Consumptive Use of irrigation

Chapter 2 – Con ons sumptive us use 

Topics to be covered in this chapter:  Ef f ecti ectiv ve rain rainff all all and and Irriga Irrigatition on eff ef f icien icienc cies  Cons onsumptive us use an and Estimation of wat water requ equiremen ementts of crops  Delta and Duty  Irrigat Irrigation ion Water Water  Field ield capacity

Chapter 2 – Som ome e defi def inition ons s 

Ef f ect ective Rainf all (Re): Precipitation falling during the growing period of a crop that is available to meet the evapot evapotrranspiration (ET) need needs s of the crop crop is called effective rainfall. It is that part of rainfall which is available to meet ET that is needed for a crop Re = R - Rr - Dr wher where e, R = Preci ecipitation, on, Rr = Surface runoff and Dr = Deep percolation


Ef f ect ective Rainf all (Re): Precipitation falling during the growing period of a crop that is available to meet the evapot evapotrranspiration (ET) need needs s of the crop crop is called effective rainfall. It is that part of rainfall which is available to meet ET that is needed for a crop Re = R - Rr - Dr wher where e, R = Preci ecipitation, on, Rr = Surface runoff and Dr = Deep percolation

Chapter 2 – Ef f ec ecttive Rainf all 

Factors affecting Re: Rainf all cha characteri eristics (in (intens ensity, fr f requ equenc ency and duration  Land slope  Soil characteristics  Ground water level characteri eristics (ET rate, root root dept epth, st stage of  Crop cha growth, ground cover)  Land management practices i.e bunding, terracing etc etc whi which may redu educe ru runoff of f and increas ease Re 

Chapter 2 – Ef f ec ecttive Rainf all

Chapter 2 – Ef f ec ecttive Rainf all 

Factors affecting Re: Carryover of soil oil mois oisture (fr (f rom previou evious s seas eason) on)  Surface and sub-surface in and out flows  Deep percolation etc 



Generally a percentage of total rainfall is taken as effective rainfall


Consumptive Irrigation Requirement (CIR): Irrigation water required in order to meet the evapotranspiration needed for crop growth  CIR = Cu - Re 

Chapter 2 – Some definitions 

Net Irrigation Requirement (NIR): 

NIR = ETc - Re - Ge - ∆SW

where, ETc = Consumptive use by crop Re = Effective rainfall Ge = Ground water contribution ∆SW = Stored soil-moisture

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Field Irrigation Requirement (FIR): It is the amount of water required to be applied to the field  FIR = NIR + water application losses = NIR/ Ea  where, Ea = water application efficiency 


Gross Irrigation Requirement (GIR): It is the amount of water required at the head of a canal  GIR = FIR + conveyance loss= FIR/ Ec  Where, Ec = conveyance efficiency 

Chapter 2 – Irrigation Efficiencies Efficiency of water conveyance (ƞc): It is the ratio of the water delivered into the fields from the outlet point of the channel, to the water pumped into the channel at the starting point  Efficiency of water application (ƞa): It is the ratio of the quantity of water stored into the root zone of the crops to the quantity of water actually delivered into the field 

Chapter 2 – Irrigation Efficiencies Efficiency of water storage (ƞs): It is the ratio of the water stored in the root zone during irrigation to the water needed in the root zone prior to irrigation  Efficiency of water use (ƞu): It is the ratio of the water beneficially used including leaching water, to the quantity of water delivered 

Chapter 2 – Irrigation Efficiencies 

Uniformity coefficient or water distribution efficiency (ƞd): The effectiveness of irrigation may also be measured by its water distribution efficiency, which is defined below: ƞd = (1 - d/D)

where, ƞd = water distribution efficiency

D = Mean depth of water stored during irrigation d = Average of the absolute values of deviations from the mean

Chapter 2 – Irrigation Efficiency 

Problem 1

The depths of penetrations/ waters along the length of a border strip at points 30 meters apart were measured. Their values are 2.0, 1.9, 1.8, 1.6 and 1.5 meters. Compute the distribution efficiency Solution: Mean Depth, D = (2.0+1.9+1.8+1.6+1.5)/ 5 = 1.76 m Values of deviations from the mean are (2.0-1.76), (1.9-1.76), (1.8-1.76), (1.6-1.76), (1.5-1.76) = 0.24, 0.14, 0.04, -0.16, -0.26

