5/10/2013
EEE 118: Energy Conversion
Dr. Mongkol Konghirun
Department of Electrical Engineering
King Mongkut’s University of Technology Thonburi
Chapter 7
Induction Motors
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7.1 Induction Motor Construction
Induction Motor Construction
As seen in previous Chapter, the amortisseur
(damper) windings on a synchronous motor could
develop a starting torque without the necessity of
supplying an external field current to them.
Induction motor is a machine with only amortisseur
(damper) windings. It is called induction machines
because the rotor voltage (which produces the rotor
current and the rotor magnetic field) is induced in the
rotor windings rather than being physically connected
by wires (no DC field current is required).
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Induction Motor Construction
Stator: same as a synchronous machine.
Rotor: there are two types.
1. Cage rotor: consists of a series of conducing bars laid into
slots carved in the face of rotor and shorted at either end of
large shorting rings.
2. Wound rotor: consists of a complete set of three-phase
windings that are mirror images of the windings on the stator.
The three-phase rotor windings are usually Y-connected. The
rotor windings are shorted through brushes riding on the slip
rings.
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Induction Motor Construction
Cage rotor
Wound rotor
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7.2 Basic Induction Motor
Concepts
The Development of Induced
Torque in an Induction Motor
A three-phase set of currents is flowing in the
stator windings, producing the stator magnetic
field, Bs (rotating in a counterclockwise direction).
The speed of the magnetic field’s rotation is given
by
nsync = 120fe/P
rpm
…(7-1)
This BS passes over the rotor bars and induces a
voltage in them. The induced voltage at rotor is
given by
eind = (v × B) • L
…(1-45)
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The Development of Induced
Torque in an Induction Motor
Since the rotor assembly is inductive,
the peak rotor current lags behind the
peak rotor voltage.
The rotor current flow produces a
rotor magnetic field, BR.
The interaction between BR and BS
causes the induced torque given
below,
τind = k (BR × BS)
…(4-58)
The net magnetic field, Bnet,
produces the resulting voltage ER.
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Could the Induction Motor’s Rotor
Rotate at Synchronous Speed ?
If the induction motor’s rotor is turning at synchronous speed, then
the rotor bars would be stationary relative to the magnetic field.
As a result, there would be no induced voltage, eind = 0. Then,
there is no rotor current, causing no rotor magnetic field, BR = 0.
Finally, there is no induced torque, τind = 0.
Thus, the induction motor can speed up to near-synchronous
speed, but it can never exactly reach synchronous speed.
However, both rotor and stator magnetic fields, BR and BS rotate
together at synchronous speed, nsync while rotor itself turns at a
slower speed.
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The Concept of Rotor Slip
The voltage induced in a rotor bar of an induction motor depends
on the speed of rotor relative to the speed of the magnetic field.
Such speed is known as slip speed (nslip), defined as the difference
between synchronous speed and rotor speed.
nslip = nsync – nm
…(7-2)
Another term used to describe the relative motion is slip (s).
s = nslip /nsync
= (nsync – nm)/nsync
= (ωsync – ωm)/ ωsync
…(7-3)
…(7-4)
…(7-5)
When nm = nsync, s = 0
nm = 0,
s=1
The mechanical speed in terms of the synchronous speed and slip,
nm = (1 – s)nsync
…(7-6)
ωm = (1 – s)ωsync
…(7-7)
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The Electrical Frequency on
the Rotor
An induction motor works by inducing voltages and currents in the rotor of
the machine, and for that reason it has sometimes been called a rotating
transformer.
Unlike transformer, the secondary part of induction motor is rotating. Also,
there is the air-gap between primary and secondary windings in induction
motor. The secondary windings are shorted circuit as well.
What is the rotor frequency (fr) ?
At nm = nsync rpm, fr = 0, and slip s = 0.
At nm = 0 rpm, fr = fe, and slip s = 1.
