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.
ASME Section I and Section VIII Fundamentals – Part 3
ASME SECTION VIII – Division 1, Division 2, Division 3:
The ASME Section VIII, rules for fired or unfired pressure vessels, is divided into three divisions applicable to design, fabrication, inspection, testing, and certification. The formulae and allowable stresses presented in this sketch are only for Division 1, the main code. It contains mandatory and nonmandatory appendices detailing supplementary design criteria, nondestructive examination and inspection acceptance standards.
The use of the U, UM and UV Code symbol stamps are also included.
4.1 – SECTION VIII - Thin Cylindrical Shells:
The formulae in ASME Section VIII, Division 1, paragraph UG-27, used for calculating the wall thickness and design pressure of pressure vessels, are: a) Circumferential Stress (longitudinal welds):
When, P < 0.385SE:
Example 5 – Thin Cylindrical Shells:
A vertical boiler is constructed of SA-515-60 according to Section VIII-1. It has an internal diameter of 96 in. and an internal design pressure of 1,000 psi at 450 F°. The corrosion allowance is 0.125 in., and joint efficiency is E = 0.85. Calculate the required thickness of the shell. (Consider SA-515-60 = 15,000 psi – allowable stress, ASME Section II, Table 1A, Div.1).
Since P < 0.385SE = P < 6,545 psi. Thus, as 1,000 psi < 6,545 psi, use equation 1.3:
t = PR… + C / SE – 0.6P
Considering the internal radius = (48 in.) and corrosion allowance = 0.125 in.
t = 1,000 x 48 …………. + 0.125 / 2(15,000)(0.85) – 0.6(1,000)
t = 2.052 in
SECTION VIII - Thick Cylindrical Shells:
For internal pressures higher than 3,000 psi, special considerations as specified in paragraph U-1 (d). As the ratio of t/R increases beyond 0.5, an accurate equation is required to determine the thickness. The formulae in ASME Appendix 1, Supplementary Design Formulas is used for calculating thick wall and design pressure, are:
a) For longitudinal welds:
When, P > 0.385SE:
b) For circumferential welds:
When, P > 1.25SE:
Where: R = Design Radius (in.):
Z = Dimensionless Factor.
Example 6 – Thick Cylindrical Shells:
a) When P > 0.385SE:
Calculate the required shell thickness of an accumulator with a design pressure of P = 10,000 psi, R = 18 in., S = 20,000 psi, and E = 1.0. Assume a corrosion allowance of 0.125 in.
Since P > 0.385SE = P > 7,700 psi. Thus, as 10,000 psi > 7,700 psi, use equation 1.7:
t = R (Z ½ - 1) =
Z = (20,000).(1.0) + 10,000 / (20,000).(1.0) – 10,000 = 30,000/10,000 = 3
t = (18) (3½ – 1) + 0.125 = 8.08 in.
Example 7 – Thick Cylindrical Shells:
b) When P < 0.385SE:
Calculate the required shell thickness of an accumulator with a design pressure of P = 7,650 psi, R = 18 in., S = 20,000 psi, and E = 1.0. Assume corrosion allowance = 0.
Since P< 0.385SE = P < 7,700 psi. Thus, as 7,650 psi < 7,700 psi, use equation 1.3:
t = PR… + C / SE – 0.6P =
t = 7,650 x 18………… + 0 / (20,000)(1.0) – 0.6(7,650)
t = 8.9 in.
Example 8 – Comparison between Equation 1.3 and Equation 1.7:
Calculate the shell thickness, comparing the equation 1.3 with another answer using equation 1.7.
t = R (Z ½ - 1) =
Z = (20,000)(1.0) + 7,650 = 27,650 / (20,000)(1.0) - 7,650 = 27,650 / 12,350 = 2.24
t = (18 + 0) (2.24 ½ - 1) = 8.9 in.
ASME I and ASME-ANSI B31 – Pressure Piping - Minimum Wall Thickness:
According to ASME Section I and ANSI B31, the minimum thickness of piping under pressure is:
t (min)= Minimum wall thickness required (in);
P = Design pressure (psig);
D = Outside diameter of pipe (in);
S = Allowable stress in pipe material (psi);
E = Longitudinal joint factor - E = 1.0 for seamless pipe, E = 0.85 for ERW pipe;
C = Corrosion allowance, typically 0.05 in.
y = Wall thickness coefficient in ASME according to Table 304.1.1 for ferritic steels, where:
y = 0.4 for T <= 900 ºF;
y = 0.5 for 900 < T <= 950 ºF;
y = 0.7 for 950 < T <= 1000 ºF
ASME SECTION VIII – Reinforcement Wall Thickness Plate:
The standard design method uses an increased wall thickness plate at the equator line of the vessel to support the additional stresses caused by the attachment of the legs. The formula for calculation the wall thickness of a segmented plate of to be welded in a vessel or spherical shell is:
t = PL…… + C / 2SE – 0.2P
L = Di/2
t = Minimum Design Wall Thickness (in);
P = Design Pressure (psi); Di = Inside Diameter of Sphere (in);
L = Sphere Radius (in);
E = Tube Welding Factor (1.0 for seamless pipe; 0.85 = for welded pipe);
C = Corrosion Allowance (0 for no corrosion; 0.0625 in. commonly used; 0.125 in. maximum);
S = Maximum Allowable Stress According to ASME Section II, Table 1A.
