1

LECTURE NOTES FOR

EE 162

ELECTROMECHANICAL ENERGY

CONVERSION & TRANSFORMERS

COURSE LECTURER: DR. PHILIP YAW OKYERE

2

Course Outline

EE 162 ELECTRICAL MACHINES I (3 0 3)

Principles of Electromechanical Energy Conversion:

Force and torque as rate of change of energy with position. Basic Transducers: Single

Excitation, Alignment Force and Torque; Double Excitation, Alignment and Interaction

Forces and torque

Transformers:

Construction; Basic theory; Phasor Diagram; Equivalent Circuits; No-load and Short-

Circuit Tests; Voltage Regulation; Efficiency; Cooling methods; Polarity; Polyphase

transformer Connections; Per-Unit Calculation; Parallel Operation of Transformers;

Auto transformers; Tap-Changing transformers; Instrument Transformers.

3

CHAPTER ONE Transformers

1. Introduction

The transformer transfers electrical energy from one circuit to another via the medium of

a pulsating magnetic field that links both circuits. The widespread development of ac

power systems is principally due to the transformer. It enables us to produce and transmit

power at economical voltages and to distribute it safely in factories and homes. In low-

power low-current electronic and control circuits, it is used to provide impedance

matching between a source and its load for maximum power transfer, to isolate one

circuit from another, to isolate direct current while maintaining ac continuity between

two circuits and to provide reduced ac voltages and currents for protection, metering,

instrumentation and control.

2. Principle of operation of a transformer

The transformer is a straight-forward application of Faradays Law of Electromagnetic Induction. Consider the general arrangement of a single-phase transformer shown in Fig.

1. An alternating voltage applied to coil 1, causes an alternating current to flow in the coil

and this current produces an alternating flux in the iron core. A portion of the total flux

links the second coil. The alternating flux induces a voltage in the second coil. If a load

should be connected to the coil, this voltage would drive a current through it. Energy

would then be transferred through the medium of magnetic field from coil 1 to coil 2. The

combination of the two coils is called a transformer. The coil connected to the source is

called the primary winding (or the primary) and the one connected to the load is called

the secondary winding (or the secondary).

V1

I1

E1 N1 N2 E2

I2

Load

Coil 2Coil 1

mSoft iron

Fig. 1 An elementary transformer

3. Polarity and terminal markings of a transformer Voltage E1 is induced in coil 1 and voltage E2 in coil 2. These voltages are in phase.

Suppose at any given instant when the primary terminal 1 is positive with respect to

primary terminal 2, the secondary terminal 3 is also positive with respect to secondary

terminal 4 (See Fig. 2). Then terminals 1 and 3 are said to have the same polarity. To

4

indicate that their polarities are the same, a dot is placed beside primary terminal 1 and

secondary terminal 3. Alternatively, letters of the same suffix, A1 (for the high-voltage

winding) and a1 (for the low-voltage winding) say can be used. Current I1 entering coil 1

through the dotted terminal 1 and current I2 entering coil 2 through the dotted terminal 3

create fluxes in the same direction.

1

2

3

4

A1

A2

a1

a2

1

2

3

4

A1

A2

a1

a2

(a) Magnetic Circuit (b) Electric Circuit

Fig. 2 Polarity and terminal markings

4.0 Ideal transformer

An ideal transformer has no losses, no leakage flux and its core is infinitely permeable.

An ideal transformer is shown in Fig. 3. The mutual flux m is confined to the iron.

V1

I1

E1 N1 N2 E2

I2

m

V2

S

(I1 = 0 on no load)

Fig. 3 An ideal transformer

4.1 Properties of an ideal transformer

(a) Emf equation and voltage ratio (see Fig. 3): With the primary connected to an ac

source V1, an alternating flux m is produced in the core. Let the flux be expressed as

sinmax weberstm (1)

The induced emf e1 as indicated in the figure is given by

tNtdt

dNN

dt

de m cossin max1max111

Or

90sinmax11 tNe (2)

Hence

5

max1max1

12

2

2

fNNE

Or

max11 44.4 fNE (3)

Similarly

max2

max2max2

2 44.42

2

2

fN

fNNE

(4)

From (3) and (4), we obtain

2

1

2

1

N

N

E

E (5)

The ratio a = N1/N2 is called the turns ratio. A step-up transformer has a < 1 and a step-

down transformer has a > 1. In an ideal transformer, the applied voltage V1 and the

induced voltage E1 must be identical. Hence we may write

11 EV (6.a)

And

max11 44.4 fNV (6.b)

Equation (6.b) indicates that for a given frequency, number of turns and voltage, the peak

flux max must remain constant.

(b) Current ratio and power equation: On no load I1 = 0. Now if a load is connected

across the secondary terminals (i.e. switch S is closed) current I2 flows through the load.

This current produces mmf N2I2 which if it acted alone would by Lenzs law, cause the mutual flux to reduce. Since when V1 is fixed the flux max is also fixed, the primary develops mmf N1I1 which is such that

2211 ININ (7.a)

Or

aII 21 (7.b)

In ideal transformer the secondary voltage

22 EV (8)

remains constant since E2 is fixed when the max is fixed. It can be deduced from above equations that for an ideal transformer

2211 IVIV (9)

That is there are no reactive and active losses in an ideal transformer.

6

(c) Phasor diagram of an ideal transformer (Fig. 4)

E1

E2

m

E1, V1

E2, V2 I1

I2

m

(a) No load (b) Load is resistive inductive

Fig. 4 Phasor diagram of an ideal transformer

Example 1

An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is

connected to a 200-V, 50-Hz source. The load across the secondary draws a current of 2

A at a power factor of 0.80 lagging. Calculate (a) the rms value of the primary current (b)

the flux linked by the secondary winding (c) Draw the phasor diagram.

Solution

AIIININa 502225090)( 112211

WbfN

Vb 01.0

905044.4

200

44.4)(

1

1max

VEN

NEc 5000)( 1

1

22

36.9(0.8)cos is and between angle phase The -122 IV

V2=E2 = 5000 V

V1=E1= 200 V I2 = 2 A

I1 = 50 A

36.9

m

Example 2

A 200-kVA, 6600-V / 400-V, 50-Hz 1-ph transformer has 80 turns on the secondary.

Calculate (a) the approximate values of the primary and secondary currents (b) the

approximate number of primary turns and (c) the maximum value of the flux.

7

Solution

Aa 3.306600)1000200(current primary load-Full)(

A500400)1000200(current secondary load-full and

1320400)660080()( 1 Nb

)]4([0225.0508044.4

400)( max fromWbc

(d) Impedance ratio (see Fig. 5): The impedance seen by the source

ZaIEaaIaEIEIVZe2

22

2

221111

2ZaZe (10)

V1 E1 E2 Z

I2I1

a : 1

Fig. 5 impedance ratio

(e)Equivalent circuits of an ideal transformer: From (10), we can represent the

transformer in Fig. 5 by equivalent circuit shown in Fig. 6.a. We may also write

aVaEEZI 1122

Or

21 ZI

a

V (11)

and then represent the transformer by an equivalent circuit shown in Fig. 6.b

V1 a

2Z

I1 I2

a

V1Z

(a) Circuit referred to the primary side (b) circuit referred to the secondary side

Fig. 6 Equivalent circuits of ideal transformer

8

Example 3 Calculate the voltage V and current I in the circuit of Fig. 7, knowing that the ideal

transformer has a primary to secondary turns of 1:100 (i.e. a = 1/100).

I5

V1=10V 40k

20k

V

Fig. 7 See Ex. 3

Solution:

We shall shift all impedances to the primary side

10V

5 2

4 aVV

Va

VVvoltageactualTheVIRV

AZ

VI

XXRZ

e

CLe

8008100.842

25

10

534

1

2222

5.0 Practical single-phase transformer

The windings of a practical transformer have both resistance and leakage inductance. The

core is also imperfect: it has a core loss and finite permeability. The core loss has two

components: hysteresis loss and eddy current loss.

Hysteresis loss

When ferromagnetic material is subjected to alternating magnetization, the energy put

into the magnetic field when the flux is increasing is not completely given back when the

flux dies away but a certain portion is wasted. The area of the hysteresis loop gives the

value of the energy loss taking place for each complete cycle of magnetization. The loop

area is found experimentally to vary as xBmax up to moderate values of flux density (1.0

2.0 Wb/m2)

9

The index x is named after Steinmetz and is about 1.6 though it may be higher. In

practice the hysteresis loss, for simplicity, is often taken as proportional to 2maxB

If f = frequency of magnetization, the power wasted in magnetic hysteresis 3W/mHh fWP Or

32

max

6.1

max W/mfBfBPh (12)

The hysteresis constant depends upon the magnetic material.

Eddy-current loss

When a changing magnetic flux permeates any mass of metal, eddy currents are induced.

The eddy currents cause the metal to heat. Eddy currents are also induced in a core

revolving in a stationary or constant magnetic field. If iron or metal subjected to

alternating magnetization is built up of laminations insulated from each other, the eddy

current loss is reduced.

