Transcript Chapter 7 - 3 phase Induction Motor

Review: Electromagnetism
Magnetic field due to an electric current
When a conductor carries an electric current, a magnetic field is
produced around that conductor.
Right Hand Rule
Magnetic field of a solenoid
If a coil is wound on a steel rod and connected to a supply, the
steel becomes magnetized and behaves like a permanent
magnet.
Grip Rule if solenoid is gripped with the right hand, with fingers
pointing in the direction of current, then the thumb outdtretched
parallel to the axis of the solenoid points in the direction of the
magnetic field inside the solenoid.
Solenoid with a steel core
Force on a conductor carrying current across a magnetic field
F (force on conductor, N) = B (flux density, T) x L (length, m) x I (current
through conductor, A)
F  BIL
Left Hand Rule: First finger: Flux
Second finger: Current
Thumb: Mechanical Force
Electromagnetic induction
When a conductor cuts or is cut by a magnetic flux, an emf is
generated in the conductor and a magnitude of the generated emf is
proportional to the rate at which the conductor cuts or is cut by the
magnetic flux. Ex : transformer
Direction of induced e.m.f
Two methods:
a) Fleming’s right hand rule - If the first finger of the right hand be
pointed in the direction of the magnetic flux and if the thumb be
pointed in the direction of motion of the conductor relative to the
magnetic field, then the second finger, held at right angles to both
the thumb and the first finger, represents the direction of the emf.
b) Lenz’s law
Fleming’s right hand rule
INDUCTION MACHINE
The induction machine is the most rugged and the most widely used
machine in industry. Like dc machine, the induction machine has a stator
and a rotor mounted on bearings and separated from the stator by an air
gap. However, in the induction machine both stator winding and rotor
winding carry alternating currents. The induction machine can operate both
as a motor and as generator
As motors, they have many advantages. They are rugged, relatively
inexpensive, and require very little maintenance. They range in size from a
few watts to about 10,000 hp. The speed of an induction motor is nearly
but not quite constant, dropping only a few percent in going from no load to
full load.
The main disadvantages of induction motors are:
· The speed is not easily controlled.
· The starting current may be five to eight times full-load current.
· The power factor is low and lagging when the machine is lightly loaded
INDUCTION MOTOR CONSTRUCTION
Two different types of induction motor which can be placed in stator:
a) squirrel cage rotor
b) wound rotor
Squirrel Cage rotor
Wound rotor
Types of rotor
Squirrel cage rotor – consists of conducting bars embedded in
slots in the rotor magnetic core, and these bars are short
circuited at each end by conducting end rings. The rotor bars
and the rings are shaped like squirrel cage.
Wound rotor – carries three windings similar to the stator
windings. The terminals of the rotor windings are connected to
the insulated slip rings mounted on the rotor shaft. Carbon
brushes bearing on these rings make the rotor terminals
available to the user of the machine. For steady state operation,
these terminals are short circuited.
Squirrel Cage Rotor
Rotor bars (slightly skewed)
End ring
Wound Rotor
• Most motors use the squirrel-cage rotor because of the
robust and maintenance-free construction.
• However, large, older motors use a wound rotor with three
phase windings placed in the rotor slots.
• The windings are connected in a three-wire wye.
• The ends of the windings are connected to three slip rings.
• Resistors or power supplies are connected to the slip rings
through brushes for reduction of starting current and
speed control
Induction Motor Components
BASIC INDUCTION MOTOR CONCEPT
A three phase set of voltages has been applied to the stator, and three
phase set of stator currents is flowing. These produce a magnetic field
Bs, which is rotating in a counterclockwise direction .
The speed of the magnetic field’s rotation is
nsync 
120 f e
P
The rotating stator fields Bs induces a voltage in the
rotor bars.
eind  (vxB)  I
The rotor voltage produces a rotor current flow,
which lags behind the voltage because of the
inductance of the rotor.
The rotor current produces a rotor magnetic field BR
lagging 900 behind itself, and BR interacts with Bnet
to produce a counterclockwise torque in the
machine
 ind  kBR xBs
THE CONCEPT OF ROTOR SLIP
The voltage induced in a rotor depends on the speed of the rotor
relative to the magnetic field.
Slip speed is defined as the difference between synchronous speed and
rotor speed
nslip  nsync - nm
where
nslip
nsync
nm
= slip speed of the machine
= speed of the magnetic fields
= mechanical shaft speed of motor
Slip is the relative speed expressed on a per unit or a percentage basis
s
n slip
n sync
x 100%
s
n sync - nm
n sync
x 100%
CONTINUED…
In term angular velocity (radians per second, rps)
 sync -  m
s
x 100%
 sync
If the rotor turns at synchronous speed, s = 0
while if the rotor is stationary/standstill, s = 1.
nm  (1 - s)n synx
 m  (1 - s) synx
THE ELECTRICAL FREQUENCY CONCEPT
Like a transformer, the primary (stator) induces a voltage in the secondary
(rotor) but unlike a transformer, the secondary frequency is not necessary
the same as the primary frequency.
If the rotor of a motor is locked, then the rotor will have same frequency as
the stator.
The rotor frequency can be expressed
f r  sf e
P
fr 
(n sync - nm )
120
EXERCISE 1
A 208V, 10hp, 4 pole, 60Hz, Y connected induction motor has full load
slip of 5%.
Calculate,
i) nsync (Ans:1800rpm)
ii) nm (Ans: 1710rpm)
iii) fr at the rated load (Ans: 3 Hz)
iv) Shaft torque at the rated load (Ans: 41.7Nm)
THE EQUIVALENT CIRCUIT OF AN INDUCTION
MOTOR
R1 = Per phase stator winding resistance
jX1 = Per phase stator leakage reactance
VP = Per phase terminal voltage
RC = Core resistance
jXM = Magnetizing reactance
ER = Rotor voltage
RR = Rotor resistance
jXR = Per phase rotor leakage reactance
THE ROTOR CIRCUIT MODEL
The greater the relative motion between
rotor and the stator magnetic fields, the
greater the resulting rotor voltage and rotor
frequency. The largest relative motion
occurs when the rotor is stationary. (locked
rotor or blocked rotor condition)
jXR0 = blocked rotor rotor resistance
ER
IR 
RR  jX R
ER
IR 
RR  jsX R0
IR 
E R0
RR
 jX R0
s
ER0 = Locked rotor voltage
THE FINAL EQUIVALENT CIRCUIT
In an ordinary transformer, the voltages, currents, and impedances
on the secondary of the device can be referred to the primary side by
means of the turn ratio of the transformer:
VP  VS'  aVS
IS
IP  I 
a
'
S
Z S'  a 2 Z S
The transformed rotor voltage becomes
E1  E R'  aeff E R 0
The rotor current becomes
I
I2  R
aeff
Referred rotor resistance and
reactance
The rotor impedances becomes

