1. TL2101 Mekanika Fluida I
Benno Rahardyan
Pertemuan

2. Tujuan Instruksional (TIK)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 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 + gaya pada bidang terendam 5
Aplikasi kekekalan energi
Aplikasi kekekalan energi dalam aplikasi di bidang TL
Latihan menggunakan prinsip kekekalan eneri khususnya dalam bidang air minum
UTS -

3. Pipes are Everywhere!
Owner: City of Hammond, IN Project: Water Main Relocation Pipe Size: 54"

4. Pipes are Everywhere! Drainage Pipes

5. Pipes

6. Pipes are Everywhere! Water Mains

7. 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 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.

9. 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.

10. 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

11. 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/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…

13. 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 =
Cross section 2 is ____________ from cross section 1!
downstream
irreversible
Point to point or control volume?
Why a? _____________________________________
V is average velocity, kinetic energy

14. 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

15. head (w.r.t. reference pressure)
EGL (or TEL) and HGL
elevation
head (w.r.t. datum)
pressure
head (w.r.t. reference pressure)
velocity
head
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 _____ ____ __________ ______
downward
lower than reference pressure

16. Energy equation Why is static head important? Energy Grade Line
velocity head
Hydraulic G L
static head
pressure head
Why is static head important? z
elevation
pump
z = 0
datum

17. 1 2 1 2 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/γ Z

18. 2 1 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/γ 2 Q
P/γ Z 1 Z

19. Bernoulli Equation Assumption
_________ (viscosity can’t 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

20. Pipe Flow: Review 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? __________ _______.
dimensional analysis

21. Example Pipe Flow Problem
cs1
D=20 cm
L=500 m
Find the discharge, Q.
100 m
valve
cs2
Describe the process in terms of energy!

22. Flow Profile for Delaware Aqueduct
Rondout Reservoir
(EL. 256 m)
70.5 km
West Branch Reservoir
(EL m)
Sea Level
(Designed for 39 m3/s)
Need a relationship between flow rate and head loss

23. 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?

24. 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

25. Dimensionless Parameters
Reynolds Number
Froude Number
Weber Number
Mach Number
Pressure/Drag Coefficients
(dependent parameters that we measure experimentally)

26. Problem solving approach
Identify relevant forces and any other relevant parameters
If inertia is a relevant force, than the non dimensional Re, Fr, W, M, Cp numbers can be used
If inertia isn’t relevant than create new non dimensional force numbers using the relevant forces
Create additional non dimensional terms based on geometry, velocity, or density if there are repeating parameters
If the problem uses different repeating variables then substitute (for example wd instead of V)
Write the functional relationship

27. Friction Factor : Major losses
Laminar flow
Hagen-Poiseuille
Turbulent (Smooth, Transition, Rough)
Colebrook Formula
Moody diagram
Swamee-Jain

28. Laminar Flow Friction Factor
Hagen-Poiseuille
Darcy-Weisbach

29. 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
Inertial
viscous
pressure
Reynolds
Pressure coefficient
diameter, length, roughness height
l/D
e/D
Other repeating parameters?

30. 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

31. Laminar Flow Friction Factor
Hagen-Poiseuille
Darcy-Weisbach
Slope of ___ on log-log plot -1

32. Viscous Flow in Pipes

33. Viscous Flow: Dimensional Analysis
Two important parameters!
R - Laminar or Turbulent
e/D - Rough or Smooth
Where
and

34. Laminar and Turbulent Flows
Reynolds apparatus
inertia
damping
Transition at R of 2000

35. 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 v v v
Pipe Entrance
Non-Uniform Flow
Need equation for entrance length here

36. 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
Turbulent flow
Laminar flow

37. 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

38. Laminar Flow through Circular Tubes
Different geometry, same equation development (see Streeter, et al. p 268)
Apply equation of motion to cylindrical sleeve (use cylindrical coordinates)

39. 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
Q = VA =
Vpa2

40. Laminar Flow through Circular Tubes: Diagram
Shear
Velocity
Laminar flow
Shear at the wall
True for Laminar or Turbulent flow

41. Laminar flow Continue *vz *vz*dQ dQ=vz2πrdr but
Momentum is
Mass*velocity (m*v)
Momentum per unit volume is
*vz
Rate of flow of momentum is
*vz*dQ
dQ=vz2πrdr
but
vz = constant at a fixed value of r
Laminar flow

42. Laminar flow Continue
Hagen-Poiseuille

43. The Hagen-Poiseuille Equation
cv pipe flow
Constant cross section
h or z
Laminar pipe flow equations

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
Soal-Penyelesaian Hidraulika II, 1995

46. Air mengalir melalui pipa berdiameter 150 mm dan kecepatan 5,5 m/det
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

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 m2/det. Hitung tekanan di titik A.

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 m2/det. Apabila tekanan pada ujung atas adalah p=10 kPA ditanyakan tekanan di ujung bawah.

51. 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)
instantaneous
velocity
mean
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,) uz úz
Uz average ur úr
Ur average p P’
p average
Time

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

55. 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

57. 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

58. 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)

60. Log Law for Turbulent, Established Flow, Velocity Profiles
Integration and empirical results
Laminar
Turbulent
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)

65. Pipe Flow Energy Losses
Horizontal pipe
Dimensional Analysis
Darcy-Weisbach equation

66. 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

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
or f = 64/Re Moody chart

68. Friction Factor for Smooth, Transition, and Rough Turbulent flow
Smooth pipe, Re>3000
Rough pipe, [ (D/)/(Re√ƒ) <0.01]
Transition function for both smooth and rough pipe

69. 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)

70. Moody Diagram 0.10 0.08 0.06 0.05 0.04 friction factor 0.03 0.02 0.01
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
0.01
smooth
1E+03
1E+04
1E+05
1E+06
1E+07
1E+08 R

71. Fanning Diagram
f =16/Re

72. Swamee-Jain no f L hf Each equation has two terms. Why? 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
no f L hf
Each equation has two terms. Why?

