chapter-1-introduction-to-machinery-principles.pdf

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Chapter 1 - Introduction to Machinery Principles
ELECTRICAL POWER SYSTEMS AND MACHINES (University of Sunderland)
Studocu is not sponsored or endorsed by any college or university
Chapter 1 - Introduction to Machinery Principles
ELECTRICAL POWER SYSTEMS AND MACHINES (University of Sunderland)
Downloaded by Tarak BENSLIMANE (tarak.benslimane@univ-msila.dz)
lOMoARcPSD|6847370

EEE4323 Electrical Machines – Introduction to Machinery Principles
Electrical Machines
Motor Generator
Electrical Mechanical Mechanical Electrical
energy energy energy energy
We will study the following machines:
 Induction motor
 Synchronous generator and motor
 DC motor
We will also look into transformers – useful in electrical power
distribution.
BUT….
Firstly, we need to look at the basic concepts of electrical
machines:
 Rotational motion and Newton’s Law
 Magnetic field and magnetic circuits
 Principles behind motor, generator and transformer action
 The Linear DC machine
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1
Updated 16 February 2011
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Chapter 1: Intro. to Machinery Principles
1.1. Rotational Motion
Machines rotate on a fixed shaft.
 = Angle of rotation measured from a fixed reference
point.
Unit: radians (rad) or degrees ().
 = Angular velocity
It is analogous to linear velocity, v. Therefore,
(1.1)
Unit: radians per second (rads-1
).
Angular velocity can also be expressed in terms of other
units.
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2

0
ω=
dθ
dt
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f = Angular velocity in revolutions per second.
(1.2)
n = Angular velocity in revolutions per minute.
(1.3)
 = Angular acceleration
It is analogous to linear acceleration, a. Hence,
(1.4)
Unit: radians per second squared (rads-2
).
Torque, T is produced when a force exerts a twisting action on a
body. Unit: Newton-meters (Nm).
Definition:
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3
Perpendicular distance
|T| = Force applied between line of force and
axis of rotor
f =
ω
2π
n=60f =
60ω
2π
=
30
π
α=
dω
dt
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F
r
Direction of
torque
x
T = (force) * (perpendicular distance)
= F  r sin (180 – θ)
= F  r sin θ
= F  x
Work, W is produced from the application of force, F through a
distance, r.
For linear motion:
W=∫F dr
For rotational motion, work = application of torque T through an
angle 
W=∫T dθ
(1.5)
However, when the torque applied is constant,
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4
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W=T∫dθ=Tθ (1.6)
Unit: Joules (J).
Power, P is the rate of doing work.
(1.7)
Unit: Watts (W).
Applying this definition for rotating bodies, and assuming torque is
constant,
P=
dW
dt
=
d
dt
(Tθ)=T(dθ
dt )=Tω
(1.8)
Equation (1.8) is very important!
It describes the mechanical power on the shaft of a motor or
generator.
Real, Reactive and apparent Power in AC Circuits
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5
P=
dW
dt
P=Tω
Z̄=Z∠θ°
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Real power, P = power supplied to load.
Unit: Watts (W)
Reactive power, Q = energy that is stored and released in the
magnetic field of inductor or electric field of capacitor
Unit: Volts-ampere reactive (VAr)
Apparent power, S = power that “appears” to be supplied to load
if phase angle differences between voltage and current are ignored.
Unit: Volt-amperes (VA)
1.2. Newton’s Law of Rotation
For an object moving in a straight line, Newton’s Law is given by:
F=ma
Where:
F = net force applied to the object
m = mass of object
a = resulting acceleration of object
In analogy, Newton’s Law of rotation for a rotating body is given
by:
(1.9)
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6
T=Jα=J
dω
dt
P=VI (cosθ)
Q=VI (sinθ)
S=VI
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(Unit: Nm or kgm2
rads-2
)
Where:
T = net torque applied to the object
J = moment of inertia (unit: kgm2
)
 = resulting angular acceleration of object (unit: rad/s2
)
1.3. The Magnetic Field
The conversion of energy from one form to the other in motors,
generators and transformers is through the presence of the
magnetic field.
