I
DEDICATION
========================================
To my loving and esteemed father Idrees Ahmed Butt and mother
Rakshanda Butt who have really enlightened my future by the dint of
their keen interest, devotion and heart felt prayers.
AUTHOR
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 copperloss in primary. Hence the no-load primary input current Io is not at 90 degree behind
V1 but lags by an angle θ 0 which is less than 90 degree. No-load primary input power
Wo= V1 Io cos θ 0 . 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 V1 . 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 θ 0
(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 θ 0
Obviously Io is the vector sum of Iw and Iu hence
Io= ( Iu 2 + Iw 2 )
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 I 2 is
set up. The magnitude of I 2 is determined bye the characteristic of the load. The
secondary current sets up its own mmf(= N 2 I 2 ) and hence its own flux φ2 which is in
opposition to the main primary flux φ , which is due to I o . The opposing secondary
flux φ2 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 ( I 2 ' ) to flow in the primary.
The current I 2 ' is known as load component of primary current. This current is in
phase opposition to current I 2 . The additional primary mmf N1I 2 ' sets up a flux φ2 '
which opposes φ2 (but is in the same direction as φ ) and is equal to it in magnitude.
Thus the magnetic effects of secondary current I 2 get neutralized immediately by
additional primary current I 2 ' . The whole process is illustrated in the following
figure.
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
φ2 = φ2 '
N 2 I 2 = N1I 2 '
I 2 ' = N 2 / N1 × I 2 =K I 2
Hence when transformer is on load, the primary winding has two currents I o and
I 2 ' (which is antiphase with I 2 and K times its magnitude). The total primary current
is the vector sum of I o and I 2 ' . In the following we show the vector diagrams of a
loaded transformer.
In fig(a) current I 2 ' is in phase with E 2 (for non-inductive loads). In Fig(b) it is
lagging behind E 2 (for inductive loads).
If we neglect I o as compared to I 2 ' as shown in fig(c), then φ1 = φ2 and thus
N1I 2 ' = N1 I1 = N 2 I 2
I1 / I 2 = N 2 / N1 = K
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 V2 is equal to the vector difference of the
secondary induced emf E2 and I 2 R2 where R2 is the resistance of the secondary
winding.
V2 = E2 - I 2 R2
(b) Similarly primary induced emf E1 is equal to the vector difference of V1 and I1R1
R1 is the resistance of the primary winding.
E1 = V1 - I1R1
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 V1 applied to
the primary must, therefore, have a component I1 X 1 (where X 1 is leakage reactance of
primary) to balance the primary leakage emf. In the secondary, similarly, an emf of self
induction I 2 X 2 (where X 2 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.
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
Z1 = ( R12 + X 12 )
And the secondary impedance is given by
Z 2 = ( R22 + X 2' )
V1 = E1 + I1 ( R1 + jX 1 ) = E1 + I1Z1
E2 = V2 + I 2 ( R2 + jX 2 ) = V2 + I 2 Z 2
.
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 θ1 between V1
and I1 gives the power-factor angle of the transformer.
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 heatresistant 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 singlephase 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 buttjoined 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 multiplelayer 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 disconnected 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 capacitancegrading 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 reentrant 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 onload 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 tapcharger.
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 reentrant 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 voltagecontrol 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, tapchanger 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
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:
HV, MV, and LV
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 g = (1 / 3 ) 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 =
se
3Un
=
3se
se
=
3Ug 3Ug
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 =
se
Un → rated line voltage
3Un
Phase current Ig = In
3=
se
3Un
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. 4 p1
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
Categories
of
transformer
Al winding
Cold-
Hot-rolled
Cold- rolled
Hot-rolled
si-steel
Si-steel
Si-steel
50-54
56-60
53-57
60-64
48-52
54-58
51-55
58-62
rolled
steel
Two-
winding
transformers
Three-winding
transformers
Co winding
si-
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/mt1
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 / e1t
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=
4.5et × 10 − 3
At
Where
B in Tesla.