<< locked-rotor or blocked rotor
In other words, the rotor frequency can be expressed as
fr = sfe
…(7-8)
= [(nsync– nm)/nsync] fe
But nsync = 120fe/P, so
fr = [(nsync– nm)P/(120fe)] fe
= (P/120)(nsync – nm)
…(7-9)
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Example Problem
Example 7-1 on page 387
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7.3 Equivalent Circuit of an
Induction Motor
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The Transformer Model of an
Induction Motor
Since the operation of induction motor relies on the induction of
voltages and currents in its rotor circuit from the stator circuit, just like
transformer, the equivalent circuit of an induction motor turns out to be
very similar to one of transformer.
R1 = stator resistance
X1 = stator leakage reactance
RC = core losses resistance
Xm = magnetizing reactance
RR = rotor resistance
XR = rotor leakage reactance
aff = effective turns ratio
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The Rotor Circuit Model
The induced voltage at rotor (ER) depends on the relative motion
between the stator magnetic field speed and rotor speed.
At nm = nsync rpm, fr = 0, slip s = 0, and ER = 0
At nm = 0 rpm, fr = fe, and slip s = 1, and ER = ER0 (max)
Thus,
ER = sER0
fr = sfe
XR0
…(7-10)
…(7-8)
The rotor resistance, RR, is a constant, independent to
the slip.
However, the rotor reactance, XR, is dependent to the
rotor frequency.
XR = ωrLR = 2πfr LR
= 2π(sfe) LR
= s (2πfeLR)
= blocked-rotor rotor reactance
= s XR0
…(7-11) 16
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The Rotor Circuit Model
What is the rotor current ?
IR = ER/(RR+jXR)
IR = ER/(RR+jsXR0)
…(7-12)
Since ER = sER0 ,
IR = ER0/(RR/s +jXR0)
…(7-13)
where ER0 is constant-voltage source.
ZR,eq = RR/s +jXR0
= function of slip
…(7-14)
At very low slips, the resistive term RR/s >> XR0, so the rotor resistance is
predominates and the rotor current varies linearly with slip.
At very high slips, the reactive term XR0 >> RR/s, so the rotor current
approaches a steady-state value as the slip becomes very large.
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The Final Equivalent Circuit
To produce the final per-phase equivalent circuit for an induction motor, it is
necessary to refer the rotor part of the model over to the stator side, using
effective turns ratio, aeff. This is similar to the transformer case.
Transformer:
VP = VS’ = aVS
IP = IS’ = IS/a
ZS’ = a2ZS
…(7-15)
…(7-16)
…(7-17)
Induction motor:
E1 = ER’ = aeffER0
I2 = IR/aeff
Z2 = aeff2(RR/s +jXR0)
Making the following definitions:
R2 = aeff2RR
X2 = aeff2XR0
…(7-18)
…(7-19)
…(7-20)
…(7-21)
…(7-22)
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The Final Equivalent Circuit
Finally, the per-phase equivalent circuit referred to the stator can be
constructed.
• For cage rotors, the RR, XR0, and aeff are impossible to determine directly.
• However, it is possible to make measurements that will directly give the
referred rotor resistance and reactance, R2 and X2 even though RR, XR0, and
aeff are not known separately. Using the locked-rotor test.
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7.4 Power and Torque in
Induction Motors
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Losses and the Power-Flow
Diagram
PAG = Air gap power,
transferring from stator to
rotor across the air gap.
Pconv = Power converted from
electrical to mechanical form.
Core losses occurs both stator and rotor circuits. However, the core losses at
rotor are very tiny compared to the stator core losses. Because the rotor
frequency (fr) is small.
