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# 7.4: Recycle Systems

In the previous section we studied systems in which there were incomplete chemical reactions, such as combustion reactions that yielded both carbon dioxide and carbon monoxide. We must always consider the consequences of a desirable, but incomplete, chemical reaction. Do we simply discard the unused reactants? And if we do, where do we discard them? Can we afford to release carbon monoxide to the atmosphere without recovering the energy available in the oxidation of $$\ce{CO}$$ to produce $$\ce{CO2}$$? What is the impact on the environment of carbon monoxide? Even if we can achieve complete combustion of the carbon monoxide ($$\ce{CO}$$), what do we do with the carbon dioxide ($$\ce{CO2}$$)?

Incomplete chemical reactions demand the use of recycle streams because we cannot afford to release the unused reactant into our ecological system, both for environmental and economic reasons. A typical reaction might involve combining two species, $$A$$ and $$B$$, to form a desirable product $$C$$. The product stream from the reactor will contain all three molecular species and we must separate the product C from this mixture and recycle the reactants as indicted in Figure $$\PageIndex{1}$$. In a problem of this type, one would want to know the composition of the product stream and the magnitude of the recycle stream. The molar flow rate of the recycle stream depends on the degree of completion of the reaction in the catalytic reactor and the degree of separation achieved by the separator. The actual design of these units is the subject of subsequent courses on reactor design and mass transfer, and we will introduce students to that process in Chapter 9. For the present we will concern ourselves only with the analysis of systems for which the operating characteristics are known or can be determined from the information that is given

The analysis of systems containing recycle streams is more complex than the systems we have studied previously, because recycle streams create loops in the flow of information. In a typical recycle configuration, a stream generated far downstream in the process is brought back to the front end of the process and mixed with an incoming feed. This is indicated in Figure $$\PageIndex{1}$$ where we have illustrated a unit called a mixer that combines the feed stream with the recycle stream.

In addition to mixers, systems with recycle streams often contain splitters that produce recycle streams and purge streams. In Figure $$\PageIndex{2}$$ we have illustrated a unit called an “ammonia converter” in which the feed consists of a mixture of nitrogen and hydrogen containing a small amount of argon. The ammonia produced in the reactor is removed as a liquid in a condenser, and the unconverted gas is recycled to the reactor through a mixer. In order to avoid the buildup of argon in the system, a purge stream is required and this gives rise to a splitter as illustrated in Figure $$\PageIndex{2}$$.

If the argon is not removed from the system by the splitter, the concentration of argon will increase and the efficiency of the reactor will decrease. The characteristics of splitters are discussed in detail in Sec. 7.3.1.

There are several methods for analyzing recycle systems. One approach, which we discuss first, is an extension of the method used previously. A set of control volumes is constructed and Axioms I and II are applied to those control volumes. Because more than one control volume is required, the most appropriate choice of control volumes is not always obvious. In previous chapters, we have discussed a set of rules or guidelines to be used in the construction of control volumes. Here we repeat those rules with the addition of one new rule that is important for the analysis of systems with recycle streams.

Rule I. Construct a primary cut where information is required.

Rule II. Construct a primary cut where information is given.

Rule III. Join these cuts with a surface located where $$\mathbf{v}_{A} \cdot \mathbf{n}$$ is known.

Rule IV. When joining the primary cuts to form control volumes, minimize the number of new or secondary cuts since these introduce information that is neither given nor required.

Rule V. Be sure that the surface specified by Rule III encloses regions in which volumetric information is either given or required.

Rule VI. When joining the primary cuts to form control volumes, minimize the number of redundant cuts since they provide no new information.

If one follows these rules carefully and if one pays careful attention to the associated degree-of-freedom analysis, the solution of many problems is either easy or impossible.

## Mixers and splitters

Mixers and splitters are a natural part of recycle systems, and one must pay careful attention to their properties. In Figure $$\PageIndex{3}$$ we have illustrated a mixer in which $$S$$ streams are joined to form a single stream, and we have illustrated a splitter in which a single stream is split into $$S$$ streams.

Both accumulation and chemical reaction can be neglected in mixers and splitters since these devices consist only of tubes joined in some convenient manner. This means that they can be analyzed in terms of the steady form of Equation $$(7.1.1)$$ that simplifies to

$\int_{\mathscr{A}}c_{A } \mathbf{v}_{A} \cdot \mathbf{n} dA = 0 , \quad A = 1, 2,...N \label{30}$

For the mixer shown in Figure $$\PageIndex{3}$$, this result can be expressed as

Mixer (species balances): $- \sum_{i = 1}^{i = S}(x_{A} )_{i} \dot{M}_{i} + (x_{A} )_{ o} \dot{M}_{ o} = 0 , \quad A = 1, 2,...N \label{31}$

This result also applies to the splitter shown in Figure $$\PageIndex{3}$$, and for that case the macroscopic balance takes the form

Splitter (species balances): $- (x_{A} )_{ o} \dot{M}_{ o} + \sum_{i = 1}^{i = S}(x_{A} )_{i} \dot{M}_{i} = 0 , \quad A = 1,2,...,N \label{32}$

The physics of a splitter require that the compositions in all the outgoing streams be equal to those in the incoming stream, and we express this idea as

Splitter (physics): $(x_{A} )_{i} = (x_{A} )_{ o} , \quad i = 1,2,...,S , \quad A = 1, 2,...,N \label{33}$

In addition to understanding the physics of a splitter, we must understand how this constraint on the mole fractions is influenced by the constraints that we apply in terms of our degree of freedom analysis. Table $$\PageIndex{1}$$ indicates that all streams cut by a control surface are required to satisfy the constraint on the mole fractions given by

$(x_{A} )_{i} + (x_{B} )_{i} + (x_{C} )_{i} + ..... + (x_{N} )_{i} = 1 , \quad i = 0, 1, 2,....,S \label{34}$

This means that only $$N-1$$ mole fractions can be specified in the outgoing streams of a splitter. To illustrate how these constraints influence our description of a splitter, we consider Stream #2 of the splitter illustrated in Figure $$\PageIndex{3}$$. For that stream, Equation \ref{33} provides the following $$N-1$$ equations:

$(x_{A} )_{2} = (x_{A} )_{ o} \label{35a}$

$(x_{B} )_{2} = (x_{B} )_{ o} \label{35b}$

$(x_{C} )_{2} = (x_{C} )_{ o} \label{35c}$

$\dots \dots \dots \nonumber$

$(x_{N-1} )_{2} = (x_{N-1} )_{ o} \label{35n}$

When we impose Equation \ref{34} for both Stream #2 and Stream #0 we obtain

$1 - (x_{N} )_{2} = 1 - (x_{N} )_{ o} \label{36}$

and this leads to the result that the mole fractions of the $$N^{th}$$ component must be equal.

$(x_{N} )_{2} = (x_{N} )_{ o} \label{37}$

This indicates that we should impose Equation \ref{33} on only $$N-1$$ of the components so that our degree of freedom representation of the splitter takes the form

Splitter (degree of freedom / mole fractions):

$(x_{A} )_{i} = (x_{A} )_{ o} , \quad i = 1,2,...,S , \quad A = 1, 2,...,N-1 \label{38}$

For many situations, a splitter may be enclosed in a control volume, as illustrated in Figure $$\PageIndex{3}$$. In those situations, we will impose the macroscopic species balance given by Equation \ref{32}, and we must again be careful to understand how this effects our degree of freedom analysis. In particular, we would like to prove that the splitter condition indicated by Equation \ref{38} need only be applied to $$S-1$$ streams when we make use of Equation \ref{32}. To see that this is true, we first note that Equation \ref{32} can be summed over all $$N$$ species to obtain the total molar balance given by

$- \dot{M}_{ o} + \left(\dot{M}_{1} + \dot{M}_{2} + \dot{M}_{3} + .... + \dot{M}_{S} \right) = 0 \label{39}$

Next, we rearrange Equation \ref{32} in the form

$- (x_{A} )_{ o} \dot{M}_{ o} + \sum_{i = 1}^{i = S-1}(x_{A} )_{i} \dot{M}_{i} + (x_{A} )_{S} \dot{M}_{S} = 0 , \quad A = 1,2,...,N \label{40}$

and then apply the constraints indicated by Equation \ref{38} for only $$S-1$$ of the streams leaving the splitter in order to obtain

