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

Question 4 of 7: Sanitary Sewer Design, Runoff Control, and Flood Frequency 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 4: Sanitary Sewer Design, 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) Sanitary Sewer Sizing at 80% Full

Given. Circular PVC sewer, partial flow:

Given data
QuantitySymbolValue
Peak design flow$Q_d$5 m³/s
Proportional depth$y/D$0.80
Bedding slope$S$0.05 (5%)
Manning's roughness (PVC)$n$0.013
Velocity limits$V$$0.8 < V < 7$ m/s

Find. The required pipe diameter $D$, and whether the resulting velocity satisfies the stated limits.

Approach. At $y/D=0.80$ the flow subtends a fixed central angle $\theta$; express the partial-flow area $A$, wetted perimeter $P$ and hydraulic radius $R$ in terms of $\theta$ and $D$, substitute into Manning's equation, and solve for the diameter that delivers $Q_d = 5$ m³/s.

  1. Central angle at 80% full. For $y/D = 0.80$: $\theta = 2\cos^{-1}(1-2y/D) = 2\cos^{-1}(-0.60) = 4.4286\ \text{rad}$ (253.7°).
  2. Partial-flow geometry (in terms of $D$). $$A = \frac{r^2}{2}(\theta-\sin\theta), \qquad P = r\theta, \qquad R = \frac{A}{P}, \qquad r=\frac{D}{2}.$$
  3. Solve Manning's equation for $D$. $Q_d = \tfrac{1}{n}AR^{2/3}S^{1/2}$ is a monotonic function of $D$ alone (since $\theta$ is fixed); solving numerically for $Q_d=5$ m³/s gives $$D = \boxed{983\ \text{mm}}\ (0.9825\ \text{m}), \qquad A = 0.650\ \text{m}^2,\ R = 0.299\ \text{m}, \qquad V = Q_d/A = \boxed{7.69\ \text{m/s}}.$$
  4. Check against the velocity limits and select a commercial size. $V=7.69$ m/s marginally exceeds the stated 7 m/s ceiling. Rounding up to the next standard commercial PVC size, $D=1000$ mm, does not fix this — at $y/D=0.80$ it carries $Q=5.24$ m³/s at $V=7.78$ m/s, an even higher velocity (partial-flow velocity is not monotonically decreasing with diameter at fixed $y/D$; it is essentially flat-to-rising in this size range). Because $Q_d$ and $S$ are fixed by the problem, no pipe-diameter choice alone brings $V$ under 7 m/s at this slope.
Check: the 5% bedding slope is unusually steep for a gravity sanitary sewer and is the root cause of the velocity exceedance — it is not a sizing error. In practice this reach would need a flatter grade over part of its length or an energy-dissipating drop structure/manhole to bring the operating velocity within the erosion/scour limit; select $D=1000$ mm (next standard size above the exact 983 mm) as the specified pipe and flag the velocity for the drop-structure design that follows.
QuantityValue
Exact required diameter983 mm (0.9825 m)
Velocity at exact diameter, 80% full7.69 m/s (exceeds 7 m/s cap)
Specified commercial diameter1000 mm
Velocity/capacity at 1000 mm, 80% full7.78 m/s at 5.24 m³/s — still exceeds cap (flag for drop structure)

(ii) On-Site and Off-Site Stormwater Runoff Control

On-site vs. off-site control, 25-year municipal O&M perspective
SystemDescriptionAdvantageDisadvantage
On-site (e.g. permeable pavement / bioretention at each lot)Distributed source control installed on individual private parcelsTreats/attenuates runoff at the source, reducing peak flow and pollutant load entering the piped system and downstream infrastructure sizingHundreds of small facilities on private property are difficult for the municipality to inspect and maintain consistently over 25 years — clogging or failure on any one lot often goes unnoticed and the maintenance burden falls on many separate owners
Off-site (e.g. a centralized regional detention/wet pond)A single municipally-owned facility serving the tributary catchmentOne asset to inspect, budget and maintain over the design life (economy of scale, clear ownership and access)Requires dedicated land and a conveyance network to bring flow to the facility, and provides no control at all until runoff has already left the individual sites (no source-level treatment)

(iii) Flood-Frequency Curve Fitting

The standard method is to treat the 15 annual-maximum instantaneous flows as a sample from a probability distribution and fit a flood-frequency distribution to them, most commonly the Log-Pearson Type III distribution (the method recommended in Canadian and US flood-frequency practice) or, for a more symmetric record, the Gumbel (Extreme Value Type I) distribution. The procedure is: (1) transform each annual maximum to $\log_{10}Q$; (2) compute the sample mean, standard deviation and skew of the log-transformed series; (3) read (or compute via the Wilson–Hilferty approximation) the frequency factor $K_T$ corresponding to the sample skew and the desired return period $T$; (4) recover the flood magnitude from $\log_{10}Q_T = \overline{\log Q} + K_T\,s_{\log Q}$. This lets the fitted curve be extrapolated beyond the 15-year record to estimate rare events (50- or 100-year floods) that have not necessarily been observed.

Illustrative application to the Wavy River record ($N=15$):

Log-Pearson III statistics (illustrative)
StatisticValue
Mean of $\log_{10}Q$2.743
Standard deviation of $\log_{10}Q$0.1245
Skew coefficient, $C_s$0.107
Frequency factor $K_{50}$ / $K_{100}$2.111 / 2.405
Estimated $Q_{50}$≈ 1013 m³/s
Estimated $Q_{100}$≈ 1102 m³/s
Check: a 15-year record is short for extrapolating to 50- or 100-year events (the true sampling uncertainty on $Q_{50}$/$Q_{100}$ from only 15 points is large); these values illustrate the METHOD and should be treated as indicative, not design-final, without a longer record or regional skew adjustment.