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A stream with a flow rate of 5 m3/s is having an ultimate BOD of 30 mg/litre. A wastewater discharge of 0.20 m3/s having BOD5 of 500 mg/litre joins the stream at a location and instantaneously gets mixed up completely. The cross-sectional area of the stream is 40 m2 which remains constant. BOD exertion rate constant is 0.3 per day (logarithm base to e). The BOD (in mg/litre round off to two decimal places) remaining at 3 km downstream from the mixing location, is ________.
Correct answer is '49.57'. Can you explain this answer?
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Verified Answer
A stream with a flow rate of 5 m3/s is having an ultimate BOD of 30 mg...
t = d/v where, 
Lt = L0e–k × t
= 53.6e –0.3 × 0.26
= 49.57 mg/l
Lt = L0e–k × t
= 53.6e –0.3 × 0.26
= 49.57 mg/l
Most Upvoted Answer
A stream with a flow rate of 5 m3/s is having an ultimate BOD of 30 mg...
The problem can be solved using the concept of BOD (Biochemical Oxygen Demand) and the principle of dilution and mixing of fluids.
Given data:
Flow rate of stream (Q1) = 5 m3/s
Ultimate BOD of stream (BODu1) = 30 mg/litre
Flow rate of wastewater discharge (Q2) = 0.20 m3/s
BOD5 of wastewater discharge (BOD52) = 500 mg/litre
Cross-sectional area of the stream (A) = 40 m2
BOD exertion rate constant (k) = 0.3 per day
To find the BOD remaining at 3 km downstream from the mixing location, we will use the concept of dilution and mixing of fluids.
Step 1: Calculate the initial BOD of the mixture
The initial BOD of the mixture can be calculated by considering the BOD of the stream and the wastewater discharge based on their flow rates and BOD values.
Initial BOD of mixture = (Q1 * BODu1 + Q2 * BOD52) / (Q1 + Q2)
Substituting the given values:
Initial BOD of mixture = (5 * 30 + 0.20 * 500) / (5 + 0.20)
= (150 + 100) / 5.20
= 50 mg/litre
Step 2: Calculate the BOD remaining after 3 km downstream
The BOD remaining after 3 km downstream can be calculated using the first-order decay equation for BOD.
BOD remaining = BOD initial * e^(-k * t)
Where:
BOD initial = Initial BOD of mixture
k = BOD exertion rate constant
t = time (in days)
Since the flow rate is given in m3/s, we need to convert it to m3/day. There are 24 hours in a day, so we multiply the flow rate by 24.
Flow rate of stream (Q1) = 5 m3/s * 24 * 60 * 60 = 432,000 m3/day
Now, we can calculate the BOD remaining after 3 km downstream.
BOD remaining = 50 * e^(-0.3 * 3 * 1000 * 1000 / 432,000)
Simplifying further:
BOD remaining = 50 * e^(-6.94)
Calculating the value:
BOD remaining ≈ 49.57 mg/litre (rounded to two decimal places)
Therefore, the BOD remaining at 3 km downstream from the mixing location is approximately 49.57 mg/litre.
Given data:
Flow rate of stream (Q1) = 5 m3/s
Ultimate BOD of stream (BODu1) = 30 mg/litre
Flow rate of wastewater discharge (Q2) = 0.20 m3/s
BOD5 of wastewater discharge (BOD52) = 500 mg/litre
Cross-sectional area of the stream (A) = 40 m2
BOD exertion rate constant (k) = 0.3 per day
To find the BOD remaining at 3 km downstream from the mixing location, we will use the concept of dilution and mixing of fluids.
Step 1: Calculate the initial BOD of the mixture
The initial BOD of the mixture can be calculated by considering the BOD of the stream and the wastewater discharge based on their flow rates and BOD values.
Initial BOD of mixture = (Q1 * BODu1 + Q2 * BOD52) / (Q1 + Q2)
Substituting the given values:
Initial BOD of mixture = (5 * 30 + 0.20 * 500) / (5 + 0.20)
= (150 + 100) / 5.20
= 50 mg/litre
Step 2: Calculate the BOD remaining after 3 km downstream
The BOD remaining after 3 km downstream can be calculated using the first-order decay equation for BOD.
BOD remaining = BOD initial * e^(-k * t)
Where:
BOD initial = Initial BOD of mixture
k = BOD exertion rate constant
t = time (in days)
Since the flow rate is given in m3/s, we need to convert it to m3/day. There are 24 hours in a day, so we multiply the flow rate by 24.
Flow rate of stream (Q1) = 5 m3/s * 24 * 60 * 60 = 432,000 m3/day
Now, we can calculate the BOD remaining after 3 km downstream.
BOD remaining = 50 * e^(-0.3 * 3 * 1000 * 1000 / 432,000)
Simplifying further:
BOD remaining = 50 * e^(-6.94)
Calculating the value:
BOD remaining ≈ 49.57 mg/litre (rounded to two decimal places)
Therefore, the BOD remaining at 3 km downstream from the mixing location is approximately 49.57 mg/litre.
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A stream with a flow rate of 5 m3/s is having an ultimate BOD of 30 mg/litre. A wastewater discharge of 0.20 m3/s having BOD5 of 500 mg/litre joins the stream at a location and instantaneously gets mixed up completely. The cross-sectional area of the stream is 40 m2 which remains constant. BOD exertion rate constant is 0.3 per day (logarithm base to e). The BOD (in mg/litre round off to two decimal places) remaining at 3 km downstream from the mixing location, is ________.Correct answer is '49.57'. Can you explain this answer?
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A stream with a flow rate of 5 m3/s is having an ultimate BOD of 30 mg/litre. A wastewater discharge of 0.20 m3/s having BOD5 of 500 mg/litre joins the stream at a location and instantaneously gets mixed up completely. The cross-sectional area of the stream is 40 m2 which remains constant. BOD exertion rate constant is 0.3 per day (logarithm base to e). The BOD (in mg/litre round off to two decimal places) remaining at 3 km downstream from the mixing location, is ________.Correct answer is '49.57'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A stream with a flow rate of 5 m3/s is having an ultimate BOD of 30 mg/litre. A wastewater discharge of 0.20 m3/s having BOD5 of 500 mg/litre joins the stream at a location and instantaneously gets mixed up completely. The cross-sectional area of the stream is 40 m2 which remains constant. BOD exertion rate constant is 0.3 per day (logarithm base to e). The BOD (in mg/litre round off to two decimal places) remaining at 3 km downstream from the mixing location, is ________.Correct answer is '49.57'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stream with a flow rate of 5 m3/s is having an ultimate BOD of 30 mg/litre. A wastewater discharge of 0.20 m3/s having BOD5 of 500 mg/litre joins the stream at a location and instantaneously gets mixed up completely. The cross-sectional area of the stream is 40 m2 which remains constant. BOD exertion rate constant is 0.3 per day (logarithm base to e). The BOD (in mg/litre round off to two decimal places) remaining at 3 km downstream from the mixing location, is ________.Correct answer is '49.57'. Can you explain this answer?.
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