Chapter 12. Absorption and Stripping

Thus far, we have talked almost entirely about distillation. There are other unit operations that are very useful in processing chemicals and pollution control. Absorption is a unit operation in which one or more components of a gas stream are removed by being taken up (absorbed) in a nonvolatile liquid (solvent). Liquid solvent must be added as a separating agent. Absorption is one method used to remove CO₂ from natural gas and flue gases so that CO₂ is not added to the atmosphere where it is probably one cause of climate change (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. In this operation the gas stream (stripping agent) must be added as a separating agent.

What is the separating agent for distillation? Energy.

Absorption can be either physical or chemical. In physical absorption gas is removed because it has greater solubility in solvent than other gases. An example is the removal of butane and pentane (C₄ and C₅) from a refinery gas mixture with heavy oil. In chemical absorption gas to be removed reacts with the solvent and remains in the solution. An example is the removal of CO₂ or H₂S by reaction with NaOH or monoethanolamine (MEA). The reaction can be either irreversible (as with NaOH) or reversible (as with MEA). In 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 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), which must be taken into account. Chemical absorption systems are discussed in more detail by Kister et al. (2008) and Kohl and Nielsen (1997).

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 equipment to be quite similar. This is indeed the case; both absorption and stripping are operated in packed and tray towers. Tray towers can be designed following an adaptation of the McCabe-Thiele method. Packed towers can be designed using HETP or preferably mass transfer analysis (see Chapter 16).

In both absorption and stripping a separate phase is added as a separating agent. Thus, columns are simpler than distillation columns because normally there are no reboilers or condensers. Figure 12-1 is a schematic of a typical absorption column. Solute entering with insoluble carrier gas in the inlet gas stream is absorbed into nonvolatile solvent.

A gas treatment plant often has both absorption and stripping columns as shown in Figure 12-2. In this operation solvent is continually recycled. After absorption, saturated solvent is heated, which changes equilibrium behavior so that solvent can be stripped. A very common type of gas treatment plant removes CO₂ and/or H₂S from refinery gas or natural gas. In this case MEA or other amine solvents in water are used as solvent, and steam is used as stripping gas (for more details see Kohl and Nielsen, 1997). Both random packings and sieve trays, but not structured packings, are commonly used for these acid gas systems (Kister, 2006).

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. Determine column diameter for an absorber or stripper for staged 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 matrix solution methods for nonisothermal multicomponent absorption and stripping

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

9. Use a process simulator for simulation of absorbers, strippers, and gas treatment plants

1. Carrier gas is insoluble.

2. Solvent is nonvolatile.

3. System is isothermal and isobaric.

The Gibbs phase rule is

F = C – P + 2 = 3(solute, solvent, carrier gas) – 2(vapor and liquid) + 2 = 3

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

where H_B is Henry’s law constant in atm/mole fraction, H_B = H(p, T, composition); x_B is mole fraction B in liquid; and p_B is partial pressure of B in vapor (Franses, 2014; Smith and Harvey, 2007). Henry’s law is valid only at low concentrations of B. Since partial pressure is defined as

Henry’s law becomes

Equation (12-1b) will plot as a straight line if H_B is a constant. Equilibrium data for absorption are given by Hwang (1981), Hwang et al. (1992a, b), 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, values for CO₂, CO, and H₂S are shown in Table 12-1 (Perry et al., 1963). 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, more gas is absorbed at higher pressures. This phenomenon is commonly taken advantage of to make carbonated beverages. When a bottle or can is opened, pressure drops and 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 25.0°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. Obviously, Henry’s law is valid only when x is less than the solubility limit.

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

If the Arrhenius relationship is followed, a plot of ln H versus 1/T will be a straight line.

Although the units of H in Tables 12-1 and 12-2, atm/(mole fraction), are fairly common, other units are also used.

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

Unfortunately, there are other forms of the linear equilibrium equation that are also known as Henry’s law. The basic solubility form of Henry’s law is

where H_sol,B = 1/H_B and has units (mole fraction)/atm (Smith and Harvey, 2014). The solubility form is also written as

where units on H_sol,B are concentration units/pressure units (e.g., [kmol/m³]/atm or [lb mol/ft³]/atm). Before you use any values of Henry’s law constant, you need to check which form is being used. If given, units provide sufficient clues to determine the form of Henry’s law that was used.

In Section 12.6 we convert equilibrium data to the concentration units required for calculations. If mole or mass ratios are used, equilibrium must be converted into ratios.

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

2. Isothermal system.

3. Isobaric system.

4. Negligible heat of absorption.

When gas and liquid streams both are dilute, these assumptions are usually satisfied. In concentrated systems these assumptions are not valid, and the McCabe-Thiele procedure should not be used.

If the solute mole fraction in the feed, y_B,N+1, is very low, then transferring most or even all solute to liquid will have very little effect on the overall vapor flow rate, V. If the solute mole fraction in liquid, x_N, remains low after transfer of solute, then transferring the solute has little effect on overall liquid flow rate, L. Thus, we can assume that L and V are both constant, and the operating line on a McCabe-Thiele diagram will be straight. Using a mass balance envelope around top of the absorption column shown in Figure 12-1, we can write solute B mass balance for constant L and V.

We dropped the subscript B because this mass balance plus L = constant and V = constant are the only mass balances needed.

Solving for y_j+1 we obtain the equation for the McCabe-Thiele operating line.

This operating line is a straight line with a slope of L/V and a y-intercept of [y₁ – (L/V)x₀]. All possible passing streams with compositions (x_j, y_j+1) must lie on the operating line. This includes passing streams at the absorber top (x₀, y₁) and passing streams at the absorber bottom (x_N, y_N+1).

The procedure for solving a dilute absorption problem is:

1. Plot y versus x equilibrium data.

2. For a design problem, typically x₀, y_N+1, y₁, and L/V are known. Point (x₀, y₁) is on the operating line, and the slope is L/V. Plot the operating line.

3. Start at stage 1, and step off stages by alternating between equilibrium and operating lines.

This procedure is illustrated in Figure 12-3 for Example 12-1.

Note that the operating line is above the equilibrium line. This occurs because solute is being transferred from gas to liquid. In distillation the more volatile component (MVC) was transferred from liquid to gas, and the operating line was below the equilibrium curve.

For stripping we know x₀, x_N, y_N+1, and L/V. Since (x_N, y_N+1) is a point on operating line, we can plot the operating line and step off stages (Figure 12-5).

Note that the operating line is below the equilibrium curve because solute is transferred from liquid to gas. Stripping is similar to the stripping section of a distillation column. A maximum L/V ratio can be defined; this ratio corresponds to the minimum amount of stripping gas. Start from 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, whereas the reverse is true in absorption. Thus, the top of the column is on the right side (larger mole fractions) in Figure 12-5, but on the left side (smaller mole fractions) in Figure 12-3. Stripping often has large temperature changes; thus, this calculation method is often appropriate only for very dilute systems.

Murphree efficiencies can be used on these diagrams. Efficiencies for absorption and stripping are often quite low (see Section 12.6).

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

5. Equilibrium line is straight.

Assumption 5 is reasonable for very dilute solutions and agrees with Henry’s law, Eq. (12-1b), if m = H_B/p_tot and b = 0.

