tsdos phasor transformer

8/15/2019 TSDOS Phasor Transformer

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Phasor, Transformer, andTransmission Lines

Wei-Jen Lee, Ph.D., P.E.Professor and Director

Energy Systems Research Center

The University of Texas at Arlington

1

Text Book: J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.

Overbye, “Power Systems Analysis and Design”, Fifth Edition.Cengage Learning


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Introduction

2


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A Typical Power System

3


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Components of a Modern Power

System

• Every power system has four major

components

– Generation facilities

– Power delivery system

–Loads

– Monitoring, protection, and control system

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Challenges of Power System

Operation

• Every Generation facility has its own limit.

• Some generation facilities are difficult tocontrol.

• Delivery systems also have limitations.

• Loads are seldom constant

• Since the power systems are dynamic innature, it is difficult, if not impossible, to

maintain optimal conditions of a power

system.

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Notation - Power

• Power: Instantaneous consumption of energy

• Power Units Watts – voltage x current for dc (W) kW – 1 x 103 Watt MW – 1 x 106 Watt GW – 1 x 109 Watt

• Installed U.S. generation capacity is about900 GW ( about 3 kW per person)

• Installed generation capacity in the ERCOT is

around 95GW (Including 10GW wind

generation)6


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Power System Examples

• North American Interconnected System

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• North American Interconnected System

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• MicroGrid (Isolated power system)

• Intentional islanding operation of a power system(Military campus, remote villages, and etc)

• Airplanes and Spaceships

• Ships and submarines

• Automobiles: dc with 12 volts standard• Battery operated portable systems

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AC versus DC Power System

• With the flip of a switch at 3 p.m. on Sept. 4,

1882 at Pearl Station, Thomas Edisonushered in the age of commercial electrical

power systems.

• It was a DC MicroGrid and served 240

customers within one square miles.• Tesla's patents and theoretical work formed

the basis of modern alternating current

electric power (AC) systems.11


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Concept of Phasor

• Since it is relative easy to change the voltage

level, majority of the power systems are ACsystem (either 50Hz or 60Hz).

• Since the relationship of the AC system is

based upon the phase difference between

two quantities, this evolves the concept of phasor.

• Goal of phasor analysis is to simplify the

analysis of constant frequency AC systems. 12


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Instantaneous and RMS Values

• The voltage and current of an AC system can

be expressed as:

• The DC equivalent of an AC sinusoidal signal

is shown below: (RMS value)

)cos(*)(

)cos(*)(

max

max

i

v

t I t i

t V t v

2

)(1

|| max

0

2 V dt t v

T

V V

T

RMS

13


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Phasor Representation

• Euler’s Identity:

• Phasor notation is developed by rewriting theEuler’s Identity:

sincos jej

]Re[||2)(

)cos(||2)(

)( vt j

v

eV t v

t V t v

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Phasor Representation

• The RMS cosine-reference voltage and

current phasors are:

i

j

v

j

I e I I

V eV V

i

v

||||

||||

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V/I Relationships of R, L, and C

16


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Advantages of Phasor Analysis

0

2 2

Resistor ( ) ( )

( )Inductor ( )

1 1Capacitor ( ) (0)

C

Z = Impedance

R = Resistance

X = Reactance

XZ = =arctan( )

t

v t Ri t V RI

di t v t L V j LI

dt

i t dt v V I j C

R jX Z

R X R

Device Time Analysis Phasor

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Complex Power

max

max

max max

( ) ( ) ( )

v(t) = cos( )

(t) = cos( )

1

cos cos [cos( ) cos( )]2

1( ) [cos( )

2

cos(2 )]

V

I

V I

V I

p t v t i t

V t

i I t

p t V I

t

Power

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Complex Power

max max

0

max max

1

( ) [cos( ) cos(2 )]2

1( )

1 cos( )2

cos( )

= =

V I V I

T

avg

V I

V I

V I

p t V I t

P p t dt T

V I

V I

Power Factor

Average P

Angle

ower

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Complex Power

*

cos( ) sin( )

P = Real Power (W, kW, MW)

Q = Reactive Power (var, kvar, Mvar)

S = Complex power (VA, kVA, MVA)

Power Factor (pf) = cos

If current leads voltage then pf is leading

If current

V I V I

V I

S V I j

P jQ

lags voltage then pf is lagging

For simplicity, we will use V and I as RMSvalues of voltage and current

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Example

• Instantaneous, real, and reactive power; power

factor.• The voltage v(t)=141.4*cos(t) is applied to a load consisting of a 10-W resistor in parallel with an inductive reactance XL =L =3.77 W. Calculate the instantaneous power absorbed by theresistor and by the inductor. Also calculate the real and reactive

power absorbed by the load, and the power factor

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Example


factor.• The V/I relationship of resistor

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Example


factor.• The V/I relationship of inductor

23


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Example


factor.• The load voltage is

• The resistor current is

• The inductor current is

• The total load current is• The Instantaneous power absorbed by the resistor is

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Example


factor.

