- 1 -

Lecture 1 & 2

DIRECT CURRENT MACHINE:

Fig. 1: Fluxes in a typical field system

The dotted lines represent the path of the main magnetic fluxes and full lines (1, 2, 3 & 4) represent leakage flux. The useful flux shown enters the

armature to produce e.m.f and mechanical forces. Electrical Machine: An electrical machine is an electromechanical device or Dynamo-electric

machine or more briefly dynamo which converts mechanical energy to electrical energy or vice versa.

The machine which converts mechanical energy into electrical energy is known as Generator, while the one converting electrical energy in to

mechanical energy is called Motor. However, electrical machine are

reversible capable of operating both as a generator and a motor. There are two types of electrical machines according to the nature of the

e.m.f and current in the armature conductors. They are: (a) Alternating Current (A.C) Machines. (b) Direct Current (D.C) Machine.

The first type has the current and voltage at the armature terminals alternating in direction where as the second type has unidirectional current

and voltage at the armature terminals. However, most of the DC machines except homopolar machines are basically similar to AC machines where

currents and e.m.fs in the armature conductors are alternating, but furnished with a special device – commutator which makes the current and e.m.f at the armature terminal unidirectional.

- 2 -

CONSTRUCTIONAL FEATURES AND ARMATURE WINDINGS OF D.C. MACHINES.

1. POLES: Pole cores are usually not laminated and are made of cast iron steel.

In some machines, pole cores are laminated and composed of electrical grade steel sheets of about 1mm thickness and insulated with respect to each other.

Solid poles are commonly bolted to the yoke. Laminated poles maybe secured in place by dovetail joint. As shown in diagram below

The flux density in the body of the pole core is much higher than that permitted in the air gap. Hence the pole face must have greater area than

in the pole core. The increased area is secured by means of pole shoes bolted or dovetailed

to the core. The pole face or pole shoes are always laminated to avoid heating and eddy losses. 2. Yoke.

It is part of the frame and carries the flux from pole to pole. It is made of cast iron or rolled steel in larger sizes

3. Armature Core. The armature core is the rotating part of D.C machine. The armature

conductors are placed in the slots punched on armature laminations

Fig. 2: One-piece of armature punching

The slots are usually parallel to the axis of the armature, but are also

sometimes skewed at a small angle to the axis to avoid vibrations Of the teeth.

The armature iron must be laminated and the laminations of thickness 0.35 to 0.5mm are insulated. Otherwise pole flux will induce e.m.fs and currents in armature iron causing eddy current losses.

4. Magnetic circuit According to ampere's law the line of the magnetic field strength along a

closed path is equal to the sum of the ampere turns with which this is linked. That’s,

∫ = NIHdl . Where, H is magnetic intensity.

- 3 -

Now taking one complete flux path through the central axes of two adjacent

poles of fig.1, the ampere-turns to be placed per pair of poles in the field winding for a certain pole flux is determined by the following equation:

∫ =Hdl Hyly+2Hplp+2Hgg+2Htlt+Hclc=2NI

Where the suffices y, p, g, t and c stands for yoke, pole, gap, tooth and core. H is the ampere-turns per unit length for the concerned part of the

magnetic circuit, l is the length of concerned part of the magnetic circuit per pole, g is the air gap and N is the number of turns per pole in the field

winding.

5. Commutator The end of the armature coils are joined to the commutator consisting of a

small number of wedge-shaped segments or bars of hard-drawn copper or drop-forged copper. The copper bars are separated from each other by sheet

of mica and insulated from their supporting rings by mica cones and gaskets. The bars are dovetail-fixed

6. Armature Windings There are two types of windings – Lap and Wave. (a). Lap windings:

Fig. 3: Simple lap winding

In fig 3 above, 1-12 and 3-14 are the coil sides of two adjacent coils, 12 and

14 being bottom coil-sides are shown in dotted. The end of the coil 1-12 and the beginning of the coil 3-14 are connected together through the

commutator segment 2.This is done by connecting the end of coil-side 12 and beginning of coil-side 3 to the same commutator segment 2. NB: It is important that coil side 12 and coil side 3 which are connected

should be under poles of different polarity so that the emfs of the corresponding coils are always additive.

b) Wave: The addition of emfs of consecutive coils can also be achieved by simple wave winding shown in fig 4 below.

Here ends of the coil-side 12 are connected to the beginning of the coil-side 21 which is situated under the next North Pole (N-pole). Two steps are needed from beginning of one coil to the beginning of the

next coil. The steps are one forward and one backwards for lap where as both are forward for wave.

- 4 -

Fig. 4: Simplex wave winding

WINDING PITCHES

a) Back pitch (Yb) - the distance between the top and bottom coil-sides of a coil. Thus in fig 3 and 4 Yb =12-1=11. b) Front pitch (Yf ) - the distance between the two coil-sides of the two coils

connected to the same commutator segment. In fig 3, Yf =12-3=9

In fig 4, Yf =21-12=9 It may be noted that both Yb and Yf are odd, because the two connected coil sides are from two different layers of slots, that is, if the number of one of

the coil-side is odd, the number of the other is even. c) Average pitch (Yav) - is defined as the mean of Yb and Yf.

That is Yav= (Yb+Yf)/2 d) Commutator pitch (Yc) - is the distance expressed in terms of commutator segments between the two commutator segments to which the

ends of a coil are connected. In simplex lap winding fig.(3), Yc=3-2=1, whereas in simplex wave winding fig.4, Yc=11-1=10.

For simplex lap winding Yc is always equal to one commutator interval.

COIL,COIL-SIDE,TURN,CONDUCTOR

A coil can have one or more number of turns, but it has only two coil sides. Let symbol (T) to represent the number of conductor per coil-side, so (T) is

also the number of turns per coil and the number of conductors in the coil will be (2T). To obtain satisfactory arrangement of the end connections, dc armature

windings are universally of the two-layer type. Each coil thus consists of an upper coil side placed at top of one slot and a lower coil-side situated at

bottom of another slot. The distance between them is known as coil-span.

