A textbook of electrical technology P7

A Textbook of Electrical Technology (Volume-2) is an updated and multicolored version of the first volume by authors B. L. Theraja and A. K. Theraja. Titled AC and DC Machines, this book is a detailed study on the core concepts of Electrical Technology. The book is ideal as reference or examination study material for engineering, diploma or polytechnic students in India. The course covered within the book is also beneficial for international examinations such as City & Guilds, I.E.E., and I.E.R.E. The book has detailed diagrams, solved numerical problems, and answers to questions asked in University examination papers. This volume has been made especially for students and has been written with a view to simplify the subject matter and visually appeal to its young readers. The book has additional chapters on Elements Of Electro-Mechanical Energy Conversion and Special Machines, which are included, keeping in mind the latest developments in the field.

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292 ELECTRICAL TECHNOLOGY Example 5 A six pole lap wound D.C armature has 70 slots with 20 conductors/slot The ratio of pole arc to pole pitch is 0.68 The diameter of bore of the pole shoe is 0.46 m The length of pole shoe is 0.3 m If the air gap flux density is 0.3 Wb /m? and the e.m.f induced in the armature is

500 V, find the speed at which it runs

Solution Number of poles, p= 6

Number of slots

Conduetors/slot

Total number of conductors,

Ratio of pole are to pole pitch

Diameter of bore of the pole shoe = 0.46 m

Length of the pole shoe

Air gap flux density,

Exm4 induced,

Speed of rotation, N:

Pole arc Đole piieh ~ 968

zs Pole are = 0.68 x pole pitch

is of greater advantage to use direct current (for chemical and metallurgical plants, electric traction, ete.)it is generally obtained by converting A.C to D.C with the help of converters

of ionic or machine types In the latter case wide use is made of such installations as motor generator sets in which A.C motor is coupled to a D.C generator on a common shaft

@ As primary sources of power, D.C generators are mainly used in self-contained plants such as automobiles and air planes, for electric are welding, train car lighting, in sub-marines, etc

Classification

According to method of excitation D.C generators are classified as follows :

1 Separately excited generators, 2, Self-excited generators

Separately excited generators:

‘These are those generators whose field magnets are energised from an independent external source of D.C current Such a generator is shown in Fig 15

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ROTATING MACHINES 293

Separate source of excitation

Self-excited generators :

These are those generators whose field magnets are energised by the current produced by the generators themselves Due to residual magnetism, there is always present some flux in the poles

‘When the armature is rotated, some e.m.f and hence some induced current produced which is partly

or fully passed through the field coils thereby strengthening the pole flux

Self excited generators can be divided, in accordance with how the field winding is connected into generators, as follows :

() Shunt wound generators

Gi Series wound generators

Gif) Compound wound generators :

(a) Short shunt

(6) Long shunt

(@® Shunt wound generators : Refer Fig 16

In these generators the field windings are connected

across or in parallel with the armature conductors,

and have the full voltage of the generator across

them

Important relations : Refer Fig 16 faa

where J,,= shunt field current, 7ạ=armaturecurrent, se 2 foe

I (or I) = load current, R, = armature resistance,

Ry, = shunt field resistance, E,= generated e.m., and

terminal voltage

(ii) Series wound generators : Refer Fig 17 In this

case, the field windings are joined in series armature

conductors As they carry full load current, they consist of rela-

tively few turns of thick wire or strip The use of such generators

is limited to special purposes (as boosters etc.)

Fig 17 Series wound generator

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294 ELECTRICAL TECHNOLOGY Important Relations (see Fig 17):

iii) Power developed = Ef

(iv) Power delivered = VI

(iii) Compound wound generators It is a combination of a few series and a few shunt windings and be either short shunt or long shunt as shown in Figs 18 and 19 respectively Important Relations:

(a) Short shunt compound wound (See Fig 18) :

(0) Power developed = Exly

(vi) Power delivered = VI

Fig 18 Short shunt compound wound generator Fig 19, Long shunt compound wound generator (6) Long shunt compound wound (See Fig 19) :

I= Tt ley

ý ~ laRa= L„R„ = Ey ~ LU(RẠ + R„)

(iv) Power developed = Ely

(0) Power delivered = VI

Example 6 4-pole lap wound shunt generator supplies to 50 lamps of 100 watts, 200 V each

The field and armature resistances are 50 Q and 0.2 Q respectively Allowing a brush drop of 1 V each

brush, calculate the following :

(i) Armature current Gi) Current per path

(iii) Generated emf (iv) Power output of D.C armature

Solution Number of poles,

‘Total lamp load,

~ Terminal voltage,

Field resistance,

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Armature current, J, =/+Iy,=25+4=29A (Ans.)

