fluid-mechanics-benno-new.ppt

7/27/2019 fluid-mechanics-benno-new.ppt

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TL2101

Mekanika Fluida I

Benno Rahardyan

Pertemuan


2/131

Mg Topik Sub Topik Tujuan Instruksional (TIK)

1 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, tekananhidrostatik Mengerti prinsip-2tekanan statitka

2 Pengenalan jenisaliran 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

3 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

4 Idem Idem + gaya pada bidang terendam Idem

5 Aplikasi

kekekalan

energi

Aplikasi kekekalan energi dalam

aplikasi di bidang TL

Latihan menggunakan

prinsip kekekalan

eneri khususnya

dalam bidang air

minum

- -


3/131

Pipes are Everywhere!

Owner: City ofHammond, IN

Project: Water Main

Relocation

Pipe Size: 54"


4/131


Drainage Pipes


5/131

Pipes


6/131


Water Mains


7/131

Types of Engineering

Problems How big does the pipe have to be to

carry a flow ofxm3/s?

What will the pressure in the waterdistribution system be when a firehydrant is open?


8/131

FLUID DYNAMICS

THE BERNOULLI EQUATION

The laws of Statics that we have learned cannot solveDynamic Problems. There is no way to solve for the flow

rate, or Q. Therefore, we need a new dynamic approach

to Fluid Mechanics.


9/131

The Bernoulli Equation

By assuming that fluid motion is governed only by pressure and

gravity forces, applying Newtons second law, F = ma, leads us tothe Bernoulli Equation.

P/g+ V2

/2g + z =constant along a streamline(P=pressure g =specific weight V=velocity g=gravity z=elevation)

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


10/131

Free Jets

The velocity of a jet of water is clearly related to the depth of waterabove the hole. The greater the depth, the higher the velocity. Similarbehavior can be seen as water flows at a very high velocity from the

reservoir behind the Glen Canyon Dam in Colorado


11/131

Closed Conduit Flow Energy equation

EGL and HGL

Head loss

major losses

minor losses

Non circular conduits


12/131

The Energy Line and the Hydraulic Grade Line

Looking at the Bernoulli equation again:

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

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

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

P/ = 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

22


13/131

Conservation of Energy Kinetic, potential, and thermal

energy

hL=

Ltp hhzg

Vphz

g

Vp 2

2

22

21

2

11

1

22

g

g

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 2

V


14/131

Energy Equation

Assumptions

hp g

p1

g

1V1

2

2g

z1 hp p2

g

2V2

2

2g

z2 ht hL

hydrostatic

density

Steady

kinetic

Pressure is _________ in both cross sections

pressure changes are due to elevation only

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

Constant ________at the cross section

_______ flow


15/131

EGL (or TEL) and HGL

velocity

head

elevationhead (w.r.t.

datum)

pressurehead (w.r.t.

reference pressure)

zg

VpEGL

2

2

g

z

pHGL

downward

lower than reference pressure

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

The decrease in total energy represents the head loss or

energy dissipation per unit weight EGL and HGL are coincident and lie at the free surface for

water at rest (reservoir)

If the HGL falls below the point in the system for which it

is plotted, the local pressures are _____ ____ __________


16/131

Energy equation

z = 0

pump

Energy Grade Line

Hydraulic G Lvelocity head

pressurehead

elevation

datum

z

2g

V2

g

p

Ltp hhzg

Vphzg

Vp

2

2

2221

2

11122

g

g

static headWhy is static

head important?


17/131

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

Q

Measures theStatic Pressure

Measures theTotal Head

1 2

Z

P/

V2/2gEL

HGL

12

1: Static Pressure Tap

Measures the sum of theelevation head and the

pressure Head.

2: Pilot Tube

Measures the Total Head

EL : Energy Line

Total Head along a systemHGL : Hydraulic Grade line

Sum of the elevation andthe pressure heads along a

system


18/131

The Energy Line and the Hydraulic Grade Line

Q

Z

P/

V2/2g

EL

HGL

Understanding the graphical approach ofEnergy Line and the Hydraulic Grade line is

key to understanding what forces aresupplying the energy that water holds.

V2/2g

P/

Z

1

2

Point 1:

Majority of energystored in the water is in

the Pressure Head

Point 2:Majority of energy

stored in the water is inthe elevation head

If the tube wassymmetrical, then the

velocity would beconstant, and the HGL

would be level


19/131

Bernoulli Equation

Assumption

constp

g

Vz

g2

2

density

Steady

streamline

Frictionless_________ (viscosity cant be asignificant parameter!)

Along a __________

______ flow

Constant ________

No pumps, turbines, or head lossWhy no ____________

Does direction matter? ____

Useful when head loss is small

point velocity

no


20/131

Pipe Flow: Review

2 2

1 1 2 21 1 2 2

2 2

p t L

p V p Vz h z h h

g g

g g

dimensional analysis

We have the control volume energyequation for pipe flow.

