Lecture 6 ce 1005 Irrigation and flood control by Rabindra Ranjan Saha, PEng, WUB

1
Method -2
Hargreaves Class A Pan evaporation method :
Evapotranspiration (consumptive use) is related to pan
evaporation by a constant K, called consumptive use
coefficient. The coefficients used in agriculture modeling to
estimate evaporation and transpiration in crop irrigation .
i.e.
Evapotranspiration (Et or Cu)
K =
Pan evaporation (Ep)
i.e. Et or Cu = K . Ep
Consumptive use coefficient K is different for different crops
and is also different for the same crop of different places.
Lecture-6

2
Computation of Consumptive Water-Use Coefficients for the
Great Lakes Basin and Climatically Similar Areas
Consumptive-use coefficient (%) = (Water consumed ÷ Water
withdrawn) × 100
Class A Pan evaporation ( Ep ) measurements
Evaporation, Ep can be measured experimentally directly
measuring the quantity of water evaporated from the standard
class Pan. This pan is 1.2 m in diameter, 25 cm deep and bottom
is raised 15 cm above the ground surface.
Presentation-6(contd.)

3
The depth of water is to be kept in a fixed range such that the
water surface is at least 5 cm and never more than 7.5 cm,
below the top of the pan. Take the reading after each one hour
and then calculate the average rate of evaporation.
25 cm
15 cm EGL
5cm
Class –A Pan
1.20 m dia
Evaporation by sunlight
Lecture-6(contd.)

4
Example 2-6
(a) Determine the pan evaporation from the following data for the
month of April, using Christiansen method.
Latitude: 150 19′ N, Elevation + 449 meters
Mean temperature = 31.80 C
Mean wind velocity at 0.5 m above the ground 183 km per day
Mean relative humidity = 40%
Mean sunshine percent = 89 %
Use tables for extra- terrestrial radiation
(b) What is the consumptive use for the Month of April for a crop
having a consumptive use co-efficient equal to 0.80.

5
Solution:
Given,
Latitude: 150 19′ N, Elevation + 449 meters
Mean temperature = 31.80 C
Mean wind velocity at 0.5 m above the ground 183 km per day
Mean relative humidity = 40%
Mean sunshine percent = 89 %
From the extra-terrestrial table-1 the radiation (R) for Latitude:
150 19 ‫׳‬N, is 47.30 cm for the month of April
(a) As we know,
Pan Evaporation equation, Ep = 0.459 R .Ct Cw.Ch . Cs. Ce
Using Christiansen method, as follows:
Ct = Co efficient for temperature in 0C, and is given by
= 0.393 + 0.02796 Tc + 0.0001189 Tc
2
= 0.393 + 0.02796 x 31.8 + 0.0001189x (31.8)2
= 1.403
Lecture-6(contd.)

6
Cw = Co efficient for wind velocity and is given as
=0.708 +0.0034 W -0000038 W2
= 0.708 + 0.0034 183 – 0.0000038 1832 = 1.200
Ch = Co efficient for relative humidity and is given by
= 1.250-0.0087H +0.75× 104×H2 - 0.85 × 10-8 × H4
= 1.250 – 0.0087 ×40+ 0.75 ×104×402 - 0.85 × 10-8 × 404
= 1.000
Cs = Co efficient for percent of possible sunshine and is given by
= 0. 542 + 0.008 × S– 0.78 × 10-4 xS2 + 0.62 × 10-6 ×S3
= 0. 542 + 0.008 × 89 – 0.78 × 10-4 x892 + 0.62 × 10-6 ×893
= 1.073
Ce = Co efficient of elevation and is given by
= 0.97 + 0.00984E = 0.97 + 0.00984 ×449/100
= 1.014

7
(a) Pan evaporation (Ep )
We know,
Ep = 0.459 R .Ct Cw.Ch . Cs. Ce
Putting the value of coefficients in the above equation
we get
Ep = 0.459 ×47.30 ×1.403 ×1.20 × 1.00 × 1.073×1.014
= 39.80 cm Ans.

