Fitzgerald A.E. Electric Machinery

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166 CHAPTER 3 ElectromechanicaI-Energy-Conversion Principles

Capacitance C-

Conducting

plates of area A

Figure

3.34 Capacitor plates (Problem 3.20).

3.20 A capacitor (Fig. 3.34) is made of two conducting plates of area A separated

in air by a spacing x. The terminal voltage is v, and the charge on the plates

is q. The capacitance C, defined as the ratio of charge to voltage, is

q E0A

C _._ _

13 X

where E0 is the dielectric constant of free space (in SI units E0 =

8.85 X 10 -12 F/m).

a. Using the results of Problem 3.19, derive expressions for the energy

Wild(q, x) and the coenergy W~d(V, x).

b. The terminals of the capacitor are connected to a source of constant

voltage V0. Derive an expression for the force required to maintain the

plates separated by a constant spacing x = 8.

3.21 Figure 3.35 shows in schematic form an

electrostatic voltmeter,

a capacitive

system consisting of a fixed electrode and a moveable electrode. The

moveable electrode is connected to a vane which rotates on a pivot such that

the air gap between the two electrodes remains fixed as the vane rotates. The

capacitance of this system is given by

EoRd(ot-

101)

C(O)

= (101 _< c~)

Fixed

electrode

Mow

electrode

Depth d into

paper

Figure 3.35

Schematic electrostatic

voltmeter (Problem 3.21).

3.11 Problems 167

2, N 2 turns

eable element

.........

~, l, N 1 turns

Figure 3.36

Two-winding magnetic

circuit for Problem 3.22.

A torsional spring is connected to the moveable vane, producing a torque

Tspring = --

K (0 - 00)

a. For 0 _< 0 _< or, using the results of Problem 3.19, derive an expression

for the electromagnetic torque Tnd in terms of the applied voltage Vdc.

b. Find an expression for the angular position of the moveable vane as a

function of the applied voltage Vdc.

c. For a system with

R -- 12 cm, d -- 4 cm, g -- 0.2 mm

ot-zr/3rad, 00-0rad, K--3.65N.m/rad

Plot the vane position in degrees as a function of applied voltage for

0 _< Vdc _< 1500 V.

3.22 The two-winding magnetic circuit of Fig. 3.36 has a winding on a fixed yoke

and a second winding on a moveable element. The moveable element is

constrained to motion such that the lengths of both air gaps remain equal.

a. Find the self-inductances of windings 1 and 2 in terms of the core

dimensions and the number of turns.

b. Find the mutual inductance between the two windings.

c. Calculate the coenergy

Wt~d(il, i2).

d. Find an expression for the force acting on the moveable element as a

function of the winding currents.

3.23 Two coils, one mounted on a stator and the other on a rotor, have self- and

mutual inductances of

LI1 -- 3.5 mH

L22 --

1.8 mH

L12 --

2.1cos0 mH

where 0 is the angle between the axes of the coils. The coils are connected in

168 CHAPTER 3 ElectromechanicaI-Energy-Conversion Principles

series and carry a current

i = ~/-2I sin

cot

a. Derive an expression for the instantaneous torque T on the rotor as a

function of the angular position 0.

b. Find an expression for the time-averaged torque Tavg as a function of 0.

c. Compute the numerical value of Tavg for I -- 10 A and 0 = 90 °.

d. Sketch curves of

Tavg versus

0 for currents I = 5, 7.07, and 10 A.

e. A helical restraining spring which tends to hold the rotor at 0 - 90 ° is

now attached to the rotor. The restraining torque of the spring is

proportional to the angular deflection from 0 -- 90 ° and is -0.1 N. m

when the rotor is turned to 0 - 0 °. Show on the curves of part (d) how

you could find the angular position of the rotor-plus-spring combination

for coil currents I -- 5, 7.07, and 10 A. From your curves, estimate the

rotor angle for each of these currents.

f. Write a MATLAB script to plot the angular position of the rotor as a

function of rms current for 0 < I < 10 A.

(Note that this problem illustrates the principles of the dynamometer-type ac

ammeter.)

3.24 Two windings, one mounted on a stator and the other on a rotor, have self- and

mutual inductances of

LI1 -

4.5H

L22 --

2.5H

L12 --

2.8cos0 H

where 0 is the angle between the axes of the windings. The resistances of the

windings may be neglected. Winding 2 is short-circuited, and the current in

winding 1 as a function of time is il = 10 sin wt A.

a. Derive an expression for the numerical value in newton-meters of the

instantaneous torque on the rotor in terms of the angle 0.

b. Compute the time-averaged torque in newton-meters when 0 = 45 °.

c. If the rotor is allowed to move, will it rotate continuously or will it tend

to come to rest? If the latter, at what value of 00 ?

3.25 A loudspeaker is made of a magnetic core of infinite permeability and circular

symmetry, as shown in Figs. 3.37a and b. The air-gap length g is much less

than the radius r0 of the central core. The voice coil is constrained to move

only in the x direction and is attached to the speaker cone, which is not shown

in the figure. A constant radial magnetic field is produced in the air gap by a

direct current in coil 1,

i l = I1.

