Chapter 9. Batch Distillation

1. Explain the operation of simple and multistage batch distillation systems

2. Discuss the differences between batch and continuous operation

3. Derive and use the Rayleigh equation for simple binary batch distillation

4. Solve problems for constant-mole, batch distillation

5. Solve problems in batch steam distillation with a single volatile organic component

6. Use the McCabe-Thiele method to analyze multistage binary batch distillation for:

a. Batch distillation with a constant reflux ratio.

b. Batch distillation with a constant distillate composition.

c. Inverted batch distillation.

7. Solve problems for simple, multicomponent batch distillation.

8. Determine the operating time and energy requirements for a batch distillation

Batch distillation is a much older process than continuous distillation. Mesopotamian clay distillation pots have been dated to around 3500 B.C.E. (RT, 2007), and alchemists in Alexandria used simple batch retorts in the first century A.D. (Davies, 1980). Batch distillation was developed to concentrate alcohol by Arab alchemists around 700 A.D. (Vallee, 1998). It was adopted in Western Europe, and the first known book on the subject was Hieronymus Brunschwig’s Liber de arte distillandi, published in Latin in the early 1500s. This book remained a standard pharmaceutical and medical text for more than a century. The first distillation book written for a literate but not the scholarly community was Walter Ryff’s Das New gross Distillier Buch, published in German in 1545 (Stanwood, 2005). This book included a “listing of distilling apparatus, techniques, and the plants, animals, and minerals able to be distilled for human pharmaceutical use.” Kockmann (2014) thoroughly reviews the history of batch distillation. Advances in batch distillation have been associated with its use to distil alcohol, pharmaceuticals, coal oil, petroleum oil, and fine chemicals.

Batch distillation is versatile. A run may last from a few hours to several days. Batch distillation is used when the plant does not run continuously and batches must be completed in one or two shifts (8 to 16 h). It is often used when the same equipment distills several different products at different times. If distillation is required only occasionally, batch distillation is the usual choice.

Equipment can be arranged in a variety of configurations (Sorensen, 2014). In simple batch distillation (Figure 9-1), the vapor is withdrawn continuously from the reboiler. The system differs from flash distillation in that there is no continuous feed input, and the liquid is drained only at the end of the batch. A similar alternative is constant-mole batch distillation in which solvent is fed continually to the still pot to keep the number of moles constant (see Section 9-4).

In a multistage batch distillation, a staged or packed column is placed above the reboiler as in Figure 9-2. Reflux is returned to the column. In the usual operation, distillate is withdrawn continually (Diwekar, 1995; Luyben, 1971; Mujtaba, 2004; Pratt, 1967; Robinson and Gilliland, 1950; Sorensen, 2014) until the column is shut down and drained. In an alternative method (Treybal, 1970), no distillate is withdrawn; instead, the composition of liquid in the accumulator changes. When the distillate in the accumulator is of the desired composition and in the desired amount, both the accumulator and the reboiler are drained. Luyben (1971) indicated that the usual method should be superior; however, the alternative method may be simpler to operate.

Another alternative is called inverted batch distillation (Diwekar, 1995; Pratt, 1967; Robinson and Gilliland, 1950) because bottoms are withdrawn continuously while distillate is withdrawn only at the end of the distillation (see Figure 9-8 and Problems 9.C2 and 9.E3). In this case the charge is placed in the accumulator, and a reboiler with a small holdup is used. Inverted batch distillation is seldom used, but it is useful when quite pure bottoms product is required. Additional configurations include middle-vessel, multivessel, extractive, and hybrid (Sorensen, 2014).

The mass balance on any component, typically the light component is,

For a multicomponent feed Eq. (9-2) can be written for C-1 of the C components. The feed into the column is F kmoles of mole fraction x_F of the more volatile component (MVC). At the end of the batch there are W_final kmoles of mole fraction x_fin in the still pot. The symbol W is used since the material left in the reboiler is often a waste. D_total is the total kmoles of distillate of average concentration x_D,avg. Equations (9-1) and (9-2) are applicable to simple batch and normal multistage batch distillation. Some minor changes in variable definitions are required for inverted batch distillation. If fractions are collected, D_total is split into a series of distillate fractions (see Section 9.2.2).

or

Expanding Eq. (9-4),

Then rearranging

and integrating

which is

The minus sign comes from switching the limits of integration. Equation (9-7) is a form of the Rayleigh equation that is valid for both simple and multistage and for binary and multicomponent batch distillation. Of course, to use this equation we must relate x_D to x_W and do the appropriate integration. This is covered in sections 9.3 and 9.6. Time does not appear explicitly in the derivation of Eq. (9-7), but it is implicitly present since W, x_W, and usually x_D are all time dependent (see Section 9.8).

Once W_final has been determined D can be determined from the overall mass balance

and the average distillate concentration can be found from a mass balance on the MVC.

The amount of distillate collected during this period is

And the mole fraction of the component in this distillate fraction is

The initial amount in the still pot for the second period (essentially the feed for this period) is W₁ with mole fraction x₁. Then the Rayleigh equation, distillate amount, and distillate mole fraction become

The fractions continue until the last fraction n:

Equations (9-9) are valid for both simple binary batch and multistage binary batch distillation. If these equations are written for C-1 components, they are also valid for multicom-ponent systems.

If fractions with sharp cuts are desired, you can either add a large number of stages to the column or take a waste, or offcut, between the desired product fractions, as shown in Figure 9-3 (Sorensen, 2014). The offcuts are usually reprocessed in the next batch. Equations (9-9) are still valid, but some of the fractions are products, and some are offcuts. The taking of offcuts is explored in Problem 9.D26 for simple binary batch distillation.

where y and x are in equilibrium with equilibrium expression y = f(x,p). For any given equilibrium expression, Eq. (9-10) can be integrated analytically, graphically, or numerically.

The general graphical or numerical integration procedure for Eq. (9-10) is:

1. Plot or fit the y-x equilibrium curve.

2. At a series of x values, find y – x.

3. Plot 1/(y – x) vs. x or fit it to an equation.

4. Graphically (Figure 9-4) or numerically integrate from x_F to x_fin.

5. From Eq. (9-10), find the final charge of material W_final in the still pot:

where the area is shown in Figure 9-4.

6. The amount of distillate, D, and the average distillate concentration, x_D,avg, can be found from the mass balances Eqs. (9-8a) and (9-8b).

The Rayleigh equation can also be integrated numerically. One convenient method for doing this is to use Simpson’s rule (e.g., see Mickley et al., 1957, pp. 35–42; or Larsson et al., 2014, pp. 42–52). If the ordinate in Figure 9-4 is called g(x), then the simplest form of Simpson’s rule divides the interval into two equal parts with δx = (x_F – x_fin)/2.

which is often written for batch distillation as

Terms are shown in Figure 9-4. Simpson’s rule is exact if g(x) is cubic or lower order. More accuracy can be obtained by dividing the interval into a larger, even number of parts of equal size. For example, if we use n equal subintervals, then δx = (x_F – x_fin)/n, and Simpson’s rule becomes

Note that n has to be an even integer. For smooth curves, such as in Figure 9-4, Simpson’s rule is quite accurate (see Example 9-1). For more complex shapes, Simpson’s rule may be more accurate if the integration is done in two or more separate pieces (see Example 9-2). Alternatively, half intervals can be used in regions where more accuracy is needed (Larsson et al., 2014). Simpson’s rule is also the MATLAB command quad (for quadrature) (Pratap, 2006). Other integration formulas that are more accurate can be used.

