ebvf4103 (chapter 6) fluid mechanics for civil engineering

5/21/2018 EBVF4103 (Chapter 6) Fluid Mechanics for Civil Engineering

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Subject Matter Expert/Author:Assoc. Prof. Dr Othman A. Karim (OUM)

Faculty of Engineering andTechnical Studies

Copyright ODL Jan 2005 Open University Malaysia

1

FLUID MECHANICS FORCIVIL ENGINEERINGTUTORIAL 4 UNIT 3: Application ofFluid MechanicsChapter 6: Flow in Pipelines

Assoc. Prof. Dr Othman A. KarimEBVF4103 Fluid Mechanics for Civil EngineeringJan 2005


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2Copyright ODL Jan 2005 Open University Malaysia

SEQUENCE OF CHAPTER 6

IntroductionObjectives

6.1 Pipe Flow System

6.2 Types of Flow

6.2.1 Laminar Flow

6.2.2 Turbulent Flow

6.3 Energy Losses due to Friction

6.3.1 Friction Losses in Laminar Flow

6.3.2 Friction Loss in Turbulent Flow

6.4 Minor Losses

6.4.1 Losses due to Pipe Fittings

6.4.2 Sudden Enlargement

6.4.3 Sudden Contraction

6.5 Energy Added and Extracted6.6 Pipe Flow Analysis

6.6.1 Simple Pipeline

6.6.2 Pipes in Series

6.6.3 Pipes in Parallel

6.6.4 Pipe Network

Summary


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Introduction

In considering the convenience and necessities in every daylife, it is truly amazing to note the role played by conduits intransporting fluid.

For example, the water in our homes is normally conveyed

through pressure pipelines, from the distribution system, sothat it will be available when and where we want it.

Moreover, virtually all of this water leaves our homes asdilute wastes through sewers, another type of conduits. Oilis often transferred from their source by pressure pipelines

to refineries while gas is conveyed by pipelines into adistribution network for supply.

Thus, it can be seen that the fluid flow in conduits is ofimmense practical significance in civil engineering.


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Objectives

1. Differentiate between laminar and turbulent flows inpipelines.

2. Describe the velocity profile for laminar and turbulentflows.

3. Compute Reynolds number for flow in pipes.

4. Define the friction factor, and compute the friction losses inpipelines.

5. Recognize the source of minor losses, and compute minor

losses in pipelines.

6. Analyze simple pipelines, pipelines in series, parallel, andsimple pipe networks.


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5

6.1 Pipe Flow System

This chapter introduces the fundamental theories of flow inpipelines as well as basic design procedures.

In this chapter, the pipeline system is defined as a closed conduitwith a circular cross-section with water flows (flowing full) insideit.

It is a closed system, the water is not in contact with air (i.e. nofree surface). Flow in a closed pipe results from a pressuredifference between inlet and outlet. The pressure is affected byfluid properties and flow rate.

The following diagram gives the geometrical properties for circularpipe. In the diagram, D represents the diameter of pipe, R is thepipe radius and L is the pipe length. The cross-sectional area ofthe pipe can be calculated using A = R2.


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In hydraulic applications, energy values are often converted intounits of energy per unit weight of fluid, resulting in units of length.

When using these length equivalents, the energy of the system isexpressed in terms of head. The usual unit used is meter. Theenergy at any point within a hydraulic system is often expressed in

three parts, i.e. the pressure head (P/g), elevation head (z), andvelocity head (V2/2g) The sum of all these components is known asthe total head (H).

(6.1)

R

R

D=2R

L

R

PipeCenterline

Figure 6.2:Schematic diagram of a circular pipe

g2

Vz

g

PH

2


7/54Subject Matter Expert/Author:Assoc. Prof. Dr Othman A. Karim (OUM)



Figure 6.3:Flow in a uniform

pipeline

A line plotted of total head versus distance through a system is called thetotal energy line (TEL).

The TEL is also known as energy grade line (EGL).

The sum of the elevation head and pressure head yields the hydraulicgrade line (HGL).

In a uniform pipeline, the total shear stress (resistance to flow) is constantalong the pipe resulting in a uniform degradation of the total energy orhead along the pipeline.

The total head loss along a specified length of pipeline is referred to headloss due to friction and denoted as hf.





