Chapter 9: Batch Distillation

Continuous distillation is a thermodynamically efficient method of producing large amounts of material of constant composition. When small amounts of material or varying product compositions are required, batch distillation has several advantages. In batch distillation a charge of feed is loaded into the reboiler, the steam is turned on, and after a short startup period, product can be withdrawn from the top of the column. When the distillation is finished, the heat is shut off and the material left in the reboiler is removed. Then a new batch can be started. Usually the distillate is the desired product.

Batch distillation is a much older process than continuous distillation. Batch distillation was first 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 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.” Advances in batch distillation have been associated with its use to distill 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 the choice when the plant does not run continuously and the batch must be completed in one or two shifts (8 to 16 hours). It is often used when the same equipment distills several different products at different times. If distillation is required only occasionally, batch distillation would again be the choice.

Equipment can be arranged in a variety of configurations. In simple batch distillation (Figure 9-1), the vapor is withdrawn continuously from the reboiler.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. An alternative to simple batch distillation is constant-level batch distillation where solvent is fed continually to the still pot to keep the liquid level constant (Gentilcore, 2002).

FIGURE 9-1Simple batch distillation

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 (Barton and Roche, 1997; Robinson and Gilliland, 1950; Pratt, 1962; Luyben, 1971; Diwekar, 1995) 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 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.

FIGURE 9-2Multistage batch distillation; A) schematic, B) photograph of packaged batch distillation/solvent recovery system of approximately 400-gallon capacity. Courtesy of APV Equipment, Inc., Tonowanda, New York

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 Problem 9.C2). 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.

9.1 BINARY BATCH DISTILLATION: RAYLEIGH EQUATION

The mass balances for batch distillation are somewhat different from those for continuous distillation. In batch distillation we are more interested in the total amounts of bottoms and distillate collected than in the rates. For a binary batch distillation, mass balances around the entire system for the entire operation time are

(9-1)

(9-2)

The feed into the column is F kg moles of mole fraction x_F of the more volatile component. The final moles in the reboiler at the end of the batch is W_final of mole fraction x_W,final. The symbol W is used since the material left in the reboiler is often a waste. D_total is the total kilogram moles 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.

Usually F, x_F, and the desired value of either x_W,final or x_D,avg are specified. An additional equation is required to solve for the three unknowns D_total, W_final, and x_W,final (or x_D,avg). This additional equation, known as the Rayleigh equation (Rayleigh, 1902), is derived from a differential mass balance. Assume that the holdup in the column and in the accumulator is negligible. Then if a differential amount of material, −dW, of concentration x_D is removed from the system, the differential mass balance is

(9-3)

or

(9-4)

Expanding Eq. (9-4),

(9-5)

Then rearranging and integrating,

(9-6)

which is

(9-7)

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 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 9.2 and 9.4.

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.

9.2 SIMPLE BINARY BATCH DISTILLATION

In the simple binary batch distillation system shown in Figure 9-1 the vapor product is in equilibrium with liquid in the still pot at any given time. Since we use a total condenser, y = x_D. Substituting this into Eq. (9-7), we have

(9-8)

where y and x are now in equilibrium and the equilibrium expression is y = f(x,p). For any given equilibrium expression, Eq. (9-8) can be integrated analytically, graphically, or numerically.

The general integration procedure for Eq. (9-8) is:

1. Plot or fit 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 or numerically integrate from x_F to x_W,final. Graphical integration is shown in Figure 9-3.

FIGURE 9-3Graphical integration for simple batch distillation, Example 9-1

5. From Eq. (9-8), find the final charge of material in the still pot:

where the area is shown in Figure 9-3.

6. The average distillate concentration, x_{D avg}, can be found from the mass balances. Solving Eqs. (9-1) and (9-2),

(9-10)

(9-11)

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). If the ordinate in Figure 9-3 is called f(x), then one form of Simpson’s rule is

(9-12)

where terms are shown in Figure 9-3. Simpson’s rule is exact if f(x) is cubic or lower order. For smooth curves, such as in Figure 9-3, Simpson’s rule will be quite accurate (see Example 9-1). For more complex shapes, Simpson’s rule will be more accurate if the integration is done in two or more pieces (see Example 9-2). 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 9-3 either graphically or using Simpson’s rule. Then Eq. (9-9) gives W_final and Eq. (9-10) 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_W,final that is too high. Then every time x_W,final is decreased, the additional area is added to the area already calculated.

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

Substituting this equation into Eq. (9-8) and integrating, we obtain

(9-13)

When it is applicable, Eq. (9-13) is obviously easier to apply than graphical or numerical integration.

