Transformer

2.1 Introduction

The transformer works on the principle of electromagnetic induction. A transformer is an electrical device, having no moving parts, which by mutual induction transfers electric energy from one circuit to another at the same frequency, usually with changed values of voltage and current. It consists of two winding insulated from each other and wound on a common core made up of magnetic material.

The most important task performed by transformers are : (i) changing voltage and current levels in electric power systems (ii) matching source and load impedances for maximum power transfer in electronic and control circuitry (iii) electrical isolation ( isolating one circuit from another).

Alternating voltage is connected across one of the winding called primary winding. The second winding is called the secondary winding. An alternating flux is produced in the core with an amplitude depending on the primary voltage, frequency and number of turns. This mutual flux links the secondary winding. In both the windings emf is induced by electromagnetic induction.

2.2 Working Principle

A transformer can be defined as a static device which helps in the transformation of electric power in one circuit to electric power of the same frequency in another circuit. The voltage can be raised or lowered in a circuit, but with a proportional increase or decrease in the current ratings.

The main principle of operation of a transformer is mutual inductance between two circuits which is linked by a common magnetic flux. A basic transformer consists of two coils that are electrically separated , but are magnetically linked through a path of low reluctance.

Consider two coils 1 and 2 wound on a simple magnetic circuit as shown in Fig. 2.1. These two coils are insulated from each other and there is no electrical connection between them. Let T1 and T2 be the number of turns in coils 1 and 2 respectively. When a source of alternating voltage V1 is applied to coil 1, an alternating flux Φm is produced in the magnetic circuit. The mean

path of this flux is shown in Fig .2.1 by dotted line. This alternating flux induces an emf E1 in the primary winding by self Induction.

Fig .2.1. Two winding Transformer

The emf (e2) is induced in the secondary winding by mutual induction.

The frequency of the induced emf in secondary is the same as that of the supply voltage. The magnitude of the emf induced in the secondary winding will depend upon its number of turns.

In a transformer, if the number of turns in the secondary winding is less than those in the primary winding and the output ( secondary) voltage is less than its input ( primary) voltage , it is called as step down transformer ( Fig. 2.2 (a)). When the number of turns in the secondary winding is higher than the primary winding and the output ( secondary) voltage is greater than its input ( primary) it is called as step-up transformer ( Fig .2.2 .(b)).

Fig.2 .2 .(a) .Step-down ransformer Fig. 2.2 . (b).Step-up Transformer

2.3 Construction of Transformer

A single phase transformer consists of primary and secondary windings

put on a magnetic core. Magnetic core is used to confine flux to a definite path.

Transformer cores are made from the laminated thin sheets of high grade

silicon steel. The laminations are insulated from one another by heat resistant

enamel insulation coating. L type, I type and E- type laminations are used. The

laminations are built up into stack and the joints in the laminations are staggered

to minimize air gaps. The laminations are tightly clamped.

Main constructional parts of transformer

.

Fig. 2.3 .a Fig. 2.3.b

The transformer has no moving parts. The main components of a transformer are

i) The magnetic core.ii) Primary and secondary windingsiii) Insulation of windingsiv) Expansion tank or conservator.v) Lead and tapping for coils with their supports, terminals and

terminal insulators. vi) Tank, oil, cooling arrangement, temperature gauge, oil gauge.vii) Buchholtz relayviii) Silica gel breather.

Magnetic core Magnetic circuit consists of an iron core. The transformer core is

generally laminated and is made out of a good magnetic material like silicon

steel. The thickness of laminations or stampings varies from 0.35 mm to 0.5 mm.

The laminations are insulated from each other by coating then with a thin coat of

varnish.

Various types of stampings and laminations employed in the construction

of transformers are shown in Fig . The joints are staggered to avoid continuous

gap causing increase in magnetizing current. If the joints are not staggered, the

core will have less mechanical strength and during operation there would be

undue humming noise. After arranging the laminations they are bolted together.

The two types of transformer cores are :1. Core type 2. Shell type

a) Core type transformers

In the core-type transformer, the magnetic circuit consists of two vertical

legs or limbs with two horizontal sections, called yokes. To keep the leakage

flux to a minimum half of each winding is placed on each leg of the core as

shown in fig. The low voltage winding is placed next to the core and the high-

voltage winding is placed around the low voltage winding to reduce the

insulating material required. The coils used usually are of cylindrical type and

are usually wound. For transformers of higher rating stepped core with circular

cylindrical coils are used. For transformers of smaller rating, rectangular coils

with core of square or rectangular cross section are used. Insulating cylinders

are used to separate windings from the core and from each other. In a core type

construction the winding surrounds the core. A few examples of single phase

and three phase core type constructions are shown in Fig. 4.

Fig.(a) Fig.(b)

Fig.2.4. Core type transformer

b) Shell type Transformer

Fig. 2.5 Shell type transformer

In general, the core-type has a longer mean length of core and a short

mean length of coil turn. The core type also has a smaller cross-section of iron

and so will need a greater number of turns of wire, since in general not as high a

flux may be reached in the core. However core type is better adopted for some

high-voltage service since there is more room for installation. The shell –type

has better provision for mechanically supporting and bracing the coils. This

allows better resistance to the very high mechanical forces that develop during a

high-current short circuit.

The choice of core -or- shell type construction is usually one of cost, for

similar characteristics can be obtained with both types. Both core and shell

forms are used , and selection is based upon many factors such as voltage

rating, kVA rating, weight, Insulation stress, mechanical stress and heat

distribution.

b) Spiral Core Transformer

Fig.2.5.a) .Spiral core type

The typical spiral core type is shown in Fig.2.5.a. The core is assembled either of a continuous strip of transformer steel wound in the form of a circular or elliptical cylinder or a group of strips assembled to produce the same elliptical shaped core. By using this construction the core flux always follows along the grain of the iron. Cold –rolled steel of high silicon content enables the designer to use higher operating flux densities with lower loss per kg. The higher flux density reduces the weigh per kVA.

Windings

Windings form another important part of transformers. In a two winding

transformer two windings would be present. The one which is connected to a

voltage source and creates the flux is called as a primary winding. The second

winding where the voltage is induced by induction is called a secondary. If the

secondary voltage is less than that of the primary the transformer is called a step

down transformer. If the secondary voltage is more than primary voltage , it is

a step up transformer. A step down transformer can be made a step up

transformer by making the low voltage winding its primary. Hence it may be

more appropriate to designate the windings as High Voltage (HV) and Low

Voltage (LV) windings..

Insulation

The insulation used in the case of electrical conductors is varnish or

enamel in dry type of transformers. In larger transformers to improve the heat

transfer characteristics the conductors are insulated using un-impregnated paper

or cloth and the whole core-winding assembly is immersed in a tank containing

transformer oil. The transformer oil thus has dual role. It acts an

i. Insulator ii. Coolant. The porous insulation around the conductor

helps the oil to reach the conductor surface and extract the heat. Oil ducts are

used as part of insulation between windings. The oil used in the transformer

tank should be free from moisture or other contamination.

Insulating oil

The oil used in transformer protects the insulation sheet from dirt and

moisture and removes the heat produced in the core and coil. It also acts as

insulating medium. The oil must posses the following properties.

i) High dielectric strength.

ii) Free from inorganic acid, alkali and corrosive sulphur to prevent

injury to the conductor or insulation.

iii) Low viscosity to provide good heat transfer.

iv) Free from sludging under normal operating condition.

v) Good resistance to emulsion so that the oil may throw down any

moisture entering the tank instead of holding it in suspense.

Expansion Tank or Conservator

A small auxiliary oil tank may be mounted above the transformer and

connected to main tank by a pipe. Its function is to keep the transformer tank

full of oil despite expansion or contraction of the coil with the change in

temperature. A small pipe connection between the gas space in the expansion

tank, and the cover of the transformer tank, permits the gas above the oil in the

transformer to pass into the expansion tank. So that the transformer tank will be

completely filled with the oil.

Temperature Gauge

Every transformer is provided with a temperature gauge to indicate hot

oil or hottest spot temperature. It is a self contained weather proof unit, made of

alarm contacts. It is dial operated by bourdon gauge connected to a thermometer

bulb located in the reign of hottest oil.

Oil Gauge

Every transformer is provided with an oil gauge to indicate the oil

present in the tank. The oil gauge may be provided with an alarm contact which

gives an alarm when the oil level dropped beyond permissible height due to oil

leaker due to any other reason.

Buchholz Relay

The first warning when a fault is present may be given by the presence of

bubbles in the oil. If the transformer is fitted with a conservator and there are no

pockets in which gas can collect, the gas bubbles will rise up the pipe joining

the conservator to the tank. It is possible to mount gas operated relay in this pipe to

give an alarm in case of minor fault and to disconnect the transformer from the

supply mains in case of severe fault.

Breather

The simplest method to prevent the entry of the moisture inside the

transformer tank is to provide chambers known as breather. The breather is

filled with some drying agent, such as calcium chloride or silica gel. Silica gel

or calcium chloride absorbs moisture and allows dry air to enter the transformer

tank. The drying agent is replaced periodically as routine maintenance. The

whole of the transformer tank and portion of conservator use filled with oil. The

breather is connected on one side of the conservator. Thus a small surface area

of transformer oil is exposed to the atmosphere through the breather.

Bushings

Connections from the transformer windings are brought out by means of

bushing. Ordinary porcelain insulators can be used up to a voltage of 33kV.

Above 33kV, capacitor and oil filled type of bushing are used. Bushing is fixed

on the transformer tank.

Cooling Arrangement in Transformer

The various methods of cooling employed in a transformer are

a) Oil immersed natural cooled transformers

b) Oil immersed forced air cooled transformer

c) Oil immersed water cooled transformers

d) Oil immersed forced oil cooled transformers

e) Air blast transformers

a) Oil immersed natural cooled transformers

In this type, the core and coils are immersed in insulting oil contained in

an iron tank. The heat produced in the core and winding is conduct by the

circulation of oil to the surface which dissipates heat to surroundings. In

the transformers of larger output, the dissipation surface is increased by

provided large number of tubes in its sides. The oil not only keeps the

windings cool but also provides additional insulation.

b) Oil immersed forced air cooled transformer

In this type, the core and winding are immersed in oil and cooling in

increased by forced air over the cooling surfaces. The air is forced over

external surfaces such as tank, tubes and radiators by means of fan

mounted external to the transformer.

c) Oil immersed water cooled transformers

In this type, the core and windings are immersed in oil and cooling is

increased by circulation of cold water through the tubes immersed in the

oil.

d) Oil immersed forced oil cooled transformers

In this type, the core and windings are Immersed in oil and cooling is

achieved by forced oil circulation. In this method of cooling forced oil

circulation is obtained by a centrifugal pump which is located at either the

oil inlet or outlet. The pump motor used for cooling is designed to operate

totally immersed in the cooling oil being circulated.

e) Air blast transformers

Here the transformer is cooled by a forced circulation of air through core

and windings. It is used in substations located in tickly populated places

where oil is considered a fire hazard. The air supplied is filtered to avoid

dust entered the ventilating ducts.

Fig 2.6.a Fig.2.6 b

Fig. 2.6.c

2.5 EMF Equation of a Transformer

Consider a transformer arrangement as shown in fig 2.7.(a)

Let , N1 = Number of turns in primary

N2 = Number of turns in secondary

∅m = Maximum flux in the core in webers = Bm X A

f = Frequency of alternating current input in hertz (HZ)

Fig 2.7.a.

B m = Maximum value of flux density in the core in wb/m2

A = Area of the core in m2

V1 = Supply voltage across primary in volts

V2 = Terminal voltage across secondary in volts

I1 = Full load primary current in amperes

I2 = Full load secondary current in amperes

E1 = Emf induced in the primary in volts

E2 = Emf induced in the secondary in volts

Fig.2.7 .b

As shown in figure above, the core flux increases from its zero value to

maximum value ∅m in one quarter of the cycle , that is in ¼ frequency second.

Therefore, average rate of change of flux = ∅m/ ¼ f = 4f ∅m Wb/s

Now, rate of change of flux per turn means induced electro motive force in volts.

Therefore, average electro-motive force induced/turn = 4f ∅m volt

If flux ∅ varies sinusoidally, then r.m.s value of induced e.m.f is obtained by

multiplying the average value with form factor.

Form Factor = r.m.s. value/average value = 1.11

Therefore, r.m.s value of e.m.f/turn = 1.11 X 4f ∅m = 4.44f ∅m

Now, r.m.s value of induced e.m.f in the whole of primary winding

= (induced e.m.f./turn) X Number of primary turns

Therefore,

E1 = 4.44f N1Øm = 4.44fN1BmA ------ (2.1)

Similarly, r.m.s value of induced e.m.f in secondary is

E2 = 4.44f N2 Øx = 4.44fN2BmA ------- (2.2)

In an ideal transformer on no load,

V1 = E1 and V2 = E2 , where V2 is the terminal voltage

Voltage Transformation Ratio (K)

From the above equations we get

E2/ E1 = V2/ V1= N2/N1 = K -=---- (2.3)

This constant K is known as transformation ratio.

(1) If N2>NA1, that is K>1 , then transformer is called step-up transformer.

(2) If N2<1, that is K<1 , then transformer is known as step-down transformer.

Again for an ideal transformer,

Input V1 = output V1

V1I1 = V2I2

Or, I2/I1 = V1/V2 = 1/K

Voltage ratio = E2

E1 = K

Current ratio = I 2

I 1 =

1K

Hence, currents are in the inverse ratio of the (voltage) transformation ratio.