$- (x_{A} )_{ o} \dot{M}_{ o} + (x_{A} )_{ o} \sum_{i = 1}^{i = S-1}\dot{M}_{i} + (x_{A} )_{S} \dot{M}_{S} = 0 , \quad A = 1,2,...,N \label{41}$

This result can be used with the total molar balance given by Equation \ref{39} to arrive at the condition

$- (x_{A} )_{ o} \dot{M}_{S} + (x_{A} )_{S} \dot{M}_{S} = 0 , \quad A = 1,2,...,N \label{42}$

which obviously this leads to

$(x_{A} )_{S} = (x_{A} )_{ o} , \quad A = 1,2,...,N \label{43}$

Here we see that we have derived, independently, one of the conditions implied by Equations \ref{38} and this means that the splitter constraints indicated by Equations \ref{38} can be expressed as

Splitter (degree of freedom / mole fractions / species balances):

$(x_{A} )_{i} = (x_{A} )_{ o} , \quad i = 1,2,...,S-1 , \quad A = 1, 2,...,N-1 \label{44}$

Often it is more convenient to work with species molar flow rates that are related to mole fractions by

$x_{A} = {\dot{M}_{A} / \dot{M}} = {\dot{M}_{A} / \sum_{B = 1}^{B = N}\dot{M}_{B} } \label{45}$

Use of this result in Equation \ref{44} provides the alternative representation for a splitter.

${(\dot{M}_{A} )_{i} / \sum_{B = 1}^{B = N}(\dot{M}_{B} )_{i} } = {(\dot{M}_{A} )_{ o} / \sum_{B = 1}^{B = N}(\dot{M}_{B} )_{ o} } , \quad i = 1,2,...,S-1 , \quad A = 1, 2,...,N-1 \label{46}$

To summarize, we note that the physical conditions associated with a splitter are given by Equations \ref{33}. When we take into account the constraint on mole fractions given by Equations \ref{34}, we must simplify Equations \ref{33} to the form given by Equations \ref{38} in order to be consistent with our degree of freedom analysis. When we further take into account the species mole balances that may be imposed on a control volume around the splitter, we must simplify Equations \ref{38} to the form given by Equations \ref{44} in order to be consistent with our degree of freedom analysis.

Example $$\PageIndex{1}$$: Splitter calculation

In this example we consider the splitter illustrated in Figure $$\PageIndex{4}$$ in which Stream #1 is split into Streams #2, #3 and #4, each of which contain the three species entering the splitter. We will use $$(\dot{M}_{A} )_{j}$$, $$(\dot{M}_{B} )_{j}$$ and $$(\dot{M}_{C} )_{j}$$ to represent the molar flow rates of species $$A$$, $$B$$ and $$C$$ in the $$j^{th}$$ stream and we will use $$\dot{M}_{1}$$ to represent the total molar flow rate entering Stream #1. The degree-of-freedom analysis given in Table $$\PageIndex{1}$$ indicates that we have five degrees of freedom, and there are several ways in which the splitter problem can be solved.

Table $$\PageIndex{1}$$: Degrees-of-freedom for a four-stream, three component splitter

Stream Variables $$N=12$$ $$M=4$$ 16 $$N=3$$ $$M=4$$ 0 4 11 5

### Example $$\PageIndex{1.1}$$: Specify five molar flow rates.

As an example of this case, we consider the following molar flow rates:

$(\dot{M}_{A} )_{1} = 10 \ { mol/hr} \tag{1a}\label{a1a}$

$(\dot{M}_{B} )_{1} = 25 \ { mol/hr} \tag{1b}\label{a1b}$

$(\dot{M}_{C} )_{1} = 65 \ { mol/hr} \tag{1c}\label{a1c}$

$(\dot{M}_{B} )_{3} = 4 \ { mol/hr} \tag{1d}\label{a1d}$

$(\dot{M}_{B} )_{4} = 2 \ { mol/hr} \tag{1e}\label{a1e}$

The development given in Sec. 4.6 can be used to construct the general relations given by

$(\dot{M}_{A} )_{i} = (x_{A} )_{i} \dot{M}_{i} , \quad A = 1,2,3 , \quad i = 1,2.3,4 \tag{2a}\label{a2a}$

$\dot{M}_{i} = (\dot{M}_{A} )_{i} + (\dot{M}_{B} )_{i} + (\dot{M}_{C} )_{i} , \quad i = 1,2.3,4 \tag{2b}\label{a2b}$

and from the data given in Eqs. \ref{a1a}, \ref{a1b} and \ref{a1c} we have

$(x_{A} )_{1} = 0.1 , \quad (x_{B} )_{1} = 0.25 \quad (x_{C} )_{1} = 0.65\tag{3}\label{a3}$

This indicates that the conditions for Stream #1 are completely specified, and on the basis of Equation \ref{33} all the mole fractions in the other streams are determined.

Equation \ref{a2a}-\ref{a2b} can be expressed in the form

$\dot{M}_{i} = \frac{(\dot{M}_{A} )_{i} }{(x_{A} )_{i} } , \quad A = 1,2,3 , \quad i = 1,2.3,4\tag{4}\label{a4}$

and this can be used for Stream #3 to obtain

$\dot{M}_{3} = \frac{(\dot{M}_{B} )_{3} }{(x_{B} )_{3} } = \frac{(\dot{M}_{B} )_{3} }{(x_{B} )_{1} } = \frac{4{ \ mol/h}}{ 0.25} = { 16 \ mol/h} \tag{5a}\label{a5a}$

Given the total molar flow rate in Stream #3, we can use Equation \ref{a2a} to obtain

$(\dot{M}_{A} )_{3} = (x_{A} )_{3} \dot{M}_{3} = (x_{A} )_{1} \dot{M}_{3} = (0.1)({ 16 \ mol/h}) = 1.6 \ mol/h \tag{5b}\label{a5aa}$

$(\dot{M}_{C} )_{3} = (x_{C} )_{3} \dot{M}_{3} = (x_{C} )_{1} \dot{M}_{3} = (0.65)( 16 \ mol/h ) = 10.4 \ mol/h \tag{5c}\label{a5b}$

indicating that all the molar flow rates in Stream #3 are determined.

Directing our attention to Stream #4, we repeat the analysis represented by Eqs. \ref{a5a}-\ref{a5b} to obtain

$\dot{M}_{4} = \frac{(\dot{M}_{B} )_{4} }{(x_{B} )_{4} } = \frac{(\dot{M}_{B} )_{4} }{(x_{B} )_{1} } = \frac{2{ \ mol/h}}{ 0.25} = { 8 \ mol/h} \tag{6a}\label{a6a}$

$(\dot{M}_{A} )_{4} = (x_{A} )_{4} \dot{M}_{4} = (x_{A} )_{1} \dot{M}_{4} = (0.1)({ 8 \ mol/h}) = 0.8{ \ mol/h} \tag{6b}\label{a6b}$

$(\dot{M}_{C} )_{4} = (x_{C} )_{4} \dot{M}_{4} = (x_{C} )_{1} \dot{M}_{4} = (0.65)({ 8 \ mol/h}) = 5.2{ \ mol/h} \tag{6c}\label{a6c}$

and we are ready to move on to determine all the molar flow rates in Stream #2. This requires the use of Equation \ref{32} in terms of the species molar flow rates given by

$(\dot{M}_{A} )_{1} = (\dot{M}_{A} )_{2} + (\dot{M}_{A} )_{3} + (\dot{M}_{A} )_{4} \tag{7a}\label{a7a}$

$(\dot{M}_{B} )_{1} = (\dot{M}_{B} )_{2} + (\dot{M}_{B} )_{3} + (\dot{M}_{B} )_{4} \tag{7b}\label{a7b}$

$(\dot{M}_{C} )_{1} = (\dot{M}_{C} )_{2} + (\dot{M}_{C} )_{3} + (\dot{M}_{C} )_{4} \tag{7c}\label{a7c}$

and these results can be used to determine the species molar flow rates given by

$(\dot{M}_{A} )_{2} = { 7.6 \ mol/h} , \quad (\dot{M}_{B} )_{2} = { 19 \ mol/h} , \quad (\dot{M}_{C} )_{2} = { 49.4 \ mol/h}\label{a8}$

Finally we see that the total molar flow rate in Stream #2 is given by

$\dot{M}_{3} = { 76 \ mol/h} \label{a9}$

Note that if all the species molar flow rates are specified for a single stream, as they are by Eqs. \ref{a1a}, \ref{a1b} and \ref{a1c}, then the additional specifications must be in the other streams. The above example illustrates why this is the case. Once all the species molar flow rates for a given stream are specified, the mole fractions for the species in that stream are determined, and from the splitter physics the mole fractions for all the other streams are known. If all the mole fractions are known, then a single molar flow rate for a given species (or the overall molar flow rate) determines the other molar flow rates for that stream.

