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SOLUTIONS MANUAL FOR
E LETROI
AN ICA L
IAlECH
M
IIYNAAIICS
PART II: Elastic and Fluid Media
HERBERT H. WOODSON
JAMES R.MELCHER
Prepared by MARKUS ZAHN
JOHN WILEY & SONS, INC.
NEW YORK * LONDON * SYDNEY * TORONTO
SOLUTIONS MANUAL FOR
ELECTROMECHANICAL
DYNAMICS
Part III: Elastic and Fluid Media
HERBERT H. WOODSON
Philip Sporn Professor of Energy Processing
Departments of Electrical and Mechanical Engineering
JAMES R. MELCHER
Associate Professor of Electrical Engineering
Department of Electrical Engineering
both of Massachusetts Institute of Technology
Preparedby
MARKUS ZAHN
Massachusetts Institute of Technology
JOHN WILEY & SONS, INC., NEW YORK. LONDON. SYDNEY. TORONTO
A
PREFACE TO:
SOLUTIONS MANUAL
TO
ELECTROMECHANICAL DYNAMICS, PART III:
ELASTIC AND FLUID MEDIA
This manual presents, in an informal format, solutions to the problems found
at the ends of chapters in Part III of the book, Electromechanical Dynamics.
It is intended as an aid for instructors, and in special circumstances, for use
by students.
A sufficient amount of explanatory material is included such that
the solutions, together with problem statements, are in themselves a teaching
aid.
They are substantially as found in the records for the undergraduate and
graduate courses 6.06, 6.526, and 6.527, as taught at Massachusetts Institute
of Technology over a period of several years.
It is difficult to give proper credit to all of those who contributed to
these solutions, because the individuals involved range over teaching assistants,
instructors, and faculty who have taught the material over a period of more than
four years.
However, special thanks are due the authors, Professor J. R. Melcher
and Professor H. H. Woodson, who gave me the opportunity and incentive to write
this manual.
This work has greatly increased the value of my graduate educa-
tion, in addition to giving me the pleasure
of working with these two men.
The manuscript was typed by Mrs. Evelyn M. Holmes, whom I especially thank
for her sense of humor, advice, patience and expertise which has made this work
possible.
Of most value during the course of this work was the understanding of my girl
friend, then fiancee, and now my wife, Linda, in spite of the competition for time.
Markus Zahn
Cambridge, Massachusetts
October, 1969
a
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 1
where we have taken
-- = 0.
a
a
= 0.
a
Since this is a static problem, we let
x3
ý2
a
(a)
(a)
p
x
P p2-
Thus,
aT(
= pg.
S
(b)
From 11.2.32, we obtain
a6
T11 = (2G + X)
(c)
Therefore
(2G + X)
21
pg
ax1
(d)
Solving for 61, we obtain
6
1
=
2(2G+X)
1
+ C x + C
1
2
(e)
where C 1 and C2 are arbitrary constants of integration, which can be evaluated
by the boundary conditions
61(0)
=
(f)
6
and
T 1 (L)
= (2G + X)
(L) = 0
since x = L is a free surface.
pg x
1 2(2G+X)
[x
Therefore, the solution is
- 2L].
Part b
Again applying 11.2.32
(g)
(f)
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.1 (Continued)
T11= (2G+X)
a6
-ax
pg[x 1 - L)
T12 = T21 = 0
T 1 3 = T 3 1 = 0
ax 1
T22 =X
1
a61
3x1
33
(2G+X) [
1x
- L
T32 = T23 = 0
0
T11
T
0
0
T22
0
0
T33
33
PROBLEM 11.2
Since the electric force only acts on the surface at x 1 = - L, the equation
of motion for the elastic material (-L
p
2
2
at2
ax
1
x I < 0) is
.a26 =
a26 I
(2G+X)
2
from Eqs. (11.1.4) and (11.2.32),
(a)
The boundary conditions are
6 1 (0,t) = 0
a 6 1 (-L,t)
M
a61
= aD(2G+X)
at2
(-L,t) + fe
3x1
fe is the electric force in the xl direction at xl = - L, and may be found by
using the Maxwell Stress Tensor T
for discussion of stress tensor),
e
-
f
E
2
2
E2 aD
with
V
+ V I cos Wt
d + 61 (-L,t)
E
= CE E j
1
2
ij
EEkEk to be (see Appendix G
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.2 (continued)
Expanding fe to linear terms only, we obtain
2V V d cos Wt
V2
fe
EaDo
fe
d
o 102
+
d
2V2
d3 61(-L,t
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 )
6 1 (x1 ,t) = 6 1 (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
with
-- v
P
v =
p
(f)
P
We first solve for the equilibrium configuration which is time independent.
Thus
2 6 1 (x
1
)
ax1
This implies
6 1 (x 1 ) = C x +
C2
1 1
O0.
Because 61(0) = 0, C2
From the boundary condition at xl = - L ((b)
2
V ­
aD(2G+X)C1 --
aD
& (d))
0
(h)
xl
(i)
Therefore
V 2
61(x
E
Note that 6 1 (x1 = - L)
o
2d2 (2G+X)
is negative, as it should be.
For the time varying part of the solution, using (f) and the boundary condition
6(o,t) = 0
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.2 (continued)
we can let the perturbation
61 be of the form
(j)
61 (X1 ,t) = Re 5 sin kx1 ejt
Substituting this assumed solution into (b) and using (d), we obtain
+ Mw
2
A
6 sin kL = aD(2G+X)k 6 cos kL
CaDV V
EaDV
3-
2
(k)
2
d6 sin kL
Solving for 6, we have
V
AaDV
d2 Mw2 sin kL - aD(2G+X)k cos kL +
CaDV 2
30 sin kL
Thus, because 6 has been shown to be real,
61 (-L,t)
V2 L
o
Sd(2G+X)
=
-
6 sin kL cos wt
(m)
Part b
If kZ << 1, we can approximate the sinusoidal part of (m) as
£aDV V1 cos wt
61(•-L,t)
=
[
aDV 2
2 o+
aD(2G+X)
(n)
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
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.2 (Continued)
where
EaDV V
ol
cos wt
d
caDV 2
o
aD(2G+X)
L
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 x 2 direction as
2
2
a t2
aT 2
ý2
1
1x1
From(11.2.32),
T21= G
3x
Therefore, substituting (b) into (a), we obtain an equation for 62
2
a 62
a 62
G-2
t2
axI
We assume solutions of the form
j(wt-kx 1 )
2
=
Re 62 e
where from (c)we obtain
wk
-v
v
p
2
P
G
P
P
Thus we let
62 = Re
a eJ(wt-kx 1 )
b e(t+kx)
+
with k =­
v
P
The boundary conditions are
62(,t) = 6 ejWt
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.3 (continued)
and
x1
= 0
O
(g)
since the surface at x1 = 0 is free.
There fore
6
e-jk'
+ 6b ejkt=6
a
b
o
(h)
and
-jk 6a + jk 6 b = 0
(i)
Solving, we obtain
6
6
a
=6
b
=
2cosk
(J)
Therefore
62 xt)
=
co
Re
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 Wt
coskZ
1
Part b
In the limit as w gets small
6 2 (Xl,t) + Re[6
ej
t ]
(m)
In this limit, 62 varies everywhere in phase with the source.
The slab of elastic
material moves as a rigid body.
x1 =
R
mass /(x
Note from (Z) that the force per unit area at
2
d
required to set the slab into motion is T 2 1 (L,t) = pt dt•l o cos wt) or the.
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
or
(2n+l)irv
S=
2
22.
(n)
n = 0,1,2,...
(o)
6
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.3 (continued)
The lowest frequency is for n = 0
7v
2£
or low
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:
(T1 1 - a)n1 + T 1 2 n 2 + T 1 3n 3 = 0
(a)
T21n1 + (T2 2 - a)n 2 + T 2 3 n3 = 0
T31n1 + T 3 2 n 2 + (T3 3 - a)n 3 = 0
Part b
The solution for these homogeneous equations requires that the determinant of
the coefficients of the ni's equal zero.
Thus
(T1 1 - a)[(T
22
- a)(T 3 3 - a) -
(T 2 3 ) 2 ]
(b)
- T 1 2 [T1 2 (T3 3 - a) - T 1 3T 2 3]
+ T13[T12T23 - T 1 3 (T2 2 -
=
0
0)]
where we have used the fact that
(c)
Tij = Tji.
Since the Tij are known, this equation can be solved for
a.
Part c
Consider T12 = T 2 1 = To , with all other components equal to zero.
The deter­
minant of coefficients then reduces to
-a3+
20 a
for which
or
•
=
0
(d)
=
0
(e)
+T
(f)
The a = 0 solution indicates that with the normal in the x 3 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
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),
In
since a is known, and it is known that
1.
PROBLEM 11.5
From Eqs. 11.2.25 - 11.2.28, we have
1
ell
E [11
(T22 + T33)]
e
-[T 1
- v(Tr
+ T)
]
22 =E
11
22
33
e33
[T 33
- 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)
ej
=
aikaj
ek
(f)
and
For an example, we look
at eli
e =
=
e
alkaltekt
v(T
"
- 1
2
+ T
;31
[T2
This may be rewritten as
alkaltek= [(( + v)alkalTk£ -
'kTk]
kV
where we have used the relation from Eq.(8.2.23), page G10 or 439.
a
a
pr ps
=
6
ps
Consider some values of k and X where k
# Z.
Then, from the stress-strain relation in the unprimed frame,
alkaltekk
Thus
1
E
example 11.2.1.
alkal
T
Tk
2k
=
a al,
E
(1+v)Tk
l+v
E
2G
or
=
=
2G(l +v)
which agrees with Eq. (g) of
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
Eb 2 34
3p ýx4
t2
002
(a)
where & measures the bar displacement in the x 2 direction, T 2 in Eq. 11.4.26 = 0
b are free. The boundary conditions for this problem
as the surfaces at x 2 =
are that at xl = 0 and at xl1
L
(b)
and T11 = 0
T21 = 0
as the ends are free.
We assume solutions of the form
j
w
Re E(x)e
E =
t
(c)
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)
1
with
E
Now, from Eqs. 11.4.18 and 11.4.21,
(x2
-
b 2 )E
(T -b
3
2
21
3
3
x1
(e)
which implies
3E
ax3
=
0
(f)
at Xl= 0, x 1 = L
and
T11
=-
X2
2 E
2(g)
ax1
which implies
a2E =
ax 2
1
at xl1
0 and xl = L
o
(h)
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.6 (continued)
With these relations, the boundary conditions require that
-A
=
0
+ B sincL + C cosh cL + D sinh aL
=
0
-B
=
0
=
0
+C
- A cos cL
- A sin cL
+D
- B cos aL + C sinh
aL + D coshc
L
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 =
Thus,
B
=
aL =
Part b
The roots of cos B =
1
2 3p
Eb2
cosh
cosh
B
follow from the figure.
Note from the figure that the roots aL are essentially the roots 3w/2, 57/2, ...
of cos aL = 0.
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.6 (continued)
Part c
It follows from (i) that the eigenfunction is
=
A'[(sin ax 1 + sinh aex
l ) (sin aL + sinh aL)
(Z)
+ (cos aL - cosh aL)(cos ax + cosh
ax1 )
1
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.
required dependence on xl to within an arbitrary constant.
function is shown in the figure.
Then, (d) gives the
A sketch of this
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.7
As in problem 11.6, the equation of motion for the elastic beam is
Eb2
2
+
4
.
3p
S2
=
x4
0
The four boundary conditions for this problem are:
((x
I
S(xl
= 0) = 0
61(0) = -
x
2
= L)
=
0
00 -Og
61(L)'
x1 = 0
1
x2 3x
1
X13
We assume solutions of the form
(x t) = Re
(xt) (eJwt
= Re
(x)e
it
I
, and as in problem 11.6, the solutions for
E(xl) are
(x
AA sin
with a
ax1 + B cos
=
x 1 + C sinh ax1 + D cosh ax
1
1/4
02
Applying the boundary conditions, we obtain
+
B
A sin aL + B
cos
tL
A
+ C sinh aL
+
A cos aL - B
sin CL
D
+ D cosh aL
C
+ C cosh aL
+ D sinh aL
=
0
-O
= 0
=
0
=
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.
