Designing of Distribution Transformer 1

ABSTRACT

Transformer is very important and efficient electrical power equipment. Its

ability to step-up voltages has made huge blocks of electric power transformation

over long distances very economical and its ability to step-down voltages to

convenient low levels has made electric power utilization safe. Depending upon

application for which transformers are put in electric power industry, there are

divided into power transformers, distribution transformers, grounding

transformers, instrument transformers etc.

The ratings and characteristics of these different transformers are therefore

different and hence their design considerations, fabrications and testing procedures

also different. Hence, a thorough study of above aspects and actual problems that

come up at various stages in the manufacture of this important piece of electrical

apparatus in a factory before being marketed is considered worth study.

Currently many factories are engaged in manufacturing distribution

transformers. Three phase 50 Hz distribution transformers up to 100 KVA have

been designed for given ratings.

This work is mainly divided into four chapters.

The first chapter deals with the basics of transformers construction and the

theoretical aspects behind this.

The second chapter deals with the theoretical aspects in designing of a

distribution transformer.

The third chapter deals with complete design of a three phase, 50 Hz, 100

KVA distribution transformers.

The Fourth chapter deals with the C Program illustrating problem

definitions and its results.

By giving the input parameters of a distribution transformers we can obtain

the design specification of that transformer by using a C program.

1

CONTENTS

CHAPTER 1: TRANSFORMER CONSTRUCTION

1.1 Principal of Transformer Action

1.2 Rating of transformers

1.3 Transformer on Load

1.4 Transformer with Resistance & Leakage Reactance

1.5 Transformer Tests

1.6 All-day efficiency

CHAPTER 2: DESIGN THEORY

2.1 Design of Core

2.2 Choice of Flux Density

2.3 Selection of Type of winding

2.4 Design of Insulation

CHAPTER 3: OPTIMUM DESIGN PROCEDURE

3.1 Design for minimum cost

3.2 Design for minimum loss (or) maximum efficiency

3.3 Core Design

3.4 Window Design

3.5 Yoke design

3.6 Overall Dimensions

3.7 Windings

3.8 Resistance

3.9 Leakage Reactance

3.10 Regulation

3.11 Losses

3.12 Efficiency

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3.13 All day Efficiency

3.14 No-Load Current

3.15 Tank

3.16 Cooling system

CHAPTER 4: RESULTS AND DISCUSSIONS

CHAPTER 5: CONCLUSION

CHAPTER 6: C-PROGRAM

CHAPTER 7: REFERENCES

3

TRANSFORMER CONSTRUCTION

4

CHAPTER-1

TRANSFORMER CONSTRUCTION

There are two types of general transformers, the

core type and the shell type. These two types differ from each other by the

manner in which the windings wound around the magnetic core.

The magnetic core is a stack to thin silicon-steel

laminations about 0.35 mm thick for 5o Hz transformers. In order to reduce

the eddy current losses, nearly all transformers have their magnetic core

made from cold-rolled grain-oriented sheet-steel (C.R.G.O). This material,

when magnetized in the rolling direction, has low core loss and high

permeability.

In the core-type, the windings around a considerable

part of steel core as shown in fig 1.1 (a). In the shell-type, the steel core

surrounds a major part of the windings as shown in fig 1.1 (b). For a given

output and voltage rating, core-type transformer requires less iron but

more conductor material as compared to a shell-type transformer. The

vertical portions of the core are usually called limbs or legs and the top and

bottom portions are called the yoke. This means that for single- phase

transformers, core-type has two-legged core, where as shell-type has

three legged core.

In iron-core transformers, most of the flux is confined to high

permeability core. There is, however, some flux that leaks through the core

legs and non-magnetic material surrounding the core. This flux, called

leakage flux, links one winding and not he other. A reduction in this

5

leakage flux is desirable as it improves the transformer performance

considerably. Consequently, an effort is always made to reduce it. In the

core-type transformer, this is achieved by placing half of the low voltage

(L.V) winding over one leg and other half over the second leg or limb. For

the high voltage winding also, half of the winding is over one leg and the

other half over the second leg, fig 1.1(a). L.V winding is placed adjacent to

the steel core and H.V winding outside, in order to minimize the mount of

insulation required.

In the shell type transformer, the L.V and H.V windings are

wound over the central limb and are interleaved or sandwiched as shown

in fig 1.1 (b). Note that the bottom and top L.V coils are of half the size of

other coils.

In core-type transformer, the flux has a single path around the

legs or yokes, fig 1.1 (a). In the shell-type transformer, the flux in the

central limb divides equally and returns through the outer two legs as

shown in fig 1.1(b)

6

L.V. Winding

H.V-Winding

1.1 (a) CORE TYPE TRANSFORMER

Yoke

Gaps

Limb (or)leg 1.1 (b) SHELL TYPE TRANSFORMER

7

One type of laminations for the core and shell type of

transformers is illustrated in Fig 1.2 (a) and (b) respectively. The steel core

is assembled in such a manner that the butt joints in adjacent layers are

staggered as illustrated in fig 1.2 (c). The staggering of the butt joints

avoids continuous air gap and, therefore, the reluctance of the magnetic

circuit is too increased. At the same time, a continuous air gap would

reduce the mechanical strength of the core and, therefore, the staggering

of the butt joints is essential

BUTT JOINTS

1.2 (a) 1.2(b)

8

BUTTJOINTS

BUTTJOINTS

Two adjacent layers for:

(a) Core

(b) Shell type of transformers and

(c) arrangement of Butt Joints in a magnetic core

9

1.1 Principle of Transformer Action:

A transformer works on the principle of electromagnetic

induction. According to this principle, an e.m.f is induced in a coil if it links

a changing flux.

In core-type transformer, half the L.V and H.V winding is on one

limb and the other half is on the second limb. In shell-type transformer, the

L.V and H.V windings are sandwiched. However, for both these type

transformers, the schematic diagram is as shown in fig 1.3. The primary

winding P is connected to an alternating voltage source, therefore, an

alternating current I φ is connected to an alternating voltage, voltage

source, therefore, an alternating current I φ starts flowing through N1 turns.

The alternating mmf N1I sets up alternating flux φ which confined to high

permeability iron path as indicated in fig 1.3, the alternating flux induces

voltage E1 in the primary P and E2 in the secondary S. If the load is

connected across the secondary, a load current starts flowing.

In addition to the secondary winding, there may be a third

winding on the same iron core. The emf induced in the secondary winding

is usually referred to as the emf due to transformer action. Thus the

transformer action requires the existence of alternative mutual flux linking

the various windings on a common magnetic core.

During the transformer construction, first the primary and

secondary windings are wound, then the laminations are pushed through

The coil openings, layer and the steel core is prepared, the laminations are

then tightened by means of clamps and bolts.

10

Low power transformers are air cooled where as large

power transformers are immersed in oil for better cooling. In oil cooled

transformers, the oil serves as a coolant and also as an insulating medium.

For power frequency range of 25 to 400Hz, transformers

are constructed with 0.32 mm thick silicon-steel laminations. For audio

frequency range of 20 to 20,000Hz, iron core with suitable refinements is

used. For high frequencies employed in communication circuits, core is

made up of powdered ferromagnetic alloy. In special cases the magnetic

circuit of transformer is referred to as an air core transformer is primarily

used in radio devices and in certain types of measuring and testing

instruments. Cores made of soft ferrates are also used for pulse

transformers as well as for high frequency electronic transformers.

1.2 Rating Of Transformers:

The manufacturer of transformers fixes a name plate

on the transformers, on which are recorded the rated output, the rated

voltages, the rated frequency etc. of a particular transformer. A typical

name plate rating of a single phase transformer is as follows: 20 KVA,

3300/220V,50Hz. Here 20KVA is the rated output at the secondary

terminals. Note that rated output is expressed in Kilo-volt-amperes(KVA)

rather than in Kilowatts(KW).This is due to the fact that rated transformer

output is limited by heating and hence by the losses in the transformer.

These losses depend on transformer voltage (core loss) and current loss

and are almost unaffected by the load power factor. Consequently the

transformer rated output is expressed in KVA and not in KW. At zero p.f

11

load, a transformer can be made to operate at rated KVA output while

delivering zero power

For any transformer

(Rated input in KVA at the primary terminals)(Cos θ1)

= (Rated output in KVA at the secondary terminals)(Cos θ2)+Losses

Since the transformer at a very high efficiency, losses may

be ignored. Further, the primary p.f.Cos θ2 are nearly equal. Therefore, the

rated KVA marked on the name plate of a transformer, refers to both the

windings, i.e. the rated KVA of the primary winding and the secondary

winding are equal.

