electromechanical dynamics - solutions manual
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MIT OpenCourseWare
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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
OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: CreativeCommons Attribution-NonCommercial-Share Alike For more information about citing these materials or our Terms of Use, visit:http://ocw.mit.edu/terms
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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
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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
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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
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LUMPED ELECTROMECHANICAL ELEMENTS
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
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LIUPED ELECTROMECHANICAL ELEMENTS
PROBLEM 2.4 (Continued)
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
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LUMPED ELECTROMECHANICAL ELEMENTS
PROBLEM 2.4 (Continued)
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
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LUMPED ELECTROMECHANICAL ELEMENTS
PROBLEM 2.5 (Continued)
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
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LUMPED' ELECTROMECHANICAL ELEMENTS
PROBLEM 2.5 (Continued)
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
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LUMPED ELECTROMECHANICAL ELEMENTS
PROBLEM 2.6 (Continued)
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
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b I
LUMPED ELECTROMECHANICAL ELEMENTS
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
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LUMPED ELECTROMECHANICAL ELEMENTS
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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MIT OpenCourseWare
http://ocw.mit.edu
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
OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: CreativeCommons Attribution-NonCommercial-Share Alike For more information about citing these materials or our Terms of Use, visit:http://ocw.mit.edu/terms
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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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.1 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.2 (Continued)
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
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LIUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.3 (Continued)
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)
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.4 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICS
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
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- -
LUMPED-PARAMETER ELECTROMECHANICS
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
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LUMPED-PARAMETER ELECTROMECHANICS
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.8 (Continued)
+ 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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.9 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.10 (Continued)
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.
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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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.12 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.13 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.14 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.14 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.15 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.16 (Continued)
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,
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.17 (Continued)
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)
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.18 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.19 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.20 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.21 (Continued)
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'
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.21 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.22 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICS
PROBLEM 3.22 (Continued)
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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Solutions Manual for Electromechanical Dynamics
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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
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ROTATING MACHINES
PROBLEM 4.1 (Continued)
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.
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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?- )"
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ROTATING MACHINES
PROBLEM 4.2 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.2 (Continued)
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,
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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.
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ROTATING MACHINES
PROBLEM 4.4 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.5 (Continued)
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)]
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ROTATING MACHINES
PROBLEM 4.7 (Continued)
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:
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ROTATING MACHINES
PROBLEM 4.8 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.9 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.10 (Continued)
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.
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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$sin (wt+l)]
For constant phase
wt + i = constant
and
Ob dt
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ROTATING MACHINES
PROBLEM 4.11 (Continued)
Part c
The ratio of amplitudes is
B _ ( 22 2rbm -cos) + sin 1i-cos
Brfm (1+cos8) 2 + sin2 8 l+cos8
rbmas 8 0, - -0..
Brfm
Part d
=The forward wave amplitude will go to zero when 8 W. The phase sequence
has been reversed by reversing the phase of the current in the b-winding.
PROBLEM 4.12
Equation 4.1.53 is
Pe =Vasias + Vbsibs
For steady state balanced conditions we can write
ias= I cos wt; ib = I sin wt
Vas = V cos(wt+4); vbs = V sin(wt+ý)
then
Pe=
VI[coswtcos(wt+$)+sinwt sin(wt+0)]
Using trigonometric identities
= pe VI cost
Referring to Fig. 4.1.13(b) we have the vector diagram
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ROTATING MACHINES
PROBLEM 4.12 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.13 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.13 (Continued)
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
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s ta~b iboI1bI% * -
Palej anordire
-A
r ýa te )
re-ldCFIre(ýreI4
zero p4.F
0.I
0.5
11.
1j,
2.0
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ROTATING MACHINES
PROBLEM 4.14 (Continued)
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
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ROTATING MACHINES
Q
P
LLI
captb IICa"el V Problem
4.14/
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ROTATING MACHINES
PROBLEM 4.15 (Continued)
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?
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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
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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
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ROTATING MACHINES
PROBLEM 4.16 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.17 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.17 (Continued)
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)
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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
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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
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ROTATING MACHINES
PROBLEM 4.18 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.19 (Continued)
PROBLEM 4.19 (Continued)
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.
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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)
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ROTATING MACHINES
PROBLEM 4.23 (Continued)
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)
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ROTATING MACHINES
PROBLEM 4.23 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.23 (Continued)
-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
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--
ROTATING MACHINES
PROBLEK 4.24 (Continued)
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"
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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
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ROTATING MACHINES
PROBLEM 4.24 (Continued)
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
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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.
