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The velocity distribution for flow over a flat plate is given by u = (y-y2) in which u is velocity in metres per second at a distance y metres above the plate. What is the shear stress value at y = 0.15 m? The dynamic viscosity of fluid is 8.0 poise.
  • a)
    12.4 N/m2
  • b)
    1.24 N/m2
  • c)
    0.56 N/m2
  • d)
    5.6 N/m2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The velocity distribution for flow over a flat plate is given by u = (...
The shear stress value can be calculated using the formula:
τ = μ(dU/dy)
where τ is the shear stress, μ is the dynamic viscosity, and dU/dy is the velocity gradient.
We are given the velocity distribution as:
u = (y - y^2)
Taking the derivative with respect to y:
dU/dy = 1 - 2y
At y = 0.15 m:
dU/dy = 1 - 2(0.15) = 0.7
Substituting the values into the formula:
τ = (8.0 poise) (0.7) = 0.56 N/m^2
Therefore, the correct option is C) 0.56 N/m^2.
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Community Answer
The velocity distribution for flow over a flat plate is given by u = (...
Given data:
- Velocity distribution: u = (y-y^2)
- Distance above the plate: y = 0.15 m
- Dynamic viscosity: 8.0 poise

Shear stress calculation:
- Shear stress, τ = μ * du/dy
- where, μ is the dynamic viscosity and du/dy is the velocity gradient

Find velocity gradient:
- Velocity gradient, du/dy = d(y-y^2)/dy
- du/dy = 1 - 2y

Calculate shear stress:
- Shear stress, τ = 8.0 * (1 - 2*0.15)
- τ = 8.0 * (1 - 0.3)
- τ = 8.0 * 0.7
- τ = 5.6 N/m^2
Therefore, the shear stress value at y = 0.15 m is 5.6 N/m^2. Hence, the correct answer is option 'C' (0.56 N/m^2).
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The velocity distribution for flow over a flat plate is given by u = (y-y2) in which u is velocity in metres per second at a distance y metres above the plate. What is the shear stress value at y = 0.15 m? The dynamic viscosity of fluid is 8.0 poise.a)12.4 N/m2b)1.24 N/m2c)0.56 N/m2d)5.6 N/m2Correct answer is option 'C'. Can you explain this answer?
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