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7/30/2019 Chapter 8 Binded
1/57
9/12/2
2011 Sajid
Chapter
Dr Muhammad Sajid
Assistant Professor
NUST, SMME.
Reference Text:Fundamentals of Fluid
Mechanics, 6th Ed
By Munson, Young, Okiishi
and Huebsch
Email: [email protected]
Tel: 9085 6065
Fluid Mechanics - II
12-Sep-120
8 Viscous flow in pipes
Pipe flow characteristics
Fully developed laminar
& turbulent flow
Major & Minor losses
Fluid Mechanics - II
Introduction
Now, we cover fluid with internal viscousfriction attributed by the viscosityproperties and friction between the flowsand any adjacent walls.
We will look into how to analyse thelaminar and turbulent pipe flows, and tocalculate friction losses due to pipe wallsas well as pressure losses due to fittingcomponents such as valves, junctions,faucets and flow measurement apparatus.
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Fluid Mechanics - II
Pipes and ducts
Duct: A conduit with non circular crosssection.
Pipe: A conduit of circular cross section.
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Fluid Mechanics - II
Pipe system components
Pipes
Fittings / Connectors
Flow Control devices
Pumps / Turbines
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Fluid Mechanics - II
CHARACTERISTICS OF PIPE FLOW
Chapter 8. Page384
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Fluid Mechanics - II
Flow Characteristics
Pipe flow
Completely filled.
Pressure driven.
Assumption: RoundCross section
Open channel flow
Partially filled
Gravity driven
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Fluid Mechanics - II
Laminar Turbulent Flow
Flow in pipes can be divided into two differentregimes, i.e. laminar and turbulence
Experimental demonstration of flow transitionfrom laminar to turbulent flow regimes.
12-Sep-12 6
Fluid Mechanics - II
Time Dependence of Fluid velocity
x component of velocity as a function of
time at A.
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Fluid Mechanics - II
Laminar Turbulent Flow
Streak lines for small, medium and largeflow rates (Re).
12-Sep-12 8
Fluid Mechanics - II
ExampleConsider water flow in a pipe having a diameter ofD = 20 mm which is
intended to fill a 0.35 liter container. Calculate:
(a) the minimum time required if the flow is laminar,
(b) the maximum time required if the flow is turbulent.
Density = 998 kg/m3 and dynamic viscosity = 1.12103 kg/ms.
Solution:(a) For laminar flow, use Re =VD/= 2100:
Hence, the minimum time t is:
(b) For turbulent flow, use Re = VD/= 4000:
Hence, the minimum time t is:
12-Sep
-12 9
smD
V 118.0020.0998
1012.1210021003
VDV
QVt 2
4
smD
V 224.0020.0998
1012.1400040003
VD
V
Q
Vt
2
4
st 45.9118.002.01035.04
2
3
st 96.4224.002.0
1035.04
2
3
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Fluid Mechanics - II
Entrance region & fully developed flow12-Sep-12 11
Fluid Mechanics - II
Entrance length The entrance region can be represented by entrance length le, which
can be empirically determined by the following formulae for bothregimes:
Laminar:
Turbulent:
Due to different boundary layer thickness in the inviscid core, the
pressure distribution behaves non-linearly in this region and thepressure slope is not constant as shown in Fig. 8.5. However, afterthe flow is fully developed, the slope becomes constant and thepressure drop p is directly caused only by viscous effect.
By projecting the graph back towards the tank, we can estimate thepressure drop due to entrance flow. Hence, by using the Bernoulliequation with losses, the pressure value at all position along thesame pipe can be calculated.
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Re06.0D
e
61(Re)4.4
D
e
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Fluid Mechanics - II
Problem 8.6
Solution
Volume flow rate = 0.1 m3/s
Diameter, D = 20 cm
Viscosity, = 1.79x10-5
Step 1:
V = (4 x 0.1)/(D2) = 0.4/0.1256 = 3.185 m/s
Re = VD/ = 42,700 Step 2:
le = 4.4(42700)1/6 0.2 = 5.2 m
12-Sep-12 13
Fluid Mechanics - II
END OF WEEK # 1
Home Work problems. 8.2, 8.4, 8.6 & 8.8
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Fluid Mechanics - II
Fully developed laminar flow
Fully developed: the velocityprofile is the same at anycross section of the pipe.
Whether the flow is laminaror turbulent,
Flow in a long, straight,constant diameter sections ofa pipe becomes fully
developed. But the other flow properties
are different for these twotypes of flow.