(c)
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.7 (continued)
The first natural frequency is at about
tL =
3w
2
Thus
L
w2
=
1/2
(3
L2
2
Eb2
3p )
Part b
We are given that L = .5 m and b = 5 x 10
-4
m
From Table 9.1, Appendix G, the parameters for steel are:
E
% 2 x 1011 N/m2
3
p I 7.75 x 103 kg/m
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.7 (continued)
w.
Then, f
1
120 rad/sec.
= -- L
19 Hz.
2T
Part c
5
For 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
3p
•4
)xi
2
at
At x 1 = L, there is a free end, so the boundary conditions are:
(x 1 =L)
=
0
T2 1 (x1 =L)
=
0
T1
and
1
The boundary conditions at xl= 0 are
2
M
,
St
=
t)
+
2
D dx 2 +
(T 2 1 )
f
+
F
x =0
and
61(x1 = 0) =
0
The H field in the air gap and in the plunger is
S
Ni
H= -i
D
Using the Maxwell stress tensor
(e - P)o
e
2
N2 i 2
2
N2i2
2
2
2
D
)(
with i = I
O
+ i1 cos wt = I
o
+ Re i l ejWt
o )2
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC !MEDIA
PROBLEM 11.8 (continued)
We linearize
fe to obtain
N2
-2 (1 -o
-e
) I
[
2
o + 21oi1
­
cos wt]i 2
(g)
For equilibrium
N2
-F o
Thus
F
2
0
2 ((-olo )1
- 2--
12
i2- =
0
0
N2
21
o)I
=
(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
2
36
=
T 1 1 (x=L)
Thus
- Ex
E
0
0
1
3-i2
=
(x 1 = L)
(j)
0
21
From Eq. 11.4.21
(x2 -
T
21
2
2
b2
3
a
(k)
3
1
and from Eq. 11.4.16
61(x1 = 0) = - x 2 (x10
Thus
-(+
)10
= 0
=0
(
INTRODUCTION TO THE 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
+
- Al sin caL
-
- Al cos cL
+
dX
A2 sin cL + A 3 cosh
2
+MwA2+
d
dt
+ A 4 cosh aL
=0
L + A 4 sinh cL
=0
A2 cos at + A 3 sirnaL
2 a 3b3 EDA
DA
- 2
1
V = dt
=0
A3
-
2
3b3EDA
2-= D[N1
[-DI
D o(0) +
2A4
= + N Io
o)
D-ý(0)
v =-N1o( - o)jW(A 2 + A4 ) + N2 i1
or
M
PDjw
We solve Eqs. (m) for A2 and A 4 using Cramer's rule to obtain
A 2 =
~'- )(- + sin aL sinh cL - cos aL cosh cL)
(
N Ioil
- 2Mw (1 + cos aL cosh
N21 il(A
4
=
-
L)+ - (ab)3ED(cos CL sinh cL + sin aL cosh cL)
3
1o)(- 1 - cos %L cosh aL - sin cL sinh cL)
2Mw 2(1 + cos aL cosh cL) + 34 (Cab)3ED(cos cL sinh cL + sin aL cosh cL)
Thus
z(J)
=
v(j
il
+
2
2,0 (i([N
0
o
j(+ 2 + 2 cos cL cosh cL)
2
- 2Mw (1 + cos cL cosh cL)+
(ab)3ED(cos aL sinh aL+ sin aL cosh cL)
+ N2 lDjw
Part c
Z(jw)
has poles when
+ 2M 2(1 + cos aL cosh aL) =
-
3
(Cb)
ED(cos
aL sinh cL + sin aL cosh aL)
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-i)
1
and the H field below is
Ho(a - b) i(a- b +-C)
b
Using the Maxwell stress tensor, the magnetic force on the beam is
T
o
2
(H-
2
a
2
2
2
b
_
)
-
2
-
H
2
o_
4
2
(a-b)
ab
2
2p H
(a- b)
Thus, from Eq. 11.4.26, the equation of motion on the beam is
a 22
t
2
Eb 2
4
3p
x4
1
2
1H0
=
0oHo2
(d)
(a-b)bp
Again, we let
jt
A
E(x t) =
Re
t
(xl)e
(e)
with the boundary conditions
E(xl= L) = 0
ý(xO=0) = 0
6
6 1 (x1=O0) Since 61= - x2 a /ax
-
1
1
(x 1 = L) = 0
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=
3P a4
+
oHo
(a-b)bp
o
see that our solutions are again of the form
Thus we U(x) = A sin ax + B cos ax + C sinh ax + D cosh ax
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.9
(continued)
where now
1/4
S(a-b)bp
2
/Eb
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
T
3
( 3& 4Eb
\
3
2
2
or
w2
2
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
will always be real, and thus stable.
traintimposes aforce
PROBLEMI
w2 > 0,
so W
This is expected as the constant flux cons­
which opposes the motion.
11.10
Part a
We choose a coordinate system as in Fig. 11.4.12, centered at the right end of
d
1
the rod.
Because
-= T
, we can neglect fringing and consider the right end of the
0
D
1
1
, we can assume that the electrical
rod as a capacitor plate. Also, since
force acts only at xl = 0.
-
T 2 1 D dx 2
fe =
+
2
2
(b
x2 - b)
where T2
21
=
2
0
(a)
3
E
a
since the electrical force,
in equilibrium.,
Thus, the boundary conditions at x1 = 0 are
(Eq. 11.4.21)
3
fe, must balance the shear stress T 2 1 to keep the rod
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (continued)
and
T 11(0) =
- x2 E a
(0) =
0
(b)
ax 1
2
At xl = -
since the end of the rod is free of normal stresses.
L, the rod is
clamped so
=
0(-=)
0
(c)
and
an(-)
-
(-) = 0
(d)
We
use
the
Maxwell
stress
tensor to calculate
the electrical force
to be
We use the Maxwell stress tensor to calculate the electrical force to be
(e)
[
2eAV
d
IVs +
d2
'(0)]
V
The equation of motion of the beam is (example 11.4.4)
Eb2
3t
3p
a
ax
0+
(f)
1
We write the solution to Eq. (f) in the form
C(x,t) =
Re E(x)e
(g)
where
C(x) =
Al sin cx + A 2 cos
with
ex + A 3 sinh ax + A4
cosh cx
-C
Applying the four boundary conditions, Eqs. (a),
(b), (c) and (d), we obtain
the equations
- A 3 sinh a
- Al sin cat + A 2 cos at
Al cos at + A2 sin at +
+ A4 cosh cL
cosh at - A 4 sinh ct
A
2
3
3
- b DEA
+
1
2EAV
,
d12
A2
2E AV2
+
2
A
3 b3DEA3
--
+
o
-d ---
o
0
(h)
0
=
+ A4
- A2
2
=
0
2F AV v
A =
4
o
os
d
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (continued)
dq s
Now i
dt
s
q
where
c A(v - V
(V + v ) +
a
o
d-ý(0)
s
2e Av
o
)
d + E(0)
2e AV
o o
s
d
&(0)
d
Therefore
V
2E A
is =
jW
V^s do
o
d
((0)
where
(0) =
A2 + A
We use Cramer's rule to solve Eqs. (h) for A 2 and A 4 to obtain:
- EoAVo s
[cos ctsinh at -sin at cosh at]
dy
A22 = A 4
4 =
~ 2b
3
ft
3DE(l + cos tccoshca)+ 2Eo
3
q
o (cosct sinh at- sin arXcosh aR)
(2)
d3
Thus, from Eq. (k) we obtain
2
=
Z(jO)
2d
jw2E0 A i
3E AVV
+ 3(o
o 3 (cos at sinhat - sin at cosh at)
ED
(1 + cos at cosh at )
d1(C3Lb)
Part b
We define a function g(ak) such that Eq. (m) has a zero when
I
(m)
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (Continued)
3
(aL)
3 g(aL)
3U3V2AE
o o
DEb 3 d3
= (1 + cosh at cos ae)(at)
sin at cosh at - cos ct2 sinh ct2
(n)
Substituting numerical values, we obtain
3£3 V2A
­
-3
o3 o
DEb 3 d3
3 x10l (106)1o0-(8.85x 102)
10- (2.2 x 10)10- 9 10 - 9
In Figure 1, we plot (at) 3g(at) as a function of at.
1.2 x 10 1.2
3
(o)
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
a
0
0.4
i
-iiI
0.8
-I
I
1.2
1.6
Figure 1
"
2.0
at
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (Continued)
From Eq. (g)
=
at
2
k-
1.875
=
Eb t
Solving for w, we obtain
w
1
1080 rad/sec.
Part c
The input impedance of a series LC circuit is
2
1 - LCw
jwC
Thus the impedance has a zero when
2
o LC
We let w = w
+ Aw, and expand
+ j
Z(jw)
(q) in a T;aylor series around wo to obtain
+ 2j LAw
2A- =
(m) can be written in the form
z (jW)
S2jwC [1 - f(w)]
2jwCo
where f(w)
and
C
o
= 1
oA
= ­
d
For small deviations around w0
O
j
2•oW
z (j )
T
O 0Iwo
(r) (s) and (t), we obtain the relations
(q),
Thus, from
2L =
2wC0
awlw
Ot
and
0
o
1
W2L
K
now
f(w) =
where
K 31£E AV 2
3
o o
d 3 (EDb 3 )
=
1.2 x 10-3
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (Continued)
1 + cos at cosh at
sin at cosh at - cos at sinh at
and g(at) =
Thus, we can write
df(w)
K
d
d(ca)
dw
o
Now from
(g),
d(cat)
(a)
3g (at)- dw
o
d(ak)
3p
=
/4~
Io
dw
2o/,
2
0wo
d
FK
d(at)
(at) g (at)
1
- K
[ ( a Z) g ( a Z)]
K da)
K d~ccx)
d
d ( a Z)
WO
[(at) ,g(a)]
(aa)
0
since at w = o
o
(ak)
3g(at)
0b)
K .
=
(aa),
Continuing the differentiating in
d
d(ca)
1
[(a)3(at)c
L
g(at)3(a )
+
we finally obtain
(at
KI
-K
d
g(at
d(aZ) gc
I
0
(at)3 d
d()
K
-3
g(
(cc)
WO
d
d()
d(at) g()
-
=
- sinat coshat + cos aisinh at
(sin aicosh at- cos aisinh at)
(1+cos ctcosh at)(+cosatcoshca+ sinaksinh at+ sin atsinh at(sinat coshat - cos atsinhat)
= - 1
2n(at) (sin aisinh at)
(sin aicosh ca-
cos atsinh at)
cos ti cosh at)
(dd )
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.10 (Continued)
Substituting numerical values into the second term of
value much less than one at w =
(cc),
we find it to have
wo.
Thus,
d
d(at) g(a)
(ee)
-
Thus, using
(y,
(z),(aa)
and (dd), we have
(bb)
df
dw
01
4.8
(ff)
0O
Thus, from
(v)
and (w)
l
4.8 x10
L
C
­
4(1080) (8.85 x10 -)(104)
1
X10
6(102 =
1.25 x109(1080)
6.8
=
X
10
-16
1.25 x 10
farads.
henries
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.11
From Eq. (11.4.29), the equation of motion is
/ 6
1a3
2
a263
3
p
=
a26
+ ax-
3
(a)
We let
6
=
3
Re 6(x )e(t-kx)
(b)
2
Substituting this assumed solution into the equation of motion, we obtain
-pw2 6 =
k2 6
G
+xC
(c)
or
a2 6
+^2
+ (
- k2)6
=
0
(d)
2
If we let 82 =
the solutions for
G
k2
2
(e)
6 are:
6(x ) = A sin
2
8x 2 + B cos 8 x 2
(f)
The boundary conditions are
6(0) = 0
and
(g)
B = 0
This implies that
and that
6(d) = 0
8 d = nn .
Thus, the dispersion relation is
- k2
w
(
)2
(h)
Part b
The sketch of the dispersion relation is identical to that of Fig. 11.4.19.
ever, now the n=0 solution is trivial, as it implies that
6(x ) = 0
2
Thus, there is no principal mode of propagation.
