This document provides examples and solutions for calculating corrosion rates and remaining life for different courses on a corroded tower. It determines the metal loss, corrosion rate, remaining corrosion allowance, and retirement date for each course. It also discusses converting units to mils, handling partial years, long term vs. short term corrosion rates, and API 510 requirements for inspection intervals and minimum thicknesses.

1. API 510 Preparatory ClassAPI 510 Preparatory Class
Lesson 11Lesson 11
Corrosion CalculationsCorrosion Calculations

2. A 60 foot tower consisting of four (4) shell courses was
found to have varying corrosion rates in each course.
Minimum wall thickness readings were taken after 4 years
and 6 months of service. All original wall thicknesses
included a 1/8" corrosion allowance.
The top course's original thickness was .3125". The present
thickness is .3000". The second course downward had an
original thickness of .375".
During the inspection it was found to have a minimum wall
thickness of .349". The third course was measured at .440"
its original thickness was .500". The bottom course had an
original thickness of .625" and measured to be .595".
Corrosion Example Problems

3. Determine the metal loss for the top course, the corrosion
rate for the second course, the corrosion allowance
remaining in the third course, the retirement date for the
bottom course.
Corrosion Example Problems

4. Solution A:
TOP COURSE:
Metal loss equals the previous thickness minus the present
thickness.
Previous .3125"
Present -.3000"
.0125" Metal Loss
Corrosion Example Problems

5. SECOND COURSE:
Corrosion rate equals metal loss per given unit of time.
Previous .3750“
Present - .3490"
.0260"
Metal Loss = = .006"/ Per Year
Corrosion Example Problems
yearsTime
Loss
5.4
"0260.

6. THIRD COURSE:
Remaining Corrosion Allowance equals the actual thickness
minus the required thickness.
Original Thickness .500"
Original C. A. -.125"
Required Wall Thickness .375"
Actual Wall Thickness .440"
Required Wall -.375"
Remaining C.A. .065“
Corrosion Example Problems

7. BOTTOM COURSE
Remaining life equals the remaining corrosion allowance
divided by the corrosion rate.
1. Required Thickness
Original Thickness .625"
Original C. A. -.125"
Required Thickness .500“
Corrosion Example Problems

8. 2.Remaining Corrosion Allowance
Actual Wall Thickness .595“
Required Thickness -.500"
Remaining Corr. Allow .095“
3.Corrosion Rate
Original Thickness .625"
Present Thickness -.595“
Metal Loss .030"
Metal Loss .030“ = Corrosion Rate = .0067"/Yr.
Time 4.5 Years
Corrosion Example Problems

9. 4. Remaining Life
Corrosion Allowance.095“ = 14.2 Yrs. Remaining Life
Corrosion Rate .0067"/Yr.
Corrosion Example Problems

10. Corrosion
Just Milling Around
In the previous examples we calculated using decimal
fractions of an inch such as .006”. However questions
on the exam often have the format of Mils or
thousandths of and inch. So if you are not familiar with
this terminology here are some examples.
.1 = 1/10th
of an inch
.01 = 1/100th
of an inch
.001 = 1/1000th
of an inch or 1 Mil.  this is what is
used.
Take our example above .006” this translates to 6 Mils.
.0067” translates to 6.7 Mils.

11. Class Quiz # 24
Just Milling Around
1. Convert the following decimal fractions into Mils.
.195” = _____ Mils
. 023” = _____ Mils
.009” = _____ Mils.
.0011” =______Mils.
2. Convert our previous Remaining Life problem into Mils.
Corrosion Allowance .095“ = 14.2 Yrs. Remaining Life
Corrosion Rate .0067"

12. Solution
Just Milling Around
Converting the following decimal fractions into Mils.
.195” = 195 Mils
.023 = 23 Mils
.009 = 9 Mils.
.0011 = 1.1 Mils.
Corrosion Allowance 95 Mils = 14.2 Yrs. Rem. Life
Corrosion Rate 6.7“ Mils/Yr.

