Lec1-Advancne Power System Ybus

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


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.


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.


Tentative Syllabus


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.


Today’s Outline

• Overview of power systems

• Component modeling

• Network modeling

• Network solution

• IEEE test system

• Homework 1


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


OVERVIEW OF POWER SYSTEMS

Main components in power systems

One line diagram

Power system operation and control


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)


Taken from FUNDAMENTALS OF ELECTRIC POWER SYSTEMS by Xiao - Ping Zhang


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


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


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


Energy Management System


COMPONENT MODELING

Generator

Transmission line

Transformer

Load


Generator

• Simple steady-state equivalent circuit.

• Governor controls valve which, as a result, controls constant power.

• Generator field current controls constant voltage magnitude.


+

~-

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.


ii DD jQP

iDI

iDz

Transmission Line Model

• Medium-length model (Π-equivalent circuit)


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.


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,


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


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


Node Voltage Equation


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


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.


kk,busY

mk,busY

kmmk ,, busbus YY

Can Y-bus be non-symmetric?

Bus Admittance Matrix: Example


G1

G2

Load

1 2

3 4

12y

13y 24y

34y

loady

1Gz

2GzZ- impedanceY- admittance

NETWORK SOLUTION

Motivation

Triangular Factorization

Gaussian elimination


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?


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


LUY bus

Solution Procedure

• From ,

• First we find,

• Then,


LUY bus

LUVVYI bus

VLI~

UVV ~

Backward substitution

Forward substitution

Lower and Upper Triangular Matrix


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


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.


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.


NETWORK REDUCTION

Motivation

Kron reduction


Motivation

• Consider the matrix in this example.


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.


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


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/


Thanks!, Richard.

http://www.ee.washington.edu/research/pstca/



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


http://www.ee.washington.edu/research/pstca/formats/cdf.txt



14-Bus Common Data Format


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)


Date Originator’s name MVA base Year Season Case Identification

Bus Data


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


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


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


http://www.mathworks.com/help/techdoc/learn_matlab/bqr_2pl.html

http://www.mathworks.com/help/techdoc/learn_matlab/bqr_2pl.html

http://www.mathworks.com/help/techdoc/learn_matlab/f2-8955.html








Next Lecture

• Power flow equations

• Power flow problem

• Iterative solution techniques

• N-R Application to power flow problem

• Mini-project I


Documents

Lec1-Advancne Power System Ybus