--·-4 -·I

- ILIQUID FLOWIN PIPES

0 V\:V>"tf f'\

(}'> .._;t->1\ u\ (CAr"-

Chapter Objectives

After reading this chapter you should be able to:

• explain the conseNation of energy and derive

the Bernoulli's equation;

• apply the Bernoulli's equation to solve simple problems;

• explain fluid viscosity;

• evaluate the Reynolds Number;

• distmguish between laminar and turbulent flow:

• evaluate the friction losses in pipes;

• evaluate the minor losses due to pipe-finings;

• sketch thehydraulic gradient and energy gradient lines in pipe flow:

• solve pipe flow problems.

19.1INTRODUCTION

Fluid flow in circular pipes bas wide engineering applications. Incity water distribution systems and industrial hydraulic systems liquidsare generally transported from one point tO another by forcing themthrough pipes or tubes. The aim or this cha pter is to analyse andestimate the friction and shock losses as liquids flow through pipes andfittings.

Only steady-flow situations are discussed in this chapter. In a steady flow, the velocity, pressure, density, etc., of the fluid at any specifiedpoint do not change with time.

The theory of fluid flow rests on three main principles; namely, the conservation of mass.the conser:ation of energy and the constrvotion of momentum.In Sections 3.5 and 3.6 conservation of mus and conservation of energy were discussed in det il. In the case of liquids (incompressible fluids) the analysis becomes simple as the liquid density is practically constant. In this chapter we will appl y the conservation of mass and the conservation of energy to liquid now problems.

19

19.2 CONTINUITYEQUATION

The water entering one end of a garden hose is expected to flow out fromthe other end. This assumption,deduby our common sense.can beexpressed as a law of conservation.

573

•

574 Chapt<r 19

Referring to Figure 19.1, theaboveideamayberestatJ:dasfoUows:Inagivenperiodoftime,the same mass or fluid Illat enters section I will flow out or section 2.

I

C Ill

I P,

' !

I m,I

cl 8

0 dl CtOH·sec:don at ®

"

Figure 19.1 Continuity of flow

Mass

entering

section I

per second,

Mass leaving

section 2

per =ond,

Under steady

Cross•sedat<D

I ·-G)

state

or

{19.1)

where p1, p1 s density of fluid at

sections I and 2. kgtm3

AI' A1 •cross-

sectional aJU at I and

2, m1

C , C = average fluid velocity at I and 2. m/s1 1

ril = mass flow rate of fluid, kgfs

For incompressible fluids (in general gases and vapours arc compressible; liquids are incompressible) tlle density is practically constanL Equation 19.1 tllen becomes:

A1C

1 A,.C,V (for incompressible fluids) (19.2)

where V volume flow rate,ml/s

Both Equations 19.1 and 19.2 are known as the Cominuity Equotions.

.. ----•

Uquid flow illpipts

575

Example 19.1

Sea water flows through a pipe reducer asshown in Figure 19.2. 1be diameters atsections I and 2 are 50 em and 25 emrespectively.If the average velocity of the fluid atsection I is 0.5 m/s, find its velocity at section 2.

c, -I- -

<D

---''----- c, I

I<?>

d Da a;

Figure 19.2 Liquid flow through a reducer

1= 0.5 m: 2 0.25 m:C 1 = 0.5 m/s

Solution:

•

A,. =d

lt-

4

By the equation of continuit y,

Substituting the values,

ll On simplification,I

II

576 Cloapur /9

Conclusion:

The smaller lhe pipe lhe greater lhe velocity.

19.3 TOTAL READ

Generally, a unit mass (I kg) of moving fluid po ses four forms of energy:

I. Jnurnol energy due to temperature, u.2. Flow enugy due to fluid pressure, pv.3. Pountial energy due to its height or elevation = gZ.

2

4. Kinttic enugy due to lhe fluid velocity =.2

Note: In most liquid now problems lhe change in the internal energy is negligible and hence is omiuein lhe analysis.

Thus, tlte total energy possessed by a moving fluid is

c,pv + gZ+-

2 kg

Replacing the specific volume,v, by fr, the tOL11 energy is

P c,p +gZ+

Nmkg

Dividing all the lhree terms by 'g': c, Nm s2

Total energy = .!:p + Z + -2g

- --s mkg rn

Alllhe lhree terms now have the dimension of 'm':

*= pressure headZ =potential head m

c, ml

sl

- = velocity head2g -r-•m

s m

In fluid mecha nics the sum of these three terms is known as t he toto/ htad . Itrepresents !he total energy per unit weight (I N) of the fluid as an equivalent form ofpotential energy stored at an imaginary height equal to lhe total head.

2

Nm

Uquid flow in piptt 577

Erampk/9.2

ln the piping system shown in Figure 19.3 calculate the total head of the oil at point2, where the absolute pressure and the average velocity of the Ouid are 2 bar and 1.5m/s respectively.Take the density of oil as 850 kgtml.

T

7m

,..\.......,...

<D Datum

Figure 19.3 Aow through a hydraulic pipe-line

Data: p 2 =2 bar= 2 x IOl N/m2; C2 = 1..5 m/s;Z,= 7 m; p = 850 kgtm 3

Solution:

Pressu re head p 2x lOs

-aP 850x9.81

= 24 m

Potential head, Z=7m

Velocity head,-

d = 1.52

28 2x9.81

= 0.1146 m

Tow head at point 2.

