MAHALAKSHMI 2.pdf · It is a gas operated relay which mounted in between the conservator and...

R.Thiyagarajan AP/EEE

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI- 621213.

QUESTION BANK

DEPARTMENT: ECE SEMESTER – III

SUBJECT CODE: EC2201 SUBJECT NAME: Electrical Engineering

UNIT 2

TRANSFORMERS

PART – A

1. What is meant by transformation ration? [NOV/DEC 2009]

Turns ratio= 1

2

N

N

Transformation ratio= KI

I

E

E

2

1

1

2

2. What is the purpose of conducting open circuit and short circuit tests in

transformers? [NOV/DEC 2009]

The open circuit test useful to find

i) No – load loss (or) core loss ii) No – load current iii) Ro and Xo

The short circuit test useful to find

i) Full load copper loss.

ii) Equivalent resistance and reactance referred to any side.

By using above two tests, we can predetermine the

i) Efficiency of the transformer ii) Regulation of the transformer

3. Why is the core of transformer is laminated? [NOV/DEC 20010]

The core of the transformer is laminated in order to minimize eddy current loss.

4. Define voltage regulation of a transformer. [NOV/DEC 2009], [NOV/DEC

2010], [NOV/DEC 2011]

Regulation of a transformer is defined as reduction in magnitude of the terminal

voltage due to the load, with respect to the no-load terminal voltage.


% regulation= 1002

22

loadonnoV

whenloadedVloadonnoV

5. Write down the emf equation of single phase transformer. [NOV/DEC 2009]

Emf induced in primary coil E1= 4.44fФmN1 volt emf induced in secondary

Coil E2 =4.44 fФmN2.

f--freq of AC input, Ф--maximum value of flux in the core

N1, N2--Number of primary & secondary turns.

6. State condition for maximum efficiency of a transformer. [NOV/DEC 2010]

Efficiency of the transformer will be maximum when copper losses are equal to

iron losses.

Iron loss = Copper loss (or) Constant loss = Variable loss

The load current corresponding to maximum efficiency is given by

02

2R

PI i

7. Classify the type of transformer based on core construction. [NOV/DEC 2011]

1) Core type transformer 2) Shell type transformer 3) Berry type transformer

8. Draw the equivalent circuit of transformer. [APRIL/MAY 2010]

9. What is step-up transformer? [NOV/DEC 2011]


In a transformer, if the number of turns in the secondary winding is higher than the

primary winding is called step up transformer.

10. Mention the function of breather in transformer. [NOV/DEC 2011]

The function of breather is to prevent the entry of moisture to inside of the

transformer tank. The breather filled with some drying agent, such as calcium

chloride or silica gel. Silica gel or calcium chloride absorbs the moisture and allows

dry air to enter in to the transformer tank

11. What is the function of the Bucholz relay in a transformer? [NOV/DEC 2008]

It is a gas operated relay which mounted in between the conservator and

transformer tank. The Bucholz relay give an alarm in case minor fault and

disconnect the transformer from the main supply in case of severe fault.

12. Draw the no – load phasor diagram of a transformer. [NOV/DEC 2008]

13. Define all – day efficiency. [NOV/DEC 2007]

The ratio of output in kwh to input in kwh of a transformer over a 24 hour period

is known a s all – day efficiency.

hrsininputkwh

hrsinoutputkwhdayall

24

24

14. Mention two different components of core loss in a transformer. [NOV/DEC

2007]

1. Hysteresis loss

2. Eddy current loss


15. What are the different losses occurring in a transformer? [APRIL/MAY

2008]

1. Copper loss

2. Iron loss

16. Define transformer.

The transformer works on the principle of electromagnetic induction. A

transformer is an electrical device, having no moving parts, which by mutual

induction transfers electrical energy from one circuit another at the same

frequency, with changed values of voltage and current

17. Draw the phasor diagram for a transformer with inductive load.

18. What is staggering in the construction of transformers?

In transformer, the joints in alternate layers are staggered in order to avoid the

presence of narrow gaps right through the cross – section of the core.

19. Why transformer rating is expressed in terms of KVA?

Copper loss depends on current and iron loss depends upon voltage. Hence the

total loss in a transformer depends upon volt – ampere only not 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 kW.