(i) Current per path :

Current per path be a=p=4, generator being lap wound) (iii) Generated e.m , By

“By = V + IạR, + brush drop = 200 + 29x0.2+2%1=207.8V Hence, generated e.m-f = 207.8 V (Ans.)

(iv) Power output of D.C armature:

Egle _ 207.8x 29 Power output of D.C armature = Tp “ng KWE6/026kW, (An)

Example 7 A series generator delivers a current of 100 A at 250 V Its armature and series field resistances are 0.1 €3 and 0.055 & respectively Find :

Solution See Fig 21

(ii) Generated e.m

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296 ELECTRICAL TECHNOLOGY

"FExample 8 A short shunt compound generator has armature, series field and shunt field resistances of 0.06 2, 0.03 Q and 110 Q respectively It supplies 100 lamps rated at 250 V, 40 W Find the generated e.m.f Assume that contact drop /brush = 1 V

Solution See Fig 22

Armature resistance, Ry = 0.06

Series field resistance, Ry = 0.03 2

Shunt field resistance, R,, = 110 Q

‘Terminal voltage, V = 250 Volts

Solution Refer Fig 23

Armature resistance, Ry = 0.042

Series field resistance, R = 0.03 2

Shunt field resistance, R,, = 200.2

Load current, 1= 180A

= 400 + 182 x 0.04 + 182 x 0.08 + 2x 1

100 + 7.28 + 6.46 + 9 = 414.14 V Henee, generated e.m.f, = 414.74 V (Ans.)

1.4 Parallel Operation of D.C Generators

1.4.1, Reasons for paralleling D.C generators

‘The reasons for paralleling D.C generators, (especially when it is recognised that

of the word parallel means duplicater or multiple) are enumerated below :

1, Reliability The sources of power such as generators are frequently primary safety items and are therefore duplicated or paralleled for reliability

usage

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2 Continuity of service In case of break-down or routine maintenance it is frequently required that the device being worked on be isolated from its work and shut down Therefore, if power sources are paralleled, the routine or emergency operations can be performed without disturbing the load conditions This affects both safety and economy

3, Efficiency It is a known fact that many major types of machinery, such as generators, run most efficiently when loaded to their design rating Electric power costs less per kWh when the generator producing it is efficiently loaded Therefore, when the load is reduced, one or more generators can be shut down and the remaining units kept efficiently loaded

4 Added capacity The use of electricity is constantly increasing in the modern world of expanding population, goods and services When added capacity is required, the new equipment can

be simply paralleled with the old

6 In several situations (not confined to generators), the equipment available todoa particular task may not be available in a sufficiently large capacity or size in a single unit Here paralleling must be a design feature just to meet original load requirement An absolute unit to the size and

‘output eapacity of a D.C generator does not seem apparent, but in any endeavour a new largest size

is always more expensive and usually has unforeseen ‘bugs’, which may be ruinously costly Note Power sources are rarely duplicated in home or automobile service, but usually are in air craft,

‘marine, rail and industrial use

1.4.2, Requirements or paralleling D.C generators

The following are the principal types of situations where paralleling of D.C generators is required :

There are certain requirements that must be met for successful electrical paralleling in all different situations A parallel circuit is defined as one in which the same voltage exists across each unit as the paralleling point :

This is absolutely required by Kirchhoff’s voltage law

The following three conditions may be met if the generated voltages of the individual generators are not all the same, and they are paralleled :

