Power System 3 -Power Transformer EEE3233

8/14/2019 Power System 3 -Power Transformer EEE3233

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:HAPTER 2:HAPTER 2 OWER TRANSFORMEROWER TRANSFORMER

ur Diyana Kamarudin ur Diyana Kamarudin

EEE3233POWER SYSTEM


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IntroductionIntroduction A transformer is a static machinesstatic machines. The word transformer comes form the word transform. The word transformer comes form the word transform. Transformer is not an energy conversion devicenot an energy conversion device But is a device that changes AC electrical power at onedevice that changes AC electrical power at one

voltage level into AC electrical power at anothervoltage level into AC electrical power at anothervoltage level through the action of magnetic field,voltage level through the action of magnetic field,without a change in frequency.without a change in frequency.

Can raise or lower the voltage/current in ac circuitCan raise or lower the voltage/current in ac circuit

GenerationGenerationStationStation

T X 1 T X 1

DistributionsDistributions

T X 1

T X 1

Transmission SystemTransmission System

33/13.5kV33/13.5kV 13.5/6.6kV13.5/6.6kV

6.6kV/415V6.6kV/415V

Consumer Consumer


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TransformerConstruction

There are 3 basic parts of transformer:A primary coil/winding

receives energy from the ac sourceA secondary coil/winding

receives energy from primary winding &delivers it to the loadA core that supports the coils/windings.

provide a path for magnetic lines of flux


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rans ormerConstruction

The operation of transformer is based on the principal of mutual inductance

A transformer usually consists of two coils of wire wound onthe same core

The primary coil is the input coil while the secondary coil isthe output coil A changing in the primary circuit creates a changingmagnetic field

This changing magnetic field induces a changing voltage inthe secondary circuit

This effect is called mutual induction


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Transformer can be either step-up or step-downtransformer

If the output voltage of transformer is greater than theinput voltage step-up transformer

If the output voltage of a transformer is less than theinput voltage step-down transformer By selecting appropriate numbers of turns, a transformerallows an alternating voltage to be stepped up bymaking Ns more than Np

Or stepped down by making Ns less than Np

Vs NsVp Np

=


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Example 1 There are 400 turns of wire in an iron-corecoil. If this coil is to be used as theprimary of a transformer, how manyturns must be wound on the coil to

form the secondary winding of thetransformer to have a secondaryvoltage of one volt if the primaryvoltage is 5 volts?


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Example 1 (solution)


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Core characteristic: The composition of a transformer core depends on:

voltage, current, frequency, size limitations andconstruction costs

Commonly used core materials are air, soft iron, and steel Air-core transformers are used when the voltage source has

a high frequency (above 20 kHz) Iron-core transformers are usually used when the source

frequency is low (below 20 kHz) A soft-iron-core transformer is very useful where the

transformer must be physically small, yet efficient The iron-core transformer provides better power transfer

than does the air-core transformer


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A transformer whose core is constructed of laminated sheets of steel dissipates heat readily; thus it provides for the efficienttransfer of power.

The purpose of the laminations is to reduce certain losses whichwill be discussed later in this part

Hollow-core construction


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The most efficient transformer core is one that offers the bestpath for the most lines of flux with the least loss in magnetic andelectrical energy There are two main shapes of cores used in laminated-steel-core transformers:

Core-type transformers Shell-core transformers


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Shell-core transformers: The most popular and efficient transformer core Each layer of the core consists of E- and I-shaped sections of

metal These sections are butted together to form the laminations The laminations are insulated from each other and then pressed

together to form the core.


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Advantages of Shell types Transformer

Better cooling facilityLess leakage reactanceGreater mechanical strengthLess magnetising currentLess magnetic loss

D isd va n ta g e s o f S h e ll typ e sTra n sfo rm e r

M ore d ifficu lt fo r m an u factu rin g G re a te r d ifficu lty in ca rryin g ou t re p a irs


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Core - type construction: so named because the core is shaped with a hollow square

through the center the core is made up of many laminations of steel


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The basic difference between these two transformersare:

1)The core type has two limbs & shell type has three limbs.2)Core type has longer mean length of iron core & shorter

mean length of coil turn.

