3-ph induction motor
TRANSCRIPT
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3-ph Induction Motor3-ph Induction Motor
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3-ph Induction Motor3-ph Induction Motor
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Theory
1. Three-phase induction motor is the mostpopular type of a. c. motor.
2. It is very commonly used for induction
drives since it is cheap, robust, efficientand reliable.
3. It has good speed regulation and high
starting torque.4. It requires little maintenance.5. It has a reasonable overload capacity.
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The basic principle of operation is inductionand hence thename Induction Motor.
Induction is a phenomenon of an induced voltage in a coil due
to changing flux.
This flux is established by either another coil(as a generalcase) or the same coil.
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Slot
Tooth
Stator core
or Stamping
IM consists of: 1. stator2. rotor.
Rotor
Shaft
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Tooth
Stator core
or Stamping
Rotor
Shaft
The stator is the stationarypart.
and the rotor is a part.rotating
IM consists of: 1. stator2. rotor.
Slot
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Rotor
Shaft
The laminations are slottedon the inner peripheryand are insulated from each other.
The insulated stator conductorsare placed in these slots.
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Rotor
Shaft
The stator conductors are connected to form athree-phase winding.
The phase winding may be either star or deltaconnected.
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Rotor
Shaft
The clearance between the stator and the rotor iscalled an air gap.The air gap is kept as small aspossible to
a) Reduce leakage reactance& no load currentandb) Improve the power factor.
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Rotor
Shaft
The rotor is built up of thin laminationsof thesame material as that of stator.The laminated cylindrical core is mounted directly
on the shaftor a spider carried by the shaft.
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Th t t l i ti t d i t t
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Rotor
Shaft
The stator laminations are supported in a statorframeof caste ironor fabricated steel plate.
Stator frame is connected to coverings.
The coverings are rested on bearingswhich are mounted onshaft.
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Air Gap
STATOR
ROTOR
Stator Core
Rotor Core
Shaf
tRotor wdg
Stator wdg
Bearings
Stator Frame
Base
STATOR
Common constructional features forALL
rotating electrical machines
Slip rings
Brushes
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There are TWOtypes of induction motor dependingon the types of rotor.
End RingsRotor Bars(Slightly skewed)
Rotor Bars(Slightly skewed)
Fig. 1. Cage Rotor
1. Squirrel-cagerotor or simply cage rotor. (SCIM).
The slots nearly parallel to the shaft axis or skewed .
2.Phase wound or wound rotor or slip ring rotor (SRIM).
Each slot contains an uninsulated bar conductor ofaluminium or copper.
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There are TWOtypes of induction motor dependingon the types of rotor.
End RingsRotor Bars(Slightly skewed)
Rotor Bars(Slightly skewed)
Fig. 1. Cage Rotor
1. Squirrel-cagerotor or simply cage rotor. (SCIM).
At each end of the rotor, the rotor barconductors are short circuited by heavy end ringof the same material.
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There are TWOtypes of induction motor dependingon the types of rotor.
End RingsRotor Bars(Slightly skewed)
Rotor Bars(Slightly skewed)
Fig. 1. Cage Rotor
1. Squirrel-cagerotor or simply cage rotor. (SCIM).
2.Phase wound or wound rotor or slip ring rotor (SRIM).
The conductors and the end rings look like a cage of a birdor form a cage of the type which was once commonly used for
keeping squirrel;
Hence this rotor is known as the squirrel cage rotor.
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The important features of this squirrel cage rotorare:1. The skewed bar reduces harmonics.2. This gives uniform torque and less noise.
3. The locking tendency is reduced.4. It has a compact and ruggedconstruction.5. The end rings can be projected for fanning actionfor
cooling.6. It requires no slip rings.
7. It has less loss and more efficiencyas compared to SRIM
8. It is not possibleto add extra rotor resistance to changethe torque and speed. This is the only disadvantage.
Ha Ha Ha, We are important in Electrical Machines
2 Ph d d li i
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ROTOR
OR
SHAFT
Slip Rings
Brushes
StartMax
RunMin
Fig. 2. Slip Ring Rotor
2.Phase wound or wound rotor or slip ring rotor(SRIM).
Rotor windingIn delta or Star
The wound rotor consists of a slotted armature.Insulated conductors are put in the slots and connected toform a three phase distributed double layer winding
similar to the stator winding.
2 Ph d d t li i t
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ROTOR
OR
SHAFT
Slip Rings
Brushes
StartMax
RunMin
Fig. 2. Slip Ring Rotor
2.Phase wound or wound rotor or slip ring rotor(SRIM).
Rotor windingIn delta or Star
The rotor windings are connected in star or in delta.The three ends of rotor windings are brought outside therotor and connected to three insulated slip rings.
The slip rings are mounted on the shaft withbrushes resting on them.
2 Ph d d t li i t
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ROTOR
OR
SHAFT
Slip Rings
Brushes
StartMax
RunMin
Fig. 2. Slip Ring Rotor
2.Phase wound or wound rotor or slip ring rotor(SRIM).
Rotor windingIn delta or Star
The resistors enable to increase each rotorphase resistance to serve the following purposes:
1. to increase the starting torque.2. to decrease the starting current.3. to improve the starting power factor.
4. to decrease the speedof the motor.
P i i l f O ti
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Principle of Operation
R
Y
B
R Y B
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R
Y
B
R Y B
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R
Y
B
R Y B
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0 60 120 180 240 300 360
r y b
r
y
b
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0 60 120 180 240 300 360
r y b
r
y
b
C n id n n l 0o (3/2) S
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0 60 120 180 240 300 360
r y b
Consider an angle 0o
-y0={(3)/(2)}m
b0={(3)/(2)m
r
y
b -y0b0
0=(3/2)m
R2
R1Y2
Y1
B2
B1
N
S
B1=+ & B2 = Y1= & Y2 =+
R1= 0 & R2 =0
At an angle 60o
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0 60 120 180 240 300 360
r y b
At an angle 60o
-y60={(3)/(2)}m
r60={(3)/(2)m
r
y
b-y0
r60
60=(3/2)m
R1=+ & R2 = Y1= & Y2 =+
B1= 0 & B2 =0
R2
R1Y2
Y1
B2
B1
N
S
At an angle 120o
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0 60 120 180 240 300 360
At an angle 120o
-b120={(3)/(2)}m
r120={(3)/(2)m
r
y
b
-b120
r120
120=(3/2)m
R2
R1Y2
Y1
B2
B1N
S
R1=+ & R2 = Y1= 0 & Y2 = 0
B1= & B2 =+
At an angle 180o
N
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0 60 120 180 240 300 360
At an angle 180o
-b180={(3)/(2)}m
y180={(3)/(2)m
r
y
b
-b180Y180
180=(3/2)m
R2
R1Y2
Y1
B2
B1
N
S
R1=0 & R2 = 0Y1= + & Y2 = B1= & B2 =+
At an angle 240o
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0 60 120 180 240 300 360
At an angle 240o
-r240={(3)/(2)}m
y240={(3)/(2)m
r
y
b
-r240
Y240
R2
R1Y2
Y1
B2
B1N
S
R1= & R2 = +Y1= + & Y2 = B1= 0 & B2 = 0
At an angle 300o
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0 60 120 180 240 300 360
At an angle 300o
-r300={(3)/(2)}m
b300={(3)/(2)m
r
y
b
-r300
b300
300
=(3/2)m
R2
R1Y2
Y1
B2
B1S
N
R1= & R2 = +Y1= 0 & Y2 = 0
B1= + & B2 =
At an angle 360o
=(3/2) S
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0 60 120 180 240 300 360
At an angle 360
-y360
={(3)/(2)}m
b360={(3)/(2) }m
r
y
b-y360b360
360=(3/2)m
R2
R1Y2
Y1
B2
B1
N
S
B1=+ & B2 =
Y1= & Y2 =+
R1= 0 & R2 =0
Phasor is
same asthat ofangle 0o
S
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0 60 120 180 240 300 360
R2
R1Y2
Y1
B2
B1
N
S
R
Y
B
For Clockwise RYB,Flux rotate Clockwise,Poles rotate Clockwise,
R Y B
S
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0 60 120 180 240 300 360
R2
R1
B2
B1
Y2
Y1
N
S
R
B
Y
For anticlockwise phase sequence, Flux rotate anticlockwise,Poles rotate anticlockwise,
Now change thephase sequenceof motor to RBY
R Y B
With samesupply phasesequence
Rotation of Rotor
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R
Y
ROTOR
Rotor ConductorSTATOR
Three Phase Supply is givenFlux is set up in the stator and passes from
stator to rotor
BFIRSTMETHOD
Rotation of Rotor
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Rotation of Rotor
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R
Y
BFIRSTMETHOD
Rotation of Rotor
FluxDirection
Motion ofConductor
w .r. t.StationaryField
By Faradays lawof electromagnetic induction, a voltagewillbe induced in the conductor.
