The relationship is related to the properties of soil and plants that affect the movement, retention and use of water. A simple analogy: Soil – Water Reservoir Plant Roots – pump with many inlets As the rate of pumping depends on the character of the pump, the rate of extraction of water from the soil by the plant depends on the character of the soil. Soil Water Classification Gravitational water: It is the water in the large pores that moves downward freely under the influence of gravity It drains out so fast that it is not available to the crops. The time of draining out varies from one day in sandy soils to three days in clay soils. Capillary Water: It is the amount of water retained by the soil after gravitational water has drained out. It is the water in the small pores which moves because of capillary forces and is called capillary water. Capillary water is the major source of water available for the plant Hygroscopic Water Soil moisture further reduced by ET until no longer moves because of capillary forces. The remaining water which is held on particle surfaces so tightly is called hygroscopic water. Here, the water is held by adhesive force. And therefore, it is unavailable to the plant. soil water constants Field Capacity (FC) Following saturation when all macro pores are drained by gravity and drainage ceases, usually defined 2 days following saturation by rainfall. Measured as the moisture content at -5 kPa (0.05 bar or 0.5 m tension) Permanent Wilting point (PWP) The point where plants cannot extract any more water – only very small pores are filled with water. Defined as the moisture content at -1500 kPa (15 bar or 150 m tension) Total Available Water Total Available Water (TAW): the water available to crops expressed in mm/m (mm of water per meter depth of soil). TAW = (FC – PWP)*b*Dz Readily Available Water (RAW): This is the level to which the available water in the soil can be used up without causing stress in the crop. For most crops, 50 to 60% of the total available water is taken as readily available. RAW = MAD*TAW Where, MAD = maximum allowable deficit Crop Water Requirement CU is the controlling factor for irrigation scheduling. That is, CU determines the quantity of water to be added by irrigation and helps in day to day management of irrigation systems. Actually, total water demand of crops is made up of: i) Crop water use: includes evaporation and transpiration ii) Leaching requirement: a fraction of water to be added to remove salts from the root zone. iii) Losses of water due to deep seepage in canals and losses due to the inefficiency of application. ETc = Evaporation + Transpiration ETc is normally expressed in mm/day. Factors Affecting ETc: Weather parameters (To, RH, Wind, etc.) Crop Characteristics (type, variety and length of growing period) Management and Environmental aspects (control of diseases, soil salinity, etc.)

1. Points of Discussion:
Chapter 2.0
BASICS IN IRRIGATION ENGINEERING
soil-plant-water Relationship
2.1 Soil-plant-water Relationship
2.2 Crop Water Requirement
2.3 Irrigation Scheduling
2.4 Measurement of Irrigation Water

2. 1. Soil – Plant – Water Relationship
The relationship is related to the properties of soil and plants
that affect the movement, retention and use of water.
A simple analogy:
Soil – Water Reservoir
Plant Roots – pump with many inlets
As the rate of pumping depends on the character of the pump,
the rate of extraction of water from the soil by the plant depends
on the character of the soil.

3. Soil Constituents
Soil system – a complex system

4. Soil Water
 Water is influential in all soil chemical,
physical and biological processes
 Essential for plant survival and
growth
 Essential for chemical transport, etc.
30-20%
WATER
 Soil Water/moisture --- the relative water content in the soil
system.

5. 45%
MINERAL
30-20%
WATER
20-30%
AIR
Zs
Za
Zw
Zt
A
solid
air
water
The 3-Phase Soil Model

6. The 3-Phase Soil Model
Zs
Za
Zw
Zt
A = 1
solid
air
water
s
s
w
w
s
w
m
Z
Z
M
M


 

Gravimetric Water
Content:
t
w
t
w
v
Z
Z
V
V



Volumetric Water
Content:
t
s
s
t
s
b
Z
Z
V
M 
 

Dry Bulk Density
m
b
v 

 

Volumetric Water
Content:

7. Sample problem
A 100 cm3 soil is taken in the field which weighs 174 gm at the
time of sampling. The oven dry weight of the sample is 155 gm.
Assume
density of water as 1gm/cm3. Compute:
a) the soil moisture content
i) in mass basis
ii) in volume basis
b) the soil moisture content for a 120cm deep soil (in depth
basis)
c) the bulk density
The 3-Phase Soil Model

8. Soil Water Classification
Capillary Water
Gravitational Water
Hygroscopic Water
Fully Saturated
Field Capacity
Permanent Wilting Point
Complete dry

