Question

Define turbulent flow. In a commercial pipe the flow of water, mass density is
1000 kg/m² and kinematie viscosity is 1x10‐6 m²/sec having roughness 0.12 mm yields an average shear stress at pipe boundary 600 N/m². Find the value of ratio roughness to thickness of laminar sub-layer for the pipe and the nature of the
surface as per Nikuradse experiment.​

Answer 1

$\underline{\underline{\maltese\: \: \textbf{\textsf{Answer}}}}$

$\sf{Re = \frac{VD}{v}}$

$\sf{τ = μ \frac{dv}{dy}}$

$\sf{ \frac{du}{dy} = \frac{u}{y}} (Linear ,commercial \: pipe)$

$\sf{∴ τo = μ \frac{μ}{y}}$

$\sf{ \frac{τo}{e} = \frac{μ}{ρ} \times \frac{u}{y} = v \frac{u}{y} }$

$\sf{Shear \: velocity = V1}$

$\sf{V1² = \frac{τo}{ρ} }$

$\sf{= v \frac{u}{y}}$

$\sf{ \frac{u}{V1} = \frac{V1 × y}{v}= Re}$

$\sf{ \frac{V1y1}{v} = 11.6 \: at \: y = δ1(experimental \: result)}$

$\sf{ \frac{V1δ1}{v} = 11.6}$

$\sf{V1 × δ1 = 11.6 × {10}^{ - 6} m² / s}$

$\sf{V1 = \sqrt{ \frac{τo}{ρ}} = \sqrt{ \frac{600}{1000}} = 0.6 }$

$\sf{δ1 = \frac{11.6 × {10}^{ - 6}}{ \sqrt{0.6} } = 1.5 × {10}^{ - 5} m}$

$\sf{= 1.5 × {10}^{ - 2} mm}$

$\sf{ \frac{k}{δ1}= \frac{0.12}{1.5 \times {10}^{ - 2} }= 8 }$