1.3 Transformer on Load:

When the secondary is loaded, the secondary current I2

with respect to V2 is determined by the characteristics of the load. Current

I2 is in phase with V2 if load is non-inductive, it lags if load is inductive and it

leads if load is capacitive.

The secondary current set up its own m.m.f (=N2

I2) and hence its own flux Ф2 which is in opposition to the main primary flux

Ф which is due to I0. The secondary flux Ф2 weakens the primary flux Ф

momentarily, hence primary back e.m.f. E 1 tends to be reduced. For a

moment V1 gains the upper hand over E1 and hence causes more current

to flow in primary.

Let the additional primary current be I2’.It is know as load

Component of primary current. This current is antiphase with I2.The

additional primary m.m.f N1I2’ sets up its own flux Ф2’ which is in opposition

to Ф2 (but is in the same direction as Ф) and is equal to it magnitude.

Hence, the two cancel each other out. So, we find that the magnetic effects

of secondary current I2 are immediately neutralized by the additional

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primary current I2’ which is brought into existence exactly at the same

instant as I2.

Hence, wherever the load conditions, the net flux passing

thought the core is approximately the same at no-load. An impartment

deduction is that due to the constancy of core flux at all loads, the core

loss is also practically the same under all load conditions.

As Ф2= Ф2 N2I2=N1I2

I2=(N2/N1)I2 = KI2

13

Hence, when transformer is on load, the primary winding has

two currents in it; one is I0 and the other is I2’ which is anti-phase with I2

and K time in magnitude. The total primary current is the vector sum of I0

and I2’.

In fig1.3(a) are show the vector diagrams for a load

transformer when load is non-inductive and when it is inductive (a similar

diagram could be drawn for capacitive load).Voltage transformation ratio of

unity is assumed so that primary vectors are equal to the secondary

vectors. With reference to fig 1.3(a), I2 is secondary current in phase with

E2 (strictly speaking it should be V2). It causes primary current I2 which is

anti-phase with it and equal to it in magnitude (k=1). Total primary current

I1 is the vector sum of I0 and I2 and lags behind v1 by an angle Ф1.

In fig 1.3 (b) vectors are drawn for an inductive load. Here, I2

lags E2 (actually V2) by Ф2. Current I2 is again anti phase with I2 and equal

14

to it in magnitude. As before I1 is the vector sum of I2 and I0 and lags

behind V1by Ф1.

It will be observed that Ф1 is slightly greater than Ф2 But if we

neglect I0as compared to I2 as in fig 1.3(c) then Ф1= Ф2 more over, under

this assumption. N1I2=N2I1=N1I2; I2/I1=I1/I2=N2/N1=K.

It shows that under full-load conditions, the ratio of primary

and secondary current is constant. This impartment relationship is made

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the basis of current transformer. A transformer which is used with a low-

range ammeter for measuring currents in circuits where the direct

connection of the ammeter is impracticable.

1.4 Transformer with Resistance and Leakage Reactance:

16

The above figure 1.4 shows the primary and secondary windings of a

transformer with reactance taken out of the windings. The primary

impedance is given by

Z1=√ (R12+X1

2)

Similarly second impedance is given by Z2=√ (R22+X2

2)

The resistance and leakage reluctance of each winding is

responsible for some voltage drop in each winding. In primary, the leakage

reactance drop is I1X1(usually 1 or 2%)

Hence V1 = E1 + I1(R1 +j X1) = E1 + I1 Z1

Similarly, there are I2 R2 and I2 X2 drops in secondary which combine with V2

to give E2.

E2= V2+ I2(R2 +j X2)= V2+ I2 Z2

It may be noted that leakage reactance’s can also transferred from

one winding to the other in the same way as resistance.

X2’= X2/K2 and X1’= K2 X1

And X01= X1+ X2’= X1+ X2/K2 and X02= X2+ X1’= X2+ K2 X1

It is obvious that total impedance of the transformer as referred primary is

given by

Z01 =√ (R012+X01

2)

Z02 =√ (R022+X02

2)

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1.5 Transformer Tests

The transformer constants or parameters can be easily determined by two tests (i)

pen-circuit test and (ii) the short-circuit test. These tests are very economical and

convenient, because they furnish the required information without actually loading the

transformer. In fact, the testing of every large a.c.machinary consists of running two tests

similar to the open and short – circuit test of a transformer.

Open – circuit or No-load Test

The main purpose of this test is to determine on-load losses. One winding of the

transformer – whichever is convenient but usually high voltage winding – is kept open

the other is connected to a supply of normal voltage and frequency (Fig.21-29). A

wattmeter W, voltmeter V and ammeter A are connected in the low-voltage winding i.e.

primary winding in the present case. Fig.21-29.

Shows the simplified diagram whereas, Fig.21-30 shows actual connections. With

normal voltage applied to the primary, normal flux will be set up in the core, hence

normal iron losses will occur which are recorded by the wattmeter. As the primary no

load current Io (as measured by ammeter) is small (usually 2 to 10% of rated

load current) Cu loss is negligibly small in primary and nil in secondary (being open).

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Hence, the wattmeter reading represents practically the core-loss under no-load

conditions (and which is the same for all loads as pointed out in Art.21-8).

Sometimes a high resistance voltmeter is connected across the secondary. The

reading of the voltmeter gives the induced e.m.f. in the secondary winding. This helps in

finding the transformation ration K.

Short-circuit or Impedance Test

This is an economical method for determining the following:

(i) Equivalent impedance (Zo1 or Zo2), leakage reactance (Xo1 or Xo2) and resistance

(Rol or Ro2) of the transformer as referred to the winding in which the measuring

instruments are placed.

(ii) Cu loss at full-load (and at any desired load). This loss is used in calculating the

efficiency of the transformer.

(iii) Knowing Zo1 or Z02, the total voltage drop in the transformer as referred to primary

or

secondary can be calculated and hence regulation of the determined.

In this test, one winding – usually the low- voltage winding, is solidly short-

circuited by a thick conductor ( or through an ammeter which may serve the additional

purpose of indicating rated load current) as shown in Fig. 21-31.

A low voltage (usually 5 to 10% of normal primary voltage) at correct frequency

(though for Cu losses it is not essential) is applied to the primary and is cautiously

increased till full-load currents are flowing both in primary and secondary (as indicated

by the respective ammeters).

Since, in this test, the applied voltage is a small percentage of the normal voltage,

the mutual flux Ө produced in the core is also a small percentage of its normal value

(because flux is proportional to the voltage as shown in the e.m.f. equation of the

transformer in Art.21-6). Hence, core losses are very small with the result that the

19

wattmeter reading represents the full-load Cu loss or I2 R loss for the whole transformer

i.e. both primary Cu loss and secondary Cu loss. If Vsc is the voltage required to

calculate rated load current on short circuit, then

Also

1.6 All day Efficiency:

The ordinary or commercial efficiency of transformer is given by

the ratio of output in watts and input in watts. But there are certain types of

transformers whose performance cannot be judged by this efficiency.

Transformers used for supplying lighting and general network i.e.

distribution transformers have their primaries energized all twenty-four

hours, although their secondary’s supply little or no load much of the time

during the day except during the house lighting period. It means that where

as core loss occurs throughout the day, the copper loss occurs only when

the transformer is loaded. Hence, it is considered a good practice to design

such transformers so that core losses are very low. The copper losses are

relatively less important, because they depend on the load. The

performance of such transformer should be judged by all-day efficiency

which is computed on the basis of energy consume during a certain time

period, usually a day of 24 hours.

Output in KWH

All day efficiency= ----------------------------- (for 24 hours)

Input in KWH

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This efficiency is always less than the commercial efficiency of a

transformer.

To find this all day efficiency, we have to know the load cycle

on the transformer i.e. how much and how long the transformer is loaded

during 24 hours. Practical calculations are facilitated by making use of

load factor

21

DESIGN THEORY

22

CHAPTER-2

DESIGN THEORY

2.1 Design of core:

The cross section for core type of transformers may be

rectangular, square or stepped. Shell type transformers use cores with

rectangular cross section.

2.1.1 Rectangular core:

For core type distribution transformers and small power

transformers for moderate and low voltage, the rectangular stepped core

section may be used. The ratio of depth to width of core varies between

1.4 to 2. Rectangular shaped coils are used for rectangular cores.

For shell type transformer width of central limb is 2 to 3

times the depth of the core.

2.1.2 Square and stepped cores:

When circular coils are required for high voltage

distribution and power transformers square and stepped cores are used.

Circular coils are preferred because of the superior mechanical

characteristics. A transformer coil, under mechanical stresses produced by

excessive leakage flux due to short circuit, tend to assume a circular from .

