EEE Chapter 2.pdf

GNIT EEE DEPARTMENT Electrical and Electronics Engineering 30

UNIT-II

DC GENERATORS

2.1 Introduction:

An electrical machine, deals with the energy transfer either from mechanical to electrical

form or electrical form to mechanical form. This process is called electromechanical energy

conversion.

An electrical machine which converts mechanical energy (or power) into electrical

energy (or power) called an electric generator.

While an electrical machine which converts an electrical energy into mechanical energy

is called an electrical motor.

The d.c. Machines are thus classified as,

1. The D.C. generators: These machines convert mechanical input power into electrical

input power.

2. D.C. Motors: These machines convert d.c electrical power into mechanical power.

The construction of both the types’ d.c machines basically remains same.

2.1.1Working Principle of D.C. Machine as a Generator:

All generators work on the principle of dynamically induced e.m.f.

This principle is nothing but the Faraday’s law of electromagnetic induction. It states

that, ‘whenever the number of magnetic lines of force i.e. flux linking with a conductor or a coil

changes, an electromotive force is set up in that conductor or coil.

The magnitude of induced e.m.f. in a conductor is proportional to the rate of change of

flux associated with the conductor. This is mathematically given by

풆( 풎풂품풏풊풕풖풅풆) ∝ 풅∅풅풕

The relative motion can be achieved by rotating conductor with respect to flux or by rotating flux

with respect to a conductor.

So a voltage gets generated in a conductor, as long as there exists a relative motion between

conductor and the flux.

Such an induced e.m.f which is due to physical movement of coil or conductor with respect to

flux or movement of flux with respect to coil or conductor is dynamically induced e.m.f.


So a generating action requires following basic components to exist ,

i) The conductor or a coil

ii) The flux

iii) The relative motion between conductor and the flux.

2.2 Simple loop Generator:

Working principle of a DC Generator:-

An electrical generator is a machine which converts mechanical energy (or power) into

electrical energy (or power). The energy conversion is based on the principle of the production of

“dynamically” induced EMF.

As seen whenever a conductor cuts the magnetic flux in it according to “Faraday’s laws

“of electromagnetic induction. This EMF causes a current to flow if the conductor circuit is

closed.

Construction: - Fig.2.1 shown a single turn rectangular copper coil ABCD rotating about its

own axis in a magnetic field provided by either permanent magnet (or) electromagnets. The two

ends of the coil are joined to two slip-rings ‘a’ & ‘b’ which are insulated from each other and

from the central shaft.

Fig.2.1 Simple loop Generator

Two collecting brushes (of 1, 2 carbon or copper) press against the slip rings. Their

function is to collect the current induced in the coil and convey it to the external load resistance

R.

The rotating coil may be called ‘armature’ and the magnets as ‘field magnets’


Working: - Imagine the coil to be rotating clockwise direction. As the coils assumes successive

positions in the field, the flux linked with it changes. Hence an e.m.f. Is induced in it which is

proportional to the rate of change of flux linkages i.e.

When plane of the coil is at right angles to the lines of flux. i.e when it is at position 1 ,

then flux linked with the coil is maximum but rate of change of flux linkages is minimum.

It is do because in this position, the coil sides AB and CD do not cut or shear the flux;

rather they slide along them i.e. they move parallel to them. Therefore EMF induced in the coil

is zero.

As the coil continuous making, the rate of change of flux linkages increases gradually till

position 3 is reached where 휃 = 90°. Here coil plane is parallel to the line of flux. As seen, the

flux linked with the coil is minimum but rate change of flux linkages is maximum. Hence

maximum EMF is induced in the coil when in this position shown in Fig.2.2 and Fig.2.3.

In the next quarter revolution i.e., from 90° to 180° the flux linked with the coil gradually

increases, but the rate of change of flux linkages decreases. Therefore the induced EMF

decreases gradually till in position (5) of the coil. It is reduced to zero.

So we find out that in the first half revolution of the coil, no EMF is induced in it when in

position (1), maximum ‘EMF’ induced when in position (3) & no EMF induced when in position

(5). The direction of this induced EMF can be determined by applying Flemings Right Hand

Rule which gives direction from A to B & C to D. hence the direction of current from ABMLCD

i.e., current through load ’R’ flow from ‘M to L’.

In the next half revolution i.e., from 180° to 360° the variation in the magnitude of EMF are

similar to those in the first half revolution. Its value is maximum when coil is in position (7) &

풆 = 푵풅∅풅풕


minimum when in position (1).It is seen that the direction of the induced current is from D to C

& B to A as shown Fig.2.1 Hence the path of the current flow is along DCLMBA & current

through load ‘R’ is from L to M.

Therefore, we find that the current which we obtain from such a simple generator reverses its

direction after half revolution. Such a current is called an alternating current. For making the

flow of alternating current into unidirectional current in the external circuit. The slip-rings are

replaced by split rings (or) commutator shown in Fig.2.4. These slip rings made of conducting

cylinder which cut into two halves (or) segments insulated from each other by a thin sheet of

mica or some other insulating material shown in Fig.2.5.

Fig.2.6(a) shows that in the first half revolution current flows along(ABMNLCD) i.e., the

brush no.(1) in conduct with segment ‘a’ acts as the +ve end of the supply and ‘b’ as –ve end.

In the next half revolution [Fig.2.6 (b)] the direction of the induced current in the coil is reversed.

But the same time position of segment ’a’ & ‘b’ have also reversed with result that brush no.1

comes in touch with the segment which is + ve i.e., segment ‘b’ in this case. Hence the current in

the load resistance again flows from “M to L”. The waveform of the current through the

external circuit is shown in Fig.2.7.

“This current is unidirectional but not continuous like direct current”


2.3 Construction of DC Machines:

As stated earlier whether a machine is d.c. generator or a motor the construction basically

remains the same as shown in the Fig.2.8 and Fig.2.9

Fig. 2.8 Cross view of D.C. Generator

Fig. 2.9 a cross – section of typical D.C. Machine


It consists of the following parts:

2.3.1 Yoke or Magnetic frame

The outer frame of a dc machine is called the “yoke” and generally made of cast iron or

fabricated steel for large machines.

