ECE 422/522
Power System Operations & Planning/
Power Systems Analysis II
4 –Active Power and Frequency
Control
Spring 2014
Instructor: Kai Sun
1
References
• Chapter 12 of Saadat’s book
• Chapter 11.1 of Kundur’s book (understand examples)
• Chapter 4 (Frequency Control) of the EPRI Tutorial
2
Background
• The frequency of a system depends on real power
balance.
• Changes in real power affect mainly the system
frequency, while reactive power is less sensitive to
changes in frequency and is mainly dependent on
changes in voltage magnitude.
3
Frequency Deviations
• Under normal conditions, the power system frequency in a large Interconnection
(e.g. the EI) varies approximately 0.03Hz from the scheduled value
• When abnormal events, e.g. loss of a large generator unit, the frequency
experiences larger deviations.
4
Control of Frequency
• As frequency is a common factor throughout the system, a
change in real power demand at one point is reflected
through the system by a change in frequency
• In an interconnected system with two or more independently
controlled areas, in addition to control of frequency, the
generation within each area has to be controlled so as to
maintain scheduled power interchange.
• The control of generation and frequency is commonly
referred to as Load Frequency Control (LFC), which
involves
– Speed governing system with each generator
– Automatic Generation Control (AGC) for interconnected systems
5
Generator Control Loops
• For each generator, real power
(or frequency) and reactive
power (or voltage) outputs are
controlled separately by
– LFC (Load Frequency Control)
loop
– AVR (Automatic Voltage
Regulator) loop .
• The LFC and AVR controllers
are set for a particular steadystate operating condition to
maintain frequency and voltage
against small changes in load
demand.
• Cross-coupling between the LFC
and AVR loops is negligible because
the excitation-system time constant
is much smaller than the prime
mover/governor time constants
6
Speed Governing System
7
Generator Model
Initial values: P0=0T0
P=rT
P=P0+P, T=T0+T
r=0+r
P0+P=(0+r)(T0+T)
0T0+ 0T +rT0 (r T 0)
so
P=0T +rT0
Pm=0Tm+rTm0
Pe=0Te +rTe0
Pm
=0
Pm-Pe=0(Tm-Te)+r(Tm0-Te0)
=Tm-Te in per unit (0=1)
= Tm-Te
Pe
8
Consider a frequency-dependent load model
Pe=PL +Dr
PL Frequency-insensitive load change
Dr Frequency-sensitive load change
D
Load damping constant, typically at 1~2, i.e. 1~2%
change in load per 1% frequency change
Pm
Pe
Pm-Pe
=2Hsr
PL
Pm-PL-Dr=2Hsr
Pm-PL
=(2Hs+D)r
=(Ms+D)r
9

D
Relationship between Load and Frequency
D=2
10
Kundur’s Example 11.1
• A small system consists of 4 identical 500MVA generating units feeding a
total load of 1,020MW. The inertia constant H of each unit is 5.0 on
500MVA base. The load varies by 1.5% for a 1% change in frequency.
When there is a sudden drop in load by 20MW
a. Determine the system block diagram with constants H and D expressed
on 2,000MVA base
b. Find the frequency deviation, assuming that there is no speedgoverning action
11
12
=
.
.
/
13
Governor Model
• See Bergen and Vittal’s book for the
model with time constants of key parts
Classic Watt Centrifugal Governing System
Speed changer
Linkage mechanism
Speed governor
Hydraulic Amplifier
14
Governor Model
Pv
Pref
• Without a governor, the generator
speed drops when load increases
• The speed governor closes the
loop for negative feedback control
r/R
r
– For stable operation, The governor
reduces (rather than eliminate) the
speed drop due to load increase.
– Usually, speed regulation R is 5-6%
from zero to full load
– Governor output r/R is compared
to the reference set power Pref
Pg= Pref -  r/R
– Then, Pg is transformed through
the hydraulic amplifier to the steam
valve/gate position command Pv
with time constant g
 (s)
15
Pv
Turbine Model
Pm
• The prime mover, i.e. the source of mechanical power, may be hydraulic turbines
at water falls, steam turbines burning coal and nuclear fuel, or gas turbines
• The model for the turbine relates changes in mechanical power output Pm to
changes in gate or valve position PV


T is in 0.2~2.0 seconds

16
Load Frequency Control block Diagram
 (s)
17
Load Frequency Control block Diagram
 (s)
How to choose the value of
R for a stable speed
governing system?


