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ECE 476 Power System Analysis
Lecture 4: Three-Phase, Power System Operations
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
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Announcements
• Please read Chapter 4• HW 1 is due now• HW 2 is 2.44, 2.48, 2.49, 2.51
• It does not need to be turned in, but will be covered by an in-class quiz on Sept 10
• San Diego Gas & Electric is on campus for the ECE Career Fair on 9/9) (ARC Gym) and then for interviews on 9/10
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Three-Phase - Wye Connection
• There are two ways to connect 3 systems– Wye (Y)– Delta ()
an
bn
cn
Wye Connection Voltages
V
V
V
V
V
V
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Wye Connection Line Voltages
Van
Vcn
Vbn
VabVca
Vbc
-Vbn
(1 1 120
3 30
3 90
3 150
ab an bn
bc
ca
V V V V
V
V V
V V
Line-to-line
voltages are
also balanced
(α = 0 in this case)
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Wye Connection, cont’d
• Define voltage/current across/through device to be phase voltage/current
• Define voltage/current across/through lines to be line voltage/current
6
*3
3 1 30 3
3
j
Line Phase Phase
Line Phase
Phase Phase
V V V e
I I
S V I
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Delta Connection
IcaIc
IabIbc
Ia
Ib
a
b
*3
For the Delta
phase voltages equal
line voltages
For currents
I
3
I
I
3
ab ca
ab
bc ab
c ca bc
Phase Phase
I I
I
I I
I I
S V I
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Three-Phase Example
Assume a -connected load is supplied from a 3 13.8 kV (L-L) source with Z = 10020W
13.8 0
13.8 0
13.8 0
ab
bc
ca
V kV
V kV
V kV
13.8 0138 20
138 140 138 0
ab
bc ca
kVI amps
I amps I amps
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Three-Phase Example, cont’d
*
138 20 138 0
239 50 amps
239 170 amps 239 0 amps
3 3 13.8 0 kV 138 amps
5.7 MVA
5.37 1.95 MVA
pf cos 20 lagging
a ab ca
b c
ab ab
I I I
I I
S V I
j
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Delta-Wye Transformation
Y
phase
To simplify analysis of balanced 3 systems:
1) Δ-connected loads can be replaced by 1
Y-connected loads with Z3
2) Δ-connected sources can be replaced by
Y-connected sources with V3 30Line
Z
V
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Delta-Wye Transformation Proof
From the side we get
Hence
ab ca ab caa
ab ca
a
V V V VI
Z Z Z
V VZ
I
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Delta-Wye Transformation, cont’d
a
From the side we get
( ) ( )
(2 )
Since I 0
Hence 3
3
1Therefore
3
ab Y a b ca Y c a
ab ca Y a b c
b c a b c
ab ca Y a
ab caY
a
Y
Y
V Z I I V Z I I
V V Z I I I
I I I I I
V V Z I
V VZ Z
I
Z Z
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Three Phase Transmission Line
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Per Phase Analysis
• Per phase analysis allows analysis of balanced 3 systems with the same effort as for a single phase system
• Balanced 3 Theorem: For a balanced 3 system with– All loads and sources Y connected– No mutual Inductance between phases
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Per Phase Analysis, cont’d
• Then– All neutrals are at the same potential– All phases are COMPLETELY decoupled– All system values are the same sequence as sources. The
sequence order we’ve been using (phase b lags phase a and phase c lags phase a) is known as “positive” sequence; later in the course we’ll discuss negative and zero sequence systems.
