Unit III (Transformer)

8/18/2019 Unit III (Transformer)

1/23

Lecture NotesElectrical and Electronics Engineering

TRANSFORMERS

UNIT-III

Transformer is an ac machine, the main advantage of alternating currents overdirect currents is that, the alternating currents can be easily transferable fromlow voltage to high or high voltage to low.

Alternating voltages can be raised or lowered as per the requirements in thedierent stages of electrical network as generation, transmission, distributionand utilization. This is possible with a device called transformer.Defnition:

Transformer is a static piece of apparatus by means of which electrical poweris transferred from one alternating current circuit to another with the desiredchange in voltage or current and without any change in the frequency.

Princile o! oeration:t is a static machine and it works on the principle of statically induced emf. tconsists of:

• !agnetic circuit and

• "lectric circuit

Two separate electrical windings are linked through a common magneticcircuit. The two electrical windings are isolated from each other.

The coil in which electrical energy is fed is called primary winding while the

other from which electrical energy is drawn out is called secondary winding.

The primary winding has #$ number of turns while secondary winding has #%number of turns.

&hen primary winding is e'cited by alternating voltage say ( $, it circulatesalternating current $ through it. This current produces an alternating )u' *φ+which completes its path through the common magnetic core.

This )u' links with both the windings. ecause of this, it produces self inducedemf "$ in the primary winding while due to mutual induction i.e. due to )u'produced by primary linking with secondary, it produces induced emf " % insecondary winding.

These emf+s are:

"ard#a$an %ollege o! EngineeringDeart$ent o! EEE

1


2/23


dt

d N E φ

11 −=dt

d N E φ

22 −=

f now secondary circuit is closed through the load, the mutually induced emf in the secondary winding circulates current through the load. Thus electricalenergy is transferred from primary to secondary with the help of magneticcore. A voltage (% appears across the load. -ence ($ is the supply voltage,while (% is the secondary voltage when load is connected, then:

==

1

212

2

1

2

1

N

N V V or

N

N

V

V

===1

2

1

2

E

E

N

N k

Transformation ratiof k $, then (% ($, transformer is called step up transformer.f k / $, then (% / ($, transformers is called step down transformers.f k 0 $, then (% 0 ($, then transformer is called one to one transformer.

The current )owing through primary is $ and when load is connected current %)ows through secondary voltage. The power transfer from primary to

secondary remains the same. Assuming both primary and secondary powerfactor to be the same, we can write:

1ower input to primary 0 1ower output from secondary

($ $ 0 (% % 2

1

2

1

1

2

2

1

E

E

N

N

I

I

V

V ===

%onstruction o! a trans!or$er:

There are basic parts of a transformer:

$2 !agnetic core %2 &indings or coils 32 Tank or ody 425onservator Tank62 reather 72 8adiator 92 ushings


2


3/23


%ore: The transformer core is made of silicon steel or sheet steel with 4

silicon. The sheets are laminated and are coated with ;'ide layer to reduceiron losses. The thickness of lamination is grade siliconsteel laminations are used. These laminations are insulated from each other by

using insulation like varnish.

The purpose of the core is to provide magnetic path of low reluctance betweenthe two windings so that the total )u' produced by one of the winding will belinked fully with the other winding without any leakage.

&indings: A transformer has two windings. The winding which receiveselectrical energy is called 1rimary winding and the winding which deliverselectrical energy is called ?econdary winding. &indings are generally made upof -igh grade copper. The windings are provided with insulation so that onewinding may not come in contact with the other winding. =enerally cotton,1aper and ;'ide layer is used as insulating medium.

Tan' or (od): t is part which is meant to carry the transformer and the oilused in the transformer. The tank used for a transformer should be air tight sothat moisture should also not enter into the tank so as to maintain theproperties of the transformer oil.

Trans!or$er Oil: t is the most important part of a transformer which decidesthe life of a transformer. The oil that is used in a transformer should be safeguarded properly so as to have a good life for a transformer.

%onser*ator Tan' : &hen a transformer is oil @lled and self cooled the oil in

the tank is subected to heat and thus will naturally e'pand and contract dueto variations in the load current and is also subected to seasonal variations.

The conservator tank provides the means for the oil to settle down bye'panding under heavy loads.

(reat#er: Transformer oil should not be e'posed to atmosphere directlybecause it may absorb !oisture and dust from the environment and may looseits electrical properties in a very short time. To avoid this from happening abreather is provided. The breather completely prevents the moisture and dustfrom coming into contact with the oil in the conservator tank when it e'pandsor contracts.

(us#ings: The purpose of ushings is to provide proper insulation for theoutput leads to be taken out from the transformer tank. ushings are generallyof two types.a+ 1orcelain type which are used for voltage ratings uptp33kv


3


4/23


,+ 5ondenser type and ;il @lled type are used for rating above 33kv

Radiator: These are meant to increase the surface area of the tank also toprovide a path for the circulating of the transformer oil.

