Chapter 12: Absorption and Stripping

Up to now we have talked almost entirely about distillation. There are other unit operations that are very useful in the processing of chemicals or in pollution control. Absorption is the unit operation where one or more components of a gas stream are removed by being taken up (absorbed) in a nonvolatile liquid (solvent). In this case the liquid solvent must be added as a separating agent. Absorption is one of the methods used to remove CO₂ from natural gas and flue gasses so that the CO₂ is not added to the atmosphere where it helps cause global warming (Socolow, 2005).

Stripping is the opposite of absorption. In stripping, one or more components of a liquid stream are removed by being vaporized into an insoluble gas stream. Here the gas stream (stripping agent) must be added as a separating agent.

What was the separating agent for distillation? Heat.

Absorption can be either physical or chemical. In physical absorption the gas is removed because it has greater solubility in the solvent than other gases. An example is the removal of butane and pentane from a refinery gas mixture (C₄ − C₅) with a heavy oil. In chemical absorption the gas to be removed reacts with the solvent and remains in solution. An example is the removal of CO₂ or H₂S by reaction with NaOH or with monoethanolamine (MEA). The reaction can be either irreversible (as with NaOH) or reversible (as with MEA). For irreversible reactions the resulting liquid must be disposed of, whereas in reversible reactions the solvent can be regenerated (in stripper or distillation columns). Thus, reversible reactions are often preferred. Chemical absorption systems are discussed in more detail by Astarita et al. (1983), Kohl (1987), Kohl and Nielsen (1995), and Zarycki and Chacuk (1993).

Chemical absorption usually has a much more favorable equilibrium relationship than physical absorption (solubility of most gases is usually very low) and is therefore often preferred. However, the Murphree efficiency is often quite low (10% is not unusual), and this must be taken into account.

Both absorption and stripping can be operated as equilibrium stage operations with contact of liquid and vapor. Since distillation is also an equilibrium stage operation with contact of liquid and vapor, we would expect the equipment to be quite similar. This is indeed the case; both absorption and stripping are operated in packed and plate towers. Plate towers can be designed by following an adaptation of the McCabe-Thiele method. Packed towers can be designed by use of HETP or preferably by mass transfer considerations (see Chapter 15).

In both absorption and stripping a separate phase is added as the separating agent. Thus, the columns are simpler than those for distillation in that reboilers and condensers are normally not used. Figure 12-1 is a schematic of a typical absorption column. In this column solute B entering with insoluble carrier gas C in stream Y_N+1 is absorbed into the nonvolatile solvent A.

Figure 12-1 Gas absorber

A gas treatment plant often has both absorption and stripping columns as shown in Figure 12-2. In this operation the solvent is continually recycled. The heat exchanger heats the saturated solvent, changing the equilibrium characteristics of the system so that the solvent can be stripped. A very common type of gas treatment plant is used for the removal of CO₂ and/or H₂S from refinery gas or natural gas. In this case MEA or other amine solvents in water are used as the solvent, and steam is used as the stripping gas (for more details see Kohl and Nielsen, 1997 or Ball and Veldman, 1991).

Figure 12-2 Gas treatment plant

12.1 ABSORPTION AND STRIPPING EQUILIBRIA

For absorption and stripping in three component systems we often assume that

1. Carrier gas is insoluble.

2. Solvent is nonvolatile.

3. The system is isothermal and isobaric.

The Gibbs phase rule is

F = C − P + 2 = 3(A, B, and C) − 2(vapor and liquid) + 2 = 3

If we set T and p constant, there is one remaining degree of freedom. The equilibrium data are usually represented either by plotting solute composition in vapor vs. solute composition in liquid or by giving a Henry’s law constant. Henry’s law is

(12-1)

where H_B is Henry’s law constant, in atm/mole frac, H = H(p, T, composition); x_B is the mole frac B in the liquid; and p_B is the partial pressure of B in the vapor.

Henry’s law is valid only at low concentrations of B. Since partial pressure is defined as

(12-2)

Henry’s law becomes

(12-3)

This will plot as a straight line if H_B is a constant. If the component is pure, y_B = 1 and p_B = p_tot. Equilibrium data for absorption are given by Hwang (1981), Hwang et al. (1992a, b), Kohl (1987), Kohl and Nielsen (1997), Perry et al. (1963, pp. 14–2 to 14–12), Perry and Chilton (1973, p. 14-3), Perry and Green (1997, pp. 2–125 to 2–128), and Yaws et al. (2005). For example, the values given for CO₂, CO, and H₂S are shown in Table 12-1 (Perry et al., 1963). The large H values in Table 12-1 show that CO₂ and H₂S are very sparingly soluble in water. Since H is roughly independent of p_tot, this means that more gas is absorbed at higher pressure. This phenomenon is commonly taken advantage of to make carbonated beverages. When the bottle or can is opened the pressure drops and the gas desorbs, forming little bubbles. Selected Henry’s law constants for chlorinated compounds in water are listed in Table 12-2 (Yaws et al., 2005) at 25°C. These values are useful for developing processes for removal of these compounds from contaminated water by stripping. Note that these compounds are much more soluble than the gases listed in Table 12-1.

Table 12-1. Henry’s law constants, H for CO₂, CO, and H₂S in water. H is in atm/mole frac.

Table 12-2. Henry’s law constants and solubilities for chlorinated compounds in water at 25°C and 1 atm (Yaws et al., 2005)

The Henry’s law constants depend upon temperature and usually follow an Arrhenius relationship. Thus,

(12-4)

A plot of log H vs. 1/T will often give a straight line.

The effect of concentration is shown in Table 12-3, where the absorption of ammonia in water (Perry et al., 1963) is illustrated. Note that the solubilities are nonlinear and H = p_NH3/x is not a constant. This behavior is fairly general for soluble gases.

Table 12-3. Absorption of ammonia in water

We will convert equilibrium data to the concentration units required for calculations. If mole or mass ratios are used, equilibrium must be converted into ratios.

12.2 OPERATING LINES FOR ABSORPTION

The McCabe-Thiele diagram is most useful when the operating line is straight. This requires that the energy balance is automatically satisfied and liquid flow rate/vapor flow rate = constant. In order for energy balances to be automatically satisfied, we must assume that

1. The heat of absorption is negligible.

2. Operation is isothermal.

These two assumptions will guarantee satisfaction of the energy balances. When the gas and liquid streams are both fairly dilute, the assumptions will probably be satisfied.

We also desire a straight operating line. This will be automatically true if we define

and if we assume that:

3. Solvent is nonvolatile.

4. Carrier gas is insoluble.

Assumptions three and four are often closely satisfied. The results of these last two assumptions are that the mass balance for solvent becomes

(12-5)

while the mass balance for the carrier gas is

(12-6)

Note that we cannot use overall flow rates of gas and liquid in concentrated mixtures because a significant amount of solute may be absorbed which would change gas and liquid flow rates and give a curved operating line. For very dilute solutions (< 1% solute), overall flow rates can be used, and mass or mole fracs can be used for operating equations and equilibria. Since we want to use L = moles nonvolatile solvent (S)/hr and G = moles insoluble carrier gas (C)/hr, we must define our compositions in such a way that we can write a mass balance for solute B. How do we do this?

After some manipulation we find that the correct way to define our compositions is as mole ratios. Define

(12-7a)

The mole ratios Y and X are related to our usual mole fracs by

(12-7b)

Note that both Y and X can be greater than 1.0. With the mole ratio units, we have

and

Thus, we can easily write the steady-state mass balance, input = output, in these units. The mass balance around the top of the column using the mass balance envelope shown in Figure 12-1 is

(12-8)

or

Moles B in/hr = moles B out/hr

Solving for Y_j+1 we obtain

(12-9)

This is a straight line with slope L/G and intercept (Y₁ − (L/G) X₀). It is our operating line for absorption. Thus, if we plot ratios Y vs. X we have a McCabe-Thiele type of graph as shown in Figure 12-3.

Figure 12-3 McCabe-Thiele diagram for absorption, Example 12-1

The steps in this procedure are:

1. Plot Y vs. X equilibrium data (convert from fraction to ratios).

2. Values of X₀, Y_N+1, Y₁ and L/G are known. Point (X₀, Y₁) is on operating line, since it represents passing streams.

3. Slope is L/G. Plot operating line.

4. Start at stage 1 and step off stages by alternating between the equilibrium curve and the operating line.

Note that the operating line is above the equilibrium line, because solute is being transferred from the gas to the liquid. In distillation we had material (the more volatile component) transferred from liquid to gas, and the operating line was below the equilibrium curve. If we had plotted the less volatile component, that operating line would be above the equilibrium curve in distillation.

