Prof. Ch. SAI BABU,

ONLINE GATE COACHING CLASSES

BY

Prof. Ch. SAI BABU,

Professor of Electrical & Electronics Engineering

JNTUK Kakinada

Lecture 3 of Electrical Machines

Topic: Losses and Load Tests of DC Machine

Jawaharlal Nehru Technological University, Kakinada-533003,

Andhra Pradesh

1

Number

Syllabus

Detailed Syllabus

Lecture

III

DC Machines: Types of Losses and

Efficiency Calculations of Electric

Machines.

Various losses of DC

Machines, No Load and Load

Tests of DC Motor.

Predetermination of

Efficiency

Various losses of DC Machines

The losses that occur in a DC Machine is divided into five basic categories.

The various losses are Electrical or Copper losses (I2R losses), Core

losses or Iron losses, Brush losses, Mechanical losses, Stray load losses. These losses

are explained below in detail.

L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 2 2

Contents:

Electrical or Copper losses

Magnetic Losses or Core Losses or Iron

Losses

Brush Losses

Mechanical Losses

Stray Losses


Electrical or Copper Losses in dc machine

These losses are also known as Winding losses as the copper loss occurs

because of the resistance of the windings.

The ohmic loss is produced by the current flowing in the windings.

Armature copper losses = Ia2Ra where Ia is armature current, and Ra is

the armature resistance. These losses are about 30 percent of the total full

load losses.

In shunt machine, the Copper loss in the shunt field is I2shRsh, where

Ish is the current in the shunt field, and Rsh is the resistance of the shunt

field windings. The shunt regulating resistance is included in Rsh.

In a series machine, the copper loss in the series windings is I2seRse,

where, Ise is the current through the series field windings, and Rse is the

resistance of the series field windings.

In a Compound machine, both the shunt and the series field losses occur.

These losses are almost 20 percent of the full load losses.

Copper losses in the interpole windings are written as Ia2Ri where Ri is

the resistance of the interpole windings.


Magnetic Losses or Core Losses or Iron Losses in dc machine

The core losses are the hysteresis and eddy current losses. These losses

are considered almost constant as the machines are usually operated at

constant flux density and constant speed. These losses are about 20

percent of the full load losses.

Brush Losses in dc machine

Brush losses are the losses taking place between the commutator and

the carbon brushes. It is the power loss at the brush contact point. The

brush drop depends upon the brush contact voltage drop and the

armature current Ia. It is given by the equation shown below.

The voltage drop occurring over a large range of armature currents,

across a set of brushes is approximately constant If the value of brush

voltage drop is not given than it is usually assumed to be about 2 volts.

Thus, the brush drop loss is taken as 2Ia.


Mechanical Losses in dc machine

The losses that take place because of the mechanical effects of the

machines are known as mechanical losses.

Mechanical losses are divided into bearing friction loss and windage

loss.

The losses occurring in the moving parts of the machine and the air

present in the machine is known as Windage losses. These losses are very

small. Stray Losses in dc machine

These losses are the miscellaneous type of losses. The following factors are

considered in stray load losses.

The distortion of flux because of armature reaction.

Short circuit currents in the coil, undergoing commutation.

These losses are very difficult to determine. Therefore, it is necessary to assign

the reasonable value of the stray loss. For most machines, stray losses are

taken by convention to be one percent of the full load output power.


Efficiency of DC Generator Efficiency is simply defined as the ratio of output power to the input power.

Let R = total resistance of the armature circuit (including the brush contact

resistance, at series winding resistance, inter-pole winding resistance and

compensating winding resistance). The efficiency of DC generator is explained

below in the line diagram.

If Ish is small compared with Ir, then

Ia = I

Therefore,


The efficiency ȠG will be a maximum

when the denominator Dr is a

minimum.

Where,

I is the output current

Ish is the current through the shunt field

Ia is the armature current = I + Ish

V is the terminal voltage.

Total copper loss in the armature circuit = Ia2Rat

Power loss in the shunt circuit = VIsh (this

includes the loss in the shunt regulating

resistance).