How­
INTRODUCTION TO THE ELECTROMECHANICS OF ELASTIC MEDIA
PROBLEM 11.12
From Eq. (11.4.1), the equation of motion is
a 26
(a)
p -7- = (2G+X)V(V.6) - GVx(Vx6)
We consider motions
6
=
6
e(r,z,t)io
(b)
Thus, the equation of motion reduces to
p
26'.-
7
G
F
2L6
1
+
r
a r6
=
ij
Tr
a
0
(c)
We assume solutions of the form
j
'(r,z,t
) = Re 6(r)e
6
(d)
t-kz)
which, when substituted into the equation of motion, yields
Tr
66 (r )(r + \
r ar
k2) 6(r)
G
=
(e)
0
From page 207 of Ramo, Whinnery and Van Duzer, we recognize solutions to this
equation as
F,,..PW2
6(r) =
A J
2)[
'
- k2
+ BN
- k
r
1
(f)
On page 209 of this reference there are plots of the Bessel functions J and N
. Now, at r = R
We must have B = 0 as at r = 0, N goes to 1
6(R) =
(g)
0
This implies that
If we denote
J
R]
- k2
J(
ai as the zeroes of J
(a )
=
(h)
= 0
, i.e.
0
we have the dispersion relation as
2
G
2 - k2
2
(i)
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.
pLD +(fe)
where (fe)
Thus
= 0 x
(a)
is the upwards vertical component of the electric force
2
2x
(fe)
(b)
LD Solving for the pressure p, we obtain
p=
EoVo
(c)
2x
Part b
d
Because A
small.
<<1,
we approximate the velocity of the piston to be negligibly
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
EoVo2
2
1
2
2x
p
Solving for V , we have
V
p
V=
C0
(e)
xp
Part c
The thrust T on the rocket is then
T =
V dM
p dt
=
V 2 pdD
p
(f)
2dD
x
PROBLEM 12.2
Part a
The forces on the movable piston must sum to zero.
pwD - fe
=
Thus
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= 2
w
(b)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.2 (Continued)
Therefore
(c)
---
p=
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
2
PV
=
or
V
=
T
=
p
(d)
r1­
(e)
W
Part b
The thrust T is
dM
-
V
Vo2
V pdW =
I
2d
d
(f)
Part c
For I = 10 3 A
d =
.lm
im
w=
p
3
=
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, is
1
I2
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
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.3 (Continued)
V
I
(b)
at each orifice.
Part b
The thrust is
T=
dM
2V
2V2 pdw
=
(c)
2
21
0
T =
I d
(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
-
a + p[V(L)] b =
V
<+
L.
P a - P(L)b + F
(a)
where F is the x i 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
(b)
l
Bernoulli's equation gives us
1 pV22 + P =
2
o
0
Solving
1 pV aV2+ P(L)
2
obj
(c)
(c) for P(L), and then substituting this result and that of
(b) into
(a), we finally obtain
F =
P (b-a) + pV2
o
0
(-a +
2
2b
)
(d)
The problem asked for the force
on the lower wall, which is just the
negative of F.
Thus
Fwall
=
- Po(b-a)o
pV
2
(-a +
2b
b
)
(e)
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.
dipole fields.
We expect similar results here.
The solutions are then
As in Eqs. (12.2.1 - 12.2.3), we
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.5 (continued)
define
V =
- Vý
and since
Vv =
then
0
2
V 0 =
0.
Using our experience from the electromagnetic field problems, we guess a solution
of the form
A
+ Br cos
-r cos O
=
9
Then
A
(T
A
cos 9r - B
sin 9 + B sin 9)T­
Now, as r *-
V =
Vo(cos Gi
=
V
- i
sin 9)
Therefore
B
=
-V
The other boundary condition at r = a is that
Vr(r=a)
=
0
Thus
A
=
B a
2
=
V
cos(
2
-Va
Therefore
V
-
o
a
-)i
r
_ - V
r
2
a
sin 9 ( 1 + a)
Part b
vs
--
;10
i
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.5 (continued)
Part c
Using Bernoulli's law, we have
1p2
o =
2 po
1
2
a'
a
r
2a2
2+ Vo (1 + r--
2a
a-cos 20) + P
Therefore the pressure is
P =
pO
0
1 2
pV
2
o
1 cos 2 9)
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.
I pv(v'n)da =
-I
We write the momentum theorem as
Pd7a + F
where F is the force which the cylinder exerts on the fluid.
However, with our
surface far away from the cylinder
V=
Vi
S1
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.
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
r
Theta is measured clockwise from 'the x
=
(a)
axis.
Thus
V V
=
As r
+
2B cos
+ r
A cos
B
A + -)
sin
(b
(b:)
oo
V
Vo(i cos 8 -
4
A = - V
Therefore
i
sin 8)
:(c)
(d)
0
At r = -a
V (a) =
Thus
2B
--
3
-a
0
:(e)
=A= -V
Va
or
B
O
3
o
(f)
2
Therefore
3
V=
with
a3
V(l r
o
r =V
x
2
2
+x
1
)cos
a
Or - Vo( + -)3
2r
r
o
sin
(g)
(g)
0i 2
+x
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 cos 2 0
+ sin 2
(1
)2
(h)
Part c
We realize that the pressure force acts normal to the sphere in the - i
direction.
r
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.6
(continued)
at r = a
e
2
9 PV 2 sin
8
o
We see that the magnitude of p remains unchanged if, for any value of
at the pressure at e + r.
Thus, by the symmetry, the force in the x
6, we look
direction is
zero,
= f
0.
PROBLEM 12.7
'art a
We are given the potential of the velocity field as
Vo
V
xoo
a
xI
.
a
(X2 1 +
1 2)
If we sketch the equipotential lines in the x x plane, we know that the velocity dis­
1 2
tribution will cross these lines at right angles, in the direction of decreasing
potential.
Part b
-
dv
a =
=
+ (vV)v
at
(xi
=V
a
av
=
dt
+ x i
(a)
2
a)\ rfir
(b)
where
r
=
2
/x
1
+ x2
and
2
i
r
is a unit vector in the radial direction.
Part c
This flow could represent a fluid impinging normally on a flat plate, located
along the line
+ x
x
1
=
2
0.
See sketches on next page.
PROBLEM 12.8
Part a
Given that
x2
v = i V
x1
1 oa
+ Tiv
2 oa
dv
av
T
+
(a)
we have that
a
=
ax
1
(v-V)v
+ ,v
v
2.
(b)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
Acceleration V2
a=
o
(-)
a
x2
ri
r
yr
xl
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.8 (Continued)
Thus
2
a=
v
-
,
i
O T2
xi
0(
1
2
(Vc)
2 2
Part b
Using Bernoulli's law, we have
)
p
1
S=
o 2 (d)
pX2+ +
(xv
+2)
(x2
V2
1
2P
o
where 2
x+x
r=
(e)
2r 0 a2
1
2
PROBLEM 12.9
Part a
The addition of a gravitational force will not change the velocity from that
Only the pressure will change.
of Problem 12.8.
-
i
v
1 a
Therefore,
v
v
x
+i
2
-- x
2a
(a)
1
Part b
The boundary conditions at the walls are that the normal component of the velocity
Consider first the wall
must be zero at the walls.
- x
x
=
(b)
0 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
(c)
1i, v
o
- (x - x)
Then ---
v*n
2
1
a
=
(d)
0 Thus, the boundary condition is satisfied along this wall.
Similarly, along the wall
x
2
n =
and
0
(e)
1 2 + i,
(f)
+x
--
v*n =
=
1
o
a
(x + x 2 ) =
0 Thus, the boundary condition is satisfied here.
x
2
2
n =
2
1
a2
x 2i2 - xi
(g)
Along the parabolic wall
(h)
i
(Ci)
ELECTROMECHANICS OF INCOMPRESSIBLE,
INVISCID FLUIDS
PROBLEM 12.9 (Continued)
v
-
a
0n
(x x
- xx )
1 2
1 2
=
0
(j)
Thus, we have shown that along all the walls, the fluid flows purely tangential to
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
=
ax
0
x=O
- =0
= 0
(b)
y=O
To aid in the sketch of J(x,y), we realize that since at the boundary the velocity
must be purely tangential, the potential lines must come in normal to the walls.
Part b
For the fluid to be irrotational and incompressible, the potential must obey
ELECTROIECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.10 (Continued)
Laplace's equaiion
V2 0 =
0
(c)
From our sketch of part (a), and from the boundary conditions, we guess a solution
of the form
V
Vv
where -
o
2
ao (x
S=
(d)
y2)
is 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
(xi x - yi )
(e)
y
Using Bernoulli's equation, we obtain
V2
=
p
p + 2
(X2 +y
0
2
)
(f)
The net force on the wall between x=c and x=d is
x=d
z=w
I
f
z=O
f
(p - p)dxdz i
(g)
X=c
where w is the depth of the wall.
Thus
v
=
d
+
x 2dx
w
i
y
6
V
=
+
2
C
w (d'
-
c')f
(h)
Y
6
Part d
The acceleration is
V
a =
(v'V)v =
a =
4
2 -a
V
x(2
a
V
ix) - 2
a
V
y (-2
a
-- y).
y
or
(xi + yi)
(i)
or in cylindrical coordinates
= 4
4(V
a a=
-- rir
ri r
Va
(j))
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
+ P
SpVo
=
2
2
P = Po
2
[Vx(x)]
+
P
2 pV2o (1 + hh
Part c
= 0 to obtain
We linearize P around
Po - pV 2
P
o
oh
Thus
T
z
- P + P
o
=
pV2o
oh
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.11 (continued)
Thus
Tz =
with
CE
(g)
p V o
h
C =
Part d
We can write the equations of motoion of the membrane as
a2
m 3t2
=
Sa2
ax2
+ T z
(h)
=
s
+ CE (i)
=
Re ' j e (wt-kx) 2
We assume
E(x,t)
(j)
Solving for the dispersion relation, we obtain
-
m02 =
- Sk2 + C
or
(k)
2
k2 - T]
W -
M
m
=
Now, since the membrane is fixed at x = 0 and x = L, we know that
k =
ni
n =
1,2,3, ..... (m)
Now if
S(
)
- C <0 (n)
we realize that the membrane will become unstable.
So for
PV2
<-S () 2
(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.
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.12
Part a
We wish to write the equation of motion for the membrane.
325
m
m
=
3t2
a 25
S
+ p ()-po
1
ax
E
oS Vo
2
= --2 d- - ) %
2
where
Tee
o
0"
Te
+ T
V
o
-
(a)
(a)
mg
ý-)
2 dd (1 +
is the electric force per unit area on the membrane.
C(x,t) = 0, we must have
In the equilibrium
V2
E
-p (0)
=
Po -
(d)
+
mg
(b)
As in example 12.1.3
P
=
-pgy
+
C
(b), we obtain
and, using the boundary condition of
- pgy
p1 =
+
Eo
mg + Po -
2
Vo
(c)
(d)
Part b
We are interested in calculating the perturbations in p, for small deflections of
From Bernoulli's law, a constant of motion of the fluid is D, where
the membrane.
D equals
E
D =
mg
U2+
--
V
2
(d)
0
For small perturbations ix,t), the velocity in the region 0 < x<
L is
Ud
d+E
We use Bernoulli's law to write
2
21
+
+
(
P
=
g
(e)
D
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 is
2
S
a
-=
m)t 2
2-
+
a2
d
pg +
We define
VO2
2
d--­
d
d3
and assume solutions .of the form
C =-
pg +
E(x,t) =
Re
+
ej
(w t - k
pg
x)
EV
E
Cg)
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
) , then w will be imaginary, and the system is unstable.
For C > S(
Part c
The dispersion relation is thus
Cs
Sf mm
Consider first C <
2 2S
0
for
1 W
C>0
lex w for
real k
_·
~I~~
_I__·__
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.12 (Continued)
Part d
Since the membrane is not moving, one wave propagates upstream and the other
Thus, to find the solution we need two boundary conditions,
propagates downstream.
If, however, both waves had propagated down­
one upstream and one downstream.
stream, then causality does not allow us to apply a downstream boundary condition.
This is not the case here.
PROBLEM 12.13
Part a
V-v = 0, in the region 0<
Since
V
v
x
d
01
o d+~l-
x<
(a)
d
o
where d is the spacing between membranes.
pressure p
2 )]
(ý1 -
V
2
L,
Using Bernoulli's law, we can find the
right below membrane 1, and pressure p 2 right above membrane 2.
