Transcript 25471_ENERGY_CONVERSION_18

ENERGY CONVERSION ONE
(Course 25741)
CHAPTER SEVEN
INDUCTION MOTORS … (Maximum Torque…)
INDUCTION MOTOR
MAXIMUM TORQUE
• maximum power transfer occurs when:
R2/s=√RTH^2 + (XTH+X2)^2
(1)
solving (1) for slip 
smax=R2 / √RTH^2 + (XTH+X2)^2
(2)
Note: slip of rotor (at maximum torque) ~ R2 rotor
resistance
applying this value of slip to torque equation
 max 
2
3VTH
2 sync  RTH 

2
TH
R
  X TH  X 2  

2
INDUCTION MOTOR
MAXIMUM TORQUE
• This maximum torque ~ VTH ^2 (or square of
supply voltage)
& inversely related to stator Impedances & rotor
reactance
• The smaller a machine’s reactance the larger
its maximum torque
• Note: smax ~ R2 , however maximum torque is
independent of R2
• Torque-speed characteristic of a wound-rotor
induction motor shown if figure next
INDUCTION MOTOR
MAXIMUM TORQUE
• Effect of varying rotor resistance on T-ω of
wound rotor
INDUCTION MOTOR
MAXIMUM TORQUE
• as the value of external resistor connected to rotor
circuit of a wound rotor through slip rings is increased
the pullout speed decreased, however the maximum
torque remains constant
• Advantage can be taken from this characteristic of
wound-rotor induction motors to start very heavy loads
• If a resistance inserted into rotor circuit, Tmax can be
adjusted to at starting conditions
• And while load is turning, extra resistance can
be removed from circuit, & Tmax move up to near
synchronous speed for regular operation
INDUCTION MOTOR
EXAMPLE(1)
• A 2 pole, 50 Hz induction motor supplies 15kW
to a load at a speed of 2950 r/min.
• What is the motor’s slip?
• What is the induced torque in the motor in Nm
under these conditions?
• What will the operating speed of the motor be if
its torque is doubled?
• How much power will be supplied by the motor
when the torque is doubled?
INDUCTION MOTOR
EXAMPLE(1)-SOLUTION
(a) nsync= 120fe/p= 120x50/2=3000 r/min
s= 3000-2950/3000=0.0167 or 1.67%
(b) Tind=Pconv/ωm=15 / (2950)(2πx1/60)=48.6 N.m.
(c) as shown, in low slip region, torque-speed is
linear & induced torque ~ s  doubling Tind
slip would be 3.33 % 
nm=(1-s)nsync =(1-0.0333)(3000)=2900 r/min
(d) Pconv=Tind ωm=97.2 x 2900 x 2πx1/60=29.5 kW
INDUCTION MOTOR
EXAMPLE(2)
• A 460V, 25hp, 60Hz, 4-pole, Y-connected wound rotor
induction motor has the following impedances in ohms
per-phase referred to the stator circuit:
• R1 = 0.641 Ω
R2 = 0.332 Ω
• X1 = 1.106 Ω
X2 = 0.464 Ω
Xm = 26.3 Ω
– What is the max torque of this motor? At what speed and
slip does it occur?
– What is the starting torque?
– When the rotor resistance is doubled, what is the speed at
which the max torque now occurs? What is the new starting
torque?
INDUCTION MOTOR
EXAMPLE(2)-SOLUTION
VTH  V
Xm
Thevenin Voltage :
=
R  X  X 
= 266/ √(0.641)^2+(1.106+26.3)^2= 255.2 V
2
1
2
1
m
=(0.641)(26.3/[1.106+26.3])^2=0.59 Ω
XTH≈X1=1.106 Ω
(a) smax= R2 / √RTH^2 + (XTH+X2)^2
=0.332/√(0.59)^2+(1.106+0.464)^2=0.198
INDUCTION MOTOR
EXAMPLE(2)-SOLUTION
• This corresponds to a mechanical speed of :
nm=(1-s)nsync=(1-0.198)(1800)=1444 r/min
• the torque at this speed :
 max 
2
3VTH
2 sync
R
TH 


2
TH
R

 X TH
 X2  


2
= 3(255.2)^2 /
{2x188.5x[0.59+√0.59^2+(1.106+0.464)^2]}
=229 N.m.
