chapter 1.2_single phase transformer_nj

EMPS_BEE2133_NJ

CHAPTER 1: TRANSFORMER

1.2 SINGLE PHASE TRANSFORMER

EMPS_BEE2133_NJ

Review from 1.1 (Review from 1.1 (review of review of electromagnetism)electromagnetism)

From previous sub-topic, we have learned: An electromagnet is an object that acts like a magnet, but its

magnetic force is created and controlled by electricity--thus the name electromagnet

A wire with DC electric current flowing through it has a magnetic field around it

By wrapping the wire around a piece of iron, the magnetic field is increased many times

The greater the current through the wire (or the higher the voltage) the greater the strength of the electromagnet

The greater the number of turns around the iron core the greater the strength of the electromagnet.

The direction of the magnetic field is determined by the direction of the current and the direction of the turns around the iron core (right-hand rule)

EMPS_BEE2133_NJ

Learning OutcomesLearning Outcomes

At the end of the lecture, student should be able to: Understand the principle and the nature of principle and the nature of

static machines of transformerstatic machines of transformer

Perform an analysis on transformers which their principles are basic to the understanding of electrical machineselectrical machines

EMPS_BEE2133_NJ

Chapter Outline

Introduction Transformer Construction and Equations Equivalent circuit Losses, Voltage Regulation and Efficiency Open Circuit test and Short Circuit test

(measurement on transformer)

EMPS_BEE2133_NJ

1.2.1 Introduction1.2.1 Introduction A transformer is a static machinesstatic machines. The word ‘transformer’ comes form the word ‘transform’.The word ‘transformer’ comes form the word ‘transform’. Transformer is not an energy conversion devicenot an energy conversion device But is a device that changes AC electrical power at one voltage device that changes AC electrical power at one voltage

level into AC electrical power at another voltage level through level into AC electrical power at another voltage level through the action of magnetic field, without a change in frequency. the action of magnetic field, without a change in frequency.

Can raise or lower the voltage/current in ac circuitCan raise or lower the voltage/current in ac circuit

Generation Generation StationStation

TX1 TX1

DistributionsDistributions

TX1

TX1

Transmission SystemTransmission System

33/13.5kV33/13.5kV 13.5/6.6kV13.5/6.6kV

6.6kV/415V6.6kV/415V

ConsumerConsumer

EMPS_BEE2133_NJ

1.2.2 Transformer Construction There are 3 basic parts of transformer:

A primary coil/winding

receives energy from the ac source A secondary coil/winding

receives energy from primary winding & delivers it to the load

A core that supports the coils/windings.

provide a path for magnetic lines of flux

EMPS_BEE2133_NJ

The operation of transformer is based on the principal of mutual inductance

A transformer usually consists of two coils of wire wound on the same core

The primary coil is the input coil while the secondary coil is the output coil

A changing in the primary circuit creates a changing magnetic field

This changing magnetic field induces a changing voltage in the secondary circuit

This effect is called mutual induction

1.2.2 Transformer Construction

EMPS_BEE2133_NJ

1.2.2 Transformer Construction Transformer can be either step-up or step-down transformer If the output voltage of transformer is greater than the input

voltage step-up transformer If the output voltage of a transformer is less than the input

voltage step-down transformer By selecting appropriate numbers of turns, a transformer

allows an alternating voltage to be stepped up – by making Ns more than Np

Or stepped down by making Ns less than Np

Vs Ns

Vp Np

EMPS_BEE2133_NJ

Example 1 There are 400 turns of wire in an iron-core coil.

If this coil is to be used as the primary of a transformer, how many turns must be wound on the coil to form the secondary winding of the transformer to have a secondary voltage of one volt if the primary voltage is 5 volts?

