BEKP 2443

PENGANTAR KEJURUTERAAN KUASA

Lecture 7 : Power Transformers

UNIVERSITI TEKNIKAL

MALAYSIA MELAKA UNIVERSITI TEKNIKAL MALAYSIA MELAKA

Learning Outcomes:

To review basic concepts and establish terminology & notation for :-

The ideal transformer (3.1)

Equivalent circuit for practical transformers(3.2)

3 phase transformers & phase shift (3.4)

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Transformers Overview

Transformer = a device that transfers electrical

energy from one circuit to another through inductively

coupled conductors/coil.

Operation Principle : A varying current in the first or

primary winding creates a varying magnetic flux in

the transformer's core, and thus a varying magnetic

field through the secondary winding. This varying

magnetic field induces a varying electromotive force

(EMF) or "voltage" in the secondary winding by a

process called “mutual induction”.

1st development of commercial transformer started in

1885 by William Stanley

Type of Tranformers (application)

- Unit transformer

- Substation transformer

- Distribution transformer

Power Transformers

- DC-DC converter

- Switching Mode Power Supply

- Uninterruptible Power Supply

High Frequency Transformers

- Potential Transformer

- Current Transformer

Sensing (Instrument) Transformers

Type of Transformers

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5

Step-up transformer

- provides a secondary voltage that is greater than the primary voltage.

Step-down transformer

- provides a secondary voltage that is less than the primary voltage.

Isolation transformer

- provides a secondary voltage that is equal to the primary voltage.

- to isolate the power supply electrically from the power line, which serves as a protection.

Other classification of Power Transformer

Pole-mounted transformer

Current transformer (CT) power transformer (oil immersed)

Household transformer Dry type distribution transformer

IDEAL TRANSFORMER An ideal transformer has:

– no real power losses (no winding resistance)

– magnetic core has infinite permeability (no core reluctance, similar to resistance in elec circuit))

– no leakage flux (all flux is confined in the core & links both primary & secondary windings)

– No core losses

We’ll define the “primary” side of the transformer as the side that usually takes power, and the “secondary” as the side that usually delivers power.

– primary is usually the side with the higher voltage, but may be the low voltage side on a generator step-up transformer.

Physical Model of Ideal Transformer

• Made up of

inductors/coils

• Not electrically

connected.

• An AC voltage

applied to the primary

induces an AC voltage

in the secondary.

Voltage polarity and current direction

Dot Convention/standard:

Ip Is

1. If the primary voltage is positive at the dotted end of winding with

respect to undotted end, then the secondary voltage will be positive at

the dotted end also. Voltage polarities are the same with respect to the

dots on each of the core

2. If primary current of the transformer flows into the dotted end of the

primary winding, the secondary current will flow out the dotted end of

the secondary winding

The direction of the windings determines the polarity of the voltage across the secondary winding with respect to the voltage across the primary. Phase dots are sometimes used to indicate polarities.

In phase Out of phase

Voltage polarity and current direction

Equivalent Circuit of 2 winding,

1Φ Transformer

Notation: 1=primary 2=secondary

12

2 Winding, 1Φ Transformer (Voltage

Relationships Between Primary & Secondary)

From Faradays Law & Lenz’s law, the turns ratio of a transformer is equal to the voltage ratio of the component:

2

1

2

1

N

N

V

V and we

define also 2

1

N

Nat

For example:

acac VVVN

NV 30)120(

4

11

1

22

ratio turnsampereta

13

2 winding, 1Φ Transformer (current

relationships between primary & secondary)

Assuming the transformer is 100% efficient, then the power delivered & received from winding 1 to winding 2 is SAME, thus:

Therefore:

2

*

22

*

111 SIVIVS 2

1

1

2

V

V

I

I

2

1

1

2

N

N

I

I

and

14

Example 1 Consider the source, transformer, and load shown in the circuit below. Determine the rms values of the currents and voltages (a) with the switch open and (b) with the switch closed.

VrmsV 110)(1 Solution Voltage applied to the primary,

VrmsVN

NrmsV 22)110(

5

1)()( 1

1

22

(a) With the switch open, the secondary current is zero. Hence, the primary current is also zero. Why?. Because no power transfer between the two “circuit” (remember : S=VI*).

(b) With the switch closed:

AR

rmsVrmsI

L

2.210

22)()( 2

2 ArmsIN

NrmsI 44.0)2.2(

5

1)()( 2

1

21

Transformer Rating

The rating of a transformer is stated as Volt Ampere (VA) that it can transform without overheating.

The transformer rating can be calculated as either V1I1 or V2I2 where I2 is the full load secondary current.

E.g. transformer is rated at 20kVA, 480/120V, 60Hz : Means: “This transformer is capable of transferring a normal/nominal

of up to 20kVA of complex power between primary & secondary winding. Its primary & secondary winding is capable of having a nominal/normal voltage of up to 480V & 120V respectively. The whole transformer can only be placed at a system with frequency of 60Hz.”

