Question

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027). Determine the terminal speed.

Answer 1

Given:

diameter of sphere, d = 6 inches

radius of sphere, r =
`(d)/(2)` = 3 inches

density,
`\rho}` = 493 lbm/
`ft^(3)`

S.G = 1.0027

g = 9.8 m/
`m^(2)` = 386.22 inch/
`s^(2)`

Solution:

Using the formula for terminal velocity,

`v_(T)` =
`\sqrt{(2V\rho  g)/(A \rho C_(d))}` (1)

`(Since, m = V* \rho)`

where,

V = volume of sphere

`C_(d)` = drag coefficient

Now,

Surface area of sphere, A =
`4\pi r^(2)`

Volume of sphere, V =
`(4)/(3) \pi r^(3)`

Using the above formulae in eqn (1):

`v_(T)` =
`\sqrt{(2* (4)/(3) \pir^(3)\rho  g)/(4\pi r^(2) \rho C_(d))}`

`v_(T)` =
`\sqrt{(2gr)/(3C_(d))}`

`v_(T)` =
`\sqrt{(2* 386.22* 3)/(3C_(d))}`

Therefore, terminal velcity is given by:

`v_(T)` =
`(27.79)/(√(C_d))` inch/sec