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

066 0840 1190 1080 0951 1143 0898 1114 1206 451 px

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Unformatted text preview: 8 − θ 100 − θ . . . . i α i ,OX − θ There are two roots to Eq. (2), which from a spreadsheet calculation are: θ = 1.0280 and 1.2027 Eq. (9-29) is applied in the following form for each of the two values of θ, in terms of the two unknowns, dEB and Lmin. The form used is obtained by multiplying Eq. (9-29) by the total distillate rate, D, which converts distillate mole fractions to component distillate rates, and the reflux ratio, R to L. The distillate rate of PX is not an unknown because it has the same relative volatility as MX. Thus, it must have the same d/b ratio as MX, making its distillate rate as 99 kmol/h. D + Lmin = 12217 (d EB ) . 1.1448 (297) 1.00 (1) + + 12217 − 10280 11448 − 1.0280 1.00 − 10280 . . . . = 6.307(d EB ) + 2875.3 D + Lmin = 12217(d EB ) . 1.1448(297) 1.00 (1) + + 12217 − 12027 11448 − 1.2027 1.00 − 1.2027 . . . = 64.30(d EB ) − 5877.2 Exercise 9.28 (continued) Analysis: (a) (continued) Because the total distillate rate, D, is also an unknown, the following equation is also needed: D = dEB + 99 + 198 + 1 = dEB + 298 Solving these 3 linear equations, the following results are obtained: D =448.9 kmol/h, Lmin = 3378 kmol/h, and dEB = 150.9 kmol/h. Thus, the assumption that EB distributes is incorrect because only 100 kmol/h of EB is in the feed. Therefore discard the root of θ = 1.2027. Also, D now equals 100 + 99 + 198 + 1 = 398 kmol/h. From above for the θ root of 1.0280, D + Lmin = 6.307 dEB + 2875.3. Therefore, Lmin = 6.307(100) + 2875.3 - 398 = 3108 kmol/h or Rmin = Lmin/D = 3108/398 = 7.81 The operating reflux ratio = 1.1(7.81) = 8.59. In the Gilliland equation, (9-34), X = (R - Rmin)/(R + 1) = (8.59 - 7.81)/(8.59 + 1) = 0.0813. Using Eq. (9.34), Y = 0.573 = (N - Nmin)/(N + 1). Solving, for Nmin = 68.3, N = 161.3 equilibrium stages. The Kirkbride Eq. (9-36) is used to determine the feed-stage location. NR = NS zOX , F x MX , B zMX , F xOX , D 2 0.206 B D = 0.20 0.40 0.0196 0.00251 2 102.01 397.99 0.206 = 153 . Therefore, the feed stage is located at (1.53/2.53) 161.3 = stage 97.5 from the top. (b) The Edmister group method is applied by using Eq. (5-66) to estimate the b/d ratio for each component: b = AF d AC φ SE + 1 φ AE S B φ AX + 1 φ SX (3) For a total condenser, AC = the reflux ratio = 8.59 for all components. To compute values of φ for each component, assume the following values of flow rates and stages for the enriching (rectifying) section and exhausting (stripping) section, based on constant molar overflow and noting that the feed and reboiler stages are not included in these two sections. Variable Number of stages Vapor rate, kmol/h Liquid rate, kmol/h Enriching Section 97 3817 3419 Exhaustin g Section 63 3817 3919 Exercise 9.28 (continued) Analysis: (b) (continued) Because in the above table the K-values do not vary much, use arithmetic mean values for each section. These values and those for the feed stage and partial reboiler are as follows along with the corresponding values of the absorption and stripping factors, as defined by equations (5-38), (5-51), below (5-64), and (5-65), where VB/B = 3817/102 = 37.4 Variable KE AE SE = 1/AE KF AF KX AX SX = 1/AX KB SB EB 1.066 0.840 1.190 1.080 0.951 1.143 0.898 1.114 1.206 45.1 PX 0.999 0.897 1.115 1.012 1.015 1.072 0.958 1.044 1.131 42.3 MX 0.997 0.898 1.114 1.012 1.015 1.074 0.956 1.046 1.136 42.5 OX 0.870 1.030 0.971 0.884 1.161 0.940 1.092 0.918 0.996 37.3 Values of φA and φS are obtained from Eqs. (5-48) and (5-...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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