Chapter 1
•
Introduction
1.1
A gas at 20
°
C may be
rarefied
if it contains less than 10
12
molecules per mm
3
. If
Avogadro’s number is 6.023E23 molecules per mole, what air pressure does this represent?
Solution:
The mass of one molecule of air may be computed as
1
−
−
Molecular weight
28.97 mol
m
4.81E 23 g
Avogadro’s number
6.023E23 molecules/g mol
==
=
⋅
Then the density of air containing 10
12
molecules per mm
3
is, in SI units,
ρ
⎛⎞
⎛
⎞
=−
⎜⎟
⎜
⎟
⎠
12
molecules
g
10
4.81E 23
⎝⎠
⎝
3
33
molecule
mm
gk
g
4.81E 11
4.81E 5
mm
m
Finally, from the perfect gas law, Eq. (1.13), at 20
°
C
=
293 K, we obtain the pressure:
Α
−
=
⋅
32
kg
m
p
RT
4.81E 5
287
(293 K)
.
ms
K
ns
4.0 Pa
2
1.2
The earth’s atmosphere can be modeled as a uniform layer of air of thickness 20 km
and average density 0.6 kg/m
3
(see Table A6). Use these values to estimate the total mass
and total number of molecules of air in the entire atmosphere of the earth.
Solution:
Let Re be the earth’s radius
≈
6377 km. Then the total mass of air in the
atmosphere is
2
ta
v
g
a
v
g
e
m
dVol
(Air Vol)
4 R (Air thickness)
(0.6 kg/m )4 (6.377E6 m) (20E3 m)
.
Ans
ρρ
π
≈
=≈
∫
6.1E18 kg
Dividing by the mass of one molecule
≈
4.8E
−
23 g (see Prob. 1.1 above), we obtain the
total number of molecules in the earth’s atmosphere:
molecules
m(atmosphere)
6.1E21 grams
N
m(one molecule)
4.8E 23 gm/molecule
Ans.
≈
−
1.3E44 molecules
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Solutions Manual
•
Fluid Mechanics, Sixth Edition
1.3
For the triangular element in Fig. P1.3,
show that a tilted free liquid surface, in
contact with an atmosphere at pressure pa,
must undergo shear stress and hence begin
to flow.
Solution:
Assume zero shear. Due to
element weight, the pressure along the
lower and right sides must vary linearly as
shown, to a higher value at point C. Vertical
forces are presumably in balance with ele
ment weight included. But horizontal forces
are out of balance, with the unbalanced
force being to the left, due to the shaded
excesspressure triangle on the right side
BC.
Thus hydrostatic pressures cannot keep
the element in balance, and shear and flow
result.
Fig. P1.3
P1.4
The Saybolt Universal Viscometer, now obsolete but still sold in scientific catalogs,
measures the kinematic viscosity of lubricants [Mott, p. 40].
A container, held at constant
temperature, is filled with 60 ml of fluid.
Measure the time
t
for the fluid to drain from a small
hole or short tube in the bottom.
This time unit, called
Saybolt universal seconds
, or SUS, is
correlated with kinematic viscosity
ν
, in centistokes (1 stoke = 1 cm
2
/s), by the following curve
fit formula:
SUS
100
40
for
145
215
.
0
<
<
−
=
t
t
t
(
a
) Comment on the dimensionality of this equation.
(
b
) Is the formula physically correct?
(
c
) Since
ν
varies strongly with temperature, how does temperature enter into the formula?
(
d
)
Can we easily convert
from centistokes to mm
2
/s?
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 M
 Fluid Mechanics, The Land, Sixth Edition

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