# 7.5: Sequential Analysis for Recycle Systems

In Examples $$7.4.2$$ and $$7.4.3$$ we saw how the presence of a recycle stream created a loop in the flow of information. The determination of the molar flow rate of dichloroethane entering the reactor shown in Figure $$7.4.5a$$ required information about the recycle stream, i.e., information generated by the column used to purify the output stream for the process. The determination of the molar flow rate of ethylene entering the reactor shown in Figure $$7.4.6a$$ required information about the recycle stream, i.e., information generated by the absorber used to separate the ethylene oxide from the output of the reactor. For the systems described in Examples $$7.4.2$$ and $$7.4.3$$, it was possible to solve the linear set of equations simultaneously as we have done in other problems. However, most chemical engineering systems are nonlinear as a result of thermodynamic conditions and chemical kinetic models for reaction rates. For such systems it becomes impractical to solve the system of equations globally and in this section we examine an alternative approach.

In Figure $$\PageIndex{1}$$ we have illustrated a flowsheet5 for the manufacture of ethyl alcohol from ethylene. Here we see several units involving recycle and purge streams, and we need to think about what happens in those individual units in an operating chemical plant. Each unit is being monitored constantly on a day-to-day basis, and the data is interpreted in terms of a macroscopic balance around each unit. If a mass or mole balance is not satisfied, there is a problem with the unit and a solution needs to be found. The engineer or operator in charge of a specific unit will be thinking about a single control volume around that unit, and this motivates the sequential problem solving approach that we describe in this section. The sequential approach naturally leads to many redundant cuts; however, the advantage of this approach is that concepts associated with the analysis are simple and straightforward. We will use the pyrolysis of dichloroethane studied earlier in Example $$7.4.2$$ to illustrate the sequential analysis of a simple recycle system.

Example $$\PageIndex{1}$$: Sequential analysis of the pyrolysis of dichloroethane.

In this example we re-consider the pyrolysis of dichloroethane ($$\ce{C2H4Cl2}$$) to produce vinyl chloride ($$\ce{C2H3Cl}$$) and hydrochloric acid ($$\ce{HCl}$$) using a sequential analysis. The conversion of dichloroethane is given by

${ C}= \text{ Conversion of } \ce{C2H4Cl2} =\frac{- \mathscr{R}_{ \ce{C2H4Cl2} } }{ (\dot{M}_{\ce{C2H4Cl2}} )_{2} } =0.30 \tag{1}\label{a1}$

and the unreacted dichloroethane is separated from the reaction products and recycled back to the reactor as indicated in Figure $$\PageIndex{2}$$. The composition of the feed Stream #1 is 98% dichloroethane ($$\ce{C2H4Cl2}$$) and 2% ethane ($$\ce{C2H6}$$) on a molar basis and the total molar flow rate is $$\dot{M}_{1}$$. We assume that the separation column produces a sharp separation meaning that all the dichloroethane leaves in Stream #5 and the remaining components (ethane, hydrochloric acid and vinyl chloride) leave in Stream #4. Our objective in this example is to determine the recycle flow rate in Stream #5 along with the composition and flow rate of Stream #4 in terms of the molar flow rate of Stream #1. Since our calculation is sequential, we require the three control volumes illustrated in Figure $$\PageIndex{2}$$ instead of the two control volumes that were used in Example $$7.4.2$$. In the sequential approach we enclose each unit in a control volume and we assume that the input conditions to each control volume are known. For Control Volume 1 in Figure $$\PageIndex{2}$$ this means that we must assume the molar flow rate of Stream #5. This process is referred to as tearing the cycle and Stream #5 is referred to as the tear stream.

Given the input conditions for the mixer, we can easily calculate the output conditions, i.e., the conditions associated with Stream #2. This means that we know the input conditions for the reactor and we can calculate the output conditions in Stream #3. Moving on to the separator, we use the conditions in Stream #3 to determine the conditions in Stream #4 and Stream #5. The calculated conditions in Stream #5 provide the new assumed value for the molar flow rate entering the mixer, and a sequential computational procedure can be repeated until a converged solution is obtained. For linear systems, convergence is assured; however, the matter is more complex for the non-linear system studied in Example $$\PageIndex{2}$$.

Since the input information for each control volume is known, the structure for all the material balances has the form

$(\dot{M}_{A} )_{out} =(\dot{M}_{A} )_{i n} + \mathscr{R}_{ A} , \quad A \Rightarrow \ce{C2H6} , \quad \ce{ HCl},\quad \ce{C2H3Cl},\quad \ce{C2H4Cl2} \tag{2}\label{a2}$

Here the unknown quantities are on the left hand side and the known quantities are on the right hand side. Directing our attention to the first control volume illustrated in Figure $$\PageIndex{2}$$ we find

### Control Volume 1

$\begin{array} & \ce{C2H6} : && (\dot{M}_{\ce{C2H6}} )_{2} =(\dot{M}_{\ce{C2H6}} )_{1} \tag{3a}\label{a3a} \end{array}$

$\begin{array} & \ce{C2H4Cl2} : && (\dot{M}_{\ce{C2H4Cl2}} )_{2} =(\dot{M}_{\ce{C2H4Cl2}} )_{1} + (\dot{M}_{\ce{C2H4Cl2}} )_{5}^{({ o})} \tag{3b}\label{a3b} \end{array}$

$\begin{array} & \ce{HCl} : && (\dot{M}_{\ce{HCl}} )_{2} = 0 \tag{3c}\label{a3c} \end{array}$