Mechanical losses = friction loss of bearings +

friction loss at a commutator + windage loss.

Core losses = hysteresis loss + eddy current loss

Stray loss = mechanical loss + core loss

The sum of the shunt field copper loss and stray

losses may be considered as a combined fixed

(constant) loss that does not vary with the load

current I.

Therefore, the constant losses (in shunt and

compound generators) = stray loss + shunt field

copper losses.

Dr is minimum when


https://circuitglobe.com/wp-content/uploads/2015/12/efficiency-of-dc-generator-eq-8-compressor.jpg

Testing of DC Machine

When construction of DC motor or DC generator is completed. To perform its

electrical characteristics, several tests are conducted for testing of a dc machine

(Generator or Motor).

Testing of machines is used for finding Losses, Temperature rise and Efficiency.

The efficiency of a d.c. machine depends on its losses.

When losses are decrease, The efficiency is increase and it is vice-versa.

The important Tests of DC machines are classified to

Direct test method (break over test)

Swinburne Test (Indirect test Method )

Hopkinson’s-Test ( Regenerative Method )

Retardation or Running down Test


Direct test method (break over test)

In this method of testing the machine is put on full load and whole

developed power is wasted by applying the break. Direct test method is

used, for small machines. Because of in case of large motors, it is

difficult to dissipate the large amount of heat generated at the brake. It

is also called break over test.

Though this method is simple, but complication in the measurement of

mechanical power input in the case of generator and output in case of a

motor. If brake is to be applied to a series motor, the brake must be

tight before the motor is started, otherwise the armature may gets

damaged and winding burned.

The main draw back of this method is that the

accuracy in determining the mechanical power

output of the motor is limited. Alternately it is

difficult to provide full load for the large capacity

motor L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 10

10

Swinburne Test of DC Machine This method is an indirect method of testing a

DC machine.

It is named after Sir James Swinburne.

Swinburne’s test is the most commonly used and

simplest method of testing of shunt and

compound wound DC machines which have

constant flux.

In this test the efficiency of the machine at any

load is pre-determined.

We can run the machine as a motor or as a

generator.

In this method of testing no load losses are

measured separately and eventually we can

determine the efficiency.

Calculation of Efficiency

Let, I0 is the no load current (it can be

measured by ammeter A1)

Ish is the shunt field current (it can be

measured by ammeter A2)


https://www.electrical4u.com/ammeter/

https://www.electrical4u.com/electric-current-and-theory-of-electricity/

When the Machine is Motoring on Load Machine is Generating on Load

Power input = VI

Advantages of Swinburne’s Test

The main advantages of this test are:

This test is very convenient and economical as it is required very less power

from supply to perform the test.

Since constant losses are known, efficiency of Swinburne’s test can be pre-

determined at any load.

Disadvantages of Swinburne’s Test

The main disadvantages of this test are :

Iron loss is neglected though there is change in iron loss from no load to full

load due to armature reaction.

We cannot be sure about the satisfactory commutation on loaded condition

because the test is done on no-load.

We can’t measure the temperature rise when the machine is loaded. Power

losses can vary with the temperature.

In DC series motors, the Swinburne’s test cannot be done to find its efficiency

as it is a no load test. L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 12 12

Hopkinson’s Test Hopkinson’s test is another useful method of testing the efficiency of a dc

machine. Hopkinson’s test is also called back-to-back test or regenerative

test.

Two identical DC shunt machines are mechanically coupled and

electrically connected in parallel across the DC supply. By adjusting the

field excitation of the machine, one is run as a motor and the other as a

generator.

The electric power from the generator and electrical power from the DC

supply are fed to the motor. The electric power given to the motor is

converted into mechanical power, the rest going to the various motor

losses. This mechanical power is given to the generator. The electrical

power of the generator is given to the motor except that which is wasted

as generator losses.

The electrical power taken from the DC supply is the sum of motor and

generator losses. this power can be measured by digital multi-meter .