The remaining four ways in which this splitter problem can be solved are left as exercises for the students. These exercises are essential to gain a comprehensive understanding of the behavior of splitters.

## Recycle and purge streams

In this section we analyze systems with recycle and purge streams. We begin with a system analogous to the one shown in Figure $$\PageIndex{1}$$ in which there is a mixer. We then move on to a more complicated system analogous to the one shown in Figure $$\PageIndex{2}$$ in which there is a splitter.

Example $$\PageIndex{2}$$: Pyrolysis of dichloroethane with recycle

To illustrate a simple recycle system, we consider the pyrolysis of dichloroethane ($$\ce{C2H4} \ce{Cl2}$$) to produce vinyl chloride ($$\ce{C2H3} \ce{Cl}$$) and hydrochloric acid ($$\ce{HCl}$$). The pyrolysis reaction is not complete, and experimental measurements indicate that the conversion for a particular reactor is given by

${C } = \text{ Conversion of } \ce{C2H4Cl2} = \frac{- \mathscr{R}_{ \ce{C2H4} \ce{Cl2} } }{ (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{2} } = 0.30 \tag{1}\label{b1}$

The unreacted dichloroethane is separated from the reaction products and recycled back to the reactor for the production of more vinyl chloride as indicated in Figure $$\PageIndex{5a}$$. The composition of the feed Stream #1 is 98% dichloroethane ($$\ce{C2H4} \ce{Cl2}$$) and 2% ethane ($$\ce{C2H6}$$) on a molar basis. We assume that the separation column produces a sharp separation meaning that all the dichloroethane leaves in the bottom Stream #5, and the remaining components (vinyl chloride, hydrochloric acid and ethane) leave through the distillate Stream #4. Our objective in this example is to determine the recycle flow rate in Stream #5.

Our analysis of this process is based on Axioms I and II as given by Equations $$(7.1.4)$$ and $$(7.1.5)$$, and we begin with the steady form of the macroscopic mole balance to obtain

Axiom I: $\int_{\mathscr{A}}c_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = \mathscr{R}_{ A} , \quad A \Rightarrow \ce{C2H6} , \quad \ce{HCl}, \ce{C2H3} \ce{Cl}, \ce{C2H4} \ce{Cl2} \tag{2}\label{b2}$

The global rates of production within any control volume are constrained by Equation $$(7.1.4)$$ that takes the form

Axiom II $\sum_{A = 1}^{A = N}N_{JA} \mathscr{R}_{ A} = 0 , \quad J \Rightarrow { C}, { H}, \ce{Cl} \tag{3}\label{b3}$

For the case under consideration, the chemical composition matrix is given by

$\text{ Molecular Species }\to \ce{C2H6} \quad \ce{HCl} \quad \ce{C2H3} \ce{Cl} \quad \ce{C2H4} \ce{Cl2} \\ \begin{matrix} {carbon } \\ {hydrogen} \\ {chlorine } \end{matrix} \begin{bmatrix} { 2} & { 0} & {2} & {2} \\ { 6} & { 1} & {3} & {4} \\ { 0} & { 1} & {1} & {2} \end{bmatrix}\tag{4}\label{b4}$

and Equation \ref{b3} takes the form

$\begin{bmatrix} {2} & {0} & {2} & {2} \\ {6} & {1} & {3} & {4} \\ {0} & {1} & {1} & {2} \end{bmatrix} \begin{bmatrix} \mathscr{R}_{ \ce{C2H6} } \\ {\mathscr{R}_{ \ce{HCl}} } \\ {\mathscr{R}_{ \ce{C2H3} \ce{Cl}} } \\ {\mathscr{R}_{ \ce{C2H4} \ce{Cl2} } } \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix}\tag{5}\label{b5}$

Representing $$\left[N_{JA} \right]$$ in row reduced echelon form leads to

$\begin{bmatrix} {1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {1} \\ {0} & {0} & {1} & {1} \end{bmatrix} \begin{bmatrix} {\mathscr{R}_{ \ce{C2H6} } } \\ {\mathscr{R}_{ \ce{HCl}} } \\ {\mathscr{R}_{ \ce{C2H3} \ce{Cl}} } \\ {\mathscr{R}_{ \ce{C2H4} \ce{Cl2} } } \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix}\tag{6}\label{b6}$

and application of the global pivot theorem (see Sec. 6.4) provides the following constraints on the global net rates of production:

Axiom II: $\mathscr{R}_{ \ce{C2H6} } = 0 , \quad \mathscr{R}_{ \ce{HCl}} = - \mathscr{R}_{ \ce{C2H4} \ce{Cl2} } , \quad \mathscr{R}_{ \ce{C2H3} \ce{Cl}} = - \mathscr{R}_{ \ce{C2H4} \ce{Cl2} } \tag{7}\label{b7}$

Here we see that ethane acts as an inert, a conclusion that might have been extracted by intuition but has been made rigorous on the basis of Axiom II. The constraints given by Eqs. \ref{b7} apply to each control volume that we construct for the system illustrated in Figure $$\PageIndex{5a}$$, and we are now ready to construct those control volumes making use of the rules listed above. On the basis of those rules we make the following five primary cuts:

I. A cut of Stream #1 is made because the composition of Stream #1 is given. NOTE: “The composition of the feed Stream #1 is 98% dichloroethane and 2% ethane on a molar basis.”

II. A cut of Stream #5 is made because information about the composition is given for that stream. NOTE: “We assume that the separation column produces a sharp separation meaning that all the dichloroethane leaves in the bottom Stream #5, and the remaining components (ethane, hydrochloric acid and vinyl chloride) leave through the distillate Stream #4.”

III. A cut of Stream #4 is made because information about the composition is given for that stream. NOTE: “We assume that the separation column produces a sharp separation meaning that all the dichloroethane leaves in the bottom Stream #5, and the remaining components (ethane, hydrochloric acid and vinyl chloride) leave through the distillate Stream #4.”

IV. A cut of Stream #2 is required because information about the net global rate of production is given in terms of conditions in Stream #2. NOTE: The statement about the conversion requires that $$\mathscr{R}_{ \ce{C2H4} \ce{Cl2} } = - {C } (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{2} , \quad { C } = 0.30$$

V. At least one control volume must enclose the reactor since information about the net global rate of production is given. NOTE: $$\mathscr{R}_{ \ce{C2H4} \ce{Cl2} } = - {C } (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{2}$$.

The cuts based on Rules I through VI are indicated in Figure $$\PageIndex{5b}$$. Two control volumes can be

created that satisfy these rules, and these are illustrated in Figure $$\PageIndex{4}$$c where we note that there is a single redundant cut of Stream #1. Given this choice of control volumes, our next step is to

perform a degree-of-freedom analysis as indicated in Table $$\PageIndex{2}$$. There we see that we have a solvable problem.

