1
Fluid Mechanics
2
3
Chapter 1
Fluid Particle and Its Mainly Physical Properties
§ 1–1 Introduction
§ 1–2 Fluid Particle and Continuous Medium Hypothesis
§ 1–3 Density,Specific Volume and Relative Density of a Fluid
§ 1–4 Compressibility and Expansion of a Fluid
§ 1–5 Fluid Viscosity
§ 1–6 Forces acting on a Fluid
Exercises of Chapter 1
4
第一章 流体质点及其主要物理性质
§ 1–1 引言
§ 1–2 流体质点与连续介质假设
§ 1–3 流体的密度、比体积和相对密度
§ 1–4 流体的压缩性和膨胀性
§ 1–5 流体的粘性
§ 1–6 作用在流体上的力
第一章 习 题
5
The physical properties of a fluid are determined by the internal
reasons of fluid equilibrium and motion regulations,so we should
first know the concepts and the mainly physical properties of a
fluid before discussing the regulations of fluid mechanics,
Chapter 1
Fluid Particle and Its Mainly Physical Properties
§ 1-1 Introduction
6
流体的物理性质决定于流体平衡和运动规律的内部原因，
因此在没有讨论流体力学规律之前，应首先了解流体的概念和
流体的主要物理性质。
第一章 流体质点及其主要物理性质
§ 1-1 引言
7
Fluid and solid are different forms of material,They all have
the following basic attributes of material,
§ 1-2 Fluid Particle and Continuous Medium Hypothesis
1,The physical properties of fluid
1.Composed of a great deal of molecules,
2.Molecules do random movements continuously,
3.The molecule force exists between molecules,
fluid liq u id
g a s
?
?
?
8
流体与固体是物质的不同表现形式，它们都有下列物质基
本属性,
§ 1-2 流体质点与连续介质假设
一、流体的物理属性
1.由大量分子组成；
2.分子不断作随机热运动；
3.分子与分子之间存在着分子力的作用。
流体
?
?
?
气体
液体
9
But these three basic properties of materials expressing in gas,
liquid and solid have difference in quantity and quality,In the
same volume the number of gas molecules is less than that of
liquid molecules and the number of liquid molecules is less than
that of solid molecules,so the distance between gas molecules is
longer and the gravitation is less than that of others,and liquid
take the second place,
The macroscopical appearance caused by these
microcosmic differences are,
The solid has definite volume and shape,
The liquid has definite volume but have indefinite shape,
The gas has neither definite volume nor definite shape,
10
不过这三个物质基本属性表现在气体、液体
与固体方面却有量与质的差别 。同样体积内的
分子数目，气体少于液体,液体又少于固体，因
此气体分子间距离大，引力小，液体次之。
这些微观的差异导致宏观表象是,
固体有一定的体积和一定的形状；
液体有一定的体积而无一定的形状；
气体既无一定的体积也无一定的形状 。
11
Seeing from mechanics properties,solid can resist pressure,
pull and shear force,so it only occurs lesser distortion and
stops deforming in a certain degree under the action of external
force,
Liquid can’t keep a definite shape because the gravitation
between the molecules is small,so it can only resist pressure but
can’t resist pull and shear force,Liquid will deform continuously
and flow when it is acted by a shear force,
These are the remarkable differences between solid and fluid,
12
从力学性质来看，固体具有抵抗压力、拉力和剪切
力三种能力，因而在外力作用下，通常只发生较小的变
形，而且到一定程度后变形就停止。
流体由于分子间引力小，不能保持一定的形状，所
以它仅能抵抗压力而不能抵抗拉力和剪切力，当它受到
剪切力作用时，就要发生连续不断的变形既流动。
这就是固体与流体的显著区别。
13
2,The concepts of fluid particle
Definition,
Seeing from microstructure,there are big gaps between fluid
molecules and which are not continuous, But to fluid mechanics
which researching macroscopical rules it is to model the physical
bodies and is not to discuss molecular microstructure,
fluid parcel —— a physical body in the fluid whose
macroscopical scale is very small but microcosmic scale is big
enough and has its volume and mass,
fluid particle—— the least unit in fluid whose linear scale can
be ignored,
14
二、流体质点的概念
定义,
从微观结构来看，流体分子与分子之间存在着较大间
隙，是不连续的但是对于研究宏观规律的流体力学来说，一般
不需要讨论分子的微观结构，而是对流体的物理实体加以 模型
化。
流体微团 —— 流体中宏观尺寸非常小而微观尺寸又足够大
的任意一个物理实体，具有自己的体积和质量。
流体质点 —— 流体中可以忽略线性尺度的最小单元。
15
3,Continuous medium hypothesis
In this way,we can use continuous function and field theory to
study the problems about fluid movement and balance successfully,
Euler put forward the hypothesis of continuous medium
mechanical model in 1753,
a,Leave out of account the gaps of molecules,fluid is
considered to be composed of micro groups which have no gaps each
other and distribute continuously throughout the whole region where
the liquid lies in,
b,All the physical variables such as density,speed,pressure,
shear stress and temperature which represent liquid properties are
single values and continuous differentiable functions when fluid
flows continuously,So the scalar quantity fields and vector fields (
flow fields) of all kinds of physical variables come into being,
16
三、连续介质假设
这样，我们就可以顺利地运用连续函数和场论等数学工具研
究流体运动和平衡问题。
欧拉在 1753年提出连续介质力学模型的假设,
a.不考虑分子间隙，认为流体是由相互间没有间隙的微团组
成，连续分布于流体所占据的整个空间；
b.表征流体属性的诸物理量，如密度、速度、压强、切应力、
温度等在流体连续流动时是时间与空间坐标变量的单值、连续
可微函数，从而形成各种物理量的标量场和矢量场（流场）。
17
§ 1-3 Density,Specific Volume and Relative
Density of Fluid
Density,specific volume and relative density are basic concepts
representing fluid mass properties and they have not only differences
but also relations,
1,Density,specific volume
The density of a fluid on one point is,
In Figure 1-1,take a fluid micro group A in fluid randomly
whose volume is,mass is,When the micro unit is
infinitesimal and close to point p (x,y,z ) to become a particle,
the definition is,
V? m?
（ 1— 1）
dV
dm
V
m
V
????
?? 0
l i m?
18
§ 1-3 流体的密度、比容和相对密度
密度、比容和相对密度是代表流体质量性质的几个基本概
念，它们互有区别却又互有联系。
一、密度、比体积
V? ),,( zyxP
如图 1— 1，在流体中任取一个流体微团 A，其微元体积为
,微元质量为 。当微元无限小而趋近 点成为一个质
点时,定义,
m?
（ 1— 1）
一点上流体密度为
0l i mV
m dm
V dV? ??
???
?
19
the specific volume of a fluid at one point is,
Figure 1— 1 fluid micro group
),,( zyxP
m?V?V
m A
x
y
z
o
dm
dV
m
V
V ??
??
?? 0lim?
（ 1— 2）
20
一点上流体的比容为
图 1— 1流体微团
),,( zyxP
m?V?V
m A
x
y
z
o
dm
dV
m
V
V ??
??
?? 0lim?
（ 1— 2）
21
Definition,
From(1— 3） and（ 1— 4） we can obtain
Fluid specific volume is
kgmmV / 3??
（ 1— 4）
3/ mkg
V
m?? （ 1— 3）
Fluid density is
Both and are decided by the temperature and pressure at
different point, If fluid is uniform,then
? ?
uniform fluid— the mass distribution in space is even,but
and still can change with the temperature and pressure,? ?
??
1? （ 1— 5）
22
不同点上的, 均随各点的温度、压强状况而定。如
果流体是均质的，则 ? ?
? ? 均质流体 — 空间上质量分布是均匀的，但, 仍然是
可以随温度和压强而变化的。
定义,?
? 1?
（ 1— 5）
由（ 1— 3）和（ 1— 4）可知,
流体比容
kgmmV / 3??
（ 1— 4）
3/ mkg
V
m?? （ 1— 3）
流体密度
23
2,Relative density
?
?
?
? w
wwm
md ???
（ 1— 6）
Definition,
Relative density — the ratio of the mass of a liquid to distilled
water with the same volume at, Co4
In this formula,
d —— relative density of a fluid;
w —— the corresponding physical variable of
distilled water at ; Co4
24
二、相对密度
蒸馏水的相应物理量。—
流体相对密度；—
Cw
d
o4
式中
?
?
?
? w
wwm
md ???
（ 1— 6）
Co4 相对密度 — 液体质量与同体积 蒸馏水质量之比。即
定义,
25
§ 1-4 Compressibility and Expansion of a Fluid
The relative density,density and specific volume of a fluid are
changing with temperature and pressure,as gaps exist between the
molecules inside a fluid, The distances between molecules minish
and volume is compressed when pressure augments,The distances
between molecules enlarge and volume expands when the
temperature goes up,All fluids have the properties of
compressibility and expansion,
1, Compressibility
definition,
compressibility— at constant temperature the property that the
volume of a fluid contracts under pressure is called compressibility,
as shown in Figure 1-2,
26
§ 1-4 流体的压缩性和膨胀性
流体相对密度、密度、比容随温度与压强变化，其原因
是由于流体内部分子间存在着间隙。压强增大，分子间距减
小，体积压缩；温度升高，分子间距增大，体积膨胀。流体
都具有这种可压缩、能膨胀的性质。
一,压缩性
压缩性 — 在温度不变的条件下，流体在压力作用下体积
缩小的性质称为压缩性。如图 1— 2所示。
定义,
27
dp
dV
Vp
1??? （ 1— 7）
In this formula,
The degree of compressibility is expressed by volume compression
coefficient, the formula is p?; m; m
3
3
volume,of quantity changed dV
volume,original V
－
?
28
式中
3
3
m;m,
体积改变量，
原有体积
?
?
dV
V
dp
dV
Vp
1??? （ 1— 7）
p?
压缩性大小，用体积压缩系数 表示。即
29
-1
3 - 1
c ha nge d qua nt i t y of pr e ssure P a ;
c om pr e ssi on c oe f f i c i e nt of v ol um e P a ;
/ r a t e of c ha nge of v ol um e w he n the pr e s sur e
c ha nge s,m P a,
p
dp
dV dp
?
?
?
?
