Download Lecture Notes EEE 360 - Warsaw University of Technology

EEE 360
Energy Conversion and
Transport
George G. Karady & Keith Holbert
Chapter 7
Induction Motors
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360 Chapter 7 Induction Motor
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Lecture 19
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Induction Motors
Operation Principle
•
•
•
•
The three-phase stator is supplied by balanced threephase voltage that drives an ac magnetizing current
through each phase winding.
The magnetizing current in each phase generates a
pulsating ac flux.
The total flux in the machine is the sum of the three
fluxes.
The summation of the three ac fluxes results in a
rotating flux, which turns with constant speed and
has constant amplitude.
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Induction Motors
Operation Principle
•
The rotating flux induces a voltage in the shortcircuited bars of the rotor. This voltage drives
current through the bars.
•
The induced voltage is proportional with the
difference of motor and synchronous speed.
Consequently the motor speed is less than the
synchronous speed
•
The interaction of the rotating flux and the rotor
current generates a force that drives the motor.
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7.3.2 Equivalent circuit
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Induction Motors
Xsta
Vsup
Rsta
Ista
Xrot_t
Rc
Xm
Vsta
Rrot_t
Irot_t
Stator
Rrot_t(1-s)/s
Rotor
Air gap
Figure 7.20 Final single-phase equivalent circuit of a three-phase
induction motor.
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7.3.3 Motor performance
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Induction Motors
• Figure 7.21 shows the energy balance in a motor.
• The supply power is:

Psup  Re(S sup )  Re 3 Vsup I
*
sta

• The power transferred through the air gap by the
magnetic coupling is the input power (Psup) minus the
stator copper loss and the magnetizing (stator iron)
loss.
•
The electrically developed power (Pdv) is the difference
between the air gap power (Pag) and rotor copper loss.
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Induction Motors
• The electrically developed power can be computed
from the power dissipated in the second term of rotor
resistance:
Pdv  3 I rot_t
2
1 s 

 Rrot _ t

s 

• The subtraction of the mechanical ventilation and
friction losses (Pmloss) from the developed power gives
the mechanical output power
Pout  Pdv  Pmloss
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Induction Motors
• The motor efficiency:
Pout

Psup
• Motor torque:
T
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Pout
m
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Induction Motors
Air gap
Input power
Psup
Air gap
power Pag
Developed power
Pdv = 3 Irot2 Rrot (1-s)/s
Output power
Pout
Ventilation and
loss friction losses
Stator Iron loss Rotor Copper
3 Irot2 Rrot
3 Vsta2 / Rc
Stator Copper loss
3 Ista2 Rsta
Figure 7.21
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Motor energy balance flow diagram.
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7.3.4 Motor performance
analysis
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Induction Motors
1) Motor impedance
( 1  s)
Zrot_t ( s )  j Xrot_t  Rrot_t  Rrot_t 
s
Zsta
Zrot_t(s)
j  Xm  Rc
Zm 
j  Xm  Rc
Zm  Zrot_t ( s )
Vsup
Ista
Im
Zm
Irot_t
Zmot ( s )  Zsta 
Zm  Zrot_t ( s )
Figure 7.22 Simplified motor
equivalent circuit.
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Induction Motors
1) Motor impedance
( 1  s)
Zrot_t ( s )  j Xrot_t  Rrot_t  Rrot_t 
s
j  Xm  Rc
Zm 
j  Xm  Rc
Zsta
Zrot_t(s)
Zsta  j Xsta  Rsta
Ista
Irot_t
Vsup
Im
Zm
Zm  Zrot_t ( s )
Zmot ( s )  Zsta 
Zm  Zrot_t ( s )
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Figure 7.22 Simplified motor
equivalent circuit.
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Induction Motors
2) Motor Current
Zsta
Vsup 
Vmot
Ista ( s ) 
Vsup  254.034 V
3
Vsup
Vsup
Ista
Im
Zm
Irot_t
Zmot ( s )
Zm
Irot_t ( s )  Ista ( s ) 
Zm  Zrot_t ( s )
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Zrot_t(s)
Figure 7.22 Simplified motor
equivalent circuit.
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Induction Motors
3) Motor Input Power

