B31G and ILI Tolerances Exercises

Introduction

In this example we will go through several scenarios on how to use ILI results and B31G to make decisions about a pipeline. It will cover anomaly classification, time to failure (TTF), pressure testing and repair decisions as well as the use of ILI tool tolerances. After each scenario there will be an exercise to try on your own. Solutions are on the last tab.

Anomaly Classification

The classification of type of metal loss anomaly depends on the length and width relative to the wall thickness. When an ILI vendor reviews the results of an ILI tool the pipeline operator will receive a report classifying the different types of anomalies, the metal loss anomalies are placed into one of several categories. Figure 1 shows how the length and width of a metal loss anomaly affects the classification. For wall thickness less than 0.394 inches the variable “A” is equal to 0.394", for greater than that, use the actual wall thickness. Typical inline inspection tolerances is \(\pm\) 10%, 80% of the time for metal loss anomalies. It should be noted that is not always the case. Depending on the type of anomaly it can have different tolerances. Besides the normal tool tolerances there are several other key factors that can also affect the performance. Among these are the tool speed, the faster the tool the more likely the tool will under-report the depth of the anomaly. The tolerances for the various types of anomalies is listed in Table 1.

Figure 1

Figure 1

Probability of Detection Size and Accuracy
Classification POD Size (L xW) ± Accuracy
General 1.58 x 1.58 0.1
Pitting 0.79 x 0.79 0.1
Axial Grooving 1.58 x 0.79 0.15
Circ. Grooving 0.2 x 0.2 0.1
Pinhole 0.2 x 0.2 0.1
Circ. Slotting 0.2 x 0.79 NA
Axial Slotting 0.79 x 0.2 NA

Exercise 1

The following information is provided from the ILI vendor and the company GIS:

  • Anomaly Length = 3 inches
  • Anomaly Width = 1.0 inches
  • Anomaly Depth = 40%
  • Diameter = 24 inches
  • Nominal Wall Thickness = 0.250 inches
  • SMYS = X52
  1. What is the anomaly classification given the above information?
  2. Considering the tool tolerances for that type of anomaly, what other anomaly type could it be?
  3. If the anomaly classification changed, what tolerances would you use?

Tool Tolerances

Table 1 is a list of typical tolerances for MFL tools. The “Accuracy” variable is the percent of wall thickness

Probability of Detection Size and Accuracy
Classification POD Size (L xW) ± Accuracy
General 1.58 x 1.58 0.1
Pitting 0.79 x 0.79 0.1
Axial Grooving 1.58 x 0.79 0.15
Circ. Grooving 0.2 x 0.2 0.1
Pinhole 0.2 x 0.2 0.1
Circ. Slotting 0.2 x 0.79 NA
Axial Slotting 0.79 x 0.2 NA

Excercise 2

Using the information from Exercise 1 and the classification, calculate the B31G failure pressure based on the tolerances for that classification.

  • Anomaly Length = 3 inches
  • Anomaly Width = 1.0 inches
  • Anomaly Depth = 40%
  • Diameter = 24 inches
  • Nominal Wall Thickness = 0.250 inches
  • SMYS = X52

The equations for Modified B31G are provided below.
\[\begin{equation} \tag{Eq1} \sigma_f=(\sigma_y+10ksi)\frac{1-0.85\frac{d}{t}}{1-0.85\frac{d}{t}\frac{1}{M}} \end{equation}\]

\[\begin{equation} \tag{Eq2} M=\sqrt{1+0.6275\left(\frac{L}{\sqrt{Dt}}\right)^2-0.003375\left(\frac{L}{\sqrt{Dt}}\right)^4}\qquad for\space \bigg(\frac{L}{\sqrt{Dt}}\bigg)^2\leq50 \end{equation}\]

\[And\] \[\begin{equation} \tag{Eq3} M=0.032\left(\frac{L}{\sqrt{Dt}}\right)^2+3.3 \qquad for\space \left(\frac{L}{\sqrt{Dt}}\right)^2 >50 \end{equation}\]

Excercise 3

Using the same information from exercise 1 determine the depth at which failure will occur and the number of years to get there. Assume 9 mpy corrosion growth rate the length is constant. The MAOP is based on a Class 1 design factor. Failure is assumed to happen when the failure pressure equals the MAOP. The variable \(P_S\) is the pressure at \(SMYS + 10\; ksi\). This can be solved iteratively by trying different depths comparing the failure pressure each time to the MAOP or it can be solved directly using Equation 4.