If the average distillate concentration is specified, a trial-and-error procedure is required. This involves guessing the final still pot concentration, x_W,final, and calculating the area in Figure 9-4 either graphically or using Simpson’s rule. Then Eq. (9-11) gives W_final, and Eq. (9-8b) is used to check the value of x_D,avg. For a graphical solution, the trial-and-error procedure can be conveniently carried out by starting with a guess for x_fin that is too high. Then every time x_fin is decreased, the additional area is added to the area already calculated.

If the relative volatility α is constant, the Rayleigh equation can be integrated analytically. In this case the equilibrium expression is Eq. (2-22b), which is repeated here:

Substituting this equation into Eq. (9-10) and integrating, we obtain

When it is applicable, Eq. (9-13) is obviously easier to apply than graphical or numerical integration. Analytical solutions for limiting cases (constant α in this case) are also useful for checking numerical solutions.

Warning: If α is not constant, W_final calculated from Eq. (9-13) can be very inaccurate.

It is interesting to compare the simple binary batch distillation result to binary flash distillation of the same feed producing y = 0.892 mole fraction methanol. This flash distillation problem was solved previously as Problem 2.D1g. The results were x = 0.756 (compare to x_fin = 0.65), V = 16.18 (compare to D_total = 31.06), and L = 33.82 (compare to W_final = 18.94). The simple batch distillation gives a greater separation with more distillate product because the bottoms product is an average of liquids in equilibrium with vapor in the entire range from y = 0.915 (in equilibrium with x = z = 0.8) to y = 0.845 (in equilibrium with x = x_fin = 0.65), while for the flash distillation, the liquid is always in equilibrium with the final vapor y = 0.892. In general, a simple batch distillation will give more separation than flash distillation with the same operating conditions. However, because flash distillation is continuous, it is much easier to integrate into a continuous plant.

Extension of the solution method for simple binary batch distillation to collection of fractions is straightforward if the values of x_W1, x_W2, ... are specified. Equations (9-9) are converted to simple batch by replacing x_D with y, which is in equilibrium with x_W. In essence, a number of simple batch problems are solved sequentially (see Problem 9.D23).

Extension of the simple batch equations to multicomponent systems is also straightforward, although solution of the equations is significantly more difficult than solution of the binary equations. For simple multicomponent batch distillation Eq. (9-5b) becomes

The Rayleigh Eq. (9-10) can be written for each component. The equilibrium expression becomes a function of all the components, which couples all of the equations. Section 9.6 explores multicomponent simple distillation in detail.

A process that is closely related to simple batch distillation is differential condensation (Treybal, 1980). In this process vapor is slowly condensed, and the condensate liquid is rapidly withdrawn. A derivation similar to the derivation of the Raleigh equation for a binary system gives

where F is the moles of feed vapor of mole fraction y_F, and D_final is the leftover vapor distillate of mole fraction y_D,final. Differential condensation is rarely practiced.

For a constant-mole batch distillation the general mole balance is

For the total mole balance this is In – Out = 0, since the moles in the still pot are constant. Thus, if dS moles of the second solvent are added, the overall mole balance for a constant-mole batch distillation is

where dV is the moles of vapor withdrawn.

If we do a component mole balance on the original solvent (solute is assumed to be nonvolatile and is ignored), we obtain the following for a constant-mole system,

where y and x_w are the mole fraction of the first solvent in the vapor and liquid, respectively. Substituting in Eq. (9-15b), this becomes

Note that since the amount of liquid W is constant, there is no term equivalent to the last term in Eq. (9-4). Integration of Eq. (9-16b) and minor rearrangement gives us

Since vapor and liquid are assumed to be in equilibrium, y is related to x_w by the equilibrium relationship. Equation (9-17) can be integrated graphically or numerically in a procedure that is quite similar to that used for simple batch distillation. Problem 9.D17 leads you through these calculations for constant-mole batch distillation.

If constant relative volatility between the two solvents can be assumed, Eq. (2-22) can be substituted into Eq. (9-17), and the equation can be integrated analytically (Gentilcore, 2002):

Gentilcore presents a constant relative volatility sample calculation that illustrates the advantage of constant-mole batch distillation when it is used to exchange solvents. The method for obtaining accurate results for low values of x_fin outlined in the last paragraph of Example 9-1 can be used for constant-mole batch distillation by replacing Eq. (9-13) with Eq. (9-18).

The amount of new solvent required can be reduced by putting a column above the still pot. Equation (9-17) remains valid except y is now the vapor leaving the top of the column (the same mole fraction as x_dist if there is a total condenser), and y is not in equilibrium with the still-pot liquid. The procedures outlined in Section 9.6 for multistage systems can be used to determine the separation.

Batch steam distillation is usually operated with liquid water present in the still. Then both the liquid water and the liquid organic phases exert their own partial pressure. Equilibrium is given by Eqs. (8-14) to (8-18) when there is one volatile organic and some nonvolatile organics present. As long as there is minimal entrainment, there is no advantage in the separation to having more than one stage, but steam use can be reduced. For low-molecular-weight organics, vaporization efficiencies, defined as the actual partial pressure divided by the partial pressure at equilibrium, p*,

are often in the range from 0.9 to 0.95 (Carey, 1950). This efficiency is close enough to equilibrium that equilibrium calculations are usually adequate.

The mass balances for Figure 9-5 are typically written on a water-free basis. These balances are identical in form to Eqs. (9-1) and (9-2) and can be solved for W_final if x_fin is known.

On a water-free basis F, W_final, and D are kmol organics in the feed, still pot, and distillate, respectively. The mole fractions x are (moles volatile organic)/(total moles organic). With a single volatile organic, x_D,avg = 1.0 if entrainment is negligible. Then

and the amount of the organic distillate product is given by Eq. (9-8a), D = F – W_final. The Rayleigh equation can also be used and will give the same results.

At any moment the instantaneous moles of water dn_w carried over in the vapor can be found from Eq. (8-18). This becomes

The total moles of water carried over in the vapor can be obtained by integrating this equation:

During the batch steam distillation, the mole fraction of the volatile organics in the still varies, and thus, the still temperature determined by Eq. (8-15) varies. Equation (9-21b) can be integrated numerically in steps. The total moles of water required is n_w plus the moles of water condensed to heat the feed and still pot and to vaporize the volatile organics.

