10.6. ABSORPTION IN PLATE AND PACKED TOWERS

10.6A. Introduction to Absorption

As discussed briefly in Section 10.1B, absorption is a mass-transfer process in which a vapor solute A in a gas mixture is absorbed by means of a liquid in which the solute is more or less soluble. The gas mixture consists mainly of an inert gas and the solute. The liquid also is primarily immiscible in the gas phase; that is, its vaporization into the gas phase is relatively slight. A typical example is absorption of the solute ammonia from an air–ammonia mixture by water. Subsequently, the solute is recovered from the solution by distillation. In the reverse process of desorption or stripping, the same principles and equations hold.

Equilibrium relations for gas–liquid systems in absorption were discussed in Section 10.2, and such data are needed for design of absorption towers. Some data are tabulated in Appendix A.3. Other, more extensive data are available in Perry and Green (P1, P2).

10.6B. Equipment for Absorption and Distillation

1. Various types of tray (plate) towers for absorption and distillation

In order to efficiently bring the vapor and liquid into contact in absorption and distillation, tray towers of the following types are often used.

  1. Sieve tray. The sieve tray shown in Fig. 10.6-1a is very common. Essentially the same tray is used in distillation and gas absorption. In the sieve tray, vapor bubbles up through simple holes in the tray through the flowing liquid. Hole sizes range from 3 to 12 mm in diameter, with 5 mm a common size. The vapor area of the holes varies between 5 to 15% of the tray area. The liquid is held on the tray surface and prevented from flowing down through the holes by the kinetic energy of the gas or vapor. The depth of liquid on the tray is maintained by an overflow, outlet weir. The overflow liquid flows into the downspout to the next tray below.

    Figure 10.6-1. Tray contacting devices: (a) detail of sieve-tray tower, (b) detail of valve-tray tower.

  2. Valve tray. A modification of the sieve tray is the valve tray shown in Fig. 10.6-1b. This consists of an opening in the tray and a lift-valve cover with guides to keep the cover properly positioned over the opening. This provides a variable open area which is varied by the vapor flow inhibiting leakage of liquid down the opening at low vapor rates. Hence, this type of tray can operate over a greater range of flow rates than the sieve tray, with a cost of only about 20% more than a sieve tray. The valve tray is being increasingly used today (S5).

  3. Bubble-cap tray. Bubble-cap trays have been used for over 100 years, but since 1950 they have been generally superseded by sieve-type or valve trays because of their cost, which is almost double that of sieve-type trays. In the bubble tray, the vapor or gas rises through the opening in the tray into the bubble caps. Then the gas flows through slots in the periphery of each cap and bubbles upward through the flowing liquid. Details and design procedures for many of these and other types of trays are given elsewhere (B2, P2, T1). Efficiencies for the different types of trays are discussed in Section 11.5.

2. Structured packing for absorption and distillation

Structured packing has become competitive with conventional tray towers, especially in tower revamps where increased capacity and/or efficiency is desired (K1, L2). A typical corrugated-sheet packing is shown in Fig. 10.6-2 (F1). The thin corrugated-metal sheets are formed in a triangular cross-section, as shown in Fig. 10.6-2a. The vapor flow goes upward through the triangular channels, which are set at a 45° angle with the vertical. The sheets are arranged so the liquid flows downward in the opposite direction and spreads over the surfaces, as in a wetted-wall tower.

Figure 10.6-2. Typical corrugated structured packing: (a) triangular cross section of flow channel; (b) flow-channel arrangement, with vapor flowing upward, indicated by arrows, and liquid downward. [From J. R. Fair and J. L. Bravo, Chem. Eng. Progr., 86, (Jan.), 19 (1990). With permission.]


The corrugated sheets are assembled into an element whose height E, as shown in Fig. 10.6-2b, is about 20 to 30 cm tall (8–12 in.). A shorter height tends to increase the liquid and vapor spread in the horizontal plane. Each adjacent element is rotated 90° in the horizontal plane with respect to the layer below in order to spread the liquid and vapor uniformly in all radial planes. The ratios of B/h in Fig. 10.6-2a are in the range of 2/1 to 4/1. The size ranges of the triangle are typically: for B, 2.4–4.0 cm, for S, 1.7–2.6 cm, and for h, 1.2–1.8 cm (K1). Smaller sizes of the triangle mean that more sheets are present, giving a greater surface area. This results in a higher efficiency but smaller openings. However, the increased resistance to gas flow gives a lower capacity and a greater sensitivity to plugging (K1).

The open void fraction typically varies from 0.91 to 0.96 and the specific surface area from 165 to 330 m2/m3 volume (50 to 100 ft2/ft3). In many cases the packing sheet contains perforations or holes about 2–4 mm ID spaced 0.5–1.5 cm apart to help wet both the upper and lower sides of the sheet.

3. Packed towers for absorption and distillation

Packed towers are used for continuous countercurrent contacting of gas and liquid in absorption as well as for vapor–liquid contacting in distillation. The tower in Fig. 10.6-3 consists of a cylindrical column containing a gas inlet and distributing space at the bottom, a liquid inlet and distributing device at the top, a gas outlet at the top, a liquid outlet at the bottom, and a packing or filling in the tower. The gas enters the distributing space below the packed section and rises upward through the openings or interstices in the packing and contacts the descending liquid flowing through the same openings. A large area of intimate contact between the liquid and gas is provided by the packing.

Figure 10.6-3. Packed tower flows and characteristics for absorption.


