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ENERGY CONVERSION ONE (Course 25741). CHAPTER SEVEN INDUCTION MOTORS … (Maximum Torque…) . INDUCTION MOTOR MAXIMUM TORQUE. maximum power transfer occurs when: R 2 /s=√R TH ^2 + (X TH +X 2 )^2 (1) solving (1) for slip 

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energy conversion one course 25741



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


applying this value of slip to torque equation

induction motor maximum torque1
  • 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 torque2
  • Effect of varying rotor resistance on T-ω of wound rotor
induction motor maximum torque3
  • as the value of external resistor connected to rotor circuit of a wound rotor through slip rings is increasedthe pullout speed decreased,howeverthe 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, Tmaxcan 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=Tindωm=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

Thevenin Voltage : =

= 266/ √(0.641)^2+(1.106+26.3)^2= 255.2 V

=(0.641)(26.3/[1.106+26.3])^2=0.59 Ω

XTH≈X1=1.106 Ω

  • smax=R2 / √RTH^2 + (XTH+X2)^2


induction motor example 2 solution1
  • 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 / {2x188.5x[0.59+√0.59^2+(1.106+0.464)^2]} =229 N.m.

induction motor example 2 solution2

(b) starting torque of motor found by s=1

= 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 speed1
  • 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
induction motor variation in torque speed2

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 speed3

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 d1
  • 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 d2
  • 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 loadssuch as loaded pumps, compressors, and conveyors
induction motor torque speed of class a b c d3
  • 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 d4
  • NEMA Class E and F
  • Class E and Class Fare already discontinued They are low starting torque machines
  • These called soft-start induction motors
  • These are also distinguishedby 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 d5
  • T-speedCurve for Different Rotor Design
induction motor torque speed of class a b c d6
  • Basic concepts of developingvariable rotor resistance by deep rotor bars or double-cage rotors
induction motor torque speed of class a b c d7
  • 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

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 starting1
  • 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 beobtained 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)

induction motors starting2
  • 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 starting3
  • 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 starting4
  • A typical full-voltage (across-the-line) motor magnetic starter circuit
induction motors starting5
  • 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
induction motors starting6
  • 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 starting7
  • Overload protection for motoris 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 starting8
  • 3 step resistive starter
  • Similar to previous,

except that there are

additional components

present to control

Removal of starting


  • Relays 1TD, 2TD, & 3 TD are time-delay relay
induction motors starting9
  • 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 control1
  • 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 control2

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 control3
  • 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 control4
  • 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 statorbetween two phasegroups, 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 control5
  • 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 

induction motor speed control6

Figure of possible connections

of stator coils in a pole changing


  • 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:
induction motor speed control7
  • 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 control8
  • 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 control9
  • 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 neededin 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 achievedand it is more flexible for a variety of machines and application
  • Can be employed for control of speed over a rangefrom a little as 5% of base speed up to about twice base speed
induction motor speed control10
  • 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 linearlywith 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 control11
  • Variable-frequency speed control
  • family of torque-speed

characteristic curves for speed

below base speed (assuming line

voltage derated linearly with


(b) Family of torque-speed

characteristic curves for speeds

above base speed, assuming line

voltage held constant

induction motor speed control12
  • 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 control13
  • Variable-line-voltage speed control
induction motor speed control14
  • 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 meaursement1
  • 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 meaursement2
  • 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 meaursement3
  • 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 resistance1
  • 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 test1
  • Since rotor is stationary, slip s=1. & R2/s equal R2 (small value)
  • Since R2 & X2so 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 test2
  • 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 test3

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


  • Where X’1 and X’2 are stator & rotor reactances at test frequency
induction motor locked rotor test4
  • Rotor resistance R2can be found:


  • 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 X’LR = 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 test5
  • Table summarizes this experience
  • In normal practice does not matter how XLR is divided, since reactance appears as X1+X2 in all torque equations
induction motor equivalent cct parameters example
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: 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 example1
INDUCTION MOTOREquivalent CCT Parameters-Example
  • sketch per phase equivalent circuit of this


(b) Find slip at pullout torque, and find the value of pullout torque

  • Solution :
  • 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 example2
INDUCTION MOTOREquivalent 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 example3
INDUCTION MOTOREquivalent 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 example4
INDUCTION MOTOREquivalent 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]=


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
induction generator1
  • As torque applied to shaft by prime mover increasespower produced increases
  • Fig. of last slide shows that there is a max. possible induced torque in Gen. mode
  • This torque namedpushover torqueof Gen.
  • if prime mover applies a torque greater than pushover torque to shaft of induction Gen., it will overspeed
  • As a generator, induction machine has severe limitations
  • Since itlacksa separate field circuit, it can not produce reactive power
induction generator2
It consumes reactive power, & an external source of reactive powershould be connected to it to maintain stator magnetic field

This source of reactive poweralso should controlterminal voltage of Gen.

