Barreras para el desarrollo e integración de las fuentes no convencionales de

The method leading to Equation11.27for computing the number of stages required to carry out a specified separation depends on the fact that all of the equations involved are linear in x_Aand y_A. Nonlinearities can occur because the equilibrium relation is not linear or because changes in mass flow rates resulting from interphase transfer must be accounted for. Although the full set of nonlinear equations would normally be solved using a digital computer for precise results, a number of simple graphical procedures have been developed for rapid estimation, such as might be needed in preliminary design considerations. We shall discuss here the McCabe-Thiele method, a graphical method that is suitable when the assumption of constant mass flow rates is valid but the equilibrium is not linear. The McCabe-Thiele method is very useful for visualizing the computational process and for getting an intuitive sense of the details of the separation that is difficult to obtain from a table of numbers at the end of a computerized numerical calculation.

The development of the graphical method of solving the extractor equations centers around the choice of a control volume. Instead of doing the mass balance on the single stage n, the control volume is chosen to comprise stages n to N, as shown inFigure 11.9. Conservation of mass applied to component A in this control volume then becomes

0= FxA,n−1+ SyAS− FxAN− SyAn, (11.28a) or

yAn= xA,n−1− xAN+ yAS. (11.28b) In addition, we have the equilibrium relation,

yA= f (xA). (11.29)

y_{AS −} ΛxAN

y_AS y_A

x_AN

x_A

x_AF

6

5 4

3

1 2

Eq. 11.30 Eq. 11.29

Equilibrium

Operating y_A = f(x_A)

y_A = ΛxA – ΛxAN + yAS

7 9 8

Figure 11.10. Calculation of the number of stages required for separation using the McCabe-Thiele method.

The construction is then carried out as shown onFigure 11.10.The equilibrium line, Equation11.29, is drawn, as is the operating line,

y_A= xA− xAN+ yAS. (11.30) If we set n= 1, then xA,n− 1is x_AF. According to Equation11.28,y_A1is the value of the operating line when x_A = xAF(point 1). y_A1is in equilibrium with x_A1, so the equilibrium line, Equation 11.29, must give x_A1for y= yA1(point 2). Given x_A1we find yA2from the operating line (point 3), then xA2from the equilibrium line (point 4), and so on. We ultimately reach xAN in this manner. The number of steps on the diagram corresponds to the number of stages required. Four stages would be needed inFigure 11.10,since x_ANlies between the third and fourth step.

EXAMPLE 11.7 For the equilibrium data shown inFigure 11.3,how many equi-librium stages are required to reduce the acetone mass fraction in the aqueous phase from 0.30 to 0.03 using = 0.68 and pure solvent (yAS= 0)?

The construction is shown inFigure 11.11. Between two and three (hence three) stages are needed. The equilibrium line is nearly linear in this range, so Equation11.27should provide a close approximation to the solution. In Example 11.6 a separation ratio of 0.10 was found to require 2.2 (hence three) stages for this system using the constant value K= 1.5.

y_AS = 0 y_A

x_AN = 0.03

x_A

x_AF 0.1

0.1 0.2 0.3

Operating line Eq. 11.30 Equilibrium line

Eq. 11.29

0.2 0.3 0.4

Figure 11.11. Calculation of the number of stages required to reduce xA from 0.3 to 0.03 using = 0.68 and the equilibrium data inFigure 11.3.

EXAMPLE 11.8 What is the minimum flow rate of pure solvent that can be used to reduce the acetone mass fraction of an aqueous phase from 0.30 to 0.03?

The solvent flow rate is contained in the ratio = F/S, which is the slope of the operating line in the McCabe-Thiele method. The smaller S is, the larger is the slope.Figure 11.12shows a sequence of operating lines with increasing slope. As increases, the number of stages increases. The number of stages goes to infinity as the operating line intersects the equilibrium line at xAF, and for greater than this value the construction cannot be carried out. Thus, the minimum solvent flow corresponds to the slope such that the two lines intersect at x_AF. For the given data and three-component system this is

= F

Smin = 1.52, Smin= 0.66F.

The desired separation thus requires that S be greater than 0.66F. Beyond that point an economic balance between capital investment for the larger number of stages and the cost of handling larger volumes of solvent in the smaller number of stages must be carried out to arrive at the final design.

If the equilibrium line is nearly straight, as in the acetone-water-trichloroethane system, the calculation in Example 11.7 can be carried out algebraically with ease.

0 0.1 0.2 0.3

y_A

x_AN = 0.003 0.4

0.1 0.2 0.3

Equili brium line

Λ = 1.0

Λ = 0.68 Λ = 1.52

x_A

Figure 11.12. Calculation of the minimum solvent flow rate necessary to reduce x_Afrom 0.30 to 0.03.

