Fluid Mechanics, 5
th
Ed. Kundu, Cohen, and
Dowling
Exercise 1.1. Many centuries ago, a mariner poured 100 cm
3
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 cup 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. [Some possibly useful facts: M
w
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. One cup is ~240 ml.]
Solution 1.1. To get started, first list or determine the volumes involved:
d
= volume of water dumped = 100 cm
3
,
c
= volume of a cup ≈ 240 cm
3
, and
V = volume of water in the oceans =
4
R
2
D
,
where, R is the radius of the earth, D is the mean depth of the oceans, and
is the oceans'
coverage fraction. Here we've ignored the ocean volume occupied by salt and have assumed that
the oceans' depth is small compared to the earth's diameter. Putting in the numbers produces:
V 4
(6.37 10
6
m)
2
(3.8 10
3
m)(0.71) 1.376 10
18
m
3
.
For well-mixed oceans, the probability P
o
that any water molecule in the ocean came from the
dumped water is:
P
o
(100 cm
3
of water)
(oceans' volume)
d
V
1.0 10
4
m
3
1.376 10
18
m
3
7.27 10
23
,
Denote the probability that at least one molecule from the dumped water is part of your next cup
as P
1
(this is the answer to the question). Without a lot of combinatorial analysis, P
1
is not easy
to calculate directly. It is easier to proceed by determining the probability P
2
that all the
molecules in your cup are not from the dumped water. With these definitions, P
1
can be
determined from: P
1
= 1 – P
2
. Here, we can calculate P
2
from:
P
2
= (the probability that a molecule was not in the dumped water)
[number of molecules in a cup]
.
The number of molecules, N
c
, in one cup of water is
N
c
240cm
3
1.00g
cm
3
gmole
18.0g
6.023 10
23
molecules
gmole
8.03 10
24
molecules
Thus,
P
2
(1 P
o
)
N
c
(1 7.27 10
23
)
8.0310
24
. Unfortunately, electronic calculators and modern
computer math programs cannot evaluate this expression, so analytical techniques are required.
First, take the natural log of both sides, i.e.
ln(P
2
) N
c
ln(1 P
o
) 8.0310
24
ln(1 7.27 10
23
)
then expand the natural logarithm using ln(1–
) ≈ –
(the first term of a standard Taylor series
for
0
)
ln(P
2
) N
c
P
o
8.0310
24
7.27 10
23
584
,
and exponentiate to find:
P
2
e
584
10
254
... (!)
Therefore, P
1
= 1 – P
2
is very-very close to unity, so there is a virtual certainty that the next cup
of water you drink will have at least one molecule in it from the 100 cm
3
of water dumped many
years ago. So, if one considers the rate at which they themselves and everyone else on the planet
uses water it is essentially impossible to get a truly fresh cup to drink.
剩余570页未读，继续阅读
资源评论
