TL2101 TL2101 Mekanika Fluida IMekanika Fluida I

Benno RahardyanBenno Rahardyan

Pertemuan Pertemuan

Mg Topik Sub Topik Tujuan Instruksional (TIK)

11 Pengantar Definisi dan sifat-sifat fluida, berbagai jenis fluida yang berhubungan dengan bidang TL

Memahami berbagai kegunaan mekflu dalam bidang TL

Pengaruh tekanan Tekanan dalam fluida, tekanan hidrostatik

Mengerti prinsip-2 tekanan statitka

22 Pengenalan jenis aliran fluida

Aliran laminar dan turbulen, pengembangan persamaan untuk penentuan jenis aliran: bilangan reynolds, freud, dll

Mengerti, dapat menghitung dan menggunakan prinsip dasar aliran staedy state

Idem Idem Idem

33 Prinsip kekekalan energi dalam aliran

Prinsip kontinuitas aliran, komponen energi dalam aliran fluida, penerapan persamaan Bernoulli dalam perpipaan

Mengerti, dapat menggunakan dan menghitung sistem prinsi hukum kontinuitas

44 Idem Idem + gaya pada bidang terendam

Idem

55 Aplikasi kekekalan energi

Aplikasi kekekalan energi dalam aplikasi di bidang TL

Latihan menggunakan prinsip kekekalan eneri khususnya dalam bidang air minum

UTS - -

Pipes are Everywhere!Pipes are Everywhere!

Owner: City of Hammond, INProject: Water Main RelocationPipe Size: 54"

Pipes are Everywhere!Pipes are Everywhere!Drainage PipesDrainage Pipes

PipesPipes

Pipes are Everywhere!Pipes are Everywhere!Water MainsWater Mains

Types of Engineering Types of Engineering ProblemsProblems How big does the pipe have to be How big does the pipe have to be

to carry a flow of to carry a flow of xx m m33/s?/s? What will the pressure in the What will the pressure in the

water distribution system be water distribution system be when a fire hydrant is open?when a fire hydrant is open?

FLUID DYNAMICSTHE BERNOULLI EQUATION

The laws of Statics that we have learned cannot solve Dynamic Problems. There is no way to solve for the flow rate, or Q. Therefore, we need a new dynamic approach

to Fluid Mechanics.

The Bernoulli Equation

By assuming that fluid motion is governed only by pressure and gravity forces, applying Newton’s second law, F = ma, leads us to the Bernoulli Equation.

P/ + V2/2g + z = constant along a streamline

(P=pressure =specific weight V=velocity g=gravity z=elevation)

A streamline is the path of one particle of water. Therefore, at any two points along a streamline, the Bernoulli equation can be applied and, using a set of engineering assumptions, unknown flows and pressures can easily be solved for.

Free Jets

The velocity of a jet of water is clearly related to the depth of water above the hole. The greater the depth, the higher the

velocity. Similar behavior can be seen as water flows at a very high velocity from the reservoir behind the Glen Canyon Dam

in Colorado

Closed Conduit FlowClosed Conduit Flow Energy equationEnergy equation EGL and HGLEGL and HGL Head lossHead loss

– major lossesmajor losses– minor lossesminor losses

Non circular conduitsNon circular conduits

The Energy Line and the Hydraulic Grade LineLooking at the Bernoulli equation again:

P/γ + V2/2g + z = constant on a streamline This constant is called the total head (energy), H

Because energy is assumed to be conserved, at any point along the streamline, the total head is always constant

Each term in the Bernoulli equation is a type of head.

P/γ = Pressure Head

V2/2g = Velocity Head

Z = elevation head

These three heads, summed together, will always equal H

Next we will look at this graphically…

Conservation of Conservation of EnergyEnergy Kinetic, potential, and Kinetic, potential, and

thermal energythermal energy

hL =

Ltp hhzg

Vphz

gVp 2

22

22

1

21

11

22

hp =

ht =

head supplied by a pumphead given to a turbine

mechanical energy converted to thermal

Cross section 2 is ____________ from cross section 1!downstream

Point to point or control volume?Why ? _____________________________________

irreversible

V is average velocity, kinetic energy 2V

Energy Equation Energy Equation AssumptionsAssumptions

hp

p1

1

V12

2g z1 hp

p2

2

V22

2g z2 ht hL

hydrostatic

densitySteady

kinetic

Pressure is _________ in both cross sectionsPressure is _________ in both cross sections– pressure changes are due to elevation onlypressure changes are due to elevation only

section is drawn perpendicular to the section is drawn perpendicular to the streamlines (otherwise the _______ energy streamlines (otherwise the _______ energy term is incorrect)term is incorrect)

Constant ________at the cross sectionConstant ________at the cross section _______ flow_______ flow

EGL (or TEL) and HGLEGL (or TEL) and HGL

velocityhead

elevationhead (w.r.t.

datum)

pressurehead (w.r.t. reference pressure)

zg

VpEGL

2

2

zγp

HGL

downward

lower than reference pressure

The energy grade line must always slope ___________ The energy grade line must always slope ___________ (in direction of flow) unless energy is added (pump)(in direction of flow) unless energy is added (pump)

The decrease in total energy represents the head The decrease in total energy represents the head loss or energy dissipation per unit weightloss or energy dissipation per unit weight

EGL and HGL are coincident and lie at the free EGL and HGL are coincident and lie at the free surface for water at rest (reservoir)surface for water at rest (reservoir)

If the HGL falls below the point in the system for If the HGL falls below the point in the system for which it is plotted, the local pressures are _____ ____ which it is plotted, the local pressures are _____ ____ __________ ________________ ______

Energy equationEnergy equation

z = 0

pump

Energy Grade LineHydraulic G

L

velocity head

pressure head

elevation

datum

z

2g

V2

p

Ltp hhzg

Vphz

gVp 2

22

22

1

21

11

22

static headWhy is static

head important?

The Energy Line and the Hydraulic Grade LineLets first understand this drawing:

Q

Measures the Static

Pressure

Measures the Total Head

12

Z

P/γ

V2/2gEL

HGL

12

1: Static Pressure Tap

Measures the sum of the elevation head and

the pressure Head.

2: Pilot Tube

Measures the Total Head

EL : Energy Line

Total Head along a system

HGL : Hydraulic Grade line

Sum of the elevation and the pressure heads

along a system

The Energy Line and the Hydraulic Grade Line

Q

Z

P/γ

V2/2gEL

HGL

Understanding the graphical approach of Energy Line and the

Hydraulic Grade line is key to understanding what forces are

supplying the energy that water holds.

V2/2g

P/γ

Z

1

2

Point 1:

Majority of energy stored in the water is in the Pressure Head

Point 2:

Majority of energy stored in the water is in the elevation head

If the tube was symmetrical, then the

velocity would be constant, and the

HGL would be level

Bernoulli Equation Bernoulli Equation AssumptionAssumption

constp

g

Vz

2

2

density

Steady

streamline

Frictionless _________ (viscosity can’t be a _________ (viscosity can’t be a significant parameter!)significant parameter!)

Along a __________Along a __________ ______ flow______ flow Constant ________Constant ________ No pumps, turbines, or head lossNo pumps, turbines, or head loss

Why no Does direction matter? ____Useful when head loss is small

point velocityno

Pipe Flow: ReviewPipe Flow: Review

2 21 1 2 2

1 1 2 22 2p t L

p V p Vz h z h h

g g

dimensional analysis

We have the control volume energy We have the control volume energy equation for pipe flow.equation for pipe flow.

