Emc510 s emec_lect_notes_sem-1_2016_lecture-2

EMC510S
ELECTRICAL MACHINES 215
PRINCIPLES OF
ELECTRO-MECHANICAL
ENERGY CONVERSION
LECTURE 2
Compiled by Mr. Kalaluka Kanyimba. M.Eng. (P/SYST.) B.Eng. (EMP) IETFeb-16

2
3. THE ENERGY CONVERSION
PROCESS IN A MACHINE
• An electromagnetic machine is one that links an
electrical energy system to another energy system by
providing a reversible means of energy flow in its
magnetic field. The magnetic field is therefore the
coupling between the two systems and is the mutual link.
• The energy transferred from one system to the other is
temporarily stored in the field and then released to the
other system.
• Usually the energy system coupled to the electrical
energy system is a mechanical one; the function of a
motor is to convert electrical energy into mechanical
energy while a generator converts mechanical energy
into electrical energy. Converters transfer electrical
energy from one system to another as in the transformer.
• KK - 2016

3
THE ENERGY CONVERSION
PROCESS IN A MACHINE (CONT.)
• Those electrical machines that operate at very low power
levels are often termed transducers, particularly when
providing 'signals' with which to activate electronic
control devices.
• An electromagnetic system can develop a
mechanical force in two ways:
(i) By alignment.
(ii) By interaction.
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4
3.1. FORCE OF ALIGNMENT
• The force of alignment can be illustrated by the
arrangements shown in Figure 3.1.
• This force acts in any direction that will increase the
magnetic energy stored in the arrangement or shorten
the magnetic path and reduce the reluctance.
• In the first case, it will try to bring the poles together
since this decreases the reluctance of the air-gap in the
magnetic circuit and hence will increase the flux and
consequently the stored energy.
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5
Figure 3.1. Force of alignment (a) Force of attraction; (b) lateral force of
alignment.
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FORCE OF ALIGNMENT (CONT.)

6
• In the second case, the resultant force tries to achieve
greater stored magnetic energy by two component
actions:
(i) By attraction of the poles towards one another as
before.
(ii) By aligning the poles laterally.
• If the poles move laterally, the cross-sectional area of
the air-gap is increased and the reluctance is reduced
with consequent increase in the stored magnetic energy
as before. Both actions attempt to align the poles to the
point of maximum stored energy, i.e. when the poles are
in contact with maximum area of contact.
NB: The force of alignment does not necessarily act in the
direction of the line of flux.
• KK - 2016

7
• Figures 3.2 and 3.3 show two examples of the
application of the force of alignment.
• In Figure 3.2, when the coil is energized, a flux is set up
in the relay core and the air-gap. The surfaces adjacent
to the air-gap become magnetized and are attracted,
hence pulling the armature plate in the direction
indicated.
• In Figure 3.3, the rotating piece, the rotor, experiences
radial forces in opposite directions, thereby cancelling
one another out. The rotor also experiences a torque
due to the magnetized rotor and pole faces attempting to
align themselves.
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8
(a) Electromagnetic relay. (b) Reluctance motor.
Figure 3.2. Two examples of the application of the force of alignment
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9
3.2. FORCE OF INTERACTION
• There are many applications involving the force of
interaction to give rise to rotary motion. These include
the synchronous and induction machines as well as
the commutator machines.
• The principle of interaction is illustrated in Figure
3.4(a). The magnetic field due to a current-carrying
conductor interacts with the magnetic field due to
magnet poles to produce a force on the conductor. In
Figure 3.4(a)-(iii) the two fluxes interact in such a way as
to distort the flux and cause flux reduction and flux build-
up as shown. Due to the ‘elastic’ nature of magnetic flux
lines, a force is imparted on the conductor. The direction
of the force may be determined by ‘Fleming’s left-hand-
rule for a motor’.
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10
FORCE OF INTERACTION (CONT.)
Figure 3.4(a) The principle of interaction
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(i) Uniform magnetic
field, due to magnet
poles, of flux density,
B.
(ii) Flux distribution
due to current-carrying
conductor.
(iii) Production of force
on conductor due to
the interaction of the
two magnetic fields of
(i) and (ii).

11
• Figure 3.4(b)-(i) shows the production of torque and
hence rotation of a coil pivoted at its centre due to the
force of interaction.
• Figure 3.4(b)-(ii) illustrates the interaction principle
applied to a simple rotating machine.
• By passing current through the coil, it experiences a
force on each of the coil sides resulting in a torque about
the axis of rotation.
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12
Figure 3.4(b) The principle of interaction in a simple rotating machine.
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(ii) Rotary machine illustrating
the force of interaction.
(i) Force of interaction in a coil.

