimated. Then the area control error (ACE) is computed from the system frequency and tie line flows. The ACE is multiplied with the desired gain k. Then the result is allocated to the various units of the system with prespecified allocation rules. The output of the AGC system is the desired unit output for each unit in the system, Pdes i , i = 1, 2,....n. These signals are transmitted to the plant RTUs which set the unit power set point appropriately. Then the primary generation control loop (local plant control) will track the desired unit output. The allocation rules can be designed in such a way as to incorporate economic dispatch, system security, scheduling function, etc. This function is discussed next.



10.7.1 Allocation Rules The control scheme of a modern control center is illustrated in Figure 10.21. This scheme is a two level control structure. The two levels are: (a) the primary generator control loop which is local for each generating unit, and (b) the secondary generation control loop. The objectives of this scheme are:

• The primary AGC loop is designed with feedback concepts for stable operation. • The secondary AGC loop (reset) provides the set points Pdes for each generating

unit. The set points are inputs to the primary AGC loop. The primary control loop is designed with feedback concepts for stable operation and minimal oscillations. The primary control loop thus tracks the commands, Pdesi,, coming from the secondary generation control loop. The commands, Pdesi, are reset every AGC cycle. Specifically, the desired outputs for each unit are computed every AGC cycle for which a nonzero ACE exists. The computation of the desire unit output is performed as follows. An economic dispatch algorithm is performed every few minutes. Within this computation, a linearized model of the economic dispatch solution is defined. Specifically, let be the total electric load (including net interchange) at which the economic dispatch has been performed. Let , be the economic dispatch solution. Obviously,

oloadP

niPobi ...,,2,1, =

),( o

bio

loado

loadi

obi PPqPP +=∑

The unit i output is referred to as the economic base points. The economic base points are computed for a specific total system load, net interchange and system topology. Thus:

obiP

)( o

loadio

bi PfP = The economic base points at other load levels can be approximately computed with a linearized model

loadload

biobibi P

PP

PP ∆∂∂

+=

The computation of above linearized model has been covered in Chapter 9. Recall that the derivative above was named the economic participation factor. It signifies the amount of load to be picked by unit i when the electric load changes by one unit, while maintaining economic operation. Thus at any instant of time, the AGC scheme has the following information:



(a) The previous economic dispatch solution: o

load Plevel loadat n 1,2,....,=i ,obiP

(b) The participation factors ai , i = 1,2,......, n. Now every time the system is scanned for data (every 1 to 6 seconds), the following are computed: (a) Actual unit outputs Pgi , i = 1,2,......, n. (b) Area Control Error (ACE) The above information is used at each cycle of the AGC to compute the following commands for each generating unit: (1) unit economic set point ACEaPP i

obibi +=

(2) desired output of the unit

∫∞−

++=t

iio

bidesi dACEbACEaPP ττ )(

where bi is an allocation factor for unit i for good regulation. Finally, the issued commands to the units will be lower or raise the plant set point by

: ∆PCi gidesiCi PPP −=∆ The allocation rules are illustrated in Figure 10.22



RegulatingAllocation

ai

EconomicAllocation

bi

LinearizedEconomic DispatchModel

ACE

iPgi

Pdes1

Pdes2

Pdes3

Figure 10.22 Allocation Rules of the Secondary Generation Control Loop 10.7.2 Past Practices and Regulations Because power systems are interconnected, a disturbance in one system will be felt with variable degree in any other system. Specifically:

• A change in load in any one system will produce a generation change in all other

systems.

• A change in generation in one system will produce a generation change in all other systems.

The practice of utilities with respect to the load-frequency control problem has been dictated by the spirit of cooperation and mutual assistance. It appears that this is the best alternative. Thus, utilities set the control problem in such a way as to:

• Maintain constant frequency • Import or export energy according to schedules • Assist neighboring utilities if need arises until the neighbor solves its problem.