Normally Gen’s voltageis maintained by external power systemto which it is connected

one advantage of induction Gen. is its simplicity,

an induction Gen.does not need a separate field circuit & does not have to be driven continuously at a fixed speed

induction generator3
as long as machine’s speed is greater than nsync

(of power system) it will function as a generator

The greater the torque applied to its shaft (up to certain point) the greater its resulting output power

Since it does not require a complicated regulation,this Gen. is a good choice for windmills, heat recovery systems, & similar supplementary power sourcesattached to an existing power system

In such applications, P.F. correction can be provided by capacitors, & generator’s terminal voltage can be controlled by external power system

induction generator4

Induction Generator Operating Alone

  • Induction machine can function as an isolated generator independent of any power system

as long as capacitors available to supply reactive power required by Gen. & by attached loads

Such an isolated induction generator is shown

induction generator5
  • Magnetizing current IM required by an induction machine as function of terminal voltagefound by running machine as motor at no load & measuring its armature current as a function of terminal voltage
  • Such a magnetization curve is shown: 
  • To achieve a given voltage in an induction Gen., external capacitors must supply magnetization current corresponding to that level
induction generator6
  • Since reactive current a capacitor can produce is directly proportional to voltage applied, locus of all possible combinations of voltage & current through a capacitor is straight line
  • Plot of voltage versus current for a given frequency
induction generator7
  • Having a 3 phase set of capacitors connected across terminals of induction generator, its no-load voltage found by intersection of Gen. magnetization curve & capacitor’s load line
  • Figure shows how no-load terminal voltage vary with size of capacitors 
induction generator8
  • When an induction Gen. first start to turn,residual magnetism in its field cct. produces a small voltage
  • The small voltage produces a capacitive current flow, which increases the voltage
  • Further increasing the capacitive current, & so forth until voltage fully built up
  • If no residual flux exist in rotor, then its voltage will not build up & it must be magnetized by momentarily running it as motor
induction generator9
  • Most serious problem with an induction generator is that its voltage varies wildly with changes in load (specially reactive load)
  • Typical terminal characteristics of an induction generator operating alone with a constant parallel capacitance shown in Figure

Note:in case of inductive loading

voltage collapses very rapidly

induction generator10
  • This happens (the collapse of voltage)because the fixed capacitors should supply both generator & load
  • & any reactive power diverted to load moves generator back along its magnetization curve, causing a major drop in generator voltage
  • Therefore it is very difficult to start an induction motor on a power system supplied by an induction generator(special techniques required to increase effective capacitance during starting & then decrease during normal operation)
induction generator11
  • Due to nature of induction machine torque-speed characteristic, an induction generator frequency varies with changing load;however since torque-speed characteristic is very steep in normal operating range, total frequency variation is limited to 5%
  • This amount of variation is quite acceptable in many isolated or emergency generator application
induction generator12
  • Applications:
  • Used since early twentieth century, however by 1960s & 1970s they had largely disappeared
  • Induction Gen. has made a comeback since oil price shocks of 1973.
  • With energy costs so high, energy recovery became an important part of economics of most industrial processes
  • They require very little in way of control systems & applications
  • Due to simplicity & small size per kW output, these generators are favored (for windmills)
induction motor ratings
  • Nameplate:

1- output power


3-Current (current limit is based on max. acceptable heating in motor’s windings, & power limit set by combination of voltage , current ratings with P.F. & efficiency)

4-Power factor


6-Nominal efficiency

7- NEMA design class

8-Starting code

appendix three phase power measurement
Appendix: three phase power Measurement
  • 3 phase power measurement
  • Active power measurement in a 3 phase system with neutral (four wire circuit):

three wattmeter, on in each phase while voltage element connected between each phase & neutral

appendix three phase power measurement1
Appendix: three phase power Measurement
  • Active power measurement in a 3 phase system without access to neutral (three wire circuit)
  • P=IV cos (30-θ)+IV cos(30+θ)=√3VIcosθ