ENERGY CONVERSION ONE (Course 25741)

1
ENERGY CONVERSION ONE (Course 25741)

• CHAPTER SEVEN
• INDUCTION MOTORS (Maximum Torque)

2
INDUCTION MOTORMAXIMUM TORQUE

• maximum power transfer occurs when
• R2/svRTH2 (XTHX2)2
  (1)
• solving (1) for slip ?
• smaxR2 / vRTH2 (XTHX2)2
  (2)
• Note slip of rotor (at maximum torque) R2
  rotor
• 
  resistance
• applying this value of slip to torque equation?

3
INDUCTION MOTORMAXIMUM TORQUE

• This maximum torque VTH 2 (or square of
  supply voltage)
• inversely related to stator Impedances
  rotor reactance
• The smaller a machines 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

4
INDUCTION MOTORMAXIMUM TORQUE

• Effect of varying rotor resistance on T-? of
  wound rotor

5
INDUCTION MOTORMAXIMUM 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

6
INDUCTION MOTOREXAMPLE(1)

• A 2 pole, 50 Hz induction motor supplies 15kW to
  a load at a speed of 2950 r/min.
• What is the motors 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?

7
INDUCTION MOTOREXAMPLE(1)-SOLUTION

• (a) nsync 120fe/p 120x50/23000 r/min
• s 3000-2950/30000.0167 or 1.67
• (b) TindPconv/?m15 / (2950)(2px1/60)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) PconvTind ?m97.2 x 2900 x 2px1/6029.5 kW

8
INDUCTION MOTOREXAMPLE(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 ? R2 0.332 ?
• X1 1.106 ? X2 0.464 ? Xm 26.3 ?
• 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?

9
INDUCTION MOTOREXAMPLE(2)-SOLUTION

• Thevenin Voltage
  
• 266/ v(0.641)2(1.10626.3)2 255.2 V
• (0.641)(26.3/1.10626.3)20.5
  9 O
• XTHX11.106 O
• smax R2 / vRTH2 (XTHX2)2
• 0.332/v(0.59)2(1.1060.464)20.198

10
INDUCTION MOTOREXAMPLE(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
• 3(255.2)2 / 2x188.5x0.59v0.59
  2(1.1060.464)2 229 N.m.

11
INDUCTION MOTOREXAMPLE(2)-SOLUTION

• (b) starting torque of motor found by s1
• 3x255.22 x 0.332 / 188.5x(0.590.332
  )2(1.1060.464)2104 N.m.
• (c) rotor resistance is doubled, ? s at Tmax
  doubles
• smax0.396 , and the speed at Tmax is
• nm(1-s)nsync(1-0.396)(1800)1087 r/min
• Maximum torque is still
• Tmax229 N.m. and starting torque is
• Tstart3(255.2)(0.664) / (188.5)(0.590.664
  )2(1.1060.464)2 170 N.m.

12
INDUCTION MOTORVARIATION IN TORQUE-SPEED

• Discussion

13
INDUCTION MOTORVARIATION IN TORQUE-SPEED

• Desired Motor Characteristic
• Should behave like the high-resistance
  wound-rotor curve at high slips, like the
  low-resistance wound-rotor curve at low slips

14
INDUCTION MOTORVARIATION IN TORQUE-SPEED

• Control of Motor Characteristics by Cage
  Rotor Design
• Leakage reactance X2 represents the referred form
  of the rotors leakage reactance (reactance due
  to the rotors 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 bars 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.

