Chapter 11: Economics and Energy Conservation in Distillation

There are an estimated 40,000 distillation columns in the USA, which have a combined capital value in excess of $8 billion. These columns are estimated to have at least a 30-year life, and they are used in more than 95% of all chemical processes. The total energy use of these columns is approximately 3% of total U.S. energy consumption (Humphrey and Keller, 1987). Thus, estimating and, if possible, reducing capital and operating costs of distillation are important. Because they are a major energy user, saving energy in distillation systems is particularly important in a time of high and uncertain energy costs.

11.1 DISTILLATION COSTS

Now that we have considered the design of the entire column we can explore the effect of design and operating parameters on the cost of operation. A brief review of economics will be helpful (for complete coverage, see a design or economics text such as Peters et al., 2003; Turton et al., 2003; Ulrich, 1984; or Woods, 1976).

Capital costs can be determined by estimating delivered equipment costs and adding on installation, building, piping, engineering, contingency, and indirect costs. These latter costs are often estimated as a factor times the delivered equipment cost for major items of equipment.

(11-1)

where the Lang factor ranges from approximately 3.1 to 4.8. The Lang factor for a plant processing only fluids is 4.74 (Turton et al., 2003). Thus, these “extra” costs greatly increase the capital cost.

Costs of major equipment are often estimated from a power law formula:

(11-2)

Some equipment will not follow this power law. The appropriate size term depends on the type of equipment. For example, for shell and tube heat exchangers such as condensers, the size used is the area of heat exchanger surface. The exponent has an “average” value of 0.6, but varies widely. For shell and tube heat exchangers the exponent has been reported as 0.41 (Seider et al., 2004), 044 (Turton et al., 2003), and 0.48 (Rudd and Watson, 1968). Thus,

(11-3)

As the area becomes larger, the cost per square meter decreases.

The exponent in Eq. (11-02) is usually less than 1. This means that as size increases, the cost per unit size decreases. This will be translated into a lower cost per kilogram of product. This “economy of scale” is the major reason that large plants have been built in the past. However, there is currently a trend toward smaller, more flexible plants that can change when the economy changes.

The current cost can be estimated by updating published sources or from current vendors’ quotes. The method for updating costs is to use a cost index.

(11-4)

The Marshall and Stevens equipment cost index or the Chemical Engineering magazine plant cost index are usually used. Current values are given in each issue of Chemical Engineering magazine. The total uninstalled equipment cost will be the sum of condenser, reboiler, tower casing, and tray costs. The total capital cost is then found from Eq. (11-01). The capital cost per year equals the capital cost times the depreciation rate. Note that cost estimates can easily vary by up to 35%.

The final bare module cost C_BM of distillation equipment can be estimated from charts and equations. The calculation starts with the base purchase cost C_p for systems built of carbon steel and operating at ambient pressure, and then adjusts this cost with factors for additional costs. The charts that we need to estimate the base purchase cost C_p of distillation systems are shown in Figures 11-1 to 11-3 (Turton et al., 2003). Since the shell of distillation columns is a vertical vessel, Figure 11-1 can be used to determine C_p = C_p° × (Volume) for towers. The base purchase costs for sieve and valve trays and tower packing are shown in Figure 11-2, C_p = C_p° × (Area) for trays and C_p = C_p° × (Volume) for packing. In the heat exchanger costs (Figure 11-3), distillation systems commonly use kettle type reboilers and fixed tube sheet shell and tube exchangers, C_p = C_p° × (Area). The Chemical Engineering Plant Cost Index (CEPCI) listed on these figures is the CEPCI at the time the figures were prepared.

Figure 11-1 Purchased costs of process vessels (Turton et al., 2003), reprinted with permission, copyright 2003 Prentice Hall

Figure 11-2 Purchased costs of packing, trays, and demisters (Turton et al., 2003), reprinted with permission, copyright 2003 Prentice Hall

Figure 11-3 Purchased costs for heat exchangers (Turton et al., 2003), reprinted with permission, copyright 2003 Prentice Hall

All types of equipment are affected by increased costs due to expensive materials, and the material factors F_m for these increased costs in distillation systems are given in Table 11-1 (Turton et al., 2003). Pressure effects on the costs of process vessels and heat exchangers are included through the pressure factor F_p. The pressure factor for process vessels is given by (Turton et al., 2003),

(11-5)

Table 11-1. Material factors F_m for equipment in distillation systems (Turton et al., 2003)

which is valid for a vessel wall thickness >0.0063 m. In this equation p = absolute pressure in bars and D = vessel diameter in meters. If F_p < 1 set F_p = 1.0. For vacuum operation with p < 0.5 bar, use F_p = 1.25. Further limitations on the use of this equation are discussed by Turton et al. (2003, p. 925). The pressure factor for condensers and kettle type reboilers can be determined from (Turton et al., 2003),

(11-6)

In this equation p = absolute pressure in bars and the range of validity of the equation is from 6 < p < 141 bar. If p φ 6, set F_p = 1.0.

Sieve and valve tray and packing costs do not depend directly on pressure although they depend indirectly on pressure since column diameter depends on pressure. Sieve tray costs depend on the number of trays ordered through the quantity factor F_q, which can be calculated from the following equation (Turton et al., 2003).

(11-7)

where N is the number of trays. This equation is valid for N < 20. For N φ 20 set F_q = 1.0. Packing costs also depend on the amount of packing, which is inherently included in Figure 11-2.

The final bare module cost C_BM for vertical process vessels is given by (Turton et al., 2003),

(11-8)

For condensers and kettle type reboilers the final bare module cost is,

(11-9)

For sieve and valve trays the final bare module cost is determined from,

(11-10)

where N = number of trays. For packings the final bare module cost is

(11-11)

The use of these graphs and equations is illustrated in Example 11-1.

The total operating costs per year can be determined as

(11-12)

For most continuous distillation columns, the electricity costs are modest, and the labor costs are the same regardless of the values of operating variables.

11.2 OPERATING EFFECTS ON COSTS

We can use the methods developed in Chapters 4, 6, and 7 to calculate the number of equilibrium stages, N_equil, required. Then N_actual = N_equil/E_o. The height of a staged column is

(11-13)

The column diameter is found using the methods in Chapter 10. Equation (10-16) shows that for higher pressures the diameter will be somewhat reduced. Conversely, for vacuum operation the diameter will be increased. Increases in tray spacing increase K, which also increases u_op, and thus, the diameter will decrease while the column height increases.