EXAMPLE 9-1 Simple Rayleigh distillation

We wish to use a simple batch still (one equilibrium stage) to separate methanol from water. The feed charge to the still pot is 50 moles of an 80 mole % methanol mixture. We desire an average distillate concentration of 89.2 mole % methanol. Find the amount of distillate collected, the amount of material left in the still pot, and the concentration of material in the still pot. Pressure is 1 atm. Methanol-water equilibrium data at 1 atm are given in Table 2-7 in Problem 2.D1.

Solution

A . Define. The apparatus is shown in Figure 9-1. The conditions are: p = 1 atm, F = 50, x_F = 0.80, and x_D,avg = 0.892. We wish to find x_W,final, D_tot, and W_final.

B Explore. Since the still pot acts as one equilibrium contact, the Rayleigh equation takes the form of Eqs. (9-8) and (9-9). To use these equations, either a plot of 1/(y − x)_equil vs. x is required for graphical integration or Simpson’s rule can be used. Both will be illustrated. Since x_W,final is unknown, a trial-and-error procedure will be required for either integration routine.

C Plan. First plot 1/(y − x) vs. x from the equilibrium data. The trial-and-error procedure is as follows:

Guess x_W,final

Integrate to find

Calculate W_final from Rayleigh equation and x_D,calc from mass balance.

Check: Is x_D,calc = x_D,avg? If not, continue trial-and-error.

Do it. From the equilibrium data the following table is easily generated:

These data are plotted in Figure 9-3. For the numerical solution a large graph on millimeter graph paper was constructed.

First guess: x_W,final = 0.70.

From Figure 9-3,

Then, W_final = F exp (−Area) = 50 e^−0.7044 = 24.72

D_calc = F − W_final = 25.28

The alternative integration procedure using Simpson’s rule gives

is the value of 1/(y−x) calculated at x_W,final.

Then for Simpson’s rule, W_final = 24.79, D_calc = 25.21, and x_D,calc = 0.898. Simpson’s rule appears to be quite accurate. W_final is off by 0.3%, and x_D,calc is the same as the more exact calculation. These values appear to be close to the desired value, but we don’t yet know the sensitivity of the calculation.

Second guess: x_W,final = 0.60. Calculations similar to the first trial give Area = 1.2084, W_final = 14.93, D_calc = 35.07, x_D,calc = 0.885 from Figure 9-3, and x_D,calc = 0.884 from the Simpson’s rule calculation. These are also close, but they are low. For this problem the value of x_D is insensitive to x_W,final.

Third guess: x_W,final = 0.65. Calculations give Area = 0.971, W_final = 18.94, D_calc = 31.06, x_D,calc = 0.891 from Figure 9-3 and x_D,calc = 0.890 from the Simpson’s rule calculation of the area, which are both close to the specified value of 0.892.

Thus, use x_W,final = 0.65 as the answer.

E . Check. The overall mass balance should check. This gives: W_final x_W,final + D_calc x_D,calc = 39.985 as compared to Fx_F = 40. Error is (40−39.985) / 40 × 100, or 0.038%, which is acceptable.

F . Generalize. The integration can also be done numerically on a computer using Simpson’s rule or an alternative integration method. This is an advantage, since then the entire trial-and-error procedure can be programmed. Note that large differences in x_W,final and hence in W_final cause rather small differences in x_D,avg. Thus, for this problem, exact control of the batch system may not be critical. This problem illustrates a common difficulty of simple batch distillation—a pure distillate and a pure bottoms product cannot be obtained unless the relative volatility is very large. Note that although Eq. (9-13) is strictly not applicable since methanol-water equilibrium does not have a constant relative volatility, it could be used over the limited range of this batch distillation.

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,

(9-14)

where F is the moles of vapor fed of mole fraction y_F, and D_final is the left over vapor distillate of mole fraction y_D,final.

9.3 CONSTANT-LEVEL BATCH DISTILLATION

One common application of batch distillation (or evaporation) is to switch solvents in a production process. For example, solvents may need to be exchanged prior to a crystallization or reaction step. Solvent switching can be done in a simple batch system by charging the still pot with feed, concentrating the solution (if necessary) by boiling off most of the original solvent (some solvent needs to remain to maintain agitation, keep the solute in solution, and keep the heat transfer area covered), adding the new solvent and doing a second batch distillation to remove the remainder of the original solvent. If desired the solution can be diluted by adding more of the desired solvent. Although this process mimics the procedure used by a bench chemist, a considerable amount of the second solvent is evaporated in the second batch distillation. An alternative is to do the second batch distillation by constant-level batch distillation in which the pure second solvent is added continuously during the second batch distillation at a rate that keeps the moles in the still-pot constant (Gentilcore, 2002). We will focus on the constant-level batch distillation step.