$\begin{array} & \ce{C2H3Cl} : && (\dot{M}_{\ce{C2H3Cl}} )_{2} =0 \tag{3d}\label{a3d} \end{array}$

Here we have used $$(\dot{M}_{\ce{C2H4Cl2}} )_{5}^{({ o})}$$ to identify the first assumed value for the molar flow rate of dichloroethane in the tear stream. Moving on to Control Volume 2, we obtain

### Control Volume 2

$\begin{array} & \ce{C2H6} : && (\dot{M}_{\ce{C2H6}} )_{3} =(\dot{M}_{\ce{C2H6}} )_{2} + \mathscr{R}_{ \ce{C2H6} } \tag{4a}\label{a4a} \end{array}$

$\begin{array} & \ce{C2H4Cl2} : && (\dot{M}_{\ce{C2H4Cl2}} )_{3} =(\dot{M}_{\ce{C2H4Cl2}} )_{2} + \mathscr{R}_{ \ce{C2H4Cl2} } \tag{4b}\label{a4b} \end{array}$

$\begin{array} & \ce{HCl} : && (\dot{M}_{\ce{HCl}} )_{3} =(\dot{M}_{\ce{HCl}} )_{2} + \mathscr{R}_{ \ce{HCl}} \tag{4c}\label{a4c} \end{array}$

$\begin{array} & \ce{C2H3Cl} : && (\dot{M}_{\ce{C2H3Cl}} )_{3} =(\dot{M}_{\ce{C2H3Cl}} )_{2} + \mathscr{R}_{ \ce{C2H3Cl}} \tag{4d}\label{a4d} \end{array}$

The global rates of production are constrained by Axiom II that provides the relations

Axiom II: $\mathscr{R}_{ \ce{C2H6} } = 0 , \quad \mathscr{R}_{ \ce{HCl}} = - \mathscr{R}_{ \ce{C2H4Cl2} } , \quad \mathscr{R}_{ \ce{C2H3Cl}} = - \mathscr{R}_{ \ce{C2H4Cl2} } \tag{5}\label{a5}$

which can be expressed in terms of the conversion (see Equation \ref{a1}) to obtain

$\mathscr{R}_{ \ce{C2H6} } = 0 , \quad \mathscr{R}_{ \ce{HCl}} = { C} (\dot{M}_{\ce{C2H4Cl2} } )_{2} , \quad \mathscr{R}_{ \ce{C2H3Cl} } = { C} (\dot{M}_{\ce{C2H4Cl2}} )_{2} \tag{6}\label{a6}$

These representations for the global rates of production can be used with Eqs. \ref{a4a} - \ref{a4d} to obtain

### Control Volume 2

$\begin{array} & \ce{C2H6} : && (\dot{M}_{\ce{C2H6}} )_{3} =(\dot{M}_{\ce{C2H6}} )_{2} \tag{7a}\label{a7a} \end{array}$

$\begin{array} & \ce{C2H4Cl2} : && (\dot{M}_{\ce{C2H4Cl2}} )_{3} =(\dot{M}_{\ce{C2H4Cl2}} )_{2} (1 - { C}) \tag{7b}\label{a7b} \end{array}$

$\begin{array} & \ce{HCl} : && (\dot{M}_{\ce{HCl}} )_{3} =(\dot{M}_{\ce{HCl}} )_{2} + { C} (\dot{M}_{\ce{C2H4Cl2}} )_{2} \tag{7c}\label{a7c} \end{array}$

$\begin{array} & \ce{C2H3Cl} : && (\dot{M}_{\ce{C2H3Cl}} )_{3} =(\dot{M}_{\ce{C2H3Cl}} )_{2} + { C} (\dot{M}_{\ce{C2H4Cl2}} )_{2} \tag{7d}\label{a7d} \end{array}$

This completes the first two steps in the sequence, and we are ready to move on to Control Volume 3 shown in Figure $$\PageIndex{2}$$. Application of Equation \ref{a2} to Control Volume 3 provides the following mole balances:

### Control Volume 3

$\begin{array} & \ce{C2H6} : && (\dot{M}_{\ce{C2H6}} )_{4} =(\dot{M}_{\ce{C2H6}} )_{3} \tag{8a}\label{a8a} \end{array}$

$\begin{array} & \ce{C2H4Cl2} : && (\dot{M}_{\ce{C2H4Cl2} } )_{5}^{(1)} =(\dot{M}_{\ce{C2H4Cl2}} )_{3} \tag{8b}\label{a8b} \end{array}$

$\begin{array} & \ce{HCl} : && (\dot{M}_{\ce{HCl}} )_{4} =(\dot{M}_{\ce{HCl}} )_{3} \tag{8c}\label{a8c} \end{array}$

$\begin{array} & \ce{C2H3Cl} : && (\dot{M}_{\ce{C2H3Cl}} )_{4} =(\dot{M}_{\ce{C2H3Cl}} )_{3} \tag{8d}\label{a8d} \end{array}$