The outer frame or yoke serves double purpose.

It provides mechanical support for poles and acts as a protecting cover for the whole

machine and

It carries the magnetic flux produced by the poles.

In small generators where cheapness rather than the weight is the main consideration, yoke is

made of cast iron. But for large machines usually cast steel or rolled steel is employed. The feet

and terminal box etc…, are welded to the frame afterwards. Such yokes possess sufficient

mechanical strength and have high permeability.

It provides a return path of low reluctance to the magnetic flux produced by the poles.

2.3.2 Poles

Each pole is divided into two parts namely, a) Pole core and b) Pole shoe.

This is shown in the Fig. 2.10

Fig.2.10 Pole structure

The pole cores which support the field windings are mounted on the inside circumference

of the yoke. The core is made of cast steel or sheet steel laminating high permeability.

The pole faces are shaped to fit the curvature of the armature as shown in fig. and is

known as the shoe of the pole. Pole shoe enlarges the area of armature core to come

across the flux , while necessary to produce larger induced e.m.f . To achieve this , pole

shoe has been given particular shape.

The pole core is usually smaller cross section than the pole shoe (pole face).

Due to following reasons

(a) The reduced cross section of the pole core requires less ‘cu’ for the field winding.


(b) The large pole shoe area increases the flux per pole entering the armature, due to the

reduction in air gap reluctance.

(c) Pole shoe provide mechanical strength and support to the field winding.

2.3.3 Field winding

The field winding is wound on the pole core with a definite direction

a) Functions : To carry due to which pole core , on which the field winding is placed

behaves as an electromagnet, producing necessary flux. As it helps in producing the

magnetic field i.e. exiting the pole as an electromagnet it is called field winding or exiting

winding.

b) Choice of material: It has to carry current hence obviously made up of some conducting

material. So aluminium or copper is the choice. But field coils are required to take any

type of shape and bend about pole core and copper has good flexibility i.e. it can bend

easily. So copper is the proper choice.

Field winding is divided into various coils called field coils. These are connected in series

with each other and wound in such a direction around pole cores , such that alternate ‘N’ and

‘S’ are formed.

The total number of poles is denoted by P.

2.3.4 Armature The armature is further divided into two parts namely,

i) Armature core and ii) Armature winding

i) Armature core: Armature core is cylindrical in shape mounted on the shaft. It consists of

slots on its periphery and the air ducts to permit the air flow through armature which

serves cooling purpose.

a) Functions:

1. Armature core provides house for armature winding i.e. armature conductors.

2. To provide a path of low reluctance to the magnetic flux produced by the field

winding.

b) Choice of material; as it has to provide a low reluctance path to flux, it is made up of

magnetic material like cast iron or cast steel.

It is made up of laminated construction keep eddy current loss as low as possible. A

single circular lamination used for the construction of the armature is shown in Fig.

2.11


ii) Armature winding: Armature winding is nothing but the interconnection of the

armature conductors, placed in the slots provided on the armature core periphery.

When the armature is rotated in case of generator, magnetic flux gets cut by

armature conductors and e.m.f gets induced in them.

a) Functions:

1. Generation of e.m.f. takes place in the armature winding in case of generators.

2. To carry the current supplied in case of d.c. Motors.

3. To do the useful work in the external circuit.

b) Choice of material: As armature winding carries entire current which depends on

external load, it has to be made up of conducting material, which is copper.

2.3.5 Commutator

The basic nature of e.m.f. induced in the armature conductors is alternating. This needs

rectification in case of d.c. Generator, which is possible by device called commutator.

a) Functions:

1. To facilitate the collection of current from the armature conductors.

2. To convert internally developed alternating e.m.f. to unidirectional (d.c) e.m.f.

3. To produce unidirectional torque in case of motors.

b) Choice of material: As it collects current from armature, it is also made up of copper

segments.

It consists of number of wedge shaped segments (or) bars made of high conductivity that

is assembled to form a cylinder.

These segments are insulated from each other by thin layers of mica. Each commutator is

connected to one end of the coil. The numbers of commutator segments are decided by the


number of coils. The commutator along with the brush gear helps to convert the ac voltage

induced in the armature conductors into unidirectional voltage across the brushes.

2.3.6 Brushes and Brush gear

Brushes are stationary and resting on the surface of the commutator

a) Functions: To collect current from the commutator and make it available to the

stationary external circuit.

b) Choice of material: Brushes are normally made up of soft material like carbon.

To avoid wear and tear of commutator , the brushes are made up of soft material like

carbon.

Bearings: Ball bearings are usually used as they are more reliable. For heavy duty

machines roller bearings are preferred.

2.4 Types of Armature winding

The number of armature conductors is connected in a specific manner to give armature winding.

According to way of connecting the conductors, the armature winding has two types,

a) Lap winding

b) Wave winding

In lap type, the connections overlap each other as the winding proceeds as shown in Fig.2.12

Fig.2.12 Lap winding

Due to the this, number of parallel paths in which conductors are divided is P where P = number

of poles in the machine.

A=P = Number of parallel paths for lap

Large number of parallel paths indicate high current capacity of machine hence lap winding is

preferred for high current rating generators.

In wave type, the winding travels ahead avoiding the overlapping as shown in the Fig.2.13 in a

progressive fashion.


Fig.2.13 Wave winding

Due to this, the armature conductors always get divided into two parallel paths, irrespective of

number of poles.

A=2 =Number of parallel paths for wave

2.5. DC Generator construction parts and their purpose and material used

S

No.

Parts Purpose Material used

1 Yoke Provides mechanical support for the poles acts as a

protecting cover for the whole machine carries the

magnetic flux.

Cast iron, cast steel

(or) rolled steel.

2 Pole cores Consists field winding Cast iron (or) cast

steel laminating.