Thus
1
2
1
2
2 p
SpVo + p
+ p
(b)
and
1
SpV
1
+ p
pv
2
+ p
(c)
Thus
=
p
p2
Po
2
0
+
(d)
d
We may now write the equations of motion of the membranes as
a2
1
a2
O
m
(pS1 - po) =
+
S
=
SX2
22 2
+ P.- P 2
= S
V
+
x
2
2
t2
a2 1
a2
St2
S
am
2
& )
d1 _pVO(-
C
(e)
(e)
2)
df)
Assume solutions of the form
S j(wt-kx)
1
=
Re
2
=
Re 52 e
l eiwt-kx) (g)
Substitution of these assumed solutions into our equations of motion will yield the
dispersion relation
V2
=
=
m
2
-Sk
SV
S-Sk2c
= - Sk2
+
(-
2)
2
(h)
+ PV-­
d
These equations may be rewritten as
2
1
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.13 (Continued)
2 + Sk2
[
L
+
0O
2
a
d
2[
V
2
dj
Vm
+ Sk
2
o
2d
-
]
m
0
=
=
0
d
For non-trivial solution, the determinant of coefficients of 5
zero.
2
Thus
aO2
aw
m
2
pV 2
od
2
+ Sk
2
must be
2
Od)
or
and 5
V2­
V2
+ Sk2
1
dOQ
pV 2
+
(k)
d
If we take the upper sign (+) on the right-hand side of the above equation, we obtain
2pV 2 !/2
S=
S
We see that if V
V2
o
k
2
)
-2Vd]
is large enough, w can be imaginary.
This can happen when
Sk2d
2p
(m)
Since the membranes are fixed at x=O and x=L
k =
n =
(n)
1,2,3, .....
So the membranes first become unstable when
2
S( -) d
V 2
o
L
(0)
2P
For this choice of sign (+),
=
-
2
so we
w
excite the odd mode. If we had
taken the negative sign, then the even mode would be excited
E1
= E2
However, the dispersion relation is then
w
m
and then we would have no instability.
Part b
The odd mode is unstable.
.
----- i
•--­
or
ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUIDS
PROBLEM 12.14
Part a
The force equation in the y direction is
3y
=
- pg
(a)
)y
Thus
(b)
- Pg(y-0)
p =
where we have used the fact that at y= 5,
the pressure is zero.
Part b
V*v = 0 implies
av
ax
3v
x +-
0
3y
(c)
Integrating with respect to y, we obtain
av
v
y
x y
+
ax
(d)
C
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
Part c
(e)
x (y+a)
v
y
ax
The x-component of the force equation is
av
x 3
a
at
or
av
x
at
(f)
=
x
3-x
x
(g)
=-gx
Part d
At y = 6,
v
at
y
Thus, from part (b), at y =
(h)
3v
= -
at
3x
x (C+a)
(i)
However, since C << a, and v and v are small perturbation quantities, we can
x
y
approximately write
av
at
a ax
ax
Part e
Our equations of motion are now
(j)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.14 (Continued)
av
-=a
.
at
and
ax
x
3v
x
__
xf
S- g
If we take
(k) and 3/at of
3/ax of
32 V
x
---
ag
(R) and then simplify, we obtain
x
We recognize this as the wave equation for gravity waves, with phase velocity
/ag
V
P
PROBLEM 12.15
Part a
As shown in Fig. 12P.15b, the H field is in the - i
I,
I
I
Insl
=
direction with magnitude:
0o
of integration (a)
for the MST
2N
If we integrate the MST along the surface defined in the above figure, the only
contribution will be along surface (1),
n=
1
2 'o
s7 220o0
=
-77
so we obtain for the normal traction
1
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.
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.15 (continued)
1 olo 2
0o+ 8T
Thus
Pint
(d)
Within the fluid, the pressure p must obey the relation
=
or
p =
-
pg
(e)
- pgz + C
(f)
Let us look at the point z = z , r= R .
0O
02
-
p =
pgzo + C =
1 Voo 0
+
P
There
(g)
0
2
Therefore C =
pgzo +
p
+
-
(h)
-~
o
Now let's look at any point on the interface with coordinates zs, r
s
Then, by Bernoulli's law,
P
+
1
1
1olo
2
+
PR
oo
1
pgzo
1
olo
i
-
+
2
Po
+
s
o
Thus, the equation of the surface is
pgzs+
2
-8
pgzs (i)
02
1
(J)
pgzo +
=
s
0
Part c
The total volume of the fluid is
[Ro 2
V =
-
( )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.
R
Thus V
=
r [R
-
1)2 a =
2
o
oo
o0
2
(+
10
2 !r
rdrdz r=r
0
(a)
z=O
where
r
is that value of r when z = o, or
o
2
1
r
(m
pgz +
Evaluating this
oo0
integral, and equating to V, will determine
zo0
OF INCOMPRESSIBLE,
ELECTRORECHANICS
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
Here
V =
V
+ vB) =
wd
(-
y
(b)
-
y
IR + V
(c)
o
vBw - V
Thus
O
I =
w
R +
Zda
The electric power out is
= VI=
P
(IR + Vo)I
v Bw - Vo
= V + R(vBw
.... - V)
From equations
(e)
(12.2.23 - 12.2.25)
we have
Ap = p(0) - p(a) =
M
B
Thus, the mechanical power in is
Bw(vBw-V )v
P
Mi
=
(Apwd)v =
(g)
w
R + ida
Plots of PE and PM versus v specify the operating regions of the MHD machine.
P>e
e
PM
>
<
0
e
e
0P e <
0
0
M>
0 PM <
0
Generator
Brake
Pump
P >
e
0
>
0
SPM
Generator
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.17
Part a
The mechanical power input is
Jdf
L
PM
(a)
Vpv 0 dxdydz
z=O y=0 x=O
The force equation in the steady state is
- Vp + fe = 0
(b)
fe =
(c)
where
Thus
- J B
yo
J YB
PM= Lf I
(d),
dxdydz
z=0 y=O x=O
Now
0oo
y
y
(-
+ vB)
=a(E
J
y
(e)
+ vB )
oo
Integrating, we obtain
ov2 B Lwd - oB v VLd
P=
P
0o
V2
VV
oc
oc
R
Ri
Part b
1
R.
V)V
(Voc
oc
P
out
=
Defining
(f)
oo
out
P
we have
(V
oc
- 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
V =
V
Voc
oc
2(l+a)
maximizes P
out
For this value of V, n equals
11
Plottlng
1
1
(h)
2 (1+2a)
1
n vs. -- gives
a
.5
1
a
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
Voc I
V =
(i)
+ala
We choose the negative sign, since V< Voc for generator operation.
+ 2a - 2
r =
Plotting
n vs. aa ,9 we
­
(j)
a(l+a)
obtain
3
2
4
5
6
7 8
910
PROBLEM 12.18
From Fig. 12P.18, we have
E
-
w
i
and
J=
([
i
y
y
V
+ vB] =
w
I -­
--
LD
i
y
The z component of the force equation is
3z
or
LD
v
IB
Ap = Pi - P=
D =
Solving for v, we obtain
v =
IB
D(1
)v
We thus obtain
p(l
)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.18 (Continued)
Thus, we have
I
LDa
or
V
- + B(1
w
IB
)v
DApo
o
2VW
w +
V =I
LD(
(f)
- v Bw
(g)
DAp
Thus, for our equivalent circuit
w
vowB 2
t
O
i
and
DApo
LDo
V
oc
=
- v wB
(i)
o
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 =
o
Within the fluid
(a)
d
IL
J =
Zd
VL
(-
=
w
+ v
H )Ti
00
(b)
Z
Simplifying, we obtain
Si
v
L [d
d
N
Ni
NV
S=
Ova NI
(c)
L
0o
d
w
For VL to be independent of IL, we must have
VpoNL
d
or
N
L
=
d
1
9d
(d)
1
Zavy 0
(e)
PROBLEM 12.20
We define coordinate systems as shown below.
MHD #2.
MHD # 1
ELECTROMILCHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.20 (Continued)
Now, since
V*v = 0, we have
=
vwd
vwd
2 2 2
1 1 1
In system (2),
I
V
=
J2
-
y2
AP2
= p(0)
+
v B)i
W2
2 d2
2
(a)
y2
2
and
I2B
-
p(_)
(b)
d
In system (1),
=
1
i
I
--y 1 id
V
-w
(
and
Ap
0
p( +)
=
-
) =
p(Y'
v B)
(c)
IB
d
(d)
= 0
By applying Bernoulli's law at the points x
xt =
(right before'IdD system 1) and at
+ (right after MHD system 1), we obtain
1 p2
or
+ p
(0)
p(0_) =
=
I
v2
+ p(L
(e)
p (+)
(f)
Similarly on MHD system (2):
p 2 (0_)
Now,
Now
=
P2 (2+)
=
Vp'd
(g)
0
C
Applying this relation to a closed contour which follows the shape of the channel,
x
we obtain
x =O
C
1
x =£
1
+
p (z1
)-
From
2
x =
p1(+
- p2 ( 2+)
2
+
) + p2
2+
(2­
(h)
(f) and (g) we reduce this to
bp
or
1 (0)
-
Vp-dZ
x =O
+ p2(O_)-
= 0
1
2-
Vpdt +
1+
p (0+)
- P2(O+) +
x
2
-
VpdZ +
V p-dZ +
Vp-dZ
x =
0
2
II
12
d
1
+p A
=
-I 1
d
2
0
(i)
(1)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.20 (Continued)
Thus, we may express v
as
I
- +
S(+
(k)
We substitute this into our original equation for J
(a),
to obtain
2
I
V
ad
22
w
wd
I
wd
£do
2
2
()
w
12
This may be rewritten as
w
V
2
= -
The Thevenin
wd
2 a
d
2* 2
2
d
11
2
R
eq
1V
oc
12
equivalent circuit is:
2
V
d
1 1
wkd d
2
v2
+
where
V
oc
=-
1
V
d
1
and
R
2
eq
_l
jd2
2
__
2 "d2
Z J
PROBLEM 12.21
For the MHD system
II
V
a1--V
-
and
H)
(a)
IpoH
ap
=
P1 -
P2 =
(b)
+
Now, since
IVp'dk
=
0
(c)
C
we must have
Ap =
kv =
poH
oo
LO(
D
VoH )
oo
(d)
Solving for v, we obtain
v =
POH LaV
D°
V0
D[k+(poHi ) 2 Lo]
00
(e)
ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUIDS
PROBLEM 12.22
Part a
We assume that the fluid flows in the +x
direction with -velocity v.
Thus
S
J•=
i
I
V
d
o(
3
+
VoHo) i
3
00
(a)
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
pIpoHo
L
p
-ax
L w pg
.(b)
= 0
-Thus
p=
For Ap =
Then
-
x
+ pg
Lw
p(O) - p(L) = 0
ILIoVH
o
Lw
For the external circuit shown,
V
-IR
+ V
0
Solving for I we get
V
T
+
OLw
v+ ooH
0 R
- pgLw
oPH
d
Solving for the velocity, v, we get
gL•
+
H 0
R
o
d
d
~oo
For v > 0, then
V <P
0
oH o
+
RL
w)
Part b
If the product V I > 0, then we are supplying electrical power to the fluid.
part (a),
(f) and (h), Vo is always negative, but so is I.
V I is positive.
From
So the product
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.23
Since the electrodes are short-circuited,
=
z Zd
In the upper reservoir
J
i
pi
=
OvB i
o z
(a)
(b)
Po + pg(hI - y) while in the lower reservoir
P2 =
Po + pg(h
2
-
(c)
y)
The pressure drop within the MHD system is
Ap =
IB
p(O) - p(R) =
(d)
-
Integrating along the closed contour from y=h through the duct to y=h
1
then back to y=h
Vp-d
Thus
2
, and
we obtain
=
pg(h
I
=
0
1
-
- pg(h
- h )+
2
1
-
d
h )d
(e)
(f)
2
B
BR
and so
and so
pg(h
I
atdB
1
-
o9 B
o
h )
2
(g)
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, w
e have
vA =
h
v
e
wD (b)
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.
