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/g + V2/2g + z = constant along a streamline
(P=pressure
g =specific weight
V=velocity g=gravity z=elevation)
A streamline is the path of one particle of water. Therefore, at any two
points along a streamline, the Bernoulli equation can be applied and,
using a set of engineering assumptions, unknown flows and pressures
can easily be solved for.
Free Jets
The velocity of a jet of water is clearly related to the depth of water
above the hole. The greater the depth, the higher the velocity. Similar
behavior can be seen as water flows at a very high velocity from the
reservoir behind the Glen Canyon Dam in Colorado
Closed Conduit Flow



Energy equation
EGL and HGL
Head loss
– major losses
– minor losses

Non circular conduits
The Energy Line and the Hydraulic Grade Line
Looking at the Bernoulli equation again:
P/γ + V2/2g + z = constant on a streamline
This constant is called the total head (energy), H
Because energy is assumed to be conserved, at any point along
the streamline, the total head is always constant
Each term in the Bernoulli equation is a type of head.
P/γ = Pressure Head
V2/2g = Velocity Head
Z = elevation head
These three heads, summed together, will always equal H
Next we will look at this graphically…
V12
p2
V22
 1
 z1  h p    2
 z2  ht  hL
g
2g
g
2g
p1
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  gh
section is drawn perpendicular to the streamlines
(otherwise the _______
kinetic energy term is incorrect)
Constant ________at
the cross section
density
Steady flow
_______
V12
p2
V22
 1
 z1  h p 
 2
 z 2  ht  hL
g
2g
g
2g
p1
EGL (or TEL) and HGL
EGL 
pressure
head (w.r.t.
reference pressure)




p
g

V2
2g
z
velocity
head
elevation
head (w.r.t.
datum)
p
HGL   z
γ
downward (in
The energy grade line must always slope ___________
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
lower than
pressure
is plotted, the local pressures are _____
____reference
__________
______
Energy equation
velocity head 
static head
Why is static
head important?
Energy Grade Line
Hydraulic G L
V2
2g
p
pressure
g
head
z
elevation
pump
z=0
datum
p1
V
p
V
 1
 z1  h p  2   2
 z2  ht  hL
g
2g
g
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 a streamline
__________
Steady flow
______
Constant density
________
No pumps, turbines, or head loss
z 
V
2
2g

p
g
 const
Why no ? ____________
point velocity
no
Does direction matter? ____
Useful when head loss is small
Pipe Flow: Review



We have the control volume energy
equation for pipe flow.
We need to be able to predict the
relationship between head loss and flow.
How do we get this relationship?
__________analysis
_______.
dimensional
V12
p2
V22
 1
 z1  hp 
 2
 z2  ht  hL
g
2g
g
2g
p1
Example Pipe Flow
Problem
cs1
Find the discharge, Q.
100 m
D=20 cm
L=500 m
valve
cs2
Describe the process in terms of energy!
V12
p2
V22
 1
 z1  H p    2
 z2  Ht  hl
g
2g
g
2g
p1
V22
z1 
 z2  hl
2g
a
V2  2 g z1  z2  hl
f
Flow Profile for Delaware
Aqueduct
Rondout Reservoir
(EL. 256 m)
70.5 km
West Branch Reservoir
(EL. 153.4 m)
V12
p2
V22
 1
 z1  H p 
 2
 z2  H t  hl
g
2g
g
2g
p1
Sea Level
(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  ra

F=ma F  ra
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
M  rl

M
f 2 2
LT
3
Ll
T
l
V
Substitute for the dimensions of specific force
V2
fi  r
l
Dimensionless
Parameters

Reynolds Number

Froude Number

Weber Number
V2
fi  r
l
r Vl
Re =
m
V
Fr =
gl
W
V 2 lr
fu  
V
l2
fg  r g
f 

2
l

r c2
 Mach Number
f Ev =
V
l
M
c (Dp + r g Dz )
 Pressure/Drag Coefficients
2  p  C  2Drag
d
Cp 
2
2
r
V
A
rV
– (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 wd 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
gD 2 hl
V
32 L
hl 
Hagen-Poiseuille
32LV
rgD 2
LV2
hl  f
D 2g
32LV
LV2
f
2
D 2g
rgD
64 64
f

rVD R
Darcy-Weisbach
Pipe Flow: Dimensional
Analysis
What are the important forces?
pressure Therefore
Inertial
______,viscous
______,________.
________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
rV 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?
D