Chapter 2 – Irrigation Efficiency Solution contd: Absolute values of these deviations fromt he mean are 0.24, 0.14, 0.04, 0.16, and 0.26 Average of these absolute values of deviations from the mean, d = (0.24+0.14+0.04+0.16+0.26)/ 5 = 0.168 m So water distribution efficiency, ƞd =[1-(d/ D)] = [1- (0.168/ 1.76)] = 0.905 x 100 = 90.5%


Problem 2 1.0 cumec of water is pumped into a farm distribution system, 0.8 cumec is delivered to a turn-out, 0.9 kilometer from the well. Compute the conveyance efficiency. Solution: By definition, ƞc = Output/input x 100 = 0.8/1.0 x 100 = 80%

Chapter 2 – Irrigation Efficiency Problem 3 10 cumec of water is delivered to a 32 hectare field, for 4 hours. Soil probing after the irrigation indicates that 0.3 meter of water has been stored in the root zone. Compute the water application efficiency.

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Solution: Volume of water supplied by 10 cumec of water applied for 4 hours = (10 x 4 x 60 x 60) m 3 = 144000 m3 = 14.4 x 104 m3 = 14.4 m x 104 m2 =14.4 hactare-meter

Chapter 2 – Irrigation Efficiency Solution contd: So, Input = 14.4 ha-m Output = 32 hectares land is storing water upto 0.3 m depth So, Output = 32 x 0.3 ha-m = 9.6 ha-m Water application efficiency, ƞa = Output/ Input x 100 = (9.6/ 14.4) x 100 = 66.67%


Problem 4 A stream of 130 liters per sec was diverted from a canal and 100 liters per sec were delivered to the field. An area of 1.6 ha was irrigated in 8 hrs. Effective depth of root zone was 1.7 m. Runoff loss in the field was 420 m3. Depth of water penetration varied linearly from 1.7 m at the head end of the field to 1.1 m at the tail end. Available moisture holding capacity of the soil is 20 cm per meter depth of soil. It is required to determine the (a) water conveyance efficiency; (b) water application efficiency; (c) water storage efficiency and (d) water distribution efficiency. Irrigation was started at a moisture extraction level of 50% of the available moisture.

Chapter 2 – Irrigation Efficiency Solution: (a) Water conveyance efficiency, ƞc = (Water delivered to the fields/ Water supplied into the canal at the head) x100 = (100/ 130) x 100 = 77% (b) Water application efficiency ƞa = (Water stored in the root zone during irrigation/ Water delivered to the field) x 100 Water supplied to the field during 8 hrs @ 100 liters per sec = 100 x 8 x 60 x 60 liters = 2.88 x 10 6 liters

Chapter 2 – Irrigation Efficiency Solution contd: = 2.88 x 106 / 103 m3 = 2880 m3 Runoff loss in the field = 420 m3 So, Water stored in the root zone = 2880 – 420 m3 = 2460 m3 Hence, Water application efficiency ƞa = (2460/ 2880) x 100 = 85.4%

Chapter 2 – Irrigation Efficiency Solution contd: (c) Water storage efficiency ƞs = (Water stored in the root zone during irrigation/Water needed in the root zone prior to irrigation) x 100 Moisture holding capacity of soil = 20 cm per m length x 1.7 m height of root zone = 34 cm Moisture already available in root zone at the time of start of irrigation = (50/100) x 34 = 17 cm Additional water required in root zone = 34 – 17 = 17 cm

Chapter 2 – Irrigation Efficiency Solution contd: Amount of water required in root zone = Depth x plot area = (17/ 100) m x (1.6 x 104) m2 = 2720 m3 But actual water stored in root zone = 2460 m3 So water storage efficiency ƞs = (2460/2720) x 100 = 90 % (d) Water distribution efficiency, ƞd = [1 – (d/D)]

Mean depth of water stored in the root zone, D = (1.7+1.1)/2 = 1.4 m

Chapter 2 – Irrigation Efficiency Solution contd: Average of the absolute values of deviations from the mean, d = [(1.7-1.4)+(1.1-1.4)]/ 2 = (0.3+0.3)/ 2 = 0.3 m So, water distribution efficiency, ƞd = [1 – (d/D)] = [1 – (0.30/1.4)] = 0.786 x 100 = 78.6%