An analytical solution for absorption is easily derived for the special case shown in Figure 12-6, in which the operating and equilibrium lines are parallel. The distance between the operating and equilibrium lines, Δy, is constant. To go from outlet to inlet concentrations with N stages, we have

since each stage causes the same change in vapor composition. Δy can be obtained by subtracting equilibrium Eq. (12-9) from operating Eq. (12-6).

For the special case shown in Figure 12-6, L/V = m (lines are parallel), Eq. (12-10b) becomes

Combining Eqs. (12-9) and (12-11)

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

Figure 12-6 and resulting Eq. (12-12) are for a special case. The general case is shown in Figure 12-7. Now Δy_j varies from stage to stage. Δy_j values can be determined from Eq. (12-10b). Equation (12-10b) is easier to use if we replace x_j with equilibrium Eq. (12-9):

Then

Subtracting Eq. (12-14a) from (12-14b) and solving for (Δy)_j+1 we obtain

Equation (12-15) relates change in vapor composition from stage to stage to (L/mV), which is called the absorption factor. If either 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 Δy_j values shown in Figure 12-7. Thus,

Applying Eq. (12-15)

The summation in Eq. (12-16b) can be calculated. The general formula is

Then Eq. (12-16b) is

If L/(mV) > 1.0, divide both sides of Eq. (12-16b) by (L/[mV])^N–1, and do the summation mV/L replacing L/(mV). The resulting equation will still be Eq. (12-18). From Eq. (12-15), Δy₁ = Δy₀L/(mV) where is shown in Figure 12-7. The vapor composition, , is the value that would be in equilibrium with the inlet liquid, x₀. Thus,

Removal of Δy₁ from Eq. (12-18) gives

Equation (12-20) 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 1.0 to both sides of Eq. (12-20) and rearrange, we have

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

where L/(mV) ≠ 1.0. Equations (12-21) and (12-22) 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 for L/(mV) ≠ 1.0 can be developed. Several alternative forms in terms of the gas phase composition are

where

Alternative forms in terms of the liquid phase composition are

where

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

Forms for systems with three phases in which two phases flow cocurrently and countercurrently to a third phase are developed by Wankat (1980). Kremser equations for columns with multiple sections are developed by Brian (1972, Chapter 3) and by King (1980, pp. 371–376) and for reboiled absorbers by Hwang et al. (1992a).

When the assumptions required for the derivation are valid, the Kremser equation has several advantages over stage-by-stage calculations. 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 use of the Kremser equation is not. Because calculations can be done faster, effects of varying y₁, x₀, L/V, m, and so on are easy to determine. The major disadvantage of the Kremser equation is that it is accurate only for dilute solutions in which 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.

The optimum value of mV/L for absorption is approximately 0.7 and for stripping is approximately 1.4 (Woods, 2007). For an absorber removing most solute, y₁ ≈ 0, and the operating line with the minimum L/V is essentially collinear with the equilibrium line. Thus, (L/V)_min = m, and

For stripping with most solute removed, x_N ≈ 0, and the operating line with minimum V and a maximum slope is collinear with the equilibrium line. Then

which is valid for (Hp/µ) > 0.000316. O’Connell used a solubility form of Henry’s law, Eq. (12-4c), and his Henry’s law constant H is in (lb mole B/ft³)/atm, pressure, p, is in atm, and liquid viscosity, µ, is in centipoise (cP). The relationship between O’Connell’s H, which is called H_sol,B in Eq. (12-4c) and the volatility form of H_B in Eqs. (12-1) and in Tables 12-1 and 12-2 is

where the liquid density of solution is in lbm/ft³ and H_B is atm/mole fraction. Since the units of Eq. (12-37) are a little tricky, I suggest you write them out, including expanding mole fractions to be sure you understand them.

More detailed estimates of the efficiency can be made using a mass transfer analysis (see Chapter 16). Correlations for very dilute strippers are given by Hwang (1981). As a first estimate, Seider et al. (2009) recommend overall efficiencies of 30.0% for absorbers and 50.0% for strippers.

and if we also assume that:

4. Solvent is nonvolatile.

5. Carrier gas is insoluble.

Assumptions 4 and 5 are often closely satisfied. The results of these last two assumptions are that the mass balance for solvent becomes

while the mass balance for carrier gas is

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 changes gas and liquid flow rates and gives a curved operating line. Since S = moles of nonvolatile solvent/h and G = moles of insoluble carrier gas (C)/h are constant, the operating line will be straight if we can define our compositions so that we can write a mass balance for solute B with flow rates S and G. How do we do this?

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

Mole ratios Y and X are related to our usual mole fractions by

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

and

Thus, we can easily write a 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 mol B in/h = mol B out/h, or

Solving for Y_j+1 we obtain

Equation (12-40b), the operating line for concentrated absorption, is a straight line with slope S/G and intercept (Y₁ – [S/G]X₀) on a McCabe-Thiele graph plotted with ratios Y versus X (shown later in Figure 12-9). Note that this graph looks similar to Figure 12-3, except different variables are used.

Steps in this procedure are very similar to those used for dilute systems:

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

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

3. Slope is S/G. Plot the operating line.

4. Step off stages by alternating between the equilibrium curve and the operating line.

Equilibrium data must be converted to ratio units, Y versus X. These values can be greater than 1.0, since Y = y/(1 – y) and X = x/(1 – x). Y = X line has no significance in absorption. As usual stages are counted at the equilibrium curve. A minimum S/G ratio can be defined (see Figure 12-9 in step D of Example 12-3).

Since solution of Eq. (12-39b) for y_j is

substituting Eq. (12-41b) in (12-41a)

or

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

Note that the difference between the total flow rates of passing streams is constant (see Figure 12-10).

Both total flows, V_j and L_j, will be largest where Y_j and X_j are largest. In absorbers, concentrations and flow rates are highest at the column bottom; therefore, you design absorber diameters there. In strippers, flow rates are highest at the top of the column, so you design diameters at the top. Kister et al. (2008) and Zenz (1997) discuss absorber and stripper design details.

An order of magnitude estimate of column diameter can be made quite easily (Reynolds et al., 2002). Superficial gas velocity (velocity in an empty column) is typically in the range of v = 0.9144 to 1.8288 m/s (3.0 to 6.0 ft/s). Volumetric flow rate of the gas is V/ρ_m where ρ_m, molar gas density, can be estimated from the ideal gas law; the required cross sectional area is A_c = V/(ρ_m v); and the column diameter is . Substituting an average superficial gas velocity of 1.37 m/s (4.5 ft/s) with other appropriate values gives an estimate of the column diameter:

where V is in kmol/s, T is in K, and p is in kPa.

To see what additional assumptions are required consider Gibbs phase rule for a system with three solutes, 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. To represent equilibrium as a single curve or in a linear form such as Eq. (12-9), four of these degrees of freedom must be specified. Constant temperature and pressure utilize two degrees of freedom. Two additional degrees of freedom can be specified by assuming that 5) solutes are independent of each other; in other words, equilibrium for any solute does not depend on amounts of other solutes present. In addition, analysis must be done in terms of mole or mass fractions with constant total flow rates. Assumption 5 and constant total flow rates both require dilute solutions. Ratio units will not work because the ratio calculation

requires solute concentrations that are unknown.

