Performance Characteristics The equivalent circuits can be used to predict the

performance characteristics of the induction machine. The important performance characteristics in the steady

state are:- efficiency- power factor- stator current- starting torque- maximum torque (pull-out), etc

21

22

22

2

'''

XXs

RR

VII

thth

thth

Performance Characteristics

Eqn. X

Performance Characteristics

At low slip

At high slip

*eqn. X

th

thth

Rs

R

XXs

RR

2

12

'&

'

• At low slip, torque proportional to slip s•At high slip, torque inverse proportional to slip

Note if approximate circuit is used to get the equation of torque, then

s

R

LLs

RR

VPT

sRIPP

LLs

RR

VI

synsynsyn

oe

ago

syn

2

221

22

21

21

222

221

22

21

12

m

33

torque,)(developed neticElectromag

phase),(per rotor at thePower

drawn becan abovecircuit thelarge, is L Since

T – N Single frequency characteristic

.0speed, syncat ,1 standsillat ;

: thatNote

ssssyn

msyn

emPLUGGING

TORQUE(+)

MOTORING

emT(max torque orpull-out torque)

esT (starting torque)

SLIP,s

TORQUE(-)

e

rated slip0 unity slip

(standstill)

GENERATING

e

m

e

m

zero slip(sync.speed)

SPEED-Ns

2

• Select one frequency ()•Select V1• Varies s ( at particular s, get T)•Repeat for other s to get T

T – N Single frequency characteristic

CURRENT TORQUE

operating point(rated torque)

rated slip

rated current

POWERFACTOR

SLIP

1.00

Standstill synchronous speed

EFFICIENCY

T – N Single frequency characteristic

As slip is increased from zero (synchronous), the torque rapidly reaches the maximum. Then it decreases to standstill when the slip is unity.

At synchronous speed, torque is almost zero. At standstill, torque is not too high, but the current is

very high. Thus the VA requirement of the IM is several times than the full load. Not economic to operate at this condition.

Only at “low slip”, the motor current is low and efficiency and power factor are high.

Maximum TorqueDifferentiate eqn.

dT/ds, and equate to zero

Varying R2

Increase R2,

increase slip max, increase staring torque

If R1 small

If R1 small

Maximum Torque

21

22

22 )'(' XXRSR

ththm

• Maximum air gap power transfer occurs at impedance matching principle – Another approaches

• Rext to be added to produce Tmax at starting, ie at s = 1 is

21

22

22 )'(' XXRS

RRthth

m

ext

R ext

Current and Power Factor

Stator current vs. speed

Power factor vs. speed

Efficiency

Efficiency vs. speed

sPag

(1-s)Pag

Internal efficiency *To get Max. efficiency, s must be very low

Power Flow

0 < s < 1

0 < s

s > 0

Example 4*A three-phase 460 V, 1740 rpm, 60 Hz 4-pole wound rotor induction star

connected motor has the following parameter/phase:R1 = 0.25 , R2’ = 0.2 , X1 = X2’= 0.5 , Xm = 30 The rotational losses are 1700 W. With the rotor terminal short circuited, find:

a) i) Starting current when started on full load ii) Starting torque

b) i) Full load slip ii) Full-load current iii) Full-load power factor iv) Ratio of starting current to full load current v) Full-load torque vi) Internal efficiency and motor efficiency at full load

c) i) Slip at maximum torque ii) Maximum Torqued) How much external resistance/phase should be connected in the rotor

circuit so that maximum torque occurs at start?l

Sen 241Sol- pg13

Example 5A three-phase 460 V, 60 Hz 6 -pole wound rotor induction motor

drives a constant load of 100 N-m at speed of 1140 rpm when the rotor terminal is short-circuited. It requires to reduce speed to 1000 rpm by inserting resistance in rotor circuit.

Determine the value of resistance if the rotor winding resistance / phase is 0.2 ohms. Neglect rotational losses. The stator to rotor turn ratio is unity.

l

TL

N2 N1

Since the developed

torque Tm = load torque

TL

2

2

1

2

SRR

SR ext

Sen 244

Sol_pg14R2

R2 +R2ext

15

By changing the impedance (R) connected to the rotor circuit, the speed/current and speed/torque curves can be altered.

Used primarily to start a high inertia load or a load that requires a very high starting torque across the full speed range with relatively low current from zero speed to full speed

Wound Rotor

Slip ring

Example 6 The following test results are obtained from a three-phase, 100hp,

460 V, eight-pole, star connected squirrel-cage induction machine.No load test: 460 V, 60 Hz, 40 A, 4.2 kWBlocked-rotor test: 100 V, 60 Hz, 140 A, 8 kW

Average dc resistance between two stator terminals is 0.152 Ω.

(a) Determine the parameters of the equivalent circuit. (0.076 Ω, 0.195 Ω, 6.386 Ω, 0.195 Ω, 0.062 Ω).

(b) The motor is connected to a three-phase , 460 V, 60 Hz supply and runs at 873 rpm. Determine the input current, input power, air gap power, rotor copper loss, mechanical power developed, output power, and efficiency of the motor.( 127.9/-27o A, 90.82 kW, 87.09 kW, 2.613 kW, 84.48 kW, 80.64 kW, 88.79 %)

Sen 282 (pb 5.6)Sol_pg16

Classes of Squirrel-Cage Motor To meet the various starting and running requirements of

a variety of industrial applications, several standard ( T vs. N) designs of squirrel-cage motors are available from manufacturer’s stock.

