Ultrasonic imaging algorithm for the health monitoring of pipes

Abstract

This paper presents a numerical study where guided ultrasonic waves (GUWs) are processed using a new imaging algorithm for the health monitoring of pipes. The numerical model mimics the generation and detection of guided waves from the transducers of two annular arrays located inside a pipe. The fastest mode of the detected signals is processed using the continuous wavelet transform and the Hilbert transform to extract two damage-sensitive features. The estimation of the features for the healthy condition is formulated in terms of an optimization problem solved with a competitive optimization algorithm. Finally, a probabilistic approach is used to create an image of the pipe to reveal the presence of possible structural anomalies. With respect to most GUW-based imaging methods, the proposed approach is baseline-free which means that data from pristine pipes are not necessary. A commercial finite element code is utilized to mimic the operation of the two arrays each made of four ultrasonic transducers in contact with the interior wall of a pipe. While one of the transducers acts as transmitter, the other seven act as sensors, and this is repeated for all the elements of the rings. The time waveforms associated with all the possible actuator–sensor pairs are processed using the algorithm proposed here. To demonstrate the advantages of the proposed approach, the findings are compared to the results obtained using a conventional pitch-catch approach. The results are promising and future studies should focus on the experimental validation of the methodology.

Appendix: Application of the competitive optimization algorithm

First, we define the features following the method presented in Sect. 4.2. For example, the results associated with the CWT analysis using Eq. (16) applied to the 3rd sensing path (actuator–sensor path length equal to 200 mm) are

$$\begin{aligned} F_{1}^{3} & = 1.076 \times 10^{ - 12} ,F_{2}^{3} = 2.911 \times 10^{ - 12} ,F_{3}^{3} = 2.284 \times 10^{ - 12} ,F_{4}^{3} = 2.825 \times 10^{ - 12} \\ F_{5}^{3} & = 1.079 \times 10^{ - 12} ,F_{6}^{3} = 2.964 \times 10^{ - 12} ,F_{7}^{3} = 2.292 \times 10^{ - 12} ,F_{8}^{3} = 2.815 \times 10^{ - 12} . \\ \end{aligned}$$

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We have eight features because they refer to paths AE, BF, CG, DH and vice versa. Then, the objective function for the same paths is determined applying Eq. (21):

$$f(e_{1}^{3} ,e_{2}^{3} , \ldots ,e_{8}^{3} ,F_{h}^{3} ) = \left\| {\left[ {\begin{array}{*{20}c} {e_{1}^{3} } & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & {e_{2}^{3} } & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & {e_{3}^{3} } & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {e_{4}^{3} } & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & {e_{5}^{3} } & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & {e_{6}^{3} } & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & {e_{7}^{3} } & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & {e_{8}^{3} } \\ \end{array} } \right]\left\{ {\begin{array}{*{20}c} {1.076} \\ {2.911} \\ {2.284} \\ {2.825} \\ {1.079} \\ {2.964} \\ {2.292} \\ {2.815} \\ \end{array} } \right\}10^{ - 12} - F_{h}^{3} \left\{ {\begin{array}{*{20}c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{array} } \right\}} \right\|^{2} .$$

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$$\begin{aligned} f(e_{1}^{3} ,e_{2}^{3} , \ldots ,e_{8}^{3} ,F_{h}^{3} ) = & \left( {1.076 \times 10^{ - 12} \times e_{1}^{3} - F_{h}^{3} } \right)^{2} + \left( {2.911 \times 10^{ - 12} \times e_{2}^{3} - F_{h}^{3} } \right)^{2} \\ & + \left( {2.284 \times 10^{ - 12} \times e_{3}^{3} - F_{h}^{3} } \right)^{2} + \left( {2.825 \times 10^{ - 12} \times e_{4}^{3} - F_{h}^{3} } \right)^{2} + \left( {1.079 \times 10^{ - 12} \times e_{5}^{3} - F_{h}^{3} } \right)^{2} \\ & + \left( {2.964 \times 10^{ - 12} \times e_{6}^{3} - F_{h}^{3} } \right)^{2} + \left( {2.292 \times 10^{ - 12} \times e_{7}^{3} - F_{h}^{3} } \right)^{2} + \left( {2.815 \times 10^{ - 12} \times e_{8}^{3} - F_{h}^{3} } \right)^{ + 2} \\ \end{aligned}$$

(33)