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
IR
y=
LD
L(- + v P H
w
eoo
0
Solving
(e)
)
for I, we obtain
Ho
ve po
1=
1
-
(f)
1 + ­R
-L
aLoD w
Vp'dZ = 0, we have
Now, since
Ap -
pgh = 0
Thus, using (d),
(g)
(f) and (g),
- pgh +
we obtain
1
Do
ve
0
(h)
Using (b), we finally obtain
dh
dt
(i)
+ sh = 0
where
D
pg w
h=
1
, until time
10 e
Numerically
s =
RD
7.1 x 10
-3
T, when the valve closes
at h =
5.
, thus
T
u
()
100 seconds.
For our approximations to be valid, we must have
vh
pg
<<
(k)
or
s 2h <<g.
Also, we must have
2
h2
<< I pgh
or
s2 h <<g
1
(a)
Our other approximation was
pL
<
which implies from (f) that
(m)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.24 (continued)
(o
<<
psL
Ho) 2
0o0
D
L
(n)
+
i
Substituting numerical values, we see that our approximations are all reasonable.
Part b
From (b) and (f)
pH
oo
I- =
A
wd
-
until t =
I =
0.
h
]DIt
650 x 10
e
- s
t
100 seconds, where I =
amperes.
-325 x 103
amperes.
Once the valve is closed,
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.25
Part a
Within the MHD system
-1i
LD
J =
(-
= -
p(O) -
Ap =
and
3
V
iHo)iH
w
p(-L ) =
where V = -iR + V
0
3
00
D''D
We are considering static conditions (v=0) so the pressure in tank 1 is
p1 =
-pg(x2 - h ) + po
and in tank 2 is
- h ) + Po
- pg(x
=
2
2is
2 the atmospheric
pressure,
p
where po is the atmospheric pressure,
thus
V
i =
(e)
o
1
R
­
w[
w
aL D
Now since jVp.dk = 0, we must have
+
io H
C
- pgh2
+
+ pgh
we obtain
Solving in terms of V
0
- h )wD
pg(h
21
oo
For h
2
= .5 and h
1
(f)
= 0
ar
1
= .4 and substituting for the given values of the parameters,
we obtain
= 6.3 millivolts
V
Under these static conditions, the current delivered is
pg(h 2 -hl)D
i =
=
o
210 amperes
o
and the power
delivered is
2
g(h2
o = LP
eP =
V
e
H
Hh
1
1
R]
wL D+ w
w
w
1.33
watts
Part b
and h2 around their equilibrium values h10 and h20 to obtain
We expand h
h
h
=
1
+Ah
10
h2 = h2
2
200
1
+ Ah a
2
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.25 (Continued)
Since the total volume of the fluid remains constant
Ah
= - Ah
2
1
Since we are neglecting the acceleration in the storage tanks, we may still write
PI=
- pg(x
1
- h ) + po
1
2
(h)
P2
= 2g(x2
- h2) + PO
-
Within the MHD section, the force equation is
av
S-at
= - VpMD
V MHD +
+
i1oHo
L(D
LID
Integrating with respect to x
(i)
, we obtain
LD
p(O) - p(-L) =
APMHD =
-o
pL
()
l
The pressure drop over the rest of the pipe is
APpipe =
Again, since
2
dv
dt
Vp-dk = 0, we have
C
pg(h
- h2 ) +
pipe =
pMHD +
0
(k)
For t > 0 we have
2V
w
V
1
cL D
0oHo
R
w
()
and substituting into the above equation, we obtain
pg(h- h)
w
ýv
- p(L+ý) ~
V11 0H
1ol)t0
o
1
o 0
(m)
[L 1D + ;i-
Ah .
We desire an equation just in
From the V-v = 0, we obtain
2
dAh
vwD =
2
A
(n)
Mlaking these substitutions, the resultant equation of motion is
d 2 Ah2
dt2
(o Ho)2
1
dAh
2gwdAh 2
dt
(L + L2)A
R
V •pH
p(L +L2 )A
(o)
-•D +
58
ELECTROMECHANICS OF INCOMPRESSIBLE,
INVISCID FLUIDS
PROBLEM 12.25 (continued)
Solving, we obtain
V0
Ah
st
and B
1
+ B e
1
D+ R
2pgwd
2
where B
H
=
s 2tt
+ B e
2
(p)
are arbitrary constants to be determined by initial conditions
2
and
[(H2]
S
-
2
'
gwd
L))
2
1
2
H
o
L +L
+pL
w
aL D
21
D
D1 1
ID
(q)
(L + L )A
2
R
w
Substituting values, we obtain approximately
-1
s
=
-
.025
sec.
=
-
.94
sec.
1
s
2
The initial conditions are
Ah (t=O)
and
dAh
2
dt
=
0
(t=0)
=
0
Thus, solving for B
B
and B 2 we have
=
1
1
=
w
R
2pgwD[ 11
1 D
000
2pgwD[
(r)
.051
2
- VooH
B
-
+ ---l
R
LID w
(R1
3
= + 1.36
_
x
10
sI)
Thus
h (t) =
h
2
From (£)
20
+ Ah (t)
=
2
.55 + 1.36 x 10
3
e
.94t
-
.051e
-025t
(s)
we have
2Vo
i
w
R
R
VoH
1
1
(t)
w +aL
D
Substituting numerical values, we obtain
st
i = 420 - 2.08 x 10
s
(Bzsze
= 420 - 268 (e-* 025t
- e94t)
s2
+ B2 s 2 e
)
(u)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.25 (continued)
11h_
(t)
2
It
i
2
1
4
5
50
100
150
200
i(t)
420 -
----
210 -
1
I
I
I
1
2
3
4
LI/
I
5
-
1
50
--
100
--
1
150
I-.---------­
I
1
200
250
L
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:
32Ah
2
<<
1
gh
2
and
+
ý-t
(v*V)vds
L
transition
region
Substituting values, we find the first ratio to be about .001, so there our approx­
imation is good to about .1%.
.3
/A-
.15
2
Pd
L
In the second approximation
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
d
(a)
2
We integrate Newton's law across the length £ to obtain
av
p
3t
i
Apo
=
p(O)
- p(X)
+ JBZ =
-
v
v +
d
B
(b)
i2
v + Np
d2
Ap
v
0
Thus
Ap
+
Nji
v
2
d Pz
pzv °
Tt
Nil
12 sin 2
t
2d 2Pt
12 (1 - cos
2 wt)
(c)
Solving, we obtain
v=
ov cos 2wt + 2w sin 2w()
\P
ot
0Ap
t o
Npo12
0
Ap
2
Ao
Part b
(d)
+
4 W2
\ pvo
The ratio R of ac to dc velocity components is:
R =
/v
[
Po
(e)
+
4W2
/2
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.27
Part a
The magnetic field in generator (1) is upward, with magnitude
Ni p o
Nmip o
B
=
I
a
a
and in generator (2) upward with magnitude
Nilpo
B
Ni2
m=io
a
2
+
o
a
We define the voltages V I and V 2 across the terminals of the generators.
Applying Kirchoff's voltage law around the loops of wire with currents i
1
and i
2
we have
d
V
1
and
dX
+ Nm d t
+N
dt
dX1
Nm j
dX2
V + N 2
dt
+ i RL
0
+ i 2RL=
0
where
A
=
A
=
1
2
B wb
B,1b
B wb
2
From conservation of current we have
and
i,
VI
abo
w
I
1 ++ VB
VB2
i2
V 2
=
-
-w
aba
+ VB
2
Combining these relations, we obtain
wbp
di
w
(N + N)+
i
w+
m
dt
a
ab
wbp
(N2 + N2
m
)
W
a
di 2
di 2 + i
a
ab
dt
+
pw
N]
+
a
w
a
VN i
m 2
N wVi
a
= 0
1
Part b
We combine these two first-order differential equations to obtain one second-
order equation.
di
+a
a1
UC
di2
L
.)
,1,UC +ai.~
whe re
a
=
1
N2)Wbo] 2
(N2+ WmVJ
a
wN Vp
L=
',
0
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.27 (continued)
a
=
a
=
/NNV
2 ýb-w + R2abo
VN
3
0
tN a
2
2)
+ N )
mV
oiw
a
If we assume solutions of the form
i
2
=
Aest
(9)
Then we must have
as
1
2
+ aas + a
2
3
- -Ia2
a
2
2
= 0
(m)
4a a
­
1
3
2 a
For the generators to be stable, the real part of s must be negative.
Thus
a
2
>
0
for stability
which implies the condition for stability is
NV
Sw
Part c
When a
ab
+
>
+
=Sab
a
(n)
= 0
2
wo Nv
aba
L
a
(o)
then s is purely imaginary, so the system will operate in the sinusoidal steady state.
Then
a
-a
"
N
m
V
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
=
8
or
f
v"
80 Hz.
500 rad/sec.
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.28
Part a
The magnetic field within the generator is
po N i
B=
--
i
w
2
The current through the generator is
J
i
+
VB)T
(
Solving
thefor
voltage across the channel we obtain
v,
Solving for v, the voltage across the channel, we obtain
D
v
VJ°Nw Gx
D
i
w
We apply Faraday's law around the electrical circuit to obtain
1p N
1
v + L
idt +
iR
d
\
N'
'd
We assume that i =
this equation we finally obtain
_ pNDV \
n
/2Rw
- ­
dt
w
Differentiating and simplifying
Ai+2
dt2
diji
= -
J
A2
2
+2 ,LT..
LJ
..
,.+
d
?-TOA(
dC
N
i
0
=
st
Re I es
Substituting this assumed solution back into the differential equation, we obtain
Sw
s
D
oNDV
-
+
+
oN d
oLw
w
)
+
w
PoNz£dC
=
0
Solving, we have
iRLw
S
NDV'
D-
S+ N Z­
2
D
OLw
4
o NDV )2
w
w
SN LdC
o
For the device to be a pure ac generator, we must have that s is purely imaginary, or
O
S(
L=
NDV
•w
D)
N2d
YLw
w
Part b
The frequency of operation is then
w
pN £dC
PROBLEM 12.29
Part a
The current within the MHD generator is
J =
- i
Ld y
YiV
(-+vB=
0
)1
w
o y
(h)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.29 (continued)
The pressure drop along the channel is
where V is the voltage across the channel.
iB
0
+
av
p at
Ap = Pi - Po
(b)
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
(c)
0
V + 2iR =
Since the pressure drop is maintained constant, we solve for v to obtain
2aR + _
dt !v
B at
Ud
w
In the state
steady
In the steady state
i =
=
d
+ 2R
w
Ap
B
Ap
2R
1 + -B
w
aid
v=
and
+ OvB
o
d
Ap
Part b
For t > 0, the differential equation for v is
(F +
Bod 2-
+ avB
=
S
+
R
d
Ap
O
The general solution for v is
where
YR +2)
T =
+o-t/T
d
+
-
V=
d
We evaluate A by realizing that at t = 0, the velocity must be continuous.
Therefore
1
v =
R
+
id
d ap
Rd
+ - -+w B Ap
B
w
Ap (I + pY" Rd
T w B
,Ap(1
+
=
/
-t/)
e
e
- t/T
d
Bo
dBR
A~kl wi.+!1d
PROBLEM 12.30
Part a
The magnetic field in the generator is
J Ni
B
=
d
The current within the generator is
S
=d
0 (
w
+ vB)
(d)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.30 (continued)
The pressure drop in the channel is
where V is the voltage across the channel.
Ap =
)
(1 -
Ap
Pi - Po=
(c)
--
=
o
Applying Faraday's law around the external circuit, we obtain
d(NB~w)
V + i(RL + RC) =
Lw
dt
oN
di
(d)
d Using (a), (b), (c) and (d), the differential equation for i is then
LN 2 dUN]
(PoN 2
N
oN 2 di
RL+ RC
1
d
dt
w•d
d
o
d AP
o
In the steady state, we have
FRL+ RC
2
w
i2 = iN
+
2 oNvo1
1
O-
dAPo
d
(f)
V
The power dissipated in RL is
P =
i 2 RL
1.5 x 10 , then
For P =
i
2
= .6
e
xl0
(amperes)2
Substituting in values for the parameters in (f),
2
.6 x
08 =
we obtain
10 10-6 N - 6.3 x l
(.125 + 2.5
N2
(4 x 10
-
N)40
10g)
)
Rearranging (g), we obtain
N 2 - 102N + 2.04 x 103
or
N =
= 0
75, 27
The most efficient solution is that one which dissipates the least power in the coil's
resistance.