Consider, for example, a large utility with a total generation at a specific instant of time of 15,000 MW. Assuming that the per-unit droop characteristic of all units is 0.05, then the system frequency response is approximately 5,000 MW/Hz. Further assume that the frequency bias is set equal to the system frequency response, or 5,000 MW/Hz. In this case, if the system frequency decreases to 59.9 Hz, then this large utility would deliver 500 MW to other systems. Now assume that the entire interconnection has a total generation of 300,000 MW (for example the eastern interconnection). Assuming similar AGC parameters for the entire interconnection, then the total amount of power that would be picked up by the interconnected system for a frequency change of -0.1 Hz would be 500 MWx 20 or 10,000 MW. Alternatively, one can state that in order for the system frequency to drop by 0.1 Hz, a 10,000 MW generation efficiency must exist. These values are typical for this system. Today the technology is advanced and the control centers sophisticated. Yet, the real time of the system, measured with the integral of the frequency, may drift. If this is the case, then the interconnected systems may change their scheduled frequency by a small amount until the problem is corrected. For example, assume that the real time of an interconnected system has advanced by 3 seconds. This can happen if the system frequency remains above the nominal frequency for a prolong time. In this case, it may be selected that the scheduled frequency of the system shall be 59.99 for the next 300 minutes. This action should bring the system real time back on schedule. Normally this is achieved as follows: The interconnected systems charge one company with the responsibility of monitoring the real time. When the real time shifts, then this system will issue a command to the rest of the systems to change their scheduled frequency for a specified period of time. Another operational problem is the fact that, due to system parameters and load characteristics, a particular system may have better regulation than others. To insure fairness, it has been agreed that a utility must bring the area control error to zero at least once every 10 minutes. Present practice is such that this requirement is comfortably met by most systems. As an example, consider Figure 10.23, illustrating the ACE for the Southern System on January 27, 1983 from 15:05 to 15:09. The ACE did not exceed 40 MW in this period and crossed zero at least once every minute. There are instances and periods of the time for which a system experiences large values of control error. Such cases may be morning hours during which the electric load is increasing at a fast pace. In this case, the generation control will issue large changes of individual unit generation. Unit response may limit these commands. In this case, the alternatives are:

• Allow the system frequency to drop. • Allow scheduled interchange to lag. • Increase the number of participating and regulating units.



Finally, present practices are such that emphasis is placed on economic scheduling. Such scheduling is incorporated in the AGC with a hierarchical structure as has been illustrated in Figure 10.1.

Figure 10.23 Recorded Area Control Error for the Southern System on January 27, 1983 from 15:05 to 15:05

(Courtesy of Georgia Power Company) 10.7.3 New Practices and Regulations The deregulation brought about another issue. Many independent power producers that they do not have an incentive to participate in the load-frequency control problem. Yet, the system cannot operate without the load-frequency regulation. In an independent system operation, the load-frequency control problem becomes an ancillary service. Units participating in the AGC regulation function are appropriately compensated. This market approach to load-frequency regulation has not worked well. Studies has shown that if the percentage of generating units participating in the load-frequency control is decreased [???] then the system will experience higher frequency deviations under usual system disturbances. A number of incentives and mandatory participation rules have been devised to address this issue. For the purpose that each region participates in the load-frequency control and assumes its fair share of the task, NERC has issued new control performance standards. These standards are referred to as CPS1 (Control Performance Standard 1), CPS2 (Control



Performance Standard 2), and DCS (Disturbance Control Standard). These standards have been introduced as mandatory as of January 1, 1998. Each control area must abide by these standards. The definition of the standards follow. Control Performance Standard 1 (CPS1): This standard is defined as follows:

( )( )∑

= −∆

=600,525

1 10600,52511

i

ii ACEfCPS

β

where: β is the are frequency bias is the one minute (minute i) average frequency deviation if∆ is the one minute (minute i) average ACE of the control area iACE Note that CPS1 expresses the one year average of above quantity. The standard states that CPS1 should not exceed the value e1 defined as the one year average of the one minute interconnection frequency error, i.e. , 2

11 ε≤CPS where:

∑=

∆=600,525

11 600,525

1i

ifε

Control Performance Standard 2 (CPS2): This standard is defined as follows:

∑=

=6

110,6

12j

jACECPS

where: is the ten minute average ACE of the control area 10,iACE Note that CPS2 expresses the one hour average of above quantity. The standard states that CPS2 should not exceed a certain limit referred to it as L10, i.e. )10)(10(65.12 1010 siLCPS ββε −−=≤ , where:

∑=

∆=560,52

110 560,52

1i

ifε

iβ is the area frequency bias sβ is the system frequency bias



Disturbance Control Standard (DCS): This standard states that the control area ACE must return to zero within 10 minutes (later increased to 15 minutes) for any disturbance that greater or equal 8% of the control area peak load or any single contingency.

