convection process

8/21/2019 Convection Process

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3.0 SYLLABUS CONTENT

3.1` Convection Principlesheat transfer coefficient

3.2 Convection boundary layer theory

3.3 Forced convection over exterior surface (laminarflow)

3.4 Forced convection in turbulent flow (Reynolds

analogy.

3.5 Principle of dynamic similarity applied to forcedconvection

3.6 Free convection and laminar profile over vertical

plates


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Convection Principles(Nusselt Number)

Nusselt Number

Developed by Wilhelm Nusselt

(1882-1957) from Germany

In convection analysis, it iscommon practice to non-

dimensionalized the governing

equations and combine the

variables, which group together indimensionless numbersto reduce

the number of variables.


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The Nusselt number is a non-dimensionalized h,

defined as:

k

hLNu c Lc - Characteristic Length

k - Thermal conductivity of fluid


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Since:

Heat transfer by conduction occurs when the fluid

is motionless and

Heat transfer by convection occurs when the fluidinvolves some motion.

In either case, the heat flux is the rate of heat

transfer per unit time per unit surface area.

L

Tkq

Thq

cond

conv


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Taking the ratio of these two equations:

Thus Nu represents the enhancement of heat transfer through a

fluid layer as a result of convection relative to conduction

across the same fluid layer. The larger Nu, the more effective

the convection.

Nu= 1 for a fluid layer, represents pure conduction.

Nu

k

LhTh

q

q

LTkcond

conv


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Dynamic viscosity ()The shear force per unit arearequired to drag on layer of fluid with unit velocity passed

another layer a unit distance away from the fluid.

Kinematic viscosity ()The ratio of dynamic viscosityto density.

Convection Principles(Viscosity)

dydu


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3.2 Convection boundarylayer theory


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Convection Principles(Velocity Boundary Layer)

Velocity boundary development on a flat plate:

The boundary layer thickness () is normally defined

as where:

uu 99.0


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The dashed line, divides the flow

over the plate into two regions:

Boundary layer region

In which the viscous effects and

velocity changes are significant.

Inviscid flow region

In which the friction effects arenegligible and the velocity

remains constant.

uy

x

Heated Surface

Boundary

Layer

Inviscid

Flow


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Flow regions in velocity boundary of a flat plate:


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Comparison of a laminar and turbulent velocity

boundary layer profile:


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Convection Principles(Thermal Boundary Layer)

Likewise there is a thermal

boundary layer

No temperature jump condition

Because velocity of the fluid

is zero at the point of contact

with the solid surface, the

fluid and solid surface must

have the same temperature

at the point of contact.

y

x

T

Heated Surface


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Convection Principles(Thermal Boundary Layer)

Thermal boundary development on a flat plate:

The thickness of the thermal boundary layer (t) at any location

along the surface is defined as the distance from the surface at

which:

T=T-Ts=0.99(T-Ts)

Ts+0.99(T-Ts)


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Convection Principles(Prandtl Number)

Prandtl Number

Developed by Ludwig Prandtl (1875-1953) of

Germany.

The relative thickness of the velocity and

thermal boundary layers is best described by

a dimensionless Prandtl number (below):

k

C

HeatofyDiffusivitMolecular

MomentumofyDiffusivitMolecular

p

Pr


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Prandtl Number ?

The Prandtl numbers of gases are about 1, which

indicates that both momentum and heat dissipate

through the fluid at about the same rate.

Heat diffuses very quickly in liquid metals (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,


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Convection Principles(Reynolds Number)

Reynolds Number

Derived by Osbourne Reynolds (1842-1912)

of Britain

The transition from laminar to turbulent flow

depends on the surface geometry, surface

roughness, free stream velocity, surface

temperature, and type of fluid (among other

things).

However, the flow regime primarily dependsupon the ratio of inertia forces to viscous

forces in a fluid. This is a dimensionless

quantity, known as Reynolds number (Re).


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The Reynolds number is defined as:

LVLV

ForcesViscousForcesInertia Re

Vupstream velocityLcharacteristic length

=

kinematic viscosity of fluid


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A large Re (inertia forces large) Means that the viscous forces cannot contain random and

rapid fluctuations (turbulent).

A small Re (viscous forces large)

Keeps the fluid in-line (laminar).

The Reynolds number where the flow becomes turbulent is

called the critical Reynolds number (Recrit)

LVLV

ForcesViscous

ForcesInertia

Re


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For flow over a flat plate, the generally accepted

value of Recritis:

Flat Plate:

where: xcrit= Distance between the leading edgeof the plate to the transition point

from laminar to turbulent flow takes place.

5

105Re

critcrit

xu


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3.3 Forced convection over an

exterior surface(laminar and turbulent flow)


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External Flow

The convection equations for an external flow can be

derived from the conservation of mass, conservation

of energy, and the conservation of momentum

equations.


