1.1 Types of Electrical machines

Basically the electrical machines are classified based on type of supply, they

are associated with. A machine associated with DC supply is called DC machine.

The DC machine can be a generator producing DC emf. or a DC motor using DC

supply to produce the mechanical energy.

A machine with AC supply is called an AC machine. The AC machines are

further classified as the synchronous machines and asynchronous machines. The

synchronous machines are further divided as synchronous generators which

generators are producing AC emf while the AC machines using AC supply and run

at a constant speed called synchronous speed are called synchronous motors. While

AC machines using an AC supply but running at sub synchronous speed which is

slightly less than the synchronous speed are called Induction machines. Depending

upon whether machines uses single phase or three phase AC supply, these are

further classified as single phase AC machines and three phase AC machines.

The various types of electrical machines are shown in fig 1.1

Fig.1.1 Types of Electrical Machines

1.11 Construction of rotating machines

It is known that there is relative motion between a conductor and the flux,

the EMF is induced in the conductor. This is the principle of generator. While

whenever a current carrying conductor is placed in a magnetic field then it

experiences a mechanical force, which is the principle of a motor. Hence every

rotating machine must possess following parts,

1. Stationary member called stator 4. Rotating member called rotor

2. Shaft 5. Slip ring, brush assembly

3. Bearings

Now these are mechanical parts which machine must possess. In addition to these a

machine has

1. An arrangement of winding which is used as a primary source of flux when

current is passed through it. This is called field winding or exciting winding.

2. An arrangement of conductors to form a winding in which EMF is induced.

This is called an armature winding.

The current flowing through a field winding, used to produce main flux is called

magnetizing current, exciting current or field winding.

The current flowing through the armature winding varies as the load on the

machine varies. So it is called a load current or an armature current.

The current in the field winding is always DC There is an appropriate air gap

between a stator and a rotor of the machine, In most of machines armature winding

is placed on the stator while the field winding is placed on the rotor from practical

convenience point of view.

Both stator and rotor are made up of high grade magnetic material such as

silicon steel. It provides low reluctance path to the flux. The construction used for

stator and rotor is laminated so as to keep eddy current losses as low as possible.

The general construction of an electrical machine is shown in the fig.1.2

Fig. 1.2 General construction of an electrical machine

In practice, it is required to feed in or take out power from the rotor which is

a rotating member of the machine. Thus a communication between a rotation

member and the stationary outside device is necessary. This is achieved using slip

ring and brush assembly. The slip rings are connected to the rotor winding and

rotate along with the rotor.

Thus slip rings behave as rotating winding terminals. The brushes are used

which are stationary and resting against slip rings. Thus brushes behave as

stationary terminals of rotating winding. The power can be fed in or taken out from

the bushes.

The slip rings and brush assembly is shown in fig.1.3

Fig.1.3 concept of slip ring and brush

The radial and axial ventilating air ducts are also provided in the machines for

cooling purpose. For large machines instead of air, hydrogen is used as a

ventilating medium. A closed circuit ventilating system is preferred for large

capacity machines. The hydrogen is circulated over the machine parts and is cooles

with the help of water cooled heat exchangers.

Though this is a common construction of the rotating electrical machines,

depending on whether machine is AC or DC the construction changes.

1.12 DC machines

In a DC machine the field winding is on stator while the armature winding

is on rotor. Consider an elementary DC machine with 2 pole and single armature

coil. The construction is shown in fig.1.4

The poles are divided in to two parts, pole core and pole shoe. The pole has been

given a specific shape shoe enlarges the area of armature core to come across the

flux, to have large induced EMF. The pole core carries the field winding.

Fig.1.4 .An elementary of 2 pole machine

The armature is also divided into two parts, armature core and armature

winding. The core is cylindrical in shape mounted on shaft. It consists of slots on

its periphery and armature winding is placed in these slots. The fig shows a single

armature coil from convenience point of view. When armature is rotated the

armature winding cuts the flux produced by field winding and an alternating EMF.

is induced in it.

But a DC output is required from DC machine. To convert alternating (AC)

emf. to DC, commutator is used. For simplicity, in the Fig. 1.4 the commutator is

shown in the form of two segments. The armature conductors are connected to

commutator. The commutator is rotating along with the armature. The two

commutator segments are insulated from each other. When armature conductor

changes its position, the commutator also changes its position. But the brushes

resting on commutator are stationary. So one brush always makes contact with that

commutator segment which is under which is under N pole i.e. +ve while other

always makes contact with that commutator segments which is under S pole i.e. –

ve. So though armature and commutator reverse their positions, the brushes which

are stationary behaves as +ve and –ve brushes permanently. Thus rectification is

possible and DC output voltage is available across the brush terminals. In a DC

motor, the commutator converts the alternating torque into unidirectional torque.

To have commutating action possible, the armature is rotor and field is stator in a

DC machine. The main frame is made up of magnetic material to provide low

reluctance path to the flux produced by the field winding.

In a motoring action along with the field, the DC supply is given to the

armature winding through brushes. When a current carrying conductor is placed in

a magnetic field, it experiences a force. When all such conductors experience a

force, a torque is exerted on the armature and it starts rotating. The unidirectional

torque is possible due to commutator. So whether a DC machine is a generator or

motor, its construction remains same. The air gap under the poles in a DC machine

is uniform.

1.13 Synchronous Machines

The synchronous machines are the AC machines. The AC producing

synchronous machines are called alternating generators or alternators. As these

machines produce AC voltage, commutator is absent. Another important difference

between AC and DC machines is that in AC machines the field winding is on the

rotor while the armature is on the stator. As the field winding is rotating, the DC

voltage is supplying to it with the help of slip ring and brush assembly.

Construction of Synchronous Machine

The overall rotating field construction is simple. The stationary armature

has a core and the slots to hold the armature winding as shown in the fig 1.5. It

uses a laminatedf construction. This keeps eddy current losses to a minimum value.

The grade magnetic material is used to keep down the hysteresis losses.

Fig 1.5. Section of stator stamping

The field winding is on the rotor and there are two rotor constructions used

in AC machines.

1. Salient pole: In this construction, poles

are projected out from the surface of the rotor as

shown in fig 1.6. The poles are built with thick

steel laminations. The poles are bolted to the

rotor. The pole face has been given specific

shape. The air gap becomes non uniform due to

such poles. These are used for low speeds due

Fig.1.6.Salient pole rotor to mechanical strength.

3. Smooth Cylindrical rotor:

This is also called non-salient pole construction or round rotor type of

construction. The rotor consists of smooth solid steel cylinder, having number

of slots to accommodate the field. The slots are covered at the top with the help

of steel wedges. The unslotted portions of the cylinder itself act as the poles as

shown in fig.1.7

Fig.1.7. Smooth cylindrical rotor

The poles are non-projecting and provide uniform air gap. As mechanical

strength of such rotors is high, these are used for high speed alternators.

The figure 1.8 shows cross-sectional view of synchronous machine in which the

rotor carries the field poles. The rotor is salient pole type rotor.

Fig.1.8. Cross sectional view of synchronous machine

The rotor carries field winding which is excited by a DC supply and designed for

fixed P number of poles. The stator carries 3 phase armature winding which is

wound for same number of poles as that of the field winding. When the rotor

rotates, an alternating EMF of frequency ‘f’ Hz is induced in the stator winding.

For the synchronous machines, there exists a fixed relationship between the

number of poles P, the frequency f Hz of induced EMF and the speed which is

called synchronous speed Ns rpm, This relationship is given by,

Ns = 120 fP

In case of synchronous motor, when 3 phase supply of ‘f’ Hz is supplied to the

armature, a magnetic field is produced which is rotating in nature having the

synchronous speed Ns? It produces the effect of magnet rotating at a speed Ns.