，
，
¨
t h e p h y s i c a l m e a n i n g o f,p?
the relative rate of change of a fluid volume as increasing per unit pressure under constant temperature,
The reciprocal of fluid compression coefficient is called fluid
elastic modulus,which is represented by E,
p
E ?1?
30
.Pam/;Pa
P a ;
1-3
1-
率，压强改变时的体积变化
体积压缩系数，
压强改变量，
?
?
?
dpdV
dp
p?
的物理意义：p?
当温度不变时每增加单位压强所产生的流体体积相对
变化率。
流体压缩系数的倒数称为流体的弹性模量，用 E表示。
p
E ?1?
31
Figure 1— 2 the volume compression of
a fluid under the condition of constant temperature
p
V
T
dV
dpp ?
dVV ?
T
The compression ratio of a gas under the condition of constant
temperature can obtain from gaseousness equation ( let T=C),
ppV
TmR
p
TmR
dp
d
V
gg
p
1)1()(1
2 ???????
（ 1— 8）
32
图 1— 2流体在等温下的体积压缩
p
V
T
dV
dpp?
dVV ?
T
气体的等温压缩率亦可由气体状态方程（令 ）求得。 CT?
ppV
TmR
p
TmR
dp
d
V
gg
p
1)1()(1
2 ???????
（ 1— 8）
33
as shown in Figure 1-3,
is inversely proportional to P p ?
in the b o u n d o f g a se o u sne ss e q u a ti o n,
d if f ic u l t t o c o m p r e ss
e a sy to c o m p r e ss
p
p
p
p
?
?
? ? ?
? ? ?
2,Expansion
1Pa/ ?p?
Pa/p
CT ?
o
Figure 1— 3 the curve of
compression ratio of a
gas
The property that the temperature of fluid rises and the volume
augments under constant pressure is called expansion,as shown in
Figure 1— 4,
definition,
34
成反比。如图 1— 3 所示。 与 p p ?
压缩容易
压缩困难
的范围内：在气体状态方程式适用
???
???
p
p
p
p
?
?
二、膨胀性
1Pa/ ?p?
Pa/p
CT ?
o
图 1— 3气体压缩率曲线
在压力不变的条件下,流体温度升高，其体积增大的性
质称为膨胀性。如图 1— 4所示。
定义,
35
dT
dV
Vt
1?? (1— 9）
In this formula
the degree of dilatability is represented by volume dilatation
coefficient, t?
./m c h a n g in g et e m p e r a t u r
w h e n v o lu m eof c h a n g e of r a t e t h e; v o lu m eoft c o e f f ic ie n d ila t a t io n t h e
,re t e m p e r a t uofq u a n t it y c h a n g e d t h e
3
1
K
dT
dV
K
KdT
t
，
—
，—；—
?
?
36
膨胀性的大小用体积膨胀系数 表示。即
t?
dT
dV
Vt
1?? （ 1— 9）
式中
./m;
3
1
K
dT
dV
K
KdT
t
率，温度变化时的体积变化—
体积膨胀系数，—；温度改变量，—
?
?
37
The relative rate of change of a fluid volume as increasing
per unit temperature under constant pressure,
The physical meaning of,
t?
（ 1— 10）
, 5 1 Figurein shown T toalproportion inversely is
1 1
c) plet (equation
sgaseousnes from obtained becan t coefficien dilatation gas the
—,
） （
as
T Vp
mR
p
T mR
dT
d
dT
dV
V
t
g g
t
?
? ? ? ? ?
?
38
当压强不变时每增加单位温度所产生的流 体体积相
对变化率。
所示。—成反比，如图与
）（
）求得：状态方程式（令气体膨胀系数可由气体
51
11
T
TVp
mR
p
TmR
dT
d
dT
dV
V
Cp
t
gg
t
?
? ????
?
（ 1— 10）
t? 的 物 理 意 义,
39
The compression coefficient of water
pressure（ at） 5 10 20 40 80
0.538 0.536 0.531 0.528 0.515
Table 1— 1
)/m(10 29 ?? ?p?
The expansion coefficient of water
temperature 1～ 10 10～ 20 40～ 50 60~70 90~100
0.14 0.15 0.42 0.55 0.72
Table 1— 2
)/1(10 4 Cot ???
)( Co
40
水的压缩系数
压强（ at） 5 10 20 40 80
0.538 0.536 0.531 0.528 0.515
表 1— 1
)/m(10 29 ?? ?p?
水的膨胀系数
温度 1～ 10 10～ 20 40～ 50 60~70 90~100
0.14 0.15 0.42 0.55 0.72
表 1— 2
)/1(10 4 Cot ???
Co
41
Example 1—1 Calculate the relative rate of change of water
density when the water pressure was increased from 5at to
10at under normal temperature,
Solution,
this problem is the application of compression coefficient
equation of a liquid under normal temperature,
Solution 1,
94
1
th e n
0, 5 3 8 1 0 ( 1 0 5 ) 9, 8 1 1 0 0, 0 2 6 4 %
p
p
d
dp
d
d p
?
?
?
?
?
?
?
?
? ? ? ? ? ? ? ?
42
例 1—1 水在常温下由 5at增加到 10at时，求水的密度的相对
变化率。
解,此题是液体在常温下压缩性系数公式的应用。
解法一,
%0 2 6 4.01081.9)510(105 3 8.0
1
49 ????????
?
?dpd
dp
d
p
p
?
?
?
?
?
?
则
43
0 2 1
21
0
9
0
0
to in te g r a te th e f o r m u l a o f
l n l n ( )
the n e x p [ ( ) ]
e xp [ 0.5 38 10 ( 10 5 ) 98 10 0]
1.0 00 26 4
n d 0,0 0 0 2 6 4 0,0 2 6 4
so th e r e l a tiv
p
p
p
d
dp
pp
pp
a
?
?
?
? ? ?
?
?
?
??
?
?
?
? ? ?
??
? ? ? ?
?
－
＝ ＝ ％
e r a te o f c h a n g e o f th e d e n sity o f w a t e r is 0,0 2 6 4 %
Solution 2,
44
％。为故水密度的相对变化率
％＝＝
－
所以
则
有
进行积分对
0 2 6 4.0
0 2 6 4.00 0 0 2 6 4.0
0 0 0 2 6 4.1
]9 8 1 0 0)510(10538.0e x p [
)](e x p [
)(lnln
0
0
9
12
0
120
?
??
?
?
?
???
?
?
?
?
????
??
???
?
?
pp
pp
dp
d
p
p
p
解法二,
45
In the applicable range of gaseousness equation,???
tT ?
Figure 1— 4 the volume dilatation of a fluid
under constant pressure
P
dTT ?
dVV ?
dV
T
V
P
1/ ?Kt?
KT /
CP ?
o
Figure 1— 5 the expansion coefficient of gas
46
在气体状态方程式的适用范围内,???
tT ?
图 1— 4流体在定压下的体积膨胀
P
dTT ?
dVV ?
dV
T
V
P
1/ ?Kt?
KT /
CP ?
o
图 1— 5气体膨胀系数
47
3,Incompressible fluid
The fluid whose compression coefficient and expansion
coefficient are zero is called incompressible fluid,
definition,
The volume of this kind of fluid doesn’t minish while being
pressed and doesn’t expand while being heated so v and d are
all constant,It is much easier to study the laws of balance and
motion,
，?
48
三、不可压缩流体
d，,??
这种流体受压体积不减小，受热体积不膨胀，
因而 均为常数，讨论其平衡和运动规律自
然简单得多。
压缩系数和膨胀系数为零的流体叫做不可压缩
流体。
定义,
49
§ 1-5 Viscosity of a Fluid
Viscosity is the capability of a fluid to resist deformation and is
a inherent attribute of a fluid,
1,Newtonian law of internal friction
Because of the gravitation between the molecules of fluid and
the momentum transfer brought from heat motion of molecules the
viscosity of fluid comes into being and brings the internal friction
also,
In 1686 Newton brought forward the law of internal friction
The property that the shear stress comes into being in fluid
when fluid flows is called the viscosity of a fluid,
definition,
50
§ 1-5 流体的粘性
粘性是流体抵抗变形的能力，它是流体的固有属性。
一、牛顿内摩擦定律
流体粘性是由于流体分子间的引力以及分子热运动产生动量
交换，而形成的并因此产生内摩擦力。
1686年牛顿提出牛顿内摩擦定律。
流体运动时内部产生切应力的性质，叫做流体的粘性。
定义,
51
experimental approach,
As shown in Figure 1— 6,Suppose that there are
two big enough parallel flat boards and the symbol h
which represents the distance between them is very
small,The space inside two boards is full of generic
homogeneous fluid,The upper board is fixed and the
lower one moves along a line as a properly constant
speed under the shear force F,The area A is so big that
the influence of the borders of boards can be neglected,
52
实验方法,
0u
如图 1— 6所示。 设有两个足够大，相距 h 很小的平行平板。
中间充满一般的均质流体，上板固定，下板在切向力 F 作用下以
不大的速度 作匀速直线运动。平板面积 A足够大，以至于可
忽略平板边缘的影响。
53
dy
duAT ??? （ 1-11）
in the thin layer whose thickness is, Newton considered
that the value of internal friction T inside a fluid（ that is to say
the value of shear force F ) has relations with the properties of
fluid,
dy
du
dy
moreover it is directly proportional to velocity gradient
and contact area A,but have nothing to do with the pressure
on the contact surface,The law is
definition,
Research an infinitely thin fluid layer,velocity of flow is u
where the coordinate is y,velocity of flow is where the
coordinate is, it is obviously that the velocity gradient is
duu ?
dyy ?
54
dy
duAT ??? （ 1-11）
取无限薄的流体层进行研究，坐标为 y处流速为 u，坐标
为 处流速为,显然在厚度为 的薄层中速度梯度
为,牛顿认为：液层间内摩擦力 T 的大小（也就是切向力
F 的大小）与液体性质有关，并与速度梯度 和接触面积 A
成正比，而与接触面上压力无关。即
dy
du
dy
du
dyy? duu? dy
内容,
55
The physical meaning,
The shear stress is directly proportional to velocity gradient,
In the formula（ 1— 11） the symbol is used to make
and positive, It is to say that when choose the
positive sign and when choose the minus sign,
T ?
0?dydu
0?dydu
?
