Question 5 of 7

Problem 5 (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) Good locations for a water storage reservoir (6 marks)

Two good locations: (1) near the hydraulic centre of demand, at a controlling ground elevation within the pressure zone (e.g. on high ground overlooking the served area), and (2) at the downstream end of a long transmission main, ahead of a large or remote demand cluster. Three reasons: (a) a reservoir near the demand centre equalizes pumping/treatment output against the diurnal demand curve — filling during low-demand hours and discharging during peaks — so upstream mains and the treatment plant can be sized to the average rather than the instantaneous peak; (b) a reservoir at a controlling elevation sets and stabilizes the hydraulic grade line for its zone, improving pressure uniformity and damping transient (water-hammer) pressure swings from pump starts/stops; and (c) locating a reservoir at the end of a long main or near a remote cluster provides a local emergency/fire-flow reserve and continued service during an upstream main break or treatment-plant outage, without depending on instantaneous transmission capacity from the source.

(ii) Wastewater collection system components (6 marks)

(a) Sanitary forcemain. A forcemain is a pressurized pipe downstream of a sewage lift/pumping station, used where gravity flow is impractical (flat topography, a summit to cross, or excessive excavation depth); unlike a gravity sewer it flows full under pump-generated pressure rather than at partial depth under gravity, so it must be sized for pressure-pipe hydraulics (Hazen-Williams/Darcy-Weisbach head loss, surge/water-hammer protection, air-release valves at high points) rather than self-cleansing open-channel velocity criteria.

(b) Sanitary drop manhole structure. A drop manhole accommodates a large elevation difference between an incoming sewer invert and the manhole's outgoing invert (e.g. where a lateral must connect into a much deeper trunk sewer) by piping the incoming flow down an external or internal drop pipe to the lower outgoing invert, rather than letting the flow cascade down the manhole's open interior; this prevents the high-velocity, splashing, corrosive/odour-generating turbulence (and the associated risk of hydrogen-sulfide generation and structure/worker safety hazards) that an unconfined vertical drop would cause.

(iii) Pump head, R-1 to R-2 (8 marks)

Given.

Flow rate $Q=150\ \text{L/s}=0.15\ \text{m}^3/\text{s}$; static lift $= 70-30 = 40$ m.

Find. Pump head $H_{pump}$ required to deliver 150 L/s from R-1 to R-2.

Approach. Energy balance from R-1 to R-2: pump head equals the static lift plus the sum of Hazen-Williams friction losses in Pipe 1 and Pipe 2 (both carry the same $Q$, in series).

1. Hazen-Williams friction loss (SI form), each pipe. $$h_f = 10.67\,\frac{L\,Q^{1.852}}{C^{1.852}D^{4.87}}$$ $$h_{f1} = 10.67\times\frac{500\times0.15^{1.852}}{60^{1.852}\times0.30^{4.87}} = \boxed{28.5\ \text{m}}$$ $$h_{f2} = 10.67\times\frac{1200\times0.15^{1.852}}{80^{1.852}\times0.25^{4.87}} = \boxed{97.5\ \text{m}}$$
2. Pump head = static lift + total friction loss. $$H_{pump} = (z_{R2}-z_{R1}) + h_{f1}+h_{f2} = 40+28.5+97.5 = \boxed{166\ \text{m}}$$