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Current Category » Water Management Including Micro Irrigation

Quantity of Irrigation Water or How Much to Irrigation

The net quantity of water to be applied depends upon magnitude of moisture deficit in the soil, leaching requirement and expectancy of rainfall. When no rainfall is likely to be received and soil is not saline, net quantity of water to be applied is equal to the moisture deficit in the soil i.e. the quantity required to fill the root zone to filed capacity. The moisture deficit (d) in the effective root zone found out by determining the field capacity moisture content and bulk density of each layer.

Problem: Find out the net quantity of irrigation water to be applied to wheat field with the following moisture status.

 Sr. No. Depth of Soil Layer ( cm ) Moisture % on oven dry basis Apparent specific gravity g/cc Field Capacity Actual 1 0 - 15 25.0 16.4 1.39 2 15 - 30 24.0 17.8 1.47 3 30 - 60 22.3 19.2 1.51 4 60 - 90 22.2 20.5 1.53

Solution: Moisture deficit in the different layers will be as follows,
25.0 – 17.8
1. First Layer = ----------------- X 1.39 X 15 = 1.79
100
24.0 – 17.8
2. Second Layer = ----------------- X 1.47 X 15 = 1.36
100
22.3 – 19.2
3. Third Layer = ------------------ X 1.81 X 30 = 1.40
100
22.2 – 20.5
4. Forth Layer = ----------------- X 1.53 X 30 = 0.78
100

Therefore, net quantity of water to be applied is 5.33 cm to fill the root zone to field capacity again.

Requirement of irrigation water:

Units for water measuring:

Water is measured under two conditions. Water at rest measured in units of volume such as liter, cubic meter, hectare meter. Water in motion is expressed in rate off low units such as liters per hour and meters per day.

Liter: One liter is equivalent to 0.22 imperial gallons or 0.0353 cubic feet or 1/1000cubic meters.

Cubic meters: A volume of water equal to that of one cubic meter in length, one meter in breadth and one meter in thickness. One cubic meter of water = one kilo liters or 100 liters or 220 gallons or 35.3 cubic feet or one ton (approx)

Gallon: A gallon is 0.1602 cubic foot. One gallon of water weighs about 10Ib

Cubic foot: A volume of water equal to that of a cube 1 foot in length, 1 foot in breadth and 1 cubic foot in thickness. One cubic foot of water = 28.37 liters or 6.23 gallons or 0.0283 cubic meters or 0.028 ton.

Hectare centimeters: A volume of water necessary to ci\over an area of one hectare (10,000 aq. meters) surface to a depth of one centimeters (1hectare centimeter = 100 cubic meters = 100,000 liters)

Acre inch: the volume of water necessary to cover one-acre (43,560 sq. feet) surface to a depth of one inch. One hectare inch = 3630 cubic feet or 101 ton.

Acre foot: The volume of water necessary to cover one acre to a depth of one foot.
(One acre feet = 43,560 cubic feet)

Cusec: It is the quantity of water flowing at the rate of one cubic foot per second. As one cubic feet of water weighs about 62.4 Ib or 28.37 kg. One cusec of water flowing for one hour is equal to 62.4 Ib X 60 X 60 = 22464 gallons, 101 tons, or one- hectare inch (28.37 liter X 60 X 60 = 101952 liters or one-acre inch.

Duty of water: It denoted the number of acres covered by one causes of water flowing continuously through out the growing season of a crop. It is therefore varies with kind of crop, season, nature of soil, method of irrigation and method of cultivation.

Delta: It is the total depth of water required by a crop.

Problem: A pump with an average discharge of 15 liters/second irrigates one-hectare wheat crop in 12 hours. What is an average depth of irrigation?

Solution:
Discharge in 12 hours = 15 X 60 X 60 X 12
= 648000 liters
= 648 M^3
Volume of water (Cu. m)
Depth of irrigation (cm)             = -------------------------------------- X 100
Area of land (sq. m)

648
=      ---------------      X 100
10,000

= 6.48 cm

Problem: Wheat crop requires 40 cm of irrigation water during 120 days irrigation period. How much land can be irrigation with a flow of 20 liters per second for 12 hours a day?

Solution:
20 X 60 X 60 X 12 X 120
Total discharge during irrigation period = ------------------------------------- M3
1000

= 1, 03,680 M3

40
Irrigation requirement per hectare      =   ------- X 10,000 M3
100

= 1, 03,680 M3

40
Irrigation requirement per hectare      = ----------- X 10,000 M3
100

= 4000 M3

Volume of available water
Area irrigated                    = -------------------------------  X 10,000 M3
Volume of water required/ha (M3)

1, 03,680
=  ---------------
40

= 25.92 hectare land can be irrigated

Current Category » Water Management Including Micro Irrigation