Electrical Drive Systems 324 - Stellenbosch...

Electrical Drive Systems 324DC Motors & Generators

Dr. P.J. Randewijk

Stellenbosch UniversityDep. of Electrical & Electronic Engineering

Stephan J. Chapman

Chapter 8 (5th Edition)

1 / 48

Outline

1 Chapter 8 – DC Motors and Generators8.1 – Introduction to DC Motors8.2 – The Equivalent Circuit of a DC Motor8.3 – The Magnetization Curve of a DC Machine8.4 – Separately Excited and Shunt DC Motors8.5 – The Permanent-Magnet DC Motor8.8 – DC Motor Starters8.10 – DC Motor Efficiency Calculations8.11 – Introduction to DC Generators8.12 – The Separately Excited DC Generator8.13 – The Shunt DC Generator

2 / 48

8.1 Introduction to DC Motors

The speed regulation (SR) of a DC motor is defined as

SR =nm,nl − nm,fl

nm,fl× 100% (8–2)

with nl ≡ no-load & fl ≡ full-load

The different types of DC motor that exist are:Ê separately excitedË shuntÌ permanent-magnetÍ seriesÎ compound

3 / 48

8.2 DC Motor Equivalent Circuit

The equivalent electrical circuit of the DC machine lookas follows,

A2

−EA=Kφωm

+

RA

LA

A1

ωm

F2

−LF

+

RF

F1

φ =⇒

LA, RA & EA is the Thevenin equivalent circuit of thearmature, andRF & LF is the equivalent impedance of the field

4 / 48

mailto:[email protected]

8.2 DC Motor Equivalent Circuit (cont.)

The dynamic model, including that for the mechanicalshaft, will look as follows:

−EA=Kφωm

+−

LAdiAdt

+

LA

−iARA

+

RA

iA

−+ VT

τload

τb=bωm

b

τJ=Jdωm

dt

J

ωm

τind=KφiAA2

A1

5 / 48


For the mechanical side, we can thus write:

τind = τb + τJ + τload

= bωm + Jdωm

dt+ τload

For steady state operation, i.e. dωmdt =0 (no acceleration),

the above equation simplifies to

τind = bωm + τload

For no-load operation, the only load is the viscousfriction component, so that at no-load

τind = bωm

6 / 48


All the DC machine’s mechanical (i.e. windage & friction)losses and core/iron (i.e. eddy & hysteresis) losses canusually be modelled a viscous frictional losses. . .

Pwindage & friction + core = PNL

= Pviscous

= bω2m

For the electrical side, we can thus write (KVL):

VT = EA + Vbrush + iARA + LAdiAdt

For steady state operation, i.e. diAdt =0 (with iA=IA

constant), the above equation simplifies to

7 / 48


VT = EA + Vbrush + IARA

If we were to ignore the brush losses, the aboveequation simplifies to

VT = EA + IARA

The steady state equivalent circuit of the DC machinecan thus be simplified to:

A2

−EA=Kφωm

+

RA

A1

ωm

F2

−LF

+

RF

F1

φ =⇒

8 / 48

8.3 DC Machine Magnetisation Curve

The flux produced by the field of the DC machine willdepend on the MMF of the field circuit,

F = NF IF

And the total reluctance, R of the DC machine, so thatwe can write

φ =FR

=NF IFR

=NF VF

RRF

with VF the voltage applied to the field circuit and RF theresistance of the field circuit

9 / 48

8.3 DC Machine Magnetisation Curve (cont.)

With an increase in field current, the flux in the DCmachine would increase linearly if saturation were to beignoredThe relationship would ultimately be determined by themagnetic core’s B–H relationship

10 / 48


If we were to drive the DC machine at a constant speed(e.g. ωm=ω0) whilst varying the field currentThe internal generated voltage will exhibit the sameB–H relationship as shown on the previous slide. . .

