4.2 Ellipsoidal & Hemispherical Head Calculations

Ellipsoidal & Hemispherical Head Calculations

The current API 510 Body of Knowledge requires internal-pressure calculations for 2:1 semi-ellipsoidal heads and hemispherical heads. It also requires candidates to determine a head's part MAWP and decide whether a measured head thickness is acceptable. Appendix 1 formulas for nonstandard heads are expressly outside the calculation scope. Accordingly, this section concentrates on the two named geometries and does not make torispherical or flat-head sizing part of the study target.

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1. Geometry Comes Before Substitution

A 2:1 semi-ellipsoidal head has a major-to-minor axis ratio of 2:1. Its inside depth is approximately one-fourth of its inside diameter. The standard geometry factor is therefore K = 1.0. The formula uses the head's inside diameter D, not its radius.

A hemispherical head is one-half of a sphere. Its inside crown radius L equals one-half of its inside diameter. The formula uses the inside radius L, not diameter. Because pressure creates equal membrane stress in every direction, a hemisphere is structurally efficient and normally needs about half the pressure thickness of a 2:1 ellipsoidal head having the same diameter, pressure, material, and joint efficiency.

Where P is internal pressure at the component, S is allowable stress at the evaluation temperature, E is the applicable joint efficiency, and t is pressure-retaining metal in the corroded condition. Keep units consistent.

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2. Corroded Dimensions and Corrosion Allowance

API 510 problems commonly switch between an existing corroded condition and a new-design condition. Read the wording before touching the calculator.

• For an existing head evaluated at its current measured thickness, use the current corroded inside dimension and do not subtract the historical corrosion allowance a second time.
• If a problem gives an original inside diameter and asks for a required thickness including a future allowance CA, first use the corresponding corroded dimension: D_corroded = D_original + 2CA for diameter, or L_corroded = L_original + CA for radius.
• Calculate pressure thickness with the corroded dimension, then add CA to obtain a nominal required thickness for future service.
• For a part-MAWP calculation, use only the thickness available for pressure after deducting the future corrosion allowance from the measured or nominal thickness when the problem requires that allowance to remain.

The diameter-versus-radius distinction is a frequent error. Adding a radial corrosion allowance once to a diameter is also wrong; loss on both sides increases the corroded inside diameter by 2CA.

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3. Worked Required-Thickness Example

An accumulator has an original inside diameter of 72.0 in. Both heads are seamless. Design pressure is 250 psi, allowable stress is 20,000 psi, E = 1.00, and future corrosion allowance is 0.125 in.

2:1 Ellipsoidal Head

Corroded inside diameter:

D = 72.0 + 2(0.125) = 72.250 in.

Pressure thickness:

t = [250(72.250)] / [2(20,000)(1.00) - 0.2(250)]

t = 18,062.5 / 39,950 = 0.4521 in.

Required nominal thickness including future allowance:

t_required = 0.4521 + 0.125 = 0.5771 in.

Hemispherical Head

Corroded inside radius:

L = 36.0 + 0.125 = 36.125 in.

Pressure thickness:

t = [250(36.125)] / [2(20,000)(1.00) - 0.2(250)]

t = 9,031.25 / 39,950 = 0.2261 in.

Required nominal thickness including future allowance:

t_required = 0.2261 + 0.125 = 0.3511 in.

The pressure portion of the hemispherical result is one-half of the ellipsoidal value because L = D/2 and the denominators are identical. The same fixed corrosion allowance is then added to each.

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4. Reversing the Formula for Part MAWP

When actual thickness is known, remove any required future corrosion allowance to obtain t before calculating part MAWP. Then use the inverse formula for the correct geometry. Do not simply divide test pressure or design pressure by a safety factor.

For example, a 2:1 ellipsoidal head has D = 60.0 in., S = 18,000 psi, E = 0.85, and 0.500 in. of pressure-retaining thickness:

P = [2(18,000)(0.85)(0.500)] / [60.0 + 0.2(0.500)]

P = 15,300 / 60.1 = 254.6 psi

That is the head's pressure-only part MAWP at the stated temperature. A completed vessel is governed by its weakest component. For a lower component in a liquid-filled vertical vessel, subtract the static head acting at that elevation when expressing the allowable pressure at the top of the vessel.

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5. Exam-Safe Workflow

1. Identify the geometry: 2:1 ellipsoidal or hemispherical.
2. Decide whether the problem asks for pressure thickness, nominal thickness with CA, or part MAWP.
3. Put D in the ellipsoidal equation or L in the hemispherical equation.
4. Use allowable stress at the stated temperature and the applicable joint efficiency.
5. Treat dimensions and thickness consistently as corroded or uncorroded.
6. Include static head at the evaluated component when P must represent local pressure.
7. Keep unrounded values through the calculation and round only the reported result.

Common distractors arise from using radius in the ellipsoidal equation, diameter in the hemispherical equation, adding CA only once to a diameter, adding CA before rather than after the pressure-thickness calculation, or using nominal thickness rather than pressure-retaining thickness in a MAWP calculation.