EMPS_BEE2133_NJ

Example 1 (solution)

EMPS_BEE2133_NJ

1.2.2 Transformer ConstructionCore characteristic:

The composition of a transformer core depends on:

voltage, current, frequency, size limitations and construction costs

Commonly used core materials are air, soft iron, and steel

Air-core transformers are used when the voltage source has a high frequency (above 20 kHz)

Iron-core transformers are usually used when the source frequency is low (below 20 kHz)

A soft-iron-core transformer is very useful where the transformer must be physically small, yet efficient The iron-core transformer provides better power transfer than does the air-core transformer

EMPS_BEE2133_NJ

1.2.2 Transformer Construction A transformer whose core is constructed of laminated sheets of steel

dissipates heat readily; thus it provides for the efficient transfer of power. The purpose of the laminations is to reduce certain losses which will be

discussed later in this part

Hollow-core construction

EMPS_BEE2133_NJ

1.2.2 Transformer Construction The most efficient transformer core is one that offers the best path for the most lines of flux with the least loss in magnetic and electrical energy There are two main shapes of cores used in laminated-steel-core transformers:

Core-type transformers Shell-core transformers

EMPS_BEE2133_NJ

1.2.2 Transformer ConstructionCore - type construction:

so named because the core is shaped with a hollow square through the center

the core is made up of many laminations of steel

EMPS_BEE2133_NJ

1.2.2 Transformer ConstructionShell-core transformers:

The most popular and efficient transformer core

Each layer of the core consists of E- and I-shaped sections of metal

These sections are butted together to form the laminations

The laminations are insulated from each other and then pressed together to form the core.

EMPS_BEE2133_NJ

1.2.2 Transformer ConstructionTypical schematic symbols for transformers:

The bars between the coils are used to indicate an iron core

Frequently, additional connections are made to the transformer windings at points other than the ends of the windings

These additional connections are called TAPS

When a tap is connected to the center of the winding, it is called a CENTER TAP

EMPS_BEE2133_NJ

1.2.2 Transformer Construction An ideal transformer is a transformer which has no loseswhich has no loses, i.e. it’s

winding has no ohmic resistance, no magnetic leakage, and therefore no I2R and core loses.

However, it is impossibleimpossible to realize such a transformer in practice.

Yet, the approximate characteristic of ideal transformer will be approximate characteristic of ideal transformer will be used in characterized the practical transformer.used in characterized the practical transformer.

VV11 VV22

NN1 1 : N: N22

EE11 EE22

II11 II22

VV11 – Primary Voltage – Primary Voltage

VV22 – Secondary Voltage – Secondary Voltage

EE11 – Primary induced Voltage – Primary induced Voltage

EE22 – secondary induced Voltage – secondary induced Voltage

NN11:N:N22 – Transformer ratio – Transformer ratio

EMPS_BEE2133_NJ

1.2.2 Transformer ConstructionNo-load condition:

is said to exist when a voltage is applied to the primary, but no load is connected to the secondary

Because of the open switch, there is no current flowing in the

secondary winding. With the switch open and an ac voltage applied to the primary,

there is, however, a very small amount of current called EXCITING CURRENT flowing in the primary

EMPS_BEE2133_NJ

1.2.2 Transformer ConstructionWith-load condition:

When a load device is connected across the secondary winding of a transformer, current flows through the secondary and the load

The magnetic field produced by the current in the secondary interacts with the magnetic field produced by the current in the primary

This interaction results from the mutual inductance between the primary and secondary windings.

EMPS_BEE2133_NJ

1.2.2 Transformer Equation Faraday’s Law states that,

If the flux passes through a coil of wire, a voltage will be induced in the turns of wire. This voltage is directly proportional to the rate of change in the flux with respect of time.

If we have NN turns of wire,

dt

tdEmfV indind

)(

dt

tdNEmfV indind

)(

Lenz’s Law

EMPS_BEE2133_NJ

1.2.2 Transformer Equation

For an ac sources, Let V(t) = Vm sint

i(t) = im sint

Since the flux is a sinusoidal function; Then:

Therefore:

Thus:

tt m sin)(

tNdt

tdNEmfV

m

mindind

cos

sin

maxmindind fNNEmfV 2(max)

maxmm

rmsind fNfNN

Emf 44.42

2

2)(

m mB x A

EMPS_BEE2133_NJ

1.2.2 Transformer Equation For an ideal transformer,

In the equilibrium condition, both the input power will be equaled to the output power, and this condition is said to ideal condition of a transformer.

