Materials of construction and corrosion allowance

How to choose materials for process equipment (corrosion resistance, strength at temperature, fabricability, life cost), the forms of corrosion, and how a corrosion allowance is set from corrosion rate and design life.

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Why it matters

Every vessel, column and exchanger thickness calculation starts with two choices: which material, and how much extra metal to add for the corrosion it will suffer. Choose a material the process fluid attacks and the equipment leaks within months; choose an expensive alloy where carbon steel would do and the plant is uneconomic. The corrosion allowance is the bridge between a stress calculation done for "new" metal and a vessel that is still safe at the end of its design life.

Key ideas

What drives material selection. The designer weighs, roughly in this order:

  • Corrosion resistance to every fluid the equipment sees, including start-up, cleaning and upset conditions, not just normal operation.
  • Mechanical properties at the design temperature: tensile and yield strength fall as temperature rises; above roughly 400 °C creep strength governs for carbon steel. At low temperature the concern is brittle fracture, so impact (Charpy) toughness must be proven.
  • Fabricability: weldability, formability of heads, availability of plate, pipe and forgings in the needed sizes.
  • Cost over the life: material + fabrication + maintenance + downtime, not purchase price alone.

Common materials.

  • Carbon steel (boiler-quality plate such as IS 2002 or SA-516 Gr 70): the default for non-corrosive or mildly corrosive service; cheap, strong, easy to weld.
  • Low-alloy steels (Cr–Mo grades): better high-temperature and hydrogen-service strength.
  • Austenitic stainless steels: 304 (about 18 % Cr, 8 % Ni) and 316 (adds about 2–3 % Mo for better resistance to chloride pitting). The chromium forms a thin passive oxide film. The "L" grades (304L, 316L, carbon ≤ 0.03 %) resist sensitisation, the precipitation of chromium carbide at grain boundaries when the steel is held roughly between 425 and 815 °C (for example during welding), which leaves chromium-depleted zones open to intergranular corrosion.
  • Nickel alloys, titanium, copper alloys: for severe service (hot chlorides, seawater, strong acids).
  • Non-metallics and linings: FRP, PTFE/PVDF, rubber and glass-lined steel give excellent chemical resistance but limited temperature and pressure capability. Clad plate (a thin alloy layer bonded to carbon steel) gives alloy resistance at near carbon-steel cost.

Forms of corrosion a designer must recognise: uniform (general) attack, which a corrosion allowance can handle; and localised forms — pitting, crevice corrosion, galvanic corrosion between dissimilar metals in contact, intergranular attack, erosion–corrosion at high velocity, and stress-corrosion cracking (for example austenitic stainless steel in warm chloride solutions, typically above about 60 °C). Extra thickness does not protect against localised attack; only correct material choice, design detail or environment control does.

Corrosion allowance (CA). The stress calculation gives the minimum thickness needed to carry the load. Because uniform corrosion removes metal over time, the code requires the designer to add a corrosion allowance so that the calculated minimum still remains at the end of the design life. CA is estimated from the expected corrosion rate × design life, from plant experience, or from a client/code specification. For carbon steel in mild service a value of a few millimetres is usual; for stainless steel in a fluid it resists, CA is often small or zero. Take the governing value from your code book or the project specification rather than a rule of thumb.

From calculated to ordered thickness. Calculated thickness + corrosion allowance (+ any allowance for thinning during forming of heads or mill under-tolerance) is then rounded up to the next available plate thickness. A minimum thickness for handling rigidity also applies, which is stated in the code.

Allowable stress (design stress) is not the tensile strength. Codes derive it from tensile strength, yield strength and, at high temperature, creep data, each divided by a safety factor; the factors and the tabulated values differ between IS 2825 and ASME Section VIII — always read the allowable stress for the actual material and design temperature from the code tables.

