Question:medium

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:

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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).
Updated On: Feb 19, 2026
  • D = b x d
  • D = b / d
  • D = 8.64 \(\frac{b}{d}\)
  • D = 8.64 \(\frac{d}{b}\)
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The Correct Option is C

Solution and Explanation

Step 1: Define the terms Duty (D), Base Period (b), and Delta (d).
- Duty (D): Land area (hectares) irrigable by 1 cubic meter per second (cumec) of water flowing continuously for the entire base period.
- Base Period (b): Total days irrigation is supplied to a crop.
- Delta (d): Total depth of water (meters) required by a crop during the entire base period.
Step 2: Derive the relationship.
Consider a discharge of 1 cumec over 'b' days.
Total water volume, 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 irrigates 'D' hectares to a depth of 'd' meters.
Volume 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 volumes:
\[ 10000 \cdot D \cdot d = 86400 \cdot b \]\[ D = \frac{86400}{10000} \frac{b}{d} = 8.64 \frac{b}{d} \]
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