3 Pole shoe Spreads out the flux in the air gap reduces the

reluctance supports the exciting coils.

Cast iron (or) cast

steel laminating.

4 Inter poles Located in b/w main poles to reduce the armature

reaction produces MMF & armature current

Cast iron (or) cast

steel laminating

4 Brushes To collect current from the commutator and delivers it

to the load.

Carbon (or) graphite

(or) cu.

5 Brush holders To support the brushes Metal

7 Pole coils To electro magnetize the poles and produce the

necessary flux.

Cu wire or strip

8 End bells &

bearings

To enclose the m/c support the armature shaft Ball bearings (or)

roller bearings

9 Armature core Rotating part of the m/c support the armature shaft Steel laminating

10 Armature

winding

Placed in the armature slots and are insulated from each

other to carry armature current

Cu

11 Commutator Convert ac voltage induced in armature conducting to

dc voltage in the external circuit

Wedge shaped ‘cu’

segments.


2.6 E.M.F. Equation of a D.C Generator.

The DC Generator operates on the principle of the “Faraday’s law of electromagnetic

induction”. It’s states that E.M.F. will be induced in the armature conductor when the armature is

rotated such as to cut the magnetic flux.

Let ∅ = useful flux per pole in Wb.

P = total no. of poles

Z = total no. of armature conductors

= No. of slots x No. of conductors /slot

A = No. of parallel paths

N = Speed of the armature in rpm.

E = E.M.F. induced in the armature conductor.

Generated E.M.F. Eg = e.m.f. generated by any one of the parallel paths i.e E

Average e.m.f. generated /conductor = ∅ volts (n =1)

Now , flux cut/ conductor in one revolution d∅= ∅P Wb

No. of revolutions /second = N/60 ( time for one revolution , dt=60/N second)

Hence , according to Faraday’s law of electromagnetic induction,

E.M.F. generated /conductor = ∅ = ∅ 푣표푙푡

This is the e.m.f. induced in one conductor. Now the conductors in one parallel path are

always in series. There are total Z conductors with A parallel paths , hence number of

conductors are always in series and e.m.f remains same across all the parallel paths.

Total e.m.f can be expressed as,

푬 = ∅푷푵ퟔퟎ

풁 푨 풗풐풍풕풔

This is nothing but the e.m.f equation of a d.c. generator.

So

푬 =∅푷푵ퟔퟎ

풁 푨 풗풐풍풕풔

푬 = ∅푵풁ퟔퟎ

풗풐풍풕풔 for lap type as A=P

푬 = ∅푷푵풁ퟏퟐퟎ

풗풐풍풕풔 for wave type as A=2


Solved Examples:

Example 2.1: Calculate the e.m.f. generated by a 6 pole lap wound armature with 65 slots and 12

conductors per slot, when driven at 1000 r.p.m. the flux per pole is 0.02 Wb.

Solution: Given data P=6 , ∅ = 0.02Wb , N= 1000 r.p.m , A=P as lap wound

Z= Slots x Conductors /slot = 65 x 12 = 780

퐸 = ∅ = . = 260푉

Answer : Generated e.m.f =260V

Example 2.2: The armature of 8 pole D.C. generator has 960 conductors and runs at 400 r.p.m.

The flux per pole is 40mWb.

a) Calculate the induced e.m.f , if armature is lap wound.

b) At what speed it must be driven to generate 400 V, if the armature is wave connected.

Solution: P=8, N=400 r.p.m , ∅ = 40푚푊푏 , Z=960

a) Lap wound , A=P=8

퐸 = ∅푍푁푃60퐴 =

40푋10 푋8푋400푋96060푋8 = 256푉

b) Wave connected , A=2 , Eg= 400V

퐸 = ∅푍푁푃

60퐴 푖. 푒. 400 = 40푋10 푋8 푋 푁 푋960

60푋8

N = 156.25 r.p.m

Example 2.3: A d.c generator generates an e.m.f of 520V. It has 2000 armature conductors ,

flux/pole of 0.013Wb , speed of 1200 r.p.m and the armature winding has 4 parallel paths . Find

the number of poles.

Solution: Given data

Generated e.m.f =520V, Z=2000 , flux/pole = 0.013Wb and

Speed =1200 r.p.m

Number of parallel paths = A =4 .

Number of poles =?

We have Generated e.m.f = 퐸 = ∅

푃 = ∅

푃 = .

= 4

Answer : Number of poles =P=4


2.7 Symbolic Representation of D.C. Generator

The armature is denoted by a circle with two brushes. Mechanically it is connected to another

device called prime mover. The two ends of armature are denoted as A1 - A2. The field winding

is shown near armature and the two ends are denoted as F1 - F2. The representations of field vary

little bit, depending on the type of generator.

The symbolic representation is shown in the Fig. 2.14. Many times an arrow (↑) is indicated near

armature. This arrow denotes the direction of current which induced e.m.f. will set up, when

connected to an external load.

Fig.2.14.Symbolic representation of D.C. generator

Key point: Every practical generator needs a prime mover to rotate its armature. Hence to avoid

complexity of the diagram, prime mover need not be included in the symbolic representation of

generator.

2.8 Types of D.C. Generators

2.8.1 Methods of Excitation

The magnetic field required for the operation of a d.c. generator is produced by an

electromagnet. This electromagnet carries a field winding which produces required magnetic flux

when current is passed through it.

Key point: The field winding is also called exciting winding and current carried by the field

winding is called an exciting current.

Thus supplying current to the field winding is called excitation and the way of supplying the

exciting current is called method of excitation.

There are two methods of excitation used for d.c. generators,

1. Separately excitation 2. Self excitation.

Depending on the method of excitation used, the generators are classified as,


1. Separately excited generator 2. Self excited generator.

In separately excited generator, a separate external d.c. supply is used to provide exciting current

through the field winding.

The d.c. generator produces d.c. voltage. If this generated voltage itself is used to excite the field

winding of the same d.c. generator, it is called self excited generator.

