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3.1 Corrosion Averaging & Minimum Thickness Evaluation

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API 510 Pressure Vessel Inspector Exam

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1 Chapter 1: Introduction, Scope & Certification Requirements

1.1 Exam Structure, Administration & Study Strategy 1.2 API 510 Code Scope, Exclusions & Jurisdictional Boundaries 1.3 Roles, Responsibilities & Inspector Qualification

2 Chapter 2: Inspection Planning, Data Evaluation & Remaining Life

2.1 Corrosion Rates, Remaining Life & Corrosion Allowance Calculations 2.2 Inspection Intervals, On-Stream Inspection & Interval Deferrals 2.3 Condition Monitoring Locations (CMLs) & Data Analysis

3 Chapter 3: Vessel Evaluation, Corrosion Averaging & Fitness-for-Service

3.1 Corrosion Averaging & Minimum Thickness Evaluation 3.2 Pitting Evaluation & Corroded Area Limits 3.3 Fitness-for-Service (API 579-1) & Risk-Based Inspection (API 580)

4 Chapter 4: ASME Section VIII Div 1 Design & Thickness Calculations

4.1 Internal Pressure Design & Cylindrical Shell Calculations 4.2 Ellipsoidal & Hemispherical Head Calculations 4.3 Spherical Shells, Static Head & Vessel MAWP 4.4 Openings, Nozzle Reinforcement & External Pressure Basics

5 Chapter 5: ASME Section VIII Fabrication, Welds & Joint Efficiency

5.1 Weld Joint Categories (UW-3) & Joint Efficiencies (UW-12) 5.2 Radiographic Examination (UW-11, UW-51, UW-52) 5.3 Post-Weld Heat Treatment (PWHT) & Impact Testing (UCS-56, UCS-66)

6 Chapter 6: Welding Inspection & Metallurgy (API 577 & ASME Sec IX)

6.1 Welding Procedures (WPS/PQR) & Essential Variables 6.2 Welder Performance Qualification (WPQ) & Testing 6.3 Welding Processes, Metallurgy & Weld Discontinuities (API 577)

7 Chapter 7: Nondestructive Examination (ASME Section V)

7.1 Radiographic Examination Techniques & Image Quality Indicators (Article 2) 7.2 Ultrasonic Thickness Measurement (Article 23, SE-797) 7.3 ASME Section V Articles 1, 6 & 7: General Requirements, PT & MT

8 Chapter 8: Damage Mechanisms (API RP 571)

8.1 Atmospheric Corrosion & Corrosion Under Insulation 8.2 HTHA, High-Temperature H2/H2S Corrosion & Sulfidation 8.3 Amine, Caustic & Chloride Stress Corrosion Cracking 8.4 Wet H2S Damage, Brittle Fracture & Erosion-Corrosion 8.5 Boiler-Water, Steam-Condensate & Hydrochloric Acid Corrosion 8.6 Caustic Corrosion & Naphthenic Acid Corrosion

9 Chapter 9: Pressure Vessel Inspection Practices (API RP 572 & API RP 578)

9.1 Internal & External Vessel Inspection Practices (API 572) 9.2 Structural Components, Internal Trays, Linings & Skirts 9.3 Material Verification Programs & Positive Material Identification (API 578)

10 Chapter 10: Pressure-Relieving Devices (API RP 576)

10.1 PRD Types, Construction & Operating Principles 10.2 PRD Causes of Failure, Shop Testing & Overhaul 10.3 PRD Inspection Intervals, Records & Rupture Disks

11 Chapter 11: Repairs, Alterations, Rerating & Pressure Testing (API 510 & PCC-2)

11.1 Repair vs. Alteration Authorization, Execution & Methods 11.2 PCC-2 Core Welded Repairs: Articles 101, 201, 202, 211 & 212 11.3 PCC-2 Special Welding: Articles 209, 210, 215 & 216 11.4 PCC-2 Excavation, Flange, Exchanger & Test Articles 304, 305, 312, 501 & 502 11.5 Vessel Rerating Requirements & Engineering Calculations 11.6 Pressure Testing: Hydrostatic, Pneumatic & NDE in Lieu of Testing

3.1 Corrosion Averaging & Minimum Thickness Evaluation

Key Takeaways Explain This Topic AI

For vessels with inside diameter D ≤ 60 in. (1500 mm), the critical evaluation length L is the lesser of D/2 or 20 in. (500 mm); for D > 60 in., L is the lesser of D/3 or 40 in. (1000 mm).

