RGPV BE 104 Unit II

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Prof M D Dutt HOD EX Department SRCT Bhopal

READING MATERIAL FOR B.E STUDENTS

OF RGPV AFFILIATED ENGINEERING COLLEGES

SUBJECT BASIC ELECTRICAL AND ELECTRONICS

Professor MD Dutt

Addl General Manager (Retd)

BHARAT HEAVY ELECTRICALS LIMITED

Professor(Ex) of EX Department

Bansal Institute of Science and Technology

KOKTA ANANAD NAGAR BHOPAL

Presently Head of The Department ( EX)

Shri Ram College Of Technology

Thuakheda BHOPAL

Sub Code BE 104 Subject Basic Electrical & Electronics

UNIT II Magnetic circuits and Transformers

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RGPV Syllabus

BE 104 BASIC ELECTRICAL & ELECTRONICS ENGINEERING

UNIT II

MAGNETIC CIRCUITS AND TRANSFORMER

Review of laws of magnetism flux and their relation. Analysis of magnetic circuit and

single phase transformer, Basic concepts and construction feature of transformer.

Voltage , current and impedance transformation,EMF equation, equivalent circuits

and phasor diagrams, Voltage regulation, Losses and efficiency, Open circuit test,

Short circuit test.

INDEX

S No Topic Page

1 Review of laws of magnetism flux and their relation 3,4,5

2 Analysis of magnetic circuit and single phase transformer 6,7,8

3 Basic concepts and construction feature of transformer 9,10,11

4 Voltage , current and impedance transformation EMF equation 11,12,13

5 Equivalent circuits and phasor diagrams 13, to 17

6 Voltage regulation 18,

7 Losses and efficiency 19,20,21

8 Open circuit test 22,23

9 Short circuit test. 23,24

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REVIEW OF LAWS OF ELECTROMAGNETISM M.M.F FLUX AND THEIR

RELATION

The space around the poles of a magnet is called magnetic field. The force in the space

around a magnet can be pictured by examining the pattern made by iron fillings. These

chain of iron fillings to the assumption that the region (field) contains invisible lines of

force. The total number of lines of force surrounding a magnet , is called the total flux.

The lines of flux of N and S pole attract each other

The lines of same pole that is N N gives the lines which repel each other

LINE OF INDUCTION

Lines of flux travels from N pole to S pole and continues to travel through the magnet and

finally they reach the N pole again forming closed curves. The portion of the curves within

the magnetic material are called LINES OF INDUCTION

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MAGNETIC FLUX

Magnetic flux is the total number of lines of force comprising the magnetic field . It is

represented by ᶲ and is measured in Weber.

MAGNETIC FLUX DENSITY

Magnetic flux density is defined as the magnetic flux passing through per unit area of

material through a plane right angles to the direction of flux. This is also known as magnetic

induction It is represented by B.

B= ᶲ/a Magnetic flux density is scalar quantity.

RELATION BETWEEN MAGNETIC FIELD INTENSITY H AND INDUCTION

DENSITY

The field intensity H is the because of the flux density B ‘s effect . Thus flux density can be

assumed to be proportional to the field intensity in a magnetic field i.e free space

Β =µ0 H

B = Webers per square meter

H= in newton per weber ( amper turn/meter

µ0 = is the magnetic space constant

for free space the value of µ0 = 4π 10¯7 H/M

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MAGNETO MOTIVE FORCE

M.MF of the magnetic circuit is defined as the magnetic potential that derives or tends to

derive flux around the magnetic circuit and is analogous to the e.m.f in an electric circuit. It

is measured is ampere turns AT.

MAGNETIC FIELD INTENSITY

M.M F per unit length ( along the path of magnetic flux) is called the magnetic field

intensity H is given by

H = M.M.F /Length AT/mtr

RELUCTANCE

It is the name given to that property of material which opposes the creation of flux in it, It is

analogous to resistance of an electric circuit. It is measured in ampere turn/Wb

PERMEABILITY

It is the measure of receptiveness of material of having magnetic flux developed in it.

