Irrigation Water Conveyance and Application Methods
Subtopic: Design of irrigation channels and underground pipelines
Question Statement:
Furrows of 120 m length with 0.5 % slope are made at 90 cm spacing. The maximum non-erosive stream flow rate is applied in a furrow that takes 1.0 hour to reach the lower end. Then this flow rate is reduced to half of its size and, subsequently, continued for another 1.0 hour. The average depth of applied water is ________ cm.
Official Answer Key:
6.00 to 6.00
Step-by-Step Solution & Explanation
Maximum non-erosive stream size (Criddle's formula): q_max = 0.60/S, with q in L s⁻¹ and S in %.
q_max = 0.60/0.5 = 1.2 L s⁻¹.
Advance phase: 1.2 L s⁻¹ × 3600 s = 4320 L.
Wetting (cut-back) phase: 0.6 L s⁻¹ × 3600 s = 2160 L.
Total volume applied = 6480 L = 6.48 m³.
Area served by one furrow = 120 × 0.90 = 108 m².
Average depth = 6.48/108 = 0.06 m = 6.0 cm.
Section 6 — Agricultural Process Engineering
9 questions · 14 marks
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