Ssup ( s)  3  Vsup Ista ( s)
Psup ( s)  Re Ssup ( s)

Pfsup ( s ) 
4)

Psup ( s )

Qsup ( s)  Im Ssup ( s)
Ssup ( s )

Motor Output Power and efficiency

Pdev ( s )  3  Irot_t ( s )

2
 Rrot_t 
( 1  s)
s
Pmech ( s)  Pdev ( s)  Pmech_loss
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 ( s ) 
Pmech ( s )
Psup ( s )
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Induction Motors
s  0.1  %  0.2%  100  %
40
Pmax
Pmech( s )
30
hp
Pmotor
20
hp
10
Operating Point
0
0
20
smax
40
60
80
100
s
%
Figure 7.24 Mechanical output power versus slip.
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Induction Motors
5. Motor Speed
rpm 
1
f
nsy 
p
min
nsy  1800 rpm
2
nm ( s)  nsy ( 1  s)
m ( s)  2 nm ( s)
6. Motor Torque
Tm ( s ) 
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Pmech ( s )
m ( s )
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Induction Motors
s  0.5%  0.6%  80%
200
150
Tm( s )
N m
100
50
0
0
20
40
60
80
s
%
Figure 7.26 Torque versus slip.
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Induction Motors
s  0.5%  0.6%  80%
200
175
Tmax
Tm( s ) 150
N m
Trated
N m
125
100
75
50
Operating Point
25
0
200
400
600
800
1000
1200
1400
n m( s )
rpm
1600
1800
nmax
Figure 7.26 Torque versus speed.
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Induction Motors
7.3.4.5.Motor Starting torque
• When the motor starts at s = 1,
• The ventilation losses are zero and the friction
loss is passive. The negative friction loss does not
drive the motor backwards.
• The mechanical losses are zero when s = 1
• This implies that the starting torque is calculated
from the developed power instead of the
mechanical output power.
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Induction Motors
7.3.4.5. Motor Starting torque
Tm_start ( s ) 
Tm_start ( s ) 
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
3  Irot_t ( s )

2
 Rrot_t 
( 1  s)
s
2   nsy  ( 1  s )

3  Irot_t ( s )

2
Rrot_t

s
2   nsy
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Induction Motors
• The resistances and reactance in the equivalent
circuit for an induction motor can be
determined by a series of measurements. The
measurements are:
– No-load test. This test determines the magnetizing
reactance and core loss resistance.
– Blocked-rotor test. This test gives the combined
value of the stator and rotor resistance and
reactance.
– Stator resistance measurement.
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Induction Motors
7.3.6.1.No-load test
• The motor shaft is free
• The rated voltage supplies the motor.
• In the case of a three-phase motor:
– the line-to-line voltages,
– line currents
– three-phase power using two wattmeters are
measured
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Induction Motors
7.3.6.1. No-load test
Xsta
Fig 7.29 Equivalent
motor circuit in noload test
Rsta
Ino_load
Vno_load
Rc
Xrot_t
Xm
Rrot_t
Irot_t= small
Vsta
Rrot_t(1-s)/s
s~0
Stator
Rotor
Air gap
Fig. 7.30 Simplified
equivalent motor
circuit in no-load test
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Ino_load
Vno_load_ln
360 Chapter 7 Induction Motor
Rc
Xm
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Induction Motors
7.3.6.1.No-load test
2
Vno_load_ln 
Pno_load_A 
Vno_load_ln
RV

