Page 1
CHAPTER 5 Principles of Convection
5-1 INTRODUCTION
5-2 VISCOUS FLOW
]15[ dy
du
viscosityDynamic
boundary layer : the region of flow that develops from the leading edge of a plate in which the effects of the viscosity are observed
The outside boundary of a boundary layer is usually chosen as the point where the velocity of flow is 99 percent of the free-stream value.
Shear stress:
Page 2
Three regimes of boundary-layer flow
1. Laminar flow
2. Transitional flow
3. Turbulent flow
Renolds number
For most analytical purposes, the critical number for the transition is usually taken as 5105
5105
xu
v
xu
]25[Re
v
xux
The transition occurs when
Page 3
The critical Re for transition is strongly dependent on the surface roughness condition and the “turbulent level”of the free-stream.
The normal range for the beginning of transition is between :65 10105 to
For very large disturbances present in the flow, transition may begin with Renolds number as low as
510Foe flows that are very free from fluctuation, the transition may not start until
6102Re The transition is completed at Re twice the value at the transition begin.
Page 4
The relative shape for the velocity profiles in laminar and turbulent flow
The laminar profile is approximately parabolic
Structure of turbulent profile :
Laminar sublayer that is nearly linear.
Turbulent portion which is relatively flat in comparison with the laminar profile.
Page 5
The physical mechanism of viscosity in fluids
In laminar flow, the viscosity is attributed to the exchange of momentum between different laminas by the movement of molecules.
For a gas T
In turbulent flow, the momentum exchange between different layers is caused by the macroscopic movement of fluid chunks. We can expect a larger viscous-shear in turbulent flow than in laminar flow, due to which the velocity profile is flat in a turbulent boundary layer.
Page 6
]35[2300Re v
dumd
4000Re2000 d
]45[
Aum m
Flow in a tube
The critical Re
The range of Re for transition is
]55[
muA
mGvelocityMass
]65[Re Gd
d
Continuity relation in a tube is
Re based on mass velocity is defined as
Page 7
]75[2
1 2
aconstg
vp
c
]75[0 bg
VdVdp
c
]85[2
1
2
1 222
211 WkV
gciQV
gi
c
]95[ pvei
5-3 INVISCOUS FLOWThe Bernoulli equation for flow along a stream results:
In differential form,
The energy equation for compressible fluid
i is the enthalpy defined by
Page 8
2
1
0
2
11 M
T
T
)1/(20
2
11
Mp
p
)1/(120
2
11
M
]105[ RTga c
]115[/045.20 smTa
RTp Tce v Tci pKmolkgJ
MR
./5.8314
CkgkJaircCkgkJaircKkgJR vpair00 /718.0,/005.1,/287
Equation of state of fluid
Relations applicable to reversible adiabatic flow:
a
VM
Page 9
5-4 LAMINAR BOUNDARY LAYER ON A FLAT PLATE
dmVd
F xx
Assumptions:
1. Incompressible and steady flow
2. No pressure variation in the direction perpendicular to the plate.
3. Constant viscosity
4. Viscous-shear in y direction is negligible.
1. Newton’s law of motion
Which applies to a system of constant mass.
2. Force balance
directionin x flux momentumin increse xF
Which applies to a elemental control volume fixed in space.
Two methods to study motion of fluid.
Page 10
dydxx
uu
dxdyy
vv
dxdyy
vvdydx
x
uuvdxudy
vdx
udy
0
y
v
x
u
Mass continuity equation
Mass in the left face is
Mass out of the left face is
Mass in the bottom face is
Mass out of the top face is
Mass balance on the element is Mass continuity equation
Page 11
dyuudyu 2
vdx
dydxx
uu
2
udy
Derivation of momentum equation
Mass in the left face is
Momentum flux in the left face is
Momentum flux out of the left face is
Mass in the bottom face is
vudxMomentum flux in the x direction entering the bottom face
dxdyy
uudy
y
vv
Momentum flux in the x direction leaving the top face
dxdyy
vv
Mass out of the top face is
Page 12
dxdyx
p
dxy
u
dyy
u
yy
udx
Pressure forces on the left and right faces are andpdy dydxxpp ])/([
Net pressure force in the direction of motion is
Viscous-shear force on the bottom face is
Viscous-shear force on the top face is
x
p
y
u
y
uv
x
uu
2
2
dxdyy
u2
2
forceshear -sNet viscou
dyudydxx
uudxdy
x
pdxdy
y
u 22
2
2
vudxdxdyy
uudy
y
vv
Balancing force and momentum in x direction gives
Final result
——Momentum equation
Page 13
dxdyudx
ddyu
HH
0
2
0
2
dxudydx
dudy
HH
00
dxudydx
du
H
0
dxudydx
dudxdyu
dx
d HH
00
2
Hudy
0
H
dyu0
2
[d]
[c]
[b]
[a]
Mass flow through plane 1
Momentum flow through plane 1
Momentum flow through plane 2
Mass flow through plane 2
Carried momentum in x direction by the flow through plane A-A
The net momentum flow out of the control volume is
Integral momentum equation of the boundary layer.
