S:CHRISSISupplIWMTMcover · The text is substantially the work of Dr Melvyn Kay, of Silsoe College, UK, with additional technical input from N. Hatcho of the Land and Water Development

Small-scale pumped irrigation - energy and cost iii

PREFACE

This manual is about reducing the costs involved in small-scale pumped irrigation schemes.Too often, schemes are designed and constructed with thought given only to the immediate costof constructing the scheme and of buying and installing equipment. Little or no attention isgiven to operating costs, with the result that some schemes may well be cheap to install but verycostly to run. When water is pumped, every litre has a real cost because of the energy needed.If more water is pumped than is needed or is pumped inefficiently, then operating costs can risesignificantly because of the additional energy which is wasted.

Ways of approaching scheme design and equipment selection are described so as to takeaccount of the operating costs. Simple examples are used to show how this can be done, andhow true comparisons can be made between different designs. Guidelines are given, based onexperience in many developing countries, so that sound practical choices can be made.

The manual is not just for those starting a new scheme. It is also for those who wish toevaluate and improve existing schemes, and practical ways of reducing operating costs byimproving the efficiency of water use and pumping are described.

The readership envisioned is that group of people with some practical experience insmall-scale irrigation but who have little or no technical or engineering knowledge and wish tobe able to advise farmers on appropriate equipment selection and its proper and efficient use.

Although not numbered in the same series as the FAO/ILRI Irrigation Water ManagementTraining Manuals, this particular publication is seen as being complementary to that series,and, as a consequence, numerous cross-references are made in the text to the various volumesof the Training Manuals series.

The text is substantially the work of Dr Melvyn Kay, of Silsoe College, UK, with additionaltechnical input from N. Hatcho of the Land and Water Development Division, FAO, Rome.The text was edited and prepared by Thorgeir Lawrence for publication by FAO.

Any comments on the text as it stands and any suggestions for potential improvementsthat could be included in subsequent editions are welcomed, and should be addressed to:

Water Resources, Development and Management Service , AGLWFAOViale delle Terme di CaracallaI-00100 ROMA,Italy

iv

Small-scale pumped irrigation - energy and cost v

CONTENTSPage

1. INTRODUCTION 11.1 Small-scale irrigation 11.2 Problems 11.3 Solutions 21.4 Making choices 2

2. SOME BASIC CONCEPTS 52.1 Introduction 52.2 Pressure 52.3 Discharge 72.4 Energy 92.5 Power112.6 Efficiency 13

3. CHOOSING A NEW IRRIGATION SYSTEM 153.1 Introduction 153.2 Water sources 183.3 Pumps and power units 18

3.3.1 Pump types 183.3.2 Pump Characteristics 223.3.3 Pump selection 233.3.4 Power units 243.3.5 Efficiency 26

3.4 Distribution systems 293.4.1 Open channels 293.4.2 Pipelines 313.4.3 Distribution efficiency 35

3.5 Methods of irrigation 373.5.1 Surface irrigation 373.5.2 Sprinkler irrigation 393.5.3 Trickle irrigation 403.5.4 Selecting an irrigation method 40

3.6 System capacity 413.6.1 Crop water requirements 423.6.2 Peak scheme water demand 443.6.3 Seasonal scheme water demand 45

3.7 Peak power and energy demand 453.8 Costs 46

3.8.1 Capital cost 473.8.2 Operating cost 473.8.3 Overall cost 493.8.4 Effects of changes 523.8.5 Some general conclusions 533.8.6 Some practical considerations 54

vi

4. CASE STUDY - 1 55Introduction 55

4.1 Options available 554.2 Scheme water demand 564.3 Peak power and energy demand 574.4 Overall costs 594.5 Conclusions 594.6 Guidelines 62

5. CASE STUDY - 2 635.1 Options available 635.2 Scheme water demand 635.3 Overall power and energy demand 645.4 Overall costs 655.5 Conclusions 675.6 Guidelines 70

6. IMPROVING EXISTING SCHEMES 716.1 Introduction 716.2 Inefficient water use 726.3 Inefficient equipment 736.4 Effect of inefficiency 746.5 Evaluating a scheme 746.6 Obtaining data 76

6.6.1Observing and questioning 766.6.2Some basic data 76

ANNEX 79

Small-scale pumped irrigation - energy and cost vii

TABLES

Page 1. Energy content of fuels and foods 10 2. A guide to selecting centrifugal pumps 21 3. Pump selection for small-scale schemes 23 4. Indicative values of distribution efficiency (%) 35 5. Typical field application efficiencies for irrigation methods 37 6. Typical sprinkler data 39 7. Factors affecting selection of irrigation method 41 8. Indicative values for crop water needs and growing periods 43 9. Useful life of irrigation system components 4710. Indicative maintenance and repair costs 4811. Capital recovery factors (CRF) 5012. EAC values for pumps at various discount rates 5113. EAC values for pumps for different life expectancies 5114. Changing the distribution system and its effects on energy and cost 5215. Calculating scheme water demand 5616. Overall power and energy demands 5717. Overall cost comparisons 5818. Calculating scheme water demand 6419. Overall power and energy demands 6420. Overall cost comparisons 6521. Efficiency of surface irrigation methods 73

viii

FIGURESPage

1. Making choices , the design process 4 2. Relationship between force and pressure 5 3. Measuring pressure in a pipe 6 4. Calculating discharge 7 5. Measuring discharge 8 6. Energy conversion - analagous systems in people and machines 10 7. Illustration of the problem considered in Example 3 11 8. Relationship between rate of energy use and power 12 9. Graph relating flow, static head and power 1310. Choosing irrigation system components 1511. Components of a typical irrigation scheme 1612. Typical axial flow pump 1913. The radial flow (centrifugal) pump 2014. Typical mixed flow pump 2115. Pump characteristics of the three pump types 2216. Pump selection based on head and discharge parameters 2417. Manufacturer’s data for a centrifugal pump 2518. Efficiency of components of pumping plant 2619. Suction lift limitations 2820. Energy demand for open channel distribution 2921. Channel design: dimensions and drop structures 3022. Pipe system and its energy demand 3223. Hydraulic gradient 3324. Nomograph relating pipe diameter, discharge, head loss and velocity 3425. Basin, border and furrow irrigation 3626. Sprinkler irrigation 3827. Hose-pull sprinkler system 3928. Trickle irrigation 4029. Peak and seasonal scheme water demands 4230. The concept of water requirements in mm 4331. Relationship between pipe size and seasonal energy cost 5332. Effect on EAC values of reducing pump efficiency 6033. Effect on EAC values of changing interest rate 6034. Effect on EAC values of a greater depth to the groundwater table 6135. Effect of reducing pumping efficiency on EAC values 6636. Effect of greater depth to groundwater on EAC values 6637. Effect of increasing scheme size on capital and operating costs 6838. Evaluating irrigation scheme performance 7139. System efficiency value ranges 7440. The relationship between efficiency and seasonal operating costs 75

Small-scale pumped irrigation - energy and cost 1

Chapter 1

Introduction

1.1 SMALL-SCALE IRRIGATION

Small-scale irrigation is an important aspect of irrigation development in many countries.Approximately half of the irrigated area in Sub-Saharan Africa, for example, is irrigated in thisway. It involves individual or small groups of farms, organized and managed by farmers,usually independent of government resources. This type of development has often provedsuccessful in places where the larger-scale, primarily government controlled, projects havenot. This is not to say that small-scale is therefore better than large-scale farming, or indeedthat small-scale is more simple to develop. It is a different approach to irrigated farming, withits own challenges. Irigation development requires careful design, construction and managementto be successful. It is, perhaps, in the management element that the key difference lies. In asmall system there are no tiers of management, as in the large-scale schemes. Farmers alonedecide when to irrigate and how much water to apply; start and stop the pumps; and generallyrun the entire scheme with the help of the family or local community.

Small-scale farming can be highly productive in terms of yield per hectare of land. Theenergy input into large-scale schemes can be up to 15 times greater than that required for small-scale farming for the same output of crops produced. This is in sharp contrast to large-scaleschemes where the ratio is normally less than 4. Thus, on a national or regional scale, whenconsidering the use of commercial fuel in agriculture, which in many countries is both scarceand expensive, the small-scale approach can be an attractive one.

1.2 PROBLEMS

Despite their apparent attractiveness in terms of potential productivity, small-scale schemesare, however, not always as efficiently run as they could be. Most schemes rely on pumping tosupply their water needs and are often designed on the basis of minimum investment cost, withlittle or no thought given to the effect that this might have on operating costs over many years.For example, a farmer may purchase a cheap pump which runs at a very low level of efficiency.The energy cost may be considerable and it may require much servicing and spare parts. If thefarmer were to purchase a better and more appropriate pump then more money might be spentinitially but there should be much more money saved over the years through reduced fuel(energy) costs and maintenance. Similar issues arise when selecting other components of anirrigation system.

An equally important issue to consider is how well the scheme is managed once it is operating.The most appropriate system design and selection will be of little use in the hands of aninexperienced or unskilled irrigator. Good equipment is no substitute for good managementand, here too, considerable savings in energy and operating costs can be made by ensuringgood equipment and water management practices.

Introduction2

1.3 SOLUTIONS

This manual describes ways of approaching scheme design and equipment selection whichtake account of both investment and operating costs, and, in particular, emphasise the significanceof energy costs.

Some basic concepts need to be understood about water flow, energy and power, and, forthose who have little or no knowledge of these, they are described in Chapter 2.

In Chapter 3 the basic components of an irrigation scheme are described together with waysof choosing between different pumps, distribution systems and methods of irrigation. Theremay be many different ways of irrigating a farm and a basis for comparison and selection isneeded. Cost is often the dominant factor. Thus the idea of cost effectiveness is introduced,showing that both capital costs and operating costs must be considered when selecting equipment,and that the one affects the other. This is demonstrated in Chapters 4 and 5, where two contrastingcase studies show how the principles and practices of Chapter 3 can be applied.

Many small-scale irrigation schemes are already in operation, and one question here mightbe how to get the best results from what is already there. Chapter 6 examines ways of lookingat existing schemes to determine energy use and operating costs, and to find ways of reducingthem through improved efficiency of equipment and water use.

1.4 MAKING CHOICES

Much of this manual is about the process of design — the process of making logical choicesbetween systems of irrigation and equipment (Figure 1). It is important to realize at the outsetthat there is unlikely to be just one ideal choice; there may be many alternatives, any one ofwhich might be quite appropriate. The job of the designer is to present the options available inrelation to good irrigation practice, water availability, equipment, its reliability and cost. Thefarmer can then choose the system which he or she feels is most appropriate.

The design process

A preliminary design is usually done first. This is often done quickly in order to establish theoptions available. Once a choice has been made, work then proceeds to a detailed designwhich details every nut and bolt to be purchased and every canal and structure to be constructed.

To undertake a preliminary design, basic information is needed about the land and crops tobe irrigated. However, accurate details about land areas and crops may not be necessary at thisstage. To understand this it is important to realize what preliminary design is about. It is toestablish the maximum capacity or size of the system to be constructed and the choices availableto the farmer. The system capacity must be enough to satisfy the maximum amount of waterneeded by the crops, and there are simple ways of assessing this without detailed knowledge ofthe cropping. Clearly the answer will not be exact but great accuracy is not needed at this stage.Remember that when a scheme is operating it will run for most of the time at well below itsmaximum capacity. It may only run at full capacity for a very short period when the crops arematuring and need most water. It is very much like designing and using a car. It may bedesigned to operate at a maximum speed of 150 km/h, but most drivers would travel well belowthis speed and only use the maximum speed occasionally. Thus whether the maximum speed is150 or 160 km/h is not really very critical to the overall use of the car if it otherwise meets allthe demands made upon it by the driver. If the actual maximum performance is less than


150 km/h, then the car will still get there — it will just take a little longer. In the same way thecapacity of an irrigation system need not be determined with great accuracy as long as thecapacity will meet most, if not all, of the operating demands that the farmer will make. If thecapacity falls a little short of demand then the difference can be made up by running the systemfor a longer period.

A further aspect of design is considering “How will the final cost of the scheme be affectedby the decisions made during the design process?”. If, for example, the crop water requirementis changed by 10%, or a channel is increased in size by 20%, does this significantly affect theoverall cost of the scheme? If it does, then this figure needs to be chosen with considerablecare. If it does not, then such accuracy is not needed. A good designer will concentrate on theimportant factors which will have significant effects on the outcome. The inexperienced designerwill need to experiment a little to determine which are the critical factors in the design process.

A final aspect of design, which the inexperienced designer may not realize at first, is thatthere are no formulae which can help with the initial decision making. For example, there is noformula which would show that a pipe should be used instead of an open channel. This is amatter of choice, which may eventually be decided by cost or some other constraint. Thedesigner would thus consider both options, prepare a preliminary design for each one, and thensee which was best. Several designs may be done in this way before the best one can be chosen.In other words, the designer will often choose what seems to be appropriate and then set aboutproving that the choices made are indeed the best. This is where an experienced designer canbe invaluable. On the basis of past experience of similar situations the designer may well beable to greatly simplify the design process because he or she may have a very good idea of whatwill be the best solution. Unfortunately, the inexperienced designer must go through a morerigorous process to arrive at the best solution. This manual is to help the inexperienced designer,and to try and pass on some of the experience of others in order to shorten and simplify thedesign process.

Cost

Cost will be an important factor when making choices. In this manual typical costs are used todemonstrate the selection process, but the reader must take great care when using conclusionsdrawn from this because local costs may vary considerably from those shown. The reader isthus encouraged to go through the design process using local costs and to make judgementsbased on local solutions. Throughout the text the unit of currency used is the United Statesdollar ($US).

Introduction4

FIGURE 1Making choices - the design process


Chapter 2

Some basic concepts

2.1 INTRODUCTION

This chapter provides a guide to some of the basic principles which affect energy needs insmall-scale irrigation. SI units — the International Metric System — are used throughout thetext. Reference is made to other units where appropriate, because it is an unfortunate fact oflife that many different systems are in use in irrigation, and sometimes it can be confusing andlead to serious mistakes.

The fundamental units in the SI systems are:

Measurement Unit SymbolLength metre mVolume cubic metre m3

Mass kilogramme kgForce newton N

2.2 PRESSURE

Pressure is a commonly used term, but it does have a special meaning in hydraulics. It describesthe force exerted by water on each square metre of some object submerged in water. It may bethe bottom of a tank, the side of a dam, or a pipeline.

Pressure is normally measured in kilonewtons per square metre (kN/m2). An alternative tothis in irrigation is the ‘bar’, where 1 bar is equal to 100 kN/m2. Pressure is calculated by:

Thus pressure is force per unit area (Figure 2).

Pressure (kN/m2) =force (kN)area (m2)

FIGURE 2Relationship between force and pressure

Some basic concepts6

Pressure measurement

Pressure in pipes can be measured using abourdon gauge (Figure 3). Inside the gauge isa curved tube of oval section, which tries tostraighten out when the system is underpressure. The tube is linked to a pointer whichmoves across a graduated scale and indicatespressure. Irrigators normally measure pressurein the field using these gauges as they are robustand simple to use.

However, engineers often refer to pressureas a head of water in metres (m) rather than apressure in kN/m2. If the bourdon gauge wasreplaced with a long vertical tube, the waterpressure in the pipe would cause water to riseup the tube. The height of this water columnis a measure of the pressure in the pipe. Forexample, a pressure of 3 bar in the pipe wouldresult in water rising to a height of 30 m in thetube. Thus, engineers may refer to the pressureas 3 bar or 30 m head of water.

EXAMPLE 1

Calculate the pressure when a force of 10 kN is applied to an area of 2 m2. - We know that Pressure = force / area, so P = 10 / 2 = 5 kN/m2.If the area is increased to 4 m2, what will be the nre pressure? - P = 10 / 4 = 2.5 kN/m2.Thus the force has remained the same but the pressure is reduced by spreading the force over

A typical operating pressure for a sprinkler system is 3 bar pressure, or 300 kN/m2. Thismeans that every square metre of the inside of the pipes and pump has a uniform force of300 kN acting on it. Other common units of pressure are kilogrammes-force per squarecentimetre (kgf/cm2) and pounds-force per square inch (lbf/in2).

