What should be the maximum average velocity of water in a tube of diameter 0.5cm . So the flow is laminar .The viscosity of water is 0.00125newton per metre square second?
Answers
Answer is 57.5 cm/sec.
Reynolds's formula for fluid flow:
Re = row * v * D / μ
Re = Reynolds's number < 2300 for laminar stream lined non-turbulent flow. row = density = 1000 kg/m^3
μ = dynamic or kinematic viscosity = 0.00125 Pa-s.
D = inner diameter = 0.005 meter.
So v = Re * μ / (row* D)
v < 57.5 cm/s.
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...