First Course on Electric Drives Electric Drives: An

88 Pages·2007·2.19 MB·English

by  Ned Mohan

First Course on Electric Drives
Electric Drives: An Integrated Approach Chapter 1
Introduction to Electric
Ned Mohan
Drive Systems
Oscar A. Schott Professor of Power Electronics and Systems
Department of Electrical and Computer Engineering
University of Minnesota
Minneapolis, MN 55455
USA
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Trend : Adjustable Speed Drives
History
(ASDs)
Constant frequency AC -essentially constant pump speed
(cid:134)
(cid:134) Inefficient -Heat generated in pump and throttling valve Outlet
Not amenable to automation
(cid:134)
Adjustable
Speed Drive
(ASD)
Inlet
Constant
frequency Adjustable frequency Adjustable Pump
AC electric converter speed
Throttling
Valve
Outlet
Driven at appropriate speed
(cid:134)
MOTOR
Inlet (cid:139) High Efficiency
Constant Pump
frequency ⇒ Essentially
AC Constant Speed
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What is an Electric-Motor Drive?
World Market and Scope
Electric
Drive
Power
Processing Motor Load
fixed Unit (PPU)
form
Elec(truitcil iStyo)urce adjfuosrtmable psopseietido n/
Sensors
Controller measured Power
speed/ position Signal
input command
(speed / position)
Role of Electric Drive: Efficient conversion of power
(cid:138)
from electrical to mechanical
(cid:138)healthy growth rate (cid:138)A very wide range of
Role of PPU: Delivers appropriate form of frequency to
(cid:138)
motor (as required by the load) ~ 25% per year speed and power
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Typical Applications
Factors for Growth
Process Industry -agitators, pumps, fans, compressors
(cid:134)
Machining -planers, winches, calendars, chippers, drill presses,
(cid:134)
Technical Advances in sanders, extruders, grinders, mills, presses
(cid:134)
Heating and Air Conditioning -blowers, compressors
(cid:139) Linear ICs and Digital Signal Processors (cid:134)
Paper and Steel Industry -hoists, rollers
(cid:134)
(cid:139) Power devices (cid:134) Transportation -elevators, trains, automobiles
Textile -looms
(cid:134)
Packaging -shears
Market Needs (cid:134)
(cid:134)
Food -conveyors, fans
(cid:134)
(cid:139)$20 Billion market in 1997 Oil, Gas , Mining -compressors, pumps, cranes, shovels
(cid:134)
Residential -heat pumps, freezers, washing machines
(cid:139) 25% Growth rate (cid:134)
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Application for Energy Conservation Energy Conservation in Pumps
(cid:134)Heat Pumps and air-conditioners -cycled operation
throttling
ideal actual valve outlet Pressure (H)
loss ∆H Q2 A ( H2+ ∆H , Q2) ( H1, Q1)
Compressor motor H2+∆H H2
Output Q2 Pump curve at
pump inlet D C B full speed
full speed
Throttled ( H2, Q2)
t System Pump curve at
no throttling outlet Curve reduced speed
ON OFF S Apdeejuds Dtarbilvee valve H2 Q2 UnStyhsrtoetmtled thPrroetstlsiunrge vaaclrvoes,s ∆H
30% improvement in efficiency by running (ASD) Curve
(cid:138) compressor at appropriately reduced speed reduced pump inlet Q2 0 Q2 Q1 Flow Rate (Q)
speed
using an ASD
(cid:139)Throttling introduces extra pressure drop,∆H
(cid:139)ASD reduces pump speed to match load requirement
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Energy Conservation in Blower Harnessing Wind Energy
Systems
Constant
Variable frequency
100 Outlet damper V awriianbdle V sapreiaebdle F r e AquCency P rPoocwesesring AC
Variable generator Unit
80 speed Utility
Power Consumption 60 Inlet vane Wind turbine
(% of full flow rate)
40
Electric drive
20
0
100 90 80 70 60 50 40 30
% Flow Rate
(cid:134)Relative power consumption using three methods
to reduce blower flow rate
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Hybrid Electric Vehicles Multi-disciplinary Nature of Electric
Drives
Machine
Utility Theory Power
Interaction Electronics
IC Continuously
Variable
Engine Transmission Electric Control
Sensors Drives
Theory
Electric
Motor Mechanical Real-time
system
modelling DSP control
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Summary
Evolution of Power Processing Unit
•What is an electric drive? Draw the block diagram and
explain the roles of its various components.
120 200 •What has been the traditional approach to controlling flow
% 104680000 SWiezeig (hvtolume) Relative unit 11055000 CFuonmcptioonnesnts •Wrathea ti na rteh et hper omcaejsosr i nddisuasdtvrya?ntages which can be overcome
20
0 0 by using adjustable speed drives?
