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![Page 1: Three phase induction motor - Ústav přístrojové a ...elektro.fs.cvut.cz/en/SSem/2141025.old/AM_ThreePhase.pdf · Induction Motors • The single-phase induction motor is the most](https://reader031.fdocuments.us/reader031/viewer/2022030402/5a78e1a57f8b9a83238dd02c/html5/thumbnails/1.jpg)
Induction Motor
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Induction Motors
• The single-phase induction motor is the most frequently used motor in the world
• Most appliances, such as washing machines and refrigerators, use a single-phase induction machine
• Highly reliable and economical
Figure 1 Single-phase induction motor.
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Induction Motors
• For industrial applications, the three-phase induction motor is used to drive machines
• Figure 2 Large three-phase induction motor. (Courtesy Siemens).
Housing
Motor
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Induction Motors
Figure 3Induction motor
components.
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Induction Motors• The motor housing consists of three parts:
– The cylindrical middle piece that holds the stator iron core,
– The two bell-shaped end covers holding the ball bearings.
– This motor housing is made of cast aluminum or cast iron. Long screws hold the three parts together.
– The legs at the middle section permit the attachment of the motor to a base.
– A cooling fan is attached to the shaft at the left-hand side. This fan blows air over the ribbed stator frame.
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Induction Motors
Figure 4 Stator of a large induction motor. (Courtesy Siemens).
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Induction Motors• The iron core has cylindrical
shape and is laminated with slots
• The iron core on the figure has paper liner insulation placed in some of the slots.
• In a three-phase motor, the three phase windings are placed in the slots
• A single-phase motor has two windings: the main and the starting windings.
• Typically, thin enamel insulated wires are used
Figure 5 Stator iron core without windings
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Induction Motors• A single-phase motor has
two windings: the main and the starting windings
• The elements of the laminated iron core are punched from a silicon iron sheet.
• The sheet has 36 slots and 4 holes for the assembly of the iron core.
Figure 6 Single-phase stator with main windings.
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Induction Motors
Figure 7 Stator iron core sheet.
• The elements of the laminated iron core are punched from a silicon iron sheet.
• The sheet has 36 slots and 4 holes for the assembly of the iron core
![Page 10: Three phase induction motor - Ústav přístrojové a ...elektro.fs.cvut.cz/en/SSem/2141025.old/AM_ThreePhase.pdf · Induction Motors • The single-phase induction motor is the most](https://reader031.fdocuments.us/reader031/viewer/2022030402/5a78e1a57f8b9a83238dd02c/html5/thumbnails/10.jpg)
Induction Motors
Figure 8 Stator and rotor magnetic circuit
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Induction Motors
Squirrel cage rotor. • This rotor has a laminated iron core with slots,
and is mounted on a shaft.• Aluminum bars are molded in the slots and the
bars are short circuited with two end rings. • The bars are slanted on a small rotor to reduce
audible noise. • Fins are placed on the ring that shorts the
bars. These fins work as a fan and improve cooling.
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Induction MotorsRotor bars (slightly skewed)
End ring
Figure 9 Squirrel cage rotor concept.
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Induction Motors
Figure 10 Squirrel cage rotor.
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Induction MotorsWound rotor. • Most motors use the squirrel-cage rotor because of the
robust and maintenance-free construction.• However, large, older motors use a wound rotor with
three phase windings placed in the rotor slots.• The windings are connected in a three-wire wye.• The ends of the windings are connected to three slip rings. • Resistors or power supplies are connected to the slip rings
through brushes for reduction of starting current and speed control
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Induction Motors
Figure 11 Rotor of a large induction motor. (Courtesy Siemens).
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Operating principle
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Induction Motors• This two-pole motor has three
stator phase windings, connected in three-wire wye.
• Each phase has 2 × 3 = 6 slots. The phases are shifted by 120°
• The squirrel cage rotor has short-circuited bars.
• The motor is supplied by balanced three-phase voltage at the terminals.
• The stator three-phase windings can also be connected in a delta configuration.
A+
A-
B+
B-
C+
C-
A BC
Figure 11 Connection diagram of a two-pole induction motor with
squirrel cage rotor.
