UNIT 2 Emi1 Final

8/13/2019 UNIT 2 Emi1 Final

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ELECTRICAL MACHINES 1/J.GOPI/MIET/EEE 2

or radiators

FACULTY INCHARGE COURSEMATERIALCOORDINATOR HOD

5. State the various methods of cooling of power transformers.

i) oil natural-air forced

ii) oil natural-water forced

iii)oil forcediv)water forced

6 . Why is the core of a transformer laminated? (MAY 2011)

The cores of a transformer laminated in order to reduce the eddy current losses.

The eddy current losses are proportion to the square of thickness of laminations. This

apparently implies that the thickness of laminations should be extremely small in order to

reduce the eddy current losses to a minimum.

7. What is the cause of noise in transformer?

The cause of noise in the transformer is mainly because of magnetostriction effect and

also loosening of stampings and mechanical forces produced during working.

8.How the leakage reactance of the transformer is reduced?

In transformers the leakage reactance is reduced by interleaving the high voltage and low

voltage winding.

9.What is step down transformer?

The transformer used to step down the voltage from primary to secondary is called as step

down transformer. (Ex: 220/110V).


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10.Draw the noload phasor diagram of a single phase

transformer.

11. Why is an auto-transformer not used as a distribution transformer?

The autotransformer cannot provide isolation between HV and LV side. Due to open

circuit in the common portion, the voltage on the load side may soot up to dangerously

high voltage causingdamage to equipment. This unexpected rise in the voltage on LV side

is potentially dangerous. Hence the autotransformer cannotbe used as distribution

transformer.

12. Give the basic principle behind the working of transformer.(DEC2008)

The transformers works in the principle of mutual inductionbetween two coils which are

electrically isolated but magnetically coupled.

13. What are the conditions for parallel operation of transformer?(AUT2006,2010)

In order that the transformers work satisfactorily in parallel, the following conditions

should be satisfied:

* Transformers should be properly connected with regard to their polarities.

*The voltage ratings and voltage ratios of the transformers should be the same.

* The per unit or percentage impedances of the transformers should be equal.

* The reactance/resistance ratios of the transformers should be the same.


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14. Why is transformer rated in kVA?(AUC 2005, 2007, 2010,2011)

The copper loss of a transformer depends on current and iron loss on voltage. Hence, total

transformer loss depends on volt-ampere (VA) and not on phase angle between voltage and

current i.e., it is independent of load power factor. Thatswhy rating of transformers arein

kVA and not in kW.

15. Classify the transformer according to the construction.

Depending upon the manner in which the primary and secondary are wound on the core,

transformers are oftwo types viz., (i) core-type transformer and (ii) shell-type transformer.

16. What is transformation ratio?(AUC NOV 2010)

It is the ratio in which the voltage to be transformed (stepped up or down) from primary to

secondary of a transformer. It is given by,

Where, K is the transformation ratio.

17.Draw the exact equivalent circuit of a transformer.(DEC 2005,2007,MAY 2009,DEC

2011)


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22. Name the factors on which hysteresis loss depends (MAY 2012)

1. Frequency

2. Volume of the core

3. Maximum flux density.

23. What is All day efficiency?(AUC MAY 2010)

The ratio of output in kWh to the input in kWh of a transformer over a 24-hour period is known

as all-day efficiency i.e.,

24. Define efficiency of transformer.(NOV 2003, MAY 2006)

25.What is a transformer?

Transformer is a static device which,

a) Transfers electric power from one circuit to another,b) It does so without a change in frequency,c) It accomplishes this by electromagnetic induction,d) Where the two electric circuits are in mutual inductive influence of each other.

26.Briefly explain the principle of operation of transformers.

A transformer consists of two coils which are in mutual inductance. When AC

supply is given to one of the coils, an alternating flux is set up, which is linked with the

second coil. Due to this alternating flux there is a mutually induced emf produced in the

second coil. If the second coil is closed, current flows in it and so electric energy is

transferred magnetically from the first coil to the second coil.

27.What are the parts of a transformer?(AUT MAY 2009)

The transformer has mainly following parts.

e) Primary windingthe coil to which the AC supply is givenf) Secondary windingthe coil from which output is taken and given to load.g) Laminated Core this acts as a mechanical support to the coils as well as

provides magnetic path for the flux.


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28.What are the types of core in transformer?