9. Gravitational water:
 It is the water in the large pores that moves downward freely under
the influence of gravity
 It drains out so fast that it is not available to the crops.
 The time of draining out varies from one day in sandy soils to three
days in clay soils.
Capillary Water:
 It is the amount of water retained by the soil after gravitational
water has drained out.
 It is the water in the small pores which moves because of capillary
forces and is called capillary water.
 Capillary water is the major source of water available for the plant
Soil Water Classification

10. Hygroscopic Water
 Soil moisture further reduced by ET until no longer moves
because of capillary forces.
 The remaining water which is held on particle surfaces so
tightly is called hygroscopic water.
 Here, the water is held by adhesive force. And therefore, it is
unavailable to the plant.
Soil Water Classification

11. soil water constants
Field Capacity (FC)
 Following saturation when all macro pores are drained by gravity and
drainage ceases, usually defined 2 days following saturation by
rainfall.
 Measured as the moisture content at -5 kPa (0.05 bar or 0.5 m
tension)
Permanent Wilting point (PWP)
 The point where plants cannot extract any more water – only very
small pores are filled with water.
 Defined as the moisture content at -1500 kPa (15 bar or 150 m
tension)

12. Available Water (AW) – the soil water available to the plant
which is defined as: AW= θFC - θPWP
22
5
4
20
29
6
6
12
29
0 5 10 15 20 25 30 35 40 45 50 55
Clay
Loam
Sand
Soil
Type
Volumetric Water Content (%v/v)
Unavailable water Available water Drainage water
Available Water

13. Water Release Characteristic Curve
-1 -10 -100 -1000
clay
sandy loam
pure sand
0.00
30.00
20.00
10.00
50.00
40.00
60.00
Volumetric
water
content,
%
v/v
Soil water potential (kPa)
FC PWP
Available
water
(sand)

14. -1 -10 -100 -1000
clay
sandy loam
pure sand
0.00
30.00
20.00
10.00
50.00
40.00
60.00
Volumetric
water
content,
%
v/v
Soil water potential (kPa)
FC PWP
Available
water
(clay) Water Release Characteristic Curve

15. -1 -10 -100 -1000
clay
sandy loam
pure sand
0.00
30.00
20.00
10.00
50.00
40.00
60.00
Volumetric
water
content,
%
v/v
Soil water potential (kPa)
FC PWP
Available
water
(sandy
loam) Water Release Characteristic Curve

16.  Unique relationship between any
soil and its water content at a
particular suction (or tension)
 General shape is the same for all
soils but the curvature differs from
soil to soil
 Largest pores drain first and air
occupies the space
 Water in the small pores are held
under tension
Volumetric
water
content,
%
v/v
-1
-10 -100 -1000
clay
sandy loam
pure sand
0.00
30.00
20.00
10.00
50.00
40.00
60.00
Soil water potential (kPa)
FC PWP
Soil water Potential - the energy status of Water. i.e. the ease with which the
water is released from the soil or the work required to release water.
Water Release Characteristic Curve

17. Total Available Water
Total Available Water (TAW): the water available to crops
expressed in mm/m (mm of water per meter depth of soil).
TAW = (FC – PWP)*b*Dz
Readily Available Water (RAW):
 This is the level to which the available water in the soil can be
used up without causing stress in the crop.
 For most crops, 50 to 60% of the total available water is taken as
readily available.
RAW = MAD*TAW
Where, MAD = maximum allowable deficit

18. Note: Irrigation application not expected till soil water reaches the PWP.
Total Available Water
TAW
MAD
TAW
NIR
RAW
wp
fc
c
fc

















  z
wp
fc
wp
fc
c
fc
D
NIR
RAW 











 






19. As a rule of thumb, two-third of TAW is easily accessible to plants
and on the average, only three-fourth of the root zone is most
effective. Thus,
RAW or NIR = 2/3 (FC - pwp). ¾ Drz
RAW = ½ (FC - pwp) . Drz
= ½ TAW
= 0.50 TAW
Or, MAD = 0.50 for most crops; NIR = net irrigation requirement
Total Available Water

20. Total Available Water

21. 2.4 Crop Water Requirement
 Irrigation water applied to crop is lost due to Consumptive use
(CU), runoff and deep percolation.
CU = ET loss + water for metabolic activities.
ET = > 99% of total water uptake
Metabolic activities = <1% of total water uptake
 Thus, CU  ETcrop
 CU is the most important single factor whose value must be known
before planning for utilization of available irrigation water.
 Losses due to DP and runoff can be controlled but consumptive is
unavoidable.

