Fundamental Principles & Methods

Competency Training and Certification Program in Electric Power Distribution System Engineering

Certificate in

Power System Modeling and Analysis Training Course in

Fundamental Principles and Methods in Power System Analysis

U. P. NATIONAL ENGINEERING CENTER NATIONAL ELECTRIFICATION ADMINISTRATION

Training Course in Fundamental Principles and Methods in Power System Analysis

Course Outline 1. Circuit Conventions and Notations 2. Power System Representation 3. Per Unit Quantities 4. Symmetrical Components 5. Network Equations and Methods of Solution

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

2

Training Course in Fundamental Principles and Methods in Power System Analysis

Circuit Conventions & Notations

Voltage and Current Directions

Double Subscript Notation

Voltage, Current and Phasor Notation

Complex Impedance and Phasor Notation

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

3

Training Course in Fundamental Principles and Methods in Power System Analysis

4

Circuit Conventions & Notations Voltage and Current Directions Polarity Marking of Voltage Source Terminals: Plus sign (+) for the terminal where positive current comes out

+

I

+ Specification of Load Terminals: Plus sign (+) for the terminal whereVs positive current enters Specification of Current Direction: Arrows for the positive current (i.e., from the source towards the load)

U. P. National Engineering Center National Electrification Administration

ZA

I

a

VA I

b

-

I +

VB

ZB -

o

n

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Circuit Conventions & Notations Double Subscript Notation The letter subscripts on a voltage indicate the nodes of the circuit between which the voltage exists. The first subscript denotes the voltage of that node with respect to the node identified by the second subscript.

ZA

a

+ +

VS = Vao

VA I = Iab

-

b

-

+

Zb

Vb = Vbn -

o U. P. National Engineering Center National Electrification Administration

n

The current direction is from first subscript to the second subscript .

Vao - IabZA - Vbn = 0 Vao - Vbn Iab = ZA

Competency Training & Certification Program in Electric Power Distribution System Engineering

5

Training Course in Fundamental Principles and Methods in Power System Analysis

6

Circuit Conventions & Notations Voltage, Current and Phasor Notation 1

V=

v 0.5

i 0 -4

π

-2

0

π

2

4

π

π

Vm ∠0o volts 2

Im I= ∠ − θ o amperes 2

-0.5

-1

v = Vmε i = I mε

jω t j (ω t − θ )

V

θ

ε

j ωt

j =

= cos ωt + j sin ωt

I

−1

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

7

Circuit Conventions & Notations Complex Impedance and Phasor Notation + +

VS = Vmε

VL

i(t)

jω t

R (Resistance)

-

L (Inductance)

-

Applying Kirchoff’s voltage law,

di ( t ) = Vmε Ri ( t ) + L dt

jω t

The first order linear differential equation has a particular jω t i ( t ) = K ε solution of the form .

Hence, RK ε

jω t

+ j ω LK ε

U. P. National Engineering Center National Electrification Administration

jω t

= Vmε

jω t

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

8

Circuit Conventions & Notations Complex Impedance and Phasor Notation Solving for the current

Vm ε i( t ) = R + jω L

jω t

Dividing voltage by current to get the impedance,

v( t ) Z = = i( t )

V m ε jω t Vm ε R + jω L

jω t

= R + jω L

Therefore, the impedance Z is a complex quantity with real part R and an imaginary (j) part ωL U. P. National Engineering Center National Electrification Administration

ωL

Z

θ R

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Circuit Conventions & Notations Complex Impedance and Phasor Notation + +

-

-

For Capacitive Circuit,

1 Z = R − j( ) ωC Z= |Z|ejφφ

VL

or

.1 ωC

C (Capacitance) R

φ

VS = Vmε

i(t)

jω t

R (Resistance)

Z

Z = |Z|(cosφ φ + jsinφ φ) or Z = |Z|∠φ ∠φ

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

9

Training Course in Fundamental Principles and Methods in Power System Analysis

10

Circuit Conventions & Notations Complex Impedance and Phasor Notation Z= |Z|ejφφ

or

Z = |Z|(cosφ φ + jsinφ φ) or Z = |Z|∠φ ∠φ

v = 141.4 cos(ω ωt + 30°) volts i = 7.07 cos(ω ωt) amperes Vmax = 141.4 |V| = 100 V = 100∠ ∠30 Imax = 7.07

|I| = 5

100 ∠ 30 Z = = 20 ∠ 30 5 ∠0

I = 5∠ ∠0

Z = 20(cos 30 + j sin 30) = 17.32 + j10 U. P. National Engineering Center National Electrification Administration

10

20 30°

17.32

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation

Electrical Symbols

Three-Line and Single-Line Diagram

Equivalent Circuit of Power System Components

Impedance and Admittance Diagram

Bus Admittance Matrix

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

11

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System representation Electrical Symbols G

Generator

Switch

Circuit Breaker or

Transformer Fuse Transmission or Distribution Line Bus U. P. National Engineering Center National Electrification Administration

Node

Competency Training & Certification Program in Electric Power Distribution System Engineering

12

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System representation 3-phase wye neutral ungrounded 3-phase delta connection 3-phase wye neutral grounded Current Transformer Potential Transformer

U. P. National Engineering Center National Electrification Administration

V

Voltmeter

A

Ammeter

R

Protective Relay

Competency Training & Certification Program in Electric Power Distribution System Engineering

13

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation Three Line Diagram

A

B

C

Circuit Recloser

Main Bus

Circuit Breaker

The three-line diagram is used to represent each phase of a threeTransformer phase power system.

CTs

R R

Distribution Lines

Relays

R

R U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

14

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation Single Line Diagram The three-line diagram becomes rather cluttered for large power systems. A shorthand version of the three-line diagram is referred to as the Single Line Diagram.

Bus CB

Transformer Distribution Line Recloser

R CT and Relay

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

15

Training Course in Fundamental Principles and Methods in Power System Analysis

16

Power System Representation Equivalent Circuit of Power System Components: Generator a Ia

Za

R a + jX

s

Ea

+

Eb Zb

Ec

Zc

Ib

Ia

Eg b

Ic

-

+

Va -

c

Three-Phase Equivalent U. P. National Engineering Center National Electrification Administration

Single-Phase Equivalent Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

17

Power System Representation Equivalent Circuit of Power System Components: Transformer Primary A B C

Y11 Y12 Y13 Y14 Y15 Y16

Secondary a

Y21 Y22 Y23 Y24 Y25 Y26 Y31 Y32 Y33 Y34 Y35 Y36

b

Y41 Y42 Y43 Y44 Y45 Y46

c

Y51 Y52 Y53 Y54 Y55 Y56 Y61 Y62 Y63 Y64 Y65 Y66

6x6 Admittance Matrix Core Loss

Three-Phase Equivalent U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

18

Power System Representation Equivalent Circuit of Power System Components: Transformer R H + jX H

v Iex

+

r VH

Rc

a 2 R X + ja 2 X X

jX m

r IH

+

r aV X

RT = R H + a 2 R X 2 XT = X H + a X X ZT

-

-

Single-Phase Equivalent

U. P. National Engineering Center National Electrification Administration

+

+

-

-

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation Equivalent Circuit of Power System Components: Transmission & Distribution Lines T&D Lines can be represented by an infinite series of resistance and inductance and shunt capacitance.

Δl

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

19

Training Course in Fundamental Principles and Methods in Power System Analysis

20

Power System Representation Equivalent Circuit of Power System Components: Distribution Lines

Equivalent π-Network

A B C

1/2

Yaa

Yab

Yac

Yba

Ybb

Ybc

Yca

Ycb

Ycc

Capture Unbalanced Characteristics U. P. National Engineering Center National Electrification Administration

Zaa

Zab

Zac

Zba

Zbb

Zbc

Zca

Zcb

Zcc

1/2

a b c Yaa

Yab

Yac

Yba

Ybb

Ybc

Yca

Ycb

Ycc

Three-Phase Equivalent Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

21

Power System Representation Equivalent Circuit of Power System Components: Long Transmission Lines Length = Longer than 240 km. (150 mi.)

Z ' = Z c sinh γ l

+• Vs

γl Y' 1 = tanh 2 2 Zc

IS

-•

Characteristic Impedance

ZC =

z y

U. P. National Engineering Center National Electrification Administration

• + Y' 2

IR

VR

• -

Propagation Constant

γ=

zy

Single-Phase Equivalent Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

22

Power System Representation Equivalent Circuit of Power System Components: Medium-Length Transmission Lines Length = 80 – 240 km. (50 - 150 mi.)

Z = (r + jx L )l

+• Vs

-•

IS

Y 1 / ωc = 2 2

• + Y 2

IR

VR

• -

Single-Phase Equivalent U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation Equivalent Circuit of Power System Components: Short Transmission Lines Length = up to 80 km. (50 mi.)

+•

Z = (r + jx L )l

Vs

-•

Is = IR

• + VR

• Single-Phase Equivalent

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

23

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation Single Phase Equivalent of Balanced Three-Phase System c

Ic

c

Eco = |E| ∠120° V o

a

Ia

ZR a

n

Eao = |E|∠0° V

ZR

Ebo= |E| ∠240° V b

ZR

b

Ib U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

24

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation a

o

Ia

a

Eao = |E|∠ 0° V

E∠0 Ia = = I∠ − θ Z R ∠θ

n

ZR

o

n

ZR

b c

Ebo= |E| ∠240° V Eco = |E| ∠120° V

E∠240 Ib = = I∠( 240 − θ ) Z R ∠θ

ZR

Ib

b c

Ic ZR

o

E∠120 Ic = = I∠( 120 − θ ) Z R ∠θ

n

ZR U. P. National Engineering Center National Electrification Administration

Note: Currents are Balanced

Competency Training & Certification Program in Electric Power Distribution System Engineering

25

Training Course in Fundamental Principles and Methods in Power System Analysis

26

Power System Representation Single Phase c Representation of a Balanced Three-Phase System b

Ic Eco = |E| ∠120° V

c

ZR

Ia a

o

a

n

Eao = |E|∠0° V

ZR

Ebo= |E| ∠240° V Ia

a

Ia

Eao = |E|∠ 0° V o U. P. National Engineering Center National Electrification Administration

ZR b

a

ZR

n Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

27

Power System Representation Impedance and Admittance Diagrams a Bus

Gen

Line

1 2 3 4

a b c

1-3 2-3 1-4 2-4 3-4

c

1

b

3

2

4 Single Line Diagram U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

28

Power System Representation Impedance and Admittance Diagrams 0

Ea

Ea

za 1

za

Eb

zc

Generator

1

zb 3

zd

0

zf

z13 1

Ec

2

ze zh

zg

3 Line 4 0

U. P. National Engineering Center National Electrification Administration

Impedance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation Impedance and Admittance Diagrams

Zg Eg

VL

VL

+

-

IL

Is

Zp

IL

Eg = ISZP Zg = Zp

The two sources will be equivalent if VL and IL are the same for both circuits.

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

29

Training Course in Fundamental Principles and Methods in Power System Analysis

30

Impedance and Admittance Diagrams a Bus

Gen

Line

1 2 3 4

a b c

1-3 2-3 1-4 2-4 3-4

c

1

b

3

2

4 Single Line Diagram U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

31

Impedance and Admittance Diagrams 0

Ea

Ea

za 1

za

Eb

zc

Generator

1

zb 3

zd

0

zf

z13 1

Ec

2

ze zh

zg

3 Line 4 0

U. P. National Engineering Center National Electrification Administration

Impedance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Impedance and Admittance Diagrams Equivalent Sources

Zg Eg

VL

VL

+

-

IL

Is

Zp

IL

Eg = ISZP Zg = Zp

The two sources will be equivalent if VL and IL are the same for both circuits.

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

32

Training Course in Fundamental Principles and Methods in Power System Analysis

33

Impedance and Admittance Diagrams 0

I1 = Ea/za y01 = 1/za I2 = Eb/zb y02 = 1/zb

I1

y01 1

y13

I3 = Ec/zc y03 = 1/zc

I3

y03 3

y14 = 1/zf

y23 = 1/ze

y24 = 1/zg y34 = 1/zh

U. P. National Engineering Center National Electrification Administration

y02 2

y23

y14

y13 = 1/zd

I2

y24 4

Admittance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Nodal Voltage Equations Applying Kirchoff’s Current Law

at node 1: I 1 = V 1 y 01 + (V 1 − V 3 ) y 13 + (V 1 − V 4 ) y 14

at node 2: I 2 = V 2 y 02 + (V 2 − V 3 ) y 23 + (V 2 − V 4 ) y 24 at node 3:

I 3 = V 3 Y03 + (V 3 − V 2 ) y 23 + (V 3 − V 4 ) y 34 + (V 3 − V 1 ) y 13

at node 4: 0 = (V 4 − V1 ) y 14 + (V 4 − V 2 ) y 24 + (V 4 − V 3 ) y 34 U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

34

Training Course in Fundamental Principles and Methods in Power System Analysis

Nodal Voltage Equations Rearranging the equations,

I 1 = V1 ( y 01 + y13 + y14 ) − V 3 y13 − V 4 y14 I 2 = V 2 ( y 02 + y 23 + y 24 ) − V 3 y 23 − V 4 y 24 I 3 = V 3 ( y 03 + y 23 + y 34 + y 13

0 = V 4 ( y 14 + y 24 + y 34

)−

) − V 2 y 23

− V 4 y 34 − V 1 y 13

V 1 y 14 − V 2 y 24 − V 3 y 34

In matrix form,

0 - y13 - y14 ⎡I1 ⎤ ⎡y01 + y13 + y14 ⎤⎡V1 ⎤ ⎥ ⎢ ⎢I ⎥ ⎢ ⎥ 0 y02 + y23 + y24 - y23 - y24 ⎢ 2⎥ = ⎢ ⎥⎢V2 ⎥ ⎢I3 ⎥ ⎢ ⎥⎢V3 ⎥ - y13 - y23 y03 + y23 + y34 + y13 - y34 ⎢ ⎥ ⎢ ⎥⎢ ⎥ y y y y y y − − − + + 14 24 34 14 24 34 ⎦⎣V4 ⎦ ⎣0 ⎦ ⎣ U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