Since the power input from the DC supply is equal to the power required

to supply the losses of the two machines. this test can be carried out with

a small amount of power.

By adjusting the field strengths of the machines, any load can be put on

the machines. Therefore, we can measure the total loss of the machines at

load. L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 13

13

Armature copper losses in the generator = I2² rg

Armature circuit copper losses in the motor = (I1 + I2)² rm

Total power drawn by the armature circuit of the motor = V x I1 watts

Let the sum of iron losses and mechanical losses of each machine be Wc,

then

V x I1 = 2Wc + I2 ² rg + (I1 + I2)² rm

Thus, Wc = ½ [ V x I1 – I2 ² rg – (I1 + I2) ² rm ]

Efficiency of Motor

Shunt field copper losses of the motor = V x I3

Hence, total losses of the motor = Wc + (I1 + I2) ² rm + V x I3

Total power input to the motor, P1 = V x (I1 + I2 + I3)


Advantages of Hopkinson’s test

Power drawn from the supply is low.

Both the DC machines are operating under loaded conditions, can such

stray load losses are taken into account.

The temperature rise and the commutation conditions can be checked

under rated load conditions.

Efficiency at different loads can be determined.

Disadvantage of Hopkinson’s test

The necessity of two practically identical machines for performing the

test.

It is difficult to operate the machines at rated speed because field

currents vary widely.


Retardation Test or Running Down on D.C Machines

Retardation test is also called as running down test.This is

very efficient way to find out stray losses in dc shunt motors.

In this test we get total stray losses nothing but combination of mechanical

(friction & windage) and iron losses of the machine.

The circuit diagram of retardation test on dc machines shown

below.A1,A2 are armature terminals.


Procedure of Retardation Test on D.C Machines

The main points in the retardation or running down test are discussed below,

1. Now start the dc machine normally,run the machine slightly above the rated

speed by adjusting resistance.

2. After achieving above the rated speed just cutoff the power supply to the

armature,but keeping field normally excited.

3. Now wait for some time to fall down of speed below rated,then using the

tachometer note down the values of speed (in rpm) and time (in sec).

4. The armature consequently slows down and the amount of kinetic energy

present in the armature is used to supply the rotational or stray losses which

includes iron, friction and winding loss.

If I is the amount of inertia of the armature ans is the angular velocity.

Kinetic energy of armature = 0.5 Iω².

Rotational losses, W = Rate of change of kinetic energy.

I=Moment of inertia of the armature.


https://2.bp.blogspot.com/--YTP89Y-tJU/Vsw4X9jOPaI/AAAAAAAAA5E/ci9up_3RO40/s1600/Untitled.png

I=Moment of inertia of the armature.

In retardation test of dc machines, the

rotational losses are given by

We have the formula of stray losses in retardation test of

machines, but here moment of inertia (I) of the armature is

unknown.


Determination of dN/dt in retardation test of dc machines

The voltmeter (V1) across the armature will give the value of

back e.m.f. of the motor.

When motor is cut off from the supply, the speed decrease in

speed is noted with the help of stop.

At any point C corresponding to normal speed, a tangent AB

is drawn. Then the value obtained from below can be

substituted in the expression for W which can give the

rotational looses.

dN/dt=OA(in rpm)/OB(in seconds)


https://4.bp.blogspot.com/-z6gRfpEDMgE/Vsw7LkKl13I/AAAAAAAAA5Y/_vjBtS74gK8/s1600/curve.jpeg

Using Flywheel in retardation test

In this method we use the fly wheel whose moment of inertia is I1 to find the

I value. In first case retardation test is performed with armature alone and

dN/dt1 is determined. In next case, flywheel is employed on the shaft,

change in speed, dN/dt2 is noted.Addition of fly-wheel will not materially

affect the rotational losses.


(b) Without using Flywheel in retardation test

Without using flywheel, I is eliminated from the expression by an

experiment. First, retardation test is performed with armature alone.

The rotational losses are given by;

W = 0.011 IN dN/dt1

Next the motor is loaded with a known amount of power W' with a

brake.

