Notes Electromechanics

8/4/2019 Notes Electromechanics

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Chapter 1

Electromechanical EnergyConversion

These notes were used to teach a course in Electromechanical Energy Conversionin conjunction with the course textbook: Electric Machinery (6th Ed.) by A.E. Fitzgerald, C. Kingsley, and S. D. Umans, McGraw-Hill, NY, 2003. (ISBN0-07-366009-4). In some Chapters of these notes there are references to guresin that textbook.

1.1 Conversion Process

Electric or Magnetic Fields in device couple between electro and mechan-

ical domains. Various conversion devices operate on similar physical principles

Structure of a specic device depends on its function.

1.2 Device Categories

Transducers

for measurement and control

linear operating conditions for input-output (usually)

relatively small signals e.g.: microphones; pickups; sensors; loudspeakers

Force-producing Devices

for actuation and holding- limited range of motion

1


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2 CHAPTER 1. ELECTROMECHANICAL ENERGY CONVERSION

e.g.: solenoids; relays; electromagnets

Continuous Energy-Conversion Equipment

for energy transfer in large quantities (continuous)

e.g.: motors; generators

1.3 Purposes of Analysis

understand process

develop design/optimization methods for devices with specic require-ments

develop models for performance evaluation of system components

1.4 Scope of Present Treatment

Magnetic elds only are considered for conversion medium.

Transducers and force-producing devices are considered rst.

Continuous energy-conversion devices are considered later.

For electric-eld systems analytical methods are similar.

1.5 Force in Electric and Magnetic Fields (Lorentz)

In simultaneous electric and magnetic elds, Lorentz force on a chargedparticle is

F = q(E+ v B) (1.1)

whereF = force (newtons)

q = charge (coulombs)

E = electric eld strength (volts per meter)

B = magnetic eld strength (teslas)

v = velocity of particle relative to magnetic eld (meters per second).

Note that superposition of electric and magnetic forces occurs.In a pure electric eld B = 0; so by equation 1.1 the force is

F = qE (1.2)

which acts in the direction of the electric eld irrespective of any particle motion.


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1.6. FORCE BY CHARGE AND CURRENT DENSITIES 3

In a pure magnetic eld E= 0; so by equation 1.1 the force is

F = q(v B) (1.3)

which acts in the direction determined by the vector cross product v B.Recall that the cross product v B denes a vector whose direction is

perpendicular to the plane of vectors v and B and whose magnitude is

jv Bj = jvj jBj sin ; (1.4)

where :

= angle between vectors v and B. The sense of direction for v Bfollows from a right-hand screw oriented along the perpendicular. Rotate v intoB to turn the screw, whereby the vector v B points in the direction of advanceof the screw. Alternatively use the equivalent right-hand rule. I.e., point thethumb and index nger in the directions of v and B; respectively. Then v Bemanates outward from the palm.

1.6 Force by Charge and Current Densities

The following results apply on a "microscopic scale", meaning a volume Vis very small compared to everyday dimensions but large compared to atomicdimensions. So "V ! 0" in a limit expression must be interpreted withinthese constraints rather than in the true mathematical sense.

Theorem 1FV = (E+ v B) (1.5)

where

Fv = force density:

= force per unit volume (newtons per cubic meter)

:

= charge density (newtons per cubic meter).

Proof. Consider a small element of volume

V = A jlj ; (1.6)

of cross sectional area A and length jlj as shown in Fig. 1, which containscharge

q= V: (1.7)

Assume that this elemental volume is moving with velocity v; so

F = q(E+ v B) = V(E+ v B) (1.8)

by substitution of equation 1.7 into equation 1.1. This gives

FV:

= limV!0

F

V= (E+ v B): (1.9)


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Theorem 2

FV = J B (1.10)

is the magnetic force density where

J = current density (amperes per square meter).

Proof. Specialize Theorem 1 for E = 0 :

FV = (v B) = (v) B: (1.11)

It remains only to show that J = v where

J:

=i

A: (1.12)

Now for the volume element V the charge density is

:

=q

V(1.13)

so

q = V = A jlj (1.14)

where the last expression follows by equation 1.6; and the velocity is

v =l

t(1.15)

ifV moves as part of a current in a continuous charge distribution : (In this

discussion both i and l are vectors normal to the cross section A:) This volumetransports charge q through a boundary surface with area A in time t. Thiscorresponds to a current i with magnitude

jij:

=q

t=

A jlj

t; (1.16)

where the last expression follows by equation 1.14; so

jJj:

=jij

A=

A jlj

At=

jlj

t= jvj =) J = v. (1.17)

where the last step before the arrow follows from equation 1.15.

Remark 3 Equation 1.10 for magnetic force density corresponds to equation1.3.

The results in Theorems 1 and 2 are useful for analysis in cases where alarge number of charged particles are in motion. The net force on a body canbe determined by integrating FV over its volume.


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1.6. FORCE BY CHARGE AND CURRENT DENSITIES 5

Example 4 (Example 3.1- Fitzgerald et al)

See Fig. 2 (RE: Fitzgerald (p. 114) Fig. 3.2 )A single-turn coil (rectangular loop on non-magnetic rotor) resides in a uni-form magnetic eld B0 = 0:02 T.

Coil sides are length l = 0:3 m. located at radius R = 0:05 m. from rotoraxis.

Coil plane is inclined at angle with respect to plane surface normal to eld.Find torque on coil when coil current I = 10 A.

Solution 5 Apply equation 1.10 to determine the force on each coil side. Forone side the force F is the integral of the magnetic force density vector FVover the volume of the wire. FV is constant over a side because J has constantmagnitude and direction over the length of the wire on that side and B0 isuniform. Hence the integral reduces to

F = V (FV) == V (J B0) = lA (J B0) = (JA B0) l

where V = lA is the volume of the wire with A as its cross-sectional area. Thisgives

F = (I B0) l

because I = JA (where I inherits the vector property of J with respect to direc-tion). By evaluation of the above equation for wire 1

F1 = (jIj jB0j l)bxwherebx is a unit vector coincident with thexaxis, because the currentI directedinto the paper and the vertical eld B0 are perpendicular. Likewise for wire 2

F2 = (jIj jB0j l)bxbecause the current I is directed out of the paper. The moment or torque aroundthe rotor axis is

= 1 + 2

where1 = R F1 = [jRj (jIj jB0j l)sin ]bz

and2 = (R) F2 = [jRj (jIj jB0j l)sin ]bz

with bz a unit vector perpendicular to and out of the paper. I.e., the forces F1and F2 are a couple which produces the net torque

= [2 jRj (jIj jB0j l)sin ]bz

which evaluates to

= [2(0:05) (10)(0:02) (0:3)sin ]bz = [0:006 sin]bz N m .Note that the translational forces on the ends which close the loop cancel andalso produce no torque around the rotor axix because both are parallel to it.


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1.7 Magnetic Force in Electromechanical Devices

All forces due to magnetic elds in electromechanical energy conversion devicesderive from the Lorenz force given by equation 1.10. Direct computation of suchforces by means of this equation requires the integration of the force density Fvover the volume of some component in a conversion device. Here integrationcan be interpreted as a sum of innitesimal constituent forces; i.e., superpositionapplies. This integration requires a detailed knowledge of the magnetic eld Band the current density J as functions of position. The direct integration is astraightforward problem only if B and J are simple functions, as in Example3.1 where both are uniform.

In general the functions for the eld B and current density J in a devicecomponent are not simple. This situation may be due in part to the presence ofnecessary magnetic material which can be analyzed in terms of equivalent surfacemagnetization curents due to the net eect of magnetic dipole alignment. Insuch cases the forces do not act only on elements carrying free current. Thisoften occurs by design to meet performance objectives.

In such general cases, the B eld computation (e.g., due to the stator currentsin a motor) invokes a form such as the Biot-Savart equation:

B =0

4

ZV

J(x0; y0; z0) baRR2

dV0: (1.18)

(Note that this equation follows from the dierential form of Amperes law,assuming the displacement current is negligible.) Likewise computation of J(e.g., for the rotor currents in a motor) becomes complex. Finally there is therequired integration of the force density Fv over the component volume. Clearlythis approach is feasible only by means of numerical algorithms.

The energy method is an alternative approach which can be used to deter-mine the net external eect of the detailed internal Lorenz force distribution. Itapplies under the reasonable assumption that device components are rigid (non-deformable) bodies and enables some simplifying assumptions used for magneticcircuits. In many situations results are suciently accurate and application isfar simpler than the direct method. The energy method is presented in the nextsectiion.

1.8 Energy Method

The energy method derives from conservation of energy in a system where

some energy is stored in a magnetic eld. I.e., the magnetic eld is part of anenergy conversion device and couples the electrical and mechanical portions ofthe system. This is shown conceptually in Fig. 3 where a lossless magneticenergy storage medium appears, to which a pair of electrical terminals and apair of mechanical terminals are attached. This can be generalized to includemultiple electrical and mechanical terminal pairs. Note that any electrical and


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1.9. MAGNETIC ENERGY BY STATE VARIABLES 7

mechanical loss mechanisms must be extracted analytically from a real mag-netic device to yield the lossless magnetic energy storage model. Of course loss

mechanisms can simply be ignored if such does not introduce signicant errors.The energy conservation principle for the lossless magnetic storage medium

isdWelec = dWmech + dWfld (1.19)

where the following represent dierential changes:

dWelec:

= electrical energy input = id = (ei) dt

dWmech:

= mechanical energy output = (ffld) dx

dWfld:

= magnetic stored energy.

