Question 1 of 7: Sanitary Sewer Design, Off-Site Runoff Control and Flood-Frequency Analysis

Problem 1: Sanitary Sewer Design, Off-Site Runoff Control and Flood-Frequency 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) Required Sanitary Sewer Diameter

Given.

Find. The pipe diameter $D$ that conveys $Q$ flowing full at slope $S$, then check $V$ against the 0.7–10 m/s envelope.

Approach. For a circular pipe flowing 100% full, $A=\pi D^2/4$ and the hydraulic radius is exactly $R=D/4$; substitute both into Manning's equation and solve for $D$ directly, then back-calculate $V=Q/A$ to check the velocity constraints.

1. Write Manning's equation for a full circular pipe. $$Q=\frac{1}{n}A R^{2/3}S^{1/2} = \frac{1}{n}\left(\frac{\pi D^2}{4}\right)\left(\frac{D}{4}\right)^{2/3}S^{1/2} = K\,D^{8/3},$$ where $K=\dfrac{1}{n}\cdot\dfrac{\pi}{4}\cdot\left(\dfrac{1}{4}\right)^{2/3}S^{1/2}$ collects every quantity that does not depend on $D$.
2. Evaluate the coefficient $K$. $$K=\frac{1}{0.020}\cdot\frac{\pi}{4}\cdot(0.25)^{2/3}\cdot\sqrt{0.06} = \boxed{3.817\ \text{s}^{-1}\text{m}^{-2/3}}.$$
3. Solve for $D$. Rearranging $Q=KD^{8/3}$, $$D=\left(\frac{Q}{K}\right)^{3/8}=\left(\frac{100}{3.817}\right)^{3/8}=\boxed{3.40\ \text{m}}.$$
4. Back-calculate the full-flow velocity and check the constraints. $$A=\frac{\pi D^2}{4}=\frac{\pi(3.40)^2}{4}=9.09\ \text{m}^2,\qquad V=\frac{Q}{A}=\frac{100}{9.09}=\boxed{11.0\ \text{m/s}}.$$ The lower bound (self-cleansing, $V>0.7$ m/s) is comfortably met, but the upper bound ($V<10$ m/s, protecting the concrete crown/invert from abrasion) is not met — the full-flow velocity is about 10% over the 10 m/s ceiling.

(ii) Off-Site Stormwater Runoff Control Systems

Two common off-site (regional, end-of-pipe) stormwater control systems are the dry detention pond and the wet retention pond (stormwater wetpond). A dry pond is normally empty and only fills temporarily during a storm, releasing the stored volume through a low-flow outlet (orifice/riser) over 24–48 hours to attenuate the peak; it provides quantity control but limited water-quality treatment because there is no permanent pool for particulates to settle into between storms. A wet pond maintains a permanent pool sized to the "water quality volume," so incoming storm flow displaces (and partially mixes with) standing water, giving both peak attenuation and pollutant removal through extended settling, some biological uptake, and thermal buffering if vegetated.

From a 25-year municipal O&M perspective, the dry pond is cheaper to build and mow, but its outlet structure (trash rack, low-flow orifice) is prone to clogging, and because there is no permanent pool, sediment removal requires periodic excavation with heavy equipment and dewatering. The wet pond costs more up front (larger footprint, engineered embankment, safety benching) and needs regular removal of accumulated sediment from the permanent pool (dredging, typically every 10–20 years) plus shoreline/vegetation management, but it delivers materially better water quality performance and, if naturalized, lower nuisance/odour risk than a poorly maintained dry pond.

Two recommendations for long-term viability: (1) fund a dedicated stormwater utility fee (rather than general tax revenue) sized to cover scheduled sediment removal, structural inspection and outlet-structure maintenance over the full 25-year design life, since deferred maintenance is the leading cause of pond failure; and (2) require an as-built survey and a maintenance/inspection covenant registered against the property at construction, with mandatory inspection intervals (e.g., annual outlet inspection, 5-year bathymetric survey of sediment accumulation) so degradation is caught and budgeted for before it becomes a capacity or safety failure.

(iii) Flood-Frequency Curve-Fitting Method

Given. 12 years (1940–1951) of annual instantaneous maximum flow on the French River, QC: 430, 500, 650, 750, 480, 350, 650, 750, 750, 600, 550, 500 m³/s.

Find. The method used to fit these annual-maximum data to a frequency (probability) distribution so that the flood magnitude for a given return period ($T=25$, 50, 100 yr) can be read off the fitted curve.

This is a classic flood-frequency analysis problem. The data are first treated as an annual-maximum series (one value per year, the largest instantaneous flow that year — already the case here). The series is then fit to a probability distribution commonly used for hydrologic extremes — the Gumbel (Extreme Value Type I) distribution or the Log-Pearson Type III distribution (the latter is Environment Canada/USGS standard practice) — using the method of moments: compute the sample mean $\bar{X}$ and standard deviation $S$ (and, for Log-Pearson III, the skew of the log-transformed data). For the Gumbel method, the flood magnitude for return period $T$ is $$X_T=\bar{X}+K_T\,S,$$ where $K_T=\dfrac{y_T-\bar{y}_n}{S_n}$ is the frequency factor, $y_T=-\ln\!\left[\ln\!\left(\dfrac{T}{T-1}\right)\right]$ is the Gumbel reduced variate for return period $T$, and $\bar{y}_n$, $S_n$ are the reduced mean/standard deviation tabulated as a function of sample size $N$ (Gumbel/Chow tables).

As a cross-check, the data can also be plotted directly: rank the $N=12$ values in descending order and assign each an empirical exceedance probability using a plotting-position formula — most commonly the Weibull formula $P=\dfrac{m}{N+1}$ (equivalently $T=\dfrac{N+1}{m}$), where $m$ is the rank (1 = largest). Plotting $X$ against $T$ on Gumbel (or log-normal) probability paper and comparing the theoretical fitted line against the plotted points visually confirms goodness of fit before extrapolating to the 25-, 50- and 100-year floods, which lie beyond the 12-year record and are therefore extrapolated, not observed.