Question 5 of 6: Gradually varied flow — profile classification
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 16-Civ-A5 Hydraulic Engineering — December 2019 · 3 hours · closed book (one aid sheet) · six questions, complete any five (all six solved here) · each question 20 marks, equal-value parts.
Reference texts: Chin, Water-Resources Engineering (Pearson); Chow, Open-Channel Hydraulics (McGraw-Hill); Chaudhry, Open-Channel Flow (Springer); Wylie & Streeter, Fluid Transients in Systems (Prentice-Hall); Munson, Young & Okiishi, Fundamentals of Fluid Mechanics (Wiley); Roberson, Cassidy & Chaudhry, Hydraulic Engineering.
Governing relations supplied on the exam cover sheet. Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with slope $S=\Delta h/L$ (SI, $Q$ in m³/s, $D$ in m); Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$; Darcy–Weisbach $\Delta h = 0.0826\,\tfrac{fL}{D^5}Q^2$. Unless stated, local losses and velocity head are neglected, diameters are nominal, and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Note on question/figure labels. The transient plots on pages 2–3 are printed as “Figure Q4–A / Q4–B” but belong to Question 1 (the 450 mm and 750 mm pipe responses); “Figure 1” (valve main) is the figure for Question 3 and “Figure 2” (pipe grid) is for Question 4. Labels are given as printed; the figures are matched to their questions by content.
Given. Wide rectangular channel: $b=150\ \text{m}$, $n=0.023$, $S_0=0.001$, $Q=50\ \text{m}^3/\text{s}$; a downstream constriction sets the control depth (≈0.4–0.45 m, Figure 3). Unit discharge $q=Q/b=0.333\ \text{m}^2/\text{s}$.
Find. Critical depth $y_c$, normal depth $y_n$, the slope class and the profile type; justify with the classification table.
Figure 5-1. Mild-slope backwater: subcritical flow ($y_n\gt y_c$) backed up by the downstream constriction rises above normal depth toward the control, then decays asymptotically to $y_n$ upstream — an M1 profile (positive water-surface slope). (Vertical scale exaggerated.)
Approach. Compute $y_c$ from the unit discharge, $y_n$ from Manning, compare them to fix the slope class, check the Froude number to confirm the regime, and match the observed profile shape (depth increasing toward the downstream control) against the classification table.
Critical depth. For a rectangular channel $y_c=\big(q^2/g\big)^{1/3}$ with $q=50/150=0.333\ \text{m}^2/\text{s}$:
$$y_c=\left(\frac{0.333^2}{9.81}\right)^{1/3}=\boxed{0.225\ \text{m}.}$$
Normal depth. Manning $Q=\tfrac{1}{n}A\,R^{2/3}S_0^{1/2}$ with $A=b\,y$, $R=by/(b+2y)$. Solving $50=\tfrac{1}{0.023}(150y)\!\left(\tfrac{150y}{150+2y}\right)^{2/3}\!\sqrt{0.001}$ (wide channel, $R\approx y$) gives
$$\boxed{y_n=0.428\ \text{m}.}$$
Slope class. Since $y_n=0.428\ \text{m} \gt y_c=0.225\ \text{m}$, the slope is mild ($S_0 \lt S_c$; the critical slope computed at $y_c$ is $S_c=0.0086 \gt 0.001$). Normal flow is therefore subcritical.
Regime check. At the control depth the Froude number $Fr=\dfrac{Q}{b\,y\sqrt{g\,y}}$ is $Fr(0.40)=0.42$ and $Fr(0.45)=0.35$ — both $\lt 1$, subcritical, consistent with a mild slope.
Profile classification. A constriction on a subcritical mild-slope channel acts as a downstream control that backs water up: the depth at the constriction exceeds normal depth and the surface rises in the flow direction (positive slope, Figure 3), decaying to $y_n$ far upstream. That is the row $y \gt y_n \gt y_c$, subcritical, positive surface — a
$$\boxed{\textbf{Type M1 (mild) backwater profile.}}$$
Check: prose vs figure depth. The stem states a downstream depth of ~0.4 m, essentially equal to the computed normal depth (0.428 m), while Figure 3 shows the surface rising to ~0.45 m at the constriction. The classification is governed by the positive (rising toward downstream) surface slope of a constriction-controlled subcritical mild channel, which is unambiguously M1; the small 0.40–0.45 m reading discrepancy does not change the class. Were the control instead below $y_n$ (e.g. a free overfall), the same channel would show an M2 drawdown.