From Equations 11.29 and 11.30 we see that the two lines intersect when y_A= KxAF= (xAF− xAN)+ yAS, from which it follows that the maximum value of , corresponding to the minimum value of S, is given by

max= K 1− scN

− yAS

x_AF− xAN

. (11.31)

When yAS= 0 the same result is obtained from Equation11.27by setting 1 –μ + μscNto zero so that N goes to infinity, reflecting the fact that an infinite number of stages would be required to achieve the required separation at the minimum solvent flow rate.

EXAMPLE 11.9 Repeat the calculation of the minimum solvent flow rate in Exam-ple 11.8 with the assumption that the equilibrium line is linear with K= 1.5.

From Equation11.31, with scN = 0.1 and yAS= 0, we obtain max = 1.67, hence Smin= 0.60F. It is obvious fromFigure 11.12that taking the equilibrium slope as constant with an equilibrium curve that is concave will always lead to too low an estimate of the minimum solvent flow rate.

11.6 Concluding Remarks

The use of the equilibrium stage is not a serious restriction, since everything done in this chapter can be reformulated in nearly the same form for nonequilibrium situations by using the stage efficiency E_fintroduced in Section 10.4.2. The essence of the analytical design question is expressed in Example 11.8: A certain minimum solvent flow rate is required for a given separation. Beyond the minimum there is a trade-off between costs associated with handling larger volumes and costs associated with adding additional stages. Given the capital and operating costs, the computation of an optimum follows in an identical manner to that for reactors inChapter 7. The same comments apply to other separation processes, such as gas absorption, ion exchange, and distillation. Indeed, with appropriate changes of nomenclature to account for different physical processes, the equations and techniques developed in this chapter carry over to these and other separation processes almost without change. In particular, the logarithmic dependence of the number of stages on the desired degree of separation, as expressed in the Kremser Equation, is a common feature of all separations processes.

Selection of the solvent is a crucial issue in extraction, especially for high-value-added products like pharmaceuticals, which may be produced in small amounts.

One interesting line of research undertaken by some chemical engineers in recent years has been to employ quantum mechanics to determine the interaction energy between complex molecules and potential solvents in order to evaluate solubility, utilizing computer codes that obtain approximate solutions to the time-independent Schr ¨odinger equation. This approach will become more predictive when interac-tion energies obtained in this way are combined with statistical mechanical calcula-tions that properly account for entropic contribucalcula-tions to the free energy of solva-tion. (These issues require a grounding in statistical thermodynamics and quantum mechanics, of course, both of which are areas of science utilized by many chemical engineers.)

The engineering art in separation begins with the choice of a process. We cannot deal with that crucial aspect of design here, for it is highly coupled with specific expe-rience and with a thorough understanding of separations technologies. For difficult separations, in which a large number of stages is required, calculation uncertainty can be compensated for at minimal extra cost by adding extra stages, and this pro-vides a useful safety factor if throughput must subsequently be increased because, say, of inaccurate market forecasts. The design of the configuration of a stage that will provide equilibrium, or at least high efficiency, is the significant engineering problem. This may involve the arrangement of baffling to induce adequate phase contact, the construction of efficient and inexpensive mechanical agitation for devel-oping interfacial area, the design of phase separators that minimize the entrainment of one phase in the other, and so on. These considerations are beyond our analytical treatment here, and they still depend, to a large extent, on the designer’s skill and appreciation of the particular system at hand.

Bibliographical Notes

The subject matter in this chapter is addressed in the unit operations books cited at the end ofChapter 10,as well as books specifically concerned with separations processes. Two recent examples of the latter are

Henley, E. J., and J. D. Seader, Separation Process Principles, 2nd Ed., Wiley, New York, 2005.

Wankat, P. K., Separation Process Engineering, 2nd Ed., Prentice-Hall, Engle-wood Cliffs, NJ, 2006.

The classic treatment of extraction is

Treybal, R. E., Liquid Extraction, McGraw-Hill, New York, 1951; reprinted by Pierce Press, 2008.

PROBLEMS

The equilibrium data inTable 11.P1 for the distribution of acetic acid in water and isopropyl ether phases are used for Problems 11.1 through 11.4. x_Adenotes the mass fraction of acid in the aqueous phase and y_Athe mass fraction of acid in the organic phase.

Table 11.P1. Equilibrium distribution of acetic acid between water (x) and isopropyl ether (y). Data of D. F.Othmer, R. E. White, and E. Trueger, Industrial & Engineering Chem., 33, 1240–1248 (1941).

xA 0.007 0.014 0.029 0.064 0.133 0.255 0.367

yA 0.002 0.004 0.008 0.019 0.048 0.114 0.216

11.1. It is necessary to process 100 kg/min of a water stream containing 15 percent