Page 14
dxudydx
dudxudyu
dx
ddxudy
dx
du
HHH
000
dxudydx
dudxdyuu
dx
d HH
00
ddd ddd
0
y
w y
udxdx
By the use of or
The pressure force on plane 1 is
The pressure force on plane 2 is
pH
Hdxdxdpp ])/([
The shear force at wall is
Setting the force on the element equal to the net increase in momentum gives
]155[00
HHw udy
dx
duudyuu
dx
dH
dx
dp
——Integral momentum equation of the boundary layer.
dxudydx
dudxdyu
dx
d HH
00
2
Page 15
dx
duu
dx
dp 0 165
0
0
y
w y
udyuuu
dx
d
175
For constant pressure , from Bernoulli equation
Integral momentum equation of the boundary layer becomes
0cg
VdVdp
One obtains
Page 16
0
0
y
w y
udyuuu
dx
d 175
0u at 0y
uu at y
a b
0y
uat y c
Boundary conditions are
02
2
y
uat 0y d
Evaluation of boundary layer thickness
3
2
1
2
3
yy
u
u 195
u
y
u
dyyyyy
udx
d
y 2
3
2
1
2
31
2
1
2
3
0
3
0
32
Appling boundary conditions obtains
Inserting the expression into equation [5-17]
Carrying out the integration leads to
u
udx
d
2
3
280
39 2
For constant-pressure condition
suppose
34
2321 yCyCyCCu
dxu
dxu
d
13
140
13
140
Separation of variables leads to
Page 17
constu
x
13
140
2
2
205
]205[64.4 u
x
00 xat , so that
In terms of Renolds number
2/1Re
64.4
xx
a215
215 xu
x2/1Re
2/1Re
0.5
xx
Exact solution
Page 18
5-5 ENERGY EQUATION OF THE BOUNDARY LAYER
Assumptions:
1. incompressible steady flow
2. Constant viscosity ,thermal conductivity, and specific heat
3. Negligible heat conduction in the direction of flow
Energy convected in left face
+ energy convected in bottom face
+ energy conducted in bottom face
+net viscous work done on element
= energy convected out right face
+ energy out top face
+heat conducted out top face
Page 19
2
2
2
y
u
cy
T
y
Tv
x
Tu
p
[ 5-12]
[ 5-22]
dxy
u
The viscous shear force over dx
0
y
v
x
u
dxdyy
u2
dyy
u
The distance through which the force moves in respect to the control volume dxdy is
The net viscous energy delivered to the element is
Energy balance corresponding to the quantities shown in figure 5-6 is
Using
And dividing by gives pc
——Energy equation of the laminar boundary layer.
y
v
x
uT
y
Tv
x
Tucp dxdy
y
udxdy
y
Tkdxdy
2
2
2
Page 20
1Pr
Pr
1
~
~
~~
2
2
2
22
22
2
Tc
u
k
cv
T
u
c
u
cy
u
c
T
y
Tthatso
yanduu
p
p
p
pp
[ 5-24]
[ 5-23]
Order-of-magnitude analysis
2
2
2
2
22
0.1012.02931005
707.0Pr
1
29320
/70
y
uv
y
uv
x
uu
y
T
y
Tv
x
Tu
Tc
u
atmp
K℃T
smu
p
[ 5-25]
[ 5-26]
A striking similarity between [5-25] and [5-26]
Page 21
TT
yTk
hw
wall
ty
[ 5-27]
[ 5-28]
[ 5-29]
[ a]
[ b]
[ c]
[ d]
At y=0
At
ty At
At y=0
5-6 THE THERMAL BOUNDARY LAYER
wally
Tkq
A
q
''
TThq w''
TT
y
T
TT w
0
02
2
y
T
2. Definition of h
3. Temperature distribution in the thermal boundary layer
1. Thermal boundary layer
Boundary conditions
Page 22
3
2
1
2
3
ttw
w yy
TT
TT
[ 5-30]
Conditions (a) to (d) may be fitted to a cubic polynomial
H
puTdyc 0
dxuTdycdx
duTdyc H
pH
p 00
dxdyudx
d H0
dxdyudx
dTc Hp 0
dxdydy
duH
0
2
Energy convected in +viscous work within element +heat transfer at wall=energy convected out
The energy convected through plane 1 is
The energy convected out through plane 2 is
The mass flow through plane A-A
The energy carried with is
The net viscous work done within element is
Heat transfer at wall
ww x
Tkdxdq
4. Integral energy equation of the boundary layer
Page 23
[ 5-32] wH
p
H
dy
Tdy
dy
du
cudyTT
dx
d]0
2
0
Combining the above energy quantities gives
ty
H
tt
HH
y
T
dyyyyy
dx
du
udydx
dudyTT
dx
d
2
3]
2
1
2
3
2
1
2
31
0
0
33
00
——integral energy equation of the boundary layer.