As L/D → ∞ (total reflux), the number of stages approaches a minimum that minimizes the column height, but the diameter goes to infinity. As L/D → (L/D)_min, the number of stages and the height become infinite while the diameter becomes a minimum. Both these limits will have infinite capital costs. Thus, we expect an optimum L/D to minimize capital costs. Column height is independent of feed flow rate, while diameter is proportional to F^1/2 and (L/D)^1/2.

Pressure effects on distillation columns are extremely important. Equation (10-16) shows that the diameter of staged columns is proportional to (1/p)^1/2. Operating at higher pressures reduces the column diameter although the height may increase since relative volatility typically decreases as pressure increases. Usually the effect of pressure on diameter is significantly larger than its effect on column height. Based on Figure 11-1 we would expect the base purchased cost of the column to decrease. Typically the pressure factor F_p = 1.0 below approximately 5 bar and rises modestly for pressures below about 20 bar. If the column is made of an expensive material (high F_m), cost factors for pressure increases are accelerated since the bare module factor depends on the product of F_p and F_m, Eq. (11-08). Since sieve tray purchase cost C_p decreases as diameter decreases (Figure 11-2), we would expect the purchased cost per tray to drop although a few more trays may be required. Keller (1987) reported on detailed economic analyses and found that the net result is that bare module costs of distillation columns (shell and trays) generally decrease up to a pressure of approximately 6.8 atm (100 psia). This result assumes that there is no thermal degradation at these pressures. Compounds that degrade must often be processed under vacuum (at lower pressures the boiling points and hence column temperatures are lower) even though the columns are more expensive. The effect of pressure on purchase costs above 6.8 atm need to be determined on a case-by-case basis.

Condenser and reboiler sizes depend on Q_c and Q_R. These values can be determined from external mass and energy balances around the column. Equations (03-14) and (03-16) allow us to calculate Q_c and Q_R for columns with a single feed. Note that Q_c and hence Q_R both increase linearly with L/D and with F. The amount of cooling water required is easily determined from an energy balance on the cooling water.

(11-14)

where θT_w = T_w,hot − T_w,cold. A water condenser can easily cool to 100°F (O’Brien and Schultz, 2004). The cooling water cost per year is

(11-15)

This will be a linear function of L/D and of F if cost per kilogram is constant.

In the reboiler the steam is usually condensed from a saturated or superheated vapor to a saturated liquid. Then the steam rate is

(11-16a)

In many applications, H_steam = H_{saturated vapor}, and

(11-16b)

where λ is the latent heat of vaporization of water at the operating pressure. The steam rate will increase linearly with L/D and with F. The value of λ can be determined from the steam tables. Then the steam cost per year is

(11-17)

Note that Q_c and hence Q_R depend linearly on L/D and F; thus, increases in L/D or F linearly increase cooling water and steam rates.

Although detailed design of condensers and particularly reboilers is specialized (Ludwig, 2001; McCarthy and Smith, 1995) and is beyond the scope of this book, an estimate of the heat transfer area is sufficient for preliminary cost estimates. The sizes of the heat exchangers can be estimated from the heat transfer equation

(11-18)

where U = overall heat transfer coefficient, A = heat transfer areas, and θT is the temperature difference between the fluid being heated and the fluid being cooled. Use of Eq. (11-18) is explained in detail in books on transport phenomena and heat transfer (e.g., Chengel, 2003; Geankoplis, 2003; Griskey, 2002; Ludwig, 2001; Greenkorn and Kessler, 1972; and Kern, 1950). For condensers and reboilers, the condensing fluid is at constant temperature. Then

(11-19)

where T_hot = condensing temperature of fluid or of steam, and T_cold,avg = (1/2)(T_cold,1 + T_cold,2). For a reboiler, the cold temperature will be constant at the boiling temperature, and θT = T_steam − T_bp. The values of the heat transfer coefficient U depend upon the fluids being heated and cooled and the condition of the heat exchangers. Tabulated values and methods of calculating U are given in the references. Approximate ranges are given in Table 11-2.

Table 11-2. Approximate heat transfer coefficients

Source: Greenkorn and Kessler (1972)

If we use average values for U and for the water temperature in the condenser, we can estimate the condenser area. With the steam pressure known, the steam temperature can be found from the steam tables; then, with an average U, the area of the reboiler can be found. Condenser and reboiler costs can then be determined. Since area is directly proportional to Q, which depends linearly on L/D and F, the heat exchanger areas increase linearly with L/D or F.

If the column pressure is raised, the condensation temperature in the condenser will be higher. This is desirable, since θT in Eqs. (11-18) and (11-19) will be larger and required condenser area will be less. In addition, higher pressures will often allow the designer to cool with water instead of using refrigeration. This can result in a large decrease in cooling costs because refrigeration is expensive. With increased column pressure, the boiling point in the reboiler will be raised. Since this is the cold temperature in Eq. (11-18), the value of θT in Eqs. (11-18) and (11-19) is reduced and the reboiler area will be increased. An alternative solution is to use a higher pressure steam so that the steam temperature is increased and a larger reboiler won’t be required. This approach does increase operating costs, though, since higher pressure steam is more expensive.

The total operating cost per year is given in Eq. (11-12). This value can be estimated as steam costs [(Eq. (11-17)] plus cooling water costs [Eq. (11-15)]. A look at these equations shows the effects of various variables; these effects are summarized in Table 11-3.

Table 11-3. Effect of changes in operating variables on operating costs

The capital cost per year is

(11-20)

The individual equipment costs depend on the condenser area, reboiler area, tower size, and tray diameter. Some of the variable effects on the capital costs are complex. These are outlined in Table 11-4. The net result of increasing L/D is shown in Figure 11-4; capital cost goes through a minimum.

Table 11-4. Effect of changes in design variables on capital costs

Figure 11-4 Effect of reflux ratio on costs

The total cost per year is the sum of capital and operating costs and is also illustrated in Figure 11-4. Note that there is an optimum reflux ratio. As operating costs increase (increased energy costs), the optimum will shift closer to the minimum reflux ratio. As capital cost increases due to special materials or very high pressures, the total cost optimum will shift toward the capital cost optimum. The optimum L/D is usually in the range from 1.05 (L/D)_min to 1.25 (L/D)_min.

The column pressure also has complex effects on the costs. If two pressures both above 1 atm are compared and cooling water can be used for both pressures, then total costs can be either higher or lower for the higher pressure. The effect depends on whether tower costs or tray costs dominate. If refrigeration would be required for condenser cooling at the lower pressure and cooling water can be used at the higher pressure, then the operating costs and the total costs will be less at the higher pressure. See Tables 11-3 and 11-4 for more details.