From the ideal transformer circuit, note that,

Hence, substitute in (i)

………………… (i)

2211 VEandVE

max

max

fNE

fNE

22

11

44.4

44.4

1 2

2 1

V I

V I

Input power = output power

1 1 2 2V I V I

EMPS_BEE2133_NJ

1.2.2 Transformer Equation

aI

I

N

N

E

ETherefore

1

2

2

1

2

1,

‘aa’ = Voltage Transformation RatioVoltage Transformation Ratio; which will determine whether the transformer is going to be step-up or step-down

EE11 > E > E22For a >1For a >1

For a <1For a <1 EE11 < E < E22

Step-up

Step-down

EMPS_BEE2133_NJ

1.2.2 Transformer Equation Transformer rating is normally writtenwritten in terms of

Apparent PowerApparent Power. Apparent power is actually the product of its rated its rated

current and rated voltagecurrent and rated voltage.

2211 IVIVVA

Where, I1 and I2 = rated current on primary and secondary winding.

V1 and V2 = rated voltage on primary and secondary winding.

** ** Rated currents are actually the full load currents in Rated currents are actually the full load currents in transformertransformer

EMPS_BEE2133_NJ

Example 1 1.5kVA single phase transformer has rated voltage

of 144/240 V. Finds its full load current.

Solution:Solution:

AI

AI

FL

FL

6240

1500

42.10144

1500

2

1

EMPS_BEE2133_NJ

Example 2 A single phase transformer has 400 primary and

1000 secondary turns. The net cross-sectional area of the core is 60m2. If the primary winding is connected to a 50Hz supply at 520V, calculate:

The induced voltage in the secondary winding The peak value of flux density in the core

EMPS_BEE2133_NJ

N1=400 V1=520V A=60m2 N2=1000 V2=?

a) We know that,

b) Emf,

2

520

1000

400

V

2

1

2

1

V

V

N

Na VV 13002


2

21

/0976.0

)60)()(400)(50(44.4520

44.4

1300,520,

44.4

44.4

mmWbB

B

ABfNE

VEVEknown

ABfN

fNE

m

m

m

m

m

EMPS_BEE2133_NJ

Example 3

A 25kVA transformer has 500 turns on the primary and 50 turns on the secondary winding. The primary is connected to 3000V, 50Hz supply. Find:

a) Full load primary current

b) The induced voltage in the secondary winding

c) The maximum flux in the core

EMPS_BEE2133_NJ


VA = 25kVA N1=500 V1=3000V N2=50 V2=?

a) We know that,

b) Induced voltage,

c) Max flux,

AV

VAI

IVVA

FL 33.83000

1025 3

11

VI

IEE

AI

I

I

N

Na

3003.83

33.83000

3.8350

33.8500

2

112

2

1

2

2

1

mWb

fNE

27

)50)(50(44.4300

44.4

EMPS_BEE2133_NJ

Equivalent circuit:

Often given to explain the operation of a complicated or unfamiliar device

If a circuit is truly an equivalent circuit, the original device can be removed from a system & replaced with its equivalent circuit without changing the behavior or performance of the system

For purposes of analysis the transformer may be represented by a 1:1 turns ratio equivalent circuit

This circuit is based on the following assumptions:

• Primary and secondary turns are equal in number. One winding is chosen as the reference winding; the other is the referred winding.

1.2.3 Equivalent Circuit of Practical Transformer

EMPS_BEE2133_NJ

• Core loss may be represented by a resistance across the terminals of the reference winding.

• Core flux reactance may be represented by a reactance across the terminals of the reference winding.

• Primary and secondary IR and IX voltage drops may be lumped together; the voltage drops in the referred winding are multiplied by a factor derived at the end of this section, to give them the

correct equivalent value. • Equivalent reactance and resistances are linear.



