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

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Unformatted text preview: ( 0.1830 ) − 0.1505 = 0.682 = 68.2% Exercise 12.5 Subject: Reciprocal rate functions for a five-component system. Given:. Eqs. (12-31) and (12-32) for a general C-component system. Find: Expanded forms of Eqs. (12-31) and Eq. (12-32). Analysis: From Eq. (12-31): P R11 = z1 z z z z + 2+ 3+ 4+ 5 P P P P P k15 k12 k13 k14 k15 P R22 = z2 z z z z + 1+ 3+ 4+ 5 P P P P P k25 k 21 k 23 k 24 k 25 P R33 = z z z z z3 + 1+ 2+ 4+ 5 P P P P P k 35 k 31 k 32 k 34 k 35 P R44 = z4 z z z z + 1+ 2+ 3+ 5 P P P P P k45 k 41 k 42 k 43 k 45 P R12 = − z1 1 1 −P P k12 k15 P R13 = − z1 1 1 −P P k13 k15 1 1 −P P k14 k15 P R21 = − z2 1 1 −P P k 21 k 25 P R23 = − z2 1 1 −P P k 23 k 25 P R24 = − z2 1 1 −P P k 24 k 25 P R31 = − z3 1 1 −P P k 31 k 35 P R32 = − z3 1 1 −P P k 32 k 35 P R34 = − z3 1 1 −P P k 34 k 35 P R41 = − z4 1 1 −P P k 41 k 45 P R42 = − z4 1 1 −P P k 42 k 45 P R43 = − z4 1 1 −P P k 43 k 45 P R14 = − z1 Exercise 12.6 Subject: Mass-transfer rates and tray efficiencies from data for a perfectly mixed tray. Given: Vapor-liquid traffic, component mole fractions, and mass-transfer coefficients for a tray in a column separating methanol (1), water (2), and acetone (3) at 14.7 psia from Example 12.1, but with the designations 1, 2, 3 changed to see if different results are obtained. Find: (a) Component molar diffusion rates. (b) Component mass-transfer rates. (c) Murphree vapor tray efficiencies. Analysis: Because mass-transfer coefficients are given for the gas phase, work with the values of yi ,n and yiI,n . Because rates instead of fluxes are given, the equations developed in this section are used with rates rather than fluxes. (a) Compute the reciprocal rate functions, R, from Eqs. (12-31) and (12-32), assuming linear mole fraction gradients such that zi can be replaced by ( yi + yiI ) / 2 . Thus, with the new designations for 1, 2, 3, the values from Example 12.1 become: z1 = 0.4654 z2 = 0.2100 z3 = 0.3246 z z z 0.4654 0.2100 0.3246 V R11 = 1 + 2 + 3 = + + = 0.000479 h / lbmol k13 k12 k13 1955 2797 1955 z z z 0.2100 0.4654 0.3246 V R22 = 2 + 1 + 3 = + + = 0.000388 h / lbmol k23 k 21 k 23 2407 2797 2407 V R12 = − z1 1 1 1 1 = −0.4654 − = −0.0000717 h / lbmol − 2797 1955 k12 k13 V R21 = − z2 1 1 1 1 = −0.2100 − = −0.0000122 h / lbmol − 2797 2407 k 21 k 23 In matrix form: RV = 0.000497 0.0000717 0.0000122 0.000388 From Eq. (12-29), by matrix inversion, κ V = RV −1 = 2021 − 3735 . −63.6 2589 Exercise 12.6 (continued) Analysis: (continued) Because the off-diagonal terms in the above 2 x 2 matrix are much smaller that the diagonal terms, the effect of coupling in this example is small, approximately 10%. From Eq. (12-27) with units of JV in lbmol/h V J1 V J2 = κV 11 κV 12 y1 − y1I κV 21 κV 22 I y2 − y2 V V I J1V = κ11 ( y1 − y1I ) + κ12 ( y2 − y2 ) = 2021( 0.4631 − 0.4677 ) − 373.5(0.2398 − 0.1802) = −31.6 lbmol/h I J = κV ( y1 − y1I ) + κV2 ( y2 − y2 ) 21 2 V 2 = −63.6(0.4631 − 0.4677) + 2589(0.2398 ...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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