We need to be able to predict the We need to be able to predict the relationship between head loss and relationship between head loss and flow.flow.

How do we get this relationship?How do we get this relationship? __________ _______. __________ _______.

Example Pipe Flow Example Pipe Flow ProblemProblem

D=20 cmL=500 m

valve

100 mFind the discharge, Q.

Describe the process in terms of energy!Describe the process in terms of energy!

cs1cs1

cs2cs2

p Vg

z Hp V

gz H hp t l

11

12

12

222

22 2

p Vg

z Hp V

gz H hp t l

11

12

12

222

22 2

zV

gz hl1

22

22 z

Vg

z hl122

22 V g z z hl2 1 22 a fV g z z hl2 1 22 a f

Flow Profile for Flow Profile for Delaware AqueductDelaware Aqueduct

Rondout Reservoir(EL. 256 m)

West Branch Reservoir(EL. 153.4 m)

70.5 km

Sea Level

(Designed for 39 m3/s)

2 21 1 2 2

1 1 2 22 2p t l

p V p Vz H z H h

g g

Need a relationship between flow rate and head loss

1 2lh z z

Ratio of ForcesRatio of Forces

Create ratios of the various forcesCreate ratios of the various forces The magnitude of the ratio will The magnitude of the ratio will

tell us which forces are most tell us which forces are most important and which forces could important and which forces could be ignoredbe ignored

Which force shall we use to Which force shall we use to create the ratios?create the ratios?

Inertia as our Inertia as our Reference ForceReference Force

F=maF=ma Fluids problems (except for statics) Fluids problems (except for statics)

include a velocity (include a velocity (VV), a dimension of ), a dimension of flow (flow (ll), and a density (), and a density ())

Substitute Substitute VV, , ll, , for the dimensions MLT for the dimensions MLT

Substitute for the dimensions of specific Substitute for the dimensions of specific forceforce

F a F

a f f

ML T2 2

L l T M

fi

lV

l3

Vl

2

Dimensionless Dimensionless ParametersParameters

Reynolds NumberReynolds Number

Froude NumberFroude Number

Weber NumberWeber Number

Mach NumberMach Number

Pressure/Drag CoefficientsPressure/Drag Coefficients

– (dependent parameters that we measure experimentally)(dependent parameters that we measure experimentally)

ReVlrm

=

FrV

gl=

2

2C p

p

V

lV

W2

cV

M

AVd

2

Drag2C

2fu

V

l

fg g

2f

l

2

fvE

cl

r=

2

fi

V

l

( )p g zrD + D

Problem solving Problem solving approachapproach

1.1. Identify relevant forces and any other relevant Identify relevant forces and any other relevant parametersparameters

2.2. If inertia is a relevant force, than the non dimensional If inertia is a relevant force, than the non dimensional ReRe, , FrFr, , WW, , M, CpM, Cp numbers can be used numbers can be used

3.3. If inertia isn’t relevant than create new non If inertia isn’t relevant than create new non dimensional force numbers using the relevant forcesdimensional force numbers using the relevant forces

4.4. Create additional non dimensional terms based on Create additional non dimensional terms based on geometry, velocity, or density if there are repeating geometry, velocity, or density if there are repeating parametersparameters

5.5. If the problem uses different repeating variables then If the problem uses different repeating variables then substitute (for example substitute (for example d instead of V)d instead of V)

6.6. Write the functional relationshipWrite the functional relationship

Friction Factor : Major Friction Factor : Major losseslosses Laminar flowLaminar flow

– Hagen-PoiseuilleHagen-Poiseuille Turbulent (Smooth, Transition, Turbulent (Smooth, Transition,

Rough) Rough) – Colebrook FormulaColebrook Formula– Moody diagramMoody diagram– Swamee-JainSwamee-Jain

Laminar Flow Friction Laminar Flow Friction FactorFactor

L

hDV l

32

2

L

hDV l

32

2

2

32gD

LVhl

2

32gD

LVhl

gV

DL

hl 2f

2

g

VDL

hl 2f

2

gV

DL

gDLV

2f

32 2

2

gV

DL

gDLV

2f

32 2

2

RVD6464

f

RVD6464

f

Hagen-Poiseuille

Darcy-Weisbach

Pipe Flow: Pipe Flow: Dimensional AnalysisDimensional Analysis What are the important forces?What are the important forces?

______, ______,________. Therefore ______, ______,________. Therefore ________number and _______________ .________number and _______________ .

What are the important geometric What are the important geometric parameters? _________________________parameters? _________________________– Create dimensionless geometric groupsCreate dimensionless geometric groups

______, ____________, ______ Write the functional relationshipWrite the functional relationship

C p f

Re, ,

l

D D

Inertial

diameter, length, roughness height

Reynolds

l/D

viscous

/D

2

2C

Vp

p

Other repeating parameters?

pressurePressure coefficient

Dimensional Dimensional AnalysisAnalysis How will the results of dimensional How will the results of dimensional

analysis guide our experiments to analysis guide our experiments to determine the relationships that determine the relationships that govern pipe flow?govern pipe flow?

If we hold the other two dimensionless If we hold the other two dimensionless parameters constant and increase the parameters constant and increase the length to diameter ratio, how will Clength to diameter ratio, how will Cpp change?change? ,Rep

DC f

l D

f ,Rep

DC f

l D

2

2C

Vp

p Cp proportional to l

f is friction factor

, ,Rep

lC f

D D

Hagen-Poiseuille

Darcy-Weisbach

Laminar Flow Friction Laminar Flow Friction FactorFactor

2

32lhD

VL

f 2

32 LVh

gD

2

f f2

L Vh

D g

2

2

32f

2

LV L V

gD D g

64 64f

ReVD

Slope of ___ on log-log plot

f 4

128 LQh

gD

-1

Viscous Flow in Viscous Flow in PipesPipes

Two important parameters!Two important parameters!R - Laminar or TurbulentR - Laminar or Turbulent

/D/D - Rough or Smooth - Rough or Smooth

R,

Df

lD

C p

R,

Df

lD

C p

2

2C

Vp

p 2

2C

Vp

p

VDRVDR

Viscous Flow: Viscous Flow: Dimensional AnalysisDimensional Analysis

Where and

Transition at R of 2000Transition at R of 2000

Laminar and Turbulent Laminar and Turbulent FlowsFlows Reynolds apparatusReynolds apparatus

VD

R VD

Rdampingdampinginertiainertia

Boundary layer Boundary layer growth: Transition growth: Transition lengthlength

Pipe Entrance

What does the water near the pipeline wall experience? _________________________Why does the water in the center of the pipeline speed up? _________________________

v v

Drag or shear

Conservation of mass

Non-Uniform Flowv

Need equation for entrance length here

Images - Laminar/Turbulent FlowsImages - Laminar/Turbulent Flows

Laser - induced florescence image of an incompressible turbulent boundary layer

Simulation of turbulent flow coming out of a tailpipe

Laminar flow (Blood Flow)

Laminar flowTurbulent flow

http://www.engineering.uiowa.edu/~cfd/gallery/lim-turb.html

Laminar, Laminar, Incompressible, Incompressible, Steady, Uniform FlowSteady, Uniform Flow Between Parallel PlatesBetween Parallel Plates Through circular tubesThrough circular tubes Hagen-Poiseuille EquationHagen-Poiseuille Equation ApproachApproach

– Because it is laminar flow the shear Because it is laminar flow the shear forces can be quantifiedforces can be quantified

– Velocity profiles can be determined Velocity profiles can be determined from a force balancefrom a force balance