11 / 48


With the relationship between the internal generatedvoltage and speed given by

EA = Kφωm (7–38)

The value of Kφ for each value of the field current (IFcan be determined from the graphAs long as IF remains constant, Kφ will remain constantas Kφ is only a function of field current, i.e. Kφ = f (IF )Thus with IF constant, we can determine EA at adifferent speed directly from (7–38)Alternatively, with IF constant, if we know EA at onespeed,

EA|1 = Kφωm|1

12 / 48


We can also determine EA at a different speed, by usingthe following relationship

EA|2 = Kφωm|2

=EA|1

ωm|1ωm|2

= EA|1nm|1(

2π60 )

nm|2(2π60 )

= EA|1nm|1

nm|2

of

nm|2 = nm|1EA|1

EA|2(8–9)

13 / 48

8.4 SEPEX and Shunt DC Motors

For a Separately Excited (SEPEX) DC Machine, thefield circuit is connected to a separate supplyFor a Shunt DC Machine, the field circuit is connected inshunt / parallel with the armature circuit

The Terminal Characteristic of a Shunt DC Motor

+ A SEPEX DC Machine’s characteristics will be thesame. . .We are interested in the output speed versus torque ofthe DC machine, i.e. torque as a function of speed. . .

Ê Write down the Kirchoff’s voltage law (KVL) equation forthe armature circuit

Ë Make use of (7–39) & (7–49) to substitute EA & IAÌ Rewrite the equation so that we have ωm as a function of

τind

14 / 48

8.4 SEPEX and Shunt DC Motors (cont.)

Thus:

VT = EA + IARA (8–3)

= Kφωm +τind

KφRA (8–6)

resulting in the following equation

ωm =VT

Kφ− RA

(Kφ)2 τind (8–7)

which is the equation for a straight line, see Fig. 8–6 (a)+ We will ignore the effect of armature reaction, see Fig.

8–6 (b)

Nonlinear Analysis of a Shunt DC Motor – ignore

15 / 48


Speed Control of Shunt DC Motors

Speed control of SEPEX and Shunt DC Motors canaccomplished by:

Ê Inserting an external resistor in series with the armaturecircuit

Ë Adjusting the terminal voltage applied to the armaturecircuit

Ì Inserting an external resistor in series with the fieldcircuit in order to change the field current and hence thefield flux

16 / 48


INSERTING A RESISTOR IN SERIES WITH THEARMATURE CIRCUIT

An external variable resistor / rheostat is inserted inseries with the armature circuit

A2

−EA=Kφωm

+

RA

A1

ωm

F2

−LF

+

RF

F1

φ =⇒−+VF −

+ VT

RA|ext

17 / 48


Equation (8–7) can thus be rewritten as

ωm =VT

Kφ−

RA + RA|ext

(Kφ)2 τind

The y–intersect is not affected by RA|ext , only the slopeof the graph as RA|ext is adjusted (see next slide)

The advantage of rheostatic control is4 simple and easy to implement

The disadvantage of rheostatic control is8 it is usually manual control, i.e. difficult to automate8 not very energy efficient, i.e. additional IA|ext

2RA losses

18 / 48


0.00 p.u.

0.05 p.u.

0.10 p.u.

0.15 p.u.

0.20 p.u.

0.25 p.u.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Torque [p.u.]

Spe

ed[p

.u.] RA|ext

19 / 48


Sometimes it is more convenient to express torque as afunction of speed

τind =Kφ

(RA + RA|ext)VT −

(Kφ)2

(RA + RA|ext)ωm

Now the x–intersect is constant, but again with achanging slope as RA|ext is adjusted (see next slide)Specifically were one has load with a known torque vs.speed characteristic (e.g. a pump or a fan, were weknow τload=k2ωm)This is done in order to determine the operating point ofthe system, i.e. were the load’s characteristic curveintersects the motor’s characteristic curve

20 / 48


0.00

p.u.

0.05

p.u.

0.10

p.u.

0.15

p.u.

0.20

p.u.

0.25

p.u.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Speed [p.u.]

Torq

ue[p

.u.]

