chapter 2: heat conduction...

Chapter 2: Heat Conduction

Equation

Yoav PelesDepartment of Mechanical, Aerospace and Nuclear Engineering

Rensselaer Polytechnic Institute

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

ObjectivesWhen you finish studying this chapter, you should be able to:

• Understand multidimensionality and time dependence of heat transfer, and the conditions under which a heat transfer problem can be approximated as being one-dimensional,

• Obtain the differential equation of heat conduction in various coordinate systems, and simplify it for steady one-dimensional case,

• Identify the thermal conditions on surfaces, and express them mathematically as boundary and initial conditions,

• Solve one-dimensional heat conduction problems and obtain the temperature distributions within a medium and the heat flux,

• Analyze one-dimensional heat conduction in solids that involve heat generation, and

• Evaluate heat conduction in solids with temperature-dependent thermal conductivity.

Introduction

• Although heat transfer and temperature are

closely related, they are of a different nature.

• Temperature has only magnitude

it is a scalar quantity.

• Heat transfer has direction as well as magnitude

it is a vector quantity.

• We work with a coordinate system and indicate

direction with plus or minus signs.

Introduction ─ Continue

• The driving force for any form of heat transfer is the

temperature difference.

• The larger the temperature difference, the larger the

rate of heat transfer.

• Three prime coordinate systems:

– rectangular (T(x, y, z, t)) ,

– cylindrical (T(r, f, z, t)),

– spherical (T(r, f, q, t)).

Classification of conduction heat transfer problems:

• steady versus transient heat transfer,

• multidimensional heat transfer,

• heat generation.

Introduction ─ Continue

Steady versus Transient Heat Transfer

• Steady implies no change with time at any point

within the medium

• Transient implies variation with time or time

dependence

Multidimensional Heat Transfer

• Heat transfer problems are also classified as being:

– one-dimensional,

– two dimensional,

– three-dimensional.

• In the most general case, heat transfer through a

medium is three-dimensional. However, some

problems can be classified as two- or one-dimensional

depending on the relative magnitudes of heat transfer

rates in different directions and the level of accuracy

desired.

• The rate of heat conduction through a medium in

a specified direction (say, in the x-direction) is

expressed by Fourier’s law of heat conduction

for one-dimensional heat conduction as:

• Heat is conducted in the direction

of decreasing temperature, and thus

the temperature gradient is negative

when heat is conducted in the positive x-

direction.

(W)cond

dTQ kA

dx (2-1)

General Relation for Fourier’s Law of

Heat Conduction• The heat flux vector at a point P on the surface of

the figure must be perpendicular to the surface,

and it must point in the direction of decreasing

temperature

• If n is the normal of the

isothermal surface at point P,

the rate of heat conduction at

that point can be expressed by

Fourier’s law as

(W)n

dTQ kA

dn (2-2)

General Relation for Fourier’s Law of

Heat Conduction-Continue

• In rectangular coordinates, the heat conduction

vector can be expressed in terms of its components as

• which can be determined from Fourier’s law asn x y zQ Q i Q j Q k

x x

y y

z z

TQ kA

x

TQ kA

y

TQ kA

z

(2-3)

(2-4)

Heat Generation• Examples:

– electrical energy being converted to heat at a rate of I2R,

– fuel elements of nuclear reactors,

– exothermic chemical reactions.

• Heat generation is a volumetric phenomenon.

• The rate of heat generation units : W/m3 or Btu/h · ft3.

• The rate of heat generation in a medium may vary

with time as well as position within the medium.

• The total rate of heat generation in a medium of

volume V can be determined from

(W)gen gen

V

E e dV (2-5)

One-Dimensional Heat Conduction

Equation - Plane Wall

xQ

Rate of heat

conduction

at x

Rate of heat

conduction

at x+Dx

Rate of heat

generation inside

the element

Rate of change of

the energy content

of the element

- + =

,gen elementEx xQ D

elementE

t

D

D

(2-6)

• The change in the energy content and the rate of heat

generation can be expressed as

• Substituting into Eq. 2–6, we get

,

element t t t t t t t t t

gen element gen element gen

E E E mc T T cA x T T

E e V e A x

D D DD D

D

,element

x x x gen element

EQ Q E

tD

D

D(2-6)

(2-7)

(2-8)

x x xQ Q D(2-9)

gene A x D t t tT TcA x

t D

DD

1gen

T TkA e c

A x x t

(2-11)

• Dividing by ADx, taking the limit as Dx 0 and Dt 0,

and from Fourier’s law:

The area A is constant for a plane wall the one dimensional

transient heat conduction equation in a plane wall is

gen

T Tk e c

x x t

Variable conductivity:

Constant conductivity:2

2

1 ;

geneT T k

x k t c

1) Steady-state:

2) Transient, no heat generation:

3) Steady-state, no heat generation:

2

20

gened T

dx k

2

2

1T T

x t

2

20

d T

dx

The one-dimensional conduction equation may be reduces

to the following forms under special conditions

(2-13)

(2-14)

(2-15)

(2-16)

(2-17)


Equation - Long Cylinder

rQ

Rate of heat

conduction

at r

Rate of heat

conduction

at r+Dr

Rate of heat

generation inside

the element

Rate of change of

the energy content

of the element

- + =

,gen elementEelementE

t

D

Dr rQ D

(2-18)

• The change in the energy content and the rate of heat

generation can be expressed as

• Substituting into Eq. 2–18, we get

,

element t t t t t t t t t

gen element gen element gen

E E E mc T T cA r T T

E e V e A r

D D DD D

D

,element

r r r gen element

EQ Q E

tD

D

D(2-18)

(2-19)

(2-20)

r r rQ Q D(2-21)

gene A r D t t tT TcA r

t D

DD

1gen

T TkA e c

A r r t

(2-23)

• Dividing by ADr, taking the limit as Dr 0 and Dt 0,

and from Fourier’s law:

Noting that the area varies with the independent variable r

according to A=2prL, the one dimensional transient heat

conduction equation in a plane wall becomes

1gen

T Trk e c

r r r t

10

gened dTr

r dr dr k

The one-dimensional conduction equation may be reduces

to the following forms under special conditions

1 1geneT Tr

r r r k t

1 1T Tr

r r r t

0d dT

rdr dr


Constant conductivity:

1) Steady-state:



(2-25)

(2-26)

(2-27)

(2-28)

(2-29)


Equation - Sphere

2

2

1gen

T Tr k e c

r r r t

2

2

1 1geneT Tr

r r r k t


Constant conductivity:

(2-30)

(2-31)

General Heat Conduction Equation

x y zQ Q Q

Rate of heat

conduction

at x, y, and z

Rate of heat

conduction

at x+Dx, y+Dy,

and z+Dz

Rate of heat

generation

inside the

element

Rate of change

of the energy

content of the

element

- + =

x x y y z zQ Q QD D D ,gen elementE elementE

t

D

D(2-36)

Repeating the mathematical approach used for the one-

dimensional heat conduction the three-dimensional heat

conduction equation is determined to be

2 2 2

2 2 2

1geneT T T T

x y z k t

2 2 2

2 2 20

geneT T T

x y z k

2 2 2

2 2 2

1T T T T

x y z t

2 2 2

2 2 20

T T T

x y z

Two-dimensional

Three-dimensional

1) Steady-state:



Constant conductivity: (2-39)

(2-40)

(2-41)

(2-42)

Cylindrical Coordinates

2

1 1gen

T T T T Trk k k e c

r r r r z z t

f f

(2-43)

Spherical Coordinates

2

2 2 2 2

1 1 1sin

sin singen

T T T Tkr k k e c

r r r r r tq

q f f q q q

(2-44)

Boundary and Initial Conditions

• Specified Temperature Boundary Condition

• Specified Heat Flux Boundary Condition

• Convection Boundary Condition

• Radiation Boundary Condition

• Interface Boundary Conditions

• Generalized Boundary Conditions

Specified Temperature Boundary

Condition

For one-dimensional heat transfer

through a plane wall of thickness

L, for example, the specified

temperature boundary conditions

can be expressed as

T(0, t) = T1

T(L, t) = T2

The specified temperatures can be constant, which is the

case for steady heat conduction, or may vary with time.

