Turbulent Pipe Flow – Roughness vs laminar sublayer thickness Water (ρ = 1000 kg/m^3, ν = 1.0×10^-6 m^2/s) flows in a commercial pipe with equivalent sand roughness k_s = 0.12 mm. If the average wall shear stress is τ_w = 600 N/m^2, compute k_s / δ′, where δ′ is the laminar sublayer thickness.
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A0.25
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B0.50
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C6.0
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D8.0
Answer
Correct Answer: 8.0
Explanation
Introduction / Context:In turbulent flow near walls, a viscous (laminar) sublayer exists adjacent to the boundary. If the roughness height protrudes beyond this sublayer, the flow behaves as transitionally rough or fully rough. Comparing the equivalent sand roughness k_s with the sublayer thickness δ′ helps decide the roughness regime and informs friction factor estimation.
Given Data / Assumptions:
- ρ = 1000 kg/m^3, ν = 1.0×10^-6 m^2/s.
- k_s = 0.12 mm = 1.2×10^-4 m.
- Average wall shear stress τ_w = 600 N/m^2.
- Laminar sublayer thickness δ′ ≈ 11.6 * ν / u_* (hydraulic smooth-wall estimate), where u_* is friction velocity.
Concept / Approach:
Compute the friction velocity u_* = sqrt(τ_w / ρ). Then compute δ′ = 11.6 * ν / u_. Finally, form the ratio k_s / δ′ to compare roughness to the sublayer thickness.
Step-by-Step Solution:
Friction velocity: u_ = sqrt(τ_w / ρ) = sqrt(600 / 1000) = sqrt(0.6) ≈ 0.7746 m/s.Laminar sublayer thickness: δ′ = 11.6 * ν / u_* = 11.6 * 1.0×10^-6 / 0.7746 ≈ 1.496×10^-5 m.Form ratio: k_s / δ′ = (1.2×10^-4) / (1.496×10^-5) ≈ 8.02 ≈ 8.0.Verification / Alternative check:
Since k_s/δ′ ≫ 1, the boundary is in the fully rough regime, consistent with high shear stress and commercial roughness magnitude.
Why Other Options Are Wrong:
0.25 and 0.50 imply hydraulically smooth behavior (k_s << δ′); 6.0 underestimates the ratio from the computed values.
Common Pitfalls:
Confusing kinematic viscosity with dynamic viscosity, or forgetting to convert k_s to metres; using τ in Pa but ρ in inconsistent units leads to erroneous u_*.
Final Answer:
8.0