3.1 Introduction
3.2 The Approximate Method
3.3 Cropping Pattern
3.4 The Calculation Method
3.5 Converting from SIN_{net} to SIN_{op}
Chapter 2 outlined some of the basic aspects of water availability for irrigation. This chapter deals with the irrigation water need of an entire irrigation area, SIN_{gross}, and the supply of irrigation water to the entire irrigation area, SIN_{op}. The SIN_{gross} is the continuous flow of water required for good crop production during the irrigation season. SIN_{op} is the flow of water actually applied to the irrigation area, continuously or at intervals after taking the operational criteria into account.
Because there is always some water stored in the soil (rootzone of the crop), or on the field (in case of paddy rice), SIN_{gross} can be supplied at intervals without affecting crop production. However, over the long run, the supplied volumes of water should equal the needed volumes of water.
The SIN_{gross} may be determined quickly, just to get a rough idea of the volume of water needed. This quick method, which is explained in section 3.2 of this manual, is called the approximate method and is suitable for preliminary planning only, or when there is a lack of reliable data. The second, the calculation method is more accurate and thus suitable for the detailed planning of sound water management. The calculation method, explained in the section 3.4, uses fairly detailed information on the irrigation area's cropping system. (The concept of a cropping system will be introduced in section 3.3.)
The approximate method of determining the SIN_{gross} assumes a constant average irrigation need, IN_{net}, for the entire season. The most simplified standard value of this IN_{net} is 1 litre per second per hectare. This is equivalent to a daily water requirement of 8.6 mm (see Annex II). When the daily water requirement is 4.3 mm, the irrigation need would be 0.5 l/s.
Like most rules of thumb, this rule should be applied with caution. The proxy values of IN_{net} in hot and dry climates can be three times as great as in humid climates. Typical values are presented in Table 1.
Table 1 APPROXIMATE AVERAGE IN_{net} VALUES FOR DIFFERENT CLIMATES AND RICE
Humid tropical climate 
0.5 l/s/ha 
Monsoon climate wet season 
0.5 l/s/ha 
Monsoon climate dry season 
1.0 l/s/ha 
Semiarid climate wet season 
1.0 l/s/ha 
Semiarid climate dry season 
1.5 l/s/ha 
Arid climate 
1.5 l/s/ha 
Rice 
1.5 l/s/ha 
The approximate values of the irrigation need for an entire area, SIN_{net} may be calculated by multiplying the IN_{net} with the area, Area (in hectares). Therefore, the equation representing the approximate Net Scheme Irrigation Need is:
SIN_{net} (l/s) = Area (ha) x IN_{net} (l/s/ha) 
EXAMPLE: The area of a specific irrigation scheme on which several crops are grown is 50 ha. The estimated net irrigation requirement is 1 l/s/ha. Thus the SIN_{net} for the entire scheme would be: 50 (ha) x 1 (l/s/ha) = 50 (l/s) 
The net scheme irrigation need, SIN_{net} is the amount of water needed to meet crop water needs of an entire scheme minus the effective rainfall. Water lost during delivery must be added before the gross scheme irrigation need, SIN_{gross}, can be determined.
The SIN_{gross} is calculated by dividing SIN_{net}_{ }by the overall scheme irrigation efficiency (e), which is the product of the conveyance efficiency (e_{c}) and the field application efficiency (e_{a}). The subject of irrigation efficiencies is discussed at length in Annex 1 in Training Manual 4, "Irrigation Scheduling".
The present series of training manuals is mainly concerned with small irrigation schemes and tertiary units of larger schemes which use gravity irrigation methods. Gravity irrigation of sandy soils is not widespread. Therefore, a conveyance efficiency of about 85% and an application efficiency of about 60% have been selected, which result in an overall area irrigation efficiency of about 50%. This value for the overall efficiency will be used in all further calculations in this manual. It should also be kept in mind that the values mentioned above are only indicative values.