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Example Problem
Example 7-2 on page 395
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Power and Torque in an
Induction Motor
I1 = Vφ/Zeq
…(7-23)
where
Zeq = R1+jX1+1/[GC-jBM+1/(R2/s +jX2)]
…(7-24)
GC = 1/RC (Core losses conductance)
BM = 1/XM (Magnetizing susceptance)
Stator copper losses:
PSCL = 3I12R1
…(7-25)
Core losses:
Pcore = 3E12GC
…(7-26)
Air-gap power:
PAG = Pin – PSCL – Pcore
…(7-27)
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Power and Torque in an
Induction Motor
Air-gap power (consumed at R2/s):
PAG = 3I22R2/s
…(7-28)
Rotor copper losses:
PRCL = 3I22R2
= s PAG
…(7-30)
…(7-32)
Converted (developed mechanical) power:
Pconv = PAG – PRCL
= 3I22R2/s – 3I22R2
= 3I22R2(1 – s)/s …(7-31)
= PAG – PRCL
= PAG – s PAG
= (1 – s)PAG
…(7-33)
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Power and Torque in an
Induction Motor
Output power:
Pout = Pconv – PF&W – Pmisc
Induced (developed) torque:
τind = Pconv/ωm
= (1 – s)PAG/[(1 – s)ωsync]
= PAG/ωsync
…(7-34)
…(7-35)
…(7-36)
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Separating the Rotor Copper Losses and the
Power Converted in an Induction Motor’s
Equivalent Circuit
The air-gap power (PAG) consists of
two parts of powers:
1. PRCL: Rotor copper losses
2. Pconv : Converted power
Rotor resistance associated with Pconv:
Rconv = R2/s – R2
= R2(1/s – 1)
= R2(1 – s)/s
…(7-37)
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Example Problem
Example 7-3 on page 399
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7.5 Induction Motor TorqueSpeed Characteristics
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Induction Motor Torque-Speed
Characteristics
How does the torque of an induction motor changes
as load changes ?
How much torque can an induction motor supply at
starting conditions ?
How much does the speed of an induction motor
drop as its shaft load increases ?
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Induced Torque from a
Physical Standpoint
The BS is produced by I1.
The Bnet is produced by IM and so Bnet ∝ E1.
Light load
1. Rotor speed is nearly synchronous speed (i.e.,
slip and rotor frequency are very small).
2. Then, ER is also very small (since ER = sER0
and fr is small).
3. Then, IR is also small.
4. Since fr is small, the rotor reactance (X2) is
also small. Rotor impedance is so resistively
dominant.
5. So, the IR lags ER a little bit (nearly in phase).
6. Then the IR produces the small BR.
7. The induced torque, τind = kBRBnet sinδ (at
this time, δ ≈ 90o, so τind ≈ kBRBnet).
8. Since BR is small, τind is also small, just large
enough to overcome the motor’s rotational
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losses.
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Induced Torque from a
Physical Standpoint
The BS is produced by I1.
The Bnet is produced by IM and so Bnet ∝ E1.
Heavy load
1. Rotor speed is slower (i.e., slip and rotor
frequency are increased).
2. Then, ER is also stronger (since ER = sER0 and
fr is increased).
3. Then, IR is also larger.
4. Since fr is larger, the rotor reactance (X2) is
also larger. Rotor impedance is so reactively
dominant.
5. So, the IR lags ER more.
6. Then the IR produces the large BR.
7. The induced torque, τind = kBRBnet sinδ (at
this time, δ > 90o, reducing the τind)
8. Since the effect of BR is larger, the overall τind
is thus increased, large enough to supply the
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motor’s load torque.
Induced Torque from a
Physical Standpoint
In summary, when the motor takes more load torque, the induced torque
depends on three following factors:
τind = kBRBnet sinδ
•
BR ∝ IR. And IR increases with increasing slip (decreasing speed). See
Figure 7-16a.
•
Bnet ∝ E1. However, E1 is considered constant because its decrease as load
increases is very small. See Figure 7-16b.
•
sin δ.
δ = θR + 90o
…(7-38)
sin δ = sin (θR + 90o) = cos θR = power factor of the rotor
The rotor power-factor angle can be calculated from the rotor impedance
as
θR = tan-1(XR/RR) = tan-1(sXR0/RR)
…(7-39)
PFR = cos θR = cos (tan-1(sXR0/RR))
…(7-40)
See Figure 7-17c. As slip increases (load increases), the rotor impedance
tends to be the inductive load.