35

Training Course in Fundamental Principles and Methods in Power System Analysis

Nodal Voltage Equations ⎡I1 ⎤ ⎢ ⎥ ⎢I2 ⎥ ⎢I3 ⎥ = ⎢ ⎥ ⎢M ⎥ ⎢I ⎥ ⎣ n⎦

⎡ Y 11 ⎢ ⎢ Y 21 ⎢ ⎢ Y 31 ⎢ ⎢M ⎢ ⎢Y n 1 ⎣

Y 12 Y 22 Y 32 M Yn2

Y 13 L Y 1 n ⎤ ⎡V 1 ⎤ ⎥⎢ ⎥ ⎥ V2 ⎥ Y 23 L Y 2 n ⎥⎢ Y 33 L Y 3 n ⎥ ⎢V 3 ⎥ ⎥⎢ ⎥ M M ⎥⎢ M ⎥ ⎥ Y n 3 L Y nn ⎥ ⎢V ⎥ ⎦⎣ n⎦

[ I ] = [Ybus][V] Yii = self-admittance, the sum of all admittances terminating on the node (diagonal elements) Yij = mutual admittance, the negative of the admittances connected directly between the nodes identifed by the double subscripts U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

36

Training Course in Fundamental Principles and Methods in Power System Analysis

Power System Representation

[YBUS] =

⎡ Y 11 ⎢ ⎢ Y 21 ⎢ ⎢ Y 31 ⎢ ⎢M ⎢ ⎢Y n 1 ⎣

Y 12 Y 22 Y 32 M Yn2

Y 13 L Y 1 n ⎤ ⎥ Y 23 L Y 2 n ⎥ ⎥ Y 33 L Y 3 n ⎥ ⎥ M M ⎥ ⎥ Y n 3 L Y nn ⎥ ⎦

Ybus is also called Bus Admittance Matrix

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

37

Training Course in Fundamental Principles and Methods in Power System Analysis

Per Unit Quantities

The Per Unit System

Per Unit Impedance

Changing Per Unit Values

Consistent Per Unit Quantities of Power System

Advantages of Per Unit Quantities

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

38

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Per Unit Value Per Unit Value

=

Actual Value Base Value

Per-unit Value is a dimensionless quantity Per-unit value is expressed as decimal

Percent U. P. National Engineering Center National Electrification Administration

=

Actual Value 100

Competency Training & Certification Program in Electric Power Distribution System Engineering

39

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Per Unit Value Per Unit Power

Per Unit Voltage

Per Unit Current

Per Unit Impedance

=

Actual Value of Power

=

Actual Value of Voltage

=

Actual Value of Current

=

Actual Value of Impedance

U. P. National Engineering Center National Electrification Administration

Base Power

Base Voltage

Base Current

Base Impedance Competency Training & Certification Program in Electric Power Distribution System Engineering

40

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Per Unit Calculations PU Voltage PU Current

= PU Impedance

PU Power = PU Voltage x PU Current

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

41

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Per Unit Calculations I

Example: +

+ Zline = 1.4 ∠75° Ω Zload = 20 ∠30° Ω

Vs = ?

-

2540 ∠0° V

Determine Vs

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

42

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Per Unit Calculations Choose: Base Impedance = 20 ohms (single phase) Base Voltage = 2540 volts (single phase) PU Impedance of the load = 20∠30°/20 = ______ p.u.

PU Impedance of the line = 1.4∠75°/20 = ______ p.u.

PU Voltage at the load = 2540∠0° /2540 = ______ p.u.

Line Current in PU = PU voltage / PU impedance of the load = ______ / ______ = ______ p.u.

PU Voltage at the Substation = Vload(pu) + IpuZLine(pu) = ________ + _______ x _______ = _______ p.u.

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

43

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Per Unit Calculations 1.0 ∠-30° p.u.

+

0.07 ∠75° p.u.

1.05∠2.70°

+

1.0∠30° p.u.

-

1.0∠0° p.u.

-

The magnitude of the voltage at the substation is 1.05 p.u. x 2540 Volts = _______ Volts U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

44

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Establishing Base Values 1. Base values must satisfy fundamental electrical laws (Ohm’s Law and Kirchoff’s Laws) 2. Choose any two electrical parameters • Normally, Base Power and Base Voltage are chosen 3. Calculate the other parameters • Base Impedance and Base Current

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

45

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Establishing Base Values For a Given Base Power and Base Voltage, Base Power Base Current = Base Voltage (Base Voltage)2

Base Voltage Base Impedance =

= Base Current

U. P. National Engineering Center National Electrification Administration

Base Power

Competency Training & Certification Program in Electric Power Distribution System Engineering

46

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Establishing Base Values For Single Phase System,

For Three Phase System,

Ibase =

Pbase(1φ) -----------Vbase(1φ)

Ibase =

Zbase =

Vbase(1φ) -----------Ibase(1φ)

Vbase(LN) Zbase = -----------Ibase(L)

[Vbase(1φ)]² = -----------Pbase(1φ) U. P. National Engineering Center National Electrification Administration

Pbase(3φ) -----------√3Vbase(LL)

[Vbase(LL)]² = -----------Pbase(3φ)

Competency Training & Certification Program in Electric Power Distribution System Engineering

47

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Establishing Base Values • Base Values can be established from Single Phase or Three Phase Quantities

Base MVA 1φ Base kV 1φ =

1 = Base MVA 3φ 3 Base kV LL 3

• Base MVA is the same base value for Apparent, Active and Reactive Power • Base Z is the same base value for Impedance, Resistance and Reactance U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

48

Training Course in Fundamental Principles and Methods in Power System Analysis

The Per Unit System Establishing Base Values Example: Base kVA3φφ

= 30,000 kVA = 30 MVA

Base kVA1φφ

= 10,000 kVA = 10 MVA

Base kVLL

= 120 kV

Base kVLN

= 69.282 kV

Base Z

= (69.282)2/10 = 480 ohms

= (120)2/30 = 480 ohms 30 x1000 Base Current = 3 ( 120 )

Base Z

= 144.34 Amps U. P. National Engineering Center National Electrification Administration

10 x1000 Base Current = 69.282 = 144.34 Amps Competency Training & Certification Program in Electric Power Distribution System Engineering

49

Training Course in Fundamental Principles and Methods in Power System Analysis

Per Unit Impedance Generators

Manufacturers provide the following impedance in per unit: 1.

Armature Resistance, Ra

2.

Direct-axis Reactances, Xd”, Xd’ and Xd

3.

Quadrature-axis Reactances, Xq”, Xq’ and Xq

4.

Negative Sequence Reactance, X2

5.

Zero Sequence Reactance, X0

}

Positive Sequence Impedances

The Base Values used by manufacturers are: 1.

Rated Capacity (MVA, KVA or VA)

2.

Rated Voltage (kV or V)

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

50

Training Course in Fundamental Principles and Methods in Power System Analysis

Per Unit Impedance Transmission and Distribution Lines

R( X

pu )

L ( pu )

= =

X C ( pu ) = U. P. National Engineering Center National Electrification Administration

R(立

)

Z

base

X

L( 立 )

Z base

X C( 立 ) Z base

Competency Training & Certification Program in Electric Power Distribution System Engineering

51

Training Course in Fundamental Principles and Methods in Power System Analysis

Per Unit Impedance Transformers

The ohmic values of resistance and leakage reactance of a transformer depends on whether they are measured on the high- or low-tension side of the transformer

The impedance of the transformer is in percent or per unit with the Rated Capacity and Rated Voltages taken as base Power and Base Voltages, respectively

The per unit impedance of the transformer is the same regardless of whether it is referred to the high-voltage or lowvoltage side

The per unit impedance of the three-phase transformer is the same regardless of the connection U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

52

Training Course in Fundamental Principles and Methods in Power System Analysis

Per Unit Impedance Example A single-phase transformer is rated 110/440 V, 2.5 kVA. The impedance of the transformer measured from the lowvoltage side is 0.06 ohms. Determine the impedance in per unit (a) when referred to low-voltage side and (b) when referred to high-voltage side

Solution 0.110 2 Low-voltage Zbase = 2.5 / 1000 PU Impedance, Zpu = U. P. National Engineering Center National Electrification Administration

= ______ ohms = ______ p.u.

Competency Training & Certification Program in Electric Power Distribution System Engineering

53

Training Course in Fundamental Principles and Methods in Power System Analysis

Per Unit Impedance If impedance had been measured on the high-voltage side, the ohmic value would be 2

⎛ 440 ⎞ Z = 0 . 06 ⎜ ⎟ = _______ ⎝ 110 ⎠

ohms

High Voltage, Zbase =

= _______ ohms

PU Impedance, Zpu =

= _______ p.u.

Note: PU value of impedance referred to any side of the transformer is the same U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values Example: Consider a three-phase transformer rated 20 MVA, 67 kV/13.2 kV voltage ratio and a reactance of 7%. The resistance is negligible. a) What is the equivalent reactance in ohms referred to the high voltage side? b) What is the equivalent reactance in ohms referred to the low voltage side? c) Calculate the per unit values both in the high voltage and low voltage side at 100 MVA. U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values SOLUTION: a) Pbase = 20 MVA Vbase = 67 kV (high voltage) ( kV)² Zbase = ( MVA)

= ________ ohms

Xhigh = Xp.u. x Zbase = _______ x _______= _______ ohms

U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values SOLUTION: b) Pbase = 20 MVA Vbase = 13.2 kV (low voltage) Zbase =

(

kV)²

(

MVA)

= ________ ohms

Xlow = Xp.u. x Zbase = _______ x _______= _______ ohms

U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values c) Pbase = 100 MVA Vbase,H = 67 kV (67)² Zbase,H = = ________ ohms 100 ohms Xp.u.,H = = ______ p.u. ohms Vbase,L = 13.2 kV (13.2)² Zbase,L = = _______ ohms 100 ohms Xp.u.,L = = ______ p.u. ohms

Note that the per unit quantities are the same regardless of the voltage level.

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Training Course in Fundamental Principles and Methods in Power System Analysis

59

Changing Per-Unit Values Three parts of an electric system are designated A, B and C and are connected to each other through transformers, as shown in the figure. The transformer are rated as follows: A-B B-C SOURCE

10 MVA, 3φ, 13.8/138 kV, leakage reactance 10% 10 MVA, 3φ, 138/69 kV, leakage reactance 8% A

B A-B

C B-C

LOAD 300 Ω/ φ PF=100 %

Determine the voltage regulation if the voltage at the load is 66 kV. U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

60

Changing Per-Unit Values Solve using actual quantities XAB=10%

XBC=8%

Vc

VA A-B 13.8/138 kV

U. P. National Engineering Center National Electrification Administration

300 Ί/ φ PF=100%

B-C 138/69 kV

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values SOLUTION USING PER UNIT METHOD: Pbase = 10 MVA VA,base = 13.8 kV VB,base = 138 kV VC,base = 69 kV ZLOAD,p.u. =

(69)² ZC,base = -------- = _____ ohms 10

---------- = ______ + j _____ p.u.

VC,p.u. = -------------- = _______ p.u.

Ip.u. = ---------------- = _______ p.u.

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values VA = _______ + ( ________ ) x ( ________ + ________ ) = ________ + j ________ p.u. = _________ p.u.

VNL - VL V.R. = ---------------- x 100% VNL V.R. = ------------------------ x 100% =

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values Consider the previous example, What if transformer A-B is 20 MVA instead of 10 MVA. The transformer nameplate impedances are specified in percent or per-unit using a base values equal to the transformer nameplate rating.

The PU impedance of the 20 MVA transformer cannot be added to the PU impedance of the 10 MVA transformer because they have different base values The per unit impedance of the 20 MVA can be referred to 10 MVA base power U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values Convert per unit value of 20 MVA transformer, Pbase = 20 MVA (Power Rating) Vbase,H = 138 kV (Voltage Rating) (138)² Zbase,H = ---------- = _______ ohms 20 Xactual,H = 0.10 p.u. x _______ ohms = _______ ohms At Pbase = 10 MVA (new base) (138)² Zbase,H = ---------- = 1904.4 ohms 10 95.22 Xp.u.(new) = ---------- = 0.05 p.u.

1904.4 U. P. National Engineering Center National Electrification Administration

The per unit impedance of the 20MVA and 10 MVA transformer can now be added.

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values Note that the transformer can have different per unit impedance for different base values (i.e., the actual ohmic impedances of the equipment is independent of the selected base values), then

Zactual = Zpu1 • Zbase1

Zactual = Zpu2 • Zbase2

Z pu 1 ⋅ Z base 1 = Z pu 2 ⋅ Z base 2 Z base Z pu 2 = Z pu 1 ⋅ Z base U. P. National Engineering Center National Electrification Administration

1 2

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Training Course in Fundamental Principles and Methods in Power System Analysis

66

Changing Per-Unit Values Recall:

Z base

( base =

voltage

base Power

(kV

Then,

Z

pu 2

= Z

LL , base 1

MVA

pu 1

(kV

Z pu 2

3 φ , base 1 2 LL , base 2

⎛ kV LL ,base 1 = Z pu 1 ⎜⎜ ⎝ kV LL ,base 2

U. P. National Engineering Center National Electrification Administration

)

2

MVA or,

)

2

)

3 φ , base 2

⎞ ⎟ ⎟ ⎠

2

⎛ MVA 3 φ ,base 2 ⎜ ⎜ MVA 3 φ ,base 1 ⎝

⎞ ⎟ ⎟ ⎠

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values

Z p .u .( new ) = Z p .u .( given ) x

[ Vbase( given ) ] [ Vbase( new ) ]

2

2

x

Pbase( new ) Pbase( given )

Example A three-phase transformer is rated 400 MVA, 220Y/22 Δ kV. The impedance measured on the low-voltage side of the transformer is 0.121 ohms (approx. equal to the leakage reactance). Determine the per-unit reactance of the transformer for 100 MVA, 230 kV base values at the high voltage side of the transformer.

