Chapter Solutions, 14
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14.1 They lower the surface tension of the final rinse, which makes the film more likely
to drain completely from the surface of the dishes, instead of breaking up into droplets,
which will leave marks on the dishes as they dry, and leave behind their dissolved salts.
When a uniform film dries it also leaves behind the salts, but they are uniformly
distributed, and thus harder to see than those left by discrete drops. Also, by lowering
the surface tension the agent makes the residual film of liquid which dries on the surface
thinner, thus leaving less solids.
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14.2* The vertical component of the surface force is Fvertical = D cos so that
hg l D reported
A
=
=
=
= 1.155 A
4cos
cos
0.886
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14.3 The figure is sketched below. A weight pulls down on the ring, the torsion
balance pulls upward. The setting of the torsion balance is slowly reduced, until the
weight pulls the ring down through the interface.
Wire to torsion balance
less dense fluid
ring
more dense fluid
weight
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14.4
Fv = D cos
= vertical = verticalo = 1.015 vertical
cos
cos10
It makes no difference which way the film is slanted.
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Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 14, page 1
14.5*
1 cm 3
g
m
0.998 3 9.81 2
Vdrop l g 20
cm
kg
N s2
s
D=
=
= 2.14 mm = 0.084 in
N
1000 g kg m
0.07274
m
I measured two burettes in our laboratory, finding 0.105 and 0.135 in diameters. This is
close, but far from an exact match . Since at equilibrium the drop size is proportional to
the diameter, one would expect 16 and 13 drops/cc from these two burettes.
From Eq 14.7 we see that the drop volume is proportional to / , so that
28.9 0.998
Vbenzene benzene water
=
=
= 0.405
Vwater
water benzene 72.74 0.8785
We would expect 20/0.405 = 49 drops per cc for benzene.
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14.6
From the solution to Prob 14.4
1
1
= 1.5 ;
= arc cos
= 48o
cos
1.5
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Correction factor =
14.7* For a small piece of film with length x, Fup = Fdown ; 2 x = x z y g
z =
2
2 A 1
B
; but y = x so that z =
which is a hyperbola.
gy
A
g B x
This only works fairly well with microscope slides, because they are too small. If one
can get the larger pieces of flat glass that were used in the lantern slides of old, they work
much better.
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14.8 On the side to the right in Fig 14.15 we get the value of R1 from the general
equation of the curvature of any surface in x-y coordinates,
dy 2 3 / 2
1 +
dz
R1 =
2
dy
dz2
y
On the other side, R2 is given by R2 =
. If we draw a tangent to the curve, where
cos
the two radii meet, and extend it to its intersection with the y axis, then that triangle will
be similar to the small triangle shown on the figure. If we let the vertical side of it be 1
arbitrary unit, then the side at right angles will be 1(dz/dy) and by Pythagorean theorem
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 14, page 2
cos =
1
dy 2 1 / 2
1 +
dz
dy 2 1 / 2
y
= y 1 +
so that R2 =
dz
cos
dy 2 3 / 2
1 +
dy 2 1 / 2
dz
Equating these two values of R we find
= y 1 + or
2
dz
dy
2
dz
dy 2 3 / 2
1 +
dy 2
d 2 y dz
y 2=
= 1 +
dy 2 1 / 2
dz
dz
1 +
dz
d2 y 1
z
dy
z
z
, then
cosh
= sinh and
2=
dz
a
a
dz
a
a
2
2
z1
z
z
z
z2
cosh = 1 + sinh
Substituting a cosh cosh = 1 + sinh ;
aa
a
a
a
a
This equation is an identity, proven in calculus books.
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Now, if we assume that y = a cosh
14.9
1
1
1
1
+ = +
;
R1 R2 cyl R1 R2 sphere
1 1 2
+ = ;
D R1
2
R1 = D
The shape is like this
, not like this
!
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14.10 The assumed shape of the gas liquid interface is as sketched below. The distance
between the pates is D.