Impedance Transformation/Reflection

2

2

2

I

VZ

2

11

2

12

1

I

V

Z

NN

NN

2

2

2

2

2

1

1

12

I

VZaZ

N

NZ t

The phasor current and voltage in the secondary are related to the load impedance by

Then,

The impedance seen by the source (referred to primary & secondary) are:

2

11

2

1

2

2

21

I

V

ta

ZZ

N

NZ

Secondary impedance referred to primary:

Primary impedance referred to secondary :

Example of Reflection of impedances & equivalent circuit

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2

2

2

2

2

1

1

1'

2I

VZaZ

N

NZ t

2

11

2

1

2

2

2"

1I

V

ta

ZZ

N

NZ

Secondary impedance referred to primary:

Primary impedance referred to secondary :

Example 2 (Glover, pg100)

A single phase , 2 winding transformer is rated at 20kVA, 480/120V, 60Hz. A source connected to the 480V winding supplies an impedance load connected to the secondary winding. When the secondary voltage is at 118V, the load absorbs 15kVA at 0.8pf lagging.(assume ideal transformer & select 118 V as reference) Calculate:

a) Voltage applied across the primary winding (V1)

b) The actual load impedance (Z2)

c) The load impedance referred to the primary winding (Z’2)

d) The real & reactive power supplied at the primary winding

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Solution Step1 : draw & label equivalent diagram (picture tells a 1000 words!!) &

write down all information given!.

a :

b:

c :

d:

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60Hzf 3.

kVA )8.0(cos15lag 0.8pf @ 15S .2

4120

480 .1

:given info

1

22

2

1

SkVA

N

Nat

oo

tVaV 0472)0118(421

o

oo

o

I

VZ

Z

VI

V

SI

IIVS

87.3693.0

A87.3612787.36127*

kVA 87.3615)0118(

2

22

2

22

2

22

*

2

0*

222

oo

tZaZ 87.3688.14)87.3693.0(42

2

'

2

Var 9000Q

W12000P

jQP900012000kVA 87.361512

jSS o

Badariah Bais KKKF163 Introduction to EE Sem II 2006/07 20

Example 3

Consider the circuit shown below. Find the phasor currents and voltages at both primary & secondary winding. Also, find the real & reactive power delivered to the load.

Solution Step1: Impedance at the secondary: )2010( jZL

Step 2: Load Impedance reflected/referred at the primary:

)20001000()2010(

1

1022

2

12'jjZ

N

NZaZ LLtL

Step 3: Total impedance

(all referred at primary) 452828)20002000(200010001000

'

1 jjZRZ LT

Badariah Bais KKKF163 Introduction to EE Sem II 2006/07 21

Step 4: Primary current and voltage (refer equivalent circuit referred to primary):

452828TZ

AZ

VI

T

S

453536.0

452828

010001

)20001000)(453536.0(11 jZIV'

L

V 43.186.790)43.632236)(453536.0(

Step 5: Secondary current and voltage (refer original circuit): AII

45536.3)453536.0(

1

10

N

N1

2

12

VVV

43.1806.79)43.186.790(

10

1

N

N1

1

22

)20001000('

jZL

Step 6: Real & Reactive Power delivered to the load:

222

*

2221

jQP

4.63279)45536.3)(43.1806.79(

S

IVSS

original circuit New equivalent circuit with load referred to primary

Rule of thumb to find primary & secondary winding

current & voltages

If given the turns ratio & you know the voltage or current of winding 1, you can always find V&I for winding 2 by only using the turns ratio relationship.

If only given : turn ratio, impedances for winding 1 & 2, & source voltage (Vs) of winding 1, then…

To calculate V & I at winding 1

1. Reflect all impedance from winding 2 to 1

2. use Ohms law to calculate the V & I of winding A using the total impedance

(Total impedance = impedance at winding A plus the ones reflected at winding A)

To calculate V & I at winding 2

1. Convert V &I at winding A to winding B using only the turns ratio relationship

To calculate power

Power at winding A = power at winding B

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Real/Practical Transformer

Real transformers

– have losses

– have leakage flux

– have finite permeability of

magnetic core

Real power losses

– resistance in windings (i2 R)

– core losses due to eddy

currents and hysteresis

Transformer Core losses

1. Eddy currents in the core

arises because of changing

flux in core. Eddy currents

are reduced by laminating

the core.

2. Hysteresis losses are

proportional to area of

BH curve and the

frequency. These losses

are reduced by using

material with a

“thin” BH curve

Equivalent Circuit of real transformer

Equivalent Circuit of real transformer

Phenomena physics represented by equivalent circuit

Proportional to voltage square

Core loss current is a current proportional to the voltage applied to the core that is in phase with the aplied voltage, so it is modeled by a resistance Rc or Gc

connected with the primary winding

- Core (eddy current) losses

- Copper losses Proportional to current square

Copper loss are resistive losses in the primary and secondary windings and are modeled by resistor R1 and R2 connected to the primary and secondary winding so respectively

- Hysterisis current losses Proportional to excitation frequency

The high excitation frequency is due to magnetic saturation in the transformer core

- Leakage flux Flux component which links winding 1 but does not link winding 2

Magnetization flux is flux produced by magnetization current

- Magnetization flux Represented by magnetization inductance

Equivalent circuit of single phase 2 winding transformer & parameter definition

R1 = primary winding resistance I2R losses (real power loss)

