Energy is required, by definition, to do work; the rate at which it is used is defined as power (see Annex I for detailed definitions of units and their relationships). A specific amount of work can be done quickly using a lot of power, or slowly using less power, but in the end the identical amount of energy is required (ignoring "sideeffects" like efficiencies).
The cost of pumping or lifting water, whether in cash or kind, is closely related to the rate at which power is used (i.e. the energy requirement in a given period). Since there is often confusion on the meaning of the words "power" and "energy", it is worth also mentioning that the energy requirement consists of the product of power and time; for example, a power of say, 5kW expended over a period of say, 6h (hours), represents an energy consumption of 30kWh (kilowatthours). The watt (W), and kilowatt (kW) are the recommended international units of power, but units such as horsepower (hp) and footpounds per second (ft.lb/s) are still in use in some places. The joule (J) is the internationally recommended unit of energy; however it is not well known and is a very small unit, being equivalent to only 1 Ws (wattsecond). For practical purposes it is common to use MJ (megajoules or millions of joules), or in the world outside scientific laboratories, kWh (kilowatthours). lkWh (which is one kilowatt for one hour or about the power of two horses being worked quite hard for one hour) is equal to 3.6MJ. Fuels of various kinds have their potency measured in energy terms; for example petroleum fuels such as kerosene or diesel oil have a gross energy value of about 36MJ/litre, which is almost exactly 10kWh/litre. Engines can only make effective use of a fraction of this energy, but the power of an engine will even so be related to the rate at which fuel (or energy) is consumed.
The hydraulic power required to lift or pump water is a function of both the apparent vertical height lifted and the flow rate at which water is lifted.
In other words, power needs are related pro rata to the head (height water is lifted) and the flow rate. In reality, the actual pumping head imposed on a pump, or "gross working head", will be somewhat greater than the actual vertical distance, or "static head", water has to be raised. Fig. 1 indicates a typical pump installation, and it can be seen that the gross pumping head, (which determines the actual power need), consists of the sum of the friction head, the velocity head and the actual static head (or lift) on both the suction side of the pump (in the case of a pump that sucks water) and on the delivery side.
Fig. 1 Typical pump installation
The friction head consists of a resistance to flow caused by viscosity of the water, turbulence in the pump or pipes, etc. It can be a considerable source of inefficiency in badly implementedwater distribution systems, as it is a function which is highly sensitive to flow rate, and particularly to pipe diameter, etc. This is discussed in more detail in Section 2.1.4.
The velocity head is the apparent resistance to flow caused by accelerating the water from rest to a given velocity through the system; any object or material with mass resists any attempt to change its state of motion so that a force is needed to accelerate it from rest to its travelling velocity. This force is "felt" by the pump or lifting device as extra resistance or head. Obviously, the higher the velocity at which water is propelled through the system, the greater the acceleration required and the greater the velocity head. The velocity head is proportional to the square of the velocity of the water. Therefore, if the water is pumped out of the system as a jet, with high velocity (such as is needed for sprinkler irrigation systems), then the velocity head can represent a sizeable proportion of the power need and hence of the running costs. But in most cases where water emerges from a pipe at low velocity, the velocity head is relatively small.
The general principle that:
applies to any water lifting technique, whether it is a centrifugal pump, or a rope with a bucket on it. The actual power and energy needs are always greater then the hydraulic energy need, because losses inevitably occur when producing and transmitting power or energy due to friction. The smaller the friction losses, the higher the quality of a system. The quality of a system in terms of minimizing losses is defined as its "efficiency":
using energy values in the equation gives the longerterm efficiency, while power values could be used to define the instantaneous efficiency.
A truly frictionless pumping system would in theory be 100% efficient; i.e. all the energy applied to it could reappear as hydraulic energy output. However, in the real world there are always friction losses associated with every mechanical and hydraulic process. Each component of a pumping system has an efficiency (or by implication, an energy loss) associated with it; the system efficiency or total efficiency is the product of multiplying together the efficiencies of all the components. For example, a small electrically driven centrifugal pump consists of an electric motor, (efficiency typically 85%), a mechanical transmission (efficiency if direct drive of say 98%), the pump itself (optimum efficiency say 70%) and the suction and delivery pipe system (say 80% efficient). The overall system efficiency will be the product of all these component efficiencies. In other words, the hydraulic power output, measured as (static head) x (flow) (since pipe losses have been considered as a pipe system efficiency) will in this case be 47%, derived as follows:
The efficiency of a component is generally not constant. There is usually an operating condition under which the efficiency is maximized or the losses are minimized as a fraction of the energy throughput; for example a centrifugal pump always has a certain speed at a given flow rate and head at which its efficiency is a maximum. Similarly, a person or draft animal also has a natural speed of operation at which the losses are minimized and pumping is easiest in relation to output.
Therefore, to obtain a pumping system which has a high overall efficiency depends very much on combining a chain of components, such as a primemover, transmission, pump and pipes, so that at the planned operating flow rate and static head, the components are all operating close to their optimum efficiencies  i.e. they are "well matched". A most important point to consider is that it is common for irrigation systems to perform badly even when all the components considered individually are potentially efficient, simply because one or more of them sometimes are forced to operate well away from their optimum condition for a particular application due to being wrongly matched or sized in relation to the rest of the system.
The complete irrigation system consists not only of a water source and water lifting mechanism and its primemover and energy supply, but then there must also be a water conveyance system to carry the water directly to the field or plots in a controlled manner according to the crop water requirements. There may also be a field distribution system to spread the water efficiently within each field. In some cases there could be a water storage tank to allow finite quantities of water to be supplied by gravity without running the water lifting mechanism. Fig. 2 indicates the key components of any irrigation system, and also shows some examples of common options that fulfil the requirements and which may be used in a variety of combinations.
Fig. 2 Key components of an irrigation system
Most of the irrigation system components influence the hydraulic power requirements. For example, if pipes are used for distribution, even if they transfer water horizontally, pipe friction will create an additional resistance "felt" at the pump, which in effect will require extra power to overcome it. If open channels are used, extra power is still needed because although the water will flow freely by gravity down the channel, the input end of the channel needs to be high enough above the field to provide the necessary slope or hydraulic gradient to cause the water to flow at a sufficient rate. So the outlet from the pump to the channel needs to be slightly higher than the field level, thus requiring an increased static head and therefore an increased power demand.
For the same reason the secondary or field distribution system will also create an additional pumping head, either because of pipe friction, or if sprinklers are used then extra pressure is needed to propel the jets of water. Even open channels or furrows imply extra static head because of the need to allow for water to flow downhill.
The power needed is the product of head and flow, and any losses that cause water to fail to reach the plants also represent a reduction in effective flow from the system. Such losses therefore add to the power demand and represent a further source of inefficiency. Typical water losses are due to leakage from the conveyance system before reaching the field, evaporation and percolation into the soil away from crop roots.
Therefore, in common with the primemover and the water lifting device an entire irrigation system can be subdivided into stages, each of which has a (variable) efficiency and a discrete need for power, either through adding to the actual pumping head or through decreasing the effective flow rate due to losses of water (or both).
Most components have an optimum efficiency. In the case of passive items like pipes or distribution systems this might be redefined as "costeffectiveness" rather than mechanical efficiency. All components need to be chosen so as to be optimized close to the planned operating condition of the system if the most economical and efficient system is to be derived. The concept of "costeffectiveness" is an important one in this connection, since most irrigation systems are a compromise or tradeoff between the conflicting requirements of minimizing the capital cost of the system and minimizing the running costs. This point may be illustrated by a comparison between earth channels and aluminium irrigation pipes as a conveyance; the channels are usually cheap to build but require regular maintenance, offer more resistance to flow and, depending on the soil conditions, are prone to lose water by both percolation and evaporation. The pipe is expensive, but usually needs little or no maintenance and involves little or no loss of water.
Because purchase costs are obvious and running costs (and what causes them) are less clear, there is a tendency for small farmers to err on the side of minimizing capital costs. They also do this as they so often lack finance to invest in a better system. This frequently results in poorer irrigation system efficiencies and reduced returns then may be possible with a more capitalintensive but better optimized system.
The proper design of water conveyance systems is complex, and numerous textbooks deal with this topic in detail. It is therefore only proposed here to provide an outline of the basic principles so far as they are important to the correct choice and selection of water lifting system. Useful references on this subject are [3] and [8].
i.Channels
When water is at rest, the water level will always be horizontal; however, if water flows down an open channel or canal, the water level will slope downwards in the direction of flow. This slope is called the "hydraulic gradient"; the greater the frictional resistance to flow the steeper it will be. Hydraulic gradient is usually measured as the ratio of the vertical drop per given length of channel; eg. lm per 100m is expressed as 1/100 or 0.01. The rate of flow (Q) that will flow down a channel depends on the cross sectional area of flow (A) and the mean velocity (v). The relationship between these factors is:
For example, if the cross sectional area is 0.5m^{2}, and the mean velocity is lm/s, then the rate of flow will be:
Table 2 SUGGESTED MAXIMUM FLOW VELOCITIES, COEFFICIENTS OF ROUGHNESS AND SIDE SLOPES, FOR LINED AND UNLINED DITCHES AND FLUMES
Type of surface 
Maximum flow velocities 
Coefficients of roughness 
Side slopes or shape 