N =
Thus, we choose
27
Part b
Substituting numerical values into (e), using N = 27, we obtain
(6 x 107 )i + 13
(1.27 x 107) di
dt
(h)
= 0
or, rewriting, we have
dt
1.27 xlO'
=
di
i(6 x10 7 -
()
) 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
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 i 2 to obtain
6 x 10
1+(13.3
12
7
-9.4t)
(k)
7.75 X 10
5.5 ,10 3
-4
-
10
I
'25
seconds
t
Part c
For N = 27, in the steady state, we use (f) to write
RL+ RC
P = i2RL
=
1
+w cd
PNvoN
dj dAPoRL
2 ON)
or
P =
a
-
where
2
dA
a
=
2
R+
1
d
N
and dAP
o
2
2
1oNvo
1.47 x 108
VO
1
2.85 x 1O
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.30 (continued)
P
1.5 X10
6
RL
PROBLEM 12.31
Part a
With the switch open, the current through the generator is
= 0 =
_
d
Iy =
(-
w
+ vB
B
)
(a)
y
where V is the voltage across the channel.
In the steady state, the pressure drop
in the channel is
Ap
iB
Pi -o
B = 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
id
2d
- oRL i + avB
w
o
=
and
iB
t
d+-
Ap =
p t
= (e)
v
Ap
(1
)
(f)
AP
(g)
Obtaining an equation in v, we have
v
t
aPo
GB
vwL-o
T
7+
w
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.31 (continued)
Solving for v we obtain
v =
- t/T
Aev-t/'T
+
_
_AP
o
where R i
(APo + Bow
vo
RL+ Ri)
w
atd
and where
Ap
wB
RL + R
'o
at t = 0, the velocity must be continuous.
Therefore,
APo
A
= v. ­
Ap
0o
w B
o+
\vo
RL+ Ri
Now, the current is
wB o v
i=
RL + R i
Thus
1=
(wB
p
RL
o
0 +
v
V
APo
B(1
wB
0
e)
)+e
-v
e-t/T
0i
RL +
0
V
Vo
W"o1
"
Ap
TR.+L)
Vo o
w
!+R.
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.32
The current in the generator is
i
zd
V
w
( Y-
i
=
(a)
vB)
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
pt
Iv
-
=
i
JB
4t
B
(b)
d
or, using (a),
3V
w
dt a
d=
3
3i
Tt
=d
I
+-
B2
dpk
i
(c)
Integrating, we have
00
w
+ B 2w
I
-+ -+
i
dia
dp2
V =
Defining
(d)
idt
w
R. =
1
atzd
C=
p2d 2
wB
i
and
i
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
C d
i =
=
dt
V
- w
ud
+
w
vB
0 0
(a)
or
dV
c
dt
+
o'd V
wC C
adv
oB
C
o
(b)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.33 (Continued)
The solution for V C is
VC =
with
=
T
vBw(l - e
-t/
I )
(c)
wC
, where we have used the initial condition that at t = 0, the vol­
atd
tage cannot change instantaneously across the capacitor.
t +
The energy stored as
m, is
2
(VW
C
Part b
The pressure drop along the fluid is
iBo
2
Ap = = B2v G£e
-t/T
d
0oo (e)
The total energy supplied by the fluid source is
Wf =
Ap vodwdt
-=I(v
B )2 awde- t / T dt - UT(v B )2 Twdet/T
000
I
C(wv B )2 Wf :
(f)
(g)
Part c
We see that the energy supplied by the fluid source is twice that stored in the
The rest of the energy has been dissipated by the conducting fluid.
capacitor.
This
dissipated energy is
(h)
J VC idt 0
wd
=
+ (v0 B )2w(
2
=
Oidw(VBB)
idw(v BBo)2
0
=
aTdw(v B )2 0o
- e-t/T)rdet/T dt
-t/T+ 1 e-2t/T
(i)
2
Therefore
W =
d 1 C(voBow)
oo
2
2
(j)
Thus
fluid
elec
dissipated
As we would expect from conservation of energy.
(k)
ELECTROMECHANICS OF INCOMPRESSIBLE. INVISCID FLUIDS
PROBLEM 12.34
The current through the generator is
= c( - - vB )
i
Zd
w
(a)
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
av
p
(b)
- V(p+U) + J x 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
+ xb )
(c)
Realizing that x = x
a
b
and
(d)
dxa
d
v
dt
We finally obtain
d2X
aB2 2 dx
a+
cB v
a
0
2
pt dt
dt
x
a
=
o
wp
k
(e)
2
We assume the transient solution to be of the form
^ st
xa =
x e
(f)
Substituting into the differential equation, we obtain
aB2
o. s
o
s2+
2g
(g)
0
=
Solving for s, we obtain
2
p
o=B
p9,
2-
2
/B2
2
t-
(h)
-22.
Substituting the given numerical values, we obtain
s=
- 29.4
s2 =
-.665
(i)
In the steady state
aB OV21
x
wp 2g
a
1
.075 meters
(j)
Thus the general solution is of the form
x
a
=
.075 + A es
1
t
+ A eS2t
2
where the initial conditions to solve for A
(k)
and A2 are
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.34 (continued)
x a(t=0) =
a
.0
d~-(t=0)
= 0
.075 s
Thus,
A
A2 =
.075 s
1
1
.0765
cad A
2
_
=
-
=
.00174
Thus, we have:
.075 +
=
x
4
.00174e 29. t -. 0765e
-
.665t
a
Xa
1
2
3
Now the current is
V
Z do( V -
i =
B
dx
d-)
0 ft)
dtt
- Bo (s I A l esl
t
+ s 2 A2 eS2)]
=
91dY[
=
100 - 2 x 103(sI AeSlt +
=
100(1 + e
2A 2 e
(m)
t
2 ) amperes
)
- e
Sketching, we have
i
100
t
1
2
3
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.35
The currents I
1
and I
2
are determined by the resistance of the fluid between
the electrodes. Thus
I
VooDx
=
I
(a)
w
and
V ODy
I
2
(b)
w
The magnetic field produced by the circuit is
poN-
B=
-
or
-
o
=
(I2 - I )i (c)
N
z VoD(y- x)i 2
(d)
From conservation of mass,
y =
Thus (L - x)
N
-
(e)
Vo oD
2
B
(f)
(L- 2x)1
The momentum equation is
av
at =
(g)
+ J x B
-V(p+U)
Integrating the equation along the conduit's length, we obtain
Now
(h)
-pg(y-x) - J o BL p -at (2L + 2a) =
ax
v =
(i)
at
so we write:
2p(L + a) -!-T +
(oNV
GDL
Pg + J
(2x -
)=
0
(j)
We assume solutions of the form
^
x =
Re x e
st
+
Thus
S 2 + -s+
(L + a)
L
(k)
2 NVo D
J L
o
pw (L + a)
0
(C)
Defining
w22
o
+ +
(L +a)
ODJ- L
pw (L+a)
NV
OOg
(mO
we have our solution in the form
L
A sin w t + B cos w t + 2
o
o
Applying the initial conditions
x
= x(O) =
L
and dx(O)
de
we obtain
x =
2 (1 + cos wot)
=
0
(n)
(0)
(p)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.36
As from Eqs. (12.2.88 - 12.2.91), we assume that
v = iev
0
B=
Biz + TOB 0
JI=
iTJ
rr
+TJ
(a)
z z
E= TE
+ TzE
z
rr
As derived in Sec. 12.2.3, Eq. (12.2.102), we know that the equation governing Alfven
waves is
aZv2
B 2 a2v
0
.p0
(b)
z_ -
For our problem, the boundary conditions are:
at z = 0
E = 0
r
at z = Z
v
=
Re[Ore j
(c)
As in section 12.2.3, we assume
Re[A(r)v
=
v
0
(d)
(z)ejet]
Thus, the pertinent differential equation reduces to
2V
dv
0
2,
+ kv
dz-Vwhere
=
0
(e)
k = W0
The solution is
=
v0
C 1 cos kz + c 2 sin kz
(f)
Imposing the boundary condition at z = 2, we obtain
A(r)[C
cos kZ + C 2 sin k
i]
=
r
(g)
We let
A(r)
(h)
and thus
C1 cos kZ + C2 sin kZ
R =
Now
E =
- VB 0
(i)
(j)
Thus, applying the second boundary condition, we obtain
v (z=0) = 0
or
C
Thus
C
=
(k)
0
OR
2
sin kZ
Now, using the relations
Er =
r
- vOB
0 o
(m)
ELECTROMECHANICS OF INCOMPRESSIBLE,
INVISCID FLUIDS
PROBLEM 12.36 (continued)
E
= 0
aE
(n)
aE
r
z
3B
=
(o)
ar
az
1
at
Be =
J
az
Po
1
3(rB6 )
or
(p)
r
= Jz
ar
(q)
we obtain
v
=
Re
Ssin
B
k.
a
= Re •
(r)
(s)
j]
sin kz ej
Re F 2 2Bo kk
1j
wsin k
=
J
z
ej t]
j
cos kz eil
nrB
k
?rBo k 2
sin k2kt
F
=ejRel
r
sin kz
cos kz e j
(t)
et
(u)
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
az
yz
B
at
(b)
C
Now, since the fluid is perfectly conducting,
E'
or
E
y
=
E + vx
=
vB
xo
B
=
0
(c)
(d)
Substituting, we obtain
3B
tXx
yv
Bo
Bzx
(e)
The x component of the force equation is
av
Pat
where
aT
X
xz
=
az
B
0
(f)
ELECTROMECHANICS OF INCOMPRESSIBLE, INVISCID FLUIDS
PROBLEM 12.37 (continued)
Thus
X
Thus
B
JB
x
O
P at
(h)
a
Bz
0
Eliminating Bx and solving for vx, we obtain
a2v
B2
o
92 v x
x (i)
or eliminating and solving for Hx, we have
32i
2
B2
B
o
x
H
(j)
where
B
(k)
pH
=
Part b
The boundary conditions are
(0)
Re Vejwt
vx(-k,t) =
E (0,t) =
0 +
=
vx(O,t)
0
(m)
We write the solution in the form
v
x
A ej ( wt - k z) + B ej(wt+kz)
=
(n)
where
k"o
Applying the boundary conditions, we obtain
S(,t)
Re I =
x
in k
sin kZ
ejt
(o)
Now
o az
x
at
x(p)
or
- BoVk cos kz
sin
Thus
=Re
Hx =
k2
=J•o
=
Hx
B Vk cos kz
Re
jw0po° in
k
(q)
t
ejet
(r)
Part c
From Maxwell's equations
3Hx
VxH= i
Thus
= J
(s)
Thus
Jjw
Re
[B
BVk2 sin kzR
sin
Rt
e
(t)
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.
ELECTRO:ECHANTCS OF COMPRESSIBLE, INVISCID
FLUIDS
PROBLE4 13.1
In static equilibrium, we have
= 0
-Vp - pgi,
Since p = pRT, (a) may be rewritten as
RT
Solving,
- + pg = 0
dx
we obtain
p = p
-L x
RT
1
a
e
PROBLEM 13.2
Since the pressure is a constant, Eq. (13.2.25) reduces to
Ov dv = dz
JB
(a)
y
where we use the coordinate system defined in Fig. 13P.4.
Now, from Eq. (13.2.21)
we obtain
Jy =
(Ey + vB)
(b)
If the loading factor K, defined by Eq. (13.2.32) is constant, then
Thus, J
y
Then
KvB =
(c)
avB(1-K)
=
pv dv
dv
or
+ E
SP dv
jdz
avB (1-K)
_-
=
(d)
2
(e)
B
GB2(1-K)
_
=
-
o(I-K)
Ai
A(z)
(f)
From conservation of mass, Eq. (13.2.24), we have
PiviAi =
pA(z)v
(g)
Thus
PiviAi
v 2
dv
-a(l-K)Bi Ai
dz =
(h)
Integrating, we obtain
- a(l-K)Bi
9n v =
z + C
(i)
-a
v=
whereZ
d
vie
d
Oi
vi
Pi Vi
a(l-K)B
v = v. at z = 0.
