SINGLE PHASE TRANSFORMERS 2012 13 2Q

TRANSFORMERS

Transmission of energy is generally divided in two parts; first is transmissionINTRODUCTION

over long distances at high voltages, which is supported by PowerTransformers. The second part is distribution of the energy from substations tothe various users; this is supported by Distribution Transformers in varioushierarchies.

1

TRANSFORMERS

220 kV 400 kV220 kV, 400 kV

132 ,66, 45 kV

3, 6, 10, 15, 20, 25, 30 kV

LV: 230/400V

LV: 230/400V

http://www.oncor.com/images/content/grid.jpg

2

TRANSFORMERSRATED VOLTAGE (kV)Overhead power lines

3 Utility frequency: 50 Hz (60 Hz in some countries)

INTRODUCTION

6 3rd 10

category 15

Utility frequency: 50 Hz (60 Hz in some countries)

Use of transformers made it possible to transmit the powereconomically hundreds of kilometers away from the

20* 25 30

economically hundreds of kilometers away from thegenerating station.

The success of the current AC power system is in grand2nd 45

Category 66*

1st 110

extend due to power, transmission and distributiontransformers.

Generators voltages: from 6 to 25 kV 1st 110

category 132* 150

i l 220*

Generators voltages: from 6 to 25 kV.

Transmission voltages(>100 km) : 220 and 400 kV

P i di ib i l ( 100 k ) 132 66 45 kV special 220* category 400*

Reglamento sobre condiciones técnicas y garantías de seguridad en líneas eléctricas de

Primary distribution voltages(<100 km): 132, 66, 45 kV

Medium voltage distribution(<10 km): 3, 6, 10, 15, 20 kV

3

garantías de seguridad en líneas eléctricas de alta tensión 2008

ITC-LAT 07 * Suggested

TRANSFORMERS

Transformer: static device that transfers electrical energy from one circuit toanother by electromagnetic induction without the change in frequency It

INTRODUCTION

another by electromagnetic induction without the change in frequency. Itchanges the relationship between voltage and current in primary andsecondary sides.

It is made of steel laminations wrapped with two coils of wire.

Winding connected to source named “Primary”Winding connected to source named Primary

Winding connected to load named “Secondary”

4

TRANSFORMERS

The transformer is an electromagnetic conversion device which has a primaryand a secondary windings. It accomplishes: V1I1 ≈ V2I2.

The primary and secondary windings are not connected electrically, butcoupled magnetically.

Machines with higher efficiency, up to 99.7%

Step-up transformer: U2>U1 I2<I1 Step-down transformer: U1>U2 I1<I2 Isolation transformer: U1=U2 I1=I2

5

TRANSFORMERS

Rating: The nominal value of any electrical, thermal, mechanical, or environmental quantity assigned to define the operating conditions under which

t hi t l t i d i t i t d t ia component, machine, apparatus, electronic device, etc., is expected to give satisfactory service.

Rated freq enc (f ) The freq enc of the alternating c rrent for hich aRated frequency (fn): The frequency of the alternating current for which a device is designed.

Rated continuous current (I ): The current expressed in amperes root meanRated continuous current (In): The current expressed in amperes, root mean square RMS, that the device can carry continuously under specified service conditions without exceeding the allowable temperature rise.

Rated voltage: The RMS voltage, at rated frequency, which may be impressed between the terminals of the device under standard operating conditions.

6

TRANSFORMERS

Power loss generate heat

Dry transformers: for low powerDry transformers: for low power

Oil-filled transformers: for large power, the transformer is completely submerged in an oil tank The oil actuates as a coolant and an insulator Mineral oil (froman oil tank. The oil actuates as a coolant and an insulator. Mineral oil (from distillation of petrol) and silicone oil (most actual)

Dry-type transformers encapsulated in epoxy resin: for installations that requireDry type transformers encapsulated in epoxy resin: for installations that require high security, specially in indoor. They don’t spread fire. High voltage windings are completely encapsulate in epoxy resin.

