Trapezoidal Channel — Normal and Critical Depth: 18-Env-B2 Water Resources, December 2018, Q2

Question 2: Trapezoidal Channel — Normal and Critical Depth (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

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The source exam gives the discharge, bottom width, and side slope but does not state the channel’s longitudinal slope S; no slope value appears anywhere on it. Normal depth cannot be found from Manning’s equation without one. Per the exam’s own Note 1 (“if doubt exists…submit a clear statement of any assumptions made”), this solution assumes S = 0.001 (the same mild municipal grade explicitly given for the sewer in Question 4) and n = 0.013 for the stated concrete lining (troweled-finish concrete, Chow Table 5-6). The critical-depth answer in part (b) does not depend on either assumption.

Given. A trapezoidal, concrete-lined channel carrying a steady discharge, with the bottom width and side slope shown below (slope and roughness as assumed above).

Given data
Quantity	Symbol	Value
Discharge	Q	15 m³/s
Bottom width	b	4.5 m
Side slope (H:V)	m	1.5
Manning roughness (assumed, concrete)	n	0.013
Longitudinal slope (assumed)	S	0.001

Find. (a) the normal depth y_n; (b) the critical depth y_c.

Trapezoidal channel cross-section: bottom width b, depth y, side slope m (horizontal run per unit vertical rise).

Approach. Solve Manning’s equation iteratively for the uniform (normal) flow depth using the assumed slope and roughness, then locate the critical depth independently from the Froude-number-unity condition, which needs neither.

Set up the trapezoidal geometry and Manning’s equation. For depth y: area, wetted perimeter, and hydraulic radius are $$A(y) = (b + m\,y)\,y, \qquad P(y) = b + 2y\sqrt{1+m^{2}}, \qquad R(y) = A(y)/P(y)$$ and uniform flow satisfies $Q = \dfrac{1}{n}A\,R^{2/3}S^{1/2}$.
Solve for normal depth by iteration. Substituting b = 4.5 m, m = 1.5, n = 0.013, S = 0.001 and searching for the y that gives Q = 15 m³/s converges to $$A_n = 7.04\ \text{m}^2, \quad P_n = 8.59\ \text{m}, \quad R_n = 0.819\ \text{m}, \quad V_n = Q/A_n = 2.13\ \text{m/s}$$ $$y_n = \boxed{1.14\ \text{m}}$$
Locate the critical depth from Fr = 1. Critical flow in a non-rectangular section occurs where $Q^{2}T/(gA^{3}) = 1$, with top width $T(y) = b + 2m\,y$. Solving this (independent of n and S) for the same Q, b, m gives $$A_c = 5.51\ \text{m}^2, \quad T_c = 7.30\ \text{m}, \quad V_c = Q/A_c = 2.72\ \text{m/s}$$ $$y_c = \boxed{0.934\ \text{m}}$$ Substituting back confirms $V_c/\sqrt{gA_c/T_c} = 1.00$.
Compare the two depths. Since y_n (1.14 m) exceeds y_c (0.934 m), the assumed mild slope produces subcritical normal flow, which is the expected regime for a gently-graded, concrete-lined municipal channel.

Final Results
Quantity	Value
Normal depth, y_n (assumed S, n)	1.14 m
Velocity at normal depth, V_n	2.13 m/s
Critical depth, y_c	0.934 m
Velocity at critical depth, V_c	2.72 m/s
Flow regime at normal depth	Subcritical (y_n > y_c)

Concept & Theory

Topic: Manning’s equation and critical flow in a trapezoidal channel.

Key relations: $Q = \tfrac{1}{n}AR^{2/3}S^{1/2}$ (uniform/normal flow); critical flow at $\mathrm{Fr} = V/\sqrt{gA/T} = 1$, equivalently $Q^2T/(gA^3)=1$.

Why this works: Normal depth is where the gravity component of weight along the slope exactly balances boundary friction, so it depends on S and n; critical depth is a purely geometric/energetic condition (minimum specific energy for the given Q) and is independent of both.

Common pitfall: Using the horizontal-to-vertical side-slope ratio backwards (m is horizontal run per unit vertical rise, not the reverse), and forgetting that the wetted perimeter excludes the free top width — only the sloped sides and the bottom are wetted.

Source: Chow, Open-Channel Hydraulics, ch. 5 (uniform flow) and ch. 3 (critical flow, specific energy).

Question 2 of 6: Trapezoidal Channel — Normal and Critical Depth

Question 2: Trapezoidal Channel — Normal and Critical Depth (20 marks)