CPD1B17 Notes1 - Fundamental Principles & Methods

7/26/2019 CPD1B17 Notes1 - Fundamental Principles & Methods

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NATIONAL ELECTRIFICATION ADMINISTRATION

U. P. NATIONAL ENGINEERING CENTERU. P. NATIONAL ENGINEERING CENTERU. P. NATIONAL ENGINEERING CENTER

Distr ibution System Modeling and Analysis

Competency Train ing and Certif ication Program in Electri c Power System Engineer ing

Fundamental Principles andMethods in Power System Analysis


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Competency Training & Cert i f icat ion Program

in Electr ic Power System Engineering

U. P. National Eng ineering Center

National Electr i f icat ion Adminis trat ion



Fundamental Principles and Methods in Power System Analysis

Course Outline

1. Circuit Conventions and Notations

2. Power System Representation

3. Per Unit Quantities4. Symmetrical Components

5. Network Equations and Methods of Solution


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Voltage and Current Directions

Double Subscript Notation

Voltage, Current and Phasor Notation

Complex Impedance and Phasor Notation

Circuit Conventions & Notations


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Polarity Marking of Voltage SourceTerminals:

Plus sign (+) for the terminal wherepositive current comes out

Specification of Load Terminals:Plus sign (+) for the terminal wherepositive current enters

Specification of Current Direction:

Arrows for the positive current (i.e.,from the source towards the load)

Vs ZB

+

-

++

-

-

VA

ZA

VB

a b

o n

I

I

I

I

Voltage and Current Directions



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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.

+

-

+ -VA

ZAa b

o n

VS= Vao Vb = VbnI= Iab Zb

+

-

Double Subscript Notation


The current direction

is from first subscript

to the second

subscript .

Vao- IabZA - Vbn= 0

Iab =Vao - Vbn

ZA

F d l P i i l d M h d i P S A l i


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Voltage, Current and Phasor Notation


tjmVv tsinjtcos

tj

voltsV

V m 0

2

-1

-0.5

0

0.5

1

-4 -2 0 2 4

v

i

1j

V

I tjmIi

amperes

I

Im

2

F d t l P i i l d M th d i P S t A l i
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Vibrations/Animations/string2.gif


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+

-

+

-

VLi(t)

R (Resistance)

L (Inductance)

tjmS VV

The first order linear differential equation has a particular

solution of the form . tjK)t(i

tj

mVdt

)t(diL)t(Ri Applying Kirchoffs voltage law,

tj

m

tjtj VLKjRK Hence,


F d t l P i i l d M th d i P S t A l i


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Complex Impedance and Phasor NotationSolving for the current

Dividing voltage by current to get the impedance,

tjm

LjR

V)t(i

LjR

LjR

V

V

)t(i

)t(vZ

tjm

tj

m

Therefore, the impedance Z is a complexquantity with real part Rand an imaginary (j)

part L

L

R




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+

-

+

-

VLi(t)

R (Resistance)

C (Capacitance)

tjmS VV

For Capacitive Circuit, .

)

C

1(jRZ


Z= |Z|ej or Z = |Z|(cos + jsin) or Z = |Z|

R

Z

1

C


1Fundamental Principles and Methods in Power System Analysis


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1








v = 141.4 cos(t + 30 ) volts

i = 7.07 cos(t) amperesVmax = 141.4 |V| = 100 V = 10030

Imax = 7.07 |I| = 5 I = 50

Complex Impedance and Phasor NotationZ= |Z|ej or Z = |Z|(cos

+ jsin

) or Z = |Z|

10

17.32

30

3020

05

30100Z

1032.17)30sin30(cos20 jjZ




11/253

1








Electrical Symbols

Three-Line and Single-Line Diagram

Equivalent Circuit of Power System

Components

Impedance and Admittance Diagram

Bus Admittance Matrix

Power System Representation



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1








Generator

Transformer

Circuit B reaker

Transmission o r

Dist r ibut ion Line

Bus

or

G Switch

Node

Fuse

Electrical Symbols

Power System representation



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1








3-phase wye neutral grounded

3-phase delta connection

Ammeter

Voltmeter

3-phase wye neutral ungrounded

Protective Relay

V

A

R

Current Transformer

Potential Transformer

Power System representation



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Three Line DiagramThe three-line diagram is used to represent each phase of a three-phase power system.

