Module 1

 Basic Introduction of Electric Drives

 Components of Drives
 Dynamic equation
 Multi-quadrant operation of hoist
 Thermal Modelling
 Drives AND Electric Drives Definition

1. Motor control is required in large number of industrial and domestic applications such
as transportation systems, rolling mills, paper machines, textile mills, machine tools,
fans, pumps, robots, and washing machines.
2. Systems employed for motion control are called drives and may employ any of the
prime movers.
3. Drives employing electric motors are known as electric drives.

OR

• Systems employed for getting the required motion and their smooth control are
called Drives.
• Drives require prime movers like Diesel or petrol engines, gas or steam turbines,
hydraulic motors or electric motors. These prime movers deliver the required
mechanical energy for getting the motion and its control.
• Drives employing Electric motors as prime movers for motion control are called
Electric Drives.
 Block Diagram of
ELECTRIC Drives
 Motor
1. Due to presence of commutator and brushes dc motors have a number of
disadvantages as compared to ac motors (induction and synchronous motors): higher
cost, weight, volume and inertia for the same rating, need for frequent maintenance,
unsuitable for explosive and contaminated environments and restrictions on maximum
voltage, speed and power ratings.
2. Squirrel-cage induction motor, which costs nearly one-third of a dc motor of the same
rating, is extremely rugged, requires practically no maintenance and can be built for
higher speeds, torques and power ratings. Wound-rotor motors are more expensive than
squirrel-cage motors.
3. Their maintenance needs, although more than squirrel-cage motors, are much less
compared to dc motors. They are also available in high power ratings.
4. Wound field and permanent magnet synchronous motors have a higher full load
efficiency and power factor than induction motors.
5. Wound field motors can be designed for a higher power rating than induction motors.
However, compared to squirrel-cage induction motors they have higher cost and
size for the same rating and require more maintenance.
6. The permanent magnet synchronous motors have all the advantages of squirrel-cage
induction motors except that they are available in lower power ratings.
7. Because of numerous advantages of ac motors described above, ac drives have
succeeded in replacing dc drives in a number of variable speed applications.
 Motor
1. Brushless dc motor is somewhat similar to a permanent magnet synchronous
motor, but has lower cost and requires simpler and cheaper converter. It is
being considered for low power high speed drives and for servo applications,
as an alternative to dc servo motor which has been very popular so far.

2. Recently, stepper motor has become popular for position control and
switched reluctance motor drive for speed control.
Sources :
1. may be of 1 phase and 3 phase. 50 Hz AC supply is the most common type of electricity
supplied in India, both for domestic and commercial purpose.
2. Synchronous motors which are fed 50 Hz supply have maximum speed up to 3000 rpm,
and for getting higher speeds higher frequency supply is needed.
3. Motors of low and medium powers are fed from 400 V supply, and higher ratings like
3.3 kV, 6.6 kV, 11 kV etc are provided also.
4. Also DC source is available from battery or Solar panel
Power Modulators –
are the devices which alter the nature or frequency as well as changes the intensity of
power to control electrical drives. Roughly, power modulators can be classified into
three types,
1. Converters,
2. Variable impedance circuits,
3. Switching circuits.

As the name suggests, converters are used to convert currents from one type to
another type. Depending on the type of function, converters can be divided into 5
types –
 AC to DC converters
 AC regulators
 Choppers or DC-DC converters
 Inverters
 Cycloconverters
 AC to DC converters (Rectifier) are used to obtain fixed DC supply from
the AC supply of fixed voltage.

You can explain in detail all the converter by following text book
 AC Regulators are used to obtain the regulated AC voltage, mainly
autotransformers or tap changer transformers are used
 Choppers or DC-DC converters are used to get a variable DC voltage. Power
transistors, IGBT’s, GPO’s, power MOSFET’s are mainly used for this purpose.
 Inverters are used to get AC from DC, the operation is just opposite to that of AC
to DC converters. PWM semiconductors are used to invert the current.
 Cycloconverters are used to convert the fixed frequency and fixed voltage AC into
variable frequency and variable voltage AC. Thyristors are used in these converters to
control the firing

You can explain in detail all the converter by following text book
• Variable Impedance circuits are used to controlling speed by varying the
resistance or impedance of the circuit. But these controlling methods are used in low cost
DC and ac drives. There can be two or more steps which can be controlled manually or
automatically with the help of contactors. To limit the starting current inductors are used
in AC motors.

