Question 4 of 7: Sanitary Sewer Design, Runoff Control, and Flood Frequency

Problem 4: Sanitary Sewer Design, Runoff Control, and Flood Frequency (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 75% Full

Given. Circular concrete sewer, partial flow:

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

Approach. At $y/D=0.75$ 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 75% full. For $y/D=0.75$: $\theta = 2\cos^{-1}(1-2y/D) = 2\cos^{-1}(-0.50) = 4.1888\ \text{rad}$ (240°).
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 $$D = \boxed{1141\ \text{mm}}\ (1.141\ \text{m}), \qquad A = 0.823\ \text{m}^2,\ R = 0.344\ \text{m}, \qquad V = Q_d/A = \boxed{6.08\ \text{m/s}}.$$
4. Check against the velocity limits. $V=6.08$ m/s exceeds the stated 5 m/s ceiling.
5. Does a bigger pipe fix it? Re-solving for the flow depth a larger commercial pipe would actually run at while still carrying $Q_d=5$ m³/s (so $y/D$ drops below 0.75) gives essentially the same or a slightly worse velocity: 1200 mm → 6.16 m/s at $y/D=0.68$; 1500 mm → 6.23 m/s; 2000 mm → 6.12 m/s; 3000 mm → 5.86 m/s — even an impractical 3 m sewer is still over the cap. The exact diameter that would give $V=5.00$ m/s at $D=1141$ mm and $y/D=0.75$ requires the slope to drop to $S\approx2.03\%$, well below the given 3%.

(ii) Minor vs. Major Runoff Control Systems

Minor and major stormwater runoff control systems differ in three principal respects. (1) Design storm frequency. The minor system is sized for frequent, moderate storms (typically the 2- to 10-year event) so that everyday runoff is conveyed underground without nuisance surface flooding; the major system is intended — by grading and route planning rather than pipe capacity — to safely convey rarer, larger storms (e.g. the 100-year event) that exceed the minor system's capacity. (2) Physical form. The minor system is a closed, engineered conduit network (storm sewers, catch basins, manholes); the major system largely reuses existing above-ground infrastructure (road profiles, swales, floodplains, designated overland flow routes). (3) Cost and visibility. The minor system is expensive per unit capacity and hidden from view, requiring dedicated maintenance access; the major system is comparatively low-cost to provide (it is mostly a grading/land-use decision made at the subdivision-design stage) but highly visible when it activates, since it is meant to flow overland during extreme events.

Example — minor system: a piped storm sewer network beneath a residential street, sized for the 5-year storm. Example — major system: the same street's road crown and curb-and-gutter profile, graded to carry the 100-year storm's excess overland to a park or watercourse without entering buildings.

(iii) Probability Frequency Hydrograph Analysis — Return Period Example

Probability-frequency (flood-frequency) analysis fits a statistical distribution — most commonly Log-Pearson Type III, the Canadian/US standard — to a series of annual maximum instantaneous flood discharges, then reads off the discharge associated with any chosen return period $T_R$. As an example: given 20 years of annual peak flows for a river, the analyst (1) ranks the 20 values and assigns each a plotting-position exceedance probability (e.g. Weibull $P=m/(N+1)$); (2) fits the Log-Pearson III distribution (mean, standard deviation and skew of $\log_{10}Q$); (3) reads the frequency factor $K_T$ for the desired $T_R$ (e.g. $T_R=100$ years, $P=0.01$) from tables keyed to the sample skew; and (4) recovers the design flood as $\log_{10}Q_{100} = \overline{\log Q} + K_{100}\,s_{\log Q}$. The resulting $Q_{100}$ is then the discharge that has a 1% chance of being equalled or exceeded in any given year — used directly to set a floodplain elevation, a bridge waterway opening, or a spillway design flow, extrapolating well beyond the 20 years actually observed.