The fluid according with the internal friction law is called
Newtonian fluid,or else which is called non-Newtonian fluid,
In the formula（ 1— 11） is a coefficient which has
relation to the kinds of fluid and temperature, It is called
dynamical stickiness coefficient (viscosity) and its unit is, sPa?
?
56
符合这样内摩擦定律的流体称为牛顿型流体，否则称为
非牛顿型流体。
式（ 1— 11）中 号是为, 永为正值而设的，即当
时取正号，当 时取负号。
? T ? 0?
dy
du
0?dydu
物理意义,
切应力与速度梯度成正比。
式（ 1— 11）中 是与流体种类、温度有关的系数，称
为动力粘性系数（粘度），单位 。
?
sPa?
57
2,Velocity gradient
0u
0u
0?u
x
y
u
duu?
Figure 1— 6
velocity distributing rule
dy
Choose a micro rectangular unit surface in the moving fluid,
deformed as a parallel quadrangle after,as shown in Figure ( 1— 7）
,dt
0
y
x
udt
dtduu )( ?
?d
Figure 1— 7
velocity gradient
dudt
58
二、速度梯度
dt 在运动流体中取一矩形微元面，经 时间变形运动成为一平
行四边形（图 1— 7）有,
0u
0u
0?u
x
y
u
duu?
图 1— 6速度分布规律
dy
0
y
x
udt
dtduu )( ?
?d
图 1— 7速度梯度
dudt
59
dt
d
dt
dtg
dt
dt
dy
du
dy
du ?? ???? )( （ 1— 13）
so d u d d y d t?? ? ?? ? ? ?
（ 1— 14）
Then,
a,the velocity gradient is equal to the velocity of shear
distortion of a fluid parcel,
b,the shear stress is directly proportional to the velocity of
shear distortion in a fluid,
60
dt
d
dt
dtg
dt
dt
dy
du
dy
du ?? ???? )( （ 1— 13）
dt
d
dy
du ???? ????故 （ 1— 14）
则有,
a.速度梯度等于流体微团的剪切变形速度。
b.流体中的切应力与剪切变形速度成正比。
61 ? v
It is the ratio of dynamical viscosity to density,If the fluid
density is six to one we can’t judge the magnitude of viscosity
according to but according to
3,Kinematic viscosity coefficient
?? /?v （ 1— 15）
Definition,
where the units of are, v s/m2
? ?v When researching fluid motion we often use the ratio of to
density which is called kinematic viscosity coefficient,It is
represented by
?
v
?
the physical meaning of, ?
the physical meaning of, v
It is the shear stress under unit velocity gradient,We can judge
the magnitude of viscosity of the same kind of fluid from, ?
62
三、运动粘性系数
?? /?v （ 1— 15）
的物理意义,?
的物理意义,v
? 单位速度梯度下的切应力，根据 的大小可直接判断
同种流体粘性的大小。
? ?
v
在研究流体运动时，常常使用 与密度 的比值,
称为运动粘性系数。以 表示,
定义,
s/m2 运动粘性系数 的单位是 。 v
?
动力粘度与密度之比。如果流体密度相差很多，不能
根据 的大小判断粘性的大小，而应根据 判断粘性的大
小。
v
63
4,Relations between viscosity and temperature
The fluid viscosity is directly proportional to press but its change
is very little and can be neglected normally,
The fluid viscosity is influenced strongly by temperature,The
gas viscosity is caused mainly by the momentum transfer which
coming from the thermal movement of molecules, So when the
temperature goes up the thermal movement of molecules becomes
strong and viscosity augments,The liquid viscosity is caused
mainly by the attractive forces between molecules,So when the
temperature goes up the momentum of molecules enlarges and
attractive forces increase and viscosity declines,
5,Property of surface tension
Definition,
Because of the attractive forces between molecules,the tiny
tension which the free surface of fluid endured is said to be surface
tension,
64
四、粘性与温度的关系
流体的粘性大小与压力成正比，但变化很小，一般不予考
虑。
流体的粘性受温度的影响很大，气体的粘性主要是由分子
热运动产生的运动的动量交换引起的，因此当温度升高时，分
子热运动加剧，粘性增大。流体的粘性主要由分子之间的引力
产生，当温度升高时，分子动量增加，引力减小，粘性下降。
五、表面张力特性
定义,
由于分子间的引力，在液体自由表面上能承受微小的张
力称表面张力。
65
Capillarity,put a thin glass tube with two open ends up straightly
into a liquid,The liquid in the tube will rise or fall under the action
of surface tension which is said to be capillarity,
Because the weight of the liquid column equals to the vertical
component of the accessional press of the surface tension,then
????? c o s4 2 dhd ?
so
dh ?
?? c o s4?
In the formula,
,a n g l ec o n t a c t
m;/ t e n s i o ns u r f a c e oft c o e f f i c i e n;m/l i q u i d of w e i g h t s p e c i f i c
m; t u b eg l a s s ofd i a m e t e r t h e
m;s u f a c e l i q u i d of r i s e t h e
3
—
，—
，—
，—
，—
?
?
?
?
?
d
h
66
由于重力与表面张力产生的附加压力在垂直方向分力平衡，
所以有,
????? c o s4 2 dhd ?
则
dh ?
?? c o s4?式中,
接触角。—
表面张力系数，—
液体重度，—
玻璃管直径，—
液面上升高度，—
?
?
?
m;/;m/
m;
m;
3
?
?
d
h
毛细管现象：将两端开口的细玻璃管竖立在液体中，在表面张力
作用下，管中液体将上升或下降一个高度，称为毛细管现象。
67
? ?
h
?
?
h
Figure 1— 8 capillarity
Obtain from experiment,the contact angle of water and glass is
when temperature is ; the contact
angle of hydrargyrum and glass is C
o20N / m0 7 2 8.0,9~3 ?? ?? oo
。N/m51.0,140~139 ?? ?? oo
hydrargyrum water
68
水
? ?
h
汞 ?
?
h
图 1— 8 表 面 管 现 象
试验测得,时水与玻璃的接触角
汞的接触角
Co20 N / m0 7 2 8.0,9~3 ?? ?? oo
。N / m51.0,140~139 ?? ?? oo
69
33
wa ter
35
6 2 6 3
3
5
9 9 8,2 k g / m,1,2 0 5 k g / m
1,0 0 2 1 0 P a s,1,8 1 1 0 P a s
1,0 0 3 1 0 m / s,1 5,0 1 0 m / s
1,0 0 2 1 0
5 5,3 4 t i m es
1,8 1 1 0
998.2
1.205
air
w a ter
air
w a ter
air
w a ter
air
w a ter
air
??
??
??
?
?
?
?
??
??
?
?
? ? ? ?
??
?
?
＝ ＝
＝ ＝
＝ ＝
＝ ＝ （ ）
＝
-6
-6
8 2 8,3 9 t i m es
1 5,0 1 0
1 4,9 6 t i m es
1,0 0 3 1 0
air
w a ter
?
?
?
?
＝ （ ）
＝ ＝ （ ）
Example 1—2 At the same temperature, Try to
demonstrate that at for water and gas which one is easy to
flow.（ use data to explain）
空气水 ?? ?
Co20
solution,when the temperature is the following data are
checked from the physical properties tables of water and air,
Co20
70
例 1—2 在相同温度下,试论证在 时，水和空
气相比，哪种流体易于流动（用数据说明）
空气水 ?? ? Co20
解,在 时，从水和空气物理特性表中查取 Co20
（倍）＝＝
（倍）＝＝
（倍）＝＝
＝＝
＝＝
＝＝
水
空气
空气
水
空气
水
空气水
空气水
空气水
1 4, 9 6
101, 0 0 3
101 5, 0
39.828
205.1
2.998
34.55
1081.1
10002.1
s/m100.15,/sm10003.1
sPa1081.1,sPa10002.1
k g / m205.1,k g / m2.998
6-
6-
5
3
3626
53
33
?
?
?
?
??
????
?
?
??
??
?
?
?
?
?
?
??
??
??
71
The example shows that it can’t judge the fluidness of a fluid
directly from but from at the same temperature, Because
that the excludes the influence of fluid density and only keeps
the movement property parameters,Which is the meaning of
importing the viscosity coefficient,
? ?
?
conclusion,At the same temperature, Air is hard to
flow compared with water,
w g ? ? ?
72
结论,在相同温度下，,空气与水相比较，空气不
易于流动。
水空气 ?? ?
此例题说明：在相同温度下，只从 值的大小不能直接
判断流体的流动性，而应由 值的大小才能够直接判别流体
的流动性。因为 值排除了流体密度的影响，只保留其运动
特征参数，这也是引入粘性系数的意义所在。
?
?
?
73
6,Concept of ideal fluid
Definition,
In fact all fluids have viscosity,The purpose to put forward
the concept of ideal fluid is to research the fluid movement rules
and simplify the deduce of the theory equations greatly,
Suppose that the viscosity of a fluid doesn’t exist,that is to
say,The fluid is said to be ideal fluid or inviscid fluid,0v ?
74
六、理想流体的概念
0v? 假定不存在粘性，即其粘度 的流体为理想流
体或无粘性流体。
定义,
实际上，一切流体都具有粘性，提出理想流体的概念在
于研究流体运动规律时，对理论方程的推导大为简化。
75
§ 1-6 Forces on a Fluid
x
y
Z
W??
V?
A?
O
Figure 1— 8 body force
and surface force
F??
T??
mF
??
IF
??RF
??
Each fluid particle is acted by all kinds of forces no matter it
keeps moving or balance,According to the behaviors of the forces
there are two classes of forces, body force and surface force,As
shown in Figure 1— 8,Choose a random fluid parcel to research
which volume is, V?
76
§ 1-6 作用在流体上的力
流体每一质点无论处于运动或平衡状态，都受到各种力。按
力的表现形式分为质量力和表面力两类。如图 1— 8所示。取体积
为 的任意微团进行研究。 V?
x
y
Z
W??
V?
A?
O
图 1— 8质量力与表面力
F??
T??
mF
??
IF
??RF
??
77
1,Body force
The force which is directly proportional to the mass of a fluid
parcel and acts on the mass center is called body force,
Definition,
Body force
gravitation
Linear motion inertia force
centrifugal inertia force
--- ---
2
?
??
??
??
