1
Mechanics of Fluid
2
3
Chapter 4
Similarity Principle and Dimensional Analysis
§ 4–1 Introduction
§ 4–2 Similarity Principle
§ 4–3 Similarity Criterion
§ 4–4 Model Laws
§ 4–5 Law and Application of
Dimensional Analysis
Chapter 4 Exercise
?
4
第四章 相似原理和量纲分析
§ 4–1 引言
§ 4–2 相似原理
§ 4–3 相似准则
§ 4–4 模型规律
§ 4–5 定理和量纲分析的应用
第四章 习 题
?
5
Chapter 4 Similarity Principle and Dimensional Analysis
§ 4-1 Introduction
Experiment is the basis of developing theories as well as the
yardstick of proving theories,sometimes science and technology
problems can’t be solved without the cooperation of experiments,
a,model experiment of engineering,For the purpose of
forecasting the flow situation of large-scale machine or water power
engineering which are being built,
b,exploratory observing experiment,For the purpose of
searching unknown flow laws,the theoretical foundation to direct
these experiments are similarity principle and dimension analyse,
There are mainly two kinds of experiments in engineering
fluid mechanics,
6
第四章 相似原理和量纲分析
§ 4-1 引言
实验既是发展理论的依据又是检验理论的准绳,解决科技
问题往往离不开实验手段的配合。
a、工程性的模型实验。目的在于预测即将建造的大型机械
或水工结构上的流动情况;
b、探索性的观察实验。目的在于寻找未知的流动规律,指
导这些实验的理论基础就是相似原理和量纲分析。
工程流体力学中的实验主要有两种,
7
§ 4-2 Similarity Principle
When we design or manufacture some complex and huge
hydraulic machines,built water power engineering as well as
search some complicated hydraulic phenomena,we often design
and made a model which is reduced size according to similarity
principle,Carrying out simulation experiment,deduce the flow
situation and relative data of practicality through observing the
flow situation of model,
The basic theory which analyze and study the
similitude relation between model and real object is called similarity
principle
Real object is called prototype too,
defination,
8
§ 4-2 相 似 原 理
当设计制造某些复杂而庞大的水力机械,建造水利工程
以及研究某些复杂的水力现象时,往往要根据相似原理,设
计制造缩小了尺寸的模型。进行模拟实验,通过对模型的流
动状况观测来推断实物的流动状况及有关数据。
分析研究模型和实物间的相似关系的基本理论称为相似
理论。
实物又称为原型。
定义,
9
1,Geometry similitude
Model parameter adds a subscript m and prototype
parameter adds a subscript n to express,
Linear scale
m
n
l l
l?? ( 4— 1)
Surface scale
2
2
2
l
m
n
m
n
l
l ?? ??
?
??
?
( 4— 2)
Volume scale 3
3
3
l
m
n
m
n
l
l
V
V ?? ???
?
( 4— 3)
The corresponding geometrical linear dimensions of prototype
are in proportion to those of model and corresponding geometrical
angles are that called geometry similitude,
defination,
10
一、几何相似
模型参数加脚码 m,原型参数加脚码 n 表示。
线性比尺为
m
n
l l
l?? ( 4— 1)
面积比尺为
2
2
2
l
m
n
m
n
l
l ?? ??
?
??
?
( 4— 2)
体积比尺为 3
3
3
l
m
n
m
n
l
l
V
V ?? ???
?
( 4— 3)
原型与模型中对应的几何线性尺寸成比例,对应的几何角
度相等,称为几何相似。
定义,
11
2,Kinematic similitude
Time scale
m
n
t t
t?? ( 4— 4)
Velocity scale
t
l
m
m
n
n
m
n
t
l
t
l
?
?
?
?
? ? ??? ( 4— 5)
Acceleration scale 2
2
2
t
l
m
m
n
n
m
n
t
l
t
l
a
a
?
?
? ? ??? ( 4— 6)
The corresponding parameters of motion of prototype such as
velocity,acceleration are consistent in direction and in proportion to
magnitude to those of model called kinematic similitude,
defination,
12
二, 运动相似
时间比尺为
m
n
t t
t?? ( 4— 4)
速度比尺为
t
l
m
m
n
n
m
n
t
l
t
l
?
?
?
?
? ? ??? ( 4— 5)
加速度比尺为 2
2
2
t
l
m
m
n
n
m
n
t
l
t
l
a
a
?
?
? ? ??? ( 4— 6)
原型与模型中对应的运动参数如速度、加速度方向一致,
大小成比例,称为运动相似。
定义,
13
3,Dynamic similarity
Density scale
m
n
?
