Question

what should be the maximum average velocity of water in a tube of diameter 0.5 cm so that the flow is laminar?viscocity of water is 1.25×_10×10×10​

Answer 1

Diameter of tube (d) = 0.5 cm = $\sf 5 \times 10^{-3} \ m$

Viscosity of water $\sf (\eta)$ = $\sf 1.25 \times 10^{-3} \ N/m^2s$

Density of water $\sf (\rho)$ = 1000 kg/m³

Reynolds Number

$\boxed{\boxed{ \bf{R_e = \dfrac{\rho vd}{\eta}}}}$

v → Average velocity

In general flow is laminar if $\sf R_e$ is less than 2000

$\rm \leadsto \dfrac{\rho vd}{\eta} \leqslant 2000 \\ \\ \rm \leadsto \dfrac{ \cancel{1000} \times v \times 5 \times \cancel{ {10}^{ - 3} }}{1.25 \times \cancel{ {10}^{ - 3} }} \leqslant 2 \cancel{000 } \\ \\ \rm \leadsto 4v \leqslant 2 \\ \\ \rm \leadsto v \leqslant \dfrac{2}{4} \\ \\ \rm \leadsto v \leqslant 0.5 \: m {s}^{ - 1}$

$\therefore$ Maximum average velocity of water in tube so that the flow is laminar = 0.5 m/s

Answer 2

Given : -

• velocity of water in a tube of diameter 0.5 cm so that the flow is lamina

• viscocity of water is 1.25×_10×10×10

To Find : -

• what should be the maximum average velocity

Solution. : -

Reynolds's formula for fluid flow:

Re = row × v × D / μ

Re = Reynolds's number < 2300 for laminar stream lined non-turbulent flow.

density = 1000 kg/m^3

μ = dynamic

inematic viscosity = 0.00125

inner diameter = 0.005 meter.

inner diameter = 0.005

inner diameter = 0.5 m/s

Maximum average of velocity of water is 0.5 m/s