Metres 
Feet 

UNLINED DITCHES 

Sand 
0.30.7 
1.02.5 
0.0300.040 
3:1 
Sandy loam 
0.50.7 
1.72.5 
0.0300.035 
2:1 to 2 1/2:1 
Clay loam 
0.60.9 
2.03.0 
0.030 
11/2:l to 2:1 
Clays 
0.91.5 
3.05.0 
0.0250.030 
1:1 to 11/2:1 
Gravel 
0.91.5 
3.05.0 
0.0300.035 
1:1 to 11:1 
Rock 
1.21.8 
4.06.0 
0.0300.040 
1/4:1 to 1:1 
LINED DITCHES  
Concrete  
Castinplace 
1.52.5 
5.07.5 
0.014 
1:1 to 1 1/2 :1 
Precast 
1.52.0 
5.07.0 
0.0180.022 
11/2:1 
Bricks 
1.21.8 
4.06.0 
0.0180.022 
11/2:1 
Asphalt  
Concrete 
1.21.8 
4.06.0 
0.015 
1:1 to 11/2:1 
Exposed membrane 
0.91.5 
3.05.0 
0.015 
11/2:1 to 1:1 
Buried membrane 
0.71.0 
2.53.5 
0.0250.030 
2:1 
Plastic  
Buried membrane 
0.60.9 
2.03.0 
0.0250.030 
21/2:11 
FLUMES  
Concrete 
1.52.0 
5.07.0 
0.0125 

