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Canal design

Intermediate Irrigation and Water Resources Engineering Reviewed 24 Oct 2026 2 min read

An irrigation canal carries water from a headworks to distributaries and field channels. Design must carry the required discharge without eroding the bed and banks and without silting up. In alluvial soils, which are erodible and carry sediment, this balance is delicate, and Indian engineers developed regime theories from long observation of stable canals in the Indus and Ganga plains.

Kennedy's theory, from the Upper Bari Doab canal, proposed a critical velocity related to depth at which a channel neither silts nor scours. Lacey later proposed that a channel in regime, with its own natural sediment, adjusts its wetted perimeter, hydraulic radius and slope to the discharge and a silt factor depending on grain size. Lacey's equations give perimeter, velocity, hydraulic radius and slope directly from discharge and silt factor. They remain useful for alluvial canals and for scour estimates at bridges.

Diagram: Indiacon original, not to scale.

Lined canals and those in non-erodible material are designed by Manning's equation for uniform flow, usually with trapezoidal sections and side slopes suited to the soil or lining. Freeboard, the height of bank above full supply level, protects against waves and surges. Banks need adequate top width for inspection roads. Structures along the canal include falls, regulators, outlets, cross-drainage works and escapes. Seepage losses, which can be large in unlined canals, drive the case for lining.

Key formulas
  • Lacey: P = 4.75 √Q; f = 1.76 √d_mm
  • Lacey velocity: V = (Q f² / 140)^(1/6)
  • R = (5/2) V² / f; S = f^(5/3) / (3 340 Q^(1/6))
  • Manning: Q = (1/n) A R^(2/3) S^(1/2)
Common mistake

Designing an alluvial canal for a much smaller discharge than it will actually run at, which makes it scour or silt quickly.

Solved example

Problem. Design an alluvial canal for Q = 30 m³/s with silt of mean size 0.3 mm using Lacey.

  1. f = 1.76 × √0.3 = 0.964.
  2. V = (30 × 0.929 / 140)^(1/6) = (0.199)^(1/6) = 0.764 m/s.
  3. A = 30 / 0.764 = 39.3 m²; P = 4.75 × √30 = 26.0 m; R = 2.5 × 0.764² / 0.964 = 1.51 m (check A/P = 1.51).
  4. S = 0.964^(5/3) / (3 340 × 30^(1/6)) = 0.941 / (3 340 × 1.763) = 1 in 6 260.

Answer. P ≈ 26 m, R ≈ 1.5 m, V ≈ 0.76 m/s, slope ≈ 1 in 6 300

Quick quiz

1What does regime theory describe?

A stable alluvial channel that neither silts nor scours on balance.

2What does Lacey's silt factor depend on?

Mean sediment size.

3What is freeboard?

Height of the bank above full supply level.

Code note
  • IS 7784 (series) (India): Design of cross-drainage works such as aqueducts and syphons.

Clauses and documents to read

  • IS 7112:2002 (criteria for design of cross-section for unlined canals in alluvial soil)
  • IS 10430:2000 (criteria for design of lined canals)
  • IS 7784 (Part 1) (design of cross-drainage works, cited by number)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • S. K. Garg, Irrigation Engineering and Hydraulic Structures, Khanna Publishers, latest edition.
  • G. L. Asawa, Irrigation and Water Resources Engineering, New Age International, latest edition.

Free and official sources

All references for this subject →

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