What this module covers
This is the most predictable calculation area. You are given thickness readings over time and asked for a corrosion rate, the remaining life and the next interval. The arithmetic is simple. Marks are lost by using the wrong rate, the wrong thickness or the wrong interval rule. The BOK says these formulas may be closed book, so learn them by heart.
All numbers below use practice circuit C-1 (invented for this course): a CML that read 0.500 in. in 2010, 0.440 in. in 2018 and 0.414 in. in 2026, with a required thickness of 0.300 in.
Metal loss and corrosion rates
Metal loss is the earlier reading minus the later reading at the same location. API 570 7.1.2 gives two rates:
- Long-term (LT): from the first reading to the latest. CR = (t first - t actual) / years between.
- Short-term (ST): from the previous reading to the latest. CR = (t previous - t actual) / years between.
Worked calculation: corrosion rate and remaining life from thickness readings.
| Step | Long term (2010 to 2026) | Short term (2018 to 2026) |
|---|---|---|
| Metal loss | 0.500 - 0.414 = 0.086 in. | 0.440 - 0.414 = 0.026 in. |
| Years | 16 | 8 |
| Rate | 0.086 / 16 = 0.0054 in./yr | 0.026 / 8 = 0.0033 in./yr |
The governing rate is normally the higher, here 0.0054 in./yr. A change in service can justify using the short-term rate, but only with a stated reason.
Two slips to avoid: subtracting in the wrong order, and miscounting the years between readings.
Remaining life and retirement thickness
Remaining life (API 570 7.2) is the margin over the rate: RL = (t actual - t required) / CR.
For C-1: (0.414 - 0.300) / 0.0054 = 0.114 / 0.0054 = 21.1 years.
Work the other way when asked for the retirement thickness: t required = t actual - (CR x RL). Projecting forward, the thickness at the next inspection is t actual - (CR x interval).
If a rate is given in mils per year, divide by 1,000 first. 8 mpy is 0.008 in./yr.
Setting thickness and visual external intervals
API 570 6.3.3 and Table 1 set the longest interval. For thickness measurement the interval is the shorter of half the remaining life and the Table 1 maximum for the class.
Worked calculation: inspection interval from remaining life and the class maximum. Use the practice maxima of 5 years for class 1, 10 years for class 2 and 10 years for class 3 on thickness measurement:
| Circuit | Remaining life | Half life | Class maximum | Interval |
|---|---|---|---|---|
| C-1, class 2 | 21.1 years | 10.6 years | 10 years | 10 years |
| C-1 as class 1 | 21.1 years | 10.6 years | 5 years | 5 years |
| Short life, class 2 | 3.0 years | 1.5 years | 10 years | 1.5 years |
Visual external inspection has its own maximum by class, and it is not set by the thickness data. Injection points have a shorter maximum of their own.
What changes an interval
- A faster corrosion rate or a smaller margin shortens the interval. A change in service can do both.
- Higher consequence classes carry shorter maxima.
- Damage that thickness readings do not catch, such as CUI or cracking, needs its own inspection added.
- A risk-based program can justify intervals but needs a documented basis and controls.
- An overdue inspection is recorded and scheduled. It does not renew itself.
- Thin spots count. A pit can leave too little wall at one place even when the general loss is small, so record the thinnest value, its size and its location.
Where required thickness is concerned, both the pressure and the structural needs must be met, so the larger requirement governs.
Check yourself
The three questions below cover a long-term rate, a remaining life and how class limits an interval.
Where to look in your own books
- API 570 7.1.2: corrosion rate determination
- API 570 7.2: remaining life
- API 570 6.3.3 and Table 1: inspection intervals
- API 570 7.5: MAWP of corroded pipe, used in the next module
- The worked formulas and practice figures in this module: every figure here is invented practice data