--- title: "Stable canal: Q 32, d 30 mm" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:stable-canal-q32-d30mm-phi37" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/stable-canal-q32-d30mm-phi37 site: BCT Engineering --- # Stable canal: Q 32, d 30 mm Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2072 Magh, 5 marks - 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-canal-q32-d30mm-phi37 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4 ## Question A canal is to be designed to carry a discharge of 32 cumecs. The bed slope is kept 1 in 1500. The soil is coarse alluvium having a grain size of 30 mm. Assuming the canal is trapezoidal and to be unlined with unprotected banks. Determine a suitable section for the canal. Assume φ = 37°. ## Short answer Shields analysis gives maximum stable depth 3.11 m (side-controlled). Using Manning's, required section: bed width 1.6 m, depth 3.11 m, side slope 2H:1V. Depth up to 3.11 m is stable, so wider shallower alternate possible. ## Answer Given: $Q = 32\ \text{m}^3/\text{s}$, $S = 0.000667$ (1 in 1500), particle size $d = 30$ mm, $\phi = 37^\circ$. The banks are unprotected, so side stability governs; a side slope of 2H:1V (26.6° with the horizontal) is assumed. Take $\gamma = 9.81$ kN/m³ and specific gravity of particles 2.65. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.03^{1/6}}{21.1} = 0.0264$ (Strickler). #### Critical shear stress of the bed (Shields) $$ \tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.03 = 22.82\ \text{N/m}^2 $$ #### Depth limited by the bed $$ \gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.02282}{9.81\times 0.000667} = 3.49\ \text{m} $$ #### Depth limited by the side slope Reduction factor for the side: $$ K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 26.6^\circ}{\sin^2 37^\circ}} = 0.669 $$ The maximum stress on the side is $0.75\gamma DS$, so: $$ D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.669\times 0.02282}{0.75\times 9.81\times 0.000667} = 3.11\ \text{m} $$ The smaller value governs: **stable depth $D = 3.11$ m** (side controlled). #### Bed width from Manning's equation Using $D = 3.11$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 1.58$ m; adopt $B = 1.6$ m. Check: $A = 24.32$ m², $R = 1.568$ m, $V = 1.319$ m/s, $Q = 32.08$ m³/s. **Answer:** Stable depth $D \approx 3.11$ m, bed width $B \approx 1.6$ m, side slope 2.0H:1V. This section is deep and narrow. Because every depth up to 3.11 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 is tau_c for 30 mm particles? A. 38.04 N/m^2 B. 2.28 N/m^2 C. tau_c = 0.047 x 1.65 x 9.81 x 0.03 = 22.82 N/m^2 D. 30.43 N/m^2 Answer: C. tau_c scales linearly with particle size d. (Key point: Shields stress) 2. Which value governs and what is it? A. D_bed = 3.49 m B. D_side = 3.49 m C. D = 1.6 m D. D_side = 3.11 m (smaller than D_bed = 3.49 m) Answer: D. Side particles are reduced by K = 0.669 and loaded to 0.75 gamma D S. (Key point: Governing limit) 3. How is Manning's n found when not given? A. From Strickler: n = d^(1/6)/21.1 = 0.0264 B. From n = 0.015 for concrete C. From n = 1.76 sqrt(d) D. From n = V/R Answer: A. Strickler relates n to particle size. (Key point: Roughness) ## Close variants set in other exams - [Stable canal: Q 40, d 5 cm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Astable-canal-q40-d5cm-phi37) — Irrigation and Drainage Engineering, chapter 4, asked 2× ## Related in Irrigation and Drainage Engineering - [Canal in alluvium, Q 64, d 0.5 mm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Aalluvial-canal-q64-d05mm) — Irrigation and Drainage Engineering, chapter 4 - [Stable alluvial canal, Q 50, d 0.5 mm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Aalluvial-canal-q50-sketch) — Irrigation and Drainage Engineering, chapter 4 - [Stable depth: Q 20, d 40 mm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Astable-depth-q20-d40mm) — Irrigation and Drainage Engineering, chapter 4, asked 2× - [Stable depth: Q 50, slope 1:1000](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Astable-depth-q50-d50mm-slope1000) — Irrigation and Drainage Engineering, chapter 4, asked 2× - [Canal in alluvium, Q 40, d 0.5 mm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Aalluvial-canal-q40-d05mm) — Irrigation and Drainage Engineering, chapter 4 - [Regime canal for 90 cumecs](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Aregime-canal-q90-d02cm) — Irrigation and Drainage Engineering, chapter 4 ## Related in other subjects - [Tractive force design in coarse alluvium, 20 m³/s](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atractive-force-trapezoid-20m3s-d75-2cm) — Hydraulics, chapter 10 - [Economical rectangular condition and trapezoidal bed slope](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atrapezoidal-channel-bed-slope) — Hydro Power, chapter 1 - [Tractive force design for 15 cumecs, 5 cm particles](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atractive-force-15-cumecs-5cm-particles) — Hydraulics, chapter 10, asked 2× - [Tractive force design for 10 cumecs, 3 cm particles](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atractive-force-10-cumecs-3cm-particles) — Hydraulics, chapter 10, asked 2×