For 5 mL sample in a 300 mL bottle, what is P?

5/295 = 0.017 of dilution water

How is the measured BOD calculated?

(D1 - D2)/P = (9.2 - 5.3)/0.0167 = 234 mg/L

Why is K25 used to get L from the 7-day test?

The test was incubated at 25°C, so K = 0.1 at 20°C must be corrected to 25°C

Because K is always taken at 25°C

2075 Baisakh · 8 marks

BOD of Bagmati sample at 30°C

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.

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.

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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BOD of Bagmati sample at 30°C

Step 1: BOD measured in the test (7 days at 25°C)

Step 2: Ultimate BOD

Step 3: 4-day BOD at 30°C

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