Solution Manual of Fluid Mechanics 4th Edition.

Chapter 01 • Introduction
1.1 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,
⎛
 = 1012 molecules ⎞⎛ 4.81E−23 g ⎞
⎝| ⎠⎝ ⎠
mm3
|| molecule |

g = 4.81E−5 kg
= 4.81E−11
mm3 m3


Finally, from the perfect gas law, Eq. (1.13), at 20C = 293 K, we obtain the pressure:
⎛ kg ⎞ ⎛ m2 ⎞
p =  RT = | 4.81E−5 | | 287 s2K |(293 K) = 4.0 Pa ns.
3
m
⎝ ⎠⎝ ⎠


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

mt = ∫  dVol = avg (Air Vol)  avg 4 Re2 (Air thickness)
= (0.6 kg/m3 )4 (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:

N =
m(atmosphere)
=
6.1E21 grams  1.3E44 molecules Ans.
molecules
m(one molecule) 4.8E−23 gm/molecule

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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 , velocity V, and surface tension Y may be combined into
a dimensionless group. Find the combination which is proportional to . This group has a
customary name, which begins with C. Can you guess its name?

Solution: The dimensions of these variables are {} = {M/LT}, {V} = {L/T}, and {Y} =
{M/T2}. We must divide  by Y to cancel mass {M}, then work the velocity into the
group:
{L F
{  F { M /LT F {T F hence multiply by {V} = ;
= = ,
{ } { 2 } { } { }
% Y J % M /T J % L J %T J
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:
 
l = 1.26 = 1.26  (RT) (1)

  (RT) p

, Chapter 1 • Introduction 3

where the latter form follows from the ideal-gas law,  = 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”:
{MF {MF { L2 F
{R} = { 2 }
{l} = {L} {} = { LT } {} = { 3 } T {T} = {}
L
% J % J % J
Therefore the above formula (first form) may be written dimensionally as
{M/LT}
{L} = {1.26?} = {1.26?}{L}
{M/L }[{L /T  }{}]
3 2 2


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  = 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.
1/2
(0.0832)[(287)(293)]


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 m.


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) ∫ p dy; (c) 2p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
ft3 ⎛ 1728 gal ⎞⎛ 1 day ⎞ gal
Q = 65340 | || |  340 min Ans. (a)
day ⎝ 231 ft3 ⎠⎝1440 min ⎠

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Similarly, 1850 m3 = 1.85E6 liters. Then a metric unit for this water usage is:
Q = ⎛1.85E6 L ⎞⎛ 1 day ⎞
 21
L Ans. (b)
| day || 86400 sec | s
⎝ ⎠⎝ ⎠


1.8 Suppose that bending stress  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 .

Solution: We are given that  = 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,
{ MF
{} = {y}{fcn(M,I)}, or: = {L}{fcn(M,I)}
%{ LT2 }J { M F
or: the function must have dimensions {fcn(M,I)} =
%{ L2T2 J}
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  is proportional to M also. Now we
have reduced the problem to:
{ M F= { ML2 F
= = −4
yM fcn(I), or { 2 } {L}{ 2 }{fcn(I)}, or: {fcn(I)} {L }
%LT J T
% J
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
 =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:  = (75 MPa)(6894.8) =
10880 lbfin2. Substitute the given data into the proposed formula:

 = 10880 lbf = C My = C (2900 lbfin)(1.5 in) , or: C  1.00 Ans.

in2 I 0.4 in4
The data show that C = 1, or  = My/I, our old friend from strength of materials.