The eddy current loss in a laminated core is given by 32

max

22 W/m/)( BtkfPe (13.a)

Where

t = thickness of the laminations

= resistivity of the material k = a constant depends on the waveform of the alternating flux

For a given thickness and waveform, (13.a) reduces to 32

max

2 / mWBfkPe (13.b)

Example 4

In a transformer core of volume 0.16 m3 the total iron loss was found to be 2170 W at 50

Hz. The hysteresis loop of the core material, taken to the same maximum flux density,

had an area of 9.0 cm2 when drawn to scales of 1 cm = 0.1 Wb/m

2 and 1 cm = 250 AT/m.

Calculate the total iron loss in the transformer core if it is energized to the same

maximum flux density, but at a frequency of 60 Hz.

Solution

Hysteresis loss = area x scale factors = 9 x 0.1 x 250 = 225 J/m3

At 50 Hz, hysteresis loss = 225 x 50 x 0.16 = 1800 W

Therefore Eddy-current loss = 2170 1800 = 370 W At 60 Hz, hysteresis loss = 1800 x (60/50) = 2160 W and

Eddy-current loss = 370 x (60/50)2 = 533 W

Therefore total iron loss = 2160 + 533 = 2693 W

Example 5

For the same maximum flux density, the total core loss in a core is 500 W at 25 Hz and

1400 W at 50 Hz. Find the hysteresis and eddy-current losses for both frequencies.

10

Solution

Since Bmax is constant, the losses can have the following forms:

AfPh ,2BfPe and

2BfAfPPP ehc

For a frequency of 25 Hz, the core loss = 500 = A(25) + B(25)2

For a frequency of 50 Hz, the core loss = 1400 = A(50) + B(50)2

Solving the two equations, we obtain A = 12, B = 0.32 and the individual losses

Ph = 300 W, Pe = 200 W at 25 Hz and Ph = 600 W, Pe = 800 W at 50 Hz

5.1 Phasor diagram on no load

To furnish the power loss in the core (core loss), a small current must be drawn from the

source. This current Ip must be in phase with induced voltage E1. Also to create the

mutual flux m, a magnetizing current Im in phase with m and lagging 90 behind E1 must be drawn to produce the required mmf.

Im

IoIp

E2

V1=E1

Fig. 8 Phasor diagram for practical transformer on no load

We note that

(i) the no-load current Io taken by the primary is the phasor sum of Ip and Im

(ii) the difference between the value of the applied voltage V1 and that of the induced emf

E1 is only about 0.05% when the transformer is on no load so the two can be considered

to be equal

(iii) Ip is very small compared with Im. Therefore the no-load power factor is very low.

Example 4

A 1-ph transformer has 480 turns on the primary and 90 turns on the secondary. The

mean length of the flux path in the iron core is 1.8 m and the joints are equivalent to an

air gap of 0.1 mm. If the peak value of the flux density is to be 1.1 T when a voltage of

2200 V at 50 Hz is applied to the primary, find (a) the cross-sectional area of the core (b)

the secondary voltage on no load (c) the primary current and power factor on no load.

Assume the value of the magnetic field strength for 1.1 T in iron to be 400 A/m, the

corresponding iron loss to be 1.7 W/kg at 50 Hz and the density of the iron to be 7800

kg/m3.

11

Solution

Wba 0206.05048044.42200)( maxmax

20187.01.10206.0 core of area sectional-Cross m

Vb 5.412480

902200 load noon voltageSecondary )(

Ammfc 7208.1400 coreiron for the required ofvaluePeak)(

g

o

g

gg lB

lHAmmf

5.870001.0104

1.1 gapair for the ofvaluePeak

7

A5.8075.87720 produce torequired mmf totalofvaluePeak max

AN

mmftotalofvaluePeak682.1

480

5.807current gmagnetizin of Peak value

1

AIm 19.12

682.1 sinusoidal be it to assuming valuerms Its

30337.00187.08.1iron of Volume m kg26378000337.0density volumeiron of Mass

W4477.1263 lossiron And

AI p 203.02200447

AIo 21.1203.019.1current load No22

laggingNo 168.021.1203.0II factor power load op

5.2 Mutual and leakage fluxes in a transformer (see Fig. 9) The actual flux linking a coil can be considered to have two components: the mutual flux

m linking both coils and the leakage flux l. The reluctance of the paths of the leakage flux is almost entirely due to the long air paths and is therefore practically constant.

Consequently, the value of the leakage flux is proportional to the current in the coil.

V1

I1

Ep l1 l2 Es V2

I2

m

Fig. 9 Ttransformer possessing two leakage fluxes and a mutual flux

The secondary induced voltage Es is composed of two parts:

(a) a voltage El2 induced by leakage flux l2 given by El2 = 4.44fN2l2,max (b) a voltage E2 induced by mutual flux m given by E2 = 4.44fN2max

12

Similarly, the primary induced voltage is composed of El1 =4.44fN1l1,max and El1 = 4.44fN1max

We can segregate the four voltages E1, E2, El1 and El2 by rearranging the transformer

circuit as shown in Fig. 10. The rearrangement of the transformer circuit makes it clear

that El1 and El2 are voltage drops across reactances. These reactances called leakage

reactances are given by 222111 and IEXIEX ll

V1

I1

E1 E2

I2

m

Es

El2 Ep

El1

l1 l2

Fig. 10 Separating out the various induced voltages

5.3 Equivalent circuit of a practical transformer

The behaviour of a practical transformer may be conveniently considered by assuming it

to be equivalent to an ideal transformer and then allowing for the imperfections of the

actual transformer by means of additional circuits or impedances inserted between the

supply and the primary winding and between the secondary winding and the load. The

complete equivalent circuit of this transformer is shown in Fig. 11. R1 and R2 are

resistances of the primary and secondary windings. X1 and X2 are the leakage reactances.

The reactance Xm is such that it takes a reactive current Im (i.e. the magnetizing current)

of the actual transformer. The core loss is accounted for by the resistor Rm which takes

the component Ip of the primary current.

R1X1

Rm

Ip

Io

Im

Xm E1 E2

I21

I2X2 R2

Z

Ideal

transformer

V1 V2

Fig. 11 Complete equivalent circuit of a practical transformer

Ideal transformer equations still apply: 2212 INNI and 2121 NNEE

13

5.4 Equivalent circuit referred to the primary side (Fig. 12a)

R1X1

Rm

Ip

Io

Im

Xm E1

I21 X2

1

R21

Z1 V2

1

Fig. 12.a.Exact equivalent circuit referred to the primary side

Let 21 NNa (14.a)

Then

2

2

2

2

212 )( XaXNNX , 22

2

2

212 )( RaRNNR , ZaZNNZ22

21 )( (14.b)

22212 )( aVVNNV

, aII 22

(14.c)

It is worth noting that for a practical transformer, 1212 XXandRR

5.5 Equivalent circuit referred to the secondary side

R11 X1

1

Rm1

Ip1

Io1

Im1

Xm1

I2 X2 R2

Z V2a

1V

I11

Fig. 12.b.Exact equivalent circuit referred to the secondary side

etcaRRNNR ,)( 2112

121

(15.a)

etcaIIorININ pppp

12 (15.b)

14

5.5 Approximate equivalent circuits

The exact equivalent circuit of the transformer is too exact for most practical

applications. Consequently, we can simplify it to make calculations easier. Approximate

equivalent circuits commonly used for power transformer calculations are given in Fig.

13

V1

I1

Rm

Ip

Io

Im

Xm

I21 Xe1=X1+X2

1

Z1 V2

1

Re1=R1+R21

(a) Voltage drop in the primary leakage impedance due to exciting or

no load current Io is neglected.

V1

I1 = I21 Re1

Xe1

Z1

V21

(b) Exciting current is neglected entirely. Note that primary rated or

full load current is at least 20 times larger than Io. This circuit may

be used when I1 = 1.15xfull load current

V1

I1 = I21 Xe1

Z1

V21

(c) For transformers above 500 kVA, Xe is at least 5 times

greater than Re. This circuit can be used to calculate

voltage regulation of such transformers.

Fig.13 Approximate equivalent circuits

15

5.5 The complete phasor diagrams for loaded conditions

They are shown in Fig. 15

Fig. 15 Complete phasor diagrams for loaded conditions (drawn in two parts)

5.6 Rating of transformers To keep the transformer temperature at an acceptable level, limits are set to both the

applied voltage (this determines the iron loss at a given frequency) and the current drawn

by the load (this determines copper loss). The limits determine the rated voltage and rated

current of transformers.