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

9-15a

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-level batch distillation is

9-15b

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-level system,

9-16a

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

9-16b

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,

(9-17)

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 will lead you through these calculations for constant-level batch distillation. Problem 9.E4 considers the dilution step followed by a simple 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),

(9-18)

Gentilcore (2002) presents a constant relative volatility sample calculation that illustrates the advantage of constant-level batch distillation when it is used to exchange solvents.

9.4 BATCH STEAM DISTILLATION

In batch steam distillation, steam is sparged directly into the still pot as shown in Figure 9-4. This is normally done for systems that are immiscible with water. The reasons for adding steam directly to the still pot are that it keeps the temperature below the boiling point of water, it eliminates the need for heat transfer surface area and it helps keep slurries and sludges well mixed so that they can be pumped. The major use is in treating wastes that contain valuable volatile organics. These waste streams are often slurries or sludges that would be difficult to process in an ordinary batch still. Compounds that are often steam distilled include glycerine, lube oils, fatty acids, and halogenated hydrocarbons (Woodland, 1978). Section 8.3 is a prerequisite for this section.

FIGURE 9-4Batch steam distillation

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 to having more than one stage. For low-molecular-weight organics, vaporization efficiencies, defined as the actual partial pressure divided by the partial pressure at equilibrium, p*,

(9-19)

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

The system shown in Figure 9-4 can be analyzed with mass balances on a water-free basis. The mass balances are Eqs. (9-1) and (9-2), which can be solved for W_final if x_W,final is given.

(9-20)

With a single volatile organic, x_D = 1.0 if entrainment is negligible. Then,

(9-21)

and the flow rate of the organic distillate product is

(9-22)

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

(9-23)

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

(9-24)

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 (9-24) 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 vaporize the volatile organics.

9.5 MULTISTAGE BATCH DISTILLATION

The separation achieved in a single equilibrium stage is often not large enough to both obtain the desired distillate concentration and a low enough bottoms concentration. In this case a distillation column is placed above the reboiler as shown in Figure 9-2. The calculation procedure will be detailed here for a staged column, but packed columns can easily be designed using the procedures explained in Chapter 10.

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 will 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,

(9-25a)

(9-25b)

(9-25c)

In these equations V, L and D are now molal flow rates. These balances are essentially the same equations we obtained for the rectifying section of a continuous column except that Eqs. (9-25) 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. (9-25a) and (9-25b) and solving for y_j+1, we obtain the operating equation for CMO:

(9-26)

At any specific time Eq. (9-26) 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 be continuously changing.

9.5.1 Constant Reflux Ratio

The most common operating method is to use a constant reflux ratio and allow x_D to vary. This procedure corresponds to a simple batch operation where x_D also varies. The relationship between x_D and x_W can now be found from a stage-by-stage calculation using a McCabe-Thiele analysis. Operating Eq. (9-26) is plotted on a McCabe-Thiele diagram for a series of x_D values. Then we step off the specified number of equilibrium contacts on each operating line starting at x_D to find the x_W value corresponding to that x_D. This procedure is shown in Figure 9-5 and Example 9-2.

FIGURE 9-5McCabe-Thiele diagram for multistage batch distillation with constant L/D, Example 9-2

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 Eq. (9-12) 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. (9-9), x_{D avg} from Eq. (9-10), and D_total from Eq. (9-11). If x_{D avg} is specified, a trial-and-error procedure will again be required.

EXAMPLE 9-2 Multistage batch distillation

We wish to batch distill 50 kg moles of a 32 mole % ethanol, 68 mole % water feed. The system has a still pot plus two equilibrium stages and a total condenser. Reflux is returned as a saturated liquid, and we use L/D = 2/3. We desire a final still pot composition of 4.5 mole % ethanol. Find the average distillate composition, the final charge in the still pot, and the amount of distillate collected. Pressure is 1 atm.

Solution

A . Define. The system is shown in the figure

Find W_final, D_total, x_D,avg.

B and C . Explore and Plan. Since we can assume CMO, a McCabe-Thiele diagram (Figure 2-2) can be used. This will relate x_D to x_W at any time. Since x_F and x_W,final are known, the Rayleigh Eq. (9-7) or (9-9) can be used to determine W_final. Then x_D,avg and D_total can be determined from Eqs. (9-10) and (9-11), respectively. A trial-and-error procedure is not needed for this problem.

