Question

With what terminal velocity will an air bubble $0.4mm$ in diameter rise in a liquid of viscosity $0.1Ns\,{m^{ - 2}}$ and specific gravity 0.9? Density of air is $1.29kg\,{m^{ - 3}}$ .

Answer 1

When an object falls freely under gravity in a fluid or gaseous system, the object attains a constant velocity after moments of fall. That constant velocity is the maximum velocity with which an object falls is called terminal velocity of substance. This terminal velocity is achieved when falling force is balanced by viscous force. Viscous force is the resistance offered to flow of fluid. This equilibrium of forces can be related to yield terminal velocity in terms of density, radii, viscosity and specific gravity.

Complete step by step solution:
A bubble inside water is acted upon by three types of forces i.e. Force due to gravity (downward), Drag due to viscosity (downward), Upward buoyancy.
${F_g} = \dfrac{4}{3}\pi {r^3}\rho g$ ….. Force due to gravity
${F_d} = 6\pi \eta rv$ ……….Viscous drag
${F_b} = \dfrac{4}{3}\pi {r^3}\sigma g$ …….Upward buoyancy
Where,
$\rho$ = density of air; $\sigma$ = density of fluid; $\eta$ = viscosity of fluid; $v$ = terminal velocity; $r$ = radius of bubble; $g$ = acceleration due to gravity.
Equating all three equations we get:
(Downward Forces = Upward Forces)
$\dfrac{4}{3}\pi {r^3}\rho g + 6\pi \eta rv = \dfrac{4}{3}\pi {r^3}\sigma g$
Evaluating above equation to get value of $v$ ,
$v = \dfrac{2}{9}\dfrac{{{r^2}\left( {\sigma - \rho } \right)g}}{\eta }$
The above evaluated equation is known as Stoke's Law.
Now we have an equation to calculate the value of terminal velocity in terms of density, radii, viscosity and specific gravity. If we look again to question we will get to know that every other variable are provided within question itself, hence
Putting values in Stoke's Law
$r = \dfrac{{0.4 \times {{10}^{ - 3}}}}{2}$ ………(As in the question length of diameter is mentioned)
If we look then we will find that specific gravity is only mentioned for fluid. To find density we must multiply specific gravity with density of water in $kg\,{m^{ - 3}}$
$\sigma = 0.9 \times 1000$
$v = \dfrac{2}{9}\dfrac{{{{\left( {2 \times {{10}^{ - 4}}} \right)}^2}\left( {900 - 1.29} \right)\left( {9.8} \right)}}{{0.1}}$
Solving resulted equation yield
$v = 7.829 \times {10^{ - 4}}\,m\,{s^{ - 1}}$
The terminal velocity with which bubble will rise is, $v = 7.829 \times {10^{ - 4}}\,m\,{s^{ - 1}}$

Note:
One must never forget that an air bubble is rising inside the water/fluid system. As the movement of the bubble is in upward direction so the acting viscous force/drag will be in opposite direction to it i.e. downward. Besides this if we directly apply the values in Stoke's Law then the result will be negative. This negative sign confirms that the direction of the air bubble is opposite as assumed.