Chapter 5. Introduction to Multicomponent Distillation

Chapter 5 is the bridge between binary and multicomponent distillation. Binary distillation problems can be solved in a straightforward manner using a stage-by-stage calculation that can be done either on a computer or graphically using a McCabe-Thiele diagram. When additional components are added, the resulting multicomponent problem becomes significantly more difficult, and solution may not be straightforward. In this chapter we consider why multicomponent distillation is more complex than binary distillation, look at profile shapes typical of multicomponent distillation, and solve a limited selection of multicomponent distillation problems with stage-by-stage calculations. In Chapter 6 matrix calculation methods are applied to multicomponent distillation, and approximate methods are developed in Chapter 7.

1. Explain why multicomponent distillation is trial and error

2. Make appropriate assumptions and solve external mass balances

3. Explain flow, temperature, and composition profiles for multicomponent distillation

4. Do stage-by-stage calculations for distillation with either no heavy non-keys or no light non-keys

Now if we add a third component, we increase the degrees of freedom to 9. Nine variables that would most likely be specified for the design of a ternary distillation column are listed in Table 5-1. Comparing this table with Tables 3-1 and 3-2, we see that the additional degree of freedom is used to completely specify feed composition. If there are four components, there are 10 degrees of freedom, and the tenth degree of freedom is used to completely specify feed composition.

In multicomponent distillation neither distillate nor bottoms composition is completely specified because there are not enough degrees of freedom to allow complete specification. This inability to completely specify distillate and bottoms compositions has a major impact on calculation procedures. Components that have their distillate and bottoms fractional recoveries specified (e.g., component 1 in distillate and component 2 in bottoms in Table 5-1) are called key components. The more volatile key is called the light key (LK) and the less volatile the heavy key (HK). Other components are non-keys (NK). If a non-key is more volatile (lighter) than the light key, it is a light non-key (LNK); if it is less volatile (heavier) than the heavy key, it is a heavy non-key (HNK).

External balance equations for the column shown in Figure 5-1 are similar to binary equations. These equations are the overall balance equation,

the component balance equations,

and the energy balance,

and mole fractions must sum to 1.

For a ternary system, Eqs. (5-2) can be written three times, but these equations add to give Eq. (5-1). Thus only three of the four equations [Eqs. (5-2) and Eq. (5-1)] are independent.

For a single-feed column, we can solve either of Eqs. (5-2) simultaneously with Eq. (5-1) to obtain results analogous to binary Eqs. (3-3) and (3-4).

If feed, bottoms, and distillate compositions are specified for any component, we can solve for D and B. In addition, these equations show that the ratio of concentration differences for all components must be identical (Doherty and Malone, 2001). These equations can be helpful but do not completely solve the external balances even when x_i,dist and x_i,bot are specified for one component since other mole fractions in distillate and bottoms are unknown.

How do we completely solve external mass balances for a ternary distillation? If component 1 is LK and 2 is HK, the unknowns are B, D, x_2,dist, x_3,dist, x_1,bot, and x_3,bot. There are six unknowns and five independent equations. Can we find an additional equation? Unfortunately, additional equations (energy balances and equilibrium expressions) always add additional variables (see Problem 5.A1), so we cannot start by solving external mass and energy balances. This is a first major difference between binary and multicomponent distillation.

Can we do internal stage-by-stage calculations first and then solve the external balances? To begin the stage-by-stage calculation procedure in a distillation column, we need to know all compositions at one end of the column. For ternary systems with variables specified as in Table 5-1, these compositions are unknown. To begin the analysis we would have to assume one of them. Thus internal calculations for multicomponent distillation problems are automatically trial and error, which is a second major difference between binary and multicomponent problems.

Fortunately, in many cases it is easy to make an excellent first guess that allows you to do external balances. If a sharp separation of keys is required, then almost all HNKs will appear only in the bottoms, and almost all LNKs will appear only in the distillate. The obvious assumption is that all LNKs appear only in the distillate and all HNKs appear only in the bottom. Thus,

The procedure for completing the external mass balances is illustrated in Example 5-1.

Surprisingly, a reasonably accurate solution of external mass balances based on a first guess does not guarantee that internal stage-by-stage calculations will be accurate. The problem given in Example 5-1 would be very difficult for stage-by-stage calculations. Let us explore why.

At the feed stage all components must be present at finite concentrations. If we wish to step off stages from the bottom up, we cannot use x_C3,bot = 0 because we would not get a nonzero concentration of propane at the feed stage. Thus, x_C3,bot must be a small but nonzero value. Unfortunately, we do not know if the correct value should be 10^–5, 10^–6, 10^–7, or 10^–20. Thus, the percentage of error in x_C3,bot will be large, and it will be difficult to obtain convergence of the trial-and-error problem. If we try to step off stages from the top down, x_C3,dist is known accurately, but x_C6,dist is not. Thus, when both HNKs and LNKs are present, stage-by-stage calculation methods are difficult, and other design procedures (see Chapter 6) should be used for accurate results.