(2-46)

Specified Heat Flux Boundary

Condition

dTq k

dx

Heat flux in the

positive x-

direction

The sign of the specified heat flux is determined by

inspection: positive if the heat flux is in the positive

direction of the coordinate axis, and negative if it is in

the opposite direction.

The heat flux in the positive x-

direction anywhere in the medium,

including the boundaries, can be

expressed by Fourier’s law of heat

conduction as

(2-47)

Two Special Cases

Insulated boundary Thermal symmetry

(0, ) (0, )0 or 0

T t T tk

x x

,2

0

LT t

x

(2-49) (2-50)

Convection Boundary Condition

1 1

(0, )(0, )

T tk h T T t

x

2 2

( , )( , )

T L tk h T L t T

x

Heat conduction

at the surface in a

selected direction

Heat convection

at the surface in

the same direction=

and

(2-51a)

(2-51b)

Radiation Boundary Condition

Heat conduction

at the surface in a

selected direction

Radiation exchange

at the surface in

the same direction=

4 4

1 ,1

(0, )(0, )surr

T tk T T t

x

4 4

2 ,2

( , )( , ) surr

T L tk T L t T

x

and

(2-52a)

(2-52b)

Interface Boundary Conditions

0 0( , ) ( , )A BA B

T x t T x tk k

x x

At the interface the requirements are:

(1) two bodies in contact must have the same

temperature at the area of contact,

(2) an interface (which is a

surface) cannot store any

energy, and thus the heat flux

on the two sides of an

interface must be the same.

TA(x0, t) = TB(x0, t)

and

(2-53)

(2-54)

Generalized Boundary ConditionsIn general a surface may involve convection, radiation,

and specified heat flux simultaneously. The boundary

condition in such cases is again obtained from a surface

energy balance, expressed as

Heat transfer

to the surface

in all modes

Heat transfer

from the surface

In all modes=

Heat Generation in SolidsThe quantities of major interest in a medium with heat

generation are the surface temperature Ts and the

maximum temperature Tmax that occurs in the medium

in steady operation.

The heat transfer rate by convection can also be

expressed from Newton’s law of cooling as

(W)s sQ hA T T

gen

s

s

e VT T

hA

Rate of

heat transfer

from the solid

Rate of

energy generation

within the solid=

For uniform heat generation within the medium

(W)genQ e V

-

Heat Generation in Solids -The Surface

Temperature

(2-64)

(2-65)

(2-66)

(2-63)

Heat Generation in Solids -The Surface

TemperatureFor a large plane wall of thickness 2L (As=2Awall and

V=2LAwall)

,

gen

s plane wall

e LT T

h

For a long solid cylinder of radius r0 (As=2pr0L and

V=pr02L)

0

,2

gen

s cylinder

e rT T

h

For a solid sphere of radius r0 (As=4pr02 and V=4/3pr0

3)

0

,3

gen

s sphere

e rT T

h

(2-68)

(2-69)

(2-67)

Heat Generation in Solids -The maximum

Temperature in a Cylinder (the Centerline)

The heat generated within an inner

cylinder must be equal to the heat

conducted through its outer surface.

r gen r

dTkA e V

dr

Substituting these expressions into the above equation

and separating the variables, we get

222

gen

gen

edTk rL e r L dT rdr

dr kp p

Integrating from r =0 where T(0) =T0 to r=ro2

0

max, 04

gen

cylinder s

e rT T T

kD (2-71)

(2-70)

Variable Thermal Conductivity, k(T)

• The thermal conductivity of a material, in general, varies with temperature.

• An average value for the thermal conductivity is commonly used when the variation is mild.

• This is also common practice for other temperature-dependent properties such as the density and specific heat.

Variable Thermal Conductivity for

One-Dimensional Cases

2

1

2 1

( )T

T

ave

k T dTk

T T

When the variation of thermal conductivity with

temperature k(T) is known, the average value of the thermal

conductivity in the temperature range between T1 and T2

can be determined from

The variation in thermal conductivity of a material

with can often be approximated as a linear function

and expressed as

0( ) (1 )k T k T

the temperature coefficient of thermal conductivity.

(2-75)

(2-79)

Variable Thermal Conductivity

• For a plane wall the

temperature varies linearly

during steady one-

dimensional heat conduction

when the thermal conductivity

is constant.

• This is no longer the case

when the thermal conductivity

changes with temperature

(even linearly).

Chapter 3: Steady Heat

Conduction





• Understand the concept of thermal resistance and its limitations, and develop thermal resistance networks for practical heat conduction problems,

• Solve steady conduction problems that involve multilayer rectangular, cylindrical, or spherical geometries,

• Develop an intuitive understanding of thermal contact resistance, and circumstances under which it may be significant,

• Identify applications in which insulation may actually increase heat transfer,

• Analyze finned surfaces, and assess how efficiently and effectively fins enhance heat transfer, and

• Solve multidimensional practical heat conduction problems using conduction shape factors.

Steady Heat Conduction in Plane

Walls

1) Considerable temperature difference

between the inner and the outer

surfaces of the wall (significant

temperature gradient in the x

direction).

2) The wall surface is nearly isothermal.

Steady one-dimensional modeling approach is

justified.

• Assuming heat transfer is the only energy interaction

and there is no heat generation, the energy balance can

be expressed as

or

The rate of heat transfer through the

wall must be constant ( ).

0

0wallin out

dEQ Q

dt

Rate of

heat transfer

into the wall

Rate of

heat transfer

out of the wall

Rate of change

of the energy

of the wall- =

Zero for steady

operation

, constantcond wallQ

(3-1)

• Then Fourier’s law of heat conduction for the wall

can be expressed as

• Remembering that the rate of conduction heat transfer

and the wall area A are constant it follows

dT/dx=constant

the temperature through the wall varies linearly with x.

• Integrating the above equation and rearranging yields

, (W)cond wall

dTQ kA

dx (3-2)

(3-3)1 2, (W)cond wall

T TQ kA

L

Thermal Resistance Concept-

Conduction Resistance

• Equation 3–3 for heat conduction through a

plane wall can be rearranged as

• Where Rwall is the conduction resistance

expressed as

(3-4)1 2

, (W)cond wall

wall

T TQ

R

(3-5) ( C/W)wall

LR

kA

Analogy to Electrical Current Flow• Eq. 3-5 is analogous to the relation for electric current

flow I, expressed as

Heat Transfer Electrical current flow

Rate of heat transfer Electric current

Thermal resistance Electrical resistance

Temperature difference Voltage difference

(3-6)1 2

eR

V VI


Convection Resistance

• Thermal resistance can also be applied to convection

processes.

• Newton’s law of cooling for convection heat transfer

rate ( ) can be rearranged as

• Rconv is the convection resistance

conv s sQ hA T T

(3-7) (W)sconv

conv

T TQ

R

(3-8)1

( C/W)conv

s

RhA


Radiation Resistance

• The rate of radiation heat transfer between a surface and

the surrounding

(3-9)

4 4 ( ) (W)s surrrad s s surr rad s s surr

rad

T TQ A T T h A T T

R

(3-10)1

( /W)rad

rad s

R Kh A

2 2 2 (W/m K)( )

radrad s surr s surr

s s surr

Qh T T T T

A T T

(3-11)


Radiation and Convection Resistance

• A surface exposed to the surrounding might involves convection and radiation simultaneously.

• The convection and radiation resistances are parallel to each other.