V1 = primary supply voltage

V2 = 2nd terminal (load) voltage

E1 = primary winding voltage

E2 = 2nd winding voltage

I1 = primary supply current

I2 = 2nd winding current

I1’ = primary winding current

Io = no load current

Ic = core current

Im = magnetism current

R1= primary winding resistance

R2= 2nd winding resistance

X1= primary winding leakage reactance

X2= 2nd winding leakage reactance

Rc= core resistance

Xm= magnetism reactance

VV11

II11 RR11XX11

RRCC

IIcc

XXmm

IImm

IIoo

EE11 EE22

VV22

II11’’

NN11: N: N22RR22

XX22

LoadLoad

II22

1.2.3 Equivalent Circuit of Practical TransformerSingle Phase transformer (referred to Primary)Single Phase transformer (referred to Primary) Actual MethodActual Method

22

22

2

2

12 '' RaRORR

N

NR

22

22

2

2

12 '' XaXORX

N

NX

a

II

aVVORVN

NVE

22

2222

1'21

'

'

VV11

II11 RR11 XX11

RRCC

IIcc

XXmm

IImm

IIoo

EE11 EE22 VV22

II22’’ NN11: N: N22

RR22’’ XX22

’’

Load

II22

1.2.3 Equivalent Circuit of Practical TransformerSingle Phase transformer (referred to Primary)Single Phase transformer (referred to Primary) Approximate MethodApproximate Method

VV11

II11 RR11XX11

RRCC

IIcc

XXmm

IImm

IIoo

EE11 EE22 VV22

II22’’ NN11: N: N22RR22

’’ XX22’’

Load

II22

22

22

2

2

12 '' RaRORR

N

NR

22

22

2

2

12 '' XaXORX

N

NX

a

II

aVVORVN

NVE

22

2222

1'21

'

'

1.2.3 Equivalent Circuit of Practical TransformerSingle Phase transformer (referred to Primary)Single Phase transformer (referred to Primary) Approximate MethodApproximate Method

22

22

2

2

12 '' RaRORR

N

NR

22

22

2

2

12 '' XaXORX

N

NX

'

'

2101

2101

XXX

RRR

2222

1'2 ' aVVORV

N

NV

In some application, the excitation branch has a small current compared to load current, thus it may be neglected without causing serious error.

VV11

II11RR0101 XX0101

aVaV22

Single Phase transformer (referred to Secondary)Single Phase transformer (referred to Secondary) Actual MethodActual Method


21

11

2

1

21 ''

a

RRORR

N

NR

a

VVORV

N

NV 1

111

21 ''

21

11

2

1

21 ''

a

XXORX

N

NX

I1’ R1’X1’

RC’

Ic

Xm’

Im

Io

I2 R2

X2

VV22

a

V1

EMPS_BEE2133_NJ

1.2.3 Equivalent Circuit of Practical Transformer)Single Phase transformer (referred to Secondary)Single Phase transformer (referred to Secondary) Approximate MethodApproximate Method

21

11

2

1

21 ''

a

XXORX

N

NX

2102

2102

'

'

XXX

RRR

II11’’ RR0202 XX0202

a

V1

21

11

2

1

21 ''

a

RRORR

N

NR

a

VVORV

N

NV 1

111

21 ''

11 ' aII

Neglect the excitation branch,

VV22

EMPS_BEE2133_NJ

For the parameters obtained from the test of 20kVA 2600/245 V single phase transformer, refer all the parameters to the high voltage side if all the parameters are obtained at lower voltage side.

Rc = 3.3, Xm =j1.5, R2 = 7.5, X2 = j12.4

Example 4

EMPS_BEE2133_NJ

Given : Rc = 3.3, Xm =j1.5, R2 = 7.5, X2 = j12.4

i) Refer to H.V side (primary)

R2’=(10.61)2 (7.5) = 844.65,

X2’=j(10.61)2 (12.4) = j1.396k

Rc’=(10.61)2 (3.3) = 371.6,

Xm’=j(10.61)2 (1.5) = j168.9


EMPS_BEE2133_NJ

1.2.4 Power Factor

Power factor = angle between current and voltage angle between current and voltage = cos

V

I

= -ve

Lagging

V

I

= +ve

Leading

VI

= 1

unity

Power factor always lagging for real transformer

Power factor can never be greater than unity (θ=1)

1 01 01 1 2( )( )V R jX I aV

EMPS_BEE2133_NJ

Example 5

A 10 kVA single phase transformer 2000/440V has primary resistance and reactance of 5.5 and 12 respectively, while the resistance and reactance of secondary winding is 0.2 and 0.45 respectively. Calculate:

i) The parameter referred to high voltage side and draw the equivalent circuit

ii) The approximate value of secondary voltage at full load of 0.8 lagging power factor, when primary supply is 2000V.

