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2. Incompressible In viscid Fluids
Depth W into paper
Fig. 12P.3 = 0
12.3. A magnetic rocket is shown in Fig. 12P.3. A current source 10 (distributed over the
depth W) drives a circuit composed in part of a movable piston. This piston drives an
incompressible, inviscid, nonconducting fluid through two orifices, each of height d and
depth W. Because D >> d, the flow is essentially steady.
(a) Find the exit velocity V as a function of 4.
(b) (b) What is the thrust developed by the rocket? (You may assume that it is under
static test.)
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3. For Section 7.2: Prob. 7.2.1 In Sec. 3.7, ci is defined such that in the conservative
subsystem, Eq. 3.7.3 holds. Show that ai satisfies Eq. 7.2.3 with p+ai. Further, show that
if a "specific" property Si is defined such that Bi pai, then by virture of conservation of
mass, the convective derivative of Si is zero.
For Section 7.6:
Prob. 7.6.1 Show that Eq (hb of Tah1P 7_6_2 is correct
Prob. 7.6.2 Show that Eqs. (j) and (2) from
Table 7.6.2 are correct.
Prob. 7.6.3 A pair of bubbles are formed with
the tube-valve system shown in the figure.
Bubble 1 is blown by closing valve V2 and
opening Vl. Then, Vl is closed and V2 opened so that
the second bubble is filled. Each bubble can be
regarded as having a constant surface tension y.
With the bubbles having the same initial radius Eo, when t = 0, both valves are opened
(with the upper inlet closed off). The object of the following steps is to describe the
resulting dynamics.
(a) Flow through the tube that connects the bubbles is modeled as being fully developed
and viscous dominated. Hence, for a length of tube k having inner radius R and with a
viscosity of the gas the volume rate of flow is related to the pressure difference by
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4. The inertia of the gas and bubble is ignored, as is that of the surrounding air. Find an
equation of motion for the bubble radius E1 .
(b) With the bubbles initially of equal radius Eo , there is a slight departure of the radius
of one of the bubbles from eauilibrium. What hapDens?
(c) In physical terms, explain the result of (b).
For Section 7.8:
Prob. 7.8.1 A conduit forming a closed loop consists of a pair of
tubes having cross-sections with areas Ar and Ay . These are
arranged as shown with a fluid having density pb filling the lower
half and a second fluid having density Pa filling the upper half.
The object of the following steps is to determine the dynamics of
the fluid, specifically the time dependence of the interfacial
positions Er and E .
(a) Use mass conservation to relate the displacements (Er', E )
to each other and to the fluid velocities (vr, vk) on the right
and left respectively. Assume that the fluid is inviscid and has
a uniform profile over the cross-section of a tube.
(b) Use Bernoulli's equation, Eq. 7.8.5 to relate quantities
evaluated at the interfaces in the lower fluid, and in the upper
fluid.
(c) Write the boundary conditions that relate quantities across
the interfaces.
(d) Show that these laws combine to give an equation of motion
for the right interface having the form
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5. Prob. 3.10.3 Fig. P3.10.3 shows a circular cylindrical tube of
inner radius a into which a second tube of outer radius b
projects half way. On top of this inner tube is a "blob" of liquid
metal (shown inside th broken-line box) having an arbitrary
shape, but having a base radius equal to that of the inner
tube. The outer and inner tubes, as well a the blob, are all
essentially perfectly conducting on the time scale o interest.
When t=0 , there are no magnetic fields. When t=O+, the ou -
t. tube is used to produce a magnetic flux which has density
Bo z a distan 2 >> a above the end of the inner tube.
What is the magnetic flux den ity over the cross section of the
annulus between tubes a distance 2 (2 >> a) below the end
of the inner tube? Sketch the distribution of surface
current.on the perfect conductors (outer and inner tubes and
blob), indicating the relative densities. Use qualitative
arguments to state whether the vertical magnetic force on the
blob acts upward or downward. Use the stress tensor to find
the magnetic force acting on the blob in the z direction. This
expression should be exact if 2 >> a, and be written in terms
of a, b, Bo and the permeability of free space yo.
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6. Prob. 3.10.4 The mechanical configuration is as in Prob. 3.10.3. But, instead of the
magnetic field, an electric field is produced by making the outer cylinder have the
potential Vo relative to the inner one. Sketch the distribution of the electric field, and
give qualitative arguments as to whether the electrical force on the blob is upward or
downward. What is the electric field in the annulus at points well removed from the tip
of the inner cylinder? Use the electric stress tensor to determine the z-directed electric
force on the blob. + 4 .
Prob. 3.10.5 In an EQS system with polarization, the force density is not F = PpE +
PfE, where Pp is the polarization charge. Nevertheless, this force density can be used
to correctly determine the total force on an object isolated in free space. The proof
follows from the argument given in the paragraph following Eq. 3.10.4. Show that the
stress tensor associated with this force density is
Show that the predicted total force will agree with that found by any of the force
densities in Table 3.10.1. Prob. 3.10.6 Given the force density of Eq. 3.8.13, show that
the stress tensor given for this force density in Table 3.10.1 is correct. It proves helpful
to first show that
Prob. 3.10.7 Given the Kelvin force density, Eq. 3.5.12, derive the consistent stress
tensor of Table 3.10.1. Note the vector identity given in Prob. 3.10.6.