Relays

Circui

tBrea

ker

Ma

inBus

R

R

R

R

CTs Distri bution Lines

Transformer

A B C


CircuitR

ec

loser



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p y y

The three-line diagram becomes rather cluttered for large powersystems. A shorthand version of the three-line diagram is referredto as the Single Line Diagram.

Single Line Diagram

R

BusCB Transformer Distribution L ine

CT and Relay

Recloser




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p y y

Equivalent Circuit of Power SystemComponents: Generator

Ec

Eb

Ea

Ic

Ib

Ia

b

a

c

sa jXR

aI+

-

aV

+

-

gE

Za

Zb Zc

Three-Phase Equivalent Single-Phase Equivalent



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X

2

HT RaRR

X

2

HT XaXX

Equivalent Circuit of Power System

Components: Transformer

+

-

+

-

TZ

Single-Phase Equivalent


HI

+

-

HV

+

-

XVa mjXcR

exI

XX XjaRa22 HH jXR



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Equivalent Circuit of Power SystemComponents: Transmission & DistributionLines

T&D Lines can be represented by an infinite series of

resistance and inductance and shunt capacitance.

l




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Equivalent Circuit of Power SystemComponents: Distribution Lines

ZccZcbZca

ZbcZbbZba

ZacZabZaa

Equivalent-Network

A

B

C

a

b

c

YccYcbYca

YbcYbbYba

YacYabYaa

1/2

YccYcbYca

YbcYbbYba

YacYabYaa

1/2

CaptureUnbalanced

Characteristics Three-Phase Equivalent




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Equivalent Circuit of Power SystemComponents: Long Transmission Lines

+

-

Vs VR

+

-

IS IR

Length = Longer than 240 km. (150 mi.)

lsinhZ'Z c

y

zZC

Characteristic ImpedancezyPropagation Constant

2

ltanh

Z

1

2

'Y

c

2

'Y





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Equivalent Circuit of Power SystemComponents: Medium-Length

Transmission LinesLength = 80 240 km. (50 - 150 mi.)

+

-

+

-

VsIS IR

ljxrZL

2

c/1

2

Y

2

YVR





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Equivalent Circuit of Power SystemComponents: Short Transmission Lines

Length = up to 80 km. (50 mi.)

+

-

Vs VR

+

-

Is = IR


ljxrZ L




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Single Phase Equivalent ofBalanced Three-Phase System

ZR

ZR

ZR

n

Eao = |E|0 V

Ebo= |E|

240 V

Eco= |E| 120 V

a

b

c

o

Ic

b

c

Iaa

Ib




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ZR

n

Eao = |E|0 V

ao

ZRZR

n

Ebo= |E|

240 V b

o

ZR

ZR

n

Eco= |E| 120 Vc

o

c

b

a

Ic

Ib

Ia

IZ0E

IR

a

)240(IZ

240EI

R

b

)120(IZ

120EI

R

c

Note:Currents are Balanced




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Single PhaseRepresentationof a BalancedThree-PhaseSystem

Eao = |E|0 V

n

a

o

a

Ia

ZR

ZR

ZR

n

Eao = |E|0 V

Ebo= |E| 240 V

Eco= |E|

120 V

a

b

c

o

Ic

b

c

Iaa

Ia

ZR




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Impedance and Admittance Diagrams

21 3

4

bca

Bus

1

23

4

Gen

a

bc

Line

1 - 3

2 - 31 - 4

2 - 4

3 - 4

Single Line Diagram



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21 3

4

Impedance Diagram

0

0

0

1

31

Ea za

z13

zd

zcza

ze

zf zg

zb

Ea Ec Eb

Generator

Line

zh



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VL

ILZpIs

Zg

+

-

Eg

VL

IL

The two sou rces w i l l be equ ivalent i f VL

and ILare the same for both circu i ts.