After calculating the failure depth determine the time to failure (TTF) in years. Is this anomaly safe to leave until the next reassessment in 7 years?

\[\begin{equation} \tag{Eq 4} d=\left[ \frac{1.1765 M t \left(- MAOP + P_S\right)}{M P_S - MAOP}\right] \end{equation}\]

Solutions

Exercise 1 Solution

Since the wall thickness is less than 0.394 inches the “A” variable would be 0.394. Therefore the L/A (Length to A) would be \(\frac{3}{0.394} =5.1\) and the W/A (Width to A) would be \(\frac{1.1}{0.394} = 2.8\). If you plot that point on the anomaly it would be considered pitting. Once the tool tolerances are applied the anomaly could be classified as any one of 3 classifications (Pitting, General, or Axial Grooving).

Exercise 2 Solution

Since the potential tolerance limits of the tool call put it in one of several classifications, you would use the most conservative tolerances of Axial Grooving of \(\pm 15\%\). Based on this, the depth would increase from \(40\%\) to \(55\%\). Using the updated depth the calculations would be as follows.

\[\bigg(\frac{L}{\sqrt{Dt}}\bigg)^2 = \bigg(\frac{2.5}{\sqrt{24 \times 0.250}}\bigg)^2=1.5\]
Since the \(\frac{L^2}{Dt}\) is less than 50 the first version of the folias factor \(M\) is applicable. Therefore:
\[M=\sqrt{1 + 0.6275\left(\frac{3}{\sqrt{24 \times 0.250}}\right)^2-0.003375\left(\frac{3}{\sqrt{24\times 0.250}}\right)^4}=1.39\]

\[\sigma_f=(52+10)\Bigg[\frac{1-0.85\frac{0.55 \times 0.250}{0.250}}{1-0.85\frac{0.55 \times 0.250}{0.250}\frac{1}{1.16}}\Bigg]=55.22\]

The above equation will give the failure stress of the anomaly in ksi. To Convert this to a failure pressure, use the rearranged Barlow’s equation to arrive at Equation 5.

\[\begin{equation} \tag{Eq 5} P=\frac{2\,\sigma\,t}{D}=\frac{2\times 55.22 \times 0.250}{24}=1.725\ ksi\;(1725\;psi) \end{equation}\]

Exercise 3 Solution

The first step is to calculate the design MAOP using Equation 5 and a design factor for a Class 1 location = 0.72 and the operating pressure at 100% of SMYS + 10 ksi.

\[P=\frac{2\,\sigma\,t}{D}=\frac{2\times 52 \times 0.250}{24}\times 0.72=0.780\; ksi\;(780\;psi)\]
\[P_S=\frac{2\times 62 \times 0.250}{24}=1.292\; ksi\;(1292\;psi)\] The next step is to calculate the depth at failure depth using Equation 4 and the folias factor from Exercise 2.

\[d=\left[ \frac{1.1765 M t \left(- MAOP + P_S\right)}{M P_S - MAOP}\right]=\left[ \frac{1.1765(1.39)(0.250) \left(- 780 + 1292\right)}{(1.39)(1292) - 780}\right]=0.206\;in\]

After calculating the failure depth, the TTF is the difference between the current depth and failure depth divided by the CGR. Don’t forget to add in the tool tolerances.

\[TTF=\frac{(0.55*0.250)-0.206}{0.009} =7.6\; years\]

Even though the time to failure is greater than the proposed 7 year reassessment interval, the B31G equations are not perfect and the ILI tool is certainly not perfect either. So the reassessment interval is normally set at half the TTF to allow a 2:1 factor of safety. So the reassessment interval would be in 3 years (intervals are calculated based on whole years <= TTF) or this anomaly would have to be remediated to increase the reassessment interval. The reason the reassessment interval was set to 3 years and not 3.8 is because planning for tool runs is based on annual budgets and there can be considerable lead times depending on tool availability and seasonal flow conditions on a give line.