To integrate Eq. (9-21b) we must relate the moles of the volatile organic in the distillate, n_org, to its mole fraction in the still pot, x_pot = x_volatile. From a mass balance on the volatile organic very similar to that used to derive Eq. (9-20b), we obtain

This equation is valid at any time. As the mole fraction of the volatile organic in the still pot, x_pot, decreases, the moles of volatile organic collected in the distillate, n_org, increase.

For step-by-step numerical integration, Eq. (9-21b) can be written as

The total change in the mole fraction of volatile organic as x_pot goes from x_F to x_w,final can be broken up into a number of intervals, Δx_pot. In each interval n_org, x_pot,average, and Δn_w are calculated. The amount of water carried over into the distillate with the volatile organic is given by Eq. (9-22c). The calculation becomes more accurate as more intervals are used in the range from x_F to x_w,final.

A short example for the simplified case in which the temperature in the still pot does not vary much and (VP)_volatile is essentially constant will help to clarify the procedure. Suppose F = 1.0 kmol, x_F = 0.35, P_total = 760 mm Hg, (VP)_volatile = 50 mm Hg, and we want to calculate Δn_w for a change in x_pot from 0.25 to 0.20. Then from Eq. (9-22a) at x_pot = 0.25,

n_org = F(x_F – x_pot)/(1 – x_pot) = 1.0(0.35 – 0.25)/(1 – 0.25) = 0.13333

At x_pot = 0.20, n_org = 0.18750. In the interval from x_pot = 0.25 to x_pot = 0.20, x_pot,average = 0.225, and Δn_org = 0.18750 – 0.13333 = 0.05417. Then from Eq. (9-22b),

Δn_w = 0.05417[760 – (50)(0.225)]/[(50)(0.225)] = 3.605 kmol water.

If the still pot temperature varies significantly, the temperature at each value of x_pot,average needs to be determined from Eq. (8-15) so that the value of (VP)_volatile can be calculated. This is easiest to do with a spreadsheet.

During most of the batch operation, the mole fraction of the volatile organic is considerably higher than it is at the end of the batch. In continuous steam distillation the mole fraction of the volatile organic is always at its lowest value. Thus, batch steam distillation requires less steam for a given separation than continuous steam distillation.

For multistage systems x_D and x_W are no longer in equilibrium. Thus, the Rayleigh equation, Eq. (9-7), cannot be integrated until a relationship between x_D and x_W is found. This relationship can be obtained from stage-by-stage calculations. We assume that there is negligible holdup on each plate, in the condenser, and in the accumulator. Then at any specific time we can write mass and energy balances around stage j and the top of the column, as shown in Figure 9-2A. These balances simplify to

Input = Output

since accumulation was assumed to be negligible everywhere except the reboiler. Thus, at any given time t

In these equations V, L, and D are molal flow rates. These balances are essentially the same equations we obtained for the rectifying section of a continuous column except that Eqs. (9-23) are time dependent. If we can assume constant molal overflow (CMO), the vapor and liquid flow rates will be the same on every stage, and the energy balance is not needed. Combining Eqs. (9-23a) and (9-23b) and solving for y_j+1, we obtain the operating equation for CMO:

At any specific time Eq. (9-24) represents a straight line on a y-x diagram. The slope will be L/V, and the intercept with the y = x line will be x_D. Since either x_D or L/V will have to vary during the batch distillation, the operating line will change continuously.

The McCabe-Thiele analysis gives x_W values for a series of x_D values. We can now calculate 1/(x_D – x_W). The integral in Eq. (9-7) can be determined by either numerical integration, such as Simpson’s rule given in Eqs. (9-12a) to (9-12c), or by graphical integration. Once x_W values have been found for several x_D values, the same procedure used for simple batch distillation can be used. Thus, W_final is found from Eq. (9-7), x_Davg from Eq. (9-8b), and D_total from Eq. (9-8a). If x_Davg is specified, a trial-and-error procedure will again be required.

Once x_fin, D, and W_final are determined, we can calculate the values of Q_c, Q_R, and operating time (see Section 9.8).

With x_D and the number of stages specified, trial and error is required to find the initial value of L/V that gives the feed concentration x_F with the specified number of stages. The operating line slope L/V is then increased until the specified number of equilibrium contacts gives x_W = x_fin. This gives (L/V)_final. W_final is found from mass balance Eqs. (9-1) and (9-2).

The required maximum values of Q_c and Q_R and the operating time can be determined next (see Section 9.8). Note that the Rayleigh equation is not required when x_D is constant. However, if used, the Rayleigh equation gives exactly the same answer as Eq. (9-25) (see Problem 9.C1).

depends on the concentration of all components, a numerical solution is required. We previously considered multicomponent simple batch distillation in section 8.5.2 when we discussed residue curves. Although the notations and the procedures are different, the developments in Sections 8.5.2 and 9.2.1 are equivalent (see Problem 9.C4). Thus, the residue curve gives us the path of the x_i during the batch distillation, and from the required equilibrium calculation, it has also calculated the y_i values that correspond to the x_i, although the vapor values may not be reported. The profiles of x_i and y_i values are sufficient information to numerically integrate the Rayleigh equation.

The procedure is easiest to illustrate if the relative volatility is constant. For batch distillation the equilibrium expression, Eq. (5-28) given in Problem 5.C1, is

The Rayleigh equation for the light component A becomes

To numerically integrate the integral in Eq. (9-27), we start with the spreadsheet calculation of a residue curve (see Section 8.5.2 and Figure 8-B1). For each increment in k, the incremental area is

Then the calculated value of the integral is

where Δx_A,k = x_A,k – x_A,k–1, and (1/[y_A – x_A])_avg = 0.5[(1/[y_A – x_A])_k-1 + (1/[y_A – x_A])_k].

For example, if we calculate the area for k = 10 for the spreadsheet in Figure 8-B1, we have

|Δx_A,k| = |0.989960081 – 0.990021592| = 0.000061511, and

1/(y_A – x_A)_avg = 0.5[1/(0.99617266 – 0.990021592) + 1/(0.996149968 – 0.989960081)] = 162.06.

Incremental area = (0.000061511)(162.06) = 0.009969.

If the values of h and k_final from a residue curve calculation using the Euler method (Section 8.5.2) starting with the initial conditions of the batch distillation are available, then an estimate of the area of the integral in Eq. (9-27) is

where k_final is the step for which x_A = x_A,final. This equation is useful for checking the result of a numerical integration. The rationale behind Eq. (9-28c) is explored in Problem 9.C5.

If the relative volatilities are not constant, we have two choices. First, we can divide the range of the batch operation into a series of intervals and use an average value for each interval. A reasonable definition of is the geometric average for each interval:

The recursion Eq. (8-30) uses values that correspond to the interval in which they appear.

The second procedure, which is more accurate, is to do a bubble-point calculation at every time step. The use of bubble-point calculations is explored in section 5.3 in the context of a steady-state, multistage analysis. For batch distillation the bubble-point calculation would be done at each step k. In essence, the resulting spreadsheet is a combination of the spreadsheets in Appendices 5A and 9A.