• The Instantaneous power absorbed by the inductor

is

• The real power absorbed by load is

• The reactive power absorbed by load is

• The power factor is25


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Example

• Power triangle and power factor correction

• A single-phase source delivers 100 kW to a loadoperating at a power factor of 0.8 lagging. Calculate

the reactive power to be delivered by a capacitor

connected in parallel with the load in order to raise

the source power factor to 0.95 lagging.

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Example

• Power triangle and power factor correction

•Before compensation

• After compensation

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RMS and Instantaneous values

• When voltages and currents are discussed in

this text, lowercase letters such as v(t) and

i(t) indicate instantaneous values, uppercase

letters such as V and I indicate rms values,

and uppercase letters in italics such as V and

I indicate rms phasors. When voltage or current values are specified, they shall be

rms values unless otherwise indicated.

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Single-Phase and Three-Phase

Power System

• Single-phase power system

• Single-phase-three wire system• Three-phase system

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Advantages of Three-Phase Power

System

• More power per kilogram of metal from a

three-phase machine.

• Power delivered to a balanced three-phase

load is constant at all time, instead of pulsing

as it does in a single-phase system.

• More power can be delivered.• Total line losses are reduced.

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Three-Phase Power System

• A three-phase generator consists of three

single-phase generators with voltages of

equal amplitudes and phase differences of

1200.

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Balanced Three-Phase Power

System

• We can used three

single-phase system

differing in phase angle

by 1200 to represent a

balanced three-phase

system.• The current flowing to

each load can be found

as V I

Z

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System

• Therefore, the currents flowing in each

phase are.0

0

0

0

120120

240240

A

B

A

V I I

Z

V I I

Z

V I I

Z

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System

• We will form a three-phase circuit by

connecting the return path of three-single

phase load together.

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System

• The current flows through the neutral

conductor is equal to zero.

0)240()120( ooC B A N

I I I I I I I

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D and Y Connection of Three-Phase Power System

• There are two types of connections in

three-phase circuits: Y and D.

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• General Relations Between Circuit

Quantities.

3

2

3

3

3

2

3

3033*303

303

3

30

3

30

303*

3

30

S

V

S

V V

I

V

I

V Z

V

S

Z

V I I

Connection

S

V

S

V V

I

V Z

ConnectionY

o

LL LLo

LL

L

o

LL LL

LL

o

LL

o

L

o

LL LL

o

LL

L

LN

Y

D

D

D

D

D

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• Phase quantity: voltages and currents in a

given phase.

• Line quantity: voltages between the lines

and currents in the lines connected to the

generators or loads.

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Power of a Balanced Three-Phase

Power System

• For a balanced Y-connected load with the

impedance Z

= Z 0 :

0

0

( ) 2 sin

( ) 2 sin( 120 )

( ) 2 sin( 240 )

an

bn

cn

v t V t

v t V t

v t V t

0

0

( ) 2 sin( )

( ) 2 sin( 120 )

( ) 2 sin( 240 )

a

b

c

i t I t

i t I t

i t I t

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Power of a Balanced Three-Phase

Power System

• The instantaneous power is:

• The total power on the load is

0

0

( ) cos cos(2 )( ) cos cos(2 240 )

( ) cos cos(2 480 )

a

b

c

p t VI t p t VI t

p t VI t

cos3)()()()( VI t P t P t P t P cbatotal

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Power Relationship

• Phase quantities in each phase of a Y- or

D-connection: – Real Power:

– Reactive Power:

– Apparent Power:

23 cos 3 cos P V I I Z

23 sin 3 sinQ V I I Z

23 3S V I I Z

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Power Relationship

• Line quantities of a Y-connection load:

– Power consumed by a load:

– Relationship between line quantities and

phase quantities:

– Power:

3 cos P V I

3 L LL

I I and V V

3 cos LL L P V I

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One-Line Diagram and D-YConversion

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Power Transformer

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Ideal Transformer

• For an ideal transformer, the following are

assumed:

• The windings have zero resistance; therefore,

the I2R losses in the windings are zero.

• The core permeability mc is infinite, whichcorresponds to zero core reluctance.

• There is no leakage flux; that is, the entire flux

Fc is confined to the core and links bothwindings.

• There are no core losses. 46


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Ideal Transformer

• For an ideal transformer, whatever goes

in should come out:

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Ideal Transformer

• For an ideal transformer, whatever goes

in should come out:

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Example of an Ideal Transformer

• A single-phase two-winding transformer is

rated 20 kVA, 480/120 V, 60 Hz. A source

connected to the 480-V winding supplies

an impedance load connected to the 120-

V winding. The load absorbs 15 kVA at

0.8 pf lagging when the load voltage is118V:

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• Solution:

• The complex power at load is

• The load current is

• The load impedance is

• The load impedance referred to the 480-V

winding is

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Equivalent Circuit of a Practical

Transformer

• A practical transformer :

The windings have resistance.

The core permeability mc is finite.

The magnetic flux is not entirely confined to the core.

There are real and reactive power losses in the core.

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Finding Parameters of a Practical

Single Phase Transformer

• Let’s use an example to illustrate the process. A single-phase two-winding transformer is rated 20

kVA, 480/120 volts, 60 Hz.

Short-circuit test: Rated current at rated frequency is

applied to the 480-volt winding with the 120-volt

winding shorted.

