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
Conventionally, the biochemical oxygen demand (BOD) is measured for _______ days.
Consider the following statements:
1. Ammonia nitrogen is a measure of nitrogen present as ammonium hydroxide and ammonium salts. It will progressively decrease as sewage gets treated.
2. Organic nitrogen is the total nitrogenous matter in sewages excepting that present as ammonia nitrogen, nitrites and nitrates. It becomes ammonia in anaerobic decomposition and nitrites or nitrates in aerobic decomposition.
Which of the above statements is/are correct?Identify the correct relation from the following:
Where BOD, COD, and TOD are Biochemical, Chemical, and Theoretical oxygen demand, respectively.
What is the correct range of \(\frac{{BO{D_u}}}{{COD}}\) for a waste water to be considered as fully biodegradable?
(where BODu = ultimate Bio-chemical oxygen demand and COD = chemical oxygen demand)
Determine ultimate BOD for sewage having 5-day BOD at 20°C as 180 ppm. Assume the de-oxygenation constant as 0.8 per day.