D . Do it. The McCabe-Thiele diagram for several arbitrary values of x_D is shown in Figure 9-5. The top operating line is

where

The corresponding x_W and x_D values are used to calculate x_D − x_W and then 1/(x_D − x_W) for each x_W value. These values are plotted in Figure 9-6 (some values not shown in Figure 9-5 are shown in Figure 9-6). The area under the curve (going down to an ordinate value of zero) from x_F = 0.32 to x_W,final = 0.045 is 0.608 by graphical integration.

Then from Eq. (9-6):

W_final = Fe^-Area = (50) exp (−0.608) = 27.21

From Eq. (9-11): D_total = F − W_final = 22.79

FIGURE 9-6Graphical integration, Example 9-2

and from Eq. (9-10):

The area can also be determined by Simpson’s rule. However, because of the shape of the curve in Figure 9-6 it will probably be less accurate than in Example 9-1. Simpson’s rule gives

where (x_W,final + x_F)/2 = 0.1825 and 1/(x_D − x_W) = 2.14 at this midpoint.

This can be checked by breaking the area into two parts and using Simpson’s rule for each part. Do one part from x_W,final = 0.045 to x_W = 0.10 and the other part from 0.1 to x_F = 0.32. Each of the two parts should be relatively easy to fit with a cubic. Then,

Total area = 0.6196

Note that Figure 9-6 is very useful for finding the values of 1/(x_D − x_W) at the intermediate points x_W = 0.0725 (value = 2.03) and x_W = 0.21 (value = 2.23). The total area calculated is closer to the answer obtained graphically (1.9% difference compared to 4.7% difference for the first estimate).

Then, doing the same calculations as previously [Eqs. (9-9), (9-11), and (9-10)] with Area = 0.6196,

W_final = 26.91, D_total = 23.09, x_D,avg = 0.640

E . Check. The mass balances for an entire cycle, Eqs. (9-1) and (9-2), should be and are satisfied. Since the graphical integration and Simpson’s rule (done as two parts) give similar results, this is another reassurance.

F . Generalize. Note that we did not need to find the exact value of x_D for x_F or x_W,final. We just made sure that our calculated values went beyond these values. This is true for both integration methods. Our axes in Figure 9-6 were selected to give maximum accuracy; thus, we did not graph parts of the diagram that we didn’t use. The same general idea applies if fitting data—only fit the data in the region needed. For more accuracy, Figure 9-5 should be expanded. Note that the graph in Figure 9-6 is very useful for interpolation to find values for Simpson’s rule. If Simpson’s rule is to be used for very sharply changing curves, accuracy will be better if the curve is split into two or more parts. Comparison of the results obtained with graphical integration to those obtained with the two-part integration with Simpson’s rule shows a difference in x_D,avg of 0.008. This is within the accuracy of the equilibrium data.

Once x_W,final, D and W_final are determined, we can calculate the values of Q_c, Q_R and operating time (see Section 9.6).

9.5.2 Variable Reflux Ratio

The batch distillation column can also be operated with variable reflux ratio to keep x_D constant. The operating Eq. (9-26) is still valid. Now the slope will vary, but the intersection with the y=x line will be constant at x_D. The McCabe-Thiele diagram for this case is shown in Figure 9-7. This diagram relates x_W to x_D, and the Rayleigh equation can be integrated as in Figure 9-6. Since x_D is kept constant, the calculation procedure is somewhat different.

FIGURE 9-7McCabe-Thiele diagram for multistage batch distillation with constant x_D and variable reflux ratio

With x_D and the number of stages specified, the initial value of L/V is found by trial and error to give the feed concentration x_F. The operating line slope L/V is then increased until the specified number of equilibrium contacts gives x_W = x_W,final. Once x_W,final is determined, 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 9.6).

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 calculational procedure is required for this case and for multicomponent systems (Barton and Roche, 1997; Diwekar, 1995).

9.6 OPERATING TIME

The operating time and batch size may be controlled by economics or other factors. For instance, it is not uncommon for the entire batch including startup and shutdown to be done in one eight-hour shift. If the same apparatus is used for several different chemicals, the batch sizes may vary. Also the time to change over from one chemical to another may be quite long, since a rigorous cleaning procedure may be required.

The total batch time, t_batch, is

(9-27)

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.

(9-28)

D_total is calculated from the Rayleigh equation calculation procedure, 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 kmole/hr, 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,

(9-29)

We usually operate at some fraction of this flow rate such as D = 0.75 D_max. Then Eqs. (9-28) and (9-29) can be used to estimate t_op. If the resulting t_batch is not convenient, adjustments must be made.