The production of a magnetic field by a current carrying conductor
is governed by Ampere’s Law:
∮H⋅dl=I
H is the magnetic field intensity produced by the current I. In SI
units, H is measured in Ampere-turns per meter.
What if we have more than 1 conductor?
Then, use the total current passing through the closed path, i.e. N
turns each carrying the current I. Therefore,
(1.10)
(l = mean path length of the core)
The magnetic field intensity, H can be considered to be a measure
of the “effort” required by the current to create a magnetic field.
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7
Hl=NI
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The relationship between the magnetic field intensity, H and the
produced magnetic flux density, B is given by:
(1.11)
(µ= magnetic permeability of material)
The unit of magnetic flux density is Tesla (T).
In (1.11),  is the permeability of the material in which the
magnetic field is produced. It represents the relative ease of
establishing a magnetic field in a given material.
The permeability is usually written as:
where:
0 = permeability of free space = 4 x 10-7
H/m
r = relative permeability of a given material compared to
free space.
Note:
 Permeability of air = permeability of free space.
 Steels used in modern machines have r of 2000 to 6000.
Finally, we define the magnetic flux, φ present in a given area
by the following equation:
(1.12)
where dA is the differential unit of area. If the flux density, B is
uniform over the cross-sectional area A, then:
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8
B=μH
μ=μ0μr
φ=∫
A
B
¿dA
¿
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(1.13)
The unit of magnetic flux is Webers (W).
Example: The simple magnetic core
1) Obtain an expression for the magnetic field intensity, H
using Ampere’s Law.
∮H
¿dl=NI
¿
Hl=NI
H=
NI
l
2) The magnetic flux density, B of the simple core is:
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9
φ=BA
cross-sectional
area, A
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B=μH
B=
μ NI
l
3) The total flux in the core due to the current I in the winding:
φ=BA
φ=(μ NI
l )A
φ=
μ NIA
l
1.4. Magnetic Circuits
In the simple magnetic core, we find that:
Compare this to:
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10
Current I in the coil
of wire
Magnetic flux  in
the core
Produces
Voltage V in an
electric circuit
Current I flowing in
the circuit
Produces
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I
It is possible to define a magnetic circuit in which magnetic
behaviour is governed by simple equations analogous to those of
an electric circuit.
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11
N turns
Electric Circuit Magnetic Circuit
V=IR F=φ ℜ
(1.14)
V = voltage or electromotive
force (emf)
F = magnetomotive force
(mmf)
I = current  = flux
R = resistance of circuit ℜ = reluctance of circuit
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We have two new quantities for the magnetic circuit:
 magnetomotive force, F
 reluctance, ℜ
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12
φ
Right Hand Rule
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The magnetomotive force, F is equal to the effective current
flow applied to the core, i.e.
(1.15)
The mmf is measured in ampere-turns.
Similar to the voltage source, there is a polarity associated with
the mmf source. This is determined by the flux flow in the
magnetic circuit determined using the ‘right-hand rule’:
“If fingers of the right hand curl in the direction of the current
flowing in a coil of wire, the thumb will point in the direction of
positive mmf.”
Therefore, for the simple magnetic core:
The reluctance ℜ is analogous to resistance R in an electric
circuit. Its unit is ampere-turns per weber
(A. turns/Wb).
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13
F=NI
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Due to the analogy, reluctances in a magnetic circuit obey the same
rules as resistances, i.e.
1. if the reluctances are connected in series:
Req=RT =R1+R2+... Rn
2. if the reluctances are connected in parallel:
1
Req
=
1
RT
=
1
R1
+
1
R2
+...
1
Rn
In order to obtain an expression for the reluctance, we look back at
the flux expression for the simple magnetic core obtained
previously:
(1.16)
We know that in the magnetic circuit:
(1.17)
Hence, the reluctance of a material of length l and cross-sectional
area A is given by:
(1.18)
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14
φ=
μ NIA
l
φ=
F
ℜ=
NI
ℜ =
1
ℜ (N
ℜ=
l
μA
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I
Permeance, P = reciprocal of reluctance (analogous to
conductance G)
P=
1
ℜ
Magnetic circuits assist in analysing magnetic problems. However,
the analysis carried out are approximations due to the following
assumptions employed in the analysis:
1. assumptions in reluctance calculations (mean path length)
2. no leakage flux
3. no fringing effects – cross-sectional area of air gap equals
that of core.