The power rating of transformer Srated = rated voltage x rated current can be expressed in

VA, kVA or MVA depending on the size of the transformer. The rated kVA (i.e. the rated

power), frequency and voltage are always shown on the name plate. In large

transformers, the corresponding currents are also shown. We note that

Rated kVA = V1r x I1fl x10-3

= V2r x I2fl x10-3

= V2n xI2fl x10-3

=E2n x I2fl x10-3

Where

V1r = rated primary voltage

I1fl = rated primary current = primary full load current

V2r = rated secondary voltage = V2n (no load secondary voltage corresponding to the rated

primary voltage) = E2n (no load induced secondary voltage corresponding to primary

rated voltage)

I2fl = rated secondary current = secondary full load current

5.7 The turns ratio

It is given by

21 EEa

rVE 11 on no load because ZI 1o is very small. Since V E 22 on no load

16

rrVVa

21 (16.a)

And aII flfl 21 (16.b)

Example 7

A transformer is rated 10 kVA, 2400 / 240 V, 60 Hz. The parameters for the approximate

equivalent circuit of Fig. 14.a are Rm = 80 k, Xm = 35 k, Re1 = 8.4 and Xe1 = 13.7 . Determine the voltage to be applied to the primary to obtain the rated current in the

secondary when the secondary terminal voltage is 240 V. What is the input power factor?

The load power factor is 0.8 lagging.

Solution

AIaII flfl 17.4240010000122

The power factor angle 9.368.0cos 1 If we choose the load current I2 as the reference phasor, then 9.362402V

volts144019209.362400 ja

22 VV

The voltage across the equivalent leakage impedance is

volts5735017.47.134.8 jjj 2e1IZ

The primary voltage required

voltsjjj 4.37246214971955 573514401920 2e121 IZVV

mAjk

j

Rm7125.184375.24

80

14971955

1p

VI

mAjkj

j

jX m8571.557714.42

35

14971955

1m

VI

0558.00427.00187.00244.0017.4 jjj mp21 IIII Aj 50.0237.40371.02371.4

9.3750.04.37 and between angle Phase IV 11 lagging79.09.37cosfactor power Input

Example 8

A 1-ph transformer operates from a 230-V supply. It has an equivalent resistance of 0.1 and an equivalent leakage reactance of 0.5 referred to the primary. The secondary is connected to a coil having a resistance of 200 and a reactance of 100 . Calculate the secondary terminal voltage. The secondary winding has four times as many turns as the

primary.

17

Solution

Refer to approximate equivalent circuit of Fig. 14.b

25.65.12100200,100200,4

1 2

2

1 jjajN

Na ZZ

75.66.1225.65.125.01.0impedanceTotal jjj ZZe1

25.65.1275.66.12

0230

75.66.12

0230j

jand

j

22 VI

voltsj

j8719.224

2941.14

9754.13230

75.66.12

25.65.12230

2V

voltsa

VV 8998719.224422

5.8 Definition of per-unit impedances The leakage impedances Z1 and Z2 on the primary and secondary side are expressed in

per unit as follows:

2

1

1

1

1

1

1

11

r

rated

r

fl

base V

SZ

V

IZ

Z

ZZ (17.a)

2

2

2

2

2

2

2

22

r

rated

r

fl

base V

SZ

V

IZ

Z

ZZ (17.b)

rated

rbase

rated

rbase

S

VZand

S

VZ

22

2

21

1

where

(17.c)

The impedances are said to be expressed in per unit with reference to the bases V1r, Srated

in the case of Z1 and V2r, Srated in the case of Z2. The total impedance of the transformer in

per unit . 21 ZZ

Example 5: A single-phase transformer that is rated 3000 kVA, 69 kV / 4.16 kV, 60 Hz

has an impedance of 8 percent. Calculate the total impedance of the transformer referred

to (a) the primary side (b) the secondary side

Solution

1587

103000

1069)(

3

622

1

rated

rbase

S

VZa

127158708.0 111 baseee ZZZ

7685.5103000

1016.4)(

3

6222

2

rated

r

baseS

VZb

46.07685.508.0 222 baseee ZZZ

18

Alternatively

46.0127

69

16.42

1

2

1

22 ee Z

N

NZ

5.9 Voltage regulation

With the primary voltage maintained constant, the secondary terminal voltage at no load

differs from the secondary voltage under load. The voltage regulation or simply

regulation of a transformer is the change in secondary voltage which occurs when the

rated kVA output at a specified power factor is reduced to zero, with the primary voltage

maintained constant. It is usually expressed as a percentage (called percentage regulation)

or a fraction of the rated no-load terminal voltage (in per unit).

The equivalent circuit given in Fig. 14.b is used to calculate voltage regulation. The

circuit may be either referred to the primary or the secondary side. The circuit in general

form can thus be represented as shown in Fig. 16.

If the circuit is referred to the primary side, then

1121 ,, eeee XXandRRVVVE

eReX

Load VE

Fig. 16 Circuit for computing voltage regulation

Voltage regulation puV

VV

E

VE

1

21

(18.a)

If the circuit is referred to the secondary side, then

2221

2 ,, eeeen XXandRRVVa

VVE

Voltage regulation puV

VV

n

n

2

22 (18.b)

It can be shown thatn

n

V

VV

V

VV

2

22

1

21

(19)

And nVV 22 1 (20)

In general, let the load current be I lagging behind the load voltage V by . Then taking

the load voltage as the reference phasor, we can write

19

sincossincos

sincos01

eeee

ee

IRIXjIXIRV

jIIjXRjV

E

And hence

22 sincossincos eeee IRIXIXIRVE E

The second term under the root is usually negligible except at low leading power factors.

Considering the first term only gives

sincos ee IXIRVE

Or sincos ee IXIRVE (21)

The angle is negative when current is leading and positive when current is lagging.

The voltage regulation is maximum when = where

e

e

R

X1tan

0cossin

ee IXIRVE

d

dFrom

Hence ee

ee

e

ee

e

ee ZI

Z

XRI

Z

XIX

Z

RIRVE

22

max (22.a)

Where22

eee XRZ (22.b)

From (21)

sincos

E

IX

I

I

E

IR

I

I

E

VE

fl

fl

fl

fl Or

sincos ........ upupupup XIRIE

VE

(23)

Where

etc,unitperiniresistanceunitperincurrent .... EIRRIII upflup

We note the following:

(i) Usually the quantities will be referred to the secondary side (ii) With (23), it is not necessary to refer quantities from primary to secondary side, for

per-unit values of primary and secondary impedances can be added directly

(iii)The equations are correct at any current or kVA and at rated current or kVA.

(iv) At rated current or kVA, Ip.u.=1. (v) It is supposed that E is the rated voltage. When E is not the rated voltage, voltage

regulation can still be calculated using (21).

20

Example 9

A 100-kVA 1-ph transformer has 400 turns on the primary and 80 turns on the secondary.

The primary and secondary resistances are 0.3 and 0.01 respectively, and the corresponding leakage reactances are 1.1 and 0.035 respectively. The supply voltage is 2200 V. Calculate (a) the equivalent impedance referred to the primary circuit and (b)

the voltage regulation and the secondary terminal voltage for full load having a power

factor of (i) 0.8 lagging and (ii) 0.8 leading (c) the maximum voltage regulation

Solution

55.0)80400(01.03.0)()( 222

2111 RNNRRa e

975.1)80400(035.01.1 21eX

05.2975.155.0 21

22

1eZ

6.0sin8.0cos)()( ib

AI fl 45.452200101003

1

pu0336.02200

6.0975.18.055.045.45

VV n 440400

8022002

VVV n 2.4250336.01440122

puii 0154.02200

6.0975.18.055.045.45)(

VV 8.4460154.014402

.0424.02200

05.245.45regulationvoltagemaximum)(

1

11

max puV

ZIc

efl

Example 10

The primary and secondary windings of a 30-kVA, 6000-V / 230-V transformer have

resistances of 10 and 0.016 respectively. The total reactance of the transformer referred to the primary is 23 . Calculate the percentage regulation of the transformer when supplying full load current at a power factor of 0.8 lagging.

Solution

89.20)2306000(016.010)( 222

2111 RNNRRe ]

231eX

AI fl 5600010303

1

pu0254.06000

6.0238.089.205

%54.2regulationPercent

21

5.10Transformer output

The transformer supply voltage and frequency are substantially constant; the heating

therefore depends on the current taken by the load. Since the secondary voltage of the

transformer is also substantially constant it means that the heating also depends on the

load kVA. The transformer output is therefore usually quoted in kVA. The transformer

load in kVA is given by

kVA10 322 IVS (24)

Where

V2 = actual load voltage and

I2 = actual load current.]

When S is given and V2 is unknown the load current can be estimated using the

approximate equation

kVA10 322 IVS n (25)

5.11 Efficiency

The losses which occur in a transformer on load are composed of

(i) Copper losses in primary and secondary windings, namely

22

212

12221

21 ee RIRIRIRI

(ii) Iron losses in the core due to hysteresis and eddy currents. The iron losses depend on

the peak value of the mutual flux m and frequency. It is therefore independent of load current if voltage and frequency are constant.