Laminar Flow through Laminar Flow through Circular TubesCircular Tubes Different geometry, same Different geometry, same

equation development (see equation development (see Streeter, et al. p 268)Streeter, et al. p 268)

Apply equation of motion to Apply equation of motion to cylindrical sleeve (use cylindrical cylindrical sleeve (use cylindrical coordinates)coordinates)

Laminar Flow through Laminar Flow through Circular Tubes: Circular Tubes: EquationsEquations

hpdldra

u

4

22

hpdldra

u

4

22

hpdlda

u

4

2

max hpdlda

u

4

2

max

hpdlda

V

8

2

hpdlda

V

8

2

hpdlda

Q 8

4

hpdlda

Q 8

4

Velocity distribution is paraboloid of revolution therefore _____________ _____________

Q = VA =

Max velocity when r = 0

average velocity (V) is 1/2 umax

Vpa2

a is radius of the tubea is radius of the tube

Laminar Flow through Laminar Flow through Circular Tubes: Circular Tubes: DiagramDiagram

Velocity

Shear

hpdldra

u

4

22

hpdldra

u

4

22

hpdl

dr

dr

du

2

hpdl

dr

dr

du

2

hpdl

dr

dr

du 2

hpdl

dr

dr

du 2

l

hr l

2

l

hr l

2

l

dhl

40

l

dhl

40

True for Laminar or Turbulent flow

Shear at the wallShear at the wall

Laminar flow

Laminar flowLaminar flowContinueContinue

Momentum isMass*velocity (m*v)Momentum per unit volume is*vz

Rate of flow of momentum is*vz*dQdQ=vz2πrdrbutvz = constant at a fixed value of r

vz (v2rdr )z vz (v2rdr )

z dz0

Laminar flow

Laminar flowLaminar flowContinueContinue

2r zr rdz 2 (r dr)zr r dr

dzpz2rdr p

z dz2rdr g2rdrdz 0

dvz

dr

Q 2vz dr0

R R4

8p

L

p pz0 pzL gL

Hagen-Poiseuille

The Hagen-Poiseuille The Hagen-Poiseuille EquationEquation

hpdlda

Q 8

4

hpdlda

Q 8

4

hp

dldD

Q

128

4

hp

dldD

Q

128

4

lhzp

zp 2

2

21

1

1

lhzp

zp 2

2

21

1

1

2

2

21

1

1 zp

zp

hl

2

2

21

1

1 zp

zp

hl

hp

hl

hp

hl

L

hDQ l

128

4

L

hDQ l

128

4

L

hh

pdld l

L

hh

pdld l

L

hDV l

32

2

L

hDV l

32

2

cv pipe flow

Constant cross section

Laminar pipe flow equations

h or zh or z

pz

Vg

Hp

zV

gH hp t l

1

11 1

12

2

22 2

22

2 2

Prof. Dr. Ir. Bambang Triatmodjo, CES-Prof. Dr. Ir. Bambang Triatmodjo, CES-UGM :UGM :

Hidraulika I, Beta Ofset Yogyakarta, 1993Hidraulika I, Beta Ofset Yogyakarta, 1993

Hidraulika II, Beta Ofset Yogyakarta, 1993Hidraulika II, Beta Ofset Yogyakarta, 1993

Soal-Penyelesaian Hidraulika I, 1994Soal-Penyelesaian Hidraulika I, 1994

Soal-Penyelesaian Hidraulika II, 1995Soal-Penyelesaian Hidraulika II, 1995

Air mengalir melalui pipa Air mengalir melalui pipa berdiameter 150 mm dan berdiameter 150 mm dan kecepatan 5,5 m/det.Kekentalan kecepatan 5,5 m/det.Kekentalan kinematik air adalah 1,3 x 10kinematik air adalah 1,3 x 10-4-4 m2/det. Selidiki tipe aliranm2/det. Selidiki tipe aliran

turbulenaliranberartiKarena

xx

x

v

VD

reynoldsBilangan

4000Re

1035,6103,1

15,05,5Re

:

56

Minyak di pompa melalui pipa Minyak di pompa melalui pipa sepanjang 4000 m dan diameter 30 sepanjang 4000 m dan diameter 30 cm dari titik A ke titik B. Titik B cm dari titik A ke titik B. Titik B terbuka ke udara luar. Elevasi titik B terbuka ke udara luar. Elevasi titik B adalah 50 di atas titik A. Debit 40 adalah 50 di atas titik A. Debit 40 l/det. Debit aliran 40 l/det. Rapat l/det. Debit aliran 40 l/det. Rapat relatif S=0,9 dan kekentalan relatif S=0,9 dan kekentalan kinematik 2,1 x 10kinematik 2,1 x 10-4-4 m2/det. Hitung m2/det. Hitung tekanan di titik A.tekanan di titik A.

erLaaliranberartiKarena

x

x

v

VD

reynoldsBilangan

dtkmxA

QV

aliranKecepatn

mZZAbawahujung

terhadapBpipaatasujungElevasi

mkgSrelatifRapat

dtkmxvkinematikKekentalan

dtkmQaliranDebit

mLpipaPanjang

cmDpipaDiameter

AB

min2000Re

6,808101,2

3,0566,0Re

:

/566,03,0

4

04,0

:

50:)(

)(

/9009,0:

/101,2:

/04,0:

4000:

30:

4

2

3

24

3

kPap

mNp

xxp

mp

p

VV

hfzg

Vpz

g

Vp

mx

xxx

gD

vVLhf

tenagaKehilangan

A

A

A

A

A

BA

BBB

AAA

574,593

/574,593

81,990023,67

23,67

23,175000

22

23,173,082,9

4000,566,0101,23232

2

22

2

4

2

Minyak dipompa melalui pipa Minyak dipompa melalui pipa berdiameter 25 cm dan panjang 10 berdiameter 25 cm dan panjang 10 km dengan debit aliran 0,02 km dengan debit aliran 0,02 m3/dtk. Pipa terletak miring dengan m3/dtk. Pipa terletak miring dengan kemiringan 1:200. Rapat minyak kemiringan 1:200. Rapat minyak S=0,9 dan keketnalan kinematik S=0,9 dan keketnalan kinematik v=2,1x 10v=2,1x 10-4-4 m2/det. Apabila m2/det. Apabila tekanan pada ujung atas adalah tekanan pada ujung atas adalah p=10 kPA ditanyakan tekanan di p=10 kPA ditanyakan tekanan di ujung bawah.ujung bawah.

erLaaliranberartiKarena

x

x

v

VD

reynoldsBilangan

dtkmxA

QV

aliranKecepatn

NmkPapBBdiTekanan

mkgSrelatifRapat

dtkmxvkinematikKekentalan

dtkmQaliranDebit

pipaKemiringan

mLpipaPanjang

cmDpipaDiameter

min2000Re

485101,2

25,04074,0Re

:

/4074,025,0

4

02,0

:

000.1010:

/9009,0:

/101,2:

/02,0:

200:1:

000.10:

25:

4

2

2

3

24

3

kPap

mNp

xxp

mp

x

p

VV

hfzg

Vpz

g

Vp

mxz

ujungkeduaelevasiSelisih

m

x

xxxx

gD

vVLhf

tenagaKehilangan

A

A

A

A

A

BA

BBB

AAA

642,845

/642,845

81,990078,95

78,95

65,445081,9900

000.100

22

50000.10200

1

:

65,44

25,082,9

100004074,0101,23232

2

22

2

4

2

Turbulent Pipe and Turbulent Pipe and Channel Flow: Channel Flow: OverviewOverview Velocity distributionsVelocity distributions Energy LossesEnergy Losses Steady Incompressible Flow Steady Incompressible Flow

through Simple Pipesthrough Simple Pipes Steady Uniform Flow in Open Steady Uniform Flow in Open

ChannelsChannels

TurbulenceTurbulence A characteristic of the flow. A characteristic of the flow. How can we characterize turbulence?How can we characterize turbulence?