Metal  
Smooth 
1.52.0 
5.07.0 
0.015 

Corrugated 
1.21.8 
4.06.0 
0.021 

Wood 
0.91.5 
3.05.0 
0.014 
The mean velocity (v) of water in a channel can be determined with reasonable accuracy for typical irrigation channels by the Chezy Formula:
where C is the Chezy coefficient which is dependent on the roughness of the surface of the channel (n), its hydraulic radius (r), (which is the area of crosssection of submerged channel divided by its wetted submerged perimeter), and the hydraulic gradient (s) of the channel (measured in unit fall per unit length of channel).
The Chezy coefficient is found from Manning's Formula:
in this formula; K = 1 if metric units are used, or K = 1.486 if feet are used; r is the previously defined hydraulic radius and n is the Manning's Coefficient of Roughness appropriate to the material used to construct the channel, examples of which are given in Table 2. This table is also of interest in that it indicates the recommended side slopes and maximum flow velocities for a selection of commonly used types of channels, ranging from earth ditches to concrete, metal or wooden flumes. Combining the above equations gives an expression for the quantity of water that will flow down a channel under gravity as follows:
where Q will be in m^{3}/s, if A is in m^{2}, r is in metres, and K is 1.
To obtain a greater flow rate, either the channel needs to be large in cross section (and hence expensive in terms of materials, construction costs and land utilization) or it needs to have a greater slope. Therefore irrigation channel design always introduces the classic problem of determining the best tradeoff between capital cost or first cost (i.e. construction cost) and running cost in terms of the extra energy requirement if flow is obtained by increasing the hydraulic gradient rather than the cross sectional area. The nature of the terrain also comes into consideration, as channels normally need to follow the natural slope of the ground if extensive regrading or supporting structures are to be avoided.
Obviously in reality, the design of a system is complicated by bends, junctions, changes in section, slope or surface, etc. The reader wishing to study this topic in greater detail should refer to an appropriate text book on this subject.
A further point to be considered with channels is the likely loss of water between the point of entry to the channel and the point of discharge caused by seepage through the channel walls and also by evaporation from the open surface. Any such losses need to be made up by extra inputs of water, which in turn require extra pumping power (and energy) in proportion. Seepage losses are of course most significant where the channel is unlined or has fissures which can lose water, while evaporation only becomes a problem for small and medium scale irrigation schemes with channels having a large surface area to depth ratio and low flow rates, particularly under hot and dry conditions; the greatest losses of this kind occur generally within the field distribution system rather than in conveying water to the field. The main factors effecting the seepage rate from a channel or canal are:
This latter point is important [9], as any channel will leak much more when it has been allowed to dry out and then refill. Seepage decreases steadily through the season due to sediment filling the pores and cracks in the soil. Therefore, it is desirable to avoid letting channels dry out completely to reduce water losses when irrigating on a cyclic basis.
Typical conveyance efficiencies for channels range at best from about 90% (or more) with a heavy clay surface or a lined channel in continuous use on small to medium land holdings down to 6080% in the same situation, but with intermittent use of the channel. In less favourable conditions, such as on a sandy or loamy soil, also with intermittent use, the conveyance efficiency may typically be 5060% or less; (i.e. almost half the water entering the channel failing to arrive at the other end).
Methods for calculating conveyance losses have been derived and are discussed in detail in specialist references (such as [9]). For example, an approach used by the Irrigation Department in Egypt [9] uses an empirical formula attributed to Molesworth and Yennidumia:
where S will be the conveyance loss in m^{3}/s per length L
ii.Pipes
A pipe can operate like a channel with a roof on it; i.e. it can be unpressurized, often with water not filling it. The advantage of a pipe, however, is that it need not follow the hydraulic gradient like a channel, since water cannot overflow from it if it dips below the natural level. In other words, although pipes are more expensive than channels in relation to their carrying capacity, they generally do not require accurate levelling and grading and are therefore more cheaply and simply installed. They are of course essential to convey water to a higher level or across uneven terrain. As with a channel, a pipe also is subject to a hydraulic gradient which also necessarily becomes steeper if the flow is increased; in other words a higher head or higher pressure is needed to overcome the increased resistance to a higher flow. This can be clarified by imagining a pipeline with vertical tappings of it (as in Fig. 3). When no flow takes place due to the outlet valve being closed, the water pressure along the pipe will be uniform and the levels in the vertical tappings will correspond to the head of the supply reservoir. If the valve is opened so that water starts to flow, then a hydraulic gradient will be introduced as indicated in the second diagram and the levels in the vertical tappings will relate to the hydraulic gradient, in becoming progressively lower further along the pipe. The same applies if a pump is used to push water along a pipe as in the lowest diagram in the figure. Here the pump needs to overcome a resistance equal to the static head of the reservoir indicated in the two upper diagrams, which is the pipe friction head. In low lift applications, as indicated, the pipe friction head can in some cases be as large or larger then the static head (which in the example is all suction head since the pump is mounted at the same level as the discharge). The power demand, and hence the energy costs will generally be directly related to total head for a given flow rate, so that in the example, friction losses in the pipe could be responsible for about half the energy costs.
