5.7: Continuous Equilibrium Stage Processes

In Sec. 5.6 we considered systems that consisted of a single contacting process in which equilibrium conditions were assumed to exist at the exit streams. Knowing when the condition of equilibrium is a reasonable approximation requires a detailed study of the heat and mass transfer processes that are taking place. These details will be studied in subsequent courses where it will be shown that the condition of equilibrium is a reasonable approximation for many mass transfer processes. Our first example of an equilibrium stage was the batch liquid-liquid extraction process illustrated in Figure $$\PageIndex{4}$$. In that case one could repeat the extraction process to obtain an arbitrarily small value of the concentration of species $$A$$ as indicated by Equation $$(5.6.3)$$. In this section we wish to illustrated how this same type of multi-stage extraction process can be achieved for a steady-state process. In Figure $$\PageIndex{1}$$ we have illustrated an arrangement of mixer-settlers that can be used to reduce the concentration of species $$A$$ in the organic stream to an arbitrarily small value. Rather than working with the details illustrated in this figure, we will represent the

mixer-settler unit as a single box so that our counter-current extraction process takes the form illustrated in Figure $$\PageIndex{2}$$.

Here we note that the nomenclature used to identify the incoming and outgoing streams is different than that used in Figure $$\PageIndex{6}$$ for a single liquid-liquid extraction unit. In this case the number of the unit is used to identify the outgoing streams, and this simplification is necessary for an efficient treatment of a system containing $$N$$ units.

Sequential analysis-algebraic

There are a variety of problem statements associated with a system of the type illustrated in Figure $$\PageIndex{2}$$. For example, one might be given the following:

1. The number of stages, $$N$$,
2. the mole fractions for the inlet streams, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$,
3. the equilibrium relation applicable to each stage, and
4. the molar flow rates of the two phases, $$\dot{M}_{\beta }$$ and $$\dot{M}_{\gamma }$$.

Given this information one would then be asked to determine the outlet conditions, $$(y_{A} )_{1}$$ and $$(x_{A} )_{N}$$.

Alternatively, one might be given:

1. The mole fractions for the inlet streams, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$,
2. the equilibrium relation, and
3. the molar flow rates of the two phases, $$\dot{M}_{\beta }$$ and $$\dot{M}_{\gamma }$$.

In this case one would be asked to determine the number of stages, $$N$$, that would be required to achieve a desired composition in one of the outlet streams.

In the following paragraphs we develop an algebraic solution for the cascade of $$N$$ equilibrium stages illustrated in Figure $$\PageIndex{2}$$, and we illustrate how this result can be applied several different problem statements. We begin our analysis of the cascade of equilibrium stages with the single unit illustrated in Figure $$\PageIndex{3}$$ in which the obvious control volume has been identified.

For this case the mole balance for species $$A$$ takes the form

$\underbrace{(x_{A} )_{ 1} \dot{M}_{\beta } + (y_{A} )_{1} \dot{M}_{\gamma } }_{\begin{array}{l} {\text{ molar flow of species }A} \\ \text{out of the control volume} \end{array}} = \underbrace{(x_{A} )_{ o} \dot{M}_{\beta } + (y_{A} )_{2} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{into the control volume} \end{array}} \label{68}$

and we impose the special condition given by

Special Condition: $(x_{A} )_{ o} = 0 \label{69}$

For this condition Equation \ref{68} takes the form

$(x_{A} )_{ 1} \dot{M}_{\beta } + (y_{A} )_{1} \dot{M}_{\gamma } = (y_{A} )_{2} \dot{M}_{\gamma } \label{70}$

At this point our objective is to develop a relation between $$(y_{A} )_{1}$$ and $$(y_{A} )_{2}$$, and we begin by arranging Equation \ref{70} in the form

$(y_{A} )_{1} + (x_{A} )_{ 1} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right) = (y_{A} )_{2} \label{71}$

Here we note that the process equilibrium relation is given by

Process equilibrium relation: $(y_{A} )_{1} = K_{eq,A} (x_{A} )_{1} \label{72}$

and use of this result allows us to simplify the mole balance for species $$A$$ to obtain

$(y_{A} )_{1} + (y_{A} )_{ 1} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } K_{eq,A} } \right) = (y_{A} )_{2} \label{73}$

We now define the absorption factor according to

$A = \left({\dot{M}_{\beta } / \dot{M}_{\gamma } K_{eq,A} } \right) \label{74}$

in order to develop the following relation between $$(y_{A} )_{1}$$ and $$(y_{A} )_{2}$$

One Equilibrium Stage: $(y_{A} )_{1} = \frac{(y_{A} )_{2} }{1+A} \label{75}$

Here we have chosen to arrange the macroscopic mole balance for species $$A$$ in terms of the mole fraction in the $$\gamma$$-phase. However, the choice is arbitrary and we could just as well have set up the analysis in terms of $$x_{A}$$ instead of $$y_{A}$$.

