Unit III Correct Syllabus Order

8/12/2019 Unit III Correct Syllabus Order

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UnitIII ELECTROMECHANICAL ENERGY CONVERSION

3.1 INTRODUCTION

The chief advantage of electric energy over other forms of energy is the

relative ease and high efficiency with which it can be transmitted over long

distances. Its main use is in the form of a transmitting link for transporting other

forms of energy, e.g. mechanical, sound, light, etc. from one physical location to

another. Electric energy is seldom available naturally and is rarely directly utilized.

Obviously two kinds of energy conversion devices are neededto convert one form

of energy to the electric form and to convert it back to the original or any otherdesired form. Our interests in this chapter are the devices for electromechanical

energy conversion. These devices can be transducers for low-energy conversion

processing and transporting. These devices can be transducers for processing and

transporting low-energy signals. A second category of such devices is meant for

production of force or torque with limited mechanical motion like electromagnets,

relays, actuators, etc. A third category is the continuous energy conversion devices

like motors or generators which are used for bulk energy conversion and

utilization.

Electromechanical energy conversion takes place via the medium of a magnetic

or electric fieldthe magnetic field being most suited for practical conversion

devices. Because of the inertia associated with mechanically moving members,

the fields must necessarily be slowly varying, i.e. quasistatic in nature. The

conversion process is basically a reversible one though practical devices may be

designed and constructed to particularly suit one mode of conversion or the other.

The role of electricity in modern technology is that of an extremely versatile

intermediary. Although energy is seldom directly available in electrical form, and


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ultimately it is seldom required in electrical form, yet conversion of other forms of

energies into electrical energy is a common practice. The chief advantages of this

conversion are that energy in electrical form can be transmitted, controlled and

utilized with relative simplicity, reliability and efficiency. Energy conversion

devices are required first to convert energy in non-electrical form to energy in

electrical form can be transmitted, controlled and utilized with relative simplicity,

reliability and efficiency. Energy conversion devices are required first to convert

energy in non-electrical form to energy in Electrical form and then to convert

electrical energy into the desired useful form, such as light, heat, sound or

mechanical energy. Thus energy conversion devices are needed at both ends of the

electrical system. One typical example is generation of electrical energy

(conversion of non-electrical form energy into electrical form) at nuclear power

station and then transmission and distribution over lines and finally conversion into

mechanical energy by an electric motor for final use. Another example is the

conversion of sound energy into electrical energy at the talker;s end, its

transmission in electrical form over lines and then its final conversion to sound

waves at the listener's end

We are concerned here with the lector-mechanical energy- conversion

process, which takes place through the medium of the electric or magnetic field of

the conversion device. Although the various conversion devices operate on similar

principles, their structures depend upon their function. Devices for measurement

and control are frequently referred to as transducers; they usually operate under

linear input-output conditions and with relatively small signals. The many

examples include torque motors, microphones, pickups, and loudspeakers. A

second category of devices such as solenoids, electromagnets and relays. The third


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category is of continuous energy- conversion equipment and includes generators

and motors.

3.1.2 Magnetic Circuits :

Many of the everyday electric device depend on proper magnetic design as

much as they do proper electrical design.

`The material which attract a piece of iron, placed near to it is called a magnet.

It attracts the iron piece by the existence of a field which consists of imaginary

lines called as Flux-lines. This field is known as magnetic field.

This field can also be created by means of a current carrying conductor.

3.1.3 Magnetic Quantities:

The various terms involved with magnetism are grouped as along with their

characteristics.

1.Lines of magnetic flux closer to each other and having the same direction will

repel each other while those having opposite direction will attract each other.

where is the angle in wb and a is the area of cross section in m2

3.2 ENERGY IN MAGNETIC SYSTEM

Energy can be stored or retrieved from a magnetic system by means of an

exciting coil connected to an electric source. Consider, for example the magnetic

system of an attracted armature relay of Fig. 3.1.


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Assuming for the time being that the armature is held fixed at position x , all

the input energy is stored in the magnetic field. Thus

e fdW ei dt dW --------- (3.5)

where fdW is the change in field energy in time dt. when the expression for e in Eq.

(3.2) is substituted in Eq. (3.5), we have

e fdW i d d dW --------- (3.6)

where Ni the magnetomotive force (mmf)

i relationship

The relationship i or is a functional one corresponding to the

magnetic circuit which is general is nonlinear (and is also history- dependent, i.e. it

exhibits hysteresis). The energy absorbed by the field for finite change in flux

linkages for flux is obtained

2 2

1 1

( ) ( )f

W i d d

-------(3.7)

As the flux in the magnetic circuit undergoes a cycle 1 2 1 , an

irrecoverable loss in energy takes place due to hysteresis and eddy-currents in the

iron, assuming here that these losses are separated out and are supplied directly by

the electric source. This assumption renders the ideal coil and the magnetic circuit

as a conservative system with energy interchange between themselves so that the

net energy is conserved.

The energy absorbed by the magnetic system to establish flux (or flux

linkages ) from initial zero flux is

0 0( ) ( )fW i d d

-------(3.8)


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According to Eqs (3.9) and (3.10) field energy is determined by the instantaneous

values of the system states ( ( , )x or ( , )i x ) and is independent of the path following

by these state to reach the present values. This means that the field energy at any

instant is history independent.

A change in with fixed x causes electric-magnetic energy interchange

governed by the circuit equation (3.5) and the energy equation (3.8). Similarly, if x

is allowed to change with fixed energy will interchange between the magnetic

circuit and the mechanical system.

As per Eq. (3.10) the field energy is the area between the -axis and i-

curve as shown in Fig. 1.7. A new term, co-energy is now defined as'( , ) ( , )

f fW i x i W x

wherein by expressing as ( , )i x the independent variables of 'fW become I and

x- The co energy on Fig. 3.3. is shown to be the complementary area of the i

rectangle. It easily follows from Fig.3.3. that

Fig. 3.3. Field Energy and co energy

Example 3.1.

A conductor of 1m length is dragged with a velocity of 100m/sec, perpendicular to

a field of 1T. What is the value of emf induced?

l= 1m v = 100m/sec, B= 1 web/m2 = /2

e= Blv sin


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= 1 Example 3.2.

An inductive coil of 10 mH is carrying a current of 10 A . What is the energy

stored in the magnetic field.L = 10

I = 10A

Stored energy W =L

= 10 2

= 0.5 joules

Example 3.3.

The total flax of an electromagnet is 8 x 104Wb. (a) If the cross-sectional area of

the core is 5cm2, find the flux density, (b) the coil of the electromagnet has 100

turns and a current of 5A flows through it. If the length of the magnetic circuit is

50cm, find the mmf and magnetic field intensity.