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

How do we get this relationship?__________ _______.


21/131

Example Pipe Flow

Problem

D=20 cm

L=500 mvalve

100 mFind the discharge, Q.

Describe the process in terms of energy!

cs1

cs2

p Vg

z H p Vg

z H hp t l1 1 12

12

22

2

22 2g

g

zV

g

z hl12

2

2

2

V g z z hl2 1 22


22/131

Flow Profile for Delaware

AqueductRondout Reservoir(EL. 256 m)

West Branch Reservoir

(EL. 153.4 m)

70.5 km

Sea Level

(Designed for 39 m3/s)

2 2

1 1 2 21 1 2 2

2 2p t l

p V p Vz H z H h

g g

g g

Need a relationship between flow rate and head loss

1 2lh z z


23/131

Ratio of Forces

Create ratios of the various forces

The magnitude of the ratio will tell us

which forces are most important andwhich forces could be ignored

Which force shall we use to create the

ratios?


24/131

Inertia as our Reference

Force F=ma

Fluids problems (except for statics) include a

velocity (V), a dimension of flow (l), and adensity (r)

Substitute V, l, r for the dimensions MLT

Substitute for the dimensions of specific force

F a rF

a

f r f M

L T2 2

L l T M

fi

lV

rl3

r

V

l

2


25/131

Dimensionless

Parameters Reynolds Number

Froude Number Weber Number

Mach Number

Pressure/Drag Coefficients

(dependent parameters that we measure experimentally)

ReVlr

m=

Fr

V

gl=

2

2Cp

p

Vr

rlVW

2

cVM

AVd

2

Drag2C

r

2f

u

V

l

fg gr

2f

l

2

f vEc

l

r=

2

fiV

lr

( )p g zrD + D


26/131

Problem solving approach

1. Identify relevant forces and any other relevant parameters

2. If inertia is a relevant force, than the non dimensional Re, Fr,W, M, Cp numbers can be used

3. If inertia isnt relevant than create new non dimensional forcenumbers using the relevant forces

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

5. If the problem uses different repeating variables then

substitute (for example wd instead of V)6. Write the functional relationship


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Friction Factor : Major

losses Laminar flow

Hagen-Poiseuille

Turbulent (Smooth, Transition, Rough) Colebrook Formula

Moody diagram

Swamee-Jain


28/131

Laminar Flow Friction

FactorL

hDV l

g

32

2

2

32

gD

LVhl

r

g

V

D

Lhl

2f

2

g

V

D

L

gD

LV

2f

32 2

2

r

RVD

6464f

r

Hagen-Poiseuille

Darcy-Weisbach


29/131

Pipe Flow: Dimensional

Analysis What are the important forces?

______, ______,________. Therefore________number and _______________ .

What are the important geometricparameters? _________________________ Create dimensionless geometric groups

______, ______

Write the functional relationship

Cp f

Re, ,l

D D

Inertial

diameter, length, roughness height

Reynolds

l/D

viscous

/D

2

2C

V

pp

r

Other repeating parameters?

pressure

Pressure coefficient


30/131

Dimensional

Analysis How will the results of dimensional analysis

guide our experiments to determine the

relationships that govern pipe flow? If we hold the other two dimensionless

parameters constant and increase thelength to diameter ratio, how will Cp

change?, Rep

DC f

l D

f , RepD

C f

l D

2

2C

V

pp

r

Cp proportional to l

f is friction factor

, , Repl

C fD D


31/131

Hagen-Poiseuille

Darcy-Weisbach

Laminar Flow Friction

Factor232

lhDVL

g

f 2

32 LVh gD

r

2

f f2

L Vh

D g

2

2

32f

2

LV L V

gD D g

r

64 64f

ReVD

r

Slope of ___ on log-log plot

f 4

128 LQ

h gDr

-1


32/131

Viscous Flow inPipes


33/131

Two important parameters!R - Laminar or Turbulent

/D- Rough or Smooth

R,

Df

l

DCp

2

2C

V

pp

r

rVDR

Viscous Flow:

Dimensional AnalysisWhere and


34/131

Transition at Rof 2000

Laminar and Turbulent

Flows Reynolds

apparatus

rVDR

dampinginertia
http://localhost/var/www/apps/conversion/tmp/scratch_10//civil9.civil.tamu.edu/kchang$/Clips/V8_1.mov


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Boundary layer growth:

Transition length

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 Flow

v

Need equation for entrance length here


36/131

Images - 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


37/131

Laminar, Incompressible,

Steady, Uniform Flow Between Parallel Plates

Through circular tubes

Hagen-Poiseuille EquationApproach

Because it is laminar flow the shear

forces can be quantifiedVelocity profiles can be determined froma force balance


38/131

Laminar Flow through

Circular Tubes Different geometry, same equation

development (see Streeter, et al. p

268)Apply equation of motion to cylindrical

sleeve (use cylindrical coordinates)