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External Flow Equations(Conservation of Mass)

Conservation of Mass

dxx

u

u

dx

dy

dydy

dv

v

u

v

AreaUnit

y

AreaUnit

x

dxvm

dyum

1

1


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External Flow Equations(Conservation of Mass)

Rate of mass

flow into

control volume

Rate of mass

flow out of

control volume

=

dydxyvdxvdydx

xudyudxvdyu

dxdyy

vvdydx

x

uudxvdyu

0

y

v

x

u~ 2-D Continuity Equation


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External Flow Equations(Conservation of Momentum)

Conservation of Momentum

ma = Net Force

dxx

PP

P

dyy

dx

dy


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External Flow Equations(Conservation of Momentum)

In the x-direction:

In the y-direction:

volume

unitper forceBody

x

forcesshearandviscousofeffectNet

forcepressureNet

du

gy

u

x

u

x

P

y

uv

x

uu

2

2

2

2

volumeunitperforceBody

y

forcesshearandviscousofeffectNet

force

pressureNetdv

gy

v

x

v

y

P

y

vv

x

vu

2

2

2

2


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External Flow Equations(Conservation of Energy)

Conservation of Energy

dx

dy

Eheat out, y Emass out, y

Emass in, yEheat in, y

Emass in, x

Eheat in, x

Emass out, x

Eheat out, x

0 outin

EE


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External Flow Equations(Conservation of Energy)

General 2-D energy equation

For 2-D inviscid flow:

222

2

2

2

2

2

x

v

y

u

y

v

x

u

y

T

x

Tk

y

Tv

x

TuCp

2

2

2

2

yT

xTk

yTv

xTuCp


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Convection over a Flat Plate

Consider laminar flow over a flat plat. When viscousdissipation is negligible, the convection equations

reduce for steady, incompressible laminar flow (with

constant properties) over a flat plate.

x

y

T

, u

u(x,0)= 0

v(x,0)= 0

T(x,0)= Ts

dy

dx

Boundary layer


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Consider elemental control volume for force balance

in the laminar boundary layer.

Continuity:

Momentum:

Energy:

0

y

v

x

u

2

2

y

u

y

uv

x

uu

2

2

y

T

y

Tv

x

Tu


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Boundary conditions:

At x= 0: u(0,y)= u

, T(0,y)= T

At y= 0: u(x,0)= 0, v(x,0)= 0, T(x,0)= Ts

At y=

: u(x,)= u

, T(x,)= T

Define a dimensionless similarity variable:

x

uy


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Recall, that the stream function is defined as:

Dependent variable:

xv

yu

;

yu

u

xu

f


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Therefore:

fd

df

x

u

fxu

u

dx

df

u

xu

xxv

d

dfu

x

u

d

df

u

xu

yyu

2

1

2


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So:

3

32

2

2

2

2

2

2

2

fd

x

u

y

u

d

fd

x

uu

y

u

d

fd

x

u

x

u


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Convection over a Flat Plate(Momentum Equation)

Substituting these into the momentum equation and simplifying

gives:

A 3rdorder non-linear differential equation. Therefore the system

of partial differential equations is transformed into a single

ordinary differential equation by use of a similarity variable.

022

2

3

3

d

fdf

d

fdEQN 6-49

text


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Using the definitions for f and , the boundary equations in

terms of the similarity variables can be found.

1

0

00

0

d

df

d

df

f However, the transformed equation

with its similarity variable cannot besolved analytically.

Therefore, an alternative solution is

necessary.


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The non-dimensional velocity profile can be obtained by

plotting u/u

vs. . The results agree experimentally.

A value of: corresponds to:

Recall that the definition of a velocity boundary layer is when:

992.0u

u

d

df

0.5

99.0

u

u


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So substituting these values into the definition for , gives the

boundary layer thickness for a flat plate.

d

d

xuwhere

x

x

u

x

u

x

uy

Re:Re

0.50.5

5

d y;0.5

For laminar

flat plate:

EQN 6-51

text


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Energy Equation

Knowing the velocity profile, we can now solve the energy

equation.

Introduce dimensionless temperature:

Note: both Tsand T

are constant.

Convection over a Flat Plate(Energy Equation)

s

s

TT

TyxTyx

,,


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Substituting into the energy equation gives:

Again using the similarity variable, , so = ()

So the energy equation becomes:

2

2

yyv

xu

x

uy

2

2

2

2

1

dy

d

d

d

dy

d

d

df

d

df

x

u

dx

d

d

d

d

dfu

C i Fl Pl


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Since:

x

u

dy

d

x

u

uy

x

uy

dx

d

x

uy

2

1

2

1 23

C ti Fl t Pl t


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and:

yud

df

yu

u

xu

f

C ti Fl t Pl t


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Substituting these in gives:

2

2

2

2

1

dy

d

d

d

dy

d

d

df

d

df

x

u

dx

d

d

d

d

dfu

x

u

d

d

x

u

d

d

yuyux

uy

x

u

x

u

u

y

d

d

yuu

2

2

2

1

2

C ti Fl t Pl t


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x

u

d

d

x

u

yux

u

ux

u

x

u

x

u

ud

d

2

2

1

2

1

2

2

21

dd

ux

xyxu

xxu

udd

C ti Fl t Pl t


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2

2

2

d

d

yux

u

ux

u

u

x

d

d

2

2

Pr

2

d

d

yud

d

f

0Pr22

2

d

df

d

d

Pr

Prandtl number

EQN 6-58

text

C ti Fl t Pl t


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A closed form solution cannot be obtained for this boundary

layer problem, and it must be solved numerically.

