--- title: "BOD of Bagmati sample at 30°C" subject: "Sanitary Engineering" chapter: 5 key: "semester_6/sanitary_engineering:bod-thapathali-4day-30c" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/bod-thapathali-4day-30c site: BCT Engineering --- # BOD of Bagmati sample at 30°C Sanitary Engineering · Chapter 5: Characteristics and Examination of Wastewater - Asked in: 2075 Baisakh, 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/bod-thapathali-4day-30c - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-5 ## Question 5 ml of a sewage sample taken under the Thapathali bridge of Bagmati river was pipetted into a 300 ml capacity BOD bottle which was then completely filled with dilution water. The DO concentration of this mixture is tested and found to be 9.2 mg/l. Now it is kept in the incubator maintained at 25°C for a period of 7 days. The DO concentration after incubation is found to be 5.3 mg/l. Adopting base 10 value of K as 0.1/d, determine the 4 day BOD of sewage at 30°C in Bagmati river at that particular location. ## Short answer 7-day BOD at 25°C from dilution test is 234 mg/L. Using K20=0.1/day and θ=1.047, ultimate BOD=269.5 mg/L. Then 4-day BOD at 30°C (K30=0.1583/day) = 206.8 mg/L. ## Answer #### Step 1: BOD measured in the test (7 days at 25°C) Dilution: 5 mL of sample in a 300 mL bottle, so the decimal fraction is $$ P=\frac{5}{300}=0.0167 $$ $$ BOD_7^{25}=\frac{D_1-D_2}{P}=\frac{9.2-5.3}{0.0167}=234\ \text{mg/L} $$ #### Step 2: Ultimate BOD Here $K=0.1$ day⁻¹ (base 10) is taken as the rate constant at 20°C; the temperature coefficient is $\theta=1.047$. At 25°C: $$ K_{25}=0.1\times1.047^{5}=0.1258\ \text{day}^{-1} $$ $$ L=\frac{BOD_7}{1-10^{-K_{25}t}}=\frac{234}{1-10^{-0.8807}}=269.5\ \text{mg/L} $$ #### Step 3: 4-day BOD at 30°C $$ \begin{aligned} K_{30}&=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}\\ BOD_4^{30}&=L\left(1-10^{-K_{30}\times4}\right)=269.5\left(1-10^{-0.6332}\right)=206.8\ \text{mg/L} \end{aligned} $$ **Answer: 4-day BOD at 30°C = 206.8 mg/L** (ultimate BOD = 269.5 mg/L). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 5 mL sample in a 300 mL bottle, what is P? A. 5/300 = 0.0167 B. 300/5 = 60 C. 5/295 = 0.017 of dilution water D. 0.5 Answer: A. P = volume of sample / volume of mixture. (Key point: Dilution fraction) 2. How is the measured BOD calculated? A. (D1 - D2) x P = 0.065 mg/L B. (D1 - D2)/P = (9.2 - 5.3)/0.0167 = 234 mg/L C. D2/P = 317 mg/L D. (D1 + D2)/P = 870 mg/L Answer: B. BOD = DO drop divided by decimal dilution. (Key point: Measured BOD) 3. Why is K25 used to get L from the 7-day test? A. Because the bottle is 300 mL B. Because 7 days equals 25 hours C. The test was incubated at 25°C, so K = 0.1 at 20°C must be corrected to 25°C D. Because K is always taken at 25°C Answer: C. K25 = 0.1 x 1.047^5 = 0.1258; L = 234/(1 - 10^(-0.8807)) = 269.5 mg/L. (Key point: Ultimate BOD step) 4. What is the 4-day BOD at 30°C? A. About 234 mg/L B. About 270 mg/L C. About 100 mg/L D. About 207 mg/L Answer: D. With K30 = 0.1583, y4 = 269.5 (1 - 10^(-0.6332)) = 206.8 mg/L. 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