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UNIT III
Electrical Drives
Control of Electrical Drives
1 / 9

Electronic drive control is the process of controlling the output of a motor, or
the mechanical energy it produces. It involves the use of an electric drive
control unit (EDCU) to manage the drive system, high-voltage supply, and
other subsystems.
How does it work?
•The EDCU is the central processing unit for the drive system.
•It controls the motor's output and response in relation to a controlling input.
•The EDCU coordinates subsystems and auxiliary units.

What is an Electrical Drive?
When we hear “electric motor” or “electrical generator,” we usually think
their speed is controlled only by the applied voltage and source current
frequency.
However, the speed of an electrical machine can be precisely controlled using
an electrical drive.
The main advantage of an electrical drive is that the motion control can be
easily optimized.
Simply put, electrical drives control the motion of electrical machines. A
typical drive system includes one or more electric motors and a control
system to manage the motor shaft’s rotation.
Today, software makes controlling these systems easy and very accurate,
making electrical drives user-friendly.

Parts of Electrical Drive
The main parts of the electrical drives are power modulator, motor, controlling unit and
sensing units.Their parts are explained below in details.
Power Modulator – The power modulator regulates the output power of the source. It
controls the power from the source to the motor in such a manner that motor transmits the
speed-torque characteristic required by the load. During the transient operations like starting,
braking and speed reversing the excessive current drawn from the source. This excessive
current drawn from the source may overload it or may cause a voltage drop. Hence the power
modulator restricts the source and motor current.
The power modulator converts the energy according to the requirement of the motor e.g. if
the source is DC and an induction motor is used then power modulator convert DC into AC. It
also selects the mode of operation of the motor, i.e., motoring or braking.

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 other
type. Depending on the type of function, converters can be divided into 5 types -
1. AC to DC converters
2. AC regulators
3. Choppers or DC - DC converters
4. Inverters
5. Cycloconverters

Control Unit – The control unit controls the power modulator which operates at
small voltage and power levels. The control unit also operates the power
modulator as desired. It also generates the commands for the protection of
power modulator and motor. An input command signal which adjusts the
operating point of the drive, from an input to the control unit.
Sensing Unit – It senses the certain drive parameter like motor current and
speed. It mainly required either for protection or for closed loop operation.

Advantages of Electrical
Drive
•The electric drive has very large range of torque, speed and power.
•Their working is independent of the environmental condition.
•The electric drives are free from pollution.
•The electric drives operate on all the quadrants of speed torque plane.
•The drive can easily be started and it does not require any refuelling.
•The efficiency of the drives is high because fewer losses occur on it.

Disadvantages of Electrical Drive
1.The application of the drive is limited because it cannot use in a place where the
power supply is not available.
2.It can cause noise pollution.
3.The initial cost of the system is high.
4.It has a poor dynamic response.
5.The output power obtained from the drive is low.
6.During the breakdown of conductors or short circuit, the system may get
damaged due to which several problems occur.

The classification of electrical drives can be done depending upon the
various components of the drive system. Now according to the design, the
drives can be classified into three types such as single-motor drive, group
motor drive and multi motor drive. The single motor types are the very basic
type of drive which are mainly used in simple metal working, house hold
appliances etc. Group electric drives are used in modern industries because
of various complexities. Multi motor drives are used in heavy industries or
where multiple motoring units are required such as railway transport. If we
divide from another point of view, these drives are of two types: 1. Reversible
types drives 2. Non reversible types drives. This depends mainly on the
capability of the drive system to alter the direction of the flux generated. So,
several classification of drive is discussed above.

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 Switching circuits in motors and electrical drives motor are used in AC
motors. 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 from the main circuit during any
abnormal condition or faults. Sources 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.
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. Motors of low and medium powers are fed
from 400 V supply, and higher ratings like 3.3kv, ControlUnit 6.6kv, 11kv etc are provided also. -
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. So, the above discussion provides us a simple
concept about the several parts of electrical drive.

•Dynamics of Electrical Drives Definition: The dynamics of electrical drives explain how
motors and loads interact, especially when their speeds differ.
•Key Components: Important components include the polar moment of inertia (J), angular
velocity (Wm), motor torque (T), and load torque (T1).
•Fundamental Torque Equation: This equation shows that motor torque balances load torque
and dynamic torque, which is crucial during changes in motion.
•Dynamic Torque: Dynamic torque, J(dωm/dt), appears only during transient operations like
starting or stopping, indicating acceleration or deceleration.
•Impact on Motion: By analyzing dynamic torque, we can determine if the motor is accelerating
or decelerating, which is essential for efficient drive operation.

When an electric motor rotates, it is typically connected to a load with either
rotational or translational motion. The motor’s speed can differ from the load’s
speed. To understand this relationship, we use the concept of the dynamics of
electrical drives.
We can easily describe the dynamics of an electrical drive with the following
example.
Here,
J = Polar moment of inertia of motor load
Wm = Instantaneous angular velocity
T = Instantaneous value of developed motor torque
T1 = Instantaneous value of load torque referred to motor shaft
Now, from the fundamental torque equation –

For drives with constant inertia
This equation shows that motor torque balances both load torque and
dynamic torque, represented as J(dωm/dt). Dynamic torque appears only
during transient operations, like starting or stopping.

Fundamentals Torque Equations Multiquadrant
Operation
Four Quadrant Operation
1 Fundamentals Torque Equations
2 Multiquadrant Operation
3 Four Quadrant Operation of a Motor Driving a Hoist Load
Dynamics of Electrical Drives
2 /
17

Operation
Introduction
T
Tl
ωm
Motor Load
Figure 1. Equivalent motor-load system.
• A motor generally drives a load (machine) through some transmission system.
• While motor always rotates, the load may rotate or may undergo a translational
motion.
• It is convenient to represent the motor load system by an equivalent rotational
system.
• Here, J → moment of inertia of motor-load system referred to the motor shaft,
kg-m2.
𝜔m → instantaneous angular velocity of motor shaft, rad/s.
𝜏 → instantaneous value of developed motor torque, N-m.
𝜏l → instantaneous value of load (resisting) torque, referred to motor shaft, N-m.
Dynamics of Electrical Drives 3 / 17

Operation
Introduction
• Load torque includes friction and windage torque of motor.
• Motor-load system is described by the following fundamental torque equation,
𝜏 − 𝜏l =
d d𝜔m dJ
dt
(J 𝜔m) = J
dt
+ 𝜔m
dt
.
(1)
• Eq. 1 is applicable to variable inertia drives such as mine winders, reel drives,
and industrial robots.
• For drives with constant inertia, (dJ/dt) = 0. Therefore,
𝜏 = 𝜏l + J
d𝜔m
.
(2)
dt
• Eq. 2 shows that torque developed by the motor is counter balanced by a load
torque 𝜏l and a dynamic torque J(d𝜔m/dt).

Operation
Introduction
• Drive accelerates or decelerates depending on whether 𝜏 is greater or less than 𝜏l .
• During acceleration → motor should supply not only the 𝜏l but an additional
torque component J(d𝜔m/dt) in order to overcome the drive inertia.
• In drives with large inertia (such as electric trains) 𝜏 must exceed 𝜏l by a large
amount in order to get adequate acceleration.
• In drives requiring fast transient response → 𝜏 should be maintained at
the highest value and motor-load system should be designed with a lowest
possible inertia.
• Energy associated with dynamic torque J(d𝜔m/dt) is stored in the form of kinetic
energy given by J 𝜔2
/2.
m
• During deceleration → J(d𝜔m/dt) is −ve sign.
• ∴ it assists the 𝜏 and maintains drive motion by extracting energy from
stored kinetic energy.

