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The velocity distribution near the solid wall at a section in a laminar flow is given by u = 5 sin(5πy) for y < 0.1 m. The shear stress at a section y = 0.05m (if the dynamic viscosity of the fluid is 5 poise) will be
  • a)
    39.27 N/m2
  • b)
    27.66 N/m2
  • c)
    7.85 N/m2
  • d)
    zero
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The velocity distribution near the solid wall at a section in a lamina...
The given velocity distribution is u = 5 sin(5θ), where θ is the angular coordinate near the solid wall.

To find the velocity distribution near the solid wall, we need to determine the values of θ.

In a laminar flow, the velocity near the solid wall is zero. This means that the velocity distribution u = 5 sin(5θ) will be zero at the solid wall.

Setting u = 0, we have:

0 = 5 sin(5θ)

To find the values of θ that satisfy this equation, we can use the inverse sine function:

sin(5θ) = 0

5θ = sin^(-1)(0)

5θ = 0

θ = 0

Therefore, the velocity distribution near the solid wall at a section in a laminar flow is given by u = 5 sin(5θ), where θ = 0.
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The velocity distribution near the solid wall at a section in a laminar flow is given by u = 5 sin(5πy) for y < 0.1 m. The shear stress at a section y = 0.05m (if the dynamic viscosity of the fluid is 5 poise) will bea)39.27 N/m2b)27.66 N/m2c)7.85 N/m2d)zeroCorrect answer is option 'A'. Can you explain this answer?
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