Cloaptu/9

\,24 + 7 + 0.1146 = 31.11m

Col!Clusion:

In this

case the velocn:y head is

negligible compared to the pressure head andpotential head.

19.4BERNOULU'SEQUATION

In fluid flow problems the pressure. elevation andvelocity vary from point to point throughout thefluid path. By mrucing suitable assumptions it ispossible to obtain a simple equation relating theabove variables. In 1738 Daniel BemouUi developedthe following energy balance equation applicable tofluid flow: 'During a steady flow offrictionlessincompressiblefluid, the toealenergy (total l-ead) remainsconstantalong the Dow path.'This is theprinciple of conservation of energy. I.e.

-+Z+- =constant pg 2g

whereeach term represents energyper unit weight (IN) of the ftuid:

.£.. = pressure head, mpg

Z =p

o

t

e

n

t

i

a

l

h

e

a

p c

d,

cl- =velocityhead,2g

Applying the aboveconcept to the flow situation illustrat

ed in Figure 19.4 will result in an energy balanceequation.

Figu re 19.4

To define theBernoulliEquation

Liquid flow in pipts 519

Tow energy at section 1 =Tow energy aueetion 2or

Toral bead at section 1 = Tow bead at seclion 2

2 lP a C, P2 C2-+ -+Z = -+-+Z,pg 2g I pg 2g

where p.,p2 =pressure at sections 1 and 2, N/m1

C1, = average fluid velocity all and 2, m/s

Z1, =elevation of P!liniS I and 2, m

Eq uation 19.3 is known as Btrnoulli'stquotion.

(19.3)

EXLlmp19.3

ln a chemical industry a U-rube is used as a siphon 10 drain an acid lllnk as shown inFigure 19.5. If !he diameter of the rube is 10 mm. detesmine !he rate of discharge and!he lowest pressure in !he rube.The specific gravity of !he acid is 1.05.

I O.Sm

2m

1--1

AOdtank

Oatum

2

t?'

3

Figu re 19.5 Aow lhrough asiphon

Data: p1

= p.,.., = 1 bar; z, = 2 m:C1 = 0 m/sp2 = ? :Z, = 2.5 m:p1= p

01 , a 1 bar. Z,a 0; C3 =?

Solution:

(a) Applying Bernoulli's equation between poiniS I and 3:

Total head at section I=Total head at section 3

p, c,2 P1 c2

-+ - +Z = + 3 +Zpg 2 g I pg 2g )

As p = p3 the pressure head terms on bolh sides of !he equation will be cancelled.

..

1

,..

580 Chllpt<r 19

Substituting for lhe other terms,

cl0+2s 3- +0

2 x9.81

On simplification,

Rate of

discharge,

C3 =6.26m/s

V=A1

mil=-x6.26

4= 0.492 x 10"3 ml/s

= 0.492 Vs

lffim

s

(b) Applying Bernoulli's equation between points I and 2:

Total head at section I =Total head at section 2

As lhe area of cross-section of lhe tube remains constant.

<; = s 6.26 rn/s

Density of the acid, p =Specific gravity x Density of water

= 1.05x 1000

= 1050 kg/m3

Substituting lhe values,

lxiO O P1 6.262

.,.. + 2 = + + 2.51050x9.81 1050x9.81 2x 9.81

On simplification.

= 0.743 bar absolute

= 0.257 bar vacuum

--Uquid flow in pi s 581

Conclusion:

At point 2 !he pressure bead is least, but Ibis is compensated for by an increase in the potential head and velocity head.

E:xampk 19.4

In a village water supply system water is pumped to a storage tank on a small hill as shown in Figure 19.6. The pressure gauges at points I and 2 read 3.5 bar and O.S bar respectively. Neglecting all losses, determine lhe discharge lhrough !he pipe-line. The diameter of pipe at points I and 2 are IOOmm and 50mm respectively.

Figure 19.6 Water supply system

Data: p1 = 3.5 bar= 3.5x 10' N/m1;d1 = 0.1 m

p1e 0.5 bar= 0.5 x I()> N/m 1; =0.05 m

Solwion:

Area of now at section I,

Area of now ot section 2,

Applying the equotion of continuity of flow.

581 CluJpttr /9

Substituting !he values,

0.00785 X C1

0.00196 X C2

or

C, =4C1

Applying Bernoulli's equation between pointS I a.nd 2,

Total head at section I =Total head at section 2

2p,-+ -+pg 2g I

1P1 cl

=-+ -+ Zzpg 2g

SubstiiUting !he values,

3.5 x 10'2

c..,::c...,_ O O.S X 10' (4C,) 30--=._:__....::,:'-,--- + + = + +1000x9.81 2 x9.81 1000x9.81 2x9.81

On simplification,

Volume flow rate,

C1 =0.87 m/s

•

v =A,c,= 0.00785 x 0.87 m>

2m

s= 0.00683 ml/s

Nor: The pump has to produce a large pressure head to pu.sh !he water up through !he pipeline. I

:,Bernoulli 's equation (19.3) is applicable onl y for an ideal fluid flow situation. It