Operation
6 /
17

Operation
Speed Torque Conventions and Multiquadrant Operation
• Multiquadrant operation of drives →
useful to establish suitable conventions
about the signs of torque and speed.
• Motor speed is considered +ve when rotat-
ing in the forward direction.
• For drives that operate only in one
direction, forward speed will be their
normal speed.
• In loads involving up-and-down motions,
the speed of motor which causes upward
motion is considered forward motion.
I
II
III IV
ω m
T
Forward
motoring
Forward
braking
Reverse
motoring
Reverse
braking
Figure 2. Multiquadrant-operation of drives.

Operation
• For reversible drives, forward speed is chosen arbitrarily.
• The rotation in the opposite direction → reverse speed and assigned −ve sign.
• Positive motor torque is defined as the torque which produces acceleration or the
positive rate of change of speed in forward direction.
• According to Eq. 2, +ve 𝜏l is opposite in direction to the +ve motor torque.
• Motor torque is considered −ve if it produces deceleration.
• Motor operates in two modes → motoring and braking.
• Motoring → converted electrical energy to mechanical energy, which supports its
motion.

Operation
• Braking → it works as a generator converting mechanical energy to electrical
energy, and thus, opposes the motion.
• Motor can provide motoring and braking operations for both forward and reverse
directions.
• Power developed by a motor is given by product of speed and torque.
• In quadrant I → developed power is +ve → forward motoring.
• In quadrant II → developed power is −ve → machine works under braking
opposing the motion → forward braking.
• Quadrant III and IV → reverse motoring and braking, respectively.

Operation
10 /
17

Operation
I
II
III IV T
Quadrants
Tl2 Tl1 Counter
weight
Motion
Loaded
cage
Counter
weight
Motion
Empty
cage
Counter
weight
ωm
Motion
Loaded
cage
Counter
weight
ωm
Motion
Empty
cage
Load torque with
empty cage
Load torque with
loaded cage
Four Quadrant Operation of a Motor Driving a Hoist Load
τ l
ωm
ωm
τ l
ωm
τ l
τ l
Figure 3. Four quadrant operation of a motor driving a hoist load.
11 /
17

Operation
• Consider operation of a hoist in four quadrants as shown in Fig. 3.
• A hoist consists of a rope wound on a drum coupled to the motor shaft.
• One end of the rope is tied to a cage which is used to transport man or material
form one level to another level.
• Other end of the rope has a counter weight.
• Here, the weight of the counter weight is chosen to be higher than the weight of
an empty cage but lower than a fully loaded cage.
• Forward direction of the motor speed will be one which gives upward motion of
the cage.

Operation
• Load torque is shown to be constant and independent of speed.
• Gravitational torque does not change its sign even when the direction of the driving
motor is reversed.
• Load torque line 𝜏l1 in quadrants I and IV → speed-torque characteristics for the
loaded hoist.
• This torque is the difference of torques due to loaded hoist and counter weight.
• Load torque line 𝜏l2 in quadrants II and III → speed-torque characteristics for an
empty hoist.
• This torque is the difference of torques due to counter weight and the empty hoist.
• Its sign is −ve since the weight of a counter weight is always higher than that of
an empty cage.

Operation
• Quadrant-I operation of a hoist requires the movement of the cage upward, which
corresponds to the positive motor speed which is in anticlockwise direction here.
• This motion will be obtained if the motor produces +ve torque in anticlockwise
direction equal to the magnitude of 𝜏l1.
• Since the developed motor power is +ve → forward motoring operation.
• Quadrant IV operation is obtained when a loaded cage is lowered.
• Since the weight of a loaded cage is higher than that of a counter weight, it is able
to come down due to the gravity itself.
• To limit the speed of cage within safe value, motor must produce a +ve torque
𝜏 = 𝜏l2 in anticlockwise direction.
• As both power and speed are −ve → drive is operating in reverse braking.

Operation
• Operation in quadrant II is obtained when an empty cage is moved up.
• Since a counterweight is heavier than an empty cage, it is able to pull it up.
• In order to limit the speed within a safe value, the motor must produce a braking
torque equal to 𝜏l2 in clockwise (−ve) direction. Since speed is +ve and
developed power −ve → forward braking operation.
• Operation in quadrant III is obtained when an empty cage is lowered.
• Since an empty cage has a lesser weight than a counter weight, the motor should
produce a torque in clockwise direction.
• Since speed is −ve and developed power +ve → reverse motoring operation.

Loads with Rotational Motion Loads with Translational Motion Moment of Inertia Load Torques
1 Loads with Rotational Motion
2 Loads with Translational Motion
3 Moment of Inertia
4 Components of Load Torques
2 / 23

Equivalent Values of Drive Parameters
• Different parts of a load → coupled through different mechanisms
such as gears, V-belts and crankshaft.
• These parts have different speeds and different types of motions → rotational and
translational.
Loads with Rotational Motion:
• Consider a motor driving two loads, one coupled directly to its shaft and other
through a gear with n and n1 teeth.
• 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 𝜏l0, respectively.
• Let the moment of inertia, speed and torque of the load coupled through a gear be
J1, 𝜔m1 and 𝜏l1, respectively.
3 / 23

Loads with Rotational Motion
Motor
Load
Tllo
Load
Tlo
Motor
Rotational to
Linear Motion
transmission
Mass
Mt
Force
Fl
vl
J0
J0
ωm
ωm
ωm
ωml
J1
Load
Tl1
( a ) Load with rotational motions
( b ) Load with translational and rotational and linear motions
Figure 1. (a) Loads with rotational motion and (b) loads with transnational and rotational motion.
Gear
n
n1
• Now,
𝜔m1
=
n
= a1.
𝜔m n1
Where, a1 is the gear tooth ratio.
(1)
4 / 23

• If the losses in transmission are neglected, then the kinetic energy due to
equivalent inertia must be the same as the kinetic energy of various moving
parts. Thus,
J 𝜔 = J
m 0
2 2
1 1
1
2
2
2
m 1
𝜔 +
J 𝜔
2
m1
(2)
• From Eqs. 1 and 2,
1
J = J0 + a2 J1.
(3)
• Power at the loads and motor must be the same. If transmission efficiency of the
gears be 𝜂1, then 𝜏l 1 𝜔m1
𝜏l 𝜔m = 𝜏l0 𝜔m + .
(4)
𝜂1
Where, 𝜏l is the total equivalent torque referred to
the motor shaft.
• From Eqs. 1 and 4, 𝜏l = 𝜏l0 +
a 𝜏
𝜂1
1 l1
. (5)
5 / 23

• 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
1 2 m
J = J0 + a2 J1 + a2 J2 + ........... a2 Jm.
(6)
• If m loads with torques 𝜏l1, 𝜏l2, ........, 𝜏l
m are coupled through gears with teeth
ratios a1, a2, ....... am and transmission efficiencies 𝜂1, 𝜂2, ........., 𝜂m, in
addition to one directly coupled, then
𝜏l = 𝜏l1 +
a1 𝜏l1
+
a2 𝜏l2
+ ........ +
am 𝜏l
m
.
(7)
𝜂1 𝜂2
𝜂m
• If loads are driven through a belt drive instead of gears, neglecting slippage, the
equivalent inertia and torque can be obtained from Eqs. 6 and 7.
6 / 23

