Question

A storage tank for concentrated nitric acid will be constructed from aluminum to resist corrosion. The tank is to have an inside diameter of 6 m and a height of 17 m. The maximum liquid level in the tank will be at 16 m. Estimate the plate thickness required at the base of the tank. Take the allowable design stress for aluminum as 90 /^2.

Answer 1

The plate thickness required at the base of the tank is approximately 6.6 mm.

Step-by-step explanation:

To estimate the plate thickness required at the base of the tank, we need to determine the maximum pressure that the tank will experience at the base. The pressure can be calculated using the formula:

P = ρgh

where P is pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height of the liquid above the base of the tank.

Assuming a density of 1.5 g/cm³ for concentrated nitric acid, the pressure at the base of the tank can be calculated as:

P = 1.5 x 9.81 x 16 = 235.44 kPa

To determine the plate thickness required, we can use the formula for the hoop stress in a cylindrical vessel:

σ = PD/2t

where σ is the stress, P is the pressure, D is the diameter of the vessel, and t is the thickness of the vessel wall.

Rearranging the formula to solve for t, we get:

t = PD/2σ

Assuming a diameter of 6 m and a design stress of 90 N/mm², the plate thickness required at the base of the tank can be calculated as:

t = (235.44 x 6 x 1000) / (2 x 90 x π) ≈ 6.6 mm

Therefore, the plate thickness required at the base of the tank is approximately 6.6 mm.

Answer 2

To determine the plate thickness required at the base of the tank, we can use the following formula:

t = (PD)/(2SE - 0.2P)

where:

t = plate thickness

P = design pressure

D = inside diameter of the tank

S = allowable design stress for aluminum

E = joint efficiency factor

Since the tank will contain concentrated nitric acid, which is a hazardous material, it is subject to specific design and safety regulations that require a detailed engineering analysis. Therefore, we will assume a design pressure of 2.5 psi for this calculation, which is a typical value for atmospheric storage tanks.

The joint efficiency factor for welded aluminum is typically 0.70 to 0.85, depending on the welding procedure and inspection method used. For this calculation, we will assume a joint efficiency factor of 0.80.

Plugging in the given values, we get:

t = (2.56)/(2900.80 - 0.22.5)

t = 0.0682 m

Therefore, the plate thickness required at the base of the tank is approximately 0.0682 m, or 68.2 mm