Test Results: V1 =35 volts, P1 =300 W

Open-circuit test: Rated voltage is applied to 120V,

with 480V terminal open.

Test Results: I2 =12 A, P2 =200 W

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• Short-circuit test:

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• Open-circuit test:

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• Discussion

• Allocate the impedance of primary and secondary

windings.

• In general, you can divide Xeq equally for equivalent

reactance of primary and secondary windings.

• Perform DC resistance testing to establish the equivalent

DC resistance ratio between primary and secondarywindings and allocate the Req accordingly.

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Polarity of the Transformer

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Per Unit System

•Quantity in per unit =

• Quantity in percent = (Quantity in per

unit)*100

Quantityof Value Base

Quantity Actual

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Advantages

• More meaningful when comparing differentvoltage levels

• The per unit equivalent impedance of thetransformer remains the same when referredto either the primary or the secondary side

• The per unit impedance of a transformer in athree-phase system is the same, regardlessthe winding connection

• The per unit method is independent of voltage changes and phase shifts throughtransformers61


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Selection of the Base

• VA, V, I, and Z are four power quantities

•One has to select two base quantities andderive the other two.

• Usually the base voltage VbaseLN and base

complex power Sbase1 are selected for either

a single-phase circuit or for one phase of athree-phase circuit.

• The per-unit values for current and

impedance can then be derived.63


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Selection of the Base

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Per Unit Representation of a

Transformer

• A single-phase two-winding transformer is

rated 20 kVA, 480/120 volts, 60 Hz.

• The equivalent leakage impedance of the

transformer referred to the 120-volt winding,

denoted winding 2, is Zeq2=0.0525/78.13oW.

Using the transformer ratings as basevalues, determine the per-unit leakage

impedance referred to winding 2 and referred

to winding 1.

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Transformer

• Solution

• The values of Sbase, Vbase1, and Vbase2 are from the

transformer ratings

• The base impedance on the 120-volt side of the

transformer is

• The per-unit leakage impedance referred to

winding 2 is

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Transformer

• Solution

• If Zeq2 is referred to winding 1

• The base impedance on the 480-volt side of the

transformer is

• The per-unit leakage reactance referred to

winding 1 is

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Transformer

The per-unit leakage impedance remains

unchanged when referred from winding 2 to

winding 1.

Form the per unit point of view transformer

voltage ratio is 1:1 (Rated to rated)

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Base Conversion

• To convert a per-unit impedance from ‘‘old’’to ‘‘new’’ base values, use.

or

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Example One: Base Conversion

• A 50-MVA, 34.5:161 kV transformer with

10% reactance is connected to a power

system where all the other impedance values

are on a 100 MVA, 34.5 or 161 kV base. The

reactance of the transformer under new base

is:

2.0*50

100*1.0

2

)(

2

)(

)(

newbase

old base

new pu KV

KV Z

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Example Two: Base Conversion

• A generator and transformer, as shown

below, are to be combined into a single

equivalent reactance on a 100 MVA, 110 kV

(high voltage side) base.

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Example Two: Base Conversion

• The transformer is operated at 3.9 kV tap.

• New base voltage at high side is 110 kV.

• The base voltage at low side is:

110*3.9/115 = 3.73 kV

514.1364.015.1

364.073.3

9.3*

30

100*1.0

15.173.3

4

*25

100

*25.0

)()(

2

2

)(

2

2

)(

new Xfer new geneq

new Xfer

new gen

Z Z Z

Z

Z

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Three Phase Transformer

• The American standard for marking three-

phase transformers substitutes H1, H2, and H3

on the high-voltage terminals and X1, X2, and

X3 on the low-voltage terminals in place of the

polarity dots.

• The ANSI/IEEE standard for transformersstates that the high voltage should lead the low

voltage by 30o with Y-D or D-Y banks.

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• There are nine possible configurations for

three-phase transformer.

• No Phase Shift: Y – Y, YG – Y, Y – YG, YG – YG, andD – D.

• With Phase Shift: D – Y, D – YG, Y – D, and YG – D.

• When describing the configuration of atransformer, it is typical to mention the

configuration of windings closer to the source

first.

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• Schematic representation of a YG – YG three-phase transformer.

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• Schematic representation of a YG – D three-phase transformer.

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• Per unit equivalent circuit of three-phase

transformers.

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Three-Winding Transformer

• Three-winding transformers are connected to

one source and two loads.

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• Z12: per-unit leakage impedance measured

from winding 1; with winding 2 shorted and

winding 3 open

• Z13: per-unit leakage impedance measured


winding 2 open• Z23: per-unit leakage impedance measured


winding 1 open79


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• We can be used to evaluate the per-unit series

impedances Z1, Z2, and Z3 of the three-winding

transformer equivalent circuit from the per-unitleakage impedances Z12, Z13, and Z23.

Z12 = Z1 + Z2

Z13 = Z1 + Z3 Z23 = Z2 + Z3

Z1 = (Z12 + Z13 - Z23)/2

Z2 = (Z12 + Z23 – Z13)/2

Z3 = (Z13 + Z23 – Z12)/280


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Per Unit Impedance of a Single-

Phase Three-Winding Transformer

• The ratings of a single-phase three-winding

transformer are: winding 1: 300 MVA; 13:8 kV winding 2: 300 MVA; 199:2 kV

winding 3: 50 MVA; 19:92 kV

• The leakage reactances, from short-circuit

tests, are: X12: 0.10 per unit on a 300-MVA; 13:8-kV base

X13: 0.16 per unit on a 50-MVA; 13:8-kV base

X23: 0.14 per unit on a 50-MVA; 199:2-kV base

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• Winding resistances and exciting current are

neglected.

