4.1 Internal Pressure Design & Cylindrical Shell Calculations

Internal Pressure Design & Cylindrical Shell Calculations

Pressure vessel design under ASME Section VIII, Division 1 relies on the membrane theory of thin-walled shells. In a thin-walled cylindrical vessel subjected to internal pressure, the vessel wall is in a state of biaxial tension. Understanding the mathematical derivations, code limitations, and parameter interactions defined in paragraph UG-27 is paramount for any API 510 Authorized Pressure Vessel Inspector.

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1. Membrane Stress Theory in Cylindrical Shells

When a closed cylindrical vessel experiences internal pressure ($P$), the internal fluid exerts uniform outward normal pressure on the cylindrical walls and axial thrust on the end closures (heads).

+-----------------------------------------------------------------------------------------+
|                        MEMBRANE STRESSES IN A CYLINDRICAL SHELL                         |
|                                                                                         |
|               ^ Circumferential (Hoop) Stress (σ_h)                                     |
|               |   Acts tangentially around the circumference                            |
|         +-----+---------------------------------------------------+-----+               |
|         |     |                                                   |     |               |
|   <-----+=====[=========== LONGITUDINAL WELD SEAM ===============]=====+----->         |
|         |     |   (Subjected to Hoop Stress σ_h)                  |     |  Longitudinal |
|         +-----+---------------------------------------------------+-----+  Stress (σ_L) |
|               |                                                                         |
|               v Circumferential (Hoop) Stress (σ_h = P*R / t)                           |
|                                                                                         |
|   GOVERNING PRINCIPLE:                                                                  |
|   - Hoop stress σ_h is TWICE longitudinal stress σ_L (σ_h = 2 * σ_L).                   |
|   - σ_h acts across the LONGITUDINAL weld seam.                                         |
|   - σ_L acts across the CIRCUMFERENTIAL (girth) weld seam.                              |
|   - Therefore, the longitudinal seam governs the minimum required shell thickness.      |
+-----------------------------------------------------------------------------------------+

Principal Stress Equations (Thin-Wall Idealization)

1. Circumferential (Hoop) Stress ($\sigma_h$): $\sigma_h = \frac{P \cdot R}{t}$

   • Acts along the circumferential plane.
   • Tends to split the cylinder longitudinally from end to end.
   • Directly stresses the longitudinal weld seams (Category A).

2. Longitudinal (Axial) Stress ($\sigma_L$): $\sigma_L = \frac{P \cdot R}{2t} = \frac{1}{2} \sigma_h$

   • Acts along the longitudinal axis of the cylinder.
   • Tends to pull the vessel apart like a telescoping tube.
   • Directly stresses the circumferential girth seams (Category B).

Because hoop stress is twice as large as longitudinal stress ($\sigma_h = 2 \sigma_L$), the longitudinal weld seam is subjected to double the tensile stress of the circumferential seam. Consequently, circumferential stress governs the required wall thickness of cylindrical shells under internal pressure.

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2. ASME Section VIII Div 1 Shell Formulas (UG-27)

ASME Section VIII, Division 1 modifies the classical thin-wall equations by adding empirical correction terms ($-0.6P$ and $+0.4P$) to account for non-linear through-thickness stress distribution in moderately thick shells.

Circumferential Stress (Longitudinal Joints) — UG-27(c)(1)

When the thickness does not exceed one-half of the inside radius ($t \le 0.5R$) or the design pressure does not exceed $0.385 S E$ ($P \le 0.385 S E$):

$\mathbf{t = \frac{P \cdot R}{S \cdot E - 0.6 \cdot P}}$

$\mathbf{P = \frac{S \cdot E \cdot t}{R + 0.6 \cdot t}}$

Where:

• $t$ = Minimum required thickness of shell (inches or mm)
• $P$ = Internal design pressure or MAWP (psi or kPa)
• $R$ = Inside radius of shell course under evaluation in corroded condition: $R = R_{\text{original}} + CA$ (inches or mm)
• $S$ = Maximum allowable stress value from ASME Section II, Part D, Table 1A (psi or kPa)
• $E$ = Joint efficiency of the longitudinal weld seam per UW-12 (dimensionless, $0.45 \le E \le 1.0$)

Longitudinal Stress (Circumferential Joints) — UG-27(c)(2)

When the thickness does not exceed one-half of the inside radius ($t \le 0.5R$) or $P \le 1.25 S E$:

$\mathbf{t = \frac{P \cdot R}{2 \cdot S \cdot E + 0.4 \cdot P}}$

$\mathbf{P = \frac{2 \cdot S \cdot E \cdot t}{R - 0.4 \cdot t}}$

Where:

• $E$ = Joint efficiency of the circumferential girth seam per UW-12.

Outer Radius ($R_o$) Formulas (ASME Appendix 1-1)

If the outside radius ($R_o$) or outside diameter ($D_o$) is known rather than the inside radius:

[!IMPORTANT] Outer Radius Rule with Corrosion Allowance: When calculating required thickness using the outer radius formula, the outside radius $R_o$ of a vessel subject to internal corrosion does not change because metal loss occurs on the inside surface. Therefore, do not add corrosion allowance to $R_o$; instead, calculate the pressure-required thickness $t$ and then add the corrosion allowance ($t_{\text{nominal}} = t + CA$).

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3. ASME Section II Part D Allowable Stress ($S$)

The allowable stress value ($S$) represents the maximum permissible tensile stress in the material at the design metal temperature.

