Chap_1 Magnetic Circuits

8/3/2019 Chap_1 Magnetic Circuits

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Chapter No. 01

Magnetic CircuitsDr. Intesar Ahmed, Engr. Kashif Imran,

Engr. Muhammad Shuja Khan

Department of Electrical Engineering

COMSATS Institute of Information Technology

Lahore


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Introduction

Electrical machines are used to convert energy from one form to

another form.

Magnetic materials are used for the construction of electricalmachines. Therefore it is important to understand the magneticmaterials and their application (i.e. magnetic circuits and air gap) inelectrical machines.

The main advantages of using magnetic materials in the electricalmachines are, to obtain high flux density which gives large outputand efficiency, reduced size of the machine.


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Current and Magnetism Current The rate of change of charge is known as current,

and denoted byIor iand the expression is

I=dq/dt. The current flows in a closed circuit inthe presence of electromotive force.

Magnet and Magnetism Magnet is a material that can attract or repel any

other iron material. The magnet contains thenorth and south poles. Magnetism is the amountof force that is created by electric current or theamount of motion of electrons in the atom.


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Magnetic Field

Consider a current isflowing in a currentcarrying conductor from A

to B as shown in figure.Due to this current, amagnetic field is producedaround the conductor,

whose direction is given by

Right Hand Thumb Rule.


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Magnetic Flux The amount of magnetic

lines passing through anarea near the magnet iscalled the magnetic flux.

The magnetic flux can alsobe defined as the productof magnetic flux densityand the perpendicular areathat flux penetrates, it is

denoted by, = BA Wb


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Magnetic Flux Density Magnetic Flux Density is magnetic flux per unit area

and expressed as, B= /A Wb/m2

The magneto-motive force is the product of currentand the number of turns of the coil. Mathematically,

= Ni AmpereTurns


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Magnetic Field Intensity The magneto-motive force per unit length is known as

magnetic field intensity and is expressed as,

H= Ni/l AmpereTurns/mMagnetic Field Intensity represents the effort

exerted by the current to establish a Magnetic Field


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Practice Problems: 1. The dimensions of a bar magnet are 15cm x 3cm x 2cm.

Its magnetic flux density is 0.02 Wb/m2. Determine thetotal flux of the magnet.

2. The area of a bar magnet is 14cm2. If the total flux ofthe magnet is 0.05Wb, determine the magnetic fluxdensity.


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Magnetic Permeability Magnetic Permeability, , represents the relativeease of establishing a magnetic field in a givenmaterial.

Relative Permeability is the ratio of absolutepermeability of any material () to thepermeability of the free space (o). Where o = 4 x 10-7

Expression for Relative Permeability: r = /o Relative permeability of silicon steel used in

modern machines ranges from 2000 to 6000.


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Relationship of B & HMagnetic flux density (B) developed in a magnetic field

is directly proportional to the amount of appliedmagnetic field intensity (H).

B H

B = *H

( is absolute permeability of the material).

B = r*o*H


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Practice Problems:

1. The mean circumference of an iron ring is 0.6m, uniform cross

sectional area is 0.005 square meter and relative permeability is3000. The iron ring is wound by 200 turns and carries a current of5A. Determine the (a) magnetic field intensity, (b) magnetic f luxdensity and (c) total flux.

2. The mean circumference of a silicon steel ring is 800mm, uniformcross sectional area is 600mm2 and relative permeability is 5000.The steel ring is wound by 400 turns and carries a current of 8A.Determine the (a) magnetic field intensity, (b) magnetic fluxdensity and (c) total flux.


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Reluctance Reluctance

It is property of a magnetic material to oppose thedevelopment of magnetic flux.

It can also be defined as the ratio of magneto-motive forceto the flux through any cross section of a magnetic circuit.

R= /

= Ni

= BAB = H

H= Ni/l

R= l/ A


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Permeance Permeance

It is defined as the ratio of magnetic flux to the magneto-motive force through any cross section of the magnetic

circuit. Alternatively, Permeance is the reciprocal of reluctance.


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Practice Problems: 1. A current of 4A is flowing through a wire which

consists of 9 turns. The flux around the wire is 0.06wb.

Determine the (a) magnetomotive force and (b)reluctance of the circuit.

2. The magnetomotive force of a coil is 54At when it

carries a current of 6A. Determine the number of turnsof the coil and the reluctance of the circuit if the flux is0.07 Wb.


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Magnetic Flux Density for a Long

Straight Wire

H = Ni/2r

B =Ni/2r


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Practice Problems: 1. A 15A current is f lowing through a wire conductor.

The conductor is surrounded by air and is wound by 10turns. Find the magnetic flux density at a distance of

1.5cm from the conductor. 2. A long wire of 15 turns carries a current of 20A. The

conductor is surrounded by air. Find the magneticfield intensity and flux density at a distance of 0.2m

from the wire.