Fractions (Section 9.2.2) are often collected in multicomponent batch distillation. With only a single-stage, simple batch distillation does not give pure compounds unless the relative volatilities are very large. The calculation method is quite similar to the method shown in Example 9.3. For example, if a second fraction was collected for Example 9.3 when x_W,B,2 = 0.3, we would look at the results of the spreadsheet in Appendix A to find the mole fractions and the cumulative area at this concentration. The values bracketing x_W,B,2 = 0.3 are shown in Table 9.2.

The cumulative area at x_W,B,2 = 0.3 is approximately 1.4962. If we subtract this value from the cumulative area = 1.1919 at x_W,B = 0.4, we determine the area for the fraction from x_W,B = 0.4 to x_W,B = 0.3 is 0.3043. The initial charge for this fraction was W_final = 0.3036 calculated in Example 9.3. Then W(at x_W,B = 0.3) = (0.3036)exp(–0.3043) = 0.2239 kmole. (See Problem 9.D27a for an equivalent calculation and Problem 9.D27b for the completion of the problem.)

Obviously, multistage batch columns are also used for multicomponent separations. One fairly simple approach is to assume that the changes in the column are quite slow and use a steady-state analysis similar to that employed in Section 9.6 for binary separations. Examples are the Fenske-Underwood-Gilliland shortcut method or stage-by-stage calculation with bubble-point calculation on each stage. Since these analyses ignore holdup on the stages and dynamics of the stages, they are considered simple models. More complicated models include stage holdup and dynamics and may include a mass transfer analysis. Readers interested in these more advanced topics should consult Sorensen (2014) for references and Diwekar (2012) and Mujtaba (2004) for calculation details.

The total batch time, t_batch, is

The down time, t_down, includes dumping the bottoms, cleanup, loading the next batch, and heating the next batch until reflux starts to appear. This time can be estimated from experience. The operating time, t_op, is the actual period during which distillation occurs, so it must be equal to the total amount of distillate collected divided by the distillate flow rate.

D_total is calculated from the Rayleigh equation, with F set either by the size of the still pot or by the charge size. For an existing apparatus the distillate flow rate, D in kmol/h, cannot be set arbitrarily. The column was designed for a given maximum vapor velocity, u_flood, which corresponds to a maximum molal flow rate, V_max (see Chapter 10). Then from the mass balance around the condenser,

We usually operate at a fraction of the maximum flow rate, such as D = 0.75 D_max. The operating time, t_op, can be estimated from Eqs. (9-29) and (9-30). If the resulting t_batch is not convenient, adjustments must be made.

The energy requirements in the reboiler or still pot, Q_R, and in the total condenser, Q_c, can be estimated from energy balances. For a total condenser Eq. (3-13) is valid, but V₁, h_D, and H₁ may all be functions of time (if x_D varies, the enthalpies will vary). If the reflux is a saturated liquid reflux, then H₁ – h_D = λ, the latent heat of evaporation. In this case the total condenser just condenses the vapor to saturated liquid. Likewise, the still pot vaporizes a saturated liquid to a vapor. Thus,

If CMO is valid, then V₁ = V_pot, λ₁ = λ_pot, and

Since V = (1 + L/D)D, we obtain

For an existing batch distillation apparatus we must check that the condenser and the reboiler are large enough to handle the calculated values of |Q_c| and Q_R. If |Q_c| or Q_R is too large, then the rate of vaporization needs to be decreased. Either the operating time, t_op, will need to be increased or the charge to the still pot F will have to be decreased.

During operation (assume that the charge and the still pot have been heated and vapor is flowing throughout the column), the energy balance around the entire system is

If CMO is valid, Eq. (9-31c) holds, and saturated liquid h_D = h_pot, the result simplifies to

If the assumption of negligible holdup is not valid, then the holdup on each stage and in the accumulator acts like a flywheel and retards changes. A different calculation procedure is required for this case (Diwekar, 2012; Mujtaba, 2004). Batch distillation also has somewhat different design and process control requirements than continuous distillation. In addition, startup and troubleshooting are somewhat different (Sorensen, 2014).

Davies, J. T., “Chemical Engineering: How Did It Begin and Develop?” in W. F. Furter (Ed.), History of Chemical Engineering, Washington, D.C., American Chemical Society, Adv. Chem. Ser, 190, 15–43 (1980).

Diwekar, U. W., Batch Distillation: Simulation, Optimal Design and Control, 2nd ed., Taylor and Francis, Washington, D.C., 2012.

Gentilcore, M. J., “Reduce Solvent Usage in Batch Distillation,” Chem. Engr. Progress, 98 (1), 56 (Jan. 2002).

Kockmann, N., “History of Distillation,” in Gorak, A., and E. Sorensen (Eds.), Distillation: Fundamentals and Principles, Chapter 1, Elsevier, 2014.

Larsson, L., R. E. Eliasson, and M. Orych, Principles of Yacht Design, 4th ed., International Marine/McGraw-Hill Education, Camden, Maine, 2014. [Note: An unusual reference in a chemical engineering textbook; however, mathematics is universally applicable. Their description of Simpson’s rule with examples is clear.]

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Mujtaba, I. M., Batch Distillation: Design and Operation, Imperial College Press, London, 2004.

Mickley, H. S., T. K. Sherwood, and C. E. Reed, Applied Mathematics in Chemical Engineering, McGraw-Hill, New York, 1957, pp. 35–42.

Pratap, R., Getting Started with MATLAB7, Oxford University Press, Oxford, 2006.

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Robinson, C. S., and E. R. Gilliland, Elements of Fractional Distillation, 4th ed., McGraw-Hill, New York, 1950, Chapters 6 and 16.

R. T., “Distillation through the Ages,” Chem. Heritage, 25 (2), 40 (Summer 2007).

Sorensen, E., “Design and Operation of Batch Distillation,” in Gorak, A., and E. Sorensen (Eds.), Distillation: Fundamentals and Principles, Chapter 5, Elsevier, London, UK, 2014.

Stanwood, C., “Found in the Othmer Library,” Chemical Heritage, 23 (3), 34 (Fall 2005).

Treybal, R. E., Mass-Transfer Operations, 3rd ed., McGraw-Hill, New York, 1980.

Treybal, R. E., “A Simple Method for Batch Distillation,” Chem. Eng., 77 (21), 95 (Oct. 5, 1970).

Vallee, B. L., “Alcohol in the Western World,” Scientific American, 80 (June 1998).

Woodland, L. R., “Steam Distillation,” in D. J. DeRenzo (Ed.), Unit Operations for Treatment of Hazardous Industrial Wastes, Noyes Data Corp., Park Ridge, NJ, 1978, pp. 849–868.

A1. Why is the still pot in Figure 9-2B much larger than the column?

A2. In the derivation of the Rayleigh equation:

a. In Eq. (9-4), why do we have –x_D dW instead of –x_D dD?

b. In Eq. (9-4), why is the left-hand side –x_D dW instead of –d(x_DW)?

A3. Explain how the graphical integration shown in Figures 9-3 and 9-5 could be done numerically on a computer.