When rotor produces the poles, there is magnetic locking between stator rotating

field and rotor field and electromagnetic torque is produced. Thus the rotor starts

rotating along with the rotating magnetic field of stator at a speed Ns. This is the

working principle of synchronous motor is not self starting due to rotor inertia.

In both DC and synchronous machines, the power handling capacity is

decided by the voltage and current of the armature winding when the field is

excited by a DC supply. Hence such machines are called doubly excited machines.

1.14 Induction Machines

The three phase induction motors are widely used for various

industrial applications. The three phase induction motors have advantages

compared to synchronous motors such as self starting property, high power factor,

good speed regulation and control possible, robust in construction, control over the

starting torque etc.

Construction

The induction motor consists of two main parts, namely

1. The part i.e. three phase windings, which is stationary called stator.

2. The part which rotates and is connected to the mechanical load through shaft

called rotor.

The conversion of electrical power to mechanical power takes place in a rotor.

Hence rotor develops a driving torque and rotates.

Stator

The ststor has a laminated type of construction made up of stampings which are

0.4 to 0.5 mm thick. The stampings are slotted on its periphery to carry the stator

winding. The stampings are insulated from each other. Such a construction

essentially keeps the iron losses to a minimum value. The number of stampings are

stamped together to build the stator core. The built up core is then fitted in a casted

or fabricated stel frame. The choice of material for the stamping is generally

silicon steel, which minimizes the hysteresis loss. The slots on the periphery of

the ststor core carries a three phase winding, connected either in star or delta. This

three phase winding is called stator winding. It is wound for definite number of

poles.

Fig 1.9. Stator lamination

The winding when excited by a supply produces a rotating magnetic field. The

choice of number of poles depends on the speed of the rotating magnetic field

required. The radial ducts are provided for the cooling purpose. In some cases, all

the six terminals of three phase stator winding are brought out which gives

flexibility to the user to connect them either in star or delta. The figure 1.9 shows

the stator lamination.

Rotor

The rotor is placed inside the stator. The rotor core is also laminated in

construction and uses cast iron. It is cylindrical, withslots on its periphery. The

rotor conductors or windings is placed in the rotor slots. The two types of rotor

constructions which are used for induction motors are

1. Squirrel cage rotor

2. Slip ring or wound rotor

Squirrel cage rotor

The rotor is cylindrical and slotted on its periphery. The rotor consists of

uninsulted copper or aluminium bars called roto conductors. The bars are placed in

the slots. These bars are permanently shorted at each end with the help of

conducting copper ring called end ring. The bars usually brazed to the end rings to

provide good mechanical strength. The entire structure looks like a cage, forming a

closed electrical circuit. So the rotor is called squirrel cage rotor. The construction

is shown in fig 1.10

Fig. a. Cage type structure of rotor fig b. Symbolic representation

Fig 1.10 Squirrel cage rotor

As the bars are permanently shorted to each other through end ring,

the entire rotor resistance is very small. Hence this rotor is also called short

circuited rotor. As rotor resistance itself is short circuited, no external resistance

can have any effect on the rotor resistance. Hence no external resistance can be

introduced in the rotor circuit. So slip ring and brush assembly is not required for

this rotor. Hence the construction of this rotor is very simple.

Fan blades are generally provided at the ends of the rotor core. This circulates the

air through the machine while operation, providing the necessary cooling. The air

gap between stator and rotor is kept uniform and as small as possible.

1.2 Introduction of magnetic circuits

All of us are familiar with a magnet. It is piece of solid body which

posses a property of attracting iron pieces of some other metals. This is called

natural magnet. Any current carrying conductor is always surrounded by a

magnetic field. The property of such current is called magnetic effect of an electric

current. Natural magnet or an electromagnet, both have close relation with

electromotive force experienced by conductor, electric current etc. To understand

this relationship it is necessary to study the fundamental concept of magnetic

circuits.

1.21 Laws of Magnetism

There are two fundamental laws of magnetism

Law 1: It states that Like magnetic poles repel each other and unlike poles attract each other

Law 2: This law is experimentally proved by Scientist Coulomb and hence also known as Coulomb ‘s law.

The force F exerted by one pole on the other pole is

i. Directly proportional to the product of the pole strengthsii. Inversely proportional to the square of the distance between them.iii. Nature of medium surrounding the poles

Mathematically this law can be expressed

F∝ KM 1M 2

d2

Where M 1 and M 2 are pole strength of the poles while d is distance between the poles.

F= KM 1M 2

d2

Where k depends on the nature of the surroundings and called permeability.

1.22 Magnetic flux (∅)

The total number of lines of force existing in a particular magnetic field is called magnetic flux. Lines of force can be called lines of magnetic flux. The unit of flux is webers and flux is denoted by symbol (∅ ¿. The unit weber is denoted as Wb.

1 weber = 108 lines of force

Magnetic flux density (B)

It can be defined as the flux per unit area (A) in a plane at right angles to the flux

is known as flux density. Mathematically,

B = ∅A

wb

m2 or Tesla ------ (1.1)

It is shown in the Fig 1.11

Fig 1.11 Concept of magnetic flux density

1.23 Magnetic Field strength (H)

This gives quantitative measure of strongness or weakness of magnetic field.

The pole strength and magnetic field strength are different. This can be defined as

the force experienced by a unit N-pole ( i.e. N pole with 1 wb of pole strength)

when placed at any point in a magnetic field is known as magnetic field strength at

that point.

It is denoted by H and its unit is newtons per weber i.e. ( N/Wb) or amperes per

metre (A/m) or ampere turns per metre ( AT/m). The mathematical expression

For calculating magnetic field strength is

H = Ampere turnsLength

H = ¿l AT/m -------- (1.2)

1.24 Fleming’s Left Hand Rule

Fig.1.12

The direction of the force experienced by the current carrying conductor

placed in magnetic field can be determined by a rule called ‘Flemings Left Hand

Rule’. The rule states that, ‘Outstretch the three fingers of the left hand namely the

first finger, middle finger and thumb such that they are mutually perpendicular to

each other. Now point the first finger in the direction of magnetic field and the

middle finger in the direction of current then the thumb gives the direction of the

force experienced by the conductor.

The rule is explained in diagrammatic form in the fig.1.12

Apply the rule to crosscheck the direction of force experienced by a single

conductor, placed in a magnetic field shown in Fig 1.13 a,(b), (c). and d

Fig. 1.13 Direction of force experienced by conductor

1.25 Permeability

The flow of flux produced by the magnet not only depends on the magnetic field

strength but also on one important property of the magnetic material called

permeability. It is related to the medium in which magnet is placed. The force

exerted by one magnetic pole on other depends on the medium in which magnets

are placed.

For any magnetic material, there are two permeabilities,

Absolute permeability ii) Relative permeability

Absolute Permeability (µ)

The magnetic field strength (H) decides the flux density (B) to be produced

by the magnet around it, in a given medium. The ratio of magnetic flux density B

in a particular medium (other than vacuum or air) to the magnetic field strength H

producing that flux density is called absolute permeability of that medium.

It is denoted by µ and mathematically can be expressed as

µ = BH

B = µH The permeability is measured in units henries per metre denoted as H/m.

Relative permeability

Generally the permeability of different magnetic materials is defined

relative to the permeability of free space (µ0). The relative permeability is defined

as the ratio of flux density produced in a medium to the flux density produced in

free space, under the influence of same magnetic field strength and under identical

conditions.

If the magnetic field strength is H which is producing flux density B in the

medium while flux density B0 in free space then the relative permeability is defined

as,

µr = BB0

It is dimension less and has no units.