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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
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ROTATING MACHINES
PROBLEM 4.26 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.26 (Continued)
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
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ROTATING MACHINES
PROBLEK 4.27 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.27 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.27 (Continued)
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
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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
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ROTATING MACHINES
PROBLEM 4.28 (Continued)
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)
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ROTATING MACHINES
PROBLEM 4.28 (Continued)
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
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ROTATING MACHINES
PROBLEM 4.28 (Continued)
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.
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ROTATING MACHINES
FIELDP' I£.E1
lPOIAEREAL tso
(MW)
€
t
A~eMtArTeE
CU~I~
SirABlLTry LIMiT
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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)
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ROTATING MACHINES
PROBLEM 4.29 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.1 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.1 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.2 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.3 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.3 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.4 (Continued)
-- 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
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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
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LUMPED-PARAMETER ELECTROMECHANICAL DYANAMICS
I
IV II
1/
/1/I
(~)
.a\e s +C ýc
k-,ý , "v W
Figure for Prob. 5.6
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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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.8 (Continued)
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
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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.
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.9 (Continued)
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.
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JUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.10 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.11 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.11 (Continued)
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
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LUMPED-PARAMETER ELECTRGCECHANICAL DYNAMICS
PROBLEM 5.12 (Continued)
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).
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.12 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.13 (Continued)
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
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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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEK 5.14 (Continued)
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).
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.14 (Continued)
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
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(-O) 4-U
CA~)
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LUMPED-PARAMETER ELECTR(OMECHANICAL DYNAMICS
PROBLEM 5.14 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
to
-Ir tl
c,,
O
-0
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
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'
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.15 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.15 (Continued)
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)
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.16 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.18 (Continued)
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,
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.18 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.19 (Continued)
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.20 (Continued)
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).
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.23 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
(0,)
OA x---aco s%~ 0
oY\
V~xr
(b)
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.25 (Continued)
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
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--
LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.25 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
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
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.26 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.26 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
PROBLEM 5.26 (Continued)
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.
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LUMPED-PARAMETER ELECTROMECHANICAL DYNAMICS
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)
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FIELDS AND MOVING MEDIA
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.
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FIELDS AND MOVING MEDIA
PROBLEM 6.3 (Continued)
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
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FIELDS AND MOVING MEDIA
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.
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FIELDS AND MOVING MEDIA
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)
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FIELDS AND MOVING MEDIA
PROBLEM 6.6 (Continued)
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.7 (Continued)
-_ 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
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FIELDS AND MOVING MEDIA
PROBLEM 6.8 (Continued)
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.8 (Continued)
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)
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FIELDS AND MOVING MEDIA
PROBLEM 6.8 (Continued)
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
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FIELDS AND MOVING MEDIA
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.
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--
FIELDS AND MOVING MEDIA
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.12 (Continued)
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.12 (Continued)
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.
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FIELDS AND MOVING MEDIA
PROBLEM 6.12 (Continued)
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)
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FIELDS AND MOVING MEDIA
PROBLEM 6.13 (Continued)
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~ +
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FIELDS AND MOVING MEDIA
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)
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FIELDS AND MOVING MEDIA
PROBLEM 6.14 (Continued)
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.15 (Continued)
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)
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FIELDS AND MOVING MEDIA
PROBLEM 6.16 (Continued)
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
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FIELDS AND MOVING MEDIA
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*
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FIELDS' AND MOVING MEDIA
PROBLEM 6.17 (Continued)
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
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- -
FIELDS AND MOVING MEDIA
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.18 (Continued)
(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
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FIELDS AND MOVING MEDIA
PROBLEM 6.19 (Continued)
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.19 (Continued)
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.20 (Continued)
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.22 (Continued)
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
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FIELDS AND MOVING MEDIA
PROBLEM 6.23 (Continued)
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 ,
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FIELDS AND MOVING MEDIA
PROBLEM 6.23 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.1 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.2 (Continued)
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.
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.3 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.4 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.4 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.5 (Continued)
Let B (x) = Be , then
a2 jw~pa (b)
a6 (c)
The boundary conditions are
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.
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.6 (Continued)
(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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.6 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.6 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.7 (Continued)
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)
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.8 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.9 (Continued)
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)
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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.
