Workshop on hydraulic network modelling with waterCAD

Transcription

Workshop on
Hydraulic
Network Modelling with WaterCAD
16-20 October 2000
Paula Dawe
SOPAC Water Resources Unit
October 2000
SOPAC Miscellaneous Report 395
[2]
[SOPAC Miscellaneous Report 395 - Dawe]
[3]
CONTENTS
INTRODUCTION
4
PARTiCiPANTS
4
TRAINING SCHEDULE FOR WATERCAD COURSE
4
RECOMMENDATION
.4
ARISING FROM THE WATERCAD TRAINING COURSE
APPENDICES
A
Training Schedule for PWD WaterCAD Course
5
B
Background Notes on Hydraulics
9
C
Design Examples
31
D
Data for Model Calibration
.43
E
Feedback
49
[4]
INTRODUCTION
In September of 2000 a request was made to the SOPAC Water Resources Unit to perform a
week long training course on the hydraulic modelling software WaterCAD by the Fiji Public
Works Department (PWD). WaterCAD is software produced by Haestad Methods.
As part of a review of the Suva-Nausori master plan, an Australian consulting firm produced
a hydraulic model of this network. This model and the WaterCAD software used to build it
were then given to PWD without any training on how to use it. SOPAC was then approached
to provide the required training.
The training course took place during 16-20 October 2000. The venue for the workshop was
the SOPAC Headquarters, Mead Road.
PARTICIPANTS
The following from Fiji PWD participated in the workshop:
Name
1
2
3
4
5
6
7
8
9
Taito Apisarome
Timoci Turaga
Sereicocoko Yanuyanurua
Samuela Tubui
Piu Sekitoga
Viiendra Prasad
Aiav Prasad Gautam
Taito Delana
Taniela Qutonilaba
Position
Technical Assistant
Senior Engineer
Hydraulics Enqineer
Operations Enclneer
Acting Supervisor
Water Engineer
Senior Enqineer
Senior Engineer
Technical Assistant
Area
Suva Water Supply
Suva Water Supply
Suva Water Supply
Northern Division
Western Division
Suva Water Supply
Suva Water Supply
Western Division
Suva Water Supply
Phone Contact
321099
321099
385334
812044
660899
385334
385334
660899
385334
TRAINING SCHEDULE FOR WATERCAD COURSE
An outline of the workshop activities can be found in Appendix A. Material that was used in
the training course has also been appended inclUding:
•
•
•
Background notes on hydraulics (Appendix B)
Design examples (Appendix C)
Data for model calibration (Appendix D)
RECOMMENDATION
ARISING FROM THE WATERCAD TRAINING COURSE
Response from the training course was overwhelmingly positive (Appendix E). A number of
points for future note were made however. These include the following:
•
•
That future training courses in the use of AutoCAD and Maplnfo software could be
organised through SOPAC.
That the SOPAC WRU are available to assist PWD with the development of WaterCAD
hydraulic models for the Suva, Northern and Western Divisions and that an agreement to
this effect be worked out.
[5]
Appendix A:
Training Schedule for PWD WaterCAD
Course
[SO PAC Miscellaneous Report 395 - Dawe]
[6]
Day 1
1. Introduction to modelling.
2. Theory behind hydraulic models (read through notes- Background Theory for Hydraulic
Modelling):
• Energy principle
• Conservation of mass
3. Go over notes, highlighting important points
• A model is only as good as the data you put in it
• Where velocity is high, pressure is low
• As pipe diameter increases, head loss decreases
• As pipe roughness increases, head loss decreases
4. Introduction to WaterCAD network elements-looking at inputs required:
• Pipe
• Junction
• Tank
• Reservoir
• Pump
• Valve
5. Do Cybernet element tutorials- pipe, pump, reservoir, tank, valve
6. Practice linking the network elements to create a simple model.
7. Try creating a simple system with a pump, using the pump performance curve in your
notes as input data.
Day 2
8. Do tutorial on calculating model results.
9. Do tutorial on reporting model results.
10. Practice calculating and displaying model results using file Paula-PWDIExample2. wcd :
• Run model for steady state and extended period analysis
• Display results through tables, colour coding, annotation, profiles and contour maps
for the different analysis and the different time steps in the extended period
analysis
11. Simple design example- rural water supply system (Design Example 1 from notes)
• Does the total demand from the vii/age exceed the minimum flow from the source
spring?
• Are the pressures in your system reasonable?
• Are your pipe sizes reasonable?
• How do they compare with the hand worked solution?
12. Building blocks of a hydraulic model (where to find what types of data):
• user demand data
• elevation/topography
• pipe network layout
13. Do the tutorial on patterns
14. Extended period simulation- flow patterns
• Using file Paula-PWDIExample2. wed, change the flow pattern from residential to
quarry demand using the global edit function in the tabular reports for junctions
• Using the demand pattern in the Water Distribution section of the notes, create a
demand pattern for the rural example you created and analyse it for the extended
period option. Are the system pressures and flows still reasonable?
Day 3
15. Simple urban system- design, calculate, report (Design Example 2)
• Do you have the most cost effective combination of pipes?
16. Calibration of a hydraulic model:
[7)
• Data required for calibration
• Does it simulate real events
17. For the simple urban system you've just worked on, apply demand and roughness
calibration factors
• How does this affect your previous design?
• Are you taking into account leakage and the daily peaking factor?
18. Do the database tutorial.
19. Cybernet and other software:
• AutoCAD (.dxf files)
~ For creating .dxf backgrounds (Paula-PWDIRaro
Cadastral PE Sample.dwg and
Raro.dxf)
• Access (databases)
~ see handout on Creating Database Connections Between Maplnfo and Cybernet
• Excel (spreadsheet)
~ Create a database connection with Excel exporting pipe length and diameter data from
the pipe table and importing elevation data for the junction table using the simple urban
system you've created. Make sure you save the database connection you've created.
Use the pipe length and diameter data in excel to estimate the cost of the system.
• Maplnfo (GIS)
~ DTM of Rarotonga for importing elevation data (Turangi-with_elev. wcd and Turangiwo_elev.wcd) and Suva layers for creating a .dxf background (Suva.dxf)
Day 4
20.
21.
22.
23.
24.
25.
26.
27.
28.
•
•
Using the control function for pumps and valves
Using the find function to locate problems within your model
Do the scenarios tutorial
Running model scenarios:
• For calibration
• For analysis of system
Create scenarios for 10 and 20 years in the future using the simple urban system you've
created based on the given population growth rate
Investigate the different scenarios in the file Example2.wcd
Using the information provided for the Cook Islands- (pressure, flow and demand data)
calibrate the model in Turangi-with_elev.wcd
Things to remember when creating a model for a large-scale network.
Complex urban system design example- design, calculate, report.