Example 2.1

A 20 kVA, single-phase transformer has 200 turns in the primary and 40 turns in the secondary. The primary is connected to 1000V,50 Hz supply. Determine i) The secondary voltage on open circuit, ii ) The current flowing through the two winding on full load, iii ) The maximum value of flux.

Given data:

Transformer rating = 20 kVA; Number of primary turns N1 = 200 ;

V1 = 1000V

Number of secondary turns N2 = 40 ; Supply frequency f = 50 HzSolution i. Secondary voltage on open circuit

By using transformation ratio (K)

K= V 2V 1 =

N 2N 1 =

V 21000 =

40200

V2 = 40

200 ×1000 = 200V

ii. The current flowing through the two windings on full loads Full load secondary current I2

I 2 = Transformer rating (kVA)

secondary voltage (V 2) = 10× 103

200 = 100A

Full load primary current I1

I 1 = Transformer rating (kVA)Secondary voltage (V 2)

= 20× 103

200 = 20

I1= 20 A

iii. The maximum value of flux ( φm) The maximum value of flux ( φm) can be find out from the emf

equation of the Transformer

V1 = E1 = 4.44 f φm N1

Φm = V 1

4.44 f N 1 =

10004.44 × 50× 200 = 22.52 mwb

Example 2.2 . The no-load ratio required in a single-phase 50Hz transformer is 6600/300 V. If the maximum value of flux in the core is to be about 0.09 Weber, Find the number of turns in each winding.Given data:

Supply frequency f = 50Hz, Primary voltage V1 = 6600VSecondary voltage V2 = 300V, Maximum value of flux

Φm = 0.09 wb

To find The number of turns in primary and secondary windings (N1 and N2)Solution:

The primary and secondary turns can be found out from the emf equation of the transformer.

V1 = E1 = 4.44f ΦmN1 ---- (1)

V2 = E2 = 4.44f Φm N2 ---- (2)

From equation (1)

Primary turns

N1 = V 1

4.44 fΦm = 6600

4.44 × 50× 0.09 = 330

N2 = V 2

4.44 f ∅m =

3004.44 × 50× 0.09 = 15

Example 2.3. A single-phase,25 Hz transformer has 50 primary turns and 600 secondary turns. The cross sectional area of the core is 400 sq-cm. If the primary of the transformer is connected to 230 v supply. Find 1) the secondary induced emf ii) the flux density (peak) in the core.Given data:

Supply frequency f = 25 Hz , Primary turns N1 = 50 Secondary turns N2 = 600, Primary voltage E1 = 230v

Cross sectional area of the core A = 400 sq-cm = 400 × 10 -4sq-m

Solutioni.The secondary induced emf (E2)

E2

E1 =

N 2

N 1

E2 = E1

N 2

N1 = 230×

60050 = 2760 v

ii. Maximum value of flux density (Bm)

E1 = 4.44f Φm N1

∅m = V 1

4.44 f N 1 230

4.44 × 25× 50 = 0.0414 wb

∅m = 0.0414

400 ×10−4 = 1.036 wb/sq.m

Example 2.4. The emf per turn of a single phase, 6.6kV/440V, 50Hz transformer approximately 12 V. Calculate number of turns in the HV and LV windings and the net cross sectional area of the core for a maximum flux density of 1.5 T

Given dataEmf per turn = 12VPrimary voltage = 6.6 kVSecondary voltage = 440VFrequency f = 50 HzMaximum flux density Bm = 1.5 T

To findNumber of turns in the HV and LV windingsNet cross sectional area

Solution:

E1 = N1× emf induced/turn

N1 = 6.6 ×103

12 = 550 turns

N2 = E2

Emf induced / turn = 440

12 = 37 turns

E1 = 4.44 f N1 Bm A

A = E 1

4.44 fN 1 Bm = 6.6 ×103

4.44 × 50× 550 ×1.5 = 0.036 m2

Example 2.5.

A 2200/250V transformer trakes 0.5A at a power factor 0.3 on open circuit. Find

magnetizing and working componenets of no-load primary current.

Given data:Primary voltage V1 = 2200VSecondary voltageV2 = 250VNo-load current Io= 0.5ANo load power factorcosφ0= 0.3

To find

i) Magnetisingcomponent Iμ , ii) Working componentI ẉ

Solution : i) Magnetisingcomponenet

Iμ =Iosinφ0

sinφ0 =sin ( cos-1 0.3) = 0.9539

I μ = 0.5× 0.9539 I μ = 0.476 A

ii) Working componenet

I ẉ=Iocosφ0= 0.5 ×0.3

I ẉ=0.15 A

Example 2.6.

Find active and reactive components of no-load current. Ii) no-load current of a

440V/220V single phase transformer , if the input power on no-load to the high

voltage winding is 100w and power factor of no-load current is 0.3 lagging.

Given data:Primary voltage V1 = 440V, Secondary voltage V2 = 220V

No-load input power P0= 100W , No-load input power factor cosφ0= 0.3 lag

Solution:

P0=V1Iocosφ0

Io =

P0

V 1cos∅ 0 = 100

440 ×0.3 = 0.757 A

1.Active component (I ẉ)

I ẉ =Iocosφ0

= 0.757× 0.3 = 0.2271 A

Ii Reactive component ( I μ)sinφ0= 0.953

Iμ = Iosinφ0 = 0.757 × 0.953 = 0.722 A

Example 2.7.

Find (i) active component and reactive component of no-load current and (ii) no-

load current of a 230/115V single-phase transformer, if the power input on no-

load to the high voltage winding is 70W and power factor of no-load current is

0.25 lagging.

Given data:Primary voltage V1=230V,Secondary voltage V2=115V

No-load input power W0=70W,No-load input power factor cos0=0.25 lag.Solution:i.Active component (I ẉ)

Iẉ= no− loadinput power

Primary voltage = 70

230 = 0.3043 A

ii) No-load current (Io)

P0=V1Iocosφ0

Io =

P0

V 1cos∅ 0=¿

70230× 0.25 = 1.217 A

iii) Reactive component ( I μ)

________ ______________ Io = √I ẉ

2+ I μ 2 = √1.2172 – 0.30432

Io= 1.179 A

Example 2.8.

Assume the transformer of Fig. 2.8 (d) to be the ideal transformer. The

secondary is connected to a load of . Calculate the primary and secondary

side impedances, current and their pf, and the real powers. What is the

secondary terminal voltage?

Solution

The circuit model of the ideal transformer is drawn in Fig,2.9.(d)

Fig.2.9.(d)

(Secondary terminal voltage)

or Lagging

or Lagging

(Secondary power output) =

(primary power input) = (as the transformer is lossless)

= 1.732kW

or otherwise

= 1.732kW

2.6. Equivalent Circuit of Transformer

In Fig 2.10 the current flowing in the primary of the semi-ideal transformer

can be visualized to comprise two components as below:

i) Exciting current whose magnetizing component creates mutual

flux and whose core-loss component provides the loss associated

with alternation of flux.

ii) A load component which counterbalances the secondary mmf so that the mutual flux remains constant independent of load, determined only by thus

. Thus

------- (2.22)

Where

------- (2.23)

The exciting current can be represented by the circuit model of Fig. 3.7 so

that the semi-ideal transformer of Fig.3.13 is now reduced to the true ideal

transformer. The corresponding circuit (equivalent circuit) modeling the

behavior of a real transformer is drawn in Fig. 3.14(a) wherein for ease of

drawing the core is not shown for the ideal transformer.

In the equivalent circuit,

R1,X1 – Primary winding resistance and reactance in Ω

R0 – No load resistance in Ω

X0 – No load reactance in Ω

I1 – Full load primary current in A

I0 – No load primary current in A

I2'– Load component of primary current in A

Iw – working component

Fig. 2.10. Evolution of transformer equivalent circuit

The impedance on the secondary side of the transformer can now be

referred to its primary side resulting in the equivalent circuit of Fig.2.10 (b)

wherein

------ (2.24)

------ (2.25)

The load voltage and current referred to the primary side are

------ (2.26)

------ (2.27)

Therefore there is no need to show the ideal transformer reducing the

transformer equivalent circuit to the T-circuit of Fig 2.10 (c) as referred to side

1. The transformer equivalent circuit can similarly be referred to side 2 by

transforming all impedances (resistances and reactances), voltages and currents

to side 2. It may be noted here that admittances (conductances and

susceptances) are transformed in the inverse ratio squared in contrast to

impedances (resistances and reactances) which as already shown in Sec.3.4

transform in direct ratio squared. The equivalent circuit of Fig.2.10 (c) referred

to side 2 is given in Fig.2.10(d) wherein

With the understanding that all quantities have been referred to a particular isde,

a superscript dash can be dropped with a corresponding equivalent circuit as

drawn in Fig. 2.10 (e).

In the equivalent circuit of Fig 2.10 (c) if is taken as constant, the core-

loss is assumed to vary as or (Eq.(3.6)). It is fairly accurate

representation as core-loss comprises hysteresis and eddy-current loss expressed

as . The magnetizing current for linear B-H curve varies

proportional to . If inductance corresponding to susceptance is

assumed constant,

It therefore is a good model except for the fact that the saturation effect has

been neglected in which case would be a nonlinear function of It is an

acceptable practice to find the shunt parameters at the rated voltage and

frequency and assume these as constant for small variations in voltage and

frequency.

The passive lumped T-circuit representation of a transformer discussed

above is adequate for most power and radio frequency transformers. In

transformers operating at higher frequencies, the interwinding capacitances are

often significant and must be included in the equivalent circuit. The equivalent

circuit given here is valid for a sinusoidal steady-state analysis. In carrying out

transient analysis all reactances must be converted to equivalent inductances.

The equivalent circuit developed above can also be arrived at by

following the classical theory of magnetically coupled circuits [40]. The above

treatment is, however, more instructive and gives a clearer insight into the

physical processes involved.

Example 2.9.

A 20 kVA, 2000/200 V, single-phase, 50Hz transformer has a primary

resistance of 2.5Ω and reactance of 4.8 Ω . The secondary resistance and

reactance are 0.01 Ω and 0.018 Ω respectively. Find i) Equivalent reactance

referred to primary. Ii) Equivalent impedance referred to primary. Iii)

Equivalent reactance , impedance and reactance referred to secondary, iv)

Total copper loss of the transformer.

Given data:

Transformer rating = 20 kVA, Primary voltage V1 = 200V

Secondary voltage V2 = 200 V, Supply frequency f = 50 Hz

Primary resistance R1 = 2.5 Ω , Primary reactance X1 = 4.8Ω

Econdary resistance R2 = 0.01 Ω , Secondary reactance X2 = 0.018Ω

Solution:

i. Equivalent resistance referred to primary (R01

R01 = R1 +R’2 = R1 +sR 2

K2

Transformer ratio K = V 2

V 1 =

2002000 == 0.1

R01 = 2.5 + 0.01

0.12 = 3.5 𝝮

iii) Equivalentimpedance referred to primary (Z01)

X01 = X1 + X’2 = X1+ X 2

K2 = 4.8 +0.18

0.12 = 6.6 𝝮 __________ _________

Z01 = √ R012 +X012= √ 3.52 + 6.62= 7.47 Ω

iii. Equivalent resistance referred to secondary (R02)

R02 = R2 + R1K2 = 0.01 +2.5×0.12= 0.035 Ω

Equivalent reactance referred to secondary (X02)

X02 = X2 +X1K2= 0.018 + 4.8×0.12= 0.0665 Ω

Equivalent impedance referred to secondary (Z02)

__________ _____________

Z02 = √ R02 2 + X022 = √ 0.0352 + 0.0662 = 0.075 Ω

iv. Total Cu losses

P cu = I12 R01

I 1 = kVA rating

Primary voltage = 20× 103

2000 = 10A

P cu = 102×3.5 = 350W

Example 2 10.

Consider the transformer of Example (Fig. 3.8) with load impedance as

specified in Example 3.3. Neglecting voltage drops (resistive and leakage

reactive drops), calculate the primary current and its pf. Compare with the

current as calculated in Example 3.3

Solution

As per Eqs (3.28) and (3.29)

and

As calculated in Example 3.3

Further as calculated in Example 3.2

Hence

= 9.41-j7.17=11.83

Lagging

Compared to the primary current computed in Example 3.3 (ignoring the

exciting current) the magnitude of the current increases slightly but its pf

reduces slightly when the exciting current is taken into account.

In large size transformers the magnitude of the magnetizing current is 5%

or less than the full -load current and so its effect on primary current under

loaded conditions may even be altogether ignored without any significant loss

in accuracy. This is a usual approximation made in power system computations

involving transformers.

Example 2.11.

A 20 – kVA, 50 Hz, 2000/200 Distribution transformer has a leakage

impedance of in the high- voltage (HV) winding and 0.004+j 0.05

in the low-voltage (LV) winding. When seen from the LV side, the shunt

branch admittance is (0.002-j 0.0015) (at rated voltage and frequency). Draw

the equivalent circuit referred to (a) HV side and (b) LV side, indicating all

impedances on the circuit.

Solution

The HV side will be referred as 1 and LV side as 2.

Transformation ratio, = 10 (ratio of rated voltages; see Eq. (3.27))

(a) Equivalent circuit referred to HV side (side 1)

(Notice that in transforming admittance is

divided by a2)

The equivalent circuit is drawn in Fig. 2.11(a).

