electric drives professor mohammed zeki khedher lecture one

Electric Drives

Professor Mohammed Zeki Khedher

Lecture One

Introduction

In some countries nearly 65% of the total electric energy produced is consumed by electric motors.

Some Applications of Electric Drives

• Electric Propulsion

• Pumps, fans, compressors

• Plant automation

• Flexible manufacturing systems

• Spindles and servos

• Appliances and power tools

• Cement kilns

• Paper and pulp mills; textile mills

• Automotive applications

• Conveyors, elevators, escalators, lifts

1. ENERGY CONVERSION IN ELECTRIC DRIVES

1.1. ELECTRIC DRIVES - A DEFINITION

Figure 1.1. Constant speed electric drive

About 50% of electrical energy produced is used in electric drives today.Electric drives may run at constant speed (figure 1.1) or at variable speed (figure 1.2).

Figure 1.2. Variable speed electric drive

1.2. APPLICATION RANGE OF ELECTRIC DRIVES

A summary of main industrial applications and power range of electric drives is shown on figure 1.3.

Figure 1.3. Electric drives - variable speed applications

INTRODUCTION TO ELECTRIC DRIVES - MODULE 1

Overview of AC and DC drives

Energy/Cost Savings

• System efficiency can be increased from 15% to 27% by introducing variable-speed drive operation in place of constant-speed operation.

• For a large pump variable-speed drive, payback period ~ 3-5 years whereas operating life is ~ 20 years.

Electric Machines

“An engineer designing a high-performance drive system must have intimate knowledge about machine performance and Power Electronics”

Electric Machines (cont’d)

• DC Machines - shunt, series, compound, separately excited dc motors and switched reluctance machines

• AC Machines - Induction, wound rotor synchronous, permanent magnet synchronous, synchronous reluctance, and switched reluctance machines.

• Special Machines - switched reluctance machines

Electric Machines (cont’d)

All of the above machines are commercially available in fractional kW to MW ranges except permanent-magnet, synchronous, synchronous reluctance, and switched reluctance which are available up to 150 kW level.

Selection Criteria for Electric Machines

• Cost• Thermal Capacity• Efficiency• Torque-speed profile• Acceleration• Power density, volume of motor• Ripple, cogging torques• Peak torque capability


Electrical Drives

• About 50% of electrical energy used for drives

• Can be either used for fixed speed or variable speed

• 75% - constant speed, 25% variable speed (expanding)

Example on VSD application

motor pump

valve

Supply

Constant speed Variable Speed Drives

PowerIn

Power lossMainly in valve

Power out



motor pump

valve

SupplymotorPEC pump

Supply


PowerIn

Power loss

Power out



Power outPowerIn


Power out

motor pump

valve

SupplymotorPEC pump

Supply




PowerIn

Power loss

PowerIn

Power out


Conventional electric drives (variable speed)

• Bulky

• Inefficient

• inflexible


Modern electric drives (With power electronic converters)

• Small

• Efficient

• Flexible


Modern electric drives

• Inter-disciplinary

• Several research area

• Expanding

Machine designSpeed sensorlessMachine Theory

Non-linear controlReal-time controlDSP applicationPFCSpeed sensorless Power electronic converters

Utility interfaceRenewable energy

20

Controllers

Controllers embody the control laws governing the load and motor characteristics and their interaction.

Controller

Torque/speed/position commands

Torque/speed/ position feedback

Thermal andother feedback

Vc, fc, start,shut-out, signals, etc.


Components in electric drives

e.g. Single drive - sensorless vector control from Hitachi



e.g. Multidrives system from ABB



Motors• DC motors - permanent magnet – wound field• AC motors – induction, synchronous , brushless DC• Applications, cost, environment

Power sources• DC – batteries, fuel cell, photovoltaic - unregulated• AC – Single- three- phase utility, wind generator - unregulated

Power processor• To provide a regulated power supply• Combination of power electronic converters

•More efficient •Flexible •Compact •AC-DC DC-DC DC-AC AC-AC

25



Control unit• Complexity depends on performance requirement• analog- noisy, inflexible, ideally has infinite bandwidth.• digital – immune to noise, configurable, bandwidth is smaller than

the analog controller’s • DSP/microprocessor – flexible, lower bandwidth - DSPs perform

faster operation than microprocessors (multiplication in single cycle), can perform complex estimations

