Transcript Stator Voltage Control

Stator Voltage Control
AC
Variable
Voltage
Sources
Controlling Induction Motor Speed by
Adjusting The Stator Voltage
Td 
IM

Td
Vs
3 Rr' Vs2

Rr'
S s  Rs 
S

2

  X s  X r'



2



Td
Vs > Vs1 > Vs2
Ii
Xs
Xr’
Rs
Tmax
Is=Ir’
Ir’
Im
Rr’/s
Vs
Po
Pi
Stator
air
gap
Tst
Tst1
Tst2
TL
rotor
S=1
Nm =0
2 1 
S=0
s
Ns
Frequency Voltage Control
Controlling Induction Motor Speed by
Adjusting The Frequency Stator Voltage
AC
Variable
Voltage
Sources
IM
Vs

f
Td 
Td
3 Rr' Vs2

Rr'
S s  Rs 
S

2


' 2
  X s  X r  


Td
fs2 < fs1 < fs
Tmax
Ii
Xs
Xr ’
Rs
Ir’
Im
Vs
Tst2
Is=Ir’
Rr’/s
f
TL
Po
Pi
Stator
Tst1
Tst
Air
gap
rotor
S=1
 m =0
2
S=0
fs2
1
S=0
fs1
s
S=0
fs
If the frequency is increased above its rated value, the flux and torque
would decrease. If the synchronous speed corresponding to the rated
frequency is call the base speed b, the synchronous speed at any other
frequency becomes:
s   b
And :
S
b  m

1  m
b
b
The motor torque :
Td 
Td 
3 Rr' Vs2

Rr'
S s  Rs 
S

2

  X s  X r'



2



3 Rr' Vs2

Rr'
S b  Rs 
S

2


' 2
  X s  X r  


If Rs is negligible, the maximum torque at the base speed as :
3 Vs2
Tmb 
2S b X s  X r'


And the maximum torque at any other frequency is :
2
Vs
3
Tm 
2S b X s  X r'  2


Sm 
At this maximum torque, slip S is :
Rr '
 X s  X r'


2
Normalizing :
Vs
3
Tm 
2S b X s  X r'  2


3 Vs2
Tmb 
2S b X s  X r'

Tm
1
 2
Tmb 

And
Tm  2  Tmb
Example :
A three-phase , 11.2 kW, 1750 rpm, 460 V, 60 Hz, four pole, Y-connected
induction motor has the following parameters : Rs = 0.1W, Rr’ = 0.38W, Xs =
1.14W, Xr’ = 1.71W, and Xm = 33.2W. If the breakdown torque requiretment is
35 Nm, Calculate : a) the frequency of supply voltage, b) speed of motor at
the maximum torque
Solution :
Input voltage per-phase : Vs 
Base frequency :
b  2 f  2 x 3.14 x 60 377rad / s
Base Torque : Tmb 
Motor Torque :
460
 265volt
3
60Po
60 x11200

 61.11Nm
2  N m 2 x 3.14 x1750
Tm  35 Nm
a) the frequency of supply voltage :
Tm
1
 2
Tmb 

Tmb
61.11

 1.321
Tm
35
Synchronous speed at this frequency is :
s   b
s 1.321x 377 498.01rad / s
or
60 x 498.01
N s   Nb 
 4755.65 rpm
2 x
p NS
4 x 4755.65
So, the supply frequency is : f s 

158.52 Hz
120 b
120
b) speed of motor at the maximum torque :
Rr '
 X s  X r'
Rr’ = 0.38W, Xs = 1.14W, Xr’ = 1.71W and   1.321
At this maximum torque, slip Sm is :
So,
Sm 
0.38
 0.101
1.3211.14  1.71
Sm 


or,
Nm  NS (1  S )  4755.65(1  0.101)  4275rpm
CONTROLLING INDUCTION MOTOR SPEED USING
ROTOR RESISTANCE
(Rotor Voltage Control)
Equation of Speed-Torque :
Td 
3 Rr' Vs2

Rr'
S s  Rs 
S

In a wound rotor induction motor, an external
three-phase resistor may be connected to its
slip rings,
RX
Stator
RX
Rotor
Three-phase
supply
RX


