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

0 ya top ya bottom 10 ya bottom 0401 00882 0001609

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Unformatted text preview: s = 169 psia, and yA b = 0.001609. At the top, PAs = 1995 psia, yA b = 0.0882, T = 126oF, . . and P = 14.3 psia. Assumptions: All mass-transfer resistance resides in the gas phase because pure water is used. Assume air rate pertains to moisture-free air. Bulk-flow effect may be appreciable. Find: Overall mass-transfer coefficient, Ky Analysis: From a rearrangement of Eq. (3-234) and including the bulk-flow effect, nA 10 − yA LM . Ky = (1) A yA i − yA b LM However, no data are given on the packing to enable the determination of the area, A. Therefore, let a = area, A, for mass transfer per unit volume, V, of tower. Then Eq. (1) becomes, Kya = nA 10 − yA . LM V yA i − yA b LM (2) Therefore, Kya will be determined instead of Ky . The rate of mass transfer of water into the air is obtained by material balance, with the water mole fraction data and the air flow rate (water-free basis), where, nA = n yA top 1.0 - yA top − yA bottom 1.0 - yA bottom = 0.401 0.0882 0.001609 − = 0.0381 lbmol/h 1.0 - 0.0882 1.0 - 0.001609 The volume over which the data were taken = V = SH = 0.5(0.7) = 0.35 ft2 Interface mole fractions of water vapor are, yA itop = PAstop / Ptop = 1995 / 14.3 = 0.1395 . yA i bottom = PAsbottom / Pbottom = 1.69 / 14.1 = 01199 . Exercise 3.40 (continued) Analysis: (continued) y A i − yA b LM = yA itop − yA btop − yA ibottom − yAbbottom ln = yA itop − yA btop yA ibottom − yA bbottom 01395 − 0.0882 − 01199 − 0.001609 . . = 0.0802 01395 − 0.0882 . ln 01199 − 0.001609 . For the bulk-flow effect, (1.0 - yA)LM at the top = 1 − yA i top ln − 1 − yA b top = bottom = 1 − yA i top 1 − yA b . 1 − 01395 − 1 − 0.0882 = 0.886 1 − 0.1395 ln 1 − 0.0882 top (1.0 - yA)LM at the bottom = 1 − yA i bottom ln − 1 − yA b 1 − yA i bottom 1 − yA b 1 − 01199 − 1 − 0.001609 . = 0.938 1 − 0.1199 ln 1 0.001609 bottom The average value of (1.0 - yA)LM is 0.912. From Eq. (2), K ya = 0.0381(0.912) = 1.24 lbmol/h-ft3-unit mole fraction (0.35)(0.0802) For the same packing and gas and liquid velocities, this value can be used to design a large unit. Exercise 4.1 Subject: Degrees of freedom analysis for a three-phase equilibrium stage. Given: Equilibrium stage of Figure 4.35, with two feeds (one vapor, one liquid), vapor entering from stage below, liquid entering from stage above, three exiting streams (one vapor, two liquid), and heat transfer. Assumptions: Equilibrium stage Find: (a) (b) (c) (d) Analysis: List and count of variables. List and count of equations. Number of degrees of freedom. List of reasonable set of design variables. (a) With 4 streams in and 3 streams out, and heat transfer, Number of variables = NV = 7(C+3) + 1 = 7C + 22 The variables are 7 total flow rates, 7 temperatures, 7 pressures, 1 heat transfer rate and C mole fractions for each of the 7 streams. (b) The equations are: C Component material balances 1 Energy balance 2 Pressure identity equations for the 3 exiting streams 2 Temperature identity equation...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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