The effects of other variables are somewhat simpler than the effect of L/D or pressure. For example, when the design feed rate increases, all costs go up; however, the capital cost per kilogram of feed drops significantly. Thus, total costs per kilogram can be significantly cheaper in large plants than in small plants. The effects of other variables are also summarized in Tables 14-3 and 14-4.

EXAMPLE 11.1 Cost estimate for distillation

Estimate the cost of the distillation column (shell and trays) designed in Examples 10-1 to 10-3.

Solution

The number of equilibrium stages can be calculated from a McCabe-Thiele diagram or estimated by the Fenske-Underwood-Gilliland approach. We will use the latter approach. In Example 10-1, α = 2.35 was used. Then from the Fenske Eq. (07-16),

From Example 10-1, y = 0.7 when x = 0.5, which is the feed concentration. Then for a saturated liquid feed,

Since L/V = 0.8,

Using Eq. (7-42b) for the Gilliland correlation, we have

and

Subtract 1 for a partial reboiler. From Example 10-1, the overall efficiency E_o = 0.59. Then

The column diameter was 11 feet in Example 10-2 and 12 feet in Problem 10-D2. Use the larger value.

With 24 in = 2 foot tray spacing, we need (36)(2) = 72 feet. In addition, add approximately 4 feet for vapor disengagement and a liquid pool at bottom. Convert to meters since Figures 11-1 and 11-2 are in metric units:

Height = (76 ft)(1 m/ 3.2808 ft) = 23.165 m, Diameter = (12 ft) (1 m/3.2808 ft) = 3.66 m, Tray area = πD²/4 = 10.5 m², and Volume of tower = (πD²/4)(L) = 243.7 m³.

From Figure 11-1, C_p° = Cost/volume ~ $680/m³, and C_p,tower = (680)(243.7) = $166,000.

From Figure 11-2, C_p° = Tray cost/area ~ $720/m², and C_p,tray = (720)(10.5) = $7560/tray.

We now need to include the extra cost factors and calculate the bare module costs. Since the column and trays are probably of carbon steel, F_m = 1.0. At 1 atm, F_p = 1.0. Since N > 20, F_q = 1.0 for the trays. Then from Eqs. (11-08) and (11-09), respectively,

C_BM,tower = ($166,000)(2.25 + 1.82 (1.0)(1.0)) = $675,000

C_BM,trays = ($7560)(36)(1.0)(1.0) = $272,000.

Total bare module cost for tower and trays is $947,000.

This result should be compared with Problem 11.D1 for the same separation, but at a pressure of 7 bar. That column costs about 10% more.

This cost does not include pumps, instrumentation and controls, reboiler, condenser, or installation. Cost is as of September 2001. It can be updated to current cost with the cost index.

11.3 CHANGES IN PLANT OPERATING RATES

Plants are designed for some maximum nameplate capacity but commonly produce less. The operating cost per kilogram of feed can be found by dividing Eq. (11-02) by the feed rate.

(11-21)

The kg steam/hr, kg water/hr, and kW elec/hr are all directly proportional to F. Thus, except for labor costs, the operating cost per kilogram will be constant regardless of the feed rate. Since labor costs are often a small fraction of total costs in automated continuous chemical plants, we can treat the operating cost per kilogram as constant.

Capital cost per kilogram depends on the total amount of feed processed per year. Then, from Eq. (11-20),

(11-22)

Operation at half the designed feed rate doubles the capital cost per kilogram.

The total cost per kilogram is

(11-23)

The effect of reduced feed rates depends on what percent of the total cost is due to capital cost.

If the cost per kilogram for the entire plant is greater than the selling price, the plant will be losing money. However, this does not mean that it should be shut down. If the selling cost is greater than the operating cost plus administrative costs, then the plant is still helping to pay off the capital costs. Since the capital charges are present even if F = 0, it is usually better to keep operating. Of course, a new plant would not be built under these circumstances.

11.4 ENERGY CONSERVATION IN DISTILLATION

Distillation columns are often the major user of energy in a plant. Mix et al. (1978) estimated that approximately 3% of the total U.S. energy consumption is used by distillation! Thus, energy conservation in distillation systems is extremely important, regardless of the current energy price. Although the cost of energy oscillates, the long-term trend has been up and will probably continue to be up for many years. This passage, originally written in 1986, is obviously also true for the twenty-first century. Several energy-conservation schemes have already been discussed in detail. Most important among these are optimization of the reflux ratio and choice of the correct operating pressure.

What can be done to reduce energy consumption in an existing, operating plant? Since the equipment already exists, there is an incentive to make rather modest, inexpensive changes. Retrofits like this are a favorite assignment to give new engineers, since they serve to familiarize the new engineer with the plant and failure will not be critical. The first thing to do is to challenge the operating conditions (Geyer and Kline, 1976). If energy can be saved by changing the operating conditions the change may not require any capital. When the feed rate to the column changes, is the column still operating at vapor rates that are near those for optimum efficiency? If not, explore the possibility of varying the column pressure to change the vapor flow rate and thus, operate closer to the optimum. This will allow the column to have the equivalent of more equilibrium contacts and allow the operator to reduce the reflux ratio. Reducing the reflux ratio saves energy in the system. Challenge the specifications for the distillate and bottoms products. When products are very pure, rather small changes in product purities can mean significant changes in the reflux ratio. If a product of intermediate composition is required, side withdrawals require less energy than mixing top and bottom products (Allen and Shonnard, 2002).

Second, look at modifications that require capital investment. Improving the controls and instrumentation can increase the efficiency of the system (Geyer and Kline, 1976; Mix et al., 1978; Shinskey, 1984). Better control allows the operator to operate much closer to the required specifications, which means a lower reflux ratio. Payback on this investment can be as short as 6 months. Distillation column control is an extremely important topic that is beyond the scope of this textbook. Rose (1985) provides a qualitative description of distillation control systems while Shinskey (1984) is more detailed. If the column has relatively inefficient trays (e.g., bubble-caps) or packing (e.g., Raschig rings), putting in new, highly efficient trays (e.g., valve trays) or new high-efficiency packing (e.g., modern rings, saddles, or structured packing) will usually pay even though it is fairly expensive (Kenney, 1988). Certainly, any damaged trays or packing should be replaced with high-efficiency/high-capacity trays or packing (Kenney, 1988). Any column with damaged insulation should have the insulation removed and replaced. Heat exchange and integration of columns may be far from optimum in existing distillation systems. An upgrade of these facilities should be considered to determine whether it is economical.