Having developed an expression for the exit mole fraction, $$(y_{A} )_{1}$$, in terms of the entrance mole fraction, $$(y_{A} )_{2}$$, for a single equilibrium stage, we now wish to relate $$(y_{A} )_{1}$$ to $$(y_{A} )_{3}$$. To accomplish this we examine the two equilibrium stages illustrated in Figure $$\PageIndex{4}$$.

In this case the molar balance for species $$A$$ can be expressed as

$\underbrace{(x_{A} )_{ 2} \dot{M}_{\beta } + (y_{A} )_{1} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{out of the control volume} \end{array}} = \underbrace{(x_{A} )_{ o} \dot{M}_{\beta } + (y_{A} )_{3} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{into the control volume} \end{array}} \label{76}$

and we continue to impose the special condition

Special Condition: $(x_{A} )_{ o} = 0 \label{77}$

so that Equation \ref{76} takes the form

$(x_{A} )_{ 2} \dot{M}_{\beta } + (y_{A} )_{1} \dot{M}_{\gamma } = (y_{A} )_{3} \dot{M}_{\gamma } \label{78}$

At this point our objective is to determine $$(y_{A} )_{1}$$ in terms of $$(y_{A} )_{3}$$, and we begin by arranging this result in the form

$(y_{A} )_{1} + (x_{A} )_{ 2} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right) = (y_{A} )_{3} \label{79}$

Here we note that the process equilibrium relation is given by

Process equilibrium relation: $(y_{A} )_{2} = K_{eq,A} (x_{A} )_{2} \label{80}$

and use of this result allows us to simplify the mole balance for species $$A$$ to

$(y_{A} )_{1} + (y_{A} )_{ 2} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } K_{eq,A} } \right) = (y_{A} )_{3} \label{81}$

We now use Equation \ref{75} to determine $$(y_{A} )_{2}$$ so that Equation \ref{81} provides the following result

$(y_{A} )_{1} + (y_{A} )_{ 1} \left[A\left(1+A\right)\right] = (y_{A} )_{3} \label{82}$

which can be solved for $$(y_{A} )_{1}$$ to obtain

Two Equilibrium Stages: $(y_{A} )_{1} = \frac{(y_{A} )_{3} }{1+A+A^{2} } \label{83}$

We are now ready to determine $$(y_{A} )_{1}$$ when three equilibrium stages are employed as illustrated in Figure $$\PageIndex{5}$$.

In this case the molar balance for species $$A$$ can be expressed as

$\underbrace{(x_{A} )_{ 3} \dot{M}_{\beta } + (y_{A} )_{1} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{out of the control volume} \end{array}} = \underbrace{(x_{A} )_{ o} \dot{M}_{\beta } + (y_{A} )_{4} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{into the control volume} \end{array}} \label{84}$

and we continue to impose the condition

Special Condition: $(x_{A} )_{ o} = 0 \label{85}$

so that Equation \ref{84} takes the form

$(x_{A} )_{ 3} \dot{M}_{\beta } + (y_{A} )_{1} \dot{M}_{\gamma } = (y_{A} )_{4} \dot{M}_{\gamma } \label{86}$

At this point our objective is to determine $$(y_{A} )_{1}$$ as a function of $$(y_{A} )_{4}$$, and we begin by arranging this result in the form

$(y_{A} )_{1} + (x_{A} )_{ 3} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right) = (y_{A} )_{4} \label{87}$

Here we note that the process equilibrium relation is given by

Process equilibrium relation: $(y_{A} )_{3} = K_{eq,A} (x_{A} )_{3} \label{88}$

and use of this result allows us to simplify the mole balance for species $$A$$ to

$(y_{A} )_{1} + (y_{A} )_{ 3} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } K_{eq,A} } \right) = (y_{A} )_{4} \label{89}$

We now use Equation \ref{82} to eliminate $$(y_{A} )_{3}$$ so that Equation \ref{89} provides the following result

$(y_{A} )_{1} + (y_{A} )_{ 1} A\left[1+A\left(1+A\right)\right] = (y_{A} )_{4} \label{90}$

which can be solved for $$(y_{A} )_{1}$$ as a function of $$(y_{A} )_{4}$$ and the absorption coefficient.