(® Ifa generator is developing an internally generated voltage E, that is appreciably above the voltage at the paralleling point, generator action is taking place and the unit is delivering current

to the load

(ii) When a generator is producing the same voltage as that existing at the paralleling point,

no effective generating action is taking place and no current is flowing to the load The generator is said to be floating’ on the line It is noither contributing nor drawing current and is still being rotated

by its own prime-mover

(ii) Ifthe setting of the generator is so made that it develops less internal Z, than voltage at

he paralleling terminal, it will draw current from the paralleling point and will be operating as

‘motor’

The above three situations are in entire agreement with Kirchhoff’s current law, parallel circuit must be

The following are the requirements or conditions of paralleling D.C generators :

1 The polarities of the generators must be the same or the connections must be interchanged until they are

2 The voltages should be nearly if not exactly identical so that each machine will contribute

8 The change of voltage with change of load should be of the same character

8 any

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298 BLECTRICAL TECHNOLOGY

A positive regulation machine cannot usefully combine with a negative regulation machine Circulating currents would dominate the situation An exact match of characteristics is desirable but not always achieved

4 The prime-movers that drive the generators should have similar and stable rotational speed characteristics, The prime-movers should either all be such that they have constant or flat rotational characteristics or should all droop in speed with increasing load A rising speed characteristics with increasing load is unstable and will cause the affected machine to take more than its share, or even,

all of the load

Note Whenever generators are in parallel their +ve and ve terminals are respectively connected to the

+veand~ve: sof the bus-bars These bus-bars are heavy thick copper bars and they act as +veand~ve terminals: for the whole power station

Construction A D.C, motor is similar in construction to a D.C generator As a matter of fact any D.C generator will run as a motor when its field and armature windings are connected to source of direct current The field winding produces the necessary magnetic field The flow ofeurrent through the armature conductors produces a force which rotates the armature

‘Though the essential construction of D.C motor is identical to that ofa generator, theexternal

‘appearance of a motor may be somewhat different from that of a generator This is mainly duc to the fact that the frame of a generator may be partially open because it is located in relatively clean

environment and only skilled operators are present in its vicinity A motor, on the other hand, may

be operating in a rather dusty environment and only unskilled operators may be working in its vicinity Therefore, frames of motors are to a large extent closed

The body of D.C mill motors is made in two halves bolted together for easy access to the field windings and inter-poles

Applications Because of their inherent characteristics D.C motors find extensive applica- tion in :

(ii) Textile mills (iv) Printing presses

(vii) Excavators ete where precise and accurate speed control over a wide range is required Advantages The D.C motors possess the following advantages :

(@) High starting torque

(Gi) Speed control over a wide range, both below and above the normal speed

(iii) Accurate stepless speed control with constant torque

(iv) Quick starting, stopping, reversing and accelerating

Disadvantages The disadvantages of D.C motors are:

(0) High initial cost

(ii) Increased operating and maintenance costs because of the commutators and brushgear

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ROTATING MACHINES 299 1.5.2 Principle of operation of D.C motor

The principle of motor action can be stated as follows :

“Whenever a current carrying conductor is placed in a magnetic field, it experiences a force whose direction is given by Fleming’s left hand rule”

Fig 24 (a) shows the field set up by the poles

Fig 24 (b) shows the conductor field due to flow of current in the conductor

Fig 24 (c) shows the resultant field produced when the current carrying conductor wire of Fig 24 (6) is inserted in the air gap of Fig 24 (a) with the axis of the conductor at right angles to the direction of the flux

On the upper side of the conductor in Fig 24 (c) the magnetizing forces of the field and of the

‘current in the conductor are additive while on the lower side these are subtractive This explains why the resultant field is strengthened above and weakened below the conductor (wire)

‘The above experiment shows that the wire in Fig 24 (c) has a force on it which tends to move

it downward Thus the force acts in the direction of the weaker field When the current in the wire is reversed, the direction of the force is also reversed, as in Fig 24 (d)

., ZÔN : (\.j s

4

tì ress ta)

Fig 24 The principle of motor action

‘The force (F) developed in the conductor is given by the relation,

F = Bil newtons

where 3= flux density, 7 (Wb/m?),

I= current in conductor, A, and

1= exposed length of conductor, m

Now consider the magnetic field of a D.C motor in which there is no current in the armature conductors ; the lines of force will be distributed as shown in Fig 25

