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EE5702R Advance Power System Analysis:: Introduction
Panida Jirutitijaroen
Fall 2011
17/08/2011
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 1
ABOUT THIS CLASS
Assessment
Tentative syllabus
Learning outcomes
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 2
About This Class
• Instructors: – Asst. Prof. Panida Jirutitijaroen (1st half)– Assoc. Prof. Chang Che Sau (2nd half)
• Core module for MEng or PhD students majoring in Power and Energy area, seen from suffice ‘R’.
• Assume that all students have adequate background in power systems analysis equivalent to EE4501.
• Assume that all students are familiar with software used to perform analysis such as C program or MATLAB.
• Fundamental materials for power systems analysis.• Followed by case studies to discuss current research
activities. Students prepare report and discuss their findings in class.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 3
Assessment
• Homework 20%– To test your understanding in the fundamental concepts
• Case Study 10%– To apply your knowledge to analyze and evaluate related
materials in research papers
• Two Mini-Projects 40%– To be able to formulate the problem, run experiments,
analyze the results and draw conclusion based on the knowledge acquired.
• Final 30%– To test your knowledge in the subject.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 4
Tentative Syllabus
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 5
Week Topics Instructor Assignment Due
1
2:17/08 Overview of power system
operation. Network representation.
PJ Homework 1,
Homework 4
3:24/08 Power flow analysis I PJ Mini-Project 1
4:31/08 Power flow analysis II PJ Homework 2 Homework 1
5:07/09 Economic dispatch and optimal
power flow
PJ Case Study 1 Homework 2
6:14/09 Power system state estimation I PJ Homework 3
Recess:
21/09
Power system state estimation II PJ Case Study 2,
Mini-Project 2
Homework 3
7:28/09 Case study 1&2 presentation PJ
8:05/10 Transient stability I CS Mini-Project 1
9:12/10 Transient stability II CS Homework 4
10:19/10 Small-perturbation stability I CS Homework 5
11:26/10 Small-perturbation stability II CS Homework 5
12:02/11 Load-frequency stability CS Homework 6 Mini-Project 2
13:09/11 Homework 6
Learning Outcome
• Understand fundamental concepts in power system analysis, namely, power flow, optimal power flow, state estimation, transient stability, small-perturbation stability, load-frequency control.
• Apply fundamental to solve application problems.– Formulate the problem– Design the experiment– Draw conclusion
• Understand and evaluate research papers using fundamental concepts.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 6
Today’s Outline
• Overview of power systems
• Component modeling
• Network modeling
• Network solution
• IEEE test system
• Homework 1
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 7
Reading Materials
• MATLAB tutorial• Background reading
– Chapter 2 Basic Principle, “Power Systems Analysis 2nd edition” by Arthur R. Bergen and Vijay Vittal.
– Chapter 2 Fundamentals, “Power System Analysis and Design 4th edition”by J. Duncan Glover, Mulukutla S. Sarma, Thomas J. Overbye.
• Today’s material– Chapter 9.1-9.3, Network Matrices, “Power Systems Analysis
2nd edition” by Arthur R. Bergen and Vijay Vittal.
• Supplementary reading– Chapter 1 Fundamentals of Electric Power System by Xiao-Ping
Zhang
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 8
OVERVIEW OF POWER SYSTEMS
Main components in power systems
One line diagram
Power system operation and control
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 9
Main Components of a Power System
• Generation (11 – 36 KV)
• Transmission and distribution (110 – 765 KV)
• Load (0.12 – 138 KV)
– Industrial customer (23 – 138 KV)
– Commercial customer (4.16 – 34.5 KV)
– Residential customer (120 – 240 V)
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 10
Taken from FUNDAMENTALS OF ELECTRIC POWER SYSTEMS by Xiao - Ping Zhang
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 11
Balanced 3Ф Circuit
• Why 3-phase? – More efficient use of equipment
and materials: 3 conductors instead of 6.
– Saving in losses
• Any pair of voltage sources differ by 120° with equal impedance
• 2 sequences, positive and negative
• In practice, phase sequence depends on how we label the wires.
RI 2
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 12
01anV
1201bnV
1201cnV
Positive sequence,
abc
01anV
1201bnV
1201cnV
Negative sequence,
acb
A One-line Diagram
• Show the interconnections of a transmission system
– Generator
– Load
– Transmission line
– Transformer
• 3Ф circuit
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 13
Power System Operation and Control
• Power flow analysis is the fundamental tool for,
– Operational planning
– Operation control
– Security analysis
– Power system planning
• Security-constrained economic dispatch
• Optimal power flow
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 14
Energy Management System
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 15
COMPONENT MODELING
Generator
Transmission line
Transformer
Load
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 16
Generator
• Simple steady-state equivalent circuit.
• Governor controls valve which, as a result, controls constant power.
