Turbulent Flow MCQ with answer and solution

Concept

In steady pipe flow (laminar or turbulent), shear stress varies linearly with radius: zero at the centre and maximum at the wall.

Formula

τ = (-dp/dx) × r/2 ; τ₀ = (-dp/dx) × R/2 ; τ/τ₀ = r/R

Step-by-step solution

1. Take a force balance on a fluid cylinder of radius r: pressure force = shear force. 2. (-dp/dx) × πr² = τ × 2πr, so τ = (-dp/dx) × r/2. 3. This result does not depend on the type of flow, so it holds for turbulent flow too. 4. At r = 0, τ = 0. At r = R (the wall), τ = τ₀, which is the maximum. 5. So τ is maximum at the wall and decreases linearly to zero at the centre. Answer (D).

Explanation

A simple force balance gives τ = (-dp/dx) r/2 for any steady flow in a pipe. So the shear stress is directly proportional to r. It is zero at the centre and maximum at the wall, in both laminar and turbulent flow.

Why the other options are wrong

A: The shear stress is zero at the centre, not maximum. B: The variation is linear, not logarithmic. The velocity profile is logarithmic, not the shear stress. C: There is no peak midway. The value increases steadily up to the wall.

Shortcut

Shear stress is linear in r for both laminar and turbulent pipe flow: zero at the centre, maximum at the wall.

Common mistake

Mixing up the velocity profile (maximum at the centre, logarithmic in turbulent flow) with the shear stress profile (maximum at the wall, linear).