The critical length L must always be oriented along the longitudinal axis for cylindrical shells (governed by circumferential hoop stress) and along any great circle arc for spherical shells and hemispherical heads.

If a locally thinned area (LTA) does not encompass a weld and lies at least the greater of 1.0 in. (25 mm) or 2t from the weld toe, the base metal required thickness can be calculated using a joint efficiency of E = 1.0.

The calculated average thickness t_avg across critical length L must be equal to or greater than the required thickness t_min plus the corrosion allowance for the next operating period.

When an LTA encompasses or crosses a weld joint, the weld joint efficiency E must be incorporated into the minimum required thickness determination.

Last updated: August 2026

Corrosion Averaging & Minimum Thickness Evaluation

In pressure vessel in-service inspection, metal loss rarely occurs in a perfectly uniform pattern across an entire shell course or head. Instead, localized corrosion mechanisms—such as condensate grooving, acid under-deposit attack, impingement, and flow-accelerated degradation—create Locally Thinned Areas (LTAs) . When an inspector encounters an area where the measured thickness falls below the minimum required thickness ( t min t_{\text{min}} t min ​ ) established by the construction code (ASME Section VIII, Division 1), the vessel does not automatically fail inspection or require immediate retirement.

API 510 Section 7.4 establishes rigorous engineering evaluation procedures that allow inspectors to evaluate LTAs through corrosion averaging . By assessing the structural reinforcement provided by the surrounding thicker base metal and properly orienting the evaluation length along the principal stress plane, an inspector can determine whether the component remains safe for continued operation at its design Maximum Allowable Working Pressure (MAWP).

1. Mechanics of Stress and Orientation of Evaluation Length

To understand why API 510 imposes strict directional rules on thickness averaging, an inspector must understand the fundamental membrane stresses operating in thin-walled pressure vessels.

Principal Membrane Stresses in Cylindrical Shells

Under internal pressure ( P P P ), a cylindrical shell experiences two primary membrane stresses:

Circumferential (Hoop) Stress ( σ h \sigma_h σ h ​ ): σ h = P ⋅ R t \sigma_h = \frac{P \cdot R}{t} σ h ​ = t P ⋅ R ​

Acts tangentially around the circumference, tending to split the cylinder longitudinally along its length.

Longitudinal (Axial) Stress ( σ L \sigma_L σ L ​ ): σ L = P ⋅ R 2 t = 1 2 σ h \sigma_L = \frac{P \cdot R}{2t} = \frac{1}{2} \sigma_h σ L ​ = 2 t P ⋅ R ​ = 2 1 ​ σ h ​

Acts parallel to the vessel's longitudinal axis, tending to pull the cylinder apart end-to-end.

Because hoop stress is twice the magnitude of longitudinal stress, the longitudinal plane is the most highly stressed cross-section in a cylindrical vessel. Consequently, circumferential stress governs the required wall thickness for cylindrical shells under internal pressure. +-----------------------------------------------------------------------------------------+

  • | PRINCIPAL STRESS VECTORS ON A CYLINDER |
  • | |
  • | ^ Circumferential (Hoop) Stress (σ_h = P*R/t) [MAX STRESS] |
  • | | |
  • | +-----+---------------------------------------------------+-----+ |
  • | | | | | |
  • | <-----+=====[=========== CRITICAL LENGTH L AXIS ===============]=====+-----> |
  • | | | (Evaluated longitudinally along the cylinder) | | Longitudinal |
  • | +-----+---------------------------------------------------+-----+ Stress (σ_L) |
  • | | |
  • | v Circumferential (Hoop) Stress (σ_h) |
  • | |
  • | KEY CODE PRINCIPLE: The critical length L must be oriented LONGITUDINALLY along the |
  • | cylinder axis so that the hoop stress vector acts perpendicularly across L. |
  • +-----------------------------------------------------------------------------------------+
  • Code Rules for Orienting the Critical Length L L L

According to API 510 Section 7.4.2.1 :

For Cylindrical Shells: The critical length L L L must be oriented parallel to the longitudinal axis of the vessel. Averaging along the circumferential direction is prohibited for hoop stress qualification because doing so would average across the maximum tensile stress field rather than along the resisting cross-section.

