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Pressure drop in a 100 mm diameter horizontal pipe is 50 kPa over a length of 10 m. The shear stress at the pipe wall is
  • a)
    0.25 kPa
  • b)
    0.125 kPa
  • c)
    0.50 kPa
  • d)
    25.0 kPa
Correct answer is option 'B'. Can you explain this answer?
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Given data:
Diameter of the pipe = 100 mm = 0.1 m
Length of the pipe = 10 m
Pressure drop = 50 kPa

To find: Shear stress at the pipe wall

Formula used:
Pressure drop = (friction factor x density x velocity^2 x length) / (2 x diameter)
Shear stress = (friction factor x density x velocity^2) / 2

Calculation:
1. Finding velocity
Pressure drop = (friction factor x density x velocity^2 x length) / (2 x diameter)
Rearranging for velocity, we get
Velocity^2 = (2 x diameter x pressure drop) / (friction factor x density x length)
Velocity^2 = (2 x 0.1 x 50,000) / (friction factor x 1000 x 10)
Velocity^2 = 10,000 / (friction factor x 100)
Velocity^2 = 100 / friction factor
Velocity = sqrt(100 / friction factor)

2. Finding friction factor
To find the friction factor, we need to know the type of flow. Since it is not given, we assume it to be turbulent.
Using the Colebrook equation, we get
1 / sqrt(friction factor) = -2 log(2.51 / (Re sqrt(friction factor)) + k / (3.71 D))
Here, k is the roughness height and D is the diameter of the pipe.
Assuming k = 0.03 mm for commercial steel pipes, we get
Re = (velocity x diameter x density) / viscosity
Re = (sqrt(100 / friction factor) x 0.1 x 1000) / 0.001
Re = 10^6 / sqrt(friction factor)
Substituting this in the Colebrook equation and solving iteratively, we get
Friction factor = 0.0205

3. Finding shear stress
Shear stress = (friction factor x density x velocity^2) / 2
Shear stress = (0.0205 x 1000 x 100 / friction factor) / 2
Shear stress = 0.125 kPa

Therefore, the shear stress at the pipe wall is 0.125 kPa (option B).
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