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Induced Torque from a
Physical Standpoint
τpullout ≈ (2-3)τrated
τind = kBRBnet sinδ
τstarting ≈ 1.5τrated
τind decreases
as τload increases
(slip increases)
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The Derivation of the Induction
Motor Induced-Torque Equation
Induced (developed) torque:
τind = Pconv/ωm
= PAG/ωsync
…(7-35)
…(7-36)
Total (three-phase) air-gap power:
PAG = 3I22 (R2/s)
Finding τind expression as a function of slip (or rotor speed)
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The Derivation of the Induction
Motor Induced-Torque Equation
Using Thevenin’s theorem,
VTH = open-circuit voltage at
terminals
ZTH = equivalent impedance seen at
terminals while Vφ is shorted
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The Derivation of the Induction
Motor Induced-Torque Equation
Since the magnetizing reactance XM >> X1 and XM >> R1, the magnitude of the
Thevenin voltage, VTH is approximately as
VTH = Vφ [ XM/√ (R12 + (X1+XM)2) ]
≈ Vφ [XM/(X1+XM)]
…(7-41a)
…(7-41b)
The Thevenin impedance ZTH is given by
ZTH = Z1ZM/(Z1+ZM)
= (jXM)(R1 + jX1)/[(R1 + j(X1+XM))]
= RTH + jXTH
…(7-42)
…(7-43)
Since the magnetizing reactance XM >> X1 and XM >> R1, the Thevenin
impedance, ZTH is approximately as
RTH ≈ R1[XM/(X1+XM)]2
…(7-44)
XTH ≈ X1
…(7-45)
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The Derivation of the Induction
Motor Induced-Torque Equation
According to the resulting equivalent circuit in Figure 7-18c, the current I2 is
given by
I2 = VTH / (ZTH+Z2)
…(7-46)
= VTH / [RTH+R2/s +jXTH+jX2]
…(7-47)
The magnitude of this current I2 is
I2 = VTH/[√((RTH+R2/s)2 + (XTH+X2)2) ]
…(7-48)
The air-gap power is thus given by
PAG = 3I22 (R2/s)
= 3VTH2(R2/s) /[(RTH+R2/s)2 + (XTH+X2)2]
…(7-49)
Finally, the induced torque is given by
τind = PAG/ωsync
= 3VTH2(R2/s) /[ωsync((RTH+R2/s)2 + (XTH+X2)2)]
…(7-50)
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The Derivation of the Induction
Motor Induced-Torque Equation
τind = 3VTH2(R2/s) /[ωsync((RTH+R2/s)2 + (XTH+X2)2)]
…(7-50)
(Breakdown torque)
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Maximum (Pullout) Torque in
an Induction Motor
τind = 3VTH2(R2/s) /[ωsync((RTH+R2/s)2 + (XTH+X2)2)]
…(7-50)
The maximum induced torque will occur
when the power consumed by the rotor
resistor, RR/s is maximum.
The maximum power transfer occurs when the R2/s is the magnitude of source
impedance.
Zsource = RTH + jXTH + jX2
…(7-51)
R2/s =√[RTH2 + (XTH+X2)2]
…(7-52)
Slip at pullout torque, smax = R2/√[RTH2 + (XTH+X2)2]
…(7-53)
Pullout torque, τmax = 3VTH2 /[2ωsync(RTH+√(RTH2 + (XTH+X2)2)]
…(7-54)
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Example Problem
Example 7-4 on page 413
Example 7-5 on page 414
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7.6 Variations in Induction Motor
Torque-Speed Characteristics
Control of Motor Characteristics
by Cage Rotor Design
The rotor design affects
• Starting torque
• Slip at normal operation
• Starting current
• Efficiency
The torque-speed characteristic curve
could be modified by changing
• Rotor resistance R2
• Rotor leakage reactance X2
When the rotor resistance R2 increases,
• Starting torque increases
• Slip at normal operation increases
• Starting current reduces
• Efficiency reduces
Note: Pconv = (1 - s)PAG
(Higher slip, smaller Pconv
⇒ lowering the efficiency)
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Deep-Bar and Double-Cage
Rotor Designs
Laminations from typical
cage induction motor
rotors, showing the cross
section of the rotor bars.