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

For typical design problems, as shown in Figure 12-11, inlet gas and liquid compositions and flow rates are specified. Each additional solute increases the degrees of freedom for the absorber by two. These two degrees of freedom are required to specify inlet gas and liquid compositions. With temperature and pressure also specified, one degree of freedom is left, which is usually used to specify one outlet solute concentration, such as y_B,1, which fully specifies the design problem for solute B. To determine the number of stages, we can plot equilibrium data on a McCabe-Thiele diagram. When assumptions 1 to 5 are satisfied, equilibrium is usually linear. Operating equation

is the same as Eq. (12-6) and can also be plotted on the McCabe-Thiele diagram. Then number of stages is stepped off as usual (see Figure 12-12 for solute B).

The calculated number of stages for solute B has to be the number of stages for solutes A and C. Thus, concentrations of solutes A and C can be determined by solving two fully specified simulation problems. Simulation problems require a trial-and-error (or guess-and-check) 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 the same as for solute B. Point (y_A,1, x_A,0) is on the operating line.

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

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

The procedure for solute C is identical.

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 all solutes have linear equilibrium as in Eq. (12-9), the Kremser equation can be used. First solve the solute B design problem using Eq. (12-22), (12-25), or (12-26) to find N. Then use the calculated value of N to separately solve solute A and C simulation problems using equations such as Eq. (12-20), (12-21), or (12-24). Remember to use m_A and m_C when you use the Kremser equation. Note that when the Kremser equation can be used, simulation problems are not trial and error.

These 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. Computer solution methods involving simultaneous mass and energy balances plus equilibrium are required. Computer methods are discussed in Section 12.9 and this chapter’s appendix.

Absorbers, similar to flash distillation, are equivalent to very wide-boiling feeds. Thus, in contrast with distillation, a wide-boiling feed (sum rates) flowchart such as Figure 2-12 should be used. The flow rate loop is now solved first, since flow rates are never constant in absorbers. Energy balances, which require more information, are done last and are used to calculate new temperatures. 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. Initial steps are similar to those for distillation, and usually the same physical properties package is used.

Mass balances and equilibrium equations are very similar to those for distillation, and numbering of stages is again from the top down, as shown in Figure 12-14. To fit into matrix form, streams V_N+1 and L₀ are relabeled as feeds to stages N and 1, respectively. Stages 2 to N have the same general shape as for distillation (Figure 6-3). Thus, mass balances and manipulations [Eqs. (6-1) to (6-6) and (6-10) to (6-12)] are unchanged.

For stage 1 the mass balance becomes

where

and component flow rates are l₁ = L₁x₁ and l₂ = L₂x₂. These equations are repeated for each component. Equations (12-45) and (12-46) differ from Eqs. (6-7) to (6-9) since the absorber equations are for an equilibrium stage, not a total condenser.

For stage N the mass balance is

where

These equations are essentially identical to Eqs. (6-10) to (6-12).

Combining Eqs. (12-45), (6-5), and (12-47) results in a tridiagonal matrix, Eq. (6-13), with all terms defined in Eqs. (12-46), (6-6), and (12-48). There is one matrix for each component. Tridiagonal matrices can be inverted with the Thomas algorithm (Table 6-1). 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 summation equations to find new total flow rates L_j and V_j. New liquid flow rate is conveniently determined as

Vapor flow rates are determined by summing component vapor flow rates:

Convergence can be checked with

for all stages. For computer calculations, ε = 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 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, energy balances are used to solve for temperatures on each stage (King, 1980; Smith, 1963). We discuss only a multivariate Newtonian convergence procedure. For general stage j, energy balance is given as Eq. (6-20), which can be rewritten as

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

where k is the trial number. ΔT values are defined as

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

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

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

where (E_j)_k is the numerical value of the energy balance for trial k on stage j, then equations for ΔT_j can be written in matrix form.

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 ΔT_j values.

Convergence can be checked from ΔT_j. If

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), return to calculate new K values, and redo 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 this chapter’s appendix.

Consider an absorber where gas to be treated contains a carrier gas, C, and a solute, B. Solvent is a mixture of a nonvolatile solvent, S, and a reagent, R, that will react irreversibly with B:

Product RB is nonvolatile. At equilibrium, x_B (in the free form) = 0 since the reaction is irreversible, and any B that is in solution 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 reaction stoichiometry 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 is Eq. (12-5), where x is total mole fraction of B in liquid (as free B and as bound RB). Operating and equilibrium lines 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, stage efficiency is often very low because of low mass transfer and kinetic rates. If Murphree vapor efficiency E_MV [Eq. (10-2)] is used in Figure 12-15, the number of real stages can be stepped off. Murphree vapor efficiencies less than 30.0% are common in absorbers and strippers with reaction.

Since only one equilibrium stage is required, alternatives to countercurrent cascades may be preferable. A cocurrent cascade is shown in Figure 12-16A. Packed columns are normally used for the cocurrent cascade. The main advantage of cocurrent cascades is they cannot flood, so smaller diameter columns with higher vapor velocities can be used. Higher vapor velocities give higher mass transfer rates (see Chapter 15), and less packing is required (Gianetto et al., 1973). The mass balance for the cocurrent system using the mass balance envelope shown in Figure 12-16A is

Solving for y, we obtain the operating equation

Equation (12-62) plots as a straight line with a slope of –L/V (Figure 12-16B). At equilibrium, y_N = 0, and x_N can be found from the operating line shown in Figure 12-16B. When equilibrium is not attained, Murphree efficiencies or a mass transfer analysis can be used to design cocurrent systems (see Chapter 16). Cocurrent absorbers are used commercially for irreversible absorption.

For reversible chemical absorption, the higher flow rates and lack of flooding available in cocurrent absorbers are still desirable, but one equilibrium contact is rarely sufficient. Connecting a cocurrent 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.

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

Franses, E. I., Thermodynamics with Chemical Engineering Applications, Cambridge University Press, Cambridge, UK, 2014.

Gianetto, A., V. Specchia, and G. Baldi, “Absorption in Packed Towers with Concurrent Downward High-Velocity Flows—II: Mass Transfer,” AIChE J., 19 (5), 916 (1973).

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. Part 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 (September 26, 1988).

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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 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.

A5. If the volatility Henry’s law constant H follows the Arrhenius equation, will the solubility Henry’s law constant H_sol follow the Arrhenius equation? If it does, what is the relationship between E in Eq. (12-3) and E_sol in the analogous equation for H_sol?

A6. A stripper is unable to obtain the specified purity of outlet liquid. Outlet liquid concentration of impurity can be decreased (which might allow one to reach or exceed specified purity) by doing which of the following changes? Circle all correct answers. There are five correct answers—one in each pair (A, B, C, D, E). Note: Think of this as doing each change by itself, and determine whether it decreases concentration of impurity in outlet liquid.

A. a. Increase inlet gas flow rate.

b. Decrease inlet gas flow rate.

B. c. Increase inlet liquid flow rate

d. Decrease inlet liquid flow rate.

C. e. Increase number of stages.

f. Decrease number of stages.

D. g. Increase column pressure.

h. Decrease column pressure.

E. i. Increase temperature of inlet gas and inlet liquid.

j. Decrease temperature of inlet gas and inlet liquid.

A7. Explain how the single assumption that “solutes are independent of each other” can specify more than one degree of freedom.

A8. Develop your key relations chart for this chapter.

A9. A column can be either an absorber or a stripper. What would you do to change it from an absorber to a stripper?

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 (in which 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 configurations as possible.

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

C. Derivations

C1. Derive Eq. (12-43).

C2. Derive an equation that is equivalent to Eq. (12-12) for L/(mV) = 1.0 but in terms of liquid mole fractions.