For the same change in speed, dN/dt2 is noted. Then,

Since the values of W', t1 and t2 are known, the

value of W can be determined.

The electric loading in retardation test W' (or

extra power loss) is given by;

W' = average voltage x average current = V' I'a.

This is the simple article on retardation test on

dc machines.


1. A dc motor is connected in the short-shunt configuration, but the series

and the shunt windings get interchanged by mistake and the motor is

connected to the rated voltage,

a) neither of the windings get short-circuit

b) series winding gets overloaded

c) shunt winding gets overloaded

d) both of the winding get overloaded

View Answer

Answer: a

Explanation: In the short shunt configuration, the winding interchanging

does not change the configuration of the set up so none of the winding get

affected.

2. The main concern before doing hopkinson’s test for finding efficiency is

a) needs one motor and one generator is that it

b) ignores iron and stray losses

c) needs one motor

d) requires identical shunt machines

View Answer

Answer: d

Explanation: It is must that both the machines are identical as required by

the calculations.


3. Retardation test on DC shunt motor is used for finding _______

a) stray losses

b) copper losses

c) friction losses

d) iron losses

View Answer

Answer: a

Explanation: Retardation test is used for finding the stray losses.

4. The quantities needed to complete retardation test are/is

a) dw/dt and moment of inertia

b) dw/dt

c) current

d) any of the mentioned

View Answer

Answer: a

Explanation: We need the angular acceleration and the moment of inertia of

the machine.

L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 28

28

5. ______ test is used for determining the efficiency of a traction motor

a) Field

b) Retardation

c) Hopkinson

d) Swineburne’s

View Answer

Answer: a

Explanation: Field test is used for finding efficiency of a traction motor.

6. If the field current and armature current are reversed, then

a) direction of rotation remains same

b) direction of rotation reverses

c) stops

d) none of the mentioned

View Answer

Answer: a

Explanation: When both field current and armature current are reversed then

the direction will not change.


7. Mark the possible causes of overheating of commutator in a DC machine.

a) Overload

b) Restricted ventilation

c) Shorted winding

d) Any of the mentioned

View Answer

Answer: a

Explanation: All the factors can cause the overheating of commutator.

8. In the hopkinson’s test on two DC machines. Machine A has field current of

1.4A and B has field current of 1.3A. Which machine acts as motor?

a) B

b) A

c) A,B

d) None of the mentioned

View Answer

Answer: a

Explanation: Ea/Eb = 1.4/1.3

Ea= 1.077 Eb > Eb

So, the machine will act as motor. L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 30 30

9. In the Hopkinson’s test on two DC machines. Machine A has field

current of 1.4A and B has field current of 1.3A. Which machine acts as

generator?

a) B

b) A

c) A,B


View Answer

Answer: b

Explanation: Ea/Eb = 1.4/1.3

Ea= 1.077 Eb > Eb

So, the machine will act as generator.

10. Hopkinson’s test is also called regenerative test due to

a) energy of one machine is used to drive other

b) feedback

c) losses are least

d) extra motor is used

View Answer

Answer: a

Explanation: Regenerative phenomena utilizes the energy of the running

machine to tap the dynamically produced energy and use it to charge the

batter


11. The most appropriate relation to find efficiency of the generator

a) output/(output + losses)

b) (output – losses)/input

c) (output – losses)/(output + losses)


View Answer

Answer: a

Explanation: For a generator the output can be measured and so this

expression is used.

12. The efficiency in the swineburne’s test can be found.

a) True

b) False

View Answer

Answer: a

Explanation: Because the constant losses are known.


32

13. The possible assumption is/are made while doing swineburne’s test

(i) mechanical loss constant

(ii) Armature reaction neglected

(iii) Increases in flux

a) (i) and (ii)

b) (iii) and (i)

c) (i), (ii), and (iii)

d) (i)

View Answer

Answer: a

Explanation: There is decrease in flux due to positive temperature coefficient of

resistance in shunt machine.