Chapter 2 – Irrigation Efficiencies 

Efficiency 100% is not always desirable: because Over irrigation may occured  If leaching is required  It is not economical 

Chapter 2 – Consumptive use 

Consumptive use (Cu) means crop water requirement Cu = ET + water used for tissue building = 99% + 1%  So, Cu ≈ ET 

   

Estimation of Irrigation water requirement Crop Water Requirement = ETcrop Net Irrigation Requirement, NIR= ETcrop – Re – Ge – ∆SW Losses: Conveyance loss  Field channel loss  Water application loss 


GIR: 



 

At the outlet point (FIR): need to include field channel loss and water application loss At the diversion point (GIR): need to include all losses

FIR= NIR+ water losses in field channel and field = NIR/ Ea GIR = FIR + Conveyance loss = FIR/ Ec

Chapter 2 – Irrigation requirement 

Problem 5 Design water requirement at the outlet of canal diversion. Assume Ea = 0.7. Area = 1 ha. Month

ETcrop(cm)

Re (cm)

Nov

2.4

0.4

Dec

5.89

1.6

Jan

6.86

3.2

Feb

13.86

2.2

Chapter 2 – Irrigation requirement Solution Month

ETcrop(cm)

Re (cm)

NIR = ET – Re (cm)

FIR= NIR/Ea (cm)

Nov

2.4

0.4

2.0

2.86

Dec

5.89

1.6

4.29

6.13

Jan

6.86

3.2

3.66

5.23

Feb

13.86

2.2

11.66

16.66

Feb is govern, so, water required = [(1x104) x (16.66x10-2) / (28x24x60x60)] x 103 = 0.7 l/ s

Assume, conveyance efficiency = 80% GIR = 0.7/ 0.8 = 0.875 l/ s

Chapter 2 – Soil Moisture Depletion Study

Soil is sampled 2 to 4 days after irrigation and again 7 to 15 days later or just before next irrigation  Only those sampling periods are considered in which rainfall is light. This is done to minimize drainage and percolation errors 

Chapter 2 – Soil Moisture Depletion Study Depth of ground water should be such that it will not influence the soil moisture fluctuation within the root zone  Cannot be applied where water table is high 

M1i = moisture content at the time of 1st sampling in the ith layer th  M2i = moisture content at the time of 2nd sampling in the i layer 


Definition: Consumptive use (CU) or Evapotranspiration (ET) is the sum of two terms (a) Transpiration: Water entering plant roots and used to build plant tissue or being passed through leaves of the plant into the atmosphere (b) Evaporation: Water evaporating from adjacent soil, water surfaces and surfaces of leaves of the plant or intercepted precipitation

Chapter 2 – Factors affecting CU or ET 

Evaporation affected by Degree of saturation of soil surface  Temperature of air and soil  Humidity  Wind velocity  Extent of vegetative cover etc 


Transpiration affected by Climate factors: Temperature  Humidity  Wind speed  Duration and intensity of light  Atmospheric vapor pressure 


Transpiration affected by Soil factors: Texture  Structure  Moisture content  Hydraulic conductivity 


Transpiration affected by Plant factors: Efficiency of root systems in moisture absorption  Leaf area  Leaf arrangement and structure  Stomatal behavior 

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Stomata: These are pores in the leaf that allows gas exchange where water vapor leaves the plant and CO2 enters

Chapter 2 – Direct Measurement of CU/ ET (a) Tank or Lysimeter experiments: Lysimeter is a device that isolates a volume of soil or earth between the soil surface and a given depth and includes a percolating water sampling system at its bottom Limitations: Reproductin of physical conditions such as temperature, water table, soil texture, density etc is very difficult

Chapter 2 – Type of Lysimeters 

There are types of lysimeters: Non-weighing constant water table type  Non-weighing percolation type  Weighing type 

In non-weighing lysimeters, changes in water balance are measured volumetrically weekly or biweekly, no accurate daily estimates can be done  Weighing lysimeters can provide precise information on soil moisture changes for daily or even hourly 

Chapter 2 – Non-weighing constant water table type By recording amount of rainfall and amount lost through soil, the amount of water lost by ETcan be estimated

Chapter 2 – Non-weighing constant water table type Constant water level is maintained by applying water ffective rainfall (Re) and Irrigation (I) are  E measured by rain-gauges and calibrated container  Overflow (R) and Deep Percolation (D ), if any, are r measured T = I + Re - R – Dr  E  R , R, D may be zero depending on site condition e r 

Chapter 2 – Non-weighing percolation type

Chapter 2 – Non-weighing percolation type Consumptive Use (CU) is computed by adding measured quantities of irrigation water, the effective rainfall received during the season and the contribtion of moisture from the soil  Applicable for areas having high precipitation  Special arrangements are made to drain and measure the water percolating through the soil mass 

Chapter 2 – Non-weighing percolation type

Where, ET = Evapotranspiration I = Total irrigation water applied (mm) Re = Effective rainfall (mm) Mbi = Moisture content at the begining of the season in the ith layer of the soil Mei = Moisture content at the end of the season in the ith layer of the soil

Chapter 2 – Non-weighing percolation type Ai = Apparent specific gravity of the ith layer of soil Di = Depth of the ith layer of soil within root zone (mm) n = No of soil layers in the root zone Dr = Deep Percolation Apparent Sp Gr: is the ratio of the weight of a volume of a substance to the weight of an equal volume of a reference substance (ie water), or ratio of densities, and is a dimensionless quantity

Chapter 2 – Weighing type ET is determined by taking the weight of the tank and making adjustment for any rain  Provides most accurate data for short time periods 

Chapter 2 – ET using Empirical Equations 

(a) Blaney-Criddle Formula: It is used extensively  It gives good estimates of seasonal water needs under arid condition or initial condition  Limitation: not suitable for a period shorter than 1 month 

Cu = (k.p)[1.8t + 32]/ 40, k values from table7.5 (Israelsen) where, Cu = Monthly consumptive use in cm k = Crop factor, determined by experiments t = Mean monthly temperature in °C p = Monthly percent of annual day light hours that occur during the period

Chapter 2 – Blaney-Criddle Formula 

Problem 6

Wheat has to be grown at a certain place, the useful climatological conditions of which are tabulated below. Determine the evapo-transpiration and consumptive irrigation requirement of wheat. Also, determine the field irrigation requirement if the water application efficiency is 80%. Use Blaney-Criddle equation. Assume 0.8 as crop factor.


Problem 6

Month

Monthly temperature (°C) averaged over the last 5 years

Monthly percent of day timehour of the year computed from the sun-shine

Useful rainfall in cm averaged over the last 5 years

November

18.0

7.20

1.7

December

15.0

7.15

1.42

January

13.5

7.30

3.01

February

14.5

7.10

2.75

Chapter 2 – Blaney-Criddle Formula Solution: Blaney-Criddle Eq, Month

t (°C)

p (hr) Re (cm)

f = p/40(1.8t+32) cm

November

18.0

7.20

1.7

11.6

December

15.0

7.15

1.42

10.5

January

13.5

7.30

3.01

10.3

February

14.5

7.10

2.75

10.3

∑ = 8.38 ∑ = 42.7

Chapter 2 – Blaney-Criddle Formula Solution: Cu = k∑f = 0.8 x 42.7 = 34.16 cm Hence, Consumptive use, Cu = 34.16 cm Consumptive Irrigation Requirement, CIR = Cu - Re = 34.16-8.38 = 25.78 cm Field Irrigation Requirement, FIR = CIR/ Ƞa = 25.78/ 0.8 = 32.225 cm


Problem 7 Determine the volume of water required to be diverted from the head works to irrigate area of 5000 ha using the data given in the table below. Assume 80% as the effective precipitation to take care of the consumptive use of the crop. Also assume 50% efficiency of water application in the field and 75% as the conveyance efficiency of canal.


Problem 7

Month

Temp (°F)

% hrs of sunshine

Rainfall (mm)

Crop factor (k)

June

70.8

9.90

75

0.80

July

74.4

10.20

108

0.85

August

72.8

9.60

130

0.85

Sept

71.6

8.40

115

0.85

Oct

69.3

7.86

105

0.65

Nov

55.2

7.25

25

0.65

Dec

47.1

6.42

0

0.60

Jan

48.8

8.62

0

0.60

Feb

53.9

9.95

0

0.65

March

60.0

8.84

0

0.70

April

62.5

8.86

0

0.70

May

67.4

9.84

0

0.75

Chapter 2Mont –h Blaney-Criddle F ormula Temp % hrs of Rainfall Crop Factor C =kf=kpt/40 u

Solution:

(°F)

Sunshine (cm)

(k)

(cm)

(1)

(2)

(3)

(4)

(5)

(6)

June

70.8

9.90

7.5

0.80

14.02

July

74.4

10.20

10.8

0.85

16.13

August

72.8

9.60

13.0

0.85

14.85

Sept

71.6

8.40

11.5

0.85

12.78

Oct

69.3

7.86

10.5

0.65

8.85

Nov

55.2

7.25

2.5

0.65

6.5

Dec

47.1

6.42

0

0.60

4.54

Jan

48.8

8.62

0

0.60

6.31

Feb

53.9

9.95

0

0.65

8.71

March

60.0

8.84

0

0.70

9.28

April

62.5

8.86

0

0.70

9.68

May

67.4

9.84

0

0.75

12.44

55.8

124.09

Chapter 2 – Blaney-Criddle Formula Solution: Total consumptive use = 124.09 cm Useful rainfall = 80% of total precipitation = (0.80 x 55.8) cm = 44.64 cm So, Net Irrigation Requirement, NIR or CIR = Cu - Re = 124.09-44.64 = 79.45 cm Field Irrigation Requirement, FIR = NIR/ Ƞa = 79.45/ 0.5 = 158.9 cm Ƞc = Conveyance Efficiency = 75% = 0.75 So, Gross Irrigatin Requirement GIR = FIR/ Ƞc = 158.9/ 0.75 cm =211.87 cm Volume of water required for 5000 ha area = (211.87/ 100)x(5000x104) = 105.93 x 106 m3

Chapter 2 – Measurement of Evaporation 

(b) Hargreaves class A pan evaporation method: Quantity of water (Ep) evaporated from the standard class A evaporation pan is measured  Pan is 1.2 m in diameter, 25 cm deep, and bottom is raised 15 cm above the ground surface  Depth of water is maintained such that the water surface is atleast 5 cm, and never more than 7.5 cm, below the top of the pan 

Chapter 2 – Class A pan evaporimeter

Chapter 2 – Class A pan evaporimeter Evapotranspiration is related to pan evaporation by a constant k, called consumptive use co-efficient  Evapotranspiration, ET or Cu = k x Ep (Pan Evaporation)  Consumptive use co-efficient, k varies with crop type; crop growth etc 

Chapter 2 – Irrigation Scheduling 

Irrigation schedule is a decision making process involving:  When to irrigate?  How much water to apply each time?  How to apply (method of irrigation)?


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Gross Command Area (GCA): It is the surface area which can be brought under the irrigation command of a canal ie the total area within the extreme limits set for irrigation Culturable Command Area (CCA): It is the gross command area less the area of unculturable land (eg rocky, marshy, ponds, roads etc) included in the GCA






Net Command Area (NCA): CCA – Culturable land where irrigation cannot be provided due to limitation of sources Intensity of Irrigation: It is the ratio of the actual area irrigated to the CCA Crop Ratio: Ratio between different crop area to be irrigated during a year


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Available water (AW): Water contained in the soil between FC (Field Capacity) and PWP (Permanent Wilting Point) is known as the available water Total Available Water (TAW): Amount of water that is available for plants in root zone is known as Total Available Water (TAW). It is the difference in volumetric moisture content at FC and that at PWP, multiplied by root zone depth

Chapter 2 – Some definitions Management Allowable Depletion (MAD): MAD is the degree, to which water in the soil is allowed to be depleted by management decision and expressed as, MAD = f x TAW where, f = allowable depletion (in %)  Reference crop Evapotranspiration (ET o): The rate of evapotranspiration from an extensive surface of 8~15 cm tall, green grass cover of uniform height, actively growing in the ground and not short of water is known as ETo 




Crop Evapotranspiration (ETc): The depth of water need to meet the water loss through evapotranspiration of a disease free crop, growing in large fields under nonrestricting soil conditions including water and fertility and achieving full production potential under the given growing environment Crop co-efficient (kc): The ratio of crop evapotranspiration (ETc) to the reference evapotranspiration (ETo) is called Crop co-efficient (kc). so, kc = ETc / ETo


Water Requirements of Crops:  Water requirements of a crop mean the total quantity and the way in which a crop requires water from the time it is sown to the time it is harvested  Water requirements depend on: water table, crop type, ground slope, intensity of irrigation, method of application of water, place/ location, climate condition, type of soil, method of cultivation and useful rainfall


Crop Period or Base Period  The time period that elapes from the instant of its sowing to the instant of its harvesting is called the crop period  The time between the first watering of a crop at the time of its sowing to its last watering before harvesting is called the base period

Chapter 2 – Duty and Delta 

Delta: Total quantity of water required by a crop for its full growth may be expressed in ha-m or as depth. Simply, it is the depth of water to be applied over a given area for the given a base period (ie the time interval from planting to harvesting of a crop). This depth of water is called delta (∆).


Problem 8 If rice requires about 10 cm depth of water at an average interval of about 10 days, and the crop period for rice is 120 days, find out the delta for rice. Solution: No of watering required = 120/ 10 = 12 days Total depth of water required in 120 days = 10 x 12 = 120 cm So ∆ for rice = 120 cm


Problem 9 If wheat requires about 7.5 cm of water after every 28 days, and the base period for wheat is 140 days, find out the value of delta for wheat. Solution: No of watering required = 140/ 28 = 5 days Total depth of water required in 140 days = 7.5 x 5 = 37.5 cm So ∆ for rice = 37.5 cm

Chapter 2 – Duty and Delta Duty: It may defined as the number of hectares of land irrigated for full growth of a given crop by supply of 1 m3 / s of water continuously during the entire base of that crop. Simply we can say that, the area (in ha) of land can be irrigated for a crop period, B (in days) using one cubic meter or water  Factors on which duty depends Type of crop, Climatic condition, Useful rainfall, Type of soil, Efficiency of cultivation method 


Importance of Duty It helps us in designing an efficient canal irrigation system. Knowing the total available water at the head of a main canal, and the overall duty for all the crops required to be irrigated in different seasons of the year, the area which can be irrigated can be worked out. Inversely, if we know the crops area required to be irrigated and their duties, we can work out the discharge required for designing the channel


Measures for improving duty of water Duty of canal water can certainly be improved by selecting economy in use of water by considering the following precautions and practives: Precautions in field preparation and sowing:  Land to be used for cultivation should, as far as possible, be levelled  Fields should be properly ploughed to the required depth  Improved modern cultivation methods may preferably be adopted

Chapter 2 – Duty and Delta Precautions in field preparation and sowing:  Porous soils should be treated before sowing crops to reduce seepage of water  Manure fertilizers should be added to increase water holding capacity of the soil Precautions in handling irrigation supplies:  Source of irrigation water should be situated within the prescribed limits  Canals carrying irrigation supplies should be lined to reduce seepage and evaporation

Chapter 2 – Duty and Delta Precautions in handling irrigation supplies:  Irrigation supplies should be economically used by proper control on its distribution  Free flooding of fields should be avoided and furrow irrigation method may preferably be adopted, if surface irrigation is resorted  Sub-surface irrigation and Drip irrigation may be preferred to ordinary surface irrigation


Relation between Duty and Delta Let, there be a crop of base period B days and 1 m3 / s of water is applied to this crop on the field for B days. Now, volume of water applied to this crop during B days, V = (1x60x60x24xB) m3 = 86400 B m3 This quantity of water (V) matures D hectares of land or 104 D m2 of area Depth of water applied on this land = Volume/ Area = 86400 B/ 104 D = 8.64B/ D m


Relation between Duty and Delta By definition, this total depth of water is called delta So, ∆ = 8.64 B/D m = 864 B/D cm where, ∆ is in cm, B is in days and D in ha/cumec


Problem 10 Find the delta for a crop when its duty is 864 ha/ cumec on the field, base period of this crop is 120 days. Solution: Here, B = 120 days, D = 864 ha/ cumec So ∆ = 864 x 120/864 = 120 cm

Chapter 2 – Definition 

Cash Crop A cash crop may be defined as a crop which has to be en-cashed in the market for processing as it cannot be consumed directly by the cultivators. All non food crops are thus included in cash crops. Examples: Jute, Tea, Cotton, Tobacco etc.

Chapter 2 – Irrigation Water Optimum Utilization of Irrigation Water In an identical situation, yield is going to vary with the application of different quantities of water. Yield increases with water, reaches maximum value and then falls down. Quantity of



water at which yield is maximum, is called the optimum water depth

Chapter 2 – Irrigation Water 

Optimum Utilization of Irrigation Water


Optimum Utilization of Irrigation Water Optimum utilization of irrigation generally means, getting max yield with any amount of water.  Supplies of water to varies crops should be adjusted in such a fashion, as to get optimum benefit ratio, not only for the efficient use of available water and max yield, but also to prevent water-logging of the land.  To achieve economy in water use, it is necessary that farmers be acquainted with the fact only a certain fixed amount of water gives best results. 

Chapter 2 – Irrigation Water Estimating depth and frequency of irrigation on the basis of soil moisture regime concept Water or soil moisture is consumed by plants through their roots. It therefore becomes necessary that sufficient moisture remains available in the soil from the surface to the root zone depth.




Depth and frequency of Irrigation

Chapt apte er 2 – Irr rrigat igation ion Water Depth and frequency of irrigation Irrigation water should be supplied as soon as the moisture falls up to this optimum level and its quantity should be sufficient to bring the moisture content up to its field capacity, considering water application losses



Chapt apte er 2 – Irr rrigat igation ion Water 

Depth and and f requen requenc cy of Irriga Irrigatition on

Chapt apte er 2 – Irr rrigat igation ion Water Depth and frequency of irrigation Water will be utilized by the plants after the fresh irrigation dose is given, and soil moisture will start falling. It will again be recouped by a f resh esh dose ose of irrigat gation, on, as as soon as the soi soill moisture reaches the optimum level, as shown in the previous slide



Chapter 2 – Irrigation Water Permanent Wilting Point (PWP): It is that water content at which plant can no longer extract sufficient water for its growth, and wilts up. It is the point at which permanent wilting of plants take place.  Available Moisture (AM): It can be defined as the difference in water content of the soil between field capacity and permanent wilting point. 

Chapter 2 – Irrigation Water Readily available moisture: It is that portion of the available moisture which is most easily extracted by the plants, and is approximately 75 to 80% of the available moisture

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Field Capacity(FC):

Immediately after a rain or irrigation water application, when all the gravity water drained down to the water table, a certain amount of water is retained within the surfaces of soil grains by molecular attraction and by loose bonds (ie adsorption). This water cannot be easily drained under the action of gravity is called FC.  So Field Capacity is the water content of a soil after free drainage has taken place for a sufficient period. 

Chapter 2 – Field Capacity Period of free gravity drainage is generally 2 to 5 days  Field capacity water further consists of two parts:  One part is that which is attached to the soil molecules by surface tension against gravitation forces, and can be extracted by plants by capillarity. This water is called capillary water.  Other part is that which is attached to the soil molecules by loose chemicals bonds. This water which cannot be removed by capillarity is not available to the plants, and is called hygroscopic water 

Chapter 2 – Field Capacity 

Derivation

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Field capacity water (ie the quantity of water which any soil can retain against gravity) is expressed as the ratio of the weight of water contained in the soil to the weight of the dry soil retaining that water, ie

Field Capacity = (Wt of water contained in a certain vol of soil/ Wt of the same volume of dry soil) x 100

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Derivation

If we consider 1 m2 area of soil and d meter depth of root zone, The volume of soil = d x 1 = d m3 If the dry unit wt of soil = γd kN/ m3 Dry wt of d m3 of soil = γd x d kN Field Capacity, F = Wt of water retained in unit area of soil/ (γd x d) So, wt of water retained in unit area of soil = (γd x d x F) kN/ m2


Derivation

So, Volume of water stored in unit area of soil x γw = (γd x d x F) kN/ m2 Vol of water stored in unit area of soil = (γd x d x F/ γw) m So, total water storage capacity of soil in (m depth of water) = (γd x d x F/ γw) m, which is equivalent to the depth of water stored in the root zone in filling the soil up to field capacity


Problem 11 After how many days will you supply water to soil in order to ensure sufficient irrigation of the given crop, if Field capacity of soil = 28% Permanent Wilting Point = 13% Dry density of soil = 1.3 gm/ cc Effective depth of root zone = 70 cm Daily Cu of water for the given crop = 12 mm Solution: Available Moisture = FC – PWP = 28 – 13 = 15%