 C   2p
Cp proportional to l
C p  f  , Re 
p
2

 D


f   C p   f  , Re 
l 

D

l
D
f is friction factor

rV
Laminar Flow Friction
Factor
g D 2 hl
V
32 L
32 LV
hf 
r gD 2
L V2
hf  f
D 2g
32 LV
L V2
f
2
r gD
D 2g
64 64
f

rVD Re
Hagen-Poiseuille
hf 
128LQ
r 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
rVD
R

Two important parameters!
R - Laminar or Turbulent
/D - Rough or Smooth
and
 2p
Cp 
rV 2
Laminar and Turbulent
Flows

Reynolds
apparatus
rVD inertia
R

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  gh 
4 dl
umax
a2 d

 p  gh 
4 dl
a is radius of the tube
Max velocity when r = 0
a2 d
V 
 p  gh 
8 dl
Velocity distribution is paraboloid of
average velocity
revolution therefore _____________
(V) is 1/2 umax
_____________
a 4 d
Q
 p  gh 
8 dl
Q = VA = Vpa2
Laminar Flow through
Circular Tubes: Diagram
a2  r 2 d
u
 p  gh 
4 dl
du

dr
 
r d
2  dl
du
dr

 p  gh 
r d
Velocity
Shear
 p  gh 
2 dl
 ghl  True for Laminar or
  r  
 2l  Turbulent flow
Laminar flow
Shear at the wall
0  
ghl d
4l
Laminar flow
Continue
Momentum is
Mass*velocity (m*v)
Momentum per unit volume is
r*vz
Rate of flow of momentum is
r*vz*dQ
dQ=vz2πrdr
but
vz = constant at a fixed value of r
 rv z (v2rdr) z  rv z (v2rdr) z dz  0
Laminar flow
Laminar flow
Continue
2r zr r dz  2 (r  dr) zr r dr dz p z 2rdr  p z dz 2rdr  rg2rdrdz  0
dvz
 
dr
Q

R
0
p  pz 0  pz L  rgL
R 4 p
2vz dr 
8 L
Hagen-Poiseuille
The Hagen-Poiseuille
Equation
V12
p2
V22
 z1   1
 Hp 
 z2   2
 Ht  hl cv pipe flow
g1
2g
g2
2g
p1
p1
g1
 z1 
p2
g2
Constant cross section
h or z
 z 2  hl
p
 p

hl   1  z1    2  z 2 
 g1
 g2

p

hl     h 
g

Laminar pipe flow equations
a 4 d
Q
 p  gh 
8 dl
gD 4 hl
Q
128 L
gD 4 d  p 
Q
  h
128 dl  g

gD 2 hl
V
32 L
hl
dp


h




dl  g
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
Kekentalan kinematik : v  2,1x10 m / dtk
2
Rapat relatif : S  0,9 
 r  900 kg / m 3
Elevasi ujung atas pipa ( B) terhadap
ujung bawah ( A) : Z B  Z A  50m
Kecepatn aliran :
Kehilangan tenaga
32vVL 32 x 2,1x10  4 x0,566,4000
hf 

 17,23 m
2
2
gD
9,82 x0,3
pA
2
Q
0,04
V 
 0,566 m / dtk
A  x0,32
4
Bilangan reynolds :
pA
 0  0  50  17,23
pA
 67,23m
VD 0,566 x0,3

 808,6
4
v
2,1x10
Karena Re  2000 berarti aliran La min er
p A  67,23 x900 x9,81
Re 
2
V
p V
 A  z A  B  B  z B  hf
g
2g
g
2g
VA  VB
g
g
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  2,1x10  4 m 2 / dtk
Selisih elevasi kedua ujung :
Rapat relatif : S  0,9 
 r  900 kg / m 3
1
x10.000  50m
200
2
2
p A VA
pB VB