For conversion from one unit to another:

1 bar = 14.7 lbf/in2 = 1 kgf/cm2 = 100 kN/m2

FIGURE 3Measuring pressure in a pipe

In this manual both the terms pressure and head are used to mean the same thing.

Head of water in metres (m) = 0.1 x pressure (kN/m2) = 10 x pressure (bar)

It is simple to change from pressure to head of water:


Atmospheric pressure is important to the understanding of suction when pumping water(Section 3.3.3) and particularly its effects on the efficiency of pumping (Section 3.3.5).

2.3 DISCHARGE

The speed at which water flows in a pipe or channel is called the velocity and is measured inmetres per second (m/s). The discharge is the volume of water flowing along the pipe orchannel each second, and is measured in cubic metres per second (m3/s). To understand this,consider the case of water flowing in a 100 mm diameter pipe at a velocity of 1.5 m/s (Figure 4).In one second the quantity of water moving past some point in the pipe will be equal to theshaded volume shown. This volume is numerically equal to the water velocity multiplied bythe cross-sectional area of the pipe, i.e., 1.5 × 0.008 = 0.012 m3/s.

Importance of Pressure

Pressure is important to the successful operation of both sprinkler and trickle irrigation.Sprinklers must be operated at the right pressure so that the water jet breaks up properly and auniform water application is achieved (Section 3.5.2.). The right pressure is also required intrickle systems so that each emitter gives the same discharge throughout the scheme(Section 3.5.3).

Atmospheric pressure

Atmospheric pressure is the pressure of the atmosphere around us, pressing down on our bodiesand the surface of the earth. Although air seems very light, when there is a large depth, as at theearth’s surface, it creates a pressure of approximately 100 kN/m2. This is equivalent to l bar or10 m head of water.

Atmospheric pressure = 100 kN/m2 = 1 bar = 10 m head of water

In general terms:

Discharge (m3/s) = cross-sectional area of pipe (m2) x velocity of water (m/s)

FIGURE 4Calculating discharge


For most small irrigation systems the unit of discharge (m3/s) is much too large and so litresper second (l/s) is very often used. The conversion is made by multiplying by 1000.

FIGURE 5Measuring discharge

Figure 5-A

Figure 5-C

Figure 5-B

Discharge (l/s) = discharge (m3/s) x 1000


2.4 ENERGY

Energy is another word commonly used in everyday language, but in hydraulics and irrigationit has a very specific meaning: — Energy enables useful work to be done

People and animals require energy to do work. This is obtained by eating food and convertingit into useful energy for work through the muscles of the body.

In irrigation, energy is needed to lift or pump water. Water energy is supplied by a pumpingdevice driven by human or animal power, or a motor using solar, wind or fossil fuel energy.

Energy measurement

Energy is normally measured in units of watt-hours. One watt-hour is a very small amount ofenergy and so engineers tend to use a larger unit, the kilowatt-hour (kWh) instead, where1 kilowatt-hour = 1000 watt-hours.

Here are some examples of energy use which may be familiar to the reader and which willprovide some practical indication of energy use:

• A farmer working in the field uses 0.2 - 0.3 kWh every day.• An electric desk fan uses 0.3 kWh every hour.• An air-conditioner uses 1 kWh every hour.

Notice how a time period (e.g., every hour, every day) is always given when describing theamount of energy needed. The farmer using 0.2 kWh every day, for example, indicates that thisenergy must be supplied from food each day otherwise he or she would not be able to workproperly. In irrigation, energy requirements may be determined on a daily, monthly or seasonalbasis.

Discharge measurement

Discharge in a pipeline can be measured using a flow meter (Figure 5-A). The meter indicatesthe volume of water passing through the pipeline. By noting the time for a given volume ofwater to pass the discharge can be determined using the formula:

Discharge (m3/s) = volume of water (m3) / time (s)

A simple way of measuring discharge from a pipe or sprinkler is to catch the flow in abucket of known volume, measuring how long it takes to fill (Figure 5-C). The discharge iscalculated using the above formula. See Example 2.

Discharge in open channels can be measured using a weir or flume measuring structure(Figure 5-B). If no measuring structure is available, a rough guide can be obtained by estimatingthe velocity of flow using a float; measuring the cross-sectional area of the channel; andmultiplying the velocity and the area together. (See Training Manual 7: Canals)

EXAMPLE 2

A small plastic tube is connected to a sprinkler nozzle to collect water in a bucket. If the bucketholds 5 litres and it takes 15 seconds to fill, calculate the sprinkler discharge.

Discharge = volume / time = 5 / 15 = 0.33 l/s


Energy sources

Energy comes from food, in the case of animals and people, and from fossil fuel, wind andsunshine in the case of engines and motors.

Foods have energy values which our bodies convert into useful energy so that we can douseful work. In the same way fossil fuels, wind and sunshine have energy which can be convertedinto useful energy to pump water.

Table 1 gives some indication of energy values for typical foods, fossil fuels and energysources.

TABLE 1Energy content of fuels and foods

Changing energy

An important aspect of energy is that it can be changed from one form of energy to another.People and animals can convert food into useful energy to drive their muscles (Figure 6). In atypical pumping system powered by a diesel engine, the energy is changed several times beforeit is usefully used by the water. Chemical energy contained within the fuel (diesel oil in thiscase) is burnt in a diesel engine to produce mechanical energy. This is passed to the pump via

Fuel orfood

Energy Indicativeefficiency (1)

Comment

MaizeWoodDieselPetrolWindSolar

1 kWh/kg4 kWh/kg11 kWh/l9 kWh/l

0.01-41 kWh/m2

1 kWh/m2

10%10%20%10%20%5%

As animal and human consumption

Sometimes also expressed as fuel consumption(0.09 l/kWh for diesel and 0.11 l/kWh for petrol)For wind speeds from 2.5 to 40 m/s respectivelyMaximum solar energy at sea level

Note: 1. Approximate efficiency when converted to mechanical power.

FIGURE 6Energy conversion - analogous systems in people (top) and machines (bottom)


a drive shaft, and finally to the water. Thus the discharge, pressure or both can be increased. Apump can be thought of as a device for putting additional energy into a water system.

The system of energy transfer is not perfect and energy losses occur through friction betweenthe moving parts and are usually lost as heat energy (the human body temperature rises whenwork hard; an engine heats as fuel is burnt to provide power). Energy losses can be significantin pumping systems, and so can be costly in terms of fuel use. This concept is discussed furtherin Section 2.6.

Calculating energy requirement

The amount of energy needed to pump water depends on the volume of water to be pumped andthe head required and can be calculated using the formula:

Increasing either the volume of water or the head will directly increase the energy requiredfor pumping.

2.5 POWER

Power is often confused with the term energy. They are related, but they have different meanings.Energy is the capacity to do useful work whereas power is the rate at which the energy is used.

Water energy (kWh) =volume of water (m3) x head (m)

367

EXAMPLE 3

600 m3 of water is pumped each day to a tank 10 m above ground (Figure 7). Calculate theamount of energy reguired to do this.

Water energy (kWh) = (600 x 10) / 367 = 16.3 kWh.This is the energy required each day.

FIGURE 7Illustration of the problem considered in Example 3


Power is the rate of using energy and is commonly measured in kilowatts (kW). Thepower needed to pump water is called water power.

Another commonly used measure of power is horse power (HP). As it is not part of themetric units system it will not be used in this manual. However, if comparison is needed therelationship is 1 kW = 1.36 HP.

An air conditioner may have a power rating of 3 kW. This means that it uses 3 kWh ofenergy every hour. In 24 hours it will consume 72 kWh (3 kW × 24 h) of energy at the rate of3 kW every hour. Thus, power is describing the rate at which the energy is used. The greaterthe energy use rate the greater is the power need (Figure 8).

Another way of calculating power and energy is to use the pump discharge rather thanthe volume of water to be pumped.

In this case the water power required can be calculated by first using the formula:Figure 9 is a graph of this formula and from which water power can be obtained.

Energy can then be calculated from power. It is the amount of power used in a giventime period and so:

Power (kW) =energy (kWh)

time (h)

EXAMPLE 4

In Example 3 it was calculated that the water energy required each day to lift 600 m3 of waterthrough 10 m was 16.3 kWh. Calculate the water power needed to do this.To calculate water power from water energy it is necessary to know the time over which pumpingtakes place.• If pumping continues for 24 hours per day:

Water power (kW) = energy used per day (kWh) / time (h) = 16.3 / 24 = 0.68 kW.• If the pump operates only 12 h/day:

Water power = 16.3 / 12 = 1.35 kW.• If pumping is only 6 h/day:

Water power = 16.3 / 6 = 2.7 kW.Note that the water energy is the same in each case, but that the rate of using the energy - thepower - changes with the time period. More power is needed when less time is available for pumpingthe same volume of water.

FIGURE 8Relationship between rate of energy use and power

Water power (kW) = 9.81 x discharge (m3/s) x head (m)


Example 5 demonstrates this approach and shows that the results are the same whichevermethod is used to calculate power and energy.

2.6 EFFICIENCY

When pumping irrigation water it is not enough just to meet the water power and energyrequirements. Additional energy and power must be provided because losses occur in transferringfuel energy to water energy via the power unit and pump. The losses in the system are causedby friction and water turbulence and are usually expressed as efficiency. This can be expressedboth in terms of energy use and of power use.

FIGURE 9Graph relating flow, static head and power

Water energy (kWh) = water power (kW) x operating time (h)

EXAMPLE 5Referring to Example 4, if 600 m3 of water is pumped 10 m each day, calculate the water powerand energy required, using the pump discharge approach if pumping is for only 6 h/day.Discharge (m3/s) = volume (m3) / time (s) = 600 / (6 x 3600) = 0.028 m3/s.Using the above equations:Water power (kW) = 9.81 x discharge (m3/s) x head (m) = 9.81 x 0.028 x 10 = 2.7 kW.Water energy (kWh) = water power (kW) x operating time (h) = 2.7 x 6 = 16.3 kWh.These answers are the same as those obtained in the previous example, thus demonstratingthat water power and energy can be calculated using either approach.


Energy use efficiency

This provides an overall indication of the way energy is used. It would usually be assessed ona seasonal or annual basis.

Power use efficiency

This provides an assessment of the efficiency with which power is converted from the fuel tothe water, but only at the moment of measurement. The efficiency may vary over time,particularly if there is wear in the engine and pump.

A system with no friction would have an efficiency of 100% and all the energy andpower input would be transferred to the water. However, this is not the case in real life andthere are always friction losses in all the components of the power unit and pump. This isdiscussed more fully in Section 3.3.5.

Sometimes efficiencies can be very low without pump users being aware of the problem.This can result in excessive energy use and high pumping costs. This is an important aspect ofpumping and is discussed more fully in Chapter 5.

For the purposes of this manual, the efficiencies of energy and power use are assumed tobe the same. In practice this may not be the case. A seasonal assessment of energy use efficiencymay not always give the same value as power use efficiency measured only once or twiceduring the season. Note that, in calculations using efficiencies, we always use the decimalform [(efficiency in %) / 100] of the value.

Pumping plant efficiency (%) = (water energy / actual energy) x 100

Pumping plant power efficiency (%) = (water power / power input) x 100


Chapter 3

Choosing a new irrigation system

3.1 INTRODUCTION

Choosing a new irrigation system is about choosing the various components which make up thesystem. In this chapter the main components are listed, and guidance is given in how to choose,for preliminary design purposes, between the various options and component configurationsavailable.

Figure 10 illustrates the process of preliminary design and the decisions to be made.Each part of the process is described in this chapter.

Small-scale pumped irrigation systems are made up of the following components (Figure 11):

• Water source;• Pump and power unit;• Distribution system; and• Method of irrigation.

FIGURE 10Choosing irrigation system components

Choosing a new irrigation system16

FIGURE 11Components of a typical irrigation scheme


— The water source, the distribution system and the method of irrigation determinethe energy demand.

— The pump and power unit provide the energy supply.

Water source

The water source may be a river or lake (surface water) or a shallow well or borehole(groundwater). In some cases, water can be abstracted from rivers by gravity, but in manycases pumping will be needed. In the case of groundwater abstraction, pumping is essential.(See also Training Manual 6: Scheme irrigation water needs and supply.)

The amount of water abstracted and the height through which it must be lifted from theriver or borehole add to the energy demand.

Pump and power unit

The pump may be driven by a power unit such as a diesel or petrol engine, or an electric motor.In some special cases solar or wind power, or even hand or animal power, may be used toprovide the power source for the pump, but they are not so common and are generally limitedto very small irrigated plots. In this manual the primary concern is with the use of pumpsdriven by diesel or petrol engines, as these are usually the main sources of energy supplyavailable to most small-scale farmers.

Distribution system

The distribution system conveys water from the pump to the fields and may consist of pipes oropen channels. Some systems are a combination of both. The choice of distribution system hasa significant effect on the energy demand.

Method of irrigation

The method of irrigation may be surface, sprinkler or trickle irrigation. This may also affectthe choice of distribution system and is also significant in determining the energy demand.Surface irrigation may be supplied by either pipe or open channel systems. Sprinkler andtrickle irrigation systems would normally use piped distribution systems. (See also TrainingManual 5: Irrigation Methods.)

Typical systems

The most common combinations of components for an irrigation system are:

• Pump open channel surface irrigation.• Pump pipe supply surface irrigation.• Pump pipe supply sprinkler or trickle irrigation.

The first system is the most common for small-scale irrigation, although the advantages ofthe second are now being more fully realized. Sprinkle, and especially trickle, irrigation aregrowing in importance in some areas where soils are very sandy and water is scarce, or energycosts are high, or both, but surface irrigation is the dominant method and is likely to remain soin many countries for the foreseeable future.

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3.2 WATER SOURCES

Rivers and lakes

Many small irrigation schemes are located close to natural river channels and lakes and obtainwater by pumping from these sources. They provide a supply which can be seen by the farmerand be judged whether sufficient or not for the seasonal needs of the farm. Usually, the pumpingpressures , and hence energy requirements, needed to use such sources are small because thedifference in elevation between the source water level and the level of the field are usually notlarge.

Shallow groundwater

This is an ideal source of supply for farms located some distance from a river or lake. Usuallythe groundwater table is fed by seepage from a river or lake and may be only a few metersbelow ground level. This source may be less reliable than surface water because except throughpumping experience there is no easy way of assessing whether there is a sufficient reserve ofwater to ensure adequate irrigation. However, the farmer can save the cost of an expensivecanal or pipe system to bring water from a more distant surface supply.

As with surface supplies, the energy costs involved in pumping are relatively low.

Deep groundwater

This may be water which has permeated through the ground from a surface source manykilometres away or water which has been trapped in the ground by impermeable soils for manythousands of years (fossil water).

Pumping deep groundwater which may be 20 - 100 m or more below ground level can beexpensive in terms of energy use, as well as in the cost of drilling the borehole, and requiresspecial, deep borehole, pumping equipment, which may also be expensive to buy.

3.3 PUMPS AND POWER UNITS

A pump is a machine which changes fuel energy into useful water energy and needs a petrol ordiesel engine or an electric motor to drive it. In special circumstances it may also be possible touse wind or solar energy. For surface irrigation the pump lifts water from a river or groundwaterinto a channel or pipe system. For sprinkler and trickle irrigation the pump provides the energyfor the pressure and discharge needed to distribute water in the pipes to the sprinklers andemitters, in addition to the energy needed to lift water from the source.

3.3.1 Pump types

Although there are many types of pumps and water lifting devices, many are unsuited to irrigationuse. The most commonly used types are the axial flow (or propeller) pump, the radial flow (orcentrifugal) pump, and the mixed flow pump. These are looked at in detail below.