Pacross the interface 2 = / D
Fon glass = AP = A 2 / D =
3.25 in 4.25 in 2 0.000415
in / 32
lbf
in = 0.37 lbf = 1.63 N
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 14, page 3
This does not appear to be a large force, but it is proportional to 1/(thickness of the liquid
film) so that as the film drains the force becomes much larger. Placing them horizontally
makes no difference in this force. They are presumably easier to slide apart than pull
apart because this is a force perpendicular to the plates, not in the plane of the plates. To
move them in the plane of the plates one need only overcome shear resistance, which is
proportional to the velocity, and hence small at low velocities.
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14.11 (a)
Vcyl = Vsph ;
2
4
2
c
DL=
6
1
3
s
D;
3
3
2
1
Dc L
2
3 3 Dc 3
Ss
Ds
2
=
=
=
For
1
2
Sc Dc L
Dc L
3
L
3
3
Ds = Dc2 L
2
Ss = Sc ,
L=
4
D
9c
(b) by inspection, if we take the area of the ends into account, any length of cylinder can
reduce its area by changing to a sphere.
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14.12 I hope the hints are enough to allow one to fill in the blanks.
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14.13 The large area is created by extruding the fluid through an orifice, e.g. a diesel fuel
injector nozzle. The breakup of the cylindrical jet actually reduces the area, so that the
statement that breakup of a jet creates surface area is false; the extrusion of the jet creates
the surface area.
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14.14* P =
2
=
r
2(4.15 10 4 )
lbf
in = 1.66 psi = 11.4 kPa
1
(10 3 in )
2
See problem 14.17 which explores the same idea in mercury porosimeters.
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14.15 The surface tension must be greater at the top than at the bottom, to overcome the
small but significant gravity force on the film. Most often this result is accomplished by
using a liquid which is a mixture, like soapy water. The soap concentration will adjust
itself to be lower at the top of the film than the bottom, so that the liquid surface tension
is higher at the top, etc. It can also be done by a temperature or electric gradient, but that
is much less common. In the absence of temperature or electrostatic gradients, such a
film cannot exist in a pure liquid. You can verity this by taking such a ring, dipping it in
soapy water, and withdrawing it; the film forms. Then repeat with pure tap water. No
film forms.
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Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 14, page 4
14.16(a) See Prob 14.14
P =
2
=
r
2(0.07274)
N
m = 29.1 kPa = 0.29 atm
1
(105 m )
2
3
(b)
V = (105 m ) = 5.2 10 16 m 3 = 0.5picoliters The ink jet literature has
6
all drop volumes in picoliter = pL.
(c)
The observed value is 20 times that calculated above.
1/ 3
6V 1/ 3 6 10 14 m
= 2.7 10 5 m = 0.0016 in
(d)
Dsphere of = =
volume V
If, as suggested the diameter increases by 3 as the drop flattens, then each drop should
have a diameter of 0.0048 inches, or there would be 200 drops/inch. This is in
tolerable agreement with the claimed 300 dpi for such printers.
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N
2
m = 193.6 kPa = 28 psi
14.17 See Prob 14.14. For 10 microns P =
=
1
5
r
(10 m )
2
If we reduce the diameter by a factor of 10, we increase the calculated pressure by a
factor of 10, so the calculated values for 1, 0.1 and 0.01 microns are 280, 2800 and 28
000 psi.
2(0.484)
This solution assumes that the angle between the edge of the interface and the outside
mercury is 180. The mercury porosimeter literature indicates that the observed values
are less than one would calculate here, and that they can be well represented by assuming
that that angle is 140. That results in multiplying the above values by minus cos 140
= 0.766. Thus the computed values are 21 psi, 210, 2100 and 21 000. You can
download a simple introduction to this type of measurement from
www.quantachrome.com/Mercury Porosimetry/poreMaster.htm.
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Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 14, page 5