project report - final year

Project Report

SSTTUUDDYY AANNDD DDEESSIIGGNN OOFF PPOOWWEERR TTRRAANNSSFFOORRMMEERR

UUNNDDEERR SSHHOORRTT CCIIRRCCUUIITT CCOONNDDIITTIIOONNSS

Submitted by:

MANGESH DILIP ALEKAL (D050302)

GAURAV SANJIV DEWAL (D050309)

KSHITEESH VINAYAK KANEGAONKAR (D050320)

OMKAR DEEPAK LIMAYE (D050330)

KEDAR NANDAN RAJE (D050347)

Programme: B. Tech (Electrical Engineering)

Year: 2008-09

Guided by:

Dr. M. S. PANSE

Department of Electrical Engineering

Veermata Jijabai Technological Institute

(Autonomous Institute Affiliated to the University of Mumbai)

Mumbai 400019

Statement by the Candidates

We wish to state that work embodied in this report titled “Study and Design of

Power Transformer under Short Circuit Conditions” forms the Group's

contribution to the work carried out under the guidance of Dr M. S. PANSE at

the Veermata Jijabai Technological Institute. This work has not been

submitted for any other Degree or Diploma of any University/Institute.

Wherever references have been made to previous works of others, it has been

clearly indicated.

Name of Candidates Signature

MANGESH DILIP ALEKAL _____________________

GAURAV SANJIV DEWAL _____________________

KSHITEESH VINAYAK KANEGAONKAR _____________________

OMKAR DEEPAK LIMAYE _____________________

KEDAR NANDAN RAJE _____________________

CERTIFICATE

This is to certify that the following students:






have completed satisfactorily, the dissertation entitled “Study and Design of

Power Transformer under Short-Circuit Conditions” towards partial

fulfillment of the degree of Bachelor of Technology at the Veermata Jijabai

Technological Institute.

Prof. M. S. Panse Dr. B. K. Lande

Project Guide Head of Electrical Engg. Dept.

V.J.T.I., Mumbai-19

CERTIFICATE

The dissertation entitled 'Study and Design of Power Transformer under Short-

Circuit Conditions' submitted by the following students:






is found to be satisfactory and is approved for the degree of B.Tech. (Electrical

Engineering) of the University of Mumbai.

Guide Examiner

(Dr. M. S. Panse)

ACKNOWLEDGEMENT

We, the aforementioned, from B.Tech. (Electrical Engineering) take great

pleasure in presenting in presenting this report titled 'Study and Design of

Power Transformer under Short-Circuit conditions'. We would like to take

this opportunity to thank Dr. B.K. Lande, Head of the Department of Electrical

Engineering, Veermata Jijabai Technological Institute, who permitted us to

undertake this project.

We would like to convey our deepest appreciation for Dr. M. S. Panse from

the Electrical Engineering Department, whose continued support and motivation,

has allowed us to express ourselves and helped us to complete this project within

the stipulated time. We would like to thank all the college officials for their co-

operation.

A special vote of thanks goes out to Mr. Shekhar Vora, from Crompton

Greaves Ltd., whose excellent guidance and valuable inputs, helped make this

project a reality.

We would also like to express our deepest gratitude to all the people, who

have directly or indirectly, influenced the completion of this project.

CHAPTER 1 ..................................................................................................................... 1

INTRODUCTION (1) .................................................................................................... 2

CHAPTER 2 ..................................................................................................................... 2

CONSTRUCTION OF TRANSFORMER ................................................................... 4

2.1 WINDINGS ........................................................................................................ 6

2.2 NEED FOR CALCULATION OF TRANSFORMER IMPEDANCE ............ 11

2.3 IMPEDANCE CALCULATION ...................................................................... 12

2.4 RADIAL FORCES (2, 3) .................................................................................... 18

2.5 AXIAL FORCES (5) .......................................................................................... 27

2.6 EFFECT OF PRESTRESS (14, 15,16) ................................................................... 36

2.7 STRESSES IN FLITCH PLATES AND FRAMES ......................................... 37

CHAPTER 3 ................................................................................................................... 38

TRANSFORMER DESIGN PROCEDURE .............................................................. 38

3.1 DIFFERENT PARAMETERS CONSIDERED IN TRANSFORMER

DESIGN (17) ............................................................................................................. 38

CHAPTER 4 ................................................................................................................... 55

DESIGNED TRANSFORMER .................................................................................. 55

4.1 PROBLEM STATEMENT ............................................................................... 55

4.2 PARAMETERS OF DESIGNED TRANSFORMER ...................................... 56

REFERENCES ............................................................................................................... 76

ABSTRACT

The project deals with the study of the practical power transformer as used in

transformer industries and its design to successfully withstand severe short circuit

conditions.

Highlighting the basics of a power transformer (like emf equation, windings, core), we

designed a practical transformer using design specifications provided to us by

Crompton Greaves Ltd. like MVA rating of transformer, core diameter, % impedance

restrictions, core and copper loss limits, constraint on the price of the transformer.

After having successfully designed a power transformer according to the given

specifications and meeting the loss, impedance and price constraints, we proceeded to

analyzing the short circuit withstand capacity of the designed transformer.

The short circuit analysis involves calculation of first peak of short circuit current,

calculation of radial and axial stresses on the windings and on the insulation blocks

placed in between windings.

The radial and axial stresses calculated were found to be well within the stipulated

limits given in the design problem and hence, we conclude that our transformer has

been successfully passed to withstand short circuit conditions.

CHAPTER 1

INTRODUCTION (1)

In the recent development of the power systems, increase in the power plant

capacity has been a major achievement in the power industry. This power is transmitted

at a voltage which is much higher than the generated voltage as the power loss at high

voltage and low current is lesser than that at low voltage and high current. Hence, it is

necessary to step up the voltage level of the generated voltage to a much higher value

for transmission purposes in order to avoid extensive power loss. For this purpose,

power transformers are widely used in the industry.

Addition of more generating capacity and interconnections in the power systems

have contributed to an increase in the short circuit capacity of networks, making the

short circuit duty of power transformers more severe. Failure of transformers due to

short circuits is an area of major concern for transformer manufacturers and consumers.

There are continuous efforts by manufacturers and consumers to improve the short

circuit withstand performance of transformers.

The short circuit strength of a power transformer enables it to survive through

fault currents due to external short circuits in a power system network. An inadequate

strength may lead to a mechanical collapse of windings, deformation/damage to

clamping structures, or an electrical fault in the transformer itself. The internal faults

initiated by the external short circuits are dangerous as they may involve blow-out of

bushings, fire hazards and may eventually lead to bursting of the entire transformer

structure.

In recent short circuit tests on power transformers greater than 100 MVA, it has

been seen that increase of short circuit inductance beyond 1% has caused significant

deformation in windings. A much stricter control on the variations in materials and

manufacturing processes will have to be exercised to avoid looseness and winding

movements. Hence, the causes of internal short-circuits in the transformer are inherently

related to the design parameters of the transformer.

Thus, it is always desirable to design a transformer for normal working

conditions and then make variations to it in order to account for the forces developed on

transformer winding during short-circuit condition.

CHAPTER 2

CONSTRUCTION OF TRANSFORMER

A transformer is a static device that transfers electrical energy from one circuit to

another by electromagnetic induction without change in frequency.

A part of a core, which is surrounded by windings, is called a limb or leg. Remaining

part of the core, which is not surrounded by windings, but is essential for completing

the path of flux, is called as yoke. This type of construction (termed as core type) is

more common and has the following distinct advantages: viz. construction is simpler,

cooling is better and repair is easy.

Fig 2.1

Core type transformer

Shell-type construction, in which a cross section of windings in the plane of core is

surrounded by limbs and yokes, is also used. It has the advantage that one can use

sandwich construction of LV and HV windings to get very low impedance, if desired,

which is not easily possible in the core-type construction.

Fig 2.2

Shell type transformer

The core construction mainly depends on technical specifications, manufacturing

limitations, and transport considerations. It is economical to have all the windings of

three phases in one core frame. A three-phase transformer is cheaper (by about 20 to

25%) than three single-phase transformers connected in a bank. But from the spare unit

consideration, users find it more economical to buy four single-phase transformers as

compared to two three-phase transformers. Also, if the three-phase rating is too large to

be manufactured in transformer works (weights and dimensions exceeding the

manufacturing capability) and transported, there is no option but to manufacture and

supply single-phase units.

2.1 WINDINGS

The transformer consists of two coils called „windings‟ which are wrapped around a

core. The transformer operates when a source of ac voltage is connected to one of the

windings and a load device is connected to the other. The winding that is connected to

the source is called the „primary winding‟. The winding that is connected to the load is

called the „secondary winding‟.

The conducting material used for the windings depends upon the application, but in all

cases the individual turns must be electrically insulated from each other to ensure that

the current travels throughout every turn.

