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
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Scheduled
Frequency
Scheduled
Net
Interchange
Computation of Error
sch
P tie = P tie - P tie
Data Acquisition and
Processing Subsystem
sch
P gi = P gi - P gi
f =f
Measurements of
. Generator output Pgi
. Frequency
. Tie Line Flows
. Electric Load
Economic Dispatch
or
Optimal Power Flow
sch
-
f
ACE
Allocation Rules
. Regulating
. Economic
Unit Desired Output
des
P gi
Control
Commands
des
P gi
Figure 10.1 Functional Diagram of the Automatic Generation Control Loop
Page 4
Copyright © A. P. Sakis Meliopoulos – 1990-2006
New York Blackout
Wednesday, July 13, 1977
Manned Downtown Substation
Birmingham, AL
59.96
59.94
5 min
Time
59.92
N.Y. Separated
59.98
60.00
60.02
60.04
N.Y. Lost Generator
at Indian Point
Frequency(Hz)
Attempts to Pick Up
N.Y. Loadon Tie Lines
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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
Oil, Gas
.5
.05
Coal
0
30sec
0
20min
Drum Type Steam-Turbine Leading
.5
.05
0
30sec
0
20min
0
20min
Steam Unit- Boiler Leading
.05
.5
0
30sec
Hydro
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Figure 10.3 Typical Responses of the Three Types of Controllable
Generating Units
LUMPED
STORAGE
SUPER
HEATER
Turbine
Turbine
Generator
BOILER
REHEAT
COOLING
TOWER
Figure 10.4 Schematic Representation of a Drum-Type Steam Turbine
Power Plant
TRANSMISSION
SYSTEM
AND LOAD
G
PRIME
MOVER
TIE
LINE
Pg , f, V
+
E(s)
Governor
f sched
+
+
+
P sched
∆f
PSS
Vref
f
Pg
U.C.E.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
∆ Pint
+
-
L(s)
Interchange
Error
BIAS
Bf
+
+
K(s)
ACE
Pdes
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Steam
Servomotor
f
To Turbine
Pc
A
c
B
High
Pressure
Oil
Pilot
Valve
Main
Piston
Pilot
Valve
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).
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
•
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 ∆f and ∆Pc . Thus, the position of the pilot valve is determined by
the quantities ∆f and ∆Pc . With this observation and utilizing the geometry illustrated in
Figure 10.6, the mathematical model of the governor/hydraulic actuator can be
constructed.
Assume a specific operating condition ( f o , Pg o , etc.). Assume that at this operating
condition the position of point A is XA. In case of a frequency change,∆f or a set point
change, ∆Pc 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:
∆XA = a 1∆f − a 2 ∆PC
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:
'
∆XB = a 3 ∆XA + a 4 ∆XC
∆X'C = a 5 ∫ ( −∆XB )d t
∆X'C denotes the deviation of the point C position (valve position) and ∆XB 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
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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 ∆XB from above equations:
∆X C' = a5 ∫ ( −a1a3∆f + a2 a3∆PC − a4 ∆X C' )dt
Differentiation of the last equation yields:
d∆X C'
aa
aa
= a4 a5 ( − 1 3 ∆f + 2 3 ∆PC − ∆X C' )
dt
a4
a4
Define
TG =
1
a 4a 5
GG =
a 2a 3
a4
R=
a2
a1
Then
d∆X C'
1
G
1
= − ∆X C' + G ( − ∆f + ∆PC )
dt
TG
TG
R
Without loss of generality, we can assume that GG equals 1.0. This is equivalent of
scaling the parameters R and ∆PC . Then the equation reads:
d∆X C'
1
1
1
= − ∆X C' −
∆f +
∆PC
dt
TG
TG R
TG
Taking the Laplace transform of above equation and solving for ∆X'C (s)
∆X'C ( s ) = H G ( s )( −
where H G =
1
∆f ( s ) + ∆PC ( s ))
R
(10.1)
1
is the transfer function of the speed governor system.
1 + sTG
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
TG
R
time constant of the speed governing system
speed regulation due to governor action
∆ Pc
+
-
1/sTg
∆ Xc
+
1/R
-1
∆f
(a)
∆ Pc
+
1/sTg
+
-
-1
1/R
∆
∆ Xc
f
(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%.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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 ∆X'C . In the
simplest form, the turbine can be assumed to be a first order dynamic system expressed
with the following simple equation:
1
G
d∆PG
= − ∆PG + T ∆X C'
TT
dt
TT
where
G T is the gain of the system in MW per valve displacement
TT is the time constant of the turbine system.
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 ∆XC ' , or equivalently, we can introduce the variable ∆XC = G T ∆XC ' . Note
that the units of the variable ∆XC is power. The equation then becomes.
d∆PG
1
1
= − ∆PG + ∆X C'
dt
TT
TT
For reheat units with multiple turbines, the differential equations describing the response
of the turbines are more complex.
∆Pv
Steam
Chest
∆PT
Turbine
To Condenser
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
(a)
Reheater
∆Pv
Steam
Chest
HP
Turbine
∆PT
LP
Turbine
To Condenser
(b)
Penstock
'Head'
∆PT
∆Pv
(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
J
d 2θ m (t )
= Tm − Te
dt 2
where
θ m(t): the position of the generator rotor
Tm : mechanical torque applied by the turbine
Te : electromagnetic torque developed by the generator.
Let
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θ m (t ) = ω o t + δ (t )
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
ω m (t ) =
where
dθ m (t )
dδ (t )
= ωo +
dt
dt
ω o : the synchronous angular speed (mechanical)
ω m (t): the actual angular speed of the generator rotor.
Then
J
dω m (t )
= Tm − Te
dt
(10.2)
Consider the per unit inertia constant, H, of a generating unit.
1 2
Jω o
2
H=
SB
Then:
E=
1 2
Jω 0 = HS B
2
Multiplication of Equation (10.2) by ω o and substitution of J in terms of E yields:
2 E dω m
= ω oTm − ω oTe = PG − Pe
ω 0 dt
(10.3)
Note that ω m = 2πf and ω0 = 2πf o. Linearization of above equation around an operating
point, and substituting ω m and ω o yields:
2 E d∆f
= ∆PG − ∆Pe
f o dt
(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:
∆Pe = D∆f + g (t )
D is the sensitivity of the electric load to frequency variations. The load model will be
examined in the next paragraph.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
A signal flow model for the generator can be derived by taking the Laplace transform of
equation (10.4):
Tp s∆F ( s ) = ∆PG ( s ) − ∆Pe ( s )
(10.5)
where:
2 E0
fs
Figure 10.9 illustrates the signal flow diagram for the generator model.
