NivaarExam PrepOfficial exam papers ↗

18-Env-B5 Industrial & Hazardous Waste Management · May 2015

Question 11 of 18: Municipal Wastewater BOD 5 /SS Content and Cannery Population Equivalent

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: Nemerow & Dasgupta, Industrial and Hazardous Waste Treatment, 2nd ed.; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery, 5th ed.; Davis & Cornwell, Introduction to Environmental Engineering, 6th ed.; LaGrega, Buckingham & Evans, Hazardous Waste Management, 2nd ed.; CCME, Guidelines for the Management of Biomedical Waste in Canada (1992); Canadian Environmental Protection Act (CEPA), 1999; Basel Convention on the Control of Transboundary Movements of Hazardous Wastes (1989); provincial hazardous waste regulations (e.g. BC's Environmental Management Act and Hazardous Waste Regulation).

Question 11: Municipal Wastewater BOD5/SS Content and Cannery Population Equivalent (10 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.

Check: the source prints the industrial wastewater-generation factors as "10,000E/tonne production" and "100,000E/tonne production." Read literally as population-equivalents-per-tonne, this would make sub-part 11.3 circular — the question separately asks to determine the cannery's PE, which cannot already be given as raw data. It is also inconsistent with the paired mg/L concentrations, which need a flow rate (not a population count) to convert to a mass load. Read instead as 10,000 L/tonne and 100,000 L/tonne (a wastewater-generation-rate factor, "E" being a misprint of "L"), consistent with the classic textbook method for this exact problem type (flow-per-unit-production × concentration → mass load; mass load ÷ domestic per-capita rate → PE) and with this paper's other misprints (e.g. "BODS" for "BOD5," "C6HSOH" for "C6H5OH" in Question 8). This reading is what is solved below; it consumes every given number exactly once with no redundancy.

Given. Municipal population and domestic per-capita flow/strength, plus a flow-generation factor and wastewater strength for each of two industries.

Given data
QuantitySymbolValue
PopulationP10,000
Domestic flow—400 L/cap·day
Domestic BOD5 / SS (concentration)—190 / 225 mg/L
Cannery production—5,000 tonnes over a 7-month season
Cannery wastewater factor / BOD5 / SS—10,000 L/tonne; 1,200 / 700 mg/L
Textile production—2,000 kg/day = 2.0 tonnes/day
Textile wastewater factor / BOD5 / SS—100,000 L/tonne; 400 / 100 mg/L

Find. (11.1) municipal BOD5/SS mass loading and concentration with the two industries served, (11.2) the same without them, and (11.3) the population equivalent of the cannery on a BOD5 basis.

Approach. Convert each industry's production rate to a daily basis, apply its wastewater-generation factor to get flow, multiply flow by concentration to get a daily mass load, sum with the domestic load for "with industries" (dividing by total flow for the combined concentration), and express the cannery's own BOD5 load as a population equivalent using the domestic per-capita BOD5 rate.

  1. Domestic flow and load. $$Q_{dom} = 10{,}000 \times 400 = 4{,}000{,}000\ \text{L/day} = 4{,}000\ \text{m}^3/\text{day}$$ $$BOD_{dom} = 4{,}000\ \text{m}^3 \times 190\ \text{g/m}^3 = 760{,}000\ \text{g/day} = 760\ \text{kg/day}$$ $$SS_{dom} = 4{,}000 \times 225 = 900{,}000\ \text{g/day} = 900\ \text{kg/day}$$ The domestic per-capita rates implied are $760/10{,}000 = 76\ \text{g BOD}_5/\text{cap}\cdot\text{day}$ and $900/10{,}000 = 90\ \text{g SS/cap}\cdot\text{day}$.
  2. Cannery daily production, flow and load. A 7-month season is taken as $\tfrac{7}{12}\times 365 \approx 213$ days of continuous operation: $$\dot{m}_{cannery} = \dfrac{5{,}000\ \text{tonnes}}{213\ \text{days}} = 23.47\ \text{tonnes/day}$$ $$Q_{cannery} = 23.47 \times 10{,}000 = 234{,}700\ \text{L/day} = 234.7\ \text{m}^3/\text{day}$$ $$BOD_{cannery} = 234.7 \times 1{,}200\ \text{g/m}^3 = 281{,}640\ \text{g/day} \approx 281.6\ \text{kg/day}$$ $$SS_{cannery} = 234.7 \times 700 = 164{,}290\ \text{g/day} \approx 164.3\ \text{kg/day}$$
  3. Textile flow and load. $2{,}000\ \text{kg/day} = 2.0$ tonnes/day of cotton goods: $$Q_{textile} = 2.0 \times 100{,}000 = 200{,}000\ \text{L/day} = 200\ \text{m}^3/\text{day}$$ $$BOD_{textile} = 200 \times 400 = 80{,}000\ \text{g/day} = 80.0\ \text{kg/day}, \qquad SS_{textile} = 200 \times 100 = 20{,}000\ \text{g/day} = 20.0\ \text{kg/day}$$
  4. (11.1) Combined flow, mass load and concentration with industries served. $$Q_{with} = 4{,}000 + 234.7 + 200 = 4{,}434.7\ \text{m}^3/\text{day}$$ $$BOD_{with} = 760 + 281.6 + 80.0 = \boxed{1{,}122\ \text{kg/day}} \ \Rightarrow\ \dfrac{1{,}122\ \text{kg/day}}{4{,}434.7\ \text{m}^3/\text{day}} \approx \boxed{253\ \text{mg/L}}$$ $$SS_{with} = 900 + 164.3 + 20.0 = \boxed{1{,}084\ \text{kg/day}} \ \Rightarrow\ \dfrac{1{,}084\ \text{kg/day}}{4{,}434.7\ \text{m}^3/\text{day}} \approx \boxed{245\ \text{mg/L}}$$
  5. (11.2) Content without the industries (domestic only). This is simply the given residential wastewater itself: $$\boxed{BOD_{without} = 190\ \text{mg/L}\ (760\ \text{kg/day})}, \qquad \boxed{SS_{without} = 225\ \text{mg/L}\ (900\ \text{kg/day})}$$
  6. (11.3) Population equivalent of the cannery, BOD5 basis. PE expresses the cannery's own BOD5 load in terms of an equivalent number of people, using the domestic per-capita rate found in Step 1: $$PE_{cannery} = \dfrac{BOD_{cannery}}{76\ \text{g/cap}\cdot\text{day}} = \dfrac{281{,}640\ \text{g/day}}{76\ \text{g/cap}\cdot\text{day}}$$ $$\boxed{PE_{cannery} \approx 3{,}706\ \text{people}}$$
Final results
QuantityValue
BOD5, without industries (11.2)190 mg/L (760 kg/day)
SS, without industries (11.2)225 mg/L (900 kg/day)
BOD5, with industries (11.1)≈ 253 mg/L (1,122 kg/day)
SS, with industries (11.1)≈ 245 mg/L (1,084 kg/day)
Population equivalent of cannery, BOD5 basis (11.3)≈ 3,706 PE

Serving the two industries raises the municipal BOD5 concentration by roughly a third (190 → 253 mg/L) even though the cannery operates only 7 months of the year, and the cannery alone contributes an organic load equivalent to more than a third of the municipality's own resident population — a material addition a treatment plant sized on population alone would not accommodate.