– intensity of the velocity fluctuationsintensity of the velocity fluctuations– size of the fluctuations (length scale)size of the fluctuations (length scale)

meanvelocitymean

velocityinstantaneous

velocityinstantaneous

velocityvelocity

fluctuationvelocity

fluctuationR

uuu uuu uu

uu

Turbulent flowTurbulent flow

When fluid flow at higher flowrates, the streamlines are not steady and straight and the flow is not laminar. Generally, the flow field will vary in both space and time with fluctuations that comprise "turbulence

For this case almost all terms in the Navier-Stokes equations are important and there is no simple solution

P = P (D, , , L, U,)

uz

úz

Uz

average

ur

úr

Ur

average

p

P’

p

average

Time

Turbulent flowTurbulent flow

All previous parameters involved three fundamental dimensions,

Mass, length, and time

From these parameters, three dimensionless groups can be build

P

U 2 f (Re,L

D)

ReUD

inertia

Viscous forces

ReUD

inertia

Viscous forces

Turbulence: Size of the Turbulence: Size of the Fluctuations or EddiesFluctuations or Eddies

Eddies must be smaller than the Eddies must be smaller than the physical dimension of the flowphysical dimension of the flow

Generally the largest eddies are of Generally the largest eddies are of similar size to the smallest dimension of similar size to the smallest dimension of the flowthe flow

Examples of turbulence length scalesExamples of turbulence length scales– rivers: ________________rivers: ________________– pipes: _________________pipes: _________________– lakes: ____________________lakes: ____________________

Actually a spectrum of eddy sizesActually a spectrum of eddy sizes

depth (R = 500)

diameter (R = 2000)depth to thermocline

Turbulence: Flow Turbulence: Flow InstabilityInstability

In turbulent flow (high Reynolds number) the force In turbulent flow (high Reynolds number) the force leading to stability (_________) is small relative to leading to stability (_________) is small relative to the force leading to instability (_______). the force leading to instability (_______).

Any disturbance in the flow results in large scale Any disturbance in the flow results in large scale motions superimposed on the mean flow. motions superimposed on the mean flow.

Some of the kinetic energy of the flow is Some of the kinetic energy of the flow is transferred to these large scale motions (eddies).transferred to these large scale motions (eddies).

Large scale instabilities gradually lose kinetic Large scale instabilities gradually lose kinetic energy to smaller scale motions.energy to smaller scale motions.

The kinetic energy of the smallest eddies is The kinetic energy of the smallest eddies is dissipated by viscous resistance and turned into dissipated by viscous resistance and turned into heat. (=___________)heat. (=___________)

head loss

viscosityviscosity

inertiainertia

Velocity DistributionsVelocity Distributions

Turbulence causes transfer of momentum Turbulence causes transfer of momentum from center of pipe to fluid closer to the from center of pipe to fluid closer to the pipe wall.pipe wall.

Mixing of fluid (transfer of momentum) Mixing of fluid (transfer of momentum) causes the central region of the pipe to causes the central region of the pipe to have relatively _______velocity (compared have relatively _______velocity (compared to laminar flow)to laminar flow)

Close to the pipe wall eddies are smaller Close to the pipe wall eddies are smaller (size proportional to distance to the (size proportional to distance to the boundary)boundary)

constant

Turbulent Flow Velocity Turbulent Flow Velocity ProfileProfile

dy

du dy

du

dy

du dy

du

IIul IIul

dy

dulu II

dy

dulu II

dy

dulI

2 dy

dulI

2

Length scale and velocity of “large” eddies

y

Turbulent shear is from momentum transfer

h = eddy viscosity

Dimensional analysis

Turbulent Flow Turbulent Flow Velocity ProfileVelocity Profile

ylI ylI

dy

duy22

dy

duy22

2

22

dy

duy

2

22

dy

duy

dy

duy

dy

duy

dy

dulI

2 dy

dulI

2 dy

du dy

du

Size of the eddies __________ as we move further from the wall.

increases

k = 0.4 (from experiments)

Log Law for Turbulent, Log Law for Turbulent, Established Flow, Established Flow, Velocity ProfilesVelocity Profiles

5.5ln1 *

*

yu

u

u5.5ln

1 *

*

yu

u

u

0* u

0* u

dy

duy

dy

duy

Iuu * Iuu *

Shear velocity

Integration and empirical results

Laminar Turbulent

x

y

Pipe Flow: The Pipe Flow: The ProblemProblem We have the control volume We have the control volume

energy equation for pipe flowenergy equation for pipe flow We need to be able to predict the We need to be able to predict the

head loss term.head loss term. We will use the results we We will use the results we

obtained using dimensional obtained using dimensional analysisanalysis

Friction Factor : Major Friction Factor : Major losseslosses Laminar flowLaminar flow

– Hagen-PoiseuilleHagen-Poiseuille Turbulent (Smooth, Transition, Turbulent (Smooth, Transition,

Rough) Rough) – Colebrook FormulaColebrook Formula– Moody diagramMoody diagram– Swamee-JainSwamee-Jain

Turbulent Pipe Flow Turbulent Pipe Flow Head LossHead Loss ___________ to the length of the pipe___________ to the length of the pipe Proportional to the _______ of the Proportional to the _______ of the

velocity (almost)velocity (almost) ________ with surface roughness________ with surface roughness Is a function of density and viscosityIs a function of density and viscosity Is __________ of pressureIs __________ of pressure

Proportional

Increases

independent

2

f f2

L Vh

D g

square

(used to draw the Moody diagram)

Smooth, Transition, Smooth, Transition, Rough Rough Turbulent FlowTurbulent Flow

Hydraulically Hydraulically smooth pipe law smooth pipe law (von Karman, 1930)(von Karman, 1930)

Rough pipe law Rough pipe law (von Karman, 1930)(von Karman, 1930)

Transition function Transition function for both smooth for both smooth and rough pipe and rough pipe laws (Colebrook)laws (Colebrook)

1 Re f2log

2.51f

1 2.512log

3.7f Re f

D

1 3.72log

f

D

2

f2f

L Vh

D g

Pipe Flow Energy Pipe Flow Energy LossesLosses

p

hl

p

hl

R,f

Df

L

DC p

R,f

Df

L

DC p

2

2C

Vp

p 2

2C

Vp

p

2

2C

V

ghlp

2

2C

V

ghlp

LD

V

ghl2

2f

LD

V

ghl2

2f

gV

DL

hl 2f

2

g

VDL

hl 2f

2

Horizontal pipe

Dimensional Analysis

Darcy-Weisbach equation

p Vg

z hp V

gz h hp t l

11

12

12

222

22 2

Turbulent Pipe Flow Turbulent Pipe Flow Head LossHead Loss ___________ to the length of the pipe___________ to the length of the pipe ___________ to the square of the ___________ to the square of the

velocity (almost)velocity (almost) ________ with the diameter (almost)________ with the diameter (almost) ________ with surface roughness________ with surface roughness Is a function of density and viscosityIs a function of density and viscosity Is __________ of pressureIs __________ of pressure

Proportional

Proportional

Inversely

Increase

independent

Surface RoughnessSurface Roughness

Additional dimensionless group /D need to be characterize

Thus more than one curve on friction factor-Reynolds number plot

Fanning diagram or Moody diagram

Depending on the laminar region.