An audio-frequency signal

i2 =

12 COS cot is

then applied to the voice coil. Assume the voice coil to be of negligible

thickness and composed of N2 turns uniformly distributed over its height h.

Also assume that its displacement is such that it remains in the air gap

(O<x <l-h).

a. Calculate the force on the voice coil, using the Lorentz Force Law

(Eq. 3.1).

3.11 Problems 169

Voice coil

N 2 tums

positive i 2

(a) (b)

Figure

3.37 Loudspeaker for Problem 3.25.

b. Calculate the self-inductance of each coil.

c. Calculate the mutual inductance between the coils. (Hint: Assume that

current is applied to the voice coil, and calculate the flux linkages of

coil 1. Note that these flux linkages vary with the displacement x.)

d. Calculate the force on the voice coil from the coenergy W~a.

3.26 Repeat Example 3.8 with the samarium-cobalt magnet replaced by a

neodymium-iron-boron magnet.

3.27 The magnetic structure of Fig. 3.38 is a schematic view of a system designed

to support a block of magnetic material (/x --+ c~) of mass M against the force

of gravity. The system includes a permanent magnet and a winding. Under

normal conditions, the force is supplied by the permanent magnet alone. The

function of the winding is to counteract the field produced by the magnet so

that the mass can be removed from the device. The system is designed such

that the air gaps at each side of the mass remain constant at length

go~2.

Depth D into the paper

~-- d--~

ling,

rns

lOravity ,,a,.etio mate,a,

mass M

Figure

3.38 Magnetic support system for

Problem 3.27.

170 CHAPTER 3

ElectromechanicaI-Energy-Conversion Principles

Assume that the permanent magnet can be represented by a linear

characteristic of the form

Bm =/ZR(Hm- Hc)

and that the winding direction is such that positive winding current reduces

the air-gap flux produced by the permanent magnet. Neglect the effects of

magnetic fringing.

a. Assume the winding current to be zero. (i) Find the force fnd acting on

the mass in the x direction due to the permanent magnet alone as a

function of

x (0 < x < h). (ii)

Find the maximum mass Mmax that can be

supported against gravity for 0 < x < h.

b. For M = Mmax/2, find the minimum current required to ensure that the

mass will fall out of the system when the current is applied.

3.28 Winding 1 in the loudspeaker of Problem 3.25 (Fig. 3.37) is replaced by a

permanent magnet as shown in Fig. 3.39. The magnet can be represented by

the linear characteristic Bm

-- /ZR (Hm

-- Hc).

a. Assuming the voice coil current to be zero, (i2 = 0), calculate the

magnetic flux density in the air gap.

b. Calculate the flux linkage of the voice coil due to the permanent magnet

as a function of the displacement x.

c. Calculate the coenergy W~d(i2, x) assuming that the voice coil current is

sufficiently small so that the component of W~d due to the voice coil self

inductance can be ignored.

d. Calculate the force on the voice coil.

3.29 Figure 3.40 shows a circularly symmetric system in which a moveable

plunger (constrained to move only in the vertical direction) is supported by a

spring of spring constant K = 5.28 N/m. The system is excited by a

Permanent

magnet ~ f

Figure

3.39 Central core of

loudspeaker of Fig. 3.37 with winding 1

replaced by a permanent magnet

(Problem 3.28).

3.11 Problems 171

~ Spring

Plunger

~!ii!ii!ii:i!iii!iiiii!ili!ii~iiiiii!iiiiilz~iiiiiiii!i:ii:iiiiiiil,2i!:ili!ii~

g g << R1

tm X

T R3

Magnet

C/L

Figure 3.40

Permanent-magnet system for Problem 3.29.

samarium-cobalt permanent-magnet in the shape of a washer of outer radius

R3, inner radius

R2,

and thickness tm. The system dimensions are:

R1 -- 2.1 cm, R2 = 4 cm, R3 = 4.5 cm

h -- 1 cm, g = 1 mm, tm --= 3 mm

The equilibrium position of the plunger is observed to be x - 1.0 mm.

a. Find the magnetic flux density

in the fixed gap and Bx in the

variable gap.

b. Calculate the x-directed magnetic force pulling down on the plunger.

c. The spring force is of the form

fspring --

K (X0 - x). Find X0.

3.30 The plunger of a solenoid is connected to a spring. The spring force is given

by f = K0(0.9a - x), where x is the air-gap length. The inductance of the

solenoid is of the form

L = Lo(1 - x/a),

and its winding resistance is R.

The plunger is initially stationary at position x = 0.9a when a dc voltage

of magnitude V0 is applied to the solenoid.

a. Find an expression for the force as a function of time required to hold the

plunger at position a/2.

b. If the plunger is then released and allowed to come to equilibrium, find

the equilibrium position X0. You may assume that this position falls in the

range 0 _< X0 < a.