Formulas

t = t_calc + CA

  • t = nominal thickness required before rounding up to a plate size (mm)
  • t_calc = thickness from the stress formula for the given pressure and loads (mm)
  • CA = corrosion allowance (mm)

CA = r_c × n

  • r_c = uniform corrosion rate (mm/year); n = design life (years). Applies to uniform attack only.

r_c = 87.6 × W / (ρ × A × τ)

  • r_c in mm/year; W = mass loss of a test coupon (mg); ρ = metal density (g/cm³); A = exposed area (cm²); τ = exposure time (h).
  • The constant 87.6 = 8760 h/year × 10 mm/cm ÷ 1000 mg/g.

1 mpy = 0.0254 mm/year

  • mpy = mils (thousandths of an inch) per year, still common in corrosion data tables.

Worked examples

Example 1 (standard): corrosion allowance from a coupon test Given: carbon-steel coupon, ρ = 7.86 g/cm³, exposed area A = 40 cm², mass loss W = 300 mg in τ = 500 h of exposure. Design life 20 years.

  1. Corrosion rate: r_c = 87.6 × W / (ρ × A × τ) r_c = 87.6 × 300 / (7.86 × 40 × 500) = 26 280 / 157 200 = 0.167 mm/year
  2. Metal lost over the life: CA = r_c × n = 0.167 × 20 = 3.34 mm
  3. Round up to a practical value: CA = 3.5 mm.

Example 2 (GATE level): carbon steel with CA versus stainless without Given: cylindrical shell, internal design pressure P = 1.0 MPa, inside diameter D = 1500 mm, weld joint efficiency J = 0.85, thin-shell formula t_calc = P·D / (2·f·J − P). Carbon steel: allowable stress f = 118 MPa, CA = 3 mm, ρ = 7850 kg/m³, ₹70/kg. Stainless steel 304: f = 130 MPa, CA = 0, ρ = 7900 kg/m³, ₹300/kg. Available plates: 6, 8, 10, 12, 14 mm. (Allowable stresses here are given data; in practice take them from your code book.)

  1. Carbon steel: t_calc = 1.0 × 1500 / (2 × 118 × 0.85 − 1.0) = 1500 / 199.6 = 7.52 mm; t = 7.52 + 3 = 10.52 mm → choose 12 mm.
  2. Stainless: t_calc = 1500 / (2 × 130 × 0.85 − 1.0) = 1500 / 220 = 6.82 mm; CA = 0 → choose 8 mm.
  3. Shell mass per metre length (thin wall, m ≈ π·D·t·ρ): CS: π × 1.5 × 0.012 × 7850 = 444 kg/m; SS: π × 1.5 × 0.008 × 7900 = 298 kg/m.
  4. Material cost per metre: CS 444 × 70 ≈ ₹31 100; SS 298 × 300 ≈ ₹89 300.
  5. Result: carbon steel with a corrosion allowance is about 2.9 times cheaper in plate cost. Stainless is justified only if the CS corrosion rate is too high for any sensible allowance, if localised corrosion is expected, or if product contamination by iron is unacceptable.

Common mistakes

  • Adding the corrosion allowance to the diameter in the stress formula but forgetting to add it to the thickness (or adding it twice). Use the corroded dimensions in the stress calculation if the code asks, and add CA once to the thickness.
  • Assuming a corrosion allowance protects against pitting, stress-corrosion cracking or intergranular attack. It does not.
  • Using tensile strength, or a room-temperature allowable stress, at an elevated design temperature.
  • Rounding the thickness down to the nearest plate, or forgetting forming thinning in dished heads.
  • Mixing mpy and mm/year (1 mpy is only 0.0254 mm/year).
  • Choosing 304 instead of 304L/316L for welded equipment in corrosive service, ignoring sensitisation.

For GATE CH

Expect short numericals combining a thickness formula with a corrosion allowance and design life, corrosion-rate calculations from coupon mass loss, and conceptual questions on which material suits a given service (chlorides, low temperature, high temperature) or which corrosion mechanism a given situation describes. Practise carrying units through the 87.6 formula and always rounding thickness up.