13
• Each electrical machine is a system-linking device.
• At one end there is the electrical system; at the other
end is the mechanical (or other) system. In between
there is the magnetic field forming a two-way link
between them.
• If there is to be flow of energy, all three will be involved
simultaneously.
• The reaction in the electrical system, apart from the flow
of current, is the introduction of an e.m.f. into the system;
the product of e.m.f. and current gives rise to the rate of
electrical energy conversion.
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14
4.1. THE ENERGY METHOD
• Based on the principle of conservation of energy, energy
is neither created nor destroyed.
• It is merely changed in form (transformed).
• Figure 4.1(a) shows a loss-less magnetic-energy-storage
system with two terminals. It represents a magnetic-
field-based electromechanical-energy-conversion
device.
• The electrical terminal has two variables, voltage (e) and
current (i). The mechanical terminal also has two
variables, force (f) and position/displacement (x).
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4. ANALYSIS OF THE
FORCE OF INTERACTION

15
THE ENERGY METHOD (CONT.)
• The losses (loss mechanism) are separated from the
energy-storage mechanism and they are:
• Electrical: ohmic/resistive losses and
• Mechanical: friction and windage.
• Figure 4.1(b) shows a simple force-producing device
with a single coil forming the electrical terminal and a
moving plunger serving as the mechanical terminal.
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Figure 4.1(a). Loss-less magnetic-energy-storage system.

16
Figure 4.1(b). A singly-excited linear actuator.
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17
• Referring to Figure 4.1(b), at a certain time instant t, the
terminal voltage applied to the excitation winding is v, the
excitation winding current is i, the position of the movable
plunger is x and the force acting on the plunger is f with the
reference direction chosen in the positive direction of the x
axis.
• After a time interval dt, it is noticed that the plunger has
moved for a distance dx under the action of the force f.
• The mechanical work done by the force acting on the
plunger during this time interval is thus: dWm = fdx
• The amount of electrical energy that has been transferred
into the magnetic field and converted into the mechanical
work during this time interval can be calculated by
subtracting the power loss dissipated in the winding
resistance from the total power fed into the excitation
winding.
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18
• The interaction between the electrical and mechanical
terminals, i.e. the electro-mechanical energy conversion,
occurs through the medium of the stored magnetic energy.
• Assuming a loss-less system, the stored energy in the
magnetic field, Wfld, is the difference between the
electrical energy input into the magnetic system and the
mechanical energy output and is given by:
where
i.e. rate of energy change in magnetic field = Electrical
power input – mechanical power output
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... eqn. 4.1.flddW dx d dx
ei f i f
dt dt dt dt
λ
=− = −
d
e
dt
λ
=

19
• The infinitesimal change in energy is:
• Equation 4.2 shows that the force can be solved simply as a
function of the flux and the mechanical terminal position x.
• Equations 4.1 and 4.2 form the basis of the energy method.
4.1.1. Energy Balance
• The energy balance in an electro-mechanical system can be
expressed as follows:
• E.g. for motor action:
… eqn. 4.3.
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... eqn. 4.2.flddW id fdxλ= −

20
Energy Balance (cont.)
• Assuming a loss-less magnetic-energy-storage system
(Figure 4.1) and rearranging eqn. 4.2 in the form of eqn
4.3 yields:
or
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21
• Referring to Figure 4.1, 'e' is the voltage induced in
the electrical terminal by the changing stored
magnetic energy.
• It is through this reaction voltage that the external
electric circuit supplies power to the coupling magnetic
field and hence to the mechanical output terminals.
• The basic energy-conversion process is one involving
the coupling field and its action and reaction on the
electric and mechanical systems.
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Energy Balance (cont.)