Axial flow pump

An axial flow pump consists of a propeller — hence its alternative name — housed inside atube which is located below the water level (Figure 12). The tube acts as the discharge pipe,and the power unit turns the propeller by means of a long shaft running down the middle of the


pipe and this lifts the water up the pipe. This pump is very efficient for lifting large volumes ofwater at low pressure and is ideally suited to lifting water from a river or lake to provide surfaceirrigation water to a farm with open channel distribution. However, these pumps tend to bevery expensive because of the high cost of materials, particularly the drive shaft and bearings tosupport the shafted propeller. For this reason there are no small axial flow pumps manufacturedof a size suitable for the small farm of 1 - 2 ha. They tend only to be used on larger farms andfor communal schemes, where several small farms are irrigated from the same pump. They areparticularly suited to paddy rice schemes because of the large volumes of water usually neededfor this crop.

Radial flow (centrifugal) pump

Centrifugal pumps are the most common type of pump used on small schemes because they aremuch cheaper than axial pumps to buy and maintain. Small pump sets are often readily availablein most developing countries (Figure 13). They are best suited to sprinkler and trickle irrigation,where a higher pressure is needed than for surface irrigation.

To understand how a centrifugal pump works, consider first how centrifugal forces occur.Most readers will at some time have spun a bucket of water around at arm’s length and observedthat no water falls from the bucket even when it is upside down (Figure 13). Water is held inthe bucket by the centrifugal forces created by spinning the bucket. A centrifugal pump makesuse of this idea and can be thought of as many buckets all spinning around together. Thebuckets are replaced by an impeller with blades or vanes which spin at high speed inside thepump casing (Figure 13). Water is drawn into the pump from the source of supply through a

FIGURE 12Typical axial flow pump


FIGURE 13Radial flow (centrifugal) pump


short length of inlet pipe called the suction pipe. As the impeller spins, water is thrown outwardsand is collected by the pump casing and guided towards the outlet. This is called the delivery.

Some pumps have very simple impellers withstraight vanes. These tend to be inefficient becausethey create a lot of turbulence in the flow and henceenergy losses. However they are cheap to make andare used in cases where efficiency is not important.Most irrigation pumps have curved vanes so that thewater enters and leaves the impeller smoothly. Thismeans lower energy losses and higher energy useefficiency. Some impellers have side plates and arecalled closed impellers. When there is debris in thewater open impellers are used to reduce the risk ofblockage.

Centrifugal pumps can be classified into twotypes: volute pumps, and turbine (diffuser) pumps. The main difference between them is thatthe turbine type has diffuser vanes, which provide diverging passages to direct the water flow.

Centrifugal pumps are often described by the diameter of the delivery connection pipe,e.g., ‘a 50 mm pump’. Table 2 is a guide to selecting centrifugal pump sizes for different flowranges.

Mixed flow

This pump is a mixture of the axial flow and the centrifugal pump and has the advantage ofcombining the best features of both pump types (Figure 14). Mixed flow pumps are moreefficient at pumping larger quantities of water than centrifugal pumps and are more efficient atpumping to higher pressures than axial flow pumps.

They can also operate as submersible pumps, i.e., being completely below the sourcewater surface.

FIGURE 14Typical mixed flow pump

Pump size (mm) Discharge (l/s)

255075100125

0 - 55 - 1515 - 2525 - 3535 - 50

TABLE 2A guide to selecting centrifugal pumps


3.3.2 Pump Characteristics

Axial flow, centrifugal and mixed flow pumpsare designed to run at a constant speed andtheir performances are described by the fol-lowing characteristics:

• Head and discharge;• Power requirements; and• Efficiency of operation.

Typical characteristics for operating headand discharge for the three pump types appearas Figures 15-A, 15-B and 15-C. They showhow head, power and efficiency vary as thedischarge changes. For example, when thehead requirement is 120% of the design headvalue, discharge is reduced to 60%, 80% and90% of design discharge for centrifugal, mixedflow and axial flow pumps respectively.

Head and discharge

Pumps can deliver a wide range of discharges depending on the pressure required and the speedat which the pump is operated. However, there is a trade off between head and discharge. Ifmore discharge is needed the head drops, and if less discharge is needed, then the head rises. Adifferent set of curves would be obtained if the pump was running at a different speed. Thefaster it runs the greater the head and the discharge.

FIGURE 15-APump characteristics: discharge - head

FIGURE 15-BPump characteristics: discharge - power

FIGURE 15-CPump characteristics: discharge - efficiency


Power

All pumps need power to rotate the impeller. The amount of power needed depends on thespeed of the pump and the discharge that is required. The faster the pump rotates, the morepower is needed.

For axial flow pumps there is a very large power demand as the pump is starting becausethere is a lot of water and a heavy pump impeller to get moving. Once the pump is under waythe power demand drops to its normal running level.

Centrifugal pumps behave quite differently. The power demand is very low when starting,but as the discharge increases the power also gradually increases.

Mixed flow pumps operate in between these two contrasting conditions and have a moreuniform power demand over the discharge range.

Efficiency

The concept of efficiency was first developed in Section 2.6. It measures how well the mech-anical energy and power from the power unit is converted into water energy and power in thepump. The pump power efficiency is calculated by:

The efficiency generally increases to some maximum value and then falls again over thedischarge range. The maximum efficiency is usually between 30 - 80% and there is only alimited range of discharges and heads over which the pumps operate at maximum efficiency.Outside this range the pump will be less efficient and so more power and energy will be neededto operate the system. Smaller pumps tend to operate at lower efficiencies than larger onesbecause they have more friction to overcome relative to their size.

3.3.3 Pump selection

There are many pumps on the market and the designer must try to select a pump which willprovide the discharge and head needed for the scheme while the pump is operating within itsmaximum efficiency range.

Table 3 indicates the range of good operating conditions for different pump types.TABLE 3Pump selection for small-scale schemes

Note: 1. The ideal pump type, but not usually available for small-scale farming.

A large number of irrigation schemes use surface irrigation and open channel distributionpumping from shallow water supplies. This situation is ideal for axial flow pumps butunfortunately few, if any, pumps are available at a reasonable price for the small discharges

Pump power efficiency (%) = (water power output / actual power output) x 100

Irrigation system Pressure or Head (bar) (m)

Discharge(l/s)

Pump type

Surface irrigation- open channel distribution- pipe distribution- deep tube wellSprinkler systemTrickle system

0.5 51.0 10

>2.0 >202 - 6 2 - 601 - 2 10 - 20

any dischargeany dischargeany dischargeany dischargeany discharge

axial1 or mixedaxial1 or mixed

mixed or centrifugalcentrifugalcentrifugal


required on many farms. The only alternative is to use centrifugal pumps instead and acceptthat they will run at well below their peak efficiency (Figure 16).

For sprinkler and trickle irrigation much higher pressures are needed and so centrifugalpumps are ideally suited to this use and will operate more efficiently.

A typical example of pump selection using the data supplied by a manufacturer would beas follows:

3.3.4 Power units

There are two main types of power unit: internal combustion engines, and electric motors.

Internal combustion engines

Many small irrigation schemes do not have access to electricity and so rely on petrol (sparkignition) engines or diesel (compression ignition) engines to drive the pumps. These engineshave a good weight:power output ratio, and are compact in size and relatively cheap due tomass production techniques.

FIGURE 16Pump selection based on head and discharge parameters

EXAMPLE 6

A centrifugal pump is required for a small sprinkler irrigation system. The discharge required is12 l/s, at a pressure of 2 bar. Using the information supplied by the manufacturer (see Figure 17),determine the pump efficiency.If the same pump was to be used to pump water into an open channel and the pressure neededfor this was only 1 bar, show what effect this would have on the pump discharge and the efficiency.From Figure 17, the efficiency of the pump at a discharge of 12 l/s and pressure of 2 bar (20 m ofhead) is 52%. This is within the high efficiency zone of the pump.If the pressure required was only 1 bar (10 m of head) the discharge would increase to 18 l/s, butat the much reduced efficiency of only 12%.Thus, using an inappropriate pump for the surface irrigation option has a significant effect on theefficiency of pumping.


Diesel engines tend to be heavier and more robust than petrol engines and are moreexpensive to buy. However, they are also more efficient to run and if operated and maintainedproperly they have a longer working life and are more reliable than petrol. In some countriespetrol-driven pumps have needed replacing after only 3 years of operation. Diesel pumpsoperating in similar conditions could be expected to last at least 6 years. However, it must notbe forgotten that engine life is not just measured in years, it is measured in hours of operationand its useful life depends on how well it is operated and serviced. There are cases in developingcountries where diesel pumps have been in continual use for 30 years and more.

A diesel-engined pump can be up to four times as heavy as a petrol-engined pump ofequivalent power, and so if portability is important a petrol pump may be the answer.

Electric motors

Electric motors are very efficient in energy use (75 - 85%) and can be used to drive all sizes andtypes of pumps. The main drawback is the reliance on a power supply which is beyond thecontrol of the farmer, and which in many places is unreliable. Inevitably electrical powersupplies usually fail when they are most needed. Heavy demands occur when crops are needingmost water and so a power failure over several days can have disastrous consequences for acrop. When using trickle irrigation on light sandy soils, serious crop losses may well occurafter only a few days without power.

FIGURE 17Manufacturer’s data for a centrifugal pump


3.3.5 Efficiency

The efficiency of power units and pumps is very variable, and few data are available on actualfield performance of small-scale irrigation pumping installations. The data that are availableindicate that efficiencies are very low, in the range 0.5 to 8%, and that such poor levels are quitecommon.

Many of the common causes of low efficiency can be corrected at little cost once theproblem is identified, but unfortunately it is easy to run an inefficient pumping system withouteven realizing it. Any shortfall in output is simply made up by running the system for longerthan would otherwise be necessary.

Pumping efficiencies are likely to be much higher for sprinkler and trickle systems as thehead needs of these systems are more favourable to the hydraulic characteristics of centrifugalpumps.

Figure 18 shows the main components of a small pumping system and the poor efficienciesthat can commonly occur. The main reasons for inefficiency are listed below. Note that improvedefficiency can be achieved by rectifying the common faults.

• Fuel efficiency 90-100%. Fuel is often spilt or leaks from tanks, or from joints in thefuel pipeline.

• Power unit efficiencySmall petrol engines (1 kW) — 10%.Small diesel engines (1.5 - 2 kW) — 15-35%.Large diesel engines — 30-40%. (Text books normally quote 30-40% for enginesbut these are optimistic. Ageing of engine, poor quality maintenance, excessivepower consumed by cooling fans, injectors, etc., all bring down efficiency.)

Electric motors have much higher efficiencies — 75-85% — but a reliable electricitysupply may be difficult to obtain in many locations.

FIGURE 18Efficiency of components of pumping plant


• Power unit to pump transmission If the engine and pump are direct coupled, thentransmission efficiency nears 100%.

• Pump efficiency A pump running at optimum head and speed has an efficiency of between40% and 80%. Many pumps are not run at optimum head and speed, and so their efficiencycould be much lower. This is particularly true for small pumps where the frictionallosses are a higher proportion of the total power requirement.

The overall efficiency of the pumping system can be found by multiplying together theefficiencies of each component:

Note that in any calculation of this type the decimal equivalent of the percentage is used,i.e., an efficiency of 10% becomes 0.1 in the calculation, 20% becomes 0.2, and so on.

Taking the worst and best possible combinations of all the above efficiencies providessome indication of the most likely range of overall efficiencies:

This implies that the worst likely efficiency is around 3%. Even this seems good whencompared to the actual field measurements of 0.5% referred to earlier in this section!

Although an efficiency of 30% might be expected from a centrifugal pump operating asprinkler or trickle system, it is unlikely to reach this level of efficiency for surface irrigation.The best that can be achieved would be around 10%.

Peak power demandThe water power and overall efficiency of the pumping plant are used to calculate the overallpower demand.

Developing the formula from Section 2.5:

Pumping plant efficiency (%)= fuel efficiency x power unit efficiency x transmission efficiency x pump efficiency x 100

EXAMPLE 7

Worst condition = 0.9 x 0.1 x 0.9 x 0.4 x 100 = 3%Best condition = 1.0 x 0.35 x 1.0 x 0.8 x 100 = 28%

Overall power demand = water power (kW) / pumping plant efficiency

Overall power demand (kW) =9.81 x discharge (m3/s) x head (m)

pumping plant efficiency

EXAMPLE 8

A small diesel-driven pump delivers a discharge of 2 l/s when lifting water 3 m from a river.

Calculate the peak power demand when the overall efficiency of pump and power unit is 10%.

Peak power demand = (9.81 x 0.002 x 3) / 0.1 = 0.59 kW

Note that the discharge of 2 l/s must be divided by 1000 to convert it into m3/s.


Pump suction

An aspect of using centrifugal and mixed flow pumps which is not always fully understood,and which can seriously impair efficiency, is the suction side of the pump.

In cases of shallow groundwater or surface water pumping, the pump is located abovethe water surface and water has to be sucked up a short length of pipe into the pump, as shownin Figure 19. The difference in height between the water surface and the pump is called thesuction lift.

When a pump is operating it draws in water in much the same way as a person suckswater up through a drinking straw. There is a limit to how high water can be lifted in this wayand it depends on atmospheric pressure (Section 2.2). At sea level this is approximately 10 mhead of water. Sucking creates a low pressure in the drinking straw and the outside pressure ofthe atmosphere pushes down on the water surface and forces water up the straw. As atmosphericpressure is the driving force, this puts a practical limit on the height to which water can be liftedin this way.

Ideally it should be possible to lift water 10 m, but because of friction losses in the pipeand pump a practical limit is 7 m. Even at this level many pumps will have difficulty suckingwater. Considerable energy will be needed to suck the water and the pump operator may havedifficulty keeping the pump primed (i.e., keeping the pump and suction pipes full of waterwhen starting the pump). For this reason, pumps should be located so that the suction lift is lessthan 3 m if possible. If the depth to the water is greater than 3 m, then a small shelf can beexcavated and the pump located nearer to the water surface (Figure 19).

Note that these rules only apply when operating in areas close to sea level. Here theatmospheric pressure is approximately 10 m head of water. For schemes operating at higheraltitudes in mountainous regions the atmospheric pressure may be much lower than 10 m andso the suction lift will need to be reduced well below 3 m to ensure proper pump operation.

However, not all pumps suffer from suction lift limitations. Pumps designed to workbelow the water surface — submersible pumps — have no such problems.

An example of the effects of variations in suction lift on pump discharge is given by thecase of a small centrifugal pump, which delivered 6.5 l/s when operating at 3 m suction. Whenthe suction lift was increased to 8 m the discharge dropped to 1.2 l/s — a loss in flow of 5.3 l/

FIGURE 19Suction lift limitations


s, or a loss of 85% of the original discharge! Thus, at the greater suction lift the pump wouldhave to be operated considerably longer to meet water demand, and at such a low flow rate thepump may be well away from its best operating efficiency. This example was cited by Wagnerand Lanoix1 (1969).

3.4 DISTRIBUTION SYSTEMS

The distribution system conveys water from the pump to the fields. This may be by openchannels or through pipes. The choice of distribution system affects both the power and energyrequirements.

3.4.1 Open channels

The most common method of distribution is through open channels, which may be lined orunlined. Channel design affects the energy demand of the system in three ways:

• by determining the energy requirement to lift water from its source into the channels;• by influencing energy losses resulting from friction between the water and the canal; and• by influencing the extent of any additional energy required to pump water which is lost

through seepage, canal breaches and misuse.

Water will only flow downhill in open channels and so the layout of canals should ensurethat the highest point in the canal system is near to the pump and water source. In this waywater will then flow downhill under the force of gravity and out onto the fields. Sufficientpower must be provided in this case to lift water from its source into the channels (Figure 20).The head required is determined by the difference in level between the water source and thewater level in the channel. The water level in the channel at the source must be high enough toensure an adequate flow of water to the field, and must include adequate head to allow effectiveflow from the channel to the field.