Inserting (5-30) and (5-19) into (5-32) gives
5. Thermal boundary layer thickness
3
2
1
2
3
yy
u
u 195
3
2
1
2
3
ttw
w yy
TT
TT
[ 5-3
0]
Page 24
2
34280
32203
dxθ
du 5-33
5-34
23)2(
203
dxdu
dx
ddxd
u 22101
dx
ddxd
u 3222101
dxuvd
13
140
uvx
132802
Assume thermal boundary layer is thinner than the hydrodynamic boundary layer
Making substitution
/t
Neglecting gives4
or
Performing the differentiation gives
But according to page 217
and
Page 25
vaCx
14134/33
oxx atWhen the boundary condition
is applied , the final solution become
5-36
where 5-37
5-38
5-35so that we have
Noting that
vdxdx
1413243
3312 dxd
dxd
3/1Pr026.11
t
0t0
3/14/3013/1Pr
026.11
x
xt
avPr
oxx at
When the plate is heated over the entire length
Solution is
Page 26
kpc
pckav
//
Pr 5-39
5-40
5-41
5-42
5-43
5-44
23
23/ k
t
kTwT
wyTkh
3/14/301
2/13/1Pr332.0
x
x
vx
ukxh
k
xxh
xNu
3/1012/1Re3/1Pr332.0
x
x
xxNu
2/1Re3/1Pr332.0 xxNu
Substituting (5-21) and (5-36) gives
Nusselt number
Finally,
For the plate heated over its entire length
6. Prandtl number(see page 225)
7. Nusselt number
Page 27
LxL
x hdx
dxhh
L
20
0 5-45
5-46a
5-46b
or
where
5-47
LxNu
kLhNu L 2
3/12/1 PrRe664.0 LL kLhNu
Lu
LRe
2 TTT w
f
9. Average heat transfer coefficient and Nusselt number
Film temperature
Page 28
5-48
5-49
5-50
3/1Pr2/1Re453.0 xkhxxNu
TwTkxwqxNu
3/1Pr2/1Re6795.0
/00
11
L
kLwqdxLxkNuxwqL
ldxTwTL
TwT
or
For Rex Pr > 100 5-51
TWTLxhwq 23
4/13/2
Pr0468.01
3/1PrRe3387.0
xxNu
10. Constant heat flux
11. Other relationsFor laminar flow on an isothermal flat plate
For the constant-heat-flux case, 0.3387 is changed to 0.4637, and 0.0468 is replaced by 0.0207.
Page 29
0
y
v
x
uMass continuity equation
x
p
y
u
y
uv
x
uu
2
2
Momentum equation
0
0
y
w y
udyuuu
dx
d
Integral momentum equation
2
2
2
y
u
cy
T
y
Tv
x
Tu
p
Energy equation
3/1Pr026.11
t
23
23/ k
t
kTwT
wyTkh
wH
p
H
dy
Tdy
dy
du
cudyTT
dx
d]0
2
0
Integral energy equation
2/1Re
0.5
xx
2/1Re3/1Pr332.0 xxNu
3
2
1
2
3
yy
u
u
3
2
1
2
3
ttw
w yy
TT
TT
Velocity and temperature distributions
2
2
2
2
y
uv
y
uv
x
uu
y
T
y
Tv
x
Tu
A striking similarity
Results:
consg
vp
c
2
2
1
kc
WVgc
iQVg
i 222
211 2
1
2
1
Basic laws for inviscous flow
Page 30
2
2u
fCw
5-7 THE RELATION BETWEEN FLUID FRICTION AND HEAT TRANSFER
Using the velocity distribution given by equation(5-19), we have
[5-52]
wyu
w
u
w 23
The shear stress is3
2
1
2
3
yy
u
u
Making use of the relation for the boundary-layer thickness gives
53564.42
3 21
vx
uuw
Combining (5-52) and (5-53) leads to
545Re323.01
64.42
3
22
1
2
21
x
fx
uvx
uuC
Page 31
aCx
fx 545Re332.02
21
21
32
RePr332.0PrRe
x
p
x
x
x
uc
hNu
The exact solution is
[5-44] may be rewritten as
uc
hSt
p
xx
555Re332.0 213
2
xx prSt
5652
32
fxx
CprSt
By introduction of Stanton number
——Reynolds-Colburn analogy
Page 32
5-8 TURBULENT-BOUNDARY-LAYER HEAT TRANSFER
575 uuu
585 vvv
Structure of turbulent flow:
1. Laminar sublayer
2. Buffer layer
3. Turbulent
velocity fluctuation in a turbulent flow
The physical mechanism of heat transfer in turbulent flow is similar to that in laminar flow.