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Per Phase Analysis Procedure
• To do per phase analysis
1. Convert all load/sources to equivalent Y’s
2. Solve phase “a” independent of the other phases
3. Total system power S = 3 Va Ia*
4. If desired, phase “b” and “c” values can be determined by inspection (i.e., ±120° degree phase shifts)
5. If necessary, go back to original circuit to determine line-line values or internal values.
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Per Phase Example
Assume a 3, Y-connected generator with Van = 10 volts supplies a -connected load with Z = -j through a transmission line with impedance of j0.1 per phase. The load is also connected to a -connected generator with Va”b” = 10 through a second transmission line which also has an impedance of j0.1 per phase.Find
1. The load voltage Va’b’
2. The total power supplied by each generator, SY and S
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Per Phase Example, cont’d
First convert the delta load and source to equivalent
Y values and draw just the "a" phase circuit
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Per Phase Example, cont’d
' ' 'a a a
To solve the circuit, write the KCL equation at a'
1(V 1 0)( 10 ) V (3 ) (V j
3j j
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Per Phase Example, cont’d
' ' 'a a a
'a
' 'a b
' 'c ab
To solve the circuit, write the KCL equation at a'
1(V 1 0)( 10 ) V (3 ) (V j
310
(10 60 ) V (10 3 10 )3
V 0.9 volts V 0.9 volts
V 0.9 volts V 1.56
j j
j j j j
volts
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Per Phase Example, cont’d
*'*
ygen
*" '"
S 3 3 5.1 3.5 VA0.1
3 5.1 4.7 VA0.1
a aa a a
a agen a
V VV I V j
j
V VS V j
j
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Power System Operations Overview
• Goal is to provide an intuitive feel for power system operation
• Emphasis will be on the impact of the transmission system
• Introduce basic power flow concepts through small system examples
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Power System Basics
• All power systems have three major components: Generation, Load and Transmission/Distribution.
• Generation: Creates electric power.• Load: Consumes electric power.• Transmission/Distribution: Transmits electric power
from generation to load. – Lines/transformers operating at voltages above 100 kV are
usually called the transmission system. The transmission system is usually networked.
– Lines/transformers operating at voltages below 100 kV are usually called the distribution system (radial).
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Simulation of the Eastern Interconnect
Bloomington
Peoria
RockfordWaukegan
Zi on
Pl easant
Des Pl ai nes
El mhurst
I t asca
Tol l w ay
W407 ( Fermi )
Cherry Val ley
Wempleton
Paddock
Cl inton
Powerton
Ipava
D uck Creek
Brai dw ood
Shefi eld
Chiave
Munster
El ect r i c Junct i on
Pl ano
La Sal le
Lombard
Li sle
Col l ins
D resden
Lockport
East Frankfort
Goodings Grove
Tazwel l
Lake George
D unacr
Green Acres
Schahfer
Tower Rd
Babcock
Prai ri e
Racine
Michi gan Ci ty
El wood
Gulon
Sti l lwel l
O l i ve
Sycamore
Bondurant
Motezuma
Arnold
Se Pol k
H azel ton
Burr Oak
Pl ai sades
Benton Harbor
RN TOUL J
PERKN SRD
CH AMP ECH AMP W
LEVER RD
RISIN G
CH AMP TP
MAHOMET
WEED MAN
