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1-0

PowerPoint Slides to accompany

Electric MachinerySixth Edition

A.E. Fitzgerald

Charles Kingsley, Jr.

Stephen D. Umans

Chapter 1

Magnetic Circuits and

Magnetic Materials




1-1

1.1 Introduction to Magnetic Circuits




Simple magnetic circuit.Figure 1.1

1-2




Magnetic circuit with air gap.

Figure 1.2

1-3




1-4

Analogy between electric and magnetic circuits.

(a) Electric circuit, (b) magnetic circuit.Figure 1.3




1-5

Air-gap fringing fields.

Figure 1.4




1-6

Simple

synchronousmachine.Figure 1.5




1-7

1.2 Flux linkage, Inductance, and Energy




1-8

• When magnetic field varies in time an electric field is produced in space as

determined by Faraday’s Law:

Thisimagecannotcurrently bedisplayed.

S C

dt

d daBdsE ..

• Line integral of the electric field intensity E around a closed contour C is equal to the

time rate of the magnetic flux linking that contour.

dt

d N

dt

d t e

)(

• Since the winding (and hence the contour C) links the core flux N times then above

equation reduces

• The induced voltage is usually refered as electromotive force to represent the

voltage due to a time-varying flux linkage.

Faraday’s Law




1-9

t B At t c sinsin)( maxmax

t E t N t e coscos)( maxmax

maxmaxmax 2 B NA f N E c

+

-

e(t) N

The direction of emf: If the winding terminalswere short-circuited a current would flow insuch a direction as to oppose the change of fluxlinkage.

max2 B NA f E crms




1-10

(a) Magnetic circuit and (b) equivalent circuit for Example 1.3.Figure 1.6




1-11

MATLAB plot of inductance vs. relative permeability for

Example 1.5.Figure 1.7




1-12

Magnetic circuit with two windings.

Figure 1.8




1-13

1.3 Properties of Magnetic Materials




1-14

B-H loops for M-5 grain-oriented electrical steel 0.012 in thick.

Only the top halves of the loops are shown here. (Armco Inc.)Figure 1.9




1-15

Dc magnetization curve for M-5 grain-oriented electrical steel

0.012 in thick. (Armco Inc.)Figure 1.10




1-16

1.4 AC Excitation




1-17

Excitation phenomena. (a) Voltage, flux, and exciting current;(b) corresponding hysteresis loop.Figure 1.11




1-18

Exciting rms voltamperes per kilogram at 60 Hz for M-5

grain-oriented electrical steel 0.012 in thick. (Armco Inc.)Figure 1.12




1-19

Hysteresis Losses: hysteresisloss is proportional to the loop area(shaded).

Figure 1.13

cccccccc dB H l A NdB A

N

l H d iW

Eddy Current Losses:

Time-varying magnetic fields give rise to electric fields in the materialresulting in induced currents. These induced currents cause Eddy CurrentLosses. These losses can be reduced by using thin sheets of laminations ofthe magnetic material.

CORE LOSSES




1-20




1-21

Core loss at 60 Hz in watts per kilogram for M-5

grain-oriented electrical steel 0.012 in thick. (Armco Inc.)Figure 1.14




1-22

Example 1.8.The magnetic core is made from laminations of M-5 grain-oriented electrical steel. The

winding is excited with a 60 Hz voltage to produce a flux density in the steel ofB=1.5sin wt T, where w=377 rad/sec. The steel occupies 0.94 of the core cross-sectional area. The mass-density of the steel is 7.65 g/cm3. Find

a) The applied voltage,

b) The peak current,

c) The rms exciting current, and

d) The core loss.

(Note that 1 meter is 39.4 inch)

Solution:




1-23

1.5 Permanent Magnets




1-24




1-25

(a) Second quadrant of hysteresis loop for Alnico 5;

(b) Second quadrant of hysteresis loop for M-5 electrical steel;

(c) hysteresis loop for M-5 electrical steel expanded for small B. (Armco Inc.)

Figure 1.16




1-26

1.6 Application of Permanent Magnet

Materials




1-27

Magnetization curves for common permanent-magnet

materials.Figure 1.19

C © G C f




1-28

Portion of a B-H characteristic showing a minor loop

and a recoil line.Figure 1.21

C i ht © Th M G Hill C i I P i i i d f d ti di l




1-29

Example (Final Exam Question).Consider the magnetic circuit shown on the right. The permanent magnet material

has cross-sectional area Am=4 cm2 and length lm=3.45 mm. The air-gap hasthe same cross-sectional area as the magnet and its length is g=2 mm. Theiron core has infinite permeability. Assume no leakage flux, no fringing.Answer the following.

a) Express the area and the length in meters correctly (if you get the unitsincorrectly, it will be carried out to the remaining of the problem) and thenobtain the load line equation.

b) Draw the load line in the graph and give the magnetic field intensity valueand magnetic flux density value of the operating point for Samarium-Cobalt .

Copyright © The McGraw Hill Companies Inc Permission required for reproduction or display




1-30

Load Line

Copyright © The McGraw-Hill Companies Inc Permission required for reproduction or display




1-31

- 1000 - 800 - 600 - 400 - 200

0. 2

0. 4

0. 6

0. 8

1

1. 2

1. 4

3

k J / m

1 0 0

(kA/m)H,

2Wb/m

B

3

k J / m

2 0 0 3

k J/ m

2 9 4

It is desired to achieve a time-varying magnetic flux density in the air gap of the magnetic circuit inthe figure of the form Bg=Bo+B1sin wt where B0=0.5 Wb/m2 and B1=0.25 T. The dc field B0 isto be created by a Neodymium-Iron-Boron (characteristic is shown below) magnet, whereasthe time-varying field is to be created by a time-varying current. Please, answer the followingfor Ag=6 cm2, g=0.4 cm, and N=200 turns.

a) Write down the flux density of NdFeB as a function of the magnetic field intensity using thecharacteristic given

b) The magnet length and the magnet area Am that will achieve the desired dc air gap fluxdensity and minimize the magnet volume.

Example (MidTerm 1 Exam Question).




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1-32

Example:

Consider the figure below and answer the following.

a) Draw magnetic equivalent circuit,

b) Find the reluctances as a function of x,

c) Calculate the flux,

d) Calculate the flux linkage of the coil,

e) Calculate the inductance of the coil.




py g p , q p p y

1-33

• Examples and Problems




1-34

Magnetic circuit for Example 1.10.

Figure 1.18




1-35

Magnetic circuit

including both apermanent magnetand an excitationwinding.Figure 1.20




1-36

Magnetic circuit for Example 1.11.