Here we have indicated that the molar flow rate of dichloroethane ($$\ce{C2H4Cl2}$$) leaving the separator in Stream #5 is the first approximation based on the assumed value entering the mixer. This assumed value is given in Equation \ref{a3b} as $$(\dot{M}_{\ce{C2H4Cl2} } )_{5}^{({ o})}$$. Because Stream #5 contains only a single component, this problem is especially simple, and we only need to make use of Eqs. \ref{a3b}, \ref{a7b} and \ref{a8b} to obtain a relation between $$(\dot{M}_{\ce{C2H4Cl2}} )_{5}^{(1)}$$ and $$(\dot{M}_{\ce{C2H4Cl2}} )_{5}^{({ o})}$$. This relation is given by

\begin{align} & \ce{C2H4Cl2} : && (\dot{M}_{\ce{C2H4Cl2}} )_{5}^{(1)} =\left(1 - { C}\right)\left[(\dot{M}_{\ce{C2H4Cl2}} )_{1} + (\dot{M}_{\ce{C2H4Cl2}} )_{5}^{({ o})} \right] \tag{9}\label{a13} \end{align}

Neither of the two molar flow rates on the right hand side of this result are known; however, we can eliminate the molar flow rate of dichloroethane $$(\dot{M}_{\ce{C2H4Cl2}} )_{1}$$by working in terms of a dimensionless molar flow rate6 defined by

$\mathscr{M}_{ 5} =\frac{(\dot{M}_{\ce{C2H4Cl2}} )_{5} }{(\dot{M}_{\ce{C2H4Cl2}} )_{1} } \tag{10}\label{a14}$

This allows us to express Equation \ref{a13} as

$\mathscr{M}_{ 5}^{(1)} =\left(1 - { C}\right)\left[1 + \mathscr{M}_{ 5}^{({ o})} \right] , \quad { C}=0.30 \tag{11}\label{a15}$

and we can assume a value of $$\mathscr{M}_{ 5}^{({ o})}$$ in order to compute a value of $$\mathscr{M}_{ 5}^{(1)}$$. On the basis of this computed value of $$\mathscr{M}_{ 5}^{(1)}$$, the analysis can be repeated to determine $$\mathscr{M}_{ 5}^{({ 2})}$$ that is given by

$\mathscr{M}_{ 5}^{({ 2})} =\left(1 - { C}\right)\left[1 + \mathscr{M}_{ 5}^{({ 1})} \right] , \quad { C}=0.30 \tag{12}\label{a16}$

This representation can be generalized to obtain the $$(i+1)^{th}$$ value that is given by

$\mathscr{M}_{ 5}^{(i+1)} =\left(1 - { C}\right)\left[1 + \mathscr{M}_{ 5}^{(i)} \right] , \quad { C}=0.30 , \quad i=1,2,3,..... \tag{13}\label{a17}$

This procedure is referred to as Picard’s method7, or as a fixed point iteration, or as the method of successive substitution 8 and it is often represented in the form

$x_{i+1} =f(x_{i} ) , \quad i = 1, 2, 3, ...{ etc} \tag{14}\label{a18}$

To illustrate how this iterative calculation is carried out, we assume that $$\mathscr{M}_{ 5}^{({ o})} =0$$ to produce the values listed in Table $$\PageIndex{1a}$$ where we see a converged value given by $$\mathscr{M}_{ 5} =2.333$$. One can avoid these detailed calculations by noting that for arbitrarily large values of $$i$$ we arrive at the fixed point condition given by

$\mathscr{M}_{ 5}^{(i+1)} =\mathscr{M}_{ 5}^{(i)} , \quad i \to \infty \tag{15}\label{a19}$

and the converged solution for the dimensionless molar flow rate is

$\mathscr{M}_{ 5}^{(\infty )} =\frac{1-{ C}}{ C} =2.333 \tag{16}\label{a20}$

This indicates that the molar flow rate of dichloroethane ($$\ce{C2H4Cl2}$$) in the recycle stream is

$(\dot{M}_{\ce{C2H4Cl2}} )_{5} =2.333 (\dot{M}_{\ce{C2H4Cl2}} )_{1} \tag{17}\label{a21}$

In terms of the total molar flow rate in Stream #1, this takes the form

$(\dot{M}_{\ce{C2H4Cl2}} )_{5} =2.2867\dot{M}_{1} \tag{18}\label{a22}$

which is exactly the answer obtained in Example $$7.4.2$$.

A variation on Picard’s method is called Wegstein’s method9 and in terms of the nomenclature used in Equation \ref{a18} this iterative procedure takes the form10

$x_{i+1} =(1-q)f(x_{i} ) + q x_{i} , \quad i = 1, 2, 3, ...{ etc} \tag{19}\label{a24}$

in which $$q$$ is an adjustable parameter. When this adjustable parameter is equal to zero, $$q=0$$, we obtain the original successive substitution scheme given by Equation \ref{a18}.

Table $$\PageIndex{1a}$$: Converging Values for Dimensionless Recycle Flow Rate (Picard’s Method)

$$i$$ $$\mathscr{M}_{ 5}^{(i)}$$ $$\mathscr{M}_{ 5}^{(i+1)}$$
0 0.000 0.700
1 0.700 1.190
2 1.190 1.533
3 1.533 1.773
4 1.773 1.941
. . .
. . .
22 2.332 2.333
23 2.333 2.333

When the adjustable parameter greater than zero and less than one, $$0<q<1$$, we obtain a damped successive substitution process that improves stability for nonlinear systems. When the adjustable parameter is negative, $$q<0$$, we obtain an accelerated successive substitution that may lead to an unstable procedure. For the problem under consideration in this example, Wegstein’s method can be expressed as

$\mathscr{M}_{ 5}^{(i+1)} =\left(1-q\right) \left\{\left(1 - { C}\right) \left[1 + \mathscr{M}_{ 5}^{(i)} \right]\right\} + q \mathscr{M}_{ 5}^{(i)} , \quad { C}=0.30 \tag{20}\label{a25}$

When the adjustable parameter is given by $$q=- 1.30$$ we obtain the accelerated convergence illustrated in Table $$\PageIndex{1b}$$. When confronted with nonlinear systems, the use of values of $$q$$ near one may be necessary to obtain stable convergence.