If, at the lowest Reynolds numbers, the laminar portion corresponds to f =16/Re Fanning Chart

or f = 64/Re Moody chart

Friction Factor for Smooth, Transition, and Friction Factor for Smooth, Transition, and Rough Turbulent flowRough Turbulent flow

1

f4.0 * log Re* f 0.4

Smooth pipe, Re>3000

1

f4.0 * log

D

2.28

Rough pipe, [ (D/)/(Re√ƒ) <0.01]

1

f4.0 * log

D

2.28 4.0 * log 4.67

D /Re f

1

Transition function for both smooth and rough pipe

f P

L

D

2U 2

f 0.079Re 0.25

Smooth, Transition, Smooth, Transition, Rough Rough Turbulent FlowTurbulent Flow

Hydraulically Hydraulically smooth pipe law smooth pipe law (von Karman, 1930)(von Karman, 1930)

Rough pipe law Rough pipe law (von Karman, 1930)(von Karman, 1930)

Transition function Transition function for both smooth for both smooth and rough pipe and rough pipe laws (Colebrook)laws (Colebrook)

51.2

Relog2

1 f

f

51.2

Relog2

1 f

f

D

f

7.3log2

1

D

f

7.3log2

1

g

V

D

Lfh f

2

2

g

V

D

Lfh f

2

2

(used to draw the Moody diagram)

f

D

f Re

51.2

7.3log2

1

f

D

f Re

51.2

7.3log2

1

Moody DiagramMoody Diagram

0.01

0.10

1E+03 1E+04 1E+05 1E+06 1E+07 1E+08R

fric

tion

fact

or

laminar

0.050.04

0.03

0.020.015

0.010.0080.006

0.004

0.002

0.0010.0008

0.0004

0.0002

0.0001

0.00005

smooth

lD

C pf

lD

C pf

D

D

0.02

0.03

0.04

0.050.06

0.08

Fanning DiagramFanning Diagram

f =16/Re

1

f4.0 * log

D

2.28

1

f4.0 * log

D

2.28 4.0 * log 4.67

D /Re f

1

Swamee-JainSwamee-Jain 19761976 limitationslimitations

/D < 2 x 10/D < 2 x 10-2-2

– Re >3 x 10Re >3 x 1033

– less than 3% less than 3% deviation from deviation from results obtained results obtained with Moody diagramwith Moody diagram

easy to program easy to program for computer or for computer or calculator usecalculator use

5/ 2 f

3/ 2 f

1.782.22 log

3.7

ghQ D

L D ghD

L

0.044.75 5.221.25 9.4

f f

0.66LQ L

D Qgh gh

2

0.9

0.25f

5.74log

3.7 ReD

no f

Each equation has two terms. Why?

fgh

L

L hf

Colebrook Solution for Colebrook Solution for QQ

1 2.512log

3.7f Re f

D

2

f 2 5

8f

LQh

g D

2

2 5f

1 1 8

f

LQ

h g D

Re 4QD

2 5

f 2

4Re f

8

Q g Dh

D LQ

3f21

Re fgh D

L

21 2.51

4 logf 3.7 Re f

D

f2 5 2

8f

h g

D LQ

Colebrook Solution for Colebrook Solution for QQ

2

2

2 5 3f f

1 8 2.514 log

3.7 21

LQ

h g D D gh DL

5/ 2 3f f

2 2.51log

3.7 21

L Q

gh D D gh DL

5/ 2 f3

f

log 2.513.7 22

gh LQ D

L D gh D

Swamee Swamee D?D?

0.045 1/ 4 5 1/52 2 2 21.250.66

Q Q Q QD

g g Q g g

1/ 251/5 1/ 4 1/52 2 25/ 40.66

Q Q QD

g g Q g

2

f 2 5

8f

LQh

g D

25

2

8f

QD

g

25

2

64f

8

QD

g

1/51/ 4 1/52 25/ 4

2

64f

Q Q

g Q g

1/52

2

64f

8

QD

g

1/51/51/ 4 1/52 2 25/ 4

8

Q Q QD

g g Q g

1/51/ 4 1/52 2 25/ 41

f4 4

Q Q

g Q g

Pipe roughnessPipe roughnesspipe materialpipe material pipe roughness pipe roughness (mm) (mm)

glass, drawn brass, copperglass, drawn brass, copper 0.00150.0015

commercial steel or wrought ironcommercial steel or wrought iron 0.0450.045

asphalted cast ironasphalted cast iron 0.120.12

galvanized irongalvanized iron 0.150.15

cast ironcast iron 0.260.26

concreteconcrete 0.18-0.60.18-0.6

rivet steelrivet steel 0.9-9.00.9-9.0

corrugated metalcorrugated metal 4545

PVCPVC 0.120.12

d

d Must be

dimensionless! Must be dimensionless!

Solution TechniquesSolution Techniquesfind head loss given (D, type of pipe, Q)

find flow rate given (head, D, L, type of pipe)

find pipe size given (head, type of pipe,L, Q)0.044.75 5.22

1.25 9.4

f f

0.66LQ L

D Qgh gh

2

2 5

8ff

LQh

g D2

0.9

0.25f

5.74log

3.7 ReD

Re 4QD

5/ 2 f3

f

log 2.513.7 22

gh LQ D

L D gh D

Exponential Friction Exponential Friction FormulasFormulas

f

n

m

RLQh

D=

units SI

675.10

units USC727.4

n

n

C

CR

1.852

f 4.8704

10.675 SI units

L Qh

D Cæ ö=è ø

C = Hazen-Williams coefficient

range of data

Commonly used in commercial Commonly used in commercial and industrial settingsand industrial settings

Only applicable over _____ __ ____ Only applicable over _____ __ ____ collectedcollected

Hazen-Williams exponential Hazen-Williams exponential friction formulafriction formula

Head loss:Head loss:Hazen-Williams Hazen-Williams CoefficientCoefficient

CC ConditionCondition

150150 PVCPVC

140140 Extremely smooth, straight pipes; asbestos Extremely smooth, straight pipes; asbestos cementcement

130130 Very smooth pipes; concrete; new cast ironVery smooth pipes; concrete; new cast iron

120120 Wood stave; new welded steelWood stave; new welded steel

110110 Vitrified clay; new riveted steelVitrified clay; new riveted steel

100100 Cast iron after years of useCast iron after years of use

9595 Riveted steel after years of useRiveted steel after years of use

60-8060-80 Old pipes in bad conditionOld pipes in bad condition

Hazen-Hazen-Williams Williams vsvs

Darcy-Darcy-WeisbachWeisbach

1.852

f 4.8704

10.675 SI units

L Qh

D C

2

f 2 5

8f

LQh

g D

preferred

Both equations are empiricalBoth equations are empirical Darcy-Weisbach is dimensionally Darcy-Weisbach is dimensionally

correct, and ________.correct, and ________. Hazen-Williams can be considered Hazen-Williams can be considered

valid only over the range of gathered valid only over the range of gathered data.data.

Hazen-Williams can’t be extended to Hazen-Williams can’t be extended to other fluids without further other fluids without further experimentation.experimentation.

Non-Circular Conduits:Non-Circular Conduits:Hydraulic Radius Hydraulic Radius ConceptConcept

A is cross sectional areaA is cross sectional area P is wetted perimeterP is wetted perimeter RRhh is the “Hydraulic Radius” is the “Hydraulic Radius”

(Area/Perimeter)(Area/Perimeter) Don’t confuse with radius!Don’t confuse with radius!

