Applied Electronics and Electrical Machines

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School of Electrical and Computer Engineering
Applied Electronics and Electrical
Machines
(ELEC 365)
Fall 2015
DC Machines
1
DC Machines
 Key educational goals:
Develop the basic principle of operation of a dc machine. Classify the different
types of dc machines and evaluate their characteristics.
 Reading/Preparatory activities for class
i) Textbook:
Chapter 16: 16.1, 16.2, 16.3, 16.4, 16.5, 16.6, 16.7
2
DC Machines
 Questions to guide your reading and to think about ahead of time.
1. What is operating principle of a dc machine?
3. What are the different hardware components of dc machine?
4. Name one application of a dc generator.
5. What are the different types of dc motors?
6. How do you distinguish a dc series motor from a separately excited dc motor?
7. What is the effect of armature reaction?
3
Classification of motors
4
Introduction
 In a transformer energy is transferred from primary to secondary, form of
energy (electrical) is unchanged.
 In a ROTATING ELECTRIC MACHINE (M/C or m/c), electrical energy is
converted into mechanical form and vice versa:
Electric Motor  Electric energy to Mechanical Energy.
Electric Generator  Mechanical Energy to Electrical Energy.
∴ Electric machines are referred to as “ELECTRO-MECHANICAL ENERGY
CONVERTERS”.
5
Introduction
Electromagnetic conversion
Electrical system
Coupling
Magnetic Field
V, i
Mechanical system
T, n
Conversion of energy from electrical to mechanical form or vice versa results from
the following two electromagnetic phenomena:
1. When a conductor moves in a magnetic field, voltage is induced in the
conductor.
2. When a current-carrying conductor is placed in a magnetic field, the conductor
experiences a mechanical force.
Note: These two effects occur simultaneously.
7
Operating Principle of a DC Machine
The linear m/c – a simple example dc
machine consisting of a conducting
bar sliding on conducting rails.
There are four basic equations to the m/c. These equations are necessary to
understand the above m/c.
1) The equation for the FORCE on a wire in the presence of a MAGNETIC FIELD:
𝐹 =𝑖 𝑙×𝐵
The direction of the force is given by Fleming’s left hand rule .
8
Operating Principle of a DC Machine
2) The equation for the voltage induced in a wire moving in a magnetic field:
eind = (v × B).l
where v = velocity of wire, l = length of conductor in magnetic field with the vector
pointing along the direction of the wire toward the end assumed to be positive.
3) KIRCHOFF’S VOLTAGE LAW:
VB – iR – eind = 0.
VB = iR + eind
Equivalent circuit for the linear m/c.
4. NEWTON’S LAW for the bar across the tracks:
Fnet = ma
9
Fleming’s Left Hand Rule Or Motor Rule
To find the direction of force:
FORE FINGER = MAGNETIC FIELD
900
900
900
FORCE = (ilB).sin(θ)
MIDDLE FINGER=
CURRENT
Where θ is the angle between the wire and the flux density vector. This principle
is basis for MOTOR ACTION. (called Lorentz Force).
10
Fleming’s Right Hand Rule Or Generator Rule
To find the direction of induced emf: right hand rule is applicable.
FORE FINGER = MAGNETIC FIELD
900
900
900
MIDDLE FINGER =
INDUCED VOLTAGE
eind (induced emf) = vBlcos(θ)
θ = smallest angle between direction of wire l and (v × B) .
11
STARTING THE DC MACHINE
To start the m/c, close the switch.
(i) A current flows through the bar. Since the bar is initially at rest, eind = 0;
𝑖=
𝑉𝐵 − 𝑒𝑖𝑛𝑑 𝑉𝐵
=
𝑅
𝑅
(𝐾𝑉𝐿)
13
STARTING THE DC MACHINE
(ii) ∴ A current flows through a wire in the presence of a MAGNETIC FIELD and
hence a FORCE is induced on the wire.
Find = ilB
TO THE RIGHT.