For free space, vacuum or air, µr =1

According to definition of absolute permeability we can write for given H,

µ = BH in medium ------ (1.3)

µ0 = B0

H in free space ------- (1.4)

Dividing equation 1 and 2

µµ0

= BB0

But BB0

= µr

µµ0

= µr

µ = µ0µr H/m

For example if relative permeability of the iron is 1000 means it is 1000 times

more magnetic than the free space or air

1.26 Reluctance (S)

Reluctance is the property of a magnetic circuit by which the setting up of

flux is opposed. It can also be defined as the ratio of magneto motive force to the

flux.

S = ¿Φ ------------ (1.5)

The unit of reluctance is Ampere turns /wb.

1.27 Leakage coefficient or Hopkinson’s coefficient

The ratio of the total flux (∅ T) to the useful flux (∅ u) is defined as the

leakage coefficient of Hopkinson’s coefficient or leakage factor of that magnetic

circuit.

It is denoted by λ

λ = Total fluxUseful flux =

∅T

∅ u

The value of λ is always greater than 1 as ∅ T is always more than∅ u. It

generally varies between 1.1 and 1.25. Ideally its value should be 1.

1.28 Magnetic Fringing

When flux enters into the air gap, it passes through the air gap in terms of

parallel flux lines. There exists a force of repulsion between the magnetic lines of

force which are parallel and having same direction. Due to this repulsive force

there is tendency of the magnetic flux to bulge out at the edge of the air gap. This

tendency of flux to bulge out at the edge of the air gap is called magnetic fringing.

Fig.1.14 Magnetic fringing

It has following effects:

i. It increases the effective cross sectional area of the air gap

ii. It reduces the flux density in the air gap

So leakage, fringing and reluctance, in practice should be as small as possible.

Practically if fringing effect is to be considered then the correction for short gaps

are empirically made by adding one gap length to each of the two dimensions

making up its area.

Thus if area of core is ac = lc×lc then the ag = (lc +lg ¿ × ((lc +lg) is the area of cross-

section of air gap for considering the effect of fringing. Thus ag>¿ ac

And the air gap reluctance is less due to fringing.

In case of circular cross-section while calculating the area of cross –section for air

gap the diameter is considered as ( d+2lg) where lg is the air gap length and d is

the diameter of cross section of core. This takes into account the effect of fringing.

1.29 Stacking Factor

Generally magnetic cores are made up of laminations which are lightly

insulated. The laminated construction helps to keep eddy current losses to low

value. Due to stacks of laminations, the net cross sectional area occupied by the

magnetic material is less than its gross cross-sectional area. Thus the stacking

factor is defined as,

Stacking Factor = Net Cross−sectional area occupied by magneticmaterial

gross cross−sectional area

As gross cross-sectional area is higher, the stacking factor is always less than

unity.

Example 1.1

A cast iron ring of 40 cm mean length and circular cross section of 5 cm diameter

is wound with a coil. The coil carries a current of 3 A and produces a flux of 3

mWb in the air gap. The length of the air gap is 2mm. The relative permeability of

the cast iron is 800. The leakage coefficient is 1.2. Calculate number of turns of the

coil.

Fig 1.15

lt = 40 cm = 0.4 m, lg = 2×10−3 m

li = lt - lgap = 0.4 - 2×10−3 = 0.398m

I = 3A, ∅ g = 2×10−3Wb , µr = 800 , λ = 1.2

λ =∅T

∅ g--- leakage flux

1.2 = ∅T

2×10−3 ,

∅ T=¿2.4× 10−3 Wb

Now reluctance of iron path Si = li

aµ0µr

A = π4d2

= π4

52 = 19.634 ×10−4 m2

Si = 0.398

4 π ×10−7×800×19.634×10−4 = 201629.16 AT/Wb

∅ T = m .m. freluctance =

¿S i

Therefore 2.4× 10−3 = ¿

201629.16

Ni for iron path = 483.909 AT

Reluctance of air gap Sg = lgµ0a

= 2×10−3

4 π ×10−7×19.634×10−4

= 810608.86 AT/Wb

∅ s = m.m . fSg

= ¿

810608.86

Ni for air gap = 1621.2177 AT

Therefore total m.m.f required = (NI) iron + (NI)airgap

= 483.909 +1621.2177

NI = 2105.1267 i.e N× 3 = 2105.1267

N = 701.7 = 702 turns

1.2. 10 Magnetic ciruits

The magnetic circuit can be defined as the closed path traced by the

magnetic lines of force i.e.flux. Such a magnetic circuit is associated with

different magnetic quantities as m.m.f , flux, reluctance , permeability etc.

Consider simple magnetic circuit shown in fig. 2. The circuit consists of an

iron core with cross sectional area of ‘a’ m2 with a mean length of ‘l’ m. A coil of

N turns is wound on one of the sides of the square core which is excited by a

supply . This supply drives a current I through the coil. This current carrying coil

produces the flux (∅ ¿which completes its path through the core as shown in

fig.1.16 (a).

This is analoguos to simple electric circuit in which a supply i.e emf of E volts

drives a current I which completes its path through a closed conductor having

resistance R. This analogous electrical circuit is shown in fig 1.16 (b).

(a) .Magnetic circuit (b). Electrical Equivalent

Fig 1.16

The relationship between mmf, flux , reluctance

I = current flowing through the coil

N = Number of turns

∅ = Flux in webers

B = Flux density in core

µ = Absolute permeability of the magnetic material

µr= relative permeability of the magnetic material

Magnetic field strength inside the solenoid is given by,

H = ¿l AT/m ------(1.6)

Noe flux density B == µH

B = µ0

µr∋¿

l¿ wb/m2 ------(1.7)

Now as area of cross-section is ‘a’ m2total flux in core is,

∅ = Ba = µ0

µr∋¿a

l¿ wb - ------(1.8)

i.e ∅ =¿l

µ0µra

∅ = m .m. freluctance =

FS

Where NI = Magnetomotive force m.m.f in AT

S = l

µ0µra

= Reluctance offered by the magnetic path

This expression of the flux is very much similar to expression for current in electric circuit.

I = e .m . f

resistance

Example 1.2

An iron ring of circular cross sectional area of 30 cm2and mean diameter of 20 cm is wound with 500 turns of wire and carries a current of 2.09 A to produce the magnetic flux of 0.5 m Wb in the ring. Determine the permeability of the material.

A = 30 cm2 = 3×10−4 m2 , d = 20 cm, I = 2 A, ∅ = 0.5 Wb

L=πd = π ×20 = 62.8318 cm = .628318 m

S = l

µ0µra =

0.628313

4 π ×10−7×µr×10−4 = 1.6667×109

µr -----(1.9)

F = mmfS =

¿S

S = ¿∅

500×2

0.5×10−3 = 2×106 AT/Wb ------(1.10)

Equating 1.9 and 1.10

2 ×106=¿ 1.6667×109

µr

µr = 833,334

1.2.11 Series Magnetic circuit

In practice magnetic circuit may be composed of various materials of

different permeabilities, of different length and of different cross sectional areas.

Such a circuit is called composite magnetic circuit. When such parts are connected

one after the other , the circuit is called series magnetic circuit.

Consider a circular ring made up of different materials of lengths l1,l3 and l3

and with cross sectional are a1,a2 and a3 with absolute permeabilities µ1,µ2 and µ3 as

shown in fig 1.17.

Fig1.17 Series magnetic circuit

Let coil wound on ring has N turns carrying a current of I amperes. The total

m.m.f available is NI AT. This will set the flux ∅ which is same through all the

three elements of the circuit.