3 Moment of Inertia
4 Components of Load
Torques
7 / 23

Loads with Translational Motion
• Consider a motor driving two loads, one coupled directly to the shaft and the
other through a transmission system converting rotational motion to linear
motion.
• Let the moment of inertia of the motor and load directly coupled to it be J0,
𝜏l directly coupled to motor be 𝜏l0, and mass, velocity and force of load with
translational motion be M1 (kg), v1 (m/s) and F1 (N), respectively.
• If the transmission losses are neglected, then kinetic energy due to equivalent
inertia J must be the same as kinetic energy of various moving parts. Thus,
2
J 𝜔 = J
2 2
m 0
m
1 1
1
2
2
1
2
1
𝜔 + M v . (8)
or
J = J0 + M1
v1
𝜔m
2
. (9)
8 / 23

Loads with Translational Motion
• ly, power at the motor and load should be the same, thus if efficiency of
transmission be 𝜂1
𝜂1
𝜏l 𝜔m = 𝜏l0 𝜔m +
F1 v1
(10)
l l0
𝜏 = 𝜏 +
F v
1 1
𝜂1 𝜔m
. (11)
• If, in addition to one load directly coupled tot he motor shaft, there are m other
loads with translational motion with velocities v1, v2, ......vm and masses M1, M2,
......Mm, respectively, then
J = J0 + M1
v1
𝜔m
2
+ M2
v2
𝜔m
2
+ ...... + M m v
m
𝜔m
2
. (12)
and
l l0
𝜏 = 𝜏 +
F v
1 1
𝜂1 𝜔m
+
F v
2 2
𝜂2 𝜔m
+ ...... +
F v
m m
𝜂m 𝜔m
. (13)
9 / 23

3 Moment of Inertia
10 /
23

Measurement of Moment of Inertia
• Moment of inertia is calculated if dimensions and weights of various parts of the
load and motor are known.
• Measured experimentally → retardation test.
• In retardation test → drive runs at a speed slightly higher than rated speed and
then the supply to it cut off.
• Drive continues to run due to kinetic energy stored in it and decelerates due to
rotational mechanical losses.
• At any speed 𝜔m, power P consumed in supplying rotational losses is given by
d
1
2
P = J 𝜔 =
J 𝜔
m m
d𝜔m
dt 2
dt
. (14)
• From retardation test d𝜔m/dt at rated speed is obtained.
11 /
23

• Now, drive is connected to the supply and run at rated speed and rotational
mechanical power input P to the drive is measured.
• However, the rotational mechanical losses cannot be measured accurately since
the core losses and rotational mechanical losses cannot be separated.
• Hence, the retardation test on a dc separately excited motor or a synchronous
motor is carried-out with field on.
• Now, core loss is included in the rotational loss, which is the difference between
armature power input and armature copper loss.
• In case of a wound rotor IM → retardation test can be carried-out by keeping the
stator supply and opening the rotor winding connection.
12 /
23

A
P1
ωm vs t
P vs t curve
t1 t
ωm
P vs ωm
ωm1
ωm1
p1
P ,ω ωm0
Figure 2. Graphical method of determination of the equivalent moment of inertia.
• J can be determined more accurately by obtaining the speed time curve from the
retardation test and also rotational versus speed plot shown in Fig. 2.
• Using these two plots, rotational losses versus time plot can be obtained.
• Area A enclosed between the rotational loss versus t plot and the time axis
(shaded area) → kinetic energy dissipated during retardation test.
• If the initial speed of the drive during retardation test is 𝜔m0 then
2 m0
1
J 𝜔2
= A.
(15)
13 /
23

3 Moment of Inertia
14 /
23

Components of Load Torques
• Load torque 𝜏l → friction torque 𝜏f and windage torque 𝜏W .
• (i) Friction torque, 𝜏f : Friction will be present at the motor shaft and also in
various parts of the load. 𝜏f is equivalent value of various friction torques referred
to the motor shaft.
• (ii) Windage torque, 𝜏w : When a motor runs, wind generates a torque opposing
the motion.
• (iii) Torque required to do the useful mechanical work, 𝜏l : This torque depends on
the particular application. It may be constant and independent of speed. It
may depend on the position or path followed by load.
15 /
23

Torque
Tc
Tv
Ts
Ts
Tc
0
0
ωm
ωm
Tf
( a ) ( b )
Figure 3. Friction torque and its components.
• Variation of friction torque with speed as shown in Fig. 3. Its value at standstill is
much higher than its value slightly above zero speed.
• Friction at zero speed → stiction or static friction.
• Friction torque is resolved into three components.
• Component 𝜏v which varies lineraly with speed is called viscous friction and given
as,
𝜏v = B 𝜔m.
(16)
16 /
23

• Another component 𝜏c which is independent of speed → Coulomb friction.
• Third component 𝜏s accounts for additional torque present at standstill. Since 𝜏s
is present only at standstill, it is not taken into account in the dynamic analysis.
• Windage torque 𝜏w is given as
𝜏w = C 𝜔2 .
(17)
m
Where, C is constant.
• For finite speed,
m
𝜏l = 𝜏l + B 𝜔m + 𝜏c + C 𝜔2
.
(18)
m
• In many applications 𝜏c + c 𝜔2
is very small compared to B 𝜔m and neglisable
compared to 𝜏l .
m
• In order to simplify the analysis, term 𝜏c + C 𝜔2 is accounted by updating the
value of viscous friction coefficient B. Then,
dt
𝜏 = J
d𝜔m
+ 𝜏l + B 𝜔m.
(19)
17 /
23

• If there is a torsional elasticity in shaft coupling the load to the motor, an
additional component of load torque known as coupling torque, will be
present.
• Coupling torque 𝜏e is given as
𝜏e = Ke 𝜃e.
(20)
Where, 𝜃e is the torsion angle of coupling (radians) and Ke the rotational stiffness
of the shaft (N-m/rad).
• In most applications, shaft can be assumed to be perfectly stiff and 𝜏e can be
neglected.
• There is potential energy associated with coupling torque and kinetic energy with
dynamic torque.
• Exchange of energy between these two energy storage tends to
produce oscillations which are damped by viscous friction torque b 𝜔m.
18 /
23

Nature and Classification of Load Torques
Tll
ωm
0
(a) Tl α ω2
m
Tll
ωm
0
(b) High speed hoist (c) Traction load Tll
ωm
0 Tl
ωm
0
(d) Constant power load
• A low speed hoist → torque is constant and independent of the speed.
• Paper mill drive → torque is independent of speed.
• Fans, compressors, aeroplanes, centrifugal pumps, ship-propellors, high speed
hoists, traction → load torque is a function of speed.
19 /
23

• Fans, compressors, aeroplanes → windade dominates → 𝜏L ∝ 𝜔2
shown in Fig. 4 (a).
• Windage is the opposition offered by air to the motion.
• Similar nature of 𝜏L an be expected when the motion is opposed by any other fluid,
example, by water in centrifugal pumps and ship-propellors → Fig. 4 (a).
• High speed hoist → viscous friction and windage also have appreciable magnitude,
in addition to gravity → Fig. 4 (b).
• Traction load → since its heavy mass, the stictiuon is large → → Fig. 4 (c).
• Torque in a coiler drive → hyperbolic in nature → Fig. 4 (d). The developed
power is nearly constant at all speeds.
20 /
23

• Load torque → active and passive loads.
• 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 due to gravitational force, tension, compression and torsion, undergone by
elastic body, comes under this category.
• 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.