Eg = ISZPZg = Zp



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21 3

4

bca

Bus

1

23

4

Gen

a

bc

Line

1 - 3

2 - 31 - 4

2 - 4

3 - 4

Single Line Diagram




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21 3

4

Impedance Diagram

0

0

0

1

31

Ea za

z13

zd

zcza

ze

zf zg

zb

Ea Ec Eb

Generator

Line

zh




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Equivalent Sources

VL

ILZpIs

Zg

+

-

Eg

VL

IL

The two sou rces w i l l be equ ivalent i f VL

and ILare the same for both circu i ts.

Eg = ISZPZg = Zp




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21

3

4

Admittance Diagram

0

y13

y03y01

y23

y14 y24

y02

I1 I3 I2I1= Ea/zay01 = 1/za

I2= Eb/zb

y02 = 1/zbI3= Ec/zcy03 = 1/zc

y13 = 1/zd

y23 = 1/ze

y14 = 1/zf

y24 = 1/zg

y34 = 1/zh


y34



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at node 1:

144113310111 yVVyVVyVI

at node 4:

343424241414 yVVyVVyVV0

at node 2:

244223320222 yVVyVVyVI at node 3:

1313344323230333 yVVyVVyVVYVI

Applying Kirchoffs Current Law

Nodal Voltage Equations



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Rearranging the equations,

In matrix form,

4

3

2

1

342414342414

34133423032313

2423242302

1413141301

3

2

1

VV

V

V

yyyyyyy-yyyyy-y-

y-y-yyy0

y-y-0yyy

0I

I

I


4342414334224114

4343342313032231133

42432322423022

41431311413011

0 VyyyVyVyVy

VyVyyyyVyVyI

VyVyVyyyIVyVyVyyyI



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n

3

2

1

I

I

I

I

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

n

3

2

1

V

V

V

V

[ I ] = [Ybus][V]


Yii= self-admittance, the sum of all admittances terminating on thenode (diagonal elements)

Yij = mutual admittance, the negative of the admittances connecteddirectly between the nodes identifed by the double subscripts



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nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

[YBUS] =


Ybusis also called Bus Admittance Matrix



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The Per Unit System

Per Unit Impedance

Changing Per Unit Values

Consistent Per Unit Quantities of PowerSystem

Advantages of Per Unit Quantities

Per Unit Quantities



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Base Value

Actual ValuePer Unit Value

Per-unit Value is a dimensionless quantity

Per-unit value is expressed as decimal

100

Actual ValuePercent

Per Unit Value

The Per Unit System


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PU Voltage

PU Current =

PU Impedance

PU Power = PU Voltage x PU Current

Per Unit Calculations

The Per Unit System



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I

Zline= 1.4 75

Zload= 20 302540 0 V

+

-

+

-

Example:

Vs= ?

Determine Vs


The Per Unit System


h i


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Choose: Base Impedance = 20 ohms (single phase)

Base Voltage = 2540 volts (single phase)

PU Impedance of the load = 2030 /20 = ______ p.u.

PU Impedance of the line = 1.475 /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.

The Per Unit System



Th P U i S


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The magnitude of the voltage at the substation is

1.05 p.u. x 2540 Volts = _______ Volts

1.0

-30 p.u.

1.05

2.70

0.07 75 p.u.

1.0

30 p.u. 1.0

0 p.u.

+

-

+

-

The Per Unit System



Th P U it S t


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1. Base values must satisfy fundamental electricallaws (Ohms Law and Kirchoffs 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

The Per Unit System

Establishing Base Values


Th P U it S t


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Base Power

Base Current = Base Voltage

Base Voltage (Base Voltage)2

Base Impedance = =Base Current Base Power

For a Given Base Power and Base Voltage,

The Per Unit System



Th P U it S t


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For Single Phase System,

Pbase(1)------------Vbase(1

)

Ibase=

Vbase(1)------------Ibase(1)

Zbase=

[Vbase(1)]= ------------Pbase(1)

For Three Phase System,

Pbase(3)------------3Vbase(LL)