Shell type has shorter mean length of iron core &longer mean length of coil turn.3) In core type transformers the LV(low voltage) coil is wound

next to the core & HV(high voltage) coil is wound onthe LV coil after the insulation layer. In Shell type

transformers the LV & HV windings are sandwichedbetween each other.


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f


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An ideal transformer is a transformer which has no loseswhich has no loses,i.e. its winding has no ohmic resistance, no magneticleakage, and therefore no I 2R and core loses.

However, it is impossibleimpossible to realize such a transformer inpractice.

Yet, the approximate characteristic of ideal transformer willapproximate characteristic of ideal transformer willbe used in characterized the practical transformer.be used in characterized the practical transformer.

VV11 VV22

NN11 : N: N 22

EE 11 EE 22

II11 II22

VV11 Primary Voltage Primary VoltageVV22 Secondary Voltage Secondary VoltageEE 11 Primary induced Voltage Primary induced VoltageEE 22 secondary induced Voltage secondary induced VoltageNN11 :N:N22 Transformer ratio Transformer ratio


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No-load condition: is said to exist when a voltage is applied to the primary,

but no load is connected to the secondary Because of the open switch, there is no current flowing in

the secondary winding. With the switch open and an ac voltage applied to theprimary, there is, however, a very small amount of current called EXCITING CURRENT flowing in the primary


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With-load condition: When a load device is connected across the secondary

winding of a transformer, current flows through thesecondary and the load

The magnetic field produced by the current in thesecondary interacts with the magnetic field produced bythe current in the primary

This interaction results from the mutual inductancebetween the primary and secondary windings.


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Transformer Equation For an ac sources,

Let V(t) = V m sin ti(t) = i m sin t

Since the flux is a sinusoidal function;

Then:

Therefore:

Thus:

t t m sin)( =

t N dt

t d N Emf V

m

mind ind

cos

sin

=

==

maxmind ind fN N Emf V === 2(max)

maxmm

rmsind fN fN N

Emf =

=

= 44.42

22

)(

m m B x A =


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Transformer Equation For an ideal transformer,

In the equilibrium condition, both the input power will beequaled to the output power, and this condition is said toideal condition of a transformer.

From the ideal transformer circuit, note that,

1 1 2 2V I V I =

1 2

2 1

V I V I

=

(i)

2211 V E and V E ==

max

max

fN E

fN E = =

= =

22

11

44.4

44.4

Input power = output power


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Transformer Equationn

a I I

N N

E E

Therefore ===1

2

2

1

2

1,

aa = Voltage Transformation RatioVoltage Transformation Ratio ;which will determine whether the transformer is goingto be step-up or step-down

EE 22 > E> E 11For a >1For a >1

For a


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Transformer Equation Transformer rating is normally writtenwritten in terms of

Apparent Power Apparent Power. Apparent power is actually the product of its ratedits rated

current and rated voltagecurrent and rated voltage.

2211 I V I V VA == Where,

I1 and I 2 = rated current on primary and secondary

winding. V1 and V 2 = rated voltage on primary and secondarywinding.

**** Rated currents are actually the full load currents inRated currents are actually the full load currents intransformer transformer


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Example 1 1.5kVA single phase transformer has ratedvoltage of 144/240 V. Finds its full loadcurrent.

Solution:Solution:

A I

A I

FL

FL

62401500

42.10144

1500

2

1

==

==


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Example 2 A single phase transformer has 400 primary and

1000 secondary turns. The net cross-sectionalarea of the core is 60m2. If the primary winding

is connected to a 50Hz supply at 520V,calculate: The induced voltage in the secondary winding The peak value of flux density in the core


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Example 2 (solution)N1=400 V1=520V A=60m2 N2=1000

V2=?a) We know that,n

n

n

b) Emf,n

nn

n

2

520

1000

400

V =

2

1

2

1

V

V

N

N a

==V V 1300

2=

[ ]

[ ]

2

21

/0976.0

)60)()(400)(50(44.4520

44.4

1300,520,

44.4

44.4

mmWb B

B

A B fN E

V E V E known

A B fN

fN E

m

m

m

m

m

=

=

=

==

=

=


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Example 3 A 25kVA transformer has 500 turns on the

primary and 50 turns on the secondarywinding. The primary is connected to3000V, 50Hz supply. Find:

a) Full load primary currentb) The induced voltage in the secondary

windingc) The maximum flux in the core


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Equivalent Circuit of

Practical TransformerEquivalent circuit: Often given to explain the operation of a complicated or

unfamiliar device

If a circuit is truly an equivalent circuit, the original devicecan be removed from a system & replaced with itsequivalent circuit without changing the behavior orperformance of the system