Coil span is about one pole pitch (τ ).This is necessary so that emfs induced

in the conductors of the two coil-sides are always additive so as to get

minimum coil emf. Usually a slot contains more than two coil sides in two layers to reduce the number of slots in the armature for the sake of economy. The coils are

arranged side by side as shown in fig. 5 below for 2 turns per coil and six coil sides per slot, a, b are the 2 conductor of the coil side A, one of the three

coil sides of the three adjacent coils in the slot.

- 5 -

Fig. 5: Arrangement of conductors in a slot and a 2 turn coil

If the number of t-turn coils in a machine is (C) ,number of slots (S) and

number of coil sides per slot is (U),which is an even number, then total number of coil-sides (Zs)=2C=US and total number of conductors Z=tZs.

Fig. 6: Numbering of coil sides (2 turns per coil)

The coils of the armature have to be electrically connected amongst

themselves as well as with the commutator segments following a certain

repetitive pattern. For convenience of these connections or winding, the coil-sides are numbered (fig 6) where top coil-sides are numbered odd and

bottom coil sides are even numbered. SIMPLE LAP WINDING

Before calculation of winding pitches it is necessary to define progressive and

retrogressive lap winding. a) Progressive-when the winding progresses in the direction in which the

coils are wound. b) Retrogressive-when the winding progresses in the opposite direction in which the coils are wound.

Fig. 7: Progressive and Retrogressive simple lap winding

NB: In simplex lap winding, both Yb and Yf should be close to one pole

pitch so as to have additive emfs in the armature coils connected in series.

- 6 -

So, numberevenp

kZyysfb

..2

=±

=+

Where Zs = Total number of coil sides.

P = The number of poles K = Usually the integer to make Yav an even number.

Yav has to be even in the simplex lap because both Yb and Yf are odd and their difference is 2..

In calculating Yav,

1±= avb yy , 1mavf yy =

The commutator pitch 12

±=−

= fb

c

yyy

The upper signs are for progressive winding and lower for retrogressive

The retrogressive lap winding is preferred compared to the progressive lap, as the span of coils is less making these cheaper.

SIMPLEX WAVE WINDING In case of wave winding (fig.4),if the winding is traced from top coil side

1,the bottom coil side of the corresponding coil which is 1+Yb is connected to the top coil side number 1+Yb +Yf situated under the next pole of same

polarity as that of coil 1.In this way the coil side under the consecutive poles are connected till it passes through all p poles of the armature surface and the winding returns to either 3 (progressive) or Zs -1 (retrogressive),the top

coil-side next to the starting coil side 1.All the coils are connected in this manner and the winding finally ends in the starting coil-side 1.

When the winding traverses one pair of poles, the number of coil sides connected increases by Yb +Yf =2Yav ,when it passes through the whole armature with P/2 pairs of poles, the number of coil-side is to increase by

Zs ± 2 in case of simplex wave.

DUMMY COILS

It may be seen that for a wave winding ( )

p

Zy sav

2±= must be an integer.

Therefore it is not possible to have a wave winding with any number of coils.

For example: an armature core for a 4-pole armature has 65 slots and 4 coil sides can be accommodated per slot. That is, S=65 and U=4.The armature

then has coil sides Zs =4×65=260. But simplex wave winding is not possible with Zs =260 as Yav then will not be an integer. However, winding can be done 2 coil-side or 1 coil less, i.e. Zs =258.Then we may have

Yav=Yb=Yf =65. However, the extra coil called a dummy coil, is put in position to maintain the

mechanical balance of the armature, but is completely insulated from the rest of the winding.

- 7 -

Fig. 8: An armature with dummy coil

MULTIPLEX WINDING A multiplex winding is equivalent to two or more simplex windings in parallel

with one another. Thus a duplex winding has 2 such windings in parallel, triplex has 3 etc. Multiplex lap and wave windings, except a relatively few duplex winding are rarely in use. The number of parallel paths in the

armature is mp for multiplex lap and 2m for multiplex wave winding where m is the multiplicity of the winding.

IDENTICAL COILS IN WINDING (SYMMETRICAL) All armature coils should be of similar shape and size. This is only possible if

back pitch of coils Yb and number of coil sides per slot U satisfied certain relationship.

In fig (9a) here one coil has physical span more than that of others as the former has more number of slots in its span. In fig (9b) all coils are of identical span with equal length of connections. Note that in this case the coil

sides numbering 1 to 1−by fully occupy U

yb 1− number of slots. Thus the

armature to have similar and identical coils the following condition has to be

satisfied.

U

yb 1−=an integer

- 8 -

Fig. 9: Coil spans for windings

EQUALIZER RING

As seen the armature has a number of parallel paths across its terminals. If for any reason the emf generated in one path or circuit is greater than in the other, circulating currents will flow through the brushes, and if the machine is

loaded, he tendency will be to overload windings and some of the brushes. If preventive measures are not employed, even very small difference of

potential may give rise to high circulating or internal equalizing currents owing to the low resistance of the armature circuits. This results in excessive heating of the winding and brushes and sparking at the brushes. The possible

causes of unequal emfs in various paths are: 1. The air gap may vary for different poles and iron parts may have

considerable variation. 2. The armature circuit may be unsymmetrical owing to the use of a number of coils that is not an exact multiple of the number of paths.

3. The induced emfs may be unequal in different parallel paths if the positions of the conductors with respect to the poles are not similar.

The circulating currents, though a source of loss and extra heating of armature conductors are mainly objectionable because of the sparking at the brushes. Steps must be taken to prevent circulating current flowing through

the brushes, this is done by equipping the armature with equalizer rings, made up of conducting material, each ring connecting certain points in the

winding under ideal conditions have no difference of potential between them.

ADVANTAGES OF WAVE OVER LAP WINDING

1. A wave winding does not usually require equalizer rings.

2. Though wave winding requires only two brushes, it is usually fitted with as many brushes as there are poles. As such poor contact at any

brush does not impair satisfactory operation of the machine. 3. Most of the machines are wave wound. Lap windings may be used in

high power machines (above 500KW) to reduce the current per

armature path.

