42357119 Transformers

8/8/2019 42357119 Transformers

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Chapter 3.Transformers and Per-Unit Systems

(My Bin p v Hn V Tng i)

A primary function of the transformer is to convert electrical energy at one voltage levelto voltages at another level. The transformation may be to increase voltages or to

decrease voltages, depending on the application. Transformers are essential parts of mostpower systems since they are utilised to interconnect different parts of transmission and

distribution power grids that operate at different voltage levels. Transformers also are

found in almost all power supplies for small power electronic equipment such as TV sets,

laptops, chargers A key application of power transformers is to reduce the currentbefore transmitting electrical energy over long distances through conductors. Most

conductors have resistance and so dissipate electrical energy at a rate proportional to the

square of the current through them. By transforming electrical power to a high-voltage,and therefore low-current form for transmission and back again afterwards, transformers

enable the economic transmission of power over long distances. Consequently,

transformers have shaped the electricity supply industry, permitting generation to belocated remotely from points of demand. All but a fraction of the world's electrical

power has passed through a series of transformers by the time it reaches the consumer. In

this chapter, we will deal with some simplified models of transformers without discussingthe electromagnetic elements in details.

3.1 The Ideal Transformer

For analysis of transformers, it is convenient to start by using the ideal transformerrelations. Suppose we consider two coupled coils on a steel core of high magneticpermeability, a simplified model for this transformer can be shown in Figure 3.1.

i2i1N1 : N2

v1

_

+

_

+

v2

Figure 3.1 Ideal Transformer

If the flux varies sinusoidally, there will be a sinusoidal voltage generated in each turn of

each coil according to Faraday. This quantity is called volt-per-turn and occupies animportant role in transformer design. If windings 1 and 2 have N1 and N2 turns,

respectively, then:


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2

2

1

1

N

v

N

v (3.1)

One of the ideal transformer relations is that there can be no energy absorbed, stored or

lost in the device. Whatever complex power enters one winding must leave the other.

Therefore, we have*

22

*

11 iviv (3.2)

and

1

2

2

1

N

N

i

i (3.3)

The transformer also tends to transform impedances. In Figure 3.3, some impedance is

connected to one side of the ideal transformer. We can find an equivalent impedance 'Z viewed from the other side of the transformer.

Z

I2I1N1 : N2

V1

_

+

Figure 3.2 Impedance Coupling

Noting that

221

2

12 ZIVandI

N

NI

The ratio between input voltage and current is:

1

2

2

1

2

2

1

1

'

1 ZIN

NV

N

NIZV

We may derive the expression of the equivalent impedance viewed from the primary side

of the transformer:

ZN

NZ

2

2

1'

(3.4)

From the ideal transformer relations we see that the voltamperes into one winding of a

two-winding transformer must equal the voltamperes out of the second winding. Thevoltampere rating of a two winding transformer is then given as the voltampere rating of

either winding since the two are equal. In large power transformers the nameplate gives a

voltamperes (or kVA or MVA) rating for the device as well as the voltage ratings of thetwo windings. The current ratings then follow from these data since S = VI. Small

transformers, for example, those used in electronic power supplies, are often rated by


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giving the voltage and current ratings of each winding, from which the voltampere rating

would follow if desired.

No mention was made ofpower in the above statements. In a practical transformer therelative phase angle of voltage and current has almost no effect on the voltage and current

capabilities of the windings, and hence the magnitude ofS is the important factor andhow S is divided intoPand Q is immaterial to the rating.

3.2 Three-Phase Transformer

A three-phase transformer is conceptually the same as three single phase transformers.There are a number of ways of winding them, and a number of ways of interconnectingthem. On either side of a transformer connection (i.e. the high voltage and low voltage

sides), it is possible to connect transformers windings either line to neutral (wye), or line

to line (delta). Thus we will allow four connecting combinations: wye-wye, delta-delta,wye-delta, delta-wye.

