I

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AAUUTTHHOORR

II

ACKNOWLEDGEMENT

We would like to express our profound gratitude, most sincere appreciation and

special thanks to our project advisor Prof. Tahir Nadeem Malik, Engr.Abdul Samad (DGM

Hec Hattar) and Engr.Fazal (Asst.DGM Hec Hattar) . They helped us in every regard.

Without there continuous moral support and enthusiasm it would have been impossible to

complete our project.

III

Preface Power industry in on the rise. With power shortage jinx haunting the country new

power projects are bound to be started. With every power plant there comes the use of

power transformers. Doesn’t matter whether its hydro power plant or diesel power plant,

to transfer electric power over greater distances and distribute it among the consumers we

need power transformers. So this was the incentive that gave us the passion to dig deep

into the design of power transformers. That was our main aim. How an actual power

transformer is designed. What are the constraints and limitations? What is the cost? How

do we make an actual power transformer? What is the core? What is the winding? How

do we select the winding? How do we select the core diameter? How we select the type of

winding? How do we calculate the losses? All these questions were in our mind when we

started this project. And now while writing these preface we can proudly say that we have

found the answers. This thesis will help any learning engineer to take an insight to the

design world.

We have divided this book into three broad sections. The first section deals with

the elementary theory of the power transformer. It covers the core and winding in detail.

And then we have explained briefly the operation of tap changers and bucholz relay.

After knowing about the basics of transformer we come to our main topic which is

design of course. We present the design consideration section. Here we have mentioned

every single design aspect with formulas. What is the type of transformer? What are the

cooling methods? What are the line and phase currents? What are the line and phase

voltages? How we select the volt/turn? How do we balance the ampere turns? All these

topics have been covered.

After giving the design considerations we design an actual 20/26 MVA 132/11.5 KV

power transformer. All the design calculations are included with formulae and adequate

explanation. In the end there is a small section in which we have compared prices of

power transformers being made both in our country and out of country.

In the end we would thank Allah who gave us the courage to pursue our objective. And

then Prof. Tahir Nadeem Malik who has been guiding us for the whole year. He always

helped us and told us better ways of doing things. Then we would like to thank

Engr.Abdul Samad(DGM Hec Hattar) and Engr.Fazal (Asst.DGM Hec Hattar) who

provided us with the practical knowledge and concept of design

IV

TABLE OF CONTENTS

1. Introduction.........................................................................1

1.1 Introduction of power transformer...................................................1

1.2 Significance of power transformer...................................................2

1.3 Utilization of our work.....................................................................2

1.4 Methodology....................................................................................2

2. Power Transformer ---- An Overview...............................3

2.1 Application of transformers.............................................................3

2.2 Working principles transformer........................................................4

2.3 Elementary theory of an ideal transformer.......................................5

2.4 EMF equation of a transformer.........................................................6

2.5 Ideal transformer...............................................................................8

2.5.1 Transformer on No-load..................................................8

2.5.2 Transformer on load........................................................9

2.6 Transformer having winding resistance but no magnetic leakage....12

2.7 Magnetic leakage...............................................................................13

2.8 Transformer with resistance and reactance.......................................14

2.9 Transformer core...............................................................................15

2.10 The winding................................................................................21

2.10.1 Continuous disk winding................................................23

2.10.2 Helical winding...............................................................28

2.11 Transformer winding arrangement for on-load tap changing.....30

2.12 External insulation.......................................................................34

2.13 Transformer Tank, Cooler, Oil conservator................................34

2.13.1 The tank...........................................................................35

2.13.2 Cooler..............................................................................38

2.13.3 Oil Conservator................................................................40

2.14 Breather.......................................................................................41

2.15 Oil sampler...................................................................................43

2.16 Protective Devices and Instrument..............................................44

V

2.16.1 Explosion Vent................................................................44

2.16.2 Buchholz Relay................................................................45

3. Design Considerations.........................................................47

3.1 Basic Procedure for design calculation..............................................47

3.2 Evaluation of core diameter and calculation of winding turns..........48

3.3 Calculations of winding.....................................................................51

3.4 Calculation of impedance voltage......................................................57

3.5 Calculation of Winding Parameters...................................................59

3.6 Calculation of core data.....................................................................62

3.7 Calculation of tank dimensions..........................................................64

3.8 Calculation of stray losses..................................................................65

3.9 Calculation of temperature-rise..........................................................68

3.10 Calculation of axial mechanic stress of windings............................71

3.11 Calculation of stress conductor under short-circuits........................74

3.12 Calculation of weight.......................................................................76.

4. Design Calculations of 20/26 MVA, 132/11.5 KV Power

Transformer....................................................................................84

4.1 Current calculation ...............................................................................84

4.2 Core Calculation...................................................................................85

4.3 L.V Winding calculations.....................................................................86

4.4 H.V winding calculations......................................................................88

4.5 R.V winding calculations......................................................................89

4.6 Calculation of Load Losses...................................................................91

4.7 Stray Losses..........................................................................................94

4.8 No Load Losses.....................................................................................97

4.9 No Load Current...................................................................................99

4.10 Regulation of Transformer ................................................................100

4.11 Efficiency...........................................................................................102

VI

4.12 Calculation of Winding Temperature Rise.........................................103

4.13 Short Circuit Current for H.V Winding..............................................112

4.14 Highest Average temperature rise of Winding Under Short Circuit...114

4.15 Balance of ampere turns......................................................................115

4.16 Axial Mechanical Stress under Short Circuit Condition ....................117

4.17 Calculations of weight ......................................................................120

5. Problems And Failures In Power Transformers.......................124

5.1 Procurement problems of power transformers.................................................124

5.2 Problems in magnetic circuits..........................................................................126

5.3 Problems in winding.........................................................................................128

5.4 Problems due to various structural defects and to other causes .......................130

5.5 Prices of Power Transformers………………………………………………..132

6. Conclusion And Recommendations............................................138

7. References.....................................................................................142

1

Chapter 1

Introduction

1.1 Introduction

Power industry is at its peak in Pakistan due to scarcity of Electrical power

generation. The industry for construction of Power Transformer locally will grow keeping

in view the demand of power transformer. We want to have:

1) Experience of design of power transformer for which we

Perform survey/review of different design aspects.

Effects of changing design aspects on power transformer.

We cover all the following aspects of Construction.

(i) Body

(ii) Winding

(iii) Core

(iv) Other parts

2) Experiences tap changer installation and working.

2

1.2 SIGNIFICANCE OF PROPOSED PROJECT

Why were we doing this project? That’s the most important question.

Presently the power transformers are being made by a handful of companies like

HEC Hattar, Siemens and now PEL have joined in. So they need design and

maintenance engineers. If we have a detailed study of power transformers then we

will be ideally suited for these kinds of jobs. Since we already have mentioned our

project will cover all the aspects related to design.

1.3 UTILIZATION OF THIS WORK

Power transformers are very costly. Their prices are in millions of rupees. Our

work will give any engineer a full-fledged research work on these transformers. Although

we may find books on power transformers. But our research is based on practical work.

Going to construction sites watching everything being built and then writing it down. This

will provide a lot of help to anyone interested in design and construction of power

transformers.

1.4 METHODOLOGY

By methodology we mean our approach to this project. We start from the design

aspects of power transformers. For this we visit factory sites and meet the design

engineers asking them about the various designing tools and at the same time study

different books so that we can get an insight into the various design aspects. For the

construction of these transformers again we visit the factory and watch every part being

made and then being assembled in the assembly section.

So basically our work includes a lot of visits to factory where these transformers

are being designed. To pursue our project we went to HEC Hattar. We had detailed

discussions with the design engineer there. That has helped us to complete our project.

3

Chapter 2

Power Transformer --- An Overview

2.1 Application of transformers Electrical energy generated by fuel-fired (thermal) power stations usually located

near large fuel deposits and by hydroelectric stations built in regions where waterpower

resources are, available has to be transmitted to industrial canters which may lie hundreds

and thousands of kilometers away from the stations, hence the need for vast transmission

lines between the greeting plant and the consumers.

It is a well-known fact that when current is transmitted over a line, some of the

power it carries is dissipated as heat in the line conductors. This loss grows higher as the

current and the resistances of the conductors are increased. It is not economical to try to

reduce the loss by solely decreasing the conductor resistance, because this would require

a substantial increase in the cross-sectional area of the conductor entailing a large

consumption of costly nonferrous metals.

It is precisely to reduce the power loss and consumption of nonferrous metals that

the transformer is used. The transformer while leaving the transmitted power unchanged

decrease current by increasing voltage and the loss which is proportional to the square of

the current (I2R loss) is thus sharply reduced. For example, ten-fold increases in the

supply voltage reduce the power loss by a factor of 100.

At the beginning of a power transmission line the voltage is raised by step-up

transformers and at the end of the line step-down transformers to a value convenient for

the consumer (from 127 V to a few kilovolts) lower it. Electric power is distributed

among the consumer (works. Factories. Residential areas. etc) through transformer

substations the prime role in the present-day power engineering is played by power

transformers. I.e. transformers used to raise or lower voltages in the supply networks of

power systems, which serve to transmit electric power over great distances and

4

distribute it among the consumers. Power transformers are notable for their high power

capacity and operating voltage.

Since electricity has to be conveyed over thousands of kilometers to the integrated

power grid the load centers and directly to numerous minor consumers it has to be

transformed four or even five times, hence the need to install a large number of set-up and

step-down transformers. Also, it should be noted that at each transformation Stage

operating at progressively lower voltage the total capacity of power transformers is

usually greater than that at the preceding stage. Therefore, in a y power system the

installed transforming capacity is six or seven times the installed generation capacity. As

an example figure shows the layout of a transmission and distribution network.

Supply networks operating at a voltage of 220 KV and higher make wide use of

autotransformer. Such transformers have two or more windings conductively connected

so that there is some winding portion to both the primary and the secondary circular. Now

we shall consider the structural components of a transformer in greater detail.

2.2 WORKING PRINCIPLE OF A TRANSFORMER A transformer is a static piece of apparatus used for transferring power from one

circuit to another without change in frequency . it can raise or lower the voltage with the

corresponding decrease or increase in current . in its simplest form the transformer

consists of a conducting coil have mutual inductance . the primary is the winding which

5

receives electric power and secondary is the one which delivers it. The coils are wound on

laminated core of magnetic material.

The physical basis of transformer is mutual inductance between two circuits linked by

common magnetic flux through a path of low reluctance as shown in the following figure.

The two coils possess high mutual inductance if one coil is connected to a source of

alternating voltage. An alternating flux is setup in the laminated core. Most of which is

linked up with other coil. In which it produces mutually induced e.m.f(electromotive

force). According to the Faraday’s law of electromagnetic induction.

e = M di/dt

Where

e=induced emf

M= mutual inductance

If second circuit is closed a current flows in it and so electric energy is transferred from

the first coil (primary coil) to the second coil (secondary winding).

2.3 Elementary theory of an ideal transformer

An ideal transformer is one which has no losses i.e. its

windings have no ohmic resistance and there is no magnetic leakage. In other words an

ideal transformer consists of two coils which are purely inductive and wound on a loss

free core. It may, however, be noted that it is impossible to realize such a transformer in

practice yet for convenience we will first analyze such a transformer and then an actual

transformer .

6

Since the primary coil is purely inductive and there is no cutput the primary

draws the magnetizing current only. The function of this current is to merely to magnetize

the core.

It is small in magnitude and lags v1 by 90 degree. This alternating current produces an

alternating flux which is proportional to the current and hence is in phase with it. This

changing flux is linked with both the windings therefore it produces self-induce emf in

the coil. This self induce emf e1 is, at any instant, equal to and in opposition to v1. This is

also known as counter emf of the primary.

Similarly in the secondary winding as induced emf e2 is produced this is known as

mutually induced emf. This emf is in phase opposition with V1 and its magnitude is

proportional to the rate of change of flux and the number of secondary turns. Figure

shows the vectorial representations of the above quantities.

2.4 EMF EQUATION OF A TRANSFORMER Let N1 = Number of turns in primary

N2 = Number of turns in secondary

Φm = maximum flux in the core in Weber’s

Bm = flux density in Weber/sq M (Tesla)

A = Net cross-sectional area of core in sq m

f = frequency of ac input in Hz

V1 = Instantaneous value of applied voltage in primary winding in

volts

V1m = Maximum value of applied voltage involtes

The instantaneous value of counter electromotive force e1 is

e1 = N1 dΦ /dt – volt

the counter emf e1is equal and opposite to applied voltage V1, I.e.

7

V1 = N1 dΦ /dt

If the applied voltage is sinusoidal, that is

V1 = V1m sin 2лft

Then Φ = Φm sin 2лft

Hence e1 = ¯ N 1 Φm cos 2л ft X 2лf

These equation are expressed as vectors as shown in Figure where’s V1 and e1 are the

rms values of V1 and e1. To obtain the rms value of counter emf e1, divided its maximum

value given above by V2.

Then E1 = 2л / V2 f N1Φm

The cosine term has no significance except to derive the instantaneous

Values.

i.o. E1 = 4.44 f N l ф m

or E1 = 4.44 f N l B m A

Similarly rms value of emf induced in secondary is

E2 = 4.44 f N2 Bm A

In an ideal transformer

V1 = E1

and V1 = E2

Where V2 is the secondary terminal voltage

8

VOLTAGE TRANSFORMATION RATIO (K)

From transformer equations we get

E2 = N1 = K

E1 N2

This constant is known as voltage transformation ratio.

(a) If, N2 > N1, i.e, K > 1, then the transformer is called as step-up transformer.

(b) If N2 < NI, i.e, K < l, then the transformer is called as step-down transformer.

Again for an ideal transformer

Input = Output

V1 I1 = V2 I2 (neglecting Iµ)

Or I2 = V1 = 1

I2 V2 K

Where I1 and. I2 are primary and secondary currents

Hence the currents are in the inverse ratio of the transformation ratio.

2.5 IDEAL TRANSFORMER We will consider two cases

1. When such a transformer is on no-load and

2. When it is loaded

2.5.1 Transformer on No-Load

The primary input current under no-load condition has to supply (i)iron-loss in the

core i.e, hysteresis loss and eddy current loss and (ii)a very small amount of copper-

loss in primary. Hence the no-load primary input current Io is not at 90 degree behind

1V but lags by an angle 0θ which is less than 90 degree. No-load primary input power

Wo= Io cos 1V 0θ . No-load condition of an actual transformer is shown vectorially in

9

the following figure.

As seen from the figure primary current Io has two components.

(i) One in phase with 1V . This is known as active or working or iron-loss

component Iw, because it supplies the iron-loss plus a small quantity of

primary Cu-loss.

Iw=Io cos 0θ

(ii) The other component is in quadrature with V1 and is known as magnetizing

component because its function is to sustain the alternating flux in the core. It

is wattless.

Iu=Io sin 0θ

Obviously Io is the vector sum of Iw and Iu hence

Io= )( 22 IwIu +

The no load primary current Io is very small as compared to full-load primary current.

As Io is very small hence no-load primary copper-loss is negligibly small which

means that no-load primary input is practically equal to the iron-loss in the

transformer.

2.5.2 TRANSFORMER ON LOAD

When the secondary is loaded, secondary current is

set up. The magnitude of is determined bye the characteristic of the load. The

secondary current sets up its own mmf(= ) and hence its own flux

2I

2I

22IN 2φ which is in

opposition to the main primary flux φ , which is due to . The opposing secondary

flux

oI

2φ weakens the primary flux momentarily and primary back emf E1 tends to

10

reduce. For a moment V1 gains the upper hand over E1 and hence causes more

current ( ) to flow in the primary. '2I

The current is known as load component of primary current. This current is in

phase opposition to current . The additional primary mmf sets up a flux

'2I

2I '21IN '2φ

which opposes 2φ (but is in the same direction as φ ) and is equal to it in magnitude.

Thus the magnetic effects of secondary current get neutralized immediately by

additional primary current . The whole process is illustrated in the following

figure.

2I

'2I

Hence whatever may be the load conditions the net flux passing through the core is

approximately the same at no-load. Due to this reason the core-loss in also practically

the same under all load conditions. Due to this reason the core-loss is also practically

the same under all load conditions.

As

11

2φ = '2φ

= 22IN '21IN

='2I 212 / INN × =K 2I

Hence when transformer is on load, the primary winding has two currents and

(which is antiphase with and K times its magnitude). The total primary current

is the vector sum of and . In the following we show the vector diagrams of a

loaded transformer.

oI

'2I 2I

oI '2I

In fig(a) current is in phase with (for non-inductive loads). In Fig(b) it is

lagging behind (for inductive loads).

'2I 2E

2E

If we neglect as compared to as shown in fig(c), then oI '2I 21 φφ = and thus

= = '21IN 11IN 22IN

KNNII == 1221 //

It shows that under all load conditions the ratio of primary and secondary current is

constant.

12

2.6 TRANSFORMER HAVING WINDING RESISTANCE

BUT NO MAGNETIC LEAKAGE An ideal transformer was supposed to possess no resistance but an actual

transformer has primary and secondary windings with some resistances. Due to these

resistances there is some voltage drop in the two windings.

The result is that:

(a) The secondary terminal voltage 2V is equal to the vector difference of the

secondary induced emf 2E and 22RI where 2R is the resistance of the secondary

winding.

= - 2V 2E 22RI

(b) Similarly primary induced emf is equal to the vector difference of and 1E 1V 11RI

is the resistance of the primary winding. 1R

= - 1E 1V 11RI

The vector diagrams for non-inductive, inductive and capacitive loads are shown

In Fig (a),(b) and (c) respectively.

13

2.7 MAGNETIC LEAKAGE

In an ideal case it is assumed that all the flux linked with the

primary winding also links the secondary winding. But in practice it is impossible to

realize this condition as magnetic flux cannot be confined. The greater the portion of the

flux(i.e, the mutual flux) flows in the core while a small proportion(Fig) called the

leakage flux links one or the other winding but not both.

On account of the leakage flux, both the primary and secondary windings have leakage

reactance, that is, each will become the seat of an emf of self induction, of a magnitude

equal to a small fraction of the emf due to main flux. The terminal voltage applied to

the primary must, therefore, have a component (where is leakage reactance of

primary) to balance the primary leakage emf. In the secondary, similarly, an emf of self

induction (where is leakage reactance of secondary) is developed. The primary

and secondary coils in figure are shown on separate limbs an arrangement that would

result in an exceptionally large leakage. Leakage between primary and secondary could

be eliminated if the windings could be made to occupy the same space. This of course is

physically impossible but an approximation to it is achieved if the coils of primary and

secondary are placed concentrically. Such an arrangement leads to a marked reduction of

leakage reactance. If on the other hand the primary and secondary are kept separate and

widely spaced there will be much room for leakage flux and the leakage reactance will be

greater.