Table $$\PageIndex{2}$$: Degrees-of-freedom for production of vinyl chloride from dichloroethane

Stream & Net Rate of Production Variables $$N \times M = 4 \times 3 = 16$$ $$M=4$$ $$N=4$$ 24 $$4 \times 2 = 8$$ $$M=4$$ $$T=3$$ 15 $$3+1+3=7$$ 0 0 1 8 1

Since there is one degree of freedom3, we will need to work in terms of dimensionless molar flow rates. For the control volumes illustrated in Figure $$\PageIndex{5c}$$ we can express Equation \ref{b2} as

Axiom I: $\sum_{ exits} (\dot{M}_{A} )_{ exit} = \sum_{ entrances}(\dot{M}_{A} )_{ entrance} + \mathscr{R}_{ A} , \quad A = 1,2, 3, 4 \tag{8}\label{b8}$

We can utilize Eqs. \ref{a2a}-\ref{a2b} of Example $$\PageIndex{1}$$ to express the molar flow rates according to

At Entrances & Exits: $\dot{M}_{A} = x_{A} \dot{M} , \quad A = 1,2,3,4 \tag{9a}\label{b9a}$

At Entrances & Exits: $\dot{M} = \dot{M}_{A} + \dot{M}_{B} + \dot{M}_{C} + \dot{M}_{D} \tag{9b}\label{b9b}$

and in many cases it will be convenient to use Equation \ref{b9a} in the form

At Entrances & Exits: $x_{A} = \frac{\dot{M}_{A} }{\dot{M}} , \quad A = 1,2,3,4 \tag{9c}\label{b9c}$

Use of Equation \ref{b8} allows us to express the species mole balances for Control Volume 1 in the form

#### Control Volume 1

\begin{align} \ce{C2H6} & : & - (\dot{M}_{\ce{C2H6} } )_{1} + (\dot{M}_{\ce{C2H6} } )_{4} = \mathscr{R}_{ \ce{C2H6} } \tag{10a}\label{b10a} \end{align}

\begin{align} \ce{C2H4} \ce{Cl2} & : & - (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} + (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{4} = \mathscr{R}_{ \ce{C2H4} \ce{Cl2} } \tag{10b}\label{b10b} \end{align}

\begin{align}\ce{HCl} & : & - (\dot{M}_{\ce{HCl}} )_{1} + (\dot{M}_{\ce{HCl}} )_{4} = \mathscr{R}_{ \ce{HCl}} \tag{10c}\label{b10c} \end{align}

\begin{align} \ce{C2H3} \ce{Cl} & : & - (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{1} + (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{4} = \mathscr{R}_{ \ce{C2H3} \ce{Cl}} \tag{10d}\label{b10d} \end{align}

Since there are no chemical reactions taking place in the mixer illustrated in Figure $$\PageIndex{5c}$$, the mole balances Control Volume 2 take the simple forms given by

#### Control Volume 2

\begin{align} \ce{C2H6} & : & - (\dot{M}_{\ce{C2H6} } )_{1} - (\dot{M}_{\ce{C2H6} } )_{5} + (\dot{M}_{\ce{C2H6} } )_{2} = 0 \tag{11a}\label{b11a} \end{align}

\begin{align} \ce{C2H4} \ce{Cl2} & : & - (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} - (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{5} + (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{2} = 0 \tag{11b}\label{b11b} \end{align}

\begin{align} \ce{HCl} & : & - (\dot{M}_{\ce{HCl}} )_{1} - (\dot{M}_{\ce{HCl}} )_{5} + (\dot{M}_{\ce{HCl}} )_{2} = 0 \tag{11c}\label{b11c} \end{align}

\begin{align} \ce{C2H3} \ce{Cl} & : & - (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{1} - (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{5} + (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{2} = 0 \tag{11d}\label{b11d} \end{align}

We begin our analysis with Control Volume 1 and simplify the mole balances in terms of conditions on the molar flow rates. Some of those molar flow rates are zero and when we make use of this information the mole balances for Control Volume 1 take the form

#### Control Volume 1 (constraints on molar flow rates)

\begin{align} \ce{C2H6} & : & - (\dot{M}_{\ce{C2H6} } )_{1} + (\dot{M}_{\ce{C2H6} } )_{4} = \mathscr{R}_{ \ce{C2H6} } \tag{12a}\label{b12a} \end{align}

\begin{align} \ce{C2H4} \ce{Cl2} & : & - (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} = \mathscr{R}_{ \ce{C2H4} \ce{Cl2} } \tag{12b}\label{b12b} \end{align}

\begin{align} \ce{HCl} & : & (\dot{M}_{\ce{HCl}} )_{4} = \mathscr{R}_{ \ce{HCl}} \tag{12c}\label{b12c} \end{align}

\begin{align} \ce{C2H3} \ce{Cl} & : & (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{4} = \mathscr{R}_{ \ce{C2H3} \ce{Cl}} \tag{12d}\label{b12d} \end{align}

We now make use of the results from Axiom II given by Eqs. \ref{b7} and impose those constraints on global rates of production to obtain

#### Control Volume 1 (constraints on global rates of production)

\begin{align} \ce{C2H6} & : & - (\dot{M}_{\ce{C2H6} } )_{1} + (\dot{M}_{\ce{C2H6} } )_{4} = 0 \tag{13a}\label{b13a} \end{align}

\begin{align} \ce{C2H4} \ce{Cl2} & : & - (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} = \mathscr{R}_{ \ce{C2H4} \ce{Cl2} } \tag{13b}\label{b13b} \end{align}

\begin{align} \ce{HCl} & : & (\dot{M}_{\ce{HCl}} )_{4} = - \mathscr{R}_{ \ce{C2H4} \ce{Cl2} } \tag{13c}\label{b13c} \end{align}

\begin{align} \ce{C2H3} \ce{Cl} & : & (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{4} = - \mathscr{R}_{ \ce{C2H4} \ce{Cl2} } \tag{13d}\label{b13d} \end{align}

At this point we can use Equation \ref{b1} to express the global rate of production in terms of the conversion leading to

#### Control Volume 1 (global rates of production in terms of the conversion)

\begin{align} \ce{C2H6} & : & - (\dot{M}_{\ce{C2H6} } )_{1} + (\dot{M}_{\ce{C2H6} } )_{4} = 0 \tag{14a}\label{b14a} \end{align}

\begin{align} \ce{C2H4} \ce{Cl2} & : & (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} = { C} (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{2} \tag{14b}\label{b14b} \end{align}

\begin{align} \ce{HCl} & : & (\dot{M}_{\ce{HCl}} )_{4} = { C} (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{2} \tag{14c}\label{b14c} \end{align}

\begin{align} \ce{C2H3} \ce{Cl} & : & (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{4} = { C} (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{2} \tag{14d}\label{b14d} \end{align}

Moving on to Control Volume 2, we note that all global rates of production are zero and that only ethane and dichloroethane are present in this control volume. This leads to the two non-trivial species mole balances given by

#### Control Volume 2 (constraints on molar flow rates)

\begin{align} \ce{C2H6} & : & - (\dot{M}_{\ce{C2H6} } )_{1} + (\dot{M}_{\ce{C2H6} } )_{2} = 0 \tag{15a}\label{b15a} \end{align}

\begin{align} \ce{C2H4} \ce{Cl2} & : & - (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} - (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{5} + (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{2} = 0 \tag{15b}\label{b15b} \end{align}

At this point we use Equation \ref{b14b} to obtain

$(\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{5} = \frac{1-C}{C} (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} \tag{16}\label{b16}$

Since the molar flow rate of dichloroethane entering the system is not specified, it is convenient to work in terms of dimensionless molar flow rates, and this leads to

\begin{align} \ce{C2H4} \ce{Cl2} & : & (\mathscr{M}_{ \ce{C2H4} \ce{Cl2} } )_{5} = \frac{(\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{5} }{(\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} } = \frac{1-C}{C} \tag{17a}\label{b17a} \end{align}

\begin{align} \ce{C2H3} \ce{Cl} & : & (\mathscr{M}_{ \ce{C2H3} \ce{Cl}} )_{4} = \frac{(\dot{M}_{\ce{C2H3} \ce{Cl}} )_{4} }{(\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} } = 1.0 \tag{17b}\label{b17b} \end{align}

\begin{align} \ce{C2H6} & : & (\mathscr{M}_{ \ce{C2H6} } )_{4} = \frac{(\dot{M}_{\ce{C2H6} } )_{4} }{(\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} } = \frac{0.02}{0.98} \tag{17c}\label{b17c} \end{align}

\begin{align} \ce{HCl} & : & (\mathscr{M}_{ \ce{HCl}} )_{4} = \frac{(\dot{M}_{\ce{HCl}} )_{4} }{(\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{1} } = 1.00 \tag{17d}\label{b17d} \end{align}