On circular coils, these forces are radial there is no tendency for the coil to

change its shape; on

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rectangular coils the forces are perpendicular to the conductors and tend

to give the coils a circular form, thus deforming it.

With core type transformers of small sizes, simple rectangular core can

be used with either square or rectangular coils. For this purpose the cores

are squares shaped. This circle is know as the circumscribing circle being

large in comparison giving rise to higher copper loss and conductor cost.

With larger transformer cruciform cores which utilize the

space better are used. As space utilization is better with cruciform cores.

The diameter of circumscribing circle is smaller than with square

cores of the same area. Thus the length of mean turn of copper is reduced

with consequent reduction in cost of copper. It should be kept in mind that

two different sizes of laminations are used in cruciform cores, the large

transformers further steps are introduced to utilize the core space which

reduces the length of mean turn with consequent reduction in both cost of

copper and copper loss.

It would seen that we can go on introducing steps with resultant

reduction in cost of winding . However with larger number of steps a large

numbers of sizes

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of laminations have to be used. This results in higher labour charges for

shearing and assembling different types of laminations.

Thus the reduction in winding costs with a certain numbers of

steps has to be balanced with the extra labour cost. The numbers of steps

to be used for particular transformers have to be decided by the about

considerations.

2.2 Choice of Flux Density:

The value of flux density in the core determines the core area.

Higher values of flux density give smaller core area and therefore there is

a saving in cost of iron. Also with the reduction in core area the length of

mean turn of windings is also reduced. Thus there is saving in conductor

costs also. But with higher flux densities the iron loss become high,

resulting in considerable temperature gradient across the core.

High flux density necessitates a large magnetizing current which

contains objectionable harmonics.

The value of flux density to be chosen also depends upon the

service conditions of the transformers. As a distribution transformer has to

e designed for a high all day efficiency, and there fore the value of flux

density should be low in order to keep down the iron loss.

25

The usual values of maximum flux density BM for transformers using hot

rolled silicon steel are:

Distribution transformer-----------1.1to1.35 wb/m2

Power transformer------------------1.25to1.45 wb/m2

Lower values should be used for small rating transformers.

2.3 Selection of Type of Winding:

It is first necessary it select proper types of windings to be used in

the transformer. The design of the winding chosen must be such that the

desired electrical characteristics and adequate mechanical strength is

obtained.

The high voltage winding are usually of the following types;

1) Cylindrical winding with circular conductor

2) Cross over winding with either circular or small rectangular

conductors

3) Continuous disc type winding with rectangular conductors

The cylindrical and the cross-over winding are used for

transformers of ratings upto 1000 KVA and 33 KV. The disc type winding

is used for transformers of higher rating ranging from 200K VA to tens of

MVA and Voltages from 11KV upwards.

26

The low voltage are usually of the following two types:

1) cylindrical winding

2) Helical winding(usually double helical)

But these windings employ rectangular conductors. Cylindrical

windings are used for KVA rating up to 800 and voltages up to 15KV and

sometimes up to 33KV.

2.4 Design of Insulation:

During the processes of power transfer from one circuit to

another; electrical, mechanical and thermal phenomenon take place in a

transformer., the winding voltage produce an electrostatic field in the

dielectric and therefore stress the insulation, the current in the windings

and to mechanical stressing of insulation, finally the losses. In the

transformer produce temperature rise which produce thermal stresses.

Hence, the fundamental considerations in the design of

insulation of transformers may be described as those of arranging core,

windings and insulation to obtain satisfactory electrical and thermal

characteristics during the steady state as well as transient conditions. The

three basic considerations in the design of insulation are various

conditions.

2.4.1 Electrical considerations:

The basic insulation structure is primarily determined from

consideration of the magnitude and nature of voltage which appears

between different parts of the transformer i.e., voltages between individual

27

turns, between coil or layers, between windings and from winding to core

and tank.

The electrical design should also take care of the eddy current losses

in conductors and leakage reactance of windings.

a) Eddy current losses:

The winding should be designed that the stray load losses

small. The stray load loss includes eddy current loss in conductor and also

in tank walls and clamping structure. The conductor should be split into

small strips to reduce eddy current losses in conductor.

b) Leakage Reactance:

A given arrangement of core and winding determines the

leakage reactance of the windings. The leakage reactance is adjusted by

changing the winding configuration and brought within desired limits.

2.4.2 Mechanical Considerations:

The basic mechanical considerations in the design of insulation

are of two types:

a) The insulation must be capable of withstanding the mechanical stresses

imposed on it during manufacturing processes.

b) The insulation must be able to withstand the mechanical stresses which

are developed in the winding due to electromagnetic forces and

mechanical stresses produced under normal conditions of operation are

28

quite small and ordinarily are of minor importance .However, under fault

conditions, particularly dead short circuit, the electromagnetic forces may

be increased several hundred times.

2.4.3 Thermal considerations:

The thermal aspects of design of insulation are determine from

the consideration of insulating materials used, selection of maximum

operating temperatures and types of cooling method employed.

The transformer structures should be such that the losses developed in

the core and winding produce temperature rises in the various parts which

no where exceed the permissible limit both under normal over load/fault

conditions and which, in the interest of economy, approved those limits as

early as possible. The insulation of a transformer is divided into four types:

a) Major insulation

b) Minor insulation

c) Insulation relative to tank

d) Insulation between the phases

a) Major Insulation:

The insulation between windings and grounded core

and insulation between the windings of the same phase is called major

insulation.

b) Minor insulation:

Insulation between different parts of one winding i.e.

insulation between turns, coils and layers etc is called Minor insulation.

c) Insulation relative to tank:

The insulation relative to the tank is called oil barrier

insulator in oil immersed transformers. This insulation consists of oils

29

ducts, barriers and coverings. Partitions of solid insulating materials placed

inside oil ducts are called barriers

d) Insulation between the phases:

There must be provided relative insulation between the

phases to avoid the short circuit between the phases

30

DESIGN PROCEDURE

31

CHAPTER 3

DESIGN PROCEDURE

OPTIMUM DESIGN:

Transformers may be designed to make of following qualities

as minimum.

1) Total volume

2) Total weight

3) Total cost

4) Total losses

Фm

All these qualities vary with r= --------

AT

Thus we say r is a controlling factor for all these qualities.

3.1 Design for minimum cost:

Let us consider a single transformer, whose KVA output is,

Q = 2.22fBmAiKwAwδ X 10-3

=2.22fBmAiAcδ X 10-3

Assuming flux and current densities are constant, the product

Ac Ai is constant for a given transformer.

Let,

Ac Ai = M2

Фm

Now r = --------

AT

32

Фm = BmAi

AT = KwAwδ / 2 = δ / 2

r = 2 BmAi / δ Ac

=> Ai = δ * r / 2 Bm =β (say)

Where β is a function of only as Bm and δ are constants.

From above

Ai = M √ β, Ac = M / √ β

Let Ct = Total cost of transformer active materials

Ci = Total cost of Iron

Ct = Cc+ Ci

Cc = Total cost of conductor

Ct = CI gi li Ai + Cc gc lmt Ac

Where Ci & Cc are specific costs of iron & copper.

Now, Ct = CI gi li Ai M √ β + Cc gc lmt M / √ β

Differentiate Ct with respect to β

d Ct CI gi li Ai M (β)-1/2 - Cc gc lmt M (β)-3/2

-------- = ----------------------- -------------------------

d β 2 2

d Ct

For minimum cost ------ = 0

d β

33

CI gi li = Cc Gc Imt (β)-1

= Cc Gc Imt Ac / AI

CI gi Ii Ai = Cc Gc Imt Ac or Ci = Cc

Hence for minimum total cost, cost of iron must equal to the cost of the

conductor. Now, Gi == Cc

Gc == CI

Knowing the value of iron and conductor ratio of weight of iron to

conductor can be determined. This can be substituted in equation, to

determine core area which gives minimum cost of transformer

1) For minimum cost, Gi / Gc= Cc / CI

2) For minimum volume, Gi / Gc = gi / gc

3) For minimum weight, Gi = Gc

4) For minimum losses (or) maximum efficiency,

Iron loss = I2 R loss

P I = X2 P c

3.2 Design for minimum loss or maximum efficiency:

Total losses at full load = Pi + Pc

At any fraction `X’ of full load, total losses are,

= Pi + X2 Pc

If Q is out put at full load, out put at fraction X is XQ

XQ

Efficiency at output xQ, nx =------------------

XQ+Pi + X2 Pc

34

dnx

This maximum if, ----------------------- = 0

dx

Differentiating with respect to `X’

dnx (XQ + Pi + X2 Pc)Q – XP (Q + 2XPc)

------ --------------------------------------------------

dx (XQ + Pi + Pc) 2

For maximum n Nr = 0

(XQ + Pi + X2 Pc)Q - XP (Q + 2XPc) = 0

Pi = X2 Pc

So, maximum efficiency is obtained when variable losses are equal to

constant losses.