INDUCTION MOTOR
EXAMPLE(2)-SOLUTION
(b) starting torque of motor found by s=1
 start 
sync [(RTH
2
3VTH
R2

2
2
 R2 )  ( X TH  X 2 ) ]
= 3x255.2^2 x 0.332 /
{188.5x[(0.59+0.332)^2+(1.106+0.464)^2]}=104 N.m.
(c) rotor resistance is doubled,  s at Tmax doubles
smax=0.396 , and the speed at Tmax is:
nm=(1-s)nsync=(1-0.396)(1800)=1087 r/min
Maximum torque is still:
Tmax=229 N.m. and starting torque is :
Tstart=3(255.2)(0.664) /
{(188.5)[(0.59+0.664)^2+(1.106+0.464)^2]} =170 N.m.
INDUCTION MOTOR
VARIATION IN TORQUE-SPEED
Discussion:
INDUCTION MOTOR
VARIATION IN TORQUE-SPEED
• Desired Motor Characteristic
• Should behave: like the high-resistance woundrotor curve; at high slips, & like the lowresistance wound-rotor curve at low slips
INDUCTION MOTOR
VARIATION IN TORQUE-SPEED
Control of Motor Characteristics by Cage Rotor
Design:
• Leakage reactance X2 represents the referred form of
the rotor’s leakage reactance (reactance due to the
rotor’s flux lines that do not couple with the stator
windings.)
• Generally, the farther away the rotor bar is from the
stator, the greater its X2 , since a smaller percentage
of the bar’s flux will reach the stator.
• Thus, if the bars of a cage rotor are placed near the
surface of the rotor, they will have small leakage flux
and X2 will be small.
INDUCTION MOTOR
VARIATION IN TORQUE-SPEED
Laminations from typical cage induction motor, cross section of the rotor
bars:
NEMA class A – large bars near the surface
NEMA class B – large, deep rotor bars
NEMA class C – double-cage rotor design
NEMA class D – small bars near the surface
INDUCTION MOTOR
TORQUE-SPEED of CLASS:A,B,C,D
• NEMA (National Electrical Manufacturers Association)
class A
• Rotor bars are quite large and are placed near the surface of
the rotor
• Low resistance (due to its large cross section) and a low
leakage reactance X2 (due to the bar’s location near the stator)
• Because of the low resistance, the pullout torque will be quite
near synchronous speed ; full load slip less than 5%
• Motor will be quite efficient, since little air gap power is lost in
the rotor resistance. ;
• However, since R2 is small, starting torque will be small, and
starting current will be high
• This design is the standard motor design
• Typical applications : driving fans, pumps, and other machine
tools
• Principal problem: extremely high inrush current on starting,
500 to 800 % of rated
INDUCTION MOTOR
TORQUE-SPEED of CLASS:A,B,C,D
•
NEMA Class B
• At the upper part of a deep rotor bar, the current flowing is tightly
coupled to the stator, and hence the leakage inductance is small in this
region. Deeper in the bar, the leakage inductance is higher
• At low slips, the rotor’s frequency is very small, and the reactances of
all the parallel paths are small compared to their resistances. The
impedances of all parts of the bar are approx equal, so current flows
through all the parts of the bar equally. The resulting large cross
sectional area makes the rotor resistance quite small, resulting in good
efficiency at low slips.
