1 chapter 1 emt 113: v-2008 school of computer and communication engineering, unimap prepared by:...

1

CHAPTER CHAPTER 11

EMT 113: V-2008EMT 113: V-2008

School of Computer and School of Computer and Communication Engineering, Communication Engineering,

UniMAPUniMAP Prepared By: Prepared By:

Amir Razif A. b. Jamil AbdullahAmir Razif A. b. Jamil Abdullah

TransforTransformermer

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1.1 1.1 IntroductionIntroduction to Transformer. to Transformer.1.2 1.2 ApplicationsApplications of Transformer. of Transformer.1.3 1.3 TypesTypes and and Constructions Constructions of Transformers.of Transformers.1.4 General 1.4 General TheoryTheory of Transformer Operation. of Transformer Operation.1.5 1.5 Ideal Ideal Transformer.Transformer.1.6 1.6 Real Real Single-Phase Transformer. Single-Phase Transformer.1.7 1.7 ExactExact Equivalent Circuit of a Real Transformer. Equivalent Circuit of a Real Transformer.1.8 1.8 Approximate Approximate Equivalent Circuit of a Equivalent Circuit of a

Transformer.Transformer.1.9 Transformer 1.9 Transformer Voltage RegulationVoltage Regulation and and EfficiencyEfficiency..1.10 1.10 OpenOpen Circuit and Circuit and ShortShort Circuit. Circuit.1.11 1.11 Three PhaseThree Phase Transformer. Transformer.

1.0 Transformer.1.0 Transformer.

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What is Transformer ? What is Transformer ? Transformer Transformer is a device that changes ac electrical device that changes ac electrical

power at one voltage level to ac electric power at power at one voltage level to ac electric power at another voltage level through the action of another voltage level through the action of magnetic field.magnetic field.

Figure 1.1 is the block diagrams of; (a) Transformer.(b) Electric motor.(c) Generator.

Figure 1.1: Block Figure 1.1: Block Diagrams of Transformer, Diagrams of Transformer, an Electric Motor and a an Electric Motor and a Generator..Generator..

1.1 Introduction to 1.1 Introduction to Transformer.Transformer.

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Figure 1.1a: A 500 MVA Power Transformer. (left) Figure 1.1a: A 500 MVA Power Transformer. (left)

A Pole-Mount 15 kVA Distribution Transformer. (right) (Courtesy of A Pole-Mount 15 kVA Distribution Transformer. (right) (Courtesy of ABB)ABB)

Cont’d…Cont’d…

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Why do we need transformer for ?Why do we need transformer for ?

(a) (a) Step Up.Step Up. Step up transformer, it will decrease the current to keep

the power into the device equal to the power out of it. In modern power system, electrical power is generated

at voltage of 12kV to 25kV. Transformer will step up the voltage to between 110kV to 1000kV for transmission over long distance at very low lost.

- Unit transformer.

(b) (b) Step Down.Step Down. The transformer will stepped down the voltage to the

12kV to 34.5kV range for local distribution in the homes, offices and factories as low as 120V (America) and 240V (Malaysia).

- Substation transformer. – distribution transformer.

1.2 Application of 1.2 Application of Transformer.Transformer.

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Power transformers are constructed on two types two types of coresof cores;(i) Core form.(ii) Shell form.

Figure 1.2: Core Form and Shell Form.Figure 1.2: Core Form and Shell Form.

A) Core type B) Shell type

1.3 Types and 1.3 Types and Construction of Construction of Transformer.Transformer.

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Core Form. Core Form. The core form construction consists of a simple rectangular

laminated piece of steel with the transform winding wrapped around the two sides of the rectangle.

Shell Form. Shell Form.

The shell form construction consists of a three-legged laminated core with the winding wrapped around the center leg.

In both cases the core is constructed of thin laminations electrically isolated from each other in order to minimize eddy current.

The coils are usually not directly connected. The common magnetic flux present within the coils connects the

coils.

A) Core type B) Shell type


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Construction.Construction. Transformer consists of two or more coils of wire wrapped around

a common ferromagnetic core. The coils are usually not directly connected.

The common magnetic flux present within the coils connects the coils.

There are two windings;(i) Primary winding (input winding); the winding that is connected to the power source.(ii) Secondary winding (output winding); the winding connected to the loads.

Figure 1.3: A Simple Figure 1.3: A Simple Transformer.Transformer.


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Operation.Operation. When AC voltage is applied to the primary winding of the

transformer, an AC current will result i1 or i2 (current at load). The AC primary current i1 set up time varying magnetic flux

f in the core. The flux links the secondary winding of the transformer.