T)es o! Trans!or$ers:

The transformers are classi@ed based on the relative position or arrangementof the core B the windings, based on cooling and based on (oltage.

a2 ased on arrangement of the core B the windings transformers areclassi@ed as

• 5ore type

• ?hell type

5ore type Transformer

t has a single magnetic circuit. n this type, winding encircles the core, coilsused are of cylindrical type. ?uch coils are wound in helical layers withdierent layers insulated from each other by paper, cloth, mica, etc. 5ore ismade up of large number of thin laminations to reduce eddy current losses.

The windings are uniformly distributed over two limbs and hence naturalcooling is more eective.

The coils can be easily removed by removing laminations of top yoke formaintenance.

?hell type transformer


4


5/23


t has a double magnetic circuit. n this type core encircles the most part of thewinding. The core is again laminated one and while arranging the laminations,care is taken that all oints at alternate layers are staggered.

This is done to avoid narrow air gap at the oint, right through the cross sectionof the core. ?uch oints are called as overlapped. The coils are multi>layered

disc type or sandwich type coils and are placed on only one limb and aresurrounded by the core. ?o natural cooling does not e'ist.

,+ (ased on "oltage

• ?tep up

• ?tep Cown

?tep up transformer is a transformer where the ;utput (oltage isgreater than input (oltage i,e (%($ ?tep down transformer is a transformer where the ;utput (oltage isless than input (oltage i,e (%/($

Transformation 8atio 0(%D($0#%D#$ &here #%, #$ are number of turns in secondary and primary windings.

c+ (ased on cooling:

• ;il cooled

• ;il @lled water cooled

• Air cooled

EMF euation o! a trans!or$er:

1rimary winding is e'cited by a voltage, which is alternating in nature. Thiscirculates current through primary, which is also alternating and hence the )u'produced is also sinusoidal in nature.

Eet φ 0 )u' in the core

φ m 0 m F A

#$ 0 number of turns in the primary winding


5


6/23


#% 0 number of turns in the secondary winding

f 0 frequency of ac input in -z.

The )u' increases from its zero value to ma'imum value φm in one quarter of the cycle i.e., in G f second.

Average rate of change of )u' 0f

m

¼

φ

0 4 f φ m wbDsec8ate of change of )u' per turn means induced emf in volts

Average emf D turn 0 4 f φm volt.

f )u' φ varies sinusoidally then rms value of induced emf is obtained bymultiplying the average value with the form factor.

Horm factor 011.1=

valueaverage

vauerms

rms value of emfD turn 0 $.$$ ' 4 f φm 0 4.44 f φm

#ow rms value of induced emfin the whole of primary winding 0 induced emf D turn ' number of primaryturns

"$ 0 4.44 f #$ φm 0 4.44 f #$ m A

?imilarly, rms value of emf induced in secondary is:

"% 0 4.44 f #% φm 0 4.44 f #% m A

mf N

E

N

E φ 44.4

2

2

1

1 ==

That is, emfD turn is same in both primary and secondary windings.

n an ideal transformers on no load ($ 0 "$ B "% 0 (%

&here, (% is the terminal voltage.

Ideal trans!or$ers

Transformer is called ideal if it satises the following properties:

$. t has no losses

%. ts windings have zero resistance

3. Eeakage )u' is zero i.e. $


7/23


8/23


Lor "% 0 J"$ 0 3


9/23


$. The no>load primary current o is very small as compared to the full>loadprimary current. t is about $ percent of the full load current.

%. ;wing to the fact that the permeability of the core varies with theinstantaneous value of the e'citing or magnetizing current is not trulysinusiodal. As such, it should not be represented by a vector becauseonly sinusoidally varying quantities are represented by rotating vectors.

3. As o is very small, the no load primary copper loss is negligibly smallwhich means that no load ri$ar) inut is racticall) eual to t#eiron loss in t#e trans!or$er/

4. As it is principally, the core loss which is responsible for shift in thecurrent vector, angle o is 'no.n as #)steresis angle of advance.

Pro,le$:

a2 A %,%


10/23


The additional primary mmf #$ %$ sets up its own )u' φ%l which is in opposition

to φ% Lbut is in the same directions as φ2 and is equal to it in magnitude. -ence,the two cancel each outer out. ?o, the magnetic eects of secondary current %are immediately neutralized by the additional primary current %l which isbrought into e'istence e'actly at the same instant as %. 0ence .#ate*er t#eload conditions1 t#e net 2u3 assing t#roug# t#e core isaro3i$atel) t#e sa$e as at no load/

φ% 0 φ%l

#% % 0 #$ %l

221

22 kI I x

N

N I l ==

-ence, when transformer is on load the primary winding has two currents in itNone is phase with % and k times inmagnitude. The total primary current is the vector sum of inductive,inductive and when it is capacitive is shown. (oltage transformation ratio of unity is assumed so that primary vectors are equal to secondary vectors.