When designing new facilities, many energy conservation approaches can be used that might not be economical in retrofits. Heat exchange between streams and integration of processes should be used extensively to minimize overall energy requirements. These methods have been known for many years (e.g., see Robinson and Gilliland, 1950 or Rudd and Watson, 1968), but they were not economical when energy costs were very low. When energy costs shot up in the 1970s and early 1980s and again from 2004 through 2006, many energy conservation techniques suddenly became very economical (Doherty and Malone, 2001; Geyer and Kline, 1976; King, 1981; Mix et al., 1978; Null, 1976; O’Brien, 1976; Shinskey, 1984; and Siirola, 1996). These sources and many other references discussed by these authors give more details.

The basic idea of heat exchange is to use hot streams that need to be cooled to heat cold streams that need to be heated. The optimum way to do this depends upon the configuration of the entire plant, since streams from outside the distillation system can be exchanged. The goal is to use exothermic reactions to supply all or at least as much as possible of the heat energy requirements in the plant. If only the distillation system is considered, there are two main heat exchange locations, as illustrated in Figure 11-5. The cold feed is preheated by heat exchange with the hot distillate; this partially or totally condenses the distillate. The trim condenser is used for any additional cooling that’s needed and for improved control of the system. The feed is then further heated with the sensible heat from the bottoms product. The heat exchange is done in this order since the bottoms is hotter than the distillate. Further heating of the feed is done in a trim heater to help control the distillation. The system shown in Figure 11-5 may not be optimum, though, particularly if several columns are integrated. Nevertheless, the heat exchange ideas shown in Figure 11-5 are quite basic.

Figure 11-5 Heat exchange for an isolated distillation column

A technique similar to that of Figure 11-5 is to produce steam in the condenser. If there is a use for this low-pressure steam elsewhere in the plant, this can be a very economical use of the energy available in the overhead vapors. Intermediate condensers (section 4.9.4) can also be used to produce steam, and they have a hotter vapor.

Heat exchange integration of columns is an important concept for reducing energy use (Andrecovich and Westerberg, 1985; Biegler et al., 1997; Doherty and Malone, 2001; Douglas, 1988; King, 1981; Linnhoff et al., 1982; Robinson and Gilliland, 1950). The basic idea is to condense the overhead vapor from one column in the reboiler of a second column. This is illustrated in Figure 11-6. (In practice, heat exchanges like those in Figure 11-5 will also be used, but they have been left off Figure 11-6 to keep the figure simple.) Obviously, the condensation temperature of stream D₁ must be higher than the boiling temperature of stream B₂. When distillation is used for two rather different separations the system shown in Figure 11-6 can be used without modification. However, in many cases stream D₁ is the feed to the second column. The system shown in Figure 11-6 will work if the first column is at a higher pressure than the second column so that stream D₁ condenses at a higher temperature than that at which stream B₂ boils.

Figure 11-6 Integration of distillation columns

Many variations of the basic idea shown in Figure 11-6 have been developed. If a solvent is recovered from considerably heavier impurities, some variant of the multieffect system shown in Figure 11-7 is useful (Agrawal, 2000; Andrecovich and Westerberg, 1985; Doherty and Malone, 2001; King, 1981; O’Brien, 1976; Robinson and Gilliland, 1950; Siirola, 1996; Wankat, 1993). After preheating, the solvent is first recovered as the distillate product in the first column, which operates at low pressure. The bottoms from this column is pumped to a higher pressure, preheated, and fed to the second column. Since the second column is at a higher pressure, the overheads can be used in the reboiler of the low-pressure column. Thus, the steam used in the reboiler of the higher pressure column serves to heat both columns. The steam efficiency is almost doubled. Since the separation is easy, not too many stages are required, and the two distillate products are both essentially pure solvent. Multieffect systems can also have feed to the high-pressure column in addition to or instead of feed to the low-pressure column. Liquified oxygen and nitrogen from air are produced on very large scales in Linde double columns, which are multieffect distillation columns. Multieffect distillation is closely related to multieffect evaporation (Mehra, 1986).

Figure 11-7 Multieffect distillation system

The condensing vapor from overhead can be used to heat the reboiler of the same column if vapor recompression or a heat pump is used (Humphrey and Keller, 1997; King, 1981; Meili, 1990; Null, 1976; Robinson and Gilliland, 1950). One arrangement for this is illustrated in Figure 11-8. The overhead vapors are condensed to a pressure at which they condense at a higher temperature than that at which the bottoms boil. Vapor recompression works best for close-boiling distillations, since modest pressure increases are required. Generally, vapor recompression is more expensive than heat integration or multieffect operation of columns. Thus, vapor recompression is used when the column is an isolated installation or is operating at extremes of high or low temperatures. O’Brien and Schultz (2004) report that UOP (formerly Universal Oil Products, Inc.) uses heat pumps for the difficult propane-propylene separation.

Figure 11-8 Vapor recompression or heat pump system

11.5 SYNTHESIS OF COLUMN SEQUENCES FOR ALMOST IDEAL MULTICOMPONENT DISTILLATION

A continuous distillation column is essentially a binary separator; that is, it separates a feed into two parts. For binary systems, both parts can be the desired pure products. However, for multicomponent systems, a single column is unable to separate all the components. For ternary systems, two columns are required to produce pure products; for four-component systems, three columns are required; and so forth. There are many ways in which these multiple columns can be coupled together for multicomponent separations. The choice of cascade can have a large effect on both capital and operating costs. In this section we will briefly look at the coupling of columns for systems that are almost ideal. More detailed presentations are available in other books (Biegler et al., 1997; Doherty and Malone, 2001; Douglas, 1988; King, 1981; Rudd et al., 1973; Woods, 1995), in reviews (Nishida et al., 1981; Siirola, 1996), and in a huge number of papers only a few of which are cited here (Agrawal, 2000; Garg et al., 1991; Kim and Wankat, 2004; Schultz et al., 2002; Shah, 2002; Tedder and Rudd, 1978; Thompson and King, 1972). Synthesis of sequences for nonideal systems are considered in the next section.

How many ways can columns be coupled for multicomponent distillation? Lots! For example, Figure 11-9 illustrates nine ways in which columns can be coupled for a ternary system that does not form azeotropes. With more components, the number of possibilities increases geometrically. Figure 11-9A shows the “normal” sequence, where the more volatile components are removed in the distillate one at a time. This is probably the most commonly used sequence, particularly in older plants. Scheme B shows an inverted sequence, where products are removed in the bottoms one at a time. With more components, a wide variety of combinations of these two schemes are possible. The scheme in Figure 11-9C is similar to the one in part A except that the reboiler has been removed and a return vapor stream from the second column supplies boilup to the first column. Capital costs will be reduced, but the columns are coupled, which will make control and startup more difficult. Scheme D is similar to B, except that a return liquid stream supplies reflux.

