Transformer

Laboratorium Penelitian Konversi Energi ElektrikInstitut Teknologi Bandung

Agus Purwadi, Qamaruzzaman & Nana Heryana

Transformer

A A transformertransformer is a device that is a device that convert ac electric energy at convert ac electric energy at one voltage level at another one voltage level at another voltage level.voltage level.

Transformer are important to modern life

• Thomas A. Edison (1882)Power Distribution System : New York-USA ; 120 VdcProblems : Losses & Voltage Drop = inefficient

120 kVA 50 kVA 70 kVA

%58

Example :

120 V 70 kVA

Rs=0,05 ohm/10km1000 A

50 V

70V

1.

High Voltage

70 kV 70 kVA

Rs=0,05 ohm/10km1 A

0,05 V

70kV~

70 kVA 0,05 VA 70 kVA

Transformer

are important to modern power system

Low voltage High Voltage

Real Application on Power System

PM G TR MTR

380 V/20 kV 20 kV/380 V

Transformer

1. Power Frequency , 50/60 Hz

2. Radio frequency, > 30 kHz

1. Power Transformer

2. Instrument Transformer

2. Type and Construction Of Transformer

3. The Ideal Transformer

V P(t)

IP(t) IS(t)

V S(t)N P N S

+

-

+

-

S

P

N

Na a

N

N

tv

tv

S

P

S

P )(

)(Turn ratio a

The Ideal Transformer

V P(t)

IP(t) IS(t)

V S(t)N P N S

+

-

+

-

ati

ti

S

P 1

)(

)(Turn ratio a)(.)(. tINtiN SSPP

The Ideal Transformer

V P(t)

IP(t) IS(t)

V S(t)N P N S

+

-

+

-

aV

V

S

P aI

I

S

P 1

In term of phasor quantities

Power in ideal Transformer

PPPIN IVP cos SSSOUT IVP cos

SPcosSSOUT IVP

PSPS aIIandaVV /

INPP

OUT PaIa

VP cos

Power input

Power input

ideal

Turn ratio equation

Power in ideal Transformer

INPPOUT PIVP cos

INPPSSOUT QIVIVQ sinsin

INPPSSOUT SIVIVS

Active power

reactive power

Apparent power

Evaluasi

• Untuk memahami konsep trafo ideal berikan ke mahasiswa, contoh sederhana misalnya :– Cara mendapatkan tegangan sekunder y

volt jika tegangan primernya x volt.– Arus sekunder b ampere pada tegangan

y volt, cari arus primer a ampere pada tegangan X volt.

– Ambil nilai x dan y yang bedanya sangat besar

– Soal pada sistem 1 fasa dulu saja.

Impedance Transformer

L

LL I

VZ

Impedance =

ratio of the phasor voltage across it ti the phasor current

Z LV L

IL

Impedance Transformer

S

S

S

S

P

PL I

Va

aI

aV

I

VZ 2

/'

SP aVV

S

SL I

VZ

P

PL I

VZ '

aII SP /

LL ZaZ 2'

Z L

IP IS

V P V S

+

-

+

-

if

Theory of operation of real single-phase transformers

The basis transformer operation can be derived from Faraday’s law dt

deind

N

ii

1

Flux linkage N

dt

dNeind

VP(t)

IP(t) IS(t)

VS(t)N P N S

+

-

+

-

EMF equation of a transformer

dt

dNe

Faraday law of

electromagnetic induction

t cosmax

tNe sinmaxRMS Value

mm fN

NE

44,4

2

Transformer Losses

• Copper (I2R) losses• Eddy current losses• Hysteresis losses

• Leakage flux

The equivalent circuit of a

transformer

RPIP XP

Rc

Idealtransformer

+

-

PV

RSISXS

+

-

SVjXM NP NS

the model of a real transformer

The equivalent circuit of a

transformer

RP

IP

j XP

Rc

+

-

PVa2 RS

IS /a

j a2 XS

+

-

SaVjXM

Referred to primary

RP /a2

a IP j XP /a2

+

-

a/VP

RS

ISj XS

+

-

SVRC /a2 j XM /a

2

Referred to secondary

Phasor diagram corresponding – referred to primary

aV 2

aI 2R 2

I1R 1

j I1X1

ImI0 I1

I2 /a

IC

2

The equivalent circuit of a

transformer

ReqIP j Xeq

Rc

+

-

PV

IS /a

+

-

SaVjXM

Approximate transformer models

ImIh+e

Test on transformers

• Open-Circuit Test

transformer

Ip (t)wattmeter

+

-

Vp (t)