This is similar to three resistance connected in series electrical circuit and

connected to e.m.f carrying same current T through all of them.ts analogous can be

shown in fig1.18.

Fig 1.18 Equivalent electric circuit

The total resistance of the electric circuit is R1 +R2+R3. Similarly the total

reluctance of the magnetic circuit is

Total sT = S1 +S2+S3 = l1µ1a1

+l2µ2a2

+l3µ3a3

Total flux ∅ = Total m.m . fTotalreluctance =

¿sT =

¿S1+S2+S3

NI = sT∅ = (S1 +S2+S3) ∅

NI = S1∅ +S2∅+S3∅

(m.m.f)T = (m.m.f)1 +(m.m.f)2 +(m.m.f)3

The total m.m.f can be expressed as

(m.m.f)T = H 1l1+ H 2 l2+H 3 l3

Where H 1 = B1

µ1 , H 2 =

B2

µ2 , , H 3 =

B3

µ3

So far a series magnetic circuit we can remember

1. The magnetic flux through all the parts is same.

2. The equivalent reluctance is sum of reluctance of different parts.

3. The resultant m.m.f necessary is sum of the m.m.fs in each individual part

1.2.12 Parallel Circuit

In case of electric circuits, resistance can be connected in parallel. A current

through each of such resistance is different while voltage across all of them is

same. Similarly different reluctances may be in parallel in case of magnetic

circuits. A magnetic circuit which has more than one path for the flux is known as

a parallel magnetic circuit.

Consider a magnetic circuit shown in fig 1.19.a. At point A the total flux ∅ ,

divided into two parts ∅ 1 and ∅ 2.

∅ = ∅ 1+∅2

The fluxes ∅ 1 and ∅ 2 have their paths completed through ABCD and AFED

respectively.

This is similar to division of current in case of parallel connection of two

resistances in an electric circuit. The analogous electric circuit is shown in

Fig.1.19. b.

(a). Magnetic circuit (b.) equivalent Electric circuit

Fig1.19 Parallel Magnetic circuit

The mean length of path ABCD = l1

The mean length of the path AFED = l2

The mean length of path AD = l3

The reluctance of the path ABCD = S1

The reluctance of the path AFED = S2

The reluctance of the path AD = S3

The total m.m.f produced = NI AT

Flux = m .m. freluctance

m.m.f = ∅×S

For path ABCDA, NI = ∅ 1S1 + ∅ Sc

For path AFEDA, NI = ∅ 2S2 + ∅ Sc

Where s1 = L1

µa1, s2 =

L2

µa2 and sc =

Lcµac

Generally a1 =a2 = a3 = Area of cross section

Total m.m.f = m.m.f required

By central limb + m.m.f required by any one of outer limbs

NI = (NI) AD + (NI)ABCD +(NI) AFED

NI = ∅ sc + [∅ 1S1 or ∅ 2S2 ]

As in the electric circuit e.m.f across parallel branches is same, in the

magnetic circuit the m.m.f across parallel branches is same.

Thus same m.m.f produces different fluxes in the two parallel branches. For

such parallel branches,

∅ 1S1 = ∅ 2S2

Hence while calculating total m.m.f, the m.m.f of only one of the two

parallel branches must be considered.

Example 1.3

An iron ring 8 cm. mean diameter is made up of diameter 1 cm and permeability

of 900, has an air gap of 2 mm wide. It consists of winding with 400 turns carrying

a current of 3.5 A. determine

1. Mmf 2. Total reluctance 3. The flux 4. Flux density in ring

The ring and the winding is shown in fig 5

Diameter of ring d = 8 cm

Length of iron = πd – Length of air gap

l1 = π ×(8×10−2) - 2×10−3

Fig.1.20 = 0.2493 m

lg = Length of the air gap = 2 mm = 2×10−3 m

Diameter of the iron = 1 cm

∴ Area of cross section a= π4d2= π

4¿

a= 7.853×10−5m2

Area of cross section of air gap and ring is to be assumed same

i) Total m.m.f produced = N1 = 400×3.5

= 1400 AT (ampere turns)

ii) Total reluctance Sr=S i+Sg

Si=li

μo μra…given μr=900

=0.2493

4 π ×10−7×900×7.853×10−5

= 2806947.615AT/Wb

Sg=lgμoa

…given μr=1 for air

Sg =2×10−3

4 π ×10−7×7.853×10−5 = 20.2667×106 AT/Wb

Sr=¿2806947.615+20.2667×106= 23.0737×106 AT/Wb

iii. ∅ = m .m. freluctance =

¿ST =

1400

23.0737×106 = 6.067×10−6 Wb

iv. Flux density

= ∅a =

6.067×10−5

7.853×10−5 = 0.7725 Wb/m2

Example 1.4

A cost steel structure is made of a rod of square section 2.5 cm ×2.5 cm `as shown

in fig. What is the current that should be passed in a 500 turn coil on the left limb

so that a flux of 2.5 mWb is made to pass in the right limb. Assume permeability

as 750 and neglect leakage.

Fig.1.21

This is parallel magnetic circuit . its electrical equivalent is shown in the fig 1.22

The total flux produced gets disturbed in to two parts having reluctances S1 and S2

S1 = Reluctance of centre limb

S2 = reluctance of right side

Fig.1.22

S1 = l1

µ0µra1 =

25×10−2

4 π ×10−7×750×2.5×2.5×10−4 = 424.413×103 AT/Wb

S2 = l2

µ0µra2 =

40×10−2

4 π ×10−7×750×2.5×2.5×10−4 = 679.061 AT/Wb

Foe branch AB and CD , m.m.f is same.

Therefore m.m.f = ∅ 1S1 = ∅ 2S2

And ∅ 2 = 2.5 mWb

∅ 1 = ∅2S2

S1 =

2.5×10−3×679.061×103

424.413×103 = 4 mWb

∅ = ∅ 1+¿ ∅ 2 = 2.5 + 4 = 6.5 mWb

Total m.m.f required is sum of the m.m.f required for AEFB and that for either

central or side limb.

SAFEB = S2 = 679.061 ×103 AT/Wb

Therefore m.m.f for AEFB = SAFEB×∅ = 679.061 ×103× 6.5×10−3

= 4413.896 AT

Total m.m.f = 4413.896 + ∅ 1S1

= 4413.896 + 4×10−3×424.413×103

But NI = total m.m.f

I = 6111.548

500 = 12.223 A

Difference between electric circuit and magnetic circuit;-

Electric circuit Magnetic circuit

1. Path traced by the current is called electric circuit

2.MMF is driving force in electric circuit, the unit is volts

3. There is current I in circuit, measured in amperes.

4. EMF ( volts)

Current = emf

reluctance

5.Resistance R =ρlA

1. Path traced by the magnetic flux is defined as magnetic circuit

2. MMF is driving force in magnetic circuit, the unit is ampere turns

3. There is flux ∅ in circuit, measured in Webers

4.MMF (Amp-turns)

Magnetic flux = mmf

reluctance

5..Reluctance = R = lμA

6.Conductance = 1R

7.Resistance opposing the flow of

the current , unit is ohm

6 Permeance P = 1/R

7. Reluctance is opposed by

magnetic path to the flux, unit

is Ampere turns/ webers

1.3 Inductance

1.3.1 Self inductance

By Faraday’s law, changing current will produce an induced emf in the

circuit to oppose the change in flux. This phenomenon is known as self induction.

Self inductance of a circuit is the property of the circuit by which changing current

induces emf in the circuit to oppose the changing current.

Consider a coil having ‘N’ number turns. If changing current (alternating

current) is applied, the emf is induced in the coil.

