MAHALAKSHMI
ENGINEERING COLLEGE
TIRUCHIRAPALLI – 621213
QUESTION BANK WITH ANSWER
--------------------------------------------------------------------------------------------------------------Sub. Code
: EE2352
Semester
: VI
Subject
: SOLID STATE DRIVES
Unit
: IV
----------------------------------------------------------------------------------------------------------------
INDUCTION MOTOR DRIVES
1. What are the different methods of braking applied to the induction motor?
Regenerative braking
Plugging
Dynamic braking.
2. What are the different methods of speed control of IM?
Stator voltage control
Supply freq. control
Rotor resistance control
Slip power recovery control.
3. What is meant by stator voltage control?
The speed of the IM can be changed by changing the stator voltage, because
the torque is proportional to the square of the voltage.
4. Mention the application of stator voltage control.
This method is suitable for applications where torque demand reduced with
speed, which points towards its suitability for fan and pump drives.
5. Mention the applications of ac drives.
AC drives are used in a no. of applications such as fans, blowers, mill runout tables, cranes, conveyors, traction etc.
6. What are the three regions in the speed-torque characteristics in the IM?
Motoring region(0<=s<=1)
Generating region(s<0)
Plugging region (1<=s<=2) where s is the slip.
7. What is the advantage of stator voltage control method?
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The control circuitry is simple
Compact size
Quick response time
8. What is meant by soft start?
The ac voltage controllers show a step less control of supply voltage from zero
to rated voltage is this control is employed in motor then it is called soft start.
9. List the advantages of squirrel cage induction motor?
Cheaper
light in weight
Rugged in construction
More efficient
Require less maintenance
It can be operated in dirty and explosive environment
10. Define base speed.
The synchronous speed corresponding to the rated frequency is called the
base speed.
11. What is meant by frequency control of IM?
The speed of IM can be controlled by changing the supply frequency
because the speed is directly proportional to supply frequency. This method of speed
ctrl is called frequency control.
12. What is meant by V/F control?
When the frequency is reduced the input voltage must be reduced
proportionally so as to maintain constant flux otherwise the core will get saturated
resulting in excessive iron loss and magnetizing current. So maintaining constant
Voltage/Frequency ratio is called V/F control.
13. What are the advantages of V/F control?
Smooth speed ctrl
Small input current and improved power factor at low frequency start
Higher starting torque
14. What are the 3 modes of region in the adjustable-frequency IM drives
characteristics?
Constant torque region
Constant power region
High speed series motoring region
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15. What are the advantages of induction motors over D.C. motors?
The main drawback of D.C. motors is the presence of commutate and brushes,
which require frequent maintenance and make them unsuitable for explosive and dirty
environments. On the other hand, induction motors, particularly squirrel-cage are
rugged, cheaper, lighter, smaller, more efficient, require lower maintenance and can
operate in dirty and explosive environments.
16. Why the control of a three-phase indication motor is more difficult than D.C.
motors.
The control of a three-phase induction
motor, particularly when the
dynamic performance involved is more difficult than D.C. motors. This is due to
a. Relatively large internal resistance of the converter causes voltage
fluctuations following load fluctuations because the capacitor cannot be
ideally
large.
b. In a D.C. motor there is a decoupling between the flux producing
magnetizing current and torque producing armature current. They can be
independently controlled. This is not the case with induction motors.
c. An induction motor is very poorly damped compared to a D.C. motor.
17. What is indirect flux control?
The method of maintaining the flux constant by providing a voltage boost
proportional to slip frequency is a kind of indirect flux control. This method of flux
control is not desirable if very good dynamic behavior is required.
18. What is the purpose of inductance and capacitance in the D.C. link circuit?
The inductance in the D.C. link circuit provides smoothing whereas the
capacitance maintains the constancy of link voltage. The link voltage is a controlled
quality.
19. What are the effects of harmonics in VSI fed induction motor drive?
If the motor receives square wave voltages this voltage has harmonic
components the harmonics of the stator current cause additional losses and heating.
These harmonics are also responsible for torque pulsations. The reaction of the fifth
and seventh harmonics with the fundamental gives rise to the seventh harmonic
pulsations in the torque developed.
20. Where is rotor resistance control used?
Where the motors drive loads with intermittent type duty, such as cranes, ore or
coal unloaders, skip hoists, mine hoists, lifts, etc. slip-ring induction motors with speed
control by variation of resistance in the rotor circuit are frequently used. This method
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of speed control is employed for a motor generator set with a flywheel (Ilgner set)
used as an automatic slip regulator under shock loading conditions.
21. Why the static scherbius drive has a poor power factor?
Drive input power is difference between motor input power and the power fed
back. Reactive input power is the sum of motor and inverter reactive power.
Therefore, drive has a poor power factor throughout the range of its options.
22. How is super synchronous speed achieved?
Super synchronous speed can be achieved if the power is fed to the rotor
from A.C. mains. This can be made possible by replacing the converter cascade by a
cycloconverter. A cycloconverter allows power flow in either direction making the
static sherbets drive operate at both sub and supper synchronous speeds.
23. What is static Kramer drive?
Instead of wasting the slip power in the rotor circuit resistance, it can be
converted to 60 Hz A.C. and pumped back to the line. The slip power controlled
drive that permits only a sub synchronous range of speed control through a converter