(b)Equivalent circuit referred to LV side (side 2)

The equivalent circuit is drawn in Fig. 2.11(b)

Fig. 2.11.

2.6.1 Approximate Equivalent Circuit

In constant frequency (50Hz) power transformers, approximate forms of

the exact T-circuit equivalent of the transformer are commonly used. With

reference to Fig. 2.11(c),it is immediately observed that since winding

resistances and leakage reactances are very small, even under conditions

of load. Therefore, the exciting current drawn by the magnetizing branch

would not be affected significantly by shifting it to the input terminals,

i.e. it is now excited by instead of as shown in Fig.2.12 (a). It may also be

observed that with this approximation, the current through is now excited

by instead of as shown in Fig. (a). It may also be observed that with this

approximation, the current through is now rather than . Since

is very small (less than 5% of full - load current), this approximation. The

winding resistances and reactances being in series can now be combined into

equivalent resistance and reactance of the transformer as seen from the

appropriate side (in this case side 1). Remembering that all quantities in the

equivalent circuit are referred either to the primary or secondary dash in the

referred quantities and suffixes and 2 in equivalent resistance, reactance and

impedance can be dropped as in Fig.2.12(b).

` Fig.2.12. Approximate equivalent circuits of a transformer

Hence,

(equivalent resistance) =

(equivalent resistance) =

(equivalent resistance) =

In computing voltages from the approximate equivalent circuit, the

parallel magnetizing branch has no role to play and can, therefore, be ignored as

in Fig. 2.12 (b).

The approximate equivalent circuit offers excellent computational ease

without any significant loss in the accuracy of results. Further, the equivalent

resistance and reactance as used in the approximate equivalent circuit offer an

added advantage in that these can be readily measured experimentally , while

separation of and experimentally is an intricate task and is rarely

attempted.

The approximate equivalent circuit of Fig. 2.12 (b) in which the

transformer is represented as a series impedance is found to be quite accurate

for power system modeling . In fact in some system studies, a transformer may

be represented as a mere series reactance as in Fig 2.12 (c). This is a good

approximation for large transformers which always have a negligible equivalent

resistance compared to the equivalent reactance.

The suffix 'eq' need not be carried all the time so that R and X from now

onwards will be understood to be equivalent resistance and reactance of the

transformer referred to one side of the transformer.

Example 2. 12

A 2000/200V transformer has primary resistance and reactance of 2Ω and 4Ω

respectively. The corresponding secondary values are 0.025Ω and

0.04Ω.Determine i) Equivalent resistance and reactance referred to secondary

ii) Total resistance and reactance referred to secondary iii) Equivalent

resistance and reactance referred to secondary referred to primary iv) total

resistance and reactance referred to primary.

Given data:

V1 = 2000V , V2 = 200V, R1 = 2 Ω , 1 = 4Ω

R2 = 0.025Ω, X2 = 0.04Ω

To fi nd:

1) R’1 , X’1 2) R01 3) R’2 , X’2 4) X02

Solution:

1) Transformation ratio k = V 2

V 1 =

2002000 = 0.1

Equivalent resistance of primary referred to secondary

R’1 = R1 k2

= 2 ×0.12

= 0.025 Ω

Equivalent reactance of primary referred to secondary

X’1 = X1k2

= 4 ×0.12

X’1 = 0.04Ω

2) Total resistance referred to secondary

R02 = R2 + R’1 = 0.025 + 0.02

R02 = 0.045 Ω

Total resistance referred to secondary

X02 = X2 +X’1 = 0.04+0.04 = 0.08Ω

3) Equivalent resistance of secondary referred to primary

R’2 = R 2

K2 = 0.025

0.12 = 2.5 Ω

Equivalent resistance of secondary referred to primary

X’2 = X 2

K2 = 0.04

0.12 = 4𝝮4) Total resistance referred to primary (R01)

= R1 +R’2

= 2 +2.5 = 4.5Ω

Total reactance referred to primary X01

= X1 +X’2

X01= 4+4 =8Ω

2.7 Losses in transformer

In a transformer, there exists two types of losses.

i) The core gets subjected to an alternating flux, causing core losses.

ii) The windings carry currents when transformer is loaded, causing copper losses

Core or Iron losses

Due to alternating flux set up in the magnetic core of the transformer, it undergoes a cycle of magnetization. Due to hysteresis effect there is loss of energy in this process which is called hysteresis loss.

Hysteresis loss = Kh Bm1.67 f v watts

Where Kh = Hysteresis constant depends on material. Bm = Maximum flux density

f = frequency.

v = Volume of the core.

The induced e.m.f in the core tries to set up eddy currents in the core and hence responsible for the eddy current losses. The eddy current loss is given by

Eddy current loss = K e Bm2 f 2t 2 watts / unit volume

K e = eddy current constant

t = thickness of the core

The flux in the core is almost constant as supply voltage V 1 at rated frequency f is always constant. Hence the flux density Bm in the core and hence the hysteresis and eddy current losses are constant at all the loads. Hence the core or Iron losses are also called Constant losses. The iron losses are denoted by Pi.

The iron losses are minimized by using high grade core material like silicon steel which has very low hysteresis loop and by manufacturing the core in the form of laminations.

Copper losses

The copper losses are due to the power wasted in the form of I 2R loss due to resistance of the primary and secondary windings. The copper loss depends on the magnitude of the currents flowing through the windings.

Total Cu loss = I 12 R1 + I 2

2 R2 = I 12 (R1 + R2

, ) = I 22 (R2 + R1

, )

= I 12 R1 e = I 2

2 R2 e

The copper losses are denoted by Pcu. If the current through the windings is full load current, we get copper losses at full load. If the load on transformer is half then we get copper losses at half load which are less than full load copper losses.

For transformer , VA rating is V 1 I 1 or V 2 I 2. As V 1 is constant ,

Copper losses are proportional to its kVA rating.

Pcu ∝ I 2 ∝ (kVA)2

Total losses = Iron losses + Copper losses

= Pi + Pcu

2.8 Testing of the Transformers

i) Open circuit Transformerii) Ii) Short circuit test (or) Impedance test

By using these two tests we can find,

1. Circuit constants (R0,X0,R01,X01,R02and X02)2. Core loss and full load copper loss3. Predetermine the efficiency and voltage regulation.\

These tests are convenient to perform and very economical because they provide the required information without actually loading the transformer.

iii) Load testiv) Sumpner’s test

2.8.1 OPEN -CIRCUIT TEST

Figure 2.13 shows the connection diagram for the open circuit test. The

high voltage (hv) side is left open.

Fig.2.13 open circuit test

A voltmeter V, an ammeter, A, and a wattmeter W are connected in the

low-voltage(lv) side (primary in our case) which is supplied at rated voltage and

frequency. Thus, the voltmeter V reads the rated voltage V1 of the primary.

Since the secondary is open- circuited, a very small current I0, called the no-load

current, flows in the primary. The ammeter A, therefore, reads the no-load

current I0. The power loss in the transformer is due to core loss and a very small

I2R loss in the primary. There is no I2R loss in the secondary since it is open and

I2=0. Since the no-load current I0 is very small (usually 2 to 5 percent of the full-

load primary current), the I2R loss in the primary winding can be neglected. The

core loss depends upon the flux. Since the rated voltage V1 is applied, the flux

set up by it will have a normal value so that normal core losses will occur. This

core loss is the same at all loads. Therefore the wattmeter which is connected to

measure input power reads the core loss (iron loss)P i only. The readings of the

instruments in an open-circuit test are as follows:

Ammeter reading = no-load current I0

Voltmeter reading = primary rated voltage V1

Wattmeter reading = Iron or core loss P1

From these measured values the components of the no-load equivalent

circuit can be determined.

The no-load power factor,

2.8.2 Short circuit test

The Short circuit test is useful to find

i) Full load copper lossii) Equivalent resistance and reactance referred to metering side

In the short circuit (SC) test (Fig. 2.14), usually the low-voltage side is

short-circulated by a thick conductor (or through an ammeter which may serve

an additional purpose of indicating rated load current). An ammeter, a voltmeter

and a watt meter are connected on the high-voltage side. The reasons for short-

circuiting the lv side and taking measurements on the hv side are as follows:

2.14 Short circuit test on a transformer

(1)The rated current on hv side is lower than that on lv side This current can

be safely measured with the available laboratory ammeters.

(2)Since the applied voltage is less than 5 percent of the rated voltage of the

winding, greater accuracy in the reading of the voltmeter is possible when

the hv side is used as the primary.

The high voltage is supplied at the reduced voltage from a variable voltage

supply. The supply voltage is gradually increased until full-load primary current

flows. When the rated full-load current flows in the primary winding rated full-

load current will flow in the secondary winding by transformer action.

Readings of the ammeter, voltmeter and wattmeter are noted. The ammeter

reading gives the full-load primary current. The voltmeter reading gives

the value of the primary applied voltage when full-load currents are flowing in

the primary and secondary. Since the applied voltage is low (usually about 5 to

10 percent of the normal rated supply voltage), the flux produced is low.

Also, since the core loss is nearly proportional to the square of the flux, the core

loss is so small that it can be neglected. However, the windings are carrying

normal full-load currents and therefore the input is supplying the normal full-

load copper losses. Thus the wattmeter gives the full-load copper losses .The

output voltage V2 is zero because of the short circuit. Consequently, whole of

the primary voltage is used in supplying the voltage drop in the total impedance

referred to the primary

If cos = power factor at short circuit then

The readings of the instruments in a short-circuit test are as follows:

Ammeter reading = full-load primary current,

Voltmeter reading = short circuit voltage

Wattmeter reading = full-load copper loss of the transformer

From the readings of the instruments on short-circuit test, the following

calculations can be made:

Equivalent resistance of the transformer referred to primary

Equivalent impedance referred to primary

Equivalent reactance referred to primary

With short-circuit test performed only on one side the equivalent circuit

constants referred to other side can also be calculated as follows:

Efficiency from OC and SC test

From the open circuit test , we can get core loss (Pi)of the Transformer

and short circuit test, we can get full load copper loss (Pcu).

Now, we can find the full load efficiency of the Transformer at any

power factor without actually loading the transformer.

Efficiency = full load kVA × p . f

( full load kVA × p . f )+ pi+Pcu

For any load (n)

η = (n× full load kVA)× p. f(n× full load kVA × p . f )+ p i+n2 P cu

2.8.3 BACK-TO-BACK TEST

(SUMPNER'S TEST OR REGENERATIVE TEST)

In order to determine the maximum temperature rise, it is necessary to

conduct a full-load test on a transformer. For small transformers full-load test is

conveniently possible, but for large transformers full-load test is very difficult.

A suitable load to absorb full-load power of a large transformer may not be

easily available. It will also be very expensive as a large amount of every will

be wasted in the load during the test. Large transformers can be tested for

determining the maximum temperature rise by the back-to-back test. This test is

also called the Regenerative test or Sumpner's test.

The back-to-back test on single-phase transformers requires two identical

transformers. Figure2.14, shows the circuit diagram for the back-to-back test on

two identical single-phase transformers and . The primary windings of the

two transformers are connected in parallel and supplied at rated voltage and

rated frequency. A voltmeter, an ammeter and a wattmeter are connected to the

input side as shown in Fig.2.14.

Fig.2.15. Back to Back test on two identical single phase transformer

2.8.4 TEST FOR POLARITY

Polarities can be checked by a simple test requiring only voltage

measurements with transformer on no load. In this test, rated voltage is applied

on one winding, and electrical connection is made between one terminal from

one winding and one terminal from the other, as shown in Fig.2.16. The voltage

across the two remaining terminals (one from each winding) is then measured.

If this measured voltage V' is greater than the input test voltage V, the polarity

is additive. If the measured voltage V' is smaller than the input test voltage V,

the polarity is subtractive. The terminals are then marked accordingly.

2.16. Polarity test

Example 2.13.

Obtain the approximate equivalent circuit of a given 200/2000 V, single-phase

25kVA transformer having the following test results.

O.C test : 200V,6A,350W on l.v side

S.C test : 70V, 15 A, 600W on H.V side

Given data:

Primary voltage V 1=¿¿200V, Secondary voltage V 2 = 2000V, Transformer rating

= 25kVA

Solution :

O.C test ( l.v side):

Here, the instruments are connected in the low voltage side

Primary voltage V 1 = 200V , No-load input current I 0 = 6A

No-load input power P0 = 350W

P0 = V 1 I 0 cos∅ 0

No-load input power factor cos∅ 0 = P0

V 1 I 0 =

350200× 6 = 0.2916

sin∅ 0 = 0.956

Wattfull component (working component)

I w = I 0cos∅ 0 = 6× 0.2916=1.75 A

Representing the core loss

R0 = V 1

I w = 200

1.75 = 114.28 𝝮Wattless component (magnetizing component)

I μ = I 0sin∅ 0 = 6×0.956 = 5.736A Magnetizing reactance

X 0 = V 1

I μ =

2005.736 = 34.86 𝝮

S.C test (H.V side)Here, the instruments are connected in the high voltage sideShort circuit voltage V sc = 70 V, Short circuit current I sc = 15 A, Losses W sc = 600WImpedance of transformer referred to h.v sideZ02 =

V sc

I sc = 70

15 = 4.66 𝝮R02 =

W sc

I sc2 =

600

152 = 2.66 𝝮Transformation ratio K = V 2

V 1 = 2000

200 = 10 Referred to 200V (l.v side)Z01 =

Z02

K2 = 4.66

102 = 0.0466 𝝮X 01 = √Z01

2−R012 = √0.04662−0.02662 = 0.0382 𝝮

2.9 Efficiency of a Transformer

Due to the losses in a transformer , the output power of a transformer is

less than the input power supplied.

