TL2101 Mekanika Fluida I
TL2101
Mekanika Fluida I
Benno Rahardyan
Pertemuan
Mg
1
Topik
Sub Topik
Tujuan Instruksional (TIK)
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
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 Latihan menggunakan
aplikasi di bidang TL
prinsip kekekalan
eneri khususnya
dalam bidang air
minum
2
Pipes are Everywhere!
Owner: City of
Hammond, IN
Project: Water Main
Relocation
Pipe Size: 54"
Pipes are Everywhere!
Drainage Pipes
Pipes
Pipes are Everywhere!
Water Mains
Types of Engineering
Problems
How big does the pipe have to be to
carry a flow of x m3/s?
What will the pressure in the water
distribution system be when a fire
hydrant is open?
FLUID DYNAMICS
THE BERNOULLI EQUATION
The laws of Statics that we have learned cannot solve
Dynamic Problems. There is no way to solve for the flow
rate, or Q. Therefore, we need a new dynamic approach
to Fluid Mechanics.
The Bernoulli Equation
By assuming that fluid motion is governed only by pressure and
gravity forces, applying Newton’s second law, F = ma, leads us to
the Bernoulli Equation.
P/ + V2/2g + z = constant along a streamline
(P=pressure =specific weight V=velocity g=gravity z=elevation)
A streamline is the path of one particle of water. Therefore, at any two
points along a streamline, the Bernoulli equation can be applied and,
using a set of engineering assumptions, unknown flows and pressures
can easily be solved for.
Free Jets
The velocity of a jet of water is clearly related to the depth of water
above the hole. The greater the depth, the higher the velocity. Similar
behavior can be seen as water flows at a very high velocity from the
reservoir behind the Glen Canyon Dam in Colorado
Closed Conduit Flow
Energy equation
EGL and HGL
Head loss
– major losses
– minor losses
Non circular conduits
The Energy Line and the Hydraulic Grade Line
Looking at the Bernoulli equation again:
P/γ + V2/2g + z = constant on a streamline
This constant is called the total head (energy), H
Because energy is assumed to be conserved, at any point along
the streamline, the total head is always constant
Each term in the Bernoulli equation is a type of head.
P/γ = Pressure Head
V2/2g = Velocity Head
Z = elevation head
These three heads, summed together, will always equal H
Next we will look at this graphically…
p1
V12
p2
V22
1
z1 h p 2
z 2 ht hL
2g
2g
Conservation of Energy
Kinetic, potential, and thermal
energy
hp = head supplied by a pump
ht = head given to a turbine
hL = mechanical energy converted to thermal
downstream from cross section 1!
Cross section 2 is ____________
irreversible
Point to point or control volume?
V is average velocity, kinetic energy V 2
Why ? _____________________________________
Energy Equation
Assumptions
hydrostatic
Pressure is _________ in both cross sections
– pressure changes are due to elevation only
p h
section is drawn perpendicular to the streamlines
(otherwise the _______
kinetic energy term is incorrect)
Constant ________at
the cross section
density
Steady flow
_______
p1
V12
p2
V22
1
z1 h p
2
z 2 ht hL
2g
2g
EGL (or TEL) and HGL
EGL
pressure
head (w.r.t.
reference
pressure)
p
V2
2g
p
HGL z
γ
z
velocity
head
elevation
head (w.r.t.
datum)
The energy grade line must always slope ___________
downward(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
lower
than
reference
plotted, the local pressures are _____
____
__________
______
pressure
Energy equation
velocity
head
Energy Grade
Line G
Hydraulic
L
V2
2g
p
static
head
Why is static
pressur
e head
head important?
z
pump
z=0
elevatio
n
datum
p1
V
p
V
1
z1 h p 2 2
z 2 ht hL
2g
2g
2
1
2
2
The Energy Line and the Hydraulic Grade Line
Lets first understand this drawing:
Measures the
Static Pressure
Measures the
Total Head
12
12
V2/2g
EL
HGL
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
Q
P/γ
Total Head along a system
HGL : Hydraulic Grade line
Z
Sum of the elevation and
the pressure heads along a
system
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.
V2/2g
EL
Q
P/γ
V2/2g
HGL
2
P/γ
Z
1
Z
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
Bernoulli Equation
Assumption
Frictionless
_________
(viscosity can’t be a
significant parameter!)
Along astreamline
__________
Steady
______ flow
Constantdensity
________
No pumps, turbines, or head loss
Why no
point velocity
V
p
z
const
Does direction matter? ____
no
2g
Useful when head loss is small
2
Pipe Flow: Review
p1
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?
__________analysis
_______.
dimensional
2
1
2
2
V
p2
V
1
z1 h p
2
z2 ht hL
2g
2g
Example Pipe Flow
Problem
cs1
Find the discharge, Q.
100 m
valve
D=20 cm
L=500 m
cs2
Describe the process in terms of energy!
p1
V12
p2
V22
1
z1 H p 2
z2 Ht hl
2g
2g
V22
z1
z2 hl
2g
a
V2 2 g z1 z2 hl
f
Flow Profile for Delaware
Aqueduct
Rondout Reservoir
(EL. 256 m)
Sea Level
70.5 km
West Branch Reservoir
(EL. 153.4 m)
p1
V12
p2
V22
1
z1 H p
2
z2 H t hl
2g
2g
(Designed for 39 m3/s)
hl z1 z2
Need a relationship between flow rate and head loss
Ratio of Forces
Create ratios of the various forces
The magnitude of the ratio will tell us
which forces are most important and
which forces could be ignored
Which force shall we use to create the
ratios?
Inertia as our Reference
Force
F
f a
F=ma F a
Fluids problems (except for statics) include a
velocity (V), a dimension of flow (l), and a
density ()
Substitute V, l, for the dimensions MLT
M l
M
f 2 2
LT
3
L l
T
l
V
Substitute for the dimensions of specific force
V2
fi
l
Dimensionless
Parameters
Reynolds Number
Froude Number
Weber Number
r Vl
Re =
m
V
Fr =
gl
V2
fi
l
V
fu 2
l
fg g
f 2
l
2
r
c
Mach Number
fE =
V
l
M
c
( Dp +r g Dz )
Pressure/Drag Coefficients
2 p C 2Drag
d
Cp
2
2
V
A
V
V 2 l
W
v
– (dependent parameters that we measure experimentally)
Problem solving approach
1.
2.
3.
4.
5.
6.
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 d instead of V)
Write the functional relationship
Friction Factor : Major
losses
Laminar flow
– Hagen-Poiseuille
Turbulent (Smooth, Transition, Rough)
– Colebrook Formula
– Moody diagram
– Swamee-Jain
Laminar Flow Friction
Factor
D 2 hl
V
32 L
hl
Hagen-Poiseuille
32 LV
gD 2
LV2
hl f
D 2g
32 LV
LV2
f
2
D 2g
gD
64 64
f
VD R
Darcy-Weisbach
Pipe Flow: Dimensional
Analysis
What are the important forces?
pressure Therefore
______,viscous
______,________.
Inertial
________number and _______________
Pressure coefficient .
Reynolds
What are the important geometric
length, roughness height
parameters? diameter,
_________________________
– Create dimensionless geometric groups
______,
______
l/D
/D
Other repeating parameters?
Write the functional relationship
l
C p f Re, ,
D D
Cp
2 p
V 2
l
C p f , , Re
D D
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?
2 p
D
Cp proportional to l
C p f , Re C p
2
D
f C p f , Re
l
D
l
D
f is friction factor
V
Laminar Flow Friction
Factor
D 2 hl
V
32 L
32 LV
hf
gD 2
L V2
hf f
D 2g
32 LV
L V2
f
2
gD
D 2g
64 64
f
VD Re
Hagen-Poiseuille
hf
128LQ
gD 4
Darcy-Weisbach
-1 on log-log plot
Slope of ___
Viscous Flow in
Pipes
Viscous Flow:
Dimensional Analysis
D
Cp
l
f ,R
D
Where
VD
R
Two important parameters!
R - Laminar or Turbulent
/D - Rough or Smooth
and
2 p
Cp
V 2
Laminar and Turbulent
Flows
Reynolds apparatus
R
VD inertia
damping
Transition at R of 2000
Boundary layer growth:
Transition length
What does the water near the pipeline wall experience?