2

f2f

L Vh

D g=

2

f f4 2h

L Vh

R g=

2

44h

DA DR

P D

p

p= = = 4 hD R=

For a pipe

We can use Moody diagram or Swamee-Jain with D = 4Rh!

Pipe Flow Summary (1)Pipe Flow Summary (1)

Shear increases _________ with Shear increases _________ with distance from the center of the pipe distance from the center of the pipe (for both laminar and turbulent flow)(for both laminar and turbulent flow)

Laminar flow losses and velocity Laminar flow losses and velocity distributions can be derived based on distributions can be derived based on momentum and energy conservationmomentum and energy conservation

Turbulent flow losses and velocity Turbulent flow losses and velocity distributions require ___________ distributions require ___________ resultsresults

linearly

experimental

Pipe Flow Summary (2)Pipe Flow Summary (2)

Energy equation left us with the elusive Energy equation left us with the elusive head loss termhead loss term

Dimensional analysis gave us the form of Dimensional analysis gave us the form of the head loss term (pressure coefficient)the head loss term (pressure coefficient)

Experiments gave us the relationship Experiments gave us the relationship between the pressure coefficient and the between the pressure coefficient and the geometric parameters and the Reynolds geometric parameters and the Reynolds number (results summarized on Moody number (results summarized on Moody diagram)diagram)

QuestionsQuestions Can the Darcy-Weisbach equation Can the Darcy-Weisbach equation

and Moody Diagram be used for and Moody Diagram be used for fluids other than water? _____fluids other than water? _____Yes

No

Yes

Yes

What about the Hazen-Williams equation? ___

Does a perfectly smooth pipe have head loss? _____

Is it possible to decrease the head loss in a pipe by installing a smooth liner? ______

Darcy Weisbach

Major and Minor LossesMajor Losses:

Hmaj = f x (L/D)(V2/2g)

f = friction factor L = pipe length D = pipe diameterV = Velocity g = gravity

Minor Losses:Hmin = KL(V2/2g)

Kl = sum of loss coefficients V = Velocity g = gravity

When solving problems, the loss terms are added to the system at the second point

P1/γ + V12/2g + z1 = P2/γ + V2

2/2g + z2 + Hmaj + Hmin

Hitung kehilangan tenaga karena gesekan di Hitung kehilangan tenaga karena gesekan di dalam pipa sepanjang 1500 m dan diameter dalam pipa sepanjang 1500 m dan diameter 20 cm, apabila air mengalir dengan 20 cm, apabila air mengalir dengan kecepatan 2 m/det. Koefisien gesekan kecepatan 2 m/det. Koefisien gesekan f=0,02f=0,02

Penyelesaian : Penyelesaian : Panjang pipa : L = 1500 mPanjang pipa : L = 1500 mDiameter pipa : D = 20 cm = 0,2 mDiameter pipa : D = 20 cm = 0,2 mKecepatan aliran : V = 2 m/dtkKecepatan aliran : V = 2 m/dtkKoefisien gesekan f = 0,02Koefisien gesekan f = 0,02

m

xx

x

g

V

D

Lfhf

tenagaKehilangan

58,30

81,922,0

2150002,0

22

2

Air melalui pipa sepanjang 1000 m dan Air melalui pipa sepanjang 1000 m dan diameternya 150 mm dengan debit 50 diameternya 150 mm dengan debit 50 l/det. Hitung kehilangan tenaga l/det. Hitung kehilangan tenaga karenagesekan apabila koefisien karenagesekan apabila koefisien gesekan f = 0,02gesekan f = 0,02

Penyelesaian : Penyelesaian : Panjang pipa : L = 1000 mPanjang pipa : L = 1000 mDiameter pipa : D = 0,15 mDiameter pipa : D = 0,15 mDebit aliran : Q = 50 liter/detikDebit aliran : Q = 50 liter/detikKoefisien gesekan f = 0,02Koefisien gesekan f = 0,02 m

xx

xx

QDg

Lfhf

tenagaKehilangan

4,54

)015,0(81,9

100002,0802,0

8

22

552

Hitung kehilangan tenaga karena gesekan di Hitung kehilangan tenaga karena gesekan di dalam pipa sepanjang 1500 m dan diameter dalam pipa sepanjang 1500 m dan diameter 20 cm, apabila air mengalir dengan 20 cm, apabila air mengalir dengan kecepatan 2 m/det. Koefisien gesekan kecepatan 2 m/det. Koefisien gesekan f=0,02f=0,02

Penyelesaian : Penyelesaian : Panjang pipa : L = 1500 mPanjang pipa : L = 1500 mDiameter pipa : D = 20 cm = 0,2 mDiameter pipa : D = 20 cm = 0,2 mKecepatan aliran : V = 2 m/dtkKecepatan aliran : V = 2 m/dtkKoefisien gesekan f = 0,02Koefisien gesekan f = 0,02

m

xx

x

g

V

D

Lfhf

tenagaKehilangan

58,30

81,922,0

2150002,0

22

2

Air melalui pipa sepanjang 1000 m dan Air melalui pipa sepanjang 1000 m dan diameternya 150 mm dengan debit 50 diameternya 150 mm dengan debit 50 l/det. Hitung kehilangan tenaga l/det. Hitung kehilangan tenaga karenagesekan apabila koefisien karenagesekan apabila koefisien gesekan f = 0,02gesekan f = 0,02

Penyelesaian : Penyelesaian : Panjang pipa : L = 1000 mPanjang pipa : L = 1000 mDiameter pipa : D = 0,15 mDiameter pipa : D = 0,15 mDebit aliran : Q = 50 liter/detikDebit aliran : Q = 50 liter/detikKoefisien gesekan f = 0,02Koefisien gesekan f = 0,02

m

xx

xx

QDg

Lfhf

tenagaKehilangan

4,54

)015,0(81,9

100005,0802,0

8

52

5

552

ExampleSolve for the Pressure Head, Velocity Head, and Elevation Head at each point, and then plot the Energy Line and the Hydraulic

Grade Line

1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Assumptions and Hints:

P1 and P4 = 0 --- V3 = V4 same diameter tube

We must work backwards to solve this problem

R = .5’ R = .25’

1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Point 1:

Pressure Head : Only atmospheric P1/γ = 0

Velocity Head : In a large tank, V1 = 0 V12/2g = 0

Elevation Head : Z1 = 4’

R = .5’ R = .25’

1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Point 4:

Apply the Bernoulli equation between 1 and 4 0 + 0 + 4 = 0 + V4

2/2(32.2) + 1

V4 = 13.9 ft/s

Pressure Head : Only atmospheric P4/γ = 0

Velocity Head : V42/2g = 3’

Elevation Head : Z4 = 1’

R = .5’ R = .25’

1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Point 3:

Apply the Bernoulli equation between 3 and 4 (V3=V4) P3/62.4 + 3 + 1 = 0 + 3 + 1

P3 = 0

Pressure Head : P3/γ = 0

Velocity Head : V32/2g = 3’

Elevation Head : Z3 = 1’

R = .5’ R = .25’

1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Point 2:

Apply the Bernoulli equation between 2 and 3 P2/62.4 + V2

2/2(32.2) + 1 = 0 + 3 + 1

Apply the Continuity Equation

(Π.52)V2 = (Π.252)x13.9 V2 = 3.475 ft/s

P2/62.4 + 3.4752/2(32.2) + 1 = 4 P2 = 175.5 lbs/ft2

R = .5’ R = .25’

Pressure Head :P2/γ = 2.81’

Velocity Head : V2

2/2g = .19’

Elevation Head : Z2 = 1’

Plotting the EL and HGLEnergy Line = Sum of the Pressure, Velocity and Elevation

heads

Hydraulic Grade Line = Sum of the Pressure and Velocity heads

EL

HGL

Z=1’Z=1’Z=1’

V2/2g=3’V2/2g=3’

Z=4’

P/γ =2.81’

V2/2g=.19’

Pipe Flow and the Energy Equation

For pipe flow, the Bernoulli equation alone is not sufficient. Friction loss along the pipe, and momentum loss through

diameter changes and corners take head (energy) out of a system that theoretically conserves energy. Therefore, to

correctly calculate the flow and pressures in pipe systems, the Bernoulli Equation must be modified.