Since the rotor is complete, either through the end rings oran external resistance, the induced voltage causes a currentto flowin the rotor conductor.
By right-hand rulewe can determine the directionof induced
current in the conductor.
Rotation of Rotor
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R
Y
BFIRSTMETHOD
Rotation of Rotor
By this rule the direction of the induced current is outwards,represented by dot.The current in the rotor conductor produces its own
magnetic fieldwhich is opposite to stator field on right handside and addition of flux on left hand side.This flux opposes the cause of it and cause is stator rotatingmagnetic field. (Lenzs Law)
FluxDirection
Motion ofConductor
w .r. t.StationaryField
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Rotation of Rotor
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FIRSTMETHOD
Rotation of Rotor
So rotor condror rotor moves in the same dirnof stator field.
R
Y
B
By this rule the direction of the induced current is outwards,represented by dot.The current in the rotor conductor produces its own
magnetic fieldwhich is opposite to stator field on right handside and addition of flux on left hand side.This flux opposes the cause of it and cause is stator rotatingmagnetic field.
Rotation of Rotor
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R
Y
BFIRSTMETHOD
Rotat on of Rotor
If rotor moves in opposite directionthen the speeddifferencebetween rotor and stator rotating magnetic field
increases.This increases the opposition. (Not opposing the cause of it)
Rotation of Rotor
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R
Y
BFIRSTMETHOD
f
If rotor moves in same directionthen the speed differencebetween rotor and stator rotating magnetic field decreases.
This decreases the opposition. (opposing the cause of it)
Rotation of Rotor
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The current in the rotor conductor produces its own magneticfield which is opposite to stator field on right hand side andaddition of flux on left hand side.
Rotor conductor moves towards right due to tension action offlux. (Catapult)It is seen that the force acting on the conductor is in the
same directionas the direction of the rotating magnetic field.
MoreFlux
Less
Flux
SECONDMETHOD
Rotation of Rotor
R
Y
B
Rotation of Rotor
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R
Y
BTHIRDMETHOD
Rotation of Rotor
FluxDirection
Direction ofCurrent
When a conductor carrying currentis put in a magnetic field aforceis produced on it. The direction of force can be found byleft hand Rule.
It is seen that the force acting on the conductor is in thesame directionas the direction of the rotating magnetic field.
Force on Conductor
S
Rotation of Rotor
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R2
R1
Y2
Y1
B2
B1
N
S
3 ph magnetic field rotates in clockwisedirection forclockwise RYBApply RHR to rotor conductor
Motion
Flux
Mark the Poles formed in rotor
N SN S
Now consider N of stator and N of rotor, Repulsion
Now consider N of stator and S of rotor, Attraction
So rotor has to rotate in clockwise direction
FOURTHMETHOD
Rotation of Rotor
Similarly S of stator and S of rotor,
S
Rotation of Rotor
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R2
R1
Y2
Y1
B2
B1
N
S
3 ph magnetic field rotates in clockwisedirection forclockwise RYBApply RHR to rotor conductor
Motion
Flux
Mark the Poles formed in rotor
N SN S
Now consider N of stator and N of rotor, Repulsion
Now consider N of stator and S of rotor, Attraction
So rotor has to rotate in clockwise direction
FOURTHMETHOD
Rotation of Rotor
Similarly S of stator and S of rotor,
S
Rotation of Rotor
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R2
R1
Y2
Y1
B2
B1
N
S
N SN S
FOURTHMETHOD
Rotation of Rotor
Flux
Motion
3 ph magnetic field rotates in clockwisedirection forclockwise RYBApply RHR to rotor conductor
Mark the Poles formed in rotorNow consider N of stator and N of rotor, Repulsion
Now consider N of stator and S of rotor, Attraction
So rotor has to rotate in clockwise direction
Similarly S of stator and S of rotor,
SRotation of Rotor
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R2
R1
Y2
Y1
B2
B1
N
S
N S
FOURTHMETHOD
Rotation of Rotor
3 ph magnetic field rotates in clockwisedirection forclockwise RYBApply RHR to rotor conductor
Mark the Poles formed in rotorNow consider N of stator and N of rotor, Repulsion
Now consider N of stator and S of rotor, Attraction
So rotor has to rotate in clockwise direction
Similarly S of stator and S of rotor,
Torque
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R
Y
BTorque
Since the rotor conductor is in a slot on the circumference ofthe rotor, this force acts in a tangential directionto the rotorand develops a torque on the rotor.
Similar torques are produced on all the rotor conductors.
BTorque
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R
Y
BTorque
Since the rotor conductor is in a slot on the circumference ofthe rotor, this force acts in a tangential directionto the rotorand develops a torque on the rotor.
Similar torques are produced on all the rotor conductors.
Hence rotor rotates
Three-phase induction motor is self starting.
and thus
BRotor Rotation
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R
Y
Bw. r. t. StatorFlux
Rotor Current
Ns
Nr
fr=sf
s=(Ns-Nr)/Ns=0.04
sf=0.04x50=2Hz
Stator Current, f=50Hz, t=20msec
Rotor Current, f=2Hz, t=500msec
t=500msec
BRotor Rotation
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Y
B
R
B
R
Y
w. r. t. StatorFlux
Ns
Ns
In this case Nr =slip = 0;No Speed difference between Rotorand Rotating Magnetic Field.No cutting of flux;No voltage; No current;
Nr
Nr
Ns;
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BRotor Rotation
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R
Y
Bw. r. t. StatorFlux
Ns
Nr
R
Y
B Ns
Nr
As soon as speed decreases, rotor current flowsand
In this case Nr =slip = 0;No Speed difference between Rotorand Rotating Magnetic Field.No cutting of flux;No voltage; No current;
Ns;
No Rotor Flux.No torque. Speed decreases.
Torque is produced
BRotor Rotation
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R
Y
Bw. r. t. StatorFlux
Ns
Nr
R
Y
B Ns
Nr
As soon as speed decreases, rotor current flowsand
In this case Nr =slip = 0;No Speed difference between Rotorand Rotating Magnetic Field.No cutting of flux;No voltage; No current;
Ns;
No Rotor Flux.No torque. Speed decreases.
Torque is produced
BRotor Rotation
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R
Y
Bw. r. t. StatorFlux
Ns
Nr
An induction motor cannotrun at synchronous speed.
Therefore the rotor speed is slightly less thanthesynchronous speed.
Since the operation of this motor depends on the inducedvoltagein its rotor conductors, it is called an induction motor.
An induction motor may also be called as asynchronous
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y ymotor as it does not run at synchronous speed.
The difference between the synchronous speed and the
actual rotor speed is called slip speed Ns-Nr.
The slip is defined as the ratio of slip speed to thesynchronous speed, s =(Ns-Nr)/Ns.
The slip at full load varies from 2 to 5%.
