Transcript Induction Motor - ENCON - Department of ENergy CONversion

Induction Motor Drive
• Why induction motor (IM)?
–
–
–
–
Robust; No brushes. No contacts on rotor shaft
High Power/Weight, Lower Cost/Power ratios
Easy to manufacture
Almost maintenance-free, except for bearing and other
“external” mechanical parts
• Disadvantages
– Essentially a “fixed-speed” machine
– Speed is determined by the supply frequency
– To vary its speed need a variable frequency supply
• Motivation for variable-speed AC drives
–
–
–
–
Inverter configuration improved
Fast switching, high power switches
Sophisticated control strategy
Microprocessor/DSP implementation
• Applications
– Conveyer line (belt) drives, Roller table, Paper mills,
Traction, Electric vehicles, Elevators, pulleys, Airconditioning and any industrial process that requires
variable-speed operation.
• The state-of-the-art in IM drives is such that most of the
DC drives will be replaced with IM in very near future.
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Torque production (1)
• Only “squirrel-cage” IM (SCIM) is considered in
this module
• Neglecting all harmonics, the stator establishes a
spatially distributed magnetic flux density in the
air-gap that rotate at a synchronous speed, w1 :
w1 
we
p
where
we : supply frequency (in Hz)
p: pole pairs (p=1for 2 pole motor, p=2
for 4 pole motor etc)
• If the rotor is initially stationary, its conductor is
subjected to a sweeping magnetic field, inducing
rotor current at synchronous speed.
• If the rotor is rotating at synchronous speed (i.e.
equals to f1), then the rotor experience no induction.
No current is induced in the rotor.
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Torque production (2)
• At any other rotor speed, say wm, the speed
differential wi-w2 creates slip. Per-unit slip is
defined as:
w  wm
s 1
;
w1
w1 
we
p
where :
we : supply frequency
w m : rotor frequency
p : pole pair
• Slip frequency is defined as: w2=w1-wm.
• When rotor is rotating at wm., rotor current at slip
frequency will be induced.
• The interaction between rotor current and air-gap
flux produces torque.
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Single-phase Equivalent Circuit
(SPEC)
1:nS
R1
V1
L1
L2
Rm
Lm
Vm
STATOR SIDE
V2 = nSV m
R2
ROTOR SIDE
R1 : Stator resistance
L1 : Stator leakage inductance
R2 : Rotor resistance
L2 : Rotor leakage inductance
Lm : Magnetisin g inductance
v1 : Supply vol tage (phase voltage)
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SPEC, referred to stator
I1
V1
R1
L1
I2
Rm
L2
Lm
R2
S
• From previous diagram, SPEC is a dual frequency
circuit. On the stator is w1 and on the rotor wm
• Difficult to do calculations.
• We can make the circuit a single frequency type, by
referring the quantities to the stator
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Rotor current
If E1 is the back EMF in the stator phase, then
the back EMF in an equivalent rotor phase with
the same effective turns ratio will be E 2 where :
E2  sE1
At standstill , i.e when w m  0,
E2  sE1  1E1  E1
At synchronous speed, i.e whenw m  1,
E2  (0) E1  0
Hence the current in the rotor phase,
E2
sE1

R2  jsX 2 R2  jsX 2
E1

R2
 jX 2
s
Note that the quantities are now referred to
the stator, but with the rotor resistance alteration.
I2 
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Performance calculation using SPEC
I1
R1
V1
L1
Rm
I2
L2
Lm
R2
S
Pin  3V1I1 cos 
Input Power :
Note : V1 and I1 must be phase voltage and current
Stator copper loss :
Core loss :
Power across the air - gap :
Rotor copper loss :
Pls  3I12 R1
3V12
Plc 
Rm
3I 2 2 R2
Pg 
s
 Pin  Pls  Plc
Plr  3I 2 2 R2
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Performance calculation (2)
Gross output power :
Po  Pg  Plr
2
3I 2 2 R2
3
I
R (1  s )

