Chapter 5

Principles of Convection

1. Convection heat transfer:

2. Basic law: Newton’s law of cooling thA

5-1 Introduction

The thermal-energy-exchange process resulting from the contact between fluid and a solid surface at a different temperature

Fourier’s law0

yy

tA

Heat balance gives0

yy

tAtAh

Then 0

yy

t

th

1). Motive force 驱动力①Forced convection ( 强制对流 ): Outside motive force moves fluids past a surface.②Natural convection ( 自然对流 ): Motive force comes from the density difference in the fluid.

2). Phase change 有无相变Single phase heat transfer ( 单相介质传热 )Boiling heat transfer ( 沸腾换热 ):

③mixed convection ( 混合对流 ): Both forces works.

3. Classification

Condensation heat transfer( 凝结换热 )Melting heat transfer ( 熔化换热 )Solidification heat transfer ( 凝固换热 )Sublimation heat transfer ( 升华换热 )Sublimation heat transfer （凝华换热）

3) Flow regime 流动状态

Laminar heat transfer ( 层流流动换热 )

Turbulent heat transfer 湍流流动换热 ()

4) Geometric configuration 按几何形状

Flow in ducts ( 管内 ( 槽道内 ) 流动 )

Around a body ( 外部绕流 )

ConvectionHeat transfer

Singlephase

Phase change

Forced

Internalflow

In tubes

In ducts of various cross section

In infinite spaceNatural

In enclosure

Externalflow

Over a flat plateOver a cylinderOver a bank of tubesOver noncircular cylinderJet impingement

Mixed

Boiling Pool boilingFlow boiling

Condensation In-tube condensationCondensation on exterior surface

4. Governing equations

1). Postulated conditions

(1) 2-D

(2) incompressible flow

(3) Newtonian fluid

(4) Constant viscosity, thermal conductivity, and specific heat

(5) No heat generation

(6) Negligible viscous dissipation

2) Energy equation

xO

y

dyx

t

x

tx

2

2

dxdy

y

ttdy

y

vvcp

x

tdy

tdyucp

y

tdx

tdxvcp

dxx

t

x

tdy

2

2

dydxx

ttdx

x

uucp

Elementalvolume

Energy balance

Increased = Net flow + Source

Net heat conducted in the x direction dxdyx

t2

2

Net heat conducted in the y direction dxdyy

t2

2

Net heat convected in the x direction

dxdxdyx

t

x

ucdxdy

x

tucdxdy

x

utc

dxdxx

t

x

u

x

tudxtdx

x

uutdyctdyuc

dydxx

ttdx

x

uuctdyuc

ppp

pp

pp

dxdyx

tu

x

utcp

Energy balance

Change in internal energy

By the same way, the net heat convected in the y direction

dxdyy

tv

y

vtcp

t

dxdycp

dxdyy

tv

y

vt

x

tu

x

utc

dxdyy

tdxdy

x

ttdxdyc

p

p

2

2

2

2

y

tv

x

tu

y

v

x

ut

y

t

x

t

c

kt

p2

2

2

2

taD

Dt 2

2

2

2

2

y

t

x

ta

y

tv

x

tu

t

zw

yv

xu

D

D

Substaintial derivative

local derivative convection derivative

3). Summary (1) Continuity equation ( 连续性方程 ) 0

y

v

x

u

(2) Momentum equation ( 动量方程 )

2

2

2

2

2

2

2

2

y

v

x

v

y

pF

y

vv

x

vu

v

y

u

x

u

x

pF

y

uv

x

uu

u

y

x

inertial force ( 惯性力 )

body force( 体积力 )

pressure gradient

( 压力梯度 )

viscous force( 粘性力 )

(3) Energy equation ( 能量守恒方程 )

2

2

2

2

y

t

x

ta

y

tv

x

tu

t

Change in internal energy

convection conduction

(4) Convection heat transfer coefficient

0

yy

t

th

Variables: u, v, p, t, h Number of equations ： 5 Solvable h is different from u, v, p, t Nonlinear Initial condition and boundary condition Initial condition Boundary condition the 1st kind of the 2nd kind of why do not use 3rd boundary condition ？

5. The parameters influencing convection heat transfer velocity ： V h V=0 physical properties:　　　 density ， viscosity ， thermal conductivity ， specific heat temperatures of fluid and wall reference temperature ( 定性温度 ) flow regime

geometric configuration plate, tube, horizontal, vertical, internal, external

4

4

10

102200

2200

Re

Re

Re laminar ( 层流 )transition flow( 过渡流 )turbulent flow ( 旺盛 ) 湍流

tube flow( 管流 )