For Spherical Shells and Hemispherical Heads: Because membrane stress is equal in all tangential directions ( σ = P ⋅ R 2 t \sigma = \frac{P \cdot R}{2t} σ = 2 t P ⋅ R ​ ), the critical length L L L may be oriented along any great circle arc .

For Torispherical and Ellipsoidal Heads: The critical length L L L is evaluated along the arc of the head profile, with special attention to the high-stress knuckle or toroidal transition zone.

2. API 510 Critical Length ( L L L ) Formulas

The maximum length over which thickness measurements may be averaged is called the Critical Length ( L L L ) . It depends strictly on the inside diameter ( D D D ) of the vessel section containing the thinned area.

Mathematical Definition of Critical Length L L L

Vessel Inside Diameter ( D D D ) Critical Length Formula (US Customary) Critical Length Formula (SI Metric)

D ≤ 60 in. D \le 60\text{ in.} D ≤ 60 in. ( 1500 mm 1500\text{ mm} 1500 mm ) L = min ⁡ ( D 2 , 20 in. ) L = \min\left(\frac{D}{2},\; 20\text{ in.}\right) L = min ( 2 D ​ , 20 in. ) L = min ⁡ ( D 2 , 500 mm ) L = \min\left(\frac{D}{2},\; 500\text{ mm}\right) L = min ( 2 D ​ , 500 mm )

D > 60 in. D > 60\text{ in.} D > 60 in. ( 1500 mm 1500\text{ mm} 1500 mm ) L = min ⁡ ( D 3 , 40 in. ) L = \min\left(\frac{D}{3},\; 40\text{ in.}\right) L = min ( 3 D ​ , 40 in. ) L = min ⁡ ( D 3 , 1000 mm ) L = \min\left(\frac{D}{3},\; 1000\text{ mm}\right) L = min ( 3 D ​ , 1000 mm )

Diameter Transition & Length Caps Table

To avoid calculation errors on the API 510 exam, memorize how the diameter cutoff triggers both the divisor change ( D / 2 D/2 D /2 vs D / 3 D/3 D /3 ) and the upper ceiling cap ( 20 in. 20\text{ in.} 20 in. vs 40 in. 40\text{ in.} 40 in. ):

Vessel Inside Diameter ( D D D ) Divisor Calculation Upper Limit Cap Governing Critical Length ( L L L )

24 in. 24\text{ in.} 24 in. ( 600 mm 600\text{ mm} 600 mm ) 24 / 2 = 12 in. 24 / 2 = 12\text{ in.} 24/2 = 12 in. 20 in. 20\text{ in.} 20 in. 12.0 in. 12.0\text{ in.} 12.0 in. ( 300 mm 300\text{ mm} 300 mm )

36 in. 36\text{ in.} 36 in. ( 900 mm 900\text{ mm} 900 mm ) 36 / 2 = 18 in. 36 / 2 = 18\text{ in.} 36/2 = 18 in. 20 in. 20\text{ in.} 20 in. 18.0 in. 18.0\text{ in.} 18.0 in. ( 450 mm 450\text{ mm} 450 mm )

48 in. 48\text{ in.} 48 in. ( 1200 mm 1200\text{ mm} 1200 mm ) 48 / 2 = 24 in. 48 / 2 = 24\text{ in.} 48/2 = 24 in. 20 in. 20\text{ in.} 20 in. 20.0 in. 20.0\text{ in.} 20.0 in. ( 500 mm 500\text{ mm} 500 mm ) (Capped)