- 9 -

EXAMPLES:

1 The armature core of a 6-poles machine has 84 slots. The commutator has 252 segments. The windings are to have six coil – sides per slot. What

must be the front and back pitches so that the elements may be insulated in groups of three (i.e., symmetrical winding)

a. If the winding is to be a simplex lap. b. If it is to be simplex wave.

Solution:

1. The number of coil sides 5042252 == xZ s

Number of coil-sides per slot 6=u , Number of slots 846

504==

a. Simple lap:

,846

504==avy 85184 =+=by as

u

yb 1− is an integer to satisfy the

condition of symmetrical. 83=fy

b. Simple wave:

6

2504 ±=avy is not an integer. Therefore with one dummy coil of two-

sides, the number of the active coil-sides = 502.

,846

2504=

+=avy 85=by , 83=fy will make the winding symmetrical.

2. A four pole wave-connected armature has 1500 conductors to be

wound on a standard armature with 50 slots and 149 commutator segments. Calculate the winding pitches.

Solution: The number of coils C = number of commutator segments = 149. The number of coil-sides Zs = 2C = 298,

74,754

2298=

±avy

The number of coil-sides per slot ( )

650

2298=

+==

S

Zu s

There is one dummy coil with two coil-sides. For symmetrical winding

73=by , as then =−

u

yb 1an integer. So yf = 75, yav being equal to 74.

- 10 -

Lecture 3

ELECTRO-MECHANICAL AND ELECTRO MAGNETIC

PRINCIPLES OF D.C. MACHINES Every machine whether used as a generator or as a motor consists of one or

more electric circuits interlinked with magnetic circuit. EMF INDUCED IN DRUM ARMATURE

The direction of induced emf in an armature conductor can easily be determined with the help of Fleming’s right hand rule or right hand palm rule. In fig below, the total voltage available between the terminals of the brushes

will be the sum of the emfs of all the coil sides in series. When the armature rotates in a magnetic field, the instant emf induced in

any one conductor is

Ex= b x l v Where l - Axial length of the armature

v - Linear velocity of the armature Bx - Flux density at a point x where the conductor is situated at that given instant.

Fig. 1: Illustrating the addition of emfs of coil sides in series between

the brushes. Let Z be the total no. of conductors and “a” be number of parallel paths in

the armature winding.

Then each parallel path will consist of a

Z conductors connected in series.

- 11 -

To obtain the voltage across the brushes Ea, it is necessary to sum the emfs

of all conductors in series in a parallel path. Thus

a

za eeeeE ++++= ............321

vBBBBa

z .....................321 l

++++=

∑=

=a

z

x

xBv1

.l

With large number of coil sides between the brushes then

av

az

x

x xBa

ZB∑

=

=/

1

Where average value of flux density Bav is

∫=τ

τ0

1dxBB xav

Consequently

ava Ba

ZvE ...l=

Fig. 2: Schematic diagram of a 2 pole machine with drum armature.

Let D be the diameter of the armature. Then the area of the armature over

one pole pitch lxP

DA

π= , where P=number of poles.

Let Φ be flux per pole. Then

)1....(..................................................lD

P

ABav π

Φ=

Φ=

The linear velocity of the armature 2

.60

.22

..DND

rV πωω ===

Where N=speed of the armature in rpm Therefore,

- 12 -

a

ZNP

D

P

a

ZDNEa

60..

60.

Φ=

Φ=

ll

ππ

)2.(..................................................60a

ZNPEa

Φ=

Where a=m.p for a lap winding = 2m for wave winding. m=multiplicity of winding for simplex winding m=1

2. TORQUE DEVELOPED IN DRUM ARMATURE

When there is no current through the armature of the machine there is no interaction between the main field flux and armature. But as soon as

current flows through the armature, a torque is produced due to interaction between the main field flux and armature m.m.f.

The direction of the force acting on an armature conductor can easily be determined with the help of Fleming’s left hand rule or left hand palm rule.

Fig. 3: Mechanical forces and flux

Let I be the current flowing through each of the armature conductors and l

be the length of each armature conductor which is equal to the axial length of the armature. Then the force acting on any one armature conductor chosen is given by

l..IBF xx = and torque

2

...2

.D

IBD

FT xxx l==

Where D is the diameter

If there are Z number of conductors on the armature, then the net torque acting on the armature

- 13 -

2

...11

DIBTT

z

x

x

z

x

xa l∑∑==

==

∑=

=z

x

xBD

I1

.2

..l

For large no. of conductors av

Z

X

x BZB .1

=∑=

Where average value of flux density

lD

PBav π

φ=

Let Ia be the total armature current and let there be ‘a’ number of parallel

paths in the armature winding. Then the current through each armature conductor I is

a

II a=

Therefore, armature torque a

ZIPZ

D

PD

a

IT aaa π

φπφ

2..

2.. ==

ll

)3...(........................................2 a

ZIPT aa π

φ=

NB: To have an idea of the physical process of the generation of machine

torque, the magnetic flux lines produced by the field poles mostly concentrate through the teeth of the armature and not through the conductors of the slot because of less magnetic reluctance.

- 14 -

Lecture 4 3. COUNTER-TORQUE AND COUNTER EMF

a) Generator operation:

Fig. (a): Illustrating counter-torque.

In dc generator the armature is rotated by a prime-mover and emf is induced in the armature. As soon as an electrical load is connected between the

armature terminals a load current starts flowing through the armature and a torque is developed due to interaction of main field flux and armature

current. The direction of the torque is opposite to the direction of rotation as may be verified by FLEMING’S LEFT HAND RULE. Further this phenomenon also conforms to Lenz’s law as the torque opposes the very cause of its

production, that is, emf and current generated by rotation. In other words this developed torque opposes the driving torque and this is illustrated in fig.

