A device to raise and lower voltage in power systems · PDF fileA device to raise and lower...

7-Oct-11

1Lecture 07 Power Engineering - Egill Benedikt Hreinsson

Transformers in power systems

A device to raise and lower voltage in power systems

7-Oct-11


Presentation Outline• Introduction: What is a transformer?• Pictures and Examples of Transformers in Power Systems • Transformer Circuit models

– A circuit model for a single winding around a core– The ideal transformer as a part of the circuit model – The per unit system to simplify the analysis of transformers and power

systems– A circuit model for two windings around an iron core (transformer

circuit model) • 3-phase transformers• Multi-winding transformers• Auto-transformers• Control transformers • Applications of transformers in electrical power systems

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What is a transformer?• A device to raise or lower the voltage in a alternating current (AC)

power system. (High voltage is needed to transmit a lot of power)• At least 2 windings wound around a ferromagnetic core with mutual

magnetic flux and induced voltage.

p v i= ⋅

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Transformers -- Why?

PV V

X=

⋅1 2 sinδ = Power flow

P V I= cosφ = Power flow2P R I= = Power losses

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What is a transformer? (ctd.)

• A common magnetic flux, Φ in the core induces a voltage in windings according to the law of Faraday: e = dΦ/dt

•Some basic concepts for a transformer: Primary windings, secondary windings, core, leakage flux, ohmic resistance of windings, windings ratio, common magnetic flux

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Transformers -- Why?

• In an AC power systems there is the need to raise and lower the voltage (“step-up”, “step-down”) according to the power transfer need in the specific location in the power system:

• The power system is composed of disjoint zones, each with a standard voltage

• Standardized voltages in the system: 0,4 kV (400 V), 11kV, 33 kV, 66kV, 132 kV, 220 kV, 400 kV

• (RMS phase to phase!)

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Transformer symbols

Symbols for transformers in one-line diagrams

“American”

“European”

Generator symbolscircuit

breakers

Disconnect switches

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3 Separate One-Phase Transformers

One-phase high voltage inputs/ outputs

Transformer housing

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Dry transformer

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10Lecture 07 Power Engineering - Egill Benedikt HreinssonExamples of Real Transformers in Power Systems

The oil tank for the cooling medium

The cooling radiator for air cooling

Higher voltage terminal

Lower voltage terminal

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11Lecture 07 Power Engineering - Egill Benedikt HreinssonExamples of Real Transformers in Power Systems

Distribution transformer in North America

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The Basic Concepts of a Transformer

Leakage fluxLeakage flux

Iron core

Mutual flux

Primary windings

Secondary windings

LoadGenerator

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Transformer Circuit Model• A circuit model for a single winding around a

core• The ideal transformer as a part of the

circuit model • Per unit system with new units to simplify the

analysis of transformers• A circuit model for a two windings around an iron

core (transformer circuit model)

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A circuit model for one winding linking a magnetic core

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A Single Winding Around an Iron Core (1)

• (A) Consider a closed electric circuit with a single winding, wound around an iron core:

• (B) Consider a closed magnetic circuit along the closed path inside the magnetic core.

Closed path

Most of the flux Φc is within the core, but a small part of the flux Φl leaves the core

xV

−

+

cΦ1N

1I

Φ

1R

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16Lecture 07 Power Engineering - Egill Benedikt HreinssonA Circuit Model for a Single Winding Around an Iron Core

• Magnetic circuit: Apply the Amperes law to the closed path :

11INdlH∫ =⋅μBH =

ABC ⋅=Φ

CC RIN

Al

Φ==⋅Φ⋅

11μ

μ⋅=

AlR

A = A = Core cross sectional areaCore cross sectional area:: R = R = reluctancereluctance

The reluctance is low when μ is

high

xV

−

+

cΦ1N

1I

Φ

1R

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17Lecture 07 Power Engineering - Egill Benedikt HreinssonA Circuit Model for a Single Winding Around an Iron Core (3)

• Electric circuit: The sum of the voltages in the loop, is zero. Assume sinusoidal voltages and currents:

dtdN

dtdNIRV Cl Φ

+Φ

+= 11111

CNjILjIRV Φ++= 111111 ωω

1

11 I

NL lΦ=

μr is about 1000 - 5000 in the coreμr is about 1.0 outside the core

⇒ 3 orders of magnitude difference!