60 in. 60\text{ in.} 60 in. ( 1500 mm 1500\text{ mm} 1500 mm ) 60 / 2 = 30 in. 60 / 2 = 30\text{ in.} 60/2 = 30 in. 20 in. 20\text{ in.} 20 in. 20.0 in. 20.0\text{ in.} 20.0 in. ( 500 mm 500\text{ mm} 500 mm ) (Capped)

72 in. 72\text{ in.} 72 in. ( 1800 mm 1800\text{ mm} 1800 mm ) 72 / 3 = 24 in. 72 / 3 = 24\text{ in.} 72/3 = 24 in. 40 in. 40\text{ in.} 40 in. 24.0 in. 24.0\text{ in.} 24.0 in. ( 600 mm 600\text{ mm} 600 mm )

96 in. 96\text{ in.} 96 in. ( 2400 mm 2400\text{ mm} 2400 mm ) 96 / 3 = 32 in. 96 / 3 = 32\text{ in.} 96/3 = 32 in. 40 in. 40\text{ in.} 40 in. 32.0 in. 32.0\text{ in.} 32.0 in. ( 800 mm 800\text{ mm} 800 mm )

120 in. 120\text{ in.} 120 in. ( 3000 mm 3000\text{ mm} 3000 mm ) 120 / 3 = 40 in. 120 / 3 = 40\text{ in.} 120/3 = 40 in. 40 in. 40\text{ in.} 40 in. 40.0 in. 40.0\text{ in.} 40.0 in. ( 1000 mm 1000\text{ mm} 1000 mm ) (Capped)

144 in. 144\text{ in.} 144 in. ( 3600 mm 3600\text{ mm} 3600 mm ) 144 / 3 = 48 in. 144 / 3 = 48\text{ in.} 144/3 = 48 in. 40 in. 40\text{ in.} 40 in. 40.0 in. 40.0\text{ in.} 40.0 in. ( 1000 mm 1000\text{ mm} 1000 mm ) (Capped)

Important

Exam Watchpoint: Notice that for a 48 in. 48\text{ in.} 48 in. vessel, D / 2 = 24 in. D/2 = 24\text{ in.} D /2 = 24 in. , but the maximum length is capped at 20 in. 20\text{ in.} 20 in. . For a 72 in. 72\text{ in.} 72 in. vessel ( D > 60 in. D > 60\text{ in.} D > 60 in. ), the formula shifts immediately to D / 3 = 24 in. D/3 = 24\text{ in.} D /3 = 24 in. , which is under the 40 in. 40\text{ in.} 40 in. cap. Confusing these two rules is one of the most common calculation traps on the API 510 exam!

3. Thickness Averaging Procedure and Grid Requirements

When performing thickness averaging across the critical length L L L , the inspector must follow a disciplined measurement protocol:

Locate the Minimum Point: Use ultrasonic thickness gauging (UT) or profile radiography to scan the locally thinned area and locate the absolute minimum thickness reading ( t act, min t_{\text{act, min}} t act, min ​ ).

Lay Out Critical Length L L L : Center the line of length L L L over the lowest thickness reading, oriented parallel to the longitudinal axis of the vessel.

Select Measurement Spacing: Take a series of equally spaced thickness readings along line L L L . API 510 recommends taking multiple readings (typically at least 5 to 10 points spaced no more than 1 in. 1\text{ in.} 1 in. to 2 in. 2\text{ in.} 2 in. apart) to accurately characterize the profile.

Calculate Average Thickness ( t avg t_{\text{avg}} t avg ​ ): t avg = 1 N ∑ i = 1 N t i = t 1 + t 2 + t 3 + ⋯ + t N N t_{\text{avg}} = \frac{1}{N} \sum_{i=1}^{N} t_i = \frac{t_1 + t_2 + t_3 + \dots + t_N}{N} t avg ​ = N 1 ​ i = 1 ∑ N ​ t i ​ = N t 1 ​ + t 2 ​ + t 3 ​ + ⋯ + t N ​ ​

Evaluate Minimum Point Criteria: In addition to the average meeting t min t_{\text{min}} t min ​ , the lowest single reading within the averaged length must not be less than the minimum thickness limits permitted by API 510 or Fitness-for-Service Level 1 criteria (typically not less than 0.5 × t required 0.5 \times t_{\text{required}} 0.5 × t required ​ without an API 579 evaluation). +-----------------------------------------------------------------------------------------+

