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Application of an Artificial Neural Network Method for the Prediction of the Tube-Side Fouling Resistance in a Shell-And-Tube Heat Exchanger

期刊

FDMP-FLUID DYNAMICS & MATERIALS PROCESSING
卷 18, 期 5, 页码 1511-1519

出版社

TECH SCIENCE PRESS
DOI: 10.32604/fdmp.2022.021925

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Artificial neural network; fouling resistance; phosphoric acid concentration process; shell-and-tube heat exchanger

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In this study, an Artificial Neural Network (ANN) is used to predict the fouling resistance in the phosphoric acid concentration process. By selecting the best training algorithm and the optimal number of hidden neurons, the model accurately predicts the fouling resistance.
The accumulation of undesirable deposits on the heat exchange surface represents a critical issue in industrial heat exchangers. Taking experimental measurements of the fouling is relatively difficult and, often, this method does not lead to precise results. To overcome these problems, in the present study, a new approach based on an Artificial Neural Network (ANN) is used to predict the fouling resistance as a function of specific measurable variables in the phosphoric acid concentration process. These include: the phosphoric acid inlet and outlet temperatures, the steam temperature, the phosphoric acid density, the phosphoric acid volume flow rate circulating in the loop. Some statistical accuracy indices are employed simultaneously to justify the interrelation between these independent variables and the fouling resistance and to select the best training algorithm allowing the determination of the optimal number of hidden neurons. In particular, the BFGS quasi-Newton back-propagation approach is found to be the most performing of the considered training algorithms. Furthermore, the best topology ANN for the shell and tube heat exchanger is obtained with a network consisting of one hidden layer with 13 neurons using a tangent sigmoid transfer function for the hidden and output layers. This model predicts the experimental values of the fouling resistance with AARD% = 0.065, MSE = 2.168 x 10(-11), RMSE = 4.656 x 10(-6) and r(2) = 0.994.

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