The conservation principle also applies to the entire system. In this case it is

necessary only to add the termdWloss

:= loss energy (converted to heat)

to the right-hand side of equation 1.19 if all system energy storage is in magneticelds.The term dWloss, which accounts for all system loss mechanisms, includesthose which were necessarily extracted (mathematically) to yield the (ideal)lossless magnetic energy storage medium. Hence it includes:

1. ohmic electrical losses due to winding currents and magnetic-core eddycurrents;

2. core magnetization losses due to hysteresis;

3. mechanical losses due to friction and windage.

These losses (hence the term dWloss) must be represented by electricaland/or mechanical elements external to the medium. Items 1 and 2 are repre-sented by resistances connected to the electrical terminals. Item 3 is representedby dampers connected to the mechanical terminals. In fact even mass (which islossless) associated with a moving section of a magnetic core must be representedby an external mass element connected to the mechanical terminals.

1.9 Magnetic Energy by State Variables

The magnetic stored energy in the lossless medium of Fig. 3 follows by appli-cation of

dWfld = dWelec dWmech = id (ffld) dx; (1.20)

which is a simple algebraic rearrangement of equation 1.19. Here we considerenergy-conversion devices which consist of stationary and moving components sothe magnetic circuit includes moving or varying air gaps. Here signicant energyis stored in the air-gap magnetic eld which serves a "reservoir" between the


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electrical and mechanical portions of the system. This is in contrast to systemswith xed geometry (e.g., inductors and transformers), where energy stored in

gap, leakage, and core elds does not enter directly into the transformationprocess.

The magnetic stored energy is

Wfld (0; x0) =

ZP

dWfld; (1.21)

where P represents a path of integration from the initial point x = 0; = 0to the nal point x0; 0; as shown in Fig. 4. (A path P can be dened by afunction relation between and x.) Equation 1.20 is applied over the path Pnoting that i and ffld are each functions of and x :

i = i(; x) (1.22)

and

ffld = ffld(; x): (1.23)

The dependency on x enters due to moving device component(s). Magneticenergy storage is conservative (lossless) here so Wfld (0; x0) will be the sameirrespective of the chosen path P: In other words increments of electrical andmechanical energy have the same eect on the balance Wfld (; x) regardless ofthe order in which they are added or subtracted by varying and x: Hence andx are state variables, meaning that no knowledge of system history is necessaryto determine the stored energy Wfld (0; x0) under the present conditions.

The evaluation of equation 1.21 is greatly simplied by choosing P as path2 in Fig. 4.

) Wfld (0; x0) = ZP2a

dWfld + ZP2b

dWfld (1.24)

Here the variations in and x are essentially independent over P2a and P2b.Consider path P2a which starts with = 0, maintains d = 0; hence maintains = 0; this implies ffld 0 also, because there is no magnetic force without amagnetic eld. So Z

P2a

dWfld =

ZP2a

[id (ffld) dx] = 0; (1.25)

by application of equation 1.20, which reduces equation 1.24 to

Wfld (0; x0) = ZP2b

dWfld = Z00

i(; x0)d: (1.26)

Note that Wfld (0; x0) retains dependency on x0 because the current i(; x0) isdependent on the xed x0 as well as the varying during integration over thevertical path P2b.


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Magnetic stored energy is also obtained by integration of the eld energydensity over the volume V of the eld:

Wfld =

ZV

ZB0

H dB0

!dV: (1.27)

For soft magnetic material with constant permeability (i.e., B = H), thisgives

Wfld =

ZV

B2

2

dV: (1.28)

These results follow from eld theory.


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Example 6 Electromagnetic Relay

See Fig. 5 (RE: Fitzgerald- p. 119, Fig. 3.4 )

The magnetic core B H characteristic is assumed linear and lossless forsimplicity.

The resistance of the excitation coil is extracted to a lumped external resis-tance R:

Mass of moving component of armature is extracted to external (lossless)mass element connected to mechanical terminals.

Any friction loss due to armature pivot is extracted to external damperelement connected to mechanical terminals.

Electrical terminal variables are ;e; and i:

Mechanical terminal variables are displacement x and force ffld producedby the magnetic eld.

)Excitation coil, magnetic core, and armature comprise a lossless magneticenergy storage medium, in conformance with the requirements of Fig. 3.

By previous analysis of magnetic circuits, where x plays the role of gaplength, we obtain the linear relation:

= L (x) i:

I.e., L (x) is the inductance which depends only on geometry, permeability ofmagnetic material, and armature position x:

Solution 7 Now determine the stored magnetic energy as a function of andx, by applying equation 1.26:

Wfld (0; x0) =

Z00

i(; x0)d:

First solve the linear relation above for i :

) i(; x0) =

L (x0);

then substitute this into the equation above for Wfld (0; x0), which gives:

Wfld (0; x0) =

Z00

L (x0)d =

1

L (x0)

Z00

d =20

2L (x0):


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See Fig. 6 (RE: Fitzgerald- p. 122, Fig. 3.6 )Electromagnetic Relay with movable plunger is shown.Core and plunger materials: permeability is assumed innite ( ! 1).h

:= plunger height; g

:= air-gap length; assume h g:

Dimensions: d = 0:15 mm; l = 0:1 mm; g = 2:0 mm.MMF Data: i = 10 A.; N = 1000 turns.=) Calculate magnetic stored energy Wfld as a function of plunger position

(0 < x < d) :

Solution 9 Use

Wfld (; x) =2

2L (x)

derived in previous example, but convert this to function of i (given here) bysubstitution of = (L (x)) i: This gives

Wfld (i; x) =(L (x)) i2

2

where

L (x) =0N

2Agap2g

;

by previous analysis of magnetic circuits, with the gap cross-sectional area Agapa function of x :

Agap = ld

1

x

d from geometry of structure. Substitution of this last equation into the previousequation gives

L (x) =0N

2ld

1 xd

2g

and by substitution of this equation for L (x) into the equation for Wfld (i; x) :

Wfld (i; x) =

"0N

2ld

1 xd

2 (2g)

#i2:

Finally evaluation for given parameter values yields

Wfld = "4 107 (1000)2 (0:1)(0:15)2(2(0:02)) # (10)2 1 xd :=) Wfld = 236

1

x

d

J.


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1.10 Magnetic Force from Energy

Wfld (; x) is already established as a function of the two (state) variables andx: The dierential of any function of two variables can be written in terms ofits partial derivatives with respect to those variables. Applying this concept toWfld (; x) gives

dWfld (; x) =

@Wfld (; x)

@

x=constant

d+

@Wfld (; x)

@x

=constant

dx:

(1.29)Both this equation and equation 1.20 must yield the same dWfld (; x) in termsof the dierentials d and dx: Note that this requirement holds for arbitraryvalues of d and dx because and x are independent variables. This imposesthe requirement that the coecients of d must be the same in both equationsand likewise the coecients of dx must be the same in both equations. Hence

we concludei =

@Wfld (; x)@

x=constant

(1.30)

and

ffld =

@Wfld (; x)

@x

=constant

: (1.31)

Equation 1.30 provides the means to deduce the current i from the energy func-tion Wfld (; x). Equation 1.31 provides the means to deduce mechanical forceffld from the energy function Wfld (; x) and can be converted to a functionof i and x by insertion of the function relating to i: Evaluation of a partialderivative with respect to a variable requires that all remaining (in this caseone) variables be held constant. However this is a purely mathematical require-ment which imposes no requirement at all that such remaining variables be held

constant during actual operation of a device.

Example 10 Any Linear Magnetic System

By denition of a linear system

= [L (x)] i

and in this case the energy is

Wfld (; x) =2

2L (x)

by the results of Example 6. Now applying equation 1.31 to this gives:

ffld = @

@x 2

2L (x)=constant =2

2 [L (x)]2

d [L (x)]

dx

;

and substitution of = [L (x)] i into this equation gives

ffld =i2

2

d [L (x)]

dx;

which is a function of i.


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1.10. MAGNETIC FORCE FROM ENERGY 13



Solenoid Inductance=) Table given as a function of plunger position:

x (cm) 0 0:2 0:4 0:4 0:8 1:0 1:2 1:4 1:6 1:8 2:0L (mH) 2:8 2:26 1:78 1:52 1:34 1:26 1:20 1:16 1:13 1:11 1:10

Note: x = 0 is position of full retraction.

Plot solenoid force at current 0:75 A. as function of position for range 0:2 x 1:8 cm.

Solution 12 Apply

ffld =i2

2

d [L (x)]

dx;

derived in Example 10, where i = 0:75 A. is given here. Hence it is necessary

only to evaluate d[L(x)]dx

from the given table by numerical approximation.

A fourth-order polynomial t to the data for L (x) can be obtain by usingthe Matlab function polyt: I.e., we assume

L (x) = a1x4 + a2x

3 + a3x2 + a4x

1 + a5

and polyt generates the coecients ak; k = 1; 2; :::; 5; after the table data isentered. See textbook pp. 126-27 for Matlab code.

From this

d [L (x)]dx

= 4a1x3 + 3a2x2 + 2a3x + a4

can be substituted into the rst equation to obtain ffld : See Fig. 7 for plots:L (x);ffld :


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1.11 Torque from Energy

Consider a system with rotational rather than linear mechanical motion. Thederivation of magnetic eld energy in this rotational system is analogous tothat in Section 1.9 for systems with linear displacement. Here torque fld andangular displacement replace force ffld and linear displacement x; respectively,so the energy balance is

dWfld = dWelec dWmech = id (fld) d (1.32)

in place of equation 1.20. Likewise the mechanical terminal variables in Fig. 3become fld and :

We continue the analogy to Section 1.9 to obtain

Wfld (0; 0) = Z0

0

i(; 0)d: (1.33)

in place of equation 1.26. Likewise

i =

@Wfld (; )

@

=constant

(1.34)

and

fld =

@Wfld (; )

@

=constant

(1.35)

replace equations 1.30 and 1.31, respectively.