Given: = 8 x 104Wb; A = 5 x 104m2; N = 100; = 50 x 504m2; I = 5Amps

Solution:

(a)Flax Density B = =

B = 1.6 Wb/m2.

(b)mmf = NI= 100 x 5

mmf = 500 AT


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(c)Magnetic field IntensityH =

H = 1000 AT/m

Example 3.4.

The core of the magnetic mode of cast iron. Find the total flux for the same

exiting mmf.

Solution:

mmf = 318 AT

H =

= 1590 AT/m

For cast iron when H = 1590 AT/m, B = 0.5T

Flux = 0.5 x 5 x 104

= 2.5 x 104WbI =1.59 Amps.

Example 3.5.

The co-efficient of coupling between two coils is 0.85. Coil 1 has 250

turns. When the current in coil 1 is 2 Amps, the total flax in this coil 1 is 3 x104Wb. When 1is decreased from 2A to zero linearly in 2ms, the voltage

induced in coil 2 is 63.75V. Find L1, L2, M and N2.

Given: K = 0.85 1 = 3 x 104Wb


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N1= 250 V2= 63.75 V

I = 2 Amp di/dt =

Solution:

(i) =

L1= 37.5 x 103H

(ii) V2= 63.75 = M x

M = 63.75 x 103H

(iii) M = K

63.75 x 103= L2= 150 x 10

3H

(iv)

=

N2= 500

Example 3.6.


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Two coils having 30 and 600 turns are wound side by side on a closed iron

circuit of 100cm2cross-sectional area and mean length of 150cm.

(a)Estimate the self inductance of the two coils and the mutual inductance ifrof iron is 2000.

(b)Calculate the voltage induced in coil if the current changes from 0 to 10Asteadily in 0.01 sec.

Given data: N1= 30 turns N2= 600 turns

a = 100cm2= 100 x 104m

= 150 cm = 150 x 10

2

m

r= 2000= 250 A/seconds.

Solution:

(a) L = Where,

= o r

= 4x 107x 2000

= 2.5132 x 103

L1=

L1= 150.8 x 10

4H

L2=

=

L2= 60320 x 10

4H


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M = () K = 1 (assume)M = M = 3016 x 104H

(b)V2= = (3016 x 104) V2= 301.6 V

3.3 Magnetic-field System: Energy and Co-energy

3.3.1 Linear System

Fig .3.4 (a) Magnetic Circuit

Fig.3.4. (b) Characteristic of a magnetic circuit Fig.3.4. (c) Energy and co-energy


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A simple magnetic current is shown in Fig 3.4(a), with assumption that air-

gap length at the joints is negligible, and the magnetic medium is not saturated.

With A as the cross-sectional area of the core and mL as the mean length of the

path, a coil with N turns carrying a current of i amp has an mmf of F, establishing a

flux of , related by

mF P -------(3.11)

Where mP =Permeance of the Magnetic circuit

o r / mA L

with r = relative permeability of the magnetic medium

This corresponds to the following relationships:

Coil Inductance, 2 / /mL N P N i i -------- (3.12)

where = flux-linkage of the coil, in weber- turns

fldW = Energy stored in the coil = 2 2 2 21 1 1

2 2 2m mLi N P i F P

1 1( )2 2mF FP F

In this eqn., mP is the slope of the characteristic in Fig 3.4. (b). Hence, the

inductance is proportional to the slope of F plot. In Fig. 3.4. (b), for the

operating point A, the mmf is F1and the flux 1 . At the point A, the energy stored

in the field is given by eqn. below:

1 11

2fldW F

------ (3.13)

1F is due to the current 1. fldi W is given by area OATO is Fig 3.4.(b)

In Fig 3.4.(b), the origin refers to the system without magnetization.

The system can reach the point A, starting from O as the current in the coil is

increased from O to 1i .


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(a)HC' vertically, if the mechanical movement is too slow so that change of flux isslow and induced emf in the coil is negligible. This corresponds to the coil-mmf

remaining constant at F during the transition. Constant mmf means vertical

travel of the operating point from H to C.

(b)H to K horizontally and then K to C' along the characteristic corresponding to( )x dc as the displacement of the movable part. This is possible when the

motion is very fast, resulting into flux remaining constant till the operating

point traverses from H to K. Then, from K to C. the flux increases, an emf in

induced the coil and the mmf finally teaches its value of F, at the point C;

(c)In reality, the transition form H to C will be somewhere in between these twoextremes mentioned above.

However, for simplicity, one of these extreme conditions has to be accepted. In

(a) above, the mmf remains constant, In (b) above, the flux (and hence the flux-

linkage) remains constant. Let us take the case of constant-mmf. If the process has

taken a time of dt.

Electrical -energy input during the process = elecdW

= (Voltage applied to the coil) Current dt = eidt

/d dt i dt id i N d Fd area of rectangle B'HC'D

In this case, coil - resistance has been neglected.

In terms of Field Energy

At the previous operating point H, the energy stored in the magnetic field

1fldW area of 'OHB


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At the new position corresponding to the operating Point C, the field energy

stored is given by2fld

W = area of the 'OC D

The difference of these two is the change in the energy stored in the

Magnetic field = 1 ' ' ' '2flddW OA A C OA A H

1 1' ' ' ' ' '2 2OA A C A H OA HC

1 1. .2 2 elec

F d dW ----- (3.17)-

Out of the energy delivered by the source, half is stored in the magnetic

field. Where has the remaining half been utilized. Obviously, this must have been

transformed into the mechanical work done. In this case, neglecting losses, it isfinally stored in the stretched spring due to its elongation by dx.

Comparing this with the equation ()3.17,

elec mech fld dW dW dW

12mech elec

dW dW ------ (3.18)

Or 12mech fld elec

dW dW dW

Consider that a force F is operative at the displacement of X. This force is in

such a direction that x increases or the movable member is attracted towards D. In

the same direction, a displacement by dx results into the increase in the energy

stored by the spring. Relating the concerned terms,

sF k x

mechdW Mechanical work done against the force of the stretched spring

fldF dx dW

or F / ,flddW dx in this case

/ ,fldW x in general


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Alternatively, the difference in the energy stored in the spring also gives a

very useful relationship.

In the position corresponding to x dx the energy stored in spring

21 ( ) .2 sk x dx Similarly, at x , the energy =212 sk x

Difference 2 21 ( ) ( )2 s

k x dx x

2 2 21 2 . ( )2 s

k x x dx dx x

1 2. .2 s

k x dx neglecting2( )dx

. .sk x dx F dx ------ (3.19)

This difference is nothing but mechdW which is equal in magnitude to flddW and

confirms the relationship obtained earlier.

3.4 FORCE ON A MOVING CHARGE

In an electric field, the electric field strength or electric field intensity, E is

defined as the vector force on a unit charge.Obviously the force on a charged particle will be as given by the expression.