39/131


Circular Tubes: Equations

hpdl

drau g

4

22

hpdl

dau g

4

2

max

hpdl

daV g

8

2

hpdl

daQ g

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 tube


40/131


Circular Tubes: Diagram

Velocity

Shear

hpdl

drau g

4

22

hpdldr

drdu g

2

hpdl

dr

dr

dug

2

l

hr l

2

g

l

dhl4

0

g True for Laminar or

Turbulent flow

Shear at the wall

Laminar flow

L i fl


41/131

Laminar flowContinue

Momentum is

Mass*velocity (m*v)

Momentum per unit volume is

r*vzRate of flow of momentum isr*vz* dQdQ=vz2rdr

butvz= constant at a fixed value of r

rvz(v2rdr) z rvz(v2rdr) zdz 0Laminar flow


42/131

Laminar flow

Continue2rzr rdz2(r dr)zr rdrdzp z2rdr p zdz2rdr rg2rdrdz 0

dvz

dr

Q 2vzdr0R R

4

8pL

p pz 0 pzL rgL

Hagen-Poiseuille


43/131

The Hagen-Poiseuille

Equation

hpdl

daQ g

8

4

h

p

dl

dDQ

g

g

128

4

lhzp

zp

22

21

1

1

gg

2

2

21

1

1 zp

zp

hlgg

h

phl

g

L

hDQ l

g

128

4

L

hh

p

dl

d l

g

L

hDV l

g

32

2

cv pipe flow

Constant cross section

Laminar pipe flow equations

h orz

pz

V

gH

pz

V

gH hp t l

1

1

1 11

2

2

2

2 22

2

2 2g

g


44/131


45/131

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

Hidraulika I, Beta Ofset Yogyakarta, 1993

Hidraulika II, Beta Ofset Yogyakarta, 1993

Soal-Penyelesaian Hidraulika I, 1994

Soal-Penyelesaian Hidraulika II, 1995


46/131

Air mengalir melalui pipa berdiameter

150 mm dan kecepatan 5,5m/det.Kekentalan kinematik air adalah1,3 x 10-4 m2/det. Selidiki tipe aliran

turbulenaliranberartiKarena

xx

x

v

VD

reynoldsBilangan

4000Re

1035,6103,1

15,05,5Re

:

5

6


47/131

Minyak di pompa melalui pipasepanjang 4000 m dan diameter 30

cm dari titik A ke titik B. Titik Bterbuka ke udara luar. Elevasi titik Badalah 50 di atas titik A. Debit 40 l/det.

Debit aliran 40 l/det. Rapat relatifS=0,9 dan kekentalan kinematik 2,1 x10-4 m2/det. Hitung tekanan di titik A.


48/131

erLaaliranberartiKarena

x

x

v

VD

reynoldsBilangan

dtkm

xA

QV

aliranKecepatn

mZZAbawahujung

terhadapBpipaatasujungElevasi

mkgSrelatifRapat

dtkmxvkinematikKekentalan

dtkmQaliranDebit

mLpipaPanjang

cmDpipaDiameter

AB

min2000Re

6,808101,2

3,0566,0Re

:

/566,0

3,04

04,0

:

50:)(

)(

/9009,0:

/101,2:

/04,0:

4000:

30:

4

2

3

24

3

r

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

g

g

gg


49/131

Minyak dipompa melalui pipaberdiameter 25 cm dan panjang 10 km

dengan debit aliran 0,02 m3/dtk. Pipaterletak miring dengan kemiringan1:200. Rapat minyak S=0,9 dankeketnalan kinematik v=2,1x 10-4

m2/det. Apabila tekanan pada ujungatas adalah p=10 kPA ditanyakantekanan di ujung bawah.


50/131

erLaaliranberartiKarena

x

x

v

VDreynoldsBilangan

dtkm

xA

QV

aliranKecepatn

NmkPapBBdiTekanan

mkgSrelatifRapat

dtkmxvkinematikKekentalan

dtkmQaliranDebit

pipaKemiringan

mLpipaPanjang

cmDpipaDiameter

min2000Re

485101,2

25,04074,0Re

:

/4074,0

25,04

02,0

:

000.1010:

/9009,0:

/101,2:

/02,0:

200:1:

000.10:

25:

4

2

2

3

24

3

r

kPap

mNp

xxp

mp

x

p

VV

hfzg

Vpzg

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

g

g

gg


51/131

Turbulent Pipe andChannel Flow: Overview

Velocity distributions

Energy Losses

Steady Incompressible Flow throughSimple Pipes

Steady Uniform Flow in Open Channels


52/131

Turbulence

A characteristic of the flow.