Equilibrium data must be converted to ratio units, Y vs. X. These values can be greater than 1.0, since Y = y/(1 − y) and X = x/(1 − x). The Y = X line has no significance in absorption. As usual the stages are counted at the equilibrium curve. A minimum L/G ratio can be defined as shown in Figure 12-3. If the system is not isothermal, the operating line will not be affected, but the equilibrium line will be. Then the McCabe-Thiele method must be modified to include changing equilibrium curves.

For very dilute systems we can use mole fracs, since total flows are approximately constant. This is illustrated in section 12.5 on the Kremser equation.

EXAMPLE 12.1 Graphical absorption analysis

A gas stream is 90 mole % N₂ and 10 mole % CO₂. We wish to absorb the CO₂ into water. The inlet water is pure and is at 5°C. Because of cooling coils, operation can be assumed to be isothermal. Operation is at 10 atm. If the liquid flow rate is 1.5 times the minimum liquid flow rate, how many equilibrium stages are required to absorb 92% of the CO₂? Choose a basis of 1 mole/hr of entering gas.

Solution

A. Define. See the sketch. We need to find the minimum liquid flow rate, the value of the outlet gas concentration, and the number of equilibrium stages required.

B. Explore. First we need equilibrium data. These are available in Table 12-1. Since concentrations are fairly high, the problem should be solved in mole ratios. Thus, we need to convert all compositions including equilibrium data to mole ratios.

C. Plan. Derive the equilibrium equation from Henry’s law. Convert compositions from mole fracs to mole ratios using Eq. (12-7). Calculate Y₁ by a percent recovery analysis. Plot mole ratio equilibrium data on a Y−X diagram, and determine (L/G)_min and hence L_min. Calculate actual L/G, plot operating line, and step off stages.

The problem appears to be straightforward.

D. Do it. Equilibrium:

Change the equilibrium data to mole ratios with a table as shown below. (The equation can also be converted, but it is easier to avoid a mistake with a table.)

Note that x = X in this concentration range, but y ≠ Y. The inlet gas mole ratio is

Percent Recovery Analysis: 8% of CO₂ exits.

(0.1 mole in)(0.08 exits) = 0.008 moles CO₂ out

Thus,

Operating Line:

Goes through point (Y₁, X₀) = (0.008888, 0).

(L/G)_min is found as the slope of the operating line from point (Y₁, X₀) to the intersection with the equilibrium curve at Y_N+1. This is shown on Figure 12-3.

The fraction was calculated as

E. Check. The overall mass balances are satisfied by the outlet concentrations. The significant figures carried in this example are excessive compared with the equilibrium data. Thus, they should be rounded off when reported (e.g., N = 4.1). The concentrations used were quite high for Henry’s law. Thus, it would be wise to check the equilibrium data.

F. Generalize. Note that the gas concentration is considerably greater than the liquid concentration. This situation is common for physical absorption (solubility is low). Chemical absorption is used to obtain more favorable equilibrium. The liquid flow rate required for physical absorption is often excessive. Thus, in practice, this type of operation uses chemical absorption.

If we had assumed that total gas and liquid flow rates were constant (dilute solutions), the result would be in error. An estimate of this error can be obtained by estimating (L/G)_min. The minimum operating line goes from (y₁, x₀) = (0.00881, 0) to (y_N+1, x_equil,N+1). y_N+1 = 0.1 and x_equil,N+1 = y_N+1/87.6 = 0.1/87.6 = 0.0011415.

Then

This is in error by more than 10%.

12.3 STRIPPING ANALYSIS

Since stripping is very similar to absorption we expect the method to be similar. The mass balance for the column shown in Figure 12-4 is the same as for absorption and the operating line is still

Figure 12-4 Stripping column

For stripping we know X₀, X_N, Y_N+1, and L/G. Since (X_N, Y_N+1) is a point on the operating line, we can plot the operating line and step off stages. This is illustrated in Figure 12-5.

Figure 12-5 McCabe-Thiele diagram for stripping

Note that the operating line is below the equilibrium curve because solute is transferred from liquid to gas. This is therefore similar to the stripping section of a distillation column. A maximum L/G ratio can be defined; this corresponds to the minimum amount of stripping gas. Start from the known point (Y_N+1, X_N), and draw a line to the intersection of X = X₀ and the equilibrium curve. Alternatively, there may be a tangent pinch point. For a stripper, Y₁ > Y_N+1, while the reverse is true in absorption. Thus, the top of the column is on the right side in Figure 12-5 but on the left side in Figure 12-3. Stripping often has large temperature changes, so the calculation method used here may not work.

Murphree efficiencies can be used on these diagrams if they are defined as

(12-10)

For dilute systems the more common definition of Murphree vapor efficiency in mole fracs would be used. Efficiencies for absorption and stripping are often quite low.

Usually the best way to determine efficiencies is to measure them on commercial-scale equipment. In the absence of such data a rough prediction of the overall efficiency E₀ can be obtained from O’Connell’s correlation shown in Figure 12-6 (O’Connell, 1946). Although originally done for bubble-cap systems, the results can be used for a first estimate for sieve and valve trays. The data in Figure 12-6 is fit by (Kessler and Wankat, 1988)

Figure 12-6 O’Connell’s correlation for overall efficiency of bubble-cap absorbers, reprinted from O’Connell, Trans. AIChE, 42, 741 (1946), copyright 1946, AIChE

(12-11)

which is valid for (Hp/µ) > 0.000316. The Henry’s law constant H is in lb mole/(atm · ft³), pressure p in atm, and liquid viscosity µ in centipoise (cP). More detailed estimates can be made using a mass transfer analysis (see Chapter 15). Correlations for very dilute strippers are given by Hwang (1981).

12.4 COLUMN DIAMETER

For absorption and stripping, the column diameter is designed the same way as for a staged or packed distillation column (Chapter 10). However, note that the gas flow rate, G, must now be converted to the total gas flow rate, V. The carrier gas flow rate, G, is

(12-12)

Since

this is

(12-13a)

or

(12-13b)

The total liquid flow rate, L_j, can be determined from an overall mass balance. Using the balance envelope shown in Figure 12-1, we obtain

(12-14)

Note that the difference between total flow rates of passing streams is constant. This is illustrated in Figure 12-7.

Figure 12-7 Total flow rates in absorber

Both total flows V_j and L_j will be largest where Y_j and X_j are largest. This is at the bottom of the column for absorption, and therefore you design the diameter at the bottom of the column. In strippers, flow rates are highest at the top of the column, so you design the diameter for the top of the column. Specific design details for absorbers and strippers are discussed by Zenz (1997).

An order of magnitude estimate of the column diameter can be made quite easily (Reynolds et al., 2002). The superficial gas velocity (velocity in an empty column) is typically in the range v = 3 to 6 ft/sec. The volumetric flow rate of the gas is ( is the molar gas density), the required cross sectional area is and the column diameter is . Substituting an average superficial gas velocity of 4.5 ft/sec with other appropriate values gives an estimate of the column diameter.

12.5 ANALYTICAL SOLUTION: KREMSER EQUATION

When the solution is quite dilute (say less than 1% solute in both gas and liquid), the total liquid and gas flow rates will not change significantly since little solute is transferred. Then the entire analysis can be done with mole or mass fractions and total flow rates. In this case the column shown in Figure 12-1 will look like the one in Figure 12-8, where streams have been relabeled. The operating equation is derived by writing a mass balance around stage j and solving for y_j+1. The result,

(12-15)

is essentially the same as Eq. (12-09) except that the units are different.

Figure 12-8 Dilute absorber; L and V are total flow rates, y and x are mass or mole fracs.

To use Eq. (12-15) in a McCabe-Thiele diagram, we assume the following:

1. L/V (total flows) is constant.

2. Isothermal system.

3. Isobaric system.

4. Negligible heat of absorption.

These are reasonable assumptions for dilute absorbers and strippers. The solutions on a plot of y vs. x (mole or mass fraction) will look like Figure 12-2 for absorbers and like Figure 12-4 for strippers. The operating line slope will be L/V. Figures 12-9 and 12-10 show two special cases for absorbers.

Figure 12-9 McCabe-Thiele diagram for dilute absorber with parallel equilibrium and operating lines

Figure 12-10 McCabe-Thiele diagram for dilute absorber. (L/V) < m

If one additional assumption is valid, the stage-by-stage problem can be solved analytically. This additional assumption is that the

5. Equilibrium line is straight.

(12-16)

This assumption is reasonable for very dilute solutions and agrees with Henry’s law, Eq. (12-03), if m = H_B/p_tot and b = 0.

