VIDYA COLLEGE OF VIDYA COLLEGE OF ENGINEERING ,MEERUTENGINEERING ,MEERUT

ABHISHEK(EN 4th Year)

ABHISHEK(EN 4th Year)

HARSH VARDHAN(EN 4th Year)

HARSH VARDHAN(EN 4th Year)

HEMANT (EN 4th Year)

HEMANT (EN 4th Year)

CONTENT

REFRENCES

A transformer is an electrical device that transfers energy from one circuit to another purely by magnetic coupling. Relative motion of the parts of the transformer is not required for transfer of energy. Transformers are often used to convert between high and low voltages, to change impedance, and to provide electrical isolation between circuits

Transformer Basics

SINGLE PHASE

TWO WINDING TRANSFORMER CONNECTION

BASIC TRANSFORMER ACTION

508

Michael Faraday, who invented an 'induction ring' on August 29, 1831. This was the first transformer, although Faraday used it only to demonstrate the principle of electromagnetic induction and did not foresee the use to which it would eventually be put.

INVENTION

Very high speed growth of industries in our

country.

High transmission loss from generation point to distribution point as well as more demand compared to generation of electrical energy.

Progress of new projects for generation of power is very slow as compared to the growth of the requirement

NEED FOR REDUCTION IN LOSS

Transformer losses categorized as follows.

No load Loss (Excitation Loss) Hysteresis Loss Eddy current loss

Load Loss (Impedance Loss) I²R loss. Stray loss

HYSTERESIS LOSSES

MECHANICAL LOSS

This loss consist of bearing friction,brush friction and windage loss.The windage loss includes the power required to circulate air through the machine.

Hysteresis loss is that energy lost by reversing the magneticfield in the core as the magnetizing AC rises and falls and reversesdirection.

Increment in core loss caused by distortion of air-gap flux plus the increment in I2R loss due to non uniform distribution of conductor current is called stray load loss.

Increment in core loss caused by distortion of air-gap flux plus the increment in I2R loss due to non uniform distribution of conductor current is called stray load loss.

When the probe is brought in close to a conductive material, the probes changing magnetic field generates current flow in the material.

The eddy currents produce their own magnetic fields that interact with the primary magnetic field of the coil.

WHY WE USE LAMINATED CORE?

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Copper Bar(Anti Ferro Magnetic Material)

Winding

Input Wires

Embedded Type Core

The losses which occur in transformer are:(a):- Copper loss or I2 R.

(b):- Iron loss or Core loss.Copper loss Pc :- Here we will calculate copper loss for one

embedded bar similarly losses can be calculated for all other bars

Pc=I2RWhere R is resistance of coil and I is current in coil

R=ƍL/A Where ƍ is specific resistance of material, L is length

of coil and A is area of coil, so copper loss can be calculated.

Working Methodology

Iron loss or core loss Pi:- Iron loss occurs in the magnetic core of the transformer. This loss is the sum of hysteresis

loss(Ph) and Eddy current loss(Pe).Pi=Ph + Pc

Pi=Kh fBmn + Kef2B2mKh= Proportionality constant which depends upon the volume and quality of

the core material and the units used.Ke= Proportionality constant whose value depends upon the volume and resistivity of the core material,thickness of laminations and units used .

Bm=Maximum flux density in the core.f=Frequency of the alternating flux.

Separation of Hysteresis and Eddy-Current losses:- The transformer core loss

Pi has two components namely hysteresis loss P and Eddy-current loss Pe. Pi=Ph + Pc

Pi=Kh fBmn + Kef2B2mThe exponents n varies in the rage 1.5 to 2.5 depending upon the

ferromagnetic material for a given Bm the hysteresis loss varies directly as the frequency

and the Eddy current loss varies as the square of the frequency. That is,

Ph α f or Ph=afand

Pe α f2 or Pe=bf2Where a and b are constants .

Pi=af+bf2

For separation of these two losses the no load test is performed on the transformer.

However, the primary of the transformer is connected to a variable frequencyand

variable sinusoidal supply and the secondary is open circuited. Now

V=4.44fǾmTOr

V/f=4.44BmAiTFor any transformer T and Ai are constants. Therefore Bm will remain constant

If the test is conducted so that the ratio (V/f) is kept constant Pi/f=a+bf

During this test, the applied voltage V and frequency f are varied together so that

(V/f) is kept constant. The core loss is obtained at different frequencies

This graph is a straight line AB of the form y=mx+c,as shown in figure.The intercept of the straight line on the vertical axis gives a and slope of line AB gives b. Thus, knowing the constants a and b, hysteresis and Eddy current losses can be separated.

