EEM EEM 355 –Mechatronic SystemMechatronic System

AC DRIVES 2BY:

DR. ROSMIWATI MOHD MOKHTAR

Example 1

A 15 hp, 460 V, 60 Hz, 3-phase squirrel cage5 p, 4 , , 3 p q ginduction motor is having a torque speed curve givenas below.

By neglecting windage and friction losses, calculate the y g g g ,power supplied to the rotor when the machine runs

a) As a motor at 1650 rpmb) As a brake at 750 rpmc) As a generator at 2550 rpm

Motor: nm = +1650 rpm; m = +100 Nm

55.9

m

mm

nn

P

083.01800

16501800

s

ms

nnns

91001650

55.9

mmm

nP

kW851828.17

1

PP

PsPm

rm

kW 28.1755.9

kW85.18

083.011 sPr

Brake: nm = -750 rpm; m = +60 Nm

55.9

m

mm

nn

P

417.11800

7501800

s

ms

nnns

960750

55.9

mmm

nP

kW31171.4

1

PP

PsPm

rm

kW 71.455.9

kW3.11

417.111

sPr

Generator: nm = +2550 rpm; m = ??

Example 2

A 3-phase, 4-pole, 1800 rpm squirrel cage induction motori d 8 V 6 H W his rated at 208 V, 60 Hz. We want the motor to turn at a no-load speed of about 225 rpm while maintaining the sameflux in the air gap. Calculate the required voltage andf b li d hfrequency to be applied to the stator.

To run at 225 rpm, the voltage and frequency must beTo run at 225 rpm, the voltage and frequency must bereduced in the ratio of 225/1800 = 0.125E

V l i d 8 6 V Voltage required = 0.125 x 208 = 26 V

and frequency = 0.125 x 60 = 7.5 Hzand frequency 0.125 x 60 7.5 Hz

Kinetic Energy

Recall that the dynamic braking will stop motor bydissipating its stored kinetic energy into resistive load.

The kinetic energy is a form of mechanical energy is givenby the equationby the equation

2322 1048.521

21 JnmvJKE

22J = moment of inertia [kg.m2]= rotational speed [rad/s]m mass of the body [kg]m = mass of the body [kg]v = translational speed [m/s]n = rotational speed [rpm]5.48 x 10-3 = constant to take care of units [exact value = 2/1800]

Torque, Inertia and Change in Speed

The rate of change of speed depends upon theg p p pinertia, as well as on the torque. The equation thatrelates these factors is

Jtn

55.9

Jn = change in speed [rpm]= Torque [Nm]t = Interval of time during which the

torque is applied [s]torque is applied [s]J = moment of inertia [kg.m2]9.55 = constant to take care of units

[exact value = 30/]

Example 3

The flywheel having total moment of inertia 10.6y gkg.m2 turns at 60 rpm. We wish to increase its speedto 600 rpm by applying a torque of 20 Nm. For howl t th t b li d?long must the torque be applied?

55.9

tn

559

nJtJ

n

302055.9

6.106060055.9

s30

Electronic AC Drives

Static frequency changersConvert the incoming line frequency directly into the desired loado Convert the incoming line frequency directly into the desired loadfrequency.

o Cycloconverters fall into this category.o Used to drive synchronous and squirrel-cage induction motors.o Used to drive synchronous and squirrel cage induction motors.o Motor can be started, stopped, reversed ad decelerated by

regenerative braking.o The stator voltage is automatically adjusted in relation to the

frequency to maintain a constant flux in the machine.o To obtain regenerative braking,

frequency produced bycycloconverter must be slightlyless than the frequencycorresponding to the speed of themotor.

Variable speed drive using cycloconverter

Squirrel-cage induction motor fed from a 3-phase cycloconverter

Static voltage controllersgo Enable speed and torque control by varying the ac voltage.o Used with squirrel cage induction motors.

S i l ll l d f i d io Static voltage controllers are also used to soft-start inductionmotors.

o Particularly useful for a motor driving a blower or centrifugaly g gpump.

Variable speed drive using static switch

Variable-speed blower motor

Variable-voltage speed control of a squirrel-cage induction motor using back-to-back thyristors

Rectifier-inverter systems with line commutation- Rectify the incoming line frequency to dc, and the dc is reconverted to

ac by an inverter.- The inverter is line commutated by the motor it drives.y- Used to control speed of wound-rotor induction motors.- Adding a chopper to the rectifier circuit also enable for control of speed

of wound-rotor induction motorof wound rotor induction motor.

Rectifier-inverter systems with self commutation- Rectify the incoming line frequency to dc, and the dc is reconverted to

ac by an inverter.- The inverter is self commutated, generating its own frequency., g g q y- Used to control speed of squirrel cage induction motors.