N LEROY
PAXTON E
RANTOUL
GIBSON C
GIBSONCP
H OOPESTN
VERMILON
TILTN EC
BUNSONVL
W TILTON
VERML 1
VERMILON
GILMAN
WATSEKA 17GOD LN D
CLTN TAP
CLT RT54
S CLN TN
R FAL; R
1346A TP
H OLLAND
MACOMB WH AMLTNAM
H AVAN A
IPAVA
CANTON
CUBA
KICKAPOO
MASON
H AVAN A SMASON CY
TAZEWELL
H ALLOCK
ED WARD S3
CAT MOSSFARGO
KEYSTONE
E PEORIA
RSW EAST
CAT SUB2
H INES
PION EERC
RAD N OR
CAT TAP
CAT MAP
CAT SUB1
EASTERNBURLIN 1G
MACOMBN E
N IOTA
N IOTA
BRLGTN 5
SB 18 5
E MOLIN E
SB 43 5
SB 112 5
KPECKTP5
SO.SUB 5
SB 85 5
SB 31T 5
SB 28 5
SB 17 5
SB 49 5
SB 53 5
SB 47 5SB 48 5
SB A 5
SB 70 5
SB 79 5
SB 88 5
SB 71 5
BVR CH 65 BVR CH 5 ALBAN Y 5
YORK 5
SAVAN NA5
GALEN A 5
8TH ST.5
LORE 5
SO .GVW.5
SALEM N5
ALBAN Y 6
GARD E;
H 71 ;BTH 71 ; B
H 71 ; R
R FAL; B
N ELSO; R
N ELSO;RT
STERL; B
D IXON;BT
MECCORD 3
CORD O;
LEECO;BP
MARYL; B
MEND O; T
STILL;RT
B427 ;1T
LANCA; R
PECAT; B
FREEP;
ELERO;BT ELERO;RT
LENA ; RLENA ; B
H 440 ;RT
H 440 ; R
STEWA; B
H 445 ;3B
Roscoe
Pi erpont
S PEC; R
FORD A; R
H arlem
Sand Park
N WT 138
BLK 138
ROR 138
JAN 138
ALB 138
N OM 138
D AR 138
H LM 138
POT 138 MRE 138
COR 138 D IK 138
BCH 138
Sabrooke
Bl awkhawk
Al pine
E. Rockford
Charles
Belvidere
B465
Marengo
WIB 138
WBT 138 ELK 138
N LG 138
N LK GV T
SGR CK5
BRLGTN 1
BRLGTN 2
SGR CK4
UN IVRSTY
UN IV NEU
WHTWTR5
WHTWTR4
WHTWTR3
SUN 138
VIK 138
LBT 138
TICH IGN PARIS WE
ALBERS-2
Green Lake
Sand Ridge
Chicago H eights
Burnham
Lansing
F-575
F-503Gl enwood
Bl oomPark ForestMatteson
Country Cl ub H i l l s
Woodhi l lU. Park
Bl ue Island
Wi lmington
Wi l ton Center
Frankfort
N Len
Brigg
D owners Groove
N Aurora
Hanover
Bart l et t
Church
Addi son
NordiG l endal e
Romeo
Wayne
Wi l l Co.
Jol i et
D umont 345 D umont 765
Kenzie Creek
JackSr
Cook - 345 kV
Cook - 765 kV
Twin Branch
BAIN 4
SOMERS
ST RITA
BIG BEND
MUKWON GO
N ED 138
N ED 161
LAN 138
EEN 138
CASVILL5TRK RIV5D UND EE 5
LIBERTY5
ASBURY 5
CN TRGRV5
JULIAN 5
MQOKETA5
E CALMS5
GR MN D 5
D EWITT 5
SBHYC5
SUB 77 5
SB 74 5SB 90 5
SB 78 5D AVN PRT5
SB 76 5
SB 58 5
SB 52 5
SB 89 5
IPSCO 5
IPSCO 3
D ENMARK5
N EWPORT5
H WY61 5
WEST 5
9 SUB 5
H ENRYCO5
VIELE 5
TRIVERS5
CARBID E5
TRIPP
Kenda
MPWSPLIT
WYOMIN G5
MT VERN5
BERTRAM5
H AZL S 5
BLKHAWK5
ED WARD S1
JEFF 5
WAPELLO5
APANOSE5
AD AIR
H ILLSIE5
PCI 5
BEVERLY5
FAIRFAX5
H ILLS 5PARNEL 5POWESHK5
BEACON 5
EIC 5
BRD GPRT5
LUCAS 5
CRLRID G5
SB PIC 5
SB J IC 5
SB EIC 5
SB YIC 5
SB GIC 5
SB UIC 5
WASH BRN 5
UN IONTP5
EL FARM5
D R EN G 5
WTWEST 5
D RFN D RY5
MID PORT5
6 ST 5
VINTON 5
H IAWATA5
D YSART 5
TRAER 5
D R N E 5
LUND QST5
D RCOMP 5
H AMPTON 5
FRANKLN5BUTLER 5
M-TOWN 5
REASN OR5
N EWTON 5
JASPER 5
TIMBRCK5
McCook
Covert
93%B
MVA105%
B
MVA
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Small PowerWorld Simulator Case
Bus 2 Bus 1
Bus 3Home Area
204 MW
102 MVR
150 MW
150 MW 37 MVR
116 MVR
102 MW 51 MVR
1.00 PU
-20 MW 4 MVR
20 MW -4 MVR
-34 MW 10 MVR
34 MW-10 MVR
14 MW -4 MVR
-14 MW
4 MVR
1.00 PU
1.00 PU
106 MW 0 MVR
100 MWAGC ONAVR ON
AGC ONAVR ON
Load withgreenarrows indicatingamountof MWflow
Usedto controloutput ofgenerator
Direction of arrow is used to indicatedirection of real power (MW) flow
Note thepower balance ateach bus
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Power Balance Constraints
• Power flow refers to how the power is moving through the system.