EMPS_BEE2133_NJ

Example 5 (solution)R1=5.5 X1=j12 R2=0.2 X2=j0.45

i) Refer to H.V side (primary)

R2’=(4.55)2 (0.2) = 4.14,

X2’=j(4.55)2(0.45) = j9.32

Therefore,

R01=R1+R2’=5.5 + 4.13 = 9.64 X01=X1+X2’=j12 + j9. 32 = j21.32

55.4440

2000

2

1

2

1 V

V

E

Ea

V1 aV2

R01 X01

21.329.64

I1

EMPS_BEE2133_NJ


ii) Secondary voltage

p.f = 0.8

cos = 0.8

=36.87o

Full load,

From cct eqn.,

AV

VAIFL 5

2000

1010 3

1

o

oo

oo

V

Vj

aVIjXRV

8.06.422

)55.4()87.365)(32.2164.9(02000

))((0

2

2

2101011

1.2.5 Transformer Losses An ideal transformer would have no energy losses, and would

be 100% efficient

In practical, transformer energy is dissipated in the windings, core, and surrounding structures

Iron LossesIron Losses

- occur in core parameters- occur in core parameters

Copper LossesCopper Losses

- occur in winding resistance- occur in winding resistance

circuitopenccciron PRIPP 2)(

022

2012

1

22

212

1

)()(,

)()(

RIRIPreferredifor

PRIRIPP

cu

circuitshortcucopper

**Poc and Psc will be discusses later in transformer test

EMPS_BEE2133_NJ

1.2.5 Transformer Efficiency To check the performance of the device, by comparing

the output with respect to the input The higher the efficiency, the better the system

%100cos

cos

%100

%100,

22

22

cuc

lossesout

out

PPIV

IV

PP

P

PowerInput

PowerOutputEfficiency

%100cos

cos

%100cos

cos

2)(

)(

cucnload

cucloadfull

PnPnVA

nVA

PPVA

VA

Where, if ½ load, hence n = ½ ,½ load, hence n = ½ , ¼ load, n= ¼ ,¼ load, n= ¼ , 90% of full load, n =0.990% of full load, n =0.9Where Pcu = Psc

Pc = Poc

EMPS_BEE2133_NJ

1.2.5 Voltage Regulation

The voltage regulation of the transformer is the percentage change in the output voltage from no-load to full-load

Voltage Regulation can be determined based on 3 methods:

Basic Definition

Short – circuit Test

Equivalent Circuit

EMPS_BEE2133_NJ

1.2.5 Voltage Regulation

Basic Definition In this method, all parameters are being referred either

to primary or secondary side. It can be represented in either

Down – voltage Regulation

Up – Voltage Regulation

%100.

NL

FLNL

V

VVRV

%100.

FL

FLNL

V

VVRV

1.2.5 Voltage Regulation Short-circuit test

In this method, direct formula can be used.

%100

cos.

1

. V

VRV fpscsc

If referred to primary side

%100

cos.

2

. V

VRV fpscsc

If referred to secondary side

Note that:Note that:

‘–’ is for Lagging power factor‘+’ is for Leading power factor Isc must equal to IFL

Psc = Vsc Isc cos θsc

EMPS_BEE2133_NJ

1.2.5 Voltage Regulation (Equivalent Circuit )Equivalent circuit

In this method, the parameters must be referred to primary or secondary

%100

sincos.

1

.01.011

V

XRIRV fpfp If referred to

primary side

If referred to secondary side

Note that:Note that:

‘+’ is for Lagging power factor‘–’ is for Leading power factor j terms ~0

%100

sincos.

2

.02.022

V

XRIRV fpfp

EMPS_BEE2133_NJ

Example 6

Determine the Voltage regulation by using down – voltage regulation and equivalent circuit in example 1.5.

EMPS_BEE2133_NJ


By using down – voltage regulation,

We know that, V2FL=422.6V , V2NL=440V

Therefore,

%95.3

%100440

6.422440

%100.