  • | THICKNESS PROFILE ALONG CRITICAL LENGTH L |
  • | |
  • | Wall Thickness (in.) |
  • | 0.500 |--- Nominal Thickness (t_nom = 0.500") ----------------------------------- |
  • | | |
  • | 0.375 |--- Required Thickness (t_req = 0.375") ---------------------------------- |
  • | | |
  • | 0.350 | t1=0.380" t5=0.390" |
  • | | \ t2=0.340" t4=0.350" / |
  • | 0.300 | \ \ t3=0.290" / / |
  • | | *-----------*-----(MIN POINT)-----*-------------* |
  • | | * |
  • | +----------------------------------|------------------------------------> |
  • | |<- CRITICAL LENGTH L = 20.0" ->| |
  • | |
  • | AVERAGE CALCULATION: |
  • | t_avg = (0.380 + 0.340 + 0.290 + 0.350 + 0.390) / 5 = 1.750 / 5 = 0.350" |
  • +-----------------------------------------------------------------------------------------+
  • 4. Joint Efficiency ( E E E ) Rules for Local Thin Areas

One of the most powerful and tested concepts in API 510 is the application of weld joint efficiency ( E E E ) during the evaluation of local thin areas.

The Base Metal Exemption ( E = 1.0 E = 1.0 E = 1.0 )

In ASME Section VIII, Division 1, shell thickness calculations for welded vessels include a weld joint efficiency factor E E E (e.g., E = 0.85 E = 0.85 E = 0.85 for spot radiography or E = 0.70 E = 0.70 E = 0.70 for no radiography) to account for potential weld defects.

However, API 510 Section 7.4.2.2 states:

When the locally thinned area is located entirely within the base metal and does not touch or cross a weld joint, the required thickness of the thinned area may be calculated using a joint efficiency of E = 1.0 E = 1.0 E = 1.0 (seamless vessel equivalent), provided the LTA is located away from the weld by a distance of at least the greater of:

1.0 in. 1.0\text{ in.} 1.0 in. ( 25 mm 25\text{ mm} 25 mm ) , or

2 × t 2 \times t 2 × t (where t t t is the nominal wall thickness).

Rationale for E = 1.0 E = 1.0 E = 1.0

Weld joint efficiency is a penalty applied to the weld seam itself due to the possibility of volumetric or planar weld flaws. The base plate away from the weld is seamless wrought material and does not possess weld defects. Therefore, if corrosion occurs strictly in the base metal, penalizing the corroded plate with the weld joint efficiency of a remote seam is overly conservative.

Weld Inclusion Rule

If the locally thinned area lies on, across, or within the heat-affected zone (HAZ) of a weld (i.e., closer than max ⁡ ( 1.0 in. , 2 t ) \max(1.0\text{ in.}, 2t) max ( 1.0 in. , 2 t ) to the weld toe), the calculation of required thickness for that LTA must use the actual joint efficiency ( E E E ) of the affected weld seam.

Location of Locally Thinned Area (LTA) Joint Efficiency ( E E E ) to Use in t required t_{\text{required}} t required ​ Formula

In Base Metal (Distance from weld toe ≥ max ⁡ [ 1.0 in. , 2 t ] \ge \max[1.0\text{ in.}, 2t] ≥ max [ 1.0 in. , 2 t ] ) E = 1.0 E = 1.0 E = 1.0 (Seamless Equivalent)

In or Touching Longitudinal Weld Seam E = E long E = E_{\text{long}} E = E long ​ (e.g., 0.85 for spot RT, 0.70 for visual only)

In or Touching Circumferential Weld Seam E = E circ E = E_{\text{circ}} E = E circ ​ (for longitudinal stress evaluation)

5. Comprehensive Step-by-Step Calculation Example

Let us work through a complete, open-book style API 510 calculation problem demonstrating every aspect of corrosion averaging, critical length determination, joint efficiency selection, and remaining life assessment.