C3. Derive an operating equation similar to Eq. (12-40), but draw your balance envelope around the bottom of the column. Show that result is equivalent to Eq. (12-40).

C4. Derive Eq. (10-4) for dilute systems by determining N_equil and N_actual in Eq. (10-1) from appropriate forms of the Kremser equation.

C5. Occasionally it is useful to apply the Kremser equation to systems with a constant relative volatility. Where on the y versus 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.

C6. For dilute systems show that (L/V)_min for absorbers and (L/V)_max for strippers calculated from the Kremser equation agree with graphical calculations.

D. Problems

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

D1. A five-stage countercurrent absorber is used to absorb acetone from air into water at 3.0 atm pressure and 20.0°C. Total inlet gas flow rate is 100.0 kmol/h and is 0.4 mol% acetone. Inlet liquid contains 0.01 mol% acetone. Outlet gas is 0.02 mol% acetone. Assume total liquid and gas flow rates are constant. At 20.0°C Henry’s law constant for acetone in water is H = 1.186 atm/(mole fraction). Find the liquid flow rate required and the mole fraction acetone in the outlet liquid.

D2. We are absorbing hydrogen sulfide at 15.0°C into water. Entering water is pure. Feed gas contains 0.12 mol% H₂S. Recover 97.0% of H₂S in the water. The total gas flow rate is 10.0 kmol/h. Total liquid flow rate is 2000.0 kmol/h. Total pressure is 2.5 atm. You can assume that total liquid and gas flow rates are constant. Equilibrium data are in Table 12-1.

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

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

c. If L/V = M(L/V)_min, find multiplier M (M > 1.0).

d. Why is this operation not practical? What would you do to make the process practical?

D3. A stripper with one equilibrium stage is stripping 1-chloronaphthalene from water into air. Liquid feed is x_in = 2.0 × 10^–6 mole fraction 1-chloronaphthalene, and total liquid flow rate L = 100.0 kmol/h. Inlet air is pure, and total gas flow rate is V = 10.0 kmol/h. Column operates at T = 25.0°C, and Henry’s law constant is given in Table 12-2. If outlet liquid mole fraction of x_out = 0.4 × 10^–6, what is the pressure (in atm) of the stripper, and what is the value of y_out? Assume H does not depend on pressure and that both L and V are constant. Note: There are multiple approaches to solve this problem.

D4. A steam stripper is operating isothermally at 100.0°C. Entering liquid stream contains 0.02 mol% nitrobenzene in water at 100.0°C. Flow rate of entering liquid is 1.0 kmol/min. Entering steam is pure water at 100.0°C. Outlet liquid mole fraction is 0.00001 nitrobenzene. L/V = 12.0. Total liquid and gas flow rates are constant. At 100.0°C equilibrium in terms of nitrobenzene mole fraction is y = 28.0x. Find the outlet mole fraction of nitrobenzene in the vapor stream and the number of stages.

D5. Absorb ammonia from air into water at 20.0°C and 1.5 atm pressure. Inlet water is recycled from a stripper and contains 0.2 wt% ammonia. Gas flow rate is 100.0 kmol/h. Inlet gas is 3.1 mol% ammonia. Exit gas should be 0.31 mol% ammonia. Assume L/V is constant. Equilibrium data are in Table 12-3.

a. What is the minimum flow rate of water in kg/h?

b. What is the exit concentration of the liquid, and how many equilibrium stages are needed if L = 1.5 × L_min?

Note: Keep your units straight. In Table 12-3, 10.0 weight NH₃ per 100.0 weight water = 10.0 weight/110.0 weight solution = 0.090909 mass fraction.

Approach 1: Convert all units to either mass or to moles. [MW NH₃ = 17.0, MW H₂O = 18.0, MW air = 28.9]

Approach 2: Keep liquid in weight fraction and gas in mole fraction, and plot equilibrium in this form. In the NH₃ mass balance, if y is mole fraction NH₃ in gas and x is weight fraction ammonia in liquid, Vy = (kmol gas/h)(kmol NH₃/kmol gas) = kmol NH₃/h, and Lx = (kg liquid/h)(kg NH₃/kg liquid) = kg NH₃/h.

Then adjust the NH₃ mass balance with the molecular weight of NH₃ so that all terms are in kg NH₃/h. Derive the operating equation from this form of mass balance.

c. Estimate the overall efficiency from the O’Connell correlation, and estimate the number of actual stages required.

D6.* Design a stripping column to remove carbon dioxide from water by heating the water and passing it countercurrently to a nitrogen stream in a staged stripper. Operation is isothermal and isobaric at 60.0°C and 1.0 atm pressure. Feed water contains 9.2 × 10^–6 mole fraction CO₂ and flows at 100,000 lb/h. Entering nitrogen is pure and flows at 2500.0 ft³/h. Nitrogen is at 1.0 atm and 60.0°C. Outlet water concentration is 2.0 × 10^–7 mole fraction CO₂. Ignore nitrogen solubility in water, and ignore water volatility. Equilibrium data are in Table 12-1. Use a Murphree vapor efficiency of 40.0%. Find the outlet vapor composition and the number of real stages needed.

D7. We wish to absorb ammonia from an air stream using water at 0°C and 1.3 atm. Entering water stream is pure water, and entering vapor is 17.2 wt% ammonia. Recover 98.0% of the ammonia in the water outlet stream. Total gas flow rate is 1050.0 kg/h. Use a solvent rate that is 1.5 × minimum solvent rate. Temperature is constant at 0°C, water is nonvolatile, and air does not dissolve in water. Equilibrium data are in Table 12-3. Find L_min, L, and N.

D8. HCl is being absorbed from two air streams into water in a countercurrent staged laboratory absorber at 10.0°C and a pressure of 2.0 atm. Feed rate of gas feed 1 is 1.0 kmol/h of total gas, and this gas is 20.0 mol% HCl. Feed rate of gas feed 2 is 0.5 kmol/h of total gas, and this gas is 5.0 mol% HCl. Entering water is pure. Outlet gas should be 0.2 mol% HCl. Equilibrium data follow. Use optimum feed locations for both gas feeds.

a. Find the minimum water flow rate required.

b. If L = 0.4 kg water/h, find the outlet liquid concentration, the number of stages required (including a fractional number of stages), and the optimum feed location for gas feeds 1 and 2.

Note: Use ratio units. If careful with units, liquid units can be in mass and gas units in moles, which is effectively 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). Because HCl has a very large heat of absorption in water, the column will have to be well cooled to maintain temperature at 10.0°C. Commercial units are not isothermal.

Equilibrium data from Perry and Green (1997, p. 2-127):

D9. We are operating a stripper at 0.75 atm pressure and 25.0°C to strip 1,2,3-trichloropropane from water using air as carrier gas. Inlet water contains 140.0 ppm (mol) 1,2,3-trichloropropane, and water flow rate is 10,000 kg/h. Entering air is pure, and its flow rate is 20.0 kmol/h. Outlet air is 3400.0 ppm (mol) of 1,2,3-trichloropropane. On stage 2 from the top of the column a liquid sidestream is withdrawn at a flow rate of 1000.0 kg/h. Find x_side, x_N = x_out, and N.