Three Equilibrium Stages: $(y_{A} )_{1} = \frac{(y_{A} )_{4} }{1+A+A^{2} +A^{3} } \label{91}$

Here we are certainly in a position to deduce that $$N$$ equilibrium stages would produce the result given by

$$\mathbf{N}$$ Equilibrium Stages: $(y_{A} )_{1} = \frac{(y_{A} )_{N+1} }{1+A+A^{2} +A^{3} + .... +A^{N} } \label{92}$

This type of sequential analysis can be used to analyze any multi-stage process such as the one illustrated in Figure $$\PageIndex{2}$$; however, one must remember that there are three important simplifications associated with this result. These simplifications are:

1. The total molar flow rates, $$\dot{M}_{\beta }$$ and $$\dot{M}_{\gamma }$$, are constant,
2. a linear equilibrium relation is applicable, and
3. the mole fraction of species $$A$$ entering the system in the $$\beta$$-phase is zero, i.e., $$(x_{A} )_{ o} = 0$$.

This latter constraint can be easily removed and that will be left as an exercise for the student.

If the absorption factor and the number of stages are specified for the system illustrated in Figure $$\PageIndex{2}$$, one can easily calculate the mole fraction of species $$A$$ at the exit of the system, $$(y_{A} )_{1}$$, in terms of the mole fraction at the entrance, $$(y_{A} )_{N+1}$$. If $$(y_{A} )_{1}$$ is specified as some fraction of $$(y_{A} )_{N+1}$$ and the number of stages are specified, the required absorption factor can be determine using the implicit expression for $$A$$ given by8

$1+A+A^{2} +A^{3} + .... +A^{N} = \frac{(y_{A} )_{N+1} }{(y_{A} )_{1} } \label{93}$

In general, it is convenient to use this result along with the macroscopic balance around the entire cascade illustrated in Figure $$\PageIndex{2}$$. This is given by

$\underbrace{(x_{A} )_{N} \dot{M}_{\beta } + (y_{A} )_{1} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{out of the control volume} \end{array}} = \underbrace{(x_{A} )_{ o} \dot{M}_{\beta } + (y_{A} )_{N+1} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{into the control volume} \end{array}} \label{94}$

and it is often convenient to arrange this result in the form

$(x_{A} )_{N} = \frac{(y_{A} )_{N+1} - (y_{A} )_{1} }{\dot{M}_{\beta } / \dot{M}_{\gamma } } \label{95}$

Here we have continued to impose the condition that $$(x_{A} )_{ o} = 0$$ in order to keep this initial study as simple as possible.

There are several types of problems that can be explored using the analysis leading to Equation \ref{93} and Equation \ref{95}, and we list three of the types as follows:

• Type I: Given the inlet mole fractions, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$, the system parameters, and the desired value of $$(y_{A} )_{1}$$, we would like to determine the number of stages, $$N$$.
• Type II: Given the inlet mole fractions, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$, the system parameters, and the number of stages, $$N$$, we would like to determine the value of $$(y_{A} )_{1}$$.
• Type III: Given the inlet mole fractions, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$, the system parameters $$K_{eq,A}$$, $$\dot{M}_{\gamma }$$, the number of stages, $$N$$, and a desired value of $$(y_{A} )_{1}$$, we would like to determine the molar flow rate of the $$\beta$$-phase, $$\dot{M}_{\beta }$$.

Example $$\PageIndex{1}$$: Absorption of acetone from air into water

In order to extract acetone from air using water, one can apply a contacting device such as the one illustrated in Figure $$\PageIndex{7}$$. The cascade of equilibrium stages can be represented by Figure $$\PageIndex{2}$$; however, the analogous system illustrated in Figure $$\PageIndex{6}$$ is often used to more closely represent the gas-liquid contacting process under consideration. There are various problems associated with the design and operation of this unit and we begin with the first type of problem identified in the previous paragraph.

Type I: Given the inlet mole fractions, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$, the system parameters, and the desired value of $$(y_{A} )_{1}$$, we would like to determine the number of stages, $$N$$.

