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

# Analysis for a single stage and the cocurrent

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Unformatted text preview: 72 mol/L, x1 = 0.016 mol/L % extracted = y1S/ xFQ x 100% = (0.0672)(500)/(0.05)(1,000) x 100% = 67.2% (3) (b) For three crosscurrent stages, can use Eq. (5-21) in the following form, x3 1 = xF 1+ E / N 3 (4) where from Eq. (5-14), E = KS/Q = (4.2)(500)/1,000 = 2.1 and N = 3. Frm Eq. (4), Percent extraction = 1 − x3 1 × 100% = 1 − × 100% = 79.6% 3 xF (1 + 2.1/ 3) (c) ) For three-stage countercurrent extraction, Eq. (5-28) applies in the form, x3 1 1 = = = 0.058 2 3 xF 1+ E + E + E 1 + 2.1 + 2.12 + 2.13 % extraction = (1 - 0.058) x 100% = 94.2% (5) Exercise 5.13 Subject: Given: Extraction of p-dioxane (B) from water (A) with solvent (C). Extraction factor, E = 0.9 compared to 2.4 in Example 5.2. Assumptions: Water and solvent are mutually insoluble. Equilibrium stages. Find: The effect on % extraction of the number and configuration of stages. Analysis: For a single stage and the cocurrent arrangement with any number of stages, N, Eq. (5-13) applies, where the B subscript has been dropped: X (N) 1 1 = = = 0.526 (F) X 1 + E 1 + 0.9 Therefore, the % extraction = (1 - 0.526) x 100% = 47.4% for any number of cocurrent stages. For a crosscurrent cascade, Eq. (5-21) applies: X (N) 1 = (F) X 1+ E / N 1 = N 1 + 0.9 / N N (1) For a countercurrent cascade, Eq. (5-29) applies: X (N) = X (F) 1 N n=0 E = n 1 N 0.9 (2) n n=0 Results from Eqs. (1) and (2) are used to calculate % extraction from: X (N) × 100% X (F) Using Eqs. (1) to (3), the following results are obtained: % extraction = 1 − N 2 3 4 5 % extraction, crosscurrent 52.4 54.5 55.6 56.3 % extraction, countercurrent 63.1 70.9 75.6 78.7 (3) Exercise 5.13 (continued) We see that for E < 1, the % extraction increases only slowly with increasing number of stages. It is desirable to have the situation where E > 1. Exercise 5.14 Subject: Multistage absorption of a hydrocarbon gas with a high-molecular-weight oil. Given: Hydrocarbon feed gas and absorbent oil of Example 5.3 in Fig. 5.11. Absorber pressure of 400 psia and average temperature of 97.5oF, with K-values at these conditions. Assumptions: Applicability of Kremser's method. Find: % absorption of components and total gas and % stripping of oil for N = 1, 3, 10, and 30 equilibrium stages. Effect of N on each component. Analysis: The same values of the absorption factors, A, and stripping factors, S, apply. Also Eqs. (5-48) and (5-50) are used to compute the fractions not absorbed, φΑ, and not stripped, φB, respectively. Then Eq. (5-54) is used to compute the component flow rates in the exit vapor. Then an overall material balance is used to compute the component flow rates in the exit liquid: l N = l0 + υ N +1 − υ 1 Let the % absorption of a component in the entering gas = l N / l0 + υ N +1 × 100% The % stripping of the oil in the absorbent = υ 1 oil / l0 oil × 100% Using a spreadsheet, the following results are obtained for the % absorption of C1, C2, C3, nC4, and nC5, the total % absorption of these light hydrocarbons, and % stripping of oil. N 1 3 6 10 30 C1 3.01 3.10 3.10 3.10 3.10 C2...
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