1
(j)
and we evaluate the arbitrary constant b
y realizing that
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.3
Part a
We assume T, Bo, w, a, cp and c
Since the electrodes are short-
E = 0, and so
circuited,
J
are constant.
=
y
vB
(a)
o,
We use the coordinate system defined in Fig. 13P.4.
Applying conservation of energy,
Eq. (13.2.26), we have
v2) = 0, where we have set h = constant.
v
Thus, v is a constant, v = vi.
Conservation of momentum, Eq. (13.2.25), implies
2
- ViBo
io
dz
Thus, p =
(b)
(c)
- viBz + p
(d)
The mechanical equation of state, Eq. (13.1.10) then implies
v.B 2 z +i
V B 2 z
RT
RT
-
=
RT
i
(e)
From conservation of mass, we then obtain
(-
Piviwdi =
+
vi wd(z)
(f)
Thus
d(z)
=
pld
(
(g)
viBoz
Part b
Then
v.B
p(z) =
Pi
2
z
1RT
(h)
RT
i
PROBLEM 13.4
Note:
There are errors in Eqs. (13.2.16) and (13.2.31).
1
d(M 2 )
S dx
They should read:
{(Y-1)(1+y1M2 )E 3 + y[2 +(y-1)M 2 ]v 1B 2 } J3
-M
r
(13.2.16)
and
i2
M2
2 ) d(M
d
)
dx
(1M) )
(1-
[2 +
S
(
[(y-I)(I1+yM2)E
+
3
'2
+
2 }
3
(y-1)M
v1 2]
P
1 2 ypv
(y-1)M2 ]dA
A
dx
(13.2.31)
Part a
We assume that
0, y , Bo, K and.M are constant along the channel.
the corrected form of Eq. (13.2.31), we must have
Then, from
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
2
[(y-l) (l+M2)(-K) + y(2+(y-1)M )]
1-
0 =
-K)
[2 +(y-1)M ]
(a)
Now, using the relations
and
v2
M2 yRT
p
pRT
v
M2
Yp
pv
we write
(b)
B 2 0C(l-K)M 2
Thus, we.obtain
[(y-)(l+yM2 )(-K) + y(2+ (y-1)M2 )]
2 + (y - 1)M 2
1 dA
A2 dz
pVA
(c)
From conservation of mass,
pvA =
(d)
piviAi
A
Using (d), we integrate (c) and solve for
Ai
to obtain
A(z)
Ai
1
1 - ýz
(e)
where
)]B2M2 (1-K)
2
yM12 )(-K) + y(2 +(y-l)M
[(y-1)(1+
Sivi[ 2
+ (y-l)M2 ]
We now substitute into Eq. (13.2.27) to obtain
vB 2 (1-K)o
1
1 dv
S
(-M)
-v
dz
o
[(y-l)(-K) + y]
(-M2)
Yp
1 dA
A dz
(f)
A dz
Thus may be rewritten as
d
=
v dz
(-M
2
)
[(y-l)(-K)
+ y]
iiAi
(1-M2)
p viAi
A
(g)
Ai
Solving, we obtain
£n v =
or
v(z)
- 82
(1 -
=
Vi
where
8 z) + en vi
8 z)2/
(h)
1
(i)
1
i
82 =
n(l -
(-i
2
)
2
(1-K)M 2 [(y-l)(-K) +y]Bo
i vi
01
Now the temperature is related through Eq. (13.2.12), as
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
M2 yRT =
Thus
v2
(j)
2
T(z)i
(k)
From (d), we have
P(z)
Pi
vi Ai
v A
=
(
Thus, from Eq. (13.1.10)
T
Vi Ai
p(z) =
Pi
v
A
Ti
(m)
Since the voltage across the electrodes is constant,
w(
or
w(z)
Thus,
- Kv(z)B
KviBoi
Kv(z)B
Kv(z)B0
=
w(z)
wi
(n)
Vi
wi
v(s)
(o)
vi
v(z)
Then
d(z)
A(z)
di
i
Ai
(q)
()
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-)
Cl-M 2 )
But, from Eq.
--
C(y-l) (-K) + y]viB2
1l
i0
1 ­
(l-K)v. B2 z
=
ov.
i
3z
(s)
2
B
1 0
Pi
Substituting the results
A(z)
1-
Pi
(l-K)
=
1 A
A dz
(13.2.25) we know that
Pi
where
(-K)
yp
_
i
A(z)
of
b),
(-Ka)
pi into
and solvin
for
we obtain
(u)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
and so, from Eq. (13.1.10)
=
T(z)
T.
p(z) Pi
Pi p(z)
(v)
As in (p)
w(z)
wi
vi
=
v(z)
=
1
(w)
Thus
d(z)
di
A(z)
A
=
(x)
Part c
We wish to find the length Z such that
C T(i) +
]2
1 [v
C
p
z
C T(o) +
.9 =
(y)
[v(o)]
For the constant M generator of part (a), we obtain from (i) and (k)
2
1
C
C
[
Ti
p
()
]2
1
Ti
C
p T(z)
i
1 2
Vi
1
221
-28
1
/Bl
9
+1 2
(Z
2i
Reducing, we obtain
-
)
(1 -
26 /8
2
.9
(aa)
Substituting the given numerical values, we have
,1 =
We then solve
2/8
and
.396
(aa) for
2
1
=
- 7.3 x 10
-2
Z, to obtain
% 1.3 meters
k
For the constant v generator of part (b), we obtain from (s),
CT i
pI
+
)
C
v
i
2
(1 -
B )
(u) and (v)
i
+ S= 2
T
p
2
(t),
(bb)
.9
i
or
C T.
C
(1- 8 4/ )
3 1 2
v2
1
2
C T + L v
1
2
p i
=
.9
Substituting the given numerical values, we have
83
(cc)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
= B
and
.45
=
B /B
.857
Solving for Z, we obtain
Z
1.3 meters.
PROBLEM 13.5
We are given the following relations:
E(z)
wi
di
B.
E.
w(z)
d(z)
I
/2
(z)
Ai'
B(z)
0, y, and K are constant.
and that v,
Part a
From Eq.
(13.2.33),
J =
(1-K)ovB
(a)
For constant velocity, conservation of momentum yields
dp
dz
=
(1-K)ovB 2
-
(b)
Conservation of energy yields
pvC
pdz
=
-
K(1-K)G(yB)
2
(c)
Using the equation of state,
=
p
pRT
(d)
we obtain
or
T
Thus,
(-K) (1-K)ovB 2
+
dz
C
p
2
vB (1-K)
=
T
(l-K)
-K
OvBB
R
p dT
dz
T dP
-- +
T dz +
dz
(1-K)OvB 2
R
(f)
(g)
+
(g)
B.i2 (A.)
1
1
A(z)
2
and
.A. =
and
(e)
(e)
+
R
Also
2
p(z)A(z)
11
d =
T d-
Therefore
and
pc
dT d =
p dz
ovB'fdi(1-K)(-2
+ --- )
P(!)
Bp
-K(1-K)av
(i)
Pi
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.5
(continued)
and so
K(1-K)OvB2
dT
i
dz
()
pic p
Therefore
T
2
OvB i
- K(1-K) z + T
PiCp
=
Let
(k)
2
Let
-K(1-K)avBi
=
Pi Cp
(i)
Then
T =
az
C-
Ti
+ 1)
(m)
1
+ ovB (1-K)(
i
S
K
cp
i
R
dz
(n)
pi (az + Ti)
We let
S vB(-K)(
K
1
Cp
R
P.
a
Pi
e
KR
Integrating (n), we then obtain
in p =
8 in(az + Ti)
+ constant
or
p=
=z
i
-
+
1 )
8
(o)
Therefore
A
A(z)
=
()
Ti
Part b
From (m),
T()
at
T(a)
+
T
i
.8
Ti
Ti
or
-l-
.
-. 2
Ti
Now
­ K(1-K)viBi
E
Ti
Pic Ti
But
R Ti
c T
pi
2
=
(1)
Pi
=
y
pi(1­
i
y
=
2.5 x 10 6
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.5 (Continued)
Thus
Thus
- .5(.5)50(700)16
Ti
-2
8.0 x 10-2
-
.7(2.5 X 10 6)
Solving for R,
we obtain
•2x 102
z=
1.25 meters
=
8
Part c
pi
p=
( (Xz
1)
•.+
Numerically
8=
- 1
KR
1
1
(l1)K
6.
%
Thus
p(z)
.7(1 -
=
.08z)
6
Then it follows:
7
Ti(l -
=
T(z)
Pi(1 -
pRT =
p(z) =
7
= 5 x 10S(l - .08z)
.08z)
.08z)
From the given information, we cannot solve for Ti, only for
v
Pi
RT. =
I
Pi
Now
=
2
-i
s
7 x 10
yM
vi
yRT(z)
v2
yp(z)
.5
1 -
.08z
Part d
The total electric power drawn from this generator is
VI =
p
= -
E
Thus
pe
-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
= K(vB )2
w diO (-K)
=
.5(700)216(.5)50(.5)1.25
=
61.3 x 106 watts =
61.3 megawatts
86
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
T (z)
T
(z)
Ti
ST
(1 -
.08z)
1.25
p(z)
p(z) =
p(z)
7ý(1
- .08z)
1.25
p(z) =
5 x 10
5
(1 -
1.25
m
6
(z)
M2 (z) =
1.25
(1-
.5
.08z)
.08z)
7
ELECTROILECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.6
Part a
We are given that
4 Vo
E--i
x3
x
X
Lj/3
and
4
e
Eo VO
9
L'3 x
The force equation in the steady state is
dv
ix
pv
m xx d•
=
Since pe/p
pe E
=
q/m =
constant, we can write
Vd O
2 v2A
dx2Vx
Solving for v
/3
X!3
V4
=
L3
m)
we obtain
x
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
=
total
-
FdVi
x
EV 2
-16 E0 0
=-f
L8
27
-3
x
Adx­
Adx i
x
0
f total
8
=
8
9
oo
2
Ai
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
v
2 a 2(a)
a ax 2
•-22V
3-ý
=
The boundary conditions are
v(-L) =
V
v(O)
0
=
cos Wt
We write the solution in the form
(a)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.7 (continued)
Re[A ej(wt-kx )+ B ej(wt+kx l
v(x t) =
) ] (b)
a
where
a
Using the boundary condition at x 1 = 0, we can alternately write the solution as
Re[A sin kx e j
v =
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
at
-x,
a(e)'
Solving, we obtain
0Vo
oL
s sin kx sin wt =
sinkL
1
x
(f)
1
p V
0 k
k 0 sin
kL cos kx 1 sin wt
or
p
f)
-
=
(g)
where po
0 is the equilibrium density, related to the speed of sound
a, through
Eq. (13.2.83).
Thus, the total pressure is
p Vw
+ k sin kL cos kx
P= + p
and the perturbation pressure at x, =
p'(-L, t) =
sin at
(h)
- L is
00Voa
sin k cos kL sin wt sin kL
(i)
Part b
There will be resonances in the pressure if
sin kL =
0
n = 1,2,3....
n7
kL =
or
(j)
(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= i2J2 (x,
t)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.8
B =
i
[P 00
H +
3
i
H' (x 1 ,t)]
03
Conservation of momentum yields
Dv
D 1 =
Dtax
.x
+J2
o(Ho + H3)
20
1
(a)
0
Conservation of energy gives us
D
p Dt
2
1
(u +
v ) =
1
2
ax
pv
1
(b)
+ J E
22
We use Ampere's and Faraday's laws to obtain
aH'
=
- - JJ
-@
ax
and
(c)
2
P oaH
DE
a2
x
(d)
at
3x,
while
Ohm's law yields
J
Since
=
2
-+
([E
- v B
2
13
i
(e)
oo
E
=
vB
2
(f)
1 3
We linearize, as in Eq. (13.2.91), so E
2
'
v p H
100
O
Substituting into Faraday's law
av,
alH'3
Linearization of the conservation of mass yields
at
=
o
)v
(h)
Thus, from (g)
,H'
3
ooy_
p
o at
at
()
Then
Ho Po
p'
H'3
Linearizing Eq.