Determination of Allowable Stress ($S$)

• Listed in ASME Section II, Part D, Subpart 1, Table 1A (ferrous materials) and Table 1B (non-ferrous materials).
• Design Margin: Since the 1999 Addenda of ASME Section VIII Div 1, the design margin on tensile strength is 3.5 (i.e., $S = \min\left(\frac{S_u}{3.5},; \frac{2}{3} S_y\right)$ at room temperature).
• Prior to 1999, the design margin was 4.0 (i.e., $S = \min\left(\frac{S_u}{4.0},; \frac{2}{3} S_y\right)$).

Typical Allowable Stress Values at Temperature (SA-516 Gr. 70 Carbon Steel)

[!NOTE] Notice that allowable stress decreases significantly at elevated temperatures due to thermal softening and the onset of creep mechanisms. In API 510 evaluations, always verify the design metal temperature from the vessel nameplate or U-1A Form before selecting $S$.

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4. Weld Joint Efficiency ($E$) Summary (UW-12)

Joint efficiency ($E$) is a numerical factor that reduces the allowable stress to compensate for potential flaws in welded joints.

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5. Comprehensive Step-by-Step Shell Thickness Calculation

Let us work through an exam-level calculation covering all aspects of shell sizing, corrosion allowance, and joint efficiency.

Problem Statement

A horizontal flash drum is to be fabricated from SA-516 Grade 70 carbon steel plate. The vessel data is as follows:

• Design Internal Pressure ($P$): $350\text{ psig}$
• Design Temperature: $500^\circ\text{F}$
• Inside Diameter ($D_i$): $60.0\text{ in.}$ ($R_i = 30.0\text{ in.}$)
• Corrosion Allowance ($CA$): $0.125\text{ in.}$ (1/8 in.)
• Longitudinal Joint: Type 1 butt weld, Spot Radiographed per UW-52
• Circumferential Joint: Type 1 butt weld, No Radiography (visual examination only)
• Allowable Stress ($S$): $20,000\text{ psi}$ at $500^\circ\text{F}$

Required:

1. Determine the minimum required thickness ($t_{\text{circ}}$) based on circumferential stress.
2. Determine the minimum required thickness ($t_{\text{long}}$) based on longitudinal stress.
3. Determine the governing required thickness in new condition ($t_{\text{design}}$).
4. Calculate the MAWP of the new shell if nominal plate thickness selected is $t_{\text{nom}} = 0.750\text{ in.}$

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Step 1: Establish Corroded Inside Radius ($R$)

$R = R_i + CA = 30.0\text{ in.} + 0.125\text{ in.} = \mathbf{30.125\text{ in.}}$

Step 2: Determine Joint Efficiencies ($E$)

• For Circumferential Stress (acting across longitudinal joint): Type 1 Butt + Spot RT $\rightarrow \mathbf{E = 0.85}$
• For Longitudinal Stress (acting across circumferential joint): Type 1 Butt + Visual Only $\rightarrow \mathbf{E = 0.70}$

Step 3: Calculate Required Thickness for Circumferential Stress (Hoop Stress)

$t_{\text{circ}} = \frac{P \cdot R}{S \cdot E - 0.6 \cdot P}$

$t_{\text{circ}} = \frac{350 \cdot 30.125}{(20,000 \cdot 0.85) - (0.6 \cdot 350)} = \frac{10,543.75}{17,000 - 210} = \frac{10,543.75}{16,790} = \mathbf{0.6280\text{ in.}}$

Step 4: Calculate Required Thickness for Longitudinal Stress (Axial Stress)

$t_{\text{long}} = \frac{P \cdot R}{2 \cdot S \cdot E + 0.4 \cdot P}$

$t_{\text{long}} = \frac{350 \cdot 30.125}{(2 \cdot 20,000 \cdot 0.70) + (0.4 \cdot 350)} = \frac{10,543.75}{28,000 + 140} = \frac{10,543.75}{28,140} = \mathbf{0.3747\text{ in.}}$

Step 5: Determine Governing Required Thickness

Comparing $t_{\text{circ}} = 0.6280\text{ in.}$ vs $t_{\text{long}} = 0.3747\text{ in.}$, circumferential stress governs: $t_{\text{required, corroded}} = \mathbf{0.6280\text{ in.}}$

Total minimum required thickness in new uncorroded condition: $t_{\text{design}} = t_{\text{required, corroded}} + CA = 0.6280\text{ in.} + 0.125\text{ in.} = \mathbf{0.7530\text{ in.}}$

Step 6: Calculate Shell MAWP in Corroded Condition for $t_{\text{nom}} = 0.750\text{ in.}$

If nominal thickness is $0.750\text{ in.}$, the corroded thickness available is: $t = t_{\text{nom}} - CA = 0.750\text{ in.} - 0.125\text{ in.} = 0.625\text{ in.}$

$MAWP = \frac{S \cdot E \cdot t}{R + 0.6 \cdot t} = \frac{20,000 \cdot 0.85 \cdot 0.625}{30.125 + (0.6 \cdot 0.625)} = \frac{10,625}{30.125 + 0.375} = \frac{10,625}{30.50} = \mathbf{348.36\text{ psig}}$

(Since 348.36 psig < 350 psig, a standard 0.750 in. plate would require slight derating or selecting the next plate size of 0.8125 in. / 13/16 in.).

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6. Common Exam Pitfalls & Calculation Traps