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Instructor:

Kai Sun

Fall 2015

ECE 325 – Electric Energy System Components4- Transformers

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Content

• Ideal Transformer (Ch. 9)

•Practical Transformers (Ch. 10)

•Special Transformers (Ch. 11)

•Three-Phase Transformers (Ch. 12)

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Power Transformer

• Ideal transformers

– Winding resistance is negligible

– No leakage flux

– Permeability of the core is infinite

(zero magnetizing current Ip is

needed to produce flux)

– No core loss

• Real transformers:

– Windings have resistance

– Windings do not link the same

flux (having leakage flux)

– Permeability of the core is finite

– Have core losses (hysteresis

losses and eddy current losses

due to time varying flux)

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I1 I2

Ideal Transformer

• Primary winding (assuming sinusoidal flux):

Because there is no flux leakage,

1=2=m (mutual flux)

• Secondary winding:

Because there is no flux leakage

• Voltage/turns ratio a=E1/E2=N1/N2

max cosm t

1 max

1

1

x

1

ma

sin

cos 90

g N

E

de e N t

dt

t

1max 1 max 1 max 1 max2 6.28 2 4.44E fN fN fN

1 1 max4.44 90gE E fN

2 2 max4.44 90E fN 2 2 2max cos 90mde N E t

dt

E1

2max 2 max 2 max2 2 4.44E fN fN

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I1 I2

No-load and Load Conditions• Under no-load conditions, because of

infinitely permeable core, no magnetizing

current is required to produce flux m.

There are

I1=I2=0

• Under load conditions (I10, I20,)

because of the infinitely permeable core,

there is an exact mmf balance

F1=I1N1=I2N2=F2

• Phasor diagram:

– Assume power factor cos

1 2 1

2 1 2

E I Na

E I N

E1

Eg, E1

E2

I1

I2

Eg, E1

E2

I1

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Circuit Symbol for an Ideal Transformer

1 2

21

E aE

II

a

21

ab cdE aE

II

a

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Impedance Ratio

• Impedance transformation

– The impedance seen by the source (primary side) is a2 times

the real impedance (secondary side)

– Thus, an ideal transformer has the amazing ability to increase

or decrease the value of an impedance

1

1

22

2 /X

E aE

I I aZ a Z

2

2

EZ

I

1 2

1 2 /

E aE

I I a

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Shifting Impedances (SP)

a2Zp Zs

aEp Es

1/aIp Is

1

2

3

4

9

Shifting Impedances (PS)

1/a2Zp Zs

1/aEp Es

aIp Is

1

2

aI1

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Examples 9-4&9-5

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• Pm+jQm and Io are the complex power and current outputs of source Eg

under no-loading conditions. Pm is the iron loss

• Rm is a resistance causing the iron loss and resulting heat

Rm=E12/Pm If=E1/Rm

• Xm is the magnetizing reactance that measures permeability of the core.

A smaller Xm means lower permeability and needs a higher magnetizing

current Im (i.e. bigger reactive power Qm) to set up mutual flux m

Xm=E12/Qm Im=E1/jXm

Im If

If

Im

Considering an imperfect core

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Considering loose coupling and resistances of windings

f1 f2

f1 f2

Ef1 Ef2

E1 =4.44fN1m E2 =4.44fN2m

Ef1 =4.44fN1f1 Ef2 =4.44fN2f2

E1+Ef1=Ep=Eg E2+Ef2=Es

• Define winding leakage reactances

Xf1=Ef1/I1 Xf2=Ef2/I2

• Add winding resistances R1 and R2 in series

with Xf1 and Xf2, respectively

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Equivalent Circuit of a Practical Transformer

• No-load conditions

– I1=I2=0

– Ignore R1, Xf1, R2 and Xf2

– EpE1=aE2=aEs

• 10%-100% load conditions

– Ip>>Io,

– Ignore Io

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• No-load conditions

– Ignoring winding

leakage fluxes and losses

• 10%-100% load conditions

– Ignoring core reluctance

and losses

Simplified Equivalent Circuits

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Equivalent Circuits Referred to One Side

a2 a2

a a2

/a

a/a2

/a2

Xm/a2 Rm/a2

aIo

/a

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Simplified Equivalent Circuits Referred to One Side

• Under load conditions

2 2

1 2 1 2

( )

p p p

f f

Z R j X

R a R j X a X

1

2

Na

N

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Voltage Regulation

• With the primary voltage held constant at its rated value, the

voltage regulation is % change of the secondary voltage from

no-load to full load (rated)