cascade is know as static Kramer drive.
24. What are the advantages of static Kramer drive?
The static Kramer drive has been very popular in large power pump and fantype drives, where the range of speed control is limited near, but below the
synchronous speed. The drive system is very efficient and the converted power rating
is low because t has to handle only the slip power, In fact, the power rating becomes
lower with a more restricted range of speed control. The additional advantages are
that the drive system has
D.C. machine like characteristics and the control is very simple.
25. What are the causes of harmonic currents in static Kramer drive?
The rectification of slip power causes harmonic currents in the rotor, and these
harmonics are reflected to the stator by the transformer action of the machine.
The harmonic currents are also injected into the A.C. line by the inverter. As a result,
the machine losses are increased and some amount of harmonic torque is produced.
Each harmonic current in the rotor will create a reading magnetic filed and its direction
of rotation will depend on the order pf the harmonic.
26. Give the four modes of operation of a Scherbius drive
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The four modes of operation of static Scherbius drive are,
Sub synchronous motoring.
Sub synchronous regeneration
Super synchronous motoring
Super synchronous regeneration
27. How is the static Scherbius drive operated in super synchronous motoring
mode?
In super synchronous motoring mode, the shaft speed increases beyond
the synchronous speed, the slip becomes negative and the slip power is
absorbed by the rotor. The slip power supplements the air gap power for the total
mechanical power output. The line therefore supplies slip power in addition to
stator input power. At this condition, the phase sequence of slip frequency is
reversed so that the slip current – induced rotating magnetic filed is opposite to
that of the stator
PART-B
1)Explain construction
motor.(AU/2012/11/)
and
working
principle
of three phase induction
Construction:
The induction motor consists of two main parts,
 Stator
 Rotor
Stator:
the stator made up of a stamping with alternate slot and tooth. Stamping are insulated
from each other. Each stamping is 0.4 to 0.5 mm thick. The slot house the three phase winding
just like a three phase alternator. The three phase winding is called stator winding. It may be
connected either in star or delta. The stator winding is made for a fixed number of poles.
Rotor:
There are two type of rotors used in induction motor. They are,
 Squirrel cage
 Slip ring
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Squirrel cage Rotor:
This is made up of a cylindrical laminated core with slots to carry the rotor conductor.
The rotor conductor are heavy bars of copper or aluminum short circuited at both ends by end
rings. Hence the rotor is called a short circuited rotor.
Slip ring or wound Rotor:
The rotor winding may be star or delta connected, distributed winding, wound for as
many number of poles as the stator is wound for. Variable external resistance can be connected
in the rotor circuit, with the help of brushes and slip ring arrangements. By varying the external
resistance in the rotor circuit, the motor speed and torque can be controlled.
WORKING PRINCIPLE:
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Three –phase supply is given to the stator winding , due to this current flows through the
stator winding , this current is called stator current , it produces a rotating magnetic field in the
space between stator and rotor , this magnetic field rotates at synchronous speed given by,
Ns=
Where,
Ns=synchronous speed
F=supply frequency
P=number of poles for which the stator is wounded
As a result of the rotating magnetic field cutting the rotor conductors , an is emf induced in the
rotor, if the rotor winding is shorted (in cage rotor they are already shorted and in wound rotor ,
they have to be shorted externally)
Then the induced emf produces current , this current produces a rotor field (rotor emf).
The interaction of stator and rotor field developed torque , then the rotor rotates in the same
direction as the rotating maganetic field when the rotor stand still , the frequency of rotor emf is
equal to the supply frequency .
As the rotor speed up , the frequency of the rotor emf and the magnitude of rotor decrease.
The rotor tries to catch up with the rotating magnetic field however, the rotor cannot really catch
up the synchronous speed , because if it does so the relative speed would becomes zero and
then ther is no rotor induced emf , no current and hence no torque. therefore, the rotor runs at a
speed slightly less then the synchronous speed , in an induction motor the rotor speed is always
less than the synchronous speed , therefore this machine is called as synchronous machine .
The difference between synchronous speed and rotor speed is called as slip speed.
(Slip speed) = Ns - Nr
Slip, s=
So,
Nr=Ns (1-s)
% slip =
At no-load, the difference between synchronous speed and rotor speed is only about 1%.
Advantages of squirrel cage induction motor
1. Cheaper
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2. Light weight
3. Higher efficiency
4. Required less maintenance
Disadvantages of squirrel cage induction motor
1. Moderate starting torque
2. External resistance cannot be connected to rotor circuit
Applications of squirrel cage induction motor
Squirrel cage induction motors are used in lathes, fans, blowers water pumps, grinder
etc..,.
Advantages of slip ring induction motor
1. The starting torque can be controlled by varying the rotor circuit resistance.
2. The speed of the motor can be controlled by varying the rotor circuit resistance.
Disadvantages of slip ring induction motor
1. Wound – rotor machine is heavier.
2. High cost
3. High rotor inertia
2)Analysis and performance of three phase induction motor:
Per phase equivalent circuit of a three phase induction motor is shown in the fig.
and
are the stator referred values of rotor resistance Rr rotor reactance Xr. Slip is defined by
s
 ms   m
 ms
 ms 
_____________ (1)
4ππ
Rad/sec _______________ (2)
p
Since, stator impedance drop is generally negligible compared to terminal voltage V, the
equivalent circuit can be simplified to that shown in the fig:
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a)
b)
from equation ( 1 )
   ms (1  s)
_______________( 3)
From fig
I 'r =
V