Ignoring all the imperfections, connection of transformers in either wye-wye or delta-delta is reasonably easy to understand. On the other hand, the interconnections of a wye-

delta or delta-wye transformer are a little more complex. Figure 3.3 shows a delta-wye

connection, in what might be called wiring diagram form. A more schematic (and

more common) form of the same picture is shown in Figure 3.4.

ab YaYbYcc

Figure 3.3 Delta-Wye Transformer Connection

Assume that N and NY are numbers of turns. If the three individual transformers areconsidered to be ideal, the following voltage and current constraints exist:

baY

aY vvNNv

cbY

bYvv

N

Nv

acY

cY vvN

Nv (3.5)


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cYaYY

a iiN

Ni

aYbYY

b iiN

Ni

bYcYYc iiNNi

where each of the voltages are line-neutral and the currents are in the lines at the

transformer terminals.

Now, consider what happens if a -Y transformer is connected to a balanced three-phasevoltage source, so that:

3

2

3

2

Re

Re

Re

tj

c

tj

b

tj

a

eVv

eVv

eVv

Where: Re denotes the real part; V is the line-neutral (phase) voltage amplitude, an

underline beneath the variable means it is a vector. Then the complex amplitudes on the

wye side are:

63

2

31

j

Yj

YaY eV

N

Ne

N

NV

232

32

3 j

Yjj

YbY eV

N

NeeN

NV

6

5

3

2

31

j

Yj

YcY eV

N

Ne

N

NV

Two observations should be made here:

The ratio of voltages (that is, the ration of either line-line or line-neutral) isdifferent from the turns ratio by a factor of 3

All wye side voltages are shifted in phase by 30o with respect to the delta sidevoltage.

It can be proved that impedances transform across transformers by the square of the

voltage ratio, no matter what connection is used.

As an example of some of the things said above, suppose that we read from the

nameplate of a large three-phase transformer at a hydroelectric generating station the

following rating data: 40 MVA, 115/24 kV. These data now tell us other things by usingthe ideal transformer relations, for example:


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79.424/115/ 21 NN

since the voltage ratio and the turns ratio are the same under rating standards of largetransformers. Also the rated current of the high voltage winding, which we call I1 is given

by

currentratedamperes82.200

101153/1040 361

I

and36

2 10243/1040 I

= 962.25 amperes rated current for the low voltage windingIt will be noted that this latter figure forIcould also have been obtained by using the

ideal transformer relationI1/I2 = N1/N2if more convenient.

3.3 An Actual Transformer

The ideal transformer relations give very good answers to many transformer problems,as in the examples preceding this section. For some problems, however, we must takeaccount of the departures from perfection to get an adequate answer to a transformer

problem. The first imperfection we will discuss is that of the core. The core is notinfinitely permeable, it does require ampere turns to establish the flux, and in addition,

there are internal energy losses in the core when the flux varies with time.

Figure 3.4 shows the hysteresis properties of a transformer core. Each time the magnetic

field is reversed, a small amount of energy is lost due to hysteresis within the core. For a

given core material, the hysteresis loss is proportional to the frequency, and is a function

of the peak flux density (Bm

) to which it is subjected.

Figure 3.4 (a) Hysteresis Loops of Steel; (b) The Normal Magnetization Curve

Transformer losses arising from the magnetic circuit, are sometimes called iron loss.These losses are independent of the load current, and may furthermore be expressed as


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"no-load" loss. Iron losses are caused mostly by hysteresis and eddy current effects in thecore, and tend to be proportional to the square of the core flux for operation at a given

frequency.

The hysteresis loss occurs as an inherent property of the magnetic material. The internal

structure of a ferromagnetic material is organized into domains and these domains arereoriented as the magnetic flux density, B vector, goes through a cyclic change inmagnitude or direction. An internal energy loss appears as a result. The energy loss may

be minimized by suitable alloying and heat treatment of the metal. The treatment

processes may also affect the mechanical properties, however, so compromises must be

made.

If the frequency of the applied voltage were reduced but the range ofB in the core

maintained (by applying lower voltage), a similar loop would be observed but withsmaller area than that originally observed. The reason for the larger area with higher

frequency is the effect of eddy currents in the steel. The steel is a conductor and, as the

flux in the steel varies, voltages are induced within the closed contours in the material.Currents flow as a result of the voltage and an I2R loss occurs known as eddy current

loss. The loss is reduced by building the core from sheets of steel called laminations and

by increasing the resistivity of the material by alloying.

