# 5.6: Equilibrium Stages

Mass transfer of a chemical species from one phase to another phase is an essential feature of the mixing and purification processes that are ubiquitous in the chemical and biological process industries. A comprehensive analysis of mass transfer requires an understanding of the prerequisite subjects of fluid mechanics, thermodynamics and heat transfer; however, there are some mass transfer processes that can be approximated as equilibrium stages and these processes can be analyzed using the techniques presented in this text. Most students are familiar with an equilibrium stage when it is carried out in a batch-wise manner, since this is a common purification technique used in organic chemistry laboratories.

If an organic reaction produces a desirable product that is soluble in an organic phase and an undesirable product that is soluble in an aqueous phase, the desirable product can be purified by liquid-liquid extraction as illustrated in Figure $$\PageIndex{1}$$. In the first step of this process, the mixture from a reactor is placed in a separatory funnel. Water is added, the system is agitated, and the phases are allowed to equilibrate and separate. The amount of the undesirable product in the organic phase is reduced by an amount related to the volumes of the organic and aqueous phases and the equilibrium relation that determines how species $$A$$ is distributed between the two phases. If the equilibrium relation is linear, the concentrations in the aqueous phase ($$\beta$$ phase) and organic phase ($$\gamma$$ phase) can be related by6

Equilibrium relation: $c_{A\gamma } = \kappa_{eq,A} (c_{A\beta } ) \label{45}$

In this case we have used $$\kappa_{eq,A}$$ to represent the equilibrium coefficient; however, this information may given in terms of a distribution coefficient defined as

$\kappa_{dist,A} = \text{ distribution coefficient } = \frac{c_{A\beta } }{c_{A\gamma } } \label{46}$

When working with equilibrium coefficients or distribution coefficients, one must be very careful to note the definition since it is quite easy to invert these relations and create an enormous error. In addition, equilibrium relations are usually given in terms of mole fractions, but they are also given in terms of molar concentrations (as in Equation \ref{45}), and they are sometime given in terms of species densities. A clear and unambiguous definition of any and all equilibrium relations is essential to avoid errors.

The analysis of the process illustrated in Figure $$\PageIndex{1}$$ is relatively simple provided that the following conditions are valid:

1. There are negligible changes in the volumes of the organic and aqueous phases caused by the mass transfer process,
2. there is no species $$A$$ in the aqueous phase used in the extraction process, and
3. the linear equilibrium relation given by Equation \ref{45} is valid.

If the batch process illustrated in Figure $$\PageIndex{1}$$ is repeated $$N$$ times, the concentration of species $$A$$ in the organic phase is given by

$(c_{A\gamma } )_{N} = \frac{(c_{A\gamma } )_o }{\left(1 + \frac{V_{\beta } }{\kappa_{eq,A} V_{\gamma } } \right)^{N} } \label{47}$

Here we have used $$V_{\beta }$$ to represent the volume of the aqueous phase, $$V_{\gamma }$$ to represent the volume of the organic phase, and $$(c_{A\gamma } )_{N}$$ to represent the concentration of the undesirable product of the chemical reaction in the organic phase after $$N$$ extractions. Equation \ref{47} indicates that repeated batch-wise extractions can be used to reduce the concentration of species $$A$$ in the organic phase to arbitrarily small values.

To characterize the behavior of the repeated batch extraction process, it is convenient to define an absorption factor according to

$A = \left({V_{\beta } / \kappa_{eq,A} V_{\gamma } } \right) \label{48}$

so that Equation \ref{47} takes the form

$(c_{A\gamma } )_{N} = \frac{(c_{A\gamma } )_o }{\left(1 + A\right)^{N} } \label{49}$

This allows us to express two important limiting cases as

$A \to 0 , \quad (c_{A\gamma } )_{N} \to (c_{A\gamma } )_o , \quad \text{ no change occurs} \label{50a}$

$A \to \infty , \quad (c_{A\gamma } )_{N} \to 0 , \quad \text{ maximum change occurs} \label{50b}$

Here it is clear that one would like the absorption factor to be as large as possible; however, the definition given by Equation \ref{48} indicates that the value of $$A$$ is limited by the process illustrated in Figure $$\PageIndex{1}$$ and the equilibrium coefficient defined by Equation \ref{45}. The derivation of Equation \ref{47} is left as an exercise for the student as is the case in which the concentration of species $$A$$ in the original aqueous phase,$$(c_{A\beta } )_o$$, is not zero.

