Transcript AC Machine - Portal UniMAP

EMT 462
ELECTRICAL
SYSTEM
TECHNOLOGY
Chapter 2:
AC Machines
By:
En. Muhammad Mahyiddin Ramli
AC Machines: Introduction
2 major classes:
a)
Asynchronous machines / induction machines :–
Motors or generators whose field current is supplied by magnetic
induction (transformer action) into their field windings.
b)
Synchronous machines :–
Motors or generators whose field current is supplied by a separate dc
power source.
Note:
1) Induction motor has the same physical stator as a synchronous machine, with a
different rotor construction.
2) The fields circuit of most synchronous and induction machines are located on
their rotors.
Motors = ac electrical energy  mechanical energy
Generators = mechanical energy  ac electrical energy
Chap 2: AC Machines
2
AC Machinery Fundamentals
A SIMPLE LOOP IN A UNIFORM MAGNETIC FIELDS.



A rotating loop of wire within the magnetic field.
Magnetic field produced by a large stationary magnet produce-constant
and uniform magnetic field, B.
Rotation of the loop induced a voltage in the wire.
V
eind
ө

Current flows in the loop, a torque will be induced on the wire loop.
eind = Фmax ω sin ωt
• Thus, the voltage generated in loop is a sinusoid whose magnitude is
equal to the product of the flux and rotation speed of the machine
Chap 2: AC Machines
3
AC Machinery Fundamentals
THE ROTATING MAGNETIC FIELD

When two magnetic fields are present in a machine, a torque will be
created which will tend to line up the two magnetic fields.

Magnetic field is produced by the stator and rotor of an ac machine.

Then a torque will be induced in the rotor cause the rotor to turn and
align itself with the stator magnetic field.

The induced torque in the rotor would cause the rotor to constantly “
chase “ the stator magnetic field around in circle
- the basic principle of all ac motor operation.
But how the stator rotate?
Chap 2: AC Machines
4
AC Machinery Fundamentals
AC MACHINE POWER LOSSES
The efficiency of an AC machines is defined as:
Pout

X 100%
Pin

Pin  Ploss
X 100%
Pin
Four types of losses in AC machines:




Electrical or copper losses (I2R losses)
Core losses
Mechanical losses
Stray load losses
Chap 2: AC Machines
5
AC Machinery Fundamentals
VOLTAGE REGULATION AND SPEED REGULATION
VR is a measure of the ability of a generator to keep a constant voltage at
its terminals as load varies. It is defined as follow:
VR 
Vnl  V fl
V fl
X 100%
SR is a measure of the ability of a motor to keep a constant shaft speed as
load varies.
SR 
N nl  N fl
N fl
X 100%
nl   fl
SR 
X 100%
 fl
Chap 2: AC Machines
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INDUCTION
MOTOR
Chap 2: AC Machines
7
Induction Motors
Induction motors are the motor frequently encountered in industry.
It simple, rugged, low-priced and easy to maintain.
It run essentially constant speed from zero to full-load.
The speed is frequency-dependent and consequently these motors are not
easily adapted to speed control
Induction machines is called induction because the rotor voltage (which
produces the rotor current and the rotor magnetic field) is induced in the rotor
winding rather than physically connected by wires.
Chap 2: AC Machines
8
INDUCTION MOTOR CONSTRUCTION
A 3-phase induction motor has two main parts :
• A stationary stator (stationary part of the machine)
• Revolving rotor (rotating part of the machine)
The rotor is separated from the stator by a small air gap (the tolerances is
depending on the power of the motor).
Chap 2: AC Machines
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Two types of rotor which can placed inside the stator.
a) Squirrel-cage induction motor (also called cage motors)
b) Wound-rotor induction motor
a) Cage rotor
b) Wound rotor induction motor
Wound rotor induction motors more expensive- maintenance
Chap 2: AC Machines
10
Two types of rotor which can placed inside the stator.
a) Squirrel-cage induction motor (also called cage motors)
b) Wound-rotor induction motor
a) Squirrel cage – the conductors would look like one of
the exercise wheels that squirrel or hamsters run on.
Chap 2: AC Machines
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Two types of rotor which can placed inside the stator.
a) Squirrel-cage induction motor (also called cage motors)
b) Wound-rotor induction motor
b) Wound rotor – have a brushes and slip ring at the end
of rotor. (Y-connected)
Chap 2: AC Machines
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INDUCTION MOTOR CONSTRUCTION
Cage induction Motor rotor consists of a series of conducting
bars laid into slot carved in the face of rotor and shorted at
either end by large shorting ring
A wound rotor has a complete set of three-phase winding that are
mirror images of the winding on the stator.
The three phases of the rotor windings are usually Y-connected, the
end of the three rotor wires are tied to slip ring on the rotor shaft.
Rotor windings are shorted through brushes riding on the slip rings.
Wound-rotor induction motors are more expensive than the cage
induction motors, they required much more maintenance because
the wear associated with their brushes and slip rings.
Chap 2: AC Machines
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INDUCTION MOTOR CONSTRUCTION
Small cage rotor induction motor
Large cage rotor induction motor
Chap 2: AC Machines
14
INDUCTION MOTOR CONCEPT
INDUCED TORQUE IN AN INDUCTION MOTOR
The speed of the magnetic field’s rotation in a cage rotor induction motor
(Figure 7.6, Chapman) is given by:
nsync
Where
120 f e