In this case we are given the inlet mole fractions, $$(x_{A} )_{ o} = 0$$ and $$(y_{A} )_{N+1} = 0.010$$, and the system parameters, $$K_{eq,A} = 2.53$$, $$\dot{M}_{\beta } = 90$$ kmol/h and $$\dot{M}_{\gamma } = 30$$ kmol/h. For these particular values, we want to know how many stages, $$N$$, are required to reduce the exit mole fraction of acetone to $$(y_{A} )_{1} = 0.001$$. In this case we express Equation \ref{93} as

$1+A+A^{2} +A^{3} + .... +A^{N} = \frac{(y_{A} )_{N+1} }{(y_{A} )_{1} } = \frac{0.010}{0.001} = 10.0 \label{1a}\tag{1}$

in which the value of the absorption coefficient is given by $$A = {\dot{M}_{\beta } / \dot{M}_{\gamma } K_{eq,A} } = 1.186$$. This leads to the values listed in Table $$\PageIndex{1}$$ where we see that 5 stages are insufficient to achieve the desired result, $$(y_{A} )_{1} = 0.001$$. In addition, the use of 6 stages reduces the exit mole fraction of acetone in the air stream to $$(y_{A} )_{1} = 0.00081$$ which is less than the desired result. For this type of situation, it is the responsibility of the chemical engineer to make a judgment based on safety considerations,

Table $$\PageIndex{1}$$: Number of stages, $$N$$, versus $${(y_{A} )_{N+1} / (y_{A} )_{1} }$$
Number of stages, $$N$$ $$1+A+A^{2} +A^{3} +....+A^{N}$$
1 2.186
2 3.592
3 5.260
4 7.238
5 9.584
6 12.366

environmental constraints, requirements for other processing units, and economic optimization. Such matters are covered in a future course on process design, and we have alluded to some of these concerns in Sec. 1.2.

Type II: Given the inlet mole fractions, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$, the system parameters, and the number of stages, $$N$$, we would like to determine the value of $$(y_{A} )_{1}$$.

In this case we consider an existing unit in which there are 7 stages. The inlet mole fractions are given by $$(x_{A} )_{ o} = 0$$ and $$(y_{A} )_{8} = 0.010$$, and the parameters associated with the system are specified as $$K_{eq,A} = 2.53$$, $$\dot{M}_{\beta } = 90$$ kmol/h and $$\dot{M}_{\gamma } = 30$$ kmol/h. In order to determine the mole fraction in the $$\gamma$$-phase (air) leaving the cascade, we make use of Equation \ref{93} to express $$(y_{A} )_{1}$$ as

$(y_{A} )_{1} = \frac{(y_{A} )_{8} }{1+A+A^{2} +A^{3} +A^{4} +A^{5} +A^{6} +A^{7} } \label{2a}\tag{2}$

This leads to the following value of the mole fraction of acetone leaving the top of the cascade illustrated in Figure $$\PageIndex{6}$$:

$(y_{A} )_{1} = \frac{0.010}{15.67} = 0.00064 \label{3a}\tag{3}$

In addition to calculating the exit mole fraction of acetone in the air stream, one may want to determine the mole fraction of acetone in the water stream leaving the cascade. This is obtained from Equation \ref{95} which provides

$(x_{A} )_{7} = \frac{(y_{A} )_{8} - (y_{A} )_{1} }{\dot{M}_{\beta } / \dot{M}_{\gamma } } = \frac{0.010 - 0.00064}{( 90 \ kmol/h) / (30 \ kmol/h)} = 0.0031 \label{4a}\tag{4}$

Some times it is possible to change the operating characteristics of a cascade to achieve a desired result and this situation is considered in the following example.

Type III: Given the inlet mole fractions, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$, the system parameters $$K_{eq,A}$$, $$\dot{M}_{\gamma }$$, the number of stages, $$N$$, and a desired value of $$(y_{A} )_{1}$$, we would like to determine the molar flow rate of the $$\beta$$-phase, $$\dot{M}_{\beta }$$.

