Process Gas Chromatographs. Tony Waters

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Название Process Gas Chromatographs
Автор произведения Tony Waters
Жанр Отраслевые издания
Серия
Издательство Отраслевые издания
Год выпуска 0
isbn 9781119633013



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      We don't need this information now, but Equation 3.1 is used in more advanced work to optimize the carrier gas flow rate.

      The air peak location is good to know when troubleshooting because component molecules can't travel faster than the carrier gas, so no peaks from the current sample injection can elute before the air peak.

      But the air peak has a much more significant role because it shows how long the carrier gas took to pass through the column. Recall that component molecules can move along the column only when they are in the carrier gas. So, to get through the column, all component molecules must spend the same time in the carrier gas as the air peak does.

      As a way of emphasizing the point, imagine that the column had no liquid inside; where would the peaks be on the chromatogram?

      Yes, of course, all the peaks would elute together with the air peak. With no liquid present, no separation can occur. The air peak reveals the time that component molecules spend in the gas phase, and it's the same for every component.

      As an aside, realize from this imagined experiment that the time a peak spends in the liquid phase is proportional to the amount of liquid phase present in the column. In packed columns, the liquid film can slowly evaporate into the carrier gas, thereby, over time, reducing the observed peak retention times.

      An air peak tells us how long the carrier gas takes to get through the column, which is also the time that each component spends in the gas phase. The additional retention time for each peak is the time that peak spends in the liquid phase. Graph depicts the significance of an Air Peak. An air peak tells us how long the carrier gas takes to get through the column, which is also the time that each component spends in the gas phase. The additional retention time for each peak is the time that peak spends in the liquid phase.

      Notice that the air peak neatly divides the chromatogram into two time zones and makes them easy to measure:

       From injection to the air peak is the time that all components spend in the gas phase. This time is often called the “dead time” because it doesn't contribute to separation. The formal name for the time in the gas phase is the holdup time.

       From the air peak to the apex of each peak is the average time that each component spends in the liquid phase. The formal name for the time in the liquid phase is the adjusted retention time.

      These chromatogram measurements are an essential prerequisite for optimizing or troubleshooting column performance. For example; they allow the optimum setting of column temperature. We plan a second book to explore these more advanced techniques.

      The answer

      Figure 3.3 is also the correct answer for the chromatogram you were invited to draw on Figure 3.2. The four peaks are centered at 3, 4, 6, and 12 minutes from sample injection. You may find this surprising; the peaks are no longer equally spaced! Let's see how that happened.

      The air peak is just leaving the column after three minutes of traveling. Since it doesn't dissolve in the liquid phase, its time in the liquid phase is zero. One way to indicate this is to show its gas:liquid residence time ratio as 3:0. Since the air molecules are not delayed in the liquid phase, the air peak should be very narrow, and it's reasonable to draw it centered at three minutes.

      Most people find it very difficult to decide where the other peaks come out on the chromatogram, perhaps because the correct answer defies all expectations.

      Actually, it's easy to get it right. Starting from the position of each peak in Figure 3.1, there are two ways of reaching the correct conclusion about its final position on the chromatogram. Both ways are noted below.

      To locate the remaining peaks, it's easier to think about them in the following order, starting with the propane peak.

      The propane peak took three minutes to reach the center of the column, so it should take another three minutes to reach the detector, for a retention time of six minutes overall. A better argument comes from its solubility ratio. Recall that retention time is the sum of the time a component spends in the gas phase and the time it spends in the liquid phase (Equation 3.2). To reach the detector, the propane peak must spend three minutes traveling in the gas phase. But, because of its 50:50 solubility ratio, it also spends three minutes stopped in the liquid – so its residence time ratio is 3:3 for a total retention time of six minutes.

      The 1‐butene peak took three minutes to move 25 % of the column length, so it should take four times as long to reach the detector, for a retention time of twelve minutes overall. A better argument invokes its solubility ratio. To reach the detector, the 1‐butene peak must spend three minutes traveling in the gas phase. But, because of its 25:75 solubility ratio, it also spends nine minutes stopped in the liquid – so its residence time ratio is 3:9 for a total retention time of twelve minutes.

      The carbon dioxide peak took three minutes to move 75 % of the column length, so it should take one additional minute to travel the last 25 % of the column, for a retention time of four minutes overall. A better argument comes from its solubility ratio. To reach the detector, the carbon dioxide peak must spend three minutes traveling in the gas phase. But, because of its 75:25 solubility ratio, it also spends one minute stopped in the liquid – so its residence time ratio is 3:1 for a total retention time of four minutes.

      Notice that the different molecules all spend exactly three minutes in the gas phase traveling.

      If you closely followed what is going on here, you may be thinking that the pressure drop along the column must distort the predictions made in the above arguments. That's somewhat true, but way too complex to consider at this early stage.

      A practical task

      To evaluate various measures of column performance,