4. Types of random packing for absorption and distillation

Many different types of tower packings have been developed and a number are used quite often. Common types of packing which are dumped at random in the tower are shown in Fig. 10.6-4. Such packings and other commercial packings are available in sizes of 3 mm to about 75 mm. Most of the tower packings are made of materials such as clay, porcelain, metal, or plastic. High void spaces of 65–95% are characteristic of good packings. The packings permit relatively large volumes of liquid to pass countercurrent to the gas flow through the openings with relatively low pressure drops for the gas. These same types of packings are also used in vapor–liquid separation processes of distillation.

Figure 10.6-4. Typical random or dumped tower packings: (a) Raschig ring; (b) Berl saddle; (c) Pall ring; (d) Intalox metal, IMTP; (e) Jaeger Metal Tri-Pack.


Ceramic Raschig rings and Berl saddles shown in Figs. 10.6-4a and b are older types of random packing and are seldom used now (K1). Pall rings (second-generation packing) shown in Fig. 10.6-4c, are made of plastic or metal; they are much more efficient and are still used now. They have porosities or void spaces of 0.90–0.96 and areas of 100–200 m2/m3 (30–60 ft2/ft3). The latest or third-generation packings are the Intalox metal type, shown in Fig. 10.6-4d, which is a combination of the Berl saddle and the Pall ring, and the Metal Tri-Pack, shown in Fig. 10.6-4e, which is a Pall ring in spherical shape. Porosities range from 0.95 to 0.98. Many other types of new packings are available. These third-generation packings are only slightly more efficient than the Pall rings.

Stacked packings having sizes of 75 mm or so and larger are also used. The packing is stacked vertically, with open channels running uninterruptedly through the bed. The advantage of the lower pressure drop of the gas is offset in part by the poorer gas–liquid contact in stacked packings. Typical stacked packings are wood grids, drip-point grids, spiral partition rings, and others.

10.6.C. Pressure Drop and Flooding in Packed Towers

In a given packed tower with a given type and size of packing and with a definite flow of liquid, there is an upper limit to the rate of gas flow, called the flooding velocity. Above this gas velocity the tower cannot operate. At low gas velocities the liquid flows downward through the packing, essentially uninfluenced by the upward gas flow. As the gas flow rate is increased at low gas velocities, the pressure drop is proportional to the flow rate to the 1.8 power. At a gas flow rate called the loading point, the gas starts to hinder the liquid downflow, and local accumulations or pools of liquid start to appear in the packing. The pressure drop of the gas starts to rise at a faster rate. As the gas flow rate is increased, the liquid holdup or accumulation increases. At the flooding point, the liquid can no longer flow down through the packing and is blown out with the gas.

In an actual, operating tower, the gas velocity is well below flooding. The optimum economic gas velocity is about one-half or more of the flooding velocity. It depends upon a balance of economic factors including equipment cost, pressure drop, and processing variables. Pressure drop in the packing is an important consideration in design of a tower and is covered in detail below.

1. Pressure drop in random packings

Empirical correlations for various random packings based on experimental data are used to predict the pressure drop in the gas flow. The original correlation by Eckert (K1) correlated the gas and liquid flow rates and properties with pressure drop. The latest version has been replotted by Strigle (K1, S4) and is shown in Fig. 10.6-5. The line for ΔP = 2.0 in. H2O/ft has been extrapolated. The ordinate (capacity parameter) is vGG/(ρL − ρG)]0.5 ν0.05 and the abscissa (flow parameter) is (GL/GG)(ρGL)0.5, where vG is superficial gas velocity in ft/s, ρG is gas density in lbm/ft3, vG = GGG, ρL is liquid density in lbm/ft3, Fp is a packing factor in ft1, v is kinematic viscosity μL/(ρL/62.4) in centstokes, μL is liquid viscosity in cp, GL is liquid mass velocity in lbm/(s · ft2), and GG is gas mass velocity in lbm/(s · ft2).Note that this capacity parameter is not dimensionless and that only these units should be used. This correlation predicts pressure drops to an accuracy of ±11% (L2).

Figure 10.6-5. Pressure-drop correlation for random packings by Strigle. (From R. F. Strigle, Jr., Random Packings and Packed Towers, Houston: Gulf Publishing Company, 1987. With permission from Elsevier Science.)


The packing factor Fp is almost inversely proportional to packing size. This packing factor Fp is determined empirically for each size and type of packing, and some data are given in Table 10.6-1. A very extensive list of values of Fp is given by Kister (K1).

Table 10.6-1. Packing Factors for Random and Structured Packing
TypeMaterialNominal size, in.Void fraction, εSurface area, a, ft2/ft3 (m2/m3)Packing factor, Fp, ft1 (m1)Relative mass-transfer coefficient, fp
Random Packing
Raschig RingsCeramic1/20.64111(364)580(1900)1.52
   10.7458(190)179(587)1.20
   1 1/20.7337(121)95(312)1.00
   20.7428(92)65(213)0.85
Berl SaddlesCeramic1/20.62142(466)240(787)1.58
   10.6876(249)110(361)1.36
   2 32(105)45(148) 
Pall RingsMetal10.9463(207)56(184)1.61
   1 1/20.9539(128)40(131)1.34
   20.9631(102)27(89)1.14
Metal Intalox (IMTP)Metal10.9770(230)41(134)1.78
   20.9830(98)18(59)1.27
Nor-PacPlastic10.9255(180)25(82) 
   20.9431(102)12(39) 
Hy-PakMetal10.9654(177)45(148)1.51
   20.9729(95)26(85)1.07
  Plastic10.9255(180)25(82) 
   20.9431(102)12(39) 
Structured Packing
Mellapak250YMetal 0.9576(249)20(66) 
 500Y   152(499)34(112) 
Flexipac2  0.9368(223)22(72) 
 4  0.98  6(20) 
Gempak2A  0.9367(220)16(52) 
 4A  0.91138(452)32(105) 
Norton Intalox2T  0.9765(213)17(56)1.98
 3T  0.9754(177)13(43)1.94
MontzB300   91(299)33(108) 
SulzerCYWire 0.85213(700)70(230) 
  Mesh       
 BX  0.90150(492)21(69) 
Data from Ref. (K1, L2, P2, S4). The relative mass-transfer coefficient, fp, is discussed in Section 10.8B.