φm = 4.5et × 10-3
Where: φm 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:
Δ We1 = Δ U g / et ( take integer for W)
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
Total
number
disk
10
35
63
40-60
56-75
60-80
110(132)
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 = Δ WF / NF (take integer)
Where
WF – Turns per disk for tapping winding
Δ EG – 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 mediumvoltage 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 with at beginning or or final
disks
Number of parallel
1-6
1
2,3
4,5
6
conductors in radial
direction
Fractional
part
number of turns
of
n1 − 1
n1
n1 − 1
n1
≤
n1 − 2
n1
N1 - Number of axial strips for the winding.
Determination of number of disks and turns per disk
≤
n1 − 3
n1
≤
n1 − 4
n1
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 select
n1 − 1
, except those for
n1
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 -
n +1
2
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 singlerow 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
_________Radius of core
S1
_________ Distance between LV winding
R2
BD
________Inner radius of LV winding
R3
________Radial dimension of LV winding
S2
R4
BG
________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
________Outer radius of HV winding
X -2
D
________Outer diameter of HV winding
+E
________Distance between phases ( see “ Main insulation”)
Mo
________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
_______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)
+EK
_____Thickness of pressing ring
H5
_____Clearance between pressing ring and upper yoke
H6
_____Window height (take 0 or 5 as the last figure)
δ
Ho
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 (
11
0.5
0.6
0.7
0.9
1.0
1.1
1.2
11
λ
1.4
)
1.5
1.6
1.7
1.8
1.9
λ
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
11
2.0
8.0
10.0 20.00 30.00
2.5
3.0
3.5
4.0
5.0
6.0
7.0
9.0
0.84
λ
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 interleaved 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 =
49.6 fI 1W 1ΣD12 P12 K 12
%
etH 12.106
Ukx13 =
49.6 fI 1W 1ΣD13 P13 K 13
%
etH 13.106
Ul x 23 =
49.6 fI 1W 1ΣD 23 P 23 K 23
%
etH 23.106
λ 12 = R3-R1; λ 13 = R4-R1; λ 23 = R4-R2;
59
Σ D17 =
1
(a1r1 + az rz) + a12r12
3
Σ D13 =
1
(a1r1 + a3 r3) + a13r13
3
Σ D23 =
1
( azrz + a9 r3) + a23r23
3
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;
a × b ---tickness and width of b are conductor (mm)
a1 × b1 ---thickness and width of paper wrapped conductor (mm)
A=MB 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 = 2 π 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 = I p 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%=
17.T (a + b + 1.57T )
Ax
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 (
235 + θ 2
)
235 + θ 1
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
Ln
A
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 =
Po
%
10 se
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
geGF + gjGAt
10 se
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 =
Ioa + Ior
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
Flux density
14000
15000
16000
17000
Hot-
Cold-
Hot-
Cold-
Hot-
Cold-
rolled
rolled
rolled
rolled
rolled
rolled
000
1.75
0.853
2.46
1.37
3.45
1.98
2.74
3.72
100
1.81
0.902
2.55
1.41
2.046
2.83
3.83
200
1.87
0.938
2.65
1.48
2.13
2.91
3.97
300
1.93
0.99
2.75
1.54
2.19
3.01
4.09
400
1.99
1.04
2.85
1.6
2.26
3.1
4.23
500
2.06
1.095
2.95
1.66
2.34
3.19
4.37
600
2.14
1.15
3.05
1.72
2.42
3.28
700
2.22
1.2
3.15
1.795
2.5
3.39
800
2.3
1.255
3.25
1.87
2.58
3.49
900
2.38
1.31
3.35
1.94
2.65
3.6
(Gause)
Cold-rolled
(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.