8
(b)Consumptive use (Evapotranspiration )
Et = K × Ep , where,
Et = Consumptive use
Ep = Pan Evaporation
K = crop coefficient = 0.80 (given)
Now putting the respective values in the above equation
Et = 0.80 × 39.80 = 31.84 cm
Hence required value of consumptive use = 31.84 cm.
Lecture-6(contd.)

9
Table 2.2 Slope of the saturation vapor pressure vs temperature curve
at the mean air temperature in mm of mercury per 0C
Temperature 0C Saturation vapor pressure (ew)
in mm of Hg
A
(mm /0C )
15 12.79 0.80
17.5 15.00 0.85
20 17.54 1.05
22.5 20.44 1.24
25 23.76 1.40
27.5 27.54 1.61
30 31.82 1.85
32.5 36.68 2.07
35 42.81 2.35
37. 48.36 2.62
40 55.32 2.95
45 71.20 3.66
Table-2.3 : Mean monthly solar radiation at top of atmosphere, Ha in mm of
evaporable water/day
Latitudes
in degrees
Jan Feb March April May June July Aug Sept Oct Nov Dec.
North
0 14.5 15.0 15.2 14.7 13.9 13.4 13.5 14.2 14.9 15.0 14.6 14.3
10 12.8 13.9 14.8 15.2 15.0 14.8 14.8 15.0 14.9 14.1 13.1 12.4
20 10.8 12.3 13.9 15.2 15.7 15.8 15.7 15.3 14.4 12.9 11.2 10.3
30 8.5 10.5 12.7 14.8 16.0 16.5 16.2 15.3 13.5 11.3 9.1 7.9
40 6.0 8.3 11.0 13.9 15.9 16.7 16.3 14.8 12.2 9.3 6.7 5.4
50 3.6 8.0 9.1 12.7 15.4 16.7 16.1 13.9 10.5 7.1 4.3 3.0

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Table 2.4
Mean monthly values of possible sunshine hours, N
North
latitude in
degree
Jan Feb March April May June July Aug Sept Oct Nov Dec.
0 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1
10 11.6 11.8 12.1 12.4 12.6 12.7 12.6 12.4 12.9 11.9 11.7 11.5
20 11.1 11.5 12.0 12.6 13.1 13.3 13.2 12.8 12.3 11.7 11.2 10.9
30 10.4 11.1 12.0 12.9 13.7 14.1 13.9 13.2 12.4 11.5 10.6 10.2
40 9.6 10.7 11.9 13.2 14.4 15.0 14.7 13.8 12.5 11.2 10.0 9.4
50 8.6 10.1 11.8 13.8 15.4 16.4 16.0 14.5 12.7 10.8 9.1 8.1
Table-1
Extra –terrestrial radiation for different latitudes
Latitudes in
degrees -
North
Jan Feb March April May June July
45 15.621 19.990 31.953 41.072 50.317 52.146 52.426
40 19.609 23.393 35.027 42.936 50.003 52.146 52.730
35 23.546 26.670 37.871 44.450 51384 51.918 52.730
30 27.407 29.794 40.411 45.669 51.460 51.384 52.451
25 31.140 32.690 42.647 46.558 51.206 50.571 51.841
20 34.722 35.382 44.552 47.117 50.597 49.428 50.902
15 38.100 37.821 46.126 47.320 49.657 47.980 49.657
10 41.250 40.005 47.346 47.168 48.388 46.253 48.082
5 44.120 41.885 48.209 46.660 46.787 44.221 46.178
0 46.736 43.485 48.692 45.822 44.822 41.910 43.993

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Lecture-6(contd.)
Estimation of Evapotrnaspiration
Crop water need = potential evapotranspiration – actual
evapotranspiration
Evapotranspiration = { precipitation + irrigation input – runoff
– increase in soil storage – groundwater loss}
Potential Evapotranspiration(PET)
The measure of the ability of the atmosphere to remove water from
the surface through the process of evaporation and transpiration
assuming no control on water supply is called PET
PET does not depend on soil and plant factors but depends on
climatic factors.

