Furrow is a long, narrow irrigation trench made in the ground used for an optimal supply of water.
Furrows can be level and are very similar to long narrow basins. However, a minimum grade of 0.05% is recommended so that effective drainage can occur following irrigation or excessive rainfall.
Important points:
The maximum recommended furrow slope is 0.5% to avoid soil erosion. Furrows can be set when the mainland slope does not exceed 3%. Beyond this, there is a major risk of soil erosion following a breach in the furrow system. On steep land, terraces can also be constructed and furrows cultivated along the terraces.Calculation:
Given,
60 % of area during Kharif remained without water
i.e. Area got irrigated during Kharif = (100 – 60) % = 40 %
Area during Rabi remained without water = 46 %
i.e. Area got irrigated during Rabi = (100 – 46) % = 54 %
∴ Intensity of Irrigation (I.O.I) is the area of land irrigated
∴ I.O.I = 40% + 54 % = 94 %
Based on Alignment Canals are classified into 3 categories. These are:
1. Ridge Canal, 2. Contour Canal and 3. Side Slope Canal
Their characteristics are given below:
Ridge Canal ( Watershed canal) |
Contour Canal ( Single Bank Canal) |
Side Slope Canal |
Aligned along the ridge or natural watershed Line |
Aligned along the natural contour of the country |
Aligned perpendicular to the contour of the country. |
No Cross-Drainage work required |
Maximum cross drainage work is required |
No Cross Drainage work required. |
Can irrigates on both sides of the ridge and hence, a large area can be cultivated |
Can irrigate areas only on one side |
Can irrigate areas only on one side |
Concept:
For an irrigation land:
SC = Saturation capacity, FC = Field capacity, OMC = Optimum moisture content, PWP = Permanent welting point and UWP = Ultimate welting point
1. Equivalent depth of water held at field capacity (x) = S × d × Fc
2. Equivalent depth of water held at PWP (x’) = S × d × (PWP)
3. Available moisture/storage capacity of soil (y) = S × d × (Fc - PWP)
4. Readily available moisture content (dw) = S × d × (Fc - OMC)
Given:
Field capacity (F. C) = 25 % = 0.25
Permanent willing point (PWP) = 15 % = 0.15
Depth of the root zone (d) = 80 cm
Dry Unit weight of soil (γ_{d}) = 1.5 g/cc
Calculations:
The storage capacity \(= \frac{{{\gamma _d} \times d \times \left( {F.C - PWP} \right)}}{{{\gamma _w}}}\)
The storage capacity \(= \frac{{1.5 \times 80 \times \left( {0.25 - 0.15} \right)}}{1} = 12\;cm\)
Hence, the storage capacity of the soil is 12 cm.
Explanation:
Weirs are commonly used for the measurement of open channel flow rate.
Suppressed weir: A rectangular weir whose notch or opening sides are coincident with the sides of the approach channel is known as a Suppressed weir. In this type of Weir crest length (l) is equal to the width of the channel.
If the crest length of the weir is less than the width of the channel, it is known as a Contracted weir.
Note:
There are different shapes of sharp-crested weir but the most commonly used are
Concept:
Water distribution efficiency \(= {{\rm{\eta }}_{\rm{d}}} = \left[ {1 - \frac{{\rm{y}}}{{{{\rm{d}}_{\rm{w}}}}}} \right] \times 100{\rm{\;}}\left( {{\rm{in\;\% }}} \right)\)
\({{\rm{d}}_{\rm{w}}} = \frac{{{{\rm{d}}_1} + {{\rm{d}}_2} + \ldots + {{\rm{d}}_{\rm{n}}}}}{{\rm{n}}},\) d_{w} = average depth of water stored
\({\rm{y}} = \frac{{y_1 \ + \ y_2\ + \ ......\ +\ y_n}}{n}\), y = Average of absolute value of deviations
y_{1}, y_{2},......., y_{n} = absolute deviations from average depth of water stored
Calculation:
\({\rm{d_w}} = \frac{{1.5 + 1.8 + 2.1}}{3} = 1.8\)
_{y}_{1} = 0.3 m, y_{2} = 1.8 – 1.8 = 0, y_{3} = 2.1 – 1.8 = 0.3 m
\({\rm{y}} = \frac{{0.3 + 0.3 + 0}}{3} = 0.2\)
\({{\rm{\eta }}_{\rm{d}}} = \left( {1 - \frac{{0.2}}{{1.8}}} \right) = \frac{8}{9} = 0.8889\)Concept:
As per Lacey’s theory-
Hydraulic radius (R) is given by,
\(R =\frac{5}{2}\frac{{{{\rm{V}}^2}}}{{\rm{f}}}\)
Where,
V is the flow velocity
f is silt factor
Calculation:
Given,
v = 1.5 m/s, f = 1
Hydraulic mean radius,
\(R =\frac{5}{2} \times \frac{{{{\rm{V}}^2}}}{{\rm{f}}} = \frac{5}{2} \times \frac{{{1.5^2}}}{1}\)
= 5.625 m
Important Points
Following relations are used in Lacey's Theory
The velocity of flow is given by V = \({\left( {\frac{{Q{f^2}}}{{140}}} \right)^{\frac{1}{6}}}\)
Where,
f = silt factor = 1.76√d and d = average size of soil particles in mm.
