Chapter 1

Introduction to Machinery

Principles

Magnetic materials form a major part in the

construction of electrical machines.

In this chapter

• Production of a Magnetic Field

• Magnetic Circuits

• Magnetic Behaviour of Ferromagnetic Materials

• Energy Losses in a Ferromagnetic Core

Production of Magnetic Field

i

f

N Ni

Cross-sectional area

Faraday’s Law

Whenever a conductor is moved through a magnetic field, or

whenever the magnetic field near a conductor is changed, currents

flow in the conductor. This effect is called electromagnetic

induction. Faraday's law gives the magnitude and direction of the

emf produced in a conducting loop whenever the loop is moved, or

whenever the magnetic field near the loop is changed:

where l=Nf

and

B is the flux density and A is the area

f is the magnetic flux in webers (Wb)

l is the flux linkage in weber-turns

N is the number of turns

dt

de

l

BAdABA

f

Magnetic Circuits

Transformers: the magnetic circuits may be formed by ferromagnetic materials only.

Rotating Machines: the magnetic circuits may be formed by ferromagnetic materials in conjunction with an air medium.

Magnetic Circuits

The magnetomotive force (mmf) of an N-turn current carrying coil is given by

=Ni

The reluctance of a path for magnetic flux, such as the bar or iron shown in the following figure, is given by:

where,

lc = mean length of the path

m = permeability of the material

A = cross-sectional area

f

A

l

A

lc

m

Reluctance is analogous to resistance in an electrical circuit.

When the bar is not straight, the length of the path is taken to

be the length of the centerline. l is therefore considered as

the mean length of the path.

Magnetic flux, f, in a magnetic circuit is equivalent of

current in an electrical circuit. Magnetic flux, reluctance and

mmf are related by:

which is equivalent to Ohm’s Law (V=iR).

The current carrying coil is the magnetic circuit equivalent to

a voltage source in an electrical circuit. Therefore, mmf can

be considered to be analogous to a source voltage.

f

Example 1

Using the magnetic circuit concepts, analyze the toroid coil shown in

Fig. 1. The magnetic circuit is analogous to the electrical circuit shown

in Fig. 2, with a resistance connected across a voltage source.

Fig. 1 Fig. 2

Fig. 1. Coil on a toroid iron core

Fig. 2. The magnetic circuit for the toroid coil

Find the expressions for the reluctance and magnetic flux.

R

H Core centreline

i

N turn coil

f

+

R

Solution to Example 1

The mean length of the magnetic path is: l=2pR

The cross-section of the core is circular with a radius, r.

Thus, the area of the cross section is: A=pr2.

The expression for the reluctance (from Eq. 1-2):

The mmf is: =Ni

And the magnetic flux:

22

22

r

R

r

R

A

l

m

mp

p

m

R

INr

2

2m

f

R

H Core centreline

i

N turn coil

Example 2

Consider the magnetic core with an air gap as shown in Fig. 3.

The core material has a relative permeability of 6000 and a

rectangular cross-section of 2cm by 3cm. The coil has 500 turns.

Determine the current required to establish a flux density of

Bgap=0.25T in the air gap.

Fig. 3. Iron core with an air gap Fig. 4. Magnetic circuit

i

N=500

0.5 cm

8 cm

8 cm

2 cm Core thickness=3 cm

f

+ Rgap

Rcore

Solution to Example 2

The mean length of the iron core and the cross-sectional area:

lc=4*6-0.5=23.5cm, Ac=(2*10-2)*(3*10-2)=6*10-4 m2

The permeability of the core is: mcore=mrm0=6000*4p*10-7=7.54*10-3

The reluctance of the core:

Air gap cross-sectional area:

Agap=2*3 cm2=6*10-4 m2

corecore

corec

A

l

m

Solution to Example 2 (cont’d)

The permeability of air is approximately equal to that of free space:

mgap=m0=4p*10-7

The reluctance of the air gap = lgap/(mgap*Agap)= 6.63*106

The total reluctance = 5.195*104 + 6.63*106 = 6.68*106

Even though the gap is much shorter than the iron core, the reluctance of

the gap is higher than that of core because of the much higher permeability

of the iron. Most of the mmf is dropped across the air gap.

The flux is given by: f=Bgap Agap=0.25*8.75*10-4 =2.188*10-4 Wb

The flux in the core is the same as that in the air gap. However, the flux

density is higher in the core, because the area is smaller.

The mmf is given by: =f =2.188*10-4*6.68*106 = 1,461

And the required current: i = /N = 1461/500 = 2.922 A

Example 3

Consider the magnetic core with an air gap as shown in Fig. 5.

The core material has a relative permeability of 6000 and a

rectangular cross-section of 2cm by 3cm. The coil has 500 turns.

Determine the current required to establish a flux density of

Bgap=0.25T in the air gap considering fringing at the air gap.

Fig. 5. Iron core with an air gap Fig. 6. Magnetic circuit

i

N=500

0.5 cm

8 cm

8 cm

2 cm Core thickness=3 cm

f

+ Rgap

Rcore

Solution to Example 3

The mean length of the iron core and the cross-sectional area:

lc=4*6-0.5=23.5cm, Ac=(2*10-2)*(3*10-2)=6*10-4 m2

The permeability of the core is: mcore=mrm0=6000*4p*10-7=7.54*10-3

The reluctance of the core:

The flux lines tend to bow out in the air gap as shown in Fig. 5. This is

called fringing. Thus, the effective area of the air gap is larger than that of

the iron core. Customarily, we take this into account by adding the length

of the air gap to each of the dimensions of the air gap cross-section giving:

Agap=2.5*3.5 cm2=8.75*10-4 m2

corecore

corec

A

l

m

Solution to Example 3 (cont’d)

The permeability of air is approximately equal to that of free space:

mgap=m0=4p*10-7

The reluctance of the air gap = lgap/(mgap*Agap)= 4.547*106

The total reluctance = 5.195*104 + 4.547*106 = 4.552*106

Even though the gap is much shorter than the iron core, the reluctance of

the gap is higher than that of core because of the much higher permeability

of the iron. Most of the mmf is dropped across the air gap.

