TL2101 Mekanika Fluida I

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TL2101

Mekanika Fluida I

Benno Rahardyan

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Mg Topik Sub Topik Tujuan Instruksional (TIK)

1

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

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

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

4 Idem Idem + gaya pada bidang

terendam Idem

5

5 Aplikasi

kekekalan energi

Aplikasi kekekalan energi dalam

aplikasi di bidang TL Latihan menggunakan prinsip kekekalan eneri khususnya dalam bidang air minum

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-Pipes are Everywhere!

Pipes are Everywhere!

Owner: City of Hammond, IN

Project: Water Main Relocation

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Types of Engineering

Problems



How big does the pipe have to be to

carry a flow of

x

m

33

/s?



What will the pressure in the water

distribution system be when a fire

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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

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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)

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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

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Closed Conduit Flow



Energy equation



EGL and HGL



Head loss

– major lossesmajor losses – minor losses_{minor losses}

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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

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Conservation of Energy



Kinetic, potential, and thermal

energy

h_L =

L t

p z h h

g V p

h z

g V p

 



 ₂

2 2 2 2

1 2

1 1 1

2

2 _ 

 

h_p= h_t =

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 _V 2

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Energy Equation

Assumptions

h

p



p₁



1

V₁2

2g  z1 hp 

p₂



2

V₂2

2g  z2 ht  hL

hydrostatic

density Steady

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 streamlines section is drawn perpendicular to the streamlines

(otherwise the _______ energy term is incorrect) (otherwise the _______ energy term is incorrect)

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EGL (or TEL) and HGL

velocity head

elevation head (w.r.t.

datum) pressure

head (w.r.t. reference pressure)

z g

V p

EGL _ _ _

2

2 



z γ

p

HGL _ _

downward

lower than reference pressure

 The energy grade line must always slope ___________ (in 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 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 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 If the HGL falls below the point in the system for which it is

plotted, the local pressures are _____ ____ __________

______

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Energy equation

z = 0

pump

Energy Grade Line

Hydraulic G L

velocity head

pressur e head

elevatio n

datum

z

2g V2



 p

L t

p z h h

g V p

h z

g V p

 



 ₂

2 2 2 2

1 2

1 1 1

2

2 _ 

 

static

head_{Why is static}

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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

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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

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Bernoulli Equation

Assumption

const p

g V

z _ _ _



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

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Pipe Flow: Review

2 2

1 1 2 2

2 p 2 t L

p V p V

z 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 flow. relationship between head loss and flow.

 How do we get this relationship?How do we get this relationship?

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Example Pipe Flow

Problem

D=20 cm

L=500 m valve

100 m Find the discharge, Q.

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

cs₁

cs₂

p V

g z H

p V

g z H h

p t l

1

1 1 2

1 2 2 2

2

2 2

         

p V

g z H

p V

g z H h

p t l

1

1 1 2

1 2 2 2

2

2 2

         

z V

g z hl

1 2 2

2

  

z V

g z hl

1 2

2

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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 2

2 p 2 t l

p V p V

z H z H h

g g

 

         

Need a relationship between flow rate and head loss

1 2

l

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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

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Inertia as our Reference

Force

 F=maF=ma

 Fluids problems (except for statics) include a Fluids problems (except for statics) include a

velocity (

velocity (VV), a dimension of flow (_{), a dimension of flow (}ll), and a _{), and a} density (

density (__))

 Substitute Substitute VV, , ll, , __ for the dimensions MLT for the dimensions MLT

 Substitute for the dimensions of specific forceSubstitute for the dimensions of specific force F _{ }_ a _F  f a f  _{L T}M2 2

L l_ T 

M 

f_i _

l V l3

V l

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Dimensionless

Parameters

 _{Reynolds Number}_{Reynolds Number}  _{Froude Number}_{Froude Number}  _{Weber Number}_{Weber Number}  _{Mach Number}_{Mach Number}

 _{Pressure/Drag Coefficients}_{Pressure/Drag Coefficients}

– _{(dependent parameters that we measure experimentally)}_{(dependent parameters that we measure experimentally)}

Re rVl

m = Fr V gl =

 

2 2

C_p p

V       l V W 2  c V M _ A V d 2 Drag 2 C   2

f_u V

l 



f_g  _g

2 f l    2 f v E c l r = 2

f_i V

l 



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Problem solving approach

1.