1V

11XI 1X

22 XI 2X

14

2.8 TRANSFORMER WITH RESISTANCE AND

REACTANCE

The following figure

shows the primary and secondary windings of a transformer with resistance and leakage

reactances taken out of the windings.

The primary impedance is given by

)( 21

211 XRZ +=

And the secondary impedance is given by

)( '2

222 XRZ +=

11111111 )( ZIEjXRIEV +=++=

22222222 )( ZIVjXRIVE +=++=

.

The vector diagram of such a transformer for different kinds of loads is shown in the

following figure

15

In these diagrams vectors for resistive drops are drawn parallel to current vectors,

whereas reactive drops are perpendicular to the current vectors. The angle 1θ between

and gives the power-factor angle of the transformer.

1V

1I

2.9 Transformer Core The transformer core is a closed magnetic circuit built up of thin laminations of

electrical sheet steel. It is intended to concentrate the main magnetic flux linking with the

windings and consists of limbs, which carry the winding, and yokes, which close the

magnetic circuit. The core laminations are insulated from one another by a film on heat-

resistant coating or varnish or by a combination of both. There are may be two forms of

magnetic circuit the shell types and the core type.

A magnetic circuit of the shell type is branched there are two yokes per limb

which encircle the limbs on both sides. As the magnetic flux leaves a limb it branches off

into two parts therefore in shell-type transformers, the cross-sectional area of the limb is

twice that of the yokes. The limbs and yokes are rectangular in section, which necessitates

the use of rectangular disk windings. Because of the insufficient strength of such

windings in the event of short circuit complications in assembly and also somewhat

greater mass of the shell-type magnetic circuit as compared with the core-type circuits

using cylindrical windings the shell type in the soviet union is employed only for single-

phase transformers in household appliances and for some special-purpose transformers.

The core-type magnetic circuit of butt-joint or interleaved (of imprecated)

construction are used in power transformers. In such circuit two or three (depending on

the number of phase) vertical limbs are bridged over by two horizontal yokes -- the top

and the bottom one – so that a closed magnetic circuit is formed.

The core limbs and yokes are built up of separate laminations of electrical sheet steel 0.35

or 0.5 mm thick.

Limbs 1(see Figure below) and yokes 3 and 5 are stacked up separately and then butt-

joined and clamped with vertical tie-rods 4 to obtain a butt-joint core. Butt-joint cores are

easy to assemble but they suffer from a number of substantial drawbacks. At present, this

type of core construction can be found only in old transformers and in some models

manufactured in other countries.

16

Most power transformers are made of the imbricated-core type. In such cores, the

limbs and yokes laminations are interleaved (see Figure below) in one layer. The short

limb laminations are butt-joined with the long yoke laminations and in next layer the long

limb laminations are butt-joined with the short yoke laminations so as to overlap the

joints between the laminations in the preceding layer by stacking layer upon layers of

such alternately arranged laminations a core of the required thickness is obtained. Such an

assembly of core laminations is called interleaving with right-angled joints between the

laminations.

17

To speed up the assembly each layer is made two or three laminations thick. Accordingly,

the assembly is called double lamination interleaving or triple-lamination

interleaving.The arrangement of lamination in the alternating layers of an imbricated

three-limb core with right-angled joints between the laminations is shown in Figure.

The interleaving with right-angled joints between laminations was widely used for cores

made from hot-rolled steel. When using cold-rolled steel to make the most of its

properties the cores are designed and assembled in such a way as to ensure that the lines

of magnetic flux many coincide with the steel rolling direction not only in the core limbs

and yokes and vice versa (in Figure these areas are hatched).

This is achieved by making use of beveled (miter) joints between the limb and yoke

laminations (see Figure below). The miter joints reduce the magnetic circuit areas where

the lines of magnetic forces do not coincide with the steel rolling direction. Moreover

they increase the length and hence the area of the joint by factor of √2, thus reducing the

magnetic induction in the gap and consequently the exciting current.

18

The use of interleaved cores with miter joints between aminations may reduce the no-load

losses of transformers by 10 to 12% and the exciting current. By 25 to 30% how ever

such joints complicate the fabrication of laminations and core assembly therefore resort is

frequently made to some simplifications imbricated cores are made with four miter joints

(at the corners) and three right joints or a combined pattern is used where in the miter

joints at two corners of the core in one layer of laminations alternate with the right –

angled joints in the next layer (see Figure)

Transformers are also made with spatial rather than plane cores of butt-joint construction,

which are noted for the symmetrical arrangement of their limbs.

Such cores consist of two triangular yokes wound from electrical steel strip or ribbon

between which there are three stepped section limbs built up of laminations of the same

length. The chief advantage of this core is its simple construction which make possible

extensive mechanization and complete automation of production processes.

Size I transformers also use distributed three-frame wound magnetic circuits consisting of

three O-shaped cores wound from electrical steel strip and arranged in space so as to from

19

a trihedral prism. In such transformers the winding are wound directly on the core limbs

by passing wire through the opening in the adjacent O-shaped cores.

The limbs and yoke section are built up of a number of steps in order to make their shape

approximately a circle (see Figure). The steps are obtained by using laminations differing

in width. In order transformer models the yokes were made rectangular T-shaped or

cross-shaped in section.

To obtain the required electromagnetic characterstics of the core and to make it

mechanically strong the core field clamping tightness is attained. In transformers with a

capacity of up to 630 KVA, the core limbs are not clamped because they are sufficiently

rigid to ensure stable vertical position of the core. When fitting windings on the core

limbs of transformers with a capacity of 250 to 630 KV A the limbs are temporarily

compressed with screw clamps.

20

After fitting the binding, the required clamping tightness of the core limbs is ensured by

wedging them which beech wood blockers and cleats.

21

2.10 The Windings

Transformer windings differ from one another in type, number of turns, wire

grade and gauge, hand, creep age distances, and interterm insulation thickness. The higher the

voltage of the transformer the greater the number of turns in its windings, and the higher its

capacity, the heavier the wire gauge and the greater size of the windings the density, in the

windings, calculated on the basis of temperature rise. Ranges from 2.5 to 4.5 A/mm2 depending

on the capacity and design of the transformer.

One must strictly distinguish between right and left hand windings. The hand of single

layer windings is determined by the direction in which their turns have been wound during

manufacture, no matter which of their ends (upper or lower) is considered the start. In multiple-

layer windings wound during with the same conductor passing from one layer in another without

interruption, the hand alternates from layer to layer. The hand of such windings is considered to

be that of the layer whose entrance end is taken as the start of the winding.

Disk windings made in the form of flat spiral coils are considered left- or right-hand,

according as their back or front end is taken to be the start. From the figure it is clear that if the

front ends of such windings are considered the start, the first winding (the one on the left) will

then be right-hand, and the second (on the right) will be left-hand. Now if we consider the back

ends of the coils to be the start of the windings, their hand will then change to the left and right,

respectively. If a disk coil is turned upside down, its hand will be reversed: a left-hand coil will

become right –hand and a right-hand one will become left-hand.

Disk coils are usually arranged in pairs (as shown in figure). In this case, the front ends of

the coils are the entrance ones,

22

Left-hand and right-hand windings

(a) single-layer; (b) multiple-layer; (c) single disk (flat spiral) coils;

(d) Paired disk coils

and the interdisk connections made at the back. Here, the winding hand remains strictly defined,

and a winding consisting of any number of paired coils of the same hand in series will have the

same hand as the separate paired coils.

To give them mechanical strength and improve their moisture resistance, the windings

are dried and the impregnated with varnish and baked in an oven at 100 to 110 C. Recent

advances in the design, manufacture, and assembly of transformer windings have made it

possible to dispense with the winding impregnation and baking. The materially cuts down the

cost of the windings and clears production space.

The LV and HV windings are arranged on the core limbs concentrically: the L.V

windings are placed on the inside and the HV ones, on the outside (in some special transformers,

the LV and HV windings are arranged the other way round). The windings are separated by

insulating cylinders.

23

The following types of transformer windings are most widely used in the USSR and other

countries; single-, double and multiple-layer cylindrical windings, multiple-layer bobbing

windings, continuous disk windings, helical and pancake windings.

2.10.1 Continuous-Disk Winding

This consists of several series-connected flat coils or disks 1 of the same radial size (as

shown in Figure). The disks are placed one above another, with horizontal cooling ducts 2

formed between

them by pressboard spacing blocks 3. Each disk is wound with a strip conductor on the flat and

may have several turns touching in the radial direction.

Each turn of the disk may be wound with a single or several parallel conductors. The

disks in the winding shown in above Figure are wound with a single conductor per turn, and

there are six voltage-control taps 8 arranged in the middle of the winding.

24

This type has been termed “continuous-disk” because a special winding technique is used

to make the conductor pass from disk to disk without a single break in its continuity.

The winding is wound around wooden bars 6 placed axially all the way round the

periphery of a paper base laminate cylinder 7 at regular intervals, so that vertical cooling ducts 5

are formed between the winding and the cylinder. Support insulation rings 4 provide reliable

bearing surfaces for the winding.

In transformers, continuous-disk windings, as a rule, have no paper base laminate

cylinders. They are wound around wooden bars placed on a special metal cylinder (template)

which is removed after the winding is competed. The cylinders (formers) for such windings are

made from pressboard blanks immediately before fitting the winding on the core limbs.

Insulation components are made of pressboard sheets. The horizontal ducts between the disks are

formed by spacing blocks stacked up from separate pressboard strips 2 on bars 1, as shown in

Figure. If the windings are not impregnated, their mechanical strength is improved by means of

pressboard bars passed through dovetail recesses in the strips at the front of the coils.

25

Continuous-disk windings for voltages up to 35 kV are made from conductors with a two

side insulation thickness from 0.45 to 0.55 mm, and those for 110 kV use conductors with 1.2 to

1.35 mm thick insulation. In 110-kV windings, the entrance or line end coils, i.e., the two first

and two last disks, are wound with conductors of increased insulation thickness in order to

improve their electric strength. This impairs the cooling of these disks, so one has to take a

heavier conductor for them. The disks are wound separately in pairs and then soldered in series

with the disks in the main part of the winding. The number of disks in the winding is always

even, therefore its start and finish are always at the back or the front of the extreme disks. In the

former

Figure Transposition of conductors

1 – conductor passing from the front of the coil to the back; 2 – conductor passing from the back

of the coil to the front; 3 – strip

case, the entrance disks are not reversed when making the winding, while in the letter case, they

have to be reversed.

The interdisk connections in continuous-disk windings are made in free spaces between

the pacing blocks. Here, the conductor is bent on the edge and reinforced with a press-board box

taped to it, or a special pressboard strip is placed under the bent conductor.

In cases where the windings are wound with several parallel conductors per turn , the

conductors arranged farther from the coil a axis will have a greater length than those closer to the

axis .To equalize the lengths and consequently, the resistances of the parallel conductors, they

26

are transposed so that each conductor may take each possible position when passing from disk to

disk, as shown in above figure. This makes for uniform distribution of current among the parallel

conductors and, also, reduces the losses due to circulating currents caused by stray fluxes.

Parallel conductors reduce eddy-current., losses in the winding copper and facilitate the

winding of the disks, because several light-gauge conductors are used in place of a single heavy

one.

Continuous-disk windings are solid and inechanically strong. They find application in

LV, MV, and HV coils. The MV and HV coils are usually rapped in accordance with the direct

and reverse tapped-winding arrangements.

As distinct from the winding for up to 35 KV, those of the 110 –KV class and upwards

include some special structural components serving the purpose of capacitive protection. Such a

protection is necessary, because there is always a possibility that the transformer may be

subjected to dangerous overvoltages likely to cause damage to its insulation

High –frequency pulses of large amplitude and short duration, which arise in power

transmission lines as a result of lightning discharges, propagate as waves along the lines at a

speed close to that of light. For the oncoming high-frequency wave with a steep front, the

transformer may be regarded as a capacitor, for at high frequencies the inductive reactance of the

transformer grows materially and its individual winding elements (disks) become as if discon-

nected from one another; the disked then begin to act like the plates of distributed capacitors .

When the oncoming wave reaches the transformer, its input capacitance charges in a few

fractions of a micro-second, and the surge voltage impressed on the entrance coils firs drops to

zero and then sharply rises, this setting up electromagnetic oscillations in the transformer

windings, which depend on the winding inductance, capacitance, and resistance. In the

27

transformer there sets in a process of transition fro the initial, no steady distribution of voltage

over the windings to the final, steady-state voltage distribution. In the course of these process,

potential differences tens of times in excess of normal may develop between the winding disks,

and still greater ones between the turns. The reason for this is the extremely non uniform

distribution of electric charge over the windings, stemming form the presence of distributed

capacitances (between the winding disks, and between the disks and earthed parts. Such as the

core, tank, etc.) That disturbs the uniformity of electric field. Here the line end turns and coils are

most liable to insulation break-down.

To make the initial distribution of potential over the entrance parts of the winding more uniform

and to bring it closer to the final, steady –state distribution, use is made of capacitance-grading

rings. These rings increase the input capacitance of the windings and equalize the electric field of

the line –end coils and turns, thus reducing voltage gradients across them. The capacitance-

grading rings are electrically connected to the line ends of the coils and are reliably insulated

from the earthed parts of the transformer.

28

Arrangement of a capacitance-grading ring and shielding turns (electrostatic shields) on a

winding

The initial and final distribution of voltage over the entrance parts of the windings are

equalized by means of electrostatic shields made in the from of open shielding turns encircling

the five entrance coils and electrically connected to the capacitance grading rings and line ends

of the windings

Above figure shows the arrangement of the capacitance grading ring and shielding turns on a

winding. The grading ring l is pressed from several pressboard washers, and is of the ring 60 to

70 mm long is left unwrapped in order to avoid the formation of closed turn.

2.10.2 Helical Winding

In this winding, the turns follow a helical line, each turn consisting of several parallel

strip conductors touching in the radial direction (such a winding is sometime referred to as the

spiral winding).

Figure shows a singly re-entrant helical winding wound with eight parallel conductors per turn.

The insulation components of this winding are chiefly the same as in the continuous-disk

winding. Spacers 7 between the turns form horizontal oil ducts, shield bars 4 form vertical ducts

between the winding and cylinder 5.

Helical windings of transformers are wound around wooden bars placed on paper base laminate

cylinders, while those for bigger units, around bars arranged on

29

Temporary steel cylinders (templates). The end faces of the winding are made level by

gradually increasing the thickness of the spacers between the coil-end turns and insulation rings.

Since the parallel conductors in the helical winding are arranged concentrically and are at

different distances from the winding axis, the conductors closer to the axis will be shorter than

those farther away (as in the continuous –disk winding), unless some special measures are taken.

To equalize the resistances and inductive reactance’s of the parallel conductors and

reduce the losses due to circulating currents caused by stray fluxes, the conductors are transposed

three times each (see Figure). At one-fourth and three-fourths of the winding’s height, the

conductors are divided into two equal groups and the groups are transposed

30

These are group transpositions. In the middle of the winding, all the conductors are transposed.

This is general transpositions are made in the free spaces between the spacing blocks separating

the turns. As a result of the transpositions, the conductors of the helical winding (which usually

have an even count) change places in consecutive order, so that each conductor takes each

possible position, and the lengths of all the conductors are thus equalized.

To make the transitions smooth and to equalize the radial sizes of the winding turns,

special insulating wedges are placed under the conductors at placed where the transpositions are

made.

Besides the singly re-entrant helical winding considered above, doubly and quadruply re-

entrant helical windings also find application. Multiply re-entrant winding are used where the

number of parallel conductors per turn is rather great (from 18-50). The conductor transpositions

in such are more perfect, because the conductors during winding are made to continually pass

from one winding re-entry into another. Such a transposition is called distributed, or Hobart’s

31

transposition. There are also other types of transposition, for example, uniformly distributed and

DeBuda’s transposition.

Helical winding have a comparatively small number of turns; they are wound for heavy current.

2.11 Transformer Winding Arrangement for One-Load Tap

changing

Among the great variety of transformers using on-load tap changing, worthy of notice are

power transformers and autotransformers equipped with built-in switching devices which make it

possible to change taps and thus to maintain voltage within the required limits directly at the

transformer terminal business without interrupting the load. The Winding of Transformers using

on-load tap-charges differ from those of transformers equipped with no-load tap-charges in that

they have a greater number of voltage-control taps, provide for a wider voltage-control range and

are made up of two individual windings which are referred to as the excitation (or main) winding

and the regulating winding (separate winding for cross and fine voltage control may sometimes

be also include). As a rule, voltage control is carried out on the HV side; therefore Figure shows

the HV winding arrangements only. The arrangements for one phase only are shown, because all

the three phase windings of the transformer are indential. The design and operation of the on-

load tap-charges are considered elsewhere in the text and here we restrict ourselves to the

examination of transformer winding arrangements for use in conjunction with the on-load tap-

charger.

Figure shows the reverse tapped-winding arrangement. It is similar to the one examined

earlier. Like all of the winding arrangements used for on-load tap changing this arrangements

includes two windings an excitation winding 1 and a regulation winding 2. the latter is made as a

separate tapped coil and is designed BW by means of a drive mechanism the movable contact

32

(finger) of a selector switch 3 (shown schematically) of the tap-changer is moved from tap to tap

without interrupting the load current and thus the required voltage corresponding to the selected

step is obtained between the points A and X.

To extend the voltage-control range, use is frequently made of an arrangement where in

the connection of the regulating winding can be reversed i.e changed from aiding to opposing

and vice versa with respect to the excitation winding With the aiding connection of the

regulating winding (a reversing 4 is in position III-I ) the number of turns being put in operation

as the

Selector switch is shifted from position 9 to position 1 is increased (if the winding 1 and 2 are

wound in opposite directions) and in position 1 the resultant voltage of the HV winding is raised

33

by the total voltage of the regulating winding. To lower the voltage the moveable contact of the

selector switch is shifted back from position 1 to position 9.

When the reversing switch is shifted to position III-II the regulating winding and the

excitation winding are connected in series opposition and at the same time the selector switch

contact is shifted to position 1. in switching from the 1st to the 9th step the number of the

opposing turns of the regulating winding grows larger and the total voltage of the Hv winding is

reduced. With the selector switch in position 9 the voltage is decreased by the total voltage range

of the regulating winding switching in the reverse direction increase the voltage.

Thus reversing the connection of the regulating winding doubles the voltage-control

range. Such an arrangement simplifies the regulating winding but complicates the tap changer

design.

A tapped-winding arrangement which two parallel branches (see Fig) has found wide

application. It ensures a better utilization of the core windows and winding wires. The upper and

lower portions of the winding are strictly symmetrical; they are wound in opposite directions and

consist of excitation winding 1 and 1’ and regulating windings 2 and 2’, respectively. Because of

the different directions of their turns, both winding halves can be connected in parallel and

provided with common voltage-control taps.

