Chapter 2 Consumptive use
Topics to be covered in this chapter:
Effective rainfall and Irrigation efficiencies
Consumptive use and Estimation of water requirements
of crops
Delta 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 crop
Re = R - Rr - Dr
where, R = Precipitation, Rr = Surface runoff
and Dr = Deep percolation
Chapter 2 Effective Rainfall
Factors
affecting Re:
Rainfall
characteristics (intensity, frequency and
duration
Land slope
Soil characteristics
Ground water level
Crop 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
Chapter 2 Effective Rainfall
Chapter 2 Effective Rainfall
Factors
affecting Re:
Carryover
of soil moisture (from previous season)
Surface and sub-surface in and out flows
Deep percolation etc
Generally
a percentage of total rainfall is taken
as effective rainfall
Chapter 2 Some definitions
Consumptive
Irrigation
Irrigation Requirement (CIR):
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
Chapter 2 Some definitions
Field
It
Irrigation Requirement (FIR):
is the amount of water required to be applied to
the field
FIR = NIR + water application losses = NIR/Ea
where, Ea = water application efficiency
Chapter 2 Some definitions
Gross
It
Irrigation Requirement (GIR):
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 (c): 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 (a): 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 (s): 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 (u): 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 (d): The effectiveness of irrigation may
also be measured by its water distribution efficiency,
which is defined below:
d = (1 - d/D)
where,
d = 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
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, d =[1-(d/D)]
= [1- (0.168/1.76)] = 0.905 x 100 = 90.5%
Chapter 2 Irrigation Efficiency
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, c = 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.
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
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, a = Output/Input x 100
= (9.6/14.4) x 100 = 66.67%
Chapter 2 Irrigation Efficiency
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, c
= (Water delivered to the fields/Water supplied into
the canal at the head) x100
= (100/130) x 100 = 77%
(b) Water application efficiency a
= (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 106 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 a =
(2460/2880) x 100 = 85.4%
Chapter 2 Irrigation Efficiency
Solution contd:
(c) Water storage efficiency s = (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 s = (2460/2720) x 100
= 90 %
(d) Water distribution efficiency, d = [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, d = [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
Chapter 2 Consumptive use
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
= moisture content at the time of 1st sampling in the ith layer
M2i = moisture content at the time of 2nd sampling in the ith layer
M1i
Chapter 2 Consumptive use
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
Degree
affected by
of saturation of soil surface
Temperature of air and soil
Humidity
Wind velocity
Extent of vegetative cover etc
Chapter 2 Factors affecting CU or ET
Transpiration
affected by
Climate factors:
Temperature
Humidity
Wind
speed
Duration and intensity of light
Atmospheric vapor pressure
Chapter 2 Factors affecting CU or ET
Transpiration
affected by
Soil factors:
Texture
Structure
Moisture
content
Hydraulic conductivity
Chapter 2 Factors affecting CU or ET
Transpiration
affected by
Plant factors:
Efficiency
of root systems in moisture absorption
Leaf
area
Leaf arrangement and structure
Stomatal 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
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 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
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
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.
Chapter 2 Blaney-Criddle Formula
Problem
Month
Monthly
temperature
(C) averaged
over the last 5
years
November
December
January
February
18.0
15.0
13.5
14.5
Monthly
percent of day
time hour of
the year
computed from
the sun-shine
7.20
7.15
7.30
7.10
Useful rainfall
in cm
averaged over
the last 5
years
1.7
1.42
3.01
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
December
January
February
18.0
15.0
13.5
14.5
7.20
7.15
7.30
7.10
1.7
1.42
3.01
2.75
= 8.38
11.6
10.5
10.3
10.3
= 42.7
Chapter 2 Blaney-Criddle Formula
Solution:
Cu = kf = 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/a
= 25.78/0.8 = 32.225 cm
Chapter 2 Blaney-Criddle Formula
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.
