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Exercise 1.1. Many centuries ago, a mariner poured 100 cm3 of water into the ocean. As time
passed, the action of currents, tides, and weather mixed the liquid uniformly throughout the
earth’s oceans, lakes, and rivers. Ignoring salinity, estimate the probability that the next sip (5
ml) of water you drink will contain at least one water molecule that was dumped by the mariner.
Assess your chances of ever drinking truly pristine water. (Consider the following facts: Mw for
water is 18.0 kg per kg-mole, the radius of the earth is 6370 km, the mean depth of the oceans is
approximately 3.8 km and they cover 71% of the surface of the earth.)
Solution 1.1. To get started, first list or determine the volumes involved:
υ
υ
Thus,
P
2=(1−P
o)Nc=(1−7.27 ×10−23 )1.673×1023
. Unfortunately, electronic calculators and modern
computer math programs cannot evaluate this expression, so analytical techniques are required.
Exercise 1.2. An adult human expels approximately 500 ml of air with each breath during
ordinary breathing. Imagining that two people exchanged greetings (one breath each) many
centuries ago, and that their breath subsequently has been mixed uniformly throughout the
atmosphere, estimate the probability that the next breath you take will contain at least one air
molecule from that age-old verbal exchange. Assess your chances of ever getting a truly fresh
breath of air. For this problem, assume that air is composed of identical molecules having Mw =
29.0 kg per kg-mole and that the average atmospheric pressure on the surface of the earth is 100
kPa. Use 6370 km for the radius of the earth and 1.20 kg/m3 for the density of air at room
temperature and pressure.
Solution 1.2. To get started, first determine the masses involved.
m = mass of air in one breath = density x volume =
1.20kg /m3
( )
0.5 ×10−3m3
( )
=
0.60 ×10−3kg
where the factor of two comes from one breath for each person. Denote the probability that at
least one molecule from the age-old verbal exchange is part of your next breath as P1 (this is the
answer to the question). Without a lot of combinatorial analysis, P1 is not easy to calculate
directly. It is easier to proceed by determining the probability P2 that all the molecules in your
Exercise 1.3. The Maxwell probability distribution, f(v) = f(v1,v2,v3), of molecular velocities in a
gas flow at a point in space with average velocity u is given by (1.1).
a) Verify that u is the average molecular velocity, and determine the standard deviations (
σ
1,
σ
2,
σ
3) of each component of u using
σ
i=1
n(vi−ui)2
all v
∫∫∫ f(v)d3v
#
$
%&
'
(
1 2
for i = 1, 2, and 3.
b) Using (1.27) or (1.28), determine n = N/V at room temperature T = 295 K and atmospheric
pressure p = 101.3 kPa.
c) Determine N = nV = number of molecules in volumes V = (10 µm)3, 1 µm3, and (0.1 µm)3.
d) For the ith velocity component, the standard deviation of the average,
σ
a,i, over N molecules
is
σ
a,i =
σ
iN
when N >> 1. For an airflow at u = (1.0 ms–1, 0, 0), compute the relative
uncertainty,
2
σ
a,1 u1
, at the 95% confidence level for the average velocity for the three volumes
listed in part c).
e) For the conditions specified in parts b) and d), what is the smallest volume of gas that ensures
a relative uncertainty in U of one percent or less?
Solution 1.3. a) Use the given distribution, and the definition of an average:
The change of integration variable to
β
=(v1−u1)m2kBT
( )
1 2
changes this integral to:
b) From (1.27),
n V =p kBT=(101.3kPa) [1.381×10−23 J/K⋅295K]=2.487 ×1025 m−3
Exercise 1.4. Using the Maxwell molecular speed distribution given by (1.4),
a) determine the most probable molecular speed,
b) show that the average molecular speed is as given in (1.5),
c) determine the root-mean square molecular speed =
vrms =1
nv2f(v)
0
∞
∫dv
#
$
%&
'
(
1 2
,
d) and compare the results from parts a), b) and c) with c = speed of sound in a perfect gas
under the same conditions.
Solution 1.4. a) The most probable speed, vmp, occurs where f(v) is maximum. Thus, differentiate
(1.4) with respect v, set this derivative equal to zero, and solve for vmp. Start from:
3 2
"
%
0
and this matches the result provided in (1.5).
c) The root-mean-square molecular speed vrms is given by:
vrms
2=1
nv2
∞
∫f(v)dv =4
π
m
2
π
kBT
#
$
%&
'
(
3 2
v4
∞
∫exp −mv2
2kBT
*
+
,
-
.