20. Name the factors on which hysteresis loss depends.


1. Frequency 2. Volume of the core 3. Maximum flux density

21. What are the different properties of ideal transformer?

1. No winding resistance

2. No magnetic leakage flux

3. No copper loss

4. No core loss

22. What is eddy current loss?

When a magnetic core carries a time varying flux, voltages are induced in all possible

path enclosing flux. Resulting is the production of circulating flux in core. These

circulating current do no useful work are known as eddy current and have power loss

known as eddy current loss.

23. How hysteresis and eddy current losses are minimized?

Hysteresis loss can be minimized by selecting materials for core such as silicon steel

& steel alloys with low hysteresis co-efficient and electrical resistivity. Eddy current

losses are minimized by laminating the core.

24. Why the open circuit test on a transformer is conducted at rated voltage?

The open circuit on a transformer is conducted at a rated voltage because core loss

depends upon the voltage. This open circuit test gives only core loss or iron loss of the

transformer.

25. Mention the difference between core and shell type transformers?

In core type, the windings surround the core considerably and in shell type the core

surrounds the windings i.e. winding is placed inside the core.

PART – B

1. With necessary vector diagrams, discuss about transformer on no

– load and loaded conditions. [NOV/DEC 2012]


If the primary winding is connected to an alternating voltage and

secondary winding left open, then the transformer is said to be on n –

load. Let the supply voltage be ‘V1’ volts. This causes an alternating

current flow through the primary. Since secondary is open, this

current is called no load primary current (Io). This ‘Io’ establishes the

flux ‘ ’ weber in the core. Thus Io is not at 90o behind V1, but lags it

by an angle o > 90o. No load input power Po=V1Iocos o . ‘Io’ has two

components

i) Active or working or iron loss or wattful component (Iw), Which

is in phase with ‘V1’ and supplies the iron loss and negligible

amount of primary copper loss.

Iw = Io cos o ----- (1)

Where, cos o = No load power factor


ii) Reactive or magnetizing or wattles component ( I ) which is in

quadrature with V1, and its function is to sustain the flux in the

core.

I = Io sin o ----- (2)

From the above two equations we get

Io=22 IIw

2. (i) Draw the equivalent circuit of a transformer with all its

notations [NOV/DEC 2012]

Under no load condition, the primary of a transformer draws no load

current Io. It is mainly used to supply the iron loss and to produce the

flux in the core. The effect of iron loss is represented by a non –

inductive resistance Ro and the magnetizing current is represented by

Xo. Both of them are connected in parallel with primary winding. This

circuit is known as exciting branch or no – load branch.

(ii) Write a note on open circuit test on transformer. [NOV/DEC

2012]


Open-Circuit or No-Load Test:

This test is conducted to determine the iron losses (or core losses) and

parameters R0 and X0 of the transformer. In this test, the rated

voltage is applied to the primary (usually low-voltage winding) while

the secondary is left open circuited. The applied primary voltage V1

is measured by the voltmeter, the no-load current I0 by ammeter and

no-load input power W0 by wattmeter as shown in Fig. (i). As the

normal rated voltage is applied to the primary, therefore, normal iron

losses will occur in the transformer core. Hence wattmeter will record

the iron losses and small copper loss in the primary. Since no-load

current I0 is very small (usually 2-10 % of rated current). Cu losses in

the primary under no-load condition are negligible as compared with

iron losses. Hence, wattmeter reading practically gives the iron losses

in the transformer. It is reminded that iron losses are the same at all

loads. Fig. (ii) Shows the equivalent circuit of transformer on no-

load.

Iron losses, Pi = Wattmeter reading = W0

No load current = Ammeter reading = I0

Applied voltage = Voltmeter reading = V1

Input power, W0 = V1 I0 cos f0


3. (a) Deduce the equivalent circuit of the transformer from the

basic principle. [NOV/DEC 2011]


Under no load condition, the primary of a transformer draws no load

current Io. It is mainly used to supply the iron loss and to produce the

flux in the core. The effect of iron loss is represented by a non –

inductive resistance Ro and the magnetizing current is represented by

Xo. Both of them are connected in parallel with primary winding. This

circuit is known as exciting branch or no – load branch.

Equivalent circuit of a transformer referred to primary

If all the parameters are referred to primary side, we get the

equivalent circuit of transformer referred to primary. When secondary

parameters are referred to primary resistance and reactance are

divided by K2, voltage are divided by K and currents multiplied by K.

This circuit also called as exact equivalent circuit of a transformer.