Suva based on Suva.dxf- try importing elevations from Maplnfo
Lautoka based on Lau.dxf
DayS
29. Other functions of Cybernet:
• Do fire flow tutorial
• Do water quality tutorial and experiment with Example1.wcd
30. Uses of hydraulic modelling/ interpreting results:
• Identifying low pressure areas
• Design/effect of upgrades
• Leakage estimates
• Operation and maintenance
• Functioning of the network for different scenarios
• For presenting data
31. Remaining questions.
32. Course review.
[8]
[9]
Appendix B:
Background Notes on Hydraulics
[10]
Background
Theory for Hydraulic Modelling
Computer Modelling
A model is something that represents something in the real world. A computer model uses
mathematical equations to help explain and predict physical events. Modelling of water
distribution systems can allow you to determine system pressures and flow rates under a
variety of different conditions without having to go out and physically monitor your system.
Cybernet or WaterCAD will help you to do the following:
• Perform steady state, extended period and water quality simulations
• Size pipes and pumps
• Analyse for demands that vary over time
• Model tank, pump and valve behaviour
• Track chemical constituents in the water
• Make estimates of leakage from the network
• Planning upgrades to the network
It is important to remember the following when modelling:
A model is only as good as the information you put into it!
Basically, if the data you input into the model is crap your results will be crap. The
predictions of pressure and flow rates that the model produces are only as accurate as the
assumptions or data used to formulate the equations in the model. Appropriate values for
friction loss, pump performance, demand, etc. must be carefully defined before being input
into the model. When a model has been properly calibrated, predicted pressures in actual
systems have been found to be within 35 to 70 kPa of measured values.
Hydraulic Theory
Hydrodynamics
Hydrodynamics deals with the movement of fluid. There are 3 basic laws to fluid flow:
1. Conservation of mass
2. Conservation of energy
3. Newton's second law of motion (F=ma)
The first two laws appear in many different forms, depending on how the symbols are
defined, the importance of terms, the mathematical language used, etc. Basically, you will
still have the same basic equations but different constants and unit conversion terms will be
used.
For the purposes of simplification, it is often assumed that water flows as an incompressible
ideal fluid. An ideal fluid is without Viscosity and therefore can have no frictional effects
between moving fluid layers or between these layers and boundary walls. This means that
there will be no eddy formation or energy dissipation due to friction. The assumption of an
ideal fluid allows a fluid to be treated as a collection of small particles, which will support
pressure forces normal to their surfaces but will slide over one another without resistance. In
situations where friction is small, the frictionless assumption will give good results. Where
friction is large, the assumption of an ideal fluid will not provide good results.
Conservation of Momentum or Newton's Second Law
The simplest definition of momentum is: momentum = mass x velocity. A body has
momentum by the fact that it is moving. If the velocity is zero, then the momentum is zero.
[11]
The law of conservation of momentum states that a body in motion cannot gain or lose
momentum, unless some external force is applied.
Newtons 2nd law of motion is that a force is equal to the rate of change of momentum. Since
we are dealing with the movement of fluid, it only makes sense that laws of motion now apply
to the particles of that fluid. As a fluid particle moves, it is displaced from its original position
over time in the direction of motion. The velocity of this particle can be described by the
equation:
v=-
d
t
If the velocity of the particle changes over time, it will have acceleration. Unbalanced forces
acting on particles of an ideal fluid will result in the acceleration of these particles according
to Newton's 2nd law. Thus, a body cannot gain or loose momentum unless an external force
is applied. With acceleration defined, Newton's 2nd law of motion can now be applied to the
moving fluid particle. This equation is:
F
= (mv
-mvl)lt=m(v2
2
-vl)lt=ma
Another form of this equation for a moving fluid can be written as:
F
= pQ(v
2
-VI)
= pAv
2
The application of the momentum equation can be demonstrated as moving liquid
approaches a bend in a pipe. The waters tendency is to continue moving in a straight line.
To make it flow around the bend the pipe must exert a force on the water as shown in the
diagram below. Looking at a control volume in a pipe bend which changes direction in either
the horizontal or vertical plane, summing the forces on the control volume results in the
following equations:
This summing of forces can be applied to many different control volume examples, such as
in reducers and nozzles.
Conservation of Mass
The law of conservation of mass states that mass must be conserved. It can neither be
created nor destroyed. Basically, what this means is that what goes in, must come out. The
following diagram illustrates this:
The above is further explained by the following equation, otherwise known as one form of the
continuity equation:
y-X
=0
The continuity equation expresses the continuity of flow from one section of a fluid to another
as the fluid moves. Another way to express conservation of mass is by the following
equation:
p\A\v\ = P2A2V2
This equation expresses the fact that in steady flow, the mass flowrate passing all sections of
a streamtube is constant.
What are the units of pAv? Does they constitute a mass flow rate?
For fluids of constant density, the continuity equation can be expressed as follows where, Q
is designated as the volume flowrate:
Q = A\v\ = A2v2
What are the units of Q?
Conservation of Energy
The principle of conservation of energy states that energy can neither be created nor
destroyed, but can be transformed from one form to another. Energy must be conserved.
There are numerous forms of energy- mechanical energy, potential energy, heat energy,
kinetic energy, sound energy, etc. that it can be transferred to and from.
Considering a fluid streamline, the driving force tending to accelerate the fluid mass are (the
Ll can be termed the "change in" the following parameter):
1. pressure forces acting on either ends of the element
F=MM
2.
and the component of weight acting in the direction of motion
W=pgM&
The change in mass being accelerated by the action of these forces can be experssed by:
M =pi1dM
Applying Newton's 2nd law we can substitute these previous equations into
F=ma
-MM- pgM&
= (pi1dM)v-
i1v
i1d
Dividing by pM gives:
[13]
M
-+ g&+vL1v =0
p
For incompressible flow, this form of Newton's 2nd equation can be divided by g in order to
obtain Bernoulli's equation:
p
v2
-+-+z=H
pg 2g
Bernoulli's equation is an energy balance form one point in a hydraulic system to another. It
applies to all points on the streamline and thus provides a useful relationship between
pressure P, velocity v, and height above a datum, z. Units of head are in meters, but head is
just another way to express energy.
The components of Bernoulli's Equation represent different forms of energy present in a fluid.
These can be broken down as follows:
•
Pressure Head:
•
Elevation Head:
P
r
z
Energy imparted by pressure, or work done on fluid
Same as Potential Energy or energy due to gravity
2
•
Velocity Head: ~
2g
Same as Kinetic Energy or energy due to motion
The Bernoulli equation may be visualised for liquids as vertical distances. The sum of all
three terms (or total head) is the distance between the horizontal datum and the Energy
Hydraulic Grade Line (HGL):
• Sum of pressure head and elevation head
• In water open to the atmosphere (river, lake), the HGL is at the water surface
• The hydraulic gradient is the slope of the hydraulic grade line
• The height to which water will rise in a piezometer or standpipe in a pipeline, if a tapping
is made.