22.  CU is the controlling factor for irrigation scheduling.
 That is, CU determines the quantity of water to be added by
irrigation and helps in day to day management of irrigation
systems.
Actually, total water demand of crops is made up of:
i) Crop water use: includes evaporation and transpiration
ii) Leaching requirement: a fraction of water to be added to remove
salts from the root zone.
iii) Losses of water due to deep seepage in canals and losses due to
the inefficiency of application.
2.4 Crop Water Requirement

23. ETcrop (ETc)
 ETc = Evaporation + Transpiration
 ETc is normally expressed in mm/day.
 Factors Affecting ETc:
 Weather parameters (To, RH, Wind, etc.)
 Crop Characteristics (type, variety and
length of growing period)
 Management and Environmental aspects
(control of diseases, soil salinity, etc.)

24. ETcrop – Direct method
 Water balance equation – measurement of input and output
parameters.
ET = I + P – DP –(θf – θi)*Dz
WEIGHING LYSIMETER

25. ETcrop – Indirect method
 Indirect method – using empirical method.
 ETc can be derived from ETo using the equation:
ETc = Kc . ETo
Where,
ETo = reference ET which reflects the effect of climate on ETc
Kc is crop coefficient which reflects the effect of crop on ETc.
 Some common Equations to estimate ETo:
Blaney -Criddle method
Penman Equation
Radiation method
Pan Evaportation method
Penman-Monteith Equation

26. ET crop

27. Potential ET (ETo)
Some common Equations to estimate ETo:
 Blaney -Criddle method
 Penman Equation
 Radiation method
 Pan Evaporation method (fairly accurate)
 Penman-Monteith Equation (most accurate )

28. Types of Pans
ETo = Kp*Ep

29. Class A-Pan
ETo = Kp*Ep

30. Class A-Pan
ETo = Kp*Ep

31. Crop Coefficient, Kc

32. Crop Water Requirement
Example: Calculate the seasonal crop water requirement of maize
[25/35/45/30 (135)] If the average Kc is 0.65 and ETo=7mm/day.
Solution:

33. 2.5 Irrigation Scheduling
How much to irrigate?
Answer: As much as the plants have used
since the previous irrigation.
RAW = NIR = MAD*TAW
How often to irrigate?
Answer: Often enough to prevent the
plants suffering from drought.
Irrigation frequency, f = NIR/ETc
Note: Actual irrigation water = gross irrigation water, GIR = NIR/Ea

34. 2.5 Irrigation Scheduling

35. 2.5 Irrigation Scheduling

36. Example 1. How much water must be added to a field of area 3ha
to increase the volumetric water content of the top 40cm from 16%
to 28%? Assume all water added to the field stays in the top 40cm.
Ans: GIR 1440m3
Example 2. Compute the depth and frequency of irrigation required
for a certain crop with data given below.
Crop : wheat Average daily CU: 1.2cm
Drz: 1.20m Application efficiency: 75%
FC: 17% PWP: 5%
Bulk density: 1.72gm/cm3
Ans: d = 24.77cm; f = 10days
2.5 Irrigation Scheduling

37.  Irrigation interval … the number of days between two successive
irrigation applications.
 Irrigation cycle … the number of days allowed to complete one
irrigation.
Irrigation interval vs. Irrigation period
Example: If the calculated irrigation interval is 7 days and if an irrigated area is
divided into 6 sub-areas to be irrigated in shift.
The irrigation cycle is 6days.
A1 A2 A3 A4 A5 A6

38.  Design capacity --- the flow rate determined by the water
requirement, irrigation time, irrigation period and the irrigation
application efficiency.
 It is the flow rate required at the water supply source or that
required to determine channel x-sections.
Design capacity/Estimation of Discharge
a
E
H
F
n
d
A
c
Q
.
.
.

Where,
Qc is the Desired Design Capacity;
d is the Net Irrigation Depth = Readily Available Moisture;
F is the number of days required to complete the Irrigation
H is the number of hours the System is operated (hrs/day) and
Ea is the Irrigation Efficiency

39. Example: A 12-hectare field is to be irrigated with a sprinkler system. The root zone
depth is 0.9m and the field capacity of the soil is 28% while the permanent wilting
point is 17% by weight. The soil bulk density is 1.36 g/cc and the water application
efficiency is 70%. The soil is to be irrigated when 50% of the available water has
depleted. The peak evapotranspiration is 5.0mm/day and the system is to be run for
10 hours in a day.
Determine:
(i) The net irrigation depth
(ii) Gross irrigation depth
(iii) Irrigation period
(iv) Area to be irrigated per day and
(v) the required system capacity.
Ans:
i) 67.5mm; ii) 96.4mm; iii)f =13days;iv)1ha/day;
V)Qc = 89m3/hr or 25 lit/sec
Design capacity/Estimation of Discharge

40. Base, Delta and Duty
Base Period, B ... The time between first watering and last watering.
B = length of growing period
Delta,  … total depth of water required in a growing period.
Duty, D … relates volume of irrigation water and area to be
irrigated. It refers to the area irrigated for 24hrs per unit discharge
in a base period B.
 