(a). Hence this torque is called counter-torque. The magnitude of counter-torque is governed by the formula

a

ZIPT aa π

φ2

=

b) Motor Operation:

Fig. (b); Illustrating counter emf

In dc motor the armature which is under the influence of main field flux is

being fed with dc current from an external source and a torque is developed due to which the armature starts to rotate. Now due to the rotation of the

armature in the main field flux, an emf is induced in the armature which will oppose the rotation of the driving torque or in other words the driving emf of

- 15 -

the armature current as can be verified by FLEMING’S RIGHT HAND RULE

and also by Lenz’s law. This phenomenon is illustrated in fig.(b). This is why the induced emf is called counter emf or back emf and the magnitude of the

counter emf is governed by the formula

a

ZNPEa

60

φ=

RELATIONSHIP BETWEEN EMF AND DEVELOPED TORQUE

From equation (2), Ea is given by

==60

2..

260

N

a

PZ

a

ZNPEa

πφ

πφ

, but 60

2 Nπω =

Therefore, )4(........................................φωaa KE =

Again from eq. (3), Ta is given by

aa

a Ia

PZ

a

ZIPT ..

22φ

ππφ

==

aa IK φ=

Therefore, )5........(..............................aaa IKT φ=

Where ==a

PZK a π2

constant for a given machine.

So from equation (4) and (5),

ωφωφ a

a

aa

a

a I

K

IK

E

T==

Therefore, )6........(................................ aaa IET =ω

NB: Equation (6) above is the power balance equation. It shows that in the

case of a generator mechanical power supplied is equal to the electrical power developed in the armature and in the case of a motor; the electrical

power supplied is equal to the mechanical power developed in the armature.

ARMATURE REACTION

The effect of the magnetic field setup by the armature current on the

distribution of flux under main poles is known as armature reaction From fig. 1, below, it may be observed that the effect of armature current is to displace the resultant field flux in the direction of rotation of the generator

and against the direction of rotation of the motor. Note that the flux is not

displaced by the mechanical rotation of the armature but by the armature

current.

- 16 -

fig.1: Effect of armature current on magnetic field of a dc

machine.

Quantitative Analysis of Armature Field To analyse the armature field quantitatively, it is necessary to determine the Magnetising force of the armature. For this purpose the real slotted

armature is reduced to a smooth armature with a layer of conductors uniformly distributed over the armature periphery but having the same

effective air-gap length al as in the real machine.

Then armature ampere-conductors per unit length of armature periphery (A) is given by

aD

ZIA a

.π=

Where, Z = total no. of conductors a= parallel number of paths

D = Diameter (armature). The quantity A is called the armature Electric loading and is one of the most important quantities in the design of a dc machine.

The quantity A maybe measured the same way as magnetic field intensity H and may be expressed in Amperes/unit length or AT/Unit length.

Applying Ampere's law an expression for the armature cross mmf can be developed. According to Ampere’s law, the line integral of magnetic field

- 17 -

intensity over any closed line path is equal to total current enclosed by the

path. Mathematically,

∫ ∑= IdxH x

Where Hx is the magnetic field intensity at a distance x. For example take a conductor which carries a current I at right angle to the

plane of the paper as shown in fig 2 below

Fig. 2: Illustration of Ampere’s law

By virtue of circular geometrical symmetry, the flux lines are concentric circles with the conductor as the central axis.

If a unit N-pole moves once around the conductor in a circular path of radius r, then work done on it is

∫ == rHxdrH r π2

According to Ampere’s law, this must be equal to the total current enclosed by the path.

Now, total current enclosed=I or H×2π r=I

r

IH

π2= Amperes/unit length or (AT/unit length)

- 18 -

Lecture 5

Let the brushes be on the geometrical neutral line and the winding pitch is equal to pole pitch(τ ).Then, the midpoint of each armature current path is

exactly on the corresponding pole centre line and the lines of the armature

field are arranged symmetrically at both sides of the polar axis as shown in fig 3(a).

Let one of these lines be at a distance x from the polar axis. The mmf along the circuit of this line is equal to the total current enclosed by this circuit

following Ampere’s law, that is, m.m.f is equal to A×2x

Within the limits of the pole-shoe each magnetic line passes through the air

gap twice on either side of the polar axis and within the armature and pole steel. It may be assumed that the reluctance of an armature cross flux line

path is determined only by the reluctance of both air gap and the armature teeth. Therefore the above mentioned armature mmf A×2x can be considered to be

consumed by the air gap and the armature teeth at a distance x from the pole axis on either side.

So the armature m.m.f Fax=Ax can be plotted as a straight line over the armature periphery as shown in fig (b) considering the direction of flux lines. The magnitude of armature mmf is maximum at the brush axis i.e. when

2

τ=x and this maximum value ATa is known as armature ampere-turns

per pole.

Hence armature ampere-turns per pole

2

.τ

AATa =

Where A = armature electric loadingaD

ZI a

.π=

And P

DpitchPole

πτ == .

Therefore, aP

ZI

P

Dx

aD

ZIAT aa

a.22.

==π

π

At any point under the pole-shoe the magnetic flux density Bax due to

armature current will be: Bax = µ oHax where Hax is the magnetic field intensity at a distance x from the

pole-axis

Nowa

ax

ax

FH

l2= , where Fax =2.Ax

Therefore a

ax

AxBl

..0µ= where al being the length of air gap.

- 19 -

- 20 -

NB: Thus within the limits of the pole shoe the air gap flux due to armature

m.m.f is proportional to the armature m.m.f. but in the interpolar space the flux density sharply decreases owing to a large increase in the length of the

flux line in air and consequently an increase in its reluctance. Therefore the curve for flux density due to armature current Ba is saddle-shaped as shown

in fig. (b). In constant flux dc machines such as shunt machines, the flux distortion is much more prominent under heavy loads. In series and compound machines

the flux distortion is less because with the increase in armature m.m.f there is a corresponding increase in the field m.m.f.

COMMUTATION The two very important parts of the d.c. machine are commutator and its accompanying brushes. Two essential actions take place here, namely, the

passage of current from or to a moving armature to or from external circuit and the commutation process.

Commutation process involves the change from a generated alternating voltage to an externally available unidirectional voltage.

Fig.1: Reversal of current through a coil undergoing Commutation.

For simplicity it is assumed that the width of the brush is equal to the width of the Commutator bar and the thickness of the insulation between the Commutator bars is negligible. The direction of rotation of the armature is

taken from left to right.

- 21 -

Fig. 2: Illustration of commutation process.