How can we characterize turbulence?

intensity of the velocity fluctuations

size of the fluctuations (length scale)

mean

velocity

instantaneous

velocity

velocity

fluctuationtuuu

u

u


53/131

Turbulent 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, , r, L , U,)

uz

z

Uzaverage

ur

r

Ur

average

p

P

paverage

Time
http://localhost/var/www/apps/conversion/tmp/scratch_10//civil9.civil.tamu.edu/kchang$/Clips/V8_1.movhttp://localhost/var/www/apps/conversion/tmp/scratch_10//civil9.civil.tamu.edu/kchang$/Clips/V8_1.mov


54/131

Turbulent flow

All previous parameters involved three fundamental dimensions,

Mass, length, and time

From these parameters, three dimensionless groups can be build

PrU

2 f(Re,L

D)

Re rUD

inertia

Viscous forces


55/131

Turbulence: Size of theFluctuations or Eddies

Eddies must be smaller than the physicaldimension of the flow

Generally the largest eddies are of similar size

to the smallest dimension of the flow Examples of turbulence length scales

rivers: ________________

pipes: _________________ lakes: ____________________

Actually a spectrum of eddy sizes

depth (R = 500)

diameter (R = 2000)depth to thermocline


56/131

Turbulence: FlowInstability

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

Any disturbance in the flow results in large scalemotions superimposed on the mean flow.

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

Large scale instabilities gradually lose kinetic energyto smaller scale motions.

The kinetic energy of the smallest eddies isdissipated by viscous resistance and turned into heat.

(=___________)head loss

viscosity

inertia


57/131

Velocity Distributions

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

Mixing of fluid (transfer of momentum) causesthe central region of the pipe to have relatively_______velocity (compared to laminar flow)

Close to the pipe wall eddies are smaller (size

proportional to distance to the boundary)

constant

Turbulent Flow Velocity


58/131

Turbulent Flow VelocityProfile

dy

du

dydu

IIulr

dy

dulu II

dy

dulI

2r

Length scale and velocity of large eddies

y

Turbulent shear is from momentum transfer

h = eddy viscosity

Dimensional analysis

b l l l i


59/131

Turbulent Flow VelocityProfile

ylI

dy

duy 22r

2

22

dy

duyr

dy

duy

r

dy

dulI

2r

dy

du

Size of the eddies __________ as we

move further from the wall.

increases

k = 0.4 (from experiments)

Log Law for Turbulent,


60/131

Log Law for Turbulent,Established Flow, Velocity

Profiles

5.5ln1 *

*

yu

u

u

r

0* u

dy

duy

r

Iuu *

Shear velocity

Integration and empirical results

Laminar Turbulent

x

y


61/131

Pipe Flow: The Problem

We have the control volume energyequation for pipe flow

We need to be able to predict thehead loss term.

We will use the results we obtained

using dimensional analysis

F i i F M j


62/131

Friction Factor : Majorlosses

Laminar flow

Hagen-Poiseuille

Turbulent (Smooth, Transition, Rough) Colebrook Formula

Moody diagram

Swamee-Jain

T b l t Pi Fl H d


63/131

Turbulent Pipe Flow HeadLoss

___________ to the length of the pipe

Proportional to the _______ of the velocity

(almost) ________ with surface roughness

Is a function of density and viscosity

Is __________ of pressure

Proportional

Increases

independent

2

f f2

L Vh

D g

square

Smooth Transition Rough


64/131

(used to draw the Moody diagram)

Smooth, Transition, RoughTurbulent Flow

Hydraulically smoothpipe law (vonKarman, 1930)

Rough pipe law (vonKarman, 1930)

Transition function

for both smooth andrough pipe laws(Colebrook)

1 Re f 2log

2.51f

1 2.512log

3.7f Re f

D

1 3.72log

f

D

2

f2

f

L Vh

D g


65/131

Pipe Flow Energy Losses

gphl

R,f

Df

L

DCp

2

2CV

pp

r

2

2CV

ghlp

L

D

V

ghl2

2f

g

V

D

Lhl

2

f2

Horizontal pipe

Dimensional Analysis

Darcy-Weisbach equation

p V

gz h

p V

gz h hp t l

11

1

2

12

22

2

22 2g

g

T b l t Pi Fl H d


66/131

Turbulent Pipe Flow HeadLoss

___________ to the length of the pipe

___________ to the square of the

velocity (almost)________ with the diameter (almost)

________ with surface roughness

Is a function of density and viscosity

Is __________ of pressure

Proportional

Proportional

Inversely

Increase

independent


67/131

Surface Roughness

Additional dimensionless group /D needto be characterize

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

Fanning diagram or Moody diagramDepending on the laminar region.