Figure 1.22




1-37

(a) Magnetization curve for Alnico 5 for Example 1.11;

(b) series of load lines for Ag = 2 cm2 and varying of valuesof i showing the magnetization procedure for Example 1.11.Figure 1.23

(a) (b)




1-38

Magnetic circuit for Problem 1.1.

Figure 1.24




1-39

Magnetic circuit for

Problem 1.9.Figure 1.26




1-40

Inductor for Problem 1.12.

Figure 1.27




1-41

Pot-core inductor for Problem 1.15.

Figure 1.28




1-42

Inductor for Problem 1.17.

Figure 1.29




1-43

Toroidal winding for Problem 1.19.Figure 1.30




1-44

Iron-core inductor for Problem 1.20.

Figure 1.31


M ti i it f P bl 1 22



1-45

Magnetic circuit for Problem 1.22.Figure 1.32




1-46

Symmetric magnetic circuit for Problem 1.23.

Figure 1.33


Reciprocating generator for Problem 1 24



1-47

Reciprocating generator for Problem 1.24.Figure 1.34


Configuration for measurement of magnetic properties



1-48

Configuration for measurement of magnetic propertiesof electrical steel.Figure 1.35


Magnetic circuit for Problem 1 28



1-49

Magnetic circuit for Problem 1.28.Figure 1.36


M ti i it f th l d k f P bl 1 34



1-50

Magnetic circuit for the loudspeaker of Problem 1.34

(voice coil not shown).Figure 1.37


M ti i it



1-51

Magnetic circuit

for Problem 1.35.Figure 1.38


PowerPoint Slides



1-52



A.E. Fitzgerald


Stephen D. Umans

Chapter 2

Transformers

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

T f ith d



1-532-53

Transformer with open secondary.

Figure 2.4




1-54

No-load phasor diagram.

Figure 2.5


Ideal transformer and load



1-55

Ideal transformer and load.

Figure 2.6


Three circuits which are identical at terminals ab when the



1-56

transformeris ideal.

Figure 2.7


Eq i alent circ its for E ample 2 2 (a) Impedance



1-57

Equivalent circuits for Example 2.2 (a) Impedancein series with the secondary. (b) Impedance referred

to the primary.

Figure 2.8


Schematic view of mutual



1-58

Schematic view of mutualand leakage fluxes in a transformer.

Figure 2.9


Steps in thed l t f th



1-59

development of thetransformer

equivalent circuit.

Figure 2.10


Equivalent circuits for transformer of Example 2.3 referred to (a) the high-voltage sideand (b) the



1-60

and (b) thelow-voltage side.

Figure 2.11


Approximate transformer equivalent circuits.



1-61

Figure 2.12


Cantilever equivalent circuit for Example 2.4.



1-62

Figure 2.13


(a) Equivalent circuit and (b) phasor diagramfor Example 2.5.



1-63

p

Figure 2.14


Equivalent circuit with short-circuited secondary.(a) Complete equivalent circuit (b) Cantilever equivalent circuit with the exciting branch



1-64

(a) Complete equivalent circuit. (b) Cantilever equivalent circuit with the exciting branchat the transformer secondary.

Figure 2.15


Equivalent circuit with open-circuited secondary.(a) Complete equivalent circuit. (b) Cantilever equivalent circuit with the exciting branch



1-65

(a) Complete equivalent circuit. (b) Cantilever equivalent circuit with the exciting branchat the transformer primary.

Figure 2.16


(a) Two-winding transformer. (b) Connectionas an autotransformer.



1-66

Figure 2.17


(a) Autotransformer connection for Example 2.7.(b) Currents under rated load.



1-67

Figure 2.18


Common three-phase transformer connections; the transformer windings are indicated bythe heavy lines.



1-68

Figure 2.19


2.8 VOLTAGE AND CURRENT TRANSFORMERS

Used in instrumentation applications



1-69

Equivalent circuit for an instrumentation transformer. Figure 2.21

Used in instrumentation applications

765 000 Volt, 10 000 Ampers can not be measured directly

Most instruments range: Voltage (Potantial) Transformers (PT)0-120 V rms; Current Transformers (CT) 0-5 A rms

Rc (core loss resistance) neglected in equivalent circuit.

Zb is referred to as the BURDEN on that transformer.




1-70

))((ˆ

ˆ

22111

2

1

2

X j R Z Z X j R

Z Z

N

N

V

V

beq

beq

FOR PT:

)()(

11

11

X X j R X j R X j Z

m

meq

bb Z N N Z 2

21 )/(

FOR CT:

)(ˆ

ˆ

222

1

1

2

mb

m

X X j R Z

X j

N

N

I

I


Example 2.10: A 2400:120 V, 60 Hz potential transformer has thefollowing parameter values (referred to the 2400 V winding



1-71

1431 X 1642 X

1281 R

k 163m X

1412 R

g p ( g

side):

a) Assuming a 2400 V input, which ideally should produce avoltage of 120 V at the low voltage winding, calculate the

magnitude and relative phase-angle errors of the secondaryvoltage if the secondary winding is open-circuited.

b) Assuming the burden impedance to be purely resistive(Zb=Rb), calculate the minimum resistance (maximum burden)that can be applied to the secondary such that the magnitude

error is less than 0.5 percent.

c) Repeat part (b) but find the minimum resistance such that thephase-angle error is less than 1 degree.


2.9 THE PER-UNIT SYSTEM

Computations relating to machines transformers and systems



1-72

quantityof valueBase

quantityActualunit per inQuantity

basebasebase I V S basebasebase I V Z /

Computations relating to machines, transformers, and systems

of machines are often carried out in per-unit form.

Quantities are expressed as ratios to chosen BASE values.

V, I, P, Q, S, R, X, Z, G (conductance), B (susceptance), Y

can be translated.

1

22

2

21

12

base

base

base

base pu pu

S

S

V

V Z Z

Single Phase:

Base Change:


Example 2.12: The equivalent circuit for a 100 MVA, 7.97 kV:79.7kV transformer is shown in Fig. 2.22a. Convert the equivalent



1-73

g q

circuit parameters to per-unit using the transformer rating asbase.