2
• For a step load change, i.e.


1
1


1


1
1/
= /
lim  (s) = 

/
(final value theorem)
• If the load is supported by n generators


1
1
1
1
18
2
⋯
1
Saadat’s Example 12.1
19
 (s)
The open-loop transfer function is
20
Review: Stability of a Linear System
s 3 + 7.08s 2 + 10.56s + 0.8 + K = 0
Characteristic equation:
• A necessary and sufficient condition for a linear system to be stable: Poles of the
system transfer function (i.e. roots of the characteristic equation) are only in the lefthand portion of the s-plane (i.e. having negative real parts)
21
Review: Routh-Hurwitz Stability Criterion
• Characteristic equation
ansn+an-1sn-1+…+a1s+a0=0 (an>0)
s 3 + 7.08s 2 + 10.56s + 0.8 + K = 0
• Routh table:
For i>2, xij=(xi-2,j+1xi-1,1  xi-2,1xi-1,j+1)/xi-1,1
where xij is the element in the i-th row and j-th column
s3
1
10.56
7.08
0.8 + K
73.965 - K
0
s1
7.08
s0 0.8 + K
0
s2
• Routh-Hurwitz criterion: No. of roots of the equation
with positive real parts = No. of changes in sign of the
1st column of the Routh table
• Necessary and sufficient condition for a linear system
to be stable: The 1st column only has positive numbers
22
• s1 row>0 if K<73.965
• s0 row>0 since K>0
• So R=1/K>1/73.965=0.0135
Review: Root-Locus Method
When s=j3.25, Rmin=1/K=0.0135
So R>0.0135
•
•
•
•
•
i-th
j-th
-zi is the
zero and -pj is
pole
The locus of roots of 1+KG(s)H(s) begins at KG(s)H(s)’s poles and ends at its zeros as K=0
No. of separate loci = No. of poles; root loci must be symmetrical with respect to the real axis
The root locus on the real axis always lies in a section of the real axis to the left of an odd number of poles and zeros
Linear asymptotes of loci are centered at a point (x, 0) on the real axis with angle  with respect to the real axis
x=[ j=1~n(-pj) -i=1~m(-zi) ]/(n-m)
=(2k+1)/(n-m)
k=0, 1, …, (n-m-1)
23
• Closed-loop transfer function with R=0.05pu (>0.0135):
(1 + 0.2 s )(1 + 0.5s )
Dw ( s )
0.1s 2 + 0.7 s + 1
= T ( s) =
= 3
(10s + 0.8)(1 + 0.2 s )(1 + 0.5s ) + 1 / 0.05
-DPL ( s )
s + 7.08s 2 + 10.56s + 20.8