If this equation is solved for numerous values of Pr, then for

Pr > 0.6, the non-dimensional temperature gradient at the

surface is found to be (reference Table 6-3, p. 378 in text):

31

Pr332.0

0

d

d

C ti Fl t Pl t


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The temperature gradient at the surface is:

Since: then:

Therefore substituting these values in gives:

0000

y

s

y

s

y y

TTy

TTy

T

x

u

y

x

uy

x

uTT

y

Ts

y

31

Pr332.0

0

C ti Fl t Pl t


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Therefore the local convection coefficient and Nusselt number

become:

TT

TTk

TT

k

TT

qh

s

x

u

s

s

yyT

s

s

x

3

1

Pr332.00

xukhx

31

Pr332.0

C ti Fl t Pl t


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The local Nusselt number is the dimensionless temperature

gradient at the surface. This is defined as:

Thus for Pr > 0.6, the local Nusselt number for laminar flow is:

Convection over a Flat Plate(Laminar Flow)

k

xhNu xx

21

31

RePr332.0 xNu

C ti Fl t Pl t


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The local friction coefficient (CFx) can also be determined.

Since the wall shear stress is:

From Table 6-3 (pp. 378 in text) this is found to be:

0

2

2

0

d

fd

x

u

uy

u

y

wall

x

wall

u

Re

332.0 2


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Con ection o er a Flat Plate


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Therefore the local skin friction coefficient is:

21

Re664.02

2,

x

wall

xF

u

C

2,

2

1 uC xFwall



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The average heat transfer coefficient over the entire plate can be

obtained by integrating over its length:

dxh

L

h

L

x

0

1

L

k

LuL

k

xu

L

k

dx

x

u

L

kh

L

L

21

31

3

1

31

31

RePr664.0

Pr664.0

2Pr332.0

Pr332.0

0

0



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So the average Nusselt number for laminar flow over

the entire plate is:

31

PrRe664.0 5.0

L

k

Lh

Nu



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Solving numerically for temperature profile for

different Prandtl numbers, and using the definition of

the thermal boundary layer, it is determined that for

laminar flow over a flat plate:

31

31

PrPr

026.1

dd

dt



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Convection over a Flat Plate(Example 3.1)

Example 3.1a Calculate the heat transfer and the

thermal boundary layer thickness of the way along

a flat plate that is 50 m long. Liquid (Tsat = 40 C)

flows over it at 4 m/s. The plate is kept at a surface

temperature (Ts= 80 C).

50 m

x

y

40 C

4 m/s

Ts= 80C



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The first step is to calculate the mean film temperature of the

fluid flowing along the plate.

This is just the average of the surface temperature and the fluid

bulk temperature.

CCCTT

T sfilm

602

4080

2

50 mx

y

80 C

40 C



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For liquid water at 60 C from Table

50 mx

y

80 C

40 C

99.2Pr

654.0

67.4

3.983 3

CmW

sm

kg

m

kg

k



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First calculate the Reynolds number to determine whether theflow is laminar or turbulent.

Since Re < Recrit = 5x105 or 500,000 ~ Flow is laminar

Therefore:

8.527,10

67.4

5043.983Re

41

3

sm

kg

sm

m

kgmxu

m

m

mx

t

m

412.0

026.1

99.2609.0

026.1

Pr

609.08.527,10

5

Re

5

31

31

450

d

d

d



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Example 3.1b Now calculate the convective heat transfer.

First we must check to see whether the entire plate is in a

laminar boundary layer or not.

Since Re < Recrit = 5x105 or 500,000 ~ Flow is laminar over the

entire plate

3.111,42

67.4

5043.983Re

3

sm

kg

sm

m

kgmLu



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Therefore we can use the following equation to find h:

CmW

sm

kg

m

kg

sm

Cm

W

m

x

uk

x

ukh

2

33

1

31

31

619.0

5067.4

3.9834654.099.2332.0

Pr332.0Pr332.0



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Using this h, we can now find the convection heat transfer:

2

2

8.244080619.0

)(

m

W

Cm

W

s

CC

TThq



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Convection over a Flat Plate(Turbulent and Mixed Flows)

Turbulent

Completely Turbulent Flow

Mixed Laminar/Turbulent Flow


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Convection over a Flat Plate(Turbulent and Mixed Flows)

Note: if it had been found that the boundary layer was notcompletely laminar another equation for h could have been

used instead.

