2022
DOI: 10.1016/j.seppur.2022.121096
|View full text |Cite
|
Sign up to set email alerts
|

A realistic approach for determining the pore size distribution of nanofiltration membranes

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1
1
1

Citation Types

1
7
0

Year Published

2022
2022
2024
2024

Publication Types

Select...
9

Relationship

1
8

Authors

Journals

citations
Cited by 34 publications
(8 citation statements)
references
References 103 publications
1
7
0
Order By: Relevance
“…Subsequently, we compared our numerical parameters to prior empirical membrane characterizations. The pore radius of NF 270 has been reported to be between 0.43 and 0.54 nm, , and the pore size distribution has been estimated to be approximately 0.3 ± 0.1 nm, based on MWCO experiments . In comparison, DSPM-DE suggested pore radii of 0.416 and 0.461 nm, at pH 7 and 2, which were within the error of the reported estimates.…”
Section: Resultssupporting
confidence: 60%
“…Subsequently, we compared our numerical parameters to prior empirical membrane characterizations. The pore radius of NF 270 has been reported to be between 0.43 and 0.54 nm, , and the pore size distribution has been estimated to be approximately 0.3 ± 0.1 nm, based on MWCO experiments . In comparison, DSPM-DE suggested pore radii of 0.416 and 0.461 nm, at pH 7 and 2, which were within the error of the reported estimates.…”
Section: Resultssupporting
confidence: 60%
“…The oxidative debris and metal throughput for a 14 kDa membrane are higher, with the main 1 μm fraction remaining in the solution. Although some approaches have been developed to determine the pore size distribution of nanofiltration membranes [ 81 ] for graphene oxide, such techniques have not yet been validated.…”
Section: Discussionmentioning
confidence: 99%
“…When the effect of steric and hydrodynamic hindrances is taken into account, the rejection of a solute with diameter D s by a membrane with uniform pore size D p depends upon the factor λ (= D s / D p ) in the following manner: 49–51 R = 1 − (1 − ( λ ( λ − 2)) 2 )exp(−0.7146 λ 2 )for, R = 0.5, λ = 0.416.…”
Section: Methodsmentioning
confidence: 99%