2. Pressure drop in structured packings

An empirical correlation for structured packings is given in Fig. 10.6-6 by Kister and Gill (K1). They modified the Eckert correlation for random packings to better fit only the structured-packing data. An extrapolated line for ΔP = 0.05 in. H2O/ft and for ΔP = 2.0 has been added. The packing factors Fp to be used for structured packing are those given in Table 10.6-1 and references (K1, L2, P2). The units on the ordinate and abscissa of Fig. 10.6-6 are the same as those for Fig. 10.6-5.

Figure 10.6-6. Pressure-drop correlation for structured packings by Kister and Gill (K2). (From H. Z. Kister, Distillation Design, New York: McGraw-Hill Book Company, 1992. With permission.)


3. Flooding pressure drop in packed and structured packings

It is important for proper design to be able to predict the flooding pressure drop in towers and, hence, the limiting flow rates at flooding. Figures 10.6-5 and 10.6-6 do not predict flooding conditions. Kister and Gill (K2) have developed an empirical equation to predict the limiting pressure drop at flooding. This equation is

Equation 10.6-1


where ΔPflood is in in. H2O/ft height of packing and Fp is the packing factor in ft1 given in Table 10.6-1 for random or structured packing. To convert from English to SI units, 1.00 in. H2O/ft height = 83.33 mm H2O/m height of packing. This can be used for packing factors from 9 up to 60. It predicts all of the data for flooding within ±15% and most for ±10%. At a packing factor of 60 or higher, Eq. (10.6-1) should not be used; instead, the pressure drop at flooding can be taken as 2.00 in. H2O/ft (166.7 mm H2O/m).

The following procedure can be used to determine the limiting flow rates and the tower diameter.

1.
First, a suitable random packing or structured packing is selected, giving an Fp value.

2.
A suitable liquid-to-gas ratio GL/GG is selected along with the total gas flow rate.

3.
The pressure drop at flooding is calculated using Eq. (10.6-1), or if Fp is 60 or over, the ΔPflooding is taken as 2.0 in./ft packing height.

4.
Then the flow parameter is calculated, and using the pressure drop at flooding and either Fig. 10.6-5 or 10.6-6, the capacity parameter is read off the plot.

5.
Using the capacity parameter, the value of GG is obtained, which is the maximum value at flooding.

6.
Using a suitable % of the flooding value of GG for design, a new GG and GL are obtained. The pressure drop can also be obtained from Figure 10.6-5 or 10.6-6.

7.
Knowing the total gas flow rate and GG, the tower cross-sectional area and ID can be calculated.

4. Approximate design factors to use

In using random packing, the radio of tower diameter to packing size should be 10/1 or greater. This is to ensure good liquid and gas distribution. For every 3 m (10 ft) height of packing, a liquid redistribution should be used to prevent channeling of liquid to the sides. Random-packed towers are generally used only for diameters of 1.0 m (3.3 ft) or less. Tray towers less than 0.6 m (2 ft) in diameter are usually not used because of cleaning and access problems.

The start of loading in packed towers is usually at about 65–70% of the flooding velocity (L2). For absorption, the tower should be designed using about 50–70% of the gas flooding velocity, with the high value used at high flow parameters. For atmospheric-pressure distillation, values of 70–80% can be used (S5). For distillation and structured packing, 80% of flooding is often used in design (L3). For tray towers, see Section 11.5.

EXAMPLE 10.6-1. Pressure Drop and Tower Diameter for Ammonia Absorption

Ammonia is being absorbed in a tower using pure water at 25°C and 1.0 atm abs pressure. The feed rate is 1440 lbm/h (653.2 kg/h) and contains 3.0 mol % ammonia in air. The process design specifies a liquid-to-gas mass flow rate ratio GL/GG of 2/1 and the use of 1-in. metal Pall rings.

  1. Calculate the pressure drop in the packing and gas mass velocity at flooding. Using 50% of the flooding velocity, calculate the pressure drop, gas and liquid flows, and tower diameter.

  2. Repeat (a) above but use Mellapak 250Y structured packing.

Solution: The gas and liquid flows in the bottom of the tower are the largest, so the tower will be sized for these flows. Assume that approximately all of the ammonia is absorbed. The gas average mol wt is 28.97(0.97) + 17.0(0.03) = 28.61. The weight fraction of ammonia is 0.03(17)/(28.61) = 0.01783.


Assuming the water is dilute, from Appendix A.2-4, the water viscosity μ = 0.8937 cp. From A.2-3, the water density is 0.99708 gm/cm3. Then, ρL = 0.99708(62.43) = 62.25 lbm/ft3. Also, v = μ/ρ = 0.8937/0.99708 = 0.8963 centistokes.

From Table 10.6-1, for 1-in. Pall rings, Fp = 56 ft1. Using Eq. 10.6-1, ΔPflood = 0.115 = 0.115(56)0.7 = 1.925 in. H2O/ft packing height. The flow parameter for Fig. 10.6-5 is


Using Fig. 10.6-5, for a flow parameter of 0.06853 (abscissa) and a pressure drop of 1.925 in./ft at flooding, a capacity parameter (ordinate) of 1.7 is read off the plot. Then, substituting into the capacity parameter equation and solving for vG,


vG = 6.663 ft/s. Then GG = vGρG = 6.663(0.07309) = 0.4870 lbm/(s · ft2) at flooding. Using 50% of the flooding velocity for design, GG = 0.5(0.4870) = 0.2435 lbm/(s · ft2) [1.189 kg/(s · m2)]. Also, the liquid flow rate GL = 2.0(0.2435) = 0.4870 lbm/(s · ft2) [2.378 kg/(s · m2)].

