Question 3 of 5: Raw Water Intake Structures; BOD Test with Temperature Correction

Question 3: Raw Water Intake Structures; BOD Test with Temperature Correction (25 marks: a 10, b 15)

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.

a. Raw water intake structures (10 marks)

A raw water intake is the first structure in a water-supply system: it withdraws water from a surface source (lake, reservoir or river) at a controlled rate and quality and conveys it, protected from debris and biofouling, to the treatment plant. A lake intake typically consists of an intake tower standing offshore in water deep enough to avoid both surface effects (wave action, ice, floating debris, warm epilimnetic algae blooms) and the bottom sediment/anoxic hypolimnion; the tower carries multiple intake ports at different elevations, each fitted with a trash rack and bar screen, so the operator can select the port giving the best available water quality as the lake stratifies seasonally (thermocline position, algal bloom depth). Water enters the selected port, passes through the screening, and flows by gravity (or is drawn by a submerged low-lift pump) through a submerged conduit laid across the lake bed to an onshore wet well, from which low-lift pumps deliver it to the treatment plant.

Key requirements for site selection follow directly from that need to reliably capture good-quality water: (1) adequate depth and distance offshore to reach water below wave/ice action and away from shoreline contamination (marinas, storm outfalls, shallow warm/algal zones); (2) a location that avoids navigation channels, existing outfalls and predicted future shoreline development; (3) stable lake-bed/foundation conditions able to support the tower and conduit without excessive settlement or scour; (4) water quality and depth profile data (temperature, DO, turbidity, algae) through at least a full seasonal cycle, to confirm a usable intake elevation exists year-round; (5) protection from ice damage and floating debris, and accessibility for maintenance/screen cleaning; and (6) adequate raw water quantity even at the source's minimum (drought/low-lake-level) condition, with intake capacity matched to the plant's design and peak demand.

b. Standard BOD5 and ultimate BOD at a non-standard test temperature (15 marks)

Given. A raw sewage sample (3 mL) diluted to 300 mL in a BOD bottle is incubated at 15 °C (not the standard 20 °C); DO falls from 8.0 mg/L to 4.5 mg/L over 4 days, of which 5% of the observed depletion is attributed to seed already present in the sample.

Find. The standard BOD5 (5-day, 20 °C, the value the exam labels “standard”) and the ultimate BOD ($\text{BOD}_u$) of the undiluted sample.

Approach. Scale the seed-corrected DO depletion by the dilution factor to get the 4-day BOD measured at the test's actual (non-standard) 15 °C; because the incubation ran at 15 °C rather than the standard 20 °C, correct the first-order rate constant to 15 °C with the Van't Hoff–Arrhenius relation to back out the (temperature-independent) ultimate BOD, then re-apply the first-order model at 20 °C, 5 days to get the standard BOD5.

1. Dilution fraction. $$P=\frac{V_s}{V_b}=\frac{3}{300}=0.0100.$$
2. Seed-corrected 4-day depletion. Raw DO drop $\Delta DO=8.0-4.5=3.5\ \text{mg/L}$; with 5% attributed to the seed, the sewage's own share is $$\Delta DO_\text{net}=3.5\times(1-0.05)=3.325\ \text{mg/L}.$$
3. 4-day BOD at the test temperature, 15 °C. $$\text{BOD}_{4,15^\circ C}=\frac{\Delta DO_\text{net}}{P}=\frac{3.325}{0.0100}=\boxed{332.5\ \text{mg/L}}.$$
4. Rate constant at 15 °C. The exam gives no deoxygenation rate constant, so a typical value for raw domestic sewage at the standard 20 °C is assumed, $k_1(20^\circ C)=0.10\ \text{d}^{-1}$ (base-10; check assumption), corrected to the actual 15 °C test temperature with $\theta=1.135$ (Metcalf & Eddy, valid for $T \lt 20^\circ C$): $$k_1(15^\circ C)=k_1(20^\circ C)\,\theta^{(T-20)}=0.10\times1.135^{(15-20)}=0.10\times0.5309=0.05309\ \text{d}^{-1}.$$
5. Ultimate BOD. $\text{BOD}_u$ is a property of the sample's total oxidizable organic content and does not change with test temperature (only the rate $k_1$ does), so it can be recovered from the 15 °C, 4-day reading using $k_1(15^\circ C)$: $$\text{BOD}_u=\frac{\text{BOD}_{4,15^\circ C}}{1-10^{-k_1(15^\circ C)(4)}}=\frac{332.5}{1-10^{-0.2124}}=\frac{332.5}{0.3868}=\boxed{859.7\ \text{mg/L}}.$$
6. Standard BOD5 (5-day, 20 °C). Re-applying the first-order model at the standard test conditions, $$\text{BOD}_5=\text{BOD}_u\left(1-10^{-k_1(20^\circ C)(5)}\right)=859.7\times\left(1-10^{-0.50}\right)=859.7\times0.6838=\boxed{587.9\ \text{mg/L}}.$$