From the Faraday law, the emf will be induced in the secondary winding. This is known as transformer action.

The current i2 will flow in the secondary winding and electric power will be transfer to the load.

The direction of the current in the secondary winding is determined by Len’z law. The secondary current’s direction is such that the flux produced by this current opposes the change in the original flux with respect to time.


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According to the Faraday’s lawFaraday’s law of electromagnetic induction, electromagnetic force (emf’s) are induced in N1 and N2 due to a time rate of change of fM,

Where, (1.1)e = instantaneous voltage induced by magnetic field (emf), = number of flux linkages between the magnetic field and the electric circuit. f = effective flux

Lenz’s LawLenz’s Law states that the direction of e1 is such to produce a current that opposes the flux changes.

dt

dN

dt

de

dt

dNe

22 ;11

dt

dNe

Figure 1.4: Basic Figure 1.4: Basic Transformer Transformer Components.Components.

1.4 General Theory of 1.4 General Theory of Transformer Operation.Transformer Operation.

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If the winding resistance is neglected, then equation (1.1) become;

(1.2)

Taking the voltage ratio in equation (1.2) results in,

(1.3) Neglecting losses means that the instantaneous power is the same

on both sides of the transformer;(1.4)

Combining all the above equation we get the equation (1.5) where a is the turn ratio of the transformer.

(1.5)

);(111dt

dNev )(222dt

dNev

2

1

2

1

e

e

N

N

2211 ieie

1

2

2

1

2

1

i

i

v

v

N

Na


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The flux varies sinusoidally such that;

The rms value of the induce voltage is; )2sin( max ft

dt

dN

dt

dNe

= max sin t

maxmax

44.42

fN

NE

max


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Losses are composed of two parts;(a) The Eddy-Current lost. (b) The Hysteresis loss.

Eddy current lostEddy current lost is basically loss due to the induced current in the magnetic material. To reduce this lost, the magnetic circuit is usually made of a stack of thin laminations.

Hysteresis lostHysteresis lost is caused by the energy used in orienting the magnetic domains of the material along the field. The lost depends on the material used.

Figure 1.5: A Magnetic Figure 1.5: A Magnetic Hysteresis or B-H Curve of Hysteresis or B-H Curve of Core Steel.Core Steel.


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Example 1.1:Example 1.1: Transformer. Transformer.How many turns must the primary and the How many turns must the primary and the secondary windings of a 220 V-110 V, 60 Hz ideal secondary windings of a 220 V-110 V, 60 Hz ideal transformer have if the core flux is not allowed to transformer have if the core flux is not allowed to exceed 5mWb?exceed 5mWb?

Solution:Solution:For an ideal transformer with no losses,

From the emf equation, we have

.

VVE

VVE

110

220

22

11

.83)105)(60)(11.4(

110

.166)105)(60)(11.4(

220

**11.4

32

3

max

11

turnsX

N

turnsX

f

EN

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An Ideal transformer is a lossless device with an input winding and an output winding.

Zero resistance result in zero voltage drops between the terminal voltages and induced voltages

Figure 1.6 shows the relationship of input voltage and output voltage of the ideal transformer.

Figure 1.6: An Ideal Transformer and the Schematic Symbols.Figure 1.6: An Ideal Transformer and the Schematic Symbols.

1.5 The Ideal 1.5 The Ideal Transformer.Transformer.

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The relationship between voltage and the number of turns. Np , number of turns of wire on its primary side.

Ns , number of turns of wire on its secondary side.

Vp(t), voltage applied to the primary side.

Vs(t), voltage applied to the secondary side.

where a is defined to be the turns ratio of the transformer.

aN

N

tv

tv

s

p

s

p )(

)(


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The relationship between current into the primary side, Ip(t), of transformer versus the secondary side, Is(t), of the transformer;

In term of phasor quantitiesphasor quantities;

-Note that Vp and Vs are in the same phase angle. Ip and Is are in the same phase angle too.

- the turn ratio, a, of the ideal transformer affects the magnitude only but not the their angle.

)()( tINtIN sspp

atI

tI

s

p 1

)(

)(

aV

V

s

p aI

I

s

p 1


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The dot conventiondot convention appearing at one end of each winding tell the polarity of the voltage and current on the secondary side of the transformer.

If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the doted on each side of the core.

If the primary current of the transformer flow into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.


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Power supplied to the transformer by the primary circuit is given by ;

where, p is the angle between the primary voltage and the primary current.

The power supplied by the transformer secondary circuit to its loads is given by the equation;

where, s is the angle between the secondary voltage and the secondary current.