D10. Dichloromethane and chloroform are being stripped from water into air at 1.2 atm and 25.0°C. Feed water contains 1000.0 ppm (mol) of chloroform and 2000.0 ppm (mol) of dichloromethane. We want a 90.0% or better removal of both solutes—design for 90.0% removal of the solute that is more difficult to remove. Water flow rate is 200.0 kmol/h and can be assumed constant. Air flow rate is 3.0 kmol/h and can be assumed constant. Entering air is pure at 25.0°C and 1.2 atm.

a. Find the outlet mole fractions of chloroform and dichloromethane in both gas and liquid.

b. Find the number of equilibrium stages required in the stripper.

D11.* An absorption column for laboratory use has been carefully constructed so that it has exactly four equilibrium stages and is being used to measure equilibrium data. Water is used as solvent to absorb ammonia from air. Operation is isothermal at 80.0°F and 1.0 atm. Inlet water is pure distilled water. L/V = 1.2, inlet gas concentration is 1.0 mol% ammonia, and measured outlet gas concentration is 0.27 mol% ammonia. Assuming that equilibrium is of the form y = mx, calculate the value of m for ammonia. Check your result.

D12.* Read Section 13.4 on crossflow in Chapter 13 before proceeding. We wish to strip CO₂ from a liquid solvent using air as carrier gas. Since air and CO₂ mixtures are vented and crossflow has a lower pressure drop, use a crossflow system. Inlet liquid is 20.4 wt% CO₂, and total inlet liquid flow rate is 1000.0 kg/h. Outlet liquid composition is 2.5 wt% (0.025 weight fraction) CO₂. Gas flow to each stage is 25,190 kg air/h. The air contains 0.12 wt% CO₂, except 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 CO₂ is not a good idea even if it is legal.

D13. A water cleanup is stripping vinyl chloride from contaminated ground water at 25.0°C and 850.0 mm Hg using a countercurrent, staged stripper. Feed is 5.0 ppm (molar) vinyl chloride. Outlet water that contains 0.1 ppm (molar) vinyl chloride is required. Inlet air used for stripping is pure. For a liquid flow rate of L = 1.0 kmol/h, determine the following:

a. Minimum gas flow rate G_min in kmol/h.

b. If G = 2.0G_min, use a McCabe-Thiele diagram to determine the number of equilibrium stages needed (including a fractional number). Note: Scales on your y and x axes should be different. Calculate two points to plot the straight lines.

c. If G = 2.0G_min, use one form of the Kremser equation to determine the number of equilibrium stages needed (including a fractional number).

d. For parts b and c, what is the concentration of vinyl chloride in outlet gas? What would you propose doing with this gas (it cannot be vented to the atmosphere)?

e. Are the assumptions required for the solution methods satisfied?

Data are in Table 12-2.

D14. In an ammonia plant traces of argon and methane are absorbed from a nitrogen stream using liquid ammonia. Operation is at 253.2 K and 175.0 atm pressure. Feed flow rate of gas containing 0.024 mol% argon and 0.129 mol% methane is 100.0 kmol/h. We remove 95.0% of the methane. Entering liquid ammonia is pure. Operate with L/V = 1.4(L/V)_min. Total gas and liquid rates are constant. Equilibrium data at 253.2 K are:

Methane: partial pressure methane atm = 3600.0 × (methane mole fraction in liquid).

Argon: partial pressure argon atm = 7700.0 × (argon mole fraction in liquid) (Alesandrini et al., 1972).

a. Find the outlet methane mole fraction in the gas.

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

c. Find the outlet methane mole fraction in the liquid.

d. Find the number of equilibrium stages required.

e. Find the outlet argon mole fractions in liquid and gas, and the percentage of recovery of argon in liquid.

D15. We plan to treat 150.0 kmol/h of water that is saturated with carbon tetrachloride (the CCl₄ is at the solubility limit shown in Table 12-2) at 25.0°C and a pressure of 0.96 bar by stripping with pure air at 25.0°C and 0.96 bar (parts a, b, c). Remove 90.0% of the CCl₄. Assume the Henry’s law constant and solubility of CCl₄ do not depend on pressure.

a. What is the minimum value of gas flow rate V_min?

b. If we operate at V = 2.5 × V_min, what is the value of V, how many stages are needed, and what is the outlet gas mole fraction of CCl₄?

c. If we operate at V = 1.10 × V_min, what is the value of V, how many stages are needed, and what is the outlet gas mole fraction of CCl₄?

d. Your boss wants to know what the minimum gas flow rate will be if the stripper is operated under a vacuum at 0.5 bar and 25.0°C.

e. If we operate at 0.5 bar and 25.0°C and V = 2.5 × V_min, what is the value of V, how many stages are needed, and what is the outlet gas mole fraction of CCl₄?

f. If we operate at 0.5 bar and 25.0°C and V = 1.10 × V_min, what is the value of V, how many stages are needed, and what is the outlet gas mole fraction of CCl₄?

g. Explain how N can be the same in parts b and e at different pressures, but y_out is larger at the lower pressure.

h. Approximately, what is lowest pressure that can be used if temperature is 25.0°C?

D16. Argon and methane are absorbed from nitrogen into liquid ammonia at 252.3 K and 175.0 atm. Feed rate of gas is 100.0 kmol/h. Feed gas contains 0.020 mol% Ar and 0.130 mol% methane. Outlet gas contains 0.0008 mol% methane. Entering liquid ammonia is pure. Operate with L/V = 26.0. Assume L and V flow rates are constant. Equilibrium data are given in Problem 12.D14.

a. Find the outlet mole fraction of methane in the liquid.

b. Find the number of equilibrium stages required.

c. Find the outlet mole fraction of argon in the gas.

D17.* We absorb ammonia from air into water. Equilibrium data are given as y_NH3 = 1.414x_NH3 in mole fractions. A countercurrent column has three equilibrium stages. Entering air stream has total flow rate of 10.0 kmol/h and is 0.83 mol% NH₃. Inlet water stream contains 0.02 mol% ammonia. Outlet gas stream is 0.05 mol% ammonia. Find the required liquid flow rate, L.

D18. Dilute amounts of ammonia are absorbed from two air streams into water. The absorber operates at 30.0°C and 2.0 atm pressure, and equilibrium expression is y = 0.596x where y is the mole fraction of ammonia in gas and x is the mole fraction of ammonia in liquid. Solvent is pure water and flows at 100.0 kmol/h. The main gas stream to be treated has mole fraction ammonia of y_N+1 = 0.0058 (the remainder is air) and flow rate V_N+1 = 100.0 kmol/h. The second gas feed stream has mole fraction ammonia of y_F = 0.003 and flow rate of V_F = 50.0 kmol/h. We desire an outlet ammonia mole fraction in the gas stream of y₁ = 0.0004. Find:

a. Outlet mole fraction of ammonia in the liquid, x_N.

b. Optimum feed stage for feed gas of mole fraction y_F, and the total number of stages.

c. Minimum solvent flow rate, L_min.

D19. We will treat two feeds of liquid water containing carbon tetrachloride in a stripper at a pressure of 0.96 bar and a temperature of 25.0°C by stripping with pure air. The first feed is L = 100.0 kmol/h water that is saturated with carbon tetrachloride (CCl₄) at the solubility limit of x_in = 93.68 ppm (mol) at 25.0°C. This feed is input at the top of the column. The second feed is F = 75.0 kmol/h of water that contains x_F = 46.84 ppm (mol) CCl₄ at 25.0°C. We want to remove 90.0% of the total CCl₄ fed to the column. Pure air is fed to the bottom of the column, and V = 0.176 kmol/h. Assume that V is constant.

a. Find x_out.

b. Find y_out, the optimum feed location for feed 2, and N.