Oil insulated type transformer 33 kV to 400 V

Dry type transformer 33 kV to 400 V

25 KVA - 20 MVA Refrigeration: mineral oil a

low flammability synthetic fluid

40 KVA - 5 MVA Insulation: epoxy resin Ventilation

fluid Ventilation (possibility)

Kotsons Catalogue

Indoor Very safeKotsons Catalogue

7

TRANSFORMERS

Insulation Symbol Type of i l ti

Symbol

REFRIGERATION SYSTEMS

circulation

Mineral oil O Natural N

Pyralene L Forced F

Gas G

Water W

Air AAir A

Solid insulation S

ONAN: Oil Natural (transformer submerged in an oil tank; oil with natural convection).It is refrigerated by Air with Natural convection

ONAF: In this case the Air is Forced by means of a fan

8

TRANSFORMERS

ENERGY LOSSES

1. Winding losses : I2 R

2. Iron losses:2. Iron losses:2.1 Hysteresis2.2 Eddy currents (Foucault currents)

9

TRANSFORMERSENERGY LOSSES

Hysteresis loopStarting from zero we gradually increase I [A] so that H [A m] and B [T]Starting from zero, we gradually increase I [A] so that H [A m] and B [T]increase, curve 0a, the flux density reaches a value Bm for a magnetic fieldintensity Hm. If the current is now gradually reduced to zero, the flux density Bdoes not follow the original curve but moves along a curve abdoes not follow the original curve, but moves along a curve ab.

As we reduce the magnetic field intensity, the magnetic domains that werelined up under the influence of field Hm tend to retain their original orientation.This phenomenon is called hysteresis

10

This phenomenon is called hysteresis.There is a frictional resistance of the magnetic domains, so the energysupplied is dissipated as heat in the material.

TRANSFORMERS

ENERGY LOSSESHysteresis lossThe ac current causes the flux in the iron changes continuously both in valueand direction. The magnetic domains are therefore oriented first in one directionand then the other, at a rated that depends on the frequency. Thus if the fluxhas a frequency of 50 Hz, the domains describe a complete cycle every 1/50 s.In describing a hysteresis loop, the flux moves successively from +Bm, +Br,, -Bm,-Br, o, and +Bm.

Hysteresis loop. If B is expressed in teslas and H in amperes per meter, the p p ,area of the loop is the energy dissipated per cycle, in joules per kilogram.Energy lost in moving the domain walls as the magnet is poled from one extreme to the other and back again.

]/[2max KgWBfkP hhysteresisfe

11

TRANSFORMERS

ENERGY LOSSES

Hysteresis lossHysteresis loss

The magnetic material absorbs energy during each cycle and this energy isdissipated as heatdissipated as heat.

To reduce hysteresis losses, we select magnetic materials that have a narrow hysteresis loop such as the grain oriented silicon steel used in the cores ofhysteresis loop, such as the grain oriented silicon steel used in the cores of transformers.

B Soft (electrical steel)

Hard (permanent magnet)

H

12

TRANSFORMERS

ENERGY LOSSES

Eddy currents (Foucault currents)Eddy currents (Foucault currents)

Consider an ac flux that links rectangular shaped conductors. According toFaraday’s law and ac voltage is induced across its terminalsFaraday s law, and ac voltage is induced across its terminals.If the conductors are short-circuited, a substantial ac current will flow in eachone, causing the conductors heat up.

SB

dttdNte

13

Rteti

TRANSFORMERS

Eddy currents (Foucault currents)ENERGY LOSSES ]/[2

max22 KgWBdfkP ecurrentseddyfe

The solid metal plate is basically equivalent to a densely packed set ofrectangular conductors touching each other. Currents swirl back and forthi id th l t Th ll d dd t F lt t binside the plate. These are so-called eddy currents or Foucault currents, can bevery large, due to the low resistance of the plate, thus the metal plate canbecome very hot. In this regard, special care has to be taken.

dAB

The losses can be reduced by splitting the core and insulating the twosections from each other

S

sections from each other.Currently the lowest thickness available is 0.23 mm, and the popular thicknessrange is 0.23 mm to 0.35 mm for power transformers. 14

TRANSFORMERS

Transformer cores are built from thin sheets of steel.