RA|ext

21 / 48


CHANGING THE ARMATURE VOLTAGE

This requires that the voltage applied to the terminals ofthe armature circuit, VT , be variable

A2

−EA=Kφωm

+

RA

A1

ωm

F2

−LF

+

RF

F1

φ =⇒−+VF −

+ VT

22 / 48


Equation (8–7) can thus be used as is

ωm =VT

Kφ− RA

(Kφ)2 τind

We can now see that only the y–intersect is affected byadjusting VT , whilst the slope remains constant (seenext slide)

The advantage of armature terminal voltage control is4 more energy efficient that rheostatic control4 there is a “nice” linear relationship between terminal

voltage and speed (will be shown later)

The disadvantage of armature terminal voltage controlis

8 a variable voltage source is not that trivial and8 it can be expensive

23 / 48


0.50 p.e.

0.60 p.e.

0.70 p.e.

0.80 p.e.

0.90 p.e.

1.00 p.e.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Draaimoment [p.e.]S

poed

[p.e

.]

VT

24 / 48


Again sometimes it is more convenient to expresstorque as a function of speed

τind =KφRA

VT −(Kφ)2

RAωm

We can now see that only the x–intersect is affected byadjusting VT , whilst again the slope remains constant(see next slide)

25 / 48


0.5

p.e.

0.6

p.e.

0.7

p.e.

0.8

p.e.

0.9

p.e.

1.0

p.e.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Spoed [p.e.]

Dra

aim

omen

t[p.

e.]

VT

26 / 48


Yet another way of looking at armature terminal voltagecontrol is by plotting VT as a function of ωm for differentvalues of τm (see next slide)Again we start by writing down the KVL equation andthen substituting (7–38) & (7–49) into it

VT =

IA︷︸︸︷τm

KφRA +

EA︷︸︸︷Kφωm

N.B.There is a “nice” linear relationship between the speed of aDC motor and the applied Armature-Terminal Voltage

27 / 48


0.0 p.u.

0.5 p.u.

1.0 p.u.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Speed [p.u.]Te

rmin

alVo

ltage

[p.u

.]

τm

+ If RA → 0, the the three graphs will be on top of oneanother. . .

28 / 48


CHANGING THE FIELD RESISTANCE (OR FIELDVOLTAGE)

This requires that the voltage applied to the terminals ofthe field circuit, VF , be variable, or an external resistor inseries with the field circuit being used (as shown in Fig.8–7)

A2

−EA=Kφωm

+

RA

A1

ωm

F2

−LF

+

RF

F1

φ =⇒−+VF −

+ VT

29 / 48


Equation (8–7) can again be used as is

ωm =VT

Kφ− RA

(Kφ)2 τind

but now with

Kφ = f (IF )

= f (VF

RF)

We can now see that both the y–intersect and the slopeis affected by the change in flux, i.e. the change in IF(see next slide)

30 / 48


1.00 p.u.0.90 p.u.

0.80 p.u.

0.70 p.u.

0.60 p.u.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Torque [p.u.]

Spe

ed[p

.u.]

IF

31 / 48


The advantage of field current control is4 can operate above rated speed4 easier to control due to the lower current value of the

field current compared to the armature current – thepower and current rating of RF |ext or VF is much smallerthan that of RA|ext or VT

The disadvantage of field current control is8 if the field current and hence the flux goes to zero, i.e.φ→ 0, the speed of the machine can “run away”, i.e.ωm →∞

32 / 48


Again sometimes it is more convenient to expresstorque as a function of speed

τind =KφRA

VT −(Kφ)2

RAωm

but again now with

Kφ = f (IF )

= f (VF

RF)

We can again now see that both the y–intersect and theslope is affected by the change in flux, i.e. the change inIF (see next slide)

33 / 48


1.0

p.u.

0.9

p.u.

0.8

p.u.

0.7

p.u.

0.6

p.u.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Speed [p.u.]