Large water losses can easily occur in open channels. This may be due to seepagethrough the bed and sides of a channel. However, open channels, particularly unlined ones, areprone to breaching, whereupon considerable amounts of water can be lost. They are also easily

FIGURE 20Energy demand for open channel distribution

1 Wagner, E.G. & Lanoix, J.N. 1969. Water Supply for Rural Areas and Small Communities. Geneva:World Health Organization.


misused. Channels may be left open, particularly when control gates are not working properly,and water runs to waste. These features of open channels mean that considerable amounts ofwater may be pumped which are wasted, using additional energy and fuel for which there is nobenefit in terms of additional crops.

Of course channels can be lined to reduce seepage, but this requires additional capitalexpenditure. A choice must then be made between the additional cost of lining and the cost ofpumping the water which would be lost through seepage. This involves a comparison betweencapital expenditure and operating costs, which is discussed later in Section 3.8.3.

Lining canals can often seem an attractive way of reducing seepage losses. It can alsoreduce maintenance costs and improve irrigation system distribution efficiency. However, iflinings are to be successful they must be constructed with great care. A concrete lining, forexample, needs to be well vibrated as it is poured so as to be impermeable, and must be placedon channel beds and banks that have been well compacted. If settlement occurs after constructionand the lining cracks, then not only will seepage losses be high but the cost of the specialistrepairs will also be significant.

Water losses in channels for typical irrigation schemes expressed in terms of efficiencyare shown in Table 4.

Channel hydraulics

Most irrigation channels excavated in the natural soil are trapezoidal in shape and slope downhillaway from the water source. Channels usually follow the natural ground slope but if the land issteep, then drop (or fall) structures may be needed to avoid serious erosion problems (Figure21). Channels with longitudinal bed slopes of less than 1:1000 will usually avoid seriouserosion problems, but a minimum slope of greater than 1:5000 is needed to discourage siltationand plant growth problems.

Channels which are lined may be trapezoidal but can also be rectangular or semi-circular.

The main aspect of channel design is choosing the bed width and depth of flow. This canpresent some difficulties because choosing a value for one affects the other. Thus channeldesign is a little more complicated than pipe design because pipes are always circular and soonly one value is chosen — the pipe diameter. The reader must look to other texts for thedetailed design of channels, but as guidelines:

FIGURE 21Channel design: dimensions and drop structures


• Unlined channels are designed so that the velocity is low and the bed and sides are noteroded by the water. For this reason, unlined channels tend to be wide and shallow, spreadingthe flow over a large area to reduce the erosive power of the water.

• Lined channels are expensive to construct. For this reason they tend to be narrow and deepwhich ensures the minimum area of lining for a given channel carrying capacity. The velocityalso tends to be high, but this is not usually a problem as the channel is protected fromerosion by the lining.

3.4.2 Pipelines

Pipelines are often thought to be too expensive for many small irrigation schemes except whensprinkler or trickle irrigation is used, as then the use of pipes is essential. However, expensiveis a relative word and does not convey a specific meaning. It may well be that when the fulloperating advantages of pipes are considered they may be a viable alternative to open channels.

For small-scale surface irrigation schemes, recent research has shown many advantagesfor piped distribution systems:

• Very low distribution losses — even less than lined channels, as it is much easier to closeoff the flow in a pipe than in an open channel (See Table 4 for water losses in pipelinesexpressed as an efficiency).

• Less land area is taken up by buried pipes. Channels can take up 0.5-2% of the commandarea.

• Pipes can often be installed at lower cost than lined canals.

• Pipe systems can provide a more flexible and reliable system of supply.

• Reduced contact with water has potential health benefits.

Pipelines for surface irrigation usually operate at low pressures, typically around 0.5 bar(5 m of head).

Pipelines are essential for the use of sprinkler and trickle irrigation, and they need tooperate at much higher pressures (typically 2 - 6 bar for sprinkler and 1 - 2 bar for tricklesystems) and need to be strong enough to withstand up to twice the working pressure. Thereason for this is that pressure surges which are much greater than the normal working pressurecan occur in pipes, and can cause bursts. It is thus important to install a pipe with the correctpressure rating to avoid the expense of repair or even replacement of a complete system.

Energy is needed in pipe systems not only to pump water from the source to the pipe butalso to overcome the energy losses due to friction as water flows down the pipe (Figure 22). Ifsurface irrigation is used, then water can flow freely from the pipe into the field. If sprinkler ortrickle irrigation is used, then additional energy is needed to ensure the water sprays or dripsproperly.

Predicting head losses in pipes is not an exact science, and it easy to make mistakeswhen calculating them. In addition, losses can increase as the pipe ages and becomes rougherinside through continued use. For these reasons the losses in the distribution system should bekept low at the design stage by choosing pipe diameters that are large enough for friction to notdominate the operation of the system at some later date. As a guideline, energy losses in thepipes should be less than 30% of the total pumping head.


Pipeline hydraulics

Energy is lost when water flows along a pipe. This is due to friction between the flowing waterand the pipe wall. This energy loss means that the pressure near the pump will always begreater than at the far end of the pipe. The change in pressure is called the hydraulic gradient(Figure 23). Additional power and energy must be supplied by the pump to overcome thatfriction so that sufficient water is still delivered to the scheme at the right pressures.

Energy loss in pipelines can be measured as a head loss in metres (m). It depends on thefollowing factors:

• Discharge — small changes in discharge can cause very large changes in head loss.

• Pipe diameter — small changes in pipe diameter can cause very large changes in head loss.

• Pipe length — changes in pipe length cause similar changes in head loss. Increasing a pipelength from 100 m to 200 m will double the head loss.

• Pipe layout — the kinds and numbers of bends and junctions.

FIGURE 22Pipe system and its energy demand energy needed

to pressurize sprinklers


• Pipe material — it determines frictional resistance by its smoothness or otherwise.

Energy loss in pipes can be determined from information supplied by pipe manufacturers.A typical nomograph for PVC pipes is shown in Figure 24. The following examples willdemonstrate effects of discharge, pipe diameter and pipe length on the head loss.

A good guide to selecting the right pipe diameter is to keep the velocity below 1.6 m/s. Thisis good engineering practice. It ensures that head losses are low and it will help to avoid thesurge and water hammer (sudden oscillations in water pressure) problems which can causepipes to burst.

Practical considerations

• Different pipe materials have different friction characteristics. The example used in thistext is PVC. If other pipes are used, then values for friction head losses must be obtainedfrom the supplier.

• The smallest diameter pipe may be the cheapest, but it is not always the best choice. Pressurelosses can be very high and so can the cost of providing the extra energy to overcome thelosses. It may be cheaper in the long term to use a larger pipe size, which may have a highercapital cost but requires less energy in use and so has a much lower operating cost. Thisissue is discussed in detail in Section 3.8.5.

• Think long term when selecting pipes. Will more water be needed in the future? Will thesystem be extended? If so, investment now in a larger pipe size may save high energy costslater when trying to pump an increased discharge down a pipe which is too small. A commonproblem across the world is that farmers install pipelines which are too small. Many regretthe decision later when they see the potential for irrigation and wish to expand their system.

• It is not necessary to use a pipe size which is the same diameter as the pump delivery pipe.For example, a 50 mm diameter pump does not mean the farmer must use a 50 mm diameter

FIGURE 23Hydraulic gradient


pipe. The diameter is selected according to the above guidelines and if it is different fromthe pump diameter then a special section of pipe with a varying diameter (a reducer) issimply used to connect the pump to the pipeline.

• It is important to see what pipe sizes and pumps are available in the local market and todesign around this equipment. This may not always give the most efficient system from anenergy use point of view but it will mean that local support for servicing, maintenance andrepair will be available. Such an advantage may far outweigh any fuel efficiency use issues.

FIGURE 24Nomograph relating pipe diameter, discharge, head loss and velocity


3.4.3 Distribution efficiency

Water is not always distributed efficiently, and losses may occur from channels through seepage,evaporation and mismanagement of the system. In the case of open channels this may involvegates being left open when no one is irrigating, and canal banks breaching through poormaintenance. For pipe systems, there may be leakage from the joints because of poor sealingand, again, valves may not always be closed properly. However, it is likely that pipelines havea potentially higher efficiency than open channels. For design purposes, Table 4 indicatestypical values of distribution efficiency.

EXAMPLE 9

An irrigation scheme uses a 100 mm diameter pipeline, 130 m long, to deliver a discharge of 8 l/s.Determine the head loss.From Figure 24:

When discharge is 8 l/s through a pipe of 100 mm ∅ , head loss is 10 m/km.Therefore, over 130 m [= 0.13 km] head loss will be 10 x 0.13 = 1.3 m.

What will be the increase in head loss is the discharge is increased to 16 l/s?From Figure 24:


The increase in head loss is 4.8 - 1.3 = 3.5 m.Increasing discharge causes a large increase in head loss.

Determine the change in head loss if a pipe of 80 mm ∅ is used to deliver the same discharge [8 l/s]over the same distance [130 m].FromFigure 24:


Difference is 3.8 - 1.3 = 2.5 m, i.e. an increase in head loss.A decrease in pipe diameter causes an increase head loss.

Determine the change in head loss if the 100 mm ∅ pipe is used to deliver the same discharge [8 l/s] over twice the distance [260 m].FromFigure 24:


Difference is 2.6 - 1.3 = 1.3 m, an increase in head loss.An increase in pipe length causes a corresponding increase in head loss.

TABLE 4Indicative values of distribution efficiency (%)

Scheme size (ha)Earth canals Lined canals Pipes

sand loam clay

Large: >2 000 haMedium: 200 - 2 000 haSmall: <200 ha

607080

707585

808590

959595

959595


FIGURE 25Basin, forder and furrow irrigation


3.5 METHODS OF IRRIGATION

There are three methods of irrigation commonly used on small schemes (See also TrainingManual 5: Irrigation Methods):

• Surface irrigation• Sprinkler irrigation• Trickle irrigation

The main objectives of these methods are to:

• Apply an adequate amount of water to meet crop needs• Apply water uniformly across the field• Ensure there are no long-term problems (e.g., soil erosion, salinization).

3.5.1 Surface irrigation

This is the most common method used on small schemes and involves flooding water acrossthe soil surface so that it can infiltrate into the root zone and be used by the crop. Basinirrigation, border irrigation and furrow irrigation are all surface methods (Figure 25). Thechoice of surface method depends on the crop, cultivation practices, soils and topography, andfarmer preferences.

Surface irrigation is a labour-intensive method but generally requires less energy thanother methods because of the low head required for distribution.

Although surface irrigation is considered to be a simple method of irrigation this can bevery misleading. Surface irrigation design and construction is relatively simple and little or noimported specialist materials are needed. However, the proper management of the method isvery complex. The efficient use of irrigation water all depends on the skill of the farmer, whomust decide when to irrigate and how much to apply, and then provide the right discharge intothe field so that water infiltrates adequately and uniformly into the root zone. This is not aneasy task, as the soil and topographic conditions can be very variable and the farmer may nothave the necessary degree of control over the discharge and timing of the application

Potentially, surface irrigation can be very efficientif all the factors involved are under the careful controlof an experienced irrigator. More often however, thewater management skills are lacking and efficiency tendsto be low. As the designer will not know the level offield application efficiency that the farmer will achieveonce the scheme is built, a typical value is used for designpurposes (Table 5). If the actual efficiency is less thanthe typical value once the scheme is operating, then thefarmer will need to operate the system for longer eachday, or to reduce the cropped area to compensate. Thisfall in efficiency will increase the energy demand(Section 5.2).

For additional information on surface irrigation see Kay (1986)1.

TABLE 5Typical field application efficienciesfor irrigation methods

Irrigationmethod

Efficiency (%)

SurfaceSprinklerTrickle

607590

1. Kay, M. 1986. Surface Irrigation: Systems and Practice. Cranfield, UK: Cranfield Press.


3.5.2 Sprinkler irrigation

Sprinkler irrigation involves distributing water in pipes under pressure and spraying it into theair so that it breaks up into small droplets and falls to the ground like natural rainfall. Sprinklersystems are generally more efficient and use less labour than surface irrigation and can beadapted more easily to sandy and erodible soils on undulating ground. There are many types ofsprinkler system available, but the most common is a system using portable pipes (aluminiumor plastic) supplying rotary impact sprinklers (Figure 26).

An individual rotary impact sprinkler produces a circular wetting pattern with pooruniformity. To obtain good uniformity, several sprinklers are always operated close togetherso that the patterns overlap.

Pressure is an important factor in successful sprinkler operation. Typical operatingpressures range from 2 to 6 bar, and so energy requirements can be much greater than forsurface irrigation. If sprinklers are working at the pressure recommended by the manufacturerthen the distribution will be good. If the pressure is above or below this value then the distributionwill be adversely affected. The most common problem is when pressure is too low and thishappens when pump and pipes wear, increasing friction and so reducing pressure.

Typical data for rotary impact sprinklers are shown in Table 6.

It is usually assumed that sprinkler irrigation is more efficient than surface irrigation.Potentially this is the case, but it largely depends on how well the system is operated andmaintained. If pipe seals leak or burst, and if sprinklers are left running for longer than necessary,then wastage is inevitable. For design purposes, a field application efficiency of 75% is generallyused.

Traditional sprinkler irrigation is not so well suited to small farms. Typical spacings forsprinklers are 18 m × 18 m, and so they are not so flexible and adaptable to the multitude ofsmall plots usually found on many farms. An alternative which may be more applicable to

FIGURE 26Sprinkler irrigation


small farms is the use of smaller sprinklers connected to the mainline by flexible hoses (Figure27). This is often called a hose-pull system. These sprinklers have great flexibility in operationand are easily re-located around the farm.

For fuller details of the methods, their design and management the reader should refer tostandard text books and other publications3.

3.5.3 Trickle irrigation

Trickle irrigation involves dripping water onto the soil at very low flow rates (2-20 l/h) from asystem of small diameter plastic pipes fitted with outlets called emitters. Water is applied closeto the plants so that only the part of the soil volume in which the roots develop is wetted.Applications are usually frequent (every 2-3 days) and this can provide a favourable high moisturelevel condition in which the plants can flourish. Many other claims are made about the method,including increased crop yields, greater efficiency of water use, possible use of saline water,reduced labour requirements and its adaptability to poor soils. An important advantage is theease with which nutrients can be applied with the irrigation water. The relative importance ofeach of these attributes will vary depending on the situation.

A typical trickle irrigation system is shown in Figure 28.

TABLE 6Typical sprinkler data

FIGURE 27Hose-pull sprinkler system

1. Two publications for further reading are: FAO/ILRI. [1988]. Irrigation methods. Irrigation WaterManagement Training Manual 5. Kay, M. 1983. Sprinkler Irrigation: Equipment and Practice.London: Batsford.

Nozzlediameter

(mm)

Pressure(bar)

Diameter ofwetted circle

(m)

Flow(m3/h)

Application rate (mm/h) for spacings:

18 x 18 m 18 x 24 m 24 x 24 m

456810

3.03.03.04.04.5

2932354348

1.021.672.444.968.13

3.25.27.515.325.1

..3.85.711.418.9

..

..4.28.614.0


Trickle irrigation is potentially a very efficient method of applying water to crops. Fieldapplication efficiency can be as high as 90%, but like any other method it relies very much onthe skill of the irrigator to achieve this. Field measurements on trickle systems have shownapplication efficiencies as low as 25%. This was the result of poor system management ratherthan design. The farmers had not fully understood the concept of partial wetting of the rootzone and so they wasted a lot of water trying to wet up the entire area.

Because of the potentially higher efficiency and the operating pressure of only 1-2 barthis method can use less energy than sprinkler irrigation and in some cases less than surfaceirrigation.

Trickle irrigation is very adaptable to small-scale irrigation. It can be ideal for smallplots of trees and row crops requiring different amounts of water. Trickle laterals may also bemoved from one crop row to another to reduce the cost of the system.