• Flow normally occurs in the direction of the hydraulic gradient from high to low pressure,
although the hydraulic gradient may rise over short distances giving an adverse gradient
which can be overcome by the momentum of the fluid
• Negative pressures occur at any place where the pipeline rises above the hydraulic grade
line
When might an adverse pressure gradient form, trying to push the liquid back in the direction
of flow?
Energy Grade Line (EGL):
• Sum of elevation, pressure and velocity head
• Always goes down in the direction of flow
Because energy can neither be created nor destroyed, the Bernoulli equation can be further
expanded to act as an energy balance for fluid in a system. The total energy of the fluid at
one point has to equal the total energy of the fluid at a point farther down the streamline, but
relative proportions of the form the energy is in (pressure energy, elevation energy, kinetic
energy) may change. The following equation demonstrates this:
[15]
From this equation, it makes sense that when velocity increases, the sum of pressure and
elevation head must decrease. In many flow problems, elevation may vary little and the
general statement- where velocity is high, pressure is low- can be made. When liquid flows
through a pipeline, the continuity equation has to be obeyed, so any loss of energy appears a
a reduction in pressure.
For example, if water flows through a ling pipeline of constant
diameter at a constant rate, then the mean velocity must be the same at all points along the
pipeline to maintain continuity of flow. Thus any loss of energy appears as a reduction of
pressure or head.
One useful application of Bernoulli's equation is that it shows that the velocity of an ideal fluid
exiting from a small orifice under a static head varies with the square root of the head. This
can be expressed by the following equation called Torricelli's theorem:
v=~2gh
The following points should be remembered when applying Bernoullis Equation.
1. Apply the Bernoulli equation in such a way as to minimize the number of unknown
variables. If energy losses are ignored there are six variables. After the use of other
equations, such as the continuity equation, there must be only one unknown to be able to
solve the problem. You can select which 2 points to use in the analysis.
2. Many problems involve a body with a free water surface, such as a tank or reservoir.
Normally we work with gauge pressure which uses atmospheric pressure as a datum, so
if a point is selected on the water surface, the pressure is atmospherice and P=O.
3. If a pipe or nozzle discharges to the atmosphere and the jet has a constant diameter then
it can be assumed that the water pressure in the jet is the same as the surrounding
atmosphere. If gauge pressure is used, P=O.
4. With large tanks or reservoirs the velocity on the water surface can be assumed to be
zero, so v=O.
5. The datum from which elevation is measured can be taken through the lower of the two
points being used in the analysis, so either Z1 or Z2 = O.
6. Make a drawing of the hydraulic system marking in the known values and the unknown
values.
[16]
Real Fluid Flow
The flow of a real fluid is more complex than that of an ideal fluid owing to the presence of
viscosity.
Viscosity introduces resistance to motion by causing shear or friction forces
between fluid particles and between these and boundary walls. For flow to take place, work
must be done against these resistance forces, and in the process energy is converted to
heat.
The effects of viscosity cause the flow of a real fluid to occur under two very different
conditions or regimes:
• Laminar flow
• Turbulent flow
In experiments conducted by Reynold's, he discovered that for low velocities of flow in a
glass pipe, a thin filament of dye issuing from the tube did not diffuse but was maintained
intact throughout the pipe, forming a thin straight streamline.
However, as velocity was
increased in the pipe, the dye filament would waver and break up, diffusing through the
flowing water in the pipe. From this experiment, Reynold's was able to isolate a critical
velocitv from which flow chanaed from one reaime to the other.
Since surface roughness increases the turbulence in a flowing fluid and thus decreasing the
effect of viscous action, roughness contributes to energy loss within the fluid. Energy is
dissipated by the work done in continually generating turbulence by the roughness. The
energy involved in this turbulence is composed of the kinetic enrgy of the fluid mass. Energy
dissipation is therefore proportional to the square of velocity.
Velocity Profiles
The shearing stress created by viscosity effects in the fluid produce velocity profiles
characterized by reduced velocities near the boundary surfaces. This differs from the
uniform velocity distribution of an ideal fluid. Since the velocity is no longer uniform, mean
velocity is now used in calculations with real fluid flow, and a correction factor is applied to
the velocity head. This correction is expressed as follows, where a is the correction factor.
v2
a-
2g
This equation shows that head loss is not a loss of total energy, but rather a conversion of
energy to heat, part of which leaves the fluid. Friction energy or head loss is then in effect
lost from the useful total of pressure, velocity and potential energies.
The subject of pipe flow involves only those pipes in which flow is completely full. Pipes that
flow partially full such as in channels and sewers are treated as open channels. The solution
of pipe flow problems results from the application of the energy principle, equation of
continuity and the principles and equations of fluid resistance. Resistance to flow in pipes is
offered by long stretches of pipe and also by pipe-fittings, such as bends and valves.
Head Loss can be calculated using the following 3 equations:
• Darcy-Weisbach Equation (use this one for pipe systems)
• Mannings Equation
• Hazen-Williams Equation
Darcy-Weisbach
Equation
Early experiments indicated that head loss varied directly with velocity head and pipe length
(I) and inversely with pipe diameter (d). Knowing this, the following equation for head loss
was dirived where f is called the friction factor:
hL
=1
I v2
d 2g
This equation, is the basic equation for calculating head loss caused by pipe friction (not pipe
fittings) in long, straight, uniform pipes.
In words it expresses the following relationships:
hLuf
the rougher the pipe, the greater the head loss
hLul
the longer the pipe, the greater the head loss
[20]
hLaV
hLa11d
the higher the velocity, the greater the head loss
the larger the diameter, the smaller the head loss
It is found that f depends only on the Reynolds number and another dimentionless parameter
eld, called the relative roughness, where e is the height of surface roughness on the wall of
the pipe, and depends on the pipe material. Values of typical pipe roughness can be found
in the following insert.
This relationship indicates
a convenient means of presenting
experimental data on the friction factor.
The dependence of f on the Reynolds number and eld is different in laminar and turbulent
flow regimes. In laminar flow, f is only dependent on R and may be calculated from the
following equation:
f
= 64
R
Laminar flow can then be expressed as:
Q = ;rd4pghL
128J.ll
Can you derive this equation?
Within the turbulent flow regime, as velocity and R increase, it is evident that the thickness of
the laminar film will decrease and the effect of viscous friction will decrease while roughness
will become more important. In the region described as completely turbulent, f depends only
upon eld. The variation of fwith these parameters is shown on the Moody diagram.
Hazen-Williams Equation
The Hazen-Williams equation was also developed for use in the pipe-flow problems.
be expressed as follows:
v~
kCR°"' SO"
or
h
L
=
~9,f( ~
It can
J"
where:
v is the mean velocity, C is a factor dependent on relative roughness, R is the hydraulic
radius (area of flow divided by the wetted perimeter), S is the slope of the energy grade line,
and k is a factor dependent on units (0.849 for mls and m).
