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1/61

FULLY DEVELOPED PIPE AND

CHANNEL FLOWS

KUMAR DINKAR ANAND

3rd YEAR, MECHANICAL ENGG.

IIT-KHARAGPUR

GUIDANCE : PROF. S CHAKRABORTY

INDO-GERMAN WINTER ACADEMY-DECEMBER 2006


2/61

: THE OUTLINE :

Hydraulically developing flow through pipes and channels and

evaluation of hydraulic entrance length. Hydraulically fully developed flows through pipes and channels .

Hydraulically fully developed flow through non-circular ducts.

Definition of Thermally fully developed flow and analysis of thermally

fully developed flow through pipe and channels.

Analysis of the problem of Thermal Entrance: The Graetz Problem.


3/61

Fully Developed Flows

There are two types of fully developed flows :

1.) Hydraulically Fully Developed Flow

2.) Thermally Fully Developed Flow

Contd


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Hydraulically Fully Developed Flow

Definition: As fluid enters any pipe or channel , boundary layers keep on growing

till they meet after some distance downstream from the entrance region. After this

distance velocity profile doesn't change, flow is said to be Fully Developed.

Analysis of fluid flow before it is fully developed:

Velocity in the core of the flow outside the boundary layer increases with

increasing distance from entrance. This is due to the fact that through any cross

section same amount of fluid flows, and boundary layer is growing.

This means

hence

0>dxdU

0


5/61

Schematic picture of internal flow through a pipe :

Velocity Profile ,Using the boundary conditions :

1.) At

2.) At

3.) At

We get the velocity profile as :

Contd

2

)( cybyayu ++=

=y Uu =

0=u

=y 0=dydu2

)()(2

)(

yyU

yu

=

0=y


6/61

Where Free stream velocity of entering fluid

Free stream core velocity inside the tube

Core velocity of fully developed flow

Radius of pipe

Now from the principle of conservation of mass :

Hence ,

Contd

=U=U

+=

R

R

R

UrdrurdrRU

0

2

22*

rRy ==R=eU

2)/(6/1)/(3/21

1

RRU

U

+=

2

2

)/(6/1)/(3/21

)/()/(2

RR

yy

U

u

+

=


7/61

+=

0 0

2 })/1()/1(/{ dyUudx

dUUdyUuUuUdxd

w

Boundary Layer momentum integral equation:

Where, Shear stress at wall,

From Bernoulli's Equation for free stream flow through core:

Using Navier-Stokes equation at the wall

Contd

0==

y

wy

u

x

p

dx

dUU

=

1

0

2

2

=

=

yy

u

x

p


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Solving for boundary layer thickness :

Integrate momentum Integral Equation

Using the boundary condition at

For determination of Entrance Length :

putting at

We get the expression for Entrance Length as:

Contd

)(

0= 0=x

)( eL

R=eLx =

)( eL

De

D

LRe03.0=


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Analytical expression for Entrance Length :

Hence it can be observed that our expression for Entrance Length differsfrom the analytical expression due to the following reasons:

1.) We have assumed parabolic velocity profile in the boundary layer

2.)We have not used the Navier-Stokes boundary equation at wall for velocity

profile determination

3.) We are doing boundary layer analysis which gives approximate results

Contd

)( eL

De

DL Re06.0=

2)/()/(2 yyUu =

0

2

2

==

yy

u

x

p

2)/(6/1)/(3/211

RRUU

+=


10/61

Schematic picture of internal flow through a channel:

Velocity Profile

Using the boundary conditions :

1.) At

2.) At

3.) At

We get the velocity profile as :

Contd

2)( cybyayu ++=

0=y 0=u

=y Uu ==y 0=dydu

2)()(2)( yyU

yu =


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Here , Distance between the parallel plates of channel

Width of the Channel

Free stream velocity of entering fluid

Free stream velocity inside channel

Core velocity of fully developed flow

Entrance Length

Hydraulic Diameter

Contd

=D

=W

=U

=U

=eU

=eL

=HD DW

WD

P

AH 22

44 ===


12/61

From the principle of conservation of mass:

Hence when flow is fully developed

Contd

+=

)2/(

00

22*

D

UdyudyDU

)/(3/21 1 DUU =

)/(3/21)/()/(2 2

D

yy

U

u

=

)2/( D=

= UUe 5.1


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From Boundary layer momentum integral equation :

Where, Shear stress at wall,

From Bernoulli's Equation for free stream flow through core:

Using Navier-Stokes equation at the wall

Contd

+=

0 0

2 })/1()/1(/{ dyUudx

dUUdyUuUuUdxd

w

0==

y

wy

u

x

p

dx

dUU

=

1

0

2

2

=

=

yy

u

x

p


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Solving for boundary layer thickness :

Integrate momentum Integral Equation

Using the boundary condition at

For determination of Entrance Length :

putting at

We get the expression for Entrance Length as:

OR

Contd

)(

0= 0=x

)(e

L

eLx = R=

)( eL

D

e

D

L

Re025.0=HD

H

e

D

L

Re00625.0=


15/61

Analytical expression for Entrance Length :

Hence it can be observed that our expression for Entrance Length differs

from the analytical expression due to the following reasons:

1.) We have assumed parabolic velocity profile in the boundary layer

2.) We have not used the Navier-Stokes boundary equation at wall for

velocity profile determination

3.) We are doing boundary layer analysis which gives approximate results.

Contd

)( eL

De

D

LRe05.0=

2

)/()/(2 yyU

u

=

0

2

2

=

=

yy

u

x

p

)/(3/21

1

DU

U

=


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Analysis of fully developed fluid flow:

Fully Developed Flow Through a Pipe:

From Equation of continuity in cylindrical coordinates:

for an incompressible fluid flowing through a pipe

Contd

0)(1 =+

xuru

rrr


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Here, radial velocity

axial velocity

radius of pipe

No fluid property varies with ,

,at wall of the pipe

hence it is zero everywhere.

Hence Equation of continuity reduces to :

Momentum Equation in radial coordinate:

Contd

=ru=u

=a

0=r

u

,0=xu )(ruu =

,0=

r

p )(xpp =


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Momentum Equation in axial direction :

)( dr

du

rdr

d

rdx

dp

=

Solving above differential equation in (r) using the boundary conditions:

1.) Axial velocity (u) is zero at wall of pipe (r =R)

2.) Velocity is finite at the pipe centerline (r=0).

We get the fully developed velocity profile:

Contd

=

22

14 a

rxpau


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Shear Stress Distribution :

Shear stress ,

Maximum shear stress at wall ,

=

=

x

pr

dr

durx

2

=

x

pa

20

Contd

Hence it can be observed that

Shear stress decreases from

maximum to zero at pipe

centerline and then increases

to maximum again at wall.


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Volume Flow Rate :

volume flow rate ,

== xpaurdrQ

a

0

4

82

Now in a fully developed flow pressure gradient is constant ,

Hence ,( )

L

p

L

pp

x

p entexit

=

=

LpaQ 84

=

Contd


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Average Velocity :

Average velocity ,

=== x

pa

a

Q

A

QV 8

2

2

Maximum Velocity :

At the point of maximum velocity , 0=dr

du

This corresponds to core of pipe , 0=r

Hence VxpaUuu r 2

4

2

0max =

=== =

Contd


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Fully Developed Flow through Channel :

From equation of continuity within the entrance length : 0=

+

y

v

x

u

In entrance length boundary layers growing , 0xu 0v

It means flow is not parallel to walls in entrance region

Contd

)(

a


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Equation of Continuity for an incompressible fluid in fully developed region :

0=

x

u

)(yuu =

Momentum equation in y-direction (transverse direction) :

0=yp )(xpp =

Momentum equation in x-direction (along length of channel) :

=

2

2

y

u

x

p

Solving above differential equation in y using boundary conditions :

u(y)=0 at y=0 and y=a

Contd


24/61

We get the velocity profile :

= a

y

a

y

x

pau

22

2

Shear Stress Distribution :

=

=

21

ay

xpa

yu

yx Shear Stress ,

Maximum Shear Stress at walls ,

=x

pa

20

Contd


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Hence it can be observed that

Shear stress decreases from

maximum to zero at centre

of the channel and increases

to maximum again at wall.

Volume Flow Rate :

Volume flow rate per unit width of channel,3

012

1a

x

pudyQ

a

==

Contd


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Average Velocity :

2

121 a

xp

aQV

==

Average Velocity ,

Maximum Velocity:

At the point of maximum velocity , 0=

y

u

This corresponds to centre of channel ,

2

ay =

Hence , Va

x

puu

2

3

8

1 2max =

==

Contd


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Fully Developed Flow Through Non-Circular Ducts :

)( Elliptical Cross Section :

As flow is fully developed in the elliptical section pipe : 0== zy uuFrom equation of continuity for incompressible flow :

,0=

+

+

zu

yu

xu zyx 0=

xux

),( zyuu xx = Contd

12

2

2

2

=

+

bz

ay

M t E ti i di ti


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Momentum Equation in x-direction :

+

=

2

2

2

2

z

u

y

u

x

p xx

Boundary condition : on0=xu 12

2

2

2

=

+

b

z

a

y

Solution Procedure :

Use ,

Such that non zero constants and to be determined using :1c 2c

2

2

2

1),(),( zcyczyuzyu xx ++

=

1.)