The various types of d.c generators are shown in Fig.2.15.

Fig.2.15. Types of D.C. Generators

2.8.2 Separately Excited Generator

When the field winding is supplied from external, separate d.c. supply i.e. excitation of field

winding is separate then the generator is called separately excited generator. Schematic

representation of this type is shown in the Fig.2.16

The field winding of this type of generator has large number of turns of thin wire. So length of

such winding is more with less cross-sectional area. So resistance of this field winding is high in

order to limit the field current.

Fig.2.16. separately exited generator


2.8.2.1 Voltage and Current Relations

The field winding is excited separately, so the field current depends supply voltage and

resistance of the field winding.

For armature side, we can see that it is supplying a load, demanding a load current of IL at a

voltage of Vt which is called terminal voltage.

Now Ia = IL

The internally induced e.m.f. E is supplying the voltage of the load hence terminal voltage Vt

is a part of E. but E is not equal to Vt while supplying a load. This is because when armature

current Ia flows through armature winding, due to armature winding resistance Ra ohms, there is

a voltage drop across armature winding equal to Ia Ra volts. The induced e.m.f. has to supply this

drop, along with the terminal voltage Vt . To keep Ia Ra drop to minimum, the resistance Ra is

designed to be very very small. In addition to this drop, there is some voltage drop at the

contacts of the brush called brush contact drop. But this drop is negligible and hence generally

neglected. So in all, induced e.m.f. E has three components namely,

i) Terminal voltage Vt

ii) Armature resistance drop Ia Ra

iii) Brush contact drop Vbrush

So voltage equation for separately excited generator can be written as,

퐄 = Vt + Ia Ra + Vbrush + Armature reaction drop

Where 퐄 = ∅퐏퐍ퟔퟎ

퐙 퐀

= Generated e.m.f

Generally Vbrush is neglected as is negligible compared to other voltages.

2.8.3 Self Excited Generator

When the field winding is supplied from the armature of the generator itself then it is said to be

self excited generator. Now without generated e.m.f., field cannot be excited in such generator

and without excitation there cannot be generated e.m.f. so one may obviously wonder, how this

type of generator works. The answer to this is residual magnetism possessed by the field poles,

under normal condition.

Practically though the generator is not working, without any current through field winding, the

field poles posses some magnetic flux. This is called residual flux and the property is called

residual magnetism. Thus when the generator is started, due to such residual flux, it develops a

small e.m.f. which now drives a small current through the field winding. This tends to increase

the flux produced. This in turn increases the induced e.m.f. and further increases the field current


and the flux. The process is cumulative and continues till the generator develops rated voltage

across its armature. This is voltage building process in self excited generators.

Based on how field winding is connected to the armature to derive its excitation, this type is

further divided into following three types:

i) Shunt generator

ii) Series generator

iii) Compound generator.

Let us see the connection diagrams and voltage, current relations for these types of generators.

2.8.3.1 Shunt Generator

When the field winding is connected in parallel with the armature and the combination across the

load then the generator is called shunt generator.

The Fig.2.17. Shows the symbolic representation of d.c. Generator.

The field winding has large number of turns of thin wire so it has high resistance. Let R be the

resistance of the field winding.

Fig.2.17. D.C .Shunt generator

From Fig.2.17 we can write

Ia = I + I

Now voltage across load is Vt which is same across field winding as both are in parallel with

each other

∴ I = Vt

R

While induced e.m.f. E, still requires supplying voltage drop Ia Ra and brushing contact drop.

∴ E = Vt + Ia Ra + Vbrush

where E = ∅

In practice, brush contact drop can be neglected.

The output of a d.c. Generator is given by product of terminal voltage and the load current.


While the product of generated e.m.f. and the armature current is called power developed in a

d.c. generator.

Pg= power developed = E x Ia

Pout = power output = Vt x IL

2.8.3.2 Series Generator

When the field winding is connected in series with the armature winding while the supplying the

load then the generator is called series generator. It is shown in the Fig.4.5.

Field winding, in this case is denoted as S and S . The resistance of series field winding is very

small and hence naturally it has less number of turns of thick cross-section wire as shown in the

Fig. 2.18

Let Rse be the resistance of the series field winding.

Fig.2.18. Series Generator

Voltage and Current Relations

As all armatures, field and load are in series they carry the same current.

∴ Ia = I = I

Where I = current through series field winding.

Now in addition to drop Ia Ra , induced e.m.f. has to supply voltage drop across series field

winding too. This is Ise Rse i.e. Ia Rse as Ia = Ise. So voltage equation can be written as,

∴ E = Vt + Ia Ra + Ise Rse + Vbrush

∴ E = v + I (R + R ) + Vbrush

Where E = ∅ volts

2.8.3.3 Compound Generator

In this type, the part of the field winding is connected in parallel with armature and part

in series with the armature. Both series and shunt field windings are mounted on the same poles.

Depending upon the connection of shunt and series field winding, compound generator is further

classified as:


i) Long shunt compound generator and ii) Short shunt compound generator.

i) Long Shunt Compound generator

In this type, shunt field winding is connected across the entire series combination of armature

and series field winding as shown in Fig.2.19.

From the Fig.2.18, Ia = Ise

Ia = Ish +IL

Voltage across shunt field winding is Vt

퐼 = Where 푅 = Resistance of shunt field winding.

And voltage equation is

E = V + I R + I R + V h Where R = Resistance of series field winding.

Fig.2.19 Long shunt compound generator

ii) Short Shunt Compound Generator

In this type, shunt field winding is connected , only across the armature, excluding series field

winding as shown in Fig.2.20.

Fig.2.20 short shunt compound generator

For the Fig.2.20, 퐼 = 퐼 + 퐼

And Ise = IL

퐼 = 퐼 + 퐼


The drop across shunt field winding is drop across the armature only and not the total Vt , in the

case.

So drop across shunt field winding is E- IaRa.

