Question

A field channel has cultivable commanded area of 2000 hectares. The intensities of irrigation for gram and wheat are 30% and 50% respectively. Gram has a kor period of 18 days, kor depth of 12 cm, while wheat has a kor period of 18 days and a kor depth of 15 cm. The discharge (in m³/s) required in the field channel to supply water to the commanded area during the kor period is _______________.

Answer 1

Concept:

Gross command area (GCA):
It is the total area that can be irrigated by a canal so that an unlimited quantity of water is available. A canal is usually aligned along with the watershed between two drainage valley.

In other words, GCA is the total area lying between the drainage boundaries that can be irrigated by the canal system. It is measured in hectares.

GCA = CCA + UCA

Where,

CCA = Clturable command area, UCA = Unculturable command area

Duty of canal water (D)

Duty is defined as the area of the land irrigated by a unit discharge of water flowing continuously for the duration of the base period of the crop.

\({\rm{D}} = \frac{{{\rm{Area\;of\;land\;being\;irrigated}}}}{{{\rm{Rate\;of\;water\;supply}}}} = \frac{A}{Q}\)

Duty is measured in hectares/cumec (ha/cumec)

Delta of a crop (Δ):

It is defined as the total thickness of water required for complete maturity of the crop within its base period. It is measured in cm

The sugarcane crop has maximum delta, i.e Δ = 120 cm. Other crops have delta less than 120 cm

Duty and delta can be related by

\({\bf{D}} = \frac{{8.64 \times {\bf{B}}}}{{\bf{\Delta }}}\)

Where, D is in ha/cumec, B is base period or kor period in days, and Delta (Δ) is in meters

Calculation:

Given,

CCA = 2000 hectares

Gram and wheat are Rabi crops,

For Gram,

Δ_g = 12 cm = 0.12 m, B_g = 18 days

Discharge required for gram

\({Q_g} = \frac{{{A_g}}}{{{D_g}}} = \frac{{2000 \times 0.3}}{{8.64 \times \frac{{18}}{{0.12}}}} = 0.4630\frac{{{m^3}}}{s}\)

For wheat,

Δ_w = 15 cm = 0.15 m, B_w = 18 days

Discharge required for wheat,

\({Q_w} = \frac{{{A_w}}}{{{D_w}}} = \frac{{2000 \times 0.5}}{{8.64 \times \frac{{18}}{{0.15}}}} = 0.9645\frac{{{m^3}}}{s}\)

So both are grown in the same period

Total discharge = Q_g + Q_w = 1.4275 m³/s