Problem Statement

A vertical process separator has the following design and inspection data:

  • Design Pressure ( P P P ): 250 psig 250\text{ psig} 250 psig
  • Design Temperature: 400 ∘ F 400^\circ\text{F} 40 0 ∘ F

Inside Diameter ( D D D ): 48.0 in. 48.0\text{ in.} 48.0 in. (Inside Radius R = 24.0 in. R = 24.0\text{ in.} R = 24.0 in. )

Allowable Stress ( S S S ): 20 , 000 psi 20,000\text{ psi} 20 , 000 psi (SA-516 Gr. 70)

Original Longitudinal Weld Joint Efficiency ( E E E ): 0.85 0.85 0.85 (Type 1 Butt Weld, Spot Radiographed per UW-52)

Nominal Shell Thickness ( t nom t_{\text{nom}} t nom ​ ): 0.500 in. 0.500\text{ in.} 0.500 in.

Corrosion Allowance ( C A CA C A ): 0.050 in. 0.050\text{ in.} 0.050 in. required for next 5-year run

During an internal turnaround inspection, an isolated local thin area is discovered on the shell. Ultrasonic readings along a longitudinal line centered on the lowest reading yield the following 5 measurements spaced 4 inches apart (total span = 16 in. = 16\text{ in.} = 16 in. ): Readings: [ 0.320 in. , 0.290 in. , 0.265 in. , 0.295 in. , 0.330 in. ] \text{Readings: } [0.320\text{ in.},\; 0.290\text{ in.},\; 0.265\text{ in.},\; 0.295\text{ in.},\; 0.330\text{ in.}] Readings: [ 0.320 in. , 0.290 in. , 0.265 in. , 0.295 in. , 0.330 in. ]

Evaluate the acceptability of this LTA for two separate cases:

Case A: The LTA is located in the middle of the base plate, 8 in. 8\text{ in.} 8 in. away from all welds.

Case B: The LTA runs directly along the longitudinal weld seam.

Step 1: Determine the Critical Evaluation Length ( L L L )

Check the vessel inside diameter: D = 48.0 in. ≤ 60 in. D = 48.0\text{ in.} \le 60\text{ in.} D = 48.0 in. ≤ 60 in.

Apply the critical length formula for D ≤ 60 in. D \le 60\text{ in.} D ≤ 60 in. : L = min ⁡ ( D 2 , 20 in. ) = min ⁡ ( 48 2 , 20 ) = min ⁡ ( 24 in. , 20 in. ) = 20.0 in. L = \min\left(\frac{D}{2},\; 20\text{ in.}\right) = \min\left(\frac{48}{2},\; 20\right) = \min(24\text{ in.},\; 20\text{ in.}) = \mathbf{20.0\text{ in.}} L = min ( 2 D ​ , 20 in. ) = min ( 2 48 ​ , 20 ) = min ( 24 in. , 20 in. ) = 20.0 in.

The 16-inch span of readings fits completely within the allowable 20.0-inch critical length L L L .

Step 2: Calculate the Average Thickness ( t avg t_{\text{avg}} t avg ​ ) t avg = 0.320 + 0.290 + 0.265 + 0.295 + 0.330 5 = 1.500 5 = 0.300 in. t_{\text{avg}} = \frac{0.320 + 0.290 + 0.265 + 0.295 + 0.330}{5} = \frac{1.500}{5} = \mathbf{0.300\text{ in.}} t avg ​ = 5 0.320 + 0.290 + 0.265 + 0.295 + 0.330 ​ = 5 1.500 ​ = 0.300 in.

Step 3: Evaluate Case A (LTA in Base Metal Away from Welds)

Check distance from weld: Distance = 8.0 in. = 8.0\text{ in.} = 8.0 in. Threshold distance = max ⁡ ( 1.0 in. , 2 × t nom ) = max ⁡ ( 1.0 , 2 × 0.500 ) = max ⁡ ( 1.0 , 1.0 ) = 1.0 in. = \max(1.0\text{ in.},\; 2 \times t_{\text{nom}}) = \max(1.0,\; 2 \times 0.500) = \max(1.0,\; 1.0) = 1.0\text{ in.} = max ( 1.0 in. , 2 × t nom ​ ) = max ( 1.0 , 2 × 0.500 ) = max ( 1.0 , 1.0 ) = 1.0 in. Since 8.0 in. ≥ 1.0 in. 8.0\text{ in.} \ge 1.0\text{ in.} 8.0 in. ≥ 1.0 in. , the base metal exemption applies, and we can use E = 1.0 E = 1.0 E = 1.0 .

