Electromechanical Dynamics - Solutions Manual

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Solutions Manual for Electromechanical Dynamics

For any use or distribution of this solutions manual, please cite as follows:

Woodson, Herbert H., James R. Melcher. Solutions Manual for Electromechanical

Dynamics. vols. 1 and 2. (Massachusetts Institute of Technology: MIT

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LUMPED ELECTROMECHANICAL ELEENTS

PROBLEM 2.1

We start with Maxwell"s equations for a magnetic system in integral form:

•di = I Jda

B-da = 0

Using either path 1 or 2 shown in the figure with the first Maxwell

equation we find that

J*da = ni

To compute the line integral of H we first note that whenever p-*we

must have HRO if E=pH is to remain finite. Thus we will only need to know

H in the three gaps (H1,H2 and H3) where the fields are assumed uniform

because of the shortness of the gaps. Then

fH*di = H(c-b-y) + H3 x = ni

path 1

C

r.path 2

LUMPED ELECTROMECHANICAL ELEMENTS

Using the second Maxwell equation we write that the flux of B into the

movable slab equals the flux of B out of the movable slab

U H1 LD = H2 aD + UoH3bD

or

H1 L = H2a + H3 b (c)

Note that in determining the relative strengths of H1,H2 and H3 in this last

equation we have let (a-x) = a, (b-y) = b to simplify the solution. This means

that we are assuming that

x/a << 1, y/b << 1 (d)

Solving for H1 using (a), (b), and (c)

ni(y/a + x/b)

HI = (c-b-y)(y/a + x/b) + L(y/a * x/b)

The flux of B through the n turns of the coil is then

(x,y,i) = nB1LD = npoH LD

2j n (y/a + x/b)LD i

(c-b-y)(y/a+x/b) + L(y/a'x/b)

Because we have assumed that the air gaps are short compared to their

cross-sectional dimensions we must have

(c-b-y) << 1, y/a << 1 and x/b << 1L

in addition to the constraints of (d) for our expression for X to be valid.

If we assume that a>L>c>b>(c-b) as shown in the diagram, these conditions

become

x << b

y << b



LURIPED ELECTROMECHANICAL ELEMENTS

PROBLEM 2.2

Because the charge is linearly related to the applied voltages we

know that q1(v1 ,V2,e) = q 1(v 1 ,O,e) + q 1 (O,v 2 ,8)

EV V

q1(V1 ,O,) = - w + ( +

EVV2

q1 (O,V2 ,8) = Ra- w

Hence

S (n/4+6)R

1 ,2, ) = v1 (+o -2

E ()+/4-O)R

2(V1,V2,) = -Vl •~+ 2 g

PROBLEM 2.3

The device has

cylindrical symmetry

so that we assume that

the fields in the gaps

are essentially radial

and denoted as shown

in the fi-ure

Ampere's law can be

integrated around each of the current loops to obtain the relations

3




PROBLEM 2.3 (Continued)

gH1 + gHm = Nil (a)

gH2 - gHm = Ni 2 (b)

In addition, the net flux into the plunger must be zero, and so

0o(d-x)2nrH1 - 2d(2rrr)0oH m - (d+x)(2Tr)o H2 (c)

These three equations can be solved for any one of the intensities. In

particular we are interested in HI and H2, because the terminal fluxes can

be written simply in terms of these quantities. For example, the flux linking

the (1) winding is N times the flux through the air gap to the left

,1 j 0oN(d-x)(21Tr)H 1 (d)

Similarly, to the right,

X2 = poN(d+x) (27)H 2 (e)

Now, if we use the values of H1 and H2 found from (a) (c), we obtain the

terminal relations of Prob. 2.3 with

jo ' rN2d

L =o 2g

PROBLEM 2.4



LIUPED ELECTROMECHANICAL ELEMENTS


Part a

dx2

Sf = Ma =M 2i i dt

2

dxt fdx

fDAMPER -B t ; coul d Mdt

dx 1

dt

2-dx•x dx

M - f(t)-B - + f

dt2 dt coul

2dx dX1

Mdx2• +B =- f(t) - UdMg dt2dt

dxI

Id--Idt

Part b

dx

First we recognize that the block will move so that > 0, hencedt

dx

coul = - g;- > 0

Then fo r t > 0

Md2x dx

B =-l dMgdt 2

which has a solution

1dMg

-(B/M) tx(t) = - - t + c1 +c 2 e

Equating sintWlarities at t 0

2 2 I

M x(t)0) = Iod (0)

= - io(t)dt dt

S=dx = d2x = 0

Then since x(0 ) = -(0 ) -(t2

dt





Idx (0+ ; x(0+ 0

Hence x(t) u-1( t + Ud 2) (1-e-(B/M) t

dxActually, this solution will only hold until to , where dx(t o) O0,at which

point the mass will stop.

Jx

10o-

i~.

t

PROBLEM 2.5

Part a

Equation of motion

M2

+ Bdt

= f(t)

dtdt

(1) f(t) = IoUo(t)

I (B/M) t)

x(t) = u (t) (1-e(1B

6

as shown in Prob. 2.4 with Vd = 0.

(2) f(t) = F° u l(t)

Integrating the answer in (1)

x(t) =F

[t +M

(e-(B/M)t

-1)]ul(t)

Part b

Consider the node connecting the

damper and the spring; there must be no

net force on this node or it will

suffer infinite acceleration.

dx-B • + K(y-x) = 0

or

dxB/K - + x = y(t)

dt

1. Let y(t) = Auo(t)

X) x'XBdx

K dt- t>Ox=O

x(t) = C e -K/Bt t > 0

But at t = 0

B dx

K J-t(O) Ao

dxNow since x(t) and -(t) are zero fo r t < 0

dt

+ AKx(O ) AK C

-x(t) = Ul(t) e (K

/B)t all t

2. Let y(t) = Au (t)

7



LUMPED' ELECTROMECHANICAL ELEMENTS


Integrating the answer in (1)

-(K/B)tx(t) = ul(t) Yo(1-e all t

PROBLEM 2.6

Part a

k]. (.

dx

fl = B3 d ; f2= K3 (x 2 -x 3-t-Lo)

d

f3= K2 (x 1 -x 2 -t-Lo); f 4 = B2 ~(x 1 -x 2 )

f5 =Kl(h-x

1-Lo)

Part b

Summing forces at the nodes and using Newton's law

Kl(h-x1-Lo)

= K2(X1-X2-t-Lo) + B2ddt

(X1-X2)

2

+ M d xl

1 2dt

K2 (x 1-x 2 -t-Lo) + B2

d

dt

(x1-~2)

d2x2

= K3 (x 2 -x 3 -t-L o ) + M2 2dt

dx3 d2x 3

K3 (x 2-x 3 -t-L ) = f(t) + B 3 - + M 2 dt





Let's solve these equations for the special case

M1 =M2 = M3 = B2 = B3 = Lo = 0

Now nothing is left except three springs pulled by force f(t). The three

equations are now

Kl(h-x 1) = K2 (x1-x 2 ) (a)

K2 (x1-x2)= K3 (x2-x3 ) (b)

K3 (x2-x3)= f(t) (c)

We write the equation of geometric constraint

x3 + (x2-x3) + (x1-X2) + (h-x1 )-h = 0

or (h-x) = (x2-x3) + (x1-x2) + (h-x) (d)

which is really a useful identity rather than a new independent equation.

Substituting in (a) and (b) into (d)

K3 (x2-x3) K3 (x2-x3) K3 (x2-x3)

(h-x 3 ) + 2 +

3 K2 K1

= 3 (x 2 -x 3 ) 3+ 2 +1

which can be plugged into (c)

K11(K+ 2

+(h-x )=f(t)

K K K 3

which tells us that three springs in series act like a spring with

-1

K' = (- + + 1 -)K K K3 2 1



b I


73 zPROBLEM 2.7

B, 1i

dxl

f = B x f= K x

1 ldt 2 1 1

d(x2-x I )

f3 B2 dt f4 = K2 (x 2 -x 1)

Node equations:

dx1 d(x 2 -x 1 )

Node 1 B1 dt + K11 2 dt + K2 (x2-xl)

2(x dNodeNode 2 B2 2dt

x)+ K2 (x 2 -x 1) = f

To find natural frequencies let f = 0

dx l st

B + K X, = 0 Let x = e1 dt 11 1

Bl1 + K, = 0 s1 =- K1/B1

d(x 2 -x 1 ) st

B2 dt + K2 (x 2 -x 1 ) 0 Let (x 2 -x 1 ) e

= B2s + K2 0 s2 - K2/B

The general solution when f = 0 is then

-(K 1 /BI) t

e1 c 1

-(K 1 /Bl)t -(K2/B2)tele2 o (x 2 -x 1 ) + x 1 c + c2 e




PROBLEM 2.8

r ae

LtVDLJ

From the diagram, the change in ir in the time At is 1iA. Hence

di =r A8 = d8im ii A6 ddt O At 8 dt

(a)

At-*O

Similarly,

di - A dO-= lim - - =. i (b)

dt At0 r At rdt

Then, the product rule of differentiation on v gives

-di 2 didv r dr dr O6 dO d dOdv r dr + 1 - + (r -) + i (r -) (c)dt dt dt r dt

2dt dt dt d

and the required acceleration follows by combining these equations.



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LUMPED-PARAMETER ELECTROMECHANICS

PROBLEM 3.1

This problem is a simple extension of that considered in Sec. 3.2, having

the purpose of emphasizing how the geometric dependence of the electrical force

depends intimately on the electrical constraints.

Part a

The system is electrically linear. Hence, W' i Li and the force fm 2_

that must be applied to the plunger is

f _fe 1Lio

2a (+ x2a

The terminal equation can be used to write this force in terms of X

f = -fe = X2/2aL

Part b

With the current constant, the force decreases rapidly as a function of

the plunger gap spacing x, as shown by (a) and the sketch below

Z= cor1;50_'M

x

With the current constant, the drop in IH'dR! across the gap increases with x,

and hence the field in the gap is reduced by increasing x.

Part c

By contrast with part b, at constant X, the force is independent of x

ftm





With this constraint, the field in the gap must remain constant, independent

of theposition x.

PROBLEM 3.2

Part a

The terminal relations are

= S11 + S12 2

(a)

2 = S2 1q1 + S22q 2

Energy input can result only through

the electrical terminal pairs, because

the mechanical terminal pairs are

constrained to constant position. Thus,

8,

We v= dq l+ v 2 dq 2 (b )

First carry out this line integral along the contour A: from a-b, ql=

0, while

from b-c, dq2 = 0. Hence,

We = 2(0q 2 )dq2 + 1v 1(,Q 2 )dql (c)o o

and using (a),

We =f 22 2dq2 + fJ (S1 1q1 + S 1 2Q2 )dql (d)

0 0

and for path A,

1 2 2

e 2 22 2 + 1 2 1 2 + 11 (e)

If instead of path A, we use C, the roles of ql and q2 are simply reversed.

Mathematically this means 1+2 and 2+1 in the above. Hence, for path C

/ 1 2 2

e 2 SllO + S212 1 + 222 f)

(To use path B in carrying out the integration of (b), we relate q2 and ql

13





Q2

q2 = 1 q1

Then, (a) becomes,

12 2 S22 2v1 1 [Sl11 + '1 1 ]q11; v22 = [S221

+Q

q 1q1I

and, from (b), where dq2 and Q2dql/Q1

e1I[ + S12Q2 1_ Q dqS11 Q dq + [S21 S 2 2Q2 1 Q2e-o o 1 1

1 2 1 1 1 2e = 2 11 1 2 I2Q21I + 21 S221 1Q2 + 2 2Q2

Part b

The integrations along paths A, B and C are the same only if S21 = S12

as can be seen by comparing (e), (f) and (j).

Part c

Conservation of energy requires

aW aW

dW(qlq 2) = vldql + v2 dq 2d q l

qd q 2

Since ql and q2 are independent variables

e e

1 =ql v2 =q

Taking cross derivatives of these two expressions and combining gives

av8v1 av2v2

3q2 3q 1

or, from (a), S12 = S2 1 '

PROBLEM 3.3

The electric field intensity between the plates is

E = v/a



LIUMPED-PARAMETER ELECTROMECHANICS


Hence, the surface charge adjacent to the free space region on the upper

plate is

of = E v/a (b)

while that next to the nonlinear dielectric slab is

3V V

Of=a -2 + 0 Y (c)afffi a + •o a

It follows that the total charge on the upper plate is

dxE v 3 E Vo av o

q a + d(.t-x)[--3 + a ] (d)

a

The electric co-energy is

d£ v 22

W o + d(£-x)cv4

(e) qdv = v d(-x)ave 2a 4a

3

Then, the force of electrical origin is

aw' 4e e dav

f = f (f)ax 4a3

PROBLEM 3.4

Part a

The magnetic field intensity in the gap must first be related to the

excitation current. From Ampere's law,

Ni = dHd + xH (a)

where the fields Hd and Hx are directed counterclockwise around the magnetic

circuit when they are positive. These fields are further related because

the magnetic flux into the movable member must equal that ou t of it

lowbHd I lowaHx (b)

From these two expressions

= daH Ni/(i-- + x)x b (c)





The flux linked by the electrical terminals is X NU awH which in view of

(c) is N2aA = Li; L = (d)da- + x)

Part b

The system is electrically linear. Hence, Wm X2/L (See Sec. 3.1.2b)

and from (d), daS2 (--

+ x)w ( (e)m 2 N2a

Np aw

Part c

From conservation of energy fe = -~W /3x, W =Wm,x). Hence,

2 2

Part d

In view of (d) the current node equation can be written as (remember

that the terminal voltage is dX/dt)

-da +i( x)

I(t) = d + b (g)R dt N210aw

Part e

The inertial force due to the mass M must he equal to two other forces,

one due to gravity and the other fe. Hence,

2 X2dx 1

Md 2

= Mg 2 (h)

dt N2oaw

(g) and (h) are the required equations of motion, where (X,x) are the

dependent variables.




PROBLEM 3.5

Part a

From Ampere's Law

H1 (a+x) + H2 (a-x) = N1il + N212

Because fBnda = 0

S

loH1A1 oH2A2

solving for H 1

NlH + N2 2

1 A A1 1

a(l + -) + x(1 -A A2 2

Now the flux $ in each air gap must be the same because

$ = poH1 A1 jiH2 A2

=and the flux linkages are determined to be X i N1P and X2 = NO. Using

these ideas

AX = N2L(x)il + N1 N 2L(x)i2

X2 = N2N L(x)i1 + N 2L(x)i2

BoA1=where L(x)

A Aa(l +

A 2 A 2)

Part b

From part a the system is electrically linear, hence

L(x)[N2 + NIN2ii + 1 2 2

W'm = 2lil 1 2 2*N2 1 2 ]

pAwhere L(x) =

A Aa(l + ) + x(l

A2 A2



- -


PROBLEM 3.6

Part a

Conservation of energy requires that

dW = idX - fedx (a)

In addition,

aw aWdW = - dX + -2 dx (b)

ax ax

so that

w e = wi ; feW W (c)

Now if we take cross-derivatives of these last relations and combine,

ai afeS- e (d)

This condition of reciprocity between the electrical and mechanical terminal

pairs must be satisfied if the system is to be conservative. For the given

terminal relations,

ai IoX + ( )

2(e)

o 0

afe o ]/(1 + xf)2

aX a 3 a0 x

and the system is conservative.

Part b

The stored energy is

I 2 41 10 1 4

W = id = [ + i (g)_a x 2A 4 3a O




IL ,

PROBLEM 3.7

To find the co-energy from the

electrical terminal relations alone,

we must assume that in the absence

of electrical excitations there is no

#oorce of electrical origin. Then, the L ,•6JO

system can be assembled mechanically,

with the currents constrained to zero,

and there will be no contribution of " r

0 g ti L--dCco-energy in the process (see

Sec. 3.1.1). The co-energy input through the electrical terminal pairs with the

mechanical system held fixed is

Wm lldil + 2Ai2

For the path shown in the (il,i 2 ) plane of the figure, this becomes

W' = i 2 2 (O,i)di +lil (i',i2)di

o o

amd in view of the given terminal relations, the required co-energy is

c 4 a 4W' -xi +bxxii +-xim 4 x2i2 + b1X2i2il1 4 Xll1

PROBLEM 3.8

Steps (a) and (b) establish the flux in the rotor winding.

X2=IL2 om

With the current constrained on the stator coil, as in step (c), the current

ii is known, and since the flux X2 is also known, we can use the second

terminal equations to solve for the current in the rotor winding as a function

of the angular position

L

2 = 2 [IO - I( t )cos8]

2 L2 0

This is the electrical equation of motion for the system. To complete the

picture, the torque equation must be found. From the terminal relations,

the co-energy is





+ 12 12W' m ldil 12di2 = 1i2 + ili2Lm cose + i2L (c)

mJf1

1 2 2 2 11 12m222

and hence, the electrical torque is

awl

T ae= = -ili2Lm sine (d)

Now, we use this expression in the torque equation, with 12 given by (b)

and ii = I(t)

Jd26 IL2

d = - m (I I(t)cose)sine (e)-dt

2 L2 o

This is the required equation of motion. Note that we did not substitute

12 from (b) into the co-energy expression and then take the derivative with

respect to 0. This gives the wrong answer because we have assumed in using

the basic energy method to find the torque that il, 2 and e are thermodynamically

independent variables.

PROBLEM 3.9

Part a

From the terminal relations, the electrical co-energy is (Table 3.1.1)

rm =

Jdil + X2di2 (a) L~. 1

or

Wm = ax2 i+bx2xlili2

1 2 4+ Cx2i 2 (b)

Part

It follows that the required forces are

awle ml 4 2

f ,m -e = Iaxi + bx2 1i2 (c)

awl1m1f2 e = = 2bx xii + x2

42 (d)

f2 Dx 2 2112 2 2





Part c

There are four equations of motion in the dependent variables il,i 2,x1 and

x2: two of these are the electrical voltage equations, which in view of the

terminal equations for the A's, are

d 2 3 2

ilR 1 = -dt(ax1il + bX2X1i2) (e)

d 2 23

v 2 (t)-i 2 R2 dt(bx 2li + cx 2i2 ) (f)

and two are the mechanical force equations

1 4 20 =- axlil + bX2 1 2 - Kx (g)

1 4 dx20 = 2bx 2xlili2 + x22 dt (h)

PROBLEM 3.10

Part a

Because the terminal relations are expressed as functions of the current

and x, it is most appropriate to use the co-energy to find the force. Hence,

W'm •dil + X di (a)

which becomes,

1 2 .1 2(b) 2

W' = - L=i + Aim 2 0 1 2 2 o 2

b)

From this it follows that the force is,

Se 1 22

2 2 (C)

The currents ii and 12 and x will be used as the dependent variables.

Then, the voltage equations for the two electrical circuits can be written,

using thý electrical terminal equations, as

d 2el(t) iR 1 + d(Loil + Aili2x) (d)

e2'(t) i2R + d(Ai 1i2x + Loi 2) (e)

21





The equation fo r mechanical equilibrium of the mass M is the third equation

of motion2 1 22

M d2dt o 2

Ai i21 2

PROBLEM 3.11

Part a

The electrical torques are simply found by taking the appropriate

derivatives of the co-energy (see Table 3.1.1)

T= m = -M sin8cos i11 2 (a)

aw l

T2 =. = -M cosOsin* i1 2 (b)

Part b

The only torques acting on the rotors are due to the fields. In view

of the above expressions the mechanical equations of motion, written using

8,p, 11 and 12 as dependent variables, are

J - = -M sinecos$p11 2 (c)

dt2

J2 = -M cosOsin*• i (d)

dt

Remember that the terminal voltages are the time rates of change of the res

pective fluxes. Hence, we can make use of the terminal equations to write

the current node equations for each of the circuits as

l(t) = Cdt

(Llil + Mi 2 coscoscos) + iI (e)

I2(t) = G (Mi1cosOcos + L2 i) + 12

Thus, we have four equations, two mechanical and two electrical, which involve

the dependent variables 8,P, ii and 12 and the known driving functions

I1 and 12.



LUMPED PARAMETER-ELECTROMECHANICS

PROBLEM 3.12

We can approach this problem in two ways. First from conservation of

energy,

dW = Aldil + 2di2 + Xdim 1 1 2 2 3 3

aw' aW' aW'

m+ di2 + m dim ai

11

2 3

Hence, aw'm

aW'm2

aw'Mi

=atail 2 2S3 3

Taking combinations of cross-derivatives, this gives

aAI ax a32 3i1 2' •i •

a12 1Di 3 3i 2

1I

33

L12 L2 1 ; L2 3 = L3 2 ; L3 1 = L13

Another way to show the same thing is to carry ou t the integrations along

the three different paths shown

Since

wm ff 1diI + A2 di 2 + x3 di 3





these paths of integration lead to differing results. For path (a), we

have

1 2 1 2 1 2

m 2111 21li2 L22i2 L31Y13 32i2i3 2 L3313

(g)

while fo r path (b)

1 2+ + 1 L 2+1

2 +L i +L (h)m 2 L2212 L32i2i3 2 33i L2 i + L122 1 L3i31 (h)

and path (c)

W 1 2 + 1L 1 2 +L ii +L ii + L 2 +L ()m 2 L3313 2 L11 + L133il + L211 22L222 L23 3i2

These equations will be identical only if (e) holds.

PROBLEM 3.13

Part a

When 8 = 0, there is no overlap between the stator and rotor plates,

as compared to complete overlap when 8 = w/2. Because the total exposed

area between one pair of stator and rotor plates is ITR2/2, at an angle 6

the area is

A =R 8 R2 (a)

There are 2N-1 pairs of such surfaces, and hence the total capacitance is

C (2N-1)8R2 o/g (b)

The required terminal relation is then q = Cv.

Part b

The system is electrically linear. Hence, WeCv2

and

eW' (2N-1)R C vTe e o (C)T (c)

ae 2g

Part c

There are three torques acting on the shaft, one due to the torsional

spring, the second from viscous damping and the third the electrical torque.





d2 de 1 v2 (2N-I)R2EJ d = -K(e-a) - B d + (d)

dt t2 dt 2 g

Part d

The voltage circuit equation, in view of the electrical terminal equation

is simply

(2N-1)R28s vV (t)0

R ddt g

o ]+ v (e)

Part e

When the rotor is in static equilibrium, the derivatives in (d) vanish

and we can solve for O-a,

V2(2N-1)R2E

8-a = o o (f)2gK

This equation would comprise a theoretical calibration for the voltmeter if

effects of fringing fields could be ignored. In practice, the plates are shaped

so as to somewhat offset the square law dependence of the deflections.

PROBLEM 3.14

Part a

Fringing fields are ignored near the ends of the metal coaxial cylinders.

In the region between the cylinders, the electric field has the form

E = Air/r, where r is the radial distance from the axis and A is a constant

determined by the voltage. This solution is both divergence and curl free,

and hence satisfies the basic electric field equations (See Table 1.2)

everywhere between the cylinders. The boundary conditions on the surfaces of

the dielectric slab are also satisfied because there is no normal electric field

at a dielectric interface and the tangential electric fields are continuous.

To determine the constant A, note that

b Erdr = -v = Aln( ); A = -v/ln((a)

The surface charge on the inner surface of the outer cylinder in the regions

adjacent to free space is then





VE

o o (b)

In -)b

while that adjacent to regions occupied by the dielectric is

af (c)In( )b

It follows that the total charge on the outer cylinder is

q = v [L(c +E)-x(E--E )] (d)In (

Part b

Conservation of power requires

dW

v d + f dx (e)dt dt e dt

Parts c and d

It follows from integration of (c) that

2Iq__ W 1 2

eW

2 Cor W'e =Cv2 v (f)

where

C = [L(E +E)-x(Ec- )]

In() o o

Part e

The force of electrical origin is therefore

awl

feiýx

e 12

2b

(C-ECo) (g)

Part f

The electrical constraints of the system have been left unspecified.

The mechanical equation of motion, in terms of the terminal voltage v, is

d2x = 1M

d 2x-K(x-k)- 2ib

v2n(E-, ) (h)

dt2

2 In( -)a





Part g

In static equilibrium, the inertial term makes no contribution, and (h)

can be simply solved fo r the equilibrium position x.

1V2 Tr(-C )x=- 2 o o (i)

2 K bK In(-)

PROBLEM 3.15

Part a

Call r the radial distance from the origin 0. Then, the field in the gap

to the right is, (from Ampere's law integrated across the gaps at a radius r

H = i/(O-a-e)r (directed to the right)

(a)

and to the left

H = Ni/(B-a+8)r (directed to the left)

(b)

These fields satisfy the conditions that VxH =0 and VB*=0 in the gaps. The

flux is computed by integrating the flux density over the two gaps and multiply

ing by N

S=N (H + Hr)dr (c)

a

which, in view of (a) and (b) becomes,

S= Li, L = p DN21n(b) 1 (d)S a -•+ -]

Part b

The system is electrically linear, and hence the co-energy is simply

(See Sec. 3.1.2b)

W' = Li2

(e)m 2

Part c

Th e torque follows from (e) as





e 1 oDN21n(b)[ 1 1

2 (f)

2 (a+e)2 (B-a-)2

Part d

The torque equation is then

J2d-2 -K6 + Te

(g)

dt

Part e

This equation is satisfied if e=0, and hence it is possible for the wedge

to be in static equilibrium at this position.

PROBLEM 3.16

We ignore fringing fields. Then the electric field is completely between

the center plate and the outer plates, where it has the value E = v/b. The

constraints on the electrical terminals further require that v.= V -Ax.

The surface charge on the outer plates is E v/b and hence the total

charge q on these plates is

de

q = 2(a-x) bv (a)

It follows that the co-energy is

e b

and the electrical force is

aW' defe e o2 (c)

ax b

Finally, we use the electrical circuit conditions to write

dEfe do (V -AX)2 (d)

b 0

The major point to be made in this situation is this. One might substitute the

voltage, as it depends on x, into (b) before taking the derivative. This clearly

gives an answer not in agreement with (d). We have assumed in writing (c) that

the variables (v,x) remain thermodynamically independent until after the force

has been found. Of course, in the actual situation, external constraints





relate these variables, but these constraints can only be introduced with care

in the energy functions. To be safe they should not be introduced until after

the force has been found.

PROBLEM 3.17

Part a

The magnetic field intensities in

the gaps can be found by using Ampere's

law integrated around closed contours

passing through the gaps. These give

Hg = N(i + i2 )/g (a)

H1 = Nil/d (b)

H2 = Ni2/d (c)

In the magnetic material, the flux densities are

B1 3 1 od1 3 + d

d

33N 2 o2Ni

2 3 + dd

The flux linking the individual coils can now be computed as simply the

flux through the appropriate gaps. For example, the flux A1 is

AX= ND[9,Hg + x1 Hl+(£-x)B ] (f)

which upon substitution from the above equations becomes the first terminal relation.

The second is obtained in a similar manner.

Part b

The co-energy is found by integrating, first on ii with 12 = 0 and then

on i2withil fixed at its final value. Hence,





Wm= fldl + X2 di2 (g)

1 d2 1 x 4 d= (1+ g +-LoB(1 - )i + Lo

1 L 4 1 di+•1Lo i +1 L(1+d)i4 o 2 2 o g 22

Part c (*-)

The force of electrical origin follows from the co-energy functions as,

fe 1 L 4 +1 Lo 4S=- Lo i +---- 1 (h)

4 o t 1 4 t 2

PROBLEM 3.18

Part a

Assuming simple uniform E fields

in the gaps'4 Vr

=E1 = (V -Vr)/g; E2 V /d = E3 )Q~9 / 60,~jObE = E5 = Vr/d

These fields leave surface charge-densities 'on the top electrodes

01 = Eo(Vt-Vr)/g , 02 = c o Vt/d

a 3 = [a(V /d) + Co (V9/d)

04 = [a(Vr/d)2 + e ](Vr/d)

a5 = C (V /d)

These urface charge densities cause net charges on the electrodes of

3o wb owLV V

q = - (-V) + d + aw(L-x) )

owb wcL V

r (V ) + Vr + w(x-g)(d)

Part b

1 2W' =

oqdV + fo qrdVr

q2=0 q1 1

VV

Ew(b + ) 4Ew(+-)y- ow(L-x)d d2 4

SV 2 wb Va

+ EW( +f [

) o VY.r

ci(x-)d (r)4 d

awl owd Vr V

e[ )

] (pulled to side with more voltage)

PROBLEM 3.19

Part a

The rotating plate forms a simple capacitor plate with respect to the

other two curved plates. There is no mutual capacitance if the fringing fields

are ignored. For example, the terminal relations over the first half cycle

of the rotor are

(ct+O)RDov1 (a-e)RDeoV 2=-a<O<a; q ; 2 Aa)

2aRDEoV

a<<-c-a; 12 = o ; = 0 (b)

So that the co-energy can be simply written as the sum of the capacitances

for the two outer electrodes relative to the rotor.

1 2 1 2

e 2 C1Vl + 2 C2v2

The dependence of this quantity on 6 is as shown below

dAID .- /

TT a.





Part b

Thetorque is the spatial derivative of the above function

Te

Part c

The torque equation is then

Sd 2 TeJd = T

dt2

where Te is graphically as above.

PROBLEM 3.20

*Part a

The electric energy is

w

e 2q2/C (a)

whereEX

C = EA/d(l-+ C-) (b)

It follows that the force on the upper plate due to the electric field is,

aW 2ffe e 1 --.-f=f

ax 2 E A0

So long as the charge on the plate is constant, so also is the force.





Part b

The electric co-energy is

W' Cv2 (c)

e 2

and hence the force, in terms of the voltage is

aw'2 2_ffe e = 1 V2e2A

ax 2 2

ed-E (l +

c dix 0

The energy converted to mechanical form is f fe dx. The contribution to this

integral from d+c and b+a in the figure is zero. Hence,

J2/eEnergy converted to mechanical form fe( 2QX)d

EOd/d

+ddQoI 0 2

J d2 Acfe( o,x)dx =- 3 de

2c d/E0

That is, the energy 3dQ /2Ac is converted from mechanical to electrical form.

PROBLEM 3.21

Part a

The magnetic energy stored in the coupling is

W 1 X2 /

L (a)m 2

where L = L /(l + X)o a

Hence, in terms of X, the force of electrical origin is

_f=fe= m x_/2aL (b)ax o





Part b

According to the terminal equation, i depends on (X,x) according to

S= (1+ )L a0

Thus, the process represented in the X-x plane has the corresponding path

i hU i A lLLI

n t e - p ane i

b

Path c A,

At the same time, the force traverses a loop in the f-x plane which,

from (b) is,

4

LOD-- -- - ID

F'





Part d

The energy converted per cycle to mechanical form is ffedx. Hence,

Energy converted to mechanical form = fedx + fAfedx (d)

2 2= -(1 2-_x)(X2-X1 )/2aLo (e)

That is, the energy converted to electrical form per cycle is

(X2_ 1 ) (X2-X 1 )/2aLo. (Note that the energy stored in the coupling, summed

around the closed path, is zero because the coupling is conservative.)

PROBLE• 3.22

Part a

The plates are pushed apart by the fields. Therefore energy is converted

from mechanical form to either electrical form or energy storage in the

coupling as the plate is moved from Xb to Xa . To make the net conversion

from mechanical to electrical form, we therefore make the current the largest

during this phase of the cycle or, I >12

Part b

With the currents related as in part a, the cycle appears in the i-x plane

as shown

I-

a &1l,

-11

u

·- T 'V





Quantitatively, the magnetic field intensity into the paper is H = I/D so that

X = pIxh/D. Hence,

Sx n

2

m'=

2(-

D- ) I

aw' oh

fe m = 1 o 2

ax 2 D

Because the cycle is closed, there is no net energy stored in the coupling,

and the energy converted to electrical form is simply that put in in mechanical

form:

B DMechanical to electrical energy per cycle = - fedx - fedx (c)

A C

I= J -X) 2h 2 (d)

Part c

From the terminal equation and the defined cycle conditions, the cycle

in the A-x plane can be pictured as

L h1/t

--- C

7_·

The energy converted to electrical form on each of the legs is





p oI°IlX h/D oI2h

(A--+B) - Ild =- ld =- D ) (e)

foI1Xbh/D D XaXb

j II X h h/D (B-.C) - id X °

a 1 Xah D 2 1B-+C) -idX Io 2 ah XDdX oa (2 2 (f)

aXh/Doa

2

(C-D) - 1 2 d = (Xa-Xb) (g)

(D-+A) - idX = 2Db (o2 2) (h)

The sum of these is equal to (c). Note however that the mechanical energy in

put on each leg is not necessarily converted to electrical form, but can be

stored in the coupling.



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ROTATING MACHINES

PROBLEM 4.1 ýZ(R+J

S10 +art a

With stator current acting alone

the situation is as depicted at the

right. Recognizing by symmetry that

Hrs(4rw)= -H rs() we use the contour

shown and Ampere's law to get

2Hrs = J+IrNsis

g sin i'](R+g)dV'=

from which s iN oiN i cosQ

Hrrs (i) S

2g

5

and

SNsi scostS

rs 2g

Part b

Following the same procedure for rotor excitation alone we obtain

poN i cos(I-e)

Brr() 2g

Note that this result is obtained from part (a) by making the replacements

N -- P N

s ri --- i

s r

Part c

The flux density varies around the periphery and the windings are distributed,

thus a double integration is required to find inductances, whether they are found

from stored energy or from flux linkages. We will use flux linkages.

The total radial flux density is

B =rs + Brr [Ni cosi + Nri cos(*-8)]r rs rr 2g a s r r

ROTATING MACHINES


Taking first the elemental coil

on the stator having sides of angular

span dip at positions I and *+ i as

illustrated. This coil links an amount

of flux

N

dX =a2(R+g) sin(R)d

number of turns in flux linking one turn

elemental coil of elemental coil

ji N (R+g)t .+w

dA = - 0 sindi I[N i cosi' + N i cos(i '-P6d1's rr

UoNs (R+g),dA = sinV[N i ins + Nri sin(i-6)]dip

To find the total flux linkage with the stator coil we add up all of the

contributions

poN (R+g)2 wA = 0 sini[N sisnp + Nri sin(*-6)]dt

5 g rr

ioN (R+g)tIX g [- N +- Ni cosa]s g 2 ss 2 r r

This can be written as

A L i + Mi cosas S r

where

m' N2RALs =•s

ml N N RkM = osr

and we have written R+g % R because g << R.

When a similar process is carried out for the rotor winding, it yields

A = L i + Mi cosOr rr s

where •oN2Rk

orLr =

and M is the same as calculated before.

ROTATING MACHINES

PROBLEM 4.2

Part a

Application of Ampere's law

with the contour shown and use of the

symmetry condition

Hrs (~r)=-Hrs (i) yields

2H (')g = Nsis (1- ); for 0 < 4 < nrs se iT

2H (')g = Ni (-3t 2); for r < <Z fs ssa 7vZ

The resulting flux density is sketched F-or 0 :-- Y -c

I)

(,

\,I

2 LI

Part b

The same process applied to excitation of the rotor winding yields

B&,

Y1

S?- )"



ROTATING MACHINES


Part c

For calculating inductances it will be helpful to have both flux densities

and turn densities in terms of Fourier series. The turn density on the stator

is expressible as

4N

n = 4 -- sin nrs (R+g) nodd

and the turn density on the rotor is4N

n = -- 1 sin(q-9)r 7w2R nodd

and the flux densities are expressible as

4ji NsiBrs = n n2 cos niB = 2 cos n=r

nodd n gn

Brr 2 2 cos n(i-O)1nodd gnt

h t. .ToU

l UJ.~ i . lUfl d i

e aa ux ens ty isl

B =B +Br rs rr

First calculating stator flux

linkages, we first consider the

elemental coil having sides di

long and 7wradians apart

dAXs = n (R+g)di Br (ip')(R+g)f(dt']

number of flux linking one

turns turn of elemental

coil

Substitution of series for Br yields

2 [ 81 N 1881i 8N Nir

d's = n (R+g) 2 d• nodd i 2o sin ný + 1r 3 r sin n(4-0)1gn nodd r gn

The total flux linkage with the stator coil is

32 u N (R+g) Zi Ni Ni32 oNs(R+g) r sinn -s s in ný + - r rsin n(*-6) d#

r g 0 odd nodd n nodd n

ROTATING MACHINES


Recognition that

sin n* sin m(P-O)di = 0 when m # n

simplifies the work in finding the solution

321 N (R+g)£t 7N i WN i

+ rr cos nO)a s4

s g nodd 2n 2n

This can be written in the form

S= i + [ M cos nO i

nodd

where16p N2 R9,

L = gs 3 4S3g nodd n

16 NsN rRM = 4osr

n 3 4T gn

In these expressions we have used the fact that g << R to write R+g 1 R.

A similar process with the rotor winding yields

S= Li + [ M cos nO i sr rr n nodd

where 16 2

16p 4sRa 1Lr 3 4

7 g nodd n

and Mn is as given above.

PROBLEM 4.3

With reference to the solution of Prob. 4.2, if the stator winding is

sinusoidally distributed, Xs becomes

32p°N (R+g)t F Ni

s 0 g - sin[Ni sin 'P + r 3r sin n(ý-) d*

Bag o nodd n

Because Jsin t sin n(P-6)

=0 when n # 1

o

3210N (R+g) 9, Fa = 4g o sin sNi sin * + Nir sin(ý- d*

and the mutual inductance will contain no harmonic terms.

Similarly, if the rotor winding is sinusoidally distributed,



ROTATING MACHINES

PROBLEM .3 (Continued)

3210N (R+g)2. IT NiS = -s in n Ni

s 4g o nodd n3 s + Nrirsin(-8) d*Usingorthogonalhe

Using the orthogonality condition

sin ný sin(P-e)d =P0 when n#l

32oNs (R+g)2.

S 4gr --adsinsin 2n + Nr ir siný sin(e- d

l"g o Lnodd n- )I

and the mutual inductance once again contains only a space fundamental term.

PROBLEM 4.4

Part a

The open-circuit stator voltage is

dX M

vI -- cos nwis dt dt

nodd n

(M I

(t) - -sin nwt

nodd n

Part b

V Vsn _1 , s3_ 1

=n -3 4 percentV

s -n3 V

ss

=27

This indicates that uniform turn density does no t yield unreasonably high values

of harmonics.



ROTATING MACHINES


Part c

9J'ees)

PROBLEM 4.5

Given electrical terminal relations are

S= L i + Mi cosOs s r

X = Mi cos6 + L ir s rr

System is conservative so energy or coenergy is independent of path. Select

currents and 6 as independent variables and use coenergy (see Table 3.1).

Assemble system first mechanically, then electrically so torque is not needed

in calculation of coenergy. Selecting one of many possible paths of integration

for i and i we haves r



ROTATING MACHINES


i i

W'(i, i ,) = A (i',O,e)di' + Ar(is,i',)di'm s r J s s rs0 r r

1 21 2w'(ii 0r)= -!Li2 + Mi cosO + 1 Lrir

m s'r 2 s rs s 2 r r

aW'(is, r,6)Te m = - Mi i cosO

36 rs

PROBLEM 4.6

The conditions existing at the time the rotor winding terminals are short-

circuited lead to the constant rotor winding flux linkages

A = MIr o

This constraint leads to a relation between i

r

and i

s

= i(t)

MI = Mi cos8+ L io s rr

iL

[I0-i(t)cosel

r r

The torque equation (4.1.8) is valid for any terminal constraint, thus

Te = -M i i cos = - (t)[Io-i(t)cos]sin6

r

The equation of motion for the shaft is then

d26 M2

dr

2 L r

i(t)[Io-i(t)cos68]sin

dt r

PROBLEM 4.7

Part a

Coenergy is

1 2 1 2W'(isi ',6) = - L i + - L i + L ()i im s r 2 ssa 2 rr sr sr

TeaW'(i s ,i ,) ii dLsr(8)m r

Do s r dO

Te i iir [M1sinO + 3M3sin36]

Part b

With the given constraints

Te = -II rsinw t sinw t[M sin(w t+y)+ 3M sin3(wmt+Y)]



ROTATING MACHINES


Repeated application of trigonometric identities leads to:

Te = s r sin[(W+ws-W) t+y]+ sin[(m-w +w )t+y]4 m s r mflLs r

-sin[(wm +mw+r)t+y]-sin[(wm-ws )t+Y]3

3M I3I r4 sin[(3m

+-Or)

+r)t+3Y]s t+3y]+ sin[(3wm-w s

-sin[(3w s++ )t+3y]- sin[(3wm- -w )t+3y]4 r s r n s rsr

To have a time-average torque, one of the coefficients of time must equal

zero. This leads to the eight possible mechanical speeds

w +w

t = +W + and + rm - s- r m - 3

For

Wm = +(Ws - w )

MeIsIrTe s sin yavg 4 sin

For

= wm +(Wts + w )

T sin y

avg 4For (ws )

W = +m - 3

3M3I I

Te3s sin 3y

For ( + w

S= s rm - 3

3M31s1

Te = rsin 3yavg 4

PROBLEM 4.8

From 4.1.8 and the given constraints the instantaneous torque is

Te= -I M sinw t cos(w t+y)(I slinw t + I sin 3w t)

Repeated use of trigonometric identities leads to:



ROTATING MACHINES


II M .

4 Cos[ (r+•mW s) t+Y]-cos[mr+W m t+

+ cos[(rW -W•m )t-y]-cos[(w -w +s )t-y]

rIs3M4s3 cos[ (wr+w -3w )t+y]-cos[ (o +w +3w )t+y]

+ cos[(Wr-w -3ws )t-y]-cos[ (wr-m+3ws )t-y]l

For a time-average torque one of the coefficients of t must be zero. This leads

to eight values of w :m

W = + W + W and w = + w + 3wm - r- s m - r- s

For

W = +(Wr-W )m -tr s

II MTe r sl1

avg 4 cos

For

S= +(r + s)

II MTe r slTavg 4

cos y

For

W = +(W - 3w )

e IrIs3M

avg 4

For

w = +(w + 3w )m r s

II MTe Ir s3Tavg

=4

cos y

PROBLEM 4.9

Electrical terminal relations are 4.1.19-4.1.22. For conservative system,

coenergy is independent of path and if we bring system to its final mechanical

configuration before exciting it electrically there is no contribution to the

coenergy from the torque term. Thus, of the many possible paths of integration

we choose one

http://slidepdf.com/reader/full/4.1.19-4.1.22

http://slidepdf.com/reader/full/4.1.19-4.1.22



ROTATING MACHINES


ias

W' (i ia) = (i' ,0,0,0,8)di'

+ fibs bs(ias ib sO90,0)dibs0

+ iarX (i ibs ,' ,,6)di'o ar as50s ar ar

0

The use of 4.1.19-4.1.22 in this expression yields

W, i as L i' i' + L iibsm J s as as sibs s•S

o o

+ iar(L i'r +Mi cos + Mi sin)di'r ar as bs ar

-+ br(Lrbr - MiassinO + Mibscos )dibr0

Evaluation of these integrals yields

1 2 1 2 1 2 1 2W'=-Li +-L +-Li ++Lm 2 sas 2 sbs 2 r at 2 r br

+ Mia i cos8+ M b i sine

- Miasibrsin 0 + Mibsibrcos

Th e torque of electric origin is then (see Table 3.1)

T e = m(a s ibs ar'ibr'19Te

Te = -M[ asiarsin-iibsiarcos +iasibrcos+.sibs rsinO]

PROBLEM 4.10

Part a

Substitution of currents into given expressions for flux density

B =B +4BBr ra rb

1JNB o [IaCos t cos * + Ib sin wt sin 4]r 2g a b



ROTATING MACHINES


Part b

Application of trigonometric identities and simplification yield.

u·•N I I

Brr 2g[2

cos(wt-i) +2

cos(wt + 4)]

II+ j-

cos(wt-0)-

bcos(wt + 4)]

p NB = - [(Ia+Ib)cos(wt- 1)+(Ia-I )cos(wt+P)]

The forward wave is1oN(I + Ib )

B = 4g cos(wt-0)rf 4g

For constant phase on the forward wave

wt - =- constant

f dt

The backward wave is

SN(I a - Ib)Brb = cos(wt + ')

rb 4g

For

wt + ' = constant

= - WWb dt

Part c

The ratio of amplitudes is

B I -Irbm Ia b

B I+Lrfm ab

Sr 0 as I Ibrfm

Part d

When Ib - I a

Brf = 0

This has simply reversed the phase sequence.



ROTATING MACHINES

PROBLEM 4.11

Part a

Br

= Bra + Brb

z NI

Br --- [cos wt cos P + sin(wt + 8)sin*]

Part b

Using trigonometric identities

i NI

Br = [cos wt cos * + cos 8 sin wt sin q + sin B cos wt sin $]

Bo 1 1Br g cos(wt-*0+os(t-)+ cos(wt+r)

cosB cos8+ 22 cos(wt-4)- -2--2 cos(wt+8)

sinB sin8+ n sin(wt+i)- 2 sin(wt-e)]

p NI

Br 4-[

(l+cosB)cos(wt-*)-sinBsin(wt-ý)

+ (1-cosa)cos(wt+i$)+sinBsin(wrt+p)]

Forward wave is

i NI

Brf = 4g [(l+cos8)cos (wt-p)-sinasin(wt-i)]

For constant phase

wt - + = constant

and

Wf dt

Backward wave is

l NI

Brb -w-{-(•-cos) cos(wt+-)+sin$sin (wt+l)]

For constant phase

wt + i = constant

and

Ob dt



ROTATING MACHINES


Part c

The ratio of amplitudes is

B _ ( 22 2rbm -cos) + sin 1i-cos

Brfm (1+cos8) 2 + sin2 8 l+cos8

rbmas 8 0, - -0..

Brfm

Part d

=The forward wave amplitude will go to zero when 8 W. The phase sequence

has been reversed by reversing the phase of the current in the b-winding.

PROBLEM 4.12

Equation 4.1.53 is

Pe =Vasias + Vbsibs

For steady state balanced conditions we can write

ias= I cos wt; ib = I sin wt

Vas = V cos(wt+4); vbs = V sin(wt+ý)

then

Pe=

VI[coswtcos(wt+$)+sinwt sin(wt+0)]

Using trigonometric identities

= pe VI cost

Referring to Fig. 4.1.13(b) we have the vector diagram

ROTATING MACHINES


From this figure it is clear that

wLsI cos = -Efsin

(remember that 6 < 0)

VE

Then pe - sin 6e wL

which was to be shown.

PROBLEM 4.13

For the generator we adopt the notation for one phase of the armature circuit

(see Fig. 4.1.12 with current convention reversed)

A, r V

E-

The vector diagra

From the vector diagram

XI sin' = Ef cos 6-V

XI coso = Ef sin 6

Also, the mechanical power input is

EV

P = - sin 6

Eliminating 0 and 6 from these equations and solving for I yields



ROTATING MACHINES


I I - 2 (ý-) - () + 1V

Normalizing as indicated in the problem statement we define

I = rated armature current

Ifo= field current to give rated voltage

on open circuit.

Po = rated power

I(f) I 2+ 1- 2 I 2-() 2(P X 2

0 fo fo (Injecting given numbers and being careful about rms and peak quantities we have

If 2 I 2 2S= 0.431 + 1 - 2 (-) -3. 9 2 (•-)

0 V fo fo 0

Ifo = 2,030 amps

and

S ) = 3.00fo max

The condition that 6 =2 is

2

PXf = V-

f PX PX P( ) 2 = 1.98

fo mi n fo V V2 P

For unity p.f., cos c = 1, sin * = 0

Ef cos 6 = V and Ef sin 6 = IX

eliminating 6 we have

V__ fI ( )2 -1o o

of 2S0.431 F

0 fo



ROTATING MACHINES


for 0.85 p.f.

Ef sin 6 = 0.85 IX

Ef cos 6 - V = 1-(0.85)2 IX

eliminating 6, solving for I, and normalizing yields

I 2- = 0.431 [-0.527 + (--) - 0.722]

I Io fo

This is double-valued and the magnitude of the bracketed term is used.

The required curves are shown on the next page.

PROBLEM 4.14

The armature current limit is defined by a circle of radius VIo, where Io

is the amplitude of rated armature current.

To find the effect of the field current limit we must express the complex

power in terms of field current. Defining quantities in terms of this circuit

The vector diagram is

A

S E - V

I jxjxAA

2V2A VE*-V

2VEfe

P + jQ = VI* =-ix = X



s ta~b iboI1bI% * -

Palej anordire

-A

r ýa te )

re-ldCFIre(ýreI4

zero p4.F

0.I

0.5

11.

1j,

2.0



ROTATING MACHINES


If we denote the voltage for maximum field current as Efo, this expression

becomes2 VEfo VEfo

V fo foP+JQ = -j + sinS + j cos6

On a P+jQ plane this trajectory is as sketched below

1

\

P

0i t- L, 1W

The stability limit (6= 2) is also shown in the sketch, along with the armature

current limit.

The capability curve for the generator of Prob. 4.13 is shown on the next

page.

P and Q are normalized to 724 MVA.

PROBLEM 4.15

The steady state deflection ipof the rotatable frame is found by setting

sum of torques to zero

Te + TS = 0 = Te - K* i)

where Te is electromagnetic torque. This equation is solved for 4.

Torque Te is found from



ROTATING MACHINES

Q

P

LLI

captb IICa"el V Problem

4.14/

ROTATING MACHINES


awem(il'i2i3'T =

and the magnetic coenergy for this electrically linear system is

1 21 2 1 2W' = - Li + -Li + - Lim 2 1 2 2 2 3 3

+ Mili 3cos (O-)+Mi2i3sin($-9)

from which

Te = Mili3sin(4-t) - Mi2i3cos(4-1)

For constant shaft speed w, the shaft position is

0 = wt.

Then, with 1 3 = 1o as given

dX di=

-t - MI sin(wt-0)+L 1 =-ilR

and

dX 2 di 2

d--=

WMI cos(wt-p)+L dt= -i 2 R

Using the given assumptions that

di 2

IL ' <<-Ril1

and Ldt

i<<Ri2

I-diR dt

we have

0MI

il Ro sin(wt-0)

wMI

i2=- R cos(wt-9)

and the torque Te is

TMI

Te = MIo(~)[sin2 (wt-9)+cos 2 (wt-)]

Hence, from (1) 2

(MIo)KR

which shows that pointer displacement 0 is a linear function of shaft speed w

which is in turn proportional to car speed.

Suppose we had not neglected the voltage drops due to self inductance.

Would the final result still be the same?

