Workshop
on
Hydraulic
Network
Modelling
with WaterCAD
16-20 October 2000
Paula Dawe
SOPAC Water Resources Unit
[2]
[3]
CONTENTS
INTRODUCTION 4
PARTiCiPANTS 4
TRAINING SCHEDULE FOR WATERCAD COURSE 4
RECOMMENDATION ARISING FROM THE WATERCAD TRAINING COURSE .4
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 Position Area Phone Contact
1 Taito Apisarome Technical Assistant Suva Water Supply 321099
2 Timoci Turaga Senior Engineer Suva Water Supply 321099
3 Sereicocoko Yanuyanurua Hydraulics Enqineer Suva Water Supply 385334
4 Samuela Tubui Operations Enclneer Northern Division 812044
5 Piu Sekitoga Acting Supervisor Western Division 660899
6 Viiendra Prasad Water Engineer Suva Water Supply 385334
7 Aiav Prasad Gautam Senior Enqineer Suva Water Supply 385334
8 Taito Delana Senior Engineer Western Division 660899
9 Taniela Qutonilaba Technical Assistant Suva Water Supply 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
[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:
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• 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 (with_elev. wcd and
Turangi-wo_elev.wcd) and Suva layers for creating a .dxf background (Suva.dxf) Day 4
20. Using the control function for pumps and valves
21. Using the find function to locate problems within your model 22. Do the scenarios tutorial
23. Running model scenarios: • For calibration
• For analysis of system
24. 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
25. Investigate the different scenarios in the file Example2.wcd
26. Using the information provided for the Cook Islands- (pressure, flow and demand data) calibrate the model in Turangi-with_elev.wcd
27. Things to remember when creating a model for a large-scale network. 28. 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.
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[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:
d
v=-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
=
(mv2-mvl)lt=m(v2 -vl)lt=maAnother form of this equation for a moving fluid can be written as:
F
=
pQ(v2 -VI)=
pAv2The 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
i1v
-MM- pgM& = (pi1dM)v-i1d
Dividing by pM gives:
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M
-+
g&+vL1v=0p
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: P Energy imparted by pressure, or work done on fluid
r
• Elevation Head: z Same as Potential Energy or energy due to gravity 2
• Velocity Head: ~ Same as Kinetic Energy or energy due to motion
2g
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:
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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 Z1or 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
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:
I v2
hL
=1
d2g
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 the higher the velocity, the greater the head loss
hLa11d 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
=64R
Laminar flow can then be expressed as:
Q=;rd4
pg
hL128J.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. It can be expressed as follows:
v ~
kCR°"'
SO" orh
L=~9,f( ~
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 Value of C Cast iron: new 130
5 yrs old 120 10 vrs old 110 20 vrs old 90-100 30 yrs old 75-90 Concrete 120 Cement lined 140 Plastic 150 Asbestos cement 140
16 (;11 ,.\ new 8.QOmmdmmcicr duenlc iron pipe ••••ith cementmortar 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 (100)°.38 ( C )
s;
=
SIOOC
de=
dlOOC
o;
=
QlOo 100Mannings 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.
= 0.397D2/3S0.5
h=(~J2L
v F or L 066
n ~
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. Ine neaa lOSS may be expressed as:
2
HL=k~
2g
where k is the loss coefficient and is a function of changes in direction, obstructions, or changes in velocity. kis constant for a given fitting, but varies with fitting size.
Expansions in piplines, produce substantial energy losses. At abrupt enlargements energy loss can be calculated from:
HL=
(V\-V2)22g
One special case of a sudden contraction is that of a square edged pipe entrance from a large tank where V1 isO. For this situation:
2
HL=0.5
2
2g
If the entrance is bell-mouthed, kcan be taken as 0.4. The insert table on the following page gives various kvalues for different fittings.
Fitting K Entrance 0.5 Contraction 0.143 90° bend 0.18 Gate valve 0.12 Check valve 0.75 Elbow 0.39 Expansion 0.277 Bell-mouthed Entrance 0.4 Exit 1
k can also be expressed in terms of equivalent length (Vd) at a certain velocity. This is expressed as follows and demonstrates the relationship that exists between fand k:
k=f
i
d
Where:
1= pipe length
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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, henhand exit losses,
[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
[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, P2 V22-+-+z,+Epump =-+-+Z2
pg 2g pg 2g
Epump will appear as an abrupt rise in the energy line over the pump machine. Pumps
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
[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 usedFor any pipe in a loop of the system, the actual flow will differ from an assumed flow by an amounti1:
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.QxI , =0
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.kiQt8. Compute, without regard to sign, the sum
I.
kiQt-19. 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
32400 ~ max i e on/off valves 150-250 m to prevent deposition of 0.3-0.6 rn/s
~1m/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 X 5.1 1000
=
lOAmand the height of the HGL above the water level in the tank is 20 - lOA
=
9.6m which meets the design requirements of 5m minimum., • n
Three Hour Practical
iJes'3')
E
I
1
1. GeneralXQi'Y'tp
e
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 17 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 12.1m3should be in Part II and 605 + 850 X 42m3or 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 10m3each 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.