Every substance posses a certain power of conducting magnetic flux. For example iron is

better conductor for magnetic flux than air. It is the ratio of flux density B and magnetic

field strength H

µ = B/H

ANALYSIS OF MAGNETIC CIRCUITS SINGLE PHASE TRANSFORMER

Magnetic Circuits with Air Gap

Energy conversion devices which incorporates a moving element have air gaps in heir

magnetic circuits. Airgaps are also provided in the magnetic circuits to avoid saturation. The

length of air gap is Lg is equal to the distance between two magnetic surfaces. When the air

gap length is very smaller than the adjacent core faces, the magnetic flux Φ is constrained

essentially to reside in the core and air gap is continuous through magnetic circuits. Then

configuration A can be analyzed as the magnetic circuit li and air gap permeability µo and

length lg. Since the permeability of the air is constant, the air gap is linear part of the

magnetic circuit and flux density in the air gap is proportional to the m.m.f is calculated

separately for the air gap and iron portion and than added together to determine the total

m.m.f

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COMPOSITE CIRCUIT Consider a circular ring made from different material of l1, l2 and

l3 having cross sectional are a1, a2 and a3 and relative permeability µr1,µr2 and µr3 respectively

with a cut of length lg known as air gap. The total reluctance as they are joined in series.

There fore

Total reluctance = l1 + l2 + l3

µ0µr1 a1 µ0µr2 a2 µ0µr3 a3

Total M M F =Φ Xs = Φ [ l1 + l2 + l3 ]

µ0µr1 a1 µ0µr2 a2 µ0µr3 a3

Total ampere turn required = H1l1 + H2 l2 + H3 l3 + Hglg

Sum of ampere turns required for individual parts of magnetic circuit

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PARALLEL MAGNETIC CIRCUITS

In series circuit all parts of the magnetic circuit carry same flux and total ampere turns

required to create a given flux is the arithmetic sum of the ampere turns required for

individual parts of circuit .

But if the various paths of the magnetic circuit are parallel as shown in figure, the ampere

turns required for the combination is equal to the ampere turns required to create the given

flux in one path.

In circuit ABCD and AFED are in parallel, so ampere turns required to create flux Φ, in

path ABCD is equal to ampere turns required create flux Φ2 in path AFED and also equal to

the ampere turns required for both the parts.

Hence total ampere turns required for the magnetic circuit= AT for path DA +AT for path

ABCD = AT for path DA +AT for path AFED.

SELF INDUCED E.M.F

When current flowing through the coil is changed, the flux linking with its own winding

changes and due to the change in linking flux with the coil, an E.M.F is induced. This

known as self induced EMF.

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MUTUALLY INDUCED E.M.F

When ever current in coil A changes, the flux linkage with coil B changes and an EMF in

induced. This induced EMF in coil B is known as mutually induced EMF.

SINGLE PHASE TRANSFORMER

It is a static machine.

i) Transfers electric energy from one electric circuit to another electric circuit.

ii) It does not change the frequency

iii) It works on the principle of electro magnetic induction

iv) It has electric circuits which are linked by a common magnetic circuits.

STEP UP TRANSFORMER

When the transformer raises the out put voltage compare to input voltage it is called the step

up transformer.

STEP DOWN TRANSFORMER

When the transformer reduces the out put voltage compare to input voltage it is called the

step down transformer.

BASIC CONCEPTS AND CONSTRUCTION FEATURES OF TRANSFORMER

It essentially consists of two separate windings placed over the laminated silicon steel core,

The windings to which a.c supply is connected is called primary winding and the winding to

which load is connected is called secondary winding.

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When a.c supply of voltage V1 is connected to the primary winding, an alternating flux is

set up in the core. This alternating flux when links with secondary winding, an EMF is

induced in it called mutually induced EMF. The direction of this EMF is opposite to the

applied voltage.

The same alternating flux links with primary winding and produces self induced EMF E1,

The EMF E1 also acts opposite to applied voltage V1 as per LENZ’s Law.

Although there is no connection between primary and secondary winding but the electrical

power is transferred from one circuit to another circuit through mutual flux .

The induced EMF in the primary and secondary winding depends upon the rate of change of

flux linkage

That is N dΦ/dt

dΦ/dt is Same for primary and secondary windings. The induced EMF in primary winding

E1 is proportional to N1 ( Number of turns in primary windings) and secondary winding E2

α N2.