Now the voltage equation is

Now, Ise = IL hence , we can write ,

i.e.

Any of the two above expressions of can be used, depending on the quantities known while

solving the problems.

Solved Examples:

Example 2.4: A shunt generator delivers 450A at 230V and the resistance of the shunt field and

armature are 50Ω and 0.03Ω respectively . Calculate the generated e.m.f.

Solution: Given data Terminal voltage (Vt) = 230V , Ra=0.03Ω , Rsh= 50Ω and IL=450A

Generator circuit is shown in Fig.2.21.

Current through shunt field winding is Ish= = = 4.6A

Load current IL=450A

Armature current Ia = IL + Ish = 450+4.6=454.6A

Armature voltage drop Ia Ra = 454.6 x 0.03 = 13.6V

Now Eg = Terminal voltage + armature drop

= Vt + Ia Ra

= 230+13.6 = 243.6V

Answer: E.M.F generated in the armature =243.6V


Example.2.5: A long –shunt compound generator delivers a load current of 50A at 500V and has

armature, series field and shunt field resistances of 0.05Ω, 0.03Ω and 250Ω respectively.

Calculate the generated voltage and the armature current. Allow 1V per brush for contact drop.

Solution: Given data, load current = 50A , Terminal voltage =500V

Ra= 0.05Ω , Rse = 0.03 Ω and Rsh= 250 Ω

Generated voltage =? and Armature current =?

Generator circuit is shown in Fig.2.22

Current through armature and series winding is

Ia =Ise = IL + Ish = 50 +2 =52A

Voltage drop on series winding = Ise Rse = 52 x 0.03 =1.56V

Armature Voltage drop = Ia Ra = 52 x 0.05 = 2.6V

Drop at brushes = 2 x 1 =2V

Now Eg = + Ia Ra + Ise Rse +brush drop

= 500 +2.6+1.56+2 = 506.16 V

Answer : Generated Voltage (Eg) =506.16 V and Armature current = 52A

Example.2.6: A d.c. shunt generator runs at 400 r.p.m and delivers 500kW to bus bars having a

constant voltage of 400V. Assuming the field excitation to be constant at 5A. Calculate the speed

at which the generator must run if the load on it is to be reduced to 300kW. The armature

resistance is 0.015Ω .

Solution: We have

In the present problem , is given to be constant , therefore


Initial conditions :

Load current I1 = = 1250퐴

Armature current Ia = I1 +If = 1250 +5 =1255A

Generated e.m.f Eg1= Vt + Ia Ra = 400+1255 X 0.015 =418.8V

Final conditions : Load current I2 = = 750퐴

Armature current Ia = I2 +If = 750 +5 =755A

Generated e.m.f Eg2= Vt + Ia Ra = 400+755 X 0.015 =411.3V

Now =

푁 = 푁 = ..

푋400 = 393 푟. 푝.푚

Answer: Speed = 393 r.p.m


DC MOTORS 2.9 Principle and operation of DC Motor

1. Motor: An electric motor is a machine which converts electric energy into mechanical

energy.

2. Principle of operation: motor action is based on the principle that when a current carrying

conductor is placed in a magnetic field, it experiences a mechanical force. Force direction

is given by Fleming’s left-hand rule and whose magnitude is given by

F =BIL Newton

3. Constructional there is no difference between a D.C generator and D.C motor. In fact the

same d.c machine can be used

2.10 Back E.M.F in a D.C. Motor:

It is seen in the generating action, that when a conductor cuts the lines of flux, e.m.f. gets

induced in the conductor. In a d.c. motor, after a motoring action, armature stars rotating and

armature conductors cuts the main flux. So is there a generating action existing in a motor after

motoring action.

There is an induced e.m.f. in the rotating armature conductors according to Faraday’s law

of electromagnetic induction. This induced e.m.f. in the armature always acts in the opposite

direction to the supply voltage. This is according to the Lenz’s law which states that the direction

of the induced e.m.f. is always so as to oppose the cause producing it.

In a d.c. motor, electrical input i.e. the supply voltage is the cause for the armature current and

the motoring action and hence this induced e.m.f. opposes the supply voltage. This e.m.f. tries to

set up a current through the armature which is in the opposite direction to that, which supply

voltage is forcing through the conductor.

As this e.m.f. always opposes the supply voltage, it is called back e.m.f. and denoted as

E , basically it gets generated by the generating action which we have seen earlier in case of

generators. So its magnitude can be determined by the e.m.f. equation which is derived earlier.

So,

E = ∅퐏퐍퐙ퟔퟎ퐀

volts

Where all symbols carry the same meaning as seen earlier in case of generators.


Thus e.m.f. is shown schematically in the Fig. 2.23(a). so if V is supply voltage in volts

and R is the value of the armature resistance, the equivalent electric circuit can be shown as in

the Fig. 2.23(b).

2.10.1 Voltage Equation of D.C. Motor

From the equivalent circuit, the voltage equation for a d.c. motor can be obtained as,

V = E + I R + Brush drop

The brush drop is practically neglected.

Hence the armature current I can be expressed as,

I =

2.10.2 Significance of Back E.M.F.

Due to the presence of back e.m.f. the d.c. motor becomes a regulating machine i.e. motor

adjusts itself to draw the armature current just enough to satisfy the load demand.

The basic principle of this fact is that the back e.m.f. is proportional to speed, E α N.

When load is suddenly put on to the motor, motor tries to slow down. So speed of the

motor reduces due to which back e.m.f. also decreases. So the net voltage across the armature

(V − E ) increases and motor draws more armature current.

Due to the increased armature current, force experienced by the conductors and hence the

torque on the armature increases. The increase in the torque is just sufficient to satisfy increased

load demand.

When load on the motor is decreased, the speed of the motor tries to increase. Hence back

e.m.f. increases. This causes (V − E ) to reduce which eventually reduces the current drawn by

the armature. The motor speed stops increasing when the armature current is just enough to

produce the less torque required by the new load.