ROTATING MACHINES

PROBLEM 4.16

The equivalent circuit with parameter values as given is

W (L. 0)=.o3 XL * (L•-.) -j 0.3 _n

• - :s0. 1\ =I Y00 'O <

p. %

From (4.1.82) the torque is

k2 L R 2

(-) (- ) (L-)v se s s

[w(l-k

2 )Lr 2+(Rr/s)

2 M2 -_m

where k L L and s = LL trs s

Solution of (4.1.81) for I yieldss

VR 2VI

Is

s R 2 L

(-r) + [wsLr(1-k 2 ) 5s

volt-ampere input is simply (for two phases)

(VA)in = VIs

The electrical input power can be calculated in a variety of ways, the

simplest being to recognize that in the equivalent circuit the power dissipated

in,R /s (for two phases) is just ws times the electromagnetic torque, hence

in T s

Finally, the mechanical power output is

mech m

These five quantities are shown plotted in the attached graphs. Numerical constants

used in the computations are



ROTATING MACHINES

Crves foY ProbleMvA 416rEdetbon /Vlace

- - ý r

OK

I )

4-oo

200

00 2160o ZS620'SL4'

LO O. 0, o,.7 0.o .s 0, 0O3 0. . ,-Sh1,

?oc

60C

9oo

0oc

CI

i

i ~c .oLo 0.9 0.6 0.7 04 OS 0,4 0 • t- 0.1 o0 jLF

60



ROTATING MACHINES


w L = w L = wM + 0.3 = 4.80ss s r S

2 4.5k = (-) = 0.878

117e s

T 0.01 newton-meters

0.342 +2

s

0.0123.0 -+

Is 0.01 147 amps (K.0.342 +

2s

smT =0.188

PROBLEM 4.17

Part a

For ease in calculation it is useful to write the mechanical speed as

wm = (l-s)

and the fan characteristic as

T = -B3 ( 1 -s)

3

m s

With w = 120w rad/sec

Bw3 = 400 newton-meters

The results of Prob. 4.16 for torque yields

117

400(1-s)3 = s0.01

0.342 +2

s

Solution of this equation by cut-and-try for s yields:

s = 0.032

Then Pech = (400) (1-s)3w = (400)(w )(l-s)4

Pmech = 133 kilowatts into fanmech

mechinput = -s = 138 kilowattsinput 1-s

Circuit seen by electrical source is



ROTATING MACHINES


j0.3 K jo.•3

3o 3 -0

Input impedance is

Zln JO.3 + (j4.5)(3.13+j0.3) = -2.79+j15.0

in 3.13 + j4.8 3.13+j4.8

in = 100.60 - 56.80 = 43.80

Hence,

p.f. = cos in = 0.72 lagging

Part b

Electromagnetic torque scales as the square of the terminal voltage,

thus

117 2e

s 0.01 s

0.342 + 2 so2

where Vso = /W 50 0 volts peak. Th e slip for any terminal voltage is now

found from

3117

V2

400(1-s) s 0.01 (V=

0.342 + 0 2 sos

The mechanical power into the fan is

=P e 400 wa (1-s)

4

mech s

electrical power input is

Pmech

in 1-s

ROTATING MACHINES


and the power factor is found as the cosine of the angle of the input impedance

of the circuit

.'s -jC

O,/I _

These quantities are protted as required on the attached graph.PROBLE4 4.18

Part a

The solution to Prob. 4.1 can be used to find the flux densities here.

For the stator a-winding, the solution of Prob. 4.1 applies directly, thus,

the radial component of flux density due to current in stator winding a is

B () aocossra(2) 2g

Windings b and c on the stator are identical with the a winding except for the

indicated angular displacements, thus,

BrbNsaib 2v r

Bb 2 cos('P- -)

SoN i 47

Brc($)rc

22g

cos((-Cos

3)

The solution in Prob. 4.1 for the flux density due to rotor winding current

applies directly here, thus,

orr ( io)Nrr2g rcos(Q-e)trr(•) 2g

Part b

The method of part (c) of Prob. 4.1 can be used and the results of that

analysis applied directly by replacing rotor quantities by stator b-winding

quantities and 0 by 2w/3. The resulting mutual inductance is (assuming

g << R)



ROTATING MACHINES

5i1I

L1KV)PC vVF

0,7T 5 150- I

250 Iooif

0.256- S

/INLJ v

.

I 5501

.s (volrs pFEAK)

L4Mdý10iLdu o su

Ma"e, LUvesV_Y~ieC~e... _-------

oe-r Proble.-K 4A7



ROTATING MACHINES

PROBLEK 4.18 (Continued)

irp N2Ri 29L = 0o 27r

2gab cos -2g 3

j N Re Los S

ab 4g 2

where Ls is the self inductance of one stator winding alone. Note that

Lac Lab because of relative geometry.

Part c

The X-i relations are thus

L L

L L s si + Mcos=Lia sa 2 b 2 + McosOi

L L

b~- 2- ia+Ls ib 2- ic + Mcos(e- 3)irL L

s a 44wSc

2 i L + Lsic + Mcos(O- --)i

2wXr McosOia + Mcos(e- 32)ib

+ Mcos(6- )i + Li3 )ic + Lrr

where from Prob. 4.1, 2

1riN2 RL =s 2g

wyoN N R.

2g

wup N2 RL =

or

r 2g

Part d

The torque of electric origin is found most easily by using magnetic

coenergy which for this electrically linear system is

W'(iibii,) 1 L (2 + 2 + 2m Ls(i +c + c)

+ 1 L(iib + ii + iic)+Mcosoi i2 s a ac r a

2w 4w+ Mcos(- -)it + Mcos(0- -)i ic

3 rcb o3 r

The torque of.electric origin is



ROTATING MACHINES


aW'(iLbi ,i ,e)Mi(ia'sic ire m=

Tae

Te

= -Mir[i sin + ibsin(8- 7--)+ i sin(O- f)]

PROBLEM 4.19

Part a

Superimposing the three component stator flux densities from Part a

of Prob. 4.18, we have

Brs Ns [i cosa + ibcos(p- 2T)+ i cos(P- -)]rs 2 a b 3 c 3

Substituting the given currents

cos(wt- 2--)cos(W- 2)B =oNs [I cos wtcosý + I b 3 3rs 2g a

47 47 + I cos(wt- 3 -cos (P-T-)

Using trigonometric identities and simplifying yields

oNs [( a + Ib + Ic cos(trs 2g 2

+ (I + IbCOS + I cos - )cos(Wt+)

+ -(I sin + I sin 2• )sin(wt+)

Positive traveling wave has point of constant phase defined by

wt - + =constant

from which

dt

This is positive traveling wave with amplitude

yoN

rfm =4g (a b c

Negative traveling wave has point of constant phase

wt + P = constant

from which

d_ = _dt

This defines negative traveling wave with amplitude



ROTATING MACHINES



B ISoNs

(Ib )

22 Ib+

2 2

rbm 4g a2 22 2c

Part b

When three phase currents are balanced

I = I = Ia b c

and Brbm = 0 leaving only a forward (positive) traveling wave.

PROBLEM 4.20

Part a

Total radial flux density due to stator excitation is

oUNBrs

rs=

2g(iaos 21 +

+ibsin 2*)

Substituting given values for currents

UoN

rs =- (Ia cos wt cos2p+ I sin wt sin 24)

Part b

= N I + Ib I - s

Brs a b cos(wt-2*) + (a 2 )cos(wt+2q)

The forward (positive-traveling) component has constant phase defined by

wt - 24 = constant

from which

d_~*wdt 2

The backward (negative-traveling) component has constant phase defined by

wt + 21 = constant

from which

dt 2

Part c

From part b, when IIa b a -b = 0 and the backward-wave amplitude

goes to zero. When Ib = - a, Ia + Ib = 0 and the forward-wave amplitude goes

to zero.



ROTATING MACHINES

PROBLEM 4.21

Referring to the solution for Prob. 4.20,

Part a12N

Brs =-N (ia cos p + i b sin p)rs 2g a

rs = (Ia cos wt cos pI + Ib sin wt sin p$)

Part b

Using trigonometric identities yields

°N I + I I- Ib

B = _o 2a b-)cos(wt-p$) + a cos(t+p4)rs 2g 2 2

Forwardwave has

constantphase

wt - pp = constant

from which

dt p

Backward wave has constant phase

wt + pJ = constant

from which

dt p

Part c

From p'art b, when Ib = Ia , I a - I b = 0, and backward-wave amplitude goes

to zero. When Ib a, Ia + Ib = 0, and forward-wave amplitude goes to zero.

PROBLEM 4.22

This is an electrically linear system, so the magnetic coenergy is

Wi(ii B) =!(L + L cos 2)i2 +1 L 12 + Mi i cos 6m(s, r 2 2 2 r r r s

Then the torque is

TW'(i 'i 8) 2Te m r =-Mi is sin 6 - L 12 sin 20

T6 r s 2 sPROBLEM 4.23

Part a

L

L 0(1-0.25 cos 46 - 0.25 cos 88)



ROTATING MACHINES


The variation of this inductance with 0 is shown plotted below.

L

LO

I

0 lo 30 so o6 70 B0 40 80

Roroe Posi rlov 0 (ereEs)



ROTATING MACHINES


From this plot and the configuration of Fig. 4P.23, it is evident that minimum

reluctance and maximum inductance occur when 0 = 0, w/2, i,... wi,.... The

IT 2 37r 7 niinductance is symmetrical about 0 = 0, ,... and about 0 = 4' +

as it should be. Minimum inductance occurs on both sides of e = which ought

to be maximum reluctance.

The general trend of the inductance is correct for the geometry of Fig.

4P.23 but the equation would probably be a better representation if the sign

of the 86 term were reversed.

Part b

For this electrically linear system, the magnetic stored energy is

1 X

m 2 L

Wm(XO) X2(1-0.25 cos 46 - 0.25 cos 80)

wm 2L

The torque is then

awm(X,0)

e me

Te = - (sin 40 + 2sin 80)

Part c

With X = A cos wt and 0 = Ot + 6

A2

cos wt

Te o2L [sin(4Qt+46 )+

2sin(80t+8

6)]

Repeated use of trig identities yields for the instantaneous converted power

2PA

rTe 4L [sin(4it+46) + 2 sin(86t+8

6)

o

1 1+ sin(2wt + 49t + 46)+1 sin(4Ct - 2wt + 46)

+ sin(2wt + 8St + 86)+ sin(80t - 2wt + 86)]

This can only have a non-zero average value when Q # 0 and a coefficient of

t in one argument is zero. This gives 4 conditions

S1 + + W

When S = +-2 A

[e - o sin 46Savg 8L



ROTATING MACHINES


-and when 9 = +

2QA

[ 4Lsin 86vg

PROBLEM 4.24

It will be helpful to express the given ratings in alternative ways.

Rated output power = 6000 HP = 4480 KW at 0.8 p.f. this is

4480-80= 5600 KVA total0.8

or

2800 KVA per phase

The rated phase current is then

2800 x 10

Is s 3 xx 103

=933 amps rms = 1320 amps pk.

Given:

Direct axis reactance w(Lo+L 2) = 4.0 ohms

Quadrature axis reactance w(L -L2) = 2.2 ohms

wL = 3.1 ohms wL2 = 0.9 ohms

The number of poles is not given in the problem statement. We assume 2 poles.

Part a

Rated field current can be found in several ways, all involving cut-and-try

procedures. Our method will be based on a vector diagram like that of

Fig. 4.2.5(a), thus

Do"IARY Ais A

) ,) ) ýAID

MEAs UeFD.4

Ls ro V AD

os, r,•E A-S

A/.

I~ALais

--

ROTATING MACHINES


Evaluating the horizontal and vertical components of Vs we have (remember that

y < 0)

Vs cos 6 = Ef cos( + y) + uL21scos(i + 2y)

V sin 8 = Ef sin( + y) + wL2I sin(j + 2y) + wL0 I

Using trigonometric identities we rewrite these as

V cos 86= -Ef sin Y - wL I s sin 2y

Vs sin 6 = Ef cos y + WL2Icos 2y +WLo1

Next, it will be convenient to normalize these equations to Vs,wL2I

cos 6 = -ef sin y V sin 2y

WL2I wLoI

sin 8 = ef cosy + cos 2y +s s

where E

ef =EEf

f Vs

Solution of these two equations for'ef yields

WLI wL I21s os

sine- 2 cos 2y V V

s sf = co s y

wL I2s

-cos - V sin 2y

ef = sin y

For rated conditions as given the constants are:

cos 0 = p.f. = 0.8

sin 6 = - 0.6 (negative sign for leading p.f.)

WL2I wL I

V = 0.280; - s = 0.964V V

s s

Solution by trial and error for a value of y that satisfies both equations

simultaneously yields

y = - 148"



ROTATING MACHINES

PROBLEN 4.24 '(Continued)

and the resulting value for ef is

ef = 1.99

yielding for the rated field current

Ve

Ir = 24.1 amps.

where Vs is in volts peak.

Part b

The V-curves can be calculated in several ways. Our choice here is to

first relate power converted to terminal voltage and field generated voltage

by multiplying (4.2.46) by w, thus

EfV (Xd-X )V

P = tTe f f-s sin 6 - s sin 26X 2X d Xd dq

where Xd = w(Lo+L2 )

Xq = w(Lo-L 2 )

We normalize this expression with respect to V /Xd, then

PXd (Xd-X )-= - e sin 6 - sin 26

V2 f 2X

Pull-out torque occurs when the derivative of this power with respect to 6 goes

to zero. Thus pull-out torque angle is defined by

PXd (Xd-X)( = -ef cos 6 - cos 26 = 0V q

The use of (4.2.44) and (4.2.45) then yield the armature (stator) current

amplitude as

V V s E 2I s = sin 6)

2 + cos 6q d d

A more useful form is

V IX 2

Is sin 6)2 + (cos 6- ef)2d q

The computation procedure used here was to fix the power and assume values of

6 over a range going from either rated armature current or rated field current

to pull-out. For each value of 6, the necessary value of e is calculated



ROTATING MACHINES


from the expression fo r power as

PX (Xd-X)

ef

-V2V

s

+2X

q

-sin 6

sin 26

and then the armature current magnitude is calculated from

V X

I d sin 6) 2 + (cos 6 - e )2s X X fd q

For zero load power, y = 0 and 6 = 0 and, from the vector diagram given earlier,

the armature current amplitude is

SIv - EflI =s w(Lo+L2)

with pull-out still defined as before. The required V-curves are shown in the

followinggraph. Note that pull-out conditions are never reached because range of

operation is limited by rated field current and rated armature current.

PROBLEM 4.25

Equation (4.2.41) is (assuming arbitrary phase for Is )

Vs =J L I + jWL2 Is ej + JMIr er2e

With y = 0 as specified

Vs = jw(Lo+L2)I + JWMIr

The two vector diagrams required are

A .\A

v5i

Lo

•4A

IVJo

5>Ji VE_

- CAW

V3 4



ROTATING MACHINES

V- CIRVE-S Foe.RoIEM 4,Z4

ARMATuVEaupeEA/r

(AMPs I MS)

AReM~ril~E CuI~EjLv ::333 AAMeS JE'M1coo

o600

03

FIELD CU eEN I- 4 (AMPS~

e &V~i24,i AMPA.



ROTATING MACHINES

PROBLEM 4.26

Part a

From Fig. 4P.26(a) 1 j._ e

VS jXs + Y

from which the ratio of the magnitudes is

1

Asl N/ cosol•2+1 sino + xs 2

For the values Y =0.01 mho, Xs = 10 ohms

i_ 100

s^I J(100 cos) 2 + (100 sin+10)2

Then, for 0 = 0

100= 0.995

IVs V1l0,000 + 1O

and, for 4 = 45*

100 -

S 0.932

s 2+ (-•i + 10)2

Part b

It is instructive to represent the synchronous condenser as a susceptance

jB, then when B is positive the synchronous condenser appears capacitive. Now the

circuit is

'Xs

A



ROTATING MACHINES


Now the voltage ratio is

1

V Ye-+ jB 1

V +1 1 + jXYe -BXVs Ye-+ jB jxs JXs B xs

V 1

V 5-s s

Then

lsl •1-BXs+X Ysin)2+(X YCosO)2

For C = 0

J±L 1

Il (1-BXs)2 + (X Y)2

If this is to be unity

(1-BXs )2 + (X Y)2 = 1

1-BX = 1-(X Y)2

1- 1- (XsY)

B = s

for the constants given

1- l-0.01 0.005 0.0005 mho10 10

Volt-amperes required from synchronous condenser

V BVA)sc = 2B (2)(1010)(5)(10-4) = 10,000 KVA

Real power supplied to load

PLf • 1

2Y cos = i Iy fo r O = 0

Then

(VA)sc B 0.0005= 0.05

P Y 0.01L

For * = 0 the synchronous condenser needs to supply reactive volt ampere5 equal to

5 percent of the load power to regulate the voltage perfectly.

77



ROTATING MACHINES


For * = 45*

A1

IV I 2 2

- +BX - +

In order fo r this to be unity

1-BX + +- 1

X Y jX Y 2

1+ s \ 1- 21s=

B

s

For the constants given

1 + 0.0707 - V1-0.005B = = 0.00732 mho

10

Volt-amperes required from synchronous condenser

-(VA)s c = V B = (2)(1010)(7.32)(10

3) = 146,400 KVA

Real power supplied to load

P = IV12

Y cos # = for O=

450

Then

(VA)sc B2 (/2) (0.00732).... 1.04

PL Y 0.01

Thus for a load having power factor of 0.707 lagging a synchronous condenser needs

to supply reactive volt-amperes equal to 1.04 times the power supplied to the

load to regulate the voltage perfectly.

These results, of course, depend on the internal impedance of the source.

That given is typical of large power systems.

PROBLEM 4.27

Part a

This part of this problem is very much like part a of Prob. 4.24. Using

results from that problem we define



ROTATING MACHINES


Ef wIEf rI

ef, V Vs s

where V is in volts peak. ThenS

WL2 Is wL Isin 8 2 sos 2y

Vs V

e f= coscos y

wL I-cos 8 si n 2y

sef = sin y

From the constants given

cos 0 1.0; sin 8 = 0

WLo = 2.5 ohms wL2 = 0.5 ohm

Rated power

PL= 1000 = 746 KW

Armature current at rated load is

I

s

= 746,000

/2 1000

= 527 amps peak = 373 amps RMS

Then

wL I wL IS0.186;

os-- = 0.932

5 s

Using the constants

-0.186 cos 2y - 0.932e =f cos y

-1 - 0.186 sin 2yef sin y

The use of trial-and-error to find a value of y that satisfies these two

equations simultaneously yields

Y = - 127* and ef = 1.48

Using the given constants we obtain

efVs (1.48)(/2)(1000)I = 1 r wM 150

14.0 amps

For Lf/Rf very large compared to a half period of the supply voltage the field

y



ROTATING MACHINES


current will essentially be equal to the peak of the supply voltage divided by

the field current; thus, the required value of Rf is

RA r (1000)= s= N(1000) 100 ohmsRf I 14.0

r

Part b

We can use (4.2.46) multiplied by the rotational speed w to write the

output power as

E V (X-X)V

P = WTe - sin 6 - q s sin 26

L Xd 2Xd q

where

Xd = w(Lo+L 2 ) = direct axis reactance

Xq= (L -L2 ) quadrature axis reactance

With the full-wave rectifier supplying the field winding we can express

E = WMIr

=Rf

Then WM V2 (Xd-X)V2

- sin 6 - sin 26= -PL RfXd 2X xq

Factoring out V2

yields

s

2= RM sin 6 - Xdq sin 2

L sV RfXd 2XdXq

Substitution of given constants yields

746 x 103 V2 [-0.500 sin 6 - 0.083 sin 26]

To find the required curve it is easiest to assume 6 and calculate the required

Vs, the range of 6 being limited by pull-out which occurs when

aPS= 0 = - 0.500 cos6 - 0.166 cos 26

The resulting curve of 6 as a function of Vs is shown in the attached graph.

Note that the voltage can only drop 15.5% before the motor pulls out

of step.

-3



ROTATING MACHINES


6o

PULL

OUr

40.

C0

)0 Zoo 400

Rviw-ruvEo00

vo/-rs J 4&s

0 raD

VoL4 4L E

1000

L

2C0

__

Although it was not required fo r this problem calculations will show that

operation at reduced voltage will lead to excessive armature current, thus,

operation in this range must be limited to transient conditions.

81



ROTATING MACHINES

PROBLEM 4.28

Part a

This is similar to part a of Prob. 4.24 except that now we are considering

a number of pole pairs greater than two and we are treating a generator. Consider

ing first the problem of pole pairs, reference to Sec. 4.1.8 and 4.2.4 shows that

when we define electrical angles ye and 6e as

Ye = P and 6e 6

where p is number of pole pairs (36 in this problem) and when we realize that the

electromagnetic torque was obtained as a derivative of inductances with respect to

angle we get the results

V Ef p(Xd-X ) V2

Te = sf sin 6 d q s sin 26

Xd e w2XdXq e

where Xd = w(Lo+L2 ) and Xq = m(Lo-L2), and, because the synchronous speed is w/p

(see 4.1.95) the electrical power output from the generator is

V E (X -X )V2

P = = - sin 6 + 2X sin 2&

p Xd e 2XdXq e

Next, we are dealing with a generator so it is convenient to replace Is

by -I in the equations. To make clear what is involved we redraw Fig. 4.2.5(a)

with the sign of the current reversed.

1 EALIA I,-.1 A 15



ROTATING MACHINES


Now, evaluating horizontal and vertical components of Vs we have

V cos 6 - wL2Is sin 2Ye = Ef sin Ye

-Vs sin 0 = WLO + wL 2 I s cos 2ye + Ef cos Ye

From these equations we obtain

oL2 I

cos 8 - sin 2ye

ef sin y

wL I wL2I s

-sin O cos 2ef = s s e

cos Y

whereEf MI

ef V , Vs 5

with V in volts peak

I s in amps peak

w is the electrical frequency

For the given constants

cos =-p.f. = 0.850 sin e = 0.528

wL2I s wLoIs0.200 =1.00

V Vs s

and

0.850 - 0.200 sin 2yeef = sin Y

-1.528 - 0.200 cos 2ye

ef = os ye

Trial-and-error solution of these two equations to find a positive value of

Ye that satisfies both equations simultaneously yields

ye = 147.50 and ef = 1.92

From the definition of ef we have

I = U = (1.92)()(0,000) = 576 ampsr wM (120) (7) (0.125)



ROTATING MACHINES


Part b

From Prob. 4.14 the definition of complex power is

VI* P+ jQss

where V and I are complex amplitudes.s B

The capability curve is not as easy to calculate for a salient-pole

machine as it was for a smooth-air-gap machine in Prob. 4.14. It will be easiest

to calculate the curve using the power output expression of part a

VE (Xd-X )V2

P = -- sin 6 + d q sin 26Xd e 2XdX q e

the facts that

P = V I cos 8as

Q = V I sin 6ss

and that Is is given from (4.2.44) and (4.2.45) as

V 2 V E 2Is = sin 6e) + s cos 6

sX e X e d

First, assuming operation at rated field current the power is

P = 320 x 106 sin 6 + 41.7 x 106

sin 26 watts.e e

We assume values of 6 starting from zero and calculate P; then we calculate Is

for the same values of 6 frome

= 11,800 (1.50 sin 6 ) + (cos -1.92) amps peaks e e

Next, because we know P, Vs, and Is we find 6 from

Pcos 8 =

VIss

From 6 we then find Q from

Q = V I sin 8.

This process is continued until rated armature current

I = /i 10,000 amps peak

is reached.

The next part of the capability curve is limited by rated armature

current which defines the trajectory

ROTATING MACHINES


rP2 2

where V and I are rated values.s a

For Q < 0, the capability curve is limited by pull-out conditions

defined by the condition

V Ef (X -X )V2dP s f d q a cos 26= 0 cos 6 + cos 26dd X e XX e

e d dq e

To evaluate this part of the curve we evaluate ef in terms of 6e from the power

and current expressions

PX (X -X )d (XdXq) sin 26

V2 2X es

ef f sin 6e

IX 2 X 2

ef = cos6e- (Isd - ( sin 6e)

s q

For each level of power at a given power factor we find the value of 6e that

simultaneously satisfies both equations. The.resulting values of ef and 6e are

used in the stability criterion

dP V2e (Xd-X )VsdP=a f cos 6 + d q cos 26 > 0dS ee

Xdde X

dX

qe-

When this condition is no longer met (equal sign holds) the stability limit is

reached. For the given constants

- 0.25 sin 26

167 x 106 e

ef = sin 6e

I 2ef

fcos 6

e-

11,800- (1.5 sin 6

e)2

dP e cos 6 + 0.5 cos 26 > 0dd f e e-

e

The results of this calculation along with the preceding two are shown on the

attached graph. Note that the steady-state stability never limits the capability.

In practice, however, more margin of stability is required and the capability in

the fourth quadrant is limited accordingly.



ROTATING MACHINES

FIELDP' I£.E1

lPOIAEREAL tso

(MW)

€

t

A~eMtArTeE

CU~I~

SirABlLTry LIMiT

ROTATING MACHINES

PROBLEM 4.29

Part a

For this electrically linear system the electric coenergy is

1 2We(v 2,6) = 2 Co(l + cos 26)v 1

+-1

Co(l + sin 26)v22

The torque of electric origin is

e3We(vl'V2'8)

Te e a 29

= c 'v2

cos 26 - v22

sin 26)

Part b

With v = V cos wt; v2 = Vo sin wt

Te = C V2(sin2 wt cos 2e - cos 2 wt sin 28)

Using trig identities

C V2

Te -~o2[cos 26 - cos 2wt cos 26 - sin 28 - cos 2wt cosZO]

C V2 C V2

T o- (cos 26 - sin 26 ) - 2 [cos(2wt-26) + cos(2wt + 26)]

Three possibilities for time-average torque:

Case I:

Shaft sitting still at fixed angle 6Case II:

Shaft turning in positive 6 direction

6 = Wt + y

where y is a constant

Case III:

Shaft turning in negative 6 direction

=-w t + 6

where 6 is a constant.

Part c

The time average torques are:

Case 1:6 = const.

C V

<Te> -- (cos 26 - sin 28)

ROTATING MACHINES


Case II: 6 = wt + yC V

2

<Te> o cos 2y

Case III: 0 = - wt + 62

C V

<Te>_ oo cos 262

PROBLEN 4.30

For an applied voltage v(t) the electric coenergy for this electrically

linear system is

W'(v,e) = (C + C1 cos 26)v 2

The torque of electric origin is then

aW'(v,6)Te = ee = CI sin 20 v2

For v = V sin wto

Te = - C V2 sin2wt sin 2e

1 o

C V2

Te = o2(sin 20- cos 2wt cos 26)

C1V2

CV 2

Te - o=sin 28 +-- o [cos(2wt-20) + cos(2wt+26)]2 4

For rotational velocity wm we write

8 = t +ym

and then

C V2

Te

12o sin 2(wmt + y)

C1V2

+1 - {cos[2(w-wm)t-2y] + cos[2(w+wm)t + 2y]}

This device can behave as a motor if it can produce a time-average torque for

w = constant. This can occur whenm

W = + Wm



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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS

PROBLEM 5.1

Part a

The capacitance of the system of plane parallel electrodes is

C = (L+x)dEo/s (a)

1 2and since the co-energy W' of an electrically linear system is simply -CCv

(remember v is the terminal voltage of the capacitor, not the voltage of the

driving source)

fe 9W' I dEo 2- --- v (b)

ax 22

The plates tend to increase their area of overlap.

Part b

The force equation is

d2x 1 dEo 2M =-Kx + v (

dtdt2 2 s

while the electrical loop equation, written using the fact that the current

dq/dt through the resistance can be written as Cv, is

dE

V(t) = R d-(L+x)- v]+ v (d)

These are two equations in the dependent variables (x,v).

Part c

This problem illustrates the important point that unless a system

involving electromechanical components is either intrinsically or externally

biased, its response will not in general be a linear reproduction of the

input. The force is proportional to the square of the terminal voltage, which

in the limit of small R is simply V2(t). Hence, the equation of motion is

(c) with 2V

v 2 u2(t) u= (t) o (1-cos 2wt) (e)

2 1

where we have used th e identity sin2t = (1-cos 2wt). For convenience

the equation of motion is normalized





d2x 2x = aul(t)(l-cos2wt)

d2 odt

where

2 = K/M ; a = V2 d E/4sMo 0 0

To solve this equation, we note that there are two parts to the particular

solution, one a constant

x= 2

o

and the other a cosinusoid having the frequency 2w. To find this second

part solve the equation

2dx

+22 x=- Reae

2jwt

dt2 o

for the particular solution

-acos 2wtx = W

_ 4 2

o

The general solution is then the sum of these two particular solutions and the

homogeneous solution t > 0

a a cos 2wtx(t) a cos 2t + A sinw t + Bcosw t (j)

2 2 _ 2 o o

o O

The constants A and B are determined by the initial conditions. At t=0,

dx/dt = 0, and this requires that A = 0. The spring determines that the initial

position is x = 0, from which it follows that

-B = a4w2/w

2 (W2 4w

2)

o o

Finally, the required response is (t > 0)

( -) cos o tx(t) = 2

cos 2wt o

1-( 2]0

0





Note that there are constant and double frequency components in this response,

reflecting the effect of the drive. In addition, there is the response

frequency w0 reflecting the natural response of the spring mass system. No

part of the response has the same frequency as the driving voltage.

PROBLEM 5.2

Part a

The field intensit ies are defined as in the figure

t, 2

Ampere's law, integrated around the outside magnetic circuit gives

2Nli = H1 (a+x) + H2 (a-x) (a)

and integrated around the left inner circuit gives

N1il - N2i2 H1 (a+x) - H3a(b)

In addition, the net flux into the movable plunger must be zero

0 = H1 - H2 + H3 (c)

These three equations can be solved fo r H1, H2 and H3 as functions of i1 and

12 . Then, the required terminal fluxes are

A, = NlPodW(H1+H2) (d)

X2 = N2p dWH 3 (e)

Hence, we have

N o dW

1 2 2 [il6aN 1 + i22N2x] (f)

12 = [ il2N1x + i22aN2 (g)2 2- 2 i 1 2 1 2 2

3a -x

Part b

To use the device as a differential transformer, it would be

excited at a frequency such that

2w-- << T (h)

where T is a period characterizing the movement of the plunger. This means

that in so far as the signal induced at the output terminals is concerned,

the effect of the motion can be ignored and the problem treated as though x

is a constant (a quasi-static situation, but not in the sense of Chap. 1).

Put another way, because the excitation is at a frequency such that (h) is

satisfied, we can ignore idL/dt compared to Ldi/dt and write

vdA2 w2N1N2

oodWxI o sin wt

2 dt 2_x 2(3a -x )

At any instant, the amplitude is determined by x(t), but the phase remains

independent of x(t), with the voltage leading the current by 90%. By

design, the output signal is zero at x=0O and tends to be proportional to x over

a range of x << a.

PROBLEM 5.3

Part a

The potential function which satisfies the boundary conditions along

constant 8 planes is

=vO (a)

where differentiation shows that Laplaces equation is satisfied. The constant

has been set so that the potential is V on the upper electrode where 8 = i,

and zero on the lower electrode where 0 = 0. Then, the electric field is

- 1 3 _-_ v

=E V =-i0 r ;3E 6

Yri

(b)

Part b

The charge on the upper electrode can he written as a function of (V,p)

by writing

b V DEVS= DE -d r - I(T) (c)

0 fap I





Part c

Then, the energy stored in the electromechanical coupling follows as

W = Vdq = dq q) (d)

Deoln( ) DE ln( )

and hence

e aW 1 q22 I 2Doln( b)

(e)

Part d

The mechanical torque equation for the movable plate requires that the

inertial torque be balanced by that due to the torsion spring and the electric

field2 2

Jd29 1 22 a(*oo 1 2 b

dt Dc ln()

The electrical equation requires that currents sum to zero at the current node,

and makes use of the terminal equation (c).

dO dq + d qý g)dt dt dt ln (

o a

Part e

With G = 0, (t) = q(t). (This is true to within a constant, corresponding

to charge placed on the upper plate initially. We will assume that this constant

is zero.) Then, (f) reduces to

2 0Od2+ a 1 o

d2 J o (l+cos 2wt) (h)dt JDE

on(-)

a2 1 is equation has a

where we have used the identity cos t =-2(1 + cos 20t). This equation has a

solution with a constant part 2

1 o41SaDE ln ()

(i)o a

and a sinusoidal steady state part

2Q cos 2wt

J4Dol n(b)[ - (2w)2

0 Ja





as can be seen by direct substitution. The plate responds with a d-c part and a

part which has twice the frequency of the drive. As canbe

seen from the

mathematical description itself, this is because regardless of whether the upper

plate is positive or negative, it will be attracted toward the opposite plate

where the image charges reside. The plates always attract. Hence, if we wish

to obtain a mechanical response that is proportional to the driving signal, we

must bias the system with an additional source and.used the drive to simply

increase and decrease the amount of this force.

PROBLEM 5.4

Part a

The equation of motion is found from (d) and (h) with i=Io, as given in

the solution to Prob. 3.4.

2

d2 12 (N vaw)dx 1 2 o(a)

M - = Mg - Io da(o da 2

dt (- + x)

Part b

The mass M can be in static

equilibrium if the forces due to the

field and gravity just balance,

f = f

g

or

=1 2 (N2 oaw)Mg =. 2

2 o da 2(ý + x) Y X

A solution to this equation is shown

graphically in the figure. The equilibrium is statically unstable because if

the mass moves in the positive x direction from xo, the gravitational force

exceeds the magnetic force and tends to carry it further from equilibrium.

Part c

Because small perturbations from equilibrium are being considered it is

appropriate to linearize. We assume x = x +x' (t) and expand the last term

in (a) to obtain





-- I2 (N2V aw)

+ I2 (N

2oaw)

2 o

(-

da ++

X)

2 +

o (--

a +

Xo)

3x' + ... (c)

b o b 0

(see Sec. 5.1.2a). The constant terms in the equation of motion cancel out by

virtue of (b) and the equation of motion is

2d~x 2 I12 (N2 oaw)d x 2 x' = O; a - (d)dt 2

(-+ x )M

Solutions are exp + at, and the linear combination which satisfies the given

initial conditions is

Vx = ea- ea] (e)

PROBLEM 5.5

Part a

For small values of x relative to d, the equation of motion is

2 QOd2x o 1 2x 1 2x

M 2 [ (a)dt d

2d3 d d'

which reduces to

-d2x + 2x = 0 where w

2=

Qo_1(b)

dt 2 0 0 N•ed 3

The equivalent spring constant will be positive if

QolQQ > 0 (c)

rcd

and hence this is the condition fo r stability. The system is stable if the

charges have like signs.

Part b

The solution to (b) has the form

x = A cos w t + B sin w t (d)o o

and in view of the initial conditions, B = 0 and A = x .

95



LUMPED-PARAMETER ELECTRCOECHANICAL DYNAMICS

PROBLEM 5.6

Part a

Questions of equilibrium and stability are of interest. Therefore, the

equation of motion is written in the standard form

M d 2

dtx V (a)

ax

where

V = Mgx - W' (b)

Here the contribution of W' to the potential is negative because Fe = aw'/ax.

The separate potentials are shown in the figure, together with the total

potential. From this plot it is clear that there will be one point of static

equilibrium as indicated.

Part b

An analytical expression for the point of equilibrium follows by setting

the force equal to zero

2L Xav 2LX

~ Mg + 0 (c)

3x b 4

Solving for X, we have

4 1/3

x =- [ ] (d)

2L I0

Part c

It is clear from the potential plot that the equilibrium is stable.

PROBLEM 5.7

From Prob. 3.15 the equation of motion is, for small 0

J K+ DN2 In( )I2 46 (a)

dt2 2 o o )3

Thus, the system will have a stable static equilibrium at 0 = 0 if the

effective spring constant is positive, or if

221 DN b

K > - in( ) (b))3 a o



LUMPED-PARAMETER ELECTROMECHANICAL DYANAMICS

I

IV II

1/

/1/I

(~)

.a\e s +C ýc

k-,ý , "v W

Figure for Prob. 5.6



LUMPED-PARAMETER ELECTRCMECHANICAL DYNAMICS

PROBLEM 5.8

Part a

The coenergy is

W' = 1 )1(iO,x)di' + 12 X (il, i ' ,x)di' (a)

o o

which can be evaluated using the given terminal relations

' = [ i + Mil2 + L2i/( + x 3 (b)

T-1 1 M1 2 2 L2i2

If follows that the force of electrical origin is

fe = aW' 3 2 2i/

fe a[Llil+2Mil2

+2i/( + ) (c)

Part b

The static force equation takes the form

_fe Mg (d)

or, ith i2=02 and il1=I,

3 L1 1

2a X 4 Mg

[1 + --o

Solution of this equation gives the required equilibrium position X

1 1/4Xo L I-a

= [

2a

_

Mg

_ ] - 1 (f)

Part c

For small perturbations from the equilibrium defined by (e),

d2 x, 6L 12

x'M

ox 6L

X 5= f(t) (g)

dt 2 + o)a

where f(t) is an external force acting in the x direction on M.

With the external force an impulse of magnitude I and the mass initially

at rest, one initial condition is x(O) = 0. The second is given by integrating

the equation of motion form 0 to 0+

+ + +

dt 0oddtl)dtt - constant f 0x'dt I. 0= .- 0(t)dt (h)

0 0 0 0a

The first term is the jump in momentum at t=0, while the second is zero if

x is to remain continuous. By definition, the integral on the right is 0

Hence, from (h) the second initial condition is





MA ) =

Mo () = 0o x

In view of these conditions, the response is

(e - e )X'(t)

I2

where 117.x o 5~= aLl2aM'r

a = LI2/a2M (1 + o

Part d

With proportional feedback through the current 12, the mutual term in

the force equation makes a linear contribution and the force equation becomes

d2x' 6L12 "4tM- 2 X - ]x ' = f(t)"

0 dt a2 ( 1 + ao-)5- a

The effective spring constant is positive if

? X%aI > 2L I /a (1 + -- ) M

1 a

and hence this is the condition for stability. However, once initiated

oscillations remain undamped according to this model.

Part e &5ee \• )

With a damping term introduced by the feedback, the mechanical

equation becomes

d2x' 3MI4 dx'

M + - + K x' = f(t)

Sdta t e

where

K 3MIa 6L 12 3 3:,4 -GLtI

2suX 5

a

This equation hias soluttns of the form exp st, where substitution shows that

e a the fho

s = 3MIBM + (3MVI61 e (n)2aM - 2aMo, Mo

For the response to decay, K must be positive (the system must be stable withe

out damping) and 6 must be positive.

23~4 n~IT> I-Ohc-

Ic~ic~P tFo

'7'7



LUMPED-PARAMETER ELECTRCMECHANICAL DYNAMICS

PROBLEM 5.9

Part a

The mechanical equation of motion is

Md 2 x

= K(x-£ )-B+

fe (a)

dt S2 o dt

Part b

where the force fe is found from the coenergy function which is (because

1 2 1 32the system is electrically linear) W' = Li2 = Ax i

fe = 3W'= 3 Ax 2 (b)f - = Ax i (b)

ax 2

Part c

We can both find the equilibrium points X and determine if they are stable0

by writing the linearized equation at the outset. Hence, we let x(t)=X +x'(t)

and (a) and (b) combine to give

d2x' dx' 3 2 2

M - K(Xo-Po)-Kx' - B + - AI (X + 2X x') (c)dtt

2 0 0 dt 2 o + 0

With the given condition on 1o, the constant (equilibrium) part of this equation

is 3X2

X - o (d)o o 16Z

0

which can be solved for X /Z, to obtaino o

x 12/3

o 1/3 (e)

That is, there are two possible equilibrium positions. The perturbation part

of (c) tells whether or not these are stable. That equation, upon substitution

of Xo and the given value of Io, becomes

d2x' 3/2 dx'

M_2- -K[l- (

1/2)]x' - B

dt(f)

dt

where the two possibilities correspond to the two equilibrium noints. Hence,

we conclude that the effective spring constant is positive (and the system is

stable) at XO/k = 4/3 and the effective spring constant is negative (and hence

the equilibrium is unstable) at X /0o = 4.

Part d

The same conclusions as to the stability of the equilibrium noints can be

made from the figure.





T

Consider the equilibrium at Xo = 4. A small displacement to the right makes

the force fe dominate the spring force, and this tends to carry the mass

further in the x direction. Hence, this point is unstable. Similar arguments

show that the other point is stable.

PROBLEM 5.10

Part a

The terminals are constrained to constant potential, so use coenergy found

from terminal equation as

W' = qdv = -4 (l + cos 2e)V22o o

Then, since Te = aW'/ae and there are no other torques acting on the shaft, the

total torque can be found by taking the negative derivative of a potential

V =-W', where V is the potential well. A sketch of this well is as shown in

the figure.



JUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS


IV

SSa~b\C

~c~-.$ c~;~lb~;a

Here it is clear that there are points of zero slope (and hence zero torque

and possible static equilibrium) at

e= o0o 7, 3r

Part b

From the potential well it is clear that the first and third equilibria

are stable, while the second and fourth are unstable.

PROBLEM 5.11

Part a

From the terminal pair relation, the coenergy is given by

Wm (ii,i2'e)= (Lo+M cos 20)il + (Lo-M cos 20) 2 + M sin 2i ili 2

so that the torque of electrical origin is

Te = M[sin 20(i 2-i1 ) + 2 cos 26 ili 22 1 1 21

Part b

For the two phase currents, as given,

2 2 12i _ 1 I cos 2w t2 1 s

i1 2 1 sin 2w t1 2 s

so that the torque Te becomes





Te = MI2[-sin 20 cos 2w t + sin 2wst cos 20] (d)

or

Te = MI2sin(2 s t- 20) (e)

Substitution of 6 w t + 6 obtainsm

Te 2

T= - MI sin[2(wm-w)t + 2] (f)

and for this torque to be constant, we must have the frequency condition

Wm=W

s(g)

under which condition, the torque can be writtenas

Te = - MI2sin 26 (h)

Part c

To determine the possible equilibrium angles 60, the perturbations and

time derivatives are set to zero in the mechanical equations of motion.

T = MI 2sin 26 (i)

o o

Here, we have written the time dependence in a form that is convenient if

cos 260 > 0, as it is at the points marked (s) in the figure. Hence, these

points are stable. At the points marked (u), the argument of the sin function

and the denominator are,imaginary, and the response takes the form of a sinh

function. Hence, the/equilibrium points indicated by (u) are unstable.

Graphical solutions of this expression are shown in the figure. For there

to be equilibrium values of 6 the currents must be large enough that the torque

can be maintained with the rotor in synchronism with the rotating field.

(MI > T )

MAI 2r

76 maa

VIA





Returning to the perturbation part of the equation of motion with wm = us,

J 2 (Wt + 6 + 6') = T + T' - MI sin(26 + 26') (j)

dt2

m o o odt

linearization gives

J A-+ (2MI2

cos 26~)6' = T' (k)

dt 2

where the constant terms cancel out by virtue of (i). With T' = Tu o(t) and

initial rest conditions,the initial conditions are

* ( 0+

) = -o (1)dt J

6'(0+

) = 0 (m)

and hence the solution for 6'(t) is

S2MI 2 cos 266'(t) = o sin o t (n)

2MI2cos

26

PROBLEM 5.12

Part a

The magnitude of the field intensity\ (H) in the gaps is the same. Hence,

from Ampere's law,

H = Ni/2x (a)

and the flux linked by the terminals is N times that passing across either

of the gaps.

~ adN2

= i = L(x)i (b )2x

Because the system is electrically linear, W'(i,x) =1

Li

22 , and we have.

2

fe = N2ad=o i2

(c)ax 2

4x

as the required force of electrical origin acting in the x direction.

Part b

Taking into account the forces due to the springs, gravity and the

magnetic field, the force equation becomes

104



LUMPED-PARAMETER ELECTRGCECHANICAL DYNAMICS


2 N2ado

M 2 = - 2K x + Mg 2 i + f(t) (d)dt 4x 2

where the last term accounts for the driving force.

The electrical equation requires that the currents sum to zero at the

electrical node, where the voltage is dA/dt, with X given by (b).

I adN

IR dt

[ •2x

i] + i (e)

Part c

In static equilibrium, the electrical equation reduces to i=I, while

the mechanical equation which takes the form fl f2 is satisfied if

2 2N2adj o2

-2KX + M g = (f)4X

Here, f2 is the negative of the force of electrical origin and therefore

(if positive) acts in the - x direction. The respective sides of (f) are

shown in the sketch, where the points of possible static equilibrium are

indicated. Point (1) is stable, because a small excursion to the right makes

f2 dominate over fl and this tends to return the mass in the minus x direction

toward the equilibrium point. By contrast, equilibrium point (2) is

characterized by having a larger force f2 and fl for small excursions to the

left. Hence, the dominate force tends to carry the mass even further from the

point of equilibrium and the situation is unstable. In what follows, x = X

will be used to indicate the position of stable static equilibrium (1).





Part d

If R is very large, then

i : I

even under dynamic conditions. This approximation allows the removal of

the characteristic time L/R from the analysis as reflected in the

reduction in the order of differential equation required to define the

dynamics. The mechanical response is determined by the mechanical

equation (x = X + x')

M-d22 xx'

= - 2Kx' +N adpo

0I2

x' + f(t) (g)dt

22X 3

where the constant terms have been balanced out and small perturbations are

assumed. In view of the form taken by the excitation, assume x = Re x ejet

.and define Ke E 2K - N2adoI2/2X

3Then, (g) shows that

S= f/(Ke-0M) (h)

To compute the output voltage

S p 0a dN2 1 dx'

o dt i= 22 dti=I =i

orupor adN2 I

=2 x (0)o2X

Then, from (h), the transfer function is

2v w0 adN I

o oj (k)

f 2X2(Ke2jM)

PROBLEM 5.13

Part a

The system is electrically linear. Hence, the coenergy takes the

standard form

W'1L 1

2 +L ii + 11

L 122 (a)2 111 1212 2 222(a)

and it follows that the force of electrical origin on the plunger is

Sx= i + i l22 + 2i2 (b)

ax 2 1 x 1 2 3x 2 2 ax





which, for the particular terminal relations of this problem becomes

2

2-if ilix 2 i

fe L { (+ x_ 1 2x 2 ( x (c)o d d d d d dc)

Finally, in terms of this force, the mechanical equation of motion is

2d2--2 = -Kx - B T- + fe (d)

dtdt

The circuit connections show that the currents i and 12 are related to the

source currents by

i = I + i (e)1 =o -i

Part b

If we use (e) in (b) and linearize, it follows that

4L I 4L 12fe oo oo (f)

d d d

and the equation of motion is

dx dx 22 + a - +wx = - Ci (g)dt dt o

where

4L 12ao

=[K + ]/M

2o

a = B/M

C = 4L /dM

Part c

Both the spring constant and damping in the equation of motion are

positive, and hence the system is always stable.

Part d

The homogeneous equation has solutions of the form ept where

p2

+ ap +2

= 0 (h)0

or , since the system is underdamped

a 22

a) p = - 2 + J - (i)

2 - 02 p



LUMPED-PARAMETER ELECTROM4ECHANICAL DYNAMICS

PROBLE~ 5.13 (Continued)

The general solution is

CI t

x(t) = - + e [A sin w t + D cos w t] (j)2 p p

o

where the constants are determined by the initial conditions x(O) = 0 and

dx/dt(O) = 0

CI tCI

D =--o

; A =o

(k)w 2w wo po

Part e

With a sinusoidal steady state condition, assume x = Re x e and write

i(t) = Re(-jI )ej t and (g) becomes

- 2 2 )x(-w + jwa + = Cj (1)

Thus, the required solution is

RejCI et

x(t) ( 2 2 (m)

0

PROBLEM 5.14

Part a

From the terminal equations, the current ii is determined by Kirchhoff's

current law

di

G L di+ i = I + CMI sin Pt (a)G1 dt 1 2

The first term in this expression is the current which flows through G because

of the voltage developed across the self inductance of the coil, while the last

is a current through G induced bhv the rotational motion. The terms on the right

are known functions of time, and constitute a driving function for the linear

equation.

Part b

We can divide the solution into particular solutions due to the two driving

terms and a homogeneous solution. From the constant drive I we have the solution

i I = I (b)

Because sin Pt = Re(-jej t), if we assume a particular solution for the

) we havesinusoidal drive of the form i1 = Re(IeI ), we have





1

1 (jDGL1

+ 1) = - J~GMI2

(c)

or, rearranging

-OGMI 2 (GCL1 + j)

1 +( 1) 2 (d)

We now multiply this complex amplitude by ejot and take the real .part to

obtain the particular solution due to the sinusoidal drive

-GMI2l1 1 2 (QGL1 cos Pt - sin Qt) (e)

1+(PGLI)

The homogeneous solution is

-t/GL 1

t1 = Ae (f)

and the total solution is the sum of (b), (e) and (f)with the constant A

determined by the initial conditions.

In view of the initial conditions, the complete solution for il, normalized

to the value necessary to produce a flux equal to the maximum mutual flux, is

then

Llil 1e LMI Q(tGLI) 1MI 22 1+(GL ) 2

+ GL1R

2 (sin t - GGLL1I

L+(QG2L )11

cos Qt) +MI2

(g)

Part c

The terminal relation is used to find the flux linking coil 1

l GLI) 2 LI 1

MI2 I+I(GL ) M 2 t

GLIQ LIG~1R cos Rt 1

2 ( L 2 MI1+(QGLI) 1+( GL1) 2

The flux has been normalized with respect to the maximum mutual flux (MI2).

Part d

In order to identify the limiting cases and the appropriate approximations

it is useful to plot (g) and (h) as functions of time. These equations contain

two constants, QGL 1 and L I/MI 2 . The time required for one rotation is 2r/S and

GL1 is the time constant of the inductance L1 and conductance G in series. Thus,

QGL1 is essentially the ratio of an electrical time constant to the time required

for the coil to traverse the applied field one time. The quantity MI2 is the

maximum flux of the externally applied field that can link the rotatable coil and

I1 is the self flux of the coil due to current I acting alone. Thus, I1I/MI 2

is' the ratio of self excitation to mutual excitation.

To first consider the limiting case that can be approximated by a current

source we require that

L1IQGL <<

11 and GL << MI

1 MI2(i)

To demonstrate this set

LIWGL = 0.1 and -- = 1 (j)

1 MI

and plot current and flux as shown in Fig. (a). We note first that the

transient dies out very quickly compared to the time of one rotation. Further

more, -the flux varies appreciably while the current varies very little compared

to its average value. In the ideal limit (GqO) the transient would die out

instantaneously and the current would be constant. Thus the approximation of

the situation by an ideal current-source excitation would involve a small

error; however, the saving in analytical time is often well worth the decrease

in accuracy resulting from the approximation.