D20. We need to remove H₂S and CO₂ from 1000.0 kmol/h of a water stream at 0°C and 15.5 atm. Inlet liquid contains 0.0024 mol% H₂S and 0.0038 mol% CO₂. We desire a 99.0% recovery of H₂S in the gas stream. Pure nitrogen at 0°C and 15.5 atm is used to strip out H₂S and CO₂. Nitrogen flow rate is 3.44 kmol/h. A staged countercurrent stripper will be used. Assume water flow rate and air flow rate are constant. Data are in Table 12-1. 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 fraction in the outlet liquid stream.

D21. A gas-processing plant has an absorber and a stripper set up as shown in Figure 12-2, except both columns operate at 25.0°C but are at different pressures. Absorber is at 5.0 atm and stripper is at 0.2 atm. Feed to the plant is V_abs = 100.0 kmol/h of air containing y_in,abs = y_N+1,abs = 0.00098 mole fraction 1,2,3-trichloropropane. Outlet gas from absorber is y_out,abs = y_1,abs = 0.000079 mole fraction 1,2,3-trichloropropane. Inlet liquid to absorber is x_in,abs = x_0,abs = 0.00001 mole fraction 1,2,3-trichloropropane. Note that because absorber and stripper are connected, x_in,abs = x_out,stripper and x_out,abs = x_in,stripper. Entering gas in stripper is pure air. Determine the minimum liquid flow rate in the absorber and then operate with L_abs = 1.6(L_abs,min). Note that L_abs = L_stripper. In the stripper determine the minimum gas flow rate, and operate with V_stripper = 1.5(V_stripper,min). Equilibrium data are in Table 12-2.

a. Find L_abs,min, L_abs, x_out,abs, N_abs, V_stripper,min, V_stripper, y_out,stripper, and N_stripper.

b. Do mole fractions in the liquid ever exceed solubility limits?

Suggestion: Easiest solution path is to roughly sketch McCabe-Thiele diagrams to help in calculation of L_abs,min and V_stripper,min, use external balances to find x_out,abs = x_in,stripper and y_out,stripper, and use the Kremser equation to find N_abs and N_stripper. Watch your decimal points!

D22. We are absorbing n-butane from a light gas into heavy oil at 1000.0 kPa and 15.0°C. Flow rate of inlet gas is V_N+1 = 150.0 kmol/h and mole fraction n-butane in inlet gas is y_N+1 = 0.003. Inlet solvent flows at L₀ = 75.0 kmol/h and contains no n-butane, x₀ = 0. We want an exit vapor with y₁ = 0.0004 mole fraction n-butane. Use a DePriester chart for equilibrium data. Assume light gas is insoluble and heavy oil is nonvolatile.

a. Find the mole fraction of n-butane in the outlet liquid, x_N.

b. Find the sufficient number of equilibrium stages.

D23.* A complete gas treatment plant often consists of both an absorber to remove solute from air and a stripper to regenerate solvent. Some treated air is heated and is used as stripping gas. In a particular application we wish to remove obnoxious impurity from inlet air. Absorber operates at 1.5 atm and 24.0°C where obnoxious impurity equilibrium is given as y = 0.5x (units are mole fractions). Stripper operates at 1.0 atm and 92.0°C where equilibrium is y = 3.0x (units are mole fractions). Total feed air flow rate is 1400.0 mol/day, and feed gas is 15.0 mol% obnoxious impurity, and the remainder is air. We desire a treated air concentration of 0.5 mol% obnoxious impurity. Liquid flow rate into the absorber is 800.0 mol/day, and the liquid feed is 0.5 mol% obnoxious impurity. Note that this liquid is the exiting liquid from the stripper.

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

b. The stripper has four equilibrium stages. Calculate the gas flow rate of heated stripping gas (concentration is 0.5 mol% A) and the outlet gas concentration from the stripper.

D24. A stripping column with 27 actual stages has an overall efficiency of 0.2. Feed is 100.0 kmol/h of liquid water that contains 0.010 mol% CHCl₃. Stripping gas is pure nitrogen saturated with water vapor. Remove 95.0% of CHCl₃ from the water. Operation is at 0.5 atm and 25.0°C.

a. What is the mole fraction CHCl₃ in the outlet liquid?

b. What is the vapor flow rate V in kmol/h?

c. What is the mole fraction CHCl₃ in the outlet vapor?

d. What would the minimum vapor flow rate be if a large number of additional stages were added?

E. More Complex Problems

E1. A laboratory steam stripper with 11 real stages is used to remove 1000.0 ppm (wt) nitrobenzene from an aqueous feed stream that enters at 97.0°C. Flow rate of liquid feed stream is L_in = F =1726 g/h. Entering steam rate is measured as S = 99.0 g/h. Leaving vapor rate is measured as V_out = 61.8 g/h. Column pressure is 1.0 atm. Treated water is measured at 28.1 ppm nitrobenzene. Data at 100.0°C are in Problem 12.D4. Molecular weights are 123.11 and 18.016 for nitrobenzene and water, respectively. What is the overall efficiency of 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 liquid flow rate in the column.

E2. Find V_min for Problem 12.D19.

F. Problems Requiring Other Resources

F1.* Laboratory tests are being made prior to design of an absorption column to absorb bromine (Br₂) from air into water. Tests were made in a laboratory packed column that is 0.1524 m in diameter, has 1.524 m of packing, and is packed with saddles. The column was operated at 20.0°C and 5.0 atm total pressure, and the following data were obtained:

Inlet solvent is pure water.

Inlet gas is 2.0 mol% bromine in air.

Exit gas is 0.2 mol% bromine in air.

Exit liquid is 0.1 mol% 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 1.0 atm. Feed is 10.0 mol/h of a 60.0 mol% ethane, 40.0 mol% n-pentane mixture that enters at 30.0°F. Solvent is pure n-octane at 70.0°F, and solvent flow rate is 20.0 mol/h. Find the outlet compositions and temperatures. Column is adiabatic. For the first guess assume that all stages are at 70.0°F, and assume flow rates are:

L₀ = 20.0   V₁ = 6.00 mol/h

L₁ = 20.2   V₂ = 6.18

L₂ = 20.8   V₃ = 6.66

L₃ = 24.0   V₄ = 10.0

Then go through one iteration of 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 in the flow loop, use a paired simultaneous convergence routine. To do this, use 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 mass balances for a second iteration. However, for this assignment stop after new temperatures have been estimated. Use a DePriester chart for K values. Pure-component enthalpies are given in Table 2-5, Maxwell (1950), and Smith (1963). Assume ideal solution behavior to find the enthalpy of each stream.

F3. Climate change is very much in the news. Engineers need to be heavily involved to control climate change. One approach is to capture carbon dioxide and sequester it. Write a two to three page engineering analysis of the feasibility of using absorption to capture carbon dioxide. In your analysis explain why capturing carbon dioxide from flue gases of large power plants is considered to be more feasible than capturing carbon dioxide from automobile exhaust or from air. Cite appropriate references in your paper.

G. Computer Simulation Problems

G1. Feed gas is at 1.0 atm and 30.0°C and is 90.0 mol% air and 10.0 mol% ammonia. Flow rate is 200.0 kmol/h. Ammonia is absorbed at 1.0 atm using water at 25.0°C as solvent. We desire outlet ammonia in exiting air that is 0.32 mol% or less. Column is adiabatic.

a. If N = 4, what L is required (± 10.0 kmol/h)?

b. If N = 8, what L is required (± 10.0 kmol/h)?

c. If N = 16, what L is required (± 10.0 kmol/h)?