??
? ?
( 4— 7)
Mass scale
3
l
mm
nn
m
n
m V
V
m
m ??
?
??
????
( 4— 8)
Force scale
22
lam
mm
nn
m
n
F am
am
F
F ??????
??????
( 4— 9)
The forces of corresponding points of prototype and model are
consistent in direction and in proportion to magnitude that is called
dynamic similarity,
defination,
14
三, 动力相似
密度比尺为
m
n
?
??
? ?
( 4— 7)
质量比尺为
3
l
mm
nn
m
n
m V
V
m
m ??
?
??
????
( 4— 8)
力的比尺为
22
lam
mm
nn
m
n
F am
am
F
F ??????
??????
( 4— 9)
原型与模型中对应点处受力方向相同、大小成比例,称为
动力相似。
定义,
15
Unit mass scale
m
n
g g
g?? ( 4— 10)
according to( 4— 9)
22
22
mmm
nnn
m
n
l
l
F
F
??
???
That is
2222
mmm
m
nnn
n
l
F
l
F
???? ?
( 4— 11)
In formula,is a dimensionless number,called Newton
number, denoted by 22??l
F
Ne
mn NeNe ?
( 4— 12)
That is,two geometry similitude flows,if dynamic similarity,then
their Newton number must be equal; Whereas,two geometry
similitude flows whose Newton number are equal,their dynamic
similarity must be equal,So geometry similitude is only necessary
condition of similitude, and that kinematic similitude and dynamic
similarity are necessary and sufficient condition of similitude,
So formula( 4— 11) change to
16
单位质量比尺为
m
n
g g
g?? ( 4— 10)
据式( 4— 9)可写
22
22
mmm
nnn
m
n
l
l
F
F
??
???

2222
mmm
m
nnn
n
l
F
l
F
???? ?
( 4— 11)
式中,为一个无因次量,称为牛顿数,以 表示。
22??l
F Ne
mn NeNe ?
( 4— 12)
就是说,两个几何相似的流动,如果动力相似,则牛顿数必
相等;反之,牛顿数相等的两个几何相似的流动,必然是动力相
似的。故几何相似仅是相似的必要条件,而运动相似和动力相似
才是相似的充要条件。
故式( 4— 11)变为
17
§ 4-3 Similarity Criterion
1.Reynolds number
? ? ? ?? ? ??? lLuLT ??
?
?
??
?? 2
through formula( 4— 10) we know the dimension of inertia
force is, if substitution of T for F,then ? ? 22?? lF ?
2222
mmm
mmm
nnn
nnn
l
l
l
l
??
??
??
?? ?
For, so simplifing above formula we get,v?
?
?
m
mm
n
nn
v
l
v
l ?? ? ( 4— 13)
In formula, called Reynolds number Re?
v
l?
Physical meaning,
dy
duAT ??internal friction caused by viscosity,which can be
expressed from dimension,
The ratio of inertial force to viscosity force,
18
§ 4-3 相 似 准 则
一、雷诺数
? ? ? ?? ? ??? lLuLT ??
?
?
??
?? 2
由式( 4— 10)知惯性力的因次为,如用 T 替
换 F,则
? ? 22?? lF ?
2222
mmm
mmm
nnn
nnn
l
l
l
l
??
??
??
?? ?
因,故简化上式得 v?
?
?
m
mm
n
nn
v
l
v
l ?? ? ( 4— 13)
式中,称为雷诺数 数。
Re?vl? ? ?yn o ldRe
物理意义,
dy
duAT ??粘性引起的内摩擦力为,从因次上可写
惯性力与粘性力的比。
19
2,Froude number
? ? ? ?? ?? ? 33 glLgG ?? ??
substitution for F in (4— 10),then
22
3
22
3
mmm
mmm
nnn
nnn
l
lg
l
lg
??
?
??
? ?
Simplify it then
mm
m
nn
n
lglg
22 ??
? ( 4— 14)
In formala, called Froude number
Frgl ?
2?
Physical meaning,
The ratio of inertial force to gravity,
,?? gmgG ??
In those liquid which have free surface,Which having leading
effect is gravity,on dimension it is
20
二,弗劳德数
? ? ? ?? ?? ? 33 glLgG ?? ??
用以代替式( 4— 10)中的 F,则
22
3
22
3
mmm
mmm
nnn
nnn
l
lg
l
lg
??
?
??
? ?
简化后得
mm
m
nn
n
lglg
22 ??
? ( 4— 14)
式中,称为弗劳德 数。
Frgl ?
