watercad theory document

Workshop onHydraulic Network Modelling with WaterCAD

16-20 October 2000

Paula DaweSOPAC Water Resources Unit

October 2000 SOPAC Miscellaneous Report 395

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[SOPAC Miscellaneous Report 395 - Dawe]

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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


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INTRODUCTION

In September of 2000 a request was made to the SOPAC Water Resources Unit to perform aweek long training course on the hydraulic modelling software WaterCAD by the Fiji PublicWorks Department (PWD). WaterCAD is software produced by Haestad Methods.

As part of a review of the Suva-Nausori master plan, an Australian consulting firm produceda hydraulic model of this network. This model and the WaterCAD software used to build itwere then given to PWD without any training on how to use it. SOPAC was then approachedto provide the required training.

The training course took place during 16-20 October 2000. The venue for the workshop wasthe SOPAC Headquarters, Mead Road.

PARTICIPANTS

The following from Fiji PWD participated in the workshop:

Name Position Area PhoneContact1 Taito Apisarome Technical Assistant Suva Water Supply 3210992 Timoci Turaga Senior Engineer Suva Water Supply 3210993 Sereicocoko Yanuyanurua Hydraulics Enqineer Suva Water Supply 3853344 Samuela Tubui Operations Enclneer Northern Division 8120445 Piu Sekitoga Acting Supervisor Western Division 6608996 Viiendra Prasad Water Engineer Suva Water Supply 3853347 Aiav Prasad Gautam Senior Enqineer Suva Water Supply 3853348 Taito Delana Senior Engineer Western Division 6608999 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 inthe 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 ofpoints for future note were made however. These include the following:

• That future training courses in the use of AutoCAD and Maplnfo software could beorganised through SOPAC.

• That the SOPAC WRU are available to assist PWD with the development of WaterCADhydraulic models for the Suva, Northern and Western Divisions and that an agreement tothis effect be worked out.


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Appendix A:Training Schedule for PWD WaterCAD

Course

[SO PAC Miscellaneous Report 395 - Dawe]

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Day 11. 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, valve6. 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 28. 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 periodanalysis

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 patterns14. Extended period simulation- flow patterns

• Using file Paula-PWDIExample2. wed, change the flow pattern from residential toquarry 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 ademand pattern for the rural example you created and analyse it for the extendedperiod option. Are the system pressures and flows still reasonable?

Day 315. 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 roughnesscalibration 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 fromthe pipe table and importing elevation data for the junction table using the simple urbansystem 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 Turangi-

wo_elev.wcd) and Suva layers for creating a .dxf background (Suva.dxf)

Day 420. Using the control function for pumps and valves21. Using the find function to locate problems within your model22. Do the scenarios tutorial23. 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'vecreated based on the given population growth rate

25. Investigate the different scenarios in the file Example2.wcd26. Using the information provided for the Cook Islands- (pressure, flow and demand data)

calibrate the model in Turangi-with_elev.wcd27. 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

DayS29. 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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Appendix B:Background Notes on Hydraulics


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Background Theory for Hydraulic Modelling

Computer Modelling

A model is something that represents something in the real world. A computer model usesmathematical equations to help explain and predict physical events. Modelling of waterdistribution systems can allow you to determine system pressures and flow rates under avariety 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. Thepredictions of pressure and flow rates that the model produces are only as accurate as theassumptions or data used to formulate the equations in the model. Appropriate values forfriction loss, pump performance, demand, etc. must be carefully defined before being inputinto the model. When a model has been properly calibrated, predicted pressures in actualsystems have been found to be within 35 to 70 kPa of measured values.

Hydraulic Theory

HydrodynamicsHydrodynamics deals with the movement of fluid. There are 3 basic laws to fluid flow:1. Conservation of mass2. Conservation of energy3. Newton's second law of motion (F=ma)

The first two laws appear in many different forms, depending on how the symbols aredefined, the importance of terms, the mathematical language used, etc. Basically, you willstill have the same basic equations but different constants and unit conversion terms will beused.