Power output = Power input – Total losses

Power input = Power output + Total losses

= Power output + Pi + Pcu

The efficiency of any device is defined as the ratio of thr power output to

power input.

The efficiency is

𝜼 = Power outputPower input

𝜼 = Power output

Power input+Pi+Pcu

Now power output = V 2 I 2 cos∅

Where cos∅ = load power factor

The transformer supplies full load current I 2 and terminal voltage V 2.

Pcu=Copper losses on full load = I 22 R2 e

𝜼 = V 2 I 2cos∅ 2

V 2 I 2cos∅ 2+Pi+ I 22 R2 e

But V 2 I 2 = VA rating of a transformer

𝜼 = (VA rating )× cos∅

(VA rating ) ×cos∅+Pi+ I 22 R2e

% 𝜼 = (VA rating )× cos∅

(VA rating ) ×cos∅+Pi+ I 22 R2e

× 100

This is full load percentage efficiency with,

I 2 = Current Full load secondary c

But if the transformer is subjected to fractional load then using the appropriate

values of various quantities, the efficiency can be obtained.

N = Fraction by which load is less than full load = Actual loadFull load

For example, if transformer is subjected to half load then

n = Half loadFull load =

( 12)

1 = 0.5

When load changes , the load current changes by same proportion.

New I 2 = n (I 2) F.L

In general for fractional load the efficiency is

% 𝜼 = n (VA rating ) ×cos∅

n (VA rating ) ×cos∅+Pi+n2 ( Pcu ) F . L × 100

Since the transformer is a static device, there are no rotational losses such as

windage and frictional losses in a rotating machine. In a well-designed

transformer the efficiency can be as high as 99% .

Since the transformer is a static device, there are no rotational losses such as

windage and frictional losses in a rotating machine. In a well-designed

transformer the efficiency can be as high as 99% .

Example 2.14.

A 4 kVA, 200/400 V, 50 Hz, single phase transformer has equivalent

resistance referred to primary as 0.15 𝝮. Calculate,

i) The total cupper losses on full load

ii) The efficiency while supplying full load at 0.9 p.f lagging

iii) The efficiency while supplying Half load at 0.9 p.f leading

Assume total iron losses equal to 60W

Solution :

V 1 = 200 V , V 2 = 400 V, S = 4 kVA , R1 e = 0.15 𝝮 , Pi = 60 W

K = 400200 = 2

R2 e = K2 R1 e = 22 × 0.15 = 0.6 𝝮 I 2¿F.L) =

kVAV 2

= 4 × 103

400 =10 A

(i) Total copper losses on full load,

(Pcu) F.L = ( I 2¿F.L))2 R2 e = (10)2 ×0.6 = 60 W

(ii) cos∅ = 0.9 lagging and full load

% 𝜼 = n (VA rating ) ×cos∅

n (VA rating ) ×cos∅+Pi+n2 ( Pcu ) F . L × 100

% 𝜼 = 4 ×103 ×0.9

4 × 103× 0.9+60+60 ×100=¿ 96.77%

(iii) cos∅ = 0.9 leading and half load

As half load n = 0.5

(Pcu) H.L = n2 ( Pcu ) F . L = (0.5)2× 60 15 W

% 𝜼 = n (VA rating ) ×cos∅

n (VA rating ) ×cos∅+Pi+n2 ( Pcu ) F . L × 100

= 0.5 × 4 ×103 ×0.80.5 ×4 ×103 ×0.8+60+15

× 100 = 95.52%

2.9.1 CONDITION FOR MAXIMUM EFFICIENCY

The per-unit (pu) efficiency at load current I2 is

------(2.28)

------ (2.29)

Equation (1.29) shows that the efficiency varies with the load. The plot of

efficiency versus load (or load current) is shown in Fig.2.17

2.17. Efficiency Vs Load curve

It is seen that the efficiency is low at small loads and reaches a maximum value

for a certain load. The efficiency then deceases as the load is increased,

Sometimes it is desired to know under what conditions and at which value of

the load, the transformer will operate at its maximum efficiency.

At maximum efficiency

and

Since and are constants for a given load, the efficiency will be a

maximum when the denominator is a maximum.

For a minimum value of the denominator Dr

and

For a minimum Dr,

------ (2.30)

Also,

Since is positive, the expression given by Eq. (2.30) is a condition

for the minimum value of Dr, and therefore the condition for maximum value of

efficiency.

Equation (2.30) shows that the efficiency of a transformer for a given

power factor is a maximum when the variable copper loss is equal to the

constant iron (core) loss.

2.9.2 CURRENT AND kVA AT MAXIMUM EFFICIENCY

Let Full-load secondary current

Secondary current at maximum efficiency

full load VA = rated VA =

VA at maximum efficiency

For maximum efficiency, variable copper loss= constant iron loss

-----(2.31)

Equation (1.31) gives the value of current at maximum efficiency.

Multiplying both sides of Eq. (1.31) by we get

----- (2.32)

Equation (1.38.2) gives the value of VA at maximum efficiency.

Maximum efficiency

Current at maximum efficiency = (full-load current) x

At maximum efficiency

------- (2.33)

2.9.3 EFFICIENCY CURVES OF A TRANSFORMER

We have seen that maximum efficiency of a transformer occurs at the

load point where variable copper loss is equal to fixed core loss . That is

------ (2.34)

The load current at which maximum efficiency occurs is

------ (2.35)

Equations (2.34) and (2.35) enable us to predict the shape of the efficiency

curves under various load and power factor conditions. Equation (2.35) shows

that regardless of the power factor of the load, maximum efficiency occurs at

the same load (current) value as shown in Fig. 2.17.(b). The maximum

efficiency for any power factor occurs at the same load and the highest possible

efficiency occurs at unity power factor.

Fig. 2.17.(b). Effect of power factor on Efficiency

Example 2.15.

A 100 kVA, 2.2 kV/220V, 50 Hz transformer has an iron loss of 900 W and

full load copper loss of 1000 W. Determine the efficiency at full load 0.8 p.f.

Given data:

Transformer rating = 100 kVA, Primary voltage = 2.2 kV

Secondary voltage = 220 V,

Iron loss Pi = 900 W

Full load copper loss Pcufl = 1000W ,

Power factor cos∅ = 0.8

Solution:

%η = nkV A × cos∅

nkVA ×cos∅+Pi+n2 Pcufl ×100

%η = 50 ×103 ×0.8

100× 103× 0.8+900+1000 ×100

%η = 97.68%

Example 2.16.

For a 40 kVA, single phase transformer, the iron losses and full load copper

losses are 350W and 400 W respectively. Find the efficiency at unity power

factor on full load and determine the load for maximum efficiency.

%η = Transformer rating×cos∅ 2

Transformer rating× cos∅ 2+Pi+Pcu ×100

= 40 ×103 ×140 ×10 ×1+350+400

×100

%η = 98.16%

The load for maximum efficiency

= Full load kVA×√ Iron lossF . L. cu . loss

= 40×103 √ 350400

= 37.41 kVA

Example 2.17.

A 120 kVA,6000/400 V,star/star, 3 phase 50 Hz transformer has a iron loss of

1800 W. The maximum efficiency occur at ¾ full load. Find the efficiency of

the transformer at 1) full load and 0.8 2) the maximum efficiency at unity pf

Since maximum efficiency occurs at ¾ full- load, copper loss at ¾ full

l,oad equals iron loss of 1800W.

Copper loss at full load = (4/3)2 ×1800=3200 W

i) Full load output at 0.8 p.f = kVA × cos∅

= 120 × 0.8=96 kW

= 1800 +3200 = 5000W

η = Pout

Pout+total losses ×100

= 96 × 103

96 ×103+5000 ×100

η = 95.04%

ii) Maximum efficiency occurs at ¾ full load when iron loss equals

copper loss.

Total loss = 1800 ×2=3600 w

Output power at unity pf

= 34 ×120 = 90kW

Input power = 90000+3600 = 93600W

= 9000093600 ×100 = 96.15%

2.10 VOLTAGE REGULATION OF A TRANSFORMER

Majority of loads connected to the secondary of a transformer are

designed to operate at practically constant voltage. However, as the current is

taken through the transformer, the load terminal voltage changes because of the

voltage drop in the internal impedance of the transformer. The term voltage

regulation is used to identity the characteristic of the voltage change in a

transformer with loading.

The voltage regulation of a transformer is defined as the arithmetical

difference in the secondary terminal voltage between no-load and full-

rated load at a given power factor with the same value of primary

voltage for both rated load and no-load. It is expressed as either a per unit or a

percentage of the rated load voltage. Rated voltage is usually taken to be the

nameplate value.

The numerical difference between no-load and full-load voltage is called

inherent voltage regulation.

Inherent voltage regulation

------ (2.36)

Where = rated secondary terminal voltage at rated load

= no-load secondary terminal voltage with the same value of primary voltage for both rated load and no load.

The quantities in Eq. (2.36)are magnitudes, not phasors.

Per-unit voltage regulation at full load

----- (2.37)

Percent voltage regulation at full load

------ (2.38)

The conditions under which the regulation is to be figured are as follows:

(1)Rated voltage, current and frequency

(2)When regulation is stated without specific reference to the load

conditions, rated load is to be understood.

(3)Waveform o voltage should be assumed sinusoidal unless stated

otherwise.

(4)Power factor of the load should be mentioned. If the power factor is not

specified, its value is to be assumed unity.

The voltage regulation is an important measure of transformer performance.

The limits of voltage variation are specified in terms of voltage regulation. For

example, transformers in public supply systems must be so adjusted that the

voltage at the terminals of the consumers must not exceed 5%

2.10.1 VOLTAGE RGULATION IN TERMS OF PRIMARY VALUES

Per unit voltage regulation

pu voltage regulation

------ (2.39)

2.10.2 CALCULATION OF VOLTAGE REGULATION

The voltage regulation of a transformer can be calculated in terms of its

circuit parameters. The approximate equivalent circuit of the transformer

referred to the secondary is shown in Fig. 2.18

Fig.2.18. Approximate equivalent circuit of the transformer referred to secondary

By KVL,

Since depends on the power factor of the load, the regulation depends on the load power factor. In order to calculate regulation the following steps are used:

(1) Take as reference phasor

(2) Write in phasor form

For lagging power-factor

For leading power –factor

For unity power factor

Example 2.18

The primary and secondary winding resistance of a 10kVA, 6600/250V

single-phase transformer are 10𝝮 and 0.02𝝮 respectively. The equivalent

leakage reactance as referred to the primary winding is 35 𝝮.Find the full load regulation for load power factors of (i) unity (ii)0.8 lagging (iii)0.8leading

Given data:

Transformer rating = 40kVA, Primary voltageV1 = 6600V, secondary

winding resistanceR1 = 10 𝝮, secondary winding resistance R2 = 0.02 𝝮Solution:Equivalent leakage reactance as referred to the primary winding X01 = 35 𝝮Transformation ratio K = V 2

V 1 = 250

6600 = 0.0378 Equivalent resistance referred to the primary winding

R01 = R1 +R2

K2 = 10 +0.02

0.3782 = 24 𝝮Primary current I 1=kVA

V 1 = 40 ×103

6600 = 6.06 A

1. Unity power factor

% regulation = Voltage drop

V 1 ×100

Regulation = I 1 R 01

V 1 = 6.06 ×24

6600 = 2.202%

2. 0.8 power factor

% Regulation = I 1 R 01cos∅+ I 1 X 01sin∅

V 1 ×100

= (6.06×24× 0.8 )+(6.06×35×0.6)

6600 ×100

= 3.69%

3. 0.8 power factor Lead

% Regulation = I 1 R 01cos∅−I1 X 01 sin∅

V 1 ×100

= (6.06 ×24 × 0.8 )−(6.06 ×35 × 0.6)

6600 ×100

= -1.165%

Example 2.19.

A single phase transformer on full load has an impedance drop of 20 V. calculate the value of power factor when its regulation will be zero.

Given data Impedance drop I 2 Z02 = 20V, Resistance drop I 2 R02 = 10V To find:Power factor when regulation will be zero

Solution: Regulation can be zero only when power factor is leading. Then

I 2 R 02cos∅ 2−I 2 X 02sin∅ 2

V 2 = 0

I 2 X 02 = √¿¿ = √202−102 = 17.32 v

I 2 R 02 cos∅ 2−I 2 X 02 sin∅ 2 = 0

I 2 R 02 cos∅ 2=I 2 X 02 sin∅ 2

I 2 R02

I 2 X02 =

sin∅ 2

cos∅ 2 =tan∅ 2

∅ 2 = 30 °

tan∅ 2 = I 2 R02

I 2 X02 = 10

17.32 = 0.5773

∅ 2 = 30°

cos∅ 2 = cos 30° = 0.866 leading

2.11 Autotransformer (or) Variac

A transformer in which part of the winding is common to both the primary and

secondary is known as an auto transformer. The primary is electrically

connected to the secondary, as well as magnetically coupled to it.

The auto transformer differs from a conventional two winding transformer in

the way in which the primary and secondary are interrelated. In the

conventional transformer, the primary and secondary windings are completely

insulated from each other but are magnetically by a common core. In the

autotransformer the two windings primary and secondary are connected

electrically as well as magnetically, in fact a part of the single continuous

winding is common to both primary and secondary.