Once the scheme irrigation efficiency (e, expressed in percentages) is known, SIN_{gross} can be calculated with the following formula:
SIN_{gross} (l/s) = 100/e x SIN_{net} (l/s) 
The formula above shows that as the efficiency drops (becomes smaller), more water has to be supplied to satisfy the irrigation need of a scheme.
EXAMPLE: An irrigation scheme of 70 ha located in a semiarid area is irrigated during the dry season. The average IN_{net} can be estimated at 1.5 l/s/ha (see Table 1). The total net irrigation need of the scheme is: Area times IN_{net} which is 70 x 1.5 = 105 l/s. The irrigation efficiency is 50% and thus the SIN_{gross} equals 100/e x SIN_{net}, which equals 100/50 x 105, or 210 l/s. 
Operational criteria
The amount of water that is supplied to an irrigation area should be equal to the gross scheme irrigation water need.
This would mean that, if the same flow of water is supplied continuously, second after second, hour after hour, and day after day without interruption throughout the entire growing season, the SIN_{gross} which is expressed in a continuous flow of litres per second would have been completely supplied, and SIN_{op}, operational scheme irrigation need, would equal SIN_{gross}.
However, water is not always supplied continuously. If irrigation water is applied only 12 out of a possible 24 hours per day, then during these 12 hours twice as much water must be given to supply the same total. Further, if irrigation does not take place every day, more water must be supplied on irrigation days in order to maintain the same total quantity per week. This is illustrated in the following examples and in Figures 17 and 18.
EXAMPLE: An area has an irrigation need of 25 l/s. However, the water supply is available only 4 hours per day, every day. Since the farmers can irrigate only during 4 out of the 24 hours a day, the available time is only 4/24 or 1/6 of the total time. The supply should therefore be 6 times as great, or: 6 x 25, or 150 l/s. 
Figure 17  Irrigation water applied for 4 hours instead of 24 hours
EXAMPLE: Another scheme also has a continuous irrigation need of 25 l/s. Although farmers irrigate the whole day (24 hours), they only do it 2 days a week. Now the amount of time available for irrigation is 2 days out of 7, which is 2/7 or 1/3.5. The supply should therefore be 3.5 times as much during the 2 irrigation days, to supply the same total amount. The actual irrigation flow should therefore be 3.5 x 25, or 87.5 l/s. 
Figure 18  Irrigation water applied for 2 days instead of 7 days
EXAMPLE: What happens if the water requirement of 25 l/s has to be supplied in 8 hours per day and only 4 days per week? In this case, the total amount of time in which the irrigation water has to be supplied is 8/24 or 1/3 of a day during only 4/7 of a week. This leaves farmers with only 1/3 x 4/7, or 4/21 of the time in which to irrigate. During this time the actual flow should be 21/4 times the continuous requirement, or 21/4 x 25, which equals 131 l/s. 
The ratios of hours per day and days per week during which the irrigation system functions are called, in this manual, the operational criteria. By using these operational criteria, the required scheme irrigation supply can be determined with the following formula:
SIN_{op} (l/s) = SIN_{gross}/T_{op} T_{op} = d/7 x h/24 
where:
SIN_{op} = operational scheme irrigation need
T_{op }= operational criteria
d = number of irrigation days per week
h = number of irrigation hours per day
SIN_{gross} = gross scheme irrigation need
The formula above shows that the fewer the number of irrigation applications per week and the fewer the number of hours of irrigation per day, the larger the scheme's supply should be, and the larger should also be the capacity of the irrigation system.