3.31 Consider the solenoid system of Problem 3.30. Assume the following

parameter values:

L0=4.0mH a--2.2cm R=1.5~ K0=3.5N/cm

172 CHAPTER 3

ElectromechanicaI-Energy-Conversion Principles

The plunger has mass M = 0.2 kg. Assume the coil to be connected to a dc

source of magnitude 4 A. Neglect any effects of gravity.

a. Find the equilibrium displacement X0.

b. Write the dynamic equations of motion for the system.

c. Linearize these dynamic equations for incremental motion of the system

around its equilibrium position.

d. If the plunger is displaced by an incremental distance E from its

equilibrium position X0 and released with zero velocity at time t = 0, find

(i) The resultant motion of the plunger as a function of time, and

(ii)

The

corresponding time-varying component of current induced across the coil

terminals.

3.32 The solenoid of Problem 3.31 is now connected to a dc voltage source of

magnitude 6 V.

a. Find the equilibrium displacement X0.

b. Write the dynamic equations of motion for the system.

c. Linearize these dynamic equations for incremental motion of the system

around its equilibrium position.

3.33 Consider the single-coil rotor of Example 3.1. Assume the rotor winding to be

carrying a constant current of I - 8 A and the rotor to have a moment of

inertia J = 0.0125 kg. m 2.

a. Find the equilibrium position of the rotor. Is it stable?

b. Write the dynamic equations for the system.

c. Find the natural frequency in hertz for incremental rotor motion around

this equilibrium position.

3.34 Consider a solenoid magnet similar to that of Example 3.10 (Fig. 3.24) except

that the length of the cylindrical plunger is reduced to a + h. Derive the

dynamic equations of motion of the system.

CHAPTER

Introduction to Rotating

Machines

he object of this chapter is to introduce and discuss some of the principles

underlying the performance of electric machinery. As will be seen, these prin-

ciples are common to both ac and dc machines. Various techniques and approx-

imations involved in reducing a physical machine to simple mathematical models,

sufficient to illustrate the basic principles, will be developed.

4.1 ELEMENTARY CONCEPTS

Equation 1.27, e = d)~/dt, can be used to determine the voltages induced by time-

varying magnetic fields. Electromagnetic energy conversion occurs when changes in

the flux linkage ~. result from mechanical motion. In rotating machines, voltages are

generated in windings or groups of coils by rotating these windings mechanically

through a magnetic field, by mechanically rotating a magnetic field past the winding,

or by designing the magnetic circuit so that the reluctance varies with rotation of the

rotor. By any of these methods, the flux linking a specific coil is changed cyclically,

and a time-varying voltage is generated.

A set of such coils connected together is typically referred to as an armature

winding. In general, the term armature winding is used to refer to a winding or a set

of windings on a rotating machine which carry ac currents. In ac machines such as

synchronous or induction machines, the armature winding is typically on the station-

ary portion of the motor referred to as the stator, in which case these windings may

also be referred to as stator windings. Figure 4.1 shows the stator winding of a large,

multipole, three-phase synchronous motor under construction.

In a dc machine, the armature winding is found on the rotating member, referred

to as the rotor. Figure 4.2 shows a dc-machine rotor. As we will see, the armature

winding of a dc machine consists of many coils connected together to form a closed

loop. A rotating mechanical contact is used to supply current to the armature winding

as the rotor rotates.

173

174 CHAPTER 4 Introduction to Rotating Machines

Figure 4.1

Stator of a 190-MVA three-phase 12-kV 37-r/min hydroelectric generator.

The conductors have hollow passages through which cooling water is circulated.

(Brown

Boveri Corporation.)

Synchronous and dc machines typically include a second winding (or set of

windings) which carry dc current and which are used to produce the main operating

flux in the machine. Such a winding is typically referred to as

field winding.

The field

winding on a dc machine is found on the stator, while that on a synchronous machine

is found on the rotor, in which case current must be supplied to the field winding via

a rotating mechanical contact. As we have seen, permanent magnets also produce dc

magnetic flux and are used in the place of field windings in some machines.

In most rotating machines, the stator and rotor are made of electrical steel, and

the windings are installed in slots on these structures. As is discussed in Chapter 1,

the use of such high-permeability material maximizes the coupling between the coils

and increases the magnetic energy density associated with the electromechanical

interaction. It also enables the machine designer to shape and distribute the magnetic

fields according to the requirements of each particular machine design. The time-

varying flux present in the armature structures of these machines tends to induce

currents, known as

eddy currents,

in the electrical steel. Eddy currents can be a large

source of loss in such machines and can significantly reduce machine performance. In

order to minimize the effects of eddy currents, the armature structure is typically built

from thin laminations of electrical steel which are insulated from each other. This is

illustrated in Fig. 4.3, which shows the stator core of an ac motor being constructed

as a stack of individual laminations.

In some machines, such as

variable reluctance machines

and

stepper motors,

there are no windings on the rotor. Operation of these machines depends on the

4.1 Elementary Concepts 175

Figure 4.2 Armature of a dc motor.

(General Electric Company.)

Figure 4.3 Partially completed stator core for an ac motor.

(Westinghouse Electric Corporation.)