1. Lecture 21Electro Mechanical System1 Assignment 7 Page 303,304 – Problems: 13-15, 13-16, 13-17, 13-20 and 13-23 Due Date: Tuesday 22 nd March, 2011 Quiz No.4 Next Week Quiz 4

2. Lecture 21Electro Mechanical System2 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 = (n s – n)/n s where s = slip n s = 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.

3. Lecture 21Electro Mechanical System3 Example A 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 is n s = 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: n s – n = 1200 – 1140 = 60 r/min The slip is S = (n s – n)/ n s = 60/1200 = 0.05 or 5%

4. Lecture 21Electro Mechanical System4 Voltage & freq. induced in the rotor  Voltage and frequency induced in rotor, both depend upon the slip. Given by the following equations: f 2 = sf E 2 = sE oc (approx.) where f 2 = freq. of the voltage and current in the rotor [Hz] f = freq. of the source connected to the stator [Hz] s = slip E 2 = voltage induced in the rotor at slip s E oc = open-circuit voltage induced in the rotor at rest [V]  In a cage motor, open-circuit voltage E oc 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.

5. Lecture 21Electro Mechanical System5 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 standstill b)Motor turning at 500 r/min in the same direction as the revolving field c)Motor turning at 500 r/min in the opposite direction to the revolving field d)Motor turning at 2000 r/min in the same direc­tion 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 = (n s - n)/n s = (1200 – 0)/1200 = 1 The frequency of the induced voltage and induced current is: f 2 = sf = 1 X 60 = 60 Hz Example

6. Lecture 21Electro Mechanical System6 b)When the motor turns in the same direction as the field, the motor speed n is positive. The slip is: s = (n s – n)/n s = (1200 – 500)/1200 = 700/1200 = 0.583 The frequency of the induced voltage and rotor current is: f 2 = sf= 0.583 X 60 = 35 Hz c)When the motor turns in the opposite direction to the field, the motor speed is negative; thus, n = – 500. The slip is: s = (n s – n)/n s = [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 is f 2 = sf = 1.417 X 60 = 85 Hz Example

7. Lecture 21Electro Mechanical System7 d)The motor speed is positive because the rotor turns in the same direction as the field: n = +2000. The slip is: s = (n s - n)/n s = (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: f 2 = 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

8. Lecture 21Electro Mechanical System8 Characteristics of Squirrel Cage Induction Motor  Let us analyze induction motor: 1.Motor at no load 2.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.

9. Lecture 21Electro Mechanical System9 Characteristics of Squirrel Cage Induction Motor Motor 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.  I 2 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

10. Lecture 21Electro Mechanical System10 Estimating Motor Currents  The full-load current  approximate value based on empirical models I = 500P h /E where: I = full-load current P h = output power in HP E = 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