_________________________
Drag or shear
Why does the water in the center of the pipeline speed
Conservation of mass
up? _________________________
Pipe
Entrance
v
v
Non-Uniform Flow
Need equation for entrance length here
v
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
http://www.engineering.uiowa.edu/~cfd/gallery/lim-turb.html
Laminar, Incompressible,
Steady, Uniform Flow
Between Parallel Plates
Through circular tubes
Hagen-Poiseuille Equation
Approach
– Because it is laminar flow the shear
forces can be quantified
– Velocity profiles can be determined from
a force balance
Laminar Flow through
Circular Tubes
Different geometry, same equation
development (see Streeter, et al. p
268)
Apply equation of motion to cylindrical
sleeve (use cylindrical coordinates)
Laminar Flow through
Circular Tubes: Equations
a2 r 2 d
u
p h
4 dl
a2 d
umax
p h
4 dl
a is radius of the tube
Max velocity when r = 0
a2 d
V
p h
8 dl
Velocity distribution is paraboloid of
average velocity
revolution therefore _____________
(V) is 1/2 umax
_____________
a 4 d
Q
p h
8 dl
Q = VA = Vpa2
Laminar Flow through
Circular Tubes: Diagram
a2 r 2 d
u
p h
4 dl
du
dr
r d
2 dl
du
dr
Velocity
p h
r d
2 dl
Shear
Laminar flow
p h
hl True for Laminar or
r
2l Turbulent flow
Shear at the wall
0
hl d
4l
Laminar flow
Continue
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
v z (v2rdr ) z v z (v2rdr ) z dz 0
Laminar flow
Laminar flow
Continue
2rzr r dz 2 (r dr)zr r dr dzp z 2rdr p z dz 2rdr g2rdrdz 0
dvz
dr
p pz 0 pz L gL
R 4 p
Q 0 2vz dr
8 L
R
Hagen-Poiseuille
The Hagen-Poiseuille
Equation
p1
V12
p2
V22
z1 1
H p z2 2
Ht hl cv pipe flow
1
2g
2
2g
p1
1
z1
p2
2
Constant cross section
h or z
z 2 hl
p
p
hl 1 z1 2 z 2
1
2
p
hl h
Laminar pipe flow equations
a 4 d
Q
p h
8 dl
D 4 hl
Q
128 L
D 4 d p
Q
h
128 dl
D 2 hl
V
32 L
hl
d p
h
dl
L
Prof. Dr. Ir. Bambang Triatmodjo, CES-UGM :
Hidraulika I, Beta Ofset Yogyakarta, 1993
Hidraulika II, Beta Ofset Yogyakarta, 1993
Soal-Penyelesaian Hidraulika I, 1994
Soal-Penyelesaian Hidraulika II, 1995
Air mengalir melalui pipa berdiameter
150 mm dan kecepatan 5,5
m/det.Kekentalan kinematik air adalah
1,3 x 10-4 m2/det. Selidiki tipe aliran
Bilangan reynolds :
VD 5,5 x0,15
5
6
,
35
x
10
v
1,3 x10 6
Karena Re 4000 berarti aliran turbulen
Re
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.
Diameter pipa : D 30 cm
Panjang pipa : L 4000 m
Debit aliran : Q 0,04 m 3 / dtk
4
2
Kekentalan kinematik : v 2,1x10 m / dtk
Rapat relatif : S 0,9 900 kg / m 3
Elevasi ujung atas pipa ( B) terhadap
ujung bawah ( A) : Z B Z A 50m
Kecepatn aliran :
Q
0,04
V
0,566 m / dtk
A x0,32
4
Bilangan reynolds :
VD 0,566 x0,3
808,6
4
v
2,1x10
Karena Re 2000 berarti aliran La min er
Re
Kehilangan tenaga
32vVL 32 x 2,1x10 4 x0,566,4000
hf
17,23 m
2
2
gD
9,82 x0,3
2
2
p A VA
p V
z A B B z B hf
2g
2g
V A VB
pA
0 0 50 17,23
pA
67,23m
p A 67,23 x900 x9,81
p A 593,574 N / m 2
p A 593,574kPa
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.
Diameter pipa : D 25 cm
Kehilangan tenaga
Panjang pipa : L 10.000 m
Debit aliran : Q 0,02 m 3 / dtk
32vVL 32 x 2,1x10 4 x0,4074 x10000
hf
gD 2
9,82 x0,252
44,65 m
Kekentalan kinematik : v 2,1x104 m 2 / dtk
Selisih elevasi kedua ujung :
Rapat relatif : S 0,9 900 kg / m 3
1
x10.000 50m
200
2
2
p A VA
p B VB
zA
z B hf
2g
2g
V A VB
Kemiringan pipa : 1 : 200
Tekanan di B : pB 10kPa 10.000 Nm 2
Kecepatn aliran :
Q
0,02
0,4074 m / dtk
A x0,252
4
Bilangan reynolds :
V
VD 0,4074 x0,25
485
v
2,1x10 4
Karena Re 2000 berarti aliran La min er
Re
z
pA
10.000
0
50 44,65
900 x9,81
pA
95,78m
p A 95,78 x900 x9,81
p A 845,642 N / m 2
p A 845,642kPa
Turbulent Pipe and
Channel Flow: Overview
Velocity distributions
Energy Losses
Steady Incompressible Flow through
Simple Pipes
Steady Uniform Flow in Open Channels
Turbulence
A characteristic of the flow.
How can we characterize turbulence?
– intensity of the velocity fluctuations
– size of the fluctuations (length scale) u
u u u
instantaneous
velocity
mean
velocity
velocity
fluctuation
u
Turbulent flow
When fluid flow at higher flowrates,
the streamlines are not steady and
straight and the flow is not laminar.
Generally, the flow field will vary in
both space and time with fluctuations
that comprise "turbulence
For this case almost all terms in the
Navier-Stokes equations are important
and there is no simple solution
uz
úz
Uz
average
ur
úr
Ur
average
p
p
average
P = P (D, , , L, U,)
P’
Time
Turbulent flow
All previous parameters involved three fundamental dimensions,
Mass, length, and time
From these parameters, three dimensionless groups can be build
P
L
)
2 f (Re,
U
D
UD
inertia
Re
Viscous forces
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
depth (R = 500)
– rivers: ________________
– pipes: _________________
diameter (R = 2000)
– lakes: ____________________
depth to thermocline
Actually a spectrum of eddy sizes
Turbulence: Flow
Instability
In turbulent flow (high Reynolds number) the force
viscosity is small relative to the
leading to stability (_________)
force leading to instability (_______).
inertia
Any disturbance in the flow results in large scale
motions superimposed on the mean flow.
Some of the kinetic energy of the flow is transferred to
these large scale motions (eddies).
Large scale instabilities gradually lose kinetic energy to
smaller scale motions.
The kinetic energy of the smallest eddies is dissipated
by viscous resistance and turned into heat.
(=___________)
head loss
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
constant
_______velocity
(compared to laminar flow)
Close to the pipe wall eddies are smaller (size
proportional to distance to the boundary)
Turbulent Flow Velocity
Profile
du
dy
du
Turbulent shear is from momentum transfer
h = eddy viscosity
dy
Length scale and velocity of “large” eddies
l I u I
du
u I lI
dy
2
I
l
du
dy
Dimensional analysis
y
Turbulent Flow Velocity
Profile
du
2
I
l
dy
increases as we
Size of the eddies __________
move further from the wall.
dy
l I y
2
y
2
du
k = 0.4 (from experiments)
dy
2
du
y
dy
2
2
du
y
dy
du
Log Law for Turbulent,
Established Flow, Velocity
Profiles
du
y
dy
u
1
yu*
ln
5.5 Integration and empirical results
u*
Turbulent
Laminar
0
Shear velocity
u*
y
u* u I
x
Pipe Flow: The Problem
We have the control volume energy
equation for pipe flow
We need to be able to predict the
head loss term.
We will use the results we obtained
using dimensional analysis
Friction Factor : Major
losses
Laminar flow
– Hagen-Poiseuille
Turbulent (Smooth, Transition, Rough)
– Colebrook Formula
– Moody diagram
– Swamee-Jain
Turbulent Pipe Flow Head
Loss
L V2
Proportional
___________ to the length of the pipe hf f
D 2g
square of the velocity
Proportional to the _______
(almost)
Increases
________ with surface roughness
Is a function of density and viscosity
Is __________
independent of pressure
Smooth, Transition, Rough
LV
h f
Turbulent Flow
D 2g
2
f
Hydraulically smooth
pipe law (von
Karman, 1930)
Rough pipe law (von
Karman, 1930)
Transition function
for both smooth and
1
rough pipe laws
f
(Colebrook)
Re f
1
2 log
f
2.51
1
3.7 D
2 log
f
2.51
D
2 log
3.7
Re f
(used to draw the Moody diagram)
Pipe Flow Energy Losses
p1
V12
p2
V22
1
z1 hp 2
z2 ht hl
2g
2g
hl
D
f
C p
L
2 p
Cp
V 2
2 ghl D
f 2
V L
p
f ,R
D
2 ghl
Cp
V2
LV2
hl f
D 2g
Horizontal pipe
Dimensional Analysis
Darcy-Weisbach equation
Turbulent Pipe Flow Head
Loss
Proportional to the length of the pipe
___________
___________
Proportional to the square of the
velocity (almost)
Inversely
________ with the diameter (almost)
________
Increase with surface roughness
Is a function of density and viscosity
Is independent
__________ of pressure
Surface Roughness
Additional dimensionless group /D need
to be characterize
Thus more than one curve on friction factorReynolds number plot
Fanning diagram or Moody diagram
Depending on the laminar region.
