\(Capacity\;factor = \frac{{\left( {Number\;of\;days\;canal\;has\;actually\;num\;during\;a\;watering\;period} \right)}}{{Maximum\;discahrge\;capacity\;of\;canal}}\)

\(Time\;Factor = \frac{{Number\;of\;days\;canal\;has\;actually\;run\;during\;a\;watering\;period}}{{Number\;of\;days\;of\;watering\;period}}\)

Outlet factor = Duty of water at the head of a field channel.

Ring Basin Method → orchid plantation

Furrow irrigation → sugarcane, groundnut, potato, tobacco etc.

Check flooding → Jowar / Paddy

Drip irrigation → tomatoes, grapes, corn, cauliflowers, cabbage.

Concept:

Gross Command Area (G.C.A):

It is the total area, bounded within the irrigation boundary of a project, which can be economically irrigated without considering the limitation of the quantity of available water. It includes the cultivable as well as the un-cultivable area. For example, ponds, residential areas, roads, reserved forests, etc. are the uncultivable areas of the gross command area.

Culturable or Cultivable Command Area (CCA):

Culturable area is the cultivable part of the gross command area, and includes all land of GCA on which cultivation is possible. It will, thus, includes pastures and follow lands, which can be made cultivable. Obviously, it does not include uncultivable part of the gross command, like, populated areas, ponds, roads, reserved forests, usher land etc.

Intensity of irrigation (IOI): It is the fraction of the cultural command area (CCA) which needs to be irrigated for a particular crop.

Area of irrigation = IOI × CCA

∴ Intensity of irrigation is the ratio of the actual irrigated area to the total cultural command area.

The intensity of irrigation is usually around 40 to 60 percent for the nation. But for the states like Uttar Pradesh, Punjab, Haryana, and West Bengal it is 139.6%, 133.4%, 90.2%, and 73.3% respectively. More of the 100 percent of IOI is achieved by cultivating larger parts of CCA with more than one crop in a year.

Other Important Terms:

KOR period: The period for a crop during which the requirement of water is maximum.

KOR depth: Water required during the KOR period is called KOR depth.

Concept:

The duty of water goes on increasing as the water flows.

For example, in the above Figure, let C be the head of the field, B be the head of the water course or the field channel, and A be the head of the distributary.

Let the area of the field be 1700 hectares, and let 1 cumec water be required to be delivered at point C, for the growth of the crop. Thus, the duty at the head of the field will be 1700 hectares/cumec.

Assuming the conveyance losses between B and C to be 0.1 cumec (say), the discharge required at B will be 1.1 cumecs, and hence duty of water measured at B will be 1700/1.1 = 1545 hectare/cumec only.

Again, if the losses between A to B are taken to be equal to 0.2 cumec, the discharge required at the head of the distributary will be 1.1 + 0.2 = 1.3 cumecs, i e. if 1.3 cumecs are discharged at A, then cumec will 1 reach at the head of the field. Hence the duty of water at A will be 1700/1.3 = 1308 hectares/cumec only.

Thus, duty at the head of the water course (at B) is lesser than the duty at the head of the field, and is greater than the duty at the head of the distributary.

So the order will be DM < DD < DW.

Concept:

Duty (hectares/cumec), Delta (cm) and Base period (days) relationship is given by

\(D = \frac{{864 \times B}}{{\rm{\Delta }}}\)

Discharge required \(\left( Q \right) = \frac{{Area\;to\;be\;irrigated\;\left( A \right)}}{{Duty\;\left( D \right)}}\) 

Calculation:

For Rabi season:

\(D = \frac{{864 \times 4 \times 7}}{{13.5}} = 1792\) hectares/cumec

Area to be irrigated (A) = Intensity of irrigation (IOI) × Culturable command area (CCA)

Culturable command area = 0.75 × 5000 = 3750 hectares

∴ A = 0.6 × 3750 = 2250 hectares

∴ Discharge \(\left( Q \right) = \frac{A}{D} = \frac{{2250}}{{1792}} = 1.256\) cumec

For kharif season:

\(D = \frac{{864 \times 205 \times 7}}{{19}} = 795.79\) hectares/cumec

Area to be irrigation (A) = 0.3 × 3750 = 112.5 hectares

∴ Discharge \(\left( Q \right) = \frac{A}{D} = \frac{{112.5}}{{795.79}} = 1.4133\) cumec

The distributary should be designed for the maximum discharge of rabi and kharif season.

∴ The distributary should be designed for 1.4133 cumec.

Concept:

Time required for irrigation (t) is given by

\(t = 2.3 \times \frac{y}{f} \times {\log _{10}}\left( {\frac{Q}{{Q - f \times A}}} \right)\)

Where,

y = Depth of water flowing over the border strip

f = Rate of infiltration of soil (in m/hour)

A = Area of land strip to be irrigated (in m²)

Q = Discharge-image supply depth (in m³/hour)

Calculation:

Given:

Area (A) = 0.05 hectares = 0.05 × 10⁴ m² = 500 m²

Discharge (Q) = 0.025 cumec = 0.025 × 60 × 60 m³/hour = 90 m²/hour

Filtration rate (f) = 0.05 m/hour

Depth of flow (y) = 10 cm = 0.1 m

Now,

\(t = 2.3 \times \frac{y}{f} \times {\log _{10}}\left( {\frac{Q}{{Q - f \times A}}} \right)\)