1. Identify relevant forces and any other relevant parametersIdentify relevant forces and any other relevant parameters 2.

2. If inertia is a relevant force, than the non dimensional If inertia is a relevant force, than the non dimensional ReRe, , FrFr, ,

W

W, , M, CpM, Cp numbers can be used numbers can be used

3.

3. If inertia isn’t relevant than create new non dimensional force If inertia isn’t relevant than create new non dimensional force

numbers using the relevant forces

4.

4. Create additional non dimensional terms based on geometry, Create additional non dimensional terms based on geometry,

velocity, or density if there are repeating parameters

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.

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Friction Factor : Major

losses



Laminar flow

– Hagen-Poiseuille_{Hagen-Poiseuille}



Turbulent (Smooth, Transition, Rough)

– Colebrook FormulaColebrook Formula – Moody diagramMoody diagram

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Laminar Flow Friction

Factor

L h D V l   32 2  L h D V l   32 2  2 32 gD LV h_l   32 ₂

gD LV h_l    g V D L h_l 2 f 2  g V D L h_l 2 f 2  g V D L gD LV 2 f 32 2 2    g V D L gD LV 2 f 32 2 2    R VD 64 64

f     R VD 64 64

f _ _

 

Hagen-Poiseuille

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Pipe Flow: Dimensional

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

2 C

V p p



  

Other repeating parameters?

pressure

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Dimensional

Analysis

 How will the results of dimensional analysis How will the results of dimensional analysis

guide our experiments to determine the guide our experiments to determine the

relationships that govern pipe flow? relationships that 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 C length to diameter ratio, how will Cp_p

change? change?

,Re

p

D

C f

l D



 

  

 

f C_p D f ,Re

l D



   

_ _ _ _

   

2

2 C

V p p



  

Cp proportional to l

f is friction factor

, ,Re

p

l

C f

D D 

 

  

(31)

Hagen-Poiseuille

Darcy-Weisbach

Laminar Flow Friction

Factor

2

32

l

h D V

L 



f ₂

32 LV h

gD  



2

f f

2

L V h

D g



2

32

f

2

LV L V

gD D g



 

64 64

f

Re

VD  

  _{Slope of ___ on log-log plot}

f 4

128 LQ h

gD 

 

(32)

Viscous Flow in

Pipes

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

Two important parameters!

R - Laminar or Turbulent R - Laminar or Turbulent



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

   

 

 ,R

D f

l D

C_p  _

  

 

 ,R

D f

l D

C_p  C 2 ₂

V p

p

    2 ₂

C

V p p



  

 VD 

R

 VD

 R

Viscous Flow:

Dimensional Analysis

(34)

Transition at R of 2000

Laminar and Turbulent

Flows

 Reynolds apparatusReynolds apparatus

 

 VD

R  

 VD

R

damping

inertia

(35)

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 Flowv

(36)

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

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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 Because it is laminar flow the shear forces can be quantified

forces can be quantified

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

(38)

Laminar Flow through

Circular Tubes



Different geometry, same equation

development (see Streeter, et al. p

268)



Apply equation of motion to cylindrical

(39)

Laminar Flow through

Circular Tubes: Equations



p h



dl d r a u _      4 2 2



p h



dl d r a u _      4 2 2



p h



dl d a u _     4 2

max _dl



p h



d a u _     4 2 max



p h



dl d a V _     8 2



p h



dl d a V _     8 2



p h



dl d a Q _      8 4



p h



dl d a Q _      8 4

Velocity distribution is paraboloid of revolution therefore _____________ _____________

Q = VA =

Max velocity when r = 0

average velocity (V) is 1/2 u_max

Vpa2

a is radius of the tube

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Circular Tubes: Diagram

Velocity

Shear



p h



dl d r a u _      4 2 2



p h



dl d r a u _      4 2 2



p h



dl d r dr du    

2



h p dl d r dr du     2



p h



dl d r dr du      

2



h p dl d r dr du       2         l h r l 2   _        l h r l 2 

 0 h₄l_ld

    l d h_l 4 0    

True for Laminar or Turbulent flow

Shear at the wall

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Laminar flow

Continue

Momentum is

Mass*velocity (m*v)

Momentum per unit volume is

*v_z

Rate of flow of momentum is

*v_z*dQ

dQ=vz2πrdr

but

v_z = constant at a fixed value of r

v_z(v2rdr) _z  v_z(v2rdr) _z__dz  0

(42)