Time and Energy-Loss Steady State Stability
1 Time and Energy-Loss
2 Steady State Stability
2 / 13

Calculation of Time and Energy-Loss in Transient Operations
• Transient operations → starting, braking, speed change and speed reversal.
• The time taken and energy dissipation in motor during the transient operations
can be evaluated by solving Eq. 1 along with motor circuit
equations.
𝜏 = J
d𝜔m
+ 𝜏l + B 𝜔m.
(1)
dt
• When 𝜏 and 𝜏l are constants or ∝ 𝜔2 → Eq. 1 is first order linear differential
equation → can be solved analytically.
• When 𝜏 and 𝜏l is neither constant nor ∝ 𝜔2 → non-liner differential equation →
can be solved by Runga-Kutta method.
• The transient operation is considered to be over when 95 % change in speed has
taken place.
• When speed changes from 𝜔m1 to [𝜔m1 + 0.95 (𝜔me − 𝜔m1)] is considered to
3 / 13

Calculation of Time and Energy-Loss in Transient Operation
• Transient time and energy loss can also be computed with satisfactory accuracy
using steady-state torque and speed-current curves of motor and
speed-torque curve of load.
• This is because the mechanical time constant of a drive is usually very large
compared to the electrical time constant of motor.
• Consequently, electrical transients die down very fast and motor operation can
occur along the steady-state speed-torque and speed-current curves.
𝜏 (𝜔m) − 𝜏l (𝜔m)
dt =
J d𝜔m
.
(2)
Where, 𝜏 (𝜔m) and 𝜏l (𝜔m) indicate that the motor and load torques are
functions of drive speed 𝜔m.
• Time taken for drive speed to change from 𝜔m1 to 𝜔m2 is obtained by
t = J
∫ 𝜔m2
𝜔m1
d𝜔
m
𝜏 (𝜔m) − 𝜏l (𝜔m)
. (3)
4 / 13

t=shaded area
J
T(ω
m
)
-
T
l
(ω
m
)
0
ωme ωm
ωm1
ωm2
• The area between the reciprocal of the acceleration J/[𝜏 (𝜔m) − 𝜏l (𝜔m)] versus
𝜔m curve and 𝜔m-axis shown in Fig. 1.
• The transient time can be evaluated by measuring this area.
5 / 13

• When 𝜔m2 is an equilibrium speed 𝜔me → the reciprocal of acceleration will
become infinite at 𝜔me.
• Consequently, time evaluated this way will be infinite.
• ∴ transient time is computed by measuring the area between speeds 𝜔m1 and
𝜔m1 + 0.95(𝜔m2 − 𝜔m1).
• Energy dissipated in a motor winding during a transient operation is given by
∫ t
2
E = R i
dt.
(4)
0
Where, R is the motor winding resistance and i is the current flowing through it.
• The area enclosed between the curve and time axis multiplied by R → gives
energy dissipated in the motor winding.
6 / 13

1 Time and Energy-Loss
2 Steady State Stability
7 / 13

Steady State Stability
T
A
Tl
Δωm
ωm
0
Torque
T
B Tl
ωm
0
Torque
Tl
T
C
0 Torque
ωm
Tl
T
D
0 Torque
ωm
• Equilibrium speed of a motor-load system is obtained when 𝜏m = 𝜏l .
• Drive will operate in steady-state at this speed, provided it is the speed of stable
equilibrium.
• In most drives, the electrical time constant of the motor is negligible compared to
its mechanical time constant.
• ∴ during transient operation, motor is assumed to be in electrical
equilibrium, implying that steady-state speed curves are also applicable to the
transient operation.
8 / 13

• Steady state stability of equilibrium point A → stable, when the operation will
restored to it after a small departure from it due to a disturbance in the motor or
load.
• Let the disturbance cause a reduction of ∆ 𝜔m in speed.
• At new speed, 𝜏m > 𝜏l → motor will accelerate and operation will be restored to
A.
• Similarly, an increase of ∆ 𝜔m in speed caused by a disturbance will make 𝜏l >
𝜏m → deceleration and restoration of operation to point A.
• Hence, the drive is steady-state stable at point A.
• Examine equilibrium point B which is obtained when the same motor drives
another load.
• A decrease in speed causes the 𝜏l to become > 𝜏m → drive decelerates and
operating point moves away from B.
9 / 13

10 /
13
• Similarly, when working at B, an increase in speed will make 𝜏m < 𝜏l ,
which will move the operating point away from B.
• Thus, B is an unstable point of equilibrium.
• Hence, an equilibrium point will be stable when as increase in speed causes 𝜏l
to exceed 𝜏m, i.e., at the equilibrium point, the following condition is satisfied.
d𝜔m d𝜔m
d𝜏l d𝜏
.
> (5)
• Let a small perturbation in speed, ∆𝜔m, results in ∆ 𝜏 and ∆ 𝜏l
perturbations in 𝜏 and 𝜏l , respectively. Then,
(𝜏 + ∆ 𝜏) = (𝜏l + ∆ 𝜏l ) + J
d (𝜔m + ∆𝜔m)
(6)
dt
or
𝜏 + ∆ 𝜏 = 𝜏l + ∆ 𝜏l +
J
d𝜔m d∆𝜔m
dt
dt
+ J . (7)
• Substracting Eq. 1 from 11 and rearranging terms gives
dt
J
d∆𝜔m
= ∆𝜏 − ∆𝜏l .
(8)

11 /
13
• For small perturbations, the speed-torque curves of the motor and load can be
assumed to be straight lines.
d𝜏
∆𝜏 =
d𝜔m
∆𝜔m.
(9)
d𝜔m
d𝜏l
∆𝜏l = ∆𝜔m.
(10)
Where, d𝜏/d𝜔m and d𝜏l /d𝜔m are slopes of steady-state speed-torque curves of
motor and load at operating point under consideration.
• Substituting Eqs. 9 and 10 into 8 and rearranging the terms yields
J
m l
d𝜏 d𝜏
d∆𝜔
dt d𝜔m
d𝜔m
m
+ − ∆𝜔 = 0. (11)
• This is a first-order linear differential equation. If the initial deviation in speed at
t = 0 be (∆𝜔m)0, then the solution of Eq. 11 will be
∆𝜔m = ∆𝜔m0 exp −
l
1 d𝜏 d𝜏
J d𝜔m d𝜔m
– t . (12)
• An operating point will be stable when ∆𝜔m approaches zero as t approaches
infinity.

Load Equalisation
1 Load Equalisation
2 /
13

Load Equalisation
Load Equalisation
3 /
13
• In some drive applications, 𝜏l fluctuates widely within short intervals of time.
• In pressing machines, a large torque of short duration is required during pressing
operation. Otherwise, the torque is nearly zero.
• Electric hammer, steel rolling mills and reciprocating pumps drive → motor is
required to supply peak torque demanded by load, the first motor rating has to be
high.
• Secondly, the motor will draw a pulsed current from the supply.
• When the amplitude of pulsed current forms an appreciable proportion of supply
line capacity, it gives rise to line voltage fluctuations, which adversely affect
other loads connected to the line.
• In some applications, peak load demanded may form major proportion of the
source capacity itself, as in blooming mills, then load fluctuations may also
adversely affect the stability of the source.

Load Equalisation
Load Equalisation
4 /
13
Tr Tmax
T
C
B
A
ωm
ωmo
ωmr
0
Figure 1. Shapes of motor speed torque curves for
fluctuating loads.
0
t
Tlh
Tmax
Tl Tl
ω m
ω m
Tmin
Tll
T T
th
tl
Figure 2. Variation of motor and load torques and
speed for a periodic load for a drooping
motor speed-curve.

