chapter 3_induction motor cont
TRANSCRIPT
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CHAPTER 3:INDUCTION MOTOR
(CONT.)
NJFKEE, UMP
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3.5 PRINCIPLE OF OPERATIONThe AC IM is a rotating electric machinedesigned to operate from a three-phase sourceof alternating voltage.
The stator is a classic three phase stator withthe winding displaced by 120.
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Rotating Magnetic FieldWhen a 3 phase stator winding is connectedconnected to a 3 phase voltage supply, 3 phasecurrent will flow in the windingsflow in the windings, which also will inducedinduced 3 phase flux in the stator.These flux will rotate at a speed called a Synchronous Speed, nSynchronous Speed, n ss . The flux is called
Rotating magnetic FieldSynchronous speed = speed of rotating flux:
Where; p = is the number of poles, andf = the frequency of supply
p f n s 120=
3.5 PRINCIPLE OF OPERATION
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At synchronous speed, there is no difference betweenrotor speed and rotating field speed
so no voltage is induced in the rotor bars and no currentflow, hence no torque is developed.
Therefore, when running, the rotor must rotate slower than the magnetic field.
The rotor speed is just slow enough to cause the proper amount of rotor current to flow, so that the resultingtorque is sufficient to overcome windage and frictionlosses, and drive the load.
3.5 PRINCIPLE OF OPERATION
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Slip,Slip
, ssThis speed difference between the rotor and magnetic field iscalled slip:
Where; n s = synchronous speed (rpm)n r = mechanical speed of rotor (rpm)
under normal operating conditions, s= 0.01 ~ 0.05, which is verysmall and the actual speed is very close to synchronous speed.Note that : s is not negligibles is not negligible
)1( snnORn
nn s sr s
r s =
=
3.5 PRINCIPLE OF OPERATION
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Rotor SpeedRotor Sp
eed When the rotor move at rotor speed, n r (rps), the stator fluxwill circulate the rotor conductor at a speed of (n s-n r ) per
second. Hence, the frequency of the rotor is written as:
sf
pnn f r sr
=
= )(
f s f iii
ii pnn
f
nn Rotor At
i pn
f
n stator At
Note
r
r sr
p f
r s
s
p f
s
.:)()(
).....(120
)(
:
).....(
120
:
:
120
120
=
=
=
=
=
3.5 PRINCIPLE OF OPERATION
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Torque producing mechanismTorq ue producing mechanismWhen a 3 phase stator winding is connectedconnected to a3 phase voltage supply, 3 phase current will flowflowin the windingsin the windings, hence the stator is energized.
A rotating flux, is produced in the air gap. The induces a voltage E a in the rotor winding (like atransformer).The induced voltage produces rotor current, if
rotor circuit is closed.The rotor current interacts with the flux ,producing torque. The rotor rotates in thedirection of the rotating flux.
3.5 PRINCIPLE OF OPERATION
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Q: How to change the direction of rotation?
3.5 PRINCIPLE OF OPERATION
A: Change the phase sequence of the power supply.
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Note: Never use three-phase equivalent circuit. Always
use per- phase equivalent circuit.
The equivalent circuit always bases on the Y always bases on the Y connection regardless of the actual connection of connection regardless of the actual connection of the motor the motor.
IM equivalent circuit is very similar to the single-
phase equivalent circuit of transformer. It iscomposed of stator circuit and rotor circuit
3.6 EQUIVALENT CIRCUIT
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3.6 EQUIVALENT CIRCUIT
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Step2 Rotor winding is shorted(Under normal operating conditions, the rotor winding is shorted. The slip is s )
Note:the frequency of E 2 is f r =sf because rotor is rotatingrotor is rotating .
f f r
3.6 EQUIVALENT CIRCUIT
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Step3: Eliminate f 2
Keep the rotor current same:
3.6 EQUIVALENT CIRCUIT
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Step 4: Referred to the stator side
Note: X 2 and R 2 will be given or measured.