Table $$\PageIndex{1b}$$: Convergence for Dimensionless Recycle Flow Rate (Wegstein’s Method)
$$i$$ $$q$$ $$\mathscr{M}_{ 5}^{(i)}$$ $$\mathscr{M}_{ 5}^{(i+1)}$$
0 $$- 1.30$$ 0.000 1.610
1 $$- 1.30$$ 1.610 2.109
2 $$- 1.30$$ 2.109 2.264
3 $$- 1.30$$ 2.264 2.312
4 $$- 1.30$$ 2.312 2.327
5 $$- 1.30$$ 2.327 2.331
6 $$- 1.30$$ 2.331 2.333
7 $$- 1.30$$ 2.333 2.333

In the next example we return to a more complex problem associated with the production of ethylene oxide that was studied earlier in Example $$7.4.3$$. In that case we will find that a damped successive substitution process is necessary to obtain a converged solution.

Example $$\PageIndex{2}$$: Sequential analysis of the production of ethylene oxide

In Figure $$\PageIndex{3}$$ we have illustrated a process in which ethylene oxide ($$\ce{C2H4O}$$) is produced by the oxidation of ethylene ($$\ce{C2H4}$$) over a catalyst containing silver. In a side reaction, ethylene is oxidized to carbon dioxide ($$\ce{CO2}$$) and water ($$\ce{H2O}$$). The feed stream, Stream #1, consists of ethylene ($$\ce{C2H4}$$) and air ($$\ce{N2}$$ and $$\ce{O2}$$) which is combined with Stream #5 that contains the unreacted ethylene ($$\ce{C2H4}$$), carbon dioxide ($$\ce{CO2}$$) and nitrogen ($$\ce{N2}$$). The mole fraction of ethylene in Stream #2 entering the reactor must be maintained at $$0.05$$ for satisfactory catalyst operation. The conversion of ethylene in the reactor is optimized to be 70% ($${ C}=0.70$$) and the yield of ethylene oxide is 50% ($${ Y}=0.50$$). All of the oxygen in the feed reacts, thus there is no oxygen in Stream #3 and we have $$(\dot{M}_{\ce{O2} } )_{3} =0$$. The reactor effluent is sent to an absorber where all the ethylene oxide is absorbed in the water entering in Stream #8. Water is fed to the absorber such that $$(\dot{M}_{\ce{H2O}} )_{8} =100 (\dot{M}_{\ce{C2H4O}} )_{7}$$ and we assume that all the water leaves the absorber in Stream #7. A portion of Stream #4 is recycled in Stream #5 and a portion is purged in Stream #6. In this example we want to determine the fraction of $$\dot{M}_{4}$$ that needs to be purged in order that 100 mol/hr of ethylene oxide are produced by the reactor.

This problem statement is identical to that given by Example $$7.4.3$$, and it is only the procedure for solving this problem that will be changed. In this case we assume the values of the flow rates entering the mixer from the tear stream. We then update these assumed values on the basis of the sequential analysis of the four control volumes.

## Available Data

The conversion and the yield are key parameters in this problem, and we define these quantities explicitly as

${ C}= \text{ Conversion of ethylene } = \frac{-\mathscr{R}_{ \ce{C2H4}} }{ (\dot{M}_{ \ce{C2H4} })_2} = 0.70 \tag{1}\label{b1}$

${ Y}= \text{ Yield of } \left(\ce{C2H4O} {\bf /} \ce{C2H4} \right)=\frac{\mathscr{R}_{ \ce{C2H4O}} }{- \mathscr{R}_{ \ce{C2H4} } } =0.50 \tag{2}\label{b2}$

We are given that 100 mol/hr of ethylene oxide are produced by the reactor and we express this condition as

Reactor: $\mathscr{R}_{ \ce{C2H4O}} =\beta, \quad \beta = 100 \ mol/h \tag{3}\label{b3}$

The results expressed by Eqs. \ref{b1}, \ref{b2} and \ref{b3} can be used to immediately deduce that

$\mathscr{R}_{ \ce{C2H4} } =- {\beta / { Y}} \tag{4a}\label{b4a}$

$(\dot{M}_{\ce{C2H4}} )_{2} ={\beta / { CY}} \tag{4b}\label{b4b}$

In addition, we are given the following relations:

Streams #7 and #8: $(\dot{M}_{\ce{H2O}} )_{8} =\psi (\dot{M}_{\ce{C2H4O}} )_{7} , \quad \psi =100 \tag{5a}\label{b5a}$

Steam #1: $(\dot{M}_{\ce{O2} } )_{1} =\varphi (\dot{M}_{\ce{N2}} )_{1} , \quad \varphi = \left({21 / 79} \right) \tag{5b}\label{b5b}$