2? ? ?F ro u d e
物理意义,
惯性力与重力之比。
,?? gmgG ??在具有自由表面的液流中,起主要作用的为重力
在因次上为
21
3,Euler number
On dimension
? ? ? ?? ? 2PlApF ??
Substitution for F in (4— 10),then
22
2
22
2
mmm
mm
nnn
nn
l
lp
l
lp
????
?
That is
In formula,called Euler number
Eup ?2??
Physical meaning,
22
mm
m
nn
n pp
???? ?
( 4— 15)
The ratio is press to initial force,
pAF ?
When study the distributing of press or pressure of the objects
submerged in liquid,the major force is press,
22
三、欧拉数
在因次上为
? ? ? ?? ? 2PlApF ??
将其代替式( 4— 10)中的 F时,则
22
2
22
2
mmm
mm
nnn
nn
l
lp
l
lp
????
?

式中,称为欧拉 数。 Eup ?
2?? ? ?Euler
物理意义,
22
mm
m
nn
n pp
???? ?
( 4— 15)
压力与惯性力之比。
pAF ?
研究淹没在流体中的物体表面上的压力或压强分布时,
起主要作用的力为压力 。
23
Definition,
mn
mn
mn
EuEu
FrFr
ReRe ?
?
?
?
( 4— 16)
Came to be known as mechanics similarity criterion of stationary
flow of incompressible fluid,
If two flows are dynamic similarity,then they must have the
same Euler number,Reynolds number and Froude number,That is
24
定义,
mn
mn
mn
EuEu
FrFr
ReRe ?
?
?
?
( 4— 16)
称为不可压缩流体定常流动的力学相似准则。
如果两个流动成力学相似,则它们的弗劳德数、欧拉数、
雷诺数必须各自相等。于是
25
4,Mach number
When thinking about fluid compressibility,elastic force predominates F=EA
f o r c e, e l a s t i c t of o r c e i n e r t i a l of r a t i o t h em e a n i n g P h y s i c a l
n u m b e r,M a c h a l l e d,
r e c i p r o c a l I t s,t h e n
f o r m u l a a b o v e i n t o s u b s t i t u d e
,
1
so,is s o u n d ofv e l o c i t y,f l u i d lec o m p r e s s i b to
ist h a t
t h e ni t,1 0 ) f i n t o—(4i n F s u b s t i t u t e
]][[][ d i m e n s i o n O n
2
22
22
2
22
2
2
cM
a
V
V
a
V
a
aE
E
a
V
E
V
E
Vl
lE
Vl
lE
ElAEF
a
m
m
n
n
mm
m
nn
n
mmm
mm
nnn
nn
?
?
??
?
?
??
?
?
??
??
26
四、马赫数
当考虑流体压缩性时,弹性力起主要作用 F=EA
性力之比。物理意义:惯性力与弹
称为马赫数。
其倒数得
代入上式因此对可压缩流体,音速

时,则)中的—代入(
在因次上
,
,
,
1
,
104
]][[][
2
22
22
2
22
2
2
a
m
m
n
n
mm
m
nn
n
mmm
mm
nnn
nn
M
a
V
V
a
V
a
aE
E
a
V
E
V
E
Vl
lE
Vl
lE
F
ElAEF
?
?
??
?
?
??
?
?
??
??
27
5,Weber number
t e n s i o n,s u r f a c e t of o r c e i n e r t i a l of r a t i o T h e m e a n i n g P h y s i c a l
n u m b e r, W e b e r c a l l e d,r a t i o O r d e r t h e
e d u c ec a n,104 i n t o t e n s i o n s u r f a c e ngS u b s t i t u t i
2
22

)—(
We
lV
VlVl
lF
m
mmm
n
nnn
?
?
?
?
?
?
?
?
?
?
6,Archimedes Number
number,Archimedes flow difference temperature
called,flow difference temperature 2
number,Archimedes flow difference
density called,flow difference density 1
2
0
0 0
2
e
r
r
T V
T gl A
V
gl A
D ?
D ?
) (
) ( ? ?
28
五、韦伯数
面张力之比。物理意义:惯性力与表
称为韦伯数。令该比值
)中,可导出—代入(将表面张力
,
104
2
22
We
lV
VlVl
lF
m
mmm
n
nnn
?
?
?
?
?
?
?
?
?
?
六、阿基米德数
称温差流动阿基米德数)温差流动时(
数称浓度差流动阿基米德)密度差流动时(
,2
,1
2
0
00
2
e
r
r
TV
Tgl
A
V
gl
A
D
?