10.8 Summary and Discussion (to be added)



10.9 Problems Problem 10.1 Consider an electric power system supplying a 552 MW load. Assume that the composition of the load is: - 8% is a constant resistance/inductance load of power factor 0.90 at 60 Hz. - 3% is a constant resistance/capacitance load of power factor 0.90 at 60 Hz. - 89% consists of induction motors driving constant torque mechanical loads. Compute the frequency characteristic D of the electric load. Solution: ( ) ( ) 0.190.91sinθ 22 =−= The frequency characteristic D of the electric load is:

)MW/Hz(0132.860

)89.0)(552()19.0)(03.0)(552)(2()19.0)(08.0)(552)(2(=

++−=D

Problem 10.2 The total electric load of a power system is 8620 MW. It can be assumed that (a) 10% of the electric load consists of constant resistance/inductance load of power factor 0.85 at 60 Hz. (b) 90% of the electric load consists of induction motors driving constant torque mechanical loads. Compute the frequency characteristic D of the electric load. Problem 10.3 Consider the electric power system of Figure P10.3. The voltage at the terminals of the generator is maintained at 1.0 pu. It is assumed that the voltage control loop is fast enough that load variations do not affect the voltage at the terminals of the generator. The generator is rated 500 MVA and operates at frequency 60 Hz, supplying the load R1, L1, i.e., switch S1 is closed; switch S2 is open. The numerical values of the load parameters (on a 500 MVA base) are:

pu 0.21 =R , pu 2.01 =Loω , pu 0.52 =R The generator droop characteristic is 0.05 pu on the generator ratings. (a) Compute the generator set point in MW, i.e. PC. (b) At time t=0, the switch S2 is closed. Assuming that the secondary frequency control loop is open, i.e. , compute the steady state frequency of the system. 0.0=∆ CP



G

R2R2

R2.

R1 L1 L1 R1

L1 R1

S2

S1

Figure P10.3

Problem 10.4 Consider the electric power system of Figure P10.4. The voltage at the terminals of the generator is maintained constant at 1.0 pu. It is assumed that the voltage control loop is fast enough that load variations do not affect the voltage at the terminals of the generator. The generator is rated 250 MVA and operates at frequency 60 Hz, supplying an electric load of 150 MW consisting of induction motors, i.e., switch S1 is closed, switch S2 is open. The numerical value of the resistor R2 is 5.5 pu expressed on the generator rating. The unit's droop characteristic is 0.05 pu. (a) Compute the unit power set point in MWs. (b) Compute the approximate frequency characteristic of the load. (c) At time t=0, the switch S2 closes. Assuming that the secondary control loop is open, compute the steady state frequency of the system.

Inductionmotors

G

R2R2

R2.

S1

S2

Figure P10.4

Problem 10.5 The electric power system of the Figure P10.5 has the following parameters:



sec5.2,60,500 === HHzfMVAS bb puRTT GT 05.0sec,25.0sec,5.0 ===

Assume that the system operates under steady state conditions with f = fsch = 60 Hz and total electric load equal to 310 MW. (a) Compute the constant D of the load in MW/Hz. (b) At time t = 0 a resistive electric load of 20 MW is connected to the system by closing the switch S. Assume that the system is uncontrolled, i.e. 0.0)( =∆ tPC . Compute the steady state frequency of the system. (c) At time t = 0 a resistive electric load of 20 MW is connected to the system by closing the switch S. Assume that the system is controlled with the following feedback law:

. Compute the steady state frequency of the system. )( )/(0.50)( tfHzMWtPC ∆−=∆(d) Consider part (b). Compute the transient frequency of the system for t > 0. (e) Consider part (c). Compute the transient frequency of the system for t > 0

G

+

-fsch

∆f

PrimeMover

Governor

∆Pc

LoadConsisting ofInductionmotors

ResistiveLoad20 MW

S

Figure P10.5

Solution:

(a) HzMWD /167.560310

==

(b) Uncontrolled case

( )( ) HzMWR

Dtgf /8333.1716005.0

500167.51,)(=+=+=−=∆ β

β

Hzf 1164.08333.171

20−=−=∆

Hzf 8836.59=



MWR

tgPG 3986.19)(==∆

β

MWfDPD 6014.0−=∆=∆ (c) Controlled case:

fPPtgf C

C ∆−=∆∆

+−=∆ 50,)(ββ

Hzf 09016.0−=∆ Hzf 90984.59=

MWP

DR

tgP CG 5341.19)(

=∆

+=∆ββ

MWfDPD 4659.0−=∆=∆ Problem 10.6: Consider a power plant comprising two generators rated 60 Hz and 200 MVA and 630 MVA, respectively. The two generators are operating in parallel supplying electric power to a load area. Both generators have regulation parameters R equal to 0.05 pu (expressed on their respective ratings). Assume that each unit outputs 150 MW of real power when suddenly the frequency drops by 0.01 Hz. Compute the change of generator output in MW for both units, assuming that the secondary automatic generation control loop is open. Problem 10.7: Consider a power plant comprising two units rated 60 Hz and 250 MVA and 350 MVA, respectively. The two units are operating in parallel supplying electric power to an industrial area. Assume that the load consists of induction motors, each unit supplies 200 MVA of real power and the droop characteristic of the two units is 0.05 pu (expressed on their respective ratings). (a) Compute the approximate frequency characteristic of the load. (b) Compute the steady state frequency and unit output for a 20 MW change (increase) of the electric load. Assume that the secondary generation control loop is open (uncontrolled case). Problem 10.8: A single generator rated 25 kV line to line, 800 MVA, 60 Hz is supplying the electric load of an isolated electric power system. The governor droop characteristic is R=0.05 pu and the load frequency characteristic is D=5 MW/Hz. At a certain instant of time, the generator is operating at 59.95 Hz with the secondary AGC loop open, i.e., ∆PC = 0. Under these conditions the generator supplies 600 MW. (a) Compute the unit power set point, P . C(b) Assume that suddenly a 10 MW customer switches in. Compute the new steady state operating frequency of the system assuming that the power set point, PC, did not change (uncontrolled unit).



(c) What is the unit power set point change, ∆PC , that the operator must perform to bring the system to the nominal frequency of 60 Hz? Problem P10.9: A power system consists of two generating units and an electric load. The data of the two units are: Unit 1 Unit 2 Rated Power 500 MVA 750 MVA Rated Frequency 60 Hz 60 Hz Droop Characteristic 0.05 pu 0.05 pu At a given instant of time, the following are observed. The total system load is 1000 MW, the system frequency is 59.90 Hz, the unit outputs are 400 MW and 600 MW respectively, and the secondary AGC loop is open, i.e. ∆ PC1 =∆ PC2 = 0. The frequency constant of the electric load is 15 MW/Hz. The power setpoint of unit 1 is PC1 = 383.333 MW.

• Compute the power set point PC2 of unit 2. • What is the required change ∆ PC1 for unit 1 which will bring the frequency to 60

Hz? (Under the assumption that ∆ PC2 = 0). Problem P10.10: Consider an electric power system comprising an 800 MVA generator supplying a load area. At a given instant of time the system operates at 60 Hz while supplying a 600 MW electric load. The other parameters of the system are: H = 3.5 sec generator per unit inertia constant R = 0.05 pu regulation parameters ("droop" characteristic) D = 15 MW/Hz electric load frequency characteristic TT = 0.6 sec steam turbine time constant TG = 0.6 sec governor time constant. Develop the dynamical equation of the system in the form

)()()()( tcgtbutAxdt

tdx++=

where u(t) = ∆Pc (unit set point) and g(t) the electric load additions. (a) Compute the transient response of the system to a 10 MW increase in the electric load assuming that the system is not controlled, i.e. u(t)=0.0. (b) Compute the transient response of the system to a 10 MW increase of the plant set point. (c) Assume that u(t)=-15(MW/Hz)∆f(t). Compute the steady state frequency of the system for a 10 MW increase of the system load, i.e. g(t)=10 MW. (d) Assume that the system is controlled with an integral feedback control law:



∫∞−

∆−=t

dftu ττβ )(1.0)(

Compute the steady state frequency of the system for a 10 MW increase of the system load, i.e. g(t)=10 MW. (e) An engineer decides to measure the performance of the system using the following performance index:

∫∞

∆+∆=0

22 ))()(2.0(21 τττ dPfJ C

Compute the value of the performance index for the load step change of 10 MW and for the following cases: (1) system is uncontrolled, (2) system is controlled as in case (c) above, and (3) system is controlled with an integral control law as in part (d) above. Hint: Solve the differential equations of the system numerically and compute the performance index J numerically.

Problem P10.11 Consider a single unit electric power system connected to a very large power system (modeled as an infinite bus) through a 115 kV transmission line as it is illustrated in Figure P10.11. The series impedance of the line is j42.5 ohms. The voltage at the terminals of the generator and the infinite bus is maintained constant at 115 kV line to line. Other system parameter are: SB = 250 MVA (generator rated power) H = 3.5 sec D = 12 MW/Hz TG = 0.1 sec TT = 0.3 sec R = 0.05 pu The generator generates 225 MW real power and the load is 200 MW.

