Question 6 of 6: Trapezoidal channel — normal depth, critical depth, regime
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 16-Civ-A5 Hydraulic Engineering — May 2018 · 3 hours · closed book (one aid sheet) · six questions, complete any five (all solved here) · each question 20 marks, equal-value parts.
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 $\Delta h = 0.0826\,\tfrac{fL}{D^5}Q^2$; total dynamic head $TDH=H_s+H_f$. Unless stated, local losses and velocity head are neglected, diameters are nominal, and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Find. (a) design channel height with 10% freeboard; (b) critical depth at the weir; (c) upstream flow regime; (d) the specific-energy diagram showing the sub-critical→critical transition.
Figure 6a. Trapezoidal cross-section: $b=10$ m, 3H:1V sides, normal-depth water line.
Approach. Normal depth from Manning fixes the flow depth on the channel slope; adding freeboard gives the design height. Critical depth from the trapezoidal criterion $Q^2/g=A^3/T$ sets the control at the weir. Comparing the two depths classifies the regime, and the specific-energy curve $E=y+Q^2/(2gA^2)$ shows the draw-down toward critical.
Normal depth (Manning). Solve $Q=\tfrac1n A R^{2/3}S_0^{1/2}$, i.e. $3.0=\tfrac{1}{0.023}A R^{2/3}(0.0015)^{1/2}$, for $y_n$. Iteration gives
$$\boxed{y_n\approx0.348\ \text{m}} \quad(V_n=Q/A_n\approx0.76\ \text{m/s}).$$
Design height with 10% freeboard (part a).
$$H_{\text{design}}=1.10\,y_n=1.10(0.348)\approx\boxed{0.38\ \text{m}}.$$
Critical depth (part b). At critical flow $\dfrac{Q^2}{g}=\dfrac{A_c^3}{T_c}$ with $A_c=(10+3y_c)y_c$, $T_c=10+6y_c$. Solving $\dfrac{3.0^2}{9.81}=0.9174=\dfrac{A_c^3}{T_c}$ gives
$$\boxed{y_c\approx0.205\ \text{m}}.$$
Regime well upstream (part c). Compare depths: $y_n=0.348\ \text{m}\gt y_c=0.205\ \text{m}$, so the normal-depth Froude number is
$$Fr_n=\frac{V_n}{\sqrt{gA_n/T_n}}\approx0.44\lt1.$$
$$\boxed{\text{Flow well upstream is sub-critical (mild slope, } y_n\gt y_c).}$$
Specific-energy transition (part d). On the mild slope the flow sits on the upper (sub-critical) limb of the $E$–$y$ curve at $y_n=0.348\ \text{m}$ ($E_n\approx0.379\ \text{m}$). Approaching the broad-crested weir the flow accelerates and the surface draws down toward the critical control at the crest, where specific energy is minimum, $E_{\min}\approx0.302\ \text{m}$ at $y_c=0.205\ \text{m}$. The operating point slides down the upper limb to the nose of the curve:
Figure 6b. Specific-energy curve $E=y+Q^2/(2gA^2)$. The flow moves from the sub-critical point ($y_n=0.348$ m) down the upper limb to the critical nose ($y_c=0.205$ m, $E_{\min}$) at the weir.