Let

Pi = the iron losses (fixed loss) in kW and

Pc = the copper loss with full-load S kVA in kW

Then the total loss at any load xS kVA at power factor cos is Pi + x2 Pc and the

efficiency is

c

ici xPx

PS

S

PxPxS

xS

input

output

cos

cos

cos

cos2

For a given power factor, the efficiency is maximum when the expression in brackets is a

minimum. Hence for a maximum efficiency, we have

cic

i

c

i

c

i PxPorPx

PorxP

x

P

dx

dFrom

P

Px 2

200 (26)

i.e. efficiency is maximum when the copper loss, ic PPx 2 , the fixed loss or iron losses.

The efficiency of a transformer is calculated using this form of efficiency equation:

ci

ci

PxPxS

PxP

outputlosses

losses2

2

cos11

(27)

22

Example 13

The primary and secondary windings of a 500-kVA transformer have resistances of 0.42

and 0.0011 respectively. The primary and secondary voltages are 6600 V and 400 V respectively and the iron loss is 2.9 kW. Calculate the efficiency on full load at a power

factor of 0.8.

Solution

AI fl 1250400

1000500currentsecondary load-Full 2

AI currentprimary load-Full fl 8.756600

10005001

W17200011.01250load fullon losscopper Secondary 2

W241542.08.75load fullon losscopper Primary 2

kW135.4W413524151720load fullon losscopper Total cP ,

kW035.79.2135.4load fullon loss Total

kW4008.0500pf 0.8at load fullon power Output

%27.989827.0035.7400

035.71

load full on Efficiency

Example 12

Find the output, at which the efficiency of the transformer of example 11 is maximum

and calculate its value assuming the power factor of the load to be 0.8.

Solution

837.0135.4

9.2

c

i

P

Px

kVA5.418500837.0efficiency maximumat output Therefore

kW8.59.22loss total,efficiency maximumAt

kW8.3348.05.418pf0.8at power output and

%30.98983.08.3348.5

8.51efficiency maximum Therefore,

Example 13

A 400-kVA transformer has an iron loss of 2 kW and the maximum efficiency at 0.8 pf

occurs when the load is 240 kW. Calculate (a) the maximum efficiency at unity power

factor and (b) the efficiency on full load at 0.71 power factor

Solution

kW422efficiency maximumat loss Total(a)

kVA3008.0

240 efficiency maximumat kVA Output

kW3001300pfunity at efficiency maximumat power Output

23

%68.989868.03004

41pfunity at efficiency Maximum

75.0400

300max is efficiency heat which tkVA load full of fraction x The)( b

kW56.375.0

2losscopper load Full

2

2

ic

x

PP

pf at efficiency load Full %08.989808.071.040056.32

56.32171.0

5.12 All-day efficiency

Power transformers operate at substantially constant load. They are designed such that

the maximum efficiency occurs at the normal operating load. Generally their efficiency

varies little as the load varies from 50 to 130 % of its rated kVA.

Distribution transformers on the other hand supplies load which varies widely over a 24-h

period. The efficiency of these transformers is better assessed on energy basis. The

output and losses are calculated in kW hours over a 24-hour day. The all-day efficiency is

defined as

kWhinoutputkWhinlosses

kWhinlossesdayall

1 (28)

Example 14

A 200-kVA 1-ph transformer has full load copper loss of 3.02 kW and iron loss of 1.6

kW. The transformer is in circuit continuously. For a total of 8 hours, it delivers a load of

160 kW at 0.8 pf. For a total of 6 hours, it delivers a load of 80 kW at unity power factor.

For the remainder of the 24-h cycle, it is on no load. What is the all-day efficiency?

Solution

kWlossTotal

kWlossiron

kWlosscopper

62.4

60.1

02.3load),(fullkVA200kVApf,0.8kW,160At

kWlossTotal

kWlossiron

kWlosscopper

08.2

60.1

48.002.3200

80,kVA80kVAupf,kW,80At

2

kW loss Total 6.1load), noon losscopper no is (there load noOn

kWhlosstotalkWhoutputtotal

kWhlosskWh output

kWhlosskWh output

kWhlosskWh output

5.651760h,24In

16106.10100h 10For

5.12608.2480680h 6For

37862.412808160h 8For

24

%41.969641.017605.65

5.651efficiencydayAll

5.13 Open-circuit and short-circuit tests on a transformer

These two tests enable the efficiency and voltage regulation to be calculated without

actually loading the transformer.

(a) Short-circuit test: This test is used to determine the leakage impedance. During this

test, one winding is short-circuited and a reduced voltage Vsc applied to the other to cause

rated current to flow. The test circuit and the equivalent circuit are shown in Fig. 18.a and

Fig. 18.b respectively. The magnetizing branch is neglected because its current under

this condition is less than 1 % of the total.

Voltage Vsc, current Isc and power Psc measured by the instruments are used to make the

following calculations.

)()()()( 21211211

cRZXcI

PRb

I

VZa eee

sc

sce

sc

sce (29)

We note that the following:

(i) Isc need not be the rated current since the equivalent circuit is linear. However, it is desirable that it should be near to the rated value so that stray losses (they are due to

eddy currents set up in large section conductors, tank and metallic supports by

leakage fluxes) are normal.

(ii) The supply could be fed to either winding. It is often convenient on the higher voltage transformers to supply the high-voltage winding, thus using a smaller current.

Vsc which will be about 3-15 % of the rated value may also be more suitable for test

facilities.

(iii)In laboratory experiments using small transformers, the instruments positions shown

minimize measurement errors.

(iv) The circuit parameters obtained with (27) are referred to the side which the test voltage is applied.

)()()()( 21211211

cRZXcI

PRb

I

VZa eee

sc

sce

sc

sce

Fig. 18 Short-circuit test

(b) Open-circuit test or no load test: During this test, one winding is open-circuited and

rated voltage at rated frequency is applied to the other. Quite often the low-voltage side is

supplied to reduce the test voltage required for safety reasons. As with the short-circuit

25

test, the equivalent circuit parameters will be referred to the side to which the test voltage

is applied. The test circuit and equivalent circuit are shown in Fig. 19.a and b.

The following calculations can be made

2

11122

1

)()()()()(V

Vae

I

VXd

I

VRcIIIb

V

PIa

m

m

p

mpomo

p (30)

Psc and Po represent the full load copper loss and the core loss (or iron losses)

respectively. They can be used directly to calculate efficiency.

.

2

11122

1

)()()()()(V

Vae

I

VXd

I

VRcIIIb

V

PIa

m

m

p

mpomo

p

Fig. 19 Open-circuit or no-load test

Example 15

The circuit shown below was used in a test on a 3-kVA transformer. A variable voltage

supply of fixed frequency was connected to terminals A and B and two tests were

performed:

(a) The voltage was raised to normal rated voltage and the meter reading were then 200 V, 24 W, 1.2 A

(b) The terminals C and D were short-circuited and the voltage was raised until the transformer full-load current was flowing. The meter readings were then 6.4 V, 28 W,

15 A

From the results of the tests, obtain:

(i) the no-load current and its power factor (ii) the iron losses of the transformer at normal frequency and voltage (iii)the full-load copper loss

(iv) the transformer resistances, Re1 and Rm and reactances Xe1 and Xm (v) the efficiency of the transformer at full load at a power factor of 0.8

26

A

B

V

A C

D Circuit diagram for Example 13

Solution

A1.2I i o current load-No )(

W 24PP (ii) oi lossesIron

W 28PP (iii) scc losscopper load Full

:nscalculatiost circuit te-Open (iv)

AIIIAV

PI pom

op 19.112.02.112.0

200

24 2222

1

16819.1

20067.1

12.0

200 11

m

m

p

mI

VXk

I

VR

:nscalculatiost circuit te-Short

12.015

2843.0

15

4.6211 2

sc

sce

sc

sce

I

PR

I

VZ

41.012.043.0 2221211 eee RZX

%88.979788.02824)8.03000(

28241)

puEfficiencyv

Example 16

A 10-kVA 1-ph transformer has a voltage ratio 1100 / 250 V. On no load and at normal

voltage (1100 V) and frequency the input current is 0.75 A at a pf of 0.2 lagging. With

the secondary short-circuited, full-load currents flow when the primary applied voltage is

77 V, the power input being 240 W. Calculate

(a) the transformer equivalent resistance and reactance referred to the secondary side (b) the maximum value of the voltage regulation at full load and the load power factor at

this regulation

(c) the percentage of full-load current at which the transformer has maximum efficiency

27

Solution

(a) Referring the circuit to the secondary side,

V5.171100

250771

a

V

A40250

00010

2

n

rated

V

SI

438.040

5.172eZ , 15.0

40

24022e

R and 41.015.0438.0 222eX

(b) The maximum voltage drop at full load = 5.1722 efl ZI V

The maximum regulation occurs when or when

lagging342.0438.0

15.0coscos

2

2 e

e

Z

R

(c) Maximum efficiency occurs when oc PPx 2 . Therefore

%9.82or829.015.040

2.075.011002

x

5.14 Construction of transformers

Power transformers are designed so that their characteristics approach those of an ideal

transformer

(a) To attain high permeability and consequently a small magnetizing current, the core is

made of iron and the core forms a closed magnetic circuit

(b) To keep the hysteresis loss down, high-grade grain-oriented steel having a narrow

hysteresis loop is used

(c) To keep the eddy current loss down, the core is laminated and 3 to 4% silicon is added

to increase the resistivity of the steel. The laminations (about 0.4mm thick) are insulated

from each other by a thin layer of insulation, thus overall cross-sectional area is slightly

greater than the actual cross-sectional area of the iron.