If there are only LNKs or only HNKs, then an accurate first guess of compositions can be made. Suppose in Example 5-1 that we specified 99.4% recovery of propane in distillate and 99.7% recovery of n-butane in bottoms. This makes propane LK, n-butane HK, and n-pentane and n-hexane HNKs. Assuming all HNKs appear in bottoms is an excellent first guess. Then we can calculate distillate and bottoms compositions from external mass balances. Compositions calculated in bottoms are quite accurate, and we can step off stages from the bottom up and be confident the results are accurate. If only LNKs are present, stage-by-stage calculation is done from the top down.

If CMO is valid, total vapor and liquid flow rates will be constant in each column section. Total flow rates can change at each feed stage or sidestream withdrawal stage. This behavior is illustrated for a computer simulation for a saturated liquid feed in Figure 5-2 and is the same behavior we expect for a binary system. For nonconstant molal overflow, total flow rates vary within a section. This is also shown in Figure 5-2. Although both liquid and vapor flow rates may vary significantly, the ratio L/V will be much more constant.

The temperature profile decreases monotonically from the reboiler to the condenser. This is illustrated in Figure 5-3 and is again similar to behavior of binary systems. Note that plateaus have started to form where there is little temperature change between stages. When there are a large number of stages, these plateaus can be quite pronounced. They represent pinch points in the column.

The composition profiles in the column are much more complex. To study these, we first look at a computer simulation for distillation of benzene, toluene, and cumene in a column with 20 equilibrium contacts. Total flow and temperature profiles for this simulation are given in Figures 5-2 and 5-3, respectively. With a specified 99.0% recovery of benzene in distillate, the liquid mole fractions are shown in Figure 5-4.

At first Figure 5-4 is a bit confusing, but it will make sense after we go through it step by step. Since benzene recovery in distillate was specified as 99.0%, benzene is LK. Typically, the next less volatile component (LVC), toluene, is HK, which makes cumene an HNK. There is no LNK. Following the benzene curve, we see that the benzene mole fraction is very low in the reboiler and increases monotonically to a high value in the total condenser. This is essentially the same behavior as an MVC in binary distillation (see Figure 4-14). In this simulation benzene is always most volatile, so its behavior is simple.

Since cumene is HNK, we typically assume that all cumene leaves in the bottoms. Figure 5-4 shows that this is essentially true (cumene distillate mole fraction was calculated as 2.45 × 10^–8). Starting at the reboiler, the mole fraction of cumene rapidly decreases and then levels off to a plateau value until the feed stage. Above the feed stage the cumene mole fraction decreases rapidly. This behavior is fairly easy to understand. Cumene’s mole fraction decreases above the reboiler because it is the LVC. Since there is cumene in the feed, there must be a finite concentration at the feed stage. Thus, after the initial decrease there is a plateau until the feed stage. Note that the concentration of cumene on the feed stage is not the same as in the feed. Above the feed stage, cumene concentration decreases rapidly because cumene is the LVC.

The concentration profile for HK toluene is most complex in this example. The behavior of the HK can be explained by noting which binary pairs of components are distilling in each part of the column. In the reboiler and stages 1 and 2 there is very little benzene (LK), and distillation is between HK and HNK. In these stages the toluene (HK) concentration increases as we go up the column because toluene is the more volatile of the two components distilling. In stages 3 to 10 the cumene (HNK) concentration plateaus. Thus, distillation is between LK and HK. Now toluene is the LVC, and its concentration decreases as we go up the column. This causes the primary maximum in HK concentration, which peaks at stage 3. Above the feed stage, in stages 11, 12, and 13, HNK concentration plummets. The major distillation is again between HK and HNK. Since HK is temporarily the more volatile component (MVC), its concentration increases as we go up the column and peaks at stage 12. After stage 12 there is very little HNK present, and the major distillation is between benzene and toluene. Toluene concentration then decreases as we continue up to the condenser. The secondary maximum above the feed stage is often much smaller than shown in Figure 5-4. The large amounts of cumene in this example cause a larger than normal secondary maximum.

In this example HNK (cumene) causes the two maxima in HK (toluene) concentration. Since there was no LNK, LK (benzene) has no maxima. It is informative to repeat this example, except now we specify 99.0% recovery of toluene in distillate. This specification makes toluene LK, cumene HK, and benzene LNK. Results are shown in Figure 5-5. This figure can also be explained qualitatively in terms of distillation of binary pairs (see Problem 5.A12). With no HNKs, HK concentration has no maxima.

What happens for a four-component distillation if there are LKs, HKs, LNKs, and HNKs present? Since there is an LNK, we would expect the LK curve to show maxima; and since there is an HNK, we would expect maxima in the HK concentration profile. This behavior is observed in Figure 5-6 for distillation of a benzene-toluene-xylene-cumene mixture. Although secondary maxima near the feed stage are drastically repressed, the primary maxima are readily evident.