1968 Y19e8a8r 1998 1968 1983 1988Yea1993r 1998 •What are the factors responsible for the growth of the
adjustable-speed drive market?
4kW DanfossVLTR power processing unit
•How does an air conditioner work?
(Consult a handbook such as [10].)
•How does a heat pump work?
•How do ASDssave energy in air conditioning and heat
pump systems?
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Summary
(cid:134)What is the role of ASDsin industrial systems?
Chapter 2
(cid:134)There are proposals to store energy in flywheels for load
leveling in utility systems. During the off-peak period
for energy demand at night, these flywheels are charged
to high speeds. At peak periods during the day, this energy Understanding Mechanical
is supplied back to the utility. How would ASDsplay a role
System Requirements
in this scheme?
(cid:134)What is the role of electric drives in electric transportation
systems of various types?
(cid:134)List a few specific examples from the applications mentioned
in section 1-4 that you are personally familiar with.
(cid:134)What are the different disciplines that make up the study
and design of electric-drive systems?
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Motivation Systems With Linear Motion
(cid:134)How can the ASD accelerate and decelerate the load to fe M fL fM M
give desired speed profile ⇒
x u=ddxt ; a=ddut = feM−fL u=dx ; a=du = fM
dt dt M
ASD Load
(cid:139)Figure on left includes load force, f , that must be overcome
L
ω
L
(cid:139)Figure on right shows only the force, f , available to accelerate
M
the mass, M
ω(rad/sec)
L
desired speed profile Accelaration Power Input Kinetic energy
100 a= feM−fL= fMM Pe(t)=fe⋅u= fM⋅u+fL⋅u WM=12Mu2
0 1 2 3 4 5 6 7 ( )
t sec
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Rotating Systems
(cid:134)Torque in an electric drive
f f
ω
90o Motor Tem TL Load
M β
r
θ torque Mg
θ (cid:139)Tem electromagnetic torque produced by motor
(cid:139)Tem is opposed by load torque, TL
(cid:139) Torque = force radius (cid:139) The difference, T e m − T L = T J , will accelerate the system
[Nm] [N] [m] (cid:139) dω=Tem −TL=TJ
dt J J
(cid:139) Example: what torque is needed to hold M motionless
where Jis the moment of inertia
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Calculation of Moment of Inertia
Accelaration, Speed and Position, Power
J
and Energy
of a Uniform Cylinder
d(cid:65)
dr
(cid:65)
rdθ
drf dM Motor Tem ωm TL Load Tem+−Σ TJ J1eq α ∫ ωm ∫ θ
r1 ω dθ TL
θ
acceleration, α = dωm = 1 (T −T )= TJ
df=dM d v dM =ρr(cid:78)dθ d(cid:78)r d(cid:78)(cid:65) dt (Jm+JL) em L Jeq
dt arc heightlength
⇒dT=r2dM dω=ρ(r3drdθd(cid:65))dω ⇒speed, ωm(t) =ωm(0)+∫0t α(τ)dτ
dt dt
⇒position, θ(t) =θ(0)+ ∫t ω(τ)dτ
T=ρ(r∫1r3dr2∫πdθ(cid:65)∫d(cid:65))dω= (πρ(cid:65)r14)dω 0
0 0 0 dt (cid:8)2(cid:11)(cid:9)(cid:11)(cid:10)dt Power P =T ⋅ω ; P =T ⋅ω
em em m L L m
J
Jsolid =π2ρ(cid:65)r14=21Mr12 KineticEnergy W = 1Jω2
2
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Frictional Torque
(cid:134)Example: Aerodynamic
coloumb drag
T friction
f Drag power at different speeds
stiction visTcfou=sB fωriction pfL==0fL.0⋅4u6CwAv2; (Cw: drag coefficient)
0 ω
 stiction ∴powerαspeed3

coloumb Power (W)
friction Speed
(km/h)
Stiction: static component C =0.3 C =0.5
(cid:139) w w
(cid:139) Coulomb friction: dynamic component 50 0.86 kW 1.44 kW
(constant magnitude) 100 6.9 kW 11.5 kW
Viscous friction: speed dependent 150 23.3 kW 38.8 kW
(cid:139)
In general, friction is non-linear
(cid:139)
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Torsional Resonances
Mechanical - Electrical Analogy
ω ω
m L
Motor Tem TL Load
T
shaft J
L
At motor end T Jm = T −J dωm (cid:132) Torque (cid:132) Current
shaft em m
dt (cid:132) Angular Velocity (cid:132) Voltage
dω Angular Displacement Flux Linkage
At load end T = T +J L (cid:132) (cid:132)