04.0 L

X

75.3 H X

m76.0 L R

114m X

085.0 H R

7.97 kV:79.7 kV


Example 2.8: Three single-phase, 50 kV 2400:240 V transformers,each identical with an impedance of 1.42+j1.82 Ohm referred to



1-74

high voltage side is connected Wye-Delta in a three-phase 150kVA bank to step down the voltage at the load end of a feederwhose impedance is 0.15+j1 Ohm/phase. The voltage at the

sending end of the feeder is 4160 V line-to-line. On theirsecondary sides, the transformer supply a balanced three-

phase load through a feeder whos impedance is 0.0005+j0.0020 Ohm/phase. Find the line-to-line voltage at the loadwhen the load draws rated current from the transformer at a

power factor of 0.8 lagging.


PowerPoint Slides



1-75

to accompany


A.E. Fitzgerald Charles Kingsley, Jr.

Stephen D. Umans

Chapter 3

Electromechanical-Energy-

Conversion Principles


3.1 FORCES AND TORQUES IN MAGNETIC FIELD SYSTEMS



1-763-76

)( BvEF q

Lorentz Force Law:

For many charged particle

)( BvEF v

N/m3 ( coulombs/m3)

vJ Current density A/m2 BJF v

AJI Current A

BIF N/m


Example 3.1: A nonmagnetic motor containing a single-turn coil isplaced in a uniform magnetic field of magnitude B0, as shown



1-77

in Fig. 3.2. The coil sides are at radius R and the wire carriescurrent I as indicated. Find the θ-directed torque as a functionof rotor position α when I=10 A, B0=0.02 T and R=0.05 m.Assume that the rotor is of length l =0.3 m.




1-78

Very few problems can be solved using Lorentz force, wherecurrent-carrying elements and simple structures exist.

Most electromechanical-energy-conversion devices containmagnetic material and forces can not be calculated from Lorentzforce.

Thus, We will use ENERGY METHOD based on conservation ofenergy.




1-79

Electrical terminals: e and i Mechanical terminals: f fld and x

Losses separated from energy storage mechanism

Interaction through magnetic stored energy


Time rate of change of Wfld (field energy) equals to the differenceof input electrical power and output mechanical power for



1-80

dt

dx f ie

dt

W d fld

fld

p p p plossless systems.

or

dx f d iW d fld fld

Force can be solved as a function of flux linkage λ and position x.


3.2 ENERGY BALANCE

Energy neither created nor destroyed, it only changes the form.



1-81

heatto

converted

Energy

field magneticin

stored energy

inIncrease

output

energy

Mechanical

sources

electricfrom

inputEnergy

fld mechelec W d W d W d

Energy balance equation is written for motor action below

For lossless magnetic-energy-storage system

energystored magneticinchangealDifferenti:outputenergymechanicalalDifferenti:

inputenergyelectricalalDifferenti:

fld

mech

elec

W d W d

W d


3.3 ENERGY IN SINGLY-EXCITED MAGNETIC FIELD SYSTEMS

Schematic of an electromagnetic relay. Figure 3.4



1-82




1-83

i x L )(

The magnetic circuit can be described by an inductance which is afunction of the geometry and permeability of the magnetic material.

When air-gap exist in most cases Rgap>>Rcore and energy storageoccurs in the gap.

Magnetic nonlinearity and core losses neglected in practicaldevices.

Flux linkage and current linearly related.

Energy equation

Wfld uniquely specified by the value of λ and x . Thus, λ and x arecalled STATE VARIABLES.

dx f d iW d fld fld


Magnetic stored energy W fld uniquely determined by λ and x regardless of how they are brought to their final values.



1-84

Integration paths for W fld. Figure 3.5

b path

fld

a path

fld fld W d W d W 22

OR magnetic stored energy:

dV Bd H W V

B

fld

0

0

0

000 ),(),(

d xi xW fld


Example 3.2:The relay shown on the figure is made of infinitely-permeable magnetic material with a movable plunger, also of



1-85

infinitely-permeable material. The height of the plunger is muchgreater than the air-gap length (h>>g). Calculate the magneticstored energy Wfld as a function of plunger position (0<x<d) forN=1000 turns, g=2 mm, d=0.15 m, l =0.1 m, and i=10 A.


3.4 DETERMINATION OF MAGNETIC FORCE AND TORQUEFROM ENERGY



1-86

Consider any state function F(x1, x2), the total differential of Fwith respect to the two variables x1 and x2

22

11

2112

),( xd x

F

xd x

F

x x F d x x

Similarly, for energy function Wfld(λ, x)

xd x

W d W xW d fld

x

fld fld

),(

dx f d i xW d fld fld ),(


fld W

i

W f

fld



1-87

xi

x f fld

fld fld T f

Once we know the energy, current and more importantly force

can be calculated.For a system with rotating mechanical terminal

x

d T d iW d fld fld ),(

),( fld

fld W T


Example 3.4:The magnetic circuit below consists of a single-coilstator and an oval rotor. Because the air-gap is nonuniform, the

il i d t i ith t l iti d



1-88

coil inductance varies with rotor angular position, measuredbetween the magnetic axis of the stator coil and the major axisof the rotor, as

)2(cos)( 20 L L L

where where L0=10.6 mH and L2=2.7 mH. Note the second-harmonic variation of inductance with rotor angle θ.


3.5 DETERMINATION OF MAGNETIC FORCE AND TORQUE FROMCOENERGY

dfdiWd )(



1-89

dx f d i xW d fld fld ),(

Mathematically manipulated to define a new state functionknown as the COENERGY, from which force can be obtaineddirectly as a function of current.

),(),( xW i xiW fld fld

dx f di xiW d fld fld ),(

Note that energy and coenergyequal for linear systems.


xd W

id i

W xiW d

fld fld

fld

),(



1-90

dx f di xiW d fld fld ),(

xi i x fld

x

fld

i

xiW

),(

i

fld

fld x

xiW f

),(

i

fld id xi xiW 0

),(),(


In field-theory terms, for soft magnetic materals(B=0 when H=0)



1-91

dV dH BW

V

H

fld

0

0

dV dH BW V

H

H

fld

c

0

For permanent magnet materials (B=0 when H=Hc)


Effect of x on the energy and coenergy of a singly-excited device: (a)change of energy with held constant; (b) change of coenergy with i held

constant Figure 3 11



1-92

constant. Figure 3.11


Example 3.5: For the relay below, find the force on the plunger as afunction of x when the coil is driven by a controller which

produces a current as a function of x of the form



1-93

produces a current as a function of x of the form

A)( 0

d

x I xi


Example 3.6: The magnetic circuit in the figure is made of high-permeabilityelectrical steel. The rotor is free to turn about a vertical axis. The dimensionsare shown in the figure.

a) Derive an expression for the torque acting on the rotor in terms of the



1-94

a) Derive an expression for the torque acting on the rotor in terms of thedimensions and the magnetic field in the two air gaps. Assume the reluctanceof the steel to be negligible and neglect the effects of fringing.

b) The maximum flux density in the overlapping portions of the air gaps is to belimited to approximately 1.65 T to avoid excessive saturation of the steel.