Referring to the above figure, the Bernoulli equation

can be written from section 1 to section 2 as;

(6.2)f

22

22

21

11 h

g2

Vz

g

P

g2

Vz

g

P





6.2 Types of Flow

The physical nature of fluid flow can be categorized into three types, i.e.laminar, transition and turbulent flow. It has been mentioned earlier thatReynolds Number (Re) can be used to characterize these flow.

(6.3)

where = density= dynamic viscosity

= kinematic viscosity (= /)V= mean velocity

D= pipe diameter

In general, flow in commercial pipes have been found to conform to thefollowing condition:

Laminar Flow: Re





6.2.1 Laminar flow

Viscous shears dominate in this type of flow and thefluid appears to be moving in discreet layers. Theshear stress is governed by Newtons law of viscosity(equation 1.1):

(1.1)

In general the shear stress is almost impossible tomeasure. But for laminar flow it is possible tocalculate the theoretical value for a given velocity,fluid and the appropriate geometrical shape.

dy

du





Pressure loss during a laminar flow in a pipe Up to this point we only considered ideal fluidwhere there is no losses due to friction or anyother factors.

In reality, because fluids are viscous, energy islost by flowing fluids due to friction which must

be taken into account. The effect of friction shows itself as a pressure (or

head) loss. In a pipe with a real fluid flowing, theshear stress at the wall retard the flow.

The shear stress will vary with velocity of flow andhence with Re. Many experiments have been done

with various fluids measuring the pressure loss atvarious Reynolds numbers.

Figure below shows a typical velocity distributionin a pipe flow. It can be seen the velocityincreases from zero at the wall to a maximum inthe mainstream of the flow.

Figure 6.4:

A typical velocitydistribution in a

pipe flow





In laminar flow the paths of individual particles of fluid donot cross, so the flow may be considered as a series ofconcentric cylinders sliding over each other rather like thecylinders of a collapsible pocket telescope.

Lets consider a cylinder of fluid with a length L, radius r,flowing steadily in the center of pipe.

Figure 6.5:Cylindrical of fluid flowing steadily in a pipe





The fluid is in equilibrium, shearing forces equal the pressure forces.

Shearing force = Pressure force

(6.5)

Taking the direction of measurement r(measured from the center ofpipe), rather than the use of y(measured from the pipe wall), theabove equation can be written as;

(6.6)

Equatting (6.5) with (6.6) will give:

2

r

L

P

rPPArL2 2

dr

du r

2

r

L

P

dr

du

dr

du

2

r

L

P





In an integral form this gives an expression for velocity, with thevalues of r= 0 (at the pipe center) to r= R (at the pipe wall)

(6.7)

where P= change in pressureL= length of pipe

R= pipe radius

r= distance measured from the center of pipe

The maximum velocity is at the center of the pipe, i.e. when r= 0.

It can be shown that the mean velocity is half the maximum velocity, i.e. V=umax/2

Rr0r rdr2

1

L

Pu

LP

4rRu

22

r

L

P

4

Ru

2

max





Figure 6.6: Shear stress and velocity distribution in pipe for laminar flow

The discharge may be found using the Hagen-Poiseuilleequation,which is given by the following;

(6.8)

The Hagen-Poiseuilleexpresses the discharge Q in terms of the

pressure gradient , diameter of pipe, and viscosity of thefluid.

Pressure drop throughout the length of pipe can then be calculated

by

(6.9)

128

D

L

PQ

4

L

P

dx

dP

4R

LQ8P





6.2.2 Turbulent flow This is the most commonly occurring flow in engineering

practice in which fluid particles move erratically causinginstantaneous fluctuations in the velocity components.

These fluctuations cause additional shear stresses. In thistype of flow both viscous and turbulent shear stresses exists.

Thus, the shear stress in turbulent flow is a combination oflaminar and turbulent shear stresses, and can be written as:

where = dynamic viscosity

= eddy viscosity which is not a fluid property butdepends upon turbulence condition of flow.

dydU

turbulentarminla





The velocity at any point in the cross-section will be proportionalto the one-seventh power of the distance from the wall, which canbe expressed as:

(6.10)

where Uyis the velocity at a distance yfrom the wall, UCLis the velocityat the centerline of pipe, and Ris the radius of pipe. This equation is

known as the Prandtl one-seventh law.

Figure 6.7 below shows the velocity profile for turbulent flow in a pipe.

