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Fluid Mechanics Solutions Manual 4th Edition
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Fluid Mechanics Solutions Manual 4th Edition Complete and Comprehensive Guide
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Chapter 1 • Introduction1.1 A gas at 20°C may be rarefied if it contains less than 1012 molecules per mm3. 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
Molecular weight 28.97 mol −1
m= = = 4.81E−23 g
Avogadro’s number 6.023E23 molecules/g ⋅ mol
Then the density of air containing 1012 molecules per mm3 is, in SI units,
æ molecules öæ g ö
ρ = ç 1012 3 ÷ç 4.81E−23 ÷
è mm øè molecule ø
g kg
= 4.81E−11 3
= 4.81E−5 3
mm m
Finally, from the perfect gas law, Eq. (1.13), at 20°C = 293 K, we obtain the pressure:
æ kg ö æ m2 ö
p = ρ RT = ç 4.81E−5 3 ÷ ç 287 2 ÷ (293 K) = 4.0 Pa Αns.
è m øè s ⋅K ø
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/m3 (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 Re be the earth’s radius ≈ 6377 km. Then the total mass of air in the
atmosphere is
m t = ò ρ dVol = ρavg (Air Vol) ≈ ρavg 4π R 2e (Air thickness)
= (0.6 kg/m 3 )4π (6.377E6 m)2 (20E3 m) ≈ 6.1E18 kg Ans.
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:
m(atmosphere) 6.1E21 grams
N molecules = = ≈ 1.3E44 molecules Ans.
m(one molecule) 4.8E −23 gm/molecule
,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 pa,
must undergo shear stress and hence begin
to flow.
Solution: Assume zero shear. Due to Fig. P1.3
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.
1.4 The quantities viscosity µ, velocity V, and surface tension Y may be combined into
a dimensionless group. Find the combination which is proportional to 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/T2}. We must divide µ by Y to cancel mass {M}, then work the velocity into the
group:
ì µ ü ì M / LT ü ì T ü ìLü
í ý=í 2 ý = í ý , hence multiply by {V } = í ý;
î Y þ î M /T þ î L þ îT þ
µV
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
l = 1.26 = 1.26 √ (RT) (1)
ρ √ (RT) p
, Chapter 1 • Introduction 3
where the latter form follows from the ideal-gas law, ρ = p/RT. What are the dimensions
of the constant “1.26”? Estimate the mean free path of air at 20°C and 7 kPa. Is air
rarefied at this condition?
Solution: We know the dimensions of every term except “1.26”:
ìMü ìMü ì L2 ü
{l} = {L} {µ} = í ý {ρ} = í 3 ý {R} = í 2 ý {T} = {Θ}
î LT þ îL þ îT Θþ
Therefore the above formula (first form) may be written dimensionally as
{M/L⋅T}
{L} = {1.26?} = {1.26?}{L}
{M/L } √ [{L2 /T 2 ⋅ Θ}{Θ}]
3
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)] =
0.0832 kg/m3. From Table A-2, its viscosity is 1.80E−5 N ⋅ s/m2. Then the formula predict
a mean free path of
1.80E−5
l = 1.26 ≈ 9.4E−7 m Ans.
(0.0832)[(287)(293)]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 l, that is, greater than about 94 µm.
1.6 If p is pressure and y is a coordinate, state, in the {MLT} system, the dimensions of
the quantities (a) ∂p/∂y; (b) ò p dy; (c) ∂ 2 p/∂y2; (d) ∇p.
Solution: (a) {ML−2T−2}; (b) {MT−2}; (c) {ML−3T−2}; (d) {ML−2T−2}
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.
Solution: One acre = (1 mi2/640) = (5280 ft)2/640 = 43560 ft2. Therefore 1.5 acre-ft =
65340 ft3 = 1850 m3. Meanwhile, 1 gallon = 231 in3 = 231/1728 ft3. Then 1.5 acre-ft of
water per day is equivalent to
ft 3 æ 1728 gal ö æ 1 day ö gal
Q = 65340 ç 3 ÷ç ÷ ≈ 340 Ans. (a)
day è 231 ft ø è 1440 min ø min
, 4 Solutions Manual • Fluid Mechanics, Fifth Edition
Similarly, 1850 m3 = 1.85E6 liters. Then a metric unit for this water usage is:
æ L öæ 1 day ö L
Q = ç 1.85E6 ÷ç ÷ ≈ 21 Ans. (b)
è day ø è 86400 sec ø s
1.8 Suppose that bending stress σ 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 = 2900 in⋅lbf, y = 1.5 in, and I = 0.4 in4, the predicted
stress is 75 MPa. Find the only possible dimensionally homogeneous formula for σ.
Solution: We are given that σ = 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,
ì M ü
{σ } = {y}{fcn(M,I)}, or: í 2 ý = {L}{fcn(M,I)}
î LT þ
ì M ü
or: the function must have dimensions {fcn(M,I)} = í 2 2 ý
îL T þ
Therefore, to achieve dimensional homogeneity, we somehow must combine bending
moment, whose dimensions are {ML2T –2}, with area moment of inertia, {I} = {L4}, and
end up with {ML–2T –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:
ì M ü ì ML2 ü
σ = yM fcn(I), or í 2 ý = {L} í 2 ý{fcn(I)}, or: {fcn(I)} = {L−4 }
î LT þ î T þ
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:
My
σ =C , where {C} = {unity} Ans.
I
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) =
10880 lbf/in2. Substitute the given data into the proposed formula:
lbf My (2900 lbf ⋅in)(1.5 in)
σ = 10880 2
=C =C , or: C ≈ 1.00 Ans.
in I 0.4 in 4
The data show that C = 1, or σ = My/I, our old friend from strength of materials.
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