EXAMPLE: In an irrigation scheme, irrigation takes place 4 days per week, 9 hours per day. Thus the operational criteria, T_{op} = 4/7 x 9/24 = 3/14. The SIN_{gross} is 50 l/s. According to the formula the SIN_{op} should be: 50 (l/s) / (3/14) = 233 l/s. SAMPLE PROBLEM: Estimate, using the approximate method, the SIN_{op} for the following scheme. ASSUMPTIONS: The scheme is situated in a monsoon climate. Irrigation is practised during the dry season. The area of the scheme is 20 ha, with loamy soil. Irrigation takes place 5 days per week for 10 hours per day. The scheme uses surface irrigation methods, and the canals are made of earth. CALCULATION: Step 1: Estimate the Net Irrigation Need, IN_{net} According to Table 1, the approximate IN_{net} for this climate should be 1 l/s/ha. Step 2: Estimate the Net Scheme Irrigation Need for the entire area. SIN_{net} (l/s) = IN_{net} (l/s/ha) x Area (ha) For an IN_{net} of 1 l/s/ha and an area of 20 ha, the result is a SIN_{net} of 20 l/s. Step 3: Estimate the scheme irrigation efficiency, e. The scheme irrigation efficiency (e) can be found by multiplying the e_{c} and e_{a} values in the following formula (the following calculation is used throughout this manual for the irrigation efficiency): e= e_{c} x e_{a}/100 = 85 x 60/100 » 50% Step 4: Calculate the gross scheme irrigation need, SIN_{gross} in l/s. The SIN_{gross} can be calculated with the following formula: SIN_{gross} = SIN_{net} x 100/e In this case, SIN_{gross} = 20 x 100/50 = 40 l/s. Step 5: Calculate the operational scheme irrigation need, considering the following operational criteria. The scheme is irrigated 5 days per week for 10 hours per day, T_{op} and SIN_{op} may be calculated according to the following formula: T_{op} = d/7 x h/24 = 5/7 x 10/24 » 0.3 SIN_{op} (l/s) = SIN_{gross} / T_{op} = 40 / 0.3 = 133 l/s 
In Chapter 3.2 a quick but rough estimation was made of a scheme's irrigation requirement, by using the rule of thumb for the IN_{net}. A much more accurate prediction of a scheme's irrigation flow requirements can be made, if it is based on the scheme's cropping pattern. The cropping pattern, or cropping schedule of an irrigation area provides information, for a period of at least one season, on three important elements:
 which crops are grown
 when are they cultivated
 how many hectares of each crop are grown.
This information is written down, in detail, let us say for the area of one farmer during one year, and from this record there emerges a pattern. For example, the cropping pattern for farm A:
 onions cultivated from April 15 to September 15 on 1 ha
 potatoes from October 15 to February 15 on 1/2 ha, on the same plot as the onions
 cotton from July 10 to January 20 on 1 and 1/2 ha.
Information on the type of crop and the period of cultivation can be visualized in a line diagram.
Figure 19  Line diagram of an example farm's crop calendar
This crop calendar diagram shows that farmer A starts with onion cultivation in midApril which he then harvests midSeptember. Within one month he will plant potatoes on the same plot, which he expects to uproot in midFebruary. Farmer A has a second plot on which he grows cotton over a sixmonth period from midJuly to the end of January of the following year.
If the area of each crop has to be shown, the lines representing individual crops should be replaced by bars where the height of each bar is a measure of the area of each crop. In the example below, for instance, an area of one hectare is represented by a barheight of about 0.5 centimetre. The length of the bar indicates the period in which the crop is grown. A diagram with crop bars rather than crop lines is called a cropping pattern.
Figure 20  Bar diagram of an example farm's cropping pattern
Figure 20 shows that farmer A grew 1 and 1/2 ha of cotton from 10 July to 20 January, 1 ha of onions from midApril to midSeptember and 1/2 ha of potatoes from midOctober to midFebruary.