F QE ------ (3.20)

This force is in the same direction as the electric field intensity (for a

positive charge) and is directly proportional to both E and Q. If the charge is in

motion, the force at any point in its trajectory is then given by equation (3.20).

A charged particle in motion in a magnetic field of flux density B is found

experimentally to experience a force whose magnitude is proportional to the

product of the charge Q, its velocity v, and the flux density B, and to the sine of the

angle between the vectors v and B. The direction of the force is perpendicular to


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both v and B and is given by a unit vector in the direction of v x B. The force may,

therefore, be expressed as

F Q v B ------ (3.21)

A fundamental difference in the effect of the electric and magnetic fields on

the charged particles is now apparent, for a force which is always applied in a

direction at right angles to the direction in which the particle is proceeding can

never change the magnitude of the particle velocity. In other words, the

acceleration vector is always normal to the velocity vector. The kinetic energy of

the particle remains unchanged, and it follows that the steady magnetic field is

incapable of transferring energy to the moving charge. The electric field, on the

other hand, exerts a force on the particle which is independent of the direction in

which the particle is progressing and, therefore, affects an energy transfer between

field and particle in general.

The force on a moving particle due to combined electric and magnetic fields

is obtained easily by superposition

( )F QE Qv B Q E v B ----- (3.22)

This equation is known as the Lorentz force equation, and its solution is

required in determining electron orbits in the magnetron, proton paths in the

cyclotron, plasma characteristics in a magneto-hydrodynamic (MHD) generator,

or, in general, charged particle motion in combined electric and magnetic fields.

3.4.1 FORCES AND TORQUES IN MAGNETIC FIELD SYSTEMS

According to Lorentz force equation

( )F q E v B --------- (3.23)


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Where F is the force in newtons on a particle of charge q coulombs in the presence

of electric and magnetic fields, E being in volts per metre, B in Teslas and v is the

velocity of the particle relative to the magnetic field, in m/s.

Thus in a pure electric field system, the force is determined simply by the

charge on the particle and the electric field.

F qE ------- (3.24)

The force F acts in the direction of the electric field and is independent of

the particle motion.

In magnetic field systems the situation is somewhat more complex. Here the

force is given as

( )F q v B -------- (3.25)

and is thus determined by the magnitude of the charge on the particle and

the magnitude of induction density B as well as the velocity of the particle. In

fact the direction of force is always perpendicular to both the direction of

Fig 3.6 Right Hand Rules For Determination

of Direction of Lorentz Force


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particle motion and the direction of magnetic field. Mathematically, this is

given by the vector cross product v x B in equation (3.25). The magnitude of

this cross product is equal to the product of the magnitudes of v and B and the

sine of the angle between them; its direction can be determined from the right

hand rule, which states that when the thumb of the right hand points in tL

direction of v and the index finger points in the direction of B, the force

points in th direction normal to the palm of the hand, as depicted in fig. 3.6.

For situations where large number of charged particles are in motion,

it is convenient to rewrite equation (3.25) in terms of the current density J, in

which case the force is a force densityF = J x B N/m3 ------- (3.26)

For currents flowing in conducting media, equation (3.26) can be employed for

determination of force density acting on the material itself.

For situations in which the forces act only upon current-carrying elements

and which a of simple geometry. Equation (3.26) usually the simplest and

easiest way to determine forces acting on the system. Very few practical

situations fall into this class, however. In i--j most electromechanical energy-

conversion devices contain magnetic material; in such system! forces act directly

on the magnetic material and clearly cannot be calculated from equation (3.26)

Techniques for calculating the detailed, localized forces acting on the magnetic

material are extremely complex and need detailed knowledge of the field

distribution throughout the structure.

Fortunately, most electro-mechanical energy-conversion devices are

constructed of rigid, non-deforming structures. In these devices, it is the net

force or torque that is of importance, and the details of the localized force


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distribution are of secondary interest. For example, in a properly designed

motor the net accelerating torque acting on the rotor determines the motor's

characteristics; accompanying forces, which act to squash or make the solid

rotor oval, play no significant role and generally are not even computed.

For understanding the behavior of rotating machinery, a simple physical

picture is quite useful. Associated with the rotor structure is a magnetic field,

and similarly with the stator; one can picture them as a set of north and south

poles associated with each structure. Just as a compass needle tries to align

with the earth's magnetic field, these two sets of fields attempt to align, and

torque is associated with their displacement from alignment. Thus in a motor,

the stator magnetic field rotates ahead

of that of the rotor, pulling on it and

performing work. The opposite is true

for a generator, here the rotor does the

work on the stator.

Various techniques have been evolved for calculation of the net forces of

concern in electromechanical energy conversion. The technique usually

employed is known as the energy method and is based upon the principle of

conservation of energy. The basis for this method can be understood with

reference to fig.3.7.(a) where a magnetic field-based electromechanical energy-

conversion device is indicated schematically as a lossless magnetic energystorage system with two terminals. The electric terminal has the terminal

variables voltage e and current i and the mechanical terminal has the terminal

variables force fieldf and position x. This sort of representation is valid in

situations where the loss mechanism can be separated out from the energy-

Fig .3.7. (a) Schematic Magnetic Field

Electro-Mechanical Energy-


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storage mechanism; in these cases the electrical losses such as ohmic losses

can be included as external elements connected to the electrical terminals and

the mechanical losses such as frictional losses can be included external to the

mechanical systems.

-

Fig. 3.7.(b) shows a typical example of such a system; a simple force

producing device with a single coil forms the electrical terminal, and a

movable plunger

Fig, 3.7. (b) .Simple Force Producing Device

The interaction between the electric and mechanical terminals, i.e., the

electromechanical terminals, i.e., the electromechanical energy conversion,

occurs through the medium of the magnetic stored energy. Since the energy

storage system is lossless, it is a simple matter to write that the time rate of

change of fieldW , the stored energy in the magnetic field, is equal to the

electric power input less the mechanical power output of the energy - storage

system i.e,

field

field field

dW dx dx d dxei f i i f

dt dt dt dt dt

------- (3.27)


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d de N

dt dt

-------- (3.28)

or field fielddW id f dx

Equation (3.28) permits us to solve the force simply as a function of the flux

linkage and the mechanical -terminal position x . This result comes about as a

consequence of our ability to separate the losses out of the physical problem,

resulting in a lossless energy storage element, as in fig. 3.7 (a).

Equations (3.27) and (3.28) form the basis for the energy method. "This

technique is quite powerful in its ability to compute forces and torques in complex

situations of electromechanical energy-conversion systems. It is to be noted

here that this power comes at the expense of a detailed picture of the force-

producing mechanism. The forces themselves are produced by such well known

physical phenomena as the Lorentz force on current -carrying element, or the

interaction of the magnetic fields with the dipoles in the magnetic material.