' 2
  X s  X r  


2
3Vs2 S
Td 
 s R 'r
These resistors Rx are used to control motor starting and stopping
anywhere from reduced voltage motors of low horsepower up to
large motor applications such as materials handling, mine hoists,
cranes etc.
The most common applications are:
AC Wound Rotor Induction Motors – where the resistor is wired into the
motor secondary slip rings and provides a soft start as resistance is
removed in steps.
AC Squirrel Cage Motors – where the resistor is used as a ballast for soft
starting also known as reduced voltage starting.
DC Series Wound Motors – where the current limiting resistor is wired to
the field to control motor current, since torque is directly proportional to
current, for starting and stopping.
The developed torque may be varying the resistance Rx
The torque-speed characteristic for variations in rotor resistance
This method increase the starting torque while limiting the starting current.
The wound rotor induction motor are widely used in applications requiring
frequent starting and braking with large motor torque (crane, hoists, etc)
The three-phase resistor may be replaced by a three-phase diode rectifier
and a DC chopper. The inductor Ld acts as a current source Id and the DC
chopper varies the effective resistance:
Re  R(1  k )
Where k is duty cycle of DC chopper
The speed can controlled by varying the duty cycle k, (slip power)
Id
Ld
Stator
D1
D3
Vd
Rotor
Three-phase
supply
D5
D4
D6
D2
GTO
R
Vdc
The slip power in the rotor circuit may be returned to the supply by
replacing the DC converter and resistance R with a three-phase full
converter (inverter)
Three-phase
supply
Transformer
Id
Ld
Stator
D1
D3
Vd
Rotor
Slip Power
D5
D4
D6
Diode rectifier
D2
T1
T3
T5
T2
T4
T6
Vdc
Controlled rectifier/
inverter
Na:Nb
Example:
A three-phase induction motor, 460, 60Hz, six-pole, Y connected, wound rotor
that speed is controlled by slip power such as shown in Figure below. The
motor parameters are Rs=0.041 W, Rr’=0.044 W, Xs=0.29 W, Xr’=0.44 W and
Xm=6.1 W. The turn ratio of the rotor to stator winding is nm=Nr/Ns=0.9. The
inductance Ld is very large and its current Id has negligible ripple.
The value of Rs, Rr’, Xs and Xr’ for equivalent circuit can be considered
negligible compared with the effective impedance of Ld. The no-load of motor is
negligible. The losses of rectifier and Dc chopper are also negligible.
The load torque, which is proportional to speed square is 750 Nm at 1175 rpm.
(a) If the motor has to operate with a minimum speed of 800 rpm, determine
the resistance R, if the desired speed is 1050 rpm,
(b) Calculate the inductor current Id.
(c) The duty cycle k of the DC chopper.
(d) The voltage Vd.
(e) The efficiency.
(f) The power factor of input line of the motor.
Vs 
460
 265.58 volt
3
p 6
  2 x 60  377rad / s
s  2 x 377/ 6  125.66 rad / s
The equivalent circuit :
The dc voltage at the rectifier output is :
Vd  I d Re  I d R (1  k )
Nr
Er  S Vs
 S Vs nm
Ns
and
For a three-phase rectifier, relates
Er and Vd as :
Vd 1.65 x 2 Er  2.3394Er
Nr
E

S
V
 S Vs nm
Using :
r
s
Ns
Vd  2.3394S Vs nm
If Pr is the slip power, air gap power is :
Pg 
Pr
S
Pr
3Pr (1  S )
 S) 
Developed power is : Pd  3( Pg  Pr )  3(
S
S
Because the total slip power is 3Pr = Vd Id and
So,
Pd  TL m
(1  S )Vd I d
Pd 
 TLm  TLm (1  S )
S
Substituting Vd from
Solving for Id gives :
Vd  2.3394S Vs nm
TLs
Id 
2.3394Vs nm
In equation
:
Pd above, so
Which indicates that the inductor current is independent of the speed.
Vd  I d Re  I d R (1  k ) and equation : Vd  2.3394S Vs nm
I d R(1  k )  2.3394S Vs nm
From equation :
So,
Which gives :
S
I d R(1  k )
2.3394S Vs n m
I d R(1  k )
S
2.3394S Vs n m

I d R(1  k ) 
m  s (1  S )  s 1 

2
.
3394
V
n
s m

The speed can be found from equation :
as :

TLs R(1  k ) 
m  s 1 
2
(
2
.
3394
V
n
)
s m


Which shows that for a fixed duty cycle, the speed decrease with load
torque. By varying k from 0 to 1, the speed can be varied from minimum
value to s
m  180  / 30  83.77 rad / s
 Kvm
2
 800 
 750x 
  347.67 Nm
 1175
From torque equation : T
L
2
From equation :
TLs
Id 
The corresponding inductor current is :
2.3394Vs nm
347.67 x 125.66
Id 
 78.13 A
2.3394x 265.58 x 0.9
The speed is minimum when the duty-cycle k is zero and equation :

I d R(1  k ) 
m  s (1  S )  s 1 

2
.
3394
V
n
s m

78.13 R
83.77  125.66(1 
)
2.3394x 265.58 x 0.9
And :
R  2.3856W