12/10/2016

1

TL2101 Mekanika Fluida I

Benno Rahardyan

Pertemuan

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, tekanan

hidrostatik

Mengerti prinsip-2

tekanan statitka

2 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

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

UTS - -

Pipes are Everywhere!

Owner: City of

Hammond, IN

Project: Water Main

Relocation

Pipe Size: 54"

Pipes are Everywhere! Drainage Pipes

Pipes Pipes are Everywhere! Water Mains

12/10/2016

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Types of Engineering Problems

How big does the pipe have to be to carry a flow of x m3/s?

What will the pressure in the water distribution system be when a fire hydrant is open?

FLUID DYNAMICS

THE 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/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 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 Flow

Energy equation

EGL and HGL

Head loss

– major losses

– minor losses

Non circular conduits

The Energy Line and the Hydraulic Grade Line

Looking 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…

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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 pump

head 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 Assumptions

hp g

p1

g1

V1

2

2g z1 hp

p2

g 2

V2

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

EGL (or TEL) and HGL

velocity head

elevation head (w.r.t.

datum)

pressure head (w.r.t.

reference pressure)

zg

VpEGL

2

2

g

zγ

pHGL

downward

lower than reference pressure

The energy grade line must always slope ___________ (in direction 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 _____ ____ __________ ______

Energy equation

z = 0

pump

Energy Grade Line

Hydraulic G L velocity head

pressure head

elevation

datum

z

2g

V2

g

p

Ltp hhzg

Vphz

g

Vp 2

2

22

21

2

11

1

22

g

g

static head

Why is static

head important?

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

Q

Measures the Static Pressure

Measures the Total Head

1 2

Z

P/γ

V2/2g EL

HGL

1 2

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

V2/2g EL

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

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Bernoulli Equation Assumption

constp

g

Vz

g2

2

density

Steady

streamline

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

Along a __________

______ flow

Constant ________

No pumps, turbines, or head loss Why no ? ____________

Does direction matter? ____

Useful when head loss is small

point velocity

no

Pipe Flow: Review

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

dimensional analysis

We have the control volume energy equation for pipe flow.

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

How do we get this relationship? __________ _______.

Example Pipe Flow Problem

D=20 cm

L=500 m valve

100 m Find the discharge, Q.

Describe the process in terms of energy!

cs1

cs2

p V

gz H

p V

gz H hp t l

11

1

2

12

22

2

22 2g

g

zV

gz hl1

2

2

22

V g z z hl2 1 22 a f

Flow Profile for Delaware Aqueduct

Rondout 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

Ratio of Forces

Create ratios of the various forces

The magnitude of the ratio will tell us which forces are most important and which forces could be ignored

Which force shall we use to create the ratios?

Inertia as our Reference Force

F=ma

Fluids problems (except for statics) include a velocity (V), a dimension of flow (l), and a density (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

l

Vrl3

rV

l

2

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Dimensionless Parameters

Reynolds Number

Froude Number

Weber Number

Mach Number

Pressure/Drag Coefficients

– (dependent parameters that we measure experimentally)

ReVlr

m=

FrV

gl=

2

2C p

p

Vr

rlVW

2

c

VM

AVd

2

Drag2C

r

2fu

V

l

fg gr

2f

l

2

fvE

c

l

r=

2

fi

V

lr

( )p g zrD + D

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 isn’t relevant than create new non dimensional force numbers 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

Friction Factor : Major losses

Laminar flow

– Hagen-Poiseuille

Turbulent (Smooth, Transition, Rough)

– Colebrook Formula

– Moody diagram

– Swamee-Jain

Laminar Flow Friction Factor

L

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

Pipe Flow: Dimensional Analysis

What are the important forces? ______, ______,________. Therefore ________number and _______________ .

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

______, ______

Write the functional relationship

C p 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

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 the length to diameter ratio, how will Cp change?

,Rep

DC f

l D

f ,Rep

DC f

l D

2

2C

V

pp

r

Cp proportional to l

f is friction factor

, ,Rep

lC f

D D

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Hagen-Poiseuille

Darcy-Weisbach

Laminar Flow Friction Factor

2

32

lhDV

L

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 LQh

gDr

-1

Viscous Flow in Pipes

Two important parameters!