ENL: secondary voltage at no-load

EFL: secondary voltage at full-load

• The voltage regulation depends on the power factor of the load

on the secondary side

– If the load is capacitive, the full-load voltage may exceed the

no-load voltage, in which case the voltage regulation

becomes negative

Voltage Regulation = 100NL FL

FL

E E

E

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Measuring Transformer Impedances

• Open-circuit (no-load) test

– Neglect R1, Xf1, R2, Xf2

– Measure Pm (core loss), |Ep|,

|Io| and |Es|

2| |p

m

m

ER

P

| || |

p

f

m

EI

R

2 2| | | | | |m o fI I I

| |

| |

p

m

m

EX

I

E1=aEs

1

2

| |

| |

p

s

ENa

N E

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| || |

| |

scp

sc

EZ

I

2| |

scp

sc

PR

I

2 2| |p p pX Z R

• Short-circuit test

– Apply a low voltage ESC to the

primary side to create ISC less than

the nominal value to prevent

overheating and rapid change in

winding resistance

– Neglect Rm and Xm due to low core

flux

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Construction of a power transformer

• Core: made of iron for high permeability; laminated and high

resistive to reduce iron losses

• Windings: the primary and secondary coils are wound closely on

top of each other with careful insulation for tight coupling; the

HV winding has more turns but uses a smaller size of conductor,

so copper/aluminum of two windings are about the same

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• High-Voltage winding (H1&H2) and Low-Voltage winding (X1& X2)

EH1,H2/EX1,X2=EH/EX=NH/NX

• Polarity test:

1. Connect HV winding to a low voltage source Eg

2. Connect a jumper J between any two adjacent HV and LV terminals

3. Connect two voltmeters to as shown in Figure 10.11

4. The polarity is additive if |Ex|>|Ep| or, otherwise, is subtractive.

Standard terminal markings and polarity tests

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The Per-Unit System

• Quantity in Per-Unit = Actual quantity / Base or nominal value of quantity

• Why per-unit notations?

– Neglecting different voltage levels of transformers, lines and generators

– Powers, voltages, currents and impedances are expressed as decimal

fractions of respective base quantities

(Source: EPRI Dynamic Tutorial)

Choose 20kV,

345KV and 138kV

as base voltages

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• Four base quantities are required to completely define a per-unit system

• We need to select two independent base quantities of the four and calculate the

other two, e.g. selecting Sn and En

• For a transformer:

– Usually, Sn, Epn, Esn and the total impedance Zp (in p.u. or %) referred to the

primary side are given.

nn

n

SI

E

2

nnn

n n

EEZ

I S

pu

n

SS

S pu

n

EE

E pu

n

II

I

pu

n

ZZ

Z

nnp

np

SI

E

2

( )npnp

np

np n

EEZ

I S

nns

ns

SI

E

p np

s ns

N Ea

N E

2

( )nsns

ns

ns n

EEZ

I S

( ) (p.u.)p p npZ Z Z

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Examples 10-5, 10-6, 10-7, 10-8 & 10-10

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Autotransformer• A conventional two-winding transformer can be changed to an autotransformer by

connecting its two coils in series (note: 4 combinations).

• The connection may use a sliding contact to providing variable output voltage.

20MVA (115/69kV) McGraw-Edison Substation

Auto-Transformer (Y-Y) (Source:

http://www.tucsontransformer.com)(Source: EPRI Power System Dynamic Tutorial)

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• The primary and secondary windings have a common terminal

A, and hence are not isolated from each other

E2/E1=ECA/EBA=N2/N1 (There is always N1>N2)

I1N1=I2N2 I1(N1-N2)=(I2-I1)N2

E1I1=E2I2

• An autotransformer has a smaller size (or equivalently, a higher

kVA rating for the same size) but loses insulation between

primary and secondary windings

N1

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Conventional transformer connected as an autotransformer (Example 11-2)

• Reconnect the transformer to obtain

– 600V / 480V

– 600V / 720V

– 120V / 480V

– 120V / 720V

• I1=15000/600=25A

• I2=15000/120=125A

When a transformer connects a source (on the primary side) with a

load (on the secondary side), it is the load (current flows into

terminal +) of the primary circuit and is the source (current flows

out of terminal +) of the secondary circuit, so two winding

currents (I1 and I2) have opposite directions.