Frequency of rotor emf = f2
120
&rotorfluxrotatingbetnspeedrelativeP
120
rs NNP
120
ssNP 1sf
Speed of rotor
f2120sf1120 sN
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flux w r t itself P
P ssN
Rotor itself rotates at a speedNr = (1-s) Ns, wrt stator
Speed of rotor flux wrt stator=sNs+ Nr
=NsThus stator androtorfluxrotate at synchronous speed wrt
stator
Speed of rotor
f2120sf1120 sN
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flux w r t itself P
P ssN
Rotor itself rotates at a speedNr = (1-s) Ns, wrt stator
Speed of rotor flux wrt stator=sNs+ Nr
=NsThus stator androtorfluxrotate at synchronous speed wrt
statorN
swrt stator
Nswrt stator
N
S
Nrwrt stator
N S
Speed of rotor
f2120sf1120 sN
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flux w r t itself P
P ssN
Rotor itself rotates at a speedNr = (1-s) Ns, wrt stator
Speed of rotor flux wrt stator=sNs+ Nr
=Ns
Thus stator androtorfluxrotate at synchronous speed wrt
stator
Relative speed is zero. MMF are stationarywrt each other
This producesuniform torque
The rotor freqquantity acts at stator freqwhen referred tostator
Induction Motor & Transformer
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Induction Motor is also called as
Induction Motor TransformerStator Primary
Rotor Secondary
Generalized TransformerTherefore, equivalent circuitdiagram of transformer is
applicable to IMVoltage equation of stator, V1= -E1+I1(r1+jx1) =V1
+I1(r1+jx1)
jXmRc
I1
I0
Ic I
V1
r1x1
f1
Voltage equation of rotor,
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V2=E2I2(r2+jx2), for transformer
= sE2I2(r2+jsx2), for IM
= 0, for shorted rotor (IM)
= sE2I2(r2+jsx2), for IM
Rotor currentI2= ,22
2
jsxr
sE
at slip frequencyf2I2
sE2f
2
=sf1
jsx2
r2
Rotor currentI2= ,
22
2
jx
s
r
E
at frequencyf1I2
E2f1
jx2
sr /2
E2 = I2(r2+jsx2),
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)(' 22222
221 xjas
ra
a
IaEEE
2 2( 2 j 2),
)''
(' 22
11 jxs
rIE
I1
E1f1
Jx2
sr /'2
Complete Equivalent Circuit Diagram
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p q g
f1
sr /'2jx1 Jx2
jXmRc
r1 I1 I1
I0
Ic I
V1
Complete Equivalent Circuit Diagram
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p q g
f1
jx1Jx2
jXmRc
r1 I1 I1
I0
Ic I
V1 sr /'2
s
s1rr-
s
r22
2
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Power Balance
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jx1jx2
jXmRc
r1 I1 I1
I0
Ic I
V1
r2
s
s1r2
StatorInput
powerV1I1cos1
Stator
I12r1loss
Stator
coreloss
Rotor Input powerE2 I2cos2=I22r2/s
(Air Gap Power Pg)
Rotor
I22r2loss
Mechanical
PowerDeveloped inRotor Pm
Rotor Input power= V1I1 cos 1- I12r1- Pi
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p p 1 1 1 1 1 iAir gap power,Pg=I22r2+ Pm
=E2I2 cos 2
and
2
2
2
2
22
xs
r
EI
2
2
2
2
2
2
xs
r
sr
cos
2
2
2
2
2
22
xs
r
sr
IE
Pg
srI 222Pg
s
s1rIrI 2
2
22
2
2Pg
gg s)P(1sP Pg
P
s1
rI 22
2
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Pm
TeMechanical
s
rI 22
gPs)(1
Internal (or Gross) torquedeveloped per phase is
SecondsperRadianinSpeedRotor
RotorinDevelopedPowerInternal
Mechanical
r(m)
m
P
s(m)
g
s)(1Ps)(1
secWatts/rad/
P
s(m)
g
Tecan also be expressed as Synchronous Watts
due toPgand s.
Shaft PowerPsh=Pm -F & W Loss
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Tsh
=Te - Lost Torque due to constant loss
r(m)
sh
P
- Iron LossPi (If not considered
earlierin Equi Ckt Diagram)
=Pm -Constant Loss
Efficiency= 100%PPP
P
cuconstantm
m
0.5%is deducted from calculated efficiency
in order to consider STRAY load losses.
Torque- SlipCharacteristics
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jx1jx2
jXmRc
r1 I1 I1
I0
Ic I
V1 /sr2
Apply Thevenins Theorem at points Aand B.
Obtain VTH, ITH, RTH andXTH
r1andx1are in parallel withXm
B
A
Torque- SlipCharacteristics
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jXTHjx2RTH ITH
VTH /sr2
q p
Thevenins Equivalent Circuit Diagram
B
A
I1 =I2
Torque- SlipCharacteristics
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q p
s
rI
m
PT 22
2ss
g
e
Nms
r
xXsrR
V
m 2
22TH
22
TH
2TH
s
Slip1 0
Ns0 Speed
Te
Motoring Mode
Test
TemTop
TLUnstable
Stable (Shunt ch)
222TH2
2TH
THest r
xXrRKT
22
THTH
THem
XRR2
KT
xXX TH
2
THop r
KsT
(T-s linear ch)
KMotor torque Tein terms of Tem
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em
e
TT
sr
xX
s
rR
K 22
2TH
2
2TH
TH
Assume r1 very small, so neglected.
Xsr mt2
2
2THTH
TH
XRR2
K
XxXxXRs
r2TH2THTH
mt
2 22
)(
RTHr1, RTHis also neglected.
em
e
T
T 2X
22
mt Xs
Xs
s
Xsmt
22THTH XRR2
22TH2
2TH xXs
rR
sr2
2
mt
mt
s
s
s
s
e 2T
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mt
mtem
e
ss
s
s2
T
T
mt
mtem
s
s
s
sT
2
s
1
1
s
eTTest
Analysis of equivalent circuit
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y q
r2/s
jx1 jx2
jXmRc
r1 I1 I2
I0Ic I
V1
ss1rr
sr 222
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jx1 jx2
jXmRc
r1 I1 I2
I0Ic I
V1
r2
s
s1r2
With this, circuit is equivalent to transformer.
At standstill, s=1,
At synchronous speed, s=0,
ckt becomes equivalent to shortcircuited transformer.
ckt becomes equivalent toopen circuited transformer.
Equivalent Circuit parameters
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q pThe equivalent circuit parameters can be
determined from
1.DC resistance measurement method2.No Load Test or Running Light test or Open Circuit Test3.Blocked Rotor test or Short Circuit Test
1.DC resistance measurement methodThe DC stator resistance r1, at room temperature ismeasured by circulating a suitable DC currentmeasuring voltage dropbetween stator terminals
The hot resistance at 750is
rdc=Vdc/Idc
r1=(1.1 to 1.3) rdc
A
V
2.No Load Test or Running Light Test orOpen Circuit Test
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Open Circuit TestThis test gives
1. Core loss2. F & W loss3. No load current I0
5. Ic, Rc, I, Xm6. Mechanical faults, noise
Rated per voltage V0,with
rated freq is given to stator.
Motor is run at NO LOAD
STATOR
A
I0
VV0
R
YB
ROTOR
N
W0
P0, I0and V0are recorded
P0 = I02r1+Pc+Pfw
4. No load power factor
No load power factor is small,0 05 t 0 15
00 IV
PC os
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0.05 to 0.15
1. Ic=I0cos0 2. I=I0sin0
3.
On No load, Motor runs near to syn speedSo, s zero 1/s=or open circuit
4.
00IV
)(, 11000c
0c jxrIVEI
ER
I
EX 0m
r1
r2/s
jx1 jx2I2
jXmRc
I0I0
Ic I
V0
open
cir
cuit
provided x1is known
The F & W loss Pfw, can be obtained from thistest
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test.Vary input voltage and note input power
Input Power
Input Voltage
Pfw
Thus Pc=P0 - I02r1 - Pfw
R t i bl k d S d 0 li 1
3.Blocked Rotor test or Short Circuit Test
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Rotor is blocked, Speed = 0, slip = 1
A
Isc
V Vsc
R
YB
N
Wsc
3-ph Variac
I M
Rotor is blocked or held stationary by
belt pulleyor by hand
Low voltageis applied upto rated statorcurrent
Voltage Vsc, Current Iscand Power Pscare measured.
Since slip is 1, secondary is shortcircuited
j
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Mechanical loss =0
Rc and Xm >> r2+jx2
Therefore, Zsc
= Vsc
/ Isc
=Rsc+jXsc
This test gives copper loss
jx1 jx2
jXmRc
r1 Isc
I0
Ic I
Vsc
r2
s
s1r2
Core loss negligible
Hence omittedscscIV
Pcos scsc =0.8 to 0.9
= r1+r2Rsc= Psc/Isc2
R
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Class of motor x1 x21. Class A(normal Tstand Ist) 0.5 0.5
For wound rotormotor, x1 = x2 = Xsc /2
22scscsc RZX
r2= Rscr1
21 xx
For squirrel cagemotor,
2. Class B(normal Tstand low Ist) 0.4 0.6
3. Class C(high Tstand low Ist) 0.3 0.7
4. Class D(high Tstand high slip) 0.5 0.5
Circle Diagram of Ind Motor
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But the advantage of circle diagram is that
torque and slipcan be known from circle diagram
The circle diagram is constructed with the help of
Graphical representation
The equivalent ckt., operating ch. can be obtained
by computer quickly and accurately
1. No load test (I0& 0)
2. Blocked rotor test (Isc& sc)
g
extremities or Limits of stator current, Power,
y
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xI00
Isc
sc
1. Draw x and y axes(V1on y axis)2. Draw I0and Isc(=V1/Zsc)3. Draw parallel line to x axis from I0.
This line indicates constant lossvertically
V1
Line I0Isc isoutput line
4. Join I0 and Isc
Output line
O
y
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xI00
sc
C
Output line
L1
T
V1
5. Draw perpendicular bisector to output line
6. Draw circle with Cas a centre
7. Draw perpendicular from Isc on x axis..
8. Divide IscL1in such a way that. LossCuStator
LossCuRotor
r
'r
LT
TI
1
2
1
sc
Isc
L2O
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y
R
P
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xI00
sc
C
Output line
11. FromR, draw line parallel to output linecrossing at P & P.
P is operating point
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
12. Join O and P. Cos1is operating pf.
1
Lebel O, T , L1 and L2
13. From P draw perpendicular on x axis
O
y
R
P
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xI00
sc
C
Output line
14. Determine the following1.Constant Losses and copper losses
=Core loss + F & W loss
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
L1L2=L1L2=constant losses
no load current I0
1
O
y
R
P
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xI00
sc
C
Output line
At standstill, input power = IscL2 L1L2=Constant Loss
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
Constant loss= Stator core loss +rotor core loss (f)
F & W loss=0
1
O
y R
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
At operating point P, input power = PL2,
1
L1L2=Constant Loss
Constant loss = Stator core loss + F & W loss
Rotor core loss 0 (sf)
Thus L1L
2=L
1L
2= Constant loss
O
y
R
P
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xI00
sc
C
Output line
At standstill,Stator Cu loss=TL1
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
rotor Cu loss = IscTAt P, stator Cu loss =TL1 and
1
rotor Cu loss = OT
O
y R
P
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L22. Output Power and Torque
1
Output Power = OP
The gap betnoutput line and circle is OUTPUTPower.