CORE MATERIALS

Core steel is alloyed with silicon (Si). Silicon increases the specific electricalresistance, which again reduces the eddy current losses in the core.Increased silicon content makes the core steel brittle; therefore the content is keptbelow 3%.To minimize eddy current losses, the sheets must be insulated from eachother. Today the core steel is delivered ready insulated from the manufacturer.Th i l ti ti i thi 4The insulating coating is very thin < 4 μm.

Eddy current losses in the core steel are proportional to the square of thethickness Therefore the steel sheets have to be thin in order to reduce the nothickness. Therefore the steel sheets have to be thin in order to reduce the noload losses. Typical thickness is from 0.23 mm to 0.35 mm.

15

TRANSFORMERS

MAGNETOSTRICTION

The dominant generating source of transformer sound is magnetostriction.

Magnetostriction is the change in dimensions which takes place in certaint i l h th bj t d t h i ti fl I timaterials when they are subjected to a change in magnetic flux. In magnetic

core steel the dimensional change is in the range of 10-7 to 10-5 meters permeter length at typical induction levels.

This produces the buzzing sound commonly associated with transformers andin turn causes losses due to frictional heating in susceptible cores.

16

TRANSFORMERS

CORE MATERIALS

17Standard IEC specification for nonoriented magnetic steel sheet

TRANSFORMERS

CORE MATERIALS

Standard IEC speci cation for grain oriented magnetic steel sheet (a) normal material; (b) material with reduced

18

Standard IEC speci.cation for grain-oriented magnetic steel sheet. (a) normal material; (b) material with reducedloss; (c) high-permeability material.

TRANSFORMERS

19

TRANSFORMERS

IDEAL TRANSFORMER AT NO LOAD

Supposing ideal coil t

021 RR

Supposing ideal coil

)( tu

)(0 ti

)(2 te )(2 tu

t

)(t

0)(2 ti

tute

)(1 tu )(2)(1 te

dt

tdNte

tetu

11

11

ttdt

tdNte

tute

22

22

sin

fNNtNtetu

tt

m

m

111

2cos

sin

fNfNNU

tNtetutt

mm

m

m

222

2

222

4442cos

sin

fSBNU

fNfNNU mmm

111

1

444

44.422

2

fSBNU

fNU

m

m

22

22

44.4

44.422

fSBNU m11 44.4 m22

tet rUU

NN

EEr

20

1

2

1

2

120

TRANSFORMERS

IDEAL TRANSFORMER AT NO LOAD WITH LINEAR WITH LINEAR B-H CURVE

200

250

v1v2fluxi0

50

100

150 i0

-50

0

50

-150

-100

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-250

-200

t [s]

[ ]

21

TRANSFORMERS

No load current Hysteresis not consideredIDEAL TRANSFORMER AT NO LOAD WITH NONLINEAR B-H CURVE

u

SB

1u

tetu

tutu cos

11

max1

t0i

ttdt

tdNte

m

sin

11

22

0

TRANSFORMERS

[A]No load current Hysteresis considered

TRANSFORMER AT NO LOAD WITH NONLINEAR B-H CURVE

oad

curr

ent [

No

lo

u

SB

t [ms]1u

Core Losses

fePfe

23t0i

TRANSFORMERS

TRANSFORMER AT NO LOAD

The no-load current is not sinusoidal

T k ith h d it t b i idTo work with phasors we need it to be a sinosoid

An equivalent sinosoid is defined with the same RMS value that the real no load current to perform the calculations and in such a way that it cause theload current to perform the calculations and in such a way that it cause the same core losses:

PfefeP

010cos

IUfe

24

TRANSFORMERS


)(0 ti t

Supposing ideal coil)(1 tu

)(0

)(1 te021 RR

Supposing ideal coil

IUPPfe

1U

001 cosIUPfe

0I01

0cosIU

feII 0

I0feP 1U

I0I0

Diagram with no losses feI I1U

I

feI

25Diagram taking into account the iron losses

TRANSFORMERS


and

01 R

Leakage flux

101011 EIXjIRU d 26

TRANSFORMERS


101011 EIXjIRU d

Regarding to iron losses the voltage drop in R1 and Xd1 could be neglected (0.2 to 6 % of U1)

001 cos

EU

IEPfe

11 EU

001 cosIUPfe

27

TRANSFORMERS

TRANSFORMER UNDER LOAD

If this was so, the flux would increase, which it is no possible due to the Lenz law

td

possible due to the Lenz law.

dt

tdNte 22

28

TRANSFORMERS


dt

tdNte 22

29

TRANSFORMERS


30

TRANSFORMERS

Can I do it?


Can I do it?

31

TRANSFORMERS

A-aA’-a’


32

TRANSFORMERS


33

TRANSFORMERS


t tbt

dt

tdNtetu

systempowerthebyfixedtutu

cos

111

max1

01

tan;

loadnoIN

tconsbemust

ttdt

m sin

221101

2211

INININ

loadunderININ

220111

221101

NINININ

21

201 I

NNII

34

TRANSFORMERS

2N

TRANSFORMER UNDER LOAD t

)(2 ti tititi ')()(

22

21

201

' IINI

INNII

)(2 te )(2 tu

)(2 ti

)(1 tu

tititi ')()( 201

)(1 te

2221

22

1

22

'

'

ININ

rI

NI

t

'2012

021

201 III

rIII

NNII

t

1 t

111111 EIXjIRU d

222222 UIXjIRE d

35222 IZU

TRANSFORMERS


11 NE)(2 te )(2 tu

)(2 ti

)(1 tu

)(1 ti

)(1 teIt is always true

2

1

2

1

; EUEU

NEr et

XI

It is always true

11

2211 ;

UEr

EUEU

2E RI

22 dXI1U

11 RI

11 dXI

202 UEr et

2U22 RI

1EWhen

2I2

1I1'2I

When

nn UUthenUUIf 22011

0I

2

n

nt U

Ur2

1

36

TRANSFORMERS

EQUIVALENT CIRCUIT REFERED TO THE PRIMARY

222222

111111

UIXjIRE

EIXjIRU

d

d

222222

rEIXjIRU

UIXjIRE d

trEE 21

''' 22222222222222

211111

rUrIXjrIRrErIIrUIXjIRrE

rEIXjIRU

ttdttttdt

td

'''

''22

222

222

211111

UUXXRR

rUIrXjIrRIXjIRU ttdtd

''''''

''' 222

222

22 rUUrXXrRR ttddt

37

'''''' 2012222211111 IIIUIXjIRIXjIRU dd

TRANSFORMERSEQUIVALENT CIRCUIT REFERED TO THE PRIMARY

'2I

+

1I

'2U1U

+

1E 12' EE '21 NN

'2I1I

+

'2U

+

1U

+1E 2'E'21 NN

'X 'R

+

trII 2

2 ' trUU 22 '

1U

2dX 2dR1dX1R

'201 III

1U221 ' ErEE t

TRANSFORMERS


'2dX '2dRXR

+

trII 2

2 ' trUU 22 '

U

1dX1R

'201 III

t1U221 ' ErEE t

''''' 222221 UXIjRIE d

39

TRANSFORMERS


40

TRANSFORMERS

Exact equivalent circuit


Exact equivalent circuit

Approximate equivalent circuit, since Rfe>>R1 & R2, X>>X1&X2

41

TRANSFORMERS


Approximate equivalent circuit,

'21 II

jXRIUU ''42

'21110 IIII scsc jXRIUU 11221 ''