An analytical solution for absorption is easily derived for the special case shown in Figure 12-9, where the operating and equilibrium lines are parallel. Now the distance between operating and equilibrium lines, θy, is constant. To go from outlet to inlet concentrations with N stages, we have

(12-17)

since each stage causes the same change in vapor composition. θy can be obtained by subtracting the equilibrium Eq. (12-16) from the operating Eq. (12-15).

(12-18)

For the special case shown in Figure 12-9 L/V = m (the lines are parallel), Eq. (12-18) becomes

(12-19)

Combining Eqs. (12-17) and (12-19), we get

(12-20)

Equation (12-20) is a special case of the Kremser equation. When this equation is applicable, absorption and stripping problems can be solved quite simple and accurately without the need for a stage-by-stage calculation.

Figure 12-9 and the resulting Eq. (12-20) were for a special case. The more general case is shown in Figure 12-10. Now θy_j varies from stage to stage. The θy_j values can be determined from Eq. (12-18). Equation (12-18) is easier to use if we replace x_j with the equilibrium Eq. (12-16),

(12-21)

Then

(12-22a)

(12-22b)

Subtracting Eq. (12-22a) from (12-22b), and solving for (θy)_j+1

(12-23)

Equation (12-23) relates the change in vapor composition from stage to stage to (L/mV), which is known as the absorption factor. If either the operating or equilibrium line is curved, this simple relationship no longer holds and a simple analytical solution does not exist.

The difference between inlet and outlet gas concentrations must be the sum of the θy_j values shown in Figure 12-10. Thus,

(12-24a)

Applying Eq. (12-23)

(12-24b)

The summation in Eq. (12-24b) can be calculated. The general formula is

(12-25)

Then Eq. (12-24b) is

(12-26)

If L/mV > 1, then divide both sides of Eq. (12-24b) by (L/mV)^N−1 and do the summation in terms of mV/L. The resulting equation will still be Eq. (12-26). From Eq. (12-23), θy₁ = θy₀L/mV where

(12-27)

Removal of θy₁ from Eq. (12-26) gives

(12-28)

Equation (12-28) is one form of the Kremser equation (Kremser, 1930; Souders and Brown, 1932). A large variety of alternative forms can be developed by algebraic manipulation. For instance, if we add 1 to both sides of Eq. (12-28) and rearrange, we have

(12-29)

which can be solved for N. After manipulation, this result is

(12-30)

where L/mV ≠ 1. Equations (12-29) and (12-30) are also known as forms of the Kremser equation. Alternative derivations of the Kremser equation are given by Brian (1972) and King (1980).

A variety of forms of the Kremser equation can be developed. Several alternative forms in terms of the gas-phase composition are

(12-31)

(12-32)

(12-33)

(12-34)

where

(12-35)

Alternative forms in terms of the liquid phase composition are

(12-36)

(12-37)

(12-38)

(12-39)

(12-40)

where

(12-41)

A form including a constant Murphree vapor efficiency is (King, 1980)

(12-42)

Forms for systems with three phases where two phases flow co-currently and countercurrent to the third phase were developed by Wankat (1980). Forms of the Kremser equation for columns with multiple sections are developed by Brian (1972, Chap. 3) and by King (1980, pp. 371–376). Forms for reboiled absorbers are given by Hwang et al. (1992a).

When the assumptions required for the derivation are valid, the Kremser equation has several advantages over the stage-by-stage calculation procedure. If the number of stages is large, the Kremser equation is much more convenient to use, and it is easy to program on a computer or calculator. When the number of stages is specified, the McCabe-Thiele stage-by-stage procedure is trial-and-error, but the use of the Kremser equation is not. Because calculations can be done faster, the effects of varying y₁, x₀, L/V, m etc. are easy to determine. The major disadvantage of the Kremser equation is that it is accurate only for dilute solutions where L/V is constant, equilibrium is linear, and the system is isothermal. The appropriate form of the Kremser equation depends on the context of the problem.

EXAMPLE 12.2 Stripping analysis with Kremser equation

A plate tower providing six equilibrium stages is employed for stripping ammonia from a wastewater stream by means of countercurrent air at atmospheric pressure and 80°F. Calculate the concentration of ammonia in the exit water if the inlet liquid concentration is 0.1 mole % ammonia in water, the inlet air is free of ammonia, and 30 standard cubic feet (scf) of air are fed to the tower per pound of wastewater.

Solution

A. Define. The column is sketched in the figure.

We wish to find the exit water concentration, x₆.

B. Explore. Since the concentrations are quite low we can use the Kremser equation. Equilibrium data are available in several sources. From King (1971, p. 273) we find y_NH3 = 1.414 x_NH3 at 80°F.

C. Plan. We have to convert flow to molar units. Since we want a concentration of liquid, forms (12-39) or (12-40) of the Kremser equation will be convenient. We will use Eq. (12-39).

D. Do it. We can calculate ratio V/L,

Note that the individual flow rates are not needed.

The Kremser equation [form (12-39)] is

Where x_N = x₆ is unknown, x₀ = 0.001, m = 1.414, b = 0, , V/L = 1.43, N = 6

Rearranging,

Most of the ammonia is stripped out by the air.

E. Check. We can check with a different form of the Kremser equation or by solving the results graphically; both give the same result. We should also check that the major assumptions of the Kremser equation (constant flow rates, linear equilibrium, and isothermal) are satisfied. In this dilute system they are.

F. Generalize. This problem is trial-and-error when it is solved graphically, but a graphical check is not trial-and-error. Also, the Kremser equation is very easy to set up on a computer or calculator. Thus, when it is applicable, the Kremser equation is very convenient.

12.6 DILUTE MULTISOLUTE ABSORBERS AND STRIPPERS

Up to this point we have been restricted to cases where there is a single solute to recover. Both the stage-by-stage McCabe-Thiele procedures and the Kremser equation can be used for multisolute absorption and stripping if certain assumptions are valid. The single-solute analysis by both procedures required systems that 1) are isothermal, 2) are isobaric, 3) have a negligible heat of absorption, and 4) have constant flow rates. These assumptions are again required.

To see what other assumptions are required consider the Gibbs phase rule for a system with three solutes plus a solvent and a carrier gas. The phase rule is

F = C − P + 2 = 5 − 2 + 2 = 5

Five degrees of freedom is a large number. In order to represent equilibrium as a single curve or in a linear form like Eq. (12-16), four of these degrees of freedom must be specified. Constant temperature and pressure utilize two degrees of freedom. The other two degrees of freedom can be specified by assuming that 5) solutes are independent of each other; in other words, that equilibrium for any solute does not depend on the amounts of other solutes present. This assumption requires dilute solutions. In addition, the analysis must be done in terms of mole or mass fractions and total flow rates for which dilute solutions are also required. An analysis using ratio units will not work because the ratio calculation

involves other solute concentrations that will be unknown.

The practical effect of the fifth assumption is that we can solve the multisolute problem once for each solute, treating each problem as a single-component problem. This is true for both the stage-by-stage solution method and the Kremser equation. Thus, for the absorber shown in Figure 12-11 we solve three single-solute problems.

Figure 12-11 Dilute multisolute absorber

Each additional solute increases the degrees of freedom for the absorber by two. These two degrees of freedom are required to specify the inlet gas and inlet liquid compositions. For the usual design problem, as shown in Figure 12-11, the inlet gas and inlet liquid compositions and flow rates will be specified. With temperature and pressure also specified, one degree of freedom is left. This is usually used to specify one of the outlet solute concentrations such as y_B,1. The design problem for solute B is now fully specified. To solve for the number of stages, we can plot the equilibrium data, which are of the form

(12-43)

on a McCabe-Thiele diagram. When assumptions 1 to 5 are satisfied, the equilibrium expression will usually be linear. The operating equation

(12-44)

is the same as Eq. (12-15) and can also be plotted on the McCabe-Thiele diagram. Then the number of stages is stepped off as usual. This is shown in Figure 12-12.

Figure 12-12 McCabe-Thiele solution for dilute three-solute absorber where solute B is specified, and solutes A and C are trial-and-error

Once the number of stages has been found from the solute B calculation, the concentrations of solutes A and C can be determined by solving two fully specified simulation problems. That is, the number of stages is known and the outlet compositions have to be calculated. Simulation problems require a trial-and-error procedure when a stage-by-stage calculation is used. One way to do this calculation for component A is:

1. Plot the A equilibrium curve, y_A = f_A(x_A).

2. Guess y_A,1 for solute A.

3. Plot the A operating line,

Slope = L/V, which is same as for solute B. Point (y_A,1, x_A,0) is on the operating line. This is shown in Figure 12-12.

4. Step off stages up to y_A,N+1 (see Figure 12-12).

5. Check: Are the number of stages the same as calculated? If yes, you have the answer. If no, return to step 2.

The procedure for solute C is the same.