TURN RATIO

The ratio of number of turns in primary winding N1 and secondary winding N2 is called

turn ratio. = N1/N2

TRANSFORMATION RATIO

The ratio of secondary voltage E2 to the primary voltage E1 is called transformation ratio. It

is represented by K.

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K= E2/E1 = N2/N1

CONSTRUCTION FEATURES OF TRANSFORMER

The transformer consists of following parts

i) Magnetic Circuit core of transformer

ii) Electric circuit Primary and secondary windings

iii) Insulation of windings

iv) Tanks, cooling methods, conservator, bushings and protective relay

TYPES OF TRANSFORMER AS PER CONSTRUCTION

The type of transformer according to the core construction and the manner in which primary

and secondary windings are placed around it, there are two type of transformers.

a) CORE TYPE TRANSFORMER

b) SHELL TYPE TRANSFORMER

CORE TYPE TRANSFORMER

In a core type transformer the magnetic core is built up of lamination to form a

rectangular frame. The laminations are cut in L shape. In order to avoid high

reluctance at the joints where laminations are butted against each other, the alternate

layers are stacked differently to eliminate continuous joints. While placing primary

winding an insulation layer ( Bakelite former) is provided between core an winding.

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SHELL TYPE TRANSFORMER

The laminations are cut and formed E’s and I’s . In order to avoid high reluctance at

the joints where laminations are butted against each other, the alternate layers are

stacked differently to eliminate continuous joints.

In shell type transformer there are three limbs, The central limb carries the whole flux

while the side limbs carries half flux. The width of central limb is double than the

outer limb. Windings both primary and secondary are placed on the central limb.

VOLTAGE, CURRENT AND IMPEDANCE TRANSFORMATION

IDEAL TRANSFORMER

An Ideal transformer is one which there is no ohmic resistance and no magnetic

leakage flux, i.e all flux produced in the core links with primary as well as secondary

. Hence transformer has no copper losses and core losses. It means an ideal

transformer consists of two purely inductive coils wound on loss free core. In actual

practice it is not possible In an ideal transformer there is no power loss.

E2I2cosΦ = E1I1 cosΦ

E2I2c = E1I1

E2 /E1 =I1/ I2

E2 α N2, E1 αN1 , E2 ≡ V2 , E1≡ V1

E2 /E1 =I1/ I2 = N2/ N1 = V2/ V1

The currents are inversely proportional to the transformation ratio.

E.M.F EQUATION

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When a sinusoidal Voltage is applied to the primary winding of a transformer a

sinusoidal flux is set up in the iron core which links with primary and secondary

windings.

Let

Φm = Maximum value of flux in weber

f = supply frequency in Hz

N1 = No of turns in primary winding

N2 = No of turns in secondary winding

Flux change from Φm to – Φmin half cycle i.e 1/2f seconds

Average rate of change of flux = Φm – (– Φmin)

1

2f

= (Φm + Φm) 2f webers

= 4 Φmf

Now the rate of change of flux per turn is the average induced emf per turn in volts

Therefore Average e.m.f induced / turn = 4 Φmf volts

Form factor for sinusoidal wave =1.11 = R.M.S value/Average value

Considering form factor = 1.11

R.M.S value of induced emf /turn = 1.11X4 Φmf = 4.44 Φmf

E.M.F induced in primary winding E1 depends on number of turn of primary winding

N1

So

E1 = 4.44 ΦmfN1

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Similarly for secondary winding

E2 = 4.44 Φmf N2

In the form of flux density Bm

Φm = BmAi Ai is the iron core area

E1 = 4.44 Bm Aif N1 Volts

E2 = 4.44Bm A2 f N2 volts

TRANSFORMER ON NOLOAD

On no load a small current I0 is drawn by the primary winding when the secondary

winding is open. This current is called exciting current, magnetizing current. This

current has to supply the iron losses in the core and a very small amount of copper

loss.

The no load current I0 has two components

1) Iw is in the phase with voltage V1 called active voltage or working component it

supplies iron loss and small component of copper loss.