2.) is constant on the wall .

0),(2

=

zyux

),( zyux

Contd

Using the assumed velocity profile and solving the momentum


29/61

Using the assumed velocity profile and solving the momentum

equation using two stated conditions:

2

1),( aczyux =

, along the wall

Using Laplace maximum criteria ( Maximum and minimum of a function

satisfying Laplace equation lies on the boundary) :

over entire domain.),(2

1 constaczyux ==

We get our velocity profile as :

+

= 2

2

2

2

22

22

121),(

bz

ay

baba

xpzyux

Contd

Volumetric Flow Rate :


30/61

Volumetric Flow Rate :

Volume flow rate ,

dA

b

z

a

y

ba

ba

x

pdAzyuQ

tion tion

x

+

==

sec sec

2

2

2

2

22

22

1

2

1),(

22

33

4 ba

ba

x

pQ

+

=

Contd


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Thermally Fully Developed Flows :

)( Thermally fully developed flow through a pipe :

Contd

When fluid enters the tube with tube walls at a different temperature


32/61

When fluid enters the tube with tube walls at a different temperature

from the fluid temperature , thermal boundary layer starts growing.

After some distance downstream (thermal entry length) thermally fully

developed condition is eventually reached :

Thermally fully developed condition is different from Hydraulic

fully developed condition .

,0=xu

,0

x

T

for hydraulic fully developed flow

at any radial location for thermally fully developed

flow as convection heat transfer is occurring.

Contd

Condition for Thermally Fully Developed Flow :


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Condition for Thermally Fully Developed Flow :

Because of convective heat transfer , continuously changes

with axial coordinate x .

)(rT

Condition for fully developed thermal flow is defined as :

0

)()(

),()(=

xTxT

xrTxT

x ms

s

This means although temperature profile changes with x

But the relative temperature profile does not change with x.

)(rT

Contd

Here )(xT Surface Temperature of the pipe


34/61

Here , =)(xTs

=)(xTm Mean Temperature

Mean Temperature ( ) is defined as:)(xTm

Surface Temperature of the pipe

v

A

cv

mcm

TdAuc

T c&

=

Thermal Energy transported by the fluid as it moves past any

cross section ,

mvA

cvt

TcmTdAucE

c

&& ==

From Newtons Law of Cooling : )( mss TThq =

Since there is continuous heat transfer between fluid and walls : 0dx

dTm

Contd

From the definition of thermally fully developed flow :


35/61

0)()(

),()(

=

xTxT

xrTxT

x mss

From the definition of thermally fully developed flow :

Hence , )()()()()(

),()(0

0

xfxTxT

r

T

xTxT

xrTxT

r ms

rr

rrms

s

=

=

=

Here is radius of the pipe .)( 0r

From Fouriers heat conduction law at the wall and Newton's law of cooling:

[ ])()(00

xTxThr

Tk

y

Tkq ms

rryy

s =

=

=

==

h


36/61

Hence , )(xfk

h

Here , Local convection heat transfer coefficient=h=k Coefficient of thermal conduction (fluid)

Hence, h is infinite in the

Beginning (boundary layers

just building up), then decays

exponentially to a constant

value when flow is fully

developed (thermally )and

thereafter remains constant.

Contd

tfdx ,

fdh

h

x

Competit ion between Thermal and Velocity boundary Layers :


37/61

p y y y

This competition is judged by a dimensionless number , called

Prandtl number .

=

Pr

Where , Kinematic friction coefficient (momentum diffusivity)==

==pc

k

Thermal diffusivity

n

t

Pr

Where , = Velocity boundary layer thickness

=t Thermal boundary layer thickness

=n Positive exponent Contd

If , 1Pr


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,

1Pr

It means Velocity Boundary Layer grows faster than Thermal

boundary layer. Hence flow first hydraulically developed and

then thermally developed.

If ,

It means Thermal Boundary Layer grows faster than Velocity

boundary layer. Hence flow first thermally developed and then

hydraulically developed.

)( Hence if, and flow is said to be thermally developed it means

Flow is already hydraulically developed.

1Pr

)( Similarly if, and flow is said to be hydraulically developed it

Means flow is already thermally developed.