Many claims are made about trickle irrigation, such as that it saves irrigation water,increases yield, etc., but care should be taken in accepting such claims. Crops need a certainamount of water to grow (Section 3.6.1) and generally they are not aware of where the water iscoming from. If it comes from surface flooding, sprinkling or trickle, it makes little differenceto the plants — they respond to water. The saving in water comes from the efficiency withwhich the water can be applied and it is here that trickle has a distinct advantage. Some yieldincreases have been shown with trickle and this may be due to the favourable soil water conditionsand the nutrients added to the water.

For further detailed information reference should be made to specialist publications1.

3.5.4 Selecting an irrigation methodThe selection of an appropriate irrigation method depends on a wide range of technical, economicand social factors. The main features which would be considered are summarized in Table 7.

FIGURE 28Trickle irrigation

1. Such as: FAO. 1980. Localized irrigation. FAO Irrigation and Drainage Paper, 36.


3.6 SYSTEM CAPACITY

An irrigation scheme must be capable of supplying the water needed to grow the crops. Inother words the supply of water must be equal to the demand. The capacity to supply therequired amount of water is called system capacity.

The demand for irrigation water varies considerably throughout the growing season.For most crops only small amounts are needed in the early stages of crop growth but thendemand rises to a peak as the crop matures. An exception to this is flooded rice when largequantities of water are needed at the beginning of the season to flood the field. This initialwater demand may even exceed the peak water demand when the crop matures.

Design criteria

To design a scheme for such conditions the designer needs to know (Figure 29):

• The maximum discharge required to satisfy the peak water requirements of the scheme, i.e.,the peak scheme water demand. This is the rate at which water must flow to meet the peakdemand. It will determine the size of the pump and the distribution system and the powerneeded for the scheme. The pipes or channels must be large enough to carry this dischargeand the pump and power unit must be powerful enough to deliver the discharge at thepressure required. It is useful at this point to consider the possibility of future extensions tothe scheme. If this is a possibility, then the designer may oversize the current scheme toallow for this.

• The volume of water required over the season, i.e., the seasonal scheme water demand.This is the total amount of water needed over the growing season and the designer must besatisfied that there is enough water available to meet the total water demand for growing thecrops. From this the energy demand for pumping over the growing season can be determined.

System capacity depends on the following:

• Crop water requirements, determined by— crop type,— stage of growth, and— climatic conditions;

• Field application efficiency; and• Distribution efficiency.

TABLE 7Factors affecting selection of irrigation method

Irrigationmethod

Crop suitability Soils Labour1 Energydemand

Surface- basin- border- furrowSprinklerTrickle

AllAll except paddy riceAll except paddy rice and sown or drilled cropsAll except paddy riceRow crops; orchards

Clay, loamClay, loamClay, loamLoam, sandAll soils

0.5 - 1.51.0 - 3.02.0 - 4.01.5 - 3.00.2 - 0.5

LowLowLowHighMedium

Note. 1. Labour requirement in hours per hectare per irrigation.


These factors vary from day to day throughout the growing season and from one season toanother and so this is the reason why the designer is first interested in the maximum dailydemand for water so that a large enough system can be provided. The following explains howthe peak scheme water demand and the seasonal scheme water demand are determined for thepurposes of design.

3.6.1 Crop water requirements

This is the amount of water needed to grow a crop, and can be determined in several ways.There may be local data available from extension services or research stations, and these arecommonly used as a basis for design.

Another approach is to calculate crop water requirements using information about thecrop, the stage of growth and the rate of evapotranspiration of water from the crop. For detailsof this calculation procedure the reader should refer to FAO and other publications1.

FIGURE 29Peak and seasonal scheme water demands

1. Suggested publications are:FAO/ILRI. 1986. Irrigation water needs. Irrigation Water Management Training Manual, 3.

This is a simple, practical guide to assessing water needs for small-scale irrigation.FAO. 1977 (Rev. ed. 1984). Crop water requirements. FAO Irrigation and Drainage Paper, 24.

This is a more comprehensive and scientific approach to the whole aspect of crop waterrequirements.

FAO. 1992. CROPWAT - A computer program for irrigation planning and management. FAOIrrigation and Drainage Paper, 46.


For most design purposes, however,some estimate of crop water requirement canbe made from the information below.Table 8 provides an indication of theseasonal water needs of several importantcrops and the range of growing periods.Crop water requirements are given inmillimetres (mm), which is the normal wayof expressing crop water needs. It is thedepth of water which must be applied over aseason to grow a crop. Just how much isapplied at each irrigation and the intervalbetween successive irrigations is determinedby the ability of the soil to store the waterfor use by the crop.

Irrigation water is normally suppliedin cubic metres (m3) and so crop water re-quirements in mm are usually converted intoa volume of water needed for 1 ha of land. If more (or less) than 1 ha is to be irrigated, then theamount of water required is found by multiplying the rate per hectare by the area in hectares(Figure 30).

If the number of days over which the crop grows — the crop duration — is known, then theaverage crop water requirement per day can be calculated:

Crop water requirement can vary considerably throughout the growing season and the peakrequirement can be at least double the average value calculated above. It is the peak waterdemand which the system must be capable of supplying, and so it is this value which must beused to design the system.

Thus one way of estimating the peak crop water requirement is:

A discharge in m3/d/ha is not a very convenient unit to use for design purposes. A morecommon unit is l/s/ha, calculated by:

If no information is available on the crops, then assume peak crop water requirement tobe assume 1 l/s/ha for vegetables and cereals and 1.5 l/s/ha for paddy rice.

TABLE 8Indicative values for crop water needs andgrowing periods

Crop Crop duration(days)

Need1 (mm)

CerealsVegetablesRice (paddy)

120-14090-12090-120

450-650400-600800-1500

Note: 1. Seasonal crop water requirement

FIGURE 30The concept of water requirements in mm

Seasonal crop water requirement (m3/ha) = Seasonal crop water requirement (mm) x 10

Average crop water requirement (m3/d/ha) = Seasonal crop water requirement (m3/ha) Length of crop duration (days)

Peak crop water requirement (m3/d/ha) = average crop water requirement (m3/d/ha) x 2

Peak crop water requirement (l/s/ha) = peak crop water requirement (m3/d/ha) x 0.012

Peak crop water requirement = 1.0 l/s/ha for vegetables and cereals(as a working assumption) = 1.5 l/s/ha for paddy rice


3.6.2 Peak scheme water demand

The peak scheme water demand is the discharge in litres per second (l/s) required to meet thepeak crop water requirements, plus the losses which occur in field application and the distributionsystem. The overall loss is called irrigation efficiency and can be caslculated by:

Peak scheme water demand can be calculated from:

This discharge in l/s/ha is called the duty. The value assumes that 1 ha of land is beingirrigated and the system will be running 24 hours every day to meet the water demand. Inpractice the irrigated area may be more (or less) than 1 ha, and pumping systems are not normallyrun 24 hours a day, and may only operate during the day or for a few hours each day. To takeaccount of areas more (or less) than 1 ha and for different hours of operation, use the followingequation to calculate peak scheme water demand:

Note that peak scheme water demand is now expressed as a discharge in l/s and not as aduty in l/s/ha. Note that to change l/s to m3/s, divide by 1000.

EXAMPLE 10

Calculate the peak crop water requirement for a vegetable crop in m3/d/ha and in l/s/ha, using theindicative values given in Table 8.From Table 8:

Seasonal crop water requirement for vegetables = 400 mm = 400 x 10 = 4000 m3/ha;Crop duration = 120 days.Average crop water requirement = 4000 / 120 = 33 m3/d/ha.Peak daily crop water requirement = 33 x 2 = 66 m3/d/ha.

Converting this to l/s/ha:- Peak daily crop water requirement = 66 x 0.012 = 0.79 l/s/ha

Irrigation efficiency (%) = field application efficiency x distribution efficiency x 100

Peak water demand (l/s/ha) = Peak crop water requirement (l/s/ha) Irrigation efficiency

EXAMPLE 11

Continuing Example 10, calculate the peak scheme water demand in both l/s and m3/s for an irrigationsystem where the irrigation area is 0.5 ha and pumping will take place for 10 hours each day duringthe peak demand period. Surface irrigation will be used, with unlined canals on a sandy soil.- From the previous example, peak crop water requirement = 0.79 l/s/ha.- From Tables 4 and 5, distribution efficiency = 0.8, and field application efficiency = 0.6.Thus, peak scheme water demand (duty) = - 0.79 / (0.6 x 0.8) = 1.65 l/s/haand peak scheme water demand - (1.65 x 24 x 0.5) / 10 = 1.98 l/s, say 2 l/s [=0.002 m3/s].

Peak scheme water demand = Peak water demand (l/s/ha) x cropped area (ha) x 24 hours of operation (h)


3.6.3 Seasonal scheme water demand

The seasonal water demand of a scheme is the volume of water (in cubic metres (m3)) used overthe growing season, taking into account the water losses in the distribution system and in fieldapplication.

It can be calculated from:

3.7 PEAK POWER AND ENERGY DEMAND

Peak power demand is determined using the equation developed in Section 3.3.5, namely:

where ‘discharge’ is the peak scheme water demand expressed in m3s.

Energy demand can be calculated from peak power demand, using the equation fromSection 2.5:

This formula is useful when calculating the maximum daily energy use based on thepeak scheme water demand. It is not so suitable for calculating seasonal energy use becausethe demand for water may vary considerably from week to week. The demand is low at firstand then builds up to a peak as the crop matures. The pump will still deliver water at the samepower (i.e., same discharge and pressure) and so the hours of operation are varied in order forthe required volume of water to be provided at each stage of crop growth.

Thus, to calculate overall seasonal energy demand the total hours of operation throughoutthe season (or year) would need to be known.

Another way of approaching this is to work from the total amount of water to be pumpedin a season (or year) using the equation developed in Section 2.4.

The volume of water is the seasonal scheme water demand (Section 3.6.3). Allowing for heefficiency of the pumping plant gives the overall energy need as follows:

Seasonal scheme water demand (m3) = Crop water requirement (m3/ha) x cropped area (ha)irrigation efficiency

Continuing Example 12, calculate the seasonal scheme water demand in m3.From a previous calculation, the seasonal crop water requirement = 4000 m3//ha.Seasonal scheme watger demand = (4000 x 0.5) / (0.6 x 0.8) = 4166 m3.

Overall power demand (kW) = 9.81 x discharge (m3/s) x head (m)pumping plant efficiency

Energy demand (kWh) = Peak power demand (kW) x operating time (h)

Water energy =volume of water (m3) x head (m)

367

Seasonal energy demand (kWh) =volume of water (m3) x head (m) 367 x pumping plant efficiency


3.8 COSTS

The selection of an irrigation system cannot be done without considering the cost. The designerwill try to select the least costly system or one which meets the farmer’s requirements at a costthat can be recovered from the sale of the produce from the scheme. In other words it must befinancially worthwhile to irrigate.

The system capacity, the choice of technology and its management and maintenance,determine the overall cost of the scheme. This is not just the cost of constructing the systemand buying pumps and irrigation equipment (capital cost) but also the cost of running thesystem over many years (operating costs).

The idea of cost-effectiveness is an important one. Although the choice of irrigationsystem should involve both capital and operating costs, sometimes the choice is not an easyone. Capital costs are easily identified sums of money which must be paid out when installinga scheme. Operating costs are much less clear and are spread over many years, and so there isa tendency for farmers to choose a scheme based only on a minimum or acceptable capital cost.They may also lack the immediate cash to invest in the more expensive systems which couldsave them money in operating costs in the longer term.

Even choosing capital equipment can create difficulties. Should a farmer, for example,buy a cheap pump which may only last a few years or buy a more robust but more expensivemodel which may provide good service for many years?

Faced with a choice of using pipes or open earth canals, a farmer may opt for canals asthey are cheaper to construct. But they do require regular maintenance; they are prone toseepage problems; and are difficult to manage efficiently. Pipes, on the other hand, may bemore expensive to buy but they will need little maintenance; there should be no loss of water;and they are easier to manage.

EXAMPLE 12Continuing Example 12, a small diesel pump working at an efficiency of 10% delivers 2 l/s [= 0.002m3/s] to irrigate 0.5 ha of vegetables from a shallow well 5 m deep. Maximum daily pumping is for 10hours. Calculate the peak power required to do this, the daily energy use, and the seasonal energydemand over the season.

Peak power demand (kW) = (9.81 x 0.002 x 5) / 0.10 = 0.98 kWEnergy demand per day = 0.98 x 10 = 9.8 kWh.

For seasonal energy demand an assessment of the seasonal scheme water demand must first bemade:

Seasonal crop water requirement = 400 mmSeasonal crop water requirement (m3) = crop water requirement (mm) x 10

= 400 x 10 = 4000 m3/ha.Allowing for field application efficiency (60%) and distribution efficiency (80%):

Seasonal scheme water demand = (4000 x 0.5) / (0.6 x 0.8) = 4 166 m3Overall seasonal energy demand = volume of water (m3) x head (m)

367 x pumping plant efficiency= (4 166 x 5) / (367 x 0.1) = 568 kWh.

From the power of the pump and the seasonal energy requirement it is possible to calculate theaverage number of hours that the pump must operate daily:

Crop duration = 120 daysAverage daily hours of pumping = overall seasonal energy demand (kWh)

crop duration (days) x power (kW)This can be compared to the maximum daily pumping of 10 hours, and indicates that there areconsiderable periods when the hours of pumping will be well below the maximum.


For sprinkler or trickle irrigation, a farmer may prefer to buy small diameter pipes becausethey are cheaper than larger ones. What may not be considered is the increase in power requiredto pump water along smaller pipes because of increased friction, and the resulting, and oftenconsiderable, rise in energy use and hence the energy cost. This additional energy cost over afew years of operation may well be far greater than the cost of installing larger diameter pipes.

We now come to the key question that this manual has been gradually leading up to:

“How can the designer reconcile all these issues and come forward with sensible advicefor the farmer?”

The following method is the one that is normally used to make real cost comparisons betweensystems, but it must be stressed that it may not always provide the right answer. There are somevery practical considerations which may influence the farmer’s final choice of equipment.

One other point to bear in mind is that both capital and operating costs will vary a greatdeal from country to country. They will depend on the cost and availability of materials andlabour, and taxes or duties imposed by countries on manufactured and imported goods. Henceit is not possible to provide a universally applicable list of costings in this manual and fromwhich the designer could choose. Such costings must be prepared locally to suit localcircumstances. However, there are broad guidelines, and they are given.

3.8.1 Capital cost

This is the cost of constructing the irrigationscheme to the point where it is ready for use.It may include pumps; pipes and field equip-ment; construction of open channels; and landpreparation such as bush clearance andlevelling.

Just how long equipment lasts before itneeds replacing obviously depends on the qual-ity of the equipment, how much it is used andhow well it is maintained. Table 9 is aguideline to the useful life of equipment forsmall-scale schemes when it is properly usedand maintained according to the manufact-urer’s recommendations. Clearly if the equipment is badly treated then its useful life will beconsiderably shortened.

3.8.2 Operating cost

There are three main operating costs:• Energy• Maintenance and repair• Labour

These costs are incurred regularly throughout the useful life of the scheme and so a timeperiod needs to be set over which the costs can be assessed. Usually the operation of a schemeis similar from one season or year to the next and so a common approach is to consider costs onthe basis of one cropping season or over a full year as a suitable period.

TABLE 9Useful life of irrigation system components

Item Years

Petrol-engined pumpDiesel-engined pumpElectrically-driven pumpPipelines- on the surface- buriedSprinkler and trickle equipmentOpen channels- unlined- lined

41010

4 - 710 - 205 - 10

510


Energy

This is the cost of providing fuel to operate the irrigation system. In some cases it can be themost important of the operating costs, and needs to be considered most carefully at the designstage (see Chapter 4).

If diesel or petrol is the main fuel used, then the cost per litre can be determined from thelocal market. The scarcity of such fuels must also be considered. If there are occasionalshortages, particularly at peak pumping periods, then the farmer may have to pay a higher pricethan normal.

