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

Separation Process Principles 2n Seader& Henley Solutions Manual

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Unformatted text preview: h Benzoic acid Nitrogen Water Total 0.634 5.770 0.150 6.554 0.00519 0.20600 0.00832 0.21951 Mol fraction 0.0236 0.9385 0.0379 1.0000 Partial Pressure if no condensation, torr 17.9 713.3 28.8 760.0 Exercise 17.37 (continued) Now determine the exit temperature and the amount of BA desublimed. Assume that water does not condense. Let x = fraction of BA desublimed and T be the exiting temperature Assume a thermodynamic path of all water evaporated at 25oC and BA condensed at the final temperature (however, it doesn’t matter because BA and N2 specific heats are assumed constant regardless of the phase state). An enthalpy balance can be written as: Cooling of feed gas to T and desublimation of BA = vaporization of added water and heating of gas of added water and added N2 to T. Using cgs units, with a heat of vaporization for water = 583 cal/g at 25oC, 2,760(0.32)(177.5 – T) + 634x(134) = 150(583) + 150(0.44)(T – 25) + 1,354(0.32)(T – 105) + 2,290(0.32)(T – 25) (1) This balance must be consistent with a partial pressure of non-desublimed BA in the gas equal to the vapor pressure at T. For a set of assumed values of T, compute x and compare the partial pressure of BA to its vapor pressure, using a spreadsheet. We need an interpolating equation for the vapor pressure. A fit of the Clausius-Clapeyron equation from 96 to 105oC with Polymath, gives, s log10 ( PBA in torr ) = 9.679 − T( o 3572 C ) + 273 The spreadsheet results are: T, C x Vapor Pressure, torr kmol/h BA in vapor Partial Pressure, torr 96 97 98 99 100 101 102 103 105 0.804 0.829 0.854 0.878 0.903 0.928 0.953 0.978 1.028 1.00 1.06 1.12 1.19 1.27 1.34 1.42 1.51 1.70 0.001018 0.000889 0.000760 0.000631 0.000502 0.000372 0.000243 0.000114 -0.000145 3.59 3.14 2.69 2.23 1.77 1.32 0.86 0.40 -0.51 The vapor pressure = the partial pressure at an exit temperature of 101oC with 93% of the BA desublimed. Check to determine if any water is condensed at this temperature. The vapor pressure of water at 101oC exceeds 760 torr. Since the partial pressure of water in the exiting gas is much less than this, the assumption of no condensation of water is valid. Exercise 17.38 Subject: Desublimation on the outside of a heat exchanger tube. Given: Heat exchanger tube with a coolant flowing through the inside and desublimation occurring on the outside surface of the tube. Assumptions: Rate of desublimation controlled by heat conduction through the deposited crystal layer. Sensible heat effect is negligible compared to the latent heat of desublimation. Temperatures of the gas and the coolant assumed constant over tube length. Find: Derive (17-75), which gives the time for a crystal layer to grow from the outside tube radius, ro, to a radius, rs. Analysis: Equation (17-75) gives, t= ρc ∆H s rs2 r 1 ln s − ( rs2 − ro2 ) ro 4 kc (Tg − Tc ) 2 (1) Rate of heat released by desublimation = rate of heat conduction through the crystal layer. Therefore, ∆H s dm dr kc ALM (Tg − Tc ) = ∆H s 2πrLρc = dt dt r − ro (2) where, ∆H s = heat o...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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