Example 13 Any Linear Magnetic System with Rotational Motion

(This example is analogous to Example 10)By denition of a linear system

= [L ()] i

and in this case the energy is

Wfld (; ) =2

2L ()

analogous to the results of Example 6. Now applying equation 1.35 to this gives:

fld = @

@

2

2L ()

=constant

=2

2 [L ()]2d [L ()]

d;

and substitution of = [L ()] i into this equation gives

fld =i2

2

d [L ()]

d;

which is a function of i.


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1.11. TORQUE FROM ENERGY 15



Rotational system is shown.

Magnetic circuit consists of a single-coil stator and an oval rotor.

Coil inductance L () varies with rotor angular position :

L () = L0 + L2 cos(2) ;

where is measured between magnetic axis of stator coil and major axis ofrotor.

Note second-harmonic variation of L () ; which is consistent with fact that

L () is the same for ! + 180

(rotor rotation through 180

)Given parameters: L0 = 10:6 mH; L2 = 2:7 mH.

Find torque as function of for current of 2 A:

Solution 15 By the results of Example 13

fld =i2

2

d [L ()]

d:

Here

d [L ()]

d =

d [L0 + L2 cos(2)]

d = 2L2 sin (2)

so

fld =i2

2(2L2 sin(2)) :

Evaluation for given parameters:

fld () =(2)

2

2

2

2:7

103

sin(2)

=) fld () = 1:08 103 sin(2) N m:Note that torque acts in direction to pull rotor axis into alignment with coil axis.This alignment maximizes coil inductance.


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1.11. TORQUE FROM ENERGY 19


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1.12 Coenergy

The magnetic coenergy (complement of the magnetic energy) is related to themagnetic energy on the path P2 in Fig. 4 that led from equation 1.24 to 1.26.I.e, we assume that the position x0 is attained by following path P2a before anymagnetic energy is stored in the eld. Next magnetic energy Wfld is acquired bythe eld as the ux increases from 0 to 0 along path P2b while position is heldconstant at x0: Fig. 9 is a detailed illustration of the latter process, where vs.i represents a general function or characteristic seen at the electrical terminal-pair of the system. Each characteristic vs. i shown in Fig. 9 corresponds toa xed value of the position parameter x.

The magnetic coenergy is

W0fld (i0; x0):

=

Zi0

0

(i; x0) di; (1.36)

which is the area below the curve in Fig. 9, whereas the magnetic energy givenby equation 1.26 is the area above the same curve. (Henceforth subscripts of iand x will be omitted, as the meaning will remain clear aside from the evaluationof integrals.) Clearly

W0fld (i; x) + Wfld (; x) = i; (1.37)

i.e., these two areas add to the area of the rectangle with upper corner at thepoint (i; ) ; so

W0fld (i; x) = i Wfld (; x) : (1.38)

The last equation can be reduced to a function of only i and x because is afunction of i:

Theorem 16dW0fld (i; x) = di + (ffld) dx: (1.39)

Proof. Take the dierential of equation 1.38:

dW0fld (i; x) = d (i) dWfld (; x)

Note that d (i) = id + di and dWfld (; x) = id (ffld ) dx (by equation1.20). Substitution of these last two expressions into the above equation gives

dW0fld (i; x) = id + di [id (ffld) dx] = di + (ffld) dx:

In general

W0fld (i; x) =

ZP0

dW0fld

and it follows from equation 1.39 that the result W0fld (i; x) will be independentof the path P0 chosen in the i x plane. Note that the result in equation 1.36


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1.13. MAGNETIC FORCE FROM COENERGY 21

was obtained by choice of one such path. Hence i and x are state variables forthe coenergy, whereas and x were taken as state variables for the energy. The

general reasoning process here is analogous to that in Section 1.9. The choiceto use energy or coenergy is primarily based on mathematical convenience inthe context of a given problem.

Magnetic coenergy is also obtained by integration of the eld energy densityover the volume V of the eld:

W0fld =

ZV

ZH0

B dH0

!dV (1.40)

for soft magnetic material (where H = 0 gives B = 0). With constant perme-ability (i.e., B = H), this gives

W0fld = ZVH2

2 dV: (1.41)Equation 1.40 must be modied to

W0fld =

ZV

ZHHc

B dH0

!dV; (1.42)

whereas equation 1.27 remains unchanged, for hard magnetic material (whereH = Hc gives B = 0) as in permanent magnets. This follows because energyand coenergy are zero when B = 0 which is equivalent to H = Hc. These resultsfollow from eld theory. Equation 1.42 is general because it includes equation1.40 as the special case (soft material) where Hc = 0:

1.13 Magnetic Force from CoenergyW

0

fld (i; x) is already established as a function of the two (state) variables i andx; hence

dW0

fld (i; x) =

"@W

0

fld (i; x)

@i

#x=constant

!di +

"@W

0

fld (i; x)

@x

#i=constant

!dx:

(1.43)Matching coecients of di and dx in this equation with those in equation 1.39,we obtain

=

"@W

0

fld (i; x)

@i

#x=constant

(1.44)

and

ffld =

"@W

0

fld (i; x)

@x

#i=constant

: (1.45)

Equation 1.44 provides the means to deduce the ux from the coenergy func-tion W

0

fld (i; x). Equation 1.45 provides the means to deduce mechanical force


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ffld from the coenergy function W0

fld (i; x) and can be converted to a functionof and x by insertion of the function relating to i: The reasoning here is

analogous to that in Section 1.10 where more detail can be found.

1.14 Torque from Coenergy

Consider a system with rotational rather than linear mechanical motion. Thederivation of magnetic eld coenergy in this rotational system is analogous tothat in Section 1.12 for systems with linear displacement. Now torque fld andangular displacement replace force ffld and linear displacement x; respectively,and likewise become the mechanical terminal variables in Fig. 3. So magneticcoenergy becomes

W0fld (i0; 0):

= Zi0

0

(i; 0) di (1.46)

in place of equation 1.36 and likewise

W0

fld (i; ) = i Wfld (; ) (1.47)

replaces equation 1.38. Then analogous to Theorem 16 we have

Theorem 17

dW0fld (i; ) = di + (fld) d:


dW0

fld (i; ) = d (i) dWfld (; )

Note that d (i) = id + di and dWfld (; ) = id (fld) d (by equation1.32). Substitution of these last two expressions into the above equation gives

dW0fld (i; x) = id + di [id (fld) d] = di + (fld) d:

By analogy to Section 1.13 we obtain

i =

"@W

0

fld (i; )

@i

#=constant

(1.48)

and

fld =

"@W

0

fld (i; )

@

#i=constant

(1.49)

in place of equations 1.44 and 1.45, respectively.


23/127

1.14. TORQUE FROM COENERGY 23

Example 18 Any Linear Magnetic System

Here we repeat Example 10 by using coenergy instead of energy.By denition of a linear system

= [L (x)] i:

In this case the coenergy is, by application of equation 1.36:

W0fld (i0; x):

=

Zi00

(i; x) di =

Zi00

[L (x)] idi =

1

2

[L (x)] i20:

The force (same result as obtained in Example 10) is, by equation 1.45:

ffld =

"@W

0

fld (i; x)

@x

#i=constant

=

"@12

[L (x)] i2

@x

#i=constant

=

i2

2

dL (x)

dx

:


For the relay of Example 3.2, nd the force on the plunger if the coil is drivenby a controller which produces a current

i (x) = I0

xd

A:

Note that both the current i (x) and the force are functions of x: Based on theassumptions in Example 3.2 (which is Example 8 and Fig. 6 here) this is amagnetically linear system and the following result was obtained:

L (x) =0N

2ld

1 x

d

2g :By application of the result from Example 18

ffld =

i2

2

dL (x)

dx

=

i2

2

0N

2l

2g

:

Force as a function of x is obtained by substitution of i (x) into the previousequation:

) ffld =

I200N

2l

4g

xd

2:

Also from Example 18 the coenergy here is

W0fld (i; x) = i2

2 [L (x)] = i2

2 "0N

2ld

1 x

d2g # :which can be written correctly as a function of x alone by substitution of i (x) :

W0fld (i; x) =

"I200N

2ld

1 xd

4g

#xd

2:


24/127


However direct application of Equation 1.45 to this last expression would givean incorrect result ffld (dierent from that found above) because partial dif-

ferentiation with respect to x imposes the condition that the current i be heldconstant (at least mathematically). I.e., this condition has been violated alreadyby the substitution i (x) which renders i a function of x in the last expression.