• When Tsurr≈T∞, the radiation

effect can properly be

accounted for by replacing h

in the convection resistance

relation by

hcombined = hconv+hrad (W/m2K)

(3-12)

Thermal Resistance Network• consider steady one-dimensional heat transfer

through a plane wall that is exposed to convection on

both sides.

• Under steady conditions we have

or

Rate of

heat

convection

into the wall

Rate of

heat conduction

through the wall

Rate of

heat convection

from the wall= =

(3-13)

1 ,1 1

1 22 2 ,2

Q h A T T

T TkA h A T T

L

Rearranging and adding

1 2 wallT T Q R

,1 ,2

1 2

1 1 ( C/W)total conv wall conv

LR R R R

h A kA h A

(3-15),1 ,2

(W)total

T TQ

R

(3-16)

where

,1 ,2T T ,1 ,2( )conv wall convQ R R R

totalQ R

,1 1 ,1convT T Q R

2 ,2 ,2convT T Q R

• It is sometimes convenient to express heat transfer

through a medium in an analogous manner to

Newton’s law of cooling as

• where U is the overall heat transfer coefficient.

• Note that

(W)Q UA T D (3-18)

1 ( C/K)

total

UAR

(3-19)

Multilayer Plane Walls

• In practice we often encounter plane walls that consist

of several layers of different materials.

• The rate of steady heat transfer through this two-layer

composite wall can be expressed through Eq. 3-15

where the total thermal

resistance is

,1 ,1 ,2 ,2

1 2

1 1 2 2

1 1

total conv wall wall convR R R R R

L L

h A k A k A h A

(3-22)

Thermal Contact Resistance• In reality surfaces have some roughness.

• When two surfaces are pressed against each other, the peaks form good material contact but the valleys form voids filled with air.

• As a result, an interface contains

numerous air gaps of varying sizes

that act as insulation because of the

low thermal conductivity of air.

• Thus, an interface offers some

resistance to heat transfer, which

is termed the thermal contact

resistance, Rc.

• The value of thermal contact resistance depends on the

– surface roughness,

– material properties,

– temperature and pressure at the interface,

– type of fluid trapped at the interface.

• Thermal contact resistance is observed to decrease with decreasing surface roughness and increasing interface pressure.

• The thermal contact resistance can be minimized by applying a thermally conducting liquid called a thermal grease.

Generalized Thermal Resistance

Network• The thermal resistance concept can be used to solve

steady heat transfer problems that involve parallel

layers or combined series-parallel arrangements.

• The total heat transfer of two parallel layers

1 2 1 21 2 1 2

1 2 1 2

1 1T T T TQ Q Q T T

R R R R

(3-29)1

totalR

1 2

1 2 1 2

1 1 1 = total

total

R RR

R R R R R

(3-31)

Combined Series-Parallel Arrangement

The total rate of heat transfer through

the composite system

where

31 21 2 3

1 1 2 2 3 3 3

1 ; ; ; conv

LL LR R R R

k A k A k A hA

(3-32)1

total

T TQ

R

1 212 3 3

1 2

total conv conv

R RR R R R R R

R R

(3-33)

(3-34)

Heat Conduction in Cylinders

Consider the long cylindrical layer

Assumptions:

– the two surfaces of the cylindrical

layer are maintained at constant

temperatures T1 and T2,

– no heat generation,

– constant thermal conductivity,

– one-dimensional heat conduction.

Fourier’s law of heat conduction

, (W)cond cyl

dTQ kA

dr (3-35)

Separating the variables and integrating from r=r1,

where T(r1)=T1, to r=r2, where T(r2)=T2

Substituting A =2prL and performing the integrations

give

Since the heat transfer rate is constant

, (W)cond cyl

dTQ kA

dr (3-35)

2 2

1 1

,

r T

cond cyl

r r T T

Qdr kdT

A

(3-36)

1 2

,

2 1

2ln /

cond cyl

T TQ Lk

r rp

(3-37)

1 2,cond cyl

cyl

T TQ

R

(3-38)

Thermal Resistance with Convection

Steady one-dimensional heat transfer through a

cylindrical or spherical layer that is exposed to

convection on both sides

where

(3-32),1 ,2

total

T TQ

R

,1 ,2

2 1

1 1 2 2

ln /1 1

2 2 2

total conv cyl convR R R R

r r

r L h Lk r L hp p p

(3-43)

Multilayered

Cylinders

• Steady heat transfer through

multilayered cylindrical or

spherical shells can be handled just like multilayered plane.

• The steady heat transfer rate through a three-layered

composite cylinder of length L with convection on both

sides is expressed by Eq. 3-32 where:

,1 ,1 ,3 ,3 ,2

2 1 3 2 4 3

1 1 1 2 3 2 2

ln / ln / ln /1 1

2 2 2 2 2

total conv cyl cyl cyl convR R R R R R

r r r r r r

r L h Lk Lk Lk r L hp p p p p

(3-46)

Critical Radius of Insulation

• Adding more insulation to a wall or to the attic

always decreases heat transfer.

• Adding insulation to a cylindrical pipe or a spherical

shell, however, is a different matter.

• Adding insulation increases the conduction resistance

of the insulation layer but decreases the convection

resistance of the surface because of the increase in the

outer surface area for convection.

• The heat transfer from the pipe may increase or

decrease, depending on which effect dominates.

• A cylindrical pipe of outer radius r1

whose outer surface temperature T1 is

maintained constant.

• The pipe is covered with an insulator

(k and r2).

• Convection heat transfer at T∞ and h.

• The rate of heat transfer from the insulated pipe to the

surrounding air can be expressed as

1 1

2 1

2

ln / 1

2 2

ins conv

T T T TQ

r rR R

Lk h r Lp p

(3-37)

2

0dQ

dr

, (m)cr cylinder

kr

h

• The variation of the heat transfer rate with the outer radius of the insulation r2 is shown

in the figure.

• The value of r2 at which

reaches a maximum is

determined by

• Performing the differentiation

and solving for r2 yields

• Thus, insulating the pipe may actually increase the rate of heat transfer instead of decreasing it.

(3-50)

Q

Heat Transfer from Finned Surfaces

• Newton’s law of cooling

• Two ways to increase the rate of heat transfer:

– increasing the heat transfer coefficient,

– increase the surface area fins

• Fins are the topic of this section.

conv s sQ hA T T

Fin EquationUnder steady conditions, the energy balance on this

volume element can be expressed as

or

where

Substituting and dividing by Dx, we obtain

Rate of heat

conduction into

the element at x

Rate of heat

conduction from the

element at x+Dx

Rate of heat

convection from

the element= +

, ,cond x cond x x convQ Q QD

convQ h p x T T D

, ,0

cond x x cond xQ Qhp T T

x

D

D(3-52)

Taking the limit as Dx → 0 gives

From Fourier’s law of heat conduction we have

Substitution of Eq. 3-54 into Eq. 3–53 gives

0conddQhp T T

dx (3-53)

cond c

dTQ kA

dx (3-54)

0c

d dTkA hp T T

dx dx

(3-55)

For constant cross section and constant thermal conductivity

Where

• Equation 3–56 is a linear, homogeneous, second-order differential equation with constant coefficients.

• The general solution of Eq. 3–56 is

• C1 and C2 are constants whose values are to be determined from the boundary conditions at the base and at the tip of the fin.

22

20

dm

dx

qq (3-56)

; c

hpT T m

kAq

1 2( ) mx mxx C e C eq (3-58)

Boundary Conditions

Several boundary conditions are typically employed:

• At the fin base

– Specified temperature boundary condition, expressed

as: q(0)= qb= Tb-T∞

• At the fin tip

1. Specified temperature

2. Infinitely Long Fin

3. Adiabatic tip

4. Convection (and

combined convection

and radiation).