• Steady-state frequency deviation due to a step input:
Dwss = lim sDw ( s ) = -DPL
s0
1
1
= -0.2´
= -0.0096 p.u.
20.8
D + 1/ R
Df = -0.0096 ´ 60 = 0.576 Hz
Note: The frequency is not restored to 60Hz
(there is an offset)
24
Using the MATLAB toolbox with Saadat’s book
chp12.ex1.m
sim12ex1.mdl
Frequency deviation step response
60
pu
Freq.,
Hz
Without LFC (Open-loop)
55
50
45
0
20
40
60
t, sec
25
80
100
Saadat’s Example 12.2
Note: two generators use different MVA bases. Select 1000MVA as the common MVA base
S Dw Sbase1
S
Dw
Dw
Dw
= base1
=
= base1
Rbase1 =
DP
Sbase 2 DP / Sbase 2 Sbase 2 D P base 2
D P base1
R1 =
1000
1000
(0.06) = 0.1 pu R2 =
(0.04) = 0.08 pu
600
500
26
DPL =
Rbase1 =
Sbase1
Rbase 2
Sbase 2
90
= 0.09 pu
1000
(b) D=1.5(900+90)/1000=1.485 (frequency dependent)
(a) D=0
Dwss =
-DPL
-0.09
=
=-0.004 pu
1 1 10 +12.5
+
R1 R2
Dwss =
-DPL
-0.09
=
= -0.00375 pu
1
1
10
12.5
1.485
+
+
+ +D
R1 R2
Df = -0.004 ´ 60 = -0.24 Hz
Df = -0.00375´60 = -0.225 Hz
f = f 0 + Df = 60 - 0.24 = 59.76 Hz
f = f 0 + Df = 60 - 0.225 = 59.775 Hz
Dw
-0.004
DP1 === 0.04 pu = 40 MW
R1
0.1
DP2 =-
Dw
-0.004
== 0.05 pu = 50 MW
R2
0.08
Unit 1 supplies 540MW and unit 2
supplies 450MW at the new operating
frequency of 59.76Hz.
DP1 = -
Dw
-0.00375
== 0.0375 pu=37.5MW
R1
0.1
DP2 = -
Dw
-0.00375
== 0.0469 pu=46.9MW
R2
0.08
Unit supplies 537.5MW and unit 2 supplies 446.9MW
at the new operating frequency of 59.775Hz. The total
change in generation is 84.4MW, i.e. 5.6MW less than
90MW load change, because of the change in load
due to frequency drop.
Dw ⋅ D = -0.00375´1.485 = -0.005572 pu = -5.6MW
27
DP1 =-
Dw
R1
DP2 =-
Dw
R2
D P1
R
= 2
R1
D P2
D=1.485
D=0
Adjusting R1 and R2 may change the generation dispatch between Units 1
and 2 for economic concerns
28
Composite Frequency Response Characteristic (FRC)
• When analyzing LFCs for a multi-generator system, we may assume the
coherent response of all generators to changes in system load represent
them by an equivalent generator.
• Meq =2Heq= sum of the inertia constants of all generators