The flux is given by: f=Bgap Agap=0.25*8.75*10-4 =2.188*10-4 Wb

The flux in the core is the same as that in the air gap. However, the flux

density is higher in the core, because the area is smaller.

The mmf is given by: =f =2.188*10-4*4.552*106 = 996

And the required current: i = /N = 996/500 = 1.992 A

Example 4:

The following figure shows a electromagnet with two parallel air

gaps. Neglecting leakage, iron reluctance, and fringing at the air

gaps, calculate the flux and flux density in each leg of the magnetic

circuit. All dimensions are in inch.

I=0.251 A

N=1,000 0.02

1

0.04

6

1

1 1 2

4

The calculated flux by the magnetic circuit method is within 5%

of the actual value. These inaccuracies are due to:

It is assumed that the flux is contained in the core. A small amount of the

flux escapes the core into the surrounding low permeability air - this is

referred to as leakage flux.

The calculations we have used for reluctance rely on a measurement of

the mean path. We have assumed that the mean path lies in the center of

the core. This assumption is not necessarily true, and in particular is a

poor assumption at the corner;

Where there are air gaps in the core fringing takes place. We have

assumed that the effective area caused by fringing increases is related to

the length of the gap, this is obviously an approximation;

In ferromagnetic materials, the permeability varies according to the

amount of flux already in the core. This non-linear effect (saturation) adds

inaccuracies to the results as the reluctance calculations depend upon the

permeability.

Magnetic Behaviour of Ferromagnetic Materials

B-H Characteristic

The B-H characteristic demonstrates the magnetic behaviour of a

ferromagnetic material. Permeability m is defined by the following

equation:

B=mH

m is constant in free space but not true for iron or other ferromagnetic

materials as magnetomotive force (mmf) is applied.

i

f=BA

N

Fig. 7. Coil wounded on a ferromagnetic core.

Ni

B-H Characteristic (cont’d)

Start with 0 DC current and slowly raise current to a possible maximum

value. When flux produced in the core is plotted versus the mmf

producing it, the resulting plot looks the one shown in Fig. 8.

(AT)

f (Wb)

Transition: knee of the curve

Saturated

region

Unsaturated

region

Fig. 8. Saturation or magnetization curve

B-H Characteristic (cont’d)

A B-H Characteristic is a representation (plot) of the magnetic flux

density as a function of the applied magnetic field. A magnetization

curve is typically nonlinear. Fig. 9 shows a plot of B versus H.

H = /l = Ni/l

F = BA or B = f/A

H is directly proportional to

f is directly proportional to B

Therefore B-H characteristic has the same shape as the relationship

between f - .

H (AT/m)

B (Tesla)

Fig. 9. B-H Characteristic

Energy Losses in a Ferromagnetic Core

• Hysteresis loss

• Eddy current loss

Hysteresis Loss

• Apply a magnetic field intensity, H, to a sample iron.

• Magnetic fields of the atoms in small domains are aligned with the

applied field.

• If the applied field is reduced to zero, the domains return to their

original orientations.

• If H is increased in the reverse direction, the domains align in the

reverse orientations.

H

i

+ e

Fig. 10a. Iron sample and coil for applying H Fig. 10b. Magnetic domains

Hysteresis Loss (cont’d)

• For an ac system, if the applied field is varied through a cycle,

during some interval of time energy flows from the source to

the coil-core assembly and during some other interval of time

energy returns to the source. However, the energy flowing in

is greater than the energy returned. Therefore, during a cycle

of variation of the applied field, there is a net energy flow

from the source to the coil-core assembly. This energy loss

goes to heat the core. The loss of power in the core due to the

hysteresis effect is called hysteresis loss.

Eddy Current Loss

• This current leads to core loss for ac operation. The solid iron core itself is an electrical conductor, acting much like shorted turns. As the magnetic field changes, voltages are induced in the core causing current, known as eddy currents to circulate in the core material. As a result, power is dissipated in the core according to P = v2/R.

• The eddy current loss can be reduced in two ways:

– A high-resistivity core material can be used.

– A laminated core can be used. The core can be laminated with thin sheets of iron that are electrical insulated from one another. The orientation of the sheets is selected to interrupt the flow of current. Thus, the resistance is higher for eddy currents, and the loss is greatly reduced. Another approach is to make the core with powdered iron held together by an insulated binder.

Questions

1. Explain the terms magnetomotive force, reluctance and flux and hence

derive the expression that relates these quantities.

2. What is a ferromagnetic material? Why is the permeability of

ferromagnetic materials so high?

3. What are the relationships between the followings:

- flux density and magnetic field strength

- flux and flux density

- mmf and flux

4. What are hysteresis and eddy current losses? What can be done to

minimize them?

5. An air gap has a length of 0.1cm. What length of iron has the same

reluctance as the air gap? The relative permeability of the iron is 5000.

The cross sectional areas of the air gap and the core are the same. [0.5m]