Flow on a flat plate Re=2×105 to 3×106,

Generally Re= u∞x/ =5×105

Summary

wfp

wfp

ttckufTA

qh

ttckufq

,,,,,,

,,,,,,

character dimension ( 特征尺度 ):

The dimension used in dimensionless group

reference temperature ( 定性温度 )

The temperature to determine physical properties

5-2 viscous flow

1. Boundary layer

Boundary layer: The region of flow that from the leading ed

ge of the plate in which the effects of viscosity are observed. the region near the surface in which temperature changes remarkably (靠近壁面处流体速度发生显著变化的薄层 )

boundary layer ends or thickness ( 边界层的厚度 ): the y coordinate where the velocity becomes

99% of free-stream flow

(2) =(x) x (x) (3) (x) << x (L) << L

(4) The flow field:

undisturbed flow (or potential) regime 主流区 boundary regime ( 边界层区 )

Feature of boundary layer

(1) laminar flow ( 层流 ), turbulent flow ( 湍流 )• critical value Rec=5 × 105

• viscous (or laminar) sub-layer （层流 or 粘性底层）

2. Thermal boundary layer ( 热边界层 or 温度边界层 )

If t∞≠tw , heat transfer will occur

Definition:

the region near the surface in which

temperature changes remarkably

boundary layer ends or thickness t

the y coordinate where the temperature difference

=t-tw=0.99(t-tw)

Feature of thermal boundary layer:

similar to boundary layer

3.Comparison between flow boundary layer and thermal boundary layer

On the definition of boundary layer, u and T are used. If body force is negligible, then

uvD

Du 2

Energy equation taD

Dt 2

If =a , the two equations are the same, and

u=u(x,y,z,) & u=u(x,y,z,) have the same forms, or

The distribution of T is the same as the distribution of u,

a is significant

pp cc

aPr

4. Prandtl number ( 普朗特数 )

Momentum equation

Kinematic viscosity 运动粘度

Thermal diffusivity 热扩散率

pca

diffusionheat

diffusion momentumPr

1Pr av Momentum diffusion=heat diffusion t

Ordinary Pr=0.6~4000air Pr=0.6~1Liquid metals Pr =0.01-0.001

1Pr

1Pr

av

av t

t Momentum diffusion>heat diffusion

Momentum diffusion<heat diffusion

Temperature distribution t=t(x,y)

)( 0

0 xfy

t

tt

yt

tth

yw

y

w

∴h= h (x) is the local heat transfer coefficient at x

local heat transfer coefficient ( 局部对流换热系数 ). 　　

average heat transfer coefficient ( 平均对流换热系数 )

dxxhl

hl

0

1

5. Importance of boundary layer

(1).Reducing equations (2). Analysis of heat transfer process

ttAh w

The convection heat transfer coefficient is the averaged convection heat transfer except for special declaration.

h(x) feature

Laminar region x (x) h (x) heat conduction

Transition region disturbance h(x)

Turbulent region Rturbulent core << Rlaminar sublayer

Heat transfer enhancement by controlling boundary layer

5-3 Inviscid flow

We have known from fluid mechanics and

engineering thermodynamics.

Lx ~ Then )1(0~x y0 Then 0~y

10~x

u

From the equation of continuity

0

y

v

x

uThen 10~

y

v

or 0~v

No body force.

It is known from boundary layer feature

The order of magnitude of u, t, L are 0(1)

The order of magnitude of , , are 0(）

5-4 Boundary layer differential equations

The order of magnitude of is 0(1) ， then 20~ 20~ a

The analysis of the order of magnitude

concordance of the order of magnitude ( 数量级一致）

0

y

v

x

u

1

1

2

2

2

2

y

u

x

u

x

p

y

uv

x

uu

1

111

1 1

1

12

2

1

The momentum diffusion at x direction is neglectable

2

2

2

2

y

v

x

v

y

p

y

vv

x

vu

111

12

2

2

2

2

2

y

t

x

ta

y

tv

x

tu

1

11

t

1

12

2

1

Then we have

The heat conduction at x direction is neglectable

2

21

y

uv

dx

dp

y

uv

x

uu

2

2

y

ta

y

tv

x

tu

0

y

v

x

u

p/y0, p=p(x)dp/dx=? According to Bernoulli equation （ for plane flow Fx=

0 ）

dx

duu

dx

dp

Boundary condition

0

0

x

y

y

uu

uu

vu 0

tt

tt

tt w

const.2

1 2 gzup

Solution to the equations for laminar flow（ 1908,Blasius, 1921, Pohlhausen)