Stream #2: $(x _{\ce{C2H4}} )_{2} =\eta , \quad \eta = 0.05 \tag{5c}\label{b5c}$

Stream #3: $(\dot{M}_{\ce{O2}} )_{3} =0 \tag{5d}\label{b5d}$

We can use Equation \ref{b4b} and Equation \ref{b5c} to provide

$(x _{\ce{C2H4}} )_{2} =\frac{(\dot{M}_{\ce{C2H4}} )_{2} }{\dot{M}_{2} } =\frac{\beta / { CY} }{\dot{M}_{2} } =\eta \tag{6a}\label{b6a}$

which indicates that the total molar flow rate entering the reactor is given by

$\dot{M}_{2} =\frac{\beta }{\eta { C} { Y}} \tag{6b}\label{b6b}$

In Example $$7.4.3$$ we carefully constructed control volumes that would minimize redundant cuts, while in Figure $$\PageIndex{3}$$ we have simply enclosed each unit in a control volume and this creates four redundant cuts. As in Example $$\PageIndex{1}$$, we will solve the material balances for each control volume in a sequential manner, and we will assume a value for any variable that is unknown.

## Stoichiometry

The details associated with Axiom II (see Eqs. \ref{b7a} - \ref{b7f} through \ref{b12a} - \ref{b12d} of Example $$7.4.3$$) can be used along with the definitions of the conversion and yield to express the global rates of production in terms of the conversion and the yield. These results are given by

$\mathscr{R}_{ \ce{C2H4} } =- {\beta / { Y}} \tag{7a}\label{b7a}$

$\mathscr{R}_{ \ce{CO2} } =2(1-{ Y}) {\beta / { Y}} \tag{7b}\label{b7b}$

$\mathscr{R}_{ \ce{N2} } =0 \tag{7c}\label{b7c}$

$\mathscr{R}_{ \ce{O2} } =- (3 - {\frac{5}{2}} { Y}) {\beta / { Y}} \tag{7d}\label{b7d}$

$\mathscr{R}_{ \ce{C2H4O}} =\beta \tag{7e}\label{b7e}$

$\mathscr{R}_{ \ce{H2O}} =2(1-{ Y}){\beta / { Y}} \tag{7f}\label{b7f}$

## Mole Balances

We begin our analysis of this process with Axiom I for steady processes and fixed control volumes

Axiom I: $\int _{A}c_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA=\mathscr{R}_{ A} , \quad A = 1, 2, 3, ..., 6 \tag{8}\label{b8}$

and we note that each mole balance can be expressed as

$(\dot{M}_{A} )_{out} =(\dot{M}_{A} )_{i n} + \mathscr{R}_{ A} , \quad A = 1, 2, 3, ..., 6 \tag{9}\label{b9}$

Application of Axiom I to the first control volume leads to

### Control Volume 1 (Mixer)

$\begin{array} & ^{**}\ce{C2H4} : && (\dot{M}_{\ce{C2H4}} )_{2} =(\dot{M}_{\ce{C2H4}} )_{1} + (\dot{M}_{\ce{C2H4}} )_{5} \tag{10a}\label{b10a} \end{array}$

$\begin{array} & ^{**}\ce{CO2} : && (\dot{M}_{\ce{CO2}} )_{2} = 0 + (\dot{M}_{\ce{CO2}} )_{5} \tag{10b}\label{b10b} \end{array}$

$\begin{array} & ^{**}\ce{N2} : && (\dot{M}_{\ce{N2}} )_{2} =(\dot{M}_{\ce{N2}} )_{1} + (\dot{M}_{\ce{N2}} )_{5} \tag{10c}\label{b10c} \end{array}$

$\begin{array} & \ce{O2} : && (\dot{M}_{\ce{O2}} )_{2} =(\dot{M}_{\ce{O2}} )_{1} \tag{10d}\label{b10d} \end{array}$

$\begin{array} & \ce{C2H4O} : && (\dot{M}_{\ce{C2H4O}} )_{2} =(\dot{M}_{\ce{C2H4O}} )_{1} =0 \tag{10e}\label{b10e} \end{array}$

$\begin{array} & \ce{H2O} : && (\dot{M}_{\ce{H2O}} )_{2} =(\dot{M}_{\ce{H2O}} )_{1} =0 \tag{10f}\label{b10f} \end{array}$

Here we have marked with a double asterisk (**) the three molecular species that are contained in the recycle stream. In order to determine the portion of Stream #4 that is recycled in Stream #5, we would normally require macroscopic balances for only those components that appear in the recycle stream. However, in this particular problem, the molar flow rates of oxygen and nitrogen entering the mixer are connected by Equation \ref{b5b}, thus we require the macroscopic balances for all four species that enter the mixer. This means that only Eqs. \ref{b10a}, \ref{b10b}, \ref{b10c} and \ref{b10d} are required and we can move on to the remaining control volumes making use of only the mole balances associated with these four species.

### Control Volume 2 (Reactor)

In this case the four mole balances include a term representing the net global rate of production owning to chemical reaction.