If, at the lowest Reynolds numbers, the laminar portion

corresponds to f =16/Re Fanning Chart

orf= 64/Re Moody chart

Friction Factor for Smooth Transition and


68/131

Friction Factor for Smooth, Transition, and

Rough Turbulent flow

1

f 4.0 * log Re* f 0.

Smooth pipe, Re>3000

1

f 4.0 * log

D

2.28

Rough pipe, [ (D/)/(Re)


69/131

Smooth, Transition, RoughTurbulent Flow

Hydraulically smoothpipe law (vonKarman, 1930)

Rough pipe law (vonKarman, 1930)

Transition function

for both smooth andrough pipe laws(Colebrook)

51.2

Relog2

1 f

f

D

f

7.3log2

1

g

V

D

Lfhf

2

2

(used to draw the Moody diagram)

f

D

f Re

51.2

7.3log2

1


70/131

Moody Diagram

0.01

0.10

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

frictionf

actor

laminar

0.05

0.04

0.03

0.02

0.015

0.010.0080.006

0.004

0.002

0.0010.0008

0.0004

0.0002

0.0001

0.00005

smooth

l

DCpf

D

0.02

0.03

0.04

0.05

0.06

0.08


71/131

Fanning Diagram

f=16/Re

1

f 4.0 * log

D

2.28

1

f 4.0* log

D

2.28 4.0* log 4.67

D /

Re f1


72/131

Swamee-Jain

1976

limitations

/D < 2 x 10-2

Re >3 x 103 less than 3% deviation

from results obtainedwith Moody diagram

easy to program forcomputer orcalculator use

5 / 2 f

3/ 2 f

1.782.22 log

3.7

ghQ D

L D ghD

L

0.04

4.75 5.22

1.25 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

hf


73/131

Colebrook Solution for Q

1 2.512log

3.7f Re f

D

2

f 2 5

8f

LQh

g D

2

2 5

f

1 1 8f

LQh g D

Re 4Q

D

2 5

f 2

4Re f

8

Q g Dh

D LQ

3

f21

Re fgh D

L

2

1 2.514 logf 3.7 Re f

D

f

2 5 2

8f

h g

D LQ


74/131

Colebrook Solution for Q

2

2

2 5 3f f

1 8 2.514 log

3.7 21

LQ

h g D D gh D

L

5 / 2 3f f

2 2.51log

3.7 21

L Q

gh D D gh D

L

5 / 2 f

3f

log 2.513.7 22

gh LQ D

L D gh D

Swamee0.04

5 1/ 4 5 1/ 52 2 2 2Q Q Q Q


75/131

SwameeD?

2 2 2 21.250.66

Q Q Q QD

g g Q g g

1/251/ 5 1/ 4 1/ 5

2 2 25 / 40.66

Q Q QD

g g Q g

2

f 2 5

8f

LQh

g D

25

2

8f QDg

25

2

64f

8

QD

g

1/ 51/ 4 1/ 5

2 25 / 4

2

64f

Q Q

g Q g

1/ 5

2

2

64f

8

QD

g

1/ 51/ 5

1/ 4 1/52 2 2

5/ 4

8

Q Q QD

g g Q g

1/ 51/ 4 1/ 52 2 2

5 / 41f4 4

Q Q

g Q g


76/131

Pipe roughness

pipe material pipe roughness (mm)

glass, drawn brass, copper 0.0015

commercial steel or wrought iron 0.045

asphalted cast iron 0.12

galvanized iron 0.15

cast iron 0.26

concrete 0.18-0.6rivet steel 0.9-9.0

corrugated metal 45

PVC 0.12

d Must be

dimensionless!


77/131

Solution Techniques

find 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)0.04

4.75 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 4Q

D

5 / 2 f

3

f

log 2.513.7 22

gh LQ D

L D gh D

Exponential Friction


78/131

Exponential FrictionFormulas

f

n

m

RLQh

D=

unitsSI

675.10

unitsUSC727.4

n

n

C

CR

1.852

f 4.8704

10.675SI units

L Qh

D C

=

C = Hazen-Williams coefficient

range of data

Commonly used in commercial andindustrial settings

Only applicable over _____ __ ____collected

Hazen-Williams exponential friction

formula

Head loss:


79/131

Head loss:Hazen-Williams Coefficient

C Condition

150 PVC

140 Extremely smooth, straight pipes; asbestos

cement

130 Very smooth pipes; concrete; new cast iron

120 Wood stave; new welded steel

110 Vitrified clay; new riveted steel

100 Cast iron after years of use

95 Riveted steel after years of use

60-80 Old pipes in bad condition

Hazen-Williams1.852

f 4.8704

10.675SI units

L Qh

D C


80/131

vs

Darcy-Weisbach

D C 2

f 2 5

8f

LQh

g D

preferred

Both equations are empirical

Darcy-Weisbach is dimensionally correct,

and ________. Hazen-Williams can be considered valid only

over the range of gathered data.