12
Lecture-6(contd.)
Penman’s equation
AHn + Ea ʏ
PET =
A + ʏ
where, PET = Potential Evapotranspiration in mm per day
A = Slope of the saturation vapor pressure vs
temperature curve at the mean air temperature
in mm of mercury per 0 C (table 3.3)
Ha = Net radiation in mm of evaporable water per day
Ea = parameter including wind velocity and saturation
deficit
γ = Psychrometric constant = 0.49 mm of mercury 0C

13
The net radiation Hn may be calculated by the following equation :
Hn = Ha (1-r) [ a + b(n/N)] - б T a
4 ( 0.56 – 0.092 √ea ) [ 0.10 + 0.90 (n/N)]
where,
Ha = Incident solar radiation outside the atmosphere in mm of
evaporable water per day
a = A constant depending upon the latitude and Ø = 0.29 cos Ø
b = A constant with an average value of 0.52
n = Actual duration of bright sunshine in hours
N = Maximum possible hours of bright sunshine
r = Reflection coefficient (albedo). Usual ranges of values of r
given in table below-
Lecture-6(contd.)
Surface ranges of r values
Close ground crops 0.15 – 0.25
Bare lands 0.05 – 0.45
Water surface 0.05
Snow 0.45 – 0.95

14
Lecture-6(contd.)
The parameter Ea is estimated as
Ea = 0.35 [{ 1 + ( u2 / 160)} { ew – ea} ]
where , u2 = mean wind speed at 2 m above ground in km/day
ew = saturation vapor pressure at mean air temperature
in mm of mercury
ea = actual mean vapor pressure in the air in mm of Hg
б = Stefan – Boltzman constant = 2.01 x 10-9 mm/day
Ta = mean air temperature in degrees Kelvin = 273 + 0C
ea = actual mean vapor pressure in the air in mm of mercury.

15
Lecture-6(contd.)
Example 2-7
Estimate the daily evapotranspiration for the month of
November from a lake situated in New Delhi with the data given
below :
latitude - : 280 4′ N
Elevation : 230 m above sea level
Mean monthly temperature : 190C
Mean relative humidity : 75%
Mean observed sunshine hours : 9 h
Wind velocity at 2 m height : 85 km/day
Nature of surface cover : Close-ground green crop
Solution :
Estimation of Potential Evapotranspiration (PET)
We know, from Penman’s Equation
PET = ( AHn + Ea ʏ) / (A + ʏ )

16
Lecture-6(contd.)
Use Table 2-2,2.3, and 2.4 for different values
From table : 2-2 - A = 1.00mm/0C
(November) ew = 16.50 mm of Hg
From table – 2.3 - Ha = 9.506 mm of water/day
From Table- 2-4 - N = 10.716 h
Mean observed sunshine hours, n/N = 9 / 10.716 = 0.84
From the given data unknown values are calculated:-
ea = ew x relative humidity
ea = 16.50 × 0.75 = 12.38 mm of Hg
a = constant depending on latitude, = 0.29 Cos Ø
= 0.29 cos 280 4′ = 0.2559
b = a constant, average value = 0.52

17
б = Stefan – Boltzman constant = 2.01 × 10-9 mm/day
Ta = Mean air temperature in degrees Kelvin
= 273 + 0C
Ta = 273 + 19 = 292 kelvin
б T a
4 = 2.01 × 10-9 × 2924 = 14.613
r = Albedo for close-ground green crop is taken from
table above as 0.25
Lecture-6(contd.)