Those wishing to undertake scientifically rigorous analysis should consult a specialized hydraulics text book, (eg. [8] or [9]) but an
approximate value of the head loss through a pipe can be gained using the empirical equation [8],[10]:
Fig. 3The concept of an 'hydraulic gradient'
The head loss due to friction is expressed as an "hydraulic gradient", i.e. head per length of pipe (m per m or ft per ft).
Note: use K = 10 with metric units, (L and D in metres and Q in cubic metres per second), and K = 4.3 with L and D in feet and Q in cubic feet per second. Values of C are typically 1.0 for steel, 1.5 for concrete, 0.8 for plastics.
An easy way to estimate pipe friction is to use charts, such as Fig. 4. Reference to this figure indicates that a flow, for example, of 6 litres/second (95 US gall/minute) through a pipe of 80mm (3" nominal bore) diameter results in a loss of head per 100m of pipe of just over 2m., As an alternative method, Fig. 5 gives a nomogram (from reference [10]) for obtaining the head loss, given in this case as m/km, for rigid PVC pipe. These results must be modified, depending on the type of pipe, by multiplying the result obtained from the chart by the roughness coefficient of the pipe relative to the material for which the chart of nomogram was derived; for example, if Fig. 4 is to be used for PVC pipe, the result must be multiplied by the factor 0.8 (as indicated at the foot of the figure) because PVC is smoother than iron and typically therefore imposes only 80% as much friction head.
Account must also be taken of the effects of changes of cross section, bends, valves or junctions, which all tend to create turbulence which in effect raises the effective friction head. Ageing of pipes due to growth of either organic matter or corrosion, or both, also increases the friction head per unit length because it increases the frictional resistance and it also decreases the available cross section of flow. This is a complex subject and various formulae are given in text books to allow this effect to be estimated when calculating head losses in pipes.
The head loss due to friction in a pipeline is approximately related to the mean velocity and hence the flow rate squared; i.e.:
therefore, the total head felt by a pump will be approximately the sum of the static head, the friction head and (if the water emerges from the outlet with significant velocity) the velocity head:
^{ }i.e. (total head) = (static head) + (friction head) + (velocity head)
Fig. 4 Determination of head friction losses in straight pipes
Fig. 5 Head loss nomogram calculated for rigid PVC pipes using Blasius formula
Since the velocity of flow is proportional to the flow rate (Q), the above equation can be rewritten:
where
Fig. 6 illustrates the relationship between the total head and the flow rate for a pumped pipeline, and the pipeline efficiency which can be expressed in energy terms as:
^{ }
Fig. 6 How pipeline and efficiency vary with flow
Certain types of pump are capable of sucking water from a source; i.e. the pump can be located above the water level and will literally pull water up by creating a vacuum in the suction pipe. Drawing water by suction depends on the difference between the atmospheric pressure on the free surface of the water and the reduced pressure in the suction pipe developed by the pump. The greater the difference in pressure, the higher the water will rise in the pipe. However, the maximum pressure difference that can be created is between sea level atmospheric pressure on the free surface and a pure vacuum, which theoretically will cause a difference of level of water of 10.4m (or 34ft). However, before a drop in pressure even approaching a pure vacuum can be produced, the water will start gassing due to release of air held in solution (just like soda water gasses when released from a pressurized container); if the pressure is reduced further, the water can boil at ambient temperature. As soon as this happens, the pump loses its prime and the discharge will cease (due to loss of prime) or at least be severely reduced. In addition, boiling and gassing within the pump (known as cavitation) can cause damage if allowed to continue for any length of time.
The suction lifts that can be achieved in practice are therefore much less than 10.4m. For example, centrifugal pumps, which are prone to cavitation due to the high speed of the water through the impeller, are generally limited to a suction lift of around 4.5m (15ft) even at sea level with a short suction pipe. Reciprocating pumps generally impose lower velocities on the water and can therefore pull a higher suction lift, but again, for practical applications, this should never normally exceed about 6.5m (21ft) even under cool sea level conditions with a short suction pipe.
At higher altitudes, or if the water is warmer than normal, the suction lift will be reduced further. For example, at an altitude of 3 000m (10 000ft) above sea level, due to reduced atmospheric pressure, the practical suction lift will be reduced by about 3m compared with sea level, (and proportionately for intermediate altitudes, so that 1 500m above sea level will reduce suction lift by about 1.5m). Higher water temperatures also cause a reduction in practical suction head; for example, if the water is at say 30°C, (or 86°F) the reduction in suction head compared with water at a more normal 20°C will be about 7%.
Extending the length of the suction pipe also reduces the suction head that is permissible, because pipe friction adds to the suction required; this effect depends on the pipe diameter, but typically a suction pipe of say 80m length will only function satisfactorily on half the above suction head.
Groundwater and river water levels vary, both seasonally and in some cases due to the rate of pumping. Such changes in head can significantly influence the power requirements, and hence the running costs. However, changes in head can also influence the efficiency with which the system works, and thereby can compound any extra running costs caused by a head increase. More serious problems can arise, resulting in total system failure, if for example a surface mounted suction pump is in use, and the supply water level falls sufficiently to make the suction lift exceed the practical suction lift limits discussed in the previous section.