Ibase=

Vbase(LN)------------Ibase(L)

Zbase=

[Vbase(LL)]= ------------Pbase(3)

The Per Unit System



The Pe U it S te


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3

kVBasekVBase

MVABase3

1MVABase

LL1

31

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

Base Values can be established from Single Phase or Three

Phase Quantities

The Per Unit System



The Per Unit System


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Base kVA1 = 10,000 kVA

= 10 MVA

Base kVLN = 69.282 kVBase Z = (69.282)2/10

= 480 ohms

Base kVA3 = 30,000 kVA

= 30 MVA

Base kVLL = 120 kVBase Z = (120)2/30

= 480 ohms

Amps144.34

)120(3

1000x30CurrentBase

Amps144.34282.69

1000x10CurrentBase

Example:

The Per Unit System



Per Unit Impedance


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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

The Base Values used by manufacturers are:

1. Rated Capacity (MVA, KVA or VA)

2. Rated Voltage (kV or V)

} PositiveSequenceImpedances

Per Unit Impedance

Generators


Per Unit Impedance


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base

)(L)pu(L

Z

XX

base

)(

)pu(Z

RR

base

)(C

)pu(C

Z

XX

Per Unit Impedance

Transmission and Distribution Lines


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Per Unit Impedance


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Example

A single-phase transformer is rated 110/440 V, 2.5 kVA.The impedance of the transformer measured from the low-

voltage 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 sideSolution

Low-voltage Zbase= = ______ ohms1000/5.2

110.02

PU Impedance, Zpu = = ______ p.u.

Per Unit Impedance


Per Unit Impedance


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High Voltage, Zbase = = _______ ohms

PU Impedance, Zpu = = _______ p.u.

If impedance had been measured on the high-voltageside, the ohmic value would be

ohmsZ _______

110

44006.0

2

Note: PU value of impedance referred to any side of the

transformer is the same

Per Unit Impedance

5


Changing Per-Unit Values


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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 andlow voltage side at 100 MVA.


5




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SOLUTION:

a) Pbase = 20 MVA

Vbase = 67 kV (high voltage)

( kV)

= ________ ohms( MVA)

Zbase =

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


5




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b) Pbase = 20 MVA

Vbase = 13.2 kV (low voltage)

= ________ ohmsZbase =

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

( kV)( MVA)

SOLUTION:


5




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c) Pbase = 100 MVA

Vbase,H = 67 kV(67)

= ________ ohms100

Zbase,H =

ohms= ______ p.u.Xp.u.,H=

Vbase,L = 13.2 kV

(13.2)= _______ ohms

100Zbase,L =

= ______ p.u.Xp.u.,L=

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

ohms

ohms

ohms


5




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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 10 MVA, 3, 13.8/138 kV, leakage reactance 10%

B-C 10 MVA, 3

, 138/69 kV, leakage reactance 8%

Determine the voltage regulation if the voltage at the load is 66 kV.

SOURCEA B C

300 /

PF=100 %A-B B-C


LOAD


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6




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SOLUTION USING PER UNIT METHOD:

Pbase = 10 MVA

VA,base = 13.8 kV

VB,base = 138 kV

VC,base = 69 kV(69)

-------- = _____ ohms

10

ZC,base =

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

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

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


6




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VA = _______ + ( ________ ) x ( ________ + ________ )

= ________ + j ________ p.u.

= _________ p.u.

VNL- VL---------------- x 100%

VL

V.R.=

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

=

a g g e a ues

6




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Consider the previous example, What if transformer A-B is20 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 beadded 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

g g

6




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Convert per unit value of 20 MVA transformer,

Pbase = 20 MVA (Power Rating)

Vbase,H = 138 kV (Voltage Rating)(138)

---------- = _______ ohms20

Zbase,H =

0.10 p.u. x _______ ohms = _______ ohmsXactual,H =

At Pbase = 10 MVA (new base)(138)

---------- = 1904.4 ohms10

Zbase,H =

95.22---------- = 0.05 p.u.1904.4

Xp.u.(new) =

The per unit impedance of

the 20MVA and 10 MVA

transformer can now beadded.