The shape of the profile is said to be logarithmic.

7/1

CL

y

R

y

U

U

Figure 6.7:Velocity profile for turbulent flow





For smooth pipe:

(6.11a)

For rough pipe:

(6.11b)

In the above equations, Urepresents the velocity at a distance yfrom the pipe wall, U*is the shear velocity =

yis the distance form the pipe wall, kis the surface roughness and

is the kinematic viscosity of the fluid.

5.5yU

log75.5U

U *10

*

5.8k

ylog75.5

U

U10

*





Example 6.1Glycerin (= 1258 kg/m3, =9.60 x10-1N.s/m2) flows with avelocity of 3.6 m/s in a 150-mm diameter pipe. Determinewhether the flow is laminar or turbulent.

Solution:

Since Re=708, which is less than 2000, the flow is laminar.

VD

Re

70810x60.9

15.0x6.3x1258

Re 1





6.3 Energy Losses due to friction

When a liquid flows through a pipeline, shear stressesdevelop between the liquid and the pipe wall.

This shear stress is a result of friction, and its magnitude isdependent upon the properties of the fluid, the speed at

which it is moving, the internal roughness of the pipe, thelength and diameter of pipe.

Friction loss, also known as major loss, is a primary causeof energy loss in a pipeline system.


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6.3.1 Friction Losses in Laminar flow In laminar flow, the fluid seems to flow as several layers, one on another.

Because of the viscosity of the fluid, a shear stress is created betweenthe layers of fluid.

Energy is lost from the fluid to overcome the frictional forces produced bythe shear stress.

Energy loss is usually represented by the drop of pressure in the directionof flow.

Therefore, the frictional head loss, hf, can be written in terms of pressuredrop along the pipeline, as follows:

(6.12)

Substituting the Hagen-Poiseuille equation and applying the continuityequation, Q = VA, to the above resulted into the following expression:

(6.13)

g

Ph f

g2

V

D

L

Re

64

gD

LV32h

2

2f


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6.3.2 Friction Losses in Turbulent flow In turbulent flow, the friction head loss can be calculated by considering

the pressure losses along the pipelines.

In a horizontal pipe of diameter Dcarrying a steady flow there will be apressure drop in a length Lof the pipe.

Equating the frictional resistance to the difference in pressure forces, andmanipulating resulted into the following expression:

(6.14)

This equation is known as Darcy-Weisbach (D-W)equation, in which isthe friction factor. It should be noted that is dimensionless, and the

value is not constant

g2

V

D

Lh

2

f


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Blasius (1913) was the first to propose an accurate empiricalrelation for the friction factor in turbulent flow in smooth

pipes, namely

(6.15)

Thus, the calculation of losses in turbulent pipe flow is

dependent on the use of empirical results and the most

common reference source is the Moody diagram. A Moody

diagram is a logarithmic plot of vs. Re for a range of k/D.

A typical Moody diagram is shown in Figure 6.9.

= 0.316/Re0.25


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Figure 6.8: Moody Diagram


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Table 6.1: Typical pipe roughness

(Reference: White, 1999)

Material Roughness, k (mm)

Glass smooth

Brass, new 0.002

Concrete

Smoothed

Rough

0.04

2.0

Iron

Cast, new

Galvanised, new

Wrought, new

0.26

0.15

0.046

Steel

Commercial, new

Riveted

0.046

3


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6.4 Minor Losses

In addition to head loss due to friction, there are alwaysother head losses due to pipe expansions and contractions,bends, valves, and other pipe fittings. These losses areusually known as minor losses (hLm).

In case of a long pipeline, the minor losses maybe negligiblecompared to the friction losses, however, in the case ofshort pipelines, their contribution may be significant.


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6.4.1 Losses due to pipe fittings

where hLm= minor loss

K= minor loss coefficientV= mean flow velocity

g2

VKh

2

Lm (6.16)

Table 6.2:Typical K values

Type K

Exit (pipe to tank) 1.0

Entrance (tank to pipe) 0.5

90elbow 0.9

45elbow 0.4

T-junction 1.8

Gate valve 0.25 - 25

l f i i d


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6.4.2 Sudden Enlargement As fluid flows from a smaller pipe into a larger pipe through

sudden enlargement, its velocity abruptly decreases; causingturbulence that generates an energy loss.