2) Im is in the quadrature with the applied voltage V1 , called reactive or

magnetizing component

Working component = Iw = I0cosΦᵒ

Im = I0sinΦᵒ

No load current I0 =√Iw² + Im²

No load power P0= V1 I0cosΦᵒ

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TRANSFORMER ON LOAD

When the secondary circuit of a transformer is completed through an impedance or

load, the secondary current I2 starts flowing through secondary winding. The

magnitude of I2 and voltage V2 will depend upon the load characteristics. I2 will be in

phase with V2 or it can lag or lead the V2 . The Secondary current I2 sets up its own

m.m.f and hence create secondary flux Φ2 , which opposes main flux Φ momentarily

gap between V1 and back emf E1 increases. The I1’ primary current increase to

maintain the value of Φ and the gap between V1 and E1 reduces , until the original

value of flux Φ is achieved . The current The I1’ is in phase opposition to I2 and is

called COUNTER BALANCING CURRENT. The phasor representation is given

below.

EQUIVALENT CIRCUITS AND PHASOR DIAGRAMS

Equivalent resistance and reactance:- The two independent circuits of a transformer can be

resolved into an equivalent circuit to make the calculation simple.

Let the resistance and reactance of primary and secondary windings of transformer be R1 ,R2

, X1 and X 2 ohms and transformation ratio K

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Resistance drop in primary winding I1 R1

Reactive drop in primary winding I1 X1

Resistance drop in secondary winding I2 R2

Reactive drop in secondary winding I2 X2

REFERRED TO SENDARY SIDE Since K is the transformation ratio, resistive and

reactive drop referred to secondary side shall be K times , i.e KI1 R1 and K I1 X1

If I1 Is substituted as equal to KI2 than we get K² I2 R1 K²I2 X1

Total resistive drop = K² I2 R1 + I2 R1 = I2 (K² R1 + R1) = I2 R02

Total reactive drop = K²I2 X1 + I2 X2 = I2(K² X1 + X2) = I2 X02

From Phasor diagram

KV1 =√( V2 + I2 R02 cosΦ + I2 X02 sinΦ)² + (I2 X02 cosΦ -- I2 R02 sinΦ)²

I2 secondary current which lags V2 by angle Φ

Since the term (I2 X02 cosΦ -- I2 R02 sinΦ)² is very small to compare to (I2 R02 cosΦ + I2 X02

sinΦ)

Neglecting small value we get

KV1= V2 + I2 R02 cosΦ + I2 X02 sinΦ

V2 = KV1- I2 R02 cosΦ - I2 X02 sinΦ

If the load is pure resistive Φ=0

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V2 = KV1- I2 R02

If the load is capacitive than Φ is negative

V2 = KV1- I2 R02 cosΦ + I2 X02 sinΦ

REFERRED TO PRIMARY SIDE

Resistance drop in secondary winding I2 R2 EVIDED /K

Reactive drop in secondary winding I2 X2/K

REFERRED TO PRIMARY SIDE Since K is the transformation ratio, resistive and reactive

drop referred to PRIMARY side shall be divided by K , i.e I2 R2 /K and I1 X1 /K

If I2 Is substituted as equal to I1/K than we get I1 R2 / K² K²I1 X2/ K²

Total resistive drop = I1 R2/K² + I1 R1 = I1 (R1 + R1 / K²) = I1 R01

Total reactive drop = I1 X1 + I1 X2 / K² = I2(X1 + X2//K²) = I1 X01

From Phasor diagram

V1 =√( V2 /K+ I1 R01 cosΦ + I1 X01 sinΦ)² + (I1 X01 cosΦ -- I1 R01 sinΦ)²

I2 secondary current which lags V2 by angle Φ

Since the term (I1 X01 cosΦ -- I1 R01 sinΦ)² is very small, neglecting small value

we get

V1= V2/K+ I1 R01 cosΦ + I1 X01 sinΦ

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EQUIVALENT CIRCUITS OF TRANSFORMER

The equivalent circuit of any device can be quite helpful in predetermination of the

behavior of the device under various condition of operation.