Fig. 7 illustrates various effects on the water level of a well in a confined aquifer. The figure shows that there is a natural ground water level (the water table), which often rises either side of a river or pond since ground water must flow slightly downhill into the open water area. The water table tends to develop a greater slope in impermeable soils (due to higher resistance to flow and greater capillary effects), and is fairly level in porous soil or sand.
If a well is bored to below the water table and water is extracted, the level in the well tends to drop until the inflow of water flowing "downhill" from the surrounding water table balances the rate at which water is being extracted. This forms a "cone of depression" of the water table surrounding the well. The greater the rate of extraction, the greater the drop in level. The actual drop in level in a given well depends on a number of factors, including soil permeability and type, and the wetted surface area of well below the water table (the greater the internal surface of the well the greater the inflow rate that is possible). Extra inflow can be gained either by increasing the well diameter (in the case of a handdug well) or by deepening it (the best possibility being with a borehole).
Fig. 7 Effects of various physical conditions on the elevation of water surfaces in wells
Drawdown usually will increase in proportion to extraction rate. A danger therefore if large and powerful pumps are used on small wells or boreholes is to draw the water down to the pump intake level, at which stage the pump goes on "snore" (to use a commonly used descriptive term). In other words, it draws a mixture of air and water which in many cases causes it to lose its prime and cease to deliver. As with cavitation, a "snoring" pump can soon be damaged. But not only the pump is at risk; excessive extraction rates on boreholes can damage the internal surface below the water table and cause voids to be formed which then leads to eventual collapse of the bore. Even when a fully lined and screened borehole is used, excessive extraction rates can pull a lot of silt and other fine material out with the water and block the screen and the natural voids in the surrounding subsoil, thereby increasing the drawdown further and putting an increasing strain on the lowermost part of the bore. Alternatively, with certain subsoils, the screen slots can be eroded by particles suspended in the water, when the extraction rate is too high, allowing larger particles to enter the bore and eventually the possible collapse of the screen.
Neighbouring wells or boreholes can influence each other if they are close enough for their respective cones of depression to overlap, as indicated in Fig. 7 . Similarly, the level of rivers and lakes will often vary seasonally, particularly in most tropical countries having distinct monsoon type seasons with most rain in just a few months of the year. The water table level will also be influenced by seasonal rainfall, particularly in proximity to rivers or lakes with varying levels, (as indicated in Fig. 7).
Therefore, when using boreholes, the pump intake is best located safely below the lowest likely water level, allowing for seasonal changes and drawdown, but above the screen in order to avoid producing high water velocities at the screen.
When specifying a mechanized pumping system, it is therefore most important to be certain of the minimum and maximum levels if a surface water source is to be used, or when using a well or borehole, the drawdown to be expected at the proposed extraction rate. A pumping test is necessary to determine the drawdown in wells and boreholes; this is normally done by extracting water with a portable enginepump, and measuring the drop in level at various pumping rates after the level has stablized. In many countries, boreholes are normally pumped as a matter of routine to test their drawdown and the information from the pumping test is commonly logged and filed in the official records and can be referred to later by potential users.
The factors that impose a power load on a pump or water lifting device are clearly more complicated than simply multiplying the static head between the water source and the field by the flow rate. The load consists mainly of various resistances to flow which when added together comprise the gross pumping head, but it also is increased by the need to pump extra water to make up for losses between the water source and the crop.
Fig. 8 summarizes these in a general way, so that the advantages and disadvantages of different systems discussed later in this paper can be seen in the context of their general efficiency. The table indicates the various heads and losses which are superimposed until the water reaches the field; actual field losses are discussed in more detail in Section 2.2 which follows.
The system hydraulic efficiency can be defined as the ratio of hydraulic energy to raise the water delivered to the field through the static head, to the hydraulic energy actually needed for the amount of water drawn by the pump:
Where E_{stat} is the hydraulic energy output, and E_{g}_{ross} is hydraulic energy actually applied.
Finally, Fig. 9 indicates the energy flow through typical complete irrigation water lifting and distribution systems and shows the various losses.
Fig. 8 Factors affecting system hydraulic efficiency
Fig. 9 Energy flow through typical irrigation system (showing percentage of original energy flow that is transmitted from each component to the next).
Calculating the power requirement for water lifting is fundamental to determining the type and size of equipment that should be used, so it is worth detailing the principles for calculating it. In general the maximum power required will simply be:
where the mass flow is measured in kg/s of water. 1kg of water is equal to 1 litre in volume, so it is numerically equal to the flow in litres per second; g is the acceleration due to gravity of 9.81m/s^{2} (or 32.2 ft/s^{2}). Therefore, for example, 5 litre/sec through 10m with a system having an overall efficiency of 10% requires:
The daily energy requirement will similarly be:
eg. for 60m^{3}/day lifted through 6m with an average efficiency of 5%
Note: 60m3 = 60 000 litres which in turn has a mass of 60 000kg (= 60 tonne). Also, since 1kWh = 3.6MJ, we can express the above result in kWh simply by dividing by 3.6:
so
It follows from these relationships that a simple formula can be derived for converting an hydraulic energy requirement into kWh, as follows:
If the above calculation relates to a gasoline engine pump irrigation system, as it might with the figures chosen for the example, then we know that as the energy input is 19.6kWh/day and as gasoline typically has an energy content of 32MJ/litre or 8.9kWh/litre, this system will typically require an input of 2.2 litre of gasoline per day.
Fig. 10 illustrates the hydraulic power requirement to lift water at a range of pumping rates appropriate to the small to medium sized landholdings this publication relates to. These figures are the hydraulic output power and need to be divided by the pumping system efficiency to arrive at the input power requirement. For example, if a pump of 50% efficiency is used, then a shaft power of twice the hydraulic power requirement is needed; (pump efficiencies are discussed in more detail in Chapter 3). The small table on Fig. 10 indicates the typical hydraulic power output of various prime movers when working with a 50% efficient water lifting device; i.e. it shows about half the "shaft power" capability. The ranges as indicated are meant to show "typical" applications; obviously there are exceptions.
Fig. 10 Hydraulic power requirements to lift water
Fig. 11 Relationship between power, head and flow
These power curves, which are hyperbolas, make it difficult to show the entire power range of possible interest in connection with landholdings from less than 1ha to 25ha, even though they cover the flow, head and power range of most general interest. Fig. 11 is a loglog graph of head versus flow, which straightens out the power curves and allows easier estimation of the hydraulic power requirement for flows up to 100 l/s and hydraulic powers of up to 16kW.
Fig. 12 is perhaps more generally useful, being a similar loglog graph, but of daily hydraulic energy requirement to deliver different volumes of water through a range of heads of up to 32m. The area of land that can be covered, as an example to 8mm depth, using a given hydraulic energy output over the range of heads is also given.
Fig. 12 Relationship between energy, head and daily output (areas that can be irrigated to a depth of 8mm are shown in parentheses)
Finally, Fig. 13 is a nomograph which allows the entire procedure of calculating power needs for a given irrigation requirement to be reduced to ruling a few lines so as to arrive at an answer. The following example of the procedure is indicated and helps to illustrate the process; starting with the area to be irrigated (in the example 3ha is used), rule a line vertically upwards until it intersects the diagonal. This point of intersection gives the required depth of irrigation; 8mm is used in the example but field and distribution losses are not accounted for in this nomograph, so the irrigation demand used must be the gross and not the nett requirement. Rule horizontally from the point of intersection, across the vertical axis (which indicates the daily water requirement in cubic metres per day  240 in the example) until the line intersects the diagonal relating to the pumping head; 10 m head is used in the example. Dropping a vertical line from the point of intersection gives the hydraulic energy requirement (6.5kWh (hyd)/day). This is converted to a shaft energy requirement by continuing the line downwards to the diagonal which corresponds with the expected pumping efficiency; 50% efficiency is assumed for the example (the actual figure depends on the type of pumping system) and this gives a shaft power requirement of 13kWh/day when a line is ruled horizontally through the shaft power axis. The final decision is the time per day which is to be spent pumping the required quantity of water; 5h is used as the example. Hence, ruling a line vertically from the point of intersection to the average power axis (which coincides with the starting axis), shows that a mean power requirement (shaft power) of about 2.6kW is necessary for the duty chosen in the example. It should be noted that this is mean shaft power; a significantly higher peak power or rated power may be necessary to achieve this mean power for the number of hours necessary.
This nomograph readily allows the reader to explore the implications of varying these parameters; in the example it is perhaps interesting to explore the implications of completing the pumping in say 3h rather than 5h and it is clear that the mean power requirement then goes up to about 4.25kW.
In some cases it may be useful to work backwards around the nomograph to see what a power unit of a certain size is capable of doing in terms of areas and depths of irrigation.
The nomograph has been drawn to cover the range from 0  10ha, which makes it difficult to see clearly what the answers are for very small landholdings of under lha. However, the nomograph also works if you divide the area scale by 10, in which case it is also necessary to divide the answer in terms of power needed by 10. In the example, if we were interested in 0.3ha instead of 3ha, and if the same assumptions are used on depth of irrigation, pumping head, pump efficiency and hours per day for pumping, the result will be 0.26kW (or 260W) instead of 2.6kW as indicated. Obviously the daily water requirement from the top axis will also need to be divided by ten, and in the example will be 24m^{3}/day. Similarly, it is possible to scale the nomograph up by a factor of ten to look at the requirements for 10 to 100 ha in exactly the same way. Note that in most real cases, if the scale is changed, factors like the pump efficiency ought to be changed too. An efficiency of 50% used in the example is a poorish efficiency for a pump large enough to deliver 240m^{3}/day, but it is rather a high efficiency for a pump capable of only one tenth of this daily discharge.
Fig. 13 Nomogram for calculating power needs for a given area, depth of irrigation and head
The quantity of water needed to irrigate a given land area depends on numerous factors, the most important being:

nature of crop

crop growth cycle

climatic conditions

type and condition of soil

topography

conveyance efficiency

field application efficiency

water quality

effectiveness of water management
Few of these factors remain constant, so that the quantity of water required will vary from day to day, and particularly from one season to the next. The selection of a small scale irrigation system needs to take all of the above factors into account.
The crop takes its water from moisture held in the soil in the root zone. The soil therefore effectively acts as a water storage for the plants, and the soil moisture needs replenishing before the moisture level falls to what is known as the "Permanent Wilting Point" where irreversible damage to the crop can occur. The maximum capacity of the soil for water is when the soil is "saturated", although certain crops do not tolerate waterlogged soil and in any case this can be a wasteful use of water. In all cases there is an optimum soil moisture level at which plant growth is maximized (see Fig. 12 ). The art of efficient irrigation is to try to keep the moisture level in the soil as close to the optimum as possible.
References such as [3], [8], [10] and [11] give a more detailed treatment of this subject.
The estimation of irrigation water requirements starts with the water needs of the crop. First the "Reference Crop Evapotranspiration" ET is determined; this is a standardized rate of evapotranspiration (related to a reference crop of tall green grass completely shading the ground and not short of water) which provides a baseline and which depends on climatic factors including pan evaporation data and windspeed. A full description on the determination of ET_{o} is presented in reference [11]. Because ET_{o} depends on climatic factors, it varies from month to month, often by a factor of 2 or more. The evapotranspiration of a particular crop (ET_{crop} ) will of course be different from that of the reference crop, and this is determined from the relationship:
K_{c} is a "crop coefficient" which depends on the type of crop, its stage of growth, the growing season and the prevailing climatic conditions. It can vary typically from around 0.3 during initial growth to around 1.0 (or a bit over 1.0) during the midseason maximum rate of growth period; Fig. 13 shows an example. Therefore the actual value of the crop water requirement, ET_{crop }usually varies considerably through the growing season.
The actual nett irrigation requirement at any time is the crop evapotranspiration demand, minus any contributions from rainfall, groundwater or stored moisture in the soil. Since not all rainfall will reach the plant roots, because a proportion will be lost through runoff, deep percolation and evaporation, the rainfall is factored to arrive at a figure for "effective rainfall". Also, some crops require water for soil preparation, particularly for example, rice, and this need has to be allowed for in addition to the nett irrigation requirement.
To give an idea of what these translate into in terms of actual water requirements an approximate "typical" nett irrigation requirement under tropical conditions with a reasonably efficient irrigation system and good water management is 4 000m^{3}/ha per crop, but under less favourable conditions as much as 13 000m^{3}/ha per crop can be needed. This is equal to 4001300mm of water per crop respectively. Since typical growing cycles are in the range of 100150 days in the tropics, the average daily requirement will therefore be in the 30130m^{3}/ha range (313mm/day). Because the water demand varies through the growing season, the peak requirement can be more than double the average, implying that a nett peak output of 50200m^{3}/ha will generally be required (which gives an indication of the capacity of pumping system needed for a given area of field).
Inadequate applications of irrigation water will not generally kill a crop, but are more likely to result in reduced yield [11]. Conversely, excessive applications of water can also be counterproductive apart from being a waste of water and pumping energy. Accurate application is therefore of importance mainly to maximise crop yields and to get the best efficiency from an irrigation system.
The output from the water lifting device has to be increased to allow for conveyance and field losses; this amount is the gross irrigation requirement. Typical conveyance and field distribution system efficiencies are given in Table 3 [11] [12], from which it can be seen that conveyance efficiencies fall into the range 6590% (depending on the type of system), while "farm ditch efficiency" or field application efficiency will typically be 5590%. Therefore, the overall irrigation system efficiency, after the discharge from the water lifting device, will be the product of these two; typically 3080%. This implies a gross irrigation water requirement at best about 25% greater than the nett requirement for the crop, and at worst 300% or more.
The previous "typical peak nett irrigation" figures of 50200m^{3}/day per hectare imply "peak gross irrigation" requirements of 60600m^{3}/day; a wide variation due to compounding so many variable parameters. Clearly there is often much scope for conservation of pumping energy by improving the water distribution efficiency; investment in a better conveyance and field distribution system will frequently pay back faster than investment in improved pumping capacity and will achieve the same result. Certainly costly pumping systems should generally only be considered in conjunction with efficient conveyance and field distribution techniques. The only real justification for extravagant water losses is where pumping costs are low and water distribution equipment is expensive.
Table 3 A AVERAGE CONVEYANCE EFFICIENCY
Irrigation Method 
Method of Water Delivery 
Irrigated Area (ha) 
Efficiency (%) 
Basin for rice cultivation 
Continuous supply with no substantial change in flow 
— 
90 
Surface irrigation (Basin, 
Rotational supply based on 
3,0005,000 
88 
Rotational supply based on 
< 1,000 
70 