The insulation system must be designed to withstand the effects of lightning strikes and

switching surges to which the transformer is subjected, in addition to the normal

operating voltages. A further requirement of the insulation system is that it must

withstand the environmental conditions to which it is exposed, such as moisture, dust

etc. A variety of techniques and materials are employed to achieve the necessary

performance characteristics of the insulation system.

http://en.wikipedia.org/wiki/Electrical_conductor

2.1.1 LAYER WINDING

(a) (b)

Fig 2.3

a) Cross section of layer winding b) Transformer with layer wound coils

For low voltage, i.e. 600 Volt class windings, the winding technique used almost

exclusively is the layer winding technique, also sometimes called helical winding or

barrel winding. In this technique, the turns required for the winding are wound in one or

more concentric layers connected in series, with the turns of each layer being wound

side by side along the axial length of the coil until the layer is full. The conductors of

the winding are insulated and so between turns there will be a minimum of two

thicknesses of insulation. Between each pair of layers there will be layers of insulation

material and/or an air duct.

Low voltage windings will generally be wound top to bottom, bottom to top etc. using a

continuous conductor, until all layers are complete. High voltage windings, i.e. above

600 Volt class, may be wound in the same way, provided the voltage between layers is

not too great. To reduce the voltage stress between layers, high voltage windings are

often wound in only one direction, for example, top to bottom. When the first layer of

winding is complete, the winding conductor is laid across the completed layer from

bottom to top and then the next layer is wound, again from top to bottom. In this way,

the voltage stress between layers is halved. The conductor must, of course, have

additional insulation where it crosses the winding from bottom to top.

2.1.1.1 CHARACTERISTICS OF LAYER WOUND COILS

As stated previously, the layer wound coil requires insulation between layers, in

addition to the conductor insulation. The thickness of insulation required will depend

upon the voltage stress between layers, and comprises one or more thicknesses of the

appropriate insulation material. In practice, due to the nature of the construction of a

layer wound coil, the finished coil will have several unavoidable small air pockets

between turns and between layers. Many of these air pockets will become filled with

resin during vacuum pressure impregnation of the coil. However, it sometimes happens

that some air pockets remain and it is in these air pockets that partial discharges can

occur, greatly increasing the possibility of premature aging of the insulation and

eventual failure. Under short circuit conditions, the electromagnetic forces developed

cause transformer windings to attempt to telescope. At the same time the coil end

blocking is trying to prevent movement. The result is often that the turns of the winding

have a tendency to slip over one another, causing turn-to turn failure, due to abrasion of

the insulation as the turns rub together. A further disadvantage of the layer wound coil

is its poor impulse voltage distribution between the first few turns of the winding, due to

the high ground capacitance and the low series capacitance. A transformer winding

forms a complex network of resistance, inductance and capacitance. As far as the

impulse voltage distribution is concerned, the resistance can be ignored and at the

instant of application of the impulse wave, when very high frequencies are predominant,

the inductive elements become effectively infinite impedances. The whole structure

therefore reduces to a capacitive network. Each turn of a transformer winding is

insulated with a dielectric material and can be thought of as one plate of a multiple plate

capacitor. In addition, the combination of dielectric material and air between each turn

and ground forms further capacitive elements.

2.1.2 DISC WINDING

(a) (b)

Fig 2.4

a) Cross section of disc winding b) Transformer with disc winding

In the disc winding, the required number of turns is wound in a number of horizontal

discs spaced along the axial length of the coil. The conductor is usually rectangular in

cross-section and the turns are wound in a radial direction, one on top of the other i.e.

one turn per layer, until the required number of turns per disc has been wound. The

conductor is then moved to the next disc and the process repeated until all turns have

been wound. There is an air space, or duct, between each pair of discs. The disc winding

requires insulation only on the conductor itself, no additional insulation is required

between layers, as in the layer winding.

The disc wound high voltage winding is usually wound in two halves, in order that the

required voltage adjustment taps may be positioned at the electrical center of the

winding. In this way the magnetic or effective length of the winding is maintained,

irrespective of which tap is used, and therefore the magnetic balance between primary

and secondary windings is always close to its optimum. This is essential to maintain the

short circuit strength of the winding, and reduces the axial electromagnetic forces which

arise when the windings are not perfectly balanced.

2.1.2.1 CHARACTERISTICS OF DISC WOUND COILS

The major advantage of the disc wound coil lies in its open construction and relative

lack of insulation. For a 15kV class transformer employing a disc wound primary

winding, the number of discs will typically be in the range 36 to 48, resulting in a

relatively low voltage per disc. Since each disc is separated from the next by an air

space, the voltage stress between discs can easily be handled by the combination of

conductor insulation and air, no additional insulation being necessary. Each disc

comprises a number of turns with each turn occupying one layer, i.e. one turn per layer:

the voltage stress between layers is therefore the same as the voltage stress between

turns and again, can easily be handled by the conductor insulation. The turns of each

disc, being wound tightly together provide almost no possibility of air pockets being

present within the disc. Due to the open construction of the discs, any small air pockets

which may be present are readily filled with resin during vacuum pressure impregnation

of the coil. A properly designed and manufactured dry-type transformer disc winding

therefore displays very low values of partial discharge, typically in the range of 10 to

20 Pico coulombs. Unlike the layer wound coil, the disc wound coil provides good

impulse voltage distribution, due to its inherently low value of ground capacitance and

high series capacitance. The disc wound coil also displays excellent short circuit

strength. Each disc by itself is mechanically very strong and the complete assembly of

discs is held very securely in place. While the electromagnetic forces resulting from a

short circuit result in tendency, for the windings to telescope, high voltage turns usually

remain intact relative to each other. Instead, the complete disc has a tendency to distort

as assembly, with all the turns distorting by same amount. The transformer can often

continue to function, despite the distortion, until a convenient time arises for repair.

2.2 NEED FOR CALCULATION OF TRANSFORMER IMPEDANCE

There are basically two types of forces in windings: axial and radial electromagnetic

forces produced by radial and axial leakage fields respectively. The average stress on

windings due to the radial electromagnetic forces acting on them is given below.

For a copper conductor,

avg = 0.48 10 4 (k 2 ) 2 2

puW

R

ZH

P kg / cm 2 ……………………………… (2.1)

where,

Zpu = Per-unit impedance of the transformer

Hw = Height of winding in meters

PR = Copper loss per phase in transformer

k = asymmetry factor

For an aluminium conductor,

avg = 0.29 10 4 (k 2 ) 2 2

puW

R

ZH

P kg / cm 2 ……………………………... (2.2)

Now, the total axial compressive force acting on the windings is given by the

expression,

F a = fHZ

S

wpu

8.50 kg…………………………………………………………… (2.3)

where,

f = Frequency in kHz

S = Rated power per limb in kVA

Hence, we can see that both force components on the winding are dependent on the

impedance of the transformer. Thus, we shall perform the calculation of transformer

impedance in detail.

2.3 IMPEDANCE CALCULATION

For a power transformer, the ratio of reactance to resistance is greater than 20.

So its impedance is generally taken as equal to its reactance. Its per- unit value can be

calculated as follows:

For uniformly distributed ampere-turns along LV & HV windings with equal

heights, the leakage flux is predominantly axial except at winding ends, where there‟s

fringing. This is because leakage flux finds a shorter path to return via limb or yoke.

Hence the equivalent height (Heq) can be obtained by dividing winding height (Hw) by

Rogowski factor KR, which is less than one.

Fig 2.5

Heq= Hw/ KR………………………………………………………………………………………………………... (2.4)

The leakage magneto motive force (mmf) distribution across the cross section of

windings is of trapezoidal form. The mmf at any point depends on the ampere-turns

enclosed by a flux contour at that point; it increases linearly with the ampere-turns from

a value of zero at the inside diameter of LV winding to the maximum value of one per-

unit (total ampere-turns of LV or HV winding) at the outside diameter. In the gap

between LV and HV windings, since flux contour at any point encloses full LV (or HV)

ampere-turns, the mmf is of constant value. The mmf starts reducing linearly from the

maximum value at the inside diameter of the HV winding and approaches zero at its

outside diameter. The core is assumed to have infinite permeability requiring no

magnetizing mmf, and hence the primary and secondary mmfs exactly balance each

other. The flux density distribution is of the same form as that of the mmf distribution.

Since the core is assumed to have zero reluctance, no mmf is expended in the return

path through it for any contour of flux.

(a) Flux Tube

(b) MMF Diagram

Fig 2.6

Hence, for a closed contour of flux at a distance x from the inside diameter of LV

winding, it can be written that

( 0

xB) H eq = (NI) x …………………………………………………………. ……. (2.5)

i.e.

B x = eq

x

H

NI )(0 …………………………………………………………………... (2.6)

For deriving the formula for reactance, let us derive a general expression for the flux

linkages of a flux tube having radial depth R and height Heq. The ampere-turns

enclosed by a flux contour at the inside diameter (ID) and outside diameter (OD) of this

flux tube are a(NI) and b(NI) respectively.

B x = eqH

0 [ { a + R

ab )( x }NI ] ………………………………………………... (2.7)

The flux linkages of an incremental flux tube of width dx placed at x are

d = N x x = N x B x A ………………………………………………………….. (2.8)

where,

A is the area of flux tube given by

A = (ID + 2x) dx …………………………………………………………....... (2.9)

d = [ { a + R

ab )( x }N ] [

eqH

0 [ { a + R

ab )( x }NI ]]

[ (ID + 2x) dx] …………………………………………………… (2.10)

Hence the total flux linkages of the flux tube are given by

= R

d0

= eqH

IN 2

0R

0

{ a + R

ab )( x } 2 ( ID + 2x ) dx …………………… (2.11)

= eqH

IN 2

0

3

R[ ( a 2 + ab + b 2 ) ID +

2

3)( 22 Rbaba -

2

2 2 aba R ]………..(2.12)

The last term in square bracket can be neglected without introducing an appreciable

error to arrive at a simple formula for the regular design use.

