Tony Waters - Process Gas Chromatographs

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A guide to the fundamentals of applied gas chromatography and the process gas chromatograph, with practical procedures for design and troubleshooting This comprehensive resource provides the theory that underpins a full understanding of the fundamental techniques of gas chromatography and the process analyzer. Without relying on complex mathematics, the book addresses hands-on applications of gas chromatographs within process industries. The author – a noted expert on the topic – details both the scientific information needed to grasp the material presented and the practical applications for professionals working in the field.
Process Gas Chromatographs:
Fundamentals, Design and Implementation
Describes practical procedures for design and troubleshooting Contains concise chapters that provide a structured course for advanced students in process engineering Reviews the fundamentals of applied gas chromatography Details the operation and maintenance of process gas chromatographs Offers a summary, and self-assessment questions, for every chapter Is written by an international expert in the field with extensive industry knowledge and teaching experience in courses on process sampling systems and gas chromatography Written for process analyzer engineers and technicians, application engineers, and industrial environmental engineers, offers an essential guide to the basics of gas chromatography and reviews the applications of process gas chromatographs in industry today.

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Plate height gives a measure of column performance that is independent of the column length and is the primary variable used for optimizing performance.

Knowledge Gained

It's obviously true to say that peaks come out of the column at different times.

It's not true to say that molecules move at different speeds inside the column.

For all injected molecules, there are only two speeds possible inside the column; stop or go.

When molecules are in the liquid phase, they cannot move along the column.

When molecules are in the gas phase, they move at full carrier gas speed.

The “air peak” doesn't dissolve in the liquid phase so it travels at the same speed as the carrier gas.

It's impossible for any peak to travel faster than the carrier gas.

No peak from the same injection can appear on the chromatogram earlier than the air peak.

To get through the column, all injected molecules must travel for the same time as the air peak does.

If the column had no liquid in it, all the peaks would elute together with the air peak.

Any additional retention time beyond the air peak time is the time a peak stopped in the liquid phase.

Separation is not caused by motion, it's due to peaks stopping for different times in the liquid.

The time that a peak stops in the liquid phase is directly proportional to its solubility in that liquid.

The spacing of peaks on the chromatogram is different than their spacing inside the column.

Peaks that migrate slowly along the column come out very much later on the chromatogram.

Retention time is the sum of time traveling in the gas phase and time stopped in the liquid phase.

Holdup time is the time in the gas phase; adjusted retention time is the time in the liquid phase.

Column operating performance can be evaluated from chromatogram measurements.

Chromatogram measurements may be made in seconds or millimeters.

You can measure the holdup time, peak retention time, and peak width on the chromatogram.

When measuring base width, triangulate the peak and extend the baseline under the peak.

Alternatively, measure the peak width at half the peak height.

Plate number is calculated from measurements of retention time and peak width.

Any change in operating conditions that increases plate number increases column separating power.

Did you get it?

Self‐assessment quiz: SAQ 03

These questions relate to gas‐liquid chromatography:

1 Q1. At what speed do component molecules move along the column when they are in the gas phase?

2 Q2. At what speed do component molecules move along the column when they are in the liquid phase?

3 Q3. What causes the separation between components?

4 Q4. What information may be derived from the retention time of an unretained peak, such as air?

5 Q5. If all the liquid phase was washed out of the column, where would all the peaks be on the chromatogram?

6 Q6. What causes the additional retention time beyond the holdup time?

7 Q7. What is the purpose of triangulating a peak?Check your SAQ answers with those given at the end of the book.

Student evaluation test: SET 03

Your instructor will provide answers to these questions.

All of the questions relate to gas‐liquid chromatography.

1 S1. Which one of the statements below best defines the retention time of a component peak? Select one:It's the average time that the component molecules spend in the column.It's the average time that the component molecules spend in the gas phase.It's the average time that the component molecules spend in the liquid phase.It's the average time that the component molecules spend in the detector.

2 S2. Which one of the statements below best explains the separation of component peaks on the chromatogram? Select one:The molecules of some components move at different speeds than others do.The molecules of some components spend less time moving than others do.The molecules of some components spend more time moving than others do.The molecules of some components spend more time not moving than others do.

3 S3. Which one of the statements below best explains the significance of the air peak? Select one:The air peak marks the time when the component molecules start to dissolve in the liquid phase.The air peak indicates the time that every component molecule spends in the gas phase.The air peak indicates the time that every component molecule spends in the liquid phase.Any peaks traveling faster than the air peak cannot be separated on that liquid phase.

4 S4. On the chromatogram shown in Figure 3.3, what percentage of the 1‐butene dissolves in the liquid phase at equilibrium?

5 S5. If the holdup time on a certain column is sixty seconds, what would be the adjusted retention time of a peak having a liquid phase solubility of 60 %?Express your answer in integer seconds.

6 S6. If the propane peak has a retention time of sixty seconds on a three‐meter column and propane has a solubility in the column liquid phase of 50 %, what is the average velocity of the carrier gas?Express your answer in m/s.

7 S7. Why do peaks that are evenly spaced in the column turn out to be unevenly spaced on the chromatogram? Combine two of the arguments given below to get the best overall explanation.Select two:All of the peaks spend the same time in the carrier gas.The later peaks are more soluble in the liquid phase than the earlier peaks are.After the air peak reaches the end of the column, the later peaks have farther to go than the earlier peaks do.The later peaks migrate through the column more slowly than the earlier peaks do.

8 S8. This question asks you to make logical deductions beyond what you learned in the chapter.Assuming that all other variables can be held constant, evaluate whether each statement listed below is true or false.Select all of the true statements and none of the false statements:The holdup time would be proportional to the column length.The retention time of a component peak would be proportional to the column length.If the carrier gas flow increased, the holdup time would increase.If the column temperature increased, the adjusted retention time of a component peak would decrease.None of the statements A−D above are true.

9 S9. This question asks you to make logical deductions beyond what you learned in the chapter.Consider a column that loses some of its stationary phase due to slow evaporation, but all other variables are held constant. Then, evaluate whether each statement listed below is true or false.Select all of the true statements and none of the false statements:The holdup time would be reduced.The retention time of all peaks would be reduced.The adjusted retention time of all peaks would be reduced.The separation of component peaks would be increased.None of the statements A−D above are true.

References

1

Figures

1 3.1 Effect of Component Solubility

2 3.2 Draw Your Own Chromatogram

3 3.3 Significance of the Air Peak

4 3.4 Typical Chromatogram Measurements

Equations

3.1 Process Gas Chromatographs - изображение 47 Average carrier gas velocity
3.2 t R= t M+ t ′ R Retention time as sum of times in gas and liquid phases
3.3 Plate number from base width 34 Plate number from width at half height - фото 48 Plate number from base width
3.4 Process Gas Chromatographs - изображение 49 Plate number from width at half height
3.5 Process Gas Chromatographs - изображение 50 Plate height (usually expressed in mm)

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