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+

600V

-

-

120V

+

+

480V

-

H1

H2

X2

X1

I2=125A

I2-I1=100A

I1=25A

+

600V

-

+

120V

-

+

720V

-

H1

H2

X1

X2

I2=125A

I2+I1=150A

I1=25A

+

120V

-

-

600V

+

-

480V

+

X1

X2

I1=25A

I2-I1=100A

I2=125A

H2

H1

|Si|=600150=90kVA |Si|=120100=12kVA

|So|=720125=90kVA |So|=48025 =12kVA

The maximum apparent power |Smax|=max(|E1|,|E2|)(|I1|+|I2|)= (|E1|+|E2|) max(|I1|, |I2|)

600V 120V

H1

H2 X2

X1

|Si|=600100=60kVA

|So|=480125=60kVA

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• Voltage transformers (VTs),

also called potential

transformers (PTs), are

installed on HV lines to

measure line-to-neutral

voltage

• Current transformers (CTs)

are used to measure or

monitor the current in a line

for system measurement and

protection

Voltage Transformers and Current Transformers

7kVA

80.5kV

50/60Hz PT

5kVA,

100A/5A

60Hz CT

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Three-Phase Transformers

• A bank of three single-phase transformers connected in Y or configurations

– Four possible combinations: Y-Y, △-△, Y-△ and △-Y

– Y: lower insulation costs, with neutral for grounding, 3rd harmonics

– △: more insulation costs, no neutral, no 3rd harmonics

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3rd harmonics problem with three-phase transformers

• A △ configuration provides a closed path for 3rd harmonics, or

in other words, all triple harmonics are trapped in the △ loop.

Distorted waveform

Fundamental

3rd harmonic3

9

, 3, 5,

7 9

, 3, 5, 7, 9

, 3, 5,

7 9

, 5, 7

, 5, 7

, 5, 7

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Three-phase transformers: - connection

3 Ip 3 Is

• Each transformer, considered

alone, acts as if it were placed in

a single-phase circuit.

• Because EH1H2 and EX1X2 are in

phase for each single

transformer, line voltages EAB

with E12, EBC with E23, and ECA

with E31 are in phase.

• A balanced three-phase load

produces balanced line currents

in line 1-2-3 and lines A-B-C

• The power rating is 3x of a

single transformer.

• One transformer may be absent

(i.e. open-delta connection)

under maintenance or

emergency conditions

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Three-phase transformers: -Y (or Y-) connection

3 Ip

• X2 terminals of 3 secondary

windings are connected to a

common neutral N.

• Because EH1H2 and EX1X2 are in

phase for each single

transformer, EAB with E1N, EBC

with E2N, and ECA with E3N are

in phase.

• There is a 30o phase shift

between the line voltages of the

incoming (EAB) and outgoing

(E12) lines

• △-Y is commonly used for

voltage step-up transformers

while Y-△ is commonly used

for voltage step-down

transformers

E1N

EN2

E2N

E3N

30o

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Three-phase transformers: Y-Y connection• No phase shift between the incoming and outgoing line voltages

• Without a configuration to trap the 3rd harmonic, it has seriously distorted voltages due to

the 3rd harmonics.

• Two solutions:

– Ground the neutrals of the primary winding and the source

– Provide each transformer with a tertiary winding, and connect 3 tertiary windings in delta

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Phase-shifting Principle

• A 3-phase system can easily shift the phase

angle of a voltage to create 2-, 6-, and 12-phase

systems used in power electronic converters

and controllers

• Phase shifting (performed by phase-shifting

transformers) is also used in power flow

control over long transmission lines

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Phase-shifting transformers

• A special type of 3-phase autotransformer that

shits the phase angle between the incoming and

outgoing lines without changing the voltage ratio.

• Twist all incoming line voltages by angle

(0~20o) without changing their magnitudes

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Calculations involving 3-phase transformers

• For any 3-phase transformer bank, assume that the primary and

secondary windings are both connected in Y configuration (if not,

transform it to Y by |ZY|=|Z|/3)

• Consider only one transformer (single phase) of the assumed Y-Y

transformer bank

– Its primary and secondary voltages are both the line-to-neutral

voltages of the incoming and outgoing lines

– Its nominal power rating and load are 1/3 of the 3-phase

transformer bank

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Examples 12-1, 12-2, 12-3, 12-6 & 12-7

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Homework Assignment #3

• Read Chapters 9&10

• Questions:

– Prove the polarity test on slide #21

– 9-4, 9-5, 9-6, 9-7, 10-19, 10-28, 10-29, 10-30, 10-31

• Due date:

– hand in your solution in the class on 10/5 (Mon) or

– to Denis at MK 205 or by email (dosipov@vols.utk.edu)

before the end of 10/5