= eqH

IN 2

0

3

R ( a 2 + ab + b 2 ) [ID + 1.5 R]……………………………………(2.13)

The last term [ID + can be taken to be approximately equal to the mean diameter

(Dm) of the flux tube (for large diameters of windings/gaps with comparatively lower

values of their radial depths).

= eqH

IN 2

0

3

R ( a 2 + ab + b 2 ) D m ……………………………………………..(2.14)

Now let,

ATD = 3

R ( a 2 + ab + b 2 ) D m ………………………………………………….....(2.15)

which corresponds to the area of Ampere-Turn Diagram.

The leakage inductance of a transformer with n flux tubes can now be given as

L = I

n

k

1

= eqH

N 2

0

n

k

ATD1

…………………………………………………. (2.16)

And the corresponding expression for the leakage reactance X is given by

X = 2 f eqH

N 2

0

n

k

ATD1

.………………………………………………….......... (2.17)

For the base impedance Zb the formula for percentage leakage reactance is

% X = bZ

X =

V

IX 100 = 2 f

VH

IN

eq

2

0

n

k

ATD1

100…………………………... (2.18)

= 2 f )/(

)(0

NVH

NI

eq

n

k

ATD1

100……………………… (2.19)

where,

V is rated voltage and

(V/N) is volts/turn of the transformer.

Substituting µ0=4 π×10-7 and adjusting constants so that the dimensions used in the

formula are in units of centimeters (Heq in cm and Σ ATD in cm2),

% X = 2.48 10 5 f )/(

)(

TurnVoltsH

sAmpereTurn

eq

n

k

ATD1

………………………………… (2.20)

After having derived the general formula, we will now apply it for a simple case of a

two winding transformer. The constants a and b have the values of 0 and 1 for LV, 1

and 1 for gap, and 1 and 0 for HV respectively. If 2,1 DD and gD are the mean diameters

and 2,1 TT and gT are the radial depths of LV, gap and HV respectively, using equation

3.12 we get

ATD = 3

1 ( T 1 D 1 ) + ( T g D g ) +

3

1( T 2 D 2 )………………………….. (2.21)

The value of Heq is calculated by equation 2.4, for which the Rogowski factor KR is

given by

)(

11

21

)21(

TTT

H

eK

g

w

TgTT

wH

R

………………………………………………… (2.22)

For taking into account the effect of core, a more accurate but complex expression for

KR can be used. For most of the cases, equation 2.22 gives sufficiently accurate results.

2.4 RADIAL FORCES (2, 3)

The radial forces produced by the axial leakage field act outwards on the outer

winding tending to stretch the winding conductor, producing a tensile stress (also

called as hoop stress)

The inner winding experiences radial forces acting inwards tending to collapse or

crush it, producing a compressive stress.

Due to the fringing of the leakage field at the ends of the windings, the axial

component of the field reduces resulting into smaller radial forces in these regions.

For deriving a simple formula for the radial force in a winding, the fringing of the field

is neglected; the approximation is justified because the maximum value of the radial

force is important which occurs in the major middle portion of the winding.

Let us consider an outer winding, which is subjected to hoop stresses. The value of the

leakage field increases from zero at the outside diameter to a maximum at the inside

diameter (at the gap between the two windings). The peak value of flux density in the

gap is

wgp

H

NIB 02

…………………………………………………………………. (2.23)

where,

NI is the r.m.s. value of winding ampere-turns

Hw is winding height in meters.

The whole winding is in the average value of flux density of half the gap value. The

total radial force acting on the winding having a mean diameter of Dm (in meters) can be

calculated by

....(2.25)................................................................................newton

2.24).........(......................................................................22

2

1

20

0

mw

r

mw

r

DH

NIF

DNIH

NIF

For the outer winding, the conductors close to gap (at the inside diameter) experience

higher forces as compared to those near the outside diameter (force reduces linearly

from a maximum value at the gap to zero at the outside diameter).

The force can be considered to be transferred from conductors with high load (force) to

those with low load if the conductors are wound tightly

Hence, averaging of the force value over the radial depth of the winding as done in the

above equation is justified since the winding conductors share the load almost

uniformly.

If the curvature is taken into account by the process of integration across the winding

radial depth the mean diameter of the winding in the above equation should be replaced

by its inside diameter plus two-thirds of the radial depth.

2.4.1 CALCULATION OF RADIAL STRESSES (4, 5)

Fig 2.7

The average hoop stress for the outer winding is calculated as for a cylindrical boiler

shell shown in figure. The transverse force F acting on two halves of the winding is

equivalent to pressure on the diameter, hence it will be given by π Dm replaced by Dm in

the above equation.

If the cross-sectional area of turn is At (in m2), the average hoop stress in the winding is

(2.27) ........................................................................................ 2

(2.26) ...............................................................................................2(

20

20

t

m

wavg

tw

mavg

A

NDp

Hp

I

ANH

DNI

Let Ir be the rated r.m.s. current and Zpu be the per-unit impedance of a transformer.

Under the short circuit condition, the r.m.s. value of current in the winding is equal to

(Ir/Zpu). To take into account the asymmetry, this current value is multiplied by the

asymmetry factor k. If we denote copper loss per phase by PR, the expression for σavg

under the short circuit condition is

(2.29) N/m 2

22

(2.28) 2

2

2

2

2

0

2

220

pu

R

w

avg

t

m

pu

r

wavg

Z

kP

Hp

A

NDp

Z

kI

Hp

Substituting the values of μ0(=4π×10-7

) and ρ (resistivity of copper at 75° = 0.0211×10-

6), we finally get

2

2

2N/m 274.4

puw

Ravg

ZH

Pk ………………………………………….. (2.30)

where,

PR is in watts and

Hw in meters.

It is to be noted that the term PR is only the DC I2R loss (without having any component

of stray loss) of the winding per phase at 75°C. Hence, with very little and basic

information of the design, the average value of hoop stress can be easily calculated.

The previous value of average stress can be assumed to be applicable for an entire

tightly wound disc winding without much error. This is because of the fact that although

the stress is higher for the inner conductors of the outer winding, these conductors

cannot elongate without stressing the outer conductors. This results in a near uniform

hoop stress distribution over the entire winding.

In layer/helical windings having two or more layers, the layers do not firmly support

each other and there is no transfer of load between them. Hence, the hoop stress is

highest for the innermost layer and it decreases towards the outer layers. For a double-

layer winding, the average stress in the layer near the gap is 1.5 times higher than the

average stress for the two layers considered together.

Generalizing, if there are L layers, the average stress in kth

layer (from gap) is [2- ((2k-

1)/L)] times the average stress of all the layers considered together.

For an inner winding subjected to radial forces acting inwards, the average stress can be

calculated by the same formulae as above for the outer winding. However, since the

inner winding can either fail by collapsing or due to bending between the supports, the

compressive stresses of the inner winding are not the simple equivalents of the hoop

stresses of the outer winding.

Thus, the inner winding design considerations are quite different, and these aspects

along with the failure modes are discussed in the further sections.

2.4.2 FAILURE MODES DUE TO RADIAL FORCES

The failure modes of windings are quite different for inward and outward radial forces.

Winding conductors subjected to outward forces experience the tensile (hoop) stresses.

The compressive stresses are developed in conductors of a winding subjected to the

inward forces. In concentric windings, the strength of outer windings subjected to the

outward forces depends on the tensile strength of the conductor; on the contrary the

strength of inner windings subjected to the inward forces depends on the support

structure provided. The radial collapse of the inner windings is common, whereas the

outward bursting of the outer windings usually does not take place.

2.4.2.1 WINDING SUBJECTED TO TENSILE STRESSES (5, 8, 9, 10, 11)

If a winding is tightly wound, the conductors in the radial direction in a disc winding or

in any layer of a multi-layer winding can be assumed to have a uniform tensile stress.

Since most of the space in the radial direction is occupied with copper (except for the

small paper covering on the conductors), the ratio of stiffness to mass is high. As

mentioned earlier, the natural frequency is much higher than the exciting frequencies,

and hence chances of resonance are remote.

Under a stretched condition, if the stress exceeds the yield strength of the conductor, a

failure occurs. The conductor insulation may get damaged or there could be local

bulging of the winding. The conductor may even break due to improper joints. The

chances of failure of windings subjected to the tensile hoop stresses are unlikely if a

conductor with a certain minimum 0.2% proof strength is used. The 0.2% proof stress

can be defined as that stress value which produces a permanent strain of 0.2% (2 mm in

1000 mm). One of the common ways to increase the strength is the use of work-

hardened conductor; the hardness should not be very high since there could be difficulty

in winding operation with such a hard conductor. A lower value of current density is

also used to improve the withstand characteristics.

2.4.2.2 WINDINGS SUBJECTED TO COMPRESSIVE STRESSES (5, 8, 9, 10, 11)

Conductors of inner windings, which are subjected to the radial compressive load, may

fail due to bending between supports or buckling. The former case is applicable when

the inner winding is firmly supported by the axially placed supporting spacers (strips),

and the supporting structure as a whole has higher stiffness than conductors (e.g., if the

spacers are supported by the core structure) as shown in figure (a).