It is interesting to compare the purities of the distillate and bottoms products in Figures 5-4 to 5-6. In Figure 5-4 (no LNK) benzene (LK) is pure in distillate, but bottoms (HK and HNK present) is clearly not pure. In Figure 5-5 (no HNK) cumene (HK) is pure in bottoms, but distillate (LK and LNK present) is not pure. In Figure 5-6 (LNK and HNK present) neither distillate nor bottoms is pure. Simple distillation columns separate feed into two fractions, and a single column can produce either pure MVC as distillate by making it the LK with no LNK, or it can produce pure LVC as bottoms by making it the HK with no HNK. If we want to completely separate a multicomponent mixture, we need to couple several columns (see Lab 6 in the appendix to Chapter 6 and Section 11.5).

There is one additional case to consider. Suppose we have the four components benzene, toluene, xylene, and cumene, and we choose cumene as HK and toluene as LK. This makes benzene LNK, but what is xylene? Xylene is an intermediate or sandwich component, which is an NK component with volatility between the two keys. Sandwich components tend to concentrate in the column’s middle since they are less volatile than LK in the rectifying section and more volatile than HK in the stripping section. Prediction of their final distribution requires a complete simulation.

If top temperature is too cold and bottom temperature is too hot to allow sandwich components to exit at the rate they enter the column, they become trapped in the column’s center and accumulate there (Kister, 2004). This accumulation can be quite large for trace components and can cause column flooding and development of a second liquid phase. The accumulation can be identified from simulation if all trace components that occur in the feed are known, accurate vapor-liquid equilibrium (VLE) correlations are available, and the simulator allows two liquid phases and one vapor phase. Unfortunately, VLE may be very nonideal, and trace components may not accumulate where we think they will. For example, when ethanol and water are distilled, there often are traces of heavier alcohols present. Alcohols with four or more carbons (butanol and heavier) are only partially miscible in water. They are easily stripped from a water phase (relative volatility >> 1), but when there is little water present they are less volatile than ethanol. Thus, they collect somewhere in the middle of the column where they may form a second liquid phase in which the heavy alcohols have low volatility. The usual solution to this problem is to install a side withdrawal line, separate intermediate components from desired components, and return desired components to the column. These heterogeneous systems are discussed in more detail in Chapter 8.

Differences in composition profiles for multicomponent and binary distillation for relatively ideal VLE with no azeotropes can be summarized as follows:

1. In multicomponent distillation key component concentrations can have maxima.

2. NK usually do not distribute. That is, HNKs usually appear only in bottoms and LNKs only in distillate.

3. NK often go through a plateau region of nearly constant composition.

4. All components must be present at the feed stage, but at that stage the primary distillation changes. Thus, discontinuities occur at the feed stage.

Understanding the differences between binary and multicomponent distillation is helpful when you are doing calculations for multicomponent distillation.

Consider a typical design problem for ternary distillation with an LK, an HK, and an HNK. Feed flow rate, composition, and temperature are specified, as are L₀/D, saturated liquid reflux, pressure, use of optimum feed stage, and recoveries of LKs and HKs in distillate and bottoms, respectively. We wish to predict the number of stages required and the separation obtained.

To start we need to assume fractional recovery (FR_HNK,bot) for HNK in bottoms. Then Dx_HNK,dist = Fz_HNK (1 – FR_HNK,bot) and Bx_HNK,bot = Fz_HNK (FR_HNK,bot). If, for example, (FR_HNK,bot) = 1.0, then Bx_HNK,bot = Fz_HNK and x_HNK,dist = 0. Once fractional recovery is assumed, we can find D and B and then L and V in the rectifying section. If CMO is assumed valid,

At the feed stage, q can be estimated from enthalpies as

or q = L_F/F can be found from a flash calculation on the feed stream. Then and are determined from balances at the feed stage:

This completes the preliminary calculations for this assumption of how HNK splits in the column.

If there is an HNK and no LNK, assumed compositions are very accurate at the column bottom but not at the top. Thus, with only an HNK present, we want to step off stages from the bottom up. The general procedure for stepping off stages up the column when CMO is valid with stages numbered from the bottom up (see Figure 5-7) is:

1. Label the partial reboiler as stage j = 0. The x_i,0 = x_i,bot.

2. Use equilibrium (bubble-point calculations) to determine y_i,j values from known x_i,j values.

3. Use mass balances (operating equations) to calculate x_i,j+l values from known y_i,j values.

4. Increment j by 1.

5. Repeat steps 2 to 4 until the feed stage is reached. Change to the rectifying section operating equations and continue.