2.11.1 SINGLE-PHASE AUTO TRANSFORMER

A Single-phase autotransformer is a one-winding transformer in which a

part of the winding is common to both high-voltage and low-voltage sides.

Consider a single winding abc of Fig. 2.1. The terminals a and c are the

high-voltage terminals. The low-voltage terminals are b and c where b is a

suitable tapping point. The portion bc of the full winding abc is common to both

high-volt-age and low-voltage sides. The winding bc is called the common

winding and the smaller winding ab is called the series winding because it is

connected in series with the common winding.

Fig 2.1. Auto transformer (step down)

A step-down autotransformer is one in which the primary voltage is

greater than the secondary voltage. The source voltage VH is applied to the full

winding abc and the load is connected across the secondary terminals bc. This

arrangement is called the step-down autotransformer as shown in Fig. 2.1

Since the transformer windings are physically connected, a different

terminology is used for the autotransformer than for other types of transformers.

= number of turns of full winding abc

= number of turns of hv side

= number of turns of full winding bc

= number of turns of the lv side

number of turns of full winding ab

= input voltage on the hv side

= input voltage on the lv side

= input voltage on the hv side

Current in the series winding =

Current in the common winding

In an autotransformer there are two voltage ratios namely circuit voltage ratio

and winding voltage ratio. The circuit -voltage ratio

------ (2.40)

The quantity is called the transformation ratio of the autotransformer,

It is seen from Eq. (1.40) that is always greater than 1.

When the load is connected across the secondary terminals a current I

flows in the common winding bc. It has a tendency to reduce the main flux but

the primary current increases to such a value that the mmf in winding ab

neutralizes the mmf in winding bc. That is,

or ----- (2.41)

----- (2.42)

The winding-voltage ratio is

----- (2.43)

or

Since the right-hand side of Eq. (2.43) is a pure number, it follows that

the current in windings ab and cb are in phase.

By KCL at point b

------ (2.44)

------ (2.45)

Since and I are in phase

------ (1.46)

Therefore Eq (2.43) becomes

or

and----- (2.47)

From Eqs, (1.43)) and (1.47)

---- (2.48)

The induced voltages in windings ab and bc are in time phase because

they are induced by the same flux.

Let

----- (2.49)

and

or ------ (2.50)

where

= two -winding transformation ratio

Equation (2.50) shows that the transformation ratio of an autotransformer

is greater than that if the same set of windings were connected as a 2-winding

transformer.

Equations (2.40) and (2.48) show that the ratio of voltages and currents in

the windings ab and bc are the same as if turns Tab formed the primary and the

turns Tbcformed the secondary of an ordinary transformer having a ratio of

transformation of . Thus, an autotransformer may be considered as an

ordinary transformer, treating winding ab as the primary and the winding bc as

the secondary. In other words, the transformer action present in the

autotransformer takes place in the windings ab and bc.

2.12 Three phase connections of Transformer

The generation of electric power is three-phase in nature and the generated

voltage is 13.2kV, 22 kV or higher. Transmission of power is carried out at high

voltages like 132kV or 400kV. Before transmission, it is required to step-up the

voltage and for this a three phase step-up transformer is required. Similarly, at

the distribution sub-station the voltage must be stepped down and it is necessary

to reduce the voltage up to 6000V, 400 V, 230V and so on.

Here a three phase step down transformer is required. Therefore it is

economical to use three phase transformers for transmission and utilization

purposes.

Three phase transformer construction is similar to single phase transformer like

shell or core type. It is shown in fig

Three phase shell type transformer has three limbs. Here we are using only I

core. Around each limb, the primary and secondary windings are placed. Due to

this, three phase flux is produced in the primary winding.

This flux is linked with secondary winding. Depending upon the number of

turns in the secondary, the secondary voltage will be stepped up or stepped

down. The primary and secondary windings can be connected either in star or

delta.

Advantages of Three Phase Transformer

1. It occupies less space for same rating, compared to a bank of three single

phase transformers.

2. It weighs less.

3.The cost is also low

4.Easy to handle

5. It can be transported easily

6. The core is of a smaller size and hence less material is required.

Three Phase Transformer Connection

The following are the most useful and commonly used transformer connections

Star – Star connection

Delta – delta connection

Star – delta connection

Delta – star connection

2.12.1 DELTA-DELTA ( ) CONNECTION

Fig, 2.12(a) shows the connection of three identical single-phase

transformers or three identical windings on each of the primary and secondary

sides of the three-phase transformer. The secondary winding the terminals

and have same polarity. The polarity of terminal a connecting and is

the same as that of A connecting and Fig. 2.12 (b) shows the phasor

diagrams for lagging power factor . Magnetizing current and voltage drops

in impedance have been neglected. Under balanced conditions, the line currents

are times the phase (winding) currents and displaced behind the phase

currents. In the configuration the corresponding line and phase voltages

are identical in magnitude on both primary and secondary sides.

Primary Secondary

Fig.2.19. Delta-Delta connection of Transformer (00 phase shift)

The secondary line-to-line voltages and are in phase with

primary line-to-line voltages and with voltage ratios equal to the

turns ratio:

The current ratios when the magnetizing current is neglected are

It is to be noted that in Fig. 2.19 each winding is drawn along the line of

the phasor of its induced voltage. The voltage and current phasors are

determined very easily from the windings drawn in this manner.

It is seen from the phasor diagram [Fig2.19(b)] that the primary and

secondary line voltages are in phase. This connection is called 00-connection.

If the connections of the phase windings are reversed on either side, we

obtain the phase difference of 1800 between the primary and secondary systems,

Such a connection is known as 1800- connection. In Fig.2.20 delta-delta

connection with 180 phase shift is shown.

Primary Secondary

Fig. 2.20 (a)

Primary Secondary

Fig 2.20 (b) --- phasor Diagram

In Fig with 1800 phase shift is shown. Here and are

connected to form delta on secondary side as shown in Fig. 2.20(a). Fig 2.20(b)

shows the phasor diagrams. It is seen that the secondary voltages are in phase

opposition to the primary voltages.

The transformers has no phase shift associated with it, an no

problem with unbalanced loads or harmonics.

Advantages of Transformation

1. The connection is satisfactory for both balanced and unbalanced

loading.

2. If a third harmonic is present, it circulates in the closed path and therefore

does not appear in the output voltage wave.

3. If one transformer fails, the reaming two transformers will continue to

supply three-phase power. This is called open-delta (or V-V) connection.

The operation of V-V connection is discussed later in this chapter.

However, the connection has the disadvantages that there is no star point

(neutral point) available. A connection is useful when neither primary nor

secondary requires a neutral and the voltages are low and moderate.

2.12.2 STAR-STAR (Y-Y) CONNECTION

Fig.2.21 tar-tar connection of transformer

(a) 00 phase shift (b) 1800phase shift

Fig. 2.21 shows the Y-Y connection of three identical single-phase

transformers or the three identical windings on each of the primary and

secondary sides of the three-phase transformer. The phasor diagrams are drawn

in the similar manner as done in connection. The phase current is equal to

the line current and they are in phase. The line voltage is times the phase

voltage. There is a phase separation of 300 between line and phase voltages.

Fig.2.21(a) shows the Y-Y connection for 00 phase shift and in Fig.2.21(b) there

is a phase shift of 1800 between primary and secondary systems.

For ideal transformers the voltage ratios are

and current ratios are

The Y-Y connection has two very serious problems:

1. If the neutral is not provided, the phase voltage tend to become

severely unbalance when the load is unbalanced. Therefore, the Y-Y connection

is not satisfactory for unbalanced loading in absence of a neutral connection.

2. The magnetizing current of any transformer is verynonsinusoidal and

contains a very large third harmonic, which is necessary to overcome saturation

in order to produce a sinusoidal flux. In a balanced three-phase system, the third

harmonic components in the magnetising currents of three primary windings are

equal in magnitude and in phase with each other. Therefore they will be directly

additive. Their sum at the neutral of a star connection is not zero. Since thre is

no path for these current components in an ungrounded star connection these

components will distort the flux wave which will produce a voltage having a

third harmonic in each of the transformers, both on the primary and secondary

sides. The third harmonic component of induced voltage may be nearly as large

as the fundamental voltage. When this voltage is added to the fundamental, the

peak voltage is nearly two times the normal value.

Both the unbalance and third harmonic problems of the Y-Y connection

can be solved using one of the following methods:

1. Solid grounding of neutrals. By providing a solid (low impedance)

connection between the star point of the primary transformer and the neutral

point of the alternator allows third harmonic currents to flow in the neutral

instead of building up large voltages. The triple-frequency currents in the

neutral wire may interfere with nearby telephone and other communication

circuits. The neutral also provides a return path for unbalanced currents due to

unbalanced loads.

2. Providing tertiary windings. When it is necessary to have a Y-Y

connection without neutral, each transformer is provided with a third winding in

addition to primary and secondary. The third winding is called tertiary. It is

connected in delta. This connection is called Y- -Y connection.

2.12.3 DELTA-STAR ( -Y) CONNECTION

A -Y connection of 3-phase transformers is shown in Fig.2.22(a). In

-Y connection, the primary line voltage is equal to the primary phase voltage

. The relationship between secondary voltages is .

Therefore, the line-to-line voltage ratio of this connection is

But

Fig.2.22. (a) Delta – star connection of transformer , phase shift 300 lead

(b) Phasor Diagram

Fig. 2.22(b) shows the phasor diagrams for the connection

supplying a balanced load at power factor logging. It is seen from the

phasor diagram that the secondary phase voltage leads the primary phase

voltage by 300. Similarly, leads by 300 and leads by 300. This

is also the phase relationship between the respective line-to-lien voltages. This

connection is called +300connection.

By reversing the connections on either side, the secondary system

voltage can be made to lag the primary system by 300 as shown in Fig.2.23. This

connection is called -300connection.

Fig. 2.23 . Delta – star connection of transformer (phase shift 300 lag).

2.12.4 STAR-DELTA CONNECTION

The connection of three-phase transformers is shown in Fig.

2.24. In this connection, the primary line voltage is equal to times the

primary phase voltage . The secondary line voltage is equal to the

secondary phase voltage . The voltage ratio of each phase is

Therefore line-to-lien voltage ratio of a connection is

Fig. 2.24. Star - Delta connection of transformer ( phase shift 300 lead).

The phasor diagrams can be drawn with the help of winding diagrams.

There is a phase shift of 300 lead between respective line-to-line voltages.

Similarly, a phase shift of 300 lead exists between respective phase voltages.

This connection is called + 300 connection.

Fig 2.25 shows the star-delta connection of transformer for a phase shift

of 300 lag. This connection is known as -300connection.

It is to be noted that a connection is simply obtained by

interchanging the primary and secondary roles in connection.

Fig.2.25 . Star - Delta connection of transformer (phase shift 300 lag).

The connection or connection has no problem with

unbalanced loads and third harmonics. The delta connection assures balanced

phase voltages on the Y side and provides a path for the circulation of the third

harmonics and their multiples without the use of a neutral wire.

Example 2.20

Determine the number of turns per phase in each winding of a 2-phase

transformer with a ratio of 20,000/2000 V at 50Hz. The high-voltage winding

is delta connected and the low voltage winding is star connected. Each core has

a gross section of 500cm2. Assume a flux density of about 1.2 Wb/m2.

SOLUTION.

Since the primary winding (hv) is connected in delta, the phase voltage is equal

to the line voltage.

and

The low voltage secondary winding is connected in star. Hence the

phase voltage is equal to times the line voltage.

Example 2.21.

An 11000/440 V, 50 Hz, 3-phase transformer is delta connected on the hv side

and the lv windings are start connected. There are to be 12V per turn and the

flux density is not to exceed 1.2 .Calculate the number of turns per phase

on each winding and the ne iron cross-sectional area of the core.

SOLUTION. Induced voltage in the primary per turn

Since the hv side is delta connected, phase voltage = line voltage

Also,

2.12.5 OPEN-DELTA OR V-V CONNECTION

If one transformer of a system is damaged or accidently opened,

the system will continue to supply 3-phase power. If this defective transformer

is disconnected and removed, as shown in Fig.2.1, the remaining two

transformers continue to function as a 3-phase bank with rating reduced to

about 58per cent of that of the original bank. This is known as opern-delta

or V-V system. Thus, in the open-delta system, two instead of three single-

phase transformers are used for 3-phase operation. Let and be the

applied voltages of the primary. The voltage induced in transformer secondary

or lv winding II is . There is no winding between points a and cm, but there is

a potential difference between a and c. This voltage may be found by applying

KVL around closed path made up of points a,b and c. Thus,

------ (2.51)

------ (2.52)

Let

and

where is the magnitude of the line voltage on the primary side.

If the leakage impedances of the transformers are negligible, then

Vbc = Vs

where is the magnitude of the secondary voltage.

Fig.2.26. open-delta ( or v-v connection )

a) common physical arrangement b) schematic diagram

Substituting the values of and in in Eq. (2.51)

= =

It is seen that is equal in magnitude to the secondary transformer

voltage and 1200 apart in time from both of them. Thus balanced 3-phase line

voltages applied to the V-V primaries produce balanced 3-phased voltages on

the secondary side if leakage impedances are negligible.

