Chromatography Forum

Hi peter
I wanted to see what the effect of high pressure was on the efficiency of a GC column. As with a given system I cannot change velocity without changing the pressure, so I have 2 variables. The data points at 80 and 4 psi are both around the "perceived " optimum velocity of helium, which is around 30 cm/s.

But I agree: it could be that running a column under 80psi, the optimum velocity will be quite different. I think it will move to the left.
The only way to run such an experiment is to change the outlet restrictions systematically and work always at the same high inlet pressure. It maybe a nice experiment to try..

Jaap,

In my last post I was still referring to the experiment where the uncoated column was put ahead OR behind the coated tube. I did not intend to convey the meaning of a sandwiched column which would be senseless.

best wishes,

Rod

OK,Rod
Sorry, I anticipated the wrong situation

This situation you refer to is almost similar as the starting experiment with the integrated guard. The thing you add also is that the separation pressures are also different. You will get the same effect as I tried to explain with the previous 2 examples: the change of pressure used in this range has no measurable effect on retention factor. The difference in retention time is caused by the difference in gas-velocities in each segment.

By the way: this concept of restriction to 0.53mm, is deeply tested with vacuum GC, showing also same retention factors.

rgds
jaap

The effect of pressure on retension factor can also be tested by using Agilent method translator software. Keep everything else the same except the inlet and outlet pressure of the column, one can see that the retention time (retetion factor) will be exactly the same if average linear velocity is kept the same. This is true whether the pressure increases or decreases.

I think that, under ideal conditions, 3 is the correct answer because the carrier gas is a compressible fluid. Otherwise (in LC, for example) the correct answer would be 1. These answers do not depend on the lengths of the two sections of the column.

It looks like the problem is mathematically simple, but logically it is a bit trickier. Here is why I think so.

Due to the gas decompression along the column, the gas velocity increases in the direction from the column inlet to the column outlet. The column consists of a coated and a bare segment. In configuration A, the bare segment is on the inlet side and the coated segment is on the outlet side. Let tM_bare_A and tM_coated_A be the hold-up times of these segments in configuration A. In configuration B, the column is reversed so that the bare segment gets on the outlet side and its hold-up time (tM_bare_B) becomes smaller. Conversely, the coated segment gets on the inlet side and its hold-up time (tM_coated_B) becomes larger. Reversing the column increases the time of the residence of a compound in one segment while reducing the time of that compound residence in another segment. This complicates a purely logical approach to the problem (i.e. solving it without the exact mathematical evaluations).

Now is the time to bring the retention to the picture. If k is the retention factor in any segment of a column then parameter (1+k) can be viewed as a magnifier of that segment's hold-up time (tM). Thus, the retention time in a segment is equal to (1+k)tM where tM is that segment's hold-up time. The compounds are retained in the coated segment – the one whose hold-up time in configuration B is larger than in configuration A. Now, it is important to recognize that the net hold-up time in configuration B is the same as the net hold-up time in configuration A. The net retention time in a column is the sum of the hold-up time in one segment (in the bare segment to be precise) and the magnified hold-up time in another segment (in the coated segment to be precise).

For any retained peak, the net retention time in configuration B is larger than it is in configuration A because the reversal of the column direction
a) does not affect the net hold-up time (the sum of the hold-up times in the coated and the bare segments),
b) magnifies by the same factor of (1+k) the hold-up time that is larger in configuration B than it is in configuration A.

Here is a more formal expression of the same logic.
If tR_A and tR_B are the net retention times in configurations A and B, respectively, then:
tR_A = tM_bare_A + (1+k)tM_coated_A
tR_B = tM_bare_B + (1+k)tM_coated_B
tR_B = tR_A when k = 0
tM_coated_B > tM_coated_A

Therefore, tR_B > tR_A when k > 0.

Outstanding explanation.

Agree completely.. answer 3 is correct.
We deal with the fact that a component has has retention time of 200s is NOT exactly in the middle of the column when 100 seconds have passed. Compressibility of carrier gas translating in different linear velocities

I explained it as follows:

If you make a simplified visualization, it will explain itself. Imagine a component with a k=4 and 10 meter uncoated Integra Guard

To make a simple model we will assume average gas velocity in the first 10m = 10 cm/s and in last 10 m the average velocity  is 20 cm/s.

A:   If the IntegraGuard  is at the FRONT, you have in the first 10m, slow moving gas part of the column, no retention.  So if the gas velocity on average is 10cm/s IN THAT SECTION,  components will need 100s time to travel through a 10m IG section. (all components)
B:  In case we put the IntegraGuard at the BACK, the gas velocity is higher on average, say 20 cm/s. So to pass this section all components take only 50 seconds.

Also for the  coated part there is a difference:
If there is 10m coated at the inlet, it takes 100 + (4 x100)  = 500 sec to pass, while at the END it takes  50 + (4x100) = 250 sec to pass similar length of the column.

So if sum this up for both situations:
In A:  we have retention time of 100  + 250   = 350 s
In B   we have retention time of   50  + 500  =  550 s

Of course for both systems we need to add some absolute time for the time it takes to travel through the “middle” part of the column, which is the same for both.

We saw this effect because at Restek we make many IntegraGuard columns and by developing a new test-procedure, columns were tested reversed They were all systematically out of specification..

thanks all for participating in this discussion!
jaap

Hi
The previous explanation may be a bit difficult to understand because there are no figures shown. If you look at the link below, you find some more details + figures. Also find a figure with the chromatograms obtained using the 2 m Rtx-5 column in front - and in the back of a 30m uncoated section.

http://blog.restek.com/?p=2394

jaap

Jaap,

Regarding the peak width difference, I always thought it has more to do with retention time and dead volume than van Deemter Equation.

When retention time is longer, the analyte diffuses more in both phases, so the peak width is wider.

It can be argued that the dead volume is the same for guard column in the front or in the back. But if the guard column is in the front, there is a focusing effect when the analyte reach the coated section of the column. If the guard column is in the back, the 30m guard column is all dead volume that will cause the peak to be wider.

My two cents.

Yes, you are correct. Retention will sure add to peak width.

The thing I was amazed was that the peaks were still that symmetrical after traveling through a 30 m guard! components must be hugely overloaded.. I would have expect more asymmetry due to the overload, but somehow the 2m coated column, did "correct" that..