D
?
?
?
29
In formula,
Archimedes number is Froude number which thinking
about flotational process,that is which corrects the gravity
similitude criterion,
Is the temperature difference between twyer and indoor
is thermodynamic temperature indoor,
0
e T
T D
30
正。重力相似准则进行了修作用的弗劳德数,即对阿基米德数是考虑浮力
为室内热力学温度
境气流温度之差为风口气流相对室内环式中:
0
eT
TD
31
§ 4-4 Model Laws
In the same physical phenomenon two similitude numbers
often can’t meet similarity relationship at the same time,Such as
Reynolds number and Froude number are not easy to meet the
conditions at the same time,
Wish Reynolds number to
meet the condition,
Wish Froude number to
meet the condition,
m
n
n
m
m
n
v
v
l
l ??
?
?
2121
???
?
???
?
???
?
???
??
m
n
m
n
m
n
g
g
l
l
?
? ?
2321 lvl
l
v ???
?
? ???
It is hard to realize and even never on technique,In fact,we
often analyze the researched flow problems deeply to find out the
main acting force which influenc flow problems,to meet a main
force similitude and ignore other minor forces similitude,
32
§ 4-4 模 型 律
两个相似准数在同一个物理现象中,常不能同时满足相
似关系。例如雷诺数和弗劳德数就不易同时满足。
欲使雷诺数相等,将有
欲使弗劳德数相等,将有
m
n
n
m
m
n
v
v
l
l ??
?
?
2121
???
?
???
?
???
?
???
??
m
n
m
n
m
n
g
g
l
l
?
?
?
2321 lvl
l
v ???
?
? ???
这在技术上很难甚至不可能做到。实际中,常常要对所研
究的流动问题作深入的分析找出影响流动问题的主要作用力,
满足一个主要力的相似而忽略其它次要力的相似。
33
example,For the pressure flow in the pipe and submerged body
ambient flow etc,as long as the Reynolds number of the flow is
not too large,commonly its similitude condition depends on
Reynolds Criterion number,
Wave motion caused by ship movement,water flow in open
channal,water flow around pier,jet from small hole in container
wall etc are mainly influenced by gravity,Similitude condition
need ensure that Froude numbers are equal,
34
例,对于管中的有压流动及潜体绕流等,只要流动的雷
诺数不是特别大,一般其相似条件依赖于雷诺准则数。
而行船引起的波浪运动、明渠水流、绕桥墩的水流、容
器壁小孔射流等主要受重力影响,相似条件要保证弗劳德数
相等。
35
§ 4-5 Application ofπTheorem and
Dimensional Analysis
1,π theorem
Assume that the function
? ?ki nnnnnfN,,,,,,321 ??? ( 4— 18)
is used to express a physical rule to be studied,Under certain
system of units,all these k+1 physical variables have definite
units and values,
π
Functional relations among physical variables included in a
physical phenomenon,if choosing a certain system of units,are
determined,If changing the system of units,the functional relations
may be influenced,If we want it not to be influenced by the
selection of system of units,it should have special structure form of
functional relations,theorem is the method which changing
dimensional functional relations into dimensionless ones,
36
§ 4-5 定理和量纲分析的应用 ?
一,定理 ?
假定用函数
? ?ki nnnnnfN,,,,,,321 ??? ( 4— 18)
表示一个需要研究的物理规律,在一定的单位制下,这
k+1个物理量都有一定的单位和数值。
?
一物理现象所包含的各物理量间的函数关系,如果选用一
定单位制,则其关系的函数式就确定了,若改变单位制则函
数关系可能受影响。要使它不受单位制选择的影响,必须具有
特殊的函数关系的结构形式。 定理就是化有量纲的函数关系
为无量纲的函数关系式的方法。
37
Choose those physical variables of great influence,such as
choosing as basic units,
321,,nnn Of which the system of units of should meet,
( 1) basic units should be independent each other;
( 2) Using these basic units can educe all necessary units of
physical variables,
321,,nnn
Because the researching problems are different,the factor which
has great influence on each problem is different,The base unit which
meets above two requirements may have a lot of combination forms,
Such as studying head loss and flow resistance and so on,its
influencing factor often can’t do without the three basic physical
variables, linear dimension,flow velocity and fluid density,
These three physical variables separately have the specialities of
geometry,kinematics and dynamics, They meet the requirements (1)
and (2),so can form a special system of units,
l ? ?