U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Changing Per-Unit Values Solution On its own base the transformer reactance is (

X=

) (

)2

(

)

= ________ pu

On the chosen base the reactance becomes X=(

)x

(

)2

(

)2

x

U. P. National Engineering Center National Electrification Administration

(

= )________ pu

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Consistent Per Unit Quantities of Power System Procedure: a)

Establish Base Power and Base Voltages • Declare Base Power for the whole Power System • Declare Base Voltage for any one of the Power System components • Compute the Base Voltages for the rest of the Power System Components using the voltage ratio of the transformers

Note: Define each subsystem with unique Base Voltage based on separation due to magnetic coupling U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Consistent Per Unit Quantities of Power System b) Compute Base Impedance and Base Current • Using the Declared Base Power and Base Voltages, compute the Base Impedances and Base Currents for each Subsystem c) Compute Per Unit Impedance • Using the declared and computed Base Values, compute the Per Unit values of the impedance by: Dividing Actual Values by Base Values Changing Per Unit Impedance with change in Base Values U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Consistent Per Unit Quantities of Power System Use Base Power = 100 MVA T1

T2

G2

L1

G1

G3 Generator 1 (G1):

300 MVA; 20 kV; 3φ; Xd” = 20 %

Transmission Line(L1):

64 km; XL = 0.5 Ω / km

Transformer 1 (T1): 3φ; 350 MVA; 230 / 20 kV; XT1 = 10 % Transformer 2 (T2): 3-1φ; 100 MVA; 127 / 13.2 kV; XT2 = 10 % Generator 2 (G2): 200 MVA; 13.8kV, Xd” = 20 % Generator 3 (G3): 100 MVA; 13.8kV, Xd” = 20 % U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

72

Consistent Per Unit Quantities of Power System T1

T2

G2

Transmission Line

G1 G3 1

XT1 2

XLINE

3

XT2 4

XG1 E1

U. P. National Engineering Center National Electrification Administration

XG2 E2

XG3 E3

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Consistent Per Unit Quantities of Power System a) & b) Establish Base Power, Base Voltages, Base Impedance, and Base Current Base Power: SubSystem

Vbase (kV)

U. P. National Engineering Center National Electrification Administration

Zbase (Ohm)

Ibase (Amp)

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Consistent Per Unit Quantities of Power System c) Compute Per Unit Impedance G1: G2: G3: L1: T1: T2: U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Advantages of Per-Unit Quantities The computation for electric systems in per-unit simplifies the work greatly. The advantages of Per Unit Quantities are: 1. Manufacturers usually specify the impedances of equipments in percent or per-unit on the base of the nameplate rating.

2. The per-unit impedances of machines of the same type and widely different rating usually lie within a narrow range.

When the impedance is not known definitely, it is generally possible to select from tabulated average values.

U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Advantages of Per-Unit Quantities 3. When working in the per-unit system, base voltages can be selected such that the per-unit turns ratio of most transformers in the system is equal to 1:1.

4. The way in which transformers are connected in threephase circuits does not affect the per-unit impedances of the equivalent circuit, although the transformer connection does determine the relation between the voltage bases on the two sides of the transformer.

5. Per unit representation yields more meaningful and easily correlated data.

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Training Course in Fundamental Principles and Methods in Power System Analysis

Advantages of Per-Unit Quantities 6. Network calculations are done in a much more handier fashion with less chance of mix-up • between phase and line voltages • between single-phase and three-phase powers, and • between primary and secondary voltages.

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Symmetrical Components

Sequence Components of Unbalanced Phasor

Sequence Impedance of Power System Components

Practical Implications of Sequence Components of Electric Currents

U. P. National Engineering Center National Electrification Administration

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78

Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor In a balanced Power System, Generator Voltages are three-phase balanced Line and transformer impedances are balanced Loads are three-phased balanced Single-Phase Representation and Analysis can be used for the Balanced Three-Phase Power System

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor In a practical Power Systems, Lines are not transposed.

Single-phase transformers used to form three-phase banks are not identical.

Loads are not balanced.

Presence of vee-phase and single phase lines.

Faults Single-phase Representation and Analysis cannot be use for an unbalanced three-phase power system.

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor Any unbalanced three-phase system of phasors may be resolved into three balanced systems of phasors which are referred to as the symmetrical components of the original unbalanced phasors, namely:

a) POSITIVE-SEQUENCE PHASOR b) NEGATIVE-SEQUENCE PHASOR c) ZERO-SEQUENCE PHASOR U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor REFERENCE PHASE SEQUENCE: abc Phase c

120째 120째

Phase a

120째

Phase b

Positive Sequence Phasors are three-phase, balanced and have the phase sequence as the original set of unbalanced phasors.

Negative Sequence Phasors are three-phase, balanced but with a phase sequence opposite to that original set of unbalnced phasors.

Zero Sequence Phasors are single-phase, equal in magnitude and in the same direction.

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor Each of the original unbalanced phasor is the sum of it’s sequence components. Thus,

Va = Va1 + Va2 + Va0 Vb = Vb1 + Vb2 + Vb0 Vc = Vc1 + Vc2 + Vc0 Where, Va1 – Positive Sequence component of Voltage Va Va2 – Negative Sequence component of Voltage Va Va0 – Zero Sequence component of Voltage Va U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor OPERATOR “a” An operator “a” causes a rotation of 120° in the counter clockwise direction of any phasor.

a = 1 ∠ 120° a² = 1 ∠ 240° a³ = 1 ∠ 0°

U. P. National Engineering Center National Electrification Administration

aV

120°

V

Operating V by a

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor The original Phasor and Positive Sequence components in terms of phase a Vc = aVa

Vb in terms of Va

120°

Vb = a² Va Vb1 = a² Va1

120°

Va

120°

Vb = a2Va

U. P. National Engineering Center National Electrification Administration

Vc in terms of Va Vc = a Va Vc1 = a Va1 Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor The Negative Sequence components in terms of phase a

Vb in terms of Va

Vb2 = aVa2

Vb2 = aVa2 120° 120°

Va2

120°

Vc2 = a2Va2

U. P. National Engineering Center National Electrification Administration

Vc in terms of Va Vc2 = a²Va2 Competency Training & Certification Program in Electric Power Distribution System Engineering

86

Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor The Zero Sequence components in terms of phase a

Vb in terms of Va Vb0 = Va0 Va0= Vb0 = Vc0

Vc in terms of Va Vc0 = Va0

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Training Course in Fundamental Principles and Methods in Power System Analysis

88

Sequence Components of Unbalanced Phasor Writing again the phasors in terms of phasor Va and operator “a”,

Va = Va0 + Va1 + Va2 Vb = Va0 + a²Va1 + aVa2 Vc = Va0 + aVa1 + a²Va2 Computing for Va0, Va1 & Va2

V a0

[

1 1 = [V a + V b + V c ] V a 1 = V a + aV b + a 2 V c 3 3 1 V a2 = V a + a 2 V b + aV c 3

[

U. P. National Engineering Center National Electrification Administration

]

]

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor Vc = 8 ∠143.1

EXAMPLE: Determine the symmetrical components of the following unbalanced voltages.

Va = 4 ∠0

Vb = 3 ∠-90 U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor For Phasor Va: 1 V a0 = ( V a + V b + V c ) 3 1 = ( 4 ∠ 0 + 3 ∠ - 90 + 8 ∠ 143.1) 3 = 1 ∠ 143.05 1 V a 1 = ( V a + aV b + a 2V c ) 3 1 = [4 ∠ 0 + (1 ∠ 120)(3 ∠ - 90) + (1 ∠ 240)(8 ∠ 143.1) ] 3 = 4.9 ∠ 18.38 U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor For Phasor Va:

Va 2

1 = ( V a + a 2Vb + aV c ) 3 1 = [4 ∠0 + (1 ∠ 240)(3 ∠ - 90) + (1 ∠120)(8 ∠143.1) ] 3 = 2.15 ∠ − 86.08

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor Components of Vb can be obtained by operating the sequence components of phasor Va.

V b0 = V a0 = 1 ∠ 143.05

= 1 ∠ 143.05

V b1 = a 2V a 1

V b2

= (1 ∠ 240)(4.9 ∠ 18.38) = 4.9 ∠ 258.38 = 4.9 ∠ - 101.62 = aV a2 = (1 ∠ 120)(2.15 ∠ - 86.08) = 2.15 ∠ 33.92

U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor Similarly, components of phasor Vc can be obtained by operating Va.

V c0 = V a0 V c1

= 1 ∠ 143.05 = aV a 1

= 1 ∠ 143.05

= (1 ∠ 1 2 0)(4.9 ∠ 18.38) = 4.9 ∠ 1 3 8.38 V c2 = a 2 V a2 = (1 ∠ 2 4 0)(2.15 ∠ - 86.08) = 2.15 ∠ 1 5 3.92 U. P. National Engineering Center National Electrification Administration

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94

Sequence Components of Unbalanced Phasor Vc1

Va0 = Vb0 = Vc0 143.05 째

Va1 18.38 째

Zero Sequence Components Vc2

Vb2 86.08 째

Vb1 Positive Sequence Components U. P. National Engineering Center National Electrification Administration

Va2 Negative Sequence Components Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor Add Sequence Components Graphically

Va1 Va0

Vc2

Va = 4 ∠0

Components of Va

Va2

Vb0 Vb1

Vb = 3 ∠-90

Vc1 Vc = 8 ∠143.1

Vb2 Vc0

Components of Vb

Components of Vc U. P. National Engineering Center National Electrification Administration

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor The results can be checked either mathematically or graphically.

V a = V a0 + V a1 + V a2 = 1 ∠ 143 . 05 + 4.9 ∠ 18.38 + 2.15 ∠ - 86.08 = 4 ∠0 V b = V b0 + V b1 + V b2 = 1 ∠ 143.05 + 4.9 ∠ - 101.62 + 2.15 ∠ 33.92 = 3 ∠ - 90 V c = V c0 + V c1 + V c2 = 1 ∠ 143 . 05 + 4.9 ∠ 138.38 + 2.15 ∠ 153.92 = 8 ∠ 143.1 U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Components of Unbalanced Phasor Ia a

Ia1 + Ia2 + Ia0

a b

b

c

c Ib

Ib1 + Ib2 + Ib0

Ic Ic1 + Ic2 + Ic0

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97

Training Course in Fundamental Principles and Methods in Power System Analysis

98

Sequence Impedance of Power System Components Sequence Networks +

+

+

Ia1 Va1

Ia2

Z1

Va2

Ia0

Z2

Va0

Z0

+ Vf -

-

Va1 = V f – I a1 Z 1

Positive Sequence

-

V =-I Z a2 a2 2

Negative Sequence

U. P. National Engineering Center National Electrification Administration

V =-I Z ao ao o

Zero Sequence

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

Sequence Impedance of Power System Components

In general,

Z1 ≠ Z2 ≠ Z0

for generators

Z1 = Z2 = Z0

for transformers

Z1 = Z2 ≠ Z0

for lines

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Competency Training & Certification Program in Electric Power Distribution System Engineering

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Training Course in Fundamental Principles and Methods in Power System Analysis

100

Practical Implications of Sequence Components of Electric Currents ZERO-SEQUENCE CURRENTS: IA0

Ia0 A

IC0 C

a B

3I0

IB0

Ic0

b c

3Io

Ib0

The neutral return (ground) carries the in-phase zerosequence currents.

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101

Practical Implications of Sequence Components of Electric Currents ZERO-SEQUENCE CURRENTS: I0 = 0

IC0

Ia0 a

C A I0 = 0 I0 = 0

B

IA0

b c

IB0

In-phase zero-sequence currents circulates in the delta-connected transformer windings.

There is “balancing ampere-turns� for the zero-sequence currents.

U. P. National Engineering Center National Electrification Administration

Ic0 3Io

Ib0

The neutral return (ground) carries the in-phase zerosequence currents.

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102

Practical Implications of Sequence Components of Electric Currents NEGATIVE-SEQUENCE CURRENTS: A three-phase unbalanced load produces a reaction field which rotates synchronously with the rotor-field system of generators.

Any unbalanced condition will have negative sequence components.

This negative sequence currents rotates counter to the synchronously revolving field of the generator.

The flux produced by sequence currents cuts the rotor field at twice the rotational velocity, thereby inducing double frequency currents in the field system and in the rotor body.

The resulting eddy-currents are very large and cause severe heating of the rotor.

U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

Training Course in Fundamental Principles and Methods in Power System Analysis

103

Network Equations and Methods of Solution

Network Equations

Matrix Representation of System of Equations

Type of Matrices

Matrix Operations

Direct Solutions of System of Equations

Iterative Solutions of System of Equations

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104

Network Equations The standard form of n independent equations: ⎡ Y 11 Y 12 Y 13 L Y 1 n ⎤ ⎡V 1 ⎤ ⎡I1 ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ V Y 21 Y 22 Y 23 L Y 2 n 2⎥ ⎢I2 ⎥ ⎢ ⎥⎢ ⎢ I 3 ⎥ = ⎢ Y 31 Y 32 Y 33 L Y 3 n ⎥ ⎢V 3 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ M M M ⎥⎢ M ⎥ ⎢M ⎢M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢I ⎥ Y Y Y L Y ⎢ n1 n2 n3 nn ⎥ V n n ⎣ ⎦ ⎣ ⎦ ⎣

⎦

[ I ] = [Ybus][V] Ypp = self-admittance, the sum of all admittances terminating on the node (diagonal elements) Ypq = mutual admittance, the negative of the admittances connected directly between the nodes identifed by the double subscripts U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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105

Network Equations a Bus

Gen

Line

1 2 3 4

a b c

1-3 2-3 1-4 2-4 3-4

c

1

b

3

2

4 Single Line Diagram U. P. National Engineering Center National Electrification Administration

Competency Training & Certification Program in Electric Power Distribution System Engineering

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106

Network Equations 0

Ea

Ea

za 1

za

Eb

zc

Generator

1

zb 3

zd

0

zf

z13 1

Ec

2

ze zh

zg

3 Line 4 0

U. P. National Engineering Center National Electrification Administration

Impedance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering

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107

Network Equations 0

I1 = Ea/za y01 = 1/za I2 = Eb/zb y02 = 1/zb

I1

y01 1

y13

I3 = Ec/zc y03 = 1/zc

I3

y03 3

y14 = 1/zf

y23 = 1/ze

y24 = 1/zg y34 = 1/zh

U. P. National Engineering Center National Electrification Administration

y02 2

y23

y14

y13 = 1/zd

I2

y24 4

Admittance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering

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108

Network Equations I 1 = V1 ( y 01 + y13 + y14 ) − V 3 y13 − V 4 y14 I 2 = V 2 ( y 02 + y 23 + y 24 ) − V 3 y 23 − V 4 y 24

I 3 = V 3 ( y 03 + y 23 + y 34 + y 13 ) − V 2 y 23 − V 4 y 34 − V 1 y 13 0 = V 4 ( y 14 + y 24 + y 34 ) − V 1 y 14 − V 2 y 24 − V 3 y 34

In matrix form,

0 - y13 - y14 ⎡I1 ⎤ ⎡y01 + y13 + y14 ⎤⎡V1 ⎤ ⎢I ⎥ ⎢ ⎥⎢V ⎥ 0 y y y y y + + 02 23 24 23 24 ⎢ 2⎥ = ⎢ ⎥⎢ 2 ⎥ ⎢I3 ⎥ ⎢ ⎥⎢V3 ⎥ - y13 - y23 y03 + y23 + y34 + y13 - y34 ⎢ ⎥ ⎢ ⎥⎢ ⎥ y14 + y24 + y34⎦⎣V4 ⎦ − y24 − y34 ⎣0 ⎦ ⎣ − y14 U. P. National Engineering Center National Electrification Administration

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Matrix Representations of System of Equations

System of n Linear Equations

In the following system of equations:

a11 x1 + a12 x 2 + a13 x 3 = y 1 a 21 x1 + a 22 x 2 + a 23 x 3 = y 2 a 31 x1 + a 32 x 2 + a 33 x 3 = y 3 x1, x 2, and x3, are unknown variables, a11, a12,‌ ‌, a33 are the coefficient of these variables and y1, y2, and y3 are known parameters.