If appears that removal of one transformer would permit the remaining

two transformers to carry two-thirds (66.7%) of the load kVA. This, however, is

not the case. If and are the rated secondary voltage and rated secondary

current of the transformers, the line current to the load of a closed delta system

is

Closed delta load VA

line voltage line current

When one transformer is removed, the transformer becomes open delta (V-

V) connected transformer. The line is in series with the windings of the

transformers and therefore secondary line current is equal to the rated

secondary current of the transformer. The VA load that can be carried by the

open delta bank without exceeding the ratings of the transformers is

Thus, is seen that the load that can be carried by the open-delta bank

without exceeding the ratings of the transformers is 57.7 per cent of the original

load carried by the bank.

of

Also in open-delta system

Thus, the VA supplied by each transformer in a V-V system is not half

(50%) of the to the total VA but it is 57.7 per cent.

If three transformers in are supplying rated load and as soon

as it becomes a V-V transformers, the current in each phase winding is

increased by times. That is, full line current flows in each of the two phase

windings of the transformers. Thus each transformer in the V-V system is

overloaded by 73.2 per cent.

Summary

1. Load in VA transformer in V-V bank

original load in bank. 2. Per cent rated VA load carried by each transformer in V-V bank,

3. Total VA rating of V-V bank

rating per transformer in V- V bank

4. Ratio of VA ratings

5. Per cent increase in VA load on each transformer when one transformer is

removed

Power Supplied by Open-Delta System

When a V-V bank of two transformers supplies a balanced 3-phase load

operating at a power factor the angle between the line voltage and line

current in one transformer is while the angle between the line voltage

and line current in the order transformer is . Therefore one transformer

operates at a power factor of cos and the other of cos and the

powers supplied by the two transformers are

The total power supplied by the transformers

At unity power factor of the load,

Therefore the power supplied by each transformer is

APPLICATIONS OF OPEN-DELTA SYSTEM

The open-delta system is used in one of the following circumstances:

1. As a temporary measure when one transformer of a system damaged

and removed for repair and maintenance.

2. To provide service in a few development area where the full growth of load

may require several years. In such cases a V-V system is installed in the

initial stage. This reduces the initial cost. Whenever the need arises at a

future date to accommodate the growth in the power demand, a third

transformer is added for operation. The addition of one transformer

increase the capacity of the total bank by 73.2 per cent.

3. To supply a combination of large single-phase and smaller 3-phase loads

Example 2.22.

A 400 kVA load at 0.7 pf lagging is supplied by three transformers

connected in Each of the transformers is rated at 200 kVA,

2300/230V. If one defective transformer is removed from service, calculate for

the V-V Connection:

a) the kVA load carried by each transformer.

b) Percent rated load carried by each transformer.

c) total kVA ratings of the transformer bank in V-V.

d) ratio of V-V bank to bank transformer ratings.

e) percent increase in load on each transformer when one transformer is

removed.

SOLUTION. (a) Load in kVA per transformer in V-V bank

original load in bank

400 = 230.9kVA

(b) Per cent rated load carried by each transformer

per cent

(c) Total kVA rating of the V-V bank

rating per transformer in

(d)

pu = 57.7%

(e) Original load kVA per transformer in

total load kVA

400 = 133.3kVA

Percent increase in load on each transformer when one transformer is removed.

Example 2.23.

A 900-kVA load is supplied by three transformers connected in delta-delta.

The primaries are connected to a 2300-V supply line, while the secondaries are

connected to a 230-V load. If one transformer is removed for repair, what load

can the remaining two transformers supply without over loading? What are the

currents in the high-and low voltage sides of the transformer windings when

connected in open delta?

Solution. Delta-delta operation

If and are the line voltage and line current respectively on hv side,

Transformer current per phase on hv side

line current = 225.9=130.4A

Transformer current per phase on lv side

Open-delta operation

When one transformer is removed from chosed delta, the currents through

the phase windings of the transformers should not exceed the rated currents to

avoid overloading. Therefore the permitted phase winding currents on hv and lv

sides are 130.4 A and 1304 A respectively. For open delta the line is in series

with the windings of the transformer and therefore, the secondary the line

current is equal to the rated secondary current of the transformer. Therefore the

VA load that can be carried by the open-delta bank without overloading is

Alternatively

It is seen that the kVA supplied by the V-V bank is times the original

closed-delta kVA.

Example 2.24.

Two 40-kVA single-phase transformers are connected in open-delta to supply

a 230-V balanced 3-phase load.

(a) What is the total load that can be supplied without overloading either

transformer?

(b)When the delta is closed by the addition of a third 40-kVA transformer,

what total load can now be supplied?

(c) Percent increase in load.

SOLUTION (a) The rated secondary transformer current

This is also the load line current. Therefore the load kVA is

b) When the delta is closed by the addition of a third 40-kVA transformer, the

bank will operate at full capacity of the individual transformers. Therfore

the load kVA supplied by the bank is

(c) Percent increase in load kVA =

2.12.6 SCOTT THREE PHASE/TWO-PHASE CONNECTION

The scott connection is the most common method of connecting two

single phase transformers to perform the 3-phase to two-phase conversion and

vice-versa. The two transformers are connected electrically but not

magnetically. One transformer is called main transformer and the other is

known as auxiliary or teaser transformer. Fig.2.27 shows the Scott transformer

connection. The main transformer is centre-tapped at D and is connected across

the lines B and C of the 3-phase side. It has primary BC and secondary

.The teaser transformer is connected between the line terminal A and the centre

tapping D. It has primary AD and secondary

Frequently identical interchangeable transformers are used forth Scott

connection, in which each transformer has a primary winding of turns and is

provided with tappings at 0.289 0.5 and 0.866

Fig. 2.27. scott connection of transformer

2.13 PARALLEL OPERATION OF TRANSFORMERS

Transformers are said to be connected in parallel when their primary

windings are connected to a common voltage supply and their secondary

windings are connected to a common load.

When the load outgrows the capacity of an existing transformer, it may be

economical to install another one in parallel with it rather than replacing it with

a single larger unit. Also, sometimes in a new installation, two units in parallel,

though more expensive, may be preferred over a single unit for reasons of

reliability - half the load can be supplied with one unit out. Further, the cost of

maintaining a spare is less with two units in parallel. However, when spare units

are maintained at a central location to serve transformer installations in a certain

region, single-unit installations would be preferred. It is, therefore, seen that

parallel operation of the transformer is quite important and desirable under

certain circumstances.

The satisfactory and successful operation of transformers connected in

parallel on both sides requires that they fulfill the following conditions:

i) The transformers must be connected properly as far as their polarities

are concerned so that the net voltage around the local loop is zero. A

wrong polarity connection results in a dead short circuit.

ii) Three-phase transformers must have zero relative phase displacement

on the secondary sides and must be connected in a proper phase

sequence. Only the transformers of the same phase group can be

paralleled. For example, Y/Y and Y/ transformers cannot be

paralleled as their secondary voltages will have a phase difference of

300. Transformers with +300 and - 300 phase shift can, however, be

paralleled by reversing the phase-sequence of one of them.

iii) The transformers must have the same voltage-ratio to avoid no-load

circulating current when transformers are in parallel on both primary

and secondary sides. Since the leakage impedance is low, even a small

voltage difference can give rise to considerable no-load circulating

current and extra I2 R loss.

iv) There should exist only a limited disparity in the pet-unit impedances

(one their own bases) of the transformers. The currents carried by two

transformers (also their kVA loadings) are proportional to their ratings

if their ohmic impedances (or their pu impedances on a common base)

are inversely proportional to their ratings or their per unit impedances

on their own ratings are equal. The ratio of equivalent leakage

reactance to equivalent resistance should be the same for all the

transformers. A difference in this ratio results in a divergence of the

phase angle of the two currents, so that one transformer will be

operating with a higher, and the other with a lower power factor than

that of the total output; as a result, the given active load is not

proportionally shared by them.

REASONS FOR PARALLEL OPERATION

The main reasons for operating transformers in parallel are as follows:

(a) For large loads it may be impracticable or uneconomical to have a single

large transformer.

(b) In substations the total load required may be supplied by an appropriate

number of transformers of standard size. This reduces the spare capacity

of the substation.

(c) There is a scope of future expansion of a substation to supply a load

beyond the capacity of the transformers already installed.

(d) If there is a breakdown of a transformer in a system of transformers

connected in parallel, there is no interruption of power supply for

essential services. Similarly, when a transformer is taken out of service

for its maintenance and inspection, the continuity of supply is maintained.

SINGLE-PHASE TRANSFORMERS IN PARALLEL

Fig. 2.28 shows the circuit diagram of two transformers A and B in

parallel.

Let a1 = turns ratio of transformer A

a2 = turns ratio of transformer B

ZA = Equivalent impedance of transformer A referred to secondary

ZB = Equivalent impedance of transformer B referred to secondary

ZL = Load impedance across the secondary

IA = Current supplied to the load by the secondary of transformer A

IB = Current supplied to the load by the secondary of transformer B

VL = Load secondary voltage

IL = Load Current

Fig. 2.28 . Two single phase transformer in parallel

By KCL, ----- (2.52)

By KVL, ----- (2.53)

----- (2.54)

Solving Eqs. (2.53) and (2.54) we get

----- (2.55)

----- (2.56)

Each of the these currents has two components; the first component represents

the transformer's share of the load current and the second component is a

circulating current in the secondary windings. Circulating currents have the

following undesirable effects:

(a) They increase the copper loss.

(b)They overload one transformer and reduce the permissible load kVA.

Equal Voltage Ratios

In order to eliminate circulating currents, the voltage ratios must be

identicated. That is

Under this condition,

------ (2.57)

------(2.58)

----- (2.59)

Equation (2.41.8) shows that the transformer currents are inversely proportional

to the transformer impedances.

Also, ------ (2.60)

Multiplying Eqs.(2.57) and (2.58) by common load voltage VL Changes

currents into volt amperes. Hence writing

= total load VA

= VA of transformer A

and of transformer B, we get

----- (2.61)

----- (2.62)

From Eqs. (2.61) and (2.62)

------ (2.63)

Thus, the volt ampere load on each transformer is inversely proportional

to its ohmic impedance. Hence to share the load in proportion to their ratings,

the transformers should have ohmic impedances which are inversely

proportional to their ratings. In terms of per-unit values, the above statement

may be expressed as follows:

The transformers should have equal per-unit impedances in order to share

the load in proportion to their volt ampere ratings.

Equation (2.60) shows that for efficient parallel operation of two

transformers, the potential differences at full load across the transformers'

internal impedances should be equal. This condition ensures that the load

sharing between them is according to the rating of each transformer. If the per

unit equivalent impedances are not equal, the transformers will not share the

load in proportion to their kVA ratings, so that the overall rating of the

transformer bank is reduced.

It is often convenient to specify the percentage or per-unit values of

resistance and leakage reactance for a transformer. In these circumstances Eqs.

(2.57) to (2.63) do not, in general, apply directly.

We have, per unit impedance =

----- (2.64)

Then

Where is the rated voltage and is the rated VA. Then

----- (2.65)

Since

Therefore ----- (2.66)

Equation (2.66) shows that, if the transformers are to share the load in

proportion to their ratings, their per-unit impedances must have the same

magnitude and that, if the transformers are to operate at the same power factor,

their per-unit impedances must have the same phase angle.

Since

If the impedance angles of both transformers are equal, that is,

then and will be in phase and the load current will be the arithmetic

sum of and .That is

But if that is, the magnitudes of the currents remain

inversely proportional to the magnitudes of the impedances, but and will

not be in phase. At rated load, the load current will be the phasor sum of

and . Since the phasor sum of and is less than the arithmetic sum of

and the load kVA (= ) is less than the sum of the transformer kVAs.

Finally both the transformers should have the same polarity while

connecting them in parallel..

Example 2.25

A 600-kVA, single-phase transformer with 0.012 pu resistance and 0.06pu

reactance is connected in parallel with a 300-kVA transformer with 0.014pu

resistance and 0.045pu reactance to share a load of 800kVA at 0.8pf lagging.

Find how they share the load (a) when both the secondary voltages are 440V

and (b) when the open-circuit secondary voltages are respectively 445V and

455V.

Solution

(a) The pu impedance expressed on a common base of 600 kVA are

The load is

From Eqs (3.86) and (3.87)

It may be noted that the transformers are not loaded in proportion to

their ratings. At a total load of 800kVA, the 300kVA transformer operates with

5% overload because of its pu impedance (on common kVA base) being less

than twice that of the 600kVA transformer.

The maximum kVA load the two transformers can feed in parallel

without any one of them getting overloaded can now be determined. From

above it is observed that the 300kVA transformer will be the first to reach its

full-load as the total load is increased. In terms of magnitudes

while the sum of the ratings of the two transformers is 900kVA. This is

consequence of the fact that the transformer impedances (on common base) are

not in the inverse ratio of their ratings.

(b) In this case it is more convenient to work with actual ohmic impedances,

Calculating the impedances referred to secondary

The load impedance must also be estimated. Assuming an output voltage on

load of 440V.

From Eqs (3.98) and (3.99)

The Corresponding kVAs are

The total output power will be

This is about 3% less than required by the load because of the

assumption of the value of the output voltage in order to calculate the load

impedance.

The secondary circulating current on no-load is

Which corresponds to about 88kVA and a considerable waste as copper-loss.

2.14 TAP CHANGING TRANSFORMERS

Voltage variation in power systems is a normal phenomenon owing to the

rapid growth of industries and distribution network. System voltage control is

therefore essential for:

i) Adjustment of consumers' terminal voltage within prescribed limits.

ii) Control of real and reactive power flow in the network.

iii) Periodical adjustment (1-10%) to check off-set load variations.

