Question

water flows at a speed 5 cm/s through a pipe of radis 2cm .the viscosity of water is 0.001 pl.the reynolds number and nature of flow are respectively ??????????????????????

Answer 1

Question 37142

Reynolds number is $\mathrm{Re} = \frac{\nu l}{\nu}$ , where $\nu$ is the mean velocity, $l$ is the typical length scale in system and $\nu$ is the kinematic viscosity. The kinematic viscosity is $\nu = \frac{\eta}{\rho}$ , where $\eta$ is the dynamical viscosity and $\rho$ is the density of the fluid.

Let us first convert given viscosity into $Pa \cdot s$ (in SI). $1P = 10 \cdot 1PI$ And $1Pa \cdot s = 10P$ , hence $1Pa \cdot s = 100PI$ so $\eta = 10^{-5}Pa \cdot s$ . Kinematic viscosity is then $\nu = 10^{-8}\frac{m^2}{s}$ .

Also knowing $l = 2cm = 2 \cdot 10^{-2}m$ and $\nu = 5\frac{cm}{s} = 5 \cdot 10^{-2}\frac{m}{s}$ , obtain $\mathrm{Re} = 100$ . Critical Reynolds value is 1000 hence the flow is laminar (calculated Reynolds value is lower than critical).