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

Because rates instead of fluxes are given the

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Unformatted text preview: as when taking into account entrainment, occlusion, backmixing, weeping, and tray hydraulics. The percentage reduction in the number of variables and equations for the rate-based model is not nearly as great as for the equilibrium model because, at least not for many components, that number is dominated by the 5C term. Exercise 12.4 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 acetone (1), methanol (2), and water (3) at 14.7 psia 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, z1 = (0.4913 + 0.5291) / 2 = 0.5102 z2 = (0.4203 + 0.4070) / 2 = 0.4136 z3 = (0.0884 + 0.0639) / 2 = 0.0762 z z z 0.5102 0.4136 0.0762 V R11 = 1 + 2 + 3 = + + = 0.000509 h / lbmol k13 k12 k13 2154 1750 2154 z z z 0.4136 0.5102 0.0762 V R22 = 2 + 1 + 3 = + + = 0.000487 h / lbmol k23 k 21 k 23 2503 1750 2503 V R12 = − z1 1 1 1 1 = −0.5102 − = −0.000055 h / lbmol − 1750 2154 k12 k13 V R21 = − z2 1 1 1 1 = −0.4136 − = −0.000071 h / lbmol − 1750 2503 k 21 k 23 In matrix form: RV = 0.000509 − 0.000055 −0.000071 0.000487 From Eq. (12-29), by matrix inversion, κ V = RV −1 = 1996 225.4 291 2086 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%. Exercise 12.4 (continued) Analysis: (continued) 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 I J1V = κ11 ( y1 − y1I ) + κV2 ( y2 − y2 ) 1 = 1996 ( 0.4913 − 0.5291) + 225.4(0.4203 − 0.4070) = −72.45 lbmol/h V I J 2 = κV ( y1 − y1I ) + κV2 ( y2 − y2 ) 21 2 = 291(0.4913 − 0.5291) + 2086(0.4203 − 0.4070) = 16.74 lbmol/h From Eq. (12-23): V V J 3 = − J1V − J 2 = 72.45 − 16.74 = 55.71 lbmol/h (b) NT = Vn+1 - Vn = 1164 - 1200 = -36 lbmol/h. From Eq. (12-19), but with diffusion and mass-transfer rates instead of fluxes, V N1V = J1V + z1 NT = −72.45 + 0.5102(-36) = -90.8 lbmol/h V V V N 2 = J 2 + z2 NT = 16.74 + 0.4136(−36) = 1.9 lbmol/h V V V N 3 = J 3 + z3 NT = 55.71 + 0.0762(−36) = 53.0 lbmol/h (c) Approximate values of the Murphree vapor tray efficiency are obtained from (12-3), with K-values at phase interface conditions: E MVi = yi ,n − yi ,n +1 / KiI,n xi ,n − yi ,n +1 Accordingly: EMV1 = ( 0.4913 − 0.4106 ) (1.507 )( 0.3683) − 0.4106 = 0.559 = 55. 9% EMV2 = ( 0.4203 − 0.4389 ) ( 0.900 )( 0.4487 ) − 0.4389 = 0.530 = 53. 0% EMV3 = ( 0.0884 − 0.1505 ) ( 0.3247 )...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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