A more perfect tapped-winding arrangement for on-load tap changing is the multiply re-

entrant helical-winding arrangement having so many reentries as there are voltage control steps

the winding turns being uniformly spaced all along the winding (see Figure). with the winding

arrangements shown in Figure and c in switching from tap to tap(especially when the amount of

control is large) a “dead” zone is formed in the regulating winding where the turns are put out of

operation and thus do not take part in producing the magnetizing force (as measured in ampere

34

turns). To equal the magnetizing forces of the primary and secondary windings the latter have to

b spread i.e. their turns have to be pushed wider apart along the winding in the region of the

regulating winding in order to avoid heavy leakage fluxes. The multiply re-entered winding

meant is free from the above shortcoming because disconnection of one or several voltage

control steps (ding re-entries) does not disturb the uniform distribution magnetizing forces along

the windings and the magnetize forces of the secondary windings thus remain equalizer. Multiply

re-entrant layer-by-layer windings are not very difficult to manufacture they are wound with

several parallel wires each of which forms an individual winding entry serving as a voltage-

control step and is provide a voltage control tap.

2.12 External Insulation

The external insulation comprising the air spaces between the live parts of the terminal

bushings and between the bushing and earthed parts of the transformer. Insulation clearances are

selected in accordance with the standard creep age distances for air. Here are some of them.

The insulation clearance between the bushing and earthed parts (explosion vent, oil

conservator, etc.) are taken at nearly the same values. In practice, these clearances are increased

by 10 to 15 mm to allow for possible size deviation in transformer assembly.

2.13 Transformer Tank, Coolers, Oil Conservator

In operation, the heat given off the core, windings, and other current carrying parts of the

transformer is transferred to the oil which surrounds them. The oil transfers the heat by

conduction and convection to the walls of the transformer tank whose external surface dissipates

it into the surroundings. Such a method of heat removal is called oil-natural cooling.

As the capacity of the transformer grows higher, the absolute power loss in it increases

and consequently, the amount of heat that the tank walls must dissipate. With natural oil

35

circulation, each square meter of the tank surface can dissipate from 400 to 450 watts of power.

If the incoming heat load on the tank surface should be greater, the temperature of the core-coil

assembly and the transformer as a whole would raise prohibitively high, this impairing its

reliability. In low-capacity transformers (25to 40 kV A), the absolute loss of power dissipate as

heat is comparatively low, so they use plain tanks. The cooling surface of larger units has to be

increased by welding steel tubes onto the tank walls or by fitting the radiators cannot provide for

adequate heat removal, a blast of air is forced onto them by means of special propeller- type fans.

This method of heat removal is called oil-natural air-blast cooling Transformers of very large

aixes use combination cooling systems, such as forced- oil-air-blast and forced-oil and water.

2.13.1 The Tank

The transformer tank is an oval or rectangular container intended for housing the core-coil

assembly of the transformer. It is arc-welded form steel sheets, all the welds being of the oil-tight

type. After manufacture, the tank is tested for tightness under a pressure of 0.5 x 10 Pa (gauge).

At the top of the tank, there is a frame with bolt holes for fastening the tank cover. The cover

closes the tank and serves as a support for mounting the oil conservator, terminal bushings, tap-

changer drive, lifting lugs, etc. To facilitate moving the transformer, the tank is provided with a

truck or undercarriages on rollers.

In major repair work, the cover has to be removed and the core-coil assembly withdrawn

from the tank. The lifting of the core-coil assembly of high-capacity transformers requires heavy

hoisting equipment, transformers, so the tanks of such trans- -formers are made with a detachable

bottom in order to ease the uncovering of the core- coil assembly. In this case, instead of

unbolting the cover and lifting the core-coil assembly, they unbolt the bottom and lift the tank,

36

leaving the core-coil assembly on its support the tank bottom, the oil being preliminarily drained

from the tank.

Transformers of up to 40 kV A generally have terminal bushings mounted on the side

walls of the tank, which also carry an oil gauge, transformer nameplate, lifting hooks, spark-gap

protector bracket, oil-drain filler plug provided with an air-bleed hole, protective hood for the

operating knob of the tap-changer, and thermometer pocket

Transformer form 63 to 1600 kV A in capacity are equipped with tubular tanks. The

tubes on the tank walls may be arranged in one, two, or three rows, depending on the transformer

capacity. At present, the transformer manufacturers in this country are putting out transformer

equipped with elliptical tubes which, as compared with round tubes, provide for higher heat

removal efficiency and can be placed closer to one another along the tank periphery, so that more

tubes can be accommodated on a given tank. Figure shows the Type TM-630/10 transformer

whose cooling surface is increased by bank 2 of elliptical tubes. The tank is filled with oil

through a globe valve 1 which also serves for draining the oil from the tank. Oil samples can be

taken through a sampler mounted the tank wall. In the tank bottom there is a hole normally

closed by a sealed –off plug (not shown in the figure) which serves for drain-in oil residue from

the during repair. The tank has four hooks 3 for lifting the transformer and a truck on four

rollers 5 for moving it horizontally the tank cover is provided with special opening and studs for

mounting and fastening the terminal bushing , tap-changer drive valve, thermometer pocket, and

pipe for connecting the tank to the oil conservator. Plates 4 welded on to the tube banks serve for

mounting the transformer nameplate and temperature indicator.

37

Figure shows the arrangement of the terminal bushings and other transformer fittings on a

transformer cover

38

2.13.2 Coolers

Transformers larger than 1600 kVA use detachable tubular radiators, so their tanks are

reinforce: by stiffening ribs and are provided with flanged ports for attaching the

Radiators and radiator valves. The radiators and valves are flange mounted and are fixed

in place by means of special steel bolts

A radiator consists of two rows of parallel tubes 1. top and bottom headers

headers 2. and flanged ports 6 which are welded into the ends of the headers and serve

for mounting the radiator on the tank. Each header carries a lifting lug 4, bracket 3 with a

hole for mechanical connection of separate radiator, while the top one is used for bleeding

air from the radiator when the transformer tank is being filled with oil. They give them

rigidity, the tubes in the parallel rows are joined by means of angle bars 7 and tubes 8

held together by bolts

39

Diagram of oil circulation in a radiator

The greater the surface of the radiator, the greater the amount of heat it can

remove from the transformer. Single sided tubular radiators are sufficient for units of 630

to 4000 kVA. In such radiators, the tubes are arranged only on one side of the headers,

and the flanged connecting ports are provided on the opposite side of the headers.

Lately, radiators with straight vertical tubes have been widely adopted. These use

round or elliptical tubes of smaller diameter and wall thickness that single or double row

tubular radiators of ordinary design. With several rows comparatively closely spaced

tubes, the straight tube radiators are lighter and more compact. Also their heat removal

efficiency is higher and they are more easy to assemble and repair, since the tubes are

straight and have the same length Straight type radiators are attached to the transformer

tank in the same way as the double row tubular radiators of ordinary design; in small

transformer, the attachment is without flanged connections, the connecting piper being

wiled directly into the tank walls.

In operation, hot oil raises to the top the tank and enters the top radiator header

(see Figure). The large cooling surface of the radiator causes the temperature of the oil to

drop. Since there is a difference in density between hot and cold oil the oil, while cooling,

descends down the radiator tubes giving away its heat to the tube walls that dissipate it

into the ambient air. Fresh portions of hot oil from the tank enter the top radiator header,

40

replacing the cooled oil which flows into the tank form the bottom header. In this way,

natural oil circulation sets in the transformer.

2.13.3 Oil Conservator

As the load on the transformer and the temperature of its surroundings vary, the

temperature of the oil filling the transformer also changes. Under the same load

conditions the temperature of the transformer oil is higher in summer than in winter.

Temperature variation cause changes in the oil volume in the transformer tank. To ensure

that the tank is always completely filled with oil, transformer having a capacity form 25

kVA upwards and working at 6 kV and over use a special expansion tank called oil

conservator.

The conservator is a metal vessel, usually cylindrical, which communicate with

the main transformer tank. Figure shows the oil conservator of a transformer. It is

installed slightly above the level of the tank cover 6. As the oil gains in temperature. It is

forced out of the tank and into the conservator via a a pipe connecting the tank to the

flanged connection 5 of the conservator; when its temperature falls, the oil flows back

into the tank. The conservator tank is mounted on the main tank cover on brackets 10 and

support plates 9.

The conservator reduces the oil surface exposed to air thus lowering the rate of

sludging and acid-production. Its capacity must be such as to ensure that it is never empty

of oil. Irrespective of any normal variations in the transformer operating conditions (from

of-circuit to full and ambient temperature (form –45C to +40C) the volume of oil in the

conservator amounting to 8 or 10% that in the main transformer tank.

One of the end walls or the conservator accommodates an oil level gauge 1 whose

glass hears three painted horizontal lines, marked – 45C,+15C, and +40C,which fix the

oil level in the inoperative transformer corresponding to the indicated ambient

41

temperatures. The opposite and wall of the conservator is made detachable to facilitate

cleaning and painting the conservator on the inside curing repair

The oil gauge server for checking the oil level in the transformer vessels, but to prevent

the ingress of atmospheric moisture and impurities, its upper part is made to communicate

with the air filled space above the oil in the conservator. Rather than directly with the

atmosphere.

2.14 Breather

This is a special air filter incorporating a dehydrating material (silica gel). It is

used to prevent the ingress of moist, contaminated air into the conservator. Depending on

the construction and size of the transformer, the breather is mounted either on the

conservator or on the main tank. At present, the breathers of transformer are built into the

conservators.

The breather consists of a metal cylinder 1 filled with silica gel 3, a wire mesh

screen 7, and a perforated cartridge 4 filled with a sight glass 6.

At the bottom to the breather there is an oil seal operating on the principle of

communicating vessels, which prevents the silica gel dehydrator from being constantly in

42

contact with the ambient air and thus continuously adsorbing moisture the oil seal also

serves to remove mechanical impurities from the air, which precipitate in the oil filling

the seal when the air passes through it.

When the oil level in the conservator drops. Fresh air is drawn into it through the

breather. The air passes the transformer oil 11 filling the seal wall 8, the wire mesh

screen, and the silica gel dehydrator which absorbs moisture from it. Then the cleaned.dry

air enters the conservator by a pipe connected to the flanged connected ion 2 of the

breather. When the oil volume in the conservator increases, the air flows in the opposite

direction and is exhausted into the atmosphere.

The oil seal is provided with several plugs. A 14 serves for filling the seal with

transformer oil, a plug 13, for draining the used oil from the seal, and a plug 10, for

draining any excess oil that might raise the oil level in the seal above the normal

(indicated by an arrow in the figure).

43

The silica-gel dehydrator in the breather is periodically changed. An indication

that the silica gel has become moist and requires replacement is the change of the colour

of the indicator silica gel from light blue to pink. The colour of the indicator silica gel is

observed through the sight glass in the cover of the indicator cartridge.

The breather uses Grade KCM silica gel in grain sizes from 2.7 to 7 mm,

impregnated with a calcium chloride solution. The indicator silica gel is additionally

impregnated with a cobaltic chloride solution. Silica gel is dried at a prior to charging the

breather with it.

2.15 Oil Sampler

Samples of transformer oil for tests are taken from the tank through a special

device (as shown in figure) mounted on the tank wall at the lower part of the transformer.

It consists of a steel body l, a stopper 2 which is free to turn in its seat in a plug 3, and a

nipple 4. as the threaded plug is screwed out of the body, the oil in the tank forces the

stopper to the right and flows out through the nipple.

44

2.16 Protective Devices and Instruments

2.16.1 Explosion Vent

Failures inside the transformer are frequently accompanied by arcing. The high

temperature of electric are causes intensive decomposition of transformer oil, the gas

evolved in the process greatly increasing the pressure inside the tank. In the event of a

short circuit the pressure inside the tank grows so high that the tank may explode and

cause an out break of fire.

To avoid damage to the transformer tank, an explosion vent (as shown in figure) is

provided. It consists of a knee-shaped tube 2 made of sheet steel 4.5 mm thick, whose top

end is closed by a diaphragm 4 with a flat, round glass. The lower end of the tube

communicates with the tank through an opening in the tank cover. Should the pressure

inside the tank grow too high, the glass will break and the gas, together with oil, will be

expelled from the tank through the tube.

The explosion vent tube at its lower end is provided with a flange I for bolting the

tube to the tank cover, and there are two flanges, 9 and 11, at the top end of the tube for

mounting the glass disk 12 sealed by rubber gaskets 7 and 8. a ring 10 welded to the

flange 9 serves for centering the gaskets. An oil tight weld is used on the joint between

the flange 9 and the tube wall 6.

Formerly, explosion vent tubes were equipped with a holed plug (at 3 in Figure) to

bleed air from the tube when filling the transformer tank with oil and to permit of the

ingress and egress of air in accordance with temperature variations in the tank, but

nowadays, to prevent any other contact of the oil with air except in the conservator, the

tube is connected to the air space of the conservator by means of a small gauge steel pipe

bolted to flange 5 on the tube. Where such plugs are still in use, they must be stopped

45

altogether and connecting pipes fitted between the explosion vent tubes and conservators.

Explosion vents are used on transformers of 1000 kV A and over.

2.16.2 Buchholz Relay

Any fault which occurs inside the transformer is generally accompanied by the

evolution of gas due to the decomposition of insulating materials (oil, paper, pressboard,

wood, etc.) under the influence of elevated temperature. With minor faults, the gas

evolution slow, the gas bubbles gradually rise to the tank cover and then enter the

conservator through the oil pipe connecting the tank to the conservator. In the case of

grave faults, oil rapidly flows through the pipe and is expelled into the conservator under

the pressure of a large amount of gas evolved.

The Buchholz relay is a gas-operated device connected between the transformer

tank and the conservator. It is fitted with alarm and tripping contacts, so that warning can

be given of incipient gas evolution and a major breakdown can be averted. On its way

from the tank to the conservator, the gas is collected in the relay housing, and after the

gas collected has reached a preset volume, the relay gives an audible or visible warning of

the fault. When there is an intensive flow of oil from the tank into the conservator, the

relay trips the circuit breaker of the transformer. Besides fault indication, this relay will

also indicate oil leakage should the conservator and pipe become empty of oil.

An analysis of a gas sample taken from the Buchhloz relay of a faulty transformer

helps to determine the nature of the fault. Usually under normal conditions, the gas

dissolved in the transformer oil has the following composition 70 to 79% nitrogen, 20 to

30% oxygen, and 0.1 to 0.2% methane; hydrogen and acetylene are absent. A sharp

change in the gas composition (for example, 50 to 70% hydrogen, 3 to 10% methane, 10

to 25% acetylene, 4 to 8% oxygen) testifies to a grave internal fault accompanied by

arcing (insulation puncture, shorted turns, flashover in the contact system of the tap-

46

changer, etc.) In the event of minor faults not accompanied by violent oil decomposition

and gas evolution, the gas composition may be as follows: 2 to 5% hydrogen, 0.5 to 1%

methane, 0.5 to 2% acetylene, 85 to 92% nitrogen and 5 to 8% oxygen. Such a gas

composition bears witness to a fault, such as shorted parallel conductors in the windings,

poor contacts in the tap connections or soldered joints, or closed paths in the magnetic

system of the core, which, if not remedied, may eventually lead to a serious trouble.

Soviet-made transformers use two types of Buchholz relay, namely,. The ITT-22

float-type relay and the PTH3-66 cup-type relay.

47

Chapter 3

Design Considerations

3.1 Basic Procedure for design calculation

a. Voltage calculation.

Determine the line voltage and phase voltage for winding (HV. MV and LV)

b. current calculation

Determine the line current and phase current for HV, MV and LV windings and

current flowing through winding.

c. Evaluation of core diameter and calculation of number of winding turns:

d. Calculation of main and long it urinal insulation distances for windings.

Chose and determine the type and the dimensions of HV, MV, and LV

windings.

e. Calculation of impedance voltage:

f. Calculation of winding data:

g. Calculation of core and lank dime nations:

h. Calculation of stray losses:

i. Calculation of temperature-rise:

j. Calculation of axial mechanical stress for windings:

k. Calculation of stresses in winding conductors:

l. Calculation of weight:

48

Line and phase voltages for three-phase transformers

For three-phase transformers or single-phase transformer used as a three phase bank

with star-connection the phase voltage equals 1/ 3 of the line voltage i.e.

u 3/1(=g ) Ug

where U g = Rated phase voltage KV:

Un = Rated line voltage KV.

For delta-connection the phase voltage equals the line voltage Ug = Un

Three-phase transformers with YN (Y) connection

Ig = In = Un

se3

= Ugse

Ugse

333

=

Where

In = Rated line current

Ig=Rated phase current

Three-phase transformer with delta connection

The calculation of current under this condition is as follows:

Line current In = Un

se3

Un rated line voltage →

Phase current Ig = In Unse

33 =

The line current phase current and tapping currents may be calculated by the

formulated listed above with the given rated power rated voltages (including the

cores.

3.2 Evaluation of core diameter and calculation of winding

turns Evaluation of core diameter

The determination of core diameter represents a very important job in the design

calculation which would affect directly the techno-economic indices of the

transformer such as raw material consumption manufacturing cost weight and

dimension for transportation etc.

49

Generally for a specific transformer the greater core diameter results in the heavier

mass of core and the more no-lead loss. But the weight of winding conductors and

their load losses would be comparatively with a dumpy configuration when the core

diameter is too small. The contrary results would be obtained.

The core diameter may be determined with an experimental formula by which the

calculation would be going on according to the design procedure. As the impedance

voltage calculated with core diameter aforesaid reveal too much deviation from their

standard values specified the core diameter should be adjusted to an appropriate

values small difference between the standard value and values calculated from the

predetermined core diameter the adjustment may be carried out by changing other

constructional dimensions.

The experimental formula for core diameter determination is.

D = k. 14 p

Where

D- core diameter:

P1 – power for each core leg KVA:

K- Experimental coefficient.

The value k varies as the frequency of source flux density and construction of core;

here the values listed in table are recommendable foe design calculation.

Table

K

Al winding Co winding Categories of

transformer Cold-

rolled si-

steel

Hot-rolled

si-steel

Cold- rolled

Si-steel

Hot-rolled

Si-steel

Two- winding

transformers 50-54 56-60 53-57 60-64

Three-winding

transformers 48-52 54-58 51-55 58-62

Calculation of power for each core leg

The power for each core leg is defied as the power for core on which the windings are

mounted, referred to a double-winding transformer.

