Page 224 - Separation process engineering
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(5-16)

                    At the feed stage, q can be estimated from enthalpies as





                                                                                                                                (5-17)

                    or q = L /F can be found from a flash calculation on the feed stream. Then   and   are determined from
                             F
                    balances at the feed stage,




                                                                                                                                (5-18)

                    This completes the preliminary calculations for this assumption of how the LNK splits in the column.

                    If there is an LNK and no HNK, the assumed compositions are very accurate at the top of the column but
                                                                                                                                -9
                    not at the bottom. (At the bottom, the mole fraction cannot be exactly zero. If we assume it is 10  when it
                                   -11
                    is actually 10 , then our relative error is very large.) Thus, with only an LNK present, we want to step
                    off stages from the top down. The general procedure for stepping off stages down the column when CMO
                    is valid is:

                     1. Set j = 1. For total condenser, y  = x    i,dist , where i is component A, B, or C, and the second subscript is
                                                            i,l
                        the stage location.

                     2. Use equilibrium to calculate x  values from known y  values for stage j.
                                                          i,j                       i,j
                     3. Use mass balances (operating equations) to calculate y  values from known x  values.
                                                                                                                 i,j
                                                                                      i,j+l
                     4. Repeat steps 2 and 3 until the feed stage is reached. Then change to the stripping section operating
                        equations and continue.

                     5. The calculation is finished when x     HK,N+1  ≥ x HK,bot  and x LK,N+1  ≤ x LK,bot

                    If the relative volatilities are constant, the equilibrium calculations become simple. We arbitrarily choose
                    component B as the reference. Then by definition,






                                                                                                                                (5-19)

                    As we step down the column, the y values leaving a stage will be known and the x values can be
                    calculated from equilibrium. Thus for component i on stage j,






                                                                                                                                (5-20)

                    Note that the first equals sign for equation (5-20) reduces to x  = x .
                                                                                               i,j
                                                                                          i,j
                    In general, both α  and K  depend upon temperature and thus vary from stage to stage. When the
                                                  B,j
                                        iB,j
                    relative volatilities are constant, only K  varies. Fortunately, because the liquid mole fractions must sum
                                                                 B,j
                    to 1.0, we can obtain an equation for K  and then remove it from Eq. (5-20). For this ternary problem,
                                                                B,j
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