A

~ Vv(t)

• Result : VOC IOC POC

Open-Circuit Test Result

Conductance of the core-loss resistor C

C RG

1

MM X

B1

MCE jBGY CC

E Xj

RY

11

OC

OCE V

IY

Susceptance of the magnetizing inductor

Admittance

Magnitude admittance

Open-Circuit Test Result

OCOC

OC

IV

PPF

.cos

OCOC

OC

IV

P

.cos 1

OC

OCE V

IY

PFV

IY

OC

OCE

1cos

Power Factor

Power Factor angle

Admittance

Test on transformers

Short-Circuit Test

transformer

Ip (t)wattmeter

+

-

Vp (t)

A

~ V

IS (t)

v(t)

Result : VSC ,ISC ,POC

Short-Circuit Test Result

Series impedanceSC

SCSE I

VZ

SCSC

SC

IV

PPF

.cos

SCSC

SC

IV

P

.cos 1

Power Factor of the current

Lagging - Current angle (-),Impedance angle (+)

SC

SC

SC

SCSE I

V

I

VZ

00Therefore,

Short-Circuit Test Result

Admittance

eqeqSE jXRZ

)()( 22SPSPSE XaXjRaRZ

Series impedance

2.2

The equivalent circuit impedances of a 20 kVA, 8000 / 240 V, 60 Hz transformer are to be determined. The open circuit test were performed on the primary side of the transformer, and the following data were taken :

Open-circuit test (on primary)

Shots-circuit test (on primary)VOC = 8000 V

IOC = 0,214 A

POC = 400 W

VOC = 489 V

IOC = 2,5 A

POC = 240 W

pI a

Is

pV sVacR mXj

ehI mI

eqR eqXj

192j4.38

kj 4.38k159

+ +

- -

The per-unit system of measurement

quantityofvaluebase

quantityactualunitperQuantity

In single-phase system : basebasebasebasebase IVSorQP ,

base

basebase I

VZ

base

basebase V

IY

base

basebase S

VZ

2)(

Voltage Regulation & Efficiency

%100,

,, xV

VVVR

loadfullS

loadfullSloadnoS

Efficiency

%100xP

P

in

out %100xPP

P

lossout

out

Voltage Regulation

Autotransformer

ISE

ILV H

V L

N SE

N CIC

IHStep-down autotransformer

SE

C

SE

C

N

N

V

V SESECC ININ

CL VV SECH VVV

SEH II CSEL III

CSE

C

H

L

NN

N

V

V

C

CSE

H

L

N

NN

I

I

Power rating -autotransformer

SE

CSE

W

IO

N

NN

S

S

SIO =Input and Output apparent powers

SW = apparent power in the

transformer windings

For example, a 5000 kVA autotransformer connecting a 110 kV system to a 138 kV system would have an NC/NSE turn of ratio of 110 : 28. Such an autotransformer would actually have windings rated at :

kVAkVANN

NSS

SEC

SEIOW 10155000

11028

28

Three-phase transformer

N P1 N S1 N P2N S2 N P3 N S3

A three-phase transformer bank composed of independent transformer

N P1 N P2 N P3

N S1 N S2 N S3

A three-phase transformer wound on a single three-legged core

Three-phase transformer connections

• Wye-wye (Yy)• Wye-delta (Yd)• Delta-wye (Dy)• Delta-delta (Dd)

Wye-wye (Yy) Connection

aV

V

V

V

S

P

LS

LP

3

3

aV

V

S

P

Wye-delta (Yd) Connection

S

P

LS

LP

V

V

V

V

3

aV

V

LS

LP 3

Delta-wye (Dy) Connection

S

P

LS

LP

V

V

V

V

3

3

a

V

V

LS

LP

Delta-delta (Dd) Connection

aV

V

V

V

S

P

LS

LP

A 50 kVA 13.800 / 208 V, Dy distribution transformer has a resistance of 1 percent and reactance of 7 percent per unit.

(a) What is the transformer’s phase impedance referred to the high-voltage side?

(b) Calculate this transformer’s voltage regulation at full load and 0,8 PF lagging, using the calculated high-side impedance.

(c) Calculate this transformer’s voltage regulation under the same conditions, using the per-unit system.

Instrument transformers

• Potential transformer (PT)• Current transformers (CT)

V W

S

A

CT

PT