50

P1 = Se1/mt

1

Where:

Se1 – power referred to a double-winding transformer (KVA), it does not need to

convert in the case of double-winding transformer, for three-winding transformers:

Se= (SG + SZ + SD)/2

SG , SZ , SD - power (KVA) for HV, MV, and LV winding respectively: Mt1 – number

of core legs on which the winding is mounted, e.g. for single-phase transformer with

two core legs, mt1 = 2: for three- phase transformer with three core legs and two side

yokes, mt1 = 3

Cross - Section of core

The cross- section of core leg and yoke are made up of a number of lamination stacks

to form a steeped periphery, for a specifies diameter, the more lamination stacks to be

encircled, the more effective cross-sectional area of core may be absorbed the

complexity of manufacturing labor force.

Calculation of winding turns

After the determination of core diameter, generally, the calculation of winding

translating with the winding without tapping (e.g. low-voltage winding) and

calculated the winding turns for HV or MV.

Calculation of volts per turns

For rated frequency = 50 HZ

et1 = B At x 103 /4.5

For rated frequency = 60 HZ

et1 = B At x 103/3.75

where

et1 – volts per turn calculated preliminarily , v :

At – core- sectional area of core m2:

B – flux density , T.

Generally, for 0.3 mm cold-rolled grain oriented silicon sheet steel (e.g. Z8H) The

flux density adopted may be 1.5-1.7T.

Preliminary calculation of low-voltage winding turns

W1D = UXI / e1

t

Where

W1D – turns of low-voltage winding calculated; (take integer for WD)

UX I – Phase voltage for LV winding, V.

51

The value of W1D calculated with formula may be not an integer, when the decimal is

rounded to an integer, the flux density may be higher slightly than that calculated with

e1t; when the decimal is carried as in adding the flux density may be lower slightly

than that calculated with: e1t

Determination of volts per turn et

et = UxI / WD

Where the value of e1 should be calculated to three places of decimals.

Calculation of flux and flux density

After the determination of the flux and flux density may be calculated used with the

following formulae at the frequency of 50 HZ.

B = At

et 3105.4 −×

Where

B in Tesla.

×= etm 5.4φ 10-3

Where: mφ in Weber

Calculation of winding turns for high-voltage or medium-voltage winding

For most of high-voltage or medium-voltage winding of corresponding tapping.

Calculate the turns of corresponding tapping according to their phase voltages, firstly,

calculate the winding turns for the max. tapping then determine the tapping turns for

each tapping voltage according to the voltage difference between two adjacent

tapings.

Turns for max, tapping is:

W1e1 = Ug1 / et ( take integer for Wg1)

Turns for each tapping:

We1 = U g / et ( take integer for W) Δ Δ

Turns for each tapping may be calculated by We1 - Δ WG Correspondingly.

3.3 Calculations of winding Determination of winding construction

The construction of winding in accordance with the power and voltage-class of the

transformer, the following description on winding construction may be used for

reference.

52

a) Layer windings The merit of this type of windings easy to manufacture and of good cooling

performer due to its axial oil-duct between layers. Generally it is adopted in the small

and medium size transformer with single or double layer for the low-voltage windings

and multi-layer for high-voltage windings, the demerit of then presents the poor

supporting stability of windings ends, and the axial height of winding is hardly to

control.

b) Helical windings.

This type of winding is easy to manufacture but is does not suit to winding with

large number of turns. It may be classified into single-row, double-row and four-row

helical winding. It is suitable for low-voltage and heavy cur-rent winding or

regulating winding.

Determination of number of disks and turns per disk

a.Determination of number of disk for outer winding (HV winding)

The total number of disk for HV winding depends on the voltage class,

longitudinal insulation and impedance voltage etc. Here, the number of disks listed in

table provides the value recommended for pre-determination.

Total disk number for HV winding

Voltage class

for winding 10 35 63 110(132)

Total disk

number 40-60 56-75 60-80

60-80

2(34-44)

Note: The number multiples 2 denoting the upper and lower part of winding connecting in

parallel.

Determination of turns per disk

For tapping winding disks:

WF = WF / NF (take integer) Δ

Where

WF – Turns per disk for tapping winding

ΔEG – turns for tapping:

NF – number of disks for tapping winding.

Turns per disk for normal part of winding

WG = W1 /(N-NF1) (take integer)

53

Where

WG – turns per disk for normal part of winding:

W1 – turns for minimum tapping voltage:

N – Total number of disks for HV winding. When the line terminal is located at the

middle of winding, take N/2:

NF1 – Number of disks for tapping winding. Generally, for 10 KV line terminal at middle,

choosing 4; for 35 KV line terminals at end of winding, choosing 8.

Turns per disk for disks with additional insulation

For the sake of additional insulation, the turns in these disks should be 1-2 turns less than

those disks in normal part of winding (e.g. the inner diameter of disk should be enlarged

for the incense from insulation or to make room for shielding wire inserting)

The total turns of high-voltage winding equal the sum of different disks times the

corresponding turns per disk. For winding with middle out-let terminal, the total number

of turns is calculated by one-half the total number of disks, because the upper half and

lower half of disks are connected in parallel.

Determination of number of disks and turns per disk for low-voltage or medium-

voltage winding

The low-voltage winding is situated in the middle or inner part of the winding

assembled on the core leg, the fractional number of turns should e adopted for the normal

disks of a continuous winding in order to avoid increasing the radial dimension of the

wining. The fractional part of fractional number of turns may be considered as Table

Table

Disks Normal

disks

Disks with at beginning or or final

Number of parallel

conductors in radial

direction

1-6 1 2,3 4,5 6

Fractional part of

number of turns 111

nn −

111

nn −

121

nn −

≤ 1

31n

n −≤

141

nn −

≤

N1 - Number of axial strips for the winding.

Determination of number of disks and turns per disk

54

a. firstly, set the number of disks as N1, take the even number for N1, its magnitude

should adapt to the outer winding:

b. The fractional part of turns per each disk is better to select1

11n

n − , except those for

disks at the beginning and the final of the winding;

where n1 is number of axial strips for the winding.

c. The fractional number of turns for disks at the beginning and final of the winding

is as follows:

For the number of parallel conuctornal part as < n - 2

1+n

d. The number of disks and turns per disks may be determined by solving the

following simultaneous equations: XW + YWY = W -------------------------------------- (1)

X + y = N ------------------------------------------------ (2)

Where

X1 & Y1 – number of disks to be determined;

W – total number of the winding;

Wx1 - Wy – turns per disks assumed by the analysis with the given total turns and

number of disks pre-determined.

By solving the equation (1) and (2) shows that the correct determination of number of

disks is most important link in the design calculation. When the obtained by solving the

simultaneous equations.

Calculation of winding height and radial dimension General technical requirements

a. when enamel wires are used as the winding conductor, take the maximum

insulation diameter of wire to calculate the winding height and radial dimension.

b. When paper-wrapped round conductors are used as the winding conductor, take

the maximum insulation diameter to calculate the winding height and radial dimension.

c. When paper-wrapped that conductors are used as the winding conductor, the

following tolerances should be added to the insulated conductor dimension for the

winding height and radial dimension.

The turn insulation of conductor is the total thickness of paper on both sides,

during calculation, the turn insulation should be added by 0.5 mm, e.g. for 0.45 mm turn

insulation, take it as 0.5mm for calculation.

55

If the winding is to be dried under pressure, the compression of dimension of turn

insulation in axial direction of winding should be considered as follows:

For 1.95 mm turn insulation – 2.0 × 0.9 = 1.8:

For 1.35 mm turn insulation - 1.4 × 0.9 = 1.26:

For 0.95 mm turn insulation – 1.0 × 0.9 = 0.90:

For 0.6 mm turn insulation – 0.65 × 0.9 = 0.0585:

For 0.45 mm turn insulation – 0.5 × 0.9 = 0.45:

As above listed, take 0.9 as the compressing factor for the turn insulation, if the

winding is to be dried not under pressure, the compressing factor should be 1.

d. The enlarged oil-duct of winding should not exceed 20 mm: if the centre

distance between two adjacent radial spacers exceeding 120 mm. the maximum height of

the enlarged oil-duct may be 24 mm.

e. For double-row and four-row helical windings, the additional height of one

turn due to transposition of conductor should be taken in consideration during the

determination of main insulation and dimension of spacers.

f. Determination of the number of radial spacers along the periphery of winding

manufacturing and dynamic stability of winding, generally, the centre distance between

the adjacent radial spacers adopted is 100 mm or more for winding with conductor with

conductor with or 8 mm or below; 120 mm or more for winding with conductor with

more than 8 mm.

g. The last figure of radial dimension of winding is taken 0.5 mm as minimum;

for the winding height, take 0 or 5 as last figure.

Determination of number of winding conductors along the winding axis for single-

row helical winding (one standard-transposition and two group-transposition):

N = number of turns + 4

Where:

N – number of conductors for calculation of winding height for double-row helical

winding (uniform transposition)

N = ( number of turns ) x2 + 2

For four-row helical winding (uniform transposition)

N = (number of turns ) x4 + 4

Or

N = ( number of turns x4 + 2 ( terminals at opposite side for continuous winding

(including capacitor shield winding and interleaved winding):

56

Where

N = number of disks (for terminals at winding ends)

N = (number of disks) x2 (for terminal at middle)

Determination of number of oil-duct along the winding axis

N1 = N – 1

Where:

N1 – number of oil-duct along the winding axis;

N – Number of conductors for winding height calculation.

If static rings are adopted, it requires the additional oil-duct. The dimension and position

of these oil-ducts should be arranged according to the requirements of longitudinal

insulation.

Calculation of winding height

H = H1 + H2

Where

H – winding after drying, mm:

H1 – overall height of insulated conductors along the winding axis, mm:

H2- overall height of insulation spacers along the winding axis after drying, mm

If the static rings are adopted, the thicnkess of ring and it, oil-duct should be

added to the winding height.

Calculation of insulation radii and window height of core

Calculation of insulation radii

R1

S1

R2

BD

R3

S2

R4

BG

_________Radius of core

_________ Distance between LV winding

________Inner radius of LV winding

________Radial dimension of LV winding

________Outer radius of LV winding

________Distance between HV and LV winding (see “ main

insulation”)

_______Inner radius of HV winding

57

_______Radial dimension of HV winding

R5

X -2

D

+E

Mo

________Outer radius of HV winding

________Outer diameter of HV winding

________Distance between phases ( see “ Main insulation”)

________Distance between core legs

RD – Mean radius of LV winding

Rr – Mean radius of Main axial oil-duct

RG – Mean radius of HV winding

Calculation of window height of core.

H1

+ H2

H3

- H2

H4

+EK

H5

H6

δ

Ho

_______Over all height of conductor

_______Over all height of spacer

_______Overall height of spacer to be compressed after dying

______Thickness of static ring and its oil-duct (need not to

calculate for winding without static ring)

_____Thickness of pressing ring

_____Clearance between pressing ring and upper yoke

_____Window height (take 0 or 5 as the last figure)

Note: Use mm for above calculation.

3.4 Calculation of impedance voltage The impedance voltage is and important performance parameter for transformers, it has

been specified in the standard of transformer performance, the calculation of impedance

voltage presents a very important job in the design calculation.

The impedance voltage under rated current and principal tapping allows a

tolerance of + 10%, however, the allowable tolerance for calculation should be kept

58

within + 2.5%, because some inevitable deviation may be revealed during test and

manufacturing.

The impedance voltage consists of two components i.e. the resistive voltage-drop

and the reactive voltage-drop, generally, the resistive voltage-drop is very small, it may

be neglected for transformer with power of 8000 KVA and above. The following

formulae are all applied to the calculation of impedance voltage with windings arranged

concentrically.

Table Value of P= f ( λ11 )

λ11

0.5 0.6 0.7 0.9 1.0 1.1 1.2 1.4 1.5 1.6 1.7 1.8 1.9

P 0.50 0.55 0.60 0.69 0.72 0.74 0.77 0.79 0.80 0.81 0.82 0.83 0.84

λ11

2.0 2.5 3.0 3.5 4.0 5.0 6.0 7.0 8.0 9.0 10.0 20.00 30.00

P 0.84 0.87 0.89 0.91 0.93 0.94 0.95 0.955 0.96 0.965 0.97 0.984 0.99

K – Coefficient of additional reactance

The distribution of ampere-turns may be non-uniform due to the presence of inter-

leaved disks and disks with tapping, thereby the additional reactance may be caused by

the radial leakage flux, it may be corrected by coefficient k. Select K=1.02 for continuous

winding; K=1.05 for helical winding these suit to the unbalance of ampere-turn within the

range of 5%.

For three-winding transformer

For three-winding transformers, the impedance voltage fore each pair of windings should

be calculated, if these three winding are of different power are of different power, the

impedance voltage should be converted to 100%.

Uk×12 = %106.

6.4912

12121211

HeKPDWfI

t

Σ

Ukx13 = %106.

6.4913

13131311

HeKPDWfI

t

Σ

Ul x 23 = %106.

6.4923

23232311

HeKPDWfI

t

Σ

λ 12 = R3-R1; λ 13 = R4-R1; λ 23 = R4-R2;

59

Σ D17 = 31 (a1r1 + az rz) + a12r12

Σ D13 = 31 (a1r1 + a3 r3) + a13r13

Σ D23 = 31 ( azrz + a9 r3) + a23r23

Where: I1 W1- rated ampere-turn for winding 1:

R1, r2, r3 – mean radius for winding 1,2,2 respectively:

R1, R2, R3, R4 – mean radius of winding respectively ( to bare conductor) :

λ 12, λ 13, λ 23 – winding of leakage flux path between winding 1 and 2, and 3.2 and 3

respectively:

A1, a2, a3 – radial dimension o winding 1, 2, 3 respectively (bare conductor to bare

conductor);

R12, r13, r23 – mean radius of oil-duct between winding 1 and 2 and 3,2 and 3 respectively;

A12, a13, a23 – width of oil-duct between winding 1 and 2, 1 and 3, 2 and 3 respectively.

L12, k13, k23 – coefficient of additional reactance between winding 1 and 2 1 and 3, 2 and

3 respectively;

p12, p13, p23 – Rogovski coefficient for winding 1 and 2, 1 and 3, and 2 and 3 respectively.

Unit for length in the formulae are all cm.

3.5 Calculation of Winding Parameters When the values of impedance voltage calculated have met the requirements of

standard or contract, the following data may be put into calculation.

The essential data to be written.

a. Connection of winding , Line-voltage and phase-voltage, V;

b. Line-current and phase-current, A;

c. Winding turns for each tapping;

d. Number of disks and turns per disks;

e. dimension and cross-sectional area of single conductor in each disk;

11 baba

××

---tickness and width of b are conductor (mm)---thickness and width of paper wrapped conductor (mm)

A = M B A

Where

A - overall cross-sectional area conductors, mm2;

A1- cross-sectional area of single conductor, mm2;

60

Mb – number of parallel conductors

c) Current density

j = I x g /A

Where

J-current density , a/mm2;

Ixg – phase current , A; A-overall cross-sectional area of conductors , mm2

Calculation of winding data

The axial and radial dimensions of winding and the radial of insulation have

already been determined, with the result that the following data may be calculated

successively.

Calculation of conductor length

The mean length of turn may be calculated as:

Ip = 2π RX

Where

Ip – mean length of turn, m;

RX – mean radius of a specific winding, m.

The overall length of conductor may calculated as:

L = Ip Wm Where

L – overall length of conductor, m;

Wm – overall number of turns for maximum tapping (for each leg).

The conductor for winding terminals have to be taken into consideration during the

calculation of conductor length, usually adding 1.5-2 m to the conductor length

calculated.

The overall conductor length for principal tapping (winding with tapings): this is

mainly for the purpose of calculation of de resistance and resister leg.

LN = LP WN

Where

LN – overall conductor length for principal tapping, m:

WN – overall number f turns for principal tapping.

Calculation of bare conductor weight

For three-phase transformers

GX = 3LAg x 10-3

Where: GX – weight of bare conductor, Kg;

61

g – Specific weight of conductor, for copper conductor g=8.9 g/cm;

The other denotations see above

Calculation of weight for insulated conductors

GC = GX 9 1+ c%)

Where: GC – weight of insulated conductor, Kg;

GX – weight of bare consenter, Kg;

C % - the percentage weight of paper to the weight of bare conductor, for flat copper

conductor wrapped with insulating paper, the value may be calculated as:

C % = Ax

TbaT )57.1(.17 ++

Where:

T– insulation thickness on each side of conductor, it equals thickness of turn insulation;

A – thickness of bare conductor, mm;

B – thickness of bare conductor, mm;

Ax – cross-sectional area of single conductor, mm;

Calculation of D.C. resistance of winding.

The d.c. resistance of winding is an important parameter for transformers, its

value for three phases reveals whether the specification of conductor adopted

being conformed to the design requirement; during manufacturing the d.c.

resistance measurement may be applied to control the winding quality and

welding joint of connection between leads and bushings, and the balance of d.c

resistances among three phases.

a) Resistively of conductor

The resistively of copper conductor at 75oC is 0.02096 kgn. Mm2/m, 1/58

ohm. Mm2/2 at 20oC.

The receptivity of copper conductor at a specifies temperature may be

calculated as follows:

P2 = P1 ( )235235

1

2

θθ

++

Where: θ 1, θ 2 – given temperatures, oC

b) d.c resistance of winding

The standard value of d.c resistance for winding is referred to the value at 75 oC

R75 oC = P75 oC ALn

62

Where: R75 oC – resistance of winding at 75 oC, π ;

P75 oC – receptivity of conductor at 75 oC π mm2 /m;

LN – length of winding conductor at rated voltage, m;

A – overall cross-sectional area of winding conductor, mm2

Calculation of resistance losses of winding

For three-phase transformer:

Pr = 3 I2 x gR

I – phase current, A:

R – d.c. resistance of winding at 75 oC

3.6 Calculation of core data

Calculation of core and yoke weight.

For three-phase core with three legs:

GF1 = 3 λ HoA x 10-4 (for core legs)

GF2 = 4 λ HoA x 10-4+ Go (for yoke)

GF = GF1 + GF2

Where: GF1, GF2, GF – weight of core legs, weight of yoke overall weight of core, kg

Ho – height of core window, mm.

Mo – centre distance between core legs, mm;

At – cross-sectional area of core leg, cm2

Ac – cross-sectional area of yoke, cm2

Go – weight of corner, kg;

λ - specific weight of silicon steel, for cold rolled brain oriented silicon sheet steel , λ =

7.65.

Calculation of no-load loss

Once the core material adopted is determined, the specific core loss (W/kg)

depends upon the flux density to be used, the specific losses for different flux densities of

core are listed in standard table.

The no-load of core may be calculated as:

Po = K1 p GF

Where

K1 – coefficient for additional core loss, 1.15 may be adopted;

P – specific core loss, W/Kg

GF – weight of core, Kg.

63

Calculation of no-load current.

The no-load current is composed of the real components and the reactive component,

usually; it is designated by the percentage of rated current.

a) The real component of no-load current.

