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18-Env-A2 Hydrology and Municipal Hydraulics Engineering · December 2014

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

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

Notes on this paper

National Exams — December 2014 — 04-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 (first five answers marked); all seven are solved below for completeness. Each question ("Problem") is worth 20 marks.

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 Analysis (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$3 m
Base width$b$8 m
Side slope (H:V)$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 $D_{50}$).

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. $$A = (b+zy)y = (8+0.25\times3)(3) = 26.25\ \text{m}^2, \qquad P = b+2y\sqrt{1+z^2} = 8+2(3)\sqrt{1.0625}=14.18\ \text{m},$$ $$R = A/P = 1.851\ \text{m}, \qquad V = \frac{1}{n}R^{2/3}S_o^{1/2} = \frac{1}{0.035}(1.851)^{2/3}(0.04)^{1/2} = 8.61\ \text{m/s},$$ $$Q = VA = (8.61)(26.25) = \boxed{226.1\ \text{m}^3/\text{s}}.$$
  2. (b) Reynolds number. Using the hydraulic diameter $D_h=4R=7.40$ m as the open-channel analogue of pipe diameter: $$Re = \frac{V(4R)}{\nu} = \frac{(8.61)(7.40)}{1.0\times10^{-6}} = \boxed{6.38\times10^{7}}.$$ Since $Re \gg 4000$, the flow is turbulent (as expected for essentially any open-channel flow at engineering scale).
QuantityValue
Flow area, $A$26.25 m²
Hydraulic radius, $R$1.851 m
Velocity, $V$8.61 m/s
Discharge, $Q$226.1 m³/s
Reynolds number, $Re$$6.38\times10^{7}$ (turbulent)

(ii) Specific Energy over a Streambed Rise

Y₁ Y₂ Δz Rock-lined trapezoidal channel bed Flow over a raised bed section (specific-energy problem)
Flow depth Y₁ upstream drops to Y₂ over the 0.6 m bed rise, matching the printed figure's dip in the water surface (subcritical approach flow accelerates over the hump).

Given. Same trapezoidal channel ($b=8$ m, $z=0.25$); $Q=20$ m³/s, $Y_1=3$ m, bed rise $\Delta z=0.6$ m, 8 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. $A_1=(8+0.25\times3)(3)=26.25\ \text{m}^2$, $V_1=Q/A_1=20/26.25=0.762\ \text{m/s}$: $$E_1 = Y_1+\frac{V_1^2}{2g} = 3+\frac{(0.762)^2}{19.62} = \boxed{3.030\ \text{m}}.$$
  2. Specific energy at the hump. $$E_2 = E_1-\Delta z = 3.030-0.600 = 2.430\ \text{m}.$$
  3. Choking check. The critical depth for this section and $Q$ is $y_c=0.853$ m, giving a minimum specific energy $E_{min}=1.268$ m. Since $E_2=2.430\ \text{m} > E_{min}=1.268\ \text{m}$, 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 $Y_2>y_c$: $$Y_2 = \boxed{2.38\ \text{m}}.$$ The depth drops from 3.00 m to 2.38 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$3.030 m
Specific energy over hump, $E_2$2.430 m
Critical depth, $y_c$ (check)0.853 m (no choking)
Downstream depth, $Y_2$2.38 m

(iii) Using the Swift River Streamflow Record

The 70-year daily-mean-flow record supports two distinct design uses. (1) Flood-probability estimation: extracting the annual maximum instantaneous (or daily) flow from each of the 70 years and fitting a flood-frequency distribution to that annual-maximum series (exactly the Log-Pearson III procedure illustrated in Problem 4(iii), but now with a far longer, more reliable 70-year record) gives the discharge associated with any chosen return period, and hence the probability that a "major flood" magnitude is equalled or exceeded in any given year. (2) Reservoir storage design: rather than annual maxima, the full daily time series is used in a mass-curve (sequent-peak / Rippl) analysis — cumulating inflow volume against a target release rate identifies the maximum cumulative deficit, which is the active storage needed to carry the watershed through its low-flow season (here, roughly July–September, at 0.5–1.0 mm/day) using surplus captured during the high-flow spring freshet (the pronounced peak near day 195, ≈4.8 mm/day). Sized this way, the reservoir both provides firm yield through the dry season and, by capturing the freshet peak, directly attenuates the downstream flood the same peak would otherwise cause.

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