73. Colebrook Solution for Q

74. Colebrook Solution for Q

75. Swamee D?

76. Pipe roughness Must be dimensionless! pipe material pipe roughness 
(mm)
glass, drawn brass, copper
0.0015
commercial steel or wrought iron
0.045
Must be dimensionless!
asphalted cast iron
0.12
galvanized iron
0.15
cast iron
0.26
concrete
rivet steel
corrugated metal 45
0.12
PVC

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)

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

79. 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

80. 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 can’t be extended to other fluids without further experimentation.
preferred

81. 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!
For a pipe
We can use Moody diagram or Swamee-Jain with D = 4Rh!

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
experimental

83. 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)

84. 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

85. Darcy Weisbach

86. Major and Minor Losses Major Losses: Minor 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 + V22/2g + z2 + Hmaj + Hmin

87. 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

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 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

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 = V3 = V4 same diameter tube
We must work backwards to solve this problem 1
γH2O= 62.4 lbs/ft3
R = .5’ 4’
R = .25’ 2 3 4 1’

92. Pressure Head : Only atmospheric  P1/γ = 0
Point 1:
Pressure Head : Only atmospheric  P1/γ = 0
Velocity Head : In a large tank, V1 = 0  V12/2g = 0
Elevation Head : Z1 = 4’ 1
γH2O= 62.4 lbs/ft3 4’
R = .5’
R = .25’ 2 3 4 1’

93. Pressure Head : Only atmospheric  P4/γ = 0
Point 4:
Apply the Bernoulli equation between 1 and = 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’ 1
γH2O= 62.4 lbs/ft3 4’
R = .5’
R = .25’ 2 3 4 1’

94. Point 3:
Apply the Bernoulli equation between 3 and 4 (V3=V4) P3/ =
P3 = 0
Pressure Head : P3/γ = 0
Velocity Head : V32/2g = 3’
Elevation Head : Z3 = 1’ 1
γH2O= 62.4 lbs/ft3 4’
R = .5’
R = .25’ 2 3 4 1’

95. Apply the Continuity Equation
Point 2:
Apply the Bernoulli equation between 2 and P2/ V22/2(32.2) + 1 =
Apply the Continuity Equation
(Π.52)V2 = (Π.252)x13.9  V2 = ft/s
P2/ /2(32.2) + 1 = 4  P2 = lbs/ft2
Pressure Head :
P2/γ = 2.81’
Velocity Head :
V22/2g = .19’
Elevation Head : Z2 = 1’ 1
γH2O= 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/γ =2.81’
V2/2g=3’
V2/2g=3’
Z=4’
HGL
Z=1’
Z=1’
Z=1’

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/γ + V12/2g + z1 = P2/γ + 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
Vehicle drag
Hydraulic jump
Vena contracta
Minor losses!

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)
diffusor angle ()
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8 20 40 60 80 KE

103. Sudden Contraction V2 V1 flow separation
losses are reduced with a gradual contraction

104. Sudden Contraction Cc A2/A1 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 0.2
0.2
0.4
A2/A1 Cc

105. Entrance Losses vena contracta
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

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 Technique: Discharge 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) constant
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 Expansion: Conservation of Energy1 2
z1 = z2
What is p1 - p2?

119. Head Loss due to Sudden Expansion: Conservation of Momentum1 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 Expansion!!! vena contracta
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 What is the head loss when a pipe enters a reservoir?
EGL
HGL V
What is the head loss when a pipe enters a reservoir?
Draw the EGL and HGL

126. Example cs1 cs2 100 m valve D=40 cm D=20 cm L=1000 m L=500 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
γoil= 8.82 kN/m3
f = .035 1
Z1 = ? 2
Z2 = 130 m
60 m
Kout=1
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. Hmaj = (fxLxV2)/(Dx2g)=(.035 x 197m x (1.58m/s)2)/(.15 x 2 x 9.8m/s2)
Pipe Flow Example
γoil= 8.82 kN/m3
f = .035 1
Z1 = ? 2
Z2 = 130 m
60 m
Kout=1
7 m
r/D = 0
130 m
r/D = 2
Apply Bernoulli’s equation between points 1 and 2: Assumptions: P1 = P2 = Atmospheric = V1 = V2 = 0 (large tank)
Z1 = m + 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 0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + Hmin
γoil= 8.82 kN/m3
f = .035 1
Z1 = ? 2
Z2 = 130 m
60 m
Kout=1
7 m
r/D = 0
130 m
r/D = 2
Z1 = m m + Hmin
Hmin= 2KbendV2/2g + KentV2/2g + KoutV2/2g
From Loss Coefficient table: Kbend = Kent = Kout = 1
Hmin = (0.19x ) x (1.582/2x9.8)
Hmin = 0.24 m

130. Pipe Flow Example 0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + 0.24m
γoil= 8.82 kN/m3
f = .035 1
Z1 = ? 2
Z2 = 130 m
60 m
Kout=1
7 m
r/D = 0
130 m
r/D = 2
Z1 = m + Hmaj + Hmin
Z1 = m m m
Z1 = 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