Values of C are found in the following table. They reflect the fact that long-term corrosion
and encrustation occurs in the pipe as it ages, increasing the pipes roughness. This effect
can be seen in the pictures below.
Description of Pipe
Valueof C
Cast iron: new
5 yrs old
10 vrs old
20 vrs old
30 yrs old
Concrete
Cement lined
Plastic
Asbestos cement
130
120
110
90-100
75-90
120
140
150
140
16
(;11 ,.\ new 8.QOmmdmmcicr duenlc iron pipe ••••
ith cement mortar lining. I~} An old. ~Omm rliami:l~.
pipe showillS severe Htl:r.:rCltlali,)It.t"o,e 111<::
reduced hut:: as wdl a~ )I~re;lsed roughn I:"'i
This equation is not applicable for low values of Reynold's number. The following nomogram
for the Hazen-Williams equation can be used to graphically solve the equation for discharge,
pipe size or energy slope given the other two variables. The following corrections can be
used for C values other than 100.
100)1.85
s; = SIOO ( C
de
= dlOO
(100)°.38
C
o; =
QlOo
( C )
100
Mannings Equation
The Manning equation is most commonly used for the analysis of flow in open channels, but
it can also be applied to pipelines. For a pipe flowing full this equation is as follows, where n
is the Manning roughness coeficient.
v
= 0.397D2/3S0.5F
n
or
h =(~J2L
L
066
~
Minor Losses
The category of minor losses in pipes includes losses incurred by change of section, bends
elbows, valves and fittings of all types. In longer pipes, minor losses can be neglected
without serious error in calculation. In shorter pipes, these losses become more important.
Minor losses usually result from rather abrupt changes (in magnitude or direction) of velocity.
Generally, an increase of velocity is associated with small head loss, but a decrease of
velocity causes large head loss because of boundary layer effects which result in flow
seperation and extreme turbulence.
eany experiments mcicateo mat minor losses vary wnn me square or veiocny.
may be expressed as:
I
ne neaa
lOSS
2
HL=k~
2g
where k is the loss coefficient and is a function of changes in direction, obstructions,
changes in velocity. k is constant for a given fitting, but varies with fitting size.
Expansions in piplines, produce substantial energy losses.
loss can be calculated from:
HL=
or
At abrupt enlargements energy
(V\-V2)2
2g
One special case of a sudden contraction is that of a square edged pipe entrance from a
large tank where V1 is O. For this situation:
2
HL=0.52
2g
If the entrance is bell-mouthed, k can be taken as 0.4. The insert table on the following page
gives various k values for different fittings.
Fitting
Entrance
Contraction
90° bend
Gate valve
Check valve
Elbow
Expansion
Bell-mouthed Entrance
Exit
K
0.5
0.143
0.18
0.12
0.75
0.39
0.277
0.4
1
k can also be expressed in terms of equivalent length (Vd) at a certain velocity.
expressed as follows and demonstrates the relationship that exists between f and k:
k=fi
d
Where:
1= pipe length
This is
[23]
d= pipe diameter
Pipe Line Problem
All steady-flow pipe line problems may be solved by the application of the Bernoulli and
continuity equations. Ususlly, the engineering problems consist of:
1. Calculation of
head loss and pressure variation from flowrate and pipe-line
characteristics
2. Calculation of flowrate from pipe characteristics and the head which produces flow
3. Calculation of required pipe diameter to pass a given flowrate between two regions of
known pressure difference
The first of these problems can be solved directly, but solution by trial is required for 2 and 3.
Trial and error solutions are necessitated by the fact that the friction factor, f, depend upon
the Reynolds number, which in turn depends upon flowrate and pipe diameter. However,
flow in rough pipes at high Reynolds numbers usually does not warrant trial and error
solutions.
There are a few points to remember when using Bernoullis equation:
1. Pipes must be flowing completely full under pressure
2. Open channels or pipes which run partially full with a free water surface that is at
atmospheric pressure are not analysed using Bernoulli
Construction of the energy and hydraulic grade lines for many problems is quite useful.
Consider a pipe line laid between 2 reservoirs having different elevations. The energy line
must start in one reservoirs surface and end in the other, using a gradual drop to repersent
head loss due to pipe friction hI, and abrupt drops to represent entrance, henh and exit losses,
hexit• The total head loss can then be expressed as:
[25]
Step 1- Estimating the level of water consumption
• the amount of water you have to supply determines how big your distribution system will
have to be- ego pipe size
• in order to estimate future water use, you have to estimate the future population you are
going to be supplying
Things that increase (t) or decrease (-/,)water usage:
• population- more people use more water
• climate- people use more water in drier, hotter climates (eg. watering gardens)
• economic level- rich people use more water than poorer people
• population density- areas where you have high concentrations of people living have a
lower water demand (eg. in apartment buildlnqs, don't have to water lawns)
• industry- industrial demands tend to be high, but it depends on the type of industry
• cost- people who pay for their water use less
• pressure- distributions systems that operate under high pressures use more water
• quality of supply- people use less water if the quality of that water is poor
• culture- some cultures use more water than others (eg. keeping pigs uses a lot of water)
Different types of users:
• domestic
• commercial (stores, bars, restaurants, hotels)
• industrial (airports, factories)
• institutional (government buildinqs, schools, hospitals, prisons)
• agricultural
Total Consumption
= domestic
use + commercial use + public use + loss and waste
Water consumption varies during the:
• year- highest during the dry season
• day- highest around 7am in the morning when people getting up and showering, lowest
from 2-4am in the morning when people are asleep
[26]
Components of Water Distribution System
Pipes~
pressurized closed conduits
Stresses Acting on Pipes:
•
Pressure of water acting on the pipe (remember that the water doesn't want to be in the
pipe and is always trying to force its way out)
Forces caused by changes in the direction of flow within the pipe
External loads like the weight of dirt on the buried pipe
Changes in velocity
•
•
•
Water Hammer
•
•
Results from the sudden stopping or slowing of flow in a pipe
The kinetic energy of the water is transferred to the pipe wall and acts to stretch, deform
and burst the pipe
Can be avoided by closing valves slowly for example
•
Low Points
•
•
•
Where the depth of the pipe below the ground surface is great
High pressures may form at low points in the distribution system
You want to break the hydraulic gradient at low points with pressure reducing valves
(PRV), overflows, auxiliary reservoirs
Place hydrants at low points in order to drain the distribution lines for maintenance
purposes, and to remove sediment
•
High Points
•
•
•
•
Should be kept below the HGL, otherwise you can get negative pressures in pipes which
leads to the accumulation of gasses that may block the flow of water through pipes
Negative pressures in pipes can create a vacuum that will actually suck water from the
ground into your pipe~ problem if you are sucking in contaminated water from a septic
tank
Flow in a pipe is possible up to around -7.5m of water, after this vaporisation of the liquid
can be expected
Use vacuum, air relief valves, or pressure sustaining valves (PSV) to release air initially
in the line or that accumulates over time, or to admit air when the line is being emptied for
maintenance purposes
Pumps
The addition of mechanical energy to moving fluid by a pump alters the basic energy balance
of the Bernoulli equation. With the addition of energy by a pump, an additional term must be
included in the equation.
r;
v2,
-+-+z,+E
pg
2g
pump
P2 V22
=-+-+Z2
pg
2g
Epump will appear as an abrupt rise in the energy line over the pump machine.
therefore, add head to hydraulic systems.