For turbulent flow (all over the plate):

For a mixed combination of laminar and turbulent flow over the

plate:

7510Re105

60Pr6.0

3

1

PrRe037.0 8.0 LNu

31Pr871Re037.0 8.0 LNu 75 10Re10560Pr6.0

L



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Example 3.2 Oil flows over a 40-m long heated plate at freestream conditions of 5 m/s and 25C. If the plate is held at 45C.

a) Determine the velocity and thermal boundary layer

thicknesses at the middle of the plate.

b) Calculate the distance where the laminar change to

turbulence flow

c) Calculate the total convection heat transfer for a 1-m

width.


40 m

u

= 5 m/s

T= 25C

Ts= 45C



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First calculate the film temperature (Tf)

From Tables for oil at 35C, the fluid properties are:

CCCTT

T sfilm

352

4525

2

3

2

255,1

2864.0

105.3

711,3Pr

4

m

kg

Cm

W

sm

k



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a) At the middle of the plate:

Since the critical Reynolds number is 5x105, then:

The flow at the mid-point of the plate is laminar.

mm

x 202

40

54

int

1086.2105.3

205Re

2

sm

sm

pomid

mxu

critpomid ReRe

int



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The hydrodynamic (or velocity) boundary layer is:

The thermal boundary layer is:

cmorm

mxx 7.18187.0

1086.2

205

Re

5

520

d

mmormm

t

8.110118.0711,3026.1

187.0

Pr026.1

31

31

dd



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b) At the end of the plate:

Since Re > Recrit the flow is turbulent at the end

The critical distance (transition point from laminar to

turbulent is:

54

10714.5105.3

405Re

2

sm

sm

end

mLu

m

ux

sm

sm

crit

crit

35

5

105.3105

Re

245



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c) Using the mixed Nu equation for a flat plate:

7.600,9 711,38711071.5037.0

Pr871Re037.0

31

31

8.05

8.0

LNu

Cm

WCmW

m

L

kNu

h

27.6840

2864.07.600,9



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The total heat transfer per is:

W

CCmm

TTAhQ

Cm

W

ss

960,54

25451407.68 2


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Example 7-1 7-2, 7-3 pp 404-407

Forced Convection


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Forced Convection(on Cylinders and Spheres)

Flows across cylinders andspheres, in general, involve

flow separation which is

difficult to handle analytically.

Thus these must be studied

empirically or experimentally

Several correlations have

been developed for the heattransfer coefficient (h).

Forced Convection


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Forced Convection(on Cylinders and Spheres)

Churchill and Bernstein developed this empiricalequation for flow over a cylinder (Eqn. 7-35 in text):

Whitaker developed this empirical equation for flow

over a sphere (Eqn. 7-36 in text):

54

85

413

2

31

21

000,282

Re

11

PrRe62.0

3.0Pr4.0

k

hD

Nucyl

4

1

32

21 4.0

PrRe06.0Re4.02

s

sphk

hDNu

Forced Convection


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Forced Convection(over Circular and Non-Circular Cylinders)

Additionally the following empirical correlations have been madeby Zukauskas and Jakob for the average Nusselt number for flow

over circular and non-circular cylinders (Table 7-1 in text):

Forced Convection


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Forced Convection(over Circular and Non-Circular Cylinders)

Forced Convection


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Forced Convection(Example 3.3)

Example 3.3 A long 10-cm diameter hexagonal steam pipewhose external surface temperature is 110C passes through

some open area that is not protected against the wind.

Determine the rate of heat loss when the air is at 1 atm

pressure and 10C and the wind is blowing across a 1-m length

of pipe at a velocity of 8 m/s.

V= 8 m/s

T

= 10C Ts=110C

10 cm

1 m

Forced Convection


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The properties of air at the average film temperatureof:

can be found from Table A-15 as:

CCCTT

T sfilm

60

2

10110

2

sm

Cm

W

k25

10896.1

7202.0Pr;02808.0

Forced Convection


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The Reynolds number is:

The Nusselt number can be determined from Table 7-1

in the text book:

45

10219.410896.1

10.08Re

2

sm

sm mDV

5.122

7202.010219.4153.0

PrRe153.03

1

31

638.04

638.0

Nu

Forced Convection


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Therefore:

The surface area of the hexagon is:

CmWCm

W

m

NuD

kh

24.345.12210.0

02808.0

2

346.0

60sin

110.03

60sin26

m

mm

LD

As

D/260

Forced Convection


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Therefore, the heat transfer is:

W

CCm

TTAhQ

Cm

W

ss

7.191,1

10110346.04.34 2

2


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Example 7-5 pp 414


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3.4 Principle of dynamic similarity

and dimensional analysis(applied to forced convection)

Non-dimensionalized


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convection equations

The continuity , momentum, and energy equations for steady,incompressible, laminar flow of a fluid with constant properties

can be non-dimensionalized by dividing all the dependent and

independent variables, as follows:

Note: the asterisks denote non-dimensional variables.

