NivaarExam PrepOfficial exam papers ↗

18-Env-A2 Hydrology and Municipal Hydraulics Engineering · May 2015

Question 7 of 7: Open-Channel Flow, Specific Energy, and Streamflow Measurement

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

Notes on this paper

National Exams — May 2015 — 04-Env-A2 / Hydrology and Municipal Hydraulics Engineering. 3 hours duration; closed book with a candidate-prepared 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 as they appear in the work book are marked); all seven are solved below for completeness. Each question ("Problem") is worth 20 marks, with sub-part weights shown in brackets.

Reference texts. Chow, Open-Channel Hydraulics; Linsley, Kohler & Paulhus, Hydrology for Engineers (3rd ed.); Walski et al., Advanced Water Distribution Modeling and Management; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.).

Problem 7: Open-Channel Flow, Specific Energy, and Streamflow Measurement (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) Rock-Lined Trapezoidal Channel — Discharge and Reynolds Number

Given. Trapezoidal channel, uniform flow:

Given data
QuantitySymbolValue
Normal depth$y$5 m
Base width$b$12 m
Side slope (H:V, as printed)$z$1:4 → $z=0.25$
Bed slope$S_o$0.04 (4%)
Check: no Manning's $n$ is given for the rock lining; a typical riprap/rock-lined-channel value $n=0.035$ is assumed (a common textbook default for moderate riprap, absent a stated $D_{50}$). A 1:4 (H:V) side slope is steeper than the 2:1–4:1 range usually recommended for stable riprap placement, but it is taken literally as printed rather than "corrected," since it is not numerically impossible; the resulting hydraulic radius (and hence velocity) is dominated by the 5 m depth and 12 m base rather than by the side slope, so the computed velocity would be similarly high under either reading.

Find. The discharge $Q$ and the Reynolds number/flow type.

Approach. Compute the trapezoidal section's area, wetted perimeter and hydraulic radius, apply Manning's equation for $V$ and $Q$, then evaluate $Re$ using the hydraulic diameter $4R$.

  1. (a) Channel geometry and discharge. $$A = (b+zy)y = (12+0.25\times5)(5) = 66.25\ \text{m}^2, \qquad P = b+2y\sqrt{1+z^2} = 12+2(5)\sqrt{1.0625}=22.31\ \text{m},$$ $$R = A/P = 2.970\ \text{m}, \qquad V = \frac{1}{n}R^{2/3}S_o^{1/2} = \frac{1}{0.035}(2.970)^{2/3}(0.04)^{1/2} = 11.81\ \text{m/s},$$ $$Q = VA = (11.81)(66.25) = \boxed{782\ \text{m}^3/\text{s}}.$$
  2. (b) Reynolds number. Using the hydraulic diameter $D_h=4R=11.88$ m as the open-channel analogue of pipe diameter: $$Re = \frac{V(4R)}{\nu} = \frac{(11.81)(11.88)}{1.0\times10^{-6}} = \boxed{1.40\times10^{8}}.$$ Since $Re \gg 4000$, the flow is turbulent (as expected for essentially any open-channel flow at engineering scale).
QuantityValue
Flow area, $A$66.25 m²
Hydraulic radius, $R$2.970 m
Velocity, $V$11.81 m/s
Discharge, $Q$782 m³/s
Reynolds number, $Re$$1.40\times10^{8}$ (turbulent)

(ii) Specific Energy over a Streambed Rise

Y₁ Y₂ Δz Trapezoidal channel bed (Problem 7i), 15 m reach Flow over a raised bed section (specific-energy problem)
Flow depth Y₁ upstream drops to Y₂ over the 0.6 m bed rise, 15 m downstream — subcritical approach flow accelerates over the hump, matching the printed figure's dip in the water surface.

Given. Same trapezoidal channel as (i) ($b=12$ m, $z=0.25$); $Q=50$ m³/s, $Y_1=2.5$ m, bed rise $\Delta z=0.6$ m, 15 m downstream, frictionless.

Find. The downstream depth $Y_2$.

Approach. Compute the upstream specific energy $E_1$, subtract the bed rise to get $E_2=E_1-\Delta z$, confirm the hump does not choke the flow (compare $E_2$ to the critical minimum specific energy), then solve $E_2=Y_2+Q^2/(2gA(Y_2)^2)$ for the subcritical root.

  1. Upstream specific energy. $A_1=(12+0.25\times2.5)(2.5)=31.56\ \text{m}^2$, $V_1=Q/A_1=50/31.56=1.584\ \text{m/s}$: $$E_1 = Y_1+\frac{V_1^2}{2g} = 2.5+\frac{(1.584)^2}{19.62} = \boxed{2.628\ \text{m}}.$$
  2. Specific energy at the hump. $$E_2 = E_1-\Delta z = 2.628-0.600 = 2.028\ \text{m}.$$
  3. Choking check. The critical depth for this section and $Q$ is $y_c=1.199$ m, giving a minimum specific energy $E_{min}=1.785$ m. Since $E_2=2.028\ \text{m} > E_{min}=1.785\ \text{m}$, the hump does not choke the flow and a subcritical solution exists.
  4. Solve for $Y_2$ (subcritical branch). Solving $E_2=Y_2+Q^2/(2gA(Y_2)^2)$ numerically on the branch $Y_2>y_c$: $$Y_2 = \boxed{1.76\ \text{m}}.$$ The depth drops from 2.50 m to 1.76 m over the hump — consistent with the printed figure's dip in the water surface for subcritical flow accelerating over a bed rise.
QuantityValue
Upstream specific energy, $E_1$2.628 m
Specific energy over hump, $E_2$2.028 m
Critical depth, $y_c$ (check)1.199 m (no choking)
Downstream depth, $Y_2$1.76 m

(iii) Streamflow Measurement: Flume, Weir, and Rating Curve

A flume with a stilling well is a specially-shaped open-channel constriction (e.g. Parshall flume) that forces flow through a critical-depth control section; a stilling well connected to the flume via a small-diameter pipe damps out short-period surface turbulence and waves so the water-surface (or head) elevation can be read or recorded steadily. Flumes are preferred where the stream carries significant sediment or debris (unlike a weir, a flume has no raised crest to trap bed load) and where head loss must be kept low, since a flume causes a much smaller afflux (upstream backwater) than an equivalent weir for the same discharge.

A weir (sharp-crested or broad-crested) is a raised, fixed overflow structure across the channel; discharge is computed from the measured upstream head above the weir crest via a calibrated head-discharge equation (e.g. $Q = C_wLH^{3/2}$ for a rectangular weir). Weirs are preferred in relatively clean, low-sediment streams with a stable channel section, where the simplicity, accuracy and low cost of a fixed structure outweigh the drawbacks of sediment trapping and the larger backwater it creates.

A rating curve is the empirical relationship between stage (water-surface elevation, continuously and cheaply recorded) and discharge (periodically measured directly, e.g. by current-meter or ADCP gauging), fitted from a scatter of paired stage/discharge observations across a range of flows. Once established for a gauging site — whether that site is a natural channel cross-section, a flume, or a weir — the rating curve lets the agency convert a continuous stage record into a continuous discharge record without measuring discharge directly at every time step; it is preferred (indeed necessary) at natural channel gauging stations without an engineered control structure, and it must be periodically re-verified/updated because the natural channel's stage-discharge relationship shifts as the bed scours or aggrades.

Back to the paper →