26

DC Motors

• Limitations:

• Advantage: simple torque and speed control without sophisticated electronics

• Regular Maintenance • Expensive motor

• Heavy motor • Sparking

27

DC DRIVES Vs AC DRIVES

DC drives:

Advantage in control unit

Disadvantage in motor

AC Drives:

Advantage in motor

Disadvantage in control unit

28

1.3. ENERGY SAVINGS PAYS OFF RAPIDLY

Consider a real case when a motor pump system of 15kW works 300 days a year, 24 hours a day and pumps 1200m3 of water per day. By on/off and throttling control, only, the system uses 0.36kWh/m3 of pumped water to keep the pressure rather constant for variable flow rate.Adding a P.E.C., in the same conditions, the energy consumption is 0.28kWh/m3 of pumped water, with a refined pressure control.Let us consider that the cost of electrical energy is 40fils/kWh.The energy savings per year S is:

S = 1200 * 300 * (0.36 -0.28) * 0.04 /year = 1152 JD/year

Now the costs of a 15kW PWM - P.E.C. for an induction motor is less than 4000JD. Thus, to a first approximation, the loss savings only pay off the extra investment in less than 4 years.

29

Costs

• Power Electronics Controller costs approximately 2 to 5 times AC motor

• Cost decreases with bigger size

30

1.4. GLOBAL ENERGY SAVINGS THROUGH P.E.C. DRIVES

So far the energy savings produced by the P.E.C. in variable speed drives have been calculated for the drive only - P.E.C. and motor.

Figure 1.5. Primary energy consumption for throttle / motor / pump system

31

Figure 1.6. Primary energy consumption for P.E.C. / motor / pump systems

32

Power consumption with flow

Load

The motor drives a load that has a characteristic torque vs. speed requirement.

In general, load torque is a function of speed and can be written as:

Tl mx

x=1 for frictional systems (e.g. feed drives)

x=2 for fans and pumps

34

General Torque EquationTranslational (linear) motion:

dt

dJT

Rotational motion:

dt

dvMF

F : Force (Nm)M : Mass (Kg )v : velocity (m/s)

T : Torque (Nm)J : Moment of Inertia (Kgm2 ) : angular velocity ( rad/s )

35

Torque Equation: Motor drives

dt

dJTTor

dt

dJTT LeLe

0 Le TT Acceleration

0 Le TT Deceleration

0 Le TT Constant speed

Te : motor torque (Nm) TL : Load torque (Nm)

36

…continue

Drive accelerates or decelerates depending on whether Te is greater or less than TL

During acceleration, motor must supply not onlythe load torque but also dynamic torque, ( Jd/dt ).

During deceleration, the dynamic torque, ( Jd/dt ), hasa negative sign. Therefore, it assists the motortorque, Te.


Elementary principles of mechanics

M

v

Fm

Ff

dtMvd

FF fm

Newton’s law

Linear motion, constant M

• First order differential equation for speed• Second order differential equation for displacement

Ma

dtxd

Mdtvd

MFF 2

2

fm

x



• First order differential equation for angular frequency (or velocity)• Second order differential equation for angle (or position)

2

2m

le dtd

Jdt

dJTT

With constant J,

Rotational motion

- Normally is the case for electrical drives

dtJd

TT mle

Te , m

Tl

J


dtd

JTT mle

For constant J,

dt

dJ m Torque dynamic – present during speed transient

dt

d m Angular acceleration (speed)

The larger the net torque, the faster the acceleration is.

0.19 0.2 0.21 0.22 0.23 0.24 0.25-200

-100

0

100

200

spee

d (r

ad/s

)

0.19 0.2 0.21 0.22 0.23 0.24 0.250

5

10

15

20

torq

ue (

Nm

)