Coldrolled
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 noload 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 =
K FmnaAxP 2
(
) %
107
Hx
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=(
235 + 75 2
) × 3.95
235 + t
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 Thickness of single bare Height
conductor perpendicular to conductor (axial)
of
single
bare
conductor (axial)
the leakage flux
Hx – Reactance height of Reactance
windings (mm)
height
of Radial width of winding
windings
M – number of conductors Continuous winding: turns Continuous
perpendicular to the leakage per
flux
disk
parallel
x
number
conducts
windings;
parallel to the leakage flux
of number of disks in balance
helical group M1 ( or m2) Helical
number
parallel single conductors
N – number of conductors Continuous
winding:
of winding: number of disks in
balance group m1 ( or m2)
winding: Continuous winding: turns
n=number of disks Helical per
disk
x
number
of
windings n= turns x number 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 = {
Kzφ 2Uk 2 Hk 3
f
× ( )2
sδ [ Hk + 2( Rδ − Rp)]2
50
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)
S δ - Inner tank-wall girth, mm;
R δ - Mean equivalent radius of rank mm;
For three- Phase transformer.
R δ = (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;
φo=BA
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 =
23
× 104
UK
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 =
PLy
,
SY
Ly – mean length of load for each phase (including the length within the bushing crossessectional 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=
K 1K 2 IWJ
K4
(+
)
100
K 3L
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 + T Δ j + T Δ y
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 Δ j – insulation correction of winding temperature – rise,
T Δ 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 Δ y – correction of temperature for oil-duct, T Δ y = 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
100 < B
≤
150
δ = a1 – a +2c
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
Ts – top oil temperature-rise, oK;
Where:
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 =
po
+
A
pk
Where
po - no – load loss, w;
Pk – loss (including additional loss), w;
A – overall effective dissipating area of tanks, m2
T Δ - value of correction for oil temperature – rise, oK.
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 ampereturn 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 highvoltage 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 highvoltage 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 -
1
(1-euπ) × [ 1-0.5(1-e-2πv) ( 1-eπu)]
πu
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 =
9.8 × 8.04 KaKl 2
P0 (Rcp) (In)2 (Wn)2 Σm-n [σm + σm-1 Cpm]
λ1011
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 =
0.98 12.8 (ky)2 (Kl)2 ( I)2 W (λ )( ml)2 pn
MPa
mn λab × 108
b. axial bending stress of conductor under short – circuit
σ2 =
0.98 12.8 ky2 Kl2 I2 W λ ml2 pn
mn a1 × 108
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 =
0.98 ( 12.8 ) (ky)2 ( Kl)2 (I)2 (W) ) (R p)P
m n Hk A1× 10 8
x N’
a. Axial bending stress of conductor under short – circuit
σ2 =
0.98 12.8 ky2Kl2 I2 W λ ml2pn
mn ab2×1010
MPa
b. Radial bending stress of conductor under short-circuit
=
0.098 × 6.4 × Ky 2 KJ 2l 2Wlp 2 p
× m’ MPa
mnHka 2b × 108
Overall stress:
=(
β3
100
)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=(
2πR
- b, cm);
N
R – outer diameter of winding, cm;
N – number of radial spacers;
B – width of spacer, cm;
Ip =
2πRp
- d, cm;
N
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=
λ
h
, v=
s
,
h
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 =
Gpe Gcu
+
7 .8 4 .5
b. Transformer with high-voltage side of 63 – 132 KV.
Gpy =
Gfe Gcu
+
7.6 3.9
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;
B – Width of tank, dm.