For the purposes of simplification, it is often assumed that water flows as an incompressibleideal fluid. An ideal fluid is without Viscosity and therefore can have no frictional effectsbetween moving fluid layers or between these layers and boundary walls. This means thatthere will be no eddy formation or energy dissipation due to friction. The assumption of anideal fluid allows a fluid to be treated as a collection of small particles, which will supportpressure forces normal to their surfaces but will slide over one another without resistance. Insituations where friction is small, the frictionless assumption will give good results. Wherefriction is large, the assumption of an ideal fluid will not provide good results.

Conservation of Momentum or Newton's Second LawThe simplest definition of momentum is: momentum = mass x velocity. A body hasmomentum by the fact that it is moving. If the velocity is zero, then the momentum is zero.


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The law of conservation of momentum states that a body in motion cannot gain or losemomentum, unless some external force is applied.

Newtons 2nd law of motion is that a force is equal to the rate of change of momentum. Sincewe are dealing with the movement of fluid, it only makes sense that laws of motion now applyto the particles of that fluid. As a fluid particle moves, it is displaced from its original positionover time in the direction of motion. The velocity of this particle can be described by theequation:

dv=-

t

If the velocity of the particle changes over time, it will have acceleration. Unbalanced forcesacting on particles of an ideal fluid will result in the acceleration of these particles accordingto Newton's 2nd law. Thus, a body cannot gain or loose momentum unless an external forceis applied. With acceleration defined, Newton's 2nd law of motion can now be applied to themoving fluid particle. This equation is:F = (mv2 -mvl)lt=m(v2 -vl)lt=ma

Another form of this equation for a moving fluid can be written as:F = pQ(v2 -VI) = pAv2

The application of the momentum equation can be demonstrated as moving liquidapproaches 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 thediagram below. Looking at a control volume in a pipe bend which changes direction in eitherthe horizontal or vertical plane, summing the forces on the control volume results in thefollowing equations:

This summing of forces can be applied to many different control volume examples, such asin reducers and nozzles.

Conservation of MassThe law of conservation of mass states that mass must be conserved. It can neither becreated nor destroyed. Basically, what this means is that what goes in, must come out. Thefollowing diagram illustrates this:


The above is further explained by the following equation, otherwise known as one form of thecontinuity equation:y-X =0

The continuity equation expresses the continuity of flow from one section of a fluid to anotheras the fluid moves. Another way to express conservation of mass is by the followingequation:p\A\v\ = P2A2V2

This equation expresses the fact that in steady flow, the mass flowrate passing all sections ofa 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, Qis designated as the volume flowrate:Q = A\v\ = A2v2

What are the units of Q?

Conservation of EnergyThe principle of conservation of energy states that energy can neither be created nordestroyed, 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 (theLl can be termed the "change in" the following parameter):

1. pressure forces acting on either ends of the elementF=MM2. and the component of weight acting in the direction of motionW=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 intoF=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 toobtain Bernoulli's equation:

p v2

-+-+z=Hpg 2g

Bernoulli's equation is an energy balance form one point in a hydraulic system to another. Itapplies to all points on the streamline and thus provides a useful relationship betweenpressure P, velocity v, and height above a datum, z. Units of head are in meters, but head isjust 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 fluidr• Elevation Head: z Same as Potential Energy or energy due to gravity

2

• Velocity Head: ~ Same as Kinetic Energy or energy due to motion2g

The Bernoulli equation may be visualised for liquids as vertical distances. The sum of allthree 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 gradientwhich can be overcome by the momentum of the fluid

• Negative pressures occur at any place where the pipeline rises above the hydraulic gradeline

When might an adverse pressure gradient form, trying to push the liquid back in the directionof 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 furtherexpanded to act as an energy balance for fluid in a system. The total energy of the fluid atone point has to equal the total energy of the fluid at a point farther down the streamline, butrelative proportions of the form the energy is in (pressure energy, elevation energy, kineticenergy) 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 andelevation head must decrease. In many flow problems, elevation may vary little and thegeneral statement- where velocity is high, pressure is low- can be made. When liquid flowsthrough a pipeline, the continuity equation has to be obeyed, so any loss of energy appears aa reduction in pressure. For example, if water flows through a ling pipeline of constantdiameter at a constant rate, then the mean velocity must be the same at all points along thepipeline to maintain continuity of flow. Thus any loss of energy appears as a reduction ofpressure or head.