The energy requirements in the reboiler or still pot Q_R and the total condenser Q_c can be estimated from energy balances on the condenser and around the entire system. 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. If we use λ_avg, Eq. (3-13) simplifies to

(9-30)

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

(9-31)

This differs from Eq. (3-15a) because during operation there is no feed and no bottoms product. For an existing batch distillation apparatus we must check that the condenser and reboiler are large enough to handle the calculated values of |Q_c| and Q_R. If |Q_c| or Q_R are 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.

Batch distillation has somewhat different design and process control requirements than continuous distillation. In addition, startup and troubleshooting are somewhat different. These aspects are discussed by Ellerbe (1979).

SUMMARY—OBJECTIVES

In this chapter we have explored binary batch distillation calculations. At this time you should be able to satisfy the following objectives:

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 batch distillation

4. Solve problems for constant-level batch distillation

5. Solve problems in batch steam distillation

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

a . Batch distillation with constant reflux ratio

b . Batch distillation with constant distillate composition

c . Inverted batch distillation

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

REFERENCES

Barton, P. and E.C. Roche, Jr., “Batch Distillation,” Section 1.3 in Schweitzer, P.A. (Ed.), Handbook of Separation Techniques for Chemical Engineers, 3rd ed., McGraw-Hill, New York, 1997.

Carey, J.S., “Distillation” in J.H. Perry (Ed.), Chemical Engineer’s Handbook, 3rd ed., McGraw-Hill, New York, 1950, pp. 582-585.

Diwekar, U.W., Batch Distillation: Simulation, Optimal Design and Control, Taylor and Francis, Washington, DC, 1995.

Ellerbe, R.W., “Batch Distillation,” in P.A. Schweitzer (Ed.), Handbook of Separation Techniques for Chemical Engineers, McGraw-Hill, New York, 1979, p. 1.147.

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

Luyben, W.L., “Some Practical Aspects of Optimal Batch Distillation,” Ind. Eng. Chem. Process Des. Develop., 10, 54 (1971).

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

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

Pratt, H.R.C., Countercurrent Separation Processes, Elsevier, New York, 1967.

Rayleigh, Lord, Phil. Mag. [vi], 4(23), 521 (1902).

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

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.

HOMEWORK

Discussion Problems>

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. Discuss the advantages and disadvantages of batch distillation compared to continuous distillation. Would you expect to see batch or continuous distillation in the following industries? Why?

a . Large basic chemical plant

b . Plant producing fine chemicals

c . Petroleum refinery

d . Petrochemical plant

e . Pharmaceutical plant

f . Cryogenic plant for recovering oxygen from air

g . Still for solvent recovery in a painting operation

h . Ethanol production facilities for a farmer

Note that continuous distillation is often used in campaigns. That is, the plant produces a product continuously but for a relatively short period of time. How does this change your answers?

A5. Assuming that either batch or continuous distillation could be used, which would use less energy? Explain why.

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 will outperform the normal simple batch (Figure 9-1). Suppose you want to batch distill a mixture similar to methanol and water where 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

c . less than

the x_d,avg 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. Batch distillation with multiple stages is easier to operate

a . At constant distillate mole fraction.

b . At constant reflux ratio.

c . At constant bottoms mole fraction.

A9. Develop a key relations chart for this chapter.

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.

C Derivations

C1. Derive Eq. (9-13).

C2. Assume that holdup in the column and in the total reboiler is negligible in an inverted batch distillation (Figure 9-8).

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

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

C3. Derive Eq. (9-14).

D Problems

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

D1. A simple batch still (one equilibrium stage) is used to separate ethanol from water. The feed to the still pot is 1.0 kmoles of a mixture that is 30 mole % ethanol. The final concentration in the still pot should be 10 mole % ethanol. The ethanol-water equilibrium data is in Table 2-1. Find:

a . The final charge of material in the still pot, W_final.

b . The amount of distillate collected, D_total.

c . The average mole fraction of the distillate, x_D,avg.

Hint: Use Simpson’s rule.

D2.* We wish to use a simple batch still (one equilibrium stage) to separate methanol from water. The feed charge to the still pot is 100 moles of a 75 mole % methanol mixture. We desire a final bottoms concentration of 55 mole % methanol. Find the amount of distillate collected, the amount of material left in the still pot, and the average concentration of distillate. Pressure is 1 atm. Equilibrium data are given in Table 2-7

D3.* We wish to use a distillation system with a still pot plus a column with one equilibrium stage to batch distill a mixture of methanol and water. A total condenser is used. The feed is 57 mole % methanol. We desire a final bottoms concentration of 15 mole % methanol. Pressure is 101.3 kPa. Reflux is a saturated liquid, and L₀/D is constant at 1.85. Find W_final, D_total, and x_D,avg. Methanol-water equilibrium data is given in Table 2-7 Calculate on the basis of 1 kg mole of feed.