While the batch extraction process illustrated in Figure $$\PageIndex{1}$$ is convenient for use in an organic chemistry laboratory, a continuous process is preferred for a large-scale commercial purification process such as we have illustrated in Figure $$\PageIndex{2}$$. There we have shown a system consisting of a mixer that provides a large surface area for mass transfer followed by a settler that separates the aqueous and organic phases. If the mixer-settler system is efficient, the two phases will be in equilibrium as they leave the settler.

In this example, we are given an equilibrium relation in the form of Henry’s law

Equilibrium relation: $y_{A} = K_{eq,A} (x_{A} ) , \quad \text{ at the fluid-fluid interface} \label{51}$

which is the mole fraction version of the equilibrium relation given earlier by Equation \ref{45}. If the flow rates of the aqueous and organic phases are slow enough and the mass transfer of species $$A$$ between the two phases is fast enough, we can use Equation \ref{51} to construct a process equilibrium relation as

Process equilibrium relation: $(y_{A} )_{3} = K_{eq,A} (x_{A} )_{4} \label{52}$

We refer to this as a process equilibrium relation because it expresses the organic phase mole fraction in Stream #3 in terms of the aqueous phase mole fraction in Stream #4. Knowing when Equation \ref{52} is a valid approximation of Equation \ref{51} requires a detailed study of the mass transfer of species $$A$$, and this is something that will take place in subsequent courses in the chemical engineering curriculum.

In a problem of this type, we wish to know how the concentrations in the exit streams are related to the concentrations in the inlet streams, and this leads to the use of the control volume illustrated in Figure $$\PageIndex{3}$$. The macroscopic mole balance for species $$A$$ takes the form

Species $$A$$: $\int_{A_{ e} }c_{A} \mathbf{v} \cdot \mathbf{n} dA = 0 \label{53}$

in which $$\mathbf{v} \cdot \mathbf{n}$$is used in place of $$\mathbf{v}_{A} \cdot \mathbf{n}$$ with the idea that diffusive effects are negligible at the entrances and exits of the system. Since the process equilibrium relation given by Equation \ref{52} is expressed in terms of mole fractions, it is convenient to express Equation \ref{53} in the form (see Sec. 4.6)

Species $$A$$: $\int_{A_{ e} }x_{A} c \mathbf{v} \cdot \mathbf{n} dA = 0 \label{54}$

When this result is applied to the control volume illustrated in Figure $$\PageIndex{3}$$, we obtain

Species $$A$$: $- (x_{A} )_{1} \dot{M}_{1} - (y_{A} )_{2} \dot{M}_{2} + (y_{A} )_{3} \dot{M}_{3} + (x_{A} )_{4} \dot{M}_{4} = 0 \label{55}$

Here we have used $$x_{A}$$ to represent the mole fraction of species $$A$$ in the aqueous phase and $$y_{A}$$ to represent the mole fraction of species $$A$$ in the organic phase.

In addition, we have used $$\dot{M}$$ to represent the total molar flow rate in each of the four streams. At the two exits of the settler, we assume that the organic and aqueous streams are in equilibrium, thus the mixer-settler is considered to be an equilibrium stage and Equation \ref{52} is applicable. When changes in the mole fraction of species $$A$$ are sufficiently small, we can use the approximations given by

$\dot{M}_{1} = \dot{M}_{4} = \dot{M}_{\beta } , \quad\dot{M}_{2} = \dot{M}_{3} = \dot{M}_{\gamma } \label{56}$

