# 2 let region 1 extend from x 08 to 098 ie xd apply

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Unformatted text preview: , Eq. (7-39) to this region. Apply this equation to the heavy component, O. Obtain the K-value for O from the top of 1.167, taking the K-value 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. (7-39) 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. (7-40) to this region. Apply this equation to the light component, P. Obtain the K-value for P from the bottom of 1.159, taking the K-value 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. (7-40) 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 .

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