Example 3.7.

A non-magnetic rotor having a single-turn coil is placed in a uniform

magnetic field of magnitude 0.8 T, as shown n fig 3.8. The coil sides are at a

radius of 0.125 m, and the coil carries a current of 12 A, as shown. Determine

the 9-directed torque as a function of rotor position ex.

Rotor may be assumed to be 0.5 m long.

Solution: The force per unit length on a wire carrying a

current of I amperes can be determined by multiplyingequation (2.9) by cross-sectional area of the wire.

Single Coil Rotor

Fig. 3.8


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As the product of the current density and the cross-sectional area of the wire is

simply the currentI, the force per unit length acting on the wire is given as

F I B

Thus, for wire 1 carrying a current of I amperes into the paper, the 9-

directed force is given as

1 sinF I B l

and for wire 2 located 180 away from wire 1 and carrying a current of I

amperes out of paper is given as

2 sinF I B l

where / is the length of rotor in metres. The torque acting on the rotor is given

by the sum of the force-moment-arm products for each wire i.e.

2 sin 2 12 0.8 0.125 0.5sin 1.2 sinT I B R l N m

3.5 SINGLY EXCITED MAGNETIC FIELD SYSTEMS

Energy in MagneticSystem. Energy can be stored or retrieved from a

magnetic system by means of an exciting coil connected to an electrical source.

Consider, for instance, the magnetic field system of an attracted armature relay

shown in fig.3.9. The resistance r of the exciting coil is shown by a series

resistance lumping outside the coil which then is considered as an ideal loss-less

coil. The magnetic flux is produced in the magnetic circuit due o flow of current

i in the exciting coil. It is assumed that all the flux so created is confined to the

iron core and, therefore, links all the N turns producing the coil flux linkages y

which is equal to N


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Fig 3.9. Electro-magnetic Relay

The instantaneous voltage equation for the electric circuit is written,

according to Kirchhoffs voltage law.

di r e i r

dt

------- (3.29)

Multiplying both sides of above equation (3.29) by i dtwe have

2i dt i r dt i d

or ( )ir i dt i d

or e i dt i d ------- (3.30)

The electrical energy input to the ideal coil due to flow of current i in time dt is

given as

elecdW e i dt i d

------- (3.31)

Assuming for the time being that armature is held fixed at position x , all the input

energy is stored in the magnetic field. Thus

elec field dW e i dt dW ------- (3.32)


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where feileddW is the change in field energy in time dt. Substituting ei dt i d from

equation (3.31)in above equation (1.69) we have

elec field dW i d i N d M d dW ------- (3.33)

where M=Ni, the mmf

The relationship i or M is a functional one corresponding to the

magnetic circuit which in general, is non-linear and also it exhibits hysteresis. The

energy absorbed by the field for finite change in flux linkages or flux is given,

from the above equation (3.33), as

2 2

1 1

( ) ( )fieldW i d M d

------- (3.34)

As the flux in the magnetic circuit undergoes a cycle 1 2 1 , an

irrecoverable loss in energy takes place due to hysteresis and eddy currents in the

iron. Assuming here that these losses are separated out and are supplied directly by

the electric source, the ideal coil and the magnetic circuit becomes a conservative

system with energy interchange between themselves so that the net energy is

conserved.

The energy absorbed by the magnetic system is establishing flux linkages

or flux from initial zero value.

0 0( ) ( )fieldW i d M d

------- (3.35)

Fig. 3.10 Current - Flux Linkages curve with variable x


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This is the energy of the magnetic field with given mechanical configuration when

its state corresponds to flux or flux linkages

The i relationship is actually the magnetization curve and this curve

varies with the configuration value x, as illustrated in fig.3.10.

The total reluctance of the magnetic path decreases with the increase in x

and air-gap between the armature and core varies with position x of the armature.

Thus i relationship can be expresses as

( , )i i x if is the independent variable

or ( , )i x if i is the independent variable.

Thus, the field energy [equation (3.35)] is in general, a function of two

variables i.e.

( , )field fieldW W x ------ (3.36)

( , )field fieldW W i x ------- (3.37)

According to above equations field energy is determined by instantaneous

values of the system states ( , )x or ( , )i x and is independent of the path followed by

these states in attaining these values.

It means that the field energy at any instant is history independent (does not

exhibit hysteresis)

A change in with fixed x causes electro-magnetic energy interchange

which is governed by equations (3.33). Similarly a change in x with fixed

causes interchange of energy between the magnetic circuit and the mechanical

system. The general case of such energy interchanges (3.35), the field energy is the

area between the -axis and i curve as illustrated in Fig.3.11. A new term, co-

energy is now defined as


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W field(i, x) = i Wfield(,x) ------ (3.38)

wherein by expressing flux linkages as ( , )i x the independent variables of fieldW

become i and x. The co energy on fig. 3.11 is shown to be complementary area of

i rectangle.

Wfield = -------- (3.39)

Fig.3.11 Field energy and co energy

3.5.1 Linear Case. Electro - mechanical energy conversion devices contain air-

gaps in their magnetic circuits so as to separate the stationary and moving

members. In most of such cases the reluctance of the air-gap is much larger than

that of the magnetic material. Thus the predominant energy storage occurs in the

air-gap, and the properties of the magnetic circuit are determined by the

dimensions of the air-gap.

Because of the simplicity of the resulting relations, magnetic non-linearity

and core losses are often neglected in the analysis of practical devices. The final

results of such approximate analysis can, if necessary, be corrected for the effects


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of these neglected factors by semi empirical methods. Thus, analysis are carried

out under the assumption that the flux and mmf are directly proportional, as in air,

for the entire magnetic circuit. With the linearity assumption the analysis is greatly

simplified.

Assuming linearity, equation (3.35) becomes

21 1 1

2 2 2fieldW i M S

------ (3.40)

Where S is the reluctance of the magnetic circuit and is equal to . .mmf N

i e

Since coil inductance is given as

,NLi i

iL

and the field energy can be expressed as

21

2field

WL

------ ( 3.41)

1 1

2 2i

L

In the linear case the inductance L is independent of current i but is a fuction

of configuration x. Thus the field energy is a special function of two independent

variables and i.e.

21

( , )2 ( )

fieldW xL x

-------- (3.42)

The field energy is distributed throughout the space occupied by the field.

Neglecting losses and assuming permeability to be constant, the energy of the field

is obtained from equation (3.35) i.e.

23

0 0

1 1/

2 2

Bfield

field

field

W Ni BW d H d B HB J m

W l N

------- (3.43)


31/58

where H is magnetic field intensity in AT/m and is equal toNi

land B is magnetic

field density in 2/Wb m or Tesla.