25/127

1.15. ENERGY AND COENERGY CONSIDERATIONS 25

1.15 Energy and Coenergy Considerations

1.15.1 Equivalence of ForcesEquations 1.31 and 1.45 must yield the same result for the force ffld whenapplied to the same device. These equations are repeated here for reference as

ffld =

@Wfld (; x)

@x

=constant

:=

lim

4x!0

4Wfld (; x)

4x

=constant

(1.50)

and

ffld =

"@W

0

fld (i; x)

@x

#i=constant

:=

"lim

4x!0

4W0

fld (i; x)

4x

#i=constant

; (1.51)

respectively, where the right-hand expressions have been added for utility. This

equivalence of forces can be explained by consideration of the i characteristicsof a typical device, as shown in Fig. 10. Note that a family of such characteristicscan be generated by variation of the position parameter x:

The device is assumed to be operating initially along the i curve shownand labelled for a particular position x: Then the position is changed from xto x + 4x; which moves the device operation onto the curve labelled x + 4x:The area enclosed between these two curves includes both j4Wfld (0; x)j and

4W0

fld (i0; x), in view of the denitions of Wfld and W0

fld : It is necessary onlyto show

lim

4x!0

4Wfld (; x)

4x

=constant

=

"lim

4x!0

4W0

fld (i; x)

4x

#i=constant

; (1.52)

i.e., the right-hand expressions in equations 1.50 and 1.51 converge to the samequantity as 4x ! 0.The equality in equation 1.52 follows by consideration of the conditions under

which the partial derivatives are dened. For equation 1.50 the limit is takenwhile = 0 is constant and i is allowed to vary from i0 to i0 4i, whichcorresponds to bounding an area (hence bounding j4Wfld (0; x)j) in Fig. 10by the two curves and the horizontal line from point a to point b: For equation1.51 the limit is taken while i = i0 is constant and is allowed to vary from 0 to04, which corresponds to bounding an area (hence bounding 4W

0

fld (i0; x))in Fig. 10 by the two curves and the vertical line from point a to point c: Thesetwo areas (hence j4Wfld (0; x)j and 4W

0

fld (i0; x)) dier by only the amountin the (roughly) triangular region bounded by the line ab; line ac; and curvesegment bc: The ratio of this dierence amount to the remaining large bounded

area goes to zero as 4x ! 0. The establishes the claimed equality.As a general rule, it will be more convenient to use energy to determine force

or torque if ux is given as a driving or regulated quantity. Likewise it will bemore convenient to use coenergy to determine force or torque if current is givenas a driving quantity. The latter case is quite common, as shown by subsequentexamples.


26/127


1.15.2 Direction of Forces

The eld force ffld acts to decrease the magnetic energy at constant ux orincrease the magnetic coenergy at constant current. This follows directly byinterpretation of equations 1.50 and 1.51, respectively. In the case of a singleelectrical terminal-pair, increase of coenergy at constant current implies thatthe force acts to increase the inductance of a device. I.e., the force pulls devicecomponents in a way which would reduce the reluctance in the correspondingmagnetic-circuit model.

1.15.3 Linear Systems

In a linear system = [L (x)] i; from which it follows by previous results thatmagnetic energy and coenergy are equal:

2

2L=

Li2

2: (1.53)

Likewise it can be shown from eld theory that energy and coenergy densitiesare equal:

B2

2=

H2

2: (1.54)

In a nonlinear system is not linearly proportional to i and equivalentlyB is not linearly proportional to H: Hence in that case the above relations donot hold. Also, examination of Fig. 10 shows that in general the energy andcoenergy in a nonlinear system are not equal.

1.16 General Solutions

A general solution for the magnetic eld is necessary in systems where an equiv-alent magnetic circuit is inadequate or innaccurate. E.g., such systems mayinclude complex geometry and/or very nonlinear magnetic materials. Generalsolutions are obtained by numerical algorithms implemented in commercial soft-ware ("eld solvers"). The nite-element method is a common example of suchalgorithms.

The energy or coenergy method is still usable even if a general eld solu-tion is required. I.e., in this case the energy and coenergy can be computed

by application of equation 1.27 and equation 1.40 or 1.42, respectively. There-upon forces are calculated by taking partial derivatives (as before), where suchderivatives must be determined by numerical approximation. E.g., commercialsoftware is available to determine magnetic coenergy for a linear-displacementactuator as a function of displacement x; from which force can be determinedas described above.


27/127

1.16. GENERAL SOLUTIONS 27


See Fig. 11 (RE: Fitzgerald- p. 135, Fig. 3.12 )Magnetic circuit with stator and rotor free to turn about axis is shown.

Magnetic materials (stator and rotor): permeability is assumed innite ( !1).

Neglect eects of fringe elds.

h:

= axial length (height perpendicular to page); g:

= air-gap length; r1:

=radiusof rotor:

Dimensions: h = 1:8mm; g = 3:0mm; r1 = 2:5 cm:

MMF (N i) Data: To be determined in solution (i.e., N and i are not speci-ed).

=) a) Derive an expression for torque acting on rotor, in terms of dimensionsand magnetic eld in the two air gaps.

=) b) Assume maximum ux density B = 1:65 T (to avoid excessive satu-ration of steel) in portion of air gap where rotor and stator overlap. Computemaximum torque.

a) In the (overlapping) air gap:

Hag =N i

2g

because the net air gap length is 2g (due to two gaps in series), Bsteel = Bag isnite, and Hsteel =

Bsteel

= 0 due to ! 1:

Coenergy:

W0

fld =

hW

0

fld

iag+

hW

0

fld

isteelwhere

hW

0

fld

iag

andh

W0

fld

isteel

are found by integration of coenergy density

(equation 1.41) over the volumes of the air gap and the steel, respectively.hW

0

fld

isteel

= 0

because H2

steel

2 =B2steel2

= 0 in view of Hsteel = 0 (above). This leaves

W0

fld =h

W0

fld

iag

=

ZV

0H

2ag

2

!dV =

0H

2ag

2

!V

where the integral is the application of equation 1.41 to the (overlapping) airgap. The integral reduces to the simple product of coenergy density

0H2ag2

and volume V because Hag is taken as constant in the (overlapping) air gap.By simple geometry

V = 2gh (r1 + 0:5g)


28/127


closely approximates the volume of a cylindrical slice where radius r1 + 0:5g istaken midway in the gap and is given in radians.

) W0

fld =

0H

2ag

2

!V =

0H

2ag

2

!(2gh (r1 + 0:5g) ) =

0 (N i)2 h (r1 + 0:5g)

4g

Torque follows by application of equation 1.49:

fld =

"@W

0

fld (i; )

@

#i=constant

=

24

@h0(Ni)2h(r1+0:5g)

4g

i@

35i=constant

=0 (N i)

2 h (r1 + 0:5g)

4g:

This is a constant torque which is present only when there is some overlap (i.e.,when 6= 0). Its sign is positive so that it acts in a direction to increase theoverlap angle ; which tends to align the rotor and stator pole faces.

b) It is necessary to determine mmf (N i) to apply the above equation fortorque. First nd Hag: Bag = 0Hag =)

Hag =Bag0

=1:65

4 [107]= 1:31

106

A= m:

Then from rst equation in part a):

N i = 2gHag = 2

3

103

1:31

106

= 7860 A turns.

Now evaluate above equation for torque:

fld =4

107

(7860)

2

1:8

102

2:5

102

+ 0:5

3

103

4(3[103])= 3:09 N m:

N.B.: Refer to Practice Problem 3.6 (Fitzgerald et al) to determine torquefld by rst deriving inductance L: Then fld follows by application of fld =hi2

2

i hdL()d

i:


29/127

1.17. MULTIPLY-EXCITED MAGNETIC FIELD SYSTEMS 29

1.17 Multiply-Excited Magnetic Field Systems

1.17.1 BackgroundElectromechanical systems with multiple electrical terminals are quite common.E.g., for power (p = vi) measurement by a wattmeter it is necessary to obtaintorque proportional to the product of two electrical variables (voltage v andcurrent i). Likewise most energy conversion devices require multiple electricalterminals to excite multiple windings which produce the magnetic elds forintended operation.

1.17.2 Methods of Analysis

The energy and coenergy methods will be extended and applied here to systemswith two electrical terminal-pairs, where analogous extension to more than two

terminal pairs is also possible. This amounts to a generalization of the approachpresented so far. A conceptual representation of a system with two electricalterminal-pairs and one mechanical terminal-pair is shown in Fig. 12. It isassumed that the mechanical portion of the system in Fig. 12 undergoes rotarymotion.

The system state is determined by three independent variables because thereare three terminal-pairs. Angular displacement is usually chosen as the me-chanical state variable. Either ux linkages 1 and 2 or currents i1 and i2 maybe chosen as electrical state variables. A hybrid set is also possible.

1.17.3 Analysis by Energy Method

The dierential energy is

dWfld (1; 2; ) = i1d1 + i2d2 (fld) d; (1.55)

for a system with two electrical terminal-pairs and one mechanical terminal-pair, analogous to equation 1.32. In this case 1; 2; and are the naturalchoice for state variables because they are the "running variables" to determineWfld (1; 2; ) by integration of equation 1.55. The currents are given by

i1 =

@Wfld (1; 2; )

@1

=constant2=constant

(1.56)

and

i2 = @Wfld (1; 2; )

@2 =constant1=constant ; (1.57)

analogous to equation 1.30. The torque is given by

fld =

@Wfld (1; 2; )

@

1=constant2=constant

(1.58)


30/127


analogous to equation 1.35.The energy is

Wfld (10 ; 20 ; 0) =

Z200

i2 (1 = 0; 2; = 0) d2+

Z100

i1 (1; 2 = 20 ; = 0) d1

(1.59)by integration of equation 1.55 over the specic path shown in Fig. 13. Anal-ogous to the path in Fig. 4 for the singly-excited case, this path begins by in-tegrating over while 1 = 2 = 0 which contributes nothing because fld = 0when there is no magnetic eld. Then this path completes by integrating over2 while 1 = 0 and nally integrating over 1: The quantity Wfld (10 ; 20 ; 0)found by integration over any other path will be the same as that computed byequation 1.59 because energy is a state function.

