# 4.7: Species Mole/Mass Balance

In this section we examine the problem of solving the $$N$$ equations represented by either Equation $$(4.1.7)$$ or Equation $$(4.1.17)$$ under steady-state conditions in the absence of chemical reactions. The distillation process illustrated in Figure $$\PageIndex{1}$$ provides a simple example; however, most distillation processes are more complex than the one shown in Figure $$\PageIndex{1}$$ and most are integrated into a chemical plant as discussed in Chapter 1. It was also pointed out in Chapter 1 that complex chemical plants can be understood by first understanding the individual units that make up the plants (see Figures $$1.3.1$$ and $$1.3.2$$), thus understanding the simple process illustrated in Figure $$\PageIndex{1}$$ is an important step in our studies.

For this particular ternary distillation process, we are given the information listed in Table $$\PageIndex{1}$$. Often the input conditions for a process are completely specified, and this means that the input flow rate and all the compositions would be specified. However, in Table $$\PageIndex{1}$$, we have not specified $$x_{C}$$ since this mole fraction will be determined by Equation $$(4.6.8)$$. If we list this mole fraction as $$x_{C} =0.5$$, we would be over-specifying the problem and this would lead to difficulties with our degree of freedom analysis.

Table $$\PageIndex{1}$$: Specified conditions
Stream #1 Stream #2 Stream #3
$$\dot{M}_{1} =1200$$ moles/hr
$$x_{A} = 0.3$$ $$x_{A} = 0.6$$ $$x_{A} = 0.1$$
$$x_{B} = 0.2$$ $$x_{B} = 0.3$$

In our application of macroscopic balances to single component systems in Chapter 3, we began each problem by identifying a control volume and we listed rules that should be followed for the construction of control volumes. For multi-component systems, we change those rules only slightly to obtain

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.

Here it is understood that $$\mathbf{v}_{A}$$ represents the species velocity for all $$N$$ species, and in Rule II it is assumed that the given information is necessary for the solution of the problem. For the system illustrated in Figure $$\PageIndex{1}$$, it should be obvious that we need to cut the entrance and exit streams and then join the cuts as illustrated in Figure $$\PageIndex{2}$$ where we have shown the details of the cut at stream #2, and we have illustrated that the cuts at the entrance and exit streams are joined by a surface that is coincident with the solid-air interface where $$\mathbf{v}_{A} \cdot \mathbf{n}=0$$.

## Degrees-of-freedom analysis

In order to solve the macroscopic balance equations for this distillation process, we require that the number of constraining equations be equal to the number of unknowns. To be certain that this is the case, we perform a degrees-of-freedom analysis which consists of three parts. We begin this analysis with a generic part in which we identify the process variables that apply to a single control volume in which there are $$N$$ molecular species and $$M$$ streams. We assume that every molecular species is present in every stream, and this leads to the generic degrees of freedom. Having determined the generic degrees of freedom, we direct our attention to the generic specifications and constraints which also apply to the control volume in which there are $$N$$ molecular species and $$M$$ streams. Finally, we consider the particular specifications and constraints that reduce the generic degrees of freedom to zero if we have a well-posed problem in which all process variables can be determined. If the last part of our analysis does not reduce the degrees of freedom to zero, we need more information in order to solve the problem. The inclusion of chemical reactions in the degree of freedom analysis will be delayed until we study stoichiometry in Chapter 6.

The first step in our analysis is to prepare a list of the process variables, and this leads to

Mole fractions: $(x_{A} )_{i} , \quad (x_{B} )_{i} , \quad (x_{C} )_{i} \quad i=1,2,3 \label{78}$

Molar flow rates: $\dot{M}_{i} , \quad i=1,2,3 \label{79}$

For a system containing three molecular species and having three streams, we determine that there are twelve generic process variables as indicated below.

 I. Three mole fractions in each of three streams 9 II. Three molar flow rates 3

For this process the generic degrees of freedom are given by

 Generic Degrees of Freedom (A) 12

In this first step, it is important to recognize that we have assumed that all species are present in all streams, and it is for this reason that we obtain the generic degrees of freedom.

The second step in this process is to determine the generic specifications and constraints associated with a system containing three molecular species and three streams. In order to solve this ternary distillation problem, we will make use of the three molecular species balances given by

Species balances: $\int_{\mathscr{A}}x_{A} c\mathbf{v} \cdot \mathbf{n} dA = 0 , \quad \int_{\mathscr{A}}x_{B} c\mathbf{v} \cdot \mathbf{n} dA = 0 , \quad \int_{\mathscr{A}}x_{C} c\mathbf{v} \cdot \mathbf{n} dA = 0 \label{80}$

along with the three mole fraction constraints that apply at the streams that are cut by the control volume illustrated in Figure $$\PageIndex{2}$$.

