10.4. MASS TRANSFER BETWEEN PHASES

10.4A. Introduction and Equilibrium Relations

1. Introduction to interphase mass transfer

In Chapter 7 we considered mass transfer from a fluid phase to another phase, which was most often a solid phase. The solute A was usually transferred from the fluid phase by convective mass transfer and through the solid by diffusion. In the present section we shall be concerned with the mass transfer of solute A from one fluid phase by convection and then through a second fluid phase by convection. For example, the solute may diffuse through a gas phase and then diffuse through and be absorbed in an adjacent and immiscible liquid phase. This occurs in the case of absorption of ammonia from air by water.

The two phases are in direct contact with each other, such as in a packed, tray, or spray-type tower, and the interfacial area between the phases is usually not well defined. In two-phase mass transfer, a concentration gradient will exist in each phase, causing mass transfer to occur. At the interface between the two fluid phases, equilibrium exists in most cases.

2. Equilibrium relations

Even when mass transfer is occurring, equilibrium relations are important to determine concentration profiles for predicting rates of mass transfer. In Section 10.2 the equilibrium relation in a gas–liquid system and Henry's law were discussed. In Section 7.1C a discussion covered equilibrium distribution coefficients between two phases. These equilibrium relations will be used in discussion of mass transfer between phases in this section.

10.4B. Concentration Profiles in Interphase Mass Transfer

In the majority of mass-transfer systems, two phases, which are essentially immiscible in each other, are present together with an interface between these two phases. Assuming solute A is diffusing from the bulk gas phase G to the liquid phase L, it must pass through phase G, through the interface, and then into phase L in series. A concentration gradient must exist to cause this mass transfer through the resistances in each phase, as shown in Fig. 10.4-1. The average or bulk concentration of A in the gas phase in mole fraction units is y_AG, where y_AG = p_A/P, and in the bulk liquid phase in mole fraction units it is x_AL.

The concentration in the bulk gas phase y_AG decreases to y_Ai at the interface. The liquid concentration starts at x_Ai at the interface and falls to x_AL. At the interface, since there would be no resistance to transfer across this interface, y_Ai and x_Ai are in equilibrium and are related by the equilibrium distribution relation

Equation 10.4-1

where y_Ai is a function of x_Ai. They are related by an equilibrium plot such as Fig. 10.1-1. If the system follows Henry's law, y_AP or p_A and x_A are related by Eq. (10.2-2) at the interface.

Experimentally, the resistance at the interface has been shown to be negligible for most cases of mass transfer where chemical reactions do not occur, such as absorption of common gases from air to water and extraction of organic solutes from one phase to another. However, there are some exceptions. Certain surface-active compounds may concentrate at the interface and cause an “interfacial resistance” that slows down the diffusion of solute molecules. Theories for predicting when interfacial resistance may occur are often obscure and unreliable.

10.4C. Mass Transfer Using Film Mass-Transfer Coefficients and Interface Concentrations

1. Equimolar counterdiffusion

For equimolar counterdiffusion the concentrations of Fig. 10.4-1 can be plotted on an x-y diagram as in Fig. 10.4-2. Point P represents the bulk phase compositions x_AG and x_AL of the two phases and point M the interface concentrations y_Ai and x_Ai. For A diffusing from the gas to liquid and B in equimolar counterdiffusion from liquid to gas,

Equation 10.4-2

where is the gas-phase mass-transfer coefficient in kg mol/s · m² · mol frac (g mol/s · cm² · mol frac, lb mol/h · ft² · mol frac) and the liquid-phase mass-transfer coefficient in kg mol/s · m² · mol frac (g mol/s · cm² · mol frac, lb mol/h · ft² · mol frac). Rearranging Eq. (10.4-2),

Equation 10.4-3

The driving force in the gas phase is (y_AG − y_Ai) and in the liquid phase it is (x_Ai − x_AL). The slope of the line PM is This means that if the two film coefficients and are known, the interface compositions can be determined by drawing line PM with a slope intersecting the equilibrium line.

The bulk-phase concentrations y_AG and x_AL can be determined by simply sampling the mixed bulk gas phase and sampling the mixed bulk liquid phase. The interface concentrations are determined by Eq. (10.4-3).