Voltage and current angles are unaffected by an ideal transformer, p – s = . The primary and secondary windings of an ideal transformer have the same power factor.

sssout IVP cos

pppin IVP cos

1.5.1 Power in an Ideal 1.5.1 Power in an Ideal Transformer.Transformer.

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The power outpower out of a transformer;- apply Vs= Vp/a and Is= aIp into the above equation gives,

** The output power of an ideal transformer is equal to the input power.

The reactive powerreactive power, Q, and the apparent powerapparent power, S; In term of phasor quantities;

-Vp and Vs are in the same phase angle. Ip and Is are in the same phase angle too.

- the turn ratio, a, of the ideal transformer affects the magnitude only but not the their angle.

cosssout IVP

inppout

pp

out

PIVP

aIa

VP

cos

cos)(

outssppin

outssppin

SIVIVS

QIVIVQ

sinsin


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Example 1.2:Example 1.2: Ideal Transformer. Ideal Transformer.Consider an ideal, single-phase 2400V-240V Consider an ideal, single-phase 2400V-240V transformer. The primary is connected to a 2200V transformer. The primary is connected to a 2200V source and the secondary is connected to an source and the secondary is connected to an impedance of 2 W < 36.9impedance of 2 W < 36.9oo, find,, find,(a) The secondary output current and voltage.(b) The primary input current.© The load impedance as seen from the primary side.(d) The input and output apparent power.(e) The output power factor.

Solution:Solution:(a) The ratio of rated terminal voltage equal to the actual turns-ratio as;

Ω

22

.


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The ideal transformer in Section 1.5 can never been made. The real transformer has many imperfections.

The real transformers consists of two or more coils of wire physically wrapped around the ferromagnetic core. The real transformer approximate the characteristic of the ideal transformer.

OperationOperation if the real transformer; (i) It consists of two coils of wire wrapped around a transformer core.(ii) The primary of the transformer is connected to an ac power source, and the secondary winding is an open-circuited.(iii) Figure 1.5 is the hysteresis of the transformer.(iv) Basic operation from the faraday law,

dt

deind

1.6 Real Single-Phase 1.6 Real Single-Phase Transformer.Transformer.

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is the flux linkage in the coil across which the voltage is being induced.

The sum of the flux passing through each turn in the coil added over all the turns of the coil is;

The average flux per turns is given by ;

And Faraday’s law can be written as ,N

N

ii

1

dt

dNeind


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Copper lossesCopper losses are resistive loses in the primary and the secondary windings of the transformer core. They are proportional to the square of the current in the windings.

Copper losses are modeled by placing a resistor RRpp in the primary circuit of the transformer and a resistor RRss in the secondary circuit.

The leakage flux in the primary windings is,

Figure 1.7 is an exact model of a real transformer. To analyze the transformer it is necessary to convert the entire circuit to an equivalent circuit at a single voltage level as in Figure 1.8.

dt

diLte

dt

diLte

ssLS

ppLP

)(

)(

1.7 1.7 Exact EquivalentExact Equivalent Circuit of Real Circuit of Real Transformer.Transformer.

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Figure 1.7: Model of a Figure 1.7: Model of a Real Transformer.Real Transformer.


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Figure 1.8: (a) The Transformer Model Referred to its Figure 1.8: (a) The Transformer Model Referred to its Primary Primary Windings (top).Windings (top).

(b) The Transformer Model Referred to its (b) The Transformer Model Referred to its Secondary Voltage Level Secondary Voltage Level (bottom).(bottom).


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Symbols used for the Exact Equivalent Circuit above;


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The major use of the exact equivalent circuit of a transformer is to determine the characteristics such as voltage regulation and efficiency.

A phasor diagram for the circuit of Figure 1.8, for lagging power factor can be obtained by using the following equations:

Based on the above equations and assuming a zero degree reference angle for V2, the phasor diagram is shown in Figure 1.9 for the exact equivalent circuit model of a transformer.

Figure 1.9: RMS Phasor Diagram for the Figure 1.9: RMS Phasor Diagram for the Exact Equivalent Circuit Model of a Exact Equivalent Circuit Model of a Transformer.Transformer.


Not Not CoveredCovered

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In the Approximate model ,the voltage drop in Rp and Xp is negligible because the current is very small. Figure 1.10 is the Approximate equivalent circuit referred to the primary side.

Figure 1.10: Figure 1.10: Approximate Transformer ModelApproximate Transformer Model Referred to the Primary Side. Referred to the Primary Side.

1.8 1.8 Approximate Approximate EquivalentEquivalent Circuit of Real Circuit of Real Transformer.Transformer.