Here we note that the change in the molar flow rates can be estimated as

$\Delta \dot{M}_{\beta } \approx \Delta (x_{A} ) \dot{M}_{\beta } , \quad\Delta \dot{M}_{\gamma } \approx \Delta (y_{A} ) \dot{M}_{\gamma } \label{57}$

in which $$\Delta (x_{A} )$$ and $$\Delta (y_{A} )$$ represent the changes in the mole fractions that occur between the inlet and outlet streams. From these estimate we conclude that

$\Delta \dot{M}_{\beta } <<\dot{M}_{\beta } , \quad \Delta \dot{M}_{\gamma } <<\dot{M}_{\gamma } \label{58}$

whenever the change in the mole fractions are constrained by

$\Delta (x_{A} )<<1 , \quad\Delta (y_{A} )<<1 \label{59}$

If we impose these constraints on the process illustrated in Figure $$\PageIndex{3}$$, we can use the approximation given by Equation \ref{56} in order to express Equation \ref{55} as

$(y_{A} )_{3} = (y_{A} )_{2} + \left[(x_{A} )_{1} - (x_{A} )_{4} \right]\left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right) \label{60}$

We now make use of the process equilibrium relation given by Equation \ref{52} to eliminate $$(x_{A} )_{4}$$ leading to

$(y_{A} )_{3} = (y_{A} )_{2} + (x_{A} )_{1} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right) - \frac{(y_{A} )_{3} }{K_{eq,A} } \left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right) \label{61}$

Here it is convenient to arrange the macroscopic mole balance for species $$A$$ in the form

$(y_{A} )_{3} \left[1 + \frac{\left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right)}{K_{eq,A} } \right] = (y_{A} )_{2} + K_{eq,A} (x_{A} )_{1} \left[\frac{\left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right)}{K_{eq,A} } \right] \label{62}$

which suggests that we define an absorption factor according to

$A = \frac{\left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right)}{K_{eq,A} } \label{63}$

Use of this expression in Equation \ref{62} and solving for $$(y_{A} )_{3}$$ leads to the following expression for the mole fraction of species $$A$$ in the organic stream ($$\gamma$$-phase) stream leaving the settler illustrated in Figure $$\PageIndex{3}$$.

$(y_{A} )_{3} = \frac{(y_{A} )_{2} }{1+A} + \frac{A}{1+A} \left[K_{eq,A} (x_{A} )_{1} \right] \label{64}$

This allows us to express two important limiting cases given by

$A \to 0 , \quad(y_{A} )_{3} \to (y_{A} )_{2} , \quad \text{ no change occurs} \label{65a}$

$A \to \infty , \quad(y_{A} )_{3} \to K_{eq,A} (x_{A} )_{1} , \quad \text{ maximum change occurs} \label{65b}$

In order to design a mixer-settler to achieve a specified mole fraction of species $$A$$ in Stream #3, one needs only to specify the absorption factor.

In the previous paragraphs, we examined a purification process from the point of view of an equilibrium stage, i.e., we assumed that the organic and aqueous streams leaving the settler are in equilibrium. The true state of equilibrium will never be achieved in a dynamic system such as the mixer-settler illustrated in Figure $$\PageIndex{3}$$. However, the assumption of equilibrium is often a reasonable approximation and when that is the case, various mass transfer systems can be successfully designed and analyzed using the concept of an equilibrium stage.

If we are confronted with the problem of species $$A$$ being transfer between a gas stream and a liquid stream, we require a contacting device that is quite different from that shown in Figure $$\PageIndex{3}$$. In this case we employ a gas-liquid contacting device of the type illustrated in Figure $$\PageIndex{4}$$. Here a gas is forced through a perforated plate and then up through a liquid stream that flows across the plate. If the gas bubbles are small enough and the liquid is deep enough, the gas will be in equilibrium with the liquid as it leaves the control volume. If this is the case, we can treat the system illustrated in Figure $$\PageIndex{4}$$ as an equilibrium stage. When the analysis is restricted to dilute solutions, one can usually employ a linear equilibrium relation, thus the mole fractions of species $$A$$ in the exiting gas and liquid streams are related by

Process equilibrium relation: $(y_{A} )_{3} = K_{eq,A} (x_{A} )_{4} \label{66}$

Under these circumstances the analysis of this system is identical to the analysis of the liquid-liquid extraction process illustrated in Figure $$\PageIndex{2}$$.