Infinitely Long Fin (Tfin tip=T)• For a sufficiently long fin the temperature at the fin

tip approaches the ambient temperature

Boundary condition: q(L→∞)=T(L)-T∞=0

• When x→∞ so does emx→∞

C1=0

• @ x=0: emx=1 C2= qb

• The temperature distribution:

• heat transfer from the entire fin

/( )cx hp kAmx

b

T x Te e

T T

(3-60)

0

c c b

x

dTQ kA hpkA T T

dx

(3-61)

Adiabatic Tip• Boundary condition at fin tip:

• After some manipulations, the temperature distribution:

• heat transfer from the entire fin

0x L

d

dx

q

(3-63)

cosh( )

coshb

m L xT x T

T T mL

(3-64)

0

tanhc c b

x

dTQ kA hpkA T T mL

dx

(3-65)

Convection (or Combined Convection

and Radiation) from Fin Tip

• A practical way of accounting for the heat loss from

the fin tip is to replace the fin length L in the relation

for the insulated tip case by a corrected length

defined as

Lc=L+Ac/p (3-66)

• For rectangular and cylindrical

fins Lc is

• Lc,rectangular=L+t/2

• Lc,cylindrical =L+D/4

Fin Efficiency• To maximize the heat transfer from a fin the

temperature of the fin should be uniform (maximized)

at the base value of Tb

• In reality, the temperature drops along the fin, and thus

the heat transfer from the fin is less

• To account for the effect we define

a fin efficiency

or

(3-69)

,max

fin

fin

fin

Q

Q

Actual heat transfer rate from the fin

Ideal heat transfer rate from the fin

if the entire fin were at base temperature

,max ( )fin fin fin fin fin bQ Q hA T T

Fin Efficiency

• For constant cross section of very long fins:

• For constant cross section with adiabatic tip:

,

,max

1 1fin c b clong fin

fin fin b

Q hpkA T T kA

Q hA T T L hp mL

(3-70)

,

,max

tanh

tanh

fin c b

adiabatic fin

fin fin b

Q hpkA T T aL

Q hA T T

mL

mL

(3-71)

Fin Effectiveness• The performance of the fins is judged on the basis of the

enhancement in heat transfer relative to the no-fin case.

• The performance of fins is expressed

in terms of the fin effectiveness fin

defined as

fin fin

fin

no fin b b

Q Q

Q hA T T

Heat transfer rate

from the surface

of area Ab

Heat transfer rate

from the fin of base

area Ab

(3-72)

Remarks regarding fin effectiveness

• The thermal conductivity k of the fin material should be as high as possible. It is no coincidence that fins are made from metals.

• The ratio of the perimeter to the cross-sectional area of the fin p/Ac should be as high as possible.

• The use of fins is most effective in applications involving a low convection heat transfer coefficient.

The use of fins is more easily justified when the medium is a gas instead of a liquid and the heat transfer is by natural convection instead of by forced convection.

Overall Effectiveness• An overall effectiveness for a

finned surface is defined as the

ratio of the total heat transfer

from the finned surface to the

heat transfer from the same

surface if there were no fins.

,

fin

fin overall

no fin

unfin fin fin

no fin

Q

Q

h A A

hA

(3-76)

Proper Length of a Fin

• An important step in the design of a fin is the

determination of the appropriate length of the fin once

the fin material and the fin cross section are specified.

• The temperature drops along

the fin exponentially and

asymptotically approaches the

ambient temperature at some

length.

Heat Transfer in Common Configurations

• Many problems encountered in practice are two- or three-dimensional and involve rather complicated geometries for which no simple solutions are available.

• An important class of heat transfer problems for which simple solutions are obtained encompasses those involving two surfaces maintained at constant temperatures T1 and T2.

• The steady rate of heat transfer between these two surfaces is expressed as

Q=Sk(T1=T2) (3-79)

• S is the conduction shape factor, which has the dimension of length.

Table 3-7

Chapter 6: Fundamentals of

Convection





• Understand the physical mechanism of convection, and its classification,

• Visualize the development of velocity and thermal boundary layers during flow over surfaces,

• Gain a working knowledge of the dimensionless Reynolds, Prandtl, and Nusselt numbers,

• Distinguish between laminar and turbulent flows, and gain an understanding of the mechanisms of momentum and heat transfer in turbulent flow,

• Derive the differential equations that govern convection on the basis of mass, momentum, and energy balances, and solve these equations for some simple cases such as laminar flow over a flat plate,

• Nondimensionalize the convection equations and obtain the functional forms of friction and heat transfer coefficients, and

• Use analogies between momentum and heat transfer, and determine heat transfer coefficient from knowledge of friction coefficient.

Physical Mechanism of Convection

• Conduction and convection are similar in that both

mechanisms require the presence of a material medium.

• But they are different in that convection requires the presence

of fluid motion.

• Heat transfer through a liquid or gas can be by conduction or

convection, depending on the presence of any bulk fluid

motion.

• The fluid motion enhances heat transfer, since it brings

warmer and cooler chunks of fluid into contact, initiating

higher rates of conduction at a greater number of sites in a

fluid.

• Experience shows that convection heat transfer strongly depends on the fluid properties:

– dynamic viscosity m,

– thermal conductivity k,

– density , and

– specific heat cp, as well as the

– fluid velocity V.

• It also depends on the geometry and the roughness of the solid surface.

• The rate of convection heat transfer is observed to be proportional to the temperature difference and is expressed by Newton’s law of cooling as

• The convection heat transfer coefficient h depends on the several of the mentioned variables, and thus is difficult to determine.

2 (W/m )conv sq h T T (6-1)

• All experimental observations indicate that a fluid in

motion comes to a complete stop at the surface and

assumes a zero velocity relative to the surface (no-slip).

• The no-slip condition is responsible for the development

of the velocity profile.

• The flow region adjacent

to the wall in which the

viscous effects (and thus

the velocity gradients) are

significant is called the boundary layer.

• An implication of the no-slip condition is that heat

transfer from the solid surface to the fluid layer

adjacent to the surface is by pure conduction, and can

be expressed as

• Equating Eqs. 6–1 and 6–3 for the heat flux to obtain

• The convection heat transfer coefficient, in general,

varies along the flow direction.

2

0

(W/m )conv cond fluid

y

Tq q k

y

(6-3)

0 2 (W/m C)

fluid y

s

k T yh

T T

(6-4)

The Nusselt Number• It is common practice to nondimensionalize the heat transfer

coefficient h with the Nusselt number

• Heat flux through the fluid layer by convection and by conduction can be expressed as, respectively:

• Taking their ratio gives

• The Nusselt number represents the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid layer.

• Nu=1 pure conduction.

chLNu

k (6-5)

convq h T D (6-6)cond

Tq k

L

D (6-7)

/

conv

cond

q h T hLNu

q k T L k

D

D(6-8)

Classification of Fluid Flows

• Viscous versus inviscid regions of flow

• Internal versus external flow

• Compressible versus incompressible flow

• Laminar versus turbulent flow

• Natural (or unforced) versus forced flow

• Steady versus unsteady flow

• One-, two-, and three-dimensional flows

Velocity Boundary Layer• Consider the parallel flow of a fluid over a flat plate.

• x-coordinate: along the plate surface

• y-coordinate: from the surface in the normal direction.

• The fluid approaches the plate in the x-direction with a uniform

velocity V.

• Because of the no-slip condition V(y=0)=0.

• The presence of the plate is felt up to d.

• Beyond d the free-stream velocity remains essentially unchanged.

• The fluid velocity, u, varies from 0 at y=0 to nearly V at y=d.

Velocity Boundary Layer

• The region of the flow above the plate bounded by dis called the velocity boundary layer.

• d is typically defined as

the distance y from the

surface at which

u=0.99V.

• The hypothetical line of

u=0.99V divides the flow over a plate into two

regions:

– the boundary layer region, and

– the irrotational flow region.