The amount of turbulence, and therefore the amount of energy, is

dependent on the ratio of the sizes of the two pipes. The minor loss (hLm)is calculated from;

(6.16a)

where is KEis the coefficient of expansion, and the values depends on the

ratio of the pipe diameters (Da/Db) as shown below.

g2

VKh

2a

ELm

Da/Db 0.0 0.2 0.4 0.6 0.8

K 1.00 0.87 0.70 0.41 0.15

Table 6.3:Values of KE vs. Da/Db

F lt f E i i d


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Figure 6.9:Flow at Sudden Enlargement

F lt f E i i d


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6.4.3 Sudden Contraction

The energy loss due to a sudden contraction can be calculated usingthe following;

(6.16b)

The KCis the coefficient of contraction and the values depends onthe ratio of the pipe diameter (Db/Da) as shown below.

g2

VKh

2b

CLm

Db/Da 0.0 0.2 0.4 0.6 0.8 1.0

K 0.5 0.49 0.42 0.27 0.20 0.0

Table 6.4:Values of KC vs. Db/Da

Figure 6.10:Flow at sudden contraction

F lt f E i i d


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Example 6.2 Water at 10C is flowing at a rate of 0.03 m3/s through a pipe. The pipe

has 150-mm diameter, 500 m long, and the surface roughness is estimatedat 0.06 mm. Find the head loss and the pressure drop throughout thelength of the pipe.

Solution: From Table 1.3 (for water): = 1000 kg/m3and =1.30x10-3N.s/m2

V= Q/A and A=R2

A= (0.15/2)2= 0.01767 m2

V= Q/A =0.03/.0.01767 =1.7 m/s

Re= (1000x1.7x0.15)/(1.30x10-3) = 1.96x105 > 2000 turbulent flow

To find, use Moody Diagram with Reand relative roughness(k/D).k/D = 0.06x10-3/0.15 = 4x10-4

From Moody diagram, 0.018The head loss may be computed using the Darcy-Weisbach equation.

The pressure drop along the pipe can be calculated using the relationship:

P=ghf= 1000 x 9.81 x 8.84P = 8.67 x 104 Pa

.m84.881.9x2x15.0

7.1x500x018.0

g2

V

D

Lh

22

f

F lt f E i i d


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Example 6.3

Determine the energy loss that will occur as 0.06 m3/s water flowsfrom a 40-mm pipe diameter into a 100-mm pipe diameter througha sudden expansion.

Solution:

The head loss through a sudden enlargement is given by;

Da/Db= 40/100 = 0.4

From Table 6.3: K = 0.70

Thus, the head loss is

g2

VKh

2

am

s/m58.3)2/04.0(

0045.0

A

QV

2a

a

m47.081.9x2

58.3x70.0h

2

Lm

Faculty of Engineering and


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6.5 Energy added and extracted

There are many occasions when energy needs to be added toa hydraulic system to overcome elevation differences,friction losses and minor losses.

A pump is a common device to which mechanical energy isapplied and transferred to the water as total head of the

pump.

The head added is called pump head (Hp), and is a functionof flow rate through the pump.

On the other hand, fluid motor or turbines are common

examples of devices that extract energy from a fluid, andthe head extracted is called head of turbine (Ht), deliver itin a form of work.



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Denoting the head loss due to friction and minor losses as HL, and

the external energy added/extracted by HE, then the Bernoulliequation may be rewritten as

(6.17)

HE= Hp(positive for pump) when the head is added to the fluid, or

HE= Ht(negative for turbine) when the head is extracted from thefluid. Note the similarity of this equation with equation (3.12). Itis often necessary to convert the total power (P) of a pump orturbine to HEor vice versa. Recall from Chapter 3, therelationship between P and HEis given by the following

P = gQHE (3.8)In a pump HE= HP, the value is positive since power is added to

the fluid. In a turbine, HE= Ht is negative and power is extracted

from the flow.

21L

22

22

E

21

11 H

g2

Vz

g

PH

g2

Vz

g

P



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The term efficiency is used to denote the ratio of the power delivered bythe pump to the fluid to the power supplied to the pump.

Because of energy loses due to mechanical friction in pump components,fluid friction in the pump, and excessive fluid turbulence in the pump,not all of the input power is delivered to the fluid. Therefore, theefficiency of a pump can be written as;

The value of pis less than 1.0.