R=
B
: (for oblate cross – section)
2
77
H
H
B
R
R
B
LZ
L
L
B1
H2
H1
Fig: 1
R3
R4
B
R2
B2
L
Fig 2
Tank with ladder – shaped top
R1
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,
(kg)
Table 1 weight of tube type radiators
Center
distance
88 – tubes
100 – tubes
120 – tubes
45o 88 – tubes
44 - tubes
Radiator
Oil
Radiator
Oil
Radiator
Oil
Radiator
Oil
Radiator
Oil
weight
weight
weight
weight
weight
weight
weight
weight
weight
weight
1400
309
108
255
96
1650
342
119
274
103
205
76
1880
372
129
421
147
305
113
220
81
2000
390
134
430
152
321
118
229
83
2285
426
145
482
166
436.6
359
130
247
89
2485
454
154
512
175
608
210
386
138
261
93
2685
481
163
243
185
644
221
2885
508
171
574
195
680
231
3000
522
176
591
200
702
239
3250
556
186
629
212
748
253
3500
590
197
667
224
96
267
3750
624
208
704
236
838
282
4000
656
218
743
248
885
296
between
connecting
flanges
(mm)
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
122
160
Yf – 100/380
130
1000
YF – 120/380
136
1160
Table 3 weight of conservators (kg)
Diameter
Length (mm)
Total weight of oil
(mm)
φ 180
φ 250
φ 310
φ 440
φ 610
Weight of oil in Weight of steel parts of
conservator
conservator
400
90
5
9
500
115
6
10
600
137
7
11
400
176
8
13
500
220
11
14
600
265
13
15
700
310
16
16
800
353
18
17
600
405
20
65
700
475
24
68
800
542
27
71
900
610
30
74
1000
678
34
77
1100
745
37
80
800
1100
52
100
900
1230
58
104
1000
1370
65
108
1200
1640
77
115
1400
1920
90
122
1600
2190
103
129
900
2370
112
132
1000
2630
124
135
80
φ 760
φ 920
φ 1080
φ 1240
1200
3150
148
140
1400
3680
173
145
1600
4200
200
150
1800
4740
224
155
2000
5260
248
160
2500
6600
300
170
2000
140
370
420
2500
10150
465
450
3000
12200
560
485
2500
14900
675
625
3000
17350
810
675
3500
20850
945
725
3000
24700
1105
800
3500
28800
1285
865
4000
32950
1465
930
3500
38000
1720
1030
4000
43500
1950
1100
4500
48900
2200
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 H δ 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 2.8
7.2
12
24
wheels),
55
90
340
30
For large transformer, dismountable bogie wheels are adopted (each group of wheel)
Load bearing for each group, ton weight of wheels 10
(each group)
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;
15
165 195
20
30
300 530
40
780
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)
ii.
KVA = 20000
Ip (H.V)
= 59.68779
KV = 145.2
IͅL(H.V)
= 45.91360
= 79.52483 Amp
85
iii.
IͅL(H.V)
KVA = 26000 KV = 132
Ip (H.V) = 65.65657 amp
iv.
113.7205 Amp
= 37.4773 amp
KVA = 20000 KV = 132
Ip (H.V) = 50.50505 amp
v.
vi.
KVA= 26000 KV = 118.8
IͅL(H.V)
Ip (H.V)
= 72.95174amp
KVA = 2000 KV = 118.8
Ip (H.V)
56.11672 amp
vii.
KVA = 26000 KV= 115
viii.
