--- title: "Is treatment necessary? Town wastewater 300 l/s" subject: "Sanitary Engineering" chapter: 6 key: "semester_6/sanitary_engineering:treatment-needed-town-300-ls" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/treatment-needed-town-300-ls site: BCT Engineering --- # Is treatment necessary? Town wastewater 300 l/s Sanitary Engineering · Chapter 6: Wastewater Disposal - Asked in: 2080 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/treatment-needed-town-300-ls - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6 ## Question The wastewater generated from a newly established town is proposed to dispose in the nearby river. The characteristics of raw wastewater and that of river are as follows: Wastewater discharge = 300 l/s; BOD5 of wastewater at 20°C = 450 mg/l; The sewage is in putrefied state; Minimum river discharge = 5.5 m³/s; BOD5 of river at 20°C = 0 mg/l; DO of river = 8.5 mg/l; Temperature in the river after mixing = 20°C; De-oxygenation constant = 0.1/day (base 10) at 20°C; Re-oxygenation constant = 0.4/day (base 10) at 20°C; Saturation DO at 20°C = 9.1 mg/l; Allowable minimum DO deficit at d/s of disposal = 4.5 mg/l. Is treatment necessary? If so, what will be the required degree of treatment? ## Short answer Streeter-Phelps analysis: mixed flow 5.8 m³/s, initial deficit 1.04 mg/L, ultimate BOD 34.04 mg/L, f=4, critical time 1.87 days, critical deficit 5.54 mg/L > allowable 4.5 mg/L, so treatment needed. Required effluent BOD₅ ≤ 362.8 mg/L, corresponding to 19.4% BOD removal efficiency. ## Answer The oxygen sag (Streeter-Phelps) analysis is applied to the mixture of river and wastewater. Constants are in base 10. #### Step 1: Conditions after mixing $$ \begin{aligned} Q_{mix}&=0.3+5.5=5.8\ \text{m}^3/\text{s}\\ DO_{mix}&=\frac{0.3(0)+5.5(8.5)}{5.8}=8.06\ \text{mg/L}\quad(\text{putrefied sewage: DO}\approx0)\\ D_0&=9.1-8.06=1.04\ \text{mg/L}\\ BOD_5^{mix}&=\frac{0.3(450)+5.5(0)}{5.8}=23.28\ \text{mg/L}\\ L_0&=\frac{BOD_5}{1-10^{-0.1\times5}}=\frac{23.28}{0.6838}=34.04\ \text{mg/L} \end{aligned} $$ #### Step 2: Critical deficit without treatment $$ f=\frac{K_2}{K_1}=\frac{0.4}{0.1}=4 $$ $$ t_c=\frac{1}{K_2-K_1}\log_{10}\left[f\left(1-\frac{(f-1)D_0}{L_0}\right)\right]=\frac{1}{0.3}\log_{10}\left[4\left(1-\frac{3\times1.04}{34.04}\right)\right]=1.87\ \text{days} $$ $$ D_c=\frac{L_0}{f}10^{-K_1t_c}=\frac{34.04}{4}\times10^{-0.1\times1.87}=5.54\ \text{mg/L} $$ Since $D_c=5.54$ mg/L is greater than the allowable deficit of 4.5 mg/L, **treatment is necessary**. #### Step 3: Required degree of treatment Find the maximum $L_0$ for which $D_c=4.5$ mg/L (with the same $D_0$, by trial using the same equations): $L_0=27.45$ mg/L. Then $$ BOD_5^{mix}=27.45\times0.6838=18.77\ \text{mg/L} $$ The effluent BOD₅ that can be allowed is: $$ BOD_5^{eff}=18.77\times\frac{5.8}{0.3}=362.8\ \text{mg/L} $$ $$ \text{Efficiency of treatment}=\frac{450-362.8}{450}\times100=19.4\% $$ **Answer: Treatment is necessary; the plant must remove about 19.4% of the BOD** (effluent BOD₅ not more than about 362.8 mg/L). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is DO of the mixture (8.06 mg/L) lower than the river DO of 8.5 mg/L? A. Because the river BOD is high B. Because the sewage is putrefied and carries about zero DO C. Because temperature rises after mixing D. Because saturation DO is 9.1 Answer: B. The putrefied sewage has DO about 0, so the weighted average (5.5 x 8.5)/5.8 falls to 8.06 mg/L. (Key point: Mixing conditions) 2. How is the need for treatment decided? A. If the BOD of the sewage exceeds 20 mg/L B. If the river flow is less than 10 m3/s C. If the calculated critical deficit exceeds the allowable deficit D. If the initial deficit D0 is greater than zero Answer: C. Untreated Dc = 5.54 mg/L is greater than the allowable 4.5 mg/L, so treatment is required. (Key point: Treatment decision) 3. What is f here and what does it represent? A. f = K1/K2 = 0.25 B. f = K2 - K1 = 0.3 C. f = D0/L0 D. f = K2/K1 = 4, the ratio of reaeration to deoxygenation constants Answer: D. f = K2/K1 = 0.4/0.1 = 4 and appears in tc and Dc = (L0/f)10^(-K1 tc). (Key point: Self-purification constant) 4. After finding the permissible mixture BOD5 (18.77 mg/L), how is the effluent BOD5 obtained? A. Multiply by 5.8/0.3 B. Divide by 5.8/0.3 C. Subtract 450 D. Multiply by 0.3/5.8 Answer: A. All the mixture BOD comes from the sewage, so the effluent BOD5 = 18.77 x (5.8/0.3) = 362.8 mg/L, giving 19.4% removal. (Key point: Degree of treatment) ## Related in Sanitary Engineering - [Is treatment necessary? 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