P
nsync = synchronous speed [r/min]
fe
= System frequency [Hz]
p
= number of poles
This equation shows that the synchronous speed increases with frequency and
decrease with the number of poles.
The three-phase of voltages has been applied to the stator, and three-phase set
of stator current is flowing . These currents produce a magnetic field BS ,
rotating counterclockwise direction.
Chap 2: AC Machines
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Basic Induction Motor Concepts
This rotating magnetic field BS passes over the rotor bars and induces a
voltage, eind in them:
eind  (v  B)  l
where,
v = velocity of the bar relative to the magnetic field
B = magnetic flux density vector
l = length of conductor in the magnetic field
•It is the relative motion of the rotor compared to the stator magnetic
field that produces induced voltage in a rotor bar. The rotor current
flow produces a rotor magnetic field, BR.
•The induce torque in the machine is given by:
 ind  kBR  BS
•The voltage induced in a rotor bar depends on the speed of the rotor
relative to the magnetic fields
Chap 2: AC Machines
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Basic Induction Motor Concepts
THE CONCEPT OF ROTOR SLIP
Slip speed is defined as the differences between synchronous speed and rotor
speed:
nslip  nsync  nm
The other term used to describe
the relative motion is slip, which
is relative speed expressed on a
per unit or a percentage basis.
The slip is defined as :
Where
s
s
nslip = slip speed of the machines
nsync = speed of the magnetic field
nm = mechanical shaft speed of motor
nslip
nsyns
 100%
nsyns  nm
nsyns
Chap 2: AC Machines
 100%
17
Basic Induction Motor Concepts
•The previous equation also can be expressed in term of angular velocity 
(radians per second) as :
sync  m
s
100%
sync
•If the rotor turns at synchronous speed, s=0 ; if the rotor is stationary (locked
or stop) , s=1. All normal motor speeds fall somewhere between those limits.
•As for mechanical speed
nm  (1  s)nsync
m  (1  s)sync
•These equation are useful in the derivation of induction motor torque and
power relationship.
Chap 2: AC Machines
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Basic Induction Motor Concepts
THE ELECTRICAL FREQUENCY ON THE ROTOR.
 The induction motor works by inducing voltages and current in the rotor
of the machine-called a rotating transformer.
 Like a transformer; primary (stator) induced a voltage in the secondary (rotor)
 Unlike a transformer, the secondary frequency not necessarily the same as
primary.
 If the rotor of a motor is locked so that it cannot move, the rotor will have
the same frequency as the stator.
 If the rotor turns at synchronous speed, the frequency on the rotor will be
zero.
 For nm=0 r/min & the rotor frequency fr=fe  slip, s = 1
nm=nsync & the rotor frequency fr=0  slip, s = 0
- For any speed in between, the rotor frequency is directly proportional to
the difference between the speed of the magnetic field nsync and the speed
of the rotor nm.
Chap 2: AC Machines
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Basic Induction Motor Concepts
THE ELECTRICAL FREQUENCY ON THE ROTOR
Since the slip of the rotor is defined as :
Then the rotor frequency can be expressed as :
s
f r  sf e
Substituting between these two equation become :
But nsync = 120fe/P, so
fr 
P
f r  (nsync  nm )
fe
120 f e
Therefore,
nsync  nm
nsync
nsync  nm
nsync
fe
P
fr 
(nsync  nm )
120
fr = frequency rotor; fe = frequency stator
Chap 2: AC Machines
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Example 2.1: Induction Motor
A 208-V, 10-hp, four-pole, 60-Hz, Y- connected induction motor has a
full-load slip of 5%.
a) What is the synchronous speed of this motor?
b) What is the rotor speed of this motor at the rated load?
c) What is the rotor frequency of this motor at the rated load?
d) What is the shaft torque of this motor at the rated load?
- Taken from Chapman’s Book, Example 7-1, pg. 387.
Chap 2: AC Machines
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The Equivalent Circuit of An Induction Motor
a) Transformer model
-
model of the transformer action-induction of voltages and
currents in the rotor circuit of an IM is essentially a transformer
operation.
as in transformer model – certain resistance, self inductance in
