 # Volume XLIII.IV: ASME Sec I & Sec VIII Fundamentals - Part 4

• Volume XLIII.IV: ASME Sec I & Sec VIII Fundamentals - Part 4

## ASME PRESSURE VESSELS

The scope of this presentation is to present basic information and understanding of the ASME code for the design of pressure vessels for the chemical and process industry as applicable in the United States and most of North and South America. For more information about our productsheavy plate & custom fabrication services or fabrication capabilities contact us today!

## ASME SECTION VIII – DIVISION 1: Dished Head Formula:

Flanged and dished heads can be formed from carbon steel, stainless steel, and other ferrous and nonferrous metals and alloys. Flanged and dished heads can be formed in a size range from 4 inches to 300 inches diameter and in thickness range of 1/8 in. to 1 ½ inches thick. The ASME Section VIII – Division 1 determines the rules for dished heads. The most common configurations are spherical, hemispherical, elliptical (or ellipsoidal) and torispherical shapes.

How the shapes are, make some confusion for beginners and even professionals users of ASME Section VIII. To cast a little light on these subjects see the resume below: a) When t < 0.356R or P < 0.665SE - (Thin Spherical or Hemispherical Heads): And ### Example 12 - Thin Spherical or Hemispherical Head:

A pressure vessel is built of SA-516-70 material and has an inside diameter of 96 in. The internal design pressure is 100 psi at 450°F. Corrosion allowance is 0.125 in. and joint efficiency is E = 0.85. Calculate the required spherical head thickness of the pressure vessel if “S” is 20,000 psi?

Since P > 0.665SE = P > 11,305 psi. Thus, as 11,305 > 100 psi, use equation 2.1.

The inside radius in a corroded condition is equal to, R = 48 + 0.125 = 48.125 in.

t = PR / 2SE – 0.2P

t = 100 x 48.125 / 2(20,000) (0.85) – 0.2(100)

t = 0.14 in.

### Example 12.1 - Thin Spherical or Hemispherical Head:

A spherical pressure vessel with an internal diameter of 120 in has a head thickness of 1 in. Determine the design pressure if the allowable stress is 16,300 psi. Assume joint efficiency E = 0.85. No corrosion allowance is stated to the design pressure.

Using t = 1.0 in., then t < 0.356R = 1.0 in. < 21.3 in., use equation 2.2.

P = 2SEt / R + 0.2t

P = 2(16,300)(0.85)(1) / 60 + 0.2(1)

P = 460 psi

The calculated pressure 460 psi is < 0.665SE (9213 psi); therefore, equation 2.2 is acceptable.

b) When t > 0.356R or P > 0.665SE – (Thick Spherical or Hemispherical Heads): ### Example 13 - Thick Hemispherical Head:

Calculate the required hemispherical head thickness of an accumulator with a design pressure of P = 10,000 psi, R = 18.0 in, S = 15,000 psi, and E = 1.0. Corrosion allowance is 0.125 in.

Since P > 0.665SE = P > 9975 psi. Thus, as 9975 psi < 10,000 psi, use equation 2.3. Y = 2 [(15,000)(1.0) + 10,000] / 2 [(15,000)(1.0)] – 10,000 =

Y = 50,000 / 20,000 = 2.5

t = R (Y1/3 – 1) = 18.0 (2.51/3 – 1) + 0,125 = 6.5 in

## Elliptical or Ellipsoidal Heads - Semi-Elliptical or Semi-Ellipsoidal Heads – 2:1:

The commonly used semi-ellipsoidal head has a ratio of base radius to depth of 2:1 (shown below).

The actual shape can be approximated by a spherical radius of 0.9D and a knuckle radius of 0.17D. The required thickness of 2:1 heads with pressure on the concave side is given below: Example 14 - Elliptical or Ellipsoidal Heads, or, Semi-Elliptical or Semi-Ellipsoidal Heads – 2:1:

Calculate a semi-elliptical head thickness considering the dimensions given below:

• Inside Diameter of Head (Di): 18.0 in;
• Inside Crown Radius (L): (18.0 x 0.9Di) in;
• Inside Knuckle Radius (ri): (18.0 x 0.17Di) in;
• Straight Skirt Length (h): 1.500 in.;
• Radius L - (18.0 x 0.9Di) = 16.20 in.;
• Radius ri - (18.0 x 0.17Di) = 3.06 in. • Material and Conditions:
• Material: SA-202 Gr. B (Room Temperature);
• Internal Pressure: 200 psi;
• Allowable Stress: 20,000 psi;
• Head Longitudinal Joint Efficiency: 0.85;
• Corrosion Allowance: 0.010 in.

Variable:

L/r = L/ri = 16.20/3.06 = 5.29 in

a) Required Thickness. (Formula 2.5):

t = PD / 2se - 0.2p + corrosion allowance

t = 200 x 18.0 + 0.010 / 2(20,000)(0.85) – 0.2(200)

t = 0.116 in. Shallow heads, commonly referred to as flanged and dished heads (F&D heads), are according to paragraph UG-32 (e), with a spherical radius L of 1.0D and a knuckle radius r of 0.06D.

The dish radius of a Flanged and Dished Head is 1.0 D and the knuckle radius is 0.06% D. The required thickness of a Torispherical F&D Head with r/L = 0.06 and L = Di, is: ### Example 15 - Torispherical Heads:

A drum is to operate at 500°F and 350 psi and to hold 5000 gallons of water. The inside radius of the Dished Torispherical Heads is 78 in. The material is SA 285 Grade A. Assume “S” = 11,200 psi and “E” = 0.85.

Solution: Dished Torispherical Heads with L = Di and r/L = 0.06, use equation 2.7.

t = 0.885 PL / SE – 0.1P =

t = 0.885 (350) (78) / (11,200)(0.85) – 0.1(350) = 2.54

### Non-Standard 80-10 - Flanged & Dished Heads: The inside radius (L) of an 80-10 is 0.8 Di and the knuckle radius (ri) is 10% of the head diameter. For the required thickness of a Non-Standard 80-10 Head, use equations 2.7 and 2.8. Designing 80-10 Torispherical Heads rather than standard shapes can be achieved by lowering the material costs. The 80-10 is typically only 66% the thickness of the standard Torispherical Heads.

The required thickness of the Conical or Toriconical Head (knuckle radius > 6% OD) shall be determined by formula using internal diameter of shell, α ≤ 30º.  • IKR – Inside Knuckle Radius (At least 60% of diameter if code – Typically 3 x THK if non code)
• HA - Half Apex
• OAH – Overall Height
• THK – Thickness
• SF – Straight Flange
• OD - Outside Diameter
• Typical Thin Out Allowance – ½” and under – add 0.03125 to the minimum thickness, Over ½” – Add 0.0625” to the minimum thickness

In the next issue we will discuss more ASME Section VIII Code

Source:  ASME International Boiler & Pressure Vessel Code

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