Surface Shear Stress• Consider the flow of a fluid over the surface of a plate.

• The fluid layer in contact with the surface tries to drag the plate along via friction, exerting a friction force on it.

• Friction force per unit area is called shear stress, and is denoted by t.

• Experimental studies indicate that the shear stress for most fluids is proportional to the velocity gradient.

• The shear stress at the wall surface for these fluids is expressed as

• The fluids that that obey the linear relationship above are called Newtonian fluids.

• The viscosity of a fluid is a measure of its resistance to deformation.

2

0

(N/m )s

y

u

yt m

(6-9)

• The viscosities of liquids decrease with temperature, whereas

the viscosities of gases increase with temperature.

• In many cases the flow velocity profile is

unknown and the surface shear stress ts

from Eq. 6–9 can not be obtained.

• A more practical approach in external flow

is to relate ts to the upstream velocity V as

• Cf is the dimensionless friction coefficient (most cases is

determined experimentally).

• The friction force over the entire surface is determined from

22 (N/m )

2s f

VC

t (6-10)

2

(N)2

f f s

VF C A

(6-11)

Thermal Boundary Layer• Like the velocity a thermal boundary layer develops when a

fluid at a specified temperature flows over a surface that is at

a different temperature.

• Consider the flow of a fluid

at a uniform temperature of

T∞ over an isothermal flat

plate at temperature Ts.

• The fluid particles in the

layer adjacent assume the surface temperature Ts.

• A temperature profile develops that ranges from Ts at the

surface to T∞ sufficiently far from the surface.

• The thermal boundary layer ─ the flow region over the

surface in which the temperature variation in the direction

normal to the surface is significant.

• The thickness of the thermal boundary layer dt at any

location along the surface is defined as the distance

from the surface at which the temperature difference

T(y=dt)-Ts= 0.99(T∞-Ts).

• The thickness of the thermal boundary layer increases

in the flow direction.

• The convection heat transfer rate anywhere along the

surface is directly related to the temperature gradient

at that location.

Prandtl Number• The relative thickness of the velocity and the

thermal boundary layers is best described by the

dimensionless parameter Prandtl number, defined

as

• Heat diffuses very quickly in liquid metals (Pr«1)

and very slowly in oils (Pr»1) relative to momentum.

• Consequently the thermal boundary layer is much

thicker for liquid metals and much thinner for oils

relative to the velocity boundary layer.

Molecular diffusivity of momentumPr

Molecular diffusivity of heat

pc

k

m

(6-12)

Laminar and Turbulent Flows

• Laminar flow ─ the flow is characterized by

smooth streamlines and highly-ordered

motion.

• Turbulent flow ─ the flow is

characterized by velocity

fluctuations and

highly-disordered motion.

• The transition from laminar

to turbulent flow does not

occur suddenly.

• The velocity profile in turbulent flow is much fuller than that in

laminar flow, with a sharp drop near the surface.

• The turbulent boundary layer can be considered to consist of

four regions:

– Viscous sublayer

– Buffer layer

– Overlap layer

– Turbulent layer

• The intense mixing in turbulent flow enhances heat and

momentum transfer, which increases the friction force on the

surface and the convection heat transfer rate.

Reynolds Number• The transition from laminar to turbulent flow depends on the

surface geometry, surface roughness, flow velocity, surface temperature, and type of fluid.

• The flow regime depends mainly on the ratio of the inertia forcesto viscous forces in the fluid.

• This ratio is called the Reynolds number, which is expressed for external flow as

• At large Reynolds numbers (turbulent flow) the inertia forces are large relative to the viscous forces.

• At small or moderate Reynolds numbers (laminar flow), the viscous forces are large enough to suppress these fluctuations and to keep the fluid “inline.”

• Critical Reynolds number ─ the Reynolds number at which the flow becomes turbulent.

Inertia forcesRe

Viscous forces

c cVL VL

m (6-13)

Heat and Momentum Transfer in

Turbulent Flow

• Turbulent flow is a complex mechanism dominated by

fluctuations, and despite tremendous amounts of research the

theory of turbulent flow remains largely undeveloped.

• Knowledge is based primarily on experiments and the empirical

or semi-empirical correlations developed for various situations.

• Turbulent flow is characterized by random and rapid fluctuations

of swirling regions of fluid, called eddies.

• The velocity can be expressed as the sum

of an average value u and a fluctuating

component u’

'u u u (6-14)

• It is convenient to think of the turbulent shear stress as

consisting of two parts:

– the laminar component, and

– the turbulent component.

• The turbulent shear stress can be expressed as

• The rate of thermal energy transport by turbulent eddies is

• The turbulent wall shear stress and turbulent heat transfer

• mt ─ turbulent (or eddy) viscosity.

• kt ─ turbulent (or eddy) thermal conductivity.

' 'turb pq c v T

' 'turb u vt

' ' ; turb t turb p t

u Tu v q c vT k

y yt m

(6-15)

• The total shear stress and total heat flux can be

expressed as

and

• In the core region of a turbulent boundary layer ─

eddy motion (and eddy diffusivities) are much larger

than their molecular counterparts.

• Close to the wall ─ the eddy motion loses its intensity.

• At the wall ─ the eddy motion diminishes because of

the no-slip condition.

turb t t

u u

y yt m m

(6-16)

(6-17) turb t p t

T Tq k k c

y y

In the core region ─ the velocity and temperature profiles

are very moderate.

In the thin layer adjacent to the wall ─ the velocity and

temperature profiles are very steep.

Large velocity and temperature gradients at the

wall surface.

The wall shear stress

and wall heat flux are much larger

in turbulent flow than they

are in laminar

flow.

Derivation of Differential Convection

Equations• Consider the parallel flow of a fluid over a surface.

• Assumptions:

– steady two-dimensional flow,

– Newtonian fluid,

– constant properties, and

– laminar flow.

• The fluid flows over the surface with a uniform free-stream velocity V, but the velocity within boundary layer is two-dimensional (u=u(x,y), v=v(x,y)).

• Three fundamental laws:

– conservation of mass continuity equation

– conservation of momentum momentum equation

– conservation of energy energy equation

The Continuity Equation

• Conservation of mass principle ─ the mass can

not be created or destroyed during a process.

• In steady flow:

• The mass flow rate is equal to: uA

Rate of mass flow

into the control volume

Rate of mass flow

out of the control volume= (6-18)

uA

u

u+∂u/∂x·dx

x,y

dx

dy

v+∂v/∂y·dy

v

Repeating this for the y direction

1u dy

The fluid leaves the control volume from the left surface at a rate of

1u

u dx dyx

the fluid leaves the control volume from the right surface at a rate of

(6-19)

The continuity equation

and substituting the results into Eq.

6–18, we obtain

1 1

1 1

u dy v dx

u vu dx dy v dy dx

x y

(6-20)

Simplifying and dividing by dx·dy

0u v

x y

(6-21)

The Momentum Equation• The differential forms of the equations of motion in

the velocity boundary layer are obtained by applying

Newton’s second law of motion to a differential

control volume element in the boundary layer.

• Two type of forces:

– body forces,

– surface forces.

• Newton’s second law of motion for the control

volume

or

(Mass)Acceleration

in a specified direction

Net force (body and surface)

acting in that direction=X

, ,x surface x body xm a F Fd (6-23)

(6-22)

• where the mass of the fluid element within the control

volume is

• The flow is steady and two-dimensional and thus

u=u(x, y), the total differential of u is

• Then the acceleration of the fluid element in the x

direction becomes

1m dx dyd (6-24)

u udu dx dy

x y

(6-25)

x

du u dx u dy u ua u v

dt x dt y dt x y

(6-26)

• The forces acting on a surface are due to pressure and

viscous effects.

• Viscous stress can be resolved into

two perpendicular components:

– normal stress,

– shear stress.