???
???
???
rmF
amF
gmW
R
?
78
一、质量力
与流体微团质量成正比并且集中作用在微团质量中心上的力
称为质量力。
定义,
质量力
重力
直线运动惯性力
离心惯性力
--- ---
2
?
??
??
??
???
???
???
rmF
amF
gmW
R
?
79
Definition,
kfjfiff zyxm ??? ???
dm
Fd
m
Ff mm
mm
??
????
?? 0
lim （ 1— 16）
The body force acting on unit mass fluid is called unit body
force,and is expressed as follows,
In the formula
—— the mass of the fluid parcel；
—— body force acting on the parcel；
m?
mF
??
)( kfjfifdmfdmFd zyxmm ???? ??????
（ 1— 17）
,, are projections of unit body force on the
axes,
xf yf zf zyx,,
80
定义,
kfjfiff zyxm ??? ???
dm
Fd
m
Ff mm
mm
??
????
?? 0
lim （ 1— 16）
式中
—— 流体微元体的质量；
—— 作用在该微元体上的质量力；
m?
mF
??
)( kfjfifdmfdmFd zyxmm ???? ??????
（ 1— 17）
,, 为单位质量力在 轴的投影。 xf
yf zf zyx,,
单位质量流体所受的质量力称为单位质量力，记作
81
2,Surface force
a,The press force along the inner normal of surface
b,The friction along tangent orientation of surface
Classified according to the act direction,
The force that is directly proportional to surface area of a fluid
and distributes on the fluid surface is called surface force,
definition,
{
82
二、表面力
a.沿表面内法线方向的压力；
b.沿表面切向的摩擦力。
按其作用方向分类,
大小与流体表面积成正比而且分布作用在流体表面上的力
称为表面力。
定义,
{
83
Then the pressure stress of every point is
A
Fp
A ?
??
??
?
0
lim （ 1— 18）
So the surface press is
ApF ??? （ 1— 19）
Shear stress of every point is
A
T
A ?
??
??
??
0
lim? （ 1— 20）
So the surface shear force is
AT ?? ??? （ 1— 21）
As shown in Figure1— 8,choose a micro area on the fluid
parcel,suppose the tiny press force acting on is and the
tiny shear force is 。
A?
A?
T??
F??
84
如图 1— 8所示，在流体微团上取微元面积,设作用在
上的微小压力,微小切力 。
A? A?
F?? T??
则各点处的压应力为,
A
Fp
A ?
??
??
?
0
lim （ 1— 18）
故表面上的压力为,
ApF ??? （ 1— 19）
各点处的切应力为,
A
T
A ?
??
??
??
0
lim? （ 1— 20）
故表面上的切力为,
AT ?? ??? （ 1— 21）
85
Exercises of Chapter 1
C
V
V
TTT
V
V
ppp
kk
kk
o4.010
0
1
0
0
4.10
0
0
0
1 0 7k3 8 02)152 7 3()()(
k P a 3.2 6 723.1 0 1)()(
???????
?????
??
?
?
?
?
1—1 The original state of air is and,
After adiabatic compression in the cylinder the volume of the
air reduces a half,What are the temperature and pressure of
the end state?
k P a3.1 0 10 ?p 150 Ct o?
solution,In the process of adiabatic compression,the process
equation is, in the formula k =1.4,k is called
adiabatic exponent of air, Uniting the state equation of ideal gas
we can obtain
C o n s tpV k ?
86
第一章 习 题
1—1 空气初始状态为,在汽缸内绝
热压缩后体积减少了一半，求终态温度和压强。
k P a3.1 0 1,15 00 ?? pCt o
解,绝热压缩时，其过程方程为,式中 k=1.4,称
为空气绝热指数。联立理想气体状态方程可得
C o n s tpV k ?
C
V
V
TTT
V
V
ppp
kk
kk
o4.010
0
1
0
0
4.10
0
0
0
1 0 7k3 8 02)152 7 3()()(
k P a 3.2 6 723.1 0 1)()(
???????
?????
??
?
?
?
?
87
solution,the exercise is the question to calculate the net increment
of water volume in the system,we can do as the formula (1— 9）,
3m25080 0 5.0
t h e n
1
????
?
?
V d TdV
dT
dV
V
t
t
?
?
?
d ia l a tio n w a te r b a n k
boiler
radiator
3m8
Co50
1—2 In heating system there is a dilatation water tank,The whole
volume of the water in the system is, The largest temperature rise
is and the dilatation coefficient is,what is the
smallest cubage of the water bank? 0 0 5.0?t?
so the smallest cubage of the dilatation water bank is 2, But
in engineering design we should pay attention to keep surplus to
ensure the safety of the system,
3 m
88
1—2 采暖系统在顶部设一膨胀水箱，系统内的水总体积为,
最大温升,膨胀系数,求该水箱的最小容积？
3m8
Co50 005.0?t?
解,该题为求解系统内水体积净增量的问题，可依（ 1— 9）式
进行求解。
3m25080 0 5.0
1
????
?
?
V d TdV
dT
dV
V
t
t
?
?
则
故膨胀水箱的最小体积应为 2立方米，但在工程设计中，应
注意按照设计规范增加一定的富裕量，以确保系统安全。
? 膨胀水箱
锅炉
散热器
89
solution,This is a simple and familiar example that Newtonian
internal friction law is applied to engineering,There are two
emphases,the one is to simply the velocity gradient in the
lubricant,that is to think the distribute rule is linear； the other is
to build the balance equation of force correctly,Since it slides at
uniform velocity,the component of gravity is equal to the
resistance of viscosity,
V
? ?
13
5
12
5kg
cm1 5kg
1—3 The bottom surface of a wood block is, thickness
is and the mass is, It slides down along a incline which is
spreaded lubricant at a speed of, The thickness of
lubricant is 1, what is the dynamical viscosity coefficient?
2cm4514 ?
m /s1?V
mm
90
1—3 一木块底面积为,厚度为,质量为,沿
着涂有润滑油的斜面以速度 等速度下滑，油层厚度,
求润滑油的动力粘性系数。
2cm4514 ? cm1 5kg
m/s1?V mm1
解,这是牛顿内摩擦定律在工程中应用的一个简单而又常见的
例子。求解此题有两个重点，一是对油层内速度梯度进行
简化，即认为是线性分布规律；二是正确列出力的平衡方
程。
由于是等速下滑，故重力分力与粘性阻力相等
V
? ?
13
5
12
5kg
91
B a l a nc e e qua ti on,g sin,e l oc it y gr a die nt
substit uti ng into 1 11 g sin
g sin
T he n
S ubstit uti ng k now n da ta into i t H e nc e 0,104 7 P a s
du V
T m v
dy
du
Am
dy
m
V
A
?
?
??
?
?
?
?
??
?
?
（ — ）
＝
,＝
Caution,When solving,all physical quantities must adopt the same
unit system,in order to avoid conversion mistakes,
92
g si n,
1 11 g si n
g si n
0.10 47 P a s
du V
Tm
dy
du
Am
dy
m
V
A
?
?
??
?
?
?
?
??
?
?
平 衡 方 程 为 速 度 梯 度 为
代 入 式 （ — ）
解 出 ＝
代 入 已 知 数 据, 解 得 ＝
注意,在解题时，所有物理量的单位必须采用相同单位制，避免
出现换算错误。
93
m5.0?H
1—4 A cone rotates around its vertical center axis at uniform
velocity,The gap between two cones is,It filled with
lubricant which, In Figure 1-2,,
,,
mm1??
s0, 1 P a??? m3.0?R
1/s 16??
Solution,This problem belongs to the application of Newtonian
internal friction law,It’s feature is that the action radius,
contact area and rotation velocity change with h,We should
find its changing rules and carry out the idea which solving
with physical methods,
Exercise Figure1— 2
?
R
? r
h
dh
H
94
1—4 一圆锥体绕竖直中心轴等速旋转，锥体与固定的外锥体之
间的隙缝,其中充满 的润滑油。已知锥体
顶面半径,锥体高度,当旋转角速度
时，求所需要的旋转力矩。
mm1?? s0, 1 P a???
m3.0?R m5.0?H 1/s 16??
解,此题属于牛顿那摩擦定律应用。该题的特点是作用半径，
液体和固壁接触面积及锥体旋转线速度都随高度变化，应逐个
找出其变化规律并贯彻物理方法解题的思想。
?
R
? r
h
dh
H
习题 1— 2图
95
In Figure,rotating torque expression of a fluid parcel
rdAdydurdAdM ????? ??
(1) Changing rule of the cone’s radius r ?ta n?? hr
(2) The expression of dA which is corresponding to dh
????? c o st a n2c o s2
dhhdhrdA ???
h r u dy du
2
tan
? ?
?
? ? ? ? ? ?
3 3 dh h dM
cos
1 tan ? ? ? ?, ?
? ? ?
Substituting above formulas into dM,
? (3)Because is very small,We can think that the changing of
velocity gradient is linear,
96
如图所示，旋转力矩的微元表达式
rdAdydurdAdM ????? ??
( 1 )锥体半径 r的变化规律 ?ta n?? hr
( 2 )对应 dh 的 dA 表达式
????? c o st a n2c o s2
dhhdhrdA ???
33
3
ta n
1
2 ta n
c o s
d u u r
h
dy
dM
d M h d h
?
??
?
? ? ?
?
? ? ?
??
? ? ?
? ? ?
（ ） 因 为 很 小, 可 把 速 度 梯 度 按 线 性 变 化 考 虑
将 上 三 式 代 入 表 达 式 中, 整 理 得
97
33
34
0
4 se e k f o r th e g e n e r a l m o m e n t
ta n ta n
2
c o s 2 c o s
substi tut ing the kn ow n d a ta int o it the r e su l t is
3 9,6 ( N m )
( in i t t a n,he nc e 31 )
H
o
d M h d h H
M
R
H
? ? ? ?? ?
??
? ? ? ?
??
??
??
?
??
（ ）
，
＝
98
) 31,t a n (
m)N( 6.39
c o s
t a n
2c o s
t a n
2
4
4
3
0
3
3
o
H
H
R
M
HdhhdMM
＝求得其中
代入已知数据，解得
）求总力矩（
??
?
?
?
? ? ?
?
?
?
?
??
?
??
??? ? ?