Fig 2.8

The latter case of buckling, termed as free buckling, is essentially an unsupported

buckling mode, in which the span of the conductor buckle bears no relation to the span

of axial supporting spacers as shown in figure (b). This kind of failure occurs mostly

with thin winding cylinders, where conductor has higher stiffness as compared to that of

inner cylinders and/or the cylinders (and the axial spacers) are not firmly supported

from inside. The conductors bulge inwards as well as outwards at one or more locations

along the circumference. There are many factors which may lead to the buckling

phenomenon, viz. winding looseness, inferior material characteristics, eccentricities in

windings, lower stiffness of supporting structures as compared to the conductor, etc.

The buckling can be viewed as a sequential chain of failures, initiated at the outermost

conductor of the inner winding and moving towards the innermost conductor facing the

core. The number of winding supports should be adequate for giving the necessary

strength to the winding against the radial forces. When the supporting structures are in

direct contact with the core, a winding can be taken as very rigidly supported. On the

contrary, if there is no direct contact (fully or partly) with the core, the winding is only

supported by the insulating cylinder made of mostly the pressboard material thereby

reducing the effective stiffness of the support structure and increasing the chances of

failure. The supports provided are effective only when the support structure as a whole

is in firm contact with the core.

A winding conductor subjected to the inward radial forces is usually modeled as a

circular loop under a uniformly distributed radial load. The critical load per unit length

of the winding conductor is given by,

]4[12

2 23

3 s

m

r Nwt

D

Ef ………………………………………………………… (2.31)

where,

E is modulus of elasticity of conductor material,

Ns is total number of axially placed supports (spacers),

w is width of conductor,

t is thickness of conductor and

Dm is mean diameter of winding

The compressive stress on the inner winding conductor is given as,

A

Df mravg

2 …………………………………………………............................ (2.32)

where,

A is area of conductor (=w t)

Substituting the value of fr, we get,

2

224

12s

s

m

savg

N

N

D

tNE ………………………………………………… (2.33)

For Ns >>1, the expression for the minimum number of supports to be provided is,

Et

DN

avgms

12 …………………………………………………............... (2.34)

The term σavg is the average value of the compressive stress (in an entire disc winding or

in a layer of a multi-layer winding)

2.5 AXIAL FORCES (5)

For a uniform ampere-turn distribution in windings with equal heights (ideal

conditions), the axial forces due to the radial leakage field at the winding ends are

directed towards the winding center as shown.

Fig 2.9

Although, there is higher local force per unit length at the winding ends, the cumulative

compressive force is maximum at the center of windings. Thus, both the inner and outer

windings experience compressive forces with no end thrust on the clamping structures

(under ideal conditions).

The inner winding being closer to the limb, by virtue of higher radial flux, experiences

higher compressive force as compared to the outer winding.

It can be assumed that 25 to 33% of force is taken by the outer winding, and the

remaining 75 to 67% is taken by the inner winding.

2.5.1 CALCULATION OF AXIAL FORCES (6)

For an asymmetry factor of 1.8, the total axial compressive force acting on the inner and

outer windings taken together is given by

conditions ideal ... kg 8.50

fHZ

SF

wpua

…………………………………….. (2.35)

where,

S is rated power per limb in kVA

Hw is winding height in meters

Zpu is per-unit impedance, and

f is frequency in Hz.

Once the total axial force for each winding is calculated, the compressive stress in the

supporting radial spacers (blocks) can be calculated by dividing the compressive force

by the total area of radial spacers. The stress should be less than a certain limit, which

depends on the material of the spacer.

2.5.2 REASONS FOR HIGH VALUE OF AXIAL FORCES (7)

The reasons for a higher value of radial field and consequent axial forces are:

Mismatch of ampere-turn distribution between LV and HV windings,

Tapping in the winding,

Unaccounted shrinkage of insulation during drying and impregnation processes, etc.

When the windings are not placed symmetrically with respect to the center-line as

shown, the resulting axial forces are in such a direction that the asymmetry and the end

thrusts on the clamping structures increase further.

Fig 2.10

It is well known that even a small axial displacement of windings or misalignment of

magnetic centers of windings can eventually cause enormous axial forces leading to

failure of transformers.

Hence, strict sizing/ dimension control is required during processing and assembling of

windings so that the windings get symmetrically placed.

2.5.3 FAILURE MODES DUE TO AXIAL FORCES (5)

There are various types of failures under the action of axial compressive forces:

If a layer winding is not wound tightly, some conductors may just axially pass over the

adjacent conductors, which may damage the conductor insulation leading eventually

into a turn-to-turn fault. In another mode of failure, if a winding is set into vibration

under the action of axial forces, the conductor insulation may get damaged due to a

relative movement between the winding and axially placed insulation spacers.

High axial end thrusts could lead to deformations of the end clamping structures and

windings. The end clamping structures play the most important role in resisting axial

forces during short circuits. They have to maintain an effective pressure on the

windings, applied usually on the clamping ring made of stiff insulating material (pre-

compressed board or densified wood). The type of insulation material used for the

clamping ring depends on the dielectric stress in the end insulation region of windings.

The densified wood material is used for lower stresses and pre-compressed board, being

a better grade dielectrically, is used for higher stresses and for complying stringent

partial discharge requirements.

When a clamping ring made of an insulating material is reinforced by the fiberglass

material, an extra strength is provided. Some manufacturers use clamping rings made of

steel material. The thickness of metallic clamping rings is smaller than that made from

the insulating material. The metallic ring has to be properly grounded with a cut so that

it does not form a short-circuited turn around the limb. The sharp edges of the metallic

ring should be rounded off and covered with a suitable insulation.

In addition to above types of failures due to the axial forces, there are two principal

types of failures, viz. bending between radial spacers and tilting.

2.5.3.1 BENDING BETWEEN RADIAL SPACERS

Under the action of axial forces, the winding conductor can bend between the radially

placed insulation spacers as shown.

Fig 2.11

The conductor bending can result into a damage of its insulation.

The maximum stress in the conductor due to bending occurs at the corners of the radial

spacers and is given by

2

0

2

max kg/cm 12I

ySFAL ……………………………………………………….. (2.36)

where,

FAL is maximum axial bending load in kg/cm. It corresponds to the most highly stressed

disc in a disc winding or turn in a helical winding (layer winding with radial spacers).

The maximum axial load may usually lie in the region of non-uniform ampere-turn

distribution (e.g., tap zone).

The maximum axial load, calculated accurately by a method such as FEM, divided by

the mean turn length (πDm) gives the value of FAL, where Dm is mean diameter of

winding in cm.

S is span between two radial spacers in cm

thspacer wid

s

m

N

D

Ns is number of radial spacers.

y is maximum distance from neutral axis for conductor in cm (i.e., half of conductor

axial width: w/2).

I0 is moment of inertia of disc or turn

12

3ntw

n being number conductors in radial direction, and

t is conductor thickness in cm.

The maximum stress in the conductor calculated should be less than the limiting value

for the type of conductor used (about 1200 kg/cm2 less than the limit for soft copper).

2.5.3.2 TILTING UNDER AXIAL LOAD (12, 13)

The failure due to tilting under the action of axial compressive forces is one of the

principal modes of failures in large power transformers. When these forces are more

than a certain limit, a failure can occur in disc winding due to tilting of conductors in a

zigzag fashion as shown.

Fig 2.12

In this mode of failure, there is turning of cross section of conductors around the

perpendicular axis of symmetry. There are two kinds of forces that resist the tilting of

the conductors. The first one is due to the conductor material, which resists being

twisted.

The second resisting force is the friction force (due to corners of conductors); during

tilting the conductors at both ends must bite into the material of the radial spacer,

producing a couple at the conductor ends which resists tilting. The two resisting forces

are usually considered separately to arrive at the critical stress and load, causing the

failure.

The tilting strength decreases inversely as the square of winding radius, suggesting that

the large windings should be carefully designed. If the conductor has sharp ends, the

frictional force resists tilting.

Actually, due to the conductor corner radius, the contribution to tilting resistance (due

to friction) reduces and this reduction should be considered. The critical strength of a

helical winding (a layer winding with radial spacers) is higher than a layer winding

(which is without radial spacers) because of the additional strength offered by the

spacers.

When a continuously transposed cable (CTC) conductor is used, although there are two

axially placed rows of conductors in one common paper covering, it cannot be assumed

that the effective tilting strength is higher. Two possible modes of failures are

described. The first type of failure (termed as cable-wise tilting), in which two adjacent

cables tilt against each other, is shown

Fig 2.14 Cable wise tilting

If the inter-strand friction is higher, the winding is forced to tilt in pairs of strands in the

CTC conductor.

For the same strand dimensions, the CTC conductor has four times greater tilting

strength; the result is obvious because the effective width of its conductor is doubled

increasing the strength by four times.

This increase in strength is valid only when the two axially placed strands in the CTC

conductor can be considered to act together under the tilting load.

In the second mode of failure (termed as strand-wise tilting), two axially placed strands

in the CTC conductor tilt against each other as shown.