At I0, o/p=0, at Isc, o/p=0
O
Pmax
Max output power=PmaxO
Slip1 0
Ns0 Speed
Pmax
TL1L2O
y R
P
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
1
Output Torque = TPThe gap betntorque line and circle is OUTPUTtorque.At I0, torque=0, but at
Isc, torque=T Isc
Pmax
=Starting torque
Tmax
Max output torque=TmaxT
2. Output Power and Torque
Slip1 0
Ns0 Speed
Tmax
OTL1L2
O
TL1L2O
y R
P
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
1
Max Power and Max Torque are not occurring at same time
Contradictionto max power transfer theorem
Pmax
Tmax
2. Output Power and Torque
OTL1L2
O
TL1L2O
y R
P
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
1
Pmax
Tmax
OTL1L2
O
TL1L2
Air gap power Pg= Input power Stator Cu loss- core loss
=PL2-TL1-L1L2
3. Slip, Power factor and Efficiency
=PT
s = rotor Cu loss/Pg =OT/PT"
""
TP
TOsmp
max
'"
'"'"
TT
TOsmt
max
O
y R
P
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
1
Pmax
Tmax
OTL1L2
O
TL1L2
3. Slip, Power factor and Efficiency
O
Power factor cos1= PL2/OP
PO/PL2Efficiency=
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y R
P
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
1
s=0
O
Pmax
Tmax
T
5. Induction Generator
O
s=1
s=
braking torque
y R
P
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
1
s=0
(Generator)
O
Pmax
Tmax
T
5. Induction Generator
O
s=1
s=
braking torque
s= -ve
G
G
OG=Gen CurrentOG=Mech I/p
L2G=Active power
OL2=reactive powerPGmax
y R
P
P
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xI00
sc
C
Output line
Torque line
V1 Isc
T
P
P
L2
L1TO
L1
L2
1
s=0
(Generator)
O
Pmax
Tmax
T
5. Induction Generator
O
s=1
s=
braking torque
s= -ve
G
G
OG=Gen CurrentOG=Mech I/p
L2G=Active power
OL2=reactive powerPGmax
Slip 0 -1Speed 2Ns
Ns0Slip
Speed
Te
1
(Double Cage Motors)High Torque Cage Motors
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Lower starting current
Lower full load speed
(Double Cage Motors)
Highrotor resistance at the time of STARTgives
Higher starting torque
Better power factor
Poor speed regulation
More rotor ohmic loss
Reduced efficiency
Advantages
Disadvantages
Lowrotor resistance at the time of STARTOn the other hand
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Higher starting current
Higher full load speed
results in
Lower starting torque
Poor power factor
Better speed regulation
Less rotor ohmic loss
More efficiency
Disadvantages
Advantages
Therefore, it may be concludedthatHighrotor resistance at starting
Lowrotor resistance at running
are the desirable features.
conditions are obtained by usingIn SRIM or Wound Rotor IM these two
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which are called as High Torque Rotors.
1. Deep Bar Rotor
yexternal resistance
But in SCIM it is achieved by special rotor arrangement
There are two types of rotors
2. Double cage Rotor or Boucherot Rotor
1. Deep Bar Rotor
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A
B
Slot Leakage Flux
More Flux lines
Less Flux lines
Air Gap
MagneticMaterial
Types of deep bar
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yp p
Parallelsided bar
Conductor
Trapezoidalbar
L-bar T-bar
Portion B has more leakage fluxlines
More leakage reactance x
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A
B
Slot Leakage Flux
More Flux lines
Less Flux lines
Air Gap
MagneticMaterial
More leakage reactance xb
Less leakage reactance xa
At the time of startRotor freq. =f
More current at top (A)
Less current at B (skin effect)
This non-uniformdistribution of current
increases resistanceof lower part (B)
So highstarting torqueis developed
Under running condition,
Freq = sf
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A
B
Slot Leakage Flux
More Flux lines
Less Flux lines
Air Gap
MagneticMaterial
Freq = sfLeakage reactances of both parts decrease
Current is distributed uniformly
Resistance decreases.
Thus better performance
at start and at normalspeed
Equivalent Circuit Diagram
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r2/s
jx1 jx2
jXmRc
r1 I1 I2
I0Ic I
V1
Same phasor diagram of induction motorBut for calculations, r2 and x2must be consideredaccordingly
At start freq f and under running freq sf
2. Double cage Rotor or Boucherot RotorTwo cages topcage less cross section -
h h b l b
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Ax
Bx
a
b
ab
-A
-B
Dumb-bell
slotting
Staggered
slotting
Types
highresistance- brass, aluminium, bronzebottom cage more cross section -lowresistance- copper
SlitHeat
conduction
a= leakage flux of A - rotorb= leakage flux of B-rotor
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Ax
Bx
a
b
ab
SlitHeat
conduction
ab= leakage flux of rotor= mutual flux of A & B
b >> a
At the time of start,
rb+jxb>> ra+jxa
More Current flows through A
Zb>> Za
ra>> rb (ra= 5 to 6 times rb)
More resistance
More starting torque
Under running condition, freq = sfxa& xbare negligible and ra>> rb
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Ax
Bx
a
b
ab
SlitHeat
conduction
Therefore, Za >> Zb
More Current flows through B
Less resistanceGood operating characteristics
jx1 jx2r1 I1 I2
I0
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Equivalent
Circuit Diagram r2/sjXmRc
I0Ic I
V1
With top cage only
With bottom cage only
r3/s
jx1 jx3
jXmRc
r1 I1 I3
I0Ic I
V1
Equivalent Circuit Diagram
For double cage
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r2/s
jx1
jx2
jXm
Rc
r1 I1
I2I0Ic I
V1
r3/s
jx3
I3
All parameters are referred to statorAir gap power, Pg =
s
rI 2
2
2s
rI 3
2
3
Te =s
gP
s
rI
s
rI
ns
32
322
2
2
1
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Torque slip characteristics
First consider top cage(more resistance)
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Slip1 0
Ns0 Speed
Te
Now consider bottom cage(less resistance)
Double cage
Bottom cage
Comparison between single cageanddouble cage IM
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1. T-s characteristics
Slip1 0
Ns0 Speed
Te
Double cage
Single cage
Wide range of torque slip chcan be obtained by choice of
top and bottom cage
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2. Starting performance
Slip1 0
Ns0 Speed
Te
Double cage
Single cage
In case of double cage,higher starting torque canbe obtained
Due to more resistance,
suitable for Direct-On-LinestartingBut there is more heating
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3. Full load performance
Slip1 0
Ns0 Speed
Te
Double cage
Single cage
Double cage has higher r2and x2,
So lower breakdown torquelow power factor
low efficiency
low full load slip
4. Circle diagramAt the time of start, rb+jxb>> ra+jxa, Zb>> Za
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More Current flows through A
x
y
V1
O
A
B
Double Cage Single Cage
ExampleA 4 pole, 50Hz, double cage induction motor has
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the following per phase parameters referred tostator.
Stator: r1= 0.5, x1= 1.5
Rotor: Top cage: r2= 2, x2= 0.6
Bottom cage: r3= 0.4, x3= 3.4
Magnetizing current is zero. The primary in delta is
energized from 400V. Calculate the starting torque and full
load torqueat 4% slip using the approximate equivalent
Circuit. Find also the pfat starting and at full load
SolutionThe equivalent circuit diagram with I0=0 is
jx
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r2/s
jx1
jx2
r1 I1
I2
V1
r3/s
jx3I30.5
1.5
0.6
2/s
3.4
0.4/s
Total impedance at the input terminal is
j42.4
j3.40.4j0.62j1.50.5Z
j2.50431.657
056.513.003
Synchronous speed=2ns =225 =50rad/sec
St t t ti t I 400 133 2A
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Stator starting current, I13.003
400
Starting torque, Test resistancerotorequivalentI
s
21
1.157133.2
50
1 2
133.2A
Nm130.68
Pf at starting Lagging0.5518cos56.51
Total impedance at full load is
j40.042.4
j3.40.040.4j0.6
0.042
j1.50.5Z j3.7698.985
022.767439 .