·Iassumes that !here is no loss or gain of energy as the fluid flows between sections 1and 2. In !he case of a real fluid there is always some energy loss due to friction as thefluid flows through pipes and fittings. On the other hand. when a fluid flows through Ia pump. energy is added to !he fluid. Similarly. a fluid flowing through a IUrbine givesaway some of itS energy 10 rotate !he machine. All lhese situations may be taken intoaccount by modifying Bernoulli's equation. I

c, z

Il

m

l.fquldflow i•pi[MI 583

Genorator

Energyinput

• 2

P\Jmp - I

t Tant

F1gure 19.7 An experimental apparatuS

Figure 19.7 shows an experimental apparatuS in a fluid mechanics labonuory. Anenergy balance for this resultS in

Totalenergy Energy Energy Energy Totalenergyat section I + added - losses - output = at section 2

l 2Pt Ct P 2 C2- + - + Z + h. = -+-+pg 2g I '" L ... pg 2g

(19.4)

J• where h., • the energy (head) input in m of the liquid columnh.,., = the energy (head) output in m of the liquid columnThL = the energy (head) loss between sections I and 2 in m of the liq uid column

IEzampl19.5

24 littes/min of oil flows through the pipe-line shown in Figure 19.8. The pressuregauge at point I reads 4 bar. What is the pressure at point 2 ? The specific gravity ofoil is 0.82. The diameters at section I and 2 are 12 mm and 25 mm respectively.The friction and other losses between pointS I and 2 add up to 2.3m of liquid column.

- Th -b

l

'

584 Clwptul9 \

2

...

30m

-=+====:::.-Oetum

Figu re 19.8 Flow through a hydra ulic pipeline

Data: p1 = 4 bar+ P..,. = S x lOS N/ml; d 1

=0.012m:

p2 = ?:<1, = 0.02Sm:

V = 24 Vmin:s.g = 0.82: l:hL = 2.3 m

Solution:

Volume Oow rate.V =24Vmin

= 24= 0.0004 ml/s

1000 x 60

Density of oil. p = 0.82 x 1000 = 820 kglml

Referring to Figure 19.8.choosing point I as datum. 2

= 0: Z,= 30m:

A - IIXO.OI2l = 1.131 X 10_. m',- 4

c, 5L., 0.0004 ., 3537 m/sA 1 (1.131x10_. )

C = 5L = 0 00041

2 (· • 0.8149 m/s

4.909 xl 0_.)

1

A

Uquid flow in plpu 585

Using the modified BemouUi's equation (19.4):l l

Pt C1 P2 C:-+ -+ = -+ -+pg 2g 1 pg 2g

Substituting the values,

Sx 5 2

+ 3.537

2

+ 0 -2.3 e ---, P2 + 0.8149 + 30

820X9.81 2x9.81 820x9.81 2x 9.81

62.16 + 0.6376 + 0 -2.3 = 8+ 0.0338+30

On simplification.

p1 = 2.450 x lOS N/m2

=2.4SObar

Conclusion:

The friction and other losses reduce the pressuredownstream.

£Jcampk 19.6

For a water treatment plant a pump is required to deliver 100m'of water per hottr.Thewater pressure at the pump inlet and exit are 0.5 bar and 3.5 bar rupectively.Neglecting thechanges in velocity and elevation, delermine the input power.

t

2

tFlaure 19.9 Water flow through a pump

z -D;.

10..:..:..-

586 Chaptu/9

Data: V = 100 mlJh0.027 8 rrl'/sp

1 = 0.5 x lOS N/m1

; p1= 3.5 x !OS N/m

Solution:

By energy balance under steady state:

Total energy Energyat section I + added

Total enttgy• at sec.tion 2

l !Pt C1 P1 C,-+ -+Z +h .= + ·+Z,pg 2g I "' pg 2g

Cancelling the potentia.l and velocity heads on both sides.

+ h

£..!.=Substituting lhe values,

pg .. pg

5 50.5xi0 +h = 3.5xl0

1000x9.81 "' 1000x9.81

On simplification, hin = 30.58 m

Le.energy input per unit weight of water = 30.58 m of water column.

Mass flow nne. .n =pv= 1000 X 0.0278

= 27.8 kg/s

Weight flow rate of water, w = rhg

= 27.8 X

9.81

= 272.5 N/s

kg m

57

Then energy input per second -w h,. = 272.5 x

30.se-Nm• Nm

= 8333 J/s = 8333 Ws s

i.e. power input = 8.3 kW

2

/

/ Liquid flow in pip.s 587

Conclusion:

The power input is proportional to the wau.r flow rate and the-pressure difference acrossthe pump.

19.5 FLUID VISCOSITY

In the case of an ideal fluid there is no intemal friction or resistance when one layerof fluid flows over another layer. Consequently, when an ideal fluid flows through apipe the velocity is uniform over the entire cross-section. That is, the velocity is the sameat all points in that section.This is illustrated in Figure 19.10.

c

Figure 19.10 Aow of an ideal fluid

In the case of a real fluid there is always some internal friction or resistance to flow.This inu.mal resistance to flow is due to what is known as viscosiry of the fluid. It isa very important fluid property in fluid dynamics. In liquids the viscosity is causedmainly because of the cohesion (anractive fon:oes) between the molecules.