To calculate the tower pressure drop at 50% of flooding, GG = 0.2435 and GL = 0.4870, the new capacity parameter is 0.5(1.7) = 0.85. Using this value of 0.85 and the same flow parameter, 0.06853, a value of 0.18 in. water/ft is obtained from Fig. 10.6-5.

The tower cross-sectional area = (1440/3600 lbm/s)(1/0.2435 lbm/(s · ft2)) = 1.6427 ft2 = (π/4)D2. Solving, D = 1.446 ft (0.441 m). The amount of ammonia in the outlet water assuming all of the ammonia is absorbed is 0.01783(1440) = 25.68 lb. Since the liquid flow rate is 2 times the gas flow rate, the total liquid flow rate is 2.0(1440) = 2880 lbm/hr. Hence, the flow rate of the pure inlet water is 2880 − 25.68 = 2858.3 lbm/s.

For part (b), using Mellapak 250Y, Fp = 20 from Table 10.6-1. The Δpflood = 0.115 = 0.115(20)0.7 = 0.936 in. water/ft. The flow parameter of 0.06853 is the same. Using Fig. 10.6-6, a capacity parameter = 1.38 at flooding is obtained. Then,


Solving, GG = 0.6615 lbm/(h · ft2). For 50% flood, GG = 0.5(0.6615) = 0.3308 lbm/(s · ft2) [1.615 kg/(s · m2)]. Also, the liquid flow rate is GL = 2(0.3308) = 0.6616 lbm/(s · ft2) [3.230 kg/(s · m2)]. The new capacity parameter is 0.5(1.38) = 0.69. Using this and Fig. 10.6-6, a pressure drop of 0.11 in. water/ft is obtained.

The tower cross-sectional area = (1440/3600)(1/0.3308) = 1.209 ft2 = (π/4)D2. Solving, D = 1.241 ft (0.3784 m). Note that the tower with structured packing uses about 25% less cross-sectional area.


10.6D. Design of Plate Absorption Towers

1. Operating-line derivation

A plate (tray) absorption tower has the same process flow diagram as the countercurrent multiple-stage process in Fig. 10.3-2 and is shown as a vertical tray tower in Fig. 10.6-7. In the case of solute A diffusing through a stagnant gas (B) and then into a stagnant fluid, as in the absorption of acetone (A) from air (B) by water, the moles of inert or stagnant air and inert water remain constant throughout the entire tower. If the rates are V' kg mol inert air/s and L' kg mol inert solvent water/s or in kg mol inert/s · m2 units (lb mol inert/h · ft2), an overall material balance on component A in Fig. 10.6-7 is

Equation 10.6-2


Figure 10.6-7. Material balance in an absorption tray tower.


A balance around the dashed-line box gives

Equation 10.6-3


where x is the mole fraction A in the liquid, y the mole fraction of A in the gas, Ln the total moles liquid/s, and Vn+1 the total moles gas/s. The total flows/s of liquid and gas vary throughout the tower.

Equation (10.6-3) is the material balance or operating line for the absorption tower and is similar to Eq. (10.3-13) for a countercurrent-stage process, except that the inert streams L' and V' are used instead of the total flow rates L and V. Equation (10.6-3) relates the concentration yn+1 in the gas stream with xn in the liquid stream passing it. The terms V', L', x0, and y1 are constant and are usually known or can be determined.

2. Graphical determination of the number of trays

A plot of the operating-line equation (10.6-3) as y versus x will give a curved line. If x and y are very dilute, the denominators 1 − x and 1 − y will be close to 1.0, and the line will be approximately straight, with a slope ≅ L'/V'. The number of theoretical trays is determined by simply stepping off the number of trays, as done in Fig. 10.3-3 for a countercurrent multiple-stage process.

EXAMPLE 10.6-2. Absorption of SO2 in a Tray Tower

A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 293 K (68°F). The entering gas contains 20 mol % SO2 and that leaving 2 mol % at a total pressure of 101.3 kPa. The inert air flow rate is 150 kg air/h · m2, and the entering water flow rate is 6000 kg water/h · m2. Assuming an overall tray efficiency of 25%, how many theoretical trays and actual trays are needed? Assume that the tower operates at 293 K (20°C).

Solution: First calculating the molar flow rates,


Referring to Fig. 10.6-7, yN+1 = 0.20, y1 = 0.02, and x0 = 0. Substituting into Eq. (10.6-2) and solving for xN,


Substituting into Eq. (10.6-3), using V' and L' as kg mol/h · m2 instead of kg mol/s · m2,


In order to plot the operating line, several intermediate points will be calculated. Setting yn+1 = 0.07 and substituting into the operating equation,


Hence, xn = 0.000855. To calculate another intermediate point, we set yn+1 = 0.13, and xn is calculated as 0.00201. The two end points and the two intermediate points on the operating line are plotted in Fig. 10.6-8, as are the equilibrium data from Appendix A.3. The operating line is somewhat curved.

Figure 10.6-8. Theoretical number of trays for absorption of SO2 in Example 10.6-2.


The number of theoretical trays is determined by stepping off the steps to give 2.4 theoretical trays. The actual number of trays is 2.4/0.25 = 9.6 trays.


