Annex II - Irrigation scheduling

1. Introduction
2. Rotational supply with limited water
3. Rotational scheduling under full water supply conditions

1. Introduction

The purpose of this annex is to illustrate rotational supply in two specific situations:

Case A when there is insufficient water to meet crop water requirements
Case B when there is sufficient water to meet crop water requirements.

These two cases have been selected because the suggested approach for situation A is quite innovative and is the result of recent findings in the study of the relationship between water and crop yield. Situation B has been selected to illustrate some of the practical problems associated with rotational distribution when based on technical calculations.

There are many other cases that could be considered in irrigation scheduling but technical literature is reasonably abundant on the topic and their consideration here is not appropriate.

2. Rotational supply with limited water

2.1 Theoretical base
2.2 Determining the rotational supply
2.3 Example
2.4 Related design considerations

The question often arises of how to modify a rotational supply during water shortages so that crop yield reductions can be kept to the minimum.

The most commonly adopted solution is to reduce the quantity given at each irrigation in proportion to the relationship:

or, more commonly, to stretch the interval between irrigations in proportion to the relationship:

However, neither of these practices is fully satisfactory since they are not in accordance with the physiological development of the plant. It is known that water stress at certain periods of crop growth adversely affects the crop yields while at other times the effect is much less significant. Therefore, water savings should be made predominantly during those periods when the plant is less sensitive to water stress and reduced to the minimum during the sensitive periods. The critical periods are different for each crop and have variable lengths of time depending on the climatic characteristics.

2.1 Theoretical base

Much research has been dedicated to the investigation of water-yield relationships during recent years. Although researchers continue to investigate the subject, some of the results already obtained offer a satisfactory level of reliability for planning purposes. FAO Irrigation and Drainage Paper No. 33 (Yield response to water) gives a description of the current methods used to estimate crop yields as a function of the water supply. One of the most practical and simple methods indicates that there is a linear relationship between relative yield decreases (1 - Ya/Ym) and relative evapotranspiration deficit (1 - ETa/ETm) for each characteristic growth period, i.e. establishment (0), vegetative (1), flowering (2), yield formation (3) and ripening (4). This constant is called "yield response factor" (ky) and its value is given by the relationship:

{1}

where:	Ya = actual yield
	Ym = maximum yield
	ky = yield response factor
	ETa = actual evapotranspiration
	ETm = maximum evapotranspiration

The values of ky for each growing period and many crops have been calculated from experimental data, and are given in Table 12. Once these values are known formula {1} can be utilized to predict yields for given deficits of evapotranspiration (which reflect the water deficits). For details of how to calculate ETa and ETm the reader is referred to the cited Irrigation and Drainage Paper.

The application of formula {1} to two or more growing periods:

Growing period 1:

Growing period 2:

gives the following relationship:

{2}

and assuming that the tolerable yield decreases (Ym₁ - Ya₂) and (Ym₂ - Ya₂) are equally distributed among the main periods, relationship {2} can be written as:

or

{3}

Since the evapotranspiration in each period (Etm₁ Etm₂) is proportional to the water supply in those periods (V₁ V₂), formula {3} can be written as follows:

where:	V1 and V2 = water applied when full supply available for growing periods
	Va1 and Va2 = water applied with limited supply for growing periods 1 and
	V1-Va1 = water deficit in growing period 1 = WS1
	V2-Va2 = water deficit in growing period 2 = WS2.
and	SW1 = water saved in period 1
	SW2 = water saved in period 2

The relationship can be finally written as!

Table 12 YIELD RESPONSE FACTOR (ky)

Crop		Vegetative period (1)			Flowering period	Yield formation	Ripening	Total growing period
		early (1a)	late (1b)	total	(2)	(3)	(4)
Alfalfa				0.7-1.1				0.7-1.1
Banana								1.2-1.35
Bean				0.2	1.1	0.75	0.2	1.15
Cabbage		0.2				0.45	0.6	0.95
Citrus								0.8-1.1
Cotton				0.2	0.5		0.25	0.85
Grape								0.85
Groundnut				0.2	0.8	0.6	0.2	0.7
Maize				0.4	1.5	0.5	0.2	1.25
Onion				0.45		0.8	0.3	1.1
Pea		0.2			0.9	0.7	0.2	1.15
Pepper								1.1
Potato		0.45	0.8			0.7	0.2	1.1
Safflower			0.3		0.55	0.6		0.8
Sorghum				0.2	0.55	0.45	0.2	0.9
Soybean				0.2	0.8	1.0		0.85
Sugarbeet
	beet							0.6-1.0
	sugar							0.7-1.1
Sugarcane				0.75		0.5	0.1	1.2
Sunflower		0.25	0.5		1.0	0.8		0.95
Tobacco		0.2	1.0			0.5		0.9
Tomato				0.4	1.1	0.8	0.4	1.05
Water melon		0.45	0.7		0.8	0.8	0.3	1.1
Wheat
winter				0.2	0.6	0.5		1.0
spring				0.2	0.65	0.55		1.15