• Normal stress should not be confused with pressure.

• Neglecting the normal stresses the net surface force

acting in the x-direction is

,

2

2

1 1 1

1

surface x

P PF dy dx dx dy dx dy

y x y x

u Pdx dy

y x

t t

m

(6-27)

• Substituting Eqs. 6–21, 6–23, and 6–24 into Eq. 6–20 and

dividing by dx·dy·1 gives

Boundary Layer Approximation

Assumptions:

1) Velocity components:

u>>v

2) Velocity gradients:

∂v/∂x≈0 and ∂v/∂y≈0

∂u/∂y >> ∂u/∂x

3) Temperature gradients:

∂T/∂y >> ∂T/∂x

• When gravity effects and other body forces are negligible the

y-momentum equation

2

2

u u u Pu v

x y y x m

(6-28)

The x-momentum

equation

0Py

(6-29)

Conservation of Energy Equation• The energy balance for any system undergoing any

process is expressed as Ein-Eout=Esystem.

• During a steady-flow process DEsystem=0.

• Energy can be transferred by

– heat,

– work, and

– mass.

• The energy balance for a steady-flow control volume can be written explicitly as

• Energy is a scalar quantity, and thus energy interactions in all directions can be combined in one equation.

0in out in out in outby heat by work by mass

E E E E E E (6-30)

Energy Transfer by Mass

• The total energy of a flowing fluid stream per unit

mass is

• Noting that mass flow rate of the fluid entering the

control volume from the left is u(dy·1), the rate of

energy transfer to the control volume by mass in the

x-direction is

2 22

2 2

p

stream

u vC T gzV

e enthalpy kinetic potential

,

1

stream xin out stream streamx xby mass x

p

p

meE E me me dx

x

u dy c T T udx c u T dxdy

x x x

(6-31)

• Repeating this for the y-direction and adding the results, the net rate of energy transfer to the control volume by mass is determined to be

• Note that ∂u/∂x+∂v/∂y=0 from the continuity equation.

in outby mass

p p

p

E E

T u T vc u T dxdy c v T dxdy

x x y y

T Tc u v dxdy

x y

(6-32)

Energy Transfer by Heat Conduction

• The net rate of heat conduction to the volume element

in the x-direction is

• Repeating this for the y-direction and adding the

results, the net rate of energy transfer to the control

volume by heat conduction becomes

,

2

2

1

xin out x x

by heat x

Q TE E Q Q dx k dy dx

x x x

Tk dxdy

x

(6-33)

2 2 2 2

2 2 2 2 in out

by heat

T T T TE E k dxdy k dxdy k dxdy

x y x y

(6-34)

Energy Transfer by Work

• The work done by a body force is determined by multiplying this force by the velocity in the direction of the force and the volume of the fluid element.

• This work needs to be considered only in the presence of significant gravitational, electric, or magnetic effects.

• The work done by pressure (the flow work) is already accounted for in the analysis above by using enthalpyfor the microscopic energy of the fluid instead of internal energy.

• The shear stresses that result from viscous effects are usually very small, and can be neglected in many cases.

The Energy Equation• The energy equation is obtained by substituting Eqs.

6–32 and 6–34 into 6–30 to be

• When the viscous shear stresses are not negligible,

• where the viscous dissipation function is obtained

after a lengthy analysis to be

• Viscous dissipation may play a dominant role in high-

speed flows.

2 2

2 2p

T T T Tc u v k

x y x y

(6-35)

2 2

2 2p

T T T Tc u v k

x y x y m

(6-36)

2 22

2u v u v

x y y x

(6-37)

Solution of Convection Equations for a

Flat Plate (Blasius Equation)• Consider laminar flow of a fluid over

a flat plate.

• Steady, incompressible, laminar flow

of a fluid with constant properties

• Continuity equation

• Momentum equation

• Energy equation

2

2

u u uu v

x y y

(6-40)

u v

x y

(6-39)

2

2

T T Tu v

x y y

(6-41)

Boundary conditions

• At x=0

• At y=0

• As y∞

• When fluid properties are assumed to be constant, the first two equations can be solved separately for the velocity components u and v.

• knowing u and v, the temperature becomes the only unknown in the last equation, and it can be solved for temperature distribution.

0, , 0,u y V T y T

,0 0, ,0 0, ,0 su x v x T x T

, , ,u x V T x T

(6-42)

• The continuity and momentum equations are solved

by transforming the two partial differential equations

into a single ordinary differential equation by

introducing a new independent variable (similarity

variable).

• The argument ─ the nondimensional velocity profile

u/V should remain unchanged when plotted against

the nondimensional distance y/d.

• d is proportional to (x/V)1/2, therefore defining

dimensionless similarity variable as

might enable a similarity solution.

Vyx

(6-43)

• Introducing a stream function y(x, y) as

• The continuity equation (Eq. 6–39) is automatically

satisfied and thus eliminated.

• Defining a function f() as the dependent variable as

• The velocity components become

; u vy x

y y

(6-44)

/

fV x V

y

(6-45)

1

2 2

x df V dfu V V

y y V d x d

x df V V dfv V f f

x V d Vx x d

y y

y

(6-46)

(6-47)

• By differentiating these u and v relations, the

derivatives of the velocity components can be shown

to be

• Substituting these relations into the momentum

equation and simplifying

• which is a third-order nonlinear differential equation.

Therefore, the system of two partial differential

equations is transformed into a single ordinary

differential equation by the use of a similarity

variable.

2 2 2 2 3

2 2 2 3 ; ;

2

u V d f u V d f u V d fV

x x d y x d y x d

(6-48)

3 2

3 22 0

d f d ff

d d (6-49)

• The boundary conditions in terms of the similarity

variables

• The transformed equation with its

associated boundary conditions

cannot be solved analytically, and

thus an alternative solution method

is necessary.

• The results shown in Table 6-3 was

obtained using different numerical approach.

• The value of corresponding to u/V=0.99 is =4.91.

0

0 0, 0, 1df df

fd d

(6-50)

• Substituting =4.91 and y=d into the definition of the

similarity variable (Eq. 6–43) gives 4.91=d(V/x)1/2.

• The velocity boundary layer thickness becomes

• The shear stress on the wall can be determined from its

definition and the ∂u/∂y relation in Eq. 6–48:

4.91 4.91

Rex

x

V xd

(6-51)

2

2

0 0

w

y

u V d fV

y x d

t m m

(6-52)

20.3320.332

Rew

x

V VV

x

m t (6-53)

Substituting the value of the second derivative of f at h=0

from Table 6–3 gives

1/ 2

, 20.664Re

/ 2

wf x xC

V

t

(6-54)

Then the average local skin friction coefficient becomes

The Energy Equation• Introducing a dimensionless temperature q as

• Noting that both Ts and T are constant, substitution

into the energy equation Eq. 6–41 gives

• Using the chain rule and substituting the u and v

expressions from Eqs. 6–46 and 6–47 into the energy

equation gives

,

,s

s

T x y Tx y

T Tq

(6-55)

2

2u v

x y y

q q q

(6-56)

22

2

1

2

df d d Vy df d d dV f

d d dx x d d dy d y

q q q

(6-57)

• Simplifying and noting that Pr=/ gives

Boundary conditions:

• Obtaining an equation for q as a function of alone confirms that the temperature profiles are similar, and thus a similarity solution exists.

• for Pr=1, this equation reduces to Eq. 6–49 when q is replaced by df/d.

• Equation 6–58 is solved for numerous values of Prandtl numbers.