Ioa = se

Po10

%

Where

Ioa – real component of no-load current;

Po – no - load loss, w;

Se – rated power of transformer, KVA.

b) The reactive component of no-load current

Ior = ko se

GAgGg tjFe

10+

Where: ior – reactive component of no-load current;

Ko – coefficient, for cold rolled steel take ko=1.3;

GF – weight of core, Kg

C – Number of lamination joint ( according to lamination diagram).

At – effective cross section of core, cm2; (with the same cross-sectional area for core leg

and yoke).

Se – rated power of transformer, KVA;

Ge – specific magnetizing power,

Gj – magnetizing power per specific area of lamination joint, VA/cm2 (see table below)

c) formula for no-load current calculation

io = IorIoa+

where: io – no-load current;

ioa – real component of n o-load current

ior – reactive component of no-load current.

Table magnetizing power per specific area of joint gj (VA /cm2)

64

13000 14000 15000 16000 17000 Flux density

(Gause) Hot-

rolled

Cold-

rolled

Hot-

rolled

Cold-

rolled

Hot-

rolled

Cold-

rolled Cold-rolled

Cold-

rolled

000

100

200

300

400

500

600

700

800

900

1.75

1.81

1.87

1.93

1.99

2.06

2.14

2.22

2.3

2.38

0.853

0.902

0.938

0.99

1.04

1.095

1.15

1.2

1.255

1.31

2.46

2.55

2.65

2.75

2.85

2.95

3.05

3.15

3.25

3.35

1.37

1.41

1.48

1.54

1.6

1.66

1.72

1.795

1.87

1.94

3.45 1.98

2.046

2.13

2.19

2.26

2.34

2.42

2.5

2.58

2.65

2.74

2.83

2.91

3.01

3.1

3.19

3.28

3.39

3.49

3.6

3.72

3.83

3.97

4.09

4.23

4.37

(1 Gs = 10-4 T)

3.7 Calculation of tank dimensions Calculation of tank height

The inner side tank height H as shown in figure may be calculated as:

H = Ho + 2Hc + Hd + Hl

H

Where

H – tank height, mn:

Ho – core window height mm:

Hc – maximum width of yoke Lamination (height of yoke) mm;

Hd – height of base plate of core, mm;

Hl – distance between core and tank cover, mm.

65

Calculation of tank width

The width of tank B as shown in Figure may be calculated as:

B = D + B1

Where:

b – width of tanks, mm:

D – Diameter of outer winding, mm:

B1 – distances between tank wall and active part of transformer from both HV and

LV sides.

Calculation of tank length

The length of tank L as shown in Figure may be calculated as:

L = D + 2Mo + B2

Where: L – length of tank, mm.

D – Diameter of outer winding, mm:

Mo – Centre distance between corn legs, mm;

B2 – Clearance along the longitudinal axis from the tank wall to windings of

phase A and phase C.

3.8 Calculation of stray losses Classification of transformer losses.

The transformer losses may be divided into two parts, first the basic losses including no-

load loss and resistance loss of winding, second, the stray losses which are caused by the

leakage flux. In windings, leads, tank, core and yoke clamping etc.

The calculation of basic losses was described already in sub-clauses.

the stray losses may be calculated by means of the coefficient K1. The method for

calculation is as follows.

Losses of eddy current and circulating current caused by the longitudinal leakage flux of

winding.

The winding is situated in the longitudinal leakage flux field, so the eddy losses

would be produced accordingly, the magnitude of such losses depend on the

dimension of winding and the leakage flux density, the leakage flux density should be

kept within 0.16 T.

It would be best, if the percentage of eddy losses caused by longitudinal leakage flux

could be kept within 20% of the resistance loss.

66

The percentage of eddy-loss means value under complete transposition to the

resistance loss.

For continuous and interleaved winding with two conductors wound in parallel. Or less

or for double – row or four – row helical winding, the transposition may be longitudinal

leakage flux may be calculated as.

K2 = 710

K (x

x

HPFmnaA )2 %

Where

K – coefficient depended on temperature: for copper conductor, K = 3.95 at 75o C K =

3.707 at 85oC:

For temperature of ToC, the value of K m ay be Calculated as:

K = ( t+

+235

75235 )2 ×3.95

Where: f – frequency, Hz;

P – Rogovski coefficient;

Ax – cross – sectional area of single conductor, mn2; other symbols see table

Table

Designation of symbols Concentric winding Sandwich windings

S or b – width of bare

conductor perpendicular to

the leakage flux

Thickness of single bare

conductor (axial)

Height of single bare

conductor (axial)

Hx – Reactance height of

windings (mm)

Reactance height of

windings

Radial width of winding

M – number of conductors

perpendicular to the leakage

flux

Continuous winding: turns

per disk x number of

parallel conducts helical

windings; number of

parallel single conductors

Continuous winding:

number of disks in balance

group M1 ( or m2) Helical

winding: number of disks in

balance group m1 ( or m2)

N – number of conductors

parallel to the leakage flux

Continuous winding:

n=number of disks Helical

windings n= turns x number

Continuous winding: turns

per disk x number of

Helical winding: number of

67

of helixes parallel single conductors

N1 ( or N2)

Note: a, d, Hx, m, n, m1, n1, m2, n2, in the Table

Calculation of stray losses:

Under the operation of transformer, the leakage flux may penetrate into steel

constructional parts (e.g yoke clamping, winding , ring and tank wall etc.) the additional

loss produced thereby is called stray loss. Owing to the complexity of leakage flux paths,

the stray loss is hardly to calculate exactly up to now the calculation is based on semi –

experimental and approximate method. For winding with concentric arrangement, stray

losses are caused by longitudinal and radial leakage flux. The stray loss in a double –

winding transformer of three – winding transformer with only one pair of winding tin

operation may be calculated with the following semi-experimental formula:

P2 = {2

322

)](2[ RpRHksHkUkKz−+ δδ

φ ×(

50f )2

Where

Uk – percentage impedance voltage at rated power.

Hk – Reactance height of winding, mm;

F – frequency, Hz;

Rp – Mean radius of main leakage flux path, mm; see calculation of impedance voltage)

Sδ - Inner tank-wall girth, mm;

Rδ - Mean equivalent radius of rank mm;

For three- Phase transformer.

Rδ = (a + b – 2mo) /4

Where

a – Length of tank;

b – Width of tank;

mo – centre distance between core legs;

φ o – Main flux in the core at rated excitation, Wb;

φ o = B A

Where

B – Flux density in the core, T;

At – cross-sectional area of core leg, m2;

Se – rated power, KVA

68

S – Actual power during operation, KVA;

Kz – Stray-loss coefficient,

By the analysis of transformer, the coefficient Kz for existing construction may be

estimated with following formula;

Kz = UK23 × 104

Calculation of losses in leads:

Current passing trough the lead causes resistance loss which pertains to the main loss in

the lead, it should be added to the additional losses for transformer. The formula for

calculation is as following.

Py = M Iy2 Ry

Where: Py – loss of lead, W;

M – number of phases;

Ly - current passing through the lead, A;

Ry – resistance of lead;

Ry = SYPLy ,

Ly – mean length of load for each phase (including the length within the bushing crosses-

sectional area of lead, mm2.

Attention: The calculation of load-loss for three-phase transformers may be conducted by a pair of

winding, i.e. load-loss between inner and outer inner and outer winding, inner and

winding leakage flux is maximum, thus for calculating the load-loss between inner and

outer winding the value of 3 time longitudinal eddy-loss of intermediate winding should

be added.

3.9 Calculation of temperature-rise Contents of temperature-rise calculation

The cooling method for transformers varies according to the power of equipment e.g.

oil-immersed nature cooling, oil-immersed forced-air cooling, forced-oil forced-air

cooling or water cooling and forced-directed oil circulation for the very large size

transformers.

The main contents of temperature-rise calculation include.

(I) Calculation of average temperature difference between winding and oil;

69

(II) Calculation of surface area of tank for dissipation;

(III) Calculation of temperature-rise for oil-immersed nature cooling oil-immersed

forced-air cooling and forced-oil circulation.

Average difference of temperature between winding and oil for continuous,

interleaved and helical windings

Winding disks are immersed in the oil, all the surfaces act as heal dissipation surface

(excluding semi-continuous and semi-helical windings), if there is no axial spacers

between winding and cylinder, the longitudinal and radial oil-ducts of the windings, the

oil flows upward from bottom of tank, a part of them passing through the radial ducts and

flowing upward also.

Calculation of heat load for winding surface

The specific heat load is calculated with winding disk possessing the maximum number

of turns.

q = )100

( 4

3

21 KLKIWJKK

+

Where

q – specific heat load for winding surface, W/m2.

K1 – coefficient, related to temperature of material, take 21.63 for copper conductor at

85o C.

I – current passing through winding disk, A;

J – current density within winding disk, A/m2;

W – Number of turns per disk, it should be carried in an integer for fractional turns: for

helical winding, W = 1;

K2 – correction factor of turn insulation:

For a1 < 1.75, K2 = 1:

For a1 > 1.75, K2 = a1/1.75a;

a – thickness of bare conductor, mm;

a1 – thickness of conductor with insulation, mm;

K3 - coverage coefficient of winding disk

K3 = 1 – number of radial spacers along girth × width of spacer / mean length of turn in

disk

K4 – percentage overall additional losses in the conductor;

L – outer girth of disk cross - section

70

For continuous and helical windings, L = 2 (ma1 + B1)

For semi-helical windings, L = ma1 + 2b1

For disk with inner turn padded with paper strips, L = 2ma1 + b1;

M – number of parallel conductors in disk;

b1 – width of conductor with insulation, mm.

If winding disks with insulating strips padded underneath the inner turn, the thickness of

these strip should not be counted for dissipating, surface; if there are additional

insulations on the outer side of disks, L should be calculated with the outermost

insulation.

Calculation of average temperature difference between winding and oil

Tx = Tx1 + TΔ j + TΔy

where

Tx1 – temperature difference between winding and oil, oK;

For oil – immersed nature cooling, inner winding; Tx1 = 0.41

For oil – immersed nature cooling outer winding: Tx1 = 0.358

For oil – immersed forced – air cooling, inner and outer winding Tx1 = 0.159

T j – insulation correction of winding temperature – rise, Δ

TΔ j = Kjq

Kj – coefficient of insulation correction, it is a function a δ see table

For disks without additional insulation, δ = a1 – 1; for disks with additional

insulation, the value of δ see Table

T y – correction of temperature for oil-duct, TΔ Δy = Pq/1550;

P – Additional coefficient,

Table Corresponding Value of K to δ

δ 0.6 0.75 1 1.4 2 2.5 3 3.5 4

Kj ( ×10 -4) 4 9 17.5 29.5 49 62 79 94 110

δ 4.4 5 6 7 8 9 10 11 12

Kj ( × 10-4) 123 139 165 200 230 262 298 326 364

Table Values of δ for disks with additional insulation

B 100mm ≤ δ = a1 – a +2c

100 < B 150 ≤ δ = (a1-a) +2c+0.2c (B-100)/50

71

150 < B≤200 δ = (a1 –a) +2.2c + 0.46c (B -150)/50

Calculation of top oil temperature-rise for transformers with oil-immersed. Nature

cooling and forced-air cooling.

The temperature-rise of top oil may be calculated as:

Ts = 1.2 Ty + T Δ

Where: Ts – top oil temperature-rise, oK;

Ty – average temperature –rise of oil, oK;

For Nature cooling,

Ty = 0.262 qr0.8

For forces – air cooling

Ty, = 0.191 qr0.8 /7;

Qr – specific heat load for tank cover, w/m2;

Qr = A

pkpo +

Where

po - no – load loss, w;

Pk – loss (including additional loss), w;

A – overall effective dissipating area of tanks, m2

T - value of correction for oil temperature – rise, oK. Δ

3.10 Calculation of axial mechanical stress of windings Significance of ampere-turn balance

The conductor size and oil-duct dimension vary as the different type of winding the

distribution of turns along the winding height may be non uniform, especially for winding

with tapping (except those with separate regulating winding). The calculation on ampere-

turn distribution aims at adjusting the ampere-turn of certain zone of winding along its

height to math the ampere-turn of another winding for balance as far as possible, it would

be best to keep the difference between them minimum for the sake of ampere-turn

balance.

The more unbalanced ampere-turn, the more radial leakage flux would be produced.

Consequently, the more radial leakage reactance, more eddy losses and larger mechanic

stress under short-circuit would be arisen. For large transformers, the leakage flux may

72

cause local over-heat in winding or constructional parts. As a result, the calculation on

balance of ampere-turn presents a very important significance despite the absolute

balance could not be obtained practically, the balance of ampere-turn should be kept at

least to meet the requirement of mechanic of mechanic strontium for winding.

The general rules for delimitation of balance zone.

a. Generally, the balance zone should be divided based on high-voltage winding.

Zones for low-voltage (or medium-voltage) winding correspond to those of high-

voltage winding, i.e the number of zones of low-voltage winding (or MV) equals

the number of zones of high-voltage winding. The height of corresponding zones

for both winding should be equal to each other as far as possible in addition, The

distribution of ampere-turn along the winding height should be as uniform as

better within each zone.

b. Within a zone, the radial dimension of winding and the main conduit for leakage

flux should be kept identical as it could be.

c. The percentage number of turns in each zone of low-voltage (or medium-voltage)

winding should correspond the percentage number of turns in those of high-

voltage winding, the difference between them should be minimum as far as

possible, otherwise, the unbalance of ampere-turn distribution would be enhanced.

d. For winding with symmetric arrangement of disks, the calculation may be

conducted in one half of the winding, winding consisting of two parallel branches,

may be considered as one combined branch.

Calculation of number of turns in each zone and height of zone.

a. Two requirement as follows must considered during the delimitation of balance

zone.

First, for winding static rings the height of zone located at the ends of winding should

exclude the thickness of static rings.

Second, for helical windings, the height of zone located at the ends of winding

should exclude the height of 2/2 turn for terminals.

b. The delimitation of balance zone should follow the related of:

The total number of turns of high-voltage winding equals the sum of number of

turns for each zone.

W1 = W11 + W12 + W13 + W14

The total number of urns of low-voltage winding( or medium-voltage winding)

equals the sum of number of turn for each zone.

73

W3 = W 31 + W32 + W 33 + W34

The overall height of high-voltage winding equals the sum of height for each

zone:

H1 = H11 + H12 + H13 + H14

The overall height of low-voltage ( or medium-voltage) winding equals the of

height for each zone.

H3 = H31 +H32 +H32 + H33

c. Calculated number of turns for each zone of HV, MV and LV winding:

Turn for zone = disks in zone × turns in each disk.

d. Calculate the height for each zone.

(Two zones are delimited by the centre of oil-duct between zones)

For a special zone:

Number of small oil-duct x height of duct = overall height of small oil duct number of

large oil-duct × height of duct = overall height of large oil-duct

+

Overall height of oil-duct in that zone

Factor for concussing (0.92)

×

Overall of oil-duct in that zone after drying +

number of conductor along axis × height of insulated conductor = overall height of

conductor

Overall height of that zone(rm)

Rogovski Coefficient.

The rogovski coefficient of axial mechanic stress and reactance for sandwich winding

calculation may be selected as follows.

Pn = 1 - uπ1 (1-euπ)×[ 1-0.5(1-e-2πv) ( 1-eπu)]

Where

U = λ / h λ = overall width of leakage flux

h =outer diameter of winding- inner diameter of winding

S = s’ + 0.03D

s; - distance between circum circle

74

D – diameter of core, cm;

Calculation of axial mechanic stress under shout- circuit for concentric winding.

The axial mechanic stress caused by the short - circuit current under maximum

radial leakage flux group may be calculated as:

Fi = 11

2

1004.88.9

λKlKa× P0 (Rcp) (In)2 (Wn)2 Σm-n [σm + σm-1 Cpm]

Where:

Fi – axial mechanic stress, N;

In - rated phase current. A:

WN – rated number of turns per phase:

Pn - Rogovski coefficient

3.11 Calculation of stress conductor under short-circuit For outer winding

a. Tensile stress of conductor under short – circuit

σ2= 810 mn

pn ml)2 )(( W I)2 ( (Kl)2 (ky)2 12.8 0.98×abλ

λ MPa

b. axial bending stress of conductor under short – circuit

σ2= 810 a1mn

pn ml2 W I2 Kl2 ky2 12.8 0.98

×

λ MPa

Overall stress in the conductor:

σ = (β1 /100 )2 (σ1 + σ2 ) MPa

For inner winding,

a. Compressing stress of conductor under short – circuit

σ2= 8 10 A1Hk n m

p)P (R ) (W) (I)2 Kl)2 ( (ky)2 ) 12.8 ( 0.98

× x N’

a. Axial bending stress of conductor under short – circuit

σ2= 010 ab2mn

pn ml2 W I2 Kl2 ky2 12.8 0.98

1×

λMPa

b. Radial bending stress of conductor under short-circuit

= ××

××82

2

104.6098.0 222

bmnHkapWlplKJKy m’ MPa

Overall stress:

= (100

3β )2 (σ1 + σ2 + σ3) MPa

75

The allowable value of overall stress for copper conductors should be kept within

156.8 MPa, it may be higher for conductors of high strength.

The designation of symbols in above formulae:

σ1, σ2, σ2, σ – stresses of conductors, Mpa

Ky – coefficient for short – circuit impulse current, generally take 1.8;

K1 – multiples of steady short – circuit current,

I – phase current of short - circuit tapping, A;

W – corresponding turns in the winding;

Rp – mean radius of winding, cm;

I = (NRπ2 - b, cm);

R – outer diameter of winding, cm;

N – number of radial spacers;

B – width of spacer, cm;

Ip = NRpπ2 - d, cm;

D – width of axial strips, cm;

P – Rogovski coefficient, see calculation of reactance;

N – parallel branches per phase;

M – number of parallel conductors in disk;

Hk – reactance height of winding, cm;

A1 – cross – sectional area for single conductors, cm2;

A – radial dimension of bare conductor ( thickness), cm;

B – axial dimension of bare conductor ( width), cm;

β 1, β 2 – current distribution coefficient for double – winding transformer β 1= β 3=100;

For three – winding transformer, the value of β 1 and β 3,;

Pn – Rogovski coefficient;

Pn – f(H,v), calculate with formula

U = hλ , v =

hs ,

S = s’ + 0.03D, where s’ and D

λ - overall width of leakage flux, cm;

S’ – distance between core and inner winding, cm;

D – core diameter, cm;

76

H – height of leakage flux group α m cm;

M’, N’, = f ( Z, 2rp/a),

Z – number of axial strips.

3.12 Calculation of weight Calculation of oil weight

Quantity of oil displaced by the active part of transformer

a. Transformer with high-voltage side of 35 KV or below:

Gpy = 5.48.7cupe GG

+

b. Transformer with high-voltage side of 63 – 132 KV.