For purposes of analysis the transformer may be representedby a 1:1 turns ratio equivalent circuit This circuit is based on the following assumptions: Primary and secondary turns are equal in number. One

winding is chosen as the reference winding; theother is the referred winding.
http://www.vias.org/eltransformers/gloss_turns_ratio.htmlhttp://www.vias.org/eltransformers/gloss_turns_ratio.html


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Practical Transformer Core loss may be represented by a resistance across theterminals of the reference winding.

Core flux reactance may be represented by a reactanceacross the terminals of the reference winding. Primary and secondary IR and IX voltage drops may belumped together; the voltage drops in the referredwinding are multiplied by a factor derived at the end of this section, to give them the correct equivalent value .

Equivalent reactance and resistances are linear.


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Equivalent Circuit of PracticalTransformer

V1 = primary supply voltageV2 = 2 nd terminal (load) voltageE1 = primary winding voltageE2 = 2 nd winding voltageI1 = primary supply currentI2 = 2 nd winding currentI1= primary winding currentIo = no load current

Ic = core currentIm = magnetism currentR

1= primary winding resistance

R2= 2 nd winding resistanceX1= primary winding leakage reactanceX2= 2 nd winding leakage reactanceR c= core resistanceXm= magnetism reactance

VV 11

II 11 R R 11XX11

R R CC

II cc

XXmm

II mm

II oo

EE 11 EE 22VV22

II 11

NN 11: N: N 22R R 22

XX 22

LoadLoad

II 22


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Equivalent Circuit of Practical TransformerSingle Phase transformer (referred to Primary)Sing le Phase transformer (referred to Primary)

Actual MethodActual Method

22

22

2

2

12 '' Ra ROR R N

N R =

=

22

22

2

2

12 '' X a X OR X

N

N X =

=a

I I

aV V ORV N N

V E

22

2222

1'21

'

'

=

=

==

VV 11

II 11 R R 11 XX 11

R R CC

II cc

XX mm

II mm

II oo

EE 11 EE 22 VV 22

II 22 NN 11: N: N 22

R R 22 XX 22

Load

II 22


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Equivalent Circuit of Practical TransformerSingle Phase transformer (referred to Primary)Sing le Phase transformer (referred to Primary)

Approximate MethodApproximate Method

VV11

II 11 R R 11XX 11

R R CC

II cc

XX mm

II mm

II oo

EE 11 EE 22 VV22

II 22 NN 11 : N: N 22R R 22

XX22

Load

II 22

22

22

2

2

12 '' Ra ROR R N

N R =

=

22

22

2

2

12 '' X a X OR X

N

N X =

= a I

I

aV V ORV N N

V E

22

2222

1'21

'

'

=

=

==


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Equivalent Circuit of Practical Transformer

Single Phase transformer (referred to Primary)Sing le Phase transformer (referred to Primary) Approximate MethodApproximate Method

22

22

2

2

12 '' Ra ROR R N

N R =

=

22

22

2

2

12 '' X a X OR X N

N X =

= '

'

2101

2101

X X X

R R R

+=+=

2222

1'2 ' aV V ORV N

N V =

=

In some application, the excitation

branch has a small current comparedto load current, thus it may beneglected without causing seriouserror.