The three diagrams shown in Figure 12-12 are often plotted on the same y−x graph. This saves paper but tends to be confusing.

If the equilibrium function for each solute is linear as in Eq. (12-16), the Kremser equation can be used. First the design problem for solute B is solved by using Eq. (12-30), (12-33), or (12-34) to find N. Then separately solve the two simulation problems (for solutes A and C) using equations such as (12-28), (12-29), or (12-32). Remember to use m_A and m_C when you use the Kremser equation. Note that when the Kremser equation can be used, the simulation problems are not trial-and-error.

These solution methods are restricted to very dilute solutions. In more concentrated solutions, flow rates are not constant, solutes may not have independent equilibria, and temperature effects become important. When this is true, more complicated computer solution methods involving simultaneous mass and energy balances plus equilibrium are required. These methods are discussed in the next section and the computer simulation is explored in the chapter appendix.

12.7 MATRIX SOLUTION FOR CONCENTRATED ABSORBERS AND STRIPPERS

For more concentrated solutions, absorbers, and strippers are usually not isothermal, total flow rates are not constant, and solutes may not be independent. The matrix methods discussed for multicomponent distillation (reread section 6.2) can be adapted for absorption and stripping.

Absorbers, like flash distillation, are equivalent to very wide boiling feeds. Thus, in contrast with distillation, a wide-boiling feed (sum rates) flowchart such as Figure 02-13 should be used. The flow rate loop is now solved first, since flow rates are never constant in absorbers. The energy balance, which requires the most information, is used to calculate new temperatures, since this is done last. Figure 12-13 shows the sum-rates flow diagram for absorbers and strippers when K_i = K_i(T, p). If K_i = K_i (T, p, x_i, x₂,… x_c) a concentration correction loop is added. The initial steps are very similar to those for distillation, and usually the same physical properties package is used.

Figure 12-13 Sum rates convergence procedure for absorption and stripping

The mass balance and equilibrium equations are very similar to those for distillation and the column is again numbered from the top down as shown in Figure 12-14. To fit into the matrix form, streams V_N+1 and L₀ are relabeled as feeds to stages N and 1, respectively. The stages from 2 to N − 1 have the same general shape as for distillation (Figure 06-3). Thus, the mass balances and the manipulations [Eqs. (06-01) to (06-06)] are the same for distillation and absorption.

Figure 12-14 Absorber nomenclature for matrix analysis

For stage 1 the mass balance becomes

(12-45)

where

(12-46)

and the component flow rates are l₁ = L₁x₁ and l₂ = L₂x₂. These equations are repeated for each component.

For stage N the mass balance is

(12-47)

where

(12-48)

This equation differs from the one for distillation, since stage 1 is an equilibrium stage in absorption, not a total condenser.

Combining Eqs. (12-45), (06-05), and (12-47) results in a tridiagonal matrix, Eq. (06-13), with all terms defined in Eqs. (12-46), (06-06), and (12-48). There is one matrix for each component. These tridiagonal matrices can each be inverted with the Thomas algorithm (Table 6-1). The results are liquid component flow rates, l_i,j, that are valid for the assumed L_j, V_j, and T_j.

The next step is to use the summation equations to find new total flow rates L_j and V_j. The new liquid flow rate is conveniently determined as

(12-49)

The vapor flow rates are determined by summing the component vapor flow rates,

(12-50)

Convergence can be checked with

(12-51)

for all stages. For computer calculations, an ε of 10⁻⁴ or 10⁻⁵ can be used. If convergence has not been reached, new liquid and vapor flow rates are determined, and we return to the component mass balances (see Figure 12-13). Direct substitution (L_j = L_j,new, V_j = V_j,new) is usually adequate.

Once the flow rate loop has converged, the energy balances are used to solve for the temperatures on each stage. This can be done in several different ways (King, 1980; Smith, 1963). We will discuss only a multivariate Newtonian convergence procedure. For the general stage j, the energy balance was given as Eq. (06-26). This can be rewritten as

(12-52)

The multivariate Newtonian approach is an extension of the single variable Newtonian convergence procedure. The change in the energy balance is

(12-53)

where k is the trial number. The θT values are defined as

(12-54)

The partial derivatives can be determined from Eq. (12-52) by determining which terms in the equation are direct functions of the stage temperatures T_j−1, T_j, and T_j+1. The partial derivatives are

(12-55a)

(12-55b)

(12-55c)

where we have identified the terms as A, B, and C terms for a matrix. The total stream heat capacities can be determined from individual component heat capacities. For ideal mixtures this is

(12-56)

For the next trial we hope to have (E_j)_k+1 = 0. If we define

(12-57)

where (E_j)_k is the numerical value of the energy balance for trial k on stage j, then the equations for θT_j can be written as

(12-58)

Equation (12-58) can be inverted using any computer inversion program or the Thomas algorithm shown in Table 6-1. The result will be all the θT_j values.

Convergence can be checked from the θT_j. If

(12-59)

for all stages, then convergence has been achieved. The problem is finished! If Eq. (12-59) is not satisfied, determine new temperatures from Eq. (12-54) and return to calculate new K values and redo the component mass balances (see Figure 12-13). A reasonable range for ε_T for computer solution is 10⁻² to 10⁻³. Computer solution methods are explored further in the appendix to this chapter.

12.8 IRREVERSIBLE ABSORPTION

Absorption with an irreversible chemical reaction is often used in small facilities for removing obnoxious chemicals. For example, NaOH is used to remove both CO₂ and H₂S; it reacts with the acid gas in solution and forms a nonvolatile salt. This is convenient in small facilities, because the absorber is usually small and simple and no regeneration facilities are required. However, the cost of the reactant (NaOH) can make operation expensive. In addition, the salt formed has to be disposed of responsibly. In large-scale systems it is usually cheaper to use a solvent that can be regenerated.

Consider a simple absorber where the gas to be treated contains carrier gas C and solute B. The solvent contains a nonvolatile solvent S and a reagent R that will react irreversibly with B according to the irreversible reaction

(12-60)

The resulting product RB is nonvolatile. At equilibrium, x_B (in the free form) = 0 since the reaction is irreversible and any B that dissolves will form product RB. Thus, y_B = 0 at equilibrium. As long as there is any reagent R present, the equilibrium expression is y_B = 0. From the stoichiometry of the reaction shown in Eq. (12-60), there will be reagent available as long as Lx_R,0 > Vy_B,N+1.

For a dilute countercurrent absorber, the mass balance was given by Eq. (12-15), where x is the total mole frac of B in the liquid (as free B and as bound RB). The operating and equilibrium diagrams can be plotted on a McCabe-Thiele diagram as shown in Figure 12-15. One equilibrium stage will give y₁ = 0, which is more than sufficient. Unfortunately, the stage efficiency is often very low because of low mass transfer rates of the solute into the liquid. If the Murphree vapor efficiency

(12-61)

is used in Figure 12-15, the number of real stages can be stepped off. Murphree vapor efficiencies less than 30% are common.

Figure 12-15 McCabe-Thiele diagram for countercurrent irreversible absorption

Since only one equilibrium stage is required, alternatives to countercurrent cascades may be preferable. A co-current cascade is shown in Figure 12-16A. Packed columns would normally be used for the co-current cascade. The advantage of the co-current cascade is that it cannot flood, so smaller diameter columns with higher vapor velocities can be used. The higher vapor velocities give higher mass transfer rates (see Chapter 15), and less packing will be required. The mass balance for the co-current system using the mass balance envelope shown in Figure 12-16A is

(12-62)

Figure 12-16 Co-current irreversible absorption; A) apparatus, B) McCabe-Thiele diagram

Solving for y, we obtain the operating equation

(12-63)

This is a straight line with a slope of − L/V, and is plotted in Figure 12-16B. At equilibrium, y_N = 0, and x_N can be found from the operating equation as shown in Figure 12-16B. When equilibrium is not attained, the system can be designed with the mass transfer analysis discussed in Chapter 15. Co-current absorbers are used commercially for irreversible absorption.

For reversible chemical absorption, the higher flow rates and lack of flooding available in co-current absorbers is still desirable, but one equilibrium contact is rarely sufficient. Connecting a co-current absorber and a countercurrent absorber in series or parallel can provide more flexibility for operation (Isom and Rogers, 1994). This combination is explored in problem 12.B3.

