Spherical Tanks and Liquefied Petroleum Gas (LPG)

A sphere is the most efficient pressure vessel because it offers the maximum volume for the least surface area and the required thickness of a sphere is one-half the thickness of a cylinder of the same diameter.

The stresses in a sphere are equal in each of the major axes, ignoring the effects of supports. In terms of weight, the proportions are similar.

When compared with a cylindrical vessel, for a given volume, a sphere would weigh approximately only half as much. However, spheres are more expensive to fabricate, so they aren't used extensively until larger sizes.

In the larger sizes, the higher costs of fabrication are balanced out by larger volumes.

Spheres are typically utilized as "storage" vessels rather than "process" vessels.

Spheres are economical for the storage of volatile liquids and gases under pressure, the design pressure being based on some marginal allowance above the vapor pressure of the contents.

Now from above you can come to know that why sphere is used for LPG storage.

I wish I could have explained the benefits of a sphere over a cylinder as accurately and succinctly as our colleague.

A few further observations

LPG is stored at ambient conditions in spheres (or bullets) or as a refrigerated liquid in cylinders

At ambient conditions (52C for The Gulf) Propane has a vapour pressure of ~17.5barg, Butane ~5barg. To store these fluids needs a pressure vessel and a sphere is the most efficient. Note spheres are only used for storage not processing vessels as their floor plan is much larger than the equivalent vertical cylinder and connecting nozzles is more expensive. In a storage area (tank farm - usually the biggest area of a refinery) space is not so critical so the most efficient storage shape is used.

Refrigerated storage is at atmospheric pressure and -45C for Propane, -12C Butane. These are not pressure containing vessels so the advantage of a sphere is lost. Also they are usually doubled walled with an insulating layer and typically 35000m3 plus (my current project has 10 off 72350m3 tanks) and building concentric spheres to this size is prohibitively expensive

When we make a comparison between spherical and cylindrical vessels from wall thicknesses' point of view which will resist the applied pressure and temperature, we shall find that the resistance of sphere is the better.

In other words, if you have two pressure vessels, the 1^st is spherical and the 2^nd is cylindrical and the both have the same inside diameter. If both the vessels fabricated from the same material and same grade, and subjected to the same conditions of pressure and temperature, by calculating the minimum required thickness for both the vessels, you will find that the 1^st vessel (spherical) has the half thickness of the 2^nd one (cylindrical). And this is also clear by comparing the equations used in calculating the shell wall thicknesses.

Sample Problem. Comparison between shell wall thicknesses for cylindrical & spherical vessels per ASME VIII-1.

Given :

• Design Pressure, P: 250 psi

• Design Temperature: 150°F

• Vessel Inside Diameter, D: 200 inch

• Shell material: ASME SA 516, Grade 70, Normalized

• Joint efficiency, E: 1.0

Find : The min. required thickness of shell, t for two cases:

Case 1. When you use a spherical vessel & Case 2. When you use a cylindrical vessel.

Data:

• Allowable tensile strength S of A 516-70 at 150 ^oF = 20 000 psi (ASME II-D, Table 1A)

• Inside radius of shell, R = D/2 = 200 / 2 = 100 inch

Case 1. Spherical shell (t < 0.356 R or P < 0.665 SE), t = PR / (2 SE – 0.2 P) ……… [UG-27 (c)(3)]

Case 2. Cylindrical shell (hoop stress) (t < 0.5 R or P < 0.385 SE), t = P R / (SE – 0.6 P) ………[UG-27 (c)(1)]

Solution:

Case 1. Spherical shell: t = PR / (2 SE – 0.2 P) = (250 psi)(100 inch) / [2(20 000 psi)(1.0) – 0.2 (250 psi)] = 0.626 inch (15.89 mm = 16 mm)

Case 2. Cylindrical shell: t = P R / (SE – 0.6 P) = (250 psi)(100 inch) / [(20 000 psi)(1.0) – 0.6 (250 psi)] = 1.259 inch ( 32 mm)

By comparing the results of thicknesses, we find that the cylindrical shells need to twice the thickness of spherical shell.

Notes.

1. The same results can be attained in a cylindrical pressure vessel with heads of hemispherical type, where you find the wall thickness of its head t_h equal almost half the wall thickness of its shell t_s.

2. In trucks used for transportation of LPG, the shell is cylindrical, where its heads are preferred to be hemispherical type (not ellipsoidal) to minimize the thickness, and inturn minimize the total weight of vessel over the truck.

3. Minimizing the thicknesses is an advantage in certain cases where we handle LPG, LNG and any other fluids in cryogenic condition, which lower thickness will lead to a pressure part with lower MDMT if compared with pressure parts characterized by its higher thicknesses.

By referring to the solved example, and from Figure UCS-66 or Table UCS-66, the material ASME SA 516 Grade 70, Normalized belong to Curve D. If the material ASME SA 516 Grade 70 is not normalized, it will belong to Curve B. And the following table summarizes the MDMT for lower and higher thicknesses. And in handling of LPG, we prefer the material with thicknesses which leads to lower MDMT.

● Note that the values of temperatures are extracted from ASME Table UCS-66, and interpolation is accepted.

● Also the conversion between ^oF and ^oC can result in some different variations other than that tabulated values.

● By comparing the results of MDMT for both curves, we find that Curve D has a lower values of MDMT than Curve B, even for the same thickness. This means that the materials belong to Curve D can resist lower temperatures, an