If, at the lowest Reynolds numbers, the laminar portion
corresponds to f =16/Re Fanning Chart
or f = 64/Re Moody chart
Friction Factor for Smooth, Transition, and
Rough Turbulent flow
P D
f
L 2 U 2
1
4.0 * log Re*
f
Smooth pipe, Re>3000
f 0.4
f 0.079Re 0.25
Rough pipe, [ (D/)/(Re√ƒ) 3 x 103
– less than 3% deviation
from results obtained
5/ 2
Q
2.22
D
with Moody diagram
ghf
L
easy to program for
computer or
calculator use
L
hf
2
no f
ghf
1.78
log
L
ghf
3.7 D
3/ 2
D
L
4.75
1.25 LQ
D 0.66
gh
f
2
5.2
0.04
L
Q
ghf
9.4
Each equation has two terms. Why?
Colebrook Solution for Q
8 LQ 2
hf f 2
g D5
2
1 1 8 LQ
f hf 2 g D 5
4Q
Re
D
4Q
2 g D5
Re f
hf
D
8 LQ 2
1 2 ghf D 3
Re f
L
1
2.51
D
2 log
3.7
f
Re f
2
1
2.51
D
4 log
f
3.7
Re f
hf g
8
f 2 5
D
LQ 2
Colebrook Solution for Q
2
1 8 LQ 2
4 log
2
5
hf g D
3.7 D
1
2.51
2 ghf D 3
L
log
3.7 D
1
2.51
2 ghf D 3
L
2
L Q
ghf D 5 / 2
5 / 2
Q
D
2
ghf
log
2.51
L
3.7 D
L
3
2 ghf D
Swamee
D?
8 LQ 2
hf f 2
g D5
2
1/ 4
1/ 5
5
1/ 5 0.04
Q Q
Q g g
2
1/ 4
2
1/ 5 1/ 25
Q 5/ 4 Q
Q
D 0.66
g
Q g
g
1/ 5
1/
5
2
2 1/ 4
2 1/ 5
Q
Q 5/ 4 Q
D
Q g
8 g
g
1/ 5
1/
4
1/
5
64 5/ 4 Q 2
Q 2
f 2
g
Q g
2
2
8
Q
D5 f 2
g
64 Q 2
D f 2
8g
5
2
5
1.25 Q Q
D 0.66
g g
2
2
1/ 5
Q 64
D f 2
8 g
2
1/ 4
1 5/ 4 Q
f
4 4 g
2
2
1/ 5 1/ 5
Q
Q g
2
Pipe roughness
pipe material
glass, drawn brass, copper
commercial steel or wrought iron
asphalted cast iron
galvanized iron
cast iron
concrete
rivet steel
corrugated metal
PVC
pipe roughness (mm)
0.0015
0.045
0.12
d Must be
0.15
dimensionless!
0.26
0.18-0.6
0.9-9.0
45
0.12
Solution Techniques
find
head loss given (D, type of pipe, Q)
0.25
2
8
LQ
f
4Q
2
hf f 2
Re
5
5.74
g
D
D
log 3.7 D Re0.9
find flow rate given (head, D, L, type of pipe)
5 / 2
Q
D
2
find
ghf
log
2.51
L
3.7 D
L
2 ghf D 3
pipe size given (head, type of pipe,L, Q)
4.75
2
LQ
1.25
D 0.66
gh
f
5.2
0.04
L
Q
ghf
9.4
Exponential Friction
Formulas
RLQ n
and hf = m
D
Commonly used in commercial
industrial settings
data
Only applicable over range
_____of__
____
collected
Hazen-Williams exponential
friction
formula
4.727
USC units
1.852
10.675 L æQ ö
hf = 4.8704
èC ø
D
Cn
SI units
R
10.675
SI units
n
C
C = Hazen-Williams coefcient
Head loss:
Hazen-Williams Coefficient
C
150
140
130
120
110
100
95
60-80
Condition
PVC
Extremely smooth, straight pipes; asbestos cement
Very smooth pipes; concrete; new cast iron
Wood stave; new welded steel
Vitrified clay; new riveted steel
Cast iron after years of use
Riveted steel after years of use
Old pipes in bad condition
Hazen-Williams
vs
Darcy-Weisbach
1.852
10.675 L Q
hf 4.8704
D
C
SI units
8 LQ 2
hf f 2
g D5
Both equations are empirical
Darcy-Weisbach is dimensionally correct,
andpreferred
________.
Hazen-Williams can be considered valid only
over the range of gathered data.
Hazen-Williams can’t be extended to other
fluids without further experimentation.
Non-Circular Conduits:
Hydraulic Radius Concept
A is cross sectional area
P is wetted perimeter
Rh is the “Hydraulic Radius”
(Area/Perimeter)
Don’tpconfuse
with
radius!
2
D
A 4
D
Rh = =
=
P pD
4
For a pipe
D =4 Rh
LV2
h f =f
D 2g
L V2
hf =f
4 Rh 2 g
We can use Moody diagram or Swamee-Jain with D = 4Rh!
Pipe Flow Summary (1)
Shear increases linearly
_________ 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 requireexperimental
___________
results
Pipe Flow Summary (2)
Energy equation left us with the elusive
head loss term
Dimensional analysis gave us the form of
the head loss term (pressure coefficient)
Experiments gave us the relationship
between the pressure coefficient and the
geometric parameters and the Reynolds
number (results summarized on Moody
diagram)
Questions
Can the Darcy-Weisbach equation and
Moody Diagram be used for fluids other
than water? Yes
_____
What
No
about the Hazen-Williams equation? ___
Does a perfectly smooth pipe have head loss?
Yes
_____
Is it possible to decrease the head loss in a
Yes
pipe by installing a smooth liner? ______
Darcy Weisbach
Major and Minor Losses
Major Losses:
Hmaj = f x (L/D)(V2/2g)
f = friction factor L = pipe length D = pipe diameter
V = Velocity g = gravity
Minor Losses:
Hmin = KL(V2/2g)
Kl = sum of loss coefficients V = Velocity g = gravity
When solving problems, the loss terms are added to the system at the
second point
P1/γ + V12/2g + z1 = P2/γ + V22/2g + z2 + Hmaj + Hmin
Hitung kehilangan tenaga karena gesekan di
dalam pipa sepanjang 1500 m dan diameter 20
cm, apabila air mengalir dengan kecepatan 2 m/
det. Koefisien gesekan f=0,02
Penyelesaian :
Panjang pipa : L = 1500 m
Diameter pipa : D = 20 cm = 0,2 m
Kecepatan aliran : V = 2 m/dtk
Koefisien gesekan f = 0,02
Kehilangan tenaga
L V2
hf f
D 2g
1500 x 2 2
0,02
0,2 x 2 x9,81
30,58 m
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
Kehilangan tenaga
8L
5
hf f
Q
g 2 D 5
8 x0,02 x1000
0,02
9,81x 2 x(0,015) 2
54,4 m
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
Kehilangan tenaga
L V2
hf f
D 2g
1500 x 2 2
0,02
0,2 x 2 x9,81
30,58 m
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
Kehilangan tenaga
hf f
8L
5
Q
g 2 D 5
8 x0,055 x1000
0,02
9,81x 2 x(0,015) 5
54,4 m
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
γH2O= 62.4 lbs/ft3
R = .5’
4’
2
R = .25’
3
4
1’
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’
2
R = .25’
3
4
1’
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’
1
γH2O= 62.4 lbs/ft3
4’
R = .5’
2
R = .25’
3
4
1’
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’
1
γH2O= 62.4 lbs/ft3
4’
R = .5’
2
R = .25’
3
4
1’
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 P2 = 175.5 lbs/ft2
Pressure Head :
P2/γ = 2.81’
1
γH2O= 62.4 lbs/ft3
4’
R = .5’
2
Velocity Head :
V22/2g = .19’
R = .25’
3
4
1’
Elevation Head :
Z2 = 1’
Plotting the EL and HGL
Energy Line = Sum of the Pressure, Velocity and Elevation heads
Hydraulic Grade Line = Sum of the Pressure and Velocity heads
V2/2g=.19’
EL
P/γ
=2.81’
V2/2g=3’ V2/2g=3’
Z=4’
HGL
Z=1’
Z=1’
Z=1’
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
Pipe Flow and the Energy Equation
Minor Losses : Hmin
Momentum losses in Pipe diameter changes and in pipe bends are called
minor losses. Unlike major losses, minor losses do not occur over the
length of the pipe, but only at points of momentum loss. Since Minor
losses occur at unique points along a pipe, to find the total minor loss
throughout a pipe, sum all of the minor losses along the pipe. Each
type of bend, or narrowing has a loss coefficient, KL to go with it.