There are two types of core available. They are,

Core typeThis has two legs. The primary winding is wounded on one leg and

the secondary winding is wounded on the second leg. This is mainly used in single

phase transformers.

Shell typeThis has three legs. Both primary and secondary windings are wound on

the same leg. This is mainly used in three phase transformers where three primary

and three secondary windings are present.

29.What is an ideal transformer?(MAY 2010)

An ideal transformer is the one which has no losses i.e., its windings have noohmic resistance, there is no magnetic leakage and hence which has no I2R and core

losses. The efficiency of an ideal transformer is 100%.

30.Give the emf equation of the transformer.(DEC 2009,2007)

The emf induced in the transformer is given by

mfNE 21 44.4 V

mfNE 12 44.4 V

where, E1 is the emf induced in the primary winding.

E2is the emf induced in the secondary winding.

N1is the number of turns in the primary winding.

N2is the number of turns in the secondary winding.

mis the maximum flux produced.

f is the frequency in Hz.


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31.What are the losses occurring in a transformer.(AUC 2003)

The losses occurring in a transformer are

a) Core loss loss occurring in the core of the transformer. This has two

components.

o Hysteresis loss The loss occurring due to the magnetizationcharacteristics of the core.

o Eddy current lossThe loss occurring due to the eddy currentproduced in the core.

b) Copper loss (I2R loss) loss occurring in the windings of thetransformer. This has two components.

o Primary copper loss The loss occurring due to the currentflowing through the primary winding.o Secondary copper loss The loss occurring due to the current

flowing through the secondary winding.

32.What is the purpose of the magnetizing current in transformer?(AUT 2012)

The magnetizing current is the component of the primary current which is

responsible for the production of flux in the core.

33.What are the components of primary no-load current?

The primary no-load current consists of mainly two components. They are,

a) Magnetizing current ( Im ) Produces flux in the core and hencemagnetizes the core.

b) Core loss component current ( Ic)compensates for the core losses.


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PART-B

1.Explain in detail of various testing methods of transformers.(NOV/DEC 2009)

Testing:

Transformer Tests

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. 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. 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


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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. . 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 copperloss 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. shows the equivalent circuit of a transformer on short circuit as referred to

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


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2.Explain in detail : (DEC 2007, AUT 2009,AUC 2010)

1) All day efficiency.2) Voltage Regulation


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For step-down autotransformer I2 > I1 (as for 2-winding transformer) so that I2 - I1 current

flows through the common portion of the winding. For step-up autotransformer, I2 < I1.

Therefore, I1 - I2 current flews in the common portion of the winding.

In an ideal autotransformer, exciting current and losses are neglected. For such an

autotransformer, as K approaches 1, the value of current in the common portion (I2 - I1 or I1 - I2)

of the winding approaches zero. Therefore, for value of K near unity, the common portion of the

winding can be wound with wire of smaller cross-sectional area. For this reason, an

autotransformer requires less copper.

Advantages and Disadvantages of autotransformers

Advantages

(i) An autotransformer requires less Cu than a two-winding transformer of similar rating.

(ii) An autotransformer operates at a higher efficiency than a two-winding transformer of similar

rating.

(iii) An autotransformer has better voltage regulation than a two-winding transformer of the same

rating.

(iv) An autotransformer has smaller size than a two-winding transformer of the same rating.

(v) An autotransformer requires smaller exciting current than a two-winding transformer of the

same rating. It may be noted that these advantages of the autotransformer decrease as the ratio of

transformation increases. Therefore, an autotransformer has marked advantages only for

relatively low values of transformation ratio .


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Disadvantages

(i) There is a direct connection between the primary and secondary. Therefore, the output is no

longer d.c. isolated from the input.

(ii) An autotransformer is not safe for stepping down a high voltage to a low voltage. As an

illustration, Fig. (7.40) shows 11000/230 V step-down autotransformer. If an open circuit

develops in the common portion 2-3 of the winding, then full-primary voltage (i.e., 11000 V in

this case) will appear across the load. In such a case, any one coming in contact with the

secondary is subjected to high voltage. This could be dangerous to both the persons and

equipment. For this reason, autotransformers are prohibited for general use.

(iii) The short-circuit current is much larger than for the two-winding transformer of the samerating.

4. Draw the equivalent circuit of transformer on load and no load conditions and draw the

phasordiagram.


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Problem-1


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Problem-2


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Problem-3


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Problem-4


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Problem-6


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Problem-7


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Problem8

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UNIT 2 Emi1 Final