s

s

TT

TTT

V

PP

V

vv

V

uu

Lyy

Lxx

*

2

*

**

**

;

;

;;

Free stream velocity

Free stream temperature

Surface temperature

Non-dimensionalized


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Introducing these variables the equations become:

Cont inui ty :

Momentum:

Energy:

0*

*

*

*

y

v

x

u

*

*

2*

*2

*

*

*

*

*

*

Re

1

dx

dP

y

u

y

uv

x

uu

L

2*

2

*

*

*

*

*

*

PrRe

1

y

T

y

Tv

x

Tu

L

Non-dimensionalized


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For a plate, the boundary conditions are:

1,1,

00,00,

1,000,1,0

****

****

******

xTxu

xTxu

yTxvyu

x*

y*

Ts

u

, T

Similarity


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87

Similarity

Where:

For a given geometry, the solutions of problems with

the same Re and Nu are similar, thus Re and Nu are

called similarity parameters.

Two physical phenomena are similar if they have the

same dimensionless forms of the governing

differential equations and boundary conditions.

PrRe LV

L

Similarity


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Similarity

A major advantage in non-dimensionalizing is thesignificant reduction in the number of similarity

parameters.

Original equations have 6 parameters: (L, V

, T,Ts, , and n)

The non-dimensionalized equations have only 2

parameters (ReLand Pr).

Similarity


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Similarity

For a given geometry, problems that have the samevalues of similarity parameters (ReLand Pr) have

identical solutions.

Fig 6-28 (text)

Similarity


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Similarity

Example: Determining the convection heat transfercoefficient (h) for flow over a given surface willrequire numerical solutions or experiments withseveral sets of:

Velocities (V) Surface lengths (L)

Wall temperatures (Ts)

Free stream temperatures (T

).

The same information can be determined with farfewer experiments or investigations by grouping thedata into the dimensionless:

Reynolds number (Re)

Prantdl number (Pr)

Similarity


91/162

91

Similarity

Fig 6-29 (text)

Similarity


92/162

92

Similarity

Another advantage is that data from a large group ofexperiments can be conveniently reported in the

terms of the similarity parameters.


93/162

93

3.5 Reynolds Analogy

Forced Convection


94/162

94

(Drag Force)

Forced Convection


95/162

95

(Reynolds Analogy)

In forced convection analysis, we are primarilyinterested in the determination of quantities of:

The coefficient of friction (CF) (to calculate the

shear stress at the wall)

Nusselt number (Nu) ( to calculate the heat

transfer rates).

Therefore, it is desirable to have a relation between

CFand Nu, so that we can calculate one when theother is available.

Forced Convection


96/162

96

(Reynolds Analogy)

Since:

The shear stress at the surface becomes:

Lyxfu Re,, **1*

Lyy

s xfL

V

y

u

L

V

y

uRe,*2

0

*

*

0 *

Forced Convection


97/162

97

(Reynolds Analogy)

L

L

L

y

V

yy

uL

V

Vsxf

xf

xf

y

uC

Re,

Re,

Re

2

Re

2

*

3

*

2

0*

*

*

2

0*

2

, 2

*

*

2

Substituting this into its definition gives the local

friction coefficient:

Forced Convection


98/162

98

(Reynolds Analogy)

Similarly, solving the energy equation for thedimensionless temperature (T*) for a given geometry

gives:

Using this definition, the convection heat transfer

coefficient (h) becomes:

Pr,Re,, **

1

*

L

yxgT

TT

kh

s

yyT

0

Forced Convection


99/162

99

(Reynolds Analogy)

Since:

Then:

localfor

s

s

x

y

L

yy

TT

TTT

**

**

*

*

yT

xTT

xyTTTT

yT sss

Forced Convection


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100

(Reynolds Analogy)

Therefore:

0*

*

*

0*

*

*

y

ys

s

y

T

L

k

y

T

TTx

TTkh

Forced Convection


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101

Substituting this into the local Nusselt numberequation gives:

We previously determined that:

Therefore:

0*

*

*

0*

*

*

y

h

y

xy

T

y

T

x

k

k

x

k

xhNu

(Reynolds Analogy)

Pr,Re,, **

1

*

LyxgT

Pr,Re,*2

0*

*

*

L

y

x xg

y

TNu

Forced Convection


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102

(Reynolds Analogy)

Note: the Nusselt number is equivalent to thedimensionless temperature gradient at the surface, and

this is why it is sometimes called the dimensionless heat

transfer coefficient (h).

Fig 6.30 (text)

Forced Convection


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103

(Reynolds Analogy)

The average friction and heat transfer coefficientsare determined by integrating the local CF,xand Nux

over the surface of the given body with respect to x*

(from 0 to 0.1), which removes the dependence on x*

and thus gives:

These relations allow experimenters to study aproblem with a minimum amount of experiments and

report their results in terms of just Re and Pr.