?? ??? 321,,nnln
38
取对所研究的问题有重大影响的几个物理量。例如取
作为基本单位。
321,,nnn
其中 单位制需满足,
( 1)基本单位应该是各自独立的;
( 2)利用这几个基本单位应该能够导出其它所需要的一
切物理量的单位。
321,,nnn
由于研究问题各不相同,对每种问题起重大影响的因素自
然也不同。满足上述两点要求的基本单位可以有很多种组合形
式。
例如研究水头损失及流动阻力等问题,其影响因素常离不
开线性尺寸,流体运动速度 及流体密度 这三个基本物理
量。这三个物理量分别具有几何学、运动学和动力学的特性。
满足要求( 1)、( 2),因而以 可以组成一
组特殊单位制。
l ? ?
?? ??? 321,,nnln
39
So all the physical variables in formula (4— 18) can be expressed
as a product of a certain power combination of the three basic units
multiplied by a dimensionless number,that is
iii zyx
ii
zyx
nnnn
nnnN
321
321
?
?
?
? ( 4— 19)
?
and
In formula the
dimensionless number
iii zyx
i
i
zyx
nnn
n
nnn
N
321
321
?
?
?
?
?
( 4— 20) and
are the values of physical variable N and under basic
system of units, So under basic system of units the basic
laws in formula (4— 18) still unchanged,only the value of each
physical variable varies,
321,,nnn
321,,nnn
40
因而式( 4— 18)中的物理量都可以表示成这三种基本单位
的一定幂次组合与一个无量纲数的乘积,即
iii zyx
ii
zyx
nnnn
nnnN
321
321
?
?
?
? ( 4— 19)
?

式中无量纲数
iii zyx
i
i
zyx
nnn
n
nnn
N
321
321
?
?
?
?
?
( 4— 20) 与
就是物理量 N与 基本单位制下的数值,因而在
基本单位制下,式( 4— 18)的规律仍然不变,只是各
物理量的数值有所改变。
321,,nnn
321,,nnn
41
so,formula (4— 18) can be written as
),,,
,,,(
321321
321
3
321
2
321
1
321
333222111
kkkiii zyx
k
zyx
i
zyxzyxzyxzyx
nnn
n
nnn
n
nnn
n
nnn
n
nnn
n
f
nnn
N
??
?
( 4— 21)
It’s not hard to find from the first three to right,the powers of
denominator are,
1
1
1
3
2
1
?
?
?
z
y
x
0
0
0
33
22
11
??
??
??
yx
zx
zy
According to formula (4— 20) can get
1321 ??? ???
42
于是,式( 4— 18)可写成
),,,
,,,(
321321
321
3
321
2
321
1
321
333222111
kkkiii zyx
k
zyx
i
zyxzyxzyxzyx
nnn
n
nnn
n
nnn
n
nnn
n
nnn
n
f
nnn
N
??
?
( 4— 21)
从右端前三项不难看出,其分母上的乘幂为
1
1
1
3
2
1
?
?
?
z
y
x
0
0
0
33
22
11
??
??
??
yx
zx
zy
根据式( 4— 20)可得
1321 ??? ???
43
so
? ?
? ?ki
ki
f
f
?????
?????
,,,,,
,,,,,,1,1,1
54
54
??
??
?
?or ( 4— 22)
In this way,Using the method of choosing a new basic unit,we
can turn the functions (4— 18 ) among original k+1 dimensional
variables into the functions (4— 22) among k+1-3 namely k-2
dimensionless variables,It’s the Buckingham theorem, Because we
often express a dimensionless variable as π,it is also called
πtheorem,
Sinceπis dimensionless,the dimension of numerator and
denominator to the right of formula (4— 20) must be equal,To each
physical variable,list its exponent functions of dimensions of
numerator and denominator (L,M.T),simultaneousness and solve,
we can confirm all dimensional numbers in formula (4— 22),
44
于是
? ?
? ?ki
ki
f
f
?????
?????
,,,,,
,,,,,,1,1,1
54
54
??
??
?
?或 ( 4— 22)
这样,运用选择新基本单位的办法,可使原来 k+1个有量
纲的物理量之间的函数式( 4— 18)变成 k+1-3个即 k-2个无量
纲数之间的函数式( 4— 22),这就是泊金汉
定理。因为经常用 表示无量纲数,故又简称 定理。
? ?B u ck i n g h a mE,
? ?
因为 是无量纲数,因而式( 4— 20)右端分子分母的量
纲必须相同,对每个物理量 列出其分子分母量纲 的
幂次方程,联立求解,即可确定式( 4— 22)中的所有量纲数。
?
i n ? ?MTL,,
45
2,Application of dimension analysis
[solution ] according to above
? ?D?D,,,,,ldfp ???
choose as basic units,so ??,,d
666555444 654
,,,zyxzyxzyxzyx dd ldd p ???????? ????? D???D?