Equivalent circuit of a transformer having transformation ratio K

E2/E1=K

E1 is the induced emf due to V1 less primary voltage drop. This voltage causes iron loss

current Ie and magnetizing current Im, these two components are represented by R0 and X0

as pure resistance and pure reactance X0

Secondary current I2 =I1’/K

Equivalent Diagram of a transformer with all secondary impedance transferred to primary

side.

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Approximate equivalent diagram of a transformer.

PHASOR DIAGRAM OF TRANSFORMER WITH RESISTIVE,REACTIVE AND

CAPACITIVE LOAD

VOLTAGE REGULATION

The way in which the secondary terminal voltage with load depends upon load

current, the internal impedance and the load power factor. The change in secondary

voltage from no load to load with primary voltage and frequency held constant, this

termed as the inherent regulation.

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The voltage characteristics depends on the voltage regulation of transformer. It is

expressed in percentage of rated terminal voltage (secondary). It is defined as the

change in magnitude of terminal voltage (secondary) when the load is thrown off

( reduced to no load) with primary voltage and frequency constant.

If V2 is the secondary voltage at full load power factor and E2 is the secondary

voltage at no load

Voltage regulation = (E2-V2)/E2

As per IS the secondary rated voltage of transformer is equal to secondary voltage at

no load.

% Voltage regulation = {(E2-V2)/E2}100

= Voltage drop / No load secondary rated voltage

The voltage drop = I2 R02 cosΦ + I2 X02 sinΦ

So % regulation = (I2 R02 cosΦ + I2 X02 sinΦ)100

E2

When the power factor is leading the percentage regulation becomes

%Regulation = (I2 R02 cosΦ - I2 X02 sinΦ)100

E2

So we can write

%Regulation = (I2 R02 cosΦ ± I2 X02 sinΦ)100

E2

+ve sign for lagging P.F

-ve sign for leading P.F

Voltage regulation on an average in transformer is 4%. From consumer point of

view, voltage regulation due to variation in load is not desirable.

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DETERMINATION OF REGULATION FROM OPENCIRCUIT AND SHORT

CIRCUIT TEST

As we know

%Regulation = (I2 R02 cosΦ - I2 X02 sinΦ)100

E2

Equivalent R02 and X02 can be easily determined from short circuit test , I2 is the

secondary current and cosΦ is the power factor either leading or lagging. No load

secondary terminal voltage is equal to emf induced E2.

Open circuit data are not required for determination of regulation

LOSSES AND EFFICIENCY

There are two type of losses in transformer.

1) Iron loss or core loss , fixed loss

2) Copper loss or ohmic loss

Iron losses:-Iron losses are due to alternating flux in the core and consists of

hysteresis and eddy current losses.

a) Hysteresis Losses:- The core of a transformer is subjected to an alternating

magnetizing force and for each cycle of e.m.f. A hysteresis loop is traced out. The

hysteresis loss is given by

Ph = ń(ßmax) ˣ f v joules per second or watts

ń = hysteresis coefficient

ßmax = flux density maximum

f = frequency, The value x varies from 1.5 to 2.5 and it depends upon the material.

b) Eddy current losses:- We have seen whenever flux linkage with closed electric

circuit changes an e.m.f induced in the circuit and current flows, If the magnetic

circuit is made up of iron and if the flux in the circuit is , current will be induced by

induction in iron circuit itself. All such currents are called eddy currents.

Pe = Ke (ßmax)² t² f² v watts

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COPPER LOSS:- These losses occur due to the ohmic resistance of the transformer

winding. If I1 and I2 are the primary and secondary currents, R1 and R2 are primary

and secondary winding resistance. =

In that case the copper losses are = I1² R1 + I2 ² R2

These losses vary in the square of load current, If the copper loss at full load is Pc

than copper loss at ½ load and 1/3 load shall be

(½)² Pc = Pc/4,

(1/3)² = Pc/9

Copper losses can be determined by short circuit test.