So back e.m.f. regulates the flow of armature current and it automatically alters the

armature current to meet the load requirement. This is the practical significance of the back

e.m.f.

At start the speed N of the motor is zero hence the back e.m.f. is also zero.

2.11 Torque equation of D.C motor :

Torque: The turning or twisting about an axis is called torque.

Torque = force ∗ radius at which this force acting = F x r

Consider a pulley of radius r meter acted by circumferential force of F Newton which causes it to

rotate at ‘N’ rpm as shown in Fig.2.24.

T = F x r (N-M)

Work done by this force in one revolution = force x distance = Fx2휋r

Power developed = = π

= π

= ( ) π = T x W

P = T x W T in Nm W in rad/sec

Power in armature = Armature torque x W

E I = T x π

But E = ∅퐙퐍 ퟔퟎ

x 퐏 퐀

∅퐙퐍 ퟔퟎ

x 퐏 퐀

x I = T x π

T = ∅ ퟐ훑

x 퐏 퐙퐀

= 0.159∅ I 퐏 퐙퐀

N-m

T = 0.159∅I 퐏 퐙퐀

N-m ∴ T ∝ ∅I

Armature Torque (T ) : Torque developed in armature is called armature torque.


Lost Torque (T ) : The torque required to overcome frictional, windage and iron losses is called

T .

Shaft Torque (T ) : the torque which is available at the shaft for doing the useful work is known

as load torque or shaft torque denoted as T .

2.12 Types of D.C motor

The classification of d.c motors is based up on how the armature winding is connected

with field winding. The different types of d.c motors are (i) series motors (ii) shunt motors (iii)

compound motors. Compound motors are further classified as short compound and long

compound motors.

Fig.2.25 Classification of DC motor

2.12.1 D.C series motors

In this type of motor, field winding is connected in series with armature winding as well as

power supply as shown in Fig.2.26

Let Rse the series field winding resistance and it is very small and it is made of small no of turns

having large cross sectional are (thick wire).

I = line current drawn from the supply

I = I = I

Dc Motor

Series Shunt compound

Differentially compound

Cummulatively Compound


Applying the KVL in the above connection diagram of d.c series motor

-V+ I R + I R + E + V = 0

V = E + I (R + R ) + V [ I = I ]

V (Voltage drop across brushes) is small

neglected V = E + I (R + R )

In d.c series motor, entire armature current is passing through the series field winding. “∅”

produced is proportional to I

∅ α I α I for series motor

2.12.2 D.C Shunt motor

In this type, the field winding is connected across (parallel) the armature winding and

combination is connected across the supply , as shown in the Fig.2.27

Let R is the shunt field winding resistance and it is quite large and is made up of large no of

turns and thin wire.

The value of armature resistance (R ) is small.

I = I + I

I = I − I

And I =

Applying KVL -V+E + I R + V = 0

V = E + I R + V

In shunt motor, flux produced by field winding is proportional to the field current I .

∅ α I

Here, the input voltage is constant and so the flux is constant. Therefore shunt motor is a

constant. Flux motor or constant speed motor.


2.12.3 D.C Compound motor :

It consist of two field windings one field winding is connected in series with armature

called series field winding and another winding connected across the armature is called shunt

field winding. It is further classified as

a) Long shunt compound motor

b) Short shunt compound motor.

a) Long shunt compound motor :

In this type of motor, the shunt field winding is connected across the combination of armature

winding and series field winding as shown in Fig.2.28

Let Rse be the resistance of series field and Rsh be the resistance of the shunt field winding.

The total current drawn from supply is IL.

So 퐼 = 퐼 + 퐼

But 퐼 = 퐼 i.e. 퐼 = 퐼 + 퐼 and 퐼 =

and 푉 = 퐸 + 퐼 푅 + 퐼 푅 + 푉 but as 퐼 = 퐼

푉 = 퐸 + 퐼 (푅 + 푅 ) + 푉

b) Short shunt compound motor :

In this type, the shunt field is connected purely in parallel with armature and the series field is

connected in series with the combination shown in the Fig.2.29.


The entire line current is passing through the series field winding .

퐼 = 퐼 and 퐼 = 퐼 + 퐼

Now the drop across the shunt field winding is to be calculated from the voltage equation.

So 푉 = 퐸 + 퐼 푅 + 퐼 푅 + 푉

But 퐼 = 퐼

푉 = 퐸 + 퐼 푅 + 퐼 푅 + 푉

Drop across the shunt field winding = 푉 − 퐼 푅 = 퐸 + 퐼 푅 + 푉

퐼 = 푉 − 퐼 푅

푅 =퐸 + 퐼 푅 + 푉

푅

2.13 Necessity of Starter:

The starter is required to start a d.c. motor but it enables us to start the motor in a desired safe

way.

At the starting instant the speed of the motor is zero, (N=0) . As speed is zero, there cannot be

any back e.m.f as E ∝ 푁 and N is zero at start.

∴ E at start = 0

The voltage equation of a d.c motor is , V = E + I R

So at start, V=I R as E =0

∴ 퐈퐚 = 퐕퐑퐚

……At start

Generally motor is switched on with normal voltage and as armature resistance is very small, the

armature current at start is very high.

So at start, motor is showing a tendency to draw an armature current which is 15 to 20 times

more than the full load current.

Such high current drawn by the armature at start is highly objectionable for the following

reasons:

1. It may affect the performance of the other equipments connected to the same line.

2. Such excessively high armature current, blows out the fuses.

3. A large armature current flowing for a longer time may burn the insulation of the

armature winding.

To restrict this high starting armature current, a variable resistance is connected in series

with the armature at start. This resistance is called starter or a starting resistance. So starter is

basically a current limiting device.


In the beginning the entire resistance is in series with the armature and then gradually

cuts off as motor gathers speed, producing the back e.m.f.

In addition to the starting resistance, there are some protective devices provided in a

starter.

2.13.1 Three Point Starter

The Fig.2.30 shows three point starter.

Fig.2.30.Three point starter.