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Matrix Representations of System of Equations The coefficients form an array

⎡ a11 ⎢ A = a21 ⎢ ⎢⎣a31

a12 a22 a32

a13 ⎤ ⎥ a23 ⎥ a33 ⎥⎦

which is called the Coefficient Matrix of the system of equations.

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Matrix Representations of System of Equations Similarly, the variables and parameters can be written in matrix form as.

⎡ x1 ⎤ ⎥ ⎢ X = x2 ⎢ ⎥ ⎢⎣ x3 ⎥⎦

and

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⎡ y1 ⎤ ⎥ ⎢ Y = y2 ⎢ ⎥ ⎢⎣ y3 ⎥⎦

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Matrix Representations of System of Equations

System of Equations in Matrix Form

The system of equations in matrix notation is

⎡ a11 ⎢a ⎢ 21 ⎢⎣ a31

a13 ⎤ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎥ ⎢ ⎥ ⎢ ⎥ a23 x2 = y 2 ⎥⎢ ⎥ ⎢ ⎥ a33 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ y 3 ⎥⎦

a12 a 22 a32

AX = Y U. P. National Engineering Center National Electrification Administration

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Matrix Representations of System of Equations Sequence Components of Unbalanced Phasor

Va = Va0 + Va1 + Va2 Vb = Va0 + a²Va1 + aVa2 Vc = Va0 + aVa1 + a²Va2 Rearranging and writing in matrix form

⎡Va ⎤ ⎡1 1 ⎢V ⎥ = ⎢1 a 2 ⎢ b⎥ ⎢ ⎢⎣Vc ⎥⎦ ⎢⎣1 a U. P. National Engineering Center National Electrification Administration

1 ⎤ ⎡Va 0 ⎤ ⎥ ⎢ ⎥ a ⎥ ⎢Va1 ⎥ a 2 ⎥⎦ ⎢⎣Va 2 ⎥⎦

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Matrix Representations of System of Equations V a0

1 = [V a + V b + V c ] 3 V a2

Va1

[

[

1 = V a + aV b + a 2 V c 3

1 = V a + a 2 V b + aV 3

c

]

]

In matrix form

⎡1 1 ⎡Va 0 ⎤ ⎢V ⎥ = 1 ⎢1 a ⎢ a1 ⎥ 3 ⎢ ⎢⎣1 a 2 ⎢⎣Va 2 ⎥⎦ U. P. National Engineering Center National Electrification Administration

1 ⎤ ⎡Va ⎤ a 2 ⎥⎥ ⎢⎢Vb ⎥⎥ a ⎥⎦ ⎢⎣Vc ⎥⎦

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Definition of a MATRIX A matrix consists of a rectangular array of elements represented by a single symbol.

[A] is a shorthand notation for the matrix and aij designates an individual element of the matrix.

A horizontal set of elements is called a row and a vertical set is called a column.

The first subscript i always designates the number of the row in which the element lies. The second subscript j designates the column. For example, element a23 is in row 2 and column 3.

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Definition of a MATRIX ⎡ a11 ⎢a ⎢ 21 [A] = [a ij ] = ⎢ a31 ⎢ ⎢ M ⎢⎣am1

a12 a22 a32 M am2

a13 a23 a33

K K K

am3

K

a 1n ⎤ a2n ⎥ ⎥ a3n ⎥ ⎥ M ⎥ amn ⎥⎦

The matrix has m rows and n columns and is said to have a dimension of m by n (or m x n).

[aij]mxn U. P. National Engineering Center National Electrification Administration

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Definition of a Vector A vector X is defined as an ordered set of elements. The components x1, X2…, Xn may be real or complex numbers or functions of some dependent variable.

⎡ x1 ⎤ ⎢x ⎥ 2 ⎢ ⎥ X = ⎢M ⎥ ⎢ ⎥ ⎣ xn ⎦ “n” defines the dimensionality or size of the vector.

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Matrices with only one row (n = 1) are called Row Vectors while those with one column (m=1) are called Column Vectors. The elements of a vectors are denoted by single subscripts as the following:

R = [r1

r2

L rn ]

⎡ c1 ⎢c 2 ⎢ C = ⎢ M ⎢ ⎣cm

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

Thus, R is a row vector of dimension n while C is a column vector of dimension m.

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Type of Matrices Square Matrix Upper Triangular Matrix Lower Triangular Matrix Diagonal Matrix Identity or Unit Matrix Null Matrix Symmetric Matrix Skew-symmetric Matrix

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Type of Matrices A square matrix is a matrix in which m = n.

For a square, the main or principal diagonal consists of the elements of the form aii; e.g., for the 3 x 3 matrix shown

⎡a11 ⎢ A = a21 ⎢ ⎢⎣a31

a12 a22 a32

a13 ⎤ ⎥ a23 ⎥ a33 ⎥⎦

the elements a11, a22, and a33 constitute the principal diagonal.

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Type of Matrices An upper triangular matrix is one where all the elements below the main diagonal are zero.

⎡u11 ⎢ U= 0 ⎢ ⎢⎣ 0 U. P. National Engineering Center National Electrification Administration

u12 u22 0

u13 ⎤ ⎥ u23 ⎥ u33 ⎥⎦

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Type of Matrices A lower triangular matrix is one where all elements above the main diagonal are zero.

⎡l11 ⎢ L = l21 ⎢ ⎢⎣l31 U. P. National Engineering Center National Electrification Administration

0 l22 l32

0⎤ ⎥ 0 ⎥ l33 ⎥⎦

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Type of Matrices A diagonal matrix is a square matrix where all elements off the diagonal are equal to zero.

Note that where large blocks of elements are zero, they are left blank.

⎡d11 ⎢ D= 0 ⎢ ⎢⎣ 0 U. P. National Engineering Center National Electrification Administration

0 d22 0

0⎤ ⎥ 0 ⎥ d33 ⎥⎦

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Type of Matrices An identity or unit matrix is a diagonal matrix where all elements on the main diagonal are equal to one.

⎡1 ⎢ I = ⎢0 ⎢⎣ 0 U. P. National Engineering Center National Electrification Administration

0 1 0

0⎤ ⎥ 0⎥ 1 ⎥⎦

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Type of Matrices The null matrix is matrix whose elements are equal to zero.

⎡0 ⎢ N= 0 ⎢ ⎢⎣0 U. P. National Engineering Center National Electrification Administration

0 0 0

0⎤ ⎥ 0 ⎥ 0 ⎥⎦

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Type of Matrices A symmetric matrix is one where aij = aji for all i’s and j’s.

⎡5 ⎢ S= 1 ⎢ ⎢⎣2

1 3 7

2⎤ ⎥ 7 ⎥ 8 ⎥⎦

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Type of Matrices A skew-symmetric matrix is a matrix which has the property aij = -aji for all i and j; this implies aii = 0

⎡0 ⎢ K= 5 ⎢ ⎢⎣− 3

−5 0 −6

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3⎤ ⎥ 6 ⎥ 0⎥⎦

a12 = −5 a 21 = +5

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Matrix Operations Addition of Matrices Product of a Matrix with a Scalar Multiplication of Matrices Transpose of a Matrix Kron Reduction Method Determinant of a Matrix Minors and Cofactors of a Matrix Inverse of a Matrix

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Addition of Matrices Two matrices A = [aij] and B = [bij] can be added together if they are of the same order (mxn). The sum C = A + B is obtained by adding the corresponding elements.

C = [cij] = [aij + bij]

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Addition of Matrices Example:

⎡1 A = ⎢ ⎣2

4 7

0⎤ 3 ⎥⎦

⎡5 B = ⎢ ⎣0

6⎤ 1 ⎥⎦

2 1

then,

⎡(1 + 5) A+ B= ⎢ ⎣(2 + 0) ⎡(1 − 5) A− B = ⎢ ⎣(2 − 0)

(4 + 2) (7 + 1) (4 − 2) (7 − 1)

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(0 + 6)⎤ ⎡6 =⎢ ⎥ (3 + 1)⎦ ⎣2 (0 − 6)⎤ ⎡− 4 =⎢ ⎥ (3 − 1)⎦ ⎣ 2

6 8 2 6

6⎤ ⎥ 4⎦ − 6⎤ ⎥ 2⎦

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Addition of Matrices Example:

⎡1+ j2 A = ⎢4 + j1 ⎢ ⎢⎣6 + j3

4 − j1 5 + j3 1 − j1

6 − j3⎤ ⎡3 + j2 1 + j1⎥ B = ⎢2 + j1 ⎢ ⎥ ⎢⎣7 − j5 8 + j9⎥⎦

2 − j1 4 + j6 5 − j4

7 + j5⎤ 5 + j4⎥ ⎥ 6 + j5⎥⎦

then,

⎡(1+ j2) +(3+ j2) A+ B= ⎢(4+ j1) +(2+ j1) ⎢ ⎢⎣(6 + j3) +(7 − j5) U. P. National Engineering Center National Electrification Administration

(4− j1) +(2− j1) (6 − j3) +(7 + j5)⎤ (5+ j3) +(4+ j6) (1+ j1) +(5+ j4) ⎥⎥ (1− j1) +(5− j4) (8+ j9) +(6 + j5)⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering

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Addition of Matrices ⎡ 4 + j4 A + B = ⎢ 6 + j2 ⎢ ⎢⎣13− j2 ⎡− 2 + j0 ⎢ A− B = 2 + j0 ⎢ ⎢⎣−1+ j8 U. P. National Engineering Center National Electrification Administration

6 − j2

13+ j2 ⎤ 9 + j9 6 + j5 ⎥ ⎥ 6 − j5 14+ j14⎥⎦ 2 + j0 −1− j8⎤ ⎥ 1− j3 − 4 − j3 ⎥ − 4 + j3 2 + j4 ⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering

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Product of a Matrix with a Scalar A matrix is multiplied by a scalar k all elements mn by k , that is,

⎡ ka11 ⎢ ka 21 ⎢ kA = Ak = ⎢ M ⎢ ⎣kam1 U. P. National Engineering Center National Electrification Administration

ka12 ka22 M kam 2

by multiplying

L ka1n ⎤ ⎥ L ka2 n ⎥ M M ⎥ ⎥ L kamn ⎦

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Product of a Matrix with a Scalar Example:

⎡4 ⎢ A= 5 ⎢ ⎢⎣6

⎡4 ⎢ B = kA = 3 5 ⎢ ⎢⎣6

3⎤ ⎥ 2 ⎥ 1⎥⎦

3⎤ ⎥ 2 ⎥ 1⎥⎦

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and

k =3

⎡ 12 ⎢ B = 15 ⎢ ⎢⎣18

9⎤ ⎥ 6 ⎥ 3 ⎥⎦

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Product of a Matrix with a Scalar Example:

⎡4 + j1 ⎢ A = 5 - j2 ⎢ ⎢⎣6 + j3

3 − j6⎤ ⎥ 2 + j5 and k = 3 ⎥ 1 − j4⎥⎦

⎡12+ j3 ⎡4 + j1 3- j6⎤ ⎢ ⎢ ⎥ B = kA=3 5 - j2 2 + j5 B = 15− j6 ⎢ ⎢ ⎥ ⎢⎣18+ j9 ⎢⎣6 + j3 1- j4⎥⎦ U. P. National Engineering Center National Electrification Administration

9 − j18⎤ ⎥ 6 + j15 ⎥ 3 − j12⎥⎦

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Multiplication of Matrices Two matrices A = [aij] and B = [bij] can be multiplied in the order AB if and only if the number of columns of A is equal to the number of rows of B .

That is, if A is of order of (m x l), then B should be of order (l x n).

If the product matrix is denoted by C = A B, then C is of order (m x n).

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Multiplication of Matrices An easy way to check whether two matrices can be multiplied.

[A ]m x l [B ]l x n

= [C ]m x n

Interior dimensions are equal multiplication is possible Exterior dimensions define the dimensions of the result U. P. National Engineering Center National Electrification Administration

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Multiplication of Matrices If the product matrix is denoted by C = A B, then C is of order (m x n). The elements cij are given by l

cij = ∑ aik bkj k =1

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for all i and j.