Approximate equivalent circuit

μ

2o

w

1o

2'

22

L'

L2

'

22

2'

22

2'

2I

VX,

I

VR,

K

VV,

K

ZZ,KII,

K

XX,

K

RR


The no – load current Io is only 1 – 3% of rated primary current. So '

2I

practically equal to I1. Due to this, equivalent circuit can be simplified

by transferring the exciting branch (Ro and Xo) to the left position of

the circuit.

R01=R1+R’2 and X01=X1+X

’1

2

01

2

0101 XRZ

The above figure shows all the parameters referred to primary.


2

02

2

0202 XRZ

The above figure shows all the parameters referred to secondary

(b) A 2200/200 V transformer draws a no load primary current of

0.6A and absorbs 400 Watts. Find the magnetizing and iron loss

current. Draw also the no load phasor diagram. [NOV/DEC 2011]

Given Data:

V1=2200V, V2=200V, Io=0.6A, Po=400W

To find? Iw, I

Solution:

Iron loss component Iw = 2200

400

V

P

1

o

Iw = 0.18 A

Magnetizing component 222

w

2

oμ 0.180.6III

μI = 0.57A

No load phasor diagram


4 . Explain the method of conducting OC and SC test on a

transformer with neat diagram. [NOV/DEC 2011]

The circuit constants, efficiency and voltage regulation of a

transformer can be determined by two simple tests (i) open-circuit test

and (ii) short-circuit lest. These tests are very convenient as they

provide the required information without actually loading the

transformer. Further, the power required to carry out these tests is

very small as compared with full-load output of the transformer.

These tests consist of measuring the input voltage, current and power

to the primary first with secondary open-circuited (open- circuit test)

and then with the secondary short-circuited (short circuit test).

Open-Circuit or No-Load Test:

This test is conducted to determine the iron losses (or core losses) and

parameters R0 and X0 of the transformer. In this test, the rated voltage

is applied to the primary (usually low-voltage winding) while the

secondary is left open circuited. The applied primary voltage V1 is

measured by the voltmeter, the no-load current I0 by ammeter and no-

load input power W0 by wattmeter as shown in Fig. ((i)). As the

normal rated voltage is applied to the primary, therefore, normal iron

losses will occur in the transformer core. Hence wattmeter will record


the iron losses and small copper loss in the primary. Since no-load

current I0 is very small (usually 2-10 % of rated current). Cu losses in

the primary under no-load condition are negligible as compared with

iron losses. Hence, wattmeter reading practically gives the iron losses

in the transformer. It is reminded that iron losses are the same at all

loads. Fig. (ii)) shows the equivalent circuit of transformer on no-

load.

Iron losses, Pi = Wattmeter reading = W0

No load current = Ammeter reading = I0

Applied voltage = Voltmeter reading = V1

Input power, W0 = V1 I0 cos f0

Short-Circuit or Impedance Test:

This test is conducted to determine R01 (or R02), X01 (or X02) and full-

load copper losses of the transformer. In this test, the secondary


(usually low-voltage winding) is short-circuited by a thick conductor

and variable low voltage is applied to the primary as shown in Fig.

(i). The low input voltage is gradually raised till at voltage VSC, full-

load current I1 flows in the primary. Then I2 in the secondary also has

full-load value since I1/I2 = N2/N1. Under such conditions, the copper

loss in the windings is the same as that on full load. There is no output

from the transformer under short-circuit conditions. Therefore, input

power is all loss and this loss is almost entirely copper loss. It is

because iron loss in the core is negligibly small since the voltage VSC

is very small. Hence, the wattmeter will practically register the full-

load copper losses in the transformer windings. Fig. (ii) shows the

equivalent circuit of a transformer on short circuit as referred to

primary; the no-load current being neglected due to its smallness.

Full load Cu loss, PC = Wattmeter reading = WS

Applied voltage = Voltmeter reading = VSC

F.L. primary current = Ammeter reading = I1


Note: The short-circuit test will give full-load Cu loss only if the

applied voltage VSC is such so as to circulate full-load currents in the

windings. If in a short circuit test, current value is other than full-load

value, the Cu loss will be corresponding to that current value.

Advantages of Transformer Tests

The above two simple transformer tests offer the following

advantages:

(i) The power required to carry out these tests is very small as

compared to the full-load output of the transformer. In case of open-

circuit lest, power required is equal to the iron loss whereas for a

short-circuit test, power required is equal to full-load copper loss.

