--- title: "Stable depth: Q 50, slope 1:1000" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:stable-depth-q50-d50mm-slope1000" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/stable-depth-q50-d50mm-slope1000 site: BCT Engineering --- # Stable depth: Q 50, slope 1:1000 Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2072 Asoj, 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-depth-q50-d50mm-slope1000 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4 ## Question A canal has to be designed to carry a design discharge as 50 m³/s. The slope of the canal is 1:1000 and passes through medium with mean particles as 50 mm. Assuming a trapezoidal section, determine the stable depth of the canal assuming angle of repose of canal bed/side particles as 36°. ## Short answer Using Shields' critical shear stress 38.04 N/m², stable depth is 3.35 m (side-controlled, K=0.649). Bed width 2.6 m from Manning’s equation, side slopes 2H:1V, n=0.0288, discharge 50 m³/s. ## Answer Given: $Q = 50\ \text{m}^3/\text{s}$, $S = 0.001000$ (1 in 1000), particle size $d = 50$ mm, $\phi = 36^\circ$. The side slope is not given, so 2H:1V (26.6° with the horizontal, flatter than $\phi$) 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 $$ \gamma D S \le \tau_c \Rightarrow D_{bed} = \frac{\tau_c}{\gamma S} = \frac{0.03804}{9.81\times 0.001000} = 3.88\ \text{m} $$ #### 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: $$ D_{side} = \frac{K\tau_c}{0.75\gamma S} = \frac{0.649\times 0.03804}{0.75\times 9.81\times 0.001000} = 3.35\ \text{m} $$ The smaller value governs: **stable depth $D = 3.35$ m** (side controlled). #### Bed width from Manning's equation Using $D = 3.35$ m and $Q = \frac{1}{n}AR^{2/3}S^{1/2}$, trial gives $B = 2.58$ m; adopt $B = 2.6$ m. Check: $A = 31.16$ m², $R = 1.772$ m, $V = 1.610$ m/s, $Q = 50.15$ m³/s. **Answer:** Stable depth $D \approx 3.35$ m, bed width $B \approx 2.6$ m, side slope 2.0H:1V. This section is deep and narrow. Because every depth up to 3.35 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. How is the critical shear stress of the bed particles found? A. tau_c = gamma D S B. tau_c = 0.75 gamma D S C. tau_c = 0.047 (gamma_s - gamma) d = 38.04 N/m^2 D. tau_c = 0.047 gamma d only Answer: C. Shields' relation gives the stress at which 50 mm particles start to move. (Key point: Shields stress) 2. Why does the side slope control the stable depth (3.35 m, not 3.88 m)? A. The side stress is larger than the bed stress B. Bed particles are finer than side particles C. Manning's n is larger on the sides D. Particles on the slope are less stable, so the allowable stress is reduced by K = 0.649, while the side stress is 0.75 gamma D S Answer: D. D_side = K tau_c/(0.75 gamma S) = 3.35 m is smaller than D_bed = tau_c/(gamma S) = 3.88 m. (Key point: Side controls) 3. Which expression gives the side reduction factor K? A. K = sqrt(1 - sin^2(theta)/sin^2(phi)) B. K = tan(theta)/tan(phi) C. K = cos(phi)/cos(theta) D. K = 1 - theta/phi Answer: A. K depends on the side slope angle theta and angle of repose phi. (Key point: Reduction factor) 4. The result B = 2.6 m with D = 3.35 m is deep and narrow. What does the answer say? A. The channel is unstable and must be redesigned with D = 3.88 m B. Every depth up to 3.35 m is stable, so a wider, shallower section is equally stable and usually preferred C. Only the section with D = 3.35 m is stable D. A shallower section would be scoured Answer: B. 3.35 m is a maximum stable depth, not a required depth. (Key point: Deep narrow section) ## Close variants set in other exams - [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× ## Related in Irrigation and Drainage Engineering - [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 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 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 - [Balancing depth and canal cross-section](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Abalancing-depth-bed8-section) — Irrigation and Drainage Engineering, chapter 3 - [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 ## 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× - [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×