Laminar flow

Continue



2r_{zr r} dz 2(r dr)_{zr r}__drdzp_z2rdr  p _z__dz2rdr  g2rdrdz 0





dv

z

dr



Q



₀R

2

v

_z

dr



R

4

8



p

L





p



p

_z_₀



p

_z__L



gL

(43)

The Hagen-Poiseuille

Equation



p h



dl d a Q _      8 4



p h



dl d a Q _      8 4        

 p h

dl d D Q    128 4        

 p h

dl d D Q    128 4 l h z p z p     ₂ 2 2 1 1 1 

 z hl

p z p     ₂ 2 2 1 1 1                   ₂ 2 2 1 1

1 _z p _z

p h_l   _               ₂ 2 2 1 1

1 _z p _z

p h_l   _       

 p h

h_l  _       

 p h

h_l  L h D Q l   128 4  L h D Q l   128 4  L h h p dl d _l           L h h p dl d _l           L h D V l   32 2  L h D V l   32 2 

cv pipe flow

Constant cross section

Laminar pipe flow equations

h or z

p

z V

g H

p

z V

g H h

p t l

1

1 1 1 2

2

2 2 2

2

2 2

(44)

(45)

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

Hidraulika I, Beta Ofset Yogyakarta, 1993

Hidraulika II, Beta Ofset Yogyakarta, 1993

Soal-Penyelesaian Hidraulika I, 1994

(46)



Air mengalir melalui pipa berdiameter

150 mm dan kecepatan 5,5

m/det.Kekentalan kinematik air adalah

1,3 x 10

-4-4

m2/det. Selidiki tipe aliran

turbulen aliran

berarti Karena

x x

x v

VD

reynolds Bilangan

4000 Re

10 35 , 6 10

3 , 1

15 , 0 5 , 5 Re

:

5 6



 

(47)



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

-4-4

m2/det.

Hitung tekanan di titik A.

(48)

(49)

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

-4-4

m2/

det. Apabila tekanan pada ujung atas

adalah p=10 kPA ditanyakan tekanan

(50)

(51)

Turbulent Pipe and

Channel Flow: Overview



Velocity distributions



Energy Losses



Steady Incompressible Flow through

Simple Pipes

(52)

Turbulence

 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)

mean velocity

instantaneous velocity

velocity fluctuation

velocity

fluctuation 

u



u



u



u



u

(53)

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, , , L, U,)

u_z

ú_z

Uz

average

u_r

ú_r

U_r average

p

P’

p average

(54)

Turbulent 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

)



Re 



UD



inertia

Viscous forces



Re 



UD



inertia

(55)

Turbulence: Size of the

Fluctuations or Eddies

 Eddies must be smaller than the physical Eddies must be smaller than the physical

dimension of the flow dimension of the flow

 Generally the largest eddies are of similar size Generally the largest eddies are of similar size

to the smallest dimension of the flow to the smallest dimension of the 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)

(56)

Turbulence: Flow

Instability

 In turbulent flow (high Reynolds number) the force 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 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 Some of the kinetic energy of the flow is transferred to

these large scale motions (eddies).

 Large scale instabilities gradually lose kinetic energy to Large scale instabilities gradually lose kinetic energy to

smaller scale motions.

 The kinetic energy of the smallest eddies is dissipated The kinetic energy of the smallest eddies is dissipated

by viscous resistance and turned into heat.

(=___________)

(=___________)_{head loss}

viscosity

inertia

(57)

Velocity Distributions

 Turbulence causes transfer of momentum from Turbulence causes transfer of momentum from

center of pipe to fluid closer to the pipe wall. center of pipe to fluid closer to the pipe wall.

