A Silting Distributary: Steepen the Slope or Line the Canal?
A 250 ha rice command is served by a 3 km unlined trapezoidal distributary with side slopes 1.5:1, Manning n = 0.025 and a bed slope of 1 in 4000, built with a 1.4 m bed width to carry 1.0 m³/s at 0.90 m normal depth. After two seasons the tail reach has silted about 25 cm and head-reach farmers over-irrigate while the tail receives almost nothing. The section is in sandy loam where Kennedy's critical velocity applies, V0 = 0.55 m^0.64 with m in metres. Measured seepage on the unlined section is 0.22 m³/s per 10⁶ m² of wetted area, desilting costs about Rs 150 per m³ of deposited silt, and the canal runs 240 days a year. Water is valued at Rs 1.2 per m³. Two proposals are on the table: steepen the bed slope to 1 in 2500, or line the full length in brick masonry (n = 0.015) at Rs 450 per m² of wetted perimeter.
1. Compute the actual velocity and discharge at 0.90 m depth, and check the section against Kennedy's critical velocity.
Read the worked answer
With b = 1.4 m and y = 0.90 m:
A = 1.4(0.9) + 1.5(0.9)² = 1.26 + 1.215 = 2.475 m²
P = 1.4 + 2(0.9)√(1 + 1.5²) = 1.4 + 3.245 = 4.645 m; R = 2.475/4.645 = 0.533 m
V = (1/0.025)(0.533)^(2/3)(1/4000)^0.5 = 40 × 0.657 × 0.01581 = 0.416 m/s
Q = 0.416 × 2.475 = 1.03 m³/s, so the section does carry the design flow.
Kennedy: V0 = 0.55(0.9)^0.64 = 0.55 × 0.935 = 0.514 m/s.
Actual velocity is only 81% of critical velocity. The canal is a silting section, which is exactly consistent with 25 cm of tail-reach deposition, and the head-tail inequity follows because the reduced tail section throttles flow. The discharge figure is irrelevant if the sediment transport regime is wrong.
2. Evaluate the two proposals quantitatively and state which you would implement first.
Read the worked answer
Slope option: to reach 0.514 m/s at the same depth and width, S = (Vn/R^(2/3))² = (0.514 × 0.025/0.657)² = 3.83 × 10⁻⁴, i.e. about 1 in 2610. The proposed 1 in 2500 is therefore just sufficient, and gives V ≈ 0.53 m/s and Q ≈ 1.31 m³/s at 0.90 m. Cost is mainly head: over 3 km the loss rises from 0.75 m to 1.2 m, which must be checked against available command level.
Lining option: n = 0.015 raises velocity to 0.69 m/s - above critical, so scour rather than silt becomes the issue, and the downstream unlined reach will receive the scoured load.
Judgement: implement the slope correction first. It is cheap, directly addresses the diagnosed regime failure, and keeps the canal within its own sedimentation balance. Lining is a capital-intensive answer to a problem whose cause is inadequate velocity; it should be considered later as a separate decision about seepage and maintenance.
3. Using the seepage and desilting figures, decide whether lining pays for itself.
Read the worked answer
Wetted area per km = 4,645 m². Seepage = 0.22 × 4,645/10⁶ = 1.02 × 10⁻³ m³/s, i.e. 0.00102 × 86,400 × 240 ≈ 21,200 m³/km/year; over 3 km about 63,500 m³, worth Rs 76,200/year at Rs 1.2/m³.
Deposited silt = 0.25 × 1.4 × 3,000 = 1,050 m³/year, costing Rs 157,500/year to remove.
Total benefit of lining ≈ Rs 234,000/year.
Cost = 4,645 × 3,000... i.e. 13,935 m² × Rs 450 ≈ Rs 6.27 million; annualised at 10% over 20 years (CRF 0.1175) = Rs 737,000/year.
Benefits are less than a third of the annualised cost, so lining is not justified on water saving plus desilting alone on this reach. The honest conclusion is to fix the hydraulic regime first, and to revisit lining only if land acquisition, weed control, groundwater recharge loss or crop-value increases at the tail are included - those, not seepage, are what can tip the balance.