Electricity, if available, will be costed at each unit of energy consumed, i.e., in kWh.

The energy cost is calculated from the seasonal energy demand (Section 3.7), the fuelconsumption of the engine (Table 1), and the cost of fuel, using the formula:

Maintenance and repair

These are difficult costs to determine and will vary greatly depending on the type of scheme.Consider, for example, a surface irrigation scheme with open earth channels which may requiresubstantial annual maintenance and repair. This may be done by the farmer and his family inbetween irrigation seasons, and so no money is actually paid to outsiders for this work. However,it can be argued that this effort does have a money value. If the farmer or his family are able tofind paid work elsewhere then they will be foregoing this income in order to do the maintenancework. It may be better in money terms to seek paid work, particularly if the farmer has skillswhich are in demand, and to pay others to do the maintenance. It all depends on localcircumstances. In many cases no money is paid out by the farmer and so maintenance is donewith little or no cash being spent outside the family.However, maintenance of pumping equipment andequipment for sprinkler and trickle irrigation may needoutside specialist help and spare parts (often imported),and so cash will be needed for this.

To allow for maintenance and repair costs at thedesign stage, a percentage of the capital cost is usuallyallocated. Table 10 gives some indication of likely cost asa percentage of the capital cost.

Labour

Labour is needed to operate irrigation systems, includingsuch jobs as pump operation and the day-to-day irrigation

Seasonal energy cost ($)= seasonal energy demand (kWh) x fuel consumption (l/kW) x cost of fuel ($/l)

EXAMPLE 14

Continuing Example 13, if the cost of diesel fuel is $ 0.35/l, and the fuel consumption is 0.09 l/kWh,calculate the seasonal energy cost.

Seasonal energy demand is 568 kWh (from Example 13).Overall seasonal energy cost = 568 x 0.09 x 0.35 = $ 18

TABLE 10Indicative maintenance and repaircosts

1 Maintenance and repair cost as apercentage of capital cost

Item Cost1

Petrol-engined pumpDiesel-engined pumpElectrically driven pumpPipelines, sprinkler andtrickle equipmentUnlined channels

10%5%1%

2%10%


of plots. The labour required will vary from system to system. Surface irrigation tends to bemore labour intensive than sprinkler or trickle irrigation. As in the case of scheme maintenance,a farmer and his family may provide the labour and so it is not something to be paid out for incash.

3.8.3 Overall cost

When a suitable irrigation system has been selected and a capital cost determined, the operatingcosts can then be calculated. From this the overall cost can be found, which is the sum of thecapital cost and the operating cost.

The designer may then consider other suitable systems to see what effect they have onthe overall cost of the scheme. From this process the designer, with the farmer, can investigatedifferent ways of irrigating and select the most appropriate system at the right level of overallcost.

Adding capital costs to operating costs to determine the overall cost is not just a matter ofsimple addition.

The capital cost is easily determined and is fixed at the time of purchasing the equipment,but how can the life of the equipment be taken into account? How can a petrol-engined pumpwith a relatively low cost, but lasting only 4 years, be compared with a diesel-engined pumpcosting much more, but lasting 10 years?

Operating costs can also be easily assessed for the coming year because the cost of fueland spare parts will be known, but prices change from year to year, and may be quite differentin 4-5 years time. Also, how many years of operation should be considered when trying tocompare a capital investment now with possible savings in operating costs in the future?

Comparing different costs

There are several ways in which both capital and operating costs can be combined for comparison,but one simple approach is to use the idea of Equivalent Annual Cost (EAC).

In order to use the EAC method, the interest rate (sometimes called the discount rate) onmoney invested locally in the bank must be known, as this affects the overall costs of systems.This rate is usually published by the bank on a regular basis. The reason for this is that thevalue of money changes each year and this needs to be taken into account when making decisionsabout costs over several years. For example, $ 100 invested in the bank now at an interest rateof 8% will have a value of $ 108 in one year’s time. Conversely, if $ 100 needs to be spent inone year’s time then its present value this year would be $ 93 — this being the amount thatwould have to be invested now to produce $ 100 in one year’s time. It is called the PresentValue (PV). For $ 100 to be spent in two years time, the present value would be $ 86.

EAC is a way of adjusting the probable costs of items to the stream of equal amounts ofpayment over a certain period (equivalent annual cost) so that they can properly be comparedwith each other.

EAC Method

The EAC method works in a slightly different way to the idea of Present Value. Rather thanconverting future running costs to present values, it converts initial capital costs to an equivalent

Overall cost = capital cost + operating cost


annual cost over the useful life of the equipment, by multiplying the capital cost by a factor — the Capital Recovery Factor (CRF) — which permits one to calculate the equal installmentsnecessary to repay a loan over a given period at a stated interest rate. To this is added theannual operating cost. This is done for each alternative option or system and the one whichshows the lowest EAC is the cheapest solution. CRF values are based on bank interest ratesand are listed in Table 11 for different interest rates and years of useful life.

Example 15 and the Case Studies of Chapters 4 and 5 show how this method is used inpractice.

Comparing the EAC of both pumps as calculated in Example 15, the diesel-enginedpump would be the cheaper solution, even though it has the higher capital cost. However, this

TABLE 11Capital recovery factors (CRF)

Interestrate %

Years

2 3 4 5 6 7 8 9 10 15 20

5 .538 .367 .282 .231 .197 .173 .155 .141 .130 .096 .0806 .545 .374 .289 .237 .203 .179 .161 .147 .136 .103 .087

7 .553 .381 .295 .244 .210 .186 .167 .153 .142 .110 .0948 .561 .388 .302 .250 ..216 ..192 .174 .160 .149 .117 .1029 .568 .395 .309 .257 .223 .199 .181 .167 .156 .124 .11010 .576 .402 .315 .264 .230 .205 .187 .174 .163 .131 .117

11 .584 .409 .322 .271 .238 .212 .194 .181 .170 .139 .12612 .592 .416 .329 .277 .243 .219 .201 .188 .177 .147 .13413 .599 .424 .336 .284 .250 .226 .208 .195 .184 .155 .14214 .607 .431 .343 .291 .257 .233 .216 .202 .192 .163 .15115 .615 .438 .350 .298 .264 .240 .223 .210 .199 .171 .160

EXAMPLE 15

Two pumps - one petrol driven and the other diesel driven - are being considered for an irrigationscheme. Based on the information given below, determine which is the cheaper pump to buy andoperate.

Diesel-engined pump Petrol-engined pumpCapital cost ($) 2000 500Useful life expectancy (years) 10 4Annual operating costs ($) 100 300Interest rate (%) 6 6

First calculate the EAC for the diesel pump:EAC of capital cost = CRF x capital cost.From Table 11: CRF = 0.136. Therefore EAC = 0.136 x 2000 = $ 272

To find the full EAC the annual operating cost must be added to this:Full EAC = 272 + 100 = $ 372.

Similarly for the petrol pump:EAC of capital cost = CRF x capital cost.From Table 11: CRF = 0.289. Therefore EAC = 0.289 x 500 = $ 145

To find the full EAC the annual operating cost must be added to this:Full EAC = 145 + 300 = $ 445.


is not the complete picture, as changes in both the interest rate and the life of the pump canaffect things quite significantly.

Changing interest rate

Using the figures for the diesel and petrol pumps to calculate the EAC values for differentinterest rates allows us to see how this would affect the choice of pump.

Table 12 is a tabulation of the abovecalculations repeated for different discountrates.

The interest rate clearly has a signif-icant effect on choice of pump. If interestrates are low then the more expensive cap-ital equipment with relatively low operatingcost is favoured. If they are high then itmay be more cost effective to choose a lessdurable pump with a lower capital cost anda relatively high operating cost.

Changing useful life expectancy

Table 13 provides a tabulation of the abovecalculations for different useful lifeexpectancies for the two pumps, based onan interest rate of 6%.

If the life expectancy of the petrolpump is extended due to good care andmaintenance then it becomes a moreattractive option. If the life expectancy ofthe diesel is reduced below 10 years thenthe cost of this option rises and it becomesless attractive.

To summarize:

— At a low interest rate: invest in low capital cost equipment with high operating cost.

— At a high interest rate: invest in high capital cost equipment with low operating cost.

— Extending the useful life of equipment reduces overall costs and may influence equipmentselection

What about other equipment?

The above example was applied only to pump selection, simply to demonstrate the principleand the process of calculation. Clearly there are many other components that together make anirrigation system (pipes, canals, structures, etc.,) and these would all need to be taken intoaccount when comparing the overall costs of different scheme.

The EAC for each scheme option would be determined in total and this would form thebasis for comparison. In Chapters 4 and 5 there are examples of how this process is applied toa scheme as a whole. In the Annex, a simple program for calculating EACs for the variousoptions is provided for those with access to a computer and Lotus 1-2-3 software.

TABLE 12EAC values for pumps at various discount rates

Interestrate (%)

EAC values ($) Choice

Diesel Petrol

61215

372454498

445465475

dieseldieselpetrol

TABLE 13EAC values for pumps for different lifeexpectancies

Petrol pump Diesel pump

Life(years)

EAC(%)

Life(years)

EAC(%)

246

572445401

6810

506422372


3.8.4 Effects of changes

A good designer will always ask “What happens if one or more of the design parameterschanges?” and “How will the change affect the design and the cost?”. If small changes canhave significant effects on cost, then these parameters need to be selected with care and accuracy.

Changing the distribution system

Suppose the distribution system used in Example 13 were changed from unlined open channelsto a pipe system. What effect would this have on the overall seasonal energy demand and theoperating cost? To investigate this, assume that the system is 100 m long and a choice of pipediameters is available.

As the distribution is now a pipe system the efficiency will be improved from 80% to95% (Table 4). This will directly reduce the volume of water to be pumped and hence theenergy demand. However, the head required to deliver the discharge will increase dependingon the energy loss in the pipe.

The cost of each option can be calculated using the above procedures. The followingTable 14 and Figure 31 summarize the results.

The results demonstrate that there are savings in energy to be made by changing from anunlined channel to a pipe system. However, remember that this does not take account of thecapital cost of the pipe and so it is difficult at this stage to say which will be cheaper overall.What is clear, though, is the dramatic rise in energy cost as the pipe diameter is reduced in size.Note the significant difference in energy cost between 50 and 40 mm and between 40 and30 mm diameter pipes. The extra energy cost here over several years of pumping may wellmake it worthwhile to buy the larger pipe.

Changing the irrigation method

The effects on seasonal energy use of changing the irrigation method can also be calculatedusing the procedures described. However, a direct comparison between the methods solely onthe basis of seasonal energy use does not in itself have any significant meaning and must beconsidered in the context of the scheme as a whole. This is done in Chapters 4 and 5, where twocontrasting case studies are examined. The reasons for this are as follows:

• Each irrigation method has a different level of field application efficiency, which meansdifferent volumes of water will be pumped.

• All three methods will require significantly different pressures. Sprinkler irrigation requiresmuch higher pressures than surface irrigation and so may require more energy. Trickle

TABLE 14Changing the distribution system and its effects on energy and cost

Distributionsystem

Seasonal schemewater demand (m3)

Head (m) Seasonal energydemand (kWh)

Seasonal energycost ($)

Unlined channels100 mm dia. pipe75 mm dia. pipe50 mm dia. pipe40 mm dia. pipe30 mm dia. pipe

416633603360336033603360

5.005.065.116.109.5016.80

5684634685588691538

181515182748


irrigation may require less energy than surface irrigation because of the greatly improvedpotential efficiency of water use.

• Pumping plant efficiency may also be different. If a centrifugal pump is used it is bettersuited to the pressure and discharge requirements of a sprinkler or trickle system than asurface system and so it is likely that it will operate at a higher efficiency.

All these factors have a direct influence on energy use. Some increase energy use whilstothers reduce it. The amount of increase or decrease largely depends on the numbers chosen.For example, sprinklers may operate between 2.5 and 4.0 bar pressure and centrifugal pumpefficiency may vary between 10 and 30%. Choosing one set of numbers for surface irrigationand other sets for sprinkler and trickle would not allow any meaningful comparison to be made.The choice between methods can only really be considered on a cost basis in the context of thewhole scheme (Chapters 4 and 5).

3.8.5 Some general conclusions

A detailed analysis involving both capital and operating costs of each option available can becarried out to determine the most cost-effective solution. However, this has been done manytimes by others and from this some conclusions can be drawn which are generally applicable tosmall-scale irrigation.

• Mains electricity, if it is available, can usually provide the cheapest method of pumping forfarms of all sizes.

• Petrol pumps are normally cheaper to buy and operate for small farms of 0 - 2 ha).• Diesel pumps are normally more cost effective for farms larger than 2 ha.

Individual farms in many countries are much less than 2 ha. Diesels can be made costeffective by grouping a number of small farms together to share a common irrigation system.

FIGURE 31Relationship between pipe size and seasonal energy cost


A detailed analysis of the cost of buying and operating pumping systems was made in anFAO (1986) publication, Water Lifting Devices1. The reader is referred to this document forfurther detailed reading.

3.8.6 Some practical considerations

Availability of equipment

In practice the designer may not always have the full range of choice of pumping systems, pipesizes, etc. For example, if only 75 mm petrol pumps are available on the local market there islittle point in specifying a 50 mm pump. The 50 mm pump may be the most cost-effectivechoice on paper, but the availability of pumps and spares may be the deciding factor.

Availability of money

The availability of cash may well determine what the farmer can afford to buy and this may notalways be the most cost effective in the long term.

If the farmer has access to good credit or is receiving a cash grant from an agency tobuild an irrigation system, then the best type of scheme to buy would be one which requires ahigh capital cost and low operating costs.

If, on the other hand, the farmer is buying from his own resources or has to borrowmoney at very high interest rates, then the best scheme might be one with a low capital cost(that the farmer can afford to buy) involving higher operating costs. In this way the farmer willavoid a large debt during construction and will be able to pay the operating costs from revenuefrom the scheme.

Availability of labour

On many small farms the farmer and his family provide the labour for operating the irrigationsystem and no direct cash payment is made for outside help. In this case the cost of labour maynot be included in the operating cost calculations. However, in some cases the farmer andsome members of his family may have employment elsewhere and have to hire labour forirrigation. In this case the cost of the labour would need to be included. This can be a significantcost and may well have an influence on the EAC values and the choice of system, particularlyas some systems are more labour intensive than others.

1 FAO Irrigation and Drainage Paper, 43.


Chapter 4

Case study - 1A small vegetable farm using surface water

or shallow groundwater

INTRODUCTION

In this chapter and the next the principles and processes described in Chapter 3 are broughttogether to show how they can be applied to two typical small-scale irrigation schemes. Thetwo examples that have been chosen are:

• a small farm growing vegetables and using surface water or shallow groundwater, which isconsidered in this chapter, and

• a group scheme of several small farms growing paddy rice and using deep groundwater,which is considered in the next chapter.

Alternative methods of irrigation and distribution will be considered to show how choicescan be made and how this selection is influenced by energy costs.

4.1 OPTIONS AVAILABLE

A farmer visits the local extension office and indicates an interest in irrigating vegetables in thedry season. The farm size is 0.5 ha, the soil is a sandy loam, and at this stage the farmer hasonly a vague idea of the cropping pattern but wishes to get the best returns from the marketplace in terms of expenditure on the scheme. There is a river nearby and the local groundwateris only 2 m below ground level, and this source of water supply is assumed to be adequate.

There are several possible design options. The main ones are:

i. Pump ð open channels ð surface irrigation.

ii. Pump ð pipelines ð surface irrigation.

iii. Pump ð pipelines ð sprinkle irrigation.

iv. Pump ð pipelines ð trickle irrigation.

Several more combinations could be added to these by choosing lined or unlined canals, ordifferent pump motors , diesel, electric or petrol. These are not considered here but would allbe part of the design process.

The design follows the general steps in Figure 1, and the more detailed steps in Figure 10.

