Problem Of
The Month
April 1997--Tank Bottom Corrosion
Oil storage tanks are expected to be leak tight for long periods of time. Unfortunately oil contains many contaminates which corrode the tank bottom. The economic issue is: 1) When to inspect the bottom (which must be done from the tank interior), and 2) When is the due date for tank replacement. The corrosion effects for this tank are believed due to Cl effects. Plan for the number of pits to grow as k1*t^2 and the depth of pits to grow as k2*t^3.
The Problem
The bottom of an oil storage tank was inspected for corrosion using ultrasonic
devices to study pit depths. The tank was marked-off in 1/4 zones. Seven measurements
were taken in each quadrant by examining the worst 7 pits based on a visual
inspection. The tank has been in service for 10 years. The maximum allowable
pit depth is 0.275 inches. The pit depth data includes some questionable
measurements as shown below:
|
Pit Depths--> |
North 0.065 |
East 0.093 |
South 0.100 |
West 0.050 |
|
Mean
= |
0.1016 |
0.1114 |
0.1140 |
0.1100 |
Notice the traditional measures of central tendency (Mean = average, Mode = the
largest number of observed data points, and Median = half of the data points
lie each side) have considerable variance and give a clue that the data has
long tails.
Questions:
1) Which distribution best fits the data? How do we decide?
Normal_____________
Log-normal__________
Weibull_____________
Distribution decision
method_______________
2) Is the data from each position significantly different from each of
the locations and can we pool the data? How do we decide?
Yes___________
No____________
How do we decide if all four data
sets are the same__________________
3) Given the data is representative of bad pits, what risks exist of pits
exceeding the maximum allowed?
Risk is
_____________________
4) If a leak occurs, the clean-up costs will be $500,000. What is our exposure
today? In 5 more years? What is the optimum replacement interval?
Risk today for a leak in
$'s______________
Risk in 5 more years in
$'s_______________
Optimum replacement
interval____________
5) When should we schedule the next inspection based on the cost of
incurring a leak?
At 10 years of age in
$'s________
At 15 years of age in
$'s________
At 20 years of age in
$'s________
At 25 years of age in
$'s________
At 30 years of age in
$'s________
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Solutions:
Answer For Question #1:
Which distribution best fits the data? How do we decide?
The (regression coefficient squared minus the critical correlation coefficient squared) = (r^2 - ccc^2) is used for deciding goodness of fit. A distribution analysis of the data shows:
|
|
North |
East |
South |
West |
|
Best Fit = |
Log-normal |
Log-normal & Normal |
Log-normal |
Weibull |
Three out of four show log-normal distributions have a better fit. (For the East data, log-normal and normal have the same value for the (r^2 - ccc^2) goodness of fit criteria--see the March 97 problem for details of how the criteria is used.) Therefore choose the log-normal distribution.
Answer For Question #2:
Is the data from each position significantly different from each of the
locations and can we pool the data? How do we decide?
WinSMITH Weibull is used to
calculate zones of uncertainty for the measure of central tendency and the
slope factor. WinSMITH Visual
is used to plot the uncertainty zones. If the uncertainty zones overlap, the
data sets are not significantly different.
Data from the East and South positions overlap each other for the slope factor and the characteristic pit depth--and they in turn are overlapped by the data from the North position. This says the data are different but not significantly different.
Likewise, data from the West position overlaps the data from the North position. Notice the zones almost overlap the East and South positions. For practical engineering purposes, we will conclude the zones overlap--remember we have questions about data measurements from all data sets except for the East which has the smallest size boundary of uncertainty.
From these overlapping boundaries, we can conclude the data is different in each of the locations but not significantly different. So for practical purposes, pool the results. Merge the data sets and treat the four data sets as one single set using the criteria that the boundaries overlap and thus no significant difference exist.
As you would suspect, the boundary for pooling all of the data into one data set will lie in the area where the four boundaries are closest together. The size of the boundary for the pooled data set will be larger than the smallest boundary but not by very much--we've got less uncertainty from the larger number of data points.
Answers For Question #3:
3) Given the data is representative of bad pits, what risks exist of pits
exceeding the maximum allowed?
A WinSMITH Weibull
probability plot of the existing data describing ten years of experience
predicts that 99.98% of the data will be less than 0.275 inches deep. Based on
the trend line, the risk of exceeding the maximum allowed pit depth of 0.275
inches is 0.01999664% as found by using the prediction feature of WinSMITH
Weibull.
Confidence limits can also be used with the prediction feature of WinSMITH Weibull software. Using 90% overall confidence limits, the right-hand limit confidence limit will occur at 95% confidence (!remember5% out of bounds on the right and 5% out of bounds on the left to achieve a 90% overall confidence interval!) and the risk grows to 0.21% for a pit depth of 0.275 inches.