In addition, the molar flow rates of the four species can be expressed in terms of the total molar flow rate entering the system, and this leads to

\begin{align} \ce{C2H4} \ce{Cl2} & : & (\dot{M}_{\ce{C2H4} \ce{Cl2} } )_{5} = 2.2867\dot{M}_{1} \tag{18a}\label{b18a} \end{align}

\begin{align} \ce{C2H3} \ce{Cl} & : & (\dot{M}_{\ce{C2H3} \ce{Cl}} )_{4} = 0.98\dot{M}_{1} \tag{18b}\label{b18b} \end{align}

\begin{align} \ce{C2H6} & : & (\dot{M}_{\ce{C2H6} } )_{4} = 0.02\dot{M}_{1} \tag{18c}\label{b18c} \end{align}

\begin{align} \ce{HCl} & : & (\dot{M}_{\ce{HCl}} )_{4} = 0.98\dot{M}_{1} \tag{18d}\label{b18d} \end{align}

These results can be used to determine the total molar flow rates in Streams #4 and #5

$\dot{M}_{5} = 2.2867\dot{M}_{1} , \quad \dot{M}_{4} = 1.98\dot{M}_{1} \tag{19}\label{b19}$

along with the composition in Stream #4 that is given by

$(x_{\ce{C2H3} \ce{Cl}} )_{4} = 0.495 , \quad (x_{\ce{C2H6} } )_{4} = 0.01 , \quad (x_{\ce{HCl}} )_{4} = 0.495 \tag{20}\label{b20}$

A direct solution of the recycle problem given in Example $$\PageIndex{2}$$ was possible because of the simplicity of the process, both in terms of the chemical reaction and in terms of the structure of the process. When purge streams are required, as illustrated in Figure $$\PageIndex{2}$$, the analysis becomes more complex.

Example $$\PageIndex{3}$$: Production of ethylene oxide with recycle and purge

In Figure $$\PageIndex{6a}$$ 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. 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.

We begin this example by constructing the equations containing information that is either given or required. To be specific, we want to determine the parameter $$\alpha$$ defined by

$\dot{M}_6 = \alpha \dot{M}_4 \tag{1}\label{c1}$

when 100 mol/h of ethylene oxide are produced by the reactor. We represent this information as

$\mathscr{R}_{\ce{C2H4O}} = \beta, \quad \beta = 100 \ mol/h \tag{2}\label{c2}$

We are also given information about the conversion and yield (see Eqs. $$(7.2.3)$$ and $$(7.2.8)$$) that we express as

${C } = \text{ Conversion of ethylene } = \frac{-\mathscr{R}_{ \ce{C2H4}} }{ (\dot{M}_{ \ce{C2H4} })_2} = 0.70 \tag{3}\label{c3}$

${ Y} = \text{ Yield of }\left(\ce{C2H4O} / \ce{C2H4} \right) = \frac{\mathscr{R}_{ \ce{C2H4O}} }{- \mathscr{R}_{ \ce{C2H4} } } = 0.50\tag{4}\label{c4}$

Other information concerning this process becomes clear when we consider the problem of constructing control volumes for the application of Axiom I. On the basis of the rules for constructing control volumes given at the beginning of this section, we make the following primary cuts of the streams indicated in Figure $$\PageIndex{6a}$$:

I. A cut of Stream #1 is made because information concerning the composition of Stream #1 is given. NOTE: “The feed stream, Stream #1, consists of ethylene ($$\ce{C2H4}$$) and air ($$\ce{N2}$$ and $$\ce{O2}$$).”

II. A cut of Stream #2 is made because a constraint on the composition is given. NOTE: “the mole fraction of ethylene in Stream #2 leaving the mixer must be maintained at $$0.05$$ for satisfactory catalyst operation.”

III. One might make a cut of Stream #3 because a constraint on the composition is given. NOTE: “All of the oxygen in the feed reacts, thus there is no oxygen in Stream #3.” However, there are other interpretations of the original statement. For example one could say that since all of the oxygen in the feed reacts, there is no oxygen in Stream #4 and in Stream #7. In addition, one could say that the molar flow rate of oxygen in Stream #2 is equal to the molar rate of consumption of oxygen in the reactor, i.e., $$(\dot{M}_{ \ce{O2} } )_{2} = - \mathscr{R}_{ \ce{O2} }$$. If we choose the second of these three possibilities, there is no need to cut Stream #3, thus we do not make a cut of Stream #3.

IV. The reactor is enclosed in a control volume because volumetric information about the reactor is given. NOTE: “The conversion of ethylene in the reactor is optimized to be 70% ($${C } = 0.70$$) and the yield of ethylene oxide is 50% ($${ Y} = 0.50$$).”

V. Cuts of Stream #4 and Stream #6 are made because the operating characteristics of the splitter are required. NOTE: “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.”

VI. A cut of Stream #7 is made because information is required, NOTE: “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.”

VII. A cut of Stream #8 is made because information is given. NOTE: “Water is fed to the absorber such that $$(\dot{M}_{\ce{H2O}} )_{8} = 100 (\dot{M}_{\ce{C2H4O}} )_{7}$$.”

The information identified in II and IV can be expressed as

$(x_{\ce{C2H4}})_2 = \eta, \quad \eta = 0.05 \tag{5}\label{c5}$

$(\dot{M}_{\ce{H2O}} )_{8} = \Psi (\dot{M}_{\ce{C2H4O}} )_{7} , \quad \Psi = 100\tag{6}\label{c6}$

The cuts and enclosures based on observations I through VII are illustrated in Figure $$\PageIndex{6b}$$ where we see one redundant cut of Stream #2 and an unnecessary cut of Stream #3. We can eliminate

the latter with the control volumes illustrated in Figure $$\PageIndex{6c}$$. There we see that we have avoided a cut of Stream #3 by enclosing the reactor and the absorber in Control Volume 1; however, we have created redundant cuts of Stream #2, Stream #4 and Stream #5.

We begin the analysis of this process with Equation $$(7.1.4)$$ that takes the form

Axiom II $\sum_{A = 1}^{A = N}N_{JA} \mathscr{R}_{ A} = 0 , \quad J \Rightarrow { C} , \quad { H} , \quad { O} \tag{7}\label{c7}$

and for the case under consideration, the atomic matrix is given by

$\text{ Molecular Species }\to \ce{C2H4} \quad \ce{O2} \quad \ce{CO2} \quad \ce{H2O} \quad \ce{C2H4O} \quad \ce{N2} \\ \begin{matrix} {carbon } \\ {hydrogen} \\ {oxygen } \\ nitrogen \end{matrix} \begin{bmatrix} { 2 } & {0} & {1 } & {0} & {2 } & {0 } \\ { 4 } & {0} & {0 } & {2} & {4 } & {0 } \\ { 0 } & {2} & {2 } & {1} & {1 } & {0 } \\ { 0 } & {0} & {0 } & {0} & {0 } & {2 } \end{bmatrix} \tag{8}\label{c8}$

Given this form of the atomic matrix, we can express Equation \ref{c7} as

$\begin{bmatrix} { 2 } & {0} & {1 } & {0} & {2 } & {0 } \\ { 4 } & {0} & {0 } & {2} & {4 } & {0 } \\ { 0 } & {2} & {2 } & {1} & {1 } & {0 } \\ { 0 } & {0} & {0 } & {0} & {0 } & {2 } \end{bmatrix} \begin{bmatrix} {\mathscr{R}_{ \ce{C2H4} } } \\ {\mathscr{R}_{ \ce{O2} } } \\ {\mathscr{R}_{ \ce{CO2} } } \\ {\mathscr{R}_{ \ce{H2O}} } \\ {\mathscr{R}_{ \ce{C2H4O}} } \\ \mathscr{R}_{\ce{N2}} \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \\ 0 \end{bmatrix}\tag{9}\label{c9}$