10


11/23


&ith reference to the @g. La2 % is secondary current in phase with "% Lit shouldbe v%2. t causes primary current %l which is anti>phase with it and equal to it in

magnitude Lk 0 $2. The total primary current $ is the vector sum of


12/23


9I


13/23


3. Eet % be the load current in line with (% Lresistive load2

4. At the tip of (%, draw A parallel to % to represent % 8% drop Lin line with(%2

6. At *+ draw 5 at right angles to A to represent % F% meeting the

vertical line at *5+ then ;5 is equal to secondary induced emf "%.

7. 1roduce *;+ backwards to *C+ such that ;C 0 ;5Dk then ;C representsthe primary induced emf.

9. 1roduce the current line % backwards to H such that ;H 0 J%. Eet *;"+ beequal to the load current *


14/23


$% 8%l 0 %% 8%

8%l 0 L%D$2% 8%.

f no load current


15/23


8$ 0


16/23


Hurther simpli@cation may be achieved by omitting


17/23


& 0 ($ o cos φo.

5os φo 0 &D($o

µ 0 o sin φo

w 0 o cos φo

Fo 0 ($Dµ B 8o 0 ($Dw.

o 0 ($ So

So 0 oD($

=o is given equation & 0 ($% =o

=o 0 &D($%

o 0 GY 2

0

2

0 −


17


18/23


S#ort-circuit or I$edance Test:

n this test, one winding usually the lo.-*oltage .inding is solidl) s#ort 6circuited ,) a t#ic' conductor/

This method is used to fnd the ollowing parameters.

$. "quivalent impedance LQo$ or Qo%2, leakage reactance LFo$ or Fo%2 andtotal resistance L8o$ or 8o%2 of the transformer as refereed to the windingin which the measuring instruments are placed.

%. 5u loss at full>load Lat any desired load2. This loss is used in calculatingthe eciency of the transformer.


19/23


Core losses:

t includes both the hysteresis loss and eddy current loss. These losses areminimized by using steel of high silicon content for the core and by using verythin laminations.

ron or core loss is found from the ;.5 test. The input of the transformer whenon no>load measures the core loss.

Hysteresis loss:

&hen a magnetic material is subected to repeated cycles of magnetizationand demagnetization it results into disturbance and there will be loss of energy and this loss of energy appears as heat in the magnetic material. Thisis called as hysteresis loss.

0)steresis loss 5 ' # ($7/= ! 3 *olu$e1 .atts

&here, J h 0 constant m 0 ma'imum )u' density f 0 frequency

The hysteresis loss can be reduced by using thin laminations for the core.

Eddy current loss:

Cue to alternating )u'es linking with the core, eddy currents get induced inthe laminations of the core. ?uch eddy currents cause the eddy current loss inthe core and heat up the core.

"ddy current loss can be reduced by selecting high resistivity material likesilicon.

The most commonly used method to reduce this loss is to use laminatedconstruction to construct the core. 5ore is constructed by stacking thin piecesknown as laminations. The laminations are insulated from each other by thinlayers of insulating material like varnish, paper, mica. This restricts the pathsof eddy currents, to respective laminations only. ?o area through whichcurrents )ow decreases, increasing the resistance and magnitude of currentsgets reduced.

Edd) current loss 5 > e ($4

! 4

t4 1

.atts

&here,J e 0 constant

m 0 ma'imum )u' density f 0 frequency t 0 thickness of the laminations

Copper loss:

This loss is due to the ohmic resistance of the transformer windings.

Total copper loss 0 I74 R7 I44 R4 5 I74 R97 5 I44 R94

5opper loss is proportional to Lcurrent2% or LJ(A2%


19


20/23


5opper loss at half the full>load is one fourth of that at full>load.

E?cienc):

ironlosslossuoutput

output

input

output

++

=.

input

losses

input

lossesinput −=

−1

5opper loss 0 $% 8loss equal to the ironloss.

$. f we are given iron loss and full load 5u loss, then the load at which twolosses would be equal

losscuL.F.

lossIron xload Full

%. "ciency at any load

100.).!

.

.

X ! ! f pkVAload full x

f pkVAload full x

i u ++−−=η


20


21/23


Trans!or$er rating in >"A:

5u loss of a transformer depends on current and iron loss on voltage. -ence,total transformer loss depends on volt>ampere L(A2 and not on phase anglebetween voltage and current i.e., it is independent of load power factor. That iswhy rating of transformer is in J(A and not in J&.

Regulation o! a Trans!or$er

L$2 &hen a transformer is loaded with a constant primary voltage, thesecondary voltage decreases because of its internal resistance andleakage reactance.

Eet, load 0 "% 0 J"$ 0 J($ because at no>load the impedance drop is negligible.

(% 0 secondary terminal voltage on full>load

The change in secondary terminal voltage from no > load to full>load isload to full>load,e'pressed as a percentage of no>load secondary voltage, is:

0 (r cos φ ± (' sin φ Lappro'imately2

where (r0 percent resistive drop 0 $


22/23


regulation 0100

1

121 x

V

V V −

f angle between ($ and (%U is neglected, then the value of numericaldierence ($ M (%U is given by L$ 8


23/23


Pro,le$ 4:

A $load J(A F loss"u#$

lossiron

..

0 $

Unit III (Transformer)

Documents