1.17.4 Analysis by Coenergy MethodThe coenergy for a system with two electrical terminal-pairs is

W0

fld (i1; i2; ):

= 1i1 + 2i2 Wfld (1; 2; ) ; (1.60)

analogous to equation 1.38 for a system with one electrical terminal-pair.

Theorem 21

dW0fld (i1; i2; ) = 1di1 + 2di2 + (fld) d: (1.61)


dW0fld (i1; i2; ) = d (1di1) + d (2di2) dWfld (1; 2; )

Note that d (i11) = i1d1+1di1; d (i22) = i2d2+2di2; and dWfld (1; 2; ) =i1d1 + i2d2 (fld) d (by equation 1.55). Substitution of these last threeexpressions into the above equation gives

dW0fld (i1; i2; ) = i1d1 + 1di1 + i2d2 + 2di2 [i1d1 + i2d2 (fld ) d]

= 1di1 + 2di2 + (fld) d:

Remark 22 This theorem is analogous to Theorem 17.

It follows from equation 1.61, by derivation analogous to that for equations1.48 and 1.49, that

1 =

"@W

0

fld (i1; i2; )

@i1

#=constanti2=constant

; (1.62)


31/127

1.17. MULTIPLY-EXCITED MAGNETIC FIELD SYSTEMS 31

2 = "@W

0

fld (i1; i2; )

@i2 # =constanti1=constant; (1.63)

and

fld =

"@W

0

fld (i1; i2; )

@

#i1=constanti2=constant

: (1.64)

Note that fld is determined directly in terms of currents- a signicant advan-tage.

The coenergy is

W0

fld (i10 ; i20 ; 0) =

Zi200

2 (i1 = 0; i2; = 0) di2+

Zi100

1 (i1; i2 = 0; = 0) di1

(1.65)by integration of equation 1.61. The path of integration is analogous to thatused to derive equation 1.59.

1.17.5 Linear Systems

For a linear system

1 = L11i1 + L12i2 (1.66)

and

2 = L21i1 + L22i2; (1.67)

where L11; L22 are self-inductances, L12; L21 are mutual inductances, and all

inductances are functions of angular position (in general). Further L21 = L12due to the physical reciprocity of magnetic-eld phenomena.

1.17.6 Application of Energy Method to Linear Systems

The immediate goal is to determine Wfld (10 ; 20 ; 0) for a linear system interms of10 ; 20 ; 0 and the given inductance parameters L11; L22; L12; L21; byapplication of equation 1.59. For this it is necessary to express i1 and i2 asfunctions of1 and 2 which are the running variables of integration in equation1.59. Hence we nd i1 and i2 by inversion of equations 1.66 and 1.67:

i1 =L221 L122

D(1.68)

and

i2 =L211 + L112

D(1.69)

where

D = L11L22 L12L21: (1.70)


32/127


Then evaluation of equation 1.59, after substitution of equations 1.68, 1.69, and1.70, gives

Wfld (10 ; 20 ; 0) =Z200

[L11 (0)] 2D (0)

d2 +Z100

[L22 (0)] 1 [L12 (0)] 20D (0)

d1(1.71)

=[L11 (0)]

220

2D (0)+

[L22 (0)] 210

2D (0)

[L12 (0)] 1020D (0)

:

The torque can be determined by application of equation 1.58, which re-quires partial dierentiation of the above expression with respect to 0. Un-fortunately partial dierentiation of this somewhat complicated expression forWfld (10 ; 20 ; 0) will generate a very complicated expression for the torque,in particular due to functions of 0 which are present in all numerators anddenominators. It turns out that the coenergy method leads to much simplerexpression of results.

1.17.7 Application of Coenergy Method to Linear Sys-tems

The immediate goal is to determine W0

fld (i10 ; i20 ; 0) for a linear system in termsof i10 ; i20 ; 0 and the given inductance parameters L11; L22; L12; L21; by appli-cation of equation 1.65. Evaluation of equation 1.65, after direct substitutionof equations 1.66 and 1.67, yields

W0

fld (i1; i2; ) =[L11 ()] i21

2+

[L22 ()] i222

+ [L12 (0)] i1i2: (1.72)

The torque is

fld = "@W0fld (i1; i2; )@

#i1=constanti2=constant

= i212 d [L11 ()]

d+ i22

2 d [L22 ()]

d+[i1i2] d [L

12 ()]d

(1.73)by direct application of equation 1.64 to equation 1.72. Clearly the coenergymethod is far more ecient than the energy method in the case of linear systemsand gives the torque directly in terms of currents.


See Figs. 14, 15 (RE: Fitzgerald- pp. 140-41, Figs. 3.15, 3.16 )Magnetic circuit with stator and rotor free to turn about axis is shown.System has two electrical terminal-pairs (rotor and stator windings) for ex-

citation.Inductances are specied directly:

L11 = (3 + cos 2)

103

;

L22 = 30 + 10 cos 2;

L12 = 0:3cos :


33/127

1.18. SYSTEMS WITH PERMANENT MAGNETS 33

Excitation Currents are: i1 = 0:8 A; i2 = 0:01 A:=) Find and plot torque fld () :

Solution 24 Torque can be computed from equation 1.73.:

fld =

i212

d [L11 ()]

d+

i222

d [L22 ()]

d+ [i1i2]

d [L12 ()]

d

=

i212

2

103

sin2

+

i222

(20sin2) [i1i2] (0:3) sin

= 1:64

103

sin2 2:4

103

sin :

Remark 25 The torque components

i21

2 d [L11 ()]

d

= i21

2 2 103

sin2and

i222

d [L22 ()]

d=

i222

(20sin2)

are due to winding currents acting separately as in a singly excited system.These torques occur because self inductances are a function of rotor position.So they are dependent on the reluctance of the magnetic path where both indi-vidual winding currents actually "see" the same magnetic path. They are calledreluctance torques and act in a direction to maximize coenergy, hence maximizeself inductances. I.e., these torques tend to align the rotor axis with the statoraxis.

The torque component

[i1i2]d [L12 ()]

d= [i1i2] (0:3)sin

is due to the mutual interaction between rotor and stator currents and is hencecalled the mutual interaction torque. This is also dependent on the reluctanceof the magnetic path and acts in a direction to maximize mutual inductance.It acts in a direction to align the rotor and stator, hence align their magneticelds.

1.18 Systems with Permanent Magnets

1.18.1 Background

Here we consider an electromechanical system or device which includes a perma-nent magnet. For simplicity we assume that the system as given has no electricalterminal-pairs but note that our method can be generalized to systems whichinclude multiple electrical terminal-pairs.


34/127


We wish to apply the energy/coenergy method so need to determine same.In the most dicult cases a general eld solution will yield Wfld or W

0

fld by

application of equation 1.27 or equation 1.42, respectively. Either of these equa-tions will integrate over volume the magnetic eld created in the system by thepermanent magnet. Note that in the absence of any electrical terminal pairWfld and W

0

fld will be functions of only displacement x and the system (in-cluding magnet) parameters. Then magnetic force ffld can be determined byapplication of equation 1.31 to Wfld or equation 1.45 to W

0

fld. The absence ofan electrical terminal pair actually simplies the partial derivatives in these lastequations.

1.18.2 Coenergy Method with Magnetic Circuit

We wish to retain the advantages that magnetic-circuit analysis provides inthose systems which are suciently uniform or simple to permit its application

without excessive error. It turns out that a "ctitious" or auxiliary windingwith an electrical terminal current expedites the analysis here, in which casecurrent is the natural choice for a state variable. Hence we choose the coenergymethod, for which current is the natural state variable and for which analysisof linear systems is most ecient (as shown previously).

The rst task is to determine W0fld as a function of displacement x; where weconsider the simple system in Fig. 16 for deniteness. Previously this was doneby integration of equation 1.39 but so far there is no state variable i: Now thereis only a state variable ffld to be integrated with respect to x; but ffld is actuallythe desired result. In addition we must account for the eect of the permanentmagnet in determination of the eld by magnetic-circuit analysis. These issuesare resolved by introduction of the ctitious winding (a mathematical artice)shown in Fig. 16. Initially the winding current if is adjusted (analytically) toa value If0 which cancels the magnetic eld in the permanent magnet (hencereduces the eld to zero in the entire system). At this point we procede to theintegration of equation 1.39 where the second state variable has become i = if,with = f:

The integration is performed over the path P1 in the ifx plane, as shown inFig. 17. This path goes from the point (if = If0; x = 0) where the system eld(hence W0fld) is zero to the point (if = 0; x = x0) where if = 0 is equivalent toremoving the ctitious winding. I.e., we seek W0fld (if = 0; x = x0) by movingon the path P1 between these two points through the (deliberately chosen)segments P1a and P1b:

) W0fld (if = 0; x = x0) = ZP1adW0fld + ZP1b

dW0fld : (1.74)

Note thatRP1a

dW0fld = 0 because f = 0 (as well as dif = 0) due to the nulleld condition. Hence

W0fld (if = 0; x = x0) =

ZP1b

dW0fld =

Z0If0

f (if; x = x0) dif (1.75)


35/127


where the term fflddx does not appear here because dx 0 on path P1b: (Notethat the development here was analogous to that which led from equation 1.24

to equation 1.26, but here the "oset" current If0 was necessary to render theintegral zero along the rst path segment.)

The result in equation 1.75 is completely general so it applies in cases whereany magnetic material in the system (including the magnet itself) is nonlinear.The magnetic force ffld due to the permanent magnet alone follows by applica-tion of equation 1.45 to W0fld (if = 0; x) ; where we note that neither W

0

fld norffld is zero.