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DESIGN

Design of 100 KVA-3-Phase, 50 Hz, Distribution transformer:

The transformer of above rating is designed, which is based on the

theoretical assumptions. But the transformer of the same rating which is

manufactured in the factory is having different values of dimensions and

other things. The errors are due to the practical considerations and

assumptions.

Design Notations :

Let,

m = Main flux, wb

B m = Maximum flux density, wb/m2

= Current density, A/m2

Agi = Gross area,m2

Ai = Net core area,m2

= Stacking factor * Gross core area

Ac = Area of copper in window, m2

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Aw = Window area, m2

D = Distance between core centers, m

d = Diameter of circumscribing circle, m

Kw = Window space factor

f = Frequency, Hz

Et = E.M.F. per turn, V

Tp,Ts = Number of turns in primary and

Secondary windings respectively

Ip,Is = Current in primary and secondary

Windings respectively, A

Vp,Vs = Terminal voltage of primary and secondary

Windings respectively, V

ap,as = Area of conductor of primary and

Secondary windings respectively, m2

37

Ii = Mean length of flux path in iron, m

Lmt = Length of mean turn of transformer windings, m

Gi = Weight of active iron, Kg

gc = Weight of copper, Kg

gi = Weight per m3 of iron, Kg

Gc = Weight per m3 of copper, Kg

Pi = Loss of iron per Kg, W

Pc = Loss of copper per Kg, W

Output Equation-VOLT PER TURN: Considering the output of one phase

KVA rating of 1-phase:

Q = VpIp 10 -3 = (4.44).f. m.Tp.Ip.10 -3

= (4.44).f. m.AT. 10 -3

The ratio of m /AT is a constant for a transformer of a given type, service

and 4.44 method of construction.

Let m /AT = r, where `r’ is a constant.

But we have individual voltage E = (4.44).f. m.T

38

Voltage per turn Et = E/T = (4.44).f. m

Q = (4.44).pim.f.AT.10-3

= 4.44 f. m.( m /r).10-3

= 4.44. m 2. {f/r}. 10-3

m = [r.103/4.44]1/2 = √Q

Voltage per turn Et = 4.44 f m

= 4.44f [r.103 / 4.44f]1/2. √Q

= k. √Q

where k = √(4.44f r.10) =√ [4.44f. m 103 / AT]

3.3 Core Design:

Taking the value of k = 0.35

Voltage per term Et = 0.35 100 =3.5V

Et = 4.44m

Flux (m) = Et/4.44f = 3.5/(4.44x50) = 0.01576 wb

Taking Bm = 1.2 wb/m

39

Net iron area = Ai = 0.01576 / 1.2 = 0.01313 = 131.2 cm

Using a cruciform core Ai = 0.56 d2

d = √ (131.3 / 0.56)

Fig. 1.2

a = 0.85 d = 13.0 cm

b = 0.53 d = 8.1 cm

3.4 Winding design:

Window space factor Kw =10/ (30 + kV)

= 10 / (30+11) = 0.243

Rating of a 3-Phase transformer in KVA,

Q = 3.3f.Bm. Kw.Aw.Ai.10-3

Taking del () = 2.3 Amps/mm2

100 = 3.33 x 50x1.2x2.3x106x0.24xAw x 0.013 x 10-3

Aw= 0.0679 M2 = 679 cm2

Taking the ratio of height to window as 2.5

40

BUTTJOINTS

Width of window, Ww = 16.56 cm

Height of window, Hw= 2.5 x 16.5 = 41 cm

Window area = 41 x 16.56 = 679 cm2

Distance between adjacent core centres:

D= Ww + d = 16.5 + 15.3 = 31.8 cm

3.5 Yoke Design:

The area of yoke is taken as 1.3 times that of limb.

Flux density in yoke = 1.2/1.3 = 0.92 Wb/m2

Net area of yoke = 1.2 x 131.3 = 157.5 cm2

Gross area of yoke = 157.5 / 0.9 = 175 cm2

Taking the section of yoke as rectangular

Depth of yoke, Dy = a = 13 cm

Height of yoke Hy = 174/13 = 13.5 cm

3.6 Overall dimensions:

Height of frame H = Hw + 2 Hy

= 41 + 2 x 13.5 = 67.5 cm

Length of frame W = 2D+a

= 2 x 32 + 13.5 = 76.6 cm

41

Depth of frame a = 13.0 cm

Fig 3.1 gives the various dimensions

3.7 Windings:

L.V. Winding:

Secondary line voltage = 433 v

Secondary phase voltage = 433 / √3 = 250 v

Number of turns Ts=Vs /Es = 250 / 3.5 = 71.426 z 72

Secondary phase current = Is = 100x x 1000 / 3 x 250

42

= 133.3 A

using a current density of 2.3 Amps/ mm2

area of secondary conductor = as = 133.3 / 2.3 = 58 mm2

using bare conductor of 9.5 x 6 mm

area of bare conductor = 57 mm2

current density in secondary windings,

dels = 133.3 / 57 = 2.33 Amps/ mm2

using paper tape of 0.25 mm over bare conductor

dimensions of insulated conductor = 10 x 6.5 mm 2

using two layers for the winding,

turns per layer = 72 / 2 = 36

Maximum voltage between layers = 2 x turns per layer x Et

= 2x 36 x 3.5 = 252 volts

this is below the limit of 300v

since the helical winding is used

space has to be provided for 36 + 1 = 37 turns

along the axial depth

Axial depth of LV winding les = 37 x axial depth of conductor

= 37 x 10 = 370 mm = 37 cm

The height of window is 41 cm. This leaves a clearance of

( 41 -37) / 2 = cm

On each of winding conductor 37 x 10.5 = 38.85

Insulation bracing etc. = 2.15

--------

= 41.00 cm

------------

using 0.5 mm pressboard cylinders between layers

radial depth of LV winding

bs = (No. of layers x Radial depth of conductor) + insulation

43

between layers

= (2 x 6) + 0.5 = 12.5 mm = 1.25 cm

diameter of circumscribing circle, d = 15.3 cm

using pressboard wraps of 1.5 mm thick as insulation between

L.V. winding and core.

Inside dia of L.V. winding = 15.3 + 0.15 = 15.5 cm

Outside dia of L.V. winding = 15.5 + 12.5 = 16.75 cm

H.V. winding:

Primary voltage = 11000v

Primary voltage = 11000v

No. of turns per phase = 11000 x 71.426 / 250 = 3143

Taking 17 layers

Turns per layer = 3143 / 17 = 185

HV winding phase current Ip = 100 x 1000 / (3 x 110000)

= 3.03 Amps

Taking current density = 2.4 Amps / mm2

Area of HV conductor = 3.03 / 2.4 = 1.26 mm 2

Dia of conductor d =√ [1.26 x 4 / (22/7)] = 1.3 mm

Using cotton covered conductors

Insulated dia = 1.70 mm

Modified area of HV conductor = (3.142 / 4) (1.300)2

= 1.32 mm2

Axia dept = No of turns per layer x dia of insulated conductor

= 186 x 1.70 = 316.2 mm =n 31.62 cm

Axial depth of HV winding = 31.62

Insulation basing etc. = 9.38

------------

41.00 cm

-------------

44

the clearance of 9.38 cm with 4.69 cm on each side sufficient

of 11 KV transformer.

This is utilized for insulation and supporting coolers

This insulation between layers is 0.3 mm thick paper

Radial depth of HV coil bp == 17 x 1.70 + 16 x 0.3

= 3.4 cm

the thickness of insulation between HV and LV winding is = 5 + 0.9 Kv

= 5 + 0.9 x 11 = 15 mm

This includes width of the oil duct.

