--- title: "Stable channel: Q 20, particle 4 cm" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:stable-channel-q20-particles-4cm" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/stable-channel-q20-particles-4cm site: BCT Engineering --- # Stable channel: Q 20, particle 4 cm Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2066 Bhadra (old course), 10 marks - Video: [sediment Transport I Design of irrigation Channels I Irrigation](https://www.youtube.com/watch?v=VA9mZIZIBlQ) — Vishwanatha Bhat - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/stable-channel-q20-particles-4cm - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4 ## Question From the data given below, design a stable irrigation canal, ensuring the stability of particles at bed as well as sides. Discharge = 20 m³/s; Bed slope = 0.001; Particle size = 4 cm; Side slope = 30°; Angle of repose of soil = 38°. ## Short answer Shields analysis yields maximum stable depth 2.41 m (side-controlled). Adopted trapezoidal section: bed width 2.0 m, depth 2.41 m, side slope 1.7H:1V. ## Answer Given: $Q = 20\ \text{m}^3/\text{s}$, $S = 0.001000$ (1 in 1000), particle size $d = 40$ mm, $\phi = 38^\circ$. Side slope $1.73H:1V$ (30.0° with horizontal); $\gamma = 9.81$ kN/m³, specific gravity of particles 2.65. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.04^{1/6}}{21.1} = 0.0277$ (Strickler). #### Critical shear stress of the bed (Shields) $$ \tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.04 = 30.43\ \text{N/m}^2 $$ #### Depth limited by the bed $$ \gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03043}{9.81\times 0.001000} = 3.10\ \text{m} $$ #### Depth limited by the side slope Reduction factor for the side: $$ K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 30.0^\circ}{\sin^2 38^\circ}} = 0.583 $$ Maximum stress on the side is $0.75\gamma DS$, so: $$ D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.583\times 0.03043}{0.75\times 9.81\times 0.001000} = 2.41\ \text{m} $$ The smaller value governs: **stable depth $D = 2.41$ m** (side controlled). #### Bed width from Manning's equation Using $D = 2.41$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 2.00$ m, adopt $B = 2.0$ m. Check: $A = 14.88$ m², $R = 1.278$ m, $V = 1.344$ m/s, $Q = 20.00$ m³/s. **Answer:** Stable depth $D \approx 2.41$ m, bed width $B \approx 2.0$ m, side slope 1.7H:1V. This section is deep and narrow. Because every depth up to 2.41 m is stable, a wider and shallower section (larger $B$, smaller $D$) carrying the same discharge is equally stable and is usually preferred in practice. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What must be checked to ensure stability of particles at both bed and sides? A. Only the bed stress B. Only Manning's V C. Take the smaller of D_bed and D_side D. The larger of D_bed and D_side Answer: C. Bed gives 3.10 m; side gives 2.41 m, so 2.41 m governs. (Key point: Bed and side checks) 2. For theta = 30 deg and phi = 38 deg, what is K? A. About 0.866 B. About 0.30 C. About 0.78 D. About 0.583 Answer: D. K = sqrt(1 - sin^2 30/sin^2 38) = 0.583. (Key point: K value) 3. What maximum stress is taken on the sides? A. 0.75 gamma D S B. 1.0 gamma D S C. 0.5 gamma D S D. 1.25 gamma D S Answer: A. The side stress is lower than the bed stress. (Key point: Side stress) 4. What section is adopted? A. D about 3.10 m, B about 2.0 m B. D about 2.41 m, B about 2.0 m, side slope 1.7H:1V C. D about 2.41 m, B about 20 m D. D about 2.0 m, B about 2.41 m Answer: B. Manning with D = 2.41 m gives B = 2.0 m, carrying 20 m^3/s. 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