1Pr

Contd

)( Usually surface conditions of pipe fixed by imposing conditions :


39/61

)( y p p y p g

1.) Surface temperature of pipe is made constant ,

2.) Uniform surface heat flux ,.constTs =

.constqs =

Constant Surface Heat Flux :

From the definition of fully developed thermal flow:

0)()(

),()(=

xTxT

xrTxT

x ms

s

dx

dT

TT

TT

dx

dT

TT

TT

dx

dT

x

T m

ms

ss

ms

ss

+

=

Contd

From Newtons Law of cooling : )( TThq =


40/61

From Newton s Law of cooling : )( mss TThq =

.constqs =As, hence,dx

dTdx

dT ms =

.constdx

dT

dx

dT

x

T ms ===

Hence using definition of thermally fully developed flowand Newtons Law:

sT

mT

.constqs =

sq

Contd

Neglecting viscous dissipation, energy equation :


41/61

eg ec g scous d ss pa o , e e gy equa o

=

+

r

T

rrr

Tvx

Tu

Assuming the flow to be both hydraulically and thermally developed :

,0=

x

u,0=v

dxdT

xT m=

,12

2

0

=

r

ruu m

Hence energy equation reduces to :

=

2

0

121

r

r

dx

dTu

r

T

rr

mm

Contd

,Vum =

dxdT

xT m=

Integrating energy equation using boundary conditions :


42/61

1.) Temperature , is finite at centre ,),( xrT 0=r

)(0

xTT srr ==We get Temperature profile :

2.) Temperature ,

+

=

2

0

4

0

20

4

1

16

1

16

32)(),(

r

r

r

r

dx

dTruxTxrT mms

From definition of mean temperature ,

v

A

cv

m

cm

TdAuc

T c

&

=

=

dx

dTruxTxT mmsm

2

0

48

11)()(

Contd

From the principle of energy conservation :Pdxqdq

=


43/61

Pdxqdq sconv =

dx

m& mT mm dTT +

x

)(pv )()( pvdpv +1=v

specific

volume,

)( pvTcdmdq mvconv += &

For an ideal gas, ,mRTpv = Rcc vp +=

Perimeter,

PdxqdTcmdq smpconv == &

( )mspp

sm TTcm

Ph

cm

Pq

dx

dT=

=

&&

Contd

DP =

=4

2Dum m

&

D

Hence combining the equations obtained by integration of energy


44/61

equation in boundary layer and conservation of energy equation :

k

Dq

dx

dTruxTxT smmsm

=

=

48

11

48

11)()(

2

0

( ))()(4811)()( xTxT

k

hDxTxT smsm =

36.41148 ===

k

hDNuD

Hence Nusselt number for fully developed flow through a circular pipeexposed to uniform heat flux on its surface is a constant ,independent of

axial location ,Reynolds number and Prandtl number .

)(

Contd

Constant Surface Temperature :


45/61

p

From the definition of fully developed thermal flow :

0

)()(

),()(=

xTxT

xrTxT

x ms

s

dx

dT

TT

TT

dx

dT

TT

TT

dx

dT

x

T m

ms

ss

ms

ss

+

=

0=dx

dTsConstant surface temperature ,

=

dx

dT

TT

TT

x

T m

ms

s

Contd

Hence it can be seen that , depends on radial coordinate.T


46/61

, p

Fully developed temperature profile for constant wall temperature

hence differs from constant surface heat flux condition.

x

.constTs =

mT

sq

Contd

Neglecting viscous dissipation, energy equation :


47/61

g g p , gy q

=

+

rT

rrrTv

xTu

Assuming the flow to be both hydraulically and thermally developed :

,0=

x

u ,0=v

,12

2

0

=

rruu m

,Vum =

= dx

dTTTTT

xT m

ms

s

Contd

Hence boundary layer energy equation becomes :


48/61

ms

smm

TT

TT

r

r

dx

dTu

r

T

rr

=

2

0

121

Above equation is solved using iterative procedure :

66.3=DNu

Contd

Fully developed thermal flow through a channel :


49/61

)( Channel walls subjected to constant heat flux :

Here we consider a channel with :

=a

=

sq

=sT

=),( xrT

=mT

=W,Depth of channel Width of channel

Heat flux at the walls

Temperature of fluid flowing through channel

Temperature at the wall

Mean Temperature or Bulk Temperature

Contd

== WP 2 perimeter


50/61

Neglecting viscous dissipation , energy equation :

2

2

y

T

y

Tv

x

Tu

=

+

== WP 2

==== aW

aWPADH 2

244 Hydraulic diameter

perimeter

Assuming the flow to be both Hydraulically and thermally developed :

,0=

x

u,0=v .const

dx

dT

x

T m ==

=

=

a

y

a

yu

a

y

a

y

x

pau m

222

62

Contd

Here, =mu Mean velocity , is defined as :


51/61

x

pa

A

udA

uc

A

c

mc

==

12

2

Now solving for boundary layer energy equation :

21

3

2

4

6126 cyca

y

a

y

dx

dTu

Tmm

++

=

Constants of integration obtained using :

1.) at ( as temperature profile is

symmetric hence has extreme

value at centre.)

