$200\ \text{m}^3/\text{s}$ (as printed on the exam)
Bedding (pipe) slope, $S$
4% = 0.04
Manning's $n$ (PVC, as printed)
0.10
Velocity envelope
$0.8\ \text{m/s} < V < 7\ \text{m/s}$
Find. Required sewer diameter $D$; confirm the velocity condition.
Approach. For a circular pipe flowing 100% full, the hydraulic radius reduces to $R=D/4$, so Manning's equation $Q=\tfrac{1}{n}AR^{2/3}S^{1/2}$ can be collapsed into a single closed-form expression for $D$ in terms of $Q$, $n$ and $S$; the resulting velocity is then checked against the stipulated $0.8$–$7\ \text{m/s}$ envelope.
Collapse Manning's equation to solve for $D$. With $A=\pi D^2/4$ and $R=D/4$,
$$Q = \frac{1}{n}\cdot\frac{\pi D^2}{4}\cdot\left(\frac{D}{4}\right)^{2/3}S^{1/2} \;\Rightarrow\; D^{8/3} = \frac{Qn}{\left(\tfrac{\pi}{4}\right)\left(\tfrac{1}{4}\right)^{2/3}S^{1/2}}$$
Substitute and solve. With $Q=200$, $n=0.10$, $S=0.04$: the bracketed constant $k=\left(\tfrac{\pi}{4}\right)\left(\tfrac{1}{4}\right)^{2/3}(0.04)^{1/2}=0.06237$, so
$$D = \left(\frac{200\times0.10}{0.06237}\right)^{3/8} = \boxed{8.71\ \text{m}}$$
Check the velocity condition.
$$A = \frac{\pi D^2}{4} = \frac{\pi(8.71)^2}{4} = 59.5\ \text{m}^2 \qquad V = \frac{Q}{A} = \frac{200}{59.5} = \boxed{3.36\ \text{m/s}}$$
Since $0.8 < 3.36 < 7$, the velocity condition is technically satisfied by this literal solution.
Quantity
Value
Required diameter, $D$
8.71 m
Full-pipe velocity, $V$
3.36 m/s (within 0.8–7 m/s)
Check: $D=8.71\ \text{m}$ is the literal, arithmetically-correct answer to the numbers exactly as printed, but an 8.7 m diameter is not a "sanitary sewer" in any practical sense — it is larger than most subway tunnels, and $n=0.10$ is roughly ten times the accepted Manning roughness for PVC ($n\approx0.009$–$0.011$). The two misprints are almost certainly $Q=200\ \text{L/s}=0.200\ \text{m}^3/\text{s}$ and $n=0.010$. Re-solving with those values through the same closed-form expression gives $D\approx0.275\ \text{m}$, i.e. a commercial 300 mm PVC pipe (full-flow $V\approx3.56\ \text{m/s}$, also inside the 0.8–7 m/s envelope) — the realistic sanitary-sewer design a candidate would actually specify. Both readings are shown so the method (collapsing Manning's equation for a full pipe) is clear regardless of which figures are correct.
(ii) On-site vs. off-site stormwater runoff control (6 marks)
An on-site control system manages runoff at or very near the point it is generated — a common example is a rooftop or parking-lot infiltration/bioretention trench (a gravel-filled, perforated-pipe trench beneath a landscaped surface depression) that intercepts roof leaders and lot sheet flow before it reaches the public storm system, reducing both peak flow and total volume leaving the individual lot. An off-site (regional) control system instead collects runoff from many upstream lots through the piped minor system and attenuates it at a single downstream facility — a common example is a municipal stormwater detention/dry pond sized to hold the combined peak inflow from an entire subdivision and release it slowly to the receiving watercourse or downstream trunk sewer at the pre-development rate.
Two key maintenance issues for the on-site infiltration trench are: (1) clogging of the infiltration surface and gravel voids by fine sediment and organic debris washed off the contributing area, which progressively reduces the design infiltration rate and must be controlled by regular sediment forebay/pretreatment cleanout; and (2) verifying long-term infiltration capacity by periodic drawdown (ponding) testing, since a trench that no longer drains within the specified 24–48 hour window has effectively failed and risks becoming a standing-water nuisance or mosquito habitat over a 25-year service life. Two key maintenance issues for the off-site detention pond are: (1) sediment and debris accumulation in the forebay and outlet control structure, which must be dredged/cleaned on a scheduled basis or the low-flow orifice will clog and defeat the outlet's rate control; and (2) vegetation and embankment/emergency-spillway integrity, since erosion of the pond banks, burrowing animals in the embankment, or overgrowth blocking the emergency spillway all threaten the structure's ability to safely pass and store the design storm over a multi-decade design life.
(iii) Fitting the annual-flood series to a frequency curve (7 marks)
Approach. Flood-frequency analysis fits the 12-year annual-maximum series to a probability distribution so the record can be extrapolated to return periods longer than the record itself (here 25, 50 and 100 years from only 12 years of data). The standard method has two complementary parts: (1) a graphical/empirical check using a plotting-position formula, and (2) an analytical fit of a frequency distribution from which $T$-year floods are read directly.
Rank the data and assign plotting positions (empirical check). Rank the $n=12$ annual maxima in descending order $m=1\ldots12$ (largest $=1$) and assign each a plotting-position return period using the Weibull formula $T=\tfrac{n+1}{m}$. The three largest values (760, 760, 760 m³/s, tied at $m=1,2,3$) plot at $T\approx13.0,\ 6.5,\ 4.3$ years respectively — these empirical points are plotted on log-probability paper (Fig. 1) and a straight line/curve is fitted through them by eye or least squares, but a 12-point record cannot itself reach $T=25$–100 years, so the fitted line must be extrapolated well beyond the data — this is the weak link the analytical distribution below is meant to strengthen.
Fit an analytical extreme-value distribution. The Gumbel (Extreme Value Type I) distribution is the standard choice for annual-maximum flood series and gives the $T$-year flood directly from the sample mean $\bar{Q}$ and standard deviation $s$:
$$Q_T = \bar{Q} + K_T\,s \qquad K_T = -\frac{\sqrt6}{\pi}\left[0.5772+\ln\!\left(\ln\frac{T}{T-1}\right)\right]$$
From the 12 values: $\bar{Q}=588.3\ \text{m}^3/\text{s}$, $s=135.6\ \text{m}^3/\text{s}$.
Evaluate the frequency factor and the design flood for each return period.
$$T=25:\ K_{25}=2.044,\quad Q_{25}=588.3+2.044(135.6)=\boxed{866\ \text{m}^3/\text{s}}$$
$$T=50:\ K_{50}=2.592,\quad Q_{50}=588.3+2.592(135.6)=\boxed{940\ \text{m}^3/\text{s}}$$
$$T=100:\ K_{100}=3.137,\quad Q_{100}=588.3+3.137(135.6)=\boxed{1014\ \text{m}^3/\text{s}}$$
These three points are marked on the fitted Gumbel curve in Fig. 1, well beyond the empirical plotting-position range — the analytical distribution, not the eye-fitted line through 12 points, is what makes that extrapolation defensible.
Fig. 1 — Weibull plotting positions for the 12-year Red River annual-maximum series (blue points) with the fitted Gumbel (EV-I) curve (red) extrapolated to the 25-, 50- and 100-year design floods.