Compute the maximum torque for r 1=2.5 cm, h=1.8 cm, and g =3 mm.


3.6 MULTIPLY-EXCITED MAGNETIC FIELD SYSTEMS

Many electromechanical devices have multiple electrical terminals.



1-95

Many electromechanical devices have multiple electrical terminals.


d T d id iW d fld fld

221121

),,(

USING ENERGY FUNCTON:



1-96Integration path to obtain W fld( 10

, 20, 0).

ff

01

0

02

00

0

102211

0

20212021 ),,(),,0(),,(

d id iW fld

For magnetically linear systems

2121111 i Li L

2221212 i Li L

D

L Li 2121221

D

L Li 2111212

21122211 L L L L D


01

002

00 1

20121022

22011

021)(

)()(

)(

)(

),,(

d D

L L

d D

L

W fld



1-97

0 00 0 )()(),,( D DW

000000 21

0

01221

0

02222

0

011021)()(

)(2)(

)(2)(),,(

D L

D L

D LW fld

21,

fld

fld

W T


dTdidiiiWd fld fld 221121 )(

USING COENERGY FUNCTON:



1-98

d T didiiiW d ff ),,(

01

0

02

00

0

102211

0

20212021 ),,(),,0(),,(

ii

fld diiiidiiiiiW

000000 2101222

02221

011021 )(

2

)(

2

)(),,( ii Li

Li

LiiW fld

2121111 i Li L

2221212 i Li L

21, ii

fld

fld

W T


d

Ld ii

d

Ld i

d

Ld iT fld

)()(

2

)(

2

1221

222211

21



1-99

nnnnn

n

n

n i

i

i

L L L

L L L

L L L

2

1

21

22221

11211

2

1

ILλ )(

For a general n electrical terminal

ILI )(2

1 T

fld W IL

Iθ d

d T T

fld

)(

2

1


Example 3.7: In the figure, the inductances in henrys are given asL11=(3+cos 2θ)x10-3; L12=0.3 cos θ; L22=30+10 cos 2θ. Find andplot the torque T

fld(θ) for current i

1=0.8 A and i

2=0.01 A.



1-100


310)sin4.22sin64.1( fld T



1-101


3.7 FORCES AND TORQUES IN SYSTEMS WITH PERMANENT MAGNETS

Special case must be taken when dealing with hard magnetic material

because magnetic flux density is zero when H=Hc not when H=0.



1-102

•Consider fictitious winding

•In normal operation, the fictitious winding carries NO current

•Current in the winding can be adjusted to zero out the field produced bypermanent magnet in order to achieve the “zero force” starting point.


dx f di xiW d fld f f f fld ),(

WdWdiW )0(



1-103Integration path for calculating W fld (i f = 0, x ) in the permanentmagnet system of Fig. 3.17. Figure 3.18

b path

fld

a path

fld f fld W d W d xiW 11

),0(

0

00

0 ),(),(),0( f I

f f f

x

f f fld f fld di xi xd x I i f xiW

0

0

),(),0(

f I

f f f f fld di xi xiW

If0 is the current to zero-out the field.


Example 3.8: The magnetic circuitis excited by a samarium-cobalt permanent magnet andincludes a movable plunger.Also shown is the fictitous



1-104

Figure 3.19

Also shown is the fictitouswinding of Nf turns carrying acurrent if which is includedhere for the sake of theanalysis. The dimensions are:Wm=2 cm, Wg=3 cm, W0=2 cm,d=2 cm, g0=0.2 cm, and D=3cm.

a) Find an expression for thecoenergy of the system as afunction of plunger position x,

b) Find an expression for theforce on the plunger as afunction of x,

c) Calculate the force at x=0 andx=0.5 cm.


A different solution for permanent magnet circuits:



1-105

d H A e

c R

d d Ni A eeq

R)(

d H Ni ceq )(


Example 3.9: Figure shows an actuator consisting of an infinitely-permeable yokeand plunger, excited by a section of NdFeB magnet and an excitation windingof N1=1500 turns. The dimensions are: W=4 cm, W1=4.5 cm, D=3.5 cm, d=8 mm,

and g0=1 mm.a) Find x-directed force on the plunger when the current in the excitation winding



1-106

) p g gis zero and x=3 mm.

b) Calculate the current in the excitation winding required to reduce the plungerforce to zero.


3.8 DYNAMIC EQUATIONS

We are interested in the operation of complete electromechanical system and not just of theelectromechanical energy conversion system around which it is built.



1-107

For Electrical Terminal:

t d

xd

xd

x Ld i

t d

id x Li Rv

)()(0

For multiple-excited system, we will have similar equation for each terminal


)( 0 x x K f K

For Mechanical Terminal:

Spring:

x

: Spring constant (N/m)



1-108

t d

xd

B f D

2

2

t d

xd

M f M

Damper:

Mass:

B

002

2

)( f x x K dt xd B

dt xd M f fld

M

B

K

fld f

0 f

: Damping constant (N.s/m)

: Mass (kg)


txdxLdtid )()()(

Dynamic Equations (Electrical and Mechanical Equations Together):



1-109

t d

t xd

xd

x Ld i

t d

t id x Lt i Rt v

)()()()()()(0

))(),(())(()()(

)( 02

2

0 t xt i f xt x K dt

t xd B

dt

t xd M t f fld

xd x Ld i f fld

)(2

2

These equations completely specify the behavior of electromechanical device. Solutionof these equations will describe the position x and the current i at any time t in thesystem.


)( 0 K T K

For Rotational Mechanical Terminal:

Torsional Spring:

: Torsional Spring constant (N.m/rad)



1-110

t d

d

BT F

2

2

t d

d

J T J

Friction:

Inertia:

B

J

002

2

)( T K dt

d

Bdt

d

J T fld

: Friction constant (N.m.s/rad)

: Inertia constant (kg.m2

/rad)


Example 3.10: Figure shows in crosssection a cylindrical solenoid magnetin which the cylindrical plunger of

mass M moves vertically in brassquide rings of thickness g and meandiameter d . The permeability of brass



1-111

p yis µ 0. The plunger is supported by aspring with K constant. Itsunstretched length is l 0. A mechanicalload force f

t

is applied to the plungerfrom the mechanical systemconnected to it. Assume thatfrictional force is linearly proportionalto the velocity with coefficient B. Thecoil has N turns and resistance R . Itsterminal voltage is v t and its current i .