When the process equilibrium relation given by Equation \ref{66} is not valid, the simplification of an equilibrium stage can no longer be applied and one must move to a smaller length scale to analyze the mass transfer process. This situation is illustrated in Figure $$\PageIndex{5}$$ where we have indicated that the mass transfer

process must be studied at a smaller scale7. The analysis of mass transfer at this scale will occur in a subsequent chemical engineering course where the equilibrium relation given by Equation \ref{66} will be applied at the gas-liquid interface. We express this type of equilibrium condition as

Equilibrium relation: $y_{A} = K_{eq,A} x_{A} , \quad \text{ at the gas-liquid interface} \label{67}$

When the form given by Equation \ref{66} represents a valid approximation, the techniques studied in this text can be successfully applied to real systems. When Equation \ref{66} is not a valid approximation, one must move to a smaller length scale and make use of Equation \ref{67}.

In the preceding paragraphs we have illustrated liquid-liquid and gas-liquid systems that can be treated as equilibrium stages. Such systems are ubiquitous in the world of chemical engineering where purification of liquid and gas streams is a major activity. In addition to the contacting devices illustrated in Figures $$\PageIndex{2}$$ and $$\PageIndex{4}$$, there are many other processes that can often be approximated as equilibrium stages and some of these are examined in the following paragraphs.

Example $$\PageIndex{1}$$: Condensation of water in humid air

On a warm spring day in Baton Rouge, LA, the atmospheric pressure is 755 mm Hg, the temperature is 80 F, and the relative humidity is 80%. A large industrial air conditioner operating at atmospheric pressure is illustrated in Figure $$\PageIndex{6}$$. It treats 1000 kg/h of air (dry air basis) and lowers the air temperature from 80 F to 15 C. The cool, dry air leaving the unit is assumed to be in equilibrium with the water that leaves the unit at 15 C. Thus the air conditioner is treated as an equilibrium stage.

Here we want to determine how much liquid water, in kg/h, is removed from the warm, humid air by the air conditioning system. In this case the obvious control volume cuts all three streams and encloses the air conditioner. If we assume that the system operates at steady state and we conclude that there are no chemical reactions, the appropriate form of the species mass balance is given by

$\int_{\mathscr{A}}\rho_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = 0 , \quad A = 1, 2, ...., N \label{1a}\tag{1}$

In problems of this type, it is plausible to neglect diffusion velocities, $$\mathbf{u}_{A} <<\mathbf{v}_{A}$$, so that Equation \ref{1a} takes the form

$\int_{\mathscr{A}}\rho_{A} \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad A = 1, 2, ...., N\label{2a}\tag{2}$

Under these circumstances we can combine the equations for nitrogen and oxygen to obtain the following mass balance for dry air:

$\int_{A_{ 1} }\rho_{\text{dry air}} \mathbf{v} \cdot \mathbf{n} dA + \int_{A_{ 2} }\rho_{\text{dry air}} \mathbf{v} \cdot \mathbf{n} dA + \int_{A_{ 3} }\rho_{\text{dry air}} \mathbf{v} \cdot \mathbf{n} dA = 0\label{3a}\tag{3}$

Here we have been careful to use the phrase “dry air” to indicate the nitrogen and oxygen since “air” would normally mean nitrogen, oxygen, and water. Since significant amounts of nitrogen and oxygen do not leave the system in Stream #3, the mass balance for dry air takes the form

Dry Air: $- (\dot{m}_{\text{dry air}} )_{1} + (\dot{m}_{\text{dry air}} )_{2} = 0 \label{4a}\tag{4}$

Water: $- (\dot{m}_{\ce{H2O}} )_{1} + (\dot{m}_{\ce{H2O}})_2 + (\dot{m}_{\ce{H2O}} )_{3} = 0 \label{5a}\tag{5}$

In which we have used the nomenclature illustrated by

$\int_{A_{ 1} }\rho_{\text{dry air}} \left|\mathbf{v} \cdot \mathbf{n}\right| dA = (\dot{m}_{\text{dry air}} )_{1}\label{6a}\tag{6}$

Equations \ref{4a} and \ref{5a} represent the fundamental balance equations associated with the system illustrated in Figure 5.6, and we need to use the information that is given to determine $$(\dot{m}_{\ce{H2O}} )_{3}$$.