2. Diffusion of A through stagnant or nondiffusing B

For the common case of A diffusing through a stagnant gas phase and then through a stagnant liquid phase, the concentrations are shown in Fig. 10.4-3, where P again represents bulk-phase compositions and M interface compositions. The equations for A diffusing through a stagnant gas and then through a stagnant liquid are

Equation 10.4-4

Now,

Equation 10.4-5

where

Equation 10.4-6

Equation 10.4-7

Then,

Equation 10.4-8

Note that (1 − y_A)_iM is the same as y_BM of Eq. (7.2-11) but is written for the interface, and (1 − x_A)_iM is the same as x_BM of Eq. (7.2-11). Using Eq. (10.4-8) and rearranging,

Equation 10.4-9

The slope of the line PM in Fig. 10.4-3 to obtain the interface compositions is given by the left-hand side of Eq. (10.4-9). This differs from the slope of Eq. (10.4-3) for equimolar counterdiffusion by the terms (1 − y_A)_iM and (1 − x_A)_iM. When A is diffusing through stagnant B and the solutions are dilute, (1 − y_A)_iM and (1 − x_A)_iM are close to 1.

A trial-and-error method is needed to use Eq. (10.4-9) to get the slope, since the left-hand side contains the terms y_Ai and x_Ai that are being sought. For the first trial (1 − y_A)_iM and (1 − x_A)_iM are assumed to be 1.0, and Eq. (10.4-9) is used to get the slope and y_Ai and x_Ai values. Then for the second trial, these values of y_Ai and x_Ai are used to calculate a new slope to get new values of y_Ai and x_Ai. This is repeated until the interface compositions do not change. Three trials are usually sufficient.

EXAMPLE 10.4-1. Interface Compositions in Interphase Mass Transfer The solute A is being absorbed from a gas mixture of A and B in a wetted-wall tower with the liquid flowing as a film downward along the wall. At a certain point in the tower the bulk gas concentration yAG = 0.380 mol fraction and the bulk liquid concentration is xAL = 0.100. The tower is operating at 298 K and 1.013 × 105 Pa and the equilibrium data are as follows: xA yA xA yA 0 0 0.20 0.131 0.05 0.022 0.25 0.187 0.10 0.052 0.30 0.265 0.15 0.087 0.35 0.385 The solute A diffuses through stagnant B in the gas phase and then through a nondiffusing liquid. Using correlations for dilute solutions in wetted-wall towers, the film mass-transfer coefficient for A in the gas phase is predicted as ky = 1.465 × 10−3 kg mol A/s · m2 · mol frac (1.08 lb mol/h · ft2 · mol frac) and for the liquid phase as kx = 1.967 × 10−3 kg mol A/s · m2 · mol frac (1.45 lb mol/h · ft2 · mol frac). Calculate the interface concentrations yAi and xAi and the flux NA. Solution: Since the correlations are for dilute solutions, (1 − yA)iM and (1 − xA)iM are approximately 1.0 and the coefficients are the same as and . The equilibrium data are plotted in Fig. 10.4-4. Point P is plotted at yAG = 0.380 and xAL = 0.100. For the first trial, (1 − yA)iM and (1 − xA)iM are assumed as 1.0 and the slope of line PM is, from Eq. (10.4-9), Figure 10.4-4. Location of interface concentrations for Example 10.4-1. A line through point P with a slope of −1.342 is plotted in Fig. 10.4-4 intersecting the equilibrium line at M1, where yAi = 0.183 and xAi = 0.247. For the second trial we use yAi and xAi from the first trial to calculate the new slope. Substituting into Eqs. (10.4-6) and (10.4-7), Substituting into Eq. (10.4-9) to obtain the new slope, A line through point P with a slope of −1.163 is plotted and intersects the equilibrium line at M, where yAi = 0.197 and xAi = 0.257. Using these new values for the third trial, the following values are calculated: This slope of −1.160 is essentially the same as the slope of −1.163 for the second trial. Hence, the final values are yAi = 0.197 and xAi = 0.257 and are shown as point M. To calculate the flux, Eq. (10.4-8) is used: Note that the flux NA through each phase is the same as in the other phase, which should be the case at steady state.

10.4D. Overall Mass-Transfer Coefficients and Driving Forces

1. Introduction

Film or single-phase mass-transfer coefficients and or k_y and k_x are often difficult to measure experimentally, except in certain experiments designed so that the concentration difference across one phase is small and can be neglected. As a result, overall mass-transfer coefficients and are measured based on the gas phase or liquid phase. This method is used in heat transfer, where overall heat-transfer coefficients are measured based on inside or outside areas instead of film coefficients.

The overall mass transfer is defined as

Equation 10.4-10

where is based on the overall gas-phase driving force in kg mol/s · m² · mol frac, and is the value that would be in equilibrium with x_AL, as shown in Fig. 10.4-2. Also, is defined as

Equation 10.4-11

where is based on the overall liquid-phase driving force in kg mol/s · m² · mol frac and is the value that would be in equilibrium with y_AG.