One useful application of Bernoulli's equation is that it shows that the velocity of an ideal fluidexiting from a small orifice under a static head varies with the square root of the head. Thiscan 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 otherequations, such as the continuity equation, there must be only one unknown to be able tosolve 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, soif 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 thenit can be assumed that the water pressure in the jet is the same as the surroundingatmosphere. If gauge pressure is used, P=O.

4. With large tanks or reservoirs the velocity on the water surface can be assumed to bezero, so v=O.

5. The datum from which elevation is measured can be taken through the lower of the twopoints 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 unknownvalues.


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Real Fluid FlowThe flow of a real fluid is more complex than that of an ideal fluid owing to the presence ofviscosity. Viscosity introduces resistance to motion by causing shear or friction forcesbetween fluid particles and between these and boundary walls. For flow to take place, workmust be done against these resistance forces, and in the process energy is converted toheat.

The effects of viscosity cause the flow of a real fluid to occur under two very differentconditions or regimes:• Laminar flow• Turbulent flow

In experiments conducted by Reynold's, he discovered that for low velocities of flow in aglass pipe, a thin filament of dye issuing from the tube did not diffuse but was maintainedintact throughout the pipe, forming a thin straight streamline. However, as velocity wasincreased in the pipe, the dye filament would waver and break up, diffusing through theflowing water in the pipe. From this experiment, Reynold's was able to isolate a criticalvelocitv from which flow chanaed from one reaime to the other.

Since surface roughness increases the turbulence in a flowing fluid and thus decreasing theeffect of viscous action, roughness contributes to energy loss within the fluid. Energy isdissipated by the work done in continually generating turbulence by the roughness. Theenergy involved in this turbulence is composed of the kinetic enrgy of the fluid mass. Energydissipation is therefore proportional to the square of velocity.

Velocity ProfilesThe shearing stress created by viscosity effects in the fluid produce velocity profilescharacterized by reduced velocities near the boundary surfaces. This differs from theuniform velocity distribution of an ideal fluid. Since the velocity is no longer uniform, meanvelocity is now used in calculations with real fluid flow, and a correction factor is applied tothe 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 ofenergy to heat, part of which leaves the fluid. Friction energy or head loss is then in effectlost 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 thatflow partially full such as in channels and sewers are treated as open channels. The solutionof pipe flow problems results from the application of the energy principle, equation ofcontinuity and the principles and equations of fluid resistance. Resistance to flow in pipes isoffered 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 EquationEarly 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 losswas dirived where f is called the friction factor:

I v2

hL =1 d 2g

This equation, is the basic equation for calculating head loss caused by pipe friction (not pipefittings) in long, straight, uniform pipes.

In words it expresses the following relationships:hLuf the rougher the pipe, the greater the head losshLul the longer the pipe, the greater the head loss


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hLaV the higher the velocity, the greater the head losshLa11d the larger the diameter, the smaller the head loss

It is found that f depends only on the Reynolds number and another dimentionless parametereld, called the relative roughness, where e is the height of surface roughness on the wall ofthe pipe, and depends on the pipe material. Values of typical pipe roughness can be foundin the following insert. This relationship indicates a convenient means of presentingexperimental data on the friction factor.