dtvd

MFF le

Combination of rotational and translational motions

r r

Te,

Tl

Fl Fe

v

M

Te = r(Fe), Tl = r(Fl), v =r

dtd

MrTT 2le

r2M - Equivalent moment inertia of the linearly moving mass


Elementary principles of mechanics – effect of gearing

Motors designed for high speed are smaller in size and volume

Low speed applications use gear to utilize high speed motors

MotorTe

Load 1, Tl1

Load 2, Tl2

J1

J2

mm1

m2

n1

n2


MotorTe

Load 1, Tl1

Load 2, Tl2

J1

J2

mm1

m2

n1

n2

MotorTe

Jequ

Equivalent Load , Tlequ

m2

221equ JaJJ

Tlequ = Tl1 + a2Tl2

a2 = n1/n2

Elementary principles of mechanics – effect of gearing

43

1.6. MOTION / TIME PROFILE MATCH

Figure 1.8. Motion / time profilea.) speed b.) position c.) required load torque

44

Figure 1.9. Required speed / time profile

T t J t T te r L

Example 1.2. The direct drive torque / time curveA direct drive has to provide a speed / time curve such as in figure 1.9. against a constant load torque of TL = 10Nm, for a motor load inertia J = 0.02 kgm2.

Neglecting the mechanical losses let us calculate the motor torque (Te) / time requirements.The motion equation for a direct drive is:

45

rr

atrad smax .

./

376 8

0 21884 2

T

Nm

Nm

Nme

1884 0 02 10 37 68 10 47 68

0 10 10

1884 0 02 10 37 68 10 27 68

. . . ;

;

. . . ;

for 0 t 0.2s

for 0.2 t 0.8s

for 0.8 t 1s

r 0 0.

For the linear speed / time (acceleration - deceleration) zones the speed derivative is:

For the constant speed (cruising) zone .Consequently the torque requirements from the motor for the three zones are:

46

Figure 1.10. Motor torque / time requirements

The motor torque / time requirements are shown on figure 1.10.

47

Example 1.3. gear - box drive torque / time curveLet us consider an electric drive for an elevator with the data shown in

figure 1.11.

Figure 1.11. Elevator electric drive with multiple mechanical transmissions and counterweight

48

The motor rated speed nn = 1550rpm. The efficiency of gearing system is = 0.8.Let us calculate the total inertia (reduced to motor shaft), torque and power without and with counterweight.First the motor angular speed m is:

rad/s 22.16260

15502n2 nm (1.12)

The gear ratios may be defined as speed ratios - t /m for J4+J5 and d /m for J6 (figure 1.11).Consequently the inertia of all rotating parts Jr, reduced to the motor shaft, (figure 1.11), is:

2

m

2d

62m

2t

54321r JJJJJJJ

222

kgm 062.2522.162

5.78

22.162

5.22005.02815

(1.13)

49

For the cabin and the counterweight, the inertia, reduced to motor shaft (Je) is:

22

2

2m

2

cwce kgm 07238.022.166

18001200

ummJ

(1.14)

Thus the total inertia Jt is:

2ert kgm 135.2507238.0062.25JJJ (1.15)

In absence of counterweight the la of energy conservation leads to:

ugmT cmem (1.16)

Consequently the motor torque, Tem, yields:

Nm 71.908.022.162

181.91200Tem

(1.17)

50

The motor electromagnetic power Pem is:

W 1471522.16271.90TP memem (1.18)

On the other hand in presence of counterweight (1.16) becomes:

ugmm'T cwcmem (1.19)

Nm 71.30

8.022.162

181.98001200'Tem

(1.20)

So the motor electromagnetic P’em is:

W 490522.16271.30'T'P memem (1.21)


Motor steady state torque-speed characteristic

Synchronous mch

Induction mch

Separately / shunt DC mch

Series DC

SPEED

TORQUE

By using power electronic converters, the motor characteristic can be change at will


Load steady state torque-speed characteristic

SPEED

TORQUE

Frictional torque (passive load) • Exist in all motor-load drive system simultaneously

• In most cases, only one or two are dominating

• Exists when there is motion

T~ C

Coulomb friction

T~

Viscous friction

T~ 2

Friction due to turbulent flow

TL

Te

Vehicle drive



Constant torque, e.g. gravitational torque (active load)

SPEED

TORQUE

Gravitational torque

gM

FL

TL = rFL = r g M sin



Hoist drive

Speed

Torque

Gravitational torque


Load and motor steady state torque

At constant speed, Te= Tl Steady state speed is at point of intersection between Te and Tl of the steady state torque characteristics

TlTe

Steady state speed

r

Torque

Speedr2r3r1


Torque and speed profile

10 25 45 60 t (ms)

speed (rad/s)

100

The system is described by: Te – Tload = J(d/dt) + B

J = 0.01 kg-m2, B = 0.01 Nm/rads-1 and Tload = 5 Nm.

