PE4 - DC Machines (3).pdf and thier applications in engineering

EN1215 Fundamentals of Integrated Engineering
EN1216 Fundamentals of Electrical & Electroninc Engineering
PE4: DC MACHINES
Dr David Clark
Image: Electricaltechnology.org
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Aims and Objectives
Aims:
▶ To identify and understand the key components of DC motors &
Generators
▶ To evaluate the key operating variables of an electric motor in both
unloaded and loaded conditions
Objectives: By the end of this section, you should be able to:
▶ Describe the basic construction of a DC motor
▶ Explain how torque (T) and back emf (E) are produced in an electric
motor
▶ Calculate the operating variables of a DC motor under a set of
prescribed conditions (Torque, speed, back emf, mechanical power
delivered, electrical power consumed, loss & efficiency)
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4.1 - What is a Motor?
Consider a conductor carrying a DC current in a static magnetic field...
In which direction will the force act upon the conductor? Recall from
electromag that the force per unit length on the wire is:
F′ = I × B
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4.2 - DC Motor: Simple Construction
A DC motor is simply a current-carrying conductor placed in a magnetic
field.
If a DC current is applied to the conductor, a torque is established due to
the accumulated Lorentz force on the moving charges in the conductor,
causing the circuit to rotate.
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4.2 - DC Motor: Simple Construction
You can construct a simple DC motor with the following items:
▶ permanent magnet
▶ enamelled copper wire
▶ 2x Paper clips
▶ battery
Link to DC Motor Video
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4.3 - DC Motor - Current Reversal
...but what happens if we allow this coil to rotate more than 90o?
If the current in the coil cannot reverse, then our motor will STALL in the
upright position. We need some means of reversing the current direction as
the machine rotates.
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4.3 - DC Motor: Current Reversal
To do this, we must use a COMMUTATOR, consisting of:
▶ a rotating metal split-ring, connected to the coil terminals
▶ a pair of carbon brushes, connected to our power source
At the commutator rotates, the direction of the current in the coil is
reversed each half-cycle
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4.4 - Developing an Equivalent Circuit
Some useful definitions:
Back EMF
A conductor passing through a magnetic field induces an emf. This
back-emf, acts in opposition to the applied voltage that causes the motor
to spin. It therefore serves to reduce the current flowing through the coils
of the motor
Torque
The tendency for a force to rotate an object about its axis. It is defined
mathematically as the product of the turning force and its radial
displacement from the axis of rotation. Think about loosening a nut &
bolt - is this easiest with a short or long handled spanner?
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4.4 - Developing an Equivalent Circuit
Consider this balance of electrical power in this simple electrical equivalent
circuit of a dc motor.
▶ i2
aRa is the power loss due to heating
in the armature coil.
▶ The residual power available to
develop mechanical torque is
therefore Eia
V = E + iaRa
∴ Via = Eia + i2
aRa
Pm = Eia
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4.5 - Steady State Mechanical Analysis
Consider a mechanical load, applying a rotational force F at some radial
distance d from the shaft. If the motor spins at speed ω...
Torque: Td = F × d [Nm]
ω = speed [rad · s−1]
Pm = Mechanical Power [W]
Pm = Eia
Pm = Tdω
∴ Eia = Tdω
10 / 36

4.6 - Relating Torque, Current, Speed & Back-EMF
From Faraday’s law, the back emf, E is proportional to the total flux,
Φ [Wb], and the speed of the machine, ω [rad · s−1
].
E = kΦω
The Lorentz force equation tells us that the developed torque, Td is
proportional to the total flux, and armature current, ia [A].
Td = kΦia
For a fixed field (e.g. permanent magnet) kΦ is a constant, its value
is dependent on the construction of the machine.
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Worked Example 4.1
A permanent magnet machine is supplied with 200Vdc and has a free
running (i.e. zero torque) speed of 800 revolutions/min (rpm). Calculate
the kΦ constant. (you may neglect frictional losses in the machine).
At no load, the armature current and the torque is zero. Also:
Eb = kΦω
First, we must convert 800 rpm to rad/sec (SI units):
800 rpm =
800 · 2π
60
= 83.776 rad/s
kΦ =
E
ω
=
200
83.776
= 2.387 Vs/rad
12 / 36