Since information is given about the humidity of the incoming air stream, we need to think about how we can connect the mass flow rates in Eqs. \ref{4a} and \ref{5a} with the humidity vaguely described by Equation $$(5.5.3)$$ or the point humidity precisely defined by Equation $$(5.5.4)$$. We start by examining the ratio of mass flow rates (water to dry air) given by

$\frac{(\dot{m}_{\ce{H2O}} )_{1} }{(\dot{m}_{\text{dry air}} )_{1} } = \frac{\int_{A_{ 1} }\rho_{\ce{H2O}} \left|\mathbf{v} \cdot \mathbf{n}\right| dA }{\int_{A_{ 1} }\rho_\text{dry air} \left|\mathbf{v} \cdot \mathbf{n}\right| dA }\label{7a}\tag{7}$

If the velocity, $$\mathbf{v} \cdot \mathbf{n}$$, is uniform across the entrance, or the species densities, $$\rho_{\ce{H2O}}$$ and $$\rho_\text{dry air}$$, are uniform across the entrance, we can follow the development in Sec. 4.5.1 to conclude that

$\frac{(\dot{m}_{\ce{H2O}} )_{1} }{(\dot{m}_{\text{dry air}} )_{1} } = \frac{\int_{A_{ 1} }\rho_{\ce{H2O}} \left|\mathbf{v} \cdot \mathbf{n}\right| dA }{\int_{A_{ 1} }\rho_\text{dry air} \left|\mathbf{v} \cdot \mathbf{n}\right| dA } = \frac{\langle \rho_{\ce{H2O}} \rangle_{1} Q_{ 1} }{\langle \rho_\text{dry air} \rangle_{1} Q_{ 1} } = \frac{\langle \rho_{\ce{H2O}} \rangle_{1} }{\langle \rho_\text{dry air} \rangle_{1} }\label{8a}\tag{8}$

In addition, if we accept the ratio of the area average species densities as our measure of the humidity we obtain

$\frac{\langle \rho_{\ce{H2O}} \rangle_{1} }{\langle \rho_\text{dry air} \rangle_{1} } = \frac{(\dot{m}_{\ce{H2O}} )_{1} }{(\dot{m}_{\text{dry air}} )_{1} } = \left( \text{ humidity }\right)_{1}\label{9a}\tag{9}$

This suggests a particular strategy for solving this mass balance problem. Dividing Equation \ref{5a} by the mass flow rate of dry air in Stream #1 leads to

$- \frac{(\dot{m}_{\ce{H2O}} )_{1} }{(\dot{m}_{\text{dry air}} )_{1} } + \frac{(\dot{m}_{\ce{H2O}} )_2}{(\dot{m}_{\text{dry air}} )_{1} } + \frac{(\dot{m}_{\ce{H2O}} )_{3} }{(\dot{m}_{\text{dry air}} )_{1} } = 0 \label{10a}\tag{10}$

and on the basis of the mass balance for dry air given by Equation \ref{4a}, we can express this result in the form

$- \frac{(\dot{m}_{\ce{H2O}} )_{1} }{(\dot{m}_{\text{dry air}} )_{1} } + \frac{(\dot{m}_{\ce{H2O}})_2 }{(\dot{m}_{\text{dry air}} )_{2} } + \frac{(\dot{m}_{\ce{H2O}} )_{3} }{(\dot{m}_{\text{dry air}} )_{1} } = 0\label{11a}\tag{11}$

