--- title: "Critical DO deficit, 300 l/s into river" subject: "Sanitary Engineering" chapter: 6 key: "semester_6/sanitary_engineering:critical-do-city-300ls-48kmday" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-city-300ls-48kmday site: BCT Engineering --- # Critical DO deficit, 300 l/s into river Sanitary Engineering · Chapter 6: Wastewater Disposal - Asked in: 2072 Asoj, 8 marks - Video: [oxygen deficit problem | DO deficit solved problem | oxygen deficit in downstream side of river](https://www.youtube.com/watch?v=4wst5Ot1YIg) — Civil Engineering Exam - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/critical-do-city-300ls-48kmday - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6 ## Question A town is discharging sewage of 300 liters/second in the river having discharge of 1300 liters/second and a velocity of 48 km/day. The BOD5 of sewage and river water are 400 mg/l and 4 mg/l respectively. The DO of sewage sample is nil. D.O. in the river water is 80% of the saturation value. The temperature of both sewage and river water is 20°C. The saturation DO at 20°C is 9.17 mg/l. Assume K1 = 0.1/day and K2 = 0.4/day (both with base 10). Calculate the critical DO deficit. ## Short answer Critical DO deficit is 18.56 mg/l at 1.88 days (90.2 km downstream). Mixed BOD5=78.25 mg/l, ultimate BOD=114.4 mg/l, initial deficit=3.21 mg/l. Minimum DO negative, stream would become anaerobic without treatment. ## Answer River DO $=0.8\times 9.17=7.336$ mg/l. Both temperatures are 20°C, so no correction is needed. $K_1=0.1$, $K_2=0.4$ (base 10), $f=4$. Flows are in l/s. #### Step 1: Conditions just after mixing $$ \begin{aligned} Q_{mix} &= Q_s+Q_r = 300 + 1300 = 1600\ l/s\\ BOD_5^{mix} &= \frac{Q_sB_s+Q_rB_r}{Q_s+Q_r} = \frac{300\times 400 + 1300\times 4}{1600} = 78.25\ \text{mg/l}\\ DO_{mix} &= \frac{Q_s DO_s+Q_r DO_r}{Q_s+Q_r} = \frac{300\times 0 + 1300\times 7.336}{1600} = 5.96\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17 - 5.96 = 3.21\ \text{mg/l} \end{aligned} $$ #### Step 2: Ultimate BOD and self-purification constant $$ \begin{aligned} L_0 &= \frac{BOD_5}{1-10^{-5K}} = \frac{78.25}{1-10^{-5\times 0.1}} = 114.44\ \text{mg/l}\\ f &= \frac{R}{K} = \frac{0.4}{0.1} = 4.00 \end{aligned} $$ #### Step 3: Critical time and critical deficit $$ \begin{aligned} t_c &= \frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_0}\right\}\right]\\ &= \frac{1}{0.1(4.00-1)}\log_{10}\left[4.00\left\{1-(4.00-1)\frac{3.21}{114.44}\right\}\right] = 1.880\ \text{days}\\ D_c &= \frac{L_0}{f}\,10^{-K t_c} = \frac{114.44}{4.00}\times 10^{-0.1\times 1.880} = 18.56\ \text{mg/l} \end{aligned} $$ **Answer: critical DO deficit $D_c = 18.56$ mg/l**, reached after $t_c=1.880$ days (about 90.2 km downstream at 48 km/day). Minimum DO $=9.17-18.56=-9.39$ mg/l. *Note:* the computed deficit is larger than the saturation DO (9.17 mg/l). In reality DO cannot fall below zero, so the stream would go anaerobic near the critical point and the effluent needs treatment before discharge. The negative value is simply what the formula gives for these data. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is BOD5 of the mixture? A. 202 mg/L B. (300 x 400 + 1300 x 4)/1600 = 78.25 mg/L C. 400 mg/L D. 4 mg/L Answer: B. BOD is flow-weighted. (Key point: Mixed BOD) 2. What is f? A. 0.25 B. 0.3 C. 4 D. 40 Answer: C. f = K2/K1 = 0.4/0.1 = 4. (Key point: Self-purification factor) 3. What is Dc? A. 3.21 mg/L B. 9.17 mg/L C. 1.88 mg/L D. 18.56 mg/L, after tc = 1.880 days Answer: D. Dc = (114.44/4) x 10^(-0.188) = 18.56 mg/L. (Key point: Critical deficit) 4. Dc exceeds the saturation DO. What does this imply? A. 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