6. The calculation is finished when x_LK,N+1 ≥ x_LK,dist and x_HK,N+1 ≤ x_HK,dist.

Stage-by-stage calculations for multicomponent distillation require bubble-point or dew-point equilibrium calculations on every stage. Stepping off stages from the bottom up a bubble-point calculation, which determines the temperature at which the liquid mixture begins to boil, is required. Pressure, p_j, and liquid mole fractions x_i,j are known. Find the temperature, T_j, at which ∑y_i,j = 1.0, where y_i,j = K_i,j(T_j)x_i,j. When stepping off stages down the column (HK, LK, and LNKs present), we use dew-point calculations (vapor starts to condense) to determine T_j for which ∑x_i,j = 1.0 with x_i,j = y_i,j/K_i,j.

Consider the distillation column shown in Figure 5-7 where all NKs are HNKs. Note that the column is numbered from the bottom up, since that is direction in which we will step off stages. With no LNK, a good first guess of concentrations can be made at the column bottom, and we can start stage-by-stage calculations by calculating the reboiler temperature and the values of y_i,R = y_i,0 from a bubble-point calculation by finding the temperature that satisfies the stoichiometric equation.

If a simple expression for K_i as a function of temperature is available, we may be able to solve Eq. (5-12) explicitly for temperature. Otherwise, a root-finding technique or a trial-and-error procedure is required. When graphs or charts are used, a trial-and-error procedure is needed. When we are finished with the calculation, the calculated y_i,j are mole fractions of vapor leaving stage j.

How can we make a good initial guess for the temperature T_j? Note that and we want . If all K_i,j > 1.0, then If all K_i,j < 1.0, then . Therefore, choose a temperature T_j so that some K_i,j are greater than 1.0 and some are less than 1.0. This procedure is similar to the procedure we used for a first guess of drum temperature in flash distillation.

How do we pick a new temperature when the previous trial was not accurate? If we look at DePriester charts or Eq. (2-28), we see that K_i are complex functions of temperature. However, functional dependence is quite similar for all K’s. Thus, change in K for a reference component and change in the value of will be quite similar when the temperature changes. By setting the ratio of new and old values of the reference component equal to the ratio of the desired new value of summation (1.0) and the calculated value of summation, we can estimate the appropriate K value for the reference component:

Temperature for the next trial is determined from the new value, K_ref,j(T_new,j). This procedure, which is again similar to the flash distillation procedure, should converge quite rapidly. As an alternative, a Newtonian convergence scheme can be used. If a spreadsheet is used, Goal Seek will find the temperature that makes for any reasonable first guess of temperature.

Once the temperature and y_i,j values are known, we can use operating equations to solve for x_i,j+1. Mass balances for Figure 5-7 in the stripping section with CMO are

and

Since we are stepping off stages up the column, solve for x_i,k+1. The operating equations are

At the feed stage switch to the enriching section with the operating equation for each component:

Continue to alternate between equilibrium (bubble point) and operating equations until

at which point the calculation is finished for the feed plate and HNK split employed. This leaves us with two unanswered questions: How do we determine the optimum feed plate, and how do we correct our initial assumption of HNK split?

The optimum feed plate is defined as a feed plate that gives the fewest total number of stages. To be absolutely sure you have the optimum feed plate location, use this definition. That is, pick a feed plate location and calculate N. Then repeat until you find the minimum total number of stages. Note that often several stages must be stepped off before feed can be input. The first legal feed stage may be optimum. This procedure sounds laborious, but, as we will see, it is not too difficult to implement on a spreadsheet (see Appendix A of this chapter). If you try to switch stages too early, the stage-by-stage calculation will eventually give negative mole fractions. With a spreadsheet, you can guard against switching stages too soon by checking that all mole fractions (x_i,j and y_i,j) are between 0 and 1 for every stage.

How do we check and correct our initial guess for splits of NK components? One approach is to use the calculated value of the HNK mole fraction to estimate the fractional recovery of HNK:

If ε is the acceptable error, then if

a new trial is required. For the next trial we can use a damped direct substitution and set

where df is a damping factor ≤ 1. If df = 1, Eq. (5-23) becomes direct substitution, which may result in oscillations.

This discussion was for calculations up the column. If only LNKs are present, then calculation should proceed down the column. Renumber the stages so that stage 1 is at the top. Now liquid mole fractions are determined from dew-point calculations, and vapor mole fractions are found from operating equations. For dew-point calculations, all y_i,j values are known. Find temperature T_j at which

Determine the temperature for the next guess by calculating the reference component K value:

Then solve for the temperature from the calculated value of K_ref [e.g., with Eq. (5-15)]. The dew-point equation for constant relative volatility is shown in Problem 5.C1b.

Operating equations are essentially the same as for binary systems. These are

in the enriching section and

in the stripping section.

Remember convergence of stage-by-stage distillation calculations is easy only when NKs are all heavy or are all light. Stage-by-stage calculations are conveniently done with a spreadsheet (see this chapter’s appendix for two examples).

