Question 2 of 7

Problem 2 (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) Empirical vs. reservoir-based runoff models (8 marks)

Empirical (black-box) models such as the Rational Method or the SCS Curve Number method: (1) relate rainfall excess to runoff through a statistically or empirically fitted relationship (a coefficient, a curve number) rather than an explicit representation of internal watershed storage; (2) are computationally simple with few parameters, calibrated directly against observed rainfall-runoff pairs for the basin or region; (3) predict peak flow (or a total runoff depth) well for the conditions they were calibrated to, but transfer poorly to ungauged basins, different storm patterns, or basins of very different size/shape than the calibration set.

Reservoir-based (conceptual/linear-reservoir) models: (1) represent the watershed's storage–outflow behaviour with one or more conceptual reservoirs obeying continuity, $\dfrac{dS}{dt} = I - O$, often with a linear storage law $S = kQ$; (2) can reproduce the full shape of the outflow hydrograph — rising limb, peak, recession — not just the peak value, because the storage routing naturally attenuates and lags inflow; (3) require calibration of a storage constant $k$ (and often multiple reservoirs in series/parallel) but generalize better to continuous, multi-event simulation.

Preferred use. An empirical model (Rational Method/SCS-CN) is preferred for a small urban catchment where only the design peak flow is needed quickly (e.g. sizing a storm sewer or culvert), because the extra complexity of storage routing buys little accuracy at that scale. A reservoir-based model is preferred for a larger or natural watershed, or wherever the full hydrograph time-distribution matters — flood routing through a reservoir or floodplain, continuous streamflow simulation, or baseflow recession analysis — because it captures storage effects that a single empirical coefficient cannot.

(ii) PVC pipe flow — velocity, Reynolds number, friction loss

Given.

Find. Average velocity $V$; Reynolds number $Re$ and flow regime; friction head loss $H_f$.

Approach. Continuity for velocity, $Re = Vd/\nu$ for flow regime, then Darcy–Weisbach with a Colebrook–White friction factor (PVC is hydraulically smooth, so $f$ is governed almost entirely by $Re$) for the friction loss.

1. Average velocity. $A = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}(0.400)^2 = 0.1257\ \text{m}^2$. $$V = \frac{Q}{A} = \frac{8.0}{0.1257} = \boxed{63.7\ \text{m/s}}$$
2. Reynolds number and regime. $$Re = \frac{Vd}{\nu} = \frac{63.7 \times 0.400}{1.00\times10^{-6}} = \boxed{2.54\times10^{7}}$$ $Re \gg 4000$, so the flow is unambiguously turbulent (fully rough/smooth-pipe turbulent regime).
3. Friction factor and head loss. With relative roughness $\varepsilon/d = 0.0015/400 = 3.8\times10^{-6}$ (essentially hydraulically smooth), solving the Colebrook–White equation $$\frac{1}{\sqrt{f}} = -2\log_{10}\!\left(\frac{\varepsilon/d}{3.7} + \frac{2.51}{Re\sqrt{f}}\right)$$ iteratively gives $f \approx 0.0078$. Applying Darcy–Weisbach, $$H_f = f\,\frac{L}{d}\,\frac{V^2}{2g} = 0.0078 \times \frac{2000}{0.400}\times\frac{63.7^2}{2(9.81)} = \boxed{8.0\times10^{3}\ \text{m}}$$

(iii) Functions of a storage reservoir in a water distribution system (6 marks)

1. Demand equalization. A distribution reservoir absorbs the difference between the (relatively constant) treatment/pumping rate and the highly variable diurnal customer demand, filling during low-demand hours and drawing down during peak hours, so the treatment plant and transmission mains can be sized to the average rather than the instantaneous peak demand.
2. Fire-flow and emergency reserve. The reservoir holds a dedicated volume for fire suppression demand and for continued supply during a treatment plant shutdown, power outage, or transmission main break, without which the system would have to rely entirely on instantaneous pumping capacity.
3. Pressure stabilization / hydraulic grade control. An elevated or ground-level reservoir at a controlling elevation sets and steadies the hydraulic grade line in its zone, damping pressure transients (including water-hammer surges) from pump starts/stops and allowing part of the zone to be served by gravity, improving reliability and reducing pumping energy.