 Mixing of fluid (transfer of momentum) causes Mixing of fluid (transfer of momentum) causes

the central region of the pipe to have relatively the central region of the pipe to have relatively

_______velocity (compared to laminar flow) _______velocity (compared to laminar flow)

 Close to the pipe wall eddies are smaller (size Close to the pipe wall eddies are smaller (size

proportional to distance to the boundary) proportional to distance to the boundary)

(58)

Turbulent Flow Velocity

Profile

dy du

  

dy du   

dy du

  

dy du   

I Iu

l

   lIuI  

dy du l

u_I _ _I dy du l

u_I _ _I

dy du l_I2

  

dy du l_I2   

Length scale and velocity of “large” eddies

y

Turbulent shear is from momentum transfer

h = eddy viscosity

(59)

Turbulent Flow Velocity

Profile

y l_I __y l_I __

dy du y2 2    dy du y2 2    2 2 2          dy du y   2 2 2          dy du y            dy du y             dy du y    dy du l_I2

  

dy du l_I2    dy du    dy du   

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

increases

(60)

Log Law for Turbulent,

Established Flow, Velocity

Profiles

5 . 5 ln 1 _* *     yu u u 5 . 5 ln 1 _* *     yu u u  ₀ *  u  ₀ *  u          dy du y             dy du y    I u u_* _u_I u_* _

Shear velocity

Integration and empirical results

Laminar Turbulent

(61)

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

(62)

Friction Factor : Major

losses



Laminar flow

– Hagen-Poiseuille_{Hagen-Poiseuille}



Turbulent (Smooth, Transition, Rough)

– Colebrook FormulaColebrook Formula – Moody diagramMoody diagram

(63)

Turbulent Pipe Flow Head

Loss

 ___________ to the length of the pipe___________ to the length of the pipe

 Proportional to the _______ of the velocity Proportional to the _______ of the velocity

(almost) (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 f

2

L V h

D g

(64)

(used to draw the Moody diagram)

Smooth, Transition, Rough

Turbulent Flow

 Hydraulically smooth Hydraulically smooth

pipe law (von pipe law (von

Karman, 1930) Karman, 1930)

 Rough pipe law (von Rough pipe law (von

 Transition function Transition function

for both smooth and for both smooth and

rough pipe laws rough pipe laws

(Colebrook) (Colebrook)

1 Re f

2 log

2.51 f

 

 _ _

 

1 2.51

2 log

3.7

f Re f

D



 

  _  _

 

1 3.7

2 log f

D



 

 _ _

 

2

f

2

f

L V h

D g

(65)

Pipe Flow Energy Losses



p h_l _ _ 



p h_l _ _ 

                

 ,R

f

D f

L D

C_p  _

              

 ,R

f

D f

L D

C_p 

2 2 C V p p     2 ₂

C V p p   

 C 2 ₂

V gh_l p  2 2 C V gh_l p  L D V gh_l 2 2 f _ L D V gh_l 2 2 f _ g V D L h_l 2 f 2  g V D L h_l 2 f 2  Horizontal pipe Dimensional Analysis Darcy-Weisbach equation p V

g z h

p V

g z h h

p t l

1

1 1 2

1 2 2 2

2

2 2

(66)

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

Inversely

Increase

(67)

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

(68)

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 f 1



 

Transition function for both smooth and rough pipe



f  P

L

D

2U2



(69)

Smooth, Transition, Rough

Turbulent Flow

 Hydraulically smooth Hydraulically smooth

pipe law (von pipe law (von

 Rough pipe law (von Rough pipe law (von

 Transition function Transition function

for both smooth and for both smooth and

rough pipe laws rough pipe laws

(Colebrook) (Colebrook)          51 . 2 Re log 2 1 f f          51 . 2 Re log 2 1 f f           D f 7 . 3 log 2 1           D f 7 . 3 log 2 1 g V D L f h_f 2 2  g V D L f h_f 2 2 

(used to draw the Moody diagram)

(70)

Moody Diagram

0.01 0.10

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

(71)

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 f 1

  

(72)

Swamee-Jain

 19761976

 limitationslimitations

 __/D < 2 x 10_{/D < 2 x 10}-2-2

– Re >3 x 10Re >3 x 1033

– less than 3% deviation less than 3% deviation from results obtained

from results obtained

with Moody diagram

 easy to program for easy to program for

computer or

calculator use

5/ 2 _f

3/ 2 _f

1.78

2.22 log

3.7

gh

Q D

L D _gh

D L                  0.04 4.75 5.2 2 1.25 9.4 f f

0.66 LQ L

D Q gh gh    _ _ _ _    _ _  _ _          2 0.9 0.25 f 5.74 log

3.7D Re

   _ _       

  no f

Each equation has two terms. Why?

f

gh L

L _h

(73)

