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18-Env-A2 Hydrology and Municipal Hydraulics Engineering · December 2019

Question 5 of 7

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2019 — 18-Env-A2 / Hydrology and Municipal Hydraulics Engineering. 3 hours duration; closed book with an 8.5×11 in double-sided aid sheet; Casio or Sharp approved calculator only. Any five questions constitute a complete paper (only the first five answers in the work book are marked); all seven Problems are solved below for completeness. Each question is worth 20 marks.

Reference texts. Linsley, Kohler & Paulhus, Hydrology for Engineers (3rd ed.); Chow, Open-Channel Hydraulics; Walski et al., Advanced Water Distribution Modeling and Management; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Guidelines for Canadian Drinking Water Quality (Health Canada); Canadian Council of Ministers of the Environment (CCME) water-quality guidelines.

Problem 5 (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.

(i) Three predictors of streamflow change: reasonableness test (7 marks)

Precipitation. This is the most direct predictor: increasing precipitation depth and/or intensity increases both the volume and peak of streamflow (more rainfall/snowmelt input to the water balance, $P=ET+Q+\Delta S$), so a reasonableness test simply checks that predicted streamflow trends track observed precipitation trends in sign and rough magnitude — a model predicting rising streamflow in a region with declining precipitation would fail this basic check unless another predictor (e.g., glacier/snowpack melt) explains the discrepancy.

Antecedent wetness. Wetter antecedent soil moisture reduces the infiltration capacity available at the start of a storm, converting a larger fraction of subsequent rainfall directly to runoff (saturation-excess mechanism) — so a model should predict higher streamflow response for the same rainfall input when antecedent wetness is high, and a reasonableness test compares predicted flow response for similar-sized storms under wet versus dry antecedent conditions; a model insensitive to antecedent wetness is missing a first-order control on runoff generation.

Population density (urbanization). Higher population density is typically associated with greater impervious cover (roads, roofs, parking) and piped drainage, which reduces infiltration and shortens time of concentration — so a model should predict higher, "flashier" peak flows (higher peak, shorter time-to-peak) for otherwise-similar storms in more densely populated/urbanized sub-basins; a reasonableness check compares the predicted hydrograph shape (not just volume) between urbanizing and rural sub-catchments, since urbanization's main signature is a sharper, earlier peak rather than simply more total runoff.

(Temperature and agriculture are the remaining two predictors; a full reasonableness test would similarly confirm temperature's role via evapotranspiration/snowmelt rate and agriculture's role via altered infiltration/soil compaction and tile drainage, but the analysis above satisfies "any three" as requested.)

(ii) Trapezoidal grass channel: uniform-flow discharge and Reynolds number (6 marks)

Given.

QuantityValue
Normal depth, $y$2.0 m
Base width, $b$7 m
Side slope (H:V)1:3, i.e. $z=1/3$
Bed slope, $S_o$4% = 0.04
Manning's $n$ (short, well-maintained grass)0.035
b = 7 my = 2.0 mH:V = 1:3Grass-lined trapezoidal channel (uniform flow)
Fig. 2 — Trapezoidal cross-section, $b=7$ m, $y=2.0$ m, side slopes 1:3 (H:V).

Find. (a) Discharge $Q$ in m³/min; (b) Reynolds number and flow type.

Approach. Trapezoidal geometry gives $A$, $P$, $R$; Manning's equation gives $Q$; open-channel $Re$ uses the hydraulic radius, $Re=VR/\nu$.

  1. Part (a) — geometry and discharge. $$A=(b+zy)y=(7+\tfrac{1}{3}\times2)(2)=\boxed{15.33\ \text{m}^2}\qquad P=b+2y\sqrt{1+z^2}=7+2(2)\sqrt{1.111}=11.22\ \text{m}\qquad R=\frac{A}{P}=1.367\ \text{m}$$ $$Q=\frac{1}{n}AR^{2/3}S_o^{1/2}=\frac{1}{0.035}(15.33)(1.367)^{2/3}(0.04)^{1/2}=107.9\ \text{m}^3/\text{s}=\boxed{6475\ \text{m}^3/\text{min}}$$
  2. Part (b) — Reynolds number. $$V=\frac{Q}{A}=\frac{107.9}{15.33}=7.04\ \text{m/s}\qquad Re=\frac{VR}{\nu}=\frac{(7.04)(1.367)}{1.0\times10^{-6}}=\boxed{9.62\times10^{6}}$$ Since open-channel flow is turbulent for $Re\gtrsim12{,}500$ (hydraulic-radius Reynolds number), this flow is strongly turbulent.
QuantityValue
Flow area, $A$15.33 m²
Hydraulic radius, $R$1.367 m
Discharge, $Q$107.9 m³/s = 6475 m³/min
Reynolds number, $Re$9.62×10&sup6; (turbulent)
Check: an H:V of 1:3 (steeper than vertical-leaning, i.e. only 1 m of horizontal run per 3 m of rise) is an unusually steep side slope for a grass-lined channel — most vegetated channels use $z\ge2$–3 for mowability and bank stability — and the resulting velocity (7.04 m/s) would erode any real grass lining (typical permissible velocities for grass channels are ≈1.5–2.5 m/s). Both the side slope and the 4% bed slope are taken exactly as printed on the exam, consistent with this subject's established convention of solving with the literal given data and flagging an atypical value rather than silently adjusting it.