Derive the dynamic equations ofmotion of the electromechanicalsystem.


EXTRA Example: A two poles VR machine is shown in figure. Stator and rotor hasinfinite permeability.

a) Find gap cross-sectional area as a function of θ.

b) Find the inductance for the machine.

) W it d th d i ti



1-112

c) Write down the dynamic equations.

d) Solve the dynamic equations to find the position of rotor as a function of timeinitially starting from θ0=25 degrees.

0 Stator Axes

R o t o r

A x e s

r

Numerical Values:

N=100 turns, g=0.0005 m, d=0.1m, r=0.04 m, J=0.05, B=0.02,θ0=30, R=0.5 ohm, E=10 Volt.



Electric Machinery



1-113

ySixth Edition


Stephen D. Umans

Chapter 4Introduction to Rotating

Machines


4.1 ELEMENTARY CONCEPT

Electromechanical energy conversion occurs when changes in the flux

linkagesλ

resulting from mechanical motion.d

te )(



1-114

Horizontalaxis

MagneticField

e(t)

dt t e )(

•Rotating the winding in magnetic

field•Rotating magnetic field through the

winding

•Stationary winding and timechanging magnetic field (Transformer

action)

Producing voltage in the coil


Armature winding: AC current carrying winding

Synchronous machine Armature winding is

Induction machine stator winding (stationary)



1-115

DC machine Armature winding is on

the rotor

Field winding: DC current carrying winding

DC machine Field winding is on the stator

Synchronous machine Field winding is on the rotor

Note: Permanent magnets produce DC magnetic flux and are usedin the place of field windings in some machines.

VRM (Variable Reluctance Machines) No windings on the rotor

Stepper Motors (non-uniform air-gaps)


4.2 INTRODUCTION TO AC AND DC MACHINES

AC Machines: Synchronous Machines and Induction Machines

Synchronous Machines:•Two-pole, single phase machine



1-1164-116

(a) Space distribution of flux density and(b) corresponding waveform of the generated voltage for the single-

phase generator.

•Rotor rotates with a constant speed

•Constraction is made such that air-gap flux density is sinusoidal

•Sinusoidal flux distribution resultswith sinusoidal induced voltage


•Four-pole, single phase machine

•a1,-a1 and a2,-a2 windings connected

in series•The generator voltage goes throught l t l l ti f



1-117

two complete cycles per revolution ofthe rotor. The frequency in hertz willbe twice the speed in revolutions per

second.

aae

p

2

602f

n pe

n: rpm

f e: Hz

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. . Yıldır ım, İTÜLecture Notes)



1-118


Hydroelectric power plant



1-119generator sets giant shaftconnecting turbine

to generator

generator

turbine

hydropower-plant-generator.swf


715 MW generator



1-120

Diameter of rotor:

16 meters

Rotating mass:

2650 ton


Field winding is a two-poledistributed winding

Winding distributed in multipleslots and arranged to produce



1-121

Elementary two-polecylindrical-rotor field winding.

sinusoidal distributed air-gapflux.

Why some synchronousgenerators have salient-pole

rotor while others havecylindirical rotors?

Answer : In salient-polemachines the number of polescan be large therefore they will

be able to operate in slow speedto produce 50 Hz voltage.


Schematic views of three-phase generators: (a) two-pole, (b) four-pole, and (c) Y connection of the windings.

Figure 4.12



1-122


Induction Machines:

•The stator winding excited

by ac current. The currentproduces a rotating magneticfield which in turn produces

t i t d t



1-123

Typical induction-motor speed-torquecharacteristic.

Figure 4.15

currents in rotor conductorsdue to induction.

•These machines mostly usedas motors.

•Rotor windings are shortcircuited (electrically) and

frequently have no externalconnections.

•Stator and rotor fluxes rotatein synchronism with eachother and that torque isrelated to the relative

displacement between them.

•Rotor does not rotatesynchronously


Typical Induction Motor



1-124


Windings placed in stator slots



1-125


Inside View of An Induction Motor



1-126


DC Machines:

Armature winding on the rotor

with current conducted from itby means of carbon brushes



1-127

Elementary dc machinewith commutator.

Figure 4.17


(a) Schematic view of fluxproduced by a concentrated,full-pitch winding in a machinewith a uniform air gap (b) The

4.3 MMF OF DISTRIBUTED WINDINGS



1-128

with a uniform air gap. (b) Theair-gap mmf produced

by current in

this winding.Figure 4.19

aag i N

F

cos2

41

Fourier Analysis

2

4

)( 1

i N

F peak ag


The mmf of one phase of a distributed two-pole,three-phase winding with full-pitch coils.

Figure 4.20

AC Machines:



1-129

a

a ph

ag

p

p

i N k

F

2cos

41

k Winding factor (usuallybetween 0.85 and 0.95)

ph N Series turns per phase


Example 4.1: The phase-a two-pole armature winding of figure below can beconsidered to consists of 8 Nc-turn full-pitch coils connected in series, with

each slot contaning two coils. There are a total of 24 armature slots, and thuseach slot is separated by 3600 /24=150. Assume angle θa is measures from the

magnetic axis of phase a such that the four slots containing the coil sideslabeled a are at 67.50, 82.50, 97.50, and 112.50. The opposite sides of each coil

are thus found in the slots found at -112.50, -97.50, -82.50, and 67.50,respectively Assume this winding to be carrying current i



1-130

respectively. Assume this winding to be carrying current ia.

a) Write an expression for the space-fundamental mmf produced by the two coils

whose sides are in the slots atθ

a=112.50

and -67.50

.b) Write an expression for the space-fundamental mmf produced by the two coils

whose sides are in the slots at θa=67.50 and -112.50.

c) Write an expression for the space-fundamental mmf of the complete armaturewinding.

d) Determine the winding factor kw for this distributed winding.


The air-gap mmf of a distributed winding on the rotorof a round-rotor generator.

r

r r r

ag

p

p

I N k

F 2cos

41



1-131


Cross section of a two-pole dc

machine.

DC Machines:



1-132


(a) Developed sketch ofthe dc machine(b) mmf wave;

(c) equivalent sawtoothmmf wave, itsfundamentalcomponent



1-133

component,and equivalent

rectangular currentsheet.

Sawtooth waveformbecause of restrictions

imposed by thecommutator.

Peak value offundamental component

81.0

82


(a) Cross section of a four-pole dc machine;(b) development of current sheet and mmf wave.