• Calculate the impedances of the per-unit

equivalent circuit using a base of 300 MVA and

13.8 kV for terminal 1

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• Solution

• Sbase = 300 MVA is the same for all three terminals.

• The specified voltage base for terminal 1 is Vbase1 =

13:8 kV.

• The base voltages for terminals 2 and 3 are then

Vbase2 = 199:2 kV and Vbase3 = 19:92 kV.

• Base Conversion

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• Solution

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Autotransformer

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Autotransformer

• Pros and Cons of autotransformer

• Pros

• The autotransformer has smaller per-unit leakage

• impedances than the usual transformer; this results in both

smaller series voltage drops.

• The autotransformer also has lower per-unit losses (higher

efficiency), lower exciting current, and lower cost if the turns

ratio is not too large.

• Cons

• The electrical connection of the windings allows transient

overvoltages to pass through the autotransformer more

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Autotransformer

• Example

A single-phase two-winding 20-kVA, 480/120-volt

transformer of is connected as a 120V/600Vautotransformer. Determine (a) the kVA rating, and

(c) the per-unit leakage impedance (The original

leakage impedance of the transformer is )

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Autotransformer

• Solution

As a normal two-winding transformer rated 20 kVA,

the rated current of the 480-volt winding is I2 = IH =20000/480 = 41.667 A. As an autotransformer, the

480-volt winding can carry the same current.

Therefore, the kVA rating SH = EHIH = 600*41.667 =

25 kVA.

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Autotransformer

• Solution

As an autotransformer, the leakage impedance in

ohms is the same as for the normal transformer,since the core and windings are the same for both.

However, the base impedances are different.

Therefore

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Transformer with Off-Nominal TurnRatio

• For a transformer that V1rated and V2rated satisfy

V1rated = at*V2rated

• The selected voltage bases satisfy Vbase1 =

b*Vbase2. Define c = at/b.

• Then it can be represented as two transformers

in series.

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• The per unit model of the transformer

• Set up equations for two-port circuit

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• Where

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• p circuit representation for real c.

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Transmission System

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Transmission System

• Transmission line design consideration

An overhead transmission line consists of

conductors, insulators, support structures, and, inmost cases, shield wires.

Other factors of consideration: Electrical,

Mechanical, Environmental, and Economic

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Conductors

• Conventional Conductors

ACSR (Aluminum conductor steel reinforced) has

been in service for more than 80 years.

In the last 15 to 20 years, the AAAC

(homogeneous all-aluminum alloy conductor) has

become quite popular, especially for National Grid

in the UK where it is now the standard conductor type employed for new and refurbished lines.

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Conductors

• High Temperature Conductors

• Research in Japan in the 1960s produced a

series of aluminum-zirconium alloys that resistedthe annealing effects of high temperatures.

• These alloys can retain their strength at

temperatures up to 230 C

• The most common of these alloys—TA1, ZTA1and XTA1—are the basis of a variety of hightemperature conductors.

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Conductors

• Emerging Conductor Technologies

ACFR: Aluminum conductor carbon fiber

reinforced from Japan makes use of the very-lowexpansion coefficient of carbon fiber.

ACCR: Aluminum Conductor Composite

Reinforced from 3M.

Also in the United States, two more designsbased on glass-fiber composites are emerging.

Aluminum conductor composite core (ACCC),

Composite reinforced aluminum conductor (CRAC).

Aluminum Conductor Steel Supported (ACSS)99


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Insulator

• Insulators for transmission lines above 69 kV

are typically suspension-type Insulators.

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Support Structures

• Transmission lines employ a variety of

support structures.

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Shield Wires

• Shield wires located above the phase

conductors protect the phase conductors

against lightning.

• They are usually high- or extra-high-strength

steel, Alumoweld, or ACSR with much

smaller cross section than the phaseconductors.

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Transmission Line Parameters

• Resistance

The dc resistance of a conductor at a specified

temperature T is

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• Resistance

In English units, conductor cross sectional area is

expressed in circular mils (cmil). One inch equals1000 mils and 1 cmil equals p/4 sq mil.

A circle with diameter D in has an area

or

where 1000 D mil = d mil104


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• Resistance

Resistivity depends on the conductor metal.

Annealed copper is the international standard for measuring resistivity r.

Conductor resistance depends on the following

factors:

Spiraling Temperature

Frequency (‘‘skin effect’’)

Current magnitude—magnetic conductors

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• Resistance

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• Inductance: Solid Cylindrical Conductor

The inductance of a magnetic circuit that has a

constant permeability m can be obtained bydetermining the following:

Magnetic field intensity H, from Ampere’s law.

Magnetic flux density B (B = mH).

Flux linkages l. Inductance from flux linkages per ampere (L = l/I).

