Question 4 of 7

Problem 4 (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) Butterfly vs. globe valves (6 marks)

A butterfly valve uses a circular disc mounted on a central shaft that rotates 90° within the pipe bore; in the open position the disc lies edge-on to the flow (low head loss, nearly full bore), and it throttles by rotating toward the closed position. It is compact, light, and quick to operate (quarter-turn), so it is favoured for large-diameter transmission and distribution mains where space and actuation torque matter and where the valve is mostly used fully open or fully closed (isolation duty), with only occasional throttling.

A globe valve closes by lowering a disc/plug straight down onto a seat inside a globe-shaped body, forcing the flow to change direction twice as it passes through. This gives it much finer, more linear control of flow at partial openings (better throttling characteristics) at the cost of a permanently higher head loss (even fully open) and a multi-turn stem that is slower to operate. It is therefore used at points where accurate flow or pressure throttling is required — e.g., a pump discharge control valve or a pressure-reducing station — rather than for simple isolation of a large main.

(ii) Wastewater collection system components (12 marks)

(a) Sanitary forcemain (6 marks). A forcemain is the pressurized pipe downstream of a sanitary pumping station that conveys wastewater under pump head (not gravity) to the point where it can re-enter a gravity sewer or reach the treatment plant. Because it runs full and pressurized, it can cross high ground or long flat stretches that a gravity sewer could not follow economically, but it must be sized to maintain a minimum scouring velocity across the pump's full operating range (to stay self-cleansing under pressure) and must be protected against water hammer/surge at pump start-stop, since a forcemain has no manholes to relieve trapped air or pressure surges the way a gravity main does.

(b) Sanitary pumping station (6 marks). A pumping station lifts wastewater from a low-lying collection area (where a continuous gravity grade to the treatment plant is not achievable without excessive excavation depth) into a forcemain or a higher gravity sewer downstream. Its critical functions are providing enough wet-well storage and duty/standby pump capacity to handle peak wet-weather flows without surcharging the upstream collection network, and providing odour and corrosion control (H&sub2;S generation is aggravated by the detention time in the wet well), since pumping stations are one of the few points in a sanitary system where wastewater is deliberately held rather than kept flowing.

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

[Figure not reproduced: Fig. 4 — Pumping system from low reservoir R-1 (WL = 20 m) to high reservoir R-2 (WL = 60 m), as printed on the exam. See the official exam paper.]

Given.

Find. Required pump head, $H_{pump}$.

Approach. Energy balance between the two (open, low-velocity) reservoir surfaces: pump head = static lift + total friction losses (velocity heads at the free surfaces and any minor losses are neglected, as no fitting loss coefficients are given). Friction loss from the Hazen–Williams equation, $H_f=10.67\,L\,Q^{1.852}/(C^{1.852}D^{4.87})$ (SI units).

1. Static lift. $$\Delta Z = 60-20=\boxed{40\ \text{m}}$$
2. Friction loss, Pipe 1. $$H_{f1}=\frac{10.67(400)(0.100)^{1.852}}{(70)^{1.852}(0.400)^{4.87}}=\boxed{1.99\ \text{m}}$$
3. Friction loss, Pipe 2. $$H_{f2}=\frac{10.67(1000)(0.100)^{1.852}}{(60)^{1.852}(0.300)^{4.87}}=\boxed{26.88\ \text{m}}$$
4. Total pump head. $$H_{pump}=\Delta Z+H_{f1}+H_{f2}=40+1.99+26.88=\boxed{68.87\ \text{m}}$$