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

Question 1 of 7: Runoff Models, Closed-Pipe Hydraulics, and Distribution Reservoirs

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 1: Runoff Models, Closed-Pipe Hydraulics, and Distribution 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) Properties of Conceptual Runoff Models

A conceptual model of runoff represents a watershed as a small network of interconnected storage elements (surface/interception storage, upper-zone soil moisture, lower-zone/groundwater storage) linked by simplified transfer or routing functions, rather than solving the full physical equations of overland and subsurface flow. Three important properties follow from this: (1) lumped or semi-distributed representation — spatial variability of soils, land cover and rainfall across the watershed is aggregated into a small number of storage compartments and parameters, trading physical detail for tractability; (2) calibration against observed records — the storage/transfer coefficients are conceptually meaningful (e.g. an infiltration capacity, a groundwater recession constant) but are normally fitted by matching simulated to observed hydrographs rather than measured directly in the field; and (3) mass-conservative bookkeeping — every model routes precipitation through the storages via a water balance, so the runoff produced is always consistent with the abstractions (interception, infiltration, evapotranspiration) the model accounts for.

A widely used example is the Stanford Watershed Model lineage (the soil-moisture-accounting engine inside HEC-HMS): rainfall is routed through interception storage, then upper-zone and lower-zone soil storages and a groundwater storage, each draining to the channel through a calibrated storage-discharge relationship. It is used operationally to generate continuous or event runoff hydrographs for reservoir operation and flood forecasting without resolving Richards' equation or overland-flow hydraulics explicitly.

(ii) Corrugated Steel Pipe — Velocity, Reynolds Number, Friction Loss

Given. Corrugated steel pipe flowing full:

Given data
QuantitySymbolValue
Pipe length$L$100 m
Pipe diameter$d$300 mm = 0.300 m
Full-flow discharge$Q$100 L/s = 0.100 m³/s
Kinematic viscosity (water)$\nu$$1.0\times10^{-6}$ m²/s

Find. The average velocity $V$, the Reynolds number $Re$ (and flow regime), and the friction head loss $H_f$.

Check: the question does not state a pipe roughness. A corrugated-steel-pipe Manning's roughness of $n = 0.024$ (standard 68×13 mm annular corrugation, uncoated — the typical design value for CSP culverts and storm drains) is assumed; corrugated pipe is far rougher than smooth concrete or plastic, so a Manning (not Colebrook) formulation is used.

Approach. Get $V$ from continuity, $Re$ from the pipe-flow definition, then Manning's equation (solved for the energy-grade-line slope $S$) to evaluate the friction head loss over the 100 m length.

  1. (a) Average velocity. Cross-sectional area and continuity: $$A = \frac{\pi d^2}{4} = \frac{\pi (0.300)^2}{4} = 0.07069\ \text{m}^2, \qquad V = \frac{Q}{A} = \frac{0.100}{0.07069} = \boxed{1.41\ \text{m/s}}.$$
  2. (b) Reynolds number. $$Re = \frac{Vd}{\nu} = \frac{(1.41)(0.300)}{1.0\times10^{-6}} = \boxed{4.24\times10^{5}}.$$ Since $Re \gg 4000$, the flow is turbulent.
  3. (c) Friction head loss via Manning's equation. For a pipe flowing full, the hydraulic radius is $R = d/4 = 0.0750$ m. Solving Manning's equation $V = \tfrac{1}{n}R^{2/3}S^{1/2}$ for the slope of the energy grade line: $$S = \left(\frac{Vn}{R^{2/3}}\right)^2 = \left(\frac{(1.41)(0.024)}{(0.0750)^{2/3}}\right)^2 = 0.0364\ \text{m/m},$$ $$H_f = S\,L = (0.0364)(100) = \boxed{3.64\ \text{m}}.$$
QuantityValue
Average velocity, $V$1.41 m/s
Reynolds number, $Re$$4.24\times10^{5}$ (turbulent)
Friction head loss, $H_f$≈ 3.64 m over 100 m

(iii) Uses of Elevated Water Reservoirs

An elevated tank or standpipe serves two important functions in a distribution system. (1) Equalizing (peaking) storage — customer demand fluctuates sharply through the day (a pronounced peak hour, low overnight demand) while pumps and treatment plants operate most efficiently at a steady rate; the elevated tank fills during low-demand hours and discharges during the peak, letting pumps run near their best-efficiency point continuously rather than cycling to chase the instantaneous demand curve. (2) Pressure maintenance and emergency reserve — the tank's elevation converts stored volume directly into distribution-system pressure (hydrostatic head) without continuous pumping, damping pressure transients, and the stored volume provides a reserve for fire flow or for a pump/power outage, keeping the system pressurized (and hence protected from backflow/contamination ingress) until pumping resumes.

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