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Induction Motors
Operation Principle• The three-phase stator is supplied by balanced three-
phase voltage that drives an ac magnetizing current through each phase winding.
• The magnetizing current in each phase generates a pulsating ac flux.
• The flux amplitude varies sinusoidally and the direction of the flux is perpendicular to the phase winding.
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Induction Motors
Operation Principle• The three-phase stator is supplied by balanced three-
phase 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 MotorsOperation Principle• The rotating flux induces a voltage in the short-
circuited 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.
• The force is proportional with the flux density and the rotor bar current
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Induction Motors• The figure shows the three
components of the magnetic field at a phase angle of –60°.
• Each phase generates a magnetic field vector.
• The vector sum of the component vectors Φa, Φb, Φc gives the resulting rotating field vector Φrot,
• The amplitude is 1.5 times the individual phase vector amplitudes, and Φrot rotates with constant speed.
A+
A-
B+
B-
C+
C-
Φb
Φa
Φrot
Φc
Φb
Φc
Figure 12 Three-phase winding-generated rotating magnetic field.
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Induced Voltage Generation
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Induction Motors
Faraday’s law• Voltage is induced in a
conductor that moves perpendicular to a magnetic field,
• The induced voltage is:
Magnetic field B into pageConductor
movingupward with
speed v
Induced voltage V
Conductor length L
v v
Figure 14 Voltage induced in a conductor moving through a magnetic field.
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Induction Motors• The three-phase winding on
the stator generates a rotating field.
• The rotor bar cuts the magnetic field lines as the field rotates.
• The rotating field induces a voltage in the short-circuited rotor bars
• The induced voltage is proportional to the speed difference between the rotating field and the spinning rotor
A+
A-
B+
B-
C+
C-
Φb
Φa
Φrot
Φc
Φb
Φc
V = B L (vsyn – v m)
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Induction Motors• The speed of flux cutting is
the difference between the magnetic field speed and the rotor speed.
• The two speeds can be calculated by using the radius at the rotor bar location and the rotational speed.
A+
A-
B+
B-
C+
C-
Φb
Φa
Φrot
Φc
Φb
Φc
mrotmot
synrotsyn
nrvnrv
ππ
2
2
=
=
( )msynrotrotbar nnBrV −π= l2
![Page 26: Three phase induction motor - Ústav přístrojové a ...elektro.fs.cvut.cz/en/SSem/2141025.old/AM_ThreePhase.pdf · Induction Motors • The single-phase induction motor is the most](https://reader031.fdocuments.us/reader031/viewer/2022030402/5a78e1a57f8b9a83238dd02c/html5/thumbnails/26.jpg)
Induction Motors• The voltage and current generation in the rotor bar require a
speed difference between the rotating field and the rotor.• Consequently, the rotor speed is always less than the magnetic
field speed. • The relative speed difference is the slip, which is calculated
using
sy
msy
sy
msy
nnn
sω
ωω −=
−=
2pfnsy =The synchronous speed is
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Motor Force Generation
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Induction Motors• The interaction
between the magnetic field B and the current generates a force
F = B L I+
B B B B
F
B
Figure 15 Force direction on a current-carrying conductor placed in a magnetic field (B) (current into the page).
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Induction Motors
Force generation in a motor• The three-phase winding
generates a rotating field;• The rotating field induces a
current in the rotor bars;• The current generation requires
a speed difference between the rotor and the magnetic field;
• The interaction between the field and the current produces the driving force.
Brotating
Force
Ir
Rotor Bar
Ring
Figure 16 Rotating magnetic field generated driving force.
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Equivalent circuit
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Induction Motors• An induction motor has two magnetically coupled circuits:
the stator and the rotor. The latter is short-circuited.
• This is similar to a transformer, whose secondary is rotating and short-circuited.
• The motor has balanced three-phase circuits; consequently, the single-phase representation is sufficient.
• Both the stator and rotor have windings, which have resistance and leakage inductance.
• The stator and rotor winding are represented by a resistance and leakage reactance connected in series
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Induction Motors• A transformer represents the magnetic coupling
between the two circuits.