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

G2. We wish to absorb two gas streams in an absorber. The main gas stream (stream A) is at 15.0°C, 2.5 atm, and has a flow rate of 100.0 kmol/h. Stream A is 90.0 mol% methane, 6.0 mol% n-butane, and 4.0 mol% n-pentane. The second gas stream (stream B) is at 10.0°C, 2.5 atm, and has a flow rate of 75.0 kmol/h. Stream B is 99.0 mol% methane, 0.9 mol% n-butane, and 0.1 mol% n-pentane. Liquid solvent fed to the absorber is at 15.0°C and 2.5 atm and has a flow rate of 200.0 kmol/h. This stream is 99.9 mol% n-decane and 0.1 mol% n-pentane. Absorber pressure is 2.5 atm. We desire an outlet gas stream that is 99.9 mol% 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 outlet gas stream. (Use “on stage” for all 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 fractions of gas stream leaving absorber

f. Outlet mole fractions of liquid leaving absorber

g. Outlet gas flow rate (kmol/h)

h. Outlet liquid flow rate (kmol/h)

i. Highest temperature in column (°C) and stage on which it occurs

G3. a. 200.0 kmol/h of a liquid feed that is 10.0 mol% isopropyl alcohol and 90.0 mol% water is stripped in a column with five equilibrium stages. Stripping gas is pure nitrogen. Column pressure can be varied between 1.0 and 5.0 atm (make feed gas and feed liquid pressures equal to column pressure). Feed gas can be between 25.0°C and 100.0°C, and inlet liquid can be between 25.0°C and 75.0°C. Design operating conditions (column pressure, inlet gas flow rate and temperature, and inlet liquid temperature) to recover 98.0% of isopropyl alcohol (recover means in gas phase), and we want as high an isopropyl alcohol concentration in outlet gas as possible. Report column pressure, feed gas flow rate and temperature, liquid feed temperature, recovery of the isopropyl alcohol, mole fractions and flow rates of leaving gas and liquid product streams, and temperature on every stage.

b. Determine the diameter of the column if it has sieve plates, uses a plate spacing of 0.60960 m, has an 85.0% flooding factor, and uses the Fair flooding calculation method. For the other values, use the default values in Aspen Plus.

G4. We wish to absorb n-butane and n-pentane from a gas stream that is at 25.0°C and 1.0 atm and has a flow rate of 100.0 kmol/h. Feed is 90.0 mol% methane, 4.0 mol% n-butane, and 6.0 mol% n-pentane. Two liquid streams are available as solvent. Liquid stream A is at 30.0°C and 1.0 atm and has a flow rate of 75.0 kmol/h. Stream A is 96.0 mol% n-nonane (9.0 carbons), 1.0 mol% n-butane, and 3.0 mol% n-pentane. All of stream A will be used. Liquid stream B is pure n-nonane. This stream is at 1.0 atm and 20.0°C. The flow rate of stream B can be adjusted to any desired value. Absorber operates at a pressure of 1.0 atm and has seven equilibrium stages. Gas should be fed on stage 7. The two liquid streams can be fed above any of the stages. Outlet gas stream should be between 99.0 and 99.1 mol% methane. Use the Peng-Robinson VLE package.

Determine the minimum flow rate of stream B and the optimum feed stages for streams A and B required to just achieve the desired methane purity of the outlet gas stream. Report the following:

a. Required flow rate of stream B (kmol/h)

b. Optimum feed stage location for stream A

c. Optimum feed stage location for stream B

d. Outlet mole fractions of gas stream leaving absorber

e. Outlet mole fractions of liquid leaving absorber

f. Outlet gas flow rate (kmol/h)

g. Outlet liquid flow rate (kmol/h)

h. Highest temperature in column (°C) and stage on which it occurs

Lab 11. Absorption and Stripping

Goals:

1. Investigate computer simulation of absorbers and strippers.

2. Work as a team to prepare a report on the design of a gas plant.

Preparation:

• As needed, review Computer Labs 1 to 6.

• Analyze Figure 12-2 in detail.

Teams will be assigned. Each team will develop in writing its own code of conduct. If there are difficulties in the functioning of the team, the first action of the team should be to have a meeting discussing the difficulties and whether team members are following the code. Students may choose to withdraw from a team and do all work alone.

Aspen Plus uses RadFrac for absorber calculations. If you want more information, once you are logged into Aspen Plus and are on the screen “Start Using Aspen Plus,” you can go to Help in the menu bar → Index (tab second from left), type absorbers in search and click RadFrac.

I. Getting Started

1. Input your components: acetone, water, and air (input air as a component), and pick the physical properties package (Do NOT list any component as a Henry component).

2. Click Simulation. 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.

Note: Many of the items below will change for each new problem statement.

3. Input conditions (T, p, molar flow rate, mole fractions) for the liquid and vapor feed streams (note that Aspen will determine if a stream is a liquid or a vapor).

4. In the Configuration window for the RadFrac block set:

Calculation type → Equilibrium

Number of stages → appropriate number for each of the following problems

Condenser → none

Reboiler → none

Valid phases → Vapor-Liquid

Convergence → Petroleum/wide boiling (Note: The instructions in Help are slightly different but will result in the same convergence setting, since Sum Rates is the same as Petroleum/wide boiling.)

5. In the Streams window supply Aspen Plus the locations of the two feed streams: Liquid feed on stage 1 and vapor feed on the last stage.

6. In the Pressure window set the pressure for the absorber. This only needs to be done for the top stage. Aspen Plus will assume the column is isobaric.

7. Go to Convergence for the block (probably easiest to do using the All Items list on the left side of the Aspen Plus screen). Click the name of the block for your absorber. Then click the arrow next to Convergence followed by clicking the box with the blue check labeled Convergence. In the Basic window, the algorithm should be listed as Sum Rates. Set the maximum iterations at 75.

II. Simulations

For each simulation, record the outlet gas and outlet liquid mole fractions, outlet flow rates, and temperatures on the top and bottom stages. Note if there is a temperature maximum and the stage location of the maximum. The purpose of parts 1 and 2 is to explore how different variables affect the operation of absorbers and strippers. Desired specification is 0.3 (or less) mol% acetone in the exiting air.