A4. Suppose you have two feeds containing methanol and water that you want to batch distil. One feed is 60.0 mol% methanol, and the other is 32.0 mol% methanol. How do you do the batch distillation to obtain the largest amount of distillate of a given mole fraction? N is constant.

a. Mix the two feeds together and batch distil.

b. Start the batch with the feed with higher methanol mole fraction and then add second feed (32.0 mol% methanol) when the still-pot concentration equals that feed concentration.

c. Do two separate batches—one for the more concentrated feed (60.0 mol%) and the other for the less concentrated feed (32.0 mol%).

d. All of the above.

e. a and b.

f. a and c.

g. b and c.

h. None of the above.

Explain your answer.

A5. Which system(s) require less energy for batch distillation than continuous distillation with the same amount of separation?

a. Simple batch compared to one-stage continuous.

b. Multistage batch compared to multistage continuous.

c. Batch steam distillation compared to continuous steam distillation.

d. All of the above.

e. a and b.

f. a and c.

g. b and c.

h. None of the above.

Explain your answer.

A6. Batch-by-Night, Inc., has developed a new simple batch with reflux system (Figure 9-1 with some of stream D refluxed to the still pot) that they claim outperforms the normal simple batch (Figure 9-1). Suppose you want to batch distil a mixture similar to methanol and water in which there is no azeotrope and two liquid phases are not formed. Both systems are loaded with the same charge F moles with the same mole fraction x_F, and distillation is done to the same value x_w,final. Will the value of x_d,avg from the simple batch with reflux be

a. greater than

b. the same as

c. less than

the x_d,avg value from the normal, simple batch distillation? Explain your answer.

A7. When there are multiple stages in batch distillation, the calculation looks like the McCabe-Thiele diagram for several

a. stripping columns.

b. enriching columns.

c. columns with a feed at the optimum location.

A8. Develop a key relations chart for this chapter.

A9. If we are doing a batch distillation of a heterogeneous azeotrope system, when would we employ a system similar to Figure 9-1, and when would we employ a system similar to Figure 9-10 (in Problem 9.D9)?

B. Generation of Alternatives

B1. List all the different ways a binary batch or inverted batch problem can be specified. Which of these will be trial and error?

B2. What can be done if an existing batch system cannot produce the desired values of x_D and x_W even at total reflux? Generate ideas for both operating and equipment changes.

B3. Develop ideas for how you would reprocess or utilize the offcuts from a batch distillation that collects fractions.

C. Derivations

C1. For a binary multistage batch distillation with constant x_dist, prove that the mass balances over the entire batch period and the Raleigh equation both give Eq. (9-25).

C2. Assume that holdup in the column and in the total reboiler is negligible in an inverted batch distillation (Figure 9-9).

a.* Derive the appropriate form of the Rayleigh equation.

b. Derive the necessary operating equations for CMO. Sketch the McCabe-Thiele diagrams.

C3. If relative volatility can be assumed constant over the change in concentration for each fraction, Eq. (9-13) can be adapted to the collection of fractions from a simple binary batch distillation. Derive the equations for a simple batch with three distillate fractions collected.

C4. Show that Eq. (9-5c) [immediately before Eq. (9-14)] is equivalent to Eq. (8-27c).

C5. If we approximate the area in Eq. (9-28b) as

then the incremental area for k = 10 (see the section immediately before Example 9.3) = (0.000061511)(162.573393759) = 0.01000005 = h = 0.01. From this result, you can deduce the approximate Eq. (9-28c).

a. Prove that this agreement of the numerical value of the estimated incremental area and the value of h is not an accident.

b. Explain how Eq. (9-28c) can be deduced.

c. Check the prediction of Eq. (9-28c) for Example 9.3.

D. Problems

*Answers to problems with an asterisk are at the back of the book.

D1. A simple batch distillation separates 0.6 kmol of a binary feed that is 70.0 mol% methanol and 30.0 mol% water. The final still pot is 10.0 mol% methanol. Find W_final, distillate D_total, and x_D,avg. VLE data are in Table 2-7. p = 1.0 atm.

D2.* A simple batch still (one equilibrium stage) separates 100 moles of a 75.0 mol% methanol and 25.0 mol% water feed. The final bottoms concentration is 55.0 mol% methanol. Find W_final, distillate D_total, and x_D,avg. VLE data are in Table 2-7. p = 1.0 atm.

D3.* A distillation system with a still pot plus a column with one equilibrium stage is used to batch distil 1.0 kmol of a 57.0 mol% methanol and 43.0 mol% water feed. A total condenser is used. The final bottoms concentration is 15.0 mol% methanol. Pressure is 101.3 kPa. Reflux is a saturated liquid, and L₀/D = 1.85. Find W_final, D_total, and x_D,avg. VLE data are in Table 2-7.

D4. A simple batch distillation (one equilibrium contact) is done of an 80.0 mol% acetone and 20.0 mol% ethanol feed. The final concentration in the still pot is 40.0 mol% acetone, and W_final = 2.0 kmole. VLE data are in Problem 4.D7. Find the feed amount F, the average mole fraction of the distillate, and the kmoles of distillate D_total collected.

D5. 3.0 kmol of feed containing 52.0 mol% water and 48.0 mol% n-butanol is charged to the still pot of a simple batch distillation system (Figure 9-1). The final still pot concentration should be 28.0 mol% water. Equilibrium data are in Table 8-2.

a. Find W_final (kmole), D_V,tot (total amount of distillate vapor collected, kmole), and y_D,avg (average mole fraction water in the vapor distillate).

b. After the distillate vapor is condensed in the total condenser, the liquid is sent to a liquid-liquid settler. Find the total amount of each distillate liquid collected, D₁ (with x_D1,water = 0.573) and D₂ (with x_D2,water = 0.975), in kmoles.

D6. A simple batch still separating a feed that is 60.0 mol% 1,2 dichloroethane and 40.0 mol% 1,1,2-trichloroethane. Pressure is 1 atm. The relative volatility is constant, α_di-tri = 2.4. The use of Eq. (9-13) is recommended.

a. The charge (F) is 1.3 kmol. The final still pot concentration is 30.0 mol% 1,2 dichloroethane. Find the final moles in still pot and the average mole fraction of distillate product.

b. Repeat part a if F = 3.5 kmol.

c. If F = 2.0 kmol and the average distillate concentration is 75.0 mol% 1,2-dicholoroethane, find final kmoles in the still pot and the final mole fraction of 1,2 dichloroethane in the still pot.

D7. A batch distillation system with a still pot and one equilibrium stage (two equilibrium contacts total) distills a feed that is 10.0 mol% water and 90.0 mol% n-butanol (see Table 8-2 for VLE data). p = 1.0 atm. The charge is 4.0 kmoles. The final still concentration is 2.0 mol% water. The system has a total condenser, and the reflux is a saturated liquid. L/D = 1/2. Find the final number of moles in the still pot and the average concentration of distillate.