Mole fraction constraints: $(x_{A} )_{i} + (x_{B} )_{i} + (x_{C} )_{i} =1 , \quad i=1,2,3 \label{81}$

We list these specification and constraints as

 I. Balance equations for three molecular species 3 II. Mole fraction constraints for the three streams 3

 Generic Specifications and Constraints (B) 6

Moving on to the third step in our degree of freedom analysis, we list the particular specifications and constraints according to

 I. Conditions for Stream #1: $$\dot{M}_{1} =1200$$ moles/hr, $$x_{A} =0.3 ,$$ $$x_{B} =0.2$$ 3 II. Conditions for Stream #2: $$x_{A} = 0.6 ,$$ $$x_{B} = 0.3$$ 2 III. Conditions for Stream #3: $$x_{A} = 0.1$$ 1

This leads us to the particular specifications and constraints indicated by

 Particular Specifications and Constraints (C) 6

and we can see that there are zero degrees of freedom for this problem.

When developing the particular specifications and constraints, it is extremely important to understand that the three mole fractions can be specified only in the following manner:

I. None of the mole fractions are specified in a particular stream.

II. One of the mole fractions is specified in a particular stream.

III. Two of the mole fractions are specified in a particular stream.

The point here is that one cannot specify all three mole fractions in a particular stream because of the constraint on the mole fractions given by Equation \ref{81}. If one specifies all three mole fractions in a particular stream, Equation \ref{81} for that stream must be deleted and the generic specifications and constraints are no longer generic.

There are two important results associated with this degree of freedom analysis. First, we are certain that a solution exits, and this provides motivation for persevering when we encounter difficulties. Second, we are now familiar with the nature of this problem and this should help us to organize a procedure for the development of a solution. We summarize our degree of freedom analysis in Table $$\PageIndex{2}$$ that provides a template for subsequent problems in which we have $$N$$ molecular species and $$M$$ streams.

Table $$\PageIndex{2}$$: Degrees-of-Freedom

Stream Variables $$N \times M = 9$$ $$M = 3$$ $$(N \times M) + M = \mathbf{12}$$ $$N=3$$ $$M=3$$ $$N+M = \mathbf{6}$$ 5 1 0 0 6 0

At this point we are certain that we have a well-posed problem and we can proceed with confidence knowing that we can find a solution.

## Solution of macroscopic balance equations

Before beginning the solution procedure, we should clearly identify what is known and what is unknown, and we do this with an extended version of Table $$\PageIndex{1}$$ given here as

Table $$\PageIndex{3}$$: Specified and unknown conditions
Stream #1 Stream #2 Stream #3
$$\dot{M}_{1} =1200$$ moles/hr ? ?
$$x_{A} = 0.3$$ $$x_{A} = 0.6$$ $$x_{A} = 0.1$$
$$x_{B} = 0.2$$ $$x_{B} = 0.3$$ ?
? ? ?

When the spaces identified by question marks have been filled with results, our solution will be complete. We begin with the simplest calculations and make use of the constraints given by Eqs. \ref{81}. These can be used to express Table $$\PageIndex{3}$$ as follows:

Table $$\PageIndex{4}$$: Unknowns to be determined
Stream #1 Stream #2 Stream #3
$$\dot{M}_{1} =1200$$ moles/hr ? ?
$$x_{A} = 0.3$$ $$x_{A} = 0.6$$ $$x_{A} = 0.1$$
$$x_{B} = 0.2$$ $$x_{B} = 0.3$$ $$x_{B} = 0.9-x_{C}$$
$$x_{c} = 0.5$$ $$x_{c} = 0.1$$ ?

This table indicates that we have three unknowns to be determined on the basis of the three species balance equations. Use of the results given in Table $$\PageIndex{4}$$ allows us to express the balance equations given by Eqs. \ref{80} as

Species $$A$$: $0.6 \dot{M}_{2} + 0.1 \dot{M}_{3} + 0 = 360 \text{ moles/hr} \label{82a}$

Species $$B$$: $0.3 \dot{M}_{2} + 0.9 \dot{M}_{3} - \underbrace{\dot{M}_{3} \langle x_{C} \rangle_{3} }_{\text{bi-linear form}} = 240 \text{ moles/hr} \label{82b}$