1-134

a

a

peak ag i p

N

F

)( a

a

peak ag i p

N

F

21

8

)(


Four Pole Stator of a DC Motor :



1-135


Rotor of a DC Motor :



1-136




1-137


Parts of a small DC motor



1-138


The air-gap mmf andradial component

of Hag for aconcentrated

4.4 MAGNETIC FIELDS IN ROTATINGMACHINERY



1-139

concentratedfull-pitch winding.

aag g

Ni H

cos

2

41

a

a phw

ag

p

p g

i N k H

2cos

4

1

Distributed winding:


Example 4.2: A four-pole synchronous ac generatorwith a smooth air gap has a distributed rotor

winding with 263 series turns, a winding factor of0.935, and an air gap of length 0.7 mm. Assumingthe mmf drop in the electrical steel to be negligible,find the rotor-winding current required to produce a



1-140

find the rotor winding current required to produce apeak, space-fundamental magnetic flux density of

1.6 T in the machine air gap.


Structure of typical salient-pole machines: (a) dc machine and (b)

salient-pole synchronous machine.

Machines with non-uniform air gaps.



1-141


Finite-element solution of the magnetic field distribution in a salient-pole dc generator. Field coils excited; no current in armature coils.

(General Electric Company .)



1-142


Flux distribution in a 4-pole salient-pole generator



1-143

Colors representthe strength of B.

Blue to Red : Theflux densityincreases


Single-phase-winding

space-fundamentalair-gap mmf:(a) mmf distribution of

a single-phase winding at

4.5 ROTATING MMF WAVES IN AC MACHINES



1-144

various times;

(b) total mmf ag1decomposed into two

traveling waves – and +;(c) phasor decomposition

of ag1.

a

a phw

ag

p

p

i N k F

2cos

41

t I i eaa cos


Simplified two-

pole three-phasestator winding.

MMF Wave of a Polyphase Winding



1-145

t I i ema cos)120cos( 0 t I i emb

)120cos( 0 t I iemc


Instantaneous phase currents under balancedthree-phase conditions.

Figure 4.30



1-146


The production of a rotating magnetic field by meansof three-phase currents.

Figure 4.31



1-147


Cross-sectional view ofan elementary three-

phaseac machine.

Figure 4.32



1-148


Voltage between the brushes in theelementary dc machine of Fig. 4.17.

Figure 4.33



1-149


Elementary two-pole machine with smooth air gap:(a) winding distribution and (b) schematic representation.

Figure 4.34



1-150


Simplified two-pole machine: (a) elementary model and(b) vector diagram of mmf waves. Torque is producedby the tendency of the rotor and stator magnetic fields

to align. Note that these figures are drawn with sr positive, i.e., with the rotor mmf wave

Fr leading that of the stator Fs.Figure 4.35



1-151


The mmf andH field of a

concentrated full-pitchlinear winding.

Figure 4.36



1-152


Typicalopen-circuit

characteristic andair-gap line.

Figure 4.37



1-153


Finite-element solutionfor the flux distributionaround a salient pole.

(General Electric

Company.)Figure 4.38



1-154


Flux-density wavecorresponding to Fig.

4.38 with itsfundamental and

third-harmoniccomponents.

Figure 4.39



1-155


Three-coil system showing components of mutualand leakage flux produced by current in coil 1.

Figure 4.40



1-156


Flux created by asingle coil side in

a slot.

Figure 4.41



1-157


Problem 4.8: (a) full-pitch coil and (b) fractional-pitch coil.

Figure 4.43



1-158


Elementary generator forProblem 4.13.

Figure 4.44



1-159


Elementary cylindrical-rotor,two-phase synchronous

machine for Problem 4.22.

Figure 4.45



1-160


Schematic two-phase,salient-pole synchronousmachine for Problem 4.24.

Figure 4.46



1-161






1-162


Stephen D. Umans

Chapter 5Synchronous Machines


5.1 INTRODUCTION TO POLYPHASE SYNCHRONOUS MACHINES

Two types:

1-Cylindirical rotor : High speed, fuel or gas fired power plants

n pn p

e120602

f



1-163

2-Salient-pole rotor : Low speed, hydroelectric power plants

To produce 50 Hz electricity p=2, n=3000 rpm

p=4, n=1500 rpm

To produce 50 Hz electricity

p=12, n=500 rpm

p=24, n=250 rpm



How does a synchronous generator work?

1- Apply DC current to rotor winding

(field winding)

2- Rotate the shaft (rotor) with constant

speed.



1-164

speed

3- Rotor magnetic field will create flux

linkages in stator coils and as a result

voltage will be produced because of

Faraday’s Law.

Why is impossible to rotate a synchronous motor when it is

connected to 50 Hz electric power?

Because before connecting to supply, the shaft speed of rotor is

zero. If the motor is two-pole, when it is connected to 50 Hzsupply it suddenly needs to rotate 3000 rpm. This is impossiblefor large synchronous motors.



How is DC current applied to the rotor?

1- Slip Rings Note: Magnetic field of rotorcan also be produced bypermanent magnets forsmall machine applications



1-165

2- Brushless Excitation System:

Excitation supplied from ac exciter and solid rectifiers. Thealternator of the ac exciter and the rectification system are on the

rotor. The current is supplied directly to the field-winding withoutthe need to slip rings.



RF F R F p

T

sin22

2

F

R

F windingfield dctheof mmf :

pole perfluxgap-air Resultant:

Steady-state torque equation



1-1665-166

Torque-angle characteristic.

F RF RF F and betweenangle phaseelectrical:



• Synchronous generatorswork in parallel with theinterconnected system.

• Frequency and voltage areconstant.



1-167

•The behivor is examinedbased on a generatorconnected to an INFINITE BUS.

Infinite bus

f : constant

V : constant

Generator


5.2 SYNCHRONOUS-MACHINE INDUCTANCES; EQUIVALENT CIRCUTS

f af cacbabaaaa iiii LLLL

f bf cbcbbbabab iiii LLLL



1-168

f cf cccbcbacac iiii LLLL f ff c fcb fba fa f iiii LLLL

al aaccbbaa L L 0

LLL

Fundamental component

Leakage flux

component fl ff ff L L

0L

Self inductances:


0021

32cosLLLLLL aaaacbbccaacbaab L L


p

Mutual inductances:



1-169

meaf faaf L cosLL

)3

2cos(LL

0

eeaf fccf

t L

)cos(LL 0eeaf faaf t L

)3

2cos(LL 0

eeaf fbbf t L

02 eemme t


f

c

b

a

fl ff eeaf eeaf eeaf

eeaf al aaaaaa

eeaf aaal aaaa

eeaf aaaaal aa

f

c

b

a

i

i

ii

L Lt Lt Lt L

t L L L L L

t L L L L L

t L L L L L

0000

0000

0000

0000

)

3

2cos()

3

2cos()cos(

)3

2cos(

2

1

2

1

)3

2cos(

2

1

2

1

)cos(

2

1

2

1




1-170

33

0 cba iii

f af aaaaal aaa ii Li L L L2

1

)( 00

For balanced system

f af aal aaa ii L L L)2

3( 0

f af a sa ii L L L

s

: Defined as synchronous inductance.