• The stator produces a rotating magnetic field that induces voltage in both windings.– A magnetizing reactance (Xm) and a resistance connected in
parallel represent the magnetic field generation.
– The resistance (Rc) represents the eddy current and hysteresislosses in the iron core
• The induced voltage is depend on the slip and the turn ratio
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Induction Motors
Stator Rotor
Xrot_m = ωrot Lrot Rrot
Irot
Vrot = s Vrot_s
Rsta Irot_t
VstaVsupIsta XmRc
Xsta = ωsy Lsta
Figure 17 Single-phase equivalent circuit of a three-phase induction motor.
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Induction Motors• In this circuit, the magnetizing reactance generates a flux that
links with both the stator and the rotor and induces a voltage in both circuits.
• The magnetic flux rotates with constant amplitude and synchronous speed.
• This flux cuts the stationary conductors of the stator with the synchronous speed and induces a 60 Hz voltage in the stator windings.
• The rms value of the voltage induced in the stator is:
2max systa
sta
NV
ωΦ=
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Induction Motors• The flux rotates with the synchronous speed and the rotor
with the motor speed.
• Consequently, the flux cuts the rotor conductors with the speed difference between the rotating flux and the rotor.
• The speed difference is calculated using the slip equation:
• The induced voltage is:
ssymsy ωωω =− )(
22
)( maxmax sNNV syrotmsyrot
rot
ωωω Φ=
−Φ=
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Induction Motors• The division of the rotor and stator induced voltage
results in:
• This speed difference determines the frequency of the rotor current
• The rotor circuit leakage reactance is:
sVsVNNV srotsta
sta
rotrot _==
sysymsyrot
rot fss
f ==−
==π
ωπ
ωωπ
ω222
sXsLLX rotsyrotrotrotmrot === ωω_
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Induction Motors• The relation between rotor current and the rotor-
induced voltage is calculated by the loop voltage equation:
• The division of this equation with the slip yields
• The implementation of this equation simplifies the equivalent circuit
)( sXjRs rotrot +== rotrot_srot IVV
⎟⎠⎞
⎜⎝⎛ += rot
rot Xjs
Rrotrot_s IV
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Induction Motors
Stator Rotor
Xsta Xrot Rrot/s
IrotVrot_s
Rsta Irot_t
VstaVsupIsta XmRc
Figure 18 Modified equivalent circuit of a three-phase induction motor.
The rotor impedance is transferred to the stator side. This eliminates the transformer
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Induction Motors
Figure 19 Simplified equivalent circuit of a three-phase induction motor.
Stator Rotor
Xsta Xrot_t Rrot_t/s
Irot_tVsta
Rsta
VsupIsta XmRc
Air gap
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Induction Motors• The last modification of the equivalent circuit is the
separation of the rotor resistance into two parts:
• The obtained resistance represents the outgoing mechanical power
[ ]trottrot
trot Rs
sRs
R__
_ 1−+=
[ ]trotR
ss
_1−
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Induction Motors
Stator Rotor
Xsta Xrot_t Rrot_t
Irot_tVsta
Rsta
VsupIsta XmRc
Air gap
Rrot_t(1-s)/s
Figure 20 Final single-phase equivalent circuit of a three-phase induction motor.
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Motor performance
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Induction Motors• Figure 21 shows the energy balance in a motor.
• The supply power is:
• 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.
( )*stasupsup IVS 3Re)Re(Psup ==
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Induction Motors
• The electrically developed power can be computed from the power dissipated in the second term of rotor resistance:
• The subtraction of the mechanical ventilation and friction losses (Pmloss) from the developed power gives the mechanical output power
⎟⎠⎞
⎜⎝⎛ −=
ssR trot
13P _
2
dv rot_tI
mlossdvout PPP −=
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Induction Motors• The motor efficiency:
• Motor torque:sup
out
PP
=η
m
outPω
=M
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Induction Motors
Input powerPsup
Stator Copper loss3 Ista
2 Rsta
Rotor Copper loss3 Irot
2 RrotStator Iron loss
3 Vsta2 / Rc
Ventilation andfriction losses
Output powerPout
Air gappower Pag
Developed powerPdv = 3 Irot
2 Rrot (1-s)/s
Air gap
Figure 21 Motor energy balance flow diagram.