primary (stator) windings; magnetization curve and etc.
-
b)Rotor circuit model
-
The greater the relative motion between the rotor and the stator
magnetic fields, the greater the resulting rotor voltage and
frequency.
Locked-rotor or blocked-rotor –the largest relative motion when the
rotor is stationary.
c) Final Equivalent circuit
- Refer the rotor part of the model over the stator side.
Chap 2: AC Machines
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The Equivalent Circuit of An Induction Motor
A) TRANSFORMER MODEL
Symbol
aeff
Description
IDEAL TRANSFORMER
STATOR
Effective turn ratio – ratio of the
conductors per phase on the stator
to the conductors per phase on the
rotor
ROTOR
Xm
Resistance losses (correspond to
iron losses, windage and friction
losses)
E1
Primary internal stator voltage
R1
Stator Resistance
ER
Secondary internal rotor voltage
X1
Stator Leakage Reactance
RR
Rotor Resistance
Rc
Magnetizing reactance
XR
Rotor Reactance
Chap 2: AC Machines
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The Equivalent Circuit of An Induction Motor
A) TRANSFORMER MODEL
•Induction motor operates on the induction of voltage and current in its
rotor circuit from the stator circuit (transformer action).
•An induction motor is called a singly excited machine, since power is
supply to only the stator circuit.
•The flux in the machine is related
to the integral of the applied
voltage E1.
•The curve of magnetomotive force
versus flux (magnetization curve)
for this machine is compared to a
similar curve for a power
transformer.
Chap 2: AC Machines
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The Equivalent Circuit of An Induction Motor
B) THE ROTOR CIRCUIT MODEL
Suppose the motor run at a slip s, meaning that the rotor speed is ns (1-s),
where ns is the synchronous speed, then this modify the values of VOLTAGE
and CURRENT on the primary and secondary side.
The frequency of the induced voltage at any slip will be given
fr = sfe
Assuming ER0 is the magnitude of the induced rotor voltage at LOCKED
ROTOR condition the actual voltage induced because of slip (s) is,
ER = sER0
The resistor is not frequency sensitive, the value of RR remain the same.
The rotor inductance is frequency sensitive (X=L=2fL) then
XR = sXR0
Figure 6 shows the equivalent circuit when motor is running at a slip (s).
Chap 2: AC Machines
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The Equivalent Circuit of An Induction Motor
B) THE ROTOR CIRCUIT MODEL
Equivalent circuit of a wound-rotor when it at locked or blocked condition
The frequency of the voltages and currents in the stator is f, but the frequency of the
voltages and currents in the rotor is sf.
Chap 2: AC Machines
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The Equivalent Circuit of An Induction Motor
B) THE ROTOR CIRCUIT MODEL
jXR=jsXR0
Then, resulting rotor equivalent circuit as below.
The rotor current flow can be found as :
IR 
ER
RR  jX R
ER = sER0
ER
IR 
RR  jsX R 0
RR
The rotor circuit model of
an induction motor
sE R 0
IR 
RR  jsX R 0
IR 
ER 0
RR
s
 jX R 0
ZReq
Chap 2: AC Machines
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The Equivalent Circuit of An Induction Motor
B) THE ROTOR CIRCUIT MODEL
Then the rotor equivalent circuit become:
jsXR0
IR 
ER 0
RR
s
ER0
 jX R 0
ZReq
RR
s
The rotor circuit model with all
the frequency (slip) effects
concentrated in resistor RR
Chap 2: AC Machines
28
The Equivalent Circuit of An Induction Motor
C) THE FINAL EQUIVALENT CIRCUIT
•Remember, in transformer, the voltages, currents and impedances on the
secondary side of the device can be referred to PRIMARY side by turn ratio of
the transformer :
VP  V ' s  aVs
IP  I 'S 
IS
a
Z 'S  a 2 Z S
•The same transformation can be used for the induction motor’s rotor circuit
by using effective turn ratio aeff :
E1  E ' R  aef fER 0
I2 
IR
a eff
Z 2  a 2 ef f (
RR
 jX R 0 )
s
Chap 2: AC Machines
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The Equivalent Circuit of An Induction Motor
C) THE FINAL EQUIVALENT CIRCUIT
The rotor circuit model that will be referred to the stator side as shown below
The per-phase equivalent circuit of an induction motor.
Chap 2: AC Machines
30
Power and Torque in Induction Motors
Input is 3 phase system of
voltages and currents.
Electrical to
mechanical power
