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Relationship between N, particle size, column length

Discussions about HPLC, CE, TLC, SFC, and other "liquid phase" separation techniques.

7 posts Page 1 of 1
Efficiency is said to be roughly related to column length, particle size by the equation:

N = L/(2Dp)

Where Dp is particle diameter, L = length. For a column straight-out-of-the-box, does/should this relationship hold true using manufacture's reported values for N, particle diameter?
This is a reasonable approximation.
A. Carl Sanchez
Not quite so. You see, different particle sizes exhibit different flow rate optima (for more on this see Van Deemter).
So, you can't just use the suggested equation with a constant flow rate, which in the suggested context is assumed.

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Dancho Dikov
What is your suggestion for approximating the expected efficiency?

Ofcourse there are many equations but each suffers from the same issue in that some assumptions are necessary.
A. Carl Sanchez
If I need/wish to calculate such a value, I'd probably normalize it, using the ratio between flow rate optimum and the intended flow rate. That would be a more fair tool for comparing different columns' performance in a given situation – i.e. everything else equal but the particle size.

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Dancho Dikov
Efficiency is said to be roughly related to column length, particle size by the equation:
The operant word in there is "roughly" :wink: .

If memory serves, the "random walk" model of peak broadening suggests that the plate height at the optimum flow rate for a perfectly packed column should be about twice the particle diameter. For small molecules on columns with small particles (say, in the 3-micron range or below), the C-term is fairly small, so the Knox (or VanDeemter) plots are fairly flat, which means that the plate height at "typical" flow rates is not that much worse than at the optimum flow (maybe 10% or so).

In practice, what we see is the net effect of the column plus "extra-column" band broadening, so reduced plate heights are typically > 2. That said, I've seen reduced plate heights down around 1.2 - 1.5 published for some of the 2.x-micron "superficially porous" material. It looks to my eye like the C-term is about what you'd expect, with the improved efficiency resulting from an anomalously small A term -- which goes to show that "the theory is not the reality"!

As a very rough approximation I've used:
N = 3000 x dp / L

where dp is the particle diameter in microns and L is the column length in cm. Yes, I know the units don't cancel; this has no theoretical validity whatever, but it does seem to indicate what we can expect from a good column on a good day (usually around 10k plates from a typical column).
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
Efficiency is said to be roughly related to column length, particle size by the equation:

N = L/(2Dp)

Where Dp is particle diameter, L = length. For a column straight-out-of-the-box, does/should this relationship hold true using manufacture's reported values for N, particle diameter?
This equation is for theoretical N for a column only.
The N value you could get is the sum effect of your column and your instrument;
The N value from the COA is, too (with a smaller bandspreading effect from instrumenta).
B/c the N is calculated with peak width.
7 posts Page 1 of 1

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