The objective function is subjected to the following constraints (Eq. 22b):

$$1.076 \times 10^{ - 12} \le F_{h}^{3} \le 2.964 \times 10^{ - 12} ,0.363 < e_{j}^{3} < 2.755.$$

(34)

To solve the function, we choose the following number of countries, empires and colonies:

$$N_{Country} = 50,N_{imp} = 5,N_{col} = N_{Country} - N_{imp} = 45.$$

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Randomly, the following countries are produced:

$$\begin{aligned} Country_{1} & = \left[ {e_{1}^{3} ,e_{2}^{3} , \ldots ,e_{8}^{3} ,F_{h}^{3} } \right] \\ Country_{1} & = \left[ {2.924,1.188,2.936,2.908,1.894,2.571,1.155,1.757,2.81 \times 10^{ - 12} } \right] \\ Country_{2} & = \left[ {2.553,2.913,2.259,0.926,2.675,2.858,2.309,2.479,2.479 \times 10^{ - 12} } \right] \\ & . \\ & . \\ Country_{50} & = \left[ {1.693,2.259,1.218,2.368,0.918,1.445,0.949,1.058,2.631 \times 10^{ - 12} } \right]. \\ \end{aligned}$$

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Next, the cost of the countries is obtained by means of the objective function in Eq. (33):

$$\begin{aligned} cost_{1} = & \left( {1.076 \times 10^{ - 12} \times 2.924 - 2.81 \times 10^{ - 12} } \right)^{2} + \left( {2.911 \times 10^{ - 12} \times 1.188 - 2.81 \times 10^{ - 12} } \right)^{2} \\ & + \left( {2.284 \times 10^{ - 12} \times 2.936 - 2.81 \times 10^{ - 12} } \right)^{2} + \left( {2.825 \times 10^{ - 12} \times 2.908 - 2.81 \times 10^{ - 12} } \right)^{2} \\ & + \left( {1.079 \times 10^{ - 12} \times 1.894 - 2.81 \times 10^{ - 12} } \right)^{2} + \left( {2.964 \times 10^{ - 12} \times 2.571 - 2.81 \times 10^{ - 12} } \right)^{2} \\ & + \left( {2.292 \times 10^{ - 12} \times 1.155 - 2.81 \times 10^{ - 12} } \right)^{2} + \left( {2.815 \times 10^{ - 12} \times 1.757 - 2.81 \times 10^{ - 12} } \right)^{2} = 73.4 \times 10^{ - 24} \\ cost_{2} = & 107.5 \times 10^{ - 24} \\ . & \\ . & \\ cost_{50} = & 38.5 \times 10^{ - 24} . \\ \end{aligned}$$

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The countries with the minimum cost are selected as imperialists. As we choose to establish one imperialist every ten colonies, the five countries with the least cost are selected. Not shown here, for the analysis conducted in this study, Country ₁₃ , Country ₂₉, Country ₂₃, Country ₄₁, and Country ₈ are selected as imperialist c ₁ to c ₅, respectively, and the remaining countries constitute the initial colonies. It should be emphasized that a new analysis may result in other countries for empires, because they are randomly generated. The cost of each imperialist is

$$\begin{aligned} c_{1} & = cost_{13} = 6.3 \times 10^{ - 24} ,c_{2} = 13.2 \times 10^{ - 24} ,c_{3} = 17.9 \times 10^{ - 24} \\ c_{4} & = 20.9 \times 10^{ - 24} ,c_{5} = 29.3 \times 10^{ - 24} . \\ \end{aligned}$$

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The normalized cost is determined by applying Eq. (3):

$$\begin{aligned} C_{1} & = 6.3 \times 10^{ - 24} - 29.3 \times 10^{ - 24} = - 23 \times 10^{ - 24} \\ C_{2} & = - 16.1 \times 10^{ - 24} ,C_{3} = - 11.4 \times 10^{ - 24} ,C_{4} = - 8.4 \times 10^{ - 24} ,C_{5} = 0. \\ \end{aligned}$$

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Equation (4) determines the normalized power of each imperialist, i.e.:

$$\begin{aligned} p_{1} & = \left| {\frac{{ - 23 \times 10^{24} }}{{ - 58.9 \times 10^{24} }}} \right| = 0.391 \\ p_{2} & = 0.273,p_{3} = 0.193,p_{4} = 0.143,p_{5} = 0.0. \\ \end{aligned}$$

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Then, the initial number of colonies belonging to each empire is computed:

$$\begin{aligned} N.C._{1} & = round\left\{ {0.391 \times 45} \right\} = 18 \\ N.C._{2} & = 12,N.C._{3} = 9,N.C._{4} = 6,N.C._{5} = 0. \\ \end{aligned}$$