The insulation between HV and LV winding is a 5 mm thick

bakelized paper cylinder. The HV winding is wound on a former 5 mm

thick and the duct is 5 mm wide, making the total insulation between HV

and LV winding as 15 mm

Inside of diameter of HV winding = outside dia of LV winding +

2 x thickness of insulation = 16.75 x 2 x 1.5 = 19.75 cm

Outside dia of HV winding = 19.75 + 2 x 3.40 = 27 cm

Clearance between windings of two adjacent limbs

= 32 - 26.55 = 5.45 cm

45

3.8 Resistance;

Mean dia of primary winding = (19.75 + 26.55) / 2 = 23.15 cm

Length of mean turn of primary winding = Lmpt

= (22/7) x 0.2315 = 0.7272 cm

Resistance of primary winding at 75˚ c

Rp = (Tp.p.Lmpt) / ap

= 3168 x 0.021 x 0.7272 / 1.32 = 36.65 ohms

Mean dia of secondary winding = ( 15.5 + 16.75) / 2

= 16.125 cm

Length of mean turn of secondary winding = Lmpt

= (22/7) x 0.16125

= 0.506 mm

Resistance of secondary winding at 75˚c

Rs= 72 x 0.021 x 0.506/ 58 = 0.013 ohms

Total resistance referred to primary

Rp = 36.65 + (3168 / 72)2 x 0.013 = 61.8 ohms

Per unit resistance of transformer er - IpRp /Vp

= 3.03 x 61.8 / 11000 = 0.01728

3.9 Leakage reactance :

Length of mean turn = Lmt = Lmpt+Lmts) / 2

= (0.7272 + 0.506) / 2 = 0.6166 m

Height of winding Lc= Lcp + Lcs) / 2

46

= (0.034+ 0.37) / 2 =0.202 m

Leakage reactance referred to primary =

Xp = 2( 22/7) f 0 Tp2 (Lmt) /LC (Ith + [(bp +bs) /3]

Xp = 2 (22/7) x 50 x 4(2217) X 10 -7 x (3143) 2 x 0.6166

0.202

= {0.0149 +[0.0125 + 0.0340]/3} = 360.56 ohms

per unit leakage reactance ex = 3.03 x 360.56 / 11000

= 0.0993

per unit impedance ez = √ (0.01702)2 + (0.0993)2

= 0.1 00079

3.10 Regulation:

per unit regulation e = er cos + ex sin

At unity power factor e = er = 0.01728

At zero power factor lagging e = ex = 0.0993

At 0.8 power factor lagging e = 0.01728 x 0.8 + 0.0993 x 0.6

47

= 0.0676

3.11 Losses:

Copper losses at 75° c = Wc = 3. Ip2. Rp

= 3 X (3.03)2 x 61.8 = 1.702 Kw

Core losses

Volume of 3 limbs = 3 x 0.41 x 0.01313 = 0.0161499

Taking density of stampings as 7.6 x 103 Kg / m3

Weight of limbs = 7.6 x 103 xO.0161499 = 123 Kg

Flux density in limbs = 1.2 Wb / m 2

Corresponding to this flux density core loss per Kg = 1.2 w

Core loss in limbs = 123 x 1.2 z148 watts

Volume of two yokes = 2x 0.775 x 0.01575 = 0.02441 m

Weight of yokes = 7.6 x 103x 0.02441 = 182.4 Kg

Flux density in the yoke = 1.0 wb / m2

48

Loss per Kg = 1.0 watts

Core loss in yokes = 1 x 182.4 = 182.4 watts

Total core loss = 147 + 182.4 = 330.4 watts

3.12 Efficiency:

For maximum efficiency X2 WC = Wi

X2 x 1702 = 332

x= √ (332 /1702) = 0.4410

Thus maximum efficiency occurs at 44.10 % of fun load

This is a good fig. for distribution transformers

Efficiency at full load and at unity power factor =

100000/ (100000 + 1702 +330.4) = 98.00%

Total losses at full load = 1702 + 330.4= 2032.4 watts

3.13 All day efficiency :

Let us assume full load 12 hours at 0.8 p.f

49

Half load 6 hours at 0.8 p.f

No load 6 hours

The total cu losses at full load = 1702 watts

The total core losses =330.4 watts

Total core losses for 24 hours

Wco =330.4x 24 =7.93 kw

Cu losses at full load for 12hours

WC1 =1702 x12 =20.424kw

Cu losses at 1/2, load for 6 hours

Wc2 = 1702/4 x6 =2.553kw

Total losses = Wc0 +Wcl+Wc2 =7.93 +20.424+2.553=30.907 kw

Total olp in kwh at 0.8 p.f=(100xO.8xlx120 +112 *100 xO.8 x6 =1200kw

Total I/p in kwh =total_o/p +losses

All day efficiency = Out put in kwh/Input in kwh

50

1200

= -----------------

1200+30.907

= 97.489%

3.14 No Load Current:

Corresponding to flux density of 1.2 wb I m2 in limbs mmf per meter

atc

= 240 Amps

Total mmf for limbs = 3 x 0.41 x 240 = 295

Corresponding to flux density of 1 wb I m2 in yokes

Aty = 120 Amps

Total mmf for yokes = 2 x 0.766 x 120 = 184Amps

Total magnetizing mmf= 295 + 184 = 479

Magnetizing mmf per phase = 480/3 = 160

Magnetizing current per phase = Im = AT0 / (√ 2 x Tp)

= 160 I (√ 2 x 343) = 0.03579Amps

51

Loss component of no-load current

Ii = Wi I (3Vp) = 330.41 (3 x 11000) = 0.01006 Amps

No-load current per phase I0 = √ (0.0359)2 + (0.01006)2

= 0.0370 Amps

No-load current as percentage of full load current

= (0.0370 / 3.03) x 100 = 1.22 %

Allowing for joints etc. No load current will be about 2.5 percent of full load

current

3.15 Tank:

Height over yoke, H = 67.S3cm

Width over 3 limbs = 2D + outer dia of HV winding

= 2 x 31.79 +27 =90.58 cm

Width over 1 limb = outer dia of HV winding = 27 cm

Allowing 5 cm at the base and about 15 cm for oil.

Height of oil level = 67.S3 + 5 +15 = 87.53cm

52

Allowing another 20 cm height for leads etc.

Height of transformer tank = 87.53 + 20 = 107.53 cm

Allowing 5 cm each side along the length

Length of transformer tank = 90.58 + (2 x 5) = 100.58 cm

In width allowing 10 cm

Width of transformer tank = 27 + 10 = 37 cm

Total heat dissipating area of tank, considering 4 walls only.

St = 2(lOO.58+37)x107.53x 10-4=2.695 m2

Total specific heat dissipation = 12.5 W/m2/ ˚C

Temperature rise q =2032.4/ (2.69Sx 12.5) =60.32˚C

This is greater than 35˚C so it is necessary to provide cooling

system in order to

dissipate the heat.

3.16 Cooling System:

Area of the plain tank St = 2(100.58+ 37)x 1O7.53x 10-4=2.695 m2

53

Let tubular cooling be provided

Let the tube area be = xSt

Total dissipating surface = (1 +x) St

= 2.7 (1+x) m2

Total loss to be dissipated = 2032.4 watts

Assume the convection is improved by 35% due to provision of tubes

Specific heat dissipation = 2032.41 {2.695(1 +x) x 35} =21.546/(1 +x) w/m2

- °c

Loss dissipated per m per °c = {(12.5+8.8) x} / (x+ 1) =21.521(1+x)

x = 1.027

Area of tubes= x st = 1.027 x 2.695 = 2.77 m2

Wall area of each tube = (22/7) dtlt =(22/7)x 0.05 xO.8 =0. 12566m2

Dia of each tube =O.05m

Length of each tube =0.8m

= (22/7) x 5 x 80 = 1256 cm2

Total no. of tubes to be provided = 2.695 / 0.12566= 21.446z22

The tubes are spaced 7.5 cm apart. Therefore in 101 cm we

will be able to accommodate 11 tubes leaving 9 cm on each side.

54

C- Program & Results

55

CHAPTER - 4

C-Program & Results

DESIGN NOTATIONS

F = Frequency of the generating system = 50 Hz

Bm = Maximum flux density inside the core.

Let assume Bm = 1.2 Wb/m² (or)Tesla.

m = Maximum flux inside the core in Wb

Agi = Gross iron area of the core

Ai = Net iron area of the core

ბ = Current density inside the transformer

Let = 2.3 Amps/mm²

= Resistivity of copper conductor

1.72 x 10 Ohm-meter

0 = Absolute permeability =

Aw = Area of the window in M

Hw = Height of the window in meters

Ww = width of the window in meters.

Kw = window space factor

Let Kw = 0.25

Tp = Number of primary turns

Ts = Number of secondary beams

Ip = Current in primary winding in amps.

56

Is = Current in secondary winding in Amps.

Vp = Terminal Voltage of Primary winding in volts

Vs = Terminal Voltage of Secondary winding in volts

ap = Area of conductor of primary in m2

a = Larger length of the cruci-form in meters.

b = Smaller length of the cruci-form in meters.

Ay = Cross sectional area of the yoke in m2

Dy = Depth of yoke in meters.

Hy = Height of the yoke in meters.

as = Cross sectional area of the primary conductor in m2

w = Total horizontal length of the core in meters.

H = Total height of the transformer core in meters.