R 
 R s 
  j(X s  X 'r )
s 

'
r
_______________( 4)
Power transferred to rotor (or air gap power)
Pg  3I '2r R 'r /s __________(5)
Rotor copper loss is
p cu  3I '2rR 'r _____________(6)
Electrical power converted into mechanical power
1 s 
Pm  p g  Pcu  3I '2r R 'r 
 ______________(7)
 s 
Torque developed by motor
T
pm
m
____________(8)
Substituting equation (3) and (7) yields
T
3
 ms
I '2r
R 'r
_______________(9)
s
Substituting in from eq (4) gives


V 2 R '2r /s
3 
T
 ms  
R 'r 

  X s  X 'r
R

 s
s 
 



 _______(10)
2



A comparison of eqn (5) and (9) suggests that
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T
Pg
 ms
_________(11)
The developed torque is a function of slip only and slip for maximum torque is given by
Sm  
R 'r

R  Xs  X
2
s

' 2
r
___________(12)
Substituting from eqn (12) into (10) yields an expression for maximum torque
T
3 
V2

2 ms  R s  R s2  (X s  X '2r ) 2


 ______________ (13)

Dividing eqn (10) by (13) and substituting in from (12) yields
 R

21  s' s m 
T
 Rs 

______________ (14)
Rs
S Sm
Tmax

2 '
Sm
s
R r Sm
Fig: Speed –Torque characteristics of induction motor
The term
Rs
s m is very small compared to I and dominating the denominator. Therefore, it can
R s'
be dropped from eqn (14)
T
2

__________________ (15)
S Sm
Tmax

Sm S
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3)Operation with unbalanced source voltage and single phasing
As supply voltage some time become unbalanced, it is useful to know the effect of
unbalanced voltages on motor performance.
As three – phase set of unbalanced voltages (Va,Vb and Vc) can be resolved in to three
phase balanced positive sequence, negative sequence and zero sequence voltages, using
symmetrical components relations:
Vp = 1/3(Va+αVb+  2 Vc)
Vn = 1/3(Va+  2 Vb+αVc)
V0 = 1/3(Va+Vb+Vc)
______________ (1)
Where α= e j120  cos120  j sin sin 120 ___________ (2)
For positive sequence voltages, equivalent circuits are same, except that V is replaced by
Vp . The positive sequence rotor current and torque are obtained by replacing V by Vp. Thus
I 'rp 
Vp
2