10.6E. Design of Packed Towers for Absorption

1. Operating-line derivation

For the case of solute A diffusing through a stagnant gas and then into a stagnant fluid, an overall material balance on component A in Fig. 10.6-9 for a packed tower is (note the notation change in Fig. 10.6-9 with inlets y1 and x2)

Equation 10.6-4


Figure 10.6-9. Material balance for a countercurrent packed absorption tower.


where L' is kg mol inert liquid/s or kg mol inert liquid/s · m2, V' is kg mol inert gas/s or kg mol inert gas/s · m2, and y1 and x1 are mole fractions A in gas and liquid, respectively. The flows L' and V' are constant throughout the tower, but the total flows L and V are not constant.

A balance around the dashed-line box in Fig. 10.6-9 gives the operating-line equation:

Equation 10.6-5


This equation, when plotted on yx coordinates, will give a curved line, as shown in Fig. 10.6-10a. Equation (10.6-5) can also be written in terms of partial pressure p1 of A, where y1/(1 − y1) = p1/(Pp1), and so on. If x and y are very dilute, (1 − x) and (1 − y) can be taken as 1.0 and Eq. (10.6-5) becomes

Equation 10.6-6


Figure 10.6-10. Location of operating lines: (a) for absorption of A from V to L stream, (b) for stripping of A from L to V stream.


This has a slope L'/V' and the operating line is essentially straight.

When the solute is being transferred from the L to the V stream, the process is called stripping. The operating line is below the equilibrium line, as shown in Fig. 10.6-10b.

2. Limiting and optimum L'/V' ratios.

In the absorption process, the inlet gas flow V1 (Fig. 10.6-9) and its composition y1 are generally set. The exit concentration y2 is also usually set by the designer, and the concentration x2 of the entering liquid is often fixed by process requirements. Hence, the amount of the entering liquid flow L2 or L' is open to choice.

In Fig. 10.6-11a the flow V1 and the concentrations y2, x2, and y1 are set. When the operating line has a minimum slope and touches the equilibrium line at point P, the liquid flow L' is a minimum at . The value of x1 is a maximum at x1max when L' is a minimum. At point P the driving forces yy*, yyi, x* − x, and xix are all zero. To solve for , the values y1 and x1max are substituted into the operating-line equation. In some cases, if the equilibrium line is curved concavely downward, the minimum value of L is reached by the operating line becoming tangent to the equilibrium line instead of intersecting it.

Figure 10.6-11. Operating line for limiting conditions: (a) absorption, (b) stripping.


The choice of the optimum L'/V' ratio to use in the design depends on an economic balance. In absorption, too high a value requires a large liquid flow and hence a large-diameter tower. The cost of recovering the solute from the liquid by distillation will be high. A small liquid flow results in a high tower, which is also costly. As an approximation, for absorption the optimum liquid flow rate can be taken as 1.2–1.5 times the limiting rate , with 1.5 usually used (S6).

For stripping or transfer of solute from L to V, where the operating line has a maximum slope and touches the equilibrium line at point P in Fig. 10.6-11b, then the gas flow is at the minimum The value of y2 is at a maximum at y2max. As in absorption, the optimum gas flow rate is taken at about 1.5 times . In liquid extraction, covered later in Section 12.7, the same conditions for stripping hold for extracting solute from feed liquid L to solvent V.

3. Analytical equations for theoretical number of steps or trays

Analytical equations to calculate the number of theoretical trays N in an absorption tower are the same as Eqs. (10.3-22) and (10.3-25) for calculating the number of stages in countercurrent stage processes. These are as follows. (Note that the notation refers to Fig. 10.6-9 and not Fig. 10.3-3.)

For transfer of the solute from phase V to phase L (absorption),

Equation 10.6-7


For transfer of the solute from phase L to phase V (stripping),

Equation 10.6-8


where A = L/mV.

If the equilibrium line and/or operating line is slightly curved, m and A = L/mV will vary. For absorption (referring to Fig. 10.6-9), at the bottom concentrated end tray, the slope m1 or tangent at the point x1 on the equilibrium line is used. At the dilute top tray, the slope m2 of the equilibrium line at point y2 is used. Then A1 = L1/m1V1, A2 = L2/m2V2, and . Also, the dilute m2 is used in Eq. (10.6-7). For stripping, at the top or concentrated stage, the slope m2 or tangent at point y2 on the equilibrium line is used. At the bottom or dilute end, the slope m1 at point x1 on the equilibrium line is used. Then, A1 = L1/m1V1, A2 = L2/m2V2, and . Again, the dilute m1 is used in Eq. (10.6-8).

EXAMPLE 10.6-3. Minimum Liquid Flow Rate and Analytical Determination of Number of Trays

A tray tower is absorbing ethyl alcohol from an inert gas stream using pure water at 303 K and 101.3 kPa. The inlet gas stream flow rate is 100.0 kg mol/h and it contains 2.2 mol % alcohol. It is desired to recover 90% of the alcohol. The equilibrium relationship is y = mx = 0.68x for this dilute stream. Using 1.5 times the minimum liquid flow rate, determine the number of trays needed. Do this graphically and also using the analytical equations.

Solution: The given data are y1 = 0.022, x2 = 0, V1 = 100.0 kg mol/h, m = 0.68. Then V' = V1(1 − y1) = 100.0(1 − 0.022) = 97.8 kg mol inert/h. Moles alcohol/h in V1 are 100 − 97.8 = 2.20. Removing 90%, moles/h in outlet gas V2 = 0.10(2.20) = 0.220. V2 = V' + 0.220 = 97.8 + 0.22 = 98.02. y2 = 0.22/98.02 = 0.002244. The equilibrium line is plotted in Fig. 10.6-12 along with y2, x2, and y1. The operating line for minimum liquid flow Lminis drawn from y2, x2 to point P, touching the equilibrium line where x1max = y1/m = 0.022/0.68 = 0.03235. Substituting into the operating-line equation (10.6-4) and solving for Lmin,


Figure 10.6-12. Operating line for minimum and actual liquid flow in Example 10.6-3.