Part e

We next consider the limiting case that can be approximated by a constant-

flux constraint. This requires that

QGL 1 >> 1 (k)

To study this case, set

CGL1 = 50 and I = 0 (1)

The resulting curves of flux and current are shown plotted in Fig. (b).

Note that with this constraint the current varies drastically but the flux

pulsates only slightly about a value that decays slowly compared to a rotational

period. Thus, when considering events that occur in a time interval comparable



(-O) 4-U

CA~)

LUMPED-PARAMETER ELECTR(OMECHANICAL DYNAMICS


with the rotational period, we can approximate this system with a constant-flux

constraint. In the ideal,limiting 6ase, which can be approached with super

conductors, G-m and X1 stays constant at its initial value. This initial value

is the flux that links the coil at the instant the switch S is closed.

In the limiting cases of constant-current and constant flux constraints

the losses in the electrical circuit go to zero. This fact allows us to take

advantage of the conservative character of lossless systems, as discussed in

Sec. 5.2.1.

Part f

Between the two limiting cases of constant-current and constant flux

constraints the conductance G is finite and provides electrical damping on

the mechanical system. We can show this by demonstrating that mechanical

power supplied by the speed source is dissipated in the conductance G. For

this purpose we need to evaluate the torque supplied by the speed source.

Because the rotational velocity is constant, we have

Tm= - Te (m)

The torque of electrical origin Te is in turn

aW'(il, i 22, )Te = (n)

Because the system is electrically linear, the coenergy W' is

W' Li + M i + L2 (o)2 1 1 1 2 2 2 2 Co)

and therefore,

Te = - M i 12 sin 6 (p)

The power supplied by the torque Tm

to rotate the coil is

Pin - T d = I sin Qt (W)il2

Part g

Hence, from (p) and (q), it follows that in the sinusoidal steady state

the average power <P. > supplied by the external toraue isin

<Pin>

=1 (r)

in 2




to

-Ir tl

c,,

O

-0

PROBLFM 5.14 (Continued)

This power, which is dissipated in the conductance G, is plotted as a function

of ~2GL 1 in Fig. (c). Notethat because

0and L

1are used as normalizing

constants, ?GL 1 can only be varied bhv varving G. Note that for both large and

small values of fGT.1 the average mechanical power dissipated in G becomes small.

The maximum in <Pin >occurs at GCL1 = 1.

PROBLEM 5.15

Part a

The coenergy of the capacitor is

= C(x) V2 = 1 (EA )V

2

e 2 2 ox

The electric force in the x direction is

aW' EAe 1 o 2

e ýx 2 2x

If this force is linearized around x = xo, V = V

f (x)e

=01

2

1 AV'o o

2

1)2E AV v

o o0

2+

2E AV x

o o0 0

3x x x

O O O

The linearized equation of motion is then

AV2

dx + (K-E

)xo) -c A

0V v + f(t) -ý0 ' = V

dt 3 2 ox x0 0

The equation for the electric circuit is

dV + R -6 (C(x)V) = V

Part b

We can keep the voltage constant if

R -- 0

In this case AV

B dx + K'x = f(t) = F ul(t); K' = K 0 3x

The particular solution is

=x(t) F/K'

The natural frequency S is the solution to

SB +K'x = 0 $ = - K'/B

Notice that sinceZ

E AV

X'/B = (K- 3)/B X(t)x0 F

Sr- /dthere is voltage V above which the

system is unstable. Assuming V isO

less than this voltage t

-x(t) = F/K' (1-e (K

/B)t)

Now we can be more specific about the size of R. We want the time

constant of the RC circuit to be small compared to the "action time" of the

mechanical system

RC(xo) << B/K'

B

R << BK'C(xo)

Part c

From part a we suspect that

RC(xo) >> Tmech

where Tmech can be found by letting R + m. Since the charge will be constant

d = 0 q = C(x )V = C(X +X)(V +V)dt 00 0 0

dC- C(xo)V + C(xo)v + Vo 4-c (xo)x

V Vx EAv (o

dC- (x

) x - +oo

ox Vxx

C(x) dxxo)X E A 2 oxo

Using this expression for induced v, the linearized equation of motion

becomes

SAV2

Adxo o o0 2

B + (K- )x - V x +f(t)dt 3 3 o

x xo o

dxB dx + Kx = f(t)

dt





The electric effect disappears because the force of a capacitor with

constant charge is independent of the plate separation. The solutions are the

same as part (a) except that K' = K. The constraint on the resistor is then

R >> 1 B/KC(xo)

PROBLEM 5.16

We wish to write the sum of the forces in the form

f f + faV

(a)1 2 3x

For x > 0, this is done by making

1 2V Kx + Fx (b)

2 o

as shown in the figure. The potential is symmetric about the origin. The largest

value of vo that can be contained by the potential well is determined by the peak

value of potential which, from (b), comes at

x = Fo/K (c)

where the potential is

V = 1 F2/K (d)2 o

Because the minimum value of the potential is zero, this means that the kinetic

energy must exceed this peak value to surmount the barrier. Hence,

SMv2 I F2/K (e)2 o 2 o

or F2

vo= (f)





PROBLEM 5.17

Part a

The electric field intensities

defined in the figure are

E2 = (v2-v1)/(d-x)/

I E\

E1

= v,/(d+x)1

Hence, the total charge on the

respective electrodes is

q= S

`2

A2E o

v 1 [. 0+Vl[d+x

+

AIC (v2-v

d-x

A

)

o

0-x]-

v2A1E o

od-x

Part b

Conservation of energy requires

vldq 1 + v2dq2 = dW + fedx




and since the charge q1 and voltage v2 are constrained, we make the

transformation v2dq2 = d(v2q2)-q2 dv2 to obtain

v1 dql-q 2 dv 2 = dW" + fedx (f)

It follows from this form of the conservation of energy equation that

fe Wfe = - • and hence W" H U. To find the desired function we integrate

(f) using the terminal relations.

U = W"= dql - q2dv 2 (g)

The integration on q1 makes no contribution since ql is constrained to be

zero. We require v2(ql=0,v 2) to evaluate the remaining integral

v2A1i Io1 1 (h)

q 2 (q 1 0,v 2) d-x 1- A2(dx ) (h)

SAl(d+x)

Then, from (g),

U 1 0 1 o 1U V 1 (i)

2 d-x A2 (d-x)

A(d+x) 1

PROBLEM 5.18

Part aI

Because the two outer plates are X

constrained differently once the switch

is opened, it is convenient to work in

terms of two electrical terminal pairs,

defined as shown in the figure. The

plane parallel geometry makes it

straightforward o compute the

terminal relations as being those for

simple parallel plate capacitors, with

no mutual capacitance.

ql 1 x (a)lEoA/a+

q2 V2 oA/a-x (b)

+,

~o 2)~ '4 00





Conservation of energy for the electromechanical coupling requires

v1dql+ v2 dq2 = dW + fedx (c)

This is written in a form where q1 and v2 are the independent variables by

using the transformation v2dq2 = d(v2q2)-q2dv2 and defining W"qW-v2q 2

v1dq1 - 2dv2 dW" + fedx (d)

This is done because after the switch is opened it is these variables that

are conserved. In fact, for t > 0,

v2 = V and (from (a))ql = VoeoA/a (e)

The energy function W" follows from (d) and the terminal conditions, as

W" = vldql- fq2 dv 2 (f)

or

c Av2

1 (a+x) 2 1 oAV2q -(g)

2 cA q1 2 a-xo

Hence, for t > 0, we have (from (e))

AV2 E AV2

1 (a+x) 1 oAV

2 2 o A 2 a-xa

Part b

e aW "The electrical force on the plate is fe W" Hence, the force

equation is (assuming a mass M for the plate)

2, Eo AV2

E AV2

dx 1 o o 1 o oM - Kx + (i)

dt a (a-x)

For small excursions about the origin, this can be written as

2 cA V2

EAV2 cA V2

dx 1 o 0 o o 01o oM-

2-Kx-

2 2+

2 2+ 3 x (j)

dt a a a

The constant terms balance, showing that a static equilibrium at the origin

is possible. Then, the system is stable if the effective spring constant

is positive.

K > c AV2/a3 (k)0 0

Part c

The total potential V(x) for the system is the sum of W" and the

potential energy stored in the springs. That is,





2

1 2 1 (a+x) AV 1E AV

o

2 2 2 o o 2 a-xa

2 2 EAV2

aKK1 oo x 1

2 ( 2 a a x(1- )

a

This is sketched in the figure for a2K/2 = 2 and 1/2 c AV2/a = 1. In additiono o

to the point of stable equilibrium at the origin, there is also an unstable

equilibrium point just to the right of the origin.

PROBLEM 5.19

Part a

The coenergy is

W' Li = i /1 - -42 2 ao

and hence the fbrce of electrical origin is

'

e dw4f = 2L iL/a[l

x o a

Hence, the mechanical equation of motion, written as a function of (i,x) is

S21,2

2 2L _id x

M = - M g +

dt a[1- -aa





while the electrical loop equation, written in terms of these same variables

(using the terminal relation for X) is

VVo

+ v = Ri +dt

- [(1- )(d)

(d)

a

These last two expressions are the equations of motion for the mass.

Part b

In static equilibrium, the above equations are satisfied by (x,v,i) having

the respective values (Xo,VoIo). Hence, we assume that

x = Xo + x'(t): v = Vo + v(t): i = Io + i'(t) (e)

The equilibrium part of (c) is then

2L 12 X 5

0 - Mg + a o/(1 - o) (f)a a

while the perturbations from this equilibrium are governed by

2x 10 L 12 x' 4 LI i'

Md

2+

X 6+

X 5 (g)

a (l- -) a(l- 0- )a a

The equilibrium part of (d) is simply Vo = I R, and the perturbation part is

L di* 4 LI

v =Ri ' + 0 d+

00

(h)X 4 dt X 5 dt

[1- •-1] al- -o]a a

Equations (g) and (h) are the linearized equations of motion for the system which

can be solved given the driving function v(t) and (if the transient is of interest)

the initial conditions.

PROBLEM 5.20

Part a

The electric field intensities, defined as shown, are

E1 = (V 1-V2 )/s; E2 = v2/s (a)

121





In terms of these quantities, the charges are

= q 1 Eo( - x)dE1 ; q2 - o( -x)dE + o( + x)dE 2 (b)

Combining (a) and (b), we have the required terminal relations

q = V1C11 - v2C12 (c)

q2 =V 1 12+

V2C22

whereEd E ado a o

C11 = - (- x); C22ii s 22 s

Edo a

C12 s (

For the next part it is convenient to write these as q1(vl,q2) and v2(v ,q 2).2

1 v1 [C1 1 C2 2 C2q 22 22

q2 C12 (d)v + v

2 C 1 C22 22

Part b

Conservation of energy for the coupling requires

v1dql + v2dq2 = dW + fedx (e)

To treat v1

and q2 as independent variables (since they are constrained to be

constant) we let vldq 1 = d(vlql)-q dvl, and write (e) as

-ql dv1 + v2dq2 = - dW" + fe dx (f)

From this expression it is clear that fe = aW"/,x as required. In particular,

the function W" is found by integrating (f)

W" = o l(,O)dv' - v 2 (Vo,q)dq2 (g)

o o

to obtain

C2 2 V OC

= 1 V2[C _ Q o 12 (h)2 o 11 C1 2 ] 2C2222 C22

Of course, C1 1, C22 and C12 are functions of x as defined in (c).




PROBLEM 5.21

Part a

The equation of motion as developed in Prob. 3.8 but with I(t)=Io=constant,

is

2 I L2 1J

dt2

d - mL2

(1-cos 6) sine (a)

dt 2

This has the required form if we define

ILm 1 2

V L (cos 0 + sin 0) (b)

as can be seen by differentiating (b) and recovering the equation of motion. This

potential function could also have been obtained by starting directly with the

thermodynamic energy equation and finding a hybred energy function (one having

il' X2,6 as independent variables). See Example 5.2.2 for this more fundamental

approach.

Part b

A sketch of the potential well is as shown below. The rotor can be in

stable static equilibrium at e = 0 (s) and unstable static equilibrium at

S= r(u).

Part c

For the rotor to execute continuous rotory motion from an initial rest

position at 0 = 0, it must have sufficient kinetic energy to surmount the peak

in potential at 8 = W. To do this,

2 2IL211 j (Lmo

2Jt

dt-- L

•> (c)

- c




PROBLEM 5.22

Part a

The coenergy stored in the magnetic coupling is simply

W'= Lo(l + 0.2 cos 0 + 0.05 cos 268)2

(a)

Since the gravitational field exerts a torque on the pendulum given by

T = (-Mg X cose) (b)p ae

and the torque of electrical origin is Te = ~W'/~8, the mechanical equation of

motion is

d ro[t 2 2 + V =0 (c)

where (because I 2Lo 6MgZ)

V = Mgt[0.4 cos e - 0.15 cos 20 - 3]

Part b

The potential distribution V is plotted in the figure, where it is evident

that there is a point of stable static equilibrium at 0 = 0 (the pendulum

straight up) and two points of unstable static equilibrium to either side of

center. The constant contribution has been ignored in the plot because it is

arbitrary.

strale

I \

C/h ~ta I




PROBLEM 5.23

Part a

The magnetic field intensity is uniform over the cross section and equal

to the surface current flowing around the circuit. Define H as into the paper

and H = i/D. Then X is H multiplied by Uo and the area xd.

p xd

-- i (a)

The system is electrically linear and so the energy is W X2 L. Then, since

fe = _ aW/ax, the equation of motion is

d2x 1 A2DM d 2 x

dt2 = f f - Kx + D (b)

2 2

Part bLet x = X + x'where x' is small and (b) becomes approximately

d2x' 1 A2D A2Dx'

Mdt

2 x2 = -K X - Kx' + (c)

o 2 2d o X3 d00

The constant terms define the static equilibrium

1 A2D 1/3X° = [ ]K- (d)

o

and if we use this expression for Xo, the perturbation equation becomes,

d22x'

M = -Kx' - 2Kx' (e)dt

2

Hence, the point of equilibrium at Xo as given by (d) is stable, and the magnetic

field is equivalent to the spring constant 2K.

Part c

The total force is the negative derivative with respect to x of V where

1 2 1 A2DV = Kx + A-D (f)

2 2jixd

This makes it possible to integrate the equation of motion (b) once to obtain

d= + 2 (E-V) (g)dt -M

The potential well is as shown in figure (a). Here again it is apparent that

the equilibrium point is one where the mass can be static and stable. The constant

of integration E is established physically by releasing the mass from static





positions such as (1) or (2) shown in Fig. (a). Then the bounded excursions of

the mass can be pictured as having the level E shown in the diagram. The motions

are periodic in nature regardless of the initial position or velocity.

Part d

The constant flux dynamics can be contrasted with those occurring at

constant current simply by replacing the energy function with the coenergy

function. That is, with the constant current constraint, it is appropriate to find

the electrical force from W' = Li2 ' where fe = W'/ax. Hence, in this case

1 2 1 oxd 2 (h)2 2 D

A plot of this potential well is shown in Fig. (b). Once again there is a point

X of stable static equilibrium given by

X 1d 2

(i)o 2 DK

However, note that if oscillations of sufficiently large amplitude are initiated

that it is now possible for the plate to hit the bottom of the parallel plate

system at x = 0.

PROBLEM 5.25

Part a

Force on the capacitor plate is simply

wa2 2fe 3W' 3 1 o (a)

f • x x 21 x

due to the electric field and a force f due to the attached string.

Part b

With the mass M1 rotating at a constant angular velocity, the force fe

must balance the centrifugal force Wm rM1 transmitted to the capacitor plate

by the string.

wa2E V2

1 oo = 2 (b)2 2 m 1

or \Ia a2 V2

= 0 (c)m 2 £3M

1

where t is both the equilibrium spacing of the plates and the equilibrium radius

of the trajectory for M1.




(0,)

OA x---aco s%~ 0

oY\

V~xr

(b)





Part c

The e directed force equation is (see Prob. 2.8) for the accleration

on a particle in circular coordinates)

d2eM 1[r d2 + 2

dr d6dt dt = 0 (d)

dt

which can be written as

d 2 dO

dt [M1r d-1= 0 (e)

This shows that the angular momentum is constant even as the mass M1 moves in and

out

2 de 2M1d m = r = . constant of the motion (f)

This result simply shows that if the radius increases, the angular velocity must

decrease accordingly

de 2dt 2 ()

r

Part d

The radial component of the force equation for M1 is

2 2

Ml[d - r-) ]= - f (h)

dt

where f is the force transmitted by the string, as shown in the figure.

S(

i grv\,

The force equation for the capacitor plate is

Mdr e(i)dt

where fe is supplied by (a) with v = V = constant. Hence, these last twoo

expressions can be added to eliminate f and obtain



--



2 a 2C 22 d w

(M

1+

+

2

- r ) = 1 o oa (j)

2. dt 1 Tr ro 0,

If we further use (g) to eliminate d6/dt, we obtain an expression for r(t)

that can be written in the standard form

2(M1 2 2 V = 0 (k)

2 dt

where M 4 2 7a2 2

V = 2 (1)2 r

2r

Of course, (k) can be multiplied by dr/dt and written in the form

d 1 dr1 S(M + V] =0 (m))(

to show that V is a potential well for the combined mass of the rotating particle

and the plate.

Part e

The potential well of (1) has the shape shown in the figure. The minimum

represents the equilibrium position found in (c), as can be seen by differentiat

ing (1) with respect to r, equating the expression to zero and solving for wm

assuming that r =£. In this example, the potential well is the result of

a combination of the negative coenergy for the electromechanical system,

constrained to constant potential, and the dynamic system with angular momentum

conserved.




PROBLEM 5.26

Part a

To begin the analysis we first write the Kirchhoff voltage equations for

the two electric circuits with switch S closed

dX

V = ilR 1 + (a)

dX20 = i2R 2 dtd),

(b)

To obtain the electrical terminal relations for the system we neglect fringing

fields and assume infinite permeability for the magnetic material to obtain*

1 = N1 ' 2=N24 (c)

where the flux $ through the coils is given by

21o wd (N1 1 + N2i2)

$ = (d)g(l + -)

We can also use (c) and (d) to calculate the stored magnetic energy as**

g(l + x) 2

W = (e)m 4 ° wd

We now multiply (a) by N1/R1 and (b) by N 2/R2, add the results and use

(c) and (d) to obtain

x 2 2NV g(l+ -) N N1V1 + (- + 2) (f)

R1 21 wd R1 R2 dt

Note that we have only one electrical unknown, the flux 0, and if the plunger is

at rest (x = constant) this equation has constant coefficients.

The neglect of fringing fields makes the two windings unity coupled. In practice

there will be small fringing fields that cause leakage inductances. However,

these leakage inductances affect only the initial part of the transient and

neglecting them causes negligible error when calculating the closing time of

the relay.

**Here we have used the equation QplPg)b

W =iL 2 +L i i2 + L i22m 2 1 1 12 1 2 2 2 2





Part b

Use the given definitions to write (f) in the form

S = (1 + ) + dt(g)

Part c

During interval 1 the flux is determined by (g) with x = xo and the

initial condition is * = 0. Thus the flux undergoes the transient

o-(1 + x-) t

SI - e 0 (h)

1+

To determine the time at which interval 1 ends and to describe the dynamics

of interval 2 we must write the equation of motion for the mechanical node.

Neglecting inertia and damping forces this equation is

K(x - Z) = fe (i)

In view of (c) (Al and X2 are the independent variables implicit in *) we can

use (e) to evaluate the force fe as

fe awm( ' x2 x) 2 )ax 41 wd

Thus, the mechanical equation of motion becomes

2K(x - t) = - (k)

41 wdo

The flux level 1 at which interval 1 ends is given by

2

K(x - - ) 4 (1)

Part d

During interval 2, flux and displacement are related by (k), thus we

eliminate xbetween (k) and (g) and obtain

F iE-x 2 d*= (1 +) - o T dt (m)

were we have used (k) to write the equation in terms of 1." This is the nonlinear

differential equation that must be solved to find the dynamical behavior during

interval 2.





To illustrate the solution of (m) it is convenient to normalize the equation

as follows

d(o) _-x o 2

o 0 ( )3- (1 +) + 1

d(- g 1 o

o

We can now write the necessary integral formally as

to d(-)

S-x 2 3,)to

o d(A)

(- ) ( ) - (1 + ) +0o

1 1(o)

where we are measuring time t from the start of interval 2.

Using the given parameter values,

o d(-o)

t

To

•o ao400 -) - 9 +

0.1

We factor the cubic in the denominator into a first order and a quadratic factor

and do a partial-fraction expansion* to obtain

(-2.23 - + 0.844)o.156

•Jt d(o ) =0 0

75.7 ( -) - 14.3 + 1

o

Integraticn of this expression yields

. . . .. .... . t •m q

Phillips, H.B., Analytic Geometry and Calculus, second edition, John Wiley

and Sons, New York, 1946, pp. 250-253.





2t 0.0295 In [3.46 ( -) + 0.654] - 0.0147 In [231 (-) - 43.5 (-) + 3.05]T0

+ 0.127 tan-1 [15.1 (--) - 1.43] - 0.0108

Part e

During interval 3, the differential equation is (g)with x = 0, for which

the solution is tT

14 = 02 +

(%o - 02)( - e 0) (s)

where t is measured from the start of interval 3 and where 2 is the value of flux

at the start of interval 3 and is given by (k)with x = 0

2

KZ = (t)41 wd

Part f

For the assumed constants in this problem

01

The transients in flux and position are plotted in Fig. (a) as functions

of time. Note that the mechanical transient occupies only a fraction of the time

interval of the electrical transient. Thus, this example represents a case in

which the electrical time constant is purposely made longer than the mechanical

transient time.




Y.iVe0Y\ Av

0,4

0.Z

o o,os oo o 0.20 o.zs

t/·t. 9.



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FIELDS AND MOVING MEDIA

PROBLEM 6.1

Part a

From Fig. 6P.1 we see the geometric relations

r' = r, e' = e - Pt, z' = z, t' = t (a)

There is also a set of back transformations

r = r', e = 8' + st', z = z', t = t' (b)

Part b

Using the chain rule for partial derivatives

, = () r) + (2j) (L ) + 2(- 3za) + () (-) (c)

at ar at f ae at' 9 ( at at

From (b) we learn that

SO , ,= , ' =0O , = 1 (d)

Hence,

at' at + 0 ao (e)

We note that the remaining partial derivatives of p are

4, = a* 2t = * (f)3r' ar ' ae' ae ' az' a

PROBLEM 6.2

Part a

The geometric transformation laws between the two inertial systems are

x1 = x - Vt, x' x2, x = x3, t' = t (a)

The inverse transformation laws are

1 = x' + Vt', x 2 x=, x 3 = x t t' (b)

The transformation of the magnetic field when there is no electric field

present in the laboratory faame is

P'= W (c)

Hence the time rate of change of the magnetic field seen by the moving

observer is

aB' =3Ba B ax + 3B 2 + B )x a3B at

1 2 3(d)




PROBLEM 6.2' (Continued)

From (b) we learn that

3x2atx' V, ax12 0, axt 3 = 0,I, at = 1 (e)

While from the given field we learn that

aB. kB cos kx

aB aB=

aBB 0 (f)

ax kBoo l -ax x 3 t C

Combining these results

aB' aB aBB' B, = V• = VVkB cos kx1 (g)

at, t, o 1which is just the convective derivative of B.

Part b

Now (b) becomes

S = x' x2 2 x + Vt, x 3 = x;, t = t' (h)'

When these equations are used with (d) we learn that

aB' S=aB , = V aB + aBt = 0 (i)

at, Tt ax2 at

aB aBbecause both and - are naught. The convective derivative is zero.

x2 at

PROBLEM 6.3

Part a

The quasistatic magnetic field transformation is

B' = B (a)

The geometric transformation laws are

x = x' + Vt' y', , Z z', t = t' (b)

This means that

' = E(t,x) = B(t', x' + Vt') = iyoB cos (wt' - k(x' + Vt'))

= i B cos[(w - kV)t' - kx'] (c)yo

From (c) it is possible to conclude that

w' = w - kV (d)

Part b

If w' = 0 the wave will appear stationary in time, although it will

still have a spacial distribution; it will not appear to move.





w' = 0 = w - kV; V = w/k = v (e)

The observer must move at the phase velocity v to make the wave appear

stationary.

PROBLEM 6.4

These three laws were determined in an inertial frame of reference, and

since there is no a priori reason to prefer one inertial frame more than

another, they should have the same form in the primed inertial frame.

We start with the geometric laws which relate the coordinates of the

two frames

r' r - v t, t = t', r = r' + v t' (a)

We recall from Chapter 6 that as a consequence of (a) and the definitions of the

operators

t =-t'- r a'' t +tr

In an inertial frame of reference moving with the velocity vr we expect the equation

to take the same form as in the fixed frame. Thus,

-v'p'iat', + p'(v'*V')v' + V'p' = 0 (c)

-' + V'*p'v' = 0 (d)at'

p', p,'(p) (e)However, from (b) these become

'a + p'(v'+vr )+Vp' = 0 (f)

+ V.p' (v'+v) = 0 (g )

p' = p'(p') (h)

where we have used the fact that v *Vp'=V*(v p'). Comparison of (1)-(3) with (f)-(h)

shows that a self consistent transformation rthat leaves the equations invariant in

form is

p' = P; p' =p; v

t- v -

r

PROBLEM 6.5

Part a

p'(r',t') P=(r,t) = p (1- -)=- o(1- C(a)

J' = p'v' = 0 (b)

Where we have chosen v =v i so thatr oz

v' v - v = 0 (c)

Since there are no currents, there is only an electric field in the primed

frame

r r'2E' (po/o - )r (d)

H O, B' = • •' = 0 (e)

Part b

p(r,t) = p(1- ) (f)

This charge distribution generates an electric field

2r

(P/ r 3air (g)

In the stationary frame there is an electric current

S= pV = po(1-r)vi

-(h)

This current generates a magnetic field

2 ir rH= oVo( --a)io (i)

Part c

5-= 5' - P'v = po(l---)Voi (j)Pvr o a oz

E= '-vrxB' = E' = (po/Eo - 3 ir (k)

SH' V xD r' r'2

r = oo 3a lie (1)

If we include the geometric transformation r' = r,(j), (k), and (1)

become (h), (g), and (i) of part (b) which we derived without using trans

formation laws. The above equations apply for r<a. Similar reasoning gives

the fields in each frame for r>a.




PROBLEM 6.6

Part a

In the frame rotating with the cylinder

E'(r') = -, I (a)r r

H' = 0, B' = u•H' = O (b)

But then since r' = r, Vr(r) = rwi

E=' - v x ' = 2'=-i (c)r r .r

V = f Ed = b dr = K In(b/a) (d)a

a

V 1 V 1 (e )

ln(b/a) r r In(b/a) r' r

The surface charge density is then

- = o 1 (f)a' = i *8 E' = - = aa r o In(b/a) a a

SE V

a'= -i *E E' = - (g )b r o In(b/a) b

Part b

3 = J' + Vy p' (h)

But in this problem we have only surface currents and charges

= '+ v ' =v a' (i)r r

awe V ) WE VO 0

K(a) 1a iBe e (n(b/a) In(b/a)

bwE V E V

K(b) =-b In(b/a) 6 In(b/a)

iB

(k)

Part c

WE VS(1)0

In(b/a) z (

Part d

S=' + v x D v x D' (m)r r

E V -+(nOx 1r'w(1n(bla) r x ir)= 6)(i (n)





we V(

in(b/a) iz

This result checks with the calculation of part (c).

PROBLEM 6.7

Part a

The equation of the top surface is

f(x,y,t) = y - a sin(wt) cos(kx) + d =0 (a)

The normal to this surface is then

Vf (b) = v ak sin(wt)sin(kx)ix + iy(b)

Applying the boundary condition n*4 = 0 at each surface and keeping only linear

terms, we learn that

Ah (x,d,t) = -ak sin(wt)sin(kx) (c)

h (x,O,t) = 0 (d)

We look for a solution for h that satisfies

V x =, V*h = 0 (e)

Let h = V V2 , = 0 (f)

Now we must make an intelligent guess for a Laplacian * using the

periodicity of the problem and the boundary condition hy •l/ay = 0 at

y = 0. Try

A cosh(ky)sin(kx)sin(wt) (g)

h = A sin(wt)[cos(kx)cosh(ky)i + sin(kx)sinh(ky) y] (h)

Equation (c) then requires the constant A to be

-ak A

sinh(kd)0od

Part b

VE - E (j)S •x ( ~-)-iy(---z)= (-)

= p0t cos(wt)[cos(kx)cosh(ky)ix

x+ sin(kx)sinh(ky)iy]y

(k)o





-_ AE = - wU cos(wt)[cos(kx)sinh(ky)]i

0 K

Now we check the boundary conditions. Because v(y=O) = 0

nx E = (n-v)B = 0 (y=0)

But E(y=O) = 0, so (m) is satisfied.

If a particle is on the top surface, its coordinates x,y,t must satisfy

(a). It follows that

Df 3fD• = - + v* V f = 0

VfSince n = -F we have that

-1 af(n.v) = i1 t awcos(wt)cos(kx)

Now we can check the boundary condition at the top surface

-- AnxiE f - o

ocos(wt)cos(kx)sinh(kd)[i

x-ak sin(wt)sin(kx)i

y]

(n.v)B = awcos(wt)cos(kx) inh(kd) Ai +ak x

poA sin(wt)sin(kx)sinh(kd)i ]

Comparing (p) and (q) we see that the boundary condition is satisfied at the top

surface.

PROBLEM 6.8

Part a

Since the plug is perfectly

conducting we expect that the current

I will return as a surface current on

the left side of the plug. Also E', H'

will be zero in the plug and the trans- I, Ntformation laws imply that E,H will then

also be zero.

Using ampere's law

-I2 ir

Also we know that

V*E = 0, Vx E - 0 0 < z < 5 (b)

We choose a simple Laplacian E field consistent with the perfectly conduct

ing boundary conditions

E =- i (c)r r

K can be evaluated from

E"ddi = dt da (d)

CS

If we use the deforming contour shown above which has a fixed left leg at z = z

and a moving right leg in the conductor. The notation E" means the electric

field measured in a frame of reference which is stationary with respect to the

local element of the deforming contour. Here

E"(z) = E(z), E"(C+A) = E'(C+A) = 0 (e)

J"*d = - E(z,r)dr = -K In(b/a) (f)

The contour contains a flux

JB-da = (E-z) %oHedr = - V I• ln(b/a)(E-z) (g)

S

So that

-K n(b/a) a = + n(b/a) d-- (h)-K = dtn(b/a) dt

= -Since v 'dt

vI

2w r r

0 <z

Part b

The voltage across the line at z = 0 is

b vpo0 IV = - Erdr = ln(b/a) (k)

a

vio

I(R + o In(b/a)) = V (1)2o

V

I = o (m)vi

R + 27r In(b/a)

V 2wR +1 (n)

vp ln(b/a)°

_VS0 < z <

R + 2 n(b/a)

H: (o)

Vo 1<z

f i 0 < z <

vy2R r rV11V+ o+(inb/a) rr

E= (p)

0 <z

Part c

Since E = 0 to the right of the plug the voltmeterreads zero. The terminal

voltage V is not zero because of the net change of magnetic flux in the loop

connecting these two voltage points.

Part d

Using the results of part (b)

SVI= In(b/a) i 1 V2=

Pin 27r 0 0Rn + lIn(b/a)

Tr

dWm H2

d= v fa H (r) 21Tr dr

R + V n(b/a)





There is a net electrical force on the block, the mechanical system that keeps

the block traveling at constant velocity receives power at the rate

1 V1o In(b/a) 1 2 2-- V2 27 [ v O In(b/a) o

2w

from the electrical system.

Part e

0H(r,I)x dr U0L(x) = I d= IT n (b/a)x

e awlm W 1 2

fe w •x ;W' =2 L(x)i

fe 1 3L 2 1 o In(bla)i2

f=fi T 2- In(b/a)T2 3x 2 2 U

The power converted from electrical to mechanical is then

v V 2

f' = f v In(b/a) [ ]e dt e 2 2w v o

R + o In(b/a)

as predicted in Part (d).

PROBLEM 6.9

The surface current circulating in the system must remain

BK = (a)

o

Hence the electric field in the finitely conducting plate is

B

E' o (b)oOs

But then

E= E' - V x (c)

= B ( - v)

os

v must be chosen so that E = 0 to comply with the shorted end, hence

v - (d)

os

PROBLEM 6.10

Part a

Ignoring the effect of the induced field we must conclude that

= 0 (a)

everywhere in the stationary frame. But then

E' =E+ V xB= xB (b)

Since the platejis conducting

J' = J aV x (c)

The force on the plate is then

F = 3 x B dv = DWd(oV x B)x B (d)

F = - DWd av B2 (e)

x o

Part b

M -v + (DWdoB )v = 0 (f)dt o

DWdoB2

to

Mv = v e (g)

Part c

The additional induced field must be small. From (e)

J' - OB v (h)oo

Hence K' = oB dv (i)o o

The induced field then has a magnitude

K 'B' o

-- =Iadv 1 ()i)<<o o

ad << 1 (k)

musttethinrlatepoorly conducting one.ery

It must be a very thin plate or a poorly conducting one.

--


PROBLEM 6.11

Part ai

The condition - << H means that the fieldW o

induced by the current can be ignored. Then th eK(I H+

magnetic field in the stationary frame is

H = -H i everywhere outside the perfect

conductors

The surface currents on the sliding conductor are such that

K1 + K2 = i/W

The force on the conductor is then

[(- +) x B W

F = x B dv = [(K1 + K)i x Bi ]WD1 2 yoz

= p H di i

Part b

The circuit equation is

dXRi +d- = V

dt odX

H dvdt oo

Since F = M dvdt

MR dv(---H d-) + (o H d)v = V

00

(o H d)V to MR

v = (1 - e )u_ (t)oo

PROBLEM 6.12

Part a

We assume the simple magnetic field

i 0 < x < x

0 x< x1

A(x) = fE*a = i

Part b

L(x) = X(x,) =i D

Since the system is linear

1 2 1lo1Wx 2W'(i,x) 1 L(x)i2 1 i

f = Wo i

ex 2 D

Part d

The mechanical equation is

Mdt

2 +B idt 2 D

The electrical circuit, equation is

dX d 0WX~ (-5-- i) = Vdtdt o

Part e

From (f) we learn that

dx o 2i = const

dt 2BD

while from (g)we learn that

PWi dx= V

D dt o

Solving these two simultaneously[DV2

dt 2 oWBEJ0

Part f

From (e)

2BD dx D 2/3 1/3 1/3= (ý (2B) V

w dt 0o

Part g

As in part (a)

i- i(t)i 3 O<X <X3 1

x < x

Part h

The surface current K is





K= i(t)(m)D 12

The force on the short is

S dv = DW ix 0 2 (n)3x

poW 2 2

2120 01(t)'l

Part i

xdVx E (o)

-7ýx - D dt 13 5t

_o di _

E2 = [D - x A + C]i (P)

=o di V(t)I

D dt W 3

Part i

Choosing a contour with the right leg in the moving short, the left leg

fixed at xl=

0'

a*d dt B*da (q)

C S

Since E' = 0 in the short and we are only considering quasistatic fields

H dx

'*dt = V(t) W x o o2 Wd H (r)oxat dt ooP Wx

= d ( i(t)) (s)

Part k

nx (Eb) = V b (t)

Here

l n dt D 3

=

b

.o

D

di

dt

V(t)

W

dx

dt

oo

D i)i2

)

(v)

dxW dx o

dt D = 0D

Part 1

Equations (n) and (e) are identical. Equations (s) and (g) are

identical if V(t) = V . Since we used (e) and (g) to solve the first parto

we would get the same answer using (n) and (s) in the second part.





Part m

diSinced= 0,dt

V(t) 4. VoE2(x) = i- = - t- iy (x)

PROBLEM 6.13

Part a

K 2 Te() + T2( ) (a)dt

Part b

ii oHio1r - 001 1

1 D2r ;1 J1 D2aR e (b)

Similarly

S oHoi 2 *F

0 0i (c)

2 D2R (c)

Part c

T = [f(r xf)dv]z = oHo(R2-Rl)il (d)

Te oHo(R2-R1)i2 (e)

Part d

1 = E1 (R2 -R 1); V2 = E2 (R2 -RI) (

Part e

= = =1 J GE H (E1+iB) = (EI+RUoHo -) (g)

E1

-ii-2aDR 0Ho

ddt

(h)

V a 2a R 1 - pH R(R -R1 ) -- (i)

2

1R2 - R

a 2aRD 2 HR(R 2R 1) dt

Part f

2oHo(Rdt

2-R1 )i0 u-l(t) (k)

v2 (t) =

K(t)= -R2-R1) 0t2

2R(poHo(R2R1))27200 2 1m)

u-(t)

t u-(t) (t)

(1)

(m)





1 (R2-R1) 2Rv (t) = [ 2RD + ( H (R2-R ) K t]ioU- (t)

1 a 2aRD 0oo 2 1 K 0 -1

Part g2

K2

- oH o (R 2-R 1 )i 1dtdt2

SoHo(R2-R1)O2aRD Hd

SR2R1) (R2-R 1) o

d2 d2 + KI dt K2 l(t)dt

2 2t

K1

= [( HoR)2 2aD(R2-R )a]/K

p H 2aDR Y

2 K

Find the particular solution

-JK2 o jt]P (, t) = R 2 e

K2 o K

-= 2o sin(wt+tan (t)

K+w

B -Kt

(t) = A -K1e + p (w,t)

We must choose A and B so that

(o0) = 0 (0) = 0

K2 K2mA =- v B = v

KIW o (K2 + 2 )1 (K~ +




PROBLEM 6.13(Continued)

Wct)~

Part h

The secondary terminals are constrained so that v2=-i2R2. Thus, (j) becomes

~dt R3 i ; R = 1 (R2-RI) =R + RD K pH (R -Rl) (w)dt RK4 2 3 0 2 RD 4 2 1

Then, it follows from (a), (d) and (e) that

di RK2 K2Ri

2 4 4 o+ i cos wt

dt KR3 2 KR3

from which it follows that

ji21 S K2RKR4 RI

0 RK2 2

2

R KR3

PROBLEM 6.14

Part a

The electric field in the moving laminations is

J' J i *E'

a a OAi

z(a)

The electric field in the stationary frame is

i E = E'-VxB (- + rwB )i (b)

1 Ni CA y z

B (c)y S

2D o12Dz•NV = (A - -)i (d)

Now we have the V-i characteristic of the device. The device is in series with a

inductance and a load resistorRt=RL+Rint.

2D p2DrN jN2aD

[R + _Y o ]i + d 0 (e)t uA S S dt

Part b

Let 2 2D ON 2aDrNw

R1= R

t+AaA

SS

, LS

(f)

1 = I e~ 1/L) t

(g)

=

Pd i/ - [e tIf

2D 2DjorNw

R = R + 2D o < 0 (h)

the power delivered is unbounded as t 4 o.

Part c

As the current becomes large, the electrical nonlinearity of the magnetic

circuit will limit the exponential growth and determine a level of stable

steady state operation (see Fig. 6.4.12).

PROBLEM 6.15

After the switch is closed, the armature circuit equation is#

diL i (a)(RL + Ra)i L + La - = GO i (a)

Since Ghif is a constant and iL(0) = 0 we can solve for the load current and

shaft torque

(RL+R a)

Gi f L

iL(t) = (R+Ra) (l-e )u_(t) (b)

Te(t) = iL(t) Gif(RL+Ra)

2, - t

(R+Rai) (l-e a )u_(t) (c)(RL+R ) 1

L





From the data given

T = La/RL+Ra= 2.5 x 10

- 3

sec (d)Gef

iL = R+R 628 amps (e)max RL+Ra

T (Gif) 1695 newton-meters (f)max RL+Ra

£ -j

428-

-1~ - t

1~7 cl~o·, - ~)

/67s~

PROBLEM 6.16

Part a

With S1 closed the equation of the field circuit is

di

Rfif + Lf dt Vf (a)

Since if(0) = 0 R

f f

if(t) =RP (1-e )u_1(t) (b)

Since the armature circuit is open

Rf

Sf C - f t

a Gif R (1-e )u-1(t) (c)





From the given data

T = Lf/Rf = 0.4 sec

Vf G6V = = 254 voltsa R

max f

to. 4- .

Part b

Since there is no coupling of the armature circuit to the field circuit

if is still given by (b).

Because S2 is closed, the armature circuit equation is

dVL

(RL+Ra)VL + La - - = RLG f (d)

Since the field current rises with a time constant

T = 0.4 sec (e)

while the time constant of the armature circuit is

T = La/RL+Ra = 0.0025 sec (f)

we will only need the particular solution for VL(t)R

RGf

t

RL G RL Vf LfVL(t) = RL+Ra i = ( a)G (1-e )ul(t) (g)

VL = RL (j-)Vf = 242 volts (h)max L a f

4?_

1!l~

fY0.4 sec




PROBLEM 6.17

The equation of motion of the shaft is

T

Jr + o = T + T (t) (a)r dt W o e

o

If Te(t) is thought of as a driving term, the response time of the mechanical

circuit is

Jro

T = = 0.0785 sec (b)T

o

In Probs. 6.15 to 6.16 we have already calculated the armature circuit time

constant to be

LT • a

-=2.5 x 10-3

sec (c)

R +R sRa+R L

We conclude that therise time of the armature circuit may be neglected, this is

equivalent to ignoring the armature inductance. The circuit equation for the

armature is then

(Ra + RL)iL = Gwif (d)

Then -- (Gif)2

T = ii = (e)Te f-Gif = RL (e) Ra +

Plugging into (a)

de =J d+ Kw T (f)r dt o

HereT (G i )2 V

K (-R - ) ; i = f (g)W R +R f R

Using the initial condition that w(0) = wo

T T0 /J)

w(t)+ (w - -)e t > 0 (h)

K o K

From which we can calculate the net torque on the shaft as

T= Jrdt = (T -KW )e u (t) (i)r dt o o 1

and the armature current iL(t)

Gi

iL(t) = (R l)(t) t >0 (j)a L*



FIELDS' AND MOVING MEDIA


From the given data

T

w = = 119.0 rad/sec = 1133 RPMfinal K

Tma x = (To-Kw) 1890 newton-m

Gi

i w0 700 amps

Gi

it = (R ) fn a 793 ampsL R +R , finalmax a L

K = 134.5 newton-meters, T = Jr/K=

0.09 sec

1

(k)

(1)

(m)

(n)

(o)

1i/33

/Ood

,÷

i8 O

713700

- -


PROBLEM 6.18

Part a

Let the coulomb torque be C, then the equation of motion is

d0

dt

Since w(0) = wt

w(t) - 0 (1- t) 0 < t <O/C) to

Part b

Now the equation of motion is

dwJ -+ Bw = 0dt

w(t) = 0 e•)

-. \

wCe

Part c

Let C = Bwo, the equation of motion is now

J

dwddt + BLc= -Bw o

B--

-JE{(t)-w + 2w e0o o

t

< t <Ji

B2





(C t )

'r = z/0

PROBLEM 6.19

Part a

The armature circuit equation is

diLRi L + La- = Gwif - Va U- (t)aL a dt f a -1

Differentiating2

dL diL dwLa - + R -d = Gi Vu (t)a dt2 a dt f dt ao

The mechanical equation of motion is

dwJ -4-

=

- Gi ir dt L fThus, (b) becomes 2 2

L diL di (Gif)a- +R -+-L---- i -Vu(t)

2 a dt J L aodt

Initial conditions are

VdiL + aiLfrom(d) = , (0 L

aand it follows from (d) that

a -atiL(t) = (- e-esinBt)ul(t)

a

whereR

a 7.5/seca

(Gif) Ra8 ('-L = 19.9 rad/sec

r a a





aLAB - 1160 amps

V

w(t) =a

-Gi Se tsin t + (e-at cos 8t-l)]

f

Va

Gi= 153.3 rad/sec I (k)

IN

Part b

Now we replace Ra by R+RL in part (a). Because of the additional

damping

Ir ~r\~

iL(t) 2L -Ya

(e )u1(t)

where R +R

= a L2L

75/sec

a

R + RL2 (Gi )y = a ) = 10.6/sec.

2L Jr La





2 Gif 1 -(a-Y)t 1 -(a+y)t

W(t)= [2Ly e + + +e ]

a r c-Y o)

iiI

AL.

. ir"(A 1.t/

vcLr

PROBLEM 6.20

Part a

The armature circuit equation is

v = R i + GIfwa aa f

The equation of motion is

dwJ i = GI idt fa

Which may be integrated to yield

w(t) G= i(t)wt J 'a





Combining (c) with (a) 2

S= Rai + ia(t) (d)a asa Jr

We recognize that

J

C = -(e)

(GI ) 2

Part b

JC = (0.5) 0.22 farads

(GIf)2

(1.5)2(1)

PROBLEM 6.21

According to (6.4.30) the torque of electromagnetic origin is

Te = Gi ifa

For operation on a-c, maximum torque is produced when if and is are in phase,

a situation assured for all loading conditions by a series connection of field

and armature. Parallel operation, on the other hand, will yield a phase relation

between if and is that varies with loading. This gives reduced performance unless

phase connecting means are employed. This is so troublesome and expensive that

the series connection is used almost exclusively.

PROBLEM 6.22

From (6.4.50) et. seq. the homopolar machine, viewed from the disk terminals

in the steady state, has the volt ampere relation

v Ri + Gwif

in(b/a) TRa 2Oad

For definition of v and i

shown to the right and with the

interconnection with the coil

snhown in rig. or.L2

1 Nia

B o 2d

Then from (6.4.52)

BoNio 22 2 o a 2 2

Gwi (b -a ) = a (b -a)f 4d

Substitution of this into the voltage equation yields for steady state (because

the coil resistance is zero).• Ni

0 = Ri + (b -a2)

asa 4d

for self-excitation with i-a 0 0

W1oN 2 2(b -a ) = -R

Because all terms on the left are positive except for w, we specify w < 0

(it rotates in the direction opposite to that shown). With this prov4sion'ithe

number of turns must be

4dR4dln(b/a)

M1lo (b2-a2) -2rodwjI p(b2-a 2)

N = 21n(b/a)

oralowj(b 2 -a 2)

PROBLEM 6.23

Part a

Denoting the left disk and magnet as 1 and the right one as 2, the flux

densities defined as positive upward are

BoN

B2 - (i+i 2)

Adding up voltage drops around the loop carrying current i we have:. M

dB dB QB 2 2dB 2 dBil+ ilRa 2 _B2a ,1

In( )where R =

a 2nah

Part b

Substitution of the expression for B1 and B2 into this voltage expression

and simplification yield

di

L d + il(R+Ra) - Gil + Gi 2 = 0

162





where

2 2 2N2na

- oN(b -a )

2Z

The equation for the circuit carrying current 12 can he written similarly as

di

L ti +2(R (+Ra)-Gi2-GOil = 0

These are linear differential equations with constant coefficients, hence, assume

i Ilest 2 I2est

Then

[Ls + RL+Ra-GG]I1 + GOI 2 = 0

[Ls + RL+R -G]lI2 - GoI1 = 0

EliminatCon of I1 yields

[Ls + RL+ R a- GSJ]2 +GS] 0+ GO I = 0

If 12 0 0 as it must be if we are to supply current to the load resistances,

then

[LI + RL+Ra-G]C + (Ga)2 = 0

For steady-state sinusoidal operation a must be purely imaginary. This requires

RL + R - G = 0

or 2 2 In(W4e--U N(b -a) RL + 2rh

G = 21G =

This is the condition required.

Part c

-When the condition of (b) is satisfied

-GSG

e+J + +JL b2

b2

PN(b -a )j• a -) 2 22 (-2 -1)Q

29oN22

2 2*2N

163 ,





Thus the system will operate in the sinusoidal steady-state with amplitudes

determined by initial conditions.With the condition of part (b) satisfied the

voltage equations show that

1 1fi2

and the currents form a balanced two-phase set.

164



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MAGNETIC DIFFUSION AND CHARGE RELAXATION

PROBLEM 7.1

It is the purpose of this problem _I

to illustrate the limitations

inherent to common conductors in _

achieving long magnetic time

constants. (Diffusion times.) For

convenience in making this point

consider the solenoid shown with

I = length

A = cross-sectional dimensions of single layer of

wire (square-cross section).

r = radius (r >> A) but r << £.

Then there are R/A turns, each having a length 2nr, and the total d-c resistance

is directly proportional to the length and inversely proportional to the area

and electrical conductivity a.

R 2ir)

27ra(A 2 )

iThe H ield in the axial direction, by Ampere's law, is H = and the flux

linked by one turn is 0 H(wR2 ) so that

=poH(r 29 = o(.r 2)i

and it follows that

L = 0o(wr2X92)

Finally, the time constant is

L 1E f po

rAc

Thus, the diffusion time (see Eq. 7.1.28) is based on an equivalent length

IVr. Consider using copper with

a = 5.9 x 107 mhos/m

A = 10 m

and find A required to give L/R = 102

A 2(L 1 (200)

R 0o (47x107)(10)(5.9x10 )

-1= 2.7 x 10 m or 27 cm

Note that to satisfy the condition that k >> r, the length must be' greater

than 10 meters also. The coil is larger than the average class-room! Of

course, if magnetic materials are used, the dimensions of the coil can be

reduced considerably, but long L/R time constants are difficult to obtain on a

laboratory scale with ordinary conductors.