28-Sep-12 17
Fluid Mechanics - II
Fully developed laminar flow
Knowledge of the velocity profile can leaddirectly to other useful information such aspressure drop, head loss, flowrate.
We begin by developing the equation forthe velocity profile in fully developedlaminar flow.
If the flow is not fully developed, a theoreticalanalysis becomes much more complex
If the flow is turbulent, a rigorous theoreticalanalysis is as yet not possible.
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Fluid Mechanics - II
Fully developed laminar flow
There are numerous ways to deriveimportant results pertaining to fully
developed laminar flow.
Three alternatives include:
From F = maapplied directly to a fluidelement,
From the NavierStokes equations of motion,
& From dimensional analysis methods.
28-Sep-12 19
Fluid Mechanics - II
F = ma Applied to a Fluid Element
Consider the motion of a cylindrical fluidelement at time twithin a pipe.
The local acceleration is zero because the flow issteady (V/t = 0), and
The convective acceleration is zero because theflow is fully developed (V.V= uu/xi = 0).
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Fluid Mechanics - II
F = ma Applied to a Fluid Element
Every part of the fluid merely flows along its
streamline parallel to the pipe walls with
constant velocity,
Velocity varies from one pathline to another.
This velocity variation, combined with the fluidviscosity, produces the shear stress.
28-Sep-12 21
Fluid Mechanics - II
F = ma Applied to a Fluid Element
If gravitational effects are neglected, thepressure is constant across any vertical crosssection of the pipe, although it varies alongthe pipe from one section to the next.
If the pressure is P1 at section (1), it is P1-Pat section (2).
A shear stress , acts on the surface of thecylinder of fluid it is a function of the radius ofthe cylinder, =(r).
We isolate the cylinder of fluid and applyNewtons second law, Fx= m ax,
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F = ma Applied to a Fluid Element
The fluid is not accelerating, so that ax= 0.
Thus, fully developed horizontal pipe flow is abalance between pressure and viscous forces
The pressure difference acting on the end of thecylinder of area rand
The shear stress acting on the lateral surface of thecylinder of area 2rl.
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Fluid Mechanics - II
F = ma Applied to a Fluid Element
This force balance can be written as
which can be simplified to give
Since neitherpnorlare functions of theradial coordinate, r, it impliesthat 2/rmustalso be independent ofr.
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F = ma Applied to a Fluid Element
That is, = Cr , where C is a constant.At the centerline of the pipe (r = 0) there is no
shear stress = 0.
At the pipe wall (r = D/2) the shear stress is amaximum, denotedw the wall shear stress.
Hence, C= 2w/Dand the shear stressdistribution throughout the pipe is a linear
function of the radial coordinate
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Fluid Mechanics - II
F = ma Applied to a Fluid Element
If the viscosity were zero there would be noshear stress, and pressure would be constantthroughout the pipe
We get a relation between
pressure drop, and
wall shear stress
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F = ma Applied to a Fluid Element
To carry the analysis further we mustprescribe how the shear stress is related to
the velocity.
For a laminar flow of a Newtonian fluid, the
shear stress is simply proportional to the
velocity gradient. =du/dy
In the notation associated with our pipe
flow, this becomes
28-Sep-12 27
Fluid Mechanics - II
F = ma Applied to a Fluid Element
The two governing laws for fully developed
laminar flow of a Newtonian fluid within a
horizontal pipe
By combining these equations & integrating
where c1 is a constant.
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F = ma Applied to a Fluid Element
Because the fluid is viscous it sticks to thepipe wall so that u = 0, at r= D/2.
Vc is the centerline velocity
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Fluid Mechanics - II
F = ma Applied to a Fluid Element
The volume flowrate through the pipe can
be obtained by integrating the velocity
profile across the pipe.
The average velocity is the flowrate divided
by the cross-sectional area,
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Fluid Mechanics - II
Poiseuille flow
These results show that for laminar pipeflow in a horizontal pipe the flowrate is directly proportional to the pressure drop,
inversely proportional to the viscosity,
inversely proportional to the pipe length, and
proportional to the pipe diameter to the fourthpower.
This flow, first determined experimentally
by Hagen in 1839 and Poiseuille in 1840,is termed HagenPoiseuille flow.
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Fluid Mechanics - II
Inclined pipes
Replace the pressure drop p, by the effectof both pressure and gravity p -l sin.
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From the NavierStokes Equations
General motion of an incompressibleNewtonian fluid is governed by
the continuity equation, and
the momentum equation
For steady, fully developed flow in a pipe,the velocity contains only an axialcomponent, which is a function of only theradial coordinate.