- 22 -

Lecture 6

1. INTERPOLES (COMPOLES OR COMMUTATING POLES)

The best and most widely used method of improving commutation is to use interpoles. The interpoles are arranged between the main poles along the

geometrical neutral axis and the brushes are also set on this axis and remain in these positions at all loads. The number of interpoles is usually equal to the number of main poles.

The polarity of the interpoles is determined from considerations that they should give the result similar to that obtained by shifting the brushes in

a machine without interpole. Therefore if a machine operates as a generator the interpoles must have the polarity of the leading poles and in the case of a motor, those of the trailing poles.

2. COMPENSATING WINDING

The interpole mmf is effective only in the commutating zone. In other words, the armature reaction effects are overcome only under the interpoles and the resultant flux under the pole face remains distorted. The most

effective way to achieve this compensation or to reduce distortion is by means of compensating windings embedded in slots in pole faces.

To obtain compensation at any load, it is necessary that the compensating winding be connected in series with the armature windings. Further, in order that armature and compensating winding mmfs oppose each other, the

direction of currents in the compensating winding must be opposite to that in the armature windings just below the pole face.

- 23 -

The Commutation period (Tc) is defined as the time required by the coil

current to change from a

I a+ toa

I a− .

The change of direction of current through a coil may occur in a manner

shown by the curves below fig3. which are called short-circuited current waves.

Fig. 3: Types of commutation

Curves

No.1- Shows that the current has been reversed too rapidly, over-reaching its final value. Such a case is known as sharply accelerated Commutation. Here, the current may reach its proper final value without a spark but it

involves very high localised current densities at the brush contact which will lead to sparking and excessive heating and detoriation of the brush.

No.2- represents a case of accelerated commutation or over-commutation where the current comes to its final value with a zero rate of change at the end of the commutation period. Usually will result in satisfactory

commutation. No.3 - indicates linear commutation. Power loss at brush contact is

minimum. No.4 - represents sinusoidal commutation which generally results in

satisfactory commutation. No.5 - is a case of retarded commutation or under-commutation where the final rate of change of current is very high. Under such conditions sparking at

the trailing edge of the brush is inevitable. No.6 - is a case of sharply retarded commutation which also results in

excessive current density under the brushes and inevitable sparking. NB: The current must change from full value in one direction to full-value in the reverse direction in very short-time interval.

The inductive property of a coil containing several turns of wire and wrapped around good magnetic material tends to oppose the reversal of current

through a coil. This property gives rise to a voltage called REACTANE VOLTAGE which tries to keep the current in the initial direction. This results in under commutation. In order to counteract this, the reactance voltage

must be balanced by an opposing emf in the coils undergoing commutation.

- 24 -

OPERATING CHARACTERISTICS OF DC GENERATOR: 1. Classification of DC generators Depending upon the mode of excitation dc generators are classified as

a). Separately excited b). Self-excited

a) Separately excited generators can have two types: (i) Excited by electromagnetic field

(ii) With permanent magnets As the latter has very limited application, only former will be considered in

separately excited generator. b) Self excited can be subdivided into three classes:

(i) Shunt (ii) Series

(iii) Compound (which has both shunt and series) Again compound generator can be classified depending upon the electrical connection of series field with respect to armature and shunt field:

(i) Short shunt (ii) Long shunt

A compound generator can be cumulative or differential according to relative magnetising effects of series and shunt field. If the shunt and the series winding are so that their magnetising effects are

in the same direction, i.e., additive, then the generator is a cumulative compound generator. Otherwise the generator is a differential compound

generator.

- 25 -

- 26 -

17 th July, 2007 EXAMPLES: 1. Estimate the reduction in speed of a generator working with constant excitation on 440V bus-bar to decrease its load from 500 to 250KW.The resistance between terminals is 0.02Ω, neglect armature reaction. Solution: Case 1:

Now 440Ia1=500×103 =1136.36A E1=V+Ia1RT=440+1136.36×0.02=462.73V Case 2:

Now 440·Ia2=250×103 =568.18A E=451.36V Again Eα Φ ·N But for this case flux Φ=const. as excitation is kept constant Eα N N2/N1=E2/E1 OR (N2-N1)/N1=(E2-E1)/E1 Therefore, ٪ change in speed= (451.36-462.73)/462.73×100 = -2.457٪ 2. A separately excited generator, when running at 1,000 rpm, supplies 200A at 125V to a circuit of constant resistance. If armature resistance is 0.04Ω

and total drop at brush is 2V, what will be the current when the speed is dropped to 800 rpm if the field current remains constant? Ignore change in

armature reaction.

- 27 -

Solution:

Here, Ra=0.04Ω, Brush drop=2V

Case 1: Ia=200A, Terminal voltage V=125V

External load resistance ( ) 1250.625

200a

VR

I= = = Ω

Therefore, E1=125+2+200×0.04=135V

N1=1000rpm Case 2:

For const. excitation, Eα N

Therefore, E2/E1=N2/N1 or E2=N2/N1×E1=800/1000×135=108V Eg2=V2+Ia2Ra+Vb but V2=Ia2R =>

Eg2 =Ia2(Ra+R) +Vb Ia= (Eg2-Vb)/(Ra+R) = (108-2)/ (0.04+0.625) =159.4A

2. Characteristics Curves and Regulation The simplest way of study and compare the several kinds of machines is to

construct characteristics curves showing the variables involved in the operation of the machine which are (i) Generated emf E

(ii) Terminal voltage V (iii) Armature current Ia

(iv) Load current I (v) Field current If or Ise (vi) Speed of rotation N rpm

Since generators usually operate at the constant rated speed, the principal group of characteristics is plotted for n=constant=rated speed.

The basic characteristics of a dc generator are: 1. External characteristics is the function

E=f (I) with constant resistance in the field circuit in the case of shunt and compound generator and If=constant in case of separately excited

generator.

2. Load characteristic is the relationship V=f(If),I=constant

When I=0, this is known as magnetising characteristic or no-load characteristic or saturation curve. This is also known as open circuit

characteristic. 3. Regulation characteristic, also known as armature characteristics, is the relationship

If= f(I) for V=constant. When V=0, regulation characteristic becomes short-circuit characteristics

If=f(Ish) where Ish is the short circuit current.