 3I 2 2  2 2
 Pg (1  s )
s
s
Power at the shaft :
Psh  Po  PFW ; PFW : friction and windage loss.
Developed (electromagnetic) torque :
Po 3I 2 2 R2 (1  s )
Te 

wm
sw m
Since
w1  w m
s
 w m  (1  s )w1 ,
w1
3 I 2 2 R2
 Te 
sw1
w
But w1  e ; w e is the supply frequency.
p
Then,

3 pI 2 2 R2
Te 
sw e
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Example calculation
• A single phase equivalent circuit of a 6-pole SCIM
that operates from a 220V line voltage at 60Hz is
given below. Calculate the stator current, output
power, torque and efficiency at a slip of 2.5%. The
fixed winding and friction losses is 350W. Neglect
the core loss.
I1
R1
X1
0.2
0.5
V1
I2
X2
0.2
Xm
R2
20
0.1
V1  220V line-to line  3
220V
 127V
3
 2.5%  0.025

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Calculation (solution)
X 1  0.5, X 2  0.2, X m  20 
R

Z in  ( R1  jX 1 )  jX m //  2  jX 2 
 s

0 .1


 j 0 .2 

  4.220 o 
 0.2  j 0.5  j 20 0.025
 0.1  j 0.2  j 20 
 0.025

V
220 3
o
I1  1 

30
.
0


20
A
Z in 4.220 o
Pin  3V1I1 cos  3(220
3)(30 )(cos 20 o )
 10,758W
Pls  3I12 R1  3(30 2 )( 0.2)  540W
Power tran sferred to rotor (neglectin g core loss)
Pg  Pin  Pls
 10,758  540  10,216W
Gross power
Po  Pg (1  s )  10,216 (1  0.025 )  9,961W
Power at the shaft
Psh  Po  PFW  9,961  350  9,611W
Output power 9611

 89 .3%
Input power 10758
Electromag netic Torque
P
Po
9611
Te  o 

wm (we1 p ) (1  s ) 2 (60 / 3)(1  0.025 )
Efficiency 
 78 .4 N .m
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Starting current
• For the previous example, Calculate the stating current
when motor is first switched on to rated applied
voltage.
Solution :
At standstill , s  1
 0.1  j 0.2 
Z in   0.2  j 0.5  j 20

 0.1  j 0.2  j 20 
 0.76
V
220 3
I1  1 
 167 A
Z in
0.76
Note that the starting current is about 5 times than
full load current.
This is common for induction motors.Care should be
taken when starting induction motors.
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Approximate SPEC
+
R1
L1
L2
R2
s
LM
V1
-
Since L m is large, the circuit above can be drawn
I2 
V1
2
R2 

2
2
 R1 
  w1  L1  L2 
s 

Power at the rotor (per phase),
R 
Po  I 2 2  2 
 s 
Electromag netic (developed ) torque,
Te 
3Po
w1

3R2V12
2


R2 
2
2
sw1  R1 
  w1  L1  L2  
s



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Single (fixed supply) frequency
characteristics
For a give frequency w1 , the torque (versus slip)
characteri stics can be shown as below.
Note that :
w1  wm
s
; at standsill s  1, at sync speed, s  0.
w1
T ORQUE(+)
we
wm
P LUGGING
MOT ORING
Tem
(max t orque or
pull-out t orque)
we
we
wm
wm
GENERATING
Tes (starting t orque)
0 unit y slip
(standstill)
rat ed slip
zero slip
w e (sync.speed)
SP EED
SLIP ,s
T ORQUE(-)
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Single frequency characteristics
CURRENT
TORQUE
operating point
(rated torque)
EFFICIENCY
POWER
FACTOR
rated current
rated slip
SLIP
0
1.0
Standstill
synchronous speed
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Single frequency characteristic
• As slip is increased from zero (synchronous), the
torque rapidly reaches the maximum. Then it
decreases to standstill when the slip is unity.
• At synchronous speed, torque is almost zero.
• At standstill, torque is not too high, but the current
is very high. Thus the VA requirement of the IM is
several times than the full load. Not economic to
operate at this condition.
• Only at “low slip”, the motor current is low and
efficiency and power factor are high.
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Typical IM Drive System
+

