Answer to Question #211693 in Mechanics | Relativity for steviosis

Question

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

Answer 1

Gives

$\rho_{air}=1.2kg/m^3$

$\eta_{air}=1.8\times10^{-5}Nsec/m^2$

d=$1.0\times10^{-5}m$

$\rho_{dust}=2.0\times10^{-3}kgm^{-3}$

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

$V_t=\frac{2\times1\times10^{-5}\times(1.2-0.002)}{9\times1.8\times10^{-5}}\times9.8$

$V_t=1.44m/sec$

Reynold's number

$R=\frac{\rho v_c d}{\eta}$

Put value

$R=\frac{1\times10^{-5} \times2.0\times10^{-3}\times1.44}{1.8\times10^{-5}}$

$R=1600$