TRANSFORMERS

EQUIVALENT CIRCUIT REFERED TO THE SECONDARY

EIXIRU

222222

111111

UIXjIRE

EIXjIRU

d

d

222221 UIXjIR

rE

dt

2222211111 UIXjIR

rIXjIRU

dt

d

2101 r

IIIt

21011 ' IrIIrI tt

4321

1221

122

1 UrrI

rXjIXj

rrI

rRIR

rU

t

t

t

dd

t

t

tt

TRANSFORMERS

111 rXjrRU d

EQUIVALENT CIRCUIT REFERED TO THE SECONDARY

I

11

211

2211

221

''' IrIXXRR

UrrI

rXjIXj

rrI

rRIR

rU

d

t

t

t

dd

t

t

tt

21011

2101

' IrIIrI

rIII

tt

t

111212 IrIXr

Rr td

tt

2222211111 ''''' UIRIXjIXjIRU dd

+

2dX 2dR'1dX'1R

rII ' 2ItrI10

trUU 22 ''1U

trII 11

trE1

t10

44

TRANSFORMERSEQUIVALENT CIRCUIT REFERED TO THE SECONDARY

2I2101 '' III

+

2U

'10I

E'1U

'1R '1dX 2X 2R

2Utr

EEE 121 ' 1U

2222211111 ''''' UIXjIRIXjIRU dd

trII 12 ' RRR ' XXX

2I

t12 212 RRR sc 212 XXX sc

2I2U

trUU 1

1' 2Z

45 scscsc jXRXXjRRZ 2221212 '' scsc jXRIUU 22221'

TRANSFORMERS

EQUIVALENT CIRCUITS SUMMARY

'XXX 1I 'RRR 1:tr

II 2'0I

211 XXX sc 211 RRR sc 2I

t

trI 2

2 ' trUU 22 '

IfeI1U 2U

' XXXII ' ' RRR1:r1I

trI 0

212 ' XXX sc trII 11 '212 ' RRR sc 1:tr1I

1U 2U2I

trU1

46

TRANSFORMERSVOLTAGE DROP

An example of how the secondary voltage

scsc jXRIUU 11221 '0'

p y gvaries with various angles of and loadcurrents for the particular values ofRcc= 0.74 % and Xcc=10%. Negativesvalues of mean inductive load current.Positive values mean capacitive loadcurrent.

Another example of how the secondary

scsc jXRIUU 11221 '0'

p yvoltage at rated current may vary atvarious values of Xcc and cos . Ingeneral secondary voltage decreases withg y gincreasing Xcc. For example at Xcc=10%and cos =0.8 the secondary voltage hasdropped to 93 % of the no load voltage.

47Xcc % scsc jXRIUU 11221 '0'

TRANSFORMERS

Secondary winding short-circuited.

SHORT-CIRCUIT TEST

y g

Primary winding is fed with an increasing variable voltage, starting from 0 Volt, until achieve I1 = I1n

I1 = I1rated = I1n

IO << I1 I2’ = I1 = I1nO 1 2 1 1n

Psc, V1sc and I1sc = I1n are measured (Qsc is deduced)

48

TRANSFORMERS

SHORT-CIRCUIT TEST

The test voltage U1sc very low, it will be few flux, so that the iron losses can be neglected, Pfe=k Bm

2.

Results from the test:

Losses in the cupper:

Circuit parameters: R1sc=R1+R2’ and X1sc=X1+X2’ 49

TRANSFORMERS

SHORT-CIRCUIT TEST

A) Why losses in the iron, Pfe, can be neglected?

50

TRANSFORMERS

A) Why losses in the iron P can be neglected?

