--- title: "Stable canal: Q 40, d 5 cm" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:stable-canal-q40-d5cm-phi37" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/stable-canal-q40-d5cm-phi37 site: BCT Engineering --- # Stable canal: Q 40, d 5 cm Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2073 Bhadra, 4 marks - Video: [Design an irrigation Channel | Example 4.6 | Chapter 4| Irrigation and Flood Control | S.K.Garg](https://www.youtube.com/watch?v=1zYuhOOF0ZU) — Saurov Nandi Majumdar - 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-q40-d5cm-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 40 cumecs. The bed slope is kept 1 in 1200. The soil is coarse alluvium having a grain size of 5 cm. Assuming the canal is trapezoidal and to be unlined with unprotected banks. Determine a suitable section for the canal. Assume φ = 37°. ## Short answer Using Shields criterion, max stable depth 4.15 m limited by side stability. Adopted section: bed width 15 m, depth 1.73 m, side slope 2H:1V, velocity 1.26 m/s. ## Answer Given: $Q = 40\ \text{m}^3/\text{s}$, $S = 0.000833$ (1 in 1200), particle size $d = 50$ 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. $n = \dfrac{d^{1/6}}{21.1} = \dfrac{0.05^{1/6}}{21.1} = 0.0288$ (Strickler). #### Critical shear stress of the bed (Shields) $$ \tau_c = 0.047(\gamma_s - \gamma)d = 0.047\times(2.65-1)\times 9.81\times 0.05 = 38.04\ \text{N/m}^2 $$ #### Depth limited by the bed $$ \gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03804}{9.81\times 0.000833} = 4.65\ \text{m} $$ #### 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: $$ D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.669\times 0.03804}{0.75\times 9.81\times 0.000833} = 4.15\ \text{m} $$ The smaller value governs: **stable depth $D = 4.15$ m** (side controlled). #### Section for the discharge At $D = 4.15$ m even a very small bed width would carry more than 40 m³/s, so the stable depth is only the **upper limit** of the depth: any flow depth up to 4.15 m is stable. To pass the discharge, choose a practical bed width $B = 15$ m; Manning's equation then gives a flow depth $$ D = 1.73\ \text{m}\ (< 4.15\ \text{m}),\quad A = 31.83\ \text{m}^2,\quad R = 1.401\ \text{m},\quad V = 1.257\ \text{m/s} $$ and the bed stress $\gamma DS = 14.1$ N/m² is below $\tau_c = 38.0$ N/m². **Answer:** Maximum stable depth $= 4.15$ m. Adopted section: $B = 15$ m, $D = 1.73$ m, side slope 2.0H:1V, $V = 1.26$ m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the side slope check important here? A. Unprotected banks ignore side stress B. The banks are unprotected, so the side-slope limit governs C. The sides carry more load than the bed D. The bed particles are finer Answer: B. D_side = 4.15 m is smaller than D_bed = 4.65 m. (Key point: Unprotected banks) 2. For theta = 26.6 deg and phi = 37 deg, what is K? A. About 0.449 B. About 0.866 C. About 0.669 D. About 1.20 Answer: C. K = sqrt(1 - sin^2 26.6/sin^2 37) = 0.669. (Key point: K value) 3. How is the final section chosen? A. Use D = 4.15 m and B from Lacey B. Use D = 4.65 m with B = 15 m C. Use D = 1.73 m with B = 4.15 m D. Take B = 15 m and use Manning to get D = 1.73 m, below the maximum 4.15 m Answer: D. The stable depth is an upper limit; the chosen depth stays below it. (Key point: Interpreting depth) ## Close variants set in other exams - [Stable canal: Q 32, d 30 mm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Astable-canal-q32-d30mm-phi37) — Irrigation and Drainage Engineering, chapter 4, asked 2× ## Related in Irrigation and Drainage Engineering - [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 - [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× - [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× - [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 - [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 - [Stable depth and bed width: Q 5, d50 3 cm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Astable-depth-bed-width-q5-d50) — Irrigation and Drainage Engineering, chapter 4 ## Related in other subjects - [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× - [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