2022
DOI: 10.1080/15320383.2022.2059444
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Evaluation of the effectiveness of bioaugmentation and biostimulation in atrazine removal in a polluted matrix using degradation kinetics

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Cited by 5 publications
(6 citation statements)
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“…The half-life of the dye was further reduced to only 6 min, showing that the catalyst more efficiently degraded the dye under UV irradiation. 33,34…”
Section: Resultsmentioning
confidence: 99%
See 1 more Smart Citation
“…The half-life of the dye was further reduced to only 6 min, showing that the catalyst more efficiently degraded the dye under UV irradiation. 33,34…”
Section: Resultsmentioning
confidence: 99%
“…The half-life of the dye was further reduced to only 6 min, showing that the catalyst more efficiently degraded the dye under UV irradiation. 33,34 A comparison of the degradation efficiency of methylene blue in the presence of various catalysts is presented in Table 2.…”
Section: Kinetics Of Photocatalytic Degradation Of Methylene Bluementioning
confidence: 99%
“…Various studies have demonstrated that the combination of a bioaugmentation and biostimulation approach yielded better results for atrazine removal as well as for other contaminants (El-Bestawy et al 2014;Aliyu et al 2023). Therefore, bioaugmentation with a consortium of selected bacteria coupled with biostimulation may be the best bioremediation technique since it is an effective, low-cost and environmentally friendly approach for the decontamination of atrazinecontaminated matrices and may also be used to remove many other TrOCs (El-Bestawy et al 2014;Aliyu et al 2023).…”
Section: Discussionmentioning
confidence: 99%
“…This model is applied by adjusting C 0 , k IORE , and N IORE to minimize the objective function S IORE . The half‐lives were determined by using the equations 7 and 8 for DT 50 (time required to decrease to half of the initial concentration) and DT 90 (time required to decrease 10 % of initial concentration) [56] DT50=[(C0/2)4pt(1-normalN)-4ptnormalC0(1-normalN)]/[normalkIORE4pt×4pt(N-1)] $\vcenter{\openup.5em\halign{$\displaystyle{#}$\cr {\rm DT}{_{50}}=[({\rm C}{_{0}}/2)\ {^{(1- {\rm N})}}- \ {\rm C}{_{0}}{^{(1- {\rm N})}}]/[{\rm k}{_{{\rm IORE}}}\ \times \ ({\rm N}\hbox{-}1)]\hfill\cr}}$ DT90=[(C0/10)4pt(1-normalN)]/[normalkIORE×(N-1)] $\vcenter{\openup.5em\halign{$\displaystyle{#}$\cr {\rm DT}{_{90}}=[({\rm C}{_{0}}/10)\ {^{(1- {\rm N})}}]/[{\rm k}{_{{\rm IORE}}}\times ({\rm N}\hbox{-}1)]\ \hfill\cr}}$ …”
Section: Methodsmentioning
confidence: 99%
“…The half-lives were determined by using the equations 7 and 8 for DT 50 (time required to decrease to half of the initial concentration) and DT 90 (time required to decrease 10 % of initial concentration). [56] : DT 50 ¼ ½ðC 0 =2Þ ð1À NÞ À C 0 ð1À NÞ �=½k IORE � ðN-1Þ�…”
Section: Single First Order Model (Sfo)mentioning
confidence: 99%