Since the core flux is proportional to the applied voltage, the iron loss can be represented

by a resistance Rc (or a conductance Gc=1/Rc) in parallel with the ideal transformer. A

core with finite permeability requires a magnetizing current IM to maintain the mutualflux in the core. The magnetizing current is in phase with the flux; saturation effects

cause the relationship between the two to be non-linear, but for simplicity this effecttends to be ignored in most circuit equivalents. With a sinusoidal supply, the core flux

lags the induced EMF by 90 and this effect can be modeled as a magnetising reactance

Xc ( or a susceptance Bc=1/Xc) in parallel with the core loss component. Rc and Xc aresometimes together termed the magnetising branch of the model. If the secondary

winding is made open-circuit, the current I0 taken by the magnetising branch represents

the transformer's no-load current. Figure 3.5 shows the use of a fictitious circuit added tothe ideal transformer to account for exciting current.

Iex

GcBc

Figure 3.5 Transformer Equivalent Circuit Accounted For Exciting Current


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The equivalent circuit of Figure 3.5 is an approximation, but it is valid for most purposes.The advantages of using a linear model far outweigh the slight error introduced by a

linear model. In any case the exciting current of a modern transformer of any size at all

is only a very small percentage of the full load current, so if there is a slight error in thelinear representation, it amounts to very little in terms of the total current passed by the

transformer. The model to use for a certain study is of course a matter of experience andengineering judgment. For heavy load or short circuit studies the exciting current branchis normally omitted entirely. For light load or no load the exciting current branch may be

included. For studies involving the wave form distortion of the exciting current the linear

model is completely inadequate.

Evaluation ofGcandBcof Figure 3.5 involves an approximation to best model the actual

transformer in some sense of a most useful model. The usual method of evaluating the

parameters is to choose GcandBc sothat the exciting current has the same rms value asthe actual exciting current and the power loss in Gc is the same as the actual core loss.

For example, suppose that a certain transformer is tested by applying rated voltage to a

10.5-kV winding with no load on the other winding, and it is observed that a current of10 amperes flows and a power of 10000 watts is drawn. We solve for Gc and Bc as

follows: P G E c2 ; 21050010000 cG ; siemen107.90

6cG ; I EY ex c ;

cY1050010 ; siemen1038.9526cY ; Y G Bc c c

2 2 2 ;

siemen1005.948 622 ccc GYB .

In addition to the core loss and the requirement for exciting current, there are other waysin which an actual transformer differs from the ideal transformer model. For one thing,

some of the applied voltage is absorbed in IR drop in the winding resistance. We can

modify the ideal transformer model to account for winding resistance by adding two

series resistors R1 and R2 in either side of the windings. Now, each winding has aresistance which, while not zero, is kept low in order to minimize copper losses and

increase efficiency.

Flux leakage results in a fraction of the applied voltage dropped without contributing to

the mutual coupling, and thus can be modeled as self-inductances Xl1 and Xl2 in serieswith the perfectly-coupled region. These series reactances (leakage reactance equivalent)

play a significant factor in transformer performance.

N2N1Xl1

Iex

Gc Bc

R1 Xl2 R2


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Figure 3.6 The Equivalent Circuit of An Actual TransformerThe secondary impedanceR2andXl2 are frequently moved (or "referred") to the primary

side after multiplying the components by the impedance scaling factor 2

2

1 )(N

N

2

2

2

1 RN

N

2

2

2

1lX

N

N

N2N1Xl1

Iex

Gc Bc

R1

Figure 3.7 Alternative Equivalent Circuit for the TransformerThe net series resistance and reactance are known simply as the impedance of the

transformer. This is an item of data that is available from the manufacturer of thetransformer. We designate the net resistance and reactance by the symbolsReqandXeqin

the figure where:

R R N N Req 1 1 22

2/

and

X X N N X eq l l 1 1 22

2/

Open and short-circuit tests:The parameters of the equivalent circuits of Figure 3.8 may be determined by thedesigner from the physical dimensions and material properties of the transformer. On the

other hand, an actual transformer may be tested electrically to determine these values.