ASME SECTION I: Dished Heads Formula:
Dished heads can be manufactured using a combination of processes, spinning & flanging, where the spherical radius is made via the spinning process and the knuckle is created under the flanging method. Flanged and dished heads can be formed in a size range from 4 in to 300 in diameter and in thickness range of 14 Gauge to 1-1/2” thick. Pressure vessel heads and dished ends are essentially the same – the end caps of a pressure vessel tank or an industrial boiler, supplied with a flanged edge to make it easier for the fabricator to weld the head to the main body of the tank.
Blank, Unstayed Dished Heads:
Paragraph PG-29.1 states that the thickness of a blank, unstayed dished head with the pressure on the concave side, when it is a segment of a sphere, shall be calculated by the following formula:
t = Minimum thickness of head (in);
P = maximum allowable working pressure (psi);
L = Concave side radius (in);
S = Maximum Allowable Working Stress (psi).
Paragraph PG-29.2 states: “The radius to which the head is dished shall be not greater than the outside diameter of the flanged portion of the head. Where two radii are used, the longer shall be taken as the value of L in the formula.”
Example 9 – Segment of a Spherical Dished Head:
Calculate the thickness of a seamless, blank unstayed dished head having pressure on the concave side. The head has an inside diameter of 42.7 in. with a dish radius of 36.0 in. The Maximum Allowable Working Pressure (MAWP) is 360 psi and the material is SA-285 A. The temperature does not exceed 480°F. State if this thickness meets Code. (Solution: Use equation 1.11).
P = 360 psi;
L = 36.0 in;
S = 11,300 psi – SA-285 A at 480°F.
t = 5 (360 x 36) / 4.8 (11,300)
t = 1.19 in.
PG-29.6 states: “No head, except a full-hemispherical head, shall be of a lesser thickness than that required for a seamless pipe of the same diameter.” Then, the minimum thickness of piping is:
t =360 x 42.7 / 2(11,300)(1.0) + 2 (0.4)(360)
t = 0.67 in. The calculated head thickness meets the code requirements, since 1.19 > 0.67.
Example 10 - Segment of a Spherical Dished Head with a Flanged-in Manhole:
Calculate the thickness of a seamless, unstayed dished head with pressure on the concave side, having a flanged-in manhole 6.0 in x 16 in. Diameter head is 47.5 in with a dish radius of 45 in. The MAWP is 225 psi, the material is SA-285-C, and temperature does not exceed 428°F.
First thing to check: is the radius of the dish at least 80% of the diameter of the shell?
Dish Radius = 45 = 0.947 > 0.8
Shell Diameter = 47.5
The radius of the dish meets the criteria, use equation 1.11:
This thickness is for a blank head. According to ASME - PG-29.3 requires this thickness to be increased by 15% or 0.125 in., whichever is greater. As 0.764 × 0.15 = 0.114 in., that is less than 0.125 in., then, the thickness must be increased by 0.125 in.
Then, the required head thickness is, t = 0.764 + 0.125 = ~0.90 in.
Seamless or Full-Hemispherical Head:
The thickness of a blank, unstayed, full-hemispherical head with the pressure on the concave side shall be calculated by the following formula:
t = Minimum thickness of head (in)
P = Maximum Allowable Working Pressure (psi)
L = Radius to which the head was formed (in)
S = Maximum Allowable Working Stress (psi)
Note: The above formula shall not be used when the required thickness of the head given by the formula exceeds 35.6% of the inside radius. Instead, use the following formula:
Example 11 – Seamless or Full-Hemispherical Head:
Calculate the minimum required thickness for a blank, unstayed, full-hemispherical head. The radius to which the head is dished is 7.5 in. The MAWP is 900 psi and the head material is SA 285-C. The average temperature of the header is 570°F.
Solution: Use equation 1.13.
P = 900 psi;
L = 7.5 in.;
S = 13,800 psi - (SA 285-C at 600°F).
t = 900 x 7.5 / 2 (13,800) – 0.2 (900) = 0.24 in.
Check if this thickness exceeds 35.6% of the inside radius = 7.5 x 0.356 = 2.67 in. It does not exceed 35.6%, therefore the calculated thickness of the head meets the code requirements.
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.
Dished Head 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:
In the next issue we will discuss ASME Section VIII Head Formulas
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