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:

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

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.

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.

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.

]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

]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

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

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.

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

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

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

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.

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

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

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

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

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

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

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]

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

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

［ 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］

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

(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

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

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

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

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

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 )

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