--- title: "Degree of treatment, industrial wastewater" subject: "Sanitary Engineering" chapter: 6 key: "semester_6/sanitary_engineering:degree-treatment-industrial-waste" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/degree-treatment-industrial-waste site: BCT Engineering --- # Degree of treatment, industrial wastewater Sanitary Engineering · Chapter 6: Wastewater Disposal - Asked in: 2074 Bhadra, 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/degree-treatment-industrial-waste - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6 ## Question You are assigned by an industry as a Sanitary Engineer to recommend the degree of treatment required for their industrial waste water. The effluent from the treatment plant is to be discharged into a river with a minimum flow of 5000 lps, a dissolved oxygen content of 7.4 mg/l and BOD of zero. In order to thrive aquatic life, it is necessary to maintain a minimum DO content of 4 mg/l in the river. A sanitary survey reveals the characteristics of industrial waste water as follows: Discharge = 2*10⁶ l/day; BOD = 5000 mg/l; DO = 0. Recommend the degree of treatment required for the plant. Assume saturation DO of 9.2 mg/l in the river after mixing with wastewater. It is equal to DO content of river before mixing. Assume any other appropriate data if required. ## Short answer Using Streeter-Phelps with base-10 constants (K=0.1, R=0.4), the permissible mixture ultimate BOD is 30.93 mg/l. After mixing with river (5 m³/s), the effluent BOD5 must be reduced from 5000 mg/l to 4589 mg/l, requiring 8.2% BOD removal. ## Answer Convert flows to m³/s: $Q_s = 2\times 10^6/(86400\times 1000) = 0.0231$ m³/s and $Q_r = 5.0$ m³/s. The river is at about 20°C, so I assume the standard constants $K=0.1$/day and $R=0.4$/day (base 10), $f=4$. Saturation DO after mixing is 9.2 mg/l, and the minimum DO to be kept is 4 mg/l. #### Step 1: Mixing and permissible deficit $$ \begin{aligned} DO_{mix} &= \frac{0.0231\times 0 + 5.0\times 7.4}{5.023} = 7.37\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.2-7.37 = 1.83\ \text{mg/l}\\ D_{c(allowed)} &= DO_{sat}-DO_{min} = 9.2-4.0 = 5.20\ \text{mg/l} \end{aligned} $$ #### Step 2: Permissible BOD of the mixture With $K=0.1$ and $R=0.4$ (base 10), $f=R/K=4.00$. The critical deficit is $$ t_c=\frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_m}\right\}\right],\qquad D_c = \frac{L_m}{f}10^{-Kt_c} $$ and $D_c$ must not exceed 5.20 mg/l. Solving these two equations for $L_m$ by trial (with $D_0=1.83$ fixed) gives $L_m=30.93$ mg/l, with $t_c=1.723$ days (check: $D_c=5.20$ mg/l). #### Step 3: Permissible BOD of the effluent $$ \begin{aligned} L_{raw} &= \frac{BOD_5}{1-10^{-5K}} = \frac{5000}{0.6838} = 7312.38\ \text{mg/l}\\ L_r &= 0.00\ \text{mg/l (river)}\\ L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{30.93\times 5.023 - 5.0\times 0.00}{0.0231} = 6712.02\ \text{mg/l} \end{aligned} $$ So the treated effluent can have $BOD_5 \approx 4589.49$ mg/l. $$ \text{Degree of treatment} = \frac{7312.38-6712.02}{7312.38}\times 100 = 8.2\% $$ **Answer: the effluent BOD5 must be brought from 5000 mg/l down to about 4589.49 mg/l, i.e. about 8.2% treatment (BOD removal) is required.** The required degree is small only because the river dilution is very large (5 m³/s against 0.023 m³/s of waste); if the river flow were lower, much higher treatment would be needed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the waste flow in m3/s for 2 x 10^6 l/day? A. 2.31 m3/s B. 0.0231 m3/s C. 23.1 m3/s D. 0.231 m3/s Answer: B. 2 x 10^6 l/day = 2000 m3/day = 0.0231 m3/s. (Key point: Flow conversion) 2. Why is only about 8.2% treatment needed despite BOD of 5000 mg/L? A. The BOD is low B. The river DO is zero C. The waste has high DO D. The river flow is very large compared with the waste flow Answer: D. Dilution by 5 m3/s against 0.023 m3/s makes the mixture BOD small; a lower river flow would need much more treatment. (Key point: Small treatment) 3. What is the maximum deficit allowed? A. 9.2 - 4 = 5.2 mg/L B. 4 mg/L C. 9.2 mg/L D. 1.83 mg/L Answer: A. The DO must stay at least 4 mg/L, so Dc max = saturation - 4. 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