Case Study - 156

4.2 SCHEME WATER DEMAND

The first step is to establish how much water is needed for the scheme. Two estimates areneeded:

- Peak crop scheme water demand (l/s) - to determine the size of the system; and

- Seasonal scheme water demand (m3) - to determine the amount of water needed over theseason.

The procedure for calculating both of these values is given in Section 3.6.

Note that when more detailed information becomes available this figure can be refined, but atpresent it is good enough for preliminary design.

Based on the above figures for crop water requirements, the scheme water demand canbe assessed by taking account of the efficiencies of the distribution system and the method ofirrigation. Different demands will result from using different systems and methods. Table 15uses the various values of efficiency taken from Chapter 3 to show the effect of these on thepeak scheme water demand and on the seasonal scheme water demand.TABLE 15Calculating scheme water demand

Notes: Columns [3] and [4] are obtained from Tables 3 and 4 respectively. Columns [5] and [6] are from formulaein Sections 3.6.2 and 3.6.3 respectively, where [5] = ([1] x 0.5 x 24) / ([3] x [4] x 6.0); and [6] = ([2] x 0.5) / ([3] x [4]). In the calculations, efficiency in % is expressed as the decimal equivalent, e.g., 85% is 0.85.

Assumptions!As the farmer is not sure about the crops to be grown and their timing assume a peak crop waterrequirement of 1 l/s/ha.To assess the seasonal crop water requirement assume an average seasonal crop water requirementfor vegetables of 500 mm.Seasonal crop water requirement (m3/ha) = 10 x seasonal crop water requirement (mm)

= 10 x 500 = 5000 m/haAssume average crop duration is 110 days.

Check!To check the peak crop water value of 1 l/s/ha use the seasonal crop water requirement as follows:

Average crop water requirement (m3/d/ha) =

= 5000/110 = 45.5 m3/d/ha.

Peak water requirement = 2 x 45.5 = 91 m3/d/ha.Peak daily crop water requirement = 0.012 x 91 = 1.09 l/s/ha.

Therefore an initial estimate of 1 l/s/ha is acceptable at this stage.

seasonal crop water requirement (m3/ha)crop duration (d)

Designoption

Crop water requirement Efficiency (%) Scheme water demandPeak daily

(l/s/ha)[1]

Seasonal(m3/ha)

[2]

Distribution

[3]

Irrigationmethod

[4]

Peak(l/s)[5]

Seasonal(m3)[6]

i.ii.iii.iv.

1.01.01.01.0

5000500050005000

85959595

6060

75\90

3.93.52.82.3

4900438535102925


Note that from Table 15 it can be seen that a lower peak scheme water demand andseasonal scheme water demand is required for the sprinkle and trickle options than for thesurface irrigation. This may lead to lower energy requirements for pumping the water andlower operating costs, but this may be offset by the higher capital costs of the system.

4.3 PEAK POWER AND ENERGY DEMAND

From the calculation of scheme water demand, an assessment can now be made of power andenergy demand (Section 3.7). The calculations are tabulated in Table 16.

TABLE 16Overall power and energy demands

Notes: Rows [1] and [2] are taken from Table 15, columns [5] and [6] respectively.Rows [5] and [6] are calculated from formulae in Section 3.7, where [5] = (9.81 x [1] x [4]) / [3]; and [6] = ([2]x [4]) / (367 x [3]).Row [7] is calculated from the formula in Section 3.8.2, where [7] = [6] x 0.11 x 0.35.Efficiency in the calculations is expressed as the decimal equivalent, e.g., 10% is 0.1.

Assumptions!- Assume that there will be a maximum of 6 hours of pumping each day.- Assume that the entire area of 0.5 ha is under cropping.

Assumptions!Assume that a small petrol-driven centrifugal pump will be used. This will not be very efficient for thesurface irrigation option, but there is unlikely to be an axial flow pump available. The centrifugal pumpshould operate at a higher efficiency for the sprinkler and trickle irrigation (Figures 16 and 18).Therefore assume efficiencies of - 10% for options (i) and (ii), surface irrigation; and

- 30% for options (iii), sprinkler, and (iv), trickle irrigation.Assume the following factors which make up the total pumping head:

Option Suction lift (m)[1]

Pressure loss indistribution (m)

[2]

Operating applicationpressure (m)

[3]

Total pumpingpressure (m)[1] + [2] + [3]

i.ii.iiiiv.

2222

0162

00

3010

233814

Pressure losses in distribution are assumed to be 1 m for Option (ii) because of the low pressurerequirement, and 20% of operating pressure for options (iii) and (iv).

Fuel: assume that fuel consumption is 0.11 l/kWh, and fuel cost to be $ 0.35/l.

Design option i. ii. iii. iv.

Scheme water demandpeak discharge (m3/s)- volume (m3)Pump- efficiency (%)- head (m)Peak power demand (kW)Energy demand (kWh)Energy cost ($)

[1][2]

[3][4][5][6][7]

0.00394 900

102

0.7626710

0.00354 385

103

1.0335814

0.00283 510

3038

3.48121147

0.00232 925

3014

1.0537214

Case Study - 158


4.4 OVERALL COSTS

Both capital and operating costs are tabulated in Table 17 to calculate the Equivalent AnnualCost (EAC) of each scheme in order to demonstrate which is the cheapest option.

4.5 CONCLUSIONS

A number of conclusions can be drawn from this example:

• Overall, Option (i) is the least-cost solution.• There is not a great difference in cost between the options , $ 125/year between the cheapest

and the most expensive alternatives. This difference is narrowed to only $ 60/year whenlabour costs are taken into account. Labour cost has the largest effect on Option (i).

• Energy costs are not significant and represent only 4 - 13% of the total annual cost. Thedominant cost is the capital equipment.

This example follows the widely accepted method of analysing costs and therefore should leadto the correct answer. But is it reasonable just to accept this without question? The answermust be “No!”. There are two reasons for this:

• Too many assumptions were made in the calculations which, if they were to be changed,might affect the final answer.

• Local practical and financial constraints may over-ride the calculation completely.

Changing assumptions

The answers obtained by calculations need to be tested by asking: “What happens if --------?”

. What happens if the pumping efficiency is lower than expected?

The assumed efficiency of pumping for Option (i) is 10%. In practice this may not be the caseand it could be as low as 3% or 1% (Section 3.3.5). The efficiencies of Options (ii), (iii) and(iv) may be similar to those selected because the pump characteristics are more suited to pipesystems. In this case a comparison of total EAC values would be as follows (See also Figure 32):

- In the example, the pumping efficiency for Option (i) was assumed to be 10%, and this gavea Total EAC of $ 240. If the pumping efficiency is less than this , 3% , Total EAC becomes$ 264. If efficiency is lower still, at only 1%, the Total EAC increases to $ 333.

- If water management practices are poor and the distribution efficiency dropped from 85%to 75%, and field application efficiency from 60% to 50% for Option (i), then the total EACwould further increase to $ 276 with 3% pumping plant efficiency.

Thus, if Option (i) were not operated as designed, which past experience indicates may verywell happen, then the overall cost of Option (i) could rise rapidly. Then Options (ii), (iii) and(iv) become more attractive financially.

. What happens if the interest ratechanges?

The effect on cost is set out in theaccompanying table (See also Figure 33).

A change in interest rate would not affectthe decision as the cost of each option rises

Case Study - 160


as the interest rate increases. This is contrary to the outcome of the example given in Section3.8.3 because the operating costs of Options (ii), (iii) and (iv) are generally higher than Option (i).

. What happens if the groundwater source is much deeper than expected?

The effect on EAC of increasing groundwaterdepth is set out in the accompanying table,where the original example, with a depth of2 m, is compared to three greater depths (Seealso Figure 34).

The costs of the different options are muchcloser together as the depth to the water sourceincreases. This is because the increased depthadds directly to the head and the energy for pumping. An increase in head from 2 to 17 m hasa significant effect on Options (i) and (ii), where pressure requirements are small, and only arelatively small effect on Options (iii) and (iv), which operate at much higher pressures.

Practical and financial constraintsLocal practical and financial constraints may very well over-ride the above calculations

completely.

Case Study - 162

• Equipment available (Section 3.8.6) may be limited so that an ideal choice cannot be made.An example is the choice of a small centrifugal pump for surface irrigation in Option (i). Itis not the ideal pump but it is usually the only type available.

• Availability of spare parts, good maintenance facilities and a regular supply of fuel.• The experience of local farmers and extension officers in irrigation. If people have the

experience, they may be able to take advantage of more technically advanced equipmentsuch as that needed for trickle or sprinkle irrigation. If not, then it is advisable to start withsurface irrigation, as that usually can be easily supported with the resources available.

• The availability of money will influence what the farmer can afford to buy and when (Section3.8.6).

What to choose?

What is clear from this example is that the choice between options is not a simple one. Toomuch can depend on local costs and constraints, and some of these issues may be difficult toresolve. There is no simple way of calculating the answer. All the above factors must bereconciled in the process of choice.

A general comment that might be drawn from the example, though, is that Option (i),which is the most common one chosen, will only be the best option if everything works asplanned. Experience in the field, however, indicates that it is unlikely to operate this waywithout a great deal of effort, because of the complexities involved in water management (Section3.5.1). Few designers set out to design for the very worst operating efficiencies and so Option(i) invariably works out as the most attractive on paper, but may not always be so in practice.

Options (ii), (iii) and (iv), with piped distribution, will be much easier to manage and aremore likely to work as designed because water management skills are built into the design andare not left to the farmer to decide.

4.6 GUIDELINES

The two case studies are treated together in Section 5.6 with regard to guidelines.


Chapter 5

Case study - 2A group scheme for paddy rice irrigation,

using deep groundwater

5.1 OPTIONS AVAILABLE

A group of 15 farmers wish to develop an irrigation scheme for the production of 2 crops ofpaddy rice each year. The total area to be irrigated is 19 ha, and water will be obtained fromgroundwater, abstracted by pumping from a depth of 20 m.

Both diesel- and electrically-driven pumps are available, and so there are four mainoptions:

i. Electric pump ð open channels ð surface irrigation.

ii. Diesel pump ð open channels ð surface irrigation.

iii. Electric pump ð pipelines ð surface irrigation.

iv. Diesel pump ð pipelines ð surface irrigation.

This Case Study follows a similar path to Case Study , 1, and so much of the explanationbetween steps is not repeated.

5.2 SCHEME WATER DEMAND

Scheme water demands for the four options are given in Table 18, based on the followingassumptions:

Assumptions!Peak daily crop water requirement = 1.5 l/s/ha.Assume seasonal water requirement for rice is 1 200 mm.As two crops will be grown each year, annual water requirement = 2 400 mm, butthis is used as the seasonal crop water requirement in the calculation.Seasonal crop water requirement (m3/ha) = 10 x seasonal water requirement (mm)

= 10 x 2 400 = 24 000 m3/ha.Field application efficiency = 60%.Distribution efficiency- for open channels = 85%.- for pipelines = 95%.Hours of operation daily (maximum) = 18 hours.

Case Study - 264


Case Study - 266


5.5 CONCLUSIONS

Three important conclusions can be drawn from this case study.

• Options (i) and (iii), involving electrically-driven pumps, are much cheaper to operatethan Options (ii) and (iv) that use diesel-engined pumps.

This is generally the case, as electric pumps operate much more efficiently than diesels.However, much depends on the price of electricity, and there is also the problem of securityof energy supply. Is the electricity source reliable? If not, then the cost advantages meanvery little and the diesel option may be the best.

• The energy cost is a much more significant part of the overall cost than was the case inCase Study , 1. It ranges here from 73% to 84% of the total cost, depending on the optionchosen. Thus any savings in energy will result in significant cash savings.

• The use of a piped distribution system costs only a little more than the open channel options,and may well improve irrigation water use efficiency through simplifying irrigation watermanagement practices.

Changing assumptions

If some of the assumptions made in the calculations are changed, what happens?

. What happens if the efficiency of the diesel pump falls below 10%, or that of the electricpump below 50%?

Note: 1. Original values from the example.

All the costs increase as efficiency levels drop, but the costs of the diesel options, (ii) and(iv), rise much more rapidly than the electric options, (i) and (iii) (Figure 35).

A fall in efficiency from 50 to 40% increases the cost of the electric options by only 15%,whereas a fall from 10 to 3% increases the cost of the diesel options by 160%!

. What happens if the groundwater source is much deeper than expected?


EAC in example (20 m) ($)EAC if 30 m deep ($)EAC if 40 m deep ($)EAC if 50 m deep ($)

13 27018 14023 01027 890

22 96032 34041 72051 100

14 13018 49022 85027 210

24 89033 28041 67050 060


Pumping efficiency(1) (%)Total EAC ($)

5013 270

1022 960

5014 130

1024 890

Pumping efficiency (%)Total EAC ($)

4015 700

541 720

4016 850

545 870


3019 770

366 730

3021 390

373 840


2027 890

1191 800

2030 480

1213 720

Case Study - 268

The figures given for the various EACs if groundwater is deeper than in the original exampleshow clearly that although all the options rise in cost for both diesel and electric pumping, theuse of a piped distribution (Options (iii) and (iv)) becomes more attractive economically as thedepth to groundwater increases (Figure 36).

Changes in the expected useful life period of the pump or of the distribution system, orchanges in the interest rate, all also affect the overall cost, but do not change the overall resultof the calculation.

Practical and financial constraints

As in Case Study , 1, any local practical or financial constraints may well over-ride the abovecalculations completely. The availability of a reliable supply of electricity would be a greatasset to the farmers, but in many countries such power supply security may not exist.

This is not to say that the supply of diesel fuel is always reliable. This also can be in shortsupply, with difficulties in delivery and on-farm storage. With shortages at critical times cancome cost increases, further increasing the operating costs.

What to choose?

There is a very clear choice between diesel and electric pumping, and the costs involved areclearly very different.

Options (iii) and (iv), based on a pipe distribution system, also appear attractive, with littleincrease in cost at the shallower pumping depths and a cost saving if deeper pumping is needed.This means that, in this example, the reduction in maintenance costs and savings in energy forthe pipe systems outweigh the lower capital cost of the open channel systems.

FIGURE 37Effect of increasing scheme size on capital and operating costs


5.6 GUIDELINESThe following general guidelines emerge from the two case studies chosen as examples.

• When dealing with small farms, energy and other operating costs are not important issues.It is the capital cost of the system and its useful life expectancy that dominate financialdecision making.

• As schemes increase in size, energy costs become more dominant in the decision makingprocess (Figure 37).

• Channel distribution with surface irrigation is usually the cheapest option when planninga new scheme. However, it will only be the best option if it works as planned. Fieldexperience indicates that it too often does not work properly, as surface irrigation is verydifficult to manage efficiently.

• Pipe distribution systems for surface irrigation should be considered as a serious option.On larger schemes it may work out to be as cheap as canal distribution, and will simplifywater management practices.

• Piped distribution, sprinkle and trickle irrigation are more likely to work as plannedbecause they are easier to manage.

• Labour costs can have a significant effect on the choice of irrigation system. Whether toinclude labour costs or not will depend on the availability of any employment opportunitiesfor the farming family away from the farm.

• Operating efficiencies, both of the pumping plant and of water use, can have a significanteffect on energy use and hence on operating cost. The effect is very pronounced at lowlevels of efficiency and when relatively large volumes of water are being pumped.

• If electricity is available and reliable, then this will usually be the cheapest source ofenergy for all sizes of irrigation schemes.

• Although bank interest rate affects the Equivalent Annual Cost of scheme options it willgenerally affect all the options in a similar manner and so is unlikely to influence the finalchoice of system.

• The results of the financial analysis are not the only factor to consider when makingchoices between different systems. Practical physical (availability of equipment, spares,and maintenance facilities) and financial (availability of credit) constraints may over-ridethe analysis.

Case Study - 270


Chapter 6

Improving existing schemes

6.1 INTRODUCTION

There are many thousands of small-scale irrigation schemes already in use and many of thesemay be operating well below their potential, either through poor design or through poor operationand maintenance.