⋯
=

⋯
29
• Frequency response characteristic (FRC) or Frequency bias factor 
=D+1/Req =|PL/f | (Unit: MW/0.1 Hz)
• FRC can be developed for any section of a power system. It relates the MW
response of the system (or section of the system) to a change in frequency.
• FRC depends on:
– The governor droop settings of all on-line units in the system.
– The condition of the power system when the frequency deviation occurs.
– The condition of the power system includes current generator output
levels, transmission line outages, voltage levels, etc.
– The frequency response of the connected load in the system.
30
FRCs of Different
Interconnections
31
Limitations of Governor Frequency Control
• Governors do not recover frequency back to the scheduled value
(60Hz) due to the required % droop characteristic.
• Governor control does not adequately consider the cost of power
production so control with governors alone is usually not the most
economical alternative.
• Governor control is intended as a primary means of frequency control.
As such governor control is course and not suited to fine adjustment of
the interconnected system frequency
• Other limitations (see Sec. 4.3 in EPRI Tutorial)
– Spinning Reserve is not considered
– Governors have dead-bands (not functioning in 600.03~0.04Hz)
– Depends on the type of Unit (Hydro: very responsive; Combustion turbine:
may or may not be responsive; Steam: varies depending on the type)
– Governors may be blocked: a generator operator can intentionally prevent
the unit from responding to a frequency disturbance
32
Automatic Generation Control (AGC)
• Adding supplementary control on
load reference set-points of
selected generators
− Controlling prime-mover power
to match load variations
− As system load is continually
changing, it is necessary to
change the output of generators
automatically
• Primary objective:
– LFC, i.e. regulating frequency to the specified nominal value, e.g. 60Hz,
and maintaining the interchange power between control areas at the
scheduled values by adjusting the output of selected generators
• Secondary objective:
– Generation dispatch, i.e. distributing the required change in generation
among generators to minimize operation costs.
• AGC is bypassed during large disturbances and emergencies, and
other emergency controls are applied.
33
AGC for an Isolated Power System
• An integral controller is added with gain KI
s(1+ t g s)(1+ tT s)
Dw( s)
=
-DPL ( s) s(2Hs + D)(1+ t g s)(1+ tT s) + KI + s / R
• Applied to the system in Example
12.1 (Example 12.3) with KI=7
34
LFC for a Two-Area System
• Generators in each area is coherent, i.e. closely coupled internally
• Two areas are represented by two equivalent generators (modeled by a voltage
source behind an equivalent reactance) interconnected by a lossless tie line
P12 =
DP12 »
Ps =
E1 E2
XT
dP12
d d12
dP12
d d12
sin d12
X T  X 1  X tie  X 2
12  1   2
Dd12 = Ps Dd12 = Ps (Dd1 -Dd2 )
d120
=
d120
E1 E2
XT
cos Dd120
P12
P12,max
Ps is the synchronizing power coefficient
Slope=Ps
P12,0
35
12,0
12
LFC with only the Primary Loop
• Consider a load change PL1 in area 1.
Both areas have the same steady-state
frequency deviation
Dw = Dw1 = Dw2
DPm1 -DP12 -DPL1 = Dw D1
DPm 2 + DP12 - 0 = Dw D2
DP12 = Dw D2 -DPm 2
• The change in mechanical power is
determined by the governor speed
characteristics
-Dw
DPm1 =
R1
DPm 2 =
-Dw
R2
• Solve  and P12
Dw =
=0
-DPL1
-DPL1
=
1
1
( + D1 ) + ( + D2 ) b1 + b2
R1
R2
1
+ D2 )DPL1
b2
R2
(-DPL1 )
DP12 =
=
1
1
b
b
+
1
2
( + D1 ) + ( + D2 )
R1
R2
-(
36
37
AGC with Frequency Bias Tie-Line Control
• The objective is to restore generation-load balance in each area
• A simple control strategy:
– Keep frequency approximately at the nominal value (60Hz)
– Maintain the tie-line flow at about schedule
– Each area should absorb its own load changes
• Area Control Error (ACE): supplementary control signal added to the
primary LFC through an integral controller
n
ACEi = å DPij + BiDw
j =1
– Bi: frequency bias factor (may or may not equal i)
– Any combination of ACEs containing Pij and  will result in