• Generator field current controls constant voltage magnitude.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 17
+
~-
aE
ii DD jQP
Gzii GG jQP
Load
• Steady-state mode, three types:-
– Constant power (kVA with pf) such as motor load.
– Constant current (A) such as welding machine with constant current source.
– Constant impedance (Ohm) such as lighting.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 18
ii DD jQP
iDI
iDz
Transmission Line Model
• Medium-length model (Π-equivalent circuit)
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 19
1I 2I
1V 2V2
y
2
y
z
+
-
+
-
2
1
2
1
1
2
1
11
2V
V
z
y
z
zz
y
I
I
Z is the series impedance of the line = R + jX (ohm)
Y is the total shunt admittance of the line = -jB (mho)
Transformer Model
• Transformer (except phase shifting transformer)
• Caution: usually the transformer parameter is given as ‘per unit’. If there is a change of base, this value needs to be adjusted accordingly.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 20
1I 2I
1V 2V
p.u.y
+
-
+
-
2
1
p.u.p.u.
p.u.p.u.
2
1
V
V
yy
yy
I
I
A Tap-Changing Transformer
• A tap-changing transformer,
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 21
2
12
2
1
V
V
ya
ya
y
a
y
I
I
1I 2I
1V 2V
+
-
+
-
a:1p.u.y
1I 2I
1V 2V
2p.u.
1
a
ay
+
-
+
-
a
yp.u.
a
ay
1p.u.
What will happen if the we have phase-shift transformer instead of a tap-changing transformer?
A Transmission Line with Transformer
• Transmission line with transformer
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 22
2
122
2
1
1
2
111
11
2
11
V
V
az
y
za
a
az
azaza
y
za
a
I
I
1I 2I
1V 2V
z
+
-
+
-
a:1
2
y
2
y
1I 2I
1V 2V
+
-
+
-
az
1
22 2
11
a
y
za
a
2
11 y
za
a
NETWORK MODELING
Node voltage equation
Bus admittance matrix
Bus admittance matrix by inspection
Example
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 23
Node Voltage Equation
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 24
qp
qp
pp Vz
Vz
yz
VVyVI
1111
qp
pq
qq Vz
yVzz
VVyVI
1122
q
p
q
p
V
V
zy
z
zzy
I
I
11
11
2
1
pI qI
pV qV1y 2y
z
+
-
+
-
Z = impedance (R+jX) and Y = admittance
Bus Admittance Matrix
• : Bus admittance matrix
• Matrix form of node voltage equation,
where = Vector of injected node current
= Vector of node voltage
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 25
q
p
q
p
V
V
zy
z
zzy
I
I
11
11
2
1
pI qI
pV qV1y 2y
z
+
-
+
-
VYI bus
I
busY
V
busY
Bus Admittance Matrix by Inspection
• Symmetric matrix:
• Diagonal entries:= Sum of the admittance of all components
connected to node i.
• Off-diagonal entries:= Negative of the admittance of all components
connected between node i and j.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 26
kk,busY
mk,busY
kmmk ,, busbus YY
Can Y-bus be non-symmetric?
Bus Admittance Matrix: Example
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 27
G1
G2
Load
1 2
3 4
12y
13y 24y
34y
loady
1Gz
2GzZ- impedanceY- admittance
NETWORK SOLUTION
Motivation
Triangular Factorization
Gaussian elimination
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 28
Motivation
• Given Ybus, nodal voltage equation is,
• Our goal is to find node voltage magnitude and angle.
• Different operating condition leads to different external sources (current injection).
• Most of the time Y bus remains constant.
• To find ‘V’, (not so) simple Y bus inversion?
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 29
VYI bus
Numerical Solution of Linear Equations
• Matrix inversion is NOT an easy job for a large dimension problem.
• Common computationally efficient algorithm:-– Triangular factorization
L is lower triangular matrix
U is upper triangular matrix.
– Applicable to square matrix, not necessarily symmetric.
– Together with Gaussian elimination
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 30
LUY bus
Solution Procedure
• From ,
• First we find,
• Then,
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 31
LUY bus
LUVVYI bus
VLI~
UVV ~
Backward substitution
Forward substitution
Lower and Upper Triangular Matrix
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 32
33
2322
131211
3231
21
00
0
1
01
001
u
uu
uuu
ll
lLUM
Work backward to find the l’s and u’s elements
Y21(1) ← Y21(0)/Y11(0)Y31(1) ← Y31(0)/Y11(0)Y22(1) ← Y22(0) – [Y21(0)Y12(0)]/Y11(0)Y23(1) ← Y23(0) – [Y21(0)Y13(0)]/Y11(0) Y32(1) ← Y32(0) – [Y31(0)Y12(0)]/Y11(0) Y33(1) ← Y33(0) – [Y31(0)Y13(0)]/Y11(0)
2nd iterationY32(2) ← Y32(1)/Y22(1)Y33(2) ← Y33(1) – [Y32(1)Y23(1)]/Y22(1)
333231
232221
131211
ull
uul
uuu
Y
Calculate new value and overwrite the original Y bus to save memory
3323321331223212311131
2313212212211121
131211
uululululul
uuluulul
uuu
M
From original Y bus
1st iteration
Triangular Factorization Algorithm
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 33
END
kk
ikik
Y
YY
kk
kiikijij
Y
YYYY
i = k+1, …, n
i, j = k+1, …, n
START, k = 1
0kkY No
Yes
k = k+1k = n
Yes
No
Y bus is n x n matrix
When will the algorithm be unstable and how to prevent it?