Load Equalisation
Load Equalisation
5 /
13
• The problems of fluctuating loads are overcome by mounting a flywheel on the
motor shaft in non-reversible drives.
• Motor speed-torque characteristic is made drooping (characteristic AC in Fig. 1).
• Alternatively, by closed loop current control torque is prevented from exceeding a
permissible value (characteristic ABC in Fig. 1).
• During high load period, 𝜏l will be much larger compared to 𝜏m.
• Deceleration occurs, producing a large dynamic torque component (J d𝜔m/dt).
• Dynamic torque and motor torque together are able to produce torque required by
the load.

Load Equalisation
Load Equalisation
6 / 13
• Due to deceleration, the motor speed falls.
• During light load period, 𝜏m exceeds the 𝜏l , causing acceleration and speed is
brought back to the original value before the next high load period.
• Variation of motor and load torques, and speed for a periodic load and for a
drooping motor speed-torque curve (AC in Fig. 1) are shown in Fig. 2.
• It shows that peak torque required from the motor has much smaller value than
the peak 𝜏l .
• Hence, a motor with much smaller rating than peak load can be used and peak
current drawn by motor from the source is reduced by a large amount.
• Fluctuations in motor torque and speed are also reduced.
• Since power drawn from the source fluctuates very little → load equalisation.

Load Equalisation
Load Equalisation
7 / 13
• In variable speed and reversible drives, a flywheel cannot be mounted on the
motor shaft, Fig. 2 as it will increase transient time of the drive by a large
amount.
• If motor is fed from a motor-generator set (Ward-Leonard Drive), then flywheel
can be mounted on the shaft of the motor-generator set.
• This arrangement equalizes load on the source, but not the load on motor.
• Consequently, a motor capable of supplying peak-load torque is required.
• Moment of inertia of the flywheel required for load equalisation is calculated as
follows:
• Assuming a linear motor-speed-torque curve in the region of interest (drooping
characteristic AC of Fig. 1)
𝜏r
𝜔m0 − 𝜔mr
𝜔m = 𝜔m0 − 𝜏.
(1)
Where, 𝜔m0, 𝜔mr and 𝜏l are no-load speed, rated speed and rated torque, respec-
tively.

Load Equalisation
Load Equalisation
8 / 13
• The slow response due to large inertia, motor can be assumed to be in
electrical equilibrium during transient operation of the motor-load system. In
that case Eq. 1 will be
dt 𝜏r
d𝜔m J (𝜔mo − 𝜔m
r ) d𝜏
dt
J = − (2)
J
d𝜔m
= − 𝜏m
d𝜏
.
(3)
dt dt
𝜏r
Where, 𝜏m = J ( 𝜔 m o − 𝜔 m r )
, is the mechanical time constant of the mot
or
.
• It is the time required for the motor speed to change by (𝜔m0 − 𝜔m
r ) when 𝜏m is
maintained constant at rated value 𝜏r .
d𝜏
𝜏m
dt
+ 𝜏 = 𝜏l .
(4)

Load Equalisation
Load Equalisation
9 / 13
• Consider now a periodic load torque, a cycle of which consists of one high load
period with torque 𝜏l
h and duration th and one light load period with torque 𝜏l
l
and duration tl (Fig. 2).
• For high load period (0 ≤ t ≤ th) solution of Eq. (2.35) is
𝜏 = 𝜏l
h (1 − e−t/𝜏m ) + 𝜏m
i
n e−t/𝜏m for 0 ≤ t ≤ th.
(5) Where, 𝜏m
i
n is motor torque at t = 0, which is also the instant when heavy
load
𝜏l
h is applied.
• If motor torque at the end of heavy load period is 𝜏m
a
x , then from Eq.
𝜏m
ax = 𝜏l
h (1 − e−th /𝜏m ) + 𝜏m
i
n e−th /𝜏m .
(6)
• Solution of Eq. 4 for the light load period (th ≤ t ≤ th + tl ) with the
initial motor torque equal to Tmax is
′
′
𝜏 = 𝜏l
l (1 − e−t /𝜏m ) + 𝜏m
a
x e−t /𝜏m for 0 ≤ t
′
≤ tl .
Where, t
′
= t − th.
(7)

Load Equalisation
Load Equalisation
10 /
13
• When operating in steady-state, motor torque at the end of a cycle will be the
same as at the beginning of the cycle.
• Hence at t
′
= tl, 𝜏 = 𝜏m
i
n. Substituting in Eq. 7 gives
𝜏m
i
n = 𝜏l
l (1 − e−tl /𝜏m ) + 𝜏m
a
x e−tl /𝜏m .
(8)
• Eq. 6,
m
th
log
𝜏 =
𝜏
− 𝜏
lh min
𝜏lh − 𝜏m
a
x
. (9)
𝜏r
• From 𝜏m = J ( 𝜔 m o − 𝜔 m r )
and Eq.
9,
𝜔
− 𝜔
m0 mr
J =
𝜏r




th
log
𝜏
− 𝜏
lh min
𝜏lh − 𝜏m
a
x




.
(10)

Load Equalisation
Load Equalisation
11 /
13
• Also from Eq. 8,
𝜏m =





tl
log max
ll
𝜏min
− 𝜏ll
𝜏 − 𝜏
 



.
(11)
𝜏r
• From 𝜏m = J ( 𝜔 m o − 𝜔 m r )
and Eq.
1
1
,
J =
𝜏r


𝜔m0 − 𝜔mr 
tl
log
𝜏
− 𝜏
max
ll
𝜏min
− 𝜏ll





.
(12)
• Moment of inertia of the flywheel required can be calculated either from Eq. 10
or 12.
• Further,
J = W R2
, kg − m2
.
(13)
Where, W is the weight of the flywheel (kg) and R is the radius (m).

Modes of Operation Speed Control C l o s e d - L o o p
Control
1 Modes of Operation
2 Speed Control and Drive Classifications
3 Closed-Loop Control of Drives
2 /
20

Control
Modes of Operation
• An electrical drive operates in three modes:
(a) steady-state,
(b) acceleration including starting, and
(c) deceleration including stopping.
• According to Eq. 1,
𝜏 = 𝜏l + J
d𝜔m
,
dt
steady-state operation takes place when 𝜏 =
𝜏l .
(1)
• The steady-state operation for a given speed is realised by the adjustment of
steady-state motor speed-torque curve such that 𝜏 = 𝜏l at this speed.
• Change in speed is achieved by varying the steady-state motor speed torque curve
so that 𝜏 equals 𝜏l at the new desired speed.
3 /
20

Control
Modes of
Operation
• In Fig. 1, when the motor parameters
are adjusted to provide speed torque
curve 1 → drive runs at the desired
speed 𝜔m1.
• Speed is changed to 𝜔2 when the mo-
tor parameters are adjusted to provide
speed-torque curve 2.
• When 𝜏l opposes motion, the
motor works as a motor
operating in quadrant
I or III depending on the direction
of rotation.
• When the load is active, it can reverse
its sign and act to assist the
motion.
ωm
ωm1
ωm2
Tl
1
2
0
Figure 1. Principle of speed control.
4 /
20

Control
Modes of Operation
• For example, when a loaded hoist is lowered or an unloaded hoist is lifted, the net
load-torque acts to assist the motion.
• Steady-state operation for such a case can be obtained by adding a mechanical
brake which will produce a torque in a direction to oppose the motion.
• The steady state operation is obtained at a speed for which braking torque equals
𝜏l .
• Drive operates
in quadrant II
or IV
depending on
the direction
of rotation.
• Disadvantages of mechanical braking: frequent maintenance and replacement of
brake shoes, lower life, braking power is dissipated as heat.
• Electrical braking → motor is made to work as a generator converting mechanical
5 /
20