What is the torque profile (torque needed to be produced) ?

Speed profile



10 25 45 60 t (ms)

speed (rad/s)

100

0 < t <10 ms Te = 0.01(0) + 0.01(0) + 5 Nm = 5 Nm

10ms < t <25 ms Te = 0.01(100/0.015) +0.01(-66.67 + 6666.67t) + 5 = (71 + 66.67t) Nm

25ms < t< 45ms Te = 0.01(0) + 0.01(100) + 5 = 6 Nm

45ms < t < 60ms Te = 0.01(-100/0.015) + 0.01(400 -6666.67t) + 5 = -57.67 – 66.67t

le TBdtd

JT



10 25 45 60

speed (rad/s)

100

10 25 45 60

Torque (Nm)

72.6771.67

-60.67

-61.67

56

t (ms)

t (ms)

Speed profile

torque profile



10 25 45 60

Torque (Nm)

70

-65

6

t (ms)

For the same system and with the motor torque profile given above, what would be the speed profile?

J = 0.001 kg-m2, B = 0.1 Nm/rads-1 and Tload = 5 Nm.

60

Torque Equation: Graphical

Te

Forwardrunning

Speed

Forwardbraking

Reverseacc.

Reverserunning

Reversebraking

Forwardacc.

61

Load Torque

Load torque, TL, is complex, depending on applications.

SPEED

TORQUE

TL = k2TL = k

TL = k

In general:


Thermal considerations

Unavoidable power losses causes temperature increase

Insulation used in the windings are classified based on the temperature it can withstand.

Motors must be operated within the allowable maximum temperature

Sources of power losses (hence temperature increase): - Conductor heat losses (i2R) - Core losses – hysteresis and eddy current - Friction losses – bearings, brush windage

63


Steady-state stability

64

1.5.1. Typical load torque / speed curves

Typical load torque / speed curves are shown on figure 1.7. They give a strong indication of the variety of torque / speed characteristics. Along such curves the mechanical power required from the motor varies with speed.

Figure 1.7. Typical load speed / torque, speed / power curves

65

1.7. LOAD DYNAMICS AND STABILITY

loadfrictioner

t TTTdt

dJ

(1.22)

WVCSfriction TTTTT (1.23)

where TS is the static friction torque (at zero speed); TC is Coulomb friction torque (constant with speed); TV is viscous friction torque (proportional to speed) and TW is windage friction (including the ventilator braking torque, proportional to speed squared):

rV 'BT 2

rW CT

(1.24)

(1.25)

66

Figure 1.13. Mechanical characteristics:a.) d.c. brush motor with separate excitation b.) induction motor

c.) synchronous motor

Figure 1.12. Components of friction torque, Tfriction

67

IV (Induction)I (Synchronous)

II (dc Shunt/Separately excited)

III (dc series excited)

TmaxTorque

Speed

Various Motor Characteristics

68

Example 1.4. D.C. brush motor drive stability.A permanent magnet d.c. brush motor with the torque speed curve: drives a d.c. generator which supplies a resistive load such that the generator torque / speed equation is r = 2TL. We calculate the speed and torque for the steady state point and find out if that point is stable.Solution:Let us first draw the motor and load (generator) torque speed curves on figure 1.14.

Figure 1.14. D.C. brush motor load match

69

The steady state point, A, corresponds to constant speed and B = 0 in (1.27). Simply the motor torque counteracts the generator braking torque:

eL TT (1.37)

Using the two torque speed curves we find:

21.0200 rA

rA

(1.38)

and thus:rad/s 476.190

2/1.01

200rA

(1.39)

andNm 238.95

2

476.190

2TT rA

LAeA

(1.40)

The static stability is met if:

Ar

L

Ar

e TT

(1.41)

In our case from the two torque / speed curves:

2

110 (1.42)

and thus, as expected, point A represents a situation of static equilibrium.