99
100
Fluid Mechanics
2
3
Chapter 1
Fluid Particle and Its Mainly Physical Properties
§ 1–1 Introduction
§ 1–2 Fluid Particle and Continuous Medium Hypothesis
§ 1–3 Density,Specific Volume and Relative Density of a Fluid
§ 1–4 Compressibility and Expansion of a Fluid
§ 1–5 Fluid Viscosity
§ 1–6 Forces acting on a Fluid
Exercises of Chapter 1
4
第一章 流体质点及其主要物理性质
§ 1–1 引言
§ 1–2 流体质点与连续介质假设
§ 1–3 流体的密度、比体积和相对密度
§ 1–4 流体的压缩性和膨胀性
§ 1–5 流体的粘性
§ 1–6 作用在流体上的力
第一章 习 题
5
The physical properties of a fluid are determined by the internal
reasons of fluid equilibrium and motion regulations,so we should
first know the concepts and the mainly physical properties of a
fluid before discussing the regulations of fluid mechanics,
Chapter 1
Fluid Particle and Its Mainly Physical Properties
§ 1-1 Introduction
6
流体的物理性质决定于流体平衡和运动规律的内部原因，
因此在没有讨论流体力学规律之前，应首先了解流体的概念和
流体的主要物理性质。
第一章 流体质点及其主要物理性质
§ 1-1 引言
7
Fluid and solid are different forms of material,They all have
the following basic attributes of material,
§ 1-2 Fluid Particle and Continuous Medium Hypothesis
1,The physical properties of fluid
1.Composed of a great deal of molecules,
2.Molecules do random movements continuously,
3.The molecule force exists between molecules,
fluid liq u id
g a s
?
?
?
8
流体与固体是物质的不同表现形式，它们都有下列物质基
本属性,
§ 1-2 流体质点与连续介质假设
一、流体的物理属性
1.由大量分子组成；
2.分子不断作随机热运动；
3.分子与分子之间存在着分子力的作用。
流体
?
?
?
气体
液体
9
But these three basic properties of materials expressing in gas,
liquid and solid have difference in quantity and quality,In the
same volume the number of gas molecules is less than that of
liquid molecules and the number of liquid molecules is less than
that of solid molecules,so the distance between gas molecules is
longer and the gravitation is less than that of others,and liquid
take the second place,
The macroscopical appearance caused by these
microcosmic differences are,
The solid has definite volume and shape,
The liquid has definite volume but have indefinite shape,
The gas has neither definite volume nor definite shape,
10
不过这三个物质基本属性表现在气体、液体
与固体方面却有量与质的差别 。同样体积内的
分子数目，气体少于液体,液体又少于固体，因
此气体分子间距离大，引力小，液体次之。
这些微观的差异导致宏观表象是,
固体有一定的体积和一定的形状；
液体有一定的体积而无一定的形状；
气体既无一定的体积也无一定的形状 。
11
Seeing from mechanics properties,solid can resist pressure,
pull and shear force,so it only occurs lesser distortion and
stops deforming in a certain degree under the action of external
force,
Liquid can’t keep a definite shape because the gravitation
between the molecules is small,so it can only resist pressure but
can’t resist pull and shear force,Liquid will deform continuously
and flow when it is acted by a shear force,
These are the remarkable differences between solid and fluid,
12
从力学性质来看，固体具有抵抗压力、拉力和剪切
力三种能力，因而在外力作用下，通常只发生较小的变
形，而且到一定程度后变形就停止。
流体由于分子间引力小，不能保持一定的形状，所
以它仅能抵抗压力而不能抵抗拉力和剪切力，当它受到
剪切力作用时，就要发生连续不断的变形既流动。
这就是固体与流体的显著区别。
13
2,The concepts of fluid particle
Definition,
Seeing from microstructure,there are big gaps between fluid
molecules and which are not continuous, But to fluid mechanics
which researching macroscopical rules it is to model the physical
bodies and is not to discuss molecular microstructure,
fluid parcel —— a physical body in the fluid whose
macroscopical scale is very small but microcosmic scale is big
enough and has its volume and mass,
fluid particle—— the least unit in fluid whose linear scale can
be ignored,
14
二、流体质点的概念
定义,
从微观结构来看，流体分子与分子之间存在着较大间
隙，是不连续的但是对于研究宏观规律的流体力学来说，一般
不需要讨论分子的微观结构，而是对流体的物理实体加以 模型
化。
流体微团 —— 流体中宏观尺寸非常小而微观尺寸又足够大
的任意一个物理实体，具有自己的体积和质量。
流体质点 —— 流体中可以忽略线性尺度的最小单元。
15
3,Continuous medium hypothesis
In this way,we can use continuous function and field theory to
study the problems about fluid movement and balance successfully,
Euler put forward the hypothesis of continuous medium
mechanical model in 1753,
a,Leave out of account the gaps of molecules,fluid is
considered to be composed of micro groups which have no gaps each
other and distribute continuously throughout the whole region where
the liquid lies in,
b,All the physical variables such as density,speed,pressure,
shear stress and temperature which represent liquid properties are
single values and continuous differentiable functions when fluid
flows continuously,So the scalar quantity fields and vector fields (
flow fields) of all kinds of physical variables come into being,
16
三、连续介质假设
这样，我们就可以顺利地运用连续函数和场论等数学工具研
究流体运动和平衡问题。
欧拉在 1753年提出连续介质力学模型的假设,
a.不考虑分子间隙，认为流体是由相互间没有间隙的微团组
成，连续分布于流体所占据的整个空间；
b.表征流体属性的诸物理量，如密度、速度、压强、切应力、
温度等在流体连续流动时是时间与空间坐标变量的单值、连续
可微函数，从而形成各种物理量的标量场和矢量场（流场）。
17
§ 1-3 Density,Specific Volume and Relative
Density of Fluid
Density,specific volume and relative density are basic concepts
representing fluid mass properties and they have not only differences
but also relations,
1,Density,specific volume
The density of a fluid on one point is,
In Figure 1-1,take a fluid micro group A in fluid randomly
whose volume is,mass is,When the micro unit is
infinitesimal and close to point p (x,y,z ) to become a particle,
the definition is,
V? m?
（ 1— 1）
dV
dm
V
m
V
????
?? 0
l i m?
18
§ 1-3 流体的密度、比容和相对密度
密度、比容和相对密度是代表流体质量性质的几个基本概
念，它们互有区别却又互有联系。
一、密度、比体积
V? ),,( zyxP
如图 1— 1，在流体中任取一个流体微团 A，其微元体积为
,微元质量为 。当微元无限小而趋近 点成为一个质
点时,定义,
m?
（ 1— 1）
一点上流体密度为
0l i mV
m dm
V dV? ??
???
?
19
the specific volume of a fluid at one point is,
Figure 1— 1 fluid micro group
),,( zyxP
m?V?V
m A
x
y
z
o
dm
dV
m
V
V ??
??
?? 0lim?
（ 1— 2）
20
一点上流体的比容为
图 1— 1流体微团
),,( zyxP
m?V?V
m A
x
y
z
o
dm
dV
m
V
V ??
??
?? 0lim?
（ 1— 2）
21
Definition,
From(1— 3） and（ 1— 4） we can obtain
Fluid specific volume is
kgmmV / 3??
（ 1— 4）
3/ mkg
V
m?? （ 1— 3）
Fluid density is
Both and are decided by the temperature and pressure at
different point, If fluid is uniform,then
? ?
uniform fluid— the mass distribution in space is even,but
and still can change with the temperature and pressure,? ?
??
1? （ 1— 5）
22
不同点上的, 均随各点的温度、压强状况而定。如
果流体是均质的，则 ? ?
? ? 均质流体 — 空间上质量分布是均匀的，但, 仍然是
可以随温度和压强而变化的。
定义,?
? 1?
（ 1— 5）
由（ 1— 3）和（ 1— 4）可知,
流体比容
kgmmV / 3??
（ 1— 4）
3/ mkg
V
m?? （ 1— 3）
流体密度
23
2,Relative density
?
?
?
? w
wwm
md ???
（ 1— 6）
Definition,
Relative density — the ratio of the mass of a liquid to distilled
water with the same volume at, Co4
In this formula,
d —— relative density of a fluid;
w —— the corresponding physical variable of
distilled water at ; Co4
24
二、相对密度
蒸馏水的相应物理量。—
流体相对密度；—
Cw
d
o4
式中
?
?
?
? w
wwm
md ???
（ 1— 6）
Co4 相对密度 — 液体质量与同体积 蒸馏水质量之比。即
定义,
25
§ 1-4 Compressibility and Expansion of a Fluid
The relative density,density and specific volume of a fluid are
changing with temperature and pressure,as gaps exist between the
molecules inside a fluid, The distances between molecules minish
and volume is compressed when pressure augments,The distances
between molecules enlarge and volume expands when the
temperature goes up,All fluids have the properties of
compressibility and expansion,
1, Compressibility
definition,
compressibility— at constant temperature the property that the
volume of a fluid contracts under pressure is called compressibility,
as shown in Figure 1-2,
26
§ 1-4 流体的压缩性和膨胀性
流体相对密度、密度、比容随温度与压强变化，其原因
是由于流体内部分子间存在着间隙。压强增大，分子间距减
小，体积压缩；温度升高，分子间距增大，体积膨胀。流体
都具有这种可压缩、能膨胀的性质。
一,压缩性
压缩性 — 在温度不变的条件下，流体在压力作用下体积
缩小的性质称为压缩性。如图 1— 2所示。
定义,
27
dp
dV
Vp
1??? （ 1— 7）
In this formula,
The degree of compressibility is expressed by volume compression
coefficient, the formula is p?; m; m
3
3
volume,of quantity changed dV
volume,original V
－
?
28
式中
3
3
m;m,
体积改变量，
原有体积
?
?
dV
V
dp
dV
Vp
1??? （ 1— 7）
p?
压缩性大小，用体积压缩系数 表示。即
29
-1
3 - 1
c ha nge d qua nt i t y of pr e ssure P a ;
c om pr e ssi on c oe f f i c i e nt of v ol um e P a ;
/ r a t e of c ha nge of v ol um e w he n the pr e s sur e
c ha nge s,m P a,
p
dp
dV dp
?
?
?
?
