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(A) The terminal velocity $v_t$ of a gas bubble rising through a liquid is given by the empirical relation:$v_t = 1.15 \left( \frac{g d^2 (\rho_f - \r

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The terminal velocity $v_t$ of a gas bubble rising through a liquid is given by the empirical relation:$v_t = 1.15 \left( \frac{g d^2 (ho_f - ho_g)}{18 \eta} ight)$ is for solid spheres, but for bubbles, the formula is often approximated as:$v_t = \frac{g d^2 ho_f}{18 \eta}$Given:Diameter $d = 2 \,mm = 2 	imes 10^{-3} \,m$Density of fluid $ho_f = 13.6 	imes 10^3 \,kg/m^3$Viscosity $\eta = 1.5 \,cP = 1.5 	imes 10^{-3} \,Pa \cdot s$$g = 10 \,m/s^2$Using the formula $v_t = \frac{g d^2 ho_f}{18 \eta}$:$v_t = \frac{10 	imes (2 	imes 10^{-3})^2 	imes 13.6 	imes 10^3}{18 	imes 1.5 	imes 10^{-3}}$$v_t = \frac{10 	imes 4 	imes 10^{-6} 	imes 13.6 	imes 10^3}{27 	imes 10^{-3}}$$v_t = \frac{0.544}{0.027} \approx 20.14 \,m/s$Rounding to the nearest provided option, the rate is $20 \,m/s$.

What is the rate at which a trapped bubble of $2 \,mm$ diameter rises through a solution of density $13.6 \times 10^3 \,kg/m^3$ and coefficient of viscosity $1.5 \,cP$ (in $\,m/s$)? Assume the density of air is negligible and $g = 10 \,m/s^2$.

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