4. Permeability of ferromagnetic materials is usually
assumed to be constant or infinite.
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15
Leakage fluxes present in a simple
magnetic core.
Fringing effects in airgap.
Hence, effective cross-
sectional area of airgap is
larger than cross-sectional
of core in reality.
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Even so, magnetic circuit analysis is the easiest tool for flux
calculations giving satisfactory results.
Example: Magnetic Circuits
A ferromagnetic core is shown in Figure P1-2. The depth of the
core is 5 cm. The other dimensions of the core are as shown in the
figure. Find the value of the current that will produce a flux of
0.005 Wb. With this current, what is the flux density at the top of
the core? What is the flux density at the right side of the core?
Assume that the relative permeability of the core is 1000.
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16
400 turns
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The magnetic circuit:
There are three regions considered for this core. The reluctances of
these regions are:
Since,
R=
l
μ0 μr A , where A= cross section area, and
µ0 = 4π x 10-7
H/m
Therefore,
ℜleft =
l1
μ0 μr A1
=
(7 .5+15. 0+7 . 5)×10−2
μ0 (1000) (5×10−2
)(10×10−2
)
=47.75
¿ kA . turns/Wb
¿
ℜright =
l2
μ0 μr A2
=
(7 . 5+15 .0+7 .5)×10−2
μ0 (1000) (5×10−2
)(5×10−2
)
=95 . 49
¿kA . turns/Wb
¿
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17
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ℜtop=ℜbottom=
l3
μ0 μr A3
=
(5 . 0+20 . 0+2.5)×10−2
μ0 (1000 )(5×10−2
)(15×10−2
)
=58 .36
¿kA .turns /Wb
¿
The total reluctance is thus
ℜT = ℜleft + ℜright + ℜtop + ℜ
bottom
¿¿ ¿¿ ¿ ¿
ℜT =47.75k+95.49k+2(58.36k )
¿ A .turns/Wb
¿
ℜT =201.6
¿kA .turns /Wb
¿
and the magnetomotive force required to produce a flux of 0.005
Wb is
F=φ ℜT =(0.005)×(201.6 k)=1008
¿ A .turns
¿
Hence, the required current is
Since, F=NI
I=
F
N
=
1008
400
=2.52
¿ A
¿
The flux density on the top of the core is
Btop=
φ
Atop
=
0.005
(15×10−2
)(5×10−2
)
=0.67
¿T
¿
The flux density on the right side of the core is
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18
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Bright =
φ
Aright
=
0.005
(5×10−2
)(5×10−2
)
=2
¿T
¿
1.5. Magnetic behaviour of ferromagnetic materials
Typical B-H curve for ferromagnetic materials:
B=μH
Slope of B-H curve = permeability, 
Clearly,   constant in ferromagnetic materials.
After a certain point, increase in mmf gives almost no increase in
flux, i.e. material has saturated.
“Knee” of curve – transition region, operation point for most
electrical machines.
Advantage: get higher B for a given value of H.
The most important ferromagnetic material is iron.
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19
 (or B)
F (or H)
knee
Saturated region
Linear
unsaturated
region
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What will happen when current changes direction (i.e. have
alternating current)?
Path BCDEF
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20
Hysteresis Loop
I
t
A
B
C
D
E
F
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A  assume flux in core is zero at t = 0
A-B  current increases, flux increases as well (as in saturation
curve seen previously).
B-C-D  Current decreases but flux traces different path
D-E-F  Current increases again but flux path doesn’t go through
A as seen before.
When MMF is applied and removed, flux path ABC is traced.
At C, F = 0 but flux  0  Residual Flux, φres
To force flux = zero (i.e. B = 0)  An amount of magnetomotive
force known as coercive magnetomotive force Fc must be applied
to the core in the opposite direction.
