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The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1000, develops a velocity of 1 m/s, when flowing full. When it is flowing half – full, the velocity of flow through the sewer will be
- a)0.5 m/s
- b)1.0 m/s
- c)2 m/s
- d)2.0 m/s
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1...
For,
Where d is depth of flow
Where d is depth of flow
v is flow velocity at depth d
D is diameter of sewer
V is flow velocity at full flow
∴ v = 1.0 m/s
Most Upvoted Answer
The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1...
Full, what is the velocity in the sewer?
We can use the continuity equation to solve this problem. The continuity equation states that the mass flow rate (ρAv) is constant along a pipe, where ρ is the density of the fluid, A is the cross-sectional area of the pipe, and v is the velocity of the fluid. Since the mass flow rate is constant, we can write:
ρ1A1v1 = ρ2A2v2
where the subscripts 1 and 2 refer to the conditions before and after the sewer is flowing half full, respectively. Since the sewer is a circular pipe with diameter d = 1.0 m, we can find the cross-sectional area as:
A = πd^2/4 = π/4 m^2
Using the given values, we have:
ρ1A1v1 = ρ2A2v2
ρ1(π/4)(1)^2(1) = ρ2(π/4)(1)^2v2
ρ1 = ρ2v2
Substituting the slope of the sewer, we have:
tan θ = 1/1000
θ = 0.057°
Using the Manning equation, we can find the velocity of the full sewer as:
v1 = (1/n)R2/3S1/2
where n is the Manning roughness coefficient, R is the hydraulic radius (equal to half the diameter for a circular pipe), and S is the slope of the sewer. Assuming a typical value of n = 0.013 for concrete pipes, we have:
R = d/4 = 0.25 m
v1 = (1/0.013)(0.25)^2/3(1/1000)^1/2 ≈ 1.16 m/s
Substituting these values into the continuity equation, we have:
ρ1A1v1 = ρ2A2v2
ρ2v2 = ρ1A1v1/(π/4)
ρ2v2 = ρ1v1
v2 = v1(ρ1/ρ2)
Using the density of water (1000 kg/m^3), we have:
v2 = 1.16(1000/ρ2)
When the sewer is half full, the cross-sectional area is half of the full area, so:
A2 = πd^2/8 = π/8 m^2
The hydraulic radius is also reduced by half, so:
R = d/8 = 0.125 m
Using the Manning equation again, we have:
v2 = (1/0.013)(0.125)^2/3(1/1000)^1/2 ≈ 0.70 m/s
Substituting this value into the continuity equation, we have:
v2 = v1(ρ1/ρ2)
0.70 = 1.16(1000/ρ2)
ρ2 ≈ 1667 kg/m^3
This density is higher than that of water, which suggests that the sewer is not flowing with water alone. It is possible that there are sediments or other materials in the sewer that increase the density. Nevertheless, the velocity of the half-full sewer is 0.70 m/s.
We can use the continuity equation to solve this problem. The continuity equation states that the mass flow rate (ρAv) is constant along a pipe, where ρ is the density of the fluid, A is the cross-sectional area of the pipe, and v is the velocity of the fluid. Since the mass flow rate is constant, we can write:
ρ1A1v1 = ρ2A2v2
where the subscripts 1 and 2 refer to the conditions before and after the sewer is flowing half full, respectively. Since the sewer is a circular pipe with diameter d = 1.0 m, we can find the cross-sectional area as:
A = πd^2/4 = π/4 m^2
Using the given values, we have:
ρ1A1v1 = ρ2A2v2
ρ1(π/4)(1)^2(1) = ρ2(π/4)(1)^2v2
ρ1 = ρ2v2
Substituting the slope of the sewer, we have:
tan θ = 1/1000
θ = 0.057°
Using the Manning equation, we can find the velocity of the full sewer as:
v1 = (1/n)R2/3S1/2
where n is the Manning roughness coefficient, R is the hydraulic radius (equal to half the diameter for a circular pipe), and S is the slope of the sewer. Assuming a typical value of n = 0.013 for concrete pipes, we have:
R = d/4 = 0.25 m
v1 = (1/0.013)(0.25)^2/3(1/1000)^1/2 ≈ 1.16 m/s
Substituting these values into the continuity equation, we have:
ρ1A1v1 = ρ2A2v2
ρ2v2 = ρ1A1v1/(π/4)
ρ2v2 = ρ1v1
v2 = v1(ρ1/ρ2)
Using the density of water (1000 kg/m^3), we have:
v2 = 1.16(1000/ρ2)
When the sewer is half full, the cross-sectional area is half of the full area, so:
A2 = πd^2/8 = π/8 m^2
The hydraulic radius is also reduced by half, so:
R = d/8 = 0.125 m
Using the Manning equation again, we have:
v2 = (1/0.013)(0.125)^2/3(1/1000)^1/2 ≈ 0.70 m/s
Substituting this value into the continuity equation, we have:
v2 = v1(ρ1/ρ2)
0.70 = 1.16(1000/ρ2)
ρ2 ≈ 1667 kg/m^3
This density is higher than that of water, which suggests that the sewer is not flowing with water alone. It is possible that there are sediments or other materials in the sewer that increase the density. Nevertheless, the velocity of the half-full sewer is 0.70 m/s.
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The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1000, develops a velocity of 1 m/s, when flowing full. When it is flowing half – full, the velocity of flow through the sewer will bea)0.5 m/sb)1.0 m/sc)2 m/sd)2.0 m/sCorrect answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1000, develops a velocity of 1 m/s, when flowing full. When it is flowing half – full, the velocity of flow through the sewer will bea)0.5 m/sb)1.0 m/sc)2 m/sd)2.0 m/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1000, develops a velocity of 1 m/s, when flowing full. When it is flowing half – full, the velocity of flow through the sewer will bea)0.5 m/sb)1.0 m/sc)2 m/sd)2.0 m/sCorrect answer is option 'B'. Can you explain this answer?.
The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1000, develops a velocity of 1 m/s, when flowing full. When it is flowing half – full, the velocity of flow through the sewer will bea)0.5 m/sb)1.0 m/sc)2 m/sd)2.0 m/sCorrect answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1000, develops a velocity of 1 m/s, when flowing full. When it is flowing half – full, the velocity of flow through the sewer will bea)0.5 m/sb)1.0 m/sc)2 m/sd)2.0 m/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The slope of a 1.0 m diameter concrete sewer laid at a slope of 1 in 1000, develops a velocity of 1 m/s, when flowing full. When it is flowing half – full, the velocity of flow through the sewer will bea)0.5 m/sb)1.0 m/sc)2 m/sd)2.0 m/sCorrect answer is option 'B'. Can you explain this answer?.
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