R' 
 R s  r   j(X S  X 'r ) 2
S 

__________________
(3)




2
2


VP R r /S
3

 _______________(4)
TP 
2
'
 ms   


R
' 2
   R s  r   (X S  X r )  


S 
  

The slip is
Sn 
  ms   m
_____________ (5)
  ms
Sn = (2-S) ______________(6)
When S is replaced by (2-S), and V is replaced by Vn , Then the following expressions are
obtained for rotor current and torque:
V
I 'rn 
_________________
'
(7)

Rr 
'
 R s 
  j(X S  X r )
S 

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



2
2


Vn R r /(2 - S)
3

 _____________ (8)
TP  
2
'
 ms   


R 
' 2
   R s  r   (X S  X r )  

S 
  

Torque has a negative sign because for negative sequence voltages the synchronous speed is the
synchronous speed is (-  ms )
The rms rotor current and torque are given by
I 'r  (I '2rp  I '2rn )1/2
T  TP  Tn
______________ (9)




2
2
2
2


VP R r /S
Vn R r /(2 - S)
3




2
2
'
'
 ms   
 



R
R
' 2
' 2
   R s  r   (X S  X r )    R s  r   (X S  X r )  




S 
S 
  
 

Positive sequence, negative sequence and the resultant speed – torque characteristics are shown
in fig:
Operation with unbalanced rotor impedance
Earlier unbalanced rotor impedance was employed in starting and speed control. It is
useful to examine the effect of unbalance on motor performance. Unbalalance rotor impedance
causes the unbalance in rotor currents. The unbalanced rotor currents can be resolved in to
positive and negative sequence components.
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For (0≤S≤0.5), the speed of negative sequence rotor field is positive, therefore interaction
between this field and stator current induced by it, produce a positive on the stator, and the fig
shows the nature of the speed torque characteristics.
Speed – torque curves with unbalanced rotor resistance
4)Explain Various method of Starting in three phase Induction Motor.
Starting arrangement is chosen based on the load requirements and nature of supply.
The following methods are employed for starting of induction motor.
Star-Delta Starter
 In this method motor designed run in delta connection and connected in star during
starting, this allows reduction in stator current and voltage by 1/√3.
 Circuit breakers CBr and CBs are closed to start the motor in star connection and when
the motor started to run in steady state speed, the circuit breakers CBs and CBr are
closed run the motor in Delta connection
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Auto- transformer starter

Reduced voltage for starting can be obtained from autotransformer.
 For a secondary to primary turns ratio of aT , motor terminal voltage and current are
reduced by aT . This reduces the current drawn from supply aT 2

Since the torque is proportional to square of the motor terminal voltage, it is
also reduced to a T2 .
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Rotor Resistance Starter

Wound – rotor motors are generally started by connecting rotor resistance in the
rotor circuit.

The highest value of resistance is chosen to limit the current at zero speed
within safe values.

C1, C2, C3 are the closing contacts to close step by step to reduce the current
when the motor accelerates and also between the maximum and minimum for
specified values.
5)Explain with a neat circuit diagram, the Braking of InductionMotor.
Following methods are employed for braking of an induction motor:
Regenerative braking
The power input to an induction motor is given by
Pin  3 VI S COSφ s
Where ФS is the angle between the stator phase voltage and stator phase current.

For motoring operation Ф S< 90  , if the rotor speed greater than the synchronous, relative
speed between the rotor conductor and air gap field is reverse.
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
Consequently, Ф S
regenerative braking

When fed from a source of fixed frequency, regenerative braking is possible only for
speed greater than the synchronous speed.
becomes greater than the 90  and power flow reverse, giving
Regenerative braking
Plugging or Reverse Voltage Braking

When phase sequence of the supply with respect to terminal is changed means, the motor
also start to rotate in reverse. Then the motor shift it operation from motoring to
plugging.

The slip for plugging is given by S n 
  ms   m
 2S
  ms
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
Since the instant switchover slip can be up to 2 and motor current is large and braking
torque is low,

A special case of braking occurs when an induction motor connected to positive sequence
voltage is driven by an active load in reverse direction. Crane hoist is one such
application.
Zero Sequence braking

In this case, three phases are connected in series across either single phase ac or dc, Such
connection is known as zero sequence connection.

With an ac supply, resultant field is stationary in space and pulsating at frequency of
supply.

However braking torque produced by this connection larger than the motoring. Motor
essentially works in regenerative braking.
TRANSIENT ANALYSIS

A rigorous analysis of operation induction motor can be done with help of d-q axis model
involving long calculation. Simple method of analysis with satisfactory accuracy for most
application obtained by using steady state torque relations.
The equation for transient operation of an induction motor drive is
J
d m
 T( m )  Tl ( m )
dt
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Starting and plugging
For starting and plugging torque equation is given by
J
2Tmax
d m

 Tl ( m )
dt
S  Sm
Sm
S
For no load operation
2Tmax
d m

dt
S  Sm
Sm
S
J
d m
ds
  ms
dt
dt
τ S
S
dt   m  m 
2  S
Sm
τm 

ds

J ms
Tmax
τ m is the mechanical time constant of the motor. It is defined as the time taken by the motor to
reach the synchronous speed from stand still under constant accelerating torque equal to the mechanical
torque of the motor.
Time required to start the induction on no load is
τ
ts  m
2
 S
0.05
  S
1
m

Sm 
ds
S 
 1

ts  m 
 1.5S m 
 4S m

Thus starting time is a function of Sm. Starting time has a minimum value of 1.22  m at Sm=0.4, When Rs
is negligible, rotor resistance required to start the motor in minimum time is

( R rm ) S  0.4( X s  X r
'
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
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Time required to stopping by plugging, when initially running at synchronous speed, can be
expressed as
tb  
 m 1  Sm
2
 
2
S

S
Sm


0.75 
ds   m 0.345S m 

Sm 


Stopping time function of Sm. It has a minimum value 1.027  m at Sm =1.47 corresponding value of rotor
resistance is

(R 'rm ) b  1.47( X s  X 'r

Time required for speed reversal by plugging when running on load is given by
tt  
m
2
0.05
 S
  S
2
m

Sm
S


1 
ds  m 3.69 S m 

Sm 


Minimum time for reversal is thus 2.88  m corresponding values of Sm is 0.52. Rotor resistance
required for speed reversal by plugging in minimum time is

(R 'rm ) r  0.52( X s  X 'r

Calculation Of Energy Losses
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The
E sr 
rotor
winding
loss
for
the
starting
can
be
written
ts
 3I
'2
r
R 'r dt
0
ts
E sr   msTdt
0
As the machine operating under no load
d m
J
T
dt
ds
 J ms
T
dt
Tdt   J msds
0
2
2
E sr    J ms
sds  12 J ms
1
Energy loss in the stator win ding, neglecting magnitizin g current
E ss 
ts
I
'2
r
Rs
0
1
2  Rs
 R'
as E ss  2 J ms 
 r




Hence, total winding loss during starting under no – load
ES 
R 
1
2 
1  S' 
J ms
2
 Rr 
Rotor winding loss during stopping by plugging under no load can be written
1
as E
'
sr
2
   J ms
sds 
2
3
2
J ms
2
6) Explain the stator voltage control of Induction motor drives.
Stator Voltage control
 By reducing stator voltage, speed of a induction motor can be reduced by an amount
which is sufficient for speed control of pump drives.
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
If stator copper loss, core loss and friction loss are ignored, then the motor efficiency is
P
given by η  m  1  S
Pg
Stator voltage control
7) Explain the closed loop speed control of VSI fed & CSI fed induction motor
drive.
VSI fed Induction Motor Drives
 Voltage source inverter allows a variable frequency supply to be obtained from a dc
supply.
 VSI can be operated as a stepped wave inverter or PWM inverter. When operated as
stepped wave inverter, transistors are switched in sequence with time difference.

Frequency of the inverter operation varied by varying the T and output of the inverter
also varied accordingly.

