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 5
ASME SECTION VIII – Main Scopes:
Minimum requirements for safe construction and operation, per Division 1, 2, and 3.
- Section VIII Division 1: 15 psig < P < 3000 psig.
- Other exclusions: Internals (except for attachment weld to vessel); Fired process heaters; Pressure containers integral with machinery; Piping systems.
- Section VIII, Division 2, Alternative Rules: 15 psig < P ≤ 10,000 psig - identical to Division 1, but the different requirements are: Allowable stress design; Stress calculations; Design; Quality control; Fabrication and inspection.
Note: The choice between Divisions 1 and 2 is mainly based on economics of materials, savings in weight and addresses cyclic conditions.
- Section VIII Division 3, Alternative Rules - High Pressure Vessels: Applications over 10,000 psi; Pressure from external source, process reaction, application of heat, combination of these; Maximum pressure limits for Division 1 or 2 or minimum limits for Division 3, not established.
Structure of ASME Section VIII, Division 1:
- Subsection A: Part UG applies to all vessels;
- Subsection B: Requirements based on fabrication method, Parts UW, UF, UB;
- Subsection C: Requirements based on material class, Parts UCS, UNF, UHA, UCI, UCL, UCD, UHT, ULW, ULT. Mandatory and Non-mandatory Appendices
Determination of Material Thickness: Yield Strength, Ultimate Tensile Strength, Creep Strength, Rupture Strength and Corrosion Resistance.
- Resistance to Hydrogen Attack: Temperature at 300 - 400°F, monatomic hydrogen forms molecular hydrogen in voids; Pressure build-up can cause steel to crack; Above 600°F, hydrogen attack causes irreparable damage through component thickness.
- Brittle Fracture and Fracture Toughness: The conditions that could cause brittle fracture are:
Typically at “low” temperature; Can occur below design pressure; No yielding before complete failure; High enough stress for crack initiation and growth; Low enough material fracture toughness at temperature; Critical size defect to act as stress concentration.
Material Groups – The Most Common Used Materials:
- Bolting: See the ASME Code Section VIII, Div. 1, for impact and nuts test for specified material specifications.
- Additional ASME Code Impact Test Requirements: For welded construction over 4 in. thick, or non-welded construction over 6 in. thick, if MDMT < 120°F; Not required for flanges if temperature -20°F; required if SMYS > 65 ksi unless specifically exempt.
Weld Joint Efficiencies, E:
Pressure Vessels Formula – ASME Section I & ASME Section VIII:
Notes: D = Shell/Head Inside Diameter, E = Weld Joint Efficiency, L = Crown Radius, P = Internal Pressure, h = Inside Depth of Head, r = Knuckle Radius, R = Shell/Head Inside Radius, S = Allowable Stress, t = Shell/Head Thickness.
Common Materials - Temperature Limits:
ASME SECTION VIII – Shell Nozzles:
Vessel components are weakened when material is removed to provide openings for nozzles or access openings. To avoid failure in the opening area, compensation or reinforcement is required. The Code procedure is to relocate the removed material to an area within an effective boundary around the opening. Figure bellow shows the steps necessary to reinforce an opening in a pressure vessel.
- x = Larger of d or Rn + tn +Tn
- y = Smaller of 2 ½ Ts, or 2 ½ Tn
- d = Diameter of circular opening (in.)
- D = Inside diameter of shell (in.)
- ts = Required thickness of shell (in.)
- Ts = Actual thickness of the shell (in.)
- Rn = Actual thickness of nozzle (in.)
- Ar = Area of required reinforcement (in.2)
- As = Area available in shell (in2)
- An = Area available om the nozzle (in2)
- Ar = (d) (ts)
- As = Larger of: d(Ts – ts) – 2Tn(Ts – ts) or 2(Ts = tn)(Ts – ts) – 2tn(Ts – ts)
- An = Smaller of: 2[2 ½ (Ts)(Tn – tn)] or 2[2 ½ (Tn)(Tn – tn)]
- Ar < (As + An): Acceptable configuration
Diameter of circular opening, d:
d = Diameter of Opening – 2 (Tn + Corrosion Allowance) =
• Required wall thickness of the nozzle (min):
tn = PR...... / SE – 0.6P =
Example 16 – Basic Shell & Nozzle:
Shell and Nozzle Data:
- Design Pressure = 700 psi
- Design temperature = 700 °F
- Nozzle Diameter = 8 in. (8.625 OD)
- Corrosion Allowance = 0.0625"
- Shell – SA 516 Gr.70;
- Head – SA 516 Gr. 70;
- Nozzle – SA 106 Gr. B;
- E = 1.0 (weld efficiency).
Required Shell Thickness:
ts = PR… / SE – 0.6P
ts = 700 x 30……..... / 16,600(1.0) – 0.6(700) = 1.3 (Use Tx = 1-1/2")
Required Head Thickness:
th = PR…… / 2SE – 0.2P
th = 700 x 30…….......... / 2(16,600)(1.0) – 0.2(700) = .64 (Use Th = 7/8")
Required Nozzle Thickness:
tn = PR… / SE – 0.6P
tn = 700 x 4.312”……..... = 0.21 (Use Tn = 1”)
- d = Nozzle Diameter– 2 (Tn + Corrosion Allowance)
- d = 8.625 – 2(1 + 0.0625) = 6.5 in.
- Ar (Reqd.) = d * ts = 6.5 x 1.3 = 8.45 in²
- As = Larger of: d (Ts – ts) - 2 Tn (Ts – ts) =
- As = 8.625 (1.5 – 1.3) – 2 (1.0) (1.5 – 1.3) = 1.325 in²
- An = Smaller of: 2[2.5 (Ts) (Tn – tn)]
- An = 2[2.5 (1.5) (1.0 – 0.21)] = 5.925 in²
- Ar < (As + An) =
- Ar < (1.325 + 5.925) = 7.25 in² < 8.45 (Reqd.)
- OBS.: Necessary to increase Ts and / or Tn to attend the premise (As + An) > Ar.
Area of required reinforcement, Ar:
Ar = d.ts.F (in²) =
- d =Diameter of circular opening, or finished dimension of opening in plane under consideration, in.
- ts = Minimum required thickness of shell when E = 1.0, in.
- F = Correction factor, normally 1.0
Example 17 – Basic Pipe Nozzle:
Following the definitions above, according to ASME Section VIII-1, calculate a basic pipe nozzle SA-53 Grade B, seamless, considering the data shown below:
- Design Pressure = 300 psig;
- Design Temperature = 200° F;
- Shell Material is SA-516 Gr. 60;
- Nozzle Diameter 8 in, Sch. 40;
- Nozzle Material is Corrosion Allowance = 0.0625";
- Vessel is 100% Radiographed;
a) Wall thickness of the nozzle (min):
tn = PR...... / SE – 0.6P =
tn = 300 x 4.312”…….. / 12,000(1.0) – 0.6(300) + 0.0625 (Corrosion Allowance) =
tn = 0.11” + 0.0625” = 0.17 in (min) – Pipe Sch. 40 is t = 0.32 in.
b) Circular opening, d:
d = Diameter of Opening – 2 (Tn + Corrosion Allowance)
d = 8.625 – 2(0.32 + 0.0625) =
d = 8.625 – 2 (0.3825) = 7.86 in
c) Area of required reinforcement, Ar:
Ar = d.ts.F (in²) =
Ar =7.86 x 0.487 x 1.0 = 3.82 in²
Source: International Boiler & Pressure Vessel Code
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