Fig 2.15 Strand wise tilting

The critical tilting load in this mode may be lower, reducing the effective overall tilting

strength. This is because the lower of the cable-wise and strand-wise strengths triggers

the axial instability. While the critical stress in the cable-wise tilting is independent of

number of strands in the cable (n), the critical stress in the strand-wise tilting is

inversely proportional to n. As the number of strands in the CTC conductor increases,

the critical load limit in the strand-wise tilting becomes lower than the cable-wise

tilting. Hence, with the increase in number of strands in the CTC conductor, the mode

of failure shifts from the cable-wise tilting to the strand-wise tilting.

The use of epoxy-bonded CTC conductor is quite common in which the epoxy coating

effectively bonds the strands increasing the resistance against the strand wise tilting.

Each strand in the epoxy-bonded CTC conductor has, in addition to an enamel coating,

a coat of thermosetting epoxy resin. The curing of this resin occurs at around 120°C

during the processing of windings. After curing, the epoxy-bonded CTC conductor

consisting of many strands can be considered as one conductor with an equivalent cross

section for the mechanical strength consideration. Thus, the possibility of strand-wise

tilting is eliminated, greatly increasing the strength of the CTC conductor against the

tilting load. The epoxy bonded CTC conductor not only reduces the winding eddy

losses but it also significantly improves the short circuit withstand characteristics.

2.6 EFFECT OF PRESTRESS (14, 15,16)

The clamping pressure applied on the windings after the completion of core winding

assembly is called as pre-stress. It has a significant impact on the response of windings

during short circuits. It increases the stiffness of windings thereby increasing their

mechanical natural frequencies.

The relationship between the natural frequency and pre-stress is highly non-linear. The

pre-stress reduces oscillatory forces acting on the insulation. The winding

displacements also decrease with the increase in the pre-stress value. The value of pre-

stress should be judiciously chosen depending upon the characteristics of core-winding

assembly. The chosen value of pre stress must get maintained during the entire life of a

transformer. This means that the insulation stability should be fully realized during the

processing of windings during manufacturing. If the natural frequency without pre-

stress is higher than the excitation frequencies, a higher pre-stress value will

significantly reduce the oscillatory forces. Contrary to this, if the natural frequency

without pre-stress is lower than the excitation frequencies, a certain value of pre-stress

will bring the natural frequency closer to the excitation frequencies leading to an

increase in the oscillatory forces. The natural frequency is reported to vary as some

function of square root of the ratio of pre-stress to maximum value of peak

electromagnetic stress in the winding.

The natural frequency of a winding may change during the short circuit period due to

changes in the insulation characteristics and ratio of pre-stress to total stress. Thus, the

natural frequency measured from the free response may be different after the short

circuit as compared to that before the short circuit. Also, during the short circuit, the

winding, which may be in resonance at some time experiencing a higher stress, may get

detuned from the resonance due to change in insulation characteristics at some other

instant.

2.7 STRESSES IN FLITCH PLATES AND FRAMES

A flitch plate should be designed for withstanding the clamping force and core winding

weight (static loads). During a short circuit, the axial forces (end thrusts) developed in

windings, act on the top and bottom frames; flitch plates help to keep the frames in

position. The stresses produced in the flitch plates are the tensile stresses and shearing

stresses. These stresses can be calculated by well-known formulae used in the structural

analysis.

The frames are subjected to stresses while lifting core-winding assembly, during

clamping of windings, or due to short circuit end thrusts. Usually, the short circuit

stresses decide their dimensions. The stresses in the frames are determined from the

calculated values of the short circuit forces acting on them and assuming the core bolt

points and locking arrangements (pins, etc.) between flitch plates and frames as support

locations.

Fig 2.16

CHAPTER 3

TRANSFORMER DESIGN PROCEDURE

3.1 DIFFERENT PARAMETERS CONSIDERED IN TRANSFORMER

DESIGN (17)

1. Core diameter selection

2. Flux density

3. Number of turns in HV and LV

4. Selection of layer and helical winding

5. Estimation of height of window

6. Width of conductor

7. Thickness of conductor

8. Number of parallel conductors (based on current density)

9. Calculation of percentage impedance based on dimensions (described for each

case)

10. Load loss and no load loss calculation

11. Percentage ratio error calculation

12. Costing of major material

3.1.1 CORE DIAMETER SELECTION

Core diameter of a transformer depends upon a number of factors like transformer

rating, percentage impedance between the windings, basic insulation level, transport

height, core fluxing requirements, type of core, and quality of core steel.

It is fairly complicated to derive a universal and exact formula for core diameter. In

practice, the core diameter is selected by the designer from similar designs already

available in the industry (usually core diameter varies from 350mm to 500mm). Based

on this, parameters like percentage impedance and losses are worked out and core

diameter is adjusted to meet the required parameters.

Influence of varying core diameter

Increase in core diameter causes:

Increase in core cross-sectional area

Increase in voltage/turn, hence reduction in number of turns

Overall weight of core steel increases

No load loss of transformer increases

Reduction in core diameter causes:

Increased copper weight

Increased load losses

The percentage reactance between the windings is directly proportional to the number

of turns and diameter of various coils and is inversely proportional to volts/turn and coil

depth.

3.1.2 FLUX DENSITY

Value of flux density is chosen to suit the required performance. Normally flux density

is chosen near the knee point of the magnetization curve. However, adequate margin

should be kept to take care for system conditions like over fluxing, frequency and

voltage variations. In certain cases, the value of flux density is reduced to limit the noise

level of transformers.

Influence of varying flux density

Increase in flux density, keeping other parameters constant causes:

Higher volts/turn

Reduction in number of turns in winding

Reduced core steel weight

Higher no load loss of transformer

Lower copper weight

Lower load losses

Reduction in the value of flux density causes increased core steel weight, lower no load

loss and increases copper weight and load loss.

3.1.3 NUMBER OF TURNS IN HV AND LV

The number of turns in HV and LV winding can be calculated by using the emf

equation of the transformer as:

E = 4.44fΦmN ……………………………………………………………………. (3.1)

E =4.44fBmAN ………………………………………………………………… (3.2)

where :

E =Voltage induced in the transformer winding (HV or LV winding)

F = Frequency in Hz

Φm = Max value of flux

Bm = Maximum flux density in Tesla

A = Net cross sectional area of the core in mm2

Therefore, after fixing the core area and flux density, the number of turns in the HV and

LV winding can be calculated by using the above equation.

Also, voltage per turn is calculated by dividing the voltage by the corresponding

number of turns (LV or HV).

3.1.4 SELECTION OF TYPE OF WINDING

The windings along with its insulations form the electric circuit of the transformer. Due

care must be taken while designing the windings to ensure its healthiness during normal

as well as fault conditions as the windings need to be electrically and mechanically

strong to withstand voltage surges as well as mechanical stresses during short-circuit

conditions. The temperature of windings at rated, over-load and short-circuit conditions

should be within limits, ensuring the proper life of transformer.

The requisite type of winding is chosen only after considering all the above mentioned

factors.

For LV, as we need to have less number of turns in the winding, generally layer

winding is used as it can be fitted within the limited height of the core structure.

For HV, we need to have a large number of turns in the winding and hence, disc

winding is preferred over layer winding taking into account the height constraints of the

core.

Sometimes, for large voltages, disc winding is chosen over layer winding in LV as well

due to height considerations.

3.1.5 HEIGHT OF CORE WINDOW

In the transformer industry, the height of transformers is restricted to 4.5 – 5 meters due

to the difficulty involved in transportation.

Taking this into account, the height of base and clamping structures, the relay

arrangement and insulation the actual height of the core structure is restricted to 3 - 3.5

m.

The designers select the height of the core structure less than the above specified value.

From this value of the core height, we determine the height of HV and LV winding.

3.1.6 WIDTH OF CONDUCTOR

After fixing the height of LV and HV winding, the width of the LV and HV winding is

calculated. The procedure of calculating the width in LV or HV is as follows:

1. First the total number of discs or turns in the winding is fixed. In case of disc

winding, the number of discs is calculated by fixing the number of turns per

disc. We usually take even number of discs for convenience of taking out the

leads.

2. Then we calculate the width of the conductor using the following formula:

Width = discs of No.

windingofHeight - (paper insulation thickness) – (wooden block

thickness) …………………………………………………………………….. (3.3)

Usually, we take paper insulation thickness as 0.5mm and wooden block of

thickness 2.8mm.

Width of conductor = discs of No.

windingofHeight - 0.5 – 2.8 ……………………..(3.4)

3. The width of the conductor should be between 6 to 17mm. These constraints are

due to the required mechanical properties of the conductor.

3.1.7 THICKNESS OF THE CONDUCTOR

The thickness of the conductor is selected between 1.7 to 3 mm. Again these restrictions

are due to the required mechanical properties of the conductor. This value of thickness

decides the number of parallel paths in the windings hence the designer can choose any

value as per the parallel path requirements.

Fig 3.1

0.25mm 0.25mm

0.25mm

0.25mm

0.25mm 0.25mm

0.25mm

0.25mm

Paper insulation

Wid

th o

f co

nduct

or

Thickness of conductor

3.1.8 NUMBER OF PARALLEL CONDUCTORS

The number of parallel paths in the given winding is calculated as follows:

1. The maximum current that the winding needs to carry is calculated.

2. The maximum current density is chosen as 3 A/mm2

3. The required cross-sectional area to carry the maximum winding current is

calculated.