There are 3 main types of pumps available on the market:
1. Centrifugal Pumps
2. Axial Flow Pumps
3. Mixed Flow Pumps
Pumps
[29]
•
•
•
if the water supply is located above the level of the water users, no pumping is required
~ this is a gravity distribution system
the steepness of the slope effects the pipe design and velocity of flows in the pipe
water will flow from a high point to a low point, but if there is a rise in between the water
must have sufficient energy to flow over this rise
Users
• how much water people use determines how big your distribution system is going to have
to be
Steps in designing a distribution system:
1. Flows to each section of the community must be estimated and designated to individual
subareas of your system
2. A system of interlocking loops must be laid out-» this ensures continuous delivery of
water even if a portion of the system is shut down for repairs
3. Flows are assigned to various nodes of the system
•
The actual design of the distribution network involves determining the size of the arterials,
secondary lines and small distribution mains required to ensure appropriate pressures,
flows, head losses and velocities in the system under a variety of design flow conditions
Design flow:
•
•
must make sure that the system operates during the worst case scenario~ maximum
daily flow + fire flow
The design of a distribution system is based on the provision of adequate pressure for
fire protection at the maximum daily flow, including fire demand
There are many solutions to the design problem of creating a distribution system-s you must
optimise (adjust parameters such as pipe size to achieve the most appropriate pressures at
nodes and velocities in pipes) to find the best solution. The following insert helps to explain
this.
Distribution system consists of a network of:
•
•
nodes-e points of flow withdrawal
links~ pipes connecting nodes
It is not reasonable to analyse a system up to every house-s individual flows can be
concentrated at a smaller number of points, commonly at pipe (or road) intersections
The usual engineering approach to the design of a looped pipe system involves laying out
the network, assigning estimated pipe sizes, and calculating resulting flows and head losses.
The pipe sizes are then adjusted as necessary to ensure the pressures at the various nodes
and the velocities in the various pipes meet the criteria.
The calculation of the flows and pressures can be performed using the Hardy Cross method.
This method is based upon the hydraulic formulas used to calculate energy losses in
elements of a system. The energy loss in any element of the pipe system may be expressed
as:
hj =kjQt
Where:
hi = energy loss in element i
OJ= flow in that element
~ =constant depending on pipe diameter, length, type and condition
[30]
x = 1.85 to 2, depending on the equation used
For any pipe in a loop of the system, the actual flow will differ from an assumed flow by an
amount i1:
Qi = Qassumed +.6.
For any loop, the sum of the head losses about the loop must be equal to zero. Thus, for
any loop:
I.k.Qx = 0
I
,
The above equation can then be solved for the correction:
.6. = _ I.hi
",h.
x"",-'
o,
The Hardy Cross procedure may be outlined as follows:
1. Disaggregate the flow to the various blocks or other sub-areas of the community
2. Concentrate the disaggregated flows at the nodes of the system
3. Add the required fire flow at appropriate nodes
4. Select initial pipe sizes
5. Assume any internally consistent distribution of flow. The sum of the flows entering and
leaving each node must be equal to zero
6. Compute the head loss in each element of the system. Conventionally, clockwise flows
are positive and produce positive head loss
7. With due attention to sign convention, compute the total head loss around each loop:
I.hi = I.kiQt
8. Compute, without regard to sign, the sum I. kiQt-1
9. Calculate the correction for each loop (Li) and apply the correction to each line in the
loop. Lines common to two loops receive two corrections with due attention to sign.
10. Repeat the procedure until the corrections calculated in step 9 are less than some
stipulated maximum. The flows and pressures in the initial network are then known.
11. Compare the pressures and velocities in the balanced network to standard criteria.
Adjust the pipe sizes to reduce or increase velocities and pressures and repeat the
procedure until a satisfactory solution is obtained
Typical Design Parameters
Fire Flows (Umin)
1890 ~ min
32 400 ~ max
i e on/off valves
150-250 m
to prevent deposition of 0.3-0.6 rn/s
~1 m/s
of streets
[31]
Appendix c:
Design Examples
_"'''''"C::::,~:'';-::t:'~'':=:::;;;''':'_''::-=:;C':=:-'_'_4
. - """-"""*-""~~~-'~~----~-~.=::---=::=:::-":"'-:::::::C7::=:_:=::'-=-~-
---":==--~--=::;::.·-------:=--::::::::::_?.::?~5~=;:::_:::-
By interpolation, frictional loss is 5.1 m/km for a flow of 2.3 lis. Therefore, total
frictional loss is
2038
1000
=
X
5.1
lOAm
and the height of the HGL above the water level in the tank is 20 - lOA
which meets the design requirements of 5m minimum.
,
•
iJes'3')
n
=
9.6m
Three Hour Practical
1.
E XQi'Y'tp Ie 1
General
This example is taken from an Indonesian project operated by CARP. It can be
used as the basis to create a worked example for a three hour practical by altering
parameters to be more representative of conditions in this country. Participants
should follow the step-by-step design procedure outlined below.
A village has a population of 850 divided into two parts, Part I with 605 persons
and Part II with 245 persons. A water source with an estimated minimum flow of
one lis is located approximately 2,300 metres from the village. The standpipes are
used for 12 hours per day. Assume that galvanized iron pipe must be used for all
pipe because of the rocky terrain. (A sketch map of the village and a ground
profile are presented in Figures 3 to 6. Design the pipelines required to serve and
plot the hydraulic gradient for each pipe length.
2.
Design Parameters
Population
The design population is the expected population in 10 years at a 2%
growth rate or 850 X 1.22 equal to 1,037 persons. For design purposes
this is rounded up to 1,050 persons.
Water Usage
It is preferable to supply the maximum amount of water possible but a
per capita supply of 100 litres per day would require 105,000 litres/day or
a flow of 1.21 lis. Since the estimated minimum flow of the source is only
1 lis this is not possible. A per capita use of 80 litres/day would require
an average daily flow of 0.97 lis and this is possible.