The dependence of f on the Reynolds number and eld is different in laminar and turbulentflow regimes. In laminar flow, f is only dependent on R and may be calculated from thefollowing equation:

f = 64R

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 ofthe laminar film will decrease and the effect of viscous friction will decrease while roughnesswill become more important. In the region described as completely turbulent, f depends onlyupon eld. The variation of fwith these parameters is shown on the Moody diagram.

Hazen-Williams EquationThe Hazen-Williams equation was also developed for use in the pipe-flow problems. It canbe expressed as follows:

v ~ kCR°"'SO" or hL = ~9,f( ~ J"where:v is the mean velocity, C is a factor dependent on relative roughness, R is the hydraulicradius (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 corrosionand encrustation occurs in the pipe as it ages, increasing the pipes roughness. This effectcan be seen in the pictures below.

Descriptionof Pipe Valueof CCast iron: new 1305 yrs old 12010 vrs old 11020 vrs old 90-10030 yrs old 75-90Concrete 120Cement lined 140Plastic 150Asbestos cement 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 nomogramfor 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 beused for C values other than 100.

(100)1.85 (100)°.38 ( C )s; = SIOO C de = dlOO C o; = QlOo 100

Mannings EquationThe Manning equation is most commonly used for the analysis of flow in open channels, butit can also be applied to pipelines. For a pipe flowing full this equation is as follows, where nis the Manning roughness coeficient.

= 0.397D2/3S0.5 h =(~J2Lv F or L 066n ~

Minor LossesThe category of minor losses in pipes includes losses incurred by change of section, bendselbows, valves and fittings of all types. In longer pipes, minor losses can be neglectedwithout 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 ofvelocity causes large head loss because of boundary layer effects which result in flowseperation and extreme turbulence.


eany experiments mcicateo mat minor losses vary wnn me square or veiocny. I ne 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, orchanges in velocity. k is constant for a given fitting, but varies with fitting size.

Expansions in piplines, produce substantial energy losses. At abrupt enlargements energyloss can be calculated from:

HL= (V\-V2)2

2g

One special case of a sudden contraction is that of a square edged pipe entrance from alarge tank where V1 is O. For this situation:

2

HL=0.522g

If the entrance is bell-mouthed, k can be taken as 0.4. The insert table on the following pagegives various k values for different fittings.

Fitting KEntrance 0.5Contraction 0.14390° bend 0.18Gate valve 0.12Check valve 0.75Elbow 0.39Expansion 0.277Bell-mouthed Entrance 0.4Exit 1

k can also be expressed in terms of equivalent length (Vd) at a certain velocity. This isexpressed as follows and demonstrates the relationship that exists between f and k:

k=fid

Where:1= pipe length


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d= pipe diameter

Pipe Line ProblemAll steady-flow pipe line problems may be solved by the application of the Bernoulli andcontinuity equations. Ususlly, the engineering problems consist of:1. Calculation of head loss and pressure variation from flowrate and pipe-line

characteristics2. Calculation of flowrate from pipe characteristics and the head which produces flow3. 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 uponthe 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 errorsolutions.

There are a few points to remember when using Bernoullis equation:1. Pipes must be flowing completely full under pressure2. 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 linemust start in one reservoirs surface and end in the other, using a gradual drop to repersenthead 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:

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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

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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 septictank

• Flow in a pipe is possible up to around -7.5m of water, after this vaporisation of the liquidcan be expected

• Use vacuum, air relief valves, or pressure sustaining valves (PSV) to release air initiallyin the line or that accumulates over time, or to admit air when the line is being emptied formaintenance purposes

PumpsThe addition of mechanical energy to moving fluid by a pump alters the basic energy balanceof the Bernoulli equation. With the addition of energy by a pump, an additional term must beincluded in the equation.r; v2, P2 V22-+-+z,+Epump =-+-+Z2pg 2g pg 2g

Epump will appear as an abrupt rise in the energy line over the pump machine. Pumpstherefore, add head to hydraulic systems.

