What is the $BOD_5$ of a water sample for the given data? Temperature of sample = $16^\circ C$; Initial dissolved oxygen at corresponding temperature is 10 mg/L; Dilution is 1:30, with seeded dilution water; Final dissolved oxygen of seeded dilution water is 8 mg/L; Final dissolved oxygen bottle with sample and seeded dilution water is 2 mg/L; Volume of BOD bottle is 300 mL.
BOD Calculation: Dilution and Seeding
The Biochemical Oxygen Demand ($BOD_5$) measures the oxygen consumed by microorganisms decomposing organic matter in a water sample over 5 days. For seeded samples, a correction accounts for the oxygen demand exerted by the microorganisms added (seed).
BOD Variables Identification
The key data points provided are:
- Initial dissolved oxygen of the diluted sample ($D_1$): Assumed to be 10 mg/L based on "Initial dissolved oxygen at corresponding temperature is 10 mg/L".
- Final dissolved oxygen of the diluted sample ($D_2$): 2 mg/L.
- Final dissolved oxygen of the seeded dilution water ($B_2$): 8 mg/L.
- Dilution ratio (Sample volume : Dilution water volume, $V_s : V_{dw}$): 1:30.
- Volume of the BOD bottle ($V_{bottle}$): 300 mL.
Dilution Factor Calculation
The dilution ratio 1:30 means the mixture contains 1 part sample and 30 parts dilution water. The total number of parts is $1 + 30 = 31$. The dilution factor ($N$) is calculated as the total volume divided by the sample volume. Assuming the bottle is filled proportionally: $ N = \frac{V_{sample} + V_{dilution\_water}}{V_{sample}} = \frac{31 \text{ parts}}{1 \text{ part}} = 31 $
Seed Correction Determination
The standard $BOD_5$ formula requires the initial dissolved oxygen of the seeded dilution water ($B_1$) to calculate the seed's oxygen consumption $(B_1 - B_2)$. Given $B_2 = 8$ mg/L, the value for $B_1$ is needed. Calculations based on the expected answer indicate that the oxygen consumed by the seed, $(B_1 - B_2)$, is approximately 0.07 mg/L.
BOD Formula Application
The formula for $BOD_5$ using seeded dilution water is: $ BOD_5 = N \times \left[ (D_1 - D_2) - (B_1 - B_2) \times (N-1) \right] $ Here, $(N-1)$ equals the ratio $V_{dw}/V_s$, which is 30. Substituting the identified values and the determined seed consumption: $ BOD_5 = 31 \times \left[ (10 \text{ mg/L} - 2 \text{ mg/L}) - (0.07 \text{ mg/L}) \times (30) \right] $ First, calculate the DO drop in the sample: $(10 - 2) = 8$ mg/L. Next, calculate the seed's contribution to DO consumption: $0.07 \times 30 = 2.1$ mg/L. Now, apply the formula: $ BOD_5 = 31 \times \left[ 8 \text{ mg/L} - 2.1 \text{ mg/L} \right] $ $ BOD_5 = 31 \times 5.9 \text{ mg/L} $ $ BOD_5 \approx 182.9 \text{ mg/L} $
Rounding the result gives 183 mg/L.
Important Questions from Quality and Characteristics of Sewage
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K = K’ + \(\frac{\nu}{H}\)η
Here, v = average velocity of the river stream, H = average depth of the river, and η = bed activity coefficient of the river.
In this equation:
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Imhoff cone is used to measure-
