Tip to find column length (of C18) for new particle size

[Sandra7]

Can someone help me in approaching the following question?

Question: A scientist is using 4.6x150 mm, 5um particle size C18 HPLC column to carry out a separation. The isocratic separation occurs at a flow rate of 1.0 mL/min (50% methanol/50% water) with a back pressure of 160 bar, take 18 min., and gives a resolution of 1.8 for the two least developed peaks [with k values of 2.7 and 2.8]. The unretained time is 1.5 min. He wants to speed up the separation by buying a similar column with 3um particles.
- If the maximum pressure of the HPLC is 300 bar, what length (to the nearest 25mm) should he purchase so that he can still run at 1.0 ml/min?

[AA]

Divide the column length by the particle size (150mm/5um) = 30000
To get the same seperattion (or at least similar) you would need at least L/dp of the the same so 100mm/3 um = 33000. So you could use a 100 mm column. The back pressure wont be much different.

[Uwe Neue]

I can send you an article that shows how this works - in more detail.

[Sandra7]

I can send you an article that shows how this works - in more detail.

That would be great. I want to understand it in more detail.

Does this forum have feature to send private message? If not, you can send me at this email address: deleted.

Thanks a lot.

[Sandra7]

Divide the column length by the particle size (150mm/5um) = 30000
To get the same separation (or at least similar) you would need at least L/dp of the the same so 100mm/3 um = 33000. So you could use a 100 mm column. The back pressure wont be much different.

Thanks a lot. I was thinking that I would need to use the new pressure value in my calculation though I also looked at that piece of information as reminding that maximum pressure cannot be beyond 300 bar when making changes to the column.

Can you (or someone) help me in estimating the time to perform the separation using the new column ( 100mm length made with the new particle size) assuming that % methanol stays the same and Kc and "beta" stay the same (though beta would increase some)?

Also to estimate the time to perform the separation if % methanol is optimized so that resolution stays the same.

[Sandra7]

Divide the column length by the particle size (150mm/5um) = 30000
To get the same separation (or at least similar) you would need at least L/dp of the the same so 100mm/3 um = 33000. So you could use a 100 mm column. The back pressure wont be much different.

Thanks a lot. I was thinking that I would need to use the new pressure value in my calculation though I also looked at that piece of information as reminding that maximum pressure cannot be beyond 300 bar when making changes to the column.

Can you (or someone) help me in estimating the time to perform the separation using the new column ( 100mm length made with the new particle size) assuming that % methanol stays the same and Kc and "beta" stay the same (though beta would increase some)?

I am thinking to do this: Find tR for both peaks using
tR = tM (1+k) since tM and both k values are given (see the original post).

The longer tR time can be considered the run time.

Also to estimate the time to perform the separation if % methanol is optimized so that resolution stays the same.

I am still thinking about this one......

[AA]

I did the calculation to be sure. If the flow rate remains the same and the column is shorter, the pressure will be increased on 3 um by about 100 bar (the actual amount depends on solvent composition and temperature).

The run time will be roughly, in this case, 2/3 of the orginal time.

Mass loading and injection volume may also need to change (less powder, less loading). This may not impact you depending on the loading you are at already.

[Uwe Neue]

Sandra7: I send you two publications in the mail. One was with the US Pharmacopeia.

[Sandra7]

Sandra7: I send you two publications in the mail. One was with the US Pharmacopeia.

Thanks. I got those. I am reading those now.

[Sandra7]

I did the calculation to be sure. If the flow rate remains the same and the column is shorter, the pressure will be increased on 3 um by about 100 bar (the actual amount depends on solvent composition and temperature).

Our lecture hasn't covered any formula that includes the pressure and so I can't use any formula containing pressure parameter.

The run time will be roughly, in this case, 2/3 of the original time.

Do you use the pressure information to calculate for the runt time that you got? For me, all I can think of is that the mobile phase velocity would remain the same when the shorter column with smaller particle size is used. So, calculate for mobile phase velocity (u) for the first column using u=L/tM= 150mm/1.5 min= 100 ml/min. Then plug this u value for column of 100mm and get the tM for shorter column which is 1 min.

Since Kc (Distribution coefficient) and beta (phase volume ratio) are said not to change, k (retention factor) values for each compound compound would not change either. Using k = (tR-tM)/tM, I can find tR of the 2nd peak (of k=2. for the new column and get 3.8 min.

Mass loading and injection volume may also need to change (less powder, less loading). This may not impact you depending on the loading you are at already.

[AA]

Do you use the pressure information to calculate for the runt time that you got?

No, pressure has nothing to do with the run time.

You are making this way too complicated. The run time will be the ratio of new column length to old column lenght, simple as that.

New Run time = Old run time X (new column length/old column length).

[Sandra7]

Do you use the pressure information to calculate for the runt time that you got?

No, pressure has nothing to do with the run time.

You are making this way too complicated. The run time will be the ratio of new column length to old column length, simple as that.

New Run time = Old run time X (new column length/old column length).

Wow.. didn't think of that obviously for lack of experience. I really appreciate what I am learning here. Thanks

The other way I tried was by calculating linear velocity by keeping setting up the ratio of reduced the reduce velocity the same and calculating for the new velocity that's increase when particle size was decreased. Then calculate for retention time of compound for the bigger k value after reading Uwe's article (pasted below). But, any formula not taught in class cannot be used in the quiz we will have next week.

Uwe's article:

"Isocratic separations
To scale the separation perfectly,
I need to reach the same plate
count, N. This can be done on a well
designed HPLC system by changing
the column length L and the particle
size, dp, in proportion to each other,
while maintaining the same reduced
plate height, h:

N= L/H=L/dp x h

dp = particle size

The same reduced plate height,
h, is achieved if the same reduced
velocity, υ, is used. The reduced
velocity is defined as:

υ = (u x dp)/ Dm

Dm is the diffusion coefficient of our
analyte, which remains the same in
the same mobile phase and we do
not need to worry about this value.
u is the linear velocity. In order to
maintain the exact same reduced
velocity, I must increase the linear
velocity as I decrease the particle
size, or decrease the linear velocity
as I increase the particle size."

[Sandra7]

Do you use the pressure information to calculate for the runt time that you got?

No, pressure has nothing to do with the run time.

You are making this way too complicated. The run time will be the ratio of new column length to old column lenght, simple as that.

New Run time = Old run time X (new column length/old column length).

Thanks. In the above, there's no change in the solvent. I am wondering what the run time would be at this new column length if % methanol is optimized so that the resolution remains the same.

Change in L means change in N. Retention factor or capacity factor (k) would change with the change in solvent. Assuming that the same % change occur to the k value of the last eluting peak, how do I approach this question?