Electrical machines can be overloaded as long their temperature does not exceed the temperature limit

Accurate prediction of temperature distribution in machines is complex – hetrogeneous materials, complex geometrical shapes

Simplified assuming machine as homogeneous body

p2p1 Thermal capacity, C (Ws/oC)

Surface A, (m2)Surface temperature, T (oC)Input heat power

(losses)

Emitted heat power(convection)

Ambient temperature, To



Power balance:

21 ppdtdT

C

Heat transfer by convection:

)TT(Ap o2

Cp

TC

Adt

Td 1

Which gives:

/th e1A

pT

AC

, where

With T(0) = 0 and p1 = ph = constant ,

, where is the coefficient of heat transfer



t

T

t

/te)0(TT

T

/th e1A

pT

Heating transient

Cooling transient

Aph

)0(T



The duration of overloading depends on the modes of operation:

Continuous duty Short time intermittent duty Periodic intermittent duty

Continuous duty

Load torque is constant over extended period multiple

Steady state temperature reached

Nominal output power chosen equals or exceeds continuous load

T

t

Ap n1

p1n

Losses due to continuous load



Short time intermittent duty

Operation considerably less than time constant,

Motor allowed to cool before next cycle

Motor can be overloaded until maximum temperature reached

t1




Ap s1

maxT Ap n1

t

T

p1

p1n

p1s

t1




t

T

/ts1 e1A

pT

maxT Ap n1

/ts1n1 1e1A

pA

p /ts1n1

1e1pp1

/tn1

s1

te11

pp

1



Periodic intermittent duty

Load cycles are repeated periodically

Motors are not allowed to completely cooled

Fluctuations in temperature until steady state temperature is reached




p1

t

heating coollingcoolling

coolling

heating

heating




Example of a simple case – p1 rectangular periodic pattern

pn = 100kW, nominal powerM = 800kg= 0.92, nominal efficiencyT= 50oC, steady state temperature rise due to pn

kW911

pp n1

Also, C/W180

509000

Tp

A o1

If we assume motor is solid iron of specific heat cFE=0.48 kWs/kgoC, thermal capacity C is given by

C = cFE M = 0.48 (800) = 384 kWs/oC

Finally , thermal time constant = 384000/180 = 35 minutes




Example of a simple case – p1 rectangular periodic pattern

For a duty cycle of 30% (period of 20 mins), heat losses of twice the nominal,

0 0.5 1 1.5 2 2.5

x 104

0

5

10

15

20

25

30

35


Torque-speed quadrant of operation

T

12

3 4

T +ve +vePm +ve

T -ve +vePm -ve

T -ve -vePm +ve

T +ve -vePm -ve


4-quadrant operation

m

Te

Te

m

Te

m

Te

m

T

• Direction of positive (forward) speed is arbitrary chosen

• Direction of positive torque will produce positive (forward) speed

Quadrant 1Forward motoring

Quadrant 2Forward braking

Quadrant 3Reverse motoring

Quadrant 4Reverse braking


Ratings of converters and motors

Torque

Speed

Power limit for continuous torque

Continuous torque limit

Maximum speed limit

Power limit for transient torque

Transient torque limit

1.8. MULTIQUADRANT OPERATION

These possibilities are summarised in Table 1.1 and in figure 1.16.

Table 1.1.

Mode ofoperation

Forwardmotoring

Forwardregenerative braking

Reversemotoring

Reverseregenerative braking

Speed, r + + - -

Torque, Te + - - +

Electricpower flow

+ - + -

4Q OPERATION

SPEED

TORQUE

I

III

II

IV

TeTe

Te Te

FMFB

RMRB

F: FORWARD R: REVERSE M : MOTORING B: BRAKING

4Q OPERATION: LIFT SYSTEM

Counterweight Cage

Motor

Positive speed

Negative torque


Convention:

Upward motion of the cage: Positive speed

Weight of the empty cage < Counterweight

Weight of the full-loaded cage > Counterweight

Principle:

What causes the motion?

Motor : motoring P =T= +ve

Load (counterweight) : braking P =T = -ve


You are at 10th floor, callingfully-loaded cage from gnd floor

You are at gnd floor, callingempty cage from 10th floor

You are at 10th floor, callingempty cage from gnd floor

You are at gnd floor, callingFully-loaded cage from 10th floor

Torque

Speed

FMFB

RM RB

DC MOTOR DRIVES

Principle of operation

Torque-speed characteristic

Methods of speed control

Armature voltage control

Variable voltage source

Phase-controlled Rectifier

Switch-mode converter (Chopper)

1Q-Converter

2Q-Converter

4Q-Converter

Figure 1.16. Electric drives with four quadrant operation

electric drives professor mohammed zeki khedher lecture one

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