Colebrook Solution for Q

1 2.51

2 log

3.7

f Re f

D      _  _   2

f 2 5

8 f LQ h g D   2 2 5 f

1 1 8 f

LQ h _ g D 

Re  4Q

D   2 5 f 2 4 Re f 8

Q g D

h D LQ     3 f 2 1

Re f gh D

L   2 1 2.51 4 log

f 3.7 _{Re f}

D     _  _   f

2 5 2

8

f h g

D LQ

(74)

Colebrook Solution for Q

2

2 5 ₃

f _f

1 8 2.51

4 log

3.7 ₁ ₂

LQ

h g D D _{gh D}

L

 



 

5 / 2 ₃

f _f

2 2.51

log

3.7 ₁ ₂

L Q

gh D D _{gh D}

L

 



 

5 / 2 _f

3 f

log 2.51

3.7 2

2

gh L

Q D

L D gh D

 



 



 _  _

(75)

Swamee

D?

0.04

5 1/ 4 5 1/ 5

2 2 2 2

1.25

0.66 Q Q Q Q

D

g g Q g g

    _ _{ } _ _ _{ } _    _ _{ } _  _ _{ } _                  1/ 25

1/ 5 1/ 4 1/ 5

2 2 2

5/ 4

0.66 Q Q Q

D

g g Q g

            _ _  _ _  _ _   _   _   _     _ 2

f 2 5

8 f LQ h g D   2 5 2 8 f Q D g    2 5 2 64 f 8 Q D g    1/ 5

1/ 4 1/ 5

2 2

5/ 4 2

64

f Q Q

g Q g

     _ _ _ _    _ _  _ _            1/ 5 2 2 64 f 8 Q D g _ _ _    _      1/5 1/ 5

1/ 4 1/ 5

2 2 2

5/ 4

8

Q Q Q

D

g g Q g

    _ _           __ _ _ _  _ _  _  _   _        _ _    1/ 5

1/ 4 1/ 5

2 ₂ ₂

5/ 4

1 f

4 4

Q Q

g Q g

(76)

Pipe roughness

pipe material

pipe material pipe roughness pipe roughness __ (mm) (mm) glass, drawn brass, copper

glass, drawn brass, copper 0.00150.0015 commercial steel or wrought iron

commercial steel or wrought iron 0.0450.045 asphalted cast iron

asphalted cast iron 0.120.12 galvanized iron

galvanized iron 0.150.15

cast iron

cast iron 0.260.26

concrete

concrete 0.18-0.60.18-0.6

rivet steel

rivet steel 0.9-9.00.9-9.0

corrugated metal

corrugated metal 4545

PVC

PVC 0.120.12



d  d

Must be dimensionless!

(77)

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 5.2

2

1.25 9.4

f f

0.66 LQ L

D Q gh gh    _ _ _ _    _ _  _ _          2 2 5 8 f f LQ h g D   2 0.9 0.25 f 5.74 log

3.7D Re

   _ _         

Re _ 4Q

D  

5 / 2 f

3 f log 2.51 3.7 2 2 gh L Q D

L D gh D

(78)

Exponential Friction

Formulas

f

n m

RLQ h

D

=

    



units SI

675 .

10

units USC

727 .

4

n n

C C R

1.852

f 4.8704

10.675

SI units

L Q h

D C

æ ö =

è ø

C = Hazen-Williams coefcient

range of data



Commonly used in commercial and

industrial settings



Only applicable over _____ __ ____

collected



Hazen-Williams exponential

friction formula

(79)

Head loss:

Hazen-Williams Coefficient

C

C ConditionCondition 150

150 PVCPVC 140

140 Extremely smooth, straight pipes; asbestos cementExtremely smooth, straight pipes; asbestos cement 130

130 Very smooth pipes; concrete; new cast ironVery smooth pipes; concrete; new cast iron 120

120 Wood stave; new welded steelWood stave; new welded steel 110

110 Vitrified clay; new riveted steelVitrified clay; new riveted steel 100

100 Cast iron after years of useCast iron after years of use 95

95 Riveted steel after years of useRiveted steel after years of use 60-80

(80)

Hazen-Williams

vs

Darcy-Weisbach

1.852

f _4.8704

10.675

SI units

L Q h

D C

 

 _{ }

  2

f ₂ ₅

8

f LQ

h

g D 



preferred

 Both equations are empiricalBoth equations are empirical

 Darcy-Weisbach is dimensionally correct, Darcy-Weisbach is dimensionally correct,

and ________. and ________.