KVA = 20000
IͅL(H.V)
IͅL(L.V) = Ip (L.V) =
KV= 11.5
4.2 Core Calculation
IͅL(L.V) = Ip (L.V) =
First we calculate core diameter
D=K
P = Power = 26000 KVA
K = Experimental co-officient = 53
D = Diameter = 53
From Table D = 510/522
Here core Dia = 510
= 411.374
126.3561amp
= 97.19702amp
= 1305.314amp
= 1004.087amp
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
Volt/turn
Area = 1796.45
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
2
11
0.45
0.45
2.45
11.45
Bare
ZB
With installation
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
= (102×3) × 0.92
= 1119.35
= 281.52
Total = 1460.87
Reactance Height = 1460.87 – 4.45 – 3
Actual Height (mm) = 1460
= 1446.42 mm
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
3.85
11.55
With installation
Cross – section area = 24.95
A/
=
A/
=
=
2.0242
= 2.6315
Disc No
disc type
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
turn/disc
= 97.334
= 97.5
Height = 11.55 × 88
(5×87) × 0.92
= 1016.4
= 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
zb 1.95
A/
= 2.055708
A/
= 2.672421
2.36
11.8
1.95
1.95
4.31
13.75
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
R12 means radius of oil duct = 337 +
30.2
= 306
R1 means radius of L.V winding = 275 +
36.2
= 362
4305
= 435.75
R2 means radius of H.V winding = 387 +
R23
= 484 +
= 509.5
R23
= 534.5 +
= 543.5
50.95
54.35
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
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)
= 9.61
= 386.7543 cm
Width of lea kaage flux = 48405-275-7.25 = 20.855 cm
H=
= 1428.21
= 142.821 cm
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
= Ip
Mean length / turn (m) =
= 1923.428
Total length (m)
= 1.923429
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
Resistance (0 hm)
= 196.189758
R=
= 197.6898
92
resistivity of copper at 75
C
R=
= 0.02096 0 hm
/m
= .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
Total length
L = Ip × Wm
2.739
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
↑
↑
Window
width of
+
50
↑
foot pad
+
190 = 2980
↑
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×
= 7922.03598 Kg
GF =
+ 1462.9
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)
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
4.9 No Load Current
íoa =
%
íoa =
of Mo load loss
%
íoa = 0.071554
íor =
GF = weight of core = 15095.8
se patrol Power
100
From Tables
Ge = 1.34
Gi = 3.01
A = 17.9645
C=8
íoR =
íoR =
íoR = 0.317434
Io=
Io=
I o = 0.32539 theoretical Value
Io =
Io =
Io = 0.140043
4.10 Regulation of Transformer
Ur =
%
= loaded losses
Se = Saled Power
=
Ur (%) = 0.495934
× Pr
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(
=1(
)+
(Ur cos cf – Ur sin cf
) +10.07877 (0.6)+
(0.6)
= 6.44001+0.301511
= 6.745524
= 6.444001
= 0.301512
At full load.
(10.07877 (0.8) – 0.495934
102
At half load.
= 0.5 (
) +10.07877 (0.6)+
(0.6)
= 3.22206 + 0.150756
= 3.372762
4.11 EFFICENCY
K=
= No-load losess
= load losses
K=
K = 0.379844
Cos = 0.8
Eff =
Eff =
Eff =
Eff =
= 0.9939
P = Power
(10.07877 (0.8) – 0.495934
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
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
C
104
=
= 1.050972
1×
= 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
W = 24
=
J = 2.249167
J = 2.923917
×100
107
=
7.645
L=2(
+
1+
)
L = 207.9
At ONAN
Q=
Q = 452.3106
At ONAF
Q=
Q = 769.4049
At ONAN
=
= 0.41
= 16.07111
At ONAF
=Kjq
Kj = 9 ×
From table
At ONAN
=9×
=0.40708
× 452.3106
= 1.076451
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 ×
Zs =
=
Zs = 2.88 ×
Zt = short circuit impedance of transformer
Zt =
Sn = rated power of transformer MAV = 20
In MAV
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
44+11.26
1021.5 turns
52 turns
504.24 Rise height
52×11.45
545.4 turns heights
116
43×5 = 215
52×3
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
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-
= 156 oil dacts height
3=95 oil duct
L.V (AmpUnbalance Balance
Turn)
Avg Height
Turn)
H.V – L.V
×100
×100
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
=
=
-B
1=
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
= 0.2495 L= 20.855
Pn = 0.5
= 0.25
= 1.02
I= 1.05314
= 0.2
= 1.1
∑x = 15.2644
W= 102
= 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
= 11
= 93.575
=
L=
–6
L=
L= 19.026-5
L= 14.02627
–5
= 142.821
I = 65.65657
W = 2018
∑x = 15.2644
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
Th → 0.25dm
Wr →1.35dm
Gr = 353.2907 Kg
Wr
Th
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
127
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 × 10 6 )/( 4 K f × A × f × N)
Where
B m =maximum flux density in the core in teslas;
E =applied voltage of the winding considered;
128
K f =form factor the e.m.f wave;
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
129
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.