DC Shunt Motor

3-Phase Induction Generator

Fuse

Input DC Voltmeter Generator output

Voltmeter

FrequencyMeter

Starter

Rehostate in motor field winding

Core

Ammeter

Wattmeter

Setup for Practical

Rh

Rh

M G

N

Y

BV

R

A

V 0

field

v 1

Circuit Diagram of Setup

Wattmeter

f

Frequency meter

DC Shunt MotorGenerator

NEC

EC

Rheostat

Rectifier

Calculation

For Non Embedded Core

R=Resistance of Winding=80.9 ohms

Pc=Copper loss=I^2*R= (1.06^2*80.9)= 89.99 W

Pi= Pt – Pc

Pi=Input Power(Variable)

Pt=Total Power

(Pi)1=125-89.99=35.01 W

(Pi/f)1=.778 W/Hz

(Pi)5=147-89.99=57.01 W

(Pi/f)5=1.14 W/Hz

From Graph

B=.00724

He= bf2=.00724*502=18.1 W

25.4W

For Embedded Core

R=Resistance of Winding=80.9 ohms

Pc=Copper loss=I^2*R= (1.06^2*80.9)= 89.99 W

Pi= Pt – Pc

Pi=Input Power(Variable)

Pt=Total Power

(Pi)1=112-89.99=22.01 W

(Pi/f)1=.489 W/Hz

(Pi)5=140- 89.99=50.01 W

(Pi/f)5=1 W/Hz

From Graph

B=.0062*502

He= bf2=.0062*502= 25.5W

S.NO. FREQUENCY F (hz)

CURRENTA(amper)

WATTMETER (watts)

GENRATED VOLTAGE(volts)

POWER(Pi)

Pi/F

01 45 1.05 125 340 35.01 .778

02 46 1.05 128 352

03 47 1.05 133 358

04 48 1.05 135 369

05 50 1.05 147 386 57.01 1.14

Table for Non Embedded Core

S.NO. FREQUENCYF (hz)

CURRENTA(amper)

WATTMETER (watts)

GENRATED VOLTAGE

(volts)

POWER(Pi)

Pi/F

01 45 1.07 112 336 22.01 .489

02 46 1.07 117 345

03 47 1.07 121 354

04 48 1.07 128 363

05 50 1.07 140 388 5o.01 1.00

Table for Embedded Core

OUTPUT OF PROJECT

45 46 47 48 49 500.4

0.6

0.8

1

1.2

1.4

Frequency

Po

wer

inpu

t/cyc

le

Non Embedded Core Embedded Core

Slopeb1=0.007

a1

a2

Slope b2=0.0102

CONCLUSION

CHALLENGES FACED

Easy to fabricate

Increased Transformer Efficiency

Reduced Humming Loss

Economy

Easy to fabricate

Increased Transformer Efficiency

Reduced Humming Loss

Economy

How eddy current loss minimize by using laminated core in transformer ?

Member since: June 26, 2007

Assume that a changing magnetic flux is passing through a certain square cross sectional area of the transformer core. Look at a loop of current enclosing that flux. The power dissipated in that particular loop is proportional to the square of the area enclosed by that loop (A) divided by the length of the path (L). If you divide that square into two rectangles by laminating the core, the area enclosed in the loop will be cut in half while the length will be reduced to 3/4 of the original length. The result will be two loops of current with a total power dissipation of 2X(.5A)^2/.75L. That makes the sum of the power dissipated in the two smaller loops two thirds of the power dissipated in the original loop. More laminations reduce the dissipation even more.

About me: WILLIUM TERROSE I am located in the USA central time zone (UTC -6)

• Dr.P.S. BIMBHRA Electrical Machines, Macmillan. ISBN 0-333-19627-9.

•Heathcote, MJ (1998). J&P Transformer Book, 12th ed., Newnes. ISBN 0-7506-1158-8.

•Hindmarsh, J. (1984). Electrical Machines and their Applications, 4th ed., Pergamon. ISBN 0-08-030572-5.

•Shepherd,J; Moreton,A.H; Spence,L.F. (1970). Higher Electrical Engineering, Pitman Publishing. ISBN 0-273-40025-8.

References

Vote of Thanks

Mr. A.K. Singha

Mr. Ramveer Singh

THANK YOU

THANK YOU