$\begin{array} & ^{**}\ce{C2H4} : && (\dot{M}_{\ce{C2H4}} )_{3} =(\dot{M}_{\ce{C2H4}} )_{2} + \mathscr{R}_{ \ce{C2H4} } \tag{11a}\label{b11a} \end{array}$

$\begin{array} & ^{**}\ce{CO2} : && (\dot{M}_{\ce{CO2}} )_{3} = (\dot{M}_{\ce{CO2}} )_{2} + \mathscr{R}_{ \ce{CO2} } \tag{11b}\label{b11b} \end{array}$

$\begin{array} & ^{**}\ce{N2} : && (\dot{M}_{\ce{N2}} )_{3} = (\dot{M}_{\ce{N2}} )_{2} \tag{11c}\label{b11c} \end{array}$

$\begin{array} & \ce{O2} : && (\dot{M}_{\ce{O2}} )_{3} = (\dot{M}_{\ce{O2}} )_{2} +\mathscr{R}_{ \ce{O2} } \tag{11d}\label{b11d} \end{array}$

### Control Volume 3 (Absorber)

No chemical reaction occurs in this unit, thus the four balance equations are given by

$\begin{array} & ^{**}\ce{C2H4} : && (\dot{M}_{\ce{C2H4}} )_{4} =(\dot{M}_{\ce{C2H4}} )_{3} \tag{12a}\label{b12a} \end{array}$

$\begin{array} & ^{**}\ce{CO2} : && (\dot{M}_{\ce{CO2}} )_{4} = (\dot{M}_{\ce{CO2}} )_{3} \tag{12b}\label{b12b} \end{array}$

$\begin{array} & ^{**}\ce{N2} : && (\dot{M}_{\ce{N2}} )_{4} = (\dot{M}_{\ce{N2}} )_{3} \tag{12c}\label{b12c} \end{array}$

$\begin{array} & \ce{O2} : && (\dot{M}_{\ce{O2}} )_{4} = (\dot{M}_{\ce{O2}} )_{3} =0 \tag{12d}\label{b12d} \end{array}$

Here we have made use of the information that all of the oxygen in the feed reacts, and we have used the information that ethylene ($$\ce{C2H4}$$), carbon dioxide ($$\ce{CO2}$$), nitrogen ($$\ce{N2}$$) or oxygen, ($$\ce{O2}$$) do not appear in Stream #7 and Stream #8.

### Control Volume 4 (Splitter)

This passive unit involves only three molecular species and the appropriate mole balances are given by

$\begin{array} & ^{**}\ce{C2H4} : && (\dot{M}_{\ce{C2H4}} )_{5} +(\dot{M}_{\ce{C2H4}} )_{6} =(\dot{M}_{\ce{C2H4}} )_{4} \tag{13a}\label{b13a} \end{array}$

$\begin{array} & ^{**}\ce{CO2} : && (\dot{M}_{\ce{CO2}} )_{5} +(\dot{M}_{\ce{CO2}} )_{6} = (\dot{M}_{\ce{CO2}} )_{4} \tag{13b}\label{b13b} \end{array}$

$\begin{array} & ^{**}\ce{N2} : && (\dot{M}_{\ce{N2}} )_{5} + (\dot{M}_{\ce{N2}} )_{6} = (\dot{M}_{\ce{N2}} )_{4} \tag{13c}\label{b13c} \end{array}$

## Sequential Analysis for $$\ce{O2}$$

At this point it is convenient to direct our attention to the molar balances for oxygen and carry out a sequential analysis to obtain

C.V. 1: $(\dot{M}_{\ce{O2}} )_{2} =(\dot{M}_{\ce{O2}} )_{1} \tag{14a}\label{b14a}$

C.V. 2: $(\dot{M}_{\ce{O2} } )_{3} = (\dot{M}_{\ce{O2} } )_{2} -(3 - \frac{5}{2} { Y}) {\beta / { Y}} \tag{14b}\label{b14b}$

C.V. 3: $(\dot{M}_{\ce{O2}} )_{3} = 0 \tag{14c}\label{b14c}$

These results allow us to determine the oxygen flow rate entering the mixer as

$(\dot{M}_{\ce{O2} } )_{1} = (3 - {\frac{5}{2} Y}) {\beta / Y } \tag{15}\label{b15}$

Here it is important to recall Equation \ref{b5b} and use that result to specify the molar flow rate of nitrogen ($$\ce{N2}$$) in Stream #1 as

$(\dot{M}_{\ce{N2} } )_{1} = (3 - {\frac{5}{2} Y}) {\beta / \varphi { Y}} \tag{16}\label{b16}$

We are now ready to direct our attention to the molar flow rates of ethylene ($$\ce{C2H4}$$),carbon dioxide ($$\ce{CO2}$$) and nitrogen ($$\ce{N2}$$) in the recycle stream and use these results to construct an iterative procedure that will allow us to calculate the fraction $$\alpha$$.

## Algebra

We begin by recognizing that the splitter conditions can be expressed as

$\begin{array} & \ce{C2H4} : && (\dot{M}_{\ce{C2H4}} )_{5} =(1-\alpha )(\dot{M}_{\ce{C2H4}} )_{4} \tag{17a}\label{b17a} \end{array}$

$\begin{array} & \ce{CO2} : && (\dot{M}_{\ce{CO2}} )_{5} =(1-\alpha )(\dot{M}_{\ce{CO2}} )_{4} \tag{17b}\label{b17b} \end{array}$

$\begin{array} & \ce{N2} : && (\dot{M}_{\ce{N2}} )_{5} =(1-\alpha )(\dot{M}_{\ce{N2}} )_{4} \tag{17c}\label{b17c} \end{array}$

and we will use these conditions in our analysis of ethylene ($$\ce{C2H4}$$), carbon dioxide ($$\ce{CO2}$$), nitrogen ($$\ce{N2}$$). Directing our attention to the ethylene flow rate leaving the reactor, we note that the use of Equation \ref{b4a} - \ref{b4b} in Equation \ref{b11a} leads to