2.2 Determining the rotational supply

The rotational supply which minimizes reductions in crop yield can be determined by applying formula {4}. The main steps in the calculation procedure are as follows:

a. Basic data
b. Critical periods for water deficits
c. Amounts of water to be saved
d. Available water per month
e. Case a: Determination of the extended irrigation interval
f. Case b: Determination of the depth of irrigation

a. Basic data needed are:	Net irrigation requirements
	Available annual water supply
	Cropping pattern
	Irrigation interval under full supply (it can also be calculated if soil moisture parameters are known)

b. The critical periods for water deficits are given in Table 12. High ky values indicate a sensitive period for water shortage and savings should be reduced in that period otherwise a considerable yield decrease will occur.

c. The total amount of water to be saved is the difference between needed supply and available supply. In no case should the saving be greater than 50 percent of the full supply, because under such conditions formula {4} may not be valid. The allocation of the savings over the periods can be calculated according to formula {4}, whereas the savings per months can be calculated by the following relationship:

{5}

where:	WS month = water to be saved in a given month
	WS period = water to be saved in the chosen growing period
	V month = water applied in the month under full supply
	V period = water applied in the chosen period under full supply

d. The available water (Va) per month is the difference between the water applied under full supply (V) and water savings (WS) under limited supply.

e. Case a: consists in extending the irrigation interval but keeping the depth of irrigation constant. The extended interval can be calculated taking into account the interval under full supply, which is determined through the traditional procedures and the ratio V/Va or:

(limited supply) = (full supply). V/Va

After determining the optimum irrigation interval, some practical changes may be introduced in scheduling in order to facilitate the operation, e.g. the introduction of a fixed interval throughout a part of the irrigation season.

f. Case b: consists in keeping the initial (full supply) interval but reducing the depth of irrigation. Although this is not a very practical measure, it is done in some instances. The volume of each irrigation is calculated by the relationship;

{7}

2.3 Example

a. Basic data:

Cropping pattern: sugarcane (ratoon) throughout the year

Crop water requirements: ETm = 21700 m³/ha year distributed as indicated in Table 13

Available water supply: 17400 m³/ha year

Irrigation interval, under full supply: as given in Table 13.

Length of growing periods: establishment - 30 days; vegetative - 180 days; yield formation - 60 days

b. Critical periods:

Table 12 indicates that the ripening period (ky₃ = 0.1) is the most appropriate for making water savings, followed by yield formation (ky₂ = 0.5) and then the vegetative period (ky₁ = 0.75).

c. Amount of water to be saved per period:

The total amount to be saved per period is 21700 - 17400 = 4300 m³/ha year.

The total amount of water to be saved in each period can be determined according to formula {4}.

Assuming that the savings should be made only in the two more favourable periods (ripening and yield formation), the corresponding water savings would be:

WS₂ + WS₃ = 4300 m³/ha

{8}

V2 = 3000 + 2200 + 2000 = 7200 m3/ha	ky2 = 0.5 (Table 12)
V3 = 1000 + 600 = 1600 m3/ha	ky =0.1 (Table 12)

Replacing these values in {8} and solving the equations results in:

WS = 2260 m³/ha
WS = 2039 m³/ha

However, WS₃ (2260 m³) is greater than V₃ (1600 m³) which is not possible. This indicates that water saving cannot be limited to growth periods (2) and (3). The savings should be made over a longer period and include period (1) as well. Since period (1) is the least appropriate for undertaking water savings, only the last three months (May, June and July) have been included. Hence the distribution of the water savings is recalculated as follows:

WS₁ + WS₂ + WS₃ = 4300 m³/ha

where:	V3 = 1600 m3	ky =0.1
	V2 = 7200 m3	ky =0.5
	V1 = 2500 + 3000 + 3000 = 8500 m3	ky - 0.75

replacing values and solving the equations:

WS₂ = 1484 = 1500 m³/ha
WS₃ = 1648 = 1600 m³/ha
WS₁ = 1168 = 1200 m³/ha

d. Amount of water to be saved per month (M):

(see Table 13)

e. Volume of water to be applied monthly:

May	2500 - 353 = 2147 m3/ha
June	3000 - 424 = 2576 m3/ha

(see Table 13)

f. Irrigation interval (Case a):

(see Table 13)

The distribution of the water according to these intervals leaves the amounts of each irrigation the same as they were initially under full supply conditions. The proposed method only stretches the irrigation intervals during those periods when the plant suffers least and therefore yield reductions are minimized.

g. Amount per irrigation (Case b):

As said in the main text, one can also maintain the irrigation interval and reduce the amount per irrigation gift. The application of this solution is not advisable because farmers do not understand why they receive different amounts of water every time and it is susceptible to malpractices. Nevertheless, to complete the example, the procedure is also illustrated.

Table 13 EXAMPLE OF ROTATIONAL SUPPLY WITH LIMITED WATER

¹ (0) = establishment, (1) = vegetative period, (2) = yield formation, (3) = ripening

² Figures rounded off to a full day

The irrigation dose is therefore:

(see Table 13)

2.4 Related design considerations

The benefits of the methodology just described are not limited to conditions when supplies are limited, but it can as well be used advantageously under conditions of full supply.

The advantages can best be illustrated by an example. Let us assume that there is one irrigation scheme that can irrigate 10000 ha, under full supply but the area could be increased if water were available. For reasons of simplicity it will be assumed that the only crop grown is sugarcane. Average annual yield (full supply) is 80 t/ha. The net crop water requirements are 21700 m/ha year and the gross irrigation requirements, with a 50 percent overall efficiency, are 43400 m³/ha year. We will suppose that the management decides to save 20 percent of the crop water requirements and utilize the water saved to extend the irrigated area. The consequences of such a decision are analysed below.

a. Amount of water to be saved:

at the farm level

at project level

b. Expected reduction in yield due to water shortage:

The reduction in yield can be calculated according to the formula {l}¹, i.e.

¹ For greater detail and information on how to calculate expected yields when water supply is limited, the reader is referred to FAO Irrigation and Drainage Paper No. 33, Chapter IV.

where:	ETa = 21700 - 4300 = 17400 m3/ha
	ETm =21700 m3/ha
	ky = 1.25
	Ym = 80 t

replacing values in the equation one obtains:

Ya = 60.2 t/ha

c. Incremental irrigated area and production:

The water saved will allow irrigation of:

The sugarcane production arising from this additional area would be:

2471 ha x 60.2 t/ha = 148705 t

d. Total production:

10000 ha x 60.2 t/ha + 148705 = 750705 t

whereas the production under full supply conditions was:

10000 ha x 80 t/ha = 800000 t

This measure has permitted to increase the irrigated area by nearly 25 percent and consequently those who benefit from it. If the intention is to maximize the number of beneficiaries, while keeping a reasonable income per farmer, the above methodology can be used to design the system. However, it should be noted that under such assumption the farmers will never be able to obtain the potential yield of the crop.

The suggested procedure can be improved by using some more sophisticated mathematical calculations or the trial and error method, if the economic optimum is desired.

3. Rotational scheduling under full water supply conditions

There are many books and references illustrating how to calculate rotational supply according to the technical principles involved. However, such calculations can lead to water distribution patterns which are often impractical and therefore rarely applied. Simplification of the calculated irrigation scheduling is often necessary but oversimplification can lead to wasting large amounts of water. In order to visualize some of these problems, an example has been worked out in detail.

Example:

a. Basic data

The exercise is to calculate the rotational schedule for a farm with the following characteristics.

a1) irrigable area: 3,7 ha

a2) cropping pattern by type of soil:

tomato	1.5 ha in soil A1
tomato	0.8 ha in soil A2
maize	0.25 ha in soil A1
pepper	0.50 ha in soil A1
pepper	0.50 ha in soil A2
beans	0.15 ha in soil A1
Total	3.7 ha

a3) hydraulic characteristics of soils:

soil A1: W	=	water holding capacity	=	93 mm/m
		depth of soil	=	0.8 m1
soil A2: W	=	water holding capacity	=	87 mm/m
		depth of soil	=	0.8 m1
fraction of available soil water = p			=	0.65

¹ Depth of soil limited by hard clay pan.

a4) ET crop² (in mm):

² Figures correspond to a real location, Sansirisay in Guatemala.

