Question

With what terminal velocity will an air bubble 0.8 mm in diameter rise in a liquid of viscosity 0.15 N-s//m^(2) and specific gravity 0.9? Density of air is 1.293 kg//m^(3).

Answer 1

Answer:

0.8*0.15*0.9

__________

1.293

answer is your responsibility

Answer 2

Given:

air bubble diameter (d)=.8mm

viscosity of liquid = .15$\frac{Ns}{m^{2} }$

specific gravity = .9

density of air = 1.293 $\frac{kg}{m^{3} }$

To find :

Terminal velocity of air bubble

Solution :

Terminal velocity of a air bubble is defined as the maximum velocity attained by a bubble as it falls or rises through  fluid.

terminal velocity = v = $\frac{2r^{2}g(p-p') }{9n}$    (derived from equation $w = F_{T} +F_{v}$)

where r is radius of bubble

g is acceleration due to gravity

p is density of fluid

p' is density of air

$n$ is dynamic  viscosity of fluid

substitute the values given to the formula of terminal velocity.

v = 2×$\frac{.0004^{2}9.8(900-1.293) }{9( .15)}$

v = .20 $\frac{m}{s}$

The terminal velocity of air bubble in the liquid is .20 metre per seconds.