• At all times in the simulation the total power flowing into any bus MUST be zero!
• This is know as Kirchhoff’s law. And it can not be repealed or modified.
• Power is lost in the transmission system.
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Basic Power Control
• Opening or closing a circuit breaker causes the power flow to instantaneously(nearly) change.
• No other way to directly control power flow in a transmission line.
• By changing generation or load, or by switching other lines, we can indirectly change this flow.
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Modeling Consideration – Change is Not Really Instantaneous!
• The change isn’t really instantaneous because of propagation delays, which are near the speed of light; there also wave reflection issues– This will be addressed more in Chapters 5 and 13
Red is the vs end, green the v2 end
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Transmission Line Limits
• Power flow in transmission line is limited by heating considerations.
• Losses (I2 R) can heat up the line, causing it to sag.• Each line has a limit; Simulator does not allow you
to continually exceed this limit. Many utilities use winter/summer limits.
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Overloaded Transmission Line
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Interconnected Operation
• Power systems are interconnected across large distances. For example most of North America east of the Rockies is one system, with most of Texas and Quebec being major exceptions
• Individual utilities only own and operate a small portion of the system, which is referred to an operating area (or an area).
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Operating Areas
• Transmission lines that join two areas are known as tie-lines.
• The net power out of an area is the sum of the flow on its tie-lines.
• The flow out of an area is equal to
total gen - total load - total losses = tie-flow
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Area Control Error (ACE)
• The area control error is the difference between the actual flow out of an area, and the scheduled flow.– There is also a frequency dependent component that we’ll
address in Chapter 12
• Ideally the ACE should always be zero.• Because the load is constantly changing, each
utility must constantly change its generation to “chase” the ACE.
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Automatic Generation Control
• Most utilities use automatic generation control (AGC) to automatically change their generation to keep their ACE close to zero.
• Usually the utility control center calculates ACE based upon tie-line flows; then the AGC module sends control signals out to the generators every couple seconds.
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Three Bus Case on AGC
Bus 2 Bus 1
Bus 3Home Area
266 MW
133 MVR
150 MW
250 MW 34 MVR
166 MVR
133 MW 67 MVR
1.00 PU
-40 MW 8 MVR
40 MW -8 MVR
-77 MW 25 MVR
78 MW-21 MVR
39 MW-11 MVR
-39 MW
12 MVR
1.00 PU
1.00 PU
101 MW 5 MVR
100 MWAGC ONAVR ON
AGC ONAVR ON
Net tie flow is close to zero
Generationis automaticallychanged to matchchange in load
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MISO Real-Time ACE
https://www.misoenergy.org/MARKETSOPERATIONS/REALTIMEMARKETDATA/Pages/ACEChart.aspx
Previouslyindividualutilities didtheir ownACE calculations;now we arepart of MISO,which doesone for the
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MISO Real-Time ACE
• MISO's real-time ACE is available online (along with lots of other data)
https://www.misoenergy.org/MARKETSOPERATIONS/REALTIMEMARKETDATA/Pages/ACEChart.aspx