15
INDUCTION MOTORVARIATION 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

16
INDUCTION MOTOR TORQUE-SPEED of CLASSA,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 bars
  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

17
INDUCTION MOTOR TORQUE-SPEED of CLASSA,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 rotors 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

18
INDUCTION MOTOR TORQUE-SPEED of CLASSA,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 deep-bar 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

19
INDUCTION MOTOR TORQUE-SPEED of CLASSA,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 bars
  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

20
INDUCTION MOTOR TORQUE-SPEED of CLASSA,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

21
INDUCTION MOTOR TORQUE-SPEED of CLASSA,B,C,D

• T-speed Curve for Different Rotor Design

22
INDUCTION MOTOR TORQUE-SPEED of CLASSA,B,C,D

• Basic concepts of developing variable rotor
  resistance by deep rotor bars or double-cage
  rotors

23
INDUCTION MOTOR TORQUE-SPEED of CLASSA,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
  LltL1ltL2ltL3

24
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 R10.641O, R20.3O
• X10.75 O, X20.5 O , XM26.3 O
• 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
• R2o3.2 O X2o0.5 O (of outer-cage)
• R2i0.4 O X2i3.3 O (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,

25
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

26
INDUCTION MOTORSSTARTING

• 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 motors rated power on
  effective rotor resistance at starting conditions

27
INDUCTION MOTORSSTARTING

• 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)
• In example for code letter A factor of
  kVA/hp is between 0-3.15 (not include lower bound
  of next higher class)

28
INDUCTION MOTORSSTARTING

• 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 Sstart15 x 5.6 84 kVA
• the starting current is thus
• ILSstart / v3 VT 84 / v3 x 208 233 A
• Starting current may be reduced by a starting
  circuit
• a- inductor banks
• b- resistor banks
• c-reduce motors terminal voltage by
  autotransformer

29
INDUCTION MOTORSSTARTING

• 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

30
INDUCTION MOTORSSTARTING

• A typical full-voltage (across-the-line) motor
  magnetic starter circuit

31
INDUCTION MOTORSSTARTING

• 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 de-energized,
  M contacts open, stopping motor

32
INDUCTION MOTORSSTARTING

• 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)

33
INDUCTION MOTORSSTARTING

• 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

34
INDUCTION MOTORSSTARTING

• 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

35
INDUCTION MOTORSSTARTING

• 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

36
INDUCTION MOTORSPEED 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 motors speed
• - Varying synchronous speed
• - Varying slip

37
INDUCTION MOTORSPEED 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

38
INDUCTION MOTORSPEED CONTROL

• 1- method of consequent poles
• relies on the fact that number of poles in
  stator windings can easily changed by a factor of
  21, 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

39
INDUCTION MOTORSPEED 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

40
INDUCTION MOTORSPEED 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

41
INDUCTION MOTORSPEED 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 high-speed 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 ?

42
INDUCTION MOTORSPEED CONTROL

• Figure of possible connections
• of stator coils in a pole changing
• motor
• constant-torque Connection torque capabilities
  of motor remain approximately constant in both
  high-speed low-speed connection
• Constant-hp connection power capabilities of
  motor remain approximately constant in
• Fan torque connection

43
INDUCTION MOTORSPEED CONTROL

• Major Disadvantage of consequent-pole method of
  changing speed speeds must be in ratio of 21
• 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

44
INDUCTION MOTORSPEED 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

45
INDUCTION MOTORSPEED 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

46
INDUCTION MOTORSPEED 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 v3 VLIL cos? should be
  decreased if terminal voltage decreased
• Figures (7-42 )

47
INDUCTION MOTORSPEED CONTROL

• Variable-frequency speed control
• 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

48
INDUCTION MOTORSPEED 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

49
INDUCTION MOTORSPEED CONTROL

• Variable-line-voltage speed control

50
INDUCTION MOTORSPEED 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

51
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

52
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

53
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), PSL3I12 R1

54
INDUCTION MOTOR CCT MODEL PAR. MEAURSEMENT

• PinPSCLPcorePFWPmisc3 I12 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
• ZeqVf/I1,nl X1XM
• if X1 known by another fashion, XM can be
  determined

55
INDUCTION MOTORDC 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

56
INDUCTION MOTORDC 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 R1VDC/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

57
INDUCTION MOTORLOCKED 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

58
INDUCTION MOTORLOCKED ROTOR TEST

• Since rotor is stationary, slip s1. 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

