Unformatted text preview: 2 4(0.977) = ( 0.977  0.0412 ) 0.408 92.34 1/ 2 ( 6,150 ) ( 0.54 ) 1/ 3 (1.80)(92.3) = 0.14 m (0.059 10 4 )(191) At the bottom of the column, vapor rate = V =437 lbmol/h and liquid rate = L = 612 lbmol/h. Therefore, V/L = (437)/(612) = 0.714 mV From Eq. (7.51), HOG = HG + HL L Therefore, values of HOG depend on the slope of the equilibrium line, which varies widely from the top of the column to the bottom. At the top, assuming HG and HL are constant, mV H L = 0.23 + 1.63(0.154) m = 0.23 + 0.25 m L At the bottom, assuming HG and HL are constant, H OG = H G + H OG = H G + mV H L = 0.14 + 0.714(0.126) m = 0.14 + 0.09 m L (d) The packed height is given by Eq. (6127), lT = HOGNOG = HETP Nt Apply these equations in a stagebystage fashion, using the equilibrium stages in Fig. 7.15. For each stage, convert the value of HOG to HETP with Eq. (7.53) as in Table 7.7 for Example 7.9, HETP = HOG ln/(1) Analysis: (d) (continued) Exercise 7.53 (continued) Using a spreadsheet, the following results are obtained: Rectifying Section:
Stage m 1 0.47 2 0.53 3 0.61 4 0.67 5 0.72 6 0.80 Total height = = mV/L...
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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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