The starter is basically a variable resistance, divide into number of sections. The contact

points of these sections are called studs and brought out separately shown as OFF, 1, 2, … up

to RUN.

There are three main points of this starter:

1. ‘L’ → Line terminal to be connected to positive of supply.

2. ‘A’ → To be connected to the armature winding.

3. ‘F’ → To be connected to the field winding.

Point ‘L’ is further connected an electromagnet called Overload Release (OLR). The

second end of ‘OLR’ is connected to a point where handle of the starter is pivoted. This handle is

free to move from its other side against the force of the spring. This spring brings back the

handle to the OFF position under the influence of its own force.


Another parallel path is derived from the stud ‘1’, given to the electromagnet called No

Volt Coil (NVC). The NVC is further connected to terminal ‘F’. The starting resistance is

entirely in series with the armature.

The OLR and NVC are the two protecting devices of the starter.

Initially the handle is in the OFF position. The d.c. supply to the motor is switched on.

Then handle is slowly moved against the spring force to make a contact with stud No. 1.

At this point, field winding gets supply through the parallel path provided to starting resistance,

through NVC. While entire starting resistance comes in series with the armature and armature

current which is high at start, gets limited.

As the handle is moved further, it goes on making contact with studs 2, 3, 4 etc.., cutting

out the starting resistance gradually from the armature circuit. Finally when the starter handle is

in ‘RUN’ position, the entire starting resistance gets removed from the armature circuit and

motor starts operating with normal speed.

Functions of no volt coil:

1. The supply to the field winding is derived through NVC. So when field current flows, it

magnetizes the NVC. When the handle is in the ‘RUN’ position, soft iron piece

connected to the handle gets attracted by the magnetic force produced by NVC. Design of

NVC is such that it holds the handle in ‘RUN’ position against the force of the spring as

long as supply to the motor is proper. Thus NVC holds the handle in the ‘RUN’ position

and hence also called hold on coil.

2. Whenever there is supply failure or if field circuit is broken, the current through NVC

gets affected. It loses its magnetism and hence not in a position to keep the soft iron piece

on the handle, attracted. Under the spring force, handle comes back to OFF position,

switching off the motor. This prevents the damage of the motor caused due to accidental

starting.

3. NVC performs the similar action under low voltage conditions and protects the motor

from such dangerous supply condition as well.

Action of over load release:

The current through the motor is taken through the OLR which is an electromagnet.

Under overload condition, high current is drawn by the motor from the supply which

passes through OLR.


Below this magnet, there is an arm which is fixed at its fulcrum and normally resting in

horizontal position.

Under overloading, high current through OLR produces enough force of attraction to

attract the arm upwards. Normally magnet is so designed that up to a full load value of current,

the force of attraction produced is just enough to balance the gravitational force of the arm and

hence not lifting it up.

At the end of this arm, there is a triangular iron piece fitted. When the arm is pulled upwards the

triangular piece touches the two points which are connected to the two ends of NVC. This shorts

the NVC and voltage across NVC becomes zero due to which NVC looses bits magnetism. So

under the spring force, handle comes back to the OFF position, disconnecting the motor from the

supply.

Thus motor gets saved from the overload conditions.

2.14 Losses in D.C. machine: The various losses in a d.c. machine whether it is a motor or a generator are classified into three

groups as:

1. Copper losses

2. Iron or core losses

3. Mechanical losses.

2.14.1 Copper Losses:

The copper losses are the losses taking place due to the current flowing in a winding. There are

basically two windings in a d.c. machine namely armature winding and field winding. The

copper losses are proportional to the square of the current flowing through these windings. Thus

various copper losses can be given by,

Armature copper loss = I Ra

Where Ra = Armature winding resistance

And Ia = Armature current

Shunt field copper loss = I Rsh

Where Rsh = Shunt field winding resistance

And Ish = Shunt field current

Series field copper loss = I Rse

Where Rse = Series field winding resistance

And Ise = Series field current.


In a compound d.c. machine, both shunt and series field copper losses are present. In addition to

the copper losses, there exists brush contact resistance drop. But this drop is usually included in

the armature copper loss.

2.14.2 Iron or Core Losses:

These losses are also called magnetic losses. These losses include hysteresis loss and eddy

current loss.

The hysteresis loss is proportional to the frequency and the maximum flux density Bm in the air

gap and is given by,

Hysteresis loss = B . f V watts

= Steinmetz hysteresis coefficient

Where V = Volume of core in 푚

f = Frequency of magnetic reversals.

This loss is basically due to reversal of magnetization of the armature core.

The eddy current loss exists due to eddy currents. When armature core rotates, it cuts the

magnetic flux and e.m.f. gets induced in the core. This induced e.m.f. sets up eddy currents

which cause the power loss. This loss is given by,

Eddy current loss = K B 푓 푡 V watts

Where K = Constant

t = Thickness of each lamination

V = Volume of core

f = Frequency of magnetic reversals.

The hysteresis loss is minimized by selecting the core material having low hysteresis coefficient.

While eddy current loss is minimized by selecting the laminated construction for the core.

These losses are almost constant for the d.c. machines.

2.14.3 Mechanical Losses:

These losses consist of friction and windage losses. Some power is required to overcome

mechanical friction and wind resistance at the shaft. This loss is nothing but the friction and

windage loss. The mechanical losses are also constant for d.c. machine.

The magnetic and mechanical losses together are called stray losses. For the shunt and

compound d.c. machines where field current is constant, field copper losses are also constant.

Thus stray losses along with constant field copper losses are called constant losses. While the


armature current is dependent on the load and thus armature copper losses are called variable

losses.

Thus for a d.c. machine,

Total losses = Constant losses + Variable losses

The power flow and energy transformation diagrams at various stages, which taken place in a

d.c. machine are represented diagrammatically in Fig.2.31 (a) and Fig.2.31 (b).