D8. 10.0 kmol of a feed with x_F = 0.4 (mole fraction methanol) and the remainder water is sent to a batch distillation system with a large still pot that is an equilibrium contact, a distillation column that acts as two equilibrium contacts (total three equilibrium contacts), and a total condenser. p = 1.0 atm. We operate with a constant x_D = 0.8 as we increase L/D. The batch operation is continued until L/D = 4.0. CMO is valid. VLE data are in Table 2-7. Find:

a. x_w,final.

b. W_final and D_total.

c. Initial L/D.

D9.* We wish to batch distil a mixture of 1-butanol and water. Since this system has a heterogeneous azeotrope (see Chapter 8), we will use the system shown in Figure 9-10. The bottom liquid layer from the liquid-liquid separator (97.5 mol% water) is removed as product, and the top liquid layer (57.3 mol% water) is returned to the still pot. Pressure is 1 atm. The feed is 20.0 kmol and 40.0 mol% water. Data are given in Problem 8.D2.

a. If the final concentration in the still pot is x_W,final = 0.28, find D_total in kmoles.

b. If the batch distillation is continued what is the lowest value of x_W,final that can be obtained while still producing a distillate x_D = 0.975?

D10. 4.0 kmoles of a feed that is 0.60 mole fraction acetone and 0.40 mole fraction ethanol is batch distilled in a system with a still pot (an equilibrium contact), a column that acts as one equilibrium stage (total two equilibrium contacts), a total condenser, and reflux to the column. p = 1.0 atm, the reflux ratio is L/D = 1.0, CMO is valid, and x_W,final = 0.20. VLE data are in Problem 4.D7. Find W_final, x_dist,avg, and D_total.

D11. A constant-mole batch distillation is used to change from a pure n-butanol solvent to a solvent that is 60.0 mol% n-butanol and 40.0 mol% water. If the initial charge is W = 2.0 kmol, how much water, S, must be added to achieve the desired solvent concentration? Do the integration with Simpson’s rule, dividing the entire integration from 0.6 to 1.0 first into one step and then into two steps. Compare the answers.

D12. A nonvolatile pharmaceutical is dissolved in a solution that is 90.0 mol% acetone and 10.0 mol% ethanol. A constant volume batch distillation is used to switch the solvent to 20.0 mol% acetone and 80.0 mol% ethanol without diluting the concentration of solute. The charge to the still pot is W = 1.5 kmol. VLE data are in Problem 4.D7. What are the values of the amount (kmol) of pure solvent ethanol added, the kmol of acetone that are evaporated, the kmol of ethanol evaporated, and the average mole fraction of acetone in the distillate?

D13. A simple batch distillation (Figure 9-1) is separating 8.00 kmol of a feed that is 40.0 mol% water and 60.0 mol% n-butanol. The batch distillation is continued until the still pot contains 0.080 mole fraction water. VLE data are in Table 8-2, Problem 8.D3.

a. Find W_final, D_total, and x_D,avg.

b. After settling, the final distillate product is two liquid phases. What are the mole fractions and the amounts (kmol) of each liquid phase?

Note: Problem 9.D28 shows that Figure 9-10 gives a better separation.

D14.* A simple steam distillation is being done in the apparatus shown in Figure 9-5. The organic feed is 90.0 mol% n-decane and 10.0% nonvolatile organics. The system is operated with liquid water in the still. Distillation is continued until the organic layer in the still is 10.0 mol% n-decane. F = 10 kmole. Pressure is 760 mm Hg.

a. At the final time, what is the temperature in the still?

b. What are values of W_final and D_total?

c. Estimate the moles of water passed overhead per mole of n-decane at the end of the distillation.

Data: Assume that water and n-decane are completely immiscible. Vapor pressure data for nC₁₀ are in Example 8-2. Vapor pressure data for water are listed in Problem 8.D10.

D15. 12.0 kmole of a mixture 0.40 mole fraction water and 0.60 mole fraction n-butanol at 1.0 atm pressure is batch distilled in a system with an equilibrium still pot and a column with one equilibrium stage (total two equilibrium contacts). Vapor from the column is sent to a total condenser and then to a liquid-liquid settler. The aqueous phase from the settler (0.975 mole fraction water) is removed as the distillate product. The organic phase from the settler (0.573 mole fraction water) is refluxed to the distillation column. (The equipment is a modification of Figure 9-10 with a column between the still pot and the condenser.) D_total = 3.9 kmol. Equilibrium data are in Table 8-2 (in Problem 8.D3).

a. Find W_final and x_W,final.

b. Find the approximate value of the reflux ratio L/D at end of the batch distillation.

D16.* 10.0 kmol of a 40.0 mol% methanol and 60.0 mol% water mixture is processed in a normal batch distillation system with a still pot that acts as an equilibrium stage and a column with two equilibrium stages (total of three equilibrium contacts). The column has a total condenser, and reflux is a saturated liquid. The column is operated with a varying reflux ratio so that x_D is held constant. We desire a final still-pot concentration of 8.0 mol% methanol, and distillate should be 84.0 mol% methanol. Pressure is 1 atm, and CMO is valid.

a. What initial external reflux ratio, L₀/D, must be used?

b. What final external reflux ratio must be used?

c. How much distillate product is withdrawn, and what is the value of W_final?

D17. A nonvolatile solute is dissolved in 1.0 kmol of methanol. We wish to have the solute in 1.0 kmol of solution that is 99.0 mol% water and 1.0 mol% methanol. Because the solution is already concentrated, a first batch distillation to concentrate the solution is not required. VLE data (ignore the effect of the solute) are in Table 2-7. Do a constant-mole batch distillation from x_M,initial = 1.0 (pure methanol) to x_M,final = 0.01. Find the moles of water added during the constant-mole batch distillation and the moles of water evaporated with the methanol in the distillate. Results can be compared with dilution followed by simple batch distillation in Problem 9.E4.

Note: Since x_M,final = 0.01 is quite small, 1/y is quite large at this limit. To use Simpson’s rule divide the integral into at least two sections.

D18. A differential condensation [see Eq. (9-14)] is done for a binary mixture of ethanol and water. The feed is 0.50 kmol of vapor that is 0.10 mole fraction ethanol. The differential condensation is continued until the vapor remaining is y_D,final = 0.50 mole fraction ethanol. Find the values of D_final and the condensate C_total and the average mole fraction of ethanol in the condensate, x_C,avg. Note that this operation can only be approximated in practice (Treybal, 1980). Compare the result to the solution to Problem 9.D1.

D19. Batch steam distillation is being done to recover octanol from nonvolatile organics. Assume that water and octanol, and water and the nonvolatile organics are completely immiscible. The water layer is pure. The organic layer in the still pot contains 0.60 mole fraction octanol and 0.40 mole fraction nonvolatile organics. The still pot temperature is 90.0°C. The Antoine equations (T is in °C, and VP is the vapor pressure in mm Hg) are:

Octanol: log₁₀ (VP) = 6.8379 – 1310.62/(T + 135.05)

Water: log₁₀ (VP) = 8.68105 – 2164.42/(T + 273.16)

What is the pressure of the still pot?