xV

−

+

cΦ1N

1I

Φ

1R

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RINjILjIRV 1

21

11111ωω ++=

Ohmic losses and Ohmic losses and heating of heating of windingswindings Magnetization Magnetization

of the coreof the coreLeakage fluxLeakage flux

CNjILjIRV Φ++= 111111 ωωCRIN Φ=11

Magnetic circuit: Electric circuit:

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Core MagnetizationCore Magnetization

21

mN

Xω

=ℜ

21 1

1 1 1 1 1j N I

V R I j L Iω

ω= + +ℜ

or: 1 1 1 1 1 1V R I j L I Eω= + +

1 1 1 1 1 1 mV R I j L I jI Xω= + +or:

1V

−

+

1I1R 1j Lω

21

mN

jX jω

=ℜ

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• We need to account for an imperfect core!• The following circuit model accounts for winding resistance, leakage flux,

core magnetization, eddy current losses and hysteresis losses

Hysteresis and eddy current losses

Leakage reactanceHeating of windings

A Circuit Model for a Single Winding Around an Iron Core (6)

Core magnetization

1V

−

+

1I1R 1j Lω

mjX cR

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The Ideal Transformer (IT) As a Mathematical Circuit Element

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The Ideal Transformer (IT)

aNN

EE

==2

1

2

1

I2I1

E1 E2The ideal

transformer

* *1 1 1 2 2 2S E I E I S= ⋅ = ⋅ =

2

1

I aI

=

….is a circuit element (“black box”) with the following characteristics:

The IT is lossless:

I2I1

E1 E2

Another symbol for the IT:

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The Ideal Transformer (2)

• No hysteresis losses• No eddy current losses• No leakage flux• No ohmic losses in windings • Magnetic permeability μ = ∞⇔ Reluctance R = 0• No magnetizing current for core magnetization

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Right Hand Rules

Current direction

Magnetic field

N

S

Current direction

Magnetic field

N

S

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The Ideal Transformer - Characteristics

The currents must be in opposite directions to The currents must be in opposite directions to ensure a magnetic balance in the core. ensure a magnetic balance in the core. ---- The The current induced on the secondary windings current induced on the secondary windings ““tries tries toto”” counteract the flux produced by the primary counteract the flux produced by the primary current.current.

Lenz law: The induced current will appear in such a Lenz law: The induced current will appear in such a direction that it opposes the change that produced direction that it opposes the change that produced it!it!

Heinrich FriedrichLenz1804-1865

1I 2I

1V

−

+

2V

−

+

1N 2N 2Z

cΦ1Φ 2Φ

1V

−

+

2V

−

+

rμ → ∞

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The “Dot Rule” of Transformer Windings

• The voltage at the “dot” side of the windings reaches its maximum at the same instance

• Currents entering the “dotted side” of the windings encircle the flux in the core in the same direction

1I 2I

1V

−

+

2V

−

+

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Transformer Circuit ModelsThe electrical circuits representing transformers in

electrical power systems

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Winding direction (1)• We start by drawing the core with 2 windings and opposite

winding directions and apply Ampere’s law to a closed path within the core

2211 ININdlH −=⋅∫1I

2I

cΦ

1 1 2 2C

Cl

N I N IA μ⋅Φ

= − = ℜΦ⋅

We get a minus sign atI2 but a plus sign I1 due to the direction of both windings with the right hand rule. This gives:

1V−

+

−

+

1N

2NZ'2V

1 1CdV N

dtΦ

=

'2 2

CdV NdtΦ

= −

We also get:

1 1 CV j Nω= Φ

'2 2 CV j Nω= − Φ

or:

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Winding direction (2)

• This tells us that the current difference in and out is a kind of magnetization current, I0 , that is:

1I

2I

cΦ'2

1 2 2 01 1

CN

I I I IN N

ℜ= + Φ = +

We get the following phasordiagram where1V

−

+

−

+

0I CΦ

1V1 1 CV j Nω= Φ

meaning that the voltage is perpendicular to the flux

Z'2V

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Winding direction (3)• We develop this phasor diagram more closely and get current

phasors that form a phase angle with the voltage depending on Z .

1I

2I

cΦ

'21 2 2 0

1 1C

NI I I I

N Nℜ

= + Φ = +

1V−

+

'2V−

+

0I CΦ

1V0I

1I'2I

Z

I0 is a “small” current and the voltage V’2 is induced in the lower winding according to Lenz law, or in the opposite direction relative to V1Then V2 will be in the same direction as V1

'2V2V

−

+

2V

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Winding direction (4)• We can now redraw the winding diagram to define dot

location.