31

213

12

1

PrRe332.0332.0 xx x

k

a

v

v

xu

x

kh

or PrRe332.0 31

21

k

xhNu x

x

Nux is Nusselt number

The mean of Nux is

31

21

0664.0

1PrReNudx

LNu x

L

5-5 Integral equations of boundary layer1. Assumption: P2121) The fluid is impressible and the flow is steady2) There are no pressure variations in the direction perpendicular

to the plate3) The viscosity is constant4) The viscous-forces in the y direction are negligible

5) Negligible viscous dissipation

dyuH

02

dxdyudx

ddyu

HH

0

2

0

2

2. The integral momentum equation

Increase in momentum flux in x direction xF

The momentum flux entering the left face

The momentum flux leaving the right face

Momentum theorem

The net momentum in the x direction H

dyudx

ddx

0

2

On mass flow through wall surface

The conservation of mass

The mass flow through plane 1 H

udy0

The mass flow through plane 2 dxudydx

dudy

HH

00

The mass flow entering plane AA dxudydx

d H

0

Carrying momentum with it

Hudy

dx

ddxu

0

The pressure force difference between plane 1 and 2 is

dxdx

dpHpHHdx

dx

dpp

The theorem of momentum is expressed as

dxdx

dpHdxudy

dx

ddxudyu

dx

ddx w

HH

00

2

HHH

udydx

duudy

dx

duudyu

dx

d000

dx

dpHudy

dx

duudyu

dx

ddyu

dx

dw

H HH

0 00

2

dx

dpHudy

dx

duudyuu

dx

dw

HH

00

From Bernoulli equation constup 2

2

1

dx

duu

dx

dp

We know

Substituting into the above eq.

That is

w

HH

dx

duHuudy

dx

duudyuu

dx

d

00

w

HHdyuu

dx

duudyuu

dx

d

00

∵H>, and u=u∞ when y>

∴ wdyuu

dx

duudyuu

dx

d

00

3. Integral energy equation

Energy balance

Energy convected in + heat transfer at wall=energy convected out

dyuH

0

From boundary differential eq. 2

2

2

2

y

t

x

t

Negligible heat conduction in x direction

Energy balance

dxtudydx

dctudycq

tdyuc

HH

pp

H

p

wAA

00

01

21 0

From momentum eq. we know the mass flow entering plane AA is

dxudydx

d H

0

The energy with it dxudydx

dtc

H

pAA

0

Heat transfer at wall0

y

w y

tdx

00

00

y

H

p

H

p y

tkdxudy

dx

ddxtctudy

dx

ddxc

0

0

y

H

p y

Hudytt

dx

dc

The integrand id zero for y>t since T=T∞

pca

0

0

y

y

taudytt

dx

d t

4. Boundary conditions0,0,,0,0

2

2

2

2

y

t

y

uttuy w

0,,

y

uuuy

5. ApplicabilityThe title of section 5-4 is laminar boundary layer on a flat plate. Do we use the condition of laminar? No!The differential and integral equations can be used for the laminar and turbulent situations

2

2

y

uv

y

uv

x

uu

2

2

y

ta

y

tv

x

tu

0,,

y

ttty t

1. Velocity boundary layer on plate (Von Karman 1921)

constu∵0

y

w dy

du

0

00

y

dy

dudyuu

dx

dudyuuu

dx

d

The velocity profile is unknown, the profile is chosen as a polynomial

32 dycybyayfu Boundary conditions

u u y

u y

,

0 , 0

0

02

2

y

u

y

u

00

auy

022

2

cy

u

5-6 Solution Integral equations of boundary layer

udbu 3

22 303 dbdbu

udd 333 32

ud

ub

2

3

So 3

2

1

2

3

yy

u

u

uu

dy

du

y

w 2

3

2

3

0

Substituting into integral eq., integrating gives

2

3

280

39 2

u

dx

du

u

vx64.4

xdx

u

vd

00 13

140

xx Re

64.4

v

xux

Re

x

w

u

xu

vu

vx

uu

u

Re

323.0323.0

64.42

3

2

3

2

soluyionexact Re

0.5

xx

u

vx

13

2802

x

wf u

CRe

646.02

Fanning friction coefficient （范宁摩擦系数）

meanfL

L

L

ffm CdxCL

C 2Re

292.110

2. Thermal boundary layer on plate(Γ.Η.Κружилин 克鲁齐林 1936 ）

The unknown profile of T is chosen as

wtthygyfye 32

Boundary conditions

0,,

0,0,02

2

yy

yy

t

Since the same boundary conditions, the profile is the same

xfC

Re664.0 solution exact

3

2

1

2

3

tt

yy

Energy integral eq.

