Choice of Electrical Drives

➢ Steady state operating requirements

➢Speed-torque characteristics
➢Speed regulation
➢Speed range
➢Efficiency
➢Duty cycle
➢Quadrants of operation
➢Speed fluctuations
➢ Transient operation requirements
➢ Requirements related to the source
➢ Capital and running cost, maintenance needs, life
➢ Space and weight restrictions if any.
➢ Environment and location
➢ Reliability

1 Rijil Ramchand 02-01-2025

 Dynamics of Electrical Drives

Lecture 2 (02-01-2025)

2 02-01-2025
 Fundamental Torque Equations
Elementary principles of mechanics – Linear Motion
v m/sec
xm Newton’s law

𝐝(𝐌𝐯)
𝐅𝐦 − 𝐅𝐟 =
𝐝𝐭
Fm M Ff
(Newton) (Newton)

Linear motion, constant M

𝐝𝐯 𝐝𝟐 𝐱
𝐅𝐦 − 𝐅𝐟 = 𝐌 =𝐌 = 𝐌𝐚
𝐝𝐭 𝐝𝐭 𝟐

➢ First order differential equation for speed

➢ Second order differential equation for displacement

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 Fundamental Torque Equations
Elementary principles of mechanics – Rotational Motion

➢ Normally is the case for electrical drives

Tl 𝐝(𝐉𝝎𝒎 )
𝐓𝐞 − 𝐓𝐥 =
𝐝𝐭
Te , m With constant J,
𝐝𝛚𝐦 𝐝𝟐 𝛉
𝐓𝐞 − 𝐓𝐥 = 𝐉 =𝐉
𝐝𝐭 𝐝𝐭 𝟐

➢ First order differential equation for angular frequency (or velocity)

➢ Second order differential equation for angle (or position)

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 Fundamental Torque Equations
Equivalent motor-load system

Motor Load
𝛚𝐦
𝐓𝐞 𝐓𝐥

𝐝(𝐉𝝎𝒎 )
𝐓𝐞 − 𝐓𝐥 =
𝐝𝐭
𝐝𝛚𝐦 𝐝𝐉
𝐓𝐞 − 𝐓𝐥 = 𝐉 +𝛚𝐦 𝐝𝐭
𝐝𝐭
𝐝𝐉
For drives with constant inertia, 𝐝𝐭=0

𝐝𝛚𝐦
𝐓𝐞 = 𝐓𝐥 + 𝐉
𝐝𝐭

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 Fundamental Torque Equations
Equivalent motor-load system

𝐝𝛚𝐦
𝐓𝐞 = 𝐓𝐥 + 𝐉
𝐝𝐭

Load Torque
Electromagnetic Torque
Dynamic Torque

➢ Electromagnetic torque – developed torque by the motor

➢ Load Torque
➢ Dynamic Torque – Present during transient operations
➢ Acceleration
➢ Deceleration

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 Fundamental Torque Equations
Equivalent motor-load system
𝐝𝛚𝐦
𝐓𝐞 = 𝐓𝐥 + 𝐉
𝐝𝐭
𝐝𝛚𝐦 𝐝𝛚𝐦
𝛚𝐦 𝐓𝐞 = 𝛚𝐦 𝐓𝐥 + 𝛚𝐦 𝐉 𝐩𝐃 = 𝐩𝐥 + 𝐉
𝐝𝐭 𝐝𝐭

Driving Power Load Power Change in KE

➢ A step change in speed requires an infinite driving power

➢ Therefore  is a continuous variable

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 Fundamental Torque Equations
A drive system that require fast acceleration must have
➢ large motor torque capability
➢ small overall moment of inertia

As the motor speed increases, the kinetic energy also increases.

During deceleration, the dynamic torque changes its sign and thus
helps motor to maintain the speed. This energy is extracted from
the stored kinetic energy:

J is purposely increased to do this job !