12.9 SUMMARY—OBJECTIVES

In this chapter we studied absorption and stripping. At the end of this chapter you should be able to achieve the following objectives:

1. Explain what absorption and stripping do and describe a complete gas treatment plant

2. Use the McCabe-Thiele method to analyze absorption and stripping systems for both concentrated and dilute systems

3. Design the column diameter for an absorber or stripper for staged columns and packed columns

4. Derive the Kremser equation for dilute systems

5. Use the Kremser equation for dilute absorption and stripping problems

6. Solve problems for dilute multicomponent absorbers and strippers both graphically and analytically

7. Use the matrix solution method for nonisothermal multicomponent absorption or stripping

8. Discuss how irreversible absorption differs from reversible systems, and design irreversible absorbers

REFERENCES

Astarita, G., D.W. Savage, and A. Bisio, Gas Treating with Chemical Solvents, Wiley, New York, 1983.

Ball, T. and R. Veldman, “Improve Gas Treating,” Chem. Engr. Progress, 87 (1) 67 (Jan. 1991).

Brian, P. L. T., Staged Cascades in Chemical Processes, Prentice Hall, Upper Saddle River, New Jersey, 1972.

Hwang, S. T., “Tray and Packing Efficiencies at Extremely Low Concentrations,” in N. N. Li (Ed.), Recent Developments in Separation Science, Vol. 6, CRC Press, Boca Raton, FL, 1981, pp. 137–148.

Hwang, Y. L., G. E. Keller II, and J. D. Olson, “Steam Stripping for Removal of Organic Pollutants from Water. 1. Stripping Effectiveness and Stripping Design,” p. 1753; “Part 2, Vapor-Liquid Equilibrium Data,” p. 1759, Ind. Eng. Chem. Research, 31, (1992 a, b).

Isom, C. and J. Rogers, “Sour Gas Treatment Gets More Flexible,” Chem. Engr., 147 (July 1994).

Kessler, D. P. and P. C. Wankat, “Correlations for Column Parameters,” Chem Engr., 72 (Sept. 26, 1988).

King, C. J., Separation Processes, McGraw-Hill, New York, 1971.

King, C. J., Separation Processes, 2^nd ed., McGraw-Hill, New York, 1981.

Kohl, A. L., “Absorption and Stripping,” in R. W. Rousseau (Ed.), Handbook of Separation Process Technology, Wiley, New York, 1987, Chap. 6.

Kohl, A. L. and R. B. Nielsen, Gas Purification, 5^th ed., Gulf Publishing Co., Houston, 1997.

Kremser, A., Nat. Petrol. News, 43 (May 30, 1930).

O’Connell, H. E., “Plate Efficiency of Fractionating Columns and Absorbers,” Trans. AIChE, 42, 741 (1946).

Perry, R. H., C. H. Chilton and S. D. Kirkpatrick (Eds.), Chemical Engineer’s Handbook, 4^th ed., McGraw-Hill, New York, 1963.

Perry, R. H. and C. H. Chilton (Eds.), Chemical Engineer’s Handbook, 5^th ed., McGraw-Hill, New York, 1973.

Perry, R. H. and D. W. Green (Eds.), Perry’s Chemical Engineers’ Handbook, 7^th ed., McGraw-Hill, New York, 1997.

Reynolds, J., J. Jeris and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, Wiley, New York, 2002.

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

Socolow, R. H., “Can We Bury Global Warming?” Scientific American, 49–55 (July 2005).

Souders, M. and G. G. Brown, “Fundamental Design of High Pressure Equipment Involving Paraffin Hydrocarbons. IV. Fundamental Design of Absorbing and Stripping Columns for Complex Vapors,” Ind. Eng. Chem., 24, 519 (1932).

Yaws, C. L., P. K. Narasimhan, H. H. Lou, and R. W. Pike, “Solubility & Henry’s Law Constants for Chlorinated Compounds in Water,” Chem. Engr., 50 (Feb. 2005).

Zarzycki, R. and A. Chacuk, Absorption: Fundamentals & Applications, Pergamon Press, Oxford, 1993.

Zenz, F. A., “Design of Gas Absorption Towers,” in P. A. Schweitzer (Ed.), Handbook of Separation Techniques for Chemical Engineers, 3^rd ed., McGraw-Hill, New York, 1997, Section 3.2.

HOMEWORK

A . Discussion Problems

A1. How can the direction of mass transfer be reversed as it is in a complete gas plant? What controls whether a column is a stripper or an absorber?

A2. Why is the Murphree efficiency often lower in chemical absorption than in physical absorption? (What additional resistances are present?)

A3. After reviewing Chapter 10, outline the method of determining the column diameter for an absorber or stripper in:

a. A plate column

b. A packed column

A4. As the system becomes dilute, L/G→L/V, Y → y, and X → x. At what concentration levels could you safely work in terms of fractions and total flows instead of ratios and flows of solvent and carrier gas? What variable will this depend on? Explore numerically. See also Problem 12-C3.

A5. How can the gas plant in Figure 12-2 be made thermodynamically more efficient with heat exchange? Refer to Chapter 11 for information on heat exchange in distillation.

A6. Explain the significance of the curves in Figure 12-7. Sketch a plot of Lx and Vy vs. stage location.

A7. Equation (12-16) has an extra b term that doesn’t appear in Henry’s law. In Example 12-2, b = 0. When might it be useful to have a nonzero b term?

A8. Explain how the single assumption that “solutes are independent of each other” can specify more than one degree of freedom.

A9. Develop your key relations chart for this chapter.

B . Generation of Alternatives

B1. The Kremser equation can be used for more than just determining the number of stages. List as many types of problems (where a different variable is solved for) as you can. What variables would be specified? How would you solve the equation?

B2. Many other configurations of absorbers and strippers can be devised. For example, there could be two feeds. Generate as many as possible.

B3. You want to use both co-current and countercurrent absorbers in a process. Sketch as many ways of doing this as you can think of. What are the advantages and disadvantages of each method?

C . Derivations

C1. Derive Eq. (12-30) starting with Eq. (12-29).

C2. Derive Eq. (12-39). Follow the procedure used to derive Eqs. (12-28) and (12-29) except work in terms of θx. Thus, start by writing the equilibrium and operating equations in form x = ….

C3. Plot a graph of ratio units (Y or X) vs. fraction units (y or x) to use for converting units.

C4. Derive an operating equation similar to Eq. (12-09), but draw your balance envelope around the bottom of the column. Show that the result is equivalent to Eq. (12-09).

C5. Complete the derivation of Eq. (12-58) by doing the steps outlined in the derivation.

C6. Derive Eq. (10-04), which relates the overall efficiency to the Murphree vapor efficiency, for dilute systems by determining N_equil and N_actual in Eq. (10-01) from appropriate forms of the Kremser equation.

C7. Occasionally it is useful to apply the Kremser equation to systems with a constant relative volatility. Where on the y vs. x diagram for distillation can you do this? Derive the appropriate values for m and b in the two regions where the Kremser equation can be applied.

C8. For dilute systems show that (L/V)_min for absorbers and (L/V)_max for strippers calculated from the Kremser equation agrees with a graphical calculation.

D . Problems

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

D1. Acetone is being absorbed from air using water as the solvent. Operation is at 1 atm and is isothermal at 20°C. The total flow rate of entering gas is 100 kmoles/hr. The entering gas is 12 mole % acetone. Pure water is used as solvent. The water flow rate is 150 kmoles/hr. We desire an outlet gas concentration of 1 mole %. The equilibrium data are:

Mole frac acetone in liquid, x	mole frac acetone in gas, y
0	0
0.0333	0.0395
0.0720	0.0826
0.1170	0.1124

a. Find the mole ratio of acetone in the outlet liquid.

b. Find the number of equilibrium stages required including a fractional number.

Work in mole ratio units.

D2. We are absorbing hydrogen sulfide at 15°C into water. The entering water is pure. The feed gas contains 0.0012 mole frac hydrogen sulfide and we want to remove 97% of this in the water. The total gas flow rate is 10 kg moles/hr. The total liquid flow rate is 2,000 kg moles/hr. Total pressure is 2.5 atm. You can assume that total liquid and gas flow rates are constant. Equilibrium data is:

Partial pressure of hydrogen sulfide (atm) = 423 x

x is the mole frac of hydrogen sulfide in the water. (Perry’s 7^th ed., p. 2–127).

a. Calculate the outlet gas and liquid mole fracs of hydrogen sulfide.

b. Calculate the number of equilibrium stages required using a McCabe-Thiele diagram.

c. If L/V = M x (L/V)_min, find the multiplier M (M>1).

d. Why is this operation not practical? What would an engineer do to develop a practical process?