Similarly, energy losses in a turbine are produced by mechanical andfluid friction. Therefore, not all the power delivered to the motor isultimately converted to power output from the device. The efficiency of

a turbine is defined as;

Here again, the value oftis less than 1.0.

in

p

inp

PgQH

PP

pumptopliedsupPowerfluidtodeliveredPower

t

o

t

ot

gQH

P

P

P

fluidbypliedsupPower

motorfromoutputPower



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Example 6.4

Calculate the head added by the pump when the water systemshown below carries a discharge of 0.27 m3/s. If the efficiencyof the pump is 80%, calculate the power input required by thepump to maintain the flow.



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Solution:Applying Bernoulli equation between section 1 and 2

(1)

P1= P2= Patm= 0 (atm) and V1=V20Thus equation (1) reduces to:

(2)

HL1-2= hf+ hentrance+ hbend+ hexit

From (2):

21L

22

22

p

21

11 H

g2

Vz

g

PH

g2

Vz

g

P

21L12p HzzH

g2

V4.39

14.05.04.0

1000x015.0g2

VH

2

2

21L

81.9x2

V4.39200230H

2

p



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The velocity can be calculated using the continuityequation:

Thus, the head added by the pump: Hp= 39.3 m

Pin= 130 117 Watt 130 kW.

s/m15.2

2/4.0

27.0

A

QV

2

in

pp

P

gQH

8.0

3.39x27.0x81.9x1000gQHP

p

pin



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6.6 Pipe Flow Analysis

Pipeline system used in water distribution, industrialapplication and in many engineering systems mayrange from simple arrangement to extremely complexone.

Problems regarding pipelines are usually tackled bythe use of continuity and energy equations.

The head loss due to friction is usually calculated usingthe D-Wequation while the minor losses are computed

using equations 6.16, 6.16(a) and 6.16(b) depending onthe appropriate conditions.



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6.6.2 Pipes in Series

When two or more pipes ofdifferent diameters orroughness are connected insuch a way that the fluid

follows a single flow paththroughout the system, thesystem represents a seriespipeline.

In a series pipeline the total

energy loss is the sum of theindividual minor losses andall pipe friction losses.

Figure 6.11:Pipelines in series



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Referring to Figure 6.11, the Bernoulli equationcan be writtenbetween points 1 and 2 as follows;

(6.18)where P/g = pressure head

z = elevation headV2/2g= velocity head

HL1-2 = total energy lost between point 1 and 2Realizing thatP1=P2=Patm, andV1=V2, then equation (6.14) reduces to

z1-z2= HL1-2

Or we can say that the different of reservoir water level is equivalent to thetotal head losses in the system.

The total head losses are a combination of the all the friction losses and thesum of the individual minor losses.

HL1-2= hfa + hfb+ hentrance+ hvalve+ hexpansion+ hexit.

Since the same discharge passes through all the pipes, the continuityequation can be written as;

Q1= Q2

21L

22

22

21

11 H

g2

Vz

g

P

g2

Vz

g

P



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42


6.6.3 Pipes in Parallel

A combination of two ormore pipes connectedbetween two points sothat the discharge

divides at the firstjunction and rejoins atthe next is known aspipes in parallel. Herethe head loss betweenthe two junctions is thesame for all pipes.

Figure 6.12:Pipelines in parallel



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43


Applying the continuity equation to the system;

Q1 = Qa + Qb= Q2 (6.19)

The energy equation between point 1 and 2 can be writtenas;

The head losses throughout the system are given by;

HL1-2=hLa= hLb (6.20)

Equations (6.19) and (6.20) are the governing relationshipsfor parallel pipe line systems. The system automaticallyadjusts the flow in each branch until the total system flowsatisfies these equations.

L

2

222

2

111 Hg2

Vzg

Pg2

Vzg

P



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44


6.6.4 Pipe Network A water distribution system consists of complex interconnected pipes, service

reservoirs and/or pumps, which deliver water from the treatment plant to theconsumer.

Water demand is highly variable, whereas supply is normally constant. Thus,the distribution system must include storage elements, and must be capable

of flexible operation. Pipe network analysis involves the determination of the pipe flow rates and

pressure heads at the outflows points of the network. The flow rate andpressure heads must satisfy the continuity and energy equations.