Class A (deep-bar)
Class B (deep-bar)
Different laminations
provide different designs of
rotor resistance (R2) and
rotor leakage reactance
(X2).
Class C (double-cage)
Class D (deep-bar)
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Deep-Bar and Double-Cage
Rotor Designs
Recall: rotor leakage
reactance is the reactance
due to the rotor flux lines
that do not couple with
the stator windings.
Deep bar: Classes A, B, D
Double-cage: Class C
1. Two sets of bars
buried in the
laminations.
2. Most expensive among
cage rotors, but
cheaper than woundrotor.
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Induction Motor Design
Classes
The National Electric Manufacturers
Association (NEMA) in USA and the
International Electrotechnical Commission
(IEC) in Europe have defined a series of
standard designs
Class A: Standard motor design. Normal
starting torque. Normal starting current. Low
slip (less than 5% at full load).
Class B: Normal starting torque. Lower
starting current. Low slip (less than 5% at full
load).
Class C: High starting torque. Low starting
current. Low slip (less than 5% at full load).
Class D: Very high starting torque. Low
starting current. High slip.
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7.8 Starting Induction Motors
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Starting Induction Motors
• Induction motors do not present the starting problems that synchronous motors do.
• However, the reduction of starting current (causing a dip in power system voltage)
by avoiding the simply connecting them to the power line is necessary.
• To estimate the starting current, all cage motors have a code letter (not same as
design class).
To determine the starting current,
compute the apparent power
Sstart = (rated horsepower)(code letter factor)
…(7-55)
from the nameplate data.
Then, the starting current is
computed as
Istart = Sstart /(√3 VT)
…(7-56)
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Example Problem
Example 7-7 on page 430
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Induction Motor Starting
Circuits
Ways to reduce the starting current:
1. Insert extra inductors or resistors into the power line during starting.
2. Use autotransformer (or variac) to reduce the terminal voltage during starting.
(a) Close 1 and 3: Vmotor = VT/2
(b) Open 1 and 3: Vmotor = 0
(c) Close 2: Vmotor = VT
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7.9 Speed Control of Induction
Motors
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Induction Motor Speed Control
by Pole Changing
nsync = 120fe/P
rpm
…(7-1)
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Speed Control by Changing
the Line Frequency
nsync = 120fe/P
Speeds below
the rated speed
Speeds above
the rated speed
rpm
…(7-1)
The motor speed can be increased
by increasing the line frequency.
However, the voltage applied to the
motor must be accordingly
increased up to the rated voltage at
the rated frequency.
The applied voltages for the motor
speed above the rated speed must
be kept constant at rated value. This
resign is so called “field weakening”.
Flux ∝ Vapplied/fe
…(7-57)
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Speed Control by Changing
the Line Voltage
τind = 3VTH2(R2/s) /[ωsync(RTH +R2/s)2 + (XTH+X2)2]
…(7-50)
Reducing the line voltage,
• Reducing the rotor speed.
• Reducing the torque.
• Reducing the efficiency (slip increased).
Note: The poor speed regulation respect
to change in load. This method is
sometimes used on small motors driving
fans.
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Speed Control by Changing
the Rotor Resistance
τind = 3VTH2(R2/s) /[ωsync(RTH +R2/s)2 + (XTH+X2)2]
…(7-50)
Increasing the rotor resistance,
• Reducing the rotor speed.
• Reducing the torque.
• Reducing the efficiency (slip increased).
Note: The poor speed regulation respect
to change in load.
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7.11 Determining Circuit Model
Parameters
Determining Circuit Model
Parameters
IEEE Standard 112:
Describe the detailed
methods of determining
circuit model parameters.
The tests are analogous to
the short-circuit and opencircuit tests in a transformer.