g g




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Zactual= Zpu1 Zbase1 Zactual= Zpu2 Zbase2

2base2pu1base1pu ZZZZ

2base

1base1pu2puZ

ZZZ

Note that the transformer can have different per unitimpedance for different base values (i.e., the actual ohmic

impedances of the equipment is independent of the selected

base values), then

g g




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Recall:

Powerbase

voltagebase

Z

2

base

2,3

2

2,

1,3

2

1,

12

base

baseLL

base

baseLL

pupu

MVA

kV

MVA

kV

ZZ

Then,

or,

1base,3

2base,3

2

2base,LL

1base,LL

1pu2pu MVA

MVA

kV

kV

ZZ

g g




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A three-phase transformer is rated 400 MVA, 220Y/22 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 thetransformer.

Example

g g

)given(base

)new(base

2

)new(base

2

)given(base

)given.(u.p)new.(u.pP

P

x]V[

]V[

xZZ




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Solution

On its own base the transformer reactance is

On the chosen base the reactance becomes

= ________ puX =( )

( )2

( )

X = ( ) x x = ________ pu( )2

( )2( )


Consistent Per Unit Quantitiesf P S t


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of Power System

Procedure:a) Establish Base Power and Base Voltages

Declare Base Power for the whole PowerSystem

Declare Base Voltage for any one of the PowerSystem components

Compute the Base Voltages for the rest of thePower System Components using the voltageratio of the transformers

Note: Def ine each subsystem wi th uni que Base Voltage based on separation due to

magnetic coupl ing




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b) Compute Base Impedance and Base Current

Using the Declared Base Power and BaseVoltages, compute the Base Impedances andBase Currents for each Subsystem

c) Compute Per Unit Impedance Using the declared and computed Base Values,

compute the Per Unit values of the impedanceby:

Dividing Actual Values by Base Values

Changing Per Uni t I mpedance with change in BaseValues

of Power System




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Generator 1 (G1): 300 MVA; 20 kV; 3; Xd = 20 %Transmission Line(L1): 64 km; XL= 0.5 / km

Transformer 1 (T1):3; 350 MVA; 230 / 20 kV; XT1= 10 %

Transformer 2 (T2):3-1; 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 %

T1

L1

T2

G1

G2

G3

of Power System

Use Base Power = 100 MVA




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E1

XT1

E2 E3

T1

Transmission Line

T2

G1

G2

G3

XLINE XT2

XG1 XG2 XG3

of Power System

1 2 3

4


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Consistent Per Unit Quantitiesof Po e S stem


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c) Compute Per Unit Impedance

of Power System

G2:

T1:

G1:

G3:

L1:

T2:


Advantages ofPer-Unit Quantities


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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.




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3. When working in the per-unit system, base voltages can beselected 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 three-

phase circuits does not affect the per-unit impedances ofthe 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.

Per Unit Quantities




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6. Network calculations are done in a much more handierfashion with less chance of mix-up

between phase and line voltages

between single-phase and three-phase powers, and

between primary and secondary voltages.

Per Unit Quantities


Symmetrical Components


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Sequence Components of Unbalanced Phasor

Sequence Impedance of Power SystemComponents

Practical Implications of SequenceComponents of Electric Currents


Sequence Components ofUnbalanced Phasor


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U a a ced aso

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 canbe used for the Balanced Three-Phase PowerSystem




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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 Analysiscannot be used for an unbalanced three-phase

power system.




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Any unbalanced three-phase system of phasorsmay be resolved into three balanced systems ofphasors which are referred to as the symmetricalcomponents of the original unbalanced phasors,namely:

a) POSITIVE-SEQUENCE PHASOR

b) NEGATIVE-SEQUENCE PHASOR

c) ZERO-SEQUENCE PHASOR




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Phase a

Phase b

Phase c

120

120

120

REFERENCE PHASE SEQUENCE: abc

Zero Sequence Phaso rs are sing le-ph ase, equal

in magnitude and in the same direct ion.