The earliest systematic method of network analysis (Hardy-Cross Method) isknown as the head balance or closed loop method. This method is applicable

to system in which pipes form closed loops. The outflows from the system aregenerally assumed to occur at the nodes junction.

For a given pipe system with known outflows, the Hardy-Crossmethod is aniterative procedure based on initially iterated flows in the pipes. At eachjunction these flows must satisfy the continuity criterion, i.e. the algebraicsum of the flow rates in the pipe meeting at a junction, together with anyexternal flows is zero.



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45


Assigning clockwise flows and their associated head losses are

positive, the procedure is as follows: Assume values of Q to satisfy Q = 0.

Calculate HLfrom Q using HL= K1Q2.

If HL= 0, then the solution is correct.

If HL0, then apply a correction factor, Q, to all Q and repeatfrom step (2).

For practical purposes, the calculation is usually terminated when HL< 0.01 m or Q < 1 L/s.

A reasonably efficient value of Q for rapid convergence is given by;

(6.21)

QH

2

HQ

L

L



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y g gTechnical Studies

46


Example 6.5 A pipe 6-cm in diameter, 1000m long and with = 0.018 is

connected in parallel between two points M and N withanother pipe 8-cm in diameter, 800-m long and having =0.020. A total discharge of 20 L/s enters the parallel pipe

through division at A and rejoins at B. Estimate thedischarge in each of the pipe.



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47


Solution:Continuity: Q = Q1+ Q2

(1)

Pipes in parallel: hf1= hf2

Substitute (2) into (1)

0.8165V2+ 1.778 V2= 7.074

V2= 2.73 m/s

22

221

2 V)08.0(4

V)06.0(4

02.0

074.7V778.1V 21

)2(V8165.0V

V08.0

800x020.0V

06.0

1000x018.0

gD2

VL

gD2

VL

21

22

21

2

222

21

211

1



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48


Q2= 0.0137 m3/s

From (2):

V1= 0.8165 V2= 0.8165x2.73 = 2.23 m/s

Q1 = 0.0063 m3/s

Recheck the answer:

Q1+ Q2= Q0.0063 + 0.0137 = 0.020

(same as given Q OK!)

73.2x)08.0(4

VAQ 2

222



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49


Example 6.6 For the square loop shown, find the discharge in all the

pipes. All pipes are 1 km long and 300 mm in diameter,with a friction factor of 0.0163. Assume that minor lossescan be neglected.



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Technical Studies

50


Solution: Assume values of Q to satisfy continuity equations all at nodes.

The head loss is calculated using; HL= K1Q2

HL= hf + hLm

But minor losses can be neglected: hLm= 0 Thus HL= hf

Head loss can be calculated using the Darcy-Weisbach equation

g2

V

D

Lh

2

f



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Technical Studies

51


First trial

Since HL> 0.01 m, then correction has to be applied.

554'K

Q'KH

Q554H

3.0x4

Qx77.2

A

Q77.2H

81.9x2Vx

3.01000x0163.0H

g2

V

D

LhH

2L

2

L

22

2

2

2

L

2

L

2

fL

Pipe Q (L/s) HL(m) HL/Q

AB 60 2.0 0.033BC 40 0.886 0.0222

CD 0 0 0

AD -40 -0.886 0.0222

2.00 0.0774

Faculty of Engineering andh i l S di


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Technical Studies

52


Second trial

Since HL 0.01 m, then it is OK.

Thus, the discharge in each pipe is as follows (to the nearest integer).

s/L92.120774.0x2

2

QH2

H

Q L

L

Pipe Q (L/s) HL(m) HL/Q

AB 47.08 1.23 0.0261

BC 27.08 0.407 0.015CD -12.92 -0.092 0.007

AD -52.92 -1.555 0.0294

-0.0107 0.07775

Pipe Discharge (L/s)

AB 47

BC 27

CD -13

AD -53

Faculty of Engineering andT h i l St di


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Technical Studies

53


Summary

Overall, chapter 6 has introduced basic designprocedures of a flow in pipelines and more elaborationon the type of flows.

A detail calculation of pipe flow analysis which

consists of simple pipeline, pipe in series, pipe inparallel and pipe network were discussed.

The pipe network used the Hardy Cross method and atthe end, the discharge of water of each pipe can beknown.

Faculty of Engineering andT h i l St di


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Technical Studies

54

Thank You

ebvf4103 (chapter 6) fluid mechanics for civil engineering

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