The above expression for energy density is important from the point of viewof design wherein the capability of the material is to be fully utilized in arriving at

the gross dimensions of the device.

For linear case it easily follows from equation (3.39) or fig. 3.11 that the co

energy is numerically equal to energy i.e.

21 1 1

2 2 2field fieldW W i M S

------- (3.44)

Also in terms of coil inductance

2

0 0

1

2

i i

fieldW di Li di Li N

N i Lii

or in general

21( , ) ( )2

fieldW i x L x i ------- (3.45)

The expression for co energy density Is given as

0

H

fieldW BdH ------- (3.46)

For linear case the above expression becomes

22 31 1 1

/2 2 2

field

BW BH H J m

------ (3.47)

The fact that field energy can be expressed in terms of circuit parameter L,

clears the way for electric circuit approach to the analysis of electrical machines

i.e. the generalized theory of electrical machines. Thus the field energy approach

serves as the physical basis for the generalized theory of electrical machines.


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General Analysis of Electro - magnetic System (Field Energy and Mechanical

Force)

Consider once again the attracted armature relay excited by an electrical

source, as illustrated in fig. 3.9. The field develops a magnetic force

mch field dW F dx ------ (3.48)

This energy is drawn from the field by virtue of change dx in field

configuration. According to the principle of energy conservation

Mechanical energy output= Electrical energy input = increase in field energy

or field elec fieldF dx dW dW

or field fieldF dx i d dW -------- (3.49)

It may be seen that fieldF dx is the gross mechanical output and a part of it

will be lost in mechanical friction.

From equation (3.48) we have

( , )field fieldW i W i x

So ( ) ( , )field fielddW d i dW i x

field fieldW Wi d di di dx

i x

------- (3.50)

Substituting for fielddW from equation (3.50)in equation (3.49) we have

field field

field

W WF dx id id di di dx

i x

orfield field

field

W W

F dx di dxi x

--------- (3.51)

Because the incremental changes di and dx are independent and di is not

present in the left-hand side of above equation (3.51), its coefficient on the right

hand must be equal to zero


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. . 0fieldW

i ei

orfieldW

i

---------- (3.52)

Thus field from equation (3.51) becomes

' ( , )field

field

W i xF

x

---------- (3.53)

The above expression for magnetic or mechanical force produced applies

when i is an independent variable i.e. when it is a current excited system.

( , )field fieldW W x

and field fieldfieldW W

dW d dxx

--------- (3.54)

Substituting fielddW from above equation (3.54) in equation (3.49) we have

field field

field

W WF dx i d d dx

x

orfield field

field

W W

F dx dx i dx

-------- (3.55)

Since d the independent differential is not present on the left-hand side of the

above equation (3.56) its coefficient on the right hand must be equal to zero

. . 0fieldW

i e i

or( , )

fieldW xi

------- (3.57)

Hence from equation (3.57)

( , )field

field

W xF

x

-------- (3.58)


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In the above form of expression for the magnetic force, is independent

variable i.e. it is a voltage controlled system as voltage is derivative of .

Equations (3.50) and (3.58) are equivalent and can both be used to calculate

the force for a given system, user preference and convenience often dictate the

choice.

In linear systems where inductances are specified it is more convenient o use

co energy for finding the force produced - refer equation (3.50).

In case of voltage controlled system, the current can be determined by

writing the necessary circuit equations.

In a rotary case, the equations (3.50) and (3.54) for force will apply fortorque by replacing x with angular rotation

3.5.2 Direction of Mechanical Force Developed. From equation (3.54) it is

obvious that mechanical force developed fieldF is positive when( , )

fieldW x

x

is

negative. It means that stored energy of the field is reduced with the increase of x

while flux linkages are kept fixed. In the particular case of electro-magnetic

relay shown in fig.1.18, as the armature moves towards left (i.e. as x increases), the

field energy for fixed flux linkages is reduced. This is due to reduced air gap.

Thus it can be concluded that the mechanical force developed by the field acts in a

direction to reduce field energy or in other words the system seeks a position of

minimum field energy. Similarly from equation (3.51) it follows that the system

seeks a position of maximum co energy.

Also in case of magnetic relay shown in fig,. 3.11 the mechanical force

developed acts in a direction to increase x thereby reducing the reluctance of the

magnetic circuit and increasing the inductance of coil.


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Determination of Magnetic or Mechanical Force. (a) Non-Linear Case.

Mechanical force developed by the magnetic field is given by partial derivatives of

co energy or energy [Equations (3.54) and (3.58). In the general non-linear case,

the derivative must be determined numerically or graphically by assuming a small

increment .x Thus

' '

tan

tan

field field

field i cons t field

cons t

W WF or F

x x

------ (3.59)

The above two expressions will give slightly different numerical values of

fieldF because of finite x . Obviously, fieldF is the same in each case as 0x .

(b) Linear Case (Neglecting Reluctance of Iron Path)

21( , ) ( )2

fieldW i x L x i

or 21

( )2

fieldW i L x

and mechanical force developed,

'

21 ( )

2

field

field

W L xF i

x x

------- (3.60)

From above equation it is obvious that the mechanical or magnetic force

developed acts in a direction to increase the exciting coil inductance, as already

mentioned. Also From equation (3.43)

21( , )

2 ( )fieldW x

L x

So

2( , ) 1 ( )

2 ( )

field

field

W x L x

F x L x x

-------- (3.61)

Since iL

, equations (3.60) and (13.61) are equivalent.

Also from equation (3.41), 21

2fieldW S


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21 ( )

2

field

field

W S xF

x x

------- (3.62)

It is worth mentioning here that and do not differ as independent

variables because these are related by a constant ( )N N . From above equation

(3.62) it is obvious that the force acts in a direction to reduce the magnetic system

reluctance, as already mentioned.

Also from equation (3.41)

1( , ) ( )

2fieldW x i x

or1 ( )

2

field

field

W i xF

x x

------- (3.63)

Similarly in a rotary case

( , )field

field

W iT

------ (3.64)

The above cases are applicable to motion through a limited distance or

rotation through a definite angle and are the equations for singly excited systems

such as in electro-magnetic relays, reluctance motors. In such systems the electro-

mechanical energy conversion takes place through only one source (ac or dc)and

that is why they are called singly excited magnetic field systems.

3.5.3 Mechanical Work Done. Figure 3.12(a) shows the magnetic circuit of a

relay initially with air gap between the core and the armature. On the closure of

switch S, the exciting current i increases from zero to 1V

ir

where r is the

resistance of the coil. As explained earlier, the energy stored in the magnetic field

is given by the hatched area oabo in fig. 3.12(b).


