Irrigation Methods Module 6

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Irrigation Methods Module 6

• Lesson 4
• Irrigation scheduling methods
• Simple calculation method

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Why calculation method?

• Irrigation schedule by this is method is based
  on
• Estimated depth (in mm) of the irrigation
  applications
• Calculated irrigation water need of the crop over
  the growing season
• Unlike the estimation method the simple
  calculation method is based on calculated
  irrigation water needs.
• Influence of the climate, i.e. temperature and
  rainfall, is more accurately taken into account
• More accurate than the result of the estimation
  method

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Steps to determine the irrigation schedule

• Step 1 Estimate the net and gross irrigation
  depth (d) in mm.
• Step 2 Calculate the irrigation water need (IN)
  in mm, over the total growing season
• Step 3 Calculate the number of irrigation
  applications over the total growing season
• Step 4 Calculate the irrigation interval in days

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1. Estimate the net and gross irrigation depth

• Best determined locally by checking how much
  water is given per irrigation application with
  the local irrigation method and practice
• If no local is available, use Table below to
  estimate the net irrigation depth (d net), in mm
  (here net irrigation depth is assumed to depend
  only on the root depth of the crop and on the
  soil type)

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• Not all water which is applied is used by the
  plants. Part of the water is lost through deep
  percolation and runoff.
• The gross irrigation depth (d gross), in mm,
  takes into account the water loss during the
  irrigation application and is determined using
  the following formula

d gross gross irrigation depth in mmd net
net irrigation depth in mmea field application
efficiency in percent
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• If reliable local data are available on the field
  application efficiency, these should be used. If
  such data are not available, the following values
  for the field application efficiency can be used

Problem Tomatoes are grown on a loamy soil in
furrows system. Estimate the gross irrigation
depth 3 minutes!!!
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Step 2. Calculate the irrigation water need

• Assume that the irrigation water need (in
  mm/month) for tomatoes, planted 1 February and
  harvested 30 June, is as follows
• The irrigation water need of tomatoes 67 110
  166 195 180 ) 718 mm
• If no data on irrigation water needs are
  available, the estimation method should be used

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Estimation method

• Schedules are given for 3 soil types and 3
  different climates
• Based on calculated crop water needs during peak
  period and root depth (No rains)
• Assumes maximum possible net application depth is
  70 mm
• Climate (See next slide)
• 1 low ETo - lt 4 - 5 mm/day.
• 2 Medium ETo 6 - 7 mm/day.
• 3 high ETo 8 - 9 mm/day

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Step 3. Calculate the number of irrigation
applications over the total growing season

• The number of irrigation applications over the
  total growing season can be obtained by dividing
  the irrigation water need over the growing season
  (Step 2) by the net irrigation depth per
  application (Step 1). Example of Tomato crop
• If the net depth of each irrigation application
  is 40 mm (d net 40 mm Step 1), and the
  irrigation water need over the growing season is
  718 mm (Step 2), then a total of (718/40 ) 18
  applications are required.

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Step 4 Calculate the irrigation interval (INT)
in days

• Now we have the following details
• Total of irrigation application 18 (step3)
• . The total growing season 5 months
• Means 5x30 150 days
• 18 applications for 150 days. Hence the average
  application interval
• 150/18 8.33 days/application
• Rounded up to 8 days interval per
  application (Irrigation schedule!!!)

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Adjusting the Simple Calculation Method for the
Peak Period

• Crop should not suffer from undue water shortage
  in the months of peak irrigation water need
• Given example the interval is 8 days (step 4),
  while the net irrigation depth is 40 mm (step1).
  Thus every 30 days (or each month) 30/8 x 40 mm
  150 mm water is applied. Also using step 2
  details we have

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• Inferences
• The total net amount of irrigation water applied
  (750 mm) sufficient to cover the total irrigation
  water need (718 mm).
• In February and March too much water has been
  applied, while in April, May and June, too little
  water has been applied.
• For tomato the peak period is April, May and
  June which normally coincides with the growth
  stages
• So we can refine the simple calculation method by
  looking only at the months of peak irrigation
  water need and basing the determination of the
  interval on the peak period only

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Example of peak period adjustment

• We consider only the peak period requirement and
  we get from Step 2
• The total irrigation water need from April to
  June (90 days) is 541 mm, while the net
  irrigation depth is 40 mm. Thus 541/40 13.5
  (rounded 14) applications are needed
• 90 days and 14 application. Hence interval
  between application 90/14 6.4 days (or 6 days)