The induced emf is proportional to the rate of change of current

E∝didt

E = Ldidt

Where L is the inductance (i.e constant of proportionality).

From Faraday’s law, the induced emf is

V = N dΦdt

Equating these two equations.

Ldidt = N

dΦdt

L = N dΦdi

If the permeability is constant, then

L = NΦi ------- (1.11)

The inductance is defined as the ratio of total magnetic flux linkage to the current

through the coil.

1.3.2 Mutual Inductance

Consider two coils 1 and 2 magnetically coupled together as shown in Fig.

1.23. The changing current i1 produces a fluxΦ1. If a second coil is placed near the

first coil, some of the flux links coil 2 says Φ12 .

Fig 1.23.Changing current in coil 1 produces changing magnetic flux in coil 2.

Fig.1.24.Changing current in coil 2 produces changing magnetic flux in coil 1.

The induced emf in coil is given by

v2 = N2 dΦ12

dt

Since flux Φ12 is produced by first coil currenti1, the induced emf v2 in coil is

proportional to the rate of change of currenti1.

v2 ∝d i1dt

v2 = M d i1dt

Where, M is the mutual inductance between the two coils.

Equating these two equations

M d i1dt

= N2 dΦ12

dt

M = N 2d∅ 12

dt

If the permeability is constant,

M = N 2∅ 12

i1

= N2 Φ12

i 1

Similarly, if the flux Φ12 is produced by second coil currenti2, the induced emf v1 in coil 1 is proportional to the rate of change of current i2

v1 ∝ di2dt

v1 = M di2dt

From Faraday’s law

v1 = N1 dΦ21

dt

Equating these two equations

M = N1 dΦ21

di2

If the permeability is constant, then

M = Φ21

i2--------- (1.12)

The mutual inductance between two coils is defined as the ratio of induced magnetic flux linkage in coil to the current through in other coil.

1.3.3 Co efficient of coupling

The fraction of the total flux produced by one coil linking a second coil is called the coefficient of coupling (K)

K = Φ12

Φ1 =

Φ21

Φ2

The mutual inductance is given by

M = N 2Φ12

i1 and M =

N 1Φ21

i2

M 2 = (N 2Φ12

i1 ) (

N 1Φ21

i2 )

Substitute the value of Φ12 and Φ21 interms of K

M 2 = (N 2KΦ1

i1 ) (

N 1KΦ2

i2 )

= K2(N 2Φ1

i1 ) (

N 1Φ2

i2 )

M 2 = K2 L1L2 L1=[N2Φ1

i1, L1=

N 1Φ2

i2]

M = K √L1L2 ----------- (1.13)

K = M

√L1 L2 --------- (1.14)

Since Φ12<Φ1 or Φ21<Φ2, the M ¿ √L1L2

i.e K is always less than one. K¿1

1.4 EMF INDUCED IN A COIL ROTATING IN A MAGNETIC FIELD

Consider a single turn rectangular coil rotating in counter-clockwise

direction with angular velocity radians per second in a

uniform magnetic field of flux density B teslas , as

illustrated in fig. 1.25 Fig.1.25

When the coil is rotated in a uniform magnetic field with uniform angular

velocity, an emf is induced in its coil sides. The magnitude of emf induced in the

coil depends on (i) the flux density of the main magnetic field, B teslas (ii) the

effective length of the coil side in the magnetic field, l metres (iii) number of truns

on the coil, N (iv) velocity of the coil side perpendicular to the filed metres/

second and is given by the expression

volts ------- (1.15) ( two coil sides are connected in series)

The coil sides do not always cut the lines of flux in perpendicular direction

while rotating in a magnetic field. The component of velocity of the coil

perpendicular to the field is

Fig.1.26

where is the peripheral velocity of the coil and is the angle between the

conductor and line perpendicular to the magnetic axis (fig.1.26).

So the emf equation for the generated voltage

becomes

sin volts ------ (1.16)

The magnitude of induced emf can be determined if the values of N,B,l,v

and are known.

The value of the induced emf varies in magnitude and direction according to

the instantaneous position of the coil relative to the magnetic field. The cross-

sectional view of the coil and its different position at different instant are shown in

fig. 1.26..

Fig.1.27

When the coil is in vertical position as shown in fig. 1.27. (a), (c) and (e), no

emf is induced in the coil sides as the coil sides are moving parallel to the magnetic

field and the rate of currint flux is zero. When the coil is horizontal and moving at

right angles to the magnetic field, as shown in fig. 1.27 (b) and (d), the magntidue

of emf induced in the conductors A and B is maximum because the rate of cutting

flux is maximum. The direction of emf induced in conductor A at instant shown in

fig,. 1.27 (b) is outward and at instant shown in fig. 1.27.(d) is inwards as

determined by Fleming's right hand rule. The direction of emf induced in

conductor B is opposite to that of conductor A, as obvious from figs. The wave

shape of the emf induced in the coil is also shown in fig 1.27

From the above discussion as well as from equation (1.16) it is obvious that the

waveform of the induced emf is sinusoidal

1.4.1 Equation of induced emf

Let us consider a coil with

N = N0 of turns

∅ 1 = Initial flux linked with the coil in wb

∅ 2 = Final flux linked with the coil in wb

T = time taken in seconds for the change of flux.

Initial flux linkage = N∅ 1

Final flux linkage = N∅ 2

Magnitude of induced emf E = Rate of change of flux linkage.

= N ∅ 1−N ∅ 2

t

= N ¿¿¿ volts ------- (1.17)

The instantaneous value of induced emf can be obtained as

e = ddt (N∅ ) = N

d∅dt volts ------- (1.18)

Applying lenz’s law the induced emf at any instant is

e = −Nd∅dt ------- (1.19)

Classification:

As emf is induced in a coil or conductor whenever there is a change in flux.

This change in flux linkage can be brought in two ways.

i) By moving the conductor in a stationary magnetic field. This class of

induced emf is known as dynamically induced emf.

ii) By moving or changing the magnetic field and keeping the coil or

conductor stationary. This type of emf is known as static.

1.4.2 Dynamically induced emf

Consider a stationary magnetic field of flux density B wb/m2. The direction

of this field is as shown in fig.1.28. In this field a conductor is placed.

Fig.1.28. Dynamically induced emf

Let l be the effective length of conductor in the field in meter. Let the

conductor be moved in the direction shown as moton 1. i.e at right angle to the

field. In a small time interval of dt seconds it moves a distance of dx meters.

Area swept by the motion of the conductor = l dx m2

Magnetic flux linked by the conductor = Flux density × Area

= B l dx wb ------- (1.20)

Let us consider that the conductor has a single turn.

The flux linkage in the conductor is 1

𝝋 = B l dx ×1 -------- (1.21)

Rate of change of flux linkage = B l dxdt ------- (1.22)

= B l v

This rate of change of flux linkage is nothing but the induced emf

e = B l v volts -------- (1.23)

Where v= dxdt = linear velocity of the conductoe

let us consider that the conductor is moved with the same velocity

v m/sec in an inclined direction making an angle ' θ’ to the direction of

the field. Then the induced emf in the conductor is reduced by a value as

e = B l v sinθ volts ------- (1.24)

1.4.3 Statically induced emf

This is the emf induced in stationary conductor with changing flux.

This can be classified into two different categories as follows.

i) Self induced emf :

If a single coil carries a current, a flux will be set up in it. If the current

changes the flux will also change and hence emf will be induced in the coil. This

kind of emf is called ‘self induced emf”.

Therefore the self induced emf is the emf induced in a circuit when the flux

linking in it changes, the flux being produced by current in the same circuit.