130
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.
131
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
132
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
133
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, Tapchanger 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:
134
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 100 o C
alarm operates, at 105 o C tripping occurs.
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.
135
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
1. M/s TBEA,
Shenyang
China
CONTRACT NO.
& DATE
T/F RATING
UNIT RATE
ABB-01-2006
13/06/2007
Lot-1:
Auto transformers
525/ 3 : 231 / 3 : 23 KV,
600MVA(3x200 MVA)
Auto transformer bank
512/ 3 : 220 / 3 : 11 KV,
450MVA(3x150MVA)
Auto transformer bank
US$5,348,160
QTY
ON
ORDE
R(NOS.
)
1
US$3,653,940
1
Lot-2:
2. M/s Xian
Electric
Engg:Co.
China
ADB-02-2007
13/07/2007
500 & 220KV
Auto-Transformers
Lot-1
500 / 3 : 220 / 3 : 11 KV
237MVA
US$ 946,000
Bank
Lot-2
525/ 3 : 220 / 3 : 11 KV
600MVA
US$ 1,684,000
Bank
Lot-3
500/ 3 : 220 / 3 : 22 KV
450MVA
US$ 1,320,000
Bank
220:132:11KV,138MVA
US$ 1,684,000
Bank
3. China
National
Electric
Wire &
Cable
ADB-02-2007
Lot-4:
Dated:17/07/2007
4. M/s Jiangsu
China
ADB-02-2007
13-07-2007
220KV Auto-Transformers
136
5. M/s Jiangsu
Hoping
China
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
Tender WOR-18
dt: 11/11/2006
3-Phase Auto
Transformers Complete
with all accessories.
US$ 1,120,859
04 No’s
US$ 3,75,000
19 No’s
US$ 3,69,713
25 No’s
Rs.29.8 Million
19 No’s
Rs.22.0 Million
05 No’s
06 No’s
Lot-4
6. M/s Iran
Transfo
IRAN
7. M/s Iran
Transfo
IRAN
STG-6-21 Lot-2 dt:
11/09/2006
STG-6-21 Lot-2
dt:11-09-2006
220/132KV, 125/160MVA
Power Transformers with
extended creepage
bushings.
132/11.5 KV,20/26 MVA
Power Transformers with
extended creepage
bushings.
132/11.5 KV,20/26 MVA
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
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
137
LESCO PROCUREMENT STATUS OF POWER TRANSFORMERS
(2006-2008)
Sr Name of firm
Contract/PO no. T/F Rating
Unit Rate
Qty
no
& Date
on
order
(No’s)
Contract dated 01- 20/26 MVA
US$ 318,912/- 10
1 M/s Iran
06-2006
132/11.5 KV
Transformer
IRAN
Contract dated 14- 20/26 MVA
Rs.24,000,000/- 5
2 M/s HEC Hattar
06-2006
132/11.5 KV
PO#2343 dated
20/26 MVA
Rs.29,800,000/- 10
3 M/s HEC Hattar
09-01-2007
132/11.5 KV
PO#2552 dated
20/26 MVA
Rs.29,468,000/- 3
4 M/s PEL
31-10-2007
132/11.5 KV
LAHORE
Under Award
20/26 MVA
Rs.29,800,000/- 7
5 M/s Siemens
132/11.5 KV
Karachi
PO#2637 dated
31.5/40MVA
Rs.38,475,000/- 2
6 M/s PEL
21-04-2008
132/11.5 KV
LAHORE
138
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.
139
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.
140
•
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.
141
•
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.
142
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