For such conditions, the left-hand side ofmomentum Eq. is zero.
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Fluid Mechanics - II
From the Navier
Stokes Equations
The NavierStokes equations become.
In polar coordinates
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From Dimensional Analysis
We assume that the pressure drop in thehorizontal pipe, is a function of
the average velocity of the fluid in the pipe, V,
the length of the pipe, l
the pipe diameter, D, and
the viscosity of the fluid, .
The density or the specific weight of the
fluid are not important parameters.
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Fluid Mechanics - II
From Dimensional Analysis
There are five variables that can be describedin terms of three reference dimensions M, L, T.
This flow can be described in terms of, k r = 5 3 = 2 dimensionless groups.
These are
The value ofCmust be determined by theory orexperiment. For a round pipe, For ducts of other cross-sectional
shapes, the value ofC is different
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FULLY DEVELOPED TURBULENT
FLOW
Section 8.3
Page 399
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Fluid Mechanics - II
Fully developed turbulent flow
Turbulent pipe flow is more likely to occur
than laminar flow in practical situations,
A considerable amount of knowledge about
the topic has been developed, the field of
turbulent flow still remains one of the least
understood area of fluid mechanics.
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Transition from Laminar to Turbulent Flow
Reynolds number must be less thanapprox. 2100 for laminar flow and greater
than approx. 4000 for turbulent flow.
28-Sep-12 39
Fluid Mechanics - II
Transition from Laminar to Turbulent Flow
Its irregular, random nature is the
distinguishing feature of turbulent flows.
The character of many of the important
properties of the flow (pressure drop, heat
transfer, etc.) depends strongly on the
existence and nature of the turbulentfluctuations or randomness indicated.
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Transition from Laminar to Turbulent Flow
Mixing, heat and mass transferprocesses are enhanced in turbulentflow compared to laminar flow.
The macroscopic scale of therandomness in turbulent flow is veryeffective in transporting energy andmass throughout the flow field,thereby increasing the various rateprocesses involved.
Laminar flow, is very small but finite-sized fluid particles flowing smoothlyin levels, one over another.
The only randomness and mixingtake place on the molecular scale andresult in relatively small heat, mass,and momentum transfer rates.
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Fluid Mechanics - II
Turbulent shear stress
Axial component of velocity, u =u(t), at agiven location in turbulent pipe flow is.
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Turbulent shear stress
The fundamental difference betweenlaminar and turbulent flow lies in the
chaotic, random behavior of the various
fluid parameters.
Such flows can be described in terms of
their mean values (denoted with an
overbar) on which are superimposed the
fluctuations (denoted with a prime).
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Fluid Mechanics - II
Turbulent shear stress
Thus, ifu u(x, y, z, t) is the xcomponent ofinstantaneous velocity, then its time mean
(ortime average) value, , is;
The time interval, T, is considerably longerthan the period of the longest fluctuations
And considerably shorter than any
unsteadiness of the average velocity
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Turbulent shear stress
Can the concept of viscous shear stressfor laminar flow (=du/dy) to that ofturbulent flow by replacing u, theinstantaneous velocity, by , the timeaverage velocity ?
The shear stress in turbulent flow is not merelyproportional to the gradient of the time-average velocity: d/dy.
It also contains a contribution due to therandom fluctuations of the components ofvelocity.
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Fluid Mechanics - II
Turbulent shear stress
The shear stress for turbulent flow in terms ofa new parameter called the eddy viscosity, .
The eddy viscosity changes from oneturbulent flow condition to another and fromone point in a turbulent flow to another.
The turbulent process could be viewed as therandom transport of bundles of fluid particlesover a certain distance, lm, the mixing length,from a region of one velocity to another regionof a different velocity.
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Turbulent shear stress
By the use of some ad hoc assumptions and physicalreasoning, the eddy viscosity is then given by.
The problem is shifted to determining the mixing length, lmwhichis not constant throughout the flow field.
Near a solid surface the turbulence is dependent on the distancefrom the surface.
Thus, additional assumptions are made regarding how themixing length varies throughout the flow.
There is no general model that can predict the shear stress
throughout an incompressible, viscous turbulent flow. It is impossible to integrate the force balance equation toobtain the turbulent velocity profile as was done for laminarflow.
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Fluid Mechanics - II
Turbulent Velocity Profile An often-used correlation is the
empirical power-law velocity profile.nis a function of the Reynoldsnumber, typically from 6 to 10. The power-law profile cannot be valid
near the wall, since according to thisequation the velocity gradient is infinitethere.