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• Over the total growing period of 150 days, this
  means 150/6 25 applications, each 40 mm net and
  thus in total 25 x 40 1000 mm
• 5 month 1000 mm. 1 month 200 mm
• Water wasted in Feb and March. So reduce it to
  arrive at the follow table

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Summary of peak adjusted intervals

• Feb-March
• d net 40 mmd gross 65 mmInterval 8 days
  (Step 4)
• April-May-June
• d net 40 mmd gross 65 mmInterval 6 days

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Problem

• Determine the irrigation schedule for groundnuts
  grown (duration 130 days) in loamy soil under
  furrow irrigation whose irrigation needs are
• Based on the total growing period.
• Based on the months of peak irrigation water
  need.
• Based on a combination of the two schedules above
  (1 and 2).

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Step 1 Determine the net and gross irrigation
depth (d) in mm

• Table 2 shows that groundnuts have a medium root
  depth. Grown on a loamy soil, the net irrigation
  depth (d net) will thus be approximately 40 mm
  (Table 1).
• The gross irrigation depth (d gross) can be
  calculated using the following formula
• The field application efficiency (ea) is 60 and
  the net irrigation depth (d net) is 40 mm.
• Thus

Back to problem page
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Step 2 Calculate total irrigation water need
(IN)

• Already given

Back to problem page
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Step 3 Calculate the number of irrigation
applications

• The number of applications equals the seasonal
  irrigation water need (Step 2) divided by the net
  irrigation depth (Step 1). Thus the number of
  applications is 527/40 13.2 rounded 13
  applications.

Back to problem page
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Step 4 Calculate the irrigation interval in days

• A total of 13 applications is given during the
  total growing period of 130 days. The interval is
  thus 130/13 10 days.
• IN SUMMARY
• The irrigation schedule for groundnuts, based on
  the total growing period is
• d net 40 mmd gross 65Interval 10 days
• The comparison of the irrigation water required
  (IN) and the irrigation water applied (d net) is
  given below

Back
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2. Irrigation schedule based on months of peak
irrigation water need
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Step 1 Estimate the net and gross depth (d)

• The net and gross depth (d) are calculated in the
  same way as in Answer 1.Thus
• d net 40 mmd gross 65 mm (rounded)

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Step 2 Calculate the irrigation water need at
peak period

• Peak irrigation water need are September and
  October
• Data given in the question
• The months of peak irrigation water need are
  September and October, and during these two
  months the IN (159 170) 329 mm.

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Step 3 Calculate the number of irrigation
applications during the peak months

• The number of applications is 329/40 8.2,
  rounded 8 applications.

Step 4 Calculate the irrigation interval in days
8 applications are given during the peak months
September and October i.e. during 60 days the
interval 60/8 7.5 rounded 7 days.
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Summary of answer

• The irrigation schedule for groundnuts, based on
  the months of peak irrigation water need is
• d net 40 mmd gross 65 mmInterval 7 days
• The comparison of the irrigation water required
  (IN) and the irrigation water applied (d net) is
  given below

Back to question
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3. Based on a combination of the two schedules
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• It is possible to combine the two schedules
  obtained in
• Answer 1
• Answer 2
• Using peak period requirement and answer 1

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Summary of answer

• July, August, November
• d net 40 mmd gross 65 mminterval 10
  days
• September, October
• d net 40 mmd gross 65 mminterval 7 days

Schedules can be determined by trial and error too
Back to question
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Conversion of mm/day into litres/sec.ha

• What is flow required (litres/ second) to cover
  one hectare are for any given required depth of
  water?
• Conversion formula
• 8.64 mm/day 1.0 litre/sec.hectare
• Means if you give 1 litre per second in a day of
  24 hours it will be 8.64 mm rain in one hectare

Or simply use the table
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• Problem
• Determine the continuous water flow when the
  gross irrigation depth is 64 mm and the interval
  is 8 days, for an area of 50 ha.
• 3 minutes

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Adjusting the irrigation schedule to actual
rainfall

• The estimation method determine the irrigation
  schedule can only be used when no significant
  rainfall occurs during the growing season
• The simple calculation method is based on the
  average irrigation water need of the crop which
  is the average crop water need minus the average
  effective rainfall (not actual)
• If the farmer can take water according to his
  requirement (based on rains) how can he adjust?

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Example of Rain adjusted irrigation

• A crop water need (CWN) was 8 am/day and a net
  irrigation depth (d net) of 45 mm.

As soon as the accumulated deficit exceeds the d
net ( 45 mm), irrigation water is supplied. Note
that the "deficit" can never be positive maximum
zero.