 Hazen-Williams can be considered valid only Hazen-Williams can be considered valid only

over the range of gathered data. over the range of gathered data.

 Hazen-Williams can’t be extended to other Hazen-Williams can’t be extended to other

(81)

Non-Circular Conduits:

Hydraulic Radius Concept



A is cross sectional area



P is wetted perimeter



R

_h_h

is the “Hydraulic Radius”

(Area/Perimeter)



Don’t confuse with radius!

2

f

2

f

L V h

D g

=

2

f f

4 _h 2

L V h

R g

=

2

4

h

D

A D

R

P D

p

= = = ₄

h

D = R For a pipe

(82)

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

(83)

Pipe Flow Summary (2)

 Energy equation left us with the elusive Energy equation left us with the elusive

head loss term head 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

(84)

Questions



Can the Darcy-Weisbach equation and

Moody Diagram be used for fluids other

than water? _____

Yes

No

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

(85)

(86)

Major and Minor Losses

Major Losses:

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

f = friction factor L = pipe length D = pipe diameter V = Velocity g = gravity

Minor Losses:

H_min = K_L(V2/2g)

K_l = 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 + z

(87)

 Hitung kehilangan tenaga karena gesekan di 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 x x

x g V D

L f hf

tenaga Kehilangan

58 , 30

81 , 9 2 2 , 0

2 1500 02

, 0

2

2 2

 

(88)

Air melalui pipa sepanjang 1000 m dan Air melalui pipa sepanjang 1000 m dan

diameternya 150 mm dengan debit 50 l/det. diameternya 150 mm dengan debit 50 l/det.

Hitung kehilangan tenaga karenagesekan Hitung kehilangan tenaga karenagesekan

apabila koefisien gesekan f = 0,02 apabila koefisien gesekan f = 0,02

Penyelesaian : Penyelesaian :

Panjang pipa : L = 1000 m Panjang pipa : L = 1000 m

Diameter pipa : D = 0,15 m Diameter pipa : D = 0,15 m

Debit aliran : Q = 50 liter/detik Debit aliran : Q = 50 liter/detik

Koefisien gesekan f = 0,02 Koefisien gesekan f = 0,02

(89)

 Hitung kehilangan tenaga karena gesekan di 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 x x

x g V D

L f hf

tenaga Kehilangan

58 , 30

81 , 9 2 2 , 0

2 1500 02

, 0

2

2 2

 

(90)

Air melalui pipa sepanjang 1000 m dan Air melalui pipa sepanjang 1000 m dan

diameternya 150 mm dengan debit 50 l/det. diameternya 150 mm dengan debit 50 l/det.

Hitung kehilangan tenaga karenagesekan Hitung kehilangan tenaga karenagesekan

apabila koefisien gesekan f = 0,02 apabila koefisien gesekan f = 0,02

Penyelesaian : Penyelesaian :

Panjang pipa : L = 1000 m Panjang pipa : L = 1000 m

Diameter pipa : D = 0,15 m Diameter pipa : D = 0,15 m

Debit aliran : Q = 50 liter/detik Debit aliran : Q = 50 liter/detik

Koefisien gesekan f = 0,02 Koefisien gesekan f = 0,02

(91)

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 = V₄same diameter tube We must work backwards to solve this problem

R = .5’

(92)

1

2

3 4

1’ 4’

γ_H2O= 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’

(93)

1

2

3 4

1’ 4’

γ_H2O= 62.4 lbs/ft3

Point 4:

Apply the Bernoulli equation between 1 and 4 0 + 0 + 4 = 0 + V₄2/2(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’

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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’

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1

2

3 4

1’ 4’

γ_H2O= 62.4 lbs/ft3

Point 2:

Apply the Bernoulli equation between 2 and 3

P₂/62.4 + V₂2/2(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 _ P

2 = 175.5 lbs/ft2

R = .5’

R = .25’

Pressure Head :

P₂/γ = 2.81’

Velocity Head :

V₂2/2g = .19’

Elevation Head :

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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’

Z=4’

P/γ =2.81’

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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 + z

1 = P2/γ + V22/2g + z2 + Hmaj + Hmin

Major losses: H_maj

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.

H_maj

Energy line with no losses

Energy line with major losses

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Pipe Flow and the Energy Equation

Minor Losses : H_min

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