Tp =
∆ Pg(s)
+
1/sTp
∆ f(s)
∆ 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:
2
PD =
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3V ph R
R 2 + ( 2πfL) 2
(10.6)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
(b)
(a)
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 Voph
yields:
PD ≅ PD ( f o ) +
∂PD
∂P
( f − f o ) + D (V ph − V pho )
∂f
∂V ph
where
∂PD
2P ( f )
X L2
=D=− D o
2
∂f
fo
(R2 + X L )
6V R
∂PD
= 2 ph 2
∂V ph ( R + X L )
X L = 2πf o L
Constant Resistance/Capacitance Load: The electric power absorbed by such an
electric load (Figure 10.10b) is:
2
PD =
3V ph R
1 2
R2 + (
)
2πfC
(10.7)
Linearization of equation (10.7) around the operating frequency fo and voltage Voph
yields:
PD ≅ PD ( f o ) +
Copyright © A. P. Sakis Meliopoulos – 1990-2006
∂PD
∂P
( f − f o ) + D (V ph − V pho )
∂f
∂V ph
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
where
∂PD
2 PD ( f o )
X C2
=D=
2
∂f
fo
(R2 + X C )
6V R
∂PD
= 2 ph 2
∂V ph ( R + X C )
XC =
1
2πf oC
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
1− s
is the power which is converted into
10.11b. The power absorbed at the resistor r2
s
mechanical power. s is the slip defined with ω m = (1 − s )ω e (for a two pole machine). For
simplicity of analysis, the shunt term is neglected. The electric power absorbed is:
Pe =
3V ph2 ( r1 +
r2
)
s
( x1 + x2 ) 2 + ( r1 +
r2 2
)
s
The mechanical power output is
2
V ph
1− s
)(
)
Pm = 3r2 (
r2 2
s
2
( x1 + x2 ) + ( r1 + )
s
Let T = f ( ω m ) 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:
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Torque
Torque-Speed Characteristic
Mechamical Load Torque
Speed
Network
Frequency
(a)
r1
x1
x2
r2
r2
bm
gm
1-s
s
(b)
Figure 10.11 Induction Motor Models, (a) Torque-Speed Characteristic,
(b) Equivalent Circuit
2
(1 − s )ω e f (ω m ) = 3r2 (
V ph
1− s
)(
)
r2 2
s
2
( x1 + x2 ) + ( r1 + )
s
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 ω e to
ω 'e . 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
Pm − Pm'
∆P ( f ) =
2π∆f
ω e − ω e'
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Pm − Pm'
2 π is the sensitivity of the load to the frequency and will be
ω e − ω e'
designated with D. Thus
The quantity
∆P ( f ) = D∆f
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
Pe =
3V ph2
r2
s=
3V ph2 ω e − ω m
⋅
r2
ωe
(10.7)
The mechanical power output is
3V ph2
3V ph2 ω m
(1 − s ) s =
(ω e − ω m )
Pm =
r2
r2 ω e2
Let T = f ( ω m ) be the torque speed characteristic of the mechanical load. Then
3V ph2 ω m
ω m f (ω m ) =
(ω e − ω m )
r2 ω e2
or
3V ph2
ω e2
f (ω m ) =
(ω e − ω m )
r2
Solution of above expression provides the slip s. Now assume a drop in the frequency of
the network, ω e → ω e + dω e . Differentiation of above equation yields:
2ω e (ω e − ω m ) − ω 2e
ω 2e
ω 2e
∂f
f
(
)
d
f
(
)
ω
ω
ω
+[
+
] dω m = 0
m
e
m
2
2
(ω e − ω m )
(ω e − ω m )
(ω e − ω m ) ω m
or
dω m =
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( 2ω eω m − ω e2 ) f (ω m )
ω e2 f (ω m ) + ω e2 (ω e − ω m )
∂f
∂ω m
⋅ dω e
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
This relationship suggests that a decrease in electrical frequency causes a decrease in
rotating speed.
In general, the term ω 2e (ω e − ω m )
∂f
is very small compared to the term ω 2e f ( ω m ) and
ωm
thus it can be omitted. Then
dω m =
2ω m − ω e
ωe
⋅ dω e
The incremental change of the electric load is now computed upon differentiation of
Equation (10.7).
3V ph2 ω m
1
dPe =
( 2 dω e − dω m )
r2 ω e
ωe
Upon substitution of d ω m
dPe =
3V ph2 ω e − ω m
dω e Pe
dω e = Pe
= df
2
r2
ωe
fo
ωe
Thus the sensitivity D of induction motor power to frequency changes is approximately
D=
Pe
fo
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
∆Pe = ∆PD = g (t ) + D∆f + DV ∆V
In this chapter we shall be concerned with changes in system frequency only. For this
reason we shall neglect the term D v ∆V . Figure 10.12 illustrates the variation of the
electric load versus frequency of the various electric load types discussed.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
P
fo
f
(a )
P
fo
f
(b )
P
fo
f
(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:
f
f
f
d∆f
= o ∆PG − o D∆f − o g (t )
dt
2E
2E
2E
d∆PG
1
1
=−
∆PG +
∆X C
dt
TT
TT
d∆X C
1
1
1
=−
∆X C +
( − ∆f + ∆PC )
dt
TG
TG
R
Page 22
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
In compact matrix notation:
dx
= Ax + Bu
dt
where
⎡ ∆f ⎤
⎡ g (t ) ⎤
x = ⎢ ∆PG ⎥ , u = ⎢
,
⎥
⎢
∆PC ⎥⎦
⎣
⎣⎢ ∆X C ⎦⎥
⎡ Df
⎢− o
⎢ 2E
A=⎢ 0
⎢
⎢ 1
⎢−
⎢⎣ RTG
fo
2E
1
−
TT
0
⎤
0 ⎥
⎡ fo
⎥
⎢− 2 E
1 ⎥
, B=⎢ 0
⎢
TT ⎥
⎥
⎢ 0
1
⎢⎣
− ⎥
TG ⎦⎥
⎤
0 ⎥
0 ⎥
1 ⎥
⎥
TG ⎥⎦
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.
