Question

) Evaluate the Reynold’s number at terminal velocity in air at 20oC, of a spherical dust particle of radius 1.0 X 10-5m and density 2.0 X 10-3Kgm-3

Answer 1

Answer:

Explanation:

9

Answer 2

The Reynold's number is 1.93 × 10⁻⁵

Explanation:

Given:

Radius of the dust particles = 1 × 10⁻⁵ m

Density of the dust particles = 2 × 10⁻³ kg/m³

To find out:

The Reynold's number

Solution:

We know that terminal velocity is given by

$v_t=\frac{2r^2(\rho-\sigma)}{9\eta}g$

Taking

Density of air

$\rho=1.2$ kg/m³

Viscosity of air

$\eta=1.8\times 10^{-5}$ N-s/m²

$\sigma=2\times 10^{-3}$ kg/m³

$r=1\times 10^{-5}$ m

Thus,

$v_t=\frac{2\times (10^{-5})^2(1.2-0.002)\times 9.8}{9\times 1.8\times 10^{-5}}$

$\implies v_t=1.45\times 10^{-5}$ m/s

Now Reynold's number is

$R_e=\frac{Dv_t\rho}{\eta}$

$\implies R_e=\frac{2\times 1\times 10^{-5}\times 1.45\times 10^{-5}\times 1.2}{1.8\times 10^{-5}}$

$\implies R_e=1.93\times 10^{-5}$

Hope this answer is helpful.