This phenomenon is known as Hysteresis.
Why does hysteresis occur?
Need to look at the atomic structure before, during and after the
presence of an external magnetic field.
Before magnetic field is applied:
Within a metal, there are small
regions  called domains
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21
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In each domain, atoms are aligned with a small magnetic field.
But each domain field are
randomly aligned in the
ferromagnetic material.
Appear to have no flux
When external magnetic field is applied:
Domains pointing in direction of field grows.
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22
Example of metal structure
before the presence of a
magnetic field.
Before external field applied After external field is applied
Direction of external field
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Domains in other directions realign to follow external field.
Hence, magnetic field increases!
As more domains align, the total magnetic flux will maintain at a
constant level, i.e. saturation is reached
When magnetic field is removed:
Domains will try to revert to its random state.
But some remain aligned.
Net magnetic field ≠ 0 (remaining residual flux)
Hence, to change alignment such that net field = 0,
must apply energy!
Example:
1. Apply magnetomotive force in the opposite direction, i.e.
coercive mmf, Fc.
2. Exert a large mechanical shock
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23
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3. Heat up the material
So, turning domains in a ferromagnetic structure requires energy.
Energy losses in ferromagnetic core
Two types of losses:
1. Hysteresis loss – energy required to accomplish the
reorientation of domains during each cycle of ac current
applied to the core.
Trajectory of flux built-up in material is different for
increasing and decreasing current applied, i.e. hysteresis
loop.
Every cycle of AC current will drive the material around the
hysteresis loop once.
Energy loss  area enclosed in hysteresis loop.
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24
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2. Eddy current loss – produced by induced currents in the
material. (We will come back to this)
Both losses cause heating of core material and needs to be
considered in machine or transformer design.
Since both occur within the metal core, these losses are lumped
together and called core losses.
Let’s recap….
FACT:
1. Current-carrying wire produces a magnetic field, B.
2. Existence of ferromagnetic material (mainly iron) increases B
and provides easy path for magnetic flux flow.
Electrical machines (motors or generators) and transformers are
devices made up of iron and windings carrying current.
The basic principles behind the operation of these devices are
caused by the effect of magnetic field on its surroundings:
 Effect 1: Presence of a coil of wire in a time-changing
magnetic field induces voltage (transformer action)
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25
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 Effect 2: Force is induced on a current-carrying wire in the
presence of magnetic field (motor action)
 Effect 3: A moving wire in presence of a static magnetic field
induces voltage (generator action)
Lets look closer at each of these effects.
1.6. Effect 1: Faraday’s Law
“Flux  passing through a turn of coil induces voltage eind in it that
is proportional to the rate of change of flux with respect to time.”
Faraday’s Law in equation form:
or for a coil having N turns:
(1.19)
Negative sign – induced voltage acts to oppose the flux producing
it (Lenz’s Law).
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26
eind=−N
dφ
dt
eind=−
dφ
dt
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Assumption for eq. (1.19): same flux present in each of N turns.
Not true in reality (due to leakages)!
Rewrite Faraday’s Law:
(1.20)
where  is flux linkage of the coil:
(1.21)
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27
eind=−N
dλ
dt ,
where λ=Nφ
λ= Σ
i=1
N
φi
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Units: Weber-turns.
Faraday’s Law is the basis of transformer action, i.e. have static
coils (or conductors) in a varying magnetic field.
But Faraday’s Law also applies if you have:
 Moving conductor in a stationary field
 Moving conductor in a varying field
Back to eddy current losses…
Cause: The time-varying flux also induces voltage, hence swirls
of currents to flow, within the ferromagnetic core.
Effect: heat is dissipated by the swirls of current flowing within
the resistive core. Energy loss  size of current paths.
Solution: Lamination of ferromagnetic core, i.e. break up core
into thin strips, separated by insulation to limit the areas in which
eddy currents can flow.
1.7. Effect 2: Induced force on a current carrying wire
Charges moving in a magnetic field experience a force.