Figure 11-9 Sequences for the distillation of ternary mixtures; no azeotropes. Component A is most volatile, and component C is least volatile

The scheme in Figure 11-9E uses a side enricher while the one in part F uses a side stripper to purify the intermediate component B. The stream is withdrawn at the location where component B has a concentration maximum. These schemes are often used in petroleum refineries. Figure 11-9G illustrates a thermally coupled system (sometimes called Petyluk columns). The first column separates A from C, which is the easiest separation, and the second column then produces three pure products. The system will often have relatively low energy requirements, but it will be more difficult to start up and control. This system may also require an excessive number of stages if either the A-B or B-C separations are difficult. Sometimes this scheme can be achieved in a single “divided wall” column (Schultz et al., 2002). This four-section column has a vertical wall between two parts of the column in the middle two sections. The feed enters on the left side of the wall and the two sections on the left side of the wall do the same separation done by the first column in Figure 11-9G. At the top of the wall a mixture of A and B would spill over the wall and go into the rectifying and the top intermediate sections. At the bottom of the wall a mixture of B and C would flow under the wall and into the stripping and bottom intermediate sections. Thus, the two intermediate sections on the right side of the wall do the separation done in the middle two sections of the second column in Figure 11-9G. Divided wall columns can be designed as if they were three simple columns; for example with the Fenske-Underwood-Gilliland approach (Muralikrishna et al., 2002). Compared to Figure 11-9G, divided wall columns have been reported to have savings up to 30% for energy costs and 25% for capital costs (O’Brien and Schultz, 2004). Note that this arrangement appears to be somewhat sensitive to upsets. Another variant is shown in Figure 11-9H, where the A-C separation is so easy that a flash drum can be used instead of the first column.

The scheme in Figure 11-9I is quite different from the others, since a single column with a side stream is used (Tedder and Rudd, 1978). The side stream cannot be completely pure B, although it may be pure enough to meet the product specifications. This or closely related schemes are most likely to be useful when the concentration of C in the feed is quite low. Then at the point of peak B concentration there will not be much C present. Methods using side streams to connect columns (Doherty and Malone, 2001) and for four-component separations (Kim and Wankat, 2004) can also be used.

These sequences are only the start of what can be done. For example, the heat exchange and energy integration schemes discussed in the previous section can be interwoven with the separation scheme. Agrawal (2000) shows multieffect distillation cascades for ternary separations. These schemes are currently the most economical methods to produce oxygen, nitrogen, and argon from air by cryogenic distillation. Obviously, the system becomes quite complex.

Which method is the best to use depends upon the separation problem. The ease of the various separations, the required purities, and the feed concentrations are all important in determining the optimum configuration. The optimum configuration may also depend upon how “best” is defined. The engineer in charge of operating the plant will prefer the uncoupled systems, while the engineer charged with minimizing energy consumption may prefer the coupled and integrated systems. The only way to be assured of finding the best method is to model all the systems and try them. This is difficult to do, because it involves a large number of interconnected multicomponent distillation columns. Shortcut methods are often used for the calculations to save computer time and money. Unfortunately, the result may not be optimum. Many studies have ignored some of the arrangements shown in Figure 11-9; thus, they may not have come up with the optimum scheme. Conditions are always changing, and a distillation cascade may become nonoptimum because of changes in plant operating conditions such as feed rates and feed or product concentrations. Changes in economics such as energy costs or interest rates may also alter the optimality of the system. Sometimes it is best to build a nonoptimum system because it is more versatile.

An alternative approach to design is to use heuristics, which are rules of thumb used to exclude many possible systems. The heuristic approach may not result in the optimum separation scheme, but it usually produces a scheme that is close to optimum. Heuristics have been developed by doing a large number of simulations and then looking for ideas that connect the best schemes. Some of the most common heuristics listed in approximately the order of importance (Biegler et al., 1997; Doherty and Malone, 2001; Douglas, 1988; Garg et al., 1991; King, 1981; Thompson and King, 1972) include:

1. Remove dangerous, corrosive, and reactive components first.

2. Do not use distillation if α_LK-HK < α_min, where α_min ~ 1.05 to 1.10.

3. Remove components requiring very high or very low temperatures or pressures first.

4. Do the easy splits (large α) first.

5. The next split should remove components in excess.

6. The next split should remove the most volatile component.

7. Do the most difficult separations as binary separations.

8. Favor 50:50 splits.

9. If possible, final product withdrawals should be as distillate products.

Two additional heuristics not listed in order of importance that can be used to force the designer to look at additional sequences are:

10. Consider side stream withdrawals for sloppy separations.

11. Consider thermally coupled and multieffect columns, particularly if energy is expensive.

There are rational reasons for each of the heuristics. Heuristic 1 will minimize safety concerns, remove unstable compounds, and reduce the need for expensive materials of construction in later columns. Heuristic 2 eliminates the need for excessively tall columns. Since very high and very low temperatures and pressures require expensive columns or operating conditions, heuristic 3 will keep costs down. Since the easiest split will require a shorter column and low reflux ratios, heuristic 4 says to do this when there are a number of components present and feed rates are large. The next heuristic suggests reducing feed rates as quickly as possible. Removing the most volatile component (heuristic 6) removes difficult to condense materials, probably allowing for reductions in column pressures. Heuristic 7 forces the lowest feed rate and hence the smallest diameter for the large column required for the most difficult separation. Heuristic 8 balances columns so that flow rates don’t change drastically. Since thermal degradation products are usually relatively nonvolatile, heuristic 9 is likely to result in purer products. The purpose of heuristics 10 and 11 is to force the designer to think outside the usual box and look at schemes that are known to be effective in certain cases.

Each heuristic should be preceded with the words “All other things being equal.” Unfortunately, all other things usually are not equal, and the heuristics often conflict with each other. For example, the most concentrated component may not be the most volatile. When there are conflicts between the heuristics, the cascade schemes suggested by both of the conflicting heuristics should be generated and then compared with more exact calculations.

EXAMPLE 11.2 Sequencing columns with heuristics

A feed with 25 mole % ethanol, 15 mole % isopropanol, 35 mole % n-propanol, 10 mole % isobutanol, and 15 mole % n-butanol is to be distilled. Purity of 98% for each alcohol is desired. Determine the possible optimum column configurations.

Solution

A, B, C. Define, explore, plan. With five components, there are a huge number of possibilities; thus, we will use heuristics to generate possible configurations. Equilibrium data can be approximated as constant relative volatilities (King, 1981) with n-propanol as the reference component: Ethanol, α = 2.09, Isopropanol, α = 1.82, n-propanol, α = 1.0; isobutanol, α = 0.677; n-butanol, α = 0.428.