(iii) ∴ Bar will accelerate to the right (by Newton’s Law). As velocity of bar
increases,
eind = vBl
(POSITIVE UPWARD)
(iv) This voltage reduces the current flowing through the wire (bar):
𝑖=
𝑉𝐵 − 𝑒𝑖𝑛𝑑 ↑
𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒 (↓)
𝑅
14
STARTING THE DC MACHINE
(v) ∴ Force induced Decreases. [since F = (i ↓)lB ]. The result is finally a steady state
speed (vss) is reached where the net force on the bar is zero.
i.e. F = (i ↓)lB until F = 0.
At that point eind = VB , i = 0 and the bar moves at a constant speed (NO-LOAD):
𝑉𝐵
𝑣𝑠𝑠 =
[𝑠𝑖𝑛𝑐𝑒 𝑉𝐵 = 𝑒𝑖𝑛𝑑 = 𝑣𝑠𝑠 𝐵𝑙]
𝐵𝑙
15
THE DC LINEAR MACHINE AS A MOTOR
DC Linear m/c as a motor: force applied in the opposite direction of motion.
Assume that the linear m/c is initially running at the no-load steady-state
conditions. Now if an external load is applied to it in a direction opposite to the
direction of motion, what happens?
16
THE DC LINEAR MACHINE AS A MOTOR
(i) Due to application of Fload, net force on the bar will be in the direction opposite
to the direction of motion,
Fnet = Fload – Find
(ii) The resulting acceleration
𝑎=
𝐹𝑛𝑒𝑡
𝑚
is negative, so the bar slows down, i.e. v ↓ (v decreases).
(iii) ∴ eind = (v ↓)(Bl) falls, ∴
(iv) Induced force Find = (I ↑)lB increases until│ Find │=│ Fload │and bar runs at a
lower speed (v).
𝑖=
𝑉𝐵 − 𝑒𝑖𝑛𝑑 ↓
↑
𝑅
(v) Electrical power, Pconv = eind.i is converted to mechanical power Find.v, and the
m/c is acting as a motor. [Behavior is similar to a DC shunt motor].
17
THE DC LINEAR MACHINE AS A GENERATOR
Assume again that the linear dc m/c is initially running at no-load steady-state
conditions. Now apply a force in the direction of motion.
DC Linear m/c as a generator: force applied in the direction of motion.
18
THE DC LINEAR MACHINE AS A GENERATOR
(i) Due to application of 𝐹𝑎𝑝𝑝 ; 𝐹𝑛𝑒𝑡 is in the direction of motion.
(ii) acceleration 𝑎 =
𝐹𝑛𝑒𝑡
𝑚
is positive, so the bar speeds up (𝑣 ↑).
(iii) 𝑒𝑖𝑛𝑑 = 𝑣 ↑ (𝐵𝑙) increase ( 𝑒𝑖𝑛𝑑 > 𝑉𝐵 ),
𝑒𝑖𝑛𝑑 ↑ −𝑉𝐵
𝑖=
↑.
𝑅
(iv) 𝐹𝑖𝑛𝑑 = 𝑖 ↑ 𝑙𝐵 increases until 𝐹𝑖𝑛𝑑 = 𝐹𝑙𝑜𝑎𝑑 and bar runs at a higher
speed v.
(v) Amount of mechanical power = 𝐹𝑖𝑛𝑑 . 𝑣, is converted to electrical power,
𝑃𝑐𝑜𝑛𝑣 = 𝑒𝑖𝑛𝑑 . 𝑖 and the m/c is acting as a generator.
[Behavior is similar to a DC shunt generator].
19
SUMMARY
(1) Same m/c acts as both generator and motor. Only difference is whether the
externally applied force is in the direction of motion (generator) or opposite the
direction of motion (motor).
(2) eind > VB (GENERATOR).
eind < VB (MOTOR).
(3) Whether motor or generator, Find (motor action) and eind (generator action) are
present at all times.
(4) Whether m/c was motor (moved slowly) or generator (moved rapidly), it
always moved in the same direction.
20
Basics Structure of DC Machines
21
Basics Structure of DC Machines
22
Basics Structure of DC Machines
Armature of a DC Machine
23
24
Operating Principle of Rotating DC Machine
26
Operating Principle of Rotating DC Machine
27
Action of a Commutator
28
Action of a Commutator
Generally, the commutator segments are copper bars insulated from one another
and from the shaft. The brushes contain graphite that lubricates the sliding contact.