The magnitude if this induced emf is

e = Nd ∅dt ------- (1.25)

As per Lenz’s law this will oppose the cause producing it which is nothing but the

change in flux

e = - Nd ∅dt --------- (1.26)

We know that the self inductance is

L = N ∅I -------- (1.27)

or LI = N ∅ \

If the current changes by an amount di the corresponding change in flux will be d∅

Therefore Ldi = Nd∅

Ldidt = N

d∅dt -------- (1.28)

- Ldidt

= -N d∅dt

------- (1.29)

Comparing Eqn

e = - Ldidt

where

L = N2

S

ii) Mutually induced emf :

It is the emf induced in one circuit due to change of flux, being produced by current in another circuit.

The mutually induced emf in a second coil due to the change in flux in a first coil is

em,2 = λ21

t =

N 2∅ 2

t ------- (1.30)

At any instant em,2 = -N2

d∅ 1

dt ------- (1.31)

And the mutual inductance between two coils is given by

` M = N 2∅ 2

I 1 ------- (1.32)

MI 1 = N2∅2

At any instant

M di1dt

= N2 d∅ 1

dt -------- (1.33)

- M didt = - N2

d∅ 2

dt-------- (1.34)

Comparing equations (1.31) and (1.34)

em,2 = - M di1dt

-------- (1.35)

Example 1.5

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 θ = π/2

e = Blv sinθ

= 1×1×100=100volts

Example 1.2.

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

in the magnetic field.

L = 10×10−3

I = 10A

Stored energy W = 12 LI 2

= 12×10×10−3×(10)2

= 0.5 joules

Example 1.6

The total flax of an electromagnet is 8 x 10–4 Wb. (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: ϕ = 8 x 10–4 Wb; A = 5 x 10–4 m2; N = 100; ℓ = 50 x 50–4 m2; I = 5

Amps

Solution:

(a) Flax Density B = ∅A

= 8 x10−4Wb5 x10−4m2

B = 1.6 Wb/m2.

(b)mmf = NI

= 100 x 5

mmf = 500 AT

(c) Magnetic field Intensity

H = mmfl

= 50

50x 10−2

H = 1000 AT/m

Example 1.7.

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

mmf.

Solution:

mmf = 318 AT

H = 318

20x 10−2 = 1590 AT/m

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

Flux = 0.5 x 5 x 10–4

ϕ = 2.5 x 10–4 Wb

I =1.59 Amps.

Example 1.8.

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 x 10 –4 Wb.

When 1 is 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 ϕ 1 = 3 x 10–4 Wb

N1 = 250 V2 = 63.75 V

I = 2 Amp di/dt = 2

2x 10−3

Solution:

(i) L1=N1∅ 2

i1

= 250x 3 x10−4

2

L1 = 37.5 x 10–3H

(ii) V2 = Md idt

63.75 = M x 2

2x 10−3

M = 63.75 x 10–3 H

(iii) M = K√L1L2

63.75 x 10–3 = 0.8√ (37.5 x10−3 )L2

L2 = 150 x 10–3 H

(iv) L1

L2

=N1

2

N22

37.5x 10−3

150x 10−3 = 250

N22

N2 = 500

Example 1.9

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

of 100cm2 cross-sectional area and mean length of 150cm.

(a) Estimate the self inductance of the two coils and the mutual inductance if

r of iron is 2000.

(b)Calculate the voltage induced in coil if the current changes from 0 to 10A

steadily in 0.01 sec.

Given data: N1 = 30 turns N2 = 600 turns

a = 100cm2 = 100 x 10–4 m

ℓ = 150 cm = 150 x 10–2 m

r = 2000didt = 250 A/seconds.

Solution:

(a) L = N2μal

Where,

= o r

= 4 x 10–7 x 2000

= 2.5132 x 10–3

L1 = N 1

2μal

=(30)2 x2.5132 x10−3 x100 x 10−4

150 x 10−2

L1 = 150.8 x 10–4 H

L2 = N 12μal

= (600)2 x 2.5132x 10−3 x100 x10−2

150 x10−2

L2 = 60320 x 10–4 H

= (√L1L2 )K−Y K = 1 (assume)

M = √150 x60320 x10−3

M = 3016 x 10–4 H

(b)V2 = Md idt = (3016 x 10–4) ( 10

0.01 )V2 = 301.6 V

1.5 Torque in round rotor machines

In a three phase machine, when a balanced supply is fed to a balanced

winding a revolving flux is developed. When such machined carry currents both in

the stator and rotor windings revolving flux is produced in both the parts. These

fluxes try to interact with each other end hence tends to align themselves. But this

is possible only if there persists no relative velocity (or) speed between the two

fluxes. This is not possible if both the fluxes are at synchronous as they are not

standstill with respect to each other. Hence it is conventional that the electrical

machines of various types are designed so as produce interacting magnetic fields.

To obtain an expression for the torque developed, the following assumptions are to

be considered.

1. The rotor is considered to be smooth cylindrical (wound rotor) so as to

maintain an uniform air gap.

2. Reluctance of the iron-path is negligible.

3. Stator and rotor mmfs are sinusoidal waves.

4. The type of ac windings is that of distributed windings.

5. The leakage flux affects the net voltage at the terminals of the machine.

6. The mutual flux links both the stator and rotor windings.

Let F1 and F2 be the peak value of sinusoidal fluxes in the stator and rotor

respectively.

α = Angle between the positive peaks of the stator and rotor fluxes.

Fair−gap = the mmf access the air gap

G = Radial air-gap length

H = Field intensity along radial path.

Fair−gap = Hg ------ (1.36)

By taking t h evectorial resultant of F1 and F2

FT = F1 +F2 = √F12+F2

2+2F1F2cos α

FT2=F1

2+F22+2 F1 F2 cos α ------ (1.37)

The expression for co- energy density is given by

W f' = BdH =1/2 µH 2 = ½ B

2

µ------ (1.38)

Neglecting the reluctance of iron path, this can be modified as

W f' = ½ µ0 H 2 ----- (1.39)

Average value H = H t

√2

Equation (1.39) can be written as W f' = ½ µ0

H t2

2

W f' = µ0

H t2

4----- (1.40)

Volumeof air gap=πDlg

G = air gap length

L = length of the machines

Total co-energy = co-energy density × volume

= µ0 H t

2

4 × πDlg

Co-energy = π/4 µ0 H t2Dlg -----(1.41)

Substituting for the peak value of resultant field intensity as

HT=F t

g =

mmflengt h

Equation 6 co-energy = π/4 µ0 F t

2

g2 Dlg

W f' = π/4

µ0Dl

g F t

2 -----(1.42)

Substituting equation 1.37 in Eq. (1.42), the expression for co-energy

becomes modified as

W f' = π/4

µ0Dl

g {F1

2+F22+2F1F2 cosα}

Torque developed in the machines is given as

T = + ∂W f

'

∂ x = - π/4

µ0Dl

g {F1F2sinα }

T = - π2 µ0Dl

g F1F2sinα ----- (1.43)

The negative sign indicates that the torque developed in the direction to reduce the

angle α. Hence this helps is aligning the two fields.

1.6 Hysteresis and eddy current loss

When magnetic circuits are subjected to time-varying flux densities, there

are two causes of power loss in the form of heat in the iron core. These losses are

significant in determining the heating, rating, and efficiency of rotating electrical

machines, transformers, and ac operated devices.

The first loss is associated with the phenomenon of hysteresis and is an

expression of the fact that when ferromagnetic material is involved, not all the

energy of the magnetic field is returned to the circuit when the MMF is removed. It

is known as hysteresis loss. When the flux varies from +B, to -B, at the frequency

f, the hysteresis loss per unit volume of material may be shown to be proportional

to the area of the hysteresis loop and to the number of loops traversed per second.