Minor
Losses
Minor Losses
We previously obtained losses through an
expansion using conservation of energy,
momentum, and mass
Most minor losses can not be obtained
analytically, so they must be measured
Minor losses are often expressed as a loss
coefficient, K, times the velocity head.
hK
High R
C p f geometry, R
2 p
Cp
V 2
Cp
2 ghl
V2
hl C p
V
2
2g
V2
2g
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
potential
thermalenergy
Kinetic energy ________ and _________
Hydraulic jump
drag
Examples – Vehicle
________________________________
Minor losses!
__________________________________________
Vena
contracta
Losses can be minimized by gradual transitions
Minor Losses
Most minor losses can not be obtained
analytically, so they must be measured
Minor losses are often expressed as a
loss coefficient, K, times the velocity
head. High Re
C p = f ( geometry, Re )
Cp
2 ghl
V2
hl C p
V2
2g
V2
hl =K
2g
Head Loss due to Gradual
Expansion (Diffusor)
2
V1 V2
hE K E
2g
2
2
V2 A2
hE K E
1
2 g A1
KE
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80
diffusor angle ()
Sudden Contraction
2
1
V2
hc
1 2
C
2g
c
V1
V2
flow separation
Ac
losses are reduced with a gradual contraction
Cc
A2
Sudden Contraction
Cc
1
0.95
0.9
0.85
0.8
0.75
0.7
0.65
0.6
0
F
I
1
V
h G 1J
HC K2 g
2
c
c
2
2
0.2
0.4
0.6
A2/A1
Qorifice CAorifice 2 gh
0.8
1
Entrance Losses
Losses can be
K e 1 .0
reduced by
accelerating the
flow gradually and K 0.5
e
eliminating the
vena contracta
K e 0.04
he K e
V2
2g
Head Loss in Bends
High pressure
Head loss is a
function of the ratio of
Possible
the bend radius to the
separation
R from wall
pipe diameter (R/D)
Velocity distribution
D
returns to normal
Low pressure
several pipe diameters
V2
hb K b
downstream
2g
Kb varies from 0.6 - 0.9
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)
hv K v
V2
2g
Solution Techniques
Neglect minor losses
Equivalent pipe lengths
Iterative Techniques
Simultaneous Equations
Pipe Network Software
Iterative Techniques for
D and Q (given total head
loss)
Assume all head loss is major head
loss.
Calculate D or Q using Swamee-Jain
equations
Calculate minor losses
Find new major losses by subtracting
minor losses from total head loss
Solution Technique: Head
Loss
Can be solved directly
hminor K
Re
V
2
hminor K
2g
4Q
D
f
8Q 2
g 2 D 4
0.25
2
5.74
log
3.7 D Re 0.9
hl h f hminor
hf f
8
LQ 2
g 2 D 5
Solution Technique:
Discharge or Pipe Diameter
Re
Iterative technique
Set up simultaneous equations in Excel
4Q
D
hminor K
f
0.25
2
5.74
log
0.9
3.7 D Re
8Q 2
g 2 D 4
hl h f hminor
hf f
8
LQ 2
g 2 D 5
Use goal seek or Solver to
find discharge that makes the
calculated head loss equal
the given head loss.
Example: Minor and
Major Losses
Find the maximum dependable flow between the
reservoirs for a water temperature range of 4ºC to 20ºC.
Water
25 m elevation difference in reservoir water levels
Reentrant pipes at reservoirs
Standard elbows
2500 m of 8” PVC pipe
1500 m of 6” PVC pipe
Sudden contraction
Gate valve wide open
Directions
Assume fully turbulent (rough pipe law)
– find f from Moody (or from von Karman)
Find total head loss
Solve for Q using symbols (must include
minor losses) (no iteration required)
Obtain values for minor losses from notes or
text
Example (Continued)
What are the Reynolds number in the two
pipes?
Where are we on the Moody Diagram?
What value of K would the valve have to
produce to reduce the discharge by 50%?
What is the effect of temperature?
Why is the effect of temperature so small?
Example (Continued)
Were the minor losses negligible?
Accuracy of head loss calculations?
What happens if the roughness
increases by a factor of 10?
If you needed to increase the flow by
30% what could you do?
Suppose I changed 6” pipe, what is
minimum diameter needed?
Pipe Flow Summary (3)
Dimensionally correct equations fit to the
empirical results can be incorporated into
computer or calculator solution techniques
Minor losses are obtained from the pressure
coefficient based on the fact that the
pressure coefficient is _______
constantat high
Reynolds numbers
Solutions for discharge or pipe diameter often
require iterative or computer solutions
Loss Coefficients
Use this table to find loss coefficients:
Expansion:
Conservation of Energy
1
2
V12
p2
V22
z1 1
Hp
z2 2
H t hl
1
2g
2
2g
p1
p1 p2
hl
V22 V12
hl
2g
p1 p2
V12 V22
2g
z1 = z2
What is p1 - p2?
Head Loss due to Sudden Expansion:
Conservation of Momentum
A2
A1
x
2
1
M1 M 2 W Fp Fp Fss Apply in direction of flow
1
2
M 1 x M 2 x Fp Fp
1x
M 1 x V12 A1
Neglect surface shear
2x
M 2 x V22 A2
V12 A1 V22 A2 p1 A2 p2 A2
p1 p2
V22 V12
g
A1
A2
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 )
Head Loss due to
Sudden Expansion
hl
Energy
p1 p2
hl
V2
V1
g
2
V1 V2
hl
2g
V22 V12
2
1
A1 V2
Mass A V
2
1
2g
Momentum p1 p2
V22 V12
V12 V22
A1
A2
g
2
2
V V
2g
2
1
V22 2V1V2 V12
hl
2g
2
V
A1
hl
1
2g
A2
2
A1
K 1
A2
Contraction
EGL
V22
hc K c
2g
HGL
Expansion!!!
V1
V2
vena contracta
losses are reduced with a gradual contraction
Questions:
In the rough pipe law region if the flow rate
is doubled (be as specific as possible)
– What happens to the major head loss?
– What happens to the minor head loss?
Why do contractions have energy loss?
If you wanted to compare the importance of
minor vs. major losses for a specific pipeline,
what dimensionless terms could you
compare?
Entrance Losses
Losses can be
reduced by
accelerating the
flow gradually and
eliminating the
vena contracta
reentrant
K e 1.0
K e 0.5
K e 0.04
V2
he K e
2g
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?
_____
2
V
hv K v
2g
How can K be greater than 1?
Questions
EGL
HGL
What is the head
loss when a pipe
enters a reservoir?
V
V2
Draw the EGL and
HGL
2g
2
A
K 1 1
A2
V12
p2
V22
z1 1
Hp
z2 2
H t hl
1
2g
2
2g
p1
cs1
Example
valve
100 m
cs
2
D=40 cm
D=20 cm
L=1000 m
L=500 m
Find the discharge, Q.
What additional information do you need?
V22
100m =
+hl
Apply energy equation
2g
S-J on small pipe
How could you get a quick estimate?Use
_________________
Or spreadsheet solution: fnd head loss as function of Q.
Pipe Flow Example
1
Z1 = ?
γoil= 8.82 kN/m3
f = .035
60 m
2
Kout=1
7m
r/D = 0
Z2 = 130 m
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.
Pipe Flow Example
1
Z1 = ?
γoil= 8.82 kN/m3
f = .035
60 m
2
Kout=1
7m
r/D = 0
Z2 = 130 m
130 m
r/D = 2
Apply Bernoulli’s equation between points 1 and 2:
Assumptions: P1 = P2 = Atmospheric = 0
V1 = V2 = 0 (large tank)
0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin
Hmaj = (fxLxV2)/(Dx2g)=(.035 x 197m x (1.58m/s)2)/(.15 x 2 x 9.8m/s2)
Hmaj= 5.85m
Pipe Flow Example
1
Z1 = ?
γoil= 8.82 kN/m3
f = .035
60 m
Z2 = 130 m
Kout=1
7m
r/D = 0
2
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
Hmin = (0.19x2 + 0.5 + 1) x (1.582/2x9.8)
Hmin = 0.24 m
Kout = 1
Pipe Flow Example
1
Z1 = ?