，
，
¨
t h e p h y s i c a l m e a n i n g o f,p?
the relative rate of change of a fluid volume as increasing per unit pressure under constant temperature,
The reciprocal of fluid compression coefficient is called fluid
elastic modulus,which is represented by E,
p
E ?1?
30
.Pam/;Pa
P a ;
1-3
1-
率，压强改变时的体积变化
体积压缩系数，
压强改变量，
?
?
?
dpdV
dp
p?
的物理意义：p?
当温度不变时每增加单位压强所产生的流体体积相对
变化率。
流体压缩系数的倒数称为流体的弹性模量，用 E表示。
p
E ?1?
31
Figure 1— 2 the volume compression of
a fluid under the condition of constant temperature
p
V
T
dV
dpp ?
dVV ?
T
The compression ratio of a gas under the condition of constant
temperature can obtain from gaseousness equation ( let T=C),
ppV
TmR
p
TmR
dp
d
V
gg
p
1)1()(1
2 ???????
（ 1— 8）
32
图 1— 2流体在等温下的体积压缩
p
V
T
dV
dpp?
dVV ?
T
气体的等温压缩率亦可由气体状态方程（令 ）求得。 CT?
ppV
TmR
p
TmR
dp
d
V
gg
p
1)1()(1
2 ???????
（ 1— 8）
33
as shown in Figure 1-3,
is inversely proportional to P p ?
in the b o u n d o f g a se o u sne ss e q u a ti o n,
d if f ic u l t t o c o m p r e ss
e a sy to c o m p r e ss
p
p
p
p
?
?
? ? ?
? ? ?
2,Expansion
1Pa/ ?p?
Pa/p
CT ?
o
Figure 1— 3 the curve of
compression ratio of a
gas
The property that the temperature of fluid rises and the volume
augments under constant pressure is called expansion,as shown in
Figure 1— 4,
definition,
34
成反比。如图 1— 3 所示。 与 p p ?
压缩容易
压缩困难
的范围内：在气体状态方程式适用
???
???
p
p
p
p
?
?
二、膨胀性
1Pa/ ?p?
Pa/p
CT ?
o
图 1— 3气体压缩率曲线
在压力不变的条件下,流体温度升高，其体积增大的性
质称为膨胀性。如图 1— 4所示。
定义,
35
dT
dV
Vt
1?? (1— 9）
In this formula
the degree of dilatability is represented by volume dilatation
coefficient, t?
./m c h a n g in g et e m p e r a t u r
w h e n v o lu m eof c h a n g e of r a t e t h e; v o lu m eoft c o e f f ic ie n d ila t a t io n t h e
,re t e m p e r a t uofq u a n t it y c h a n g e d t h e
3
1
K
dT
dV
K
KdT
t
，
—
，—；—
?
?
36
膨胀性的大小用体积膨胀系数 表示。即
t?
dT
dV
Vt
1?? （ 1— 9）
式中
./m;
3
1
K
dT
dV
K
KdT
t
率，温度变化时的体积变化—
体积膨胀系数，—；温度改变量，—
?
?
37
The relative rate of change of a fluid volume as increasing
per unit temperature under constant pressure,
The physical meaning of,
t?
（ 1— 10）
, 5 1 Figurein shown T toalproportion inversely is
1 1
c) plet (equation
sgaseousnes from obtained becan t coefficien dilatation gas the
—,
） （
as
T Vp
mR
p
T mR
dT
d
dT
dV
V
t
g g
t
?
? ? ? ? ?
?
38
当压强不变时每增加单位温度所产生的流 体体积相
对变化率。
所示。—成反比，如图与
）（
）求得：状态方程式（令气体膨胀系数可由气体
51
11
T
TVp
mR
p
TmR
dT
d
dT
dV
V
Cp
t
gg
t
?
? ????
?
（ 1— 10）
t? 的 物 理 意 义,
39
The compression coefficient of water
pressure（ at） 5 10 20 40 80
0.538 0.536 0.531 0.528 0.515
Table 1— 1
)/m(10 29 ?? ?p?
The expansion coefficient of water
temperature 1～ 10 10～ 20 40～ 50 60~70 90~100
0.14 0.15 0.42 0.55 0.72
Table 1— 2
)/1(10 4 Cot ???
)( Co
40
水的压缩系数
压强（ at） 5 10 20 40 80
0.538 0.536 0.531 0.528 0.515
表 1— 1
)/m(10 29 ?? ?p?
水的膨胀系数
温度 1～ 10 10～ 20 40～ 50 60~70 90~100
0.14 0.15 0.42 0.55 0.72
表 1— 2
)/1(10 4 Cot ???
Co
41
Example 1—1 Calculate the relative rate of change of water
density when the water pressure was increased from 5at to
10at under normal temperature,
Solution,
this problem is the application of compression coefficient
equation of a liquid under normal temperature,
Solution 1,
94
1
th e n
0, 5 3 8 1 0 ( 1 0 5 ) 9, 8 1 1 0 0, 0 2 6 4 %
p
p
d
dp
d
d p
?
?
?
?
?
?
?
?
? ? ? ? ? ? ? ?
42
例 1—1 水在常温下由 5at增加到 10at时，求水的密度的相对
变化率。
解,此题是液体在常温下压缩性系数公式的应用。
解法一,
%0 2 6 4.01081.9)510(105 3 8.0
1
49 ????????
?
?dpd
dp
d
p
p
?
?
?
?
?
?
则
43
0 2 1
21
0
9
0
0
to in te g r a te th e f o r m u l a o f
l n l n ( )
the n e x p [ ( ) ]
e xp [ 0.5 38 10 ( 10 5 ) 98 10 0]
1.0 00 26 4
n d 0,0 0 0 2 6 4 0,0 2 6 4
so th e r e l a tiv
p
p
p
d
dp
pp
pp
a
?
?
?
? ? ?
?
?
?
??
?
?
?
? ? ?
??
? ? ? ?
?
－
＝ ＝ ％
e r a te o f c h a n g e o f th e d e n sity o f w a t e r is 0,0 2 6 4 %
Solution 2,
44
％。为故水密度的相对变化率
％＝＝
－
所以
则
有
进行积分对
0 2 6 4.0
0 2 6 4.00 0 0 2 6 4.0
0 0 0 2 6 4.1
]9 8 1 0 0)510(10538.0e x p [
)](e x p [
)(lnln
0
0
9
12
0
120
?
??
?
?
?
???
?
?
?
?
????
??
???
?
?
pp
pp
dp
d
p
p
p
解法二,
45
In the applicable range of gaseousness equation,???
tT ?
Figure 1— 4 the volume dilatation of a fluid
under constant pressure
P
dTT ?
dVV ?
dV
T
V
P
1/ ?Kt?
KT /
CP ?
o
Figure 1— 5 the expansion coefficient of gas
46
在气体状态方程式的适用范围内,???
tT ?
图 1— 4流体在定压下的体积膨胀
P
dTT ?
dVV ?
dV
T
V
P
1/ ?Kt?
KT /
CP ?
o
图 1— 5气体膨胀系数
47
3,Incompressible fluid
The fluid whose compression coefficient and expansion
coefficient are zero is called incompressible fluid,
definition,
The volume of this kind of fluid doesn’t minish while being
pressed and doesn’t expand while being heated so v and d are
all constant,It is much easier to study the laws of balance and
motion,
，?
48
三、不可压缩流体
d，,??
这种流体受压体积不减小，受热体积不膨胀，
因而 均为常数，讨论其平衡和运动规律自
然简单得多。
压缩系数和膨胀系数为零的流体叫做不可压缩
流体。
定义,
49
§ 1-5 Viscosity of a Fluid
Viscosity is the capability of a fluid to resist deformation and is
a inherent attribute of a fluid,
1,Newtonian law of internal friction
Because of the gravitation between the molecules of fluid and
the momentum transfer brought from heat motion of molecules the
viscosity of fluid comes into being and brings the internal friction
also,
In 1686 Newton brought forward the law of internal friction
The property that the shear stress comes into being in fluid
when fluid flows is called the viscosity of a fluid,
definition,
50
§ 1-5 流体的粘性
粘性是流体抵抗变形的能力，它是流体的固有属性。
一、牛顿内摩擦定律
流体粘性是由于流体分子间的引力以及分子热运动产生动量
交换，而形成的并因此产生内摩擦力。
1686年牛顿提出牛顿内摩擦定律。
流体运动时内部产生切应力的性质，叫做流体的粘性。
定义,
51
experimental approach,
As shown in Figure 1— 6,Suppose that there are
two big enough parallel flat boards and the symbol h
which represents the distance between them is very
small,The space inside two boards is full of generic
homogeneous fluid,The upper board is fixed and the
lower one moves along a line as a properly constant
speed under the shear force F,The area A is so big that
the influence of the borders of boards can be neglected,
52
实验方法,
0u
如图 1— 6所示。 设有两个足够大，相距 h 很小的平行平板。
中间充满一般的均质流体，上板固定，下板在切向力 F 作用下以
不大的速度 作匀速直线运动。平板面积 A足够大，以至于可
忽略平板边缘的影响。
53
dy
duAT ??? （ 1-11）
in the thin layer whose thickness is, Newton considered
that the value of internal friction T inside a fluid（ that is to say
the value of shear force F ) has relations with the properties of
fluid,
dy
du
dy
moreover it is directly proportional to velocity gradient
and contact area A,but have nothing to do with the pressure
on the contact surface,The law is
definition,
Research an infinitely thin fluid layer,velocity of flow is u
where the coordinate is y,velocity of flow is where the
coordinate is, it is obviously that the velocity gradient is
duu ?
dyy ?
54
dy
duAT ??? （ 1-11）
取无限薄的流体层进行研究，坐标为 y处流速为 u，坐标
为 处流速为,显然在厚度为 的薄层中速度梯度
为,牛顿认为：液层间内摩擦力 T 的大小（也就是切向力
F 的大小）与液体性质有关，并与速度梯度 和接触面积 A
成正比，而与接触面上压力无关。即
dy
du
dy
du
dyy? duu? dy
内容,
55
The physical meaning,
The shear stress is directly proportional to velocity gradient,
In the formula（ 1— 11） the symbol is used to make
and positive, It is to say that when choose the
positive sign and when choose the minus sign,
T ?
0?dydu
0?dydu
?