• At high slips (starting conditions), the reactances are large compared
to the resistances in the rotor bars, so all the current is forced to flow
in the low-reactance part of the bar near the stator. Since the effective
cross section is lower, the rotor resistance is higher. Thus, the starting
torque is relatively higher and the starting current is relatively lower
than in a class A design (about 25% less)
• Applications similar to class A, and this type B have largely replaced
type A
• Pullout Torque greater than or equal 200% of rated load torque
INDUCTION MOTOR
TORQUE-SPEED of CLASS:A,B,C,D
• NEMA Class C
• It consists of a large, low resistance set of bars buried
deeply in the rotor and a small, high-resistance set of
bars set at the rotor surface. It is similar to the deepbar rotor, except that the difference between low-slip
and high-slip operation is even more exaggerated
• At starting conditions, only the small bars are effective,
and the rotor resistance is high. Hence, high starting
torque. However, at normal operating speeds, both
bars are effective, and the resistance is almost as low
as in a deep-bar rotor
• Used in high starting torque loads such as loaded
pumps, compressors, and conveyors
INDUCTION MOTOR
TORQUE-SPEED of CLASS:A,B,C,D
• NEMA class D
• Rotor with small bars placed near the surface of the
rotor (higher-resistance material)
• High resistance (due to its small cross section) and a
low leakage reactance X2 (due to the bar’s location
near the stator)
• Like a wound-rotor induction motor with extra
resistance inserted into the rotor
• Because of the large resistance, the pullout torque
occurs at high slip, and starting torque will be quite
high, and low starting current (starting T, 275% Trated)
• Typical applications : extremely high-inertia type loads
INDUCTION MOTOR
TORQUE-SPEED of CLASS:A,B,C,D
• NEMA Class E and F
• Class E and Class F are already discontinued
They are low starting torque machines
• These called soft-start induction motors
• These are also distinguished by having very low
starting currents & used for starting-torque
loads in situations where starting current were a
problem
INDUCTION MOTOR
TORQUE-SPEED of CLASS:A,B,C,D
• T-speed Curve for Different Rotor Design
INDUCTION MOTOR
TORQUE-SPEED of CLASS:A,B,C,D
• Basic
concepts of
developing
variable
rotor
resistance
by deep
rotor bars or
doublecage rotors
INDUCTION MOTOR
TORQUE-SPEED of CLASS:A,B,C,D
•
•
•
•
Basic Concept continued; (Last Figure)
In Fig (a): for a current flowing in the upper part of
the deep rotor bar, the flux is tightly linked to the
stator, and leakage L is small
In Fig (b): current flowing at the bottom part of the
bar, the flux is weakly linked to the stator, and
leakage L is large
Fig (c): Resulting equivalent circuit
Since all parts of rotor bar are in parallel electrically,
bar represents a series of parallel electric circuits,
upper ones have smaller inductance & lower ones
larger inductance :
L<L1<L2<L3
INDUCTION MOTOR
EXAMPLE-3
• A 460 V, 30 hp, 60 Hz, 4 pole, Y connected induction motor has
two possible rotor designs:
- A single cage rotor &
- A double-cage rotor (stator identical for both designs)
• Single-cage modeled by: R1=0.641Ω, R2=0.3Ω
X1=0.75 Ω, X2=0.5 Ω , XM=26.3 Ω
• Double-cage; modeled as tightly coupled high resistance outer
cage in parallel
with a loosely coupled, low-resistance inner cage , stator
magnetization resistance & reactances identical
R2o=3.2 Ω X2o=0.5 Ω (of outer-cage)
R2i=0.4 Ω X2i=3.3 Ω (of outer-cage)
Calculate torque-speed characteristics associated with two
rotor designs solution by MATLAB
Results: double-cage: slightly higher slip, smaller Tmax, higher
Tstarting,
TRENDS IN INDUCTION MOTOR
DESIGN
• Smaller motor for a given power output, great saving (modern
100 hp same size of 7.5 hp motor of 1897)
• However not necessarily increase in efficiency (used since
electricity was inexpensive)
• New lines of high efficiency induction motors being produced by
all major manufacturers using some the following techniques;
1- More copper in stator windings; reduce copper losses
2- rotor & stator length increased to reduce B in air gap
(decreasing saturation and core loss)
3-More steel in stator, greater amount of heat transfer
4- using special high grade steel with low hysteresis loss in
stator
5- steel made of especially thin guage & high resistivity toreduce
eddy current loss
6-rotor carefully machined to produce uniform air gap, reducing
stray load losses
INDUCTION MOTORS
STARTING
• An induction motor has the ability to start directly,
however direct starting of an induction motor is not
advised due to high starting currents, may cause dip in
power system voltage; that across-the-line starting not
acceptable
• for wound rotor, by inserting extra resistance can be
reduced; this increase starting torque, but also
reduces starting current
• For cage type, starting current vary widely depending
primarily on motor’s rated power & on effective rotor
resistance at starting conditions
INDUCTION MOTORS
STARTING
• To determine starting current, need to calculate the
starting power required by the induction motor.