shaft L L
dt Moment of Inertia Capacitance
(cid:132) (cid:132)
T
(θ −θ ) = shaft Spring Constant 1/Inductance
m L (cid:132) (cid:132)
K
θm andθL :angularrotationat thetwoendsof theshaft (cid:132) Damping Coefficient (cid:132) 1/Resistance
(cid:139) If K →∞, θ =θ (cid:132) Coupling Ratio (cid:132) Transformer ratio
m L
( J a n d J can be treated as one inertial mass )
M L
(cid:139)Finite K may lead to resonances
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Coupling Mechanisms
Electrical Analogy of Motor & Load
Required when
(cid:134)
ωm ωL (cid:139) a (rotary) motor is driving a load which requires linear
Motor Tem TL Load (translational) motion
JL (cid:139) motors prefer higher rotational speed than that required
J
m
by the load
ωM 1/K ωL ωm (cid:139) the axis of rotation needs to be changed
Tem TJM Tshaft TJL TL Tem TJ TL
J J
M L (cid:134)Types
J =J +J
eq M L
(cid:139)Conveyor belts (belt and pulley)
Finite shaft stiffness Infinite shaft stiffness (cid:139)Rack and pinion or a lead-screw type of arrangement
(cid:139)Gear mechanisms
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Conversion between Linear and Rotary Gears
Systems
T r
f Tem 1 1
M uL Jm=motor inertia MotJor ωM
r M = mass of load M r T2 TL
ω r = pulley radius 2 Load
m
ω
Motor Tem du L JL
f = M + f (cid:139) Basic relationships: radius, speed, torque
L
J dt
m u = rω Equal speeds at gear surfaces ⇒r1ωM = r2ωL
m
dω Power transferred across gears ⇒ωMT1 = ωLT2,
T = r fd=ωr2M dtm +rfLdω ⇒rr21 = ωωML = TT21 & (cid:8)Te(cid:11)m(cid:11)−(cid:11)J(cid:9)M(cid:11)dω(cid:11)dtM(cid:11)(cid:10)ωωML = (cid:8)TL(cid:11)+(cid:11)J(cid:9)L(cid:11)ddω(cid:11)tL(cid:10)
T = J m + r2M +r f T T
em m L 1 2
(cid:8)(cid:11)(cid:9)d(cid:11)t(cid:10) (cid:8)(cid:11)(cid:11)(cid:9)dt(cid:11)(cid:11)(cid:10) (cid:139)Geared up: speed increased, torque decreasedωL >ωM; T2<T1; r2<r1
required toaccelerate due toload (cid:139) Geared down: speed decreased, torque increased
motor ωL <ωM; T2>T1; r2 >r1
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Gears (cont’d)
Types of Loads
(cid:139) Equivalent Inertia
Centrifugal loads
 ω 2dω ω 
Tem = Jm+JLωmL  dtm+ωmLTL
(cid:8)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:10) Fan
Jeq
ω 2 r 2
⇒ Jeq = Jm+JLωmL = Jm+JLr21
Constant Torque loads
(cid:139)Optimum gear ratio (to minimize T )
em
2
r  r  J
Jm =  1 ⋅JL ⇒  1 = m
r2opt. r2opt. JL
Hoist
and (Tem)opt.= 2Jmddωtm =2Jmrr21opt.ddωtL
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Types of Loads
Four-Quadrant Operation
Squared power loads
ω
m
(2) (1)
ω =+ ω =+
Compressor m m
T =− T =+
em em
ω p=− p=+
T m
em T
Load em
Constant power loads Motor ω =− ω =−
m m
Power T =− T =+
em em
p=+ p=−
(3) (4)
Winder
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Summary
Dynamic Operation
(cid:137)What are the MKS units for force, torque, linear velocity,
How the operating point changes with time
(cid:134) angular velocity, speed, and power?
Important for High Performance Drives (cid:137)What is the relationship between force, torque, and power?
(cid:134)
(cid:137)Show that torque is the fundamental variable in controlling
Speed change: rapid and without any oscillations
(cid:134) speed and position.
Requires good controller design (cid:137)What is the kinetic energy stored in a moving mass and a
(cid:134)
rotating inertia?
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Summary
Chapter 3
(cid:134)What is the mechanism for torsionalresonances?
(cid:134)What are the various types of coupling mechanisms? Review of
(cid:134)What is the optimum gear ratio to minimize the torque
Basic Electric Circuits
required from the drive to accelerate a load?
(cid:134)What are the torque-speed and the power-speed profiles
for various types of loads?
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