Page 314 - Applied Process Design For Chemical And Petrochemical Plants Volume II
P. 314

Packed Towers                                           303

           Therefore, bubble point = 46°F                         Refer to Figure 9-30 for plot for this example.
              abottoms = 1.35/0.98 = 1.38                         Refer to Figure 9-29 for diagram of loading. From a mate-
             Dew point on overhead: at 443 psia                   rial balance using (L/D)  = 6.0; theoretical trays = 20
            Composition            K@lj0F      y/K                  (a) Feed = 24,000 lb/day
             Ethylene       0.90     1.05     0.85'7                (b) Distillate product: 13,900 lb/day
             Ethane        0.10      0.7j     0.138                 (c) Reflux  into  column: 83,400 lb/day  @  6:l  reflux,
                            1 .oo             0.995                    L/D

                                                                    (d) Total gross overhead: 97,300 lb/day
           Dew point on overhead: 15°F                              (e) Total bottoms out: 10,100 lb/day
           Bubble point on overhead: Try 10°F at 445 psia

           Composition      a  - K          -                    Top of Column
                                             Kx
           Ethylene         0.90    1.00    0.90
           Ethane           - 0.80          0.08                       (28) (445) (492)  = 2.47 lb/ft3 @ 430 psig & 10°F
                            0.10
                            1 .oo           0.98                  "'  = (359) (14.7) (470)
             Bubble point > 10°F < 15°F E 12°F                    PL = 0.39 (62.4) = 24.3 lb/ft3
                  = 1.05/0.73 = 1.44                               Liquid viscosity = 0.07 cp
                                                                   From Figure 421C:
                  I
            aavg = ;atop abottom = J(1.38) (1.44) = 1.41
                                                                  L/G = dx 83,400/97,300\/2.47/24.3  = 0.2'7
                                                                              =
           Minimum Trays at Total Reflux
                                                                 Read:
             Fenske Equation:
                                                                     G2 FpO.l
            N+1= log (xlk /Xhk  )D (xhk ixlk )B                   PG (PL - PG  gc   = 0.067 (flooding, avg.)
                         1%  Qavg
                                                                 For average loading condition, read ordinate = 0.030
                - log (0.90/0.10) (0.95/0.05)                    For average flooding: (from chart) using 1-in. metal Pall rings,
                -
                         log 1.41                                with F = 56.
                                                                   0.067 = G2 (56) (0.07)'.'/(32.2)  (24.3 - 2.47) (2.47)
                = 14.8 theoretical trays                           G = 1.648, ft/sec/ft2  for average flooding
                                                                   Vloading = 1.648/2.47 = 0.667 ft/.SeC
           Minimum Rejlux
           Gilliland Plot:






                    -  -  0.90 [1+ 0.41 (0.55)] - 0.775  = 3.23
                         (6.41) (0.55) (0.45)

            Use actual L/D  = 6 : 1
           Theoretical Plates us. Reflux
           Gilliland Plot
            m            Theoretical Plates                                       Theoretical trays = 20

             3.23             Infinity
             4.0               28.4
             5.0               22.4
             6.0               20.0
             7.0               18.6
             8.0               17.8
           Infinity            14.2                                        Figure 9-30.  Gilliland Plot for Example 9-1.
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