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Unformatted text preview: , Eq. (739) to this region. Apply this equation to the heavy component, O. Obtain the Kvalue for O from the top of 1.167, taking the Kvalue for P = 1.00. Therefore, KO = 1/1.167 = 0.857. Therefore, the absorption factor, A = L/KV = 0.915/0.857 = 1.067. Other quantities needed in Eq. (739) are: xo = 1  (xD)P = 1  0.98 = 0.02 y1 = xo = 0.02 From Eq. (2), for xN = 0.8, yN+1 for P = 0.8153. Therefore, for O, yN+1 = 1  0.8153 = 0.1847. log NR = 1 1 + 1 A A logA yN +1  xo K y1  xo K log = 1 1 + 1 1.067 1067 . 0.1847  0.02(0.857) 0.02  0.02(0.857) log1.067 = 23.6 Let region (5) extend from x = 0.04 (i.e. xB) to 0.20. Apply the Kremser equation, Eq. (740) to this region. Apply this equation to the light component, P. Obtain the Kvalue for P from the bottom of 1.159, taking the Kvalue for O = 1.00. Therefore, KP = 1.159. Therefore, the absorption factor in the stripping section is A = L / KV = 10535 / 1159 = 0.9085 . Other . . values needed in Eq. (740) are x1 = xB = 0.04 and xN+1 = 0.20. Therefo...
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This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .
 Spring '06
 selebi
 Distillation, pH, Mass Transfer

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