∆f
Hp(s)
1/R
D
-
∆ PG
+
HG(s)
∆Pc
D∆f
+
HT(s)
-
g(t)
Figure 10.13 Signal Flow Diagram of the Governor, Turbine, Generator and
Load Model
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 23
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Prime
mover
G
g(t)
+
f
Governor
Pd
fsch
Pc
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, ∆PC , and changes to the electric load, g(t). Quantities of
interest are the mechanical power output of the turbine, ∆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:
⎡ Df
⎢− 0
⎡ ∆f ⎤ ⎢ 2 E
d ⎢
∆PG ⎥ = ⎢ 0
⎢
⎥ ⎢
dt
⎣⎢ ∆x C ⎦⎥ ⎢
1
⎢−
⎣⎢ RTG
f0
2E
1
−
TT
0
⎤
0 ⎥
⎡
⎥ ⎡ ∆f ⎤ ⎢ −
1 ⎥⎢
∆PG ⎥ + ⎢
⎢
⎥ ⎢
⎥
TT
∆
x
⎢
⎥
C
⎣
⎦⎥ ⎢
1
⎥
−
⎣⎢
TG ⎦⎥
f 0 g (t ) ⎤
2E ⎥
⎥
0
∆PC ⎥
⎥
TG ⎦⎥
(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
∆f = −
∆PG =
Page 24
g (t )
β
+
∆PC
β
∆P
1 g (t )
+D C
R β
β
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
∆X C =
∆P
1 g (t )
+D C
β
R β
where
β = D+
1
R
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
∆PC = 0, and
g (t ) ≠ 0
The following relationships are valid in this case
∆f = −
g (t )
β
∆PG = −
(10.9)
1
1
∆f =
g (t )
R
Rβ
(10.10)
1
1
∆f =
g (t )
R
Rβ
(10.11)
∆X C = −
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):
∆PC = g (t )
∆PG = ∆PC
∆X C = ∆PC
These cases are illustrated in Figure 10.15.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
∆ Pg
∆ f1
∆ f=0
slope = 1
∆ Pc
(a)
∆ Pg
slope = -1/R
∆f
∆ Pc1
(b)
∆ Pc=0
∆ Pc2
Figure 10.15 Steady State Response of Governor/Turbine System
(a) ∆PG versus ∆PC , parameter ∆f ,
(b) ∆PG versus ∆f , parameter ∆PC
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:
a
∆f = −
1
D+
R
Page 26
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
R = 0.05
D+
60
= 0.00375 Hz / MW
800
1
1 800 MW
= 15 MW / Hz +
(
) = 281 .66 MW / HZ
R
0.05 60 Hz
∆f = −
∆PG =
10
Hz = −0.0355 Hz
281 .66
a
R(D +
1
)
R
=
10 MW
= 9.47
(0.00375)(281.66)
∆PD = D∆f = −0. 53MW
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
∆f =
b
1
D+
R
= 60.0 − 59.9645 = 0.0355 Hz
Upon solution we obtain: b = 10 MW. Note that:
∆PG =
D
1
D+
R
b = 0.53 MW
∆PD = D∆f = 0.53
MW
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
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 27
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
d∆f
f
f
f
= − o D∆f − o g (t ) + o ∆PG
dt
2E
2E
2E
0 = − ∆PG + ∆X C
0 = − ∆X C −
1
∆f + ∆PC
R
Elimination of the variables ∆XC and ∆PG from above equations yields:
d∆f
f
1
f
f
= − o ( D + ) ∆f − o g (t ) + o ∆PC
dt
2E
R
2E
2E
The response of the system to a step change of the electric load g(t) = a u-1(t) is
∆f s ( t ) = −
a
β
(1 − e −αt )
where
β =D+
α=
Page 28
1
R
is the area frequency response
fo β
2E
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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
α
d∆ f s ( t )
= − ∫ g (t − τ ) e −ατ dτ
β
dt
0
t
∆f ( t ) = g ( t ) *
f
-
a
Time
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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 29
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Solution: The model of the system upon substitution of numerical values is:
d∆f
= −0.107143∆f + 0.0107143∆PG − 0.0107143g (t )
dt
d∆PG
= −1.666∆PG + 1.666∆X C
dt
d∆X C
= −1777.77∆f − 6.666∆X C + 6.666∆PC
dt
(a) Assuming TG → 0 and TT → 0 the last two equations yield:
0 = −∆PG + ∆X C
0 = −266.67∆f − ∆X C + ∆PC
Upon solution of these equations:
∆X G = −266.67∆f + ∆PC
∆X C = −266.67∆f + ∆PC
Upon substitution in the first equation
d∆f
= −2.9643∆f + 0.0107143( ∆PC − g (t ))
dt
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.
Page 30
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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
0.000
System_Frequency (Hz)
-8.267 m
-16.53 m
-24.80 m
-33.07 m
-41.33 m
-49.60 m
-57.87 m
-66.13 m
13.84
Unit_Mechanical_Power (MW)
12.11
10.38
8.652
6.921
5.191
3.461
1.730
0.000 p
00:44:38
00:44:39
00:44:40
00:44:41
00:44:42
00:44:43
00:44:44
00:44:45
00:44:46
00:44:47
00:44:48
00:44:49
00:44:50
00:44:51
00:44:52
Tuesday, November 27
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:
d∆f
= −0.107143∆f + 0.0107143∆PG − 0.0107143g (t )
dt
d∆PG
= −1.666∆PG + 1.666∆X C
dt
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 31
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
d∆X C
= h( −1777.77∆f − 6.666∆X C + 6.666∆PC )
dt
where
x < 0.833
⎧ x,
⎪
h( x ) = ⎨ 0.833,
x > 0.833
⎪− 0.833, x < −0.833
⎩
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:
⎡ − f o g 1 ( t ) / 2 E1 ⎤
dx1 (t )
⎥
0
= A1 x1 (t ) + ⎢
⎥
⎢
dt
⎢⎣ ∆PC1 / ∆TG1 ⎥⎦
⎡ − f o g 2 (t ) / 2 E 2 ⎤
dx 2 (t )
⎥
0
= A2 x 2 (t ) + ⎢
⎥
⎢
dt
⎣⎢ ∆PC 2 / ∆TG 2 ⎦⎥
where
⎡ ∆f 1 ( t ) ⎤
x1 (t ) = ⎢ ∆PG1 ⎥
⎥
⎢
⎣⎢ ∆X C1 ⎦⎥
⎡ ∆f 2 (t )⎤
x 2 (t ) = ⎢ ∆PG 2 ⎥
⎥
⎢
⎢⎣ ∆X C 2 ⎥⎦
Page 32
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
∆f 1 = ∆f 2 = ∆f = −
g 1 (t )
β1
+
∆PC1
β1
=−
g 2 (t )
β2
+
∆PC 2
β2
g (t ) = g1 (t ) + g 2 (t ) = −( β 1 + β 2 )∆f + ∆Pc1 + ∆Pc 2
1 g 1 (t )
1
= − ∆f
R1 β 1
R1
1
∆X C 1 = − ∆f
R1
1
∆PG 2 = −
∆f
R2
1
∆X C 2 = −
∆f
R2
∆PG1 =
The above results are illustrated in Figure 10.17.
