 # What should be the maximum average velocity in a tube of diameter 2.0 cm so that the flow is laminar? The viscosity of water is 0.001N/s2.I am not understanding how do i start this question.Thank you for your help. 12 years ago

Hi

The formula is $\mathrm{Re} = {{\rho {\bold \mathrm V} D} \over {\mu}} = {{{\bold \mathrm V} D} \over {\nu}} = {{{\bold \mathrm Q} D} \over {\nu}A}$

• ${\bold \mathrm V}$ is the mean fluid velocity (SI units: m/s)
• L is a characteristic linear dimension, (traveled length of fluid, or hydraulic radius when dealing with river systems) (m)
• μ is the dynamic viscosity of the fluid (Pa·s or N·s/m² or kg/m·s)
• ν is the kinematic viscosity (ν = μ / ρ) (m²/s)
• ${\rho}\,$ is the density of the fluid (kg/m³)
• Q is the volumetric flow rate (m³/s)
• A is the pipe cross-sectional area (m²).
• D is the hydraulic diameter of the pipe (m).
• Here the formula used is (1)

Now

When the Reynolds number is less than 2000, flow will be described as laminar
When the Reynolds number is greater than 4000, flow will be described as turbulent
When the Reynolds number is in the range of 2000 to 4000 the flow is considered transitional

So we have

2000 = 103 (v) 2*10-2/0.001

=> v = 0.1 m/s

Note density of water = 103 kg/m3

Thanks

Anurag Kishore