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

# An example is the extraction of caffeine from coffee

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Unformatted text preview: Reflux ratio = 7 and bottoms flow rate is set at 197 mol/s. Chemical equilibrium is achieved at stages 4 through 11. RK EOS for vapor-phase fugacities. Assumptions: Use UNIFAC method for liquid-phase activity coefficients. Find: Distillate and bottoms products, and comparison of results with those of Example 11.9. Analysis: The ChemCad program, with the SCDS model and reactive distillation option, was used to make the equilibrium stage calculations. To do this, it is assumed that the required chemical equilibrium constant can be computed by taking the ratio of the forward to the backward rate constant. From Eqs. (1) and (2) in Example 11.9, K = 1.375 x 10-5 exp(42014/(RT)) (1) From Fig. 11.38(a), all of the reaction takes place at a temperature of about 350 K. From Eq. (1), at T = 350 K, with R = 8.314 J/mol-K, K = 25.6. In ChemCad, let lnK = A. Therefore, A = 3.24. Input data for ChemCad, included: No. of stages = 17, with M fed to stage 10, and the butenes fed to stage 11. Estimated temperatures in K were 350, 340, 400, and 420 for stages 1, 2, 16, and 17. A damping factor of 0.25 was used with a maximum of 50 iterations. A single reaction was specified for the equilibrium option and stages 1-3 and 12-17 were disabled. The problem was very slow to converge, taking almost the maximum number of iterations. The overall material balance is as follows: Mol/s: Component Feed 1 Feed 2 Distillate Bottoms Methanol 215.50 0.00 36.28 0.12 Isobutene 0.00 195.44 13.58 2.76 n-butene 0.00 353.56 338.39 15.17 MTBE 0.00 0.00 0.15 178.95 Total: 215.50 549.00 388.40 197.00 The % conversion of isobutene to MTBE = 91.64%. This compares to 95.6% in Example 11.9. The profiles are shown on the next three pages and compare favorably to Figs. 11.38(a to c). Exercise 11.24 (continued) Analysis: (continued) Exercise 11.24 (continued) Analysis: (continued) Exercise 11.24 (continued) Analysis: (continued) Exercise 11.25 Subject: Supercritical extraction of 1 mol/s of 10 wt% ethanol in water by 3 mol/s of carbon dioxide at 305 K and 9.86 MPa in a staged contactor. Given: 10 equilibrium stages. K-values from Example 11.10. Find: Flow rates and compositions of the exiting extract and raffinate. Compare results to those of Example 11.10. Analysis: The ChemCad program with the Tower Plus model was used to make the calculations, as in Example 11.10. Isothermal conditions were employed by allowing heat transfer at each stage. The polynomial K-value option was used to supply the constant K-values given in Example 11.10 (0.115 for ethanol, 0.00575 for water, and 34.5 for CO2) The calculations converged in 7 iterations. The overall material balance was as follows: Component Carbon dioxide Ethanol Water Total: Mol/s: Feed 0.0000 0.0417 0.9583 1.0000 Solvent 3.0000 0.0000 0.0000 3.0000 Raffinate 0.0287 0.0277 0.9422 0.9986 Extract 2.9713 0.0140 0.0161 3.0014 These results are almost identical to those for Example 11.10, which used just 5 equilibrium stages. This should not be surprising! Supercritical extraction is analogous to stripping or liquid-liquid extraction. Thus, an approximate calculati...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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