SHORT-CIRCUIT TEST

A) Why losses in the iron, Pfe, can be neglected?

cosmax1 td

tutu

44.4sin 11

111

mm SBfNUttdt

tdNtetu

%5%5

11 nmn

BUUBtutu

%5%5

2

11

nmnsc

BkP

BUU

4001

400105.0 1

221

nfenmnmscfe

mfe

UPBkBkUP

BkP

01 scfe UP 51

TRANSFORMERS

SHORT-CIRCUIT TEST

11 cos ccnscsc

PIUP

21

1

scscsc

n

scsc

tgPQX

IPR

21

21

1nn

sc IIX

52

TRANSFORMERS

SHORT-CIRCUIT TEST

What happens if we do this test with another U1 in such a way that does notWhat happens if we do this test with another U1sc in such a way that does not produce the rated current ?

scnnsc RIIP 12

11

scscsccunsc RIIPthenIIIf 12

1111 2

1

111

n

scnscsccu I

IIPIP

scn

scsc

nsc

sccu

RIRI

IPIP

12

1

12

1

1

1

1 n

53

TRANSFORMERS

SHORT-CIRCUIT TEST

22211

11111

XscRscscXscRscsc

nscnscsc

UUUUUU

IjXIRU

111 ccnscsc

scscscscscsc

UZI

UU

1

11

1

1

11

scnRscRsc

nnsc

URI

UU

UU

1

11

1

1

11

n

scn

n

XscXsc

nn

UXI

UU

UU

IUP222

XscRscsc

ccnscsc IUP cos11

54

TRANSFORMERS

SHORT-CIRCUIT TEST

(%)CC (%)XCC (%)RCC XCCRCC

Transformers < 1000 kVA 3 ÷ 6 2,5 ÷ 6 1,1 ÷ 2,5 1,2 ÷ 6

Transformers > 1000 kVA 6 ÷ 13 5 ÷ 13 0,4 ÷ 13 3 ÷ 30

55

TRANSFORMERS

NO LOAD TEST

No load conditions = open circuit I << I No load conditions = open circuit I0 << In The rated voltage is applied to one winding and other winding is kept

open (usually LV winding is supplied, while HV is kept open).p ( y g pp , p p )

Po, V1o = V1n and I1o are measured, Qo is deduced from Po

101 II 02 I

101 UU n 20U

56

TRANSFORMERS

NO LOAD TEST

Results from the test:

Losses in the iron:Losses in the iron:

Circuit parameters: Rfe and Xµ 57

TRANSFORMERS

NO LOAD TEST

22 UU

010100

0101010100

sincoscos

IUQIUIUP n

00

10

0

10

tgPUX

PURfe

110 EU

210UR

210

12

00

10fe

UX

RIPR

58

12

000 dXItgPX

TRANSFORMERS

What happens if we do this test with another U10 which is NOT U1 ?

NO LOAD TEST

What happens if we do this test with another U10 which is NOT U1n ?

2

102

01

nhmfe

nfe

UkPBkP

PUP

2

1

UPUP

2

111

2111

hnfe

hnfe

UkUUP

UkUUP

1

101

n

fe UPUP

21

1

0 nh

hf

UkP

59

TRANSFORMERS

TEST SUMMARY

NO LOAD TEST

2

U

NO LOAD TEST

210

210 UU

1

101

nfe U

UPUP00

10

0

10

tgPUX

PURfe

SHORT-CIRCUIT TEST

111 ccnsc ZIU

111

1

11

1

1

scnRscRsc

n

ccn

n

scsc

RIUUU

U

2'

211

11 cos

scsc

ccnscsc

PRRR

IUP

2

1

111

n

scnscsccu I

IIPIP111

11

scnXscXsc

nnRsc

UXI

UU

UU

22'

211

21

211

scscscsc

nsc

ItgP

IQRRX

I

60

1 n

22211

XscRscsc

nn UU

11 nn II

TRANSFORMERS

FERRANTI EFFECT

nUU 22

61

TRANSFORMERS

The efficiency of a transformer is defined as the ratio of useful output power to