D = Distance between the centers of adjacent lims in meters.

Lp = Length of conductor in primary winding in meters.

57

Ls = Length of conductor in secondary winding in meters.

Rp = Resistance of primary winding in ohms.

Rs = Resistance of secondary winding in ohms

R01 = Total resistance referred to primary in ohms.

er = Per unit resistance of transformer.

Lmt = Length of meom turn in meters.

Xp = Leakage resistance refer to primary in ohms.

ex = Per unit resistance.

ez = Per unit in pedance.

R = Regulation of transformer.

C = Total copper losses of the transformer in watts.

Cp = Primary copper losses in watts.

Cs = Secondary copper losses in watts.

Vi = Volume of the core in transformer in m3

Vy = Volume of the yoke in transformer in m3

58

V = Volume of the iron in transformer in m3

Wi = Total iron losses in watts.

E = Efficiency of transformer.

L = Total losses of transformer in watts.

Th = Total weight of the tank in meters.

TL = Total length of the tank in meters.

Tw = Total width of the tank in meters.

St = Area of plain tank in m2

T = Temperature rise of the transformer in C

N = Total No. of tubes.

At = Area of tube in m2

Wco = Total core losses in 24 hours in watts.

Wc1 = Total copper losses at full load for 12 hours.

Wc2 = Total copper losses at half full load for 6 hours in watts.

59

Wt = Total copper losses in 24 hours in kwh

Qo = total output of the transformer for 24 hours in kwh

Ae = All day efficiency of distribution from transformer.

Dmp = Mean dia. of primary winding in meters.

Dms = Mean dia. of secondary winding in meters.

Dip = Inside dia. of primary winding in meters.

Dop = Outside dia. of primary winding in meters

Dis = Inside dia. of secondary winding in meters

Dos = Outside dia. of secondary winding in meters

Ith = Thickness of insulation between H V and L V winding

Lcs = Arial depth of L.V winding in meters.

Lc = Height of winding in meters.

bp = Axial depth of H V coil in meters let assume bp = 0.034m

bs = Radial depth of L V coil in meters let assume bs = 0.0125m

60

C-Program

# include<stdio.h>

# include<conio.h>

# include<math.h>

# define f 50

# define Bm 1.2

# define del 2300000

# define dels 2300000

# define delp 2400000

# define bp 0.034

# define bs 0.0125

# define rho 0.000000021

# define Abs 0.000001256

# define di 7600

# define S 1.25

# define cospi 0.8

# define sinpi 0.6

void main()

{

FILE * babu;

float Agi,Ai,Pim,Et,Q,Vp,Vs,d,a,b;

float Aw,Ww,Hw;

float Ay,Dy,Hy;

float Tp,Ts,Ip,Is,ap,as;

float W,H,D,Lp,Ls,Rp,Rs,R01,er;

float Lmt,Xp,ex,ez,R;

float C,Cp,Cs;

float Vi,Vy,V,Wc,Wi,E,L;

61

float Th,Tl,Tw,St,T;

float X,N,At;

float Wc0,Wc1,Wc2,Wt,Q0,AE;

float Dmp,Dms,Dip,Dop,Dis,Dos;

float Ith,Lcs,Lc,Kw;

clrscr();

textbackground(GREEN);

textcolor(RED);

cprintf(“\n Enter volt per turn:”);

cscanf(“%f”,&Et);

cprintf(“\n Enter capacity of transformer :”);

cscanf(“%f”,&Q);

cprintf(“\n Enger Primmary voltage : “);

cscanf(“%f”,&Vp);

cprintf(“\n Enter Secondary Voltage: “);

cscanf(“%f”,&Vs);

/*printf(“\n Enter volt per turn, capacity of transformer, Primmary

voltage and Secondary Voltage”);

scanf(“%f%f%f%f”,&Et,&q,&Vp,&vs);

*/

babu=fopen(“result.txt”,”w+”);

fprintf(babu,”Volt per turn:%f\n”,Et);

fprintf(babu,”Capaicty of Transformaer: %f\n”,Q);

fprintf(babu,’Primary voltage:%f\n”,Vp);

fprintf(babu,”enter secondary Voltage:%f\n”,Vs);

Pim = Et/(4.44*f);

Ai=Pim/Bm;

Kw = (10/(30+Vp*0.001));

62

d=sqrt(Ai/056);

a=0.85*d;

b=0.53*d;

Aw=Q/(3.33*f*1.2*dels*Kw*Ai*0.004);

Ww=sqrt(Aw/2.5);

Hw=2.5*Ww;

Ay=1.2*Ai;

Agi=Ay/0.92;

Dy=a;

Hy=Agi/dy;

Ts=(Vs/(1.732*Et));

Tp=(Vp/Vs)*Ts*1.732;

Ip=(Q*1000)/(3*Vp);

Is=(Q*1000*1.732)/(3*Vs);

Ap=(Ip/delp)

As=(Is/dels);

D=Ww+d;

W=(2*D+a);

H=(2*Hy+Hw);

Dis=d+0.0015;

Dos=Dis+bs;

Dmp=(Dip+Dop)/2;

Dms=(Dis+Dos)/2;

Ith=((5+(0.9*Vp*0.001))*0.001);

Dop=dip+2*bp;

Lcs=(((Ts/2)+1)*10)*0.001;

Lc-(Lcs+bp)/2;

Dip=Dos+(2*Ith);

Lp=Tp*3.14*Dmp;

63

Ls=Ts*3.14*Dmms;

Rp=(rho*Lp)/ap;

Rs=(rho*Ls)/as;

R01=Rp+(Rs*(Vp/Vs)*(Vp/Vs)*3);

Er=(Ip*Rp)/Vp;

Lmt=((Lp/Tp)+(Ls/Ts))/2;

Xp=(2*3.14*f*Abs*Tp*Tp)*(Lmt/Lc)*(Ith+(bp+bs)/3;

Ex=(3.03*Xp)/Vp;

Ez=sqrt(er*er+ex*ex);

R=er*cospi+ex*sinpi;

Cp=Ip*Ip*Rp;

Cs=Is*is*Rs;

C=3*Ip*Ip*R01;

Vi=3*Ai*Hw;

Vy=2*Ay*W;

V=Vi+Vy;

Wc=(Vi+Vy)*di;

Wi=(Vi*1.2+Vy)*di;

L=C+Wi;

E=(Q*1000)/((Q*1000)+L);

E=E*10;

Th=H+4;

T1=(2*D+dop)+0.1;

Tw=Dop;

St=2*(Tw+T1)*Th;

T=L/(st*12.5);

X=(((L/(St*35))-12.5)/8.8);

At=3.14*0.05*0.8;

N=X*St/At;

64

Wc0=wi*24;

Wc1=C*12;

Wc2=C*0.25*6;

Wt=(Wc0+Wc1+/wc2)/1000;

Q0=(Q*cospi*12)+(0.5*Q*cospi*6);

AE=(Q0*100)/(Wt+Q0);

textbackground(MAGENTA);

textcolor(YELLO);

printf(“\n CORE DESIGN”);

printf(“\n ~~~~~~~~~~~~~~~~~~~~~~”);

cpintf(“n\ Net iron area of the core %f m2“,Ai);

cpintf(“n\ Gross iron area of the core %f m2“,Agi);

cpintf(“n\ Larger length of the cruci-form %f m“,a);

cpintf(“n\ Smaller length of the cruci-form %f m“,b);

getch();

printf(“\n\n DESIGN OF WINDOW”);

printf(“\n~~~~~~~~~~~~~~~~~~~~~”);

printf(“n\ Area of window %of m2”,Aw);

printf(“n\ Width Area of window %of m”,Ww);

printf(“\n Height of window %fm’,Hw);

getch();

printf(“\n\n DESIGN OF YOKE”);

printf(“\n~~~~~~~~~~~~~~~~~”);

printf(“\n Cross sectional Area of the yoke %f m2”,Ay);

printf(“\n Depth of the yoke %f m”,Dy);

printf(“\n Height of the yoke%f m”,Hyderabad);

getch();

printf(“\n\n WINDING DESIGN”);

printf(“\n~~~~~~~~~~~~~~~~~~~~”);