R - Laminar or Turbulent

/D - Rough or Smooth

R,

Df

l

DCp

2

2C

V

pp

r

rVDR

Viscous Flow: Dimensional Analysis

Where and

Transition at R of 2000

Laminar and Turbulent Flows

Reynolds apparatus

rVDR

damping inertia

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

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 flow Turbulent flow

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

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Laminar, Incompressible, Steady, Uniform Flow

Between Parallel Plates

Through circular tubes

Hagen-Poiseuille Equation

Approach – Because it is laminar flow the shear

forces can be quantified

– Velocity profiles can be determined from a force balance

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)

Laminar Flow through 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

Laminar Flow through Circular Tubes: Diagram

Velocity

Shear

hpdl

drau g

4

22

hpdl

dr

dr

dug

2

hpdl

dr

dr

dug

2

l

hr l

2

g

l

dhl

40

g True for Laminar or

Turbulent flow

Shear at the wall

Laminar flow

Laminar flow Continue

Momentum is

Mass*velocity (m*v)

Momentum per unit volume is

r*vz

Rate of flow of momentum is

r*vz*dQ

dQ=vz2πrdr

but

vz = constant at a fixed value of r

rvz(v2rdr)z rvz(v2rdr)

zdz 0

Laminar flow

Laminar flow Continue

2r zr r dz 2(r dr)zr rdrdzp

z2rdr p

zdz2rdr rg2rdrdz 0

dvz

dr

Q 2vzdr0

R

R4

8

p

L

p pz 0 pzL rgL

Hagen-Poiseuille

12/10/2016

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The Hagen-Poiseuille Equation

hpdl

daQ g

8

4

h

p

dl

dDQ

g

g

128

4

lhzp

zp

2

2

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 or z

pz

V

gH

pz

V

gH hp t l

1

1

1 11

2

2

2

2 22

2

2 2g

g

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

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

turbulenaliranberartiKarena

xx

x

v

VD

reynoldsBilangan

4000Re

1035,6103,1

15,05,5Re

:

5

6

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

12/10/2016

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

x

x

v

VD

reynoldsBilangan

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

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

g

g

gg

Turbulent Pipe and Channel Flow: Overview

Velocity distributions

Energy Losses

Steady Incompressible Flow through Simple Pipes

Steady Uniform Flow in Open Channels

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

fluctuation t

uuu

u

u

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

úz

Uz

average

ur

úr

Ur

average

p

P’

p

average

Time

Turbulent flow

All previous parameters involved three fundamental dimensions,

Mass, length, and time

From these parameters, three dimensionless groups can be build

P

rU2 f (Re,

L

D)

Re rUD

inertia

Viscous forces

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Turbulence: Size of the Fluctuations or Eddies

Eddies must be smaller than the physical dimension 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

Turbulence: Flow Instability

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

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

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

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

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

viscosity

inertia

Velocity Distributions

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

Mixing of fluid (transfer of momentum) causes the 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 Profile

dy

du

dy

du

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

Turbulent Flow Velocity Profile

ylI

dy

duy22r

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, 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

12/10/2016

11

Pipe Flow: The Problem

We have the control volume energy equation for pipe flow

We need to be able to predict the head loss term.

We will use the results we obtained using dimensional analysis

Friction Factor : Major losses

Laminar flow

– Hagen-Poiseuille

Turbulent (Smooth, Transition, Rough)

– Colebrook Formula

– Moody diagram

– Swamee-Jain

Turbulent Pipe Flow Head Loss

___________ 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

(used to draw the Moody diagram)

Smooth, Transition, Rough Turbulent Flow

Hydraulically smooth pipe law (von Karman, 1930)

Rough pipe law (von Karman, 1930)

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

1 Re f2log

2.51f

1 2.512log

3.7f Re f

D

1 3.72log

f

D

2

f2

f

L Vh

D g

Pipe Flow Energy Losses

g

phl

R,f

Df

L

DCp

2

2C

V

pp

r

2

2C

V

ghlp

L

D

V

ghl

2

2f

g

V

D

Lhl

2f

2

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

Turbulent Pipe Flow Head Loss

___________ 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

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12

Surface 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

Rough Turbulent flow

1

f 4.0 * log Re* f 0.4

Smooth pipe, Re>3000

1

f 4.0 * log

D

2.28

Rough pipe, [ (D/)/(Re√ƒ) <0.01]

1

f 4.0 * log

D

2.28 4.0 * log 4.67

D /

Re f1

Transition function

for both smooth and

rough pipe

f P

L

D

2rU 2

f 0.079Re0.25

Smooth, Transition, Rough Turbulent Flow

Hydraulically smooth pipe law (von Karman, 1930)

Rough pipe law (von Karman, 1930)