,0=dy

dT

2

ay =

Contd

2.) at the wall , ayy == &0sTT =


52/61

sTc = 2

Hence we obtain the temperature profile :

+=

126126

3

2

4ay

a

y

a

y

dx

dTuTT mms

From the definition of mean temperature,

v

A

cv

mcm

TdAuc

T c&

=

=

dx

dTauTT mmsm

2

140

17

Contd

From conservation of energy method ( similar to case of pipe):


53/61

( )mspp

sm TTcm

Phcm

Pqdx

dT =

=&&

Hence combining temperature profile and conservation of energy :

( )smp

msm TT

cm

PhauTT =

&

2

140

17

Using , ( ),aWum m=& ,2WP = ,pc

k

= aDH 2& =

( )

17

1402==

k

ahNu

HD

Contd

Thermal Entrance : The Graetz Problem


54/61

ur

x0

0T

0T

0T

wT

0r

Problem Statement:

Fluid initially at a uniform temperature

enters into a pipe at a surface temperature different than

the fluid. Flow assumed to be Hydraulically developed .

Contd...

wT

Prandtl number of fluid is high , hence thermal entrance starts far

d t

Prandtl number of fluid is high , hence thermal entrance starts fard t


55/61

downstream.

Flow already hydraulically developed.

downstream.

Flow already hydraulically developed.

Here , Uniform temperature of fluid before thermal entrance=0T

=wT Uniform surface temperature of walls

=),( rxT Fluid temperature in thermal entrance region

As the flow is hydraulically fully developed : ,122

0

=

r

ruu m

Neglecting viscous dissipation, boundary layer energy equation :

=

r

T

rrx

Tu

Contd

Boundary Conditions :

1 ) at ,0x 0TT =


56/61

1.) at

2.) at

,0x0TT

,0x wTxrT =),( 0

Solution :

Solution done with the help of non dimensional variables.

,0

*

TT

TTT

w

w

= ,

0

*

r

rr =

PrRe0

*

d

xx =

Here , ,Re 0

mud=k

cp

==Pr

Hence energy equation reduces to :

( )

=

*

**

*2***

*

1

2

r

Tr

rrrx

T

Contd

Boundary condition in terms of non-dimensional variables :


57/61

,1)0,(

**

=rT 0),1(**

=xTSolving the energy equation using variable separation method :

)()(),( **** xgrfxrT =Using ,the particular solutionin energy equation

We obtain :

( ).

1

2

2**

*

const

frr

ffr

g

g==

+=

Hence ,*22exp xCg =

012**2* =++ frrffr

Contd

Hence the particular solution will be :

**2***


58/61

)()2exp(),(

**2***

rfxCxrT nnnn =

From the principle of linearity and superposition :

=

= =n

n nnn rfxCxrT 0**2***

)()2exp(),(

1)0( =nf

0)1( =nf

, for simplicity

, using the boundary condition 0),1( ** =xT

Using the other condition ,

=

===n

n nn rfCrT 0 *** )(1)0,(

To be determined using theory of orthogonal

functions.

,nC

Contd

Using theory of orthogonal functions :


59/61

=

1

0

*2**

1

0

***

)1(

)1(

2

2

drfrr

drfrr

C

n

n

n

Now the rest of the problem is numerically solved for Nusselt Number :

( )

( )

=

*22

*2

2exp)1(2

2exp)1(

xfC

xfCNu

nnnn

nnn

x

contd

: KEY QUESTIONS :


60/61

IF FLOW THROUGH A PIPE OR CHANNEL IS SAID TO BE

HYDRAULICALLY FULLY DEVELOPED DOES THIS IMPLY

THERMALLY FULLY DEVELOPED AND VICE-VERSA ????

IF TWO PLATES IN THE CHANNEL ARE MAINTAINED AT

DIFFERENT TEMPERATURES THEN WHAT WILL BE THE

CRITEREA FOR THERMALLY FULLY DEVELOPED FLOW ????

THANK YOU FOR YOUR COOPERATION


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