NL

FLNL

V

VVRV

EMPS_BEE2133_NJ

By using equivalent circuit,

I1=5A R01=9.64 X01 = 21.32 V1=2000V, 0.8 lagging p.f

%12.5

%1002000

)6.0(32.21)8.0(64.95

%100sincos

.1

.01.011

V

XRIRV fpfp


EMPS_BEE2133_NJ

Example 7

A short circuit test was performed at the secondary side of 10kVA, 240/100V transformer. Determine the voltage regulation at 0.8 lagging power factor if :

Vsc =18V

Isc =100

Psc=240W

EMPS_BEE2133_NJ


,

100100

10000

2

2

scFL

FL

II

AV

VAI

Check,

Hence, we can use short-circuit method,

%100

cos.

2

. V

VRV fpscsc

EMPS_BEE2133_NJ

o

scsc

scsc

scscscsc

ofp

fpscsc

IV

P

IVP

thatKnow

Hence

fpGiven

V

VRV

34.82)100)(18(

240cos

cos

cos

,

87.368.0cos,

8.0.

%100cos

.

1

1

1.

2

.

%62.12

%100100

87.3634.82cos18.

oo

RV


EMPS_BEE2133_NJ

Example 8

The following data were obtained in test on 20kVA 2400/240V, 60Hz transformer:

Vsc =72V Isc =8.33APsc=268W Poc=170W

The measuring instrument are connected in the primary side for short circuit test. Determine the voltage regulation for 0.8 lagging p.f. (use all 3 methods), full load efficiency and half load efficiency.

EMPS_BEE2133_NJ


.72.786.34.6364.8

64.833.8

72

4.63)33.8)(72(

268cos

cos

cos

,

87.368.0cos,

8.0.

%100cos

.

0101

1

1

1.

2

.

sideprimarytoconnectedbecausejXRjZ

I

VZ

IV

P

IVP

thatKnow

Hence

fpGiven

V

VRV

osc

sc

scsc

o

scsc

scsc

scscscsc

ofp

fpscsc

EMPS_BEE2133_NJ

%68.2%100

2400

)6.0(72.7)8.0(86.32400

20000

%100sincos

.,.2

%68.2%1002400

87.364.63cos72.

%100cos

.,.1

1

.01.011

1

.

V

XRIRVcircuitEquivalent

RV

V

VRVmethodCircuitShort

fpfp

oo

fpscsc


EMPS_BEE2133_NJ

%68.2

%100240

58.233240

%100.

79.058.233

240

24004.6364.887.36

2400

2000002400

,.3

2

2

20111

NL

FLNL

o

ooo

V

VVRV

VV

V

aVZIV

DefinationBasic


EMPS_BEE2133_NJ

%12.97%100)268()5.0(170)8.0)(20000)(5.0(

)8.0)(20000)(5.0(

%34.97%100)268()1(170)8.0)(20000)(1(

)8.0)(20000)(1(

2)(

2)(

loadhalf

loadfull


EMPS_BEE2133_NJ

1.2.4 Measurement on Transformer There are two test conducted on transformer.

Open Circuit TestOpen Circuit Test Short Circuit testShort Circuit test

The test is conducted to determine the parameter of to determine the parameter of the transformerthe transformer.

Open circuitOpen circuit test test is conducted to determine magnetism parameter, Rc and XmRc and Xm.

Short circuitShort circuit test test is conducted to determine the copper parameter depending where the test is performed. If performed at primary, hence the parameters are RR0101

and XX0101 and vice-versavice-versa.

Normally, measurement at higher voltage sidehigher voltage side

From a given test parameters,

m

ocm

c

occ

mc

ococm

ococc

ococ

ococ

ococococ

I

VX

I

VR

XandRThen

II

II

Hence

IV

P

IVP

,

,,

sin

cos

,

cos

cos

1

Rc Xm

Voc

Ic Im

Voc

Ioccosoc

Ioc

Voc

Ic

Im

Iocsinoc

oc

Note:If the question asked parameters referred to low question asked parameters referred to low voltage sidevoltage side, the parameters (Rc and Xm) obtained

need to be referred to low voltage sideneed to be referred to low voltage side

1.2.5 Open-Circuit Test

EMPS_BEE2133_NJ

1.2.5 Short-Circuit Test Normally, measurement at lower voltage sideat lower voltage side

If the given test parameters are taken on primary taken on primary side, Rside, R0101 and X and X0101 will be obtained & will be obtained & vvice-versa.