• A Code Letter designated to each induction motor,
which can be seen in figure 7-34, may represent this.
(The starting code may be obtained from the motor
nameplate)
IL
S start

3VT
In example: for code letter A; factor of kVA/hp is
between 0-3.15 (not include lower bound of next
higher class)
INDUCTION MOTORS
STARTING
• EXAMPLE: what is starting current of a 15 hp, 208 V,
code letter F, 3 phase induction motor?
• Maximum kVA / hp is 5.6  max. starting kVA of this
motor is Sstart=15 x 5.6 = 84 kVA
the starting current is thus:
IL=Sstart / [√3 VT] = 84 / [√3 x 208] = 233 A
• Starting current may be reduced by a starting circuit:
a- inductor banks
b- resistor banks
c-reduce motor’s terminal voltage by autotransformer
INDUCTION MOTORS
STARTING
• Autotransformer
starter:
• During starting 1 & 3
closed, when motor is
nearly up to speed; those
contacts opened & 2
closed
• Note: as starting current
reduced proportional to
decrease in voltage,
starting torque decreased
as square of applied
voltage, therefore just a
certain reduction possible
if motor is to start with a
shaft load attached
INDUCTION MOTORS
STARTING
• A typical full-voltage (across-the-line) motor
magnetic starter circuit
INDUCTION MOTORS
STARTING
• Start button pressed, rely coil M energized, &
N.O. contacts M1,M2,M3 close
• Therefore power supplied to motor & motor
starts
• Contacts M4 also close which short out starting
switch, allowing operator to release it (start
button) without removing power from M relay
• When stop button pressed, M relay deenergized, & M contacts open, stopping motor
INDUCTION MOTORS
STARTING
• A magnetic motor starter circuit has several built-in
protective features:
1- short-circuit protection
2- overload protection
3- under-voltage protection
• Short-circuit protection provided by fuses F1,F2,F3
• If sudden sh. cct. Develops within motor causes a
current (many times greater than rated current) flow;
these fuses blow disconnecting motor from supply
(however, sh. cct. by a high resistance or excessive
motor loads will not be cleared by fuses)
INDUCTION MOTORS
STARTING
• Overload protection for motor is provided “OL” relays
which consists of 2 parts: an over load heater, and
overload contacts
• when an induction motor overloaded, it is eventually
damaged by excessive heating caused by high
currents
• However this damage takes time & motor will not be
hurt by brief periods of high current (such as starting
current)
• Undervoltage protection is also provided by controller
If voltage applied to motor falls too much, voltage
applied to M relay also fall, & relay will de-energize
The M contacts open, removing power from motor
terminals
INDUCTION MOTORS
STARTING
• 3 step resistive starter
• Similar to previous,
except that there are
additional components
present to control
Removal of starting
resistors
• Relays 1TD, 2TD, & 3
TD are time-delay
relay
INDUCTION MOTORS
STARTING
• Start button is pushed in this circuit, M relay energizes
and power is applied to motor as before
• Since 1TD, 2TD, & 3TD contacts are all open the full
starting resistor in series with motor, reducing the
starting current
• When M contacts close, notice that 1 TD relay is
energized, however there is a finite delay before 1TD
contacts close, cutting out part of starting resistance &
simultaneously energizing 2TD relay
• After another delay, 2TD contacts close, cutting out
second part of resistor & energizing 3TD relay
• Finally 3TD contacts close, & entire starting resistor is
out of circuit
INDUCTION MOTOR
SPEED CONTROL
• Induction motors are not good machines for
applications requiring considerable speed control.