2A/mm 3 ofdensity current maximum

windingby the carriedcurrent maximumarea Required ……………...(3.5)

4. The number of parallel paths are calculated by dividing this required area by the

area of one conductor.

thicknessconductor widthconductor

area Requiredpaths parallel ofNumber

………(3.6)

5. The number of parallel paths calculated by above formula is approximated to

the next integer. E.g. if we get number of parallel paths as 3.3 we approximate it

to 4.

Fig 3.2

1 2 3 4

3.1.9 CALCULATION OF %Z BASED ON DIMENSIONS

In the impedance derivation part we have seen that %Z is calculated from the formula:

ATDturnVoltsH

turnsAmperefZ

eq )/(

)(1048.2% 5 ……………………………………... (3.7)

where,

2cmin is ATD

and cmin is eqH

The term ATDis nothing but area under the ampere-turn diagram. This area is

different for maximum, minimum and normal (100%) tap positions.

For maximum tap position,

8).......(3......................................... )1428.0(3

1 )1428.03(

3

1

)1428.01428.01(3

1)111(

3

1

3

1

2

33

2

2

2211

22

11

DTDT

DTDTDTATD

gg

gg

where,

1T - Width of LV winding

1D - Mean diameter of LV winding

2T - Width of HV winding

2D - Mean diameter of HV winding

3T - Width of tapping winding

3D - Mean diameter of tapping winding

1gT - Width of air gap between LV and HV winding

1gD - Mean diameter of air gap between LV and HV winding

2gT - Width of air gap between HV and tapping winding

2gD - Mean diameter of air gap between HV and tapping winding

For minimum tap position,

112211 )(3

1gg DTDTDTATD …………………………………………… (3.9)

Similarly for normal tap,

..(3.10).................................................. )21.0(333

1 )21.03(

223

1

)21.01.01(223

1)111(

113

1

113

1

DTg

Dg

T

DTg

Dg

TDTATD

The Ampere-Turns and Voltage per turn are calculated by multiplying LV current and

LV turns.

3.1.10 COPPER LOSS AND NO LOAD LOSS (CORE LOSS) CALCULATIONS

Mainly, there are two types of losses in a transformer viz. core losses (No-load) losses

and copper losses.

Core (No-Load) Losses:

Transformer no-load losses occur due to flow of main flux in the core.

The core loss of a transformer depends upon grade of steel, frequency, flux

density, type & weight of core. Since all these things remain constant once the

transformer is designed, the core losses are practically constant at all the loads.

To calculate the core loss, we find the volume of the core. Area of core is

already fixed. We fix the length of core such that there is sufficient gap

(=10mm) between 2 HV windings of adjacent phases. Accordingly, volume is

calculated.

Weight of the core is calculated by multiplying the volume of the core by

density of the core material. Core losses are given as 1.6 watts per kg weight of

the core. Accordingly, total core losses are computed.

Copper Losses:

Load losses comprise of ohmic losses and eddy-current losses. But eddy-current

losses are about 10% of ohmic losses. Hence generally they are neglected.

The load-loss of a transformer is a function of temperature and is generally

expressed at a reference temperature of 75 degree C.

The resistance R of winding at 75 degree C is calculated as

R = ρNL/A………………………………………………………………………… (3.11)

R = resistivity of conductor at 75 degree C

L = length of winding

N = number of turns

A = area of conductor.

Area of conductor = wt……………………………………………………………. (3.12)

where,

t = Thickness of conductor

w = Width of conductor

Now the length of winding is calculated as follows:

Consider a winding with „m‟ number of discs, „n‟ number of turns per disc, and

„p‟ number of parallel paths. Now, mean thickness of one such conductor = n *

p * (0.5+ t) / 2, where the thickness of paper insulation is 0.5 mm.

Hence, length of one such conductor per disc is

2Π[(distance between core centre & LV winding)+{n x p x (t + 0.5)/2}]…… (3.13)

Hence total length of all such conductors throughout all the discs is

m*n* Π *length of one conductor…………………………………………(3.14)

But there are „p‟ parallel paths of conductors. Hence while calculating the

resistance of winding, we divide this total length by p.

Once the resistance of a winding is calculated, ohmic loss for that winding is

calculated by multiplying the value of resistance by the square of the current

through the respective winding. The procedure is same for LV, HV and tapping

windings.

In case of minimum tap, the winding resistance is small as compared to

maximum tap. But the current is maximum. Hence copper loss is more in case

of minimum tapping than maximum tapping.

The above procedure gives the per phase copper loss. Hence in case of 3-phase

transformer, the total copper loss is calculated by multiplying the above value

by 3.

The total losses in a transformer are the sum of the above two losses. Increase in losses

implies reduction in efficiency and hence increase in cost. Increment of 1kW in cost of

transformer by Rs. 1 lakh

Increment of 1kW in No-load losses implies increment in cost of transformer by Rs. 5

lakh.

3.1.11 CALCULATION OF RADIAL STRESS ON HV WINDING

1. Total copper loss is calculated for HV winding. From this we calculate the

copper loss per phase , RP .

2. We have fixed the winding height in the beginning of the design.

3. Per unit impedance is calculated.

Using the above data we calculate the radial stress in HV winding is calculated as

22 /*)2(*94.4 puwRavg ZHPk …………………………………………... (3.15)

where,

k is the asymmetry factor considered to take into account symmetry.

3.1.12. CALCULATION OF AXIAL STRESSES ON INNER AND OUTER

WINDING

For an uniform ampere-turn distribution in windings with equal heights (ideal

conditions), the axial forces due to the radial leakage field in both inner and outer

winding are given by

)**/()*8.50( fHZSF wpua …………………………………………… (3.16)

where,

S =Rated power per limb in kVA

Hw = Winding height in meters

Zpu = Per unit impedance

f = frequency in Hz

In the absence of detailed analysis, it can be assumed that 25 to 33% of force is taken by

the outer winding, and the remaining 75 to 67% is taken by the inner winding.

For inner winding, we approximate the respective percentage to 70%

Hence,

Axial forces on inner winding = 70 % of total axial forces

For outer winding, we approximate the above percentage to 30 %

Hence,

Axial forces on outer winding = 30 % of total axial forces

Calculation of number of spacers

The spacers are inserted to provide the necessary strength to the winding against the

radial forces.

The minimum number of spacers or supporters provided is given by

)/*12(*)/( EtDN avgms ………………………………………………….(3.17)

avg = the average value of the compressive stress (in an entire disc winding)

mD = mean diameter of the winding.

t =thickness of the conductor

E =modulus of elasticity of conductor material

According to standard specifications, the number of spacers are always kept in a

multiple of 4, 8 or 12.

Number of blocks

Number of blocks inserted between the two adjacent discs is always equal to the

number of spacers inserted.

Calculation of width of a block

The total width of all the blocks should be approximately 30 % to 35 % of the

circumference of the corresponding winding, which is calculated by considering the

mean diameter of the winding.

Mathematically,

Width of a block = (30 % of circumference of the winding)/ (no of blocks)………(3.18)

According to international standards,

Widths of a block commercially available are 35 mm, 45 mm and 55 mm.

After calculating width of the block, block of closest dimensions is selected.

Calculation of axial stress

Axial forces acting on each block are obtained by using axial force calculated before

and area of all the blocks.

Mathematically this can be shown as

Axial stress = (Axial force on the blocks)/ (Total area of blocks)………………... (3.19)

The procedure for calculating the axial stress is same for LV, HV and tap winding.

CHAPTER 4

DESIGNED TRANSFORMER

_____________________________________________________________________

4.1 PROBLEM STATEMENT

Design a transformer having following parameters:

Power rating 25 MVA

Voltage ratio 33/11 kV with +5% to -10% regulation in the step of 1.5%

Vector group 11DYN

Operating frequency 50Hz

No load loss 20kW

Copper losses 80 kW at 25 MVA

%X = 12%

Maximum current density in any conductor should be less than 2A/mm 3

Maximum flux density in the core = 1.7 T

Tolerance on impedance %10 .

Tolerance on losses and current density: no positive tolerance.

Radial stress on windings should be less than 80 MPa

Axial stress on the insulation block should be less than 10 MPa

4.2 PARAMETERS OF DESIGNED TRANSFORMER

1. Current in LV and HV:

First we calculate the current in both LV and HV winding. Since the HV winding is star

connected we take the phase voltage for current calculation.

(4.3).............................................................................57.757113

MVA 25 current

(4.2)......................................................................37.437053.193

MVA 25 current HV

(4.1) ................................................................................ 053.193

33kV voltageHV

AkV

LV

AkV

kV

2. Core cross-sectional area:

Then we fix the core cross-sectional area and hence the core diameter. For the given

power rating, the core diameter is usually selected between 375mm to 450mm.

Core area =141500 2mm ……………………………………………………………. (4.4)

Core diameter = 424.43 mm…………………………………………………………(4.5)

3. Number of turns in LV and HV:

Using the equation,

E = 4.44fBmAN…………………………………………………………………… (4.6)

Putting f = 50 Hz, A = 141500 2mm , Bm = 1.7 T, we calculate LV and HV number of

turns.