/dv
.
Exercise 1.5. By considering the volume swept out by a moving molecule, estimate how the
mean-free path, l, depends on the average molecular cross section dimension
d
and the
molecular number density n for nominally spherical molecules. Find a formula for (the ratio
of the mean-free path to the mean intermolecular spacing) in terms of the nominal molecular
volume ( ) and the available volume per molecule (1/n). Is this ratio typically bigger or smaller
than one?
Solution 1.5. The combined collision cross section for two spherical molecules having diameter
d
is
π
d
2
. The mean free path l is the average distance traveled by a molecule between
collisions. Thus, the average molecule should experience one
collision when sweeping a volume equal to
π
d
2l
. If the molecular
ln1 3
d3
Exercise 1.6. Compute the average relative speed,
vr
, between molecules in a gas using the
Maxwell speed distribution f given by (1.4) via the following steps.
a) If u and v are the velocities of two molecules then their relative velocity is: vr = u – v. If the
angle between u and v is
θ
, show that the relative speed is: vr = |vr|
=u2+v2−2uv cos
θ
where
u = |u|, and v = |v|.
b) The averaging of vr necessary to determine
vr
must include all possible values of the two
speeds (u and v) and all possible angles
θ
. Therefore, start from:
vr=1
2n2vrf(u)
all u,v,
θ
∫f(v)sin
θ
d
θ
dvdu
,
and note that
vr
is unchanged by exchange of u and v, to reach:
vr=1
n2u2+v2−2uvcos
θ
θ
=0
π
∫
v=u
∞
∫
u=0
∞
∫sin
θ
f(u)f(v)d
θ
dvdu
c) Note that vr must always be positive and perform the integrations, starting with the angular
one, to find:
vr=1
3n2
2u3+6uv2
uv
v=u
∞
∫
u=0
∞
∫f(u)f(v)dvdu =16kBT
π
#
$
%&
'
(
1 2
=2v
.
Solution 1.6. a) Compute the dot produce of vr with itself:
In u-v coordinates, the integration domain covers the first
quadrant, and the integrand is unchanged when u and v are
swapped. Thus, the u-v integration can be completed above the
v!dv!
u = v!
Use the variable substitution:
α
= mv2/2kBT so that d
α
= mvdv/kBT, which reduces the v-
integration to:
vr=1
3n2u3+6ukBT
m
α
!
"
#$
%
&
mu2kBT
∞
∫
u=0
∞
∫4
π
m
2
π
kBT
!
"
#$
%
&
3 2
exp −
α
{ }
kBT
md
α
f(u)
udu
Exercise 1.7. In a gas, the molecular momentum flux (MFij) in the j-coordinate direction that
crosses a flat surface of unit area with coordinate normal direction i is:
MF
ij =1
Vmvivjf(v)d3v
all v
∫∫∫
where f(v) is the Maxwell velocity distribution (1.1). For a perfect
gas that is not moving on average (i.e., u = 0), show that MFij = p (the pressure), when i = j, and
that MFij = 0, when i ≠ j.
Solution 1.7. Start from the given equation using the Maxwell distribution:
The first integral is equal to
2kBT m
( )
3 2
π
2
( )
while the second two integrals are each equal to
2
π
kBT m
( )
1 2
. Thus:
Exercise 1.8. Consider the viscous flow in a channel of width 2b. The channel is aligned in the
x-direction, and the velocity u in the x-direction at a distance y from the channel centerline is
given by the parabolic distribution
u(y) = U01−y b
( )
2
[ ]
.
Calculate the shear stress
τ
as a
function y,
µ
, b, and Uo. What is the shear stress at y = 0?
Exercise 1.9. Hydroplaning occurs on wet roadways when sudden braking causes a moving
vehicle’s tires to stop turning when the tires are separated from the road surface by a thin film of
water. When hydroplaning occurs the vehicle may slide a significant distance before the film
breaks down and the tires again contact the road. For simplicity, consider a hypothetical version
of this scenario where the water film is somehow maintained until the vehicle comes to rest.
a) Develop a formula for the friction force delivered to a vehicle of mass M and tire-contact area
A that is moving at speed u on a water film with constant thickness h and viscosity
µ
.
b) Using Newton’s second law, derive a formula for the hypothetical sliding distance D traveled
by a vehicle that started hydroplaning at speed Uo
c) Evaluate this hypothetical distance for M = 1200 kg, A = 0.1 m2, Uo = 20 m/s, h = 0.1 mm, and
µ
= 0.001 kgm–1s–1. Compare this to the dry-pavement stopping distance assuming a tire-road
coefficient of kinetic friction of 0.8.