∆ Pg
slope=-1/R1
slope=-1/R2
∆ Pg1
∆ Pg2
f
f
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
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
Ri = Ru
f0
Si
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:
S1
∆f
Ru f 0
S
= − 2 ∆f
Ru f 0
∆PG1 = −
∆PG 2
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.
Page 34
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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
d∆f
= −a∆f − Cg (t ) + b∆PC
dt
where:
fo
β
2H
C = fo/2E
b = fo/2E
a=
Assume a feedback control law for the minimization of the frequency deviation of the
form (proportional feedback control):
∆PC = −k∆f (t )
In this case the system response will obey the equation
d∆f
= −(a + bk ) ∆f − Cg (t )
dt
(10.12)
For any change g(t) in the electric load, the steady state frequency of the system will have
an error equal to
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
∆f ss = −
Cg
a + bk
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:
t
∆PC = − k ∫ ∆f (τ )dτ
−∞
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
∫
∆f (τ )dτ =
−∞
t
t
∫ f (τ )dτ − ∫ f
−∞
sch
dτ
−∞
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
x (t ) =
∫ ∆f (τ )dτ
−∞
Upon differentiation of above equation and substitution in equation (10.2), the model
becomes:
dx(t )
= ∆f (t )
dt
d∆f
= −a∆f − bkx(t ) − Cg (t )
dt
The steady state response is obtained by substituting zero for the derivatives yielding
0 = ∆f ( t )
0 = − a ∆f − bkx ( t ) − Cg ( t )
Page 36
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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
t
∆PC = − k
∫
∆f (τ )dτ
−∞
Upon differentiation
d∆PC
= −k∆f
dt
Incorporating above equation into the rest of the model, one obtains
⎡ ∆f ⎤ ⎡ − Df o / 2 E
⎢
⎥ ⎢
0
d ⎢ ∆PG ⎥ ⎢
=
dt ⎢ ∆X C ⎥ ⎢ − 1 / RTG
⎢
⎥ ⎢
⎣ ∆PC ⎦ ⎣ − K
f o / 2E
0
− 1 / TT
0
1 / TT
− 1 / TG
0
0
0 ⎤ ⎡ ∆f ⎤ ⎡ − f o
0 ⎥ ⎢ ∆PG ⎥ ⎢
⎥+⎢
⎥⎢
1 / TG ⎥ ⎢ ∆X C ⎥ ⎢
⎥ ⎢
⎥⎢
0 ⎦ ⎣ ∆PC ⎦ ⎣
/ 2 E g(t)⎤
⎥
0
⎥
0
⎥
⎥
0
⎦
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.
Steam
Line
G
g(t)
f
Pd
+
Governor
-K
Copyright © A. P. Sakis Meliopoulos – 1990-2006
fsch
Page 37
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
fo
f
f
∆PG − o D∆f − o a = 0
2E
2E
2E
−
1
1
∆PG +
∆X C = 0
TT
TT
−
1
1
1
∆X C +
( − ∆f + ∆PC ) = 0
TG
TG
R
− k∆f = 0
Solution of these equations yields:
∆f = 0
∆PG = a
∆X C = a
∆PC = a
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
Page 38
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
t
(a)
∆PC = −0.1 ∫ β ∆f (τ )dτ
−∞
t
(b)
∆PC = −0.2 ∫ β ∆f (τ )dτ
−∞
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.
d∆f
= −0.107143∆f + 0.0107143∆PG − 0.0107143g (t )
dt
d∆PG
= −1.666∆PG + 1.666∆X C
dt
d∆X C
= −1777.77∆f − 6.666∆X C + 6.666∆PC
dt
d∆PC
= −27.6666∆f
dt
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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 39
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Program XfmCtr - Page 1 of 1
c:\books\md_psa\examples\psa-ch10-ex10-3a - Nov 27, 2001, 00:52:57.000000 - 10000.0 samples/sec - 800000 Samples
-3.806 u
System_Frequency__Hz_ (V)
-13.18
-26.36
-39.54
-52.72
-65.90
14.66 k
Unit_Mechanical_Power__MW_ (V)
11.73 k
8.796 k
5.864 k
2.932 k
4.297 u
9.849 k
Unit_Set_Point__MW_ (V)
7.879 k
5.909 k
3.940 k
1.970 k
-8.188 u
00:53:00
00:53:10
00:53:20
00:53:30
00:53:40
00:53:50
00:54:00
Tuesday, November 27
00:54:10
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.
d∆f
= −0.107143∆f + 0.0107143∆PG − 0.0107143g (t )
dt
d∆PG
= −1.666∆PG + 1.666∆X C
dt
Page 40
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
d∆X C
= −1777.77∆f − 6.666∆X C + 6.666∆PC
dt
d∆PC
= −55.3333∆f
dt
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.
Program XfmCtr - Page 1 of 1
c:\books\md_psa\examples\psa-ch10-ex10-3b - Nov 27, 2001, 01:01:35.000000 - 10000.0 samples/sec - 800000 Samples
6.802 u
System_Frequency__Hz_ (V)
-13.03
-26.05
-39.08
-52.10
-65.13
15.44 k
Unit_Mechanical_Power__MW_ (V)
12.35 k
9.265 k
6.177 k
3.088 k
-2.579 u
9.999 k
Unit_Set_Point__MW_ (V)
7.999 k
5.999 k
3.999 k
2.000 k
11.89 u
01:01:40
01:01:50
01:02:00
01:02:10
01:02:20
01:02:30
01:02:40
Tuesday, November 27
01:02:50
Figure E10.3b. Transient Response of the System for k=0.2
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 41
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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.
2.
3.
4.
5.