F L stator current, I1 A9.743400
F L t T 8 4854001
2
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F L torque, Tefl 8.4859.743
400
50
1
Nm91.05
F L Pf Lagging0.922cos22.760
The starting torque is higher than F L torqueF L pf is better than starting pf
Example
A 6hp, 220V, 50Hz, 6 pole, 3-ph, star connected
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IM gave following test data:
No load test: 220V, 6A, 475W (Line values)Blocked rotor test: 110V, 27A, 1930W (Line values)
Calculate from the circle diagram, for full load condition,
line current, power factor, torque, slip and efficiencyAlso determine the maximum output, maximum torque, slipAlso determine the maximum output, maximum torque, slip
for maximum torque and starting torque.
The stator copper loss at standstill is twice the rotor
copper loss.
17.44A, 0.83 lag, 45.41Nm, 0.046, 79%
8.89hp, 73.14Nm, 0.134, 32.75Nm
Induction motors find wide application in
Starting of 3-ph IM
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Induction motors find wide application in
electrical drives because of
1.constructional advantages2.robustness3. cheapness
In industry, in order to attainhigh productivity
high quality products
it is required to PROPERLY
STARTCONTROL SPEED
STOP the motor
Depends on
Starting of SCIM
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Depends on
1.Size of motor
2.Type of load and3.Capacity of supply lines
Thereare principally TWO methods
1. Fullvoltage starting or Direct-on-Linestarting
2. Reducedvoltage starting
1. Fullvoltage starting or Direct-on-Linestarting
Simplest
inexpensiveAs the name indicates IM is directly connected to
supply mains
pply
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Fig: Direct- On-Line starting
At starting m/c has low pf
This large current may not harmrugged SCIM
but may cause objectionable voltage dropwhich may affect otherequipments
3-ph
Sup
Stator
Rotor
current 5 to 10times F L current
For example dimmingof lamps and tube lights
when refrigerator motor starts in the home.
Therefore, this method is applicable up to 3 HP.If starting timeis more, motor may burn.
Th i f i
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Te s
r
Is
22
2
1
The equation of torque is
Test
Tefl flfl
st
sr
r
I
I
/
1/
2
2
2
2
2
2
fl
fl
st sI
I 2
2
2
If no load current is neglected, then similar to transformer
I1st
I2st=
Rotor TurnsStator turns
I1stx Stator turns= I2stx Rotor turns
I1st= I2stx Ratio of Rotor turns to Stator turns
I1st= I2stx Ratio of Rotor turns to Stator turns
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I1st
I1fl=
fl
2
1fl
1st sII
At the time of start, I1st=Isc
1st 2st f
I1fl= I2flx Ratio of Rotor turns to Stator turns
I2st
I2fl
TestTefl
fl
2
1fl
sc sII
2. Reducedvoltage startingIn SCIM, it is not possible to change the rotor
i t
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Therefore, the starting current can only be reduced by
reducing the statorterminal voltage
There are three methods of reducing voltage
1.Stator reactor(or resistor) starting
resistance.
2.Auto-transformer starting
3.Star-delta starting
1.Stator reactor(or resistor) starting
Three reactors(or resistors) are connected in series withstator winding.The voltage applied to stator wdg is less than the voltage
drop across reactor.
Rotor1V 1xV
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Fig: Stator Reactor starting
The value of x is less than one.
stator and motor is started.
As speed increase, reactor is cut outin steps.
and finally no reactoris in circuit.
Stator
Rotor
Initially more reactoris in circuit, less voltageapplied to
Speed is equal to operating speed
Stator Reactor1V3
1
In place of reactors resistors can be used
Rotor1V 1xV
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Fig: Stator Reactor startingReactors are more costlythan resistors.
1. Low power lossin reactors
2. Effective reductionin voltage applied to stator
Stator
Rotor
But reactors are preferred due to
Stator Reactor1V3
1
V1
I1
VLV1
I1
VR
Vr
Rotor1V 1xV
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Fig: Stator Reactor starting
The starting current is
Stator
Rotor
The short circuit current is
Stator Reactor1V3
1
111st ZxVI /
11sc ZVI /
sc1st xII
fl
2
1fl
1st sII
Test
Teflfl
2
fl
sc2 sIIx
Thus,Starting torque with reactor starting
Starting torque with DOL starting
=x2
2.Auto-transformerstarting
1V
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Fig: Auto-transformer starting
Stator
Rotor
The fraction of xV1is applied to the stator wdg atstarting.
1V
1
xV
1. Voltage is changed by transformer action
2. So power loss and input current are less.
xV1
LIscst xII
As speed increases, gradually voltage is increased
Finally full voltage is applied to the motor.
Advantagesand not by dropping voltage as that of reactor
2.Auto-transformerstarting
1V
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Fig: Auto-transformer starting
Stator
Rotor
The stator starting current is
1V
1
xV
sc11st xIzxVI /
xV1
LIscst xII
For auto-transformer, input VA= output VA
ILV1=Ist (xV1)
IL=xIstIL=x2Isc
Therefore, line current at
inputis x2times the DOLcurrent.
fl
2
1fl
1st sI
I
Test
Tefl
fl
2
fl
sc2 sI
Ix
Thus,
2.Auto-transformerstarting
1V V
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Fig: Auto-transformer starting
Stator
Rotor
Line current at input due to auto-transformerstarting
1V
1
xVxV1
LIscst xII
Line current at input due to stator reactorstarting =x
Stator
Rotor
1V
1xVxV
1
LIscst xII
2.Auto-transformerstarting
1V V
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Fig: Auto-transformer starting
Stator
Rotor
Line current at input due to auto-transformerstarting
1V
1xVxV1
LIscst xII
Line current at input due to stator reactorstarting =x
Starting torque with auto transformerstartingStarting torque with DOLstarting =x
2
Starting torque with auto transformerstartingStarting torque with stator reactorstarting =1
3.Star-Deltastarting
For delta, 6 terminals are required.For star, 3terminals of stator wdg are required.
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Stator
Rotor
q
TPDT
R Y B
2- Run
1- Start - Star
- Delta
Fig.: Star-Deltastarting
Now make delta
Connection.
At starting TPDT to 1, wdg in star
Motor rotates.Reduced voltage is applied to wdg = VL/3
Th st ti t is
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Stator
Rotor
TPDT
R Y B
2- Run
1- Start - Star
- Delta
Fig.: Star-Deltastarting
The starting current is
1Lst.y zVI 3/
L.yI StartingLine current
Now TPDT to 2- Delta
Line voltage appliedto wdg. Motor runs at rated speed
At starting TPDT to 1, wdg in star
Motor rotates.Reduced voltage is applied to wdg = VL/3
Th st rtin curr nt is
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Rotor
Stator
TPDT
R Y B
2- Run
1- Start - Star
- Delta
Fig.: Star-Deltastarting
The starting current is
1Lst.y zVI 3/
L.yI StartingLine current
Now TPDT to 2- Delta
Line voltage appliedto wdg Motor runs at rated speed
At starting TPDT to 1, wdg in star
Motor rotates.Reduced voltage is applied to wdg = VL/3
The starting current is
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At starting, if, wdg in deltaThe starting current is
1Lst.d zVI / sc.dI
st.dL.d II 3
st.dst.y II3
1
Starting line current with Y-starter
Thus Ist.yin star is one third of that current in delta.
= Ist.y3 Ist.d
= 13Starting line current with stator in
The starting current is
1Lst.y zVI 3/
L.yI StartingLine current
Now TPDT to 2- Delta
Line voltage appliedto wdg Motor runs at rated speed
This shows that T in star is one third of starting torque
Starting torque with Y-startingStarting torque with stator in
= (V1/3)2
V12= 1
3
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In case of auto-transformer, if turn ratio x = 1/3
Then starting line current and is starting torque are
This shows that
This shows that Tst.yin star is one third of starting torquein delta.
Star delta starting is equivalent to auto transformer
reduced to one third of their values with delta.
if auto transformer turn ratio x=1/3=0.58 or 58% tapping
This method is cheap, effective and used extensivelyUsed for tool drives, pumps, motor-generator set.