Significant disadvantage of dc machines?
30
EMF EQUATION OF A GENERATOR
Let
φ = flux/pole
Weber (Wb)
Z = Total number of armature conductors.
= (No. of slots) × (No. of conductors/slot)
P = Number of Generator poles (Even number)
a = Number of parallel paths in the armature winding.
𝐸𝑎 = Average voltage induced
N = Armature speed
(V)
Revolutions per second (rps)
Ω = Angular velocity, = 2πN
 Speed regulation :
(Rads/s.)
Ω𝑛𝑜−𝑙𝑜𝑎𝑑 −Ω𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑
Ω𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑
∗ 100%
33
EMF EQUATION OF A GENERATOR
 The voltage induced in armature:
𝐸𝑎 = 𝐾𝜙Ω
𝑉
Where K is a machine constant which depends on the design parameters of the
machines.
Where 𝐾 =
𝑍𝑃
2𝜋𝑎
 Torque equation
𝑇𝑑 = 𝐾𝜙𝐼𝑎
Nm.
𝑬𝒂 𝑰𝒂 is the air-gap power of the machine.
𝐸𝑎 𝐼𝑎 = (KɸΩ) 𝐼𝑎
𝑇𝑑
= (𝐾𝜙Ω)
𝐾𝜙
= (𝑇𝑑 ) Ω = 𝑃𝑑
Watts
Watts
Where Ω is in Rads/s.
This is the gross mechanical power in Watts.
34
CLASSIFICATION OF DC MACHINES
Based on excitation.
I.
Permanent magnet:
•
No field windings.
•
Usually, fractional horsepower rating m/c.
•
With the development of permanent magnet materials, higher power
available.
35
Separately Excited DC Machine
II.
(1) Separately excited: No electrical interconnection between the field and
armature winding.
V
Vt = Ea + IaRa
= KφΩ + IaRa
Ω = Vt
− Ia Ra
Kφ
rads/sec.
separately excited DC motor.
36
Shunt excited
(2) Shunt excited (Self): Parallel connection between the field and armature
winding.
Shunt excited DC m/c.
38
Series excited
(3) Series excited: Series field winding is connected in series with the armature
and therefore, carries the same current as the armature.
Series excited DC m/c.
39
Compound excitation
(4) Compound excitation:
Short shunt: Cumulative Compound
Long Shunt: Differential Compound.
40
MAGNETIZATION CURVES
constant speed
Circuit diagram to vary the field current (in turn flux) to obtain the
magnetization curve.
Relation between Speed and backemf:
Air-gap line approximates the magnetization curve.
41
Armature Reaction (AR)
In a DC machine, two kinds of magnetic
fluxes are present; 'armature flux' and 'main
field flux'. The effect of armature flux on the
main field flux is called as armature
reaction.
• AR is the magnetic field produced by the
armature current
• AR aids the main flux in one half of the
pole and opposes the main flux in the
other half of the pole.
• However due to saturation of the pole
faces the net effect of AR is demagnetizing.
42
COMPENSATING WINDINGS
One way to reduce the effects of armature reaction is to connect winding called
“compensating winding" in series with armature windings so that it’s mmf is
proportional to the armature mmf.
These pole face windings are so arranged that
the mmf produced by currents flowing in these
windings opposes the armature mmf. These pole
face windings are expensive. Therefore they are
only used in large machines or in machines that
are subjected to abrupt change of armature
current
43
Interpoles
Another way to reduce the effects of armature reaction is to place small auxiliary
poles called "interpoles" between the main field poles. The interpoles have a few
turns of large wire and are connected in series with the armature. Interpoles are
wound and placed so that each interpole has the same magnetic polarity as the
main pole ahead of it, in the direction of rotation.