See Fig. 18 (RE: Fitzgerald- p. 145, Fig. 3.19 )Magnetic circuit including permanent magnet (samarium-cobalt) with mov-

able plunger is shown.

MMF Data: Fictitious winding (Nf turns with current if) is included foranalytic purposes.

Core and plunger materials: permeability is assumed innite ( ! 1);neglect eld fringing eects

x:

= plunger displacement (movable gap); g0:

= xed air-gap length; d =magnet length; D = depth of core and plunger/

Dimensions: d = 2:0 cm; g0 = 0:2 cm; D = 3:0 cm.;Widths: Wm = 2:0cm; Wg = 3:0cm; W0 = 2:0 cm:

=) a) Derive coenergy W0

fld as a function of plunger position x:=) b) Derive force on plunger as a function of plunger position x:=) c) Evaluate force at x = 0 and x = 0:5 cm:

Solution 27 a) Represent the DC magnetization curve (RE: Fitzgerald- p. 36,Fig. 1.19) of samarium-cobalt by the linear approximation

Bm = R

Hm H

0

c

= RHm + Br

where Br = RH0

c with R = 1:050; H0

c = 712 kA/m,jBrj = 0:94T. Notethat the apparent coercivity H

0

c adjusted for the slight downward bend in theB H curve is somewhat larger than the actual coercivity.

The other ux densities are

Bg = 0Hg

in the movable gap andB0 = 0H0

in the xed gap.Flux continuity requires

BmWmD = BgWgD = B0W0D


36/127


and by Amperes law (for mmf):

Nfif = Hmd + Hgx + H0g0:

There are six unknowns Bm; Bg; B0; Hm; Hg; H0 and six equations (wherethe ux continuity equation counts as two). In matrix terms the system issparse so it is simplest to reduce to a three-by-three system by combining someequations. I.e.,

R

Hm H

0

c

Wm = 0HgWg

by substitution of the rst and second equations into the rst "half" of thefourth equation. In a similar vein,

0HgWg = 0H0W0

by substitution of the second and third equations into the last "half" of thefourth equation. The last three equations form the three-by-three system whichcan be written in matrix form:24 RWm 0Wg 00 0Wg 0W0

d x g0

3524 HmHgH0

35 =24 RH0cWm0

Nfif

35 :This is solved for the vector of unknowns Hm; Hg; H0 by matrix inversion andHm is substituted into the rst equation to yield:

Bm =R

Nfif H

0

cd

d + Wm R0

xWg

+ g0W0

:

The ux linkage is

f = NfWmDBm = [NfWmD]

24 R

Nfif H0

cd

d + Wm

R0

xWg

+ g0W0

35

where f = 0 when

if = If0 =H

0

cd

Nfif=

Brd

RNf:

Then the coenergy is

W0fld (if = 0; x = x0) = Z0H0

cd

Nfif

[NfWmD]24 R Nfif H0

cd

d + WmR0

xWg

+ g0W0

35 dif=

24 WmD (Brd)22R

hd + Wm

R0

xWg

+ g0W0

i35 ;


37/127


by application of equation 1.75.

b) The force is

ffld =

264 W2mD (Brd)220Wg

hd + Wm

R0

xWg

+ g0W0

i2375 ;

by application of equation 1.45 to W0fld (if = 0; x = x0) ; where the negative signindicates that the force acts to decrease the movable gap by pulling the plungerinward.

c) Evaluation of the above expression for ffld by substitution of knownparameters gives

ffld = 115N at x = 0cm

and

ffld = 85:8 N at x = 0:5 cm:


38/127


1.18.3 Equivalent of Magnet with Linear Material

Claim 28 Consider a section of linear, hard magnetic material:

Bm = R

Hm H

0

c

(1.76)

with area A and length d: With respect to the external magnetic circuit which itfaces, this section can be replaced by:

1. a section of soft, linear magnetic material with the same permeability R(B = RH) and same geometry;

2. and an equivalent winding of

N i = H0

cd ampere-turns. (1.77)

I.e., this will result in same ux in the external magnetic circuit.

Proof. By Amperes law

Hmd + Fe = 0 (1.78)

where Fe represents the mmf developed at the "terminals" of the external mag-netic circuit which the magnet faces.

) Hm =Fe

d

: (1.79)

The ux produced in the external magnetic circuit by the permanent magnet is

= ABm = Ah

R

Hm H

0

c

iHm=

Fed

= RA

H

0

c Fed

:

Now consider the replacement:

= RA

N i

d

Fed

by magnetic-circuit analysis with application of linear superposition. This equa-

tion gives the same result as the previous equation because Nid = H0

c byequation 1.77 for the mmf of the equivalent winding.

Remark 29 Note the analogy between the previous theorem for magnetic cir-cuits and Thevenins theorem for electrical networks.


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See Fig. 19 (RE: Fitzgerald- p. 149, Fig. 3.22 )Magnetic circuit including permanent magnet (neodymium-iron-boron) withmovable plunger is shown.

MMF Data: Excitation winding (N1 = 1500 turns with current i1) is in-cluded.

Core and plunger materials: permeability is assumed innite ( ! 1);neglect eld fringing eects

x:

= plunger displacement (movable gap); g0:

= xed air-gap length; d =magnet length; D = depth of core and plunger/

Dimensions: d = 8mm; g0 = 1mm; D = 3:5 cm.Widths: W = 4:0cm; W1 = 4:5 cm:=) a) Find x directed force on plunger when current i1 = 0:=) b) Find current i1 required to reduce plunger force to zero.

Solution 31 a) Represent the DC magnetization curve (RE: Fitzgerald- p. 36,Fig. 1.19) of neodymium-iron-boron by the linear approximation

B = R

H H

0

c

= RH+ Br

where Br = RH0

c with R = 1:050; H0

c = 940 kA/m,jBrj = 1:25T. ApplyClaim 28 to replace magnet by section of linear material with permeability Rand equivalent winding of

N i = H0

cd =

9:4

105

8

103

= 7520 ampere-turns.

The equivalent magnetic circuit after this replacement is shown in Fig. 19b,where the two mmf sources are in series with the variable gap, xed gap, andmagnet reluctances

Rx = x0W1D

;

R0 =g0

0W D;

and

Rm =d

RW D;

respectively. For i1 = 0 this is equivalent to a single-winding system driven by(N i)equiv where

W0

fld =Li2

2=

(N i)2equiv2 [Rx + R0 + Rm]

:

The force on the plunger is

ffld ="@W0fld

@x

#iequiv=constant

=(N i)2equiv

[Rx + R0 + Rm]2

dRxdx

=(N i)2equiv

(0W1D) [Rx + R0 + Rm]2


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which gives

ffld = 703N

upon substitution of known parameters.

b) This occurs by setting the net mmf to zero:

(N i)equiv + N1i1 = 0

which gives

i1 =(N i)equiv

N1=

7520

1500= 5:01 A:

The direction of the current cannot be determined because the direction ofmagnetization of the magnet is not given. This current must be applied in thedirection which reduces net ux to zero.


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44


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Chapter 2

Rotating Machines

2.1 Introduction

Certain basic principles are common to both AC and DC rotating machines,whereby a physical machine can be reduced to a simple mathematical model.The key principle is electromagnetic induction (Faradays law):

e =d

dt(2.1)

where = N (2.2)

and

:

= ZSB dS: (2.3)Electromechanical energy conversion occurs when mechanical motion producesddt

6= 0. In rotating machines there are three methods to obtain ddt

6= 0 me-chanically:

1. Rotation of windings through a B eld;

2. Rotation of a B eld past windings;

3. Variation of reluctance in the main magnetic ux path by rotor rotation,where rotor has geometry designed for that purpose.

varies periodically by any of these methods due to the cyclic nature ofmechanical motion in a rotating machine.

2.2 General Rotating Machine

The general machine (a generator or a motor) consists of a stator (stationarymember) and a rotor (rotating member) where the stator partially or totally

45


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46 CHAPTER 2. ROTATING MACHINES

encloses the rotor, as shown in Fig. 1. Stator and rotor cores usually consistof magnetic material to concentrate and shape the B elds produced by their

respective windings.An armature winding is a set of interconnected coils through which the bulk

of machine energy passes, whether energy is electrical output (generator) orelectrical input (motor). The eld winding produces the operating ux of themachine and usually carries DC current. It may be replaced by permanentmagnets in some machines.

Torque is produced by the interaction of stator and rotor magnetic eldsand acts to align them. In a generator mechanical torque is applied and theaforementioned torque opposes rotation. In a motor electrical energy (armaturecurrent) is applied and the aforementioned torque aids rotation.

2.2.1 AC Machines

The armature winding is on the stator and the eld winding is on the rotoralmost invariably. This corresponds to method 2 in Section 2.1 to produceddt

6= 0.

Synchronous Machine

The eld winding is excited by DC current and requires rotating electrical con-tacts (slip rings) to feed it.

Induction Machine

The eld winding is shorted and no external source is applied directly to it so

no rotating electrical contacts are required. Rather AC induction excites it bygeneralized transformer action between the stator and rotor, as explained later.

Variable Reluctance Machine

There is no eld winding. A varying B eld is produced by a non-uniformair-gap reluctance with respect to rotor angle. Time-varying armature (stator)currents are applied.