[Example 4—1] head loss of flowing in pipe on the way,
?
pD
?
D
From actually observe,In pipe flowing,the pressure difference
caused by friction on the way has relation to the following
factors,pipe diameter d,average velocity in pipe,dynaflow
viscosity, pipe line length l, degree of roughness of pipe wall
,fluid density,Try to solve the head loss on the way,?
46
二、量纲分析的应用
[解 ] 根据题意知
? ?D?D,,,,,ldfp ???
选择 作为基本单位,于是 ??,,d
666555444 654
,,,zyxzyxzyxzyx dd ldd p ???????? ????? D???D?
[例题 4—1]管中流动的沿程水头损失。
?pD
? D
根据实际观测知道,管中流动由于沿程摩擦而造成的压强
差 与下列因素有关:管路直径 d,管中平均速度,流体动
力粘度,管路长度 l,管壁的粗糙度,流体的密度 试求
水中流动的沿程水头损失。
?
47
Dimensions of each physical variable are as follows,
dimension
Physical
variable d
L
?
1?LT
?
3?ML
pD ?
11 ?? TML
l
L L
D
21 ?? TML
First analyze the dimension of,Since its dimension of numerator
and denominator should be equal, pD
? ? ? ? ? ? yzyxzzyx TLMMLLTLTML ??????? ?? 33121
0,2,1 ??? xyz
so
??? 2
pD?
solve it
48
各物理量的量纲如下,
量纲
物理量
d
L
?
1?LT
?
3?ML
pD ?
11 ?? TML
l
L L
D
21 ?? TML
首先分析 的量纲,因为其分子分母的量纲应该相同,所以 pD
? ? ? ? ? ? yzyxzzyx TLMMLLTLTML ??????? ?? 33121
解得
0,2,1 ??? xyz
所以
??? 2
pD?
49
Second analyze the dimension of,in a similar way ?
? ? ? ? ? ? 44444444 33111 yzyxzzy TLMMLLTLTML x ??????? ??
Resolve it 1,1,1 444 ??? xyz
so
??
??
d?4
In a similar way
dd
l D??
65,??
Substitute all π into (4— 22),we can obtain
???????? D?D ddldfP,,2 ?????
for the head loss flowing in pipe
order
?
??? d
v
d ??Re then
?????? D?D? ddlfggPh f,,Re1
2?
?
g
ph
f ?
D?
50
其次再分析 的量纲,同理有
?
? ? ? ? ? ? 44444444 33111 yzyxzzy TLMMLLTLTML x ??????? ??解得
1,1,1 444 ??? xyz
所以
??
??
d?4同理可得
dd
l D??
65,??
将所有 值代入式( 4— 22)可得 ?
???
?
???
? D?D
dd
l
df
P,,
2 ??
?
??
因为管中流动的水头损失
g
Ph f
?
D?

?
??? d
v
d ??Re 则
?????? D?D? ddlfggPh f,,Re1
2?
?
51
Chapter 4 Exercise
4— 1 Under layer flow condition,flow through a small equilateral
triangle hole (side length is b,hole length is L),its volume flow rate Q
is the function of dynamic-viscosity coefficient,the pressure drop
on unit length and b,Try to alter the relation into
nondimensional formula,If double b,how the Q will change?
?
Lp /D
solution,when the number of unknown parameters is less then 4,
we can obtain its accurate function whit dimensional analysis
method,It is called Rayleigh method,
) r a t e f l o w t oe q u i v a l e n t ( b e s o l v e d, be
t t oc o e f f i c i e n o n a ln o m d im e n s i is,f o r m u l ai n
)( ipr e l a t i o n s h le x p o n e n t i ao u t w r i t e2
0),,,( f o r mf u n c t i o n i m p l i c i t o u t w r i t e1
Q
b
L
p
QQ
b
L
p
Qf
???
?
?
D
?
?
D
)(
)(
52
第四章 习 题
4— 1 在层流情况下,流过一小等边三角形截面的孔(边长为 b,
孔长为 L )的体积流量 Q 为动力粘性系数,单位长度上的压降
及 b 的函数。试将此关系写成无因次式。若 b 加倍,流量有
何变化?
?
Lp /D
解:对未知参数小于等于 4 个时,可直接用量纲分析法求得
其准确的函数关系,并称之为雷利法。
此处相当于流量系数)为待求的无因次系数(式中
写出指数关系式)(
写出隐函数形式)(
)( 2
0),,,( 1
Q
b
L
p
QQ
b
L
p
Qf
???