44
EQUIVALENT CIRCUIT OF A DC GENERATOR
J = Polar moment of inertia (Kg.m²)
𝑣𝑓 = 𝑅𝑓 𝑖𝑓 + 𝐿𝑓
𝑑𝑖𝑓
𝑣𝑡 = 𝐾𝜙Ω − 𝐿𝑎
V
𝑑𝑡
𝑑𝑖𝑎
𝑑𝑡
− 𝑅𝑎 𝑖𝑎
𝑑Ω
V
𝑇𝑠ℎ𝑎𝑓𝑡 = 𝐾𝜙𝑖𝑎 + 𝐽 𝑑𝑡 + 𝑇𝑙𝑜𝑠𝑠
N.m.
These equations are used under
transient conditions.
Under STEADY-STATE:
𝑣𝑓 = (𝑅𝑓 )(𝐼𝑓 )
V
𝑣𝑡 = 𝐾𝜙Ω − 𝑅𝑎 𝐼𝑎
Equivalent circuit of a separately excited generator.
V
𝑇𝑠ℎ𝑎𝑓𝑡 = 𝐾𝜙𝐼𝑎 + 𝑇𝑙𝑜𝑠𝑠
N.m.
45
PERFORMANCE OF GENERATORS
(a) External characteristics [potential difference (pd) Vt versus load current, iL].
(b) Efficiency.
(c) Voltage Regulation = [(No Load Voltage) – (Full Load Voltage)]/[Full load Voltage].
1) Separately Excited Generator
Separately excited DC generator. Re is the external resistance added in the field circuit.
46
PERFORMANCE OF GENERATORS (Separately Excited)
Steady-state:
𝐼𝑓 =
𝑉𝑓𝑠
(𝑅𝑒 +𝑅𝑓 )
𝐸𝑎 = 𝐾𝜙Ω
A
V
𝑉𝑡
𝐼𝐿
Ω
𝐼𝐿 = 𝐼𝑎
A
𝐸𝑎 = 𝑉𝑡 + 𝐼𝑎 𝑅𝑎
V
𝑉𝑡 = 𝐸𝑎 − 𝐼𝑎 𝑅𝑎 = 𝐸𝑎 − 𝐼𝐿 𝑅𝑎
V
𝑅𝐿 =
𝐼𝑎 = 𝐼𝐿 =
𝐸𝑎
(𝑅𝑎 +𝑅𝐿 )
=
𝐾𝜙Ω
(𝑅𝑎 +𝑅𝐿 )
A
External characteristic of a separately excited generator.
47
PERFORMANCE OF GENERATORS (Shunt)
2) Shunt Generator (self-excited)
When m/c is driven under NO LOAD:
A small Ear due to residual magnetism appears at the terminals. This voltage
circulates a current through field winding which in turn increases generated voltage
… (cumulative build up) … until a voltage Ea is reached, where: Ea = If.Rf = Vt
(Assume Ra is neglected)
Schematic of shunt dc machine.
Voltage buildup in a shunt dc generator.
48
PERFORMANCE OF GENERATORS (Shunt)
Under NO LOAD:
Ea = (Ra + Rfc + Rfw)If
Vt = (Rfc + Rfw)If
Vt ≅ Ea = F(If)
Shunt generator under no-load.
When the m/c is LOADED:
Vt = (RL)IL
Vt = Ea - (Ia Ra)
Ia = IL + If
Shunt generator with a load.
49
PERFORMANCE OF GENERATORS (Shunt)
Effect of field resistance.
50
PERFORMANCE OF GENERATORS (Shunt)
A shunt generator can not build up if:
(a) There is no residual magnetism,
(b) If the field is connected in a wrong-way around, opposing the permanent
magnetism, &
(c) The value of the field resistance is greater than a value called critical
resistance.
51
Example 1
 A shunt generator delivers 450 A at 230 V and the resistances of the shunt field
and armature are 50 Ω and 0.03 Ω, respectively. Calculate the generated emf.
Field current, 𝐼𝐹 =
230
50
= 4.6 𝐴
Load current, IL = 450 A with
load voltage, Vt = 230 V.
∴ Armature current,
Ia = IL + IF = 450 + 4.6 = 454.6 A.
∴Generated emf, Ea = Vt + IaRa = 230 + (454.6)(0.03) = 243.6 V.