PROBLEM 7.2

Part a

Our solution will parallel the one in the text, only now the B field will

be trapped in the slab until it diffuses away. The fundamental equations are

VxB =pJi= paE; -3

VxVxB = V(V.B) - V~B - aVxi =- pa

Because V.B = 0,

2no

or in one dimension

a2B aB1 x x (a)pa z2 at

at t = 0+

0 z<0

S B0 < z<d

0 z > d

This suggests that between 0 and d, we can write Bx(z) as

B W()= a sin(-nz ) 0 < z < d

n= l n

To solve for the coefficients an, we take advantage of the orthogonal property

of the sine functions.d ood7rz d nTz mTrz

B() sin( )dz = a sin )sin( )dz

o n=l o

But 2dB

B uTzd mdz 0 m odB ()sin( )dz = B sin( -)dx =

d o dm even

Alsod ad n

nz niza sin(-~--)sin( )dz 2

o 0 n m

Hence,

4Bm odd

0 m even'm 1mITLOB (t=O,z) =

-4 Bosin( n0< z< d

x nlnf o

n odd

We assume that fo r t > 0, O < z < d

B (t,z) - I -B sin(n--)ex nlniT 0 dn=1

n odd 2

Plugging into (a) we find that !.T-) = an Let's define T = po( )

as the fundamental diffusion time. Thenc 2

B (t,z) = --- B sin(nz-- n t)e O<z<dx nlnf o d

n=lt>O > 0

n odd

Part b

2nVxB - 1 3B x 4Bo~ no -n t/T3 i = -~T cos(--e O<z<d11 z y Y od n I

t>On odd

PROBLEM 7.3

Part a

If we neglect the capacitance of the block, the current we put in at t=O

will have to return by means of the block. This can be seen from the magnetic

field system equation

VxH = J (a)

which implies

V3 = o0 (b)

or "what goes in must come out".

If the current penetrated the block at t=0+

there would be a magnetic field

within the block at t=0+, a situation we cannot allow since some time must

elapse (relative to the diffusion time) before the fields in the block can change

significantly.

We conclude that the source current returns as a surface current down the

left side of the block. This current must be

K = -I /D (c)y o

where y is the upwards vertical direction. The current loop between x = - L

and x = 0 thus provides a magnetic field

-I /D -L~x<O

H (t=O+) = (d)

S0 O<x

where z points out of sketch.

Part b

As t + = the system will reach a static state with input current I /D per

unit length. The current will return uniformly through the block. Hence,I

J (x)= (e)y Dd

.-S (x)

4

- Z I (Y.)j M -5 5-P-

Part c

As a diffusion problem this system is very much like the system of

Fig. 7.1.1 of the text except for the fact that here diffusion occurs on only

one side of the block instead of two. This suggests a fundamental diffusion

time constant of

22

where we have replaced the term d2 by (2d) 2 in Eq. 7.1.28 of the text.

PROBLEM 7.4

Part a

This is a magnetic field system characterized by a diffusion equation.

With B =ReB (x)ez z





2Ad2B

1

dx2

z = B (a)a z

d ax

Let B (x) = Be , then

a.2 jpa (b)

or

a + (1 + j), 6 = \j (c)

The boundary conditions are

B (x=O) = -pi/D (d)

Sz('( ) -o

which means that we use only the (-) sign

p -X/6 6(WtB (x,t)=-Re D e e (e)

Part b

VxB P 3 (f)

or

aB

ax y

so that

^ . j(Wt- xRe--J = -Re e e (h)yPart

Part c





Part d

The electric fild is given byE

Vx E = waJBz z

Ey

x=O) =-22D

(1+j)i (j)

Faraday's law (Eq. 1.1.23, Table 1.2, Appendix E) written for a counter-clockwise

contour through the source and left edge of the block, gives

jw 0o(Ld)

V+Ed i (k)y D

where from (j)

A1 d1Ed =- (1 (+)y a-

Hence, assuming that V = i[R(w)+jwL(w)], (don't confuse the L's)

R(0) = ( ) = (m)

L(w) = + (n)

Thus, as w-+ the inductance becomes just that due to the free-space portion

of the circuit between x=0 and x=-L. The resistance becomes infinite because

the currents crowd to the left edge of the block.

In the opposite extreme, as w+O, the resistance approaches zero because

the currents have an infinite x-z area of the block through which to flow.

Similarly, the inductance becomes large because the x-y area enclosed by the

current paths increases without limit. At low frequencies it would be

necessary to include the finite extent of the block in the x direction in the

analysis to obtain a realistic estimate of the resistance and inductance.

PROBLEM 7.5

Part a

This is a magnetic field system characterized by a diffusion equation.

Place origin of coordinates at left edge of block, x to right and z out of paper.

With B = ReB (x)ex x

82B1 z J

Sax 2 (a)=jwBB

Let B (x) = Be , then

a2 jw~pa (b)

a6 (c)


A

Bz(x=0) = - 1 - (d)

Bz(x=•) = 0 (e)

because all of the current Io(t) is returned through the block. Thus the'

appropriate linear combination of solutions to satisfy the boundary conditions

is

B (x,t) = Re I sinh[a(x- jeJ•SR D sinh (at)

where a is a complex quantity, (c). The current is related to Bz by

aB

Vx = a i = VJ = VJyy (g)ax y yy

From (f) and (g),

jIa cosh[a(x-R)l]ejo t

(h)

y D sinh at

Part b

The time average magnetic force on the block is given by

fReDd fa J (x)B *(x)dxf

= Re Dd Y 2 (i)x 2

where we have taken advantage of the identity

^ jWt A jW t 1 ^^

<Re Ae ReBe > =-12

Re AB*

to integrate the force density (JxB)x over the volume of the block. Note that

a detailed calculation is required to complete (i), because a in (f) and (h)

is complex.

This example is one where the total force is more easily computed using

the Maxwell stress tensor. See Probs. 8.16, 8.17 and 8.22 for this approach.




PROBLEM 7.6

As an example of electromagnetic phenomena that occur in conductors at

rest we consider the system of Fig. 7.1.1 with the constant-current source

and switch replaced by an alternating current source.

i(t) = I cos Wt (a)

We make all of the assumptions of Sec. 7.1.1 and adopt the coordinate system

of Fig. 7.1.2. Interest is now confined.to a steady-state problem.

The equation that describes the behavior of the flux density in this

system is Eq. 7.1.15

2a B1 x xa a2 at (b)

and the boundary conditions are now, at z = 0 and z =d,

B = B cos at =[Re Be0Jit (c)

where

) NIB = -- (d)

o w

The boundary condition of (c) coupled with the linearity of (b) lead us to

assume a solution

Bx = Re[B(z)eijt] (e)

We substitute this form of solution into (b), cancel the exponential factor,

and drop the Re to obtain

d 2BB j poa B (f)2

dz

Solutions to this equation are of the form

rzB(z) = e (g)

where substitution shows that

r = + = + (l+) (h)

It is convenient to define the skin depth 6 as (see Sec. 7.1.3a)

6 2 (i)

We use this definition and write the solution, (g) as





(1+j) -(+j)B(z) = cle + c2e (j)

The boundary conditions at z = 0 and z = d (c) require that

B0 c + c 2

o 1 2

Solution of these equations for cl and c 2 yields

Bo1 - e (C Bo[l D (k)

1 D

-B1l e+

C2 0 D (Z)

where D = 2(cos sinh + J sin coshd

We now substitute (k) and (Z) into (J); and, after manipulation, obtain

B(z) = Bo[f(z) + J g(z)] (m)

where

M d d N d df(z) = V cos sinh -+ sin cosh

F 6 6F 6 6

N d d M d dg(z) = f cos E sinh sin d cosh

F 6 6 F 6 6

z z d-z d-zM = cos sinh + cos (- z sinh (-z

N = sin cosh + sin d-z cosh (d-z6 6 Ci::

2d 2d 2d 2dF = cos -sinh2 + sin2 cosh

6 .6 6 6

Substitution of (m) into (e) yields

Bx= B

nm(z)cos[wt + 8(z)] (n)

where

Bm(Z) = Bo \[f(z)]2 + [g(z)] (o)

-0(z) = tan1 g(z) (P)

f(z)

It is clear from the form of (n) that both the amplitude and phase of the

flux density vary as functions of z.

To illustrate the nature of the distribution of flux density predicted

-9

by this set of equations the maximum flux density is plotted as a function of

position for several values of d/6 in the figure. Recalling the definition

of the skin depth 6 in (i),we realize that for a system of fixed geometry

and fixed properties t/vw, thus, as increases, the frequency of the excitation

increases. From the curves of the figure we see that as the frequency increases

the flux density penetrates less and less into the specimen until at high

frequencies (- >> 1) the flux density is completely excluded from the conductor.

At very low frequencies (d << 1) the flux density penetrates completely and is

essentially unaffected by the presence of the conducting material.

It is clear that at high frequencies ( >> 1) when the flux penetrates

very little into the slab, the induced (eddy) currents flow near the surfaces.

In this case it is often convenient, when considering electromagnetic phenomena

external to the slab, to assume oa-x and treat the induced currents as surface

currents.

It is informative to compare the flux distribution of the figure for a

steady-state a-c problem with the distribution of Fig. 7.1.4 for a transient

problem. We made the statement in Sec. 7.1.1 that when we deal with phenomena

having characteristic times that are short compared to the diffusion time constant,

the flux will not penetrate appreciably into the slab. We can make this statement

quantitative for the steady-state a-c problem by defining a characteristic time

as

1c W

We now take the ratio of the diffusion time constant given by Eq. 7.1.28 to this

characteristic time and use the definition of skin depth in (i).

T 2 d 2 (q)

T 27

Thus, for our steady-state a-c problem, this statement that the diffusion time

constant is long compared to a characteristic time is the same as saying that

the significant dimension d is much greater than the skin depth 6.

The current distribution follows from the magnetic flux density by using

Ampere's law;

S= 1aBx (r)

y 0 az

-10-10





Thus the distribution of IJyl is somewhat as shown in the figure for B .

The instantaneous J has odd symmetry about z = 0.5 d.y

j

w

j

f

r\

Fo~j PO

4ý.- cx

o O•. 0,4, 0.6 0.08 /,.

PJsTfli1u7IloT OI lnIUX DENSITý WITH SI•N EFFEdCT

-11




PROBLEM 7.7

Part a

Assume the resistors in the circuit model each have approximately their

D.C. resistance

RS

RDD.C.

a1- oAD

(a)

The inductance is the "loop" of metal

Depth D

i atL = (b)

D

Hence the time constant involved is

- = o (c)2R 2

The equivalent length in the diffusion time is A/S >> A.

Part b

By adding the vacuum space of region 2 we have increased the amount of

magnetic field that must be stored in the region before equilibrium is reached

while the dissipation is confined to the two slabs. In the problem of Fig.

7.1.1, the slab stores a magnetic field only in a region of thickness A, the

same region occupied by the currents , while here the magnetic field region is

of thickness t.

Part c

-12

Since diffusion in the slabs takes negligible time compared to the main problem,

each slab could be modeled as a conducting sheet with

S= (aA)• (d)

In region 2

VxH = 0 or H = Ho(t) z = - K2 (t)i (e)

From

fEd9 = - -- JB n da (f)

we learn that

a[ Kl(t)-K 2 (t)] = + [poat K2 (t)] (g)

Since Ko(t) = Kl(t) + K2 (t) we know that

I dK2 (t)

K (t) = U-l(t) = 2K2 (t) + apl0 A dt (h)

The solution.is therefore

I -t/T 01 A9.

K2 (t) (l-e ) t) (i)

and,because K2 = - Ho, the magnetic field fills region (2) with the time

constant T.

PROBLEM 7.8

As in Prob. 7.7, the diffusion time associated with the thin conducting

shell is small compared to the time required for the field to fill the region

r < R. Modeling the thin shell as having the property

K = A (a)

and assuming that

H 1 (t)tz = [Ho-K(t)] z (b)

-13

We can use the induction equation

•Ei-

BEn da (c)

to learn that, becauseH ° = constant fo r t > 0

2_R K(t) = - R21 dK t (d)Aa o dt

The solution to (d) is

-t/Tu •RAR

K(t) = H e-t/Tu (t); T (e)

and from (b), it follows that

Hi(t) = H0-K(t) = H (1-e-t)u_(t) (f)

The H field is finally distributed uniformly for r < a, with a diffusion time

based on the length A¶A.

PROBLEM 7.9

Part a

V x E = - (a)at

V x B = oE (b)

So

V xV x B = - (c)at

But

Vx(VxB) =V(V.B)- V2B = - V2 (d)

So

at

Part b

Since B only has a z component

aB

V2B = Ipa z (e)z at

Incylindrical

coordinates

V2 1 3 (3 2

82

lVar +12 (f)r r 9r 2 z2

Hr a= aBz

Here B = B (r,t) soz z

-14



- -



r (r + B (g)

Part c

We want the magnetic field to remain finite at r = 0, hence C2 = 0.

Part d

At r = a

B(a,t) = 0HR o - C Jo(40--a a) = poHo (h)

Hence if CI # 0

Jo(V- a) = 0 (i)

Part e

Multiply both sides of expression for B(r,t=0) = 0 by rJo ( j r/a) and

integrate from 0 to a. Then,

a a2

pH r J (v r/a)dr = p H 2 J(V) Q)

o CiJo(V i r/a)r Jo(j r/a)dr = C - J2(v) (k)

o i=1 o1 0 J22 1

from which it follows that

C (o)j v1jJl(v)

The values of vj and J l (vj) given in the table lead to the coefficients

C C C

1

211HF2.802; 2

H.535, 3

)2pH0.425 (m)

Part f2

2

= 2 = 0.174 Uooa2 (n)

V1

T (0.174)(4rx 10-7 )

10(25) x 10

4

-- 4.35 x 10

7seconds (o)

-15




PROBLEM 7.10

Part a

aE

VxE = iz 3x at

Bz z

00 (a)

D 4.

VxB = - iy Dx

= -a o = vo0(E -U B )iy

(b)

VxVxB = V(V*B) - V2B

a2B DE aDB= x2 ;z = 0pa0(a- Y U )-z (c)

DEBut - = 0 from (a), so

ax

a2B aB

2Z x

o

aU

ax

(d)

ax

Part

At

x = 0 B = - K (e)

at x = L B =0 (f)z

Part c

Let

axBz (x) = C e ,

then

a(a-o aOU) = 0 (g)

a = 0, a = i oU (0)

Using the boundary conditions

joGU(x-L)

z 0 1-e-1 0UL

Note that as U+O

B (x) =- PoK ( ) (3)

zas expected.

-16




PROBLEM 7.11

Part a

F - JxB = - J B i

yxzR z/Z R z/1

oI2 R e (em -1)

i (a)2 R z

w (e m -1)2

Part b 2

Part b Fwd dz = P1 d

(b)z fz 2w

This result can be found more simply by using the Maxwell Stress Tensor by methods

similar to those used with Probs. 8.16 and 8.17.

Part cThe power supplied by the velocity source is

pI 2 dU I2d RP = - fU= o 1d m (c)

U - f 2w wtR 2

The electric field at the current source is

J (zn=)

Ey(z=j) = y - UB (z=£) (d)Cxy

RI m

(e)a1w R

(e m -1)

Power supplied by the current source is then

-VgI = + E dI = + 2d Rf)g y OW R

Power dissipated in the moving conductor is then

I2d Rm eRm + 1(g)P d

=PU

+VI 2 Rm (g)

~d~'U g aRw2 R

which is just what is obtained from

Pd= wd "- dx (h)o

-17




PROBLEM 7.12

If a point in the reference frame is outside the block it must satisfy

Vx= - (a)

= = 0 and VxB 0 (b)

Since the points outside the block have J=O, and uniform static fields (for

differential changes in time),(a) and (b) are satisfied.

Points inside the block must satisfy

1 aB x

=J (c)o az y

S2B DB aB1 x + x = - V a x (d)oa •z2 at

Since these points see

3B B 2x o aB

J aVB , 2 O0 (e)y o' az a' z

DB Bx

=t - v -- and Vyioa = Iat . 0

these conditions are satisfied.

Points on the block boundaries are satisfied because the field quantities

F and B are continuous.

PROBLEM 7.13

Part a

Because V*.BO the magnetic flux lines run in closed loops. The field

lines prefer to run through the high 1 material near the source, hence very

few lines will close beyond the edge of the material at z=O. Currents in

the slab will tend to remain between the pole pieces.

Part b

a2B aB aB1 _-= + v (a)

Saz2 at az

j(wt-kz)Let B (z,t) = C e , then

k - jpoVk + jpaw = 0; (b)

A quadratic equation with roots

-18





k = Ilj[Y ++[ ] (c)

or in terms of R = VaVL and 6 =/m Wia

R R 2 2

k ')L = j + j () + J2 (~) (d)

From Fig. 7.1.16 of the text we see that

k+ =k + +j k k +kk i

where

-k= k+ >0 and k >-k > 0 (e)r r i i

Tomeet

the boundary condition of part (a) we must have

B (z,t) = C[ - e -jkZeijt (f)y

Using the boundary condition at z = - L

B +

B (z,t)= (eJz _e-Jk) ejt (g)y =(ek +L e jkL)

Part caB

x az x x/Ato

+ tJ JB°/ (k+ e-jk z -K- e- z)e j(1)

x +

(ejk L-ejk L)

Part d R

Asw+-O ký -O,k + Lj

By B (Rm/L) z-RB-R - ((-e (m)

(1-e m)

B /L (R /L)zJ = R e (k)

x -R m(1-e M)

-19



MAGNETIC DIFFUSION AN D CHARGE RELAXATION


-C

As the sketch Fig. 7.1.9 of the text suggests, we could realize this problem by

placing a current sheet source

B

R=-- po i x e-S tK 0

across the end z = - L and providing perfect conductors to slide against the

slab at x = O,D. The top view of the slab then appears as shown in the

figure.

---- ~I $I~P-~D·CI - -_ --- Li~m

C-uen /7/,S~ e.(? '

Výba -6 - - -

If0 'b t

Note from (j) and (k) that as Rm0,m the current density Jx is uniform and By is

a linear function of z. This limiting case is as would be obtained with the

given driving arrangement.

PROBLEM 7.14

Part a

Since J' =

-20





= 1 (a)zK cos(kUt-kx)

= IzKocos(wt-kx); w = kU

Part b

The track can be taken as large in the y direction when it is many skin

depths thick

L = track thickness >> = o koa0 (b)

In the track we have the diffusion equation

1 V2i 3B (c)pa at

o

or, with B = Re B exp j(wt-kx),

_- _ k = jB (d)pa 2 x x

o ay

Let B (y) = C ery ,

then

x2

1 2 k- a jW + (e)

wpa U oO

a=k 1+JS ; S = = (f)

k2 k

Since the track is modeled as infinitely thick

a yB = C e e

j (wt-kx) (g)

x

The gap between track and train is very thin; thus,

- i x B--- = K ej(wt -

kx) i (h)y p o z

which yields

eay ej(t-kx) (i)

We must also have VBE = 3Bx/ax + 3B y /y = 0 or

B= Jk Bx(x,y,t ) (j)y a x

-21

To compute the current in the track we note that

3B aB

VxB = i z x - = o

= - (jS

k))

--Bx

(x,y,t)ia 0j

Part c

The time average force density in the track is (see footnote, page 368)

<F > = 1 Re(J B*)y 2 z x

Hence the time average lifting force per unit x-z area on the train is

<T> - <F >d y = - Re 0 1-J B* dy

y _m 2 zx

1 2T 0oK

See Fig. 7.1.21 of the text for a plot of this lifting force.

Part d

The time average force density in the track in the x direction is

<F > Re(J B*)x 2 z y

The force on the train in the x direction is thenoo 1<

>= - <F >dy = - Re J B dy

fj K2 -

K2o00 S

4 2Iý+S ReVl1jS

The problem is that this force drags the train instead of propelling it in the

x direction. (See Fig. 7.1.20 of the text for a plot of the magnitude of

this drag force). To make matters worse, if the train stops, the magnetic

levitation force becomes zero.

-22

PROBLEM 7.15

Part a

Let the current sheet lie in the plane y = - s. In the region -s<y<O

we have the "diffusion equation"

V2B = 0 (a)z

If B (x,y,t) = Bz(y)eJ(t-kx) this equation yieldsz z

22B

S= k2B (b)a2 z

Hence we can conclude that

B = [A cosh k(y+s) + B sinh k(y+s)]e J t- k z) (c)z

At y = - s we have'the boundary condition

iy x Bz = PoK cos(wt-kz)ix (d)

Thus

B = [ K0 cosh k(y+s) + B sinh k(y+s)]e w t-kz ) (e)

Since V*- = aB /ay + aB z/az = 0 we must have

=By [j( 0K0 sinh k(y+s) + B cosh k(y+s))]e j (t-kz) (f)

In the conductor the diffusion equation is

1 2- aB 3BVB =- + V- (g)11a at az

Then

32B- = (Jo(a-kV) + k2 )B (h)

ay

which suggests a solution

Sao(w-kV)Bz(y) C e-ay ,V = k/l+jS, S 2 (i)

k

Since V-B = 0 in the conductor too, we must have

B B' (j)B =-jkB zi) a z

As the boundary y = 0 we must have

Byl = By2, Hz1 = Hz2 (k)

Note that.

-23

cosh ks Bz2 + J sinh ks By2

= p K (cosh2ks - sinh2ks) = oIK (1)

Then we must also have

0Ko = cosh ks BZ 1 + j sinh ks Byl

k= C (cosh ks + - sinh ks) (m)

It follows that the B field for y>O is

oK k

k " y + iz)e-e (n)S 0 (_j + ay eJ(t-kz)

cosh ks + sinh ks

Comparing with Eq. 7.1.91 of Sec. 7.1.4 of the text we see that it is only

necessary to replace

K

K by k

cosh ks + sinh ks

starting with Eq. 7.1.90. The average forces depend on the magnitude, not the

phase, of Ko, which is reduced by this substitution.

Part b

We note that if ks << 1

K0 =K

k o(0)

cosh ks + - sinhks

a

which shows that the results of Sec. 7.1.4 are valid when ks << 1.

Part c

When ks *

K0 -- 0

kcosh ks + - sinh ks

a

No fields will then be present in the conductor.

PROBLEM 7.16

Part a

Because the charge needs time to move through the conductor, at t=0+ there

is only free charge on the plates. The electric fields are directed in the

negative vertical direction and satisfy

-24



MAGNETIC DIFFUSION AN D CHARGE RELAXATION


Eb + E a = V (a)

at the interface at t=O+

EE = EE (b). o g

Hence at t=0+V V

E = , E = (c)E 8 g E 0b + --- a o

E0-b + ao £

Part b

As t+a the charge on the interface excludes the fields from the conducting

liquid, hence

V

E = 0 Eg =0 (f)

Part c

The charge on the interface at any time is

Of = cE - E E (g)

Conservation of charge requires

do

- oEE= (h)

The voltage across the plates is Vo for t>O

V° = E b + E a (i)o 9. g

Solving g, h, i we find that the charge obeys

(e + E b/a) dof = E

a dt+ a

f= -

aV

o (j)

E+ £ b/a

Le t T= , then

cV

f =00

(- e-t/T

) t> 0 (k)

e AV

q A0 (1 e - t/T) t > 0 (9)f i a

-25

PROBLEM 7.17

Part a

In the inner sphere

oi PfE Pf + ý = 0 (a)

o

So we find that

-ai/ t Pf(r,t) = P (r)e , t > 0 r < Ri (b)

A similar equation holds for the charge in the outer sphere, but it has no initial

charge distribution at t = 0, so

Pf(r,t) 0, t > 0 R <r<Ro (c)

Part b

Let R

Q = f 4wr22

po(r)dr (d)

o

Also define

aA = the surface charge density at r = Ri

"B = the surface charge density at r = Ro

The field at R is, by Gauss' lawo

E(Ro) ( )4re R

oo

Then, conservation of charge requires that the electric field at r = Ro obey

a E(Ro) + o E (R)- 0 (f)

o o/g)tE(R') e , t > 0 (g)

4re Roo

We can thus conclude that

a = Q2 (1- e ), t > 0 (h)4wR

Since charge is conserved we now know that

SAA 2

(eJrR - e0 ), t > 0 (i)

i

-26

Part c

PROBEM17 (ontiued

I

sAA;tY~/QO

YO-- 0 Tr &Xr.I 44 ~7; ý=&xr

~t~k-rgi7

PROBLEM 7.18

Part a

At the radius b

e[E(b ) - E(b-)] = f (a)

aaf ao[E(b+)-E(b-)]= - = (-E (b)[E(b )-E(b)]

For t < 0 when the system has come to rest

V*3 = (o/c)V.E = - t- = 0 (c)

For cylindrical geometry this has the solution

E + A i ; V = +b Edr = A In(b/a) (d)r J r

a

-27

then

E(r=b )= +bn+ n(b/a) b

t = 0 (e)

E(r=b+ ) = 0

Since E(b+) - E(b-) = af/E it cannot change instantaneously, so

+ V

E(b ) - E(b-) = -o(b/a) e

-0/) t, t > 0 (f)

Because there is no initial charge between the shells, there will be no charge

between the shells for t > 0, thus

SCl (t)+ a<r 0 (g)+ r b<r<c

r

The battery adds the constraint

V° = C1lln(b/a) + C21ln(c/b) (h)

while (f) becomes

C - C = 0 e t1 2 In(b/a)

(i)

Solving (h) and (i) for C1 , C2

C 0 (1- e ) t (j)2 In c/a (1

Vo ln(c/b) e-(/E)Ct) (k)

1 In c/a ( + ln(b/a)

Part b

a= 4 (E(b+ ) - E(b-)) = b In(b/a) t (V)

Part cInc/b , = 27 e

'b - 2a ' b in c/b

In(b/a) 2CSCb Ra

=2ro ' Ca In b/a

cc,

-28




PROBLEM 7.19

While the potential v is applied the system reaches an equilibrium. During

this time

V*J = of=

-(a)

in the bulk of the liquid. If the potential V is applied for many time constants

(T=C/a) any.charge in the fluid decays away. For t>O if the fluid is

incompressible (V'v = 0) and J = oE + p fv we know that

V-3 =(a/E)Pf + vVpf = tf(b)

But in a frame moving with the particles of fluid

d •- f - vVof =--(+lE)pf- + (c)

Pf(t) = Pf(t=O)e( l E) t t > 0 (d )

where Pf(t) is the local charge seen by a moving particle. But for all fluid

particles

Pf(t= 0) = 0 (e)

Hence the charge remains zero everywhere for t > 0.

Now draw a volume around the upper sphere big enough to enclose it for a

few seconds even though it is moving.

-da =- f f dV (f)

S V

Now because pf = 0 in the fluid

3 = aE, fJ*da =(a/e) Eida =(a/E)Q(t) (g)

S s

Then

(a/)Q(t) = - dJ P dV = - d Q(t) (h)

V

which has solution

Q(t)= Q e-t/;T =/a

-29

PROBLEM 7.20

Part a

We can use Gauss' law

c o E da = Pf dV (a)

S V

to determine the electric field if we note that there is no net charge in the

system, which means that

E= E = 0 x<0 and x>3d (b)x x

FoE (x) = - dx = Q x (c)ox o D2d D2 d t

There is no charge in the middle region so

E = ---- d<x<2d; t= 0 (d)

x D2

o

In the region 2d<x<3d

Eo(E (x) - Ex(2d)) = ddx = - 2 (x-2d) (e)2d D2d D2 d

E (x) Q_ (3d-x) 12d<d<3d

x D2C d lt = 0

~tx

1ZE,

As t-*o all the charge on the lower plate relaxes to the surface x = d, while the

charge on the upper plate relaxes to the surface x = 2d. The electric field then

looks like

Qe6 .





Part b

Each charge distributioncan be thought of as made up of many thin charges

sheets; any two such sheets,

one located somewhere in the top conductor, one located somewhere in the bottom

conductor, attract each other with a force

AQ1 AQ2AF= (g)

22E Do

which is independent of their separation, hence the net attractive force between

plates does not change with time. At t-m

there is a surface charge

a =- - x = 2dD

(h)

B = + 9- x= d

D

and the force per unit area Tx is simply that found for a pair of capacitor plates

having separation d and supporting surface charge densities + Q. (See Sec. 3.1.2b).

2T

x= Q D-

t > 0 (i)2c D

0

This force can be easily seen to be constant from the viewpoint taken in Chapter 8,

where the force on the lower plate can be found from the Maxwell Stress Tensor.

The only contribution comes from Txx = - E2 evaluated at x = d, and thus

Txx(x = d) = Tx as given by (i) regardless of t. Problem 8.23 is worked out

following the stress-tensor approach.

PROBLEM 7.21

Part a

If the electric field beyond the plates is zero the conservation of charge

equation

J'da = t fPdV = -t eEdda (a)

S V S

-31

MAGNETIC DIFFUSION OF CHARGE RELAXATION


becomes

GE^EI

(x)=

-j

WE E (X)

That is, the equation for Exis as given by (f) of Example 7.2.3, with £ now

a function of x.

^ = I ^IA

Ex() A(o+Jw) 02 £2[ol + - x+JW(£+ -- x)1

From Coulomb's law

d El dE do I

Pf = x= -(J dx dx d+

+

(jw£ + 0)2

E2 E2 02£2

pfA (1 + T- x)(JW -- + i -)

I02 0 2 2

£2[(01 + - x) + JW(E 1 + x)] [(01+ r- x)+JW(E+ 7 x)]1 it 1 Z

Part b

Consider the effect of a small change in E alone

02 = 0; 2/l << 1

then

Pf 12 2 (f)

A2(jw£1+01)

It is seen from (f) that in the presence of conduction the gradient of C causes

free charge to be stored in the bulk of the fluid. This effect is highly

dependent on frequency, being greatest at zero frequency and disappearing when

the cycle time is short compared to the relaxation time of the material.

PROBLEM 7.22

Part a

In the fluid the consitutive law for conduction is

J = ovE + PfV

Since the given velocity distribution has the property

-32





V'v = 0 (b)

*V p=

o

pf + U 2Lx Pf

pf

(c)

a

V*j = V*(cE) +

or

S+ u = --(d)

The charge is relaxing in the frame of the moving fluid. The solution has the

form ox J(t- x

Pf= Re p e e 2 2 (e)

= 0 elsewhere in the channel

where y = 0 is the channel center. Note that (e) satisfies the boundary condition

xat x = 0 and states that a charge at x at time t has been decaying U seconds

(since it left the source) and was dumped in the channel at time

t' = tU

Substitution of (e) into (d) verifies that it is a solution.

Part b

From (e) it is clear that the wave length of the sinusoidally (and decaying)

charge stream is 2fTU/w. Thus, the wave length can be altered simply by changing

w. One technique for measuring the flow velocity would consist in measuring

the voltage induced across the resistance R (as shown in the figure) as a function

of the frequency. With the distance between electrode centers d equal to 1/2 wave

length, a peak in the output signal would be expected. If we call the frequency

at which this peak occurs wip, then

4 4 o A t 1

-33





27UW-- = 2d

p

or

dwU = --

Tr

Thus, a determination of w gives U. There are, of course, problems with this

approach. For example, there would be lesser peaks in the output at harmonic

frequencies that could be mistaken for the desired peak. Alternatives are to use

the decay rate, but such techniques are vulnerable to conductivity variations

which are likely to be large.

PROBLEM 7.23

Part a

Current is carried by the conductor because of normal conduction and also

because of convection of a net charge.

J = GE + pv

Also

(j-ply)pf/E = V(-P

VoE =

But

ap fV.J = - = 0 in steady state

V*v = '.(U ix ) = 0 also, so that

V'VPf U apfvfVE =a a ax

The solution to this last equation is

p = p e

xi.e., the charge relaxes in the conductor; the time T = is a measure of how long

since the charge left the source at the first screen.

Part b

Let

Ex(x=O) = Eo

aE o(0 a

()- x = p(x)

-- Poo -(Pq)

eax e

-34




PROBLEM 7.23 (Continued) a X

Pf(x)x PU U

E (x) dx+ = E + (1-e

Note that since J (x=0) = OE + poU Vx 0 0 RA

Sx

V poU E U

Ex(X) RA

We must finish the problem to know V

Part c

Sv 2 aE

£U

V = - Ex(x)dx = + P )oo0a (1-e

2

V= ( O ( l-e1+

RAG

PROBLEM 7.24

Part a

The model for this problem is similar to that used in Example 7.2.6 of the

text. Each ring induces a charge on the stream having opposite polarity to its

potential. Thus, conservation of charge for the can at potential v3 (under

the ring at potential v1 ) is

dv v

-C nv = C d+ v (a)

1 1dt

R

Similarly, for the other two cans,

dv v-Ci nv2 =-C d + (b)

dv v

-C i nv 3 = C +2 (c)

To solve these three equations, we assume solutions of the form

^ stv i = vi e (d)

and the complex amplitudes vi are governed by the conditions that follow from

substitution of (d) into (a)-(c)

7cn 0 (C s +)

= 0 (e)

1 Cs)0 (Cs + -) Cin





The solution for s is

s =--+ Cin + j 2 ]r3RC C 2 2

Part b

Thus, the system is unstable if

1 Cn

RC 2C (g)

Part c

In particular, from (g), the system is self-excited as

1 CinI = n(h)

R 2

Part dThe frequency of oscillation under condition (h) follows from (f) and (h),

as

cIn /r2C RC

PROBLEM 7.25

The crucial quantities in the respective systems are the magnetic diffusion

time (Eq. 7.1.28) and the charge relaxation time (Eq. 7.2.11) relative to the

period of excitation T = 1/f. The conductivities required to make these

respective times equal to the excitation period T are

a =i2 T/Po d2 (a)

a = S/T (b)

In terms of the given numbers,

a = (3.14)2(10-5)/(4)(3.14 x 10- 7)(10 - 4

(c)= 7.85 x 10 mhos/m

and

a = (81)(8.85 x 10-12)/10 - 5 = 7.16 x 10- 5 mhos/m (d)

For the change in depth to have a large effect on the inductance, the

conductivity must be greater than that given by (c). Thus, the magnetic device

would not be satisfactory. By contrast, (d) indicates that the conductivity

of the electric apparatus is more than sufficient to make a change in

capacitance with liquid depth apparent even if c=c o . Both devices would be

attractive for this application only if the conductivity exceeded that given

by (c).

-36




PROBLEM 7.26

This problem depends on the same physical reasoning as used in connection

with Prob. 7.25. There are two modes in which either device can operate.

Consider configuration (a): the inductance can change either because of the

magnetization of the water, or because of currents induced in the water. However,

water is only weakly magnetic and so the first mode of operation is not attractive.

Moreover, the frequency is too low to induce appreciable currents, as can be seen

by comparing the magnetic diffusion time to the period of excitation. Hence,

configuration (a) does not represent an attractive approach to the engineering

problem.

On the other hand, configuration (b) can operate either because of a change

in capacitance between the electrodes due to the change in position of the

polarized liquid (at high frequencies) or due to a change in position of a

perfectly conducting liquid (low frequencies). As the calculations of Prob.

7.26 show, it is this last mode of operation that is appropriate in this case.

PROBLEM 7.27

Part a

Because we have changed only a boundary condition,the potentials in regions

(a) and (b) are still of the general form

# = A sinh kx + B cosh kxa(a)

Ob = C sinh kx + D cosh kx

There are now four boundary conditions:

ýa(d) = V (b)

a(o) = o) (0) c)

a( ) (0)

Xat +V -Ve (- C az + C ax(d)

- b(O) (0)

b(-f) = 0 (e)

Only boundary condition (e) is new; it has replaced the assumption that

must go to zero as x + - -.

Solving for A, B, C and D we find that

-37





a = Re 2[(l+jS E/E )sinh kx + jS tanh kf cosh kx]ej( t - k z) (f)a A

v j(ft-k)zOb =

Re [{jS sinh kx + jS tanh kf cosh kx]ej- k (g )

where A = (1+jS c/co)sinh kd + JS tanh kf cosh kd.

Part b

If

Ifk >> 1

tanh kf + 1 (h)

A comparison shows that in this limit the results agree with Sec. 7.2.4 if

we note that

kxe = cosh kx + sinh kx (i)

PROBLEM 7.28

Part a

The regions between the traveling wave electrodes and the moving sheet are

free space, and therefore the fields are governed by

V2 = 0 (a)

where

H= - VD (b)

Moreover, solutions that have the same (z-t) dependence as the imposed

traveling wave potentials, and that satisfy (a) are

Oa = Re[Alcosh kx + A2sinh kx]ej(wt-k x ) (c)

b=

Re[Blcosh kx + B2sinh kx]ej (

wt-kz

) (d)

The constants A ,A2,B,B 2 must be adjusted to make these solutions satisfy

the boundary conditions

a = V at x= c (e)

=bV at x =-c (f)

aa b at x= o (g)

aEa

+ U- ) aEa cE)+E a (h)

Part b

The symmetry requires that

-38

Ya(x,Zt) = 4b(-X,Z,t) (i)

and this implies that A1

= B1, A

2= - B

2 .The boundary conditions become

A cosh kc + A2 sinh kc Vo ()

JS (2A2) = A1 (k)

where

S = (w-kU)Eo/kas (9)

Thus,

A = B = 2j SVo/(sinh kc + 2j S cosh kc) (m)

and

A 2 =- B

2= V/(sinh kc + 2j S cosh kc) (n)

Part b

A section of the sheet can be enclosed by a thin volume of small area in

the y-z plane to give the force per unit area as

Tz= 2T

zXa (x = 0) (o)

where the symmetry has been used to set

Ta b (p.)zx zx

Thus, the time average force per unit area is

<T > = ReE 0 (0)E (0)1 (q)

and from (m) and (n),

<T > = Re[Eo(-jk)A*(-k)A 2 ] (r)

z 2 2

= Re o (s)sinh2kc+4S2cosh2ke

2tok2Vo 2S((t)

(sinh2kc+4S2cosh2kc)

It follows from (t) that the maximum occurs as

S' = 1 tanh kc (u)2

or ok

= kU + -- tanh kc (v)

o

-39





Part c

Note that if S is held fixed at the value given by (u), the force per unit

area remains fixed. Thus, as as 4 0, the velocities of the potential wave and

the sheet must become equal to retain the force at a constant value

w - kU (w)

-40



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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES

PROBLEM 8.1

The identity to be verified is

V-(iýA) = 4IV.A + R-*VI

First express the identity in index notation..

aA

I = + APA ] ýx m x m axm m m

The repeated subscript indicates summation. Thus, expanding the first term on

the left yields:

3Am

"x+ A

m ax0iV.A + A.V (c)

m m

PROBLEM 8.2

We wish to show that

B-v(ipA) = iBviA + AB.V

First, the identity is expressed in index notation, conssidering the mth

component of this vector equation. Note that the equat:io n relates two vectors.

(B.[V(pA)]) m = (B*.[VAI) m + A B.Vm (b)

Now, consider each term separately

(BE[V(1A)])m = Bk (A m ) = A m Bk axk +PBk

aA

(iB*[VA]) =B B

Am*V = AmBk[Vi k = AmBk

The sum of (d) and (e) give (c) so that the identity is verified.

PROBLEM 8.3

Part a

aik is the cosine of the angle between the x axis and the xk axis

(see age 435). Thus for our geometry

/1 1aik

o o

-41

-4





Now, we ma y apply the transformation law for vectors (Eq. 8.2.10)

Ai = aikAk (b)

where the components of A in the (xl,x2 ,x3) system are given as

Al = 1; A2 = 2; A 3 - 1 (c)

Thus:

A' =alk =allA + al2A 2 + a13A3 (d)

A' = 1/2 + /Y (e)

A a2kAk 2+

1 (f)

A3 a3kAk = - 1 (g)

Using matrix alegbra, we can write a more concise solution. That is:

A1 a11 12 13 A1yl

A2 = a 2 1 a2 2a23 A

2 - ! + 1) (h)

L3j La31 3 2 a 3 3 3 1)

Part b

The tensor aik is associated with coordinate transforms involving the

direction of force while the tensor a is associated with coordinate trans

forms involving the direction of the area normal vectors. The tensor

transformation is (Eq. 8.2.17), page 437;

T'j = aik ajTk£ (i)

For example,

T11 = allk Tkt = allallT1 + al2allT21 + al3allT31

+ alla12T12 + a12a12T22 + a13a12T32 (

+ alla13T13 + a12a13T23 + a13a13T33

Thus:

T11= + (k)11 4 4

-42





Similarly

T2 3 +/3

12 2 4

T' = 013

3+r

21 2 4

5 3/22 4 2

T' = 023

T' = 0

T' = 0

T'3 133

Written in matrix algebra, the problem is solved below:

T' T' T] a a11 12 13 11 12 T12 a21 a31

Tl T' T2'a aT22 a22 a3221 T22 T23 21 22

T'31

T'32 33 a 31 a32 T32 a23 a33

Note that the third matrix on the right i. the transpose of aij. Matrix

multiplication of (t) gives

7 6 3

+ 2 ) (-2

+ r34

3 5 3r

ij

PROBLEM 8.4

thThe m component of the force density at a point is (Eq. 8.1.10)

F =

i dax

Thus in the 11 direction,

-43





2 2

aF aT123T

F+

+ T = ( x I-0 o x +0 = 0

X21 2 3 a a

Similarly in the 12 and 13 directions we find

aT aT aT2

F= (axi 22 x1 =0

2 1 a2 X3

aT aT aTFj3 =

3x11

+x2

32

+x3

3

Hence, the total volume force density resulting from the given stress tensor is

zero.

PROBLEM 8.5

I__.e i~> in region (1) E=Eo 3 1 i )

in region (2) E = 0

ii>

I 0b)

3

Tij =E EiE- EoEEk

Thus in region (2)

Tij = 10]

in region (1)

-44





5 E2 3 2

8 o0 2 oo

3 E2 5 2

Tij 2 oo 8 o0 Cc)

0 3 E 2

8 oo

The total contribution to the forces found by integrating the stress tensor

over surface (c) is zero, because surface (c) lies in region (2) where the

stress tensor is zero. By symmetry the sum of contributions to the force

resulting from integrations over the two surfaces perpendicular to the x3 axis

is zero.

Now let us note the fact that:

area (a) = 2 (d)

area (b) = 3 (e)

Thus:

fi = Tij n da (f)

f = fT 1 1 da + fT12da + fT1 3da

(b) (a)

= 8 o

E23) + 32

Eoo

E2 (2) (g)

f= 4 E2(h)

1 8 o0o

f2 = fT21da + fT2 2da + fT2 3da

(b) (a)

2 oo 8 oo3 2 5

f2

3 -4 o

E2

o (j)

f-3 fT 3 1 da + fT 3 2 da + fT 33da

=0 (k)

Hence, the total force is:

7 2 + 1 2 )84E E i i + 3 cEo i()8 0 0 oo 2

-45




PROBLEM 8.6

Part a

At point A, the electric field intensity is a superposition of the imposed

field and the field due to the surface charges; E = (/0f/ )1 . Thus at A,

aE= i(E ) + i (E + --) (a)

x 0 y 0 Eo

while at B,

E= ix (Eo) + Cy(Eo) (b)

Thus, from Eq. 8.3.10, at A,

12 afT = So[E -(E + ) ] CE (E + -) 0ij

0 0

af a 2

CE e( - 0-) [(E +-) E] (c)

S a 20 0 E E2+(E0+ -- ) I

o ý

while at B the components are given by (c) with of + 0.

Part b

In the x direction, because the fields are independent of x and z,

fx = cb-a)[(TxyA -(T y D = (b-a)DEo f (d)

or simply the area multiplied by the surface charge density and x component

of electric field intensity.

In the y direction

2of

f = (b-a)(T - T )D = (b-a)D[E f + 2] (e)y jA Y B 0o

Note that both (d) and (e) could be found by multiplying the surface

charge density by the average electric field intensity and the area, as

shown by Eq. 8.4.8.

-46



FIELD DESCRIPTION OF MAGNETIC AND ELECT IC FORCES

PROBLEM 8.7

I , = L

(4)

Before finding the force, we must calculate the H field at xl = L. To find

this field let us use

- 0nda =J

over the dotted surface. At x1 = + L,

H(xl=L) = Hoi1 (b)

over surface (4) H = 0, and over surface (2), H is in the 1idirection, where

n = 12. Thus over surface (2) B'n = 0.

Hence, the integral in (a) reduces to

- I H0da + pfoH(xl = + L)da = 0

(1) (3)

- oHoa + oHb = 0 per unit depth

Thus:

H(x, = + L) = ýa/b)Ho i I

Hi o kHk

Hence, the stress tensor over surfaces (1), (2) and (3) is:

-OH2

0 0

Tij o -j 2

2 1.

1o 2T 0 - H 0

ij 2 1

o 2- 0 2

H1

-47





over surface (4)

T = [01

Thus th e force in th e 1 direct ion is

=fl Tij n-da

fl=-f T 11da+ f T1 1da+ f T1 2 da

(1) (3) (2)

Thus, since the last integral makes no contribution,

11o 2 Vo 2 a 2 0o 2

f 1 - Ho (a) + -2 2 H2o )b2 *b =o

a b - 1}

Since Tij = 0 over surface (4) there is no contribution to the force from

this surface. and by symmetry, there is no contribution to the force from th e

surfaces perpendicular to th e x 3 axis. Thus, th e force per unit depth in

1 direction is (k).

PROBLEM 8.8

The appropriate surface of integrat ion is shown in th e figure

'I)

tI ~

-II

The stresses acting in the x direction on the respective surfaces are as

shown. Because the plates are perfectly conducting, all shear stresses

required to complete the integration of Eq. 8.1.17 vanish. The only

contributions are from surfaces (i), (ii), (iii) and (iv), where the fields

-48





are known to be

E =

V

ia y

V

-i=a y

(i) ;E

(ii) ; E

=

=

V

b

V

-s•ib

iy

y

(iii)

(iv)

(a)

Thus,

f = (T1 1) ad + (T11) ad (T ) bd (T11) bd

i ii iii iv

= dV2 1 -]

oob a

The plate tends to be drawn to the right, where the fields are greater.

(b)

(c)

PROBLEM 8.9

A- - - )Z.

The volume enclosing the half of the plate is arbitrary so lo6ng as it is

defined so that it does not include additional charge. Thus the volume shown

in the figure encloses no more than the desired distribution of charge. More

over, surfaces (i) and (ii pass through the fringing fields half way

between the plates where by symmetry there is no x2 component of E. Thus surfaces

(i) and (iii) support no shear stress T2 1 . There is no field at surface (iv)

and hence the only contribution is from surface (i), where the square of the

field is known to be

2

E2 V2o (a)1 2

s

1 2and it follows that because T22 on (i) is - 2 CeoE and the normal vector is

negative

4wse V2f o o (b)2 22

s

-49



------------ -----------



The fringing field tends to pull the end of the plate in the + x2

direction.

PROBLEM 8.10

'1 I:

;I(7 IB~

Part a

Consider the surface shown in Figure 1. The total force in the x

direction is:

f= f T da - T da + da TT f T da (a)

1,3 5,7 4 8 2,6

The first four integrals disappear because:

T = CE E = 0 on 1, 3, 5 and 7 because we are nextxy xy

to the conducting plates (Ex = 0)

T = 0 an 4 and 8 because the E field = 0 therexx

Hence

f= T da E2 da (b)x xx 2 y

2,6 2,6

where Tij is evaluated using Eq. 8.3.10.

E = (c)y s

and hence:

-50





f = ESd2( da= - -- v

2(d)

x -2 s

2,6

Part b

The coenergy of the system is

W' = C(x)v2 (e)

2

where C(x) = 2(a-x)d(f)s

Thus, (see Sec. 3.1.2b)

f = =W' 2 3C(x)v2 = ----e

v2

(g)x ax 2 ax s

which is the same value determined in part (a).

Part c

The equation of motion of the plate is:

2M

dt2 x2 + K(x-a) = f = -- V (h)

x s o

When the system reaches equilibrium with the switch closed,

K(X0-a) =-dE

s o(i)

thus

Xo

= asK

V2

o()

After the switch is opened,

M 2 + K(x-a) = - dc v2(t) (k)

dt sdt

The electrical circuit is like an R-C circuit with time varying elements

+÷v R(x)

v + R(x)i(t) = 0 (M)

dv + R(x) d [C(x)v] = 0 (m)

dv dC(x) dxv + R(x)C(x) - + R(x) dxv = 0 (n)

dt dx dt

-51





where:

R(x) s and C(x) 2d(a-x)E (o)2ad(a-x) s

Hence

v + d-- -x a dt La (a-x) d ]

v = 0 (p)

Part d

Dropping the inertial term from (h) leaves:

K(x-a) = - _c v2(t) from (k) (q)S

But we may write the identity

1 dx 1 d= K(x-a) (r)

(a-x) dt K(x-a) dt

and then, from (q)

1 dx s d d 2

(a-x) dt dev2

t) dt s

1 d 2 2 dv

2 dt v t v dt (s)

Substituting back into (p) we have

v+ E -dv + -2E dv= (t)a dt a dt

Solving we find

v = V e-(/3E) t (u)o

and substituting back into (q),

2a

x = a -•dE

V2

e3 E

(v)sK o

Along relaxation time is consistent with neglecting the inertial terms, as

then x(t) varies slowly.

Part e

Proceed as in (c), and record the time constant T of a-x(t) by measuring

the mechanical displacement. Then,

= 22- (w)

a 3

-52





This problem should raise questions as to the appropriate form of Tij

used in (b). Note that the surface of integration encloses liquid as well

as the plate. We want only the force on the plate, so our calculation is

correct only if there is no net force on the enclosed liquid. The electrical

force density in the liquid is given by Eq. 8.5.45. There is no free charge

or gradient of permittivity in the bulk of the liquid and hence the first

two of the three contributions to this force density vanish in the liquid.

However, there remains the electrostriction force density. Note that it is

ignored in our calculation because the electrostriction term was not included

in the stress tensor (we used Eq. 8.3.10 rather than 8.5.46). Our reason for

ignoring the electrostriction is this: it gives rise to a force density that

takes the form of the gradient of a pressure. Hence, it simply alters the

distribution of liquid pressure around the plate. Because each element of the

liquid is in static equilibrium and can give way to motions of the plate without

changing its volume, the "hydrostatic pressure" of the liquid is altered by

the electric field so as to exactly cancel the effect of the electrostriction force

density. Hence, to correctly include the effect of electrostriction in integrat

ing the stresses over the surface, we must also include the hydrostatic pressure

of the liquid. If this is done, the effect of the electrostriction will cancel

out, leaving the force on the plate we have derived by two alternative methods

here.

PROBLEM 8.11

B

(C-2)I

L_ _ _ - -- -

-53

I





First, le t us note the E fields on each of the surfaces of the figure

over surfaces (1), (3), (5) and (7), E1 = 0 (a)

over surface

(6) E2 = -a = 0 (b)

V

(4) E2 - E1 =0 (c)

V

(2) E2 = - E1 = 0 (d)c

From Eq. 8.3.10,

Tij= oEE -E F F (e)

ij oij 2 0ok

Hence, over surfaces (1), (3), (5) and (7)

T12 = 0 (f)

and over surfacesE V 2

(6) T11 = ( (g)S v 2

(4) T11 = - (h)

C V 2

(2) T11 ) (i

Now;

f= fTij nfda = T1 lnlda + T12n2da + fT1 3 n 3 da (j)

IT 1 3n 3 da = 0 because the problem is two dimensional. (k)

Let us consider each of the other integrals:

fT1 2 n 2 da = 0(a)

because the surfaces which have normal n2 are (1), (3), (5) and (7) and by(f) we have shown that T12 = 0 over these surfaces. Also, we get no

contribution to the force over surface (8), because E + 0 faster than the

area 4 m.

