Sovent Front

[tjgr00me]

Good morning all.

I am posting to ask is it possible to determine (as an estimate) when the solvent front should be expected of the column (by only going off the dimensions of the column and the flow rate, i know the dead time of a system would effect this and column particle size but is why i only want an estimate)?

I always thought i knew how to work this out as the equation i was given a while ago is:

((pi x 0.7 x (radius of the column in mm)2 x column length in mm)) / flow rate in ul.

Now i am not sure if this is correct as i cant find any literature with this equation in it.


But for example i have a kinetex column of dimensions 100mm x 2.10 being used with a flow rate of 0.4ml/min
so to fill this into my equation:

((3.1416 x 0.7 x (1.05)2 x 100)) / 400 = 0.61min

As mentioned this is how i have always gone about it in the past but now i would like to know if i am in the right ball park or if i am way off.
If i am way off is there a way to determine solvent front with only column dimensions and flow rate.

I hope this makes sense.

Thank you in advance to all who reply.

Terry

[ym3142]

Terry,

why do you think your cal is not good any more? Or just not sure.

I always do similar way for ball park.

[tjgr00me]

Thank you for you reply.

I just wanted confirmation to be honest because i have only ever known one other person to use this calculation (the person i learnt it off in fact).

If my current calculation is correct to get a ball park figure then i will continue to use it.

It was just reassurance that i was in the right area, and if i wasnt was there any other ways out there.

[HW Mueller]

Dolan and Snyder in "Troubleshooting LC Systems, page 86 in my 1989 version give

to ~ 0.5Ld^2

L = column length in cm
d = column inner diamter in cm

(seems to be oK only when flow rate is 1mL/min)

[Bryan Evans]

Does the column come with certificate of analysis? If so, you can use data from there.

[tom jupille]

Look at it this way:

If the column were an empty cylinder, the internal volume would be
V = (pi/4) x L x dc^2 where
L = length and
dc = column internal diameter

Of course, the column is not empty, but of that total volume, about 25-30% is between the particles (it calculates out to 26% for closest-packed spheres) and about 35-40% is inside the pores. That means that about 65% of the cylinder volume actually contains mobile phase. 65% of pi/4 is approximately 0.5, so
Vm = 0.5 x L x dc^2 (approximately)

That estimate probably has about +/- 15% uncertainty.

If the solid core in the superficially porous packings is about half the diameter of the particle, then it will occupy about one eighth of the volume (volume is propotional to the diameter cubed), so those packings would be toward the low end of the above estimate.