Tank Load Case 2 Calculations

LC2 calculations takes place on the Equipment. The LC2 calculations (also known as buckling calculation) is to determine the number of wind Girders (a.k.a. stiffeners) required for structural integrity of the tank. Relevant properties are Wind Speed (WS), Design Vacuum (V), Operating Temperature (OT) and tank Diameter (D). The basic approach of the buckling calculation is to calculate transformed Shell height (H_E). Then a Max H_E and an H_p is calculated. The Max H_E is the Maximum of all the Course sections (i.e., “pseudo-Course”). 

The number of Required Girders =

For each stiffener added, the Course is split into more pseudo-Courses, which will be calculated independently, as if they were actual Courses. The Course Heights (CH) are being modified for each pseudo-Course to reflect the original Course split.

Where:

T_min = Minimum thickness

T_act = Actual calculated thickness

Since the calculated thickness is a function of time, to calculate the RL we end up with a third-degree polynomial function, which is solved with bisectional optimization. The starting point is the last inspection date of the Governing Course (GC), while the end point is the worst (closest to LID) fail date of all Courses.

In the calculations a Last Governing Date (LGD) is used, which is the latest inspection date of the GC. This means that during the calculations, we only take those inspection histories that are the same or older as the latest (newest) inspection history of the GC.

For example, if Course 2 is the GC and has inspection dates 2000, 2003, 2005 and 2007, then the LGD is 2007. If Course 1 has inspection dates 2005, 2007 and 2009 then the inspection history of 2007 is used and not the one for 2009. Note: there is no check on the current date. If the current date is 2010, and the Governing Course has an inspection entry for 2012, the LGD is 2012.

The fail date per Course is calculated by taking the maximum of the inspection time failure date and the install time failure date. The failure date is the date when the projected Wall Thickness is 0. The inspection failure time is the lowest (worst case) Wall Thickness of any inspection that is before, or at the same time as the LGD, divided by its CR. The install failure time is simply the Thickness as built (TAB) divided by the CR.

Starting with the initial points, a bisectional optimization is used to converge two points to the location where y = 0, or at least closer than one day. This is taken as the RL point.

Additionally, a failsafe has been built in to ensure that the conversion does not take more than 100 iterations. After this the algorithm will bail out, to avoid an endless loop. Each time the algorithm finds a new point in time, the statistics must be calculated for all pseudo-Courses.

T_min is the lowest T_act of all the pseudo-Courses.

T_act is the lowest thickness of the inspections before the LGD.

The Max H_E is highest H_E of all pseudo-Courses or the total H_E of all (whatever is highest).

The Calculation result is the number of Required Girders =

The H_p calculation depends on the Compliance Code:

H_p API Calculation

Where:

Allowance = 0.2468

Which gives:

H_p DEP Calculation

Where:

E = 1, unless 100 ≤ OT ≤ 300, then

Which gives:

Corrosion Rates (CRs)

The final aspect of the T_act calculations are the CRs used to calculate when a Course has corrosion failure. The CR used for the pseudo-Courses depends on the Extrapolation type. For this purpose, the LC2 calculation uses a Base Corrosion Rate (BCR) and a Ratio of incr. CR / BCR (R).

Extrapolation Types:

Even: This is the simplest scenario. All pseudo-Courses use the BCR.

Top: All Courses use the BCR, except for the top Course, this uses BCR × R.

Middle: In this case, the tank is split up into three sections, the Top, Middle and Bottom. All pseudo-Courses in the Top and Bottom parts will use the BCR, the pseudo Courses in the middle section will use BCR × R.

Usually this will not fit nicely with the existing (pseudo) Courses. For example, if you have a tank of 6 meters (high), 3 Courses, each 2 meters, this will give Top part (Course 3), Bottom part (Course 1) and the Middle part (Course 2). However, if you have a tank of 10 meters (high) and 5 Courses, each 2 meters, you will get 3 sections of 3.33333 meters. So, the Top Section is Course 5 plus part of Course 4, the Bottom section is Course 1 and part of Course 2, and the Middle section consists of parts of Courses 2 and 4 and Course 3.

Since BCR or BCR×R is used, it cannot be simply decided what CR to use for Courses 2 or 4. This is solved by also splitting these Courses into pseudo Courses. However, if a split is calculated within a certain tolerance from the height, it is not seen as a split. The tolerance is set at 15 cm. For example, if Course 2 is 2 meters high and a split is calculated at 1.90 meters, there will be no split, but if it the split is calculated at 1.84m, it will be split.