Q is discharge , Area of the section is given by A = Q/V
Perimeter is given by, P = 4.75√Q
Bed slope(S), is given by S = \(\frac{{{{\rm{F}}^{\frac{5}{3}}}}}{{3340{{\rm{Q}}^{\frac{1}{6}}}}}\)
Duty:
The duty means the area of land that can be irrigated with the unit volume of irrigation water or it is the area of land expressed in hectare that can be irrigated with unit discharge i.e. 1 m3/s flowing throughout the base period, expressed in days. It is expressed as ha/m3/s or ha/cumec.
∴ \(\text{Duty}=\frac{\text{Area}}{\text{Discharge}}\)
From above it can be concluded that
1. Duty is more for more area if discharge is constant.
2. Duty is less if discharge is more for a given area.
Note:
Mathematically, Duty is given as:
\(D = \frac{{8.64B}}{{\rm{\Delta }}}\)
Where, D = Duty (in ha/m3/s), B = Base period (in days), Δ = Amount of water required by plant (in m)
Calculation:
Delta, \(\Delta ~=\frac{2\times 400\times {{10}^{6}}}{40000\times {{10}^{4}}}=2~m\)
Duty in hectares / cumec, \(D = \frac{{8.64\times125}}{{\rm{2 }}} = 540\; ha/cumec\)
Temporary spurs or bunds are temporary structures constructed every year after floods. A bund is a structure made to project flow from a riverbank into a stream or river with the aim of deflecting the flow away from the side of the river on which the groyne is built.
Important Points:
Bunds are temporary in nature whereas Weirs and barrages are permanent in nature.
In a weir, the water overflows the weir, but in a dam, the water overflows through a special place called a spillway. Weirs have traditionally been used to create mill ponds.
A barrage is a weir that has adjustable gates installed over top of it, to allow different water surface heights at different times.Concept:
For the given data, the most suitable cross drainage work is syphon aqueduct.
There are mainly two conditions for syphon aqueduct-
Full Supply Level (FSL): It is the elevation of water surface when then the channel is carrying the discharge at its peak capacity.
High flood level (HFL): It is the maximum elevation of the water surface in the river when it is carrying maximum discharge during high flood time.
Calculation:
Bed level of Canal = + 211 m,
Bed level of Drain = + 209.4 m
High flood level(HFL) = Depth of flow at High Flood Level + Bed level of Drain
High flood level(HFL) = 2.2 m + 209.4 m = 211.6 m
Hence,
(i) Bed level of Canal > Bed level of Drain
(ii) High flood level > Bed level of Canal
Important Points
The drainage water intercepting the canal can be classified in the following ways:
i) By passing the canal over the drainage
a) Aqueduct
b) Syphon aqueduct
ii) By passing the canal below the drainage
a) Super passage
b) Syphon
Explanation:
Canal head work
Functions
Depending upon the stages (reaches) of the river, the location of head work is:
Trough stage or alluvial stage
In this stage, the cross-section of the river is made up of alluvial sand and silt. The bed slope and velocity are small. It is Most suitable because
Both rocky and delta stage is not suitable for canal head work.
The gross commanded area for a distributary is 6000 hectares, 80% of which is culturable irrigable. The intensity of irrigation for Kharif season is 25%. The area to be irrigated in Kharif season is ____ hectares.
Concept:
Culturable land = percentage of culturable × Gross area
Area to be irrigated = IOI × percentage of cultivable land × Gross Area
Where,
IOI is intensity of irrigation
Calculations:
Given, Gross area = 6000 hectares.