VV 11

II 11R R 0101 XX0101

aVaV 22


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Equivalent Circuit of PracticalTransformer

Single Phase transformer (referred to Secondary)Sing le Phase transformer (referred to Secondary) Actual MethodActual Method

21

11

2

1

21 '' a

R ROR R

N N

R =

= a

V V ORV

N N

V 1111

21 '' =

=

21

11

2

1

21 ''

a

X X OR X

N

N X =

=

I1 R 1X1

R C

Ic

Xm

Im

IoI2 R 2

X2

VV 22

aV 1


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Practical Transformer )Single Phase transformer (referred to Secondary)Sing le Phase transformer (referred to Secondary) Approximate MethodApproximate Method

21

11

2

1

21 ''

a

X X OR X

N

N X =

=

2102

2102

'

'

X X X

R R R

+=+=

II 11 R R 0202 XX 0202

aV 1

21

11

2

1

21 '' a

R ROR R

N N

R =

=

aV

V ORV N N

V 1111

21 '' =

=

11 ' aI I =

Neglect the excitation branch,

VV 22


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Example 4 For the parameters obtained from the test of

20kVA 2600/245 V single phase transformer,refer all the parameters to the high voltageside if all the parameters are obtained atlower voltage side.Rc = 3.3 , X m =j1.5 , R 2 = 7.5 , X 2 =

j12.4


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Example 4 (solution)Given : R c = 3.3 , X m =j1.5 , R 2= 7.5 , X 2 = j12.4

i) Refer to H.V side (primary)

R2=(10.61) 2 (7.5) = 844.65 ,

X2=j(10.61) 2 (12.4) = j1.396k

Rc=(10.61) 2 (3.3) = 371.6 ,Xm =j(10.61) 2 (1.5) = j168.9


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Example 5 A 10 kVA single phase transformer 2000/440V

has primary resistance and reactance of 5.5and 12 respectively, while the resistanceand reactance of secondary winding is 0.2

and 0.45 respectively. Calculate:i) The parameter referred to high voltageside and draw the equivalent circuit

ii) The approximate value of secondaryvoltage at full load of 0.8 lagging powerfactor, when primary supply is 2000V.


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Example 5 (solution)R1=5.5 X 1=j12 R 2 =0.2 X 2=j0.45

i) Refer to H.V side (primary)

R2 =(4.55) 2 (0.2) = 4.14 ,X2 =j(4.55) 2(0.45) = j9.32

Therefore,

R01 =R 1+R 2 =5.5 + 4.13 = 9.64X01 =X 1 +X 2 =j12 + j9. 32 = j21.32

55.4440

2000

2

1

2

1 ====V

V

E

E a

V1 aV 2

R01 X01

21.329.64

I1


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Example 5 (solution)ii) Secondary voltage

p.f = 0.8cos = 0.8

=36.87 o

Full load,

From cct eqn.,

o

oo

oo

V

V j

aV I jX RV

8.06.422

)55.4()87.365)(32.2164.9(02000

))((0

2

2

2101011

=++=

++=

AV VA I FL 52000

10103

1==


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Transformer Efficiency To check the performance of the device, by comparing the

output with respect to the input The higher the efficiency, the better the system

%100cos

cos

%100

%100,

22

22 ++=

+=

=

cuc

lossesout

out

P P I V I V

P P P

Power Input Power Output

Efficiency

%100cos

cos

%100cos

cos

2)(

)(

++=

++=

cucnload

cucload full

P n P nVAnVA

P P VAVA

Where, if load, hence n = , load, hence n = , load, n= , load, n= ,

90% of full load, n =0.990% of full load, n =0.9Where P cu = P scP c = P oc


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Voltage Regulation The voltage regulation of the transformer is the

percentage change in the output voltage fromno-load to full-load

Voltage Regulation can be determined based on 3methods: Basic Definition Short circuit Test Equivalent Circuit


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Voltage Regulation

Basic Definition In this method, all parameters are being referred

either to primary or secondary side. It can be represented in either

Down voltage Regulation

Up Voltage Regulation

%100. = NL

FL NL

V V V

RV

%100. = FL

FL NL

V V V

RV

Voltage Regulation


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Voltage Regulation

Short-circuit test In this method, direct formula can be used.

( )%100

cos.

1

. =V

V RV f p sc sc

If referred to primaryside

( )%100

cos.

2

. =V

V RV f p sc sc

If referred to secondaryside

Note that:Note that:

is for Lagging power factor + is for Leading power factor Isc must equal to IFL

P sc = V sc I sc cos sc

V l R l i


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Voltage Regulation( Equivalent Circuit )

Equivalent circuit In this method, the parameters must be referred to

primary or secondary

[ ]%100

sincos.

1

.01.011

= V X R I

RV f p f p If referred to

primary side

If referred tosecondary side

Note that:Note that: + is for Lagging power factor is for Leading power factor

j terms ~0

[ ]%100

sincos.