= 59.24 kg mol/h. Using 1.5, L' = 1.5(59.24) = 88.86. Using L' = 88.86 in Eq. (10.6-4) and solving for the outlet concentration, x1 = 0.02180. The operating line is plotted as a straight line through the points y2, x2 and y1, x1 in Fig. 10.6-12. An intermediate point is calculated by setting y = 0.012 in Eq. (10.6-5) and solving for x = 0.01078. Plotting this point shows that the operating line is very linear. This occurs because the solutions are dilute.

The number of theoretical trays obtained by stepping them off is 4.0 trays. The total flow rates are V1 = 100.0, V2 = V'/(1 − y2) = 97.8/(1 − 0.002244) = 98.02, L2 = L' = 88.86, and L1 = L'/(1 − x1) = 88.86/(1 − 0.02180) = 90.84. To calculate the number of trays analytically, A1 = L1/mV1 = 90.84/(0.68)(100) = 1.336, A2 = L2/mV2 = 88.86/(0.68)(98.02) = 1.333. Using the geometric average, A = 1.335. Using Eq. (10.6-7),


N = 4.04 theoretical steps, which agrees closely with the 4.0 steps obtained graphically.


4. Film and overall mass-transfer coefficients in packed towers

As discussed in Section 10.5, it is very difficult to measure experimentally the interfacial area A m2 between phases L and V. Also, it is difficult to measure the film coefficients and and the overall coefficients and . Usually, experimental measurements in a packed tower yield a volumetric mass-transfer coefficient that combines the interfacial area and mass-transfer coefficient.

Defining a as interfacial area in m2 per m3 volume of packed section, the volume of packing in a height dz m (Fig. 10.6-9) is S dz, and

Equation 10.6-9


where S is m2 cross-sectional area of tower. The volumetric film and overall mass-transfer coefficients are then defined as


5. Design method for packed towers using mass-transfer coefficients

For absorption of A from stagnant B, the operating-line equation (10.6-5) holds. For the differential height of tower dz in Fig. 10.6-9, the moles of A leaving V equal the moles entering L:

Equation 10.6-10


where V = kg mol total gas/s, L = kg mol total liquid/s, and d(Vy) = d(Lx) = kg mol A transferred/s in height dz m. The kg mol A transferred/s from Eq. (10.6-10) must equal the kg mol A transferred/s from the mass-transfer equation for NA. Equation (10.4-8) gives the flux NA using the gas-film and liquid-film coefficients:

Equation 10.4-8


where (1 − yA)iM and (1 − xA)iM are defined by Eqs. (10.4-6) and (10.4-7). Multiplying the left-hand side of Eq. (10.4-8) by dA and the two right-side terms by aS dz from Eq. (10.6-9),

Equation 10.6-11


where NA dA = kg mol A transferred/s in height dz m (lb mol/h).

Equating Eq. (10.6-10) to (10.6-11) and using yAG for the bulk gas phase and xAL for the bulk liquid phase,

Equation 10.6-12


Equation 10.6-13


Since V' = V(1 − yAG) or V = V'/(1 − yAG),

Equation 10.6-14


Substituting V for V'/(1 − yAG) in Eq. (10.6-14) and then equating Eq. (10.6-14) to (10.6-12),

Equation 10.6-15


Repeating for Eq. (10.6-13), since L = L'/(1 − xAL),

Equation 10.6-16


Dropping the subscripts A, G, and L and integrating, the final equations are as follows using film coefficients:

Equation 10.6-17


Equation 10.6-18


In a similar manner, the final equations can be derived using overall coefficients:

Equation 10.6-19


Equation 10.6-20


In the general case, the equilibrium and the operating lines are usually curved, and , , and vary somewhat with total gas and liquid flows. Then Eqs. (10.6-17)–(10.6-20) must be integrated numerically or graphically. The methods for doing this for concentrated mixtures will be discussed in Section 10.7. Methods for dilute gases will be considered below.

10.6F. Simplified Design Methods for Absorption of Dilute Gas Mixtures in Packed Towers

Since a considerable percentage of the absorption processes include absorption of a dilute gas A, these cases will be considered using a simplified design procedure.

The concentrations can be considered dilute for engineering design purposes when the mole fractions y and x in the gas and liquid streams are less than about 0.10, that is, 10%. The flows will vary by less than 10% and the mass-transfer coefficients by considerably less than this. As a result, the average values of the flows V and L and the mass-transfer coefficients at the top and bottom of the tower can be taken outside the integral. Likewise, the terms (1 − y)iM/(1 − y), (1 − y)*M/(1 − y), (1 − x)iM/(1 − x), and (1 − x)*M/(1 − x) can be taken outside and average values at the top and bottom of the tower used. (Often these terms are close to 1.0 and can be dropped out entirely.) Then Eqs. (10.6-17)–(10.6-20) become

Equation 10.6-21


Equation 10.6-22


Equation 10.6-23


Equation 10.6-24


Since the solutions are dilute, the operating line will be essentially straight. Assuming the equilibrium line is approximately straight over the range of concentrations used, (yyi) varies linearly with y and also with x:

Equation 10.6-25


where k and b are constants. Therefore, the integral of Eq. (10.6-21) can be integrated to give the following:

Equation 10.6-26


where (yyi)M is the log mean driving force:

Equation 10.6-27


Similarly, Eq. (10.6-23) gives

Equation 10.6-28


If the term (1 − y)iM/(1 − y) is considered 1.0, then by substituting Eq. (10.6-26) into (10.6-21) and doing the same for Eqs. (10.6-22)–(10.6-24), the final results are as follows:

Equation 10.6-29


Equation 10.6-30


Equation 10.6-31


Equation 10.6-32


where the left side is the kg mol absorbed/s · m2 (lb mol/h · ft2) by material balance and the right-hand side is the rate equation for mass transfer. The value of V is the average (V1 + V2)/2 and of L is (L1 + L2)/2.