• Switching circuits in motors and electrical drives are used for running the motor
smoothly and they also protects the machine during faults. These circuits are used for
changing the quadrant of operations during the running condition of a motor. And these
circuits are implemented to operate the motor and drives according to predetermined
sequence, to provide interlocking, to disconnect the motor from the main circuit during
any abnormal condition or faults.

Control Unit –

Choice of control unit depends upon the type of power modulator that is used. These are
of many types, like when semiconductor converters are used, then the control unit
consists of firing circuits, which employ linear devices and microprocessors.
Advantage of Electrical drives.
1. These drives are available in wide range torque, speed and power.
2. The control characteristics of these drives are flexible. According to load requirements
these can be shaped to steady state and dynamic characteristics. As well as speed control,
electric braking, gearing, starting many things can be accomplished.
3. They can operate in all the four quadrants of speed torque plane, which is not applicable
for other prime movers.
4. They do not pollute the environment.

Dynamics of Electric Drive

 Dynamics of Motor Load System
A motor generally drives a load (Machines) through some transmission system. While
motor always rotates, the load may rotate or undergo a translational motion.

J = Moment of inertia of motor load system referred to the motor shaft kg / m2

ωm = Instantaneous angular velocity of motor shaft, rad/sec.
T = Instantaneous value of developed motor torque, N-m
Tl = Instantaneous value of load torque, referred to the motor shaft N-m
Motor-load system can be described by the following fundamental torque equation.
T  Tl  rateof changeof angular momemntum

d  J  dJ d equation applicable to variable

T  Tl   J  eq.1 inertia drives such as mine winders,
dt dt dt reel drives, Industrial robots.

d
 eq.2   J  k   0 
 dJ
T  Tl  J The motor equation applicable to
dt  dt  constant inertia drives
 d
T  Tl  J
dt
d
if T  Tl   0  accelerating motor 
dt
d
if T  Tl   0  deaccelerating motor 
dt
d
if T  Tl   0  equilibrium state 
dt
Stability in drives refers to the ability of a drive to maintain equilibrium and
return to equilibrium after a disturbance.
A drive is in equilibrium when the motor torque equals the load torque.

Stable equilibrium
1. When a drive returns to equilibrium after a disturbance, it is in stable
equilibrium.
2. For a stable equilibrium, the change in load torque with respect to speed must
be greater than the change in motor torque with respect to speed.
 If the speed decreases, the motor torque must exceed the load torque.
 If the speed increases, the load torque must exceed the motor torque.

Detail described in the class

 Graphical Analysis of steady state stability

A decrease in speed causes the

A decrease in speed causes the motor load torque to become greater than the
torque to become greater than the load torque, motor torque, electric drive decelerates
electric drive accelerates and operating point and operating point moves away from
moves toward point A. (acceleration) point B.

Similarly when working at point A and Similarly when working at point B

increase in speed will make load torque greater and increase in speed will make motor
than the motor torque, which will move the torque greater than the load torque,
operating point toward point B which will move the operating point
(deacceleration) away from point B
 Practice Problems

The torque speed characteristic of motor and load torque are shown in Figure . The
load torque is equal to motor torque at point P, Q, R and S . Find out the stable
operating points for both the cases.

Practice more problem from G K dUBEY

 Mathematical Analysis of steady state stability
Let the equilibrium of the torques and speed is T,Tl ,  and the small deviations are T , Tl ,
After the displacement from the equilibrium state the torque equation becomes,
d    
T  T   Tl  Tl   J
dt
d   d   
T  Tl  J  T  Tl  J
dt dt
If we assume that these increments are so small that may be expressed as linear functions of the
 dT   dTl 
change in speed, then T     , T    
 d  
l
 d  
 dT   dTl  d    1   dT   dTl   d 
          
    
J   d   d  
J
 d   d  dt dt
1    dTl   dT   
d  1   dTl   dT        t

    0    mo e
J    d   d   
   
dt J   d   d  
  dT   dT  
if   l       0  systemconvergetowards zero (exp onential decay )
 d    d  
 dT   
dT
Condition for stabiity  l    
 d    d  
 Practice Problems

A drive has the following equation.

Obtain equilibrium points and determine steady state stability.

T   1  2m  and Tl   3 m

Practice more problem from G K dUBEY

Equivalent Drive parameters

 Different parts of a load

may be coupled through
different mechanisms,
such as gears, V-belts and
crankshaft.
 These parts may have
different speeds and
different types of
Motions such as
rotational and
translational.