In this case we consider an existing unit in which there are 6 stages. The inlet mole fractions are given by $$(x_{A} )_{ o} = 0$$ and $$(y_{A} )_{7} = 0.010$$, and the specified parameters associated with the system are $$K_{eq,A} = 2.53$$ and $$\dot{M}_{\gamma } = 30$$ kmol/h. The desired value of the mole fraction of acetone in the exit air stream is $$(y_{A} )_{1} = 0.0005$$. We begin the analysis with Equation \ref{93} that leads to

$1+A+A^{2} +A^{3} +A^{4} +A^{5} +A^{6} = \frac{(y_{A} )_{7} }{(y_{A} )_{1} } = \frac{0.010}{0.0005} = 20.0 \label{5a}\tag{5}$

Since $$\dot{M}_{\beta }$$ is an adjustable parameter, we need only solve this implicit equation for the absorption coefficient in order to determine the molar flow rate. Use of one of the iterative methods described in Appendix B leads to

$A = 1.342 \label{6a}\tag{6}$

and from the definition of the absorption coefficient given by Equation \ref{74} we determine the molar flow rate of the $$\beta$$-phase to be

$\dot{M}_{\beta } = A \dot{M}_{\gamma } K_{eq,A} = 101.9 \ { kmol/h} \label{7a}\tag{7}$

Adjusting a molar flow rate to achieve a specific separation is a convenient operational technique provided that the auxiliary equipment required to change the flow rate is readily available.

Sequential analysis-graphical

The algebraic solution to the cascade of equilibrium stages illustrated in Figure $$\PageIndex{2}$$ can also be represented in terms of a graph. In this case we consider the cascade illustrated in Figure $$\PageIndex{7}$$ in which the

index $$n$$ is bounded according to $$1\leq n\leq N$$. The macroscopic mole balance for species $$A$$ associated with this control volume can be expressed as

$\underbrace{(x_{A} )_{n} \dot{M}_{\beta } + (y_{A} )_{1} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{out of the control volume} \end{array}} = \underbrace{(x_{A} )_{ o} \dot{M}_{\beta } + (y_{A} )_{n+1} \dot{M}_{\gamma } }_{\begin{array}{l} \text{ molar flow of species }A \\ \text{into the control volume} \end{array}} \label{96}$

In this case we want to develop an expression for $$(y_{A} )_{n+1}$$ thus we arrange Equation \ref{96} in the form

$(y_{A} )_{n+1} = (x_{A} )_{n} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right) + \left[(y_{A} )_{1} -(x_{A} )_{ o} \left({\dot{M}_{\beta } / \dot{M}_{\gamma } } \right)\right] , \quad n = 1,2,...,N \label{97}$

This is sometime called the “operating line” and it can be used in conjunction with the “equilibrium line”

$(y_{A} )_{n} = K_{eq,A} (x_{A} )_{n} , \quad n = 1,2,...,N \label{98}$

to provide a graphical representation of the solution developed using the sequential analysis. To be precise, we note that Eqs. \ref{97} and \ref{98} represent a series of operating points and equilibrium points; however, the construction of operating lines and equilibrium lines is a useful graphical tool.

To illustrate the graphical construction associated with Eqs. \ref{97} and \ref{98} we consider the Type I problem discussed in Sec. 5.7.1 and repeated here as

Type I: Given the inlet mole fractions, $$(x_{A} )_{ o}$$ and $$(y_{A} )_{N+1}$$, the system parameters, and the desired value of $$(y_{A} )_{1}$$, we would like to determine the number of stages, $$N$$.

In this case we are given the inlet mole fractions, $$(x_{A} )_{ o} = 0$$ and $$(y_{A} )_{N+1} = 0.010$$, and the system parameters, $$K_{eq,A} = 2.25$$, $$\dot{M}_{\beta } = 97.5$$ kmol/h and $$\dot{M}_{\gamma } = 30$$ kmol/h. For these particular values, we want to know how many stages, $$N$$, are required to reduce the exit mole fraction of acetone to $$(y_{A} )_{1} = 0.001$$. The equilibrium line associated with Equation \ref{98} is given directly by

Equilibrium line: $y_{A} = 2.25 x_{A} \label{99}$

while the construction of the operating line associated with Equation \ref{97} requires that we assume the value of $$(y_{A} )_{1}$$ in order to obtain

Operating line: $y_{A} = 3.25 x_{A} + 0.001 \label{100}$

This approach is comparable to our use of Equation \ref{93} in Example $$\PageIndex{1}$$ where the analysis of a Type I problem required that we specify $$(y_{A} )_{1}$$ in our search for the number of stages associated with $$1+A+...+A^{N}$$.

Given the operating line and the equilibrium line, we can construct the operating points and the equilibrium points and this is done in Figure $$\PageIndex{8}$$. Some of these points are identified by numbers such as $$\boxed{4}$$ while others are identified by solid dots such as $$\bullet$$. In order to connect the graphical analysis shown in Figure $$\PageIndex{8}$$ with the sequential analysis given in Sec. 5.7.1, we recall Figure $$\PageIndex{3}$$ in the form given below by Figure $$\PageIndex{9}$$. There we have clearly identified Points #1, #2 and #3 in the graphical analysis with those same pairs of values in the sequential analysis.

