Try a full lesson: Corrosion rates, remaining life, inspection intervals and joint efficiency

What this module covers

This is the most predictable calculation area on the exam. You are given thickness readings and asked for metal loss, a corrosion rate, the remaining life, the remaining corrosion allowance and the next inspection interval. You are also asked how joint efficiency, E, is found. The BOK says the formulas and the interval rules may be closed book, so learn them by heart.

All numbers below use practice tank T-21 (invented for this course): a shell course that read 0.500 in. at construction in 2006, 0.452 in. in 2016 and 0.428 in. in 2026, with a minimum allowable thickness of 0.320 in.

Metal loss and corrosion averaging

Metal loss is the earlier reading minus the later reading at the same location. Always subtract in that order and keep the units.

Corrosion averaging. A shell is not judged by its single thinnest reading when the loss is spread over an area. API 653 Section 4 lets you average the readings over a set length (the critical length taught in module t-m3) and compare that average with the minimum. Pits are handled by their own rules. Know which number a question wants: the single lowest reading, the average over the length, or a pit depth.

Corrosion rate and remaining life

The BOK gives both relations, and RP 575 7.1 and 7.2 discuss them:

• Corrosion rate = (t previous - t actual) / years between the two readings
• Remaining life = (t actual - t minimum) / corrosion rate
• Remaining corrosion allowance (RCA) = t actual - t minimum

Worked calculation: corrosion rate and remaining life from shell readings (tank T-21).

Use the rate the question tells you to use. If it does not say, the higher rate is the safer basis, here 0.0036 in./yr.

RCA = 0.428 - 0.320 = 0.108 in. Remaining life = 0.108 / 0.0036 = 30.0 years.

The BOK also says you must be able to "compensate for the corrosion allowance": if a stem gives a required thickness and a corrosion allowance to keep, add the allowance to the minimum before you subtract.

If a rate is given in mils per year, divide by 1,000 first: 4 mpy is 0.004 in./yr.

Setting external and internal inspection intervals

API 653 Section 6 sets a maximum interval for each inspection type. This course uses these practice relations, where N is the corrosion rate in in./yr and RCA is in inches:

Worked calculation: intervals for tank T-21 (N = 0.0036 in./yr, RCA = 0.108 in.).

Now suppose a faster rate of 0.0090 in./yr. External: 0.108 / 0.036 = 3.0 years. Ultrasonic: 0.108 / 0.018 = 6.0 years. The cap no longer governs; the margin does.

Internal intervals are driven by the bottom. The interval is set so that the bottom will still be above its minimum thickness at the next internal inspection (module t-m4 works that calculation). Table 6.1 lists features that support a longer initial interval, and a documented risk-based assessment is an alternative route.

What changes an interval: a faster rate, a smaller margin, a change of service, damage that thickness readings do not catch, settlement, and new repairs. An overdue inspection is recorded and scheduled; it does not renew itself.

Joint types, radiography and joint efficiency

E appears in the shell thickness and fill height relations, so a wrong E carries through the whole problem. API 653 4.3.3.1 points to Tables 4.2a and 4.2b for E, and the BOK marks both tables as ones that may be asked without the documents. To read the right entry you need:

1. The construction standard and its edition or date, from the nameplate or records. Tanks of unknown origin have their own entry.
2. The joint type, for example a double butt-welded joint or a lap-welded joint.
3. The radiography performed: full, spot or none. API 650 Section 8.1 and Figure 8.1 describe where spot radiographs are taken on new tanks, and API 653 Section 12 covers radiography of repairs.

Corrosion that lies well away from any weld is evaluated with the plate value instead of the joint value; read the distance rule in your copy of Section 4.

Worked calculation: where E enters (practice values). Tank T-21's records show its edition, its joint type and spot radiography. Suppose your reading of the table gives E = 0.85 for that weld, and the plate value applies to a corroded spot far from the welds. A fill height calculated for the corroded spot uses the plate value, and one calculated across a vertical weld uses 0.85. The same thickness supports less liquid height when the lower E applies.

Common traps

• Subtracting readings in the wrong order, or counting the years wrong (2016 to 2026 is 10 years).
• Using t minimum where the stem asks for RCA, or forgetting to add a required corrosion allowance.
• Quoting the capped interval when the margin gives a shorter one.
• Using the whole thickness instead of the margin over the minimum.
• Reading E for the wrong edition or ignoring the radiography actually done.

How to study this module

Work each calculation twice: once slowly, writing every step and unit, then timed with only a calculator. Keep one worked example of each type on a single page: rate, remaining life, RCA, external interval and ultrasonic interval. When an answer looks odd, check units and the order of subtraction before the arithmetic.

Check yourself

The three questions below cover a corrosion rate, an external interval and which E applies to an area away from welds.

Where to look in your own books

• API 653 Section 4: metal loss and corrosion averaging, and 4.3.3.1 for where E enters
• API 653 Tables 4.2a and 4.2b: joint efficiencies
• API 653 Section 6: inspection intervals, including Table 6.1
• API 653 Section 12 and API 650 Section 8.1 and Figure 8.1: radiography
• API RP 575 7.1 and 7.2: corrosion rate and remaining life
• Every figure in this module is invented practice data