00

yy

taudytt

dx

d t

or 0

0

y

yaudy

dx

d t

Inserting the temperature distribution and velocity distribution into the equation, and set

/t

42

0 280

3

20

3

uudyt

2

3

2

3

0

tyy

If <1, 4 is negligible. Then

a

dx

du

2

10

1

or adx

d

dx

du

223 210

1

From integral momentum eq.

dxu

vd

13

140

Substituting into the eq. adx

d

u

vx

u

vu

33

13

280

3

2

13

140

10

1

13

3 Pr14

13

14

13

3

4 v

a

dx

dx

General solution 4/33

Pr14

13 Cx

u

vx

13

2802

Boundary condition 0,0,0 txx

0Pr14

134/3

0

3

0

x

Cxx

So

4/3

031

4/3

031

3 1Pr1Pr14

13

x

x

x

xt

Local heat transfer coefficient at x0=0

31

21

0

PrRe332.064.4

Re

2

3

2

3x

x

ywx xxy

t

tth

31

21

PrRe332.0 xx

hxNu

31

21

0PrRe664.02

1

Lhdxh

Lh L

L

x

31

21

PrRe664.0Nu

The reference temperature is ttt wm 2

1

4/30Pr14

13xC

506.0ity Applicabil Pr

The result agrees well with experimental data

Constant heat flux constant wall temperature

31

21

PrRe453.0 xx

hxNu

)(

tt

xqNu

w

wx

3/12/100 PrRe6795.0

/1)(

1

x

wL

x

wL

ww

Lqdx

Nu

xq

Ldxtt

Ltt

)(2

3 tthq wLxw

Other relations for very wide range of Pr numberIsothermal flat plate

100for Pr/0468.01

PrRe3387.04/13/2

3/12/1

PrReNu x x

x

For the constant heat flux case 0.3387 is changed to 0.4637 0.0468 is changed to 0.0207

5-7 The relation between fluid friction and heat transferFanning friction coefficient

2/1332.02 x

wf Reu

C

Nusselt Number

PrRe332.0 31

21

xhNu x

x

or3/22/1332.0

PrReuc

h

PrRe

NuSt x

p

x

x

x

2/13/2 332.0 xReStPr

Then 2/3/2fCStPr

This is called Reynolds-Colburn analogy

which can be applied to turbulent flow

turbulent ： complex, nonsteady-state, 3-D

Navier-Stokes are also applicable

research method

5-8 Turbulent-boundary-layer heat transfer

direct simulation ( 直接模拟 )

Maelstrom simulation ( 大旋涡模拟 )

Reynolds time average

Analogy

Experiment

1. Momentum transfer and heat transfer

Time average and fluctuation( 脉动值 )

'ttt

vvv

uuu

For steady flow the mean value of fluctuations must be zero over an extended period 0' tvu

Turbulent shear stress or Reynolds stress ( 湍流切应力 ) The average of fluctuation production is not zero. The turbulent lump goes down at -v’, its mass is -v’, the momentum carried with it is -v’u’. The net exchanged momentum is called turbulent shear stress. Its mean value is

vut define

2N/mdy

duvu Mt

M eddy viscosity （湍流粘度） eddy diffusivity for momentum ( 湍流动量扩散率 )

'tvcq pt

definedy

dtctvcq Hppt '

2mW

Eddy diffusivity of heat ( 湍流热扩散率 ) H

M and H are not properties of matter.

H

Mt Pr Turbulent Prandtl number 6.1~0.1tPr

dy

duv Mtl

dy

dtacqqq Hptl

Prantdtl mixing length y

uKy

y

ulM

22

The turbulent lump also carries heat down the plane aa. The heat (called turbulent heat flux) carried with it is -cpv’T’.