c no_load_ln
 120.089 V
Pno_load_A
Vno_load
3
Pno_load
Qno_load_A 
SPno_load_A95 V
I
Wno_load_ln no_load
no_load_A
3
2
2
Sno_load_A  Pno_load_A
2
Vno_load_ln
Xm 
Qno_load_A
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Induction Motors
7.3.6.2. Blocked-rotor test
• The rotor is blocked to prevent rotation
• The supply voltage is reduced until the motor current is
around the rated value.
• The motor is supplied by reduced voltage and reduced
frequency. The supply frequency is typically 15 Hz.
• In the case of a three-phase motor:
– the line-to-line voltages,
– line currents
– three-phase power using two wattmeters are measured
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Induction Motors
7.3.6.2. Blocked-Rotor test
Xsta
Figure 7.31 Equivalent
motor circuit for
blocked-rotor test
Vblocked
f = 15 Hz
Rsta
Iblocked
Rc
Xrot_t
Xm
Rrot_t
Irot_t
Vsta
Stator
Rotor
Air gap
Re = Rsta + Rrot_t
Figure 7.32 Simplified
equivalent motor circuit
for blocked-rotor test
Vblocked_ln
Xe = Xsta + Xrot_t
Iblocked
f = 15 Hz
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Induction Motors
7.3.6.2. Blocked-Rotor test
Vblocked_ln 
Pblocked_A 
Vblocked
Vblocked_ln
 21.939 V
Pblocked_A
3
Re 
2
Iblocked
Pblocked
Pblocked_A  160 W
3
The stator resistance was measured directly
Rrot_t  Re  Rsta
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Induction Motors
7.3.6.2. Blocked-Rotor test
The magnitude of the
motor impedance
The leakage reactance
at 15 Hz
The leakage reactance
at 60 Hz
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Zblocked 
Xe_15Hz 
Vblocked_ln
Iblocked
2
2
Zblocked  Re
60Hz
Xe  Xe_15Hz 
15Hz
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Numerical Example
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Induction Motors
A three -phase 30hp, 208V, 4 pole, 60Hz, wye connected induction
motor was tested, the obtained results are:
No load test, 60 Hz
VnL  208V
PnL  1600W
InL  22A
Blocked Rotor test, 15Hz
Vbr  21V
Pbr  2100W
Ibr  71A
DC test
Vdc  12V
Idc  75A
Motor rating:
Pmot_rated  30hp
p  4
Vmot_ll  208V
fbr  15Hz
pfmot  0.8
f  60Hz
Calculate :
Calculate
a) The: equivalent circuit paramete rs
The equivalent
circuit
parameters
b)a)Motor
rated current
and
synchronous speed
b)
Motor
rated
current
and
synchronous speed
Draw the equivalent circuit
Draw the equivalent circuit
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Induction Motors
No load test , Determine the core losses Rc and magnetizing
reactance Xm
PnL_1F 
Single phase values:
Rc 
VnL_ln2
YnL 
InL
VnL_ln 
3
YnL  0.183 S
VnL_ln
1
2
YnL 
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3
VnL
Rc  27.04 
PnL_1F
Xm  i
PnL
 1 
R 
 c
2
Xm  5.573i 
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Induction Motors
Block Rotor test,
Neglect the magnetizing branch. Consider only the Xsta+Xrot and
Rsta+ Rrot
Xbr
Xsta  Xrot
Single phase values:
Rbr
Rsta  Rrot
Pbr_1F 
Pbr
3
Vbr_ln 
Vbr
3
Resistance value is:
Rbr 
Pbr_1F
2
Rbr  0.139 
Ibr
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Induction Motors
Zbr 
Vbr_ln
Zbr  0.171 
Ibr
2
Xbr_15Hz  i Zbr  Rbr
2
Xbr_15Hz  0.099i 
The reactance at 60 Hz is:
Xbr_60Hz  Xbr_15Hz 
Xbr  i
60
Xbr_60Hz  0.398i 
15
 Z 2  R 2   60Hz
br 
 br
f
Xbr  0.398i 
br
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Induction Motors
Determination of R1 and R2 and X1 and X2
Xsta 
Xbr
2
Xrot  Xsta
Xsta  0.199i 
Y connected motor
Rsta 
Vdc
2Idc
Rsta  0.08 
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Rrot  Rbr  Rsta
Rrot  0.059 
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Induction Motors
a) The equivalent circuit parameters
a) The equivalent circuit parameters
jX sta
jX rot
R sta
Ista
Irot
Vsup_ln
Rrot (1-s) / s
Im
Ic
Stator
R rot
Rc
jXm
Rotor
Air gap
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Induction Motors
C) Motor rated current and sy nchronous speed
Srated 
Pmot_rated
Srated  27.964 kV A
pfmot
Srated
Imot_rated 
n synch 
3 Vmot_ll
1
Imot_rated  77.62 A
rpm 
n synch  30 Hz
n synch  1800 rpm
min
f
p
2
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