IDC
+
VDC
IM

Modulation Index,
BLOCK DIAGRAM
Supply
Rectifier and Filter 3-phase Inverter
n
CIRCUIT
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IM
16
Variable speed characteristics
• For variable speed operation, the supply is an
inverter.
• The frequency of the fundamental AC voltage will
determine the speed of IM. To vary the speed of IM,
the inverter fundamental frequency need to be
changed.
• The inverter output frequency must be kept close to
the required motor speed. This is necessary as the
IM operates under low slip conditions.
• To maintain constant torque, the slip frequency has
to be maintained over the range of supply
frequencies.
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Variable voltage, variable
frequency (VVVF) operation
• In order for maximum torque production,
motor flux should be maintained at its
rated value.
   m sin w1t
But the back emf is :
d
e1  N
 Nw1 m cos w1t
dt
In RMS,
1
E1 
Nw11 m  4.44 f1 N1 m
2
E1
or
 4.44 N1 m
f1
Therefore, in order to maintain t he motor flux,
the  E1 f1  ratio has to be kept constant.
This is popularly known as the constant Volt/Hertz
operation
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Constant Torque region
• Hence for VVVF operation, there is a need to
control the fundamental voltage of the inverter if its
frequency (and therefore the frequency of the IM)
need to be varied.
• To vary the fundamental component of the inverter,
the MODULATION INDEX can be changed.
• The rated supply frequency is normally used as the
base speed
• At frequencies below the base speed, the supply
magnitude need to be reduced so as to maintain a
constant Volt/Hertz.
• The motor is operated at rated slip at all supply
frequencies. Hence a “constant torque” region is
obtained.
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Constant Torque Region
f1 f
2
T ORQUE(+)
f1  f 2  f 3  f 4  f 5
f3
f4
f5
rat ed t orque
0
SP EED
rat ed slip
SLIP ,s
T ORQUE(+)
rat ed t orque
0
SLIP ,s
SP EED
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Constant Power region
• Above base speed, the stator voltage
reaches the rated value and the motor
enters a constant power region.
• In this region, the air-gap flux decreases.
This is due to increase in frequency
frequency while maintaining fixed
voltage.
• However, the stator current is maintained
constant by increasing the slip. This is
equivalent to field weakening mode of a
separately excited DC motor.
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Constant Power region
T ORQUE(+)
rat ed t orque
0
SLIP ,s
Base speed
SP EED
T ORQUE(+)
"FIELD WEAKENING"
CONST ANT T ORQUE
REGION
0
SLIP ,s
Base speed
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SP EED
22
VVVF Summary
CONST ANT POWER
CONST ANT T ORQUE
Electromagnetic torque,eT
T erminal (supply)
voltage, V1
slip frequency,fs
slip,s
0
Base speed
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SPEED
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Examples
• A three-phase 4-pole, 10 horsepower, 460V
rms/60Hz (line-to line) runs at full-load
speed of 1746 rpm. The motor is fed from
an inverter. The flux is made to be constsnt.
Plot the torque-speed graphs for the
following frequency: 60Hz, 45 Hz, 30Hz,
15Hz.
• A three-phase induction motor is using a
three-phase VSI for VVVF operation. The
IM has the following rated parameters:
•
•
•
•
voltage:
frequency:
slip (p.u)
pole pair
415V (RMS)
50Hz
5%
2
– If the inverter gives 415V (RMS) with
modulation index of 0.8, calculate the required
modulation index if the motor need to be
operated at rotor mechanical speed of 10Hz.
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