• The normal operating range of a typical induction
motor is confined to less than 5% slip, and the speed
variation is more or less proportional to the load
• Since PRCL = s PAG , if slip is made higher, rotor
copper losses will be high as well
• There are basically 2 general methods to control
induction motor’s speed:
- Varying synchronous speed
- Varying slip
INDUCTION MOTOR
SPEED CONTROL
• nsync= 120 fe / p
• so the only ways to change nsync is (1) changing
electrical frequency (2) changing number of
poles
• slip control can be accomplished, either by
varying rotor resistance, or terminal voltage of
motor
• Speed Control by Pole Changing
• Two major approaches:
1- method of consequent poles
2- multiple stator windings
INDUCTION MOTOR
SPEED CONTROL
1- method of consequent
poles
relies on the fact that
number of poles in stator
windings can easily
changed by a factor of 2:1,
with simple changes in coil
connections
- a 2-pole stator winding for
pole changing. Very small
rotor pitch

• In next figure for windings
of phase “a” of a 2 pole
stator, method is illustrated
INDUCTION MOTOR
SPEED CONTROL
• A view of one phase of
a pole changing
winding
• In fig(a) , current flow
in phase a, causes
magnetic field leave
stator in upper phase
group (N) & enters
stator in lower phase
group (S), producing 2
stator magnetic poles
INDUCTION MOTOR
SPEED CONTROL
• Now, if direction of current flow in lower phase group reversed,
magnetic field leave stator in both upper phase group, & lower
phase group, each will be a North pole while flux in machine
must return to stator between two phase groups, producing a
pair of consequent south magnetic poles (twice as many as
before)
• Rotor in such a motor is of cage design, and a cage rotor
always has as many poles as there are in stator
• when motor reconnected from 2 pole to 4 pole , resulting
maximum torque is the same (for :constant-torque connection)
half its previous value (for: square-law-torque connection used
for fans, etc.), depending on how the stator windings are
rearranged
• Next figure, shows possible stator connections & their effect on
torque-speed
INDUCTION MOTOR
SPEED CONTROL
• Possible connections of stator coils in a pole-changing
motor, together with resulting torque-speed
characteristics:
(a) constant-torque connection : power capabilities
remain constant in both high & low speed connections
(b) constant hp connection: power capabilities of
motor remain approximately constant in both highspeed & low-speed connections
(c) Fan torque connection: torque capabilities of motor
change with speed in same manner as fan-type loads
Shown in next figure 
INDUCTION MOTOR
SPEED CONTROL
Figure of possible connections
of stator coils in a pole changing
motor
(a) constant-torque Connection:
torque capabilities of motor
remain approximately
constant in both high-speed
& low-speed connection
(b) Constant-hp connection:
power capabilities of motor
remain approximately
constant in …
(c) Fan torque connection:
INDUCTION MOTOR
SPEED CONTROL
• Major Disadvantage of consequent-pole method of changing
speed: speeds must be in ratio of 2:1
• traditional method to overcome the limitation: employ multiple
stator windings with different numbers of poles & to energize
only set at a time
Example: a motor may wound with 4 pole & a set of 6 pole
stator windings, then its sync. Speed on a 60 Hz system could
be switched from 1800 to 1200 r/min simply by supplying power
to other set of windings
• however multiple stator windings increase expense of motor &
used only it is absolutely necessary
• Combining method of consequent poles with multiple stator