The viscosity or a fluid is a measure of the internal resistance exhibited as one layer of fluid moves relative to another layer as shown in Figure 19.11. The upper layerof the fluid needs some fon:oe, F. to overcome the internal resistanCe. The viscosity of a fluid is defined as the shear stress per unit velocity gradient.

c c

_[7

Figure 19.11 One fluid layer moves relative to another

\.

588 Chapterl9

Let thesliding area of the upper layer be A. Then theshear stressis £..Referring toFigure

A

19.11, the velocity gtadient is. As per the definition, the viscosityis

F-+-=._X-A Y A C

Viscosity is denotedby thesymbolll(greek len.er mu).Several other tennssueh ascoq]icitnJof viscosiry, dynamic viscosity or absolutt viscosity also refer to the fluidviscosity. The unit for viscosity is Ns/m 2• Sometimes the uniiS Poist or t:entiPoise arealso used.

I centiPoise= JO·lNs/ml

In dealillg with problems in fluid now we shall frequently come across the ratio betweenabsolute viscosity and density; this ratio is called the kinematic viscosity and is denotedby the symbol"f (greek letter n u).

Kinematic viscosity,3 2

=r-•-m kg s

(19..5)

Thus the unit for kinematic viscosity is m2/s. Sometimes the uniiS Stookr centiStoke are also used.

I centiStoke= J0-6 m2/s

19.6 REYNOLDS NUMBER

In 1883 Osbourne Reynolds, an English scientist, classified the flow of fluids into twocategories, laminar flow and turbultnt flow, depending upon the characteristics ofthe fluid now.

When the velocity in a pipe is small the fluid molecules move in an orderly fashionas adjacent layers and there is no mixing. In flow through circular pipes the nowpauern constitutes a series of thin shells that are sliding over one another.This conditionis known as laminar flow.This is illustrated in Figure 19.12. At the centre of the pipethe fluid velocity reaches iiS maximum whilst near the pipe wall the velocity is zero.

I

I

Figun 19.12 Laminar now through a circular pipe I

c r v

Ns m m

Liquid flow lnplpts 589

When the velocity is relatively high, eddies ate fonned and lhe mixing of fluidparticles occurs. This situation is known as turbu/t11t flow.The fluid panicles have arandom motion which is transverse to the main flow direction..This IUibulence causesthe velocity of lhe fluid panicles to average out across the cross-section of the pipe.This is illustrated in Figure 19.13.

,ssssssssss ssss sss

ssssssssssssss sss•

-.. c

Fi.gure19.13 Turbulent flow through a circulllt pipe

Reynoldsconducted a seriesof experiments with different pipe diameters and a varietyof fluids. He concluded that the type of flow is dependent on:I. The average velocity of lhe fluid, C.2. The pipe diameter, d.3. The viscosity of lhe fluid·.4. The density of lhe fluid, p.

He combined lhese four variables to fonn a dimensionless parameler known as theReynolds Numbtr (NR).

(19.6)We know

y =p

then

N = Cd

R y(197. )

where p c density of the fluid, kg/m 1

C = average velocity of lhe fluid, mjs d = diameter of the pipe. mI' = dynamic viscosity of the fluid, Ns/m1

y a lcinematic viscosity of the fluid, m1/s

\

590 Cllapru19

In pipe flow, laminar flow exists when Nil is 2000 or less; turilulent flow existsif Nil is 3000 or more. If Nil is between 2000 and 3000 the flow type can not bedetermined and is called transient flow.In practice, pipe flow is generally turilulenL

\Example19.7

In an oil rer111ery diesel oil flows through a 30 em diameter pipe at the rate of700 ml/h. If the kinematic viscosity of the oil is 2.4 Jo-5 m2/s, determine the natureof the flow.

. 700 2Data: d = 0.3 m; V =3600 = 0.1944 mlfs: y = 2.4 10"l m /s

Solution.:

2

A = lt X 0 3

4

vC =A

= 0.07068m2

-m3 -::-y •m-5

= 0·1944 = 2.751 m/s0.07068

By Equation 19.7:

2.751x0.3; -i

2.4x 10

= 34 385

Conclusion:

Since NR is greater than 3000 the flow is turbulent.

!!lmss m

19.7 FRICTION LOSS IN PIPEFLOW

As we have seen, when a fluid flows through a pipe, the fluid panicles near the pipe wallhave a relatively low velocity and the ones near the cenlte (far from the surface) movewith relatively high velocity. Because of this relative motion and the viscosity of thefluid, shear stresses are produced. This viscous action causes energy dissipation whichis usually referred to as pipe friction loss.

Based on experimental results, Henry Darcy in 1857 developed the following

-·-

m s

I

I Liquid flow in pipts 591

empirical equation to detennine fric.tion losses in pipes. ILaccuracy is sufficient for most engineering calculations:

2

"r=fL£..d 2g

•where h

1= friction loss in m of liquid column

L = length of straight pipe. md = pipe diameter, mC = average velocity of fluid, m/sf = friction factor

(19.8)

The value of thefrictionfactor,f,dependson lheReynolds numberandrelativeroughness of the pipe.(Relative roughness is the ratioof surface imperfection tothe innerdiameter of lhe pipe.)In 1944 L. F. Moody developed a chan relating thesevariables (Figure 19.14).