If the moving charges are a current flowing in a conductor, a force
acting on the conductor is observed.
General equation for the force induced on the conductor:
(1.22)
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28
l = Length of wire in
the direction of
current flow
B= Magnetic flux
density vector
I= Magnitude of
current in wire
F=I(l×B)
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Hence, force magnitude:
(1.23)
( = angle between conductor and the flux density vector)
Direction of force: Fleming’s left hand rule
For motor application
Or
Note: In the book, the right hand is used but with different fingers
representing different quantities. Do not mix them up! Stick to one
convention and remember only it!
Example: A conductor placed on rails connected to a DC voltage
source in a constant magnetic field.
Since all vectors are
perpendicular:
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29
, B First finger:
SeCond finger:
ThuMb :
F=IlB(sinθ)
I
+
Vdc
-
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F direction to the left
(based on LH Fleming’s Rule)
The force can also be observed as result of the interaction between
the original field and the field created by the current in the
conductor.
This effect is basis of electric motor action, i.e. torque (force)
produced to move the motor.
In electrical motors, construction is such that the windings (i.e.
current) and magnetic field are all acting in perpendicular
directions.
Why? To achieve maximum force!
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30
I
F
F=IlB(sinθ)
F=IlB(sin 90o
)
F=IlB
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1.8. Effect 3: Induced voltage on a moving wire
Now, take the same conductor on rails from the example above.
But, take off the DC voltage source and connect a voltmeter
instead. (Note: The conductor is still placed in a constant magnetic
field region.)
Then, move the conductor to the right. What do we get?
Voltage is induced in the system!
We know:
And in the example:
 φ=BA=B(lx) , where A – total surface area perpendicular
to flux in motion of conductor

eind=N
d(Blx)
dt
=(NBl)
dx
dt
=NBlv
, if N=1
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31
V
+
eind
-
l
I
Velocity, v
Moves x distance
eind=N
dφ
dt
A
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 hence eind=Blv (for generator)
Easy because B is constant and all are in perpendicular
directions!
General equation for the induced voltage:
(1.24)
Note: The value of l is dependent upon the angle at which the wire
cuts through the magnetic field. Hence a more complete formula
will be as follows:
(1.25)
where  = angle between the conductor and the direction of the
(v×B) vector.
Direction of force: Fleming’s right hand rule
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32
First finger: B (Field)
SeCond finger: I (Current)
ThuMb: v (Motion)
B = Magnetic flux
density vector
v = Velocity vector
of the wire
l = Length vector
of wire
eind=(v×B)⋅l
eind =(v×B)l cosθ
eind=(vBsinθ)(lcosθ)
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This effect is basis of generator action, i.e. induction of voltages
in a moving wire located in a magnetic field.
1.9. The Linear DC machine
It operates on the same principles and exhibits the same behaviours
as real generators and motors.
Construction: Conducting bar placed on a pair of smooth,
frictionless rails in a constant, uniform magnetic field.
To investigate its behaviour, 4 basic equations are required:
1. Force production on a wire in the presence of a magnetic field:
(1.26)
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33
F=I(l×B)
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2. Voltage induced on a wire moving in a magnetic field:
(1.27)
3. Kirchoff’s voltage law for the machine:
(1.28)
4. Newton’s law for the bar lying across the rails:
(1.29)
The fundamental behaviour of the simple DC machine will be
examined through three cases.
Case 1: Starting the Linear DC machine
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34
eind=(v×B)⋅l
−VB+IR+eind=0
VB=IR+eind
Fnet =ma
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1. The switch is closed and current is allowed to flow in the bar.
From Kirchoff’s voltage law:
(1.30)
Note: eind = 0 because the bar is at rest.
2. With current flowing downwards in the bar, force is produced
on it.
Direction of movement: Right (use Fleming’s LH Rule)
3. Based on Newton’s law, bar will accelerate to the right. This
motion will cause voltage to be induced across the current-
carrying bar.