Torq

ue[p

.u.] IF

34 / 48


TORQUE AND POWER LIMITS OF A DC MOTOR

In all the graphs shown in die previous slides, therewere a “dot” in the middle. . .This is the rated operating point of the DC machine

Which occurs when,4 the rated terminal voltage is applied,4 the rated field current is drawn and4 the rated armature current is drawn,

So that the DC machine will be+ developing its rated torque and+ developing its rated power and+ running at rated (or base) speed

35 / 48


Below rated / base speed, speed control is done byarmature terminal voltage controlarmature external resistance (rheostatic) control

Above rated / base speed, speed control is done byfield current control

Electrical motors different somewhat from InternalCombustion Engines (ICEs) in terms of their

“maximum developed torque” and“maximum developed power”

Defined in Chapman as

τmax = KφIA|max (8–14)

and

Pmax = τmaxωm (8–15)

36 / 48


I prefer (and it is less confusing) to rather refer to it asrated torque and rated speed. . .So that I would redefine (8–14) & (8–15) to

τrated = KφIA|rated (8–14’)

and

Prated = τratedωm|rated (8–15’)

The interesting thing about electrical motors is that itcan develop more than its rated torque. . .It however implies that more than the rated armaturecurrent needs to be drawn. . .Which could result in the motor burning out . . .

37 / 48


For an ICE, this is not possible

Diesel Engine

PM Alternator

2,000-3,600 r/min Three Phase, 250-450 Vac

Diode Rectifier/

Boost Converter

350-450 Vdc Selectable Output Inverter

50/60/400 Hz

Common Mode and Transverse Mode Filter

120 Vac, Single Phase 120/240 Vac, Dual Phase 120/208 Vac Three Phase

Bidirectional Buck/Boost Converter

28 Vdc

28 Vdc Auxiliary Loads

Two 12-V Batteries in

Series

Power Supply for Controller and Power Electronics

1

Functional block diagram of power conversion components in generator set.

conditions. It is also possible to control the engine to run where it is most audibly quiet, at its least-polluting operating point (from an emissions point of view), or at its most reliable, stiffest point such that it is less sensitive to load transients. This article describes a proof-of-concept development for a 7.5-kW gen-set in a family of military gen-sets in the 5- to 60-kW range.

Gen-Set Operation A block diagram of the electronic power conversion system for the proof-of-concept gen-set developed at the Oak Ridge National Laboratory (ORNL) is shown in Figure 1 [6], [7]. The military gen-set uses an internal combustion (IC) diesel engine to drive a radial-gap permanent magnet (PM) alternator at variable speed. The speed of the engine is determined from a user-selectable interface that allows the engine to run at its most efficient operating point for a given load and ambient thermal conditions. The variable frequency, variable voltage produced by the PM alternator is diode-rectified to dc voltage, and an inverter is used to produce selectable-frequency, controllable ac voltage.

The user is allowed to select single-phase 120-V, dual-phase 120/240-V, or three-phase 120/208-V. Each of these voltage configurations can be generated at 50, 60, and 400 Hz such that the unit can be compatible with equipment produced from around the world or for aerospace applications. The power-conversion system also incorporates a bidirectional dc-dc converter that can charge 24-V batteries that are used to start the IC engine and to power auxiliary loads [8]. The converter can also draw power from the batteries to help maintain the dc link during severe load transients.

Engine and Alternator Description Each gen-set size was determined by selecting an advanced diesel engine that had a high power-to-weight ratio. For the smallest gen-set, an air-cooled Ruggerini MD 191 rated at 13 kW was used as the prime mover. The engine

Torq

ue (

N m

) or

Pow

er (

kW)

⋅

Engine Speed (r/min)

Fue

l Con

sum

ptio

n (k

g/kW

h)

50 45 40 35 30 25 20 15 10

5 0 1,800 2,200 2,600 3,000 3,400 3,800

0.40 0.38 0.36 0.34 0.32 0.30 0.28 0.26 0.24 0.22 0.20

Torque

Fuel Consumption

Power

2

Peak torque, power, and fuel consumption for the diesel

engine in the proof-of-concept gen-set.

3

PM alternator mounted to gen-set engine. 49

IEE

E IN

DU

ST

RY

AP

PL

ICA

TIO

NS

MA

GA

ZIN

E •

M

AR

|A

PR

20

03

•

WW

W.IE

EE

.OR

G/

IAS

38 / 48


For a DC motor, the maximum (or rated) torque–speedand power–speed curves, looks as follows