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Thus, the colonies are randomly assigned to a certain empire based on the empire’s number of colonies N.C.To begin the assimilation process, the distances between the colonies and the imperialists are calculated to determine the new position of the colonies. For example, the distance l ₁ between the Imperialist ₁ and its first colony is

$$\begin{aligned} l_{1} & = Imperialist_{1} - Colony_{1} \\ l_{1} & = [ - 1.211, - 0.176, - 1.571, - 1.793, - 0.648, - 1.205, - 0.592, - 0.800, - 2.46 \times 10^{ - 14} ], \\ \end{aligned}$$

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and Δl ₁ is (see Eq. 6):

$$\begin{aligned} \Delta l_{1} & = \gamma \times l_{1} \times rand[01] = 2 \times l_{1} \times rand[01] \\ \Delta l_{1} & = [ - 1.994, - 0.244, - 0.996, - 3.407, - 0.045, - 1.057,0.452, - 1.225, - 3.23 \times 10^{ - 14} ]. \\ \end{aligned}$$

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The constant γ must be greater than one. We choose γ = 2 by trial and error method. This number affects the number of iterations but it does not influence the final results. Therefore, the new position of the colony is

$$\begin{aligned} Colony_{1}^{new} & = Colony_{1} + \Delta l_{1} \\ Colony_{1}^{new} & = [0.931,0.945, 1.941, 0.499, 1.849, 1.513, 1.607, 0.532,2.77 \times 10^{ - 12} ]. \\ \end{aligned}$$

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The assimilation process is iterated for each colony belonging to the Imperialist ₁ and then repeated for the other imperialist-colonies systems. Once the assimilation process is completed, the cost of each colony is computed and then the total cost of an empire is determined using Eq. (7). For the example presented in this appendix, we have

$$\begin{aligned} T.C._{1} & = 6.3 \times 10^{ - 24} + 0.02\frac{{\sum\nolimits_{k = 1}^{18} {f\left( {Colony_{k} } \right)} }}{18} = 6.76 \times 10^{ - 24} \\ T.C._{2} & = 13.54 \times 10^{ - 24} ,T.C._{3} = 18.24 \times 10^{ - 24} ,T.C._{4} = 21.14 \times 10^{ - 24} ,T.C._{5} = 29.3 \times 10^{ - 24} . \\ \end{aligned}$$

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For the competition process, the normalized total cost of an empire is computed

$$\begin{aligned} N.T.C._{1} & = 6.76 \times 10^{ - 24} - 29.3 \times 10^{ - 24} = - 22.54 \times 10^{ - 24} \\ N.T.C._{2} & = - 15.76 \times 10^{ - 24} ,N.T.C._{3} = - 11.06 \times 10^{ - 24} ,\\N.T.C._{4} &= - 8.16 \times 10^{ - 24} ,N.T.C._{5} = 0. \\ \end{aligned}$$

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And the possession probability of the empires is calculated based on Eq. (9).

$$\begin{aligned} p_{{p_{1} }} & = \left| {\frac{{ - 22.54 \times 10^{ - 24} }}{{ - 57.52 \times 10^{ - 24} }}} \right| = 0.392 \\ p_{{p_{2} }} & = 0.274,p_{{p_{3} }} = 0.192,p_{{p_{4} }} = 0.142,p_{{p_{5} }} = 0. \\ \end{aligned}$$

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Thus, the vector P (see Eq. 10) is

$${\mathbf{P}} = \left[ {0.392,0.274,0.192,0.142,0} \right].$$

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Next, the vector R, which is formed by uniformly distributed random numbers comprised between 0 and 1, is generated:

$${\text{R}} = \left[ {0.876, 0.301,0.226,0.923,0.438} \right].$$

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Finally, the vector D is computed as:

$${\text{D}} = {\text{P}} - {\text{R}} = \left[ { - 0.484, - 0.027, - 0.034, - 0.781, - 0.438} \right].$$

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Referring to vector D, the weakest colony, i.e. the colony with the smallest number, of the Imperialist ₄ is handed to the Imperialist ₂. Then, the assimilation process is started again. Finally, the algorithm is terminated after 100 iteration at which the feature F ³ _h  = 2.447 × 10⁻¹² is found.

This process is repeated for other sensing paths to find the feature value that is associated with the pristine pipe. Once the other feature values are computed the application of the COA is terminated.