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

51.2

Relog2

1 f

f

D

f

7.3log2

1

g

V

D

Lfh f

2

2

(used to draw the Moody diagram)

f

D

f Re

51.2

7.3log2

1

Moody Diagram

0.01

0.10

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

fric

tio

n f

acto

r

laminar

0.05

0.04

0.03

0.02

0.015

0.01 0.008 0.006

0.004

0.002

0.001 0.0008

0.0004

0.0002

0.0001

0.00005

smooth

l

DCpf

D

0.02

0.03

0.04

0.05

0.06

0.08

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

Swamee-Jain

1976

limitations

/D < 2 x 10-2

– Re >3 x 103

– less than 3% deviation

from results obtained

with Moody diagram

easy to program for computer or calculator use

5/ 2 f

3/ 2 f

1.782.22 log

3.7

ghQ D

L D ghD

L

0.044.75 5.2

21.25 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

12/10/2016

13

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 8

f

LQ

h g D

Re 4Q

D 2 5

f 2

4Re f

8

Q g Dh

D LQ

3

f21Re f

gh D

L

2

1 2.514 log

f 3.7 Re f

D

f

2 5 2

8f

h g

D LQ

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

3

f

log 2.513.7 22

gh LQ D

L D gh D

Swamee D?

0.045 1/ 4 5 1/5

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

QD

g

25

2

64f

8

QD

g

1/51/ 4 1/5

2 25/ 4

2

64f

Q Q

g Q g

1/52

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/ 41f

4 4

Q Q

g Q g

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 0.18-0.6

rivet steel 0.9-9.0

corrugated metal 45

PVC 0.12

d Must be

dimensionless!

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 Formulas

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 C

æ ö=

è ø

C = Hazen-Williams coefficient

range of data

Commonly used in commercial and industrial settings

Only applicable over _____ __ ____ collected

Hazen-Williams exponential friction formula

12/10/2016

14

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-Williams vs

Darcy-Weisbach

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 empirical

Darcy-Weisbach is dimensionally correct, and ________.

Hazen-Williams can be considered valid only over the range of gathered data.

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

Non-Circular Conduits: Hydraulic Radius Concept

A is cross sectional area

P is wetted perimeter

Rh is the “Hydraulic Radius” (Area/Perimeter)

Don’t confuse with radius!

2

f2

f

L Vh

D g=

2

f f4 2h

L Vh

R g=

2

44

h

DA D

RP D

p

p= = = 4 hD R=

For a pipe

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

Pipe Flow Summary (1)

Shear increases _________ with distance 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

Pipe Flow Summary (2)

Energy equation left us with the elusive head loss term

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

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

Questions

Can the Darcy-Weisbach equation and Moody Diagram be used for 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? ______

12/10/2016

15

Darcy Weisbach

Major and Minor Losses

Major Losses:

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

f = friction factor L = pipe length D = pipe diameter V = 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 dalam pipa sepanjang 1500 m dan diameter 20 cm, apabila air mengalir dengan kecepatan 2 m/det. Koefisien gesekan f=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

m

xx

x

g

V

D

Lfhf

tenagaKehilangan

58,30

81,922,0

2150002,0

2

2

2

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

Penyelesaian : Panjang pipa : L = 1000 m Diameter pipa : D = 0,15 m Debit aliran : Q = 50 liter/detik Koefisien gesekan f = 0,02

m

xx

xx

QDg

Lfhf

tenagaKehilangan

4,54

)015,0(81,9

100002,0802,0

8

22

5

52

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

m

xx

x

g

V

D

Lfhf

tenagaKehilangan

58,30

81,922,0

2150002,0

2

2

2

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

Penyelesaian : Panjang pipa : L = 1000 m Diameter pipa : D = 0,15 m Debit aliran : Q = 50 liter/detik Koefisien gesekan f = 0,02

m

xx

xx

QDg

Lfhf

tenagaKehilangan

4,54

)015,0(81,9

100005,0802,0

8

52

5

5

52

12/10/2016

16

Example Solve for the Pressure Head, Velocity Head, and Elevation Head at each

point, and then plot the Energy Line and the Hydraulic Grade Line

1

2 3 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

2 3 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

2 3 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

2 3 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

2 3 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 HGL Energy 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’

12/10/2016

17

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 Equation

Minor 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 Losses

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

Most minor losses can not be obtained analytically, 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 2

2C

V

pp

r

2

2C

V

ghlp

g

Vh pl

2C

2

High R

Head Loss: Minor Losses

potential thermal

Vehicle drag Hydraulic jump Vena contracta Minor losses!