Equations (10.6-29)–(10.6-32) can be used in slightly different ways. The general steps to follow are discussed below and shown in Fig. 10.6-13.

1.
The operating-line equation (10.6-5) is plotted as in Fig. 10.6-13 as a straight line. Calculate V1, V2, and Vav = (V1 + V2)/2; also calculate L1, L2, and Lav = (L1 + L2)/2.

2.
Average experimental values of the film coefficients and are available or may be obtained from empirical correlations. The interface compositions yi1 and xi1 at point y1, x1 in the tower are determined by plotting line P1M1, whose slope is calculated by Eq. (10.6-33):

Equation 10.6-33


Equation 10.6-34


If terms (1 − x)iM and (1 − y)iM are used, the procedure is trial and error, as in Example 10.4-1. However, since the solutions are dilute, the terms (1 − x1) and (1 − y1) can be used in Eq. (10.6-34) without trial and error and with only a small error in slope. If the coefficients kya and kxa are available for the approximate concentration range, they can be used, since they include the terms (1 − x)iM and (1 − y)iM. For line P2M2 at the other end of the tower, values of yi2 and xi2 are determined using Eq. (10.6-33) or (10.6-34) and y2 and x2.

3.
If the overall coefficient is being used, and are determined as shown in Fig. 10.6-13. If is used, and are obtained.

4.
Calculate the log mean driving force (yyi)M by Eq. (10.6-27) if is used. For , (yy*)M is calculated by Eq. (10.6-28). Using the liquid coefficients, the appropriate driving forces are calculated.

5.
Calculate the column height z m by substituting into the appropriate form of Eqs. (10.6-29)–(10.6-32).

Figure 10.6-13. Operating-line and interface compositions in a packed tower for absorption of dilute gases.


EXAMPLE 10.6-4. Absorption of Acetone in a Packed Tower

Acetone is being absorbed by water in a packed tower having a cross-sectional area of 0.186 m2 at 293 K and 101.32 kPa (1 atm). The inlet air contains 2.6 mol % acetone and outlet 0.5%. The gas flow is 13.65 kg mol inert air/h (30.1 lb mol/h). The pure water inlet flow is 45.36 kg mol water/h (100 lb mol/h). Film coefficients for the given flows in the tower are = 3.78 × 102 kg mol/s · m3 · mol frac (8.50 lb mol/h · ft3 · mol frac) and = 6.16 × 102 kg mol/s · m3 · mol frac (13.85 lb mol/h · ft3 · mol frac). Equilibrium data are given in Appendix A.3.

  1. Calculate the tower height using..

  2. Repeat, using..

  3. Calculate and the tower height.

Solution: From Appendix A.3 for acetone–water and xA = 0.0333 mol frac, pA = 30/760 = 0.0395 atm or yA = 0.0395 mol frac. Hence, the equilibrium line is yA = mxA or 0.0395 = m (0.0333). Then, y = 1.186x. This equilibrium line is plotted in Fig. 10.6-14. The given data are L' = 45.36 kg mol/h, V' = 13.65 kg mol/h, y1 = 0.026, y2 = 0.005, and x2 = 0.

Figure 10.6-14. Location of interface compositions for Example 10.6-4.


Substituting into Eq. (10.6-4) for an overall material balance using flow rates as kg mol/h instead of kg mol/s,


The points y1, x1 and y2, x2 are plotted in Fig. 10.6-14 and a straight line is drawn for the operating line.

Using Eq. (10.6-34) the approximate slope at y1, x1 is


Plotting this line through y1, x1, the line intersects the equilibrium line at yi1 = 0.0154 and xi1 = 0.0130. Also, = 0.0077. Using Eq. (10.6-33) to calculate a more accurate slope, the preliminary values of yi1 and xi1 will be used in the trial-and-error solution. Substituting into Eq. (10.4-6),


Using Eq. (10.4-7),


Substituting into Eq. (10.6-33),


Hence, the approximate slope and interface values are accurate enough.

For the slope at point y2, x2,


The slope changes little in the tower. Plotting this line, yi2 = 0.0020, xi2 = 0.0018, and = 0.

Substituting into Eq. (10.6-27),


To calculate the total molar flow rates in kg mol/s,


For part (a), substituting into Eq. (10.6-29) and solving,


For part (b), using an equation similar to Eq. (10.6-27),


Substituting into Eq. (10.6-30) and solving,


This agrees with part (a) quite closely.

For part (c), substituting into Eq. (10.4-25) for point y1, x1,


The overall mass-transfer coefficient at point y1, x1 is calculated by substituting into Eq. (10.4-24):


Substituting into Eq. (10.6-28),


Finally, substituting into Eq. (10.6-31),


This is in agreement with parts (a) and (b).


10.6G. Design of Packed Towers Using Transfer Units

1. Design for concentrated solutions

Another and in some ways more useful design method for packed towers is the use of the transfer-unit concept. For the most common case of A diffusing through stagnant and nondiffusing B, Eqs. (10.6-17)–(10.6-20) can be rewritten as

Equation 10.6-35


Equation 10.6-36


Equation 10.6-37


Equation 10.6-38


where

Equation 10.6-39


Equation 10.6-40


Equation 10.6-41


Equation 10.6-42


where (1 − y)iM is defined by Eq. (10.4-6), (1 − x)iM by Eq. (10.4-7), (1 − y)*M by Eq. (10.4-25), and (1 − x)*M by Eq. (10.4-27). The units of H are in m (ft). HG is the height of a transfer unit based on the gas film. The values of the heights of transfer units are more constant than the mass-transfer coefficients. For example, is often proportional to V0.7, so then . The average values of the mass-transfer coefficients, (1 − y)iM, (1 − y)*M, (1 − x)iM, and (1 − x)*M, must be used in Eqs. (10.6-39)–(10.6-42).