Adjustment is normally carried out by off-circuit tap changing, the

common range being 5% in 2.5% steps. Daily and short-time control or

adjustment is carried out by means of on-load tap changing gear.

Besides the above, tapping are also provided for one of the following

purposes.

i) For varying the secondary voltage.

ii) For maintaining the secondary voltage constant with a varying

primary voltage.

iii) For providing an auxiliary secondary voltage for a special purpose,

such as lighting.

iv) For providing a low voltage for starting rotating machines.

v) For providing a neutral point, e.g. for earthing.

The principal tapping is one to which the rating of the winding is related. A

positive tapping means more, and a negative tapping implies less turns than

those of the principal tap. Tap changing may be achieved in one of the three

conditions, viz.

i) Voltage variation with constant flux and constant voltage turn

ii) With varying flux

iii) a mix of (i) and (ii). In (i) the percentage tapping range is same as the

voltage variation.

Location

The taps mage be placed on the primary or secondary side which partly

depends on construction. If tappings are near the line ends, fewer bushing

insulators are required.

If the tappings are placed near the neutral ends, the phase-to-phase

insulation conditions are eased.

For achieving large voltage variation, tappings should be placed near the

centres of the phase windings to reduce magnetic asymmetry. However, this

arrangement cannot be put on LV windings placed next to the core (as in core

type transformer) because of accessibility and insulation consideration. The HV

winding placed outside the LW winding is easily accessible and can, thus, be

tapped easily.

It is not possible to tap other than an integral number of truns and this

may not be feasible with LV side tappings. For example 250V phase winding

with 15 V/turn cannot be tapped closer than 5%. It is therefore essential to tap

the HV windings which is advantageous in a step-down transformer.

Fig. 2.34 Location of transformer tappings

Some of the methods of locating tappings are depicted in Fig.2.34(a) and (b).

Axial mmf unbalance is minimized by thinning out the LV winding or by

arranging parts of the winding more symmetrically. For very large tapping

ranges a special tapping coil may be employed.

Tap changed causes changes in leakage reactance, core loss, loss and

perhaps some problems in parallel operation of dissimilar transformers.

2.14.1 No-load (off-load or off-circuit) tap changing

The cheapest method of changing the turn ratio of a transformer is the use

of off-circuit tap changer. As the name indicates, it is required to deenergize the

transformer before changing the tap. A simple no-load tap changer is shown in

Fig.2.35. It has eight studs marked one to eight. The winding is tapped at eight

points. The face plate carrying the suitable studs can be mounted at a convenient

place on the transformer such as upper yoke or located near the tapped positions

on the windings. The movable contact arm A may be rotated by handwheel

mounted externally on the tank.

If the winding is tapped at 2% intervals, then as the rotatable arm A is

moved over to studs 1,2;2,3;……6,7;7,8 the winding in circuit reduces

progressively by it from 100% with arm at studs (1,2) to 88% at studs (7,8).

The stop F which fixes the final position of the arm A prevents further

anticlockwise rotation so that stud 1 and 8 cannot be bridged by the arm.

Adjustment of tap setting is carried out with transformer deenergized. For

example, for 94% tap the arm is brought in position to bridge studs 4 and 5. The

transformer can then be switched on.

Fig. 2.35. No- load tap changer

To prevent unauthorized operation of an off-circuit tap changer, a mechanical

lick is provided. Further, to prevent inadvertent operation, an electromagnetic

latching device or microswitch is provided to open the circuit breaker so as to

deenergize the transformer as soon as the tap changer handle is moved; well

before the contact of the arm with the stud (with which it was in contact) opens.

2.14.2 On-load Tap Changing

On-load tap changers are used to change the turn ratio of transformer to

regulate system voltage while the transformer is delivering load. With the

introduction of on-load tap changer, the operating efficiency of electrical system

gets considerably improved. Nowadays almost all the large power transformers

are fitted with on-load tap changer. During the operation of an on-load tap

changer the main circuit should not be opened to prevent (dangerous) sparking

and not part of the tapped winding should get short-circuited. All forms of on-

load tap changing circuits are provided with an impedance, which is introduced

to limit short-circuit current during the tap changing operation. The impedance

can either be a resistor or centre-tapped reactor. The on-load tap changers can in

general be classified as resistor or reactor type. In modern designs the current

limiting is almost invariably carried out by a pair of resistors.

On-load tap changing gear with resistor transition, in which one winding

tap is changed over for each operating position, is depicted in Fig.3.69. The

figure also shows the sequence of operations during the transition from one tap

to the next (adjoining) (in this case from tap 4 to tap 5). Back-up main

contractors are provided which short-circuit the resistor for normal operation.

To ensure that the transition once started gets completed, an energy

(usually a spring device) storage is provided which acts even if the auxiliary

power supply happens to fail. In resistor - aided tap changing the current break

is made easier by the fact that the short- circuit resistor causes the current to be

opened to have unity power factor.

Fig. 2.36 Simple switching sequence for on-load tap changing

On-load tap changer control gear can be from simple push-button

initiation to complex automatic control of several transformers operating in

parallel. The aim is to maintain a given voltage level within a specified

tolerance or to raise it with load to compensate for the transmission line voltage

drop. The main components are an automatic voltage regulator, a time delay

relay prevents unwanted initiation of a tap change by a small transient voltage

fluctuation. It may be set for a delay unto 1 min.

At present tap changers are available for the highest insulation level of

1475kV (peak) impulse and 630kV power frequency voltage. Efforts are

underway to develop tap changers suitable for still higher insulation levels.

More compact tap changers with high reliability and performance are being

made by employing vacuum switches in the diverter switch. Also, now

thyristorized tap changers are available for special applications where a large

number of operations are desired.

2 MARK Questions

1. What is the function of Transformer?

Transformer are energy converting devices, converting AC electrical

energy with one level of voltage and current , to AC electrical energy with

another level of voltage and current.

2. What is meant by step –up and step- down Transformer?

When transformation ratio K>1, The voltage available in the secondary U2

will be greater than the input voltage U1, the Transformer is called as

STEP-UP Transformer.

When transformation ratio K<1, The secondary voltage available in the load

U2 will be less than the input voltage U1, the Transformer is called as

STEP-DOWN Transformer.

3.Distinguish between step –up and step- down Transformer.

In step -up Transformer, the output voltage is greater than the input voltage

with K>1. In step - down Transformer, the output voltage is less than the

input voltage with K<1.

4. List out the general applications of Transformer,

1.Stepping – up of voltage 2. Stepping – down of voltage 3. Instrument

extension 4.Electrical isolation 5.Impedance mat6ching 6.Link between AC

and DC system.

5. Enumerate the various kinds of Transformers.

By numbers of phases – single and the three phases

By relative position of winding and core – core and shell

type

By number of windings / phase – One winding (Auto

transformer), Two winding and three winding .

By power rating – small, medium and high capacity.

By voltage rating – low, medium and high voltage.

By methods of cooling – AN,AB,ON,OB,OF,OW< OFW

etc.

By sensitive condition – normal. Instrument, isolation,

welding etc.

6. How are Transformers classified according to number of windings.

One winding per phase- Auto transformer – part of single winding act as

primary and part of it act as secondary.

Two windings per phase – one primary winding and one secondary

winding is provided in each phase. The KVA rating of both windings

are equal.

Three winding Transformer per phase – one primary winding and two

secondary windings at two different voltage ratings are provided in each

phase.The KVA rating of primary winding is the sum of KVA ratings of

the two secondary windings.

7.Why the central limb of shell type single – phase Transformer has al most

twice the cross section as that of outer limbs?

Single- phase shell type Transformer has two magnetic circuits.The total flux Φ

passing through the central limb divide into 2 halves in the yoke, one towards

left and the other towards right and complete their path via the outer limbs and

the other yoke. As the centre limb carry twice the flux, its cross section is

almost twice as that of outer limbs.

8. Distinguish between single – phase core and shell – type Transformer.

A single – phase core type Transformer has 2 limbs, 2 yokes and one window.

Around each limb half of LV coil and half of HV coil are housed. A single –

phase shell type Transformer has 3 limbs, 1 centre limb and 2 outer limbs, 2

yokes and 2 windows. The entire LV and HV coils are housed around the centre

limb.

9.What is the purpose of constructing Transformer core by silicon content steel

laminations?

Steel and steel alloy offers less resistance to the path of magnetic field. Addition

of 3 to 5 percent silicon with steel reduces Hysteresis loss incurred in the core.

Laminations reduces eddy current loss.

10. Why is Transformer core laminated?

Laminating the core reduces the eddy –current loss incurred in the core.

11. What is the need for stepped core in Transformer?

In Transformers of medium and large capacity, only circular coils are used

which are mechanically stronger. It is more economical to house circular shape

coils around stepped core. For the same area of iron core, required by magnetic

flux the diameter of the circumscribing circle get reduced with increase in

number of steps. Reduction of diameter of circle reduces the length of mean

turn of the winding around the core which ultimately reduce the volume of the

copper conductor and also the resistance of the coil.

12. What is the function of Bucholtz relay

Automatic monitoring of oil level is provided by Bucholtz relay fitted along

the tube connecting conservation tank and Transformer tank. If the oil level

comes down below this level without notice, the relay will initially given an

alarm and if not attende, the relay will cause tripping off power supply to the

Transformer.

13. What are the advantages of 3 – phase Transformer over 3 numbers of single

– phase Transformers ?

A single 3-phase transformer occupies less space than installing 3

numbers of 1- phase Transformers of equal capacity.

Cost of single 3 – phase Transformer is less than 3 numbers of 1 – phase

Transformers.

It is enough to install only one Transformer.

14. What is staggering in the construction of transformer ?

In transformer, the joints in the alternate layers are staggered in order to avoid

the presence of narrow gaps right through the cross- section of the core.

15.Classify the transformer according to the construction.

1. Core type Transformer

2. Shell type Transformer

3. Berry type Transformer

16.What are the two components in transformer’s no load current?

1. Active working component (Iw)

2. Reactive or magnetizing component (Iu)

17 .Why Transformer rating is expressed in terms of KVA?

Copper loss depends on current and iron loss depends on voltage. Hence the

total loss in a transformer depends upon volt-ampere (VA) only and not only on

the phase angle between voltage and current i.e it is independent of load power

factor. That is why the rating of a transformer is given in KVA and not in KW.

18. Give the conditions to be satisfied for parallel operation of transformer.

1. The voltages rating of both primaries and secondaries must be the same. i.e

transformation ratios are same.

2. The transformer polarities must be connected properly, otherwise dead short

circuit ocuurs

3. The ratio of the equivalent resistance to equivalent reactance of the

transformers should be equal.

4. The equivalent impedances should be inversely proportional to the respective

KVA ratings of the transformer.

19. State the condition for maximum efficiency of a transformer. Then what is

the corresponding output current?

Iron loss = copper loss or constant loss = Variable loss

Hence efficiency of a transformer will be maximum when copper losses are

equal to iron losses.

The load current corresponding to maximum efficiency is given by

I2 √ Pi/R02

20. Define “all day efficiency” of a transformer.

The ratio of output kwh to input in kwh of a transformer over a 24 hour period

Is k known as all-day efficiency.

ήall -day = Kwh output∈24 hoursKw hinput∈24 hours

21.Define regulation and efficiency of a transformer.

The regulation of a transformer is defined as reduction in magnitude of the

terminal voltage due to load with respect to the no-load terminal voltage.

% regulation =¿V 2 on no−load∨−¿V 2 w hen loaded

¿V 2 on no−load∨¿¿

22. Name the factors on which hysteresis loss depends

1. frequency

2. Volume of the core

3. Maximum flux density

23. Why the open circuit test on a transformer is conducted at rated voltage?

The open circuit test on a transformer is conducted at rated voltage because

core loss depends upon the voltage. This open circuit test gives onl;y core loss

or iron loss of the transformer.

24. Explain why the wattmeter in OC test on a transformer reads core loss and

that in SC test reads copper loss at full load..

In open circuit test , the transformer secondary is open. The transformer is

operated at rated voltage . The iron loss occur in transformer core. Therefore the

wattmeter in OC test on transformer reads core loss only.

In short circuit test, the transformer secondary winding is short circuited. Here

the transformer os operate at rated current. Here the input voltage is low. The full

load current depends upon the copper loss. Therefore the wattmeter in SC test on

transformer reads copper loss only.

25. Why is the short circuit test on a transformer performed on HV side?

The short circuit test normally conduct on transformer HV side and LV side is

short circuited, because the high voltage side, the current rating is low. So we

have to use normally available meter range.

26. What is the use of load test of a transformer?

Load test is helpful to determine the following

1. Efficiency of the transformer

2. Regulation of the transformer

27. What are the advantages of Sumpner’s test?

1. The power required to carry out the test is small.

2. The transformers are tested under full-load conditions

3. The iron-losses and full load copper losses are measured simultaneously.

4. The temperature rise of the transformers can be noted.

28.What is the use of open circuit test and short circuit test ina transformer?

The open circuit test is useful to find

i. No load loss (or) core loss

ii . No load current

iii. R0 and X0

The short circuit test is useful to find

i. Full load copper loss

ii. Equivalent resistance and reactance referred to any side.

By using above two test, we can predetermine the

i. Efficiency of the transformer

ii. Regulation of the transformer

30. How can the iron loss be minimized in a transformer?

The iron loss in a transformer is made up of a hyteresis loss and eddy current

loss. Hysteresis loss can be minimized by using steel of high silicon content for

the transformer core. The eddy current loss can be minimized by using very

thin laminations of transformer core.