E:rampu 19.8Water flows in a 30 em diameter cast iron pipe roofeuglahtnievses 0.0008. If the waterflow rate is200litre/S findthe frictionhead loss per lOOm of pipe.Takelhe dynamicviscosityof water as 1.49 x J0 3 Nslm2•

Data: d = 0.3 m;

L =lOOm;

. 200 3V=200Us=- =0.2m /s;1000

11 = 1.49 .x 10·3 Nslm 2

Solmion:

ltd2

0 32

A = - = 1t x · -= 0.07CJ7 m'4 4

C = - = - - =2.829 m/S A0.0707

Density of water. p = I 000 kglm3

By Equation 19.6:

NR =p Cd

11

IOOOx2.829x0.3

= 569 000 = 5.69 X J(}l

v 0.2

-

=

(

I Liquid flow in pipes 593

From Moody's chan (Figure 19.14) forNR = 5.69x lOS and relative roughness ofO.OOO 8, the friction factor f = 0.019. ·

By Darcy's equation (19.8).

L c2

h = f--r d 2g

0.019x l00X2.8292

0.3x2x9.81

2 2.58m of water column

Note: The smaller the diameter of the pipe the greater the friction loss.

19.8 MINOR LOSSES IN Pll'ES

Apan from the friction losses in pipes there will be losses d ue to fluid flow in variouspipe-finings such as valves, reducers. bends.etc.In long pipes these losses may be smallcompared to frictional loss and hence are known as minor losses. They are essentiallydue to change in velocity either in direction or magnitude, or both.The sudden changesin velocity produce eddies (see Figure 19.15) and energy is dissipated. As these effects are confmed to shon distanceS in the vicinity of the disturbance, these losses are also referred to >S loco/losses.

Minor losses are in general proportional to the k.inetic energy of the fluid. Inmathematical form,

c2 ho. m2g

c2h

.,=K

2g (1 9.9)

where K, the constant of proponionali y, is known as the loss coefliciem and its valuehas to be evaluated for each source of loss.

19.8.1 Losses dut tosudden enlargement

When the pipe diameter increases abruptly as shown i n Figure 19.15. the fluidexperiences 'shock '. Th is causes the formation of eddies. and consequentlysome energy is lost due to increased local 10rbulence.This head loss can be evaluated using the continuity.momentum and energy principles.

--

594 Chopteci9.·.

::Jflv- "1:::::

8 c,-------- c,- p,-; ---

t= "'

Cross·section of smallor pipo

-------- r--[').-F=

I :<D @

Cross-sectionof larger pipe

Figu re 19.15 Aow lhrough an abrupt pipe expansion

The head loss is given by

h = K-m 2g

11- A )where !he loss coefficient K = ( and C1= velocity in the smaller pipe. (19.10)

19.8.2 Exitloss

When a pipe discharges into a large reservoir as shown in Figure 19.I 6a, some energy is dissipated which can be evaluated using Equation 19.10.

C-:----- --

(bl Free e)(i1

Figure 19.16 Pipe discharging into a large reservoir

In this case A >> A •1 1

On substitution,

(i.e.K=l) (19.11)

Eddies

c:

Uquid flow i•pip<s 595

Thus all the kinetic energy is dissipated by mixing and rurbulence.Even if the discharge isfree as in Figure 19.16b, all the kinetic energy is still losl.

Note:When we say energy is lost. it is not lost in realit)'. It isconverted into thennal energy causing a small temperarure rise in the fluid in the reservoir.

E:uunpu 19.9

Crude oil flows through a tOO mm diameter pipe at the rate of 40 litrels. lf the pipe suddenly enlarges to 200 mm estimate the lossof head due to this abrupt change of section.

c, --+-_.,_. ., c)

Figure 19.17 An abrupt increilSe in flow cross-section

Data: d1 =0.1 m; d, =0.2

m;

. 40V --1000

= 0.04 m)

Solwion:

2 lnd2 n x0.2

A,=-=4 4

= 3 1 ..1x 10''m 2

By Equation

19.10: Loss

coc:fficient ,

C V O.04 = 5.093 m/s1 = A, = (7.854 X 10-l)

2= (• - 7.854 X 10- ) )

31.4xi O -'

= /s

596 Chilptu19.

=0.5625

Then head loss,c>h =K L2g

0..562 Sx 5.0931

= 2x9.81

= 0.7436 m of oil column

A,Nou: The grr.ater the ratio -the greater the loss.

A,

BXDmple 19.10

In an experimentAl apparatus (Figure 19.18) to measure the sboclc losses due to suddenexpansion, the following observations were made:

d1= 10mm; =20mm

(..:.:..;:

l...._

p:...:..:. =20 mm of water column = 0.02m

pg

water flow rate, V = S Vmin

Find the loss coefficient K and compare it with the theoretical value.

Figure 19.18 ExperimentAl apparatustmeasure the head loss due to sudden expansion

1td21 0 012

Data: A 1 =- = "x ·

4 42

=7.854xl()·Sm2

d 0 0 2

A = !:2. = It X ' z = 31.4 X 1O•S m22 4 4

v = 5

=8.333 x IO's m/'s1000x60

ISolution:

•

Uquid flow in pifHS 597

c, =-A,

58.333xto"=

-$7.854xlO

=1.06

1m/S

cl = vAt

8.333x 10"5

;:;

-331.4xl

0

=0.2653m/

s

Applying Bernoulli's equation between points I and 2:

Po

Pt-+-+Z -h =

c; c;

v

-+pg 2g Im pg 2g

Rearranging,

(p

p)

(Cl Cl )

h = I- l+ I-l + ( 7 )

..pg 2g

=.02 + 1.0617

= ,02

Substitutingvalues.