2

.02.022 =V

X R I RV f p f p


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Example 6 Determine the Voltage regulation by using

down voltage regulation and equivalentcircuit in example 5.


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Example 6 (solution) By using e quivalent circuit,

I1=5A R 01 =9.64 X 01 = 21.32 V 1=2000V, 0.8 lagging p.f

[ ][ ]

%12.5

%1002000

)6.0(32.21)8.0(64.95

%100sincos.1

.01.011

=

+=

=V

X R I RV f p f p


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Example 7 A short circuit test was performed at the

secondary side of 10kVA, 240/100Vtransformer. Determine the voltageregulation at 0.8 lagging power factor if :

Vsc =18V Isc =100 Psc=240W


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Example 7 (solution)( )

o

sc sc

sc sc

sc sc sc sc

o f p

f p sc sc

I V P

I V P that Know

Hence

f pGiven

V V RV

34.82)100)(18(

240cos

cos

cos,

87.368.0cos,

8.0.

%100cos.

1

1

1.

2

.

=

=

=

=

===

=

( )

%62.12

%100100

87.3634.82cos18.

=

=oo

RV


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Example 8 The following data were obtained in test on 20kVA

2400/240V, 60Hz transformer:

Vsc =72V Isc =8.33APsc=268W Poc=170W

The measuring instrument are connected inthe primary side for short circuit test. Determine

the voltage regulation for 0.8 lagging p.f. (use all3 methods), full load efficiency and half loadefficiency.


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Example 8 (solution)( )

.72.786.34.6364.8

64.833.8

72

4.63)33.8)(72(

268cos

cos

cos,

87.368.0cos,

8.0.

%100cos

.

0101

1

1

1.

2

.

side primarytoconnected because jX R j Z

I V

Z

I V P

I V P that Know

Hence

f pGiven

V

V RV

o sc

sc

sc sc

o

sc sc

sc sc

sc sc sc sc

o f p

f p sc sc

+=+==

===

=

=

=

=

===

=


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57/75


( )

%68.2

%100240

58.233240

%100.

79.058.233

2402400

4.6364.887.36240020000

02400

,.3

2

2

20111

=

=

=

=

+

=

+=

NL

FL NL

o

ooo

V

V V

RV

V V

V

aV Z I V

Defination Basic


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Open circuit testq is conducted to determine magnetism

parameter, Rc and Xm and core lossesRc and Xm and core losses.q Also known as no-load test.


61/75

With the secondary open, the primary voltage wasincreased from zero to rated voltage, where therated voltage is the name plate stamp.

A digital multimeter was used as an ammeter tomeasure the open circuit current. A wattmeter

was used to measure the open circuit power. Thepower measured was the power dissipated in R m ,the core losses.

Pin(W) = core loss+ copper loss

Copper loss is neglected because of small no loadcurrent.

Wattmeter only shows the reading of core loss.


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Open-Circuit Test

( )22

1

,

,,

sin

cos

,

cos

cos

mcoc

m

ocmc

occ

mc

ococm

ococc

ococ

ococ

ococococ

I I I

I V

X I V

R

X and RThen

I I

I I

Hence

I V P

I V P

+===

==

=

=

Rc XmVoc

Ic Im

Voc

Ioc cos oc

Ioc

Voc

Ic

Im

Ioc sin oc

oc

Note:If the question asked parameters referred to lowquestion asked parameters referred to lowvoltage sidevoltage side , the parameters (R c and X m) obtainedneed to be referred to low voltage sideneed to be referred to low voltage side


63/75

Short circuit test

is conducted to determine the copperparameter depending where the test isperformed. If performed at primary, hencethe parameters are RR0101 and XX0101 and vice-vice-versaversa.


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Short-Circuit Test Normally, measurement at lower voltage sideat lower voltage side If the given test parameters are taken on primary side,taken on primary side,

RR 0101 and Xand X 0101 will be obtained &will be obtained & vvice-versa.