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Magnetic Flux Density in a

Toroidal Coil

H = Ni/2(R-r)

B =Ni/2(R-r)

= Nir2/2(R-r)


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Practice Problems:

1. An iron (relative permeability is 4000) torodial coil iswound by 200 turns and having the inner diameter of 7cmand the outer diameter of 10cm. The cross-sectional area is

0.005m2. If the current 5A flows through the coil, find the(a) magnetic field strength, (b) flux density, and (c) f lux.

2. An iron (relative permeability is 5000) torodial coil is

wound by 200 turns and having the mean length of500mm. The cross-sectional area is 0.0008m2. If the currentof 2A flows through the coil, find the (a) magnetic fieldstrength, (b) flux density, and (c) f lux.


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Magnetic CircuitsA magnetic circuit is a closed path followed by themagnetic flux. The magnetic circuit is divided into twotypes.

Series Magnetic Circuit: The Magnetic circuit in which, same flux flows

through it.

Parallel Magnetic Circuit:

The Magnetic circuit in which flux divides into two ormore parts.


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Series Magnetic Circuits


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Series Magnetic Circuits Magnetic circuit and their equivalent circuit are shown

in figure. Where, l1 and l2 are mean length of twomaterials, A1 and A2 are the cross-sectional area of twomaterials and 1 and 2 are the absolute permeability oftwo materials. Current flowing through the left sectionof the core having N turns is I.


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Series Magnetic CircuitsNi= H1l1 + H2l2 = R

Where R= R1+R2 = B1A1 = B2A2


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Practice Problems: 1. The dimension of the magnetic circuit is shown inthe following figure. The relative permeability of themagnetic material is 750. Calculate the field intensity.

2. Determine the magnetic flux of the magnetic circuit

as shown in the figure. The dimensions of the length,width and height are 0.3m, 0.02m and 0.3mrespectively. The relative permeability of a magneticmaterial is 1100.


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Total f lux is divided into two or more sections (seefigure). The magnetomotive force is connected to theleft side and induces the flux and this flux is divided

into 1 and 2 for the mid and right section of the core.The magnetic field intensity for left, right and middleare H, H1

, and H2 and length are l,l1 and l2.


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Parallel Magnetic Circuit: The magnetic potential drop around the equivalentcircuit is, - Hl = H1l1= H2l2 Total flux is, =

1

+ 2

As we know, Hl = R

Using above equations, magnomotive force can bewritten in form of,

- R= R11= R22 = R+ R11= R+R22


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Practice Problems: Following problems are for a magnetic circuit with two windows. Itsmean window width is 22cm and window height is 10cm.

1. The cross sectional area of the above mentioned magnetic structureis 5cm2. The left limb is wound by 350 turns and the flux in the rightlimb is 4Wb. Determine the current. Assume the relative permeabilityis 300.

2. The cross sectional area of the magnetic structure is 6cm

2

. The leftlimb is wound by N turns and the flux in the right limb is 4mWb.Determine N if the 3A current flows in the coil. Assume the relative

permeability is 550.


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Comparison b/w Magnetic and

Electric Circuits 4. The magnetomotive force is equal to the product offlux and reluctance. The voltage is equal to the productof current and circuit resistance.

5. The magnetic circuit reluctance is defined as R=l/A. The electrical circuit resistance is defined asR= (l/A).

6. The magnetic field intensity is defined as H=/l and

the electric field intensity is defined as E= V/d


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Air Gap in Magnetic Circuits

The gap between two parts of magnetic bar is called AirGap. This gap is filled by non magnetic material. Inmachines, rotor moves freely inside the stator through a

small air gap. Generally, f luxes cross directly from one barto other bar at the middle of the air gap, whereas at theedges of an air gap flux bends outward and therebyincreases the effective area of the gap which reduces the

magnetic f lux density. This reduction process is known asfringing as shown in figure. The fringing will be greater forlonger air gap.


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Calculation of Magnetic Circuits with Air

GapConsider the mean length, permeability and crosssectional area of the core are lc, c and Ac and for airgap lag, ag and Aag respectively. In the composite

circuit, the core and the air gap are connected in series.As a result same flux will f low through the core and airgap and equivalent circuit can be considered as seriescircuit as shown in following figure.


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Air Gap in Magnetic Circuits


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Air Gap in Magnetic Circuits The magnetic f lux densities for the core and air gap

are,

Bc

= c

/Ac

and Bag

= ag

/Aag

,

Total magneto-motive force,

t = c+ ag = Hclc+ Haglag t = Rtwhere, Rt = Rc+ Rag


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Air Gap in Magnetic Circuits If the permeability is

very high then thereluctance of the core is

very small and it can beneglected. Therefore,total magneto-motiveforce is,

ag = Rag


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Practice Problems: 1. The mean length cross sectional area and the air gap

length of a iron ring are 35cm, 15cm2 and 0.6 mmrespectively. The ring wound by 400 turns and carries a

current of 2A, which produces 2.5mWb flux. Determinethe reluctance and the relative permeability of the ironring.