P1/γ + V12/2g + z1 = P2/γ + V2

2/2g + z2 + Hmaj + Hmin

Major losses: Hmaj

Major losses occur over the entire pipe, as the friction of the fluid over the pipe walls removes energy from the system. Each

type of pipe as a friction factor, f, associated with it. Hmaj

Energy line with no losses

Energy line with major losses

1 2

Pipe Flow and the Energy EquationMinor Losses : Hmin

Momentum losses in Pipe diameter changes and in pipe bends are called minor losses. Unlike major losses, minor losses do not occur over the length of the pipe, but only at points of momentum loss. Since Minor losses occur at unique points along a pipe, to find the total minor loss throughout a pipe,

sum all of the minor losses along the pipe. Each type of bend, or narrowing has a loss coefficient, KL to go with it.

Minor Losses

Minor LossesMinor Losses

We previously obtained losses through We previously obtained losses through an expansion using conservation of an expansion using conservation of energy, momentum, and massenergy, momentum, and mass

Most minor losses can not be obtained Most minor losses can not be obtained analytically, so they must be analytically, so they must be measuredmeasured

Minor losses are often expressed as a Minor losses are often expressed as a loss coefficient, K, times the velocity loss coefficient, K, times the velocity head.head. g

VKh

2

2

g

VKh

2

2

R,geometryfC p R,geometryfC p 2

2C

Vp

p 2

2C

Vp

p

2

2C

V

ghlp

2

2C

V

ghlp

g

Vh pl

2C

2

g

Vh pl

2C

2

High RHigh R

Head Loss: Minor Head Loss: Minor LossesLosses

potential thermalVehicle drag Hydraulic jump

Vena contracta Minor losses!

Head loss due to Head loss due to outlet, inlet, bends, elbows, valves, pipe size outlet, inlet, bends, elbows, valves, pipe size changeschanges

Flow expansions have high lossesFlow expansions have high losses– Kinetic energy decreases across expansionKinetic energy decreases across expansion– Kinetic energy Kinetic energy ________ and _________ energy ________ and _________ energy– Examples – ________________________________ Examples – ________________________________

____________________________________________________________________________________ Losses can be minimized by gradual Losses can be minimized by gradual

transitionstransitions

Minor LossesMinor Losses

Most minor losses can not be Most minor losses can not be obtained analytically, so they obtained analytically, so they must be measuredmust be measured

Minor losses are often expressed Minor losses are often expressed as a loss coefficient, K, times the as a loss coefficient, K, times the velocity head.velocity head.

2

2l

Vh K

g=

( )geometry,RepC f=

2

2C

V

ghlp

g

Vh pl

2C

2

High Re

Head Loss due to Head Loss due to Gradual Expansion Gradual Expansion (Diffusor)(Diffusor)

g

VVKh EE

2

221

g

VVKh EE

2

221

diffusor angle ()

00.10.20.30.40.50.60.70.8

0 20 40 60 80

KE

2

1

22

2 12

AA

gV

Kh EE

2

1

22

2 12

AA

gV

Kh EE

Sudden ContractionSudden Contraction

losses are reduced with a gradual contractionlosses are reduced with a gradual contraction

g

V

Ch

c

c

21

1 22

2

g

V

Ch

c

c

21

1 22

2

2A

AC c

c 2A

AC c

c

V1V2

flow separation

Sudden ContractionSudden Contraction

0.60.650.7

0.750.8

0.850.9

0.951

0 0.2 0.4 0.6 0.8 1

A2/A1

Cc

hC

Vgc

c

FHG

IKJ

11

2

2

22

Q CA ghorifice orifice 2

g

VKh ee

2

2

g

VKh ee

2

2

0.1eK 0.1eK

5.0eK 5.0eK

04.0eK 04.0eK

Entrance LossesEntrance Losses

Losses can be Losses can be reduced by reduced by accelerating the accelerating the flow gradually flow gradually and eliminating and eliminating thethevena contracta

Head Loss in BendsHead Loss in Bends Head loss is a Head loss is a

function of the ratio of function of the ratio of the bend radius to the the bend radius to the pipe diameter (R/D)pipe diameter (R/D)

Velocity distribution Velocity distribution returns to normal returns to normal several pipe several pipe diameters diameters downstreamdownstream

High pressure

Low pressure

Possible separation from wall

D

g

VKh bb

2

2

g

VKh bb

2

2

Kb varies from 0.6 - 0.9

R

Head Loss in ValvesHead Loss in Valves Function of valve type and Function of valve type and

valve positionvalve position The complex flow path The complex flow path

through valves can result in through valves can result in high head loss (of course, high head loss (of course, one of the purposes of a one of the purposes of a valve is to create head loss valve is to create head loss when it is not fully open)when it is not fully open)

g

VKh vv

2

2

g

VKh vv

2

2

Solution TechniquesSolution Techniques

Neglect minor lossesNeglect minor losses Equivalent pipe lengthsEquivalent pipe lengths Iterative TechniquesIterative Techniques Simultaneous EquationsSimultaneous Equations Pipe Network SoftwarePipe Network Software

Iterative Techniques Iterative Techniques for D and Q (given for D and Q (given total head loss)total head loss) Assume all head loss is major head Assume all head loss is major head

loss.loss. Calculate D or Q using Swamee-Calculate D or Q using Swamee-

Jain equationsJain equations Calculate minor lossesCalculate minor losses Find new major losses by Find new major losses by

subtracting minor losses from total subtracting minor losses from total head loss head loss

Solution Technique: Solution Technique: Head LossHead Loss Can be solved directlyCan be solved directly

minorfl hhh minorfl hhh

g

VKhminor

2

2

g

VKhminor

2

2

5

2

2

8

D

LQ

gfh f

5

2

2

8

D

LQ

gfh f

2

9.0Re

74.5

7.3log

25.0

D

f

2

9.0Re

74.5

7.3log

25.0

D

f

42

28

Dg

QKhminor

42

28

Dg

QKhminor

D

Q4Re

D

Q4Re

Solution Technique:Solution Technique:Discharge or Pipe Discharge or Pipe DiameterDiameter Iterative techniqueIterative technique Set up simultaneous equations in Set up simultaneous equations in

ExcelExcel

minorfl hhh minorfl hhh

42

28

Dg

QKhminor

42

28

Dg

QKhminor

5

2

2

8

D

LQ

gfh f

5

2

2

8

D

LQ

gfh f

2

9.0Re

74.5

7.3log

25.0

D

f

2

9.0Re

74.5

7.3log

25.0

D

f

D

Q4Re

D

Q4Re

Use goal seek or Solver to find discharge that makes the calculated head loss equal the given head loss.

Example: Minor and Example: Minor and Major LossesMajor Losses

Find the maximum dependable flow between the Find the maximum dependable flow between the reservoirs for a water temperature range of 4ºC reservoirs for a water temperature range of 4ºC to 20ºC. to 20ºC.