22
4.1.2. Energy in a Singly-Excited Linear Actuator
• In electro-mechanical energy conversion systems, the
magnetic circuits have air-gaps between the stationary
and the moving members in which a considerable
amount of energy is stored, in the magnetic field.
• This field acts as the energy-conversion medium and its
energy is the reservoir between the electric and
mechanical systems.
• Consider the electromagnetic relay of Figure 4.2.
• The predominant energy storage occurs in the air-
gap and the properties of the magnetic circuit are
determined by the dimensions of the air-gap.
• Since the magnetic energy storage system is loss-less, it
is known as a conservative system.
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23
Energy in a Singly-Excited
Linear Actuator (cont.)
Figure 4.2. Schematic of an electromagnetic relay (singly-excited linear
actuator).
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24
• The energy equations associated with Figure 4.2 are
derived as follows:
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fld
( ) ... eqn. 4.5.
N
(recall: L = )
I
... eqn. 4.6.
dW ... eqn.4.6. (eqn. 4.2.)
mech
L x I
N LI
dW fdx
id fdx
l
l
l
=
F
Þ = F =
=
= -
• Wfld is uniquely specified by the values of λ and x,
however, if the armature is stationary, i.e.
mechanical force = electromagnetic force
then, from eqn. 4.6, the magnetic energy stored in the air
gap is dWfld = idλ … eqn. 4.7.

25
• From eqn. 4.5 and, dλ = L(x)di … eqn. 4.8
and substituting this in eqn. 4.7 yields,
… eqn. 4.9.
• If ‘V’ is the volume of the magnetic field/circuit, from the
B/H curve of the magnetic material, energy per unit
volume is
• Therefore, the total energy in the magnetic field is
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2
2
)(
2
1
)(
2
1
)()(
ixLW
CixLidixLWdixiLdW
fld
fldfld
=⇒
+==⇒= ∫
0
B
HdB∫
( )0
' ... eqn. 4.10.
B
fld
V
W HdB dV= ∫ ∫

26
• Substituting H = B/µ in eqn. 4.10 yields,
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2
... eqn. 4.11.
2
fld
V
B
W dV
µ
 
=  
 
∫
NB: Wfld is the total energy (in joules) stored in the magnetic
field and wfld is the energy stored in the magnetic field per unit
volume (joules/m3).

27
4.1.3. Force and Torque Calculations from Energy
4.1.3.1. Singly-excited Linear Actuator
•The stored magnetic energy Wfld is determined uniquely by
the values of the independent variables λ and x.
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fld
( , ) ... eqn. 4.12 (eqn. 4.2.)
dW ( , ) ... eqn. 4.13.
fld
fld fld
x
dW x id fdx
W W
x d dx
x λ
λ λ
λ λ
λ
= −
∂ ∂
= +
∂ ∂
• Comparing eqns. 4.12 and 4.13:
... eqn. 4.14. and ... eq. 4.15.fld fld
x
W W
i f
x λ
λ
∂ ∂
= = −
∂ ∂

28
Singly-excited Linear Actuator (cont.)
• Once Wfld is known as a function of λ and as a function
of i(λ, x), eqn. 4.15 can be used to solve for the
mechanical force, f(λ, x).
• The partial derivative is taken while holding the flux
linkages λ constant.
• For linear magnetic systems for which λ = L(x)I, the
force can be found as follows:
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( )
.17.4....
)(
2
1
.16.4....
)(
)]([2)(2
1
)(2
1
)(
2
1
,
2
2
22
2
2
eqn
dx
xdL
if
eqn
dx
xdL
xLxLx
f
xL
ixLxWfld
=⇒
⋅=





∂
∂
−=⇒
==
λλ
λ
λ
λ

29
Force and Torque
Calculations from Energy (cont.)
• 4.1.3.2. Singly Excited Rotating Actuator
• The singly-excited linear actuator (Figure 4.2) mentioned
above becomes a singly excited rotating actuator if the
linearly movable plunger is replaced by a rotor, as
illustrated in Figure 4.4.
• Through a derivation similar to that for a singly-excited
linear actuator, the torque acting on the rotor can be
expressed as the negative partial derivative of the
energy stored in the magnetic field against the angular
displacement or as the positive partial derivative of the
co-energy against the angular displacement (see section
4.1.4.).
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30
Singly Excited Rotating Actuator (cont.)
Figure 4.4. A singly excited rotating actuator
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31
• For the system with a rotating mechanical terminal, the
mechanical terminal variables become the angular
displacement θ and the torque T.
• The energy equation is therefore:
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( , ) ... eqn. 4.18.
( , )
and ... eqn. 4.19.
fld
fld
dW id Td
W
T
λ θ λ θ
λ θ
θ
= −
∂
= −
∂
• For linear magnetic systems for which λ = L(θ)i:
2
1
( , ) ... eqn. 4.20.
2 ( )
fldW
L
λ
λ θ
θ
=