(d) To keep the leakage fluxes low, the windings are arranged as shown in Fig. 18 in the

case of 1-ph transformers. Two main forms of magnetic circuits are used: core- and shell-

type arrangements. The core-type construction has the primary and secondary windings

distributed over both core legs in order to reduce the amount of copper (i.e. length of

turns is reduced).

The transformer windings are carefully insulated from each other and from the core.

Winding resistances R1 and R2 are also minimized to minimize copper losses.

28

L L

H H

L LH H

H L L H H L L H

windingsconcentric

on with constructi type-Core (a)

windingssandwiched

on with constructi type-Shell (b)

LL

Fig. 20 Common Single-phase transformer construction

5.15 Polarity tests

The four terminals of a single-phase transformer may be mounted so that it has either

additive or subtractive polarity as shown in Fig. 21

A2 a1

A1 a2

A2

a1

a2

A1

polarity additive (a) polarity esubtractiv (b)

Fig. 21Two standard ways of mounting transformer terminals

If it becomes necessary to determine whether a transformer has additive or subtractive

polarity then either of the following polarity tests may be used:

(a) Using a low-voltage ac source: Referring to Fig. 22,

(i) Connect the HV winding to ac source

(ii) Connect a jumper J between any two adjacent HV and LV terminals (1, 3)

(iii) Connect a voltmeter (V2) between the other two adjacent HV and LV terminals (2, 4)

(iv) Connect a voltmeter (V1) across the HV winding.

Reading of voltmeter V2 = E1 E2 = V1 E2 if 1 and 3 possess the same polarity = E1 + E2 = V1 + E2 if they do not

Therefore if V2 is lower than V1 then the polarity is subtractive and if V2 is higher than V1

then the polarity is additive

1 3

2 4

V1V2

J

Fig. 22 Polarity test using ac source

29

(b) Using dc source: Referring to Fig. 23,

(i) Connect a dc source to LV winding

(ii) Mark the terminal connected to the positive side of the source a2

(iii) Connect a dc voltmeter across the HV winding

(iv) Close the switch and observe the movement of the pointer of the voltmeter

(v) If the pointer moves upscale, the transformer terminal connected to the (+)

terminal of the voltmeter is marked A2 and the other marked A1.

a2

V+

-

Fig. 23 Polarity test using dc source

5.15 Parallel Operation of single-phase transformers

Parallel connection of several transformers is widely used in electrical systems for the

following reasons:

(a) In many cases, the amount of power to be transformed is greater than that which can be built into one transformer

(b) Frequently, the growth of load requires that the installed transformers supply an output greater than their total kVA capacity. Additional transformers are then

installed to run in parallel with the existing transformer.

(c) It is sometimes found desirable to supply a load through two or more units in order to reduce the cost of the spare unit required to ensure continuity of service in case of

damage.

The following conditions must be fulfilled when operating two or more single-phase

transformers in parallel:

(a) The polarity should be the same. The polarity can be either right or wrong. A wrong polarity results in a severe short circuit. Terminals of the same markings are

connected together to ensure correct polarity. See Fig. 24. If the polarity markings

are either incorrect or not present, the polarity of the incoming transformer can be

checked by connecting a voltmeter across the paralleling switch

(b) The voltage ratio should be the same. This is to avoid no-load circulating current and also over-loading on one transformer when the paralleled transformers are loaded.

(c) The per-unit impedances should be equal in magnitude and have the same angle. When they are equal in magnitude, the transformers share kVA loads in proportion to

their respective ratings. If both their magnitudes and angles are the same they will not

only share kVA loads in proportion to their respective ratings but also the combined

load kVA will be the algebraic sum of the kVA carried by each transformer. If the

angles are different, the resultant kVA capacity of the paralleled group will be slightly

smaller than the sum of their individual ratings if none should be overloaded. It is not

very necessary that the angles should be the same

30

A2 a2

A1 a1

Tx1

Tx2

A2

A1

a2

a1

Load

Fig. 24 Connection ensuring correct polarity

5.16 Load sharing of parallel-connected transformers The equivalent circuit of two transformers in parallel feeding a common load ZL is shown

in Fig. 25. The voltage ratios are supposed to be equal and the magnetizing branch is

neglected. The circuit is referred to the secondary side but it may also be referred to the

primary side.

a

V1

1I

2I

1Z

2Z

I

VLZ

Fig. 25 Equivalent circuit of two transformers in parallel

Z1 is the impedance of transformer 1 referred to the secondary side and Z2 is the

impedance of transformer 2 referred to the secondary side.

The voltage drops across the impedances are the same. Therefore

)( 21

212211

ZZ

ZZIZIZI

Or

IZZ

ZIandI

ZZ

ZI

21

12

21

21

Multiplying by the terminal voltage V gives

21

21

ZZ

ZSS

(31.a)

21

12

ZZ

ZSS

(31.b)

31

Where

loading 1r transforme11 VIS (32.a)

loading 2r transformeS 22 VI (32.b)

load combinedVIS (32.c)

Equations (31) hold for per-unit impedances provided that all are expressed with

reference to a common base power. The following equation can be used to obtain a new

per unit value with reference to a new base power.

old

base

new

baseold

pu

new

puS

SZZ (33)

Example 17

A 500-kVA transformer (Transformer 1) is connected in parallel with a 250-kVA

transformer (Transformer 2). The secondary voltage of each is 400 V on no load. Find

how they share a load of 750 kVA at power factor of 0.8 lagging if

pujZ

andpupujZc

pupujZ

andpupujZb

pupujZ

andpupujZa

62.6005095.00444.0025.0

69.7805099.005.001.0)(

69.7805099.005.001.0

69.7805099.005.001.0)(

44.6904272.004.0015.0

69.7805099.005.001.0)(

2

1

2

1

2

1

Solution

If all the impedances are referred to a base power of 500 kVA, then only the impedances

of transformer 2 will change.

old

pu

old

puold

base

new

baseold

pu ZZS

SZ 2

250

500Znewpu

Case (a)

69.7805099.005.001.01 jZ and

44.6908544.008.003.004.0015.022 jjZ

Further

90.721360.013.004.008.003.005.001.021 jjjZZ kVAS 9.367508.0cos750loadkVATotal 1

kVAZZ

ZSS

36.40471

9.72136.0

44.6908544.09.36750

21

21

lagging 762036.40cos offactor power at 471 .kVA

32

Similarly

11.31281

9.72136.0

69.7805099.09.367502S

lagging 8560 offactor power at 281 .kVA

Remark: Transformer 1 with larger per-unit impedance is under-loaded whereas

transformer 2 with lower per-unit impedance is overloaded.

Case (b)

pujjZ 69.7810198.01.002.005.001.022 pujZZ 69.7815297.015.003.021

kVAS

9.36500

69.7815297.0

69.7810198.09.367501

lagging 80 offactor power at 500 .kVA

kVA

9.36250

69.7815297.0

69.7805099.09.36750S2

laggging 80 offactor power at kVA 250 .

Remark: Load shared in proportion to transformer ratings. Arithmetic sum of loadings is

equal to the combined load. A shorter approach can be used on recognizing that

pupu ZZ 21

Case (c)

pujjZ 62.601019.00888.005.00444.0025.022 62.661512.01388.006.021 jZZ

9.42505

62.661512.0

62.601019.09.367501S

lagging 7320 offactor power at 505 .kVA

82.24253

62.661512.0

69.7805099.09.367502S

lagging 910 offactor power at kVA 253 .

Remark: Transformers are slightly overloaded when the combined load is equal to the

sum of individual kVAs

33

6.0 Three-phase transformers

These are required to transform 3-phase power. The three-phase transformer may be

either of the following:

(a) A three-phase transformer bank: This consists of three identical single-phase

transformers having their windings externally connected for three-phase working. The

single-phase transformers retain all their basic single-phase properties such as current

ratio, voltage ratio and the flux in the core. The kVA capacity of the bank is the sum of

their individual ratings. See Fig.26.

(b) A three-phase transformer unit: This is a single unit of special construction for three-

phase working. Modern large transformers are usually of the three-phase three-legged

core type shown in Fig.27. A leg carries the primary and secondary windings of a phase.

The windings are internally connected. For a given total capacity, 3-phase units are much

cheaper in capital cost, lighter, smaller and more efficient.

Tx1

Tx2

Tx3

1L

2L

3L

1L

2L

3L

2A

1a1A

2a

2A

2A

1A

1A

2a

2a

1a

1a

ers transformphase-single of connectionStar -Delta 9 Fig.