Maintain constant frequency close to scheduled value
Maintain accurate real time
Maintain net interchange power equal to scheduled values
Minimize equipment wear
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:
Pei = PDi + ∑ Pij
j
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
∆PDi = g i (t ) + Di ∆f i
The power flow in the transmission line i,j, Pij can be represented with the following
model:
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Pij =
ViV j
x ij
sin(δ i − δ j )
(10.13)
Linearization of Equation (10.13) around an operating point yields:
∆Pij =
ViV j cos(δ i − δ j )
x ij
( ∆δ i − ∆δ j ) = C ij ( ∆δ i − ∆δ j )
where Cij is given by:
C ij =
ViV j
x ij
Upon substitution, one obtains
d∆f i
f
f
f
f
= o ∆PGi − o Di ∆f i − o g i (t ) − o
dt
2Ei
2Ei
2Ei
2Ei
∑C
ij
( ∆δ i − ∆δ j )
j
In addition, the phase angle δ i of the voltage Vi is dependent on the frequency of the
system. Specifically
θ i (t ) = 2πf i t = ω s t + δ i
Upon differentiation
2πf i = ω s +
dδ i
dδ
= 2πf o + i
dt
dt
Upon rearrangement and substituting ∆fi = fi − f o
d∆δ i
= 2π∆f i
dt
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Vi e
jδi
V 3e
Ptie i, v
Area 1
Area 3
Tie Lines
V 2e
Area 4
jδ 3
jδ 2
Area 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:
d∆f i
f
f
f
f
= o ∆PGi − o Di ∆f i − o g i (t ) − o
dt
2Ei
2Ei
2Ei
2Ei
∑C
ij
( ∆δ i − ∆δ j )
j
d∆δ i
= 2π∆f i
dt
d∆PGi
1
1
=−
∆PGi +
∆X Ci
dt
TTi
TTi
d∆X Ci
1
1
1
=−
∆X Ci +
( − ∆f i + ∆PCi )
dt
TGi
TGi
Ri
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.
Page 44
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
dx
= Ax + Bu + Cg (t )
dt
where
⎡ ∆PC1 ⎤
⎢ ∆P ⎥
u = ⎢ C2 ⎥
⎢ ... ⎥
⎥
⎢
⎣ ∆PCm ⎦
⎡ g 1 (t ) ⎤
⎢ g (t ) ⎥
g (t ) = ⎢ 2 ⎥
⎢ ... ⎥
⎥
⎢
⎣ g m (t )⎦
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:
f
f
f
f
d∆f 1
1
= − o ( D1 + ) ∆f 1 − o g 1 (t ) + o ∆PC1 − o C12 ∆δ
dt
2 E1
R1
2 E1
2 E1
2 E1
f
f
f
f
d∆f 2
1
= − o ( D2 +
) ∆f 2 − o g 2 (t ) + o ∆PC 2 − o C 21 ∆δ
dt
2E2
R2
2E 2
2E 2
2E2
The feedback control law for this system is given by the equations:
∆PC1 = K 01C12 ∆δ + K1 ∆f 1
∆PC 2 = K 02 C 21 ∆δ + K 2 ∆f 2
Upon substitution of above expressions into the equations:
f
f
d∆f 1
= − o ( β 1 − K1 )∆f 1 + o (1 − K 01 )C12 ∆δ
dt
2 E1
2 E1
f
f
f
d∆f 2
= − o ( β 2 − K 2 )∆f 2 + o (1 − K 02 )C 21 ∆δ − o g (t )
dt
2E2
2E2
2E2
The modeling procedure is illustrated with an example of a two area system.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
Rating`
Per Unit Inertia Constant (H)
Unit Output
Turbine Time Constant
Governor Time Constant
Regulation
Load Frequency Characteristic
Unit 1
250MVA
2.5 sec
180 MW
0.2 sec
0.25 sec
0.05 pu
5 MW/Hz
Unit 2
300MVA
2.8 sec
220 MW
0.1 sec
0.2 sec
0.05 pu
6 MW/Hz
G1
Area 1
G2
Area 2
Figure E10.4. A Simplified Two Area System
Solution: The equations for area 1 are:
f
f
f
f
dx1
= − o D1 x1 + o x 2 − o C12 ( x 4' − x 8' ) − o g 1 (t )
dt
2 E1
2 E1
2 E1
2 E1
dx 2
1
1
=−
x2 +
x3
dt
TT 1
TT 1
dx 3
1
1
1
=−
x1 −
x3 +
u1
dt
R1TG1
TG1
TG1
dx 4'
= 2πx1
dt
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
The equations for area 2 are:
dx 5
f
f
f
f
= − o D2 x 5 + o x 6 − o C12 ( x 8' − x 4' ) − o g 2 (t )
dt
2E2
2E2
2E 2
2E 2
dx 6
1
1
=−
x6 +
x7
dt
TT 2
TT 2
dx 7
1
1
1
=−
x5 −
x7 +
u2
dt
R2 TG 2
TG 2
TG 2
dx 8'
= 2πx 5
dt
where
x 1 = ∆f 1
x 2 = ∆PG1
x 3 = ∆X E 1
x 4' = ∆δ 1
x 5 = ∆f 2
x 6 = ∆PG 2
x 7 = ∆X E 2
x 8' = ∆δ 2
u1 = ∆PC1
u 2 = ∆PC 2
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
d ( x 4' − x8' )
= 2π ( x1 − x 5 )
dt
or
dx 4
= 2π ( x1 − x 5 )
dt
Equations 4 and 8 must be substituted with above equation.
equations represent the model of the two area system.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
The resulting seven
Page 47
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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.
V1V 2 sin(δ 1 − δ 2 )
= 30 MW
x12
Above equation yields the solution
δ 1 − δ 2 = 5.85890 or 0.10225 rads.
The constant C12 is:
C12 =
V1V 2 cos(δ 1 − δ 2 )
= 292.35 MW
x12
or
C12, pu = 2.9235
Upon substitution of numerical values, the seven equations read:
dx1
= −0.24 x1 + 0.048 x 2 − 14.032 x 4 − 0.048g 1 (t )
dt
dx 2
= −5 x 2 + 5 x 3
dt
dx 3
= −333.33x1 − 4 x 3 + 4u1
dt
dx 4
= 2π ( x1 − x 5 )
dt
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
dx 5
= −0.21426 x 5 + 0.035713x 6 + 10.44 x 4 − 0.035713g 2 (t )
dt
dx 6
= −10 x 6 + 10 x 7
dt
dx 7
= −500 x 5 − 5 x 7 + 5u 2
dt
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:
t
∆PCi = − k i ∫ ( ∑ ∆P ij (τ ) + Bi ∆f i (τ ))dτ
−∞
(10.14)
j
where ki is the control loop gain and Bi is the so-called frequency bias, ∆Pij (t ) is the
deviation of the net power interchange from the scheduled value and ∆f i (t ) 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 ki 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.