In addition, it cannot be precisely validnear the centerline because it does notgive d/dr =0 at r =0.
However, it does provide areasonable approximation to themeasured velocity profiles acrossmost of the pipe.
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Turbulence modeling
It is not yet possible to theoretically predictthe random, irregular details of turbulentflows.
One can time average the governing NavierStokes equations to obtain equations for theaverage velocity and pressure.
The resulting time-averaged differentialequations contain not only the desiredaverage pressure and velocity as variables,
but also averages of products of thefluctuationsterms of the type that one triedto eliminate by averaging the equations!
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Fluid Mechanics - II
Chaos and turbulence
Chaos theory, which is quite complex and iscurrently under development, involves thebehavior of nonlinear dynamical systems andtheir response to initial and boundaryconditions.
The flow of a viscous fluid, which is governed
by the nonlinear NavierStokes equations,may be such a system.
It may be that chaos theory can provide theturbulence properties and structure directlyfrom the governing equations.
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Dimensional Analysis of pipe flow
Turbulent flow can be a very complex, difficulttopic, most turbulent pipe flow analyses arebased on experimental data and semi-empiricalformulas.
These data are expressed conveniently indimensionless form.
It is often necessary to determine the head loss,hL, that occurs in a pipe flow so that the
following equation, can be used in the analysisof pipe flow problems.
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Fluid Mechanics - II
Dimensional analysis of pipe flow
The overall head loss for the pipe system
hL, consists of
the head loss due to viscous effects in the
straight pipes, termed the major loss and
denoted hL major, and
the head loss in the various pipe components,termed the minor loss and denoted hL minor,
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Dimensional analysis of pipe flow
Major losses the pressure drop and head loss in a pipe are
dependent on the wall shear stress, w,
between the fluid and pipe surface.
Difference b/w laminar and turbulent flow is:
the shear stress for turbulent flow is a function of
the density of the fluid,
the shear stress for laminar flow, is independent of
the density, leaving the viscosity, as the onlyimportant fluid property.
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Fluid Mechanics - II
Dimensional analysis of pipe flow
Major losses
the pressure drop, pfor steady,incompressible turbulent flow in ahorizontal round pipe of diameter Dis:
the pressure drop for laminar pipeflow is found to be independent ofthe roughness of the pipe,
but it is necessary to include thisparameter when consideringturbulent flow.
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Dimensional analysis of pipe flow
Major losses A relatively thin viscous sublayer is
formed in the fluid near the pipe wall inturbulent flow
Thus for turbulent flow the pressure dropis expected to be a function of the wallroughness.
relatively small roughness elementshave completely negligible effects onlaminar pipe flow.
For pipes with very large wallroughness such as that in corrugatedpipes, the flowrate may be a function
of the roughness. We will consider only typical constant
diameter pipes with relativeroughnesses in the range
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Fluid Mechanics - II
Dimensional analysis of pipe flow
Major losses
The pressure drop, pcan beexpressed in terms of k r = 4
dimensionless groups.
This result differs from that used for
laminar flow in two ways.
the pressure term is made dimensionless bydividing by the dynamic pressure, rather
than a characteristic viscous shear stress,
we have introduced two additional
dimensionless parameters, the Reynolds
number, and the relative roughness, which
are not present in the laminar formulation.
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Dimensional analysis of pipe flow
Major lossesAssume that the pressure drop should
be proportional to the pipe length. Thisway the l/Dterm can factored out.
We defined friction factor as:
Thus for horizontal pipe flow.
And For laminar fully developed flow, f = 64/Re
For turbulent flow, the functional
dependence of the friction factor on theReynolds number and the relativeroughness, is a rather complex one thatcannot, be obtained from a theory
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Fluid Mechanics - II
Dimensional analysis of pipe flow
Major losses
Join energy equation with expression of
pressure drop. We get:
This is the DarcyWeisbach equation, it is validfor any fully developed, steady, incompressible
pipe flow, horizontal or not.
In general with Vin= Vout, the energy eq gives
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Dimensional analysis of pipe flow
Major Losses It is not easy to determine the functional dependence of the
friction factor on the Reynolds number and relative roughness.
Much of this information is a result of experiments conducted byNikuradse in 1933 and amplified by many others since then.
One difficulty lies in the determination of the roughness of thepipe. Nikuradse used artificially roughened pipes produced by gluing sand
grains of known size onto pipe walls to produce pipes with sandpaper-type surfaces.