Used up to rating of 3.3kV,
After this voltage, m/c becomes expensive for delta winding
that the supply current during starting of IM does not
ExampleDetermine the % tappingof the auto-transformer so
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exceed 1.5 times full load current. The shortcircuit current
on normal voltage is 4.5 times the full load currentand the
full load torque.
that the supply currentduring starting of IM does not
Solution
full load slip is 3%.Calculate the ratio of starting torque
Stator
Rotor1V
1xVxV1
LIscst xII
IL=1.5IFLIL=1.5IFL
Isc=4.5IFL
IL/Isc=0.333
In auto-transformer IL/Isc=x2 x=0.577
Hence % tapping is 57%
fl
2
1fl
1st sII
Test
Teflfl
2
fl
sc2 sIIx
Now
0 04 50 2
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Stator
Rotor1V
1xVxV1
LIscst xII
0.034.50.333 2
0.202
its full load current, the stator of which is arranged for star
ExampleThe short circuit line current of a 6hpIM is 3.5times
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delta starting. The supply voltage is 400V, full load effnis
82%and full load power factor is 0.85% (lag).
Neglect magnetizing current.
ts fu a curr nt, th stat r f wh ch s arrang f r star
Solution
Calculate the line currentat the instant of starting.
6hp IM,
Star-delta starting
Isc=3.5IFL
Isc (line) =3.5 IFL
Voltage =400V
FL=82%, pf=0.85 (lag)
P=3 VLILcos
0.854003
1
0.82
7466I
LIFL=
=9.26A (line current fordelta)
=5.34A (phase currentfor delta)
=18.73AIsc=3.5IFL=3.5x5.34
A h i f i d i i
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At the instant of starting, motor wdg is in star
For star, line current is equal to phase current.ILat the instant of start =18.73A for delta (400V)
ILat the instant of start =18.73/3 A for star (400/3)=10.81A
All the methods used for SCIM can also be usedf SRIM
Starting of SRIM
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for SRIM
But the justification of name SR is done if this motor isstarted by external rotor resistance.
This is the simplest and cheapest method
It increases starting torque
It decreases starting currentIt is possible to produce starting torque = maximum torqueIt improves starting power factor.
Rotorin Star
Stator
Rotor
in Delta
Rotor wdg is connected to slip ring, which isStarting of SRIM
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connected to external resistancethrough brushes.
At starting, entire external resistance is connected to rotorwinding. As the motor accelerates, resistance is cut outin
steps so that torque remains maximumduring acceleration.
At operating point, this resistance is fully cut offandand slip rings are shorted.
Rotorin Star
Stator
Rotor
in Delta
Calculation of starter resistance stepsn+1nn-1
R
3
R
2
R
1
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Consider one phase with rotor resistance r2
Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1R3
R2
R2R1
R1
Let R1, R2, R3, Rnbe step resistances
Let 1, 2, 3, n, n+1 be studsof starter
Let R1, R2, R3, Rnbe totalresistances of respective studs
R1=R1+ R2+ R3 +
+ Rn+ r2 R2=R2+ R3+ R4 +
+ Rn+ r2
R3=R3+ R4+
+ Rn+ r2 Rn=Rn+ r2
Rn+1
= r2
n-section starter orn-step starterorn+1stud starter
Calculation of starter resistance stepsn+1n
R
n-1
R
3
R
2
R
1
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2. During start up, load torqueremains constant
Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1R3
R2
R2R1
R1
1. No load current is neglected
3. Input current fluctuates between I1maxand I1min.
Some assumptionsare made
I1max
t
I1min
Calculation of starter resistance stepsn+1n
R
n-1
R
3
R
2
R
1OFF ON
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The input current shoots up to I1max
Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1R3
R2
R2R1
R1
When supply is given, handle is at stud 1
I1max
t
I1min
s1
-Off-total rotor resistance is R1
2TH2
2
1
1TH
11max
Xxs
'RR
VI
statortoreferredareparametersSpeed increases
n+1n
R
n-1
R
3
R
2
R
1
R
OFF
ON
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The input current shoots up to I1max
Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1R3
R2
R2R1
R1
When supply is given, handle is at stud 1
I1max
t
I1min
s1
-Off-total rotor resistance is R1
2TH2
2
1
1TH
11max
Xxs
'RR
VI
statortoreferredareparametersSpeed increases
Current decreases to I1min
Slip decreases to s2
s2
N
t
2TH2
2
2
1TH
11min
Xxs'R
R
VI
n+1n
R
n-1
R
3
R
2
R
1
R
OFF ON
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Resistance decreases, current increases to I1max
Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1
R3
R2
R2R1
R1
I1max
t
I1min
s1
2TH2
2
2
2TH
11max
Xxs
'RR
VI
Now speed increasess2
R1
cut
Handle moves from stud 1 to stud 2
with same speed and slip=s2
n+1n
R
n-1
R
3
R
2
R
1
R
OFF
ON
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Resistance decreases, current increases to I1max
Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1
R3
R2
R2R1
R1
I1max
t
I1min
2TH2
2
2
2TH
11max
Xxs
'RR
VI
Now speed increases
R1
cut
Handle moves from stud 1 to stud 2
with same speed and slip=s2N
t
2
TH2
2
3
2
TH
11min
Xxs
'R
R
VI
Current decreases to I1min
Slip decreases to s3
s3s1 s2
n+1n
R
n-1
R
3
R
2
R
1
R
OFF ON
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Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1
R3
R2
R2R1
R1
I1max
t
I1min
2TH2
2
3
3TH
11max
Xxs
'RR
VI
Now speed increases
R1
cut
s3s1 s2
Resistance decreases, current increases to I1max
Handle moves from stud 2 to stud 3
with same speed and slip=s3
R2
cut
n+1n
R
n-1
R
3
R
2
R
1
R
OFF
ON
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Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1
R3
R2
R2R1
R1
I1max
t
I1min
2TH2
2
3
3TH
11max
Xxs
'RR
VI
Now speed increases
R1
cut
s3s1 s2
Resistance decreases, current increases to I1max
Handle moves from stud 2 to stud 3
with same speed and slip=s3
R2
cut
N
t
2TH2
2
4
3TH
11min
Xxs
'RR
VI
Current decreases to I1min
Slip decreases to s4
s4
n+1n
R
n-1
R
3
R
2
R
1
R
OFF ON
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Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1
R3
R2
R2R1
R1
I1max
t
I1min
2TH2
2
4
4TH
11max
Xxs
'RR
VI
Thus I1max& I1minare obtained
R1
cut
s3s1 s2
Resistance decreases, current increases to I1max
Handle moves from stud 3 to stud 4
with same speed and slip=s4
R2
cut
s4
R3
cut
n+1n
R
n-1
R
3
R
2
R
1
R
OFF ON
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Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1
R3
R2
R2R1
R1
1
1
sR '
sm=sflif I1max=Ifl
From all I1maxequations, we can write
2
2
sR '
n
n
3
3
sR
sR '.......'
msr
sR
1n
1n 2'
flsr2
Entire resistance is cut off.
From all I1minequations, we can write
2
1
s
R '
3
2
s
R '
1n
n
4
3
s
R
s
R '.......
'
(1)
m
s
Rn'(3)
mm s
rr
s
sR1,equFrom 1
n 22
1'
(2)
n+1n
R
n-1
R
3
R
2
R
1
R
OFF ON
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Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1
R3
R2
R2R1
R1
1
2
ss
Equn 1/3 gives
2
3
ss
n
1n
3
4
ss
ss .......
nssm
'
'
1
2
R
R
2'
3
R
R '
(4)
(5)'
'
n
1n
R
R '
2
nR
r )(, assume
'' 12 RR '' 23 RR '
2
1R
'' 11n RR n
'21
Rnr (6)
From equn 2,m
n
sr 2r2
m
n s
or n
ms /1
(7)
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n+1n
R
n-1
R
3
R
2
R
1
R
OFF ON
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Rotor
Rnr
2
Rn-1
Rn
R3
Rn-1
R3
R2
R2R1
R1
Steps for the design of starter1. Calculate s
m2. Calculate 3. Calculate R1
m
2
s
r
4. Calculate R1= R1(1-)5. Calculate R2= R1
6. Calculate Rn= n-1R1
R1 is referred to stator
In order to calculate actualvalue of rotor resistance
transfer R1to rotor
ExampleDesign a 5-sections of a 6-studsstarter for a 3-ph slip ringIM. The full load slip is 2%and the max starting current
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is limited to twicethe full load current. Rotor resistance per
phase is 0.03. (assume no of turns equal)
SolutionFull load current
2TH22
2TH
11FL
XxsrR
VI
For small slip, r2/s >> RTHand X srV
I2
11FL
Now starting current I1maxis TWICE the full load current
m2
11max sr
VI m2
11FL sr
VI 2m
2
1
2
1 srV(0.02)
rV 2 0.04sm
Here no of sections are 5 = 0.525R1=r2/sm
1/nms )(
= 0.75
65
R5r2
4
R4R
3
R3R R
2
R2R
1
R1
OFF ON0.356 0.187 0.098 0.052 0.027
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The resistance of various elements are
R1 =
R2=
R1(1-) = 0.075(1-0.525) = 0.356
R1= 0.525x0.356= 0.187
R3=2R1= R2 = 0.525x0.187= 0.098
3
R1=R3 = 0.525x0.098= 0.052
R4=
4R1= R4 = 0.525x0.052= 0.027R5=
0.03r2 =
Rotor
r2Rn Rn-1R3 R2 R1
0.03
Total resistance R1=0.075
In industrial application, for wide speed range
and for smooth speed control DC motors are preferred
Speed control of IM
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and for smooth speed control, DC motors are preferred
But it is required to convert AC to DC.The capital cost of DC motor is higher than AC motor.