ACE i (t ) = ∑ ∆Pij (t ) + Bi ∆f i (t )
j
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Page 49
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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.:
∑ ∆P
+ Bi ∆f i = 0, i = 1, 2, 3 and 4
ij
j
Summation of above four equations yields
∑ ∆P
1, j
j = 2 , 3, 4
+
∑ ∆P
j =1, 3, 4
2, j
+
∑ ∆P
+
∑ ∆P
+
j =1, 2 , 4
3, j
∑ ∆P
+ B1 ∆f 1 + B 2 ∆f 2 + B3 ∆f 3 + B 4 ∆f 4 = 0
∑ ∆P
=0
j =1, 2 , 3
4, j
Note that (by definition)
∑ ∆P
1, j
j = 2 , 3, 4
+
∑ ∆P
j =1, 3, 4
2, j
+
j =1, 2 , 4
3, j
j =1, 2 , 3
4, j
and that at steady state the frequency of any system will be equal to the frequency of any
other system:
∆f 1 = ∆f 2 = ∆f 3 = ∆f 4 = ∆f
thus
∆f = 0
and
∑ ∆P
ij
=0
for every system i
j
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
Page 50
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
a
∆f 1 = ∆f 2 = −
β1 + β 2
1
a
∆PG1 =
R1 β 1 + β 2
1
a
∆PG 2 =
R2 β 1 + β 2
a
∆PD1 = − D1
β1 + β 2
a
∆PD 2 = − D2
β1 + β 2
∆P12 = −
β2
β1 + β 2
a
1
R1
1
β 2 = D2 +
R2
β 1 = D1 +
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, ∆PC1 = b ,
∆PC 2 = 0 , is computed to be:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
b
β1 + β 2
∆f 1 = ∆f 2 =
D1 + β 2
b
β1 + β 2
1
=−
b
R2 ( β 1 + β 2 )
∆PG1 =
∆PG 2
∆P12 =
β2
β1 + β 2
b
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:
t
∆PC1 = − k 1 ∫ ( ∆P12 (τ ) + B1 ∆f 1 (τ ))dτ
−∞
t
∆PC 2 = − k 2 ∫ ( ∆P21 (τ ) + B 2 ∆f 2 (τ ))dτ
−∞
In this case the steady state response to a step load change in area 1,
g 1 (t ) = a ,
g 2 (t ) = 0
is
∆f 1 = ∆ f 1 = 0
∆PG1 = a
∆PG 2 = 0
∆P12 = 0
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.
Page 52
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
(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
∆f = −
0.5
1 1
+
0.05 + 0.06 +
1.2 1
∆PG1 = 0.2144
= −0.257289
∆PG 2 = 0.2573
∆PD1 = −0.0128
∆PD 2 = −0.015437
∆Ptie1, 2 = −0.2727
An accounting of the power is illustrated in Figure E10.5.
∆PG1 = +0.2144
∆PG2 = +0.2573
∆Ptie = -0.2727
g(t) = +0.5
∆Pd1 = -0.0128
a)
∆Pd2 = -0.0154
Figure E10.5 Accounting of Power Change
(b) Regulated system.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
∆f = 0
∆PG1 = 0.5
∆PG 2 = 0.0
∆PD1 = 0
∆PD 2 = 0
∆Ptie1, 2 = 0
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
dx(t )
= Ax(t ) + bu(t ) + cg (t )
dt
where u(t) = ∆PC (t), the unit set point and g(t) is the electric load disturbance.
(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:
∆PC (t ) = −∆Ptie (t ) − β∆f (t )
(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
∆PC (t ) = −0.1 ∫ ( ∆Ptie (τ ) + β∆f (τ ))dτ
−∞
Page 54
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
G
Tie 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:
f
f
f
f
d∆f
= − o D∆f + o ∆PG − o C1∞ ∆δ − o g (t )
dt
2E
2E
2E
2E
d∆PG
1
1
=−
∆PG +
∆X C
dt
TT
TT
d∆X C
1
1
1
=−
∆X C +
( − ∆f + ∆PC )
dt
TG
TG
R
d∆δ
= 2π∆f
dt
The constant C 1∞ is computed from the given data as follows:
P1∞ =
115 2
sin δ = −50 MW
30
Solution of above equations for δ yields: δ = -6.51o or –0.1137 rads. Thus
C1∞ =
115 2
cos δ = 437.98 MW
30
Next the parameters of this problem are converted into a consisted set of units:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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E = HS B = 1050 MW . sec
R = 0.0086 Hz / MW
D = 5 MW / Hz
TT = 0.55 sec onds
TG = 0.15 sec onds
β = 121.279 MW / Hz
Upon substitution of all numerical values:
− 12.514⎤
0
⎡ − 0.1428 0.0286
⎡- 0.0286⎤
⎡ 0 ⎤
⎢
⎥
⎢
⎥
⎢ 0 ⎥
− 1.818 1.818
0
0
0
dx (t ) ⎢
⎥ x(t) + ⎢
⎥ g (t ) + ⎢
⎥ ∆PC (t )
=
− 6.666
0
0 ⎥
⎢ − 775.19
⎢ 0 ⎥
⎢6.666⎥
dt
⎢
⎥
⎢
⎥
⎢
⎥
0
0
0 ⎦
⎣ 6.2832
⎣ 0 ⎦
⎣ 0 ⎦
where
x(t ) = [∆f (t ) ∆PG (t ) ∆X C (t ) ∆δ (t )]
T
(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):
∆f ss = 0
∆PGss = 0
∆δ = −0.02283
∆Ptie = −0.02857 pu or - 10 MW
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):
∆f ss = 0
∆δ = −0.011415
∆PG = 5.0 MW
∆X C = 5.0 MW
∆Ptie = −5.0 MW
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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
x5 (t ) = ∆PC (t )
Upon differentiation of the feedback law equation
dx 5 (t )
= −43.798∆δ (t ) − 12.1279∆f (t )
dt
Above equation is added to the previous system model. The resulting equations are:
− 12.514
0
0 ⎤
⎡− 0.0286⎤
⎡ − 0.1428 0.0286
⎥
⎢
⎥
⎢
0
0
0
0
− 1.818 1.818
⎥
⎢
⎥
⎢
dx (t )
= ⎢ − 775.19
0
0
0
6.666⎥ x(t ) + ⎢
− 6.666
⎥ g (t )
dt
⎥
⎢
⎥
⎢
0
0
0
0
0 ⎥
⎥
⎢
⎢ 6.2832
⎥⎦
⎢⎣
⎢⎣− 12.1279
0
0
0
0 ⎥⎦
− 43.798
Upon solution of above equations under steady state conditions, i.e. all derivatives equal
zero:
∆PG = 10 MW
∆X C = 10 MW
∆PC = 10 MW
∆δ = 0
∆Ptie = 0
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
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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(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:
∞
1
J = ∫ ( x T (τ )Qx (τ ) + u T (τ ) Ru (τ ))dτ
20
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
Objective Function
•
•
B=Constant
Integral Gain , k
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
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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
J=
Subject to
1
( x T (τ )Qx (τ ) + u T (τ ) Ru (τ ))dτ
2 ∫0
dx(t )
= Ax(t ) + Bu(t )
dt
x(0) = x 0
The solution of above optimization problem is obtained with the solution of the
appropriate Riccatti Equation [???]. Specifically, the solution is given with:
u (t ) = − kx (t )
where:
k = R −1 B T P
and P is determined from the solution of the Riccatti Equation
PA + AT P − PBR −1 B T P = −Q
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.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System
Pdes1 + UCE1
L1(S)
Pdes2 + UCE2
Pdes3 +
-
UCE3
Tie Line
Gov
Mover 1
G1
PG1
G2
PG2
Gov
Mover 2
L2(s)
L3(s)
Gov
Mover 3
Tie Line
G3
PG3
ESTIMATION
f
Tie Flows
ACE
Calculation
Allocation
Rules
ACE
K(s)
Every 1 to 6 sec.
