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

# Concentration to 40 wt naoh at a pressure of 37 psia

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Unformatted text preview: 0.130 = 1.508 kmol/h = 184 kg/h. The ratio of mass of entering gas mixture to entering BA mass = Mr = 28.02 ( 0.992 ) + 122.12 ( 0.008 ) 122.12 ( 0.008 ) Sensible heat + heat of fusion per gram of BA desublimed = = 29.45 M r CPg (Tin − Tout ) g + ∆H s = 29.45(0.32)(130 − 80) + 134 = 605 cal/g 1.316(605) From (17-75), time = t = 1.4(40) Number of tubes = 184 (14.2 ) 1.94 = 1,350 ( 2.5 ) 2 2 ln 2.5 1 − ( 2.52 − 1.252 ) = 14.2 h 1.25 4 Exercise 17.37 Subject: Bulk-phase desublimation of benzoic acid from N2 by injection of water followed by two steps of nitrogen quench. Given: 3 m3/h of gas at 1 atm and a temperature of 10oC above the dew point and containing 6.4 mol% benzoic acid (BA) and 93.6 mol% N2. 150 cm3/h of water at 25oC is added. Then the gas is further cooled in two steps with N2 at 1 atm, first with 1.5 m3/h at 105oC, and secondly with 2.0 m3/h at 25oC. The following are properties of BA: Melting point = 122.4oC Specific heat of solid and gas = 0.32 cal/g-oC Heat of sublimation = 134 cal/g. Therefore, desublimation is exothermic at 134 cal/g Vapor pressure of BA: 96 105 119.5 132.1 146.7 162.6 172.8 Temp., oC Vapor 1 1.7 5 10 20 40 60 Pressure, solid solid solid liquid liquid liquid liquid torr Assumptions: Equilibrium. Desublimation occurs in the bulk gas. Adiabatic process. Find: The final gas temperature and the fractional yield of solid benzoic acid crystals. Analysis: A plot of the vapor pressure data, using a spreadsheet is shown on the following page. Note the change in slope at the melting point. MW of BA = 122.12 MW of H2O = 18.02 MW of N2 = 28.01 First find the entering gas temperature by computing the dew point temperature. Partial pressure of BA in entering gas = 0.064(760) = 49 torr. From the vapor pressure curve, this is in the liquid region with a temperature of 167.5oC. Therefore, the entering gas temperature is 167.5 + 10 = 177.5oC Next, calculate the mass flow rate of the entering gas, using SI units. MW of entering gas = M = 0.064(122.12) + 0.936(28.01) = 34.03 Temperature = 177.5 + 273.2 = 450.7 K Pressure = 1 atm = 101.3 kPa Therefore, from the ideal gas law, ρ= PM (101.3) ( 34.03) = = 0.920 kg/m3 RT ( 8.314 )( 450.7 ) Entering gas flow rate = 3(0.920) = 2.76 kg/h or 2.76/34.03 = 0.0811 kmol/h This gives 0.064(0.0811) = 0.00519 kmol/h or 0.634 kg/h of BA and 2.76 – 0.634 = 2.126 kg/h of N2 Exercise 17.37 (continued) Vapor Pressure, torr 100 10 1 80 90 100 110 120 130 140 150 160 170 180 Temperature, C Water at a flow rate of 150 cm3/h or 0.150 kg/h is now added. Now add 1.5 m3/h of N2 at 105oC and 1 atm. From the ideal gas law, PM (101.3) ( 28.02 ) ρ= = = 0.903 kg/m 3 RT ( 8.314 )( 378.2 ) Therefore the mass flow rate of N2 added = 1.5(0.903) = 1.354 kg/h Next, add 2.0 m3/h of N2 at 25oC and 1 atm. From the ideal gas law, PM (101.3) ( 28.02 ) ρ= = = 1.145 kg/m3 RT ( 8.314 )( 298.2 ) Therefore the mass flow rate of N2 added = 2.0(1.145) = 2.290 kg/h Thus, the final exiting combined gas and solid is as follows: Component kg/h kmol/...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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