γoil= 8.82 kN/m3
f = .035
60 m
2
Kout=1
7m
r/D = 0
Z2 = 130 m
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
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
Mekanika Fluida I
Benno Rahardyan
Pertemuan
Mg
1
Topik
Sub Topik
Tujuan Instruksional (TIK)
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
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 Latihan menggunakan
aplikasi di bidang TL
prinsip kekekalan
eneri khususnya
dalam bidang air
minum
2
Pipes are Everywhere!
Owner: City of
Hammond, IN
Project: Water Main
Relocation
Pipe Size: 54"
Pipes are Everywhere!
Drainage Pipes
Pipes
Pipes are Everywhere!
Water Mains
Types of Engineering
Problems
How big does the pipe have to be to
carry a flow of x m3/s?
What will the pressure in the water
distribution system be when a fire
hydrant is open?
FLUID DYNAMICS
THE BERNOULLI EQUATION
The laws of Statics that we have learned cannot solve
Dynamic Problems. There is no way to solve for the flow
rate, or Q. Therefore, we need a new dynamic approach
to Fluid Mechanics.
The Bernoulli Equation
By assuming that fluid motion is governed only by pressure and
gravity forces, applying Newton’s second law, F = ma, leads us to
the Bernoulli Equation.
P/ + V2/2g + z = constant along a streamline
(P=pressure =specific weight V=velocity g=gravity z=elevation)
A streamline is the path of one particle of water. Therefore, at any two
points along a streamline, the Bernoulli equation can be applied and,
using a set of engineering assumptions, unknown flows and pressures
can easily be solved for.
Free Jets
The velocity of a jet of water is clearly related to the depth of water
above the hole. The greater the depth, the higher the velocity. Similar
behavior can be seen as water flows at a very high velocity from the
reservoir behind the Glen Canyon Dam in Colorado
Closed Conduit Flow
Energy equation
EGL and HGL
Head loss
– major losses
– minor losses
Non circular conduits
The Energy Line and the Hydraulic Grade Line
Looking at the Bernoulli equation again:
P/γ + V2/2g + z = constant on a streamline
This constant is called the total head (energy), H
Because energy is assumed to be conserved, at any point along
the streamline, the total head is always constant
Each term in the Bernoulli equation is a type of head.
P/γ = Pressure Head
V2/2g = Velocity Head
Z = elevation head
These three heads, summed together, will always equal H
Next we will look at this graphically…
p1
V12
p2
V22
1
z1 h p 2
z 2 ht hL
2g
2g
Conservation of Energy
Kinetic, potential, and thermal
energy
hp = head supplied by a pump
ht = head given to a turbine
hL = mechanical energy converted to thermal
downstream from cross section 1!
Cross section 2 is ____________
irreversible
Point to point or control volume?
V is average velocity, kinetic energy V 2
Why ? _____________________________________
Energy Equation
Assumptions
hydrostatic
Pressure is _________ in both cross sections
– pressure changes are due to elevation only
p h
section is drawn perpendicular to the streamlines
(otherwise the _______
kinetic energy term is incorrect)
Constant ________at
the cross section
density
Steady flow
_______
p1
V12
p2
V22
1
z1 h p
2
z 2 ht hL
2g
2g
EGL (or TEL) and HGL
EGL
pressure
head (w.r.t.
reference
pressure)
p
V2
2g
p
HGL z
γ
z
velocity
head
elevation
head (w.r.t.
datum)
The energy grade line must always slope ___________
downward(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
lower
than
reference
plotted, the local pressures are _____
____
__________
______
pressure
Energy equation
velocity
head
Energy Grade
Line G
Hydraulic
L
V2
2g
p
static
head
Why is static
pressur
e head
head important?
z
pump
z=0
elevatio
n
datum
p1
V
p
V
1
z1 h p 2 2
z 2 ht hL
2g
2g
2
1
2
2
The Energy Line and the Hydraulic Grade Line
Lets first understand this drawing:
Measures the
Static Pressure
Measures the
Total Head
12
12
V2/2g
EL
HGL
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
Q
P/γ
Total Head along a system
HGL : Hydraulic Grade line
Z
Sum of the elevation and
the pressure heads along a
system
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.
V2/2g
EL
Q
P/γ
V2/2g
HGL
2
P/γ
Z
1
Z
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
Bernoulli Equation
Assumption
Frictionless
_________
(viscosity can’t be a
significant parameter!)
Along astreamline
__________
Steady
______ flow
Constantdensity
________
No pumps, turbines, or head loss
Why no
point velocity
V
p
z
const
Does direction matter? ____
no
2g
Useful when head loss is small
2
Pipe Flow: Review
p1
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?
__________analysis
_______.
dimensional
2
1
2
2
V
p2
V
1
z1 h p
2
z2 ht hL
2g
2g
Example Pipe Flow
Problem
cs1
Find the discharge, Q.
100 m
valve
D=20 cm
L=500 m
cs2
Describe the process in terms of energy!
p1
V12
p2
V22
1
z1 H p 2
z2 Ht hl
2g
2g
V22
z1
z2 hl
2g
a
V2 2 g z1 z2 hl
f
Flow Profile for Delaware
Aqueduct
Rondout Reservoir
(EL. 256 m)
Sea Level
70.5 km
West Branch Reservoir
(EL. 153.4 m)
p1
V12
p2
V22
1
z1 H p
2
z2 H t hl
2g
2g
(Designed for 39 m3/s)
hl z1 z2
Need a relationship between flow rate and head loss
Ratio of Forces
Create ratios of the various forces
The magnitude of the ratio will tell us
which forces are most important and
which forces could be ignored
Which force shall we use to create the
ratios?
Inertia as our Reference
Force
F
f a
F=ma F a
Fluids problems (except for statics) include a
velocity (V), a dimension of flow (l), and a
density ()
Substitute V, l, for the dimensions MLT
M l
M
f 2 2
LT
3
L l
T
l
V
Substitute for the dimensions of specific force
V2
fi
l
Dimensionless
Parameters
Reynolds Number
Froude Number
Weber Number
r Vl
Re =
m
V
Fr =
gl
V2
fi
l
V
fu 2
l
fg g
f 2
l
2
r
c
Mach Number
fE =
V
l
M
c
( Dp +r g Dz )
Pressure/Drag Coefficients
2 p C 2Drag
d
Cp
2
2
V
A
V
V 2 l
W
v
– (dependent parameters that we measure experimentally)
Problem solving approach
1.
2.
3.
4.
5.
6.
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 d instead of V)
Write the functional relationship
Friction Factor : Major
losses
Laminar flow
– Hagen-Poiseuille
Turbulent (Smooth, Transition, Rough)
– Colebrook Formula
– Moody diagram
– Swamee-Jain
Laminar Flow Friction
Factor
D 2 hl
V
32 L
hl
Hagen-Poiseuille
32 LV
gD 2
LV2
hl f
D 2g
32 LV
LV2
f
2
D 2g
gD
64 64
f
VD R
Darcy-Weisbach
Pipe Flow: Dimensional
Analysis
What are the important forces?
pressure Therefore
______,viscous
______,________.
Inertial
________number and _______________
Pressure coefficient .
Reynolds
What are the important geometric
length, roughness height
parameters? diameter,
_________________________
– Create dimensionless geometric groups
______,
______
l/D
/D
Other repeating parameters?
Write the functional relationship
l
C p f Re, ,
D D
Cp
2 p
V 2
l
C p f , , Re
D D
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?
2 p
D
Cp proportional to l
C p f , Re C p
2
D
f C p f , Re
l
D
l
D
f is friction factor
V
Laminar Flow Friction
Factor
D 2 hl
V
32 L
32 LV
hf
gD 2
L V2
hf f
D 2g
32 LV
L V2
f
2
gD
D 2g
64 64
f
VD Re
Hagen-Poiseuille
hf
128LQ
gD 4
Darcy-Weisbach
-1 on log-log plot
Slope of ___
Viscous Flow in
Pipes
Viscous Flow:
Dimensional Analysis
D
Cp
l
f ,R
D
Where
VD
R
Two important parameters!
R - Laminar or Turbulent
/D - Rough or Smooth
and
2 p
Cp
V 2
Laminar and Turbulent
Flows
Reynolds apparatus
R
VD inertia
damping
Transition at R of 2000
Boundary layer growth:
Transition length
What does the water near the pipeline wall experience?