FIGURE 9-8Inverted batch distillation

D4. We wish to do a simple batch distillation (1 equilibrium contact) of a mixture of acetone and ethanol. The feed charge to the still pot is 80 mole % acetone. The final concentration in the still pot will be 40 mole % acetone. The final amount of material in the still pot is 2.0 kg moles. Vapor-liquid equilibrium (VLE) are in problem 4.D7. Find the feed rate F, the average mole fraction of the distillate, and the kmoles of distillate collected.

D5. A simple batch distillation system with a still pot is used to distill 0.7 kmole of a mixture of n-butanol and water. The system has a total condenser and distillate is removed. The feed is 20 mole % water and 80 mole % n-butanol. We desire a final still pot that is 4 mole % water. The equilibrium data is in Table 8-2. Use Simpson’s rule to find the final moles in the still pot. Use a mass balance to determine the average distillate mole fraction.

D6. We have a simple batch still separating 1,2 dichloroethane from 1,1,2 trichloroethane. Pressure is one atmosphere. The relative volatility for this system is very close to constant at a value of 2.4. The use of Eq. (9-13) is recommended.

a . The charge (F) is 1.3 kg mole. This feed is 60 mole % dichloroethane. We desire a final still pot concentration of 30 mole % dichloroethane. Find the final moles in the still pot and the average mole fraction of the distillate product.

b . Repeat part a if the feed charge is 3.5 kg moles.

c . If the feed charge is 2.0 kg moles, the feed is 60 mole % dichloroethane, and we want a distillate with an average concentration of 75 mole % dicholoroethane, find the final moles in the still pot and the final mole fraction of dichloroethane in the still pot.

D7. We will use a batch distillation system with a still pot and one equilibrium stage (2 equilibrium contacts total) to distill a feed that is 10 mole % water and 90 mole % n-butanol (see Table 8-2 for VLE data). Pressure is one atmosphere. The charge is 4.0 kg moles. We desire a final still concentration that is 2.0 mole % water. The system has a total condenser and the reflux is returned as a saturated liquid. The reflux ratio L/D = 1/2. Find the final number of moles in the still and the average concentration of the distillate.

D8. A mixture of acetone and ethanol is being batch distilled in a system with a still pot plus one equilibrium stage (two equilibrium contacts total). The feed is 0.50 mole fraction acetone. The external reflux ratio is constant at a value of L/D = 1.0. Pressure is one atmosphere. We desire the final amount in the still pot to be 1.5 kmole. The final mole fraction in the still pot should be 0.20 mole fraction acetone. Assume CMO. Equilibrium is listed in problem 4.D7. Find

a . The amount of feed F (in kmoles) that should be added.

b . The amount of distillate and the average mole fraction of acetone in the distillate.

D9.* We wish to batch distill a mixture of 1-butanol and water. Since this system has a heterogeneous azeotrope (see Chapter 8), we will use the system shown in the figure. The bottom liquid layer with 97.5 mole % water is removed as product, and the top liquid layer, which is 57.3 mole % water, is returned to the still pot. Pressure is 1 atm. The feed is 20 kg moles and is 40 mole % water. Data are given in Problem 8-D2.

a . If the final concentration in the still pot should be x_W,final = 0.28, what is the final amount of distillate D_total collected, in kg moles?

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. We plan to batch distill a mixture of ethanol and water. The feed contains 0.52 mole fraction ethanol and we wish to continue the distillation until x_w,final = 0.20 mole fraction ethanol. The initial charge is F = 10.0 kgmoles. The batch system has a still pot, which acts as an equilibrium contact and a column with one equilibrium stage (total of 2 equilibrium contacts). The distillate vapor is condensed in a total condenser and we use an external reflux ratio . VLE is in Table 2-1 Find: W_final, D_total, and x_D,avg

D11. We will separate 2.0 kg moles of a mixture of n-butane and n-hexane in a simple batch still. The feed is 40 mole % n-butane and 60 mole % n-hexane. Pressure is 101.3 kPa. The final still pot is 5 mole % n-butane. Do a bubble point calculation at the feed condition and a bubble point calculation at the final conditions. K values can be obtained from the DePriester charts. Calculate relative volatilities at these two conditions. Then calculate the geometric average relative volatility for use in Eq. (9-13). Determine the final amount in the still pot, the amount of distillate collected, the average distillate mole fraction, and the temperature of the still pot when the batch is finished.

D12. Acetone-ethanol VLE data are given in problem 4.D7. We have 3.0 kgmoles of a mixture of acetone and ethanol that we wish to separate in a simple batch still. The feed is 48 mole % acetone. The final still pot should be 16 mole % acetone. Find the final amount in the still pot, the amount of distillate collected, and the average distillate mole fraction.