conversion
Output is
mechanical.
PSCL – Losses In Stator Windings / I2R
PRCL=I2R
Pcore – Hysteresis & Eddy Current
The power flow diagram of an induction motor – shows the relationship between
the input electric power and output mechanical power.
Chap 2: AC Machines
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Example 2.2: Power-in in Induction Motors
An induction motor draws 60A from a 480 V, 60 Hz, 50-hp, three-phase line at
a power of 0.85 lagging. The stator copper losses are 2000W and the rotor
copper losses are 700W. The rotational losses include 600W of friction and
wind-age, 1800W of core and negligible of stray load losses. Calculate the
following quantities:
(a)The air-gap power,
(b)The power converted, Pconv
(c)The output power,
(d)The efficiency of the motor
PAG = Pin – PSCL - PCORE
PIN = √3 VTIL cos θ
= √3 (480V)(60A)(0.85)
= 42.4 kW
Solution:
(a)The air-gap power,
PAG = Pin – PSCL – PCORE
= 42.4 kW – 2 kW – 1.8 kW
= 38.6 kW
Chap 2: AC Machines
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(cont’d)
(b)The power converter, Pconv
PCONV = PAG – PRCL
= 38.6 kW – 700 W = 37.9kW
(c) The output power.
POUT = PCONV – PF&W – PMISC
= 37.9 kW – 600 W – 0 W = 37.3 kW
(d) The efficiency,
  P out X 100 %
P in
= 37.3 kW x 100% = 88%
42.4 kW
Chap 2: AC Machines
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Power and Torque in Induction Motors
The per-phase equivalent
circuit of an induction motor
Input current
I1 
V
Z eq
Where
Z eq  R1  jX1  [( Rc  jX m ) //(
Chap 2: AC Machines
R2
 jX 2 )
s
34
Torque-speed characteristics
(a)
(b)
a) A typical induction motor torque-speed characteristic curve.
b) Showing the extended operating ranges (braking region and generator region)
Chap 2: AC Machines
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Torque-speed characteristic
1.
2.
3.
4.
5.
6.
7.
The induced torque of the motor is zero at synchronous speed.
The torque speed curve is nearly linear between no-load and full-load. In
this range, the rotor resistance is much larger than the rotor reactance. So,
the rotor current increasing linearly.
There is maximum possible torque that cannot be exceeded (called pullout
torque or breakdown torque) is 2-3 times the rated full-load torque.
Starting torque on motor is slightly larger than full-load.
The torque on the motor for a given slip varies as the square of the
applied voltage.
If the rotor of the induction motor is driven faster than synchronous
speed, then the direction of the induced torque in the machine reverse
and become generator.
If motor turning backward, relative to the direction of the magnetic field,
the induced torque will stop the machine very rapidly and will try to
rotate it in the other direction (called plugging).
Chap 2: AC Machines
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Speed Control of Induction Motors
1.
By pole changing
2.
By line frequency control
3.
By line voltage control
4.
By changing the rotor resistance
Chap 2: AC Machines
To vary the synchronous
speed which is the speed
of the stator and rotor
magnetic field
To vary the slip of the
motor for a given load
37
Induction Motor Ratings
•The most important
ratings:
•Output power
•Voltage
•Current
•Power factor
•Speed
•Nominal frequency
*Refer Chapman pg. 465
*Note: 1 h.p = 746 Watts
Chap 2: AC Machines
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SYNCHRONOUS
MACHINES
Chap 2: AC Machines
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INTRODUCTION (Review)
Transformer – energy transfer device.
(transfer energy from primary to secondary)
- form of energy remain unchanged. (Electrical)
(DC/AC) Machines – electrical energy is converted to mechanical or vice versa.
Motor operation
The field induced voltage, E permits the motor to draw power from the
line to be converted into mechanical power. This time, the mechanical
output torque is also developing. The induced voltage is in opposition to
the current flow-called counter emf.
Chap 2: AC Machines
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INTRODUCTION (Review)
Generator operation
The field induced voltage, E is in the same direction as the current and is called
the “generated voltage”. The machine torque opposes the input mechanical
torque that is trying to drive the generator, and it is called the counter torque.


