Question 5 of 6: Classify the water-surface profile at a constriction
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 16-Civ-A5 Hydraulic Engineering — May 2019 · 3 hours · closed book (one aid sheet) · six questions, complete any four (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 $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
The Question 3 pipe specification used below is reconstructed from Figure 1. Check it against your copy of the paper.
Question 5: Classify the water-surface profile at a constriction (20 marks)
Given. Wide rectangular channel, unit discharge $q=Q/b=50/150=0.333\ \text{m}^2/\text{s}$.
Given data — Question 5
Quantity
Symbol
Value
Bottom width
$b$
150 m
Manning roughness
$n$
0.023
Bed slope
$S_0$
0.001 m/m
Discharge / unit discharge
$Q,\ q$
50 m³/s, 0.333 m²/s
Downstream control depth
$y$
0.4 m
Find. The profile classification (letter and number), justified by $y_c$, $y_n$, the slope type, and the Froude number.
Figure 5-1. On a mild slope ($y_c\lt y_n$) a downstream constriction backs water up: the surface rises above normal depth toward the control, the signature of an M1 profile.
Approach. Compute critical depth from the unit discharge, normal depth from Manning, and compare. Their order fixes the slope class (mild vs steep); the position of the actual surface relative to $y_n$ and $y_c$ fixes the region number. The nature of the control (a constriction) settles whether the surface rises or draws down.
Critical depth. For a rectangular channel $y_c=\left(q^2/g\right)^{1/3}$:
$$y_c=\left(\frac{0.333^2}{9.81}\right)^{1/3}=\boxed{0.225\ \text{m}.}$$
Normal depth (Manning). Solving $Q=\tfrac{1}{n}A R^{2/3}S_0^{1/2}$ with $A=by$, $R=by/(b+2y)$ for $y$ (the wide-channel value, $R\approx y$, gives $y=\left(qn/S_0^{1/2}\right)^{3/5}$; refining with the full $R$):
$$\boxed{y_n=0.428\ \text{m}.}$$
Slope type and flow regime. Since $y_n=0.428\ \text{m}\gt y_c=0.225\ \text{m}$, the bed is a mild slope ($S_0\lt S_c$) and uniform flow would be subcritical. At the control depth $y=0.4\ \text{m}$ the velocity is $v=q/y=0.833\ \text{m/s}$ and
$$Fr=\frac{v}{\sqrt{gy}}=\frac{0.833}{\sqrt{9.81(0.4)}}=0.42\ (\lt 1),$$
confirming subcritical flow.
Region and classification. A constriction is a downstream control that chokes the flow and backs water up, so the surface climbs above normal depth as it approaches the throat (Figure 3 rises to $\approx 0.45$ m at the constriction, above $y_n=0.428$). With the surface in the range $y\gt y_n\gt y_c$, subcritical, and a rising (positive) trend, Table 1 gives an
$$\boxed{\textbf{M1 profile (backwater curve).}}$$
The stated 0.4 m is essentially at normal depth (the near-tie is why the profile must be judged by its shape): a rising surface at a constriction is M1, whereas a free overfall on the same mild channel would draw the surface down toward $y_c$, an M2 profile.