32
and
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2 2 2
2
1 1 ( ) ( )
... eqn. 4.21.
2 ( ) 2 ( ( )) 2
dL i dL
T
L L d dλ
λ λ θ θ
θ θ θ θ θ
 ∂
=− = = 
∂  

33
ANALYSIS OF THE FORCE OF
INTERACTION (CONT.)
4.2. MULTIPLY-EXCITED MAGNETIC FIELD SYSTEMS –
DOUBLY EXCITED ROTATING ACTUATOR
• Many electromechanical devices have multiple electrical
terminals, e.g:
(i) Measurement systems:
- Torque proportional to two electrical quantities.
- Power as a product of voltage and current.
(ii) Energy conversion systems:
- multiply-excited magnetic field systems, e.g. direct-
current (d.c.) and synchronous machines.
• Consider a doubly excited rotating actuator and its
equivalent energy storage system, shown in Figure 4.7.
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34
DOUBLY EXCITED
ROTATING ACTUATOR (CONT.)
A doubly excited actuator
and its equivalent energy
storage system.
Figure 4.7. Multiply-
excited magnetic energy
storage system.
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35
DOUBLY EXCITED
• The differential energy functions can be derived as
follows. First three independent variables are selected:
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{ } { } { } { }1 2 1 2 1 2 1 2, , , , , , , , or , ,i i i iθ λ λ θ θ λ θ λ
fld e mdW dW dW= −
where
1 2
1 1 2 2 1 2, , ande m
d d
dW e i dt e i dt e e dW Td
dt dt
λ λ
θ=+ = = =
hence,
1 2 1 1 2 2
1 2 1 2 1 2
1 2
1 2
( , , ) (analogous to eqn. 4.18)
( , , ) ( , , ) ( , , )
fld
fld fld fld
dW i d i d Td
W W W
d d d
λ λ θ λ λ θ
λ λ θ λ λ θ λ λ θ
λ λ θ
λ λ θ
= + −
∂ ∂ ∂
= + +
∂ ∂ ∂

36
• Therefore, comparing the corresponding differential terms,
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DOUBLY EXCITED
• In a magnetically-linear system:
1 11 1 12 2 2 21 1 22 2 12 21; ; (or, in general, )ij jiL i L i L i L i L L L Lλ λ=+ =+ = =
• Therefore, the torque acting on the rotor can be calculated as

37
• For linear displacement:
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DOUBLY EXCITED
• Due to the salient structure of the rotor, the self inductance of
the stator is a function of the rotor position and the first term
on the right hand side of the torque expression of eqn. 4.46 is
non zero for that dL11/dθ is not equal to zero.
• Similarly, the second term on the right hand side is non zero
because of the salient structure of the stator.
• Therefore, these two terms are known as the reluctance
torque components.

38
• The last term in the torque expression, however, is only
related to the relative position of the stator and rotor and
is independent of the shape of the stator and rotor poles.
• For a magnetically linear system with linear
displacement, ‘x’ replaces ‘θ’.
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DOUBLY EXCITED

39
5. ANALYSIS OF THE FORCE OF ALIGNMENT
5.1. A SIMPLIFIED ANALYSIS OF THE FORCE OF
ALIGNMENT
• It will be recalled that magnetic field energy is given by
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• All the above expressions for the energy depend on the flux
and the m.m.f. being directly proportional, i.e. the
inductance is constant.
• This is generally limited to the case of air, which is the most
important one in electrical machines.
• Sometimes the energy density can be of greater importance.
• The energy stored is proportional to the shaded area due to
the B/H curve of the magnetic material.

40
A SIMPLIFIED ANALYSIS OF
THE FORCE OF ALIGNMENT (CONT.)
• In the case of an air-gap, the B/H characteristic is
straight (Figure 5.1) and the energy stored is given by:
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1
volume of air-gap
2
fldW BH= ×
Figure 5.1. Stored-energy
diagram (B/H curve in air).
• If the air-gap has a
cross-sectional area, A,
and is of length, l,
1 1
2 2
fldW BH Al F= × = Φ
• The stored energy
density is thus given by
2
3
0 0
1
/ ... eqn. 5.1.
2 2 2
fld
B B
w BH B joules m
µ µ
= = =

41
• Figure 5.2 shows the force of alignment between two poles of
a magnetic circuit.
• The flux in the air-gap is Φ and fringing is assumed to be non-
existent.
• The uniform flux density in the air-gap is given by:
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B
A
Φ
=
Figure 5.2. Force of alignment
between two poles.
• It is assumed that the poles
can be separated by a small
distance dx without there
being a change in the flux
and the flux density.
• Since there is a mechanical force experienced by the poles,
the mechanical work done is
... eqn. 5.2.mdW Fdx=