Calculate minimum required thickness ( t min t_{\text{min}} t min ​ ) using the ASME Section VIII Div 1 circumferential stress formula (UG-27(c)(1)): t min = P ⋅ R S ⋅ E − 0.6 ⋅ P t_{\text{min}} = \frac{P \cdot R}{S \cdot E - 0.6 \cdot P} t min ​ = S ⋅ E − 0.6 ⋅ P P ⋅ R ​ t min = 250 ⋅ 24.0 ( 20 , 000 ⋅ 1.0 ) − ( 0.6 ⋅ 250 ) = 6 , 000 20 , 000 − 150 = 6 , 000 19 , 850 = 0.3023 in. t_{\text{min}} = \frac{250 \cdot 24.0}{(20,000 \cdot 1.0) - (0.6 \cdot 250)} = \frac{6,000}{20,000 - 150} = \frac{6,000}{19,850} = \mathbf{0.3023\text{ in.}} t min ​ = ( 20 , 000 ⋅ 1.0 ) − ( 0.6 ⋅ 250 ) 250 ⋅ 24.0 ​ = 20 , 000 − 150 6 , 000 ​ = 19 , 850 6 , 000 ​ = 0.3023 in.

Total required thickness including future corrosion allowance ( C A = 0.050 in. CA = 0.050\text{ in.} C A = 0.050 in. ): t required = t min + C A = 0.3023 + 0.050 = 0.3523 in. t_{\text{required}} = t_{\text{min}} + CA = 0.3023 + 0.050 = \mathbf{0.3523\text{ in.}} t required ​ = t min ​ + C A = 0.3023 + 0.050 = 0.3523 in.

Comparison for Case A:

Measured t avg = 0.300 in. t_{\text{avg}} = 0.300\text{ in.} t avg ​ = 0.300 in.

Minimum required base metal thickness t min = 0.3023 in. t_{\text{min}} = 0.3023\text{ in.} t min ​ = 0.3023 in.

Since t avg < t min t_{\text{avg}} < t_{\text{min}} t avg ​ < t min ​ , the vessel cannot continue operating at 250 psig without either derating the MAWP, performing an API 579 Level 2 FFS assessment, or executing a weld buildup repair.

Calculate the maximum allowable working pressure of the thinned base metal ( E = 1.0 E = 1.0 E = 1.0 ): M A W P LTA = S ⋅ E ⋅ t avg R + 0.6 ⋅ t avg = 20 , 000 ⋅ 1.0 ⋅ 0.300 24.0 + ( 0.6 ⋅ 0.300 ) = 6 , 000 24.18 = 248.14 psig MAWP_{\text{LTA}} = \frac{S \cdot E \cdot t_{\text{avg}}}{R + 0.6 \cdot t_{\text{avg}}} = \frac{20,000 \cdot 1.0 \cdot 0.300}{24.0 + (0.6 \cdot 0.300)} = \frac{6,000}{24.18} = \mathbf{248.14\text{ psig}} M A W P LTA ​ = R + 0.6 ⋅ t avg ​ S ⋅ E ⋅ t avg ​ ​ = 24.0 + ( 0.6 ⋅ 0.300 ) 20 , 000 ⋅ 1.0 ⋅ 0.300 ​ = 24.18 6 , 000 ​ = 248.14 psig

Step 4: Evaluate Case B (LTA in Longitudinal Weld Seam)