 zA 

 z B  hf
g
2g
g
2g
VA  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
Re 

 485
v
2,1x10  4
Karena Re  2000 berarti aliran La min er
z
pA
0
pA
 95,78m
g
g
10.000
 50  44,65
900 x9,81
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
t
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 = P (D, , r, L, U,)
p
average
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,
rU
D
rUD
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
leading to stability (_________)
is small relative to
inertia
the force leading to instability (_______).
Any disturbance in the flow results in large scale
motions superimposed on the mean flow.
Some of the kinetic energy of the flow is transferred
to these large scale motions (eddies).
Large scale instabilities gradually lose kinetic energy
to smaller scale motions.
The kinetic energy of the smallest eddies is
dissipated by viscous resistance and turned into heat.
head loss
(=___________)
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
h = eddy viscosity
dy
  rlI u I
u I  lI
du
dy
  rl
2
I
Turbulent shear is from momentum transfer
du
dy
Length scale and velocity of “large” eddies
Dimensional analysis
y
Turbulent Flow Velocity
Profile
 
du
  rl
lI  y
  r y
2
du
k = 0.4 (from experiments)
dy
 du 
  r y  
 dy 
 
2
2
 du 
 y 
 dy 
r
 

du
dy
increases as we
Size of the eddies __________
move further from the wall.
dy
2
2
I
2
Log Law for Turbulent,
Established Flow, Velocity
Profiles
 du 
 y 
 dy 
r
 

u
u*

1

u* 
ln
yu*

0
r
 5.5 Integration and empirical results
Shear velocity
Laminar
Turbulent
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
independent of pressure
 Is __________

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 


Re f 
 3.7
(used to draw the Moody diagram)
Pipe Flow Energy Losses
V12
p2
V22
 1
 z1  hp    2
 z2  ht  hl
g
2g
g
2g
p1
hl  
p
g


 D

f   C p   f  , R 

D 
L
2 ghl
 2p
Cp 
Cp 
2
rV
V2
2 ghl D
f 2
V L
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 rU 2

1
 4.0 * log Re*
f
Smooth pipe, Re>3000

f  0.4
f  0.079Re 0.25

Rough pipe, [ (D/)/(Re√ƒ) <0.01]
Transition function
for both smooth and
rough pipe

1
D
 4.0 * log  2.28

f


1
D
D/
 4.0 * log  2.28  4.0 * log4.67
1


f
Re f 

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
 2 log

 2.51
f
1




 3.7 D 

 2 log


  
f
1
 D
2.51

 2 log

 3.7
Re f





(used to draw the Moody diagram)
Moody Diagram
0.10
0.08
 D
f  Cp 
l  0.06

0.05
0.04
0.03
friction factor
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.004
0.03
laminar
0.002
0.02
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
R
1E+06
1E+07
1E+08

D
Fanning Diagram


1
D
D/
 4.0 * log  2.28  4.0 * log4.67
 1

f
Re f




f =16/Re
1
D
 4.0 * log  2.28

f
Swamee-Jain


1976
limitations
f
0.25
  
5.74  
log  3.7 D  Re0.9  

 
 /D < 2 x 10-2
– Re >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
hf
2
no f

 
ghf
1.78
log 

L
ghf
 3.7 D
3/ 2
D

L


1.25  LQ 
D  0.66  


 ghf 
2
4.75






5.2 0.04
 L  
 Q 
 
 ghf  
9.4
Each equation has two terms. Why?
Colebrook Solution for Q
1
2.51 
 D
 2 log 


3.7
f
Re f 

8 LQ 2
hf  f 2
 g D5
1
2.51 
 D
 4 log 


f
3.7
Re f 

2
1 1 8 LQ

f hf  2 g D 5
4Q
Re 
D
4Q
Re f 
 D
hf
 2 g D5
8 LQ 2
1 2 ghf D 3
Re f 

L
hf g
8
f 2 5
 D
LQ 2
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

Q
D5 / 2
2
 
ghf
log 
 2.51
L
 3.7 D

L
3 
2 ghf D 
Swamee
D?