For problems where both LKs and HNKs are present, Lewis and Matheson in 1932 and Thiele and Geddes in 1933 calculated from both column ends and matched compositions at the feed stage (see Smith, 1963). Unfortunately, closure can be very difficult. When there are both LNKs and HNKs, other calculation methods such as the matrix method discussed in Chapter 6 are preferable. Perhaps surprisingly, separations with an HNK or LNK and a sandwich component converge reasonably well (see Problem 5.A8).

Elliott, J. R., and C. T. Lira, Introductory Chemical Engineering Thermodynamics, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 2012.

Kister, H. Z., “Component Trapping in Distillation Towers: Causes, Symptoms and Cures,” Chem. Engr. Progress, 100, (8), 22 (August 2004).

Smith, B. D., Design of Equilibrium Stage Processes, McGraw-Hill, New York, 1963.

Speight, J. (Ed.), Lange’s Handbook of Chemistry, 70th Anniversary Edition, 16th ed., McGraw-Hill, New York, 2005.

A1. Explain why the external mass balances cannot be solved for a ternary distillation system without an additional assumption. Why are the equations for the following not useful?

a. External energy balance

b. Energy balance around the condenser

c. Equilibrium expression in the reboiler

A2. Show for the simulation illustrated in Figure 5-2 that L/V is more constant than either L or V when CMO is not valid. Explain why this is so.

A3. Define the following:

a. Heavy key

b. Heavy non-key

c. Sandwich component (see Problem 5.A6)

d. Optimum feed stage

e. Minimum reflux ratio

A4. We are distilling a mixture with two HNKs, an HK, and an LK. Sketch the expected concentration profiles (x_i vs. stage location).

A5. How would you introduce Murphree vapor efficiencies into stage-by-stage calculations?

A6. A distillation column is separating methane, ethane, propane, and butane. We pick methane and propane as the keys, which means that ethane is a sandwich component.

a. Show the approximate composition profiles for each of the four components. Label each curve.

b. Explain in detail the reasoning used to obtain the profile for ethane.

A7. We are distilling a mixture that is 10.0 mol% methanol, 20.0 mol% ethanol, 30.0 mol% n-propanol, and 40.0 mol% n-butanol. Methanol is most volatile and n-butanol is least volatile. The feed is a saturated liquid. We desire to recover 98.0% of the ethanol in the distillate and 97.0% of the n-propanol in the bottoms. The column has a total condenser and a partial reboiler. The feed rate is 100.0 kmol/h. Pressure is 1.0 atm. L/D = 3.0.

1. The column is hottest at:

a. The condenser.

b. The feed plate.

c. The reboiler.

2. In the rectifying (enriching) section:

a. Liquid flow rate > vapor flow rate.

b. Liquid flow rate = vapor flow rate.

c. Liquid flow rate < vapor flow rate.

3. Comparing the stripping section to the rectifying section:

a. Liquid flow rate in stripping section > liquid flow rate in rectifying section.

b. Liquid flow rate in stripping section = liquid flow rate in rectifying section.

c. Liquid flow rate in stripping section < liquid flow rate in rectifying section.

4. The HK is:

a. Methanol.

b. Ethanol.

c. n-propanol.

d. n-butanol.

5. If you were going to do external mass balances around the column to find B and D, the best assumption to make is:

a. All of the methanol and n-propanol are in distillate.

b. All of the methanol is in distillate, and all of the n-butanol is in bottoms.

c. All of the ethanol is in distillate and all of the n-propanol is in bottoms.

d. All of the n-propanol and n-butanol are in bottoms.

A8. The chapter closes with the sentence, “Perhaps surprisingly, separations with a HNK or LNK and a sandwich component converge reasonably well.” Explain why.

A9. It is often suggested that when the corresponding NK component is present, key components should be withdrawn as sidestreams at the location where their concentration maximum occurs. If there is an LNK, can a pure LK be withdrawn as a sidestream? Why or why not? Can a pure LNK be obtained at the top of the column? Why or why not?

A10. Develop a key relations chart for this chapter. You will probably want to include sketches.

A11. In Figure 5-4, a 99.0% recovery of benzene does not give a high benzene purity. Why not? What would you change to also achieve a high benzene purity in distillate?

A12. Explain Figure 5-5 in terms of the distillation of binary pairs.

A13. In Figure 5-4 HNK and HK concentrations cross near the column bottom, and in Figure 5-5 LK and LNK concentrations do not cross near the top of the column. Explain when concentrations of HK and HNK and LK and LNK pairs do and do not cross.

A14. Figure 5-6 shows distillation of a four-component mixture. What would you expect the profiles to look like if xylene were LK and cumene HK?

A15. 1 to 5. Determine whether the following multicomponent distillation problems can or cannot be solved with a stage-by-stage calculation, and if a stage-by-stage calculation can be used, indicate the direction in which you should step off stages.

1. You have two LNKs, an LK, and an HK.

a. Step off stages from the top down.

b. Step off stages from the bottom up.

c. Step off stages in either direction.

d. Stage-by-stage calculation probably will not converge.