Use of Equation \ref{9a} leads us to a “humidity balance” equation given by

$\left( \text{ humidity }\right)_{1} = \left( \text{ humidity }\right)_{2} + \frac{(\dot{m}_{\ce{H2O}} )_{3} }{(\dot{m}_{\text{dry air}} )_{1} }\label{12a}\tag{12}$

Our objective in this example is to determine how much liquid water is removed from the air, and Equation \ref{12a} provides this information in the form

$(\dot{m}_{\ce{H2O}} )_{3} = (\dot{m}_{\text{dry air}} )_{1} \left[\left( \text{ humidity }\right)_{1} - \left( \text{ humidity }\right)_{2} \right]\label{13a}\tag{13}$

Here it becomes clear that the solution to this problem requires that we determine the humidity in Streams #1 and #2. This motivates us to make use of Equation $$(5.5.5)$$ that we list here as

$\text{ humidity } = \frac{MW_{\ce{H2O}} p_{\ce{H2O}} }{MW_{\text{dry air}} \left(p-p_{\ce{H2O}} \right)}\label{14a}\tag{14}$

In addition, we are given information about the percent relative humidity defined by Equation $$(5.5.6)$$ and listed here as

$\% \text{ relative humidity } = \frac{p_{\ce{H2O}} }{p_{\ce{H2O} , vap} } \times 100\label{15a}\tag{15}$

In Stream #1 the percent relative humidity is 80% and this provides

$\left(\% \text{ relative humidity }\right)_{ 1} = \frac{\left(p_{\ce{H2O}} \right)_{ 1} }{\left(p_{\ce{H2O} , vap} \right)_{ 1} } \times 100 = 80\label{16a}\tag{16}$

which allows us to express the partial pressure of water vapor in Stream #1 as

$\left(p_{\ce{H2O}} \right)_{ 1} = 0.80\times \left(p_{\ce{H2O} , vap} \right)_{ 1}\label{17a}\tag{17}$

The vapor pressure of water, in both Stream #1 and Stream #2, can be determined by Antoine’s equation (see Appendix A3) that provides

$\left(p_{\ce{H2O} , vap} \right)_{ 1} = p_{\ce{H2O} , vap} (T = 80 \ { F}) = 3.484 \ { kPa}\label{18aa}\tag{18a}$

$\left(p_{\ce{H2O} , vap} \right)_{ 2} = p_{\ce{H2O} vap} (T = 15 \ { C}) = 1.705 \ { kPa}\label{18ba}\tag{18b}$

Returning to Equation \ref{17a} and making use of Equation \ref{18aa} leads to

$\left(p_{\ce{H2O}} \right)_{ 1} = 0.80\times \left[p_{\ce{H2O} , vap} (T = 80 \ { F})\right] = 2787 \ { Pa}\label{19a}\tag{19}$

and we are now ready to determine the humidity in Stream #1.

We are given that the total pressure in the system is equivalent to 755 mm Hg and this leads to a pressure given by

$p = \frac{755 \ mm Hg} { 760 \ mm Hg/atm} \times \frac{ 101.3\times 10^{ 3} \ Pa}{atm} = 100,634 \ Pa \label{20a}\tag{20}$

and from Equation \ref{14a} we obtain the humidity in Stream #1 according to

$( \text{ humidity })_{ 1} = \frac{MW_{\ce{H2O}} \left(p_{\ce{H2O}} \right)_{1} }{MW_{\text{dry air}} \left[p-\left(p_{\ce{H2O}} \right)_{1} \right]} \\ = \frac{ 18.015 \ g \ \ce{H2O}/mol} { 28.84 \text{ g dry air/mol}} \times \frac{ 2795 \ Pa}{100,634 \ { Pa} - 2795 \ { Pa}} = 0.0178 \ g \ \ce{H2O} / \text{g dry air} \label{21a}\tag{21}$

We are given that Stream #2 is in equilibrium with Stream #3, thus the relative humidity in Stream #2 is 100% and the partial pressure of water vapor can be determined as