18
Lecture-6(contd.)
Penman’s Equation:
Hn = Ha (1-r) [ a + b(n/N)] - б T a
4 ( 0.56 – 0.092 √ea ) [ 0.10 + 0.90 (n/N)]
Putting the respective values in the above equation-
Hn = 9.506(1-0.25) × (0.2559 + (0.52 ×0.84)) – 14.613 × ( 0.56 – 0.092
√ 12.38) × (0.10 + (0.90×0.84))
Hn = 1.990 mm of water /day
Parameter including the wind velocity and saturation deficit-
Ea = 0.35 1 + { u2 / 160} { ew – ea}
Similarly putting respective values in this equation:
Ea = 0.35 [1 + { 85 / 160}] [{16.5 – 12.38}]=2.208 mm/day

19
Lecture-6(contd.)
We know , Potential Evapotranspiration
PET = (A Hn + Ea ʏ )/ (A + ʏ)
Where, ʏ = Phychrometric constant=0.49 mm of Hg./0C
Putting the values of A , Hn , Ea and ʏ in the above
equation
PET = (1× 1.990 + 2.208 × 0.49)/ (1+ 0.49)
PET = 2.06 mm/day

20
Lecture-6(contd.)
Example 2-8
A reservoir with a surface area of 250 hectares had the following
average values of parameters during a week: Water temperature =
200C , Relative humidity = 40% ; wind velocity at 1.0 m above ground
level = 16 km/h. Estimate the average daily evaporation from the lake
and the volume of water evaporated from the lake during that one
week.
Solution:
Given,
Surface area = 250 hectares
Temperature = 200C
Relative humidity = 40% = 0.40
Wind velocity at 1.0 m above the ground level = 16 km/h
Estimate Daily evaporation (= EL) and volume of
evaporation in 7 days.

21
Lecture-6(contd.)
We know,
EL = KM [( ew – ea) {1 + (u9 / 16 )}]
where,
KM = a coefficient
= 0.36 for large deep waters and 0.50 for small,
shallow waters (Mayer’s coefficient)
ew = saturated vapor pressure
ea = actual vapor pressure = ew x relative humidity
u9 = the wind velocity at a height of 9.0 m above ground level
uh = C (h) 1/7
where, C = wind velocity at 1 meter above ground level (u1)
h = height in meter above the ground level
u9 = u1 (9) 1/7 = 16.0 (9) 1/7 = 21.90 km/h

22
Lecture-6(contd.)
From Table2-3:
ew = 17.54 mm of mercury(Hg)
ea = 0.40 × 17.54 = 7.02 mm of Hg
Putting the respective values in the above equation
EL = KM [( ew – ea) ] [1 + (u9 / 16 )]
EL = 0.36 [( 17.54 – 7.02) ] [1 + (21.90 / 16 )]
EL = 8.97 mm/day
Volume of evaporation for 7 days
= (7 × 8.97 × 250 × 10,000) /1000 =156,975 m3

23
Lecture-6(contd.)
Methods of Irrigation
Broadly four (4) methods
(1) Surface irrigation
(2) Sub-surface irrigation
(3) Sprinkler irrigation
(4) Trickle or Drip irrigation
(1) Surface irrigation
Where water is applied and distributed over the soil surface
by gravity. It is by far the most common form of irrigation
throughout the world and has been practiced in many areas
virtually unchanged for thousands of years.
Traditional Method of irrigation

24
Photograph of Surface irrigation
Lecture-6(contd.)

25
Lecture-6(contd.)
Types of Surface irrigation method : Five (5) types
a. Uncontrolled (or wild or free) flooding method
b. Border strip method
c. Check method
d. Basin method
e. Furrow method

26
Lecture-6(contd.)
(a) Uncontrolled /Wild or Free Flooding:
Flooding method of irrigation is in use since centuries.
Flooding method consists in applying the water by flooding
the land of rather smooth and flat topography. In modern
irrigation practice several flooding methods have been
developed. In free flooding method water is applied to the
land from field ditches without any check or guidance to the
flow. The method is very clear from Fig. 6.1.
From the main or field ditch, laterals are taken across the
fields at a spacing of 15 m to 45 m. The laterals run along the
contours. The water flows out through the openings in
laterals to flood the field. On the other side of the field a
drain ditch is provided to take away excess water. To divert
the water from the main to the laterals generally earthen
dams are used. Sometimes steel shutters may also be used.