steady-state restoration of the tie line flow and frequency deviation
(the integral control action reduces each ACEi to 0)
– What composition of ACE signals should be selected is more
important from dynamic performance considerations.
38
Comparing different Bi’s in ACE signals
• Consider a sudden load increase in Area 1:
Bi=i=D+1/Ri
ACE1 = DP12 + b1Dw =
b2
-DPL1
(-DPL1 ) + b1
= -DPL1
b1 + b2
b1 + b2
ACE 2 = -DP12 + b2Dw = -
b2
-DPL1
(-DPL1 ) + b2
=0
b1 + b2
b1 + b2
Load change is taken care of locally
Coefficient of
(1=2=20)
B1=k1, B2=k2
b2
-DPL1
kb + b
=-DPL1 1 2
ACE1 =DP12 + kb1Dw =
(-DPL1) + kb1
b1 + b2
b1 + b2
b1 + b2
ACE2 =-DP12 + kb2Dw =-
b2
(k -1)b2
-DPL1
(-DPL1) + kb2
=-DPL1
b1 + b2
b1 + b2
b1 + b2
k=2
k=1
k=1/2
1.5
1
0.75
0.5
0
-0.5
What does k1 mean? (k>1: the generator is more active in dynamics)
39
Bi=i=D+1/Ri
1~0
2~0
Pref1
Pm1>0
Pref2
Pm2~0
P12~0
=0
Bi=2i
1
2
In practice, only selected units participate in
AGC, i.e. receiving supplementary control
signals (ACE)
Pm1
Pm2
40
P12
NERC Balancing Authority
• A Balancing Authority (BA) is a part of an interconnected power system
that is responsible for meeting its own load.
• Each BA operates an AGC system to balance its generation resources
to its load requirements.
– The generation resources may be internal or purchased from other
BAs and transferred over tie-lines between BAs.
– Similarly, load requirements may include internal customer load,
losses, or scheduled sales to other BAs.
41
• The control center is the headquarters of the BA, where the AGC
computer system is typically located. All the data collected by the
AGC system is processed in the control center.
• Based on the gathered data, the AGC signals are transmitted from the
control center to the various generators currently involved in
supplementary control to tell the generators what generation levels to
hold (adjust the generator set-points).
• It is not necessary for the AGC system to regulate the output of all the
generators in a BA. Most BAs have policies which require that as
many units as needed are under control and able to respond to the
BA’s continual load changes. Those units that receive and respond to
AGC signals are called regulating units. The number vary from a few
for a small BA to 40~50 for the largest BA
42
NERC Balancing Authorities
• The EI is composed of
approximately 90 BAs,
which range in load size
from over 130GW peaks to
BAs that serve no load but
simply use their generation
for meeting interchange
responsibilities.
• The WI (WECC) is
composed of approximately
30 BAs with a distribution
similar to the Eastern
Interconnection.
• The ERCOT and Hydro
Quebec are each operated
as single BAs.
43
AGC for more than two areas
• By means of ACEs, the frequency bias tie-line control scheme
schedules the net import/export for each area, i.e. the algebraic sum of
power flows on all the tie lines from that area to the others
44
Influences from reserves
• Sufficient or insufficient spinning reserve
– Normal conditions: each area has sufficient generation reserve to
carry out its supplementary control (AGC) obligations to eliminate
the ACE
– Abnormal conditions: one or more areas cannot fully eliminate the
ACE due to insufficient generation reserve; thus, there will be
changes in frequency and tie-line flows (under both supplementary
control and primary control)
• Operating reserve resources
– Spinning reserve: unloaded generating capacity (Pref,max-Pref), interruptible
load (controlled automatically)
– Non-spinning reserve: not currently connected to the system but can be
available within a specific time period, e.g. 15 minutes. Examples are such
as combustion turbines while cold standby and some interruptible load.
Each BA shall carry enough operating reserves.
45
Kundur’s Example 11.3
Spinning reserve:
1,000 of 4,000MW
Spinning reserve:
1,000 of 10,000MW
B1=250MW/0.1Hz
B2=500MW/0.1Hz
(losing some spinning reserve)
46
Notes on AGC
• In an interconnect system, all generators with governors
may respond to a generation/load change due to either
f/R0 or Pref 0