Due to this reasons, preference is given to IM.
IM is 1. Cheaper 2. Robustin construction
3. More economical to operate and maintain.The disadvantages are 1. Restricted small range of speed.
2. Lowoperating pf.
The operating speed of IM is given bynr=ns(1-s)Thus speed can be controlled by
varying syn speed ns=2f/pwhich can be controlled by1. Pole changing 2. freq changing
changing SlipThus speed can also be controlled by
which can be controlled by
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3. Varying line voltage
4. Varying rotor circuit resistance
5. Slip PowerControl
Methods 1, 2 and 3 are carried out from stator side.
Methods 4 and 5 are carried out from rotor side.
Methods 1, 2 and 3are applicable to SCIM and SRIM
Methods 4 and 5 are applicable only to SRIM
a) Cascading of Induction Motors
b) Rotor Voltage Injection
Since no of poles can be changed only by evenb th d t l i t ti b t t d
1. Pole changing methodSpeed is inverselyproportional to no of poles.
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numbers, the speed control is not continuousbut a stepped
one.However, this is very economicalmethod and motor can
operate on quite hardcharacteristics.
The no of poles can be changed by three methods.
1. Use of Multiple Stator Winding
Stator has separatewindings with differentno of poles.
This is not suitable for wide rangeof speed control
Winding having less no of poles is used for high speedandhaving moreno of poles is used for low speed.
TPDTis used for changeover. The unused winding is kept
open to avoid circulating current and heating.
to produce different no of poles for differentd
2. Use of Consequent Pole TechniqueNo multiple stator wdg but wdg is re-connected
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speed.
This method is suitable for SCIM sincerotor wdg adjustautomatically.
For WRIM, rotor wdgmust be designed for same no of poles.
Consider 4 coils of one phase
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12 3
4
Up=oUt of the plane-coming toward U=DOT
dowN=iNthe plane-going away from you=CROSS
Upwardflux
Downwardflux
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1 2 34
If bis connected to c
ab
cd
a b
c d
N
flux
S N S N S N SS
Series connection, 8 poles, low speed
a b
c d Parallel,8 poles,low speed
8 poles
OR
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1 2 34
Now connect bto d
ab
cd
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1 2 34
ab
cd
N S N S
4 poles
Now connect bto d
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1 2 34
Conclusions
ab
cd
N S N S
4 poles
In wdg if current flows from a to b and from c to d,8 poles ( Low speed)are produced
In wdg if current flows from a to b and from d to c,
4 poles ( High speed) are produced
Connection of three phases can be in delta or star
bd
B bc
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a
b
c
Series Delta, 8 poles, Low speed
Y
= ad
R
R
Y
B
a
bd
c
Series Delta, 4 poles, High speed
Y
B
= a
bd
c
R
RY
B
For parallel delta,voltage of wdg
increases
a
bc
S i St 8 l L d
R
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d
Series Star, 8 poles, Low speed
YB a
bd
c
Series Star, 4 poles, High speed
YB
R
ac
Parallel Star, 8 poles, Low speed
R
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bd
, p , p
YB
a
b
d
cParallel Star, 4 poles, High speed
YB
R
This is an elegant and flexiblemethod ofpole changing
3. Pole Amplitude Modulation TechniquePoles are changed by modulationtechnique.
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pole changing
Consider sinusoidally distributed MMF wavesproduced bybalanced 3-ph wdg.
FR ,
2
PsinF FY ,
3
2-
2
PsinF FB
3
2
2
PsinF
where P is no of poles and is the mech angleLet these MMF waves be modulated by another sinusoidal
modulating MMF waves.
FMR ,
2
PsinF
MM FMY ,
3
2-2
PsinF
MM FMB ,
3
2
2
P
sinFM
M
Modulated meansproduct of these two waves
Therefore, modulated MMF waves are given by
FR
2
PsinF2PsinF MM= FR x RMR
2
PPcos
2
P-PcosFF
2
1 MMM
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FY
3
2
-2PsinF32
-2PsinFMM
FB
Consider first termThese are co-phasal in space, hence net MMF is zero.
222
= FY x RMY
3
4-
2
PPcos
2
P-PcosFF
2
1 MMM
= FB x RMB
3
2-
2
PPcos
2
P-PcosFF
2
1 MMM
The second term will produce rotating magnetic field in
opposite direction depending on no of poles P+PM
.
Again if the modulating MMFs areFMR ,
2
PsinF MM FMY ,
3
2
2
PsinF MM
F
2
P
sinF M
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FMB ,
3
2
sinFM
Then, modulated MMFwaves are given by
FR
FY
FB
2
PPcos
2
P-PcosFF
2
1 MMM
2
PPcos3
2-2
P-PcosFF2
1MMM
2
PPcos
3
2
2
P-PcosFF
2
1 MMM
The second terms are co-phasal in space, hence net MMF iszero.
The first term will produce rotating magnetic field in
direction of rotation, depending on no of poles P-PM.
Thus four speeds are possible1. Speed depending on no of poles P
2. Speed depending on no of poles PM
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p p g p M
3. Speed depending on no of poles P+PM4. Speed depending on no of poles P-PM
Therefore, smooth steplessspeed control can be
2. Variation of supply frequencySpeed is directlyproportional to frequency (2f/p)
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possible by smooth variation of frequency.The voltage equation is V=4.44fmTphkw
If f is changed, for constant voltage, flux will change.
So, characteristicwill change.In order to retain high hardness ch, flux should be
maintained constant
The resultant air gap flux is m fVK4.44T1 wph
Thus in order to avoid saturation, V/f is maintained constant.
The variable frequency can be obtained by1. Rotating frequency changer
2. Adjustable frequency generator
3 S lid i f h
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3. Solid state static frequency changer
a). Cycloconverter(output freq lessthan input freq)
b). Rectifier-Inverter(V/f control)(used)
Torque-slip characteristics
V f, Reactance f,speed f
1. Slip at max torque,smt2THTH
2
jxjXR
r
Neglecting stator impedance,smt2
2
x
r
Thus slip at max torque is inversely proportional to freq.
2
2
Lf2
r
mNr
xXrR
V
322
2TH
2
2TH
2
TH
s
2. The starting torque
Test
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22
2TH
2TH
s
rxX
V
n2
3
3f
1K
Thus Testis inversely proportional to f3, if V is constant.
Neglect RTH+r2Test
2
22TH2
2TH r
lLf2
V
2f
P
2
3
Test
The Testis inversely proportional to f, if V/fis constant.
22
THTH
2TH
s XRR2
V
33. The maximum torque
Tem
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2TH
2
TH
s xX2
V
n2
3
2f
1K
Thus Temis inversely proportional to f2, if V is constant.
Neglect RTHTem
2TH
2
TH
lLf22
V
2f
P
2
3
Tem
The Temis unchanged, if V/fis constant.
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The torque-slip characteristicsIf V/f is constant
1. smtisinversely proportional to freq.
2 T i i l t f
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3. Temis independent of f.
2. Testisinversely to f.
Slip1 0Ns0 Speed
Te
f1 f1
smt1
f2f2r2 >TL
N1N2
N3
N4
Thus rotor speeddecreases withincrease in r2
Advantages:1. Simple 2. It is possible to obtain max Test.
3. Low starting currentDisadvantages
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g
1. Speed below syn speedcan only be obtained2. Reduced effnat low speed due to more rotor ohmic loss3. Poorspeed regulation
With negligible r1, the slip at max torque is Xr
s 2mt
This can be written as ,XRs 2mt R2=r2+ additional resistance
The value of smt can be obtained from ,
s
s
ss
2
T
T
mt
mt
emt
e
mt2 sXR
Under normaloperating condition, for load TL1,2
1THL1 r
sKT
For load TL2,2
2THL2 R
sKT
2
2
2
1
L2
L1
r
R
s
s
T
T
L2
L1
1
222 T
T
s
srR
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there is power loss.