1RRp

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The voltage in the primary series impedance (r1 + jx1) is small, even at full load. Also, the no load current (I0) is so small that its effect on the voltage drop in the primary series impedance is negligible.

Therefore, it matters little if the shunt branch of Rc in parallel with Xm is connected before the primary series impedance or after it. The core loss and magnetizing currents are not greatly affected by the move.

Connecting the shunt components right at the input terminals has the great advantage of permitting the two series impedance to be combined into one complex impedance.

The equivalent impedance for the circuit in Figure 1.11 is;


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The value of this equivalent impedance of a particular transformer depends on whether the model used is referred to the primary or secondary.

Figure 1.11: Figure 1.11: Approximate CircuitApproximate Circuit Model of a Transformer Model of a Transformer Referred to the Referred to the SecondarySecondary. .

The equivalent impedance for the circuit in Figure 1.11 is;


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Voltage regulationVoltage regulation is a measure of the change in the terminal voltage of the transformer with respect to loading. Therefore the voltage regulation is defined as:

At no load, Vs = Vp/a and the voltage regulation can also be express as;

In the per-unitper-unit system;

For ideal transformer VR = 0. It is a good practice to have as small voltage regulator as possible.

%100,

,,

fls

flsnls

V

VVVR

%100,

,

fls

flsp

V

Va

V

VR

%100,,

,,,

pulfs

puflspup

V

VVVR

1.9 Transformer Voltage 1.9 Transformer Voltage Regulation and Efficiency.Regulation and Efficiency.

34

Figure 1.12: Example of Transformer Voltage Regulation.Figure 1.12: Example of Transformer Voltage Regulation.


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Transformer EfficiencyTransformer Efficiency, efficiency of a transformer is defined as follows;

For Non-Ideal transformer, the output power is less than the input power because of losses. These losses are the winding or I2R loss (copper losses) and the core loss (hysteresis and eddy-current losses).

In term of the total losses, PPlosseslosses,, the above equation may be expressed as;

The winding or copper loss is load dependent, whereas the core loss is constant and almost independent of the load on the transformer.

1

2

P

P

PowerInput

PowerOutput

corecopperlosses

losses

PPP

P

PP

P

P

PP

2

2

2

2

1

\11


36

The efficiency can also be obtained by using the per-unit system.


37

Example 1.3:Example 1.3: Transformer Voltage Transformer Voltage Regulation.Regulation.

A 10 kVA, 2400V/240V, single-phase transformer A 10 kVA, 2400V/240V, single-phase transformer has the following resistance and leakage has the following resistance and leakage reactances: rreactances: r11=30 =30 , x, x11=15 =15 , r, r22=0.03 =0.03 , x, x22 = = 0.15 0.15

Use the approximate circuit model to find the Use the approximate circuit model to find the voltage regulation when the load power factor is; a) voltage regulation when the load power factor is; a) 0.8 lagging and b) 0.8 leading0.8 lagging and b) 0.8 leading

Solution:Solution:

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Solution:Solution:

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Open Circuit Test.Open Circuit Test. The open circuit test is conducted by applying

rated voltage at rated frequency to one of the windings, with the other windings open circuited. The input power and current are measured, Figure 1.13 .

For reasons of safety and convenience, the measurements are made on the low-voltage (LV) side of the transformer.

1.10 Open Circuit and 1.10 Open Circuit and Short Circuit.Short Circuit.

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Figure 1.13: Equivalent Circuit of theFigure 1.13: Equivalent Circuit of the Open-Circuit TestOpen-Circuit Test..

The high voltage (HV) side is open, the input current is equal to the no load current or exciting current (I0), and is quite small.

The voltage drops in the primary leakage reactance and winding resistance may be neglected and so may the primary loss (I12r1).

The input power is almost equal to the core loss at rated voltage and frequency.


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oc is the angle by which Io_LV lags Voc. The core loss current, Ic is in phase with Voc while Im lags Voc by 90°. Then;

The core-loss current, Ic, may be found from above equation, then Rc_LV may be calculated by the equation below,

The magnetization current Im is given by the above equation or may be found from Ioc and Ic using


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Short Circuit Test.Short Circuit Test. The short-circuit test is used to determine the

equivalent series resistance and reactance. One winding is shorted at its terminals, and the

other winding is connected through proper meters to a variable, low-voltage, high-current source of rated frequency, Figure 1.14.

The source voltage is increased until the current into the transformer reaches rated value. To avoid unnecessary high currents, the short-circuit measurements are made on the high-voltage side of the transformer.