80% of land is cultural irrigable.
Intensity of irrigation = 25 %.
Area to be irrigated = 0.25 × 0.8 × 6000
= 1200 hectares.
Explanation:
Whenever the available natural ground slope is steeper than the designed bed slope of the channel, the difference is adjusted by constructing vertical drops or falls in the canal beds at suitable intervals.
The main aim of constructing fall is to destroy the surplus energy build-up due to the difference in Energy head at the drop location.
Various types of famous falls are: Ogee falls, Trapezoidal notch falls, Well type falls, Sarda Type falls, etc.What is the variation in duty of water from the head of a main canal (M) to that in the field (F)?
Duty: It is the number of hectares of land irrigated for full growth of a given crop by a supply of 1 cumec of water continuously during the entire base period of that crop.
Duty of water changes from place to place, it will be maximum at the field and minimum at the head of the main canal.
Duty is the area that can be irrigated by the discharge of 1 cumec of water.
At the head of the canal, there are numerous losses to occur later which requires more amount of water to irrigate a particular field. However, if considered on the field, all losses have already occurred and a lesser amount of water is required to irrigate the same considered area.
The agricultural crop year in India is from July to June. The Indian cropping season is classified into two main seasons-(i) Kharif and (ii) Rabi based on the monsoon. The kharif cropping season is from July – October during the south-west monsoon and the Rabi cropping season is from October-March (winter). The crops grown between March and June are summer crops. The terms ‘kharif’ and ‘rabi’ originate from Arabic language where Kharif means autumn and Rabi means spring.
The kharif crops include rice, maize, sorghum, pearl millet/bajra, finger millet/ragi (cereals), arhar (pulses), soyabean, groundnut (oilseeds), cotton etc.
The rabi crops include wheat, barley, oats (cereals), chickpea/gram (pulses), linseed, mustard (oilseeds) etc.
Explanation:
Garret's diagram
It gives the graphical method of designing the channel dimensions based on Kennedy's Theory.
The diagram has discharge plotted on the abscissa and the ordinates on the left indicate the slope and on the right water depth in the channel and critical velocity (V_{o}).
Kennedy's theory
Lacey proposed that rugosity coefficient is dependent on grade and density of boundary material against the idea of Kutter and Manning, who defined rugosity coefficient to be dependent solely on surface roughness.
He also established relationship between rugosity coefficient (N) and silt factor (f) as below:
N α f^{1/4 }and f = 1.76 \(\sqrt d\)
Where, d is the diameter average silt particle in ‘mm’.
However, Lacey considered the rugosity coefficient for ‘very good’ condition appropriate to the grain size and considered a value of 0.0225 for most of the conditions.Concept:
i) Water Application Efficiency (ηa)
\(\boxed{{\eta _a} = \frac{{{W_s}}}{{{W_f}}} \times 100}\)
where, Ws = Water stored in the root zone
Wf = Water delivered to the field
ii) Water Conveyance Efficiency, (ηc)
\(\boxed{{\eta _c} = \frac{{{W_f}}}{{{W_r}}} \times 100}\)
where, Wf = Water delivered to the field
Wr = Water delivered from the reservoir
It includes losses from canals in form of seepage, evaporation etc.
iii) Water use Efficiency (ηu)
\(\boxed{{\eta _u} = \frac{{{W_u}}}{{{W_f}}} \times 100}\)
where, Wu = Water use consumptive.
Wf = Water delivered to the filed
iv) Water storage Efficiency (ηu)
\(\boxed{{\eta _u} = \frac{{{W_{s'}}}}{{{W_\eta }}} \times 100}\)
where, Ws’ = Actual water stored in the root zone
Wη = Water needed to be stored to bring the water content up to field capacity.
Calculation:
The volume of water in root zone,
\({V_r} = 0.4 \times 30 \times {10^4}\)
= 12 × 104m3
The volume of water supplied,
VS = 12 × 5 × 3600
= 2.16 × 105m3
Water application efficiency,
\(\eta = \frac{{{V_r}}}{{{V_s}}} \times 100 = 55.55\;\%\)
The Correct Answer is Crop Rotation:
Diversion head works like Weir or barrage is constructed across a perennial river to raise water level and to divert the water to canal, is known as diversion head work Flow of water in the canal is controlled by canal head regulator. It controls the entry of silt into canals and provides some poundage by creating small pond. However, the most appropriate will be option ‘1’.