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For simplicity, assume that the conductor

is sufficiently long that end effects are neglected,

is nonmagnetic (m = m0 = 4p x10-7 H/m)

has a uniform current density (skin effect is neglected)

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Ampere’s law states that

The magnetic field inside the conductor

Select the dashed circle of radius x < r

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Flux Density B

The differential flux dF per-unit length of conductor inthe cross-hatched rectangle of width dx shown in

previous figure (dF = Bxdx Wb/m)

Computation of the differential flux linkage dl in therectangle is tricky since only the fraction (x/r)2 of the total

current I is linked by the flux.

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Integrating the dl from x = 0 to x = r determines the totalflux linkages lint inside the conductor.

The internal inductance Lint per-unit length of conductor due to this flux linkage is then

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The magnetic field outside the conductor

Select the dashed circle of radius x > r shown in Figureas the closed contour that encloses the entire current I.

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The magnetic field outside the conductor

Integrating dl between two external points at distancesD1 and D2 from the conductor center gives the external

flux linkage l12 between D1 and D2.

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The total flux lP linking the conductor out to

external point P at distance D is the sum of theinternal flux linkage and the external flux linkage

from D1 = r to D2 = D.

where

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The flux linkage lkPm, which links conductor k out

to P due to Im is:

Assume Dkm>>r

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Rearrange the equation:

Replace IM with

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• Inductance: Single-phase two-wire

where

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• Total inductance

• If conductors X and Y are the same

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• Inductance: Three-phase three-wire with

equal spacing

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• Inductance: Composite conductors

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• Since only the fraction (1/N) of the total conductor

current I is linked by this flux, the flux linkage lk of (the current in) subconductor k is

• The total flux linkage of conductor x is

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• Using ln A = ln A and ∑ ln Ak = ln Π Ak, we can

rewrite the previous equation as

• The Inductor

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Dxy is called the geometric mean distance or GMD

between conductors x and y. Dxx is called the geometric mean radius or GMR of

conductor x

Similarly, for conductor y,.

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• Example

For a single-phase multiple conductor circuit

shown below, find GMR, GMD, and Inductance

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• Solution

For N=3 and M=2’, GMD and GMR can be

expressed as:

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• Solution


expressed as:

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• Solution


expressed as:

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• Solution


expressed as:

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• Solution


expressed as:

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• Transposition

If the spacings between phases are unequal, then

balanced positive-sequence flux linkages are notobtained from balanced positive-sequence

currents. Balance can be restored by exchanging

the conductor positions along the line, a technique

called transposition.

134


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• Transposition

The total flux linking the phase a conductor while it

is in position 1 is

Similarly, the total flux linkage of this conductor

while it is in positions 2 and 3 is

135


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• Transposition

The average of the above flux linkages is

136


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• Transposition

The average inductance of phase a is

Defining

Then,

137


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• Bundle conductors

If the conductors are stranded and the bundle

spacing d is large compared to the conductor outside radius, each stranded conductor is first

replaced by an equivalent solid cylindrical

conductor with GMR = DS. Then the bundle is

replaced by one equivalent conductor with GMR =DSL.

138


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• Bundle conductors

Two-conductor bundle.

Three-conductor bundle

Four-conductor bundle

139


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• Series Impedances: Three-phase line with

neutral conductors and earth return

• Each earth return conductor carries the negative of its overhead conductor current, has a GMR

denoted Dk’k’, distance Dkk’ from its overhead

conductor, and resistance Rk’ given by:

where r is the earth resistivity in ohm-meters and f is frequency in hertz.141


142/248


• Series Impedances: Three-phase line with

neutral conductors and earth return

• Earth resistivities and 60-Hz equivalent conductor distances are shown in the table. It is common

practice to select r = 100 Wm when actual data areunavailable

142


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• Series Impedances calculation

First, renumber the overhead conductors from 1 to

(3 + N), beginning with the phase conductors, thenoverhead neutral conductors.

The sum of the currents in all the conductors is

zero

the flux linking overhead conductor k is

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• Series Impedances calculation

Convert the flux linkage in matrix format: l = L I . Where

When k = m, Dkk is the GMR of conductor k. When

k ≠ m, Dkm is the distance between conductors k

and m144


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• Circuit representation of a 1-meter section

series-phase impedances

145


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• Using this circuit, the vector of voltage drops

across the conductors is

where L is given in previous slide and R is a (3 + N) x (3 + N) matrix

of conductor resistances.146


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• Our objective now is to reduce the (3 + N)

equations to three equations

147


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• The diagonal elements of this matrix are

• The off diagonal elements of this matrix are

• Partition the matrix

where

148


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• Solve the matrix

• ZP is a 3x3 matrix

149


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• If the line is completely transposed

where

150

Electric Field and Voltage: Solid


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Cylindrical Conductors

• The capacitance between conductors in a

medium with constant permittivity e can be

obtained by determining the following: Electric field strength E, from Gauss’s law

Voltage between conductors

Capacitance from charge per unit volt (C = q/V)

151



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• Gauss’s law states that the total electric fluxleaving a closed surface equals the total charge

within the volume enclosed by the surface.• The normal component of electric flux density

integrated over a closed surface equals the

charge enclosed:

where D ┴ denotes the normal component of electric flux density, E ┴

denotes the normal component of electric field strength, and ds denotes

the differential surface area.152



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• Electric field lines originate from positive charges

and terminate at negative charges.

