TL2101 Mekanika Fluida I

1
TL2101 Mekanika Fluida I

• Benno Rahardyan

Pertemuan
2
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 - -
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Pipes are Everywhere!
Owner City of Hammond, INProject Water Main
RelocationPipe Size 54"
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Pipes are Everywhere!Drainage Pipes
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Pipes
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Pipes are Everywhere!Water Mains
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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?

8
FLUID DYNAMICSTHE BERNOULLI EQUATION
The laws of Statics that we have learned cannot
solve Dynamic Problems. There is no way to solve
for the flow rate, or Q. Therefore, we need a new
dynamic approach to Fluid Mechanics.
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The Bernoulli Equation
By assuming that fluid motion is governed only by
pressure and gravity forces, applying Newtons
second law, F ma, leads us to the Bernoulli
Equation. P/g V2/2g z constant along a
streamline (Ppressure g
specific weight Vvelocity ggravity
zelevation) 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.
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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 behavior can be seen as water flows at a
very high velocity from the reservoir behind the
Glen Canyon Dam in Colorado
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Closed Conduit Flow

• Energy equation
• EGL and HGL
• Head loss
• major losses
• minor losses
• Non circular conduits

12
The Energy Line and the Hydraulic Grade Line
Looking at the Bernoulli equation again P/?
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/?
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

head supplied by a pump
hp
head given to a turbine
ht
mechanical energy converted to thermal
hL
downstream
Cross section 2 is ____________ from cross
section 1!
irreversible
Point to point or control volume?
Why a? _____________________________________
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Energy Equation Assumptions
hydrostatic

• 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

kinetic
density
Steady
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EGL (or TEL) and HGL
pressure head (w.r.t. reference pressure)
velocity head
downward

• 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
  _____ ____ __________ ______

lower than reference pressure
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Energy equation
Energy Grade Line
velocity head
Hydraulic G L
static head
pressure head
Why is static head important?
elevation
z
pump
z 0
datum
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The Energy Line and the Hydraulic Grade Line
Lets first understand this drawing
Measures the Total Head
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
Measures the Static Pressure
1
2
1
2
EL
V2/2g
HGL
Q
P/?
Z
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The Energy Line and the Hydraulic Grade Line
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.
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
EL
V2/2g
V2/2g
HGL
P/?
2
Q
P/?
Z
1
Z
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Bernoulli Equation Assumption

• _________ (viscosity cant be a significant
  parameter!)
• Along a __________
• ______ flow
• Constant ________
• No pumps, turbines, or head loss

Frictionless
streamline
Steady
density
Why no a? ____________
point velocity
Does direction matter? ____
no
Useful when head loss is small
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Pipe Flow Review

• 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? __________
  _______.

dimensional analysis
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Example Pipe Flow Problem
cs1
D20 cm L500 m
Find the discharge, Q.
100 m
valve
cs2
Describe the process in terms of energy!
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Flow Profile for Delaware Aqueduct
Rondout Reservoir (EL. 256 m)
70.5 km
West Branch Reservoir (EL. 153.4 m)
Sea Level
(Designed for 39 m3/s)
Need a relationship between flow rate and head
loss
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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 ratios?

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Inertia as our Reference Force

• Fma
• 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

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Dimensionless Parameters

• Reynolds Number
• Froude Number
• Weber Number
• Mach Number
• Pressure/Drag Coefficients
• (dependent parameters that we measure
  experimentally)

26
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 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

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Friction Factor Major losses

• Laminar flow
• Hagen-Poiseuille
• Turbulent (Smooth, Transition, Rough)
• Colebrook Formula
• Moody diagram
• Swamee-Jain

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Laminar Flow Friction Factor
Hagen-Poiseuille
Darcy-Weisbach
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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

viscous
pressure
Inertial
Pressure coefficient
Reynolds
diameter, length, roughness height
l/D
e/D
Other repeating parameters?
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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?

Cp proportional to l
f is friction factor
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Laminar Flow Friction Factor
Hagen-Poiseuille
Darcy-Weisbach
-1
Slope of ___ on log-log plot
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Viscous Flow in Pipes

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Viscous Flow Dimensional Analysis

• Two important parameters!
• R - Laminar or Turbulent
• e/D - Rough or Smooth

Where
and
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Laminar and Turbulent Flows

• Reynolds apparatus

inertia
damping
Transition at R of 2000
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Boundary layer growth Transition length
What does the water near the pipeline wall
experience? _________________________ Why does
the water in the center of the pipeline speed up?
_________________________
Drag or shear
Conservation of mass
Pipe Entrance
Non-Uniform Flow
Need equation for entrance length here
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Images - Laminar/Turbulent Flows
Laser - induced florescence image of an
incompressible turbulent boundary layer
Laminar flow (Blood Flow)
Simulation of turbulent flow coming out of a
tailpipe
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