27
Lecture-6(contd.)
Advantages of Free flooding method
(1) Useful for newly established farm
(2) Cheap and can be successfully used where water supply is plenty
(3) Suited for irregular surface
(4) No land area is utilized for water distribution
(5) Labor requirement is mnimum

28
Disadvantages
i. It is the most inefficient methods of irrigation as only
about 20 per cent of water is actually used by plants and the
rest is being lost as run off, seepage and evaporation
ii. Leveling of land increases cost of cultivation.
iii. This method is unsuitable for crops that are sensitive to
water logging.
iv. Crop growth is not uniform as the water distribution by
this method is very uneven.
v. There is a possibility of soil erosion.
Lecture-6(contd.)

29
Lecture-6(contd.)
Photograph of level basin flood irrigation
Photograph of Gated pipe supply system

30
Lecture-6(contd.)
(b) Border Strip Method:
The field is divided into number of strips. The width of strip varies
from 10 to 15 meters and length varies from 90 m to 400 m.
Strips are separated by low embankments or levees. The
arrangement is as shown in Fig. 6.2.
Levees

31
Lecture-6(contd.)
The water is diverted from the field channel into the strips. The
water flows slowly towards lower end, wetting the soil as it
advances. The surface between two embankments should
essentially be level. It helps in covering the entire width of the
strip. There is a general surface slope from opening to the lower
end. The surface slope from 2 to 4 m/1000 m is best suited.
When the slope is steeper, special arrangement is made to
prevent erosion of soil. In this method it is possible to maintain
more discharges successfully. The discharge may vary from
0.015 to 0.30 cumec depending upon the kind of soil, nature of
crop, size of strips etc.
This method is suitable on the fields where soil is sufficiently
capable to absorb the water.
This method requires high initial cost.

32
Lecture-6(contd.)
Relationship between the discharge (Q) though the
supply ditch and the time required in border irrigation
method:
t = 2.3 [(y/f )log 10 { Q/(Q – fA)}]
Where, Q = Discharge through the supply ditch m3/h
y = Depth of water flowing over the border strip in m
f = Rate of infiltration capacity of soil in m/h
A = Area of land strip to be irrigated in m2
t = Time in hour required to cover the given area A
Again maximum area that can be irrigated is given by the
following equation-
Amax = Q / f

33
Lecture-6(contd.)
Example 3-1
Determine the time required to irrigate a strip of land of 0.04
hectares in area from a tube – well with a discharge of 0.02 cumec.
The infiltration capacity of the soil may be taken as 5 cm/ hr and
the average depth of flow on the field as 10 cm. Also calculate the
maximum area that can be irrigated from this tube- well.
Solution
Given,
Area of strip of land (A) = 0.04 hectare = 0.04 ×10000 m2
f = Infiltration capacity= 5 cm/hr = 0.05 m/hr
Discharge (Q) = 0.02 m3/s = 0.02 x 60x60 m3 /hr = 72 m3 /hr
Average depth of flow, y = 10 cm = 0.10 m
To be determined time t to irrigate a strip of land
And maximum area (Amax)

34
Lecture-6(contd.)
(i) Determination of time
We know , time required to irrigate the land is as
t = 2.3 [(y/f )log 10 {Q/(Q – fA)}]
Now putting the respective given values in this equation
t = 2.3 [(0.10/0.05 )log 10 {72/(72 – 0.05×400)}]
t = 0.65 hr. = 39 minutes
(ii) Calculation of maximum area
We know,
Amax = Q / f = (72 m3 /hr) / (0.05 m/hr) = 1440 m2

Lecture 6 ce 1005 Irrigation and flood control by Rabindra Ranjan Saha, PEng, WUB

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Lecture 6 ce 1005 Irrigation and flood control by Rabindra Ranjan Saha, PEng, WUB