R2 = secondary winding resistance I2R losses (real power loss)

X1= primary winding leakage reactance I2X loss (reactive power loss)

X2= secondary winding leakage reactance I2X loss (reactive power loss)

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Equivalent circuit of single phase 2 winding transformer

Im= magnetizing

Ic= core loss current

Ie = excitation current

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)usceptanceinductor(sshunt 2

1

11

2

11

N

RBwhere

EjBIN

NII

cm

mm

1)( EjBGIII mcmce

Equivalent Circuit of Transformer

Another version of equivalent circuits: All losses take into account & secondary impedances all referred to primary

windings

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Another version of equivalent circuits: All losses take into account & secondary impedances all referred to

primary windings

'

2

'

2

'

21

'

22

2

2

12

2

11

IZVE

IZN

NV

N

NE

2

2

2

12

2

2

1'

2

'

2

'

2

'

2

XN

NjR

N

NZ

jXRZ

Another version of equivalent circuits: neglect core losses (ignore shunt branch) & secondary impedances all referred

to primary windings

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Another version of equivalent circuits: neglect core losses (ignore shunt branch) & ignore resistive winding losses with

secondary impedances all referred to primary windings

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Equivalent Circuit (referred to primary)- only if (R1, X1 << Rc, Xm)

Approximate equivalent circuit referred to the primary.

(R1, X1 << Rc, Xm)

- Move the excitation branch close to V1

Equivalent Circuit (referred to secondary) Approximate equivalent circuit referred to the secondary.

(R1, X1 << Rc, Xm)

- Move the excitation branch close to V1

Simplified circuit referred to one side.

-Neglect the excitation branch

(very high permeability core and very small core loss)

Method to determine equivalent Circuit parameters for real transformer

PARAMETERS

to find the shunt admittance (Y)

Data given : I1/ Io and P1/ Po, you must know that V1 = rated voltage

no load

Io = open circuit current

Open-circuit test (no load)

No load

to find the series impedances (Z)

Data given : V1/ Vsc & P1/ Psc, you must know that I1/ Isc = rated current

Short-circuit test

Transformer Performance - Efficiency of a transformer in percent is given by %100

x

powerinput

poweroutput

- Voltage Regulation – the change in the magnitude of the secondary terminal voltage from no-load to full-load.

100egulation2

22x

V

VVR

nl

- Primary no-load.

- Secondary no-load.

100egulation'

2

'

21x

V

VVR

100egulation2

2

'

1x

V

VVR

Refer to primary

Refer to secondary

Example 4

Data obtained from short-circuit and open-circuit tests of a 20 kVA, 480/120 V 60 Hz transformer are as follows. Determine the parameter of the equivalent circuit.(solution: refer Glover, P 105~106)

Result of Short-circuit test: V1 = 35V & P1 = 300W Result of Open-circuit test: I2 = 12A, & P2 = 200 W Questions: 1. From the short circuit test, find the equivalent series impedances (neglect shunt

admittances) 2. From open circuit test, determine the shunt admittances (Ym) referred to

winding 1. neglect the series impedances.

solution

Part a: step 1 draw equivalent circuit of SC test & write down all info given:

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OR

82.01728.0XR

82.0RZX

XR84.042

35

I

VZ

173.042

300

I

PRRIP

42480

20

V

SII

eqeq

2

eq

2

eqeq

2

eq

2

eq

1

1eq

22

sc

eqeq

2

sc

rated 1,

ratedSC1

jjZ

AV

kVA

eq

Given : V1 = 35V P1 = 300W

solution Part b: step 1 draw equivalent circuit of OC test & write down all info given:

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OR

o

m

m

mm

m

mcom

c

cmco

c

ot

jjY

ZYjYremember

I

VXAIII

AR

VIjIII

P

VR

AIN

NIIVa

02.820063.000619.0000868.0BG

00619.0B

000868.0G

X

1B ,

R

1G ,

1 where,BG :

6.161,97.2

42.01152

480,

1152200

480

312480

120,480120

120

480VV

mc

m

c

m

m

c

cmc

122

1

22

1

2

1

2121

Given : I2 = 12A P2 = 200W

Example 5

A single phase transformer has 2000 turns on the primary winding and 500 turns on the secondary. Winding resistances are R1 = 2Ω and R2= 0.125 Ω; leakage reactances are X1 = 8Ω and x2 = 0.5Ω. The resistance load on the secondary is 12Ω.

(a)If the applied voltage at the terminals of the primary is 1000V, determine V2 at the load terminal of the transformer, neglecting magnetizing current.

(b) Compute the percent voltage regulation

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solution

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3 phase transformer

connections & phase shift

Three-phase Transformer

a. Three one-phase transformers are composed to be a three-phase transformer bank

b. Three-phase transformer wrapped around single three-legged core

Three-Phase Transformer

Connection technique

Connect the end of a winding with the

begining of another winding

Delta

Wye (star)

Connect the end of each winding at a

common point

a

b

c

b

a

c

Three-Phase Transformer

Three-Phase Transformer Connections.

End of Lecture 7

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