59
INDUCTION MOTORLOCKED 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

60
INDUCTION MOTORLOCKED ROTOR TEST

• Pv3 VT IL cos
  ?
• So locked-rotor P.F. found as
• 
  PF cos? Pin / v3 VT IL
• Impedance angle is ?acos P.F.
• Magnitude of total impedance
• 
  ZLR Vf/I1VT/v3 IL
• Angle of total impedance is ?, therefore,
• 
  ZLRRLRjXLR ZLR cos ? j ZLR sin?
• Locked-rotor resistance RLRR1R2
• While locked-rotor reactance
• 
  XLRX1X2
• Where X1 and X2 are stator rotor reactances
  at test frequency

61
INDUCTION MOTORLOCKED ROTOR TEST

• Rotor resistance R2 can be found
• R2RLR-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
• XLRfrated/ ftest
  X1X2
• No simple way for separation of stator rotor
  contributions
• Experience, shown motors of certain design have
  certain proportions between rotor stator
  reactances

62
INDUCTION MOTORLOCKED ROTOR TEST

• Table summarizes this experience
• In normal practice does not matter how XLR is
  divided, since reactance appears as X1X2 in all
  torque equations

X1X2 as function of XLR X1X2 as function of XLR
Rotor Design X1 X2
Wound rotor 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
63
INDUCTION MOTOREquivalent 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 VDC13.6 IDC 28.0 A
• No load test
• VT208 V f
  60 Hz
• IA8.12 A, IB8.2 A, IC8.18 A Pin420 W
• Locked rotor test
• VT25 V
  f15 Hz
• IA28.1 A, IB28.0 A, IC27.6 A
  Pin920 W

64
INDUCTION MOTOREquivalent CCT Parameters-Example

• sketch per phase equivalent circuit of this
• motor
• (b) Find slip at pullout torque, and find the
  value of pullout torque
• Solution
• from dc test ?R1VDC/2IDC 13.6/2x28
• 
  0.243 O
• from no load test IL,av8.128.28.18/38.17
  A
• Vf,nl208/v3 120
  V
• therefore znl120/8.1714.7 O
  X1XM
• when X1 is known, XM can be found

65
INDUCTION MOTOREquivalent CCT Parameters-Example

• The copper losses
• PSCL3I12 R13 X 8.172 x 0.243 O 48.7 W
• No load rotational losses are
• ProtPin,nl PSCL,nl420 -48.7 371.3 W
• from locked-rotor test
• IL,av28.128.027.6/327.9 A
• Locked rotor impedance is ZLR25/v3x27.90.5
  17 O
• Impedance angle
• ?acos 920/v3 x 25 x 27.9acos 0.76240.4 ?
  
• ? RLR0.517cos 40.40.394 O R1R2
• since R10.243 O ? R20.151 O
• The reactance at 15 Hz XLR0.517 sin
  40.40.335O

66
INDUCTION MOTOREquivalent CCT Parameters-Example

• Equivalent reactance at 60 Hz
• XLR frated/ftest XLR60/15 x 0.335 1.34 O
• For design class A, this reactance divided
  equally between rotor stator
• X1X20.67 O
• XMZnl-X114.7-0.6714.03 O
• Per phase equivalent cct shown below

67
INDUCTION MOTOREquivalent CCT Parameters-Example

• (b) for this equivalent cct. Thevenin equivalent
  found as follows
• VTH114.6 V, RTH0.221 O, XTH0.67 O
• ? slip at pullout torque is
• smax R2/v RTH2(XTHX2)2
• 0.151/v0.2432(0.670.67)20.11111.1
• Maximum torque of this motor is given by
• Tmax 3 VTH2/2?syncRTHvRTH2(XTHX2)2
• 3 x 114.6 2 /2x188.5x0.221v0.2212(0.2x0.67)
  266.2 N.m.

68
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