• A solid cylindrical conductor withradius r and with charge q coulombs

per meter (assumed positive in the

figure), uniformly distributed on the

conductor surface.

• For simplicity, assume that theconductor is (1) sufficiently long that

end effects are negligible, and (2) a

perfect conductor (that is, zero

resistivity, r = 0).153



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• Inside the perfect conductor, Ohm’s law gives Eint = rJ = 0. That is, the internal electric field Eint is

zero.• To determine the electric field outside the

conductor, select the cylinder with radius x > r

and with 1-meter length as the closed surface for

Gauss’s law.

• Due to the uniform charge distribution, the

electric field strength Ex is constant on the

cylinder154



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• There is no tangential component of Ex, so the

electric field is radial to the conductor.

• For a conductor in free space

155



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• Concentric cylinders surrounding the conductor

are constant potential surfaces. The potential

difference between two concentric cylinders atdistances D1 and D2 from the conductor center

is.

156



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• Now apply this equation to the array of M solid

cylindrical conductors shown below:

157



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• Assume that each conductor m has a charge qm

C/m uniformly distributed along the conductor.

The voltage Vkim between conductors k and i dueto the charge qm acting alone is

• Using superposition, the voltage Vki betweenconductors k and i due to all the charges is

where Dmm = r m when k = m or i = m

158

T i i i P


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• Capacitance: Single-phase two-wire line

• Assume that the

conductors are energizedby a voltage source such

that conductor x has a

uniform charge q C/m and,

assuming conservation of charge, conductor y has an

equal quantity of negative

charge -q.

159

T i i Li P t


160/248



• Using Dxy = Dyx = D, Dxx = r x, and Dyy = r y.

• For 1-meter length, the capacitor between conductors

is

160

T i i Li P t


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• If the two-wire line is supplied by a transformer with a

grounded center tap, then the voltage between eachconductor and ground is one-half of the line to line

voltage (Assume r x = r y).

161

T i i Li P t


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• Capacitance: Three-phase, three-wire and

equal spacing

• To determine the positive-sequence capacitance,assume positive-sequence charges qa, qb, qc such

that qa + qb + qc = 0. Using Daa = Dbb = r, and Dab =

Dba = Dca = Dcb = D, we can rewrite the equation

162

T i i Li P t


163/248


• Capacitance: Three-phase, three-wire and equal

spacing

• Similarly,

163

T i i Li P t


164/248



spacing

• For balanced positive-sequence voltages

• Therefore,164

T i i Li P t


165/248



spacing

• Therefore,

and

165

T i i Li P t


166/248


• Capacitance: General forms

• For a complete transpose line, the line to neutral

capacitance can be expressed as:

where

166

T i i Li P t


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• For a three-phase balanced system with bundle

conductors:• qa + qb + qc = 0 for a balanced system. Assume that the

conductors in each bundle, which are in parallel, share the

charges equally.

• Also assume that the phase spacings are much larger than

the bundle spacings.

167

T i i Li P t


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conductors:• Using the following configuration (Three-phase, two

conductors per bundle) as example.

168



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conductors:

169



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• For transposed line

where

• Similarly

• Deq is the geometric mean of the distances between

phases170



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• Shunt Admittances: Line with neutral conductors

and earth return

Single conductor and earth plane

• The effect of the earth plane isdescribed as follows. Consider asingle conductor with uniform charge

distribution and with height H above

a perfectly conducting earth plane,

as shown in Figure. When the

conductor has a positive charge,an equal quantity of negative charge is induced on the earth.

The electric field lines will originate from the positive charges on

the conductor and terminate at the negative charges on the

earth. Also, the electric field lines are perpendicular to the

surfaces of the conductor and earth

171



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• Shunt Admittances: Line with neutral conductors

and earth return

Earth place is replaced by image conductor

• Now replace the earth by theimage conductor which has the

same radius as the original

conductor, lies directly below the

original conductor with conductor

separation H11 = 2H, and has anequal quantity of negative charge.

Therefore, the voltage between

any two points above the earth is

the same in both figures.172

Example: Capacitance on single-phase line


173/248

phase line

• If a 20 miles of single-phase line operating at

20kV and 60 Hz consists of two 4/0 12-strand

copper conductors with 5 ft spacing betweenconductor centers, determine the line-to-line

capacitance in F and the line-to-line admittance

in S.

173

Example: Capacitance on single-phase line


174/248

phase line

• Solution

• The outside radius of a 4/0 12-strand copper

conductor is 0.023 ft.

174

Example: Capacitance on single-phase line (Consider earth)


175/248

phase line (Consider earth)

• For the same single phase line as the previous

example with18-ft average line height. Determine

the effect of the earth on capacitance. Assume aperfectly conducting earth plane.

175



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• Solution

• The earth plane is replaced by a separate image

conductor for each overhead conductor with proper charge.

176



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• Solution

• The voltage between conductors x and y is

177



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• Solution

• The line to line capacitor is

• Using D = 5 ft, r = 0.023 ft, Hxx = 2H = 36 ft, and Hxy =

36.346 ft.