Control
Modes of Operation
• Even when electrical braking is employed, mechanical brakes may also be provided
to ensure the reliable operation of the drive.
• Mechanical brakes → employed to hold the drive at stand-still since many braking
methods are not able to produce torque at stand-still.
• Acceleration and deceleration modes → transient operations.
• Drive operates in acceleration mode whenever an increase in its speed is required.
• Hence, the motor speed-torque curve must be changed so that 𝜏 > 𝜏l .
• Time taken for a given change in speed → depends on the inertia of motor-load
system and the amount by which 𝜏 exceeds 𝜏l .
• ↑ in 𝜏 is accompanied by an ↑ in motor current, i.
6 /
20

Control
Modes of Operation
• Restrict i within a value that is safe for both motor and power modulator.
• In applications involving acceleration periods of long duration → current must
not be allowed to exceed the rated value.
• When acceleration periods are of short duration → current higher than the rated
value (is allowed during acceleration).
• In closed-loop drives requiring fast response → i may be intentionally forced to
the maximum value to achieve high acceleration.
• Torque developed by an ac motor for a given current → function of motor control
method employed.
• In high-performance drives, methods that produce high torque per ampere of
motor current are employed.
7 / 20

Control
Modes of
Operation
ωm2
1
2
ωm
Motoring
Electric braking
3
4
5
A
B
C
T
-T
E1
D1
ωm1 D2
D3
E3
ωm3
E2
Deceleration
0
Figure 2. Speed transition paths (1 to 5 are motor speed torque curves).
8 / 20

Control
Modes of Operation
• Fig. 2 shows the transition from operating point A at speed 𝜔m1 to operating
point B at a higher speed 𝜔m2, when 𝜏 is held constant during acceleration.
• The path consists of A D1 E1 B. In Fig. 2, 1 to 5 are motor speed-torque curves.
• Starting is a special case of acceleration where a speed change from 0 to a desired
speed takes place.
• When starting takes place at no-load or light loads → methods with low starting
torque can be employed.
• When the motor must start with substantial 𝜏l (around rated torque) or when
fast start is required → methods with high starting torque must be used.
• In applications, the motor should accelerate smoothly (without any jerk) → starting
torque can be increased steplessly from its zero value → soft start.
9 / 20

Control
Modes of Operation
• Motor operation in deceleration mode is required when a decrease in its speed is
required.
• When 𝜏l > 𝜏 → deceleration occurs.
• In those applications where 𝜏l is always present with substantial magnitude,
enough deceleration can be achieved by simply reducing 𝜏 to zero.
• In those applications where 𝜏l may not always have a substantial amount or
where simply reducing 𝜏 to zero does not provide enough deceleration,
mechanical brakes may be used to produce the required magnitude of
deceleration.
• Alternatively, electric braking may be employed. Now both 𝜏 and 𝜏l oppose the
motion, thus producing larger deceleration.
• During electric braking motor current tends to exceed the safe limit.
• When electric braking may persist for long periods, maximum current is usually
restricted to the rated value.
10 /
20

Control
Modes of Operation
• When electric braking occurs for short durations, maximum current is allowed to
exceed the rated value.
• Higher the braking torque → greater the deceleration.
• In high-performance closed loop schemes, motor current may be intentionally
forced to the maximum permissible value during deceleration.
• Figure 2 shows paths followed during transition from point A at speed 𝜔m1 to a
point C at a lower speed 𝜔m3.
• When deceleration is carried out using electric braking at a constant braking
torque
→ operating point moves along the path A D3 E3 C.
• When sufficient 𝜏l is present or when mechanical braking is used → operation
takes place along the path A D2 E2 C.
11 /
20

Control
Modes of
Operation
• Stopping is a special case of deceleration where the speed of a running motor is
changed to zero.
• All the discussions about deceleration are applicable to stopping also.
• In applications requiring frequent, quick, accurate or rapid emergency stops, the
electric braking is usually employed.
• It allows smooth and quick stops without subjecting the mechanical parts to unduly
large stresses, e.g. in suburban electric trains quick stops are required.
• Use of electric braking allows a smooth stop, and increases the life of track
and wheels allowing a substantial saving in cost.
12 /
20

Control
13 /
20

Control
Speed Control and Drive Classifications
• Drives where the driving motor runs at a nearly fixed speed → constant speed of
single speed drives.
• Multi-speed drives → which operate at discrete speed settings.
• Drives needing stepless change in speed and multispeed drives → variable speed
drives.
• Multi-motor drive → when a number of motors are fed from a common converter
or when a load is driven by more than one motor.
• A variable speed drive is called constant torque drive → if the drive’s maximum
torque capability does not change with a change in speed setting.
• Constant torque → refers to the maximum torque capability of the drive and not
to the actual output torque, which may vary from no load to full load torque.
14 /
20

Control
Speed Control and Drive Classifications
• Ideally, for a given speed setting, the motor speed should remain constant as 𝜏l is
changes from no load to full load.
• In practice, speed drops with an ↑ in 𝜏l .
• Quality of a speed control system is measured in terms of speed-regulation which
is defined as
Full load speed
Speed regulation =
No load speed − Full load speed
× 100. (2)
• If open-loop control fails to provide the desired speed regulation, drive is operated
as a closed-loop speed control system.
15 /
20

Control
16 /
20

Control
Closed-Loop Control of Drives
Controller Motor Load
Current
Sensor
Converter
Controller Motor Load
Current
Sensor
+
V*
If
Converter
-
I
0 I
max
Threshold logic
circuit
Figure 3. Block diagram of current limit control.
• Feedback loops in an electrical drive may be provided to satisfy one or more of
the following requirements:
(i) Protection
(ii) Enhancement of speed of response
(iii) To improve steady-state accuracy
Current-Limit Control
• Current-limit control scheme → to limit the converter and motor current below a
safe limit during transient operations.
• As long as the current is within a set maximum value → feedback loop does not
affect the operation of the drive.
17 /
20

Control
Current-Limit Control
• During a transient operation, if current exceeds the set maximum
value, the feedback loop becomes active and current is forced below the set
maximum value, which causes the feedback loop to become inactive again.
• If the current exceeds set maximum value again, it is again brought below it by
the action of feedback loop.
• Thus, the current fluctuates around a set maximum limit during the transient
operation until the drive condition is such that the current does not have a
tendency to cross the set maximum value.
• Example → during starting, current will fluctuate around the set maximum value.
• When close to the steady-state operation point, current will not have tendency to
cross the maximum value.
• Consequently, the feedback loop will have no effect on the drive operation.
18 /
20

Closed-Loop Torque Control Closed-Loop Speed Control Speed Sensing Current Sensing PLL Control Closed-Loop Position Control
1 Closed-Loop Torque Control
2 Closed-Loop Speed Control
3 Speed Sensing
2 /
23

Closed-Loop Torque Control
Torque
Controller
Converter Motor Load
Torque
sensor
+
- T
T*
Figure 1. Block diagram of closed-loop torque control.
• Closed-loop torque control scheme → battery operated vehicles, rail cars and
electric trains.
• Driver presses the accelerator to set torque reference 𝜏∗. Through closed- loop
control of torque, the actual 𝜏 follows 𝜏∗.
• Speed feedback loop is present through the driver. By applying pressure on the
accelerator, driver adjusts the speed depending on traffic, road condition, his liking,
car condition and speed limit.
3 /
23

