What is the terminal velocity of an air bubbl | vedclass

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(B) The terminal velocity $V_T$ of a spherical bubble rising in a liquid is given by the formula: $V_T = \frac{2(\rho - \sigma) r^2 g}{9 \eta}$ Where:

Vedclass · Accepted Answer

$V_T = 0.5 \ mm \ s^{-1}$

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(B) The terminal velocity $V_T$ of a spherical bubble rising in a liquid is given by the formula: $V_T = \frac{2(ho - \sigma) r^2 g}{9 \eta}$ Where: $ho$ is the density of the liquid $(900 \ kg \ m^{-3})$,$\sigma$ is the density of the air $(1.293 \ kg \ m^{-3})$,$r$ is the radius of the bubble $(d/2 = 0.5 \ mm = 0.5 	imes 10^{-3} \ m)$,$g$ is the acceleration due to gravity $(9 \ m \ s^{-2})$,$\eta$ is the coefficient of viscosity $(0.85 \ N \ s \ m^{-2})$. Substituting the values: $V_T = \frac{2(900 - 1.293) 	imes (0.5 	imes 10^{-3})^2 	imes 9}{9 	imes 0.85}$ $V_T = \frac{2(898.707) 	imes 0.25 	imes 10^{-6}}{0.85}$ $V_T \approx 0.528 	imes 10^{-3} \ m \ s^{-1} \approx 0.5 \ mm \ s^{-1}$.

What is the terminal velocity of an air bubble of $1.0 \ mm$ in diameter rising in a liquid of viscosity $0.85 \ N \ s \ m^{-2}$ and density $900 \ kg \ m^{-3}$? (Air density $= 1.293 \ kg \ m^{-3}$, $g = 9 \ m \ s^{-2}$)

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