windings a 4 –speed motor can be developed
Example: with separate 4 & 6 pole windings, it is possible to
produce a 60 Hz motor capable of running at 600, 900, 1200,
and 1800 r/min
INDUCTION MOTOR
SPEED CONTROL
•
•
•
Speed Control by Changing Line
Frequency
Changing the electrical frequency will change
the synchronous speed of the machine
Changing the electrical frequency would also
require an adjustment to the terminal voltage
in order to maintain the same amount of flux
level in the machine core. If not the machine
will experience
(a) Core saturation (non linearity effects)
(b) Excessive magnetization current
•
•
•
•
INDUCTION MOTOR
SPEED CONTROL
Varying frequency with or without adjustment to the terminal
voltage may give 2 different effects :
(a) Vary frequency, stator voltage adjusted – generally vary
speed and maintain operating torque
(b) Vary Frequency, stator voltage maintained – able to
achieve higher speeds but a reduction of torque as speed is
increased
There may also be instances where both characteristics are
needed in the motor operation; hence it may be combined to
give both effects
With the arrival of solid-state devices/power electronics, line
frequency change is easy to achieved and it is more flexible
for a variety of machines and application
Can be employed for control of speed over a range from a
little as 5% of base speed up to about twice base speed
INDUCTION MOTOR
SPEED CONTROL
• Running below base speed, the terminal voltage should be
reduced linearly with decreasing stator frequency
• This process called derating, failing to do that cause saturation
and excessive magnetization current (if fe decreased by 10% &
voltage remain constant flux increase by 10% and cause
increase in magnetization current)
• When voltage applied varied linearly with frequency below base
speed, flux remain approximately constant, & maximum torque
remain fairly high, therefore maximum power rating of motor
must be decreased linearly with frequency to protect stator cct.
From overheating
• Power supplied to : √3 VLIL cosθ should be decreased if
terminal voltage decreased
• Figures (7-42 )
INDUCTION MOTOR
SPEED CONTROL
•
Variable-frequency speed
control
(a) family of torque-speed
characteristic curves for speed
below base speed (assuming line
voltage derated linearly with
frequency
(b) Family of torque-speed
characteristic curves for speeds
above base speed, assuming line
voltage held constant
INDUCTION MOTOR
SPEED CONTROL
• Speed control by changing Line Voltage
• Torque developed by induction motor is
proportional to square of applied voltage
• Varying the terminal voltage will vary the
operating speed but with also a variation of
operating torque
• In terms of the range of speed variations, it is
not significant hence this method is only
suitable for small motors only
INDUCTION MOTOR
SPEED CONTROL
• Variable-line-voltage speed control
INDUCTION MOTOR
SPEED CONTROL
• Speed control by changing rotor resistance
• In wound rotor, it is possible to change the
torque-speed curve by inserting extra
resistances into rotor cct.
• However, inserting extra resistances into rotor
cct. seriously reduces efficiency
• Such a method of speed control normally used
for short periods, to avoid low efficiency
INDUCTION MOTOR
CCT MODEL PAR. MEAURSEMENT
• Determining Circuit Model Parameters
• R1,R2,X1,X2 and XM should be determined
• Tests (O.C. & S.C.) performed under precisely
controlled conditions
Since resistances vary with Temperature & rotor
resistance also vary with rotor frequency
• Exact details described in IEEE standards 112
• Although details of tests very complicated, concepts