2.2.2 DC Machine

The armature winding is on the rotor and the eld winding is on the stator inmost cases, which is a reversal of positions with respect to an AC machine. This

corresponds to method 1 in Section 2.1 to produced

dt 6= 0. The DC interfacebetween excitation or output and the periodic ("AC") induced armature (speed)voltage is performed by a commutator on the rotor. The commutator switchescurrent in synchronism with rotation; it is the underlying reason for placingthe armature on the rotor. An exception is the so-called brushless DC motorin which the rotor is a permanent magnet. The armature of this machine is


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2.3. STATOR AND ROTOR CORES 47

on the stator and driven by electronic switching circuitry which emulates acommutator.

2.3 Stator and Rotor Cores

Material for these cores is usually magnetically soft and highly permeable elec-trical steel, with windings placed in core slots. This core material increasescoupling between windings and the B eld strength (hence stored magnetic en-ergy) in a machine. It also allows the designer to shape and distribute the Belds. Core losses due to hysteresis and eddy currents must be considered, solaminated construction is used to reduce eddy currents. Eddy currents mayoccur in both stator and rotor of variable reluctance machines. Refer to Figs.4.1, 4.2, 4.3 in textbook.

2.4 Some Machine Type-Names

1. DC

2. Synchronous

3. Permanent magnet (PM)

4. Induction

5. Variable Reluctance

6. Hysteresis

7. Brushless

2.5 Similarity of Physical Principles

The following machines are apparently quite dierent but share a common prin-ciple with regard to the interaction between stator and rotor elds.

2.5.1 DC Machine

Both the stator and rotor B elds are xed in space, hence at a xed angu-lar displacement with respect to each other. The latter eld is xed due tocommutator action. Torque is produced by the tendency of these elds to align.

2.5.2 AC Induction Machine

Both the stator and rotor B elds rotate together in space and remain at a xedangular displacement with respect to each other. Actually the stator eld canbe analyzed as a superposition of forward and backward rotating waves, where


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the backward wave is zero in a 3-phase machine but the same as the forwardwave in a single-phase machine. Torque is produced by the tendency of these

elds to align, just as in a DC motor.

2.6 Dierentiation Rule for Integrals (Leibnitz)

Leibnitz rule is stated here for reference because it will be used to apply Fara-days law to ux integrals. Given

F(z) =

Zb(z)a(z)

f(x; z) dx (2.4)

where both the integrand f(x; z) and the limits a (z) and b (z) may be functionsof a parameter z. Also, in general the integrand f(x; z) may contain another

integral. We have

dF(z)

dz=

db (z)

dz

f(b (z) ; z) dx

da (z)

dz

f(a (z) ; z) +

Zb(z)a(z)

@f (x; z)

@zdx

(2.5)which is essentially a superposition of eects due to changes in the upper limit,lower limit, and integrand with respect to changes in a parameter z: This isrestated for evaluation at a specic value z = z0 to show that the derivativeoperator can be moved to precede the integral in the last term:

dF(z)

dz

z=z0

=

db (z)

dz

z=z0

f(b (z0) ; z0) dx

da (z)

dz

z=z0

f(a (z0) ; z0)+@

@z

Zb(z0)a(z0)

f(x; z) dx:

(2.6)

2.7 Gap Field Distribution in Rotating Machines

Clearly a gap between the rotor and the stator is necessary to permit rotorrotation. The B eld produced by the eld winding continues through thegap and into armature slots where it is accessible to the armature winding.Coincidentally a prescribed distribution of the B eld in the gap, with respectto the axis of the eld winding, is necessary to achieve the desired output voltagewaveform in a generator or desired torque characteristic in a motor. It turns outthat nearly all magnetic eld energy is concentrated in the gap volume becausethe H eld is very small in highly permeable cores. For a given energy in the

gap volume, the distributed nature of the B eld reduces the tendency towardmagnetic saturation at points in the cores where B attains its extreme values.In other words, the magnetization is spread to attain a more or less uniformdegree of magnetic-dipole alignment throughout the core volume. Analyticallythis magnetization can be represented by equivalent surface currents on thecores, as previously described.


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2.8. MULTIPOLE FIELD DISTRIBUTION 49

The B eld is normal to the rotor and stator surfaces due to boundaryconditions on the magnetic core materials. Hence B is directed radially in the

case of a uniform circular gap. For analytic purposes we assume that the Beld is essentially constant across the gap from rotor to stator if the gap issmall relative to the radius of the rotor. Further we assume that this B eldmay vary with angle (in cylindrical coordinates) around the circumference ofthe gap but does not vary with linear (axial) displacement along the rotor orstator. These principles apply to a B eld irrespective of whether it is producedby current in a eld winding or an armature winding.

2.8 Multipole Field Distribution

Consider the radially directed spatial B eld around the circumference of thegap due to an instantaneous current in either a eld winding or an armaturewinding. At any position on the circumference specied by angle ; the eld B isspecied by its magnitude and sense (i.e., radially directed outward or inward).Note that around the circumference B undergoes more or less complete reversalsin direction because a winding sets up circulating currents. The number of poles(Npoles) in B is by denition the number of maxima which occur in jBj duringa complete (360 ) revolution on the circumference of the gap. Npoles is alwaysan even number because no magnetic monopoles exist.

Npoles is determined only by the winding pattern if the gap is uniform (e.g.,round rotor and round stator with distributed winding). N

poleswill be depen-

dent primarily on the design of the pole faces if the gap is non-uniform (e.g.,salient-pole rotor or stator with concentrated winding). Refer to Figs. 4.4, 4.5,4.6, and 4.7 for illustrations of two-pole and four-pole machines including spatialdistributions of their B elds. In a machine with a large number of pole-pairseach pair "spans" a small arc and contributes a small or narrow segment of theB eld in a prescribed manner, by design of the winding distribution or poleface geometry. Generally one-half subcycle of the B eld occurs in each of thesenarrow segments.

In most multipole machines, by design the magnetic elds of the stator androtor will have the same number of poles to achieve optimal performance. Hencethe machine output, whether it be voltage in a generator or torque in a motor,can be taken as the aggregate of the outputs with respect to all the magnetic

poles.

An analogy can be drawn with bar magnets and horseshoe magnets. Asimple bar magnet has two poles hence represents a two-pole winding. Barmagnets can be curved into horseshoe magnets to form a four-pole or eight-polemachine as shown in Fig. 3.


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2.9 Induced Voltage by Application of Faradays

LawRecall Faradays law: I

C

E ds = d

dt

ZS

B da = d

dt(2.7)

where S may be chosen as any surface for which the curve C is a boundary andthe total ux is

:

=

ZS

B da: (2.8)

Hence the induced voltage in path C is

e =IC

E ds = d

dt

and the ux linkage to an Nturn winding on this path is

:

= N (2.9)

so the induced voltage in the winding is

e = Nd

dt=

d

dt: (2.10)

We apply equation 2.10 to a rotor (armature) winding consisting of a singlerectangular loop (N = 1) turning through a B eld produced in the gap by thestator (eld) winding, as shown in Fig. 2. This corresponds to method 1 inSection 2.1 to produce d

dt6= 0. Here the "curve" C is taken as the rectangular

loop. The surface S is taken as the cylindrical half-shell whose base is the planarsection enclosed by the rectangular loop and whose cyclindrically curved surfacecoincides with the circular path of the loop conductors around the gap. Thecurve C and this surface S are shown in Fig. 2. This choice of S expeditesevaluation of the surface integral in equation 2.8 because the gap B eld iseverywhere normal to this surface. (By conservation of ux, the same resultwould be obtained for the surface integral by choosing S as the aforementionedplanar section but its evaluation would be complex.)

The plane of the rectangular loop is oriented at an arbitrary angle with

respect to the vertical as shown in Fig. 2. Thus its upper and lower wires boundthe cylindrical half-shell at angles and + ; respectively. This gives

:

=

ZS

B da =

+Z

B

0; t

ld0; (2.11)


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2.10. OPTIMAL GAP FIELD 51

where l is the length of the loop, for the integral over the surface of the cylindricalhalf-shell. Application of Leibnitz rule from equation 2.6, where we now take

z = t; yields

e (t0) =

d

dt

t=t0

= l

0BBBBBB@d

dt

(t0)+Z(t0)

B

0; t

d0

| {z } transformer voltage term

+ [B ( (t0) + ) B ( (t0))]

d

dt

t=t0 | {z }

speed voltage term

1CCCCCCA :

The speed voltage can be expressed as

espeed (t0) = l [B ( (t0) + ) B ( (t0))] ! (t0) (2.12)

= l

r v [B ( (t0) + ) B ( (t0))] | {z } Lorentz force densitywhere !

:= d

dt= v

r: Note that the Lorentz force density term appears in equation

2.12 because this produces the E elds down the length of the wires. By designof machine windings the B eld is normally anti-symmetric around the gap:

B ( (t0) + ) = B ( (t0)) ; (2.13)

soespeed (t0) = l [2B ( (t0))] ! (t0) (2.14)

and nally if ! is constant

espeed (t) = !l [2B (!t)] : (2.15)

Thus we conclude that the waveform espeed (t) is simply a scaled version ofB () where = !t; for a generator this can be taken as the output voltage:This coincides with the classic "cutting of ux" interpretation; see Example 4.4in textbook.

Analogously we can apply equation 2.10 to a stator (armature) windingconsisting of a single rectangular loop (N = 1) where a rotating B eld isproduced in the gap by the rotor (eld) winding. This corresponds to method2 in Section 2.1 to produce d

dt6= 0. Now the speed voltage espeed (t) induced

in the stator can be determined simply by taking the rotating rotor as thespatial frame of reference. Hence the eld appears stationary in time and canbe denoted by B (), with relative to the rotor axis. Now the stator windingappears to be rotating with respect to the rotor eld and equation 2.15 appliesdirectly.