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.13 (Continued)
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)
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MAGNETIC DIFFUSION AN D CHARGE RELAXATION
PROBLEM 7.13 (Continued)
-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' =
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.14 (Continued)
= 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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.14 (Continued)
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.
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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.
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.15 (Continued)
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
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MAGNETIC DIFFUSION AN D CHARGE RELAXATION
PROBLEM 7.16 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.17 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.18 (Continued)
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<b
r
Er = C2 (t) t > 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,
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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 .
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.20 (Continued)
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
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MAGNETIC DIFFUSION OF CHARGE RELAXATION
PROBLEM 7.21 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.22 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.22 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.24 (Continued)
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).
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.27 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.28 (Continued)
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
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MAGNETIC DIFFUSION AND CHARGE RELAXATION
PROBLEM 7.28 (Continued)
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)
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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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.3 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.3 (Continued)
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,
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.4 (Continued)
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)
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.5 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
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.
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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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.7 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.8 (Continued)
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
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------------ -----------
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.9 (Continued)
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:
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.10 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.10 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.10 (Continued)
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_ _ _ - -- -
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.11 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.11 (Continued)
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.
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.12 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.12 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.13 (Continued)
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 =•
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FIELD DESCRIPTION OF AGNETIC AND ELECTRIC FORCES
PROBLEM 8.13 (Continued)
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.
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.13 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.13 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.14 (Continued)
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.
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.15 (Continued)
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:
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.15 (Continued)
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,
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.16 (Continued)
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
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VFIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.17 (Continued)
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)
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.17 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.17 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.18 (Continued)
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)
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.19 (Continued)
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).
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.20 (Continued)
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
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.21 (Continued)
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 - <B B (y=o)> (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-
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
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 .
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
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;
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.25 (Continued)
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.
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FIELD DESCRIPTION ON MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.26 (Continued)
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,
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.26 (Continued)
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,
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.27 (Continued)
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.
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Solutions Manual for Electromechanical Dynamics
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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.
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SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.2 (Continued)
-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)
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.3 (Continued)
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
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SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.5 (Continued)
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:
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.6 (Continued)
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,
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.6 (Continued)
(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)
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.7 (Continued)
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):
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.8 (Continued)
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.9 (Continued)
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.10 (Continued)
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.11 (Continued)
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.11 (Continued)
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.
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.12 (Continued)
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.
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U
SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.13 (Continued)
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.14 (Continued)
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
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SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
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).
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.17 (continued)
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)
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.18 (continued)
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
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SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.19 (continued)
'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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.20 (continued)
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
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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
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SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.22 (continued)
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)
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.24 (Continued)
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.24 (continued)
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.25 (continued)
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
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y
SIMPLE ELASTIC CONTINUA
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.26 (continued)
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
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.27 (continued)
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.
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SIMPLE ELASTIC CONTINUA
PROBLEM 9.27 (continued)
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.
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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
<P> =
[(ffklcoshjkflZ)2
+ (2Bw sinhj)kJi)2]
where it is convenient to use the identity
<Re Aejwt ReBejt> = 1 AB*2
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECITANICAL CONTINUA
PROBLEM 10.2 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.3 (continued)
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 ^ ^ ^ ^
-<P> [5+ * - 5_ _*]
Part b
For k purely imaginary
k = j5
with 3 real, we can write F(x,t) in the form
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DYNAMICS IN ELECTROMECHANICAL CONTINUA
PROBLEM 10.5 (continued)
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
<P> =- +* _ - + 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+
<P> = E
Thus, substitution gives
<P> 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.
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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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.7 (continued)
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
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DYNANICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.7 (continued)
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
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DYNAMICS OF ELECTROMECIIANICAL CONTINUA
PROBLEM 10.7 (continued)
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.
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.8 (continued)
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
The boundary conditions are
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.9 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.10 (continued)
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
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PROBLEM 10.11 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.12 (continued)
(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.
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.12 (continued)
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.
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.13 (continued)
,-, 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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.14 (continued)
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
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PROBLEM 10.15 (continued)
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.
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
The boundary conditions are
=(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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.17 (continued)
(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
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PROBLEM 10.18 (continued)
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.
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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.