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(a) Magnetic circuit (b) CurrentFlux Linkages Curve

Fig 3.12. Electro magnetic relay in open position

In fig. 3.13(a), the armature is assumed in the closed position i.e. the air gap g is

zero. On closure of switch S, the exciting current again rises from zero to 1V

ir

whereas the flux linkage increases from zero to 2 . It is to be noted here that the

flux linkages 2 will be greater than the flux linkages 1 , because reluctance of the

magnetic path is considerably reduced for the given mmf. The energy stored is the

magnetic field is now given by the hatched area ocdo( fig.3.13(b)).

With the armature in open position, as the magnetic field builds up,

magnetic pull between the core and the armature develops. Under the action of the

magnetic pull (a mechanical force), the armature moves to the closed position.

During the armature movement, from its open position, the reluctance of the

magnetic path decreases, thereby increasing the flux linkages from its initial value

of 1 . Because of these increments in flux linkages, an emf, known as counter emf,

is induced in the coil. The induced emf opposes the flow of exciting current i, i.e.,


38/58

Current V EiCoil impedance

(a) Magnetic circuit (b) Current-Flux linkage Curve

Fig 3.13. Electro magnetic relay in closed position

The magnitude of counter emf induced in the exciting coil depends how fast the

armature moves. We will consider the two extreme cases. One, the current is

allowed to establish and gap g is reduced to zero very slowly (i.e. at almost zerospeed). Second when the armature is held in position till the current i establishes

and then the armature is allowed to move and reduce the gap g to zero in zero time

(i.e. instantly). Neither of these two would be followed in practice.

(a)Slow Movement of Armature.With the armature in the open position, the exciting current is 1i the flux

linkages is 1

and the operating point is a. When the armature is allowed to movevery slowly, the flux changes at negligibly low rate, the induced emf is, therefore,

negligible, the current 1i stays substantially constant 1V

ir

. Operating point a

must move in such a way that 1i is maintained. Thus it moves along the vertical


39/58

dotted line ac in fig3.14. Consider the significance of the various areas in fig.3.14.

Area 'o a a b o to the left of the original magnetization curve (i.e. area 'o a a b o )

represents the energy abstracted from the electrical source and absorbed by the

magnetic field during the initial excitation of the coil while the armature is in open

position. For simplicity, neglect hysteresis and eddy currents. This energy then is

stored in the magnetic field. After the armature has moved into the final

magnetization curve (i.e. area 'o a c d b o ). Consequently the increase in energy

stored in the field,

fieldW = Magnetic energy stored in the closed position

- Magnetic energy stored in the open position= Area 'o a c d b o - area 'o a a b o ------- (3.65)

While the armature is moving, the flux linkages is increasing from 1 to 2

and an emf is induced in the coil. The corresponding energy abstracted from the

electrical source during this time period .

2

21 1 2 1

( )elec

W i d i

Area

'a c d b a o ----- (3.66)

Substituting equations (3.65) and (3.66) in equation

d Wfield= eidt =d Wfield+d Wmechwe have

elec field mechW W W

or Area 'a c d b a o = (Area 'o a c d b o - area 'o a a b o + mechW

mechW = Area' '

a c d b a o + area'

o a a b o - area'

o a c d b o

= area ' 'o a c a o ------ (3.67)

The above equation (3.67) shows that the mechanical work done (or energy

converted into mechanical form) equals the area enclosed between the two


40/58

magnetization curves at open and closed positions and the vertical i locus during

the slow movement of the armature. This is illustrated by cross batched area in

fig.3.14.

Fig.3.14. Mechanical Work Done Fig.3.15 . Mechanical Work Done

With Slow Armature Movement With Slow Armature Movement

(Non-Linear case) (Linear case)

It is to be noted here that out of the total electrical energy supplied during

slow armature movement, some energy gets stored in the magnetic field and the

remaining is converted into mechanical energy.

Linear Case. Again electrical supplied, 1 2 1( )elecW i . For linear case, it follows

from the geometry of the fig. 3.15 that

'

1 2 1

1 1( )

2 2field field mech elecW W W i W

------ (3.68)

It means that half the electrical energy input gets stored in the field and theother half is output as mechanical energy.

(b) Instantaneous Movement of Armature. When the armature moves to close

the air gap instantaneously (i.e. in zero time), change in flux linkages will be


41/58

negligible and will remain constant at 1 . This is because the flux linkages with an

inductive circuit cannot change abruptly. The operating point, therefore, moves

along the horizontal dotted line from a to a', and after the armature has closed, the

operating point moves along the closed position magnetization curve a'c as

illustrated in fig.3.16(a), because the final operating point has to be c.

(a) Non-linear Case (b) Linear Case

Fig. 3.16. Mechanical work done with instantaneous Armature movement

During the instantaneous movement of armature from open position (operating

point a ) to the closed position (operating point a ) in fig. 3.16 we have

Electrical energy supplied1

2

0elec

W i d

Change in the magnetic stored energy,

fieldW Area'o a b o area

' 'o a a b o

But elec field mechW W W

i.e. 0 = elec mechW W

or mech field W W [area'o a b o area ' 'o a a b o ]


42/58

= area 'o a a b o area 'o a b o = area 'o a b o ---- (3.69)

From the above equation (3.69) it is obvious that the mechanical work done

is equal to the area enclosed between the two magnetization curves at open and

closed positions and the horizontal i locus during the instantaneous movement

of the armature. Since there is no electrical input, the mechanical energy output is

at the expense of the field energy stored.

i.e. Mechanical energy output = Reduction in the energy stored in the field

For linear case [fig. 3.16(b)]

1 1 21 ( )2

mechW i i ------ (3.70)

( C) Transient Movement of Armature.The armature movement will neither

be too slow nor too fast, but will be somewhere in between the two extreme

limits discussed above. The flux linkage-current ( i ) relationship is a general

path from a to c as illustrated in figs. 3.17 (a) and 3.17 (b)

In general case

elecW Area 'b a a c d

fieldW Area 'o a c d o - area' 'o a a b o

and mech elec field W W W

=Area 'b a a c d - area ' 'o a a b o + area ' 'o a a b o

= Area ' 'o a a b o ----- (3.71)


43/58

(A)Non-Linear Case (B) Linear CaseFig.3.17 .Mechanical Work Done During Transient Movement of Armature

Equation (3.71) again shows that the mechanical work done is equal to

the area enclosed between the two magnetization curves at open and closed

positions and the i locus during the transient movement of armature.

Since ac is a general movement, the area 'o a c d o which represents the

mechanical energy output has to be computed graphically or numerically.