(ii) These tests enable us to determine the efficiency of the

transformer accurately at any load and p.f. without actually loading

the transformer

(iii) The short-circuit test enables us to determine R01 and X01 (or R02

and X02). We can thus find the total voltage drop in the transformer as


referred to primary or secondary. This permits us to calculate voltage

regulation of the transformer.

5. (a) List the various parts of transformer. Also explain the

constructional features of a transformer [MAY/JUNE 11]

Various parts of transformer

1. Transformer tank

2. Transformer oil

3. Core

4. Winding

5. Conservator

6. Breather

CONSTRUCTION OF A TRANSFORMER

We usually design a power transformer so that it approaches the

characteristics of an ideal transformer. To achieve this, following design

features are incorporated:


(i) The core is made of silicon steel which has low hysteresis loss and

high permeability. Further, core is laminated in order to reduce

eddy current loss. These features considerably reduce the iron

losses and the no-load current.

(ii) Instead of placing primary on one limb and secondary on the

other, it is a usual practice to wind one-half of each winding on

one limb. This ensures tight coupling between the two windings.

Consequently, leakage flux is considerably reduced.

(iii) The winding resistances R1 and R2 are minimized to reduce I2R

loss and resulting rise in temperature and to ensure high efficiency.

TYPES OF TRANSFORMERS:

Depending upon the manner in which the primary and secondary

are wound on the core, transformers are of two types viz.,

(i) core-type transformer and

(ii) shell-type transformer.

(I) Core-type transformer.

In a core-type transformer, half of the primary winding and

half of the secondary winding are placed round each limb as

shown in Fig. This reduces the leakage flux. It is a usual

practice to place the low-voltage winding below the high-

voltage winding for mechanical considerations.


(II) Shell-type transformer.

This method of construction involves the use of a double magnetic

circuit. Both the windings are placed round the central limb (See Fig,

the other two limbs acting simply as a low-reluctance flux path. The

choice of type (whether core or shell) will not greatly affect the

efficiency of the transformer. The core type is generally more suitable

for high voltage and small output while the shell-type is generally

more suitable for low voltage and high output.


6. A 100 KVA transformer has iron loss of 720 Watts and full

load copper loss of 2100 Watts. Determine its efficiency at

full load at unity power factor and 0.8 power lagging.

[MAY/JUNE 11]

Given data:

Transformer rating = 100 kVA, Iron loss Pi=720 W

Full load copper loss Pcufl = 2100 W, Power factor cos =0.8

To find:

Full load efficiency

100PPcosΦnkVA

cosΦnkVAη%

cufli

Here n = 1 (for full load)

10021007208.0101001

8.0101001%

3

3

% 96.59

7. (a) Draw the phasor diagram for an ideal transformer with

resistive and inductive load. [MAY/JUNE 11]

An ideal transformer is one that has

(i) No winding resistance


(ii) No leakage flux i.e., the same flux links both the windings

(iii) No iron losses (i.e., eddy current and hysteresis losses) in the

core

Although ideal transformer cannot be physically realized, yet

its study provides a very powerful tool in the analysis of a

practical transformer. In fact, practical transformers have

properties that approach very close to an ideal transformer.

Consider an ideal transformer on no load i.e., secondary is open-

circuited. Under such conditions, the primary is simply a coil of pure

inductance. When an alternating voltage V1 is applied to the primary,

it draws a small magnetizing current Im which lags behind the applied

voltage by 90°. This alternating current Im produces an alternating

flux which is proportional to and in phase with it. The alternating

flux links both the windings and induces e.m.f. E1 in the primary

and e.m.f. E2 in the secondary. The primary e.m.f. E1 is, at every

instant, equal to and in opposition to V1 (Lenz’s law). Both e.m.f.s E1

and E2 lag behind flux by 90°. However, their magnitudes depend

upon the number of primary and secondary turns. Fig. (ii) shows the


phasor diagram of an ideal transformer on no load. Since flux is

common to both the windings, it has been taken as the reference

phasor. The primary e.m.f. E1 and secondary e.m.f. E2 lag behind the

flux by 90

V1 and 180° out of phase with it.

7. The emf induced per turn of a single phase 11000 V/ 440 V,

50Hz single phase transformer approximately 11 volts.