The pressure drop needed to produce a desired flowrate was measuredand the data were converted into the friction factor for the correspondingReynolds number and relative roughness.
The tests were repeated numerous times for a wide range of Re and /Dto determine the f=(Re, /D ) dependence.
In commercially available pipes it is possible to obtain a measureof the effective relative roughness of typical pipes and thus toobtain the friction factor.
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Fluid Mechanics - II
Dimensional analysis of pipe flow Major losses
Typical roughness values for various pipe surfaces are shown alongwith the functional dependence offon Re and called the Moodychart in honor of L. F. Moody, who, along with C. F. Colebrook,correlated the original data of Nikuradse in terms of the relativeroughness of commercially available pipe materials.
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Dimensional analysis of pipe flow
Major losses The turbulent portion of the Moody chart is represented
by the Colebrook formula
In fact, the Moody chart is a graphical representation ofthis equation, which is an empirical fit of the pipe flowpressure drop data.
A difficulty with its use is that for given conditions it is notpossible to solve forfwithout some sort of iterativescheme.
It is possible to obtain an equation that adequatelyapproximates the Colebrook / Moody chart relationshipbut does not require an iterative scheme.
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Fluid Mechanics - II
Major Losses - Summary The head loss due to viscous effects in straight
pipes, termed the major loss and denoted hL major,
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The Typical roughness values for various pipe surfaces areshown along with the functional dependence offon Re and calledthe Moody chart.
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Example 8.5
Air flows through a 4mm diameter plastictube with an average velocity of V=50m/s.
Determine the pressure drop in a 0.1m sectionof the tube if the flow is laminar.
Repeat the calculations if the flow is turbulent.
Solution
= 1.23 kg/m3 & = 1.79x10-5 Re=13,700
For laminar flow, f = 64/Re = 0.00467
Pressure drop from , is p = 0.179kPa
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Fluid Mechanics - II
Example 8.5
For plastic tube = 0.0015mm and
/D = 0.0015/4 = 0.000375
With Re = 13700, f = 0.028 from Moody Chart
Pressure drop from , p = 1.076 kPa
Alternately, from
And the pressure drop, p = 1.076 kPa
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Solved Problem
A horizontal cast iron pipe of 8cm diametertransporting water at 20C has a pressuredrop of 500 kPa over 200m. Estimate the flow rate using the Moody diagram
for Re = 1x104, 1x105 & 1x106.
Solution: The relative roughness, /D = 0.26/80 = 0.00325
The friction factor from Moody chart is f= 0.0256
The head loss, hL = p/= 500000/9800 = 51
The average velocity, from is,V =3.92m/s
Flowrate, Q = AV, = x 0.04 x 3.92 = 0.0197m3/s.
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Fluid Mechanics - II
PROBLEMS
8.42, 8.45, 8.50, 8.58, 8.60, 8.62 & 8.70.
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In addition to straight pipes most pipingsystems consist of valves, bends, tees, etc
which add to the overall head loss of the
system.
Such losses are generally termed minor
losses, denoted as hL minor.
How to determine the various minor losses
that commonly occur in pipe systems?
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Minor losses
Fluid Mechanics - II
A valve provides a means toregulate the flowrate bychanging the geometry ofthe system.
With the valve closed, theresistance to the flow is
infinitethe fluid cannot flow. With the valve wide open the
extra resistance due to thepresence of the valve may ormay not be negligible.
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Minor losses example: Valve
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An analytical method to predict the head lossfor components of piping system is notpossible.
The head loss information is given indimensionless form and based onexperimental data.
The most common method to determinehead losses or pressure drops is to specify
the loss coefficient, kL
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Loss Coefficient
Fluid Mechanics - II
Its value depends on geometry of component.
It may also depend on fluid properties.
In many cases Re is large enough that flowthrough the component is dominated by inertiaeffects, with low viscous effects.
Here pressure drops and head losses correlatedirectly with the dynamic pressure.
Thus, in many cases the loss coefficients forcomponents are a function of geometry only
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Loss Coefficient
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Head loss through a component is given interms of the length of pipe that would producethe same head loss.
The head loss of the pipe system is the sameas that produced in a straight pipe whoselength is equal to the pipes of the original
system plus the sum of the additionalequivalent lengths of all of the components ofthe system.
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Equivalent length
Fluid Mechanics - II
Loss coefficient at flow entrance8-Oct-12
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Loss coefficient at flow exit8-Oct-12
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Fluid Mechanics - II
Loss coefficient in sudden expansion
In this case the loss coefficient can be
calculated from analytical means.