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 Torque-speed quadrant of operation

➢ Quadrant of operation is defined

1
T -ve

2
T +ve
 +ve  +ve by the speed and torque of the
Pm -ve Pm +ve motor
➢ Most rotating electrical machines
can operate in 4 quadrants
T ➢ Not all converters can operate in
4 quadrants

3 4
T -ve T +ve
 -ve  -ve
Pm +ve Pm -ve

9 02-01-2025
Courtesy: Dr. Nik Rumzi Nik Idris, Dept. of Energy Conversion, UTM
 Torque-speed quadrant of operation


m Te
m
Te • Quadrant of operation is
defined by the speed and torque
Quadrant 1 of the motor
Quadrant 2 Forward
Forward motoring • Most rotating electrical
braking machines can operate in 4
Te T quadrants
Te • Not all converters can operate
m in 4 quadrants
m
Quadrant 4
Quadrant 3
Reverse
Reverse
braking
motoring

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Courtesy: Dr. Nik Rumzi Nik Idris, Dept. of Energy Conversion, UTM
 Loads with rotational motion

m m1
Motor Load 1, n1
Te Tl1
J2

m2
n2 Load 2,
J1 Tl2

ωm2 n1
= = a2 a2 is the gear tooth ratio
ωm1 n2

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 Loads with rotational motion

If the losses in transmission are neglected, then the KE due to equivalent

MOI must be the same as KE of various moving parts

𝟏 𝟏 𝟏
𝐉𝐞𝐪𝐮 𝛚𝐦 𝟐 = 𝐉𝟏 𝛚𝐦𝟏 𝟐 + 𝐉𝟐 𝛚𝐦𝟐 𝟐
𝟐 𝟐 𝟐

𝐉𝐞𝐪𝐮 = 𝐉𝟏 + 𝐚𝟐𝟐 𝐉𝟐 ,

𝛚𝐦𝟏 𝛚𝐦𝟏
𝐚𝐬 =𝟏 𝐚𝐧𝐝 = 𝐚𝟐
𝛚𝐦 𝛚𝐦

m
Motor Equivalent
Te Load , Tlequ

Jequ

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 Loads with rotational motion

Power at the motor and loads must be the same. If transmission efficiency of
the gears is 𝛈𝟐 , then

𝐓𝐥𝟐 𝛚𝐦𝟐
𝐓𝐥𝐞𝐪𝐮 𝛚𝐦 = 𝐓𝐥𝟏 𝛚𝐦𝟏 +
𝜼𝟐

𝐓𝐥𝟐 𝐚𝟐
𝐓𝐥𝐞𝐪𝐮 = 𝐓𝐥𝟏 +
𝜼𝟐

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 Loads with rotational motion
If in addition to load directly coupled to the motor with MOI J0 there are m
other loads of MOIs J1 , J2 , J3 …… Jm and gear teeth ratios a1, a2,…am, then
the equivalent MOI Jequ is given by

𝐉𝐞𝐪𝐮 = 𝐉𝟎 + 𝐚𝟐𝟏 𝐉𝟏 + 𝐚𝟐𝟐 𝐉𝟐 + ……. + 𝐚𝟐𝐦 𝐉𝐦

If in addition to load directly coupled to the motor with torque Tl0 there are m
other loads with torques Tl1, Tl2, Tl3, …….. Tlmcoupled through gears with gear
teeth ratios a1, a2,…am and transmission efficiencies 𝜼1, 𝜼2,…𝜼m then the
equivalent torque Tlequis given by

𝐚𝟏 𝐓𝐥𝟏 𝐚𝟐 𝐓𝐥𝟐 𝐚𝐦 𝐓𝐥𝐦

𝐓𝐥𝐞𝐪𝐮 = 𝐓𝒍𝟎 + + 𝛈 + …..+
𝛈𝟏 𝟐 𝛈𝐦

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 Problem - 1
A motor drives two loads. One has rotational motion. It is coupled to the
motor through a reduction gear with a = 0.1 and efficiency of 90%. The load
has a MOI of 10 kgm2 and a load torque of 10Nm. Other load has translational
motion and consists of 1000kg weight to be lifted up at an uniform speed of
1.5m/s. Coupling between this load and the motor has an efficiency of 85%.
Motor has a MOI of 0.2 kgm2 and runs at a constant speed of 1420rpm.
Determine the equivalent MOI referred to the motor shaft and power
delivered by the motor.

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