1I

2I

cΦ

1V−

+

'2V−

+Z2V

−

+

1I 2IcΦ

1V−

+2V

2I

Z−

+

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Elementary Transformer Model

Iron core Leakage fluxLeakage flux

Mutual flux

Primary windings

Secondary windings

I1 I2

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Fluxes and Circuits in a Transformer

1V 2V

2LΦLeakage flux L2: Leakage flux L2:

CΦ

1LΦLeakage flux L1:Leakage flux L1:

I1 I2

N1 N2

Electrical CircuitA Magnetic Circuit

Electrical Circuit

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34Lecture 07 Power Engineering - Egill Benedikt Hreinsson1 Magnetic Circuit Model and 2 Electrical Circuits Models

dtdN

dtdNIRV CL Φ

+Φ

+= 11

1111

22 2 2 2 2

CL ddV R I N Ndt dt

ΦΦ= − − +

The equations for the primary and secondary electrical circuits for the general waveform of voltages and currents:

The Amperes-law for the closed path of the magnetic circuit in the iron core:

2211 ININdlH −=⋅∫(The + or - between the last 2 factors according to the definition of

current directions)

Resistance in the windings

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The Basic Transformer Model


2 2 2 2 2 2 CV R I j L I j Nω ω= − − + Φ

Primary windings using phasors and defining the leakage reactances L1 :

1

111 I

NL LΦ=

2 22

2

LNLIΦ

=

dtdN

dtdNIRV CL Φ

+Φ

+= 11

1111

22 2 2 2 2

L Cd dV R I N Ndt dtΦ Φ

= − − +

Secondary windings using phasors and defining the leakage reactances L2:

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36Lecture 07 Power Engineering - Egill Benedikt HreinssonCircuit Model based on the 2 Winding Circuits


2 2 2 2 2 2 CV R I j L I j Nω ω= − − + Φ

1 1 1 1 1 1

2 2 2 2 2 2

V R I jX I EV R I jX I E

= + += − − +

2

1

2

1

NN

EE

=CNjE Φ= 22 ω

CNjE Φ= 11 ω

R1 jωL1 R2 jωL2

V1 V2E2

I1 I2

E1

The Idealtransformer

Idea

ltra

nsfo

rmer

1 1

2 2

X LX L

ωω

==

We need to account for the magnetization current I0 from:

'21 2 2 0

1 1C

NI I I I

N Nℜ

= + Φ = +

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1V

−

+

1I 1ljX

mjb−cg2

2 11

NE E

N=

−

+

1E

−

+

1R

Álag:Z

2R 2ljX1N 2N

An idealtransformer

2V

−

+

22 2

1

NI I

N′ =

cI

0I

mI

The Transformer Circuit Model (A)

0'202

1

21 IIII

NNI +=+=

121

0 ENRjI

ω−=

21N

Rbm ω=

Here the current Ic is added due to the eddy current and hysteresis losses

i. e. I0 = Ic + Im

111111 EILjIRV ++= ω

2 2 2 2 2 2V R I j L I Eω= − − +

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The Transformation Ratio

1 1

2 2

E N aE N

= =

The symbol a is called the transformation ratio or the turns ratio of the transformer

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The Magnetic Circuit of the CoreAgain, the Amperes-law for the closed path in the iron core gives:

lHININdlH ⋅=−=⋅∫ 2211

average value of the H vectorH = −

length of the closed pathl =

μBH =

A = Core cross sectional areaR = Core reluctance

ABC ⋅=ΦC

C RININA

lΦ=−=

⋅Φ⋅

2211μ

'21 2 2 0

1 1C

NI I I I

N Nℜ

= + Φ = +

The magnetization current is I0

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Definition of Reluctance, R

μ⋅=

AlR

A = core areaμ = the magnetic permeability of the core: l= the length of the closed path in the core

μ = 1000 - 10000 in the core. μ ≈ 1 in air. If μ= ∞ it leads to R=0 This means that we have a “perfect” core.

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Deriving the Core Magnetization Current

2211 ININR C −=Φ

21 2

1 1

CRNI IN N

Φ= +

CNjE Φ= 11 ω

1 1

1 1C

E Ejj N Nω ω

Φ = = −

21 2 12

1 1

N RI I j EN Nω

= −

121

mRI j ENω

= −21 2 0

1

NI I IN

= + ..where:

0 c mI I I= +

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An impedance referred to the primary (or secondary) on an ideal transformer

Let us examine this circuit

Z2

I2I1

V1 V'1 V2

Z1 I2I1

V1 V'1

22

2

V ZI

= 12

VV

a= 2 1I I a= ⋅

211 2

1

V Z a ZI

= = ⋅

This leads to the following equivalent circuits, where the impedance can be “moved” or referred from the primary to the secondary side by multiplying by the factor a2 :

Z1

I1

V1Z2

I2I1

V1 V'1 V2

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Equivalent Transformer Circuits

Z I2I1

V1 V2

Z/a2I2I1

V1 V2

An impedance can be “referred to”either side of the transformer!

