{'fraction_non_alphanumeric': 0.027624309392265192, 'fraction_numerical': 0.011049723756906077, 'mean_word_length': 5.104868913857678, 'pattern_counts': {'":': 0, '<': 0, '<?xml version=': 0, '>': 0, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 0, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Design of Intelligent Network to Predicate Phenol Removal from Waste Water by Emulsion Liquid Membrane

This research introduced Intelligent Network's proposed design for predicting efficiency in the removal of phenol from wastewater by liquid membrane emulsion. In the inner phase of W / O emulsions, phenol extraction from an aqueous solution was investigated using emulsion liquid membrane prepared with kerosene as a membrane phase, Span 80 as a surfactant, and NaOH as a stripping agent. Experiments were conducted to investigate the effect of three emulsion composition variables, namely: surfactant concentration, membrane phase to-internal (VM / VI) volume ratio, and removal phase concentration in the internal phase, and two process parameters, feed phase agitation speed at organic acid extraction rates, and emulsion-to-feed volume ratio (VE / VF). More than 98% of phenol can be extracted in less than 5 minutes. This article describes compares the performance of different learning algorithms such as GD, RB, GDM, GDX, CG, and LM to predict the efficiency of phenol removal from wastewater through the liquid emulsion membrane. The proposed neural network consisted of (7, 11, 1) neurons in the input , hidden and output layers respectively feed forward ANN with various types of back propagation training algorithms were developed to model the emulsion liquid membrane removal of phenols. The values predicted for the neural network model are found in close agreement with the results of the batch experiment using MATLAB program with a correlation coefficient ( R2) of 0.999 and Mean Squared Error (MSE) of 0.004.