## Mathematics Of Mass Transfer

Between liquid and gas phases, the transfer of mass from one phase to the other must pass through the interfacial boundary surface. Call the concentration of the solute at this surface as [yj referred to the gas phase. The corresponding concentration referred to the liquid phase is [xj . [xi] and [ yi] are the same concentration of the solute only that they are referred to different basis; in effect, they are equal. Because they are equal and because the thickness of the interface is zero, xi and yi must be in equilibrium with respect to each other.

Consider the process of absorption. If [y] is the concentration in the bulk gas phase, the driving force toward the interfacial boundary is [ y] - [ yi] and the rate of mass transfer is k,([y] - [yj), where ky is the gas film coefficient of mass transfer. For this rate of mass transfer to exist, it must be balanced by an equal rate of mass transfer at the liquid film. The liquid phase mass transfer rate is kx([xi] - [x]), where kx is the liquid film coefficient of mass transfer and [x] is the bulk concentration of the solute in the liquid phase. Thus, ky([y] - [yi]) = kx([Xi] - [x]) (9.1)

What is known is that [xi] and [ yi] are in equilibrium. But, determining these values experimentally would be very difficult. Thus, instead of using them, use [ x* ] and [ y* ], respectively. [ x* ] is the concentration that [x] would attain if it were to reach equilibrium value. By parallel deduction, [ y* ] is also the concentration that [ y] would attain if it were to reach equilibrium value. The corresponding driving forces are now [y] - [ y* ] and [ x* ] - [x], respectively. To comprehend the physical meaning of the driving forces, refer to Figure 9.8.

As shown, [xi], [yi] is on the equilibrium curve. The equilibrium curve is the relationship between the concentration [x] in the liquid phase and the concentration [ y] in the gas phase when there is no net mass transfer between the phases. For any given pair of values [x] and [ y] in the liquid and gas phase, respectively, the point ([xi], [yi]) represents the "distance" that ([x], [y]) will have to "move" to attain their equilibrium values concurrently at the equilibrium curve. This distance, represented by the line segment ([x], [ y] ^ [xi], [ yi]), is the actual driving force for mass transfer;

FIGURE 9.8 Relationship among the various mole fractions.

however, as mentioned before, ([xj, [yj) is impossible to determine experimentally. Thus, locate the point ([x], [y* ]) to view the surrogate driving force in the gas film and locate point ([x* ], [y]) to view the surrogate driving force in the liquid film.

As seen, [y] - [y* ] is greater than [y] - [yj; however, it is not the true driving force for transfer. Also [x* ] - [x] is greater than [xj - [x], but, again, it is not the true driving force for transfer. When the transfer equation is written, however, it is prefixed with a proportionality constant. This situation is therefore taken advantage of by using a different proportionality constant for the case of the surrogate driving forces. Thus, using Ky as the proportionality constant for the gas-side mass transfer equation in the surrogate situation, ky (y - yi) = Ky (y - y*) (9.2)

On the liquid side, using Kx as the proportionality constant, kx([Xi] - [x]) = Kx([x*] - [x]) (9.3)

Ky and Kx are called overall mass transfer coefficients for the gas and liquid sides, respectively. To differentiate, ky and kx are called individual mass transfer coefficients for the respective sides.

It is instructive to determine the equation relating the overall and the individual mass transfer coefficients. Equation (9.2) may be rearranged to obtain

J_ = [y]-[y*] = ([y]-[yi]) + ([yi]-[y*]- = 1 [yj - [y*] (94)

Ky ky kx

A parallel derivation for Kx yields

Kx kx mky mKy

The coordinates [xi] and [ yi] are the coordinates of the point ([xi], [ yi]) on the equilibrium curve. On the other hand, the coordinates [x] and [ y] are coordinates of point ([x], [y]) representing the concentration [x] in the liquid phase and the concentration [y] in the gas. Because the phases are not in equilibrium, ([x], [y]) is not on the equilibrium curve. The various values of the pair ([x], [ y]) in the liquid and gas phases can be plotted; this plot is called an operating line (see figure). Any ([x], [y]) pair is called an operating point. For the operating point ([x], [y]) the corresponding points on the equilibrium curve are ([x], [y*]) and ([x*], [y]), based on the surrogate equation. Point ([x*], [y*]) can only exist if ([x], [y]) is on the equilibrium curve.

The slope between ([x], [y*]) and ([xi], [ yj) is ([ yi] - [y*])/([xj - [x]); this is equal to m. Thus, m is the slope of the equilibrium curve if it is a straight line.

Parallel derivations may be performed for the stripping operation; the results are similar. The only difference is that there will be interchange of subscripts and superscripts. Thus, the following analogous equations will be obtained: kx ([x] - [xi]) = ky([yi] - [y]), kx([x] - [xi]) = K([x] - [x*]) and ky([yJ - [y]) = Ky([y1 - [y]). Refer to the figure to visualize that the mass flow is from the liquid phase to the gas phase.

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