Generally, the magnetic field in a machine forms the energy link
between the electrical and mechanical systems.
The magnetic field performs two functions:


Magnetic attraction and repulsion produces mechanical torque (motor
operation)
The magnetic field by Faraday’s Law induces voltages in the coils of
wire. (generator operation)
Chap 2: AC Machines
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SYNCHRONOUS MACHINES
CONSTRUCTION
Have an outside stationary part, (stator)
The inner rotating part (rotor)
The rotor is centered within the stator.
Air gap - the space between the outside of
the rotor and the inside of the stator
Origin of name: syn = equal, chronos = time
Synchronous machines are called ‘synchronous’ because their mechanical shaft
speed is directly related to the power system’s line frequency.
Chap 2: AC Machines
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SYNCHRONOUS MACHINES
CONSTRUCTION
STATOR
 The stator of a synchronous
machine carries the armature or
load winding which is a threephase winding.
 The armature winding is formed
by interconnecting various
conductors in slots spread over the
periphery of the machine’s stator.
 When current flows in the winding,
each group produces a magnetic pole
having a polarity dependent on the
current direction, and a magnetomotive
force (mmf) proportional to the current
magnitude.
Chap 2: AC Machines
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SYNCHRONOUS MACHINES
CONSTRUCTION
ROTOR
 2 types of rotors
- cylindrical (or round) rotors
- salient pole rotors.
Salient pole rotor less expensive than round rotors
and rotate at lower speeds
 The rotor carries the field winding. The
field current or the excitation current is
provided by an external dc source.
 Synchronous machine rotors are simply
rotating electromagnets built to have as
many poles as are produced by the stator
windings.
 Dc currents flowing in the field coils
surrounding each pole magnetize the rotor
poles. The magnetic field produced by the
rotor poles locks in with a rotating stator
field, so that the shaft and the stator field
rotate in synchronism.
Chap 2: AC Machines
44
SYNCHRONOUS MACHINES
1) Generator
The rate of rotation of the magnetic fields in the machine is related to
the stator electrical frequency, given as:
nm P
fe 
120
f e  electrical frequency, in Hz
nm  mechanical speedofmagnetic field , in r / min
P  number of poles
The internal generated voltage of a synchronous generator is given as,
EA  K
This equation shows the magnitude of the voltage induced in a given stator
phase.
Chap 2: AC Machines
45
Equivalent Circuit of a synchronous
Generator
The per phase equivalent circuit
Chap 2: AC Machines
46
Synchronous Generator – voltage regulation