D3. We wish to strip a dilute amount of hydrogen sulfide from water using air. The system is operated warm and at lower pressure to increase the removal of hydrogen sulfide. The stripper is at 45°C and 0.98 atm. The inlet liquid mole frac is 7.00E-6 hydrogen sulfide. Inlet liquid flow rate is 100 kg moles/hr. We desire to remove 98% of the hydrogen sulfide. The inlet gas has y = 0.00. Inlet gas flow rate is 0.20 kgmoles/hr. You can assume that total liquid and gas flow rates are constant. Equilibrium data is:

partial pressure of hydrogen sulfide (atm) = 745 x

x is the mole frac of hydrogen sulfide in water. (Perry, 7^th ed., p. 2–127)

a. Calculate the outlet gas and liquid mole fracs of hydrogen sulfide.

b. Calculate the number of equilibrium stages required using a McCabe-Thiele diagram.

c. If L/V = M x (L/V)_max, find the multiplier M (M < 1).

d. Why would this column be difficult to operate?

D4. We have a steam stripper operating isothermally at 100°C. The entering liquid stream contains 0.0002 mole frac nitrobenzene in water at 100°C. Flow rate of entering liquid is 1 kmole/minute. The entering steam is pure water at 100 C. We desire an outlet liquid mole frac of 0.00001 nitrobenzene. L/V = 12.0. You can assume that total liquid and gas flow rates are constant. At 100°C equilibrium in terms of nitrobenzene mole frac is y = 28 x. Find the outlet mole frac of the nitrobenzene in the vapor stream and the number of stages.

D5. *A packed column 3-inches in diameter with 10 feet of Intalox saddle packing is being run in the laboratory. P is being stripped from nC₉ using methane gas. The methane can be assumed to be insoluble and the nC₉ is nonvolatile. Operation is isothermal. The laboratory test results are:

Equilibrium data can be approximated as Y = 1.5X. Find the HETP for the packing.

D6. *We wish to design a stripping column to remove carbon dioxide from water. This is done by heating the water and passing it countercurrent to a nitrogen stream in a staged stripper. Operation is isothermal and isobaric at 60°C and 1 atm pressure. The water contains 9.2 × 10⁻⁶ mole frac CO₂ and flows at 100,000 lb/hr. Nitrogen (N₂) enters the column as pure nitrogen and flows at 2500 ft³/hr. Nitrogen is at 1 atm and 60°C. We desire an outlet water concentration that is 2 × 10⁻⁷ mole frac CO₂. Ignore nitrogen solubility in water and ignore the volatility of the water. Equilibrium data are in Table 12-1. Use a Murphree vapor efficiency of 40%. Find outlet vapor composition and number of real stages needed.

D7. We wish to absorb ammonia from an air stream using water at 0°C and a total pressure of 1.30 atm. The entering water stream is pure water. The entering vapor is 17.2 wt % ammonia. We desire to recover 98% of the ammonia in the water outlet stream. The gas flow rate is 1050 kg/hour. We want to use a solvent rate that is 1.5 times the minimum solvent rate. Assume that temperature is constant at 0°C, water is nonvolatile, and air does not dissolve in water. Equilibrium data are available in Table 12-3.

D8. 10.9 kg moles/hr of a concentrated aqueous solution of ammonia is to be processed in a stripping column. The feed contains 0.90 kg moles of ammonia/hr and 10 kg moles of water/hr. We desire an outlet water stream that is 0.010 mole frac NH₃. The stripping gas is pure air at a flow rate of 9 kg moles/hr. Operation is at 30°C and 745 mm Hg. Assume the air is insoluble, the water is nonvolatile and the stripper is isothermal. Equilibrium data are in Table 12-3. Molecular weights: ammonia = 17, water = 18, air = 29.

a. Find the number of equilibrium stages required including the fractional number.

b. Find the minimum air flow rate.

Note: This should be analyzed as a concentrated NOT dilute system.

Note: Watch your units! This problem can be solved in mole ratios, weight ratios, or (with care) in mixed ratio units.

D9. Problems 12.D2 and 12.D3 satisfy the criteria for using the Kremser equation.

a. Repeat problem 12.D2b, but use the Kremser equation. Compare your answer with the McCabe-Thiele solution.

b. Repeat problem 12.D3b, but use the Kremser equation. Compare your answer with the McCabe-Thiele solution.

D10. *A stripping tower with four equilibrium stages is being used to remove ammonia from wastewater using air as the stripping agent. Operation is at 80°F and 1 atm. The inlet air is pure air, and the inlet water contains 0.02 mole frac ammonia. The column operates at L/V = 0.65. Equilibrium data in mole fracs are given as y = 1.414x. Find the outlet concentrations.

D11. *An absorption column for laboratory use has been carefully constructed so that it has exactly 4 equilibrium stages and is being used to measure equilibrium data. Water is used as the solvent to absorb ammonia from air. The system operates isothermally at 80°F and 1 atm. The inlet water is pure distilled water. The ratio of L/V = 1.2, inlet gas concentration is 0.01 mole frac ammonia, and the measured outlet gas concentration is 0.0027 mole frac ammonia. Assuming that equilibrium is of the form y = mx, calculate the value of m for ammonia. Check your result.

D12. *Read the section on cross flow in Chapter 13 before proceeding. We wish to strip CO₂ from a liquid solvent using air as the carrier gas. Since the air and CO₂ mixtures will be vented and since cross flow has a lower pressure drop, we will use a cross-flow system. The inlet liquid is 20.4 wt % CO₂, and the total inlet liquid flow rate is 1,000 kghr. We desire an outlet liquid composition that is 2.5 wt % (0.025 wt frac) CO₂. The gas flow to each stage is 25,190 kg air/hr. In the last stage a special purified air is used that has no CO₂. Find the number of equilibrium stages required. In weight fraction units, equilibrium is y = 0.04x. Use unequal axes for your McCabe-Thiele diagram. Note: With current environmental concerns, venting the CO₂ is not a good idea even if it is legal.

D13. We plan to absorb a feed that is 15 mole % HCl in air with water at 10°C and a pressure of 3 atm. The total feed gas flow rate is 100 kg moles/hr. We want to remove 98% of the HCl from the air. The absorber is well cooled and thus, is isothermal. Assume that the air is insoluble and the water is nonvolatile. The inlet water is pure. Total flow rates are not constant. Operate with a (water flow rate)/(air flow rate) = 1.4(water flow rate)/(air flow rate)_min. Equilibrium data from Perry’s 7^th ed., p. 2–127:

Note: You need to use ratio units. However, if careful, the liquid units can be in mass and the gas units in moles (this is the form of the equilibrium data). Derive the operating equation and external mass balances to determine where to include the molecular weight of HCl (36.46).

a. Find the outlet amount of HCl in the gas.

b. Find (L/G)_min and L/G.

c. Find the outlet concentration of HCl in the liquid.

d. Find the number of equilibrium stages needed.

D14. In an ammonia plant we wish to absorb traces of argon and methane from a nitrogen stream using liquid ammonia. Operation is at 253.2 K and 175 atm pressure. The feed flow rate of gas is 100 kgmoles/hr. The gas contains 0.00024 mole frac argon and 0.00129 mole frac methane. We desire to remove 95% of the methane. The entering liquid ammonia is pure. Operate with L/V = 1.4(L/V)_min. Assume the total gas and liquid rates are constant. The equilibrium data at 253.2 K are:

Methane: partial pressure methane atm = 3600(methane mole frac in liquid)

Argon: partial pressure argon atm = 7700(argon mole frac in liquid)

Reference: Alesandrini et al., Ind. Eng. Chem. Process Design and Develop., 11, 253 (1972)

a. Find outlet methane mole frac in the gas.

b. Find (L/V)_min and actual L/V.

c. Find outlet methane mole frac in the liquid.

d. Find the number of equilibrium stages required.

e. Find the outlet argon mole fracs in the liquid and the gas, and the % recovery of argon in the liquid.

D15. *We want to remove traces of propane and n-butane from a hydrogen stream by absorbing them into a heavy oil. The feed is 150 kg moles/hr of a gas that is 0.0017 mole frac propane and 0.0006 mole frac butane. We desire to recover 98.8% of the butane. Assume that H₂ is insoluble. The heavy oil enters as a pure stream (which is approximately C₁₀ and can be assumed to be nonvolatile). Liquid flow rate is 300 kg moles/hr. Operation is at 700 kPa and 20° C. K values can be obtained from the DePriester charts. Find the equilibrium number of stages required and the compositions of gas and liquid streams leaving the absorber.

D16. We wish to strip CO₂ out of water at 20°C and 2 atm pressure using a staged, countercurrent stripper. The liquid flow rate is 100 kgmoles per hour of water and the initial CO₂ mole frac in the water is 0.00005. The inlet air stream contains no CO₂. A 98.4 % removal of CO₂ from the water is desired. The Henry’s law constant for CO₂ in water at 20°C is 1420 atm.

a. Find the outlet CO₂ mole frac in the water.

b. Find V_min (for N = ∞).

c. If there are 7 equilibrium stages, find V and the outlet mole frac CO₂ in the gas.