Calculate the number of turns in high voltage and low

voltage windings and the net cross sectional area of the core

for maximum flux density of 1.5 Tesla. [MAY/JUNE 11]

Given data:

Emf per turn = 11 V, Primary voltage = 11000 V, Secondary = 440

V, Frequency f = 50 Hz, Maximum flux density Bm = 1.5 T.


To find:

Number turns in the HV and LV windings, Net cross sectional

area

Solution:

E1 = N1 X emf induced/turn

N1 = 11000/11 =1100 turn

N2 = E2/Emf inducedper turn =440/11

N2 = 40 turns

E1 = 4.44 f N1 Bm A

A = E1/4.44 f N1 Bm = 1.51100504.44

11000

A = 0.030 cm2

8. Derive the emf equation of a transformer. [NOV/DEC 2010]

EMF Equation of Transformer:

Since applied voltage is alternating in nature, the flux established is

also an alternating one. It is clear that the flux is attaining its

maximum value in one quarter of the cycle i.e., T/4 sec where ‘T’ is

the time period in second.


We know that T= 1/f

Average rate of change of flux = wb/seconds

4f1

Φm

Form factor = 11.1..

valueAverage

valueSMR

RMS value of induced emf/turn =(1.11) x (4f x m )

RMS value of induced emf in the entire primary winding

E1= 4.44f m xN1

RMS value of induced emf in the entire primary winding

E2= 4.44f m xN2

9. A single phase transformer is rated at 10 KVA, 50 Hz. The

secondary rated voltage is 2400 V and the turn’s ratio is 10. The

resistance and leakage reactance as referred to secondary are8.4


and 13.7 respectively. Find voltage regulation at full – loaf and

power factors of 0.8 lagging, 0.8 leading and unity. [NOV/DEC

2010]

Given data:

Transformer rating = 10 kVA

Frequency = 50 Hz

Turns ratio (K) = 10

Resistance referred to secondary R01 = 8.4 Ω

Reactance referred to secondary X01 = 13.7 Ω

To find:

Full – load voltage regulation

At 0.8 Lagging =?

At 0.8 Leading =?

At unity =?

Solution:

Secondary current (I2) =2V

kVA= 166.4

2400

10000A

Full – load voltage regulation at 0.8 lagging p.f

cos = 0.8


= cos-1

(0.8) = 36.86

Sin (36.86) = 0.6

% Voltage regulation = 100V

)sinXΦcos(RI

2

02022

= 1002400

0.6)13.70.8(8.44.166

= 2.59 %

Full – load voltage regulation at 0.8 leading p.f

% Voltage regulation = 100V

)sinXΦcos(RI

2

02022

= 1002400

0.6)13.70.8(8.44.166

= - 0.260 %

Full – load voltage regulation at unity p.f

% Voltage regulation = 100cos

2

022

V

RI

= 2400

14.8167.4

= 1.45 %

10. A 30 KVA, 2000/200V single phase, 50Hz transformer has a

primary resistance of 3.5 ohms and reactance of 4.5 ohms. The

secondary resistance and reactance are 0.015 ohms and 0.02 ohms


respectively. Find the equivalent resistance, reactance and

impedance

(i) Referred to primary

(ii) Referred to secondary side [APRIL/MAY 2010]

Given data:

Transformer rating = 30 kVA, Primary voltage V1 = 2000 V

Secondary voltage V2 = 200 V, R1 = 3.5 Ω, X1 = 4.5 Ω, R2 = 0.015 Ω,

X2 = 0.02 Ω

To find: R01, X01, Z01, R02, X02, Z02

Transformation ratio K = V2/ V1=200/2000 =0.1

R01 = R1+ R!2 = R1+R2/K

2

R01 = 3.5 + 0.015/0.1

2

R01 = 5 Ω

X01 = X1 + X!2 = X1 + X2/K

2

X01 = 4.5 + 0.02/0.12

X01 = 6.5 Ω

Z01 = 2

01

2

01 XR

Z01 = 22 5.65

Z01 =8.2 Ω


R02 = R2+ R!1 = R2+R1 K

2

R02 = 0.015 + 3.5 X 0.12

R02 = 0.05 Ω

X02 = X2 + X!1 = X2 + X1 K

2

X02 = 0.02 + 4.5 X 0.12

X02 = 0.065 Ω

Z02 = 2

02

2

02 XR

Z01 = 22 065.005.0

Z01 = 0.082 Ω

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