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x  since ethanol is the more volatile component.) Vapor stream y  is a passing stream relative to x            1
                                                                                                2
                        1
                      and can be determined from the operating Eq. (4-12).



                      Stream x  is in equilibrium with y . From Figure 2-2 we obtain x  = 0.11.
                                                             2
                                                                                                2
                                2
                      Since stage 2 is the feed stage, use bottom operating Eq. (4-14) for y .
                                                                                                    3



                      Stream x  is in equilibrium with y . From Figure 2-2, this is x  = 0.02. Since x  = x  (in mole
                                3                            3                              3                  3    B
                      fraction), we are finished.

                      The third equilibrium contact would be the partial reboiler. Thus the column has two equilibrium
                      stages plus the partial reboiler.
                      E. Check. This is a small number of stages. However, not much separation is required, the external
                         reflux ratio is large, and the separation of ethanol from water is easy in this concentration range.
                         Thus the answer is reasonable. We can check the calculation of L/V with mass balances.

                         Since V  = L  + D,
                                       0
                                  1




                         Since L , V , and D, are the same composition, L /D and L /V  have the same values in mass and
                                 0
                                                                                0
                                                                                               1
                                                                                           0
                                      1
                         molar units. We can check the equilibriom calculation with Eq. (2.B-2). For example, for y  = 0.61
                                                                                                                                1
                         we obtain x  = 0.385.
                                      1
                      F. Generalizations. We should always check that CMO is valid. Then convert all flows and
                         compositions into molar units. The procedure for stepping off stages is easily programmed on a
                         spreadsheet (Burns and Sung, 1996). We could also have started at the bottom and worked our way
                         up the column stage by stage. Going up the column we calculate y values from equilibrium and x
                         values from the operating equations.

                         Note that L/V < 1 and  /  > 1. This makes sense, since we must have a net flow of material
                         upwards in the rectifying section (to obtain a distillate product) and a net flow downwards in the
                         stripping section. We must also have a net upward flow of ethanol in the rectifying section (Lx <
                                                                                                                                    j
                         Vy ) and in the stripping section ( x <  y ). These conditions are satisfied by all pairs of
                                                                    j
                                                                          j+1
                            j+1
                         passing streams.

                    The Lewis method is obviously much faster and more convenient than the Sorel method. It is also easier
                    to program on a computer or in a spreadsheet. In addition, it is easier to understand the physical reasons
                    why separation occurs instead of becoming lost in the algebraic details. However, remember that the
                    Lewis method is based on the assumption of CMO. If CMO is not valid, the answers will be incorrect.

                    If the calculation procedure in the Lewis method is confusing to you, continue on to the next section. The
                    graphical McCabe-Thiele procedure explained there is easier for many students to understand. After
                    completing the McCabe-Thiele procedure, return to this section and study the Lewis method again.

                    4.3 Introduction to the McCabe-Thiele Method

                    McCabe and Thiele (1925) developed a graphical solution method based on Lewis’ method and the
                    observation that the operating Eqs. (4-12b) and (4-14) plot as straight lines (the operating lines) on a y-x
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