Storage Requirements
A per capita use of 80 litres/ day means that the average daily usage is 80
X 1,050 or 84,000 litres. Because the galvanized iron pipe is so expensive
the smallest pipe possible is used. In order to reduce the size of the main
pipe, storage will be located in the village. The recommended storage is
then one half of 84,000 litres or 42m3• Based on the present population
distribution 245 + 850 X 42m3 or 12.1m3 should be in Part II and 605 +
850 X 42m3 or 29.9m3 should be in Part I. Based on the village sketch it
has been decided that the water will be distributed to five public
reservoirs, three in Part I and two in Part II. With this layout, the water
will be under village control. Costs will be reduced because the main pipe
does not have to convey peak flows of water. The public reservoirs are to
be situated on high points to obtain any acceptable pressure head at the
tank outlet.
The three reservoirs in Part I will be 10m3 each for a total of 30m3
(rounded up from 29.9m3). The two reservoirs in Part II will be 6m3 each
for a total of 12m3 (rounded down from 12.1m3). With this distribution of
reservoirs, no one has to walk more than 100 meters to obtain water.
4J4¥f'¥ff&;T¥S,,:¥·J,:·;,(t,};!.')7·~~·?~':'"
\,,,,," - '-'
17
·\Sj1-":":"'< ,
.~.;
-,<"--.,
_-_ •• h~~
Number of Faucets
The number of persons per faucet should be between 30 and 100 so the
number of faucets for Part I should be between 6 and 20 and for Part II
between 2 and 8. In order to accomodate future demand, a higher number
is preferable. Thus 6 faucets at each of the three reservoirs in Part I and 4
faucets at each of the reservoirs in Part II give a total of 26. The average
number of persons per faucet (based on the future population of 1050) is
42 for Part I and 37 for Part II. Half of the faucets can be placed on one
side of the reservoir and half on the opposite side. Thus, one area can be
used by females and the other by males.
Design Flows
The path of the pipeline is sketched in Figure 3. The water will flow
continuously into the reservoirs so the design flows will be the same as the
average daily flows. The peaking factor is therefore 1. At the projected
per capita use of 80 IIday the average daily flow is 0.97 lis but the spring
has an estimated minimum flow of 1.0 lis. Therefore, 1.0 lis will be used
in designing the pipeline.
18
From the source to the junction at point A the design flow used is 1.0 lis.
At point A this flow is divided with 0.71 lis flowing to Part I and
reservoir B. The remainder of 0.29 lis will flow to Part II and reservoir E.
At reservoir B 0.23 lis is taken and the remainder of 0.48 lis flows to
reservoir C. At C 0.24 lis is taken and the remainder of 0.24 lis flows to
reservoir D. At reservoir E 0.14 lis is taken and the remainder of 0.15 lis
flows to reservoir F. The design flows are noted on the pipeline route and
profile.
4*~
.;tno'
,.",\,;,/:'.,,:
.. ,
"':',
.:
:1t>t~:4d;VW§-FifiPjJ1
'.
,> '
'.
. (d'~'
... H"H'
·»., ..
•..
;>~·... _.,..'"'.H"
•.•.•:.";.·.; ..;"".,,, •.;~."'~",.,·.,J"i,"
•.... , ..••
,. _"""""'i,,,,,, •.,,..,~";"'""',,
••••u~
change diameters in such a way that cost is reduced at each step. The worked
problem below indicates how this might be done. It is important to note however,
that the preferable approach is to use a computer, even for branched networks,
which is described in the notes for the second submodule.
()
Worked Example
-
e 5 I Q..J n
Ey. QM \e '1
P
F"igure 1 sows
h
communi WIt
. hI'present popu anon 0 f 1,000 that
a street map 0 f aa commumty
is to be served entirely by public standpipes. The purpose of this example is to
illustrate the four steps of design using a conventional approach that employs a
desk calculator.
..
Figure 1: Street Map of Worked Example
Background
Data
Present Population
Average per capita flow
Peaking factor
Town growth rate
Design period
Unaccounted for losses
Number of persons per standpipe
Maximum level of elevated tank
Minimum level of elevated tank
Ground elevation, all nodes
M;n;mnm llllowahle nressure
1,000
100
3
2
20
20
100
16
12
0
5
l/c/d
%/year
years
0/0
m
m
m
m\~
••••
i>
.
!"mlli!1!lm~~'
""Fmi::)«~~~i!
"';:k'
L"'"
'.>
·f:::lWt:iilliliih!I~~i.:·~
---------------------_
.
§co]
Layout
With 1,000 persons in town and 100 persons per standpipe, there will be 10
standpipes. If the population is more or less evenly distributed throughout the
town, the standpipes should be evenly spaced. They should be located along
streets where users have easy access, near street comers if possible. Figure 1 shows
one arrangement of the standpipes, which are denoted by node numbers.
This network has a single source at the elevated tank (node 11). The task now is to
connect the standpipe nodes to the source node keeping total pipe length as short
as possible, laying pipes in streets, and selecting routes where the greatest number
of houses are located so that in the future when the system is upgraded to
individual connections, these houses will be able to connect. Figure 2 shows the
resulting layout. This network has 1 source node (No. 11), 10 demand nodes (Nos.
1-10),3 junction nodes (Nos. 12-14), and 13 pipes. In a branched network, the
number of pipes is always 1 less than the number of nodes.
Figure 2: Network Layout of Worked Example
Flows
The peak hourly design flow for this network is calculated in the slide show and
above. It is 557 mt/d which is equivalent to 6.5 litres/second.
Since this is a small town with only domestic demand to be served and since each
standpipe serves the same number of persons, the flow at each demand node (Nos.
1-10) is assumed to be identical, namely 0.65 litres/second, and the inflow at node
No. 11 is 6.5 litres/second. Note that the network is being designed to meet peak
hourly demands.
With a takeoff flow of 0.65 Ips at each demand node, it is an easy matter to
calculate the flows in the 13 pipes of the network. It is preferable to start the
110.'"
DlIIiiiii
~_'fl)f7nl~il\'1~~MiIIC'~h~~"'9M''"~~",.~~~''C"~''=n_~f'''oI''W
,
__
~_~'''''
••••__
'''"'...,''''n_ •••...,~_n.c.,.,.-
@A]
, .•• '_.. ,,1.' ,-,: •..
,~~.";'~~i~~(~~~
,~.j·~."•..•
calculations at the terminal ends and work toward the source. For example, the
flow in pipe No. 13 is 0.65, in No. 12 it is 1.30, etc. The list of pipe flows and
their lengths is as follows:
Pipe No.
Length (m)
Flow (Ips)
1
1~
~~
2
3
4
5
6
7
8
9
10
11
12
13
210
225
95
110
165
155
lW
1~
155
140
235
160
5.85
1.95
1.30
0.65
0.65
1.30
Q~
Q~
1.95
0.65
1.30
0.65
Pressures
From the Background Data, the water elevation in the tank varies between 12 and
16 m. At the time of peak hourly demand, the level should be approximately
midway in the tank. Hence, the inlet pressure is 14 m.