Area
time
e
Disch
Area
Volume 


 arg
D
B
864


 
4
10
24
60
60
1






 D
B
Where,
 = delta in cm,
B = base period in days, and
D = duty in ha/m3/s

41. Base, Delta and Duty
Example 1. If a crop requires 20cm of water every 5days. Compute
the discharge required to irrigate 3000ha.
Solution:
Here, delta = 20cm; B = 5days
Using the relation, Delta = 864xB/D, D = 864*5/20 = 216 ha/cumec
Therefore, the required discharge to irrigate 3ha = 3000ha/483.84
= 13.89 cumec

42. Base, Delta and Duty
Example 2. A channel is to be designed for irrigating 5000ha. The
water requirement of the crop to be irrigated is 25cm every 2weeks.
Determine the discharge of the channel for which it is to be designed.
Solution: Here, delta = 25cm; B = 2weeks = 14days
Using the relation, Delta = 864xB/D,
D = 864*14/25 = 483.84 ha/cumec
Area = 5000ha
Required discharge of channel = 5000ha/483.84 = 10.33 cumec
So, the channel is to be designed for the maximum discharge of
10.33 cumec

43. Other Irrigation project related terms
Gross Command Area (G.C.A) --- the total area included in an
irrigation project (both cultivated and uncultivated).
Culturable Command area (CCA) --- cultivable part of the GCA.
Unculturable Command area --- uncultivable part of the GCA.
Intensity of Irrigation (I) --- a ratio of the cultivated area to the
CCA
100


CCA
CA
I

44. Base, Delta and Duty
Problem 1. The total command area of an irrigation project is 15,000ha,
where 750ha is uncultivable. The area covered with crop A is 6000ha and
that of crop B is 4000ha. The duty of crop A is 3000ha/cumec and the duty
of crop B is 4000ha/cumec. Find a) the design discharge of a channel
assuming 10% transmission loss; b) intensity of irrigation
Ans: a) QA = 2.2cumec and QB = 1.1cumec; b) IA= 42.1%; IB=28.07%
Problem 2. The command area of a channel is 4000ha. The intensity of
irrigation of a crop is 70%. The crop requires 60cm of water in 15 days,
when the effective rainfall is recorded as 15cm during that period. Assume
total losses as 15% and find:
a) the duty at the head of the field
b) the duty at the head of the canal
c) Discharge at the head of the canal

45. Solution
 Verified Answer The command area of a channel is 4000 hectares.
The intensity of irrigation.
Depth of water = 60cm
Effective rainfall = 15 cm
Depth of irrigating water = 60 – 15 = 45 cm ⇒ Delta = 45 cm = 0.45 m
B = 15 days
From relation, ∆= 8.64×B/D
⇒ Duty(D) = 8.64×15/0.45 = 288ha/cumec
Due to loss of water, duty at head of channel is reduced
Here losses are 15%
So, duty at head of channel = 288 × 85/100 = 244.80 ha/cumecs

46. Quiz
Q1. The gross commanded area of an irrigation project is 6000ha, 80%
of which is culturable irrigable.
The intensity of irrigation for winter season is 50% and that for autumn
season is 25%.
If the average duty at head of the distributary is 2000ha/cumecs for
winter season and 1000ha/cumecs for autumn season,
find out the discharge required.

47. 2.6 Measurement of irrigation water
Irrigation Amount … expressed in depths of water (usually mm or cm).
It refers to the volume of water per unit command area. i.e.
Where,
d = equivalent depth of water in a soil layer
L = depth (thickness) of the soil layer
L
A
AL
d v
v





48. 2.6 Measurement of irrigation water
Discharge/flow rate … expressed in units of m3/sec or lit/sec.
Discharge in small canals may be estimated by:
 Volume method
 Velocity-area method (e.g. float method)
Float method

49. Direct method of measuring Discharge -Weirs
2
/
3
)
2
.
0
(
0184
.
0 H
H
L
Q 

2
/
3
0186
.
0 LH
Q
2
/
5
0138
.
0 H
Q
Rectangular Weir
Triangular Weir
Trapezoidal Weir

50. Direct method of measuring Discharge -Weirs

51. Direct method of measuring Discharge -Flumes
Parshall flume
Cutthroat flume