Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 18-Env-B2 / Water Resources. 3 hours duration; open-book exam (any non-communicating calculator permitted). Six questions are printed; the first five as they appear in the answer book constitute a complete paper and are marked, each worth 20 marks. All six are solved below for completeness.
Reference texts. Chow, Open-Channel Hydraulics; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); Linsley, Kohler & Paulhus, Hydrology for Engineers; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Freeze & Cherry, Groundwater; Fisheries Act, Canadian Environmental Protection Act, 1999; Ontario Water Resources Act and Clean Water Act, 2006 (used here as a representative province); CCME, Canada-Wide Strategy for the Management of Municipal Wastewater Effluent.
Question 3: Sluice Gate and Hydraulic Jump (20 marks)
The channel width is not given anywhere in the source. The solution below is therefore carried per unit width of channel, using the unit discharge q (m²/s); the part (b) energy-loss rate is reported as power per metre of channel width (W/m), which is the quantity that is actually determined by the given data.
Given. Reservoir (headwater) depth upstream of the gate y0 = 0.90 m (negligible approach velocity); tailwater depth downstream of the jump y2 = 0.388 m; rectangular channel; no head loss through the gate; the jump forms immediately after the vena contracta, so its depth y1 equals the gate-opening depth.
Find. (a) the vena-contracta depth y1; (b) the rate of energy dissipation through the jump, per unit channel width.
Approach. Conserve specific energy across the gate (reservoir to vena contracta, no losses) and conserve momentum across the jump (vena contracta to tailwater); the two relations share the unknowns y1 and the unit discharge q and are solved simultaneously.
Energy equation, reservoir to vena contracta. With negligible approach velocity in the reservoir and no head loss through the gate,
$$y_0 = y_1 + \dfrac{q^{2}}{2g\,y_1^{2}}$$
Momentum (sequent-depth) equation across the jump. For a rectangular channel,
$$y_2 = \dfrac{y_1}{2}\left(\sqrt{1+8\,\mathrm{Fr}_1^{2}}-1\right), \qquad \mathrm{Fr}_1 = \dfrac{q}{\sqrt{g}\,y_1^{1.5}}$$
Solve the two equations simultaneously. Substituting y0 = 0.90 m and y2 = 0.388 m and solving numerically for the physical (shallow, supercritical) root gives
$$q = 0.204\ \text{m}^2/\text{s}, \qquad V_1 = q/y_1 = 4.08\ \text{m/s}, \qquad \mathrm{Fr}_1 = 5.83$$
$$y_1 = \boxed{0.0500\ \text{m}\ (50\ \text{mm})}$$
Fr1 ≫ 1 confirms the approach flow really is supercritical, as required for a jump to form downstream.
Specific energy before and after the jump. With V2 = q/y2 = 0.526 m/s,
$$E_1 = y_1 + \dfrac{V_1^{2}}{2g} = 0.900\ \text{m}, \qquad E_2 = y_2 + \dfrac{V_2^{2}}{2g} = 0.402\ \text{m}$$
(E1 reproduces y0 exactly, confirming energy conservation through the gate). The head lost in the jump is $\Delta E = E_1-E_2 = 0.498\ \text{m}$.
Rate of energy dissipation. Power dissipated per unit width is $P' = \gamma\,q\,\Delta E$, with $\gamma = 9.81\ \text{kN/m}^3$:
$$P' = \boxed{997\ \text{W/m}\ (\approx 1.00\ \text{kW per metre of channel width})}$$