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

Flow expansions have high losses

– Kinetic energy decreases across expansion

– Kinetic energy ________ and _________ energy

– Examples – ________________________________

__________________________________________

Losses can be minimized by gradual transitions

Minor Losses

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

Minor losses are often expressed as a loss coefficient, K, times the 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 Gradual Expansion (Diffusor)

g

VVKh EE

2

2

21

diffusor angle ( )

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0 20 40 60 80

KE

2

1

2

2

2 12

A

A

g

VKh EE

12/10/2016

18

Sudden Contraction

losses are reduced with a gradual contraction

g

V

Ch

c

c

21

1 2

2

2

2A

AC c

c

V1 V2

flow separation

Sudden Contraction

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

0 0.2 0.4 0.6 0.8 1

A2/A1

Cc

hC

V

gc

c

FHG

IKJ

11

2

2

2

2

Q CA ghorifice orifice 2

g

VKh ee

2

2

0.1eK

5.0eK

04.0eK

Entrance Losses

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

vena contracta

Head Loss in Bends

Head loss is a function of the ratio of the bend radius to the pipe diameter (R/D)

Velocity distribution returns to normal several pipe diameters downstream

High pressure

Low pressure

Possible

separation

from wall

D

g

VKh bb

2

2

Kb varies from 0.6 - 0.9

R

Head Loss in Valves

Function of valve type and valve position

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

g

VKh vv

2

2

Solution Techniques

Neglect minor losses

Equivalent pipe lengths

Iterative Techniques

Simultaneous Equations

Pipe Network Software

12/10/2016

19

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

loss.

Calculate D or Q using Swamee-Jain equations

Calculate minor losses

Find new major losses by subtracting minor losses from total head loss

Solution Technique: Head Loss

Can be solved directly

minorfl hhh

g

VKhminor

2

2

5

2

2

8

D

LQ

gfh f

2

9.0Re

74.5

7.3log

25.0

D

f

42

28

Dg

QKhminor

D

Q4Re

Solution Technique: Discharge or Pipe Diameter

Iterative technique

Set up simultaneous equations in Excel

minorfl hhh

42

28

Dg

QKhminor

5

2

2

8

D

LQ

gfh f

2

9.0Re

74.5

7.3log

25.0

D

f

D

Q4Re

Use goal seek or Solver to

find discharge that makes the

calculated head loss equal

the given head loss.

Example: Minor and Major Losses

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

Water

2500 m of 8” PVC pipe

1500 m of 6” PVC pipe Gate valve wide open

Standard elbows

Reentrant pipes at reservoirs

25 m elevation difference in reservoir water levels

Sudden contraction

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 include minor losses) (no iteration required)

Obtain values for minor losses from notes or text

Example (Continued)

What are the Reynolds number in the two 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?

12/10/2016

20

Example (Continued)

Were the minor losses negligible?

Accuracy of head loss calculations?

What happens if the roughness increases by a factor of 10?

If you needed to increase the flow by 30% what could you do?

Suppose I changed 6” pipe, what is minimum diameter needed?

Pipe Flow Summary (3)

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

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

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

constant

Loss Coefficients

Use this table to find loss coefficients:

Head Loss due to Sudden Expansion: 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 is p1 - p2?

Apply in direction of flow

Neglect surface shear

Divide by (A2 g)

Head Loss due to Sudden Expansion: 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

1

2

11 AVM x r 2

2

22 AVM x r

22212

2

21

2

1 ApApAVAV rr

g

A

AVV

pp 2

12

1

2

2

21

g

A1 A2

x

Energy

Head Loss due to Sudden Expansion

g

VVpphl

2

2

2

2

121

g

g

A

AVV

pp 2

12

1

2

2

21

g

1

2

2

1

V

V

A

A

g

VV

g

V

VVV

hl2

2

2

2

11

22

1

2

2

g

VVVVhl

2

2 2

121

2

2

g

VVhl

2

2

21

2

2

1

2

1 12

A

A

g

Vhl

2

2

11

A

AK

Momentum

Mass

12/10/2016

21

Contraction

V1 V2

EGL

HGL

vena contracta

g

VKh

cc2

2

2

losses are reduced with a gradual contraction

Expansion!!!

Questions:

In the rough pipe law region if the flow rate is 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 specific pipeline, what dimensionless terms could you compare?

Entrance Losses

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

Ke 0.5

Ke 1.0

Ke 0.04

he Ke

V 2

2g

reentrant

Head Loss in Valves

Function of valve type and valve position

The complex flow path through valves often results in high head loss

What is the maximum value that Kv can have? _____

hv Kv

V 2

2g

How can K be greater than 1?

Questions

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

Draw the EGL and HGL

V

g

V

2

2

EGL

HGL

2

2

11

A

AK

Example

D=40 cm L=1000 m

D=20 cm L=500 m

valve 100 m

Find the discharge, Q. What additional information do you need? Apply energy equation How 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

g

Vz

p

22

2

222

2

2

2

111

1

1 g

g

cs1

cs2

2

21002

l

Vm h

g= +

12/10/2016

22

Pipe Flow Example

1

2 Z2 = 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

2 Z2 = 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

2 Z2 = 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 + 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

1

2 Z2 = 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 ekivalen

Digunakan untuk menyederhanakan sistem yang ditinjau

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

Pipa ekivalen dapat dinyatakan melalui ekivalensi l,D,f