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ELECTRICAL TECHNOLOGY

Fig 25 Distribution of lines of force in a motor due to magnetic field only

If, now, the armature carries current, each of its conductors will produce a magnetic field which, when super-imposed on the main field, causes a distribution of magnetic lines as shown in Fig 26

Fig 26, Distribution of lines of force in a motor, on load, due to the armature and magnetic field

The magnetic field is said to be distorted,

straight paths ince the lines of force no longer follow approximately

These lines of force have the property of tending to shorten themselves, so that they may be regarded as being in tension Each conductor in Fig 26 will

experience « force like that exerted on a stone ina catapult Since

these conductors are embedded in slots in the armature, the

latter is caused to rotate in a clockwise direction

1.5.3, Back or counter EMF

Refer Fig 27 Ina D.C motor when the armature rotates,

the conductors on it cut the lines of force of magnetic field in

which they revolve, so that an e.m.f is induced in the armature

as in a generator The induced e.m.f acts in opposition to the

current in the machine and, therefore, to the applied voltage, 50

thatitis customary to refer to this voltage as the ‘back e.nuf That

this is so can be deduced by Lenz’s law, which states that the

direction of an induced e.m.f is such as to oppose the change

causing it, which is, of course, the applied voltage

‘The magnitude of the back or counter e.m.f can be calcu-

lated by using formula for the induced e.m.f in a generator, and

tO) Motion ~~”,

Armature Fig 27 Motoring operation

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it is important in the case of the motor, to appreciate that this is proportional to the product of the flux and the speed Thus if E}, denotes the back e.m.f,, 9 the flux and N the speed, we may write,

Ey=kon where k is a number depending on nature of armature winding

The value of back e.m.f (Ep) is always less than the applied voltage, although difference small when the machine is running under normal conditions It is the difference between these two quantities which actually drives current through the resistance of the armature circuit If this resistance is represented by R,, the back e.m.f by E,, and the applied voltage by V, then we have

V=Eạ +I,l„

where J, is the current in the armature circuit,

1.5.4, Comparison between motor and generator action

— Fig 28 (b) shows a generator action, where a mechanical force moves a conductor in upward direction inducing an e.m.f in the direetion shown When a current flows as a result of this e.m.f, there is a current-carrying conductor existing in a magnetic field ; hence motor action occurs Shown as a dotted line in Fig 28 (b), the force developed as a result of motor action opposes the motion which produced it

{o) Left-hand motor rule (b) Right-hand generator rule

Fig 28, Comparison of motor and generator uction

‘Thus it ean be stated categorically that in rotating electric machines generator action and motor action occur simultaneously Hence the same dynamo may be operated either as a motor or a generator or both (as in dynamo motor or synchronous converter)

© Fig 29 presents a more graphic representation in terms of rotational elements, which compares the elementary motor and generator for the same direction of rotation and shows, the electric circuits of each which is self-explanatory

It may be noted that when a dynamo is operating as a motor, the generated e.m.f is always

less than the terminal voltage (that produces motor action) and it opposes the armature current On the other hand when a dynamo is operating as a generator, the armature current is in the same

direction as the generated e.m.f., and the generated e.m.f E, exceeds the terminal voltage V applied

across the load This distinction between generator and motor, in which the armature-generator voltage aids or opposes the armature current, respectively give rise to the following basic armature circuit equations

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Direction of applied Directicn of induces

voltage and current oirection of rototion YoltOge ANd CUFFEME Direction of

‘ond torque produced ermature rotation

Ig —— IR, lạRs Retordeg - — - Eg=VelaRo,

> {motor Iriving torque » Retar

vey sttion Driving torque » Retard

Fig 29 Elementary motor action versus generator action

where V = applied voltage (measurable terminal voltage) across the armature,

Ey = back or counter e.m.f developed in the armature of the motor,

E, = generated e.m.f developed in the generator armature, and

1,Rq = armature voltage drop due to a flow of armature current through an armature of a

given resistance, Ry

1.6.8 Torque developed in a motor

When the field of a machine (of the type described as generator) is excited and a potential difference is impressed upon the machine terminals, the current in the armature winding reacts with

the air-gap flux to produce a turning moment or torque which tends to cause the armature to revolve Fig 30 illustrates production of torque in a motor