Apply continuity, momentum & energy
equations in control volume.
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Loss coefficient in sudden expansion8-Oct-12
76
3 33 = 33 3 3 3 = 33 3 3 = 3 3
+
2 =
3 +
32 +
3 3 +
2
32 +=
3 2 =
= 333 =
3
3
2 = 1 3
2 = 1 3
= 2
Fluid Mechanics - II
Loss coefficient in conical diffuser
Diffuseris a device shaped
to decelerate a fluid.
Losses can be reduced if
expansion is gradual.
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For small angles, the diffuser is long and most of
the head loss is due to the wall shear stress.
For moderate or large angles, the flow separates
from the walls and the losses are due mainly to
dissipation of the kinetic energy of the jet leaving
the smaller diameter pipe.
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Loss coefficient in conical diffuser
Losses ina diffuser
NOTE
Typical
results
only.
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Flow through a diffuser is very complicated and may
be strongly dependent on the area ratio specific
details of the geometry, and the Reynolds number.
Fluid Mechanics - II
Losses in bends
The losses are due
to the separated
region of flow near
the inside of the
bend, and
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The swirling secondary flow that occurs fromthe imbalance of centripetal forces as a
result of the curvature of the pipe centerline.
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Losses in miter bends
Miter bends are usedwhere space is too
limited for smooth bends.
The losses in miter
bends can be reduced by
using guide vanes that
direct the flow with less
unwanted swirl anddisturbances.
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Fluid Mechanics - II
Loss coefficient for pipe components8-Oct-12
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Loss coefficient for pipe components8-Oct-12
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Fluid Mechanics - II
Loss coefficient for valves8-Oct-12
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Loss coefficient for valves8-Oct-12
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Fluid Mechanics - II
Example 8.6
Air at STP is to flowthrough test sections(5) and (6) with avelocity of 200 m/s.
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85
Flow is driven by a fan that increases the
static pressure by the amount p1 - p9. neededto overcome head losses experienced by the
fluid as it flows around the circuit.
Find p1 - p9 and the power supplied to thefluid by the fan.
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Example 8.6
Fan provides energy toovercome the head loss.
Energy eq b/w 1 and 9.
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86
z1 = z9, V1 = V9A
Power is
Loss coefficients
Section 6 to 4 (clockwise) is a diffuser with KL = 0.6
Section 4 has KL = 4.0,
Section 4-5 is nozzle, KL=0.2.
At corners, KL = 0.2.
Total head loss is.
Fluid Mechanics - II
Example 8.68-Oct-12
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Non circular conduits
With slight modification many round piperesults can be carried over, to flow in conduits
of other shapes.
Regardless of the cross-sectional shape,
there are no inertia effects in fully developed
laminar pipe flow.
The friction factor can be written as f= C/Reh
Cdepends on the shape of the duct, and Rehis the Reynolds number, based on the
hydraulic diameter Dh.
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Fluid Mechanics - II
Non circular conduits
The hydraulic diameteris four times the
ratio of the cross-sectional flow area
divided by the wetted perimeter, P, of the
pipe.
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Example 8.7
Air at T = 50 C and Patm flows from furnacethrough an 20cm dia pipe with V = 3 m/s.
It then passes into a square duct whose sideis of length a, with smooth surfaces (= 0)
The unit head loss is the same for the pipeand the duct.
Determine the duct size, a.
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92
Plan
Determine the head loss per unit length for thepipe, and then size the square duct to give the
same value. Solution
Find Viscosity and Re
From Re and /D= 0, find friction factorf= 0.022
Fluid Mechanics - II
Example 8.7
Air properties from appendix
Re = 34,100
Friction factor, f= 0.022
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93
Head loss per unit length is .0505
Same for the square duct, i.e. .0505 =
Where Dh is
And in duct, Vs = Qpipe/Aduct = 0.09/a.
Three equations and three variables.
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PIPE FLOW EXAMPLES
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Fluid Mechanics - II
Pipe systems
The main idea involved is to apply the energy
equation between appropriate locations within
the flow system,
The head loss written in terms of the friction
factor and the minor loss coefficients.
Two classes of pipe systems: those containing a single pipe (whose length may
be interrupted by various components),
those containing multiple pipes in parallel, series,
or network configurations.
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Single pipe
The three most common types of problems are. Type I: Determine the necessary pressure difference
or head loss from the desired flowrate or averagevelocity.
