chapter 8 binded

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


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.

12-Sep

-12 1


2/57

9/12/2


Pipes and ducts

Duct: A conduit with non circular crosssection.

Pipe: A conduit of circular cross section.

12-Sep-12 2


Pipe system components

Pipes

Fittings / Connectors

Flow Control devices

Pumps / Turbines

12-Sep

-12 3


3/57

9/12/2


CHARACTERISTICS OF PIPE FLOW

Chapter 8. Page384

12-Sep-12 4


Flow Characteristics

Pipe flow

Completely filled.

Pressure driven.

Assumption: RoundCross section

Open channel flow

Partially filled

Gravity driven

12-Sep

-12 5


4/57

9/12/2


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


Time Dependence of Fluid velocity

x component of velocity as a function of

time at A.

12-Sep

-12 7


5/57

9/12/2


Laminar Turbulent Flow

Streak lines for small, medium and largeflow rates (Re).

12-Sep-12 8


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


6/57

9/12/2


Entrance region & fully developed flow12-Sep-12 11


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.

12-Sep

-12 12

Re06.0D

e

61(Re)4.4

D

e


7/57

9/12/2


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


END OF WEEK # 1

Home Work problems. 8.2, 8.4, 8.6 & 8.8

12-Sep

-12 14


8/57

9/28/2


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



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.

28-Sep

-12 18


9/57

9/28/2



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


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).

28-Sep

-12 20


10/57

9/28/2



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



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,

28-Sep

-12 22


11/57

9/28/2



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.

28-Sep-12 23



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.

28-Sep

-12 24


12/57

9/28/2



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

28-Sep-12 25



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

28-Sep

-12 26


13/57

9/28/2



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



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.

28-Sep

-12 28


14/57

9/28/2



Because the fluid is viscous it sticks to thepipe wall so that u = 0, at r= D/2.

Vc is the centerline velocity

28-Sep-12 29



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,

28-Sep

-12 30


15/57

9/28/2


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.

28-Sep-12 31


Inclined pipes

Replace the pressure drop p, by the effectof both pressure and gravity p -l sin.

28-Sep

-12 32


16/57

9/28/2


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.

28-Sep-12 33


From the Navier

Stokes Equations

The NavierStokes equations become.

In polar coordinates

28-Sep

-12 34


17/57

9/28/2


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.

28-Sep-12 35


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

28-Sep

-12 36


18/57

9/28/2


FULLY DEVELOPED TURBULENT

FLOW

Section 8.3

Page 399

28-Sep-12 37


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.

28-Sep

-12 38


19/57

9/28/2


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



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.

28-Sep

-12 40


20/57

9/28/2



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.

28-Sep-12 41


Turbulent shear stress

Axial component of velocity, u =u(t), at agiven location in turbulent pipe flow is.

28-Sep

-12 42


21/57

9/28/2



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).

28-Sep-12 43



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

28-Sep

-12 44


22/57

9/28/2



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.

28-Sep-12 45



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.

28-Sep

-12 46


23/57

9/28/2



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.

28-Sep-12 47


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.

28-Sep

-12 48


24/57

9/28/2


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!

28-Sep-12 49


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.

28-Sep

-12 50


25/57

9/28/2


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.

28-Sep-12 51


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,

28-Sep

-12 52


26/57

9/28/2



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.

28-Sep-12 53



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.

28-Sep

-12 54


27/57

9/28/2



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

28-Sep-12 55



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.

28-Sep

-12 56


28/57

10/8/2



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

8-Oct-12

57



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

8-Oct-12

58


29/57

10/8/2



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.

8-Oct-12

59


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.

8-Oct-12

60


30/57

10/8/2



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.

8-Oct-12

61


Major Losses - Summary The head loss due to viscous effects in straight

pipes, termed the major loss and denoted hL major,

8-Oct-12

62

The Typical roughness values for various pipe surfaces areshown along with the functional dependence offon Re and calledthe Moody chart.


31/57

10/8/2


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

8-Oct-12

64


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

8-Oct-12

65


32/57

10/8/2


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.

8-Oct-12

66


PROBLEMS

8.42, 8.45, 8.50, 8.58, 8.60, 8.62 & 8.70.

8-Oct-12

67


33/57

10/8/2


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?

8-Oct-12

68

Minor losses


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.

8-Oct-12

69

Minor losses example: Valve


34/57

10/8/2


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

8-Oct-12

70

Loss Coefficient


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

8-Oct-12

71

Loss Coefficient


35/57

10/8/2


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.

8-Oct-12

72

Equivalent length


Loss coefficient at flow entrance8-Oct-12

73


36/57

10/8/2


Loss coefficient at flow exit8-Oct-12

74


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.

8-Oct-12

75


37/57

10/8/2


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


Loss coefficient in conical diffuser

Diffuseris a device shaped

to decelerate a fluid.

Losses can be reduced if

expansion is gradual.

8-Oct-12

77

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.


38/57

10/8/2


Loss coefficient in conical diffuser

Losses ina diffuser

NOTE

Typical

results

only.

8-Oct-12

78

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.


Losses in bends

The losses are due

to the separated

region of flow near

the inside of the

bend, and

8-Oct-12

79

The swirling secondary flow that occurs fromthe imbalance of centripetal forces as a

result of the curvature of the pipe centerline.


39/57

10/8/2


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.

8-Oct-12

80


Loss coefficient for pipe components8-Oct-12

81


40/57

10/8/2


Loss coefficient for pipe components8-Oct-12

82


Loss coefficient for valves8-Oct-12

83


41/57

10/8/2


Loss coefficient for valves8-Oct-12

84


Example 8.6

Air at STP is to flowthrough test sections(5) and (6) with avelocity of 200 m/s.

8-Oct-12

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.


42/57

10/8/2


Example 8.6

Fan provides energy toovercome the head loss.

Energy eq b/w 1 and 9.

8-Oct-12

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.


Example 8.68-Oct-12

87


43/57

10/12/2


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.

12-Oct-12

90


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.

12-Oct

-12

91


44/57

10/12/2


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.

12-Oct-12

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


Example 8.7

Air properties from appendix

Re = 34,100

Friction factor, f= 0.022

12-Oct

-12

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.


45/57

10/12/2


PIPE FLOW EXAMPLES

12-Oct-12

94


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.

12-Oct

-12

95


46/57

10/12/2


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.

12-Oct-12

96


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.

12-Oct

-12


47/57

10/12/2


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


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

12-Oct

-12


48/57

10/12/2


PIPE NETWORKS (MULTIPLE PIPE

SYSTEMS)

12-Oct-12

10

0


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.

12-Oct

-12

10

1


49/57

10/12/2


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.

12-Oct-12

10

2


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.

12-Oct

-12

10

3


50/57

10/12/2


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.

12-Oct-12

10

4


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

12-Oct

-12


51/57

10/12/2


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

12-Oct-12


Pipe Loops

Pipe loops are common in waterdistribution systems.

Pipe network divided into loopswith nodes

12-Oct

-12

11

1


52/57

10/12/2


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

12-Oct-12

11

2


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

12-Oct

-12

11

3


53/57

10/12/2


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.

12-Oct-12

11

4


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

12-Oct

-12

11

5


54/57

10/12/2


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

12-Oct-12

11

6


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


55/57

10/12/2


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


Solution Calculate the head loss in each pipe

12-Oct

-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


56/57

10/12/2


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.

12-Oct-12

12

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


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

12-Oct

-12

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.xls


57/57

10/12/2


END OF CHAPTER

12-Oct-12

12

2

chapter 8 binded

Documents