To use the heuristics it is useful to determine the relative volatilities of all adjacent pairs of compounds: α_E-IP = 2.09/1.82 = 1.15, α_IP-nP = 1.82/1.0 = 1.82, α_nP-IB = 1.0/0.677 = 1.48, α_IB-nB = 0.677/0.428 = 1.58.

Since the easiest separation, isopropanol—n-propanol, is not that much easier than the other separations, heuristic 4 can probably be ignored.

D. Do it.

Case 1. Heuristics 6 and 9 give the direct sequence

(Reboilers and condensers are not shown.) This will certainly work, but it is not very inventive.

Case 2. Heuristic 7 is often very important. Which separation is most difficult? From the list of relative volatilities of adjacent pairs of compounds ethanol-isopropanol is the hardest separation. If we also use heuristic 8 for column A and heuristic 5 and 8 for column C, we obtain the scheme shown in the figure.

Naturally, other alternatives are possible.

Case 3. Heuristic 11 can be used to generate an entirely thermally coupled system (see Problem 11-A17). This would be difficult to operate. However, we can use heuristics 7, 8, and 11 to obtain a modification of case 2 (see figure).

Case 4. Heuristic 11 can also be used to develop a system with one or more multieffect columns. If we use heuristic 7 to do the ethanol—isopropanol separation by itself, one option is to pressurize the liquid feed to column D in case 2 and operate column D at a higher pressure. This multieffect arrangement is shown below. There are many other options possible for using multieffect columns for this separation.

Other systems can be generated, but one of the four shown here is probably reasonably close to optimal.

E. Check. Finding the optimum configuration requires a simulation of each alternative. This can be done for cases 1 and 2 using the Fenske-Underwood-Gilliland approach. For cases 3 and 4 the thermally coupled and multieffect columns are more complex and probably should be simulated in detail.

F. Generalize. It is likely that one of these designs is close to optimum. Because of the low relative volatility between ethanol and isopropanol, heuristic 7 is important. Use of the heuristics does avoid having to look at several hundred other alternatives.

These heuristics have been developed for systems that have no azeotropes. When azeotropes are present, the methods developed in Chapter 8 should be used.

11.6 SYNTHESIS OF DISTILLATION SYSTEMS FOR NONIDEAL TERNARY SYSTEMS

In the previous section we developed heuristics for synthesis of distillation sequences for almost ideal systems; unfortunately, many of these heuristics do not apply to nonideal systems. Instead, we must use a different set of operational suggestions and the tools developed in section 8.5, distillation and residue curves. The purpose of the operational suggestions is to first develop a feasible separation scheme and then work to improve it.

Heuristics for nonideal systems have not been formalized and agreed upon to the same degree as for ideal systems. The following operational suggestions are from Biegler et al. (1997), Doherty and Malone (2001), and common sense.

Operational Suggestions: Preliminary.

1. Obtain reliable equilibrium data and/or correlations for the system.

2. Develop residue curves or distillation curves for the system.

3. Classify the system as A) almost ideal; B) nonideal without azeotropes; C) one homogeneous binary azeotrope without a distillation boundary; D) one homogeneous binary azeotrope with a distillation boundary; E) two or more homogeneous binary azeotropes with possibly a ternary azeotrope; F) heterogeneous azeotrope, which may include several binary and ternary azeotropes. Although solutions for cases D to F are beyond the scope of this introductory treatment, they will be discussed briefly.

A. Almost ideal. If the system is reasonably close to ideal (Figure 08-7), rejoice and use the heuristics in Section 11.5.

B. Nonideal systems without azeotropes. These systems are often similar to ideal, and a variety of column sequences will probably work. However, it may be easier to do the most difficult separation with the non-key present instead of the binary separation recommended for ideal mixtures. Doherty and Malone (2001) present a detailed example for the acetaldehyde-methanol-water system.

1. Generate the y-x curves for each binary pair. If all separations are relatively easy (reasonable relative volatilities and no inflection points causing tangent pinches for one of the pure components) use the ideal heuristics in Section 11.5.

2. If one of the binary pairs has a small relative volatility or a tangent pinch, determine if this separation is easier in the presence of the third component. This can be done by generating distillation curves or y-x “binary” equilibrium at different constant concentrations of the third component. If the presence of the third component aids the separation, separate the difficult pair first. The concentration of the third component can be adjusted by recycling it from the column where it is purified. Note that this approach is very similar to using the third component as an extractive distillation solvent for separation of close boiling components (Section 8.6).

C. One homogeneous binary azeotrope, without a distillation boundary. The residue curve maps look like Figure 8-11a or 8-11c.

1. If the binary azeotrope is between the light and intermediate components (Figure 8-11c), the situation is very similar to using extractive distillation to separate a binary azeotrope, and the heavy component is used instead of an added solvent. In general, the flowsheet is similar to Figure 08-13 except the heavy component product is withdrawn where the makeup solvent is added. If there is sufficient heavy component in the feed, a heavy recycle may not be required. The residue curve map will be similar to the extractive distillation residue map, Figure 08-14, but since the feed contains all three components point F will be inside the triangle.

2. If the binary azeotrope is between the heavy and light components (Figure 8-11a), a separation can be obtained with an intermediate recycle as shown in Figures 11-10a and 11-10b. If there is sufficient intermediate component in the feed, the intermediate recycle may not be required. Separation using the flowchart in Figure 11-10a is illustrated in Example 11-3.

Figure 11-10 Distillation cascades for ternary feed with binary azeotrope between light L and heavy H components; A) Indirect sequence, B) direct sequence

D. One homogeneous binary azeotrope with distillation boundary. There can either be a maximum boiling azeotrope (Figure 08-8) or a minimum boiling azeotrope (Figure 8-11b).

1. If the distillation boundary is straight, complete separation of the ternary feed is not possible without addition of a mass-separating agent (Doherty and Malone, 2001).

2. If the distillation boundary is curved, complete separation of the ternary feed may be possible. Distillation boundaries can often be crossed by mixing a feed with a recycle stream. An example of the synthesis of feasible flowsheets is given by Biegler et al. (1997) for the system in Figure 08-8. Detailed solution of this case and the remaining two cases is beyond the scope of this section.

E. Two or more homogeneous binary azeotropes, with possibly a ternary azeotrope. These systems are messy, and there are invariably one or more distillation boundaries. If there is a single curved distillation boundary, it may be possible to develop a scheme to separate the mixture without addition of a mass-separating agent. If there are only two binary azeotropes and no ternary azeotropes, look for a separation method (e.g., extraction) that will remove the component that occurs in both azeotropes.