• For Pr>0.6, the nondimensional temperature gradient at the surface is found to be proportional to Pr1/3, and is expressed as

2

22 Pr 0

d df

d d

q q

(6-58)

0 0, 1 q q

1/3

0

0.332 Prd

d

q

(6-59)

• The temperature gradient at the surface is

• Then the local convection coefficient and Nusselt number become

and

• Solving Eq. 6–58 numerically for the temperature profile for different Prandtl numbers, and using the definition of the thermal boundary layer, it is determined that

00 0 0

1/3 0.332Pr

s s

y y y

s

dT d dT T T T

dy y d dy

VT T

x

q q

(6-60)

0 1/30.332Pr

ysx

s s

k T yq Vh k

T T T T x

(6-61)

1/3 1/ 20.332Pr Re Pr>0.6xx

h xNu

k (6-62)

1/3Prtd d

Nondimensional Convection Equation

and Similarity

Continuity equation

x-momentum equation

Energy equation

• Nondimensionalized variables

0u v

x y

(6-21)

2

2

u u u Pu v

x y y x m

(6-28)

2 2

2 2p

T T T Tc u v k

x y x y

(6-35)

* * * * * *

2 ; y ; u ; v ; P ; s

s

T Tx y u v Px T

L L V V V T T

• Introducing these variables into Eqs. 6–21, 6–28, and

6–35 and simplifying give

Continuity equation

x-momentum equation

Energy equation

with the boundary conditions

* *

* *0

u v

x y

(6-64)

* * 2 * ** *

* * *2 *

1

ReL

u u u Pu v

x y y x

(6-65)

* * 2 ** *

* * *2

1

Re PrL

T T Tu v

x y y

(6-66)

* * * * * * * *0, 1 ; ,0 0 ; , 1 ; ,0 0u y u x u x v x

(6-67)

* * * * * *0, 1 ; ,0 0 ; , 1T y T x T x

• For a given type of geometry, the solutions of

problems with the same Re and Nu numbers are

similar, and thus Re and Nu numbers serve as

similarity parameters.

• A major advantage of nondimensionalizing is the

significant reduction in the number of parameters.

• The original problem involves 6 parameters (L, V, T,

Ts, , ), but the nondimensionalized problem

involves just 2 parameters (ReL and Pr).

L, V, T, Ts,, Nondimensionalizing

6 parameters

ReL, Pr

2 parameters

Functional Forms of the Friction and

Convection Coefficient

• From Eqs. 6-64 and 6-65 it can be inferred that

• Then the shear stress at the surface becomes

• Substituting into its definition gives the local

friction coefficient,

* * *

1 , ,ReLu f x y (6-68)

*

**

2*

0 0

,Res L

y y

u V u Vf x

y L y L

m mt m

(6-69)

* * *

, 2 2 32 2

2,Re ,Re ,Re

2 2 Re

sf x L L L

L

V LC f x f x f x

V V

t m

(6-70)

• Similarly the solution of Eq. 6-66

• Using the definition of T*, the convection heat

transfer coefficient becomes

• Substituting this into the Nusselt Number relation

gives

* * *

1 , ,Re ,PrLT g x y (6-71)

* *

* *0

* *

0 0

y s

s s y y

k T y k T T T k Th

T T L T T y L y

(6-72)

*

**

2*

0

,Re ,Prx L

y

hL TNu g x

k y

(6-73)

• It follows that the average Nu and Cf depends on

• These relations are extremely valuable:

– The friction coefficient can be expressed as a function of

Reynolds number alone, and

– The Nusselt number as a function of Reynolds and Prandtl

numbers alone.

• The experiment data for heat transfer is often

represented by a simple power-law relation of the

form:

3 4Re ,Pr ; ReL f LNu g C f (6-74)

Re PrL

m nNu C (6-75)

Analogies Between Momentum and

Heat Transfer

• Reynolds Analogy (Chilton─Colburn Analogy) ─ under some conditions knowledge of the friction coefficient, Cf, can be used to obtain Nu and vice versa.

• Eqs. 6–65 and 6–66 (the nondimensionalized momentum and energy equations) for Pr=1 and ∂P*/∂x*=0:

x-momentum equation

Energy equation

• which are exactly of the same form for the dimensionless velocity u* and temperature T*.

* * 2 ** *

* * *2

1

ReL

u u uu v

x y y

(6-76)

* * 2 ** *

* * *2

1

ReL

T T Tu v

x y y

(6-77)

• The boundary conditions for u* and T* are also identical.

• Therefore, the functions u* and T* must be identical.

• The Reynolds analogy can be extended to a wide range of

Pr by adding a Prandtl number correction.

* *

* *

* *

0 0y y

u T

y y

(6-78)

,

Re (Pr=1)

2

Lf xC Nu (6-79)Reynolds analogy

1/ 2

, 0.664Ref x xC 1/3 1/ 20.332Pr Re Pr>0.6xNu (6-82)

1 3

,

RePr 0.6 Pr 60

2f x xC Nu (6-83)

Chapter 7: External Forced

Convection




ObjectivesWhen you finish studying this chapter, you should be

able to:

• Distinguish between internal and external flow,

• Develop an intuitive understanding of friction drag and pressure drag, and evaluate the average drag and convection coefficients in external flow,

• Evaluate the drag and heat transfer associated with flow over a flat plate for both laminar and turbulent flow,

• Calculate the drag force exerted on cylinders during cross flow, and the average heat transfer coefficient, and

• Determine the pressure drop and the average heat transfer coefficient associated with flow across a tube bank for both in-line and staggered configurations.

Drag and Heat Transfer in External

flow• Fluid flow over solid bodies is responsible for numerous

physical phenomena such as

– drag force• automobiles

• power lines

– lift force• airplane wings

– cooling of metal or plastic sheets.

• Free-stream velocity ─ the velocity of the fluid relative to an immersed solid body sufficiently far from the body.

• The fluid velocity ranges from zero at the surface (the no-slip condition) to the free-stream value away from the surface.

Friction and Pressure Drag• The force a flowing fluid exerts on a body in the flow

direction is called drag.

• Drag is compose of:

– pressure drag,

– friction drag (skin friction drag).

• The drag force FD depends on the

– density of the fluid,

– the upstream velocity V, and

– the size, shape, and orientation of the body.

• The dimensionless drag coefficient CD is defined as

21 2

DD

FC

V A (7-1)

• At low Reynolds numbers, most drag is due to friction

drag.

• The friction drag is also proportional to the surface area.

• The pressure drag is proportional to the frontal area and to

the difference between the pressures acting on the front

and back of the immersed body.

• The pressure drag is usually dominant for blunt bodies

and negligible for streamlined bodies.

• When a fluid separates from a body,

it forms a separated region between

the body and the fluid stream.

• The larger the separated region, the

larger the pressure drag.

Heat Transfer

• The phenomena that affect drag force also affect heat transfer.

• The local drag and convection coefficients vary along the surface as a result of the changes in the velocity boundary layers in the flow direction.

• The average friction and convection coefficients for the entire surface can be determined by

,

0

1L

D D xC C dxL

(7-7)

0

1L

xh h dxL

(7-8)

Parallel Flow Over Flat Plates• Consider the parallel flow of a fluid over a flat plate of

length L in the flow direction.

• The Reynolds number at a distance

x from the leading edge of a flat

plate is expressed as

• In engineering analysis, a generally accepted value for

the critical Reynolds number is

• The actual value of the engineering critical Reynolds

number may vary somewhat from 105 to 3X106.