ACE =
P tie + B
f
Security,
Economic Dispatch,
Etc.
Every
Several
Minutes
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.
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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.
o
Specifically, let Pload
be the total electric load (including net interchange) at which the
economic dispatch has been performed. Let Pbio , i = 1, 2, ..., n , be the economic dispatch
solution. Obviously,
∑P
o
bi
o
o
= Pload
+ q( Pload
, Pbio )
i
The unit i output Pbio 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:
o
Pbio = f i ( Pload
)
The economic base points at other load levels can be approximately computed with a
linearized model
Pbi = Pbio +
∂Pbi
∆Pload
∂Pload
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:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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(a) The previous economic dispatch solution:
o
Pbio , i = 1,2,...., n at load level Pload
(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
Pbi = Pbio + a i ACE
(2) desired output of the unit
t
Pdesi = Pbio + a i ACE + bi
∫ ACE (τ )dτ
−∞
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 :
∆PCi = Pdesi − Pgi
The allocation rules are illustrated in Figure 10.22
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ACE
Regulating
Allocation
Pdes1
bi
i
Pgi
Economic
Allocation
Pdes2
ai
Linearized
Economic Dispatch
Model
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.
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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:
•
•
•
Page 64
Allow the system frequency to drop.
Allow scheduled interchange to lag.
Increase the number of participating and regulating units.
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
525, 600
(∆f i )( ACE i )
1
CPS1 =
∑
525,600 i =1
− 10β
where: β is the are frequency bias
∆f i is the one minute (minute i) average frequency deviation
ACEi is the one minute (minute i) average ACE of the control area
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.
CPS1 ≤ ε12 ,
where:
ε1 =
525, 600
1
∑ ∆f i
525,600 i =1
Control Performance Standard 2 (CPS2): This standard is defined as follows:
CPS 2 =
1 6
∑ ACE j ,10
6 j =1
where: ACEi ,10 is the ten minute average ACE of the control area
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.
CPS 2 ≤ L10 = 1.65ε 10 ( −10β i )( −10β s ) ,
where:
ε 10 =
1 52,560
∑ ∆f i
52,560 i =1
βi is the area frequency bias
β s is the system frequency bias
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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)
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
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:
(sinθ)2 = 1 − (0.9)2 = 0.19
The frequency characteristic D of the electric load is:
D=
− (2)(552)(0.08)(0.19) + (2)(552)(0.03)(0.19) + (552)(0.89)
= 8.0132 ( MW/Hz)
60
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:
R1 = 2.0 pu ,
ωo L1 = 0.2 pu ,
R 2 = 5.0 pu
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. ∆PC = 0.0 , compute the steady state frequency of the system.
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
S2
R2
R2
R2.
S1
R1
G
L1
L1
L1
R1
R1
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.
S2
R2
R2
R2.
S1
G
Induction
motors
Figure P10.4
Problem 10.5 The electric power system of the Figure P10.5 has the following
parameters:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
S b = 500 MVA,
f b = 60 Hz, H = 2.5 sec
TT = 0.5 sec, TG = 0.25 sec, R = 0.05 pu
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. ∆PC (t ) = 0.0 . 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:
∆PC (t ) = −50.0( MW / Hz) ∆f (t ) . Compute the steady state frequency of the system.
(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
Prime
Mover
Load
Consisting of
Induction
motors
G
S
Governor
Resistive
Load
20 MW
+
∆Pc
∆f
fsch
Figure P10.5
Solution:
(a) D =
310
= 5.167 MW / Hz
60
(b) Uncontrolled case
∆f = −
g (t )
β
, β =D+
1
500
= 5.167 +
= 171.8333 MW / Hz
(0.05)(60)
R
20
= −0.1164 Hz
171.8333
f = 59.8836 Hz
∆f = −
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g (t )
= 19.3986 MW
Rβ
∆PD = D∆f = −0.6014 MW
∆PG =
(c) Controlled case:
∆f = −
g (t )
β
+
∆PC
β
, ∆PC = −50∆f
∆f = −0.09016 Hz
f = 59.90984 Hz
∆P
g (t )
∆PG =
+ D C = 19.5341 MW
Rβ
β
∆PD = D∆f = −0.4659 MW
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, PC.
(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).
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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(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:
Rated Power
Rated Frequency
Droop Characteristic
Unit 1
500 MVA
60 Hz
0.05 pu
Unit 2
750 MVA
60 Hz
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
R = 0.05 pu
D = 15 MW/Hz
TT = 0.6 sec
TG = 0.6 sec
generator per unit inertia constant
regulation parameters ("droop" characteristic)
electric load frequency characteristic
steam turbine time constant
governor time constant.
Develop the dynamical equation of the system in the form
dx (t )
= Ax(t ) + bu(t ) + cg (t )
dt
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:
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
t
u (t ) = −0.1 ∫ β ∆f (τ )dτ
−∞
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:
∞
J =
1
(0.2∆f 2 (τ ) + ∆PC2 (τ )) dτ
2 ∫0
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.
G
115kV
Infinite Bus
Figure P10.11
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
(a) Develop the dynamic equations of the system in the form:
x& (t ) = Ax (t ) + bu (t ) + cg (t )
where u(t) = ∆PC unit set point and g(t) = electric load disturbance.
(b) Compute the steady state frequency and generator power for a 10 MW change of the
electric load assuming ∆PC = 0.