Determination of Hysteresis Loss. Consider a ring of specimen

circumference I meters, cross sectional area a metrer2 and having N turns of an

insulated wire. Let the current flowing through the coil be of I amperes.

Fig. 1.29

Let the flux density at this instant be B.

Total flux through the ring, webers,.

When the current flowing through the solenoid alters, the flux produced in

he iron ring also alters, so the emf is induced, whose value is given by

According to Lenz's law this induced emf will oppose the flow of current,

therefore, in order to maintain the current I in the coil, the source of supply must

have an equal and opposite emf.

Hence applied emf,

Energy consumed in short time dt, during which flux density has changed

joules

Thus total work done or energy consumed during one complete cycle of

magnetization.

Now is the volume of the ring and HdB is the area of elementary strip of

B-H curve shown in Fig. 1.30 and is the total area enclosed by hysteresis

loop.

Energy consumer per cycle = Volume of ring area of the hysteresis loop

------- (1.44).

This energy expended in taking specimen through a magnetic cycle is wasted and

since it appears as heat, it is termed as hysteresis loss.

For determination of hysteresis loss, the area of the loop is measured in units

of H and B i.e. in AT/m and Wb/m2. To measure the area of the loop in units of H

and B, the area of the loop is measured in square metres may be multiplied by the

scales of H and B.

For example if one metre represents xAT/n on H-axis and y on B-

axis, then hysteresis loss,

= Area of hysteresis loop in m2× x× yjoules/ /Cycle -----(1.45)

Hysteresis loss can also be determined by Steinmetz method, according to which

hysteresis loss per cubic metre per cycle of magnetization of a magnetic material

depends upon (i) the maximum value of the flux density and (ii) the magnetic

quality of the material.

Hysteresis loss ∝ (Bmax) 1.6 J/ m2 / cycle = η (Bmax) 1.6 J/ m2/cycle.

= η (Bmax) 1.6 f v J/s or watts -----(1.46)

Where is a constant for a given specimen and a given range of flux density

and is known as Steinmetz hysteresis coefficient. f is the frequency of reversals of

magnetization and v is the volume of magnetic material in m3.

1.6.2 Eddy current Losses. The second loss arises from the fact that the core

itself is composed of conducting material, so that the voltage induced in it by the

varying flux produces circulating currents in the iron. These are called eddy

currents and are accompanied by loss in the core called the eddy current loss.

Since the eddy currents depend upon the rate of change of flux as well as the

resistance of the path, it is reasonable to expect this loss to vary as the square of

both the maximum flux density and the frequency. Eddy current loss is given by

the expression

watts or joules/ second ------(1.47)

Where is the eddy current coefficient and depends upon the type of core

material. is the number of complete magnetization cycles per second, is the

maximum flux density in teslas , t is the thickness of laminations in metres

and v is the volume of core material in m3.

To increase the core resistance and thereby minimize eddy currents,

magnetic cores subjected to alternating fluxes are assembled from thin sheets with

an insulating layer (surface oxide or varnish) between successive laminations.

Eddy currents also have a magnetic effect, tending to make the flux density at the

centre lower than at the surface. This effect, known as screening effect, is

negligible in properly laminated cores at power frequencies but may be significant

at higher frequencies.

Hysteresis and eddy current losses, taken together, are known as core or iron

loss. Core loss is present in dc machines as well as ac machines, for the rotor iron

of a rotating dc machine contains flux which varies cyclically in both magnitude

and direction. Many machines are operated at constant voltage and constant

frequency or speed and, as a result, have substantially constant core losses

regardless of load being supplied. The operating voltage and frequency of a power

transformer, for example, are normally constant, the maximum flux and flux

density are, therefore, constant, and the core loss remains the same whether the

transformer is loaded or not.

Example 1.10.

Calculate the loss of energy caused by hysteresis in one hour in 50kg of iron if the peak

flux density reached in 1.3 T and the frequency is 25Hz. Assume Steinmetz

coefficient as 628 and density of iron as 7.8 x 103 kg/m3. What will be the area of

B-H curve of this specimen if 1cm = 12.5 AT/m and 1cm = 0.1 T .

Hysteresis loss/second = ῃ (Bmax) 1.6 fv = 628 (1.3) 1.6 ×25× 50

7.8×103 = 153J/s

Solution:

Hysteresis loss per hour = 153×60×60 = 5,50,800 Joules

Hysteresis loss per m3 per cycle as per Stemintz law

= = ῃ (Bmax) 1.6 = 628 × (1.3) 1.6

Also Hysteresis loss per m3 per cycle = xy × area of B-H loop

Area of B-H loop = Hysteresis loss/m3 /cycle

Xy =

628×(1.3)1.6

12.5×0.1

= 764 cm2

1.7 A.C operation of Magnetic circuits

In many applications and machines such as transformers and a.c machines ,

the magnetic circuits are excited by a.c supply. In such an operation, inductance

plays vital role even in steady state operation through in d.c . It acts as a short

circuit. In such a case the flux is determined by the AC voltage applied and the

frequency. Thus the exciting current has to adjust itself according to the flux so

that every time B-H relationship is satisfied

Fig.1.31

Consider a coil having N turns wound on iron core as shown in Fig.1.31

The coil carries an alternating current I varying sinusoidally. Thus the flux ∅ produced by the exciting current I is also sinusoidally varying with time.

Therefore ∅ = ∅m sinωt ------(1.48)

∅m = Maximum value of flux in core.

ω = 2π f where f is frequency in Hz

According to Faraday’s law, as flux changes with respect to coil, the e.m.f gets induced in the coil given by

e = Nd∅dt = N ddt (∅m sinωt )

e = N ∅mω cosωt V -------(1.49)

Em = Maximum Value = n ∅m ω

E = r.m.s value = Em√2

= N ∅mω

√2

E = N ∅m2π f

√2 = 4.44 f N ∅m -------(1.50)

∅m = AcBm

Where Ac = Area of cross-section of core.

Bm = Maximumflux density in Wb/m2

The sign of e.m.f induced must be determined according to Lenz’s law, opposing

the changes in the flux. The current and the flux are in phase as current produces

flux instantaneously. Now induced e.m.f. is cosine term and thus leads the flux and

current by90°. This is called back e.m.f as it opposes the applied voltage. The

resistance drop is very small and is neglected in most of the electromagnetic

device.

1.8 Energy stored under A.C operation

The instantaneous electric power input into the magnetic circuit through the coil

terminals are given by

P= e i but e = dλdt

P = I dλdt ---4

The power is rate of change of energy hence energy input which gets stored in the

magnetic field during the interval t 1 to t 2 is,

W t = ∫t1

t2

pdt = ∫t1

t2

idλdtdt

W t = ∫λ1

λ2

i dλ ------(1.50)

Thus W f is the increase in the field energy as the flux linkages of the coil change

λ1 ¿ λ2 during the interval t 1 to t 2.

Now, H c l c = m.m.f. = Ni

I = H c lcN

Using the equation 5

W f = ∫λ 1

λ 2

( H c lcN ) dλ ------(1.51)

But λ = N∅ and ∅ = Bc aC i.e λ = N Bc aC

The flux density of the core is changing from B1 to B2 as the flux linkages change

from λ1 to λ2

W f = ∫B 1

B 2

(H c lcN ) N aC d Bc

W f = ac lc∫B1

B2

H c dBc ------(1.52)

This is the field energy in terms of field quantities.

Now ac lc = Volume of the core

W f per unit volume = ∫B 1

B 2

H cdBc J/m3

This is called field energy density.

