A spherical air bubble rises in water. At a depth of 9 ft, its diameter is 0.01 in. What is its diameter just as it reaches the free-surface where the pressure is 1 atmosphere?
Work:
p1 = 10.9 lb/in^2, d1 = 0.01 in, r1 = 0.005 in, h1 = 9 ft
p2 = 1 atm
pressure p = force F/area A
p2 = F/A2
p2 = (specific weight sw)*(fluid volume V2)*(height h2)/area A2
p2 = (sw)*(V2)*(h2)/(pi*radius r2^2)
r2^2 = (sw)*(V2)*(h2)/pi*p2
r2 = SQRT(sw*v2*h2/pi*p2), where SQRT = square root
d2 = diameter of air bubble at the free-surface
d2 = 2*SQRT(sw*V2*h2/pi*p2)
or...
p2 < p1
V2 > V1
V1 = (4/3)*pi*r1^3
V2 = (4/3)*pi*r2^3
increase in volume of the sphere (IV) = V1*(change in pressure/p2)
IV = (4/3*pi*r1^3)*(p2-p1/p2)
IV = (4/3*pi*0.005^3)*(3.77 lb/in^2/14.7 lb/in^2)
IV = (1.34 x 10^-7 + 5.23 x 10^-7) in^3
total volume (V2) of sphere at the free-surface
V2 = (1.34 x 10^-7 + 5.23 x 10^-7) in^3
V2 = 6.62 x 10^-7 in^3 at the free-surface
V2 = 4/3*pi*r2^3
(6.62 x 10^-7 in^3) = 4/3*pi*r2^3
r2^3 = (6.62 x 10^-7 in^3)/4*pi
r2 = CURT[3*(6.62 x 10^-7 in^3)/4*pi], where CURT = cubed root
d2 = 2*r2
d2 = 2*[CURT(1.58 x 10^-7 in^3)]
Solution
d2 = 2*SQRT(sw*V2*h2/pi*p2), or
d2 = 2*[CURT(1.58 x 10^-7 in^3)]