	Tomato	Maize	Pepper	Beans
Oct	65	66	65	31
Nov	91	85	92	92
Dec	124	124	125	103
Jan	52	125	50	34
	332	400	332	260

b. Depth of irrigation water

The depth of irrigation to be applied is given by the formula:

d = p. W. D

{9}

where:	p = fraction of available soil water
	W = water holding capacity in mm/m
	D = rooting depth in m
	d = depth of irrigation (mm)

replacing values gives:

soil A1	d1	=	0.65 x 93 x 0.83	=	48 mm
soil A2	d2	=	0.65 x 87 x 0.83	=	45 mm

³ It is assumed that the rooting depth is limited by the hard pan and therefore equal to the soil depth.

c. Irrigation intervals

The irrigation intervals (I) are calculated with the following relationship:

{10}

where ETcrop is mm/day and I in days.

The irrigation intervals for tomato in soil A₁ are calculated as follows:

For the next interval:

The application of formula {l0} to the following month gives a discontinuity in the intervals because October cannot be divided into equal intervals by the 22 day period. Therefore, in order to have a continuous irrigation schedule, the remaining days in October must also be considered:

2.10 mm/day x (31 - 23) days + 3.03 x 6 days = 48

d =10.3 days

therefore I₂ = 8 + 10.3 » 18 days

The next interval is all within November, i.e.:

The next interval is in part of November and December:

d = 8.97 » 9 days

I₄ = 4 + 9 = 13 days

The other intervals are calculated similarly and they are reflected in the bar diagram of Figure 15.

It has been assumed that crops are planted at the beginning of the 1st, 2nd or 3rd 10 day period in order to extend the harvest period and to suit labour availability.

A farmer who has free access to a water supply (well, continuous flow in the canal, etc.) could follow the irrigation schedule given for each crop in Figure 15. Such a schedule would allow him to use the water in a highly efficient manner, but it would be highly impractical since the schedules of the crops do not coincide, nor even those of the same crop in different soil. Hence the farmer would have to repeat the irrigation on many days (30 times in our example), and this would certainly conflict with other agricultural operations. Furthermore, it is difficult for the average farmer to keep track of such schedules.

If we now consider that a tertiary canal serves a large number of farms (sometimes 100 or more) of different sizes and with slight variations in the soils, planting dates and crops, the application of a rotational schedule in the canal to satisfy the individual schedules on every farm would mean that it would become almost impossible to operate.

Other alternatives must be sought, and one of them is to introduce some simplification in the irrigation scheduling, so that it becomes more manageable.

Fig. 15 Example of irrigation scheduling under the rotational supply method

d. Simplification of the irrigation schedule

The simplifications to be introduced in a calculated irrigation schedule are, to a great extent, a matter of personal judgement. A compromise must be found between an oversimplification of the schedule (which sometimes bears little relation to the actual needs of the plant), and the calculated one which is often impractical.

In the example three levels of simplification have been considered which could correspond to reality.

First level of simplification

A common irrigation schedule with variable intervals is adopted for all crops. The fact that some crops are planted earlier than others is taken into consideration in the irrigation schedule.

To calculate the common schedule the following steps are taken:

- The intervals are fitted to the planting dates (in the example, every 10 days).

- For a given month the shortest interval falling within that month is selected. For intervals falling between two months the shortest of those is adopted.

The common irrigation schedule is represented by the line h) in Figure 15.

The procedure can either be applied to a farm or canal serving many farms.

Second level of simplification

This consists of introducing slight modifications in the "common irrigation schedule" so that intervals are as similar as possible in duration. In the example, the last five intervals have a duration of 13, 11, 11, 12 and 12 days and they have been replaced by 5 intervals of 11 days. The advantage of such a simplification is that the intervals can be more easily remembered by the farmers and the operation of the canal is more regular.

Third level of simplification

This consists of adopting a constant interval - which may or may not include the planting dates - throughout the irrigation season. The length of such an interval is the shortest of the "calculated" ones (11 days in the example).

The application of such a constant interval, keeping the depth of each irrigation constant, leads to great water infiltration losses because the soil cannot hold that much water. Alternatively, the depth given in each irrigation can be different to fit the water holding capacity of the soil, but again the system may not be easily understood by the farmer. Experience shows that farmers become more skillful in managing a constant depth of irrigation water rather than a variable one.

Therefore, in most cases, the first and second levels of simplification are the most practical approach to irrigation scheduling under the rotational method.