Current-fed frequency converter

Voltage-fed frequency converter

Pulse-width modulation systemsy- Relatively new in industry.- Enable variable speed induction drives ranging from p g g

zero speed and up.

Example 4

A 3-phase, squirrel-cage induction motor has a fullload rating of 25 hp, 480 V, 1760 rpm, 60 Hz. The 3independent windings each carry a rated current of20 A. The cycloconverter is connected to a 3-phase,20 A. The cycloconverter is connected to a 3 phase,60 Hz line and generates a frequency of 8 Hz.Calculate the approximate value of the following:

) h ff i l h i dia) The effective voltage across each windingb) The no-load speed) Th d t t d tc) The speed at rated torque

d) The effective current in the windings at ratedtorquetorque

Answer 4

a) The flux in the motor should remain the same at all frequencies.C l f f f 8 H h l h i diConsequently, for a frequency of 8 Hz, the voltage across the windingsmust be reduced in proportion. Thus, the effective voltage across eachwinding is

V 64480608

E

b) The full load speed at 60 Hz = 1760 rpm. Consequently, this is a 4-pole motor. Thus, its synchronous speed is

60120120 f

The no load speed at 8 Hz is

rpm 18004

60120120

pfns

The no-load speed at 8 Hz isrpm 2401800

608

n

c) When the motor is operating at 60 Hz, the slip speed at rated torque iis

4017601800

msslip nnn

Consequently, the slip is again 40 rpm when the motor developsrated torque at 8 Hz. Therefore, the speed at rated torque

rpm40

q , p q

d) Because the flux in the motor is the same at 8 Hz and 60 Hz it

rpm 20040240 n

d) Because the flux in the motor is the same at 8 Hz and 60 Hz, itfollows that rated torque will be developed when the current in thestator windings reaches its rated value, namely 20 A. Effective current in the windings at rated torque = 20 A Effective current in the windings at rated torque = 20 A.

Example 5

A Y-connected squirrel-cage induction motor has thefollowing ratings and parameters:

400 V, 50 Hz, 4-pole, 1370 rpm, Rs = 2 , Rr’=3 , Xs = Xr’ = 3.5

Motor is controlled by a voltage source inverter at constantV/f ratio. Inverter allows frequency variation from 10 to 50/ q y 5Hz. Determine,a) Speed for the frequency of 30 Hz and 80% of full load

ttorqueb) Frequency for a speed of 1000 rpm and full load torque

a) The speed at 50 Hz = 1370 rpm. Consequently, this is a 4-pole motor.Thus its synchronous speed at 50 Hz isThus, its synchronous speed at 50 Hz is

rpm 15004

50120120

pfns

At 50 Hz, drop in speed from no-load to full load torque,= 1500 – 1370 = 130 rpm.

p

5 37 3 p

Drop in speed from no-load to 80% of full load torque,= 0.8 x 130 = 104 rpm.

Synchronous speed at 30 Hz = 30/50 x 1500 = 900 rpm.

Motor speed for frequency 30 Hz and having 80% of full load torque =900-104 = 796 rpm

b) Frequency for a speed of 1000 rpm and full load torqueb) Frequency for a speed of 1000 rpm and full load torque

Drop in speed from no load to full load torque = 130 rpm.

Synchronous speed = 1000 + 130 = 1130 rpm.

Frequency,

120 f

Hz673711304

120

s

s

pnf

pfn

Hz 67.37120120

f

Let say the regenerative braking operation isy g g pemployed, determine;

a) Speed for the frequency of 30 Hz and 80% of fullload torque

b) Frequency for a speed of 1000 rpm and full loadtorque

c) What will be the answer to (a) to (b), if the drivek d d b kworks under dynamic braking?

Answer 5

a) The speed at 50 Hz = 1370 rpm. Consequently, this is a 4-pole motor.Thus, its synchronous speed at 50 Hz isThus, its synchronous speed at 50 Hz is

rpm 15004

50120120

pfns

Increase in speed from no-load to full load torque,= 1500 – 1370 = 130 rpm.

Increase in speed from no-load to 0.8 of full load torque,= 0.8 x 130 = 104 rpm.

Synchronous speed at 30 Hz = 30/50 x 1500 = 900 rpm.

M t d f f q 30 H d h i 80% f f ll l d t q Motor speed for frequency 30 Hz and having 80% of full load torque =900+104 = 1004 rpm

b) Frequency for a speed of 1000 rpm and full load torque

Synchronous speed = 1000 – 130 = 870 rpm.

Frequency,120

s pfn

Hz 29120

8704120

spnf

c) In both regenerative and dynamic braking, motor works as agenerator. The two braking methods only differ in the way brakingenergy is disposed off (in former it is transferred to the source and inlatter it is dissipated in a resistor) Hence answer will be the samelatter it is dissipated in a resistor). Hence, answer will be the same.