2. The cross sectional area of a magnetic structure is 2cm2and is wound by 200 turns. The core is made up of iron andthe relative permeability of the core is 4000. Determine thecurrent flowing through the coil if the total f lux of the

circuit is 2.5mWb.


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Practice Problems: 3. The mean length cross sectional area and the air gap

length of a iron ring are 55cm, 25cm2 and 0.8 mmrespectively. The ring is wound by 1200 turns and carries a

current of 2A, which produces 6.5mWb flux. Determinethe reluctance and the relative permeability of the ironring.

4. The cross sectional area of a magnetic structure is 2.5cm2and is wound by 1000 turns. The core is made up of ironand the relative permeability of the core is 4000.Determine the current flowing through the coil if the total

flux of the circuit is 3.5mWb.


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Electromagnetic Force on a Conductor

When a current carrying conductor is placed in amagnetic field it experiences a force, known aselectromagnetic force or Lorentz force.

F = BiI

Where, B is the magnetic flux density, l is thelength of the conductor in meter and i is the

current flowing through the conductor. The forceis greater if the length of the conductor increases.


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Practice Problems: 1. The conductor carries a current of 150A. The lengthof the conductor is 2m and is placed in a magneticfield whose magnetic flux density is 0.35T. Determine

the force. 2. The magnetic force of a 3m conductor is 200N.The

magnetic f lux density of the conductor is 0.5T.Determine the magnitude of the current


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Forces B/W Two Parallel Conductors If two conductors are carrying currents in same

direction, the fluxes are in opposite direction incommon region (see figure 1). The total reduction in

flux shows the force of attraction. However, if the conductors are carrying current in

opposite direction the net flux is increased, whichindicates the force of repulsion (see figure 2).


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Where d distance b/w two conductors,

Conductors carrying currents I1 and I2 respectively,the flux density at the conductor 1 due to conductor 2

is B =oI1/2d

If the length of the conductor is l then force eitherattraction or repulsion is,

F = BlI2

Finally,

F =oI1I2/2d


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Magnetic Material and B-H Curve Ferromagnetic materials: Materials whose

permeability thousands times greater than freespace i.e. iron, nickel, steel and cobalt.

Diamagnetic materials: Materials whosepermeability less than free space i.e. copper, goldand silver.

Paramagnetic materials: Materials havepermeability slightly greater than free space i.e.magnesium, molybdenum, and lithium.


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Hysteresis Loop It is a loop that is obtained by variation ofmagnetic flux density with magnetic fieldintensity. Consider an un-magnetized core toobtain this loop. The relationship of magnetic fieldintensity,

H = Ni/l

For zero value of current, h= 0 and the curve willstart from the origin of axes. By increasing thevalue of current the value of field intensity is alsoincreased accordingly.

H = Ni l


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BH Graph


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The new value of magnetic field intensity at pointa is Ha (see figure). Again, increase the value ofcurrent until the magnetic field intensity reachesat the saturation point and at this point the valueof magnetic field intensity is Hb.

Further increase in current will increase magnetic

field intensity but the magnitude of magnetic fluxdensity will be same. It means magnetic materialhas reached its saturation region.


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In the reverse way, if the magnetic field intensitydecreases slowly, the magnetic flux density will alsodecrease. This decreasing value of magnetic field

intensity and magnetic flux density are greater thanthe previous values and the curve will move by anotherpath. There will be some magnetic flux densityremaining if the magnetic flux intensity is reduced to

zero.


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It means that the core sustained some flux density(OC). This remaining part of the magnetic flux densityis known as retentivity of the material. Further

increase in magnetic field intensity in the reversedirection leads to the magnetic flux density becomingzero at point D. This value OD is necessary todemagnetize the magnetic material.


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This amount of magnetic field intensity is knownas coercive force. The path will continue up to thepoint E if we increase the magnetic field intensity.The complete hysteresis curve will be obtained ifwe continue the other steps.


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Core Losses There are two types of losses in the magnetic fieldknown as hysteresis and eddy current losses, andcombination of these two losses is known as core

losses.


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Hysteresis Loss In general, the hysteresis loss can be expressed as,

Ph= KhVcvolBnf

Where k

h,

depends on the properties of the magneticmaterial and volume of the core. The value of theconstant n varies b/w 1.5 to 2.5.


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Eddy Current Loss The core of the machine normally offers conductivepath for the time varying flux. This time varying fluxinduces a voltage in the body of the core. As a result of

this voltage, a small magnitude of current will flowthrough the core. This current is known as eddycurrent. The power loss due to this current is known aseddy current loss.


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The eddy current loss can be expressed as,

Pe= KeVcvolt2B2f2

Heret is the thickness of the lamination. Vcvolis core volume

Chap_1 Magnetic Circuits

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