=0.0337 m of walercolumn

h = K 1

.. 2gz,-

d

0.033

S98 Ch.Dpur/9

...,.." ..

whichgiv

K =0.589

By Equation 19.10, the theoretical value of K is

,

.

(

1_

7

.

8

5

4

x

t

o

-5

)31.4xlo-s

= 0.5625

Conclusion:

The experimental value is very close to the theoretical value and the difference couldbe due toexperimental errors.

19.8.3 Loss d ue IO sudden contracllon

Let usconsider an abrupt contraction of a pipe from area A 1toasshown in Figure 19.19. TheDow converges up to the vena contracta(sectioo with the least Dow area)at x-x and thenexpands to the pipe area-

0

'

"

>

"

'

!

)

Ic,

c,

I0

I<D

A

,

<VA,

A,

Figu re 19.19 Flow through a

2

n abrupt pipe conl!action

It is possible to measure A, which can be expressed as a function of ·

w

h

e

r

e

c

,

i

s

k

n

o

w

n

a

s

t

h

e

c

o

e

f

f

i

c

i

e

n

t

o

f

c

o

n

l

n

l

c

t

i

o

n

.

B

y

t

h

e

e

q

u

a

t

i

o

n

o

f

c

o

n

t

i

n

u

i

t

y

,

or

c

.

=

c

,c,

I

Uquid flo"' in pi s 599

It is generally assumed that all of the energy loss occurs when the jet expands fromA, to Az· By Equation 19.10,the head loss is

h =(•- ),c!"' A2 2g

h = (c, -cS"' 2g

where K =(cl, -I r

,c,= 1 -1 _1

c, 2g

c;=K-

2g

(19.12)

C, =velocity in the smaller pipe

The value of c,is a function of A 1

and A,- Experimental analysis gives

c, =0.62 + 0.38(::J (19.13)

The above expression is valid if the ratioA,- is between 0.1 and I. When anA,

approximate value issuffic.ient c,may be taken as0.62and then K =0.375.1!thecontraction is notsudden but isgradual then thevalueofK will besmall and the headloss will be minimal.

19.8.4 Entranceloss

A poorly designed inlet to a pipe can cause an appreciable head loss. For variouscommon inlet conditions the values of K are shown in Figure 19.20. It can be seen thata slight rounding off will reduce the loss drastically.

------L -

-----i_

------L

) , K • OB

( )

_Fo•Figure 19.20 Different pipe inlet conditions and values of Joss coefficient.

600 Chapta19

For a sharp entrance, provided the pipe does not prouude into the reservoir.

h.,=0.5d2g

(19.14}

E;uunpld9.II

Lubricating oil flows through a 10 mm diameter rube. If the flow passage suddenly reduces 10 aS mm diameter.find tlie shock loss.The oil flow rate is 10 litre/min.

c, --t-- --11--4"' C,

II a>I

<D

Figure 19.21 A reducer

1 1

Data: A = ltd,

= nx 0.01

= 7.854xl0"5

m2

1 - - -4 4

2ltd=-=

0.005 • 2- -= 1.964x IO"'m

4 4

V = 101000X60

Solution:

16.7 x 10"5 m'Js

. _,C,== 16.7xl0 -l = 8.48

m/sA2 1.964x 10

By Equation 19.13, the coefficient of contraction.

c< "0.62 + 0.38(::J: 0.62 + 0.38(1.%4 X

IO

_,)2

= 0.62 + 0.024

7.854 X 10-l

nx

Liquid flow inpipu 601

:0.644

By Equation 19.12, the loss coefficient,

2

: ( 1- -1) :0.3050.644

Then head loss,

1

: 0.305 8 48

2x9.81

: 1.12 m of oil column

Nott: The shock Joss due 10 conraelion is always smallu compared 10 the loss due toa corresponding enlargemenL

19.8.5 Losses in pipe-fittings

Invariably a pumping sys1em will have connections which change the size and directionof the pipe. Pipe-fillings such as valves and elbows which constric,t the flow passagesor change the direction of flow cause additional energy losses. It is often convenientto exptess these losses as equivalent 10 the friction loss in a specific lengili of straightpipe of the same diameler. TI1e eq uivalent lengths exptessed as a ratio to the pipediameler for typical finings are given in Table 19.1.Table 19.1 Typical equivalent length for some selectedfinings

Finings

Globe valve (fully open)Gate valve (fully open)(one-quarteropen)

Check valve (fully open) Check valve (with strainer)90 degree standard elbow90 degree long radius elbow Standard T (with flow through ru n) Standard T (wi th flow through branch)

Equivalent length ratioLid

20010

1000ISO40030202060

·

602 Chapttr 19

EX{lltlple19.12

The di.schargpipe from a untrifugal pump is 35 mm in <tiameter and 20 m long. A fully open gale valve and a standard 90" elbow are connecas shown in Figure 19.22. Olivoil of specific gravity 0.92 has to be pumped at the rat of 4 titre/s. Take f 0.032. Find the total losses.

d•3SmmL =20 m

Valve

tFigure 19.22 Details of the discharge pipe line

Data: d= 0.035m; L= 20m; V = 4 Vs

Solwion:

Aowrate, v = 4Vs

= 4-= 0.004 m3/s1000

Velocity in the delivery pipe,

c =-

:...:. .c...,1.. = 4.16 mts35

!lX --4

Length of pipe2:0m.