It is the effective inductance seen by phase aunder steady state balanced conditions.


af a

saa f af a

saaa

aaa et d id Li R

t d id

t d id Li R

t d d i Rv )L(

)(L 0ff L

5.2.4 EQUIVALENT CIRCUTS

Terminal voltage for phase a



1-171

)cos(L 0eeaf af t L

)sin( 0ee f af eaf t I Le 2

L f af e

af

I E

0

2

L

ˆ e j f af e

af e

I

j E

In complex form:


5.2.4 EQUIVALENT CIRCUTS

af a saaa E I X j I RV ˆˆˆˆ a saaaf a I X j I R E V ˆˆˆˆ

Motor: Generator:



1-172

Synchronous Reactance


Synchronous-machine equivalent circuit showing air-gap and leakage componentsof synchronous reactance and air-gap voltage.

Figure 5.4



1-173


Open-circuit characteristic of a synchronous machine.

Figure 5.5



1-174


Typical form of an open-circuit core-loss curve.

Figure 5.6



1-175


Open- and short-circuit characteristics of a synchronous machine.

Figure 5.7



1-176


Phasor diagram forshort-circuitconditions.

Figure 5.8



1-177


Open- andshort-circuitcharacteristics showingequivalent magnetization

line for saturatedoperating conditions.

Figure 5.9



1-178


Typical form of short-circuit load lossand stray load-loss curves.

Figure 5.10



1-179


(a) Impedance interconnecting two voltages;(b) phasor diagram.

Figure 5.11



1-180


Equivalent-circuit representation of a synchronous machine connected to anexternal system.

Figure 5.12



1-181


Example 5.6. (a) MATLAB plot of terminal voltage vs.for part (b). (b) MATLAB plot of E af vs. power for part (c).

Figure 5.13



1-182


Equivalent circuits and phasor diagrams for Example 5.7.

Figure 5.14



1-183


Characteristic form of synchronous-generator compounding curves.

Figure 5.15



1-184


Capability curves of an 0.85power factor, 0.80short-circuit ratio,hydrogen-cooled turbinegenerator. Base MVA is

rated MVA at 0.5 psighydrogen.

Figure 5.16



1-185


Construction used forthe derivation of asynchronous generatorcapability curve.

Figure 5.17



1-186


Typical form of synchronous-generator V curves.

Figure 5.18



1-187


Losses in a three-phase,45-kVA, Y-connected,220-V, 60-Hz,six-pole synchronous

machine (Example 5.8).Figure 5.19



1-188


Direct-axis air-gap fluxes in a salient-polesynchronous machine.

Figure 5.20



1-189


Quadrature-axis air-gap fluxes in a salient-pole synchronous machine.

Figure 5.21



1-190


Phasor diagram of a salient-pole synchronous generator.

Figure 5.22



1-191


Phasor diagram for a synchronous generator showing the relationship between thevoltages and the currents.

Figure 5.23



1-192


Relationships between componentvoltages in a phasor diagram.

Figure 5.24



1-193


Generator phasor diagram for Example 5.9.

Figure 5.25



1-194


Salient-pole synchronous machine and series impedance: (a) single-line diagram and(b) phasor diagram.

Figure 5.26



1-195


Power-angle characteristic of a salient-pole synchronous machine showing thefundamental component due to field excitation and the second-harmonic componentdue to reluctance torque.

Figure 5.27



1-196


(a) Single-line diagram and (b) phasor diagram for motor of Example 5.11.

Figure 5.28



1-197


Schematic diagram of athree-phase permanent-magnet ac machine. Thearrow indicates thedirection

of rotor magnetization.

Figure 5.29



1-198




A.E. Fitzgerald


Stephen D. Umans



1-199

Chapter 6

Polyphase Induction

Machines


6.1 INTRODUCTION TO POLYPHASE INDUCTION MACHINES



1-200

Squirrel-Cage Wound Rotor

Two types of motor:


6.1 INTRODUCTION TO POLYPHASE INDUCTION MACHINESHow does an induction motor work?

1. Apply AC three-phase current to stator winding to produce rotatingmagnetic field.

2. Rotating magnetic field induces voltages in rotor windings resultingwith rotor currents.

3. Then, rotor currents will create rotor magnetic field.

4. Constant speed stator magnetic field will drag rotor magnetic field.

ns: Synchronous speed (the speedf t t t ti fi ld i )

n : Rotor speed (rpm).



1-201

sf 120

p

n s ns

n

of stator rotating field in rpm).

SLIP : It is defined as the differencebetween synchronous speed and therotor speed divided by synchronous

speed.

s

s

n

nn

s


er s f f


s sr n snnn

The speed of rotor magnetic field with respect to rotor is

The relative motion of stator flux and the rotor conductors inducesvoltages of frequency (f r is called slip frequency)



1-202

sn sn )1(

sm s )1(

The rotor speed

Mechanical angular velocity



Breakdown torque



1-203

Typical induction-motor torque-speed curve for constant-voltage,constant-frequency operation.6-203


6.2 CURRENTS AND FLUXES IN POLYPHASE INDUCTIONMACHINE



1-204

Developed rotor winding of an induction motor with its flux-density and mmfwaves in their relative positions for (a) zero and (b) nonzero rotor leakage

reactance.


Reactions of a

squirrel-cage rotorin a two-pole field.

Figure 6.6

6.2 CURRENTS AND FLUXES IN POLYPHASE INDUCTIONMACHINE



1-205


Stator equivalent circuit for a polyphase induction motor.

6.3 INDUCTION MOTOR EQUIVALENT CIRCUIT



1-206

Counter emf generated by the resultant air-gap flux


Rotor equivalent circuit for a polyphase induction motor at slipfrequency.


22ˆˆ E s E s

22ˆˆ I I s



1-207

2

2

2

2

2 ˆ

ˆ X j

s

R

I

E Z


Single-phase equivalent circuit for a polyphase induction motor.