Worked Example 4.2
If the armature resistance is 2Ω, calculate the current taken from the supply
and the motor’s speed when it is called upon to deliver a Torque of 3Nm,
the supply voltage being maintained at 200Vdc.
Ra = 2 Ω
In SI units, the Torque constant is the same as the Speed constant:
Td = kΦia → 3 = 2.387 · ia ∴ ia = 1.257 A
To calculate the speed:
V = E + iaRa
200 = E + 1.257 · 2 → E = 197.486 V
ω =
E
kΦ
=
Eia
Td
=
197.486 × 1.257
3
= 82.74 rad/s
13 / 36

Worked Example 4.3
For the same motor, calculate the current and speed at Torques of:
a) 10 Nm
b) 20 Nm
a)
Td = kΦia → 10 = 2.387 · ia
ia = 4.189 A
200 = E + 4.189 · 2 → E = 191.62 V
ω =
Eia
Td
=
191.62 × 4.189
10
= 80.27 rad/sec
b)
ia = 8.379 A
ω =
Eia
Td
=
183.243 × 8.379
20
= 76.77 rad/sec
14 / 36

4.7 - Building Useful Machines - Maximising Φ
To produce a useful amount of torque, we need to maximise the flux Φ in
the machine. The magnetic circuit should be constructed of a high µr
material (steel), with the smallest practical air gap to minimise reluctance:
Evolution of a motor magnetic circuit from a gapped core
Note how the core laminations are shaped to minimise the length of the gap
between the rotor (the movoing part of the circuit) and the stator (the static
part of the circuit)
15 / 36

4.7 - Building Useful Machines - Maximising Φ
Ideally the armature windings would be as far as possible from the rotor
axis to maximise torque, but placing them on the surface of the rotor iron
would require a very large air gap...
Solution: Armature windings are laid in slots in the rotor, minimising the
reluctance of the magnetic circuit
16 / 36

4.7 - Building Useful Machines - Uniform Torque
To keep the torque constant, the armature windings are arranged uniformly
around the circumference of the rotor. Opposing conductors are energised
using a segmented commutator:
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Building Useful Machines - Field Control
While a permanent magnet (PM) can be used to produce the required
magnetic field, it is more common to see field windings on the stator:
▶ Φ is tunable by adjustment of the field current
▶ We don’t need to worry about demagnetisation
but...
▶ Less efficient than a PM machine - you have copper losses in two sets
of windings!
18 / 36

4.7 - Building Useful Machines - Minimising Losses
Eddy current heating in the core is minimised through the use of
laminations. This minimises the extent of induced current loops in the core,
and I2R losses are reduced accordingly
▶ enables construction of extremely efficient machines
▶ very important for AC motors and generators due to the
presence of alternating fields (more in the next section)
19 / 36

Summary of DC Motors
▶ Knowledge of the Lorentz force and Faraday’s law of induction allow
us to develop an understanding of the operating principles behind
electric motors.
▶ To build a practical DC motor, we require a commutator to reverse the
direction of current every half cycle in order to maintain unidirectional
rotation.
▶ The electrical and mechanical characteristics of a simple DC motor
can be analysed with some knowledge of the operating conditions.
▶ Through careful design, powerful and efficient machines can be build
to meet all manner of applications
20 / 36

4.8 - DC Generators
Consider a moving conductor in a static magnetic field
....In which direction is the current induced in this conductor? Recall from
your electromagnetics lectures that the force acting on a charge moving
with respect to a magnetic field is...
F = q(v × B)
21 / 36

4.8 - DC Generators
Electrical generators come in a wide range of configurations and sizes, but
all exploit the same basic principle.
The fundamental requirements of all generators are:
▶ A magnetic field system
▶ A conductor (or group of conductors)
▶ Motion of the conductor(s) with respect to the magnetic field
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4.8 - DC Generators
Consider the DC motor met previously
The Motor converts Electrical energy from the source into Mechanical
energy to drive the load, plus losses in the form of Heat and Sound.
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4.8 - DC Generators
What if we were able to reverse this process?
Consider the same machine, but rather than applying a source voltage
across the electrical terminals we apply a turning force to the rotor shaft.
This source of Mechanical energy is termed a prime mover. It could be a
rotary handle, a water wheel, steam or wind turbine, or even a motor!
If you place a voltmeter across the open terminals, what would you expect
to see?
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4.8 - DC Generators
The DC motor becomes a generator...
The Generator converts Mechanical energy from the source into Electrical
energy, plus losses in the form of Heat and Sound.
25 / 36