Gpy = 9.36.7cufe GG

+

Where

Gpy – weight of oil displaced by the active part of transformer, Kg;

GFe – weight of silicon sheet steel, Kg;

GCu – weight of copper conductor with insulation, Kg;

Quantity of oil to be filled into empty tank.

for drum type tank

GKY = 0.9 HAd

Where:

GKy – oil capacity of empty tank, Kg

H – Tank height, dm:

Ad – cross – sectional area of tank, dm2

Ad = L B – 0.8584 R2

L – Length of tank, dm;

R – Radius of round corner. Dm; R =2B : (for oblate cross – section)

B – Width of tank, dm.

77

H

H

BR R

B

LZ L L

Fig: 1

B2

B1

H1 H2

R2

B R4

R1

R3

L

Fig 2 Tank with ladder – shaped top

78

Weight of oil in the tank

GNy = Gky – Gpy

Where

Gny – Weight of oil in the tank, Kg;

GKY - weight of oil to be filled in the empty tank, Kg;

GPY – weight of oil replaced b y the transformer active part, Kg;

Weight of oil for radiators and coolers

Gey = Ne Gey

Where

Gey – weigh of oil in radiator or cooler, Kg;

Ne – number of radiator or cooler;

Gey – Weight of oil in each radiator or cooler, kg.

For weight of oil in oblate tube radiators, see Table 1: those for coolers, see Table 2; and

those for panel type radiators,

Table 1 weight of tube type radiators (kg) 88 – tubes 100 – tubes 120 – tubes 45o 88 – tubes 44 - tubes Center

distance

between

connecting

flanges

(mm)

Radiator

weight

Oil

weight

Radiator

weight

Oil

weight

Radiator

weight

Oil

weight

Radiator

weight

Oil

weight

Radiator

weight

Oil

weight

1400

1650

1880

2000

2285

2485

2685

2885

3000

3250

3500

3750

4000

309

342

372

390

426

454

481

508

522

556

590

624

656

108

119

129

134

145

154

163

171

176

186

197

208

218

421

430

482

512

243

574

591

629

667

704

743

147

152

166

175

185

195

200

212

224

236

248

436.6

608

644

680

702

748

96

838

885

210

221

231

239

253

267

282

296

255

274

305

321

359

386

96

103

113

118

130

138

205

220

229

247

261

76

81

83

89

93

79

4250 690 229 781 260 931 311

Table 2 weight of single unit cooler (Kg)

Forced – oil forced – air cooler

Type of cooler Weight of oil (gey) Weight of cooler (ge)

Yf – 80/100

Yf – 100/380

YF – 120/380

122

130

136

160

1000

1160

Table 3 weight of conservators (kg)

Diameter

(mm)

Length (mm) Total weight of oil Weight of oil in

conservator

Weight of steel parts of

conservator

φ 180 400

500

600

90

115

137

5

6

7

9

10

11

φ 250

400

500

600

700

800

176

220

265

310

353

8

11

13

16

18

13

14

15

16

17

φ 310

600

700

800

900

1000

1100

405

475

542

610

678

745

20

24

27

30

34

37

65

68

71

74

77

80

φ 440

800

900

1000

1200

1400

1600

1100

1230

1370

1640

1920

2190

52

58

65

77

90

103

100

104

108

115

122

129

φ 610 900

1000

2370

2630

112

124

132

135

80

1200

1400

1600

1800

2000

2500

3150

3680

4200

4740

5260

6600

148

173

200

224

248

300

140

145

150

155

160

170

φ 760 2000

2500

3000

140

10150

12200

370

465

560

420

450

485

φ 920 2500

3000

3500

14900

17350

20850

675

810

945

625

675

725

φ 1080 3000

3500

4000

24700

28800

32950

1105

1285

1465

800

865

930

φ 1240 3500

4000

4500

38000

43500

48900

1720

1950

2200

1030

1100

1170

Calculation of tank weight

Weight of tank cover

For tank with flat cover,

Gg = 7.85 Ag δ g

Where

Gg – weight of tank cover, Kg;

Ag – area of tank cover, dm2.

δ g – Thickness of tank Cover, dm;

7.85 – specific weight of steel plate, kg/dm3.

81

Weight of tank bottom

Gd = 7.85 Ad δ d

Where

Gd – weight of tank bottom, kg;

Ad – area of tank bottom, dm2;

δ d – Thickness of tank bottom, dm.

Weight of tank wall

Gb = 7.85 Ib Hδ b

Where

Gb – weight of tank wall, kg;

Ib – girth of tank, dm;

H – Height of tank, dm;

δ b – Thickness of tank wall, dm.

Weight of oblate tube.

Gw = gw Iw

Where

Gw – weight of oblate tube, Kg;

Gw – weight of oblate tube per meter, Kg/m; Gw = 1.525 Kg/m;

Lw – overall length of tube, m.

Weight of panel type radiators

Gp = Np gp

Where:

Gp – weight of panel type radiators

Np – number of units of radiator;

Gp – weight of each unit of radiator, kg;

Weight of tank

Tube type tank

Gx = 1.15 (Gg + Gd + Gb + Gw)

Where

Gx – weight of tank, Kg:

Gg – weight of tank cover, Kg:

Gd – weight of tank bottom, Kg:

Gb – weight of tank wall, Kg:

82

Gw – weight of oblate tube, Kg.

Calculation of weight of accessories

Gf = Gs + Gt + Gz + Gxc + Gfs + Gj

Where

Gp – weight of accessories, kg.

Gs – Weight of radiator or coolers,

Gs = gs N – weight of each radiator or cooler, Kg: N – number of radiator or coolers:

Gt – weight bushings, Gt = Nt gt, Nt – number of HV, MV, and IV bushing, gl – weight of

each bushing:

Gz – weight of conservator, Kg;

Gxc – weight of bogie wheels, see table 5; for large transformer wheels are dismountable,

weight listed in the Table are unit weight, it should be multiple by the number of group

adopted.

Gfs – weight of fan motors for forced – air cooling,

Gfs = 50 Nfs – number of radiators with forced – air cooling, 50 – each radiator equipped

with 2 sets of fan motors, each weighing 25 Kg.

Gj – weight of oil – regenerators, Kg

Table 5 weight of wheels (Kg)

For small and medium size transformers, bogie being welded underneath the tank

bottom ( four wheels)

Load bearing (four wheels), lon weight of wheel ( for

wheels),

2.8

30

7.2

55

12

90

24

340

For large transformer, dismountable bogie wheels are adopted (each group of wheel)

Load bearing for each group, ton weight of wheels

(each group)

10

165

15

195

20

300

30

530

40

780

Calculation of total weight

G = Gqx + Gx + Gf + Gy

Where

G – total weight of transformer, Kg;

Gqx – weight of active part, Kg:

Gx – weight of tank, Kg;

Gf – weight of accessories, Kg;

83

Gy – weight of oil, Kg;

Weight of accessories dismounted

Gcf = Gp + Gz + Gj

Where: Gcf – weight of accessories dismounted, Kg;

Gp – weight of radiators or coolers, Kg:

Gz – weight of conservator, Kg;

Gj – weight of oil – regenerators, Kg;

Weight of transformer for delivery

Gu = G – Gty – Gf

Where: Gu – weight of transformer for delivery, Kg:

G – total weight of transformer, Kg;

Gty – weight of oil to be delivered separately, Kg:

Gf – weight of accessories, Kg.

84

Chapter 4

DESIGN CALCULATIONS OF 20/26 MVA, 132/11.5 KV

POWER TRANSFORMER

Power Transformer

Connection Symbol = DYn 11 Impedance Voltage = 10%

Winding Temp Rise = Top Oil Temp Rise =

4.1 Current calculation

i. KVA= 26000 KVA = 145.2

IL(H.V) = 103.3823 Amp

Ip (H.V) = 59.68779

ii. KVA = 20000 KV = 145.2 IͅL(H.V) = 79.52483 Amp

Ip (H.V) = 45.91360

85

iii. KVA = 26000 KV = 132 IͅL(H.V) 113.7205 Amp

Ip (H.V) = 65.65657 amp

iv. KVA = 20000 KV = 132 = 37.4773 amp

Ip (H.V) = 50.50505 amp

v. KVA= 26000 KV = 118.8 IͅL(H.V) 126.3561amp

Ip (H.V) = 72.95174amp

vi. KVA = 2000 KV = 118.8 IͅL(H.V) = 97.19702amp

Ip (H.V) 56.11672 amp

vii. KVA = 26000 KV= 115 IͅL(L.V) = Ip (L.V) = = 1305.314amp

viii. KVA = 20000 KV= 11.5 IͅL(L.V) = Ip (L.V) = = 1004.087amp

4.2 Core Calculation

First we calculate core diameter

D = K

P = Power = 26000 KVA

K = Experimental co-officient = 53

D = Diameter = 53 = 411.374

From Table D = 510/522

Here core Dia = 510

86

Inner dia of cylinder = 522

Area = 1796.45

B (T) → flux density = 163

Op = B x A

= 163 x 1796.45

=29.28

Thickness of lamination = 0.3 mm

Lamination factor = 0.95

Volt / Turn

B (T) = 1.63 Area = 1796.45

Volt/turn ELT

=

= 65.0714 volt/turn

4.3 L.V Winding calculations

KVA = 20000 I (amp) = 1004.87

KVA = 26000 I (amp) = 1305.314

It is Y connected

K.V = 115.5

Now to calculate num of turns

=

WD → num of turns (take integer)

U × g → Phase voltage (volts)

87

=

WD = 102 → take integer

NOW

E/T = =

E/T = 65.0934

CONDUCTOR SIZE

Width Height

Bare 2 11

ZB 0.45 0.45

With installation 2.45 11.45

Cross- section area ( ) = 21.64 519.36

Num of conductors in parallel = 24

A/ 1.933317 cross – section area ( ) 519.36

A/ 2.513312 cross – section area ( ) 519.36

Width (mm) = 24 (2.451.05) × 1.03

= 61.8

= 62 mm

Height (mm) = 103× 11.45 = 1119.35

= (102×3) × 0.92 = 281.52

Total = 1460.87

Reactance Height = 1460.87 – 4.45 – 3 = 1446.42 mm

Actual Height (mm) = 1460

88

4.4 H.V winding calculations

No of turns = = 2027.855

= 2028

It is D connected so line and phase voltage are equal

Conductor size bare 2.5 10.2

ZB 1.35 1.35 1.35

With installation 3.85 11.55

Cross – section area = 24.95

A/ = = 2.0242

A/ = = 2.6315

Disc No disc type turn/disc

44 A 22-14/16

32 B 22–15/16

8 C 23-15/16

4 D 24

Width = (3.85+0.05) × 24 × 1.04 = 97.334

= 97.5

Height = 11.55 × 88 = 1016.4

(5×87) × 0.92 = 400.2

Height = 1410

Total height = 1410+50 ← electric ring = 1460

89

4.5 R.V winding calculations

KV = 1015.385 turn = 15.59889 = 16

V/T = 65.0934

Conductor size bare 2.36 11.8

zb 1.95 1.95 1.95

4.31 13.75

A/ = 2.055708

A/ = 2.672421

Four row helical type winding

Width = (4.31+0.5)×4×1.03 17.9632

4×17×13.6 = 924.8

4×17×(5×92) = 312.8

Total = 1240

Actual Height = 1240

Total Height = 1240+52 = 1292

Impedance voltage (insulation distances)

Radius of core 255

6

Inner radius of L.V winding

Outer radius of L.V

Distance B/W H.V and L.V

Inner radius of H.V

90

Outer of H.V

Inner radius of R.V

Outer radius of R.V

Outer of dia meter Distance B/W phases

R1 means radius of L.V winding = 275 + = 306

R12 means radius of oil duct = 337 + = 362

R2 means radius of H.V winding = 387 + = 435.75

R23 = 484 + = 509.5

R23 = 534.5 + = 543.5

Now radial dimension of winding A = 62.005

= 61.5 =6.15

Width of oil duct between winding 1 and 2 = 50.95

Radial dimension of winding 2 = 97.5 – 1.4 = 96.1 = 9.61

ED = (A1×R1+A2×R2) + (A12×R12)

ED = (6.15×30.6+9.61×43.575) + (5.095×36.2)

= 202.31525 + (5.095×36.2) = 386.7543 cm

Width of lea kaage flux = 48405-275-7.25 = 20.855 cm

H = = 1428.21 = 142.821 cm

30.2

36.2

4305

50.95

54.35

91

Rogovski Co-efficient + = = 6.848286 = .955(table)

Now

UK =

= = 10.09097

UK = = 13.11826

4.6 Calculation of Load Losses “PK”

L.V winding data:

Inner dia (mm) = 550

Outer dia (mm) = 674

Mean length / turn (m) = = Ip

= 1923.428 = 1.923429

Total length (m) L = Ip Wn

Ip = mean length / turn

Wn = over all num of turns

Conductor for winding terminals have to be taken into considerations during calculation

of conductor length usually adding 1.5-2m.

L = 1.923429 × 102 = 196.189758 = 197.6898

Resistance (0 hm) R=

92

resistivity of copper at 75 C = 0.02096 0 hm /m

R= = .007978

Weight of bare conductor (Kg) G:x = 3 L.A.g ×

G = specific weight of conductor

For copper G = 8.9 g/

Gx = 3×(197.6898)×(519.36)×(8.9) ×

Gx = 2741.347 (Kg)

Weight of insulated conductor Gc= Gx (1+c%)

Here c% = percentage weight of paper to weight of bare conductor. It may be calculated

ad follows

C % =

£ → insulation thickness on each side of conductor, it equal ½ thickness of turn

insulation

A → thickness of bare conductor (mm)

B → width of bare conductor (mm)

Ax → cross –sectional area of single conductor ( )

=

=

=2.4

Weight of insulated conductor = 2806.05 (kg).

93

Resistive loss at ONAN (W) = R =24579.77

Resistive loss at ONAF (W)= R =41539.81

Eddy current loss in % = = %

K co-efficient depended on temp for copper K = 3.8 at 75 C

F = frequency = 50 H2

M = num of conductor perpendicular to flux

N = number of bare conductor parallel to flux

A = width of bare conductor

Ax = cross – sectional area ( )

Eddy loss in % = = (

= (13413697.35)

=

Eddy loss in % = 5.097205

Eddy current loss at

ONAN = 1252.881

Eddy loss current loss at

ONAF = 2117.37

H.V Winding Data

Inner Dia (mm) = 774

Outer Dia (mm) = 969

Mean length / turn = 2739 2.739

Total length L = Ip × Wm

94

= 5555.296

R = = = 4.75373

Weight of bare conductor = Gx = 3 Lag ×

= 3 (5555.296)×(42.95)×8.9×

= 3800.744 kg

Weight if in sulated conductor Gc = G × (1+c%)

= 3944.308 kg

Resistive loss at ONAN (w) = = 36376.88

Resistive loss at ONAF (w) = = 61476.92

Resistive loss at ONAN (-10%) =

Resistive loss at ONAF (-10%) =

Eddy current loss in% = (

= 7.646 %

Eddy current loss at ONAN (w) = 2781.031

Eddy current loss at ONAF (w) = 4699.943

(-10%) Eddy current loss at ONAN (w) = 3433.372

(-10%) Eddy current loss at ONAF (w) = 5802.398

(+10%) Eddy current loss at ONAN (w) = 2298.373

(+10%) Eddy current loss at ONAF (w) = 3884.25

4.7 Stray Losses

Tank length (mm) = 4850

Tank width (mm) = 2180

95

Tank height (mm) = 1740 + 1000 + 50 + 190 = 2980

↑ ↑ ↑ ↑

Window width of foot pad lifting

Circumference of tank = 2 (L + B)

= 2 (4108.52 + 1030)

= 10237.06 mm

= 1023.706 cm

= × × )

UK= % in pendance at rated power = 10.09097 13.11826

HK = reactance at height if winding (mm) = 142.821

F = 50

Rp = mean radis if leakage flux path = 36.2

Rs = mean equivalent radius of tank

Rs =

A – length of tank = 3860

B – width of tank = 1630

Mo = distance b/w core legs = 1175

=

= 732.5 mm

Rs = 78 cm

Cpo = main flux = B×A = 29.28214

B = flux density

96

A + = cross – sectional area of core

Kz ⇾ stray loss co-efficient

Kz = ×

For 20 MVA For 26 MVA

For 10 % impedance For 10 % impedance

Kz = 2.19 Kz = 2.19

Ss ⇾ circumference = 1023.706

For 20 MVA

P2 =

=

=

= 10565.495 Watt

For 26 MVA

P2 =

=

=

= 17857.88 watt

Stray losses at 26 MVA = 17857.88 watt

97

4.8 No –load losses

Winding height = 1460

Lower = 140

Upper = 60

Press = 60

Gap = 20

Window height = 1740

For three phase core with three legs

= 3r Ho A for core

= 4r Mo A + Go for yoke

GF = +

⇾ Weight of core legs

⇾ weight of yoke

GF ⇾ overall weight

Ho ⇾ height of window (mm) = 1740

Mo ⇾ centre distance between = 1175

A ⇾ cross – sectional area of core = 1796.45

Go ⇾ weight of silicon steel for cold rolled grain oriented

R = 7.65

Now = 3 7.65 1740 1796.45 ×

= 7173.76 Kg

= 4×7.65×1175×1796.45× + 1462.9

= 7922.03598 Kg

GF =

98

GF = 15095.8 Kg

Now

No load losses Po = k1 p GF

= co-efficient for additional core loss = 1.2

P = specific core loss = 1.027 w/Kg

Po = 102×1.027×15095.8

Po = 18604.06 (watt)

99

Total Losses (W)

4.9 No Load Current

íoa = % of Mo load loss

íoa = % se patrol Power

íoa = 0.071554

íor =

GF = weight of core = 15095.8

26 MVA 26 MVA 20 MVA 20 MVA 20 MVA

(14 Tap) (-10 %) (14 Tap) (-10 %) (1 Tap)

H.V Resistive loss 61476.92 75897.43 36376.88 44909.72 30063.53

H.V. Resistive loss 4699.993 5802.398 2781.031 3433.372 2298.37

L.V. Resistive loss 41539.81 41539.81 24579.7 24579.77 24579.77

L.V. Resistive loss 2117.37 2117.37 1252.881 1252.881 1252.881

Lead loss 1250 1250 1250 1250 1250

Stray loss 17858.86 17858.86 17858.86 17858.86 17858.86

Total load losses 128942.9 255565.9 76807.94 85993.12 70011.94

No load losses 18604.06 18604.06 18604.06 18604.06 18604.06

Total loses 147547 163069.9 95412 10745972 88616

100

From Tables

Ge = 1.34

Gi = 3.01 A = 17.9645

C = 8

íoR =

íoR =

íoR = 0.317434

I o =

I o =

I o = 0.32539 theoretical Value

Io =

Io =

Io = 0.140043

4.10 Regulation of Transformer

Ur = % = loaded losses

Se = Saled Power

= × Pr

Ur (%) = 0.495934

101

= 10.09097

= ?