There are 3 main types of pumps available on the market:1. Centrifugal Pumps2. Axial Flow Pumps3. Mixed Flow Pumps


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• 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 system2. 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 repairs3. 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 mustoptimise (adjust parameters such as pipe size to achieve the most appropriate pressures atnodes and velocities in pipes) to find the best solution. The following insert helps to explainthis.

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 beconcentrated 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 outthe 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 nodesand 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 inelements of a system. The energy loss in any element of the pipe system may be expressedas:hj =kjQt

Where:hi = energy loss in element iOJ= flow in that element~ =constant depending on pipe diameter, length, type and condition


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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 anamount i1:

Qi = Qassumed +.6.

For any loop, the sum of the head losses about the loop must be equal to zero. Thus, forany 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 community2. Concentrate the disaggregated flows at the nodes of the system3. Add the required fire flow at appropriate nodes4. Select initial pipe sizes5. Assume any internally consistent distribution of flow. The sum of the flows entering and

leaving each node must be equal to zero6. Compute the head loss in each element of the system. Conventionally, clockwise flows

are positive and produce positive head loss7. 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 theloop. 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 somestipulated 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 theprocedure until a satisfactory solution is obtained

Typical Design Parameters

Fire Flows (Umin) 1890 ~ min32 400 ~ max

i e on/off valves 150-250 mto 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, totalfrictional loss is

2038 X 5.11000

= lOAm

and the height of the HGL above the water level in the tank is 20 - lOA = 9.6mwhich meets the design requirements of 5m minimum.

, • n Three Hour PracticaliJes'3')

E I 1 1. GeneralXQi'Y'tp e

This example is taken from an Indonesian project operated by CARP. It can beused as the basis to create a worked example for a three hour practical by alteringparameters to be more representative of conditions in this country. Participantsshould follow the step-by-step design procedure outlined below.

A village has a population of 850 divided into two parts, Part I with 605 personsand Part II with 245 persons. A water source with an estimated minimum flow ofone lis is located approximately 2,300 metres from the village. The standpipes areused for 12 hours per day. Assume that galvanized iron pipe must be used for allpipe because of the rocky terrain. (A sketch map of the village and a groundprofile are presented in Figures 3 to 6. Design the pipelines required to serve andplot 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 purposesthis is rounded up to 1,050 persons.

Water Usage

It is preferable to supply the maximum amount of water possible but aper capita supply of 100 litres per day would require 105,000 litres/day ora flow of 1.21 lis. Since the estimated minimum flow of the source is only1 lis this is not possible. A per capita use of 80 litres/day would requirean 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 80X 1,050 or 84,000 litres. Because the galvanized iron pipe is so expensivethe smallest pipe possible is used. In order to reduce the size of the main 17pipe, storage will be located in the village. The recommended storage isthen one half of 84,000 litres or 42m3• Based on the present populationdistribution 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 ithas been decided that the water will be distributed to five publicreservoirs, three in Part I and two in Part II. With this layout, the waterwill be under village control. Costs will be reduced because the main pipedoes not have to convey peak flows of water. The public reservoirs are tobe situated on high points to obtain any acceptable pressure head at thetank 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 eachfor a total of 12m3 (rounded down from 12.1m3). With this distribution ofreservoirs, 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 thenumber of faucets for Part I should be between 6 and 20 and for Part IIbetween 2 and 8. In order to accomodate future demand, a higher numberis preferable. Thus 6 faucets at each of the three reservoirs in Part I and 4faucets at each of the reservoirs in Part II give a total of 26. The averagenumber of persons per faucet (based on the future population of 1050) is42 for Part I and 37 for Part II. Half of the faucets can be placed on oneside of the reservoir and half on the opposite side. Thus, one area can beused by females and the other by males.