?
?
D
?
?
D
53
3 1 1 1 2 2
3 w r it e o ut d im e ns iona l f or m [ ] [ ] [ ]
4 e xp r e ss a l l d im e ns ion of p hy sic a l v a r ia bl e s w it h
ba sic d im e ns ion ( ) [ ] [ ] [ ] [ ]
( 5 ) so l v e w it h dim e ns ion,
p
Qb
L
L,M,T L T M L T M L T L
? ? ?
? ? ?
?
? ? ?
? ? ? ? ?
D
?
?
( )
( )
、,
4
c on so na nc e the or y
, 3 2 1
, 0 he n 1
, 1 2 4
6 se tt l e f or m ul a s the n ( )
w he n d ou bl e d f l ux inc r e a se s 16 tim e s,
L
Mt
T
bp
QQ
L
b
? ? ? ?
? ? ?
? ? ?
?
? ? ? ? ??
??
?? ??
??
? ? ? ?
??
D
?
= -


( )

54
倍。倍时,流量增加增加当
整理方程)(


=-

、、根据量纲和谐原理求
表示各物理量的量纲以基本量纲()(
写出量纲式)(
16 1
)( 6
4
1
1
2 1,
0,
23,
)5(
][][][][
) 4
][][][ 3
4
221113
b
L
pb
QQ
T
M
L
LTMLTMLTL
L,M,T
b
L
p
Q
D
?
?
?
?
?
?
?
?
?
?
?
????
??
????
?
D
?
?????
?
?
?
?
??
??
???
???
?
???
???
55
4— 2 A flyer moving in stationary air is the same as airflow flowing
around fixity,According to measure,when circulating flow the flow
resistance D has relation to transect linear dimension l,average velocity
of flow,air density,dynamic viscosity, try to analyze the
resistance formula of ambient flow,v ? ?
i
so l u tion 1 w r ite o u t im p l ic it f u n c tion o f r e sista n c e (,,,)
c ho ose ba si c v a r ia b l e s,,
2 w r it e ou t e a c n d im e n sion,a n d
D f l v
lv
??
?
??
?,( )
( )
4 4 4
2 1 3 1 1
4
4
21
[ ] [ ],[ ] [ ],[ ] [ ],[ ] [ ],[ ] [ ]
,
3 sol v e a nd
T o [ ] [ ] [ ]
x y zx y z
xy
D M L T l L v L T M L M L T
D
l v l v
M L T L L T
??
?
??
??
??
?
? ? ? ? ?
??
? ? ? ? ?
??
?
( )

3
[]
z
ML
?
56
4— 2 飞行物体在静止空气中运动相当于气流绕固定物体流动。
根据测定,绕流时流体所受到的阻力 D 与物体横断面线性尺寸 l,
气流平均流速,气体密度,动力粘度 有关,试分析绕流阻力
公式。
v ? ?
zyx
zyxzyx
MLLTLM L T
vlvl
D
TMLMLLTvLlM L TD
vl
vlfD
][][][][
3
,
][][,][][,][][,][][,][][
2
,,
),,,( 1
312
4
4
11312
i
444
???
?????
?
??
?????
?
:对
和)解出(
及)写出各物理量纲和(
选择基本量
写出阻力隐函数式)(解:
?
??
?
?
?
?
?
??
??
?
??
57
)(,e q u a t i o nc r i t e r i o n as W r i t e4
So
1
2
2
r e s u l t
2,
31,
1,
][][][][, to
So
1
2
2
r e s u l t
2,
31,
1,
22
4
4
4
4
4
444
4
3111
4
22
444
?
?
?
?
?
?
?
?
?
lv
f
vl
D
lv
z
y
x
yT
zyxL
zM
MLLTLMML
vl
D
z
y
x
yT
zyxL
zM
zyx
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
?
????
)(
58
)( 4
1
2
2
2,
31,
1,
][][][][,
1
2
2
2,
31,
1,
22
4
4
4
4
4
444
4
3111
4
22
444
?
?
?
?
?
?
?
?
?
lv
f
vl
D
lv
z
y
x
yT
zyxL
zM
MLLTLMML
vl
D
z
y
x
yT
zyxL
zM
zyx
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
???
?
????