=

=

=

= 10.07877

Given

K ( A + full load = 1

K ( A half load = ½

Cos = 0.8

Sin = 06

A+ full load

= K ( ) + (Ur cos cf – Ur sin cf

= 1 ( ) +10.07877 (0.6)+ (10.07877 (0.8) – 0.495934

(0.6)

= 6.44001+0.301511

= 6.745524 At full load.

= 6.444001

= 0.301512

102

At half load.

= 0.5 ( ) +10.07877 (0.6)+ (10.07877 (0.8) – 0.495934

(0.6)

= 3.22206 + 0.150756

= 3.372762

4.11 EFFICENCY

K =

= No-load losess P = Power

= load losses

K =

K = 0.379844

Cos = 0.8

Eff =

Eff =

Eff =

Eff = = 0.9939

103

Efficiency at full load = 0.9939

Efficiency at full load (%) = 99.39

4.12 Calculation of winding temperature

L.V Winding

Q = specific head loss fort winding surface. (w)

Q = ( )

= co-efficient selated to temperature of manual taree 22.1 for copper at C

T= current passing through winding for 26 MVA = 1305.314 A

T = current through winding for20 MVA = 1004.87

S = current density = 1305.314/519.36 = 2.513312

S = current density = 1004/519.36 = 1.93317

W = no of turns for disk = 1

= coverage co-efficient of winding disk.

= 1-

= 1-

= 0.750446

= percentage overall additional losses in conductor

= × 100

104

=

1× = 1.050972

= co section factor for turn insulation = 1

L = outer girth of disk crises section

L = 2 ( )

M = no of passable conductor = 24

= thickness of conductor with insulation = 2.45

= width of conductor with insulation = 11.45

L = 2 (24 (2.45) + 11.45)

L = 140.5

At ONAN

Q =

Q = 427.5946

At ONAF

Q =

Q = 722.68052 w/

T×1= temperature between winding and oil

105

At ONAN

T×1= 0.41q 0.6

T×1 = 0.41 (427.5946)0.6

T× 1= 15.53894

At ONAF

T × 1 = 0.159 (Q) 0.7

T× 1 = 15.93577

T+J = insulation correction of winding temperature –rise

T+ j= Kj Q

Kj = co-efficient of insulation value from table.

Kj = 4×

At ONAN

T j = 4× ×427.6244

T j = 0.1717049

TJ (At ONAF) = 4× ×722.6852

TJ (At ONAF) = 0.289074

Correction of temperature rise for oil.

Try = Pq / 1500

P = additional co-efficient sec value from Graph.

P = 2

T&y At (ONAN) =

T&y (At ONAN) = 0.57016

106

T&y (At ONAF) =

T&y (At ONAF) = 0.96358

Average temperature difference between winding and oil =

= + +

For ONAN

= 15.53894+0.17105+0.57016

= 16.28015

For ONAF

= 15.94636+0.289074+0.96358

= 17.19901

H.V Winding

Q = ( )

= 22.1 = 1

= 1- ( ) I = 50.50505

I = 65.65657

= 0.706338 J = 2.249167

W = 24 J = 2.923917

= ×100

107

= 7.645 1+ = 1.076451

L = 2 ( + )

L = 207.9

At ONAN

Q =

Q = 452.3106

At ONAF

Q =

Q = 769.4049

At ONAN

=

= 0.41

= 16.07111

At ONAF

= K j q

Kj = 9 × From table

At ONAN

= 9 × × 452.3106

=0.40708

108

At ONAF

= 9 × × 764.4049

= 0.681964

=

P = 4 From Graph

At ONAN

=

= 1.206162

At ONAF

= = 2.038413

At ONAN

= + +

= 16.07111+0.40708+1.206162

= 17.68435

At ONAF

= 16.58531+0.687964+2.038413

= 19.31169

Calculation of Temperature Rise of Top oil

Ts = 1.2 Ty +Ta

Ts = total oil temperature rise ok

109

Ty = average temperature of oil

Td = value of correction for oil temperature rise

Ty = 0.262 Q + 0.8 A+ ONAN

Ty = 0.191 Q + 0.8 A+ ONAF

Now Q+ = Po + Px/A

Po = no load losses

Px = load losses

A = overall dissipating area of tank

(cover area) = 4.85 = 4.85×1.63-0. - +1.465×.55-.55×1/2

= 8.38

(wall area) = (4.25+ ×0.3×4+3.085+ +0.55+0.915+1.58) ×2.93

= 39.73792

(Radiator dissipating area at ONAN)= 8×28.2824

= 226.2592

(Radiator dissipating area at ONAN)= 8×30.6724

= 245.3792

At ONAN

A = + +

A = 8.38+39.73792+226.2592

A = 274.3771

110

At ONAF

A = + +

A = 8.38+39.73792+245.3792

A = 295.903

At ONAN

Q = Q = 391.2461 w/

At ONAF

Q = Q = 566.7156 w/

No of radiator = 8

Specified by table

A = 2250

Dissipating area of radiator At ONAF = 30.6724

At ONAN

= 0.262

= 0.262×

= 31.06428

At ONAF

= 0.262 - 7

111

= 0.262× -7

= 23.45969

(Center of head generator) = 1345

(Center of head dissipating) = 1695

=

=

=

Ta = 6 for both OA ONAN and ONAF for value of TA see graph between Ty and A

At ONAN

Ts = 1.2 Ty + Td

Ts = 1.2×31.06428+6

Ts = 43.27713

At ONAF

Ts = 1.2 Ty + Td

Ts = 1.2 ×23.45969+6

Ts = 34.15163

112

Winding Temperature in L.V

DGC (ONAN) =

= 16.28015+43.27713

T = 47.34443

At ONAF

T = 17.9901+23.45969

T = 40.6587

4.13 Short Circuit Current for H.V Winding

Z =

Zs=

Us = rated voltage of System = 145

S = Short circuit appaunt Power of = 1×

System in MAV

Zs =

Zs = 2.1025 ×

Zt =

= Impedance Voltage at rated current = 10.00839

= Rated voltage of winding = 132

Sn = Rated Power of Transformer = 20

Zt =

113

Zt =

Zt = 87.19309

I =

Z =

U = 132

I (Short circuit current) = 8.74×

X = 9.991412

I = 8.740187 Amps

For L.V Winding

I = K Amps

And Zs = Short circuit impedance of system

Zs =

Us = Rated voltage of system = 12

S = short circuit apparent Power of system = 5 × In MAV

Zs = =

Zs = 2.88 ×

Zt = short circuit impedance of transformer

Zt =

Sn = rated power of transformer MAV = 20

114

= Impedance voltage all rated current = 10.00839

Un = Rated voltage of winding = 11.5

Zt =

Zt =

Zt = 0.661805

I (short circuit current) =

I =

I = 10.02809 K Amps

As Per IEC Design Manual Short Current

K =

Zs =

4.14 Highest Average temperature rise of Winding Under

Short Circuit Condition

= + 2 ( ) /

= 105 due to class A

T is the short circuit current density = 3.164934×K

T = 3.22×

115

For 2 Sec

= 105+2(105+235)/ ((101000 2) -1

= 105 +

= 105+114.25906

= 1.19×

For 3 Sec

= 1.27×

4.15 Balance of Amp Rise Turns

H.V Side L.V side

We made three zones.

Zone 1 Zone 2

22A = 503.25 turns 25 turns

22×11.36 = 252.12 (Bit now 11.55 25.5×11.45=291.975 Turns height

22×5 = 110 oil Ducts Height 25×3= 75 ducts

Total Height = 362.12 Total height = 366.975

110×0.92 = 101.2 = 75×0.92 = 69

Oil compression = 110-101.2= 8.8 Oil compression = 75-0.92 = 6.926

Actual heights = 353.32 Actual heights = 360.975

Zone 2 Zone 2

32 B+8c+4d 1021.5 turns 52 turns

44+11.26 504.24 Rise height 52×11.45 545.4 turns heights

116

43×5 = 215 52×3 = 156 oil dacts height

Total heights = 719.24 Total heights = 751.4

215×0.92 = 197.8 156×0.92 = 143.52

Oil compression = 215-197.8 = 17.2 Oil compression = 156-143.52 = 12.48

Actual heights = 702.04 Actual heights=

Zone 3 H.V side Zone 3 L.V side

22 A = 503.25 turns 25 turns

11.26×22 = 252.12 disc Height 25.5×11.45 = 291.97

22×5= 110 oil duct height 25 3=95 oil duct

Total height = 362.12 Total height = 366.975

Oil compression 110×0.92 = 101.2 Oil compression = 6

Actual height 110-101.2 = 8.8 Actual height = 360.975

H.V (Amp-

Turn)

L.V (Amp-

Turn) Unbalance Balance Avg Height

×100

×100

H.V – L.V

24.8151 24.5098 0.30529 24.6624 357.1475

50.36982 50.98038 -0.61056 50.6751 720.48

24.81509 24.5098 0.30529 24.66844 357.1475

100 total 100 total 1.9× =0 100 1434.775

117

4.16 Axial Mechanical Stress under Short Circuit Condition

S = S+0.03×D

S = Distance between core and inner winding = 20

D = core diameters =

S= 20+0.03×510

S = 35.3 3.53 Cm

N = 143.4775 (take from prevision Page)

V = S/N =

V = 0.025

L = 20.255 U = L/n

U = 0.141172

From Phase = Pn (UV) = 0.5

∑x = ( + ( ))

= 0.30529(24.66255+(50.6749× ))

∑x = 15.26448

Arial Mechanical Stress Under Short Circuit Condition

=

=

=18547.7 New turn

Xy = Co-efficient for short circuit impairs = 1.8

118

K1 = multiple of steady short circuit current = 11

I = Phase current if short circuit tapping for L.V

I = 1305.314

For H.V

I = 65.65657

W = no of turns

(H.V) W = 2028 (L.V) W = 102

Rp = mean radius of winding

1= H.V 2 = L.V

= =

1 = - B

N = no of radial spares = 16

D = diameter of winding

B = width of spares

Ip = – d

M = no of passable conductor in disk

N = passable blanch per phase

= reactance height of winding = 142.821

= cross sectional area of single conduction

A = radial dimention of base conductor

B = radial dimention of base conductor

K = overall width of leakage flour

Ky = 1-8 P = 0.955 = 142.821 W = 2028

119

K1 = 11 Rp1 = 43.575 = 65.65657 ∑x = 15.2644

= 0.2495 L= 20.855 Pn = 0.5

= 0.25 = 1.02 I= 1.05314 W= 102

= 0.2 = 1.1 = 0.95 = 0.05

= 30.475

(L.V) MN = 24 (H.V) MN = 1

= 0.2164

=

=

= 512.3717 KG/

Where

Ky = 1.8 P = 0.955 = 142.821 W = 2018

= 11 = 93.575 I = 65.65657 ∑x = 15.2644

=

L = – 6

L = – 5

L= 19.026-5

L= 14.02627

120

=

=325.0127

=

= 2× - 4

= 9.2339

=

= = =8.944709

=

=

= = 209.594

4.17 Calculation of weight

Weight of active part = (Gfe + Gfu) 1.15

Gfe ⇾ weight of silicon steel sheet

Gfu ⇾ weight of copper conductor with insulation.

121

= (15095.8+7377.2063)1.15= 25843.96 Kg

Weight of Oil

A ⇾ weight of oil display by active part

= +

= +

= 1986.2895+1891.591

Gpy = 3877.881 Kg

Weight of oil in empty tank

Gky = 0.895×H×Ad

H = thank height ⇾ dm

Ad ⇾ cross – sectional area of tank

L ⇾ length of tank

R = Radius of round corner

B = width of tank

GKY = 0.895 × 838×29.3

= 22350.3 Kg

Weight of oil in tank = Gky – Gpy

= 18472.42 Kg

Weight of oil in turrets = 600 Kg

Weight of oil in conservator = 900×3350 = 950 From table

Volume of oil per meter of of pipe

V = 0.814153 Kg/ Meter

Total length of pipe / radiator = 318.98 m

122

Weight of oil /radiator = 259.6986 Kg

Total weight of oil/radiator= 275.686 Kg

Number of radiators = 8

Weight of oil in radiators = 2205.589 Kg

Total weight of oil = 22228.01 Kg

Weight Of Tank:

Weight of Tank cover

Gg = 7.85 Ag Th

Th → Thickness of Tank cover (Dm)

Ag →Area of Tank cover (Dm )

7.85→ Specific weight of steel plate

Gg = 7.85(0.16)(1027.675)

= 1290.76

Weight Of Tank Bottom:

Gd = 7.85 Ab Th

Ab →Area of tank bottom

Th →Thickness of bottom

Gd= 7.85 (1027.675)(0.1)

= 806.7249

Weight of tank wall:

Gw= 7.85 Iw Hw Th

Th →thickness=0.1 dm

Hw →height = 29.8 dm

Iw →Girth = 133.34874 dm

Gw = 3119.427 Kg

123

Weight of Rim:

Gr = 7.85 Ir Wr Th

Th → 0.25dm

Wr →1.35dm

Gr = 353.2907 Kg

124

Chapter 5

PROBLEMS AND FAILURES IN POWER TRANSFORMERS

This chapter deals with the problems occurring in power transformers. We have included

three types of problems. First all the procurement problems faced by Wapda. And then

there are problems and failures arising in the magnetic circuit, windings, insulation and

due to structural defects. And then finally we have problems during the actual operation

of the power transformers.

5.1 PROCUREMENT PROBLEMS OF POWER

TRANSFORMERS

Power Transformers are mostly supplied by foreign manufacturers but it is quite

encouraging to note that now three Pakistani firms i.e. SIEMENS, HEC & PEL have also

started manufacturing power transformers of different ratings including 132/11 KV 20/26

MVA transformers.

Power transformers are normally procured through international competitive bidding

(ICB). As the said transformers are very expensive therefore these are mostly procured

against World Bank and ADB Loans. Presently procurement of power transformers is

carried out by NTDC WAPDA and all Distribution Companies (DISCOs) independently.

Procurement is very specialized job and this is the reason that various problems arise

during the procurement process. Some of the main problems are listed below:

1. Tendering process comprises of preparation of tenders, floating of tenders in news

papers, Tender opening, Tender Evaluation, Tender approval by the competent

authority and finally award of purchase order/contract agreement. Due to delicate

tendering process, some times tenders are not timely finalized causing delay in the

125

award of purchase order / contract agreement which ultimately delays the supply of

transformers / material.

2. Due to preoccupation of important international manufacturer, some times tender

participation by the bidders becomes very poor and healthy competition is not held.

This leads to higher quoted prices and delays the procurement process.

3. It also happen that during the period from “Tender opening date to tender finalization

date” prices of transformers constituents i.e. copper, transformer oil and steel etc

increase abruptly and bidders show their inability to accept the PO/Contract at their

quoted prices and demand escalation in quoted prices. This factor also delays the

procurement process considerably.

4. Some times bidders do not quote their prices according to the requirement / clauses of

tender documents due to which it becomes difficult for the procurement agencies to

finalize the tender.

5. Some times offer submitted by the bidders is responsive both technically and

commercially but offered delivery period is more than that of bidding document,

which also creates lot of problems and causes delay in award of PO/Contract.

6. During sale period of tender documents, amendments to the bidding documents are

issued late by the purchaser due to which bidders can not prepare their bids properly

and participation response becomes poor. This does not bring the healthy competition.

7. Some times due to lack of funds with the purchaser, tenders for procurement for

power transformers are scrapped and procurement process is delayed.

8. It also some times happens that purchaser changes the specifications during the

tendering process and does not give sufficient time to bidders to prepare their bids

according to revised specification. This causes less participation of bidders and

ultimately higher prices are quoted by the bidders

9. Power transformers are very expensive and mostly procured against World Bank and

ADB Loans. It has been often observed that due to late approval of donor banks

procurement process gets delayed.

10. Although power transformers are procured through international tenders but still there

are chances that the bidders mix up with each other and quote higher prices through

pool system. In such cases it also becomes very difficult for purchaser to process the

tenders.

126

11. Late opening of letter of credit (LC) by the purchaser also delays the procurement

process.

12. There are lots of accessories which are to be supplied with power transformers but

some time due to certain transportation problems, power transformers are delivered in

store with fewer accessories which are detected through survey process. This also

creates complications and hampers the procurement process.

13. Some times tenders are not process/finalized by the purchased during the validity of

the bid (120 days) and ultimately tenders are to be scrapped.

After explaining the procurement problems we move on the next section. Which are

the problems and failures occurring inside the transformer.

5.2 PROBLEMS IN MAGNETIC CIRCUIT

1. Cases have occurred in core-type transformers of breakdown around the bolts

inserted cores and yokes for the purpose of clamping the laminations. This type of

fault has the effect of causing local short-circuits between the laminations which

produce intense local eddy currents, if in addition two or more of the core bolts

break down, heavy currents are liable to circulate in these bolts as they form a

short-circuited turn through which flux passes. This failure is most serious when

one of the core bolts situated at the ends of the limb breaks down together with an

adjacent bolt in the yokes as the path between the two bolts carriers almost the full

value of flux passing from the core to yoke, or vice versa. The path formed by the

bolts and the outer clamping plates if of low impedance and the amount of heat

generated by failures of this type is sometimes sufficient to distort the whole core.

The heat generated may also char the coil insulation and cause a short circuit

between turns of adjacent winding.

It is now common practice to clamp the limb laminations of large power

transformers with insulated bands. This method of construction eliminates limb bolt

failures.

2. Unless special precautions are taken to lock effectively the core-clamping bolts

and the bolts tying together the core structure, vibration will be setup which will

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tend to weaken the core insulation and produce failures similar to those outlined

above.

3. During manufacture the edges of the core and yoke laminations may have become

burred, due to the continued use of worn tools. Unless proper supervision is given

to the cutting and punching processes the resulting burrs will produce local short

circuits in the iron laminations and eddy currents with consequent abnormal

heating will occur.

4. It is important to note that no metallic fillings or small turnings are present

between the laminations as these are also liable to produce intense local eddy

currents and excessive local heating of the core in the finished transformer.

5. High flux-density in the magnetic circuit often results in larger magnetizing

current “in-rushes” occurring when switching a transformer into circuit on no

load. While this current in-rush generally dies down rapidly large electromagnetic

forces are created while the heavy current lasts and the windings are there by

strained. The phenomenon becomes more severe the nearer the transformer is

located to the generating source, in repeated switching-in may ultimately cause

movement of the windings. The effects of switching-in are dealt with more fully

in the sections on switching and transient phenomena.

