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A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do = 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.
- a)10 mg/l
- b)16.65 mg/l
- c)6.65 mg/l
- d)9.65 mg/l
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A large stream has a reoxygenation constant of 0.45 per day and at a p...
Time required for certain amount of D.O at 36 km downstream is given by
t = 0.3472 days
Oxygen deficit after time t
Dt = 2.35 mg/l
D.O at 36 km downstream = 12 - 2.35
= 9.65 mg/l
Most Upvoted Answer
A large stream has a reoxygenation constant of 0.45 per day and at a p...
Given data:
Reoxygenation constant (k1) = 0.45 per day
Initial dissolved oxygen (Do) = 0 mg/L
Ultimate demand for oxygen (U) = 19 mg/L
Deoxygenation constant (k2) = 0.2 per day
Distance downstream (d) = 36 km = 36,000 m
Velocity of the stream (v) = 1.2 m/s
To calculate the dissolved oxygen (DO) at 36 km downstream, we need to consider the reoxygenation and deoxygenation processes in the stream.
1. Reoxygenation Process:
The reoxygenation process replenishes the dissolved oxygen in the stream. The rate of reoxygenation is given by the equation:
dDO/dt = k1 * (U - DO)
where dDO/dt is the rate of change of dissolved oxygen with respect to time.
2. Deoxygenation Process:
The deoxygenation process consumes the dissolved oxygen in the stream. The rate of deoxygenation is given by the equation:
dDO/dt = -k2 * DO
Since the stream is saturated with oxygen at the point of discharge (Do = 0), we can assume the deoxygenation process dominates initially.
3. Calculation of DO at 36 km downstream:
We can solve the differential equation for the deoxygenation process to find the dissolved oxygen at any point downstream.
Integrating the equation, we get:
ln(DO) = -k2 * t + C
where C is the integration constant.
Using the initial condition at the point of discharge (Do = 0), we can determine the integration constant C:
ln(0) = -k2 * 0 + C
C = ln(0)
C = -∞ (negative infinity)
Now, let's substitute the values and calculate the dissolved oxygen at 36 km downstream:
ln(DO) = -k2 * t - ∞
ln(DO) = -k2 * t
Solving for DO:
DO = e^(-k2 * t)
Substituting the values:
DO = e^(-0.2 * 36,000 / 1.2)
Calculating the value:
DO ≈ 9.65 mg/L
Therefore, the dissolved oxygen at 36 km downstream is approximately 9.65 mg/L.
Hence, the correct answer is option 'D'.
Reoxygenation constant (k1) = 0.45 per day
Initial dissolved oxygen (Do) = 0 mg/L
Ultimate demand for oxygen (U) = 19 mg/L
Deoxygenation constant (k2) = 0.2 per day
Distance downstream (d) = 36 km = 36,000 m
Velocity of the stream (v) = 1.2 m/s
To calculate the dissolved oxygen (DO) at 36 km downstream, we need to consider the reoxygenation and deoxygenation processes in the stream.
1. Reoxygenation Process:
The reoxygenation process replenishes the dissolved oxygen in the stream. The rate of reoxygenation is given by the equation:
dDO/dt = k1 * (U - DO)
where dDO/dt is the rate of change of dissolved oxygen with respect to time.
2. Deoxygenation Process:
The deoxygenation process consumes the dissolved oxygen in the stream. The rate of deoxygenation is given by the equation:
dDO/dt = -k2 * DO
Since the stream is saturated with oxygen at the point of discharge (Do = 0), we can assume the deoxygenation process dominates initially.
3. Calculation of DO at 36 km downstream:
We can solve the differential equation for the deoxygenation process to find the dissolved oxygen at any point downstream.
Integrating the equation, we get:
ln(DO) = -k2 * t + C
where C is the integration constant.
Using the initial condition at the point of discharge (Do = 0), we can determine the integration constant C:
ln(0) = -k2 * 0 + C
C = ln(0)
C = -∞ (negative infinity)
Now, let's substitute the values and calculate the dissolved oxygen at 36 km downstream:
ln(DO) = -k2 * t - ∞
ln(DO) = -k2 * t
Solving for DO:
DO = e^(-k2 * t)
Substituting the values:
DO = e^(-0.2 * 36,000 / 1.2)
Calculating the value:
DO ≈ 9.65 mg/L
Therefore, the dissolved oxygen at 36 km downstream is approximately 9.65 mg/L.
Hence, the correct answer is option 'D'.
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A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer?
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A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer? 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 large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer?.
A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer? 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 large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer?.
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A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do= 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.a)10 mg/lb)16.65 mg/lc)6.65 mg/ld)9.65 mg/lCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.
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