Hence the calculation of the force reduces tc

-54





f l i= T6) da T (4) da - T) da2 (m)

(6) (4) (2)2

E DVD o 1 1 1(n)f 0+ 0 (n)1 2 a b c

Note: by symmetry, there is no contribution to the force from the surfaces

perpendicular to the x3 axis.

PROBLEM 8.12

Part a

,, P

77

T10-~ / - / / / / / / ,- , ,, / -

S, -- 7 -

From elementary field theory, we find that

wx2 - wxl/a= o sin -'- e (a)

o a

satisfies V25 = 0 in the region between the plates and the required boundary

conditions. The distribution of E follows from

E = - V4 (b)

Hence,

w - wxla wx wx2- 0o 1 'r2 - os2E -- e In a i - cos -- (c)

a a l a 2

The sketch of the E field is obtained by recognizing that E is directed

perpendicular to contours of constant 4.

-55





Part b

To find the force as the bottom plate, we use surface (2). E = 0 every

where except on the upper side where the normal n = 12 (d)

and the field is

o$ - Txl/a

= - - e ia 2

Hence,

fl = ITi n da = 0

f2 = T2 j n da = T2 2 n 2 da 2

per unit x3, this reduces to

f2 FT 2 2 dIx

1

2.2 1

1 1 ao

but, T EE = -E 0 e22 2 o 2 2

a

and thus

2

f2 1dx

2a

2 4a

-56





Part c

On the top plate, use surface (1). Only the sign of the normal changes,

and the result is

fl =02

2 4a

or the force is equal and opposite to that on the bottom plate.

PROBLEM 8.13

Part a

Tij EiEj - EKEk

Hence:2V 2

223a

2V 2

T21 ooE2E13a

x 1 x 2

Part b

Consider the surface of integration shown in the figure.

D

O

f2 =T 2jnjda = f22T2 1n1da + f22T22 n2 da + f~l3/3da

(2)(3) (1)(4) by symmetry

-57





Let us look at each of these integrals separately

T22daT22n2da - 22da

(1)(4) (1) (4)

over surface (1), E 0 * T220 and hence, the integral is merely:

Sxl=a E° 2V 20

- T22 da 4 = - (- ) (x2 - xl) wdx1Jx =-a 2 3a

x 2 = 2a2a

E V244 o0o

T7 a

Thus,

SV2wSnda 44 oo

S22n2 =27 a

(1)(4)

Let us now evaluate:

I T 21nlda

(2)(3)

Consider the surface shown.

in this region field = 0

hence, no contribution to the

integral over this area.

i Z•r =•

-58



FIELD DESCRIPTION OF AGNETIC AND ELECTRIC FORCES


Thus;x2 =a5d 2V 2

IT21da =x 2a

awx2dx2

(3) x2a 3axl=a

0V2w2 oo

9 a

Over surface (2), we have essentially the same thing, except n = - i

and xl = - a. Hence:

E V2wI 2 oo

ST21da2 a9a(2)

Therefore, the total force in the f2 direction is

E V2w56 00

2 27 a

Part c

0

= + T12n2da + f1T daf1 Tllnlda

(2)(3) (1)(4) by symmetry

f-T1 2da4over (1) we get

I T12n2da

(1)(4) (4) 0 as before

S 2V02 fa x x2wdx1 = 00 3a2 -a X2WdX

x2=2a

Now, over surfaces, 2 and 3

Tl1 1n

da = - T 11da2Tllda 3 0

(2)(3) (2)

because,

T1 1 12 T1113

hence fl = 0.

-59





Part d

a = n E0 (o)

at the lower surface of the movable conductor. The functional relation,

f(xl 2), for the lower surface if the movable conductor is given as

f(x1x2 ) = 4a2 + x - x2 = 0 (p)

the outward unit normal to this surface is

n ~Vf(xlx2) xl I - i(q)

l'2 [ 2

at x 2 = 4a + x1

S1/22 x 24 2 /2

of Eo[n l E + n2 E2 ] = 32 x2 4a 2 2xl

3a 2 J4a +2x

The surface force density (see Eq. 8.4.8) is equal to:

-E + E•

f 2

where, Eb = field just below the charge sheet

Ea = field just above the charge sheet

Since

a = 0, T = 1 a (t)

thus 2 2 2 1/2

e 2V 2 x 4a + xo o0

T 2- 2 ) + x2 x1il- 2i2 4a2 + 2 (u)

3a x2 4a + 2x1 J

To find the total force, the surface force density must be integrated over the

surface. Hence, we find

V 2 a 1/2 1/2

2- 0 fx2 dx12 = 2c a j2x 2 + 4a2 + 4a2} (v)3a -a

If the student wishes, he may carry out this integral, but the complexity of

the integration shows the value of the stress tensor in calculating such a

-60





force. We realize that by using the stress tensor, we have essentially

carried out this difficult integral by an integration by parts.

PROBLEM 8.14

Part a

V

i - xY (a)1 2

a

= - V (b)

hence, V

E ( 2 x2) + 2 - xl) (c)a a

and, from Eq. 8.3.10

Ti = CEE - 6j 1ýE (d)T ijhetress tensor becomes2

Thus, the stress tensor becomes:

V 2 V 2

(-) - (x2-x ) (-) o(xlx2) 0a a

V V 2 E°Tij

- (X1 X2 ) (- ) -2- (x-x2) 02 (e)

a a

V 2 )0 -(-) -•xx 2

a

Part b

Consider the surface shown, bounded by the line segment x2 = 2a, x2 = a,

and xl = a/2 and x1 = a.

XK

1

-61





As before, because the geometry and fields are two-dimensional, the force in

the13 direction is zero. Also, since along surface (1)

,

= constant, thenthe E field = 0, and hence Tij = 0 along this surface. Thus the calculation

of the force on AB reduces to:

fl = - T11da - T12da (f)

(2) (3)

f2 = - T2 1da - f T22 da (g)

(2) (3)

V 2 2a 2 a +

fl 2 oD x2 - (2) ]dx (h)a

a/2

and henceV 2

fl - Eo=- ()aa

Da3 17[*•] (i)

Similarly:

V 2a 1 a 2 2f

2 a Dox(x-a

2 x2d2 2 a/2 l)dxl (J)

and hence

V 2f

2

- oD

o

a3 (- )48

(k)

a

Thus, 2vo 17 31

f = - E D [i - +i 4

o a 1 12 48(-)

2

PROBLEM 8.15

Part a

The E field in the laboratory frame is zero since the two perfectly

conducting plates are shorted. This can be seen by integrating E around a

fixed contour through the block and short and recognizing that the enclosed

flux is constant. Hence,

E'E+vx E , E 0 (a)

and thus

E' v x B = - V1 oHi2 (b)

Therefore we may now calculate J in the moving block.

-62





J' aE' - ap VHo 2

Thus:

F Jx ap 2 VH2i0 o 1

- (3 x B)dV =- i oVH (abD) 10 0 1volume

Part b

The closed surface of integration is shown in the figure below.

All,

""~~ "I'I

' -

IX (

I i A

XI

Since the field is uniform everywhere, the only non-zero components of the stress

tensor are the diagonal elements

T =T 1 H2 T1 H2

11 22 2 oo 33 2 oo

Thus

f1 = i T1da3 - f Tllda2

(3) (2)

= H2 bD - H2bD = 02•o 2 o

Similarly

f2 = T22da - T22da4 =0

(1) (4)

f3 = T 33da5 - T 33da6 =0

(5) (6)

Hence:

Part c

The magnetic field strength and the current density are inconsistant. The

quasi-static magnetic field cannot be uniform and irrotational in a region where

a finite current density exists. The Maxwell stress tensor was developed with

the aid of Ampere's Law (quasi-static) which relates current density and magnetic

field rotation.

S=VxH (k)

F V J x H = o(VxH)x H (1)

For this case, we have assumed that

V x H = 0 (m)

In the limit of small magnetic Reynold's number, (Rm << 1), the motion does not

appreciably affect the field, and the answer found in part a is a good

approximation. There are some problems more easily handled with the stress tensor.

This problem illustrates that in other cases it is easiest to use the force

density J x B directly. Note that we could compute the field induced by J and

then use the Maxwell stress tensor and the self-consistent fields to find the

same force as given by (e).

PROBLEM 8.16

To find the force on the block, we will use the stress tensor over the

surface shown in the figure. Note that the surface is just outside the block.

X,

-64





In the region to the left of the block

I

and to the right H=0=D o 13 ,

Thus:

f n T12n2da + T1 3 da (a)11nda + 3 n

but, since

H1 = H2 = 0; T12 = T13 = 0 (b)

hence,

f = - Tl1da5 + f T1dal (c)

(5) (1)

on surface (5), J 12T o o (d)

11 2 2 dD

on surface (1)

T11 f 0 (e)

therefore 2 2

f1 + 2o2 .Dd = + 2D

(f)(f)D

Similarly, f2 reduces to

f2 = T22da2 - T22da6 (g)

2 6

But, since T22 is a function of xl alone (1 is a function of x l alone) the

two surface integrals are identical, and hence f2 = 0. Similar reasoning

shows that f = 0 and thus the total force is

- ~oodf d

2D i

PROBLEM 8.17

Part a

2-BV =po- (a)

o atAssume a solution of the form:

H = Re [H (x)e iWt (b)

e-z

- joZ o HZ (c)2

ax

-65



VFIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES


Try

KxH (x) = H eKx

where

K2

= •0 °

and hence

K = + (1+j)

Let us define the skin depth as:

6 2

And thus [ (+j)

-H= e 2He iz

Because the skin depth 6 is assumed to be small, and the excitation is on

left,

Hz(large x) + 0 which implies 12 = 0

Hence, - x(l+j)

H(xlt) = H e ejtz

But, our boundary condition at x = 0 is

H(x=O,t) =ReHe = - Re - e iH(Ot)z D z

and thus

(x•) D e +J)ejt i

3H - -K(l+j)

J=Vx H=- ((6)iyx

-D 6

e ejt iy

Part b

= f x E dV= JfxVRdV

f Re dV] + ReL 2 e2jWt dV]

Now, solving each of these integrals:

S2x

2 dV = oaD (1 (1+j) e dx

T Da

I (1+j)i4 D

xx

(d)

(e)

(f)

(g)

(h)

the

(i)

(j)

(k)

(M)

(m)

(n)





2xfixi e 2 2xT(1+j)

1 o•a

2 2jwt (p)4 D x

Hence, taking the real part, the force as in equation (n) is:

S1 oa12(1 + cos 2wt)i (q)

4 D x

Part c

Using the Maxwell stress tensor, we choose the surface shown in the

figure,

O

f = T jnjda = Txxn da + T n da (r)

(1)(3) (2)(4)

Along surfaces (2) and (4), Hx = 0 along the interface between the perfect

conductors and the finite conductivity block. Thus,

T = oHHy = 0 (s)

At surface (3), the field is zero since all current filaments complete a

closed loop circuit with the source through the block. Hence

T = 0 on surface (3) (t)xx

Therefore the calculation of the force reduces to

f =- f T da (u)

T o H2 (v)xx 2 z

And thus,

aDoH

f = o H2 (w)x 2 z

-67

where the field H z is evaluated on surface 1, i.e. x = 0 and is simply given

by the boundary condition (j). Thus it follows

al

a=4D 2 (1 + cos 2wt}ix (x)

which checks with (q). Note that the distribution of J and H, as found in

part (a), are not required to find the total force in this problem. Even more,

(x) is not limited to 6 << x block d1•lension, while the detailed integration is.

Note: We have made use of the rule for products, namely of:

a(t) = Re[Aejwt ]

=Ae +

2A*e

2

b(t) = Re[Be ] = 22

then

-AB* + A*B ABe

2jwt + A*B*e

2jwt

a(t)b(t) = 44

+44

AB* AB 2jwt= Re[ --

2* + Re[-

2e t

avg. value time varying part

PROBLEM 8.18

Choose the surface shown in the figure.

r -----/ 0j r- - - - -

I-. -- -

f fTijnjda = f Tlnlda+ T2n2da + T3n3da (a)

3,4 5,6 1,2

Since the plates are perfectly conducting, E1 = 0 at surfaces (5) and (6)

and .hence T12 = 0 on surfaces (5) and (6). Surfaces (1), (2), (3) and (4)

are far from the body so

-68





V

E= i (b)d z

at each of them, and thus, on surfaces (1) and (3), T1 3 0=. Therefore,

fl= - 11da3 + f Tllda (c)

(3) (4)

T(3) T ) o (V (d)11 11 2

and a3 = a 4 (areas). Hence,

fl = 0 (e)

PROBLEM 8.19

Part a

Since the system is electrically linear,

S= BI + Br (a)

where BZ and Br are respectively the fields from the left and right wires.

The force on a unit length of the right wire is

=r x B da = Jr x , da + x BE da (b)

but,

Jr x Br da = 0 (c)

ane hence,

f = J x Bi da (d)

Since, we don't need the fields near the wire,

P I x2 1- (xl+a) 2

9 2 (xl+a) + x2 e

0l -x2 1 + (x 1-a)12S

r 2 (x -a)

Hence,

fr = rx B da

2p o1 (2a)i

r 2w1 2(2a)

22+x 2 j

(f)

- I 3 x Bz (xx (g)1 a, x2=0)

pl2

4wa 1 (h)

-69





Part b

I

I /

Along the symmetry plane of the surface shown in the figure

- o (-2a)(i)

=2w 2 2 2(a +x 2 )

The terms of Tij go as B2, but B2a - and the surface area goes as 27R on surface

(2), hence the contributions of the stress tensor will vanish on surface (2)as

R-o; we need only compute the integral on surface (1). Because H1 0 in the

plane xl = 0

f= f-T1 1da = 2 dx2

o Ia2 dx 222 2

- -j (a +x2)

Solving this integral, we find

Io2

f P (k)1 4ra

also

f2 f3= 0 (Y)

since

T21 T = 031 (m)

and hence the total force is that of (k) and it agrees with that determined

in part (a).

-70




PROBLEM 8.20

ioL7

\~

4-21 0| • | !

- X.IF- O .j~ Ir

/

0 /

I~/

Part a

Use the contour indicated in the figure. At infinity the fields will go

to zero, and hence there will be no contribution to the force from the semi

circular part of the area, i.e. surface (2).

Along the line x2 = 0, E2 = 0 by symmetry and

2 XE1 = ( )sin8

o

2 2 2r = a + X

x1 x

sinG =r 2

Henceo x1

X X1E1 = oe a2+x2

f2 = T2jnda = T21nda + t22n2da + T23n3da

(1) (1) (1)

first and last integrals = 0, n 1 and n3 = 0 on surface 1

2

T 2 2)o2 2

22 2 1 2 c ft (a +x )2

-71





Thus

f2=- 2

f2

2c

SXl2

dxl

x2 22o (a +x )

f2

2 P4a a

Part b

From electrostatics,

f = AE

From the figure, we see that

E(x2=a) = 2 (2a)

Hence,2

4we 0 a 2

which is the same as we obtained using the stress tensor - (see equation (h)).

PROBLEM 8.21

Part a

From Eq. 8.1.11,

BBXy 0lo

0Tij 2L0 x y

0 1 2 20 - (-Bx -By)

2v x y0

where the components of B are given in the problem.

Part b

The appropriate surface of integration, which is fixed with respect to the

fixed frame, is shown in the figure.

We compute the time average force,

tod.h iUbL rL U i I LUL U fan ence contr ut. ons rom SUL aces

(1) and (3) cancel. Fields go to

zero on surface (2), which is at

y-. Thus, there remains the stress

Ion surface (4). The time average

value of the surface force density T + c c~ea

-72

is independent of x. Hence,

T = - <T (y=O)> (b)y yy

T <-B2

+ B2> (c)y 21 p x y

Observe that

^ -JkUt

<Re A -e

jkUt Re B e kUt>

22Re A B* (d)

where B* is complex conjugate of B, and (c) becomes

T R-kx (-jklj 0K°) jkx (jki K ) jk x

e-Ty Re-( Ko

ekX

) (p K )+ e e

0

2 k2 (e)

=4

(1aa*

) (f)

Finally, use the given definition of a to write (f) as

T = - - (g)y 4 U 2

S1+ (-10-)

Note that T is positive so that the train is supported by the magnetic field.Y

However, as U-O (the train is stopped) the levitation force goes to zero.

Part c

For the force per unit area in the x direction;

1T - (h)

x 21 ° x y

1 Re[VK e j k x

(jk0 -jk (i) Re Kek K ejkx M2V a* o0

Thus •K2 V aU

Tx 0 Re j 1- jSaU 2 1/2

2[1 + (-ý) I

As must be expected, the force on the train in the x directions vanishes as

U-O. Note that in any case the force always tends to retard the motion and

hence could hardly be used to propel the train.

The identity sin(e/2) = + /(l - cosO)/2 is helpful in reducing (j) to

the form

- K2 p rU 2

T=° 0 ( 1+ (-e- 1) (k)

11 0oU--2 1/22-

-73




PROBLEM 8.22

This problem makes the same point as Probs. 8.16 and 8.17, with the

additional effect of material motion included. Regardless of the motion,

with the current constrained as given, the magnetic field intensity is zero

to the right of the block and uniform into the paper (z direction) to the left

of the block, where

I

H= i 0 (a)zd

The only contribution to an integration of the stress tensor over a surface

enclosing the block is on the left surface. Thus

f = ds T = - ds1 H (b)x xx 2 0oz

2I

ds 1o 0) (c)

The magnetic force is to the right and independent of the magnetic Reynolds

number.

PROBLEM 8.23

In plane geometry, a knowledge of the charge on the upper plate is equivalent

to knowing the electric field intensity on the surface of the plate. Thus, the

surface charge density on the upper plate is

I t Ia = I coswt dt sin wt (a)

f A o a)0

and

EX

(x=a) =Qf

E AE

I0W

sin wt (b)

o o

Now, we enclose the upper plate with a surface just outside the electrode

surface. The only contribution to the integration of Eq. 8.1.17 using the

stress tensor 8.3.10 is

A 2

f = - AT (x=a)=

o E2(x=a) (c)x xx 2 x

which we can evaluate from (b) as

AE I 22

fx o2 ( ow)sin20t (d)o

The force of attraction between the conducting slab and upper electrode is not

dependent on 01 or ao .

-74




PROBLEM 8.24

The force on the lower electrode in the x direction is zero, as can be seen

by integrating the Maxwell stress tensor over the surface shown.

t

The fields are zero on surfaces (2), (3) and (4). Hence, the total force per

unit depth into the paper is

f f= T dx (a)

where contributions from surfaces in the plane of the paper cancel because the

problem is two-dimensional. Moreover, by symmetry the electric field intensity

on the surface (1), even in the fringing regions, is in the y direction only

and T = C E E in (a) is zero. Thus, the total x directed force is zero.xy oxy

PROBLEM 8.25

The force density in the dielectric slab is Eq. 8.5.45. Not only is the

first term zero, but because the block moves as a rigid body (we are interested

only in the net force giving rise to a rigid body displacement) the last term,

which originates in changes in volume of the material, does not give a

contribution. Hence, the force density is

= E.Ev- (a)

2

and the stress tensor is

6

T = EEIE - - c k E (b)

Note that, from (a), the force density in the xl direction is confined to the

right edge of the block,.where it acts as a surface force. Thus, we obtain the

total force by simply integrating over a surface that encloses the right edge;

-75





fl = aD E +1 (Eb) (c)o(E2

where a and b are to the right and left of the right edge of the slab. Also

Ea =2

E2

= - V/a.

0Hence (c) becomes

V 2

fl 2 (o) (ECo) (d)d)

The force acts to the right, as could be computed by the energy method.

PROBLEM 8.26

Part a

The force density for polarizable materials is:

- 1 1 -

F =- E*E VC + - V(E*E p -) (a)2 2 ap

The second term on the right side represents electrostriction. Note that

this is a case where the material volume must change, and hence the effect of

electrostriction is important. Sincd free space and the elastic bulk are homogeneous,

changes in permittivity and ac/ap occur only at the boundary where

the permittivity is discontinuous. The upper and lower elastic bulk surfaces

are constrained by the plates. Thus only the xl component of force is pertinent.

Since the left-hand edge is fixed, any stress arising from the discontinuity in

permittivity at that boundary is counterbalanced by the rigidity of the wall.

Therefore, all of the force arises at the right-hand boundary which is free to

move.

The closed surface of integration is shown in the figure.

-76

FIELD DESCRIPTION ON MAGNETIC AND ELECTRIC FORCES


1 9eTij EE i J 2 ij - p EkEk (b)

1Since a/c << 1 and b 2- a the field at the dielectric interface is essentially

uniform.V

- 0oE-= - i

2 aW (c)

The relevant components of the stress tensor are:

S 2 1 ae 2T - 2 + p E2 (d)11 2 2 2 ap 2

T12 = E1EE2 =0 (e)

f = n11nda + T 22da (f)

(1)(3) (4)Hence

f fT11da3 - fTldal

(3) (1)

ES 2

VT)

2(aD) -

E 1 V(1

2) (aD) + P

V 2(aD) (g)

Thus;

(E- Eo)V2 D V2 D

00 r 0(h) 1

12a

o2 d ( a

)

Part b

In order to use lumped parameter energy methods, the charge on the upper

plate will be found. The permittivity of the dielectric bulk is a junction of

the displacement of the rightýhand edge. That is, if mass conservation is to

hold,

po abD = (po + Ap)aD(b+F) (i)

where

P = Po + Ap, Ap = 0 if 5 = 0 (j)

Thus, if Ap << po and E << b, to first order

Ap = -p (k)-Po b

(see Eqs. 8.5.9 and 8.5.10)

Furthermore, to first order, using a Taylor series,

-77





a p Po 2E

1 +p 1 b ap

Also, the electric field will be assumed as uniform everywhere between the

plates. Hence; in the block

V

-2a + Ap] } (m);-T 22 [2 l

to the right of the block

V

D=-12a-E (n)2 a 0 C2 (o

By employing Gauss's law, we find the charge on the upper plate as:

V p0 V

q = (P - p E}(b+)D + Eo--)(c-b-Q)D(o)

dw = fqdv + f dx (p)

integrating we find

w' = i(a 11 E 22 a o(c-b-)D b ap (b+ý)D + 1 (q)

Thus,Thus, (E-E)V2D V2

fe

eoec v=V

-- 02a

o 1

2

o

a o

_E

pr)(PD

o

Second order terms have been dropped in the co-energy expression (alternatively,

first order terms can be dropped in the force expression).

Part c

If the result of part (a) is written for p = po + Ap, where po >> Ap,

then the answers to part (a) and (b) are identical to first order. This

should be expected since the lumped parameter approach assumed a value for

permittivity which was correct only to first order.

PROBLEM 8.27

The surface force density is

T = [Ta - Tbn (a)

m mn mn n

For this problem, we require m = 1 and n = 12. Thus

T 1 = (T 2 - Tb2 ) (b)

From Eq. 8.5.46,

-78





T1 =EoEEa EEbEb (c)

Note that E2 = E2 (see Eq. 6.2.31). Moreover, because there is no free charge

coEa = CE1 (see Eq. 6.2.33). Thus, (c) becomes

T1 = 2 ol - EEl] f 0 (d)Ea[EoE I

That the shear surface force density is zero in the x3 direction follows the

same reasoning.

PROBLEM 8.28

The force density, Eq. 8.5.45, written in component form, is

F = E ac a 1 ac (a)i i ax 2 Ekk +a EkEk ) (a)

The first term can be rewritten as two terms, one of which is in the

desired form

a i 1 De 1 E) (b)

i -x (i j ax 2 k k ax ax 2 k k -5p (b)

Because V x = O, aDE/ax j = Ej/Dx i so that the second term can be rewritten

and combined with the third. (Note the j is a dummy summation variable.)

a a i a i

Finally, we introduce 6 (see Eq. 8.1.7) to write (c) in the required form

aTF= (d)i ax

where

Tij = C E - (E-p ) (e)This is identical 8.5.46.o E.

This is identical to Eq. 8.5.46.

-79



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SIMPLE ELASTIC CONTINUA

PROBLEM 9.1

The equation of motion for a static rod is

d260 = E d

dx2+ F where F = pg

x x

We can integrate this equation directly and get

26(x) = - () + Cx + D,

E 2

where C and D are arbitrary constants.

Part a

d6The stress function is T(x) = E x and therefore

dx

T(x) = - pgx + CE. (c)

We have a free end at x = Z and this implies T(x=2)=O. Now we can write the

stress as

T(x) = - pgx + pg£. (d)

The maximum stress occurs at x = 0 and is T = pg£. Equating this to themax

maximum allowable stress, we have

2 x 109 = (7.8 x 103 )(9.8)

hence

t = 2.6 x 104 meters.

Part b

From part (a)

T(x) = - pgx + pg£ (e)

The fixed end at x = 0 implies that D = 0, so now we can write the displacement

2

6(x) =- •) +--E x)

Part c

6(R) = -g2

E 2 E 2E

For 2 = 2.6 x 10 meters, 6(£.) = 129 meters. This appears to be a large

displacement, but note that the total unstressed length is 26,000 meters.

-80




PROBLEM 9.2

Part a

The equation of motion for a static rod is

2

0 = E22 + pg (a)

dx

If we define x' = x-L1, we can write the solutions for 6 in rod I and in rod 2

asapg 2

61 (x) = E ) + C2 + D1 (b)

andp2 g ,2

62(x) - + C2 x' + D2 (c)

d6where C1,C2,D1, and D2 are arbitrary constants. Since T = E 1we can also write

the tensions,

T1(X ) = - plgx + EI C (d)

and

T2(x') = - 2 gx' + E2C 2 (e)

We must have four boundary conditions to evaluate the constants and they are:

6 (X=O) = 0, (f)

6 2 (x'=0) = 61 (x=L 1) (g)

0 = - A 1T(x=L1 )+A2T2 (x'=0) + mg, (h)

and

0 =- A2T2(x'=L 2) + Mg + fex (i)

where fe is found using the Maxwell stress tensor

x 2

o 0oe x 2d2

where we assume d >> 6(2) (x'=L2).

Equations (f), (g), (h) and (i) serve to define the constants of integration.

Substitution of (b)-(e) shows that

DI

= 0 (k)

plg L

E + C1L1 + D1 - D2 =0 (£)1

-81





-A1[-PlgL 1 + E1C1 ] + A2[E2C2] + mg = 0 (m)

E A V2

-A2 -P2L2 + E2C2] + Mg + o M_2d 0 (n)

2d

Solution of these expressions, beginning with (n), gives

C2=

g+

2d2 +P2gL 2A2 A E2 (o)

and hence

C1 = g + gL1 A1 + A2E2C2] A1E

SoA

V2

= {[(M+m) + 1 L1A1 + P2L2A2 2 AE (p)

12 2d2

L1 1M1A1 +oA+M2

D=2 AE

{[ClL)++2 P

22L2A2 2d2

0 (q)

D1 = 0 (r)

Thus, (b) and (c) are determined.

PROBLEM 9.3

Part a

Longitudinal displacements on the rod satisfy the wave equation

2 2

p = E and the stress T = E -(a)2 2 ax

at ax

We can write 6(xt) = Re[6(x)e ] for sinusoidal excitations. 6(x) can be

written as6(x) = C1sin ýx + C2 cos fx where = ww/p7E. The two constants are

found from the boundary conditions

2

M 2 6(£,t) = - AT(k,t) + f(t) (b)

at

6(0,t) = 0. (c)

These conditions become

2 ds-M2 6(£k) - AE - (k) + f 0 (d)

dx o

and

S(0) = 0 (e)

-82





for sinusoidal xcitations.

Now we find C2 = 0 and

f

Co (f)

AE~cos8t- MW sin8t2

Hence,

6(x,t) = sin~ 2 Re[foet (AE$cos 8t-Mw sinat

and

T(x,t) = E EaEcosx Re[f wt] (h)

x AE~cos$Z-Mw sin 2 0

Part b

At x = 2,

6(t,t) = Re[foe et (i)

AEMcotaS-Mw2

where 8cot8£ = wv 7T cot (wt/r7E).

For small w, cot(wt/pE) + 1 and

6(1,t) MAE f(t) (j)MW

This equation is as used to describe a mass on the end of a massless spring:

2

Mdx = - Kx + f(t) (k)o dt

2

and H Mto

x = Ref[eJWt],

- M x = - Kx fo, (0)

or

x f(t) (m)K-M w

Comparing (j) and (2)we note that

K = AE and t = Z. (n)

Our comparison is complete and since M >> pAt we can use the massless spring model

with a mass M = M on the end.0

-83




PROBLEM 9.4

A response that can be represented purely as a wave traveling in the negative

x direction implies that there be no wave reflection at the left-hand boundary.

We must have

v(O,t) + 1 T(O,t) = 0 (a)

as seen in Sec. 9.1.1b.

This condition can be satisfied by a viscous damper alone:

AT(O,t) + Bv(O,t) = 0 (b)

Hence, we can write

B = ArpE

M = 0 (c)

K = 0.

PROBLEM 9.5

Part a

At x = k the boundary condition is

360 = - AT(£,t) - B 2- (£,t) + f(t) (a)

Part b

We can write the solution as

6(x) CI sin Bx + C2 cos 8x, (b)

where a = w . At x = 0 there is a fixed end, hence 6(x=O) = 0 and C2 0.

At x = Z our boundary condition becomes

F = JwB6(x=2) + AE d•x (x6=), (c)o dx

or in terms of C1;

Fo = wBC1 sin 8£ + AESC1 cos at (d)

After solving for C1, we can write our solution as

F sinSx(x) = o (e)

AE~cosaB+jwBsini(

Part c

For w real and B>O, 6 cannot be infinite with a finite-applied force, because

the denominator of 6(x) can never be zero.

Physically, B>O implies that the system is damped and energy would be

-84





dissipated for each cycle of operation, hence a perfect resonance cannot occur.

However, there will be frequencies which will maximize the amplitude.

PROBLEM 9.6

First, we can calculate the force of magnetic origin, fx, on the rod. If

we define 6(9,t) to be the a.c. deflection of the rod at x = Z, then using

Ampere's law and the Maxwell stress tensor (Eq. 8.5.41 with magnetostriction

ignored) we find

=f 2 (a)X 2[d-6(ý,t).]

This result can also be obtained using the energy methods of Chap. 3 (See

Appendix E, Table 3.1). Since d >> 6(t,t), we may linearize f :

2oAN22AN22

fx

-2d

2 +d3 S(£,t) (b)

The first term represents a constant force which is balanced by a static deflection

on the rod. If we assume that this static deflection is included in the

equilibrium length X, then we need only use the last term of fx to compute the

dynamic deflection 6(£,t). In the bulk of the rod we have the wave equation;

for sinusoidal variations

6(x,t) = Re[6(x)e

1j t ](c)

we can write the complex amplitude

6

(x) as

6(x) = C sin Bx + C cos ýx (d)

where a = U4. At x = 0 we have a fixed end, so 6(o) =0 and C2 = 0. At x = Z

the boundary condition is

0 = f - AE (£,t), (e)x x

or AN2 2p AN d

0 = 6(x9=) - AE (x=9) (f)3 dx

Substituting we obtain

p AN2 2

d3 C1 sin 8£

=C1 AEa cos BZ (g)

Our solution is6(x) = C1 sin ax and for a non-trivial solution we must have

C1 j 0. So, divide (g) by C1 and obtain the resonance condition:

-85





i AN212

( 3 ) sinBU = AEB cos B$

d

Substituting B = w and rearranging, we have

3~(N212 = tan( .-

-o1N212t

which, when solved for w, yields the eigenfrequencies. Graphically, the first

two eigenfrequencies are found from the sketch.

I II I

StE~A 3/~44 ,N

1Ejc.A3'

;IE-

Notice that as the current I is increased, the slope of the straight line decreases

and the first eigenfrequency (denoted by wl) goes to zero and then seemingly

disappears for still higher currents. Actually w1 now becomes imaginary and can

be found from the equation

oL ( IfwlI I) = tanh( ol Z)10N212Z

Just as there are negative solutions to (i), -wl, -w2 " etc., so there are now

solutions + JIwlI Thus, because wl is imaginary, the system is unstable,

-86

(amplitude of one solution growing in time).

Hence when the slope of the straight line becomes less than unity, the system

is unstable. This condition can be stated as:

STABLEEdEd > 1 (k)

S0N2 2

or

UNSTABLEEd

2 < 1 (a)

pN2 2

PROBLEM 9.7

Part a

6(x,t) satisfies the wave equation

2 2

p 2-- E 2 (a)

at ax

acand the stress is T = E 2-. We can write

ax

6(x,t) = Re[•6(x)eJ t

] (b)

and substitution into the wave equation gives

d2ýd + B2 6 = 0. (c)

dx

For x > 0 we have,

6(x) = C1 sin ax + C2 cos ax (d)

and

Ta(X) = C1ER cos ax - C2Ea sin 8x (e)

and for x < 0 we have,

6b(x) = C3 sin ax + C4 cos Bx (f)

and

Tb(x) = C3ER cos ax - C4 EB sin ax(g)

Part b

There are four constants to be determined; thus we need four boundary

conditions. At the right end (x=L), we have

6 (x=L) = 0 (h)

-87





and the left end,

6

b (x-L) = 6oe

There are two conditions at the middle (x=O),

6a(= 0+ ) = 6^(x=O

and

(i)

(1)

-Mo2

6a(x=O) = ATa(x= +

) - ATb(x=O) - 4K6a(x0) (k)

Part c

Solving for C1 ,C 2 ,C 3 , and C4 we obtain

-j

-6 AEBe cotBL

c o()1 sin6L(4K+2AEBcotBL-M 2 )

6 AEBe

C2 sinBL(4K+2AEBcotSL-M 2)

(m)

11E 7-j -j

6 AE~e cotBL 6 e= o o (n)

3 sinL (4K+2AEBcot L-Mw2) sRL)

C =4C2 (o)

Thus, (b), (e), and (g) with these constants give the desired stress distribution.

PROBLEM 9.8

In terms of the complex amplitudes, (k) and (r) become

LI

T'(0) = i (R) - text (a)

andLLI

T'(Z) = o i' (r) - text (b)aA

where i - Gv0

Equation (t) without the approximation becomes

^ GLo (P+p) ^ L Iv = - j + j 6 (c)o 1-11 o a o

Using the steady-state solutions for the rod, we can solve for T(x) in terms of

the boundary values T(O) and T(P):

-88





T(x) T(O) sin[k(k-x)]sin[kt]

+ T() sin[kx]sin[k2]

(d)

then

1 [ cos[k(-x)] cos[kx] (e)

sin[kt] sin[k]

From (a) and (b), this becomes

)_ 1 lo_ 1 o ^ cos[kt~6(M) = 60 i v (f)

o a•aA i sin[kt] aA o sink f)

Thus, in view of (c) solved for 60, we obtain the system function

H(w) =

i 2 wGL (1+1) 2i cosfkt]+j/p ()(a oI ) sin[kP]- o AC G(+a Isin[k)]

o o

(g)

PROBLEM 9.9

Part a

First of all, y(t) = 6(-L,t) where6 (x,t) = Re[6(x)eJet]. We can write the

solution for 6 as 6(x) = C1sinBx + C2cosBx, where = w/pE. The C2 is zero

because of the fixed end at x = 0(6(0) = 0). At the other end we have

2

M

2

(-L,t) = A2 E-xL

(-L,t) +fe(t) (a)

Bt

Using the Maxwell stress tensor, (or the energy method of Chap. 3) we find

Ae N2 [ -(t)]2 [I( + I(t) 2 (b)

[d-D+6(-Lgt)] [d-D-6(-L,t)]2

which when linearized becomes,

fe(t) - C I(t) - C 6(-L,t), (c)

where

2N2p AI 2N21oAI 2

CI 02 ;C

(d-D)2 'y (d-D)3

Our boundary condition (a)becomes

2 d6 ^-Mw 6(-L) = A2 E dx (-L) - C I - C 6(-L) (d )

Solving for C1 we obtain

-89

CI

CI 2 , (e)A2EcosSL - (M -C )sinBL

and we can write our solution as

y(t) = Re[-C sinBL ejWt]. (f)

Part b

The transducer is itself made from solid materials having characteristics

that do not differ greatly from those of the rod. Thus, there is the question

of whether the elastic response of the transducer materials is of importance.

Under the assumption that the rod and transducer are constructed from materials

having essentially the same elastic properties, the assumption that the yoke

and plunger are rigid, but that the rod supports acoustic waves,is justified

provided the rod is long compared to the largest dimension of the transducer,

and that an acoustic wavelength is long compared to the largest transducer

dimension. (See Sec. 9.1.3).

PROBLEM 9.10

Part a

At the outset, we can write the equation of motion for the massless plate:

-aT(l,t) + fe (t) = M (£,t) 0 (a)at

Using the Maxwell stress tensor we find the force of electrical origin fe(t)

to be

(V0 - v(t))

6 (b)(d6 (L, t)'Zý"to'-5

Since v(t) << Vo and 6(£,t) << d, we can linearize fe(t):

2E AV2 2E AVfe(t) [oaV (,t) + v(t) (c)

Recognizing that T(£,t) = E ýx (£,t) we can write our boundary condition at

x = R in the desired form:

2E AV2 2c AV

ax (t) 0 6(,t) + o2 v(t) (d)d d

-90





Longitudinal displacements in the rod obey the wave equation and for an

jWt

assumed form of 6(x,t) = Re[6(x)et] we can write

6

(x) = C1sinax + C2cosax,where 8 = wr/7E. At x = 0 we have a fixed end, thus 6(x=0) = 0 and C2 = 0.

From part (a) and assuming sinusoidal time dependence,we can write our boundary

condition at x = 2 as

2c AV2

2E AV

aE L(Z)dx

=3

d

6() +2

d

o (e)

Solving

2E AV VC1 o o (f)

2c AV

2

aEd28cos - o o sin S£d

Finally, we can write our solution as

2eoAVo jwt ]6(x,t) = FAV 2 Re[Vet (g)

2S£

2 AVaEd

2cos - d sinat

PROBLEM 9.11

Part a

For no elastic wave reflection at the right-hand boundary we must have a

boundary condition of the form

v(0,t) + 1 T(0,t) = 0 (a)

(from Sec. 9.1.1b). Since v(O,t) = -6 (0,t), we can write

-- (O,t) = T(O,t) (b)

If we write the boundary condition at x = 0 for our example we obtain

0 = - ST(O,t) + fe(t), (c)x

or for perturbations

0 = - ST(0,t) + fe (t) (d)a.c.

Combining (b) and (d)

fe (t) = - - (0,t) (e)pEa.c. at

-91

and since 365/t (O,t) = dys/dt,

dy_

fec (t) = - SPa.c. dt

The perturbation electric force can be found using the Maxwell stress tensc

(using a surface of integration similar to that illustrated by Prob. 8.10):

E V2D E V2D 2E V Dve o oo + o s

ft) -- +x a a a

2E V Dv

where we associate fe (t) s

a.c. a

Equation (f) now becomes

2eo2

oDv

s = S/pf -dy

(Oh)a edt

Now that we have dealt with the force balance we can write the circuit equat ions.

The capacitance of the

+. V. device is found to be

C =

Note that q = Cv and i=d-.

The basic circuit equation is

v+iR = V =v+R =v+RC d+vd (i)o dt dt dt

Substituting, we obtain

2E DR

V = v + RC +dv o dy

o dt a dt

and for perturbation quantities,

dv 2E DV R dy ss o o

O = v + RC + 0s o dt a dt

Since w << v >> RC dv /d t and now we have

RC s R o

2E DV R dy

0 v +s a dt





Equations (h) and (t) must be satisfied simultaneously and this can occur

only if

2E DV0R aSp (m)o = (m)a 2c VD

oo

Finally from (m) we have the condition on the d.c. voltage,

Vo= S- (n)1/2V

o-

2E D[SvR l (n)

PROBLEM 9.12

Part a

Note that there is no mutual capacitance between the two pairs. We can find

the capacitance of the left-hand pair of plates to be

d( - y2) Eod(' + y2)c = + (a)2 h h

The current 12 can be found from 12 = dq2 /dt = d(VoC2)/dt = V dC2/dt,

and upon substitution of C2 we obtain

i (E-E )Vdd]dy (b)

2 h dt

If we solve for y2

in terms of vs

our job will be done.

Define the y-axis from left to right with y= 0 at yl

= 0. Assume all

constant forces (with v = 0) to be balanced and consider only the perturbations.s

If we assume for the rod 6(y,t)=

Re[i(y)ej t ]

then we can write

6(y) = C1 sin By + C2 cos By (c)

where = w/plE . (We have assumed that the electrical forces act only on the

surfaces of the rod. This is evident from the form of the force density, Eq.

8.5.45, if the effect of electrostriction can be ignored.) At y = 0 there is

no perturbation force and for a.c. deflections we have a free end condition:

Ad6

0 = T(O,t) E - (y = 0) = 0 (d)dy

This forces C1 to be zero. At y = I we can write the boundary condition as

0 = - hdT(t,t) + fe (t)a.c.

-93





Using the Maxwell stress tensor (or energy methods, as in Sec. 8.5.4)

fe~t)

(Ec-o)dfe(t) = 2h (V + v )2 (e)

Linearizing and ignoring the d.c. term we have

(E-E )V d

fe

(t) =o o

v .la.c. h s

From the boundary condition for complex amplitudes we obtain

(Ec-Eo)V d0 = - hdE T () + h v (f)

dy h f

Substituting and solving for C2;

-(E-E o)Vo ^2 v . (g)

S ks2E sin

Recognizing that Y2(t) = 6(0,t), we can now write

h Eh sin f3,

Since 2 =

-(CE-cE)Vodh

dY2 , we have

h dt 22

i2 = Re v et (i)

2 L 3 E sin t sv

Finally, we can write

2 o22

Y(jw) ^ - 3(j)v h ER sinr3

s

Part b

The poles can be found from

h 3Ea sin Pt = 0 (k)

where 8 = wp7iE. The lowest nonzero frequency can be found from

sin(wlJpEj = 0 to be

Note that the A)= 0 is a pole because the rod is free to translate slowly between

the plates.

-94



U


PROBLEM 9.13

Part a

The flux for the left-hand transducer is

2poN

X = 2 2wR(a-6(0,t))i£2,

and for the right-hand one,

2N

Xr = 2I7R(a+6(L t))i r(b)

For this electrically linear situation we have W21

2Li 1

Xi and f=

awlm

Hence we find, to linear terms

ff -

N2

°g

2rR(I

2

o+ 21 i)

o(c)

and, because ir = I - out

2

f = N2 rR(I - 21 Gv )r g 0 o out

Part b

For the left-hand transducer, an acceptable stress-tensor surface is shown

below ,

T Ac eC

and the mirror-image is acceptable for the right-hand transducer. Application

of f = f Txjnda to the two surfaces yields the same result as in part (a).

S

-95

Part c

The wave equation holds in the rod for 6(x,t). Assuming 6 = Re[6(x)eJWt],

we have 6(x) = C1 sin B x + C2 cos 8x, where = w/pE. At x = 0, f= -T(0,t)(rrR)

which yieldsN2I

211i0N 0 A1

T(0) = Rg = C I,ci

which in turn implies C1 =E . At x = k, fr = T(L,t)(,R ), which will yield C2.

The only other relation we need is the electrical circuit equation, which

we can find from

out dt Eze;G

to be

IoL1 ji 6(L)

v (e)out a(l+j GII1w)

where L1 = N2(27Ra)/g.

Finally we can write G(w) as

S=out jIoL1CI

= aE~sin8L(l+jGL1w)-jwGC IoLcosf(

Part d

1If G << so that the self inductance of the output transducer is negligible

1and the system is matched so that a/iE = G C IoL 1 we have

Vout JIoILICI,out=oa I C (g)

I aJrE [sinSL-jcosBL]

and

Vout o L 1 =0 a(h)

PROBLEM 9.14

Part a

With no perturbations and no volume force in the rod we know that the

stress, T(x1), will be constant. At x1 = 0,

0 = - AT(x = 0) + fe 2 (a)SV A

where, using the Maxwell stress tensor, fe = Hence,2

2d

-96





C V2A

T(x1 212Ad

Part b

The velocity of the wave will be v = p-•nd the transit time will be

td = L/vp

Using Table 9.1 we have

t 1 = 1.96 x 104

sec.d 5100

Part c

This part is similar to Prob. 9.11, where our condition for no reflection

fe (t) = - AA/ip-- (O,t)a.c. at

Using the Maxwell stress tensor

A v2 cA V2 E AVfe Ao - olo

+ ol o v'

2d 2d2

d2

where v = v' + V . Here, we ignore the effect on fe of the change in d resultingo

from the motion of the plate.

Writing the circuit equation we have

iR + v = V = R + v = R C dC

o dt d-t dt

The capacitance C is

ooA 1o = 1 + oal•

d-6(0,t) d d2 6(0,t)

Our equation becomes

0 = v' + RA1 d+RV E

21 6 ,t)

d dt o atd

and since

SEoA1R dv'

d dt '

we have

-97




PROBLEM 9.14 (continued)

RV e A

v' = 22

a t(0,t) (g)

d

Now we can use this result to write fe = EAV v'/d2, and the condition

a.c. o o

that this force take the form of (c) requires

A/p$E d = RV2 2A , (h)o o1

or equivalently

R = A/CpSd4 (:)

E2A2V2

PROBLEM 9.15

Part a

We have from the problem statement

ip(z+Az) - P(z) = 3TAz .

If we take the limit Az + 0, then we obtain

1 3IT =az

Part b

We can write the equation of motion directly as

2-

(JAz) • = T(z+Az,t) - T(z,t).

at

Dividing by Az we have

j aý = T(z+Az,t)-T(z,t)2 Az

at

Taking the limit Az + 0 we obtain

at2 az

Part c

Substituting the result of part (a) into the result of part (b) we get

9 2

at2 az

-9s




PROBLEM 9.16

Part a

We seek to write Newton's law for motions in the z direction of a slice

of the material having x thickness dx. In our situation the mass is

?2 2 padzdx, where the acceleration is a 6 /3t . The net force due to the stress is

F = [Tx(x+dx) - T x()]a dz (a)

and3-6S z dx ad z = [T (x+dx) - T (x)]a dz (b)

2zx zx

Finally) in the limit dx + 0 we have

32 6 3TZ ZX C)p z= zx (c)

t2 3x

Part

The shear strain, ezx , is defined so that it is proportional to

6 (x+dx) - 6 (x) normalized to the distance between points dx. If T =2G ezx

then in the limit dx + 0 T = G 36 /3x if we definezx z

1 ze (d)zx 2 ax

The 1/2 is included to subtract out rigid body rotation, a point that is

important in dealing with three-dimensional motions (see Chap. 11, Sec.

11.2. la).

Part c

From part (a),

a26 3T

P 2 z 3x zx (e)

Using the result of part (b) we have

p-326

=326

; Cf)

3t 3x

the wave equation for shear waves with the propagational velocity

v = P P

-99




PROBLEM 9.17

Part a

Conservation of mass implies: net mass out per unit time = time rate of

decrease of stored mass

[pv + a(pv) AxA - (pv)A = - [p(Ax)A] (a)ax at

As Ax - 0, we have

a ap (b)a (pv) + 0 (b)jx at

If we write p = p + p'(x,t) and v = v(x,t) then we obtain by substitution

p av+-a(pv) (c)

o ax ax at

Retaining only first-order terms we have

pav ap (d)o ax at

as desired.

Part b

Conservation of momentum implies:

time rate of increase of stored momentum = net momentum in

per unit time + externally applied force

-C(pvAxA) = - [pv2+ (v Ax]A +(pv2)A + pA - (p + ax A x)A (e)

at ax

as Ax + 0, we have

2a(pv) = a(pv 2) (f)

at ax ax

Expanding we have

p(a + v 2) + v (a(Pv) + ) ()

this term is zero

by conservation of

mass

-100




PROBLIEM 9.17 (continued)

Finally we have

p ( + vper = - (h)

Substituting the perturbation quantities and retaining only the first order

terms we obtain

yv= _0o •t ax

Part c

In terms of perturbation quantities we can write

p' = a2 0'

where

a

o

Substitution for p' yields the two equations

av ap'

Po ax = t

and

-a 2 ' = °avy*

Combiningxeo obtaint

Combining we obtain

2 2

at

= ax

3x

(scalar wave equation)

Part d

If we substitute v = Re[v(x)ej t ]

in the above equation we obtain

dv(x) 2

d2

+-• v(x) = 0dx a

which has solutions of the form

v(x)=C sin(- x) + C2 cos(a x).

-101





A rigid wall at x = 0 imples that v(x=n) = 0. The drive at x = Z and the

equations of part (c) imply that

v (p)

dx 2a po

at x = 2.

The solution for v is

j~o sin( x)

v() 0 (q)

apo cos((a

)

and we can now obtain v(x,t): fo r p realo

p sin(01 x)=

v(x,t) a sin wt. (r)

apo cos (- )a

PROBLEM 9.18

We can calculate the values of d6+/da and d6-/d8 for three regions of the

x-t plane as defined below.