6. If the voltage applied to the transformer must be increased appreciably on account

of the needs of the system load, the frequency must also be increased in order to

avoid high magnetic saturation of the core. An increase of voltage must not be

accompanied by a decrease of frequency, as if it is, core saturation may occur with

resultant increase iron loss and abnormal core heating. The inter dependence of

voltage, frequency and flux will easily be seen from the following formula.

B m =(E × )/( 4 610 fK × A × f ×N)

Where

B m =maximum flux density in the core in teslas;

E =applied voltage of the winding considered;

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=form factor the e.m.f wave; fK

A =cross-sectional area of core in mm(square);

N = number of turns in the winding considered;

F = frequency, Hz;

7. In older transformers ageing of the core plates may be found to have taken place.

This is due to a deterioration of the material of the laminations and results in an

increase iron loss and rise in temperature of the transformer. This may eventually

lead to a partial or complete destruction of the coil insulation and to sludging of

oil.

5.3 PROBLEMS IN WINDING

1) A short circuit between adjacent turns of a coil-usually high-voltage winding- may

be caused by the presence of sharp edges on the copper conductors. If the

transformer vibrates when on load, or if the windings are subjected to repeated

electromagnetic shocks through short-circuit or switching –in, these sharp edges

will cut through the insulation and allow adjacent turns to make metallic contact.

2) A short circuit between turns may result from dislodging of one or more turns of a

coil caused by a heavy external short-circuit across the windings. Breakdown may

not occur immediately the turns are displaced, but should the transformer vibrate

while on load due to looseness of core bolts, or should it receive heavy

electromagnetic shocks, abrasion of the insulation between adjacent dislodged

turns may produce a breakdown.

3) Transformers on large systems are now usually fitted with adjustable coil

clamping for the purpose of taking up shrinkage of the insulation which may

occur under service. Unless this adjustment of coil supports is performed very

carefully by an experienced workman, so that the correct pressure is applied to the

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windings, some of the conductors may become dislodged, and a short-circuit

between turns may occur as stated above.

4) Short-circuits between turns are almost bound to occur sooner or later should the

moisture penetrate the insulation of the coils. Breakdown from this cause is

rendered more imminent still if the coils have been insufficiently impregnated.

5) Drying out a transformer on site may be undertaken by an engineer not fully

conversant with the operation, and due to inexperience the process may be unduly

shortened. If normal voltage or a test voltage is applied while the insulation

resistance of the windings is still low, the insulation between adjacent turns is

liable to fail due to the presence of moisture vapour.

6) If a transformer is subjected to more or less rapidly fluctuating loads, the

expansion and contraction of the winding conductors alternately increases and

decreases the mechanical pressure on the insulation between turns. As the

dielectric strength of most insulation decreases with increasing mechanical

pressure, the windings become more susceptible to failure should they be

subjected to electrical or magnetic shocks.

7) Badly made joints between coils may overheat on load, and local carbonization of

the coil may occur. The heat generated at the joint will probably be transmitted to

a length of the conductor of each coil, and this may partially carbonize the

insulation round the conductors and eventually result in a short-circuit between

turns.

8) Short-circuits between turns, breakdown of windings to earth, and puncture of

insulators may take place due to the following transient phenomena:

a. Concentration of voltage on the terminal end coils when switching in or

when lightning surges reach the transformer. Owing to the change of surge

impedance at the transition point between transformer and line, the

phenomenon of reflected and transmitted voltage and current wave occurs,

which may produce high voltage rises in the transformer windings.

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b. The excessive voltages set up by surges may be accentuated at open-ended

tappings, at any point of surge impedance in windings, for instance, at the

termination of conductor reinforcement where employed, at the space

between series coils, and at the neutral or mid-point. Care should be taken

to insulate the conductors in these regions very thoroughly, in order to

eliminate so far as possible the short circuits between turns.

c. When switching out of circuit an inductive winding, such as a transformer

primary with the secondary open-circuited, the magnetizing current and

consequently the magnetic flux, tends to collapse instantaneously. While

for various reasons the flux does not do this, it does sometimes drop at a

much more rapid rate than that corresponding to the normal cyclic rate of

change, and as a result high voltage rises are sometimes produced. The

rapid cooling of the interrupting arc, particularly during the last half-cycle

has been found to augment this effect.

9) Sustained heavy overloads produce high temperatures throughout the transformer.

The coil insulation becomes brittle, and in time probably flakes off the conductor

in places, so producing the short-circuits between turns.

5.4 PROBLEMS IN THE INSULATION

1. Moisture entering the oil as a result of breathing action greatly reduces its

dielectric strength, so that breakdowns from coils or terminal leads to tank or core

structure may take place. The greatest danger, however, is to the inter-turn

insulation of the coils, as referred to previously.

2. Deterioration of the oil may occur as a result of prolonged overloading of the

transformer. Excessive accelerates the formation of sludge, water and acids.

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3. Dielectrics having different specific permittivities are often used in series, and

unless there thicknesses are correctly proportioned they may be subjected to

abnormally high dielectric stresses. For instance the insulation between high

voltage and low voltage transformer windings usually consists of paper, solid

insulation and oil. Except in very high voltage transformers, the effect of the paper

covering can be ignored, so that we only have to consider solid insulation with a

permittivity of 5, and oil with a permittivity of 2. The total voltage across two

such dielectrics in series divides up so that the voltages across equal thicknesses

of each are inversely proportional to the permittivity’s and unless the respective

thicknesses are proportional so that the voltage gradient across each is within the

safe working limit, first one and then the other dielectric will fail, due to corona

discharge and overheating.

4. Corona may take place from sharp conducting edges or small diameter conductors

if the surface voltage gradient is high.

5. Earth shields placed between primary and secondary winding have been

responsible for numerous breakdowns. Their presence produces a concentration of

dielectric stress at their edges which tends to stress insulation locally, while a

breakdown at one point from the high voltage windings to the shield has often

resulted in almost completely destroying the high voltage winding on the limb

concerned.

6. Narrow oil ducts reduce the serviceable life of a transformer. Adequate cooling

cannot be obtained, the oil insulation becomes brittle in time, and a fault between

turns follows.

7. Careless workmanship may result in metallic material on the coils and on surfaces

which are depended upon for creepage clearances. The possible effect of these

will be obvious.

8. During the course of time the oil level may fall. Unless the oil is topped up to the

correct working level, the transformer will overheat due to restriction of oil

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circulation. This applies particularly to transformers having tubular tanks, should

the oil flow the top of cooling tubes.

9. Short-circuits between phases may occur if there is insufficient clearance between

the phases. This may sometimes be aggravated by the insertion of pressboard

barrier between phases if the presence of the barrier upsets the distribution of

dielectric stress to such an extent as to put too high a stress across the oil spaces

and across the barrier.

5.5 PROBLEMS DUE TO VARIOUS STRUCTURAL

DEFECTS AND TO OTHER CAUSES

1. Transformer tanks sometimes give trouble on account of bad and porous

welding and leaky fittings, all of which may lead to oil leakage and

consequent overheating and breakdown of the transformer if not attended to

immediately. Rough handling in transit is also responsible for a large

proportion of the leaky tank troubles.

2. If a transformer fitted witch a protective gas relay is not filled with oil to the

correct level, maloperation of the relay will result. In the event of fault inside

the transformer tank this lack of protection might result in a major breakdown

of the transformer.

3. Deposits of ordinary dust, coal dust and salt spray on the surfaces of bushing

insulators often cause flashover to take place. The remedy is obvious.

4. Transformers operating in parallel should preferably possess the same turns

ratio, the same percentage impedances, and the same ratios of resistance to

reactance voltage drops. If any on these factors are different one transformer

may be overloaded and may be damaged.

5. When housing transformers care must be taken to provide sufficient space

around them to enable the tank to dissipate the losses. If a transformer is

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placed to near to another unit or to the walls of a chamber the tank will

become lagged and the temperature of the transformer will increase, so

endangering the coil insulation and the condition of oil. It is essential to

provide adequate ventilation for all transformers.

6. Puncture of high-voltage capacitor bushings has occurred owing to a

deterioration of the paper insulation consequent upon the allowance of too

high a dielectric stress per step.

Now we explain how the different protective devices trip owing to these

problems. This piece of information has been gathered by the S.E (Kot Lakhpat

Grid Station, Lahore).

1. Differential trip:

When a fault occurs in differential zone i.e. between C.T. on HV side and C.T.

on LV side differential relay will operate and differential zone will be isolated

on tripping of HV and LV side breakers. Equipment included in differential

zone is transformer, lightning arrestor. Fault may be inside/outside the

transformer.

2. Main Buchholz Trip:

When a fault occurs inside the transformer (main body) due to insulation

failure of winding/sparking, ionization of oil produce gases. These gases

escape towards Buchholz relay and conservator and close the buchholz relay

contacts for tripping.

3. Tap Changer Buchholz Trip:

When there is some fault in tap-changer (diverter-unit) compartment, Tap-

changer buchholz trips. Cause may be the low dielectric strength of oil, loose

connections, reduced value of resistors etc.

4. Oil Temperature Trip:

When oil temperature exceeds preset value of temperature gauge, trip signal is

passed to the related breaker and tripping occurs.

5. Winding Temperature Trip:

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When the winding temperature exceeds pre-set value on temperature gauge

due to overloading etc.

6. Overload Trip:

Tripping occurs on transformer due to over-loading (over-loading value is set

on C.T. ratio)

7. Pressure Relief Valve:

When a gas pressure of certain value develops in the main body of transformer

due to any fault i.e. short-circuit of winding, failure of major insulation or

excessive oil filling. Pressure relief valve operates and gives tripping signal.

8. Alarms/Warnings:

• Main Buchholz Alarm:

At low oil-level in main body of transformer, a first stage alarm warns

about the problem.

• Oil Temperature/Winding Temperature Alarm:

Normally oil and winding temperature is set on two stages, at first

stage alarm operates and at second stage tripping occurs. Let at C

alarm operates, at C tripping occurs.

o100o105

9. R.E.F Relay (Restricted Earth Fault):

When the fault is of low intensity and very close to the neutral, this relay

operates to save the transformer.

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5.5 PRICES OF POWER TRANSFORMERS Here we have performed a little survey regarding the priced of power transformers. Both national and international manufacturers have been added. STG NTDC PROCUREMENT OF POWER TRANSFORMERS (2006-2008)

NAME OF FIRM CONTRACT NO. & DATE

T/F RATING UNIT RATE QTY ON ORDER(NOS.)

ABB-01-2006 13/06/2007 Lot-1:

Auto transformers 525/ 23:3/231:3 KV, 600MVA(3x200 MVA) Auto transformer bank

US$5,348,160 1 1. M/s TBEA, Shenyang China

Lot-2: 512/ 11:3/220:3 KV, 450MVA(3x150MVA) Auto transformer bank

US$3,653,940 1

ADB-02-2007 13/07/2007

500 & 220KV Auto-Transformers

Lot-1

500 11:3/220:3/ KV 237MVA

US$ 946,000

Bank

Lot-2

525/ 11:3/220:3 KV 600MVA

US$ 1,684,000

Bank

2. M/s Xian Electric Engg:Co. China

Lot-3

500/ 22:3/220:3 KV 450MVA

US$ 1,320,000

Bank

3. China National Electric Wire & Cable

ADB-02-2007 Lot-4: Dated:17/07/2007

220:132:11KV,138MVA

US$ 1,684,000

Bank

4. M/s Jiangsu China

ADB-02-2007 13-07-2007

220KV Auto-Transformers

136

Lot-2 & Lot-3

220:132:11KV,160MVA

US$ 1,203,600

14 No’s

Lot-4

220:132:11KV,250MVA

1,890,000

05 No’s

5. M/s Jiangsu Hoping China

Tender WOR-18 dt: 11/11/2006 Lot-4

3-Phase Auto Transformers Complete with all accessories. 220/132KV, 125/160MVA

US$ 1,120,859

04 No’s

Power Transformers with extended creepage bushings.

6. M/s Iran Transfo IRAN

STG-6-21 Lot-2 dt: 11/09/2006

132/11.5 KV,20/26 MVA

US$ 3,75,000 19 No’s

Power Transformers with extended creepage bushings.

7. M/s Iran Transfo IRAN

STG-6-21 Lot-2 dt:11-09-2006

132/11.5 KV,20/26 MVA

US$ 3,69,713

25 No’s

8. M/s Siemens Pakistan

STG-6-21 Lot-3 & 4 dt:14/12/2006

Power Transformers with extended creepage bushings. 132/11.5 KV,20/26 MVA

Rs.29.8 Million

19 No’s

9. M/s Siemens and M/s HEC

STG-6-21 Dt:25/11/2006

Power Transformers with extended creepage bushings. 132/11.5 KV,20/26 MVA

Rs.22.0 Million

05 No’s 06 No’s

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LESCO PROCUREMENT STATUS OF POWER TRANSFORMERS (2006-2008) Sr no

Name of firm Contract/PO no. & Date

T/F Rating Unit Rate Qty on order (No’s)

1 M/s Iran Transformer IRAN

Contract dated 01-06-2006

20/26 MVA 132/11.5 KV

US$ 318,912/- 10

2 M/s HEC Hattar Contract dated 14-06-2006

20/26 MVA 132/11.5 KV

Rs.24,000,000/- 5

3 M/s HEC Hattar PO#2343 dated 09-01-2007

20/26 MVA 132/11.5 KV

Rs.29,800,000/- 10

4 M/s PEL LAHORE

PO#2552 dated 31-10-2007

20/26 MVA 132/11.5 KV

Rs.29,468,000/- 3

5 M/s Siemens Karachi

Under Award 20/26 MVA 132/11.5 KV

Rs.29,800,000/- 7

6 M/s PEL LAHORE

PO#2637 dated 21-04-2008

31.5/40MVA 132/11.5 KV

Rs.38,475,000/- 2

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CONCLUSION AND RECOMMENDATIONS

The Power Transformers are one of the most important components of the power system

which play a very important role in facilitating transfer of power to end users. The failure

rate of power transformers has been comparatively very high in some parts of the world, causing

not only immense loss by way of avoidable expenditure on its replacement by the utilities but

also resulting in inconvenience to the ultimate consumers and huge losses in production. The

failure analysts quote a host of reasons behind the failure of power transformers, which cover all

the aspects of deficiencies in design, manufacturing, material quality, abused operation, poor

maintenance practices, non-substandard techniques and consumables adopted/used during

manufacturing, shortcomings in testing and commissioning, inconsistency of the design

parameters with the practical field environment, abnormal operating conditions, etc. While the

manufacturers are blamed for supplying poor quality of transformers, the users are blamed for

lack of maintenance and not providing proper protection and frequent overloading of the

transformers etc. Some experts even point out towards the phenomenal decrease in the safety

margins in design of power transformers due to fierce competition thereby increasing chances of

failures during grid disturbance as well as voltage & frequency fluctuations during normal

working. The funds sometimes becoming scarce, the replacement of large number on

failed/burnt transformers is posing a serious problem thus compounding the gravity of the

situation. Unless this situation is properly and urgently tackled, it is felt that the continuity of

power supply to large sections of consumers would be threatened for comparatively longer

periods and more frequently, may lead to disastrous consequences. There may be multiple other

factors, dictated by present day realities, due to which the transformer failure rate continues to

be very high particularly in developing countries. However, considering the very high stakes

involved, all concerned viz. Designers, researchers, manufacturers, utility engineers etc. are

required to come together for working out an acceptable course of action in correcting the

situation.

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Transformer is a global product. In present day’s market, “reliability” is the key word. The

transformer being vital link, its reliability dictates the quality of the power supply. And the

construction of the transformer is a very complicated process. From calculations on paper we

make an actual transformer. The first role is played by the designer who calculates currents,

voltages, losses, stresses. It is the designer who determines the type of windings to be used.

The amount of oil being used. Then comes the role of construction where the engineers must

make sure that each and every part is assembled correctly. Even a small bulge in windings

can lead to the failure of transformer. After construction comes the role of testing engineers.

He has to make sure that the transformer passes each and every test. After the intensive tests

the transformer is ready for transportation.

Recommendations

Now we present some recommendations that may prevent the transformer from being

damaged and ensures the reliability of the transformer.

• Transformers should withstand the vibrations and shocks during transportation

from factory to installation sites. The hydraulic road trailers for transporting large

transformers should be used. Physical inspection of active part prior to installation

may be an important step to ensure that there are no damages during transit.

• During installation, utmost care is necessary to avoid moisture/dust entry inside

transformer. Dry air or nitrogen shall be fed into transformers for this purpose.

• Minimum duration for vacuuming period, hot oil circulation and settling time

shall be given as per the recommendations of manufacturer.

• During operations of the transformer, it needs to be ensured that the active part is

in dry condition.

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• Leak proof transformers can be manufactured by taking care of the specifications

/ selection of sealing gaskets, design of joints, appropriate leak tests at factory site

etc.

• Moisture content in transformer insulation needs to be controlled in order to

extend the life of transformers. Transformers are required to be adequately dried

out at factory and proper precautions are taken at site during installation so as to

ensure that moisture content in Transformer is below 0.5 % at the time of

commissioning.

• Life cycle cost involving procurement cost, operation & maintenance and

refurbishment costs over the useful life of transformers may be considered during

procurement of the transformers. Condition based maintenance is the best

technique for optimizing maintenance costs.

• Capacitance and Tan-delta, moisture content monitoring and recovery voltage

parameters are required to be regularly checked for prolonged life of the

transformers.

• Preventive maintenance and refurbishment of aged components along with

replacement of old oil with new oil would enhance the life of transformer.

• Copper sulphide in transformers/reactors is dangerous for performance. The oil

should either be diluted by adding passivators or by replacing the same to

minimize failure risk.

• Site drying of transformers may be carried out by using dry pure Nitrogen by

vacuuming and heating methods so that moisture is effectively eliminated.

• During drying/repair, all windings must be equally compressed otherwise ampere

turns unbalancing will increase failure risk.

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• Careful handling and testing of bushings at site will ensure better performance.

Capacitance and tan delta of bushings shall be regularly monitored for checking

variation in order to ensure better maintenance, if any.

• Proper Protection of transformers needs to be ensured particularly in outgoing

feeders to minimize failure during repeated line tripping faults. Extensive

insulation co-ordination and adequate protection with recommended settings

based on system faults must be provided.

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References

• Transformers (second edition) By Bharat Heavy Electricals Limited.

• The J and P Transformer Book.

• Transformer Design And Applications By William M. Flanagan.

• Modern Transformer Design Theory By D. C. Cox

• Principals Of Electronic Transformer Design By Alfred Still.

WEBSITES REFERENCES

• http://www.tpub.com

• http://www.physlink.com

• http://www.butlerwinding.com/core-types/

• http://www.bcae1.com/trnsfrmr.htm

• http://www.softbitonline.com