- 28 -

There are also other characteristics, such as characteristics with variable

speeds, but these are auxiliary use. For generator the terminal voltage at full load is generally different from no

load. Voltage regulation is frequently expressed as a % and calculated by dividing the fall of voltage from no load to full load by the voltage that is

considered to be the normal voltage of the machine (generally the full load rated voltage) and multiplying the result by 100

%100xV

VV

FL

FLNL −

STARTING, SPEED CONTROL AND BRAKING OF D.C MOTORS At starting the dc motors draw heavy current from the supply as sometimes is required for the speed and hence, the counter emf to build up.

In industrial applications to stop the driven unit quickly in an exact position or to reverse the direction of rotation quickly, electrical braking is preferred

over the mechanical brakes. METHODS OF DC MOTOR STARTING

The starting characteristics of a dc motor determine the operation of starting from the moment the motor begins running to the moment when steady-state operation is attained and including the starting current in the armature

(Ist), the starting torque (Tst) and the duration of starting (tst). At the instant of starting the speed of the motor is zero, therefore back emf

E=0.Consequently, for the armature circuit the emf equation is V=0+IaRa for shunt motor V=0+Ia(Ra+Rse) for series and compound

If rated voltage is applied at the terminals of the motor, the starting current is

a

ratedst

R

VI = for shunt motor.

sea

ratedst

RR

VI

+= for both compound and series motors.

Since Ra and (Ra+Rse) are much small, the motor draws very high current from the supply mains during starting. Such large current surge at the initial period of starting may result

in: a. Heavy sparking at the commutator and even flashovers.

b. Deterioration of the insulation of the armature winding due to overheating. c. The necessity of properly designing the supply mains feeding the motor since otherwise very high voltage drop may occur.

d. Production of a large dynamic torque on the motor shaft during starting this may damage the power transmission parts of the motor or the load.

In view of this, it is advisable to limit the armature current at starting to a

safe value that can be handled safely. The following methods of motor starting are used: 1. Direct-on-line (DOL) or full-voltage starting

2. Starting by means of additional resistance connected in series with the armature (armature resistance starting)

3. Controlled voltage starting

- 29 -

1. DOL: This method has the advantage of having the simplest form of starting

apparatus and starting operation. With proper designed motors and systems it is possible to accomplish DOL

starting of motors of rating up to 6KW with current surges of 6 to 8 times the rated value. 2. Resistance starting using starting resistors.

In this method of starting, the armature current is limited to a value that can be commutated safely by inserting suitable resistances in the armature

circuit.

Since at starting instant the back emf is zero (E=0) the current equals

( )∑+=

sta

iRR

VI Where ∑ =stR total resistance of all the

starting resistors.

If the motor torque developed by this current Ii is greater than static torque, the motor will start to rotate and a back emf will be developed proportional

to the speed N. In practical cases, mostly Ii=I1 where Ii=maximum permissible armature current at starting.

- 30 -

3. Controlled voltage starting

In heavy power installation the starting resistors become bulky and considerable loss of energy occurs in these resistors, especially with frequent

starts. Therefore non-resistance methods of starting by varying the voltage supplied to the motor becomes necessary in some installations. These include

starting by employing a booster machine or generator-motor set or by connecting two or more motors in series as for example in electric traction. Nowadays dc supply obtained from controlled rectifiers is used in place of

generator motor set. COMPOUND AND SHUNT MOTOR STARTERS

a) THREE-POINT STARTER It is called three point starters since only three terminals L, A and F are

available for the starter. One of the supply lines is connected directly to one

armature terminal and one field terminal shorted together. It makes no connection whatsoever with the starting box.

HM - hold up magnet

HM - is also known as no-voltage release S - Starting arm

OLR - Overload release C - Soft iron

A major problem with the three point starter arises due to the series connection of the shunt field winding and the hold up magnetic. If the motor

is required to be operated at a higher speed than the normal speed by the field control, the field current has to be reduced. At a certain value of reduced shunt field current, the pull of magnet HM may become less than the

spring force. In that case, the starting arm returns to the OFF position and the motor stops. Thus a three point starter is not suitable for the applications

where wide range of operating speed by field flux control is required. This problem is overcome in a Four-point starter.

- 31 -

2. FOUR-POINT STARTER

In a four-point starter the hold-up magnet (HM) is connected in series with a high resistance (R) and this series combination is connected to the line and

common armature and field terminal. The other components of the four point starter are the same as those of a three-point starter.

The function of the high resistance R is to prevent short-circuits of the supply mains when the overload release operates. If HM gets short-circuited by contacts of OLR due to overload of the running motor, the current through R

is limited by its own resistance and starting resistance. When the line is without any voltage the hold-up magnet is de-energised and allows the starting arm to spring back to the OFF position.

NB: Four-point starter permits the change of field current by field rheostat without affecting the current through HM and thus without affecting the pull

of the hold-up magnet. In view of this, the four-point-starter is more popular in practise.

SPEED CONTROL OF DC MOTOR

From the formulae φKRIV

N aa−= for shunt, there are three methods of speed

control: a) By varying the resistance in the armature circuit b) By varying the excitation current and hence the field flux of the motor

c) By varying the voltage applied to the armature terminals.

- 32 -

24th July 2007

PARALLEL OPERATION OF SHUNT GENERATORS

Let’s assume that generator I is already working under load, producing voltage U across the bus-bar. To connect generator II to the same bus-bar,

two conditions must be satisfied:

i) The (+) and (-) terminals of the generator being connected should be joined to the like terminals of the common bus-bars.

ii) The emf of the generator to be connected should be practically equal to the voltage U.

These conditions are met as follows:

1. Generator II is run at the required speed and then, without exciting it, one of the knife-switches, for example the left-hand one(S3), is closed. If a

voltmeter is connected across the open knife switch (S4), it will measure the voltage U (the influence of the residual magnetization field of generator II being disregarded).