3 Speed Sensing
4 /
23

Closed-Loop Speed Control
Speed
Controller
Motor Load
Current
Sensor
+
ω*
Δωm
ωm
Current
Limiter
Current
Controller
Converter
Speed
Sensor
Current
Reference
I*
I
+
-
-
I
ωm
(b)
(a)
Figure 2. Block diagram of closed-loop speed control.
• Fig. 2 → closed-loop speed control scheme widely used in
electrical drives.
• It employs an inner current control loop within an outer speed-
loop.
• Inner current control loop is provided to limit the converter and motor current or
motor torque below a safe limit.
• In some schemes, the current is controlled directly or indirectly. For example,
in a variable frequency IM drives the current is controlled by controlling the
slip.
Current
Limiter
5 /
23

Closed-Loop Speed Control Operation
• An increase in reference speed 𝜔∗ produce a positive error ∆ 𝜔m.
m
• Speed error is processed through a speed controller and applied to a current
limiter, which saturates even for a small speed error.
• Consequently, limiter sets the current reference for inner current control loop
at a value corresponding to the maximum allowable current.
• Drive accelerates at the maximum allowable current (and in some cases 𝜏m
a
x ).
• When close to the desired speed, limiter desaturates.
• Steady-state is reached at the desired speed (with some steady-state error) and at
current for which 𝜏 = 𝜏l .
• A decrease in reference speed produces a negative speed error.
• Current limiter saturates and sets current reference for inner current loop at a value
corresponding to the maximum allowable current.
6 /
23

Closed-Loop Speed Control Operation
• Consequently, the drive decelerates in braking mode at the maximum allowable
current.
• When close to the required speed, current limiter desaturates. The operation is
transferred from braking to motoring.
• Drive then settles at a desired speed and at current for which 𝜏 = 𝜏l .
• In those drives where the current I does not have to reverse for braking operation,
current limiter will have the input-output characteristic shown in Fig. 2 (b).
• In those drive applications where 𝜏l is able to provide enough decelerating torque,
electric braking need not be used. Then also current limiter has the characteristic
shown in Fig. 2 (b).
• Current and speed controllers → proportional and integral (PI), proportional and
derivative (PD) or proportional, integral and derivative (PID) controller, depending
on steady-state accuracy and transient response requirements.
7 /
23

3 Speed Sensing
8 /
23

Speed Sensing
• Sensing of speed is required for the implementation of closed-loop speed control
schemes.
• Speed is usually sensed → tachometers coupled to the motor shaft.
• A tachometer is an ac or dc generator with a high order of linearity between its
speed and output voltage.
• A dc tachometer is built with a permanent magnetic field with silver brushes to
reduce contact drop between brush and commutator.
• Typical voltage outputs are 10 V per 1000 rpm. The tachometer output voltage
consists of a ripple whose frequency depends on its speed.
• At low speeds → adequate filtering is done by a filter with a large enough time
constant to affect the dynamics of the drive.
• Special large diameter tachometers with a large number of commutator segments
are sometimes built to overcome this problem.
9 /
23

Speed Sensing
• Tachometers are available to measure speed with an accuracy of ± 0.1 %.
• Tachometer should be coupled to the motor with a torsionally stiff coupling →
natural frequency of the system consisting of rotor and tachometer lies well beyond
the bandwidth of the speed control loop.
• When very high-speed accuracies are required (computer peripherals and paper
mills) → digital tachometers are used.
• A digital tachometer employs a shaft encoder which gives a f ∝ 𝜔.
• In dc drives, speed can be sensed without a tachometer when field current or flux
is held constant, Eb ∝ 𝜔.
• Eb is measured by deducting a signal equal to its armature resistance drop from
motor terminal voltage.
• Method is inexpensive and provides speed measurement with an accuracy of ± 2
% of base speed.
10 /
23

Selection of Motor Power Rating Classes of Motor Duty
2 Classes of Motor Duty
Selection of Motor Power Rating
7 / 15

Classes of Motor Duty
Various classes of motor duty are:
• Continuous duty.
• Short time duty.
• Intermittent periodic duty.
• Intermittent periodic duty with starting.
• Intermittent periodic duty with starting and braking.
• Continuous duty with intermittent periodic loading.
• Continuous duty with starting and braking.
• Continuous duty with periodic speed changes.
8 / 15

t
TL
0
0
ᶿ
t
t
TL
0
0
ᶿ
t
t
TL
0
0
ᶿ
t
0
t
TL
0
ᶿ
t
0
t
TL
0
ᶿ
t
(a) (b) (c)
Starting
Running
Rest
(d) (e)
Figure 2. Some classes of motor duty.
9 / 15

Continuous Duty [Fig. 1 (a)]:
• It denotes the motor operation at a constant load torque for a duration long
enough for the motor temperature to reach steady-state value.
• This duty is characterized by a constant motor loss.
• Examples of continuous duty → paper mill drives, compressors, conveyers,
centrifugal pumps and fans.
Short Time Duty [Fig. 1 (b)]:
• Time of drive operation is considerably less than the heating time constant and
machine is allowed to cool off to ambinent temperature before the
motor is required to operate again.
• In this operation, the machine can be overloaded until temperature at the end of
loading time reaches the permissible limit.
• Examples → crane drives, drives for household appliances, turning bridges,
sluice-gate drives, and valve drives.
10 /
15

Intermittent Periodic Duty [Fig. 1 (c)]:
• It consists of periodic duty cycles, each consisting of a period of running at a
constant load and a rest period.
• Neither the duration of running period is sufficient to raise the temperature to a
steady-state value, nor the rest period is long enough for the machine to cool off to
ambient temperature.
• Examples → pressing, cutting and drilling machine drives.
Intermittent Period Duty with Starting [Fig. 1 (d)]:
• This is intermittent periodic duty where heat losses during starting cannot be
ignored.
• Thus, it consists of a period of starting, a period of operation at a constant load
and a rest period with operating and rest periods being too short for
the respective steady-state temperatures to be attained.
• Examples → metal cutting and drilling tool drives, drives for fork lift trucks, mine
hoist etc.
11 /
15

Intermittent Periodic duty with Starting and Braking [Fig. 1 (e)]:
• It consists of a period of starting, a period of operation with a constant load, a
braking period with electrical braking and a rest period; with operating and rest
periods being too short for the respective steady state temperatures to be attained.
• Examples → billet mill drive, manipulator drive, ingot buggy drive, schrewdown
mechanism of blooming mill, drives for electric suburban trains and mine hoist.
Continuous Duty with Intermittent Periodic Loading:
• It consists of periodic duty cycles, each consisting of a period of running at a
constant load and a period of running at no load, with normal voltage across the
excitation winding.
• Again the load period and no load period being too short for the respective
temperatures to be attained.
• This duty is distinguished from the intermittent periodic duty by the fact that a
period of running at a constant load is followed by a period of running
at no load instead of rest.
• Examples → pressing, cutting, shearing and drilling machine drives.
Selection of Motor Power Rating 12 / 15

Continuous Duty with Starting and Braking:
• Consists of periodic duty cycle, each having a period of starting, a period of
running at a constant load and a period of electrical braking; there is no period of
rest.
• Example → main drive of a blooming mill.
Continuous Duty with Periodic Speed Changes:
• Consists of periodic duty cycle, each having a period of running at one load
and speed, and another period of running at different speed and load; again both
operating periods are too short for respective steady-state temperatures to be
attained.
• Further, there is no period of rest.
13 /
15

Determination of Motor
Rating
Short Time Duty Intermittent Periodic Duty Frequency of Operation of Motor Subjected to Intermittent Loads
1 Determination of Motor Rating
2 Short Time Duty
2 / 24