Difficulty: there is no completely adequate theory to predict turbulent-flow behavior
Page 33
595 uvt
0'' uv 0'' vu
Shear stress giving rise to velocity fluctuations in turbulent flow
'v ''uv
605 dy
duuv Mt
y
uyuyu
y
uyuyu
615
y
uu
6252
2
y
u
y
uvu Mt
Eddy Viscosity and Mixing Length ( 湍流粘度与混合长度 )
ydiffusiviteddy or sity eddy visco—M
Mean free path and Prandtl mixing length
Prandtl postulated:
6352
y
uM
Page 34
6352
y
uM
645 Ky
2
22
y
uyKw
655ln1
CyK
u w
665
y
uv M
675
w
uu
685
v
yy w
6951
v
dydu
M
cyu
705 yu
For laminar sublayer
0~M
Prandtal’s hypothesis
Nondimensional coordinates
In the near-wall region
Universal velocity profile
becomestconsgAssu )665(,tanmin
0,0:. yatuCB
wt
Page 35
yKy
u w 11
yK wM
715 KyvM
725ln1
cyK
u
yu
05.3ln0.5 yu
5.5ln5.2 yu
Laminar sublayer : 0<y+<5Buffer layer : 5<y+<30Turbulent layer : 30<y+<400
745
y
Tc
A
qHp
turb
755
y
Tc
A
qHp
For fully turbulent region
1/ M
From equation [5-65], we have
or
Substituting this relation along with equation (5-64) into equation (5-63) gives
Substituting this relation into Eq (5-69) for and integrating gives
1/ M
For fully turbulent region For regions where both molecular and turbulent energy transport are important
Universal velocity profile
6951
v
dydu
M
6352
y
uM
645 Ky
Page 36
765Pr H
Mt
Turbulent Heat Transfer Based on Fluid-Friction Analogy
1. Fluid-friction analogy
2. The local skin-friction coefficient over a flat plate:
5/1Re0592.0 xfxC
75 10Re105 xfor
584.2)Re(log370.0 xfxC
97 10Re10 xfor
3. Average-friction coefficient for a flat plate:
LLf
AC
Re)Re(log
455.0584.2
910Re xfor
LL
f
AC
ReRe
074.05/1 710Re xfor
4. Local turbulent heat transfer coefficient5/13/2 Re0296.0Pr xxSt
584.23/2 )Re(log185.0Pr xxSt
75 10Re105 xfor
97 10Re10 xfor
Table 5-1
5652
32
fxtx
CprSt
A simpler formula for lower Reynolds number is
8940334017421055
10310105103Re 6655
A
crit
Page 37
2Pr 3/2 fCSt
15/13/2 Re871Re037.0Pr LLSt
Pr)/(ReLNuSt
)871Re037.0(Pr 8.03/1 LLk
LhNu
dxhdxh
Lh turb
L
xcritlam
xcrit
0
1
3/1584.2 Pr]871)Re(logRe228.0[ LLL
k
LhNu
5. Average heat transfer coefficient over the entire laminar-turbulent boundary layer
75 10Re105Re Lcrit andfor
From , the above equation can be rewritten as
Alternatively,
for the laminar portion2
13
2
Re332.0
xx prSt
5/13/2 Re0296.0Pr xxStOne obtains
597 105Re10Re10 critx andfor
6. Equation suggested by Whitaker4/1
8.043.0 )9200(RePr036.0
wLLNu
5.326.0
105.5Re102
380Pr7.065
w
L
constTxxw
NuNu
04.1
Constant Heat Fluxfor the turbulent portion
)871Re037.0(Pr 8.03/1 LLk
LhNu
For higher Reynolds number,using equation (5-79), one obtains
Page 38
7/1
y
u
u
2
2uC f
w
2
5/1
0296.0
u
xuw
5/17/1
0
7/1
0296.01
xu
dyyy
dx
d
]905[)0296.0(7
72 5/1
5/1
xudx
d
5/1Re381.0 xx
u
xat crit
5105
2/15 )105(0.5 critlam x
5/45/4
5/1
4
5)0296.0(
7
72critlam xx
u
15/1 Re256.10Re381.0 xxx
75 10Re105 xfor
TURBULENT BOUNDARY LAYER THICKNESS1. Velocity profile in a turbulent boundary layer
2. Shear stress at wall
5/1Re0592.0 xfxC 75 10Re105 xfor
So that
0
0
y
w y
udyuuu
dx
d 175
3. Integrating the integral momentum equation
Integrating and clearing terms gives
The first case: The boundary layer is fully turbulent from the leading edge of the plate:
lam
The second case: The boundary layer follows a laminar growth pattern up to and a turbulent growth thereafter
5105Re crit
Integrating [5-90] leads to
Combining the above various relations gives
Page 39
(a) Semilog scale
(b) log scale
Boundary-layer thickness for atmospheric air at u=30m/s.