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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)

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Laminar Flow through Circular Tubes Equations
a is radius of the tube
Max velocity when r 0
Velocity distribution is paraboloid of revolution
therefore _____________ _____________
average velocity (V) is 1/2 umax
Vpa2
Q VA
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Laminar Flow through Circular Tubes Diagram
Laminar flow
Shear at the wall
True for Laminar or Turbulent flow
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Laminar flowContinue

• Momentum is
• Massvelocity (mv)
• Momentum per unit volume is
• ?vz
• Rate of flow of momentum is
• ?vzdQ
• dQvz2prdr
• but
• vz constant at a fixed value of r

Laminar flow
42
Laminar flowContinue
Hagen-Poiseuille
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The Hagen-Poiseuille Equation
cv pipe flow
Constant cross section
h or z
Laminar pipe flow equations
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• 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

• 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

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• 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 S0,9 dan kekentalan
  kinematik 2,1 x 10-4 m2/det. Hitung tekanan di
  titik A.

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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 1200.
  Rapat minyak S0,9 dan keketnalan kinematik
  v2,1x 10-4 m2/det. Apabila tekanan pada ujung
  atas adalah p10 kPA ditanyakan tekanan di ujung
  bawah.

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Turbulent Pipe and Channel Flow Overview

• Velocity distributions
• Energy Losses
• Steady Incompressible Flow through Simple Pipes
• Steady Uniform Flow in Open Channels

52
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
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,)

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Turbulent flow
All previous parameters involved three
fundamental dimensions, Mass, length, and time
From these parameters, three dimensionless
groups can be build
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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
56
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. (___________)

viscosity
inertia
head loss
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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
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Turbulent Flow Velocity Profile
Turbulent shear is from momentum transfer
h eddy viscosity
Length scale and velocity of large eddies
Dimensional analysis
y
59
Turbulent Flow Velocity Profile
increases
Size of the eddies __________ as we move further
from the wall.
k 0.4 (from experiments)
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Log Law for Turbulent, Established Flow, Velocity
Profiles
Integration and empirical results
Turbulent
Laminar
Shear velocity
y
x
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 using
  dimensional analysis

62
Friction Factor Major losses

• Laminar flow
• Hagen-Poiseuille
• Turbulent (Smooth, Transition, Rough)
• Colebrook Formula
• Moody diagram
• Swamee-Jain

63
Turbulent Pipe Flow Head Loss
Proportional

• ___________ 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

square
Increases
independent
64
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)

(used to draw the Moody diagram)
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Pipe Flow Energy Losses
Horizontal pipe
Dimensional Analysis
Darcy-Weisbach equation
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Turbulent Pipe Flow Head Loss
Proportional

• ___________ 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
independent
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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
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Friction Factor for Smooth, Transition, and Rough
Turbulent flow
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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)

(used to draw the Moody diagram)
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Moody Diagram
0.10
0.08
0.05
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
friction factor
0.006
0.03
0.004
laminar
0.002
0.02
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
smooth
1E03
1E04
1E05
1E06
1E07
1E08
R
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Fanning Diagram
f 16/Re
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Swamee-Jain

• 1976
• limitations
• ?/D lt 2 x 10-2
• Re gt3 x 103
• less than 3 deviation from results obtained with
  Moody diagram
• easy to program for computer or calculator use

no f
L
hf
Each equation has two terms. Why?
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Colebrook Solution for Q
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Colebrook Solution for Q
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Swamee D?
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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
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0.12
PVC
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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)

78
Exponential Friction Formulas

• Commonly used in commercial and industrial
  settings
• Only applicable over _____ __ ____ collected
• Hazen-Williams exponential friction formula

range of data
C Hazen-Williams coefficient
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Head lossHazen-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

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Hazen-Williams vs Darcy-Weisbach

• 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 other fluids
  without further experimentation.

preferred
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Non-Circular ConduitsHydraulic Radius Concept

• A is cross sectional area
• P is wetted perimeter
• Rh is the Hydraulic Radius (Area/Perimeter)
• Dont confuse with radius!

