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Chapter 1 • Introduction
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?
1.1
A gas at 20
°
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,
molecules
g
12
ρ
=
10
4.81E 23 molecule
−
3
mm
g
kg
=
4.81E 11
−
=
4.81E 5
−
3
3
mm
m
Finally, from the perfect gas law, Eq. (1.13), at 20
°
C
=
293 K, we obtain the pressure:
2
kg
m
p
=
ρ
RT
=
4.81E 5
−
287
(293 K)
=
4.0 Pa
Α
ns
.
3
2
m
s
⋅
K
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 A-6). Use these values to estimate the total mass
and total number of molecules of air in the entire atmosphere of the earth.
Solution:
Let R e be the earth’s radius
≈
6377 km. Then the total mass of air in the
atmosphere is
2
m
=
ρ
dVol
=
ρ
(Air Vol)
≈
ρ
4
π
R (Air thickness)
t
avg
avg
e
3
2
=
(0.6 kg/m )4
π
(6.377E6 m) (20E3 m)
≈
6.1E18 kg
Ans
.
Dividing by the mass of one molecule
23 g (see Prob. 1.1 above), we obtain the
total number of molecules in the earth’s atmosphere:
≈
4.8E
−
m(atmosphere)
6.1E21 grams
N
=
=
≈
1.3E44 molecules
Ans.
molecules
m(one molecule)
4.8E
−
23 gm/molecule
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2
Solutions Manual • Fluid Mechanics, Fifth 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 p a ,
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
excess-pressure triangle on the right side
BC. Thus hydrostatic pressures cannot keep
the element in balance, and shear and flow
result.
Fig. P1.3
, velocity V , and surface tension Y may be combined into
a dimensionless group. Find the combination which is proportional to
1.4
The quantities viscosity
µ
µ
. This group has a
customary name, which begins with C . Can you guess its name?
Solution:
The dimensions of these variables are {
µ
}
=
{M/LT}, {V}
=
{L/T}, and {Y}
=
{M/T 2 }. We must divide
µ
by Y to cancel mass {M}, then work the velocity into the
group:
µ
MLT
/
T
L
=
=
,
hence multiply by
{ }
V
=
;
2
Y
L
T
MT
/
µ
V
finally obtain
=
dimensionless.
Ans
.
Y
This dimensionless parameter is commonly called the Capillary Number .
1.5
A formula for estimating the mean free path of a perfect gas is:
µ
µ
=
1.26
=
1.26
√
(RT)
(1)
ρ
√
(RT)
p
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3
Chapter 1 • Introduction
where the latter form follows from the ideal-gas law,
RT. What are the dimensions
of the constant “1.26”? Estimate the mean free path of air at 20
ρ
=
p
/
°
C and 7 kPa. Is air
rarefied at this condition?
Solution:
We know the dimensions of every term except “1.26”:
2
M
M
L
{} {L} { }
=
µ
=
{ }
ρ
=
{R}
=
{T} { }
= Θ
3
2
LT
L
T
Θ
Therefore the above formula (first form) may be written dimensionally as
{M/L T}
⋅
{L}
=
{1.26?}
=
{1.26?}{L}
3
2
2
{M/L }
√
[{L /T
⋅ Θ
}{ }]
Θ
Since we have {L} on both sides, {1.26}
{unity}, that is, the constant is dimensionless.
The formula is therefore dimensionally homogeneous and should hold for any unit system.
=
For air at 20
°
C
=
293 K and 7000 Pa, the density is
ρ
=
p
/
RT
=
(7000)/[(287)(293)]
=
m 3 . From Table A-2, its viscosity is 1.80E
s/m 2 . Then the formula predict
0.0832 kg
/
−
5 N
⋅
a mean free path of
1.80E 5
−
=
1.26 (0.0832)[(287)(293)]
≈
9.4E 7 m
−
Ans.
1/2
This is quite small. We would judge this gas to approximate a continuum if the physical
scales in the flow are greater than about 100 ,
that is, greater than about 94
µ
m.
1.6 If p is pressure and y is a coordinate, state, in the {MLT} system, the dimensions of
the quantities (a)
2
y 2 ; (d)
∂
p
/∂
y ; (b)
p dy ; (c)
∂
p
/∂
∇
p .
(a) {ML −
T −
}; (b) {MT −
}; (c) {ML −
T −
}; (d) {ML −
T −
2
2
2
3
2
2
2
Solution:
}
1.7 A small village draws 1.5 acre-foot of water per day from its reservoir. Convert this
water usage into (a) gallons per minute; and (b) liters per second.