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.~.;.., "':',
-,<"--., ,.",\,;,/:'.,,: .:
_-_ •• h~~ 4*~ .;tno'
:1t>t~:4d;VW§-FifiPjJ1
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.
'. i>
!"mlli!1!lm~~'
,> ' '. .
""Fmi::)«~~~i!
. (d'~' "';:k' L"'" '.>
... H"H' •..·»., ..;>~·..._.,..'"'.H" •.•.•:.";.·.; ..;"".,,, •.;~."'~",.,·.,J"i," •.... , ..•• ,. _"""""'i,,,,,, •.,,..,~";"'""',,••••u~ ••••
·f:::lWt:iilliliih!I~~i.:·~
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
()
-e5IQ..J
n
F"igure 1 hsows a street map 0faa commumty WIt present popu anoncommuni . hI' 0f 1,000 thatEy.
QM\e
'1
is to be served entirely by public standpipes. The purpose of this example is toP
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 1,000
Average per capita flow 100 l/c/d
Peaking factor 3
Town growth rate 2 %/year
Design period 20 years
Unaccounted for losses 20 0/0
Number of persons per standpipe 100 Maximum level of elevated tank 16 m Minimum level of elevated tank 12 m
Ground elevation, all nodes 0 m
---_
.§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
,
•••...,~_n.c.,.,.-@A]
, .•• '_.. ,,1.' ,-,: •..,~.j·~."•..•,~~.";'~~i~~(~~~
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~ ~~ 2 210 5.85 3 225 1.95 4 95 1.30 5 110 0.65 6 165 0.65 7 155 1.30 8 lW Q~ 9 1~ Q~ 10 155 1.95 11 140 0.65 12 235 1.30 13 160 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 1,2,3,4,6 860 2 1,2,3,4,5 805 6 1,2,7,8 650 7 1,2,7,9 705 8 1,2,10,11 670 10 1,2,10,12,13 925
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
=
3.7 X 10-6C 1)2.63(H/L)OS4 Q ::::flows, Ips C = roughness co-efficient = 130 D=
diameter, mm H/L=
hydraulic gradient=
0.00973Substituting these values into this equation yields: Q
=
3.94 x 10-5D2.63Rearranging by solving for diameter: Q
=
47.28 QO.38Using 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:
Exact
Pipe No. Diameter Rounded mm
1 96 100 2 93 100 3 61 50 4 52 50 5 40 38 6 40 38 7 52 50 8 40 38 9 40 38 10 61 50 11 40 38 12 52 50 13 40 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:
....,., '-~' '~;. ':"~~"~:<h"~"__'-~;~.""",,;_;~i_"~\i.,.,;,{.:~.'~i;.)..~"~:,~-~;:.-;ii,i~:kW~,i:,~<~'~~~~liiiWliiilG •• ~
.
H = 1.39 X 106 QI.86L/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. 1 2 3 4 6
Q(Ips) 6.50 5.85 1.95 1.30 0.65
L (m) 165 210 225 95 165
D (mm) 100 100 50 50 38
H (m) 1.33 1.39 5.73 1.14 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 5m. 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:
Pipe No. Diameter
1 100 8 2 100 3 75 4 50 5 38 6 38 7 50 8 38 9 38 10 75 11 38 12 50 13 38
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 (P1m) Length (m) Cost (103P)
38 300 870 261
50 440 485 213
75 800 380 304
100 1310 375 491
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 9 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?
rttaJ
Appendix
D:
Data for Model Calibration
r:_J
1
>
COOK ISLANDS
Utility
Profile
MINISTRY OF WORKS,
ENVIRONMENT
AND PHYSICAL PLANNING
(Water Supply Division)
Address P. O. Box 102, Rarotonga, Cook Islands Telephone : (682) 20034
Fax: (682) 21134
Head : 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 4,265
Staff 15
Annual O&M Costs NZ$405,700 US$275,181 Annual Collections' NZ$ 9,500 US$ 6,444 Annual Billings! NZ$ 9,500 US$ 6,444 Annual Capital Expenditure NZ$360,000 US$244,184
(Average over last 5 years) Expenditure Per Connection US$57.25/connection
Source of Investment Funds 72.2% commercial loan; 27.8% externally-funded government grant
1 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: 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.
I. As seen by Management II. Consumers' Opinion 1) Improve water resources management in the island. 1) Improve water quality. 2) Development of water supply master plan for Rarotonga. 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 24-hour 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%.
[~]
Appendix
E:
Feedback
~]
Taniela Qutonilaba Everything were OK. Nothin to com lain about.
Taito Apisarome 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.
Ajay Prasad Gautam 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. Samuela Tubui 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. Timoci Turaga 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.
Sereicocoko Yanuyanurua 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.