LV number of turns = 205.98……………………………………………………... (4.7)

HV number of turns = 356.78…………………………………………………….. (4.8)

The number of turns is approximated as follows:

Approximated LV turns = 206…………………………………………………… (4.9)

Approximated HV turns = 357………………………………………………….... (4.10)

Since it is required to have voltage regulation on HV side from 90% to 105% we

calculate the maximum and the minimum number of turns corresponding to these

voltage levels.

(4.11) .....................................................................322....... turnsHV minimumApprox

375 turnsHV maximumApprox

321.1 turnsHV minimum Calculated

374.62 turnsHV maximum Calculated

Since we are using linear tapping, in the HV winding there are only 322 turns

(corresponding to 90% voltage) and the rest of the turns are present in tapping winding.

4. Number of discs in LV and HV:

We fix the number of turns per disc in both LV and HV. Also, as explained in the

procedure, we add around 4-5 more discs to the calculated value to account for

dropping of turns at the crossover during actual winding process. The fact that it is

beneficial from manufacturing point of view that the number of discs should be even is

considered.

For LV winding:

Number of turns per disc = 2

Number of discs = 206/2 = 103

Actual number of discs = 105 …………………………………………………… (4.12)

For HV winding:


Number of discs = 322/3 = 107.3

Actual number of discs = 110 …………………………………………………… (4.13)

Tapping winding:

Tapping winding turns = 375-322 = 53


Number of discs = 53/3 = 17.66

Actual number of discs = 20……………………………………………………… (4.14)

5. Height of core structure:

As explained in the design procedure, we select the height of the transformer as

2450mm. This height satisfies all the constraints present on the height of core.

Height of core structure=2450mm

Height of core window = 2450- 2×core diameter

= 2450-2×424.42

= 1601.14 mm………………………………………… (4.15)

We keep a gap of 100 mm on both side of the winding so as to maintain the oil flow.

Therefore the height of the LV and HV winding is calculated as follows:

Height of LV and HV winding = 1601.14 – 200 =1401.14 mm……………….. (4.16)

6. Width of conductor:

As explained in the design procedure we calculate width of the conductor of LV and

HV as follows:

9.918mm2.80.5106

1401.142.80.5

LVin discs of no

windingofheight conductor LV of Width

9.437mm2.80.5110

1401.142.80.5

HVin discs of no

windingofheight conductor HV ofWidth

Similarly,

…………….(4.17)

7. Thickness of conductor and number of parallel paths:

We select the thickness of conductor between 1.7 – 3 mm.

Thickness of LV winding = 2.31 mm

(4.19) 8 HVin paths parallel ofnumber Actual

005.7

45.2437.93

37.437

thicknessconductor widthconductor A/mm 3 ofdensity current maximum

windingHVby carriedcurrent maximum HVin paths parallel ofNumber

mm 2.45 windingHV of Thickness

12 LVin paths parallel ofnumber Actual

4.18)( 022.11

31.2918.93

57.757

thicknessconductor widthconductor A/mm 3 ofdensity current maximum

windingLVby carriedcurrent maximum LVin paths parallel ofNumber

2

2

8. Tapping winding:

We carry the dimensions of HV to tapping winding.

Width of tapping winding conductor = 9.437mm

Thickness of tapping winding conductor = 2.45mm……………………………. (4.20)

For this design, we have 8 parallel paths in HV winding. For the tapping winding, it is

not desirable to have tapping on 8 parallel paths together.

Hence we divide 8 parallel paths in 4 parts and take 2 parallel paths together. We

construct 20 discs with 3 turns per disc and 2 parallel paths i.e. 2 parallel conductors.

Then we construct 3 more such sets of 20 discs each. There is 2.8 mm wooden block

between two consecutive discs.

Total height of taping winding

(4.21) .......................4.1018

)8.25.043.9(204

mm

9. Calculation of %Z based on dimensions:

We calculate the %Z for three positions of tapping viz. Maximum tap, Minimum tap

and Normal tap.

1T - Width of LV winding

1D - Mean diameter of LV winding (ID for LV+1.5

1T )

2T - Width of HV winding

2D - Mean diameter of HV winding (ID for HV+1.52

T )

3T - Width of tapping winding

3D - Mean diameter of tapping winding (ID for tapping+1.5

3T )

1gT - Width of air gap between LV and HV winding

1gD - Mean diameter of air gap between LV and HV winding

(ID for air gap between LV and HV winding+1.51gT )

2gT - Width of air gap between HV and tapping winding

2gD - Mean diameter of air gap between HV and tapping winding

(ID for air gap between HV and tapping winding+1.52gT )

For the given design we take the gap between core and LV winding as well as the gap

between LV and HV winding as 20 mm whereas we take the gap between HV and

tapping winding as 10 mm.

ID for LV = core diameter + 2(gap between core and LV winding)

= 424.42 + 2(20)

= 464.42 mm………………………………………………………... (4.22)

(4.23) .............. 44.67

31.2212

0.5)conductor of (thicknessdiscper turnspaths parallel ofnumber 1

mm

T

1D = 464.42 + 1.5×67.44

= 565.59 mm………………………………………………………………. (4.24)

Similarly we calculate rest of the terms and get the following values:

2

T = 70.8 mm 3T = 17.7 mm 1gT = 20 mm

2gT = 20 mm

1gD = 629.3 mm 2gD = 795.9 mm

2D = 745.5 mm

3D = 827.45 mm

Following diagram explains all the dimensions:

Fig 4.1

LV

HV T

A

P

20mm 67.44mm 20mm 70.8mm 20mm 17mm

1018.34mm

1401.14mm

For maximum tap position,

Fig 4.2

(4.25) mm 46028.27

)1428.0(3

1 )1428.03(

3

1

)1428.01428.01(3

1)111(

3

1

3

1

2

2

33

2

2

2211max

22

11

DTDT

DTDTDTATD

gg

gg

LV

HV

TAP

1

0.1429

NI

Distance

For minimum tap position,

Fig 4.3

(4.26) mm 42894.63

)(3

1

2

2211min 11

gg DTDTDTATD

LV

HV

TAP

Distance

NI 1

Similarly for normal tap,

Fig 4.4

(4.27) mm 44840.28

)21.0(333

1 )21.03(

223

1

)21.01.01(223

1)111(

113

1

113

1

2

DTg

Dg

T

DTg

Dg

TDTATD

LV

HV

TAP

1

0.1

NI

Distance

Then we calculate the %Z from the following formula:

ATDturnVoltsH

turnsAmperefZ

eq )/(

)(1048.2% 5 …………………………………… (4.28)

where,

2cmin is ATD

and cmin is eqH

Ampere-turns = 156060.6 AT

Volts/turn = 53.39 V ……………………………………………………………… (4.29)

For %Z min,

2

2min cm10

42894.63 ATD

cm9640.010

14.1401

0.9640 factor Rogowsky

eqH

Putting these values, we get,

695.10% min Z …………………………………………………………………. (4.30)

For %Z max,

2

2min cm10

46028.27 ATD

cm9577.010

14.1401


eqH


4.11%max

Z ………………………………………………………………….. (4.31)

For %Z ,

2

2cm

10

44840.28 ATD

cm9577.010

14.1401


eqH


10.11% Z ……………………………………………………………………… (4.32)

10. Width of transformer

The horizontal width of transformer can be calculated using the dimensions of windings

used to determine the parameters of the ATD (Ampere-turn diagram).

Thus, we get,

Width of transformer arrangement = 2588.92 mm……………………………… (4.33)

11. Core Losses

We wish to determine the core losses of the transformer. We know that losses in the

core material used are 1.6 watts per kilogram weight of material. Thus, we wish to find

the weight of the core. Since we know the density of core material, we can find its

weight by determining the volume of core material.

This volume can be determined by calculating the volumes of the vertical and

horizontal limbs of the core (which are cylinders). Since we know the core diameter, we

can determine the volume of the core structure.

Volume of core = 1.29*109 mm

3

Density of core material = 7.65*106 kg/mm

3

Weight of core structure = Volume*Density

= 9689.48 kg ………………………………………….(4.34)

Thus,

Core losses = 9689.48*1.6

= 15.79 KW ………………………………………………………. (4.35)

This value of core losses is well within the limits prescribed for the design (20 KW).

12. Copper Losses

Now, let us determine the copper losses in this transformer.

Copper Loss = (Current)2 x Resistance ……………………………………………(4.36)

For this, we must determine the resistance of the windings in the transformer. This can

be determined using the formula,

Resistance = (ρl)/a

where,

ρ = Resistivity of conductor material (copper) = 1.68x10-8

Ω-m

l = Length of conductor

a = Cross-sectional area of conductor

Hence, we must determine the length of winding. We compute the LV and HV winding

lengths by considering the circumference of the winding. We have used the disc type of

winding for both LV and HV windings. Also, we have used many conductors in parallel

to account for the current density, in the case of each winding. Hence, we evaluate the

length of one conductor in a single disc of the winding as the middle conductor in the

winding assuming that all conductors in one disc have the same length. We then

multiply this length with the number of conductors in each disc and then the total

number of discs, to obtain the total length of conductors for that winding. We can then

obtain the resistance of the winding and further, the copper losses of the transformer

using the pre-stated formulae.

