Ac Drives 01.ppt

8/16/2019 Ac Drives 01.ppt

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The stator winding are supplied with balanced three-phase AC voltage,

which produce induced voltage in the rotor windings. It is possible toarrange the distribution of stator winding so that there is an effect of

multiple poles, producing several cycle of magnetomotive force (mmf) or

field around the air gap.

The speed of rotation of field is called the synchronous speed ωs , which

is defined by

p s

ω ω

2=

!s is syncronous speed "rad#sec$

%s is syncronous speed "rpm$

p is numbers of poles

! is the supply fre&uency "rad#sec$

f is the supply fre&uency "'$

%m is motor speed p

f

N s120

=

or


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The rotor speed or motor speed is : )1( S sm −=ω ω

here * is slip, as defined as S

mS S ω

ω ω −= +r

S

mS

N

N N S

−=

The motor speed


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Equivalent Circuit Of Induction Motor Equivalent Circuit Of Induction Motor

here

s is resistance per-phase of stator winding

r is resistance per-phase of rotor winding

s is leaage reactance per-phase of the

winding stator

s is leaage reactance per-phase of the

winding rotor

m is magnetiing reactance

m is Core losses as a reactance


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Performance Characteristic of

Induction Motor

Stator copper loss : s scu s R I P 23=

'2')(3 r r cur R I P =Rotor copper loss :

m

s

m

mc

R

V

R

V

P

22

33 ≈=Core losses :


5/16

S

R I P r r g

'

2')(3=

)1()(3'

2' S S

R I P P P r r cur g d −=−=

)1( S P P g d −=

- Power developed on air ap !Power fropm stator to

rotor throuh air ap" :


Induction Motor

- Power developed #y motor :

or

- Torque of motor :

m

d d

P T ω

=

s

g

S

g P

S

S P

ω ω =

−−

=)1(

)1(or

m

d d

N P T π 2

60

=or


6/16

m s si I V P φ cos3=

g cu sc P P P ++=Input power of motor :


Induction Motor

load nod o P P P −=

g cu sc

load nod

i

o

P P P

P P

P

P

++−

==η

Output power of motor :

Efficiency :


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)( cu sc g P P P +>>

load nod P P >>

S P

S P

P

P

g

g

g

d

−=

−

=≈ 1)1(

η

If

and

so, the efficiency can calculated as


Induction Motor


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)(

)()(

''

''

r smr

s

r smr sm

i

X X X jS

R R

S

R R jX X X X

Z

++++

+++−=

Total Impedance of this circuit is


Induction Motor

m

The rotor current is

( ) 2

1

2'

2'

'

++

+

=

r sr

s

sr

X X S

R R

V I


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( )

++

+

=2'

2'

2'3

r sr

s s

sr d

X X S

R RS

V RT

ω

Tor&ue 2 speed Characteristic


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Three reion operation :

$% Motorin :

&% Reeneratin :

'% Pluin :

10 ≤≤ S

0


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*tarting speed of motor is ωm 3 4 or * 3 5,


Induction Motor

*tarting tor&ue of motor is

( )

++

+

=2

'

2'

2'

3

r sr

s s

sr st

X X S

R R

V RT

ω

*lip for the ma6imum tor&ue *ma6 can be found by setting 0=dS T d d

*o, the slip on ma6imum tor&ue is

( ) ( )

[ ]2

12'2

'

max

r s s

r

X X R

RS

++±=


13/16

( )

+++=

2'2

2

max

2

3

r s s s s

s

X X R R

V T

ω


Induction Motor

Tor&ue ma6imum is

And the ma6imum regenerative tor&ue can be found as

( )

+++−= 2'2

2

max2

3

r s s s s

s

X X R R

V

T ω

here the slip of motor s 3 - *m


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( )

++

+

=2

'

2'

2'3

r s

r

s s

sr d

X X S

R RS

V RT

ω Speed(Torque Characteristic :

( )2

'2'

+>>+

S

R R X X r sr s

( )2'2'3

r s s

sr d

X X S

V RT

+=

ω

( )2'2'

3

r s s

sr st

X X

V RT

+=ω

)or the hih Slip S% !startin"

So* the torque of motor is :

+nd startin torque !slip S,$" is :


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( ) s

r

r s

RS

R X X >>


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Stator -oltae Control

Controlling Induction 8otor *peed by

Ad9usting The *tator :oltage

( )

++

+

=2

'

2'

2'3

r sr

s s

sr d

X X S R RS

V RT

ω

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