Equivalent length of a straight pipe: I-for a fully open gate valve = I Od = 10 x 0.035 = 0.35m

- for 90"standard elbow= 30d = 30 x 0.035 = 1.05m

0

I

Uquid flow in pipts 603

Total equivalent length,

20 +0.35 + 1.05 = 21.4 m

Friction losses,

f L C =O.Ol2 21.4x 4.16d 2g 0.03Sx2x9.81

a 17.25 m

Losses due to pipe and fittings = 17.25 m of liquid column.

Loss at pipe exit._c_l = 4.1 6 l =0.88m2g 2x9.81

Tolllllosses on delivery side,

thl.= 17.25 + 0.88= I8. I 3 m

Note: The pump has to input enough energy (pressure) to overcome this loss in additionto the venical lif t.

19.9 ENERGY GRADIE,_'T AND HYDRAULIC GRADIENTLINES

In Section I 9.3 we saw that the total head at any point is givenby

P cl-+-+ Zpg 2g

When the valuesof the tollll head at everjpoint are ploned along theOow path, asshownin Figure 19.23, the resulting graph is called an tnergy gradient lint.It is customary touse the gauge pressure in the above expression.

-

,_ ----

l·b Loss It entrancec-d loss due to sudden con1raction•• lOS$ ll o-..t.

•t vdraulic oredlonl line

Figure 19.23 Energy a nd hydraulic gradient Jines

1

-- -

Clwpter 19

In the reservoir the fluid velocity is negligible and therefore (.E... + Z) is equal to the pgelevation of the free surface. Then there is a sudden drop (a -b) which is due to theentrance loss. As the liquid flows along, the total energy decreases gradually due tothe pipe frictiCin toss.There is a stepped drop (c-d) due to shock at the pipe contraction.Then there is a gradual drop (with greater slope for smaller pipe diameter) due toincreased pipe friction.Finally there is a sudden drop due to the exit loss (e - f).

Another useful graph is the hydraulic gradient line which is the plot of thesum (.E... +Z)pg

along the pipe.This line is therefore paralle-l to the energy gradient line and located below2

it at a venical distance equal to .£._. Also, it is wonhwhile to note that the venical height2g

of the hydraulic gradient line measured above any point on the centreline of the pipe

indicates the static pressure head .E....pg

19.10 PIPES IN SERIES

== c,r-

1) ti c, v-

1---- L,- - ----'- 1--'-- L,--- --1

Figure 19.24 Two pipes in series

When two or more pipes of different diameters are connected in series, the total headloss in the pipeline is the sum of the frictional loss in each pipe plus local shocklosses.Insteady flow the discharge through the pipes is the same.For the seriesconnectionshown in Figure 19.24 the total head loss is:

:EhL = Friction loss +in pipe I

Friction loss + Shock lossin pipe 2 at the contraction

2 2 2r,L,c, r2L2C2 C2

--+- --+K- d12g dl2g 2g

where C 1 = - and C, = - .A1 A2

:=;:: #r

v v

Uquid flow In pip<S 605

E:mmple 19.13

Two reservoirs are connected by a pipe whose total length is 36 m. From the upper reservoir the pipe is 250 ITUll in diameter for a length of 12 m and the remaining 24 m is125 mm in diameter. The entrance and exit of the pipe are sharp and the change ofsection is sudden. The difference in levels of the water in the two reservoirs is 10 m. The friction coefficient f is 0.06 for both pipes, and for !he scdden contrllCtion, K= 0.3.Find !he rate of now.

A$= 1--------------- ---------

IOm

_l-08Figure 19.25 Pipes in series connecting two reservoirs

Data: Pipe 1:d 1 = 0.25m; L

1 = 12m

Pipe 2: <1, = 0.125m; t, = 24m

Z= 10m; f=0.06; K = 0.3; V =?

Solwion:

By continuity of now, A 1 C1 = A2 C2

ltd1'-

4

Sim plifying, 0.252 X C1 = 0.1252 X C2

orc, = 0.25 c,

Loss of head at entrance,

c'0.5

2g =0.5.<:.0. :-2_5:-C.2:..)'2x 9.81

I

1

xc

606 CIIIJptu19

Friction loss in the larger pipe,

Ll 2

=0.06x l2x(0.2SC1 C1f-

dt 2g

Head loss due to sudden contraction,

c2K--2

2g

2)

0.25x2x9.81

c;=0.30x -

2g

Friction loss in the sm:lller pipe,

f ci -

d2 2g=0.06 X

2A lxc;

0.125x2x9.8l

=0.5872C/

Head loss at exit, ci c2g 2x9.81

=0.051 !

Applying Bernoulli's equation between points A a.nd 8,

Considering a datum through the point B.

which gives

Then volume Oow rate,

C2 = 3.88 mls

v --4·c2

0 1252

: It X ·4

X 3.88

=--

nd

Uquidflow in pipes 607

Note; All the potential energy of the water in the first reservoir is lost as it flowsthrough the pipe-line.