Length of LV winding = 344253 mm

Resistance of LV winding = R1 = 0.021 Ω

Length of HV winding = 736283 mm

Resistance of HV winding = R2 =0.067 Ω

Length of tapping winding = 158094 mm

Resistance of tapping winding = R3 =0.014 Ω

Copper losses in LV winding = (ILV)2 * R1

= 36.21 KW ……………………………………. (4.37)

For HV winding, we will have three important values of losses, one for the normal

100% tap conditions and two others, which are limits of losses for the maximum and

minimum tapping conditions.

Copper losses in HV winding for normal tap = (IHV)2 * [R2 + (0.333*R3)]

= 43.87 KW……………………. (4.38)

Copper losses in HV winding for minimum tap (90%) = (IHV)2 * R2

= 47.37 KW …………. (4.39)

Copper losses in HV winding for maximum tap (105%) = (IHV)2 * [R2 + R3]

= 42.29 KW………... (4.40)

Total copper losses of transformer,

For normal tap = 80.09 KW

For minimum tap = 83.59 KW

For maximum tap = 78.51 KW ………………………………………………… (4.41)

Since the transformer will usually be operated at normal tap, the limit on the copper

losses (80 KW) can be restricted to the normal tapping condition alone.

13. Cost of core and winding materials:

Further, we shall determine the costing of the transformer. We must compute the cost of

core as well as the cost of conductors used in order to obtain a rough estimate of the

material cost of the transformer.

a. Cost of core material

We have calculated the weight of core earlier. Using the market cost of core

material as Rs.165/kg, we get,

Cost of core material = Rs.16,28,465 ………………………………………... (4.42)

b. Cost of conductor material

We must also determine the weight of conductor material. This can be done by

finding the volume of the conductor material used (since we know density of

copper). The volume can easily be found out since we know cross-sectional area of

conductor in each winding and also its length.

Volume of LV conductors = 94.64*106 mm

3

Volume of HV conductors = 136.19*106 mm

3

Volume of tapping conductors = 29.22*106 mm

3

Total volume of conductors = 260.06*106 mm

3

Density of copper = 8.96*106 kg/mm

3

Weight of conductors = Volume*Density = 2330.17 kg

Cost of copper = Rs.200/kg

Cost of conductor material = Rs.4,66,034…………………………………….. (4.43)

Total cost = Cost of core material + Cost of conductor material

= Rs.20,94,500 ………………………………………………………... (4.44)

14. Radial stresses on windings:

We have to ensure that the transformer that has been designed can withstand the forces

imparted on the winding in the case of the occurrence of a short circuit. For this, we

consider an asymmetry factor of 1.8 for the transformer. Since we know the copper loss

and height for each winding and the percent impedance of the transformer, we can

easily determine the radial stress. These values are found to be within permissible

limits.

Radial stress on LV winding = 21.48 MPa

Radial stress on HV winding = 26.01 MPa………………………………. (4.45)

15. Number of spacers:

Since there is a compressive radial force on LV winding, we calculate number of

spacers using the value of stress on inner winding. The same number of spacers is used

for HV winding as well, since there is no compressive force on it.

Number of Spacers = 11.85 ≈ 16………………………………………………... (4.46)

The number of spacers should always be taken as multiples of 4. The reasons as to why

this number is rounded off to 16 are specified during discussion of axial stresses.

16. Axial stresses on windings and insulation blocks:

We always design a transformer to withstand axial forces on winding equal to one-third

of the maximum value of force. Now, since we know the value of power rating of

transformer, height of winding, percentage impedance of transformer and frequency of

voltage to be supplied, we can easily determine the total axial force on winding. We

assume that 70% of this force is applied to LV winding while the remaining 30% is

applied to HV and tap windings.

Total axial force on windings = 533273 N

Axial force on LV winding = 373291 N

Axial force on HV and tap windings = 159982 N ……………………………. (4.47)

Every insulation block between windings is attached to its corresponding vertical spacer

for ease in assembly. Hence the number of spacers is equal to the number of insulation

blocks.

Also, we design these blocks such that they cover only 30-35% of the area underneath

the winding. This is to allow enough area for the oil to pass through over the conductors

to cool the winding.

Firstly, we take into account 30% of the circumference of each winding, which is taken

around the mean radius of the winding. (This will account for 30% of the area since

thickness for winding and thickness for block is same)

30% of circumference along mean radius of LV winding = 501.27 mm

30% of circumference along mean radius of HV winding = 669.26 mm

30% of circumference along mean radius of tap winding = 790.37 mm

…………………… (4.48)

Insulations blocks are available with widths of 35, 45 and 55 mm only. Hence this

constraint is imposed on their selection. We divide the circumference obtained with the

number of insulation blocks to get the desired width of block. This width is then

rounded off to the nearest specification of block.

Calculated width of insulation block for LV winding = 31.329 mm

Calculated width of insulation block for HV winding = 41.828 mm

Calculated width of insulation block for tap winding = 49.39 mm

……………………… (4.49)

Using this and knowing the thickness of windings, we calculate the area of block under

the winding and calculate the stress on the insulation blocks as,

Stress = (Axial force on winding) / (Total Area of blocks under winding) ………. (4.50)

According to previous calculations, the number of spacers should have been rounded

off to 12. However, such a selection in the design causes the stresses on insulation block

to rise above the permissible limit of 10 MPa (even for the maximum block width of 55

mm). Hence, we choose number of spacers as 16. The width of the block is dependent

on two factors; area coverage under the winding, and the stress on block for a particular

width.

Area of block under LV winding = 0.0377 m2

Area of block under HV winding = 0.0396 m2

Area of block under tap winding = 0.0099 m2 ….……………………………… (4.51)

Stress on insulation blocks in LV winding = 9.88 MPa

Stress on insulation blocks in HV winding = 3.36 MPa

Stress on insulation blocks in tap winding = 2.69 MPa ……………………… (4.52)

All winding insulation blocks used are of 35 mm width.

REFERENCES

1. Bergonzi, L., Bertagnolli, G., Cannavale, G., Caprio, G., Iliceto, F., Dilli, B.,

and Gulyesil, O. Power transmission reliability, technical and economic issues

relating to the short circuit performance of power transformers, CIGRE 2000,

Paper No. 12–207.

2. Sollergren, B. Calculation of short circuit forces in transformers, Electra,

Report no. 67, 1979, pp. 29–75.

3. Salon, S., LaMattina, B., and Sivasubramaniam, K. Comparison of assumptions

in computation of short circuit forces in transformers, IEEE Transactions on

Magnetics, Vol. 36, No. 5, September 2000, pp. 3521–3523.

4. Bertagnolli, G. Short circuit duty of power transformers–Zthe ABB approach,

Golinelli Industrie Grafiche, Italy, 1996.

5. Waters, M. The short circuit strength of power transformers, Macdonald and

Co. Ltd., London, 1966.

6. Waters, M. The measurement and calculation of axial electromagnetic forces

in concentric transformer windings, Proceedings IEE, Vol. 101, Pt. II, February

l954, pp. 35–46.

7. Norris, E.T. Mechanical strength of power transformers in service, Proceedings

IEE, Vol. 104, February 1957, pp. 289–306.Short circuit strength: requirements, design

and demonstration, IEEETransactions on Power Apparatus and Systems, Vol. PAS-89,

No. 8, November/December 1970, pp. 1955–1969.

8. Steel, R.B., Johnson, W.M., Narbus, J.J., Patel, M.R., and Nelson, R.A.

Dynamic measurements in power transformers under short circuit conditions,

CIGRE 1972, Paper No. 12–01.

9. Thomson, H.A., Tillery, F., and Rosenberg, D.U. The dynamic response of

low voltage, high current, disk type transformer windings to through fault

loads, IEEE Transactions on Power Apparatus and Systems, Vol. PAS-98, No.

3, May/June 1979, pp. 1091–1098.

10. Saravolac, M.P., Vertigen, P.A., Sumner, C.A., and Siew, W.H. Design

verification criteria for evaluating the short circuit withstand capability of

transformer inner windings, CIGRE 2000, Paper No. 12–208.

11. Kojima, H., Miyata, H., Shida, S., and Okuyama, K. Buckling strength analysis

of large power transformer windings subjected to electromagnetic force under

short circuit, IEEE Transactions on Power Apparatus and Systems, Vol. PAS-

99, No. 3, May/June 1980, pp. 1288–1297.

12. Patel, M.R. Dynamic stability of helical and barrel coils in transformers against

axial short circuit forces, Proceedings IEE, Vol. 127, Pt. C, No. 5, September

1980, pp. 281–284.

13. Patel, M.R. Instability of the continuously transposed cable under axial short

circuit forces in transformers, IEEE Transactions on Power Delivery, Vol. 17,

No. 1, January 2002, pp. 149–154.

14. Madin, A.B. and Whitaker, J.D. The dynamic behavior of a transformer winding

under axial short circuit forces: An experimental and theoretical investigation,

Proceedings IEE, Vol. 110, No. 3, 1963, pp. 535–550.

15. Gee, F.W. and Whitaker, J.D. Factors affecting the choice of pre-stress applied

to transformer windings, IEEE Summer General Meeting, Toronto, Canada,

1963, Paper No. 63–1012.

16. Dobsa, J. and Pirktl, E. Transformers and short circuits, Brown Boveri Review,

Vol. 52, No. 11/12, November/December 1965, pp. 821–830.

17. TRANSFORMERS by BHEL, 1987 Tata - Mcgraw Hill Publications.

project report - final year

Documents