2. Now let us begin to excite generator II. If the polarity does not coincide with that of the bus-bar, then both generators are connected in series, and

the voltmeter measures the sum U+EaII. In this case the generator must

not be connected to the buses, as this would correspond to the short

circuiting of both machines. If the polarity of connected generator coincides with that of the bus-bars, the

voltmeter shows this by decreasing its indication since the difference U-EaII

is being measured. When this difference vanishes the right-hand knife switch may be closed, thus connecting generator II to the bus-bars.

- 33 -

If the emf EaII of generator II is exactly equal to the voltage U of the

mains then it follows from equation Ea=U+IaRa that the current of

generator II=0.

0=−

=aII

aIIII

R

UEI

To load the generator II, it is necessary to increase the mechanical power delivered to it from the prime-mover. This can be done either by directly

manipulating the speed regulator of the generator prime mover, thus

increasing its speed, or by increasing the excitation current fIIi of

generator II.

PARALLEL OPERATION – EXTERNAL CHARACTERISTIC

Let us assume for simplicity that the out-put of generator I and II are equal

i.e. PI=PII and that their prime movers have the same speed n=const.

Let the curve 1 and 2 be the external characteristics of generator I & II,

operating independently with the same no load voltage Uo and the curve 3

be the total external characteristic

U=f(Ie)=f(II+III) where Ie=II+III is the current delivered by the

generators to the power line. If the generators operate in parallel, then with an increase of load the voltage across both generators should decrease by

the same value

∆U=Uo-U=OF-OD=OF-Aa

- 34 -

Distribution and transfer of load

From the fundamental equation aaa RIUE += for the emf of a dc generator

we have

URIERIE aIIIIaIIaIIaI =−=−

If Re is the resistance of the external circuit, then

( ) eIII RIIU +=

Solving the equation in respect to current II and III we obtain

( )( ) aIIaIaIIaIe

eaIIaIIeaI

IRRRRR

RERREI

++

−+=

( )( ) aIIaIaIIaIe

eaIaIeaII

IIRRRRR

RERREI

++

−+=

Hence

( )( ) aIIaIaIIaIe

aIaIIaIIaIe

RRRRR

RERERU

++

+=

It follows from the above formulas that for a given resistance RaI, RaII and Re

the distribution of load currents between the generators depends on the emfs EaI and EaII i.e. on the generator speed nI and nII and their resultant fluxes

Φ I and Φ II as E=KΦn

- 35 -

EFFICIENCY AND TESTING OF DC MACHINES

I. Classification of losses: The losses in dc machines may be grouped as follows:

(a) The core losses due to: (i) Hysteresis in the armature core and teeth (ii) Eddy current in the armature core and teeth and pole faces

(b) The ohmic losses in: (i) The armature winding

(ii) The field winding (iii) The brushes and contacts

(c) The mechanical losses due to

(i) Bearing friction (ii) Brush friction

(iii) Windage (d Stray load losses II. No-load and load losses

Some of the losses stated above occur when the machine is not loaded while the rest of the losses appear when the machine is loaded.

a) No-load losses: (i) Iron losses in the armature due to the main flux (ii)Pole face losses due to slot openings

(iii) Windage losses (iv) Bearing friction losses

(v) Brush friction losses (vi) Copper losses in the conductors due to main flux (vii) Eddy current losses in the steel wire bands of the armature due to main

flux b) Load losses:

(i) Copper losses in all winding in series with the armature (ii) Skin effect losses in the armature conductors (iii) Brush contact copper losses

(iv) Eddy current losses in the armature conductors due to flux distortion (v) Copper losses in the commutating coils

(vi) Iron losses due to flux distortion

III. Efficiency of dc machine The efficiency of any device is defined as the ratio of power output to power input

)1......(...................................................

.

InputPower

OutputPowerEfficiency =

(i) )2......(...............................

...

InputPower

lossesInputPowermotorofEfficiency

−=

(ii) )3....(.....................

...

LossesOutputPower

OutputPowergeneratorofEfficiency

+=

- 36 -

IV. Maximum efficiency Thus if the output is Po, the constant losses are Pc and the variable losses are

KPo2, where K is a constant, the overall efficiency

)4..(............................................................)( 2

00

0

KPPP

P

C ++=η

The condition for maximum efficiency can be obtained by differentiating η

with respect to Po and equating the result to zero. Thus

( ) ( )

( )

20 0 0 0

22

0 0

1 20

c

oc

P P KP P KPd

dP P P KP

η + + − += =

+ +

Or )5....(..................................................20KPPc =

Or

and the maximum efficiency )6........(..............................)2( 0

0max

cPP

P

+=η

Let, the rated output of machine be P and the maximum efficiency occurs at

an output x.P. Then, the variable losses at rated output=KP2 and variable

losses at maximum efficiency 2 2 2( ) ( . )K xP x K P= =

If the constant losses are Pc, then according to the condition for maximum efficiency

2 2( . ) cx K P P=

Or

)7....(........................

.tan2 speedratedatlossesVariable

lossestCons

KP

Px c ==

Hence, the load at which maximum efficiency occurs is equal to

OutputRatedxoutputratedatlossesVariable

lossestCons..

....

.tan

Constant losses=Variable losses

- 37 -

I DIRECT METHOD—(BRAKE TEST) The efficiency of small motors may be determined directly by a brake test.

The load is varied by varying the weight on the scale pan. The actual load torque is given by

T= (W1-W2) R Where R=1/2(diameter of the pulley diameter of the rope) =effective radius of the brake

The mechanical power output of the motor is given by

P0=Wt=2π N T

The electrical power input can be calculated by measuring voltage V and the

input line current I Then, the electrical power input is given by

Pi=VI And efficiency of the motor is given by

η=iP

P0 ×100=VI

nTπ2×100

II. INDIRECT METHOD

In this method, the no-load machine losses are measured by a suitable test and then the copper losses are calculated from measured values of the

various resistances, in order to calculate the machine efficiency. The simplest

method of measuring the no load losses of a dc machine is Swinburne’s method.

SWINEBURNE’S METHOD This test consists in determining the losses and hence calculating the

efficiency. The test can be applied only to shunt and compound machines. The machine is run, on no-load, as a motor and the applied and input current

is measured.