D20. 10.0 kmole of a 40.0 mol% methanol and 60.0 mol% water mixture is batch distilled. The system consists of a column with two equilibrium stages, a still pot (an equilibrium contact), and a total condenser. The system is operated with a reflux ratio L/D = 0.515. The final concentration of methanol in the still pot is x_W,final = 0.075. CMO is valid. If you use Simpson’s rule, break the area into at least two parts. Find W_final, D_total, and x_dist,avg.

D21. Repeat Problem 9.D20 but for a distillation column with 20 stages.

Note: Think smart and this problem is significantly less work than Problem 9.D20.

D22. 3.0 kmol of a 62.0 mol% methanol and 38.0 mol% water are batch distilled in a system with a still pot and a column with one equilibrium stage (two equilibrium contacts total). Distillate concentration is constant at 85.0 mol% methanol. The final still pot concentration is 45.0 mol% methanol. Reflux is a saturated liquid, and L/D varies. Assume CMO.

a. Find D_total and W_final (kmoles).

b. Find the final value of the external reflux ratio L/D.

D23.* 1.2 kmol of a 10.0 mol% ethanol and 90.0 mol% water feed is processed in a simple batch distillation system that consists of a still pot, a total condenser, and a fraction collector. We collect three fractions with x_W1 = 7.0 mol% ethanol, x_W2 = 4.0 mol% ethanol, and x_W3 = x_W,final = 1.0 mol% ethanol. Find W₁, W₂, and W_final; D₁, D₂, and D₃ in kmole; and mole fractions of ethanol x_D1,avg, x_D2,avg, and x_D3,avg. p = 1.0 atm.

D24. A constant-mole batch distillation is being done to replace methanol with water as solvent for a nonvolatile solute currently dissolved in methanol. The system has a still pot and a condenser and operates with no reflux. The initial state has W_initial = 10.0 kmol, and x_{w,MeOH,initial} = 1.00. The final state has W_final = 10.0 kmol, x_w,MeOH,final = 0.10, and x_{w,water,final} = 0.90 mole fraction. Pressure = 1.0 atm.

a. Find S, the kmol of water added, and D_total = kmol of distillate.

b. Find the average distillate mole fraction of methanol, x_{dist,MeOH,avg}.

c. If the heating rate is constant at 100 kW, how many hours does the constant mole batch distillation require (assume water is added as a saturated liquid)?

D25. 1.5 kmol of feed one that is 40.0 mol% methanol and 60.0 mol% water and 1.0 kmol of feed two that is 20.0 mol% methanol and 80.0 mol% water will be processed in a simple batch distillation. The following three approaches have been proposed:

a. Mix the two feeds together, and do the batch distillation.

b. Do a batch distillation of feed one until the still pot concentration is 20 mol% methanol (same as feed two), add feed two, and then complete the batch distillation.

c. Do a batch distillation of feed one and a separate batch distillation of feed two. Add the two distillate products and the two still-pot products.

If we want x_{W,final,total} = 0.10, do calculations for each method. Determine W_final, D_total, and x_D,average, and compare the three methods. Equilibrium data are in Table 2-7. Logically, parts b and c should give the same result. If they do not, there is probably a numerical error from the use of Simpson’s rule. Try dividing an area into two parts for more accuracy.

D26. In Problem 9.D23 fractions were collected from ethanol mole fractions x_F = 0.10 to x_W1 = 0.07, from x_W1 to x_W2 = 0.04, and from x_W2 to x_W3 = x_W,final = 0.01. The solution is given under Answers to Selected Problems. Now collect a first product fraction (A) from x_F = 0.10 to x_WA = 0.085, a second product fraction (B) from x_WOC1 = 0.07 to x_WB = 0.05, and a third product fraction (C), which is identical to the current fraction, 3. To obtain these product fractions we need two offcuts: OC1 from x_WA = 0.085 to x_WOC1 = 0.07, and OC2 from x_WB = 0.05 to x_WOC2 = 0.04.

a. Find W and D (in kmol) and the mole fractions of ethanol x_D,avg for each fraction.

b. How would you reprocess the offcuts in the next batch?

D27. The continuation of Example 9.3 in Table 9.2 looks at removing a second fraction with x_W,A,2 = 0.30.

a. An alternative equivalent calculation is to use the entire cumulative area and the original feed F = 1 kmole. Show that this alternate calculation of W(at x_W,benzene = 0.30) gives the same result.

b. When x_W,benzene = 0.30, what are the values of x_W,toluene and x_W,cumene?

c. For the fraction collected from x_W,benzene = 0.40 to x_W,benzene = 0.30, determine the values of D_total, x_{D,benzene,avg}, x_{D,toluene,avg}, and x_D,cumene,avg.

D28. Repeat Problem 9.D13 except use Figure 9-10 as long as the distillate is in the two-phase region and then convert to simple batch distillation by bypassing the liquid-liquid settler. Compare the amounts and concentrations of the three products to Problem 9.D13.

D29. A 10.0 mol% ethanol 90.0 mol% water feed is batch distilled at 1.0 atm. The batch distillation system consists of a still pot plus a column with the equivalent of nine equilibrium contacts and a total condenser. External reflux ratio = 2/3. The initial charge to the still pot is 1000.0 kg. The final still pot concentration is 0.004 mole fraction ethanol. Find W_final, D_total, and x_D,avg. Note: Stepping off ten stages for a number of operating lines sounds like a lot of work. However, after you try it once, you will notice that there is a shortcut to determining a relationship between x_D and x_W.

E. More Complex Problems

E1. We are doing a single-stage, batch steam distillation of 1-octanol. The unit operates at 760 mm Hg. The batch steam distillation is operated with liquid water present. The distillate vapor is condensed, and two immiscible liquid layers form. The feed is 90.0 mol% octanol, and the rest is nonvolatile organic compounds. The feed is 1.0 kmol. We desire to recover 95% of the octanol. Vapor pressure data for water are given in Problem 8.D15. For small ranges in temperature, these data can be fit to an Antoine equation form with C = 273.16. The vapor pressure equation for 1-octanol is in Problem 8.D11.

a. Find the operating temperature of the still at the beginning and end of the batch.

b. Find the amount of organics left in the still pot at the end of the batch.

c. Find the kmoles of octanol recovered in the distillate.

d. Find the kmoles of water condensed in the distillate product. To do this, use the average still temperature to estimate the average octanol vapor pressure, which can be assumed to be constant. To numerically integrate Eq. (9-23b) relate x_org to n_org with a mass balance around the batch still.

e. Compare with your answer to Problem 8.D11. Which system produces more water in the distillate? Why?