TRANSFORMER EFFICIENCY :- The efficiency of transformer is defined as the

ratio of out put power to the input power. Both are to be measured in same unit either

in watts or Kw

Therefore ƞ = output output

Input input + losses

ƞ = output power

output power +losses

= V2 I2 Cosϕ2

V2 I2 Cosϕ2 + Pi + Pc

V2 Secondary voltage

I2 Secondary current

ϕ2 Power factor of load

Pi = Iron loss = eddy current loss + hysteresis loss

Pc = Copper loss = I2² Res

If x is the fraction of load than the efficiency at x

Ƞx = Xoutput

Xoutput + Pi + x² Pc

CONDITION FOR MAXIMUM EFFICIENCY :- The terminal voltage V2

is approximately constant, Thus for a given P.F efficiency depends on load current I2

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ƞ = V2 I2 Cosϕ2

V2 I2 Cosϕ2 + Pi + Pc

In the above expression numerator is constant and efficiency will be maximum when

denominator is minimum . Thus the maximum condition is obtained by

differentiating the quantity in the denominator w.r.t I2

d (V2 I2 Cosϕ2 + Pi/ I2 + I2 Res) = 0

d I2

0 = - Pi/ I2² + Res

Pi = I2²Res

Iron loss = copper loss

I2 =√ Pi/Res

If x is the fraction of full load KVA at which the efficiency of transformer is Maximum

Then copper loss = x²Pc

Iron loss = Pi

x²Pc = Pi

x = √Pi/Pc

therefore output KVA corresponding Maximum efficiency

= full load kva √iron loss/copper loss at FL

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ALL DAY EFFICIENCY

The load on certain transformer fluctuates throughout the day , The distribution

transformers are energized for 24 hours, but they deliver very light loads for the major

portion of the day. Thus iron loss occur for the whole day but copper loss occurs only when

transformers are loaded. Hence transformer cannot be judged for the commercial efficiency.

All day efficiency is the ratio of Kwh output to Kwh input for 24 hours

Ƞ allday = output in kwh ( 24hrs)

Input in kwh

The find out all day efficiency , we must know the load cycle of the transformer.

OPEN CIRCUIT TEST

PURPOSE OF THIS TEST IS TO DETERMINE

•Core loss or Iron loss Or Magnetic loss (Pi)

•No load current (Iо)

•Shunt branch parameters R о and X о

One of the winding is kept open.

Rated voltage at rated frequency is applied to other(LV) winding.

A voltmeter, wattmeter, and an ammeter are connected in LV side of the transformer.

Ammeter > Reads No-Load Current, I о

Voltmeter > Reads Applied Voltage, V о

Wattmeter> Reads No-Load Input Power, W о or P о

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DETERMINATION OF EQUIVALENT CIRCUIT CONSTANTS THROUGH NO-

LOAD TEST

No load power factor, CosΦо= W о / V о I о

Core loss component, Iw = I о CosΦ о

Magnetising component, Im = I о SinΦ о

Core Loss, Pi = No load power (W о)

Core loss resistance, R о = V о / Iw = V о / I о CosΦ о

Magnetising reactance, X о = V о / Im = V о / I о SinΦ о

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SHORT CIRCUIT / IMPEDANCE TEST ON TRANSFORMER

PURPOSE OF THIS TEST IS TO DETERMINE

•Zо₁ or Z о ₂ – Total impedance referred to either primary or secondary side

•R о ₁ or R о ₂ - Total resistance referred to either primary or secondary side

•X о ₁ or X ₂ - Total reactance referred to either primary or secondary side

•Full load cu loss I ₂ ² R о ₂

*In this test one of the winding is short circuited by thick conductor.

* Current rating of HV side is low compared with LV side.

* Power input gives total cu loss at rated load.

* Unity power factor wattmeter is used for measuring power in SC test.

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DETERMINATION OF EQUIVALENT CIRCUIT CONSTANTS THROUGH LOAD

TEST

SC power factor, CosΦsc = Wsc / Vsc Isc

Resistance of transformer referred to primary side ,

Rо ₁ = Wsc / (Isc)2

Reactance of transformer referred to primary side ,

X о ₁ = Z о ₁ SinΦsc = 𝐙 о ₁ 𝟐−𝐑 о ₁ 𝟐

Impedance of transformer referred to primary side,

Z о ₁ = Z о ₁ Cos Φsc= Vsc / Isc

RGPV BE 104 Unit II

Engineering

Transcript of RGPV BE 104 Unit II