14. Swineburne’s test is applicable to those machines in which flux is practically

_______

a) constant

b) linear

c) non-linear


View Answer

Answer: a

Explanation: Because the constant losses are known in advance of the test is

conducted L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 33

33

15. Similarity between Hopkinonson’s and field’s test is that

a) both need similar mechanically coupled motor

b) both need similar electrically coupled motor

c) regenerative power

d) negligible power

View Answer

Answer: a

Explanation: The machines have to be mechanically coupled together.

16. Inspite of heavy initial investments, dc motors are used due to

a) flexibility and ease of control

b) lower losses

c) improved power factor of the system

d) all of the mentioned

View Answer

Answer: a

Explanation: The control of dc machines is very much simplified when

compared to other machines. So this makes it very useful when compared.


17. Separately excited dc generators are used in

a) Ward leonard system of speed control

b) Hopkinson’s testing

c) Voltage control


View Answer

Answer: a

Explanation: Separately excited dc generators are used in Ward leonard

system of speed control.

18. Maximum torque in dc series motor is limited by

a) commutation

b) heating

c) field control


View Answer

Answer: a

Explanation: Commutation is the process which reduces the induced emf and

so the torque.


19. Hoists, cranes and battery powered vehicles use _________ motors in

the locomotive.

a) dc series

b) dc shunt

c) induction

d) reluctance

View Answer

Answer: a

Explanation: Hoists, cranes require large starting torque which can be

provided by dc series motor.

20. There is an application which required pulsating loads, punch presses.

The most preferred machines would be

a) compound dc machine

b) series dc machine

c) shunt dc machine


View Answer

Answer: a

Explanation: It is compound machine whose flux can varied higher as

well as lower in order to have variable torques.


21. The manufacturer has mentioned a medium starting torque and 15%

speed regulation. The most appropriate motor for his requirement is

a) dc shunt motor

b) induction motor

c) differential motor


View Answer

Answer: a

Explanation: The dc shunt motor provides a medium values of the torque.

22. Centrifugal pumps, fans-blowers use

a) shunt as well as induction motor

b) only shunt motors

c) only induction motor


View Answer

Answer: a

Explanation: Centrifugal pumps, fans-blowers use shunt as well as

induction motor. L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 37

37

23. Most commercial compound dc generator are normally supplied by

manufacturers as over compound machines because

a) degree of compounding can be adjusted by diverters acrosss series

field

b) they have ideally best for HVDC

c) cost effective than shunt

d) zero percent regulation

View Answer

Answer: a

Explanation: It is usually over compounded so that degree of

compounding can be adjusted by diverters across series field.

24. Which of the following are applied with differential compound

motor?

a) Hoist

b) Cranes

c) Drilling machines


View Answer

Answer: a

Explanation: Practically it is never used as the speed can rise very high. L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 38 38

25. Most commercial compound dc generator are normally supplied by

manufacturers as ______ compounded machines.

a) over

b) under

c) level


View Answer

Answer: a

Explanation: Most commercial compound dc generator are normally

supplied by manufacturers as over compounded.

26. Long back ago, magnetic casette were in use, having permanent

magnet dc motors which had

a) magnets on stator and armature on rotor

b) magnet on rotor and armature on stator

c) electronic commutation


View Answer

Answer: a

Explanation: PMDC motors have field fixed on the stator and rotor has

coils. L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 39

39

27. If a dc shunt motor is stopped by forcing starter handle back to OFF

position, then

a) dangerous sparking occurs at last stud

b) dangerous sparking occurs at all stud

c) it stops normally

d) heavy sparking

View Answer

Answer: a

Explanation: The sparking will occur at the last stud not all the studs of

the starter.

28. Interpoles are used to start the dc motor above the base speed.

a) True

b) False

View Answer

Answer: b

Explanation: Interpoles do not reduce the flux across the windings,

hence the speed does not rise above base speed.


