- 81 Views
- Uploaded on
- Presentation posted in: General

Performance Characteristics

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

- 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

Eqn. X

*eqn. X

At high slip

At low slip

- 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

-Ns

2

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

- 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.

Differentiate eqn. dT/ds, and equate to zero

If R1 small

Increase R2, increase slip max, increase staring torque

If R1 small

Varying R2

- 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

R ext

Stator current vs. speed

Power factor vs. speed

(1-s)Pag

sPag

Efficiency vs. speed

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

Internal efficiency

0 < s < 1

0 < s

s > 0

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 Torque

d) How much external resistance/phase should be connected in the rotor circuit so that maximum torque occurs at start?

l

Sol- pg13

Sen 241

N2

N1

A 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

Since the developed torque Tm = load torque TL

R2 +R2ext

R2

Sol_pg14

TL

Sen 244

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

- 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 kW

Blocked-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 %)

Sol_pg16

Sen 282 (pb 5.6)

- 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)

- Characterized by normalstarting 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.

- 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.

- 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.

- 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.

- 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

Pole changing

Synchronous speed change with changes number of poles (change the stator winding/coil connection)

Discrete step change in speed/ expensive

Given a load T–N characteristic,

the steady-state speed can be changed

by altering the profile of T–N of the motor:

Rotor Resistance control

For wound rotor only

Variable line voltage (amplitude), variable frequency

. Most popular method

. Using power electronics converter

. Operated at low slip frequency

Variable line voltage (amplitude), frequency fixed

Torque V2

E.g. using 3-phase autotransformer (variac) or solid state controller

Slip becomes high as voltage reduced – low efficiency

V= 1pu

V= 0.71pu

Fan (TL) load

V= 0.25pu

Auto Transformer Voltage Control

Solid State Voltage Control

Closed Loop Operation Voltage Control

Open Loop Control Scheme

Closed Loop Control Scheme

n

Control both V and freq, f

IM

Supply

3-phase pwm Inverter

Rectifier and Filter

PWM Inverter

VSI

Rectifier

3-phase supply

IM

C

f

Pulse Width Modulator

Ramp

V

+

s*

- Question 4
- 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 1

SESI 2007/2008 eg 5.4 pg 241

- QUESTION 4
- (a) Explain the working principle of a three-phase induction machine on the basis of magnetic fields.
- Show through a power flow diagram, how electrical power input is converted into mechanical power output in an induction motor.
- (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 VRth = 0.25 W
- Xth = 0.5 WX2 = 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 1

SESI 2008/2009

Sol_pg31