G115kV

Infinite Bus

Figure P10.11



(a) Develop the dynamic equations of the system in the form: )()()()( tcgtbutAxtx ++=&

where u(t) = ∆ unit set point and g(t) = electric load disturbance. PC (b) Compute the steady state frequency and generator power for a 10 MW change of the electric load assuming ∆ = 0. PC(c) Compute the steady state frequency and generator power for a 10 MW change of the electric load assuming the following feedback control law: )()/(350)( tfHzMWtPP tieC ∆−∆−=∆ (d) Compute the steady state frequency and generator power for a 10 MW change of the electric load assuming the following feedback control law:

∫∞−

∆+∆−=∆t

tieC dfHzMWPtP τττ ))()/(350)((1.0)(

(e) Consider case (b) above. Compute the system transient frequency for t>0. (f) Consider case (c) above. Compute the system transient frequency for t>0. (g) Consider case (d) above. Compute the system transient frequency for t>0. Problem P10.12. Consider a two area system, each area comprising one unit. The two areas are interconnected with an 80 mile long, 115 kV line of series impedance equal to j0.8 ohms per mile. The parameters of the two units are: Unit 1 Unit 2 Rated Power 250 MVA 400 MVA Per Unit Inertia Constant 3.0 sec 3.5 sec Droop Characteristic, R 0.05 pu 0.05 pu Turbine Time Constant 0.0 0.0 Governor Time Constant 0.0 0.0 Load Frequency Constant 5 MW/Hz 10 MW/Hz Area Load 100 MW 350 MW Area Generation 200 MW 250 MW (a) Develop the dynamic equations of the system in the form

)t(CgBu)t(Axdt

)t(dx++=

where: uPP

tg t

C

C=⎡

⎣⎢

⎤

⎦⎥

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⎡

⎣⎢

⎤

⎦⎥

∆∆

∆∆∆δ

1

2 2 x(t) =

ff g(t) =

g1

21( )( )



Provide the numerical values of the matrices A, B, and C (Use a power base of 100 MVA). (b) Assume that u1(t) = P∆ C1(t) = -20 (MW/Hz) ∆ f1(t) , u2(t) = ∆ PC2(t) = 0, g1(t) = 0 and g2(t) = 10 MW. Compute the steady state frequency of the system assuming that prior to the load g2(t) increase, the frequency was 60 Hz. (c) Compute the transient response of the system for the conditions of question (b). Problem P10.13: Consider a two area power system interconnected with a 230 kV transmission line of series impedance ohmsjz 2.65= . Additional data for this system are: Area 1 Area 2 Rated Power 2500 MVA 2000 MVA Per unit inertia constant (H) 4.5 seconds 5.0 seconds Generation 2050 MW 1750 MW Electric load 1600 MW 2200 MW Frequency characteristic of load 20 MW/Hz 18 MW/Hz Turbine time constant 0.3 seconds 0.32 seconds Governor time constant 0.2 seconds 0.2 seconds Regulation Constant 0.05 pu 0.05 pu

(a) For the defined operating conditions, compute the linearized tie line power flow equation )( 211212 δδ ∆−∆=∆ CP . (b) Write the dynamical equations of the two area power system explicitly (substitute numerical values for the parameter variables). Assume uncontrolled case. (c) Assume that the load of area 1 increased by 100 MW. Compute the new generation and frequency for both systems at steady state conditions and the new power flow on the tie line. Solution: (a) The linearized model is:

( ) ( ) 002

01

02

01

20

12 685.33450sin2.65

230=−⇒=−= δδδδP

( ) ( ) 121.675cos2.65

230 02

01

2

12 =−= δδC

)(121.675 2112 δδ ∆−∆=∆P (b) The dynamical equations are:

( )211111 8.1)(002667.00533.0002667.0 δδ ∆−∆−−∆−∆=

∆tgfP

dtfd

G



( ) ( )2121 283.6 ff

dtd

∆−∆=∆−∆ δδ

111 333.3333.3 CG

G XPdtPd

∆+∆−=∆

1111 0.5667.41660.5 CC

C PfXdtXd

∆+∆−∆−=∆

( )212222 025.2)(003.0054.0003.0 δδ ∆−∆+−∆−∆=

∆tgfP

dtfd

G

222 125.3125.3 CG

G XPdtPd

∆+∆−=∆

2222 0.533.33330.5 CC

C PfXdtXd

∆+∆−∆−=∆

(c) In this case:

0.0,0.0,0.100)( 211 =∆=∆= CC PPtg Upon substitution and solution:

Hzfff 06502.021 −=∆=∆=∆

MWfR

PG 1833.541

11 =∆−=∆

MWfR

PG 3467.431

22 =∆−=∆

MWfDPD 3.111 −=∆=∆ MWfDPD 17.122 −=∆=∆

MWfP 517.44212 −=∆=∆ β Problem 10.14. Consider a simplified electric power system consisting of two generating units which operate in parallel and supply an electric load. The ratings of the two units are 60 Hz and 400 MVA and 800 MVA respectively. The droop characteristic of each unit is 0.05 per unit referred on the respective ratings of each unit. At a certain instant of time, the two units supply a total electric load of 568 MWs and the frequency is exactly 60 Hz. The frequency sensitivity of the load is D=9 MW/Hz. (a) Based on the information provided above, can you determine the real power output of each unit? (b) Assume that the electric load suddenly increases by 25 MWs. Assume also that the two units remain uncontrolled, i.e. the set point for each unit, Pc1 and Pc2, remain constant. Compute the change in the system frequency and the change in the real power output of each unit.



(c) Assume again the electric load change of 25 MWs. Now assume that only the 800 MVA unit is controlled with an integral control law, i.e. . Unit 1 is

not controlled, i.e. . Determine the steady state frequency of the system and the real power output change for each unit.

∫ ∆−=∆ ττ dfkPC )(2

0.01 =∆ CP

Solution: (a) No, the initial set points for the two units are unknown.

(b) )Hz/MW(00375.0800

)60)(05.0(),Hz/MW(0075.0400

)60)(05.0(21 ==== RR

)Hz(061125.09

00375.01

0075.01

25f −=++

−=∆

)MW(30.1600375.0

)061125.0(P),MW(15.80075.0

)061125.0(P g2g1 =−

−=∆=−

−=∆

Notice that, the total generation increase is 24.45 (MW). The load change is: )MW(55.0)061125.0(9.0)(=fPLoad −=−∆=∆ D (c) (MW) 25.0=P(MW), 0.0=P),Hz(0.0=f g2g1 ∆∆∆ Problem 10.15. Consider a single area power system consisting of a single 250 MVA, 60 Hz generator. At a given instant of time the system operates at 60.05 Hz, the total load is 200 MW. Assume a sudden load change of 25 MW. What will be the steady state frequency of the system for the following two cases: (1) If the system is uncontrolled, i.e. 0.0)( =∆ tPC , and (2) If the system is controlled with the following law: )()/(0.75)( tfHzMWtPc ∆−=∆ . In case (2) what will be the unit set point at steady state? System Data Are: Rated Power 250MVA Rated Frequency f = 60 Hz Inertia Constant H = 2.5 sec Regulation R = 3.6 Hz/250MW Electric Load constant D = 5 MW/Hz. Solution:



)MW/Hz(140R1D =+=β

ββ

)(tPg(t)f c∆+

−=∆

(1) : 0.0=(t)Pc∆

)Hz(82143.59140

2560fff 0ss =−

+=∆+=

(2) f(t)-12(MW/Hz)=(t)Pc ∆∆

140

f12140

25f ∆−+

−=∆

17859.00.0857143)+f(1 −=∆ )Hz(1645.0=f −∆ )Hz(8355.59fff 0ss =∆+= Problem 10.16. Consider a single area power system consisting of a single 350 MVA, 60 Hz generator. At a given instant of time the system operates at 60.08 Hz, the total load is 250 MW. 1. Assume the system is uncontrolled. At a certain instant of time, a sudden load change (positive or negative) occurs and the frequency of the system returns to the nominal frequency of 60 Hz. Compute the load change and the unit set point. 2. Assume that the unit is controlled with the following proportional feddback law:

)()/(0.125)( tfHzMWtPc ∆−=∆

At a certain instant of time, a sudden load change (positive or negative) occurs and the frequency of the system returns to the nominal frequency of 60 Hz. Compute the load change and the unit set point. System Data Are: Rated Power 350MVA Rated Frequency f = 60 Hz Inertia Constant H = 2.5 sec Regulation R = 3.0 Hz/350MW Electric Load constant D = 6 MW/Hz.


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