• For load increase or generation loss, only generators with
spinning reserves may increase their outputs up to their
maximum output limits (by either governors or AGC) (see
EPRI tutorial Sec. 4.4.2: “Spinning reserves consist of
unloaded generating capacity that is synchronized to the
power system. A governor cannot increase generation in
a unit unless that unit is carrying spinning reserves. An
AGC system cannot increase a unit’s MW output unless
that unit is carrying spinning reserves.”)
• For load decrease, all generators may reduce their
outputs as long as higher than their minimum output limits
47
Spinning reserve:
1,000 of 4,000MW
Spinning reserve:
1,000 of 10,000MW
B1=250MW/0.1Hz
B2=500MW/0.1Hz
ACEi = BiDf + DPij
=0 with sufficient reserve
or 0, otherwise
Pref1
DPmi -DPLi = Di Df + DPij
Pref2
DPGi
Without supplementary control (AGC):
-å i DPL ,i = (å i 1 / Ri + å i Di )Df
= (1 / R + D )Df
48
Spinning reserve:
1,000 of 4,000MW
1000
Spinning reserve:
1,000 of 10,000MW
B1=250MW/0.1Hz
B2=500MW/0.1Hz
Loss of 1,000MW load
Online generators with
active governor control
49
Spinning reserve:
1,000 of 4,000MW
1000
322.56
B1=250MW/0.1Hz
Spinning reserve:
1,000 of 10,000MW
B2=500MW/0.1Hz
Loss of 1,000MW load
50
Spinning reserve:
1,000 of 4,000MW
1000
1000
Spinning reserve:
1,000 of 10,000MW
B1=250MW/0.1Hz
B2=500MW/0.1Hz
Loss of 1,000MW load
51
Spinning reserve:
1,000 of 4,000MW
1000
1000
?
Spinning reserve:
1,000 of 10,000MW
B1=250MW/0.1Hz
B2=500MW/0.1Hz
Loss of 500MW
generation carrying part
of spinning reserve
(losing some spinning reserve)
833.33-500=333.33MW
52
Spinning reserve:
1,000 of 4,000MW
1000
1,937.50
B1=250MW/0.1Hz
Spinning reserve:
1,000 of 10,000MW
B2=500MW/0.1Hz
Loss of 2,000MW
generation, not carrying
spinning reserve
Only held for the area with
sufficient spinning reserve
53
Spinning reserve:
1,000 of 4,000MW
B1=250MW/0.1Hz
Spinning reserve:
1,000 of 10,000MW
X
B2=500MW/0.1Hz
54
Spinning reserve:
1,000 of 4,000MW
B1=250MW/0.1Hz
Spinning reserve:
1,000 of 10,000MW
X
B2=500MW/0.1Hz
55
Frequency response following the loss of a generator
56
Impact of Abnormal Frequency Deviations
• Prolonged operation at frequencies above or below 60Hz can damage
power system equipment.
• Turbine blades of steam turbine generators can be exposed to only a
certain amount of off-frequency operation over their entire lifetime.
• Steam turbine generators often have under- and over-frequency relays
installed to trip the unit if operated at off-frequencies for a period
A typical steam turbine can be operated,
under load, for 10 minutes over the lifetime
at 58Hz before damage is likely to occur to
the turbine blades
57
Frequency Decay Due to Generation Deficiency
• Severe system disturbances can result in cascading outages and
isolation of areas to form electrical islands.
• If such an islanded area does not have sufficient generation (and
spinning reserve), it will experience a frequency decline, which is
largely determined by frequency sensitive characteristics of loads.
58
Underfrequency Load Shedding
• In many situations, the frequency decline may lead to tripping of
steam turbine generators by underfrequency protective relays, thus
aggravating the situation further
• Underfrequency Load Shedding (UFLS) is a protection program that
automatically trips selected customer loads once frequency falls
below a specific value.
• The intent of UFLS is not to recover the frequency to 60 Hz but rather
to arrest or stop the frequency decline. Once UFLS has operated,
manual intervention by the system operators is likely required to
restore the system frequency to a healthy state.
59
• A typical UFLS setting for a North American utility may include
three steps conducted by under-frequency relays, e.g.,
– shedding 10% of the load at 59.3 HZ
– shedding 10% additional load at 59.0 HZ, and
– shedding 10% more at 58.7Hz
60
UFLS and Automatic Load Restoration in the
Western Interconnection
Maximum delay
Minimum waiting time
61
Homework
• Problems 12.3 and 12.5~12.10 in Saadat’s book (3rd ed.,
Page 619), due by April 1st (Tue) in class
62