Hazen-Williams cant be extended to otherfluids without further experimentation.

Non Circular Conduits:


81/131

Non-Circular Conduits:Hydraulic Radius Concept

A is cross sectional area

P is wetted perimeter

Rhis the Hydraulic Radius(Area/Perimeter)

Dont confuse with radius!

2

f2

f

L Vh

D g=

2

f f4 2h

L VhR g

=

2

44

h

DA DR

P D

p

= = = 4 hD R=For a pipe

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


82/131

Pipe Flow Summary (1)

Shear increases _________ withdistance from the center of the pipe (for

both laminar and turbulent flow) Laminar flow losses and velocity

distributions can be derived based on

momentum and energy conservation Turbulent flow losses and velocity

distributions require ___________

results

linearly

experimental


83/131


Energy equation left us with the elusivehead loss term

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

Experiments gave us the relationshipbetween the pressure coefficient and the

geometric parameters and the Reynoldsnumber (results summarized on Moodydiagram)


84/131

Questions

Can the Darcy-Weisbach equation andMoody Diagram be used for fluids other

than water? _____YesNo

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? ______


85/131

Darcy Weisbach

Major and Minor Losses


86/131

Major 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 thesecond point

P1/ + V12/2g + z1 = P2/ + V2

2/2g + z2 + Hmaj + Hmin


87/131

Hitung kehilangan tenaga karena gesekan di

dalam pipa sepanjang 1500 m dan diameter20 cm, apabila air mengalir dengankecepatan 2 m/det. Koefisien gesekanf=0,02

Penyelesaian :

Panjang pipa : L = 1500 m

Diameter pipa : D = 20 cm = 0,2 m

Kecepatan aliran : V = 2 m/dtk

Koefisien gesekan f = 0,02 mxxx

g

V

D

Lfhf

tenagaKehilangan

58,30

81,922,02150002,0

2

2

2


88/131

Air melalui pipa sepanjang 1000 m dan

diameternya 150 mm dengan debit 50 l/det.Hitung kehilangan tenaga karenagesekanapabila koefisien gesekan f = 0,02

Penyelesaian :Panjang pipa : L = 1000 mDiameter pipa : D = 0,15 mDebit aliran : Q = 50 liter/detik

Koefisien gesekan f = 0,02m

xx

xx

QDg

Lfhf

tenagaKehilangan

4,54

)015,0(81,9

100002,0802,0

8

22

5

52


89/131

Hitung kehilangan tenaga karena gesekan di

dalam pipa sepanjang 1500 m dan diameter20 cm, apabila air mengalir dengankecepatan 2 m/det. Koefisien gesekanf=0,02

Penyelesaian :

Panjang pipa : L = 1500 m

Diameter pipa : D = 20 cm = 0,2 m

Kecepatan aliran : V = 2 m/dtk

Koefisien gesekan f = 0,02 mxxx

g

V

D

Lfhf

tenagaKehilangan

58,30

81,922,02150002,0

2

2

2


90/131

Air melalui pipa sepanjang 1000 m dan

diameternya 150 mm dengan debit 50 l/det.Hitung kehilangan tenaga karenagesekanapabila koefisien gesekan f = 0,02

Penyelesaian :Panjang pipa : L = 1000 mDiameter pipa : D = 0,15 mDebit aliran : Q = 50 liter/detik

Koefisien gesekan f = 0,02m

xx

xx

QDg

Lfhf

tenagaKehilangan

4,54

)015,0(81,9

100005,0802,0

8

52

5

5

52

Example


91/131

Solve for the Pressure Head, Velocity Head, and Elevation Head at eachpoint, and then plot the Energy Line and the Hydraulic Grade Line

1

23 4

1

4

H2O= 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


92/131

1

23 4

1

4

H2O= 62.4 lbs/ft3

Point 1:

Pressure Head : Only atmospheric P1/ = 0

Velocity Head : In a large tank,V1 = 0 V12/2g = 0Elevation Head : Z1= 4

R = .5R = .25

Point 4:


93/131

1

23 4

1

4

H2O= 62.4 lbs/ft3

Apply the Bernoulli equation between 1 and 40 + 0 + 4 = 0 + V4

2/2(32.2) + 1

V4 = 13.9 ft/s

Pressure Head : Only atmospheric P4/ = 0

Velocity Head :V42/2g = 3

Elevation Head : Z4= 1

R = .5R = .25

Point 3:


94/131

1

23 4

1

4

H2O= 62.4 lbs/ft3

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

P3 = 0

Pressure Head : P3/ = 0

Velocity Head :V32/2g = 3

Elevation Head : Z3= 1

R = .5R = .25

Point 2:


95/131

1

23 4

1

4

H2O= 62.4 lbs/ft3

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

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

Apply the Continuity Equation

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

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

2

R = .5R = .25

Pressure Head :P2/= 2.81

Velocity Head :V22/2g = .19

Elevation Head :Z2= 1

Plotting the EL and HGLE Li S f h P V l i d El i h d


96/131

Energy Line = Sum of the Pressure, Velocity and Elevation heads

Hydraulic Grade Line = Sum of the Pressure and Velocity heads

EL

HGL

Z=1Z=1Z=1

V2/2g=3V2/2g=3

Z=4

P/=2.81

V2/2g=.19

Pipe Flow and the Energy EquationFor pipe flow, the Bernoulli equation alone is not sufficient. Friction loss


97/131

p p , qalong 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 pipesystems, the Bernoulli Equation must be modified.

