Question

The duty of canal water (hectare per cm per second) 'D', base period of crop 'b' in days and delta 'd' in meter are related as:

• A. D = b x d
• B. D = b / d
• C. D = 8.64 \(\frac{b}{d}\)
• D. D = 8.64 \(\frac{d}{b}\)

Hint:

The relationship \( \Delta = \frac{8.64 B}{D} \) (with \(\Delta\) in meters, B in days, D in ha/cumec) is one of the most fundamental formulas in irrigation engineering. Always be careful with the units, as they determine the constant (e.g., it becomes 864 if Delta is in cm).

Answer 1

The Correct Option is C

Step 1: Define the terms Duty (D), Base Period (b), and Delta (d).
- Duty (D): The area of land (in hectares) that can be irrigated with a unit discharge (1 cubic meter per second, or cumec) of water flowing continuously for the entire base period.
- Base Period (b): The total period in days for which irrigation is supplied to a crop.
- Delta (d): The total depth of water (in meters) required by a crop during the entire base period.
Step 2: Derive the relationship.
Consider a discharge of 1 cumec flowing for 'b' days.
Total volume of water supplied, V = Discharge \(\times\) Time \[ V = (1 \text{ m}^3/\text{s}) \times (b \text{ days}) = 1 \times (b \times 24 \times 60 \times 60) \text{ m}^3 = 86400 \cdot b \text{ m}^3 \] This volume of water irrigates an area 'D' hectares to a depth of 'd' meters.
Volume of water used on land = Area \(\times\) Depth \[ V = (D \text{ hectares}) \times (d \text{ meters}) = (D \times 10^4 \text{ m}^2) \times (d \text{ m}) = 10000 \cdot D \cdot d \text{ m}^3 \] Equating the two expressions for volume: \[ 10000 \cdot D \cdot d = 86400 \cdot b \] \[ D = \frac{86400}{10000} \frac{b}{d} = 8.64 \frac{b}{d} \]