Pr,ReRe 34 LLF gNuandfC

Forced Convection


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104

(Reynolds Analogy)

The experimental data for heat transfer is oftenrepresented (with reasonable accuracy) by a simple

power law relation of the form:

Where m and n are constant exponents (normally between 0

and 1), and the value of C depends on geometry.

nmLCNu PrRe

Forced Convection


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105

(Reynolds Analogy)

Summary, so far (Fig 6-31 in text book):

Forced Convection(R ld A l )


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106

(Reynolds Analogy)

Now if we simplify the momentum and energyequations by assuming:

Pr = 1 (which is approximately true for gases)

(true when u = u= V= constant)0*

*

x

P

For Pr = 1, the

thermal andvelocity boundary

layers coincide



107/162

107

(Reynolds Analogy)

The equations then become:

Momentum:

Energy:

Note: These two equations are exactly in the same

form for u* and T*.

2*

*2

*

*

*

*

*

*

2*

*2

*

*

*

*

*

*

Re

1

Re

1

y

T

y

Tv

x

Tu

y

u

y

uv

x

uu

L

L



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108

Since the boundary conditions are also identical:

Recall:

Then:

(Reynolds Analogy)

1,1,00,00,

1,01,0

****

****

****

xTxu

xTxu

yTyu

0*

*

*

0*

*

*

yy y

Tyu Equation *



109/162

109

(Reynolds Analogy)

Since as previously derived:

Rearranging these equations gives:

*

*

*

*

Re

2

y

T

L

khand

y

uC

F

x

y

xF

y

NukLh

yTandC

yu

0*

*

*,

0*

*

*

2Re



110/162

110

(Reynolds Analogy)

Therefore substituting these values into Equation*gives:

0*

*

*

0*

*

*

yy y

T

y

u

2

2

Re

,

,

xF

x

xF

x

CSt

or

CNu

Reynolds Analogyfor

Pr = 1

Forced Convection(St t N b )


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111

(Stanton Number)

Reynolds Analogy can also beexpressed in terms of the Stanton

number (St).

This was derived by Sir ThomasEdward Stanton (1865-1931) from

England

PrRe

Nu

VC

hSt

P



112/162

112

(Reynolds Analogy)

Reynolds Analogy is important because it allows us

to determine the heat transfer coefficient (h) for fluids

where Pr = 1, from knowledge of the friction

coefficient (which is easier to measure).

Forced Convection(Chilt C lb A l )


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113

(Chilton-Colburn Analogy)

However, the Reynolds number is of limited usebecause of the restrictions:

Pr = 1

Therefore it is desirable to have an analogy that is

applicable over a wide range of Pr.

This is done by adding a Prandtl number correction.

0*

*

x

P

Forced Convection(Chilton Colburn Analogy)


114/162

114


Recall as previously derived:

Taking their ratio and rearranging give the relation

known as the Chilton-Colburn analogy or the modifiedReynold's analogy:

21

31

21

RePr332.0Re664.0, xxxxF NuandC

HLx

xFjNu

C

1,RePr

2

31

32

Pr2

,

VC

hCj

p

xxF

H

For 0.6 < Pr < 60

Colburn j-factor

Forced Convection(Chilton Colburn Analogy)


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115


The Chilton-Colburn Analogy is derived using: Laminar flow

Over a flat plate ( )

However, experimental studies however show that it is also

approximately applicable to turbulent flow over a surface in

the presence of pressure gradients.

For laminar flow it is not applicable unless it is a flat plate,

therefore it cannot be applied to laminar flow in a pipe.

Also the analogy above can be used for local or average

quantities.

0

x

P

Forced Convection(Example 3 4)


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116

(Example 3.4)

Example 3.4 Laminar flow profileover a vertical plate. A 2 x 3 m plate

is suspended in a room and subject

to air flow parallel to its surfaces

along its 3 m side. The total dragforce acting on the plate is 0.86 N.

Determine the average heat transfer

coefficient (h) for the plate:

The properties of air at 1 atm (Table A-15 in

text book) at Tfilm= 20C:

3 m

2 m

Air FlowT

= 15C

V

= 7 m/s

7309.0Pr

007.1;204.13

kg

kJC

m

kgp

Ts=25C



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117

(Example 3.4)

Set L= 3 m ~ Characteristic length

Since both sides of the plate are exposed to the air (and

considering the thickness negligibly small) the total surface area

is:

212322

2

mmm

LwAs



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118

For all flat plates:

Drag = Frict ion Force

Therefore:

(Example 3.4)

00243.0712204.1

86.022222

3

sm

m

kgs

Fm

N

VA

DC

221

VACDF sFfriction



119/162

119

(Example 3.4)

Then from the modified Reynolds analogy (Chilton-Colburn) the average heat transfer coefficient (h) can

be calculated:

Cm

W

CkgJ

sm

m

kg

pF CVCh

2

32

3

32

7.127309.0

10077204.1

2

00243.0

Pr2


120/162

120

3.6 Convection in an

internal flow

Internal Flow


121/162

121

Internal flow relates to flow through fixed conduitssuch as pipes or ducts.