If the generator operates at a terminal voltage VT while supplying a load
corresponding to an armature current Ia, then;

In an actual synchronous machine, the reactance is much greater than the
armature resistance, in which case;

Among the steady-state characteristics of a synchronous generator, its
voltage regulation and power-angle characteristics are the most important
ones. As for transformers, the voltage regulation of a synchronous generator
is defined at a given load as;
Chap 2: AC Machines
47
Phasor diagram of a synchronous generator
The phasor diagram is showing the relationship among the voltages
within a phase (Eφ,Vφ, jXSIA and RAIA) and the current IA in the
phase.
Unity P.F (1.0)
Chap 2: AC Machines
48
Phasor diagram of a synchronous generator
Lagging P.F
Leading P.F.
Chap 2: AC Machines
49
Power and Torque in Synchronous Generator
In generators, not all the mechanical power going into a synchronous
generator becomes electric power out of the machine
The power losses in generator are represented by difference between output
power and input power shown in power flow diagram below.
Pconv
Chap 2: AC Machines
50
Losses in Synchronous Generator
Rotor
- resistance; iron parts moving in a magnetic field causing
currents to be generated in the rotor body
- resistance of connections to the rotor (slip rings)
Stator
- resistance; magnetic losses (e.g., hysteresis)
Mechanical
- friction at bearings, friction at slip rings
Stray load losses
- due to non-uniform current distribution
Chap 2: AC Machines
51
Synchronous Generator
 The input mechanical power is the shaft power in the generator given by equation:
 The power converted from mechanical to electrical form internally is given by:
 The real electric output power of the synchronous generator can be expressed
in line and phase quantities as:
and reactive output power:
Chap 2: AC Machines
52
Synchronous Generator
In real synchronous machines of any size, the armature resistance
RA is more than 10 times smaller than the synchronous reactance XS
(Xs >> RA). Therefore, RA can be ignored
Simplified phasor diagram with armature resistance ignored.
Chap 2: AC Machines
53
SYNCHRONOUS MACHINES
2) MOTOR
Equivalent circuit of synchronous motor:
Equivalent circuit
Chap 2: AC Machines
54
Power and Torque in Synchronous Motor
The power-flow diagram of a synchronous machines
Chap 2: AC Machines
55
Example 3.3 : Synchronous Generator.
A three-phase, wye-connected 2500 kVA and 6.6 kV generator operates at fullload. The per-phase armature resistance Ra and the synchronous reactance, Xd,
are (0.07+j10.4). Calculate the percent voltage regulation at:
(a) 0.8 power-factor lagging, and
(b) 0.8 power-factor leading.
Chap 2: AC Machines
56
Solution:
Chap 2: AC Machines
57
Success is the sum of small efforts,
repeated day in and day out…
- Robert Collier
Chap 2: AC Machines
58