Chapter 2 Blaney-Criddle Formula
Problem
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
0.60
Jan
48.8
8.62
0.60
Feb
53.9
9.95
0.65
March
60.0
8.84
0.70
April
62.5
8.86
0.70
May
67.4
9.84
0.75
Chapter 2Month
Blaney-Criddle
Formula
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
0.60
4.54
Jan
48.8
8.62
0.60
6.31
Feb
53.9
9.95
0.65
8.71
March
60.0
8.84
0.70
9.28
April
62.5
8.86
0.70
9.68
May
67.4
9.84
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/a
= 79.45/0.5 = 158.9 cm
c = Conveyance Efficiency = 75% = 0.75
So, Gross Irrigatin Requirement GIR = FIR/c = 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)?
Chapter 2 Some definitions
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
Chapter 2 Some definitions
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
Chapter 2 Some definitions
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 (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
Chapter 2 Some definitions
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
Chapter 2 Some definitions
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
Chapter 2 Some definitions
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 ().
Chapter 2 Duty and 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
Chapter 2 Duty and Delta
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
Chapter 2 Duty and Delta
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
Chapter 2 Duty and Delta
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
Chapter 2 Duty and Delta
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
Chapter 2 Duty and Delta
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
Chapter 2 Duty and Delta
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
Chapter 2 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.
Chapter 2 Irrigation Water
Depth
and frequency of Irrigation
Chapter 2 Irrigation 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
Chapter 2 Irrigation Water
Depth
and frequency of Irrigation
Chapter 2 Irrigation 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
fresh dose of irrigation, as soon as the soil
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
Chapter 2 Irrigation Water
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
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
Chapter 2 Field Capacity
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 = d kN/m3
Dry wt of d m3 of soil = d x d kN
Field Capacity, F = Wt of water retained in unit area
of soil/(d x d)
So, wt of water retained in unit area of soil
= (d x d x F) kN/m2
Chapter 2 Field Capacity
Derivation
So, Volume of water stored in unit area of soil x w
= (d x d x F) kN/m2
Vol of water stored in unit area of soil
= (d x d x F/ w) m
So, total water storage capacity of soil in (m depth of
water) = (d x d x F/ w) m, which is equivalent to
the depth of water stored in the root zone in filling
the soil up to field capacity
Chapter 2 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:
Lets 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, d / w = (d x g)/(w x g) = (d x g)/(w x g)
= 1.3/1 = 1.3 gm/cc [w = 1 gm/cc]
Chapter 2 Field Capacity
Solution contd:
Depth of water stored in root zone between two limits
= (d x d/ w) 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
Chapter 2 Field Capacity
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
=(d x d/ w) x [FC PWP], (since w =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/a = 10.8/0.8 = 13.5 cm
Quantity of water needed at the canal outlet
= FIR/ a = 13.5/0.85 = 15.55 cm
Chapter 2 Field Capacity
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.
Chapter 2 Field Capacity
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
Wheat
Solution
A
contd.
D
Q
Qtota
l
210
00
0.1
259
2
8.1
900
0
0.1
259
2
3.47 11.5
7
210
00
0.1
259
2
8.1
900
0
0.1
259
2
3.47 11.5
7
210
00
0.1
259
2
8.1
900
0
0.1
259
2
3.47 11.5
7
900
0
0.1
259
2
3.47 3.47
900
0
0.15 172
8
5.21 5.21
Sugarcane
Chapter 2 Field Capacity
Rice
Wheat
Solution
Sugarcane
Qtota
contd.
D
360
00
0.25 103
6
34.7
5
34.7
5
360
00
0.3
864
41.6
7
41.6
7
360
00
0.15 172
8
20.8
3
20.8
3
360
00
0.15 172
8
20.8
3
20.8
3
Chapter 2 Field Capacity
Solution
contd.
Sample calc, D = 8.64 x B/ = 8.64x30/0.25
= 1036.8 ha/cumec
Q1 = 36000/1036 = 34.75 cumec
Canal must be designed for the max capacity = 41.67
Design capacity of canal at offtake = 41.67/0.8
= 52.08 cumec (Ans)