Water

2500 m of 8” PVC pipe

1500 m of 6” PVC pipeGate valve wide open

Standard elbows

Reentrant pipes at reservoirs

25 m elevation difference in reservoir water levels

Sudden contraction

DirectionsDirections

Assume fully turbulent (rough pipe Assume fully turbulent (rough pipe law)law)– find f from Moody (or from von Karman)find f from Moody (or from von Karman)

Find total head lossFind total head loss Solve for Q using symbols (must Solve for Q using symbols (must

include minor losses) (no iteration include minor losses) (no iteration required)required)

Obtain values for minor losses from Obtain values for minor losses from notes or textnotes or text

Example (Continued)Example (Continued)

What are the Reynolds number in the What are the Reynolds number in the two pipes?two pipes?

Where are we on the Moody Diagram?Where are we on the Moody Diagram? What value of K would the valve have to What value of K would the valve have to

produce to reduce the discharge by produce to reduce the discharge by 50%?50%?

What is the effect of temperature?What is the effect of temperature? Why is the effect of temperature so Why is the effect of temperature so

small?small?

Example (Continued)Example (Continued)

Were the minor losses negligible?Were the minor losses negligible? Accuracy of head loss calculations?Accuracy of head loss calculations? What happens if the roughness What happens if the roughness

increases by a factor of 10?increases by a factor of 10? If you needed to increase the flow If you needed to increase the flow

by 30% what could you do?by 30% what could you do? Suppose I changed 6” pipe, what is Suppose I changed 6” pipe, what is

minimum diameter needed?minimum diameter needed?

Pipe Flow Summary (3)Pipe Flow Summary (3)

Dimensionally correct equations fit to Dimensionally correct equations fit to the empirical results can be the empirical results can be incorporated into computer or calculator incorporated into computer or calculator solution techniquessolution techniques

Minor losses are obtained from the Minor losses are obtained from the pressure coefficient based on the fact pressure coefficient based on the fact that the pressure coefficient is _______ that the pressure coefficient is _______ at high Reynolds numbersat high Reynolds numbers

Solutions for discharge or pipe diameter Solutions for discharge or pipe diameter often require iterative or computer often require iterative or computer solutionssolutions

constant

Loss Coefficients Use this table to find loss coefficients:

Head Loss due to Sudden Head Loss due to Sudden Expansion:Expansion:Conservation of EnergyConservation of Energy

1 2

ltp hHg

Vz

pH

gV

zp

22

22

222

22

111

1

1

lhgVVpp

2

21

2221

gVVpp

hl 2

22

2121

z1 = z2

What is p1 - p2?

Apply in direction of flow

Neglect surface shear

Divide by (A2 )

Head Loss due to Sudden Head Loss due to Sudden Expansion:Expansion:Conservation of Momentum Conservation of Momentum

Pressure is applied over all of section 1.Momentum is transferred over area corresponding to upstream pipe diameter.V1 is velocity upstream.

sspp FFFWMM 2121

1 2

xx ppxx FFMM2121

12

11 AVM x 22

22 AVM x

222122

212

1 ApApAVAV

gAA

VVpp 2

121

22

21

AA11

AA22

xx

Energy

Head Loss due Head Loss due to Sudden to Sudden ExpansionExpansion

g

VVpphl

2

22

2121

gAA

VVpp 2

121

22

21

1

2

2

1

VV

AA

gVV

gVV

VVhl 2

22

211

221

22

g

VVVVhl 2

2 2121

22

gVV

hl 2

221

2

2

12

1 12

AA

gV

hl

2

2

11

AA

K

Momentum

Mass

ContractionContraction

V1 V2

EGL

HGL

vena contracta

gV

Kh cc 2

2

2

losses are reduced with a gradual contractionlosses are reduced with a gradual contraction

Expansion!!!

Questions:Questions:

In the rough pipe law region if the flow In the rough pipe law region if the flow rate is doubled (be as specific as rate is doubled (be as specific as possible)possible)– What happens to the major head loss?What happens to the major head loss?– What happens to the minor head loss?What happens to the minor head loss?

Why do contractions have energy loss?Why do contractions have energy loss? If you wanted to compare the importance If you wanted to compare the importance

of minor vs. major losses for a specific of minor vs. major losses for a specific pipeline, what dimensionless terms could pipeline, what dimensionless terms could you compare?you compare?

Entrance LossesEntrance Losses Losses can be Losses can be

reduced by reduced by accelerating the accelerating the flow gradually flow gradually and eliminating and eliminating the the vena vena contractacontracta

Ke 0.5

Ke 1.0

Ke 0.04

heKe

V 2

2g

reentrant

Head Loss in ValvesHead Loss in Valves Function of valve type Function of valve type

and valve positionand valve position The complex flow path The complex flow path

through valves often through valves often results in high head results in high head lossloss

What is the maximum What is the maximum value that value that KKvv can can have? _____have? _____

hv Kv

V 2

2g

How can K be greater than 1?

QuestionsQuestions

What is the What is the head loss when head loss when a pipe enters a a pipe enters a reservoir?reservoir?

Draw the EGL Draw the EGL and HGLand HGL

V

g

V

2

2

EGLHGL

2

2

11

AA

K

ExampleExample

D=40 cmL=1000 m

D=20 cmL=500 m

valve100 m

Find the discharge, Q.What additional information do you need?Apply energy equationHow could you get a quick estimate? _________________Or spreadsheet solution: find head loss as function of Q.

Use S-J on small pipe

ltp hHg

Vz

pH

gV

zp

22

22

222

22

111

1

1

cs1

cs2

22100

2 l

Vm h

g= +

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?γoil= 8.82 kN/m3

f = .035

If oil flows from the upper to lower reservoir at a velocity of 1.58 m/s in the 15 cm diameter smooth pipe, what is the elevation of the oil surface in the

upper reservoir?

Include major losses along the pipe, and the minor losses associated with the entrance, the two bends,

and the outlet.

Kout=1

r/D = 0

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?γoil= 8.82 kN/m3

f = .035

Kout=1

r/D = 0

Apply Bernoulli’s equation between points 1 and 2:Assumptions: P1 = P2 = Atmospheric = 0 V1 = V2 = 0 (large

tank)

0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin

Hmaj = (fxLxV2)/(Dx2g)=(.035 x 197m x (1.58m/s)2)/(.15 x 2 x 9.8m/s2)

Hmaj= 5.85m

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?γoil= 8.82 kN/m3

f = .035

Kout=1

r/D = 0

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + Hmin

Hmin= 2KbendV2/2g + KentV2/2g + KoutV2/2g

From Loss Coefficient table: Kbend = 0.19 Kent = 0.5 Kout = 1

Hmin = (0.19x2 + 0.5 + 1) x (1.582/2x9.8)

Hmin = 0.24 m

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?γoil= 8.82 kN/m3

f = .035

Kout=1

r/D = 0

0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + 0.24m

Z1 = 136.09 meters

Pipa ekivalenPipa ekivalen

Digunakan untuk Digunakan untuk menyederhanakan menyederhanakan sistem yang sistem yang ditinjauditinjau

Ciri khasnya adalah memiliki Ciri khasnya adalah memiliki keserupaan hidrolis dengan keserupaan hidrolis dengan kondisi nyatanya kondisi nyatanya Q, hf sama Q, hf sama

Pipa ekivalen dapat dinyatakan Pipa ekivalen dapat dinyatakan melalui ekivalensi l,D,fmelalui ekivalensi l,D,f