Design Flows

The path of the pipeline is sketched in Figure 3. The water will flowcontinuously into the reservoirs so the design flows will be the same as theaverage daily flows. The peaking factor is therefore 1. At the projectedper capita use of 80 IIday the average daily flow is 0.97 lis but the springhas an estimated minimum flow of 1.0 lis. Therefore, 1.0 lis will be usedin 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 andreservoir 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 toreservoir C. At C 0.24 lis is taken and the remainder of 0.24 lis flows toreservoir D. At reservoir E 0.14 lis is taken and the remainder of 0.15 lisflows to reservoir F. The design flows are noted on the pipeline route andprofile.

'. 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 workedproblem 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 IQ n F" 1 h f a communi . hI' f..J igure sows a street map 0 a commumty WIt present popu anon 0 1,000 that

Ey. 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 1,000Average per capita flow 100 l/c/dPeaking factor 3Town growth rate 2 %/yearDesign period 20 yearsUnaccounted for losses 20 0/0Number of persons per standpipe 100Maximum level of elevated tank 16 mMinimum level of elevated tank 12 mGround elevation, all nodes 0 mM;n;mnm llllowahle nressure 5 m\~

---------------------_ .

§co]

Layout

With 1,000 persons in town and 100 persons per standpipe, there will be 10standpipes. If the population is more or less evenly distributed throughout thetown, the standpipes should be evenly spaced. They should be located alongstreets where users have easy access, near street comers if possible. Figure 1 showsone 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 toconnect the standpipe nodes to the source node keeping total pipe length as shortas possible, laying pipes in streets, and selecting routes where the greatest numberof houses are located so that in the future when the system is upgraded toindividual connections, these houses will be able to connect. Figure 2 shows theresulting 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, thenumber 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 andabove. 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 eachstandpipe 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 nodeNo. 11 is 6.5 litres/second. Note that the network is being designed to meet peakhourly demands.

With a takeoff flow of 0.65 Ips at each demand node, it is an easy matter tocalculate the flows in the 13 pipes of the network. It is preferable to start the

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@A], .•• '_.. ,,1.' ,-,: •.. ,~.j·~."•..•,~~.";'~~i~~(~~~

calculations at the terminal ends and work toward the source. For example, theflow in pipe No. 13 is 0.65, in No. 12 it is 1.30, etc. The list of pipe flows andtheir lengths is as follows:

Pipe No. Length (m) Flow (Ips)

1 1~ ~~2 210 5.853 225 1.954 95 1.305 110 0.656 165 0.657 155 1.308 lW Q~9 1~ Q~

10 155 1.9511 140 0.6512 235 1.3013 160 0.65

Pressures

From the Background Data, the water elevation in the tank varies between 12 and16 m. At the time of peak hourly demand, the level should be approximatelymidway in the tank. Hence, the inlet pressure is 14 m.

The minimum target pressure is 5 m. Because the network is flat, this pressureshould occur at the terminal nodes of the network. That is, if the network is welldesigned, 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 minimumhydraulic gradient. This means finding the longest branch and dividing its lengthinto 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 bedesignated by the terminal node numbers. For example, branch No. 6 includespipes No. I, 2, 7 and 8, and branch No.8 includes pipes No. I, 2, 10 and 11. Thebranches with their pipe numbers and total lengths are:

Branch Pipe Numbers Total Length (m)

6

1 1,2,3,4,6 8602 1,2,3,4,5 8056 1,2,7,8 6507 1,2,7,9 7058 1,2,10,11 670

10 1,2,10,12,13 925

The branch with terminal node No. 10 is longest, and its average hydraulicgradient is 9 m: 925 m = 0.00973.