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Multiplication of Matrices Example:

then

⎡ a11 ⎢ A = ⎢a21 ⎢⎣a31

a12 ⎤ ⎥ a22 ⎥ and a32 ⎥⎦ 3 x 2

⎡ a11 ⎢ C = A B = a 21 ⎢ ⎢⎣ a31 2

cij = ∑ aik bkj k =1

⎡b11 b12 ⎤ B=⎢ ⎥ ⎣b21 b22 ⎦ 2 x 2

a12 ⎤ ⎡b11 ⎥ a22 ⎢ ⎥ ⎣b21 a32 ⎥⎦

b12 ⎤ ⎥ b22 ⎦

c 11 = a 11 b 11 + a 12 b 21

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Multiplication of Matrices 2

cij = ∑ aik bkj k =1

⎡ (a11b11 + a12b21 ) (a11b12 + a12b22 ) ⎤ ⎢ ⎥ C = AB = ⎢( a21b11 + a22b21 ) ( a21b12 + a22b22 )⎥ ⎢⎣ (a31b11 + a32b21 ) ( a31b12 + a32b22 )⎥⎦

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Multiplication of Matrices Example:

then

⎡1 4⎤ ⎢ ⎥ A = ⎢2 5⎥ ⎢⎣3 6⎥⎦ 3 x 2

and

⎡1 ⎢ C = AB = ⎢ 2 ⎢⎣ 3

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⎡7 8 ⎤ B=⎢ ⎥ ⎣9 0 ⎦ 2 x 2

4⎤ ⎡7 ⎥ 5⎥⎢ 9 ⎣ 6 ⎥⎦

8⎤ ⎥ 0⎦

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Multiplication of Matrices ⎡ (1x7 + 4 x9) (1x8 + 4 x0) ⎤ ⎥ ⎢ C = AB = ⎢(2 x7 + 5x9) (2 x8 + 5x0)⎥ ⎢⎣(3x7 + 6 x9) (3x8 + 6 x0)⎥⎦3 x 2 ⎡ 43 C = ⎢⎢ 59 ⎢⎣ 75 U. P. National Engineering Center National Electrification Administration

8 ⎤ 16 ⎥⎥ 24 ⎥⎦ 3 x 2

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Multiplication of Matrices Example:

⎡1+ j2 4− j1 6 − j3⎤ ⎡3+ j2 2− j1 7 + j5⎤ A= ⎢4+ j1 5+ j3 1+ j1⎥ B= ⎢2+ j1 4+ j6 5+ j4⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣6 + j3 1− j1 8 + j9⎥⎦ ⎢⎣7 − j5 5− j4 6 + j5⎥⎦ c 11 = (1 + j 2 )(3 + j 2 ) + (4 − j 1 )(2 + j 1 ) + (6 − j 3 )(7 − j 5 ) = 35 − j 41

c 12 = (1 + j 2 )(2 − j 1 ) + (4 − j 1 )(4 + j 6 ) + (6 − j 3 )(5 − j 4 ) = 44 − j 16

c 13 = (1 + j 2 )(7 + j 5 ) + (4 − j 1)(5 + j 4 ) + (6 − j 3 )(6 + j 5 ) = 72 + j 42

c21 = (4 + j1)(3 + j 2 ) + (5 + j 3 )(2 + j1) + (1 + j1)(7 − j 5 ) = 29 + j 24

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Multiplication of Matrices c 21 = (4 + j 1 )(3 + j 2 ) + (5 + j 3 )(2 + j 1 ) + (1 + j 1 )(7 − j 5 ) = 29 + j 24 c 22 = (4 + j 1 )(2 − j 1 ) + (5 + j 3 )(4 + j 6 ) + (1 + j 1 )(5 − j 4 ) = 20 + j 41 c 23 = (4 + j 1 )(7 + j 5 ) + (5 + j 3 )(5 + j 4 ) + (1 + j 1 )(6 + j 5 ) = 37 + j73

c 31 = (6 + j 3 )(3 + j 2 ) + (1 − j 1)(2 + j 1) + (8 + j 9 )(7 − j 5 ) = 116 + j 43 c 32 = (6 + j 3 )(2 − j 1) + (1 − j 1)(4 + j 6 ) + (8 + j 9 )(5 − j 4 ) = 101 + j 15 c 33 = (6 + j 3 )(7 + j 5 ) + (1 − j 1)(5 + j 4 ) + (8 + j 9 )(6 + j 5 ) = 39 + j 144

⎡ 35 − j 41 [A] x[B] = ⎢⎢ 29 + j 24 ⎢⎣116 + j 43 U. P. National Engineering Center National Electrification Administration

44 − j16 20 + j 41 101 + j15

72 + j 42 ⎤ 37 + j73 ⎥ ⎥ 39 + j144⎥⎦

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Transpose of a Matrix If the rows and columns of an m x n matrix are interchanged, the resultant n x m matrix is the transpose of the matrix and is designated by AT.

For the matrix

The transpose is

⎡ a11 A=⎢ ⎣a12 ⎡ a11 AT = ⎢a21 ⎢ ⎢⎣ a31

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a21 a22

a31 ⎤ a32 ⎥⎦ 2 x 3

a12 ⎤ a22 ⎥ ⎥ a32 ⎥⎦ 3 x 2

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Transpose of a Matrix Example:

then,

⎡1 A=⎢ ⎣2 ⎡1 T ⎢ A = 3 ⎢ ⎢⎣5

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3 4

5⎤ 6 ⎥⎦ 2 x 3 2⎤ ⎥ 4 ⎥ 6 ⎥⎦ 3 x 2

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Transpose of a Matrix Example:

⎡1 + j 4 A=⎢ ⎣2 − j 5

3 − j6 4 + j1

5 + j 2⎤ ⎥ 6 − j 3⎦

then,

⎡1 + j 4 T ⎢ A = 3 − j6 ⎢ ⎢⎣5 + j 2 U. P. National Engineering Center National Electrification Administration

2 − j 5⎤ ⎥ 4 + j1 ⎥ 6 − j 3⎥⎦

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Determinant of a Matrix Determinant of a 2 x 2 Matrix Two simultaneous equations:

a11 x1 + a12 x2 = y1 a21 x1 + a22 x2 = y2

(1) (2)

In Matrix Form

⎡ a11 ⎢a ⎣ 21

a12 ⎤ ⎡ x1 ⎤ ⎡ y1 ⎤ =⎢ ⎥ ⎥ ⎢ ⎥ a22 ⎦ ⎣ x2 ⎦ ⎣ y 2 ⎦

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Determinant of a Matrix The solutions of two simultaneous equations can be obtained by eliminating the variables one at a time.

Solving for x2 in terms of x1 from the second equation and substituting this expression for x2 in the first equation, the following is obtained:

a22x2 = y2 − a21x1

y2 a21 x2 = − x1 a22 a22 U. P. National Engineering Center National Electrification Administration

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Determinant of a Matrix substituting x2 and solving for x1

y 2 a21 a11 x1 + a12 ( − x1 ) = y1 a22 a22 a11 a22 x1 + a12 y 2 − a12 a21 x1 = a22 y1 ( a11 a22 − a12 a21 ) x1 = a22 y1 − a12 y2 a22 y1 − a12 y2 x1 = a11 a22 − a12 a21 U. P. National Engineering Center National Electrification Administration

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Determinant of a Matrix Then, substituting x1 in either equation (1) or (2), x2 is obtained

a11 y2 − a21 y1 x2 = a11a22 − a12a21

The expression (a11a22 – a12a21) is the value of the determinant of the coefficient matrix A, denoted by |A|.

a11 a12 | A| = a21 a22 U. P. National Engineering Center National Electrification Administration

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Determinant of a Matrix The determinant of A is can be determined by reducing the size of the original matrix by eliminating rows.

For example:

⎡ 1 [ A ] = ⎢⎢ − 1 ⎢⎣ − 6 1 | A |= − 1 −6

4

2⎤ 1 ⎥⎥ 2 ⎥⎦

1

2

− 2 4

1 2

1 − 2

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Determinant of a Matrix The determinant of A is 1 1 2

| A |= − 1 −6

| A |= ( −1 ) 1 1+1

− 2 4

1 2

−2

1

4

2

+ ( −1 ) 1 1+2

−1

1

−6

2

+ ( −1 ) 2 1+3

−1

−2

−6

4

| A |= ( +1)(1)[( −2)( 2) − (1)( 4) ] + ( −1)(1)[( −1)( 2) − (1)( −6) ] + ( +1)( 2)[( −1)( 4) − ( −2)( −6) ] = (1)[− 4 − 4 ] − (1)[− 2 + 6 ] + ( 2)[− 4 − 12 ] | A |= −44 U. P. National Engineering Center National Electrification Administration

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Determinant of a Matrix Minors and Cofactors of a Matrix The determinant obtained by striking out the ith row and jth column is called the minor element aij.

Example:

a11 a 21 a31

a12 a22 a32

a13 a12 a 23 = a32 a33

a13 a33

The minor of a 21 = ( a 12 a 33 − a 32 a 13 ) U. P. National Engineering Center National Electrification Administration

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Determinant of a Matrix Minors and Cofactors of a Matrix The cofactor of an element aij designated by Aij is

Aij = ( − 1 ) Example:

i+ j

(minor

of a ij )

A 21 = ( − 1) 2 +1 (the min or of a 21 ) = (-1) 3 (the min or of a 21 )

A 21 = - 1 (the minor of a 21 )

Since the minor of a 21 = ( a 12 a 23 − a 32 a 13 )

∴ the cofactor of A21 = − 1( a 12 a 33 − a 32 a 13 ) U. P. National Engineering Center National Electrification Administration

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Determinant of a Matrix Minors and Cofactors of a Matrix

⎡1 ⎢- 1 A = Example: ⎢ ⎢⎣- 6

1 -2 4 1

A11 = ( −1 )1+1 − 1

1

2⎤ 1⎥ ⎥ 2⎥⎦

−2

−6

4

1 A12 = ( −1 )1+2 − 1 −6

1 −2 4

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2 1 = 1[( −2 )( 2 ) − ( 1 )( 4 )] = −8 2 2 1 = −1[( −1 )( 2 ) − ( 1 )( −6 )] = −4 2 Competency Training & Certification Program in Electric Power Distribution System Engineering

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Determinant of a Matrix Minors and Cofactors of a Matrix 1 1 2 A13 = ( −1 )1+3 − 1 − 2 1 = 1[( −1 )( 4 ) − ( −2 )( −6 )] = −16 −6 4 2

A21 = ( −1 )2+1

A22 = ( −1 )2+2

1 −1

1 −2

−6

4

1 −1

1 −2

−6

4

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2 1 = −1[( 1 )( 2 ) − ( 4 )( 2 )] = 6 2

2 1 = 1[( 1 )( 2 ) − ( 2 )( −6 )] = 14 2 Competency Training & Certification Program in Electric Power Distribution System Engineering

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Determinant of a Matrix Minors and Cofactors of a Matrix

1 A23 = ( −1 )2+3 − 1 −6 1 A31 = ( − 1 )3 + 1 − 1 −6 1 A32 = ( −1 )3+ 2 − 1 −6

1 −2 4

2 1 = −1[( 1 )( 4 ) − ( 1 )( −6 )] = −10 2

1 −2 4 1 −2 4

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Determinant of a Matrix Minors and Cofactors of a Matrix

1 A33 = ( −1)3+3 − 1 −6

1 −2 4

2 1 = 1[(1 )( −2 ) − ( −1 )(1 )] = −1 2

Therefore the cofactors of matrix A are:

A11 = −8

A21 = 6

A31 = 5

A12 = −4

A22 = 14

A32 = −3

A13 = −16

A23 = −10

A33 = −1

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Determinant of a Matrix Minors and Cofactors of a Matrix and in matrix form:

⎡ A11 Cofactors of A = ⎢⎢ A21 ⎢⎣ A31

A12 A22 A32

A13 ⎤ A23 ⎥⎥ A33 ⎥⎦

⎡− 8 Cofactors of A = ⎢⎢ 6 ⎢⎣ 5

−4 14 −3

− 16 ⎤ − 10 ⎥⎥ − 1 ⎥⎦

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Inverse of a Matrix Division does not exist in matrix algebra except in the case of division of a matrix by a scalar. However, for a given set of equations.

a 11 x 1 + a 12 x 2 + a 13 x 3 = y 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = y 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = y 3 or in matrix form [AX] = [Y]. It is desirable to express x1, x2, and x3 a function of y1, y2, and y3, i.e..

[X] = [BY], where B is the inverse of A designated

by A-1.

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Inverse of a Matrix If the determinant of A is not zero, the equations can be solved for x ’s as follows;

A 11 A 21 A 31 x1 = y1 + y2 + y3 | A| | A| | A|

x2

A 12 A 22 A 32 = y1 + y2 + y3 | A| | A| | A|

A 13 A 23 A 33 x3 = y1 + y2 + y3 | A| | A| | A| where A11, A12, ‌, A33 are cofactors of a11, a12,,a33 and |A| is the determinant of A. U. P. National Engineering Center National Electrification Administration

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Inverse of a Matrix Thus,

B = A -1

⎡ A 11 ⎢| A | ⎢ A 12 ⎢ = ⎢| A | ⎢ A 13 ⎢ ⎣| A |

A 21 | A | A 22 | A | A 23 | A |

A 31 ⎤ | A |⎥ ⎥ A 32 ⎥ or | A |⎥ A 33 ⎥ ⎥ | A |⎦

+ A -1 A = | A|

A+ is called the adjoint of A. It should be noted that the elements of adjoint A+ are the cofactors of the elements of A, but are placed in transposed position.

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Inverse of a Matrix Example: Get the inverse of A

⎡1 A = ⎢- 1 ⎢ ⎢⎣- 6

1 -2 4

2⎤ 1⎥ ⎥ 2⎥⎦

A

-1

A+ = | A|

the Adjoint of A is

⎡A11 A+ = ⎢A12 ⎢ ⎢⎣A13

A21 A22 A23

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A31⎤ ⎡ − 8 A32 ⎥ = ⎢ − 4 ⎥ ⎢ A33 ⎥⎦ ⎢⎣−16

6 14 −10

5⎤ − 3⎥ ⎥ −1⎥⎦

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Inverse of a Matrix Hence, the inverse of matrix A is

6 5⎤ ⎡ −8 ⎥ ⎢ −4 14 3 − ⎥ ⎢ + ⎢⎣− 16 − 10 − 1⎥⎦ A −1 A = = A − 44

A

−1

⎡ −8 1 ⎢ = − ⎢ −4 44 ⎢⎣− 16

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⎡ −−448 ⎢ = ⎢ −−444 ⎢⎣ −−16 44

6 −44 14 −44 −10 −44

5 −44 −3 −44 −1 −44

⎤ ⎥ ⎥ ⎥⎦

5 ⎤ ⎥ 14 − 3⎥ − 10 − 1⎥⎦ 6

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Kron Reduction Method ⎡ y1 ⎤ ⎡ a11 ⎢ y ⎥ ⎢a ⎢ 2 ⎥ = ⎢ 21 ⎢ y3 ⎥ ⎢a31 ⎢ ⎥ ⎢ ⎣ 0 ⎦ ⎣a41

⎡ y1 ⎤ ⎡a11 ⎢ y ⎥ ⎢a ⎢ 2 ⎥ = ⎢ 21 ⎢ y3 ⎥ ⎢a31 ⎢ ⎥ ⎢ ⎣ 0 ⎦ ⎣a41

a12 a22

a13 a23

a 32 a42

a33 a43

a12

a13

a22 a32 a42

a23 a33 a43

a14 ⎤ ⎡ x1 ⎤ a24 ⎥ ⎢ x2 ⎥ ⎥⎢ ⎥ a34 ⎥ ⎢ x3 ⎥ ⎥⎢ ⎥ a44 ⎦ ⎣ x4 ⎦

a14 ⎤⎡ x1 ⎤ ⎥ ⎢ ⎥ a24 x2 ⎥⎢ ⎥ a34 ⎥⎢x3 ⎥ ⎥⎢ ⎥ a44 ⎦⎣x4 ⎦

The four equations can be reduced to three equations by Kron Reduction Method since the independent variable of the fourth equation is zero.

