Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 18-Env-B2 / Water Resources. 3 hours duration; open-book exam (any non-communicating calculator permitted). Six questions are printed; the first five as they appear in the answer book constitute a complete paper and are marked, each worth 20 marks. All six are solved below for completeness.
Reference texts. Chow, Open-Channel Hydraulics; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); Linsley, Kohler & Paulhus, Hydrology for Engineers; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Freeze & Cherry, Groundwater; Fisheries Act, Canadian Environmental Protection Act, 1999; Ontario Water Resources Act and Clean Water Act, 2006 (used here as a representative province); CCME, Canada-Wide Strategy for the Management of Municipal Wastewater Effluent.
The source does not state the pipe material; n = 0.013 (concrete sewer pipe, Metcalf & Eddy typical design value) is assumed and used consistently in both parts. The minimum permissible (self-cleansing) velocity is taken as 0.6 m/s, the standard Canadian municipal design minimum. The stated 450 m length, and the 0.03 m invert drop across each manhole (spaced every 90 m), fix the profile’s hydraulic-grade-line detail but are not needed to size the pipe or check its velocities — the design is governed entirely by the given slope, flows, n, and the d/D criterion.
Given.
Given data
Quantity
Value
Longitudinal slope, S
0.001
Peak wastewater flow, Qpeak
90 m³/min = 1.50 m³/s
Minimum design flow, Qmin
15 m³/min = 0.250 m³/s
Roughness (assumed, concrete)
n = 0.013
Minimum permissible velocity (assumed)
0.6 m/s
Find. (a) the required commercial pipe diameter so the peak flow runs at no more than half-full; (b) the velocity at minimum flow and the flow depth at peak flow, checked against their respective limits.
Flow depth in the selected 1800 mm pipe at minimum and peak design flow (shaded).
Approach. Size the pipe from the property that a circular section flowing exactly half full has the same hydraulic radius (and hence the same velocity) as flowing full, then check both operating extremes against the selected commercial diameter using the standard central-angle partial-flow relations.
Use the half-full property to convert the sizing target to a full-pipe capacity. At d/D = 0.5, R = D/4 exactly equals R at full pipe, so Vhalf = Vfull and Qhalf = Qfull/2. To carry the peak flow at exactly half-full, the pipe must be able to run full at
$$Q_{full} = 2\,Q_{peak} = 3.00\ \text{m}^3/\text{s}$$
Solve Manning’s equation (full-flowing) for D. With $A=\pi D^{2}/4$ and $R=D/4$,
$$Q_{full} = \dfrac{1}{n}\left(\dfrac{\pi D^{2}}{4}\right)\left(\dfrac{D}{4}\right)^{2/3}S^{1/2}$$
solving numerically gives D = 1.675 m. Rounding up to the next standard commercial (reinforced-concrete) pipe size,
$$D = \boxed{1800\ \text{mm}}$$
At this diameter, Qfull = 3.64 m³/s and Vfull = 1.43 m/s.
Partial-flow depth and velocity at minimum flow. For a circular pipe, expressing depth by the central angle θ subtended at the pipe centre, $d/D = \tfrac{1}{2}(1-\cos(\theta/2))$, $A=\tfrac{D^2}{8}(\theta-\sin\theta)$, $R=\tfrac{D}{4}\left(1-\dfrac{\sin\theta}{\theta}\right)$. Solving for the θ that gives Qmin = 0.25 m³/s in the 1800 mm pipe:
$$d/D = 0.178\ (\text{depth} = 0.320\ \text{m}), \qquad V_{min} = \boxed{0.818\ \text{m/s}}$$
Since 0.818 m/s exceeds the assumed 0.6 m/s self-cleansing minimum, the pipe is self-cleansing at minimum flow.
Depth check at peak flow. Repeating the same partial-flow solution for Qpeak = 1.50 m³/s in the same 1800 mm pipe:
$$d/D = \boxed{0.448\ (\text{depth} = 0.806\ \text{m})}$$
This is below the 0.5 (half-diameter, 0.90 m) design limit set in part (a) — the actual peak-flow depth clears the limit with margin because the diameter was rounded up from the exact 1.675 m requirement to the next commercial size, 1800 mm.