For the test, we apply rated voltage to the left side of the transformer of Figure 3.8(a)with the right side open-circuited. Since the output current is zero, the current through

ReqandXeqis zero from properties of an ideal transformer. We thus "see" only the shuntbranch and determine Gc andBc from the instrument readings.

To determine the series impedances Req and Xeq, we short-circuit one side, and it isconvenient to use the model of Figure 3.8(b) in this case. In order to limit the short-circuit

current, this test must be conducted at reduced voltage. With a short circuit onN2, the

voltage is zero on this winding and also is zero across N1according to the properties of anideal transformer. As a result we can ignore the shunt exciting current branch and we

"see" onlyReqandXeq.


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Xeq

Iex

Gc Bc

Req

(a)

Xeq

Iex

Gc Bc

Req

(b)

Figure 3.8 Simple Equivalent Circuit for the Transformer

3.4 Introduction to Per-Unit SystemsPer-unit systems are nothing more than normalizations of voltage, current, impedancepower, reactive power, and apparent power (volt-ampere). These normalizations ofsystem parameters provide simplifications in many network calculations. As we willdiscover, the transmission system and several portions of the distribution system are

operated at voltages in the kV range. This results in large amounts of power being

transmitted in the range of kilowatts to megawatts, and kilovoltamperes tomegavoltamperes. As a result, in analysis, it is useful to scale, or normalize quantities

with large physical values. This is commonly done in power system analysis and isreferred to as the per-unit system. This helps in understanding how certain types of

system behave. The numerical per-unit value of any quantity is its ratio to the chosen

base quantity of the same dimensions. Thus a per-unit quantity is a normalizedquantity

with respect to a chosen base value.

Normalization of Voltage and Current:

The basis for the per-unit system of notation is the expression of voltage and current as

fractions of base levels. Thus the first step in setting up a per-unit normalization is to

select base voltage and current.


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Consider the simple situation in Figure 3.9. For this network the complex amplitudes of

voltage and current are: IZV (an underline beneath the variable means it is a vector).

We start by defining two base quantities, Vbase for voltage and Ibase for current. In many

cases, these will be chosen to be nominal or rated values. For generating plants, for

example, it is common to use the rated voltage and rated current of the generator as base

qualities. In other situations, such as system stability studies, it is common to use astandard, system wide base system.

_

+v1

i1

_

+V Z

Figure 3.9 Example

The per-unit voltage and current are then simply:

basebase I

Ii

V

Vv ,

With IZV , we have

izV

IZ

I

I

V

IZv

base

base

basebase

Where the per-unit impedance is:

basebase

base

Z

Z

V

IZz

This leads to a definition for a base impedance for the system:

base

basebase

I

VZ

And there is also a base power, which for a single phase system is:

basebasebase IVP

where Vbase andIbase are expressed in RMS (Root Mean Square). It is interesting to notethat, as long as normalization is carried out in a consistent way, there is no ambiguity inper-unit notation. That is, peak quantities normalized to peak base will be the same, in

per-unit, as RMS quantities normalized to RMS bases. This advantage is even more

striking in polyphase systems.

Three Phase Systems:


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In power system calculations the nominal voltage of lines and equipment is almostalways known, so the voltage is a convenient base value to choose. The apparent power

(volt-ampere - S) is usually chosen as a second base. In equipment this quantity is

usually known and makes a convenient base. The choice of these two base quantities willautomatically fix the base of current, impedance, and admittance. In a system study, the

volt-ampere base can be selected to be any convenient value such as 100 MVA, 1000MVA, etc.

The same volt-ampere base is used in all parts of the system. One base voltage in a

certain part of the system is selected arbitrarily. All other base voltages must be related

to the arbitrarily selected one by the turns ratio of the connecting transformers.