This chapter shows what can be done to improve such schemes. It involves evaluatingscheme performance and comparing it with what can be reasonably expected of the scheme andidentifying aspects which are in need of improvement (Figure 38)

What performance can be reasonably expected from a scheme?

To answer this question, a thorough understanding of Chapter 3 is needed. The design is basedon what can be reasonably expected with good design, equipment selection and irrigationmanagement practice. The actual performance of the scheme can then be measured and comparedwith these criteria. Where performance falls below expectation, then ways of improving

FIGURE 38Evaluating irrigation scheme performance

Improving existing schemes72

performance can be identified. Such improvements can involve either changes in design andequipment, or changes in the way the equipment is used. There are two main areas of potentialinefficiency in a scheme:

• Inefficient use of water, and• Inefficient equipment.

6.2 INEFFICIENT WATER USE

One of the major problems in irrigation is the proper management of the irrigation system.This involves:

• Scheduling irrigation water on the farm.• Managing the irrigation method.• Managing the distribution system.

Scheduling irrigation

The farmer must decide: When to irrigate the crop and how much water to apply?

This must be decided for each crop and implies that the farmer understands about cropwater use; about water storage in the soil around the crop roots; is able to measure how muchwater is needed; and can measure and apply water to meet the demand.

Few peasant farmers have this knowledge and often think that more water means morecrops. This is true up to a point, if it is properly applied, but beyond that point additional watercan have a bad effect on the crop as well as wasting both water and energy.

Irrigation method

Surface irrigation is the most common method of irrigation and can be the most inefficient.This is not because it is a poor technology, it is because it is difficult to manage properly. It isoften chosen for its simplicity but its success relies entirely on the skill of the farmer irrigating,requiring the right discharge into the right shape and size of field that has been properly graded,and so forth. Farmers do not always have these skills, and so the method tends to be inefficient.Table 21 shows some of the common problems of surface irrigation and the effect they canhave on efficiency.

Sprinkler and trickle irrigation are much less common and are considered to be sophisticatedtechnologies. However, what it not understood by many people is that several aspects of watermanagement are built into their design, making them much easier for the farmer to use. Thedesigner has already chosen pipe and sprinkler sizes, trickle emitters and a pump so that they allmatch. The farmer has only to lay out the system as recommended by the designer and to turnthe system on and off to achieve a reasonably efficient irrigation.

Thus the choice of irrigation method in relation to skills available can have an importantbearing on water use efficiency.

Distribution system

If the farmer is working on a scheme sharing the supply with others, additional decisions willbe needed regarding the distribution system, namely “How to get the right amount of waterthrough the distribution system to the right place at the right time?”


For small schemes this is not a problem, but where channels are several kilometres longand serve several farmers, the management of water distribution becomes more difficult andless efficient. In shared systems, farmers at the tail end of a scheme tend to get less water thanthose at the head. This may be a result of mismanagement, the additional losses through seepagein the distribution system, or both. This is a common problem, referred to as the top-endertail-ender problem.

The levels of efficiency referred to in Tables 4 and 5 are for well-managed systems.Efficiencies can be much lower than these values when there are seepage problems, poormaintenance and poor liaison among the farmers using the same system.

6.3 INEFFICIENT EQUIPMENT

This may be the result of poor design, for example the selection of pipes which are too small orthe wrong type of pump or power unit.

However, a lot of inefficiency is the result of poor operation and maintenance. It doesnot seem to be a part of human nature to maintain equipment, but to run it until it stops, and thenfix it! It is thus difficult to instil in people the importance of regular maintenance for theoperating efficiency of equipment.

In Section 3.3.5 the factors which affect pumping efficiency are described. Improvementscan be made by rectifying the common faults. The efficiency that can be expected from adiesel or petrol driven pump is about 28% at best and, at worst, 3%. Efficiency from an electricpump will be much higher, at around 50%, but usually the electricity supply is not reliableenough for irrigation.

Also in Section 3.3.5 are details of how the siting of pumps in relation to the water sourcecan affect efficiency. This is a common problem with surface water and shallow groundwaterpumping. Providing water is coming out of the pump, then everyone assumes it is satisfactorilyworking. Little thought is given to the significant drop in efficiency that occurs when thesuction lift is greater than 3 m.

Remember that however well a scheme is designed, it is only as good as its operator.Technology cannot compensate for a poorly trained operator and a poorly managed scheme.

TABLE 21Efficiency of surface irrigation methods

Irrigation methodBasin Border Furrow

For design purposesWell-managed method

Common problems(1)

Poor land preparationDifferent soil typesFixed irrigation schedulesDischarge too lowNo return flow system

6090

subtract10 - 20 5 - 1010 - 2010 - 15

...

6080

subtract10 - 20 5 - 1010 - 2010 - 15

...

6090

subtract10 - 20 5 - 1010 - 2010 - 1520 - 40

Note: If common problems exist then the figures shown should be subtracted from the well-managedefficiency value to give an indication of the actual efficiency that might actually be achievedwhen the scheme is operating.


6.4 EFFECT OF INEFFICIENCY

Just what effect does poor efficiency haveon performance and cost? Its main effectis on the operating cost of the scheme andin particular on the cost of the additionalenergy or fuel needed, as was demon-strated by the case studies of Chapters 4and 5. The range of typical values ofefficiency in different parts of a schemeare summarized in Figure 39. The overallefficiency of a scheme can be found bymultiplying together the efficiency ofeach component. It is often surprising tosee that when the efficiencies of waterapplication, distribution and pumpingplant are taken into account, the overallefficiency of energy use can be a low as0.4% (i.e., 99.6% of the energy input iswasted!) and at best only 45%.

Figure 40 demonstrates the effect ofoverall efficiency on the operating costof a small scheme.

In general, low efficiencies can resultin significant increases in energy andhence operating costs. Conversely, at lowoverall efficiencies, which includes mostsmall irrigation schemes, significantsavings in operating cost can be made bymaking small improvements.

6.5 EVALUATING A SCHEME

This can be done for an individual farmor for a scheme involving several farms.The approach is the same and involvesan assessment of the efficiency. If theefficiency is below what is expected thena more detailed examination of the farmor scheme can be made to determinewhere water is being wasted.

The simplest and quickest assessmentto make is the efficiency of water use foran entire farm or scheme. This meansmeasuring the amount of water going intoa scheme , water supply , and assessingthe water used by crops , water demand.An overall measure of the farm or schemeefficiency can be made as follows:

FIGURE 39System efficiency value ranges


This value is a combination of the distribution and the water application efficiency. Only if theoverall assessment is not satisfactory will it be necessary to go further to assess the twocomponents separately, using the relationships:

Typical values for efficiencies in good, well managed scheme are given in Tables 4 and 5, anda graph showing how seasonal operating costs are reduced as efficiency improves is shown asFigure 40.

Farm (or scheme) efficiency (%) =

water demand (m3) x 100water supply (m3)

Water supply reaching the farm (fields) (m3) x 100Water supply from the source to the scheme (m3)Water distribution efficiency (%) =

Water required by the crop (m3) x 100water supply reaching the farm (field) (m3)

Field application efficiency (%) =

FIGURE 40The relationship between efficiency and seasonal operating costs


6.6 OBTAINING DATA

One of the main difficulties in making evaluations of this kind is having sufficient informationcollected over a long enough time period to make meaningful assessments.

The approaches described below are ways around this problem because to set up extensivedata collection on most schemes may well place impossible burdens on farmers and localextension staff.

6.6.1 Observing and questioning

A very effective evaluation can be made by just walking through a farm (or scheme) observingwhat is happening and asking questions. This is the way good detectives set about solvingproblems.• Observe the pumping unit in operation and check for problems such as vibration, noise or

smoky exhaust.• Ask the farmers if they think they are getting enough water. Those who share a supply and

are short of water will no doubt complain. There may be enough water supplied foreveryone, but it maybe shared out inequitably. Look out for the top-ender , tail-enderproblem.

• Ask about the schedule used for each crop and the hours of pumping needed to irrigate.• Check the drainage system at the lowest part of the farm to see if water is flowing out. This

may be wastage. If the supply is shared then look to see from which farm the wastage iscoming.

• Walk along the canals to look for signs of seepage or leakage at structures; weeds or debrisobstructing flow; or siltation.

• For underground pipe systems, look for signs of leakage, such as wet patches on the soilsurface above the pipes; or for water still flowing in the pipes when all the outlets aresupposedly closed.

• On paddy rice schemes check the depth of water in the fields to see if it conforms with goodpractice for that stage of the crop’s growth.

• On sprinkler and trickle systems, look to see if the operating pressure is as recommended bythe manufacturer. Also look for leakage from the pipes and for blockages in the sprinklernozzles and trickle emitters.

• Particularly on shared schemes, check to see that gates are not left open when a farmer is notirrigating and that farmers arrive and leave their farm in time with the water being suppliedto them.

6.6.2 Some basic data

If some basic measurements can be made, then a more detailed assessment can be carried out.These may be measurements made on a brief visit to a farm (or scheme) or measurements madeover a much longer period of time. Clearly the longer the period of data collection the moreaccurate the assessment will be.

The evaluator would determine:

• Pump discharge. For a small pump this can be measured by catching the water in a largebucket (Section 2.3). For a larger pump, a weir can be temporarily installed in the channeljust downstream of the pump or a flow meter installed in a pipe system.


• Sprinkler or trickle discharge. For schemes of this type, the discharge can be measured atthe sprinkler or trickle emitter by catching water in a large bucket (Section 2.3).

• Amount of fuel or electricity used in a season. This will help to determine whether theinefficiency is in water distribution or in the pump.

• Number of irrigation days in a season and the average number of hours of pumping eachday. These would be estimates obtained from the farmer in the absence of records. Acheck on this would be to see how much fuel had been used during the season and tocalculate the number of running hours, based on fuel consumption, for comparison withthe farmers’ estimates.

• Seasonal crop water requirements. A rough guide can be obtained from Table 8, but moreuseful and accurate data may be available from local research stations.

The following example shows how both the water use efficiency and the pumping plant can beevaluated.

EXAMPLE 16

A small irrigation scheme is growing vegetable crops in the dry season. Water is being pumped froma river nearby and is distributed in an unlined open channel system, and surface irrigation is used forfield application.

Evaluate the water use and pumping plant efficiencies of the scheme using the following data:Scheme design:

Irrigated area = 2 ha, with full cropping on entire area.Seasonal crop water requirement = 500 mm.Distribution system efficiency = 85%.Field application efficiency = 60%.Irrigation efficiency = 0.85 x 0.60 = 0.51 [=51%].It uses a petrol-driven centrifugal pump, with a plant efficiency of 10% and fuel consumption

of 0.11 l/kWh.

Data collected from site:Pump discharge = 13 l/s.Number of irrigation days = 100 d.Hours of daily pumping (average) = 6 h.Pump suction lift = 4 m.Seasonal cost of fuel = $ 285, at a cost of $ 0.35/l.

What can be expected?If the scheme was performing as designed:

Crop water requirement (m3/ha) = 10 x crop water requirement (mm) = 10 x 500 = 5 000 m3/ha.

Seasonal scheme water demand (m3) = crop water requirement (m3/ha) x cropped area (ha)Irrigation efficiency

= (5 000 x 2) / 0.51 = 19 608 m3.

Seasonal energy demand = volume of water (m3) x head (m)367 x pumping plant efficiency

= (19 608 ´ 4) / (367 ´ 0.1) = 2 137 kWh.Overall energy cost =

overall energy demand (kWh) x fuel consumption (l/kWh) x cost of fuel ($/l)= 2 137 x 0.11 x 0.35 = $ 82.


EXAMPLE 16 (continued)

What actually happens?

Check the amount of water supplied to the scheme:Seasonal scheme water supplied (m3) =

pump discharge (m3/s) x daily operating hours (h) x days of irrigation (d) x 3 600= 0.013 x 6 x 100 x 3 600 = 28 080 m3.

The amount of water delivered is in excess of the requirement of 19 608 m3. Therefore theactual irrigation efficiency, which is (5 000 x 2) / 28 080 = 0.36 [=36%], is much lower thanexpected.

A detailed visual inspection of the scheme can now be made to locate where water is beingwasted.

To investigate how efficient the pump is, first calculate the amount of energy actually expendedin pumping 28 080 m3.

The actual energy supplied = overall energy cost ($) fuel consumption (l/kWh) x cost of fuel ($/l)

= 285 / (0.35 x 0.11) = 7 402 kWh

To calculate the efficiency of the pumping plant:Seasonal energy demand = volume of water (m3) x head (m)

367 x pumping plant efficiency

Thus, pumping plant efficiency = volume of water (m3) x head (m) 367 x seasonal energy demand

= (28 080 x 4) / (367 x 7 402) = 0.04 [=4%]

This figure confirms that the pump is operating below its expected efficiency of 10%.The same calculation can be done by comparing the fuel cost actually spent with the fuel costexpected. If we assume 10% pumping plant efficiency:

Expected seasonal energy requirement = (28 080 x 4) / (367 x 0.1) = 3 060 kWh.Fuel cost expected =3 060 x 0.11 x 0.35 = $ 118.

The expected cost of fuel is much less than the $ 285 actually spent on fuel. This means thatthe pumping plant is also operating at a lower efficiency (calculated as (118 x 0.10) / 285 = 0.04[=4%]) than expected.

This may be due to any , or many , of the reasons discussed in Section 3.3.5. One point toconsider, though, is the suction lift of the pump. This is looking critical at 4 m and may belowering the efficiency. If the pump was located within 3 m of the water surface then the pumpefficiency might improve.


AnnexCost Comparison of Different System

OptionsA Lotus 1-2-31 application

A.1 INTRODUCTIONThis annex is to help those who have access to a computer with Lotus 1-2-3 software, and thathave some knowledge of spread-sheet operation.

The spread-sheet (see next page), when entered as a 1-2-3 program, can provide quickcalculation of total cost for 4 different options, in the same way as discussed in Chapters 4 and5. The calculation used for this spread-sheet example is the case study in Chapter 4 (Table 17).

A.2 PROCEDURE FOR PROGRAM INPUTTo input the spread-sheet, just input all the text in the appropriate cells, except for equations inshaded areas (columns C, D, E and F at rows 20, 21, 27, 32, 33, 46, 47, 50, 51, 52, 53, 55, 57and 61). To input equations, type the formula as described in “( Equation in Lotus )”, which isalso shaded, in appropriate rows in column C. The equation can be copied to D, E and Fcolumns by using the “Copy” command (/ Copy).

A.3 CALCULATIONInput appropriate values into cells where equation or text has not been entered. The calculationis automatically performed.

To test “What happens if ..... ?”, just change the value you want to assess, e.g., interestrate can be changed to 10% or 15% by inserting 10 or 15 instead of 5. The result is calculatedautomatically.

A.4 GRAPHIC PRESENTATION ( STACK-BAR )The calculated result can also be viewed graphically.

When the data input is finished, carry out the following steps, using the indicated Lotus1-2-3 commands:

Step 1: / Graph X (Select the range C, D, E and F-12 as X axis)Step 2: [/ Graph] A (Select the range C, D, E and F-21 as A)Step 3: [/ Graph] B (Select the range C, D, E and F-33 as B)Step 4: [/ Graph] C (Select the range C, D, E and F-47 as C)Step 5: [/ Graph] D (Select the range C, D, E and F-53 as D)Step 6: [/ Graph] Type Stack-BarStep 7: [/ Graph] View

If you want to add the labour component to the graph, do Step 8 and repeat Step 7.Step 8: [/Graph] E (Select the range C, D, E and F-59 as E)

1 Lotus and 1-2-3 are registered trademarks of Lotus Development Corporation.This program is based on Release 3.1.

Documents

S:CHRISSISupplIWMTMcover · The text is substantially the work of Dr Melvyn Kay, of Silsoe College, UK, with additional technical input from N. Hatcho of the Land and Water Development