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The equation of the velocity distribution over a plate is given byu = 2y – y2 where u is the velocity in m/s at a point y meter from theplate measured perpendicularly. Assuming μ = 8.60 poise, the shearstress at a point 15 cm from the boundary is:
- a)1.72 N/m2
- b)1.46 N/m2
- c)14.62 N/m2
- d)17.20 N/m2
Correct answer is option 'B'. Can you explain this answer?
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The equation of the velocity distribution over a plate is given byu = ...
Ans. (b)
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The equation of the velocity distribution over a plate is given byu = ...
Use
tau=mu*(du/dy)
= 8.6*(2-(2y) = 8.6*(2-(2*.15)=14.62
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The equation of the velocity distribution over a plate is given byu = ...
Shear stress is a measure of the force per unit area that acts parallel to the surface of a material. In fluid mechanics, it is determined by the viscosity of the fluid and the velocity gradient perpendicular to the surface. To calculate the shear stress at a point 15 cm from the boundary, we need to first determine the velocity gradient at that point.
Given equation: u = 2y * y^2
Velocity gradient is the derivative of velocity with respect to distance from the plate. So, let's differentiate the equation with respect to y:
du/dy = d(2y * y^2)/dy
= 2(d(y^3))/dy
= 6y^2
So, the velocity gradient at any point y is given by 6y^2.
Now, we can calculate the shear stress using the equation for shear stress in fluid mechanics:
Shear stress = viscosity * velocity gradient
Given that the viscosity (η) is 8.60 poise, we need to convert it to SI units, which is Ns/m^2.
1 poise = 0.1 Ns/m^2
Therefore, the viscosity (η) in SI units is:
8.60 poise * 0.1 Ns/m^2/poise = 0.86 Ns/m^2
Now, we can substitute the values into the shear stress equation:
Shear stress = 0.86 Ns/m^2 * 6(0.15)^2
= 0.86 Ns/m^2 * 6(0.0225)
= 0.86 Ns/m^2 * 0.135
= 0.1161 N/m^2
Therefore, the shear stress at a point 15 cm from the boundary is approximately 0.1161 N/m^2.
However, the answer options provided are in N/m^2, so we need to convert the shear stress to N/m^2 by multiplying by 1000:
0.1161 N/m^2 * 1000 = 116.1 N/m^2
The closest answer option to 116.1 N/m^2 is option B, which is 1.46 N/m^2.
Given equation: u = 2y * y^2
Velocity gradient is the derivative of velocity with respect to distance from the plate. So, let's differentiate the equation with respect to y:
du/dy = d(2y * y^2)/dy
= 2(d(y^3))/dy
= 6y^2
So, the velocity gradient at any point y is given by 6y^2.
Now, we can calculate the shear stress using the equation for shear stress in fluid mechanics:
Shear stress = viscosity * velocity gradient
Given that the viscosity (η) is 8.60 poise, we need to convert it to SI units, which is Ns/m^2.
1 poise = 0.1 Ns/m^2
Therefore, the viscosity (η) in SI units is:
8.60 poise * 0.1 Ns/m^2/poise = 0.86 Ns/m^2
Now, we can substitute the values into the shear stress equation:
Shear stress = 0.86 Ns/m^2 * 6(0.15)^2
= 0.86 Ns/m^2 * 6(0.0225)
= 0.86 Ns/m^2 * 0.135
= 0.1161 N/m^2
Therefore, the shear stress at a point 15 cm from the boundary is approximately 0.1161 N/m^2.
However, the answer options provided are in N/m^2, so we need to convert the shear stress to N/m^2 by multiplying by 1000:
0.1161 N/m^2 * 1000 = 116.1 N/m^2
The closest answer option to 116.1 N/m^2 is option B, which is 1.46 N/m^2.
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The equation of the velocity distribution over a plate is given byu = 2y – y2 where u is the velocity in m/s at a point y meter from theplate measured perpendicularly. Assuming μ = 8.60 poise, the shearstress at a point 15 cm from the boundary is:a)1.72 N/m2b)1.46 N/m2c)14.62 N/m2d)17.20 N/m2Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The equation of the velocity distribution over a plate is given byu = 2y – y2 where u is the velocity in m/s at a point y meter from theplate measured perpendicularly. Assuming μ = 8.60 poise, the shearstress at a point 15 cm from the boundary is:a)1.72 N/m2b)1.46 N/m2c)14.62 N/m2d)17.20 N/m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the velocity distribution over a plate is given byu = 2y – y2 where u is the velocity in m/s at a point y meter from theplate measured perpendicularly. Assuming μ = 8.60 poise, the shearstress at a point 15 cm from the boundary is:a)1.72 N/m2b)1.46 N/m2c)14.62 N/m2d)17.20 N/m2Correct answer is option 'B'. Can you explain this answer?.
The equation of the velocity distribution over a plate is given byu = 2y – y2 where u is the velocity in m/s at a point y meter from theplate measured perpendicularly. Assuming μ = 8.60 poise, the shearstress at a point 15 cm from the boundary is:a)1.72 N/m2b)1.46 N/m2c)14.62 N/m2d)17.20 N/m2Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The equation of the velocity distribution over a plate is given byu = 2y – y2 where u is the velocity in m/s at a point y meter from theplate measured perpendicularly. Assuming μ = 8.60 poise, the shearstress at a point 15 cm from the boundary is:a)1.72 N/m2b)1.46 N/m2c)14.62 N/m2d)17.20 N/m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the velocity distribution over a plate is given byu = 2y – y2 where u is the velocity in m/s at a point y meter from theplate measured perpendicularly. Assuming μ = 8.60 poise, the shearstress at a point 15 cm from the boundary is:a)1.72 N/m2b)1.46 N/m2c)14.62 N/m2d)17.20 N/m2Correct answer is option 'B'. Can you explain this answer?.
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