The value of projecting the risk of exceeding the allowed pit depth is to get facts on the table. We need facts rather than emotion about risk . Emotional reactions to pit depths can quickly reach hysterical magnitude that the "sky is falling"--when in fact, it is not.
Now the next important issue is to make forecasts of pit depth as time increases. The straight line on the probability chart helps make predictions into future years using the relationship that pit depth increases as k2*t^3. Using the characteristic pit depth and the corrosion depth with time for the 10 year history we can solve the corrosion pit depth equation for k2 as 0.1059 = k2*(10)^3 where k2 = 0.1059/1000 = 0.0001059.
Answers For Question #4:
If a leak occurs, the clean-up costs will be $500,000. What is our
exposure today? In 5 more years? What is the optimum replacement interval?
The forecasted characteristic pit depth (i.e., MuAL) for corrosion at various times is given in the following table along with the failure risks of exceeding the trend line depth of 0.275 inches (this is similar to drawing parallel lines to the data trend line and each new line passes through the characteristic values of pit depth shown in the table below and % risks are judged against exceeding the 0.275 inch deep pit limit. The risks for a spill in monetary values are (% risk of failure)*($ exposure = $500,000) are also shown:
|
Time (years) Char.Pit Depth (in.) Failure Risks (%) Spill Risk ($'s) |
5 0.0132 0.00000 0 |
10 0.1059 0.019996 50 |
11 0.1409 0.569999 100 |
12 0.1830 6.18 30,900 |
13 0.2327 26.37 131,850 |
14 0.2906 58.18 290,900 |
15 0.3574 83.86 419,300 |
16 0.4338 95.77 478,850 |
17 0.5203 99.21 496,050 |
18 0.6176 99.89 499,450 |
Note the pit depth is calculated from the equation (pit depth) = k2*t^3 = 0.0001059*t^3. Usually the maximum pit depth is associated with a minimum wall thickness. The minimum wall thickness has a safety allowance which exceeds the "drop dead value" and it is not always necessary to add "belts and suspenders" to the nominal calculations.
The optimum replacement interval depends on the replacement costs (which is not given), costs for an unplanned outage, and the mean time to failure--so we can't calculate the optimum replacement interval. Suppose the planned replacement cost is $200,000. Suppose the unplanned failure is $700,000 ($200,00 for replacement equipment and $500,000 for cost of the spill). These costs are shown in the following table which can be plotted to arrive at the minimum interval.
|
Time (years) Failure Risks (%) Planned ($'s) Unplanned ($'s) Total ($'s) Total ($'s/yr) |
5 0.00000 200,000 0 200,000 40,000 |
10 0.019996 199,960 140 200,100 20,010 |
11 0.569999 198,860 3,990 202,850 18,441 |
12 6.18 187,640 43,260 230,900 19,242 |
13 26.37 147,260 184,590 331,850 25,527 |
14 58.18 83,640 407,260 490,900 35,064 |
15 83.86 32,280 587,020 619,300 41,287 |
16 95.77 8,460 670,390 678,850 42,428 |
Notice the planned costs for each time period are $200,000*reliability and unplanned costs are $500,000*unreliabilty---of course the total costs are the sum of these elements. The costs per year are found by dividing the years of service into the total costs to get $'s/yr--almost everyone understands a plot of $'s/yr vs years of service.
Why not use the optimum replacement equation build into WinSMITH Weibull? Remember the data shown above does not provide an age-to-failure based on actual failure data. Thus we don't have a specific probability curve to plug into the WinSMITH Weibull equation driven by detailed ages to failure. This means we must infer the results rather than relying on factual age-to-failure facts.
Answers For Question #5:
When should we schedule the next inspection based on the cost of incurring a
leak?
The answer depends on how much risk we can afford and the replacement cost. This helps set the exposure which drives the drop dead last time to inspect--for a practical answer, plan to inspect at the mid-point of the interval between now (10 years) and the replacement time so that enough time exists to handle the uncertainty. For the assumptions given in A4, the optimum time for replacement is between 11 and 12 years which says make another inspection at 11 years (and have the replacement parts on order!!!).
Comments:
The action plan is dependent upon the corrosion model for the rate at which pit depths increase. If the model was linear--the action plan would be substantial different than for the cubic model for pit depth. Technical solutions are interesting but marry the technical details with costs and the appetite for acceptance/rejection of business risks and strategies for the business. You'll never have enough data for a risk free decision.
By the way:
Thanks to Mr. Ashok H. Rakhe who works with Hoechst Celanese in Corpus Christi, TX for finding an error in the last table of answer #4 on April 2, 1997 concerning the Total ($'s/yr) which has been corrected as shown above.
Thanks to Mr. Bob Hart who works Conoco Refinery in
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