Type II: Determine the flowrate from the applieddriving pressure or head loss.
Type III: Determine the diameter of the pipe neededfrom the the pressure drop and the flowrate.
We assume the pipe system is defined in terms of the length of
pipe sections used
the number of elbows, bends, and valves needed toconvey the fluid is known.
the fluid properties are given.
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Fluid Mechanics - II
Threaded elbows 90K 1.5
Globe valve
open, K 10
Faucet
K
2
Q
1.75 m
5.25 m
3.5 m
3.5 m
3.5 m 3.5 m
(1)
(2)
97
Example Water flows from the ground floor to the second level in a
three-storey building through a 20 mm diameter pipe(drawn-tubing, = 0.0015 mm) at a rate of 0.75 liter/s.
The water exits through a faucet of diameter 12.5 mm.
Calculate the pressure at point (1).
all losses are neglected,
the only losses included
are major losses, or
all losses are included.
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Example
mhhggzVVp 12212212
1
98
From the modified Bernoulli equation, we can write
In this problem,p2= 0,z1 = 0. Thus,
The velocities in the pipe and out from the faucet are respectively
The Reynolds number of the flow is
LghgzVpgzVp 22
221
2
112
1
2
1
smD
Q
A
QV
smD
Q
A
QV
631.6012.0
1075.044
387.2020.0
1075.044
2
3
2
22
2
2
3
2
11
1
546,421012.1
)020.0)(387.2)(998(Re
3
Vd
12-Oct-12
Fluid Mechanics - II
The roughness d= 0.0015/20 = 0.000075. From the Moody chart, 0.022 (or,0.02191 viathe Colebrook formula). The total length of the pipe is
Hence, the friction head loss is
The total minor loss is
The pressure at (1) is
Example
m71.6)81.9(2
387.2
02.0
21)022.0(
2
22
1 g
V
dfhf
99
m2175.1)5.3(425.5
m
m
94.1123.571.6
23.5)81.9(2
387.2210)5.1(4
2
22
1
mf
m
hhh
g
VKh
Pa205
23.571.681.99985.35.381.9998387.2631.69982
1
2
1
22
12
2
1
2
21
k
hhggzVVp m
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Fluid Mechanics - II
PIPE NETWORKS (MULTIPLE PIPE
SYSTEMS)
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0
Fluid Mechanics - II
Multiple pipe systems
The governing mechanisms for the flow inmultiple pipe systems are the same as for thesingle pipe systems.
But because of the numerous unknownsinvolved, additional complexities may arise insolving for the flow in multiple pipe systems.
The simplest multiple pipe systems can beclassified into series or parallel flows.
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1
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Pipes in series
Every fluid particle that passes through thesystem passes through each of the pipes. Thus, The flowrate is the same in each pipe, and
The head loss is the sum of the head losses in eachof the pipes.
The friction factors will be different for each pipebecause the Re and will be different.
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2
Fluid Mechanics - II
Pipes in series (Problem types) Type I:
If flowrate is known, the pressure drop or head loss canbe determined from given equations.
Type II: If the pressure drop is given and flowrate is required, an
iteration scheme is needed.
None of the friction factors, are known, so solution mayinvolve more trial-and-error attempts.
Type III: If the pressure drop is given and pipe diameter is to be
determined, iterations are needed as in Type II.
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Pipes in parallel
A fluid particle traveling from A to B may takeany of the paths available, with The total flowrate equal to the sum of the flowrates in
each pipe.
The head loss experienced by any fluid particletraveling between A and B is the same, independentof the path taken.
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4
Fluid Mechanics - II
Each point in thesystem can onlyhave one pressure
The pressurechange from 1 to 2by path amustequal the pressurechange from 1 to 2by path b
A
p1
V1
2
2gz1
p2
V2
2
2gz2 hL
A
aa
Lhzg
Vz
g
Vpp
2
2
2
1
2
112
22
B
1 2
B
bb
Lhzg
Vz
g
Vpp
2
2
2
1
2
112
22
Pipe networks
10
9
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hLa hLb
a
b
1 2Pressure change by path A
Pipe networks
Assumptions
Pipe diameters are constant or K.E. is small
Model withdrawals are occurring at nodes so Vis constant between nodes
Or sum of head loss around loop is zero.
B
bb
A
aa
LL hzg
Vz
g
Vhz
g
Vz
g
V 2
2
2
1
2
1
2
2
2
1
2
1
2222
11
0
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Fluid Mechanics - II
Pipe Loops
Pipe loops are common in waterdistribution systems.