Fig 30, Produetion of torque in a D.C motor

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‘When the brushes are on the neutral axis, all the armature conductors lying under the north pole carry currents in a given direction, while those lying under south pole carry currents in the reverse direction The commutator (just as in a generator) serves to reverse the current in each armature coil at the instant it passes through the neutral axis, so the above relation is always

maintained as the armature rotates

All conductors under the north pole carry inward-flowing currents which react with the air sap flux to produce down-ward acting forces and a counter clockwise torque Similarly the conductors

under the south pole carry outward-flowing currents which produce upward-acting forces These forces also give rise to counter clockwise torques If the air-gap flux is assumed to be radially directed

at all points, each of the force acts tangentially and produces a turning moment equal to the force

multiplied by its lever arm—the radial distance from the centre of the conductor to the centre of the shaft

‘Magnitude of torque developed by each conductor

= BllrNm

Ifthe motor contains Z conductors, the total torque developed by the armature

where B= gap density, T (Wb/m")

7 = armature current in a conductor, A

1 = active length of each conductor, m

+ = average lever arm ofa conductor or the average radius at which conductors are placed, m

Z = total number of armature conductors

It is more convenient to express 7, in terms of armature current Jq, total flux per pole ¢ and number of poles p

1x @

where a= number of parallel paths,

and A= the cross-sectional area of flux path at radius r

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304 ELECTRICAL TECHNOLOGY Electrical equivalent of mechanical power developed by the armature also

is lost to overcome the iron and friction losses The torque which is available for useful work is known

as shaft torque 7,, It is so called because it is available at the shaft The horse power obtained by using shaft torque is called brake horse power (B.H.P.)

jc) = Teh X28N BLP (metric) = “Toe

BALP (metric) x 735.5

60

where N’= speed of armature in r.p.m

The difference 7, - Ta is known as lost torque (i.e torque lost in iron and friction losses)

= 0.159 x 7am and friction losses Nm,

số

1.5.6 Mechanical power developed by motor armature

Refer Fig 31 The voltage V applied across the motor armature has to (i) overcome back e.m.f

Ey and (i) supply the armature ohmic drop Ï,ll,

V= + I,R, This is known as voltage equation ofa motor.

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Multiplying both sides by J, we get

Via = Bile + ba? Ra

Here Vi, = electrical input to the armature,

ExJq = electrical equivalent of mechanical power ?,, developed in the armature, and 1,2Ry = copper loss in the armature

‘The power available at the pulley for doing

useful work is somewhat less than the mechanical

power developed by the armature

This is evident, since there are certain =

mechanical losses (such as bearing and windage tae

Tie friction and iron losses) that must be supplied by `"

the driving power of the motor

Condition for maximum power We

know that, the mechanical power developed by the

in the form of heat and taking other losses, such as mechanical and magnetic, into consideration, the efficiency of the motor will be below 50 per cent

1.5.7 Types of D.C motors

‘There are three main types of motors characterised by the connection of the field winding in relation to the armature These are :

1, Shunt wound motor or the shunt motor, in which the field winding is connected in parallel

with the armature

2 Series motor, in which the armature and field windings are connected in series

3 Compound motor, which has two field windings, one of which is connected in parallel with the armature and the other in series with it

Shunt wound motor Fig 32 shows the connections of a shunt motor From these connections

it may be observed at once that the field current is constant, since it is connected directly to the supply which is assumed to be at constant voltage Hence the flux is approximately constant and, since also the back e.m.f is almast constant under normal conditions the speed is approximately constant This

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is not strietly true, but nevertheless, it is usual for all practical purposes to regard the shunt motor asaconstant speed machine It is, therefore, employed in practice for drives, the speeds of which are required to be independent of the loads The speed can, of course, be varied when necessary and this

is done by the inclusion of a variable resistor in series with the field winding, as shown in Fig 32

in Fig 33, and is called a divertor compound motor

Refer Fig 34 The compound motor has a shunt

field winding in addition to the series winding so that

the number of magnetic lines of force produced by each

of its poles is the resultant of the flux produced by the

shunt coil and that due to the series coil The flux so

produced depends not only on the current and number of

turns of each coil, but also on the winding direction of

the shunt coil in relation to that of the series coil When

the two fluxes assist each other the machine is a

cumulative compound motor, while if they oppose

each other, it is said to be a differential compound

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91 = flux in the first case

‘No, Jag and Q = corresponding quantities in the second case

gE Using the above retain ie Ney ) we get

For shunt motor :

Applying the same equation in this case also, we get

‘The speed regulation of a D.C motor is defined as follows :

“The change in speed when the load on the motor is reduced from rated value to zero, expressed 4s percent of the rated load speed.”

