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18-Env-A2 Hydrology and Municipal Hydraulics Engineering · December 2013

Question 7 of 7: Trapezoidal Channel Uniform Flow, Specific Energy Over a Bed Rise, and V-Notch Weir Calibration

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

Notes on this paper

National Exams — December 2013 — 04-Env-A2 / Hydrology and Municipal Hydraulics Engineering. 3 hours duration; closed book with a candidate-prepared 8½×11 in double-sided aid sheet; Casio or Sharp approved calculator only. Any five questions constitute a complete paper (first five answers marked, 20 marks each, 100 marks total); all seven are solved below for completeness.

Reference texts. Davis & Cornwell, Introduction to Environmental Engineering (6th ed.) — hydrology, stormwater management and water-demand chapters; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — sanitary sewer hydraulics, Manning/Harmon design formulas; MWH’s Water Treatment: Principles and Design (3rd ed.) — distribution systems, pipe-network analysis and pump selection; Chow, Open-Channel Hydraulics — Manning's n tables and specific-energy theory; Chow, Maidment & Mays, Applied Hydrology — flood-frequency analysis; CCME water quality guidelines — cold-water fishery thermal protection.

Problem 7: Trapezoidal Channel Uniform Flow, Specific Energy Over a Bed Rise, and V-Notch Weir Calibration (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) Trapezoidal Channel: Discharge and Reynolds Number

Given.

Given data
QuantitySymbolValue
Normal depth$y$6 m
Base width$b$12 m
Side slope (H:V)$z$1:4 $\Rightarrow z=1/4$
Bed slope$S_0$0.05 (5%)
Lining—concrete
b = 12 m y = 6 m side slope z = 1/4 (H:V = 1:4) water surface Concrete-lined trapezoidal channel, S₀ = 5%
Trapezoidal channel cross-section: base $b=12$ m, side slope $z=1/4$, normal depth $y=6$ m.

Find. (a) $Q$ in m³/s; (b) $Re$ and flow type.

Approach. Compute the trapezoidal geometry ($A$, wetted perimeter $P$, hydraulic radius $R=A/P$), select a Manning's n appropriate to a concrete lining, apply Manning's equation for $Q$, then classify the flow with $Re=VR/\nu$.

Check: Manning's n is not printed on this page of the source; $n=0.015$ is used, a standard textbook design value for finished concrete (Chow's Open-Channel Hydraulics table 5-6, typical range 0.011–0.017). At $z=1/4$, this side slope (H:V = 1:4) is unusually steep — steeper than the 1.5:1–3:1 (H:V) typical of gravity-stable lined channels — and combined with the 5% bed slope, produces a very high computed velocity/Froude number (supercritical, $Fr\approx4.6$) well outside typical municipal channel practice; both values are used here exactly as printed on the exam.
  1. Geometry. $$A=(b+zy)y=(12+\tfrac14\times6)\times6=\boxed{81.0\ \text{m}^2},$$ $$P=b+2y\sqrt{1+z^2}=12+2(6)\sqrt{1+\tfrac{1}{16}}=\boxed{24.37\ \text{m}},\qquad R=\frac{A}{P}=\boxed{3.32\ \text{m}}.$$
  2. Manning's equation. $$Q=\frac{1}{n}AR^{2/3}S_0^{1/2}=\frac{1}{0.015}(81.0)(3.32)^{2/3}(0.05)^{1/2}=\boxed{2689\ \text{m}^3/\text{s}}.$$
  3. Velocity, Froude number and Reynolds number. $$V=\frac{Q}{A}=\frac{2689}{81.0}=33.2\ \text{m/s},\qquad Fr=\frac{V}{\sqrt{gA/T}}=4.56\ (\textbf{supercritical}),$$ $$Re=\frac{VR}{\nu}=\frac{33.2\times3.32}{1.0\times10^{-6}}=\boxed{1.10\times10^{8}}\ \Rightarrow\ \textbf{turbulent}.$$
QuantityValue
Flow area, $A$81.0 m²
Hydraulic radius, $R$3.32 m
Discharge, $Q$2689 m³/s
Mean velocity, $V$33.2 m/s (supercritical, $Fr=4.6$)
Reynolds number, $Re$1.10×10&sup8; — turbulent

(ii) Depth of Flow Over the Bed Rise (Specific Energy)

Given. Same channel ($b=12$ m, $z=1/4$); $Q=20$ m³/s; upstream normal depth $y_1=3$ m; bed rise $\Delta z=1$ m over the 8 m reach; frictional losses negligible.

y₁ = 3.00 m y₂ = 1.98 m Δz = 1.0 m bed (y₁ section) raised bed water surface (dips over the rise)
Longitudinal profile over the bed rise: subcritical flow responds to $\Delta z$ with a drop in depth, not a rise.