39 / 48


+ Below rated / base speed,Ê the applied armature terminal voltage increases linearly

with speed,Ë but the armature current can be at its maximum / rated

value, even from standstillÌ thus the maximum power increases linearly with speedÍ where as the maximum torque the DC motor can

develop will be constant

+ Above rated / base speed,Ê the applied armature terminal voltage can’t increase any

more,Ë neither can the armature current,Ì thus the maximum power stays constant (it can’t

increase)Í and hence the maximum torque the DC motor can

develop, will decrease hyperbolically (k = xy )

40 / 48

8.5 The Permanent-Magnet DC Motor

A Permanent-Magnet (PM) DC motor uses PMs togenerate the flux in the motorThus the flux is fixed and hence Kφ is constantFor PM DC motors, manufacturers sometimes refer toKφ as motor or machine constant, Km

Advantages of PM DC Motors arewe don’t have to worry about the fieldKm = Kφ = constant

Disadvantages of PM DC Motors arePM DC motors are expensivewe can’t operate the PM DC motor above rated speedbecause we can’t reduce the flux in the machine

Ignore rest of this section. . .

41 / 48

8.8 DC Motor Starters

From the KVL equation for the armature circuit,

IA =VT − EA

RA

We can see that at start-up,the speed ωm=0, and thus the back-EMF, EA=0coupled with the fact that armature resistance, RA, isusually very smallwe can see that the armature current, IAand hence the torque developed at standstill, τind, will beextremely high

+ which could lead to “something breaking” on themechanical side

Hence a starting resistor in series with armature circuitor a lower starting armature terminal voltage is required

Ignore rest of this section. . .42 / 48

8.10 DC Motor Efficiency Calculations

+ Have a look again at Chapman Section 7.7

The efficiency of a DC motor at a certain operatingpoint, is given by

η =Pout

Pin× 100%

=Pconv − Pno-load

Parmature + Pfield× 100%

=(τind − bωm)ωm

VT IA + VF IF× 100%

+ Don’t forget about the losses in the field winding. . .+ Chapman doesn’t calculate the viscous friction

coefficient, b, and assumes the no-load losses(Pmech + Pcore) stays constant and is not affected byspeed

43 / 48

8.11 Introduction to DC Generators

For DC motor, speed regulation is important, but for aDC generator (similar to a transformer) voltageregulation (VR) is important

VR =Vnl − Vfl

Vfl× 100% (8–39)

with nl ≡ no-load & fl ≡ full-load

The different types of DC generators that exist are:Ê separately excitedË shuntÌ permanent-magnetÍ seriesÎ cumulatively compoundÏ differential compound

44 / 48

8.12 The Separately Excited Generator

The equivalent circuit for the SEPEX DC generatorlooks exactly the same as for a SEPEX DC motorThe only difference is that a motor is electrically drivenand mechanically loadedWhere as a generator is mechanically driven andelectrically loaded

A2

−EA=Kφωm

+

RA

A1

ωm

F2

−LF

+

RF

F1

φ =⇒−+VF Rload

−VT

+

IL

45 / 48

8.12 The Separately Excited Generator (cont.)

Also for a motor we were interested in the speed vs.torque, or torque vs. speed characteristicsWhere as for a generator, we are interested in theterminal voltage, VT vs. the load current, IL,characteristicsfrom the KVL equation for the armature circuit and withIA = IL

VT = EA − IARA (8–41)

Furthermore, with EA = Kφωm, with the flux being keptconstant, i.e. IF being kept constant, and the rotationalspeed, ωm, constantAnd hence the terminal voltage VT will vary linearly withthe load current, IL – see Fig. 8–45 (a)

46 / 48

8.12 The Separately Excited Generator (cont.)

Control of the Terminal VoltageAlso with EA = Kφωm, it follow that the terminal voltageof the SEPEX generator can be controlled by, either

Ê Changing the speed of rotationË Changing the field current

Nonlinear Analysis of a Separately Excited DC Generator– ignore (i.e. ignore the effect of armature reaction)

47 / 48

8.13 The Shunt DC Generator

+ The only thing that is important to understand here, isthat the voltage buildup during starting of the shunt DCgenerator occurs due to the remanent or residualmagnetic flux in the core of the machine. . .

48 / 48

Electrical Drive Systems 324 - Stellenbosch...

Documents

Transcript of Electrical Drive Systems 324 - Stellenbosch...