> 10,000 

Rotational supply based on 
< 1,000 
65 

> 10,000 
B AVERAGE FARM DITCH EFFICIENCY
Irrigation Method 
Method of Delivery 
Soil Type and Ditch Condition 
Block Size (ha) 
Efficiency (%) 
Basin for rice 
Continuous 
Unlined: Clay
to heavy clay 
up to 3 
90 
Surface 
Rotation or 
Unlined: Clay to heavy clay 
<20 
80 
Irrigation 
Intermittent 
Lined or piped 
>20 
90 
Rotation or 
Unlined: Silt clay 
<20 
6070 

Intermittent 
Lined or piped 
>20 
80 

Rotation or 
Unlined: Sand, loam 
<20 
55 

Intermittent 
Lined or piped 
>20 
65 
C AVERAGE APPLICATION EFFICIENCY
Irrigation Method 
Method of Delivery 
Soil type 
Depth of Application (mm) 
Efficiency (%) 
Basin 
Continuous 
Clay 
>60 
4050 
Heavy clay 

Furrow 
Intermittent 
Light soil 
>60 
60 
Border 
Intermittent 
Light soil 
>60 
60 
Basin 
Intermittent 
All soil 
>60 
60 
Sprinkler 
Intermittent 
Sand, loam 
<60 
70 
A summary of the procedure so far outlined to arrive at the gross crop irrigation water requirement is given in Fig. 14.
Fig. 14 Rate of crop growth as a function of soil moisture content
In order to specify a water lifting system the following basic information is needed:
Having determined the daily application required by the plants, a further consideration is the "intake rate" as different soil types absorb water at different rates, (see Table 4). Too rapid a rate of application on some soils can cause flooding and possible loss of water through runoff. This constraint determines the maximum flow rate that can usefully be absorbed by the field distribution system. For example, some silty clay soils can only take about 7 l/sec per hectare, but in contrast sandy soils do not impose a serious constraint as they can often usefully absorb over 100 l/s per hectare. Obviously lower rates than the maximum are acceptable, although the application efficiency is likely to be best at a reasonably high rate in most cases, and farmers obviously will prefer not to take longer than necessary to complete the job.
Taking account of the above constraint on flow rate, it is then possible to calculate how many hours per day the field will require irrigating, for example by using the nomogram given in Fig. 15
Table 4 AVERAGE INTAKE RATES OF WATER IN mm/hr FOR DIFFERENT SOILS AND CORRESPONDING STREAM SIZE 1/sec/ha
Soil Texture 
Intake Rate mm/hr 
Stream size q 1/sec/ha 
Sand 
50 (25 to 250) 
140 
Sandy loam 
25 (15 to 75) 
70 
Loam 
12.5 (8 to 20) 
35 
Clay loam 
8 (2.5 to 5) 
7 
Silty clay 
2.5 (0.03 to 5) 
7 
Clay 
5 (1 to 15) 
14 
Fig. 15 Example of a crop coefficient curve for corn planted in midMay at Cairo, Egypt; e.g. initial stage is 8.4mm/day with irrigation frequency of 7 days