D17. *We wish to absorb ammonia from air into water. Equilibrium data are given as y_NH3 = 1.414x_NH3 in mole fracs. The countercurrent column has three equilibrium stages. The entering air stream has a total flow rate of 10 kg moles/hr and is 0.0083 mole frac NH₃. The inlet water stream contains 0.0002 mole frac ammonia. We desire an outlet gas stream with 0.0005 mole frac ammonia. Find the required liquid flow rate, L.

D18. Dilute amounts of ammonia are to be absorbed from two air streams into water. The absorber operates at 30°C and 2 atm pressure, and the equilibrium expression is y = 0.596 x where y is mole frac of ammonia in the gas and x is mole frac of ammonia in the liquid. The solvent is pure water and flows at 100 kmoles/hr. The main gas stream to be treated has a mole frac ammonia of y_N+1 = 0.0058 (the remainder is air) and flow rate of V_N+1 = 100 kmoles/hr. The second gas stream is input as a feed in the column with a mole frac ammonia of y_F = 0.003 and flow rate of V_F = 50 kmoles/hr. We desire an outlet ammonia mole frac in the gas stream of y₁ = 0.0004. Find:

a. The outlet mole frac of ammonia in the liquid, x_N.

b. The optimum feed stage for the gas of mole frac y_F, and the total number of stages needed.

c. The minimum solvent flow rate, L_min.

D19. You have a water feed at 25°C and 1 atm. The water has been in contact with air and can be assumed to be in equilibrium with the normal CO₂ content of air (about 0.035 volume %) (Note: This dissolved CO₂ will decrease the pH of the water.) We wish to strip out the CO₂. A counter current stripping column will be used operating at 25°C and 50 mm Hg pressure. The stripping gas used will be nitrogen gas saturated with pure water at 25°C. Assume the nitrogen is insoluble in water. Assume ideal gas (vol % = mole %). Data: See Table 12-1. We want to remove 95% of the initial CO₂ in the water.

a. Calculate the inlet and outlet mole fracs of CO₂ in water.

b. Determine (L/G)_max. If L = 1 kgmolehr, calculate G_min (corresponds to (L/G)_max).

c. If we operate at G = 1.5 × G_min find the CO₂ mole frac in the outlet nitrogen and the number of equilibrium stages needed.

D20. We will strip acetone from water into an air stream. Operation is isothermal at 20°C and at a total pressure of 278 mm Hg. The entering liquid is 16 mole % acetone. The entering liquid flow rate is 10 kg moles/hr. We desire the exiting liquid to be 1 mole % acetone. The entering air is pure. Assume that the water is nonvolatile and the air is insoluble. Equilibrium data at 20°C:

a. Find the minimum flow rate of air.

b. If the air flow rate is 4 kgmole/hr, find the number of equilibrium stages.

D21. We are studying the stripping of dilute quantities of nitrobenzene from water using steam. The stripping column and all streams are at 100°C and 1 atm. We have a column that has 4 real stages. The entering liquid water contains 0.00035 mole frac nitrobenzene. The entering steam is pure water vapor. We operate at L/V = 9.5. The exiting gas stream is 0.00292 mole frac nitrobenzene. The equilibrium expression is y = 28 x (y and x are mole fracs nitrobenzene). Find the value of the Murphree vapor efficiency.

D22. We need to remove H₂S and CO₂ from 1,000 kg mole/hr of a water stream at 0°C and 15.5 atm. The inlet liquid contains 0.000024 mole frac H₂S and 0.000038 mole frac CO₂. We desire a 99% recovery of the H₂S in the gas stream. The gas used is pure nitrogen at 0°C and 15.5 atm. The nitrogen flow rate is 3.44 kg moles/hr. A staged countercurrent stripper will be used. Assume water flow rate and air flow rate are constant. At 0°C: H_H2S = 26,800 (units are atm) and H_CO2 = 728 (units are atm). Note: Watch your decimals.

a. Determine the H₂S mole-fractions in the outlet gas and liquid streams.

b. Determine the number of equilibrium stages required.

c. Determine the CO₂ mole frac in the outlet liquid stream.

D23. *A complete gas treatment plant often consists of both an absorber to remove the solute and a stripper to regenerate the solvent. Some of the treated gas is heated and is used in the stripper. This is called stream B. In a particular application we wish to remove obnoxious impurity A from the inlet gas. The absorber operates at 1.5 atm and 24°C where equilibrium is given as y = 0.5 x (units are mole fracs). The stripper operates at 1 atm and 92°C where equilibrium is y = 3.0 x (units are mole fracs). The total gas flow rate is 1,400 moles/day, and the gas is 15 mole % A. The nonsoluble carrier is air. We desire a treated gas concentration of 0.5 mole % A. The liquid flow rate into the absorber is 800 moles/day and the liquid is 0.5 mole % A.

a. Calculate the number of stages in the absorber and the liquid concentration leaving.

b. If the stripper is an already existing column with four equilibrium stages, calculate the gas flow rate of stream B (concentration is 0.5 mole % A) and the outlet gas concentration from the stripper.

E . More Complex Problems

E1. A laboratory steam stripper with 11 real stages is used to remove 1,000 ppm (wt.) nitrobenzene from an aqueous feed stream which enters at 97°C. The flow rate of the liquid feed stream is L_in = F = 1726 g/hr. The entering steam rate was measured as S = 99 g/hr. The leaving vapor rate was measured as V_out = 61.8 g/hr. Column pressure was 1 atm. The treated water was measured at 28.1 ppm nitrobenzene. Data at 100°C is in problem 12.D21. Molecular weights are 123.11 and 18.016 for nitrobenzene and water, respectively. What is the overall efficiency of this column?

Note: A significant amount of steam condenses in this system to heat the liquid feed to its boiling point and to replace significant heat losses. An approximate solution can be obtained by assuming all this condensation occurs on the top stage, ignoring condensation of nitrobenzene, and adjusting the liquid flow rate in the column.

F . Problems Requiring Other Resources

F1. Laboratory tests are being made prior to the design of an absorption column to absorb bromine (Br₂) from air into water. Tests were made in a laboratory-packed column that is 6 inches in diameter, has 5 feet of packing, and is packed with saddles. The column was operated at 20°C and 5 atm total pressure, and the following data were obtained:

Inlet solvent is pure water.

Inlet gas is 0.02 mole frac bromine in air.

Exit gas is 0.002 mole frac bromine in air.

Exit liquid is 0.001 mole frac bromine in water.

What is the L/G ratio for this system? (Base your answer on flows of pure carrier gas and pure solvent.) What is the HETP obtained at these experimental conditions? Henry’s law constant data are given in Perry’s (4th ed) on pages 14–2 to 14–12. Note: Use mole ratio units. Assume that water is nonvolatile and air is insoluble.

F2. *(Difficult) An absorber with three equilibrium stages is operating at 1 atm. The feed is 10 moles/hr of a 60 mole % ethane, 40 mole % n-pentane mixture and enters at 30° F. The solvent used is pure n-octane at 70° F, solvent flow rate is 20 moles/hr. We desire to find all the outlet compositions and temperatures. The column is insulated. For a first guess assume that all stages are at 70° F. As a first guess on flow rates, assume:

L0 = 20.0	V1 = 6.00 moles/hr
L1 = 20.2	V2 = 6.18
L2 = 20.8	V3 = 6.66
L3 = 24.0	V4 = 10.0

Then go through one iteration of the sum rates convergence procedure (Figure 12-13) using direct substitution to estimate new flow rates on each stage. You could use these new flow rates for a second iteration, but instead of doing a second iteration of the flow loop use a paired simultaneous convergence routine. To do this, use the new values for liquid and vapor flow rates to find compositions on each stage. Then calculate enthalpies and use the multivariable Newtonian method to calculate new temperatures on each stage. You will then be ready to recalculate K values and solve the mass balances for the second iteration. However, for purposes of this assignment stop after the new temperatures have been estimated. Use a DePriester chart for K values. Pure-component enthalpies are given in Maxwell (1950) and on pages 629 and 630 of Smith (1963). Assume ideal solution behavior to find the enthalpy of each stream.

F3. Global warming is very much in the news. Engineers will need to be heavily involved in methods to control global warming. One approach is to capture carbon dioxide and bury it. Read Socolow’s (2005) article. Write a one page engineering analysis of the feasibility of using absorption to capture carbon dioxide. In your analysis explain why capturing carbon dioxide from the flue gases of large power plants is considered to be more feasible than capturing carbon dioxide from automobile exhaust or from air.