29. A 200 V dc motor has external resistances of Ra and Rf in armature and

field circuits respectively. The starting current is reduced when

a) Ra is maximum and Rf minimum

b) Rf is maximum and Ra minimum

c) Ra is minimum and Rf minimum

d) Ra is maximum and Rf maximum

View Answer

Answer: a

Explanation: When Ra is maximum and Rf minimum then the current to the

armature is reduced. And hence the sparkings are reduced while starting.

30. A d.c. generator has been provided tappings on the armature winding at

intervals of 120 degrees from the side of commutator. The connection will be

a) delta connected alternator

b) star connected alternator

c) star connected induction motor


View Answer

Answer: a

Explanation: Dc generator if has displaced by 120 degrees physically then it

will be act as ac machine. L E C T U R E 3 - L O S S E S A N D L O A D T E S T S O N D C M A C H I N E S 41

41

31.


32.


33.


34.


35.


36.


37.


38.

Ans : C


39.


40.

Ans : C


41.

Ans : C


53

Few exercise Problems

What will happen when the D.C motor is connected to

A.C supply then?

1.D.C motor will run at rated speed

2.D.C motor will burn

3.D.C motor will run at slow speed

4.Both 2 & 3.

1.

54

If Ta be the armature torque and Ia be the armature

current then which of the following relation is valid for DC

series motor before saturation?

1.Ta ∝ Ia

2.Ta ∝ Ia2

3.Ta ∝ 1/Ia

4.Ta ∝ 1/Ia2

2.

55

Which DC motor is preferred for Elevator?

1.Differentially compound motor

2.Series motor

3.Shunt Motor

4.Cumulative compound motor

3.

56

In the DC machine, the fractional pitch winding

is used

1.To reduce the Harmonic in generated EMF

2.Improve Cooling

3.Increase EMF

4.To reduce the copper losses

4.

57

A 4 pole, lap wound d.c motor has 540 conductors and

230 V supply. The armature resistance is 0.8Ω. Its

speed found to be 1000 r.p.m. the flux per pole is 25

mWb. The Induced e.m.f.and Armature current will be

1.250 V, 10 A

2.300 V, 6.25 A

3.225 V, 10 A

4.225 V, 6.25 A

5.

58

A short shunt compound generator supplies a load current of 100

A at 250 V. The generator has the following winding resistances:

shunt field = 130Ω, Armature 0.1 Ω and series field = 0.1 Ω . Find

the Emf generated if the brush drop is 1 V per brush.

1.272.2 V

2.262.2 V

3.272.0 V

4.262.0 V

6.

59

For any queries/clarifications/,suggestions...feel free to

contact through...

[email protected]

62

Solutions for exercise Problems

What will happen when the D.C motor is connected to A.C supply then?

1.D.C motor will run at rated speed

2.D.C motor will burn

3.D.C motor will run at slow speed

4.Both 2 & 3.

Answer 4. Either 2 & 3

Explanation:

Series Connection:

•In DC machine, T = kφi

where k = number of poles and the number of turns in the winding.

Φ = flux per pole

i = current.

•If saturation of core is neglected, then Flux is proportional to field current.

•So Torque= k *field current*armature current.

•For DC series machine, Armature current = Field Current,

Therefore, Torque = k × Ia2

•If we apply AC then both, during positive as well as negative half cycle, torque is

positive from above relation. So, torque is unidirectional, but due to AC, it will be

pulsating.

Parallel connection:

•But, In the case of parallel connection, it won’t rotate at all and will start

humming and will create vibrations, as a torque produced by positive and

negative cycle will cancel out each other. D.C. motor will be heated

1.

63

If Ta be the armature torque and Ia be the armature current then which of the

following relation is valid for DC series motor before saturation?

1.Ta ∝ Ia

2.Ta ∝ Ia2

3.Ta ∝ 1/Ia

4.Ta ∝ 1/Ia2

Answer 2. Ta ∝ Ia2

Explanation:

•We know that torque is directly proportional to the product of armature

current and field flux, Ta ∝ ɸ.Ia.

•In DC series motors, field winding is connected in series with the armature, i.e.

Ia= If. Therefore, before magnetic saturation of the field, flux ɸ is directly

proportional to Ia.