F. Heterogeneous azeotrope, which may include several binary and ternary azeotropes. An example was shown in Figure 8-12 for an azeotropic distillation scheme. The residue curve map developed by a process simulator probably will not show the envelope of the two-phase region, and this region will have to be added. Since liquid-liquid separators can cross distillation boundaries, there is a good chance that a separation can be achieved without adding an additional mass-separating agent. Distillation boundaries can be crossed by mixing, decanting, and reaction. If possible, use components already in the feed as an extractive distillation solvent or entrainer for azeotropic distillation. Also explore using components in the feed as extraction solvents. These systems are discussed in detail by Doherty and Malone (2001), and a process example is discussed by Biegler et al. (1997).

EXAMPLE 11.3 Process development for separation of complex ternary mixture

We have 100.0 kg moles/hr of a saturated liquid feed that is 50.0 mole % methanol, 10.0 mole % methyl butyrate and 40 mole % toluene. We want to separate this feed into three pure products (99.7⁺% purities). Develop a feasible distillation cascade for this system. Prove that your system is feasible.

Solution

A. Define. We want to develop a sequence of distillation columns including recycle that we claim can produce 99.7⁺% pure methanol, methyl butyrate, and toluene. Proof requires running a process simulator to show that the separation is achieved. Note that optimization is not required.

B. Explore. These components are in the Aspen Plus data bank and residue curves were generated with Aspen Plus using NRTL (Figure 11-11) (obviously, other process simulators could be used). Since there is one minimum boiling binary azeotrope between methanol (light component) and toluene (heavy) component without a distillation boundary, this residue curve map is similar to Figure 8-11a. We expect that the flowchart in either Figure 11-10a or 11-10b will do the separation.

C. Plan. The fresh feed point is plotted in Figure 11-11. Since the methyl butyrate concentration in the fresh feed is low, this point is close to the binary toluene-methanol line. If we don’t recycle intermediate (methyl butyrate) we may have a problem with the binary azeotrope. Thus, for a feasible design, it is safer to start with recycle of intermediate.

Is Figure 11-10a or 11-10b likely to be better? Probably either one will work. Comparing the boiling points, the separation of methanol from methyl butyrate is probably much simpler than separating methyl butyrate from toluene. Thus, the second column is likely to be considerably smaller if we use the flowchart in Figure 11-10a, although the first column will be larger. We will use Figure 11-10a and leave exploration of Figure 11-10b to Problems 11-D6 and 11-G1.

D. Do it. Use the flowchart in Figure 11-10a with recycle of intermediate. Arbitrarily, pick a recycle rate of 100 kg moles/hr. (The purpose of this example is to show feasibility. This initial assumption would be varied as the design is polished.) Since the methyl butyrate needs to be 99.7⁺% pure, we will assume the recycle stream is pure. Mixing the fresh feed and the recycle stream and using the lever arm rule, we find point M as shown in Figure 11-11. This combined feed can then be split into a distillate that contains essentially no toluene and a bottoms product that is 99.7⁺% pure toluene. The distillate product can then be separated in the second column into a 99.7⁺% pure methanol distillate and a 99.7⁺% pure methyl butyrate bottoms. Thus, the separation appears to be possible.

Figure 11-11 Residue curves and mass balances for methanol, toluene, and methyl butyrate distillation. Key: M = methanol, MB = methyl butyrate, T = toluene

Proof: The mixed feed to the column is 55% methyl butyrate, 20% toluene and 25% methanol. Flow rate is 200 kg mole/hr, and it was assumed to be a saturated liquid. The system in Figure 11-10a was simulated on Aspen Plus using NRTL for equilibrium. For the feasibility study pure methyl butyrate (instead of the recycle stream) and fresh feed were mixed together and input on the same stage of the first column. After some trial-and-error, the following results were obtained.

Column 1: N = 81 (including total condenser and partial reboiler), N_feed = 41, D = 160 kmoles/hr, L/D = 8, p = 1.0 atm.

Distillate mole fractions: Methanol = 0.3125, Methyl butyrate = 0.6869907, Toluene = 0.00050925.

Bottoms mole fractions: Methanol = 6.345 E -35, Methyl butyrate = 0.002037, Toluene = 0.997963.

Column 2: N = 20 (including total condenser and partial reboiler), N_feed = 10, D = 50 kmoles/hr, L/D = 1.5, p = 1.0 atm.

Distillate mole fractions: Methanol = 0.99741, Methyl butyrate = 0.002315, Toluene = 0.0002753.

Bottoms mole fractions: Methanol = 0.0011751, Methyl butyrate = 0.998207, Toluene = 0.00061559.

Thus, the separation is feasible.

E. Check. The residue curve map plotted from the process simulator agrees with the map in Doherty and Malone (2001). The predicted distillate composition for the first column determined from the mass balance calculation on Figure 11-11 is 31% methanol and 69% methyl butyrate. This agrees quite well with the results from the simulator. The literature states that Figure 11-10a should be successful for separating this type of mixture. Thus, we are quite confident that the process is feasible.

F. Generalization. The development of a feasible process is the first major step in the design. We would also need to do a preliminary design on Figure 11-10b to make sure that process is not better (see Problems 11.D6 and 11.G1). We need to optimize the better process to find appropriate values for the recycle rate, and optimum values for N, N_feed, and L/D for each column. We should also try the processes without recycle of MB to see if either of these processes are feasible (see Problems 11.D7a and 11.G2). In the first column the use of separate feed locations for the fresh feed and the recycle stream should be explored. If neither process is clearly better than the other, we would optimize both processes to determine which is more economical.

11.7 SUMMARY—OBJECTIVES

In this chapter we have looked briefly at the economics of distillation and the effects of changing operating variables. At the end of this chapter you should be able to satisfy the following objectives:

1. Estimate the capital and operating costs for a distillation column

2. Predict the effect of the following variables on column capital and operating costs:

a. Feed rate

b. Column pressure

c. External reflux ratio

3. Estimate the effects that external factors have on capital and operating costs; external factors would include:

a. Energy costs

b. The general state of the economy

4. Discuss methods for reducing energy in distillation systems; develop flowsheets with appropriate heat exchange

5. Use heuristics to develop alternative cascades for the distillation of almost ideal multicomponent mixtures.

6. Use distillation or residue curves to develop a feasible separation scheme for nonideal mixtures.

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HOMEWORK

A. Discussion Problems

A1. If valve trays cost more than sieve trays, why are they often advertised as a way of decreasing tower costs?