/

Referring to equations from text,

Region A:

d6+ 1 m d6 S- 0da 2 v ' d0

and

vT E m

2 vp

9.1.23 and 9.1.24, 9.1.27 and 9.1.28:

(a)

(b)

-102





Region B:

d6+ d6- 1 Vmd- d6 2 v

E mT

2vP

Region C:

d6+ ds= 0 and T = 0.

de dB

Plott ing T(x,t) in the x-t plane we have

-103

PROBLEM 9.19

We can find d6+/da and d6-/d3 fo r four regions of the x-t plane:

DD

a

-0

<1D

Referring to equations from the text 9.1.23, 9.1.24 and 9.1.27, 9.1.28 we have,

Region A:

d6+ 1 T(a) d6- 1 T(R)

da 2 E ' dB 2 E (a)

Region B:

d6+ 1 T(a) d6

da 2 E ' dB

Region C:

d6+ d6- 1 T(a)da 0' df 2 E (c)

Region D:

d6+ d6

da dB=•0 (d)

We can use these values in equation 9.1.23 and 9.1.24 from text and make the

-104





'F'/~ E

PROBLEM 9.20

Part a

The free end at x = 0 implies that T(0,t) = 0 and using equations 9.1.23

through 9.1.26 we can easily find that velocity pulses "bounce off" x = 0

boundary with the same sign and magnitude. For the x-t plane we can indicate

the values for v(x,t):

/P

-105





Part b

We can make use of part (a) if we use superposition. Consider the super

position of boundary and initial conditions; a free end, T(O,t) = 0 with the

initial conditions in part (a) and the T(O,t) as shown in Fig. 9.P20b with

initial conditions on T and v zero. Since the system is linear, we can add

the velocities that result from the two situations and thus have the net

velocity. For the response to the second set of conditions we have

o-/-VP

Add this velocity set to the set in part (a) and we obtain:

vtY

-106

SIMPLE -ELASTIC CONTINUA

PROBLEM 9.21

Part a

With the current returned on th e inside surface the field in the air gap,

is H = I(t)z D

/a

/ '

and the force per unit area acting on the inside surface is

2

Tx=2o1 Ia2 , (a)

The force is f =-T aD = I (t) and the boundary condition at x = - £x x 2 D

is

326 1 oa 2S--

2(-£,t) =

2 D(t) + AT(-R,t) (b)

at

Part b

The current will flow on the surface when the time T is much shorter than

the characteristic diffusion time Td over the length b:

Tdd> > T or oabo >> T (c)

Part c

In order to ignore the mass M, the inertial term must be small compared

to AT(-£,t). For t < T, 6_ = 0 on the rod, and from Eqs. 9.1.23 and 9.1.24,

E a6T(-Z,t) = - (-Y,t) (d)

v Tt

Thus M26

t 2p

or

M << AE T/v (f)p

Our boundary condition In part (a) now becomes:

a0 = 12 D

2 (t)+ AT(-£,t) (g)

Since there is a fixed end at x = 0 we know that a stress wave traveling

in the +x direction will reflect at x = 0 with the same wave returning in the

-x direction. To satisfy the condition v(O,t) = 0, Eq. 9.1.23 shows that

d6 /da = d6 /dB at x = 0. Thus, from Eq. 9.1.24, the stress is twice that

-107




PROBLE1I 9.21 (continued)

initiated at the left end

pa

T -=oa 12 (h)r DA o

PROBLFM 9.22

Part a

We have W = W' and U = C + U' where W' and U' are perturbations from

equilibrium. Rewriting the equations we have

3W' aw' U' + K BU'+ (-') W + _ = 0 (a)

(C+U')3 ax

and

+ aU'w' aU'- (C+U') -L + (W') - = 0 (b)

at ax axNeglecting all second-order perturbation terms we have

aW' K 3U'+ (1 + -) • = o (c)

-u'

+((C)

w'= 0 (d)

at ax

Part b

Multiplying the above two equations by and x' respectively, we have

329' K 2U'e)+ (1 + = 0 (e)

t2

C3

and

a2U' 32W

,

+ (C) 2-= o (f)axat (C)

Eliminating U' we obtain

aw2' K a22a'= c(1 +

3)

2 (8)

at C ax

which is the familiar wave equation with wave velocity v = C(I + K)3

We can write the solution as 14' = ReIW(x)e jt] where

N(x) = C1 sin Ox + C2 cos Bx (h)

with = wu/vp

At x = 0, W = W' = 0 and hence C 2 = 0. At x = - L, W =W' = Wo cos wt, or equiv

alently t"(-L) = Wo, hence C =-Wo/sinpL. Upon substitution we find that

-108





the solution is

W sin3x

W = W' = -sin BL

cos Wt . (i)

PROBLEM 9.23

Part a

This part is similar to Prob. 9.24 with two simplifications:

V = 0 ando

the mass is M/unit width (Mw) instead of 2M. The two separate relations yielding

the natural frequencies are

and sin (L ) = 0 (a)

and

m = tan (wL (b)

(a) yields wL /~7/S = nir where n = 1, 2, ... and corresponds to solutionsm

which are "odd", or ý(x) = - F(-x). (b) can be solved graphically and corresponds

to solutions which are "even", or ý(x) = F(-x).

Part b

The effect of raising M is to reduce the eigenfrequencies of the "even"

modes. The "odd" solutions predicted by (a) are independent of the mass M.

This is physically reasonable since there is a node at the mass,and since

the mass doesn't move there is no inertial force. For the "even" solutions

predicted by (b), we notice that if M = 0 we have essentially the natural

frequencies of a membrane of length 2L. As M + -, the system responds like

two different membranes of length L. The infinite mass acts like a

rigid boundary.

PROBLEM 9.24

Part a

We can use the Maxwell Stress Tensor to find the forces of electric

origin. If fe corresponds to the force due to the upper electrode andu

f corresponds to the force due to the lower electrode, then we have:

C V2AA

f(t)u

=oo

2[d-_(O,t)]2i

y(a)

-109

E V2A

(t) = - 0 2 (b)

2[d+ý(0,t)]

2 2 32Our equation for the membranes is a 22C/~t = S a and if we assume

E=Re[Q(x)ejwt1, then we can writem ax2

Z(x) = C1 sin 8x + C2 cos fx (c)

for x > 0 and

Z(x) = C3 sin ýx + C4 cos 8x (d)

for x < 0 where 3 = wr7T7S.m

Our boundary condition will yield the four constants. We have

&(x = - L) = 0

.(x = L) = 0 (e)

(x = ) = (x= 0 )

and

- (0-) + fe(t) + f(t) (f)2M

22t2

(0,t) = Swxx

(0+ )u

which reduces to

2 2E 0V2A

-2Mw E(0) = Sw- (0) - (0) + 2+ d () (g)g)0)

Idx dx d3

after we linearize [fe(t) + ef(t)]. Substituting, we immediately find C2 = C4 .

Writing the remaining equations we have

0 = - C3 sin ýL + C2 cos fL (h)

0 = C1 sin BL + C2 cos aL (i)

0 = SwB C1 + - + 2Mw2 2 - Sw C3 )

If we eliminate the constants by setting the determinant of the coefficients

C1 , C2 , and C3 equal to zero, we obtain two separate relations:

SwIsin 8L

= 0 and S = tan BL. (k)

E V2Ao o + 2M

3

d

-110





Substituting for a we have

SwW m-sin(WL = 0 and 2 = tan wL

E V2AS+ M 2

d

The first relation implies that wLv/-7 = nfl where n = 1, 2, .... The secondm

relation can be solved graphically.

Part b

As V is increased from V = 0, the lowest natural frequency decreases.o o

When V approaches the valueo

CSSwd'

the lowest natural frequency approaches zero; as Vo is further increased, there

will be an imaginary solution for w and the system will be unstable.

PROBLEM 9.25

Part a

m = W' =io 2The force of th e lower2 plunger is f - . By symmetry the upper

i = (I + i ) = I + 21 i and i = (Io-il) = I - 21 i . Hence the totalmagnetic force isu o 1 o o

magnetic force is

2L I i 2L , G= o1 oo 3(

a a ax

Writing the force balance on the tip of the wire at x = - Z we have

2L IG

ff (-+,t) +X o o ag(o,t) = 0

ax a x

Part b

Away from the ends

M =2f 2

2 22at x

and if E=Re[(x)ejm t

] then

((x) = C1 sin Ox + C2 cos fx

where = . = 0 implies that = 0. From part (a) we havew7/mT E(O,t) C2

-111





d 2L I G

f (-+) + d (o) =. (e)dx a dx

Upon substitution we obtain

2L I

fB C cos 6k + ooa GBC = 0 f)1 a '1

Since C1 must be finite for a finite response, we have

2L I

fB cos B + oo G = 0, (g)a

or - 2L IG

f cos Wk - + 00 0 (h)If a

(We have ruled out one solution, because it is trivial.) A graphical solution

of (h) is shown in the figure.

Part c

If G = 0, then

S = (-n+l) (i)f 2

with n = 0, 1, 2,...

Part d

From the figure, wl increases toward wl•7"/m/f = T and co2 decreases toward

the same value. They come together at G = af/2LT I and seemingly disappear

afif G >

2L I0O

-112



y


PROBLEM q.25 (continued)

Part e

If JGI > 2L ' then (h) has imaginary solutions for w, hence the systemO2O00

will be unstable:

PROBLEM 9.26

Part a

First of all we. notice that y(t) = C(-L,t). For th e membrane

l t ma -- = _il and if F=Re[l(x)ejO then ,(x) = Clsinfx + C2 cos(x wherem at 3x

8 = Jo~m/S. At x = 0, ((x=O) = 0 and therefore C = 0. At x = - L, we can write

the boundary condition

M (-L,t) = SD - (-L,t) + fm(t) (a)

at2 ýx y

We can find fe(t) using Ampere's Law and the Maxwell stress tensor

1 AN 2 ( + I(t))2 (I- (t)) 2

fe(t) o o_ - o (h)

t) 2 d-D- (-L,Qt)) (d-D+C(-L,t))

Since I >> I(t) and (d-D) >> ,(-L,t) then we can linearize:

SI 12I

fe (t) 2N2A I(t) + U-Lt) (c)

y (d-D (d-D)3 2N2A2N2A oI o

Substitution of (c) into (a) and definition of C E 2 and

2NA•I 2 (d-D)

C E gives

Y (d-D))

M2 (-SD,t)) (-L) t)+ Clt) + C -l) (e)

Cx or in complex

form,= SD ax (-L) + CII + C •(-L) (e)Mw C(-L)After solving for C 1, we can write

C I sin 8x I

((x) = (f)

(Mw2 +C )sin(L-SD( cos 8L

or finally

-113





C sin BL IT

Sf~sf~. (ttC si 3

SDB cos T.- (MW +C )sin BL

=where y(t) Ie[y ejwt].

Part b

To find the resonance frequencies we look at the poles of y/I. This amounts

to finding the zeros of the denominator of y/I. We have

SD C w,. - [MW2

C ]sinb L (h )os +

is-

USDmWV.m

= tan(wL V)MW +C

We can represent the solution graphically:

I

I

PROBLEM 9.27

Part a

The boundary condition may be obtained by applying force equilibrium using

the following diagram, s 3C

slope

slopeax

-114

thus

F(t) =f (0-) - r+(

Part b

For the odd solution, EQ(x,t) = - r(-X,t) and it follows that

a5, agr

= 0. This implies that the odd solution is not excited by the force F(t).ax ax

Part c ag a r

For the even solution, Q(x,t) = Fr(-x,t), we have - = - and the boundary

condition (a) from part (a) becomes

F(t) = - 2f -x at x = 0

For O<x<£ we have

m = f •

at2 ax 2

with E(x,t) = 0 at x = k.

For t < 0, this reducss to

2a2 = 0ax

and we obtain

F0

o x((x) = 2f (1 - ) fo r O<x<£

Part d

We now have a combined transient and driven response, as discussed in Sec. 9.2.1.

By contrast with the developments of that section, we now have a boundary condition

at x = 0 on the slope 3&r/ax (see (b) of part (c)). Our program is: (E5Hr in the

following)

i. Find the driven sinusoidal steady-state response, This satisfies the boun

dary conditions:

F cos wt = - 2f (O,t) (f)o ax

((£,t) = 0

ii. Find normal modes, which satisfy homogeneous boundary conditions;

T- (0,t) = 0ax

E((,t) =oThe sum of these modes takes the form of a Fourier series.

-115

iii. Superimpose (i) and (ii) and use the initial conditions found in

parts (a)-(c) to evaluate the arbitrary coefficients.

The driven response is of the form

S=Re(C1 sin Bx + C2 cos Bx)eJt; = (j)

a linear combination which satisfies (g)

= ReC 3 sin ý(x-2 )ejWt (k)

while (f) evaluates C3 and the driven response is

F sin 8(x-k)ejWt5 = - Re 0(2

2fM cos B(

The normal modes are in this lossless case the resonances of the drivenresponse and occur as cos ýZ = 0. Thus

Sk = (2n+)r, n = 0, 1, 2, 3... (m)

and the total solution for O<x<R is

F sinO(x-) + jWnt -jn t2nf= cos +0 os t + [A e + A e ]sin[(-2 - (x-)] (n)

n=0

The coefficients An and A- are evaluated by requiring thatn n

o(xo F osinS(x-.)_+_o + 2n+l 2,2n+1

(x,0) • - 2f cos + (An + A )sin[ 2( ) (x-k)] (o)n=O

and

__ S + 2n+l 7(x,) = 0 = jw nAnA j- An]sin[.(-)T (x-2)] (p)

n=O

This last condition is satisfied if A+ = A-. The A+'s follow from (o) by usingn n n

the orthogonality of the functions sin[(2n+1/2)j (x-Z)] and sin[(2m+l /2) (x-Z),

m # n, over the interval R.

-116



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DYNAMICS OF ELECTROMECHANICAL CONTINUA

PROBLEM 10.1

Part a

At x = 0, the net force on an incremental length of the string has to be zero.

-2B - f n = 0at ax

This is the required boundary condition at x = 0.

Part b

The power absorbed by the dashpots is the product of force 2B C/a3t and the

velocity 3&/3t. T us

P = 2B (Q

If we solve Eq. 10.1.6 for

E(x,t) = Re[( ej(wt-k x )

and assume that w < we we get

ý(x,t) = Re {[A I sinhlklx + A 2 coshlklx]ejWt}

where k L 1/2

We can calculate A 1 and A2 using the boundary condition of part (a) and the

boundaiy condition at x =

= Re E ejwtk(-Z,t)

We then get j2o2•j0 2Bw

Al [f k cosh k Z + jw2B sinh kjkj

A2 [flklcoshlkl1 + jw2B sinh kI]1

If we plug these values into the expression for power,.and then time average,

we have

B(f kljom)2

 =

[(ffklcoshjkflZ)2

+ (2Bw sinhj)kJi)2]

where it is convenient to use the identity

<Re Aejwt ReBejt> = 1 AB*2

-117

PROBLEM 10.2

Part a

We use Eq. (10.1.6)

a• v2 32a 2 2 Ib

t2 s x2 c c mat2 s ax2

Assume solutions ( = Ref(Ae- j

kx + Bejkx)ejtt]. The dispersion equation is:2 2

k2 = ck2 = d 2

vS

Now use the boundary conditions, which require

A ejk + Be- j k

=

-jk[A - B] = 0

(i) Wd < W c (below cutoff)

=d cosh eax= cos wdt

cosh atd

a 2 v

(ii) Wd > Wc (above cutoff)

Fsd

cos x

((x,t) = os t cos Wdt

ý 2 cos 8d

Part b Lti

CL)d =0

WII

x

-118

DYNAMICS OF ELECTROMECITANICAL CONTINUA


i

e

Sd < Wc

,(

Cd >•c

d e i

Part c

The string might be attached to a massless (friction ess) slider at

x = 0, so that the end would be free to move in the transverse direction.

inForce e0uilibrium for the increment or length at x U LI•en requ•L•

E/3ax = 0 at x = 0.

PROBLEM 10.3

Part a

From Eq. 10.1.10 we have

2k 21/2

k 2v

s

with our solution of the form

E(x,t) = Re(A 1 e j(t-kx) + A2 e J(wt+kx)

We have the boundary conditions

-119

x

PROBLEM 10.3(continued)

-x (O,t) = 0ax

and

x-(-Z,t) = 0.ax

From the first boundary condition, we obtain

Al = A2 ; E(x,t) = Re A3 cos kx ej w

t

From the second boundary condition, we obtain

sin kZ = 0

This implies that

nITk = ; n = 0,1,2,3...

Note that by contrast with the case where the ends are fixed, n = 0 is a valid

(nontrivial) and crucial solution. It corresponds to an eigenmode which is

simply a rigid body translation.

From Eq. 10.1.7

2 2 2 2S=-k v + 00

s c

Therefore, the eigenfrequencies are

'22 1/2

-- c + vs

For th e n = 0 mode, w = + ) .-- c

Part b

With I as in Fig. 10.1.q, we have the same equations as in part (a) if we

replace w2

by -02. Therefore, fo r this case, the eigenfrequencies are

c c

_ 2 2 1/2W =

VS)- 11/

Part c

With I as in Fig. 10.1.9, the IxB force is destabilizing, as a small

perturbation from x = 0 tends to increase this force. If w in part (b) became

imaginary, the equilibrium 1,= 0 would become unstable as the solutions are

unbounded in time. This will happen as

2

S v W < 0Ss C

-120

or in terms of the current

n11 I )2

b (T vs

Note that any finite current makes the n = 0 mode unstable, since for this mode

there is no elastic restoring force.

PROBLEM 10.4

Multiply the system equation by ,

ma - a 2f• _ Ib * + * F(x,t)at 2 at 2 at at

at ax

Proper substitution of partial differential identities yields:

a m • 2 4f o2 + Ib 2] f , F(x,t)

-t •2t 2•x 3x 2 x x• t at

PROBLEM 10.5

We have that

F(x,t) = Re ( e( x) + e

Part a

For k real, we might write this in the form

1 j1at-kx) e - j (wt-kx)5(x,t) = 2 + e+* e

+ e (tkx) + * e-j (wt+kx)

From rob. 10.4 we have that the power carried by the string is

P = - f a aax at

If we do the indicated differentiations, then substitute into this expression,

and then time averagewe will obtain

<>= fwk ^ ^ ^ ^

- [5+ * - 5_ _*]

Part b

For k purely imaginary

k = j5

with 3 real, we can write F(x,t) in the form

-121

DYNAMICS IN ELECTROMECHANICAL CONTINUA


1 ^t eJO t+x + e-Jwt+Bx + ejwt-8x + _,e-jt-8x5(x,t) = ( et + * e + ( e + *

1

If we again substitute into our expression for power and average over time

we obtain

 =- +* _ - + E_*l2 + +

From (b), we see that it is possible to have a net power flow from two evanescent

waves, but not from a single evanescent wave. Suppose that a single evanescent

wave did carry power away from the driving source. This would correspond physicall

to a string driven at the left and infinite to the right. With Wd< c, the

response as x *o becomes vanishingly small; clearly there can be no power flow at

x + oo . Yet, there is no mechanism for power absorption by the string and so there

can be no power flow into the string from the drive. With a dissipative load, a

second evanescent wave is established, decaying to the left, and the conditions

for power flow are met.

PROBLEM 10.6

From the dispersion relation, we calculate:

r[ w2 1/2V - = v 1

g ak s 2

Now, assuming a single forward traveling wave:

S= E+ cos[wt - k(w)x]

Then: 2 2(m0 fk

2Ibn 2

<W> = + 4+

 = E

Thus, substitution gives

 fkw/2

<W> 2 2

2-1/2= vs - =Vg

which is the desired relation. This result is of some general significance, but

has been shown here for a particular case.

-122

DYNAMICS OF ELECTROMECIANICAL CONTINUA

PROBLEM 10.7

Part a

The equations of motion for th e membranes are

2 1

a = S + Tm ýt2

21

gm 2Fx =S 2F2 + T 2

where T1 and T 2 ar e th e transverse magnetic forces/area. If th e membranes extend

a distance w into the paper, and if we define regions 1, 2, and 3 as the top,

middle, and bottom regions respectively in Fig. 10P.7, the flux in each region is

11 o 1 w(d- 1)

=2 PoHl2 w(d+ý1- 2 )

3 = l o113 w(d+E 2 )

where ll , H2 , and H 3 are the magnetic field intensities within each region. Since

the flux is conservedwhen E1 = '2=

0 we have

A = =- plio wd A = + lH wd

Therefore, I d

1 od-•q ixHod i

1 d+E- 2 x

and

1 2andH do i

3 d+F, x

We will use the Maxwell stress tensor to calculate T 1 and T 2, using a pill-box

volume enclosing a section of surface on each membrane.

We then obtain

TI o° 2 2

T= 2 [11 - H ]

and

T2= 2 232 2 2[ - H3

Substituting th e expression for the H fields, and real izing that (1< <

d and

E2< <

d, we finally obtain for th e forces

-123





T 1P~oH 2

(2C11

- F)2

T d1 d

and PIt12 (2 ,2

T 2 0 0 2

2

Our equations of motion are then

2

am

1 E= - S

21

2

~ 12o o

d(2Fi-F2

1 2

and 2

32 32 0 10

m t2 x2 d 2 1

Part b

We assume that

F = Re F1 eJ(rlt-kx)

and

= Re F2e j

(tAt-kx)F2

We can substitute these functions into the equations of motion from part (a),

and solve for the relation between w and k such that the 2 equations of motion

are consistent. This dispersion relation is

2'),10 p H22 2 0 01oo

-( + Sk + = +m d - d

We see that the dispersion equation factors into two dispersion relations. If we

substitute this relation back into the equations of motion from part (a), we see

that we obtain even and odd solutions.

The dispersion relation

9

2 Sk' li ~k 0 0a +

a adm m

yields

F• =F1.

The dispersion relation

2 Sk'+ 0 0h) +

0 Oadm m

yields FL = - E12

-124



DYNANICS OF ELECTROMECHANICAL CONTINUA


Plotting w versus k, we obtain

k real

- - - - k imaginery

From the plot we see that the lowest frequency for which we have propagation

(k real) for the even mode is

ce ad

For the odd mode, the cut off frequency is

= /Wco -(md 1

od

Part d

We are given the boundary conditions that at x = 0

i= 0 2 = 0

and at x =

= - F = Re o ejt1 '2 o

-125



DYNAMICS OF ELECTROMECIIANICAL CONTINUA


From the boundary condition, we see that our solution is purely odd. Therefore

2 1/2

m 0ok= mk S Sd

We assume a solution of the form

Sl(x,t) = - C 2 (x,t) = Re{A 1 ej (

wt-kx) + A2 ej(wftkx)}

Evaluating A1 and A2 through the boundary conditions,we obtain

'oA = - A = -Jk

1 2 jk2_ -jk2

Therefore 0[e -jkx _ e+jkxlejt

(x,t)=-2(x,t)= Re [ejk _ e-jk Ik k

For w = 0, k is pure imaginary. We define k = j3 , with 8 real with value

(OO2 1/2

Sd

Therefore

Y1(xt) 0- 0inhWfx

sinh 8£

A sketch appears below.

-126

PROBLEM 10.8

Part a

The given equations follow by writing out Maxwell's equations and assuming

E and H have the given directions and dependences.

Part b

The force equation for an incremental volume element is

ývF i mn - (a)

x e ýt

where F is the force density due to electrical forces on the electrons

F = - i en E (b)x ex

Thus, 3v

-en E = mn (c)ex e at

Part c

As the electrons move, they give rise to the current density

J - en v (linearized) (d)x e x

Part

Assume ej (wt-kx) dependence and (c) and (d) require

2en

J =-jj eE (e)x - m X°

o[W2 x

where = e2n /me is called the plasma frequency. (See page 600)

p e o

=k22 - 2; c =

i1 (g)

c ACoIýo

Part e

We have a dispersion which yields evanescent waves below the plasma (cutoff)

frequency. Below this frequency, the electrons respond to the electric field

associated with theuave in such a way as to reflect rather than transmit

an incident electromagnetic wave.

Part f

Waves impinging upon a boundary between free space and plasma will be totally

reflected if the wave frequency w < wP . The plasma frequency for the ionosphere

-127

is typically

f % 10 MHp z

This result explains why AM broadcasts (500 KH! < f < 1500 KH ) can commonly bez z

monitored all over the world, whereas FM (88 MH < f < 108 MH ) has a range

limited to "line-of-sight".

PROBLEM 10.9

In the regions

x < - 2 and x > 0

the equation of motion for the string is

2F, 2 a2t2 s ýx

at ax~

In the region -Z < x < 0, this equation is modified due to the magnetic force to

2 22 v2 2(j

at s x2 c

If we assume

-F(x,t) = Re { e (w

tkx)

}

and substitute back into the equations of motion we obtain the dispersion

relations

2 2W1/2

k = + [W 2 - < x < 0v

V

- v


at x=- =

at x = 0 F and - must be continuous.ax

lie assume that

(x,t) = Re {[A e- • x + B e+x] ejwt} fo r -£ < x < 0

-128

where = 1/2

2C2

2 1/2

for w < w= s

c

U(x,t) = Re b e-jk b x e j

3t} fo r x > 0

where

b s

Using the above boundary conditions, we obtain

0o(1 + jkb)- 0A =2(1 cosh 1£ + jkb sinh 12 )

%o(1 - jk b )

2(1 cosh 3,+ jkh sinh 1R)

Bu t F = A + B

Therefore

co

[cosh 1£ + sinh 1S]a

Part b

As 0

-b

As £-+

-+ 0

Eo

PROBLEM 10.10

Part a

The equation of motion for the string is

2 2

mt-;2E2

_ 2+ S - mg

at ax

-129

where, for small deflections F in the "l/r" field from Q,

S qQ [1 + E12n7 d d

O

In static equilibrium, 5 = 0 and from (a)

qQ= Pdco'mg

Part b

The perturbation equation of motion remains;

m2

= f2

+ (2

at ax 2 d0

Assume ej (

wt-k x

) dependence and (c) requires (vs = /f7n)

2 v2k2 qQ

s 22nTd e m

0

or from (b),

2 = v2k _ g

7(ý1(11~e s d

The boundary conditions require k = nw/9,, and for stability the most critical

mode is n = 1; thus

v2()2 >

s d(e)

m < gfd

()2(f)

(f)8

Part c

Increase f, d, or decrease R.

PROBLEM li.11

m F f + S-mg

at ax

where S (IxB) andB = , r the radial distance from the fixed wire.

r= o ad 2 r

Therefore S = 2or2trr

For static equilibrium

-130





ooS = mg =

2rF,

Therefore

27rmgo

I =•olo

Note that I I > 0 for the required equilibrium.o

Part b

The force per unit length is linearized to obtain the perturbation equation.

O 10S27Therefore

Therefore

mS2

= f 2 11ooI

00St

3x

2

Part c

Assuming ej (

wt-kx) solutions, the dispersion relation is

2 22 oo-m = f k

2 2o

Solving fo r w, we obtain

w = k2 S1/2IoJ 1/2X 'jjI

+

As long as I I > 0 th e equilibrium will always he stable as w will always be

real. Note that this condition is required for the desired static equilibrium

to exist.

PROBLEM 10.12

The equation of motion is given as

,2, ,2,

m d_ = f d2L + p,

3t, 3x'

Part a

Boundary conditions follow from force equilibrium fo r the ends of the wire

-131





(i) -2KE(O,t) + f (, = 0 (b)ax

(ii) 2((,t) + f (xt) = 0 (c)

Part b

The dispersion relation follows from (a) as

w2 = v2 k2 - ; v = V7TF (d)s m s

where solutions have been assumed of the following form:

E = Re[(A sin kx + B cos kx)ejO)

t ](e)

Application of the boundary conditions yields a transcendental equation for k:

tan k = 4Kf(f)

f2k2_4K2

where, from (d),

k = 1 a2 + P/m (g)

s

Thus, (f) is the desired equation for the natural frequencies.

-132





Part c

As K + 0, the lowest root of graphic solution goes to k + 0, for which

stability criterion is:

P

0 >m

PROBLEM 10.13

Part a

This problem is very similar to that of problem 10.7. Using the same

reasoning as in that problem, we obtain

m

a2

1_2 = S

a2I

2 +

E V2

0 (2tl-62)

ax d

2 2

2a 2 S 7+ (22 -_1)m at ax2 Dxd 3 2

Part b

Assuming sinusoidal solutions in time and space, the dispersion relation is

V2

2E V2 2 2 o o oo

-a(w + Sk 03 + 3m 3 -- d

d d

We have a dispersion relation that factors into two parts. The odd mode,

S= - 2 has the dispersion relation

W 2 3E 2] 1/2

m ad

The even mode, E = E2 has the dispersion relation

V2] 1/2

[Sk22

am

m

Part c

A plot of the dispersion relation appears below.

-133





,-, I

Coecog

Part d

The lowest allowed value of k is k = since the membranes are fixed at

x = 0 and x = L. Therefore the first mode to go unstable is the even mode.

This happens as

3C V2

206D

Sd L2

7r

J2 Sd 31/2

0 L2 Eo0

-134




PROBLEM 10.14

The equation of motion.is

a22 v2 a2

2v Ca

2 2 (a)at s ax at

Part a

The dispersion for this system is:

jv - v2k2 _ 22

= 0 (b)s

We may solve for w,

w = j () - v2k2 (c)

J[a + Y

We assume solutions of the form:

- (a+yn ) t - (a-Yn ) ntxE(x,t) Ref I [Ane + B e ]sin nx (d)

n odd

Now, we may use the initial condition on- to relate A and B . Thus we obtain:

•U(x,t) = Re{( A [e - •- e le sin (e)

n odd a nn

Now, we apply the initial condition on E(x,t = 0) to determine A .

U(x,O) = d An sin (f)

n odd LYn-aJ

I iA'sin nl

n odd

The coefficient A' is determined from a Fourier analysis of the displacement:n

4EA' = --

o(g )

n n• '

So that:

y -an 4E

An n 0 (h)

n

Part b

There is one important difference between this problem and the magnetic

diffusion problems of Chap. VII. While magnetic diffusion is "true diffusion"

and satisfies the normal diffusion equation, the string equation is basically a

wave equation modified by viscosity. Hence, we note (c) that especially the

- 135





higher modes in the solution to this problem have sinusoidal time dependence as

well as decay. Magnetic diffusion as discussed in Chap. 7 exhibits no such

oscillation, because there is no mathematical analog to the inertia of the

string. If we had included the effects of electromagnetic wave propagation

(displacement current) the analogy would be more complete.

PROBLEM 10.15

From Chap. 10, page 588, Eqs. (e) and (f) we have

dE+ (vs-U) DE 1 @

da 2v ax 2v ;tS S

dE_ (v + U) _ 1 •d--- 2v x +

2v- •S S

Since -C (x = 0) = 0, we have the following relations in the three regions.axRegion

Region 1

d ++ V

o0 d&

da 2v ' d

Region 2

dF+ d_ V0

d ' dB 2vs

Region 3

d+ Vo d_ V

da -v ' 2vs s

-136





In the other regions, the derivatives are zero. From Eq. 10.2.10 on page 586,

d( dS

~x de+ d_

we have

V

H- =- o-2 [u (B)-u 1 (B-b) - ul(a) + u_l(-b)ax 2v -1

s

Integrating with respect to x, we obtain

V

F(x,t) = l[u-2(0) - u_ ( (-b) - u_2() + u- 2 (-b)]

s

A sketch of this deflection is shown in the figure.

-1 37




PROBLEM 10.16

Part a

The equation of motion is simply

2 2m = f

22 axat

The dispersion equation follows as:

(m-kU)2

= v2k222

s

Where solutions are assumed of the form:

E(x,t) = Re{(E+ejix + ý_e-jýX)e j (wt-ax)}

The boundary conditions are both applied at x = 0, because string is moving at

a "supersonic" velocity.

((x,t) = ol{cos 6x cos[wt-ax] - U sin ýx sin[wt-ax]}

Part b

A

IP

C• • I I • I \ .•

U1 Ix

W/d

-138



- -


PROBLEM 10.17

We use Eq . 10.2.9

2 2 32E

-t + U-) -v . 2

Assuming sinusoidal solutiors in time and space we obtain the dispersion relation

2 22(w-kU) = k v

s

Thus

SW(U + v s )

U+ v U2 2

s

We le t

wU

U2 _ v

s

WvU2 -v2

s

Therefore, k = a + and

E(x,t) = Re[A e- j ( a- 8 ) x + B e- j (a + a) x ] e j wt


=(x 0) = 0 which implies A = - B

E(x = - 9) = Eo

Therefore

E(x,t) = Re A[e j(a 8 )x_ e-j(a++)x]eJwt

= Re A2j sin Bx ej(wt-ax)However,

E(-t,t) = Re E ejwt

Therefore

U(x,t) = - si sin Bx cos[wt-a(x+t)]

Part b

For 5 = 0 at x = 0 and at x = - a we must have a = n7rw/

Wvs niT

U2_v22

U-v

or

-139

(U2 - v )2W=

vs

These are the natural frequencies of the wire.

Part c

The results are meaningful only for IUI < Ivs . If this inequality were not

true, we would not be able to use a downstream boundary condition to determine

upstream behavior and arrive at a result that would be obtained by "turning the driv

on". That is, if U vs the predictions are not consistent with causality.

PROBLEM 10.18

Part a

In the limit of wavelength short compared to the radius, we may "unwrap" the

system:

2 2m + U a) (a)

az

Now let z + RO, U -RQ. Then, it follows that

a + 0 a 2 f a (b)

where Q = f/(R 2)

Part b

The initial conditions are

ac/at(e,t = 0) = 0 (c)

(,t = 0) = 0 < e < d/4

o- --

(d), elsewhere

Solutions take the form

S= +(a) + S_() (e)

where

-140

a= - ts

= 8 O -30 t

s

Because aE/at (t = 0) = 0,

d+ d= - s -3sd (f)

s dc s dB

Also,

dE+ dý= - (t=0) =

da_ +-

d0(0

[u (0) uo(/4)]

0(g)

Thus, from (f) and (g),

dE+ 3da 5o[u (0) - Uo(O/4)]; on a (h)

dSE- 1= -aT o[uo(0) - u (7/4)]; on B

The solution in the 0-t plane follows from

dE+ + dE_a -+ -d (J)(j)

and an integration at constant t on e. The result is shown in the figure. Note that

the characteristicsthat leave the interval 0 < 0 < 27W atO= 27 r reappear at e = 0 to

account for the reentrant nature of the rotating wire.

-141




PROBLEM 10.19

In the moving frame we can write

at 2 fa2 + F(x',t') (a)at,2 ax,2

and so from Prob. 10.4, we can write

= aw' aP'n -" + -

Pin at' ax' ()

where

P' = F (c)in at

2 2W' = m (,) 1 f ( 2 (d)

'

P = - f (e)ax ' at'

But a a d a a U a+ax ax at at ax

Therefore (c)-(e) become

a aP'in = F( + U ý-)) (f)

W'= m(- + U )+ 2 (g)

P' = - f (a + U ) (h)-ax at ax

The conservation of energy equation, in terms of fixed frame coordinates,becomes

p aW' aW' P'P -- +U - + (i--)in at ax ax

atS+

ax(P + W'U)

(j)

If we let

P P'in in

W = W' (k)

P = P' + W'U

we can write=P. aw + a

in at ax

which is the required form.

-142




PROBLEM 10.20

The equation of motion is given by Eq. 10.2.33, and hence the dispersion

equation is 10.2.36;

k = n + jy (a)

where

=n WdU/(U 2-v

y = v (U2_2

)k22 /(U

2 -2 )

Solutions are assumed of the form

= Re[A sinh yx + B cosh yx]ei(0t-nx) (b)

Boundary conditions require;

B = 0o (c)

A = jnEo/y (d)

Thus

= Re Eo[y sinh yx + cosh yx]ej

(t-nx) (e)

The deflection has an envelope with an essentially exponentially increasing

dependence on x, with the instantaneous deflection traveling in the + x

direction.

PROBLEM 10.21 f

.'

souL6B e6•'

" 7NrH606//#j 7 peRO•fL•L

Part a

The equation of motion is

a a 2 2am t + U x) = S 2 mg + T (a)

ax

E V2

with T -- o -o V2

[1+ 22 (d-_)2 2 o d

2+3

For equilibrium, E = 0 and from (a)

2SV

oS= mg (b)

2d

or

Vo [ /J (c)

L 0

-143

http://slidepdf.com/reader/full/10.2.33






Part b

With solutions of the form ej(wt- k x)

E V2

2 S 2 oo

(w-kU) k 0d)

om a d 3

m

the dispersion relation is

(d)

Solving for k, we obtain

For U >

k = wU

S/a• ,m

S 2 U2 S o+ - U - ( )

m

2 S(U

2 _ )

m

and not to have spatially growing waves

--

(e)

Sa

m

-2(U2 S>a

m

,e V2o3

a d 3m

> 0 (f)

or

w > (U- i-) 3 (g)

POOBLEM 10.22

Part a

Neglecting the curvature of the system, as in Prob. 10.18, we write:

am

a( t

RP( •-

R a

2= S

R2282

+ Tr

(a)

where the linearized perturbation force/unit area is

2e V2

Tr (----a

Therefore, the equation of motion is

(b)

+ 0 a 2 2n 2 + m2 () (c)

s

2m

c=

2

aRm

2E V2

oo3

a

2

RS

-144




PROBLEM 10.22

Part b

(W -)2 2 (m2m2) (d)

ioW + = 2 + ( 2 Q2 ) 2m2

( mi) = -

(f)

n2 _ 2

-145




PROBLEM 10.22

Part c

Because the membrane closes on itself it can be absolutely unstable

regardless of 0 relative to s. Allowed values of m are determined by the

requirement that the deflections be periodic in 6; m = 0, 1,2,3,... Thus,

from (e) any finite me will lead to instability in the m = 0 mode. Note

however that this mode does not meet the requirement that wavelengths be

*short compared to R.

PROBLEM 10.23

We may take the results of Prob. 10.13, replacingiI-by a + U andat ax

replacing w by w-kU.

Part aThe equations of motion

2 a2 E V2

a m + U x ) = S --2 o3 (2 1 -2)ax d

a2E E V2

a a Sam it + U -x)ý2= S

ax'2 + 0

d(22 1

Part b

The dispersion relation is biquadratic, and'factors into

2e V2 e V2

-a (w-ldM) + Sk2 + 03 (c)d d

The (+) signs correspond to the cases 1 E2 and 1l= ý2 respectively, as will

be seen in part (d).

Part c

The dispersion relations are plotted in the figure for U > S-/jm.m

-146





A· U

and

1% ý 0;..

z

-147





Part d

Let =1 2. Then (a) and (b) become

2 825 E V2=

a(-a

+ U-a

)S

1+

oo1E

ax d

S+ 2 22 oo

V2

2 (e)ax 3 2

ax d

These equations are identical for 5i = 2; the dispersion equation is (c) with

the minus sign. Now let 1 =- E2 and (a) and (b) require

a2 I3E V

2

am (-a +U a ) 2 E = S

ax2 3(f)

ax dam 2

a22 3E V2

2+ 00

m(- + 3x E2=

Sax

x2d

3 2

These equations are identical for = - 2; the dispersion equation is (c)

with the + sign.

Part e#%A ,

E (0,t) = Re E ejWL = - )(0,t)A1

ac 1 a12= = 0 at x = 0

ax ax

The odd mode is excited. Hence, we u,se the + sign in (c)

38 V2

-am(w-kU)2 + Sk

2 S =0

d3Z

3E V

k2 (S-0mU2) + 20

mWkU - a

mw2

ddo033

O 0

Solving for k, we obtain

wU

where a = U2_v2

s 3 2 2 2 1/23E V (U -V )

[v22 oo s

d3s a

ad

8= 2m2

U - vs

-148

2with v = S/a m .s m

Therefore

E1 = Re {[A e-J(a+a)x + B e-j(a-8)x ]ejWt

Applying the boundary conditions, we obtain

A = 2()28

(c±+B)•B = W

21

Therefore, if • is real

(o)

(p)

(q)(q)

S (x,t) x = cos 8x cos(wt-ax)- sin Ox sin(wt-ax)

Part f

We can see that 8 can be imaginary, for which we will have spatially

growing curves. This can happen when

3E V2

22 oo 2 2w s d 3 (U2 v ) < 0

m

or d3 2 v2

v2 m s

0 v

Part g

With V = 0 and v > v ;o s

(r)

(s)

(t)

% f- F A I A I

• J

-149





Amplifying waves are obtained as (t) is satisfied;

PROBLEM 10.24

Part a

The equation of motion for the membrane is:

at 2 =S Fx2 21+T

where

T =T = 2E V2 E/S

3

z zz

The equation may be rewritten as follows:

-150





A2 = 2 (c)2

at2

ax2

ay2

where 2E V2

2 _ oo

c 3Ss

Assume solutions of the following form:

j(wt-k x-k .y)E(x,y,t) = Re[E e ] (d)

The dispersion is:

w2 = v2[k2 +k 2 - k2 (e)

s x y c

The mode which goes unstable first is the lowest spatial mode:

k k 'T (f)x a- y b

Instability occurs at

2 2

k = ar + (b (g)k2c a)

or,

3 2 2 1/2

Vo = [2(a[ +()] (h)

o

Part b

The natural frequencies follow from Eq. (e) as

2 2 1/2mn ( + _ k2 ] (i)

sn b+[a c

Part c

We superimpose eigensolutions to obtain the membrane motion for t > 0.

The solution that already satisfies the initial condition on velocity is

E(x,y,t) = Emn sin marx sin nb- cos mnt ()

mn

where m and nare odd only, since the initial condition on ((x,y,t=0) requires

no even modes. Now use the principle of orthogonality of modes. Multiply

(j)by sin(pwx/a) sin(qfy/b) and integrate over the area of the membrane.

The left hand side becomes

-151



.DYNAMICS OF ELECTROMECHANICAL CONTINUA

PROBLEM 10.24 (continued

fb fa1b f:[(x,y,t=o)sin pIx sin dx dy (k)

0b 0a

= ba J u (x- a) o( b)sin plx sin dx dyooJ o 2 a b

0 o

Thus, (j) reducesto

J ) (pq0)

which makes it possible to evaluate the Fourier amplitudes

4J

(m)mn a

The desired response is (j) with Emn given by (m).

4J(x,y,t) = (o ) sin sin n cos

'ab' sin-si b mnmn

(odd)

Note that the analysis is valid even if the lowest mode(s) is (are) unstable,

for which case:

cos w t + cosh a tpq pq

PROBLEM 10.25

The equation of motion is (see Table 9.2, page 535):

2

am

322t 2 S ( 2

x2 y2(a)

j(wt-k x-k y)With solutions of the form 5 = Re C e x yy, the dispersion equation is

W +v k2 + k 2(b)

-- s x y

A particular superposition of these solutions that satisfies the boundary

conditions along three of the four edges is

f A sin nry sink (x-b) cos w t (c)a x o

where in view of (b),

w2 22

n(i)2 (d)o s x a

Thus, there is a solution for each value of n, and

-152





I==A sin k (x-b)cos w t sin nay (e)n= 1 n n o an=1

where, from (d)

k 2 2 1/2

kn - (f)

At x = 0, (e) takes the form of a Fourier series

(y=0O) = I -A sin k b cos w t sin ng)n=1

This function of (y,t) has the correct dependence on t. The dependence on y

is made that of Fig. 10P.25 by adjusting the coefficients An as is usual

in a Fourier series. Note that because of the symmetry of the excitation abouty = a/2, only odd values of n give finite A . Thus

n

12 - y sin y dy + a/2 a (a-y)sin n- y dy (h)a a/2

= - A sin knb sin n•y sin 'Y dyo

Evaluation of the integrals gives

4E a A a sin k bSsin ()= - n n (i)

(mW)2

2

Hence, the required function is (e) ith k given by (f) and A given byn n

solving (i)

A = sin (')/sin k b (1)n m 2 2 n

PROBLEM 10.26

The force per unit length is#o0 x H, where H is the magnetic field intensity

evaluated at the position of the wire. That is,

S = p1I[H iy - H i ]. (a)

To evaluate H and H at ui + vi note that H(0,0) = 0. By symmetryx y x y

H (0O,y) = 0 and therefore aH /ay (0,0) = 0. Then, V*B = 0 requires thaty y

aH /Dx(0,0) = 0. Thus, an expansion of (a) about the origin gives

-153



--



EaH aH iI vi - -- ui (b)0 ay y ax x

Note that because Vx H = 0 at the origin, aHx/ y = a~ /lx. Thus, (b)

becomes

aH

S oI a [-ui + vi ] (c)

and 3H /lx(0,0) is easily computed becausey

(x,) o 1 1

Io2x

Hy

(x,O) =2w

[a-x a+x

]2w

[2

] (d)

Thus,

S 2 [-ui x + vii y (e)

Ira

It is the fact that V x H = 0 in the neighborhood of the origin that requires

that the contributions to (e) be negatives.

Part b

(i) Assume u = Re[u e( t - k z ) (f)

Then

2 2k2 2 2 f 2 IbtVsVp,k + =-m

(g)

The w-k plots are sketchedin the figure

M

)(~~e~~oh

&,

-154





Y,Coý%?ýeAX C-)

40'r 'rec,9t

-------- civ

Then

v = Re[v ej(t-kz)]

2 22 22 = v2k2 -

(h)

and the w-k plots are as shown

LtA ee6A-ZOL^Sýor VQcj

-155

PRCOrP\P) Cj

o r·ec

21 04 6j,

direction, it must destabilize motions in the other direction.

Part c

Driven response is found in a manner similar to that for Prob. 10.2.

Thus for

w < 4b (cutoff)

u(z,t) =-

u sinh a x

sinh ct

cos o to

(j)

v(z,t) = -

v sin k x

rsin k t

V

sin w to

(k)

u(z,t)

v(z,t)

=

=

-

-

u sin k x

isin k •

u

v sin k x

sink

v

cos wuto

sin t

()

(m)

where Vs 211/2

k

Uk=[w;

2

o -

2

2 1/2

Wb

~l/

-156

k= 2 +21/2

Part d

We must suppress instability of lowest natural mode in v.

22 T > 2

2 2

II <

for evanescent waves

2 23o %

Thus, from (n) and (p), 2 < v (w/9)2.

o

Part e

U4

if

'p.

-2

64

< 00</1o,

I~fl

-1 57

7




PROBLEM 10.26 (continued)L,

z0o

b/m

The effect of raising the current is summarized by the W-k plot, with

complex k plotted for real w.

As I is raised the hyperbola moves

outward. Thus, k increases and kv u

decreases to zero and becomes imaginary.

Thus, wavelengths for the v deflection

shorten while those for u lengthen to

infinity and then deflections decay.

Note that v waves shorter than A=2k

will not be observed because of

instability.

PROBLEM 10.27

Part a

We may take the results of problem 10.26 and replace -by + U ýz in the

differential equations, and w by w-kU in the dispersion equations.

Therefore, the equations of motion are

2 2u

m(- + U ) 2 u = f- 2- Ibu (a)

at az a

m( + U ) v = f az2 + Ibv (b)

Part b

For the x motions, the dispersion relation is

-m(w-k2

= - fk 2 - Ib (c)

We let

-158

-41





Ib 2

-f = V 2m s

Therefore w=kU + 2v + 2

or solving for k

wU +2v +

22(U

2-V 22 )

k=

k - )2U v

The w-k plot for x motions is sketched as

-W'0

For real w, we have only k real. For real k, we have only real w.

For the y motions, we obtain

222 2 2 2WU + V v v2 -ss

Tks f r

Thus for real w, the sketch is

-159





Tj Ul S

/p

(Ar (01 23

while fo r real k, the w-k plot is

-160





Part c

Since the wire is traveling at a "supersonic" velocity, we cannot impose

a downstream boundary condition to determine upstream behavior.

We are given

u(O,t)i + (O,t)i = u cosw t i + v sinw t i (h)

and the boundary conditions

(o,t) = o0 (o,t) = 0 (i)

We let

WU

2 2

s

2 2 2_+A, (U v)

b

U2_v2 s (U2_v )s(j)

2w2v

2-

(2(U

2- v)

(u2

(U2v)

- v )s

For the x otions, the allowed values of k are

kI = a + 8 with w = wo

k2 = - (k)

Therefore

u = Re Al e- k z + A2 e- Jk2]ejo ()

Applying the boundary conditions and simplying, we obtain

u = u Re[(C1 sin ýz + cos ýz) ej ( t-Z ) ] (m)

o

For the y otions, the allowed values of k are

k3 = a + y (n)

k4 = a - y

Therefore

v = -v0

Re[(-y- sin yz + j cos yz)e (w

t-az) ] (o)

-161





Part d

~N000

I - "I(- - 'I

what

As long as U > vs this is the form of u, no matter/the value of I (as long

as I > 0). As the magnitude of I increases, B increases but a remains

unchanged.

-1 62

-1





This is the form of v, as long as

2 2 2 2 _p2-Svs - (U ) > 0 (p)

As I is increased, we reach a value whereby this inequality no longer holds. At

this point y becomes imaginary and we have spatial growth.

,·

- ~ ~7N

As I is increased beyond the critical value, v will begin to grow exponentially

with z.

Part e

To simulate the moving wire, we could use a moving stream of a conducting

liquid such as mercury. We would introduce current onto the stream at the

nozzle and complete the circuit by having the stream strike a metal plate at

some downstream postion.

PROBLEM 10.28

Part a

A simple static argument establishes the required pressure difference.

The pressure, as a mechanical stress that occurs in'a fluid, always acts

on a surface in the normal direction. The figure shows a section of length

Az from the membrane. Since the volume which encloses this section must be

in force equilibrium, we can write

-163

2R(Az)[pi - po] = 2S(Az) (a)

CL

I

ltp

. .4--P

1<

where we have summed the forces acting on the surfaces. It follows that the

required pressure difference is

P - p = R (b)

Part b

To answer this question, and other questions concerning the dynamics of

the circular membrane, we must include in our description a perturbation

displacement from the equilibrium at r = R. Hence, we define the membrane

surface by the relation

r = R + E(e,z,t) (c)

The pressure difference p -po is a force per unit area acting on the membranein the normal direction. It is the surface force density necessary to counter

act a mechanical force per unit area T

mT = - S (d)

m R

which acts on each section of the membrane in the radial direction. We wish

now to determine the mechanical force acting on each section, when the surface

is perturbed to the position given by (c). We can do this in steps. First,

consider the case where 5 is independent of 6 and z, as shown in the figure.

Then from (d )

Tm R +

S S-1RR 2

(e)

where we have kept only the linear term in the expansion of T about r = R.

When the perturbation ý depends on 6, the surface has a tilt, as shown.

We can sum the components to S acting on the section in the radial direction

as

-164





lim S 1 3 - 1 (f),

AORARm 6

G 0+ A R

2

Similarly, a dependence on z gives rise to a radial force on the section due to

the mechanical tension S,

S 3z lim

Ale-0 Az 2

In general, the force per unit area exerted on a small section of membrane under

the constant tension S from the adjacent material is the sum of the forces given

by (e), (f) and (g),

T = S(- 1 + + + 2)2 (h)Tm = ( +2 R 2

It is now possible to write the dynamic force equation for radial motions.

In addition to the pressure difference pi-Po acting in the radial direction,

we will include the inertial force density om/( 2ý/at2) and a surface force

density Tr due to electric or magnetic fields. Hence,

2 1 1 a) (i2• S(- - + + 2 -+ 2 ) + T + p ()-P

m at R 2 2 z r io

Consider now the case where there is no electromechanical interaction.