D13. A batch distillation system to separate water from n-butanol is set up with a still pot, a column containing two stages, a total condenser, and a liquid-liquid separator. The water layer from the liquid-liquid separator (0.975 mole fraction water) is taken as the distillate product. The organic layer from the liquid-liquid separator (0.573 mole fraction water) is returned to the column as reflux. VLE are in Table 8-2. The feed to the batch system is 10.0 kg. The feed is 0.32 mole fraction water. Operation is at 1.0 atm., and CMO is valid.

a . If we desire a final still pot that is 0.08 mole fraction water, find D_total and W_final.

b . What is the internal reflux ratio, L/V, at the start of the batch distillation?

c . What is the lowest value of x_W,final that can be obtained with this batch distillation system?

D14.* A simple steam distillation is being done in the apparatus shown in Figure 9-4. The organic feed is 90 mole % n-decane and 10% nonvolatile organics. The system is operated with liquid water in the still. Distillation is continued until the organic layer in the still is 10 mole % n-decane. F = 10 kg moles. Pressure is 760 mmHg.

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

b . What is W_final? What is 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 is in Example 8-2. Vapor pressure data for water are listed in Problem 8.D10.

D15. We wish to batch distill 100 kgmoles of a mixture of n-butanol and water. The system consists of a batch still pot plus 1 equilibrium stage. The system is at one atmosphere. The feed is 48 mole % water and 52 mole % butanol. The distillate vapor is condensed and sent to a liquid-liquid settler. The water rich product (0.975 mole fraction water) is taken as the distillate product and the butanol rich layer (0.573 mole fraction water) is refluxed to the column. We desire a final still pot mole fraction of 0.08 water. Energy is added at a constant rate to the still pot; thus, V = constant. Note that the distillate product is a constant mole fraction. The reflux ratio increases as the distillate vapor mole fraction decreases during the course of the batch distillation. Equilibrium data are given in Table 8-2.

a . Find the final amount of material in the still pot and the amount of distillate collected.

b . Find the initial and final internal reflux ratios (L/V values).

D16.* We wish to do a normal batch distillation of methanol and water. The system has 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 returned as a saturated liquid. The column is operated with a varying reflux ratio so that x_D is held constant. The initial charge is F = 10 kg moles and is 40 mole % methanol. We desire a final still-pot concentration of 8 mole % methanol, and the distillate concentration should be 85 mole % methanol. Pressure is 1 atm and CMO is valid. Equilibrium data is given in Tale 2-7.

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 final amount, W_final, left in the still pot?

D17. A nonvolatile solute is dissolved in 1.0 kg moles 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 kg mole of solution that is 99.0 mole % water and 1.0 mole % methanol. This can be done either with a constant-level 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. Do a constant-level 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-level 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. Straightforward application of Simpson’s rule from x_M,initial = 1.0 to x_M,final = 0.01 will not be accurate. The integral needs to be divided into at least two sections.

D18. We will use a simple batch still (one equilibrium contact—the still pot) to separate 1.5 kmoles of a mixture of n-pentane and n-octane at a pressure of 101.3 kPa. The feed is 35 mole % n-pentane. We desire a final still pot concentration of 5 mole % n-pentane. Find the final amount in the still pot, the amount of distillate collected, and the average distillate mole fraction.

D19. A simple batch still is being used to process 1.0 kmole of a mixture that is 40 mole % water and 60 mole % n-butanol. The final still concentration desired is 16 mole % water. Pressure is 1.0 atm. Equilibrium data are available in Table 8-2. Find the final amount in the still pot, the amount of distillate collected, and the average distillate mole fraction. If all of the distillate liquid is collected in a stirred tank and then allowed to settle, is it one or two phases?

D20. To take advantage of the heterogeneous azeotrope, the batch distillation system shown in problem 9.D9 is used for 2.5 kmoles of feed that is 52 mole % water and 48 mole % n-butanol. Operation is at 1.0 atm. A final still concentration of 24 mole % water is desired. The distillate is always 97.5 mole % water. Equilibrium data are available in Table 8-2.

a . What are W_final and D_total?

b . What is the initial external reflux ratio L/D?

c . What is the final external reflux ratio L/D?

D21. We plan to batch distill 3.0 kmoles of a mixture of acetone and ethanol. The feed is 48.0 mole % acetone. We desire a final still pot concentration of 16.0 mole % acetone. The batch distillation system consists of the still pot and a column with one equilibrium stage. The reflux is returned as a saturated liquid and the external reflux ratio is constant at L/D = 1. Pressure is 1.0 atm. VLE data are in problem 4.D7. Find the final amount in the still pot, the amount of distillate collected, and the average distillate mole fraction. Compare your results with the simple batch system in problem 9.D12.

