2022
DOI: 10.1016/j.fuel.2022.124815
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Catalyst characterization and catalytic evaluation of 3wt%Al-KIT-6 toward biomass-derived γ-valerolactone decarboxylation to butene

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Cited by 4 publications
(3 citation statements)
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“…When the reaction time was 150 min, the butene yield increased to 86 %. The satisfactory catalytic activity of ASA(Si/Al=20) catalyst was obtained when the react time was 180 min, and the butene yield was 90 %, which is 10 % and 6 % higher than that over the 3 wt %Al‐KIT‐6 [13a,b] and 2.23 wt %Al‐SBA‐15 catalysts [13a,b] published by our group at same reaction condition. When the reaction time was extended to 210 min, the catalytic activity did not change significantly.…”
Section: Resultsmentioning
confidence: 56%
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“…When the reaction time was 150 min, the butene yield increased to 86 %. The satisfactory catalytic activity of ASA(Si/Al=20) catalyst was obtained when the react time was 180 min, and the butene yield was 90 %, which is 10 % and 6 % higher than that over the 3 wt %Al‐KIT‐6 [13a,b] and 2.23 wt %Al‐SBA‐15 catalysts [13a,b] published by our group at same reaction condition. When the reaction time was extended to 210 min, the catalytic activity did not change significantly.…”
Section: Resultsmentioning
confidence: 56%
“…Besides, all resulting by‐products are liquids (>C 5 ). The gaseous products of the reaction are only butene and CO 2 [13b] . Therefore, the butene yield ( Y (C4) %) and the GVL conversions ( Conv .%) can be calculated according to the mass conservation, as shown in equations (1) and 2: boldY()C4%=nactual4ptbutene4ptoutputntheoretical4ptbutene4ptoutput=(m0-m1)/boldMGVLboldm0/boldMGVL=100.12·(boldm0-boldm1)100.12·boldm0×100% $\vcenter{\openup.5em\halign{$\displaystyle{#}$\cr {{\bf Y}}_{\left({\bf C}4\right)}{\rm \char37 }={{{\bf n}\left(actual{\rm \ }butene{\rm \ }output\right)}\over{{\bf n}\left(theoretical{\rm \ }butene{\rm \ }output\right)}}\hfill\cr ={{({{\bf m}}_{0}-{{\bf m}}_{1})/{{\bf M}}_{{\bf G V L}}}\over{{{\bf m}}_{0}/{{\bf M}}_{{\bf G V L}}}}={{100.12\cdot{}({{\bf m}}_{0}-{{\bf m}}_{1})}\over{100.12\cdot{}{{\bf m}}_{0}}}\times 100\hskip0.17em{\rm \char37 }\ \hfill\cr}}$ boldCboldoboldnboldv.%=(boldm0-boldm2)m0×100% $\vcenter{\openup.5em\halign{$\displaystyle{#}$\cr {\bf C o n v}.…”
Section: Methodsmentioning
confidence: 99%
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