$(\dot{M}_{\ce{C2H4}} )_{3} ={\beta \left(1-{ C}\right) / { CY}} \tag{18}\label{b18}$

Moving from the reactor to the absorber, we use Equation \ref{b12a} with Equation \ref{b18} to obtain

$(\dot{M}_{\ce{C2H4}} )_{4} ={\beta \left(1-{ C}\right) / { CY}} \tag{19}\label{b19}$

and when this result is used in the first splitter condition given by Equation \ref{b17a} we obtain

$(\dot{M}_{\ce{C2H4}} )_{5} =(1-\alpha ){\beta \left(1-{ C}\right) / { CY}} \tag{20}\label{b20}$

Here we must remember that $$\alpha$$ is the parameter that we want to determine by means of an iterative process. At this point we move on to the mole balances for nitrogen ($$\ce{N2}$$) and make use of Eqs. \ref{b10c}, \ref{b11c}, \ref{b12c}, \ref{b16} and \ref{b17c} to obtain

$(M_{\ce{N2}} )_{5} =\frac{\left(1 - \alpha \right)}{\alpha } \frac{\beta \left(3 - {\frac{5}{2} Y}\right)}{\varphi { Y}} \tag{21}\label{b21}$

Finally we consider the mole balances for carbon dioxide ($$\ce{CO2}$$) and make use of Eqs. \ref{b7b}, \ref{b10b}, \ref{b11b}, and \ref{b17b} to obtain the following result for the molar flow rate of carbon dioxide in the recycle stream.

$(\dot{M}_{\ce{CO2}} )_{5} =\frac{\left(1 - \alpha \right)}{\alpha } \frac{2 \beta \left(1-{ Y}\right)}{ Y} \tag{22}\label{b22}$

At this point we consider a total molar balance for the mixer

$\dot{M}_{1} + \dot{M}_{5} =\dot{M}_{2} \tag{23}\label{b23}$

and note that the total molar flow rate of Stream #2 was given earlier by Equation \ref{b6b}. This leads to the mole balance around the mixer given by

Mixer: $\dot{M}_{1} ={\beta / \eta { C} { Y}} \dot{M}_{5} \tag{24}\label{b24}$

and we are ready to develop an iterative solution for the recycle flow and the parameter $$\alpha$$.

At this point we begin our analysis of the tear stream, i.e., Stream #5 that can be expressed as

$\dot{M}_{5} =(\dot{M}_{\ce{C2H4}} )_{5} + (\dot{M}_{\ce{CO2}} )_{5} + (\dot{M}_{\ce{N2}} )_{5} \tag{25}\label{b25}$

Given these values, we make use of Equation \ref{b24} to express the molar flow rate entering the mixer in the form

$\dot{M}_{1} ={\beta / \eta { CY}} - (\dot{M}_{\ce{C2H4}} )_{5} + (\dot{M}_{\ce{CO2}} )_{5} + (\dot{M}_{\ce{N2}} )_{5} \tag{26}\label{b26}$

Representing the total molar flow rate of Stream #1 in terms of the three species molar flow rates leads to

$\dot{M}_{1}^{(1)} =(\dot{M}_{\ce{C2H4}} )_{1}^{(1)} + (\dot{M}_{\ce{O2}} )_{1} + (\dot{M}_{\ce{N2}} )_{1} \tag{27} \label{b26b}$

in which the molar flow rates of oxygen ($$\ce{O2}$$) and nitrogen ($$\ce{N2}$$) are specified by Eqs. \ref{b15} and \ref{b16}. Use of those results provides

$\dot{M}_{1} =(\dot{M}_{\ce{C2H4}} )_{1} + (3 - {\frac{5}{2} Y}) {\beta \left(1+\varphi \right) / \varphi { Y}} \tag{28}\label{b27}$

and substitution of this result into Equation \ref{b26} yields

\begin{align} (\dot{M}_{\ce{C2H4}} )_{1} = && \beta / \eta CY - (3 - \frac{5}{2} Y) \beta \left(1+\varphi \right) / \varphi Y \nonumber\\ && - \left[(\dot{M}_{\ce{C2H4}} )_{5} + (\dot{M}_{\ce{CO2}} )_{5} + (\dot{M}_{\ce{N2}} )_{5} \right] \tag{29}\label{b28} \end{align}

Here we can use Eqs. \ref{b4b} and \ref{b10a} to obtain

$\beta / { CY} =(\dot{M}_{\ce{C2H4}} )_{1} + (\dot{M}_{\ce{C2H4}} )_{5} \tag{30}\label{b29}$

which allows us to express Equation \ref{b28} as

${\beta / { CY}} ={\beta / \eta { CY} - (3 - {\frac{5}{2}} { Y}) {\beta \left(1+\varphi \right) / \varphi { Y}} } - \left[(\dot{M}_{\ce{CO2}} )_{5} + (\dot{M}_{\ce{N2}} )_{5} \right] \tag{31}\label{b30}$

At this point we return to Eqs. \ref{b20}, \ref{b21} and \ref{b22} in order to express the molar flow rates of carbon dioxide ($$\ce{CO2}$$) and nitrogen ($$\ce{N2}$$) in the tear stream as