4.9 - Equivalent Circuit of the DC Machine
Compare the equivalent circuits of a simple motor and generator:
Motor: Generator:
▶ In the Motoring case, V > E, and power flows left to right
▶ In the Generating case, E > V and power flows right to left
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4.10 - DC Generator: Pulsed “DC” output
What is the effect of using a split ring commutator on the polarity of the
induced EMF?
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4.11 - General Construction of Machines
DC motors and DC generators are known collectively as “machines”. Their
construction is identical, comprising:
▶ Field System: Generates the magnetic field. This could be a
permanent magnet or a field winding on the stator.
▶ Armature: The rotating electric circuit that carries current “across”
the applied field, generating an EMF.
▶ Split Commutator: In a motor, allows continuous rotation. In a
generator, allows “direct current” to be generated.
▶ Brushes: Provide a conductive path from the fixed electric terminals
to the moving armature conductor(s).
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4.12 - Generating Alternating Currents
Question: how would you construct an AC generator?
The principle is identical, all that we need to do is to change the type of
commutator that we use:
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4.10 Permanent Magnet vs. Wound-Field Machines
The flux Φ in the machine can be produced in a number of ways, as we have
already seen. The simplest machine from an analytical perspective is the
permanent magnet (PM) machine, but we can also produce the flux
electromagnetically.
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4.13 Permanent Magnet vs. Wound-Field Machines
Permanent Magnet Machines
✓ Simple to analyse
✓ More Efficient
X More Expensive
Wound Field Machines
✓ Cheaper
✓ More Flexible
X More Complex Analysis
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4.14 Wound Field Machine Configurations
The Separately Excited Machine gives us independent control of both the
armature current and machine flux. It is the most versatile but requires two
power supplies
Machine flux is proportional to field current Φ = kf If
The Torque developed is T = kaΦIa = kIf Ia
The Back EMF is E = kaΦω = kIf ω (k = kf ka) is the motor constant
And the Armature voltage Va = E + IaRa = kIf ω + TRa/kIa
Rearranging in terms ω we get ω = (1/kIf )Va − (1/kIf )2RaT
Where If = Vf /Rf
32 / 36

In the Series Connected Machine the armature current is routed through
the field winding
In this configuration Ia = If
Machine flux is still proportional to field current Φ = kf Ia
The Torque developed is T = kaΦIa = kI2
a
The Back EMF is E = kaΦω = kIaω (k = kf ka) is the motor constant
And the Supply voltage Vs = E + IaRa = ω
√
kT + (Ra + Rf )
p
T/k
Our torque speed characteristic is of the form T ∝ (1/ω)2
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In the Shunt Connected Machine the armature and field are fed in parallel
from the same supply:
Machine flux is proportional to field current Φ = kf If
The Torque developed is T = kaΦIa = kIf Ia
The Back EMF is E = kaΦω = kIf ω (k = kf ka) is the motor constant
And the Supply voltage Vs = E + IaRa = kIf ω + TRa/kIf
Rearranging in terms ω we get ω = (1/kIf )Vs − (1/kIf )2RaT
Where If = Vs/Rf , giving us ω = Rf /k − (RaR2
f /k2V2
s )T
Thus our torque-speed characteristic is of the form T ∝ −ω
34 / 36

Series DC Machine
T ∝ 1/ω2
✓ Huge starting Torque, suited to
traction and lifting applications
X Speed varies widely with load
torque, not suited to constant
speed applications
X Some load must always be
connected at turn-on - why?
Shunt DC Machine
T ∝ −ω
✓ Can maintain near constant
speed over a wide load range.
Suited to constant speed
applications e.g. conveyors
✓ Speed is easily controllable by
varying either Vs or If
X Low starting torque
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Summary
▶ DC motors and DC generators are identical in their construction and
collectively known as DC machines
▶ DC motors supply direct current to an armature. The Lorentz force
resulting from the motion of the electrons within the static magnetic
field produces a torque that turns the rotor shaft.
▶ DC generators need a prime mover, for example steam or wind, to
rotate the conductor in a magnetic field. The resulting Lorentz force
on the electrons in the conductor gives rise to an electric current when
a load is connected across the terminals.
▶ Wound field machines can be configured to give different performance
characteristics to suit a particular application.
▶ We will look at evaluating a DC machines’s behaviour in the lab
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