Posit ive Sequence Phasors are

three-phase, balanced and have

the phase sequence as the

origin al set of unbalanced

phasors.

Negative Sequence Phasors are three-phase, balanced but with a

phase sequence opposite to that original set of unbalnced

phasors.




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Va = Va1 + Va2 + Va0

Vb = Vb1 + Vb2 + Vb0

Vc= Vc1 + Vc2 + Vc0

Each of the original unbalanced phasor is the sum

of its sequence components. Thus,

Where,

Va1Positive Sequence component of VoltageVa

Va2

Negative Sequence component of VoltageVaVa0Zero Sequence component of VoltageVa




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a = 1

120

OPERATORa

An operator a causes a rotation of 120 in the

counter clockwise direction of any phasor.

V

120

aV

Operating V by a

a = 1 240

a = 1 0




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Vb in terms of Va

Vb = a VaVb1 = a Va1

Vc in terms of Va

Vc = a VaVc1 = a Va1

Va

Vc= aVa

120

120

120

Vb= a2Va

The original Phasor and Positive Sequencecomponents in terms of phase a


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Vb in terms of Va

Vb0

= Va0

Vc in terms of Va

Vc0 = Va0

Va0= Vb0= Vc0

The Zero Sequence components interms of phase a




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Writing again the phasors in terms of phasor Va

and operator a,

Va = Va0 + Va1 + Va2Vb = Va0 +aVa1 + aVa2Vc= Va0 + aVa1 + aVa2

Computing for Va0, Va1& Va2

c2ba1acba0a VaaVV3

1VVVV

3

1V

cb2

a2a aVVaV31V




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EXAMPLE:Determine the symmetricalcomponents of the followingunbalanced voltages.

Vc= 8 143.1

Vb = 3 -90

Va= 4 0


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For Phasor Va:

86.082.15

143.1)120)(8(190)-240)(3(10431

)aVVaV(3

1V cb

2

a2a


C t f V b bt i d b ti th



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Components of Vbcan be obtained by operating thesequence components of phasor Va.

33.922.1586.08)-120)(2.15(1

aVV

101.62-4.9

258.384.918.38)240)(4.9(1

VaV

143.051143.051

VV

a2b2

1a

2

b1

a0b0


Si il l t f h V b bt i d b



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Similarly, components of phasor Vccan be obtained byoperating Va.

3.92512.15

86.08)-0)(2.1542(1

VaV

8.38314.9

18.38)0)(4.921(1

VaV

143.051143.051

VV

a2

2

c2

1ac1

a0c0


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Components of Vc

Components of Va

Components of Vb

Add Sequence Components Graphically

Va0

Vb0

Vc0

Vc1

Va1

Vb1

Va2

Vb2

Vc2

Vc= 8 143.1

Vb = 3 -90

Va= 4 0


h l b h k d i h h i ll



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The results can be checked either mathematically or

graphically.

143.18153.922.15138.384.905.1431

VVVV

90-3

33.922.15101.62-4.9143.051VVVV

04

86.08-2.1518.384.905.1431

VVVV

c2c1c0c

b2b1b0b

a2a1a0a




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Ia

b

a

c

b

a

c

Ib

Ic

Ia1 + Ia2+ Ia0

Ib1 + Ib2+ Ib0

Ic1

+ Ic2

+ Ic0


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Sequence Impedance ofPower System Components


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In general,

Z1 Z2 Z0 for generators

Z1 = Z2 = Z0 for transformers

Z1 = Z2 Z0 for l ines


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ZERO-SEQUENCE CURRENTS:


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ZERO-SEQUENCE CURRENTS:

Ia0

3Io

Ic0

Ib0

b

a

c

The neutral return (ground )

carries the in-phase zero-

sequence currents.

In-phase zero-sequence cu rrents

circu lates in the delta-con nected

transform er windings .

There is balancing ampere-turns

for the zero-sequence currents .

IA0

IC0

IB0

B

A

C

I0 = 0

I0 = 0

I0 = 0



NEGATI VE SEQUENCE CURRENTS


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NEGATI VE-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.