The most significant design variable in these motors is the effective resistance of the rotor cage circuit ( for wound rotor)

Class A Motors Characterized by normal starting torque, high starting

current and low operating slip. Low rotor circuit resistance and therefore operate

efficiently with a low slip (0.005<s<0.015) at full load. Suitable for applications where the load torque is low at

start (such as fan or pump) so that full speed is achieved rapidly, thereby eliminating the problem of overheating during starting.

In larger machines, low voltage starting is required to limit the starting current.

Class B Motors Characterized by normal starting torque, low starting

current and low operating slip. The starting current is about 75 % of that for class A. The starting current is reduced by designing for relatively

high leakage reactance by using either deep-bar rotors or double- cage rotors.

The full load slip and efficiency are as good as those of class A motors.

Good general-purpose motors and have a wide variety of industrial applications. Suitable for constant speed drives.

Examples are drives for fans, pumps, blowers, and motor-generator sets.

Class C Motors Characterized by high starting torque and low starting

current. A double-cage rotor is used with higher rotor resistance

than is found in class B motors. The full-load slip is somewhat higher and the efficiency

lower than for class A and class B motors. Class C motors are suitable for driving compressors,

conveyors, crushers, and so forth.

Class D Motors Characterized by high starting torque, low starting current and

high operating slip. The torque-speed characteristic is similar to that of a wound-rotor

motor with some external resistance connected to the rotor circuit. The full-load operating slip is high (8 to 15 %), and therefore the

running efficiency is slow. The high losses in the rotor circuit require that the machine be large

(and hence expensive) for a given power. Suitable for driving intermittent loads requiring rapid acceleration

and high impact loads. In the case of impact loads, a flywheel is fitted to the system which

delivers some of its kinetic energy during the impact.

Speed Control Pole Changing Line Voltage Control Line Frequency Control Constant-slip Frequency Operation Closed-loop Control Constant-Flux Operation Constant-current Operation Rotor Resistance Control Rotor Slip Energy Recovery

Speed control of induction machineGiven a load T–N characteristic, the steady-state speed can be changed by altering the profile of T–N of the motor:

Pole changing Synchronous speed change with changes number of poles (change the stator winding/coil connection)Discrete step change in speed/ expensive

Variable line voltage (amplitude), variable frequency

. Most popular method

. Using power electronics converter

. Operated at low slip frequency

Variable line voltage (amplitude), frequency fixedTorque V2

E.g. using 3-phase autotransformer (variac) or solid state controllerSlip becomes high as voltage reduced – low efficiency

Rotor Resistance control For wound rotor only

Variable line voltage, fixed frequency

0 20 40 60 80 100 120 140 1600

100

200

300

400

500

600To

rque

w (rad/s)

Fan (TL) load

V= 1pu

V= 0.71pu

V= 0.25pu

Auto Transformer Voltage Control Solid State Voltage Control

Closed Loop Operation Voltage Control

Rotor resistance Control

Open Loop Control Scheme Closed Loop Control Scheme

Typical IM Drive System - Variable voltage, variable frequency

IDC

Modulation Index,

+

VDC

+

IM

n

IM

Supply Rectifier and Filter 3-phase pwm Inverter

Control both V and freq, f

PWM Inverter

Variable voltage, variable frequencyConstant V/f

0 20 40 60 80 100 120 140 1600

100

200

300

400

500

600

700

800

900

Torq

ue

50Hz

30Hz

10Hz mfV

fE

VVVF, Constant V/f – open-loop

VSIRectifier3-phase supply

IM

Pulse Width

Modulators

*

+

Rampf

C

V

Example Final

Question 4(a)Explain briefly three methods for controlling the speed of an

induction motor. (6 marks)

(b) Draw a typical torque-speed characteristic of an induction motor and label key quantities. (3 marks)

(c) A three-phase, 415 V, 1450 rpm, 50 Hz, four-pole wound-rotor induction motor has the following parameters per phase:

R1 = 0.25 , R2’ = 0.2 X1 = X2’ = 0.5 , Xm = 30

The rotational losses are 1700 W. With the rotor terminals short-circuited, determine:

(i) Starting current when started direct on full voltage. (4 marks)(ii) Starting torque. (4 marks)

(iii) Full-load current. (4 marks)(iv) Full-load torque. (4 marks)

SEMESTER 1SESI 2007/2008

eg 5.4 pg 241

ExampleQUESTION 4  

 (a) Explain the working principle of a three-phase induction machine on the basis of

magnetic fields. 

(b)Show through a power flow diagram, how electrical power input is converted into mechanical power output in an induction motor.

(c)(c) A 3 phase , 415 V, 1450 rpm, 50 Hz, four-pole wound rotor induction motor has the following Thevenin’s equivalent circuit parameters per phase:

Vth = 236 V Rth = 0.25 WXth = 0.5 W X2 = 0.5 W R2’ = 0.2 W

The motor drives a constant load of 100 Nm at rated speed when the rotor terminals are short-circuited. Neglect rotational losses. 

 (i) Draw the Thevenin’s equivalent circuit of induction machine.    

(ii) How much external resistance per phase should be connected in the rotor circuit so that maximum torque occurs at start-up?

   (iii) It is required to reduce the speed of the motor to 1400 rpm by inserting resistance in the rotor circuit. How much external resistance per phase should be connected in the rotor circuit?

   (iv)Draw torque-speed characteristics of the motor and load to show the conditions in (iii) with and without external rotor resistance.

  

SEMESTER 1SESI 2008/2009 Sol_pg31