•Hence, before magnetic saturation Ta ∝ Ia2.

•After magnetic saturation of the field poles, flux ɸ is independent of armature

current Ia. Therefore, the torque varies proportionally to Ia only, T ∝

Ia.Therefore, after magnetic saturation, Ta-Ia curve becomes a straight line.

2.

64

Which DC motor is preferred for Elevator?

1.Differentially compound motor

2.Series motor

3.Shunt Motor

4.Cumulative compound motor

Answer 4. Cumulative compound motor

Explanation:

•Compound motor finds application where high starting torque

or pulsating load are required.

•In the elevator, the load is always pulsating and high starting

torque is required to run the elevator.

•This high starting torque is provided by the series winding of

the compound motor and maximum speed at no load is limited

by shunt winding of the compound motor.

3.

65

In the DC machine, the fractional pitch winding is used

1.To reduce the Harmonic in generated EMF

2.Improve Cooling

3.Increase EMF

4.To reduce the copper losses

Answer 1. To reduce the Harmonic in generated EMF

Explanation:

•In full pitched coils, as one conductor of a turn in a coil cuts N pole, the other

conductor of the same turn cuts S pole resulting in the production of induced

emf(E), i.e phase angle is 180 degree

•In short-pitched coils, both conductors of the same turn in a coil don’t cut the

respective poles simultaneously. so phase angle is slightly less than 180 degree As a

result, the magnitude of induced emf gets reduced to E × Cos(nα/2)

Where E × Cos(nα/2) is called a pitch factor

For eliminating 3rd harmonic from Generated EMF,

Cos(3α/2) = 0

3α/2 = π/2

α = π/3 = 60°

4.

66

A 4 pole, lap wound d.c motor has 540 conductors and 230 V supply. The armature

resistance is 0.8Ω. Its speed found to be 1000 r.p.m. the flux per pole is 25 mWb. The

Induced e.m.f.and Armature current will be

1.250 V, 10 A

2.300 V, 6.25 A

3.225 V, 10 A

4.225 V, 6.25 A

Answer. D. 225 V, 6.25A

Explanation:-

Given Data

Number of Poles P = 4

The number of parallel Path A = 4 (In Lap wound the number of brush required by

this winding is always equal to the number of poles.)

Total number of armature conductor Z = 540

Flux per Pole Φ = 20 mWb = 25 x 10-3 Wb,

Armature current Ia =?

Back EMF Eb =?

(i) Back EMF of DC motor

= (PΦZN)/(60A) = (25 x 10 -3 x 4 x 1000 x 540)/(60 x 4) = 225 V

Eb = 225 V

(ii) Armature current

From the voltage equation the back EMF of DC motor is

Eb = V − IaRa

225 = 230 − 0.8Ia

Ia = 6.25 A

Hence the back EMF is 225 V and Armature current is 6.25 A

5.

67

A short shunt compound generator supplies a load current of 100 A at 250 V. The

generator has the following winding resistances:

shunt field = 130Ω, Armature 0.1 Ω and series field = 0.1 Ω . Find the Emf

generated if the brush drop is 1 V per brush.

1.272.2 V

2.262.2 V

3.272.0 V

4.262.0 V

Answer.1. 272.2 V

Explanation:-

Voltage = 250 V

Load current (IL) = 100 A

Shunt field (Rf) = 130Ω

Armature field (Ra) = 0.1Ω

Series field (Rse) = 0.1Ω

Brush voltage Drop = 1V

Now voltage drop in the field winding = ILRse

= 100 x 0.1 = 10 V

For short shunt compound generator, the voltage drop across series field winding

gets added with voltage drop across armature winding

Voltage drop across field winding + Voltage drop across armature winding

V1 = 250 + 10 = 260 V

Field current If = V1/Rf = 260/130 = 2 A

So armature current

Ia = IL + If = 100 + 2 = 102 A

Generated EMF

E = V1 + IaRa

= 260 + 102×0.1

E = 270.2 Volt

6.

Prof. Ch. SAI BABU,

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