Then Tr = 0, and static equilibrium requires that (b) hold. Hence, the constant

terms in (i) cancel, leaving the perturbation equation

326 ( 1__2 2

m2 2 2 2at R R ae2 z

-165

Parts c & d

This equation is formally the same as those that we have encountered

previously (see Sec. 10.1.3). However, the cylindrical geometry imposes

additional requirements on the solutions. That is, if we assume solutions

having the form,

((wt+ me)( = Re (z()e

jRe(k)

the assumed dependence on 0 is a linear combination of sin me and cos me.

If the displacement is to be single valued, m must have integer values. Other

wise we would not have E((,z,t) = C(B + 27,z,t).

With the assumed dependence on 6 and t, (j) becomes,

2"d

E + k2 ==()dz

where 2a

2 1 (-m2) m

2 SR

=The membrane is attached to the rigid tubes at z 0 and z = R. The

solution to (Z) which satisfies this condition is

= A sink x (m)n

where

nWk = -- , n = 1,2,3,...

n P,

The eigenvalue kn determines the eigenfrequency, because of (Z).

2 n 2 (m2- 1) S

n R2 aR m

To obtain a picture of how these modes appear, consider the case where A is

real, and (m) and (k) become

C((,z,t) = A sin Tx cos mO cos wt (o)

The instantaneous displacements for the first four modes are shown in the

figure.

There is the possibility that the m = 0 mode is unstable, as can be seen

from (n), where if

2 1 )n7 2 (P)

(i) <R 2

we find that the time dependence has the form exp + Iwit. The first mode to

meet this condition for instability is the n = 1 mode. Hence, it is not possible

-166




Mv'

m=1

Vi = i

k=2.

tj =

-167

DYNAMICS OFp ELECTROMECHANICAL CONTINUA


to maintain the uniform cylindrical shape of the static equilibrium if

R7t/£ < 1 (q)

This condition for static instability is easily understood if we remember

that in the m = 0, n = 1 mode, there are two perturbation surface forces on a

small section of the membrane surface. One of these is the perturbation part

of (e) and arises because of the curvature in static equilibrium. This force

acts in the same direction as the displacement, and hence tends to produce

static instability. It is counteracted by a restoring force proportional to

the second derivative in the z direction, as given by (g). Condition (q) is

satisfied when the effect of the initial curvature predominates the stiffness

from the boundaries.

Part e

With rotation, the dispersion becomes:

(w-mp)2

= [Ms

12_-m2 ]c

with

2 SoR

2

aRm

2E V 22 oo R

m = 3c S

a

Because there is no z dependence (no surface curvature in the z direction) the

equilibrium is unstable in the m = 0 mode even in the absence of an applied

voltage.

PROBLEM 10.29

The solution is of the form

= (a) + (_(8) (a)

where

= x + y

We are given that at x = 0

• dl+ dld-_ + d A[U-(y)-U- (y-a)] (b)

and that

-168

which implies that

d+--r = 0 =ay do

We therefore have

d+ Ad = [u(-a)

and

d_ Ad 2 [U-0()

dE+

d0

- u l(-C-a)]

- U 1(8-a)]

(c)

(e)

Then = - + _ _(Y-X)- u (y-x-a)Ty do dR 2 -1 -1

Integrating with respect to y, we obtain

+ U-l(x+y)-ul(x+y-a)}1

(f)

where u_2 is

( {-u 2 (Y-x) + u_2(y-x-a) + u2(Y+X)

a ramp function; that is u_2(y-b) is

- u_2(y+x-a)

defined as

(g)

maJ

Part b

The constraint represented by Fig. 10P.29 could be obtained'by ejecting

the membrane from a slit at x = 0 that is planar, but tilted over the range

0 < y < a. Thus, the membrane would have no deflection ý at x = 0, but would

have the required constant slope A over the range 0 < y < a, and zero slope

elsewhere.

-169





-170




PROBLEM 10.30

For this situation, the governing equation is (10.4.15) of the text.

(M -1)2

a22 a

22

ax2 ay2

2Here M = 2; so we have the equation:

2 22 = (b )

ax av

The characteristics are determined from equations (10.4.17) and (10.4.18) to be:

a= x-y

B= x+y

The x-y plane divides into regions A...F, as shown in the sketch. Tracing

back on the characteristics from points in regions A, D, F... shows that in

these regions C = 0; the characteristics originate on "zero" boundary conditions.

At points in region B, only the C+

characteristic originates on finite data;

+(a) = o' _(B) = 0 and hence

E = co in region B

-171





In C, deflections are determined by waves, both originating from the initial

data. Hence E+(a) = o,' but E_(() is determined by the reflection of an incident

wave on the boundary at y = THence = 5o and. - 0_()

S= 0 in region C

In region E only the _((3) wave is finite because the +(a) wave originates

on zero conditions and

E = - 0oin region E

The deflection has the stationary appearance shown in the figure.

PROBLEM 10.31

From equation 10.4.30, we have

kB I2 k2v2 oS=k

S- m

We define

IBo

2mvs

BI \2 2

2myS

vS

-172







The four allowed branches of k as a function of w are therefore k = + kl,

and + k2, where

k = a + (d)

k, = - a + B (e)

The sketch shows complex w for real k. Note however, that only real values of

k are given if w is real and hence the solid lines represent the plot of complex

k for real w.

C3,

C3;

-173




PROBLEM 10.32

The effect of the longitudinal convection is accounted for by replacing w

in Eq. 10.4.3 by w-kU (see for example page 594). Thus,

o(w-kU)2

= k22 kB I

s)- m (a)

This expression can be solved to give

B I B IBI B 1B2(2wU +---) + 40U ++

k = - s2 2 m (b)2(U - v2)

The sketch of complex k for real oiis made with the help of the following

observations: Consider the modes that are represented by -Bo

1) Asymptotes for branches are k = w/(U + vs) as w + o.

2) As w is lowered, the (-B ) branches become complex as

4v2-2 + 4BU I+ 0

or at the frequencies

BI

2v2 m -- --V s

Thus, for U > vs there is a lower as well as an upper positive frequency at

which k switches from real to complex values.

In this range of complex k, real k is

BIk = (2wU - )/2(1I2 - v

2)

or a straight line intercepting the k = 0 axis at

BIo

2Um

3) as w + 0,

2 2k + 0 and k - + B I/m(U2 _ v)-o 5

where the - sign goes with the unstable branches.

4) As w + - m the values of k are real and approach the asymptotes

k = W/(U + vs).

-174





Similar reasoning gives the modes represented in (b) by +B . Note that these

modes have a plot obtained by replacing w + - w and k - - k in the figure.

-175



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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA

PROBLEM 11.1

Part a

We add up all the volume force densities on the elastic material, and with

the help of equation 11.1.4, we write Newton's law as

2 1 = 11 (a)P p2 x p (a)

a 1 a

where we have taken a a = 0. Since this is a static problem, we let

a ý2 x3

-- = 0. Thus,

aT(

S = pg. (b)

From 11.2.32, we obtain

a6

T11 = (2G + X) (c)

Therefore

(2G + X) 2 1 pg (d)ax1

Solving for 61, we obtain

61

=

2(2G+X) 1+ C

1x + C

2(e)

where C1 and C2 are arbitrary constants of integration, which can be evaluated

by the boundary conditions

61(0) = (f)

and 6

T1 (L) = (2G + X) (L) = 0 (g)

since x = L is a free surface. Therefore, the solution is

pg x

1 2(2G+X) [x - 2L]. (f)

Part b

Again applying 11.2.32



a6

T11= (2G+X) -ax pg[x 1 - L)

T12 = T21 = 0

T13 = T31 = 0

ax 1

T22 =X 1

a6133 3 x 1 (2G+X) [ - L

1x

T32 = T23 = 0

0 0T11

T 0 T22 0

0 T3333

PROBLEM 11.2

Since the electric force only acts on the surface at x1 = - L, the equation

of motion for the elastic material (-L xI < 0) is from Eqs. (11.1.4) and (11.2.32),

2 2.a26 a26 I

p = (2G+X) 2 (a)

at2 ax

1


6 1 (0,t) = 0

a 61(-L,t) a61M = aD(2G+X) (-L,t) + fe

at2 3x1

fe is the electric force in the xl direction at xl = - L, and may be found by

1using the Maxwell Stress Tensor T = CE Ej 2 ij EEkEk to be (see Appendix G

for discussion of stress tensor),

fe

-E

E22

aD2

with

V + VI cos Wt

E + 61 (-L,t)

- -



Expanding fe to linear terms only, we obtain

V2

2V Vd cosWt 2V

2

fe EaDo o 102 3fe

+d 61(-L,t (d)

d d d

We have neglected all second order products of small quantities.

Because of the constant bias Vo, and the sinusoidal nature of the

perturbations, we assume solutions of the form

^ J(wt-kx1 )

61 (x1 ,t) = 61 (x1 ) + Re 6e(e)

where

<< 6 1 (x 1 ) << L

The relationship between w and k is readily found by substituting (e) into (a),

from which we obtain

k = + ý-- v

with vp

=P

(f)

P

We first solve for the equilibrium configuration which is time independent.

Thus

2 6 1 (x1 )

ax1

This implies

6 1 (x1 ) = C1x 1 + C2

Because 61(0) = 0, C2 O0.

From the boundary condition at xl = - L ((b) & (d))2 V

aD(2G+X)C1 - aD 0 (h)

Therefore

V2

61(x E o xl (i)22 (2G+X)

Note that 61(x1 = - L) is negative, as it should be.

For the time varying part of the solution, using (f) and the boundary condition

6(o,t) = 0

we can let the perturbation 61 be of the form

61 (X1 ,t)=

Re 5 sin kx1 ejt(j)

Substituting this assumed solution into (b) and using (d), we obtain

+ Mw2

6 sin kL = aD(2G+X)kA

6 cos kL (k)

CaDV V EaDV2

2 3 d6 sin kL

Solving for 6, we have

AaDV V

CaDV

2

d2 Mw2 sin kL - aD(2G+X)k cos kL + 30 sin kL

Thus, because 6 has been shown to be real,

V2

L

61 (-L,t) = o - 6 sin kL cos wt (m)

Sd(2G+X)

Part b

If kZ << 1, we can approximate the sinusoidal part of (m) as

£aDV V1 cos wt

61•-L,t) = 2o+ aDV2

(n)[ aD(2G+X) EaDV2

We recognize this as a force-displacement relation for a mass on the end of a

spring.

Pairt

We thus can model (n) as

/6

L4





where

EaDV V cos wtol

d

caDV 2

aD(2G+X) oL

3

We see that the electrical force acts like a negative spring constant.

PROBLEM 11.3

Part a

From CL1.1.4),we have the equation of motion in the x2 direction as

2 22 aT

a t 2 1

ý 2 1x1

From(11.2.32),

T21= G 3x

Therefore, substituting (b) into (a), we obtain an equation for 62

2 a 62a 62 G-2t2

axI

We assume solutions of the form

j(wt-kx1 )

2= Re 62 e

where from (c)we obtain

wk-v

2v

P

GPP

p

Thus we le t

+ e(t+kx)

62 = Re a eJ(wt-kx 1 )

with k =vP


62(,t) = 6 ejWt





and

x1 O 0

(g)

since the surface at x1 = 0 is free.

There fore

6 e-jk' + 6b ejkt=6a b o (h)

and

-jk 6a + jk6b = 0 (i)

Solving, we obtain

6

6 =6 =

a b 2cosk (J)Therefore

62 xt) = Re co cos kx 1 ej cos kx cos wt (k)

and

T21(Xl,t)=-Re LFc sin kx1 ej ()

G6 k

cosk sin kx 1 cos WtcoskZ 1

Part b

In the limit as w gets small

6 2 (Xl,t) + Re[6 e j t ](m)

In this limit, 62 varies everywhere in phase with the source. The slab of elastic

material moves as a rigid body. Note from (Z) that the force per unit area at2

dx1 = R required to set the slab into motion is T21 (L,t) = pt dt•l o cos wt) or the.

mass /(x 2 -x 3 ) area times the rigid body acceleration.

Part c

The slab can resonate if we can have a finite displacement, even as 6 - 0.

This can happen if the denominator of (k) vanishes

cos kZ = 0 (n)

or (2n+l)irvS= 2

22.

n = 0,1,2,...(o)

6





The lowest frequency is for n = 0

7v

or low 2£

PROBLEM 11.4

Part a

We have that

Ti = Tijnj ' ijnj

It is given that the Tij are known, thus the above equation may be written as

three scalar equations (Tij - a6ij)n = 0, or:

(T11 - a)n1 + T12n2 + T13n3 = 0

T21n1 + (T2 2 - a)n2 + T2 3n3 = 0 (a)

T31n1 + T32n2 + (T33 - a)n3 = 0

Part b

The solution for these homogeneous equations requires that the determinant of

the coefficients of the ni's equal zero.

Thus

(T11 - a)[(T 22 - a)(T 3 3 - a) - (T 2 3 ) 2 ]

- T1 2 [T12 (T33 - a) - T13T 23] (b)

= 0)] T13[T12T23 - T13(T22 - 0

where we have used the fact that

Tij = Tji. (c)

Since the Tij are known, this equation can be solved for a.

Part c

Consider T12 = T21 = To , with all other components equal to zero. The deter

minant of coefficients then reduces to

-a3+ 2 a= 0 (d)

0

for which = 0 (e)

or • +T (f)

The a = 0 solution indicates that with the normal in the x3 direction, there is

no normal stress. The a = t To solution implies that there are two surfaces

where the net traction is purely normal with stresses To, respectively, as




PROBLE' 11.4 (continued)

found in example 11.2.1. Note that the normal to the surface for which the shear

stress is zero can be found from (a), since a is known, and it is known that

In 1.

PROBLEM 11.5

From Eqs. 11.2.25 - 11.2.28, we have

1

ell E [11 (T22 + T33)]

e -[T1 - v(Tr + T) ]

22 =E 2 2 33 11

e33 [T33 - v(T + T 22 )]

and

eij 2G ,

These relations must still hold in a prim ed coordinate system, where we can use

the transformations

T = aika Tk£ (e)

and

ej = aikaj ek (f)

For an example, we look at eli

e = - 1 " + T(T 2e = alkaltekt [T2 ;31

This may be rewritten as

alkaltek= [(( + v)alkalTk£ kV'kTk]

where we have used the relation from Eq.(8.2.23), page G10 or 439.

a a = 6pr ps ps

Consider some values of k and X where k # Z.

Then, from the stress-strain relation in the unprimed frame,

Tkalkaltekk = alkalT2k =

a al,

E (1+v)Tk

Thus 1 l+v

2G E

or E = 2G(l +v) which agrees with Eq. (g) of

example 11.2.1.

http://slidepdf.com/reader/full/-11.2.28






PROBLEM 11.6

Part a

Following the analysis in Eqs. 11.4.16 - 11.4.26, the equation of motion for

the bar is

Eb2 34 02 (a)t2 3p ýx4

where & measures the bar displacement in the x2 direction, T2 in Eq. 11.4.26 = 0

as the surfaces at x 2 = b are free. The boundary conditions for this problem

are that at xl = 0 and at xl1 L

T21 = 0 and T11 = 0 (b)

as the ends are free.


E = wt (c)e E(x)e

j

As in example 11.4.4, the solutions for Z(x) are

V(x) = A sin ax 1 + B cos ax1 + C sinh ax 1 + D cosh ax1 (d)

with 1

E

Now, from Eqs. 11.4.18 and 11.4.21,

(x2 - b2)E 3

(T -b 3 (e)21 2 3

x1

which implies

3E = 0 (f)

ax3

at Xl= 0, x1 = L

and

11=-

2E

2(g)T X 2

ax1

which implies

a2 E = o (h)

ax 2

1

at xl1 0 and xl = L









With these relations, the boundary conditions require that

-A +C = 0- A cos cL + B sincL + C cosh cL + D sinh aL = 0

-B +D = 0

- A sincL - B cos aL + C sinh aL + D cosh L = 0

The solution to this set of homogeneous equations requires that the determinant

of the coefficients of A, B, C, and D equal zero. Performing this operation, we

obtain

cos cL cosh cL = 1

Thus,

B = aL = 2 3pEb2

Part b

The roots of cos B = coshcosh B

follow from the figure.

Note from the figure that the roots aL are essentially the roots 3w/2, 57/2, ...

of cos aL = 0.



INTRODUCTION TO TH E ELECTROMECHANICS OF ELASTIC MEDIA


Part c

It follows from (i) that the eigenfunction is

= A'[(sin ax 1 + sinh aex) (sin aL + sinh aL)

(Z)

+ (cos aL - cosh aL)(cos ax1 + cosh ax1)

where A' is an arbitrary amplitude. This expression is found by taking one of

the constants A ... D as known, and solving for the others. Then, (d) gives the

required dependence on xl to within an arbitrary constant. A sketch of this

function is shown in the figure.




PROBLEM 11.7

As in problem 11.6, the equation of motion for the elastic beam is

2 Eb2

4+ . = 0

S2 3p x4

The four boundary conditions for this problem are:

((x I = 0) = 0 S(xl = L) = 0

61(0) 00-Og

= - 2 x 61(L)' x2 3xx1 = 0 1

1 X13


(xt) (eJwt Re(x t) = Re (x)e , and as in problem 11.6, the solutions for (c)

it I

E(xl) are

(x AA sin ax1 + B cos x1 + C sinh ax1 + D cosh ax1

with a = 02 1/4


B + D = 0

-OA sin aL + B cos tL + C sinh aL + D cosh aL = 0

A + C = 0

A cos aL - B sin CL + C cosh aL + D sinh aL = 0

The solution to this set of homogeneous equations requires that the determinant

of the coefficients of A, B, C, D, equal zero. Performing this operation, we

obtain

cos aL cosh aL = + 1

To solve for the natural frequencies, we must use a graphical procedure.





The first natural frequency is at about

3wtL=

2

Thus

w2 L =

1/22

(3 Eb2

L2 3p )

Part b

-4We are given that L = .5 m and b = 5 x 10 m

From Table 9.1, Appendix G, the parameters for steel are:

E % 2 x 1011 N/m2

p I 7.75 x 103 kg/m3





w. 120 rad/sec.

Then, f = -- L 19 Hz.1 2T

Part c

5For the next higher resonance,

Therefore, f 52 f 1 53 Hz.

PROBLEM 11.8

Part a

As in Prob. 11.7, the equation of motion for the beam is

2 Eb2 4

2 3p •4at )xi

At x1 = L, there is a free end, so the boundary conditions are:

T1 1 (x1=L) = 0

and T2 1 (x1=L) = 0

The boundary conditions at xl= 0 are

M2 ,

t) = + (T 2 1 ) D dx 2 + f + F

St 2x =0

and

61(x1 = 0) = 0

The H field in the air gap and in the plunger is

S NiH= -i

D

Using the Maxwell stress tensor

(e - P)oN2 i 2 2

N2i2e 2 2 2 2 o )2

)(

with i = I + i1 cos wt = I + Re i l ejWtO o



INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC !MEDIA


We linearize fe to obtain

-e N2

2 I

-2 (1 -o ) [ o + 21oi1

cos wt]i2 (g)

For equilibrium

N2- 2 ( 0 2 12 00o -2-- (-olo )1 i2 =

Thus - N2

F = o)I 21(h)

Part b

We write the solution to Eq. (a) in the form

(xj,t) = Re U(xl) e

where, from example 11.4.4

E(xl ) = Al sin caxl + A2cos C x + A3sinh aIx + A4 cosh cxx (i)

with

Now, from Eqs. 11.4.6 and 11.4.16

36 2 0

T1 1 (x=L) = E - Ex 0 (j)1

Thus 3-i2 (x 1 = L) = 0

21

From Eq. 11.4.21

(x2 - b2

3T

21 2 2a 3 (k)

1

and from Eq. 11.4.16

61(x1 =0) =

-x

2(x10

=0 (

Thus -(+ )10 = 0




PROBLEM 11.8(continued)

Applying the boundary conditions from Eqs. (b), (c), (d) to our solution of Eq. (i),

we obtain the four equations

Al + A3=0

- Al sin caL -A2 cos at + A3 sirnaL + A4 cosh aL =0

- Al cos cL + A2 sin cL + A3 cosh L + A4 sinh cL =0

2 3b3 DA 2 2 3b3EDA 2A4 )2 a EDA1 +MwA2+ M = + N Io o

dX d - 2-= D[N1V = dt dt [-D- D o(0) + D-ý(0)

or v =-N1o( - o)jW(A 2 + A4 ) + N2 i1 PDjw

We solve Eqs. (m) for A2 and A4 using Cramer's rule to obtain

N Ioil ~'- )(- + sin aL sinh cL - cos aL cosh cL)

=2

- 2Mw (1 + cos aL cosh L)+ - (ab)3ED(cos CL sinh cL + sin aL cosh cL)3

N21 il(- 1o)(- 1 - cos %L cosh aL - sin cL sinh cL)

A =4

- 2Mw2(1 + cos aL cosh cL) + 3

4(Cab)

3ED(cos cL sinh cL + sin aL cosh cL)

Thus

z(J) = v(j + 2 ,0 (i- o j(+ 2 + 2 cos cL cL) ([N cosh

il - 2Mw2 (1 + cos cL cosh cL)+ (ab)3ED(cos aL sinh aL+ sin aL cosh cL)

+ N2lDjw

Part c

Z(jw) has poles when

+ 2M 2(1 + cos aL cosh aL) = - (Cb) ED(cos aL sinh cL + sin aL cosh aL)




PROBLEM 11.9

Part a

The flux above and below the beam must remain constant. Therefore, the H

field above is

H (a- b)

a (a- b- ) 1

and the H field below is

Ho(a - b) i-

b (a- b +-C)

Using the Maxwell stress tensor, the magnetic force on the beam is

T 2 2 _ 2 2 42 (H-2 2 a b

) - -2

Ho_

(a-b) ab2

2p H

(a- b)

Thus, from Eq. 11.4.26, the equation of motion on the beam is

2Eb

2 4 1H0a 2 = 0oHo2

(d)

t 2 3p x4 (a-b)bp1

Again, we letA jt

E(x t) = Re (xl)et (e)

with the boundary conditions

ý(xO=0) = 0 E(xl= L) = 0

61 (x1=O0) 6

1 (x 1= L) = 0

Since 61= - x2 a /ax 1 from Eq. 11.4.16, this implies that:

- (xl=0)= 0 and (x= L) = 0

Substituting our assumed solution into the equation of motion, we have

2A

2 Eb2 4= oHo o

3P a4 + (a-b)bp

Thus we see that our solutions are again of the form

U(x) = A sin ax + B cos ax + C sinh ax + D cosh ax









where now 1/4

Sa-b)bp /Eb 2

Since the boundary conditions for this problem are identical to that of problem 11.7,

we can take the solutions from that problem, substituting the new value of a. From

problem 11.7, the solution must satisfy

cos aL cosh aL = 1 (k)

The first resonance occurs when

3 T

2

( 3& 4Eb \ 2or 2 3

w2 0213po

L4

(a-b)bp

Part c

The resonant frequencies are thus shifted upward due to the stiffening effect

of the constant flux constraint.

Part d

We see that, no matter what the values of the system parameters w2 > 0, so W

will always be real, and thus stable. This is expected as the constant flux cons

traintimposes aforce which opposes the motion.

PROBLEMI 11.10

Part a

We choose a coordinate system as in Fig. 11.4.12, centered at the right end of

d 1the rod. Because -= T , we can neglect fringing and consider the right end of the

0 D 1

rod as a capacitor plate. Also, since 1 , we can assume that the electrical

force acts only at xl = 0. Thus, the boundary conditions at x1 = 0 are

- T 2 1 D dx2 + fe = 0 (a)

2 2(b - 32 b)

where T2 = E (Eq. 11.4.21)21 2 a 3

since the electrical force, fe, must balance the shear stress T 2 1 to keep the rod

in equilibrium.,







and

T 11(0) = - x2 Ea (0) = 0 (b)2

ax 1

since the end of the rod is free of normal stresses. At xl = - L, the rod is

clamped so

= 0(-=) (c)

andan(-) - (-) = 0 (d)WeseMaxwell stresshe

tensor to calculate the electrical force to be

We use the Maxwell stress tensor to calculate the electrical force to be

(e)

2eAV [ V '(0)]

d2 IVs + d

The equation of motion of the beam is (example 11.4.4)

Eb2 a 0+(f)

3t 3p ax1

We write the solution to Eq. (f) in the form

C(x,t) = Re E(x)e (g)

where

C(x) = Al sin cx + A2cos ex + A3 sinh ax + A4 cosh cx

with -C

Applying the four boundary conditions, Eqs. (a), (b), (c) and (d), we obtain

the equations

- Al sin cat + A2 cos at - A3 sinh a + A4 cosh cL = 0

Al cos at + A2 sin at + A cosh at - A4 sinh ct 0 (h)

- A 2 + A4 = 0

2 2E AV2 2F AV v2 3 3 2EAV 2 + o o o os- b DEA

+ , A2+

3 b3DEA3 -- A =1 d12 -- d --- 4 d





dq sNow i

s dtc A(v - V )

where qs d-ý(0)

(Vo

+ va

) +

d + E(0)

2e Av 2e AVo s o o

&(0)d

d

Therefore

2E A V

is = jW od V^s do ((0)

where

(0) = A2 +

We use Cramer's rule to solve Eqs. (h) for A2 and A4 to obtain:

- EoAVo s [cos ctsinh at -sin at cosh at]

dyq22

= A44=~ 2b 3 3DE(l + cos tccoshca)+ 2Eo o

ft

(cosct sinh at- sin arXcosh aR)(2)

3d3

Thus, from Eq. (k) we obtain

2i 3E AVV

+ 3(o o (cos at sinhat - sin at cosh at)Z(jO) = 2d

Lb) I (m)jw2E0 A d1(C3

3ED (1 + cos at cosh at )

Part b

We define a unction g(ak) such that Eq. (m) has a zero when





3 3U3V2AE

(aL) 3g(aL) = (1 + cosh at cos ae)(at) o o

sin at cosh at - cos ct2 sinh ct2 DEb3

d3

(n)

Substituting numerical values, we obtain

3£3V2A -3

3 o 3 x10l (106)1o0-(8.85x 102 ) 1.2 x 10 - 3 (o)

DEb 3d3 10- (2.2 x 10)10- 9 10 - 9 1.2

In Figure 1, we plot (at) 3g(at) as a function of at. We see that the solution to

Eq. (n) first occurs when (at) g(ae) 2 0. Thus, the solution is approximately

at = 1.875

(at) g(at)

3.0

2.8

2.4-

2.0

1'.6

1.2

0.8

0.4

I i -I Ia

-iiI "0 0.4 0.8 1.2 1.6 2.0 at

Figure 1





From Eq. (g)

at = 2 k- = 1.875Eb t

Solving for w, we obtain

w 1 1080 rad/sec.

Part c

The input impedance of a series LC circuit is

21 - LCw

jwC

Thus the impedance has a zero when

2

o LC

We let w = w + Aw, and expand (q) in a T;aylor series around wo to obtain

Z(jw) + j 2A- = + 2j LAw

(m) can be written in the form

z (jW)S2jwC [1 - f(w)] where f(w) = 12jwCo oA

and C = o d

For small deviations around w0O

2•oW (j ) jT O 0Iwo

Thus, from (q), (r) (s) and (t), we obtain the relations

2L =

2wC0 awlwOt 0

o

and 1

W2L

Know f(w) =

31£E AV 2where K 3 o o = 1.2 x 10-3

d3(EDb

3)





1 + cos at cosh atand g(at) =

sin at cosh at - cos at sinh at

Thus, we can write

df(w) d K d(cat)

dw o d(ca) (a)g (at)- dw

o

Now from (g),

d(ak) = 3p /4~

dw Io 2o/,0wo

d FK 1 - K d

d(at) (at) g (at) [ ( a Z) g (a Z)] d (a Z) WO

K da) [(at) ,g(a)] (aa)K d~ccx)

0

since at w = oo

(ak)3g(at) = K . 0b)

Continuing the differentiating in (aa), we finally obtain

d [(a)3(at)c 1 g(at)3(a ) + (a t d g(atd(ca) L -K K I d(aZ) gc

I

0

-3 (at)3 d

(cc)K d() g(

WO

d - sinat coshat + cos aisinh at=d() g()

d(at) (sin aicosh at- cos aisinh at)

- (1+cos ctcosh at)(+cosatcoshca+ sinaksinh at+ sin atsinh at- cos ti cosh at)

(sinat coshat - cos atsinhat)

= - 1 2n(at) (sin aisinh at)(dd )

(sin aicosh ca- cos atsinh at)





Substituting numerical values into the second term of (cc), we find it to have

value much less than one at w = wo.

Thus,

d

d(at) g(a) - (ee)

Thus, using (y, (z),(aa) (bb) and (dd), we have

df01 4.8 (ff)

dw

0O

Thus, from (v) and (w)

lL 4(1080) (8.85 x10 -)(104) = 1.25 x 10 henries

4.8 x10

C X110 6(10

2= 6.8 X 10

-16farads.

1.25 x109(1080)




PROBLEM 11.11

From Eq. (11.4.29), the equation of motion is

a263 / 26 a26

3 1a3p = + ax- (a)

We let

6 = Re 6(x )e(t-kx) (b)3 2

Substituting this assumed solution into the equation of motion, we obtain

-pw2 6 = G k26 +xC (c)

ora26 +^2

+ ( - k2)6 = 0 (d)2

If we let 82 =G

2 k2

(e)

the solutions for 6 are:

6(x ) = A sin 8x + B cos 8 x (f)2 2 2


6(0) = 0 and 6(d) = 0 (g)

This implies that B = 0

and that 8 d = nn .

Thus, the dispersion relation is

w - k2 ( )2(h)

Part b

The sketch of the dispersion relation is identical to that of Fig. 11.4.19. How

ever, now the n=0 solution is trivial, as it implies that

6(x ) = 02

Thus, there is no principal mode of propagation.






PROBLEM 11.12

From Eq. (11.4.1), the equation of motion is

a 26p -7- = (2G+X)V(V.6) - GVx(Vx6) (a)

We consider motions

6 =6e(r,z,t)io (b)

Thus, the equation of motion reduces to

p 26'.- G F 2L6 + 1 a r6 = 0 (c)7 r Tr ij


) '(r,z,t= Re 6(r)ej t-kz) (d)

which, when substituted into the equation of motion, yields

6 (r )(r + \ k2) 6(r) = 0 (e)Tr r ar G

From page 207 of Ramo, Whinnery and Van Duzer, we recognize solutions to this

equation as

F,,..PW2k2

2)[ 1

6(r) = A J - + BN - k r (f)

On page 209 of this reference there are plots of the Bessel functions J and N

We must have B = 0 as at r = 0, N goes to - . Now, at r = R1

6(R) = 0 (g)

This implies that

J( - k2 R] = 0 (h)

If we denote ai as the zeroes of J , i.e.

J (a = 0

we have the dispersion relation as2

G2 - k2

2(i)



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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS

PROBLEM 12.1

Part a

Since we are in the steady state (C/Dt= 0), the total forces on the piston

must sum to zero. Thus

pLD +(fe) x = 0 (a)

where (fe) is the upwards vertical component of the electric force

(fe) 2x2

LD (b)

Solving for the pressure p, we obtain

EoVop= (c)

2x

Part b

dBecause A <<1, we approximate the velocity of the piston to be negligibly

small. Then, applying Bernoulli's equation, Eq. (12.2.11) right below the piston

and at the exit nozzle where the pressure is zero, we obtain

1 2 EoVo2

2x2 p

Solving for V , we have

V V= C (e)p xp

Part c

The thrust T on the rocket is then

T = V dM = V 2 pdD (f)p dt p

2dD

x

PROBLEM 12.2

Part a

The forces on the movable piston must sum to zero. Thus

pwD - fe = 0 (a)

where fe is the component of electrical force normal to the piston in the direction

of V, and p is the pressure just to the right of the piston.

fe= (b)2 w





Therefore

p= --- (c)

Assuming that the velocity of the piston is negligible, we use Bernoulli's law,

Eq. (12.2.11), just to the right of the piston and at the exit orifice where the

pressure is zero, to write

PV2

= p (d)

or r1

Part bV =

W(e)

The thrust T is

T = VdM

-VoV

2pdW =

I 2d

d (f)

Part c

For I = 103A

d = .lm

w= im

3

p= 103 kg/m

the exit velocity is

V = 3.5 x 10 2 m/sec.

and the thrust is

T = .126 newtons.

Within the assumption that the fluid is incompressible, we would prefer a dense

material, for although the thrust is independent of the fluid's density, the ex

haust velocity would decrease with increasing density, and thus the rocket will

work longer. Under these conditions, we would prefer water in our rocket, since

it is much more dense than air.

PROBLEM 12.3

Part a

From the results of problem 12.2, we have that the pressure p, acting just to

the left of the piston, is1I2

P 2 (a)2w

The exit velocity at-each orifice is-obtaineu by using Bernoulli's law just to the

left of the piston and at either orifice, from which we obtain

V I (b)

at each orifice.

Part b

The thrust is

T= 2VdM

= 2V2pdw (c)

21 I2d

T = 0 (d)w

PROBLEM 12.4

Part a

In the steady state, we choose to integrate the momentum theorem, Eq. (12.1.29),

around a rectangular surface, enclosing the system from -L < x <+ L.

- V a + p[V(L)] b = P a - P(L)b + F (a)

where F is the xi component force per unit length which the walls exert on the

fluid. We see that there is no x, component of force from the upper wall, therefore

F is the force purely from the lower wall.

In the steady state, conservation of mass, (Eq. 12.1.8), yields

V(t) = V a l (b)

Bernoulli's equation gives us

1 pV22 + P = 1pV

aV2+ P(L)

(c)2 o 0 2 obj

Solving (c) for P(L), and then substituting this result and that of (b) into

(a), we finally obtain

F = P (b-a) + pV2 (- a + ) (d )0 o 2 2b

The problem asked for the force on the lower wall, which is just the

negative of F.

Thus

F = - Po(b-a)- pV 2(-a + b

)(e)

wall o 2b 2

PROBLEM 12.5

Part a

We recognize this problem to be analogous to a dielectric or high-permeability

cylinder placed in a uniform electric or magnetic field. The solutions are then

dipole fields. We expect similar results here. As in Eqs. (12.2.1 - 12.2.3), we



--



define

V = - Vý

and since

Vv = 0

then V20 = 0.

Using our experience from the electromagnetic field problems, we guess a solution

of the form

= -A

cos O + Br cos 9r

Then

A A(T cos 9r - B sin 9 + B sin 9)T

-

Now, as r *

V = V = Vo(cos Gi - i sin 9)

Therefore

B = -V

The other boundary condition at r = a is that

Vr(r=a) = 0

Thus

A = B a2

-Va2

Therefore

V V - a _ - V sin 9 ( 1 + a)

= cos( -)i ao r r

Part b

vs

;1 0 i





Part c

Using Bernoulli's law, we have

1p2 1 2 a' 2a2 po o = 2+ Vo (1 + r- a-cos 20) + P

Therefore the pressure is

P = p-1pV

2r

2a12

cos 2 9)0 2 o r r

Part d

We choose a large rectangular surface which encloses the cylinder, but the

sides of which are far away from the cylinder. We write the momentum theorem as

I pv(v'n)da = -I Pd7a + F

where F is the force which the cylinder exerts on the fluid. However, with our

surface far away from the cylinder

V= ViS1

and the pressure is constant

P = Po.

Thus, integrating over the closed surface

F=

The force which is exerted by the fluid on the cylinder is -F, which, however, is

still zero.




PROBLEM 12.6

Part a

This problem is analogous to 12.5, only we are now working in spherical co

ordinates. As in Prob. 12.5,

V = - V

In spherical coordinates, we try the solution to Laplace's equation

B= Ar cos 0 + - Cos 8 (a)

rTheta is measured clockwise from 'the x axis.

Thus

V = 2B B (b A cos + r cos A + -) sin (b:)

As r+

oo

V 4 Vo(i cos 8 - i sin 8) :(c)

Therefore A = - V (d)0

At r = -a

V (a) = 0 :(e)

Thus 2B-- =A= -V3 O-a

or Va 3

B o2

(f)

Therefore 3

a3 a

V= V (l - )cos Or - Vo( + -) sin i (g)2r

3 0 (g)o r r o

with 2 2 2r = V x +x +x

1 2 3

Part b

=At r a, 0 = n, and =

we are given that p = 0

At this point

V = 0

Therefore, from Bernoulli's law

p = - 2 - )2 cos20 + sin

2 (1 )2 (h)

Part c

We realize that the pressure force acts normal to the sphere in the - ir

direction.





at r = a

9 PV 2 sin2 e

8 o

We see that the magnitude of p remains unchanged if, for any value of 6, we look

at the pressure at e + r. Thus, by the symmetry, the force in the x direction is

zero,

= f 0.

PROBLEM 12.7

'art a

We are given the potential of the velocity field as

V o Vxo- o

a x I . a (X2 1+

1 2)

If we sketch the equipotential lines in the x x plane, we know that the velocity dis1 2

tribution will cross these lines at right angles, in the direction of decreasing

potential.

Part b

- dv ava = = + (vV)v

dt at

=V (xi + x i 2 (a)

a = a)\ rfir (b)

where

r = /x 2+ x

2and i is a unit vector in the radial direction.

1 2 r

Part c

This flow could represent a fluid impinging normally on a flat plate, located

along the line

x + x = 0. See sketches on next page.1 2

PROBLEM 12.8

Part a

Given thatx 2 x1

v = i V + Tiv (a)1 oa 2 oa

we have that

- dv ava T + (v-V)v

= ax + ,v v (b)

1 2.




Acceleration x2

V2o

a= (-) ria r

yr

xl





Thus2 - , xi (Vc)

a= v i 0(O T2 1 2 2 2

Part b


p 1 ) (xv + pX2+ (d)

S= o 2 (x2 +2)

V2

1 2r (e)

o 2P 0 a2

where

r= x+x 2

1 2

PROBLEM 12.9

Part a

The addition of a gravitational force will not change the velocity from that

of Problem 12.8. Only the pressure will change. Therefore,

- v vv i x +i -- x (a)

1 a 2 2a 1

Part b

The boundary conditions at the walls are that the normal component of the velocity

must be zero at the walls. Consider first the wall

x - x = 0 (b)

We take the gradient of this expression to find a normal vector to the curve. (Note

that this normal vector does not have unit magnitude.)

n i 1i, (c)

Then v--- ov*n - (x - x) = 0 (d)

a 1 2

Thus, the boundary condition is satisfied along this wall.

Similarly, along the wall

x + x = 0 (e)2 1

n = 12 + i, (f)

and -- ov*n = a (x + x2) = 0 (g)

Thus, the boundary condition is satisfied here. Along the parabolic wall

x2

2 a2

(h)2 1

n = x 2i2 - xi i (Ci)





v -a

0n (x1x

2- xx

1 2) = 0 (j)

Thus, we have shown that along all the walls, the fluid flows purely tangentialto

these walls.

PROBLEM 12.10

Part a

Along the lines x = 0 and y = 0, the normal component of the velocity must be

zero. In terms of the potential, we must then have

= 0ax

x=O

- =0 = 0 (b)y=O

To aid in the sketch of J(x,y), we realize that since at the boundary the velocity

must be purely tangential, the potential lines must come in normal to the walls.

Part b

For the fluid to be irrotational and incompressible, the potential must obey



ELECTROIECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS


Laplace's equaiion

V20 = 0(c)

From our sketch of part (a), and from the boundary conditions, we guess a solution

of the form

V

S= o (x2

y2) (d)a

Vvwhere -

ois a scaling constant. By direct substitution, we see that this solution

satisfies all the conditions.

Part c

For the potential of part (b), the velocity is

=v -V = 2 -a (x ix - yi y (e)

Using Bernoulli's equation, we obtain

V2

p = p + 2 0 (X2+y

2) (f)

The net force on the wall between x=c and x=d is

z=w x=d

f I f (p - p)dxdz i (g)

z=O X=c

where w is the depth of the wall.

Thus v d

= + w x 2dx i

6 y

2V C

= + w (d ' - c')f (h)6 Y

Part d

The acceleration is

V V V V

a = (v'V)v = 2 - x(2 i ) - 2 y (-2 --y).a a x a a y

or

a = 4 (xi + yi)(i)

or in cylindrical coordinates

a = 4(V rir )a = -- ri (j)Va r



ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUTDS


PROBLEM 12.11

Part a

Since the V*v = 0, we must have

V h = v x(x)(h - &)OX

IL h

v (x)= V (1 + )

Part b


1 1 2

SpVo + P = 2 [Vx(x)] + P

P = Po 2 2 pV2o (1 + h

Part c

We linearize P around = 0 to obtain

P Po - pV 2o oh

Thus

T - P + P = pV2oz o oh

Thus Tz = CE (g)

with p VoC =

h

Part d

We can write the equations of motoion of the membrane as

a2 = Sa2 + T (h)m 3t

2ax2 z

= s 2 + CE (i)

We assume

E(x,t) = Re 'j

e (wt-kx) (j)

Solving for the dispersion relation, we obtain

- m02 = - Sk2 + C (k)

or 2

W - k2 - T]

= m M

Now, since the membrane is fixed at x = 0 and x = L, we know that

nik = n = 1,2,3, ..... (m)

Now if

S( ) - C < 0 (n)

we realize that the membrane will become unstable.

So for

PV2 2<S () (0)

we have stability.

Part e

As ý increases, the velocity of the flow above the membrane increases, since

the fluid is incompressible. Through Bernoulli's law, the pressure on the membrane

must decrease, thereby increasing the net upwards force on the membrane, which

tends to make E increase even further, thus making the membrane become unstable.

PROBLEM 12.12

Part a

We wish to write the equation of motion for the membrane.

325 a 25 Te (a)= S + p ()-po + T - mg (a)m 0 " 3t 2 ax 1

S V 2 E Vwhere o o o o

Te = --2 d- -) % 2 dd(1 + ý-)

is the electric force per unit area on the membrane.

In the equilibrium C(x,t) = 0, we must have

E V2

-p (0) = Po - (d) + mg (b)

As in example 12.1.3

P = -pgy+

C

and, using the boundary condition of (b), we obtain

Eo Vo 2

p1 = - pgy+

mg + Po - (d) (c)

Part b

We are interested in calculating the perturbations in p, for small deflections of

the membrane. From Bernoulli's law, a constant of motion of the fluid is D, where

D equals

E V 2

D = U2+ mg -- 0 (d)

For small perturbations ix,t), the velocity in the region 0 < x< L is

Ud

d+E

We use Bernoulli's law to write

21 2+

P( +

g = D (e)

Since we have already taken care of the equilibrium terms, we are interested only in

small changes of pi, so we omit constant terms in our linearization of pi.

Thus () = - pg +

U

(f)

Thus, our linearized force equation is2

a -= S + 2- pg EV

m)t2

a2 d pg + E Cg)

We define

2 VO2

C =- pg + + d--d d

3

and assume solutions .of the form

E(x,t) = Re ej(w t - k

x)

ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUIDS


from which we obtain the dispersion relation

2= a (h)

Since the membrane is fixed at x=0O and at x=L

k . n = 1,2, 3, ..... (i)

If C <0, then w is always real, and we can have oscillation about the equilibrium.2

For C > S( ) , then w will be imaginary, and the system is unstable.

Part c

The dispersion relation is thus

Cs2 2S

Sf mm

Consider first C < 0

for

1 W

C>0

lex w for

real k

_· ~I~~ _I__·__

Part d

Since the membrane is not moving, one wave propagates upstream and the other

propagates downstream. Thus, to find the solution we need two boundary conditions,

one upstream and one downstream. If, however, both waves had propagated down

stream, then causality does not allow us to apply a downstream boundary condition.

This is not the case here.

PROBLEM 12.13

Part a

Since V-v = 0, in the region 0< x< L,

vV

od V (ý1 - 2)]

01

x d+~l- 2 o d (a)

where d is the spacing between membranes. Using Bernoulli's law, we can find the

pressure p right below membrane 1, and pressure p2 right above membrane 2.

Thus

1 2 1 2SpVo + p 2 p + p (b)

and

1 1 2SpV + p pv + p (c)

Thus

0 2p = p2 Po

+d (d)

We may now write the equations of motion of the membranes as

a2 1 a2 a 2 1 V & )

am S (pS1 = x+ 1 C(e)t2 + - po) S d (e)

a2 2 2 22 2 _pVO(- 2)

m 2t2 = SX2 + P.- P2 = S df)

Assume solutions of the form

S j(wt-kx)1 = Re l eiwt-kx) (g)

2 = Re 52 e

Substitution of these assumed solutions into our equations of motion will yield the

dispersion relation

V2

= -Sk + (- 2)

SV2

(h)

S-Sk2c + PV-m 2 = - Sk2 d 2 1

These equations may be rewritten as





[ 2 + Sk2 + L = 0

Vm dj 2 V 2d o

0O a

2

+ Sk2

- ] = 0d 2[ m d

For non-trivial solution, the determinant of coefficients of 5 and 5 must be1 2

zero. 2 2

V2Thus aO2 + Sk2

V2Od) dOQ

or pV2

pV2

-2

+2

o- + (k)w Skm d d

If we take the upper sign (+) on the right-hand side of the above equation, we obtain

2p V 2

!/2S= S k2

-2Vd] )

We see that if V is large enough, w can be imaginary. This can happen when

V2 Sk2d (m)o 2p

Since the membranes are fixed at x=O and x=L

k = n = 1,2,3, ..... (n)

So the membranes first become unstable when2

S( -) dV 2 L (0)o 2P

For this choice of sign (+), = - 2 w If we hado we excite the odd mode.

taken the negative sign, then the even mode would be excited

E1 = E2

However, the dispersion relation is then

wmand then we would have no instability.

Part b

The odd mode is unstable.

. •--

---- ior

PROBLEM 12.14

Part a

The force equation in the y direction is

3y = - pg (a))y

Thus

p = - Pg(y-0) (b)

where we have used the fact that at y= 5, the pressure is zero.

Part b

V*v = 0 implies

av 3v

axx +-

3y0 (c)

Integrating with respect to y, we obtainav

v x y+ C (d)

y ax

where C is a constant of integration to be evaluated by the boundary condition at

y = -a, that

v (y= -a) = 0

since we have a rigid bottom at y = -a.

Thus 3v

v x (y+a) (e)

Part cy ax

The x-component of the force equation is

avx 3 a = (f)

at x 3-or

avx x (g)

at =-gx

Part d

At y = 6,

v (h)y at

Thus, from part (b), at y =

3v= - x (C+a) (i)

at 3xHowever, since C << a, and v and v are small perturbation quantities, we can

x yapproximately write

ava ax (j)at ax

Part e

Our equations of motion are now





av

. -=a xat ax

and

3vx __

S- g xf

If we take 3/ax of (k) and 3/at of (R) and then simplify, we obtain

32V

x x--- ag

We recognize this as the wave equation for gravity waves, with phase velocity

V /agP

PROBLEM 12.15

Part a

As shown in Fig. 12P.15b, the H field is in the - i direction with magnitude:

II, I 0o

Insl = 2N of integration (a)for the MST

If we integrate the MST along the surface defined in the above figure, the only

contribution will be along surface (1), so we obtain for the normal traction

20o0n=

12 o s72

= -771

Part b

Since the net force on the interface must be zero, we must have

Tn Pint o = 0

where pint is the hydrostatic pressure on the fluid side of the interface.





Thus 1 olo2

Pint 0o+ 8T (d)

Within the fluid, the pressure p must obey the relation

= - pg (e)

orp = - pgz + C (f)

Let us look at the point z = z , r= R . There0O 02

1 Voo 0p = - pgzo + C = P

+(g)

0

Therefore 2

C = pgzo + p+

- -~h)

o

Now let's look at any point on the interface with coordinates zs, rs

Then, by Bernoulli's law,

1 1olo2

1 oo2

P + 1 PR + pgzo - 8 + Po +pgzs (i)

o s

Thus, the equation of the surface is

1 olo2 1 02

pgzs+ i - = pgzo + (J)

s 0

Part c

The total volume of the fluid is

V = [Ro

2

- ( )2 ]a . (k)

We can find the value of z0 by finding the volume of the deformed fluid in terms of

z0, and then equating this volume to V. 2

R o oo (+ 10Thus

o02!r

V = r[R - 1)2 a = 2 rdrdz (a)

r=r 0 z=O

where

r is that value of r when z = o, or

o2 1

r (m

pgz + oo0

Evaluating this integral, and equating to V, will determine zo0

ELECTRORECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS

PROBLEM 12.16

We do an analysis similar to that of Sec. 12.2.1a, to obtain

E = - i (a)yw

and

J = i (-V

+ vB) = - (b)y wd y

Here V = IR + V (c)o

Thus vBw - VO

I =w

R +Zda

The electric power out is

P = VI= (IR + Vo)I

R(vBw - V) v Bw - Vo

(e) V + ....

From equations (12.2.23 - 12.2.25)

we have

MAp = p(0) - p(a) = B

Thus, the mechanical power in is

Bw(vBw-V )vP = (Apwd)v = (g)

Mi wR +

ida

Plots of PE and PM versus v specify the operating regions of the MHD machine.

P>e 0 e<

0P e < 0 P > 0e e e

PM> M

>0 PM < 0 >0 SPM 0

Generator Brake Pump Generator

http://slidepdf.com/reader/full/12.2.1a




PROBLEM 12.17

Part a

The mechanical power input is

PM L Jdf Vpv 0dxdydz (a)

z=O y=0 x=O

The force equation in the steady state is

- Vp + fe = 0 (b)

where

fe = - J B (c)yo

Thus YBPM= Lf I J dxdydz (d),

z=0 y=O x=O

NowJ =a(E + vB) (- + vB ) (e)y y 0oo y oo

Integrating, we obtain

P= ov2B Lwd - oB v VLd

P 0 o oo (f)

V 2 VVoc oc 1 V)V

Ri R R. (Voc oc

Part bPout

Defining = outP

we have

(Voc

- V)V - aV2(g)

S(Voc - V)Voc

First, we wish to find what terminal voltage maximizes Pout. We take

Pout

= 0 and find that

VVoc

V = oc maximizes P2(l+a) out

For this value of V, n equals

1 111 (h)2 (1+2a)

1Plottlng n vs. -- gives

a

.5

a

1

PROBLEM 12.17(Continued)

Now, we wish to find what voltage will give maximum efficiency, so we take

-- = 0

Solving for the maximum, we obtain

V = Voc I +ala (i)

We choose the negative sign, since V< Voc for generator operation. We thus obtain

r = + 2a - 2 a(l+a) (j)

Plotting n vs. a , we obtaina 9

2 3 4 5 6 7 8 910

PROBLEM 12.18

From Fig. 12P.18, we have

E - iw y

andV I -

J= i [ + vB] = -- iy w LD y

The z component of the force equation is

3z LD

IB vor Ap = Pi - P= D = p(l )

Solving for v, we obtain

IBv = D(1 )v





Thus, we have

I V IB

LDa w+ B(1

DApo)vo

(f)

or 2VWV =I w + - v Bw (g)

LD( DAp

Thus, for our equivalent circuit

w vowB2

t O

i LDo DApoand

Voc

= - vo

wB (i)

We notice that the current I in Fig. 12P.18b is not consistent with that of Fig.

