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The Reynolds number for flow of oil through a 5 cm diameter pipe is 1700. The kinematic viscosity, v = 1.02 × 10−6 m2⁄s. What is the velocity at a point 0.625 cm away from the wall?
Correct answer is 'Range: 0.02 to 0.03'. Can you explain this answer?
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The Reynolds number for flow of oil through a 5 cm diameter pipe is 1...
Note: r is taken from the center and not from the wall, so for a section 0.625 cm from wall has r = R − 0.625 = 1.875 cm] u0.625 = 0.0297 = 0.03 m⁄s
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The Reynolds number for flow of oil through a 5 cm diameter pipe is 1...
Given data:
Diameter of pipe (D) = 5 cm = 0.05 m
Reynolds number (Re) = 1700
Kinematic viscosity (v) = 1.02 × 10−6 m²/s
Distance from the wall (y) = 0.625 cm = 0.00625 m
We are required to find the velocity (v) at a point 0.625 cm away from the wall.
Formula:
Reynolds number (Re) = (ρ × v × D) / v
Where,
ρ = Density of fluid
v = Velocity of fluid
D = Diameter of the pipe
v = Kinematic viscosity of fluid
Calculations:
ρ is not given in the problem statement. However, we can assume that the fluid is oil, and its density is approximately 850 kg/m³.
ρ = 850 kg/m³
Substituting the given values in the Reynolds number formula:
1700 = (850 × v × 0.05) / 1.02 × 10^-6
v = (1700 × 1.02 × 10^-6) / (850 × 0.05)
v = 0.04 m/s
Now, we need to calculate the velocity at a distance of 0.625 cm from the wall. For this, we can use the following formula:
v(y) = v(max) × [1 - (y/D)]^2
Where,
v(max) = Velocity of fluid at the centerline of the pipe
y = Distance from the wall
D = Diameter of the pipe
To find v(max), we can use the following formula:
v(max) = (Re × v) / (D × 2)
Substituting the given values:
v(max) = (1700 × 1.02 × 10^-6) / (0.05 × 2)
v(max) = 0.017 m/s
Now, substituting all the values in the velocity formula:
v(y) = 0.017 × [1 - (0.00625/0.05)]^2
v(y) = 0.023 m/s
Therefore, the velocity at a point 0.625 cm away from the wall is in the range of 0.02 to 0.03 m/s.
Diameter of pipe (D) = 5 cm = 0.05 m
Reynolds number (Re) = 1700
Kinematic viscosity (v) = 1.02 × 10−6 m²/s
Distance from the wall (y) = 0.625 cm = 0.00625 m
We are required to find the velocity (v) at a point 0.625 cm away from the wall.
Formula:
Reynolds number (Re) = (ρ × v × D) / v
Where,
ρ = Density of fluid
v = Velocity of fluid
D = Diameter of the pipe
v = Kinematic viscosity of fluid
Calculations:
ρ is not given in the problem statement. However, we can assume that the fluid is oil, and its density is approximately 850 kg/m³.
ρ = 850 kg/m³
Substituting the given values in the Reynolds number formula:
1700 = (850 × v × 0.05) / 1.02 × 10^-6
v = (1700 × 1.02 × 10^-6) / (850 × 0.05)
v = 0.04 m/s
Now, we need to calculate the velocity at a distance of 0.625 cm from the wall. For this, we can use the following formula:
v(y) = v(max) × [1 - (y/D)]^2
Where,
v(max) = Velocity of fluid at the centerline of the pipe
y = Distance from the wall
D = Diameter of the pipe
To find v(max), we can use the following formula:
v(max) = (Re × v) / (D × 2)
Substituting the given values:
v(max) = (1700 × 1.02 × 10^-6) / (0.05 × 2)
v(max) = 0.017 m/s
Now, substituting all the values in the velocity formula:
v(y) = 0.017 × [1 - (0.00625/0.05)]^2
v(y) = 0.023 m/s
Therefore, the velocity at a point 0.625 cm away from the wall is in the range of 0.02 to 0.03 m/s.
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The Reynolds number for flow of oil through a 5 cm diameter pipe is 1700. The kinematic viscosity, v = 1.02 × 10−6 m2⁄s. What is the velocity at a point 0.625 cm away from the wall?Correct answer is 'Range: 0.02 to 0.03'. Can you explain this answer?
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The Reynolds number for flow of oil through a 5 cm diameter pipe is 1700. The kinematic viscosity, v = 1.02 × 10−6 m2⁄s. What is the velocity at a point 0.625 cm away from the wall?Correct answer is 'Range: 0.02 to 0.03'. 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 The Reynolds number for flow of oil through a 5 cm diameter pipe is 1700. The kinematic viscosity, v = 1.02 × 10−6 m2⁄s. What is the velocity at a point 0.625 cm away from the wall?Correct answer is 'Range: 0.02 to 0.03'. 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 The Reynolds number for flow of oil through a 5 cm diameter pipe is 1700. The kinematic viscosity, v = 1.02 × 10−6 m2⁄s. What is the velocity at a point 0.625 cm away from the wall?Correct answer is 'Range: 0.02 to 0.03'. Can you explain this answer?.
The Reynolds number for flow of oil through a 5 cm diameter pipe is 1700. The kinematic viscosity, v = 1.02 × 10−6 m2⁄s. What is the velocity at a point 0.625 cm away from the wall?Correct answer is 'Range: 0.02 to 0.03'. 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 The Reynolds number for flow of oil through a 5 cm diameter pipe is 1700. The kinematic viscosity, v = 1.02 × 10−6 m2⁄s. What is the velocity at a point 0.625 cm away from the wall?Correct answer is 'Range: 0.02 to 0.03'. 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 The Reynolds number for flow of oil through a 5 cm diameter pipe is 1700. The kinematic viscosity, v = 1.02 × 10−6 m2⁄s. What is the velocity at a point 0.625 cm away from the wall?Correct answer is 'Range: 0.02 to 0.03'. Can you explain this answer?.
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