Question 5 of 7

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) Partial-flow (85% full) sanitary sewer diameter (8 marks)

Given.

Find. Required pipe diameter $D$; confirm the velocity condition.

Approach. For a circular pipe flowing partly full, the flow area $A$ and wetted perimeter $P$ (and hence hydraulic radius $R$) are fixed fractions of $D^2$ and $D$ respectively once $y/D$ is fixed, via the central angle $\theta=2\cos^{-1}(1-2y/D)$. Substituting those fractions into Manning's equation collapses it to $Q=C\,D^{8/3}$ for a constant $C$, which is solved directly for $D$.

1. Central angle and geometry ratios at $y/D=0.85$. $$\theta = 2\cos^{-1}(1-2(0.85)) = 2\cos^{-1}(-0.70) = 4.692\ \text{rad}$$ $$\frac{A}{D^2}=\frac{\theta-\sin\theta}{8}=0.7130 \qquad \frac{P}{D}=\frac{\theta}{2}=2.346 \qquad \frac{R}{D}=\frac{A/D^2}{P/D}=0.3039$$
2. Collapse Manning's equation and solve for $D$. $$Q = \frac{1}{n}\left(\frac{A}{D^2}D^2\right)\left(\frac{R}{D}D\right)^{2/3}S^{1/2} = \underbrace{\frac{1}{n}\left(\frac{A}{D^2}\right)\left(\frac{R}{D}\right)^{2/3}S^{1/2}}_{C}\;D^{8/3}$$ With $n=0.014$, $S=0.05$: $C=\tfrac{1}{0.014}(0.7130)(0.3039)^{2/3}(0.05)^{1/2}=5.135$, so $$D = \left(\frac{Q}{C}\right)^{3/8} = \left(\frac{5}{5.135}\right)^{3/8} = \boxed{0.990\ \text{m} \approx 990\ \text{mm}}$$
3. Check the velocity condition. $$A = 0.7130\,D^2 = 0.698\ \text{m}^2 \qquad V = \frac{Q}{A} = \frac{5}{0.698} = \boxed{7.16\ \text{m/s}}$$ Since $1 \lt 7.16 \lt 8$, the velocity condition is satisfied — but only just, with little margin below the 8 m/s scour/erosion ceiling.

(ii) Off-site stormwater control system (6 marks)

Selecting a regional (off-site) dry detention pond as the example off-site quantity-control system, three important design features are: (1) adequate storage volume and outlet control structure sized to attenuate the combined peak inflow from the entire tributary catchment down to an acceptable release rate (often the pre-development peak) for the design storm; (2) an emergency spillway sized for a storm larger than the design storm, so the facility fails safely (controlled overflow) rather than by embankment overtopping/breach; and (3) maintenance access to the low-flow outlet structure and forebay, since off-site facilities collect sediment and debris from a large contributing area and must be reachable by maintenance equipment.

One advantage of an off-site system compared to an on-site system is economy of scale — a single regional facility is usually cheaper to build and maintain than many small on-site facilities providing the same total storage, and it centralizes inspection/maintenance effort. One disadvantage is that it does nothing to reduce runoff volume or velocity in the upstream piped network between each lot and the facility, so upstream pipes must still be sized for the full uncontrolled peak flow, and a single point of failure (blocked outlet, embankment breach) affects the whole contributing catchment rather than just one lot.

(iii) Method for fitting the French River flood series to determine the 100-year flood (6 marks)

Approach. With only 15 years of annual-maximum data, the 100-year flood must be found by extrapolating a fitted probability distribution well beyond the length of the observed record, not by reading it directly off the data. The standard method has two complementary steps.

1. Rank the data and assign empirical plotting positions. Rank the $n=15$ annual-maximum flows in descending order $m=1,\ldots,15$ (largest $=1$) and assign each a return period by a plotting-position formula — most commonly the Weibull formula $T=(n+1)/m$. These $(T, Q)$ pairs are plotted on log-probability (or Gumbel-probability) paper as a visual/graphical check of how well an extreme-value distribution fits the data, but with $n=15$ the largest plotting position reached is only $T\approx(15{+}1)/1=16$ years — far short of the required 100-year return period, so the empirical points alone cannot answer the question.
2. Fit an analytical extreme-value distribution and extrapolate. The Gumbel (Extreme Value Type I) distribution is the standard choice for annual-maximum flood series; it is fitted using only the sample mean $\bar{Q}$ and standard deviation $s$ of the 15 values, and gives the $T$-year flood directly from a tabulated frequency factor $K_T$: $$Q_T = \bar{Q} + K_T\,s \qquad K_T = -\frac{\sqrt6}{\pi}\left[0.5772+\ln\!\left(\ln\frac{T}{T-1}\right)\right]$$ This closed-form relationship is what makes the extrapolation from a 15-year record out to $T=100$ years defensible: the fitted line drawn from $\bar{Q}$ and $s$ is compared against the plotted empirical points from Step 1 as a goodness-of-fit check, and then read (or computed) at $T=100$.
3. Illustrative computation (for reference only — the question asks only for the method). From the 15 tabulated values, $\bar{Q}=480.0\ \text{m}^3/\text{s}$ and $s=137.3\ \text{m}^3/\text{s}$; at $T=100$, $K_{100}=3.137$, giving $$Q_{100} = 480.0 + 3.137(137.3) = \boxed{911\ \text{m}^3/\text{s}}$$