42
• It is also assumed that the magnetic core is ideal, i.e. it is of
infinite permeability and therefore requires no m.m.f. to create
a magnetic field.
• The stored magnetic energy will therefore be contained
entirely in the air-gap.
• The air-gap has been increased by a volume A.dx, yet, since
the flux density is constant, the energy density must remain
unchanged.
• There is, therefore, an increase in the stored energy
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2
0
... eqn. 5.3.
2
fld
B
dW Adx
µ
= ×

43
• Since the system is ideal and the motion has taken place
slowly from one point of rest to another, this energy must be
due to the input of mechanical energy, i.e.
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2 2
0 0
... eqn. 5.4.
2 2
m fld
B B A
dW dW Fdx Adx F
µ µ
= ⇒ = ⇒ =
• A more important relationship, however, may be derived from
the above argument, i.e.
... eqn. 5.5.
fld
m fld
dW
dW Fdx dW F
dx
= = ⇒ =
• That is, the force is given by the rate of change of stored
field energy with distortion of the arrangement of the
ferromagnetic poles.

44
5.2. FORCE OF ALIGNMENT BETWEEN PARALLEL
MAGNETIZED SURFACES
• The arrangement of Figure 5.3 is now considered.
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ANALYSIS OF THE
flddW
f
dx
=
Figure 5.3. Force of alignment
between parallel magnetized
surfaces.
• If ‘V’ is the volume of the air-
gap, then
• Therefore,
2
0
... eqn. 5.6.
2
flddW B
f A
dx µ
= =

45
FORCE OF ALIGNMENT BETWEEN
PARALLEL MAGNETIZED SURFACES (CONT.)
• Now the case where the poles are laterally displaced is
considered (Figure 5.4).
• A force of alignment would be experienced trying to align the
poles.
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Figure 5.4. Lateral force
between magnetized
surfaces. Let the depth
be l.
• Ignoring the effect of leakage
flux, the area covered by the
gap is xl, and the gap length is
lg.
• The air-gap volume is given by
• Therefore,
2
0
... eqn. 5.7.
2
fld
g
dW B
f ll
dx µ
= =

46
• The polarity of the expression of eqn. 5.7 indicates that
the force tries to align the poles by increasing the cross-
sectional area of the air-gap, thereby decreasing the
reluctance.
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FORCE OF ALIGNMENT BETWEEN
PARALLEL MAGNETIZED SURFACES (CONT.)

47
5.3. ROTARY MOTION – THE RELUCTANCE MOTOR
•For rotary motion linear displacement x (m) is replaced by
angular displacement θ (rads) and linear speed u (dx/dt)
(m/s) by angular speed ω (dθ/dt) (rad/s).
•Angular speed (rotational speed) may also be expressed
in rev/s (n) and its relationship with ω is given by
ω = 2πn.
•The torque of a rotating machine is given by
T = dWfld /dθ.
•A simple machine that demonstrates a torque of alignment
is the reluctance motor, shown in Figure 5.5.
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ANALYSIS OF THE

48
ROTARY MOTION –
THE RELUCTANCE MOTOR (CONT.)
Figure 5.5. Simple
reluctance torque machine.
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• If the rotor is displaced through
a given angle, it experiences a
torque which tries to align it with
the stator poles.
• The field energy density
(joules/m3) in the air-gaps is
given by: 2
02
fld
B
w
µ
=
• It will be assumed that only the
gap energies need to be
considered. The total energy is
therefore given by:

49
• Referring to the previous equation, ‘V’ is the gap volume, l is
the depth of the rotor and ‘A’ is the gap area through which
the pole flux passes.
• Therefore,
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ROTARY MOTION –
THE RELUCTANCE MOTOR (CONT.)
• Such a machine would not produce continuous rotation, but it
can be adapted to do so.
• The resulting machine is termed a reluctance motor.
• Any machine with saliency of the type illustrated in Figure 5.5.
will produce a significant torque in this way.

50
HOME WORK
From the text book:
“Hughes Electrical and Electronic Technology (10th edition),
Revised by John Hiley, Keith Brown and Ian McKenzie
Smith, (Pearson) Prentice Hall”
Study the following examples thoroughly:
35.1, 35.2 and 35.3.
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