Percent speed regulation = 29.1884

II load si full load speed

Shunt Motors Example 10 Determine the torque developed when a current of a 30 A passes through the armature of a motor with the following particulars : lap winding, 310 conductors, 4-pole, pole-shoes 16.2 em long subtending an angle of 60° at the centre, bore radius 16.2 cm, flux density in air gap 0.7 tesla

Solution Number of poles, p4

Number of armature conductors, Z=310

Flux density in air gap, B=0.7 tesla

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= 28.45 Nm

Hence, torque developed = 28.45 N-m (Ans.)

Example 11 A 230 V D.C shunt motor takes 32 A at full load Find the back e.m.f on full load if the resistances of motor armature and shunt field windings are 0.2 ohm and 115 ohms

respectively

Solution Supply voltage, V = 230 Volts

Armature resistance, Ry = 0.2 ohm

Shunt field windings’ resistance, Ry, = 115 ohms

Hence, back e.m-f on full load = 224 V (Ans.)

Example 12 The power input to a 230 volts D.C shunt motor is 8.477 kW The field resistance

is 230 Q and armature resistance is 0.28 Q Find the input current, armature current and back e.m.f

(P-TU,, May, 2001) Solution Given : Pi, = 8.477 kW ; Ry, = 230 © ; R, = 0.282

‘s Armature current, I_=1- lay

= 35.86 A (Ans.) Fig 37

= 280 ~ 36.86 x 0.28 = 219.96 V, (Ans.) Example 13 A six-pole lap-connected 230 V shunt motor has 410 armature conductors It takes 41 A on full load The flux per pole is 0.05 weber The armature and field resistances are 0.1 ohm and 230 ohms respectively Contact drop per brush = 1 V

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310

or

ELECTRICAL TECHNOLOGY

Determine the speed of motor at full load

Solution Number of poles,

‘Number of parallel paths,

Number of armature conductors,

Full load current,

Flux per pole,

Armature resistance,

Shunt field resistance,

Contact drop/brush

p=6 a=6

Speed of motor on full load, N

Shunt field current,

Armature current,

Back e.m.f, on full load,

‘We know that,

Hence, speed of motor

on full load = 655.6 r.p.m (Ans.)

224 =

Example 14 A 250 volt d.c shunt motor, on no load, runs at 1000 rpm and takes § A The field and armature resistances are 250 ohms and 0.25 ohm respectively Calculate the speed when the

‘motor is loaded such that it takes 41 A if the armature reaction weakens the field by 3%

Solution Given : 'Y = 250 V ; Nụ = 1000 r.p.m ; Í = 5 Á ; đụ, = 250 9; IP.T.U., May, 2001]

= =0.250,1=41A,0 (: ita |® 0.97 = 2(1- )g=

Speed of the motor, N:

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ROTATING MACHINES 3H 'Example 1õ A 120 uolt d.e shunt motor has an armature resistance of 0.2 ohms and a field

resistance of 60 ohms The full-load line current is 60 A and full-load speed is 1800 rp.m Ifthe brush

contact drop is 3 V, find the speed of the motor at half-load (P.T.U,, June, 2000) Solution Given : V = 120 V; Ry = 0.2.2; Ry, = 602

1, = 60 A, Nj = 1800 r.p.m ; Brush contact drop = 3 V

‘Speed of the motor at half-load, Nz:

V120 Ian Rt go TA Full load armature current,

In, =T,-1, = 60-2=58A V~ Iq, Rg brush contact drop

Armature current at half-load, a2 =

Badk e.mf at halEload, -Eyp = V~ Iy,R, - brush contact drop

= 120-29x0.2-3=111.2V

'® Example 16 A 4-pole 500 V shunt motor takes 7A on no-load, the no-load speed being 750

rpm It has a shunt field current of 2 A Caleulate the full-load speed of the motor if it takes 122 A

Solution Supply voltage, V= 600 Volts

No-load current, l=7A

No-load speed, No = 750 r.p.m

Shunt field current, J„=2A

Full-load current, 1= 122A

Armature resistance, L„ = 0.2 ohm

Contact drop/ brush =1V

= 00 ~ 120 x0.2~ 9 x 1 = 500-24-2=474V

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) Speed regulation

Armature reaction weakens the field by 1.5% on full-load

Solution Supply voltage, V = 440 Volts

Number of poles,

Number of parallel paths,

Armature resistance,

Flux per pole on no-load, 0

Number of armature conductors, Z = 750

Contact drop per brush,

No-load input current,

Shunt field current,

437.5 x60 x4 40.05 x 750 ~ 700 FP

Hence, no-load speed ‘= 700 r.p.m (Ans.)

Gi) Full-toad speed :

Armature current, J, = —I,4 = 160-10 = 140A

Ey = V-1,Rq~ brush contact drop

=440~140x 0.1~2x 1= 494V

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9 =(1-0.015) [Since armature reaction weakens the

field by 1.5 per cent]

Hence, full-load speed = 688.78 r.p.m, (Ans.)

(iii) Percentage speed regulation

no-load speed - full-load full-load speed

700 - 688.73

“Ga 3100 = 1.637 Hence, percentage speed regulation = 1.637% (Ans.)

Example 18 A 250 V shunt motor takes a line current of 60 A and runs at 800 r.p.m Its

armature and field resistances are 0.2 ohm and 125 ohms respectively Contact dropfbrush = 1 V Catculate :

@ No-load speed if the no-load current is 6A

Gi) The percentage reduction in the flux per pole in order that the speed may be 1000 r.p.m

when the armature current is 40 A

Neglect the effects of armature reaction

At no-load:

Tyo = Ip~ Ian 26 -2=4A

Eyo = V~IyoRq— brush drop

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ROTATING MACHINES 317 (ii) Speed of motor,

"Example 23 A six-pole, lap-wound 400 V series motor has the following data : Number of

armature conductors = 920, flux! pole = 0.045 Wb, total motor resistance = 0.6 ohm, iron and friction losses = 2 RW If the current taken by the motor is 90 A, find :

() Total torque ; ii) Useful torque at the shaft ;

Gili) Power output ;

(iv) Pull at the rim of a pulley of 40 cm diameter connected to the shaft

Solution Number of poles, p= 6

Supply voltage, V= 400 Volts

‘Number of armature conductors, Z = 920

Motor resistance, Ry = 0.6 ohm

Iron and friction losses = 2 kW or 2000 W

Current taken by the motor, J, =90A

Radius of the pulley, = 40/2 = 20 em or 0.2 m

Using the relation, By =V~I,Rn = 400-90 x06

(ii) Useful torque, Tusetut !

Toa = Iron and friction loss = 2000 W Thay = 2000% 60, 2000 x 60

aN ~ nx 501

Ti = Ta ~ Tat = 692.4 — 38.11

= 554.29 Nm Hence, useful torque = 854.29 Nm (Ans.)

= 38.11 Nm

Trang 23

318 ELECTRICAL TECHNOLOGY (iii) Power output :

2mN \ _ 554.29 x 2m x 501 Power output - ant a } 554.29 x 2n-< 501

= 29084.4 or 29.08 kW

Hence, power output = 29.08 kW (Ans.)

(iv) ICF isthe pull at the rim of the pulley and ris the radius,

Torqueat the shaft, Tyagi PT

Hence, pull at the rim of the pulley = 2771.45 N (Ans.)

Example 24 A 240 V series motor takes 40 A when giving its rated output at 1500 r.p.m Its

resistance is 0.3 Q Find what resistance must be added to obtain rated torque (i) at starting and (ii)

at 1000 r.p.m

Solution Given : V = 240 volts ; I (= J,) = 40 A, N = 1500 r.p.m., R = 0.8

G) Resistance to be added to obtain rated torque at starting, Raga?