Always refer the rotor side parameters to stator side.R
c represents core loss (of stator side).
3.6 EQUIVALENT CIRCUIT
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IEEE recommended equivalent circuit
Note:R
c is omitted. The core loss is lumped with therotational loss.
3.6 EQUIVALENT CIRCUIT
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IEEE recommended equivalent circuit
Note: can be separated into 2 PARTSNote: can be separated into 2 PARTS
Purpose :Purpose :to obtain the developed mechanicalto obtain the developed mechanical
I1 1 R1 X
m X
'2 X
'2 R
s s R 1'21V
s
R2
s s R
R s
R )1(22
2 +=
3.6 EQUIVALENT CIRCUIT
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Analysis of Induction MachinesFor simplicity, let assume
II ss=I=I 11 , I, I R R =I=I 22
(s=stator, R=rotor)
[ ] Rm sTotal
s s s
cmm
cmcm
R R
R
Z Z Z Z
jX R Z
neglected R jX Z neglected R jX R Z
jX s
R Z
//
;
;;//
;''
+=+=
===
+=
ZZRRZZmm
ZZss
VVs1s1
IIs1s1 IIm1m1 IIR1R1
T
s
s Z
V I
1
1=
3.6 EQUIVALENT CIRCUIT
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=
=
=
m
RM m
R
RM R
sT
m R RM
Z
V I
Z
V I Hence
V Z
Z Z V
Rules Dividing Voltage
11
11
11
,
//
,
ZZRRZZmm
ZZss
VVs1s1
IIs1s1 IIm1m1 IIR1R1
11
11
,
s Rm
m R
s Rm
Rm
I Z Z
Z I
I Z Z Z
I
Rules Dividing Current
+=
+=
OR
Note : 1hp =746WattNote : 1hp =746Watt
Analysis of Induction Machines
3.6 EQUIVALENT CIRCUIT
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Example 11. A 4 poles, 3 Induction Motor operates from a
supply which frequency is 50Hz. Calculate:a. The speed at which the magnetic field is rotatingb. The speed of the rotor when slip is 0.04c. The frequency of the rotor when slip is 3%.d. The frequency of the rotor at standstill
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2. A 500hp, 3 6 poles, 50Hz Induction Motor has a speed of 950rpm on full load. Calculate the slip.
Example 2
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3. If the emf in the rotor of an 8 poles Induction Motor has afrequency of 1.5Hz and the supply frequency is 50Hz.Calculate the slip and the speed of the motor.
Example 3
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4. A 440V, 50Hz, 6 poles, Y connected induction motor israted at 135hp. The equivalent circuit parameters are :
R s=0.084 R R=0.066 Xs=0.2 XR=0.165 s = 5% X m=6.9
Determine the stator current, magnetism current and rotor current.
SolutionGiven V=440V, p=6, f=50Hz, 135hp
]63.9685.32,76.1344.170,34.246.177[ A A Aooo
Example 4
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Example 4 (Cont 1 st
Method)
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Example 4 (Cont 2 nd
Method)
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Power Flow Diagram
PP inin (Motor)(Motor)
PP inin (Stator)(Stator)
PP corelosscoreloss(P(P cc ))
PP air Gapair Gap (P(P agag ))
PP
developeddevelopedPP mechanicalmechanicalPP convertedconverted
(P(P mm ))
PP out,out, PP oo
PP stator copper stator copper loss,loss, (P(P scuscu ))
PP rotor copper rotor copper lossloss (P(P rcurcu ))
PP windage, friction, etcwindage, friction, etc (P(P --
Given)Given)
cos3 s s I V
s R
I R R'
'3 2
2
3
c
RM
RV ''3
2 R R R I
s s
R I R R1
''3 2
W hp 7461 =
s s R I 23
PP inin (Rotor)(Rotor)
3.6 EQUIVALENT CIRCUIT
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Ratio:Ratio:
Pag P rcu Pm
s R
I R
R
''3
2
''32
R R R I
s s
R I R R1
''32
s
11
1 s
1
1 s s1
Ratio makes the analysis simpler to find the value of theparticular power if we have another particular power. For example:
s
s
P
P
m
rcu
=
1
Power Flow Diagram
3.6 EQUIVALENT CIRCUIT
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Efficiency
Watt xW hp x P
I V P otherwise
P P P
P P P givenare P if
P P
out
s sin
mo
lossesino
losses
in
out
746746
cos3,
,
%100
===
==
=
3.6 EQUIVALENT CIRCUIT
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Example 5 (Cont from Ex
4)Calculatei. Stator Copper Lossii. Air Gap Power
iii. Power converted from electrical to mechanical power iv. Output power v. Motor efficiency
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Torque-Equation
Torque , can be derived from power equation in term of mechanical power or electrical power .