2. Equimolar counterdiffusion and/or diffusion in dilute solutions

Equation (10.4-2) holds for equimolar counterdiffusion, or, when the solutions are dilute, Eqs. (10.4-8) and (10.4-2) are identical:

Equation 10.4-2

From Fig. 10.4-2,

Equation 10.4-12

Between the points E and M the slope m' can be given as

Equation 10.4-13

Solving Eq. (10.4-13) for (y_Ai − ) and substituting into Eq. (10.4-12),

Equation 10.4-14

Then, on substituting Eqs. (10.4-10) and (10.4-2) into (10.4-14) and canceling out N_A,

Equation 10.4-15

The left-hand side of Eq. (10.4-15) is the total resistance based on the overall gas driving force and equals the gas film resistance plus the liquid film resistance .

In a similar manner, from Fig. 10.4-2,

Equation 10.4-16

Equation 10.4-17

Proceeding as before,

Equation 10.4-18

Several special cases of Eqs. (10.4-15) and (10.4-18) will now be discussed. The numerical values of and are very roughly similar. The values of the slopes m' and m'' are very important. If m' is quite small, so that the equilibrium curve in Fig. 10.4-2 is almost horizontal, a small value of y_A in the gas will give a large value of x_A in equilibrium in the liquid. The gas solute A is then very soluble in the liquid phase, and hence the term in Eq. (10.4-15) is very small. Then,

Equation 10.4-19

and the major resistance is in the gas phase, or the “gas phase is controlling.” The point M has moved down very close to E, so that

Equation 10.4-20

Similarly, when m" is very large, the solute A is very insoluble in the liquid, becomes small, and

Equation 10.4-21

The “liquid phase is controlling” and x_Ai ≅ . Systems for absorption of oxygen or CO₂ from air by water are similar to Eq. (10.4-21).

3. Diffusion of A through stagnant or nondiffusing B

For the case of A diffusing through nondiffusing B, Eqs. (10.4-8) and (10.4-14) hold and Fig. 10.4-3 is used:

Equation 10.4-8

Equation 10.4-14

We must, however, define the equations for the flux using overall coefficients as follows:

Equation 10.4-22

The bracketed terms are often written as follows:

Equation 10.4-23

where K_y is the overall gas mass-transfer coefficient for A diffusing through stagnant B and K_x the overall liquid mass-transfer coefficient. These two coefficients are concentration-dependent. Substituting Eqs. (10.4-8) and (10.4-22) into (10.4-14), we obtain

Equation 10.4-24

where

Equation 10.4-25

Similarly, for ,

Equation 10.4-26

where

Equation 10.4-27

It should be noted that the relations derived here also hold for any two-phase system, where y stands for one phase and x for the other phase. For example, for the extraction of the solute acetic acid (A) from water (y phase) by isopropyl ether (x phase), the same relations will hold.

EXAMPLE 10.4-2. Overall Mass-Transfer Coefficients from Film Coefficients Using the same data as in Example 10.4-1, calculate the overall mass-transfer coefficient , the flux, and the percent resistance in the gas and liquid films. Do this for the case of A diffusing through stagnant B. Solution: From Fig. 10.4-4, = 0.052, which is in equilibrium with the bulk liquid xAL = 0.10. Also, yAG = 0.380. The slope of chord m' between E and M from Eq. (10.4-13) is, for yAi = 0.197 and xAi = 0.257, From Example 10.4-1, Using Eq. (10.4-25), Then, using Eq. (10.4-24), Solving, = 8.90 × 10−4. The percent resistance in the gas film is (484.0/868.8)100 = 55.7% and in the liquid film 44.3%. The flux is as follows, using Eq. (10.4-22): This, of course, is the same flux value as was calculated in Example 10.4-1 using the film equations.

4. Discussion of overall coefficients

If the two-phase system is such that the major resistance is in the gas phase, as in Eq. (10.4-19), then to increase the overall rate of mass transfer, efforts should be centered on increasing the gas-phase turbulence, not the liquid-phase turbulence. For a two-phase system where the liquid film resistance is controlling, turbulence should be increased in this phase to increase rates of mass transfer.

To design mass-transfer equipment, the overall mass-transfer coefficient is synthesized from the individual film coefficients, as discussed in this section.

Generally, when the major resistance to mass transfer is in the gas phase, the overall mass transfer coefficient or the film coefficient is used. An example would be absorption of ammonia from air to water. When the major resistance is in the liquid phase, as in absorption of oxygen from air by water, or is used.