--- title: "Design stable channel in 8 mm gravel" subject: "Hydraulics" chapter: 10 key: "semester_4/hydraulics:stable-channel-8mm-gravel-design" source: https://www.bctengineer.com/q/semester_4/hydraulics/stable-channel-8mm-gravel-design site: BCT Engineering --- # Design stable channel in 8 mm gravel Hydraulics · Chapter 10: Flow in mobile boundary channel - Asked in: 2082 Kartik, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/stable-channel-8mm-gravel-design - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10 ## Question Design a stable non-erodible straight channel to carry 15 m³/s of clear water through a 8 mm bed of rounded gravel. A longitudinal slope of 0.0009 and side slope of 2H:1V are to be adopted. Take angle of repose as 32°. ## Short answer Critical shear τc=0.06(1.65×9810×0.008)=7.77 N/m^2, side factor K=0.536, depth limited by sides to 0.63 m. With Strickler's n=0.021, Manning's equation yields base width 22.5 m. Section: b=22.5 m, y=0.63 m, side slope 2H:1V. ## Answer **Given:** $Q = 15\ \text{m}^3/\text{s}$, gravel $d = 8$ mm (rounded), $S_0 = 0.0009$, side slope 2H : 1V ($z = 2$), $\phi = 32^\circ$. **Assumptions:** the Shields value $\tau_* = 0.06$ (fully rough), $\gamma_s = 2.65\gamma$; maximum shear on the bed $\gamma y S_0$ and on the sides $0.75\gamma y S_0$; Manning's $n$ from Strickler's formula $n = 0.047\,d^{1/6}$ ($d$ in m). **Step 1: Critical shear stress on a level bed** $$ \tau_c = 0.06(\gamma_s - \gamma)d = 0.06(1.65)(9810)(0.008) = 7.77\ \text{N/m}^2 $$ **Step 2: Side slope factor** $$ \theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{\sin^2 26.57^\circ}{\sin^2 32^\circ}} = \sqrt{1 - \frac{0.2}{0.2808}} = 0.536 $$ **Step 3: Limiting depth** - Bed: $\gamma y S_0 \le \tau_c \Rightarrow y \le \dfrac{7.77}{9810 \times 0.0009} = 0.880$ m - Sides: $0.75\gamma y S_0 \le K\tau_c \Rightarrow y \le \dfrac{0.536 \times 7.77}{0.75 \times 9810 \times 0.0009} = 0.629$ m The side slope governs: **$y = 0.63$ m**. **Step 4: Base width.** Strickler: $n = 0.047(0.008)^{1/6} = 0.021$. Manning's equation with $y = 0.629$ m, $z = 2$: $$ Q = \frac{1}{n}AR^{2/3}S_0^{1/2}, \quad A = (b + 2 \times 0.629)(0.629) $$ Trial gives $b = 22.5$ m: $A = 14.93\ \text{m}^2$, $P = 25.27$ m, $R = 0.591$ m, and $$ Q = \frac{1}{0.021}(14.93)(0.591)^{2/3}(0.03) = 15.0\ \text{m}^3/\text{s} \checkmark $$ **Step 5: Checks** $$ V = \frac{15}{14.93} = 1.00\ \text{m/s},\quad Fr = \frac{V}{\sqrt{gA/T}} = 0.41 < 1\ \text{(subcritical)} $$ **Answer: Trapezoidal section with $b \approx 22.5$ m, $y \approx 0.63$ m, side slope 2H : 1V, $S_0 = 0.0009$.** The channel is wide and shallow because the small gravel cannot withstand more shear stress. Add a freeboard (about 0.3 m). ## 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 8 mm gravel found? A. tau_c = 0.06 (gamma_s - gamma) d = 7.77 N/m^2 B. tau_c = gamma y S0 C. tau_c = 0.75 gamma y S0 D. tau_c = K x 7.77 Answer: A. Using Shields tau* = 0.06 with gamma_s = 2.65 gamma gives 7.77 N/m^2. (Key point: Critical shear) 2. What is K for 2H:1V sides and phi = 32 degrees? A. About 1.0 B. About 0.54 C. About 0.20 D. About 0.80 Answer: B. theta = 26.57 degrees, sin^2 = 0.2, sin^2(32) = 0.2808, K = sqrt(1 - 0.712) = 0.536. (Key point: Side factor) 3. Which condition governs the depth and what is it? A. Bed, y = 0.88 m B. Side slope, y = 0.88 m C. Side slope, y = 0.63 m D. Bed, y = 0.63 m Answer: C. Bed allows 0.88 m but sides allow only 0.63 m; the smaller governs. (Key point: Governing depth) 4. Why is the final section wide (b about 22.5 m) and shallow? A. Manning's n is large B. The side slope is steep C. Flow is supercritical D. Small gravel resists little shear, so depth is limited and width must carry the flow Answer: D. Limited depth forces a large base width to carry 15 m^3/s. 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