It is desired that air tanks for scuba diving be neutrally buoyant when empty. (

Question

It is desired that air tanks for scuba diving be neutrally buoyant when empty.

(a) A tank is designed to contain 50 standard cubic feet of air when Ölled to a pressure of 3000 psig at an ambient temperature of 80 F. Calculate the interior volume of the tank. A standard cubic foot of air occupies one cubic foot at standard temperature and pressure (T = 59 F and p = 2116 lb/ft2 ). If the interior length of the tank is 1.25 ft, what is the inner diameter of the tank? You may assume that the tank is a cylinder with circular cross section.

(b) The density of aluminum is 2700 kg/m3 . If the above tank is made of aluminum, what should be the wall thickness of the tank in order for it to be neutrally buoyant?

Explanation / Answer

a. given

pressure inside the cylinder = P = 3000 psig = 2.0684*10^7 Pa

temperature, T = 80 F = 299.817 K

amount of air 50 standard cubit feet = 1.41584 m^3

density of standard air = 1.225 kg/m^3

so mass of aire inside cylinder, m = 1.225*1.41584 = 1.734404 kg

number of moles, n = 1734.404/28.97 = 59.8689 moles

using PV = nRT

V = nRT/P = 0.00721148 m^3 = 440.07151 in^3 [ R = universal gas constant, 8.31 ]

this is the internal volume of the cylinder

let innerdiameter of tank be d

then

pi*d^2*l/4 = V = pi*d^2(1.25*12/4) = 440.07151

d = 6.111 in

b. density of aluminium, rho = 2700 kg/m^3

let thisckness of the sheet be t

then volume of aluminium used, V' = surface area * thickness = [2*pi*d^2/4 + pi*d*l]t

V' = [pi*6.111^2/2 + pi*6.111*1.25*12]*t = 346.65*t in^3

so for neutral buoyancy

rho*V' = rho'*V [ where rho' is density of water]

2700*346.65*t = 1000* 440.07151

t = 0.41 in

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