. -

&oJUsing the Hazen Williams equation and this gradient, the diameter of each pipe inthe network can be calculated based on its design flow. The Hazen Williamsequation is:

Q = 3.7 X 10-6C 1)2.63(H/L)OS4

Q ::::flows, IpsC = roughness co-efficient = 130D = diameter, mmH/L = hydraulic gradient = 0.00973

Substituting these values into this equation yields:

Q = 3.94 x 10-5D2.63

Rearranging by solving for diameter:

Q = 47.28 QO.38

Using this equation, the diameter of each pipe can be calculated. For many of thepipes, the diameter will not be a commercial size. Hence, the exact diameter willhave to be rounded up rather than down, but this is a matter of judgement. Theresulting diameters in mm are:

ExactPipe No. Diameter Rounded mm

1 96 1002 93 1003 61 504 52 505 40 386 40 387 52 508 40 389 40 38

10 61 5011 40 3812 52 5013 40 38

This is an initial estimate of diameters. Note that all of them were rounded downto commercial sizes except those for pipes No. 1 and 2. If all diameters had beenrounded up including pipes No.3 to 13, there would be no question about thefeasibility of this design; all node pressures would be above the minimum targetvalue because the gradient would be less than 0.00973 which was used to calculatethe 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 targetpressure of at least 5 m is not obtained at each terminal node, then diametersshould 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 bechecked 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) asa 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.86 L/D4.87

Now, using the initial rounded diameters, pipe flows and pipe lengths, theheadloss 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 0.65L (m) 165 210 225 95 165D (mm) 100 100 50 50 38H (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 is11.68 m. Hence, with a pressure at the inlet of 14 m, the pressure at terminal nodeNo.1 is 14.0 - 11.68 = 2.34 m, which is below the minimum pressure target of 5m. Hence the diameters for this branch are infeasible.

Now let's check the branch with terminal node No. 10; the pipes in this branch areNos. 1,2, 10, 12 and 13. Using the same approach, the total headloss is calculatedto be 11.51 m, which means that the pressure at node No. 10 would be14.0 - 11.51 = 2.49 m. Again, the design is infeasible because the pressure isbelow the minimum target of 5 m. Both of these infeasibilities occurred because ofrounding eleven of the pipes down instead of up. It is important to note that thishappened despite what appears to be substantial increases that were made byrounding up pipe Nos. 1 and 2.

The question now is how can some diameters be changed to make the designfeasible? Which pipes should be adjusted? The general rule to follow in caseswhere diameters need to be enlarged is: leave pipes near the ends small and enlargethem 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 theends. This rule will generally result in lowering the cost.

In the case of these branches, it was pipes fairly close to the source that wererounded down that caused the infeasibility, namely pipes No.3 and 10. Sincediameters need to be enlarged, the above rule suggests that these are the pipeswhose diameters should be changed. Assume that the next larger commercial sizeabove 50 mm is 75 mm. Hence, it is proposed that pipes No.3 and 10 both beincreased to 75 mm. The resulting design is:

Pipe No. Diameter

1 1008 2 100

3 754 505 386 387 508 389 3810 7511 3812 5013 38

Using the Hazen Williams equation, the headloss in pipe No.3 with 75 mmdiameter 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. 1and 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 thediameters listed above. Recent prices in pesos per metre for constructing pipe inSouth Atnerica are shown below; the table also shows the length of each differentdiameter pipe in the proposed network. Total pipe cost is 1.269 million pesos.

Diameter(mm) Price (P1m) Length (m) Cost (103 P)

38 300 870 26150 440 485 21375 800 380 304100 1310 375 491

Three Hour Practical

1. The Three Hour Practical session for this submodule should be patternedafter the worked problem above except using local data. The participantscan be divided into groups of from I to 3 persons. Each group should begiven a street map of a small community and asked to design a branchednetwork. Even if the topography is hilly, it should be assumed to be flat.Each group should assume a different set of background data andproduce a design for those conditions. Because calculation will be madeby hand, the networks should have a maximum of about 25 pipes. If thereare several groups, an interesting exercise is to have each group work onthe same problem using the same map, but with different designstandards. For example, one group can assume per capita flow of tOOledand another can use 150 led. One group can use an elevated tank height of13 m and another can use 18 m. One group can assume 200 persons perstandpipe and another can assume 150. After each group does its best toproduce a least cost network design, the groups should be reassembled sothey can compare their results, especially network costs. They should payparticular attention to the effects that design pressures and flows have onpipe sizes and costs.