2.14.4 Efficiency of a D.C. Machine:

For a d.c. machine, its overall efficiency is given by,

% =

× 100

Let Pout = Total output of a machine

Pin = Total input of a machine

Pcu = Variable losses

Pi = Constant losses

Then Pin = Pout + Pcu + Pi

∴ %Efficiency = × 100 = × 100

∴ %Efficiency =

× 100


Condition for Maximum Efficiency:

In case of d.c. generator the output is given by,

Pout = VI

Pcu = Variable losses = I Ra = I Ra

Ia = I …..Neglecting shunt field current

∴ %Efficiency =

× 100 = x 100

The efficiency is maximum, when the denominator is minimum. According to maximum minima

theorem, 푑푑퐼 1 +

IRV +

PVI = 0

RV −

PVI = 0

퐼 푅 − 푃 = 0

퐼 푅 = 푃 = 푃

Thus for the maximum efficiency, the condition is,

Variable losses =Constant losses

Solved examples:

Example.2.7: A 440V, shunt motor has armature resistance of 0.8Ω and field resistance of 200Ω.

Determine the back e.m.f when given on output of 7.46kW at 85 percent efficiency.

Solution: Given data V=440, Ra =0.8Ω, Rsh = 200Ω

Power output = 7.46kW and efficiency =85%

Back e.m.f (Eb) =?

Motor input Power = = . .

=8776.4W

Motor input current = .

x440 = 19.95A

Ish = 440/200 = 2.2A

Armature current Ia = 19.95 -2.2 = 17.75A

Now , Eb =V-IaRa

Eb = 440-(17.75 x0.8) =425.8V

Answer: Back e.m.f = 425.8 V


Example 2.8: A 250V shunt motor takes a total current of 20A.the shunt field and armature is

200훺 and 0.3 훺 respectively. Determine: a) The back e.m.f. b) Gross mechanical power

developed.

Solution: V= 250V, 푅 =o.3 훺 퐼 =20A , 푅 =200 훺

퐼 = = = 1.25퐴

퐼 =퐼 − 퐼 = 20 − 1.25 = 18.75퐴

a) 퐸 = 푉 − 퐼 푅 =250-18.75× 0.3 = 244.375푉

b) 푃 = 퐸 퐼 = 244.375 × 18.75 = 4582.0312푤

Example. 2.8: A 20 kW, 250V D.C. shunt generator has armature and field resistance of 0.1 훺

and 125 훺 respectively. Calculate the total armature power developed when running 1) As a

generator delivering 20kW output 2) As a motor taking 20kW input.

Solution: 1. As a generator

푃 =20kw, 푉 = 250푣

퐼 = = ×

=80A

퐼 = = = 2A

퐼 =퐼 + 퐼 = 80 + 2 = 82퐴

퐸 =푉 + 퐼 푅 = 250 + 82 × 0.1 = 258.2

푃 =퐸 × 퐼 = 258.2 × 82 = 21.172푘푊…………………armature power developed

2. As a motor

푃 = 20푘푊, V=250V

퐼 = = × = 80퐴

퐼 = = = 2퐴

퐼 =퐼 − 퐼 = 80− 2=78A

퐸 = 푉 − 퐼 푅 =250-78× 0.1 = 242.2푉

푃 = 퐸 퐼 = 242.2 × 78 = 18.8916푘푊……………Armature power developed


Previous year’s questions: 1. Deduce the relation between the Torque and armature current of D.C. Motor.

November/December, 2012

2. A 4-pole dc generator has a lap wound armature having 400 conductors. It generates an

e.m.f of 300V when flux per pole is 0.02Wb. Find the speed of rotation of its armature.

November/December, 2012

3. A) Derive EMF equation of a D.C generator.

B) With a neat diagram, explain the operation of a three point starter used in D.C motor.

May-June, 2012, Set-1, 15 Marks

4. Derive the torque equation of a D.C motor.

May-June, 2012, Set-2,7 Marks

5. A) Derive the torque equation of a D.C motor.

B) Mention three types of DC generators and write the application of each generator

C) Explain the working principle of DC generator.


6. A) Mention three types of DC motors and write the application of each motor.

B) Derive the torque equation of a D.C motor.

C) Draw a neat sketch of three point starter.


7. Write short notes on the following:

(a) Function of a commutator in DC machines

(b) Armature reaction in DC motors and remedies

(c) Speed-torque characteristics of a DC motor

(d) Back e.m.f in DC motors.

May, 2011, Set-2, 15 Marks

8. A 6-pole, 12 KW, 240 V, DC-machine is wave connected, if the same machine is lap

Connected, all other things remain same. Calculate its voltage, current and power ratings.

May, 2011, Set-4,15 Marks

9. A) Explain the principle of operation of DC motors. What is back e.m.f in DC

Motors? What are its effects?

B) How is back e.m.f produced in a DC motor? Also derive an expression for this e.m.f.

May, 2011, Set-3,15 Marks


10. A) Draw a neat sketch of a dc generator. State the functions of each part.

B) A 110V dc generator delivers a load current of 50A. It’s Ra = 0.2Ω; Rf= 55Ω. It is

rotating with speed of 1800 rpm, has 6poles, lap – wound, 360 conductors. Calculate the

no-load voltage at the armature and flux per pole.

December, 2011, Set-1,15 Marks

11. Explain in detail the principle of operation and construction of a dc motor and also derive

the torque expression.


12. A) Derive the EMF equation of a dc generator.

B) A dc shunt motor connected to a 230V dc supply takes a line current of 12A at same

load. If the field resistance and armature resistance are 230Ω and 1Ω respectively.

Calculate back e.m.f.


13. A 4-pole, long shunt, lap wound generator supplies 25 KW at a terminal voltage of 500

V. The armature resistance is 0.03 ohms, series field resistance is 0.04 ohms and Shunt

field resistance is 200 ohms. The brush drop may be taken as 1 V. Determine:

(a) the EMF generated

(b) Cu -Losses & iron Losses

(c) Efficiency at full load

November, 2010, Set-1,15 Marks

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