behind them straightforward & will be explained here
INDUCTION MOTOR
CCT MODEL PAR. MEAURSEMENT
• No Load Test
• Measures rotational
losses & provides
information about
magnetization
current
• Test cct. shown in
(a), motor allowed
to spin freely
• Wattmeters, a
voltmeter and 3
ammeters
INDUCTION MOTOR
CCT MODEL PAR. MEAURSEMENT
• In this test, only load mechanical losses, & slip
very small (as 0.001 or less)
• Equivalent cct. shown in figure (b)
• Resistance corresponding to power conversion
is R2(1-s)/s much larger than R2 & much larger
than X2 so eq. cct. Reduces to last in (b)
• output resistor in parallel with magnetization
reactance XM & core losses RC
• Input power measured by meters equal losses,
while rotor copper losses negligible (I2
extremely small), PSL=3I1^2 R1
INDUCTION MOTOR
CCT MODEL PAR. MEAURSEMENT
• Pin=PSCL+Pcore+PF&W+Pmisc=3 I1^2 R1 + Prot
• So eq. cct. In this condition contains RC and R2(1-s)/s
in parallel with XM
• While current to provide magnetic field is large due to
high reluctance of air gap & so XM would be much
smaller than resistance in parallel with it
• Overall P.F. very small
• with large lagging current
:
|Zeq|=Vφ/I1,nl ≈ X1+XM
if X1 known by another fashion, XM can be determined
INDUCTION MOTOR
DC Test for STATOR RESISTANCE
• The locked-rotor test later
used to determine total
motor circuit resistance
• However to determine rotor
resistance R2 that is very
important and affect the
torque-speed curve, R1
should be known
• There is a dc test for
determining R1. a dc power
supply is connected to two of 3
terminals of a Y connected
induction motor
• Current adjusted to rated
value & voltage between
terminals measured
INDUCTION MOTOR
DC Test for STATOR RESISTANCE
• reason for setting current to rated value is to
heat windings to same temperature of normal
operation
• 2R1= VDC/IDC
or
R1=VDC/[2 IDC]
• With R1, stator copper losses at no load can be
determined
• rotational losses determined as difference of Pin
at no load & stator copper loss
• R1 determined by this method is not accurate,
due to neglect of skin effect using an ac voltage
INDUCTION MOTOR
LOCKED ROTOR TEST
• Third test to determine cct.
Parameters of an induction
motor is called : locked-rotor
test
• In this test rotor is locked &
cannot move
• Voltage applied to motor,
voltage, current & power are
measured
• An ac voltage applied to
stator, current flow adjusted
to full-load value
• Then, voltage, current, &
power flowing to motor
measured
INDUCTION MOTOR
LOCKED ROTOR TEST
• Since rotor is stationary, slip s=1. & R2/s equal R2
(small value)
• Since R2 & X2 so small, all input current will flow
through them rather XM and circuit is a series of
X1,R1,X2 and R2
• There is one problem with this test in normal
operation, stator frequency is line frequency (50 or 60
Hz)
• At starting conditions, rotor also at power frequency
(while in normal operation slip 2 to 4 % and frequency
1 to 3 Hz) & it does not simulate normal operation
• A compromise : is to use a frequency 25% or less of
rated frequency
INDUCTION MOTOR
LOCKED ROTOR TEST
• This acceptable for constant resistance rotors
(design class A and D)
• it leaves a lot to be desired when looking for
normal rotor resistance of a variable resistance
rotor
• a great deal of care required taking
measurement for these tests
• a test voltage & frequency set up, current flow
in motor quickly adjusted to about rated
voltage, & input power, voltage and current
measured before motor heat up
INDUCTION MOTOR
LOCKED ROTOR TEST
P=√3 VT IL cos θ
• So locked-rotor P.F. found as:
PF = cosθ= Pin / [√3 VT IL]
• Impedance angle is
θ=acos P.F.
