Question 1 of 7: Conceptual Runoff Models, Pipe Flow in a Concrete Sewer and Elevated Reservoirs

Problem 1: Conceptual Runoff Models, Pipe Flow in a Concrete Sewer and Elevated Reservoirs (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) Conceptual Runoff Models in Hydraulic Structure Design

A conceptual model of runoff represents a watershed as a small number of lumped, physically-motivated storage/routing elements whose parameters are calibrated to observed rainfall-runoff behaviour, rather than solving the full physics from first principles. Two common engineering-design examples: (1) the unit hydrograph method (e.g., SCS dimensionless unit hydrograph), which converts a design storm hyetograph directly into an inflow hydrograph used to size a culvert, detention pond outlet structure or spillway — the designer needs the peak flow and its timing, not a full physically distributed simulation; and (2) the SCS Curve Number (CN) method, a lumped conceptual loss model that converts rainfall depth to runoff depth from a single calibrated parameter (land use, soil group, antecedent moisture), routinely used to size storm sewers, culverts and detention basins across a whole municipality where a fully physically-based model would be impractical to calibrate site-by-site.

(ii) Concrete Sewer Pipe: Velocity, Reynolds Number and Friction Loss

Given.

Find. (a) $V$ in m/s, (b) $Re$ and flow regime, (c) head loss $H_f$.

Approach. Get $V$ from continuity ($V=Q/A$); get $Re$ from $Re=Vd/\nu$ to classify the flow; get the Darcy friction factor $f$ from the Swamee–Jain explicit correlation and apply Darcy–Weisbach for $H_f$.

1. Average velocity (continuity). $$A=\frac{\pi d^2}{4}=\frac{\pi(0.250)^2}{4}=0.04909\ \text{m}^2,\qquad V=\frac{Q}{A}=\frac{0.100}{0.04909}=\boxed{2.04\ \text{m/s}}.$$
2. Reynolds number. $$Re=\frac{Vd}{\nu}=\frac{2.037\times0.250}{1.0\times10^{-6}}=\boxed{5.09\times10^{5}}\ \Rightarrow\ Re\gg4000,\ \textbf{turbulent flow}.$$
3. Friction factor (Swamee–Jain). $$f=\frac{0.25}{\left[\log_{10}\!\left(\dfrac{\varepsilon}{3.7d}+\dfrac{5.74}{Re^{0.9}}\right)\right]^2}=\frac{0.25}{\left[\log_{10}(3.243\times10^{-4}+4.19\times10^{-5})\right]^2}=\boxed{0.0212}.$$
4. Head loss (Darcy–Weisbach). $$H_f=f\frac{L}{d}\frac{V^2}{2g}=0.0212\times\frac{500}{0.250}\times\frac{(2.037)^2}{2\times9.81}=\boxed{8.96\ \text{m}}\ \text{over the 500 m run}.$$

(iii) Functions of Elevated Water Reservoirs

Three functions of an elevated (or standpipe/water-tower) reservoir that reduce the need for continuous pumping: (1) peak-demand shaving — the tower fills during low-demand hours (overnight) using steadily-running, efficiently-sized pumps, then gravity-discharges during the daily peak, so the pump station is sized to the average/near-average demand rather than the much larger instantaneous peak; (2) fire-flow and emergency storage — the elevated volume supplies the large, short-duration fire flow (or covers a pump/power outage) without requiring standby pumps to start and ramp up instantaneously, and without oversizing the everyday pump station just to cover a rare event; and (3) system pressure stabilization — because the water surface elevation directly sets the hydraulic grade line for the surrounding pressure zone, the tower damps out the pressure transients that would otherwise occur every time a pump starts, stops or a large demand (fire flow, water-main break) suddenly changes, reducing the number of pump starts/stops (and associated wear/energy cost) needed to hold pressure within the distribution system's target range.