(iii) Specific energy over a bed rise, Q = 60 m³/s (7 marks)

Y₁ = 2.0 mY₂ = Y_c= 1.897 mΔz=0.7 mSection 1Crest (choked, Y=Y_c)flow
Fig. 3 — Longitudinal profile of the bed rise $\Delta z=0.7$ m; same trapezoidal cross-section ($b=7$ m, $z=1/3$) as part (ii). The upstream approach depth $Y_1=2.0$ m is very close to critical, so this bump chokes the flow.

Given. $Q=60\ \text{m}^3/\text{s}$, $Y_1=2.0$ m, bed rise $\Delta z=0.7$ m, same trapezoidal section ($b=7$ m, $z=1/3$); frictional losses negligible.

Find. Depth $Y_2$ at the raised section.

Approach. With no friction loss, specific energy referenced to the local bed falls by exactly $\Delta z$ across the rise, $E_2=E_1-\Delta z$. Before solving for $Y_2$, the crossing must be checked against the minimum specific energy (critical-flow) condition — if $\Delta z$ exceeds the available margin $E_1-E_{\min}$, no subcritical $Y_2$ exists and the crossing is choked.

  1. Approach conditions at Section 1. $$A_1=(7+\tfrac{1}{3}\times2)(2)=15.33\ \text{m}^2\qquad V_1=\frac{60}{15.33}=3.914\ \text{m/s}\qquad E_1=Y_1+\frac{V_1^2}{2g}=2.0+0.780=\boxed{2.780\ \text{m}}$$ $$\text{Froude number: } Fr_1=\frac{Q}{A_1\sqrt{gA_1/T_1}}=0.921<1\ \ (\text{subcritical, but close to critical})$$
  2. Check whether the bump chokes the flow. Solving $Q^2T/(gA^3)=1$ numerically for the critical depth of this trapezoidal section gives $Y_c=1.897$ m, with minimum specific energy $E_{\min}=2.772$ m. The available margin is only $$\Delta z_{\max}=E_1-E_{\min}=2.780-2.772=0.008\ \text{m}$$ Since the actual rise $\Delta z=0.7\ \text{m}\gg\Delta z_{\max}=0.008\ \text{m}$, the crossing is choked — no subcritical depth satisfies $E_2=E_1-\Delta z$ at this upstream depth.
  3. Depth at the raised section. A choked crossing is forced through critical depth at its highest point (the flow cannot carry more energy deficit than $E_1-E_{\min}$, so the free surface adjusts until the crest itself runs at critical depth): $$\boxed{Y_2=Y_c=1.897\ \text{m}}$$
  4. What actually happens upstream (afflux). Since $Y_1=2.0$ m cannot supply enough specific energy to clear a 0.7 m rise, water backs up until the approach section itself carries the higher specific energy $E_1'=E_{\min}+\Delta z=2.772+0.7=3.472$ m. Solving $E_1'=Y_1'+Q^2/(2gA(Y_1')^2)$ on the subcritical branch: $$Y_1'=\boxed{3.196\ \text{m}}$$ i.e., the stated $Y_1=2.0$ m upstream depth is not sustainable once the 0.7 m bump is in place — the channel would back up (afflux) to about 3.20 m upstream of the rise before flow could pass over the crest at critical depth.
QuantityValue
Specific energy at Section 1 (at stated $Y_1=2.0$ m), $E_1$2.780 m
Critical depth for $Q=60$ m³/s, $Y_c$1.897 m
Available margin, $E_1-E_{\min}$0.008 m (<< $\Delta z=0.7$ m → choked)
Depth at the raised section, $Y_2$1.897 m (= $Y_c$)
Required upstream depth to actually clear the hump, $Y_1'$3.196 m