1.25  Q   Q 
D  0.66  
  

g


  g 
8 LQ 2
hf  f 2
 g D5
1/ 5
Q 
D  0.66 


g


2
2
1/ 4


5/ 4  Q 
 


g



2
1/ 5 0.04
 Q  Q  
2
5
2
    
Q g   g  

1/ 4

1/ 5 1/ 25
 Q  
2
  
Q g  

1/ 5
1/
5
 2 

2 1/ 4
2 1/ 5 
 Q 
 Q  5/ 4  Q 

D  
 
 
 



Q  g   
 g 
 8 g   
 
1/ 5

1/
4
1/
5
64  5/ 4  Q 2 
  Q2  
f 2   
 




Q  g  

 g 

2
8
Q
D5  f 2
 g
64 Q 2
D f 2
 8g 
5
1/ 5
 Q  64 
D  
f 2

 8 g    
2
5
2
1     5/ 4  Q 
f     


4 4  
g



2
2
1/ 4

1/ 5 1/ 5
 Q  
2
  
Q g  

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)

Q
D5 / 2
2
find
 
ghf
log 
 2.51
L
 3.7 D

L
3 
2 ghf D 
pipe size given (head, type of pipe,L, Q)
2



LQ
1.25
D  0.66  

gh

 f 
4.75
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
 Cn
R
10.675

SI units
n
 C
1.852
10.675L æQ ö
hf = 4.8704
èC ø
D
SI units
C = Hazen-Williams coefficient
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
10.675L  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’tp confuse with radius!
D2
A 4
D
Rh = =
=
P pD
4
For a pipe
D = 4 Rh
LV2
hf = 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
Kehilangan tenaga
Penyelesaian :
L V2
hf  f
Panjang pipa : L = 1500 m
D 2g
Diameter pipa : D = 20 cm = 0,2 m
1500 x 2 2
 0,02
0,2 x 2 x9,81
Kecepatan aliran : V = 2 m/dtk
 30,58 m
Koefisien gesekan f = 0,02
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
Kehilangan tenaga
Penyelesaian :
L V2
hf  f
Panjang pipa : L = 1500 m
D 2g
Diameter pipa : D = 20 cm = 0,2 m
1500 x 2 2
 0,02
0,2 x 2 x9,81
Kecepatan aliran : V = 2 m/dtk
 30,58 m
Koefisien gesekan f = 0,02
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.
hK
High R
C p  f geometry , R 
 2p
Cp 
rV 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
thermal energy
– 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
1
0.95
0.9
0.85
Cc 0.8
0.75
0.7
0.65
0.6
0
F
1 I 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
Possible
separation
of the bend radius to
from wall
R
the pipe diameter
(R/D)
D
Velocity distribution
Low pressure
returns to normal
V2
hb  K b
several pipe
2g
diameters K varies from 0.6 - 0.9
b
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
V
2
hminor  K
2g
Re 
4Q
D
f 
8Q 2
g 2 D 4
0.25

 
5.74 



log

 3.7 D Re 0.9 

hl   h f   hminor
2
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

 
5.74 


log



0.9 
 3.7 D Re 

8Q 2
g 2 D 4
hl   h f   hminor
2
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
Sudden contraction
1500 m of 6” PVC pipe
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
constantat high
pressure coefficient is _______
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
g1
2g
g2
2g
p1
p1  p2
g
hl 
V22  V12

 hl
2g
p1  p2
g
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  Fp1  Fp2  Fss Apply in direction of flow
M 1x  M 2 x  Fp1 x  Fp2 x
M 1x   rV12 A1
Neglect surface shear
M 2 x  rV22 A2
 rV12 A1  rV22 A2  p1 A2  p2 A2
p1  p2
g
V V

2
2
2
1
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 g)
Head Loss due to
Sudden Expansion
hl 
Energy
p1  p2
g
Momentum p1  p2
g
hl 
V22  V12
hl 
V2
V1
g
V1  V2 
2
2g

V12  V22
A1 V2

Mass
A2
V1
2g
V V

2
2
A1
A2
g
V V

2g
2
1
2
1
2
2
V22  2V1V2  V12
hl 
2g
V 
A1 
hl 
1 

2g 
A2 
2
1
2

A1 
K  1 

A2 

2
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
Ke  1.0
Ke  0.5
Ke  0.04
V2
he  Ke
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 V
enters a
2
V
reservoir?
2g

Draw the EGL
and HGL

A 
K  1  1 
A2 

2
V12
p2
V22
 z1  1
 Hp 
 z2   2
 H t  hl
g1
2g
g2
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
Use S-J on small pipe
How could you get a quick estimate? _________________
Or spreadsheet solution: find 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