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Kron Reduction Method ⎡ y1 ⎤ [Y1 ] = ⎢⎢ y2 ⎥⎥ ⎢⎣ y3 ⎥⎦

⎡a11 [ A1 ] = ⎢⎢a21 ⎢⎣a31

a12 a22 a32

a13 ⎤ a23 ⎥ ⎥ a33 ⎥⎦

⎡ a14 ⎤ [A2 ] = ⎢⎢a24 ⎥⎥ ⎢⎣a34 ⎥⎦

⎡ x1 ⎤ [X1] = ⎢⎢x2 ⎥⎥ ⎢⎣ x3 ⎥⎦

[Y 2 ] = [0 ]

[A ] = [a

a42

a43 ]

[A4 ] = [a44 ]

[X 2 ] = [x4 ]

3

41

⎡Y1 ⎤ ⎡ A1 ⎢ 0 ⎥ = ⎢A ⎣ ⎦ ⎣ 3 U. P. National Engineering Center National Electrification Administration

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Kron Reduction Method

⎡Y1 ⎤ ⎡ A1 ⎢ 0 ⎥ = ⎢A ⎣ ⎦ ⎣ 3

A2 ⎤⎡ X1 ⎤ ⎥ ⎢ ⎥ A4 ⎦⎣X2 ⎦

[Y1 ] = [ A1 ][ X 1 ] + [ A 2 ][X 2 ] [0 ] = [A3 ][X 1 ] + [A 4 ][X 2 ]

(1) (2)

From (2)

[ A 4 ][X 2 ] = − [A3 ][X 1 ] [X 2 ] = − [A4 ]− 1 [A3 ][X 1 ] U. P. National Engineering Center National Electrification Administration

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Kron Reduction Method Substitute [X2] to equation (1)

[Y1 ] = [A1 ][X 1 ] + [A2 ](− [A4 ]−1 [A3 ][X 1 ]) Thus,

[Y1 ] = {[A1 ] − [A2 ][A4 ] [A3 ]}[X 1 ] −1

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Kron Reduction Method Example:

⎡ y1 ⎤ ⎡1 ⎢ y ⎥ ⎢8 ⎢ 2⎥ = ⎢ ⎢ y3 ⎥ ⎢3 ⎢ ⎥ ⎢ ⎣ 0 ⎦ ⎣2

2 7 4 4

3 6 5 7

4 ⎤ ⎡ x1 ⎤ 5 ⎥ ⎢x2 ⎥ ⎥⎢ ⎥ 6 ⎥ ⎢ x3 ⎥ ⎥⎢ ⎥ 8 ⎦ ⎣ x4 ⎦

⎞ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎛ ⎡1 2 3⎤ ⎡4⎤ ⎢ y ⎥ = ⎜ ⎢8 7 6⎥ − ⎢5⎥[8]−1 [2 4 7]⎟ ⎢ x ⎥ ⎟⎢ 2 ⎥ ⎢ 2⎥ ⎜⎢ ⎥ ⎢ ⎥ ⎟⎢ x ⎥ ⎢⎣ y3 ⎥⎦ ⎜⎝ ⎢⎣3 4 5⎥⎦ ⎢⎣6⎥⎦ ⎠⎣ 3 ⎦ U. P. National Engineering Center National Electrification Administration

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Kron Reduction Method ⎞ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎛ ⎡1 2 3⎤ ⎡4⎤ ⎜ ⎢ y ⎥ = ⎢8 7 6⎥ − ⎢5⎥[0.125][2 4 7]⎟ ⎢ x ⎥ ⎟⎢ 2 ⎥ ⎢ 2⎥ ⎜⎢ ⎥ ⎢ ⎥ ⎟⎢ x ⎥ ⎢⎣ y3 ⎥⎦ ⎜⎝ ⎢⎣3 4 5⎥⎦ ⎢⎣6⎥⎦ ⎠⎣ 3 ⎦

⎞ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎛ ⎡1 2 3⎤ ⎡ 0.5 ⎤ ⎜ ⎢ y ⎥ = ⎢8 7 6⎥ − ⎢0.625⎥[2 4 7 ]⎟ ⎢ x ⎥ ⎟⎢ 2 ⎥ ⎥ ⎥ ⎢ ⎢ 2⎥ ⎜⎢ ⎟⎢ x ⎥ ⎢⎣ y3 ⎥⎦ ⎜⎝ ⎢⎣3 4 5⎥⎦ ⎢⎣ 0.75 ⎥⎦ ⎠⎣ 3 ⎦ 2 3.5 ⎤ ⎞ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎛ ⎡1 2 3⎤ ⎡ 1 ⎜ ⎢ y ⎥ = ⎢8 7 6⎥ − ⎢1.25 2.5 4.375⎥ ⎟ ⎢ x ⎥ ⎢ 2⎥ ⎜⎢ ⎥ ⎢ ⎥ ⎟⎢ 2 ⎥ ⎢⎣ y3 ⎥⎦ ⎜⎝ ⎢⎣3 4 5⎥⎦ ⎢⎣ 1.5 3 5.25 ⎥⎦ ⎟⎠ ⎢⎣ x3 ⎥⎦ U. P. National Engineering Center National Electrification Administration

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Kron Reduction Method The 4x4 matrix was reduced to a 3x3matrix.

⎡ y1 ⎤ ⎡1 ⎢ y ⎥ ⎢8 ⎢ 2⎥ = ⎢ ⎢ y3 ⎥ ⎢3 ⎢ ⎥ ⎢ ⎣ 0 ⎦ ⎣2

2 7

3 6

4 4

5 7

4 ⎤ ⎡ x1 ⎤ 5 ⎥ ⎢x2 ⎥ ⎥⎢ ⎥ 6 ⎥ ⎢ x3 ⎥ ⎥⎢ ⎥ 8 ⎦ ⎣ x4 ⎦

0 − 0.50⎤ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎡ 0 ⎢ y ⎥ = ⎢6.75 4.5 1.625 ⎥ ⎢ x ⎥ ⎢ 2⎥ ⎢ ⎥⎢ 2 ⎥ ⎢⎣ y3 ⎥⎦ ⎢⎣1.50 1.0 − 0.25⎥⎦ ⎢⎣ x3 ⎥⎦ U. P. National Engineering Center National Electrification Administration

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Direct Solutions of System of Equations

Cramer’s Rule of Determinants

Matrix Inversion Method

Gaussian Elimination Method

Gauss-Jordan Method

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Solutions of System of Equations by Cramer’ s Rule The system of three linear equations in three unknowns x1, x2, x3:

a11 x1 + a12 x 2 + a13 x3 = y1 a 21 x1 + a 22 x 2 + a 23 x3 = y 2 a31 x1 + a 32 x 2 + a 33 x3 = y 3

written in matrix form as :

⎡ a11 ⎢a ⎢ 21 ⎢⎣ a 31

a12 a 22 a 32

a13 ⎤ ⎡ x1 ⎤ ⎡ y 1 ⎤ a 23 ⎥ ⎢ x 2 ⎥ = ⎢ y 2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ a 33 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ y 3 ⎥⎦

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or

AX = Y

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can be solved by Cramer’s Rule of determinants. The determinant of coefficient matrix A is

a 11

a 12

a 13

| A |= a 21

a 22

a 23

a 31

a 32

a 33

x1 can be obtained by :

x1 =

y1

a 12

a 13

y2 y3

a 22 a 32 | A|

a 23 a 33

Note that values in the numerator are the values of the determinant of A with the first column were replaced by the Y vector elements.

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Similarly, x2 and x3 can be obtained by:

x2 =

a 11 a 21

y1 y2

a 13 a 23

a 31

y3 | A|

a 33

a 11 a 21

a 12 a 22

y1 y2

a 31

a 32 | A|

y3

and

x3 =

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Solutions of System of Equations by Cramer’s Rule Example:

x 1 + x 2 + 2x 3 = 3 - x 1 − 2x 2 + x 3 = 7 - 6x 1 + 4x 2 + 2x 3 = 14

⎡ 1 ⎢- 1 ⎢ ⎢⎣ - 6

1 - 2 4

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2 ⎤ ⎡ x1 ⎤ ⎡ 3 ⎤ ⎥ ⎢ ⎥ ⎢ ⎥ 1 x2 = 7 ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 14 ⎥⎦ 2 ⎥⎦ ⎢⎣ x 3 ⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering

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Solutions of System of Equations by Cramer’s Rule ⎡1 A = ⎢- 1 ⎢ ⎢⎣- 6

x1 =

1 -2 4

2⎤ 1⎥ ⎥ 2⎥⎦

1

1

| A|= -1 −6

2

−2 4

1 = − 44 2

y1

a 12

a 13

3

1

2

y2 y3

a 22 a 32 | A|

a 23 a 33

7 14

−2 4 - 44

1 2

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x1 =

88 = = −2 − 44

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Solutions of System of Equations by Cramer’s Rule

x2 =

x3 =

a 11 a 21

y1 y2

a 13 a 23

a 31

y3 | A|

a 33

a 11 a 21 a 31

a 12 a 22 a 32 | A|

Therefore,

y1 y2 y3

1

x2 =

−1 −6

⎡ 1 ⎢− 1 ⎢ ⎢⎣ − 6 x3 =

x1 = - 2

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3

2

7 14 - 44 1

1 2

−2 4 - 44

x

2

3⎤ 7 ⎥ ⎥ 14 ⎥⎦

= -1

44 = = −1 − 44

− 132 =3 = − 44

x

3

=3

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Solutions of System of Equations by Matrix Inversion The system of equations in matrix form can be manipulated as follows:

AX = Y A AX = A Y -1

-1

IX = A Y -1

X =A Y -1

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Solutions of System of Equations by Matrix Inversion Example:

x 1 + x 2 + 2x 3 = 3 - x 1 − 2x 2 + x 3 = 7 - 6x 1 + 4x 2 + 2x 3 = 14 ⎡ 1 A = ⎢- 1 ⎢ ⎢⎣ - 6

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1 -2 4

2⎤ 1⎥ ⎥ 2 ⎥⎦

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Solutions of System of Equations by Matrix Inversion ⎡ −8 1 ⎢ -1 −4 A =− ⎢ 44 ⎢⎣− 16

6 14 − 10

⎡ −8 1 ⎢ X = A -1Y = − −4 ⎢ 44 ⎢⎣ − 16 ⎡ − 8( 3 ) + 1 ⎢ -1 X =A Y =− − 4( 3 ) + ⎢ 44 ⎢⎣− 16( 3 ) + U. P. National Engineering Center National Electrification Administration

6 14 − 10 6( 7 )

5⎤ − 3⎥ ⎥ − 1⎥⎦ 5 ⎤⎡ 3 ⎤ − 3⎥ ⎢ 7 ⎥ ⎥⎢ ⎥ − 1⎥⎦ ⎢⎣14 ⎥⎦ +

14( 7 ) + − 10( 7 ) +

5( 14 ) ⎤ − 3(14) ⎥ ⎥ − 1( 14 )⎥⎦

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Solutions of System of Equations by Matrix Inversion ⎡ 88 ⎤ ⎡ − 2 ⎤ 1 ⎢ -1 X =A Y =− 44 ⎥ = ⎢ − 1 ⎥ ⎥ ⎢ ⎥ 44 ⎢ ⎢⎣ − 132 ⎥⎦ ⎢⎣ 3 ⎥⎦ Therefore:

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Gaussian Elimination Method The following are the rules in matrix manipulation: (1) Interchange rows (2) Multiply row by constant (3) Add rows

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Gaussian Elimination Method Example:

⎡1 ⎢−1 ⎢ ⎢⎣−6

x1 + x2 + 2x3 = 3 - x1 − 2x2 + x3 = 7 - 6x1 + 4x2 + 2x3 = 14

1 −2 4

2 1 2

M M M

3⎤ 7⎥ ⎥ 14⎥⎦

Add row 1 to row 2 to get row 2.

Add 6 times row 1 to row 3 to get row 3.

⎡1 ⎢0 ⎢ ⎢⎣0

1

2

M

−1 10

3 14

M M

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3⎤ ⎥ 10 ⎥ 32⎥⎦

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Gaussian Elimination Method ⎡1 ⎢0 ⎢ ⎢⎣0

M

1

2

−1 10

3 M 14 M

3⎤ 10⎥ ⎥ 32⎥⎦

Multiply row 2 by 10 then add to row 3 to obtained row 3.

⎡1 ⎢0 ⎢ ⎢⎣0

1

2

M

−1 0

3 44

M M

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3⎤ 10 ⎥ ⎥ 132⎥⎦

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Gaussian Elimination Method

By Back Substitution:

0x1 + 0 x2 + 44 x3 = 132 0 x1 − x2 + 3( 3 ) = 10 x1 + ( −1 ) + 2( 3 ) = 3 Therefore:

x3 = 3 U. P. National Engineering Center National Electrification Administration

x2 = - 1

x1 = -2

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Gaussian Elimination Method The two phases of Gauss Elimination: forward elimination & back substitution.

The primes indicate the number of times that the coefficients and constants have been modified.

⎡a11 ⎢a ⎢ 21 ⎢⎣a31

a12 a22 a32

a13 M c1 ⎤ a23 M c2 ⎥ ⎥ a33 M c3 ⎥⎦ ⇓

⎡a11 ⎢ ⎢ ⎢⎣

a12 a'22

188

a13 M c1 ⎤ a'23 M c2' ⎥ ⎥ a"33 M c3" ⎥⎦

Forward elimination

⇓ x3 = c3" / a"33 x2 =( c2 −a23x3 ) / a22 '

'

'

Back substitution

x1 =( c1 −a12x2 −a13x3 ) / a11

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Gauss-Jordan Method From Gauss Elimination Method

⎡1 ⎢0 ⎢ ⎢⎣0

1

2

M

−1 0

3 44

M M

1

2

M

1 0

−3 44

M M

3⎤ 10 ⎥ ⎥ 132⎥⎦

Multiply row 2 by -1.