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35
VB=IR+eind
VB=IR
Find=I (l×B)
Find=IlB(sin9
Find=IlB
eind=(v×B)⋅l
eind=v(Bsin90o
)(lc
eind=vBl
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Direction of induced voltage: upwards
4. The induced voltage will cause the current flowing to be
reduced. Look back to Kirchoff’s voltage low:
5. This reduction in current will be followed by a decrease in the
force production since
Eventually, |F|=0 .
At which point: I = 0 VB= IR + eind VB = eind
And the bar will move at a constant no-load speed,
Since
eind=Blv=vB ,
v=
vB
Bl at steady state speed
(1.31)
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36
vss=
vB
Bl
VB=IR+eind
I=
VB−eind
R
F=IlB
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The linear dc machine on starting. (a) Velocity v(t) as a
function of time; (b) induced voltage eind(t); (c) current i(t); (d)
induced force Find(t)
Case 2: The Linear DC machine as a motor
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37
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Assume the linear DC machine is running at no-load and under
steady state conditions, i.e. steady state velocity of vss.
Event Outcome
1
Fload applied opposite to the
direction of motion (i.e. to left)
Fnet =Fload−Find
(
Fnet to the left)
2
Negative acceleration (
a=
Fnet
m )
Bar slows down ( i.e. v )
3 Reduction in bar speed (v )
Induced voltage (
eind ) will
decrease, since
eind=vBl
4 Reduction in induced voltage
Increase in bar current, as
I=
V B−eind
R
5 I increases
Force induced/acting on bar
increases to the right, as
F=BIl
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38
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This force will increase until it is equal in magnitude but opposite
in direction to the load force, i.e. F=Fload , which will occur at
a lower speed v.
The force F induced in the bar is in the direction of motion of the
bar and power has been converted from electrical form to
mechanical form to keep the bar moving.
The converted power is:
(1.32)
The bar is operating as a motor because power is converted
from electrical to mechanical form.
________________________________________________________________________
39
Pconsumed= Electrical power consumed
Pconv= Mechanical power
d
Pconsumed=eind I Pconv=Tind ω
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40
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Case 3: The Linear DC machine as a generator
The DC machine is assumed to operating under no-load steady
state conditions.
Event Outcome
1
Fapp applied in the
direction of motion (i.e. to
right)
Fnet is in the direction of
motion
Fnet =Find+Fapp
Find=0 , at initial
2
Positive acceleration,
a=
Fnet
m
Bar speeds up ( i.e. v )
3 Increase in bar speed (v ) Induced voltage will increase.
(
eind↑=Blv↑ )
4 When
eind increases,
eind >
VB
Current increases & reverse
original direction,
I=
eind−VB
R
________________________________________________________________________
41
lOMoARcPSD|6847370

5
I flow in reverse direction &
increases in magnitude.
LH Fleming’s Rule,
Find=IlB (to the left)
Force induced/acting on bar
increases to the left
This will continue until Find=Fapp which will cause the bar to
reach a new steady state and move at a higher speed v.
The reversal of current means that the linear DC machine is now
charging the battery, i.e. it is acting as a generator that converts
mechanical power into electric power.
The amount of power converted is:
Note:
 Same machine can act as both motor and generator.
 Difference lies in the direction of external force applied with
respect to direction of motion.
 In both operations, induced voltage and force are both
present at all times.
 Machine movement is always in the same direction.
________________________________________________________________________
42
Mechanical power
consumed
Electrical power
created
Pconv=Tω=I (eind)
lOMoARcPSD|6847370

Starting problems with the Linear DC machine
As stated previously, when the linear machine is started, there is no
induced emf present, i.e. eind=0 . Therefore, the starting current
is given by
VB=IR+eind :
(1.33)
As we can see, istart ∝1/R .
Typical machines have small R and are supplied with rated V,
therefore the istart will be very high (more than 10 times rated
current).
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43
Istart=
V B
R
=
250
0.1
=2500
¿ A
¿
lOMoARcPSD|6847370

Consequence: Possibility of severe damage to motors.
Solution: insert an extra resistance into the circuit during starting
of motor.
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44
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chapter-1-introduction-to-machinery-principles.pdf

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