6.1 Winding arrangement

The three windings, primary or secondary, can be connected in three different ways:

(a) Star connection: For this connection %583 voltagelinevoltagephase of the

line voltage. This enables the insulation of the winding to be reduced to a minimum for a

given supply voltage. Line current = phase current. It is the most economical connection

for a high-voltage winding.

(b) Delta connection: Phase voltage = Line voltage. Therefore winding must be

insulated for the full Line voltage. More turns are also required. With very high voltages

a saving of 10 % may be achieved by using star-connection rather than delta connection

on account of insulation. The saving is small, however, at voltages below 11 kV. For

delta connection 3currentlinecurrentphase so the winding cross-sectional area is

34

58 % of that required for the star connection. Therefore it is the most economical for

low-voltage winding.

(c) Zigzag (or interconnected star) connection: It is a modification of the star

connection. Each phase winding is divided into two sections and placed on two different

legs. The two sections are then connected in phase opposition. The zigzag connection is

restricted to the low-voltage winding. 15 % more turns are required for a given phase

terminal voltage compared with a normal star.

The three different winding arrangements give rise to several possible connection

combinations: star-star, star-delta, star-zigzag, delta-star, delta-delta, etc

1L 2L 3L

2A2B

1B

2C

1C

2a

1a

2b

1b

2c

1c

1L

2L

3L

1A

Fig. 27 Delta-Star connection of three-phase, three-legged core-type transformer

6.2 Phase groupings

The voltage induced in a primary phase winding is in phase with its corresponding

secondary phase winding. However, the phase angles of the primary and secondary line

voltages may differ depending on the type of winding connection. Three-phase

transformers are given symbol which indicates the type of connection used for the high-

voltage and the low-voltage winding and the phase displacement between the high-

voltage line phasor and its corresponding low-voltage line phasor. The symbol is in the

form Xxn where X (the capital letter) and x (the small letter) indicate the type of

35

connection for the high-voltage and low-voltage windings respectively and n is a clock

hour number which indicates the phase displacement. In this method of indicating the

phase shift, the high-voltage line phasor is represented by the minute hand of a clock

always set at 12 (or the zero hour) and the corresponding low-voltage line phasor by the

hour hand. Thus when n = 11 it means the clock reads 11 Oclock and the low-voltage line phasor leads by 30

o. The groups into which three-phase transformers are classified

are as follows:

Group Phase displacement Winding connections

1 zero Yyo Ddo Dzo

2 180o Yy6 Dd6 Dz6

3 30o lag Dy1 Yd1 Yz1

4 30o lead Dy11 Yd11 Yz11

6.3 Three-phase transformer connections More common transformer connections are

(a) Delta-Delta connection (Fig 28). : This connection is economical for large low

voltage transformer. This connection is not often used because there is no neutral point

and a four-wire supply cannot be given. It is used in a 3-phase transformer bank but

rarely in 3-phase transformer unit. It is possible to use this arrangement to provide 3-

phase power with one transformer removed. This connection, known as open-delta or

vee connection, can supply up to 57.7 % of the load capacity of the delta-delta

connection.

1L

2L

3L

LV phV

phI

LI

1A 2A

1B

2B1C

2C

2L

1L

3L

1a 2a

1b

2b1c

2c

phL

phL

II

VV

3

Fig. 28 Delta-delta connection

(b) Delta-Star (Fig. 29): It is commonly used to step up alternator voltage to

transmission line voltage. Another common application is in distribution service where

as a step-down transformer, the windings are not the most economical. The secondary

star point can be earthed and a four-wire supply given.

(c) Star-Delta: There is no secondary neutral and four-wire supplies cannot be given.

The main use is as a step-down transformer at the load end of transmission line.

36

2A

1B

2B1C

2C

1A

1L

3L

2L

2c

1c

1b

2b

1a 2a

phV

phI

LI 3L

1L

2L

LV

3

Lph

VV Lph II

Fig. 29 Delta-star connection

(d) Star-Star: This is economical for high-voltage transformer. However, the primary

and secondary phase voltages contain pronounced third harmonic voltages, which cause

the neutral point to oscillate at three times the line frequency. The effect of this

oscillating neutral is to cause fluctuation in the line to star-point voltage. Also if the

secondary load is unbalanced, the neutral point will be displaced and the line-to-neutral

voltages will become unequal. To stabilize the neutral, the neutral of the primary and the

neutral of the source are connected together usually by way of ground. Another way is to

provide a third delta-connected winding called tertiary winding. Although it can be used

to supply additional power, the tertiary winding generally has no external connection

Example 18

Three single-phase step-up transformers rated 40MVA, 13.2 kV / 80 kV are connected in

delta-star on a 13.2 kV transmission line. If they feed a 90MVA load, calculate the

following:

(a) the secondary line voltage

(b) the currents in the transformer windings

(c) the incoming and outgoing transmission line currents.

Assume transformers are ideal.

Solution

2.13

801)(

1

2

aV

Va

ph

ph

voltage)lineion(transmiss2.13But 1 kVVph

kVVph 80Therefore

kV138380sidesecondary on the voltageline theand

37

2phI

2phV

1phV

1phI

kV2.13

MVA b 303

90formereach transby carried load The)(

AkV

MVAI ph1 2272

2.13

30ndingprimary wi in theCurrent

AkV

MVAI ph2 375

80

30windingsecondary in theCurrent

AIc ph 3932322723 line incominginCurrent)( 1

AII Lph 375line outgoingin Current 2

Example 19

Three single-phase transformers have their primaries joined in delta to a 6600 V, three-

phase, three-wire supply. Their secondaries are connected to give a three-phase, four-wire

output at 415 V across lines. The total load on the transformers is a balanced load of 150

kW at 0.8 pf lag. If the voltage per turn on the primaries is 4, find

(a) the number of turns on the primary winding and the secondary winding

(b) the currents and voltages in all windings and lines, including the neutral wire on the

secondary side

(c) kVA load on each transformer

Assume transformers are ideal

Solution

turnsturnvolt

Va

ph1650

4

6600phaseperturnsPrimary)(

1

turnsturnvolt

Vph60

43

415phaseperturnsSecondary

2

38

kV6.61phV

1phI

1LI2LI

V415

Load2phV

kW150

laggingpf8.0

AV

PowerIIb

L

phL 2618.04153

10150

cos3Secondary)(

3

22

balanced is load because0 AI N

AN

INI

ph

ph 5.91650

26160

1

22

1

AIL 4.1635.9 .

Example 20

A three-phase 415 V load takes a line current of 800 A from a 3300 / 415 V delta/star

transformer. The 3300 V system is supplied from an 11000/3300 star/star transformer.

Draw the circuit diagram and assuming no losses, find both line and phase values of

voltages and currents in each part of the circuit. What will be the turns ratios of both

transformers?

Solution

110001 V

1phV 1phI2phI 2ph

V

32 VV

32 II

3phI 3phV4phI

4phV

AI 8004

V

V

415

4

1IArTransformeBrTransforme

Solution

Voltages:

VV

VVV ph 2403

415

3;415 444

VVVVV ph 3300;3300 333

VV

VVVV ph 19053

3300

3;3300 2232

39

VV

VV ph 63503

11000

3;11000 111

Turn ratios:

75.13240

3300

4

3

ph

ph

V

VArTransforme

33.31905

6350

2

1

ph

ph

V

VBrTransforme

Currents:

AIIAI ph 800;800 444

AAratioturns

II

ph

ph 2.5875.13

80043

AIII ph 8.1002.5833 332

AII ph 8.10022

ABratioturns

II

ph

ph 2.3033.3

8.10021

AII ph 2.3011

Check:

kVAVAIVkVAInput 5755750002.301100033 11

kVAVAIVkVAOutput 57557500080041533 44

6.4 Parallel operation of three-phase transformers (Fig. 30)

Three-phase transformers operating in parallel should have

(a) the same line voltage ratios

(b) the same per-unit impedances, i.e., they are equal in magnitude and in phase

(c) the same phase displacement between primary and secondary line voltages

(d) the same phase sequence

The last two conditions which are absolutely essential ensure that the secondary line

voltages of the transformers are in phase. When these conditions are not fulfilled a

potential difference appears across the paralleling switches S1 and S2.

From the view point of phase sequence and phase displacement, three-phase transformers

which can operate in parallel are:

40

(a) transformers of the same group. This case the terminals with the same letter must be

connected to the same line as shown in Fig. 30.

(b) transformers having -30o phase displacement (Group 3 transformers) and those having

+30o phase displacement (group 4 transformers). In this case two of the high voltage

connections and the corresponding low voltage connections are interchanged as shown in

Fig. 31.

Faulty internal connections in the transformer tank can cause the phase sequence of a

transformer to be reversed. Voltage across paralleling switches should therefore be

monitored before the switches are closed.

7. Cooling methods Cooling of transformer windings and core is provided to prevent rapid deterioration of

the insulating materials. There are several methods of transformer cooling. Each method

is described by a standard designation (or nomenclature) consisting of letter symbols.