_________________________
Drag or shear
Why does the water in the center of the pipeline speed
Conservation of mass
up? _________________________
Pipe
Entrance
v
v
Non-Uniform Flow
Need equation for entrance length here
v
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
http://www.engineering.uiowa.edu/~cfd/gallery/lim-turb.html
Laminar, Incompressible,
Steady, Uniform Flow
Between Parallel Plates
Through circular tubes
Hagen-Poiseuille Equation
Approach
– Because it is laminar flow the shear
forces can be quantified
– Velocity profiles can be determined from
a force balance
Laminar Flow through
Circular Tubes
Different geometry, same equation
development (see Streeter, et al. p
268)
Apply equation of motion to cylindrical
sleeve (use cylindrical coordinates)
Laminar Flow through
Circular Tubes: Equations
a2 r 2 d
u
p h
4 dl
a2 d
umax
p h
4 dl
a is radius of the tube
Max velocity when r = 0
a2 d
V
p h
8 dl
Velocity distribution is paraboloid of
average velocity
revolution therefore _____________
(V) is 1/2 umax
_____________
a 4 d
Q
p h
8 dl
Q = VA = Vpa2
Laminar Flow through
Circular Tubes: Diagram
a2 r 2 d
u
p h
4 dl
du
dr
r d
2 dl
du
dr
Velocity
p h
r d
2 dl
Shear
Laminar flow
p h
hl True for Laminar or
r
2l Turbulent flow
Shear at the wall
0
hl d
4l
Laminar flow
Continue
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
v z (v2rdr ) z v z (v2rdr ) z dz 0
Laminar flow
Laminar flow
Continue
2rzr r dz 2 (r dr)zr r dr dzp z 2rdr p z dz 2rdr g2rdrdz 0
dvz
dr
p pz 0 pz L gL
R 4 p
Q 0 2vz dr
8 L
R
Hagen-Poiseuille
The Hagen-Poiseuille
Equation
p1
V12
p2
V22
z1 1
H p z2 2
Ht hl cv pipe flow
1
2g
2
2g
p1
1
z1
p2
2
Constant cross section
h or z
z 2 hl
p
p
hl 1 z1 2 z 2
1
2
p
hl h
Laminar pipe flow equations
a 4 d
Q
p h
8 dl
D 4 hl
Q
128 L
D 4 d p
Q
h
128 dl
D 2 hl
V
32 L
hl
d p
h
dl
L
Prof. Dr. Ir. Bambang Triatmodjo, CES-UGM :
Hidraulika I, Beta Ofset Yogyakarta, 1993
Hidraulika II, Beta Ofset Yogyakarta, 1993
Soal-Penyelesaian Hidraulika I, 1994
Soal-Penyelesaian Hidraulika II, 1995
Air mengalir melalui pipa berdiameter
150 mm dan kecepatan 5,5
m/det.Kekentalan kinematik air adalah
1,3 x 10-4 m2/det. Selidiki tipe aliran
Bilangan reynolds :
VD 5,5 x0,15
5
6
,
35
x
10
v
1,3 x10 6
Karena Re 4000 berarti aliran turbulen
Re
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.
Diameter pipa : D 30 cm
Panjang pipa : L 4000 m
Debit aliran : Q 0,04 m 3 / dtk
4
2
Kekentalan kinematik : v 2,1x10 m / dtk
Rapat relatif : S 0,9 900 kg / m 3
Elevasi ujung atas pipa ( B) terhadap
ujung bawah ( A) : Z B Z A 50m
Kecepatn aliran :
Q
0,04
V
0,566 m / dtk
A x0,32
4
Bilangan reynolds :
VD 0,566 x0,3
808,6
4
v
2,1x10
Karena Re 2000 berarti aliran La min er
Re
Kehilangan tenaga
32vVL 32 x 2,1x10 4 x0,566,4000
hf
17,23 m
2
2
gD
9,82 x0,3
2
2
p A VA
p V
z A B B z B hf
2g
2g
V A VB
pA
0 0 50 17,23
pA
67,23m
p A 67,23 x900 x9,81
p A 593,574 N / m 2
p A 593,574kPa
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.
Diameter pipa : D 25 cm
Kehilangan tenaga
Panjang pipa : L 10.000 m
Debit aliran : Q 0,02 m 3 / dtk
32vVL 32 x 2,1x10 4 x0,4074 x10000
hf
gD 2
9,82 x0,252
44,65 m
Kekentalan kinematik : v 2,1x104 m 2 / dtk
Selisih elevasi kedua ujung :
Rapat relatif : S 0,9 900 kg / m 3
1
x10.000 50m
200
2
2
p A VA
p B VB
zA
z B hf
2g
2g
V A VB
Kemiringan pipa : 1 : 200
Tekanan di B : pB 10kPa 10.000 Nm 2
Kecepatn aliran :
Q
0,02
0,4074 m / dtk
A x0,252
4
Bilangan reynolds :
V
VD 0,4074 x0,25
485
v
2,1x10 4
Karena Re 2000 berarti aliran La min er
Re
z
pA
10.000
0
50 44,65
900 x9,81
pA
95,78m
p A 95,78 x900 x9,81
p A 845,642 N / m 2
p A 845,642kPa
Turbulent Pipe and
Channel Flow: Overview
Velocity distributions
Energy Losses
Steady Incompressible Flow through
Simple Pipes
Steady Uniform Flow in Open Channels
Turbulence
A characteristic of the flow.
How can we characterize turbulence?
– intensity of the velocity fluctuations
– size of the fluctuations (length scale) u
u u u
instantaneous
velocity
mean
velocity
velocity
fluctuation
u
Turbulent flow
When fluid flow at higher flowrates,
the streamlines are not steady and
straight and the flow is not laminar.
Generally, the flow field will vary in
both space and time with fluctuations
that comprise "turbulence
For this case almost all terms in the
Navier-Stokes equations are important
and there is no simple solution
uz
úz
Uz
average
ur
úr
Ur
average
p
p
average
P = P (D, , , L, U,)
P’
Time
Turbulent flow
All previous parameters involved three fundamental dimensions,
Mass, length, and time
From these parameters, three dimensionless groups can be build
P
L
)
2 f (Re,
U
D
UD
inertia
Re
Viscous forces
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
depth (R = 500)
– rivers: ________________
– pipes: _________________
diameter (R = 2000)
– lakes: ____________________
depth to thermocline
Actually a spectrum of eddy sizes
Turbulence: Flow
Instability
In turbulent flow (high Reynolds number) the force
viscosity is small relative to the
leading to stability (_________)
force leading to instability (_______).
inertia
Any disturbance in the flow results in large scale
motions superimposed on the mean flow.
Some of the kinetic energy of the flow is transferred to
these large scale motions (eddies).
Large scale instabilities gradually lose kinetic energy to
smaller scale motions.
The kinetic energy of the smallest eddies is dissipated
by viscous resistance and turned into heat.
(=___________)
head loss
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
constant
_______velocity
(compared to laminar flow)
Close to the pipe wall eddies are smaller (size
proportional to distance to the boundary)
Turbulent Flow Velocity
Profile
du
dy
du
Turbulent shear is from momentum transfer
h = eddy viscosity
dy
Length scale and velocity of “large” eddies
l I u I
du
u I lI
dy
2
I
l
du
dy
Dimensional analysis
y
Turbulent Flow Velocity
Profile
du
2
I
l
dy
increases as we
Size of the eddies __________
move further from the wall.
dy
l I y
2
y
2
du
k = 0.4 (from experiments)
dy
2
du
y
dy
2
2
du
y
dy
du
Log Law for Turbulent,
Established Flow, Velocity
Profiles
du
y
dy
u
1
yu*
ln
5.5 Integration and empirical results
u*
Turbulent
Laminar
0
Shear velocity
u*
y
u* u I
x
Pipe Flow: The Problem
We have the control volume energy
equation for pipe flow
We need to be able to predict the
head loss term.
We will use the results we obtained
using dimensional analysis
Friction Factor : Major
losses
Laminar flow
– Hagen-Poiseuille
Turbulent (Smooth, Transition, Rough)
– Colebrook Formula
– Moody diagram
– Swamee-Jain
Turbulent Pipe Flow Head
Loss
L V2
Proportional
___________ to the length of the pipe hf f
D 2g
square of the velocity
Proportional to the _______
(almost)
Increases
________ with surface roughness
Is a function of density and viscosity
Is __________
independent of pressure
Smooth, Transition, Rough
LV
h f
Turbulent Flow
D 2g
2
f
Hydraulically smooth
pipe law (von
Karman, 1930)
Rough pipe law (von
Karman, 1930)
Transition function
for both smooth and
1
rough pipe laws
f
(Colebrook)
Re f
1
2 log
f
2.51
1
3.7 D
2 log
f
2.51
D
2 log
3.7
Re f
(used to draw the Moody diagram)
Pipe Flow Energy Losses
p1
V12
p2
V22
1
z1 hp 2
z2 ht hl
2g
2g
hl
D
f
C p
L
2 p
Cp
V 2
2 ghl D
f 2
V L
p
f ,R
D
2 ghl
Cp
V2
LV2
hl f
D 2g
Horizontal pipe
Dimensional Analysis
Darcy-Weisbach equation
Turbulent Pipe Flow Head
Loss
Proportional to the length of the pipe
___________
___________
Proportional to the square of the
velocity (almost)
Inversely
________ with the diameter (almost)
________
Increase with surface roughness
Is a function of density and viscosity
Is independent
__________ of pressure
Surface Roughness
Additional dimensionless group /D need
to be characterize
Thus more than one curve on friction factorReynolds number plot
Fanning diagram or Moody diagram
Depending on the laminar region.