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7.3.4 Motor performanceanalysis
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Induction Motors1) Motor impedance
Vsup
Zsta Zrot_t(s)
Zm
Ista Irot_tIm
Figure 7.22 Simplified motor equivalent circuit.
Zrot_t s( ) j Xrot_t⋅ Rrot_t+ Rrot_t1 s−( )
s⋅+:=
Zmj Xm⋅ Rc⋅
j Xm⋅ Rc+:=
Zmot s( ) ZstaZm Zrot_t s( )⋅
Zm Zrot_t s( )++:=
Zsta j Xsta⋅ Rsta+:=
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Induction Motors2) Motor Current
Figure 7.22 Simplified motor equivalent circuit.
VsupVmot
3:= Vsup 254.034 V=
Ista s( )Vsup
Zmot s( ):=
Irot_t s( ) Ista s( )Zm
Zm Zrot_t s( )+⋅:=
Vsup
Zsta Zrot_t(s)
Zm
Ista Irot_tIm
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Induction Motors3) Motor Input Power
Ssup s( ) 3 Vsup⋅ Ista s( )⎯
⋅:= Psup s( ) Re Ssup s( )( ):=
Pfsup s( )Psup s( )
Ssup s( ):= Qsup s( ) Im Ssup s( )( ):=
4) Motor Output Power and efficiency
Pdev s( ) 3 Irot_t s( )( )2⋅ Rrot_t⋅1 s−( )
s⋅:=
Pmech s( ) Pdev s( ) Pmech_loss−:= η s( )Pmech s( )
Psup s( ):=
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Induction Motorss 0.1 %⋅ 0.2%, 100 %⋅..:=
0 20 40 60 80 1000
10
20
30
40
Pmech s( )
hp
Pmotorhp
s%
Pmax
smax
Operating Point
Figure 24 Mechanical output power versus slip.
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Induction Motors5. Motor Speed
rpm1
min:= nsy
fp2
:= nsy 1800rpm=
nm s( ) nsy 1 s−( )⋅:= ωm s( ) 2 π⋅ nm s( )⋅:=
6. Motor Torque
Tm s( )Pmech s( )
ωm s( ):=
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Induction Motors
Figure 25 Torque versus slip.
s 0.5% 0.6%, 80%..:=
0 20 40 60 800
50
100
150
200
Tm s( )
N m⋅
s%
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Induction Motors
Figure 26 Torque versus speed.
nmax
s 0.5% 0.6%, 80%..:=
200 400 600 800 1000 1200 1400 1600 18000
25
50
75
100
125
150
175
200
Tm s( )
N m⋅
TratedN m⋅
nm s( )
rpm
Tmax
Operating Point
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Induction Motors
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Induction MotorsMotor 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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Tm_start s( )3 Irot_t s( )( )2⋅
Rrot_ts
⋅
2 π⋅ nsy⋅:=
Tm_start s( )3 Irot_t s( )( )2⋅ Rrot_t⋅
1 s−( )s
⋅
2 π⋅ nsy⋅ 1 s−( )⋅:=
Induction Motors
Motor Starting torque
Mm
Mm
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Induction Motors
• Kloss formula
M Ms
ss
s
MAX
MAX
MAX
=+
2
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Induction MotorsCircular diagram
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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
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
No-load test
Stator Rotor
Xsta Xrot_t Rrot_t
Irot_t= smallVsta
Rsta
Vno_loadXmRc
Air gap
Rrot_t(1-s)/ss ~ 0
Ino_load
Fig 29 Equivalent motor circuit in no-load test
Fig. 30 Simplified equivalent motor circuit in no-load test
Ino_load XmRcVno_load_ln
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Induction Motors
No-load test
Vno_load_lnVno_load
3:= Vno_load_ln 120.089 V=
Pno_load_APno_load
3:= Pno_load_A 95W=
RcVno_load_ln
2
Pno_load_A:=
Sno_load_A Vno_load_ln Ino_load⋅:=