E2. We wish to use batch steam distillation at 1.0 atm to recover 1-octanol from 100.0 kg of a mixture that is 15.0 wt% 1-octanol, and the remainder consists of nonvolatile organics and solids of unknown composition. The still pot is operated with liquid water in the pot. Assume the still pot is well mixed and liquid and vapor are in equilibrium. Ninety-five percent of the 1-octanol should be recovered in the distillate. Assume that water is completely immiscible with 1-octanol and with the nonvolatile organics. Because the composition of the nonvolatile organics is not known, we do a simple experiment and boil the feed mixture under a vacuum with no water present. The result is that at 0.05 atm pressure, the mixture boils at 129.8°C.

a. Find the mole fraction of 1-octanol in the feed and the effective average molecular weight of the nonvolatile organics and the solids. (Note: This is identical to the solution of part a in Problem 8.D15.)

b. Find the kg and kmole of 1-octanol in the distillate, the kg of total organics in the waste, and the 1-octanol weight fraction and the 1-octanol mole fraction in the waste.

e. Find the initial and final values of the temperature and of VP_octanol in the still pot. A spreadsheet or MATLAB is highly recommended for finding T and VP.

d. Find the kg and kmole of water in the distillate. This requires a numerical integration of Eq. (9-23b); however, the problem is simplified in this case. Because the still-pot temperature does not change much, the value of (VP)_octanol is very close to constant, and the average value can be used.

e. Compare the solution with the solution of Problem 8.D15. Why does batch steam distillation require less steam than continuous steam distillation?

Octanol boils at about 195°C. The formula for octanol is CH₃(CH₂)₆CH₂OH, and its molecular weight is 130.23. Vapor pressure formulas for octanol and water are available in Problems 8.D11 and 8.D15, respectively.

E3. In inverted batch distillation the charge of feed is placed in the accumulator at the top of the column (Figure 9-9). Liquid is fed to the top of the column. At the bottom of the column bottoms are continuously withdrawn, and part of the stream is sent to a total reboiler, vaporized, and sent back up the column. During the course of the batch distillation, the less volatile component (LVC) is slowly removed from the liquid in the accumulator, and the mole fraction MVC x_d increases. Assuming that holdup in the total reboiler, the total condenser, and the trays is small compared to the holdup in the accumulator, the Rayleigh equation for inverted batch distillation is

ln (D_final /F) = – _xfeed∫^xd,final [(dx_d)/(x_d – x_B)]

We feed the inverted batch system shown in Figure 9-9 with F = 10.0 mole of a feed that is 50.0 mol% ethanol and 50.0 mol% water. We desire a final distillate mole fraction of 0.63. There are two equilibrium stages in the column. The total reboiler, the total condenser, and the accumulator are not equilibrium contacts. VLE data are in Table 2-1. Find D_final, B_total, and x_B,avg if the boilup ratio is 1.0.

E4. A nonvolatile solute is dissolved in 1.0 kmol of methanol. We wish to switch the solvent to water. Because the solution is already concentrated, a first batch distillation to concentrate the solution is not required. We desire to have the solute in 1.0 kmol of solution that is 99.0 mol% water and 1.0 mol% methanol. This can be done either with a constant-mole batch distillation or by diluting the mixture with water and then doing a simple batch distillation. VLE data (ignore the effect of the solute) are in Table 2-7. Dilute the original pure methanol (plus solute) with water and then do a simple batch distillation with the goal of having W_final = 1.0 and x_M,final = 0.01. Find the moles of water added and the moles of water evaporated during the batch distillation. Compare with the constant-mole batch distillation in Problem 9.D17.

E5. A simple batch distillation is used to process 1.0 kmole of methanol-water feed into three distillate fractions and a waste. The initial feed has x_F,methanol = 0.50 mole fraction. We want the average mole fractions of methanol in the distillate fractions to be x_dist,avg,1 = 0.70, x_dist,avg,2 = 0.60, and x_dist,avg,3 = 0.40. Equilibrium data is in Table 2-7.

Part a. Find W_1,final, D_total,1, x_W,final,1; W_2,final, D_total,2, x_Wfinal,2; and W_3,final, D_total,3, x_W,final,3.

To make this problem tractable, the following procedure is suggested: Calculate relative volatility values from equilibrium data at x_M = 0.5, 0.4, 0.3, 0.2, 0.15, and 0.10. Over the different ranges of still pot mole fractions expected, calculate geometric average for each range.

Solve Eq. (9-13) on a spreadsheet to calculate W_fin using the guessed value of x_W,final. Calculate x_D,avg for the fraction. Use Goal Seek to make diff = x_{D,avg,specificed} – x_D,avg,guess approach zero by changing the guess for x_W,final. If the average value of relative volatility used was for an obviously incorrect range, insert a corrected , and repeat. Do this for each fraction.

Part b. In the lab or plant, it is easier to collect fractions based on weight or time than based on mole fraction measurements. If heating is done at a rate of 20 kW, at what time in minutes should we switch from collecting the first fraction to collecting the second fraction? Required data are available in the textbook (see Appendix D in the back of book).

E6. We are separating 100 moles of a feed that is 60.0 mol% methanol and 40.0 mol% water by batch distillation in a system with a still pot and a column that has one equilibrium contact (two equilibrium contacts total). Start the operation with a constant reflux ratio of L/D = 2/3. Operate with L/D = 2/3 until x_dist = 0.70. Then keep the distillate mole fraction constant by varying L/D. When L/D = 4.0, stop the batch distillation.

a. Find x_W,final.

b. Find W_final and D_total.

c. Find x_D,avg.

H. Computer Spreadsheet Problems

H1. A mixture of benzene (A) and cumene (C) is to be distilled in a simple batch system. The initial charge of 5.0 kmol is 37.0 mol% benzene and 63.0 mol% cumene. At the system pressure of 1 atm and choosing cumene (C) as the reference component, the relative volatility is α_A–C = 10.71. Use Eq. (9-13).

a. If FR_benzene,dist = 0.75, find x_A,Wfinal, W_final, D_total, and x_A,D,avg.

b. If x_A,Wfinal = 0.05, find FR_benzene,dist, W_final, D_total, and x_A,D,avg.

c. If FR_benzene,dist = 0.982, find x_A,Wfinal, W_final, D_total, and x_A,D,avg.

H2. Run a simple batch distillation of a mixture of benzene (A), xylene (B), and cumene (C). Initial charge is 5.0 kmole that is 72.0 mol% benzene, 13.0 mol% xylene, and 15.0 mol% cumene. If x_A,Wfinal = 0.30, find W_final, x_W,B, x_W,C, D_total, x_D,A,avg, x_D,B,avg, and x_D,C,avg. Constant relative volatilities are benzene = 2.25, toluene = 1.0, xylene = 0.33, and cumene = 0.21. Note: Toluene can be the reference component despite not being part of the mixture.

H3. Run a simple batch distillation of a mixture of benzene (A), xylene (B), and cumene (C). Initial charge is 2.7 kmole that is 64.3 mol% benzene, 11.6 mol% xylene, and 24.1 mol% cumene. If x_A,D,avg = 0.800, find W_final, x_W,A, x_W,B, x_W,C, D_total, x_D,B,avg, and x_D,C,avg. Constant relative volatilities are benzene = 2.25, toluene = 1.0, xylene = 0.33, and cumene = 0.21. Note: Toluene can be the reference component despite not being part of the mixture.