Quick check

  1. A vessel needs 9.2 mm for pressure and the fluid corrodes the steel at 0.15 mm/year over a 15-year life. What thickness is ordered if plates come in even millimetres?
  2. Why is 316 preferred over 304 in chloride-containing service?
  3. Will a 3 mm corrosion allowance protect against chloride stress-corrosion cracking?
  4. Convert 10 mpy to mm/year.

Answers: 1. 9.2 + 2.25 = 11.45 mm, so order 12 mm. 2. Its molybdenum content improves resistance to chloride pitting. 3. No, extra thickness does not stop cracking; change the material or environment. 4. 0.254 mm/year.

Try answering each one aloud before you open it.

  1. 1.What is meant by 'materials of construction' in process equipment design?Concept

    It is the material chosen for every pressure-retaining and wetted part of the equipment: shell, heads, nozzles, internals, gaskets and bolting. The choice is driven first by corrosion resistance to all fluids seen (including cleaning and upsets), then by mechanical properties at the design temperature, fabricability and weldability, and life-cycle cost. Carbon steel is the default; stainless steels, nickel alloys, clad plate or non-metallic linings are used only when the service demands them.

  2. 2.Explain the term 'corrosion allowance' in the context of process equipment.Concept

    Corrosion allowance is extra thickness added to the thickness calculated for pressure and other loads, so that the required metal is still there at the end of the design life. It is usually estimated as uniform corrosion rate × design life, or taken from plant experience or the client/code specification, then the total is rounded up to an available plate. It only covers uniform thinning; it gives no protection against pitting, stress-corrosion cracking or intergranular attack.

  3. 3.Why is stainless steel commonly used in the construction of chemical reactors?Application

    Its chromium (about 18 % in 304/316) forms a thin, self-healing passive oxide film that resists many acids, alkalis and oxidising media, keeps the product free of iron contamination and is easy to clean. It also keeps good strength and toughness over a wide temperature range and welds well. It is not universal: austenitic grades suffer pitting and stress-corrosion cracking in warm chlorides, which is why 316 (with Mo), low-carbon L grades or higher alloys are chosen for such service.

  4. 4.What could happen if the wrong material is chosen for a piece of process equipment?Application

    If the wrong material is chosen, the equipment may suffer from premature failure due to corrosion, mechanical stress, or chemical attack. This can lead to leaks, contamination of the product, safety hazards, and costly downtime for repairs or replacements. Proper material selection is crucial to ensure the equipment's reliability and safety.

  5. 5.How does temperature affect the choice of materials for process equipment?Application

    Allowable stress falls as temperature rises, so thickness grows; above roughly 400 °C creep governs carbon steel and Cr–Mo or stainless grades are used. At low temperature the risk is brittle fracture, so materials with proven impact toughness (impact-tested carbon steels, austenitic stainless, aluminium) are required. Temperature also speeds up most corrosion reactions and limits non-metallics such as plastics and rubber linings.

  6. 6.Why might a corrosion allowance be unnecessary for certain materials?Application

    If the material's corrosion rate in that fluid is negligible, as with stainless steel in a medium it resists, a lined vessel or many non-metallic constructions, there is no uniform thinning to allow for, so the allowance is set to zero or a small nominal value. Leaving out the allowance makes the expensive alloy thinner and helps offset its higher price. The decision must still be based on corrosion data, because localised attack is not covered by an allowance anyway.

  7. 7.Calculate the corrosion allowance needed for a steel pipe expected to lose 0.1 mm of thickness per year over a 20-year lifespan.Numerical

    Corrosion allowance = Corrosion rate × Service life = 0.1 mm/year × 20 years = 2 mm. Therefore, an additional 2 mm should be added to the pipe's thickness to account for corrosion over its expected lifespan.

  8. 8.Explain why non-metallic materials might be chosen over metals for certain process equipment.Application

    Non-metallic materials, such as plastics and ceramics, might be chosen over metals due to their superior corrosion resistance, especially in highly acidic or alkaline environments. They are also often lighter and can be more cost-effective for certain applications. However, their mechanical strength and temperature resistance must be considered to ensure they are suitable for the intended use.

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