18-Env-A2 Hydrology and Municipal Hydraulics Engineering · December 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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.).
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.
Given. Circular PVC sewer, partial flow:
| Quantity | Symbol | Value |
|---|---|---|
| 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.
| Quantity | Value |
|---|---|
| Exact required diameter | 983 mm (0.9825 m) |
| Velocity at exact diameter, 80% full | 7.69 m/s (exceeds 7 m/s cap) |
| Specified commercial diameter | 1000 mm |
| Velocity/capacity at 1000 mm, 80% full | 7.78 m/s at 5.24 m³/s — still exceeds cap (flag for drop structure) |
| System | Description | Advantage | Disadvantage |
|---|---|---|---|
| On-site (e.g. permeable pavement / bioretention at each lot) | Distributed source control installed on individual private parcels | Treats/attenuates runoff at the source, reducing peak flow and pollutant load entering the piped system and downstream infrastructure sizing | Hundreds 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 catchment | One 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) |
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 ($N=15$):
| Statistic | Value |
|---|---|
| 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 |