In row reduced echelon form Equation \ref{c9} takes the form

$\begin{bmatrix} {1} & {0} & {0} & {\frac{1}{2} } & {1} & {0} \\ {0} & {1} & {0} & {\frac{3}{2} } & {\frac{1}{2} } & {0} \\ {0} & {0} & {1} & {-1} & {0} & {0} \\ 0 & {0} & {0} & {0} & {0} & {1} \end{bmatrix}\begin{bmatrix} {\mathscr{R}_{ \ce{C2H4} } } \\ {\mathscr{R}_{ \ce{O2} } } \\ {\mathscr{R}_{ \ce{CO2} } } \\ {\mathscr{R}_{ \ce{H2O}} } \\ {\mathscr{R}_{ \ce{C2H4O}} } \\ \mathscr{R}_{\ce{N2}} \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \\ 0 \end{bmatrix}\tag{10}\label{c10}$

and the stoichiometric constraints on the global rates of production are given by application of the global pivot theorem (see Sec. 6.4)

$\mathscr{R}_{ \ce{C2H4} } = - \frac{1}{2} \mathscr{R}_{ \ce{H2O}} - \mathscr{R}_{ \ce{C2H4O}} \tag{11a}\label{c11a}$

$\mathscr{R}_{ \ce{O2} } = -\frac{3}{2} \mathscr{R}_{ \ce{H2O}} - \frac{1}{2} \mathscr{R}_{ \ce{C2H4O}} \tag{11b}\label{c11b}$

$\mathscr{R}_{ \ce{CO2} } = \mathscr{R}_{ \ce{H2O}} \tag{11c}\label{c11c}$

$\mathscr{R}_{ \ce{N2} } = 0 \tag{11d}\label{c11d}$

### Degree of Freedom Analysis

To gain some insight into a strategy for solving this problem, we need a degree-of-freedom analysis for the control volumes illustrated in Figure $$\PageIndex{6c}$$. In this case we will use molar flow rates and global rates of production as our independent variables. Of the eight streams illustrated in Figure $$\PageIndex{6c}$$, only seven are cut by the surface of a control volume. Keeping in mind that we are dealing with six molecular species, we see that we have 42 stream variables in terms of species molar flow rates. In addition, we have six global rates of production so that the total number of generic variables is 48. The three control volumes illustrated in Figure $$\PageIndex{6c}$$, coupled with the six molecular species, gives rise to 18 generic constraints associated with Axiom I. In addition Axiom II provides 4 stoichiometric conditions (see Eqs. \ref{c11a}-\ref{c11d}), and the physics of the splitter provides $$A-1 = 5$$ constraints on the composition (see Equation \ref{44}) that we list here as

Splitter condition: $(x_{A} )_{6} = (x_{A} )_{ 4} , \quad A = 1, 2,...,5 \tag{12}\label{c12}$

This provides us with 27 generic constraints and we can move on to the matter of particular constraints.

Control Volume 1

We begin our exploration of the particular constraints with Control Volume 1 and direct our attention to Stream #1. The problem statement indicates that only ethylene ($$\ce{C2H4}$$), oxygen ($$\ce{O2}$$) and nitrogen ($$\ce{N2}$$) are present in that stream, thus we have

Stream #1: $(\dot{M}_{\ce{C2H4O}} )_{1} = (\dot{M}_{\ce{H2O}} )_{1} = (\dot{M}_{ \ce{CO2} } )_{1} = 0 \tag{13}\label{c13}$

In addition, we are given that the oxygen and nitrogen in Stream #1 are supplied by air, thus we have a constraint on the composition given by

Steam #1: $(\dot{M}_{\ce{O2} } )_{1} = \varphi (\dot{M}_{ \ce{N2} } )_{1} , \quad \varphi = \left({21 / 79} \right) \tag{14}\label{c14}$

Moving on to the other stream entering Control Volume 1, we draw upon information about the splitter to conclude that Stream #5 contains no ethylene oxide ($$\ce{C2H4O}$$), water ($$\ce{H2O}$$) or oxygen ($$\ce{O2}$$), thus we have

Stream #5: $(\dot{M}_{\ce{C2H4O}} )_{5} = (\dot{M}_{\ce{H2O}} )_{5} = (\dot{M}_{ \ce{O2} } )_{5} = 0 \tag{15}\label{c15}$

It is obvious that there is no ethylene oxide ($$\ce{C2H4O}$$) or water ($$\ce{H2O}$$) in Stream #2. However, we must be careful not to impose such a condition as a particular constraint, since this condition is “obvious” on the basis of the species mole balances and the conditions given by Eqs. \ref{c13} and \ref{c15}. At this point we conclude that there are 7 particular constraints associated with Control Volume 1 and we list this result as

#### Control Volume 1:

$\text{ particular constraints } = { 7} \tag{16}\label{c16}$

Moving on to Control Volume 2, we note that the problem statement contained the comment “the mole fraction of ethylene in Stream #2 entering the reactor must be maintained at $$0.05$$ for satisfactory catalyst operation.” This provided the basis for Equation \ref{c5} and we list this particular constraint as

Stream #2: $(x_{\ce{C2H4} } )_{2} = \eta , \quad \eta = 0.05 \tag{17}\label{c17}$

There are no particular constraints imposed on Stream #4, thus we move on to Stream #7 and Stream #8 for which the following particular constraints are imposed:

Stream #7: $(\dot{M}_{\ce{C2H4} } )_{7} = (\dot{M}_{ \ce{O2} } )_{7} = (\dot{M}_{ \ce{N2} } )_{7} = (\dot{M}_{ \ce{CO2} } )_{7} = 0 \tag{18}\label{c18}$

Stream #8: $(\dot{M}_{\ce{C2H4O}} )_{8} = (\dot{M}_{\ce{C2H4} } )_{8} = (\dot{M}_{ \ce{O2} } )_{8} = (\dot{M}_{ \ce{N2} } )_{8} = (\dot{M}_{ \ce{CO2} } )_{8} = 0 \tag{19}\label{c19}$

In addition, we have a condition involving both Stream #7 and Stream #8 that was given earlier by Equation \ref{c6} and repeated here as

Streams #7 and #8: $(\dot{M}_{\ce{H2O}} )_{8} = \Psi (\dot{M}_{\ce{C2H4O}} )_{7} , \quad \Psi = 100 \tag{20}\label{c20}$

The requirement that 100 mol/hr of ethylene oxide are produced by the reactor can be stated as

Reactor: $\mathscr{R}_{ \ce{C2H4O}} = \beta = { 100 \ mol/hr} \tag{21}\label{c21}$

while the information concerning the conversion and yield are repeated here as

Reactor: ${ C } = \text{ Conversion of ethylene } = \frac{- \mathscr{R}_{ \ce{C2H4} } }{(\dot{M}_{\ce{C2H4} } )_{ 2} } = 0.70 \tag{22}\label{c22}$

Reactor: ${ Y} = \text{ Yield of }\left(\ce{C2H4O} / \ce{C2H4} \right) = \frac{\mathscr{R}_{ \ce{C2H4O}} }{- \mathscr{R}_{ \ce{C2H4} } } = 0.50 \tag{23}\label{c23}$

Here we conclude that there are 14 particular constraints associated with Control Volume 2 and we list this result as

#### Control Volume 2:

$\text{ particular constraints} = { 14} \tag{24}\label{c24}$

Our third control volume encloses the splitter as indicated in Figure $$\PageIndex{6c}$$. The problem statement indicates that “All of the oxygen in the feed reacts, .all the ethylene oxide is absorbed,..all the water leaves the absorber” thus it is clear Stream #4 is constrained by

Stream #4: $(\dot{M}_{\ce{C2H4O}} )_{4} = (\dot{M}_{\ce{H2O}} )_{4} = (\dot{M}_{ \ce{O2} } )_{4} = 0 \tag{25}\label{c25}$

However, the splitter condition given by Equation \ref{c12} along with the particular constraint given by Equation \ref{c15} makes these three conditions redundant. Thus there are no new constraints associated with Control Volume 3.

#### Control Volume 3:

$\text{ particular constraints} = { 0} \tag{26}\label{c26}$

The total number of particular constraints is $$7+14 = 21$$, thus the total number of constraints is given by

$\text{ Total constraints: } 18 + 27 = 48 \tag{27}\label{c27}$

This means that we have zero degrees of freedom and the problem has a solution.