$(\dot{M}_{\ce{CO2} } )_{5} + (\dot{M}_{\ce{N2} } )_{5} =\frac{(\dot{M}_{\ce{C2H4}} )_{5} \left[{ CY} / \beta (1-{ C}) \right]}{1 - (\dot{M}_{\ce{C2H4}} )_{5} \left[{ CY} / \beta (1-{ C}) \right]} \left\{\frac{\beta \left(3 - \frac{5}{2} Y \right)}{\varphi { Y}} + \frac{2\beta \left(1 - { Y}\right)}{ Y} \right\} \tag{32}\label{b31}$

Substitution of this result into Equation \ref{b30} leads to an equation for the molar flow rate of ethylene ($$\ce{C2H4}$$) in Stream #5. This result can be expressed in the compact form

$\Omega - \left[\frac{(\dot{M}_{\ce{C2H4}} )_{5} }{\beta \left(1-{ C}\right) / { CY} - (\dot{M}_{\ce{C2H4}} )_{5} } \right]\Lambda =0 \tag{33}\label{b32}$

where the two parameters are given by

$\Omega =\frac{1-\eta }{\eta { CY}} - \frac{\left(3 - {\frac{5}{2}} { Y}\right)\left(1+\varphi \right)}{\varphi { Y}} , \quad \Lambda =\frac{\left(3 - {\frac{5}{2}} { Y}\right) + 2\varphi \left(1-{ Y}\right)}{\varphi { Y}} \tag{34}\label{b33}$

At this point we are ready to use a trial-and-error procedure to first solve for $$(\dot{M}_{\ce{C2H4}} )_{5}$$ and then solve for the parameter $$\alpha$$.

## Picard’s method

We begin by defining the dimensionless molar flow rate as

$x=\frac{(\dot{M}_{\ce{C2H4}} )_{5} }{\beta } \tag{35}\label{b34}$

so that the governing equation takes the form

$H(x)=\Omega - \frac{\Lambda x}{\left(1-{ C}\right) / { CY} - x} =0 \tag{36}\label{b35}$

In order to use Picard’s method (see Appendix B4), we define a new function according to

Definition: $f(x)=x + H(x) \tag{37}\label{b36}$

and for any specific value of the dependent variable, $$x_{i}$$, we can define a new value, $$x_{i+1}$$, by

Definition: $x_{i+1} =f(x_{i} ) , \quad i = 1,2,3,...,\infty \tag{38}\label{b37}$

To be explicit we note that the new value of the dependent variable is given by

$x_{i+1} = x_{i} +\left[\Omega - \frac{\Lambda x_{i} }{\left(1-{ C}\right) / { CY} - x_{i} } \right] , \quad i = 1,2,3,...,\infty \tag{39}\label{b38}$

In terms of the parameters for this particular problem

$\Omega = 37.6190 , \quad \Lambda = 15.1678 , \quad { C} = 0.7 , \quad { Y} = 0.5 \tag{40}\label{b39}$

we have the following iterative scheme

$x_{i+1} =x_{i} + \left[37.6190 - \frac{15.1678 x_{i} }{0.8571 - x_{i} } \right] , \quad i = 1,2,3,...,\infty \tag{41}\label{b40}$

By inspection, one can see that $$x < 0.8571$$ thus we choose our first guess to be $$x_{ o} =0.62$$ and this leads to the results shown in Table $$\PageIndex{2a}$$. Clearly Picard’s method does not converge for this

Table $$\PageIndex{2a}$$: Iterative Values for Dimensionless Flow Rate (Picard’s Method)
$$i$$ $$x_{i} ={(\dot{M}_{\ce{C2H4}} )_{5}^{(i)} / \beta }$$ $$x_{i+1} ={(\dot{M}_{\ce{C2H4}} )_{5}^{(i+1)} / \beta }$$
0 0.6200 – 1.4237
1 – 1.4237 45.6633
2 45.6633 98.7402
3 98.7402 151.6598
4 151.6598 204.5328
5 204.5328 257.3835
6 257.3835 ..
7 .. ..

case and we move on to the Wegstein’s method (see Appendix B5).

## Wegstein’s method

In this case we replace Equation \ref{b37} with

Definition: $x_{i+1} =\left(1-q\right)f(x_{i} ) + q x_{i} , \quad i = 1,2,3,...,\infty \tag{42}\label{b41}$

and choose a value for the parameter, $$q$$, that provides for a severely damped convergence. The results are shown in Table $$\PageIndex{2b}$$ where we see that the iterative procedure converges rapidly to the value given by

${(\dot{M}_{\ce{C2H4}} )_{5} / \beta } =0.6108 \tag{43}\label{b42}$

Use of this result in Equation \ref{b20} allows us to determine the fraction of Stream #4 that is purged and this fraction is given by

$\alpha =1 - \frac{(\dot{M}_{\ce{C2H4}} )_{5} \left({ CY}\right)}{\beta \left(1-{ C}\right)} =0.2874 \tag{44}\label{b43}$

which is identical to that obtained in Example $$7.4.3$$.

Table $$\PageIndex{2b}$$: Iterative Values for Dimensionless Flow Rate (Wegstein’s Method)
$$i$$ $$q$$ $$x_{i} ={(\dot{M}_{\ce{C2H4}} )_{5}^{(i)} / \beta }$$ $$x_{i+1} ={(\dot{M}_{\ce{C2H4}} )_{5}^{(i+1)} / \beta }$$
0 0.9950 0.7500 0.4070
1 0.9950 0.4070 0.5265
2 0.9950 0.5265 0.5938
3 0.9950 0.5938 0.6109
4 0.9950 0.6109 0.6108
5 0.9950 0.6108 0.6108