65

printf(“\n No. of turns on primary %f’,Tp);

printf(“\n No. of turns on secondary %f”,Ts);

printf(“\n Current in primary winding %f amps”, Ip);

printf(“\n Current in secondary winding %f amps”, Is);

printf(“\n Cross Sectional area of primary conductor %f m2”, ap);

printf(“\n Cross Sectional area of secondary conductor %f m2”,as);

getch();

printf(“\n \n OVERALL DESIGNING OF TRANSFORMER”);

printf(“\n ~~~~~~~~~~~~~~~~~~~~~~~~~”);

printf(“\n The total horizontal length of the core%fm”,W);

printf(“\n total height of the transformer core %fm”,H);

printf(“\n distance between the centers of adjacent limbs %f m”,D);

getch();

printf(“\n\n RESISTANCE OF WINDINGS”);

printf(“\n Length of conductor in primary winding %f m”,Lp);

printf(“\n Length of conductor in Secondary winding %f m”,Ls);

printf(“\n Resistance of Primary winding %f ohms”,Rp);

printf(“\n Resistance of secondary winding %f ohms”,Rs);

printf(“\n per Unit resistance of transformer %f ”,er);

getch();

printf(“\n\n LEAKAGE REACTANCE OF TRANSFORMER”);

printf(“\n~~~~~~~~~~~~~~~~~~~”);

printf(“\n Length of mean turn %f m”,Lmt);

printf(“\n Height of the winding %f m”,Hw);

printf(“\n Leakage reactance refer to primary %f ohms”,Xp);

printf(“\n Per Unit reactance %f”,ex);

printf(“\n Per unit impedance %f”,ez);

getch();

printf(“\n\n REGULATION OF THE TRANSFORMER”);

66

printf(“\n ~~~~~~~~~~~~~~~~~”);

printf(“\n Regulation at.8 power factor %f”,R);

printf(“\n COPPER LOSSES OF THE TRANSFORMER \n”);

printf(“\n ~~~~~~~~~~~~~~~~~~~~~\n”);

printf(“\n total copper losses of the transformer %f watts”, C);

getch();

printf(“\n\n IRON LOSSES OF TRANSFORMER”);

printf(“\n~~~~~~~~~~~~~~~~~~~~~~~~”);

printf(“\n Volume of the core in transformer %f m3”,Vi);

printf(“\n Volume of the yoke %f m3”,Vy);

printf(“\n Volume of the Iron in transformer %f m3”,V);

printf(“\n total iron losses %f watts”,Wi);

getch();

printf(“\n\n THE EFFICIENCY OF THE TRANSFORMER”);

printf(“\n~~~~~~~~~~~~~~~~~~~~~~~~”);

printf(“\n Total losses of transformer %f watts”,L);

printf(“\n Efficiency of transformer %f “,E);

getch();

printf(“\n\n THE ALLDAY EFFICIENCY OF TRANSFORMER”);

printf(“\n~~~~~~~~~~~~~~~~~~~~~~~~”);

printf(“\n\n taking full load for 12 hrs at 0.8pf lag”);

printf(“\n Taking no load for 6 hrs”);

printf(“\n Copper loss at full load %f watts”, Wc1);

printf(“\n Copper loss at half full load %f watts”, Wc2);

printf(“\n iron losses for 24 hrs %f watts”, Wc0);

printf(“\n total loss for 24 hrs %f kwh”, Wt);

printf(“\n total output for 24 hrs %f kwh”, Q0);

printf(“\n ALLDAY EFFICIENCY %f ”, AE);

getch();

67

printf(“\n\n TRANSFORMER TANK DESIGN”);

printf(“\n~~~~~~~~~~~~~~~~~~~~~~~~”);

printf(“\n Total height of the tank %f m”,Th);

printf(“\n Total length of the tank %f m”,T1);

printf(“\n Total width of the tank %f m”,Tw);

getch();

printf(“\n Temperature rise of the transformer %f m”,T);

getch();

if(T>35)

{

printf(“\n since temperature rise is>35 cooling is required\n”);

printf(“\n\n COOLING SYSTEM”);

printf(“\n~~~~~~~~~~~~~~~~~~~~~~~~”);

printf(“\n Area of plain tank %f m2”,St);

printf(“\n Area of tube %f m2”,At);

printf(“\n Number of tubes required %f ”,N);

getch();

}

else

printf(“\n since temperature rise is <35 cooling system is not required\n”);

getch();

fprintf(babu,“\n\t\t CORE DESIGN”);

fprintf(babu,“\n\t\t ~~~~~~~~~~~~”);

fprintf(babu,”\n Net iron area of the core %m2, Ai);

fprintf(babu,”\n Gross iron area of the core %m2, Agi);

fprintf(babu,”\n Larger length of the cruci-form %f m, a);

fprintf(babu,”\n smaller length of the cruci-form %f m”, b);

68

fprintf(babu,“\n\n DESIGN OF WINDOW”);

fprintf(babu,“\n ~~~~~~~~~~~~”);

fprintf(babu,”\n Area of window %f m2, Aw);

fprintf(babu,”\n width of window %f m, Ww);

fprintf(babu,”\n Height of window %f m, Hw);

fprintf(babu,“\n\n DESIGN OF YOKE”);

fprintf(babu,“\n ~~~~~~~~~~~~”);

fprintf(babu,”\n Cross sectional area of the yoke %f m2, Ay);

fprintf(babu,”\n depth of the yoke %f m, Dy);

fprintf(babu,”\n Height of the yoke %f m, Hy);

fprintf(babu,“\n\n WINDING DESIGN”);

fprintf(babu,“\n ~~~~~~~~~~~~”);

fprintf(babu,”\n No. of turns on primary %f , Tp);

fprintf(babu,”\n No. of turns on secondary %f”, Ts);

fprintf(babu,”\n Current in primary winding %f amps”,Ip);

fprintf(babu,”\n Current in secondary winding %f amps”,Is);

fprintf(babu,”\n Cross sectional area of primary conductor %f m2”,ap);

fprintf(babu,”\n Cross sectional area of Secondary conductor %f m2”,as);

fprintf(babu,“\n\n OVERALL DESIGNING OF TRANSFORMER”);

fprintf(babu,“\n ~~~~~~~~~~~~”);

fprintf(babu,”\n The total horizontal length of the core %f m , w);

fprintf(babu,”\n The total height of the transformer 0 core %f m , H);

fprintf(babu,”\n Distance between the centers of adjacent limbs %f m ,D);

fprintf(babu,”\n\n RESISTANCE OF WINDINGS”);

69

fprintf(babu,“\n Length of conductor in primary winding %f m”,Lp);

fprintf(babu,“\n Length of conductor in Secondary winding %f m”,Ls);

fprintf(babu,“\n Resistance of Primary winding %f ohms”,Rp);

fprintf(babu,“\n Resistance of secondary winding %f ohms”,Rs);

fprintf(babu,“\n per Unit resistance of transformer %f ”,er);

fprintf(babu,“\n\n LEAKAGE REACTANCE OF TRANSFORMER”);

fprintf(babu,“\n~~~~~~~~~~~~~~~~~~~”);

fprintf(babu,“\n Length of mean turn %f m”,Lmt);

fprintf(babu,“\n Height of the winding %f m”,Hw);

fprintf(babu,“\n Leakage reactance refer to primary %f ohms”,Xp);

fprintf(babu,“\n Per Unit reactance %f”,ex);

fprintf(babu,“\n Per unit impedance %f”,ez);

fprintf(babu, “\n\n REGULATION OF THE TRANSFORMER”);

fprintf(babu, “\n ~~~~~~~~~~~~~~~~~”);

fprintf(babu,“\n Regulation at.8 power factor %f”,R);

fprintf(babu,“\n COPPER LOSSES OF THE TRANSFORMER \n”);

fprintf(babu, “\n ~~~~~~~~~~~~~~~~~~~~~\n”);

fprintf(babu, “\n total copper losses of the transformer %f watts”, C);

fprintf(babu, “\n\n IRON LOSSES OF TRANSFORMER”);

fprintf(babu, “\n~~~~~~~~~~~~~~~~~~~~~~~~”);

fprintf(babu, “\n Volume of the core in transformer %f m3”,Vi);

fprintf(babu, “\n Volume of the yoke %f m3”,Vy);

fprintf(babu, “\n Volume of the Iron in transformer %f m3”,V);

fprintf(babu, “\n total iron losses %f watts”,Wi);