Rating
Determination of Motor Rating
From the point of view of calculation of motor ratings, duty cycles are broadly
classified into
• Continuous duty
• Fluctuating loads
• Short time and intermittent duty
Continuous Duty:
• The maximum continuous power demand of the load is ascertained.
• The selected motor should be the next higher power rating from commercially
available ratings.
• The motor speed should match the load speed requirements.
• The selected motor rating should match the starting torque requirement.
• It should have the ability to drive the load during the normal disturbances coming
from the supply system. This condition is generally assured by the transient and
steady-state reserve torque capacity of the motor.
3 / 24

Rating
Equivalent current, torque and power methods for fluctuating and
intermittent Loads
• This method is based on the approximation that the actual variable motor current
is replaced by the equivalent Ieq which produces the same losses as
the actual current.
• The equivalent current can be determined as follows
• Motor loss p1 consists of two components → (i) constant loss pc , which is
independent of load which consists core loss and friction loss and (ii) load
dependent copper loss.
• Therefore for fluctuating load given in fig 4.3 consists of n values of motor current
I1, I2, . . . ..In for the durations t1, t2, . . . .. tn, respectively.
• The equivalent current Ieq is given by
eq
pc + I2 R =
1 n
(pc + I2) R)t1 + (pc + I2) R)t2 + ....... + (pc + I2) R)tn
2
t1 + t2 + .........
+ tn
. (1)
or
2
c eq
p + I R =
p (t + t + ......... + t )
t1 + t2 + ......... + tn
+
c 1 2 n 1 2
n
(I2
t1 + I2
t2 + .......... + I2
tn) R
t1 + t2 + ......... + tn
.
(2)
4 / 24

Rating
Load diagram of a Fluctuating Load
Figure 1. Load diagram of a fluctuating load.
Figure 2. Load diagram of a fluctuating load.
5 / 24

Rating
I
eq =
√︄
2
1
I
t
1
2
2
2
2 n
+ I t + .......... + I tn
t1 + t2 + ......... + tn
(3)
•
∫ T
0
2 2
Integral i dt represents the area between i versus t curve and the time axis
for the duration 0 to T .
• The above analysis is based on the assumption that the heating and cooling
remain the same.
• However, if the motor is running at constant RPM then the heating and cooling
are going to be unchanged.
• If the speed varies, the constant loss will be changed marginally and if
forced ventilation is used, heating and cooling can still be assumed same without
the loss of much accuracy.
• In self-ventilating machines, colling conditions remain poor at low speeds.
• After Ieq is determined motors with next higher current rating from commercially
available ratings are selected.
6 / 24

Rating
DC Motor
• This motor is allowed to carry the larger current than the rated current for the
short duration of time.
• This condition is known as short time overload capacity of the motor.
• A normally designed dc motor is allowed to carry up to 2 times its rated current.
• Let the ratio of maximum allowable current to the rated current be
denoted by 𝜆. Then,
𝜆 ≥
Imax
.
(4)
Irated
Where, Imax is the ,maximum value of the current and Irated is the rated current of
the motor.
• If Eq. 4 is not satisfied, then the motor current rating is calculated from
𝜆
Irated ≥
Imax
.
(5)
7 / 24

Rating
Induction and Synchronous Motors
• For stable operation maximum load torque should be well within the breakdown
torque of the motor.
• If the motor rating is selected based on Eq. 2 or 3 the the above
constraint is violated.
• In the case of induction motors with normal design the ratio of breakdown to
rated torque varied from 1.65 to 3, and for synchronous motors, from 2 to
2.25.
• If the ratio between breakdown to rated torque is denoted by 𝜆
′
, then the motor
rating is based on
Trated ≥
Tmax
.
𝜆′ (6)
• When the load has high torque pulses selection of of motor rating based on this
will be large.
8 / 24

Rating
• The equivalent current method assumed constant losses to remain constant for all
operating points.
• ∴ this method must be carefully employed when these losses vary.
• This method is also not applicable to the frequency-dependent parameters in the
equivalent circuit.
• For example, in deep bar and double squirrel cage rotor motors the rotor winding
resistance and reactance vary widely during starting and braking making this
method inapplicable.
• When the torque is directly proportional to the current like dc separately excited
motor the from Eq. 2
√︄
T
eq =
1 2
1 2
2 2 2
n
T t + T t + ......... + T t n
t1 + t2 + ......... + tn
. (7)
• Eq. 7 can be employed to directly ascertain the motor torque rating.
• When the motor operates at nearly fixed speed its power will be ∝ torque; hence,
for nearly constant speed operation, the power rating of the motor can be
directly
from,
P
eq =
√︄
1 2
1 2
2 2 2
n
P t + P t + ......... + P tn
t1 + t2 + ......... + tn
. (8)
9 / 24

Rating
1 Determination of Motor Rating
2 Short Time Duty
10 /
24

Determination of Motor Rating Short Time Duty Intermittent Periodic Duty Frequency of Operation of Motor Subjected to Intermittent Loads
Short Time Duty
• In short time duty, time of motor operation is considerably less than the heating
time constant and motor is allowed to cool down to the ambient temperature
before it is required to operate again.
• If a motor with a continuous duty power rating of Pr is subjected to a short time
duty load of magnitude Pr then the motor temperature rise will be far below the
maximum permissible value theta per and the motor will be highly underutilized.
• ∴ motor can be overloaded by a factor K (K > 1) such that the maximum
temperature rise just reaches the permissible value 𝜃p
e
r .
• When the duration of the running period in a duty cycle with power K Pr is tr
then from 𝜃 = 𝜃ss 1 − e−t/𝜏 + 𝜃1 e−t/𝜏 .
11 /
24

Rating
Short Time Duty
Figure 3. 𝜃 versus t curves for short time duty loads.
−tr
𝜃p
e
r = 𝜃s
s (1 − e 𝜏 ), (9)
(10)
𝜃ss
=
1
.
𝜃per
−tr
1 − e 𝜏
• Note that 𝜃s
s is the steady state temperature rise which will be attained if
motor delivers a power (KPr ) on continuous basis, whereas the permissible
temperature rise 𝜃p
e
r is also the steady state temperature rise attained when
motor operates with a power Pr on continuous basis.
• The motor losses for powers Pr and (KPr ) be P1r and P1s, respectively.
12 /
24

Rating
• Thus Eq. 24. provides energy balance relation when the period of intermittent
loading is very small, compared to the thermal time constants of the machine.
• Applying relationship of Eq. 24 to intermittent loads with frequency starting and
braking, and short running intervals yields
Es + p1str + Eb = p1r (𝛾ts
t + tr + 𝛾tb + 𝛽ts).
Where, Es = loss of energy during starting.
Eb = loss of energy during braking.
p1s = loss of power during running
interval.
p1r = rated loss of power of the
motor.
tr = length of the running interval.
tst = length of the starting interval.
tb = length of breaking interval.
ts = length of standstill interval.
𝛾 and 𝛽 are numerical constants
based on measurements.
(25)
21 /
24

Rating
2
• The value of 𝛾 is assumed as 𝛾 = 1+𝛽
, and the speed changes from zero to running
value during starting and braking, the speed changes from zero to running value.
• ∴ effective dissipation factor can be considered as the mean of those at
running and standstill conditions.
• ts is calculated from Eq. 25. Then, the permissible frequency of switching per
hour is 3600
fmax =
tst + tr + tb + ts
(26)
• Eqs. 25 and 26 suggest that the switching frequency can be increased by reducing
loss during starting, braking and running by use of efficient method of control,
and by improving heat dissipation by use of forced ventilation.
• The most efficient methods of control for dc and ac motors are armature voltage
control and variable frequency control, respectively.
22 /
24

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