54
Example 2
 A separately excited generator is rated at 10 kW, 200 V, 1000 rpm. Rf = 80 Ω,
Ra= 0.4 Ω and Ia = 50 A. Predict the no-load voltage at 1000 rpm and the fullload voltage at 800 rpm if the field current is kept constant.
55
Example 2
At no-load, Ia = 0 and Vt = Ea = VNL
∴ VNL = Ea = VFL + IaRa = 200 + (50)(0.4) = 220 V.
If the field current is constant, the generated emf Ea is directly proportional to
speed; therefore, at 800 rpm, induced (or generated voltage) E’a is given by
The terminal voltage at 800 rpm under full-load (Ia = 50 A) is
56
EQUIVALENT CIRCUIT OF A DC MOTOR
Illustrated below for a separately excited
motor.
J=Polar moment of inertia (Kg.m²)
𝑣𝑓 = 𝑅𝑓 𝑖𝑓 + 𝐿𝑓
𝑑𝑖𝑓
𝑑𝑡
𝑑𝑖
𝑣𝑡 = 𝐾𝜙Ω + 𝐿𝑎 𝑎 + 𝑅𝑎 𝑖𝑎
𝑑𝑡
𝑑Ω
𝑇𝑙𝑜𝑎𝑑 = 𝐾𝜙𝑖𝑎 − 𝐽 − 𝑇𝑙𝑜𝑠𝑠
𝑑𝑡
V
V
N.m.
These equations are used under
transient conditions.
Under STEADY-STATE:
𝑣𝑓 = (𝑅𝑓 )(𝐼𝑓 )
𝑣𝑡 = 𝐾𝜙Ω + 𝑅𝑎 𝐼𝑎
𝑇𝑙𝑜𝑎𝑑 = 𝐾𝜙𝐼𝑎 − 𝑇𝑙𝑜𝑠𝑠
𝐸𝑎 = 𝐾𝜙Ω = F 𝑖𝑓 │
V
V
N.m.
Ω= ΩR
Equivalent circuit of a separately excited dc motor.
58
Shunt Connected DC Motor
A shunt-connected motor is similar to a separately excited dc motor except that:
Speed versus torque characteristic for a shunt motor.
59
Series Connected DC Motor
Since, φ is a function of Ia, for light loads,
speed becomes very high. Therefore,
SERIES MOTORS MUST NEVER BE STARTED
WITHOUT ANY LOAD.
Speed versus torque for a series motor.
60
Torque – Speed relationship in Series DC Motor
Assumption:
Apply Kirchhoff's Voltage Law to equivalent circuit:
61
Compound Excited DC Motor
• If the shunt and series field aid each other it is called a cumulatively excited
machine.
• If the shunt and series field oppose each other it is called a differentially excited
machine.
63
Speed – Torque characteristics of different machines
Torque – speed characteristics of different dc motors.
64
permanent-magnet DC Motor
In a permanent-magnet (PM) dc motor, the field is supplied by magnets mounted on the
stator rather than by field coils. PM motors are common on fractional or subfractional
horsepower sizes.
65
Compound Excited DC Motor
Advantages:
• No power is required to establish the field leading to better efficiency.
• Smaller
• cheaper
Disadvantages:
• Magnets can become demagnetized by overheating or because of excessive
armature currents.
• The flux density magnitude is smaller than in wound- field machines.
Consequently, the torque produced per ampere of armature current is smaller
in PM motors than in wound- field motors with equal power ratings.