The integrals on the right side of Eqs. (10.6-35)–(10.6-38) are the number of transfer units NG, NL, NOG, and NOL, respectively. The height of the packed tower is then

Equation 10.6-43


These equations are basically no different from those using mass-transfer coefficients. One still needs and to determine interface concentrations. Disregarding (1 − y)iM/(1 − y), which is near 1.0 in Eq. (10.6-35), the greater the amount of absorption (y1y2) or the smaller the driving force (yyi), the larger the number of transfer units NG and the taller the tower.

2. Design for dilute solutions

When the solutions are dilute, with concentrations below 10%, the terms (1 − y)iM/(1 − y), (1 − x)iM/(1 − x), (1 − y)*M/(1 − y), and (1 − x)*M/(1 − x) in Eqs. (10.6-35)–(10.6-38) can be taken outside the integral and average values used. Since these are quite close to 1.0, they can be dropped out. The equations then become

Equation 10.6-44


Equation 10.6-45


Equation 10.6-46


Equation 10.6-47


If both the operating and equilibrium lines are straight and dilute, the integrals in Eqs. (10.6-44)–(10.6-47) can be integrated, giving the number of transfer units:

Equation 10.6-48


Equation 10.6-49


Equation 10.6-50


Equation 10.6-51


where (yyi)M is defined in Eq. (10.6-27) and (yy*)M in (10.6-28).

When the major resistance to mass transfer is in the gas phase, as in absorption of acetone from air by water, the overall number of transfer units based on the gas phase NOG or the film NG should be used. When the major resistance is in the liquid phase, as in absorption of O2 by water or stripping of a slightly soluble solute from water, NOL or NL should be employed. This was also discussed in detail in Section 10.4D. Often the film coefficients are not available, and then it is more convenient to use NOG or NOL.

By combining the operating line with the integrals in Eqs. (10.6-46) and (10.6-47), using the equilibrium-line equation y = mx, and letting A = L/mV, different forms of the equations for absorption with NOG and for stripping with NOL are obtained:

Equation 10.6-52


Equation 10.6-53


The values of m and A to use in the above equations when operating and/or equilibrium lines are slightly curved are discussed in detail in Section 10.6E for use with Eqs. (10.6-7) and (10.6-8).

When the operating and equilibrium lines are straight and not parallel, the number of overall gas-transfer units NOG for absorption in Eq. (10.6-52) is related to the number of theoretical trays or stages N in Eq. (10.6-7) by

Equation 10.6-54


The height of a theoretical tray or stage, HETP, in m is related to HOG by

Equation 10.6-55


A detailed discussion of HETP is given in Section 11.5D. Note that if the operating and equilibrium lines are parallel (A = 1.0), then HOG = HETP and NOG = N. Equations (10.6-7) and (10.6-8) can be used to analytically calculate N, the number of theoretical steps.

EXAMPLE 10.6-5. Use of Transfer Units and Analytical Equations for Packed Tower

Repeat Example 10.6-4 using transfer units and height of a transfer unit as follows:

  1. Use HG and NG to calculate tower height.

  2. Use HOG and NOG to calculate tower height.

  3. Use Eq. (10.6-52) to calculate NOG and tower height.

  4. Using the analytical equations, calculate HETP from Eq. (10.6-55), number of theoretical steps N from Eq. (10.6-7), and tower height.

Solution: From Example 10.6-4, = 3.78 × 102 kg mol/s · m2 · mol frac, = 2.183 × 102 kg mol/s · m2 · mol frac, average V = 3.852 × 103 kg mol/s, and S = 0.186 m2.

For part (a), from Eq. (10.6-39),


From Eq. (10.6-27) in Example 10.6-4, (yyi)iM = 0.00602. Also, y1 = 0.026 and y2 = 0.005. Then, using Eq. (10.6-48),


Substituting in Eq. (10.6-44),


For part (b), using Eq. (10.6-41),


From Eq. (10.6-28) in Example 10.6-4, (yy*)*M = 0.01025. Then using Eq. (10.6-50),


From Eq. (10.6-46),


Note that the number of transfer units NOG = 2.049 is not the same as NG = 3.488.

For part (c), from Example 10.6-4, m = 1.186, and the average V = 3.852 × 103, L = 1.260 × 102, y1 = 0.026, y2 = 0.005, and x2 = 0.

Then A = L/mV = 1.260 × 102/(1.186 × 3.852 × 103) = 2.758. Substituting into Eq. (10.6-52),


This compares closely with the value of 2.049 in part (b). Also, z = HOGNOG = (0.949)(2.043) = 1.939 m.

For part (d), to calculate HETP, Eq. (10.6-55) is used:


Using Eq. (10.6-7) to calculate N,


This compares with a value of 1.35 steps obtained graphically in Figure 10.6-14:



In order to relate HOG to the film coefficients HG and HL, Eq. (10.4-24) is rewritten for dilute solutions:

Equation 10.6-56


Then, substituting HG = V/S, HL = L/S, and HOG = V/ into Eq. (10.6-56) and canceling like terms,

Equation 10.6-57


Similarly, using Eq. (10.4-26),

Equation 10.6-58


Note that often in the literature, the terms Hy, Hx, HOy, and HOx are used instead of HG, HL, HOG, and HOL; they are identical to each other.

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