P1/ + V12/2g + z1 = P2/ + V2

2/2g + z2 + Hmaj + Hmin

Major losses: Hmaj

Major losses occur over the entire pipe, as the friction of the fluid overthe 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 Equation


98/131

Minor Losses : Hmin

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

MinorLosses

i


99/131

Minor Losses

We previously obtained losses through anexpansion using conservation of energy,momentum, and mass

Most minor losses can not be obtainedanalytically, so they must be measured

Minor losses are often expressed as a loss

coefficient, K, times the velocity head.

g

VKh

2

2

R,geometryfCp 22

C

V

pp

r

2

2C

V

ghlp

g

Vh pl

2

C2

High R

H d L Mi L


100/131

Head Loss: Minor Losses

potential thermal

Vehicle drag Hydraulic jumpVena contracta Minor losses!

Head loss due tooutlet, inlet, bends, elbows, valves, pipe sizechanges

Flow expansions have high losses Kinetic energy decreases across expansion

Kinetic energy ________ and _________ energy

Examples __________________________________________________________________________

Losses can be minimized by gradual transitions

Mi L


101/131

Minor Losses

Most minor losses can not be obtainedanalytically, so they must be measured

Minor losses are often expressed as aloss coefficient, K, times the velocityhead.

2

2l

Vh K

g=

( )geometry,RepC f=

2

2C

V

ghlp

g

Vh pl

2C

2

High Re

Head Loss due to Gradual


102/131

Head Loss due to GradualExpansion (Diffusor)

g

VVKh EE

2

2

21

diffusor angle ( )

00.10.20.30.40.50.60.70.8

0 20 40 60 80

KE

2

1

2

2

2 12

A

Ag

VKh EE


103/131

Sudden Contraction

losses are reduced with a gradual contraction

g

V

Ch

cc 2

11 22

2

2A

AC cc

V1 V2

flow separation

S dd C t ti


104/131

Sudden Contraction

0.6

0.65

0.7

0.75

0.80.85

0.9

0.95

1

0 0.2 0.4 0.6 0.8 1

A2/A1

Cc

hC

V

gc

c

HG KJ

11

2

2

2

2

Q CA ghorifice orifice 2

E t L


105/131

g

VKh ee

2

2

0.1eK

5.0eK

04.0eK

Entrance Losses

Losses can bereduced byaccelerating the

flow gradually andeliminating the

vena contracta


106/131

Head Loss in Bends

Head loss is afunction of the ratio

of the bend radius tothe pipe diameter(R/D)

Velocity distributionreturns to normalseveral pipediameters

High pressure

Low pressure

Possible

separation

from wall

D

g

VKh bb

2

2

Kb

varies from 0.6 - 0.9

R


107/131

Head Loss in Valves

Function of valve type and valveposition

The complex flow path throughvalves can result in high headloss (of course, one of thepurposes of a valve is to create

head loss when it is not fullyopen)

g

VKh vv

2

2

S l ti T h i


108/131

Solution Techniques

Neglect minor losses

Equivalent pipe lengths

Iterative Techniques Simultaneous Equations

Pipe Network Software

Iterative Techniques forD d Q ( i t t l h d


109/131

D and Q (given total head

loss)Assume all head loss is major head

loss.

Calculate D or Q using Swamee-Jainequations

Calculate minor losses

Find new major losses by subtractingminor losses from total head loss

Solution Technique: Head


110/131

qLoss

Can be solved directly

minorfl hhh

g

VKhminor2

2

5

2

2

8

D

LQ

g

fhf

2

9.0Re

74.5

7.3log

25.0

D

f

42

2

8Dg

QKhminor

D

Q4Re

Solution Technique:


111/131

qDischarge or Pipe Diameter

Iterative technique

Set up simultaneous equations in Excel

minorfl hhh

42

28

Dg

Q

Khminor

5

2

2

8

D

LQ

gfhf

2

9.0Re

74.5

7.3log

25.0

D

f

D

Q4Re

Use goal seek or Solver to

find discharge that makes thecalculated head loss equal

the given head loss.

Example: Minor and


112/131

pMajor Losses

Find the maximum dependable flow between thereservoirs for a water temperature range of 4Cto 20C.