31. What are the advantages of OC and SC test of the transformer?

1. The power required to carry out these test is very small as compared to the

full load output of the transformer.

2. These tests enable us to determine the efficiency of the transformer

accurately at any load and power factor without actually loading the

transformer.

3. The short circuit test is used to determine R01 and X01 ( or R02 and X02).

By using this data , we can find out voltage drop and voltage regulation of the

transformer.

32. Mention the difference between core and shell type transformers.

In core type , the windings surround the core considerably and in shell type the

core surround the winding.

33. What is the purpose of laminating the core in a transformers ?

To reduce eddy current loss.

34. Give the emf equation of a transformer and define each term Emf induced in

primary coil E1 = 4.44 fΦ mN1 volt

Emf induced in secondary coil E2 = 4.44fΦ mN2 volt

Where f is the frequency of AC input

Φ m is the maximum value of flux in the core

N1, N2 are the number of primary and secondary turns.

35. Does the transformer draw any current when secondary is open ? Why ?

Yes,it (primary) will draw the current from the main supply in order to

magnetise the core and to supply iron and copper losses on no load . There will

not be any current in the secondary since secondary is open.

36. Define voltage regulation of a transformer (April –98)

When a transformer is loaded with a constant primary voltage , the secondary

voltage decreases for lagging power factor load, and increases for leading pf

load because of its internal resistance and leakage reactance . The change in

secondary terminal voltage from no load to full load expressed as a percentage

of no load or full load voltage is termed as regulation .

% regulation down = (0V2-V2) x 100/0V2

% regulation up = (0V2-V2) x 100/V2

37. Full load copper loss in a transformer is 1600 watts. What will be the loss at

half load ?

If x is the ratio of actual load to full load then copper loss = x2(full load copper

loss).Here Wc = (0.5)2 x 1600 = 400 watts

38. Define all day efficiency of a transformer .

It is the computed on the basis of energy consumed during a certain period ,

usually a day of 24 hrs.

ηall day = output in kWh /input in kWh for 24 hrs.

39. Why transformers are rated in kVA ?

Copper loss of a transformer depends on current and iron loss on voltage .

Hence total losses depends on Volt- Ampere and not on the power factor. That

is why the rating of transformers are in kVA and not in kW.

40. What are the typical uses of auto transformer ?

(i)To give small boost to a distribution cable to correct for the voltage drop.

(ii)As induction motor starters.

(iii)As furnace transformers

(iv)As interconnecting transformers

(v)In control equipment for single phase and 3 phase elective locomotives.

41. What are the applications of step-up and step-down transformers?

Step-up transformers are used in generating stations. Normally the generated

voltage will be either 11 kV or 22 kV. This voltage is stepped up to 110 kV or

220 kV or 400 kV and transmitted through transmission lines. (In short it may

be called as sending end). Step-down transformers are used in receiving

stations. The voltage are again stepped down to 11 kV or 22 kV and transmitted

through feeders. (In short it may be called as receiving end). Further these 11

kV or 22kV are stepped down to 3 phase 400 V by means of a distribution

transformer and made available at consumer premises. The transformers used at

generating stations and receiving stations are called power transformers

42. Explain on the material used for core construction.

The core is constructed of transformer sheet steel laminations assembled to

provide a continuous magnetic path with a minimum of air gap included. The

steel used is of high silicon content sometimes heat-treated to produce a high

permeability and a low hysteresis loss at the usual operating flux densities. The

eddy current loss is minimized by laminating the core, the laminations being

insulated from each other by light coat of core-plate vanish or by an oxide layer

on the surface .the thickness of laminations varies from 0.35 mm for a

frequency of 59 Hz and 0.5 mm for a frequency of 25 Hz.

43. When will a Bucholz relay operate in a transformer?

Bucholz rely is a protective device in a transformer. If the temperature of the

coil exceeds its limit, Bucholz relay operates and gives an alarm.

44 How does change in frequency affect the operation of a given transformer?

With a change in frequency, iron loss, copper loss, regulation, efficiency and

heating varies and thereby the Operation of the transformer is affected.

45. What is the angle by which no-load current will lag the ideal applied

voltage?

In an ideal transformer, there are no copper loss and no core loss,(i.e. loss free

core).

The no load current is only magnetizing current. Therefore the no-load current

lags behind by an angle of 90°. However the windings possess resistance and

leakage reactance and therefore the no-load current lags the applied voltage

slightly less than 90°.

46. List the advantages of stepped core arrangement in a transformer.

(i) To reduce the space effectively.

(ii) To obtain reduced length of mean turn of the windings.

(iii) To reduce I2R loss.

47. Why are breathers used in transformers?

Breathers are used to entrap the atmospheric moisture and thereby not allowing

it to pass on to the transformer oil. Also to permit the oil inside the tank to

expand and contract as its temperature increases and decreases. Also to avoid

sledging of oil i.e. decomposition of oil. Addition of 8 parts of water in 1000000

reduces the insulations quantity of oil. Normally silica gel is filled in the

breather having pink colour. This colour will be changed to white due to

continuous use, which is an indication of bad silica gel, it is normally heated

and reused.

48. What is the function of transformer oil in a transformer?

Nowadays instead of natural mineral oil, synthetic oils known as ASKRELS

(trade name) are used. They are noninflammable; under an electric arc do not

decompose to produce inflammable gases. PYROCOLOR oil possesses high

dielectric strength. Hence it can be said that transformer oil provides, (i) good

insulation and (ii) cooling.

49. A 1100/400 V, 50 Hz single phase transformer has 100 turns on the

secondary winding.

Calculate the number of turns on its primary.

We know that V1 / V2 = k = N2 / N1

Substituting in above equation 400/1100 = 100/N1

N1 = 100/400 x 1100

= 275 turns.

50. Give the method of reducing iron loss in a Transformer

The iron losses are minimized by using high-grade core material like silicon

steel having very low hysteresis loop and by manufacturing the core in the form

of laminations.

QUESTION BANK

PART-A

1. What is a Transformer?

2. Enumerate the various kinds of transformers.

3. Why is the transformer windings are divided into several coils ?

4. Draw the Phasor diagram of an ideal transformer.

5. Give the two general types of transformer.

6. What is an air – core transformer?

7. Draw the equivalent circuit of single phase transformer.

8. Name the two tests that are used to determine parameters of equivalent

circuit, voltage regulation and efficiency.

9. State the different losses in transformer.

10. Define hysteresis and eddy current losses.

11. What is Steinmetz’s constant and give its range.

12. Define efficiency and All day efficiency.

13. Enumerate the various testing of transformers.

14. What is Auto – transformer?

15. Distinguish between transformer and Auto – transformer.

16. What is tap- changer ?

17. Classify the various types of tap- changer.

18. What are the different three phase connections?

19. What is the EMF equation of a transformer?

20. Define voltage regulation.

21. What is transformation ratio?

22. At what frequency core loss and iron loss are equal?

23. What is polarity test in transformers?

24. What is equivalent resistance of transformer ? How it is calculated in

primary terms ?

25. Define all- day efficiency .

26. What is equivalent reactance of transformer ? How it is calculated in

primary terms ?

27. What are the two components of core loss ?

28. Draw the phasor diagram of inductive and capacitive load.

29. At what condition core loss and iron loss equal ?

30. Define voltage transformation ratio.

PART- B

1. Explain the operation of transformer in no load and loaded condition with

phasor diagram.

2. Draw the equivalent circuit of a transformer and derive the components

with respect to primary side.

3. Derive the EMF equation of a transformer.

4. Explain with a neat diagram the O.C and S.C test of transformer.

5. What is auto- transformer? State the application of auto- transformer.

6. Explain the parallel operation of transformer.

7. What is tap – changer and explain its various types.

8. Explain the various testing of transformers.

9. What are the various losses of the transformers and give its efficiency.

10.What is voltage regulation?

11.Draw the various three phase connections.

12.What is sumpner’s test? Draw the circuit diagram to conduct the test .

13.Explain Scott connection and explain in detail.

14.Explain Hysteresis and eddy current loss.

15.What is polarity test of transformer?

16.(a) Explain the principle and operation of auto transformer.

(b)Draw and explain the no load phasor diagram of a single phase

transformer.

17.(a) Derive the emf equation of single phase transformer.

(b) A 120kVA, 6000/400V, Y/Y, 3- phase, 50Hz transformer has a iron

loss of 1800W. The maximum efficiency occurs at ¾ full loads. Find the

efficiency of the transformer at (i) Full load and 0.8 pf (ii) The

maximum efficiency at unity pf.

18.A100 kVA, 6.6kV/415V, single phase transformer has an effective

impedance of (3+8j) referred to HV side. Estimate the full load voltage

regulation at 0.8 pf lagging and 0.8 leading pf.

19.A 100 kVA, 6.6 kV/415V, single phase transformer has an effective

impedance of (3+j8) 𝝮 referred to HV side. Estimate the full load voltage

regulation at (1) 0.8 pf lagging (2) 0.8 pf leading [ Ans : 1.653%, -

0.55%].

20.The primary and secondary windings of a 30 kVA, 6.6 kV/240V

transformer gave resistance of 10 𝝮 and 0.013 𝝮 respectively. The

leakage reactance of the windings are 17𝝮 and 0.022 𝝮. Estimate the

percentage voltage regulation of the transformer when it is delivering

full-load at 0.8 pf lagging the rated voltage. [Ans : 2.5%]

21.(a) Explain the working of auto transformer and prove that when

transformation ratio approaches unity, the amount copper used

approaches smaller value.

(b)The emf per turn of a single phase, 6.6kV/440V, 50 Hz transformer is

approximately 12V. Calculate the number of turns in the HV and LV

windings and the net cross sectional area of the core for a maximum flux

density of 1.5T.

22.Draw the circuit diagrams for conducting OC and SC tests on a single

phase transformer. Also explain how the efficiency and voltage regulation

can be estimated by these tests.

23.a) Obtain the equivalent circuit of a 200/400V,50Hz,single phase

transformer from the following testdata:

OC test: 200V,0.7A,70W on LV side

SC test: 15V, 10A, 85W on HV side.

(b)With the help of circuit diagrams, explain any two types of three phase

transformer connections.

24. Find the approximate equivalent circuit of a single- phase 400/4400 V

transformer having the following test readings.

O.C test : 400V, 5.2 A, 600W on l.v side

S.C test : 155V, 50A, 1850 W on h.v side

25.What is the sumpner’s test? Draw the circuit diagram to conduct this test

and explain its principle.

26.(a) Derive the condition for maximum efficiency in a transformer.

(b) A11000/230 V, 150 KVA, 1-phase, 50 Hz transformer has core loss

of 1.4kW and F.L cu loss of 1.6 Kw .Determine (i) The kVA load for

maximum efficiency and the value of maximum efficiency at unityp.f.

(ii) The efficiency at half F.L 0.8 pf leading.

27.Explain in detail about parallel operation of single phase transformers.

28.Data of a 500KVA, 3300/400 V, 50 Hz, single phase transformer is given

below.S.C test: 1250 W, 100 V –secondary short circuited with full load

current in it O.C test: 1000 W –with normal primary voltage.

29.Calculate the full load regulation and efficiency at a power factor of

0.8(lag).

30.(a) Derive the equivalent circuit of a single phase two winding

transformer.

(b) The maximum efficiency of a single phase 250kVA, 2000/250 V

transformer occurs at 80% of full load and is equal to 97.5% at 0.8 pf

determine the efficiency and regulation on full load at 0.8pf lagging if the

impedance of the transformer is 9 percent.

31.A 600 kVA, single phase transformer when working at unity pf has an

efficiency of 92% at full load and also at half load. Determine its

efficiency when it operates at unity power factor and 60% of full load

[Ans : 92.32%]

32.A 100kVA, 1000/10kV,50Hz, single phase transformer has an iron loss

0f 1100W. The copper loss with 5A in the high voltage winding is 400W.

Calculate the efficiencies at (i) 25% (ii)50% and (iii) 100% of normal

load for power factor of (1) 1.0 and 0.8. The output terminal voltage

being maintained at 10kV. Find also the load for maximum efficiency at

both power factors.[Ans : At upf: 94.33%,97.08%, 98.52%, : At 0.8 pf ,

93.02%,96.38%,98.15% ]

33.Explain in detail about tap changing of transformers.

34.With the help of circuit diagrams, explain any two types of three phase

transformer connections.

35.Draw and explain the phasor diagram of transformer when it is operating

under load.

36.a)Write short notes on parallel operation of transformers

b) What are the necessary conditions for parallel operation of

transformers?

37.Describe the construction of single phase transformer

38.Discuss about working principle of transformer.

39.Explain in detail about scott connection.

40.What are the tests required to draw the equivalent circuit of a Single

phase Transformer? How they are conducted?

41.Draw phasor diagram to represent conditions in a single-phase

transformer-supplying load at 1. Unity p.f , 2.Lagging p.f 3. Leading p.f

42.Explain the Back to back method of testing of two identical single phase

transformers

43.Explain the construction and principle of operation of single phase

transformer

44.Deduce the equivalent circuit of a Transformer

45.Derive the emf equation of the Transformer

46.List the losses, which occur in a loaded transformer. Deduce the

relationship between losses for maximum efficiency

47.Derive the condition for maximum efficiency of a Transformer

48.Explain the types of testing of transformer

49.Explain the Construction of 3 phase Transformer

50. Describe the various three phase transformer connections.