(13.2.71), we obtain
=
p
Dt
o
p
D'
Dt
(k)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.8 (continued)
Defining the acoustic speed
as
\ 0o/
where p
is the equilibrium pressure,
(oH
2
= P0o
p
2
we have
p'
=
as22
(p)
Linearization of convervation of momentum (a) yields
av
p o at 1 =
-
9H'
H o(m)
x3 o
oHo
T
3xl
(m
1
or, from (j) and (Z),
Iv,
ao
0
a2
(n)
-s
Po at=
Differentiating (n) with respect to time, and using conservation of mass (h), we
finally obtain
°H
S2v
2V=
2v
( a2
(0 )
Defining
0
1
2
a2
a2
p1 H
+-0
s
we have
(p)
p o
a2 v
3 v
I
1
2(V)
a2V
Part b
We assume solutions of the form
(
Re [A ej (Ct-kx )+ A e j wt+
V =
1
where
k =
1
(r)
a
V(-L,t) =
V
= - L is
cos wt =
s
V Re e jwt
(s)
s
= 0
M dViot) =
p
dt
'A'
+ poH H
oo
1=0
From (h),
)] ­
The boundary condition at x
and at x
kx
2
(j)
and
1-g__ =
a2
s
1
A (t)
1=0
(t),
P
v
at
(o
u)
ax
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.8 (continued)
H'
a
Ho
Po
Thus
dv,(0,t)
dt
M
2
aoH
=A
From (u), we solve for p'j
+
2
az-T-
I p' =A
to obtain:
x =0
Substituting into (s) and (t), we have
a 2/Poak
Mjw(A + A)
1 2
=
A(-
~)•
\
) (A
­
1
/
a'\W
A2)
Ale+jk£+ A2e-jki
Solving for A
A
1
1
and A , we obtain
2
(Mjw + Aap o)V
s
=
2(-Mw sin kZ + Aap cos k£)
(Aap
- Mjw)V
s
2(-Mw sin kZ + Aa %cos kt)
A2
Thus, the velocity of the piston is
v (O,t) =
[A
Re
j
wt
+ A ]e
S1
v (0,t)
2
Aap V
w s
+ Aaos
Wt
-Mw
Elsin kk +Aap cos kZCOWt
=
1
(aa)
PROBLEM 13.9
Part a
The differential equation for the velocity as derived in problem 13.8 is
32v
1
Ftwhere
a
2
2
a
-
=
a
2
+
s
a
a
with
=
S
a2v
1o
aX 1 2
(a)
00
Po
2
(YPo2
where p
-•oP
Po
=
0
Part b
We assume a solution of the form
p
1
p H
2
2
2
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.9 (continued)
where k =
[De J(t-kx1)]
Re
V(x ,t) =
1
­
a
We do not consider the negatively traveling wave, as we want to use this system as
a delay line without distortion.
V(-L,t) =
and at x
The boundary condition at x
=
- L is
Re V ejWt
s
= 0 is
dV
BV (O,t)+
p'(O,t)A -
=
dt
;,
(b)
A
joHoH'
From problem 13.8, (h), (j) and (9)
H'
av
pa
s
p
',
-
=t
at
o ax
and
1
-
H
=
a o
S 0
Thus, (b) becomes
ta 2
t
-BDej
+
(-)
where
=0
p'A
S
PoD(- jk)
pI
=w
x0O
j
(c)
t
2
(d)
se
w
(d)
Thus, for no reflections
2 Ap0a2
- B +
(a)
a
0
a
(e)
S
or B =
Aapo
(f)
PROBLEM 13.10
The equilibrium boundary conditions are
T[-
(L
1
(L
T[-
+ L
+ A),t]
2
+ A),t]As =
=
T
0
- PoAc
Boundary conditions for incremental motions are
T[-
(L
2) - T[-
(L
1)
1
+ L
2
+ A),t] =
+ A),t]A
v (-Ll ,t) =
3)
and 4 )
v
(O,t)
=
- p(-L
Ve[-(L
0
T
s
(t)
d
,t)Ac = M
+ A),t]
-
v
(-L ,t)
since the mass is rigid
since the wall at x=0 is fixed.
PROBLEM 13.11
Part a
We can immediately write down the equation for perturbation velocity, using
equations (13.2.76) and (13.2.77) and the results of chapters 6 and 10.
93
ELECTROMECHANTCS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.11 (continued)
We replace 3/at by 3/at + v-V
(-+
Letting
V
t
)v'
o
=
x
to obtain
a
2
sax
2V
= Re V j(t-kx)
v'
we have
(W- kV S)2 =
a2k
2
as
Solving for w, we obtain
=
W
k(V
± as )
Part b
Solving for k, we have
k =
V±a
V+a
o
s
For Vo > as, both waves propagate in the positive x- direction.
PROBLEM 13.12
Part a
We assume that
E =
i
J =
iz
z
z
i
y
B=
E (x,t)
z
J (x,t)
z
o1[H
+ H'(x,t)]
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'
and
3E
3x
(b)
J z
ax Yx =
aH'
0o----
z
at
(c)
and the constitutive law is
Jz =
z
(E + VxpoH )
z
xoo
We recognize that Eqs. (13.2.94), (13.2.96) and (13.2.97) are still valid, so
1
I3' P
at
ax
(d)
eav
(e)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM' 13.12 (continued)
and
a 2 P'
p'
(f)
S
Part b
We assume all perturbation quantities are of the form
v
Using (b),
x
=
^
Re[v e
j(wt-kx)
]
(a) may be rewritten as
(g)
+ jkp + p0HojkH
pojwv =
and (c) may now be written as
(h)
jojwH
- jkE =
Then, from (b) and (d)
o(E
- jkH =
+ 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=
(k)
kP asv
Substituting (j) and (k) back into (g), we find
2 jk(po Ho)2
v
L2
-
Lk
0)
00jk+ 'w
Thus, the dispersion relation is
k
j(MoHo)20
k,
(02 - k2a2) -_
(+ 2
=
(m)
0
+ j11w)po
We see that in the limit as a + m, this dispersion relation reduces to the lossless
dispersion relation
_-2 k2(a2 +
=
(n)
0
Part c
If a is very small, we can approximate (m) as
W2-
k2a2
_ j(
w
s fo for which we can rewrite (o) as
2
O
kr
)H = (0o)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.12 (continued)
kl
s
k2
-
O
2
2
(
=
2
, we obtain
Solv ing for k
2
0p
()02_ j
2
0
k2=
2
22a
OHo
Po
2
)
w°
0 02
2)
O+
o
a
s
2
2a s
a
s
Since a is very small, we expand the radical in (q) to obtain
2
[12
O A oo
_
2 a .
O2]
S
L
2
Thus, our approximate solutions for k
EW
are
(IIoHo)2
k2
a
(
o0
0
O
s
H )
a (11
2
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.
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.12 (continued)
Part d
If 0 is very large, then (m) reduces to
H2
W2 - k 2 a 2 -
o
0 ;
k
p0
a2 = a 2
Ow
s
H2
O
+
(w)
Po
This can be put in the form
a
2
a
where
H 2 k0
f(w,k), =-
As
(x)
f(wk)
k2=
(Y)
2
pOwa
a becomes very large, the second term in (x) becomes negligible, and so
k
2
-
2
However, it is this second'term which represents the damping in space; that is,
k
+
,a
f(w,k)
j2a
(z)
Thus, the approximate decay rate, ki, is
ki
ki
2a w
2
or
i
H2 k
o
f(w,k)a
2p
a0
a
2poaw•
k
H2
w
a
2pa
0
(aa)
ELECTROMECHANICAL COUPLING WITH VISCOUS FLUIDS
PROBLEM 14.1
Part a
We can specify the relevant variables as
v =
The x
1
v 1 (x2)
)
E =
2E2(
J=
i 2 J O
B =
iB
20
2
+ 13E3 (x
+TB
(x)
11
2
3
component of the momentum equation is
a2v 1
0 =
p sx
2
with solution
Cx +C
1 2
2
Applying the boundary conditions
V
1
V
=
I
We obtain
@
x
=0
v
@
x2 = d
2
(c)
vx
02
d
=
V
0
'1
(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
+
x B)
in the x
2
J
and
and x
o
0
(e)
directions to obtain
3
=
YGE
2
=
a(E
(f)
+ v Bo)
(g)
since no current can flow in the x
Thus
E
3
direction.
J
o
T
2
and
(h)
VVoX2B
3
(i)
d
As from Eq. (14.2.5),
d
V
=
J
o E2 dx 2 =
d
Thus, the electrical input pe per unit area in an x
(j)
- x
plane is
J2 d
p
e
=
J V =
o
0
(k)
ELECTROMECHANICAL COUPLING WITH VISCOUS FLUIDS
PROBLEM 14.1 (continued)
We see that this power is dissipated as Ohmic loss.
like a resistor from the electrical terminals.
The moving fluid looks just
The traction that must be applied
to the upper plate to maintain the steady motion is
av, Iv0
d
0
-i 2•2 x=d
=
T
2
Again we note no contribution from the magnetic forces.
The mechanical input power per unit area is then
]V2
p
m
=
T
-- d
io
(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.
electromechanical coupling.
V =
There is no
Using the parameters from Table 14.2.1, we obtain
15 millivolts
Pt
V
2.2635 x 104
.015
______________________
B
B
0
B0
U
o
and
Pt = 2.2635 x 10 4watts/m 2 , independent of B0
o
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 x 2 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
x 3 direction, there was no additional dissipated power.
However, if E3 became too
large, the fluid might experience electrical breakdown, resulting in corona arcs.
ELECTROMECHANICAL COUPLING WITH VISCOUS FLUIDS
PROBLEM 14.2
The momentum equation for the fluid is
+
-Vp +V2
+ p(v*V)v =
P
We consider solutions of the form
v =
i-z v z C(r)
-II\
P =
Pkz,*
Then in the steady state, we write the z component of (a) in cylindrical coordinates
av
as
=
z
'-
r
r
rr
r
(b)
z
Now, the left side of (b) is only a function of z, while the right side is only a
Thus, from the given information
function of r.
(c)
"=
P2 ­
S
az
L
Using the results of (c) in (b), we solve for v (r) in the form
p -p
r2 + A
2
v (r) =
z
4Lp
£n
(d)
r + B
where A and B are arbitrary constants to be evaluated by the boundary conditions
v (r = 0) is finite
I-
v zr
=_
3\
)
_
n
U
=
Thus the solution is
(p - p )
v (r) =
(r2 -
2•L
R 2)
(e)
We can also find relations between the flow rate and the pressure difference, since
R
2Wrrdr
f v
Q
=
0
PROBLEM 14.3
Part a
We are given the pressure drop Ap, the magnetic field Bo, the conductivity 0,
and the dimensions of the system.
Now
+a
i
+u
J Jddx
=
1
(E
+ v Bo)dx
V
-d
-d
where
=
E
- - is defined as the voltage across the resistor.
w
From Eq. (14.2.29), we have the solution for the velocity v, . We then perform the
V=
integrations of (a) and solve for the voltage V to obtain
ELECTROMECHANICAL COUPLING WITH VISCOUS FLUIDS
PROBLEM 14.3(continued)
(Ap)2d
V=
o
1
R
tanh M
2a0d
w
(b)
tanh M
M
where
)
B d(
M=
0
11
Then, the power pedissipated in the resistor is
A(o2d
2
2
(
tanh M)
2
V1
e
(1
R
R
1
tanh M
R
M
2
R
R
where we have defined the internal resistance R i as
w
w
R.
i
2a0,d
Part b
e
To maximize p , we differentiate (c) with respect to R, solve for that value
of R which makes this quantity zero, and then check that this value does indeed
e
maximize p . Performing these operations, we obtain
MR
=
R
max
(d)
tanh M
Part c
We must convert the given numerical values to MKS units, using the conversions
10,000 gauss =
100 cm =
and
1 Weber/meter
2
1 meter
For mercury
and
a
=
106 mhos/m
p
=
1.5 x 10 - 3 kg/m-sec.
Thus
M =
Bod(
M =
520
Then tanh M
)
=
2 x 10 2
x 10'
%1
and so
max
R
520
10
2
x
10
10 b
-
x 1 0Z
-3
2.60 x 10-
101
ohms.