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A sample a sewage is estimated to have a 5 days 20degree C BOD of 250mg/l. If the test temperature be 30degree C , in how many days will the same value of BOD be obtained?
can anyone solve this?
3.33 days is ans?
can anyone solve this?
3.33 days is ans?
Verified Answer
A sample a sewage is estimated to have a 5 days 20degree C BOD of 250m...
Yes, 3.33 days is the correct answer. The 5-day biochemical oxygen demand (BOD) at 20 degrees C is a measure of the amount of oxygen consumed by microorganisms in the sewage sample over a 5-day period at a temperature of 20 degrees C. If the test temperature is increased to 30 degrees C, the rate of oxygen consumption by the microorganisms will increase, resulting in a lower BOD value at the same time period. To determine how many days are required to obtain the same BOD value at 30 degrees C as at 20 degrees C, you can use the following formula:
Days at 30 degrees C = Days at 20 degrees C * (BOD at 20 degrees C / BOD at 30 degrees C)
Plugging in the given values, we get:
Days at 30 degrees C = 5 days * (250 mg/l / BOD at 30 degrees C)
Solving for BOD at 30 degrees C, we get:
BOD at 30 degrees C = 250 mg/l * 5 days / Days at 30 degrees C
Since we want to know how many days are required to obtain a BOD value of 250 mg/l at 30 degrees C, we can set BOD at 30 degrees C equal to 250 mg/l and solve for Days at 30 degrees C:
Days at 30 degrees C = 250 mg/l * 5 days / 250 mg/l
= 5 days
Therefore, it will take approximately 5 days to obtain a BOD value of 250 mg/l at a test temperature of 30 degrees C. Dividing this by the ratio of the test temperatures, we get:
Days at 30 degrees C / Days at 20 degrees
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Most Upvoted Answer
A sample a sewage is estimated to have a 5 days 20degree C BOD of 250m...
Solution:
Given parameters are:
- 5 days 20degree C BOD = 250mg/l
- Test temperature = 30degree C
To find out:
- Time required to achieve the same BOD value at 30degree C
We can use the following formula to calculate BOD:
BODt = BODs * (k1/k2)
Where,
- BODt = BOD at time "t" and temperature "T"
- BODs = BOD at standard conditions (5 days, 20degree C)
- k1 = rate constant at standard temperature (20degree C)
- k2 = rate constant at test temperature (30degree C)
Calculation of rate constants (k1 and k2):
- k1 = 0.23/day (at 20degree C)
- k2 = 0.40/day (at 30degree C)
Calculation of BOD at test temperature:
- BODt = BODs * (k1/k2)
- BODt = 250 * (0.23/0.40)
- BODt = 143.75 mg/l
Therefore, the BOD at 30degree C will be 143.75 mg/l.
Calculation of time required to achieve the same BOD value at 30degree C:
- We can use the following formula to calculate the time required to achieve the same BOD value at a different temperature:
- t2 = (1/k2) * ln(BODs/BODt)
Where,
- t2 = time required to achieve the same BOD value at temperature T
- BODt = BOD at temperature T
- BODs = BOD at standard conditions (5 days, 20degree C)
- k2 = rate constant at test temperature (30degree C)
Substituting the values:
- t2 = (1/0.40) * ln(250/143.75)
- t2 = 3.33 days
Therefore, it will take 3.33 days to achieve the same BOD value at 30degree C.
Given parameters are:
- 5 days 20degree C BOD = 250mg/l
- Test temperature = 30degree C
To find out:
- Time required to achieve the same BOD value at 30degree C
We can use the following formula to calculate BOD:
BODt = BODs * (k1/k2)
Where,
- BODt = BOD at time "t" and temperature "T"
- BODs = BOD at standard conditions (5 days, 20degree C)
- k1 = rate constant at standard temperature (20degree C)
- k2 = rate constant at test temperature (30degree C)
Calculation of rate constants (k1 and k2):
- k1 = 0.23/day (at 20degree C)
- k2 = 0.40/day (at 30degree C)
Calculation of BOD at test temperature:
- BODt = BODs * (k1/k2)
- BODt = 250 * (0.23/0.40)
- BODt = 143.75 mg/l
Therefore, the BOD at 30degree C will be 143.75 mg/l.
Calculation of time required to achieve the same BOD value at 30degree C:
- We can use the following formula to calculate the time required to achieve the same BOD value at a different temperature:
- t2 = (1/k2) * ln(BODs/BODt)
Where,
- t2 = time required to achieve the same BOD value at temperature T
- BODt = BOD at temperature T
- BODs = BOD at standard conditions (5 days, 20degree C)
- k2 = rate constant at test temperature (30degree C)
Substituting the values:
- t2 = (1/0.40) * ln(250/143.75)
- t2 = 3.33 days
Therefore, it will take 3.33 days to achieve the same BOD value at 30degree C.
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A sample a sewage is estimated to have a 5 days 20degree C BOD of 250mg/l. If the test temperature be 30degree C , in how many days will the same value of BOD be obtained?can anyone solve this?3.33 days is ans?
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A sample a sewage is estimated to have a 5 days 20degree C BOD of 250mg/l. If the test temperature be 30degree C , in how many days will the same value of BOD be obtained?can anyone solve this?3.33 days is ans? 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 A sample a sewage is estimated to have a 5 days 20degree C BOD of 250mg/l. If the test temperature be 30degree C , in how many days will the same value of BOD be obtained?can anyone solve this?3.33 days is ans? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample a sewage is estimated to have a 5 days 20degree C BOD of 250mg/l. If the test temperature be 30degree C , in how many days will the same value of BOD be obtained?can anyone solve this?3.33 days is ans?.
A sample a sewage is estimated to have a 5 days 20degree C BOD of 250mg/l. If the test temperature be 30degree C , in how many days will the same value of BOD be obtained?can anyone solve this?3.33 days is ans? 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 A sample a sewage is estimated to have a 5 days 20degree C BOD of 250mg/l. If the test temperature be 30degree C , in how many days will the same value of BOD be obtained?can anyone solve this?3.33 days is ans? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample a sewage is estimated to have a 5 days 20degree C BOD of 250mg/l. If the test temperature be 30degree C , in how many days will the same value of BOD be obtained?can anyone solve this?3.33 days is ans?.
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