The fluid according with the internal friction law is called
Newtonian fluid,or else which is called non-Newtonian fluid,
In the formula（ 1— 11） is a coefficient which has
relation to the kinds of fluid and temperature, It is called
dynamical stickiness coefficient (viscosity) and its unit is, sPa?
?
56
符合这样内摩擦定律的流体称为牛顿型流体，否则称为
非牛顿型流体。
式（ 1— 11）中 号是为, 永为正值而设的，即当
时取正号，当 时取负号。
? T ? 0?
dy
du
0?dydu
物理意义,
切应力与速度梯度成正比。
式（ 1— 11）中 是与流体种类、温度有关的系数，称
为动力粘性系数（粘度），单位 。
?
sPa?
57
2,Velocity gradient
0u
0u
0?u
x
y
u
duu?
Figure 1— 6
velocity distributing rule
dy
Choose a micro rectangular unit surface in the moving fluid,
deformed as a parallel quadrangle after,as shown in Figure ( 1— 7）
,dt
0
y
x
udt
dtduu )( ?
?d
Figure 1— 7
velocity gradient
dudt
58
二、速度梯度
dt 在运动流体中取一矩形微元面，经 时间变形运动成为一平
行四边形（图 1— 7）有,
0u
0u
0?u
x
y
u
duu?
图 1— 6速度分布规律
dy
0
y
x
udt
dtduu )( ?
?d
图 1— 7速度梯度
dudt
59
dt
d
dt
dtg
dt
dt
dy
du
dy
du ?? ???? )( （ 1— 13）
so d u d d y d t?? ? ?? ? ? ?
（ 1— 14）
Then,
a,the velocity gradient is equal to the velocity of shear
distortion of a fluid parcel,
b,the shear stress is directly proportional to the velocity of
shear distortion in a fluid,
60
dt
d
dt
dtg
dt
dt
dy
du
dy
du ?? ???? )( （ 1— 13）
dt
d
dy
du ???? ????故 （ 1— 14）
则有,
a.速度梯度等于流体微团的剪切变形速度。
b.流体中的切应力与剪切变形速度成正比。
61 ? v
It is the ratio of dynamical viscosity to density,If the fluid
density is six to one we can’t judge the magnitude of viscosity
according to but according to
3,Kinematic viscosity coefficient
?? /?v （ 1— 15）
Definition,
where the units of are, v s/m2
? ?v When researching fluid motion we often use the ratio of to
density which is called kinematic viscosity coefficient,It is
represented by
?
v
?
the physical meaning of, ?
the physical meaning of, v
It is the shear stress under unit velocity gradient,We can judge
the magnitude of viscosity of the same kind of fluid from, ?
62
三、运动粘性系数
?? /?v （ 1— 15）
的物理意义,?
的物理意义,v
? 单位速度梯度下的切应力，根据 的大小可直接判断
同种流体粘性的大小。
? ?
v
在研究流体运动时，常常使用 与密度 的比值,
称为运动粘性系数。以 表示,
定义,
s/m2 运动粘性系数 的单位是 。 v
?
动力粘度与密度之比。如果流体密度相差很多，不能
根据 的大小判断粘性的大小，而应根据 判断粘性的大
小。
v
63
4,Relations between viscosity and temperature
The fluid viscosity is directly proportional to press but its change
is very little and can be neglected normally,
The fluid viscosity is influenced strongly by temperature,The
gas viscosity is caused mainly by the momentum transfer which
coming from the thermal movement of molecules, So when the
temperature goes up the thermal movement of molecules becomes
strong and viscosity augments,The liquid viscosity is caused
mainly by the attractive forces between molecules,So when the
temperature goes up the momentum of molecules enlarges and
attractive forces increase and viscosity declines,
5,Property of surface tension
Definition,
Because of the attractive forces between molecules,the tiny
tension which the free surface of fluid endured is said to be surface
tension,
64
四、粘性与温度的关系
流体的粘性大小与压力成正比，但变化很小，一般不予考
虑。
流体的粘性受温度的影响很大，气体的粘性主要是由分子
热运动产生的运动的动量交换引起的，因此当温度升高时，分
子热运动加剧，粘性增大。流体的粘性主要由分子之间的引力
产生，当温度升高时，分子动量增加，引力减小，粘性下降。
五、表面张力特性
定义,
由于分子间的引力，在液体自由表面上能承受微小的张
力称表面张力。
65
Capillarity,put a thin glass tube with two open ends up straightly
into a liquid,The liquid in the tube will rise or fall under the action
of surface tension which is said to be capillarity,
Because the weight of the liquid column equals to the vertical
component of the accessional press of the surface tension,then
????? c o s4 2 dhd ?
so
dh ?
?? c o s4?
In the formula,
,a n g l ec o n t a c t
m;/ t e n s i o ns u r f a c e oft c o e f f i c i e n;m/l i q u i d of w e i g h t s p e c i f i c
m; t u b eg l a s s ofd i a m e t e r t h e
m;s u f a c e l i q u i d of r i s e t h e
3
—
，—
，—
，—
，—
?
?
?
?
?
d
h
66
由于重力与表面张力产生的附加压力在垂直方向分力平衡，
所以有,
????? c o s4 2 dhd ?
则
dh ?
?? c o s4?式中,
接触角。—
表面张力系数，—
液体重度，—
玻璃管直径，—
液面上升高度，—
?
?
?
m;/;m/
m;
m;
3
?
?
d
h
毛细管现象：将两端开口的细玻璃管竖立在液体中，在表面张力
作用下，管中液体将上升或下降一个高度，称为毛细管现象。
67
? ?
h
?
?
h
Figure 1— 8 capillarity
Obtain from experiment,the contact angle of water and glass is
when temperature is ; the contact
angle of hydrargyrum and glass is C
o20N / m0 7 2 8.0,9~3 ?? ?? oo
。N/m51.0,140~139 ?? ?? oo
hydrargyrum water
68
水
? ?
h
汞 ?
?
h
图 1— 8 表 面 管 现 象
试验测得,时水与玻璃的接触角
汞的接触角
Co20 N / m0 7 2 8.0,9~3 ?? ?? oo
。N / m51.0,140~139 ?? ?? oo
69
33
wa ter
35
6 2 6 3
3
5
9 9 8,2 k g / m,1,2 0 5 k g / m
1,0 0 2 1 0 P a s,1,8 1 1 0 P a s
1,0 0 3 1 0 m / s,1 5,0 1 0 m / s
1,0 0 2 1 0
5 5,3 4 t i m es
1,8 1 1 0
998.2
1.205
air
w a ter
air
w a ter
air
w a ter
air
w a ter
air
??
??
??
?
?
?
?
??
??
?
?
? ? ? ?
??
?
?
＝ ＝
＝ ＝
＝ ＝
＝ ＝ （ ）
＝
-6
-6
8 2 8,3 9 t i m es
1 5,0 1 0
1 4,9 6 t i m es
1,0 0 3 1 0
air
w a ter
?
?
?
?
＝ （ ）
＝ ＝ （ ）
Example 1—2 At the same temperature, Try to
demonstrate that at for water and gas which one is easy to
flow.（ use data to explain）
空气水 ?? ?
Co20
solution,when the temperature is the following data are
checked from the physical properties tables of water and air,
Co20
70
例 1—2 在相同温度下,试论证在 时，水和空
气相比，哪种流体易于流动（用数据说明）
空气水 ?? ? Co20
解,在 时，从水和空气物理特性表中查取 Co20
（倍）＝＝
（倍）＝＝
（倍）＝＝
＝＝
＝＝
＝＝
水
空气
空气
水
空气
水
空气水
空气水
空气水
1 4, 9 6
101, 0 0 3
101 5, 0
39.828
205.1
2.998
34.55
1081.1
10002.1
s/m100.15,/sm10003.1
sPa1081.1,sPa10002.1
k g / m205.1,k g / m2.998
6-
6-
5
3
3626
53
33
?
?
?
?
??
????
?
?
??
??
?
?
?
?
?
?
??
??
??
71
The example shows that it can’t judge the fluidness of a fluid
directly from but from at the same temperature, Because
that the excludes the influence of fluid density and only keeps
the movement property parameters,Which is the meaning of
importing the viscosity coefficient,
? ?
?
conclusion,At the same temperature, Air is hard to
flow compared with water,
w g ? ? ?
72
结论,在相同温度下，,空气与水相比较，空气不
易于流动。
水空气 ?? ?
此例题说明：在相同温度下，只从 值的大小不能直接
判断流体的流动性，而应由 值的大小才能够直接判别流体
的流动性。因为 值排除了流体密度的影响，只保留其运动
特征参数，这也是引入粘性系数的意义所在。
?
?
?
73
6,Concept of ideal fluid
Definition,
In fact all fluids have viscosity,The purpose to put forward
the concept of ideal fluid is to research the fluid movement rules
and simplify the deduce of the theory equations greatly,
Suppose that the viscosity of a fluid doesn’t exist,that is to
say,The fluid is said to be ideal fluid or inviscid fluid,0v ?
74
六、理想流体的概念
0v? 假定不存在粘性，即其粘度 的流体为理想流
体或无粘性流体。
定义,
实际上，一切流体都具有粘性，提出理想流体的概念在
于研究流体运动规律时，对理论方程的推导大为简化。
75
§ 1-6 Forces on a Fluid
x
y
Z
W??
V?
A?
O
Figure 1— 8 body force
and surface force
F??
T??
mF
??
IF
??RF
??
Each fluid particle is acted by all kinds of forces no matter it
keeps moving or balance,According to the behaviors of the forces
there are two classes of forces, body force and surface force,As
shown in Figure 1— 8,Choose a random fluid parcel to research
which volume is, V?
76
§ 1-6 作用在流体上的力
流体每一质点无论处于运动或平衡状态，都受到各种力。按
力的表现形式分为质量力和表面力两类。如图 1— 8所示。取体积
为 的任意微团进行研究。 V?
x
y
Z
W??
V?
A?
O
图 1— 8质量力与表面力
F??
T??
mF
??
IF
??RF
??
77
1,Body force
The force which is directly proportional to the mass of a fluid
parcel and acts on the mass center is called body force,
Definition,
Body force
gravitation
Linear motion inertia force
centrifugal inertia force
--- ---
2
?
??
??
??