In Figure $$\PageIndex{8}$$ and in Figure $$\PageIndex{9}$$ we see that Point #1 represents a point on the operating line, that Point #2 represents a point on the equilibrium line, and that Point #3 represents a second point on the operating line. To continue this type of comparison, we recall Figure $$\PageIndex{4}$$ in the form given below by Figure $$\PageIndex{10}$$.

There we have clearly identified Points #3, #4 and #5 in the graphical analysis with those same pairs of values in the sequential analysis. In Figure $$\PageIndex{8}$$ and Figure $$\PageIndex{10}$$ we see that Point #3 represents a point on the operating line, that Point #4 represents a point on the equilibrium line, and that Point #5 represents a second point on the operating line.

The sequential analysis presented in Sec. 5.7.1 has the advantage of providing a clear set of equations that describe the mass transfer process occurring in a cascade of equilibrium stages. The graphical analysis is certainly less accurate, but it can illustrate quite effectively how the system responds to changes in the operating conditions. As an example, we can reduce the molar flow rate of the $$\beta$$-phase (water) and examine the effect that will have on the separation process. For the case in which species $$A$$ is transferred from the $$\gamma$$-phase to the $$\beta$$-phase, the reduction of $$\dot{M}_{\beta }$$ will surely make the separation process less efficient.

This change is clearly indicated in Figure $$\PageIndex{11}$$ where the effect of reducing $$\dot{M}_{\beta }$$ by 21% diminishes the efficiency of the cascade significantly.

The graphical representation of a cascade of equilibrium stages is explored more thoroughly in Example $$\PageIndex{2}$$.

Example $$\PageIndex{2}$$: Graphical analysis of the absorption of acetone from air into water

Here we extend Example $$\PageIndex{1}$$ to explore the use of graphical methods to analyze the absorption of acetone from air into water. The molar flow rate of the gas mixture (air-acetone) entering the cascade (at stage $$N$$) is $$\dot{M}_{\gamma } = 30$$ kmol/h and the mole fraction of acetone is given by $$(y_{A} )_{N+1} = 0.010$$ The molar flow rate of the liquid (water) entering the cascade (at stage #1) is $$\dot{M}_{\beta } = 90$$ kmol/h and the mole fraction of acetone in this entering liquid stream is zero, i.e., $$(x_{A} )_{ o} = 0$$. In this case the slope of the operating line and the equilibrium coefficient are given by

${\dot{M}_{\beta } / \dot{M}_{\gamma } } = 3.0 , \quad K_{eq, A} = 2.53 \tag{1}\label{1b}$

and we can use these values to determine an operating line and an equilibrium line. Following the development given in Sec. 5.7.2 we construct the graph illustrated in Figure $$\PageIndex{12a}$$. There we find that 5 stages are not sufficient to produce the desired result, i.e., $$(y_{A} )_{N+1} = 0.010$$ and one is forced to make a choice of adding a sixth stage or accepting a somewhat reduced separation.

If we reduce the water flow rate from 90 kmol/h to 70 kmol/h the slope of the operating line is reduced from 3.00 to 2.33, and the situation illustrated in Figure $$\PageIndex{12a}$$ changes to that illustrated in Figure $$\PageIndex{12b}$$. An overall mole balance indicates that $$(x_{A} )_{N} = 0.00386$$ and the graphical construction indicates that slightly more than 18 equilibrium stages are required. This means that

we need 19 stages to accomplish the objective of reducing the mole fraction in the $$\gamma$$-phase to $$(y_{A} )_{1} = 0.001$$. If the water flow rate is further reduced to 67 kmol/h we see in Figure $$\PageIndex{12c}$$ that

the equilibrium line and the operating line intersect at $$x_{A} = 0.00337$$. This is referred to as a pinch point within the cascade of equilibrium stages, and an infinite number of equilibrium stages are required to reach this point. Consequently, it is impossible to reach the desired design specifications for the cascade, i.e., $$(y_{A} )_{N+1} = 0.010$$ and $$(x_{A} )_{N} = 0.0403$$.

The sequential analysis and the graphical analysis of continuous equilibrium stage processes provide a relatively clear illustration of the physical processes involved; however, more complex systems are routinely encountered in industrial practice. In those cases, powerful numerical methods are extremely useful.