The diagram can indicate more crop husbandry dates, such as crop manuring and periods of weeding and transplanting, etc. It is also possible to draw up a cropping pattern diagram for larger areas, including a group of farmers, or an entire irrigation scheme. The only difference is that with a single farm, it was supposed that the crop was sown or planted on one specific day. The same assumption cannot be made in the case of a larger irrigation scheme. In fact, there are several reasons why farmers do not all start planting on the same day on an irrigation scheme:
 labour and machinery needed to prepare the land is in short supply, and each has to wait his turn; a farmer is not ready due to other activities, so he postpones planting the crop for a while;
 also, it is impossible to supply all the fields on an irrigation scheme with the water required to start a crop on the same day. The larger the scheme's area, the longer the time it takes to serve all the fields.
The cultivation of any crop on an irrigation scheme is thus spread over a certain period of time. This is called staggered cultivation or a staggered cropping pattern.
When, for example, the first farmer starts with onion cultivation on 1 April, and other farmers follow more or less regularly on the days after until the last farmer has planted his onions at the end of April, the crop calendar for onions is staggered over one month. In a cropping pattern diagram this staggering is depicted as below.
Figure 21  Staggered cropping of onions
Each line in the diagram represents a farm, or a certain area, and the day it starts onion cultivation. Of course; when crop planting dates are staggered, so also are the end of season dates.
In a normal cropping pattern diagram the subsequent lines are simplified, forming a parallelogram bar. The lower side represents the first farmer, or the area which is planted earliest. The upper side represents the last area to be planted and cropped.
Figure 22  Parallelogram cropping pattern
As with the cropping pattern diagram for one farmer, the extent of the area cultivated for each type of crop, across the entire scheme, can be indicated with the height of the parallelogram serving as a measure. The height of the vertical axis of the diagram represents the total area of an irrigation scheme, for example a scheme of 150 ha (see Figure 23). Onions are cultivated on 60 ha and they have a growing period of 5 months. The onion planting season starts in the beginning of April and is staggered over one month. Potatoes, with a growing period of 4 months, are grown on 30 ha starting 1 October. The cultivation period is staggered over one month. Cotton, with a growth period of 190 days, is grown on 75 ha from the beginning of July. Cotton cultivation is staggered over a period of 45 days.
Figure 23  Cropping pattern for an irrigation scheme of 150 ha
An area which is empty at any given moment in the cropping pattern diagram is not cultivated at that time. For example, during the month of March nothing is being cultivated on the scheme; in midSeptember 30 ha of onions have already been harvested and 15 ha of land have been left fallow.
3.4.1 Scheme irrigation need for a simple cropping system
3.4.2 Scheme irrigation needs for a multiple cropping system
3.4.3 Determining the irrigation need for a ricebased cropping pattern
In Section 3.2, the irrigation water need and the actual scheme irrigation flow requirement were estimated using the approximate method, which is based on a rule of thumb which says that one hectare of irrigated land needs a fixed flow of water during the irrigation season.
When accurate crop and climatic data are available, a much more precise prediction may be made of the gross scheme irrigation need, SIN_{gross}, while the operational scheme irrigation need, SIN_{op}, can be finely tuned to the real irrigation need. However, when a scheme has many crops, this more precise method also requires more complex calculations. A calculation method example is presented in this section to help field technicians and extension agents understand and become familiar with the underlying principles and the basic elements involved in calculating scheme water needs.
