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

An enthalpy balance can be written as cooling of feed

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Unformatted text preview: the impurity does not occur in the solid phase. Initial impurity concentration is uniform at wo. Impurity concentration in the melt zone is in equilibrium with that in the solid phase upstream of the melt zone. Find: The concentration profile for Fe, and the average concentration, using (1) in Exercise 17.33. Analysis: For the concentration profile, use (17-73), ws = wo 1 − (1 − K ) exp − Kz l = 0.02 1 − (1 − 0.29 ) exp − 0.29 z l Use a spreadsheet to compute the concentration profile for ws as a function of z/l from 0 to 9. The results are: z/l 0 1 2 3 4 5 6 7 8 9 ws 0.0058 0.0094 0.0120 0.0141 0.0155 0.0167 0.0175 0.0181 0.0186 0.0190 The expression in Exercise 17.33 is: wavg = wo l (1 − K ) K ( z2 − z1 ) exp − z2 K zK − exp − 1 l l +1 where, here, K = 0.29 , wo = 0.02 , z1 = 0 , take l/L = 0.1, and use z2 = 0.75 z . (1) Exercise 17.35 (continued) Therefore, and, l l/z 1/ 9 = = = 0.148 z2 − z1 z2 / z − 0 0.75 − 0 z2 0.75 z = = 6.75 l z /9 Substitution into (1) gives, wavg = 0.02 (1 − 0.29 ) 0.29 ( 0.148 ) exp [ −6.75(0.29] − exp ( 0 ) + 1 = 0.0138 Exercise 17.36 Subject: Desublimation unit of heat exchanger type. Given: Feed gas of 0.8 mol% benzoic acid (BA) and 99.2 mol% N2 at 780 torr and 130oC, containing 200 kg/h. Exit temperature is 80oC with no pressure drop. Cooling water flows countercurrently through the inside of the tubes, entering at 40oC and exiting at 90oC. From Example 17.15, tubes are 1 m long and 2.5 cm in outside diameter Assumptions: Equilibrium Find: Number of tubes needed and the time to reach the maximum crystal thickness of 1.25 cm. Analysis: The following properties apply: MW of BA = 122.12, MW of N2 = 28.02 Melting point = 122.4oC Thermal conductivity of BA crystals = 1.4 cal/h-cm-oC Density of BA crystals = 1.316 g/cm3 Specific heat of solid and vapor = 0.32 cal/g-oC Heat of sublimation = 134 cal/g. Therefore, desublimation is exothermic at 134 cal/g Vapor pressure of BA: Temp., oC 96 105 119.5 132.1 146.7 162.6 172.8 Vapor 1 1.7 5 10 20 40 60 Pressure, solid solid solid liquid liquid liquid liquid torr A plot of the vapor pressure data follows: Vapor Pressure, torr 100 10 1 80 90 100 110 120 130 140 Temperature, C 150 160 170 180 Exercise 17.36 (continued) If the thickness of the dusublimate is uniform, mass of BA desublimated = π ( rs2 − ro2 ) Lρc = 3.14 (1.25 + 1.25 ) − (1.25 ) (100)(1.316) = 1,940 g = 1.94 kg 2 2 Entering gas contains 200 kg/h BA = 200/122.12 = 1.638 kmol/h Therefore, the N2 flow rate = 1.638(99.2/0.8) = 203 kmol/h Partial pressure of BA in entering gas = 0.008(780) = 6.24 torr compared to a BA vapor pressure of 9 torr at 130oC. This confirms that all entering BA is in the vapor phase. Exit gas temperature is 80oC, which is below the melting point of BA. Therefore, it will desublime as a solid. We must extrapolate the vapor pressure data to estimate the vapor pressure at 80oC. The value is approximately 0.5 torr. Therefore, the flow rate of BA in the exiting gas is 203[0.5/(780 – 0.5)] = 0.13 kmol/h Therefore, BA desublimes at the rate of 1.638 –...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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