Introduction Ac Motor3

7/22/2019 Introduction Ac Motor3

1/43

ET602AC MOTOR

CONTROL

INTRODUCTION TO AC MOTOR


2/43

At the end of this topic, you

should be able to .. Study the transfer of power in an AC

motor

Draw diagram of the power transmission ofan induction motor

Explain the power of the transmission of an

induction motor Construct equivalent circuit of an induction

motor using calculation and related formula


3/43

At the end of this topic, you

should be able to .. Know the methods of AC motor

control

Compare the characteristics of AC

motor control by varying

Voltage

Frequency

Number of poles

Rotors resistance


4/43

Study the transfer of power inan AC motor


5/43

Pin

P2

coreCopper

PC

POUT

Mechanical

Losses


6/43

Draw diagram of the

power transmission of

an induction motor


7/43

Active Power FlowInduction motor converts an electrical power supplies to it into

mechanical power. The various stages in this conversion is called

power flow in an inductor motor.

The three phase supply given to the stator is the net electrical

input to the motor. If motor power factor is cos and VL, ILare

line values of supply voltage and current drawn, then net

electrical supplied to the motor can be calculated as,


8/43

Active Power Flow

The part of this power is utilized to supply the losses in the

stator which are stator core as well as copper losses.

The remaining power is delivered to the rotor magneticallythrough the air gap with the help of rotating magnetic field.

This is called rotor input denoted as P2.


9/43

Active Power Flow

The rotor is not able to convert its entire input to the

mechanical as it has to supply rotor losses. The rotor

losses are dominantly copper losses as rotor iron losses

are very small and hence generally neglected. So rotor

losses are rotor copper losses denoted as Pc.

where I2r = Rotor current per phase in running

condition

R2

= Rotor resistance per phase.

Or can be calculated by Pc = s P2 where s = slip


10/43

Active Power Flow After supplying these losses, the remaining part of P2 is converted into

mechanical which is called gross mechanical power developed by the motor

denoted as Pm.

Now this power, motor tries to deliver to the load connected to the shaft. But

during this mechanical transmission, part of Pmis utilised to provide mechanical

losses like friction and windage.

And finally the power is available to the load at the shaft. This is called netoutput of the motor denoted as Pout. This is also called shaft power. The rating of

the motor is specified in terms of value of Pout when load condition is full load

condition.

Or can be calculated by

Pm = (1-s)/P2

usually mechanical losses is

neglected


11/43

Active Power Flow

From the power flow diagram we can

define,

= Pm

/ P2


12/43

Active Power Flow

Motor Torque:

Tm = 9.55 Pmn


13/43

Construct equivalent circuit of an induction motor using

calculation and related formula


14/43

Construct equivalent circuit of an induction motor using calculation

and related formula


15/43

Know the methods of ACmotor control


16/43

FREQUENCY

Compare the characteristics of AC motor control by varying


17/43

Varying Frequency

The best method since supply voltage and

supply frequency is varied to keep V/f

constant

Maintain speed regulation

uses power electronics circuit for frequencyand voltage controller

Constant maximum torque


18/43

Varying Frequency

The synchronous speed of an induction

motor is given by Ns = 120f/p

The synchronous speed and, therefore, thespeed of motor can be controlled by varying

the supply frequency.

The emf induced in the stator of aninduction motor is given by


19/43

Varying Frequency

Therefore, if the supply frequency is

change, E1will also change to maintain the

same air gap flux. If the stator voltage dropis neglected the terminal voltage V1 is equal

to E1in order to avoid saturation and to

minimize losses, motor is operated at ratedair gap flux by varying terminal voltage

with frequency so as to maintain (V/f) ratio

constant at rated value.


20/43

Varying Frequency

This type of control is known as constant

volt in per hertz. Thus, the speed control of

an induction motor using variable frequencysupply requires a variable voltage power

source.


21/43

Varying Frequency

T

nNL1

T

nr1nr2nr3 n

fdecreasing

nNL2nNL3


22/43

ROTOR RESISTANCE



23/43

Varying Rotor Resistance

For wound rotor only

Speed is decreasing

Constant maximum torque

The speed at which max torque occurs

changes

Disadvantages:

large speed regulation

Power loss in Rextreduce the efficiency


24/43


In wound rotor induction motor, it is possible to

change the shape of the torquespeed curve by

inserting extra resistance into rotor circuit of themachine. The resulting torquespeed

characteristic curves are shown in fig.(4).

This method of speed control is very simple. It is

possible to have a large starting torque and lowstarting current at small value of slip.


25/43


The major disadvantage of this method is that the

efficiency is low due to additional losses in

resistors connected in the rotor circuit. Because ofconvenience and simplicity, it is often employed

when speed is to be reduced for a short period

only (cranes).


26/43


27/43

NO OF POLES



28/43

No of Poles

The number of stator poles can be change

by

Multiple stator windings

Method of consequent poles

Pole amplitude modulation (PWM)


29/43

No of Poles

The methods of speed control by pole

changing are suitable for cage motors only

because the cage rotor automaticallydevelops number of poles equal to the poles

of stator winding.


30/43

No of Poles: Multiple stator windings

In this method the stator is providedwith two separate windings which are

wound for two different pole numbers.

One winding is energized at a time.Suppose that a motor has two windings

for 6 and 4 poles.


31/43

No of Poles: Multiple stator windings

For 50Hz supply the synchronous speedwill be 1000 and 1500rpm respectively. If

the full load slip is 5% in each case, the

operating speeds will be 950 rpm and 1425

rpm respectively. This method is less

efficient and more costly, and therefore,

used only when absolutely necessary.


32/43

No of Poles: Method of consequent

poles In this method a single stator winding is

divided into few coil groups. The terminals

of all these groups are brought out. The number of poles can be changed with

only simple changes in coil connections. In

practice, the stator winding is divided onlyin two coil groups. The number of poles can

be changed in the ratio of 2:1.


33/43

No of Poles: Method of consequent

poles Fig.(1) shows one phase of a stator winding

consisting of 4 coils divided into two groups

ab and cd. Group ab consists of oddnumbered coils(1,3) and connected in

series.

Group cd has even numbered coils (2, 4)connected in series. The terminals a,b,c,d

are taken out as shown.


34/43

No of Poles: Method of

consequent poles


35/43


consequent poles The coils can be made to carry current in

given directions by connecting coil groups

either in series or parallel shown in fig. (1-b) and fig.(1-c) respectively.


36/43


consequent poles With this connection, there will be a total of

4 poles giving a synchronous speed of 1500

rpm for 50 Hz system. If the current throughthe coils of group ab is reversed (fig.2),

then all coils will produce north (N) poles.


37/43


consequent poles In order to complete the magnetic path, the

flux of the pole groups must pass through

the spaces between the groups, thusinducing magnetic poles of opposite

polarity (S poles) in the interpole spaces.


38/43


39/43

VOLTAGE



40/43


41/43

Varying Voltage

Further, we also note that the starting torque

is also lower at lower voltages. Thus, even

if a given voltage level is sufficient forachieving the running torque, the machine

may not start.

This method of trying to control the speed isbest suited for loads that require very little

starting torque, but their torque requirement

may increase with speed.


42/43


43/43

THANK YOU

Date post:	09-Feb-2018
Category:	Documents
Upload:	cyril-joseph-abi
View:	214 times
Download:	0 times

Introduction Ac Motor3

Documents