没有合适的资源?快使用搜索试试~ 我知道了~
Fluid Mechanics(Fluid Mechanics Kundu-Cohen-Dowling-2012)习题答案
需积分: 5 2 下载量 47 浏览量
2022-12-22
22:21:34
上传
评论
收藏 12.54MB PDF 举报
温馨提示
试读
571页
不可否认的是答案里有些出错的地方,但这份答案也曾救我于水火之中,所以分享给有需求的人! 如果有侵权,联系本人删除!
资源推荐
资源详情
资源评论
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.
Fluid Mechanics, 5
th
Ed. Kundu, Cohen, and
Dowling
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 M
w
=
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/m
3
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/m
3
0.5 10
3
m
3
=
0.60 10
3
kg
M = mass of air in the atmosphere =
4
R
2
(z)dz
z 0
Here, R is the radius of the earth, z is the elevation above the surface of the earth, and
(z) is the
air density as function of elevation. From the law for static pressure in a gravitational field,
dP dz
g
, the surface pressure, P
s
, on the earth is determined from
P
s
P
(z)gdz
z 0
z
so
that:
M 4
R
2
P
s
P
g
4
(6.37 10
6
m)
2
(10
5
Pa) 5.2 10
18
kg
.
where the pressure (vacuum) in outer space = P
∞
= 0, and g is assumed constant throughout the
atmosphere. For a well-mixed atmosphere, the probability P
o
that any molecule in the
atmosphere came from the age-old verbal exchange is
P
o
2 (mass of one breath)
(mass of the whole atmosphere)
2m
M
1.2 10
3
kg
5.2 10
18
kg
2.3110
22
,
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 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
next breath are not from the age-old verbal exchange. 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 verbal exchange)
[number of molecules in a breath]
.
The number of molecules, N
b
, involved in one breath is
N
b
0.6 10
3
kg
29.0g/gmole
10
3
g
kg
6.023 10
23
molecules
gmole
1.25 10
22
molecules
Thus,
P
2
(1 P
o
)
N
b
(1 2.3110
22
)
1.2510
22
. 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
b
ln(1 P
o
) 1.25 10
22
ln(12.3110
22
)
then expand the natural logarithm using ln(1–
) ≈ –
(the first term of a standard Taylor series
for
0
)
ln(P
2
) N
b
P
o
1.25 10
22
2.3110
22
2.89
,
and exponentiate to find:
Fluid Mechanics, 5
th
Ed. Kundu, Cohen, and
Dowling
P
2
e
2.89
0.056
.
Therefore, P
1
= 1 – P
2
= 0.944 so there is a better than 94% chance that the next breath you take
will have at least one molecule in it from the age-old verbal exchange. So, if one considers how
often they themselves and everyone else breathes, it is essentially impossible to get a breath of
truly fresh air.
Fluid Mechanics, 5
th
Ed. Kundu, Cohen, and
Dowling
Exercise 1.3. In Cartesian coordinates, the Maxwell probability distribution, f(u) = f(u
1
,u
2
,u
3
), of
molecular velocities in a gas flow with average velocity U = (U
1
,U
2
,U
3
) is
f (u)
m
2
k
B
T
3 2
exp
m
2k
B
T
u U
2
where n is the number of gas molecules in volume V, m is the molecular mass, k
B
is Boltzmann’s
constant and T is the absolute temperature.
a) Verify that U is the average molecular velocity, and determine the standard deviations (
1
,
2
,
3
) of each component of U using
i
(u
i
U
i
)
2
all u
f (u)d
3
u
1 2
for i = 1, 2, and 3.
b) Using the molecular version of perfect gas law (1.21), determine n/V at room temperature T =
295 K and atmospheric pressure p = 101.3 kPa.
c) Determine n for volumes V = (10
m)
3
, 1
m
3
, and (0.1
m)
3
.
d) For the i
th
velocity component, the standard deviation of the average,
a,i
, over n molecules
is
a,i
=
i
n
when n >> 1. For an airflow at U = (1.0 ms
–1
, 0, 0), compute the relative
uncertainty,
2
a ,1
U
1
, 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 less than one percent?