The calculation of an individual crop's irrigation water need, IN_{net}, has already been explained in Training Manual 3, "Irrigation Water Needs". Using this calculation as a point of departure, which ends with the net irrigation need of an individual crop expressed in millimetres of water depth per month (mm/mon), this section will focus on determining the net scheme irrigation need SIN_{net} (in litres per second) of an entire irrigation scheme with a variety of crops throughout the year.
Suppose, as in the former section, that 60 ha of onions are planted on an irrigation scheme of 150 ha and that the planting of onions is spread out evenly over one month. Planting is begun on 1 April and finished on 30 April.
The irrigation water need of the earliest planted onions during the month of April is 98 mm, which is 3.3 mm/day for 30 days (see Manual 3, Part II 4.9). Onions which are not planted until April 15 do not need water during the first 15 days of April. For the remaining 15 days of April, the irrigation water need is 15 x 3.3 mm, or 49 mm. Onions which are planted on the last day of April do not need any irrigation water in April at all.
This example indicates that the irrigation water need of the onions which are planted in between the earliest and the latest planted onions, lies in between the irrigation water need of the earliest and latest planted onions. In general, the average irrigation water need of an area cropped with onions, or any other crop, is the average of the irrigation water need of the earliest (IN_{e}) and the irrigation water need of the latest (IN_{l}) planting. This is also expressed in the formula below:
_{}
This monthly irrigation need, IN_{av} , is presented as a water layer which is applied during that month on a certain area. However, water flows are not actually expressed in millimetres of water covering a certain area over a certain period of time, but in litres per second, or cubic metres per second. Therefore the IN_{av} per month should be converted into a current irrigation water flow. SIN_{net} in litres per second (net scheme irrigation need) can be calculated as follows:
_{}
or
_{}
When the monthly irrigation need of an area with staggered crop cultivation must be determined, the irrigation needs of the earliest and of the latest plantings must be calculated. The following example is based on Training Manual 3, Chapter 5, data sheets 3 and 4. Field technicians and extension agents should consult this material before proceeding with this manual.
***
SAMPLE PROBLEM: Determine the irrigation need of onions. (Following is a list of assumptions to be used in synthesizing the information in the tables below.)
ASSUMPTIONS:
Crop: 
Dry Onions 
Earliest planting: 
1 April 
Total area: 
60 ha 
Latest planting: 
30 April 
Duration of crop growth phases 
Crop factors: (K_{c})  
Initial stage: 
15 days 
K_{c} 
0.5 
Crop devt. stage: 
25 days 
K_{c} 
0.75 
Mid season: 
70 days 
k_{c} 
1.05 
Late season: 
40 days 
K_{c} 
0.85 
Potential Evapotranspiration (ET_{0}) in the example location:
Month 
Jan 
Feb 
Mar 
Apr 
May 
Jun 
Jul 
Aug 
Sep 
Oct 
Nov 
Dec 
ET_{0} (mm/day) 
4.7 
5.1 
5.2 
5.6 
5.7 
6.1 
5.8 
5.5 
5.6 
5.2 
4.3 
4.6 
Effective rainfall (P_{e}) in the example place:
Month 
Jan 
Feb 
Mar 
Apr 
May 
Jun 
Jul 
Aug 
Sep 
Oct 
Nov 
Dec 
P_{e} (mm/mon) 
1 
3 
7 
10 
12 
13 
72 
82 
16 
7 
1 
0 
CALCULATION:
Step 1: Draw the cropping pattern.
Step 2: Calculate the IN_{e} according to the method explained in Training Manual 3.
Step 3: Calculate the IN_{l} in the same way.
Step 4: Calculate the average monthly irrigation need IN_{av}.
Month 
Jan 
Feb 
Mar 
Apr 
May 
Jun 
Jul 
Aug 
Sep 
Oct 
Nov 
Dec 
IN_{e} mm/mon 