(1 mi 2
(5280 ft) 2
43560 ft 2 . Therefore 1.5 acre-ft
Solution:
One acre
=
/
640)
=
/
640
=
=
65340 ft 3
1850 m 3 . Meanwhile, 1 gallon
231 in 3
231/1728 ft 3 . Then 1.5 acre-ft of
=
=
=
water per day is equivalent to
3
ft
1728
gal
1
day
gal
Q
=
65340
≈
340
Ans
. (a)
3
day
231
ft
1440
min
min
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4
Solutions Manual • Fluid Mechanics, Fifth Edition
Similarly, 1850 m 3
=
1.85E6 liters. Then a metric unit for this water usage is:
L
1
day
L
Q
=
1.85E6
≈
21
Ans
. (b)
day
86400
sec
s
1.8
in a beam depends upon bending moment M and
beam area moment of inertia I and is proportional to the beam half-thickness y. Suppose
also that, for the particular case M
Suppose that bending stress
σ
0.4 in 4 , the predicted
stress is 75 MPa. Find the only possible dimensionally homogeneous formula for
=
2900 in
⋅
lbf, y
=
1.5 in, and I
=
σ
.
Solution:
y fcn(M,I) and we are not to study up on strength of
materials but only to use dimensional reasoning. For homogeneity, the right hand side
must have dimensions of stress, that is,
We are given that
σ
=
M
{ }
σ
=
{y}{fcn(M,I)},
or:
=
{L}{fcn(M,I)}
2
LT
M
or:
the function must have dimensions
{fcn(M,I)}
=
22
LT
Therefore, to achieve dimensional homogeneity, we somehow must combine bending
moment, whose dimensions are {ML 2 T –2 }, with area moment of inertia, {I}
{L 4 }, and
end up with {ML –2 T –2 }. Well, it is clear that {I} contains neither mass {M} nor time {T}
dimensions, but the bending moment contains both mass and time and in exactly the com-
bination we need, {MT –2 }. Thus it must be that
=
σ
is proportional to M also . Now we
have reduced the problem to:
2
M
ML
−
4
σ
=
yM fcn(I),
or
=
{L}
{fcn(I)},
or:
{fcn(I)}
=
{L
}
2
2
LT
T
We need just enough I ’s to give dimensions of {L –4 }: we need the formula to be exactly
inverse in I . The correct dimensionally homogeneous beam bending formula is thus:
M C,
I
σ =
where {C}
=
{unity}
Ans
.
The formula admits to an arbitrary dimensionless constant C whose value can only be
obtained from known data. Convert stress into English units:
σ
=
(75 MPa)
/
(6894.8)
=
in 2 . Substitute the given data into the proposed formula:
10880 lbf
/
lbf
My
(2900 lbf in)(1.5 in)
⋅
σ
=
10880
=
C
=
C
,
or:
C . 0
≈
Ans.
2
4
I
in
0.4 in
The data show that C
=
1, or
σ
=
My/I , our old friend from strength of materials.
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5
Chapter 1 • Introduction
1.9 The dimensionless Galileo number, Ga , expresses the ratio of gravitational effect to
viscous effects in a flow. It combines the quantities density
ρ
, acceleration of gravity g ,
length scale L , and viscosity
. Without peeking into another textbook, find the form of
the Galileo number if it contains g in the numerator.
µ
L 3 }, { g }
T 2 }, { L }
Solution:
The dimensions of these variables are {
ρ
}
=
{M
/
=
{L
/
=
{L}, and {
to eliminate mass {M} and then combine with g
and L to eliminate length {L} and time {T}, making sure that g appears only to the first
power:
µ
}
=
{M
/
LT}. Divide
ρ
by
µ
3
ρ
µ
ML
/
/
T
=
=
2
MLT
L
while only { g } contains { T }. To keep { g } to the 1st power, we need to multiply it by
{
{ L −
3 }.
We then make the combination dimensionless by multiplying the group by L 3 . Thus
we obtain:
} 2 . Thus {
} 2 { g }
{ T 2
L 4 }{ L
T 2 }
ρ/µ
ρ/µ
=
/
/
=
2
23
3
ρ
ρ
gL
gL
ν
3
Galileo number
==
Ga
()( )
g
L
=
=
Ans
.
2
2
µ
µ
1.10
The Stokes-Oseen formula [10] for drag on a sphere at low velocity V is:
9
1 π
22
F3 DV
=
πµ
+
ρ
VD
where D
=
sphere diameter,
µ
=
viscosity, and
ρ
=
density. Is the formula homogeneous?
Solution:
Write this formula in dimensional form, using Table 1-2:
9
1 π
2
2
{F}
=
{3
πµ
}{ }{D}{V}
+
{ }{V} {D} ?
ρ
2
ML
M
L
M
T
L
2
or:
=
{1}
{L}
+
{1}
{L } ?
2
3
2
LT
T
T
L
where, hoping for homogeneity, we have assumed that all constants (3,
π
,9,16) are pure ,
T 2 }! Therefore the Stokes-
Oseen formula (derived in fact from a theory) is dimensionally homogeneous .
i.e., {unity}. Well, yes indeed, all terms have dimensions {ML
/
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