Example 3.8

The -i characteristics of singly excited electromagnet is given by I = |2|forif air gap is 5 cm and a current of3A is flowing 0


44/58

= i

0 , l

0

i 4, l 3.63636 for x 5 cm3.63636

= * +0

Field energy = 4.848 J2. Co energy

Wf =

= di

4

=

{

} =0

Wf =

,

- = 9.69696J

3). Force

Force =

Wf = 1939.39 F =

( 1939.39)2x

At x 5 cm = 3878.78 5 Force F = 193.939N


45/58

3.6 MULTIPLY EXCITED SYSTEMS

Singly-excited devices discussed earlier, are generally employed for motion

through a limited distance or rotation through a prescribed angle. In singly-excited

devices, electromechanical conversion takes place through dc or ac source and that

is why they are called singly excited devices. Examples of singly-excited devices

are electromagnets, relays, moving-iron instruments, reluctance motors etc.

A doubly -excited magnetic system is one which has two independent

sources of excitation. Most of the electromagnetic devices belong to doubly-

excited or multiply-excited magnetic systems. In measurement systems it is often

desirable to obtain torques proportional to two electric signals, power as the

product of voltage and current is a common example. Similarly, most continuous

energy conversion devices - generators and motors need multiple excitation

(multiple windings). One group of windings is mounted on a stationary member

and the other group on a movable member. The tendency for the magnetic field

energy to change when one group of windings moves with respect to the other

gives rise to mechanical forces. The theory of numerous types of energy-

conversion devices, including most rotating machines, is based on this general

principle. DC shunt machines, synchronous machines, loud-speakers, tachom-eter

etc. fall in this category of the devices.

A simple model of a doubly- excited magnetic system is shown in fig. 1.28.

The model consists of stator iron having N1turn energized from source 1 and rotor

iron having N2turn coil energised from source 2. The mmfs produced by both the

stator and rotor windings are in the same direction and the electromagnetic torque

Te is in the counter- clockwise direction, as illustrated in fig. 3.18. Magnetic

saturation and hysteresis, for convenience may be neglected.


46/58


47/58

If the rotor is held fixed, then mechanical work done

and 0elec mech field field dW dW dW W ------ (3.74)

i.e. with rotor is held fixed, all the electrical energy input by the two supply

sources is stored in the magnetic field and from equation (3.74)

1 1 2 2 1 1 1 12 2 2 2 2 21( ) ( )field elec idW dW i d i d i d L i M i i d L i M i ---- (3.75)

Since2 2 2

/

o r

o r

N a N NL

l l a S

and 1 2 1 2 1 2

/o r

o r

N N a N N N N M

l l a S

2 2

1 21 2

1 2

;N N

L LS S

and 1 212 2112

N NM M

S

where 1 2,S S and 12S are the respective reluctances seen by the stator flux, rotor flux

and by the resultant of stator and rotor flux.

Since the rotor is held fixed (not allowed to move), the reluctances and,

therefore, inductances are constant. Thus the differential changes in inductances

1 2, ,dL dL and 12dM are all zeros and equation (3.75) is reduced to

1 1 2 2 2 1 12 2 2 21 1 1 1 2 2 2 12 1 2( , )field ldW i L di i L di i M di i M di i L di i L di M d i i

The magnetic field energy stored in establishing the currents from zero to 1i and 2i

is given as

1 2 1 2 2 2

1 1 1 2 2 2 12 1 2 1 1 2 2 12 1 20 0 0

1 1( )

2 2

i i i i

fieldW L i di L i di M d i i L i L i M i i ---- (3.76)

Let us assume that the rotor moves through a virtual displacement rd

in the

direction of magnetic torque Te, as illustrated in fig. 3.18. With the movement of

rotor, reluctances, S1, S2and S12and therefore, inductances L1, L2, M12mustchange.

Differentiation of equation (3.75) with respect to inductances and currents

provides for electrical energy input

2 2

1 1 1 12 1 2 1 1 1 2 12 2 2 2 21 2 1 2 2 1 2 12elecdW L i di M i di i dL i i dM L i di M i di i dL i i dM ---- (3.77)


48/58

Similarly from equation (3.76), the increment in field energy is

2 2

1 1 1 1 1 2 2 2 2 2 12 1 2 12 2 1 1 2 12

1 1

2 2field

dW L i di i dL L i di i dL M i di M i di i i dM ----- (3.78)

The differential mechanical work done during virtual displacement rd

mech e r dW T d

From the law of conservation of energy

elec e r field dW T d dW -----(3.79)

When the values of elecdW and fieldW are substituted from equations (3.77) and (3.78)

in equation (3.79), the four terms 1 1 1 2 2 2 12 1 2 12 2 1L i di L i di M i di M i di expressing the

electrical energy input caused by current increments 1di and 2di are balanced by

identical terms expressing the corresponding increments in field energy. These

terms, therefore, cancel, In other words, differential changes in currents have no

effect on the mechanical forces. Thus we get

2 2 2 2

1 1 2 2 1 2 12 1 1 2 2 2 2 12

1 12

2 2e ri dL i dL i i dM T d i dL i dL i i dM

-----(3.80)

or 2 21 2 121 2 1 21 12 2e r r rdL dL dM T i i i id d d ----- (3.81)

The translational equivalent of equation (3.81) is obtained if torque eT is

replaced by translational force ef and angular displacement rdby linear

displacement dx in the direction of force. The extension of equation (3.81) to a

situation involving several circuits whose self and mutual inductances depend upon

the angular position of some member should be obvious from consideration of the

line of reasoning followed in the derivation of equation (3.82). For example, with

three circuits.

2 2 2 3 23 311 2 121 2 3 1 2 2 3 3 1

1 1 1

2 2 2e

r r r r r r

dL dM dM dL dL dM T i i i i i i i i i

d d d d d d

(3.82)


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50/58


51/58

The total torque produced by the doubly excited magnetic system is given by

equation (3.81)

In fig.3.18, if rotor current 2 0I , then from equation (3.81)

2 11

12

e

r

dLT id

----- (3.88)

and if stator current 1I =0, then from equation (3.81)

2 22

1

2e

r

dLT i

d

------- (3.89)

Equation (3.88) reveals that even with rotor unexcited 2( . . 0)i e i

the torque isdeveloped. This is because reluctance S1 seen by stator flux changed with rotor

position. A change in S1causes the stator self- inductance L1 to change with r .

Similarly equation (3.89) reveals that with stator unexcited 1( . . 0)i e i the torque can

be produced because rotor inductance L2is a function of rotor position. The torque

expressions given by equation (3.88 and 3.89) are called reluctance torques. Thus

in a magnetic system shown in fig.3.19, the reluctance torque is present with anyone of the currents acting alone. The physical concepts about the development of

reluctance torques in salient pole machines is very important. In case when only

the stator is excited, then the stator flux would turn in counter-clockwise direction.