???
???
???
rmF
amF
gmW
R
?
78
一、质量力
与流体微团质量成正比并且集中作用在微团质量中心上的力
称为质量力。
定义,
质量力
重力
直线运动惯性力
离心惯性力
--- ---
2
?
??
??
??
???
???
???
rmF
amF
gmW
R
?
79
Definition,
kfjfiff zyxm ??? ???
dm
Fd
m
Ff mm
mm
??
????
?? 0
lim （ 1— 16）
The body force acting on unit mass fluid is called unit body
force,and is expressed as follows,
In the formula
—— the mass of the fluid parcel；
—— body force acting on the parcel；
m?
mF
??
)( kfjfifdmfdmFd zyxmm ???? ??????
（ 1— 17）
,, are projections of unit body force on the
axes,
xf yf zf zyx,,
80
定义,
kfjfiff zyxm ??? ???
dm
Fd
m
Ff mm
mm
??
????
?? 0
lim （ 1— 16）
式中
—— 流体微元体的质量；
—— 作用在该微元体上的质量力；
m?
mF
??
)( kfjfifdmfdmFd zyxmm ???? ??????
（ 1— 17）
,, 为单位质量力在 轴的投影。 xf
yf zf zyx,,
单位质量流体所受的质量力称为单位质量力，记作
81
2,Surface force
a,The press force along the inner normal of surface
b,The friction along tangent orientation of surface
Classified according to the act direction,
The force that is directly proportional to surface area of a fluid
and distributes on the fluid surface is called surface force,
definition,
{
82
二、表面力
a.沿表面内法线方向的压力；
b.沿表面切向的摩擦力。
按其作用方向分类,
大小与流体表面积成正比而且分布作用在流体表面上的力
称为表面力。
定义,
{
83
Then the pressure stress of every point is
A
Fp
A ?
??
??
?
0
lim （ 1— 18）
So the surface press is
ApF ??? （ 1— 19）
Shear stress of every point is
A
T
A ?
??
??
??
0
lim? （ 1— 20）
So the surface shear force is
AT ?? ??? （ 1— 21）
As shown in Figure1— 8,choose a micro area on the fluid
parcel,suppose the tiny press force acting on is and the
tiny shear force is 。
A?
A?
T??
F??
84
如图 1— 8所示，在流体微团上取微元面积,设作用在
上的微小压力,微小切力 。
A? A?
F?? T??
则各点处的压应力为,
A
Fp
A ?
??
??
?
0
lim （ 1— 18）
故表面上的压力为,
ApF ??? （ 1— 19）
各点处的切应力为,
A
T
A ?
??
??
??
0
lim? （ 1— 20）
故表面上的切力为,
AT ?? ??? （ 1— 21）
85
Exercises of Chapter 1
C
V
V
TTT
V
V
ppp
kk
kk
o4.010
0
1
0
0
4.10
0
0
0
1 0 7k3 8 02)152 7 3()()(
k P a 3.2 6 723.1 0 1)()(
???????
?????
??
?
?
?
?
1—1 The original state of air is and,
After adiabatic compression in the cylinder the volume of the
air reduces a half,What are the temperature and pressure of
the end state?
k P a3.1 0 10 ?p 150 Ct o?
solution,In the process of adiabatic compression,the process
equation is, in the formula k =1.4,k is called
adiabatic exponent of air, Uniting the state equation of ideal gas
we can obtain
C o n s tpV k ?
86
第一章 习 题
1—1 空气初始状态为,在汽缸内绝
热压缩后体积减少了一半，求终态温度和压强。
k P a3.1 0 1,15 00 ?? pCt o
解,绝热压缩时，其过程方程为,式中 k=1.4,称
为空气绝热指数。联立理想气体状态方程可得
C o n s tpV k ?
C
V
V
TTT
V
V
ppp
kk
kk
o4.010
0
1
0
0
4.10
0
0
0
1 0 7k3 8 02)152 7 3()()(
k P a 3.2 6 723.1 0 1)()(
???????
?????
??
?
?
?
?
87
solution,the exercise is the question to calculate the net increment
of water volume in the system,we can do as the formula (1— 9）,
3m25080 0 5.0
t h e n
1
????
?
?
V d TdV
dT
dV
V
t
t
?
?
?
d ia l a tio n w a te r b a n k
boiler
radiator
3m8
Co50
1—2 In heating system there is a dilatation water tank,The whole
volume of the water in the system is, The largest temperature rise
is and the dilatation coefficient is,what is the
smallest cubage of the water bank? 0 0 5.0?t?
so the smallest cubage of the dilatation water bank is 2, But
in engineering design we should pay attention to keep surplus to
ensure the safety of the system,
3 m
88
1—2 采暖系统在顶部设一膨胀水箱，系统内的水总体积为,
最大温升,膨胀系数,求该水箱的最小容积？
3m8
Co50 005.0?t?
解,该题为求解系统内水体积净增量的问题，可依（ 1— 9）式
进行求解。
3m25080 0 5.0
1
????
?
?
V d TdV
dT
dV
V
t
t
?
?
则
故膨胀水箱的最小体积应为 2立方米，但在工程设计中，应
注意按照设计规范增加一定的富裕量，以确保系统安全。
? 膨胀水箱
锅炉
散热器
89
solution,This is a simple and familiar example that Newtonian
internal friction law is applied to engineering,There are two
emphases,the one is to simply the velocity gradient in the
lubricant,that is to think the distribute rule is linear； the other is
to build the balance equation of force correctly,Since it slides at
uniform velocity,the component of gravity is equal to the
resistance of viscosity,
V
? ?
13
5
12
5kg
cm1 5kg
1—3 The bottom surface of a wood block is, thickness
is and the mass is, It slides down along a incline which is
spreaded lubricant at a speed of, The thickness of
lubricant is 1, what is the dynamical viscosity coefficient?
2cm4514 ?
m /s1?V
mm
90
1—3 一木块底面积为,厚度为,质量为,沿
着涂有润滑油的斜面以速度 等速度下滑，油层厚度,
求润滑油的动力粘性系数。
2cm4514 ? cm1 5kg
m/s1?V mm1
解,这是牛顿内摩擦定律在工程中应用的一个简单而又常见的
例子。求解此题有两个重点，一是对油层内速度梯度进行
简化，即认为是线性分布规律；二是正确列出力的平衡方
程。
由于是等速下滑，故重力分力与粘性阻力相等
V
? ?
13
5
12
5kg
91
B a l a nc e e qua ti on,g sin,e l oc it y gr a die nt
substit uti ng into 1 11 g sin
g sin
T he n
S ubstit uti ng k now n da ta into i t H e nc e 0,104 7 P a s
du V
T m v
dy
du
Am
dy
m
V
A
?
?
??
?
?
?
?
??
?
?
（ — ）
＝
,＝
Caution,When solving,all physical quantities must adopt the same
unit system,in order to avoid conversion mistakes,
92
g si n,
1 11 g si n
g si n
0.10 47 P a s
du V
Tm
dy
du
Am
dy
m
V
A
?
?
??
?
?
?
?
??
?
?
平 衡 方 程 为 速 度 梯 度 为
代 入 式 （ — ）
解 出 ＝
代 入 已 知 数 据, 解 得 ＝
注意,在解题时，所有物理量的单位必须采用相同单位制，避免
出现换算错误。
93
m5.0?H
1—4 A cone rotates around its vertical center axis at uniform
velocity,The gap between two cones is,It filled with
lubricant which, In Figure 1-2,,
,,
mm1??
s0, 1 P a??? m3.0?R
1/s 16??
Solution,This problem belongs to the application of Newtonian
internal friction law,It’s feature is that the action radius,
contact area and rotation velocity change with h,We should
find its changing rules and carry out the idea which solving
with physical methods,
Exercise Figure1— 2
?
R
? r
h
dh
H
94
1—4 一圆锥体绕竖直中心轴等速旋转，锥体与固定的外锥体之
间的隙缝,其中充满 的润滑油。已知锥体
顶面半径,锥体高度,当旋转角速度
时，求所需要的旋转力矩。
mm1?? s0, 1 P a???
m3.0?R m5.0?H 1/s 16??
解,此题属于牛顿那摩擦定律应用。该题的特点是作用半径，
液体和固壁接触面积及锥体旋转线速度都随高度变化，应逐个
找出其变化规律并贯彻物理方法解题的思想。
?
R
? r
h
dh
H
习题 1— 2图
95
In Figure,rotating torque expression of a fluid parcel
rdAdydurdAdM ????? ??
(1) Changing rule of the cone’s radius r ?ta n?? hr
(2) The expression of dA which is corresponding to dh
????? c o st a n2c o s2
dhhdhrdA ???
h r u dy du
2
tan
? ?
?
? ? ? ? ? ?
3 3 dh h dM
cos
1 tan ? ? ? ?, ?
? ? ?
Substituting above formulas into dM,
? (3)Because is very small,We can think that the changing of
velocity gradient is linear,
96
如图所示，旋转力矩的微元表达式
rdAdydurdAdM ????? ??
( 1 )锥体半径 r的变化规律 ?ta n?? hr
( 2 )对应 dh 的 dA 表达式
????? c o st a n2c o s2
dhhdhrdA ???
33
3
ta n
1
2 ta n
c o s
d u u r
h
dy
dM
d M h d h
?
??
?
? ? ?
?
? ? ?
??
? ? ?
? ? ?
（ ） 因 为 很 小, 可 把 速 度 梯 度 按 线 性 变 化 考 虑
将 上 三 式 代 入 表 达 式 中, 整 理 得
97
33
34
0
4 se e k f o r th e g e n e r a l m o m e n t
ta n ta n
2
c o s 2 c o s
substi tut ing the kn ow n d a ta int o it the r e su l t is
3 9,6 ( N m )
( in i t t a n,he nc e 31 )
H
o
d M h d h H
M
R
H
? ? ? ?? ?
??
? ? ? ?
??
??
??
?
??
（ ）
，
＝
98
) 31,t a n (
m)N( 6.39
c o s
t a n
2c o s
t a n
2
4
4
3
0
3
3
o
H
H
R
M
HdhhdMM
＝求得其中
代入已知数据，解得
）求总力矩（
??
?
?
?
? ? ?
?
?
?
?
??
?
??
??? ? ?
99
100
