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A waste water sample has an initial BOD of 200 mg/L. the first order BOD decay coefficient is 0.5/day. The BOD consumed in mg/L in 5 days is
- a)183.55
- b)200
- c)16.45
- d)30
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A waste water sample has an initial BOD of 200 mg/L. the first order ...
To calculate the BOD consumed in a waste water sample in 5 days, we can use the first-order BOD decay equation:
BODt = BOD0 * e^(-kt)
Where:
BODt = BOD at time t
BOD0 = initial BOD
k = BOD decay coefficient
t = time in days
Given:
Initial BOD (BOD0) = 200 mg/L
BOD decay coefficient (k) = 0.5/day
Time (t) = 5 days
Calculating BOD consumed in 5 days:
Using the equation mentioned above, we can substitute the given values to find the BOD consumed in 5 days (BOD5).
BOD5 = BOD0 * e^(-k*t)
BOD5 = 200 * e^(-0.5*5)
Simplifying the equation:
BOD5 = 200 * e^(-2.5)
Using the value of e^(-2.5), which is approximately 0.0821:
BOD5 = 200 * 0.0821
BOD5 = 16.42 mg/L
Therefore, the BOD consumed in 5 days is approximately 16.42 mg/L.
Therefore, the correct answer is option 'C' (16.42).
BODt = BOD0 * e^(-kt)
Where:
BODt = BOD at time t
BOD0 = initial BOD
k = BOD decay coefficient
t = time in days
Given:
Initial BOD (BOD0) = 200 mg/L
BOD decay coefficient (k) = 0.5/day
Time (t) = 5 days
Calculating BOD consumed in 5 days:
Using the equation mentioned above, we can substitute the given values to find the BOD consumed in 5 days (BOD5).
BOD5 = BOD0 * e^(-k*t)
BOD5 = 200 * e^(-0.5*5)
Simplifying the equation:
BOD5 = 200 * e^(-2.5)
Using the value of e^(-2.5), which is approximately 0.0821:
BOD5 = 200 * 0.0821
BOD5 = 16.42 mg/L
Therefore, the BOD consumed in 5 days is approximately 16.42 mg/L.
Therefore, the correct answer is option 'C' (16.42).
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Community Answer
A waste water sample has an initial BOD of 200 mg/L. the first order ...
Here,
KD = 0.5 × .434 = 0.217
Now BOD in 5 days,
(BOD)5 = (BOD)0 (1 – 10-KDt)
= 200 (1 – 10-5×.217)
= 183.55
Hence, BOD consumed = 200 – 183.55 = 16.45
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A waste water sample has an initial BOD of 200 mg/L. the first order BOD decay coefficient is 0.5/day. The BOD consumed in mg/L in 5 days isa)183.55b)200c)16.45d)30Correct answer is option 'C'. Can you explain this answer?
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