They are

(a) Letters for medium: air A, gas G, synthetic oil L, mineral oil O, solid insulation S,

water W

(b) Letters for circulation: natural N, forced F

Up to four letter symbols are used for each method for which the transformer is assigned

a rating: some big transformers are designed to have a variable rating, depending on the

method of cooling used. The order of the symbols is

S2

S1

Transformer 2

A2 a2

B2 b2

C2 c2

Transformer 1

A2 a2

B2 b2

C2 c2

C B A a b c

Fig. 30 Parallel operation of 3-phase transformers

41

(i) the medium and

(ii) the circulation of the coolant in contact with the windings; and

(iii) the medium and

(iv) the circulation of the coolant in an external heat exchanger system.

Common methods of cooling transformers are

(a) Air Cooling (Dry type transformers)

(i) AN: the ambient air as coolant and natural circulation by convection. The metallic

housing is fitted with ventilating louvers. Ratings: up to 50 kVA (low-power

transformers). Using high-temperature insulating materials (glass and silicone resins)

make ratings up to 1.5 MVA possible. The low-power transformers are used inside

buildings where the air is clean and high-power ones are for special conditions such as

those in mines.

(ii) AF: Forced air circulation is used to raise transformer loadings

(b) Oil-immersed, Oil cooling: Oil is a much better insulator than air. Consequently, it is

invariably used on high-voltage transformers

(i) ONAN: natural oil circulation and natural air flow over the tank. It is very common

for transformers rated up to 5 MVA. With radiators, it is possible to build units up to 40

MVA. (Note that limit of output is determined by tank size and cost).

(ii) ONAF: Cooling fans blow air over the radiators to enable a much bigger output from

a transformer of a given size. With this method of cooling ratings up to about 75 MVA

can be built.

S2

S1

Transformer 2

A2 a2

B2 b2

C2 c2

Transformer 1

A2 a2

B2 b2

C2 c2

C B A a b c

Fig. 31 Parallel operation of group 3 & 4 transformers

42

(iii) OFAF: Pumps are used to circulate the oil and cooling fans to blow air over

radiators. This is the usual method for transformers of 30 MVA and upward. Both OFAF

and ONAN may be used on a unit with ONAN up to 0.5 p.u. rating. Change over is

initiated automatically by temperature-sensing elements. Three-phase type OFAF step-up

transformer rated 1300 MVA installed at a nuclear power generating station is one of the

largest units ever built.

(c) Oil-immersed, Water cooling

(i) ONWF: Copper cooling coils are mounted in the tank above the level of the

transformer core, but below the oil surface.

(ii) OFWF: Oil is circulated by pump from the top of the transformer tank to an external

oil/water heat exchanger. Oil returns when cold to the bottom of the tank. Its advantages

over ONWF include

-The transformer is smaller and the tank does not have to contain the cooling coils.

-Leakage of water into oil is improbable if oil pressure is greater than that of water

This method is used for large installations. It is commonly used in generating stations,

particularly hydro stations where ample supply of water is available.

8.0 Tap-changing Transformers Most power transformers have tappings on coils brought out to terminals so that the

number of turns on one winding can be changed. The turns ratios are changed in order

(a) to maintain the secondary voltage at their rated value under the varying conditions of

load and power factor. The secondary terminal voltage may vary with changes in load

over an undesirably large range, because of changes in the impedance drop in the

transmission lines and transformers

(b) to control the flow of reactive power between two interconnected power systems or

between component parts of the same system, at the same time permitting the voltages at

specified points to be maintained at desired values.

Tappings on power transformers permit voltage adjustment within %5 . Low- and medium-power transformers usually have three taps per phase: +5%, 0, and -5%

variations in the turns ratio. Higher power ratings usually have five: +5%, +2.5%, 0, -

2.5% and -5% variations in the turns ratio. The principal tapping 0 is that to which the rating of the winding is related. A positive tapping includes more, and a negative less

turns than those of the principal tapping.

Tappings are usually fitted on the higher voltage winding to obtain tappings within fine

limits. Consider say, 11kV / 433 V, 600 kVA delta-star distribution transformer having

volts / turn = 10. On the low-voltage side, 25310433 N . Adjustment can then be

in steps of 4%. If 5 % and 10 % are required, we shall use 4 % and 8 % or 12 %.

On the high-voltage side 11001011000 N and it is possible to make adjustment in

steps of 0.09 %.

43

8.1 Changing the taps of transformers

Tap-changing may be either on-load or off-load.tap-changing:

(a) Off-load tap changing: the changes are made when the transformer is disconnected

from the primary circuit. The most common off-load tap changing transformer has

tappings inside the tank and connected to an internal switch which is operated by an

external switch handle (usually by rotary movement of a handwheel). A three-phase star

connected winding with taps made at the neutral is shown in Fig. 32.

Fig. 32 Three-phase star connected winding with taps at the neutral point

Off-load tap changing is simple and inexpensive and it is commonly used with

distribution transformers where occasional adjustments are required..

(b)On-load tap changing: Daily and short-time adjustment is generally by means of on-

load tap-changing gear. Tap changing is done without breaking the circuit. Momentary

connection must be made simultaneously to two adjacent taps during the transition, and

the short-circuit current between them must be limited by some form of impedance

known as transition impedance. Centre-tapped iron-core inductors (the reactor method) or

resistors (the resistor method) are used for this purpose.

(i) Reactor method: This method has now almost entirely been superseded by the resistor

method. It is manufactured and used only in the USA.

(ii) Resistor method: In modern designs the transition impedance is almost invariably

obtained by means of a pair of resistors. An arrangement of such a tap changer for one

phase is shown in Fig. 33. In this figure, the diverter switch and the even tap selector are

shown in the position when the T2 tap is brought in circuit.

To move to the next tap, T3, the odd tap selector should first be moved to that tap (see the

dashed lines in Fig. 33), and the diverter switch may then be rotated clockwise. The

ensuing sequence of events is as follows:

44

selector tapodd

1TS1T

3T

5T

7T

2T

4T

6T

8T

2TS

selector even tap

1R 2R

1

2 3

4

I

I

Fig. 33.a. Resistor tap changer

- contacts 3 and 4 break, contacts 1 and 4 make, Fig. 33.b

1R 2R

1

2 3

4

I

i-I2

1iI

2

1i

Fig 33.b

45

- contacts 1 and 4 break, and contacts 1 and 2 make, Fig. 33.c

1R 2R

1

2 3

4

I

Fig 33.c

The tap selectors may be moved from tap to tap only when their circuits are de-energized.

The resistors are short-time rated and it is essential to minimize their time of duty. For the

same reason, means must be adopted to ensure that it cannot be inadvertently left in the

bridging position.

On-load tap changer control gear can vary from simple push-button initiation to a

complex automatic control of as many as four transformers in parallel.

9.0 Autotransformers

An autotransformer has a single tapped winding which serves both primary and

secondary functions as shown in Fig. 34. The circuit diagrams are shown in Fig. 35.

1V

1I A

B

1N

2N

2I C

B

2V Load

1V 1N

A

B

2N

2I

2V Load

ormerautotransfdown -Step 34.a Fig. ormerautotransf up-Step 34.b Fig.

C

B

1I

9.1 Autotransformer equations If we neglect losses, leakage flux and magnetizing current then

2

1

1

2

1

2

I

I

V

V

N

Nn

9.2 Advantages and disadvantages of autotransformer over two-winding

transformer

The main advantage gained in the use of autotransformer is the saving of copper. For a

two-winding transformer and an autotransformer which can perform the same duty (they

46

1V

1I

S

C

2I C

B

2VLoad

1V

A

B

1I

S

C

2VLoad

C

BB

ormerautotransfofdiagramsCircuit35 Fig.

dingcommon win C winding;series S

should have the same voltage per turn and therefore the same flux. We can also assume

the same mean length per turn)

H

L

V

V

rtransforme winding-two in copper of Volume

ormerautotransf in copper of Volume1

Or

ormerautotransf an using

by effected copper of saving rtransforme windingtwo the in copper of Volume

V

V

H

L

sidevoltage high on voltageV

sidevoltage low the onvoltageV

Where

H

L

In practice, voltage ratios HL VV less than about 31 show little economic benefit over

two-winding transformer because of other factors such as cost of insulation.

The main disadvantage is that the primary and secondary circuits are not isolated from

each other.

Example 21

An autotransformer is required to step up a voltage from 220 to 250 V. The total number

of turns is 2000. Determine (a) the position of the tapping point (b) the approximate

value of the current in each part of the winding when the output is 10 kVA and (c) the

economy in copper over the two winding transformer having the same peak flux and the

same mean length per turn.

47

Solution:

1V

1I

2V

2I

1N

2N

17602000250

220

250

220

250

220)( 21

2

1

2

1 NNorN

N

V

Va

Position is 240 turns from one end.