2. You have an LNK, an LK, an HK, and an HNK.

a. Step off stages from the top down.

b. Step off stages from the bottom up.

c. Step off stages in either direction.

d. Stage-by-stage calculation probably will not converge.

3. You have an LK, an HK, and two HNK.

a. Step off stages from the top down.

b. Step off stages from the bottom up.

c. Step off stages in either direction.

d. Stage-by-stage calculation probably will not converge.

4. You have an LK, a sandwich component, and an HK.

a. Step off stages from the top down.

b. Step off stages from the bottom up.

c. Step off stages in either direction.

d. Stage-by-stage calculation probably will not converge.

5. You have an LK, a sandwich component, an HK, and an HNK.

a. Step off stages from the top down.

b. Step off stages from the bottom up.

c. Step off stages in either direction.

d. Stage-by-stage calculation probably will not converge.

C. Derivations

C1. If relative volatilities are constant, bubble- and dew-point calculations are simplified.

a. Show that mole fractions for a bubble-point calculation are given by

and temperature can be found from the estimated K value of the reference component:

b. For dew-point calculations show that

and temperature can be determined from the estimated K value of the reference component:

D. Problems

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

D1. The feed to a rectifying column is 32.0 mol% n-butane, 56.0 mol% n-pentane, and 12.0 mol% n-hexane. Feed rate is 100.0 kmol/h and is a saturated vapor at 3.0 bar. The column operates at 3.0 bar, has a total condenser with a saturated liquid reflux, and CMO can be assumed valid. The distillate product is 99.6 mol% n-butane and D = 20.0 kmol/h.

a. Find bottoms flow rate, reflux ratio L/D, and mole fractions of distillate and bottoms.

b. List your assumptions, if any.

D2. The feed to a distillation column with a total condenser and a partial reboiler is 19.0 wt% methanol, 31.0 wt% ethanol, 27.0 wt% n-propanol, and 23.0 wt% n-butanol. (Methanol is most volatile, and n-butanol is least volatile.). Feed rate = 12,000 kg/h. We desire a 97.8% recovery of ethanol in distillate and a 99.4% recovery of n-propanol in bottoms. Estimate D, B, and the weight fractions of all the components in distillate and bottoms.

D3.* We have a feed mixture of 22.0 mol% methanol, 47.0 mol% ethanol, 18.0 mol% n-propanol, and 13.0 mol% n-butanol. The feed is a saturated liquid, and F = 10,000 kmol/day. Recover 99.8% of methanol in distillate, and obtain a methanol mole fraction in distillate of 0.99.

a. Find D and B.

b. Find the compositions of distillate and bottoms.

D4.* 1000.0 kmol/h of a 40.0 mol% isopentane, 30.0 mol% n-hexane, and 30.0 mol% n-heptane is being distilled. We desire a 98.0% recovery of n-hexane in bottoms and a 99.0% recovery of isopentane in distillate. The feed is a two-phase mixture that is 40.0% vapor. L/D = 2.5.

a. Find D and B. List any required assumptions.

b. Find the compositions of distillate and bottoms.

c. Calculate and and and L and V assuming CMO.

d. Show schematically the expected composition profiles for isopentane, n-hexane, and n-heptane. Label the curves. Be neat!

D5. A distillation column with a total condenser and a partial reboiler is separating two feeds. The first feed is a saturated liquid, and its rate is 100.0 kmol/h. This feed is 55.0 mol% methanol, 21.0 mol% ethanol, 23.0 mol% propanol, and 1.0 mol% butanol. The second feed is a saturated liquid with a flow rate of 150.0 kmol/h. This feed is 1.0 mol% methanol, 3.0 mol% ethanol, 26.0 mol% propanol, and 70.0 mol% butanol. We want to recover 99.3% of the propanol in distillate and 99.5% of the butanol in bottoms. Make appropriate assumptions, and find distillate and bottoms flow rates and mole fractions of distillate and bottoms.

D6. We are separating a mixture of light hydrocarbons in a stripping column. The saturated liquid feed is 35.0 mol% n-butane, 45.0 mol% n-pentane, and 20.0 mol% n-hexane. The feed rate is 100.0 kmol/h, and it is at 7.0 bar. The column is also at 7.0 bar, has a partial reboiler, and CMO can be assumed valid. The bottoms product is 98.2 mol% n-hexane, and the boilup ratio is . Find bottoms flow rate B, distillate flow rate D, and mole fractions of distillate and bottoms.

D7. We are separating hydrocarbons in a two-feed column with a total condenser and a partial reboiler. Operation is at 75.0 psig. 1000.0 kg/h of saturated liquid feed 1 and 1500.0 kg/h of saturated liquid feed 2 are fed to the column. Feed 1 is 30.0 wt% ethane, 0.6 wt% propylene, 45.0 wt% propane, 15.4 wt% n-butane, and 9.0 wt% n-pentane. Feed 2 is 2.0 wt% ethane, 0.1 wt% propylene, 24.9 wt% propane, 40.0 wt% n-butane, 18.0 wt% n-pentane, and 15.0 wt% n-hexane. We desire 99.1% recovery of propane in distillate and 98.0% recovery of n-butane in bottoms. Make appropriate assumptions, and calculate distillate and bottoms flow rates and weight fractions of each component in distillate and bottoms.