⎡1 ⎢0 ⎢ ⎢⎣0 U. P. National Engineering Center National Electrification Administration

3 ⎤ ⎥ − 10 ⎥ 132⎥⎦

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Gauss-Jordan Method Divide row 3 by 44.

⎡1 ⎢0 ⎢ ⎢⎣0

1

2

M

1 0

−3 1

M M

3 ⎤ − 10⎥ ⎥ 3 ⎥⎦

Multiply row 2 by -1 then add to row 1 to get row 1.

⎡1 ⎢0 ⎢ ⎢⎣0

0 1 0

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5 −3 1

M M M

13 ⎤ − 10⎥ ⎥ 3 ⎥⎦

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Gauss-Jordan Method Multiply row 3 by -5 then add to row 1 to get row 1.

Multiply row 3 by 3 then add to row 2 to get row 2.

⎡1 ⎢0 ⎢ ⎢⎣0

0

0

M

1 0

0 1

M M

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− 2⎤ − 1⎥ ⎥ 3 ⎥⎦

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Gauss-Jordan Method Therefore:

Then,

⎡1 ⎢0 ⎢ ⎢⎣0

0 1 0

0 ⎤ ⎡ x1 ⎤ ⎡ − 2 ⎤ 0 ⎥ ⎢ x2 ⎥ = ⎢ − 1⎥ ⎥⎢ ⎥ ⎢ ⎥ 1 ⎥⎦ ⎢⎣ x 3 ⎥⎦ ⎢⎣ 3 ⎥⎦

x1 = − 2 x2 = −1 x3 = 3

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Gauss-Jordan Method

The Gauss-Jordan method is a variation of Gauss Elimination. The major differences is that when an unknown is eliminated in the GJM, it is eliminated from all other equations rather than just the subsequent ones.

In addition, all rows are normalized by dividing them by their pivot elements. Thus, the elimination steps results in an identity matrix rather than a triangular matrix.

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Gauss-Jordan Method Graphical depiction of the Gauss-Jordan Method.

The superscript (n) means that the elements of the right-hand-side vector have been modified n times (for this case, n=3).

⎡ a 11 ⎢a ⎢ 21 ⎢⎣ a 31

a 12

a 13

M

a 22 a 32

a 23 a 33

M M

c1 ⎤ c2 ⎥ ⎥ c 3 ⎥⎦

↓ ⎡1 ⎢0 ⎢ ⎢⎣0

0 1

0 0

M M

0

1

M

c 1(n) ⎤ c 2(n) ⎥⎥ c 3(n) ⎥⎦

↓ x1 x2

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194

=

c 1(n)

=

c 2(n)

x3 =

c 3(n)

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Iterative Solutions of System of Equations

Gauss Iterative Method

Gauss-Seidel Method

Newton-Raphson Method

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Iterative Solutions of System of Equations An iterative method is a repetitive process for obtaining the solution of an equation or a system of equation. It is applicable to system of equations where the main-diagonal elements of the coefficient matrix are larger in magnitude in comparison to the off-diagonal elements.

The Gauss and Gauss-Seidel iterative techniques are for solving linear algebraic solutions and the NewtonRaphson method applied to the solution of non-linear equations.

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Iterative Solutions of System of Equations The solutions starts from an arbitrarily chosen initial estimates of the unknown variables from which a new set of estimates is determined. Convergence is achieved when the absolute mismatch between the current and previous estimates is less than some pre-specified precision index for all the variables.

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Gauss Iterative Method Example:

4x 1 − x 2 + x 3 = 4 x 1 + 4x 2 + x 3 = 6 x 1 + x 2 + 3x 3 = 5

Assume a convergence index of Îľ = 0.001 and the following initial estimates:

a) x1 = x 2 = x3 = 0.0 0

0

0

b) x1 = x 2 = x 3 = 0.5 0

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0

0

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Gauss Iterative Method Solution: a) The system of equation must be expressed in standard form.

x

k +1 1

x

k +1 2

x 3k + 1

1 k k = ( 4 + x2 - x3 ) 4 1 = ( 6 - x 1k - x 3k ) 4 1 = ( 5 - x 1k - x 2k ) 3

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Gauss Iterative Method a) with x1 0 = x 2 0 = x3 0 = 0 Iteration 1 (k = 0):

1 x1 = ( 4 + 0 - 0 ) = 1.0 4 1 1 x2 = ( 6 - 0 - 0 ) = 1.5 4 1 x31 = ( 5 - 0 - 0 ) = 1.6667 3 Δ x 10 = 1 − 0 = 1 1

Δ x 20 = 1 . 5 − 0 = 1 . 5 Δ x 30 = 1 . 6667 − 0 = 1 . 6667 max

Δ x 30

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Gauss Iterative Method Iteration 2 (k = 1):

1 x1 = ( 4 + 1.5 - 1.6667 ) = 0.958333 4 1 2 x2 = ( 6 - 1.0 - 1.6667 ) = 0.833333 4 1 2 x3 = ( 5 - 1.0 - 1.5 ) = 0.833333 3 Δx11 = 0.958325 − 1 = 0.041667 2

Δx21 = 0.833333 − 1.5 = 0.66667 Δx31 = 0.833333 − 1.6667 = −0.83334 max Δx31 = 0.83334 U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 3 (k = 2):

1 ( 4 + 0.8333 - 0.8333 ) = 1.0 4 1 3 x 2 = ( 6 - 0.9583 - 0.8333 ) = 1.0521 4 1 3 x 3 = ( 5 - 0.9583 - 0.8333 ) = 1.0695 3 Δ x 12 = 1 − 0 . 958325 = 0 . 041667 x1 3 =

Δ x 22 = 1 . 0521 − 0 . 833325 = 0 . 21877 Δ x 32 = 1 . 0695 − 0 . 8333 = 0 . 23617 max Δ x 32 = 0.23617 U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 4 (k = 3):

1 x1 = ( 4 + 1.0521 - 1.0695 ) = 0.9956 4 1 x2 4 = ( 6 - 1.0 - 1.0695 ) = 0.9826 4 1 x3 4 = ( 5 - 1.0 - 1.0521 ) = 0.9826 3 Δx13 = 0.9956 − 1 = −0.0044 4

Δx23 = 0.9826 − 1.0521 = −0.0695 Δx33 = 0.9826 − 1.0695 = −0.0869 max Δx33 = 0.0869 U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 5 (k = 4):

1 x 1 = ( 4 + 0.9826 - 0.9826 ) = 1.0 4 1 5 x 2 = ( 6 - 0.9956 - 0.9826 ) = 1.0054 4 1 5 x 3 = ( 5 - 0.9956 - 0.9826) = 1.0073 3 Δ x 14 = 1 − 0 . 9956 = 0 . 0044 5

Δ x 24 = 1 . 0054 − 0 . 9826 = − 0 . 0228 Δ x 34 = 1 . 0073 − 0 . 9826 = 0 . 0247 max Δ x 34 = 0.0247 U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 6 (k = 5):

1 x1 = ( 4 + 1.0054 - 1.0073 ) = 0.9995 4 1 6 x 2 = ( 6 - 1.0 - 1.0071 ) = 0.9982 4 1 6 x 3 = ( 5 - 1.0 - 1.0054 ) = 0.9982 3 Δx15 = 0.9995 − 1 = −0.0005 6

Δx 25 = 0.9982 − 1.0054 = −0.0072 Δx 35 = o .9982 − 1.0073 = −0.0091 max Δx35 = 0.0091 U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 7 (k = 6):

1 x 1 = ( 4 + 0.9982 - 0.9982 ) = 1.0 4 1 7 x 2 = ( 6 - 0.9995 - 0.9982 ) = 1.0006 4 1 7 x 3 = ( 5 - 0.9995 - 0.9982) = 1.0008 3 Δ x 16 = 1 − 0 . 9995 = 0 . 0005 7

Δ x 62 = 1 . 0006 − 0 . 9982 = 0 . 0024 Δ x 36 = 1 . 0008 − 0 . 9982 = 0 . 0026 max Δ x 36 = 0.0026 U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 8 (k = 7):

1 x1 = ( 4 + 1.0006 - 1.0008 ) = 0.9995 4 1 8 x2 = ( 6 - 1.0 - 1.0008 ) = 0.9998 4 1 8 x3 = ( 5 - 1.0 - 1.0008 ) = 0.9998 3 Δx17 = 0.9995 − 1 = −0.0005 8

Δx27 = 0.9998 − 1.0006 = −0.0008 Δx37 = 0.9998 − 1.0008 = −0.0010 max Δx37 = 0.0010 U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method The Gauss iterative method has converged at iteration 7. The method yields the following solution.

x 1 = 0.9995 x 2 = 0.9998 x 3 = 0.9998

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Gauss Iterative Method b) with x 1 = x 2 = x 3 = 0.5 0

0

0

Iteration 1 (k = 0):

x1 1 = x21 = x31 =

Δ x 10 = Δ x 20 = Δ x 30 = max Δ x = U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 2 (k = 1):

x1 = 2

x2 2 = x3 2 =

Δx11 = Δx21 = Δx31 = max Δx 1 = U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 3 (k = 2):

x1 3 = x2 = 3

x3 3 =

Δx12 = Δx22 = Δx32 = max Δx 2 = U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 4 (k = 3):

x1 = 4

x2 4 = x3 4 =

Δx13 = Δx23 = Δx33 = max Δx 3 = U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 5 (k = 4):

x1 5 = x2 = 5

x3 5 =

Δx14 = Δx24 = Δx34 = max Δx = 4

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Gauss Iterative Method Iteration 6 (k = 5):

x1 = 6

x2 6 = x3 6 =

Δx15 = Δx25 = Δx35 = max Δx 5 = U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 7 (k = 6):

x1 7 = x2 7 = x3 7 =

Δx16 = Δx26 = Δx36 = max Δx 6 = U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Iteration 8 (k = 7):

x18 = x2 8 = x3 8 =

Δx17 = Δx27 = Δx37 = max Δx7 = U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method =

x

1

x

2

=

x

3

=

Note: Number of iterations to achieve convergence is also dependent on initial estimates U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method Given the system of algebraic equations,

a 11 x 1 + a 12 x 2 + L + a 1n x n = y 1 a 21 x 1 + a 22 x 2 + L + a 2n x n = y 2 ↓

↓

↓

↓

a 31 x 1 + a 32 x 2 + L + a nn x n = y 3 In the above equation, the x’s are unknown.

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Gauss Iterative Method From the first equation,

a 11 x 1 = y 1 − a 12 x 2 K − a 1n x n 1 x1 = ( y 1 − a 12 x 2 K − a 1n x n ) a 11

Similarly, x2, x3…xn of the 2nd to the nth equations can be obtained.

1 x2 = (b 2 − a 21 x 1 − a 23 x 3 − K a 2n x n ) a 22 ↓

↓

↓

↓

1 ( bn − a n1 x 1 − a n2 x 2 − K − a n,n -1 x n -1 ) xn = a nn U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method In general, the jth equation may be written as

1 n xj = ( b j − ∑ i = 1 a ji x i ) a jj

equation “a”

i≠ j

j = 1, 2,K n

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Gauss Iterative Method In general, the Gauss iterative estimates are:

x1 x2

xn

k +1

k +1

k +1

a 13 a 1n y1 a 12 k k k = − x2 − x 3 − ... − xn a 11 a 11 a 11 a 11 a 2n y2 a 21 a 23 k k k = − x1 − x 3 − ... − xn a 22 a 22 a 22 a 22

a n, n - 1 yn a n1 k a n2 k k = − x1 − x 2 − ... − x n -1 a nn a nn a nn a nn

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Gauss Iterative Method From an initial estimate of the unknowns (x10, x20,…xn0), updated values of the unknown variables are computed using equation “a”. This completes one iteration. The new estimates replace the original estimates. Mathematically, at the kth iteration,

k +1

xj

1 n = ( b j − ∑ i=1 a a jj i≠ j

k

ji

x )

equation “b”

i

j = 1, 2,K n U. P. National Engineering Center National Electrification Administration

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Gauss Iterative Method A convergence check is conducted after each iteration. The latest values are compared with their values respectively.

Δx = x k

k +1 j

− x

k j

equation “c”

j = 1, 2,K n The iteration process is terminated when

max | Δ x | < ε k j

k = itermax U. P. National Engineering Center National Electrification Administration

(convergen t) (non - convergent )

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Gauss-Seidel Method Solution: a) The system of equation must be expressed in standard form.

x

k +1 1

x

k +1 2

x

k +1 3

1 k k ( 4 + x2 - x3 ) = 4 1 k +1 k = ( 6 - x1 - x3 ) 4 1 k +1 k +1 = ( 5 - x1 - x2 ) 3

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Gauss-Seidel Method Example: Solve the system of equations using the Gauss-Seidel method.