Solution 1.3. a) Use the given distribution, and the definition of an average:
u
ave
u
all u
f (u)d
3
u
m
2
k
B
T
3 2
u
–
+
–
+
–
+
exp
m
2k
B
T
u U
2
d
3
u
.
Consider the first component of u, and separate out the integrations in the "2" and "3" directions.
(u
1
)
ave
m
2
k
B
T
3 2
u
1
–
+
–
+
–
+
exp
m
2k
B
T
(u
1
U
1
)
2
(u
2
U
2
)
2
(u
3
U
3
)
2
du
1
du
2
du
3
m
2
k
B
T
3 2
u
1
–
+
exp
m(u
1
U
1
)
2
2k
B
T
du
1
exp
m(u
2
U
2
)
2
2k
B
T
du
2
–
+
exp
m(u
3
U
3
)
2
2k
B
T
–
+
du
3
The integrations in the "2" and "3" directions are equal to:
2
k
B
T m
1 2
, so
(u
1
)
ave
m
2
k
B
T
1 2
u
1
–
+
exp
m(u
1
U
1
)
2
2k
B
T
du
1
The change of integration variable to
(u
1
U
1
) m 2k
B
T
1 2
changes this integral to:
(u
1
)
ave
1
2k
B
T
m
1 2
U
1
–
+
exp
2
d
0
1
U
1
U
1
,
where the first term of the integrand is an odd function integrated on an even interval so its
contribution is zero. This procedure is readily repeated for the other directions to find (u
2
)
ave
=
U
2
, and (u
3
)
ave
= U
3
. Using the same simplifications and change of integration variables
produces:
1
2
m
2
k
B
T
3 2
(u
1
U
1
)
2
–
+
–
+
–
+
exp
m
2k
B
T
(u
1
U
1
)
2
(u
2
U
2
)
2
(u
3
U
3
)
2
du
1
du
2
du
3
Fluid Mechanics, 5
th
Ed. Kundu, Cohen, and
Dowling
m
2
k
B
T
1 2
(u
1
U
1
)
2
–
+
exp
m(u
1
U
1
)
2
2k
B
T
du
1
1
2k
B
T
m
2
–
+
exp
2
d
.
The final integral over
is:
2
, so the standard deviations of molecular speed are
1
k
B
T m
1 2
2
3
,
where the second two equalities follow from repeating this calculation for the second and third
directions.
b) From (1.21),
n V p k
B
T (101.2kPa) [1.38110
23
J /K 295K] 2.487 10
25
m
3
c) From n/V from part b):
n 2.487 10
10
for V = 10
3
m
3
= 10
–15
m
3
n 2.487 10
7
for V = 1.0
m
3
= 10
–18
m
3
n 2.487 10
4
for V = 0.001
m
3
= 10
–21
m
3
d) From (1.22), the gas constant is R = (k
B
/m), and R = 287 m
2
/s
2
K for air. Compute:
2
a,1
U
1
2 k
B
T m n
1 2
1m /s
2 RT n
1 2
1m /s 2 287 295 n
1 2
582 n
. Thus,
for V = 10
–15
m
3
:
2
a,1
U
1
= 0.00369,
V = 10
–18
m
3
:
2
a,1
U
1
= 0.117, and
V = 10
–21
m
3
:
2
a,1
U
1
= 3.69.
e) To achieve a relative uncertainty of 1% we need n ≈ (582/0.01)
2
= 3.39
10
9
, and this
corresponds to a volume of 1.36
10
-16
m
3
which is a cube with side dimension ≈ 5
m.
剩余570页未读,继续阅读
资源评论
Caesarqhs
- 粉丝: 0
- 资源: 3
上传资源 快速赚钱
- 我的内容管理 展开
- 我的资源 快来上传第一个资源
- 我的收益 登录查看自己的收益
- 我的积分 登录查看自己的积分
- 我的C币 登录后查看C币余额
- 我的收藏
- 我的下载
- 下载帮助
安全验证
文档复制为VIP权益,开通VIP直接复制
信息提交成功