This section presents

finding the equivalent
moment of inertia (J) of
motor-load system and
equivalent torque
components, all referred
to motor shaft.
Let us consider a motor driving two loads,
• one coupled directly to its shaft and
• other through a gear with n and n1 teeth as shown in Figure

• Let the moment of inertia of motor and load directly coupled to its shaft be J0,
motor speed and torque of the directly coupled load be ωm and Tl0 respectively.

• Let the moment of inertia, speed and torque of the load coupled through a
gear be J1, ωm1 and Tl1 respectively.

• Now, m1 n
  a1 where a1 is the gear tooth ratio.
m n1
 If the losses in transmission are neglected, then the kinetic energy due to equivalent
inertia must be the same as kinetic energy of various moving parts. Thus
2
1 1 1  m1 
J m  J 0m 0  J1m1  J  J 0  J1    J  J 0  J1  a1 
2 2 2 2

2 2 2  m 
Power at the loads and motor must be the same. If transmission efficiency of the gears
be η1, then
Tl1m1 Tl1a1
Tlm  Tl 0m   Tl  Tl 0 
1 1
where Tl is the total equivalent torque referred to motor shaft.
If in addition to load directly coupled to the motor with inertia J0 there are m other loads with
moment of inertias J1, J2, . . . , Jm and gear teeth ratios of a1, a2, . . . am then

J  J 0  J1  a1   J 2  a2   .....  J m  am 
2 2 2

If m loads with torques Tl1, Tl2, . . . , Tlm are coupled through gears with teeth ratios a1, a2, . . .
am and transmission efficiencies η1, η2 , . . . , ηm, in addition to one directly coupled, then
Tl1a1 Tl 2 a2 Tlm am
Tl  Tl 0    ....... 
1 2 m
Practice for load with translational and rotational motion
Find equivalent Moment of inertia and equivalent torque
 Practice Problems
A motor has two loads. Load-1 has rotational motion which is coupled to the
motor through a reduction gear with gear ratio gr1 =10 and efficiency of
90%. Load-1 has a moment of inertia of 10 kg.m2 and torque of 10 N-m. Load-2
has translation motion and consists of 1000 kg weight to be lifted up at uniform
speed of 1.5 m/s. The coupling between load-2 and the motor has an efficiency
of 85%. The motor has inertia of 0.2 kg.m2 and runs at constant speed of 1420
rpm. Determine the equivalent inertia referred to the motor shaft and the power
developed by the motor

Practice more problem from G K dUBEY

 Components of Load Torques:
Friction torque TF :
Friction will be present at the motor shaft and also in various parts of the load. TF is
equivalent value of various friction torques referred to the motor shaft.
(ii) Windage torque, Tw :
When a motor runs, wind generates a torque opposing the motion. This is known as
windage torque.
(iii) Torque required to do the useful mechanical work, TL:
Nature of this Components of Load Torques depends on particular application. It may be
constant and independent of speed; it may be some function of speed; it may depend on
the position or path followed by load; it may be time invariant or time-variant; it may
vary cyclically and its nature may also change with the load’s mode of operation.
 Variation of friction torque with speed

TF  TS  TC  TV  TW
Ts : Friction at zero speed is called stiction or static friction. In order for drive to start, the
motor torque should at least exceed stiction. Ts accounts for additional torque present at
standstill. Since Ts is present only at standstill it is not taken into account in the dynamic
analysis.
Another component Tc, which is independent of speed, is known as Coulomb friction.

Viscous Component Tv which varies linearly with speed is called viscous friction and is given
by:
TV  Bm  TV  m
where B is the viscous friction coefficient.

Windage torque Tw, which is proportional to speed squared, is given by TW  Cm  Tw  m

2 2

where C is a constant.

From the above discussion, for finite speeds, TF  TS  TC  Bm  Cm2

 Classification of Load torque
1. In fans, compressors and aeroplanes, the windage dominates, consequently, load torque is
proportional to speed squared (Fig.(a)).
 Windage is the opposition offered by air to the motion. Similar nature of
Classification of Load Torques can be expected when the motion is opposed by any other
fluid, e.g. by water in centrifugal pumps and ship-propellors, giving the same characteristic
as shown in Fig. (a).
2. In a high speed hoist, viscous friction and windage also have appreciable magnitude, in
addition to gravity, thus giving the speed-torque curve of Fig. (b).
3. Nature of speed-torque characteristic of a traction load when moving on a levelled ground is
shown in Fig. (c). Because of its heavy mass, the stiction is large. Near zero speed, net torque
is mainly due to stiction. The stiction however disappears at a finite speed and then windage
and viscous friction dominate. Because of large stiction and need for accelerating a heavy
mass, the motor torque required for starting a train is much larger than what is required to run
it at full speed.
Classification of Load Torques can be broadly classified into two categories-

1. Load torques which have the potential to drive the motor under equilibrium condition
are called Active Load Torques.

 Such load torques usually retain their sign when the direction of the drive
rotation is changed.
 Torque(s) due to gravitational force, tension, compression and torsion,
undergone by an elastic body, come under this category.