2. Universal velocity profile

layerTurbulent 40030 5.5ln5.2

layerBaffer 30505.3ln0.5

sublayerLaminar 50

yyu

yyu

yyu

where /w

uu

/wy

y

3. Turbulent heat transfer based on fluid-friction analogy

4. Constant heat flux

.const.const ww04.1

Txqx NuNu

5-9 Turbulent-boundary-layer thickness

wdyuudx

duudyuu

dx

d

00

Integral momentum equation

Empirical relations 7/1

y

u

u 20296.0

u

xuw

integrating 5/1

5/1

)0296.0(7

72

x

xudx

d

Two physical situations

1. Boundary layer id fully turbulent from the leading edge of the plate.

2. It follows a laminar growth pattern up to Recrit=5X105 and a turbulent growth thereafter Situation 1 =0 at x=0 5/1381.0 xRe

x

Situation 2 = lam at x=xcrit=5105/u . lam is calculated from exact solution

21a)-(5 )105(0.5 2/15critlam

x

Integrating

5/4crit

5/4

5/1

lam 4

5)0296.0(

7

72xx

u

Combining 15/1 10256381.0 xx ReRex

5-10 Heat transfer in laminar tube flow

For fully developed flow, the analytical solution is

For constant heat flux Nud=4.364

For constant wall temperature Nud=3.66

The (non-steady) turbulence satisfies governing equation.

If the time-average value of each term is taken, the governing equations become

2

2

)(1

y

uv

dx

dp

y

uv

x

uu M

2

2

)(y

ta

y

tv

x

tu H

0

y

v

x

u

2*

*2

**

**

*

** )(

11

y

uv

ludx

dp

y

uv

x

uu M

2*

2

**

** )(

ya

yv

xu H

0 *

*

*

*

y

v

x

u

Set lyylxx /,/ **

uvvuuu /,/ **

w

w

tt

tt

Boundary condition

1,/,1:/

0,0,0:0***

***

uvvuly

vuy

For plane flow and Pr=1 u and have same form solution

5-11 Turbulent flow in tube

1. Governing equation

2. Analogy

0

*

*

0

*

*

**

yyyy

u

2221 2000

*

*

*

ReC

lu

uu

l

y

u

u

l

y

u

y

uf

w

yyy

Nul

tt

q

tt

l

y

T

y w

w

wyy

)(00

**

xf

x ReC

Nu2

Colburn-Reynolds analogy

factor 2

3/2 jjStPrC f

3. Darcy friction coefficient for tube flow

2

2mu

d

lfp

Applying momentum theorem to the control volume

4

2dpdlw

222

2824

1

4 mf

mm

w uc

ufu

d

lf

l

d

l

dp

So 4/fC f

8/2/3 fStPr 3/1

8PrRe

fNu d

For smooth tube at Re=5×103 ～ 2×105,the Darcy friction coefficient is

2.0184.0 dRefSubstituting into Reynolds analogy 3/18.0023.0 PrReNu dFor Pr=0.6 ～ 50

5-13 Similarity Principle & dimensional analysis 相似原理及其量纲分析1. Dimensionless equations

Steady-state, constant physical properties, 2-D

0

y

v

x

u

2

2

2

2

2

2

2

2

y

v

x

v

y

pF

y

vv

x

vu

y

u

x

u

x

pF

y

uv

x

uu

y

x

2

2

2

2

y

t

x

ta

y

tv

x

tu

Set2

0

0 ,,,,,

u

pP

L

yY

L

xX

tt

tt

u

vV

u

uU

Then Uuu

x

Uu

x

u

2

2

2

2

x

Uu

x

u

sinceXLx

X

Xx

1

2

2

22

2 11

XLx

X

xLXxxx

By the same way

X

U

L

u

x

u

2

2

2

2

Re

1

Y

U

X

U

X

P

Y

UV

X

UU

2

2

2

2

Re

1

Y

V

X

V

Y

P

Y

VV

X

VU

2

2

2

2

PrRe

1

YXYV

XU

If two processes have the same Pr , Re and the same boundary conditions, then they have the same solutions.

0

Y

V

X

U

aPr

LuRe

2. Concept of similarity1). Geometrical similarity——the ratios of corresponding sides of geometrical bodies are the same

constcc

c

b

b

a

al

aa

b b

cc

2). Physical similarity——the ratios of physical quantities in corresponding positions are the same

Corresponding points

lcx

x

x

x

x

x

x

x

3

3

2

2

1

1

lcy

y

y

y

y

y

y

y

3

3

2

2

1

1

ucu

u

u

u

u

u

u

u

3

3

2

2

1

1Physical quantities

Similarity: the same kinds of phenomena, equations, geometric configurations, initial conditions and boundary condition.

