Download IM ppt File

Chapter 2
Poly phase Induction Motor
Prepared by:- Jaydeepsingh Chauhan
Electrical Department
SVBIT Gandhinagar
Induction Motor
•Most popular motor today in the low and medium horsepower range
•Very robust in construction
•Speed easily controllable using V/f or Field Oriented Controllers
•Have replaced DC Motors in areas where traditional DC Motors
cannot be used such as mining or explosive environments
•Of two types depending on motor construction: Squirrel Cage
or Slip Ring
•Only Disadvantage: Most of them run with a lagging power factor
2
induction motor
Construction of Induction Motor
Stator
Squirrel Cage Rotor
Cut section Of Induction Motor
Slip ring Rotor
3
induction motor
AC Machine Stator
‘b’ phase axis
1200
1200
‘a’ phase axis
1200
‘c’ phase axis
4
induction motor
Squirrel Cage Rotor
5
induction motor
Slip Ring Rotor
•The rotor contains windings similar to stator.
•The connections from rotor are brought out using slip rings that
are rotating with the rotor and carbon brushes that are static.
6
induction motor
Currents in different
phases
of
AC
Machine
t
t
01
12
Amp
t0
t1
1 Cycle
7
t2
t3
t4
time
induction motor
a
Fc
RMF(Rotating Magnetic Field)
b’
c’
1.5
Fa
F
1
Fa
c
b
t = t0= t4
F
Fc
0.5
Fb
a’
0
Fb
-0.5
t = t0= t4
-1
-1.5
-93
F
Fb
a
c’
8
Fc
Fb
b’
c
a’
t = t1
113
216
Space angle () in degrees
a
c’
b
10
a
b’
Fa
F
c
b
Fc
t = t2
a’
b’
c’
c
b
t = t3
induction motor
Fc
a’
F
Fb
Torque Production in an Induction Motor
•In a conventional DC machine field is stationary and the current carrying
conductors rotate.
•We can obtain similar results if we make field structure rotating and current
carrying conductor stationary.
•In an induction motor the conventional 3-phase winding sets up the rotating
magnetic field(RMF) and the rotor carries the current carrying conductors.
•An EMF and hence current is induced in the rotor due to the speed
difference between the RMF and the rotor, similar to that in a DC motor.
•This current produces a torque such that the speed difference between the
RMF and rotor is reduced.
9
induction motor
Slip in Induction Motor
•However, this speed difference cannot become zero because that would stop generation of the
torque producing current itself.
•The parameter slip ‘s’ is a measure of this relative speed difference
ns  n  s  
s

ns
s
ns 
120 f1
; p # of poles
p
where ns,s,f1 are the speeds of the RMF in RPM ,rad./sec and supply frequency
respectively
n, are the speeds of the motor in RPM and rad./sec respectively
•The angular slip frequency and the slip frequency at which voltage is induced in the rotor
is given by
 2  s , f 2  sf1 , E2 s
10
N2
s
E1
N1
N1  Stator turns N2  Rotor turns
induction motor
Starting Method for Induction Motors
Therefore, 3-phase induction motors employ a starting
method not to provide a starting torque at the rotor, but
because of the following reasons;
1) Reduce heavy starting currents and prevent motor from
overheating.
2) Provide overload and no-voltage protection.
• Direct On-Line Starter (DOL)
• Star-Delta Starter
• Auto Transformer Starter
• Rotor Impedance Starter
• Power Electronics Starter
Direct On-Line Starter (DOL)
Star-Delta Starter
Auto Transformer Starter
Rotor Impedance Starter
Single Phase Induction Motor
A single phase induction motor is not self starting but requires some starting means.
If the stator winding is connected to single – phase a.c. supply, the stator winding produces a
magnetic field that pulsates in strength in a sinusoidal manner. The field polarity reverses after
each half cycle but the field does not rotate.
Double – Field Revolving Theory
Torque – slip characteristic of 1- phase induction motor
Starting of Single Phase Induction Motors
Split – phase Induction Motor
Capacitor – Start Motor
Permanent – Split Capacitor Motor
Capacitor - Start Capacitor - Run
Shaded Pole Induction Motor
Construction of Circle Diagram
Conduct No load test and blocked rotor test on the induction motor and find out the per phase
values of no load current I0, short circuit current ISC and the corresponding phase angles Ö0 and
ÖSC.
Also find short circuit current ISN corresponding to normal supply voltage.
With this data, the circle diagram can be drawn as follows.
• With suitable scale, raw vector OA with length corresponding to Io at an angle Ö0 from the
vertical axis. Draw a horizontal line AB.
• Draw OS equal to Isn at an angle ΦSC and join AS.
• Draw the perpendicular bisector to AS to meet the horizontal line AB at C.
• With C as centre, draw a portion of circle passing through A and S. This forms the circle
diagram which is the locus of the input current.
• From point S, draw a vertical line SL to meet the line AB.
• Divide SL at point K so that SK : KL = rotor resistance : stator resistance.
• For a given operating point P, draw a vertical line PEFGD as shown. then PE = output power, EF
= rotor copper loss, FG = stator copper loss, GD = constant loss (iron loss + mechanical loss).
• To find the operating points corresponding to maximum power and maximum torque, draw
tangents to the circle diagram parallel to the output line and torque line respectively. The
points at which these tangents touch the circle are respectively the maximum power point and
maximum torque point.
Efficiency line
1. The output line AS is extended backwards to meet the X-axis at O′.
2. From any convenient point on the extended output line, draw a horizontal line QT so as to
meet the vertical from O′. Divide the line QT into 100 equal parts.
3. To find the efficiency corresponding to any operating point P, draw a line from O′ to the
efficiency line through P to meet the efficiency line at T1. Now QT1 is the efficiency.
Slip Line
1. Draw line QR parallel to the torque line, meeting the vertical through A at R. Divide RQ into
100 equal parts.
2. To find the slip corresponding to any operating point P, draw a line from A to the
slip line through P to meet the slip line at R1. Now RR1 is the slip
Power Factor Curve
1. Draw a quadrant of a circle with O as centre and any convenient radius. Divide OCm into
100 equal parts.
2. To find power factor corresponding to P, extend the line OP to meet the power factor curve at
C′. Draw a horizontal line C′C1 to meet the vertical axis at C1. Now OC1 represents power
factor.