2. Load torques which always oppose the motion and change their sign on the reversal of
motion are called Passive Load Torques.
 Such torques are due to friction, windage, cutting etc.
 Different types of drives.
1. Individual Drive 2. Group Drive 3. Multi motor Drive
Individual drive
• In an electric drive system, if an individual machine is fitted with its
own motor and each operator has complete control on their
machine, then it is termed as an individual electric drive.
• Examples − Drill machine, lath machine, etc.

Multi motor drive

 A mechanism in which separate motors are used for operating
different parts of the same machine is known as a multi-motor
electric drive.
 For example, in travelling cranes, there are three motors : one for
hoisting, another for long travel motion and the third for cross
travel motors.
 Group Drive
• When several machines are organized on a single shaft and are driven by a single large
motor, this system is known as group drive.
• Group drive is also known as line shaft drive because a large sized motor (called main
motor) drives a common shaft. To the common shaft, a number of small machines are
connected by means of multi-stepped pulleys.
Advantages of group drives include the following −
1. As group drive uses a large sized single motor, so there is considerable saving in cost.
2. Group drive has low maintenance cost.
3. In case of group drive, the overload occurring for main motor is very less.
4. When the main motor operates near rated capacity, then the power factor and efficiency
of the group drive be better.
Disadvantages of group drives include the following −
• If there is a fault in main motor, then it causes stopping of all the small machines.
• In group drive, the small machines have to be installed at places which is convenient for
shafting.
• .In group drive, the speed control of individual small machine is not possible
• In case of group drive, high amount of power is wasted in power transmission (in pulleys,
etc.).
The main applications of group drives are
Grain processing industries, Food grinding mills, Paper mills, Textile mills
4 quadrant operation
You can explain in detail by following g k dubey
 Selection of Motor Power Rating

1. When a motor operates, heat is produced due to losses (copper, iron and friction)
inside the machine and its temperature rises.
2. As the temperature increases beyond ambient value, a portion of heat produced
flows out to the surrounding medium.
3. The amount of outflow of heat is a function of temperature rise of motor above the
ambient value.
4. As motor temperature rises, the heat outflow increases and equilibrium ultimately
sets in when the heat generated becomes equal to heat dissipated into the
surrounding medium.
5. Motor temperature then reaches a steady state value. Steady state temperature
depends on power loss, which in turn depends on the output power of the machine.
6. Since temperature rise has a direct relationship with the output power, it is
termed Thermal Loading on the machine

Heat applied=heat absorbed (stored) in the machine +heat dissipated

Selection of Motor Power Rating

Insulation Temperature Limit

 Let the machine, which is assumed to be a homogeneous body, and the cooling medium
has following parameters at time t:
P1 = Heat developed, joules/sec or watts.
P2 = Heat dissipated to the cooling medium, joules/sec or watts.
W = Weight of the active parts of machine, kg.
h = Specific heat, Joules per kg per °C.
A = Cooling surface, m2.
d = Coefficient of heat transfer or specific heat dissipation, joules/sec/m2/°C.
θ = Mean temperature rise, °C.
During a time increment dt, let the machine temperature rise be dθ. Since,
Heat absorbed (stored) in the machine = heat developed inside the machine – heat
dissipated to the surrounding medium

p 1 dt  whd  p2 dt
 whd  p 1 dt  p2 dt  p 1 dt  dA dt  p 1 dt  D dt
C d p1
 Cd  p 1 dt  D dt   
D dt D
C = Thermal capacity of the machine watts/degree Celsius.
D= Heat dissipation constant watts/degree Celsius


    ss 1  e
t

  i e
t

When motor temperature reach
steady state value t tends to ∞

where  
C
 ss 
p1 P1
   ss  1  e 
D
 t

D D Machine is continuously heated by
power P1.

If motor is completely cool down i.e disconnected from the supply

t
 f   ss  0    i e 