19.11 PIPES INPARALLEL

When two or more pipes are connected as shown in Figure 19.26 such that theybranch out from a single point and after equal or uneq ual lengths join at anotherpoint, then the pipes are said to be in parallel. Tn practice a parallel pipe is addedto an existing pipe to increase the discharge.

gv v

.I&. e

it, aPlpo 1L,, d,

Figu re 19.26 Pipes inparallel

Applying the conservation of mass at the junction A,

Appl ying Bernoulli's equation between points A and B (via pipe 1),

PA c Pe-+ -+Z -h - + -+ Zpg 2g A (I J g 2g

ApplyingBemoulli's equation between points A and B (via pipe 2).

PA c Ps c;- + -+Z - hn; -+ -+Z,.pg 2g A pg 2g

Compari ng the above t wo eq uations i t is seen that

--

8

C!

-

608 Clwpr<r 19

Example 19.14

A 300 mm diameter main is required for a town water supply. As pipes over 250 mmdiameter are not readily available, it was decided lO lay two parallel pipes of the samediameter instead. Find thesiz.eofthe pipes to provide thesameflowrateand incur the samehead loss as the 300 mm pipe.

d • 300nvn

v c v

v

(a) A sitlgle 300 mm pipe

.. d,

iJ ( )c,

v, d,

tb) Two parallelpipo$ of sttme $1le

Figure 19:27 Replacing a single pipe with two smaller pipes

Solwion:

Let V =the total volume flow rateC = velocity in 300 mm diameter pipe

volume flow rate in ooe of the parallel pipes1

C1

= velocity in parallel pipes

By conservation of mass.

20.3

ltX - -4

or

d2s2 xn -

I I

4

s. (A)c

If a 300 mm diameter pipe were used the head loss would be

n..d fLC2

h - ---1- d2g - 0.3x2g

When using two parallel pipes of diameter d •

c,

c

--==---

1

v

V =

·/

Uquid flow in pipu

The head losses would be the same in both cases, so

2fLC

0.3x2g

or

Substituting for C /C from (A) in (B),1

d, = 0.3(0. 5Jd = 6.075 X 10"'

d1 =0.227m

Use rwo 227 nun diameter pipes in parallel.

Conclusion:

The rwo 227 mm pipes in parallel will provide the same flow rate as a 300 mm pipe andincur the same frictional loss.

19.12 Sll'HON

A siphon is a closed conduit which conveys liquid witll a free surface from one levelto a lower level via an intermediate higher level as shown in Figure 19.28. Thisarrangement can be used to convey water from a lake over a ridge witllout any externalsource of power.

AB: \. .. \SO mAC: L • .a()OM CO: l •200m

d • SOOmmd • SOOmmd • 700mm

Figure 19.28 Siphon

r 610 Ciwpt<r/9

For a siphon to work, lhe pipe must be filled wilh water. Also the pressure at lhehighest poin1 should not drop below !he vapour pressure.olherwise !here will be vapourformation which will imerrupllhe now.

Example 19.15Two reservoits.whose surface levels differ by 5 m.are connected by a pipe-line which is500 mm in diame.ter for the first400 m and 700mm in diameter for the remaining200m.The pipe-line crosses a ridge whosesummit is2m abovelhe level of and 150m distantfrom lhe higher reservoirasshown in Figure19.28.Taking(=0.04 forbolh pipes andneglecting minor losses, find lhe rate Of now and lhe pressure a11he highest point of lhepipe-line.

Data: Pipe AC = pipe I:d 1 = 0.5m; L 1 =400m

Pipe CO = pipe 2:d,= 0.7m; L,=200m

So/urion:

By conainui1y or now,

2rtd1

-

rtd-xC2

4 I 4

Simplifying,

or <; =0.51 c,

Fric1ion loss in pipe 1.2

L1 C1hn =f- -

d1 2g 400xC2

Friction loss in pipe 2,

= 0.04 X _.:.,:_:...:.!1

O.Sx2x9.81

= 1.631 c,2

t., chn =f-- -

d2 2g

2200x(0.51C )

= 0.04 X .....,....-.:_:... :.._0.7 x2x9.81

=O.ISIS C1

xC =

1

2

,.

Liquid flow in pipes 611

Applying BernouUi 's equation be-tween poinrs Eand F,

Taking point F as the datum,

5 -< 1.631 c,2 + 0.1515 C 1 ) =

0

which gives C1

= 1.675 rn/s

1td2Then the volume flow rate. V = 1

x C4 1

2= JtXO.S X 1.675

4

=0.329m1/s

Applying Bernoulli's equation between points E and B.2

PE P o C1-+Ze -DJL=-+ Z.+

pg pg 2g

Taking the datum line through pointE,

Frictional head loss in pipe AB. 2

h =0.04x 150xl.675r 0.5x 2x9.81

= 1.716m

Substituting in Bernoulli'seq uation,

) 2J .013x!O +0 - 1.7!6 = Po + 2 + 1.675!000x9.81 t000x9.81- 2x9.81

which gives the absolute pressu re,

p0 = 63440 N/m 2 = 0.634

bar

2