For a pipe
We can use Moody diagram or Swamee-Jain with D
4Rh!
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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
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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)

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Questions

• Can the Darcy-Weisbach equation and Moody Diagram
  be used for fluids other than water? _____

Yes

• What about the Hazen-Williams equation? ___

No

• Does a perfectly smooth pipe have head loss? _____

Yes

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

Yes
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Darcy Weisbach
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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/? V12/2g z1
P2/? V22/2g z2 Hmaj Hmin
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• 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 f0,02
• 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

88

• 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

89

• 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 f0,02
• 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

90

• 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

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
Assumptions and Hints P1 and P4 0 --- V3 V4
same diameter tube We must work backwards to
solve this problem
1
?H2O 62.4 lbs/ft3
R .5
4
R .25
2
3
4
1
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Point 1 Pressure Head Only atmospheric ? P1/?
0 Velocity Head In a large tank, V1 0 ?
V12/2g 0 Elevation Head Z1 4
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
93
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/? 0 Velocity Head
V42/2g 3 Elevation Head Z4 1
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
94
Point 3 Apply the Bernoulli equation between 3
and 4 (V3V4) P3/62.4
3 1 0 3 1 P3 0 Pressure Head P3/?
0 Velocity Head V32/2g 3 Elevation Head Z3
1
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
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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 (?.52)V2 (?.252)x13.9 ? V2 3.475
ft/s P2/62.4 3.4752/2(32.2) 1 4 ? P2
175.5 lbs/ft2
Pressure Head P2/? 2.81 Velocity Head
V22/2g .19 Elevation Head Z2 1
1
?H2O 62.4 lbs/ft3
4
R .5
R .25
2
3
4
1
96
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
V2/2g.19
EL
P/? 2.81
V2/2g3
V2/2g3
Z4
HGL
Z1
Z1
Z1
97
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/? V12/2g z1 P2/? V22/2g 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.
Energy line with no losses
Hmaj
Energy line with major losses
1
2
98
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
99
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.

High R
100
Head Loss 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

potential
thermal
Hydraulic jump
Vehicle drag
Minor losses!
Vena contracta
101
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.

High Re
102
Head Loss due to Gradual Expansion (Diffusor)
103
Sudden Contraction
V2
V1
flow separation

• losses are reduced with a gradual contraction

104
Sudden Contraction
105
Entrance Losses

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

vena contracta
106
Head Loss in Bends
High pressure

• 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

Possible separation from wall
R
D
Low pressure
Kb varies from 0.6 - 0.9
107
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)

108
Solution Techniques

• Neglect minor losses
• Equivalent pipe lengths
• Iterative Techniques
• Simultaneous Equations
• Pipe Network Software

109
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

110
Solution Technique Head Loss

• Can be solved directly

111
Solution TechniqueDischarge or Pipe Diameter

• Iterative technique
• Set up simultaneous equations in Excel

Use goal seek or Solver to find discharge that
makes the calculated head loss equal the given
head loss.
112
Example Minor and Major Losses

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

25 m elevation difference in reservoir water
levels
Water
Reentrant pipes at reservoirs
Standard elbows
2500 m of 8 PVC pipe
Sudden contraction
Gate valve wide open
1500 m of 6 PVC pipe
113
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

114
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?

115
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?

116
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
117
Loss Coefficients
Use this table
to find loss coefficients
118
Head Loss due to Sudden ExpansionConservation
of Energy
z1 z2
What is p1 - p2?
119
Head Loss due to Sudden ExpansionConservation
of Momentum
A2
A1
x
1
2
Apply in direction of flow
Neglect surface shear
Pressure is applied over all of section
1. Momentum is transferred over area
corresponding to upstream pipe diameter. V1 is
velocity upstream.
Divide by (A2 g)
120
Head Loss due to Sudden Expansion
Mass
Energy
Momentum
121
Contraction
EGL
HGL
Expansion!!!
V1
V2
vena contracta

• losses are reduced with a gradual contraction

122
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?

123
Entrance Losses
reentrant

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

124
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? _____

?
How can K be greater than 1?
125
Questions
EGL
HGL

• What is the head loss when a pipe enters a
  reservoir?
• Draw the EGL and HGL

126
Example
cs1
100 m
valve
cs2
D40 cm L1000 m
D20 cm L500 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
127
Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
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.
128
Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
Apply Bernoullis equation between points 1 and
2Assumptions 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
129
Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
0 0 Z1 0 0 130m 5.85m Hmin Hmin
2KbendV2/2g KentV2/2g KoutV2/2g From Loss
Coefficient table Kbend 0.19 Kent 0.5
Kout 1 Hmin (0.19x2 0.5 1) x
(1.582/2x9.8) Hmin 0.24 m
130
Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
0 0 Z1 0 0 130m Hmaj Hmin 0 0
Z1 0 0 130m 5.85m 0.24m Z1 136.09
meters
131
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