66
Comparison
Motor Types
Separately excited
motors
DC shunt motor
DC series motor
Permanent magnet
DC motor
Characteristics
Can be controlled either by varying
the voltage applied to the field
winding or by varying the voltage
applied to the armature; can
produce high torques
Constant flux constant speed motor
creates moderate torque at start
Power
Range
Up to
100 hp
Up to
200 hp
Creates dangerously high torques at
low speeds and should be always
connected to the load; speed can be
varied
Up to
200 hp
Higher efficiency, smaller size and
simpler architecture; magnets can
become demagnetized due to
excessive use and overheating;
produces lower torque
Up to
10 hp
Applications
Traction applications, to control
the speed and torque of the
motor by changing both
armature
voltage and stator current
Machine tools like lathes, milling
machines, grinding machines,
centrifugal and reciprocating
pumps, blowers, and fans
Preferred for traction-type
loads; employed in electric
locomotives, conveyors, cranes,
elevators, trolleys
Power windows in automobiles,
computer peripherals
67
Example 5
 A 50-hp shunt-connected dc motor has the magnetization curve shown in
Figure. The dc supply voltage is VT = 240 V, the armature resistance is RA =
0.065Ω , the field resistance is RF = 10 , and the adjustable resistance is Radj =
14Ω . At a speed of 1200 rpm, the rotational loss is Prot = 1450 W. If this motor
drives a hoist that demands a torque of Tout = 250 Nm independent of speed,
determine the motor speed and efficiency.
70
Example 5
The equivalent circuit is shown in Figure. The field current is given by
Next, we use the magnetization curve
to find the machine constant K for this
value of field current.
Assuming constant torque for the rotational loss.
Thus, the developed torque is
71
Example 5
Then, applying Kirchhoff s voltage law to the armature circuit, we have
To find efficiency, we first compute the output power and the input power, given by
72
Speed Control
1 - Varying armature terminal voltage (Vt) while holding the field constant.
2 - Varying field current (If ) and therefore φ.
3 - Varying Ra (Insert resistance in series with the armature circuit.)
OR combination of these.
 Speed Control by varying Ra
By inserting additional resistance in series with armature circuit we can control the
speed of DC machine. This method is applicable for various type of DC machine.
73
Speed Control by varying Ra
• Add external resistance Rd in series with the armature.
• INEFFICIENT method due to losses in added resistance.
Speed versus Torque characteristics of a separately excited DC
motor with armature resistance control.
74
Speed Control by Variation of Terminal Voltage Vt
P0  No-load operating point.
P1  Operating point for a given load.
φ1 = Maximum (or rated) flux for which the
m/c is designed.
Speed versus Torque characteristics of a separately
excited DC motor with voltage
control.
75
Speed Control by varying If
Speed control above base (rated) value is obtained by reducing VF and therefore
If and called FIELD WEAKENING.
Speed versus Torque characteristics of a separately
excited DC motor with field control.
76
Speed Control by varying If
Vt Control: Speed control from zero to base speed (Ω𝑏 ) with rated field current.
Over this speed range, Ia must normally be limited to its rated value, and hence
developed torque will also be limited,
Since
T = KɸIa
Nm.
P = ΩT
W.
[Vt is limited to rated value because of commutator insulation and because of the
limit on the available supply, further speed increase requires field weakening].
For negligible, 𝑅𝑎 ,
1
Ω∝
𝜙
𝑟𝑎𝑑𝑠
.
𝑠
1
𝑇 = 𝐾𝜙𝐼𝑎 ∝
[since 𝐼𝑎 is limited to rated value ]
Ω
𝑃 = Ω𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑊 .
77
Early variable DC system: Ward Leonard
1- three-phase induction motor drives a dc generator.
2- Produced dc voltage supplies to the motor to be controlled.
79
Starting of DC Motors
Ea = KφΩ
∴ Ia =
Vt = Ea + IaRa
(𝑣𝑡− 𝐸𝑎 )
𝑅𝑎
Under standstill (at starting), Ω = 0, i.e., Ea = 0 and
∴ Ia =
𝑣𝑡
𝑅𝑎
Therefore, at starting, very high current (since Ra is very small) flows through the
armature windings. Therefore, limit starting current by inserting series resistance
or applying reduced voltage. As speed picks up, Ea increases, reduce the
resistance. This large current is approximately seven times greater than actual fullload current for motor. The high current would, in all probability, cause severe
damage to
the
brushes, commutator,
or
windings. Starting
resistors
are usually incorporated into the motor design to limit starting current to 125
to 200 percent of full load current .
80
Losses and efficiency of DC machines
Power flow in a dc
generator
Power flow in a
shunt-connected dc motor
81
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