Solution 1.9. a) Assume that viscous friction from the water layer transmitted to the tires is the
only force on the sliding vehicle. Here viscous shear stress at any time will be
µ
u(t)/h, where u(t)
is the vehicle's speed. Thus, the friction force will be A
µ
u(t)/h.
than the estimated stopping distance for hydroplaning.)
Exercise 1.10. Estimate the height to which water at 20°C will rise in a capillary glass tube 3
mm in diameter that is exposed to the atmosphere. For water in contact with glass the contact
angle is nearly 0°. At 20°C, the surface tension of a water-air interface is
σ
= 0.073 N/m.
Solution 1.10. Start from the result of Example 1.4.
Exercise 1.11. A manometer is a U-shaped tube containing mercury of density
ρ
m. Manometers
are used as pressure-measuring devices. If the fluid in tank A has a pressure p and density
ρ
, then
show that the gauge pressure in the tank is: p − patm =
ρ
mgh −
ρ
ga. Note that the last term on the
right side is negligible if
ρ
«
ρ
m. (Hint: Equate the pressures at X and Y.)
Solution 1.9. Start by equating the pressures at X and Y.
Exercise 1.12. Prove that if e(T,
υ
) = e(T) only and if h(T, p) = h(T) only, then the (thermal)
equation of state is (1.28) or p
υ
= kT, where k is constant.
Solution 1.12. Start with the first equation of (1.24): de = Tds – pd
υ
, and rearrange it:
$
'
υ
Now repeat this procedure using the second equation of (1.24), dh = Tds +
υ
dp.
ds =1
Tdh −
υ
Tdp =
∂
s
∂
h
%
&
' (
)
*
p
dh +
∂
s
∂
p
%
&
'
(
)
*
h
dp
.
Exercise 1.13. Starting from the property relationships (1.24) prove (1.31) and (1.32) for a
reversible adiabatic process involving a perfect gas when the specific heats cp and cv are constant.
Solution 1.13. For an isentropic process: de = Tds – pd
υ
= –pd
υ
, and dh = Tds +
υ
dp = +
υ
dp.
which is the second equation of (1.32). The remaining relationship involving the temperature is
found by using the perfect gas law, p =
ρ
RT, to eliminate
ρ
= p/RT:
Exercise 1.14. A cylinder contains 2 kg of air at 50°C and a pressure of 3 bars. The air is
compressed until its pressure rises to 8 bars. What is the initial volume? Find the final volume for
both isothermal compression and isentropic compression.
Solution 1.14. Use the perfect gas law but explicitly separate the mass M of the air and the
volume V it occupies via the substitution
ρ
= M/V:
Exercise 1.15. Derive (1.35) starting from Figure 1.9 and the discussion at the beginning of
Section 1.10.
Solution 1.15. Take the z axis vertical, and consider a small fluid element
δ
m of fluid having
volume
δ
V that starts at height z0 in a stratified fluid medium having a vertical density profile =
ρ
(z), and a vertical pressure profile p(z). Without any vertical displacement, the small mass and
its volume are related by
δ
m =
ρ
(z0)
δ
V. If the small mass is displaced vertically a small distance
ζ
via an isentropic process, its density will change isentropically according to:
Exercise 1.16. Starting with the hydrostatic pressure law (1.14), prove (1.36) without using
perfect gas relationships.
Solution 1.16. The adiabatic temperature gradient dTa/dz, can be written terms of the pressure
gradient:
where the second two equalities are mathematical manipulations that allow the introduction of
∂ρ
&
&
∂
h
!
∂
s
!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 1.17. Assume that the temperature of the atmosphere varies with height z as T = T0 +
Kz where K is a constant. Show that the pressure varies with height as
p=p0
T0
T0+Kz
!
"
#$
%
&
g KR
where
g is the acceleration of gravity and R is the gas constant for the atmospheric gas.
Solution 1.17. Start with the hydrostatic and perfect gas laws, dp/dz = –
ρ
g, and p =
ρ
RT,
eliminate the density, and substitute in the given temperature profile to find:
