Consumptive Use of irrigation

Chapter 2 – Consumptive use

Topics to be covered in this chapter: Effective rainfall and Irrigation efficienciesConsumptive use and Estimation of water requirements

of cropsDelta and Duty Irrigation Water Field capacity

Chapter 2 – Some definitions

Effective Rainfall (Re):Precipitation falling during the growing period

of a crop that is available to meet the evapotranspiration (ET) needs of the crop is called effective rainfall.It is that part of rainfall which is available to

meet ET that is needed for a cropRe = R - Rr - Dr

where, R = Precipitation, Rr = Surface runoffand Dr = Deep percolation

Chapter 2 – Effective Rainfall

Factors affecting Re:Rainfall characteristics (intensity, frequency and

durationLand slopeSoil characteristicsGround water levelCrop characteristics (ET rate, root depth, stage of

growth, ground cover)Land management practices i.e bunding, terracing

etc which may reduce runoff and increase Re



Factors affecting Re:Carryover of soil moisture (from previous season)Surface and sub-surface in and out flowsDeep 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 evapo-

transpiration needed for crop growthCIR = Cu - Re


Net Irrigation Requirement (NIR):NIR = ETc - Re - Ge - ∆SW

where, ETc = Consumptive use by cropRe = Effective rainfallGe = Ground water contribution∆SW = Stored soil-moisture


Field Irrigation Requirement (FIR): It is the amount of water required to be applied to

the fieldFIR = 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

canalGIR = 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


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


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 efficiencyD = Mean depth of water stored during irrigationd = Average of the absolute values of deviations

from the mean

Chapter 2 – Irrigation Efficiency

Problem 1The 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 efficiencySolution:Mean Depth, D = (2.0+1.9+1.8+1.6+1.5)/5 = 1.76 mValues 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


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


Problem 21.0 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%


Problem 310 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.Solution:Volume of water supplied by 10 cumec of water applied for 4 hours = (10 x 4 x 60 x 60) m3

= 144000 m3 = 14.4 x 104 m3 = 14.4 m x 104 m2

=14.4 hactare-meter


Solution contd:So, Input = 14.4 ha-mOutput = 32 hectares land is storing water upto 0.3 m depthSo, Output = 32 x 0.3 ha-m = 9.6 ha-mWater application efficiency, ƞa = Output/Input x 100= (9.6/14.4) x 100 = 66.67%


Problem 4A 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.


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 100Water supplied to the field during 8 hrs @ 100 liters per sec = 100 x 8 x 60 x 60 liters = 2.88 x 106 liters


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%


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 100Moisture holding capacity of soil = 20 cm per m length x 1.7 m height of root zone = 34 cmMoisture already available in root zone at the time of start of irrigation = (50/100) x 34 = 17 cmAdditional water required in root zone = 34 – 17

= 17 cm


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


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 mSo, water distribution efficiency, ƞd = [1 – (d/D)]

= [1 – (0.30/1.4)] = 0.786 x 100 = 78.6%


Efficiency 100% is not always desirable: becauseOver irrigation may occuredIf leaching is requiredIt is not economical


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 5Design 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

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

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

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 layerM2i = moisture content at the time of 2nd sampling in the ith 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 byDegree of saturation of soil surfaceTemperature of air and soilHumidityWind velocityExtent of vegetative cover etc


Transpiration affected byClimate factors:TemperatureHumidityWind speedDuration and intensity of lightAtmospheric vapor pressure


Transpiration affected bySoil factors:TextureStructureMoisture contentHydraulic conductivity


Transpiration affected byPlant factors:Efficiency of root systems in moisture absorptionLeaf areaLeaf arrangement and structureStomatal behavior

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 bottomLimitations: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 typeNon-weighing percolation typeWeighing 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 throughsoil, the amount of water lost by ET can be estimated

Chapter 2 – Non-weighing constant water table type

Constant water level is maintained by applying water

Effective rainfall (Re) and Irrigation (I) are measured by rain-gauges and calibrated container

Overflow (R) and Deep Percolation (Dr), if any, are measured

ET = I + Re - R – Dr

Re, R, Dr may be zero depending on site condition

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


Where,ET = EvapotranspirationI = Total irrigation water applied (mm)Re = Effective rainfall (mm)Mbi = Moisture content at the begining of the season in the

ith layer of the soilMei = Moisture content at the end of the season in the ith

layer of the soil


Ai = Apparent specific gravity of the ith layer of soilDi = Depth of the ith layer of soil within root zone (mm)n = No of soil layers in the root zoneDr = 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 conditionLimitation: not suitable for a period shorter than 1

monthCu = (k.p)[1.8t + 32]/40, k values from table7.5 (Israelsen)where, Cu = Monthly consumptive use in cm

k = Crop factor, determined by experimentst = Mean monthly temperature in °Cp = Monthly percent of annual day light hours that

occur during the period

Chapter 2 – Blaney-Criddle Formula

Problem 6Wheat 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 time hour 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


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


Solution:Cu = k∑f = 0.8 x 42.7 = 34.16 cmHence, Consumptive use, Cu = 34.16 cmConsumptive Irrigation Requirement, CIR = Cu - Re

= 34.16-8.38 = 25.78 cmField Irrigation Requirement, FIR = CIR/Ƞa

= 25.78/0.8 = 32.225 cm


Problem 7Determine 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


Solution:

Month Temp (°F)

% hrs of Sunshine

Rainfall (cm)

Crop Factor (k)

Cu=kf=kpt/40 (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


Solution:Total consumptive use = 124.09 cmUseful rainfall = 80% of total precipitation

= (0.80 x 55.8) cm = 44.64 cmSo, Net Irrigation Requirement, NIR or CIR = Cu - Re

= 124.09-44.64 = 79.45 cmField Irrigation Requirement, FIR = NIR/Ƞa

= 79.45/0.5 = 158.9 cmȠc = Conveyance Efficiency = 75% = 0.75So, Gross Irrigatin Requirement GIR = FIR/Ƞc = 158.9/0.75 cm

=211.87 cmVolume 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 measuredPan is 1.2 m in diameter, 25 cm deep, and bottom is

raised 15 cm above the ground surfaceDepth 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-efficientEvapotranspiration, 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)?


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


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


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 TAWwhere, f = allowable depletion (in %)

Reference crop Evapotranspiration (ETo): 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 non-restricting 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 harvestedWater 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 PeriodThe time period that elapes from the instant of its

sowing to the instant of its harvesting is called the crop periodThe 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 8If 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 daysTotal depth of water required in 120 days = 10 x 12

= 120 cmSo ∆ for rice = 120 cm


Problem 9If 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 daysTotal depth of water required in 140 days = 7.5 x 5

= 37.5 cmSo ∆ for rice = 37.5 cm


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 dependsType of crop, Climatic condition, Useful rainfall, Type of soil, Efficiency of cultivation method


Importance of DutyIt 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 waterDuty 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 levelledFields should be properly ploughed to the required

depth Improved modern cultivation methods may preferably

be adopted


Precautions in field preparation and sowing:Porous soils should be treated before sowing crops to

reduce seepage of waterManure fertilizers should be added to increase water

holding capacity of the soilPrecautions in handling irrigation supplies:Source of irrigation water should be situated within

the prescribed limitsCanals carrying irrigation supplies should be lined to

reduce seepage and evaporation


Precautions in handling irrigation supplies: Irrigation supplies should be economically used by

proper control on its distributionFree flooding of fields should be avoided and furrow

irrigation method may preferably be adopted, if surface irrigation is resortedSub-surface irrigation and Drip irrigation may be

preferred to ordinary surface irrigation


Relation between Duty and DeltaLet, 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 areaDepth of water applied on this land = Volume/Area

= 86400 B/104 D = 8.64B/D m


Relation between Duty and DeltaBy definition, this total depth of water is called deltaSo, ∆ = 8.64 B/D m = 864 B/D cm

where, ∆ is in cm, B is in days and D in ha/cumec


Problem 10Find 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/cumecSo ∆ = 864 x 120/864 = 120 cm

Chapter 2 – Definition

Cash CropA 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 WaterIn 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


Optimum Utilization of Irrigation Water


Optimum Utilization of Irrigation WaterOptimum 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.


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


Depth and frequency of irrigationIrrigation 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


Depth and frequency of Irrigation


Depth and frequency of irrigationWater 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 fresh dose of irrigation, as soon as the soil moisture reaches the optimum level, as shown in the previous slide


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.


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


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 daysField 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


DerivationField 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, ieField Capacity = (Wt of water contained in a

certain vol of soil/Wt of the same volume of dry soil) x 100


DerivationIf 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 kNField 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


DerivationSo, 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 11After 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/ccEffective depth of root zone = 70 cmDaily Cu of water for the given crop = 12 mm

Solution:Available Moisture = FC – PWP = 28 – 13 = 15%


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]


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 cm1.2 cm of water is utilized by the plant in 1 day10.92 cm of water will be utilized by the plant is

= 1 x 10.92 / 1.2 days = 9.1 days ~ 9 daysHence, after 9 days, water should be supplied to the crop


Problem 12Wheat 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?


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


Solution contd:Hence, 10.8 cm depth of water is the NIRQuantity of water requirement to be supplied to the

field, FIR = NIR/ƞa = 10.8/0.8 = 13.5 cmQuantity of water needed at the canal outlet

= FIR/ ƞa = 13.5/0.85 = 15.55 cm


Problem 13A 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-15cmWheat: Dec to March – 10cmSugarcane: Nov-5cm, Dec to Apr – 10cm, May-15cmWhat should be the design capacity of the main canal near the point of offtake. Assume 20% loss in conveyance.


SolutionCCA = 90000 haArea under Rice = 3/6x90000 = 45000 haArea under Wheat = 2/6x90000 = 30000 haArea under Sugarcane = 1/6x90000 = 15000 ha

Actual area under irrigation areRice = 0.8 x 45000 = 36000 haWheat = 0.7 x 30000 = 21000 haSugarcane = 0.6 x 15000 = 9000 ha


Solution contd.Rice Wheat Sugarcane Qtota

l

A ∆ D Q A ∆ D Q A ∆ D Q

J 21000

0.1 2592

8.1 9000

0.1 2592

3.47 11.57

F 21000

0.1 2592

8.1 9000

0.1 2592

3.47 11.57

M 21000

0.1 2592

8.1 9000

0.1 2592

3.47 11.57

A 9000

0.1 2592

3.47 3.47

M 9000

0.15 1728

5.21 5.21

J 0


Solution contd.Rice Wheat Sugarcane Qtota

l

A ∆ D Q A ∆ D Q A ∆ D Q

J 36000

0.25 1036

34.75

34.75

A 36000

0.3 864 41.67

41.67

S 36000

0.15 1728

20.83

20.83

O 36000

0.15 1728

20.83

20.83

N 0

D 0


Solution contd.Sample calc, D = 8.64 x B/∆ = 8.64x30/0.25

= 1036.8 ha/cumecQ1 = 36000/1036 = 34.75 cumec

Canal must be designed for the max capacity = 41.67Design capacity of canal at offtake = 41.67/0.8

= 52.08 cumec (Ans)

Documents

Consumptive Use of irrigation