By Bessi U.
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Extra info for An analytic counterexample to the KAM theorem
Valve 1, position 2; valve 2, position 1; valve 3, position 1. but now the N, and CH, (and 0, if present) are allowed to go forward to the molecular sieve column. Both columns are now isolated, with CO, and C, in column 3, the porous polymer, and 0,, N, and CH, in column 4,the molecular sieve. After elution of C,, C, and C, hydrocarbons, column 3 is reconnected for elution and measurement of CO, and C,, and finally column 4 is connected for measurement of 0,, N, and CH,. 12 is a typical chromatogram.
Among the hydrocarbons, saturates and monounsaturates (alkanes and alkenes) form the main components, with dienes and acetylenic compounds (alkynes) normally present only as traces. As the gas is subjected to different fractionations and reactions, so the relative proportions of components obviously change. Those hydrocarbon components which are not used in processing, or as fuel gas, may be added to propane or butane product streams. These products may, therefore, be totally saturated hydrocarbons, containing relatively small amounts of each other, or mainly saturated, but with significant concentrations of unsaturates.
Capillary separation. I 0 min I 10 min Fig. 14. Capillary separation. d. capillary coated with OV-101. Temperature: 35°C for 5 min, then 6"C/min to 220°C. 3 bar. Referencesp. 40 Chapter I 18 I . 6 shows a chromatogram with good resolution and peak shapes for C, to C, hydrocarbons, insufficient resolution for lighter components and long analysis time and low, broad peaks for C, and heavier components. The solution consisted of diverting groups of components onto different columns, chosen to optimize their separation at the selected operating temperature.
An analytic counterexample to the KAM theorem by Bessi U.