70

fprintf(babu, “\n\n THE EFFICIENCY OF THE TRANSFORMER”);

fprintf(babu, “\n~~~~~~~~~~~~~~~~~~~~~~~~”);

fprintf(babu, “\n Total losses of transformer %f watts”,L);

fprintf(babu, “\n Efficiency of transformer %f “,E);

fprintf(babu, “\n\n THE ALLDAY EFFICIENCY OF TRANSFORMER”);

fprintf(babu,“\n~~~~~~~~~~~~~~~~~~~~~~~~”);

fprintf(babu, “\n\n taking full load for 12 hrs at 0.8pf lag”);

fprintf(babu, “\n\n taking half full load for 6 hrs at 0.8pf lag”);

fprintf(babu, “\n\n Taking no load for 6 hrs”);

fprintf(babu, “\n Copper loss at full load %f watts”, Wc1);

fprintf(babu, “\n Copper loss at half full load %f watts”, Wc2);

fprintf(babu, “\n iron losses for 24 hrs %f watts”, Wc0);

fprintf(babu, “\n total loss for 24 hrs %f kwh”, Wt);

fprintf(babu, “\n total output for 24 hrs %f kwh”, Q0);

fprintf(babu,“\n ALLDAY EFFICIENCY %f ”, AE);

fprintf(babu, “\n\n TRANSFORMER TANK DESIGN”);

fprintf(babu, “\n~~~~~~~~~~~~~~~~~~~~~~~~”);

fprintf(babu, “\n Total height of the tank %f m”,Th);

fprintf(babu, “\n Total length of the tank %f m”,T1);

fprintf(babu, “\n Total width of the tank %f m”,Tw);

fprintf(babu,“\n Temperature rise of the transformer %f ”,T);

if(T>35)

{

fprintf(babu, “\n since temperature rise is>35 cooling is required\n”);

71

fprintf(babu, “\n COOLING SYSTEM”);

fprintf(babu,“\n~~~~~~~~~~~~~~~~~~~~~~~~”);

fprintf(babu,“\n Area of plain tank %f m2”,St);

fprintf(babu, “\n Area of tube %f m2”,At);

fprintf(babu, “\n Number of tubes required %f ”,N);

fclose(babu);

}

else

fprintf(babu,“\n since temperature rise is <35 cooling system is not

required\n”);

printf(“\n the file is generated for the Out Put as result.txt file\n”);

printf(“*********************************************\n”);

printf(“\t\t\t CLOSE WINDOW NOW \n”);

printf(“\t\t\t*************\n”);

}

72

RESULTS

Volt per turn :3.500000

Capacity of Transformer :100.000000

Primary Voltage : 11000.000000

Enter Secondary Voltage : 433.000000

CORE DESIGN

Net iron area of the core :0.013138 mÙ2

Gross iron area of the core :0.017137 mÙ2

Larger length of the cruci-form :0.130194 m

Smaller length of the cruci-form :0.081180 m

DESIGN OF WINDOW

Area of window :0.0067909 mÙ2

Width of window :0.164814 m

Height of window :0.412034 m

DESIGN OF YOKE

Cross sectional area of the yoke :0.15766 mÙ2

Depth of the yoke :0.130194 m

Height of the yoke :0.131624 m

WINDING DESIGN

73

No.of turns on primary:3142.857178

No.of turns on secondary :71.428574

Current in primary winding :3.030303 amps

Current in secondary winding :133.333328 amps

Cross sectional area of primary conductor :0.000001 m2

Cross sectional area of secondary conductor :0.000058 m2

OVERALL DESIGNING OF TRANSFORMER

The total horizontal length of the core:0.766161 m

Total height of the transformer core :0.675282 m

Distance between the centers of adjacent limbs :0.317983 m

RESISTANCE OF WINDINGS

Length of conductor in primary winding :2279.341064 m

Length of conductor in secondary winding :36.091991 m

Resistance of primary winding :37.910000 ohms

Resistance of primary secondary winding :0.013074 ohms

Per unit resistance of transformer :0.010444

LEAKAGE REACTANCE OF TRANSFORMER

Length of maen turn :0.615266 m

Height of the winding :0.412034 m

Leakage reactance refer to primary :363.275635 ohms

Per unit reactance :0.100066

Per unit impedance :0.1000609

74

REGULATION OF THE TRANSFORMER

Regulation at 0.8 power factor :0.068394

COPPER LOSSES OF THE TRANSFORMER

Total copper losses of the transformer :1741.690674 waatts

IRON LOSSES OF TRANSFORMER

Volume of the core in transformer :0.016240 m3

Volume of the yoke :0.024158 mÙ3

Volume of the iron in transformer :0.040398 m3

Total iron losses :331.712158 watts

THE EFFICIENCY OF THE TRANSFORMER

Total losses of transformer:2073.402832 watts

Efficiency of transformer :97.968712

THE ALLDAY EFFICIENCY OF THE TRANSFORMER

Taking full load for 12 hrs at 0.8pf lag

Taking half full load for 6 hrs at 0.8pf lag

Taking no load for 6 hrs.

Copper loss at full load :20900.289062 watts

Copper loss at half full load :2612.536133 watts

Iron losses for 24 hrs:7961.091797 watts

Total loss for 24 hrs :31.473917 kwh

Total output for 24 hrs :1200.000000 kwh

ALLDAY EFFICIENCY :97.444206

75

TRANSFORMER TANK DESIGN

Total height of the tank :1.075282 m

Total length of the tank :1.000937 m

Total width of the tank :0.264970 m

Temperature rise of the transformer :60.928371

Since temperature rise is>35 cooling system is required.

COOLING SYSTEM

Area of plain tank :2.722414 m2

Area of tube :0.125600 m2

Number of tubes required :22.808620

Volt per turn :2.250000

Capacity of Transformer :25.000000

Primary Voltage :11000.000000

Enter Secondary Voltage :433.000000

CORE DESIGN

Net iron area of the core:0.008446 m2

Gross iron area of the core :0.011016 m2

Larger length of the cruici-form :0.104388 m


DESIGN OF WINDOW

76

Area of window :0.026409 m2



DESIGN OF YOKE

Gross sectional Area of the yoke :0.010135 m2



WINDING DESIGN

No.of turns on primary :4888.889160

No.of turn on secondary :111.111115





OVERALL DESIGN OF TRANSFORMER

The toal horizontal length of the core :0.555564 m

Total height of the transformer core:0.468017 m






77

Resistance of secondary winding :0.066003 ohms













Volume of the core in transformer :0.006511 m3

Volume of the yoke :0.011261 m3

Volume of the iron in transformer :0.017772 m3


78



















Since temperature rise is35 cooling system is not required.

Capacity of transformer :100.000000

Primary Voltage:11000.000000

Enter secondary Voltage :250.000000

79

CORE DESIGN

Net iron area of the core:0.013138 m2

Gross iron area of the core :0.017137 m2

Larger length of the cruici-form :0.130194 m


DESIGN OF WINDOW

Area of window :0.067909 m2



DESIGN OF YOKE

Gross sectional Area of the yoke :0.015766 m2



WINDING DESIGN

No.of turns on primary :3142.856934

No.of turn on secondary :41.240513





80

OVERALL DESIGN OF TRANSFORMER

The toal horizontal length of the core :0.766161 m

Total height of the transformer core:0.675282 m






Resistance of secondary winding :0.004358 ohms













Volume of the core in transformer :0.016240 mÙ3

Volume of the yoke :0.024158 mÙ3

81

Volume of the iron in transformer :0.040398mÙ3




















Since temperature rise is>35 cooling system is required.

82

COOLING SYSTEM

Area of plain tank :2.722414 m2

Area of tube :0.125600 m2

Number of tubes required :22.808615

83

CONCLUSIONS

The design of distribution transformer is the main objective of this

project. In the design process some specifications are taken common to

all the distribution transformers for making the design easier. Here, we

used C language program for the design of distribution transformer.

In design process, by giving volt per turn, capacity of transformer

(KVA), primary voltage and secondary voltage as input parameters, we

designed core, window, yoke, windings, overall design, resistance of

winding, leakage reactance of transformer, regulation of transformer, iron

loss, copper loss, efficiency of transformer and cooling system of

transformer.

For finding regulation of the distribution transformer we assumed

power factor as 0.8 lagging, since all distribution transformers are working

more or less at this power factor only. Flux density, current density and

window space factor are assumed common to all the distribution

transformers to make C program easy. Assumed primary and secondary

winding coils are made of copper, core and yoke are made of iron.

Tubular Oil circulating cooling system is employed.

Due to time limitation, we could not study the fabrication methods

that are employed in the design of transformer. But, we have gone

through design procedures and implemented this in developing the C

language program by which we can design any distribution transformer

specifications by giving only four input parameters.

84

By using the C language program we can design up to 100 KVA

transformer due to assumption of specifications and constraints. The

same program can be useful to design desired parameter values more

than 100 KVA by altering some of assumed specifications and constraints.

85

REFERENCES

1. ELECTRICAL MACHINE DESIGN

By A.K.Sawhney

2. PERFORMANCE AND DESIGN OF AC MACHINES

By MG Say

3. ELECTRICAL MACHINARY

By Dr. B.S. Bimbhra

4. ELECTRICAL MACHINES

By I.J.Nagrath

& D.P. Kothari

86

Documents

Designing of Distribution Transformer 1