13. Time
A year has 12 parts
The next issue is how do you handle the odd months in
a year. Its all good when it has been an even number of
years. However what about 2 years and 5 months?
Five months is 5/12th
of a year. 5/12 = .416 years
rounded to 3 places. Of course 6 months is then 6/12 or
.5 years.
Example: Based on 4.416 years and a metal loss of 50
Mils we have;
Metal Loss 50 Mils = Corrosion Rate = 11.3 Mils/Yr.
Time 4.416 Years

14. Class Quiz # 25
Time
1. A vessel shell course was replaced in October 1st
of
1980. The initial or, baseline thickness was .505”. The
most recent actual thickness measured was .485” in
January 1st
1991. What is the corrosion rate, based in
Mils per year in January of 1991.

15. Solution
.505(t-initial) - .485(t-actual) = .020” = 20 Mils
Oct.1st
1990 – Oct.1st
1980 = 10 years plus November,
December of 1990 and January of 1991 so; 10 yrs. 3 Mos.
10 - 3/12th
(3/12 = .250) = 10.25 years
Therefore;
Metal Loss 20 Mils = Corrosion Rate = 1.95 Mils/Yr.
Time 10.25 Years

16. Yes, in the API 510 there are, the Long Term and Short
Term. Here is the Long Term. From API 510 6.4 page 6-3
The long term (L.T.) corrosion rate shall be calculated from
the following formula:
Corrosion rate (L.T.) =
Can there be two Corrosion Rates?
t(actual)andt(inital)between)time(years
t(actual)-t(inital)

17. The short term (S.T.) corrosion rate shall be calculated from the
following formula:
Corrosion rate (S.T.) =
Corrosion Example Problems
actualtandprevioustbetween(years)time
actualtprevioust -

18. tinitial = the thickness, in inches(millimeters), at the same
location as t actual measured at the initial installation or at
the commencement of a new corrosion rate environment.
tprevious = the thickness, in inches(millimeters), at the same
location as t actual measured during a previous inspection.
Long-term and short-term corrosion rates should be
compared to as part of the data assessment. The authorized
inspector, in consultation with a corrosion specialist, shall
select the corrosion rate that best reflects the current
process.
Corrosion Example Problems

19. Example:
A vessel shell had a second set of ultrasonic thickness
measurements after 1 year of service, the original baseline
wall thickness was 0.500”, and the second set revealed that
the shell was now at 0.489”. Five years later a third set of
wall readings were taken and the shell was measured to be
0.459”.
What value should be used in the Remaining Life
calculations? Comparing short term to long term corrosion
we find the following values. Normally the most aggressive
will be used and in this case it will be the Long Term
Corrosion Rate.
Corrosion Example Problems

20. L.T. = = 6.8 Mils/ year
S.T. = = 6 Mils/ year
The remaining life of the vessel shall be calculated from the
following formula:
Remaining life (years) = (6.4)
Corrosion Example Problems
years6
0.4590.500-
years5
0.459-0.489
ratecorrosion
requiredt-actualt

21. tactual = the actual minimum thickness, in inches determined
at the time of inspection for a given location or component.
trequired = the required thickness in inches at the same
location or component as the t actual measurement
computed by the design formulas (e.g. , pressure and
structural) before corrosion allowance and manufacturer’s….
Example: Determine Remaining life of the vessel shell
course in the example above. T required thickness of the
shell course is 0.388” (known as t minimum). Compare S.T. and
L.T. corrosion rates as follows:
S.T. rate = 0.006” or 6 Mils a year
L.T. rate = 0.0068” or 6.8 Mils a year
Corrosion Example Problems

22. Therefore we will use the most aggressive corrosion rate
found to be the Long Term Rate.
Remaining life (years)
What would be the maximum length of time before the next
inspection?
ANS: ½ the Remaining Life or 10 years whichever is less.
Therefore: 17.79/2 = 8.895 years
Corrosion Example Problems
Yrs
Mils
Mils
79.17
8.6
121
"0068.0
"121.0
yeara0.0068"
0.338"-0.459"
===

23. What would be the maximum length of time before the next
inspection? From: API 510 page 6-2
6.4 Internal An On-Stream Inspection
The period between internal or on-stream inspections shall
not exceed one half the estimated remaining life of the
vessel based on corrosion rate or 10 years, whichever is
less. In cases where the remaining safe operating life is
estimated to be less than 4 years, the inspection interval
may be the full remaining safe operating life up to a
maximum of 2 years.
ANS: ½ the Remaining Life or 10 years whichever is less.
Therefore: 17.79/2 = 8.895 years
Corrosion Example Problems