Chapter 2 – Field Capacity Solution contd: Let’s assume that the readily available moisture or optimum soil moisture level is 80% of available moisture ie Readily available moisture = 0.80 x 15% = 12% So, Optimum moisture = 28 – 12 = 16% Which means that the moisture will be filled by irrigation between 16% and 28% Now, γd / γw = (ρd x g)/(ρw x g) = (ρd x g)/(ρw x g) = 1.3/1 = 1.3 gm/cc [ρw = 1 gm/cc]

Chapter 2 – Field Capacity Solution contd: Depth of water stored in root zone between two limits = (γd x d/ γw) x [FC – OMC] = 1.3 x 0.7 x [0.28 – 0.16] = 0.1092 m = 10.92 cm Hence, water available for ET = 10.92 cm 1.2 cm of water is utilized by the plant in 1 day 10.92 cm of water will be utilized by the plant is = 1 x 10.92 / 1.2 days = 9.1 days ~ 9 days Hence, after 9 days, water should be supplied to the crop


Problem 12 Wheat is to be grown in a field having a field capacity equal to 27% and permanent wilting point is 13%. Find the storage capacity in 80 cm depth of the soil, if the dry unit weight of the soil is 14.72 kN/ m3. If irrigation water is to be supplied when the average soil moisture falls to 18%, find the water depth required to be supplied to the field if the field application efficiency is 80%. What is the amount of water needed at the canal outlet if the water loss in the water-courses and the field channels is 15% of the outlet discharge?

Chapter 2 – Field Capacity Solution: Max storage capacity or available moisture =(γd x d/ γw) x [FC – PWP], (since ρw =9.81 kN/ m3) = (14.72 x 0.8/ 9.81) x [0.27 – 0.13] = 0.168 m = 16.8 cm Since the moisture is allowed to vary between 27% and 18%, deficiency created in this fall =(14.72 x 0.8/ 9.81) x [0.27 – 0.18] =0.108 m = 10.8 cm

Chapter 2 – Field Capacity Solution contd: Hence, 10.8 cm depth of water is the NIR Quantity of water requirement to be supplied to the field, FIR = NIR/ ƞa = 10.8/ 0.8 = 13.5 cm Quantity of water needed at the canal outlet = FIR/ ƞa = 13.5/ 0.85 = 15.55 cm


Problem 13 A CCA of 90000 ha has to be irrigated by a proposed reservoir with the help of a canal system. Crops to be grown during a year are rice, wheat and sugarcane with a crop ratio 3:2:1. Intensities of irrigation are rice-80%, wheat-70% and sugarcane-60%. FIRs are as follows: Rice: July-25cm, Aug-30cm, Sept-15cm, Oct-15cm Wheat: Dec to March – 10cm Sugarcane: Nov-5cm, Dec to Apr – 10cm, May-15cm What should be the design capacity of the main canal near the point of offtake. Assume 20% loss in conveyance.


Solution CCA = 90000 ha Area under Rice = 3/ 6x90000 = 45000 ha Area under Wheat = 2/ 6x90000 = 30000 ha Area under Sugarcane = 1/ 6x90000 = 15000 ha Actual area under irrigation are Rice = 0.8 x 45000 = 36000 ha Wheat = 0.7 x 30000 = 21000 ha Sugarcane = 0.6 x 15000 = 9000 ha

Chapter 2 – Field Capacity Rice 

Solution contd. ∆ A D Q

Wheat

Sugarcane

Qtota l

A

∆

D

Q

A

∆

D

Q

J

210 00

0.1

259 2

8.1

900 0

0.1

259 2

3.47 11.5 7

F

210 00

0.1

259 2

8.1

900 0

0.1

259 2

3.47 11.5 7

M

210 00

0.1

259 2

8.1

900 0

0.1

259 2

3.47 11.5 7

A

900 0

0.1

259 2

3.47 3.47

M

900 0

0.15 172 8

5.21 5.21

J

0

Chapter 2 – Field Capacity Rice 

Wheat

Sugarcane

Qtota

Solution contd. D

Q

l

A

∆

A

J

360 00

0.25 103 6

34.7 5

34.7 5

A

360 00

0.3

864

41.6 7

41.6 7

S

360 00

0.15 172 8

20.8 3

20.8 3

O

360 00

0.15 172 8

20.8 3

20.8 3

∆

D

Q

A

∆

D

Q

N

0

D

0

Consumptive Use of irrigation

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