98 
150 
179 
102 
59 




IN_{l} mm/mon 




99 
161 
111 
83 
127 



IN_{av} mm/mon 



49 
125 
170 
107 
71 
64 



Step 5: Determine SIN_{net} in litres per second for each month with the following formula:
_{}
For example, if April's IN_{av} equals 49 mm, then the SIN_{net} equals 49 mm times the area (60 ha) divided by 260, or
_{}
The same calculation should be made and recorded on the schedule below for the months of May, June, July, August and September.
Month 
Jan 
Feb 
Mar 
Apr 
May 
Jun 
Jul 
Aug 
Sep 
Oct 
Nov 
Dec 
SIN_{net} l/s 



11 
29 
39 
25 
16 
15 



***
Please note the variations for SIN_{net} for each month in the above example. It is interesting to compare the results with the ones obtained by the approximate method. With the approximate method, only a single value for SIN_{net} can be obtained. If IN_{net} is assumed to be 1 l/s/ha, SIN_{net} for 60 ha is 60 l/s, which is about 50% greater than the peak water need (39 l/s for June) calculated above.
The data contained in the foregoing tables and results of the calculations can be summarized on Data Sheet 1 which is given in Annex III.1.
On many irrigation schemes more than one kind of crop is grown at the same time, and over the course of a year. It is therefore often necessary to calculate irrigation needs for a multiple cropping system. Once the irrigation need for a single cropping pattern has been estimated, the irrigation need for a scheme with more than one crop can be determined by adding up the irrigation requirements for each month. The way this is done is much the same as with the calculation for a simple cropping pattern, following the three steps outlined below.
Step 1: Draw the multiple cropping system diagram, as explained in Section 3.3.Step 2: Determine the scheme irrigation need for each crop separately, following the steps explained in Section 3.4.1. Use Data Sheet 1.
Step 3: Add up each month's calculated irrigation needs. The total scheme irrigation need should be expressed in litres per second.
Data Sheet 2, included in Annex III.2, is designed to calculate the scheme irrigation need for a multiple cropping system.
SAMPLE PROBLEM: Determine the scheme irrigation need based on the following assumed data. ASSUMPTIONS:  

Crop 1 
Crop 2 
Crop 3 
Name: 
onions 
cotton 
potatoes 
Area: 
60 ha 
75 ha 
30 ha 
Planting period: 
1 Apr30 Apr 
1 Jul15 Aug 
1 Oct30 Oct 
Growing period: 
150 days 
190 days 
120 days 
Climatic data are the same as the data used in the example Section 3.4.1. Scheme area op: 150 ha (see Figure 23) CALCULATION: Step 1: The multiple cropping pattern is drawn in the diagram included in Data Sheet 2 as was done in Figure 23. Note the scale of the vertical axis which indicates that the scheme area is 150 ha. Step 2: The determination of the irrigation needs for the three different crops can be done on Data Sheet 1 (see also Manual 3 Part II Chapter 5). This method was extensively explained in Section 3.4.1, and is thus not repeated here. The results are copied onto Data Sheet 2. Step 3: The net scheme irrigation need is found by simply adding up monthly irrigation needs for each crop. The results of the calculation are shown below in Data Sheet 2.

As was explained in Section 4.4 of Training Manual 3, Part II, the net irrigation needs of paddy rice consist of the crop evapotranspiration, ET_{rice}, minus the effective rainfall, P_{e}, if any. Moreover, additional irrigation water is needed to compensate for the continuous percolation losses, PERC. At the start of paddy rice cultivation, a certain amount of water must be supplied to saturate the rootzone and prepare the wet soil, called puddling. This amount of water, SAT, is about 200 mm. Finally, because rice grows well only on a submerged field, a water layer, WL, measuring about 50 mm deep must be established on the field.
The formula to determine the net irrigation need of paddy rice is written as follows:
IN_{net rice} = ET_{rice} + SAT + PERC + WL  P_{e } (mm/month)
In the same way as for other crops, one has to calculate the IN_{net}_{ }_{rice} of the earliest and the latest plantings to find the irrigation requirement for a rice crop with a staggered cultivation. The irrigation needs of a second crop after rice during the dry season, beans for instance, are determined as explained in Section 3.4.1.
SAMPLE PROBLEM: Rice is grown on an irrigated scheme of 150 ha during the wet season. Determine the scheme irrigation water need, including the saturation requirement, SAT, the establishment of a water layer, WL, and the supply needed to make up for percolation, PERC. ASSUMPTIONS: Planting period: 16 June (early)  1 August (late) Growing period: 130 days Crop factors:  