D8. A distillation column with a partial reboiler and a total condenser is being used to separate 1000.0 kmol/h of a saturated liquid feed that is 40.0 mol% benzene, 30.0 mol% toluene, and 30.0 mol% cumene. Reflux is a saturated liquid, and L/D = 2.0. Recover 95.0% of cumene in bottoms and 95.0% of toluene in distillate. Pressure is at 1 atm. Assume constant relative volatilities, α_BT = 2.25, α_TT = 1.0, and α_CT = 0.21. Find the optimum feed stage and the total number of equilibrium contacts required. Use Eq. (5-28) or (5-30) (in Problem 5.C1) for equilibrium.

D9. What is the dew point of a vapor that is 30.0 mol% n-butane, 50.0 mol% n-pentane, and 20.0 mol% n-hexane at p = 760.0 mm Hg? Use Raoult’s law to predict K values. Find vapor pressures from Antoine’s equation:

where VP is in mm Hg, and T is in °C. The Antoine constants (Speight, 2005) are

D10. What is the bubble-point temperature at 760.0 mm Hg pressure of a 40.0 mol% n-pentane and 60.0 mol% n-hexane mixture? Use Raoult’s law and Antoine coefficients from Problem 5.D9.

D11.

a. Find the dew-point temperature for a vapor mixture that is 40.0 mol% methane, 5.0 mol% ethylene, 35.0 mol% ethane, and 20.0 mol% n-hexane at a pressure of 2500.0 kPa.

b. What are the mole fractions of the first drop of liquid condensed?

Use DePriester charts.

D12. Suppose n-pentane, n-heptane, and n-octane are available so that the mole fractions of a mixture can be changed to any desired value. If system pressure is 500.0 kPa,

a. What is the highest possible bubble-point temperature?

b. What is the lowest possible bubble-point temperature?

D13.* Find the bubble-point temperature and vapor mole fractions for a mixture at 1.0 atm that is 20.0 mol% n-butane, 50.0 mol% n-pentane, and 30.0 mol% n-hexane. Use DePriester charts.

E. More Complex Problems

E1. A column with a partial reboiler and a partial condenser operates at 400.0 kPa. The saturated liquid feed flow rate is 200.0 kmol/h and is 22.0 mol% ethane, 47.0 mol% propane, and 31.0 mol% n-butane. Recover 97.0% of the ethane in distillate and 99.0% of the propane in bottoms. The reflux is a saturated liquid, and the external reflux ratio is L₀/D = 3.0. Find the optimum feed stage and the total number of equilibrium contacts required. Assume CMO, and use DePriester charts or Eq. (2-28).

E2. A column with a partial reboiler and a total condenser is distilling hydrocarbons at 7.0 bar. The saturated liquid feed is 25.0 mol% ethane, 35.0 mol% n-butane, and 40.0 mol% n-pentane, and the flow rate is 100.0 kmol/h. Recover 99.2% of the n-butane in distillate and 98.3% of the n-pentane in bottoms. The reflux is a saturated liquid, and L₀/D = 3.0. Find the optimum feed stage and the total number of equilibrium contacts required. Assume CMO, and use DePriester charts or Eq. (2-28). Use of a spreadsheet to do dew-point calculations is highly recommended.

E3. Determine the number of equilibrium stages needed for Problem 5.D6 by stepping off stages and doing a dew- or bubble-point calculation at each stage. Use either DePriester charts or Table 2-3. Report the temperature of each stage and the vapor and liquid mole fractions leaving each stage. Use of Table 2-3 constants in a spreadsheet to do dew- or bubble-point calculations will probably save you time.

H. Computer Spreadsheet Problems

H1. Do part b of this problem with a spreadsheet or with MATLAB. Although use of VBA is recommended if a spreadsheet is used, it is not required (the procedure shown in Appendix A of this chapter can be used). We have 200.0 kmol/h of a saturated liquid feed that is 35.0 mol% methanol, 40.0 mol% i-propanol, and 25.0 mol% n-propanol at 1.0 atm. We want 96.1% recovery of the methanol in distillate and 99.6% recovery of the i-propanol in bottoms. CMO is valid. The column has a total condenser and a partial reboiler. L/D = 6.0. As a first guess, assume that n-propanol does not distribute (it all exits in bottoms). The constant relative volatilities are: methanol = 3.58, i-propanol = 1.86, and n-propanol = 1.0. See Eqs. (5-28) to (5-31) in Problem 5.C1 for methods for equilibrium calculations with constant relative volatilities.

a. Find D, B, x_i,dist, and x_i,bot.

b. Step off stages, and find the optimum feed location and the total number of stages.