### Solution Procedure

We can use Eqs. \ref{c3}, \ref{c4} and \ref{c11a}-\ref{c11d} to express all the non-zero global rates of production in terms of the conversion ($${ C}$$), the yield ($${ Y}$$), and the molar flow rate of ethylene into the reactor, $$(\dot{M}_{\ce{C2H4} } )_{2}$$. These global rates of production are given by

$\mathscr{R}_{ \ce{C2H4} } = - { C} (\dot{M}_{\ce{C2H4} } )_{2} \tag{28a}\label{c28a}$

$\mathscr{R}_{ \ce{C2H4O}} = { YC} (\dot{M}_{\ce{C2H4} } )_{2} \tag{28b}\label{c28b}$

$\mathscr{R}_{ \ce{H2O}} = 2{ C} (1-{ Y}) (\dot{M}_{\ce{C2H4} } )_{2} \tag{28c}\label{c28c}$

$\mathscr{R}_{ \ce{CO2} } = 2{ C} (1-{ Y}) (\dot{M}_{\ce{C2H4} } )_{2} \tag{28d}\label{c28d}$

$\mathscr{R}_{ \ce{O2} } = { C} ({\frac{5}{2}} { Y}-3) (\dot{M}_{\ce{C2H4} } )_{2} \tag{28e}\label{c28e}$

$\mathscr{R}_{ \ce{N2} } = 0 \tag{28f}\label{c28f}$

And this completes our application of Axiom II. Directing our attention to Axiom I, we begin our analysis of the control volumes illustrated in Figure $$\PageIndex{6c}$$ with the steady form of Equation $$(7.1.1)$$ given by

Axiom I: $\int_{\mathscr{A}}c_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = \mathscr{R}_{ A} , \quad A = 1,2, ..., 6 \tag{29}\label{c29}$

We will apply this result to Control Volumes 1, 2 and 3, and then begin the algebraic process of determining 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.

#### Control Volume 1

Since there are no chemical reactions in the mixer illustrated in Figure $$\PageIndex{6c}$$, we can make use of Eqs. \ref{c29} along with Eqs. \ref{c13} and \ref{c15} to obtain the following mole balance equations:

\begin{align} \ce{C2H4} & : & - (\dot{M}_{\ce{C2H4} } )_{1} - (\dot{M}_{\ce{C2H4} } )_{5} + (\dot{M}_{\ce{C2H4} } )_{2} = 0 \tag{30a}\label{c30a} \end{align}

\begin{align} \ce{CO2} & : & - (\dot{M}_{ \ce{CO2} } )_{5} + (\dot{M}_{ \ce{CO2} } )_{2} = 0 \tag{30b}\label{c30b} \end{align}

\begin{align} \ce{O2} & : & - (\dot{M}_{ \ce{O2} } )_{1} + (\dot{M}_{ \ce{O2} } )_{2} = 0 \tag{30c}\label{c30c} \end{align}

\begin{align} \ce{N2} & : & - (\dot{M}_{ \ce{N2} } )_{1} - (\dot{M}_{ \ce{N2} } )_{5} + (\dot{M}_{ \ce{N2} } )_{2} = 0 \tag{30d}\label{c30d} \end{align}

The single constraint on theses molar flow rates is given by Equation \ref{c14} that we repeat here as

$(\dot{M}_{\ce{O2} } )_{1} = \varphi (\dot{M}_{ \ce{N2} } )_{1} , \quad \varphi = \left({21 / 79} \right) \tag{31}\label{c31}$

We cannot make any significant progress with Eqs. \ref{c30a}-\ref{c30d}, thus we move on to the second control volume.

#### Control Volume 2

In this case we make use of Eqs. \ref{c28a}-\ref{c28f} and \ref{c29} to obtain the six species mole balances. These six equations are simplified by Equation \ref{c18}, Equation \ref{c19} and Equation \ref{c25} along with the observation that there is no ethylene oxide ($$\ce{C2H4O}$$) or water ($$\ce{H2O}$$) in Stream #2. The six balance equations are given by

\begin{align} \ce{C2H4} & : & - (\dot{M}_{\ce{C2H4} } )_{2} + (\dot{M}_{\ce{C2H4} } )_{4} = - { C} (\dot{M}_{\ce{C2H4} } )_{2} \tag{32a}\label{c32a} \end{align}

\begin{align} \ce{C2H4O} & : & (\dot{M}_{\ce{C2H4O}} )_{7} = { YC} (\dot{M}_{\ce{C2H4} } )_{2} \tag{32b}\label{c32b} \end{align}

\begin{align} \ce{H2O} & : & - (\dot{M}_{\ce{H2O}} )_{8} + (\dot{M}_{\ce{H2O}} )_{7} = 2{ C} (1-{ Y}) (\dot{M}_{\ce{C2H4} } )_{2} \tag{32c}\label{c32c} \end{align}

\begin{align} \ce{CO2} & : & - (\dot{M}_{ \ce{CO2} } )_{2} + (\dot{M}_{ \ce{CO2} } )_{4} = 2{ C} (1-{ Y}) (\dot{M}_{\ce{C2H4} } )_{2} \tag{32d}\label{c32d} \end{align}

\begin{align} \ce{O2} & : & - (\dot{M}_{\ce{O2} } )_{2} = { C} (\frac{5}{2} { Y}-3) (\dot{M}_{\ce{C2H4} } )_{2} \tag{32e}\label{c32e} \end{align}

\begin{align} \ce{N2} & : & - (\dot{M}_{ \ce{N2} } )_{2} + (\dot{M}_{ \ce{N2} } )_{4} = 0 \tag{32f}\label{c32f} \end{align}

Equation \ref{c21} along with Equation \ref{c22} and Equation \ref{c23} provide the constraint

$(\dot{M}_{\ce{C2H4} } )_{2} = {\beta / { CY}} \tag{33}\label{c33}$

indicating that the molar flow rate of ethylene entering the reactor is specified. In addition, we can use Equation \ref{c32d} along with Equation \ref{c32f} and Equation \ref{c17} to determine $$(\dot{M}_{ \ce{CO2} } )_{4} +(\dot{M}_{ \ce{N2} } )_{4}$$. A summary of these results is given by

$(\dot{M}_{\ce{C2H4} } )_{2} = {\beta / { CY}} \tag{34a}\label{c34a}$

$(\dot{M}_{\ce{C2H4} } )_{4} = {\beta (1-{ C}) / { CY}} \tag{34b}\label{c34b}$

$(\dot{M}_{\ce{C2H4O}} )_{7} = \beta \tag{34c}\label{c34c}$

$(\dot{M}_{\ce{O2} } )_{2} = \beta (3-\frac{5}{2} { Y}) / { Y} \tag{34d}\label{c34d}$

$(\dot{M}_{\ce{H2O}} )_{7} = \beta \left[\Psi + 2 (1-{ Y}) / { Y} \right] \tag{34e}\label{c34e}$

$(\dot{M}_{\ce{H2O}} )_{8} = \beta \Psi \tag{34f}\label{c34f}$

$(\dot{M}_{ \ce{CO2} } )_{4} + (\dot{M}_{ \ce{N2} } )_{4} = \frac{\beta }{ CY\eta } \left[1 - \eta (1 + { C} - \frac{1}{2} { CY})\right] \tag{34g}\label{c34g}$

At this point we need information about Stream #4 and this leads us to the splitter contained in Control Volume 3.