24. (a) (the second a) For a corroded area of considerable size
in which the Circumferential Stress Govern, the least
thickness along the most critical element of the area may be
averaged over a length not exceeding the following:
1. For vessels with the inside diameters less than or equal to
60 inches one half the vessel diameter or 20 inches
whichever is less.
2. For vessels with the inside diameters greater than 60
inches one third the vessel diameter or 40 inches whichever
is less.
Example 5.7 Page 5-3
Corrosion

25. For a corroded area of considerable size in which the
Circumferential Stress Govern, the least thickness along
the most critical element of the area may be averaged over
a length not exceeding the following:
1. For vessels with the inside diameters less than or equal to
60 inches one half the vessel diameter or 20 inches
whichever is less.
2. For vessels with the inside diameters greater than 60
inches one third the vessel diameter or 40 inches whichever
is less.
Corrosion and Minimum Thickness Evaluation
5.7

26. When the area contains an opening the distance on either side
of the opening within which the thicknesses are averaged shall
not extend beyond the limits of the reinforcement as defined in
the ASME Code.
Corrosion and Minimum Thickness
Evaluation
See Fig. UG-37 and refer to reinforcement calculations for
determining the extra metal in the shell.
For Limits:
d ( t – tr
) or 2(t + tn
) ( t – tr
)
Use the Larger

27. Widely scattered pits may be ignored as long as the
following are true:
Corrosion and Minimum Thickness
Evaluation
1. The remaining thickness below the pit is greater than
one-half the required thickness. (1/2 t required)
1.0” – 0.125” = 0.875” the required thickness
Therefore no pit equal to 0.875”/2 = 0.4375” or greater!
Pit #3  0.402 < 0.4375” Acceptable.

28. 2. The total area of the pits does not exceed 7 square inches
(45 sq. centimeters) within any 8 inch (20 centimeter)
diameter circle.
We will assume the pits to be circles and use the formula Pi
x Radius Squared to determine the area of each pit.
Pit # One 3.141 x (0.170/2) 2
= 0.085 2
= 0.0226 sq.
in.
Pit # Two 3.141 x (0.330/2) 2
= 0.165 2
= 0.0855 sq. in.
Pit # Three 3.141 x (0.250/2) 2
= 0.125 2
= 0.0490 sq.
in.
Pit # Four 3.141 x (0.377/2) 2
= 0.1185 2
= 0.1116 sq.
in.
Total: 0.2678 sq. in.
less than 7 sq. inches Acceptable!
Corrosion and Minimum Thickness Evaluation

29. 3. The sum of their dimensions along any straight line within
the circle does not exceed 2 inches
Straight line for pits 1-3-4
#1. 0.170”
#3. 0.250”
#4. 0.377”
Total 0.797” < 2 inches Acceptable
When the surface at a weld with a joint factor other than 1.0
as well as the surfaces remote from the weld is corroded, an
independent calculation using the appropriate weld joint
factor must be made to determine if thickness at the weld or
remote from the weld governs the allowable working
pressure.
Corrosion and Minimum Thickness
Evaluation

30. For this calculation the surface at a weld includes 1 inch on
either side of the weld, (measured from the toe of the weld),
or twice the minimum thickness on either side of the weld,
whichever is greater.
Corrosion and Minimum Thickness
Evaluation

31. When measuring the corroded thickness of ellipsoidal
and torispherical heads the governing thickness may be
as follows:
1. The thickness of the knuckle region with the head rating
calculated by the appropriate head formula.
2. The thickness of the central portion of the dished region in
which case the dished region may be considered to be a
spherical segment whose allowable pressure is
calculated by the Code formula for spherical shells.
Corrosion and Minimum Thickness
Evaluation

32. 3. For calculating the spherical portion using the spherical
formula the following applies to find the spherical radius L:
a. L = the OD of the Torispherical head
b. L = the ID of the shell times 90% for 2 to 1 Ellipsoidal
heads
See 5.7 second e.
Corrosion and Minimum Thickness
Evaluation

33. Corrosion Over
And Out…