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FULLY DEVELOPED PIPE AND
CHANNEL FLOWS
KUMAR DINKAR ANAND
3rd YEAR, MECHANICAL ENGG.
IITKHARAGPUR
GUIDANCE : PROF. S CHAKRABORTY
INDOGERMAN WINTER ACADEMYDECEMBER 2006

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: 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 noncircular 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.

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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

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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

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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
+
=

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+=
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 NavierStokes 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 NavierStokes 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
+=

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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 ===

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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 NavierStokes 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=

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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 NavierStokes 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 ydirection (transverse direction) :
0=yp )(xpp =
Momentum equation in xdirection (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

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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 NonCircular 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 xdirection :
+
=
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

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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 :

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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

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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

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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 :

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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

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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 :

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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 :

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)( 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 =

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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 :

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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 :

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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
=

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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

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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 :

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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

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, 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 :

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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 :

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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 :

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)( 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

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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 :

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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 =

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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):

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( )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

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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

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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 =

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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 nondimensional variables :

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,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***

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)()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 :

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=
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 :

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IF FLOW THROUGH A PIPE OR CHANNEL IS SAID TO BE
HYDRAULICALLY FULLY DEVELOPED DOES THIS IMPLY
THERMALLY FULLY DEVELOPED AND VICEVERSA ????
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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THANK YOU FOR YOUR COOPERATION
THE END