AIIVb 40250

10101010)(

3

23

22

AIIV 45.45220

10101010

3

1

3

11

Therefore current in series winding is 40 A and current in common

winding A45.54045.45

%8888.0250

220)( orpu

V

VcopperinSavingc

H

L of copper used in the two-

winding transformer

9.3 Two-winding transformer connected as an autotransformer

A two-winding transformer can be changed into an autotransformer by connecting the

primary and secondary windings in series. The following rules apply whenever a two-

winding transformer is connected as autotransformer:

(a) the current in any winding should not exceed its current rating

(b) the voltage across any winding should not exceed its voltage rating

(c) rated current in one winding gives rise to rated current in the other

(d) rated voltage across one winding gives rise to rated voltage across the other

(e) if current in one winding flows from say A2 to A1, then current in the other winding

must flow from a1 to a2 and vice versa

(f) the voltages add when terminals of opposite polarity (A1 and a2 or A2 and a1) are

connected together by a jumper. The voltages subtract when A1 and a1 (or A2 and a2) are

connected together.

48

Example 22

A two-winding single-phase transformer rated 15 kVA, 600 V / 120 V, 60 Hz. We wish

to reconnect it as an autotransformer in three different ways to obtain three different

voltage ratios:

(a) 600 V primary to 480 V secondary (b) 600 V primary to 720 V secondary (c) 120 V primary to 480 V secondary Calculate the maximum load the transformer can carry in each case.

Solution

(a) The secondary voltage 120 V must be subtracted from the primary voltage to obtain

the 480 V.

%8888.0250

220)( orpu

V

VcopperinSavingc

H

L

(b) The secondary voltage 120 V must be added to the 600 V to obtain 720 V

%8888.0250

220)( orpu

V

VcopperinSavingc

H

L

49

(c) The 120 V becomes the primary of the autotransformer and the 120 V is subtracted

from the 600 V to obtain its secondary

%8888.0250

220)( orpu

V

VcopperinSavingc

H

L

9.4 Applications of autotransformers

They are mainly used for

(a) variac

(b) interconnecting power systems that are operating at roughly the same voltage (eg. 132

kV, 275 kV, 400 kV) and

(c) starting squirrel-cage induction motors.

10.0 Instrument transformers

They are used in ac circuits to serve these purposes:

(a) to make possible the measurement of high voltages with low-voltage instruments or

large currents with low current ammeters

(b) to insulate high voltage circuits being monitored from measuring circuit in order to

protect the measuring apparatus and operator

(c) to energize relays for the operation of protective and automatic control devices

The load on the secondary of an instrument transformer is called its burden and is

expressed in volt-amperes (VA). There are two types of instrument transformers: the

voltage (or potential) transformer and the current transformer

10.1 The voltage or potential transformers (VTs or PTs) The construction is similar to a power transformer. The primary is connected directly to

the power circuit either between two phases or between a phase and ground and the

secondary is connected to instruments and coils of relays. Sufficient insulation is

provided between the primary and the secondary to withstand the full line voltage as well

as the very high impulse voltage. Voltage transformers are designed to step down the

primary voltage to a nominal or rated voltage of 110 V so that standard instruments and

relays can be used. They introduce errors of two kinds into measurement being made: the

ratio errors (the ratio between input and output voltages is not constant under all

conditions of load) and the phase angle errors (the phase shift between input and output

voltages is not zero). These errors are due to the exciting current and the equivalent

50

series impedance of the transformer and they are kept low by using high quality iron

(high permeability and low loss) and operating it at low flux densities so that the exciting

current is very small. The resistance and reactance of the windings are also made very

low.

Fig. 36 shows the circuit for a potential transformer. One terminal of the secondary

winding is always earthed. The windings though insulated from each other, are

connected invisibly together by distributed capacitance between them. By earthing one

of the secondary terminals, the highest voltage between the secondary lines and earth can

never rise above that of the secondary voltage.

V

V1500

V110

PT

circuit ac H.V.

Fig. 36 Potential transformer installed on H.V. circuit

10.2 Current transformers (CTs)

Their primary consists of small number of turns connected in series with the power

circuit load. The secondary consists of a larger number of turns and it is connected to

ammeter, current coils of other instruments or current coils of relays. Current

transformers have the ratio of primary to secondary current approximately constant. The

nominal or rated secondary currents are usually 5 A or 1 A, irrespective of the primary

current rating. The transformer ratio is usually stated to include the secondary current

rating. Current transformers also introduce two errors in measurement: the ratio error

(the ratio between primary and secondary currents is not constant) and phase angle error

(the phase angle between the primary and secondary currents is not zero. The basic cause

of ratio and phase angle errors is the exciting current. To keep the exciting current small,

a high quality iron operating at very low flux densities is used as in PTs. In Cts the

secondary leakage impedance and impedances of the secondary leads and instruments

should also be very low; for any increase in these impedances increases the core flux and

therefore the exciting current. The tansformer is connected in the power circuit as shown

in Fig. 37 As in the case of PT (and for the same reasons) one of the secondary terminals

is always earthed

Current transformer secondary circuit must not be opened while current is flowing in the

primary. Without opposing ampere-turns the line current, which may be 100 to 200 times

the normal exciting current, becomes the exciting current. The iron core becomes.

51

A

AorA 51

Load

currentLoad

supplyac

AorA 51ammeter

CT

Fig. 37 Current transformer intalled in H.V. circuit

saturated and very high voltage spikes (several thousand volts) are induced across the

open-circuited secondary. These voltages are dangerous to life and to the transformer

insulation. The core when it becomes saturated can also cause excessive heating of the

core and windings. Therefore when it is desired to remove a load from the secondary

circuit, the secondary winding must first be short circuited.

When the line current exceeds 100 A we can sometimes use a toroidal or bar-primary

(N1=1) transformer shown in Fig. 38. It consists of a laminated ring-shaped core which

carries the secondary winding. The primary is composed of a single conductor that

simply passes through the centre of the ring as shown in the figure. Toroidal CTs are

simple and inexpensive and are widely used in HV and MV indoor installations.

AAI 32

AI 6001

11

N

primaryBar

Fig. 38 Toroidal or bar-primary transformer having a ratio 1000A / 5A connected to

measure a current in a line

52

Current transformers are also commonly used for the measurements of large currents

even when the circuit voltage is not dangerously high. This avoids bringing heavy leads

to the instrument panels. Whereas instrument CTs have to remain accurate up to 12 %

rated current, protection CTs must retain proportionality up to 20 times normal full load

Example 23

A potential transformer rated 14400 V / 115 V and a current transformer rated 75 A / 5 A

are used to measure the voltage and current in a transmission line. If the voltmeter

indicates 111 V and the ammeter reads 3 A, calculate the voltage and current in the line.

Solution

VV is line the on voltage The 13900115

14400111

AI is line the in currentThe 455

753

Example 24

The toroidal current transformer of Fig. 38 has a ratio of 1000 A / 5 A. The line

conductor carries a current of 600 A.

(a)Calculate the voltage across the secondary winding if the ammeter has an impedance

of 0.15 (b) Calculate the voltage drop the transformer produces on the line conductor

(c) If the primary conductor is looped four times through the toroidal opening, calculate

the new current ratio

Solution

AI secondarythe in Current (a) 36001000

52

V burden the across drop Voltage 45.015.03

mVVI

IVor

I

I

N

N

V

V b 25.245.0

1000

5)( 2

1

21

1

2

2

1

2

1

41)( 1122 xIxINIc

AIthatimpliesThis 2501

AAratiocurrentnewtheTherefore 5250

53

Exercises One

(1) A single-phase transformer has a primary winding with 1,500 turns a nd a secondary

winding with 80 turns. If the primary winding is connected to a 2300-V, 50-Hz supply,

calculate (a) the secondary voltage (b) the maximum value of the core flux. Neglect the

primary impedance

(2) A single-phase 2,300/230-V, 500-kVA, 50-Hz transformer is tested with the

secondary open-circuited. The following test results were obtained: V1 = 2,300 V, Io =

10.5 A, and Po = 2,300 W. Calculate (a) the power factor (b) the core-loss current Ip (c)

the magnetizing current Im.

(3) A 150-kVA, 2,400/240-V, 50-Hz single-phase transformer has the following

resistances and reactances: R1 =0.225 , Xl1 = 0.525 , R2 = 0.00220 and Xl2 = 0.0445 . Calculate the transformer equivalent values (a) referred to the primary (b) referred to the secondary.

(4) A 150-kVA, 2,400/240-V, 50-Hz single-phase transformer has the following

resistances and reactances: R1 =0.225 Xl1 = 0.525 , R2 = 0.00220 , Xl2 = 0.00445 , Rm = 10 k and Xm = 1.5 k . The transformer is supplying full-rated load at 0.85 lagging power factor and rated secondary terminal voltage. Calculate (a) I2 (b) Ip (c) Im

(d) Io (e) I1 (f) V1. Use the exact equivalent circuit referred to the primary side.

(5) The results of open- and short-circuit tests carried out on a 230/115-V, 60-Hz single phase transformer are

Cal