\(= 2.3 \times \frac{{0.1}}{{0.05}} \times {\log _{10}}\left( {\frac{{90}}{{90 - 0.05 \times 500}}} \right)\)

∴ t = 0.65 hour = 39 minutes

Concept:

For an irrigation land:

SC = Saturation capacity, FC (F_c) = 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 × F_c

2. Equivalent depth of water held at PWP (x’) = S × d × (PWP)

3. Available moisture/storage capacity of soil (y) = S × d × (F_c - PWP)

4. Readily available moisture content (d_w) = S × d × (F_c - OMC)

Where,

S = Specific gravity of the soil and d = soil depth

Frequency of irrigation in days = Total water to be supplied / Water consumed per day

Total water to be supplied (y_s) = S × d × (F_c – OMC)

Calculation:

Given: F_c­ = 40%, PWP = 15%, d = 100 cm, Soil density (ρ_s) = 1350 kg/m³, and water density (ρ_w) = 1030 kg/m³

Total water to be supplied = Readily available moisture content

Readily available moisture content = 0.7 × Total moisture content

Total moisture content = \(\frac{{1350}}{{1030}} \times 1 \times \left( {0.4 - 0.15} \right) = 0.328\;m\)

Total water to be supplied = 0.7 × 0.328 = 0.229 m

Frequency of irrigation = 15.29 days.

∴ The frequency of irrigation is 15 days as lower values has to be adopted as the plant with start wilting if the 16 days frequency of irrigation is adopted.

Concept:

Sodium Absorption Ration (SAR)

\({\rm{SAR}} = \frac{{{\rm{N}}{{\rm{a}}^ + }}}{{\sqrt {\frac{{{\rm{C}}{{\rm{a}}^{ + 2}} + {\rm{M}}{{\rm{g}}^{ + 2}}}}{2}} }}\)

Calculation:

Given: Na⁺ = 25, Ca⁺² = 4 and Mg⁺² = 2

\({\rm{SAR}} = \frac{{{\rm{N}}{{\rm{a}}^ + }}}{{\sqrt {\frac{{{\rm{C}}{{\rm{a}}^{ + 2}} + {\rm{M}}{{\rm{g}}^{ + 2}}}}{2}} }} = \frac{{25}}{{\sqrt {\frac{{4 + 2}}{2}} }} = \frac{{25}}{{\sqrt 3 }} = 14.43\)

Important Point:

Exchangeable sodium ratio (ESR):

\({\rm{ESR}}\left( {{\rm{Ratio}}} \right) = \frac{{{\rm{N}}{{\rm{a}}^ + }}}{{{\rm{C}}{{\rm{a}}^{ + 2}} + {\rm{M}}{{\rm{g}}^{ + 2}} + {{\rm{K}}^ + }}}\)

\({\rm{ESR}}\left( {{\rm{Percent}}} \right) = \frac{{{\rm{N}}{{\rm{a}}^ + }}}{{{\rm{N}}{{\rm{a}}^ + } + {{\rm{K}}^ + } + {\rm{C}}{{\rm{a}}^{ + 2}} + {\rm{M}}{{\rm{g}}^{ + 2}}}}\)

Concept: Depth of water stored in the root zone in every cycle of irrigation.

\({d_w} = \frac{{{\gamma _d}}}{{{\gamma _w}}} \times d\left( {Field\;capacity - MC} \right)\)

where,

γ_d = dry unit weight of soil

γ_w = unit weight of water

d = depth of the root zone

MC = current moisture content.

Calculation:

Field Capacity, F.C = 36%

M.C. = 24%

γ_d = 1.5 × 10³ kg/m³ × 9.81 = 14715 N/m³

γ_w = 9810 N/m³

d = 0.8 m

η_c = 80%

η_a = 90%

To bring the moisture content up to field capacity,

\(\therefore {d_w} = \frac{{14715}}{{9810}} \times 0.8 \times \left( {0.36 - 0.24} \right)\)

d_w = 0.144 m

Now, to bring the soil moisture content to field capacity the depth of irrigation should incorporate conveyance efficiency and application efficiency

∴ Depth of water required \(= {d_w} \times \frac{{100}}{{{\eta _c}}} \times \frac{{100}}{{{\eta _a}}}\)

\(= 0.144 \times \frac{{100}}{{80}} \times \frac{{100}}{{90}}\)

= 0.20 m

Concept:

Water distribution efficiency is given by (η_d)

\({{\rm{\eta }}_d} = \left( {1 - \frac{d}{D}} \right) \times 100\)

D = Average depth of penetration

d = S Average of absolute value of deviations from

Calculation:

\(D = \frac{{3 + 2.7 + 2.5 + 2.2 + 1.9 + 1.5}}{6}\)

D = 2.3 cm

Values of deviation from the mean is (3-2.3), (2.7-2.3), (2.2-2.3), (2.5-2.3), (1.9-2.3), (1.5-2.3)

= 0.7, 0.4, 0.1, 0.2, -0.4, -0.8

\(d = \frac{{0.7 + 0.4 + 0.1 + 0.2 + 0.4 + 0.8}}{6}\) = 0.433 metre

\({{\rm{\eta }}_d} = \left( {1 - \frac{{0.433}}{{2.3}}} \right) \times 100\)