1-208

Models the combinedeffect of rotor resistance

and shaft load


Alternative form of equivalent circuit.




1-209

Electromechanical power isequal to the power delivered

to this resistance


s

R I n P ph gap

222 2

22 R I n P phrotor

s

s R I n P P P phrotor gapmech

12

22

6.4 ANALYSIS OF THE EQUIVALENT CIRCUIT

gapmech P s P )1( gaprotor P s P mechmmech T P



1-210

P mech is not the net power but it includes the losses such as friction, windage.

Output power and torque from the shaft is

rot mech shaft P P P m

shaft

shaft

P T


6.5 TORQUE AND POWER BY USE OF THEVENIN’S THEOREM

(a) General linear network and (b) its equivalent at terminals ab byThevenin’s theorem.



1-211



Equivalent circuits with the core-loss resistance R c neglected.



1-212

)(ˆˆ

11

1,1m

meq

X X j R

X jV V

)(

)(

11

11,1

m

meq

X X j R

X j R X j Z


Induction-motor equivalent circuits simplified by Thevenin’s theorem.




1-213

s R X j Z

V

I eq

eq

22,1

,1

2

ˆ

ˆ

22,1

22,1

2

2

,1

)())((

)(1

X X s R R

s RV nT

eqeq

eq ph

s

mech


Induction-machine torque-slip curve showing braking, motor, and generatorregions.

Figure 6.14



1-214


The End of This Chapter



1-215


Computed torque, power, and current curvesfor the 7.5-kW motor in Examples 6.2 and 6.3.

Figure 6.15



1-216


Induction-motortorque-slip curves

showing effectof changing rotor-circuit resistance.

Figure 6.16



1-217


Electromechanical torque vs. speed for the wound-rotor induction motor of Example6.4 for various values of the rotor resistance R 2.

Figure 6.17



1-218


Deep rotor bar and slot-leakage flux.

Figure 6.18



1-219


Skin effect in a copper rotor bar 2.5 cm deep.

Figure 6.19



1-220


Double-squirrel-cagerotor bars and slot-

leakage flux.

Figure 6.20



1-221


Typical torque-speed curvesfor 1800-r/min general-

purpose induction motors.

Figure 6.21



1-222


Connections of a one-step starting autotransformer.

Figure 6.22



1-223


Interconnected inductionand synchronous machines

(Problems 6.7and 6.8).

Figure 6.23



1-224


Induction-motor equivalent circuits

simplified by Thevenin’s theorem.

Figure 6.13



1-225




A.E. Fitzgerald

Charles Kingsley, Jr.Stephen D. Umans

Ch t 7



1-226

Chapter 7

DC Machines


7.1 INTRODUCTION



1-227


7.1 INTRODUCTION



1-228


7.1 INTRODUCTION



1-229


7.1 INTRODUCTION



1-2307-230

ad amech i K T

Constant determined bythe design of windings

Direct axis air-gap flux perpole

Current in externalarmature circuit


Rectified coil voltages and resultant voltage between brushes in a dcmachine.

7.1 INTRODUCTION



1-231

md aa K e

Speed voltage Angular speed

mmechaa T ie

Power


Typical form of magnetization curves of a dc machine.

7.1 INTRODUCTION



1-232

0

0

m

ad a

m

a e K

e

0

0

a

m

ma ee

0

0

aa en

ne


7.1 INTRODUCTION

Equivalent Circuit (Not in the textbook):

a R

a L f R f L

ae

f i

ai



1-233

For steady-state, current is dc,therefore Lf and La can be

neglected.


7.1 INTRODUCTION



1-234

Field-circuit connections of dc machines: (a) separate excitation, (b) series,(c) shunt, (d) compound.


7.1 INTRODUCTION



1-235

Volt-ampere characteristics of dc generators.


7.1 INTRODUCTION



1-236

Speed-torque characteristics of dc motors.


EXAMPLE (Final Exam 2006) :

Assume that a 240 V self-excited shuntmotor is supplied by a line current of 102.4

A when it is loaded with a full load at aspeed of 1000 rpm. The armature-circuitresistance and the shunt-field circuit

resistance of the motor are 0.1 ohm and 100ohm, respectively. Assume that a breakingresistor of 1.05 ohm is used for dynamic

braking (breaking means that voltagesource is removed and immediately a

resistor is connected to the terminals of theDC machine) and determine the following.

Ea

Ra

Rf

Vt



1-237

a)The value of counter emf Ea.

b)The full-load torque of the motor

c)The value of the armature winding currentat the time of initial breaking.

d)The value of initial dynamic breaking(initial torque during breaking)


The End of This Chapter



1-238


Dc machine armaturewinding with commutator

and brushes.(a), (b) Current directionsfor two positions of the

armature.Figure 7.7



1-239


Waveform of current in an armature coilwith linear commutation.

Figure 7.8



1-240


Armature-mmf and flux-density distribution with brushes on neutral and only thearmature excited.

Figure 7.9



1-241


Flux with only the armature excited and brusheson neutral.

Figure 7.10



1-242


Armature,main-field,

and resultantflux-density distributions

with brusheson neutral.

Figure 7.11



1-243


Motor or generator connection diagramwith current directions.

Figure 7.12



1-244


Short-shunt compound-generator connections.Figure 7.13



1-245


Magnetization curves for a250-V 1200-r/min dc

machine. Also shown arefield-resistance lines for the

discussion of self-excitation in Section 7.6.1.

Figure 7.14



1-246


Equivalent circuit for analysis of voltage buildupin a self-excited dc generator.

Figure 7.15



1-247


Cross section of a typical permanent-magnet motor. Arrows indicate the direction ofmagnetization in the permanent magnets.

Figure 7.17



1-248


(a) Dimension definitions for the motor of Fig. 7.17.(b) approximate magnetic equivalent circuit.

Figure 7.18



1-249


Equivalent circuit of a permanent-magnet dc motor.

Figure 7.20



1-250


Section of dc machine showing compensating winding.

Figure 7.22



1-251


Schematic connection diagram of a dc machine.

Figure 7.24



1-252


Series-connected universal machine.

Figure 7.25



1-253


Typical torque-speed characteristics of a series universal motor.

Figure 7.26



1-254


1200 r/min magnetizationcurve for thedc generator

of Problem 7.4.

Figure 7.27



1-255


Series cranemotor (Problem 7.22): (a)hoisting connection and(b) lowering connection.

Figure 7.28



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machine electric

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