Network Equations and Methods ofSolution


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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


The standard form of n independent equations:

Network Equations


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p q

n

3

2

1

I

I

II

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

n

3

2

1

V

V

VV

[ 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 connecteddirectly between the nodes identifed by the double subscripts


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Network Equations


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21 3

4

Admittance Diagram

0

y13

y03y01

y23

y14 y24

y02

I1 I3 I2I1= Ea/zay01 = 1/za

I2= Eb/zby02 = 1/zb

I3= Ec/zcy03 = 1/zc

y13 = 1/zd

y23 = 1/ze

y14 = 1/zf

y24 = 1/zg

y34 = 1/zh


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Matrix Representations ofSystem of Equations


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Va = Va0 + Va1 + Va2Vb = Va0 +aVa1 + aVa2Vc= Va0 + aVa1 + aVa2

Rearranging and writing in matrix form

2a

1a

0a

2

2

c

b

a

V

V

V

aa1

aa1

111

V

V

V

Sequence Components of Unbalanced Phasor


Matrix Representations ofSystem of Equations

2b1b0 VaaVV1VVVV1V


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c

b

a

2

2

2a

1a

0a

V

V

V

aa1

aa1

111

3

1

V

V

V

cba1acba0a VaaVV3

VVVV

3

V

cb2a2a aVVaV3

1V

In matrix form


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Definition of a MATRIX

1n131211 aaaa


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mnm3m2m1

3n333231

2n232221

1n131211

ij

aaaa

aaaa

aaaa

aA

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

[ai j]mxn


Definition of a Vector

A vector Xis defined as an ordered set of elements. The

components x X X may be real or complex


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components x1, X

2, X

nmay be real or complex

numbers or functions of some dependent variable.

n

2

1

x

x

x

X

ndefines the dimensionality or size of the vector.


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An upper triangular matrix is one where all the

Type of Matrices


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33

2322

131211

u00

uu0

uuu

U

An upper triangular matrix is one where all the

elements below the main diagonal are zero.


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An identity or unit matrix is a diagonal matrix where

Type of Matrices


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100

010

001

I

all elements on the main diagonal are equal to one.


Type of Matrices


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The null matrixis matrix whose elements areequal to zero.

000

000

000

N


Type of Matrices

A symmetric matrixis one where ai j = aj ifor all isandjs.


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872

731

215

S

j j

7aa

2aa

1aa

3223

3113

2112


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Addition of Matrices

Matrix Operations


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dd t o o at ces

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



Two matrices A = [aij] and B = [bi j] can be added


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together if they are of the same order (mxn). The sumC = A + B is obtained by adding the corresponding

elements.

C= [ci j] = [ai j+ bi j]


Example:



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110625B

372041A

482666

1)(31)(70)(26)(02)(45)(1BA

then,

262

624

1)(31)(70)(2

6)(02)(45)(1BA


5j71j22j33j61j42j1

Example:



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5j64j55j7

4j56j41j2B

9j81j13j6

1j13j51j4A

5j69j84j51j15j73j6

4j51j16j43j51j21j4

5j73j61j21j42j32j1

BA

then,


2j132j64j4



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14j145j62j13

5j69j92j6BA

4j23j48j1

3j43j10j28j10j20j2

BA



A matrix is multiplied by a scalar k by multiplyingall elements mn by k, that is,


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mn2m1m

n22221

n11211

kakaka

kakaka

kakaka

AkkA


Example: 34



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3kand

16

25A

318

615912

B

16

2534

3AkB


Example:6j3j14



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3kand

4j1j36

5j2j2-5A

12j39j18

15j66j15

18j93j12

B

j4-1j36

j52j2-5

j6-3j14

3AkB


Multiplication of Matrices

Two matrices A= [ai j] and B = [bi j]can be multiplied in


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the order ABif and only if the number of columns of Aisequal to the number of rows ofB.

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

If the product matrix is denoted b

CPD1B17 Notes1 - Fundamental Principles & Methods

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Transcript of CPD1B17 Notes1 - Fundamental Principles & Methods