Irrigation Water Conveyance and Application Methods
Subtopic: Irrigation scheduling
Question Statement:
A 5 ha field under wheat crop is irrigated at 40% depletion of available moisture content. The field capacity, wilting point and bulk density of the soil in the field are 32% (on weight basis), 20% (on weight basis) and 1400 kg m⁻³ respectively. To irrigate the field in a day of 10 hours, a pump is used which lifts 270 m³ of water per hour against a total head of 20 m. If the root zone depth is 0.8 m, the volume of water required to irrigate the field in m³ will be
Options:
(A) 2700
(B) 3150
(C) 3600
(D) 4050
Official Answer Key:
A
Step-by-Step Solution & Explanation
Available water (depth) = (FC − WP) × (bulk density/density of water) × root depth = (0.32 − 0.20) × (1400/1000) × 0.8 m = 0.12 × 1.4 × 0.8 = 0.1344 m. Depth to be applied at 40% depletion = 0.1344 × 0.40 = 0.0538 m. Volume required = depth × area = 0.0538 m × 5 ha × 10,000 m²/ha ≈ 2690 ≈ 2700 m³, matching option (A).
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