In case when only the rotor is excited, the rotor flux would turn in counter-

clockwise direction. In case when only the rotor is excited, the rotor flux would

tend to follow a minimum reluctance path and this is possible only when the rotor

turns in counter-clockwise direction.

Now let the salient pole rotor be replaced by a smooth cylindrical rotor and

the excited by current 2i as before. The torque eT according to equation (3.81), is

given by the expression


52/58


53/58

and i2, In fig.3.18, with the indicated direction of currents i1 and i2, the

electromagnetic or interaction, torque is counter-clockwise, as illustrated. If the

direction of flow of current through either stator or rotor is reversed, the interaction

torque would be opposite to that shown in fig, 3.18. i.e. in clockwise direction, but

the direction of reluctance torque would still be in the same direction as before the

reversal of direction of current flow i.e. counter clock wise.

Example 3.9.

Determine approximately the maximum torque that the motor can develop when it

is connected to 230V, 50Hz supply. It is given that the exciting winding has 1,500

turns; the inductance of the winding in direct-axis position is 0.8H and when the

rotor is in quadrature - axis position, the inductance is 0.4H.

Solution : Inductance of exciting winding

when the rotor is in direct - axis position, 0.8dL H

when the rotor is in quadrature - axis position, 0.4qL H

max

2300.00069

4.44 4.44 4.44 50 1, 500

E V

fN fN

weber

2 26(1,500)

5.625 10 /0.4

q

q

NS AT Wb

L

and2 2

6(1, 500) 2.8125 10 /0.8

d

q

NS AT Wb

L

The value of the maximum torque that the motor can develop,

2 2 2( ) max max1 1

(0.00069) (5.625 2.8125) 10 0.16778 8

e av q d T S S Nm


54/58

2 mark Question and answer

1.State the principle of electromechanical energy conversion.

As electrical energy is not readily available in nature, it has to be generated , to

meet the demands of electricity. The mechanical energy is converted to electrical

energy the vice versa of which also is possible is called electro-mechanical energy

conversion which takes place through either electric field (or) magnetic field.

2. What is an electromechanical system?

The system in which the electromechanical energy conversion takes palace via themedium of a magnetic or electric field is called electromechanical system.

3. How energy is stored?

Energy can be stored of retrieved from the magnetic system by means of an

exciting coil connected to an electric source.

4. Distinguish between statically induced and dynamically induced emf.

When emf is induced in a conductor placed stationary in a field , it is called as

statically induced emf.

If emf is induced in a conductor due to relative motion between the conductor and

the field , it is called as dynamically induced emf.

5.What does speed voltage mean ?

It is that voltage generated in a coil, when there is a exist relative moment

between the coil and magnetic field.


55/58

6.Define field energy.

The energy drawn by virtue of change in the distance moved by the rotor in

electrical machines in field configuration is known as field energy.

7.How the energy stored in magnetic field?

When the moving part of any physical system is held fixed, and then the entire

electrical energy input gets stored in the magnetic field.

8. Why energy stored in a magnetic material always occur in air gap?

In a magnetic material in case of iron core (or) steel core the saturation and ageing

effects from hinderance in storage.

But in air gap as reluctance as well permeability are constant, the energy storage

takes place linearly without any complexity.

Hence energy is stored in air gap always in a magnetic medium.

9. Define mmf.

Magnetic motive force (mmf) is given by

Mmf = flux reluctanceMmf = . R Amp.turns.

10. In a solenoid coil with an inductance of 5 mH current is increasing at the rate

of 100A/sec. What is the value of induced emf?

Emf = L = 5 = 0.5v


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11. What is the expression for energy stored in magnetic field ?

W =L

Where

L is the inductance

I is the current

12. What is the energy density in the magnetic field?

Energy density w = BH

=

13. Define co energy.

Co energy is an energy used for a linear system computation keeping current as

constant. It will not be applied to the non linear systems.

14. What is the significance of co-energy?

When electric energy is fed to a coil not the whole energy is stored as magnetic

energy.

The co-energy gives the measure of other energy conversions which takes place in

a coil than magnetic energy storage.

1 fieldenergy 2 Co-energy

15.Write the expression for the principle of energy conversion.

Mechanical energy output (work done by the field force) = Electrical energy input

increased in field energy


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16. Give an example for each of single and multiple excited system.

Single excited system : Eg : Reluctance motor, single phase transformer, Relay

coil.

Multiple excited system : Eg: Alternator, Electro-mechanical transducer.

17. Write the application of single and doubly fed magnetic system.

Singly excited system are employed for motion through a limited distance (or)

rotation through a prescribed angle.

Whereas multiply excited system are used where continuous energy conversion

takes place and in case of transducers where one coil when energized produces a

proportional signal either electrical (or) mechanical.

18. Describe multiply excited magnetic field system.

The specially designed transducers have the special requirement of producing an

electrical signal proportional to forces or velocities of producing force proportional

to electrical signal. Such transducers requires two or more excitation called as

multiply excited magnetic field system.

PartB

1. Derive the expression for field energy produced in a doubly excited magnetic

field system?

2. The magnetic flux density on the surface of an iron face is 1.6 T which is a

typical saturation level value for ferromagnetic material. Find the force density on

the iron face.

3. What are the special applications where the electric field is used as a coupling

medium for electromechanical energy conversion? Also explain why electric field

coupling is preferred in such applications?


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4. Find an expression for the force per unit area between the plates of a parallel

plate condenser in terms of the electric field intensity. Use both the energy and co

energy methods. Find the value of the force per unit area when E = 3 x 106 V/m,

the breakdown strength of air.

5. Explain with neat diagram and sufficient expressions, the multiply excited

magnetic field systems.

6.Explain i - characteristics of a magnetic system. Also derive the expression for

co energy density. Assume i - relationship of the magnetic circuit is linear.

7. Explain the concept of singly excited machines and derive the expression for

the electromagnetic torque.

8. Obtain the expression for energy stored in magnetic field.

9 .A faraday copper disc of 0.3m diameter is rotated at 60 revolution per second

on a horizontal axis perpendicular to and through the centre of the disc, the axis

lying in a horizontal field of 20wb/. Determine the emf induced betweenbrushes.

10..An iron ring of relative permittivity 100 is wound uniformly with two coils of

100 and 400 turns of wire. The cross section of the ring is 4 cm and the mean

length is 50 cm. Calculate the self inductance of each of the two coils and the

mutual inductance.

11.Two identical coupled coils in series has an equivalent inductance values of

0.08 H and 0.0354H. Find the values of self inductances, mutual inductance and

coupling coefficient.

12. A three phase 16 pole alternator has a star connected winding with 144 slots

and 10 conductors per slot. Flux per pole is 0.04 and runs at 375 rpm. Calculate the

Unit III Correct Syllabus Order

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