The high-pressure air system at OSU's Aerospace Research Center is fed by a set of two cylindrical tanks. Each tank has an outer height of 50 ft and an outer diameter of 4.6 ft. The tanks are made of 0.1 ft thick steel (steel = 499 lbm/ft?) and store air at a maximum pressure of 2500 psi at - 10 °F. How much load must the support structure at the base of the tanks carry?

179000 lb

Explanation:

The supports must be able to hold the weight of the tank and the contents. Since tanks are pressure tested with water, and the supports cannot fail during testing, we disregard the air and will consider the weight of water.

The specific weight of water is ρw = 62.4 lbf/ft^3

These tanks are thin walled because

D / t = 4.6 / 0.1 = 46 > 10

To calculate the volume of steel we can approximate it by multiplying the total surface area by the thickness:

A = 2 * π/4 * D^2 + π * D * h

The steel volume is:

V = A * t

The specific weight is

ρ = δ * g

ρs = 499 lbm/ft^3 * 1 lbf/lbm = 499 lbf/ft^3

The weight of the steel tank is:

Ws = ρs * V

Ws = ρs * A * t

Ws = ρs * (2 * π/4 * D^2 + π * D * h) * t

Ws = 499 * (π/2 * 4.6^2 + π * 4.6 * 50) * 0.1 = 37700 lb

The weight of water can be approximated with the volume of the tank:

Vw = π/4 * D^2 * h

Ww = ρw * π/4 * D^2 * h

Ww = 62.4 * π/4 * 4.6^2 * 50 = 51800 lb

Wt = Ws + Ww = 37700 + 51800 = 89500 lb

Assuming the support holds both tanks

2 * 89500 = 179000 lb

The support must be able to carry 179000 lb