- 38 -

The resistance of the armature and field winding are measured. Since the

machine winding would be hot when running under normal conditions, the probable hot resistance are calculated from the above measured cold

resistances by assuming a temperature rise of about 400C.From the above data the different losses are calculated and efficiency is determined.

If V and I0 are the supply voltage and input armature current at no-load, R’a and R’sh the cold armature and field resistances, Ra and Rsh the hot resistances of armature and field, then

Motor input at no load=VI0

Field winding loss at no load='

2

shR

V

Armature copper loss at no load

= ( I0- 'shR

V ) 2R 'a

Mechanical and iron losses

=VI0- '

2

shR

V- ( I0- '

shR

V ) 2R 'a

Field winding loss at load='

2

shR

V

Armature copper loss at load

(For an armature current of Ia ampere)

=I 2a R a

Total losses=Mechanical and iron losses

+Field winding loss at load +Armature copper loss at load After the losses have been determined the efficiency can be calculated. The

stray load losses can be assumed to be 1% of the output and added to the other losses to make the calculation more accurate.

NB: This test can not be applied to series motors because the speed of a series motor being very high at no load, it is not possible to run a series motor on no-load.

III. REGENERATIVE or HOPKINSON’S METHOD

This method is a regenerative test and requires two identical machines. The connection diagram for this test on a pair of shunt machine is shown below.

- 39 -

The two machines are mechanically coupled and their field are so adjusted

that one of them acts as a motor and the other as a generator. The electric power produced by the generator is utilised by the motor and the

motor drives the generator. The power taken from the supply is that required to over come the losses only. Two identical machines of any size can be

tested under full load conditions and therefore the method is very useful for determining efficiency and also as heat run test for determining the temperature rise.

To perform the test, the machine I is started as a shunt motor through a starter and its field rheostat is adjusted so that it runs at rated speed. The

machine I will drive machine II. The switch ‘S’ is initially kept open. The excitation of machine II is gradually increased (by decreasing its field circuit resistance) till the voltmeter V1 reads zero. Then switch ‘S’ is closed. The

machine II is now floating neither taking any current from the supply nor delivering any current. Any desired load can be put on the set by adjusting

the shunt field regulators. The machine with lower excitation will act as a motor and the other machine will act as a generator. The power input to the set is only that which is required to meet the losses.

Thus with an input of only about 100KW, two 1000KW sets can be tested on full load.

At the end of the experiment when the windings have reached their final steady temperature, the resistances Ra1 and Ra2 of the armature circuit (i.e. armature winding + brush contact + interpole winding) of each machine are

measured by passing full load current through them at standstill and measuring the voltage drops with fields not excited.

The efficiency can be found as under: Input to the set=VI1

Generator armature current= gI =I2+I4

Motor armature current IM=I2+I1-I3

Motor armature current loss= I 2M 1aR

Generator armature current loss= I2g 2aR

Field winding current loss of motor=VI3

Field winding current loss of Generator=VI4

Total iron and mechanical losses of the two machines=VI1- I2M 1aR - I 2

g 2aR -

VI3-VI4

- 40 -

Iron and mechanical loss of each machine=PC=0.5(VI1- I2M 1aR - I

2g 2aR -VI3-

VI4)

Motor losses=PC+ I 2M 1aR +VI3

Motor input=V (I1+I2)

ELECTRICAL BRAKING OF DC MACHINES In practice it is often required to stop a unit driven by dc motors in an exact

position or to have the speed of the driven unit suitably controlled during its deceleration. Such requirement is satisfactorily fulfilled by electrical braking of a dc motor

which can be accomplished in three different ways. (i) Regenerative braking

(ii) Counter-current braking or plugging (iii) Dynamic braking

(i) Regenerative Braking: Regenerative braking with return of energy to the supply circuit is

obtained when under certain conditions the motor is forced to run at a speed higher than its ideal no-load speed η 0 due to which its induced emf E in the

armature exceed the applied voltage V. The motor then operates as a generator connected in parallel with the supply and the direction of the

armature. Current also reverses so that energy is returned to the supply circuit. This is

evident from expression

Ia=ar

EV −=-

ar

VE −………………… (1)

With the reversal of direction of armature current the motor torque also

reverses and a braking torque is developed so that Ta=-KaΦ Ia

(ii) Counter-current Braking or Plugging It occurs when the motor windings are connected for reverse direction of rotation at a time when the armature is still rotating in the forward direction

either under the action of an external torque or due to inertia. For counter-current braking motor reversal may be accomplished by reversing the

polarity of the applied voltage either to the armature or to the field winding. Polarity reversal of field winding is rarely practiced because it results in a longer braking time due to the relatively high inductance of the field winding

compared to that can be obtained by polarity reversal of armature winding. If the polarity of the voltage applied to the motor armature is reversed

while it is still rotating in the initial direction, then the armature induced emf E has the same polarity as before but the polarity of the supply voltage V is

reversed so that the armature circuit voltage equation can be written as

-V=E+Iara

- 41 -

Ia=-ar

EV +……………………….. (1)

Reversal of polarity of armature current produces a plugging torque which is given by

TP=KaΦ Ia=-KaΦar

EV +………….. (2)

(iii) Dynamic Braking: Dynamic braking is accomplished by disconnecting the armature from

supply and connecting an external resistance across the armature terminals. Because of this reason, this method of braking is also sometimes called

rheostat braking. Hence, the motor armature current during dynamic braking is given by

Ia=-ba rr

E

+

The reversal of armature current results in generator mode of operation and consequent production of braking torque causing deceleration. However,

dynamic braking differs from regenerative braking in the sense that in dynamic braking the kinetic energy stored in the moving parts of the motor and the driven unit is converted to heat in the braking resistor while the

same is converted to electric energy which is returned to the supply in the case of regenerative braking.

The dynamic braking torque is given by

Td=KaΦ Ia=-ba

a

rr

EK

+

Φ=-

ba

a

rr

K

+

Φ ω22

Or

ba

drr

NKT

+

Φ−=

22