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The pressure drop for a relatively low Reynolds number flow in a 600 mm diameter, 30 m long pipeline is 70 kPa. What is the wall shear stress?
  • a)
    0 Pa
  • b)
    1400 Pa
  • c)
    700 Pa
  • d)
    350 Pa
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The pressure drop for a relatively low Reynolds number flow in a 600 m...
The wall shear stress is the force per unit area acting tangentially on the surface of the pipeline. It is caused by the friction between the fluid and the pipe wall.
To calculate the wall shear stress, we can use the Darcy-Weisbach equation, which relates the pressure drop to the wall shear stress and other parameters of the flow. The equation is given as:

ΔP = (f * L * ρ * V^2) / (2 * D)

Where:
ΔP is the pressure drop
f is the friction factor
L is the length of the pipeline
ρ is the density of the fluid
V is the velocity of the fluid
D is the diameter of the pipeline

In this case, we are given the pressure drop (ΔP = 70 kPa), the diameter of the pipeline (D = 600 mm), and the length of the pipeline (L = 30 m). We need to find the wall shear stress.

To solve for the wall shear stress, we need to find the velocity of the fluid. The Reynolds number (Re) can be used to determine the flow regime. For a relatively low Reynolds number flow, the flow is typically laminar. The Reynolds number is given by:

Re = (ρ * V * D) / μ

Where:
μ is the dynamic viscosity of the fluid

Since the flow is laminar, we can assume that the friction factor (f) is 64/Re. Substituting this into the Darcy-Weisbach equation and rearranging for the wall shear stress (τ), we get:

τ = (2 * ΔP * D) / (f * L)

Substituting the given values into the equation:

τ = (2 * 70 kPa * 600 mm) / ((64/Re) * 30 m)

Simplifying the equation:

τ = (2 * 70 * 600) / (64/Re)

τ = 70 * 600 * (Re/64)

Now, since the Reynolds number is relatively low, we can assume that Re ≈ 2000. Substituting this value into the equation:

τ = 70 * 600 * (2000/64)

τ ≈ 350 Pa

Therefore, the wall shear stress is approximately 350 Pa. So, the correct answer is option D.
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Community Answer
The pressure drop for a relatively low Reynolds number flow in a 600 m...
Concept:
Shear stress at any distance ‘r’ from the center of the pipe is given by.

At r = R, i.e. at the, pipe wall, shear stress is maximum and is given by

Where, R = Radius of the pipe
∂p / x = Pressure gradient over the length of the pipe.
So, from the above, τ ∝ R
Calculation:
Given:

δp = 70 kPa, r = 300 mm, x = 30 m
Shear stress 

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