#### Control Volume 3

As indicated in Figure $$\PageIndex{6c}$$, Stream #4 contains only ethylene ($$\ce{C2H4}$$), carbon dioxide ($$\ce{CO2}$$) and nitrogen ($$\ce{N2}$$), and the appropriate mole balances are given by

\begin{align} \ce{C2H4} & : & - (\dot{M}_{\ce{C2H4} } )_{4} + (\dot{M}_{\ce{C2H4} } )_{5} + (\dot{M}_{\ce{C2H4} } )_{6} = 0 \tag{35a}\label{c35a} \end{align}

\begin{align} \ce{CO2} & : & - (\dot{M}_{ \ce{CO2} } )_{4} + (\dot{M}_{ \ce{CO2} } )_{5} + (\dot{M}_{ \ce{CO2} } )_{6} = 0 \tag{35b}\label{c35b} \end{align}

\begin{align} \ce{N2} & : & - (\dot{M}_{ \ce{N2} } )_{4} + (\dot{M}_{ \ce{N2} } )_{5} + (\dot{M}_{ \ce{N2} } )_{6} = 0 \tag{35c}\label{c35c} \end{align}

On the basis of Equation \ref{c12}, and the fact that the splitter streams contain only three species (see Figure $$\PageIndex{6c}$$), we apply Equation \ref{44} to obtain

$(x_{ \ce{CO2} } )_{6} = (x_{ \ce{CO2} } )_{4} \tag{36a}\label{c36a}$

$(x_{ \ce{N2} } )_{6} = (x_{ \ce{N2} } )_{4} \tag{36b}\label{c36b}$

At this point we make use of Equation \ref{34} to obtain the standard constraints on mole fractions when only ethylene ($$\ce{C2H4}$$), carbon dioxide ($$\ce{CO2}$$) and nitrogen ($$\ce{N2}$$) are present. This provides

$(x_{\ce{C2H4} } )_{4} + (x_{ \ce{CO2} } )_{4} + (x_{ \ce{N2} } )_{4} = 1 \tag{37a}\label{c37a}$

$(x_{\ce{C2H4} } )_{6} + (x_{ \ce{CO2} } )_{6} + (x_{ \ce{N2} } )_{6} = 1 \tag{37b}\label{c37b}$

and subtracting Equation \ref{c37a} from Equation \ref{c37b} leads to

$(x_{\ce{C2H4} } )_{6} = (x_{\ce{C2H4} } )_{4} \tag{38}\label{c38}$

We now make use of Equation \ref{45} to represent the constraints on the three mole fractions as

$(\dot{M}_{\ce{CO2} } )_{6} = \left(\dot{M}_{6} / \dot{M}_{4} \right)(\dot{M}_{ \ce{CO2} } )_{4} \tag{39a}\label{c39a}$

$(\dot{M}_{\ce{N2} } )_{6} = \left(\dot{M}_{6} / \dot{M}_{4} \right)(\dot{M}_{ \ce{N2} } )_{4} \tag{39b}\label{c39b}$

$(\dot{M}_{\ce{C2H4} } )_{6} = \left(\dot{M}_{6} / \dot{M}_{4} \right)(\dot{M}_{\ce{C2H4} } )_{4} \tag{39c}\label{c39c}$

In Equation \ref{c1} we let $$\alpha$$ be the fraction of Stream #4 that needs to be purged, i.e.,

$\alpha = \dot{M}_{6} / \dot{M}_{4} \tag{40}\label{c40}$

thus the molar flow rates in Stream #6 are given by

$(\dot{M}_{ \ce{CO2} } )_{6} = \alpha (\dot{M}_{ \ce{CO2} } )_{4} \tag{41a}\label{c41a}$

$(\dot{M}_{ \ce{N2} } )_{6} = \alpha (\dot{M}_{ \ce{N2} } )_{4} \tag{41b}\label{c41b}$

$(\dot{M}_{\ce{C2H4} } )_{6} = \alpha (\dot{M}_{\ce{C2H4} } )_{4} \tag{41c}\label{c41c}$

Use of these results with Eqs. \ref{c35a}-\ref{c35c} leads to

\begin{align} \ce{C2H4} & : & (\dot{M}_{\ce{C2H4} } )_{5} = (1-\alpha )(\dot{M}_{\ce{C2H4} } )_{4}\tag{42a}\label{c42a} \end{align}

\begin{align} \ce{CO2} & : & (\dot{M}_{ \ce{CO2} } )_{5} = (1-\alpha )(\dot{M}_{ \ce{CO2} } )_{4} \tag{42b}\label{c42b} \end{align}

\begin{align} \ce{N2} & : & (\dot{M}_{ \ce{N2} } )_{5} = (1-\alpha )(\dot{M}_{ \ce{N2} } )_{4} \tag{42c}\label{c42c} \end{align}

and we can use Equation \ref{c34b} to represent the molar flow rate of ethylene in Stream #5 as

$(\dot{M}_{\ce{C2H4} } )_{5} = \frac{\beta (1-C)(1-\alpha )}{ CY}\tag{43}\label{c43}$

At this point we need to return to Control Volume 1 in order to acquire additional information needed to determine the parameter, $$\alpha$$. Beginning with Equation \ref{c30a}, we use Equation \ref{c34a} and Equation \ref{c43} to obtain

\begin{align} \ce{C2H4} & : & (\dot{M}_{\ce{C2H4} } )_{1} = \frac{\beta }{ CY} \left[1 - \left(1-{ C}\right)\left(1-\alpha \right)\right] \tag{44a}\label{c44a} \end{align}

Moving on to Equation \ref{c30b}, we use Equation \ref{c32d} along with Equation \ref{c34a} and Equation \ref{c42b} to obtain

\begin{align} \ce{CO2} & : & (\dot{M}_{\ce{CO2} } )_{2} = \frac{2\beta \left(1-\alpha \right)\left(1-{ Y}\right)}{\alpha Y}\tag{44b}\label{c44b} \end{align}

Next we use Equation \ref{c30c} along with Equation \ref{c34d} to obtain

\begin{align} \ce{O2} & : & (\dot{M}_{\ce{O2} } )_{1} = (\dot{M}_{\ce{O2} } )_{2} = \beta (3-\frac{5}{2} { Y}) / { Y} \tag{44c}\label{c44c} \end{align}

and finally we make use of Equation \ref{c30c} along with Equation \ref{c31} to express the molar flow rate of nitrogen entering the system as

\begin{align} \ce{N2} & : & (\dot{M}_{\ce{N2} } )_{1} = \beta (3-\frac{5}{2} Y) / \varphi Y \tag{44d}\label{c44d} \end{align}

At this point we are ready to return to Equation \ref{c34g} and make use of Equation \ref{c32d} and Equation \ref{c33} to obtain

$\left(\dot{M}_{\ce{CO2}}\right)_{4}=\left(\dot{M}_{\ce{CO2}}\right)_{2}+2 \beta(1-{Y}) / {Y} \tag{45}\label{c45}$

This result can be used with Equation \ref{c44b} to provide the molar flow rate of carbon dioxide ($$\ce{CO2}$$) entering the splitter

$(\dot{M}_{\ce{N2}})_4 = 2\beta (1 - Y) / \alpha Y \tag{46}\label{c46}$

This result can be used with Equation \ref{c34g} to determine the molar flow rate of nitrogen ($$\ce{N2}$$) entering the splitter as

$\left(\dot{M}_{\ce{N2}}\right)_{4}=\frac{\beta}{CY \eta}\left[1-\eta\left(1+{C}-\frac{1}{2} CY \right)\right]-\frac{2 \beta(1-Y)}{\alpha Y} \tag{47}\label{c47}$

We now recall Equation \ref{c30d} along with Equation \ref{c32f} and combine that result with Equation \ref{c42c} to obtain4

$(\dot{M}_{\ce{N2}})_1 = \alpha (\dot{M}_{\ce{N2}})_4 \tag{48}\label{c48}$

Here we can use Equation \ref{c44d} along with Equation \ref{c47} and Equation \ref{c48} to eliminate all the molar flow ra obtain an expression in which $$\alpha$$ is the only unknown.

$\beta\left(3-\frac{5}{2} {Y}\right) / \varphi {Y}=\frac{\alpha \beta}{CY \eta}\left[1-\eta\left(1+{C}-\frac{1}{2} {CY}\right)\right]-\frac{2 \beta(1-{Y})}{Y} \tag{49}\label{c49}$

This result provides a solution for the fraction of Stream #4 that must be purged, i.e.,

$\alpha=\frac{\left(3-\frac{5}{2} {Y}\right)({C} \eta / \varphi)+2 {C}(1-{Y}) \eta}{1-\eta\left(1+{C}-\frac{1}{2} {CY}\right)} \tag{50}\label{c50}$

Given the following data:

$C = \text{ conversion } = 0.70 \nonumber$

$Y = \text{ yield } = 0.50 \nonumber$

$\eta = \text{ mole fraction of ethylene entering the reactor } = 0.05 \nonumber$

$\varphi = (\dot{M}_{\ce{O2} } )_{1} / (\dot{M}_{\ce{N2} } )_{1} = {21 / 79} = 0.2658 \nonumber$

we determine the fraction of Stream #4 that must be purged to be $$\alpha = 0.2874$$.

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