12P.18a. It should be defined flowing in the other direction.

PROBLEM 12.19

Using Ampere's law

NI + NI

H = (a)o d

Within the fluid

IL VLJ = =

Zd(-

w+ v H

00)i

Z(b)

Simplifying, we obtain

Si v NiNVS=

Ova NI0o L (c)

L [d d d w

For VL to be independent of IL, we must have

VpoNL 1

dd

9d(d)

or 1N (e)L Zavy

0

PROBLEM 12.20

We define coordinate systems as shown below.

M HD #2. M HD # 1



ELECTROMILCHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS


Now, since V*v = 0, we have

vwd = vwd

1 1 1 2 2 2In system (2),

I VJ2

=-

+v B)i (a)

2 y2 2d2 W2 2 y2

andI2B

AP2 = p(0) - p(_) d (b)

In system (1),

I V

= i --- ( -- v B) (c)1 y 1 id w

and IB

Ap = p(0

+) - p(Y' ) = d (d)

By applying Bernoulli's law at the points x = 0 (right before'IdD system 1) and at

xt = + (right after MHD system 1), we obtain

1 p2 + p (0) = I v2 + p(L (e)

or p(0_) = p (+) (f)

Similarly on MHD system (2):

p 2 (0_) = P2 (2+) (g)

Now,Now Vp'd = 0

CApplying this relation to a closed contour which follows the shape of the channel,

we obtain x 0 x = x = 02 - 2 2- 1 -

Vp-dZ V p-dZ + VpdZ + Vpdt + Vp-dZ

C x =O x =£ x =O x =1 + 1 1+ 2 + 2 2+

p (z1 )- p (0+) + p2(O_)- p1(+ ) + p2 (2-

- P2(O+) + 1 (0) - p2( 2+) (h)

From (f) and (g) we reduce this to

bp +pA = 0 (i)

orI -I112 (1)

d d1 2





Thus, we may express v asI

S(+ - + (k)

We substitute this into our original equation for J2

I V wd I

ad w wd £do w22 2 2 12

(a), to obtain

()

This may be rewritten as

w wdV = - 1 1

2 2 a d wkd d2* 2 2 12

The Thevenin equivalent circuit is:

d

11d

2

2 Req

Voc

+

1V2

v 2

where

and

Voc

1=-

dV1

Req

2jd2 2

_l

2d2 Z

__

J

PROBLEM 12.21

For the MHD system

II -V

a1--V H) (a)

and

ap

Now, since

= P1 - P 2 = +

IpoH

(b)

IVp'dk = 0 (c)

Cwe must have

Ap = kv = poH LO(oo D

- VoHoo

) (d)

Solving for v, we obtain

PO H LaV

v = D° V0D[k+(poHi )

2Lo]

00

(e)

PROBLEM 12.22

Part a

We assume that the fluid flows in the +x direction with -velocity v.

ThusS I V +J•= i o( VoHo) i (a)

3 d 00 3

where I is defined as flowing out of the positive terminal of the voltage source V .

We write the x component of the force equation as

p L pIpoHo-ax L w pg = 0 .(b)

-Thus

p= -Lw

+ pg x

For Ap = p(O) - p(L) = 0

Then ILIoVHo

Lw

For the external circuit shown,

V -IR + V0

Solving for I we get

V

T v+ ooH - pgLw0

+ R oPHOLw d

Solving for the velocity, v, we get

gL• + R o

H 0 d d

~oo

For v > 0, then

V 0, then we are supplying electrical power to the fluid. From

part (a), (f) and (h), Vo is always negative, but so is I. So the product

V I is positive.




PROBLEM 12.23

Since the electrodes are short-circuited,

J iz-Zd

= OvBoi

z(a)

In the upper reservoir

pi=

Po + pg(hI - y) (b)

while in the lower reservoir

P2 = Po + pg(h 2 - y) (c)

The pressure drop within the MHD system is

IBAp = p(O) - p(R) = - (d)

Integrating along the closed contour from y=h through the duct to y=h , and1 2

then back to y=h we obtain

Vp-d = 0 = - pg(h - h )+ - (e)1 2 d

Thus

Ipg(h

1 - h

2)d

(f)BBR

and so and so pg(h

I 1- h

2)

(g)atdB o9 B

o o

PROBLEM 12.24

Part a

We define the velocity vh as the velocity of the fluid at the top interface,

where

dh

vh - dt (a)

Since V-v = 0, e have

vA = v wD (b )h e

where ve is the velocity of flow through the MHD generator (assumed constant). We

assume that accelerations of the fluid are negligible. When we obtain the solution,

we mustcheck that these approximations are reasonable. With these approximations,

the pressure in the storage tank is

p = - pg(y-h) + Po (c)w:here po is the atmospheric pressure and y the vertical coordinate. The pressure

drop in the MHD generator is

Ap = D (d)

where I is defined positive flowing from right to left within the generator in the

end view of Fig. 12P.24.

We have also assumed that within the generator, ve does not vary with position.

The current within the generator is

I IRy= L(- + v P H ) (e)

LD w eoo0

Solving for I, we obtain

Hove po

1= 1+- (f)

1 R-L

aLoD w

Now, since Vp'dZ = 0, we have

Ap - pgh = 0 (g)

Thus, using (d), (f) and (g), we obtain

- pgh + Do 1 v 0 (h)

Using (b), we finally obtain

dh + sh = 0 (i)dt

wherepg w D RD 1

h= 10 e , until time T, when the valve closes ()

Numerically at h = 5.

s = 7.1 x 10- 3

, thus T u 100 seconds.

For our approximations to be valid, we must have

vh << pg (k)

or

s2h <<g.

Also, we must have

2 h2 << I pgh

or

1 s2h <<g

(a)

Our other approximation was

pL < (m)

which implies from (f) that

(o Ho)2psL << 0o0 (n)

L + i

Substituting numerical values, we see that our approximations are all reasonable.

Part b

From (b) and (f)

pH AI-

= oo h

wd ]D t

- 650 x 10 e- s

t amperes.

until t = 100 seconds, where I = -325 x 103 amperes. Once the valve is closed,

I = 0.




PROBLEM 12.25

Part a

Within the MHD system

-1i VJ = = - (- iHo)iH where V = -iR + V

LD 3 w 00 3 0

and Ap = p(O) - p(-L ) = D''D

We are considering static conditions (v=0) so the pressure in tank 1 is

p 1 = -pg(x2 - h ) + po

and in tank 2 is

p = - pg(x - h ) + Po

2 the atmospheric2isressure,

where po is the atmospheric pressure,

thus Vi = o (e)

1 Rw[ +

aL D w

Now since jVp.dk = 0, we must have

C io H

+ pgh + - pgh2 = 0 (f)

Solving in terms of V we obtain

0pg(h - h)wD 21 ar

oo 1

For h = .5and

h = .4and substituting for the given values of the parameters,

2 1

we obtain

V = 6.3 millivolts

Under these static conditions, the current delivered is

pg(h 2 -hl)Di = = 210 amperes

o o

and the power

delivered is 2

eP= V = Pg(h2

Hh1

w 1

+R]

1.33 wattse o L H wL D w

Part b

We expand h and h2 around their equilibrium values h10 and h20 to obtain

h = h +Ah1 10 1

h 2 2h 2 0 + Ah

2

20 a





Since the total volume of the fluid remains constant

Ah = - Ah2 1

Since we are neglecting the acceleration in the storage tanks, we may still write

PI - pg(x - h ) + po1 2 1

(h)=

P2 - - +POg(x2h2)

Within the MHD section, the force equation is

av i1oHoS- = - VpMD + L(Dat V MHD + LID (i)

Integrating with respect to x , we obtain

APMHD = p(O) - p(-L) = LD -o pL l()

The pressure drop over the rest of the pipe is

dv

APpipe = 2 dt

Again, since Vp-dk = 0, we have

C

pg(h - h2 ) + pMHD + pipe = 0 (k)

For t > 0 we have

2V

w V 0oHo1 R ()

cL D w

and substituting into the above equation, we obtain

ýv w V11 0H 1ol)t0pg(h- h) - p(L+ý) ~ 1 o o 0 (m)

[L 1D + ;i-

We desire an equation just in Ah . From the V-v = 0, we obtain2

dAh2

vwD = A (n)

Mlaking these substitutions, the resultant equation of motion is

d2Ah2 (o Ho )2 dAh 2gwdAh 2dt2 1 R dt (L + L2)A

V •pH (o)

p(L +L2 )A -•D +

58





Solving, we obtain

V0

Hst s 2 t

Ah = + B e + B e (p)

2 2pgwd D+ R 1 2

where B and B are arbitrary constants to be determined by initial conditions1 2

and

[(H2] H 2

S - o gwd (q)

2 ' L)) +pL L +L DD11 R (L + L

2)A

1 2 aL D w 21 ID w

Substituting values, we obtain approximately

-1

s = - .025 sec.1

s = - .94 sec.2

The initial conditions are

Ah (t=O) = 0

and dAh (t=0)2 = 0

dt

Thus, solving for B and B 2 we have

B = = - .051 (r)

1 2pgwD[ 11D w1 1 R 2

- Voo H 3

B 000 (R = + 1.36 x 102pgwD[ +---l 1 _

LID w sI)

Thus

3 .94t -025th (t) = h + Ah (t) = .55 + 1.36 x 10 e - .051e (s)2 20 2

From (£) we have2V

o

w V oHR 1 (t)R 1w +aL D

Substituting numerical values, we obtain

st s2

i = 420 - 2.08 x 10s

(Bzsze + B2 s 2e )

= 420 - 268 (e-*025t -e94t) (u)





11h_t)2

i1 4 5 50

It

100 150 200

420

i(t)

- ----

210 -

1

1

2

I

3

I

4

I

5

LI/

- --

50

I

100

1

--

150

1

I-.---------

200

I

250

1

L

Our approximations were made in (h) and (k). For them.to be valid, the following

relations must hold:

32Ah2

<< 1

gh2

and

+ (v*V)vds ý-t L

transition

region

Substituting values, we find the first ratio to be about .001, so there our approx

imation is good to about .1%. In the second approximation

/A- .3L Pd 2 .15

2

Here, our approximation is good only to about 15%, which provides us with an idea of

the error inherent in the approximation.

PROBLEM 12.26

Part a

We use the same coordinate system as defined in Fig. 12P.25. The magnetic field

through the pump is

Nipo -B = i (a)

d 2

We integrate Newton's law across the length £ to obtain

av Apo ip = p(O) - p(X) + JBZ = - v + B (b)

3t v d

Ap v + Np i2

v d2

Thus 0

Ap Nji Nil+ v

212 sin

2t 2 1

2(1 - cos 2 wt) (c)

Tt pzv° d Pz 2d Pt

Solving, we obtain

Npo12 ot \P ovcos 2wt + 2w sin 2w()

v= 0 0Ap Ap 2 (d)t o A o + 4 W2

Part b \pvo

The ratio R of ac to dc velocity components is:

R = /v (e)

[ Po + 4W2 /2




PROBLEM 12.27

Part a

The magnetic field in generator (1) is upward, with magnitude

Ni po Nmip oB =

I a a

and in generator (2) upward with magnitude

Nilp o Ni2 o

B m=io +2 a a

We define the voltages VI and V2 across the terminals of the generators.

Applying Kirchoff's voltage law around the loops of wire with currents i and i1 2

we have

d dXV +N + Nm d t + i RL 01 dt

and dX2dX

1

V + N - Nm j + i2RL= 02 dt

where

A = B wb1 B,1b

A = B wb2 2

From conservation of current we have

i, V I + VB I1 + VB2

abo wand

i2

V 2

- = -- + VBaba w 2

Combining these relations, we obtainwbp di w W N] pw

(N + N)+ i w+ + VN im a dt ab a a m 2

wbp di 2

(N2 + N ) di 2 + i +

w N wVi = 0m a dt ab a a 1

Part b

We combine these two first-order differential equations to obtain one second-

order equation.

di di2

a1 +a.) ,1,-

+ai. ',

= 0UC L UC ~ L

whe re

(N2+ N2)Wbo] 2a = WmVJ a

1 wN Vp





/NNV 2 2)a = 2 ýb-w + R- 0 + N )

2abo tN a m V

VN oiwa =

3 a

If we assume solutions of the form

i = Aest (9)2

Then we must have

as2

+ aas + a = 0 (m)1 2 3

- a Ia2 4a a2 2 1 3

2 a

For the generators to be stable, the real part of s must be negative.

Thus

a > 0 for stability2

which implies the condition for stability is

Sw NV

Part c ab+ > a (n)

When a = 02

woNv

+ =Sab (o)aba L a

then s is purely imaginary, so the system will operate in the sinusoidal steady state.

Then -aa

N Vm

The length b necessary for sinusoidal operation is

b= aakVWoN~ (q)

Substituting values, we obtain

b = 4 meters.

Part d

Thus, the frequency of operation is

w4000 = 500 rad/sec.8

or f v" 80 Hz.




PROBLEM 12.28

Part a

The magnetic field within the generator isN i

poB= -- i

w 2

The current through the generator is

J i (+

VB)TSolvingor the we obtainhe v,oltage across channel

Solving for v, the voltage across the channel, we obtain

D VJ°Nw v Gx w

D i

We apply Faraday's law around the electrical circuit to obtain

1pN 2

1 dijiv + L id t + iR = - d

-

J w dt

Differentiating and simplifying this equation we finally obtain

A2 /2Rw n _pNDV \

Ai+2 +2 LT.. .. d ,.+ ?-TOA(C i =dt2 \ N' 'd LJ N 0

stWe assume that i = Re I es

Substituting this assumed solution back into the differential equation, we obtain

Sw D oNDV- ) + w

s + + = 0oN d oLw w PoNz£dC

Solving, we have

iRLw D- NDV' D o NDV )2

S+N Z-OLw w

wS

2 4 SN LdCo

For the device to be a pure ac generator, we must have that s is purely imaginary, or

S(ONDV D) N2d

L = •w YLw w(h)

Part b

The frequency of operation is then

wpN £dC

PROBLEM 12.29

Part a

The current within the MHD generator is

YiVJ = i 0(-+vB= )1

Ld y w o y





where V is the voltage across the channel. The pressure drop along the channel is

iB0 + av (b)

Ap = Pi - Po p at

where we assume that v does not vary with distance along the channel. With the switch

open, we apply Faraday's law around the circuit, for which we obtain

V + 2iR = 0 (c)

Since the pressure drop is maintained constant, we solve for v to obtain

2aR + _ dt !v + OvB = + 2R Ap (d)w Ud B at o d w BIn the steadytate

In the steady state

+ 2R= 1 -- Apaid w B

andd

i = Ap

Part b

For t > 0, the differential equation for v is

(F + Bo d 2- + avB = S + R d ApO

The general solution for v is

V= - + d +o-t/T

where T = YR +2) d

We evaluate A by realizing that at t = 0, the velocity must be continuous.

Therefore

1 R d Rd - t/Tv = + ap + - -+- Ap eid w B w B

Ap (I + pY" R d -t/) dT w B Bo

-,Ap (1 + e d= / BR

A~kl wi.+!1d

PROBLEM 12.30

Part a

The magnetic field in the generator is

J Ni=

Bd

The current within the generator is

S 0 ( + vB)=d w



2



where V is the voltage across the channel. The pressure drop in the channel is

Ap = Pi - Po= Ap (1 -)

= -- (c)

oApplying Faraday's law around the external circuit, we obtain

d(NB~w) Lw N diV + i(RL + RC) = dt o d (d)

Using (a), (b), (c) and (d), the differential equation for i is then

LN2 dUN] (PoN 2

oN 2

di RL+ RC 1 N

d dt w•d d o d AP o

In the steady state, we have

FRL+ RC 1 oNvo12 w

+O- d dAPo

i2 = (f)

iN2 V

The power dissipated in RL is

P = i2RL

For P = 1.5 x 10 , then

i2

= .6 xl0

(amperes)2

Substituting in values for the parameters in (f), we obtain

.6 x 08 = (.125 + 2.5 10-6 N - 6.3 x 10 -

N)40 10g)

-N2 (4 x 10 )

Rearranging (g), we obtain

N2

- 102N + 2.04 x 103 = 0

or N = 75, 27

The most efficient solution is that one which dissipates the least power in the coil's

resistance. Thus, we choose

N = 27

Part b

Substituting numerical values into (e), using N = 27, we obtain

(1.27 x 107) di (6 x 107)i + 13 = 0 (h)

dt

or, rewriting, we have

dt di

1.27 xlO' = i(6 x107

- ) ()


9.4t + C = log 6 x10-i ()

We evaluate the arbitrary constant C by realizing that at t=O, i = 10 amps

66





Thus CThus - 13.3

We take the anti-log of both sides of (j), and solve for i2

to obtain

6 x 10 7

12 1+(13.3 -9.4t) (k)

7.75 X 10

5.5 ,10 3

- -4

10

I '25 secondst

Part c

For N = 27, in the steady state, we use (f) to write

RL+ RC 1 PNvoN

P = i2RL = +w cd dj dAPoRL

2N) or

P = a - 2 2

where dA R+ 1 1oNvo

a = - d 1.47 x 108

N VOand

dAPo 1

2 2 2.85 x 1O





P

1.5 X106

RL

PROBLEM 12.31

Part a

With the switch open, the current through the generator is

= 0 =_ d

Iy

= ( - + vB ) (a)w B y

where V is the voltage across the channel. In the steady state, the pressure drop

in the channel is

Ap Pi -oiBB = 0 = APo(l -

v) (b)

Thus, v = vo and the voltage across the channel is

V = v B w. (c)00

Part b

With the switch closed, applying Faraday's law around the circuit we obtain

V = i RL (d)

Thus

i = oRLd - i + avB (e)2d w o

and iB t v=

Ap = d+- p t Ap (1 ) (f)

Obtaining an equation in v, we have

v aPo GBt vw--o 7+ AP (g)

T w





Solving for v we obtain

-t/'T _AP

v = Aev t/T + _o where Ri

w

(APo + Bow atd

vo RL+ Ri)

and where

Ap wB

'o RL + R

at t = 0, the velocity must be continuous. Therefore,

APo=

A v. 0o Ap w B

o+\vo RL+ Ri

Now, the current is

wBo vi=

RL + R i

Thus

1= (wB p APo B(1 e) )+e -v-t/TRL o wB 0i

0 + 0

v RL +

V0

V

Vo ""o1Ap TR.+L)

Vo o

w

!+R.




PROBLEM 12.32

The current in the generator is

i Vi ( Y- (a) vB)

zd w

where we assume that the B field is up and that the fluid flows counter-clockwise.

We integrate Newton's law around the channel to obtain

Iv = ipt - JB B (b)

4t d

or, using (a),

3V w 3i B2

3 = d +- i (c)d= Ttt a dpk

I

Integrating, we have00

w + B 2wV = -+ -+ I idt (d)

dia dp2

wDefining R. = i

1 atzd

and

C= p2di wB

2

we rewrite (d) as

V = iRi + i-fidt (e)

The equivalent circuit implied by (e) is

R.

+ v

Ci

PROBLEM 12.33

Part a

We assume that the capacitor is initially uncharged when the switch is closed

at t = 0. The current through the capacitor is

dV V

i = C d=

ud - w+

vB (a)dt w 0 0

or

dV advc + o'd oB

V o (b)dt wC C C





The solution for VC is

-t/VC = vBw(l - e ) (c)

with wCT = - , where we have used the initial condition that at t = 0, the volatd

tage cannot change instantaneously across the capacitor. The energy stored as

t + m, is

2 C (VW

Part b

The pressure drop along the fluid is

iBo 2 -t/TAp = - = B2v G£e

d 0oo (e)

The total energy supplied by the fluid source is

Wf = Ap vodwdt

-=I(v B )2 awde-t/T

dt (f)

- UT(v B )2 Twdet/T I00

Wf : C(wv B )2 (g)

Part c

We see that the energy supplied by the fluid source is twice that stored in the

capacitor. The rest of the energy has been dissipated by the conducting fluid. This

dissipated energy is

wd J VC idt (h)

0

= + (v0B )2w( - e-t/T)rdet/T dt

Oidw(VBB)2

-t/T+ 1 e-2t/T

= idw(v BBo)2

= aTdw(v B )2 (i)0o 2

ThereforeW = 1 C(voBow)

2 (j)

d 2 oo

Thus

fluid elec dissipated (k)

As we would expect from conservation of energy.




PROBLEM 12.34

The current through the generator is

i = c( - - vB ) (a)Zd w o

Since the fluid is incompressible, and the channel has constant cross-sectional area,

the velocity of the fluid does not change with position. Thus, we write Newton's law

as in Eq. (12.2.41) as

avp - V(p+U) + J x B (b)

where U is the potential energy due to gravity. We integrate this expression along

the length of the tube to obtain

iB• = - - pg(x a + x b ) (c)

Realizing that xa = xb

and dxa (d)v d

dt

We finally obtain

d2X aB

22 dx cB v k

dta+ 0

pt dta

2xa

=wp

o2

(e)

We assume the transient solution to be of the form

^ stx = x e (f)

a

Substituting into the differential equation, we obtain

aB2o. s

s2+ o 2g = 0 (g)

Solving for s, we obtain

2o=Bp /B2 2

2-9, t-2 -22. (h)

Substituting the given numerical values, we obtain

s= - 29.4

s2 = -.665 (i)

In the steady state

aB V2

x

O 1

1 .075 meters (j)a wp 2g

Thus the general solution is of the form

x = .075 + A est + A eS2t (k)

a 1 2

where the initial conditions to solve for A and A2 are





x a(t=0) = .0

a

d~-(t=0) = 0.075 s .075 s

Thus, A = 1 _ 2A2 1 .0765 cad A = - = .00174

Thus, we have:

x = .075 + .00174e-29.

4t -.0765e

-.665t

aXa

1 2 3

Now the current is

V dxi = Z do( V - B d-)

0 ft)tt t= 91dY[ - Bo (s I Al esl + s 2 A2 eS2)] (m)

= 100 - 2 x 103(sI AeSlt + 2A 2e 2t) amperes

= 100(1 + e - e )

Sketching, we have

i

100

t

1 2 3




PROBLEM 12.35

The currents I and I are determined by the resistance of the fluid between1 2

the electrodes. Thus

I = VooDx (a)

I wand V ODy

I (b)2 w

The magnetic field produced by the circuit is

- poN-B= - (I2 - I )i (c)

Nor - o

= z VoD(y- x)i2 (d)

From conservation of mass,

y = (L - x) (e)

Thus -

N

Vo oDB 2 (L- 2x)1 (f)

The momentum equation is

avat = -V(p+U) + J x B (g)

Integrating the equation along the conduit's length, we obtain

p - (2L + 2a) = -pg(y-x) - J BL (h)at o

Now

v =ax (i)at

so we write:

(oNV GDL

2p(L + a) -!-T + Pg + J (2x - )= 0 (j)


^ st Lx = Re x e + (k)

Thus2 NVo D

S2 + -s- + J L 0 (C )(L + a) pw (L + a) o

Defining

NV - L22 + + OOg ODJ (mO

o (L +a) pw (L+a)

we have our solution in the form

L= x A sin w t + B cos w t + (n)

o o 2Applying the initial conditions

x(O) = L and dx(O)

de = 0 (0)

we obtain x = 2 (1 + cos wot) (p)




PROBLEM 12.36

As from Eqs. (12.2.88 - 12.2.91), we assume that

v = iev0

B= B iz + TOB 0 (a)JI= iT J +TJ

rr z z

E= TE + T Err z z

As derived in Sec. 12.2.3, Eq. (12.2.102), we know that the equation governing Alfven

waves is

aZv2 B 2 a2v0 .p0 z_- (b)

For our problem, the boundary conditions are:

at z = 0 E = 0r (c)

at z = Z v = Re[Orej

As in section 12.2.3, we assume

v = Re[A(r)v 0 (z)ejet] (d)

Thus, the pertinent differential equation reduces to

dvV0 2,

dz-V-+ kv = 0 (e)

where k = W0

The solution is

v 0 = C1 cos kz + c2 sin kz (f)

Imposing the boundary condition at z = 2, we obtain

A(r)[C cos kZ + C2 sin k ] = r (g)

We let

A(r) (h)

and thus

R = C1 cos kZ + C2 sin kZ (i)

Now E = - VB 0 (j)

Thus, applying the second boundary condition, we obtain

v (z=0) = 0

or C = 0 (k)

Thus C OR2 sin kZ

Now, using the relations

Er = - vOB (m)r 0 o





E = 0 (n)

aE aE 3Br z = (o)

az ar at1 Be = JPo az r (p)

1 3(rB6 )r = J (q)o ar z

we obtain

v = Re sin kz ej

t] (r)Ssin k.

B = Re a k il nrB cos kz ej

(s)

r =ejRel ?rBo

sink 2

k2kt sin kz ejj]

(t)

= F 2 2Bo kk cos kz e j et (u) Re

z 1j wsin k C

PROBLEM 12.37

Part a

We perform a similar analysis as in section 12.2.3, Eqs. (12.2.84 - 12.2.88).

From Maxwell's equation

VxE=

(a)t

which yields

aE

yz B (b)az at C

Now, since the fluid is perfectly conducting,

E' = E + vx B = 0 (c)

or E = vB (d )y xo

Substituting, we obtain

yv 3BBo Bz tXx

(e)

The x component of the force equation is

av aTX xz

P at = a z (f)

where B

0





Thus Thus B JB

P atX O

ax (h)

Bz0

Eliminating Bx and solving for vx, we obtain

92 B2 a2v

x o x (i)

or eliminating and solving for Hx, we have

32i B2 2 Hx o (j)

where

B = pH (k)

Part b


vx(-k,t) = Re Vejwt (0)

E (0,t) = 0 + vx(O,t)=

0 (m)

We write the solution in the form

v = A ej(wt-k z) + B ej(wt+kz) (n)

x

where

k"o


Now

xS(,t) =e

Iin k

sin kZejt (o)

or

x x(p)o az at

- BoVk cos kz

sin k2 =J•o Hx (q )

Thus

Hx

=Re

= Re

B Vk cos

jw0po°in

kz

k

t

ejet (r)

Part c

Thus

From Maxwell's equations

3HxThusxH= i = J (s)

Jjw[BBVk2

Resin kzsin R Rt

e (t)





Since V*J = 0, the current must have a return path, so the walls in the x-z plane

must be perfectly conducting.

Even though the fluid has no viscosity, since it is perfectly conducting, it

interacts with the magnetic field such that for any motion of the fluid, currents are

induced such that the magnetic force tends to restore the fluid to its original

position. This shearing motion sets the neighboring fluid elements into motion,

whereupon this process continues throughout the fluid.



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--

ELECTRO:ECHANTCS OF COMPRESSIBLE, INVISCID FLUIDS

PROBLE4 13.1

In static equilibrium, we have

-Vp - pgi, = 0

Since p = pRT, (a) may be rewritten as

RT - + pg = 0dx

Solving, we obtain-L xRT 1

p = p ea

PROBLEM 13.2

Since the pressure is a constant, Eq. (13.2.25) reduces to

Ov dv = - JBdz y

where we use the coordinate system defined in Fig. 13P.4. Now, from

we obtain

Jy = (Ey + vB)

If the loading factor K, defined by Eq. (13.2.32) is constant, then

- KvB = + E

Thus, J = avB(1-K)y

Then pv dv _- avB2(1-K)

or dv B AiSPdv =

_ GB2(1-K) =- o(I-K)

jdz A(z)

From conservation of mass, Eq. (13.2.24), we have

PiviAi = pA(z)v

Thus

PiviAi dv 2

dz = -a(l-K)Bi Ai


- a(l-K)Bi9n v = z + C

v

a

v= vie dOi viPi Vi

whereZd a(l-K)B

and we evaluate the arbitrary constant b

v = v. at z = 0.1

(a)

Eq. (13.2.21)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

y realizing that



ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS

PROBLEM 13.3

Part a

We assume T, Bo, w, a, cp and c are constant. Since the electrodes are short-

circuited, E = 0, and so

J = vB (a)y o,

We use the coordinate system defined in Fig. 13P.4. Applying conservation of energy,

Eq. (13.2.26), we have

v v2) = 0, where we have set h = constant. (b)

Thus, v is a constant, v = vi. Conservation of momentum, Eq. (13.2.25), implies

- ViBo

(c)dz io

Thus, p = - viBz + p (d)

The mechanical equation of state, Eq. (13.1.10) then impliesv.B2z +i V B

2z

RT - RT = i RT (e)

From conservation of mass, we then obtain

Piviwdi = (- + vi wd(z) (f)

Thus

(pld

d(z) = (g)viBoz

Part b

Then v.B 2 z

p(z) = Pi 1RT (h)i RT

PROBLEM 13.4

Note:

There are errors in Eqs. (13.2.16) and (13.2.31). They should read:

1 d(M2 ) {(Y-1)(1+y1M

2)E 3+ y[2 +(y-1)M 2 ]v 1B 2 } J3

S dx r M - -- (13.2.16)

and

2 d(M2 )

i d ) (1M) ) [(y-I)(I1+yM2)E 2 } 3M2 dx (1- (

+ '2 + (y-1)M v1 2] P3 1 2 ypv

[2 + (y-1)M2

]dA

S A dx (13.2.31)

Part a

We assume that 0, y , Bo, K and.M are constant along the channel. Then, from

the corrected form of Eq. (13.2.31), we must have





0 = 1- [(y-l) (l+M2)(-K) + y(2+(y-1)M )] -K) [2 +(y-1)M

](a)

Now, using the relations

v 2 M2yR T

and p pRT

we write

v M2

Yp pv

Thus, we.obtain

1 dA [(y-)(l+yM2)(-K) + y(2+ (y-1)M

2)]

A2

dz 2 + (y - 1)M2

B20C(l-K)M

2

pVA

(b )

(c)

From conservation of mass,

pvA = piviAi

Using (d), we integrate (c) and solve for A

Ai

(d)

to obtain

where

A(z)

Ai

1

1 - ýz

[(y-1)(1+ yM1)(-K) + y(2 +(y-l)M

Sivi[ 2+ (y-l)M

2]

2)]B2M

2(1-K)

(e)

We now substitute into Eq. (13.2.27) to obtain

vB 2 (1-K)o1 dv 1 o 1 dA

S (-M) [(y-l)(-K) + y] A dz-vz (-M2) Yp A dz(f)

Thus may be rewritten as

d = (-M 2 ) [(y-l)(-K) + y] i iAi Av dz (1-M2) p viAi Ai

Solving, we obtain

- 82£n v = n(l - 8 z) + en vi

or v(z) = (1 8 z)2/ 1

V i 1

i [(y-l)(-K) +y]Bo (1-K)M

2-

where 82 = (-i 2) i vi

01

(g)

(h)

(i)

Now the temperature is related through Eq. (13.2.12), as





M2yRT = v 2

(j)

Thus T(z)i 2(k)

From (d), we have

P(z) = vi Ai

Pi v A (

Thus, from Eq. (13.1.10)

p(z) = Vi Ai T

Pi v A Ti (m)

Since the voltage across the electrodes is constant,

w( - Kv(z)B (n)

or KviBoi Viw(z) = Kv(z)B wi (o)

Kv(z)B0 v(s)

Thus, w(z) vi

wi v(z)Then

d(z) A(z) i(q)di Ai w(z) ()

Part b

We now assume that a, y, Bo, K and v are constant along the channel. Then, from

Eq. (13.2.27) we have

0 = 1-) C(y-l) (-K) + y]viB2

(-K) 1 ACl-M 2 ) 1li0 yp A dz

But, from Eq. (13.2.25) we know that

(l-K)v. B2z

-- 1 = 1- 3z (s)Pi Pi

ov. B2

where = (l-K) 1 0

Pi

Substituting the results of intoA(z) _ ib),pi (-Ka) and solvin for we obtain

i A(z) (u)





and so, from Eq. (13.1.10)

T(z) = p(z) Pi (v)T. Pi p(z)

As in (p)

w(z) vi

wi=v(z)

= 1 (w)

Thus

d(z) = A(z) (x)

di A

Part c

We wish to find the length Z such that

C T(i) + [v ]2p C 1

z = .9 (y)

C T(o) + [v(o)]

For the constant M generator of part (a), we obtain from (i) and (k)

C 1 2 1 221 1 1 -28 /Bl

C Ti [ () ]2 C T i +1 2 9p Vi 1 2 p iT(z) 2 i (Z

Reducing, we obtain

- 26 /8

(1 - ) 2 .9 (aa)

Substituting the given numerical values, we have

-2,1 = .396 and 2/8

2 1= - 7.3 x 10

We then solve (aa) for Z, to obtain

k % 1.3 meters

For the constant v generator of part (b), we obtain from (s), (t), (u) and (v)

CT i ) + vpI 2 i

C T + S= 2 .9 (bb)

p i 2 i

or

(1- 8 4/ 3) 1 2C T. (1 - B ) v2C 1 = .9 (cc)

C T + L2

vp i 2 1

Substituting the given numerical values, we have

83





= B .45 and B /B = .857

Solving for Z, we obtain

Z 1.3 meters.

PROBLEM 13.5

We are given the following relations:

B(z) E(z) wi di Ai I/2

B. E. w(z) d(z) (z)

and that v, 0, y, and K are constant.

Part a

From Eq . (13.2.33),

J=

(1-K)ovB(a)

For constant velocity, conservation of momentum yields

dp = - (1-K)ovB2

(b)dz

Conservation of energy yields

pvC = - K(1-K)G(yB)2

(c)pdz

Using the equation of state,

p = pRT (d)

we obtain

dP dT (l-K) B2 (e)

T -- + p - -K OvB (e)Tdz + dz R

or

T + (-K) (1-K)ovB2

(1-K)OvB2

(f)dz C R

pThus, T = vB

2(1-K) + (g)

dz R+

(g)

AlsoB.i2 (A.)

2 1 1

A(z)

and andA. =p(z)A(z)11

=Therefore d d-

and dT =

p c d p dz

ovB'fi(1-K)(-

2 + --- ) P(!)

Bp

-K(1-K)av (i)Pi





and so

dTK(1-K)OvB2

i ()dz pic p

Therefore 2OvB i

T = - K(1-K) - z + T (k)PiCp

Let 2Let -K(1-K)avBi

=Pi Cp (i)

Then

T = Ti C-az

+ 1) (m)1

+ ovB (1-K)( K iS i cp R dz (n)

pi (az + Ti)

We let K 1

S vB(-K)( Cp R

P. aP i e

KR

Integrating (n), we then obtain

in p = 8 in(az + Ti) + constant

or

=z + 8p= i - 1 ) (o)

ThereforeA

A(z) =()

Ti

Part b

From (m),

T() at+

TT(a) i

.8T i Ti

or

-l- . -. 2Ti

Now

K(1-K)viBi2

E

Ti Pic Ti

But R Ti Pic T = = = 2.5 x 10

6

pi (1) pi(1 y i y





Thus -2hus - .5(.5)50(700)16 - 8.0 x 10-2

Ti .7(2.5 X 10 6)

Solving for R, we obtain

z= •2x 102 = 1.25 meters8

Part c

(Xzp= pi ( •.+ 1)

Numerically

8= - 1 1 1 % 6.KR (l1)K

Thus

p(z) = .7(1 - .08z) 6

Then it follows:

7 7

p(z) = pRT = Pi(1 - .08z) = 5 x 10S(l - .08z)

T(z) = Ti( l - .08z)

From the given information, we cannot solve for Ti, only for

Pi v 2

RT. = - = -i 7 x 10s

I Pi yM

Now vi v2

yRT(z) yp(z)

.51 - .08z

Part d

The total electric power drawn from this generator is

p VI = -E(z)w(z)J(z)£d(z)

= - E(z)(1-K)GvB(z)£d(z)w(z)

= - Eiwi(1-K)ovB diz

= -KvB

i

Thus pe = K(vB )2 w diO (-K)

= .5(700)216(.5)50(.5)1.25

= 61.3 x 106 watts = 61.3 megawatts

86




T (z)

T

ST (z)

Ti (1 - .08z)

p(z)

1.25

p(z) = 7ý(1 - .08z) 6

p(z) 1.25

p(z) = 5 x 105

(1 - .08z)7

m(z)

1.25

M2

(z) =(1-

.5.08z)

1.25



ELECTROILECHANICS OF COMPRESSIBLE, INVISCID FLUIDS

PROBLEM 13.6

Part a

We are given that

E--i4 Vo x

X

x3 Lj/3and

4 Eo VO

e 9 L'3 x

The force equation in the steady state is

dv

pv d• i x = p Em xx e

Since pe/p = q/m = constant, we can write

Vd O /3

v2A = V4 X!3dx2Vx m) L3

Solving for v we obtainx

Part b

The total force per unit volume acting on the accelerator system is

F = peE

Thus, the total force which the fixed support must exert is

f = - FdVitotal x

EV 2-16 E0 0 -3 Adx

=-f - x Adx i

27 L8

x

0

f 8 oototal 8 2 Ai

9 x

PROBLEM 13.7

Part a

We refer to the analysis performed in section 13.2.3a. The equation of motion

for the velocity is, Eq. (13.2.76),

•-2 a v2V 2 2(a)

3-ý = a ax 2 (a)


v(-L) = V cos Wt

v(O) = 0

We write the solution in the form







v(x t) = Re[A ej(wt-kx )+ B ej(wt+kxl ) ] (b)

where aa

Using the boundary condition at x1 = 0, we can alternately write the solution as

v = Re[A sin kx e j t I

Applying the other boundary condition at x = - L, we finally obtain

V

v(xt) = - sin kL sin kx cos wt. (d)1 sin kL 1

The perturbation pressure is related to the velocity throughEq. (13.2.74)

Pov'3 a(e)'at -x,

Solving, we obtain

0Vo f)oL sin kx sin wt = - (f)

sinkL 1 x1

or p V

p = 0 0 k cos kx sin wt (g)k sin kL 1

where po0 is the equilibrium density, related to the speed of sound a, through

Eq. (13.2.83).

Thus, the total pressure is

p Vw

P= + p + k sin kL cos kx sin at (h)

and the perturbation pressure at x, = - L is

00Voa

p'(-L, t) = sin k cos kL sin wt (i)sin kL

Part b

There will be resonances in the pressure if

sin kL = 0 (j)

or kL = n7 n = 1,2,3.... (k)

Thus

w a (a)L

PROBLEM 13.8

Part a

We carry through an analysis similar to that performed in section 13.2.3b.

We assume that

E = iE 2 (xl,t)

J= 2J2 (x, t)

http://slidepdf.com/reader/full/13.2.3b

http://slidepdf.com/reader/full/13.2.3b




PROBLEM 13.8

B = i [P H + i H' (x ,t)]3 00 03 1

Conservation of momentum yields

Dv=

D 1 .x +J 2 o(Ho + H3) (a)Dtax 1 20 0

Conservation of energy gives us

D 1 2p - (u + v ) = pv + J E (b)

Dt 2 1 ax 1 2 2

We use Ampere's and Faraday's laws to obtain

aH'-@ =

- - J

ax 2(c)

andDE P oaH

a2 x (d)3x, at

while

Ohm's law yields

J = ([E - v B i (e)2 2 13

Since-+

oo

E = vB (f)2 1 3

We linearize, as in Eq. (13.2.91), so E ' v p H2 100

Substituting into Faraday's law

av, alH'3

Linearization of the conservation of mass yields

=at o )v (h)

Thus, from (g)

,H'ooy_ 3

p at o at ()

Then

Ho Po

H'3 p'

Linearizing Eq. (13.2.71), we obtain

p o D' (k)Dt p Dt





Defining the acoustic speed

as \ 0o/ where p is the equilibrium pressure,2(oH

= P0o p 2

we have

p' = as22 (p)

Linearization of convervation of momentum (a) yields

av 9H'

p 1 = - T 3 o Ho(m)o at 3xl x oHo (m

1

or, from (j) and (Z),

Iv, ao a2 0

Po at= -s (n)

Differentiating (n) with respect to time, and using conservation of mass (h), we

finally obtain

S2v °H 2v2V= ( a (0 )

Defining 0 12

p1H

a2

a2+-0 (p)

we haves po

a 2v 3 v

1I a2V 2(V)

Part b


j ( kxV = Re [A ej (Ct-kx )+ A e wt+ )] (r)1 1 2

where k = a

The boundary condition at x = - L is

V(-L,t) = V cos wt = V Re ejwt (s)

s s

and at x = 0

M dViot) =p'A' + poH H A (t)

dt oo 11=0 1=0

From (h), (j) and (t),

1-g__ = P va2 at (o)

s ax





H'

Ho

a Po

Thus

dv,(0,t) oH2

2M dt =A a + I p' =A az-T-

From (u), we solve for p'j to obtain:

x =0

Substituting into (s) and (t), we have

a 2/Poak \Mjw(A + A) = A(- ~)• ) (A 1 A 2 )

1 2 a'\W /

Ale+jk£+ A2e-jki

Solving for A and A , we obtain1 2

(Mjw + Aapo)VA = s

1 2(-Mw sin kZ + Aap cos k£)

(Aap - Mjw)Vs

A2 2(-Mw sin kZ + Aa %cos kt)

Thus, the velocity of the piston is

wtv (O,t) = Re [A + A ]ej

S1 2

Aap V

v (0,t) = w s + Aaos Wt (aa)1 -Mwlsin kk +Aap cos kZCOWt

PROBLEM 13.9

Part a

Th e differential equation for the velocity as derived in problem 13.8 is

32v a2v1 2 1o

Ft- - a aX12 (a)

where 2 2 00a = a +

s Po

2 p H2

a (YPo2

with a = -•oP where p = p 2

S Po 0 1 2

Part b

We assume a solution of the form





V(x ,t) = Re [De J(t-kx1)] where k = 1 a

We do not consider the negatively traveling wave, as we want to use this system as

a delay line without distortion. The boundary condition at x = - L is

V(-L,t) = Re V ejWts

and at x = 0 is

dV

;, dt = p'(O,t)A - BV (O,t)+ joHoH' A (b)

From problem 13.8, (h), (j) and (9)

av H'

p p ', =t - and - =s at o ax H a o

1 S 0

Thus, (b) becomes

t a 2

-BDejt

+ (-) p'A =0 (c)S

where PoD(- jk) 2 t (d)pI =w

j w se (d)

x0O

Thus, for no reflections

2 Ap0a2

- B + (a) 0 (e)a aS

or

B = Aapo (f)

PROBLEM 13.10

The equilibrium boundary conditions are

T[- (L + L + A),t] = T1 2 0

T[- (L + A),t]As = - PoAc

Boundary conditions for incremental motions are

1) T[- (L + L + A),t] = T (t)1 2 s d

2) - T[- (L + A),t]A - p(-L ,t)Ac = M - v (-L ,t)

3) v (-Ll ,t) = Ve[-(L + A),t] since the mass is rigid

and 4 ) v (O,t) = 0 since the wall at x=0 is fixed.

PROBLEM 13.11

Part a

We can immediately write down the equation for perturbation velocity, using

equations (13.2.76) and (13.2.77) and the results of chapters 6 and 10.

93



ELECTROMECHANTCS OF COMPRESSIBLE, INVISCID FLUIDS


We replace 3/at by 3/at + v-V to obtain

(-+ V )v' = a2

2Vt o x sax

Letting v' = Re V j(t-kx)

we havea2k

2

(W- kVS)2 =

as

Solving for w, we obtain

W = k(V ± as)

Part b

Solving for k, we have

k = V±a

V+ao s

For Vo > as, both waves propagate in the positive x- direction.

PROBLEM 13.12

Part a

We assume that

E = i E (x,t)z z

J = i J (x,t)z z

B= i o1[H + H'(x,t)]y y

We also assume that all quantities can be written in the form of Eq. (13.2.91) .

Vx p' - Jz Ho (conservation of momentum (a)

linearized)

The relevant electromagnetic equations are

aH'

Yx = J (b)ax z

and3E aH'

z 0o----3x at (c)

and the constitutive law is

Jz = (E + VxpoH ) (d)z z xoo

We recognize that Eqs. (13.2.94), (13.2.96) and (13.2.97) are still valid, so eav

1 I3'

P at ax (e)




PROBLEM' 13.12 (continued)

and

p' a2P' (f)S

Part b

We assume all perturbation quantities are of the form

^ j(wt-kx)v = Re[v e ]

x

Using (b), (a) may be rewritten as

pojwv = + jkp + p0HojkH (g)

and (c) may now be written as

- jkE = jojwH (h)

Then, from (b) and (d)

- jkH = o(E + V oHo ) (i)

Solving (g) and (h) for H in terms of v, we have

vOVoH

H = (j)

- jk+po 9

From (e) and (f), we solve for p in terms of v to be

p= kP asv (k)

Substituting (j) and (k) back into (g), we find

2 jk(po Ho)2 0)

vL2 00jk+ w

-

LkThus, the dispersion relation is

k2

j(MoHo)20= 0 (m)(02 - k2a2) -_ k,

(+ + j11w)po

We see that in the limit as a + m, this dispersion relation reduces to the lossless

dispersion relation

_2 k2(a2 + = 0 (n)

Part c

If a is very small, we can approximate (m) as

W2- k2a2 _ j( 2 )H = (0o)s fo O kr

for which we can rewrite (o) as





kl = 0s - k22 O ( 2 2

Solv ing for k , we obtain

2 2

()02_ 0p 2 OHo) w° 2) 0 02

2 Po ok2= 22a

O+2

as 2as as

Since a is very small, we expand the radical in (q) to obtain

_ O A oo2 [12

2 a .S O2]

L o02

Thus, our approximate solutions for k are

EW (IIoHo)2

k2 Oa

s

( H ) 20 a (11

The wavenumbers for the first pair of waves are approximately:

while for the second pair, we obtain

+ o 0

k~+~lo~op

The wavenumbers from (u) represent a forward and backward traveling wave, both with

amplitudes exponentially decreasing. Such waves are called 'diffusion waves'. The

wavenumbers from (v) represent pure propagating waves in the forward and reverse

directions.





Part d

If 0 is very large, then (m) reduces to

H2 H2

W2- k2a2- o k 0 ; a2 = a2 + O (w)p0 Ow s Po

This can be put in the form

k2= f(wk) (x)

2 aa

where H2 k0

f(w,k), =- 2 (Y)pOwa

As a becomes very large, the second term in (x) becomes negligible, and so

2 2k -

However, it is this second'term which represents the damping in space; that is,

k + f(w,k) (z),a j2a

Thus, the approximate decay rate, ki, is

H2 k

ki f(w,k)a o a (aa)ki 2a w 2poaw•

or 2 k H2

2pa a 2p a 0 w 0



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ELECTROMECHANICAL COUPLING WITH VISCOUS FLUIDS

PROBLEM 14.1

Part a

We can specify the relevant variables as

v = v 1 (x2)

E = 2E2()2 + 13E3 (x 2

J= i 2 JO

B = iB +TB (x)2 0 11 3

The x component of the momentum equation is1 a2v 1

0 = p sx

with solution 2

V Cx +C1 1 2 2

Applying the boundary conditions

V = 0 @ x =0I 2 (c)

v @ x2 = d

We obtainvx02

V =

'1 d (d)

We note that there is no magnetic force density since the imposed current and mag

netic field are colinear. We apply Ohm's law for a moving fluid

T = oE+V (e) x B)

in the x and x directions to obtain2 3

Jo

= YGE

2 (f)

and 0 = a(E + v Bo) (g)

since no current can flow in the x direction.3

Thus J

E2 T

o(h)

and VVoX2B

3 d (i)

As from Eq. (14.2.5),

d JV = o E2dx 2 = d (j)

Thus, the electrical input pe per unit area in an x - x plane is

J2 d

pe

= Jo

=0 (k)





We see that this power is dissipated as Ohmic loss. The moving fluid looks just

like a resistor from the electrical terminals. The traction that must be applied

to the upper plate to maintain the steady motion is

av, Iv0

2•2x=d dT = -i 0

2

Again we note no contribution from the magnetic forces.

The mechanical input power per unit area is then

]V2

pm

= Tio

--d

(m)

The total input power per unit area is thus

PtPe+

P d + a (n)

The first term is due to viscous loss that results from simple shear flow, while the

second term is simply the Joule loss associated with Ohmic heating. There is no

electromechanical coupling. Using the parameters from Table 14.2.1, we obtain

V = 15 millivolts PtV

2.2635 x 104

.015

______________________ B B0B0 o

and

Pt = 2.2635 x 10 4watts/m2, independent of B0o

These results correspond to the plots of Fig. 14.2.3 in the limit as

B 1 0O

We see that the brush losses and brush voltage are much less for this configuration

than for that analyzed in Sec. 14.2.1. This is because the electrical and mechanical

equations were uncoupled when the applied flux density was in the x2 direction.

This configuration is better, because low voltages at the brush eliminate arcing,

and because the net power input per unit area is lessno matter the field strength Bo.

The only effect of applying a flux density in the x 2 direction was to cause an

electric field in the x 3 direction. However, since there was no current flow in the

x3 direction, there was no additional dissipated power. However, if E3 became too

large, the fluid might experience electrical breakdown, resulting in corona arcs.




PROBLEM 14.2

The momentum equation for the fluid is

P + -Vp +(v*V)v = +V2

We consider solutions of the form

v = i- v C(r)z z

-II\

P = Pkz,*

Then in the steady state, we write the z component of (a) in cylindrical coordinates

as av

= '- r z (b)z rr r r

Now, the left side of (b) is only a function of z, while the right side is only a

function of r. Thus, from the given information

P2 (c) "= az L

Using the results of (c) in (b), we solve for v (r) in the form

p -pv (r) = 2 r

2+ A £n r + B (d)

z 4Lp

where A and B are arbitrary constants to be evaluated by the boundary conditions

v (r = 0) is finite

v zr- = 3\

) =

_

U

n

Thus the solution is

(p - p )R 2)

v (r) = 2•L (r2 - (e)

Electromechanical Dynamics - Solutions Manual

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