(c) Compute the steady state frequency and generator power for a 10 MW change of the
electric load assuming the following feedback control law:
∆PC = −∆Ptie (t ) − 350( MW / Hz)∆f (t )
(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
∆PC (t ) = −0.1 ∫ ( ∆Ptie (τ ) + 350( MW / Hz ) ∆f (τ )) dτ
−∞
(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:
Rated Power
Per Unit Inertia Constant
Droop Characteristic, R
Turbine Time Constant
Governor Time Constant
Load Frequency Constant
Area Load
Area Generation
Unit 1
Unit 2
250 MVA
3.0 sec
0.05 pu
0.0
0.0
5 MW/Hz
100 MW
200 MW
400 MVA
3.5 sec
0.05 pu
0.0
0.0
10 MW/Hz
350 MW
250 MW
(a) Develop the dynamic equations of the system in the form
dx( t )
= Ax( t ) + Bu + Cg( t )
dt
⎡ ∆PC1 ⎤
where: u = ⎢
⎥
⎣∆PC2 ⎦
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⎡ ∆f1 ⎤
x(t) = ⎢∆f2 ⎥
⎢ ⎥
⎢⎣ ∆δ ⎥⎦
⎡ g1 (t ) ⎤
g(t) = ⎢
⎥
⎣g 2 (t )⎦
Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
Provide the numerical values of the matrices A, B, and C (Use a power base of 100
MVA).
(b) Assume that u1(t) = ∆ PC1(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 z = j 65.2 ohms . Additional data for this system
are:
Rated Power
Per unit inertia constant (H)
Generation
Electric load
Frequency characteristic of load
Turbine time constant
Governor time constant
Regulation Constant
Area 1
2500 MVA
4.5 seconds
2050 MW
1600 MW
20 MW/Hz
0.3 seconds
0.2 seconds
0.05 pu
Area 2
2000 MVA
5.0 seconds
1750 MW
2200 MW
18 MW/Hz
0.32 seconds
0.2 seconds
0.05 pu
(a) For the defined operating conditions, compute the linearized tie line power flow
equation ∆P12 = C 12 ( ∆δ 1 − ∆δ 2 ) .
(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:
2
(
230)
=
sin (δ 0 − δ 0 ) = 450
⇒ δ 10 − δ 20 = 33.685 0
1
2
65.2
(230)2 cos(δ 0 − δ 0 ) = 675.121
C12 =
1
2
65.2
∆P12 = 675 .121( ∆δ 1 − ∆δ 2 )
0
12
P
(b) The dynamical equations are:
d∆f 1
= 0.002667∆PG1 − 0.0533∆f 1 − 0.002667g 1 (t ) − 1.8(∆δ 1 − ∆δ 2 )
dt
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
d (∆δ 1 − ∆δ 2 )
= 6.283(∆f 1 − ∆f 2 )
dt
d∆PG1
= −3.333∆PG1 + 3.333∆X C1
dt
d∆X C1
= −5.0∆X C1 − 4166.667∆f 1 + 5.0∆PC1
dt
d∆f 2
= 0.003∆PG 2 − 0.054∆f 2 − 0.003g 2 (t ) + 2.025(∆δ 1 − ∆δ 2 )
dt
d∆PG 2
= −3.125∆PG 2 + 3.125∆X C 2
dt
d∆X C 2
= −5.0∆X C 2 − 3333.33∆f 2 + 5.0∆PC 2
dt
(c) In this case:
g 1 (t ) = 100.0, ∆PC1 = 0.0, ∆PC 2 = 0.0
Upon substitution and solution:
∆f 1 = ∆f 2 = ∆f = −0.06502 Hz
1
∆PG1 = −
∆f = 54.1833 MW
R1
1
∆PG 2 = −
∆f = 43.3467 MW
R2
∆PD1 = D1 ∆f = −1.3 MW
∆PD 2 = D 2 ∆f = −1.17 MW
∆P12 = β 2 ∆f = −44 .517 MW
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.
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Copyright © A. P. Sakis Meliopoulos – 1990-2006
Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
(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. ∆PC 2 = − k ∫ ∆f (τ )dτ . Unit 1 is
not controlled, i.e. ∆PC1 = 0.0 . Determine the steady state frequency of the system and
the real power output change for each unit.
Solution:
(a) No, the initial set points for the two units are unknown.
(0.05)(60)
(0.05)(60)
= 0.0075( Hz/MW), R2 =
= 0.00375( Hz/MW)
400
800
− 25
∆f =
= −0.061125( Hz )
1
1
+
+9
0.0075 0.00375
( −0.061125)
( −0.061125)
∆Pg1 = −
= 8.15( MW), ∆Pg2 = −
= 16.30( MW)
0.0075
0.00375
(b) R1 =
Notice that, the total generation increase is 24.45 (MW). The load change is:
∆PLoad = D∆f = (9.0)(−0.061125) = −0.55( MW)
(c) ∆f = 0.0( Hz ),
∆Pg1 = 0.0 (MW),
∆Pg2 = 25.0 (MW)
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. ∆PC (t ) = 0.0 ,
and (2) If the system is controlled with the following law: ∆Pc (t ) = −75.0( MW / Hz)∆f (t ) .
In case (2) what will be the unit set point at steady state?
System Data Are:
Rated Power
Rated Frequency
Inertia Constant
Regulation
Electric Load constant
250MVA
f = 60 Hz
H = 2.5 sec
R = 3.6 Hz/250MW
D = 5 MW/Hz.
Solution:
Copyright © A. P. Sakis Meliopoulos – 1990-2006
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Power System Modeling, Analysis and Control: Chapter 10, Meliopoulos
1
= 140 ( MW/Hz)
R
− g(t) ∆Pc (t )
∆f =
+
β
β
(1) ∆Pc (t) = 0.0 :
− 25
= 59.82143 ( Hz )
f ss = f 0 + ∆f = 60 +
140
(2) ∆Pc (t) = -12(MW/Hz)∆f(t)
− 25 − 12∆f
∆f =
+
140
140
∆f(1 + 0.0857143) = −0.17859
∆f = −0.1645( Hz )
f ss = f 0 + ∆f = 59.8355(Hz)
β = D+
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:
∆Pc (t ) = −125.0( MW / Hz)∆f (t )
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
Rated Frequency
Inertia Constant
Regulation
Electric Load constant
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350MVA
f = 60 Hz
H = 2.5 sec
R = 3.0 Hz/350MW
D = 6 MW/Hz.
Copyright © A. P. Sakis Meliopoulos – 1990-2006