These 2 circuits are equivalent:

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The Transformer Circuit Model (B)

All impedances referred to the primary (left) side

1V

−

+

1I 1ljX

mjb−cg

−

+

2 2V aV′ =

−

+

1R

Álag:

LZ

22 2R a R′ = 1N 2N

Fullkominnspennir

22

II

a′ =

cI

0I

mI

22 2´l ljX ja X′ =

2V

−

+

1 2E aE=

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The Transformer Circuit Model (c)

All impedances referred to the secondary(right) side

1V

−

+

11 2

ll

jXjX

a′ =

2mja b−2

ca g

+

11

VV

a′=

+

11 2

RR

a′ =

Álag:

LZ

2R1N 2N

Fullkominnspennir

1 1I aI′ =

caI

0aI

maI

2ljX

2V

2I

−

1I

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Example 10a

Gefið er 3 fasa kerfið sem sýnt er á myndinni: Spennulindirnar gefa frá sér 3-fasa samhverfa spennu þar sem spennuvísirinn 400 45

2aV = ∠ (volt). Samviðnám hvers þéttis er 2cZ j= − ohm og seríuviðnám (spanviðnám) í línunni er 0.1jX j= ohm. Viðnámið milli

núllpunktanna 1 0.01n n nZ R jX j= + = + ohm og viðnám spankennda álagsins er 1 1.0jX j= ohm.

1. Hver er spennan bV og cV bæði lengd og fasahorn ef lindin hefur samhverfa spennu? 2. Stillið upp einfasa jafngildismynd fyrir ofangreinda rás. 3. Hvert er spennan 1V yfir fasa “a” í spannkennda álaginu? (lengd og fasahorn). 4. Hver er straumurinn aI milli punkta a’ og b’ í delta tengdu þéttasamstæðunni? (Finnið bæði lengd og fasahorn). 5. Hver er spennan milli punkta a’ og b’ og milli punkta c’ og b’ í delta tengdu samstæðunni þéttasamstæðunni? (Finnið einnig hér bæði

lengd og fasahorn). 6. Hvernig breytist þessi spenna með breytingu á nZ ?

I1

VaV1

Zc Zc

Zc

jX

jX1

jX1jX1

jX

Zn=Rn+jXn

jX

Vb

Vc

+ +

++

- -

--

Ia

a

bc

a'

b'c'

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Example 10b1. Einfasa 100kVA spennir er með umsetningu

2400/240V. Hann er gerður fyrir 60 Hz og er dreifispennir sem er notaður til að lækka spennu (step down). Álagið er tengt 240 V bakvafinu og tekur 90 kVA við spankenndan aflstuðul 0.8 og er spennan inn á það í raun 230 V. Gera skal ráð fyrir ideal spenni og reikna út (a) spennu á forvafi (háspennuhlið) (b) álagsviðnám (samviðnám) á bakvafi (c) álagssamviðnám í ohm ef það er flutt yfir á forvafið og (d) raunafl og launafl sem fer inn á forvafið.

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Example 11Einfasa spennir (sbr mynd) hefur umsetningu a = 440V/220V og málraun 5 kVA. Skammhlaupsviðnám hans miðað við forvaf er R + jX þar sem R = r1 + a2r2 og X = x1 + a2x2 Þegar lágspennuhlið spennisins er skammhleypt og 35 volta spenna er tengd við forvafið, fer málstraumur um vafninga spennisins og raunaflið á forvafi er 100W. Finnið raunviðnám og spanviðnám (í ohm) á báðum hliðum spennisins ef hlutfall raunviðnáms og spanviðnáms er hið sama á báðum hliðum og sömu afltöp á báðum hliðum spennisins.

r2 + jx2 r1 + jx1

bakvaf 220 V 440 V forvaf

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Example 10a solution (a)

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Example 10a solution (b)

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Example 10a solution (c)

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Example 10b - solution

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Example 11 - solution

A device to raise and lower voltage in power systems · PDF fileA device to raise and lower...

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