5. Slip Power ControlDue to addition of rotor circuit resistance
Without power loss, slip poweris utilized by suitable slip
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power converter to control the speed.a) Cascade connection of IM
Sometimes called as tandem control
p , p p y p
Not used now-a-days
Only historical importance
MIM1
f1
AIM2
f2=s1f1
SRIM SRIM
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The actual speed is )PPP
(1P
120f
N 21
2
1
1
r
21
1
PP
120f
Speed is dependent on P1+P2
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Speed is dependent on P1+P2.If two torques are in same direction, then the scheme is
called as cumulatively cascadedIf AIM terminals are interchanged, then it is called as
differentially cascadedarrangement and its speed is
)s(1P
120fN 2
2
2r2 )s(1P
f120s2
2
11 )s(1P
120f1
1
1
12
21 PP
Ps
The actual speed is )s(1P
120fN 1
1
1r
21
1
PP
120f
This is possible only when P1 P2
b) Rotor Voltage Injection or
The air gap power is Pg=sPg+(1-s)Pg
sPgis rotor ohmic loss, wasted power, can be used for
Injection of slip frequencyemf in rotor circuit
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g , p ,
speed control. This power can be added to the shaft or
returnedto the supply.
Due to this injected voltage, motor becomes DOUBLY
excited IM. This can be used for speed and pfcontrol.
If this injected voltage, is opposite to rotor emf, speed
decreasesand motor runs at sub-synchronousspeed
If this injected voltage, is additive to rotor emf, speed
increasesand motor runs at super-synchronousspeed
The various methods for injecting slip freq emf are
1) Lablank Exciter
Armwdg
stator
Arm wdg
Stator Ring
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RotoryFrequency Converter (FC)
w g
stator
Coupled to the rotor of IM
DC armwith commutator, brusheson one end andslip ringson the other end
MIM
f
FC
Voltageregulating
device
1) Lablank Exciter
Armwdg
stator
Arm wdg
Stator Ring
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Slip rings of MIMare connected to brushes of FC
g
stator
Slip rings of FC are connected to voltage regulating device
or transformer and then to supply
R YB
MIM
f
FC
Voltageregulating
device
R YB
1) Lablank Exciter
Armwdg
stator
Arm wdg
Stator Ring
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MIM and FC are having equal no of poles
g
statorSame speed
If rotor of FC rotates in opposite direction, thenfield speed
R YB
MIM
f
FC
Voltageregulating
device
R YB
Ns-Nr= Ns-Ns(1-s)=sNsBrush frequency of FC, fb=sNsP/2= sf
1) Lablank ExciterThus at any speed, fbis equal to slip freqand suitable for injectioninto rotor circuit if IM
If brush voltage is less thanIM slip voltage, current flows
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R YB
MIM
f
FC
Voltageregulating
device
R YB
f brush voltage s less than M sl p voltage, current flows
from rotor of IM to brushes of FC to slip rings and supplyMotor runs at Sub-Synchronous speed
If brush voltage is more thanIM slip voltage, current flowsSupply- VRD- slip rings of FC -brushes of FC -rotor of IM
Motor runs at Super-Synchronous speed
Power ischanging
ConstantTorqueDrive
1) Lablank ExciterConstant Power DriveR YB
f
R YB
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MIMSynM/c
FCSynMotor
MIM Syn m/c
=equal no of polesOn one shaft=same speed, Nr
Speed of FC =Ns, due to Syn motor
Nr=Ns(1-s)freq=f(1-s)
Speed of rotating magnetic field of FC=2f(1-s)/P=Ns(1-s)If Nr & Ns are in oppositedirection,Ns-Ns(1-s)=sNs
Freq of emf across brushes of FC=sfSuitable for injectioninto rotor ckt of MIM
1) Lablank ExciterConstant Power DriveR YB
f
R YB
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MIMSynM/c
FCSynMotor
This emf can be varied by
excitation and brush angleIf this emf is more thanslip voltagePower flows from FC-Rotor of MIM-Shaft-Syn m/c (g)- FC
Nr=Ns(1-s)freq=f(1-s)
MIM operates at Super-Synchronous speedIf brush emf is less than slip voltagePower flows from Rotor of MIM- FC-Syn m/c (m)-Shaft- rotor of MIM
MIM operates at Sub-Synchronous speed
1) Lablank ExciterConstant Power DriveR YB
f
R YB
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MIMSynM/c
FCSynMotor
Power is taken and fed to same shaftTherefore, constant power drive
Nr=Ns(1-s)freq=f(1-s)
2) Kramer System
R YB
fVoltageregulating
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MIM
f
ACM
device
If brush emf is more than slip voltagePower flows from ACM-Rotor of MIM.
MIM operates at Super-Synchronous speedIf brush emf is less than slip voltage
Power flows from Rotor of MIM- ACM.MIM operates at Sub-Synchronous speedSince power is flowing from one machine to another with one
shaft, it is constant power drive.
3) Scherbius SystemR YB R YB
f
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Power changes
At Super-Synchronous speed, power flows from supply-AIM
(Motor) - ACM -rotor of MIM.
MIM
ACM AIM
Voltageregulatingdevice
Constant torque drive
At Sub-Synchronous speed, power flows from rotor of MIM
- ACM AIM (Gen) - supply.
f
4) Static Slip Power Recovery Scheme(Rectifier Inverter Scheme)Constant torque driveR YB
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from Rotor Bridge A Bridge B Transformer - Supply
Lablank, Kramer,and Scherbius methods require Auxi M/c.Here static, thyristerized ckts, two bridges A & Bare used
If rotor emf is more than bridge A voltage, then power flows
MIM
f Transformer
Bridge A Bridge B
PhaseShifter
Inductor
4) Static Slip Power Recovery Scheme(Rectifier Inverter Scheme)Constant torque driveR YB
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from SupplyTransformer Bridge B Bridge A - Rotor
IM runs at sub-synchronousspeed.Bridges A acts as a rectifier, Bridge Bacts as a Inverter
If rotor emf is less than bridge A voltage, then power flows
MIM
f Transformer
Bridge A Bridge B
PhaseShifter
Inductor
4) Static Slip Power Recovery Scheme(Rectifier Inverter Scheme)Constant torque driveR YB
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injected emf.
IM runs at super-synchronousspeed.Bridges B acts as a rectifier, Bridge A acts as a Inverter
The phase shifter is used to control the phase of the
MIM
f Transformer
Bridge A Bridge B
PhaseShifter
Inductor
In modern electrical drives, it is frequentlynecessary
Electrical Braking of IM
to stopthe motor quicklyin an exact position
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to decelarate the motorto reverse the direction of rotationof motor
The quality of the productand the productivityof an unit
are often dependent on braking
Types of braking: 1. Electrical, 2. Mechanical
In electrical braking, electrical torqueis opposite to the
direction of rotationof rotor.
Types of electrical braking:1. Regenerativebraking
2. Counter-Current braking or Plugging
3. DC Dynamicor Rheostatic or AC Dynamic Braking
Ns N N N N N
1. RegenerativebrakingRegenerative = returning
useful energy back to
linesThis braking takes place the load forcesto run the motor
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g p f m
above syn speed. For example, cranes or hoists
Slip becomes negative and IM operates as IG.
1 0
Ns0
Te 4 poles8 poles
TLSpeed
Slip
A
If regenerative braking isapplied at A
Poles are changed from 4 to 8A shifts to B
B
B shifts to CC shifts to D
C
D
From B to D, regenerative
brakingand IGoperation
Slip = (OC-OB)/OC =-ve
1. RegenerativebrakingSuitable for SCIM
It is not required to change no of poles of rotor
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1 0
Ns0
Te 4 poles8 poles
TLSpeed
Slip
A
B
C
D
+Te-Te
TL
Two quadrant operation
A
B
D
Ns N N N N N
Ns N N N N N2. Plugging:
Any two leadsare interchanged. The direction of rotating
or Counter-Currentbraking
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magnetic field getsreversed.This is a braking torquewhich stopsmotor quickly.
1 0
Ns0
Te
TLSpeedSlip
A
CD
2
-Ns
The Tegets reversed.
B
Tb=TL+T1
TL
At Csupply should be disconnectedto stop the motor.
EF
Te-Slip Ch for SCIMduring Plugging
T1
otherwise motor starts inreverse direction
Braking
Braking
Motoring
Motoring
Againg ifplugging isdone
Dto E
E to F
and F to A
Pt A to B
B to CC to D
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During plugging slip increasesSlip emfincreases.
To limit this current, extra R2can be inserted in rotor ckt.Currentincreases
Te-Slip Ch for SRIMduring Plugging
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1 0
Ns0
Te
TLSpeedSlip
A
C
D2
-Ns
B
Tb=TL+T1
TL
This R2increases Backward Torque T1.
E
H
T1
Braking
Braking
Moto