Process equilibrium relation: $\left(p_{\ce{H2O}} \right)_{ 2} = \left(p_{\ce{H2O} , vap} \right)_{ 3} = \left. p_{\ce{H2O} , vap} \right|_{T = 15 \ C} = 1705 \ Pa \label{22a}\tag{22}$

This allows us to repeat the type of calculation illustrated by Equation \ref{21a} to obtain

$( \text{ humidity })_2 = \frac{MW_{\ce{H2O}} \left( p_{\ce{H2O}} \right)_2 }{MW_{\text{dry air}} \left[p-\left(p_{\ce{H2O}} \right)_2 \right] } = 0.01076 \ g \ \ce{H2O} / \text{g dry air} \label{23a}\tag{23}$

Finally we return to Equation \ref{13a} to determine the mass flow rate of water leaving the air conditioning unit according to

$\left(\dot{m}_{\ce{H2O}} \right)_{3} = 1000 \text{ kg dry air/h} \left[0.01783 - 0.01076\right]\frac{ kg \ \ce{H2O}}{\text{kg dry air}} = 7.07 \ kg \ \ce{H2O} / h \label{24a}\tag{24}$

Example $$\PageIndex{8}$$: Use of air to dry wet solids

In Figure $$\PageIndex{7a}$$ we have illustrated a co-current air dryer. The solids entering the dryer contain 20% water on a mass basis and the mass flow rate of the wet solids entering the dryer is 1000 lbm/hr. The dried solids contain 5% water on a mass basis, and the temperature of the solid stream leaving the dryer is 65 C. The complete design of the dryer is a complex process that requires a knowledge of the flow rate of the dry air entering the dryer. This can be determined by a macroscopic mass balance analysis.

As the air and the wet solids pass through the dryer an equilibrium condition is approached and mass transfer of water from the wet solids to the air stream diminishes. As an example, we assume that the air leaving the dryer is in equilibrium with the solids leaving the dryer, and this allows us to determine the maximum amount of water that can be removed from the wet solids. For this type of approximation the dryer becomes an equilibrium stage.

To construct a control volume for the analysis of this system, we need only make cuts where information is given and required and these lead to the control volume shown in Figure 5.9b. We begin this problem with the species macroscopic mole balance for a steady-state process in the absence of chemical reaction and note that the species velocity, $$\mathbf{v}_{A}$$, can be replaced with the mass average velocity, $$v$$, at entrances and exits to obtain

$\int_{\mathscr{A}}c_{A} \mathbf{v} \cdot \mathbf{n} dA = 0 \tag{1}\label{1b}$

It is important to remember that the molar concentration can be expressed as

$c_{A} = y_{A} c , \quad \text{ gas streams}\tag{2}\label{2b}$

where $$y_{A}$$ is the mole fraction of species $$A$$ and $$c$$ is the total molar concentration. The form given

by Equation \ref{2b} is especially useful in the analysis of the air stream; however, for the wet solids stream it is convenient to work in terms of mass rather than moles and make use of

$c_{A} = \frac{\rho_{A} }{MW_{A} } , \quad \text{ wet solid streams}\tag{3}\label{3b}$

If we let species $$A$$ be water and apply Equation \ref{1b} to the control volume shown in Figure $$\PageIndex{7b}$$, we obtain

Water: $\begin{array}{l} \underbrace{\int_{A_{ 1} }\left(\rho_{\ce{H2O}} / MW_{\ce{H2O}} \right)\mathbf{v} \cdot \mathbf{n} dA }_{\begin{array}{l} \text{ molar flow rate of water} \\ \text{ entering with the solid}\end{array}}+\underbrace{\int_{A_{ 4} }\left(\rho_{\ce{H2O}} / MW_{\ce{H2O}} \right) \mathbf{v} \cdot \mathbf{n} dA }_{\begin{array}{l} \text{ molar flow rate of water} \\ \text{ leaving with the solid} \end{array}} + \\ \underbrace{ \int_{A_{ 2} }y_{\ce{H2O}} c \mathbf{v} \cdot \mathbf{n} dA }_{\begin{array}{l} \text{ molar flow rate of water} \\ \text{ entering with the air} \end{array}} + \underbrace{ \int_{A_{ 3} }y_{\ce{H2O}} c \mathbf{v} \cdot \mathbf{n} dA }_{\begin{array}{l} \text{ molar flow rate of water} \\ \text{ leaving with the air} \end{array}} = 0 \end{array} \label{4b}\tag{4}$

