Lecture 21 Electro Mechanical System 1

Assignment 7

Page 303,304 – Problems: 13-15, 13-16, 13-17, 13-20 and

13-23

Due Date: Tuesday 22nd March, 2011

Quiz No.4 Next Week

Quiz 4

Lecture 21 Electro Mechanical System 2

Slip At no-load the percent difference in speed between the rotor and

field is called slip and usually less than 0.1% of synchronous speed.

The slip s of an induction motor is the difference between the synchronous speed and the rotor speed, expressed as a percent (or per-unit) of synchronous speed.

The per-unit slip is given by the equation:

s = (ns – n)/ns

wheres = slipns = synchronous speed [r/min] n = rotor speed [r/min]

The slip is practically zero at no-load and is equal to 1 (or 100%) when the rotor is locked.

Lecture 21 Electro Mechanical System 3

ExampleA 6-pole induction motor is excited by a 3-phase, 60 Hz source. If the full-load speed is 1140 r/min, calculate the slip.

The synchronous speed of the motor isns = 120f/p = 120 X 60/6 = 1200 r/min

The difference between the synchronous speed of the revolving flux and rotor speed is the slip speed:

ns – n = 1200 – 1140 = 60 r/minThe slip isS = (ns – n)/ ns = 60/1200 = 0.05 or 5%

Lecture 21 Electro Mechanical System 4

Voltage & freq. induced in the rotor Voltage and frequency induced in rotor, both depend

upon the slip. Given by the following equations:f2 = sfE2 = sEoc (approx.)

wheref2 = freq. of the voltage and current in the rotor [Hz]f = freq. of the source connected to the stator [Hz]s = slip E2 = voltage induced in the rotor at slip sEoc = open-circuit voltage induced in the rotor at rest [V]

In a cage motor, open-circuit voltage Eoc is the voltage that would be induced in the rotor bars if the bars were disconnected from the end-rings.

In a wound-rotor motor open-circuit voltage is 1/3 times the voltage between the open-circuit slip-rings.

Lecture 21 Electro Mechanical System 5

The 6-pole wound-rotor induction motor of previous example is excited by a 3-phase 60 Hz source. Calculate the frequency of the rotor current under the following conditions:a)at standstillb)Motor turning at 500 r/min in the same direction as the revolving fieldc)Motor turning at 500 r/min in the opposite direction to the revolving fieldd)Motor turning at 2000 r/min in the same direc tion as the revolving field

We calculated synchronous speed of the motor as 1200 r/min.

a) At standstill the motor speed n = 0. The slip is:s = (ns- n)/ns = (1200 – 0)/1200 = 1

The frequency of the induced voltage and induced current is:f2 = sf = 1 X 60 = 60 Hz

Example

Lecture 21 Electro Mechanical System 6

b) When the motor turns in the same direction as the field, the motor speed n is positive. The slip is:

s = (ns – n)/ns = (1200 – 500)/1200 = 700/1200 = 0.583 The frequency of the induced voltage and rotor current is:

f2 = sf= 0.583 X 60 = 35 Hzc) When the motor turns in the opposite direction to the field, the motor speed is negative; thus,

n = – 500. The slip is:s = (ns – n)/ns = [1200 – (– 500)]/1200

= (1200 + 500)/1200 = 1700/1200 = 1.417 A slip greater than 1 implies that the motor is operating as a brake.

Frequency of the induced voltage and rotor current isf2 = sf = 1.417 X 60 = 85 Hz

Example

Lecture 21 Electro Mechanical System 7

d) The motor speed is positive because the rotor turns in the same direction as the field:

n = +2000. The slip is:s = (ns - n)/ns = (1200–2000)/1200 = –800/1200 = –0.667

 A negative slip implies that the motor is actually operating as a

generator. The frequency of the induced voltage and rotor current is:

f2 = sf = –0.667 X 60 = –40 Hz

A negative frequency means that the phase sequence of voltages induced in the rotor windings is reversed. If the phase sequence of the rotor voltages is A-B-C when the frequency is positive, the phase sequence is A-C-B when the frequency is negative.

Example

Lecture 21 Electro Mechanical System 8

Characteristics of Squirrel Cage Induction Motor Let us analyze induction motor:

1. Motor at no load2. Motor under load.3. Locked Rotor Characteristics.

Motor at no load No load current is like excitation

current in the transformer. It is composed of magnetizing

component that creates revolving flux m & is like mutual flux of transformer

A small active power is used in the windage & friction losses in the rotor and iron/ copper losses in the stator.

Considerable reactive power is needed to create the revolving flux.

To keep reactive power within acceptable limits, the air gap is made as small as mechanical tolerance will permit.

Due to reactive power the power factor at no load will be very low. It will be 0.2 for small and 0.05 for large machines.

Lecture 21 Electro Mechanical System 9

Characteristics of Squirrel Cage Induction MotorMotor under load. When motor is under load, the current in the

rotor produces mmf which tends to change the mutual flux m.

This sets up opposing flux in stator & rotor Leakage flux f1 and f2 are created. Opposing mmfs of the rotor and stator are

similar to leakage flux in transformer. Total reactive power needed to produce these fluxes will be slightly

greater than at no load. The active power(kW) absorbed increased in almost direct proportion to

mechanical load. The power factor of motor improves dramatically as the mechanical load

increases. At full load it is 0.8 to 0.9.

Locked Rotor Characteristics. Locked rotor current is 5 to 6 times of the full load current. I2 R losses are 25 to 36 times higher than normal. The rotor should never remain locked for more than few seconds. Mechanical power at stand still is zero but develops a very strong torque. Power factor is low due to reactive power needed to produce leakage flux

Lecture 21 Electro Mechanical System 10

Estimating Motor Currents The full-load current

approximate value based on empirical models

I = 500Ph/E

where:I = full-load currentPh = output power in HPE = rated line voltage

The starting current 5 to 6 times the full-load current

The no-load current 0.3 to 0.5 times the full-load current

Home Work : Page 277, Example 13.4