1. Absorber:

a. You wish to absorb acetone from air into water. The feed streams are at 1.04 bar and the column is at 1.03 bar. Inlet gas stream contains 3.2 mol% acetone, and temperature is 30.0°C. Flow of inlet gas is 100.0 kmol/h. Water flow rate is 200.0 kmol/h. Inlet water is pure, and temperature is 20.0°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 water in the outlet gas come from? Note how dilute the outlet liquid is. What can we do to increase this mole fraction?

b. Decrease water flow rate to 100.0 kmol/h, and run again. Do you meet the required acetone concentration in gas? (It should not.) But note that the outlet water is more concentrated.

c. Double the number of stages to 12 (Note: You also have to change the location of the vapor feed in the Streams tab) with L = 100.0. Does this help reduce the outlet vapor mole fraction of acetone significantly? (Probably not.)

d. Return to N = 6 with L = 100.0. Reduce the temperature of both feed gas and inlet water to 10.0°C. Did this reduction allow you to meet outlet specifications for air? (Answer depends on the VLE 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?

e. Repeat part 1d (N = 6, L = 100.0, T = 10.0°C) but with feed, solvent, and absorber at 2.0 bar (temporarily retain report for this run for simulation exercise 2). Does the outlet gas stream meet specifications? (It should).

f. From these runs write down the expected qualitative behavior (is outlet gas purer or less pure) for the following operational changes:

Increase L or increase L/G

Increase temperature of feed gas and/or solvent

Increase N

Increase pressures

2. Stripper: The same flow diagram can be used as a stripper by changing feeds and column pressure. Liquid feed to the stripper should be liquid product from the absorber (use stage 6’s liquid product mole fractions and product flow rate, kmol/h from part 1e) but at 85.0°C and 1.0 bar. Gas feed to the stripper should be pure steam at 1.0 atm. Use six stages. Be sure to put your steam (the stripping gas) on stage 6, and liquid from the absorber should be put in above stage 1. The stripper is at 1.0 bar.

a. Set steam flow rate at 20.0 kmol/h. Steam is superheated to 101.0°C. If the specification is liquid leaving should have an acetone mole fraction less than 1.0 × 10^–4, does this design satisfy the specification? (It should.) Record the acetone mole fraction in gas leaving the stripper. Note that this fraction is high enough that pure acetone could be recovered economically by distillation.

b. Repeat with saturated steam at 1.0 bar and 100.0°C (set vapor fraction = 1.0)

c. Using saturated steam (vapor fraction = 1.0) at 1.0 bar, reduce the steam flow rate to just satisfy the specification on the outlet liquid composition.

d. Try reducing the stripper and steam pressures to 0.5 bar with steam at 83.0°C and 15.0 kmol/h steam. What can you conclude about the effect of pressure?

3. Complete Gas Plant, No Recycle: Design a complete gas plant (similar to Figure 12-2 but without solvent recycle) to process air containing 3.2 mol% acetone and 0.5 mol% water at 2.0 bar and 10.0°C. Feed flow rate is 100.0 kmol/h. Absorber has six stages, pressure = 2.0 bar. The outlet mole fraction of acetone leaving in gas should be less than 0.003. Solvent fed to the absorber is pure water at 10.0°C and 2.0 bar. Flow rate is 100.0 kmol/h. If the outlet gas specification is not met, adjust operating conditions (L, T, N, or p) until it is met. Liquid exiting from the absorber should be heated (use a Heater) to 65.0°C at 0.5 bar before being fed to the stripper. Steam fed to the stripper is at 0.5 bar and is saturated steam. The stripper operates at 0.5 bar and has six stages. The acetone mole fraction leaving in 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.

a. Record your steam flow rate, flow rate and mole fractions of the outlet gas stream from the absorber, flow rate and mole fractions of the outlet liquid stream from the stripper, and flow rate and mole fractions of the outlet gas stream from the stripper.

b. What fraction of the HOTSOLV stream from the heater to the stripper is vapor? Why does this occur? Is this a problem? How cold would this stream have to be to have no vapor?

4. Complete Gas Plant with Recycle:

Absorber: 2.0 bar, N = 6, feed is 100.0 kmol/h of 3.2 mol% acetone and 0.5 mol% water in air at 2.0 bar and 10.0°C.

Stripper: 0.5 bar, N = 6, pure saturated steam at 0.5 bar is fed at the rate found in step 3.

Heater after absorber: heats liquid to 65.0°C.

Treated solvent from the stripper, which is close to pure water, should be recycled and connected to the absorber and fed above stage 1; however, first cool this stream to 10.0°C in a heat exchanger (use a Heater), and use a Pump to increase the pressure from 0.5 to 2.0 bar.

Now the tricky part. Fresh solvent has to be reduced to a flow rate at which water mass balance can be valid, and flow rate in the recycle loop must be controlled. In other words, at steady state, water in (inlet gas + makeup solvent + steam) must equal water out (process gas + concentrated gas + waste). But you do not know what waste or makeup values to use. So, after the stripper on the TRTSOLV line, put in a splitter (Fsplit in Aspen), and send TRTSOLV as feed to the splitter. Hook up one product line to the heat exchanger that cools the recycle liquid and then to the pump. Set the flow rate of this line to 100.0 kmol/h in the Specifications tab for the splitter. The other product from the splitter is WASTE and is dumped. After recycle is hooked up (Figure 12-A2), use a fresh (or makeup) solvent rate of 100.0 kmol/h, and run the simulation. Decrease the makeup solvent flow in steps (try increments of –20.0 until makeup = 20.0 and then smaller). Ultimately, you want a makeup solvent flow rate that minimizes WASTE, but must have a WASTE flow rate > 0. Often gas plants can operate with MAKEUP = 0. This gas plant will operate with MAKEUP = 0, so that is your goal (where does water come from for the waste stream?). Check that TRTGAS is less than 0.3 mol% acetone and that WASTE is less than 0.01 mol% acetone. If these conditions are not met, adjust operating conditions until they are met. Record the flow rates and the mole fractions of all streams.

Note 1: Stream WASTE also serves as a purge of liquid water. A small purge is needed to prevent nonvolatile impurities, such as salts, from building up in the solvent loop.

Note 2: How would you startup this type of gas plant?

Note 3: Although the design works, it is not optimized. Operating the absorber at 10.0°C will be expensive. The final, optimized design will probably be at a higher temperature.

Team Report:

Your team is to submit a three-page memo reporting the design of the gas plant, addressed to your instructor from the entire group of active members. The memo should have two pages of text and one page of tables and figures. Present numbers, including units, in a way that is clearly identified. Mention graphs or figures you have included in the text.

Team Report Requirements:

These requirements are very strict for the following reasons:

1. Companies want brief, to the point, but complete reports.

2. Company reports typically put details in an appendix.

3. There must be enough detail (in appendix) so that another engineer can check your work—perhaps years later.

4. A strict page count protects students from having to compete for the most complete and longest report.

5. A strict page count protects the person who grades from overly long reports.

Turn in a single team report consisting of three pages that are word processed in standard size (8½ by 11 in.). The first two pages should contain text with normal margins and 11-point Times New Roman font. Line spacing should be multiple at 1.15 with a line space between paragraphs and no indention. The third page should have figures, tables, and equations. References may be on either page 2 or 3. There should be no cover page. The first five lines of the first page should be:

1. To: Instructor’s name

2. From: List all active team members

3. Subject: Name of project

4. Date: Give date

5. (Skip a line—line is blank)

Use the following headings (you can change the wording). Note: Paragraphs start on the same line as the headings.

Summary: The first paragraph should be a summary that briefly describes the problem and your solution. (Write this paragraph last.)

Introduction: The second paragraph should describe the separation problem. Do not include excessive details.

Recommended Solution for Complete Gas Plant with Recycle: Refer to a figure on page 3 that shows a flowsheet for the recommended operating conditions and to a table with pertinent data.

Discussion: Answer the question of how you would start up this plant, and bring up any other issues that occur.

Figures and Tables: (Page 3): Include figures, tables, equations, and so forth. Give complete information (in kmol/h, mole fraction, m, kW) on your flowsheet. Numbers and letters must be 10-point Times New Roman or larger.

Additional pages: Not allowed, and will not be read.

Grading: Reports will be graded for both technical accuracy (instructor may run a simulation of your final solution) and written communication. Do a spell and grammar check, and carefully proofread before turning in the report.