If, at the lowest Reynolds numbers, the laminar portion
corresponds to f =16/Re Fanning Chart
or f = 64/Re Moody chart
Friction Factor for Smooth, Transition, and
Rough Turbulent flow
P D
f
L 2 U 2
1
4.0 * log Re*
f
Smooth pipe, Re>3000
f 0.4
f 0.079Re 0.25
Rough pipe, [ (D/)/(Re√ƒ) 3 x 103
– less than 3% deviation
from results obtained
5/ 2
Q
2.22
D
with Moody diagram
ghf
L
easy to program for
computer or
calculator use
L
hf
2
no f
ghf
1.78
log
L
ghf
3.7 D
3/ 2
D
L
4.75
1.25 LQ
D 0.66
gh
f
2
5.2
0.04
L
Q
ghf
9.4
Each equation has two terms. Why?
Colebrook Solution for Q
8 LQ 2
hf f 2
g D5
2
1 1 8 LQ
f hf 2 g D 5
4Q
Re
D
4Q
2 g D5
Re f
hf
D
8 LQ 2
1 2 ghf D 3
Re f
L
1
2.51
D
2 log
3.7
f
Re f
2
1
2.51
D
4 log
f
3.7
Re f
hf g
8
f 2 5
D
LQ 2
Colebrook Solution for Q
2
1 8 LQ 2
4 log
2
5
hf g D
3.7 D
1
2.51
2 ghf D 3
L
log
3.7 D
1
2.51
2 ghf D 3
L
2
L Q
ghf D 5 / 2
5 / 2
Q
D
2
ghf
log
2.51
L
3.7 D
L
3
2 ghf D
Swamee
D?
8 LQ 2
hf f 2
g D5
2
1/ 4
1/ 5
5
1/ 5 0.04
Q Q
Q g g
2
1/ 4
2
1/ 5 1/ 25
Q 5/ 4 Q
Q
D 0.66
g
Q g
g
1/ 5
1/
5
2
2 1/ 4
2 1/ 5
Q
Q 5/ 4 Q
D
Q g
8 g
g
1/ 5
1/
4
1/
5
64 5/ 4 Q 2
Q 2
f 2
g
Q g
2
2
8
Q
D5 f 2
g
64 Q 2
D f 2
8g
5
2
5
1.25 Q Q
D 0.66
g g
2
2
1/ 5
Q 64
D f 2
8 g
2
1/ 4
1 5/ 4 Q
f
4 4 g
2
2
1/ 5 1/ 5
Q
Q g
2
Pipe roughness
pipe material
glass, drawn brass, copper
commercial steel or wrought iron
asphalted cast iron
galvanized iron
cast iron
concrete
rivet steel
corrugated metal
PVC
pipe roughness (mm)
0.0015
0.045
0.12
d Must be
0.15
dimensionless!
0.26
0.18-0.6
0.9-9.0
45
0.12
Solution Techniques
find
head loss given (D, type of pipe, Q)
0.25
2
8
LQ
f
4Q
2
hf f 2
Re
5
5.74
g
D
D
log 3.7 D Re0.9
find flow rate given (head, D, L, type of pipe)
5 / 2
Q
D
2
find
ghf
log
2.51
L
3.7 D
L
2 ghf D 3
pipe size given (head, type of pipe,L, Q)
4.75
2
LQ
1.25
D 0.66
gh
f
5.2
0.04
L
Q
ghf
9.4
Exponential Friction
Formulas
RLQ n
and hf = m
D
Commonly used in commercial
industrial settings
data
Only applicable over range
_____of__
____
collected
Hazen-Williams exponential
friction
formula
4.727
USC units
1.852
10.675 L æQ ö
hf = 4.8704
èC ø
D
Cn
SI units
R
10.675
SI units
n
C
C = Hazen-Williams coefcient
Head loss:
Hazen-Williams Coefficient
C
150
140
130
120
110
100
95
60-80
Condition
PVC
Extremely smooth, straight pipes; asbestos cement
Very smooth pipes; concrete; new cast iron
Wood stave; new welded steel
Vitrified clay; new riveted steel
Cast iron after years of use
Riveted steel after years of use
Old pipes in bad condition
Hazen-Williams
vs
Darcy-Weisbach
1.852
10.675 L Q
hf 4.8704
D
C
SI units
8 LQ 2
hf f 2
g D5
Both equations are empirical
Darcy-Weisbach is dimensionally correct,
andpreferred
________.
Hazen-Williams can be considered valid only
over the range of gathered data.
Hazen-Williams can’t be extended to other
fluids without further experimentation.
Non-Circular Conduits:
Hydraulic Radius Concept
A is cross sectional area
P is wetted perimeter
Rh is the “Hydraulic Radius”
(Area/Perimeter)
Don’tpconfuse
with
radius!
2
D
A 4
D
Rh = =
=
P pD
4
For a pipe
D =4 Rh
LV2
h f =f
D 2g
L V2
hf =f
4 Rh 2 g
We can use Moody diagram or Swamee-Jain with D = 4Rh!
Pipe Flow Summary (1)
Shear increases linearly
_________ 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 requireexperimental
___________
results
Pipe Flow Summary (2)
Energy equation left us with the elusive
head loss term
Dimensional analysis gave us the form of
the head loss term (pressure coefficient)
Experiments gave us the relationship
between the pressure coefficient and the
geometric parameters and the Reynolds
number (results summarized on Moody
diagram)
Questions
Can the Darcy-Weisbach equation and
Moody Diagram be used for fluids other
than water? Yes
_____
What
No
about the Hazen-Williams equation? ___
Does a perfectly smooth pipe have head loss?
Yes
_____
Is it possible to decrease the head loss in a
Yes
pipe by installing a smooth liner? ______
Darcy Weisbach
Major and Minor Losses
Major Losses:
Hmaj = f x (L/D)(V2/2g)
f = friction factor L = pipe length D = pipe diameter
V = Velocity g = gravity
Minor Losses:
Hmin = KL(V2/2g)
Kl = sum of loss coefficients V = Velocity g = gravity
When solving problems, the loss terms are added to the system at the
second point
P1/γ + V12/2g + z1 = P2/γ + V22/2g + z2 + Hmaj + Hmin
Hitung kehilangan tenaga karena gesekan di
dalam pipa sepanjang 1500 m dan diameter 20
cm, apabila air mengalir dengan kecepatan 2 m/
det. Koefisien gesekan f=0,02
Penyelesaian :
Panjang pipa : L = 1500 m
Diameter pipa : D = 20 cm = 0,2 m
Kecepatan aliran : V = 2 m/dtk
Koefisien gesekan f = 0,02
Kehilangan tenaga
L V2
hf f
D 2g
1500 x 2 2
0,02
0,2 x 2 x9,81
30,58 m
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
Kehilangan tenaga
8L
5
hf f
Q
g 2 D 5
8 x0,02 x1000
0,02
9,81x 2 x(0,015) 2
54,4 m
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
Kehilangan tenaga
L V2
hf f
D 2g
1500 x 2 2
0,02
0,2 x 2 x9,81
30,58 m
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
Kehilangan tenaga
hf f
8L
5
Q
g 2 D 5
8 x0,055 x1000
0,02
9,81x 2 x(0,015) 5
54,4 m
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
γH2O= 62.4 lbs/ft3
R = .5’
4’
2
R = .25’
3
4
1’
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’
2
R = .25’
3
4
1’
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’
1
γH2O= 62.4 lbs/ft3
4’
R = .5’
2
R = .25’
3
4
1’
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’
1
γH2O= 62.4 lbs/ft3
4’
R = .5’
2
R = .25’
3
4
1’
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 P2 = 175.5 lbs/ft2
Pressure Head :
P2/γ = 2.81’
1
γH2O= 62.4 lbs/ft3
4’
R = .5’
2
Velocity Head :
V22/2g = .19’
R = .25’
3
4
1’
Elevation Head :
Z2 = 1’
Plotting the EL and HGL
Energy Line = Sum of the Pressure, Velocity and Elevation heads
Hydraulic Grade Line = Sum of the Pressure and Velocity heads
V2/2g=.19’
EL
P/γ
=2.81’
V2/2g=3’ V2/2g=3’
Z=4’
HGL
Z=1’
Z=1’
Z=1’
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
Pipe Flow and the Energy Equation
Minor Losses : Hmin
Momentum losses in Pipe diameter changes and in pipe bends are called
minor losses. Unlike major losses, minor losses do not occur over the
length of the pipe, but only at points of momentum loss. Since Minor
losses occur at unique points along a pipe, to find the total minor loss
throughout a pipe, sum all of the minor losses along the pipe. Each
type of bend, or narrowing has a loss coefficient, KL to go with it.