H2. A stripping column with a partial reboiler is processing 100.0 kmol/h of a saturated liquid feed at 300.0 kPa. The feed is 25.0 mol% n-butane, 35.0 mol% n-pentane, and 40.0 mol % n-hexane. The column is at 300.0 kPa, and CMO is valid. Bottoms flow rate is 20.0 kmol/h. The fractional recovery of n-pentane in distillate is 95.0 mol%. Assume all n-butane is in distillate, and calculate the distillate flow rate and distillate and bottoms compositions. Determine the temperature, vapor mole fractions, and liquid mole fractions leaving each stage. Manually increment the stage location (see this chapter’s appendix). Determine the total number of stages required. Use DePriester charts or Eq. (2-28) for equilibrium data. Do NOT converge on a better guess for n-butane distribution. Note: If preliminary calculations are changed, the spreadsheet in Figures 5-A1 and 5-A2 can be used.

H3. We are separating a mixture of light hydrocarbons in a stripping column. The saturated liquid feed at 4.5 bar is 27.0 mol% n-butane, 32.0 mol% n-pentane, and 41.0 mol% n-hexane. Feed rate is 150.0 kmol/h. The column is at 4.5 bar, has a partial reboiler, and CMO is valid. Bottoms product is 97.7 mol% n-hexane, and the boilup ratio is . Find bottoms flow rate B, distillate flow rate D, mole fractions of distillate and bottoms, and the number of equilibrium stages needed. Report vapor and liquid mole fractions leaving each stage and temperature of each stage. This problem is a modification of Example 5-3. The spreadsheets in Figures 5-A1 or 5-B1 can be used by changing the input conditions.

H4. Develop a spreadsheet VBA program for a rectifying column. The feed is 32.0 mol% n-butane, 56.0 mol% n-pentane, and 12.0 mol% n-hexane. The feed rate is 100.0 kmol/h of a saturated vapor at 7.0 bar. The column is at 7.0 bar, has a total condenser with a saturated liquid reflux, and CMO is valid. Distillate product is 99.6 mol% n-butane, and D = 15.0 kmol/h.

a. Find bottoms flow rate B, reflux ratio L/D, and distillate and bottoms mole fractions.

b. Find the number of stages and the temperature and mole fraction profiles in the column.

H5.

a. Find the number of stages and the temperature and mole fraction profiles in the column for Problem 5.D1. The spreadsheet VBA program for Problem 5.H4 also works for Problem 5.D1.

b. After completing part a, increase pressure to 7.0 bar, and try running the VBA program again. Why does the VBA program fail?

H6. [VBA required] Developing a spreadsheet for a complete column is not substantially more work once either a stripping or rectifying column spreadsheet has been developed.

a. Develop a program for a system with an LK, HK, and HNK (step off stages from the bottom up), doing a bubble-point calculation on every stage.

b. Find the optimum feed stage, the total number of stages, distillate and bottoms flow rates and compositions, and composition and temperature profiles for the following problem. Separate 100.0 kmol/h of a saturated liquid feed that is 30.0 mol% n-butane, 30.0 mol% n-pentane, and 40.0 mol% n-hexane. Column pressure is 14.7 psia. The fractional recovery of n-butane in distillate is 0.995, and the fractional recovery of n-pentane in bottoms is 0.997. L/D = 8.0. Use Eq. (2-28) and Table 2-3 to calculate K values.

Goal Seek is very convenient for calculating dew- and bubble-point temperatures when you are interacting directly with a spreadsheet for every calculation; however, for stepping off stages with a VBA program, automatic equilibrium calculations are needed. The method illustrated in Example 5-2 can be adapted. If the temperature does not give ∑y_i = 1.0 for a bubble-point calculation or ∑x_i = 1.0 for a dew-point calculation, an estimate of the reference K value can be determined from Eq. (5-13) or (5-14), respectively. The temperature that corresponds to the new value of K_ref will be a better estimate. This temperature is most conveniently found if you can explicitly solve for the temperature as in Eq. (5-15). Either continue until error in the summation is small enough or repeat the estimation a specified number of times (e.g., 10).

The spreadsheet in Figure 5-B1 looks quite similar to Figure 5-A1, and the results are identical. The following calculations are identical: preliminary, K values, x_i, and yop_i. The simplified spreadsheet in Figure 5-A1 uses Goal Seek to converge the dew-point calculation and the manual transfer of values for the next stage. The automated spreadsheet uses Eq. (5-14) for the dew-point calculation and Eq. (5-15) to determine a new temperature followed by an arbitrary number of iterations controlled by VBA (Table 5-B1) to converge the temperature. VBA is also used to transfer values for the next stage. The trade-off is between an easier spreadsheet to write (Figure 5-A1) versus a more difficult to write but easier to use spreadsheet (Figure 5-B1). Depending on the task, either one could be the appropriate choice.