写成准则方程)(
所以
解得

所以
解得
59
2
( R e )
2
,f o r m s t a n d a r d
i n t oi t t r a n s f o r mso f i n a l y,t c o e f f i c i e ni t w i t h m o d i f y s h o u l d w eb e c a u s e
s t r u c t u r e f o r m u l a t h ei n f l u e n c et d o n '
c o n s t a n t d e c r e a s i n go r i n c r e a s i n g )
Re
1
(o u t o r w r i t e
22
22
v
ACf
v
AD
fvlD
D
?
?
?
??
?

4— 3 Use length scale model to experiment the aerodynamic
characteristics of shells,Given that the shell’s flying velocity is 1000
m/s, air temperature is, aerodynamics viscosity coefficient is; air temperature of model is, aerodynamics
viscosity coefficient of model is,Try to obtain the wind
speed and wind pressure of model which are elastic force similitude
and viscid force similitude,
101 ??
040
sPa 102.19 6 ?? ? 010
sPa 108.17 6 ?? ?
,experiment by curve
relative get their can,we incompressible to Reith relation w having
t,coefficien resistence flow ambient called (Re) formula in
,flow ambient of formula resistence Rayleigh famous the sIt'
f C D ?,
60
的关系曲线。关,可由实验测取二者
有可压缩流体中与称绕流阻力系数,在不式中:
绕流阻力计算公式。此即为著名的雷利(
,变成标准形式:正,故将该式稍作变形因为最终有系数进行修
响公式结析,在公式中增减常数不影或写成
Re ( R e )
)
2
( R e )
2
)
Re
1
(
22
22
fC
R a y l e i g h
v
ACf
v
AD
fvlD
D
D
?
??
?
?
?
?
4— 3 用长度比例尺 的模型试验炮弹的空气动力特性,已
知炮弹的飞行速度为 1000 m/s,空气温度为,空气动力粘性
系数为 ;模型空气温度为,空气动力粘性系数为
,试求同时满足弹性力相似和粘性力相似的模型风
速和风压。
101 ??
040
sPa 102.19 6 ?? ? 010
sPa 108.17 6 ?? ?
61
Solution,This problem is the synthetic application of Mach
similarity criterion and Reynolds similarity criterion,From given
conditions,first we should solve with Mach similarity criterion
,and then solve with Reynolds similarity criterion, also we
should use ideal-gas state equation as a bridge,
?V
?p
Tp
T
p
p
s
R
T
p
T
T
VV
Rkt
k R T
V
k R T
V
a
V
a
V
MM
V
???
??
?
?
?
?
?
??
?
???
???
???
?
?
?
?
?
?
?
t h e n,1 in c e
s o lv e t oe q u a t io n s t a t e g a s-id e a l t h r o u g h 2
m /s 87.950
313
283
1000
r e s u lt,k J /k g 287,4.1 a ir o
o r t h e n
c r it e r io n s im il a r it yM a c h t h r o u g h s p e e d w in d S o lv e 1
。)(
:,
)(
?
?
?
?
?
?
?
?
62
解:此题是马赫相似准则和雷诺相似准则综合应用问题,由
已知条件,应先由马赫相似准则求,然后用雷诺相似准则求
,中间需借助理想气体状态方程作为桥梁。
?V
?p
Tp
T
p
p
R
T
p
T
T
VV
Rk
k R T
V
k R T
V
a
V
a
V
MM
V
???
??
?
?
?
?
?
??
?
???
???
???
?
?
?
?
?
?
?
,1
2
m / s 87.950
313
283
1000
,k J / k g 287,4.1
1
故因为
。求由理想气体状态方程)(
解得:对空气,
或则
由马赫准则求模型风速)(
?
?
?
?
?
?
?
?
63
82.8
3 1 3
2 8 3
10
87.09
1 0 0 0
101 9, 2
108.17
so f o r
so t h e n
t h e n ReReg h t h r o u
c r i t e r i o n R e y n o l d s f r o m S o l v e3
6
6
11
1111
???
?
??
?
??
????
?
???
?
???
?
?
?
?
?
???
??
??
pp
p
p
p
p
p
lVlV
TlV
p
p
TlVplV
p
????
?
?
?????????
?
?
?
?
?
?
???
?
?
?
???
?
)(
64
82.8
313
283
10
87.09
1000
101 9, 2
108.17
ReRe
3
6
6
11
1111
???
?
??
?
??
????
?
???
?
???
?
?
?
?
?
???
??
??
pp
p
p
p
p
p
lVlV
TlV
p
p
TlVplV
p
????
?
?
?????????
?
?
?
?
?
?
???
?
?
?
???
?
所以因为
所以得
得由
由雷诺准则求)(
65
66