chapter 5 principles of convection 1. convection heat transfer: 2. basic law: newton’s law of...
TRANSCRIPT
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!