178



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• Shunt Admittances: General form

• For the three-phase line with N

neutral conductors shown inFigure, the perfectly

conducting earth plane is

replaced by a separate image

conductor for each overheadconductor that carry opposite

charge.

179



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The voltage Vkk’ between any conductor k and its

image conductor k’ is

where Dkk = r k and Dkm is the distance between

overhead conductors k and m. Hkm is the distance

between overhead conductor k and image conductor m.180



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By symmetry, the voltage Vkn between conductor k

and the earth is one-half of Vkk’.

where (k = a; b; c; n1; n2; . . . ; nN) and (m = a; b; c; n1;

n2; . . . ; nN) Since all the neutral conductors are grounded to the

earth, Vkn = 0 for k = n1; n2; . . . ; nN.

181



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In matrix form

182



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The elements of the (3 + N) X (3 + N) matrix P are

where (k = a; b; c; n1; n2; . . . ; nN) and (m = a; b; c; n1;

n2; . . . ; nN)

The matrix can then be partitioned into

183



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The matrix can be rewritten into

or

where

184


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Electric Field at Conductor Surfaces and at Ground Level


186/248

Surfaces and at Ground Level

• When the electric field strength at a conductor

surface exceeds the breakdown strength of air,

current discharges occur.• This phenomenon, called corona, causes

additional line losses (corona loss),

communications interference, and audible noise.

• When line capacitances are determined and

conductor voltages are known, the conductor

charges can be calculated.186



187/248


• Then the electric field strength at the surface of

one phase conductor, neglecting the electric

fields due to charges on other phase conductorsand neutral wires is

where r is the conductor outside radius

187


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189/248


• For the arrangement shown in the figure, the

maximum electric field strength at the surface

of one conductor due to all charges in a bundlecan be obtained by the vector addition of

electric fields.

189



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• Two-conductor bundle (Nb = 2)

• Three-conductor bundle (Nb = 3)

• Four-conductor bundle (Nb = 4)

190


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192/248


Transmission-line heights are designed to

maintain discharge currents below prescribed

levels for any equipment that may be on theright-of-way. Table shows examples of

maximum ground level electric field strength.

192



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As shown in Figure, the ground-level electric

field strength due to charged conductor k and its

image conductor is perpendicular to the earthplane, will be:

193


194/248

Parallel Circuits on Three-PhaseLines


195/248

Lines

• Consider the double-circuit line shown below.

• Since both are connected in parallel, they have

the same series-voltage drop for each phase.Following the same procedure as described

before, we can write 2(6 X N) equations.

195


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197/248

es

• Adding IP1 and IP2 and solving for EP

• ZPeq is the equivalent 3 X 3 series phaseimpedance matrix of the double circuit line. The

matrices YB and YC account for the inductive

coupling between the two circuits.

where

197



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• Similar procedure can be used to obtain the

shunt admittance matrix.

• Following the same idea, we can write (6 + N)equations. After eliminating the neutral wire

charges, we obtain

where VP is the vector of phase-to-neutral voltages, and qP1 and qP2 are the

vectors of phase-conductor charges for lines 1 and 2. CP is a 6 X 6 capacitance

matrix that is partitioned into four 3 X 3 matrices C A, CB, CC, and CD 198



199/248

• Adding qP1 and qP2.

where

199

Transmission Lines: Steady StateOperation


200/248

p

200

Medium and Short LineApproximation


201/248

pp

• It is convenient to represent a transmission line

by the two-port network shown below.

• The relation between the sending-end andreceiving-end quantities can be written as:

201

or



202/248

pp

• A, B, C, and D are parameters that depend on

the transmission-line constants R, L, C, and G.

The ABCD parameters are, in general, complexnumbers. A and D are dimensionless. B has

units of ohms, and C has units of Siemens.

• The circuit in previous figure represents a short

transmission line, usually applied to overhead60-Hz lines less than 80 km (50 miles) long.

Only the series resistance and reactance are

included. The shunt admittance is neglected.202



203/248

pp

• The circuit applies to either single-phase or

completely transposed three-phase lines

operating under balanced conditions.• We can derive the two-port circuit and rewrite

the matrix

203



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pp

• For medium-length lines, typically ranging from

80 to 250 km (50 to 150 Miles) at 60 Hz, it is

common to lump the total shunt capacitanceand locate half at each end of the line. Such a

circuit, called a nominal p circuit.

204



205/248

pp

• Following the basic circuit theory to obtain the

ABCD parameters of the nominal p circuit.

205



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pp

• Rewrite the equations into matrix form.

• Note that for both the short and medium-lengthlines, the relation AD - BC = 1 is verified. Note

also that since the line is the same when

viewed from either end, A = D.206



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pp

• Summary of Medium and Short Line

Representation.

207



208/248

• Summary of Medium and Short Line

Representation.

208



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• Voltage Regulation

• The “percent VR” is the percent voltageregulation, |VRNL| is the magnitude of the no-

load receiving-end voltage, and |VRFL

| is the

magnitude of the full load receiving-end

voltage.

209



210/248

• In addition to voltage regulation, line loadability

is an important issue. Three major line-loading

limits are: (1) the thermal limit, (2) the voltage-drop limit, and (3) the steady-state stability limit.

