--- title: "Normal depth in compound channel" subject: "Hydraulics" chapter: 6 key: "semester_4/hydraulics:compound-channel-normal-depth-berms" source: https://www.bctengineer.com/q/semester_4/hydraulics/compound-channel-normal-depth-berms site: BCT Engineering --- # Normal depth in compound channel Hydraulics · Chapter 6: Uniform flow in open channel - Asked in: 2071 Bhadra, 6 marks - Video: [Manning’s equation to calculate discharge for a compound open channel](https://www.youtube.com/watch?v=BJZ73WWEc3M) — Fluids Explained - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/compound-channel-normal-depth-berms - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6 ## Question For given channel section shown in the figure below with bed slope = 0.00017, Manning's roughness coefficient = 0.018, discharge 8.97 m³/s, and side slope as 1:1, determine the normal depth of flow for uniform flow. [Figure: compound section with a central trapezoidal channel of bottom width 3 m and 1:1 side slopes (1 m horizontal each), and 2 m wide flood berms on both sides; depth Y measured from the berm level] ## Short answer Divide the compound section into central channel and berms. Using Manning's equation with n=0.018 and S=0.00017, trial gives depth above berm level Y≈0.82 m. Total depth from channel bottom is about 1.82 m for Q=8.97 m³/s. ## Answer #### Section (from the figure) Central channel: bottom width 3 m, side slopes 1:1, depth 1 m (1 m horizontal each side), so top width at berm level $=3+2=5$ m. Two berms (flood plains) 2 m wide each. Let $Y$ = depth of water above the berm level, so total depth $=1+Y$. The outer edges of berms are assumed vertical. ```text berm Y (above berm) berm 2 m ~~~~~~~~~~~~~~~~~~~~~~~~ 2 m ____| |____ \ / 1 m \__________________/ 3 m ``` The section is divided by vertical lines at the berm edges; those lines are not counted as wetted perimeter. $S_0=0.00017$, $n=0.018$, $Q=8.97$ m³/s. #### Subsection properties Central channel: $$ A_c=\frac{(3+5)}{2}(1)+5Y=4+5Y,\qquad P_c=3+2\sqrt2=5.828\ \text{m} $$ Each berm (width 2 m, depth $Y$, vertical outer wall): $$ A_b=2Y,\qquad P_b=2+Y $$ #### Discharge $$ Q=\frac{\sqrt{S_0}}{n}\left[A_cR_c^{2/3}+2A_bR_b^{2/3}\right],\qquad \frac{\sqrt{0.00017}}{0.018}=0.72436 $$ Trial values of $Y$: | $Y$ (m) | $Q$ (m³/s) | |---|---| | 0.6 | 6.767 | | 0.8 | 8.753 | | 0.9 | 9.815 | Solving $Q(Y)=8.97$ gives $Y=0.821$ m. #### Check at $Y=0.821$ m | Part | $A$ (m²) | $P$ (m) | $R$ (m) | $Q$ (m³/s) | |---|---|---|---|---| | Central | 8.104 | 5.828 | 1.390 | 7.312 | | Each berm | 1.642 | 2.821 | 0.582 | 0.829 | | Total | | | | 8.970 | **Answer: depth of flow above the berm level $Y\approx0.82$ m (total depth from the bed $\approx1.82$ m).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When splitting the compound section by vertical lines at the berm edges, how are those lines treated? A. Counted as wetted perimeter of both parts B. Counted as wetted perimeter of the berm only C. Counted as wetted perimeter of the main channel only D. Not counted in the wetted perimeter Answer: D. The dividing lines are imaginary, with no real boundary shear, so they are excluded from P. (Key point: Subdividing the section) 2. In the solution, Y is measured: A. From the channel bed B. Above the berm level, so total depth = 1 + Y C. From the top of the water to the berm D. As half of total depth Answer: B. The central channel is 1 m deep, so water above the berm is Y and total depth from the bed is 1 + Y. (Key point: Depth definition) 3. Which are the correct properties of the central channel and each berm? A. A_c = 4 + 5Y, P_c = 5.828 m; A_b = 2Y, P_b = 2 + Y B. A_c = 3Y, P_c = 5 m; A_b = 2Y, P_b = 2 C. A_c = 4 + 5Y, P_c = 3 m; A_b = 2Y, P_b = 4 + Y D. A_c = 8Y, P_c = 5.828 m; A_b = Y, P_b = 2 + 2Y Answer: A. The central channel is a trapezoid (area 4 m2) plus 5Y above the berm, with P = 3 + 2 sqrt2; each berm is 2 m wide with a vertical outer wall. (Key point: Subsection properties) 4. How is the normal depth found? A. Directly from Manning's formula using the total area B. By summing the discharges of the three subsections and trying Y until the total equals 8.97 m3/s (Y about 0.82 m) C. By equating the three velocities D. By taking the average of the berm and main depth Answer: B. Q(Y) = (sqrt(S0)/n)[A_c R_c^(2/3) + 2 A_b R_b^(2/3)] is solved by trial to give Y = 0.821 m. (Key point: Solving for Y) ## Related in Hydraulics - [Equivalent roughness of compound channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acompound-channel-equivalent-roughness-horton-einstein) — Hydraulics, chapter 6 - [Bed slope for critical normal depth](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Abed-slope-for-critical-flow-at-normal-depth) — Hydraulics, chapter 7 - [Program for normal depth in trapezoidal channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aalgorithm-normal-depth-trapezoidal) — Hydraulics, chapter 6 - [Boundary shear stress and compound channel flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ashear-stress-boundary-compound-channel-flow) — Hydraulics, chapter 6 - [GVF in 5 m channel, depth 0.25 m to 0.40 m](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Agvf-5m-channel-1-85-m3s-direct-step) — Hydraulics, chapter 8, asked 2× - [Contraction and bed rise, 3.5 m channel (1.2 m depth)](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acontraction-bed-rise-3-5m-wide-1-2m-depth) — Hydraulics, chapter 7 ## Related in other subjects - [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 - [Normal depth in triangular channel (1:1)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-0-0004) — Hydro Power, chapter 1, asked 2× - [Efficient rectangular section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-rect) — Hydro Power, chapter 1, asked 2×