❑Speed-Torque Characteristics of Loads

1. Type 1:- In fans, compressors and pumps,

the windage dominates, consequently, load torque is
proportional to speed squared (𝑻𝒍 ∝ 𝝎𝟐 ).
- Windage is the opposition offered by air to the motion.
- The motion in pumps is opposed by fluid like water.

2. Type 2:- in a high-speed hoist, viscous friction and

windage also have appreciable magnitude in addition
to gravity.

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3. Type 3:- Traction load when moving on a levelled ground
- 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 torques are 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.

4. Type 4:- Constant Power Drive

- There are certain situations in which it is desirable to develop
constant power over a range of speed. A typical application is
the take-up roll in a steel strip mill.

( All four types of loads operate in the 1st quadrant)

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5. Bi-Directional Speed
- An elevator is moving passengers in both directions (up and down).
- In the upward and downward directions, the motor sees the load forces 𝑭𝒍 , which is
a function of the weight of the passengers plus elevator cabin, cables, and so on.
- Since the weight and 𝑭𝒍 are unidirectional, the motor force 𝑭𝒎 is also unidirectional.
- The speed of the motor in this operation is bidirectional.

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6. Bi-directional Torque (Vehicle or Transportation drive)
- When the system motion is in the uphill direction. The force of the load is divided
into two components: one is perpendicular to the road, 𝑭 producing the frictional
force, and the other, 𝑭𝒍 , is parallel to the road and represents the load torque
exerted on the motor.
- 𝑭𝒍 always pulls the bus toward the base of the hill.
- The load torque seen by the motor is 𝑭𝒍 multiplied by the radius of the wheel. This
torque must be matched by a motor torque 𝑭𝒎 in the opposite direction to 𝑭𝒍 .

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7.Transportation Drive (Subway trains, Trolly buses, etc)

• Max speed of motor is related to max speed of the

vehicle
• 𝝎𝒃 is determined by max power that source and
converter can deliver.
• Motor will be chosen to develop 𝑻𝒎𝒂𝒙 .
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• To minimize the station-to-station time,
maximize +ve and –ve accelerations, and
speed in between.
• Maximum accelerations are dictated by
passengers comfort and also by the rate of
change of acceleration.
• Source rating is chosen to meet the
maximum power required during the
travel cycle, even though it lasts for a short
time.

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• A comparison with the previous Figure
shows that, at the cost of a very small
increase in station-to-station time, a much-
reduced maximum source power may be
used.

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❑Equivalent Values of 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 motions such as rotational.
- This section presents methods of finding the equivalent moment of inertia (J) of
motor load system and equivalent torque components, all referred to motor shaft.
- Let the moment of inertia of motor and load directly coupled to its shaft be 𝑱𝒐 , motor
speed and torque of the directly coupled load be 𝝎𝒎 and 𝑻𝒍𝒐 respectively.
- Let the moment of inertia, speed and torque of the load coupled through gears be
𝑱𝟏 , 𝝎𝒎𝟏 and 𝑻𝒍𝟏 respectively.
- 𝒏, 𝒏𝟏 are the gear teeth.

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- If the losses in transmission are neglected, the kinetic energy due to equivalent
inertia must be the same as kinetic energy of various moving parts. Thus,

- Power at the loads and motor must be the same. If the transmission efficiency of the
gears be 𝛈𝟏 .

Where 𝑇𝑙 is the total equivalent torque referred to motor shaft

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- If there are m other loads with moment of inertias 𝑱𝟏 , 𝑱𝟐 , … . . , 𝑱𝒎 and gear teeth
ratios of 𝒂𝟏 , 𝒂𝟐 , … . , 𝒂𝒎 then

- Load torques are then,

- If loads are driven through a belt drive instead of gears, then the 𝒂𝟏 , 𝒂𝟐 , … . , 𝒂𝒎 are
the ratios of diameters of wheels driven by motor to the diameters of wheels mounted
on the load shaft.

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 2
2
W v
𝑇𝑀 = TL C + + W. r. C + Tc . 𝐶 + 𝐾1 . C 2 . ω𝑚 + 𝐾2 . 𝐶 3 . 𝜔𝑚
2
g ω𝑚
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❑ Required Drive Characteristics
1. Speed Control
• At constant speed, 𝑻𝒆 = 𝑻𝒍
• Steady state speed is at point of intersection between 𝑻𝒆 and 𝑻𝒍 of the steady state
torque characteristics.
• Ideally, the motor speed should remain constant as load torque changes.
• In practical, speed drops with an increase in load torque.

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2. Speed Regulation
• For speed control, the ω-T c/c of the motor-converter should be as horizontal as
possible for any given control setting.
• Also, ensure that the s.s. torque required by the load is less than the torque limit set
by the controller. Some margin of motor torque must be available to accelerate the
system and to stabilize against overloads.

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• Speed Regulation = (NL speed –FL speed)/(FL speed)
• S.S. speeds are also specified 𝝎max /𝝎min ≈ 𝟔 ( 200 for special applications).
• For position control, motor speed must be controllable right down to standstill.

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3. Steady-state stability
• Equilibrium speed of a motor load system is obtained when motor torque equals the
load torque.
• Drive will operate in steady-state at this speed, provided it is the speed of stable
equilibrium.

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• Let the disturbance causes a reduction of ∆𝝎 in speed. At new speed, motor torque is
greater than the load torque, then motor will accelerate, and operation will be
restored to A.
• Similarly, an increase in speed will make load torque greater than the motor torque,
resulting into deceleration and restoration of operation to point A. Hence the drive is
steady state stable at point A.

Point A is stable

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• At point B, a decrease in speed causes the load torque to become greater than the
motor torque, drive decelerates, and operating point moves away from B. similarly,
an increase in speed will make motor torque to become greater than the load
torque. Which will move the operating point away from B. Thus, B is an unstable
point of equilibrium.

B is unstable C is stable D is unstable

• An equilibrium point will be stable when an increase in speed causes load torque to
exceed the motor torque, i.e. when at equilibrium point following condition is
satisfied:

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4. Speed Changes
• Acceleration and deceleration are transient operations. Drive operates in acceleration
mode when an increase is its speed is required.
• For this, motor speed-torque curve must be changed so that motor torque exceeds the
load torque. Time taken for a given change in speed depends on inertia of motor-load
system.
• Figure shows the transition from operating
point A at speed 𝝎𝒎𝟏 to operating point B
at higher speed 𝝎𝒎𝟐 , when the motor torque
is held constant during acceleration.
• The path consists of 𝑨𝑫𝟏 𝑬𝟏 𝑩.

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• Motor operation in deceleration mode is required when a decrease in speed is required.
• Deceleration occurs when load torque exceeds the motor torque. Deceleration can be
achieved by simply reducing the motor torque to zero.
• Mechanical braking may be used to produce the required magnitude of deceleration.
Alternatively, electrical braking may be employed.
• During the transition from point A at speed 𝝎𝒎𝟏 to operating point C at lower speed
𝝎𝒎𝟑 , electrical braking at constant torque is employed. The operating point moves
along the path 𝑨𝑫𝟑 𝑬𝟑 𝑪.
• When mechanical braking is used, the operation takes place along the path 𝑨𝑫𝟐 𝑬𝟐 𝑪.
• Electrical braking is usually employed in applications requiring frequent , quick,
accurate stops
• It allows smooth and quick stops without large stresses to mechanical parts.
• It is used is electric trains to allow a smooth stop without any inconvenience to
passengers and increase the life of track and wheels.

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