n P
T Hence
srad nwhereT P Power
260
,
)/(60
2,,
=
==
r
oo
r
mm
n P
T TorqueOutput
n P
T TorqueMechanical
Thus
2
60,
2
60,
,
=
=
3.6 EQUIVALENT CIRCUIT
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Note that, Mechanical torque can be written in terms of circuitNote that, Mechanical torque can be written in terms of circuitparameters. This is determined by usingparameters. This is determined by using approximation methodapproximation method
...
...
...
)1('
'3
)1('
'3
2
2
==
==
r
R R
r
mm
mr m R
Rm
s s
R I P
T
T P and s s
R I P
+=
22
2
)'()'(
'
2
)(3
R R
R
s
RM m
sX R
sR
n
V T
Torque-speed characteristic
Tm
n s
ss maxmax is the slip for Tis the slip for T maxmax to occur to occur
s=1
Tst
Tmax
s max
3.6 EQUIVALENT CIRCUIT
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+++
=
=
22
2
)'()'('
602
)(3
1,
R s R s
R
s
s st X X R R
Rn
V T
sTorqueStarting
+++
=
+=
22
2
max
22max
)'()(
1
6022
)(3
)'()(
'
R s s s s
s
R s
R
X X R Rn
V T
X R
R s
3.6 EQUIVALENT CIRCUIT
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Calculatev. Mechanical torquevi. Output torquevii. Starting torque
viii. Maximum torque and maximum slip
Example 6 (Cont from Ex
4)
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There are 3 types of speed control of 3 phase inductionmachines
i.i. Varying rotor resistanceVarying rotor resistanceii.ii. Varying supply voltageVarying supply voltageiii.iii. Varying supply voltage and supply frequencyVarying supply voltage and supply frequency
3.7 SPEED CONTROL
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Varying rotor resistance
For wound rotor onlySpeed is decreasing
Constant maximum torqueThe speed at which maxtorque occurs changesDisadvantages:
large speed regulationPower loss in R ext reduce the efficiency
T
n s ~n NL
T
n r1n r2n r3 n
n r1 < n r2 < n r3R 1R 2R 3
R 1< R 2< R 3
3.7 SPEED CONTROL
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Varying supply voltage
Maximum torque changesThe speed which at max
torque occurs is constant (atmax torque, X R=R R/s
Relatively simple method uses power electronics
circuit for voltage controller Suitable for fan type loadDisadvantages :
Large speed regulation
since ~ n s
T
n s ~n NL
T
n r1n r2n r3 n
n r1 > n r2 > n r3
V1
V2
V3
V1> V2 > V3
Vdecreasing
3.7 SPEED CONTROL
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The best method since supplyvoltage and supply frequency isvaried to keep V V / / f f constant
Maintain speed regulationuses power electronics circuitfor frequency and voltagecontroller Constant maximum torque
Varying supply voltage and supply frequency
T
n NL1
T
n r1n r2n r3 n
f decreasing
n NL2n NL3
3.7 SPEED CONTROL
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THANK YOU FOR YOUR ATTENTION!!
HAVE A NICEHOLIDAY!