Used a convergence index of Îľ = 0.001

4x

1

â&#x2C6;&#x2019; x2 + x3 = 4

x 1 + 4x

+ x3

2

x 1 + x 2 + 3x

=6 3

= 5

x 1 = x 2 = x 3 = 0.5 0

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0

0

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Gauss-Seidel Method with x 1 0 = x 2 0 = x 3 0 = 0.5 Iteration 1 (k =0):

1 ( 4 + 0.5 - 0.5 ) = 1.0 4 1 1 ( 6 - 1.0 - 0.5 ) = 1.125 x2 = 4 1 x31 = ( 5 - 1.0 - 1.125 ) = 0.9583 3 Δ x 10 = 1 − 0 . 5 = 0 . 50 x11 =

Δ x 20 = 1 . 125 − 0 . 50 = 0 . 625 Δ x 30 = 0 . 9583 − 0 . 50 = 0 . 4583 max | Δ x 20 | = 0.625 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Method Iteration 2 (k = 1):

1 x1 2 = ( 4 + 1.125 - 0.9583 ) = 1.0417 4 1 2 x2 = ( 6 - 1.0417 - 0.9583 ) = 1.0 4 1 x3 2 = ( 5 - 1.0417 - 1.0 ) = 0.9861 3 Δx11 = 1.0417 − 1 = 0.0417

Δx21 = 1 − 1.125 = −0.125 Δx31 = 0.9861 − 0.9583 = 0.0323 max | Δx21 | = 0.125 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Method Iteration 3 (k = 2):

1 x 1 = ( 4 + 1.0 - 0.9861 ) = 1.0035 4 1 3 x 2 = ( 6 - 1.0035 - 0.9861 ) = 1.0026 4 1 x 3 3 = ( 5 - 1.0035 - 1.0026 ) = 0.9980 3 Δ x 12 = 1 . 0035 − 1 . 0417 = − 0 . 0382 3

Δ x 22 = 1 . 0026 − 1 = 0 . 0026 Δ x 32 = 0 . 9980 − 0 . 9861 = 0 . 0119 max | Δ x 32 | = 0.0119 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Method Iteration 4 (k = 3):

1 x1 4 = ( 4 + 1.0026 - 0.9980 ) = 1.0012 4 1 4 x2 = ( 6 - 1.0012 - 0.9980) = 1.0002 4 1 4 x3 = ( 5 - 1.0 - 1.0012 - 1.0002) = 0.9995 3 Δx13 = 1.0012 − 1.0035 = 0.0023

Δx23 = 1.0002 − 1.0026 = −0.0024 Δx33 = 0.9995 − 0.9980 = 0.0015 max | Δx23 | = 0.0024 U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Method Iteration 5 (k = 4):

1 ( 4 + 1.0002 - 0.9995 ) = 1.0002 4 1 x 2 5 = ( 6 - 1.0002 - 0.9995) = 1.0001 4 1 5 x 3 = ( 5 - 1.0002 - 1.0001) = 0.9999 3 Δ x14 = 1 .0002 − 1 .0012 = − 0 .001 x1 5 =

Δ x 24 = 1 .0001 − 1 .0002 = − 0 .0001 Δ x 34 = 0 .9999 − 0 .9995 = 0 .0004 max | Δ x 4 | = 0.001 < ε U. P. National Engineering Center National Electrification Administration

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Gauss-Seidel Method The Gauss-Seidel Method has converged after 4 iterations only with the following solutions:

x 1 = 1.0002 x 2 = 1.0001 x 3 = 0.9999

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Gauss-Seidel Method The Gauss-Seidel method is an improvement over the Gauss iterative method. As presented in the previous section, the standard form of the jth equation may be written as follows.

1 xj = (bj − a jj

∑

n i=1 i≠ j

a ji x i )

j = 1, 2, K n

From an initial estimates (x10, x20,…xn0), an updated value is computed for x1 using the above equation with j set to 1.This new value replaces x10 and is then used together with the remaining initial estimates to compute a new value for x2. The process is repeated until a new estimate is obtained for xn. This completes one iteration.

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Gauss-Seidel Method Note that within an iteration, the latest computed values are used in computing for the remaining unknowns. In general, at iteration k,

x

k +1 j

1 = (bj − a jj

∑

n i=1

a

i≠ j

α

ji

xi )

j = 1, 2, K n where α = k

if i > j

= k + 1 if i < j After each iteration, a convergence check is conducted. The convergence criterion applied is the same with Gauss Iterative Method.

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Gauss-Seidel Method An improvement to the Gauss Iterative Method k +1

x1

k +1

x2

y1 a 12 a 1n k k = − x 2 − ... − xn a 11 a 11 a 11 a 2n k y2 a 21 k + 1 = − x 1 − ... − xn a 22 a 22 a 22

ai,i-1 k+1 ai,i+1 k+1 ain k+1 yi aij k+1 xi = − xi −...− xi-1 − xi+1 − xn aii aii aii aii aii k+1

k +1

xn

a n, n - 1 yn a n1 k + 1 k +1 = − x 1 − ... − x n -1 a nn a nn a nn

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Newton-Raphson Method Solve the non-linear equations

x12 â&#x2C6;&#x2019; 4x2 = 4 2x1 â&#x2C6;&#x2019; x2 = 2 The Newton-Raphson method is applied when the system of equations is non-linear.

Consider a set of 2 non-linear equations in 2 unknowns.

y 1 = f 1 ( x1 , x 2 )

y 2 = f 2 ( x1 , x 2 ) U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Method The system of non-linear equations can be linearized using the first order Taylor’s Series y1 = f 1 ( x 0 ) +

∂f 1 0 ∂f 0 0 ( x ) Δ x1 + 1 ( x 0 ) Δ x 2 ∂ x1 ∂x2

y2 = f 2 ( x 0 ) +

∂f 2 0 ∂f 0 0 ( x ) Δ x1 + 2 ( x 0 ) Δ x2 ∂x1 ∂x2

Where: x0 = (x10, x20) are set of initial estimates fi(x0) = the function fi (x1,x2) evaluated using the set of initial estimates.

0 ∂f i ( x ) = the partial derivatives of the function fi(x1,x2) ∂x j evaluated using the set of original estimates.

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Newton-Raphson Method The equation may be written in matrix form as follows:

⎡ y1 − f1 ( x )⎤ ⎡ = ⎢ ∂f ( x 0 ) ⎢ 0 ⎥ 2 ( ) y f x − ⎢ 2 ⎣ 2 ⎦ ⎣ ∂x1 0

∂f 1 ( x o ) ∂x 1

⎤ ⎡ Δx 0 ⎤ 1 ⎥ 0⎥ ∂f 2 ( x 0 ) ⎢ ⎥ Δx 2 ⎦ ∂x 2 ⎦ ⎣ ∂f 1 (x 0 ) ∂x 2

The matrix of partial derivatives is known as the Jacobian. The linearized system of equations may be solved for ∆x’s which are then used to update the initial estimates.

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Newton-Raphson Method At the kth iteration: k +1

= x1 + Δx1

k +1

= x2 + Δx2

x1

x2

k

k

k

k

Convergence is achieved when

y1 − f1(xk ) ≤ ε y2 − f2 (xk ) ≤ ε Where ε is pre-set precision indices.

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Newton-Raphson Method Solve the non-linear equation

x12 − 4x2 = 4 2x1 − x2 = 2 use: x1 = 1, x2 = −1 0

Solution:

0

First, form the Jacobian

f1 = x − 4 x2 2 1

f 2 = 2 x1 − x 2

∂f1 = 2 x1 ∂ x1 ∂f2 = 2 ∂ x1

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∂f1 = −4 ∂x2 ∂f 2 = −1 ∂x2

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Newton-Raphson Method In Matrix form

⎡ ∂f 1 ( x ) 0 ⎢ − y f ( x ) ⎡ 1 ⎤ ∂x1 1 ⎢ y − f ( x )⎥ = ⎢ ∂f ( x 0 ) ⎢ 2 ⎣ 2 2 0 ⎦ ⎢⎣ ∂x1 0

⎡4 − ( x1 − 4 x2 )⎤ ⎡ 2 x1 ⎢ 2 − ( 2 x − x )⎥ = ⎢ 2 ⎣ ⎣ 1 2 ⎦ 2

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∂f 1 ( x ) ⎤ ∂x2 ⎥ ⎡ Δx1 ⎤ 0 ⎥⎢ ∂f 2 ( x ) ⎥ ⎣Δx2 ⎥⎦ ∂x2 ⎥⎦ 0

- 4 ⎤ ⎡ Δx1 ⎤ ⎢ ⎥ ⎥ − 1⎦ ⎣Δx2 ⎦

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Newton-Raphson Method Iteration 0:

f 1 ( x 0 ) = 12 − 4( −1 ) = 5 , y1 = 4 ∂f 1 = 2(1) = 2 ∂x1 ∂f 1 = −4 ∂x2

f 2 ( x 0 ) = 2( 1 ) − ( − 1 ) = 3 , y 2 = 2 ∂f 2 =2 ∂x1 ∂f 2 = −1 ∂x 2

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Newton-Raphson Method The equations are:

In matrix form:

4 − 5 = ( 2 )Δx + ( −4 )Δx2 0 1

2 − 3 = ( 2 )Δx1 + ( −1 )Δx2 0

0

0

⎡ − 1⎤ ⎡ 2 ⎢ − 1⎥ = ⎢ 2 ⎣ ⎦ ⎣

− 4 ⎤ ⎡ Δ x10 ⎤ ⎢ ⎥ 0⎥ − 1 ⎦ ⎣Δx2 ⎦

Solving,

Δ x1

0

= − 0 .5

Δ x

0

= 0

2

Thus,

x1 = 1 + ( −0.5 ) = 0.5 1

x2 = −1 + 0 = −1 1

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Newton-Raphson Method Repeating the process with the new estimates, Iteration 1:

f 1 ( x1 ) = ( 0.5 )2 − 4( −1 ) = 4.25 , y1 = 4 ∂f 1 ( x1 ) = 2( 0.5 ) = 1.0 ∂x1 ∂f 1 ( x ) = −4 ∂x2 1

f 2 ( x 1 ) = 2 ( 0 .5 ) − ( − 1 ) = 2

, y2 = 2

∂f 2 ( x 1 ) = 2 ∂x1 ∂f 2 ( x 1 ) = −1 ∂x2

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Newton-Raphson Method The equations are:

In matrix form:

4 − 4.25 = Δx1 − 4 Δ x 2

1

2 − 2 = 2 Δ x1 − Δ x 2

1

1

1

⎡ − 0.25 ⎤ ⎡ 1 ⎢ 0 ⎥ = ⎢2 ⎣ ⎦ ⎣

− 4 ⎤ ⎡ Δ x1 1 ⎤ ⎢ ⎥ − 1 ⎥⎦ ⎣ Δx 2 1 ⎦

Solving,

Δ x 1 = 0 . 03571 1

Δ x 2 = 0 . 07143 1

Thus,

x 1 = 0 .5 + 0 .03571 = 0 .53571 2

x 2 = − 1 + 0 .07143 = − 0 .92857 2

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Newton-Raphson Method Repeating the process with the new estimates, Iteration 2:

f 1 ( x 2 ) = ( 0.53571 )2 − 4( −0.92857 ) = 4.001265 , y1 = 4 ∂f 1 ( x 2 ) = 2( 0.53571 ) = 1.07142 ∂x1 ∂f 1 ( x 1 ) = −4 ∂x2

f 2 ( x 2 ) = 2( 0.53571 ) − ( −0.92857 ) = 1.99999 ≅ 2 ∂f 2 ( x 2 ) =2 ∂x 1

y2 = 2

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Newton-Raphson Method The equations are:

4 − 4 . 001265 = 1 . 07142 Δ x 1 − 4 Δ x 2 2

2 − 2 .0 = 2 Δ x1 − Δ x 2 2

Solving,

2

2

In matrix form:

⎡ − 0 .001265 ⎤ ⎡1 .07142 =⎢ ⎢ ⎥ 0 2 ⎣ ⎦ ⎣

− 4 ⎤ ⎡ Δ x1 2 ⎤ ⎢ 2 ⎥ ⎥ − 1 ⎦ ⎣Δx2 ⎦

Δ x 1 = − 0 . 00018 2

Δ x 2 = 0 . 00036 2

Thus,

x 1 = 0 . 53571 − 0 . 00018 = 0 . 53553 3

x 2 = − 0 . 92857 − 0 . 00035 = − 0 . 92893 3

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Newton-Raphson Method Substituting to the original equation:

f 1 ( x1 ) = ( 0.53553 )2 − 4( −0.92893 ) = 4.0025124 , y1 = 4 3 f 2 ( x1 ) = 2( 0.53553 ) − ( −0.92893 ) = 1.99928 , y 2 = 2 3

y 1 − f 1 = − 0 . 0025 y 2 − f 2 = 0 . 00072

Therefore,

x 1 = 0 .53553 x 2 = − 0 .92893 Note the rapid convergence of the Newton-Raphson Method.

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Newton-Raphson Method The Newton-Raphson method is applied when the system of equations is non-linear.

Consider a set of n non-linear equations in n unknowns.

y 1 = f 1 ( x 1 , x 2 ,K , x n ) y 2 = f 2 ( x 1 , x 2 ,K , x n ) M y n = f n (x 1 , x 2 , K , x n ) U. P. National Engineering Center National Electrification Administration

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Newton-Raphson Method The system of non-linear equations can be linearized using Taylor’s Series

∂f1 0 ∂f1 0 ∂f1 0 0 0 0 y1 = f1 ( x ) + ( x )Δx1 + ( x )Δx2 + K + ( x )Δxn ∂x1 ∂x2 ∂xn 0

∂f 2 0 ∂f 2 0 ∂f 2 0 0 0 0 y2 = f 2 ( x ) + ( x )Δx1 + ( x )Δx2 + K+ ( x )Δxn ∂xn ∂x2 ∂x1 0

M

M

M

M

∂f n 0 ∂f1 0 ∂f1 0 0 0 0 yn = f n ( x ) + ( x )Δx1 + ( x )Δx2 + K+ ( x )Δxn ∂x1 ∂x2 ∂xn 0

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Newton-Raphson Method Where: X0 = (x10,x20, …, xn0) = set of initial estimates fi(x0) = the function fi (x1,x2, …, xn) evaluated using the set of initial estimates.

∂f i ( x 0 ) = the partial derivatives of the ∂x j function fi(x1,x2,…,xn) evaluated using the set of original estimates.

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Newton-Raphson Method The equation may be written in matrix form as follows:

⎡ y1 − f 1 ( x ) ⎤ ⎡ ⎢ ⎥ ⎢ 0 ∂f 2 ( x 0 ) ⎢ y 2 − f 2 ( x )⎥ = ⎢ ∂x1 ⎢ ⎥ ⎢ M M ⎢ ⎥ ⎢ 0 ∂f n ( x 0 ) ⎢⎣ y n − f n ( x )⎥⎦ ⎢⎣ ∂x1 0

∂f 1 ( x o ) ∂x1

∂f 1 (x 0 ) ∂x 2 ∂f 2 ( x 0 ) ∂x 2

M ∂f n ( x 0 ) ∂x 2

0 ⎤ ⎡ Δx1 ⎤ K 0 ⎥⎢ 0⎥ ∂f 2 ( x ) Δx 2 ⎥ K ∂x n ⎥ ⎢ ⎥⎢ M ⎥ M ⎥ ⎢ 0⎥ ∂f n ( x 0 ) ⎥ K ⎣Δxn ⎥⎦ ∂x n ⎦ ⎢ ∂f 1 (x 0 ) ∂x n

The matrix of partial derivatives is known as the Jacobian. The linearized system of equations may be solved for ∆x’s which are then used to update the initial estimates.

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Newton-Raphson Method At the kth iteration:

x j = x j + Δx j k =1

k

k

j = 1, 2,...,n

Convergence is achieved when

y j − f j ( x ) ≤ ε1 or

k

j = 1,2,...,n

Δx j ≤ ε 2

j = 1, 2, ..., n

k

Where ε1 and ε2 are pre-set precision indices.

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