--- title: "Stable depth: Q 20, d 40 mm" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:stable-depth-q20-d40mm" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/stable-depth-q20-d40mm site: BCT Engineering --- # Stable depth: Q 20, d 40 mm Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2079 Asoj, 6 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-depth-q20-d40mm - 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 design discharge of 20 m³/sec. The slope of the canal is 1:1500 and passes through a medium with mean particles as 40 mm. Assuming a trapezoidal section, determine the stable depth of canal assuming the angle of repose of canal materials as 36° and Manning's coefficient as 0.025. ## Short answer Use Shields criterion. Critical shear stress 30.43 N/m². Side stability governs with max stable depth 4.03 m. Adopted trapezoidal section: bed width 8 m, depth 1.57 m, side slope 2H:1V, velocity 1.14 m/s. ## Answer Given: $Q = 20\ \text{m}^3/\text{s}$, $S = 0.000667$ (1 in 1500), particle size $d = 40$ mm, $\phi = 36^\circ$, $n = 0.025$. The side slope is not given, so 2H:1V (26.6° with the horizontal) is assumed. Take $\gamma = 9.81$ kN/m³ and specific gravity of particles 2.65. #### 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 $$ \gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03043}{9.81\times 0.000667} = 4.65\ \text{m} $$ #### Depth limited by the side slope Reduction factor for the side: $$ K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{\sin^2 26.6^\circ}{\sin^2 36^\circ}} = 0.649 $$ The maximum stress on the side is $0.75\gamma DS$, so: $$ D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.649\times 0.03043}{0.75\times 9.81\times 0.000667} = 4.03\ \text{m} $$ The smaller value governs: **stable depth $D = 4.03$ m** (side controlled). #### Section for the discharge At $D = 4.03$ m even a very small bed width would carry more than 20 m³/s, so the stable depth is only the **upper limit** of the depth: any flow depth up to 4.03 m is stable. To pass the discharge, choose a practical bed width $B = 8$ m; Manning's equation then gives a flow depth $$ D = 1.57\ \text{m}\ (< 4.03\ \text{m}),\quad A = 17.49\ \text{m}^2,\quad R = 1.165\ \text{m},\quad V = 1.143\ \text{m/s} $$ and the bed stress $\gamma DS = 10.3$ N/m² is below $\tau_c = 30.4$ N/m². **Answer:** Maximum stable depth $= 4.03$ m. Adopted section: $B = 8$ m, $D = 1.57$ m, side slope 2.0H:1V, $V = 1.14$ m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the critical shear stress for the 40 mm particles? A. tau_c = 0.047 x 1.65 x 9.81 x 0.04 = 30.43 N/m^2 B. About 3.0 N/m^2 C. About 38 N/m^2 D. About 304 N/m^2 Answer: A. Using tau_c = 0.047 (gamma_s - gamma) d. (Key point: Shields stress) 2. Which depth governs and why? A. D_bed = 4.65 m, because bed stress is largest B. D_side = 4.03 m, as it is smaller than D_bed = 4.65 m C. Their average, 4.34 m D. D_bed, since the angle of repose is ignored Answer: B. The smaller of the bed and side limits governs. (Key point: Governing depth) 3. What does the stable depth 4.03 m represent? A. The exact depth the canal must have B. The minimum depth needed C. An upper limit: any flow depth up to it is stable D. The depth at which scour begins in the bed Answer: C. The discharge is carried by choosing a practical bed width, giving a shallower depth. (Key point: Meaning of stable depth) 4. With B = 8 m, what happens to the flow depth? A. D = 4.03 m exactly B. D = 8 m C. D = 0.5 m, so the bed scours D. Manning gives D = 1.57 m, below 4.03 m, with bed stress 10.3 N/m^2 below 30.4 N/m^2 Answer: D. The stress is well under the critical value, so the section is stable. (Key point: Adopted section) ## Close variants set in other exams - [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× ## Related in Irrigation and Drainage Engineering - [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× - [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× - [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 channel: Q 20, particle 4 cm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Astable-channel-q20-particles-4cm) — Irrigation and Drainage Engineering, chapter 4 - [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 - [Canal design discharge at 0, 1, 2, 3 km](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acanal-design-discharge-along-reach) — Irrigation and Drainage Engineering, chapter 4 ## Related in other subjects - [Design stable channel in 8 mm gravel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Astable-channel-8mm-gravel-design) — Hydraulics, chapter 10 - [Efficient trapezoidal section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-trap) — Hydro Power, chapter 1, asked 2× - [Cut slope design: HS 110/40, B 130/20](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Acut-slope-hs110-40) — Engineering Geology II, chapter 6 - [Settling basin design for 0.2 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-2mm) — Hydro Power, chapter 5