Inverter output line and phase voltage are given by the following Fourier series:
VAB 
2 3 
1
1
1
1

Vd Sin t  Sin 5t - Sin7t  Sin11t  Sin13t....
π
5
7
11
13


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VSI fed induction motor
Stepped wave inverter line voltage waveform
CSI fed Induction Motor Drives

CSI consist of diodes D1-D6 and capacitors C1-C2 to provide commutation for the
thyristor T1- T, which are fired in with a phase difference of 60˚ in sequence of their
numbers.

Inverter behaves as current source because of presence of large inductance Ld in dc link.
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
The fundamental component of motor phase current is I S 
6
Id
π
CSI fed induction drive
Static Rotor Resistance Control
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
Rotor resistance can also vary steplessly using the above circuit. The ac output voltage of
the rotor can be converted with Diode Bridge and it is given to parallel combination of
the resister and transistor switch.

The resistance value can be varied with help of the duty ratio of the transistor.
The RMS rotor current will be
Ir 
2
Id
3
Therefore average value of resistance between the terminals is given by
RAB= 1 δR
Power consumed by RAB is
PAB= I d2 R AB  I d2 R(1  δ)
Power consumed per phase= PAB /3  0.5R(1  δ)I 2r
Thus total rotor circuit resistance per phase will be R rT  R T  0.5R(1  δ)
8)Explain the voltage\frequency control of Induction motor drives.
Open Loop Volts/Hz Control

Open loop volts/Hz control of an induction motor is very familiar in the industry.
For adjustable speed applications, frequency control is natural.

The phase voltage command Vs* is generated from frequency command and the flux is
remain constant.
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9)with necessary diagrams explain the vector control of induction motor
drives.
Vector Control of Induction motor

Vector control of induction can be controlled like a separately excited DC motor. Vector
control applicable for both induction as well synchronous motor.

In a DC machine neglecting the armature reaction effect and field saturation, the
developed torque is given by
Te  K 't I a I f
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Te  K t ψ a ψ f  K 't I a I f
Torque Component Field Component