Since the torque remains the same in both the cases, itis obvious that the current drawn by

the motor remains constant at 40 A

(@) When the load torque is increased by 44%:

Tạ= 1447 taÏ;= 1.44 0y 1? 1.4412

Back emf Bạn = Vy + Ree) = 200 - 20 x 0.5 = 190 V

Trang 24

Mechanical charaeteristies Of major importance for industrial drive mechanisms are the mechanical characteristics, which are the relation N = (T)(where N and T stand for speed and torque

respectively) for conditions of constant voltage and resistances in the armature and field circuits

‘These also include the braking characteristics

Regulation characteristics These characteristics determine the properties of motors when their speed is controlled These include :

Nonax

“The D.C motors possess versatile and diverse regulation characteristics, and for this reason are indispensable in installations where wide-range control of speed is necessary

The characteristic curves of a motor are those curves which show relation between the following quantities :

1 Torque and armature current Le, Ta/Iq characteristic This

characteristic

2 Speed and armature current i.e., N/I, characteristic

8 Speed and torquei.e., N/T, characteristic This is also known as mechanical characteristic

‘This can be obtained from (1) and (2) above

Following relations are worth keeping in mind while discussing motor characteristics :

Ey NaF and Ty» tl

is also known as electrical

1.8.10.1 Torque-current characteristics

Shunt motor:

@ When running on no-load, a small arma-

ture current flows to supply the field and

to drive the machine against the friction

and other losses in it

is increased, the torque rises almost

proportionally to the increase in current

‘This is not quite true, because the flux

has been assumed to be constant,

whereas it decreases slightly owing to

armature reaction The effect of this is to

cause the top of the curve connecting

torque and line current to bend over as

shown in Fig 40

mined by the starting resistance, which

in turn, governs the initial current ==

switch is closed At this moment the is is Fig 40 Torque-current characteristic of speed is zero, so that the back em.f is KG,

Torque

Trang 25

zero and the starting current is given by J = V/R, where V is the supply voltage and R is

the total resistance, which includes the armature and starting resistance

Ifthe starting current is limited by heating considerations to twice the full load current, then with normal supply voltage the starting torque of a shunt motor is twice the full load torque If,

however, the supply voltage is below normal, the flux is also less than twice full load torque The

importance of this will be appreciated when the starting torque of a series motor is compared with

that of the shunt motor

Fig 41 shows the relationship between torque and current

Here the current commences at the no-load value, rises paraboli- z cally at first, but increases more slowly as the effects of armature

reaction and magnetic saturation becomes appreciable

heavy current gives rise to a very high torque, also

influences its starting characteristics In a caso of a

shunt motor, ithas already been seen, that the current

at the moment of starting may be as high as twice the

full-load value ; if we allow for the armature of the

mmagnetisation characteristic and for weakening effect

of armature reaction and assume that the flux is in-

creased to 1.5 times its full-load value, then it is

obvious that the starting torque of a series motor is

three times the full-load torque

current may still be maintained at twice full-load value 7

by cutting out some of the starting resistance, so that the high value of starting torque may still be maintained

This type of motor (series motor) is superior to shunt motor for drives in which machines have

to be started and accelerated from rest when fully loaded, as is the case with traction equipment Compound motors:

Differential compound motor Refer Fig 35 In this type of motor the two field windings (shunt and series) oppose each other On light loads, such a machine runs as a shunt motor, since the series field winding, carrying only a small current, has relatively little effect

On heavy loads, the series coils strengthen and since they are in opposition to the shunt winding, cause a reduetion in the flux and a consequence decrease in torque

In the latter case, the motor would tend to start-up in the wrong direction

It is obvious that such characteristics may cause dangerous results, so that differential

compound motors have only very limited applications in practice

Fig 42 shows the torque-current characteristic of a differential compound motor.

Tiêu đề	A Textbook of Electrical Technology P7
Trường học	XYZ University
Chuyên ngành	Electrical Technology
Thể loại	Textbook
Năm xuất bản	2023
Thành phố	Hanoi

Định dạng
Số trang	50
Dung lượng	2,63 MB