Because the LTA encompasses the weld seam, we must use the vessel's actual weld joint efficiency E = 0.85 E = 0.85 E = 0.85 : t min = P ⋅ R S ⋅ E − 0.6 ⋅ P = 250 ⋅ 24.0 ( 20 , 000 ⋅ 0.85 ) − ( 0.6 ⋅ 250 ) = 6 , 000 17 , 000 − 150 = 6 , 000 16 , 850 = 0.3561 in. t_{\text{min}} = \frac{P \cdot R}{S \cdot E - 0.6 \cdot P} = \frac{250 \cdot 24.0}{(20,000 \cdot 0.85) - (0.6 \cdot 250)} = \frac{6,000}{17,000 - 150} = \frac{6,000}{16,850} = \mathbf{0.3561\text{ in.}} t min ​ = S ⋅ E − 0.6 ⋅ P P ⋅ R ​ = ( 20 , 000 ⋅ 0.85 ) − ( 0.6 ⋅ 250 ) 250 ⋅ 24.0 ​ = 17 , 000 − 150 6 , 000 ​ = 16 , 850 6 , 000 ​ = 0.3561 in.

Comparison for Case B:

Measured t avg = 0.300 in. t_{\text{avg}} = 0.300\text{ in.} t avg ​ = 0.300 in.

Minimum required weld thickness t min = 0.3561 in. t_{\text{min}} = 0.3561\text{ in.} t min ​ = 0.3561 in.

The deficit is significantly larger ( 0.300 in. 0.300\text{ in.} 0.300 in. vs 0.3561 in. 0.3561\text{ in.} 0.3561 in. ) because the joint efficiency penalty ( E = 0.85 E = 0.85 E = 0.85 ) increases the required thickness by 17.8 % 17.8\% 17.8% .

Calculate the MAWP for the thinned weld ( E = 0.85 E = 0.85 E = 0.85 ): M A W P weld = S ⋅ E ⋅ t avg R + 0.6 ⋅ t avg = 20 , 000 ⋅ 0.85 ⋅ 0.300 24.0 + ( 0.6 ⋅ 0.300 ) = 5 , 100 24.18 = 210.92 psig MAWP_{\text{weld}} = \frac{S \cdot E \cdot t_{\text{avg}}}{R + 0.6 \cdot t_{\text{avg}}} = \frac{20,000 \cdot 0.85 \cdot 0.300}{24.0 + (0.6 \cdot 0.300)} = \frac{5,100}{24.18} = \mathbf{210.92\text{ psig}} M A W P weld ​ = R + 0.6 ⋅ t avg ​ S ⋅ E ⋅ t avg ​ ​ = 24.0 + ( 0.6 ⋅ 0.300 ) 20 , 000 ⋅ 0.85 ⋅ 0.300 ​ = 24.18 5 , 100 ​ = 210.92 psig

This calculation clearly highlights why inspectors must verify the proximity of an LTA to any weld seam.

6. Common Exam Pitfalls & Calculation Traps

Trap Description Incorrect Assumption Correct Code Rule

Averaging Direction Averaging circumferentially around the vessel Always average longitudinally on cylindrical shells because hoop stress acts across this line.

Diameter Thresholds Using D / 2 D/2 D /2 for vessels larger than 60 in. 60\text{ in.} 60 in. For D ≤ 60 in. D \le 60\text{ in.} D ≤ 60 in. , L = min ⁡ ( D / 2 , 20 ) L = \min(D/2, 20) L = min ( D /2 , 20 ) . For D > 60 in. D > 60\text{ in.} D > 60 in. , L = min ⁡ ( D / 3 , 40 ) L = \min(D/3, 40) L = min ( D /3 , 40 ) .

Joint Efficiency Misuse Always using the vessel nameplate E E E for base metal LTAs If LTA is ≥ max ⁡ ( 1 in. , 2 t ) \ge \max(1\text{ in.}, 2t) ≥ max ( 1 in. , 2 t ) from weld toe, use E = 1.0 E = 1.0 E = 1.0 .

Radius vs Diameter Plugging Inside Diameter D D D into t min t_{\text{min}} t min ​ equation UG-27 requires Inside Radius R = D / 2 R = D/2 R = D /2 .

Lowest Reading Check Accepting an average even if one point is paper-thin The lowest point cannot violate structural integrity limits (e.g., must be ≥ 0.5 t req \ge 0.5 t_{\text{req}} ≥ 0.5 t req ​ for screening).

Loading diagram... API 510 Locally Thinned Area (LTA) Evaluation Workflow