Rex

Vx Vxm

(7-10)

5Re 5 10crcr

Vxm

(7-11)

Local Friction Coefficient

• The boundary layer thickness and the local friction

coefficient at location x over a flat plate

– Laminar:

– Turbulent:

, 1/ 2

5

, 1/ 2

4.91

ReRe 5 10

0.664

Re

v x

x

x

f x

x

x

C

d

(7-12a,b)

, 1/5

5 7

, 1/5

0.38

Re5 10 Re 10

0.059

Re

v x

x

x

f x

x

x

C

d

(7-13a,b)

Average Friction Coefficient• The average friction coefficient

– Laminar:

– Turbulent:

• When laminar and turbulent flows are significant

5

1/ 2

1.33 Re 5 10

Ref L

L

C (7-14)

5 7

1/5

0.074 5 10 Re 10

Ref L

L

C (7-15)

, laminar , turbulent

0

1 cr

cr

x L

f f x f x

x

C C dx C dxL

(7-16)

5 7

1/5

0.074 1742- 5 10 Re 10

Re Ref L

L L

C (7-17)

5Re 5 10cr

Heat Transfer Coefficient

• The local Nusselt number at location x over a flat plate

– Laminar:

– Turbulent:

• hx is infinite at the leading edge

(x=0) and decreases by a factor

of x0.5 in the flow direction.

1/ 2 1/30.332Re Pr Pr 0.6xNu (7-19)

(7-20)0.8 1/30.0296Re Prx xNu 5 7

0.6 Pr 60

5 10 Re 10x

Average Nusset Number• The average Nusselt number

– Laminar:

– Turbulent:

• When laminar and turbulent flows are significant

, laminar , turbulent

0

1 cr

cr

x L

x x

x

h h dx h dxL

(7-23)

0.8 1 30.037 Re 871 PrLNu (7-24)

5Re 5 10cr

0.5 1/3 50.664Re Pr Re 5 10LNu (7-21)

(7-22)0.8 1/30.037 Re PrLNu 5 7

0.6 Pr 60

5 10 Re 10x

Uniform Heat Flux

• When a flat plate is subjected to uniform heat flux

instead of uniform temperature, the local Nusselt

number is given by

– Laminar:

– Turbulent:

• These relations give values that are 36 percent higher

for laminar flow and 4 percent higher for turbulent

flow relative to the isothermal plate case.

0.5 1/30.453Re Prx LNu (7-31)

(7-32)0.8 1/30.0308Re Prx xNu 5 7

0.6 Pr 60

5 10 Re 10x

Flow Across Cylinders and Spheres• Flow across cylinders and spheres is frequently

encountered in many heat transfer systems

– shell-and-tube heat exchanger,

– Pin fin heat sinks for electronic cooling.

• The characteristic length for a circular cylinder or sphere is taken to be the external diameter D.

• The critical Reynolds number for flow across a circular cylinder or sphere is about

Recr=2X105.

• Cross-flow over a

cylinder exhibits complex

flow patterns depending on the Reynolds number.

• At very low upstream velocities (Re≤1), the fluid completely wraps around the cylinder.

• At higher velocities the boundary layer detaches from the surface, forming a separation region behind the cylinder.

• Flow in the wake region is characterized by periodic vortex formation and low pressures.

• The nature of the flow across a cylinder or sphere strongly affects the total drag coefficient CD.

• At low Reynolds numbers (Re<10) ─ friction drag dominate.

• At high Reynolds numbers (Re>5000) ─ pressure drag dominate.

• At intermediate Reynolds numbers ─ both pressure and friction drag are significant.

Average CD for circular cylinder and

sphere

• Re≤1 ─ creeping flow

• Re≈10 ─ separation starts

• Re≈90 ─ vortex shedding

starts.

• 103<Re<105

– in the boundary

layer flow

is laminar

– in the separated

region flow is

highly turbulent

• 105<Re<106 ─

turbulent flow

Effect of Surface Roughness• Surface roughness, in general, increases the drag coefficient in

turbulent flow.

• This is especially the case for streamlined bodies.

• For blunt bodies such as a circular cylinder or sphere, however,

an increase in the surface roughness may actually decrease the

drag coefficient.

• This is done by tripping the

boundary layer into

turbulence at a lower Reynolds

number, causing the fluid to close

in behind the body, narrowing the

wake and reducing pressure drag considerably.

Heat Transfer Coefficient• Flows across cylinders and spheres, in general, involve flow

separation, which is difficult to handle analytically.

• The local Nusselt number Nuq around the periphery of a cylinder subjected to cross flow varies considerably.

Small q ─ Nuq decreases with increasing q as a

result of the thickening of the laminar boundary

layer.

80º<q <90º ─ Nuq reaches a minimum – low Reynolds numbers ─ due to separation in laminar flow

– high Reynolds numbers ─ transition to turbulent flow.

q >90º laminar flow ─ Nuq increases with increasing

q due to intense mixing in the separation zone.

90º<q <140º turbulent flow ─ Nuq decreases due to

the thickening of the boundary layer.

q ≈140º turbulent flow ─ Nuq reaches a second minimum due to flow separation point in turbulent flow.

Average Heat Transfer Coefficient• For flow over a cylinder (Churchill and Bernstein):

Re·Pr>0.2

• The fluid properties are evaluated at the film temperature

[Tf=0.5(T∞+Ts)].

• Flow over a sphere (Whitaker):

• The two correlations are accurate within ±30%.

4 55 81 2 1/3

1 42/3

0.62 Re Pr Re0.3 1

282,0001 0.4 Prcyl

hDNu

k

(7-35)

1 4

1 2 2 3 0.42 0.4Re 0.06Re Prsph

s

hDNu

k

m

m

(7-36)

• A more compact correlation

for flow across cylinders

where n=1/3 and the

experimentally

determined constants C and

m are given in Table 7-1.

• Eq. 7–35 is more accurate,

and thus should be preferred

in calculations whenever

possible.

Re Prm n

cyl

hDNu C

k (7-37)

Flow Across Tube Bank• Cross-flow over tube banks is commonly encountered

in practice in heat transfer equipment such heat exchangers.

• In such equipment, one fluid

moves through the tubes while

the other moves over the tubes

in a perpendicular direction.

• Flow through the tubes can be analyzed by considering flow through a single tube, and multiplying the results by the number of tubes.

• For flow over the tubes the tubes affect the flow pattern and turbulence level downstream, and thus heat transfer to or from them are altered.

• Typical arrangement

– in-line

– staggered

• The outer tube diameter D is the characteristic length.

• The arrangement of the tubes are characterized by the

– transverse pitch ST,

– longitudinal pitch SL , and the

– diagonal pitch SD between tube centers.

StaggeredIn-line

• As the fluid enters the tube bank, the flow area

decreases from A1=STL to AT (ST-D)L between the

tubes, and thus flow velocity increases.

• In tube banks, the flow characteristics are dominated

by the maximum velocity Vmax.

• The Reynolds number is defined on the basis of

maximum velocity as

• For in-line arrangement, the maximum velocity

occurs at the minimum flow area between the tubes

max maxReD

V D V D

m (7-39)

maxT

T

SV V

S D

(7-40)

• In staggered arrangement,

– for SD>(ST+D)/2 :

– for SD<(ST+D)/2 :

• The nature of flow around a tube in the first row resembles flow over a single tube.

• The nature of flow around a tube in the second and subsequent rows is very different.

• The level of turbulence, and thus the heat transfer coefficient, increases with row number.

• there is no significant change in turbulence level after the first few rows, and thus the heat transfer coefficient remains constant.

maxT

T

SV V

S D

(7-40)

max2

T

D

SV V

S D

(7-41)

• Zukauskas has proposed correlations whose general form is

• where the values of the constants C, m, and n depend on Reynolds number.

• The average Nusselt number relations in Table 7–2 are for tube banks with 16 or more rows.

• Those relations can also be used for tube banks with NL

provided that they

are modified as

• The correction factor F

values are given in

Table 7–3.

0.25

Re Pr Pr Prm n

D D s

hDNu C

k (7-42)

, LD N DNu F Nu (7-43)

Pressure drop • the pressure drop over tube banks is expressed as:

• f is the friction factor and c is the correction factor.

• The correction factor (c) given in the insert is used to

account for the effects of deviation from square

arrangement (in-line) and from equilateral

arrangement (staggered).

2

max

2L

VP N f

cD (7-48)

chapter 2: heat conduction...

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