A2. Develop your key relations chart for this chapter.

A3. What is the effect of increasing the feed temperature if

a. L/D is constant

b. L/D = 1.15 (L/D)_min. Note that (L/D)_min will change.

Include effects on Q_R and number of stages. Use a McCabe-Thiele diagram.

A4. Optimums usually occur because there are two major competing effects. For the optimum L/D these two major effects for capital cost are (select two answers):

a. The number of stages is infinite as L/D goes to infinity and has a minimum value at (L/D)_min.

b. The number of stages is infinite at (L/D)_min and approaches a minimum value as L/D goes to infinity.

c. The column diameter becomes infinite at (L/D)_min and approaches a minimum value as L/D approaches infinity.

d. The column diameter becomes infinite as L/D approaches infinity and has a minimum value at (L/D)_min.

A5. Working capital is the money required for day-to-day operation of the plant, and interest on working capital is an operating expense. If the feed rate drops, is the working capital cost per kilogram constant? What happens to working capital if customers are slow paying their bills? Some companies give a 5% discount for immediate payment of bills; explain why this might or might not be a good idea.

A6. How does the general state of the economy affect:

a. Design of new plants

b. Operation of existing plants

A7. Why is the dependence on size usually less than linear—in other words, why is the exponent in Eq. (11-02) less than 1?

A8. It is common to design columns at reflux ratios slightly above (L/D)_opt. Use a curve of total cost/yr vs. L/D to explain why an L/D > (L/D)_opt is used. Why isn’t there a large cost penalty?

A9. Discuss the concept of economies of scale. What happens to economies of scale if the feed rate is half the design value?

A10. Use a McCabe-Thiele diagram to explain how reducing the product concentration allows the use of a lower L/D for an existing column.

A11. Referring to Figure 11-6 if D1 is the feed to column 2, explain what conditions are necessary for this system to work.

A12. Figure 11-9I shows an arrangement that is useful when the feed concentration of the heavy component, C, is low. Sketch an arrangement for use when the light component, A, feed concentration is low.

A13. Why is the cost of packing per m³ and the cost per tray less when more packing or more trays are purchased?

A14. The use of components in the feed as solvents for extractive or azeotropic distillation or extraction is recommended even if they are not the best solvents for stand-alone separations. Explain the reasoning behind this recommendation.

A15. Residue curves and distillation curves have similar shapes but are not identical. Even though the residue curves might be misleading in some cases, they are still useful for screening possible distillation separations. Explain why.

A16. Figure 11-8 shows a heat pump system in which the distillate vapor is used as the working fluid. It may be desirable to use a separate working fluid. Sketch this. What are the advantages and disadvantages?

A17. Draw the entirely thermally coupled system (an extension of Figure 11-9G) for Example 11-2.

A18. Preheating the feed will often increase the number of stages required for the separation (F, z, x_D, x_B, L/D constant). Use a McCabe-Thiele diagram to explain why this happens.

B. Generation of Alternatives

B1. Sketch possible column arrangements for separation of a four-component system. Note that there are a large number of possibilities.

B2. Repeat Example 11-2 but for a 99.9% recovery of n-propanol. Purities can be lower.

B3. Multieffect distillation or column integration can be done with more than two columns. Use the basic ideas in Figures 11-5 and 11-6 to sketch as many ways of thermally connecting three columns as you can.

B4. We wish to separate a feed that is 10 mole % benzene, 55 mole % toluene, 10 mole % xylene, and 25 mole % cumene. Use heuristics to generate desirable alternatives. Average relative volatilities are α_BT = 2.5, α_TT = 1.0, α_XT = 0.33, α_CT = 0.21. 98% purity of all products is required.

B5. Repeat Problem 11-B4 for an 80% purity of the xylene product.

C. Derivations

C1. Show that cooling water and steam costs are directly proportional to the feed rate.

C2. For large volumes Figure 11-2 shows that packing costs are directly proportional to the volume of packing. Show that packing costs go through a minimum as L/D increases.

D. Problems

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

D1. *Repeat Example 11-1 except at a pressure of 700 kPa. At this pressure E_o = 0.73, D = 9 ft, and the relative volatility will be a function of pressure.

a. * Find (L/D)_min.

b. * Find N_min.

c. * Estimate N_equil.

d. * Estimate N_actual.

e. * Find cost of shells and trays as of Sept. 2001.

f. Update cost of shells and trays to current date.

D2. *Estimate the cost of the condenser and reboiler (both fixed tube sheet, shell and tube) for the distillation of Example 11-1. Pressure is 101.3 kPa. C_PL,C7 = 50.8 Btu/lb–mole–°F, λ_C7 = 14,908 Btu/lb–mole. Data for hexane are given in Problem 3-D6. The saturated steam in the reboiler is at 110°C. λ_steam = 958.7 Btu/lb. Cooling water enters at 70°F and leaves at 110°F. C_P,w = 1.0 Btu/lb–°F. Use heat transfer coefficients from Table 11-2. Watch your units.

D3. Determine the steam and water operating costs per hour for Problem 11-D2. Cost of steam is $20.00/1000 lb, and cost of cooling water is $3.00/1000 gal.

D4. Example 10-4 and Problem 10-D17 sized the diameter of a packed column doing the separation in Example 11-1. Suppose a 15-foot diameter column is to be used. The 1-in Intalox saddles have an HETP of 1.1 feet. Estimate the packing and tower costs. Pressure is 101.3 kPa.

D5. A 9-foot diameter column is adequate for Example 10-4 at 700 kPa. Using the same data as in Problem 11.D4 and the number of equilibrium contacts estimated in Problem 11.D1, estimate the packing and tower costs.

D6. Repeat the residue curve analysis for Example 11-3, but using the flowsheet in Figure 11-10b. Arbitrarily use a recycle flow rate of 100 kmoleshr.

D7. Repeat the residue curve analysis for Example 11-3 but with no recycle.

a. For process in Figure 11-10a.

b. For process in Figure 11-10b.

F. Problems Requiring Other Resources

F1. Look up the current cost index in Chemical Engineering magazine. Use this to update the ordinates in Figures 11-1 to 11-3.

G. Computer Problems

G1. Repeat the computer simulation proof of feasibility for Example 11-3, but using the flowsheet in Figure 11-10b. The input for the simulator should be based on the solution to Problem 11.D6. If desired, you may input the recycle stream and the fresh feed to different stages in the first column.

G2. Repeat the computer simulation proof of feasibility for Example 11-3, but with no recycle. The input for the simulator should be based on the solution to Problem 11.D7.

a. For process in Figure 11-10a.

b. For process in Figure 11-10b.