G . Computer Problems

G1. Do the following problem with a process simulator. A feed gas at 1 atm and 30°C is 90 mole % air and 10 mole % ammonia. Flow rate is 200 kmoles/hr. The ammonia is being absorbed in an absorber operating at 1 atm using water at 25°C as the solvent. We desire an outlet ammonia in the exiting air that is 0.0032 or less. The column is adiabatic.

a. If N = 4, what L is required (± 10 kmoles/hr)?

b. If N = 8, what L is required (± 10 kmoles/hr)?

c. If N = 16, what L is required (± 10 kmoles/hr)?

d. Examine your answers. Why does N = 16 not decrease L more?

G2. Solve the following problem with a process simulator. We wish to absorb two gas streams in an absorber. The main gas stream (stream A) is at 15°C, 2.5 atm, and has a flow rate of 100 kmoles/hr. Stream A is 0.90 mole frac methane, 0.06 mole frac n-butane, and 0.04 mole frac n-pentane. The other gas stream (stream B) is at 10°C, 2.5 atm, and has a flow rate of 75 kmoles/hr. Stream B is 0.99 mole frac methane, 0.009 mole frac n-butane, and 0.001 mole frac n-pentane. The liquid solvent fed to the absorber is at 15°C, 2.5 atm, and has a flow rate of 200 kmoles/hr. This stream is 0.999 mole frac n-decane and 0.001 mole frac n-pentane. Absorber pressure is 2.5 atm. We desire an outlet gas stream that is 0.999 mole frac methane or slightly higher.

Determine the total number of stages and the optimum feed stages for Streams A and B required to just achieve the desired methane purity of the outlet gas stream. (Use “on stage” for all of the feeds.) Report the following information:

a. Total number of stages required _________________________

b. Feed stage location for the solvent _______________________

c. Feed stage location for stream A _________________________

d. Feed stage location for stream B _________________________

e. Outlet mole fracs of gas stream leaving absorber________________

f. Outlet mole fracs of liquid leaving absorber ____________________

g. Outlet gas flow rate ___________________ kmoles/hr

h. Outlet liquid flow rate __________________ kmoles/hr

i. Highest temperature in column _______°C and stage it occurs on__________

G3. We wish to strip two liquid streams in a stripper. The main liquid stream (stream A) is at 3°C, 1 atm, and has a flow rate of 100 kmoles/hr. Stream A is 0.90 mole frac n-nonane (9 carbons), 0.04 mole frac n-butane, and 0.06 mole frac n-pentane. The other liquid stream (stream B) is at 30°C, 1 atm, and has a flow rate of 50 kmoles/hr. Stream B is 0.99 mole frac n-nonane, 0.001 mole frac n-butane, and 0.009 mole frac n-pentane. The gas fed to the stripper is at 30°C, 1 atm, and has a flow rate of 250 kmoles/hr. This stream is 0.999 mole frac methane and 0.001 mole frac n-butane. Stripper pressure is 1 atm. We desire an outlet liquid stream that is 0.98 mole frac n-nonane or slightly higher.

Determine the total number of stages and the optimum feed stages for Streams A and B required to just achieve the desired n-nonane purity of the outlet liquid stream. (Use “on stage” for all of the feeds.) Report the following information:

a. Total number of stages required _________________________

b. Feed stage location for the gas stream _______________________

c. Feed stage location for stream A _________________________

d. Feed stage location for stream B _________________________

e. Outlet mole fracs of gas stream leaving stripper______________________

f. Outlet mole fracs of liquid leaving stripper __________________________

g. Outlet gas flow rate ___________________ kmoles/hr

h. Outlet liquid flow rate __________________ kmoles/hr

i. Lowest temperature in column _______°C and stage it occurs on___________

CHAPTER 12 APPENDIX COMPUTER SIMULATIONS FOR ABSORPTION AND STRIPPING

This appendix follows the instructions in the appendices to Chapters 2 and 6. Although the Aspen Plus simulator is referred to, other process simulators can be used. If difficulties are encountered while running the simulator, see Appendix: Aspen Plus Separations Troubleshooting Guide that follows Chapter 17.

Lab 9. Aspen Plus uses RADFRAC for absorber calculations. If you want more information, once you are logged into Aspen Plus you can go to Help, look in the index, and click on Absorbers-RADFRAC. Note: Help may be useful for other applications of Aspen Plus.

To draw an absorber, use the RADFRAC icon. Draw a system with a vapor feed at the bottom, a liquid bottoms product, a second feed (liquid) at the top, and a vapor distillate. In the configuration window for the RADFRAC block set:

Condenser	None
Reboiler	None
Convergence	Petroleum/wide boiling

Go to convergence for the block (probably easiest to do using the flow diagram in color on the left side of the Aspen Plus screen). Click on the name of the block for your absorber. Then click on the blue check next to Convergence. In the Basic window the algorithm should be listed as Sum Rates. Set the maximum iterations at 50.

Input your components, pick the physical properties package, input conditions for the liquid and vapor feed streams (note that Aspen will determine if a stream is a liquid or a vapor), set the pressure and the number of stages for the absorber. Supply Aspen Plus the locations of the two feed streams:

In the streams window for the RADFRAC block set: Liquid feed above stage 1, and Vapor feed on the last stage.

For each simulation, determine the outlet gas and outlet liquid mole fracs, outlet flow rates, and temperatures on the top and bottom stages. Note if there is a temperature maximum.

The purpose of parts 1 and 2 is to explore how different variables affect the operation of absorbers and strippers. Desired specification is 0.003 mole frac acetone in the exiting air.

1. Absorber: a . You wish to absorb acetone from air into water. The column and feed streams are at 1 atm. The inlet gas stream is 3 mole % acetone. Flow of inlet gas is 100 kmoles/hr. Inlet gas is at 30°C. Water flow rate is 200 kmoles/hr. Inlet water temperature is 20°C. N = 6. Find the outlet concentrations and flow rates.

Results: Look at the temperature profile and note the temperature maximum. Look at the concentration profiles. Does the outlet vapor meet the acetone requirement? (It should.) Where does all the water in the outlet gas come from? Note how dilute the outlet liquid is. What can we do to increase this mole frac?

b . Decrease the water flow rate to 100 kmoles/hr and run again. Do you meet the required acetone concentration? (It shouldn’t.) But note that the outlet water is more concentrated.

c . Double the number of stages to 12 with L = 100. Does this help reduce the outlet vapor mole frac of acetone significantly? (Probably not.)

d . Return to N = 6 with L = 100. Reduce the temperature of both the feed gas and the inlet water to 10°C. Did this allow you to meet the outlet specifications for air? (Answer depends on the vapor-liquid equilibrium package you used.) Did the amount of water in the air decrease? (Why?) Do you see much change in the outlet concentration of the liquid stream?

2. Stripper. The same flowchart can be used as the stripper, or you can attach a heater and then a stripper to the liquid outlet from the absorber. The liquid feed to the stripper should be the liquid product from the absorber, but heat it to a saturated liquid (at 1 atm) first. The gas feed to the stripper should be pure steam, at 1 atm. Use 6 stages. Be sure to put your steam (the stripping gas) on stage 6 and the liquid from the absorber should be put in above stage 1.

a . Set the steam flow rate at 20 kmoles/hr. The steam should be superheated to 105°C. If the specification is liquid leaving should be < 1 E -04, does this design satisfy the spec? (It should.) What is the acetone mole frac in the gas leaving the stripper? Note that this is high enough that pure acetone could be recovered by distillation.

b . Repeat with steam at 1 atm and superheated to 101°C.

c . Repeat with saturated steam at 1 atm and 100°C. (If this won’t run, why not?)

d . Using 1 atm steam superheated to 105°C, reduce the steam flow rate to just satisfy the specification on the outlet liquid composition.

e . Try reducing the pressure of the stripper and the steam to 0.5 bar with steam at 83°C and 15 kmoles/hr of steam. What can you conclude about the effect of pressure?

3. Complete gas plant. Design a complete gas plant (similar to Figure 12-2 but without solvent recycle) to process the feed in problem 1a, but use a gas temperature of 10°C. Absorber has 6 stages. The outlet mole frac of acetone leaving in the gas should be less than 0.003 mole frac. The solvent fed to the absorber is pure water at 10°C. Flow rate is 100 kmoles/hr. The liquid outlet from the absorber should be heated and then be sent directly to a stripper (N = 6) operating at 1 atm with a gas feed of 1 atm steam superheated to 105°C. The acetone mole frac leaving in the liquid from the stripper should be less than 0.0001. Adjust the steam flow rate until your outlet is slightly below this value. The flowsheet for the gas plant should be similar to Figure 12-A1.

FIGURE 12-A1. Aspen Plus screen shot of gas plant