• Magnitude of total impedance :
|ZLR| =Vφ/I1=VT/[√3 IL]
• Angle of total impedance is θ, therefore,
ZLR=RLR+jX’LR= |ZLR| cos θ +j |ZLR| sinθ
• Locked-rotor resistance
RLR=R1+R2
• While locked-rotor reactance
X’LR=X’1+X’2
• Where X’1 and X’2 are stator & rotor reactances at test frequency
INDUCTION MOTOR
LOCKED ROTOR TEST
• Rotor resistance R2 can be found:
R2=RLR-R1
• R1 determined in dc test
• Total rotor reactance referred to stator can be found
• Since reactance ~ f  total eq. reactance at normal
operating frequency:
XLR=frated/ ftest= X1+X2
• No simple way for separation of stator & rotor
contributions
• Experience, shown motors of certain design have
certain proportions between rotor & stator reactances
INDUCTION MOTOR
LOCKED ROTOR TEST
• Table
summarizes
this experience
• In normal
practice does
not matter how
XLR is divided,
since
reactance
appears as
X1+X2 in all
torque
equations
X1&X2 as function of XLR
Rotor
Design
Wound rotor
X1
X2
0.5 XLR
0.5 XLR
Design A
0.5 XLR
0.5 XLR
Design B
0.4 XLR
0.6 XLR
Design C
0.3 XLR
0.7 XLR
Design D
0.5 XLR
0.5 XLR
INDUCTION MOTOR
Equivalent CCT Parameters-Example
• Following test data taken on a 7.5 hp, 4 pole,
208 V, 60 Hz, design A, Y connected induction
motor with a rated current of 28 A.
• Dc test result: VDC=13.6
IDC= 28.0 A
• No load test:
VT=208 V
f =60 Hz
IA=8.12 A, IB=8.2 A, IC=8.18 A Pin=420 W
• Locked rotor test:
VT=25 V
f=15 Hz
IA=28.1 A, IB=28.0 A, IC=27.6 A
Pin=920 W
INDUCTION MOTOR
Equivalent CCT Parameters-Example
(a) sketch per phase equivalent circuit of this
motor
(b) Find slip at pullout torque, and find the value of
pullout torque
• Solution :
(a) from dc test R1=VDC/[2IDC] =13.6/[2x28]=
0.243 Ω
from no load test: IL,av=[8.12+8.2+8.18]/3=8.17 A
Vφ,nl=208/√3 = 120 V
therefore:
|znl|=120/8.17=14.7 Ω =X1+XM
when X1 is known, XM can be found
INDUCTION MOTOR
Equivalent CCT Parameters-Example
• The copper losses:
PSCL=3I1^2 R1=3 X 8.17^2 x 0.243 Ω =48.7 W
• No load rotational losses are:
Prot=Pin,nl – PSCL,nl=420 -48.7 =371.3 W
• from locked-rotor test:
IL,av=[28.1+28.0+27.6]/3=27.9 A
Locked rotor impedance is: |ZLR|=25/[√3x27.9]=0.517 Ω
• Impedance angle
θ=acos 920/[√3 x 25 x 27.9]=acos 0.762=40.4 ◦
 RLR=0.517cos 40.4=0.394 Ω =R1+R2
since R1=0.243 Ω  R2=0.151 Ω
• The reactance at 15 Hz ; X’LR=0.517 sin 40.4=0.335Ω
INDUCTION MOTOR
Equivalent CCT Parameters-Example
• Equivalent reactance at 60 Hz:
XLR= frated/ftest X’LR=60/15 x 0.335 =1.34 Ω
• For design class A, this reactance divided equally
between rotor & stator:
X1=X2=0.67 Ω
XM=|Znl|-X1=14.7-0.67=14.03 Ω
• Per phase equivalent cct shown below:
INDUCTION MOTOR
Equivalent CCT Parameters-Example
(b) for this equivalent cct. Thevenin equivalent found as
follows:
• VTH=114.6 V, RTH=0.221 Ω, XTH=0.67 Ω
•  slip at pullout torque is :
• smax= R2/[√ RTH^2+(XTH+X2)^2]=
0.151/√0.243^2+(0.67+0.67)^2=0.111=11.1%
Maximum torque of this motor is given by:
Tmax= 3 VTH^2/{2ωsync[RTH+√RTH^2+(XTH+X2)^2]}=
3 x 114.6 ^2
/{2x188.5x[0.221+√0.221^2+(0.2x0.67)^2]}=66.2 N.m.
INDUCTION GENERATOR
• The torque –speed curve shown when induction
motor driven at speed greater than nsync by a
prime mover, direction of induced torque
reverses & act as a generator