2  RR
Z 2  aeff
 jX R 0 

 s

2
R2  aeff
RR
2
X 2  aeff
X R0
THE FINAL EQUIVALENT CIRCUIT
POWER FLOW DIAGRAM
When the secondary windings in an induction motor (rotor) are
shorted out, so no electrical output exists from normal induction
motors. Instead, the output is mechanical.
The relationship between input and output powers are shown below:
POWER AND TORQUE IN INDUCTION MOTOR
PSCL  3I 12 R1
Stator Copper Loss,
Pcore  3E12 GC
Core Losses,
PAG  Pin  PSCL  Pcore
Air Gap Power,
Rotor Copper Loss,
PRCL  3I 22 R2  sPAG
Developed Mech. Power,
Output power,
Pconv  Pdev  PAG  PRCL
Pout  Pconv  PF &W  Pmisc
Developed Torque,
PAG
R2
 3I
s
2
2
 ind 
Pconv
m

PAG
 sync
1- s 
 3I R2 

 s 
2
2
Separating the rotor copper losses and the power converted in
Induction motor equivalent circuit
The derivation of the induction motor induced-torque equation
The induced torque in induction motor is:
 ind 
Pconv
m
 ind 
PAG
 sync
Air gap power:
PAG  I 22
R2
s
Total Air gap power:
PAG  3I 22
Air gap power is the power crossing the gap
from the stator circuit to the rotor circuit. It is
equal to the power absorbed in the resistance
R2/s.
R2
s
Continued…
If I2 can be determined, then the air gap power and
the induced torque will be known.
To solve circuit above, we use
thevenin’s theorem:
Continued…
ZTH  RTH  jX TH
X M  X 1  R1
RTH
 XM
 R1 
 X1  X M
X M  X 1



2
X TH  X 1
jX M ( R1  jX1 )

R1  j( X 1  X M )
THEVENIN’S EQUIVALENT CIRCUIT
VTH
I2 
RTH
PAG  3I 22
R
 2  jX TH  jX 2
s
 ind 
PAG
sync

3VTH2
sync[( RTH
R2

s
3VTH2
( RTH 
R2
s
R2 2
)  ( X TH  X 2 ) 2
s
R2
s
R2 2
 )  ( X TH  X 2 ) 2 ]
s