Minor
Losses
Minor Losses
We previously obtained losses through an
expansion using conservation of energy,
momentum, and mass
Most minor losses can not be obtained
analytically, so they must be measured
Minor losses are often expressed as a loss
coefficient, K, times the velocity head.
hK
High R
C p f geometry, R
2 p
Cp
V 2
Cp
2 ghl
V2
hl C p
V
2
2g
V2
2g
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
potential
thermalenergy
Kinetic energy ________ and _________
Hydraulic jump
drag
Examples – Vehicle
________________________________
Minor losses!
__________________________________________
Vena
contracta
Losses can be minimized by gradual transitions
Minor Losses
Most minor losses can not be obtained
analytically, so they must be measured
Minor losses are often expressed as a
loss coefficient, K, times the velocity
head. High Re
C p = f ( geometry, Re )
Cp
2 ghl
V2
hl C p
V2
2g
V2
hl =K
2g
Head Loss due to Gradual
Expansion (Diffusor)
2
V1 V2
hE K E
2g
2
2
V2 A2
hE K E
1
2 g A1
KE
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80
diffusor angle ()
Sudden Contraction
2
1
V2
hc
1 2
C
2g
c
V1
V2
flow separation
Ac
losses are reduced with a gradual contraction
Cc
A2
Sudden Contraction
Cc
1
0.95
0.9
0.85
0.8
0.75
0.7
0.65
0.6
0
F
I
1
V
h G 1J
HC K2 g
2
c
c
2
2
0.2
0.4
0.6
A2/A1
Qorifice CAorifice 2 gh
0.8
1
Entrance Losses
Losses can be
K e 1 .0
reduced by
accelerating the
flow gradually and K 0.5
e
eliminating the
vena contracta
K e 0.04
he K e
V2
2g
Head Loss in Bends
High pressure
Head loss is a
function of the ratio of
Possible
the bend radius to the
separation
R from wall
pipe diameter (R/D)
Velocity distribution
D
returns to normal
Low pressure
several pipe diameters
V2
hb K b
downstream
2g
Kb varies from 0.6 - 0.9
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)
hv K v
V2
2g
Solution Techniques
Neglect minor losses
Equivalent pipe lengths
Iterative Techniques
Simultaneous Equations
Pipe Network Software
Iterative Techniques for
D and Q (given total head
loss)
Assume all head loss is major head
loss.
Calculate D or Q using Swamee-Jain
equations
Calculate minor losses
Find new major losses by subtracting
minor losses from total head loss
Solution Technique: Head
Loss
Can be solved directly
hminor K
Re
V
2
hminor K
2g
4Q
D
f
8Q 2
g 2 D 4
0.25
2
5.74
log
3.7 D Re 0.9
hl h f hminor
hf f
8
LQ 2
g 2 D 5
Solution Technique:
Discharge or Pipe Diameter
Re
Iterative technique
Set up simultaneous equations in Excel
4Q
D
hminor K
f
0.25
2
5.74
log
0.9
3.7 D Re
8Q 2
g 2 D 4
hl h f hminor
hf f
8
LQ 2
g 2 D 5
Use goal seek or Solver to
find discharge that makes the
calculated head loss equal
the given head loss.
Example: Minor and
Major Losses
Find the maximum dependable flow between the
reservoirs for a water temperature range of 4ºC to 20ºC.
Water
25 m elevation difference in reservoir water levels
Reentrant pipes at reservoirs
Standard elbows
2500 m of 8” PVC pipe
1500 m of 6” PVC pipe
Sudden contraction
Gate valve wide open
Directions
Assume fully turbulent (rough pipe law)
– find f from Moody (or from von Karman)
Find total head loss
Solve for Q using symbols (must include
minor losses) (no iteration required)
Obtain values for minor losses from notes or
text
Example (Continued)
What are the Reynolds number in the two
pipes?
Where are we on the Moody Diagram?
What value of K would the valve have to
produce to reduce the discharge by 50%?
What is the effect of temperature?
Why is the effect of temperature so small?
Example (Continued)
Were the minor losses negligible?
Accuracy of head loss calculations?
What happens if the roughness
increases by a factor of 10?
If you needed to increase the flow by
30% what could you do?
Suppose I changed 6” pipe, what is
minimum diameter needed?
Pipe Flow Summary (3)
Dimensionally correct equations fit to the
empirical results can be incorporated into
computer or calculator solution techniques
Minor losses are obtained from the pressure
coefficient based on the fact that the
pressure coefficient is _______
constantat high
Reynolds numbers
Solutions for discharge or pipe diameter often
require iterative or computer solutions
Loss Coefficients
Use this table to find loss coefficients:
Expansion:
Conservation of Energy
1
2
V12
p2
V22
z1 1
Hp
z2 2
H t hl
1
2g
2
2g
p1
p1 p2
hl
V22 V12
hl
2g
p1 p2
V12 V22
2g
z1 = z2
What is p1 - p2?
Head Loss due to Sudden Expansion:
Conservation of Momentum
A2
A1
x
2
1
M1 M 2 W Fp Fp Fss Apply in direction of flow
1
2
M 1 x M 2 x Fp Fp
1x
M 1 x V12 A1
Neglect surface shear
2x
M 2 x V22 A2
V12 A1 V22 A2 p1 A2 p2 A2
p1 p2
V22 V12
g
A1
A2
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 )
Head Loss due to
Sudden Expansion
hl
Energy
p1 p2
hl
V2
V1
g
2
V1 V2
hl
2g
V22 V12
2
1
A1 V2
Mass A V
2
1
2g
Momentum p1 p2
V22 V12
V12 V22
A1
A2
g
2
2
V V
2g
2
1
V22 2V1V2 V12
hl
2g
2
V
A1
hl
1
2g
A2
2
A1
K 1
A2
Contraction
EGL
V22
hc K c
2g
HGL
Expansion!!!
V1
V2
vena contracta
losses are reduced with a gradual contraction
Questions:
In the rough pipe law region if the flow rate
is doubled (be as specific as possible)
– What happens to the major head loss?
– What happens to the minor head loss?
Why do contractions have energy loss?
If you wanted to compare the importance of
minor vs. major losses for a specific pipeline,
what dimensionless terms could you
compare?
Entrance Losses
Losses can be
reduced by
accelerating the
flow gradually and
eliminating the
vena contracta
reentrant
K e 1.0
K e 0.5
K e 0.04
V2
he K e
2g
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?
_____
2
V
hv K v
2g
How can K be greater than 1?
Questions
EGL
HGL
What is the head
loss when a pipe
enters a reservoir?
V
V2
Draw the EGL and
HGL
2g
2
A
K 1 1
A2
V12
p2
V22
z1 1
Hp
z2 2
H t hl
1
2g
2
2g
p1
cs1
Example
valve
100 m
cs
2
D=40 cm
D=20 cm
L=1000 m
L=500 m
Find the discharge, Q.
What additional information do you need?
V22
100m =
+hl
Apply energy equation
2g
S-J on small pipe
How could you get a quick estimate?Use
_________________
Or spreadsheet solution: fnd head loss as function of Q.
Pipe Flow Example
1
Z1 = ?
γoil= 8.82 kN/m3
f = .035
60 m
2
Kout=1
7m
r/D = 0
Z2 = 130 m
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.
Pipe Flow Example
1
Z1 = ?
γoil= 8.82 kN/m3
f = .035
60 m
2
Kout=1
7m
r/D = 0
Z2 = 130 m
130 m
r/D = 2
Apply Bernoulli’s equation between points 1 and 2:
Assumptions: P1 = P2 = Atmospheric = 0
V1 = V2 = 0 (large tank)
0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin
Hmaj = (fxLxV2)/(Dx2g)=(.035 x 197m x (1.58m/s)2)/(.15 x 2 x 9.8m/s2)
Hmaj= 5.85m
Pipe Flow Example
1
Z1 = ?
γoil= 8.82 kN/m3
f = .035
60 m
Z2 = 130 m
Kout=1
7m
r/D = 0
2
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
Hmin = (0.19x2 + 0.5 + 1) x (1.582/2x9.8)
Hmin = 0.24 m
Kout = 1
Pipe Flow Example
1
Z1 = ?
γoil= 8.82 kN/m3
f = .035
60 m
2
Kout=1
7m
r/D = 0
Z2 = 130 m
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
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