The minimum target pressure is 5 m. Because the network is flat, this pressure
should occur at the terminal nodes of the network. That is, if the network is well
designed, the pressure at each of the 6 terminal nodes (Nos I, 2, 6, 7, 8 and 10)
should be about 5 m.
Diameters
Using the method in the slide show, the first task is to calculate the minimum
hydraulic gradient. This means finding the longest branch and dividing its length
into the available head, which in this case is 14 - 5 = 9 m.
This network has 6 branches, one for each terminal node. The branches can be
designated by the terminal node numbers. For example, branch No. 6 includes
pipes No. I, 2, 7 and 8, and branch No.8 includes pipes No. I, 2, 10 and 11. The
branches with their pipe numbers and total lengths are:
Branch
Pipe Numbers
Total Length (m)
6
1
2
6
7
8
10
1,2,3,4,6
1,2,3,4,5
1,2,7,8
1,2,7,9
1,2,10,11
1,2,10,12,13
860
805
650
705
670
925
The branch with terminal node No. 10 is longest, and its average hydraulic
gradient is 9 m: 925 m = 0.00973.
.
-
&oJ
Using the Hazen Williams equation and this gradient, the diameter of each pipe in
the network can be calculated based on its design flow. The Hazen Williams
equation is:
Q
=
Q
C
D
H/L
3.7
10-6C 1)2.63(H/L)OS4
X
::::flows, Ips
= roughness co-efficient = 130
= diameter, mm
= hydraulic gradient = 0.00973
Substituting these values into this equation yields:
Q
=
3.94 x 10-5 D2.63
Rearranging by solving for diameter:
Q
= 47.28 QO.38
Using this equation, the diameter of each pipe can be calculated. For many of the
pipes, the diameter will not be a commercial size. Hence, the exact diameter will
have to be rounded up rather than down, but this is a matter of judgement. The
resulting diameters in mm are:
Pipe No.
1
2
3
4
5
6
7
8
9
10
11
12
13
Exact
Diameter
Rounded mm
96
93
61
52
40
40
52
40
40
61
40
52
40
100
100
50
50
38
38
50
38
38
50
38
50
38
This is an initial estimate of diameters. Note that all of them were rounded down
to commercial sizes except those for pipes No. 1 and 2. If all diameters had been
rounded up including pipes No.3 to 13, there would be no question about the
feasibility of this design; all node pressures would be above the minimum target
value because the gradient would be less than 0.00973 which was used to calculate
the exact diameters. Because eleven of the diameters were rounded down,
however, a check must be made to verify that the proposed design is feasible.
The actual headloss should be calculated in each pipe, and if the minimum target
pressure of at least 5 m is not obtained at each terminal node, then diameters
should be adjusted until this is achieved. There are 6 branches in this network;
only two of them will be checked in this worked example. The others can be
checked as an exercise.
Consider the branch with terminal node No.1; its pipes are Nos. 1,2,3,4 and 6.
The Hazen Williams equation can be rearranged to solve for pipe headloss (H) as
a function of flow (Q), diameter (D), and length (L). The equation (for C = 130)
is:
~~~~liiiWliiilG
....,., '-~' '~;. ':"~~"~:<h"~"
__' -~;~.""",,;_;~i_"~\i.,.,;,{.:~.'~i;.)..~"~:,~-~;:.-;ii,i~:kW~,i:,~<~'
H
=
1.39
X
106 QI.86 L/D4.87
Now, using the initial rounded diameters, pipe flows and pipe lengths, the
headloss in each pipe can be calculated. For the branch with terminal node No.1,
the actual headloss would be:
Pipe No.
Q (Ips)
L (m)
D (mm)
H (m)
1
2
3
4
6
6.50
165
100
1.33
5.85
210
100
1.39
1.95
225
50
5.73
1.30
95
50
1.14
0.65
165
38
2.09
The total headloss in this branch is the sum of the individual pipe losses which is
11.68 m. Hence, with a pressure at the inlet of 14 m, the pressure at terminal node
No.1 is 14.0 - 11.68 = 2.34 m, which is below the minimum pressure target of 5
m. Hence the diameters for this branch are infeasible.
Now let's check the branch with terminal node No. 10; the pipes in this branch are
Nos. 1,2, 10, 12 and 13. Using the same approach, the total headloss is calculated
to be 11.51 m, which means that the pressure at node No. 10 would be
14.0 - 11.51 = 2.49 m. Again, the design is infeasible because the pressure is
below the minimum target of 5 m. Both of these infeasibilities occurred because of
rounding eleven of the pipes down instead of up. It is important to note that this
happened despite what appears to be substantial increases that were made by
rounding up pipe Nos. 1 and 2.
The question now is how can some diameters be changed to make the design
feasible? Which pipes should be adjusted? The general rule to follow in cases
where diameters need to be enlarged is: leave pipes near the ends small and enlarge
them near the source. In cases where diameters are too large and must be reduced,
the rule is reversed: leave pipes near the source large and reduce those nearer the
ends. This rule will generally result in lowering the cost.
In the case of these branches, it was pipes fairly close to the source that were
rounded down that caused the infeasibility, namely pipes No.3 and 10. Since
diameters need to be enlarged, the above rule suggests that these are the pipes
whose diameters should be changed. Assume that the next larger commercial size
above 50 mm is 75 mm. Hence, it is proposed that pipes No.3 and 10 both be
increased to 75 mm. The resulting design is:
8
Pipe No.
Diameter
1
2
100
100
75
50
38
38
50
38
38
75
38
50
38
3
4
5
6
7
8
9
10
11
12
13
Using the Hazen Williams equation, the headloss in pipe No.3 with 75 mm
diameter would be 0.79 m, and in pipe No. 10 and 75 mrn, the headloss would be
••
~
.
"
<
t,
"'••~.;~ ,
M;.:l.]
0.66 m. Hence, these changes would increase the pressures at terminal nodes No. 1
and 10 to 7.26 m and 5.88 m, respectively. As expected, the design is feasible.
To conclude this example, let us calculate the cost of pipe in the network using the
diameters listed above. Recent prices in pesos per metre for constructing pipe in
South Atnerica are shown below; the table also shows the length of each different
diameter pipe in the proposed network. Total pipe cost is 1.269 million pesos.
Diameter(mm) Price (P 1m)
38
50
75
100
300
440
800
1310
Length (m)
Cost (103 P)
870
485
380
375
261
213
304
491
Three Hour Practical
1.
The Three Hour Practical session for this submodule should be patterned
after the worked problem above except using local data. The participants
can be divided into groups of from I to 3 persons. Each group should be
given a street map of a small community and asked to design a branched
network. Even if the topography is hilly, it should be assumed to be flat.