2.10 Optimal Gap Field

Optimal operation of a rotating machine requires that the (Fourier) spectralcontent of the induced speed voltage waveform espeed (t) matches the spectral


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content of the required armature (generator) output or applied armature (mo-tor) input. In an AC machine this specializes to the simple requirement that

espeed (t) is a pure sinusoid; i.e., its spectrum contains no harmonics. In a DCmachine this must be interpreted by taking the intervening commutator intoaccount, so the armature itself actually sees or should provide a square wave.By equation 2.15, the gap eld waveform B () has the same spectral contentas that required in espeed (t). So B () should be a pure (fundamental) sinusoidin an AC machine or a square wave (i.e., piecewise constant) in a DC machine.

The importance of the spectral content in espeed (t) hence B () is explainedas follows. Assume that the machine is in normal steady-state operation, ro-tating at a constant angular velocity !m: Consider the electrical circuit or loopformed by the armature winding and a voltage source of the required spectralcontent connected directly to the armature terminals. By Kirchos voltagelaw (KVL), any mismatch in spectral content between espeed (t) and the voltagesource yields a voltage dierence which contains additional harmonic compo-nents with respect to !m. These harmonic components in the loop voltagewill drive corresponding current components around the loop. These currentcomponents are undesirable because they increase the RMS current value henceincrease resistive (i2R) losses and increase core eddy current losses (which areproportional to the square of frequency) but do not contribute to average electro-mechanical power transfer. The last statement follows by application of the or-thogonality principle to harmonics with respect to the fundamental component.In addition harmonics tend to introduce small periodic variations in !m:

2.11 AC Machines

2.11.1 Electrical and Mechanical FrequencyThe spatial variation in the gap B eld will be more rapid in machines with agreater number of poles:

ae =

Npoles

2

a (2.16)

where ae is the electrical angle or argument of the cyclic eld B (ae) and ais the mechanical or spatial angle around the gap. It follows that

!e =

Npoles

2

!m; (2.17)

hence

fe = Npoles2 fm; (2.18)by dierentiation of equation 2.16 with respect to time. From this last equationit follows that

fe =

Npoles

2

RP M

60

Hz; (2.19)


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2.11. AC MACHINES 53

where the term RPM60 converts mechanical revolutions per minute to fm whichis in revolutions per second. Note that fe is the electrical frequency of the speed

voltage and fm is the mechanical frequency of rotation in a given machine.

2.11.2 Synchronous Machines

Synchronous machine operation is best explained by separate consideration ofmotors and generators. An AC motor is operating synchronously if fe, theelectrical frequency of the speed voltage which it produces, equals the electricalfrequency of the voltage or current applied to its armature winding. Then fm;the mechanical frequency of rotation, is constrained or locked to the appliedfrequency by equation 2.18. Conversely an AC generator is operating synchro-nously iffm is such that fe given by equation 2.18 matches the frequency of the(possibly "innite") bus to which it is connected. Note that the inherent "fre-quency conversion" given by equation 2.18 always occurs in multipole machines(both generators and motors).

Example 1 Given a 2-pole single-phase machine. What RPM is required toobtain fe = 60 Hz ?

Solution 2 From the above equation

RP M =

2

Npoles

(60fe) =

2

2

(60 60) = 3600:

Note that the spatial B eld of this machine completes one full cycle per me-chanical revolution.

Example 3 Given a 4-pole single-phase machine. Express electrical frequencyin terms of mechanical frequency.

Solution 4 Npoles = 4; so by equation

fe =

Npoles

2

fm = fe =

4

2

fm = 2fm

2.11.3 Induction Machines

In induction machines the armature (stator) winding is the same as in synchro-nous machines. On the other hand, the eld (rotor) winding is short-circuitedso that no external excitation need be applied to it.

AC rotor current follows by induction (generalized transformer action) fromstator to rotor. E.g., consider a two-pole machine (Npoles = 2) where !m < !eso that the rotor rotation continuously slips behind rotation of the stator Beld. Hence there will be a continuosly induced rotor current of frequency

!s = !e !m (2.20)


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due to relative motion between the rotor winding and the rotating armatureeld. Note that a frequency change and electromechanical energy conversion

occur here.Induction motors are the most common and widely applied of all motors.

On the other hand induction generators are rarely found except recently in windpower applications for which they are somewhat suitable.

The squirrel-cage rotor or reasonable facsimile is a common method of con-struction for induction motors. Here the "windings" are solid aluminum barscast in slots and shorted together by common cast aluminum end rings. Conse-quently induction motors are durable, reliable, and relatively inexpensive, whichare contributing factors to their immense popularity.

Refer to Fig. 4.15 for a torque-speed curve of an induction motor. Note thatthe induction motor must operate at less sychronous speed to develop torque.Otherwise there is no slip, hence no induced rotor current.

2.11.4 Three-Phase Machines

Phase refers to the electrical phase or delay between AC currents in multiplearmature windings. A 3-phase system can be either a synchronous machineor an induction machine where there are three sets of armature windings perpole-pair; refer to Fig. 4.12 in textbook. These sets are displaced by 120 electri-cal degrees in the spatial pattern around the rotor or stator. Hence in generalthe mechanical displacement angle between sets is (2=3) = (Npoles=2) : E.g., ina two-pole machine, these sets are symmetrically displaced by 120 mechani-cal degrees. E.g., in a four-pole machine the two coils for each phase can beconnected in series (or in parallel because coil voltages are identical); see Fig.4.12b in textbook. Likewise the currents supplied to or voltages generated from

3-phase windings are symmetrically displaced by 120 electrical degrees.In a three-phase system the electrical frequency is the same as in a single-phase system. I.e., electrical frequency is determined solely by the number ofpoles in a machine. The eect of a balanced three-phase system is to produce arotating ux wave which has only a forward component, which will be shown byphasor addition of rotating waves. Hence three-phase motors possess constanttorque vs. time; i.e., there is no uctuating double-frequency torque component.Most of the worlds power systems (hence most synchronous generators) are 3-phase. The attendant lack of uctuating "reaction torque" makes it possible tobuild and operate extremely large 3-phase generators which would otherwise bedamaged by such uctuating torques.

2.11.5 Design Considerations

An AC machine (generator or motor) requires a large number of poles (Npoles)by equation 2.19 if it operates at a relatively low mechanical RPM . (Electricalpower systems operate at fe = 60Hz or fe = 50Hz in North America or Eu-rope, respectively.) Salient-pole rotors are most suitable in this case from thestandpoint of mechanical construction and economy. Rotor windings must be


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2.12. DC MACHINES 55

concentrated so the B eld will be determined only by the rotor (gap) geometryand not by a winding distribution. E.g., consider a large hydroelectric generator

where the prime mover is water power; Fig. 4.1 shows such a generator wheremechanical rotation is at 37 RPM. E.g., in a salient-pole motor more space isavailable to accomodate rotor windings.

An AC machine (generator or motor) requires a small number of poles(Npoles) by equation 2.19 if it operates at a relatively high mechanical RPM.Round rotors (hence a uniform gap) are a necessity in this case because salientpoles would incur unacceptably high stress and windage loss. Rotor windingsare distributed so the B eld will be determined only by the distribution andnot by the gap. E.g., consider a large generator where the prime mover is agas or steam turbine operating at 1800 or 3600 RPM; Fig. 4.11 shows such agenerator. In this case Npoles = 2 or Npoles = 1, respectively.

2.12 DC Machines

A DC machine may be viewed as a "commutated" version of a synchronous ACmachine. I.e., all rotating machines execute cyclic mechanical motion whichcreates a cyclic (AC) induced speed voltage in the windings to support electro-mechanical energy conversion. Such speed voltage has zero average (DC) valueso a form of rectication is required to achieve a non-zero DC value. In a con-ventional DC machine this rectication is attained by placing a commutator andhence the armature on the rotor shaft while the (DC-excited) eld winding is

on the stator. The commutator mechanically "switches" the connection of thearmature coils to external terminals in synchronism with machine rotation, bymeans of copper segments and carbon brushes. Fig. 4.17 shows an elementaryDC machine with a commutator.

The instantaneous induced speed voltage in a single turn of the rotor windingis proportional to the gap B eld at the turns particular angular position ,as shown previously. All turns between two diametrically opposed commutatorsegments are eectively in series, so it is possible to deduce the output windingvoltage as a function of time from the B eld distribution. It turns out that theideal stator B eld is piecewise constant (a square wave) to maximize the DCcomponent of the commutated (rectied) speed voltage.

The spatial orientation between the rotor and stator elds is nearly constant

during machine operation. I.e., the stator eld is inherently xed whereas therotor eld is stationary with respect to the active positive and negative com-mutator segments. This allows the rotor eld to revolve in space by only thearc subtended by the active commutator segments, which is small if segmentsare numerous. Typically by design (brush positions, etc.) the stator and rotorelds are perpendicular to each other to maximize torque.


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2.13 Design of Gap Field

2.13.1 Magnetic Circuit AnalysisAssume that a certain gap eld B () is required in the design of a given machine.The eld B () varies periodically with due to continuous rotation around

the gap, where the spatial electrical frequency isNpoles

2 electrical cycles permechanical revolution. It will be shown below by magnetic-circuit analysis thattwo options are available to design or "shape" B () :

1. Variation of the gap length g () versus by pole-face geometry;

2. Winding distribution versus to p

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