R1 = Stator resistance (Ohm)
R2 = Rotor resistance referred to stator (Ohm)
X1 = Stator leakage reactance (Ohm)
X2 = Rotor leakage reactance referred to stator (Ohm)
XM = Magnetization reactance (Ohm)
RC = Core losses resistance (Ohm)
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The No-Load Test
Applying the rated voltage at rated frequency without load torque.
During testing,
1. Speed is nearly synchronous
speed.
2. Slip is about zero.
3. R2(1-s)/s approaches infinity.
4. I2 is about zero.
No-load test provide the
following information
1. Rotational losses
(Prot = Pcore+ PF&W + Pmisc)
2. X1 + XM
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The No-Load Test
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The No-Load Test
Neglect
Neglect
Stator copper losses: PSCL = 3I12 R1
…(7-25)
Input power: Pin = PSCL + Pcore+ PF&W + Pmisc
= 3I12R1 + Prot
…(7-58)
Due to air-gap in the machine, the current established the
magnetic field is quite large, comparing with the transformer. So,
XM << [RC // R2(1-s)/s]
Neglecting R1, then
|Zeq| = Vφ/I1,nl ≈ X1 + XM
Pin ≈ Prot
…(7-60)
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The DC Test for Stator Resistance
Applying a DC voltage to the stator windings until the rated current
is reached. No induced voltage in the rotor circuit.
2R1 = VDC/IDC
R1 = VDC/2IDC
…(7-61)
Note: IEEE Standard 112 provides details when the skin effect (due
to AC current) takes into consideration. The RAC is higher than RDC.
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The Locked-Rotor (Blocked-Rotor)
Test
Applying the voltage at 25% of rated frequency until the rated
current is reached with rotor is locked or blocked (cannot move).
During testing,
1. Speed is zero.
2. Slip is one.
3. R2/s = R2.
Locked-rotor test provide
the following information
1. Copper losses
(PSCL and PRCL)
2. R2
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3. XLR = X1 + X2
Neglect
The Locked-Rotor (Blocked-Rotor)
Test
Neglect
Input power:
Pin = √3 VTIL cos θ
Power factor:
PF = cos θ = Pin/(√3 VTIL)
where θ is impedance angle.
…(7-62)
Total impedance: |ZLR| = Vφ/I1 = VT/(√3 IL)
ZLR = RLR + jXLR’
= |ZLR|cos θ + j|ZLR|sin θ
…(7-63)
…(7-64)
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The Locked-Rotor (Blocked-Rotor)
Test
Neglect
Locked-rotor resistance: RLR = R1 + R2
…(7-65)
Locked-rotor reactance: XLR’ = X1’ + X2’
…(7-66)
where X1’ and X2’ are stator and rotor leakage reactances at the
test frequency, respectively.
Rotor resistance:
R2 = RLR - R1
where R1 was found in the DC test.
…(7-67)
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The Locked-Rotor (Blocked-Rotor)
Test
Neglect
Locked-rotor reactance: XLR’ = X1’ + X2’
…(7-66)
where X1’ and X2’ are stator and rotor leakage reactances at the
test frequency (ftest), respectively.
Total leakage reactance at rated frequency:
XLR = (frated/ftest)XLR’ = X1 + X2
…(7-68)
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Separating the Stator and Rotor
Leakage Reactances
From the locked-rotor test, the XLR = X1 + X2 can be broken down
into the stator and rotor leakage reactances by using the following
rule of thumb.
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Example Problem
Example 7-8 on page 458
66
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7.13 Induction Motor Ratings
Induction Motor Ratings
Important ratings presented on the nameplate:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Output power (40 HP)
Voltage (230/460 V)
Current (97/48.5 A)
Frequency (60 Hz)
Power factor (0.827)
Speed (3565 rpm)
Nominal efficiency (93.6 %)
NEMA design class (class B)
Starting code (code G)
How many poles for this
particular induction motor ?
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5/10/2013
EEE 118: Energy Conversion
Dr. Mongkol Konghirun
Department of Electrical Engineering
King Mongkut’s University of Technology Thonburi
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