Each group should assume a different set of background data and
produce a design for those conditions. Because calculation will be made
by hand, the networks should have a maximum of about 25 pipes. If there
are several groups, an interesting exercise is to have each group work on
the same problem using the same map, but with different design
standards. For example, one group can assume per capita flow of tOOled
and another can use 150 led. One group can use an elevated tank height of
13 m and another can use 18 m. One group can assume 200 persons per
standpipe and another can assume 150. After each group does its best to
produce a least cost network design, the groups should be reassembled so
they can compare their results, especially network costs. They should pay
particular attention to the effects that design pressures and flows have on
pipe sizes and costs.
In addition to these practical examples the following points can be
considered through class discussion, or in small groups.
2.
Given a set of demand nodes, total pipe length and cost can be minimized
by linking them together with a branched network. Why then are looped
networks used? Why is it necessary to have looped networks? What are
their advantages?
3.
Suppose a network has to be designed for a target flow of X. The
engineer can assume, say, either 10 demand nodes or 20. If he assumes 20,
will the pipe diameters be larger or smaller than if he assumes 1O?In
deciding on the number of demand nodes, is it safer to assume a larger
number or a smaller number? Why is it possible that in some public
standpipe systems, the pipe diameters are larger than those in networks
that provide house connections?
4.
What are the average per capita design flows for standpipes, yard taps,
single house taps, multiple house taps? Why do they change? What are
the peaking factors for each of these different levels of service? Why are
they different?
5.
How would you conduct a study to determine the average per capita
consumption in a community? How would you collect the data to
measure the peak hourly flow? How would you measure peaking factors?
6.
Why are networks usually designed for peak hourly flow? If the network
did not have a central storage tank, but rather each house had its own
tank, would it still be necessary to design for peak hourly flow? Would it
make any difference if the tanks were equipped with flow restrictors in the
inlet?
,
~:ml1"!":!~~'~m:ffflm!'!~~\'~'''W~'~(''';>(:~!''''''''!'''i'''',,,,"',,,,,,,,,-
9
•
rttaJ
Appendix D:
Data for Model Calibration
1
r:_J
>
COOK ISLANDS
Utility Profile
MINISTRY OF WORKS, ENVIRONMENT
(Water Supply Division)
Address
Telephone
Fax:
Head
AND PHYSICAL PLANNING
P. O. Box 102, Rarotonga, Cook Islands
: (682) 20034
(682) 21134
: Mr. Nooroa Parakoti, Director
The Water Supply Division is responsible for the water supply of Rarotonga Island with a population of 11,100 including
the capital, Avarua Township. It is a division under the Ministry of Works, Environment and Physical Planning (MOWEPP)
with a water supply system that was established in 1900. The government exercises control on the number, salary and
appointment of staff, appointment of top management, budgets for O&M and development. The division has a partly
developed management information system. Development is guided by its 1995-2000 Development Plan. No annual
report is published by the Water Supply Division. The utility still do not collect any tariff from its consumers. As part of the
government reform process, consideration is being given to the utility's privatization.
No Mission Statement.
Connections
Staff
Annual O&M Costs
Annual Collections'
Annual Billings!
Annual Capital Expenditure
(Average over last 5 years)
Source of Investment Funds
1
4,265
15
NZ$405,700
NZ$ 9,500
NZ$ 9,500
US$275,181
US$ 6,444
US$ 6,444
NZ$360,000
US$244,184
Expenditure Per Connection
US$57.25/connection
72.2% commercial loan; 27.8% externally-funded government grant
Billings and collections are for new connection fees.
There is no tariff levied by the Government on water consumers
for the water supply service at present.
Notes:
I.
1 There were 50 new connections in 1996. Cost of new connection is NZ$200.00
(US$135.66)
2 About 95% of all industrial, commercial and institutional connections and 12% of
house connections are metered.
As seen by Management
1) Improve water resources management in the island.
2) Development of water supply master plan for Rarotonga.
II.
Consumers' Opinion
1) Improve water quality.
2) Better water storage facilities.
No tariff is levied on the consumers. Average monthly power bill is NZ$81.50 (US$55.28). About 80% claim to have 24hour water supply. Perception on water quality ranges from satisfactory (56%) to poor (36%) with only 8% saying quality
is good. About 75% boil, filter or do both to their drinking water. Approximately 27% said water pressure is low. Water
supply interruption was experienced by 55% of the respondents on the month prior to the survey. Leak repairs take about
3 days to be made after reporting to the utility. Overall rating of the utility is fair (52%) to good (17%).
;lf~m'
.·_.·~4'·='
-
The number of connections increased by 69% which are mostly residential users. However, UFW increased from 27% to
70"10attributed mostly to leaks in house plumbing systems and agricultural use in residential connections where only 12%
of connections are metered but comprise 98% of total connections. Staff/1,000 connections ratio improved to 3.5 from 12.6.
Funding sources also changed from purely government grant to the use of commercial loans (72.2%) and externally-funded
government grant (27.8"10).Unit production cost decreased by 13%.
108
Second Water Utilities Data Book for the Asian and Pacific Region
[~]
Appendix E:
Feedback
~]
Taniela Qutonilaba
Taito Apisarome
Ajay Prasad Gautam
Samuela
Tubui
Timoci Turaga
Sereicocoko Yanuyanurua
Everything were OK.
Nothin to com lain about.
The workshop was a great success for me. I do not have any
complaints or problems but look forward to continued backup
support until I am really familiar and can run this program
inde endentl . Thank- ou.
Good Points
-got first hand knowledge with respect to what WaterCAD is
about
-now I am fairly confident in using WaterCAD software at my
work
it would be easier for me now to relate to basic problem or
information to WaterCAD software
-course was very well organised
-good provision of tea and lunch
-good access to communicate with the course director
Note
We hope to organise more courses in future in-lieu to above.
Presentation
Excellent, a flow chart to help participants in trying to find their
way through any given exercise. This could be compiled
together as a set of notes for to help participants in their
exercise.
Facilities
Excellent, appropriate for this kind of workshop.
Participation
The examples given as exercise was quite good and they do
clearly demonstrate the strengths behind this software.
Training on Maplnfo and AutoCAD would really add some
knowledge to the use of this software
Thanks.
Presentation
Clear and easy to follow up. Very good.
Tutorials
Detailed explanation of various attributes in software. Very
good.
Model
Have now attained a very good grasp of using the WaterCAD
software.
Data Collection
Did not have tome to collect field data but this can be done by
us. A thorough explanation was done on the need and
importance of this.
Conclusion
The workshop have been a tremendous benefit to myself in
terms of m lob res onsibilities and duties.
The course was well presented and organised. I have learnt a
lot and is very beneficial to my work. I should be able to use
this software with confidence after this course. Ve well done.

Workshop on hydraulic network modelling with waterCAD

Transcription

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