Irrigation Engineering Test 1

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

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

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:

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:

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

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