2 Mark question and Answer

1. Mention the types of electrical machines.

There are three basic rotating machines types, namely

a. The dc machines

b. the poly phase synchronous machine (ac), and

c. Poly and single phase induction machine (ac) and a stationary

machine, namely Transformer

2. State Ohm’s law for magnetic circuit.

It states that the magneto motive force across the magnetic element is equal

to the product of the magnetic flux through the magnetic element and the

reluctance of the magnetic material. It is given by

MMF = Flux X Reluctance

3. Define leakage flux

The flux setup in the air paths around the magnetic material is known as

leakage flux.

4. Define magnetic reluctance

The opposition offered by the magnetic circuit for the magnetic flux path is known as magnetic reluctance. It is analogous to electric resistance.

5. Draw the typical normal magnetization curve of ferromagnetic material.

6. What is fringing?

In the air gap the magnetic flux fringes out into neighboring air paths due to

the reluctance of air gap which causes a non uniform flux density in the air gap of a

machine. This effect is called fringing effect.

7. State stacking factor.

The stacking factor is defined as the ratio of the net cross sectional area of a

magnetic core to the gross cross sectional area of the magnetic core. Due to

lamination net cross -sectional are will be always less than gross cross sectional

area. Therefore the value of stacking factor is always less than unity.

8. Mention some magnetic materials

Saturation zone

Initial nonlinear zone

Linear zone (constant μ) B(T)t H (A/m)

Alnicos, chromium steels, copper–nickel alloy, nickel, cobalt, tungsten and

aluminium.

9. What is magnetostriction?

When ferromagnetic materials are subjected to magnetizing mmf, these may

undergo small changes in dimension; this phenomenon is known as

magnetostriction

10. Define statically induced emf.

The coil remains stationary with respect to flux, but the flux through it

changes with time. The emf induced is known as statically induced emf.

11. Define dynamically induced emf.

Flux density distribution remains constant and stationary but the coil moves

relative to it.

The emf induced is known as dynamically induced emf.

12. State Fleming’s right hand rule.

Extend the thumb, fore and middle finger of the right hand so that they are

mutually perpendicular to each other. If the thumb represents the direction of

movement of conductor and the fore finger the direction of magnetic flux, then the

middle finger represents the direction of emf

13. State Fleming’s Left hand rule.

Extend the thumb, fore and middle finger of the right hand so that they are

mutually perpendicular to each other. If the forefinger represents the direction of

flux and the middle finger the direction of current, then the middle finger

represents the direction of movement of conductor.

14. What are the losses called as core loss?

Hysteresis loss and eddy current loss.

15. Define coercivity.

It is the measure of mmf which, when applied to the magnetic circuit would

reduce its flux density to zero, i.e., it demagnetizes the magnetic circuit.

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

W = 12 LI 2Where

L is the inductance

I is the current

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

Energy density w = 12 BH

= 12 μ H2

18. Define MMF.

Magnetic motive force (mmf) is given by

Mmf = flux × reluctance

Mmf = ∅ . R Amp.turns.

19. Define reluctance.

Reluctance is the ratio of magnetic circuit to the flux through

. R = mmfflux (∅ )

It is also written as

. R = lμA

Where l is the length

A is the area of cross section

is permeability

20. Mention four similarities between electric circuit and magnetic circuit;-

Electric circuit Magnetic circuit

1. Path traced by the current is called electric circuit

2.MMF is driving force in electric circuit, the unit is volts

3. There is current I in circuit, measured in amperes.

4. EMF ( volts)

Current = emf

reluctance

5.Resistance R =ρlA

1. Path traced by the magnetic flux is defined as magnetic circuit

2. MMF is driving force in magnetic circuit, the unit is ampere turns

3. There is flux ∅ in circuit, measured in Webers

4.MMF (Amp-turns)

Magnetic flux = mmf

reluctance

5..Reluctance = R = lμA

6.Conductance = 1R

7.Resistance opposing the flow of

the current , unit is ohm

6 Permeance P = 1/R

7. Reluctance is opposed by

magnetic path to the flux, unit

is Ampere turns/ webers

QUESTUION BANK

PART –A

1. What are the three basic rotating Electric machines?

2. Name the three materials used in machine manufacture.

3. What is magneto motive force?

4. Define Leakage flux and leakage inductance.

5. Explain Flux Fringing at air – gap.

6. Write short note on stacking factor and give its value.

7. Give the classification of material based on relative permeability μr .

8. Define relative permeability μr .

9. What is B-H curve and sketch it?

10. Define Faraday’s law of induction.

11. What is Lenz’s law?

12. Define statically induced emf and dynamically induced emf.

13. Write the Lorentz force equation.

14. Give few magnetic properties.

15. What is magnetic permeability?

PART - B

1. Compare the various magnetic materials?

2. Derive the expression of the flux, reluctance of the magnetic material with

air gap.

3. Derive the inductance, energy and power of a magnetic circuit with two

Windings.

4. Differentiate between Electric and magnetic circuits.

5. Explain with a neat diagram the B-H curve.

6. In a rectangular electromagnetic relay, the exciting coil has 1200 turns.

Cross sectional area of the core is A = 6 cm × 6 cm. neglect the reluctance of

the magnetic circuit and fringing effects. With coil current kept constant at

2A, derive expression for force on armature as a function of air gap of length

x. Find the work done by the magnetic field when x decreases from 1 cm to

0.5 cm by integrating the force.

7. Explain different types machines

8. Compare statically induced emf and dynamically induced emf?

9. Discuss the origin of hysteresis and eddy current losses in electrical

machines.

10.A straight conductor of 2 m length carries a current of 20A. It is lying at

right angles to a uniform magnetic flux density of 0.8 T. Find: (1) the force

developed on the conductor (2) the power required to drive the conductor at

a uniform speed of 25 m/s and (3) the emf induced in the conductor.

11.Explain the AC operation of magnetic circuit in electrical machines.

12.A coil of 400 turns is wound on a closed iron ring of mean radius 10cm and

cross-sectional area of 3cm2. Find the self inductance of the winding if r =

800.

( L = 0.768 µH)

13.Find inductance/unit length of a co-axial cable if inner and outer conductor

radius is 1 mm and 3 mm respectively. Assume r = 2. ( L = 0.439 µH).

14.Derive the expression for torque produced in Rotating machines.

15.Define self inductance and mutual inductance. Show that M=K √L1L2

16.Explain the various types of Electric machines.

17.Explain the general construction of a d,c machine.

18.Explain the general construction of a synchronous machine

19.Explain the general construction of an Induction machine.

20.State and explain the laws of magnetism

21.Define and state the following terms

(i) Magnetic flux (ii) Magnetic flux density (iii)Reluctance (iv)Relative

permeability (v)Absolute permeability

22.Explain the leakage flux and fringing

23.A steel ring of 180 cm mean diameter has a cross sectional area of 250 mm2.

Flux developed in the ring is 500 µwb when a 4000 turns coil carries certain

current. Find i. m.m.f required ii. Reluctance iii. Current in the coil Given

that the relative permeability of the steel is 1100.

24.A coil is wound uniformly with 300 turns over a steel ring of relative

permeability900, having mean circumference of 40 mm and cross sectional

area of 50mm2. If a current of 25A is passed through the coil, determine i.

m.m.f ii.reluctance of a ring iii. Flux ( Ans : 7500 AT, 707355 AT/Wb,

0.0106 Wb)

25.A iron ring of 100 cm mean circumference is made from round iron of cross

section 10cm2, its relative permeability is 800. If is wound with 300 turns,

what current is required to produce a flux of 1.1 ×10−3 Wb?. (3.647 A)