K_{c}  
Day 060 
1.1  
Midseason (40 days) 
1.15  
Last 30 days 
1.0  
SAT:  
200 mm 
 
Early 
16 May  15 June  
Late 
1 July  31 July  
PERC: 6 mm/day =180 mm/month WL:  
100 mm 
 
Early 
16 June  15 July  
Late 
1 August  30 August  
Climatic data are the same as the data used in 4.4 in Manual 3. CALCULATION: Start with Data Sheet 3 which is a modified version of Data Sheet 7 in Training Manual 3. The blank sample of this sheet is found in Annex III.3. Step 1: Draw the cropping pattern diagram with SAT and WL, and fill ET_{0}.
Step 2: Calculate the IN_{e} (mm/month) according to the method explained in 3.4.
Step 3: Calculate the IN_{l}, (mm/month) in the same way as Step 2.
Step 4: Calculate the average monthly irrigation need IN_{av} (mm/month).  
IN_{av} mm/mon 




49 
149 
146 
205 
203 
296 
171 
57  
Step 5: Determine SIN_{net} (l/s) for each month. _{}
 
SIN_{net} l/s 




28 
86 
84 
118 
117 
171 
99 
33  
Note: When cultivation does not start on the first of a month or when cultivation periods do not coincide with full months, the water requirement should be divided over the two months according to the number of days that fall in each month. For example, in establishing a water layer of 100 mm from 16 June to 15 July, i.e. during 15 days in June and 15 days in July, note that 15/30 x 100 = 50 mm is to be given in June, and 15/30 x 100 = 50 mm in July. 
The preceding sections of the present manual illustrated the calculations of the irrigation need for a simple cropping pattern. In Section 3.4.2, we calculated the net irrigation need for the entire scheme in litres per second, SIN_{net}, by adding up the irrigation needs for each crop. Now, these irrigation needs have to be converted into the gross scheme irrigation need, SIN_{gross}, and the operational scheme irrigation need, SIN_{op}. This process is explained in section 3.2, and is repeated here with the following fourstep process.
Step 1: Estimate the overall irrigation efficiency, e.
e = e_{conveyance} x e_{application }/ 100
Step 2: Determine the gross Scheme Irrigation Need, SIN_{gross}
SIN_{gross}_{ }= SIN_{net} x 100 / e
Step 3: Establish the operational criteria or the total operation time T_{op}.
T_{op }= (d/7) x (h/24)
Where:d = days of operation per 7 days of the week
h = hours of operation per 24 hours of the day.
Step 4: Determine the operational Scheme Irrigation Need, SIN_{op}.
_{}
SAMPLE PROBLEM: Determine the operational flow for the cropping pattern on page 28. The following data is assumed: SIN_{net} CALCULATION: Step 1: Estimate the scheme irrigation efficiency. e = 85 x 60 /100 » 50 % Step 2: Determine the gross Scheme Irrigation Need SIN_{gross} in litres per second for April. SIN_{gross} = SIN_{net} x 100 / e = 11 x 100 / 50 » 22 l/s. Calculate the same for the months May, June, July, August, and September in the following schedule.  
Month 
Jan 
Feb 
Mar 
Apr 
May 
Jun 
Jul 
Aug 
Sep 
Oct 
Nov 
Dec  
SIN_{gross} (l/s) 



22 
58 
78 
50 
32 
30 


 
Step 3: Establish the operational criteria, or the total operation time, T_{op}. T_{op} = days of operation per week times hours of operation per day. Step 4: Determine operational scheme irrigation need, SIN_{op}, for April. SIN_{op} = SIN_{gross} / T_{op} = 22 / 0.43 » 51 l/s Calculate the same for the months May, June, July, August, and September in the following schedule.  
Month 
Jan 
Feb 
Mar 
Apr 
May 
Jun 
Jul 
Aug 
Sep 
Oct 
Nov 
Dec  
SIN_{op} (ls) 



51 
135 
181 
116 
74 
70 