In addition to these practical examples the following points can beconsidered through class discussion, or in small groups.

2. Given a set of demand nodes, total pipe length and cost can be minimizedby linking them together with a branched network. Why then are loopednetworks used? Why is it necessary to have looped networks? What aretheir advantages?

3. Suppose a network has to be designed for a target flow of X. Theengineer 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?Indeciding on the number of demand nodes, is it safer to assume a largernumber or a smaller number? Why is it possible that in some public 9standpipe systems, the pipe diameters are larger than those in networksthat 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 arethe peaking factors for each of these different levels of service? Why arethey different?

5. How would you conduct a study to determine the average per capitaconsumption in a community? How would you collect the data tomeasure the peak hourly flow? How would you measure peaking factors?

6. Why are networks usually designed for peak hourly flow? If the networkdid not have a central storage tank, but rather each house had its owntank, 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 theinlet?

~:ml1"!":!~~'~m:ffflm!'!~~\'~'''W~'~(''';>(:~!''''''''!'''i''''-,,,,"',,,,,,,,,- •

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 includingthe 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 andappointment of staff, appointment of top management, budgets for O&M and development. The division has a partlydeveloped management information system. Development is guided by its 1995-2000 Development Plan. No annualreport is published by the Water Supply Division. The utility still do not collect any tariff from its consumers. As part of thegovernment reform process, consideration is being given to the utility's privatization.

No Mission Statement.

Connections 4,265Staff 15Annual O&M Costs NZ$405,700 US$275,181Annual Collections' NZ$ 9,500 US$ 6,444Annual Billings! NZ$ 9,500 US$ 6,444Annual Capital Expenditure NZ$360,000 US$244,184

(Average over last 5 years) Expenditure Per Connection US$57.25/connectionSource 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 consumersfor 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% ofhouse connections are metered.

I. As seen by Management II. Consumers' Opinion1) 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 qualityis good. About 75% boil, filter or do both to their drinking water. Approximately 27% said water pressure is low. Watersupply interruption was experienced by 55% of the respondents on the month prior to the survey. Leak repairs take about3 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% to70"10attributed mostly to leaks in house plumbing systemsand 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 useof commercial loans (72.2%) and externally-fundedgovernment 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 Everything were OK.Nothin to com lain about.

Taito Apisarome The workshop was a great success for me. I do not have anycomplaints or problems but look forward to continued backupsupport until I am really familiar and can run this programinde endentl . Thank- ou.

Ajay Prasad Gautam Good Points-got first hand knowledge with respect to what WaterCAD isabout-now I am fairly confident in using WaterCAD software at myworkit would be easier for me now to relate to basic problem orinformation to WaterCAD software-course was very well organised-good provision of tea and lunch-good access to communicate with the course directorNoteWe hope to organise more courses in future in-lieu to above.

Samuela Tubui PresentationExcellent, a flow chart to help participants in trying to find theirway through any given exercise. This could be compiledtogether as a set of notes for to help participants in theirexercise.FacilitiesExcellent, appropriate for this kind of workshop.ParticipationThe examples given as exercise was quite good and they doclearly demonstrate the strengths behind this software.Training on Maplnfo and AutoCAD would really add someknowledge to the use of this softwareThanks.

Timoci Turaga PresentationClear and easy to follow up. Very good.TutorialsDetailed explanation of various attributes in software. Verygood.ModelHave now attained a very good grasp of using the WaterCADsoftware.Data CollectionDid not have tome to collect field data but this can be done byus. A thorough explanation was done on the need andimportance of this.ConclusionThe workshop have been a tremendous benefit to myself interms of m lob res onsibilities and duties.

Sereicocoko Yanuyanurua The course was well presented and organised. I have learnt alot and is very beneficial to my work. I should be able to usethis software with confidence after this course. Ve well done.


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