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0.2 0.4 0.6 0.8 Mol Fraction in Liquid Fig. 9-5.

Mol Fraction in Li quid Fig. 9-6.

Mol Fraction in Li quid Fig. 9-6.

curves could be drawn on the diagrams, the problem would become similar to the stepwise procedure for a binary mixture. However, in general, such curves are not known. The Lewis and Cope method was to draw a series of equilibrium curves of the type y = Kx which at constant temperature are in general straight lines through the origin of slope K. Thus, in the present example, K = P/760, where P of any one component is a function of the temperature only. Such equilibrium curves have been drawn in for the temperatures of 105, 110, and 115°C. Starting at xw on each plot, vertical lines are drawn through this point cutting the equilibrium curves. By trial and error, temperatures are tried until the sum of the y values at the intersection of the vertical line through Xw and the equilibrium curve for the assumed temperature adds up to unity. Thus, if 115°C. is tried, the sum of the y values at the intersection of the 115°C. curve with the xw lines is 0.013 + 0.837 + 0.13 = 0.98, indicating that 115°C. is too low. By interpolation at 116°C., the sum becomes 0.013 + 0.855 + 0.132 = 1.00, indicating that this is the correct temperature, and the y values give the composition yw of the vapor in equilibrium with xw. Horizontal lines are then drawn through the yw values to the operating line, the abscissa of the intersection with the operating line being x\. Vertical lines are drawn through the Xi's; and by using the same procedure as for xw a temperature of 112.5°C. is found to give equal to unity, and the step is then completed to the operating line. In a like manner, steps are taken up the column. The same operating line is used until the feed plate is reached, and theh the change is made to the operating line for the upper portion of the column simultaneously for all three components.

A comparison of the values of these figures with those obtained in the previous algebraic calculation shows the close agreement. Actually, they have to give the same result, since they both are solutions ol the same set of equations, one being algebraic and the other graphical, Both methods have their advantages; in the algebraic method, as a rule,"higher accuracy can be obtained"than in the graphical method this iff especially true in the low-concentration region where the graphical diagrani must be greatly expanded or replotted on logarithmic paper, such as was utilized in the binary mixtures. The advantage o1 the graphical method is that it gives a visual picture of the concentration gradients and operation of the tower. The amount of labor anc time consumed i^ approximately the same for the two methods.

Numerous analytical methods based on the foregoing methods hav< been proposed to simplify the trial and error required in the Lewis anc

Matheson method. Some of these methods will be considered in a later section, but generally the stepwise method outlined above is more satisfactory. By using y = Kx/^Kx instead of making the Kx's add to exactly unity, the trial-and-error work of the Lewis and Matheson method is practically eliminated. In the example just solved, when XKx became larger than 1, the temperature was dropped, making 2Kx less than 1; and this temperature was used until %Kx again became greater than unity, and then the temperature was again dropped. Thus, no actual trial and error was required, but merely successive drops of temperature of 5 to 10°. Such calculations require only a few hours more than the simplest of the approximate methods and only two tor three such stepwise calculations at different reflux ratios together with the minimum number of plates at total reflux and the minimum reflux ratio are required to allow the construction of a curve of theoretical plates required vs. the reflux ratio. In general, the added confidence that may be placed in the stepwise calculations relative to the approximate methods more than justifies the extra work involved.

In using the stepwise method with the simplification that y = Kx/%Kx, the problem arises as to how much IZKx can differ from unity and still not appreciably affect the values of y. The justification of this simplification is that for moderate changes in temperature the percentage change in the values of if for substances that do not differ too widely is approximately the same. A little consideration will show that if all the K values change the same percentage with temperature, then the values of y calculated by such a method will be independent of the temperature chosen. This relative variation in the K values is best expressed in the relative volatility. Thus, if yA = KAxA and yB = KbXb, then yA/yB = (KA/KB)(xA/xB), and (KA/KB) is the relative volatility of A to B, aAB (see page 30). If the percentage change in both Ka and KB is the same with temperature, aAB will be a constant over this region, and a plot of aAB vs. temperature will give immediately the region over which IZKx can vary without appreciably altering the y value. Actually, the a's can be introduced into the equations, and the K's eliminated. Thus,

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