555Re332.0 213
2
xx prSt
5/13/2 Re0296.0Pr xxSt
75 10Re105 xfor
)871Re037.0(Pr 8.03/1 LLk
LhNu
Page 40
5-10 HEAT TRANSFER IN LAMINAR TUBE FLOW
constrdx
dpu
and
drdx
dprdu
ordr
dudxrrdxdpr
2
2
4
1
2
1
22
1. Velocity distribution
00 rratu
dx
dpru
4
20
0
20
2
0
1r
r
u
u
)(4
1 20
2 rrdx
dpu
5-98
Page 41
0dx
dqw
r
Trdxkdqr
2
dxr
Tucrdr p
2
drr
T
r
Tdxdrrkdq drr 2
2
)(2
dxdrr
Tr
r
Tkdxdr
r
Tucr p
2
2
r
T
r
T
rur
11
2. Energy balance analysis and temperature distribution
Net energy convected out = net heat conducted in
which may be rewritten
Page 42
constxT
00
ratr
T
constqr
Tk w
rr
0
rr
ru
x
T
r
Tr
r
20
2
0 11
120
42
0 42
1C
r
rru
x
T
r
Tr
2120
42
0 ln164
1CrC
r
rru
x
TT
01 C
cc TCthatsoratTT 20
4
0
2
0
200
4
1
4
1
r
r
r
rru
x
TTT c
assume
B.C:
Inserting Eq (5-98) into Eq (5-99)
]995[11
r
T
r
T
rur
20
2
0
1r
r
u
u 5-98
Page 43
x
Tru
r
rr
x
Tu
r
T
rrrr
44200
20
30
00
00 11
48
11
24
d
k
r
kh
364.40 k
hdNud
)(''bw TThq
0
0
0
0
2
2r
p
r
p
b
rdruc
TrdrucTT
x
TruTT cb
200
96
7
x
TruTT cw
200
16
3
0
)(rr
bw r
TkATThAq
bw
rr
TT
rTkh
0
)(
Bulk temperature
Local heat flux =
2. Bulk temperature ( 整体温度 )
1. Definition of convection heat transfer coefficient in tube flow
3. Wall temperature
4. Convection heat transfer coefficient
Page 44
5-11 TURBULENT FLOW IN A TUBE
dy
dTk
A
q
dy
dT
Ac
q
p
)1085()( dy
dT
Ac
qH
p
)1095()(
dy
du
dy
duMH
dTduAc
q
p
ww
w
A
qconst
A
q
dTducA
q b
w
m T
T
uu
upww
w
0
bwpww
mw TTcA
uq
)( bwww TThAq
givesbydividing
orassume HM
)1095()1085(
1Pr,
For laminar flow
For turbulent flow
assume
Integrating (a)
Heat transfer at wall is
( a )
Page 45
8PrRe
fNu
uc
hSt
d
d
mp
4/1Re
316.0
d
f
4/1Re0395.0PrRe
dd
dNu
4/3Re0395.0 ddNu
8Pr 3/2 f
St
3/14/3 PrRe0395.0 ddNu
4.08.0 PrRe023.0 ddNu
L
dp
Ld
dpw
0
0
20
44
)(
2
8 mw uf
2
2mu
d
Lfp
Shear stress at wall is
The pressure drop can be expressed in terms of friction factor
So that
bwpww
mw TTcA
uq
)( bwww TThAq
Heat transfer at wall is
( A )
( B )
( C )
Substituting ( B ) and ( C ) into ( A ) gives
Reynolds analogy for tube flow
( D )
( D ) is modified by Pr
A more correct relation