Triangular Factorization Example
• Find L and U of the following matrix.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 34
513
123
235
A
Ill-Conditioned Y bus Matrix
• If Ykk is very small, the algorithm may be unstable.• Can be fix by permutation. • In addition, the computation time depends on number
of non-zero elements, permutation may help to reduce this number.
• For more information, read:– FERNANDO L. ALVARADO, WILLIAM F. TINNEY, and MARK
K. ENNS, “SPARSITY IN LARGE-SCALE NETWORK COMPUTATION”, Advances in electric Power and Energy Conversion System Dynamics and Control,” Academic Press, 1991, C. T. Leondes, editor (with permission). Corrections 15 Feb 93.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 35
NETWORK REDUCTION
Motivation
Kron reduction
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 36
Motivation
• Consider the matrix in this example.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 37
G1
G2
Impedance Load
1 2
3 4
What is the current injection to node 2 and 3?
4
3
2
1
44434241
34333231
24232221
14131211
4
1
0
0
V
V
V
V
yyyy
yyyy
yyyy
yyyy
I
I
KRON Reduction
• Eliminate node with zero injection to reduce the size of Y bus matrix.
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 38
3
2
1
333231
232221
131211
2
1
0 V
V
V
yyy
yyy
yyy
I
I
2
1
33
322322
33
311321
33
321312
33
311311
2
1
V
V
y
yyy
y
yyy
y
yyy
y
yyy
I
I
2
33
321
33
313 V
y
yV
y
yV
kk
kjik
ij
new
ijy
yyyy To eliminate node k: i,j = I,2, .. ,n,
IEEE TEST SYSTEM
Test case archive
IEEE common data format
14 bus test system
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 39
Power Systems Test Case Archive
• Managed by Richard D. Christie, an Associate Professor at the University of Washington, Seattle, Washington, USA
• http://www.ee.washington.edu/research/pstca/
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 40
Thanks!, Richard.
IEEE Common Data Format
• http://www.ee.washington.edu/research/pstca/formats/cdf.txt
– Title data
– Bus data
– Branch data
– Loss zone data
– Interchange data
– Tie line data
• Data type codes:
– A - Alphanumeric (no special characters)
– I - Integer
– F - Floating point
– * - Mandatory item
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 41
14-Bus Common Data Format
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 42
Title data
Bus data
Branch data
Title Data
• Columns 2- 9 Date, in format DD/MM/YY with leading zeros. If no date provided, use 0b/0b/0b where b is blank.
• Columns 11-30 Originator's name (A)
• Columns 32-37 MVA Base (F*)
• Columns 39-42 Year (I)
• Column 44 Season (S - Summer, W - Winter)
• Column 46-73 Case identification (A)
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 43
Date Originator’s name MVA base Year Season Case Identification
Bus Data
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 44
Bus number
Name (left justify)
Type of bus, 0-load bus, 2-generator bus, 3 swing bus
Final voltage (pu) and angle (degree)
Load P (MW)
Load Q (MVAR)
Generation (MW) Generation (MVAR)
Controlled voltage (pu) Shunt susceptance B (pu)
Branch Data
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 45
Branch resistance R (pu)
Branch reactance X (pu)
Line charging B (pu) (total line charging, for each bus, this has to be divided by two)
Transformer turns ratio
Tap bus number
Z bus number
7V4V
+
-
+
-
0.978:1 j0.20912
Example:
Homework 1: 14-Bus Test System
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 46
Find Ybus of this system
MATLAB Basics
• http://www.mathworks.com/help/techdoc/learn_matlab/bqr_2pl.html
• Read
– Matrices and Arrays: http://www.mathworks.com/help/techdoc/learn_matlab/f2-8955.html
– Programming:http://www.mathworks.com/help/techdoc/learn_matlab/f4-8955.html
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 47
Next Lecture
• Power flow equations
• Power flow problem
• Iterative solution techniques
• N-R Application to power flow problem
• Mini-project I
8/11/2011 EE5702R Advance Power System Analysis:: Introduction by P. Jirutitijaroen 48