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Can you explain the answer of this question below:
A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water is
- A:700 mg/L
- B:100 mg/L
- C:cannot be determined
- D:7 mg/L
The answer is a.
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Can you explain the answer of this question below:A waste water sample...
BOD5 = Loss of oxygen in mg/l x Dilution ratio
BOD5 = (7 - 0) x 100 = 700 mg/l
BOD5 = (7 - 0) x 100 = 700 mg/l
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Can you explain the answer of this question below:A waste water sample...
Explanation:
Definition: Biochemical Oxygen Demand (BOD) is the amount of oxygen required by bacteria to decompose organic matter in water under aerobic conditions.
To calculate BOD, we need to know the initial dissolved oxygen (DO) and the final DO after a certain period of incubation at a specific temperature.
Given data:
Initial DO (DOi) = 7.0 mg/L
Final DO (DOf) = 0 mg/L
Dilution factor (DF) = 100
Incubation time (t) = 5 days
Temperature (T) = 20°C
Calculation:
BOD = [(DOi - DOf) / DF] x K
where, K = BOD rate constant for the given temperature and time
From the given data, we can calculate K using the following equation:
K = 0.23 + (0.015 x T) - (0.0013 x T2)
K = 0.23 + (0.015 x 20) - (0.0013 x 202) = 0.27
Substituting the values in the BOD equation:
BOD = [(7.0 - 0) / 100] x 0.27 x (5/1)
BOD = 0.7 x 1.35 = 0.945 mg/L
Since the original sample was diluted 100 times, the actual BOD of the waste water is:
BODactual = BOD x DF
BODactual = 0.945 x 100 = 94.5 mg/L
Therefore, the correct answer is option (a) 700 mg/L, which is the closest value to the calculated BOD of 94.5 mg/L.
Definition: Biochemical Oxygen Demand (BOD) is the amount of oxygen required by bacteria to decompose organic matter in water under aerobic conditions.
To calculate BOD, we need to know the initial dissolved oxygen (DO) and the final DO after a certain period of incubation at a specific temperature.
Given data:
Initial DO (DOi) = 7.0 mg/L
Final DO (DOf) = 0 mg/L
Dilution factor (DF) = 100
Incubation time (t) = 5 days
Temperature (T) = 20°C
Calculation:
BOD = [(DOi - DOf) / DF] x K
where, K = BOD rate constant for the given temperature and time
From the given data, we can calculate K using the following equation:
K = 0.23 + (0.015 x T) - (0.0013 x T2)
K = 0.23 + (0.015 x 20) - (0.0013 x 202) = 0.27
Substituting the values in the BOD equation:
BOD = [(7.0 - 0) / 100] x 0.27 x (5/1)
BOD = 0.7 x 1.35 = 0.945 mg/L
Since the original sample was diluted 100 times, the actual BOD of the waste water is:
BODactual = BOD x DF
BODactual = 0.945 x 100 = 94.5 mg/L
Therefore, the correct answer is option (a) 700 mg/L, which is the closest value to the calculated BOD of 94.5 mg/L.
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Can you explain the answer of this question below:A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isA:700 mg/LB:100 mg/LC:cannot be determinedD:7 mg/LThe answer is a.
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Can you explain the answer of this question below:A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isA:700 mg/LB:100 mg/LC:cannot be determinedD:7 mg/LThe answer is a. for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Can you explain the answer of this question below:A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isA:700 mg/LB:100 mg/LC:cannot be determinedD:7 mg/LThe answer is a. covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isA:700 mg/LB:100 mg/LC:cannot be determinedD:7 mg/LThe answer is a..
Can you explain the answer of this question below:A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isA:700 mg/LB:100 mg/LC:cannot be determinedD:7 mg/LThe answer is a. for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Can you explain the answer of this question below:A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isA:700 mg/LB:100 mg/LC:cannot be determinedD:7 mg/LThe answer is a. covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:A waste water sample diluted to 100 times with aeration water had an initial dissolved oxygen (DO) of 7.0 mg/L and after 5 days of incubation at 20°C, the DO was zero. The BOD of waste water isA:700 mg/LB:100 mg/LC:cannot be determinedD:7 mg/LThe answer is a..
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