Pipe network divided into loopswith nodes
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1
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Pipe Loops Basic Principles
At each node, continuity may be applied:
Around any loop, the sum of head losses
must be zero:
n
i
iq1
0
m
i
ih1
0
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2
Fluid Mechanics - II
Pipe Loops
Basic Principles
Also, in each pipe, head loss is a function
of discharge as is evident from all pipe flow
formulae
Sign Convention (very important!)
Flows intoa node are positive
Head loss clockwiseround a loop are
positive
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Pipe Loops Solution techniques
The "loop" or "head balance" method This is used when the total volume rate of
flow through the network is known but the
heads or pressures at junctions within the
network are unknown.
The "nodal" or "quantity balance" method
This is used when the heads at each flow
entry point are known and it is required to
determine the pressure heads and flows
through the network.
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4
Fluid Mechanics - II
The Loop Method1. assume values ofqi to satisfy2. calculate hfi from qi3. if then solution is correct
4. if then apply a correction factor andreturn to step 2
Correction factors can be computed from:
0iq
0fih
0fih
i
fi
fi
i
q
h
hq
2
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Fluid Mechanics - II
The Nodal Method
1. assume a value of head Hjat each junction2. calculate qifrom Hj3. if then solution is correct
4. if then apply a correction factor andreturn to step 2
Correction factors can be computed from:
0fiq
0fiq
fi
i
i
h
qqH 2
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6
Fluid Mechanics - II
Network Analysis12-Oct
-12
11
7
Find the flows in the loop given the inflowsand outflows.The pipes are all 25 cm cast iron (=0.26 mm).
A B
C D0.10 m3/s
0.32 m3/s 0.28 m3/s
0.14 m3/s
200 m
100 m
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Example Assign a flow to each pipe link
Flow into each junction must equal flow out
of the junction
A B
C D0.10 m3/s
0.32 m3/s 0.28 m3/s
0.14 m3/s
0.320.00
0.10
0.04
arbitrary
11
8
12-Oct-12
Fluid Mechanics - II
Solution Calculate the head loss in each pipe
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-12
11
9
f=0.02 for Re>200000hf 8fL
gD5
2
Q 2
339)25.0)(8.9(
)200)(02.0(8
251
k k1,k3=339
k2,k4=169
A B
C D
0.10 m3/s
0.32 m3/s 0.28 m3/s
0.14 m3/s
1
4 2
3
hf1 34.7m
hf2 0.222m
hf3 3.39m
hf4 0.00m
hfii 1
4
31.53m
g
DQ
df
g
AQ
df
g
V
Dfhf
2
4
22
221
21
21
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Solution The head loss around the loop isnt zero Need to change the flow around the loop
The clockwise flow is too great (head loss is positive)
reduce the clockwise flow to reduce the head loss
qi = -0.163
Repeat until head loss around loop is zero
Easier solution would be to use numerical tools.
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0
A B
C D
0.10 m3/s
0.32 m3/s 0.28 m3/s
0.14 m3
/s
0.157
0.163
0.263
0.123
1
4 2
3
Q0+Q
Q1 = 0.157
Q2 = -0.123
Q3 = 0.263
Q4 = 0.163
Fluid Mechanics - II
Numeric Analysis Solution techniques
Use a numeric solver (Solver in Excel) to find a change inflow that will give zero head loss around the loop, or
Use a pipe Network Analysis software.
Set up a spreadsheet as shown, initially Q is 0
Set the sum of the head loss to 0 by changing Q the column Q0+Q contains the correct flows
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12
1
Q 0.000pipe f L D k Q0 0+ hf
P1 0.02 200 0.25 339 0.32 0.320 34.69
P2 0.02 100 0.25 169 0.04 0.040 0.27
P3 0.02 200 0.25 339 -0.1 -0.100 -3.39
P4 0.02 100 0.25 169 0 0.000 0.00
31.575Sum Head Loss
http://localhost/var/www/apps/conversion/tmp/scratch_6/spreadsheets/pipe_network_analysis.xlshttp://localhost/var/www/apps/conversion/tmp/scratch_6/spreadsheets/pipe_network_analysis.xlshttp://localhost/var/www/apps/conversion/tmp/scratch_6/spreadsheets/pipe_network_analysis.xlshttp://localhost/var/www/apps/conversion/tmp/scratch_6/spreadsheets/pipe_network_analysis.xls7/30/2019 Chapter 8 Binded
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END OF CHAPTER
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