For single-phase systems or three-phase systems where the term current refers to line

current, where the term voltage refers to line to neutral voltage, and where the term volt-amperes refers to volt-amperes per phase, the following formulae relate the various

quantities:

)(

)(

NLbase

basebase

VSI

)(

2

)(

)(

2

)()(

base

NLbase

NLbasebase

NLbase

base

NLbase

baseS

V

VI

V

I

VZ

In performing per-phase analysis, the bases for the quantities in the circuit representation

are volt-amperes per-phase or kilo-volt-amperes per phase, and volts or kilovolts from

line to neutral. System specification is usually given in terms of total three-phase volt-amperes or kilo-volt-amperes or mega-volt-amperes and line-to-line volts or kilovolts.

This may result in some confusion regarding the relation between the per-unit value of

line-to-line voltage and the per-unit value of phase voltage (line to neutral voltage). In aper-phase circuit, the voltage required for the solution is the line to neutral voltage even

though a line-to-line voltage may be specified as a base. The base value of the line to

neutral voltage is the base value of the line-to line voltage divided by 3 . Since this is

also the relation between line-to-line and line to neutral voltages of a balanced three-

phase system, the per-unit value of a line to neutral voltage on the line to neutral voltage

base is equal to the per-unit value of the line-to-line voltage at the same point on the line-to -line voltage base if the system is balanced. Similarly, the three-phase volt-amperes is

three times the volt-amperes per-phase, and the base value of the three-phase volt-

amperes is three times the base value of the per-phase volt-amperes. Therefore, the per-

unit value of the three-phase volt-amperes on the three-phase volt-ampere base is

identical to the per-unit value of the volt-amperes per-phase on the volt-ampere per-phase base.

In a three-phase system, normally, a given value of base voltage is a line-to-line voltage,

and a given value of base kilo-volt-amperes or base mega-volt-amperes is the total three-

phase base.


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The values of base impedance and base current can be computed from base values ofvoltage and volt-amperes as shown earlier in the section. If the base values of volt-

amperes and voltage are specified as the volt-amperes for the total three phases and

voltage from line-to-line in a balanced three-phase system respectively, we have

)(

)3(

3 LLbase

basebase

VSI

)3(

2

)(

)(

2

)()(

33 base

LLbase

NLbasebase

LLbase

base

LLbase

baseS

V

VI

V

I

VZ

Networks With Transformers:

One of the most important advantages of the use of per-unit systems arises in the analysis

of networks with transformers. Properly applied, a per-unit normalization will cause

nearly all ideal transformers to disappear from the per-unit network, thus greatlysimplifying analysis.

+

_

V2

I2I11 : N

V1

_

+

Figure 3.10 An Ideal Transformer

To show how this comes about, consider the ideal transformer as shown in Figure 3.10.The ideal transformer imposes the constraints that:

1212

1, I

NIVNV

Note that an underline beneath the variable means it is a vector. Normalized to basequantities on the two sides of the transformer, the per-unit voltage and current are:

2

22

2

22

1

1

1

1

1

1

,

,

basebase

basebase

IIi

VVv

I

Ii

V

Vv

Note that if the base quantities are related to each other as if they had been processed by

the transformer:

1212

1, I

NINVV basebasebase

then 21 vv and 21 ii , as if the ideal transformer were not there.


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Transforming From One Base To Another:

In most instances, the per-unit impedance of a component is specified on the rated

component base which is different from the base selected for the part of the system in

which the component is located. When performing calculations, all impedances in anyone part of the system must be expressed on the same impedance base. As a result, it is

necessary to have a means of converting per-unit impedances from one base to another.

The process of changing this per-unit value of impedance to per-unit on a new base can

be done as follows:

Note that impedance in Ohms (ordinary units) is given by:

.__ newbasenewoldbaseold ZzZzZ Here, replace

base

base

base

S

VZ

2

, we can write:

newbase

newbase

newoldbase

oldbase

old S

Vz

S

Vz

_

2

_

_

2

_

This yields a convenient rule for converting from the old base to the new one:

old

newbase

oldbase

oldbase

newbase

newz

V

V

S

Sz

2

_

2

_

_

_

In other word: )()( 2

old

new

new

old

oldnew

baseVA

baseVA

baseV

baseVZUnitPerZUnitPer

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