Here we have used Eqs. \ref{2b} and \ref{3b} in order to arrange the fluxes in forms that are convenient, but not necessary, for this particular problem, and we can express those fluxes in terms of averaged quantities to obtain

Water: $- \frac{\langle \rho_{\ce{H2O}} \rangle_{1} Q_{1} }{MW_{\ce{H2O}} } + \frac{\langle \rho_{\ce{H2O}} \rangle_{4} Q_{4} }{MW_{\ce{H2O}} } - (y_{\ce{H2O}} )_{2} \dot{M}_{2} + (y_{\ce{H2O}} )_{3} \dot{M}_{3} = 0\tag{5}\label{5b}$

In this representation of the macroscopic mole balance for water, we have drawn upon the analysis presented in Sec. 4.5. Specifically, we have imposed the following assumptions:

Gas streams: $c \mathbf{v} \cdot \mathbf{n} = \text{ constant}\tag{6a}\label{6ab}$

Wet solid streams: $\mathbf{v} \cdot \mathbf{n} = \text{ constant}\tag{6b}\label{6bb}$

in which “constant” means constant across the area of the entrances and exits. The three phases contained in the wet solid streams are illustrated in Figure $$\PageIndex{7c}$$ for stream #1. The total density in these streams consists of the density of the solid, the water, and the air, and this density can be written explicitly as

$\langle \rho \rangle = \langle \rho_{solid} \rangle + \langle \rho_{\ce{H2O}} \rangle + \langle \rho_{air} \rangle\tag{7}\label{7b}$

The mass fraction of water in the wet solids is defined by

$\langle \omega_{\ce{H2O}} \rangle = \frac{\langle \rho_{\ce{H2O}} \rangle }{\langle \rho \rangle }\tag{8}\label{8b}$

and use of this representation in Equation \ref{5b} leads to

Water: $- \frac{\langle \omega_{\ce{H2O}} \rangle_{1} \dot{m}_{1} }{MW_{\ce{H2O}} } + \frac{\langle \omega_{\ce{H2O}} \rangle_{4} \dot{m}_{4} }{MW_{\ce{H2O}} } - (y_{\ce{H2O}} )_{2} \dot{M}_{2} + (y_{\ce{H2O}} )_{3} \dot{M}_{3} = 0\tag{9}\label{9b}$

Here we have identified the mass flow rates of the wet solid streams according to

$\dot{m}_{ 1} = \langle \rho \rangle_{1} Q_{ 1} , \quad \dot{m}_{ 4} = \langle \rho \rangle_{4} Q_{4}\tag{10}\label{10b}$

A little thought (see Problem 5-33) will indicate that a mass balance for the solid material leads to

Solid material: $\left[1 - \langle \omega_{\ce{H2O}} \rangle_{1} \right] \dot{m}_{ 1} = \left[1 - \langle \omega_{\ce{H2O}} \rangle_{4} \right] \dot{m}_{ 4}\tag{11}\label{11b}$

and this result can be used in Equation \ref{9b} to obtain

Water: $(y_{\ce{H2O}} )_{2} \dot{M}_{2} = (y_{\ce{H2O}} )_{3} \dot{M}_{3} + \left\{\frac{\langle \omega_{\ce{H2O}} \rangle_{4} - \langle \omega_{\ce{H2O}} \rangle_{1} }{\left[1-\langle \omega_{\ce{H2O}} \rangle_{4} \right] MW_{\ce{H2O}} } \right\} \dot{m}_{1}\tag{12}\label{12b}$

A molar balance for the air will allow us to eliminate $$\dot{M}_{3}$$ from this result and the calculation of $$\dot{M}_{2}$$ easily follows. This will be left as an exercise for the student (see Problem 5-34).