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Unformatted text preview: Chapter 1  Introduction 1.1 A gas at 20°C may be rareﬁed if it contains less than 10l2 molecules per mm}. 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 w = 4.81E—23 g m = ._
Avogadro‘s number 6.023E23 molecules/gmol Then the densit of air containin 10l2 molecules er mm3 is, in SI units,
Y 3 P p=[10‘2 MIasta—zs ——g——] mm3 molecule =4.81E——ll g 3 =4.8]E—5 k—gs
mm m Finally, from the perfect gas law, Eq. (l.l3), at 20°C = 293 K. we obtain the pressure:
kg m2
p = pRT = 4.8]E—5 —; 287 —2—— (293 K): 4.0 Pa Ans.
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/m3 (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 Rc be the earth’s radius a 6377 km. Then the total mass of air in the
atmosphere is
m, = I p dVol = pavg(Air Vol) 2 pavgtln'RRAir thickness)
= (0.6 kg/m3)41r(6.377E6 m)2(20E3 m) = 6.11518 kg Ans. Dividing by the mass of one molecule = 4.81343 g (see Prob. 1.1 above), we obtain the
total number of molecules in the earth’s atmosphere: Nmﬁleculﬂ = W = M— : 1.3E44 molecules Am.
' m(one molecule) 4.8E—23 grit/molecule Chaptert  Introduction 7 1.13 The efﬁciency 17 of a pump is deﬁned as = ____QAP
Input Power where Q is volume flow and Ap the pressure rise produced by the pump. What is n if
AP = 35 psi, Q = 40 Us, and the input power is 16 horsepower? Solution: The student should perhaps verify that QAp has units of power, so that n is a
dimensionless ratio. Then convert everything to consistent units, for example, BG: 2
s s in: ft2 #5 8800 ftlbfl's =0.Sl or 81% Ans. Similarly, one could convert to SI units: Q = 0.04 [Ill/S, Ap = 241300 Pa, and input power =
16045.7) = l 1930 W, thus h = (0.04)(24l300)l(11930) = 0.81. Ans. _______________._____—————————— 1.14 The volume ﬂow Q over a dam is
proportional to dam width B and also varies
with gravity g and excess water height H
upstream, as shown in Fig. P1.l4. What is
the only possible dimensionally homo—
geneous relation for this ﬂow rate? Solution: So far we know that
Q = B fcn(H,g). Write this in dimensional
form: 3
lQl={%}=lBltf(H.g)l =lLllftH.g)l. 2
or: lf(H.g)l So the function fcn(H,g) must provide dimensions of [LZIT ], but only 3 contains m.
Therefore g must enter in the form gm to accomplish this. The relation is now Q=Bgmfcn(H), or: {L3rri={L}{L‘”rr}{fcn(H)}, or: {rcn(n)i={L3"} 8 Solutions Manual  Fluid Mechanics, Eifth Edition ' In order for fcn(ll) to provide dimensions of {Lm}, the function must be a 3/2 power.
Thus the ﬁnal desired homogeneous relation for dam flow is: z . . .
Q = CBg" H”, where C rs a dimensmnless constant Ans.
W 1.15 As a practical application of Fig. P1.14, often termed a sharpcrested weir, civil
engineers use the following formula for ﬂow rate: Q = 3.3 8H“, with Q in ft3ls and B and H in feet. [3 this formula dimensionally homogeneous? If not, try to explain the
difﬁculty and how it might be converted to a more homogeneous form. Solution: Clearly the formula cannot be dimensionally homogeneous, because B and H
do not contain the dimension time. The formula would be invalid for anything except English units (ft, sec). By comparing with the answer to Prob. 1.14 just above, we see
that the constant “3.3” hides the square root of the acceleration of gravity. W 1.16 Test the dimensional homogeneity of the boundary—layer x—momentum equation: 1133+ vﬂuﬁst +5
p 8x '0 8y 8x pgx 07y Solution: This equation. like aﬂ theoretical partial differential equations in mechanics,
is dimensionally homogeneous. Test each term in sequence: { a e M } {aw/“1t M}
‘0 ax ’0 ay L3T L Lz'rz' ax L LZTZ { was M i {ﬁrmi M}
pg“ L3T2 L‘TZ’ ax L 1.2T2 All terms have dimension {ML”2T"2}. This e uation ma use any consistent units.
CI 3’ . W 1.17 Investigate the consistency of the HazenWiliiams formula from hydraulics: Q s 61 .913163 PET“
1.. What are the dimensions of the constant “61.9”? Can this equation be used with
conﬁdence for a variety of liquids and gases? Chaptort  Introduction. 9 Solution: Write out the dimensions of each side of the equation:  L3 1 263 {ﬂ}0.54_ M3 MILTZ 0.54
in—{T}{619HD l L —i619]{L l L The constant 61.9 has fractional dimensions: {61.9} = {L"‘5T"'°“M‘°"} Ans.
Clearly, the formula is extremely inconsistent and cannot be used with confidence
for any given fluid or condition or units. Actually, the HazenWilliams formula, still in common use in the watersupply industry, is valid only for we: flow in smooth
pipes larger than 2in. diameter and turbulent velocities less than 10 ft/s and (certain)
English units. This formula should be held at arm's length and given a vote of “No Confidence." 1.18* (“*" means “difficult"—not just a plugandchug. that is) For small particles at Drag x V
low velocities. the ﬁrst (linear) term in Stokes’ _ _ _ "® _ _ _ _ _ _ _ _ _ _ '
drag law, Prob. 1.10, is dominant, hence horizontal F = KV, where K is a constant. Suppose a particle of mass m is constrained to move horizontally from the initial position x = 0
with initial velocity V = V0. Show (a) that its velocity will decrease exponentially with
time; and (b) that it will stop after travelling at distance x = mVo/K. Solution: Set up and solve the differential equation for forces in the xdirection:
dt n KW» t ~ Kt/M
Solve V = Voem and x = I V dt = —‘3~(le""'“) Ans. (a,b) V l _
IF, = Drag = max, or: KV = mg, integrate 91 =  I r
v V o ’ “A Thus, as asked, V drops off exponentially with time, and, as t —+ on, x = mVo/K. 1.19 Marangom' convection arises when a surface has a difference in surface
tension along its length. The dimensionless Marangoni number M is a combination
of thermal diffusivity a: k/(pcp) (where k is the thermal conductivity), length scale
L. viscosity p, and surface tension difference 6Y. If M is proportional to L, find its form. ...
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