X01R01

For a case referred toPrimary side 2

012

0101

01

1

012

,

cos

cos

X R Z

I

V Z

Hence

I V

P

R I P

I V P

sc

sc

sc

sc sc

sc sc

sc sc

sc sc sc sc

+=

=

=

==


66/75

Example 9 Given the test on 500kVA 2300/208V are as follows:

Poc = 3800W P sc = 6200WVoc = 208V V sc = 95VIoc = 52.5A I sc = 217.4A

Determine the transformer parameters and drawequivalent circuit referred to high voltage side. Alsocalculate full load efficiency, half load efficiency andvoltage regulation, when power factor is 0.866 lagging.

[1392 , 517.2 , 0.13 , 0.44 , 97.74%, 97.59%, 3.04%]


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EMPS_BEE2133_NJ


A

I I

A

I I

I V P

o

ococm

o

ococc

ooc

ococococ

2.49

6.69sin5.52

sin

26.18

6.69cos5.52

cos

6.69

)208)(5.52(

3800cos

cos

1

======

=

=

=

Ioc cos oc

Ioc

Voc

Ic

Im

Ioc sin oc

oc

From Open Circuit Test,


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===

===

23.421.49

208

39.1126.18

208

m

ocm

c

occ

I V

X

I V

R

Since V oc =208Vall reading are taken on the secondary side

Parameters referred to high voltage side,

=

=

=

= =

=

21.517208

230023.4'

1392208

230039.11'

22

2

1

22

2

1

E E

X X

E E R R

mm

cc



69/75

EMPS_BEE2133_NJ


AV VA

I FL 4.217230010500 3

11

===

o sc

sc sc sc sc I V P

53.72)4.217)(95(

6200cos

cos

1 =

=

=

From Short Circuit Test,First, check theFirst, check the II scsc

Since IFL1 =Isc , all reading are actually taken on the primary side

+=

=

=

=

42.013.0

53.7244.053.724.217

95

01

j

I V

Z

oo

sc sc

sc


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Example 9 (solution)Equivalent circuit referred to high voltage side,

VV22 =aV=aV 22VV11 RR cc13921392

XXmm517.21517.21

RR 01010.130.13

XX01010.420.42


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Example 9 (solution)Efficiency,

%59.97

%1003800)5.0)(6200()866.0)(10500)(5.0(

)866.0)(10500)(5.0(

%100cos

cos

%74.97

%10038006200)866.0)(10500(

)866.0)(10500(

%100cos

cos

23

3

22

1

3

3

=

++

=

++=

=

++ =

++

=

oc sc L

oc sc FL

P P nnVAnVA

P P VAVA


72/75

Example 9 (solution)Voltage Regulation,

[ ]

[ ]

%04.3

%1002300

3053.72cos)95(

%100cos

.1

=

=

= E

V RV pf sc sc


73/75

Example 10 Data obtained from short-circuit and open-circuit test of a

75-kVA, 4600-230V, 60Hz transformer are

Open-Circuit Test Short-Circuit Test(Low-Side Data) (High-Side Data)Poc = 521W P sc = 1200WVoc = 230V V sc = 160.8VIoc = 13.04A I sc = 16.3A

Determine a) the magnetizing reactance andequivalent core-loss resistance; b) the resistance,reactance and impedance of the transformer windings;c) the voltage regulation when operating at rated loadand 0.75 power-factor lagging.


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Summary Transformers convert AC electricity from one voltage to another with

little loss of power Transformers work only with AC and this is one of the reasons why

mains electricity is AC Step-up transformers increase voltage, step-down transformers

reduce voltage Most power supplies use a step-down transformer to reduce thedangerously high mains voltage (230V in UK) to a safer lowvoltage.

The input coil is called the primary and the output coil is called thesecondary

There is no electrical connection between the two coils, instead theyare linked by an alternating magnetic field created in the soft-iron core of the transformer

The two lines in the middle of the circuit symbol represent the core.


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Summary The two lines in the middle of the circuit symbol represent the core Transformers waste very little power so the power out is (almost)

equal to the power in Note that as voltage is stepped down current is stepped up.

The ratio of the number of turns on each coil, called the turnsratio , determines the ratio of the voltages A step-down transformer has a large number of turns on its primary

(input) coil which is connected to the high voltage mains supply,and a small number of turns on its secondary (output) coil togive a low output voltage.

Date post:	30-May-2018
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Power System 3 -Power Transformer EEE3233

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