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Gas—General    241

                          Sample calculation 2                   Mol fraction of water vapor = y = P w P         (4)

             Mols of components are given:                       P w = partial pressure of water vapor. It is exactly equal to the
                                                                     saturation pressure if the gas is saturated.
           Ethane           400mols/hr
           Propane          4,800mols/hr                             MW mix =  yMW mix + (1  -  y) ¥  MW dg      (5)
           n-Butane         900mols/hr
           Total Flow       6,100mols/hr                         where w = water vapor
                                                                      dg = dry gas
           Mol % ethane = 400/6,100 = 6.56%
                                                                 m total =  m dg[1  + ( MW w ¥  P w ) ( MW dg ¥  P dg )]  (6)
                                 Table 3
                                                                   Saturated pressure of water vapor at 100°F is equal to
           Component               Mol/hr              Mol %
                                                                 0.95psia, and since the mixture is saturated, the partial pres-
           Ethane                    400                 6.56    sure is also equal to 0.95psia.
           Propane                  4,800               78.69      Calculate mol % of water vapor using Equation 4.
           n-Butane                  900                14.75
                                                                 y = 096 16  = 5 94 mol % water vapor
                                                                     .
                                                                              .
            Totals                  6,100              100.00
                                                                   Calculate MW mix using Equation 5.
             The mol % values listed in Table 4 can be used to calcu-
           late the properties of the mixture following the procedure  MW mix = 0 0594 (18 02.  ) + (1  - 0 059 ) ¥ 44 01.
                                                                                          .
                                                                          .
           detailed in Sample Calculation 1.
                                                                           .
                                                                       = 42 48
                          Sample calculation 3                     Calculate the mass flow of the mix using Equation 6.
             Mass flow of components is given.                    M total = 1 000 [1  + (18 02  ¥ 0 95 ) (44 01.  ¥ (16  - 0 95 ))]
                                                                                                        .
                                                                        ,
                                                                                  .
                                                                                       .
                                                                        ,
                                                                      = 1 026 lbm min
                                 Table 4
                        Mass Flow            Mol flow
           Component      (kg/h)    Mol wt   (kmol/h)   Mol %
           Ethane         15,000    30.070      499     11.08
           Propane       150,000    44.097    3,402     75.55
           n-Butane       35,000    58.123      602     13.37
            Totals                            4,503    100.00


           Mol flow of ethane = 15,000/30.07
                           = 499
           Mol % of ethane = 499/4,503
                         = 11.08%

             When a dry gas is saturated with water, it will be necessary
           to consider the effects of the water on the molecular weight
           of the mixture. The water will also affect the total mass flow
           of the mixture.


                          Sample calculation 4

             Given: Compress 1,000lbm/min dry CO 2 , which is initially
           water saturated. MW = 44.01, P inlet = 16psia, T inlet = 100°F.  Figure 1. Compressibility chart for very low values of reduced
             Calculate the molecular weight of the mixture and the total  pressure. Reproduced by permission of Chemical Engineering,
           mass flow for the required 1,000lbm/min of dry CO 2 .  McGraw Hill Publications Company, July 1954.
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