Separation Process Principles- 2n - Seader & Henley - Solutions Manual

Let the amount of air be 100 kmol 79 kmol n2 and 21

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Unformatted text preview: absorption with water at 25oC. Because the K-value for O2 is lower than that of N2 , the O2 is more easily absorbed. The bottoms liquid is heated to 100oC and flashed at 1 atm to release the dissolved O2. However, because of the large K-value for oxygen in the absorber, the process may require huge amounts of water to obtain a high yield of oxygen. Furthermore, the much higher concentration of nitrogen in the entering air and the fact that the relative volatility of nitrogen with respect to oxygen is not very high will make it very difficult to obtain a sharp split with absorption, as compared to distillation. Thus, the process may not be capable of either high yield or high purity of oxygen. The flash is capable of releasing any dissolved organic. Exercise 4.63 (continued) Analysis: (continued) (b) For batch absorption, the yield and purity of the oxygen, after it flashed off from the water, depends upon the amount of water used. Let the amount of air be 100 kmol (79 kmol N2 and 21 kmol O2). Consider amounts of water equal to 100, 1000, 10000, 50000, and 100000 kmol/h. Add this to the air and flash it at 100 atm and 25oC. Using the isothermal flash equations of Table 4.4 (with the exception of the energy balance, which is not needed) with the K-values of 910 for nitrogen and 450 for oxygen from the table above, the following results are obtained for a single equilibrium absorption, neglecting the mass transfer of water (stripping into the gas phase), i.e. assigning a K-value of 0 to water: Equilibrium conditions: Water, kmol kmol N2 in V kmol O2 in V kmol N2 in L kmol O2 in L Mol fraction O2 in flashed vapor 100 78.92 20.95 0.08 0.05 0.349 1,000 78.15 20.55 0.85 0.45 0.347 10,000 70.33 16.81 8.67 4.19 0.326 50,000 32.95 5.49 46.05 15.51 0.252 100,000 0.00 0.00 79.00 21.00 0.210 These results show that for small amounts of water very little absorption occurs, but the greatest increase in O2 purity occurs. If additional stages were used with small amounts of water to absorb the flashed vapor so as to reach 90 mol% purity of O2 ,the yield would be extremely small. At the highest liquid rate in the table above, all of the air is absorbed and, therefore, no separation between nitrogen and oxygen occurs. If 50,000 kmol of water is used, about 75% of the O2 is absorbed along with about 60% of the N2, with a resulting O2 mole fraction of 0.252 in the flashed vapor. If we start with 30,000 kmol of water and continue at this level using 300 kmol of water per kmol of flashed vapor for each batch absorption stage, the following results are obtained, using the isothermal flash equations for absorption at 100 atm and 25oC, followed by complete release of N2 and O2 at 1 atm and 100oC, but neglecting the vaporization of water: Equilibrium conditions: Stg. Water, kmol N2 kmol O2 kmol N2 kmol O2 y of O2 in flashed kmol in V in V in L in L vapor 1 30,000 52.07 10.27 26.93 10.73 0.285 2 11,300 17.54 5.16 9.38 5.58 0.373 3 4,490 6.02 2.62 3.36 2.96 0.468 4 1,895 2.12 1.35 1...
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