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7. Prob. 3.10.8 Total forces on objects can sometimes be found by the energy method
"ignoring" fringing fields and yet obtaining results that are "exact." This is because the
change in total energy caused by a virtual displacement leaves the fringing field
unaltered. There is a "theorem" than any configuration that can be described in this
way by an energy method can also be-described by integrating the stress tensor over
an appropriately defined surface. Use Eqs. 3.7.22 of Table 3.10.1 to find the force
derived in Prob. 2.13.2.
For Section 3.11:
Prob. 3.11.1 An alternative to the derivation represented by Eq. 3.11.7 comes from
exploiting an integral theorem that is analogous to Stokes's theorem.1
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8. � �
Thus, the combination of Eqs. 1 and 4 give a first order differential equation
describing the evolution of ξ1 or ξ2. In normalized terms, that expression is
� �
Problem 7.6.3
Mass conservation requries that
(1)
With the pressure outside the bubbles defined as p0, the pressure inside the
respective bubbles are
(2)
so that the pressure difference driving fluid between the bubbles
once the valve is opened is
.
(3)
The flow rate between bubbles given by differentiating Eq. 1 is then equal to
Qv and hence to the given expression for the pressure drop through the
connecting tubing.
(4)
Solution
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9. Thus, the velocity is a function of ξ1, and can be pictured as shown in the figure. It
is therefore evident that if ξ1 increases slightly, it will tend to further increase. The
static equilibrium at ξ1 = ξ0 is unstable. Physically this results from the fact that γ
is constant. As the radius of curvature of a bubble decreases, the pressure
increases and forces the air into the other bubble. Note that this is not what would be
found if the bubbles were replaced by most elastic membranes. The example is
useful for giving a reminder of what is implied by the concept of a surface tension.
Of course, if the bubble can not be modelled as a layer of liquid with interior and
exterior interfaces comprised of the same material, then the basic law may not apply.
In the figure, note that all variables are normalized. The asymptote comes at the
radius where the second bubble has completely collapsed.
where
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10. Figure 1: Bubble velocity versus bubble displacement from Eq. (5) (Image by MIT
OpenCourseWare.)
Courtesy of James R. Melcher. Used with permission.
Solution to Problem 7.6.3 in Solutions Manual for Continuum
Electromechanics, 1982, pp. 7.3-7.4
Problem 3.10.3
(a)
The magnetic field is “trapped” in the region between tubes. For an infinitely long pair
of coaxial conductors, the field in the annulus is uniform. Hence, because the total
flux πa2B0 must be constant over the length of the system, in the lower region
a2B0
B z =
a2 −b2
(6)
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11. (b)
The distribution of surface current is as sketched below. It is determined by the
condition that the magnetic flux at the extremities be as found in (a) and by the
condition that the normal flux density on any of the perfectly conducting surfaces
vanish.
(c)
Using the surface force density K× < B >, it is reasonable to expect the net
magnetic force in the z direction to be downward.
(d)
One way to find the net force is to enclose the “blob” by the control volume shown in
the figure and integrate the stress tensor over the enclosing surface.
�
fz = Tzj nj da
S
Contributions to this integration over surfaces (4) and (2) (the walls of the inner
and outer tubes which are perfectly conducting) vanish because there is no shear
stress on a perfectly conducting surface. Surface
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12. Figure 2: Magnetic field lines and stress tensor enclosing surface (dashed) (Image by
MIT OpenCourseWare.)
(5) cuts under the “blob” and hence sustains no magnetic stress. Hence, only
surfaces (1) and (3) make contributions, and on them the magnetic flux density is
given and uniform. Hence, the net force is
Note that, as expected, this force is negative.
Problem 3.10.4
The electric field is sketched in the figure. The force on the cap should be upward. To
find this force use the surface S shown to enclose the cap. On S1 the field is zero. On
S2 and S3 the electric shear stress is zero because it is an equi-potential and hence
can support no tangential E. On S4 the field is zero. Finally, on S5 the field is that of
infinite coaxial conductors.
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13. Thus, the normal electric stress is
and the integral for the total force reduces to
(8)
(9)
(10)
Figure 3: Electric field lines and stress tensor enclosing surface (dashed) (Image by
MIT OpenCourseWare.)
Courtesy of James R. Melcher. Used with permission.
Solution to Problem 3.10.4 in Solutions Manual for Continuum
Electromechanics, 1982, p. 3.8.
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14. (11)
The exit velocity at each orifice is obtained by using Bernoulli’s law just to the left
of the piston and at either orifice, from which we obtain
(12)
(13)
(14)
Problem 4
(a)
From the results of problem 12.2, we have that the pressure p, acting just to the left
of the piston, is
at each orifice.
(b)
The thrust is
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15. (c)In the steady state, we choose to integrate the momentum theorem, Eq. (12.1.29),
around a rectangular surface, enclosing the system for −L ≤ x1 ≤ +L.
−ρV 2 a + ρ[V (L)]2b = p0a −p(L)b + F,
(15)
0
where F is the x1 component force per unit length which the walls exert on the fluid.
We see that there is no x1 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
(17)
(18)
(19)
(16)
Bernoulli’s equation gives us
Solving (17) for P(L), and then substituting this result and that of (16) into (15), we
finally obtain
The problem asked for the force on the lower wall, which is just the negative of F.
Thus
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