Qno_load_A Sno_load_A2 Pno_load_A
2−:=
XmVno_load_ln
2
Qno_load_A:=
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Induction Motors
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
Blocked-Rotor test
Figure 32 Simplified equivalent motor circuit for blocked-rotor test
Figure 31 Equivalent motor circuit for blocked-rotor test Stator Rotor
Xsta Xrot_t Rrot_t
Irot_tVsta
Rsta
Vblocked
Iblocked XmRc
Air gapf = 15 Hz
Xe = Xsta + Xrot_tRe = Rsta + Rrot_t
f = 15 Hz
Vblocked_lnIblocked
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Induction Motors
Blocked-Rotor test
Vblocked_lnVblocked
3:= Vblocked_ln 21.939 V=
Pblocked_APblocked
3:= Pblocked_A 160W=
RePblocked_A
Iblocked2
:=
Rrot_t Re Rsta−:=
The stator resistance was measured directly
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Induction Motors
Blocked-Rotor test
The leakage reactance at 15 Hz
ZblockedVblocked_ln
Iblocked:=The magnitude of the
motor impedance
Xe_15Hz Zblocked2 Re
2−:=
The leakage reactance at 60 Hz Xe Xe_15Hz
60Hz15Hz
⋅:=
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Numerical Example
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Induction Motors
f 60Hz:=p 4:=
pfmot 0.8:=Vmot_ll 208V:=Pmot_rated 30hp:=Motor rating:
Idc 75A:=Vdc 12V:=DC test
fbr 15Hz:=Ibr 71A:=Pbr 2100W:=Vbr 21V:=Blocked Rotor test, 15Hz
InL 22A:=PnL 1600W:=VnL 208V:=No load test, 60 Hz
A three -phase 30hp, 208V, 4 pole, 60Hz, wye connected induction motor was tested, the obtained results are:
Calculate :a) The equivalent circuit parametersb) Motor rated current and synchronous speed
Draw the equivalent circuit
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Induction MotorsNo load test , Determine the core losses Rc and magnetizing reactance Xm
PnL_1FPnL
3:= VnL_ln
VnL
3:=Single phase values:
RcVnL_ln( )2
PnL_1F:= Rc 27.04Ω=
YnLInL
VnL_ln:= YnL 0.183S=
Xm i1
YnL2 1
Rc
⎛⎜⎝
⎞⎠
2−
⋅:= Xm 5.573iΩ=
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Induction MotorsBlock Rotor test,Neglect the magnetizing branch. Consider only the Xsta+Xrot and Rsta+ Rrot
Xbr Xsta Xrot+ Rbr Rsta Rrot+
Single phase values: Pbr_1FPbr
3:= Vbr_ln
Vbr
3:=
Resistance value is:
RbrPbr_1F
Ibr2
:= Rbr 0.139Ω=
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Induction Motors
ZbrVbr_ln
Ibr:= Zbr 0.171Ω=
Xbr_15Hz i Zbr2 Rbr
2−⋅:= Xbr_15Hz 0.099iΩ=
The reactance at 60 Hz is:
Xbr_60Hz Xbr_15Hz6015
⋅:= Xbr_60Hz 0.398iΩ=
Xbr i Zbr2 Rbr
2−⎛⎝
⎞⎠⋅
60Hzfbr
⋅:= Xbr 0.398iΩ=
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Induction Motors
Determination of R1 and R2 and X1 and X2
XstaXbr
2:= Xrot Xsta:= Xsta 0.199iΩ=
Y connected motor
RstaVdc
2Idc:= Rrot Rbr Rsta−:=
Rsta 0.08Ω= Rrot 0.059Ω=
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Induction Motors
a) The equivalent circuit parameters
Rotor
Air gap
Stator
sta
Ic Im
jXmRc
stajX RjX RjX R
IstaVsup_ln
Irot
rotjX RjX RjX Rrot
Rrot (1-s) / s
a) The equivalent circuit parameters
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Induction Motors
C) Motor rated current and synchronous speed
SratedPmot_rated
pfmot:= Srated 27.964kV A⋅=
Imot_ratedSrated
3 Vmot_ll⋅:= Imot_rated 77.62A= rpm
1min
:=
nsynchfp
2
:= nsynch 30Hz= nsynch 1800rpm=