Internal Flow(Non-Circular Tubes)


122/162

122

(Non-Circular Tubes)

For flow through non-circular tubes Re and Nu,

are based on the hydraulic

diameter Dh.

Where p is the perimeter, Vmis

the mean velocity, and Acis the

cross-sectional area.

hm

ch

DV

pAD

Re

4

Internal Flow(Mean Velocity)


123/162

123

(Mean Velocity)

Because the velocity varies over the cross-section itis necessary to work with a mean velocity (Vm) when

dealing with internal flows.

c

m

mc

A

mV

VAm

Internal Flow(Circular Tubes)


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124

(Circular Tubes)

In a circular tube:

Re < 2,300 laminar flow2,300 < Re < 10,000 transitional flow

Re > 10,000 turbulent flow

DV

DDp

AD

m

D

c

h

Re

444

2

Internal Flow(Entrance Region)


125/162

125

(Entrance Region)

Internal Flow Equations(Example 3 5)


126/162

126

(Example 3.5)

Example 3.5 - Temperature rise of oil in a bearing

(a) Find the temperature and velocity distributions

(b) Find the maximum temperature in the oil

(c) Find the maximum heat flux in the oil

u(y)L= 2 mm

V= 12 m/sUpper plate moving

Oil

k= 0.145 W/(mK)

= 0.8 kg/(ms)

Lower plate stationary



127/162

127

(Example 3.5)

Assumptions: Steady operating conditions

Oil is incompressible with constant properties

Body forces such as gravity are negligible

The plates are large, so no variation in the z-direction

u(y)L= 2 mm


Oilk= 0.145 W/(mK)

= 0.8 kg/(ms)

Lower plate stationary



128/162

128

(Example 3.5)

(a) Find the temperature and velocity distributions Solution:

Flow only in the x-direction v = 0

Continuity Equation:

The x-component of velocity does not change. Since also, the

flow is maintained by the upper plate and not the pressure gradient.

)(

0

0

yuu

x

u

y

v

x

u

0

0

x

P



129/162

129

(Example 3.5)

x-momentum equation:

This is a 2ndorder differential equation. So integrating twicegives:

0 0 000

xgx

P

y

u

x

u

y

vv

x

uu

2

2

2

2

02

2

y

u

21 CyCu

Internal Flow Equations(Example 3.5)


130/162

130

(Example 3.5)

The boundary conditions are:

u(0)= 0

u(L)= V= 12 m/s

Using these boundary conditions to solve for the constants C1and C2gives:

0

00

2

21

C

CC

L

V

C

LCV

1

1 0



131/162

131

(Example 3.5)

Therefore the equation becomes:

Frictional heating due to viscous dissipation in this case is

significant because of the high viscosity of oil and large plate

velocity. The plates are isothermal and there is no change in

flow direction, so the temperature changes with y only T= T(y).

VL

yu



132/162

132

0 0

0 0 0

(Example 3.5)

So the energy equation for this system is:

2

2

2

0

y

u

y

Tk

0

222

2

2

2

2

2x

v

y

u

y

v

x

u

y

T

x

Tk

y

Tv

x

TuC

p



133/162

133

(Example 3.5)

2

2

2

LV

yTk

Since:

Therefore the equation becomes:

L

V

y

u

VLyu



134/162

134

(Example 3.5)

Now integrating the equation twice:

Applying boundary conditions:

T(0) = T0

T(L) = T0

43

2

2CyCV

L

y

kT

40:0 CTy

2

3

03

2

0

2

2:

VkL

C

TLCVL

L

kTLy



135/162

135

( a p e 3 5)

Substituting these constants in to the equation gives:

2

22

0

0

22

2

2

2

22

L

y

L

y

k

VT

TVkL

yV

L

y

kT



136/162

136

( p )

(b) Find the maximum temperature in the oilThe temperature gradient is found by differentiating T(y) with

respect to y.

Now to find the maximum temperature, maximize T by setting

the above equation equal to 0.

021

2

2

L

y

kL

V

y

T

mmL

y

L

y

001.02

002.0

2

21



137/162

137

( p )

This means that the maximum temperature will occur at the mid-plane (y= 1 mm), which is not surprising since both planes are

maintained at the same temperature.

The maximum temperature at y= 1 mm is:

C

WC

k

VT

LLk

VTT

smN

CmW

sm

m

sN

LL

1191

1

145.08

128.020

8

2

2

2

0

2

2

22

2

0max

2



138/162

138

( p )

(c) Find the maximum heat flux in the oilThe heat flux at the plates is determined from the definition of a

heat flux.

2

22

2

0

0

800,28

1

1

002.02

128.0

2

212

2

m

W

W

mL

V

L

y

kL

V

kdy

dT

kq

s

mN

sm

m

sN

y



139/162

139

( p )

As a check, we can also calculate the heat flux at y= L (shouldbe equal but opposite sign).

