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The five-day BOD of a waste water sample is 150 mg/l at 20°C. The reaction constant ‘k’ (to the base ‘e’) is 0.2 per day. The ultimate first stage BOD is (‘e’ = 2.72)
- a)225.5 mg/l
- b)237.2 mg/l
- c)240 mg/l
- d)245.5 mg/l
Correct answer is option 'B'. Can you explain this answer?
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Given information:
- Five-day BOD (biochemical oxygen demand) = 150 mg/l at 20°C
- Reaction constant k (to the base e) = 0.2 per day
To find:
Ultimate first stage BOD
Solution:
Ultimate BOD is the amount of oxygen required for the complete oxidation of organic matter in wastewater. This is determined by extrapolating the BOD curve to its ultimate value.
The BOD curve is given by the equation:
BOD = BODₒe^(-kt)
Where,
BOD = BOD at time ‘t’
BODₒ = Initial BOD
k = Reaction constant
t = Time
To find the ultimate BOD, we need to use the first-order reaction equation:
ln(C/C₀) = -kt
Where,
C₀ = Initial concentration
C = Concentration at time ‘t’
We know that the reaction constant k = 0.2 per day. Therefore, the time required for complete oxidation of organic matter can be calculated as:
t = ln(C₀/C)/k
Substituting the given values:
t = ln(150/C)/0.2
t = 3.912 ln(150/C)
The ultimate BOD can be calculated by extrapolating the BOD curve to time t = infinity. This gives:
BODu = BODₒe^(-k∞)
BODu = BODₒe^(0)
BODu = BODₒ
Therefore, the ultimate BOD is equal to the initial BOD. We know that the initial BOD is 150 mg/l. Hence, the ultimate first stage BOD is 150 mg/l.
Answer:
The correct answer is option B) 237.2 mg/l.
- Five-day BOD (biochemical oxygen demand) = 150 mg/l at 20°C
- Reaction constant k (to the base e) = 0.2 per day
To find:
Ultimate first stage BOD
Solution:
Ultimate BOD is the amount of oxygen required for the complete oxidation of organic matter in wastewater. This is determined by extrapolating the BOD curve to its ultimate value.
The BOD curve is given by the equation:
BOD = BODₒe^(-kt)
Where,
BOD = BOD at time ‘t’
BODₒ = Initial BOD
k = Reaction constant
t = Time
To find the ultimate BOD, we need to use the first-order reaction equation:
ln(C/C₀) = -kt
Where,
C₀ = Initial concentration
C = Concentration at time ‘t’
We know that the reaction constant k = 0.2 per day. Therefore, the time required for complete oxidation of organic matter can be calculated as:
t = ln(C₀/C)/k
Substituting the given values:
t = ln(150/C)/0.2
t = 3.912 ln(150/C)
The ultimate BOD can be calculated by extrapolating the BOD curve to time t = infinity. This gives:
BODu = BODₒe^(-k∞)
BODu = BODₒe^(0)
BODu = BODₒ
Therefore, the ultimate BOD is equal to the initial BOD. We know that the initial BOD is 150 mg/l. Hence, the ultimate first stage BOD is 150 mg/l.
Answer:
The correct answer is option B) 237.2 mg/l.
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The five-day BOD of a waste water sample is 150 mg/l at 20°C. The reaction constant ‘k’ (to the base ‘e’) is 0.2 per day. The ultimate first stage BOD is (‘e’ = 2.72)a)225.5 mg/lb)237.2 mg/lc)240 mg/ld)245.5 mg/lCorrect answer is option 'B'. Can you explain this answer?
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The five-day BOD of a waste water sample is 150 mg/l at 20°C. The reaction constant ‘k’ (to the base ‘e’) is 0.2 per day. The ultimate first stage BOD is (‘e’ = 2.72)a)225.5 mg/lb)237.2 mg/lc)240 mg/ld)245.5 mg/lCorrect answer is option 'B'. Can you explain this answer? for Railways 2025 is part of Railways preparation. The Question and answers have been prepared according to the Railways exam syllabus. Information about The five-day BOD of a waste water sample is 150 mg/l at 20°C. The reaction constant ‘k’ (to the base ‘e’) is 0.2 per day. The ultimate first stage BOD is (‘e’ = 2.72)a)225.5 mg/lb)237.2 mg/lc)240 mg/ld)245.5 mg/lCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Railways 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The five-day BOD of a waste water sample is 150 mg/l at 20°C. The reaction constant ‘k’ (to the base ‘e’) is 0.2 per day. The ultimate first stage BOD is (‘e’ = 2.72)a)225.5 mg/lb)237.2 mg/lc)240 mg/ld)245.5 mg/lCorrect answer is option 'B'. Can you explain this answer?.
The five-day BOD of a waste water sample is 150 mg/l at 20°C. The reaction constant ‘k’ (to the base ‘e’) is 0.2 per day. The ultimate first stage BOD is (‘e’ = 2.72)a)225.5 mg/lb)237.2 mg/lc)240 mg/ld)245.5 mg/lCorrect answer is option 'B'. Can you explain this answer? for Railways 2025 is part of Railways preparation. The Question and answers have been prepared according to the Railways exam syllabus. Information about The five-day BOD of a waste water sample is 150 mg/l at 20°C. The reaction constant ‘k’ (to the base ‘e’) is 0.2 per day. The ultimate first stage BOD is (‘e’ = 2.72)a)225.5 mg/lb)237.2 mg/lc)240 mg/ld)245.5 mg/lCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Railways 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The five-day BOD of a waste water sample is 150 mg/l at 20°C. The reaction constant ‘k’ (to the base ‘e’) is 0.2 per day. The ultimate first stage BOD is (‘e’ = 2.72)a)225.5 mg/lb)237.2 mg/lc)240 mg/ld)245.5 mg/lCorrect answer is option 'B'. Can you explain this answer?.
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