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

Directions


113/131

Directions

Assume fully turbulent (rough pipe law)

find f from Moody (or from von Karman)

Find total head loss

Solve for Q using symbols (must includeminor losses) (no iteration required)

Obtain values for minor losses from notes or

text

Example (Continued)


114/131

Example (Continued)

What are the Reynolds number in thetwo pipes?

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

to produce to reduce the discharge by

50%? What is the effect of temperature?

Why is the effect of temperature so

small?

Example (Continued)


115/131

Example (Continued)

Were the minor losses negligible?

Accuracy of head loss calculations?

What happens if the roughnessincreases by a factor of 10?

If you needed to increase the flow by30% what could you do?

Suppose I changed 6 pipe, what isminimum diameter needed?



116/131


Dimensionally correct equations fit to theempirical results can be incorporated intocomputer or calculator solution techniques

Minor losses are obtained from the pressurecoefficient based on the fact that thepressure coefficient is _______ at high

Reynolds numbers Solutions for discharge or pipe diameter often

require iterative or computer solutions

constant

Loss CoefficientsUse this table to find loss coefficients:


117/131

Expansion:Conservation of Energy


118/131

Conservation of Energy

1 2

ltp hHg

Vz

pH

g

Vz

p

22

2

222

2

2

2

111

1

1

g

g

lhg

VVpp

2

2

1

2

221

g

g

VVpphl

2

2

2

2

121

g

z1 = z2

What isp1 - p2?

Head Loss due to Sudden Expansion:Conservation of Momentum


119/131

Apply in direction of flowNeglect surface shear

Divide by (A2g)

Conservation of Momentum

Pressure is applied over all ofsection 1.

Momentum is transferred overarea corresponding toupstream pipe diameter.V1 is velocity upstream.

sspp FFFWMM 2121

1 2

xx ppxxFFM

2121

1

2

11 AVM x r 22

22 AVx r

222122

212

1 ApApAVAV rr

g

A

AVV

pp 2

12

1

2

2

21

g

A1A2

x

Head Loss due toSudden Expansion


120/131

Energy

Sudden Expansion

g

VVpphl

2

22

2121

g

gA

AVV

pp 2

12

1

2

2

21

g

1

2

2

1

V

V

A

A

g

VV

g

V

VVV

hl 2

2

2

2

11

22

1

2

2

g

VVVV

hl 2

2 21212

2

g

VVhl

2

2

21

2

2

1

2

1 12

A

A

g

Vhl

2

2

11

A

AK

Momentum

Mass


121/131

Contraction

V1 V2

EGL

HGL

vena contracta

g

VKh

cc2

2

2

losses are reduced with a gradual contraction

Expansion!!!

Questions:


122/131

Questions:

In the rough pipe law region if the flow rateis doubled (be as specific as possible)

What happens to the major head loss?

What happens to the minor head loss?

Why do contractions have energy loss?

If you wanted to compare the importance of

minor vs. major losses for a specificpipeline, what dimensionless terms couldyou compare?


123/131

Entrance Losses

Losses can bereduced by

accelerating theflow gradually andeliminating thevena contracta

Ke 0.5

Ke 1.0

Ke 0.04

he Ke

V2

2g

reentrant


124/131

Head Loss in Valves

Function of valve typeand valve position

The complex flow paththrough valves oftenresults in high head loss

What is the maximumvalue that Kvcan have?_____

hv KvV2

2g

How can K be greater than 1?


125/131

Questions

What is the headloss when a pipeenters a

reservoir?

Draw the EGLand HGL

V

g

V

2

2

EGL

HGL

2

2

11

A

AK

ltp hHg

VzpHg

Vzp 22

2

222

2

2

2

111

1

1

g

g


126/131

Example

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

cs1

cs2

2

21002

l

Vm h

g= +

Pipe Flow Example


127/131

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?oil= 8.82 kN/m

3

f = .035

If oil flows from the upper to lower reservoir at a velocity of1.58 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


128/131

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?oil= 8.82 kN/m

3

f = .035

Kout=1

r/D = 0

Apply Bernoullis 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


129/131

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?oil= 8.82 kN/m

3

f = .035

Kout=1

r/D = 0

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

Hmin= 2KbendV2/2g + KentV

2/2g + 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


130/131

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?oil= 8.82 kN/m

3

f = .035

Kout=1

r/D = 0

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

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + 0.24mZ1 = 136.09 meters

Pipa ekivalen


131/131

Pipa ekivalen

Digunakan untukmenyederhanakansistem yang ditinjau

Ciri khasnya adalah memilikikeserupaan hidrolis dengan kondisinyatanya Q, hf sama

Pipa ekivalen dapat dinyatakan melaluiekivalensi l,D,f

fluid-mechanics-benno-new.ppt

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