2

22

2

800,28

1

1

002.02

128.0

2

21

2

2

m

W

W

mL

V

L

L

kL

Vk

dy

dTkq

smN

sm

m

sN

Ly

L

Correct !



140/162

140

( p )

Discussion of example

A temperature rise of 99C confirms that viscous dissipation is

very significant

T=119CL= 2 mm


Lower plate stationaryT=20C

T=20C



141/162

141

( p )

Discussion of example

Heat flux is equivalent to the mechanical energy rate ofdissipation. Therefore, mechanical energy is being convertedinto thermal energy to overcome friction in oil. This accounts forthe temperature flux.

L= 2 mm

V= 12 m/s

2

8.28

m

kWq

28.28m

kWq


142/162

142

3.7 Free (natural) convection

Free Convection


143/162

143

Hot air rises due to the buoyancyeffect.

This causes fluid motion

(possibly in a circulating pattern)

that causes natural or freeconvection

Warm air

Cold

can Cold air

Heat

Transfer

Free Convection(Volume Expansion Coefficient)


144/162

144

In heat transfer, the primary variable is thetemperature, so it is desirable to express the net

buoyancy force in terms of a temperature difference.

This requires knowledge of a property that represents the

variation of the density of a fluid with temperature at constantpressure.

This is called the volume expansion coefficient () which is

defined as:

PP TT

11



145/162

145

In natural convection studies, the condition of the fluidsufficiently far from the hot or cold surface is indicated by the

subscript

to indicate that the presence of the surface is not

felt.

In such cases, can be expressed approximately by replacing

the differential equations by differences, such as:

TTT

11

TT



146/162

146

For an ideal gas:

Thus for an ideal gas the discharge coefficientbecomes:

TR

P

TTT

RTP

RTP

P

111

Free Convection(Grashof Number)


147/162

147

The velocity and temperature for naturalconvection over a vertical plate are

shown in the figure.

As in forced convection, the

boundary layer thickness increases

in the flow direction

Unlike forced convection, the fluid

velocity (u) is 0 at the outer edge of

the boundary layer as well as the

surface of the plate.

This is expected since the fluid

beyond the boundary layer is

motionless.



148/162

148

Recall that the x-momentum equations is:

Now the momentum equation outside the boundary layer can be

obtained from this relation as a special case by setting u = 0,

giving:

gx

P

y

u

y

uv

x

uu

2

2

gx

P



149/162

149

Since:

Then the momentum equation becomes:

g

x

P

x

PxPxPP

TTgy

u

y

vv

x

uu

TTgy

u

y

v

vx

u

u

gy

u

y

vv

x

uu

2

2

2

2

2

2

EQN 9-13

in text



150/162

150

If we now non-dimensionalize this x-momentumequation, we get:

2*

*2

2

*

2

3

*

*

*

*

*

*

Re

1

Re y

uTLTTg

y

u

vx

u

uLL

cs

Grashof Number



151/162

151

The Grashof number is derived byFranz Grashof (1826-1893) from

Germany.

2

3

csL

LTTgGr



152/162

152

Gr is a measure of the relativemagnitudes of the buoyancy force

and the opposing viscous force

acting on the fluid

Free Convection(Raleigh Number)


153/162

153

Lord Raleigh (1842-1919) fromEngland derived the Raleigh Number

PrGrRa

Pr

2

3

csL

LTTgRa

Free Convection(Example 3.6)


154/162

154

Example 3.6 A 6-m long section of 8-cm diameterhorizontal hot water pipe passes through a large

room. The pipe surface temperature is 70 C.

Determine the heat loss from the pipe by natural

convection.

D= 8 cm

L= 6 m

Ts= 70 CT

= 20 C



155/162

155

Assume: Steady operating conditions

Air is an ideal gas

The local atmospheric pressure is 1 atm

From Table A-15, the properties of air are:

7241.0Pr;10749.1;02699.0sec

5 2 m

CmWk

C

CCTTT sfilm

45

2

2070

2



156/162

156

The volumetric expansion coefficient () is:

The characteristic length is the outer diameter of the

pipe:

KCTf 318

1

27345

11

mDLc 08.0



157/162

157

Therefore the Raleigh Number is:

6

25

3

3181

2

3

10869.1

10749.1

7241.008.029334381.9

Pr

2

2

sm

Ks

m

sD

mKK

DTTgRa



158/162

158

Table 9-1 in the text book gives average Nusseltnumbers for natural convection over surfaces.

For a horizontal cylinder:



159/162

159

Thus Nu is:

4.17

7241.0

559.01

10869.1387.060.0

Pr

559.01

387.060.0

2

6

2

278

169

61

278

169

61

DD

RaNu



160/162

160

Then:

The surface area of the cylinder is:

Cm

WCmW

mNu

D

kh

2869.54.1708.0

02699.0

2508.1608.0 mmmLDAs



161/162

161

Therefore the heat transfer is:

W

CCm

TTAhQ

Cm

W

ss

5.442

2070508.1869.5 2

End Of Convection SectionC C


162/162

convection process

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