3. Similarity theorem 1) Similarity nature ( 相似性质 )

Convection heat transfer0

'

'

yy

T

th

0"

"

yy

T

th

Physical similarityLtkh cy

yct

tcchh

,,, "

'

"

'

Then

0

"

"

"

yk

Lh

y

t

th

c

cc

As a result1

k

Lh

c

cc

lcL

L

y

y

Geometric

similarity "' LhLh

so

"' yhyh

uNuN The method is called similarity analysis (相似分析 )

By using similarity analysis, from momentum equation eReR From energy equation ePeP

Nature convection

From Bernoulli equation constxgp X

YO

gx

p

Body force gFx One has

ggg

x

pFx

The coefficient of volume expansion

tttt p

11

or tgg

RePrPe Peclet number 贝克利数

Buoyancy

Substituting into momentum equation

2

2

y

uvtg

y

uv

x

uu

Dimensionless form 2

2

2 Re

1

Y

U

u

tLg

X

UV

X

UU

222

2

2

3

2 ReGr

Lu

v

v

tlg

u

tlg

2

2

2 Re

1

Re Y

UGr

X

UV

X

UU

Grashof number

va

tLgPrGrRa

3

Rayleigh number

2

3

v

tLgGr

Physical significanceof dimensionless groups

2223force inertia lul

ulu

ld

dum

ululdy

duA force viscous

3gravity glmg 3lift buoyance Tlg

2pressure total pl

ul

luulRe

2

2

3

22

gl

lu

gl

uFr

2

22

2

2 lu

pl

u

pEu

lu

lu

lu

tlg

v

tlgGr

223

2

3

Conclusion ： If two phenomena are similar, the values of their dimensionless groups of the same name are equal.

2. Similarity criterion ( 相似条件 ) If the geometric configuration, initial and boundary conditions are similar, and the dimensionless groups of the same name are equal, the phenomena are similar.

3. The relation of dimensionless groups ( 相似准则之间的关系 ) The solution to governing equations must be the function of the dimensionless groups. 0,,, 21 nF

)(Re,PrfNu Known dimensionless group ( 已定准则 , 定性（型）准则 )Undetermined dimensionless group ( 待定准则 , 定性（型）准则 )

The power function of known dimensionless groups mncNu PrRe

4. Processing of experimental data

PrmRencNu lnlnlnln

5. Dimensional Analysis （量纲分析）

The process of determining appropriate dimensionless group

1).dimension: a dimension is a name given to any measurable quantity.

2). Primary dimensions and dimensional formulas

primary dimension ( 基本量纲 ) ： The choice is arbitrary. In SI system there are 7 dimensions, but 4 are used in heat transfer: length [L], time[T], mass[M] and temperature

[]

Dimensional formulas ( 导出量纲 ) of a physical quantity follows fro

m

definitions or physical laws, m/s [L]/[T]

principle of dimensional consistency ( 量纲和谐原理 ) ：

The dimensions of both sides of an equation are the same3). Dimension analysis( 量纲分析 )

6. Buckingham theorem The required number of independent dimensionless groups that can be formed by combining the physical variables pertinent to a problem is equal to the total number of these physical quantities n minus the number of primary dimensions m required to express the dimensional formulas of the n physical quantities.

Take the steady-state heat transfer without pressure gradient as a example 0),,,,,,( pckLuh

gfbgfecbfedcbagedb

gfedcba

gfp

edcba

TLM

T

M

T

L

LT

M

L

M

T

L

T

MLL

hcukL

32323

32

2

33

For to be dimensionless,The exponents of each primary dimension must be zero

0

0323

023

0

gfb

gfecb

fedcba

gedb

0...),,( 321 f

Nuk

hL1

Re2 uL

Pr3 k

cp

so PrRe,fNu

n=7 and m=4 ， there are n-m=3 independent dimensionless groups

First choose g=1, c=d=0, then a=1, b= -1, e=f=0

Second g=0, a=1, f=0, then c=1, d= -1, e=-1, b=0

Third f =1, a=g=0, then e=1, b= -1, c=d=0

gfp

edcba hcukL

7. Application of similarity principleExperiment :

Convection heat transfer in a tube

)(Re,PrfNu

),,,,,( pckdufh

Take 10 values of each variable

106 ， 100

Universality,

economy

Thank you!