X01R01

For a case referred to Primary side 010101

01

1

,

cos

cos

jXRZ

I

VZ

Hence

IV

P

IVP

scsc

sc

scsc

scsc

scscscsc

EMPS_BEE2133_NJ

Example 9 Given the test on 500kVA 2300/208V are as follows:

Poc = 3800W Psc = 6200WVoc = 208V Vsc = 95VIoc = 52.5A Isc = 217.4A

Determine the transformer parameters and draw equivalent circuit referred to high voltage side. Also calculate appropriate value of V2 at full load, the full load efficiency, half load efficiency and voltage regulation, when power factor is 0.866 lagging.

[1392, 517.2, 0.13, 0.44, 202V, 97.74%, 97.59%, 3.04%]

EMPS_BEE2133_NJA

II

A

II

IVP

o

ococm

o

ococc

ooc

ococococ

2.49

6.69sin5.52

sin

26.18

6.69cos5.52

cos

6.69)208)(5.52(

3800cos

cos

1

Ioccosoc

Ioc

Voc

Ic

Im

Iocsinoc

oc

From Open Circuit Test,


EMPS_BEE2133_NJ

23.421.49

208

39.1126.18

208

m

ocm

c

occ

I

VX

I

VR

Since Voc=208Vall reading are taken on the secondary side

Parameters referred to high voltage side,

21.517208

230023.4'

1392208

230039.11'

22

2

1

22

2

1

E

EXX

E

ERR

mm

cc


EMPS_BEE2133_NJ

AV

VAIFL 4.217

2300

10500 3

11

osc

scscscsc IVP

53.72)4.217)(95(

6200cos

cos

1

From Short Circuit Test,

First, check the First, check the IIscsc

Since IFL1 =Isc , all reading are actually taken on the primary side

42.013.0

53.7244.053.724.217

95

01

j

I

VZ

oo

scsc

sc


EMPS_BEE2133_NJ

Equivalent circuit referred to high voltage side,

VV22’=aV’=aV22VV11

RRcc

13921392 XXmm

517.21517.21

RR0101

0.130.13 XX0101

0.420.42


EMPS_BEE2133_NJ

Example 9 (solution)Efficiency,

%59.97

%1003800)5.0)(6200()866.0)(10500)(5.0(

)866.0)(10500)(5.0(

%100cos

cos

%74.97

%10038006200)866.0)(10500(

)866.0)(10500(

%100cos

cos

23

3

22

1

3

3

ocscL

ocscFL

PPnnVA

nVA

PPVA

VA

EMPS_BEE2133_NJ


Voltage Regulation,

%04.3

%1002300

3053.72cos)95(

%100cos

.1

E

VRV pfscsc

EMPS_BEE2133_NJ

Summary

Transformers convert AC electricity from one voltage to another with little loss of power

Transformers work only with AC and this is one of the reasons why mains electricity is AC

Step-up transformers increase voltage, step-down transformers reduce voltage

Most power supplies use a step-down transformer to reduce the dangerously high mains voltage (230V in UK) to a safer low voltage.

The input coil is called the primary and the output coil is called the secondary

There is no electrical connection between the two coils, instead they are linked by an alternating magnetic field created in the soft-iron core of the transformer

The two lines in the middle of the circuit symbol represent the core.

EMPS_BEE2133_NJ

Summary

The two lines in the middle of the circuit symbol represent the core

Transformers waste very little power so the power out is (almost) equal to the power in

Note that as voltage is stepped down current is stepped up. The ratio of the number of turns on each coil, called the turns

ratio, determines the ratio of the voltages A step-down transformer has a large number of turns on its

primary (input) coil which is connected to the high voltage mains supply, and a small number of turns on its secondary (output) coil to give a low output voltage.

EMPS_BEE2133_NJ

http://www.sayedsaad.com/fundmental/index_transformer.htm

THANK YOU FOR YOUR ATTENTION!!

chapter 1.2_single phase transformer_nj

Documents