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.21 (continued)
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
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PROBLEM 10.22
Part b
(W -)2 2 (m2m2) (d)
ioW + = 2 + ( 2 Q2 ) 2m2
( mi) = -
(f)
n2 _ 2
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.23 (continued)
A· U
and
1% ý 0;..
z
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.23 (continued)
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
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PROBLEM 10.23 (continued)
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
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PROBLEM 10.23 (continued)
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:
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.24 (continued)
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
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.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
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PROBLEM 10.25 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.26 (continued)
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
&,
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.26 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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/
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.26 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.27 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.27 (continued)
Tj Ul S
/p
(Ar (01 23
while fo r real k, the w-k plot is
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.27 (continued)
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)
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.27 (continued)
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.
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.27 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.28 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.28 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.28 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
Mv'
m=1
Vi = i
k=2.
tj =
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DYNAMICS OFp ELECTROMECHANICAL CONTINUA
PROBLEM 10.28 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.29 (continued)
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.
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.29 (continued)
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.30 (continued)
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
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.31 (continued)
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;
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
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).
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DYNAMICS OF ELECTROMECHANICAL CONTINUA
PROBLEM 10.32 (continued)
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.
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Solutions Manual for Electromechanical Dynamics
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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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.1 (Continued)
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
The boundary conditions are
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)
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.2 (continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.2 (continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.2 (Continued)
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
The boundary conditions are
62(,t) = 6 ejWt
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.3 (continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.3 (continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
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.
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
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.
We assume solutions of the form
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.6 (continued)
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.
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INTRODUCTION TO TH E ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.6 (continued)
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.
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INTRODUCTION TO TH E ELECTROMECHANICS OF ELASTIC MEDIA
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
We assume solutions of the form
(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
Applying the boundary conditions, we obtain
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.
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.7 (continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.7 (continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC !MEDIA
PROBLEM 11.8 (continued)
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
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INTRODUCTION TO TH E ELECTROMECHANICS OF ELASTIC MEDIA
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)
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
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
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INTRODUCTION TO TH E ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.9 (continued)
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.,
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (Continued)
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
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (Continued)
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)
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (Continued)
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)
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (Continued)
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)
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
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
The boundary conditions are
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.
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INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
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
We assume solutions of the form
) '(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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.2 (Continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.3 (Continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.5 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.5 (continued)
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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.6 (continued)
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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
Acceleration x2
V2o
a= (-) ria r
yr
xl
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.8 (Continued)
Thus2 - , xi (Vc)
a= v i 0(O T2 1 2 2 2
Part b
Using Bernoulli's law, we have
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)
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.9 (Continued)
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
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ELECTROIECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.10 (Continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUTDS
PROBLEM 12.10 (Continued)
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
Using Bernoulli's law, we have
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.11 (continued)
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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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)
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ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUIDS
PROBLEM 12.12 (Continued)
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__·__
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.12 (Continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.13 (Continued)
[ 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
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ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.14 (Continued)
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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.15 (continued)
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
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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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.18 (Continued)
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
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ELECTROMILCHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.20 (Continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.20 (Continued)
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)
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ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUIDS
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 <P + RL0 o H o w)
Part b
If the product V I > 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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.24 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.24 (continued)
(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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.25 (Continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.25 (continued)
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)
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.25 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.25 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.27 (continued)
/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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.29 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.30 (continued)
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
- ) ()
Integrating, we obtain
9.4t + C = log 6 x10-i ()
We evaluate the arbitrary constant C by realizing that at t=O, i = 10 amps
66
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.30 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.30 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.31 (continued)
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.
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.33 (Continued)
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.
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ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.34 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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)
We assume solutions of the form
^ 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)
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.36 (continued)
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
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.37 (continued)
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
The boundary conditions are
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
Applying the boundary conditions, we obtain
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)
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ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.37 (continued)
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
Integrating, we obtain
- 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
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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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
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)
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
= 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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.5 (continued)
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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.5 (Continued)
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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
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
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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)
The boundary conditions are
v(-L) = V cos Wt
v(O) = 0
We write the solution in the form
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.7 (continued)
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)
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.8 (continued)
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
We assume solutions of the form
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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.8 (continued)
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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.9 (continued)
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
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ELECTROMECHANTCS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.11 (continued)
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)
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
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
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.12 (continued)
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.
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ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.12 (continued)
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)
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ELECTROMECHANICAL COUPLING WITH VISCOUS FLUIDS
PROBLEM 14.1 (continued)
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.
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ELECTROMECHANICAL COUPLING WITH VISCOUS FLUIDS
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)