D22. A mixture that is 62 mole % methanol and 38 mole % water is batch distilled in a system with a still pot and a column with 1 equilibrium stage (2 equilibrium contacts total). F = 3.0 kmole. The system operates with a constant distillate concentration that is 85 mole % methanol. We desire a final still pot concentration that is 45 mole % methanol. Reflux is a saturated liquid and the external reflux ratio L/D varies. Assume CMO. VLE data are in Table 2-7.

a. Find D_total and W_final (kmoles).

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

D23. We wish to batch distill a mixture of ethanol and water. The feed is 10.0 mole % ethanol. Operation is at 1.0 atmosphere. The batch distillation system consists of a still pot plus a column with the equivalent of 9 equilibrium contacts and a total condenser. We operate at a constant external reflux ratio of 2/3. The initial charge to the still pot is 1000.0 kg. We desire a final still pot concentration of 0.004 mole fraction ethanol. Equilibrium data are in Table 2-1. Convert the amount of feed to kmoles using an average molecular weight. Find W_final and D_total in kmoles, and x_D,avg.

Note: Stepping off 10 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 short cut to determining the relationship between x_D and x_w.

D24. On occasion your plant needs to batch distill a mixture of acetone-ethanol. The initial charge is 10 kg moles of a mixture that is 40 mole % acetone. The batch distillation is done in a system with a still pot and 4 equilibrium stages. Your boss wants a constant distillate concentration of x_D = 0.70 mole fraction acetone. Data are available in problem 4.D7.

a. If an infinite number of stages were available, what is (L/D)_min?

b. With the 5 equilibrium contacts available, what is the lowest still pot mole fraction that could possibly be obtained (total reflux)?

c. Approximately what initial reflux ratio (L/V)_initial must be used?

d. If the largest reflux ratio that can be used in practice is L/V = 5/7, what is x_w,final?

e. What are W_final and D_total in kg moles? (The easy way is to use overall and acetone mass balances.)

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 mole % octanol and the rest is nonvolatile organic compounds. The feed is 1.0 kg mole. We desire to recover 95% of the octanol.

Vapor pressure data for water is given in problem 8.D10. For small ranges in temperature, this 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 moles of octanol recovered in the distillate.

d. Find the moles 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. (9-24) 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. A single-stage, batch steam distillation is being done of 1.2 kmoles of a feed that contains 55.0 mole % o-cymene (C₁₀H₁₄) and 45.0 mole % nonvolatile compounds. The distillation is conducted with liquid water present. The distillate vapor is condensed and two immiscible liquid layers form (o-cymene and water). Total pressure is 760 mm Hg. We desire to recover 92 % of the o-cymene.

Vapor pressure data for water is given in problem 8.D10. For small ranges of temperature, this data can be fit to an Antoine equation form with C = 273.16. The vapor pressure of o-cymene can be found from the Antoine equation,

Log₁₀ (VP) = A – B/(T+C)

With A = 7.26610, B = 1768.45, C = 224.95, and T in degrees Centigrade and VP is the vapor pressure in mm Hg. Source: Lange’s Handbook of Chemistry, 13th ed., p. 10-40, 1985.

a. Find the amount of organics in the still pot at the end of the distillation.

b. Find the moles of o-cymene recovered in the distillate.

c. Find the still operating temperatures at the beginning and end of the distillation.

d. Find the moles of water condensed in the distillate product. Use the average still temperature to estimate an average o-cymene vapor pressure, which is assumed to be constant. Numerically integrate Eq. (9-24) by relating x_org to n_org with a mass balance around the still pot.

E3. In inverted batch distillation the charge of feed is placed in the accumulator at the top of the column (Figure 9-8). 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 is slowly removed from the liquid in the accumulator and the mole fraction more volatile component x_d increases. Assuming that holdup in the total reboiler, total condenser and the trays is small compared to the holdup in the accumulator, the Rayleigh equation for inverted batch distillation is,

We feed the inverted batch system shown in the Figure 9-8 with F = 10 gmoles of a feed that is 50 mole % ethanol and 50% water. We desire a final distillate mole fraction of 0.63. There are 2 equilibrium stages in the column. The total reboiler, the total condenser and the accumulator are not equilibrium contacts. VLE 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 kg moles 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 kg mole of solution that is 99.0 mole % water and 1.0 mole % methanol. This can be done either with a constant-level 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 constant-level batch distillation in problem 9.D17.