Find. $y_2$, the flow depth 8 m downstream where the bed has risen $\Delta z=1$ m.

Approach. With friction losses negligible, apply conservation of specific energy referenced to a common datum: since the bed itself rises by $\Delta z$, $E_1=\Delta z+E_2$. First check the Froude number (sub- or supercritical), then solve $E_2=y_2+Q^2/(2gA(y_2)^2)$ for $y_2$ on the appropriate branch, confirming the bump does not choke the flow ($E_2\ge E_{c,\min}$).

  1. Upstream velocity, Froude number and specific energy. $$A_1=(b+zy_1)y_1=(12+\tfrac14\times3)(3)=\boxed{38.25\ \text{m}^2},\qquad V_1=\frac{Q}{A_1}=\frac{20}{38.25}=0.523\ \text{m/s}.$$ Top width $T_1=b+2zy_1=13.5$ m, so $Fr_1=V_1/\sqrt{gA_1/T_1}=0.10<1$: subcritical flow. $$E_1=y_1+\frac{V_1^2}{2g}=3+\frac{(0.523)^2}{2(9.81)}=\boxed{3.014\ \text{m}}.$$
  2. Specific energy after the rise. $$E_2=E_1-\Delta z=3.014-1.0=\boxed{2.014\ \text{m}}.$$
  3. Check for choking (critical depth). Solving $Q^2T_c/(gA_c^3)=1$ for this trapezoid gives critical depth $y_c\approx0.65$ m and minimum specific energy $E_{c,\min}\approx0.976$ m. Since $E_2=2.014\ \text{m} > E_{c,\min}=0.976\ \text{m}$, the bump does not choke the flow, so a subcritical solution for $y_2$ exists.
  4. Solve for $y_2$ on the subcritical branch. Solving $E_2=y_2+\dfrac{Q^2}{2g\,[(b+zy_2)y_2]^2}=2.014$ m numerically for the root $y_cdecreases ($y_2
QuantityValue
Upstream specific energy, $E_1$3.014 m
Specific energy over the rise, $E_2$2.014 m
Critical depth / min. specific energy check$y_c=0.65$ m, $E_{c,\min}=0.976$ m — not choked
Depth over the bed rise, $y_2$1.98 m
Velocity over the bed rise, $V_2$0.81 m/s

(iii) V-Notch Weir Calibration of Stage-Discharge Curves

upstream pool (head H) H θ V-notch weir plate, notch angle θ
V-notch (triangular) weir: head $H$ above the notch vertex drives discharge through the fixed-geometry opening.

Given/Find. The equation and assumptions underlying the use of a V-notch weir to calibrate a stream's stage-discharge curve.

A V-notch weir is installed at a stable, controlled cross-section and produces a discharge that depends only on the measured upstream head $H$ above the notch vertex, following (for a fully contracted, thin-plate 90° notch, SI units) $$Q=1.4\,H^{5/2},$$ or more generally $Q=\tfrac{8}{15}C_d\tan(\theta/2)\sqrt{2g}\,H^{5/2}$ for notch angle $\theta$ and discharge coefficient $C_d$. Because $Q$ depends on $H$ alone through a fixed, well-characterized geometric relationship (unlike a natural channel cross-section, which drifts with scour/deposition), the weir provides an independent, repeatable $Q$–$H$ pair at its own location; simultaneously recording the natural channel's stage at the same time as the weir's head gives a directly verified point on the channel's stage-discharge rating curve, and repeating this over a range of flows lets the rating curve be checked (and re-calibrated) without relying solely on infrequent current-meter gaugings.

Key assumptions: (1) free (unsubmerged) flow over the notch, with the downstream water surface well below the notch vertex — a submerged weir invalidates the head-discharge equation and requires a separate submergence correction; and (2) fully contracted, thin-plate flow with negligible approach velocity — the approach channel must be wide and slow enough (and the weir plate sharp-edged) that the nappe springs cleanly clear of the plate, otherwise the calibrated $C_d$ (and hence $Q$) no longer applies.

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