Te  K t ψ r I qs  K 't I qs I ds
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10) What are the various methods of speed control of 3ф induction motors and neat
diagram explain the slip power recovery scheme?
The slip power recovery system can be classified into two types
 Kramer system
 Scherbius system
These two systems can further be classified into two methods
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 Conventional method
 Static method
Kramer system:
The Kramer system is applicable only for sub synchronous speed operation. The classification
of Kramer system is
 Conventional Kramer system
 Static Kramer system
Conventional Kramer system
Fig. shows conventional Kramer system. The system consists of three phase rotary converter
and a DC motor. The slip power is converted into dc power by a rotary converter and fed to the
armature of a dc motor.
The slip ring induction motor is coupled to the dc motor. The slip rings are connected to
the rotary converter. The dc output of rotary converter is used to drive a dc motor. The rotary
converter and dc motor are excited from the dc bus bars or from an exciter. The speed of slip
ring induction motor is adjusted by adjusting the speed of dc motor with the help of field
regulator.
This system is also called the electrochemical cascade, because the slip frequency power is
returned as mechanical power to the slip ring induction motor shaft by the dc motor.If the
mechanical loses in cascade system are neglected the shaft power output of the SRIM motor is,
Pm = (1-s) Pin
Where,
Pin = input power to the stator
The slip power Ps =sPin is added to Pm by converting it to mechanical power by the dc
motor. This mechanical power is fed to the slip ring induction motor shaft. Thus, irrespective of
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the value of the slip and consequently the speed of the SRIM, the power output remains more or
less constant. Hence, it is also called the constant power cascade. This method is used only for
large motors of capacity 4000KW or above.
Advantages
 The main advantage of this method is that any speed, within the working range can be
obtained.
 If the rotary converter is over excited, it will take a leading current which compensates
for the lagging current drawn by SRIM and hence improves the power factor of the
system.
The slip power is converted into dc by a three phase diode bridge rectifier fig. this dc power
is fed to the dc motor. This dc motor is mechanically coupled to SRIM. The slip power is
converted to mechanical power and fed back to the SRIM shaft.
The torque supplied to the load is shared by SRIM and DC motor. The SRIM speed can
be controlled by controlling the field regulator (field current) of the dc motor. In this method
the speed control range is synchronous speed to around half of the synchronous speed.
In the fig. the diode bridge rectifier is replaced by thruster bridge rectifier. The speed of
SRIM can be controlled from zero to around synchronous speed, by varying the firing angle
of thirstier rectifier.
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Static Kramer system:
In the rotor resistance controlled method, the slip power is wasted in the rotor circuit
resistance. Instead of wasting the slip power in the rotor circuit resistance, it can be
converted to 50Hz ac and pumped back to the line.
Here, the slip power flows only in one direction. This method of drive is called static
Kramer drive. It is shown fig. the static Kramer drive offers speed control for sub
synchronous speed only. i.e. speed can controlled less than the synchronous speed only.
In this method, the slip power is taken from the rotor and it is rectified to dc voltage by
three phase diode bridge rectifier. Inductor Ld smoothens the ripples in the rectified voltage
Vd. this DC power is converted in to AC power by using line – commutated inverter.
The rectifier and inverter are both line commutated by alternating emfs appearing at the slip
rings and supply bus bars respectively. Here, the slip power flows from rotor circuit to
supply. This method is also called constant – torque drive.
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The static Kramer drive has been very popular in large power pump and fan type
drives, were the range of speed control is limited, but less than the synchronous
speed.This method of speed control is economical because the rectifier and inverter
have to carry only the slip power of the rotor, which is considerably less than the input
power to the stator.
SCHERBIUS SYSTEM
The scherbius system is similar to Kramer system but only difference is that in the
Kramer system the feedback is mechanical and in the scherbius system the return power is
electrical. The different types of scherbius system are:
 Conventional scherbius drive
 Static scherbius drive
CONVENTIONAL SCHERBIUS DRIVE:
Figure shows conventional method of scherbius drive. This method consists of SRIM,
rotary converter, DC motor and induction generator. Here, the rotary converter converts slip
power into DC power and the DC power fed to the DC motor.
The Dc motor is coupled with induction generator. The induction generator converters
the mechanical power into the electrical power and returns it to the supply line. The SRIM speed
can be controlled by varying the field regulator of the DC motor.
STATIC SCHERBIUS DRIVE:
For the speed control SRIM both below and the above synchronous speed, static
scherbius drive is used. This system can again be classified as
 DC link static scherbius drive
 Cycloconverter static scherbius drive
DC LINK STATIC SCHERBIUS DRIVE
This system consists of SRIM, two numbers of phase controlled bridges, smoothing
inductor and step up transformer. This system used for sub synchronous and super
synchronous speed operation.
I.
Sub synchronous speed operation
In sub synchronous speed control of SRIM, slip power is removed from the rotor circuit and
is pumped back into the ac supply. Fig. shows the dc link static scherbius system.
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In the scherbius system, when a machine is operated at sub synchronous speed, phase
controlled bridge 1 operates in the rectifier mode and bridge 2 operates in the inverter
mode. In other words, bridge 1 has firing angle less than the 90 0 whereas bridge 2 has firing
angle more than 900.
The slip power flows from rotor circuit to bridge 1, bridge2, transformer and return to the
supply.
Slip power→ rectifier (bridge1) → inverter bridge2→transformer→supply
II. Super synchronous speed operation
In super synchronous speed operation, the additional power is fed into the rotor circuit at
slip frequency. Fig. shows super synchronous speed operation of a dc link static scherbius
system. In the scherbius system, when the machine is operated at super synchronous
speed, phase controlled bridge 2 should operate in rectifier mode and bridge1 in inverter
mode. In other words, the bridge2 has firing angle less than 90 0 and bridge 1 has firing
angle more than 900. The slip power flows from the supply to transformer, bridge2 (rectifier),
bridge1 (line commutated inverter) and to the rotor circuit.
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Supply → transformer →rectifier (bridge2) → Bridge 1 (inverter) →rotor circuit
Near synchronous speed, the rotor voltage is low, and forced commutation must be employed in
the inverter, which makes the scheme less attractive. The replacement of 6 diodes by 6
thyristors increases the converter cost and also necessitate the introduction of slip frequency
gating circuit.
Difficulty is experienced near synchronous speed when slip frequency emfs are
insufficient for line or natural commutation, and special connections or forced commutation
methods are necessary for the passage through synchronism. Thus, the prevision of super
synchronous speed control unduly complicates the static converter cascade system and nullifies
the advantages of simplicity and economy.
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