EFFICIENCY OF THE TRANSFORMER

y p pinput power

out

PPPP

PP

PP 22

2

101

feUPUP

nn

cufein

indexloadII

IIC

PPPPP

1

1

2

2

21

2

I

1

01

n

fe U

nn

PCUPIU

IU

22

1022

22

12

cos

cos

1

111

n

scnscsccu I

IIPIP

Maximizing the efficiency with respect to C, it results:

scn

PCU

PIU1

022 cos

fe

PP

C maxcuP

62

TRANSFORMERS

SHORT CIRCUIT CURRENTS

nsc I

UZ 1sc

sc I

UU1

sc

nsc I

UZ1

1

sc

n

n

sc

IU

IU 1

1

1 n

scsc I

UZ1

1

nnnsc

IUUII 11

1

scn1

scscnsc U 1

1

5<ε<10% h t i it t 10 20 ti I i bt i d5<ε<10% a short circuit current 10 or 20 times I1n is obtained

63

TRANSFORMERS

TRANSFORMERS IN PARALLEL

1.-If rt A= rt B we have the following schemet A t B g

If I0T<<I2T’ then I1T=I2T64

TRANSFORMERS


''' 212121 BBSCAASC IZIZUU

65

TRANSFORMERS

BBSCAASC IZIZUU 212121 '''TRANSFORMERS IN PARALLEL

BABSCBSCAAASCASC IZIZ 221221 ''

BBSCAASC

ICIIC

IZIZ

2

2121

''

''

nSCnSC

nn

UZIZ

ICII

C

111

121

2 ';

nSCBnSCA

n

nSCSC

nSC

ICUICU

IZ

U

11

1

11

1

;

nBBnB

nSCBnAA

nA

nSCA ICI

ICI

11

11

1

1

SCBBSCAA CC 66

TRANSFORMERS

221221 '' BABSCBSCAAASCASC IZIZ '' IZIZ


22 BBSCAASC

2121 ABSCAASC IZIZ

22 ASCBSCAB

22 ABASCBSC

It can be noted that the currents in each transformer are in phase, so that it can be added arithmetically instead of vectorially.

'''''' IIIIII ...... 222222 BATBAT IIIIII

67

TRANSFORMERS

TRANSFORMERS IN PARALLEL TRAFO Sn [kVA] εsc (%)

A 100 4.5We have the following transformers, fed at rated voltageB 1000 5.5

g g

Which will be the total available aparent power (VA) ?

SCBBSCAA CC

CSSCIUIUS

CC

C

5554

10CSSCIUIUSCSSCIUIUS

BnBBBnBnBBnBB

AnAAAnAnAAnAA

ABA

BA

CC

CCCCC

81805.45.41

5.55.4

kVAkVASkVAkVAS

B

A

818818.010001001100

AB

IUSS

CC

818.05.55.5

kVAkVAkVAST 918818100 !!

nnn IUSS max

68

TRANSFORMERS

INRUSH CURRENT

tUtu

iftUtu

cos2

º0cos2

11

11

200

300

TT

dttUNddttUNd

tUdt

tdNtetu

cos2cos2

cos2

4/4/

1111

-100

0

100

u [V

]

TU

ifand

dttUNddttUNd

i

00

cos2cos20

110

11

0 0.005 0.01 0.015 0.02-300

-200

t [s]

mfN

UfN

UN

UN

UT

44.4220sin4

sin2

4 1

1

1

1

1

1

1

1

69

TRANSFORMERS

tUtu

if

º90cos2

º90INRUSH CURRENT

TTTT

dttUNddttUNd

tUtu

sin2º90cos2

º90cos22/

011

2/

0

2/

011

2/

0

11

mNU

NUUT

TNUT

ifand

2221120cos2

2cos22

00

1

1

1

11

1

1

200

300

-200

-100

0

100

u [V

]

0 0.005 0.01 0.015 0.02-300

t [s]

70

TRANSFORMERS

INRUSH CURRENT

The light grey curve is theThe light grey curve is themagnetic flux in the normalstationary condition

In practical cases if we considerer the existence of remanent flux, whichcan reach 0.5 n a instantaneous flux of 2.5 n can be obtained.Taking into account the magnetization curve of the material it means that itTaking into account the magnetization curve of the material it means that itmay demand an exciting current many hundreds of times the normalexcitation current because of core saturation.

71

SINGLE PHASE TRANSFORMERS 2012 13 2Q

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