210

Transmission Line DifferentialEquations


211/248

• Though we have used lumped model for

medium and short transmission lines, the R, L,

and C of line parameters are uniformlydistributed along the length of the line. The

circuit shown below is the better way for

transmission line representation.

211

G is usually neglected for

overhead 60-Hz lines



212/248

• From basic circuit theory:

212



213/248

• A1 and A2 are integration constants and

• Using

213

Zc is the characteristic impedance

and is the propagation constant


214/248



215/248

• Substitute these values into V(x) and I(x).

215



216/248

• Use the hyperbolic functions cosh and sinh.

216

where



217/248

• The propagation constant is a complexquantity with real and imaginary parts denoted

and ( = + j m-1

).

217



218/248

• ABCD parameters for short, medium, long, and

lossless lines.

218

Example: Exact Transmission LineModel


219/248

• A three-phase 765-kV, 60-Hz, 300-km,

completely transposed line has the following

positive-sequence impedance and admittance:

• Assuming positive-sequence operation,

calculate the exact ABCD parameters of the

line. Compare the exact B parameter with that

of the nominal p circuit.219



220/248

• Solution

220



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• Solution

221



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• Solution

222



223/248

• Solution

• The B parameter for the nominal p circuit is

223

Equivalent p Circuit for LongTransmission Line


224/248

• Many computer programs used in power

system analysis and design assume circuit

representations of components.• It is therefore convenient to represent the

terminal characteristics of a transmission line

by an equivalent circuit.

224



225/248

• The circuit shown

below is called an

equivalent p circuit. Itis identical instructure to the

nominal p circuit

except that Z’ and Y’are used instead of Zand Y.

225



226/248

• The ABCD parameters of the equivalent p circuit, which has the same structure as the

nominal p, are

226

Lossless Line


227/248

• It is easier to use lossless line (R = G = 0) to

explain the concept of surge impedance,

wavelength, surge impedance loading, voltageprofiles, and steady-state stability limit.

227

Surge Impedance


228/248

• For a lossless line,

• The characteristic impedance Zc, commonly called

surge impedance for a lossless line, is pure real -

that is, resistive. The propagation constant = j ispure imaginary228

and

where

Wavelength


229/248

• A wavelength is the distance required to

change the phase of the voltage or current by

2p radians.• For a lossless line

229

and

Wavelength


230/248

• V(x) and I (x) change phase by 2p radianswhen x = 2p/. Denoting wavelength by l.

• is the propagation velocity of voltage

and current waves along a lossless line. This isaround 3X108 m/sec for overhead lines. For

60Hz system, the wavelength is around

5000km.230

Surge Impedance Loading


231/248

• Surge impedance loading (SIL) is the power

delivered by a lossless line to a load resistance

equal to the surge impedance Zc.

231



232/248

• At SIL, the voltage profile is flat. That is, the

voltage magnitude at any point x along a

lossless line at SIL is constant.

232



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233

• The complex power flowing at any point x

along the line is



234/248

234

• The real power flow along a lossless line at

SIL remains constant from the sending end to

the receiving end. The reactive power flow iszero.

• At rated line voltage SIL of a line is V2rated/Zc.



235/248

235

• This table shows surge impedance and SIL

values for typical overhead 60-Hz three-

phase line.


236/248

Steady State Stability Limit


237/248

237

• The equivalent p circuit can be used to obtainan equation for the real power delivered by a

lossless line.

• Assume that the voltage magnitudes VS and

VR at the ends of the line are held constant.

• Also, let d denote the voltage-phase angle atthe sending end with respect to the receivingend.



238/248

238

• From circuit theory

• The complex power SR delivered to the

receiving end is



239/248

239

• The real power is



240/248

240

• Pmax represents the theoretical steady-state

stability limit of a lossless line.

• It is convenient to express the steady-statestability limit in terms of SIL



241/248

241

• Transmission line loadability curve

Maximum Power Flow


242/248

242

• For lossy lines, their ABCD parameters are

• The receiving end current is

Maximum Power Flow


243/248

243

• The complex power delivers to the receiving

end is

Maximum Power Flow


244/248

244

• The theoretical maximum real power

delivered (or steady-state stability limit)

occurs when d = Z.

Line Loadability


245/248

245

• In practice, power lines are not operated to

deliver their theoretical maximum power.

• Typical loadability is based on the voltage-drop limit VR/VS>0:95 and on a maximum

angular displacement of 30 to 35o across the

line (or about 45o across the line and

equivalent system reactances) to maintainstability during transient disturbances.

Line Loadability


246/248

246

• Note that for short lines less than 80 km long,

loadability is limited by the thermal rating of

the conductors or by terminal equipment

ratings, not by voltage drop or stability

considerations.

Reactive Compensation


247/248

247

• Inductors and capacitors are used on

medium-length and long transmission lines to

increase line loadability and to maintain

voltages near rated values.

• It can be either series or shunt

compensation.

Reactive Compensation


248/248

• Following devices are used in reactive power

compensation:

• Capacitor

• Inductor

• Static Var Compensator (FC-TCR or SC-TCR)

tsdos phasor transformer

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