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Figure 1.14: Equivalent Circuit of the Figure 1.14: Equivalent Circuit of the Short-Circuit Test.Short-Circuit Test.

Neglecting I0, the input power during this test is consumed in the equivalent resistance referred to the primary or high-voltage side, Req_HV. Then


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Almost all the major power generation and distribution systems in the world today are three-phase ac system.

Two Two ways of constructing transformer of three-phase circuit;(i) Three single phase transformers are connected in three-phase bank.(ii) Make a three-phased transformer consisting of three sets of windings wrapped on a common core.

The three-phased transformer on a common core (ii) is preferred because it is lighter, smaller, cheaper and slightly more efficient.

1.11 Three Phase 1.11 Three Phase Transformer.Transformer.

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A three-phase transformer consists of three transformers either separate or combined on one core.

There are four possible connections between the secondary and primary of a three-phase transformer.(1) Wye-Wye (Y-Y).(2) Wye-Delta (Y-D).(3) Delta-Wye (D-Y).(4) Delta-Delta (D -D).

Figure 1.15 are these four possible transformer connections.

1.11.1 Three-Phase 1.11.1 Three-Phase Transformer Connections.Transformer Connections.

46

Figure 1.15: Three-Phase Transformer Connections and Wiring Figure 1.15: Three-Phase Transformer Connections and Wiring Diagram.Diagram.


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(1) Wye-Wye Connection.(1) Wye-Wye Connection.

(2) (2) Wye-Delta Connection.Wye-Delta Connection. A three-phase transformer consists of three

transformers either separate or combined on one core.

(3)(3) Delta-Wye Connection. Delta-Wye Connection.

(4)(4) Delta-Delta Connection. Delta-Delta Connection.

YaV

V

LS

LP 3

YaV

V

LS

LP 3

aV

V

V

V

S

P

LS

LP

YYaV

V

LS

LP


48

Example 1.4:Example 1.4: Three Phase Three Phase Transformer.Transformer.What should be the ratings (voltages What should be the ratings (voltages and currents) and turns ratio of a three-and currents) and turns ratio of a three-phase transformer to transform 10 MVA phase transformer to transform 10 MVA from 230 kV to 4160 V, if the from 230 kV to 4160 V, if the transformer is to be connected:transformer is to be connected:a) wye-delta, a) wye-delta, b) delta-wye, and b) delta-wye, and c) delta-delta?c) delta-delta?

Solution:Solution:For both delta and wye connections, the line currents can be obtained as:

.

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Example 1.5:Example 1.5: Voltage Regulation at Full Voltage Regulation at Full Load.Load.

A 7200V/208V, 50kVA, three-phase distribution A 7200V/208V, 50kVA, three-phase distribution transformer is connected delta-wye. The transformer transformer is connected delta-wye. The transformer has 1.2% resistance and 5% reactance. has 1.2% resistance and 5% reactance. Find the Find the voltage regulationvoltage regulation at full load, 0.8 power at full load, 0.8 power factor lagging.factor lagging.Solution:Solution:

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Voltage Regulation.

.

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Example 1.6:Example 1.6: Transformer Efficiency. Transformer Efficiency.If the core loss of the transformer in Example 7 is If the core loss of the transformer in Example 7 is 1kW, find the 1kW, find the efficiencyefficiency of this transformer at full of this transformer at full load and 0.8 power factor.load and 0.8 power factor.

Solution:Solution:

.

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Assignment #1Assignment #1Due :Due :

1. A transformer has the following parameters;1. A transformer has the following parameters;

NN11= 1000, N= 1000, N22 = 10, I = 10, I11=200A, V=200A, V11 = 100kV = 100kV

a) Determine Ia) Determine I22 and V and V22 b) Which type of transformer is this?b) Which type of transformer is this?

2. From Example 4, Draw and label the current and voltage 2. From Example 4, Draw and label the current and voltage of the Y-Y, Y-Delta and Delta-Delta connection of the Y-Y, Y-Delta and Delta-Delta connection transformer.transformer.

3. The parameter of the equivalent circuit of a 150-kVA, 2400V/240V transformer, shown in Figure 1.7, are R1=0.2 , R2=2 m, X1=4.5 m, Rc=10 K, and Xm=1.55 K. Using the circuit referred to the primary, determine the;

(a) Voltage Regulation, (b) Efficiency of the transformer operating at rated load 0.8 lagging power factor.

Note: Assume the transformer is ideal for all cases

1 chapter 1 emt 113: v-2008 school of computer and communication engineering, unimap prepared by:...

Documents

construction of transformer

real transformer

simple transformer

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