Species $$C$$: $0.1 \dot{M}_{2} + 0 + \underbrace{\dot{M}_{3} \langle x_{C} \rangle_{3} }_{\text{bi-linear form}} = 600 \text{ moles/hr} \label{82c}$

in which the product of unknowns, $$\dot{M}_{3}$$ and $$\langle x_{C} \rangle_{3}$$, has been identified as a bi-linear form. This is different from a linear form in which $$\dot{M}_{3}$$ and $$\langle x_{C} \rangle_{3}$$ would appear separately, or a non-linear form such as $$\sqrt{\dot{M}_{3} }$$ or and $$\langle x_{C} \rangle_{3}^{2}$$. In this problem, we are confronted with three unknowns, $$\dot{M}_{2}$$, $$\dot{M}_{3}$$ and $$\langle x_{C} \rangle_{3}$$, and three equations that can easily be solved to yield

$\dot{M}_{2} = 480 \text{ mol/hr} , \quad \dot{M}_{3} = 720 \text{ mol/hr} , \quad \langle x_{C} \rangle_{3} =0.767 \label{92}$

This information can be summarized in the same form as the input data in order to obtain

Table $$\PageIndex{5}$$: Solution for molar flows and mole fractions
Stream #1 Stream #2 Stream #3
$$\dot{M}_{1} =1200$$ mol/hr $$\dot{M}_{2} =480$$ mol/hr $$\dot{M}_{3} =720$$ mol/hr
$$x_{A} = 0.3$$ $$x_{A} = 0.6$$ $$x_{A} = 0.100$$
$$x_{B} = 0.2$$ $$x_{B} = 0.3$$ $$x_{B} = 0.133$$
$$x_{C} = 0.5$$ $$x_{C} = 0.1$$ $$x_{C} = 0.767$$

The structure of this ternary distillation process is typical of macroscopic mass balance problems for multicomponent systems. These problems become increasing complex (in the algebraic sense) as the number of components increases and as chemical reactions are included, thus it is important to understand the general structure. Macroscopic mass balance problems are always linear in terms of the compositions and flow rates even though these quantities may appear in bi-linear forms. This is the case in Eqs. \ref{82} where an unknown flow rate is multiplied by an unknown composition; however, these equations are still linear in $$\dot{M}_{3}$$ and $$\langle x_{C} \rangle_{3}$$, thus a unique solution is possible. When chemical reactions occur, and the reaction rate expressions (see Sec. 8.6) are non-linear in the composition, numerical methods are generally necessary and one must be aware that nonlinear problems may have more than one solution or no solution.

## Solution of sets of equations

To illustrate a classic procedure for solving sets of algebraic equations, we direct our attention to Eqs. \ref{82a} - \ref{82c}. We begin by eliminating the term, $$\dot{M}_{3} \langle x_{C} \rangle_{3}$$, from Equation \ref{82b} to obtain the following pair of linear equations:

Species $$A$$: $0.6 \dot{M}_{2} + 0.1 \dot{M}_{3} =360 \text{ mol/hr} \label{84a}$

Species $$B$$: $0.4 \dot{M}_{2} + 0.9 \dot{M}_{3} =840 \text{ mol/hr} \label{84b}$

To solve this set of linear equations, we make use of a simple scheme known as Gaussian elimination. We begin by dividing Equation \ref{84a} by the coefficient 0.6 in order to obtain

$\dot{M}_{2} + 0.1667 \dot{M}_{3} =600 \text{ mol/hr} \label{85a}$

$0.4 \dot{M}_{2} + 0.9 \dot{M}_{3} =840 \text{ mol/hr} \label{85b}$

Next, we multiply the first equation by 0.4 and subtract that result from the second equation to provide

$\dot{M}_{2} + 0.1667 \dot{M}_{3} =600 \text{ mol/hr} \label{86a}$

$0.8333 \dot{M}_{3} =600 \text{ mol/hr} \label{86b}$

We now divide the last equation by 0.8333 to obtain the solution for the unknown $$\dot{M}_{3}$$.

$\dot{M}_{2} + 0.1667 \dot{M}_{3} =600 \text{ mol/hr} \label{87a}$

$\dot{M}_{3} =720 \text{ mol/hr} \label{87b}$

At this point, we begin the procedure of "back substitution” which requires that Equation \ref{87b} be substituted into Equation \ref{87a} in order to obtain the final solution for the two molar flow rates.

$\dot{M}_{2} =480 \text{ mol/hr} \label{88a}$

$\dot{M}_{3} =720 \text{ mol/hr} \label{88b}$

The procedure leading from Eqs. \ref{84a} - \ref{84b} to the solution given by Eqs. \ref{88a} - \ref{88b} is trivial for a pair of equations; however, if we were working with a five component system the algebra would be overwhelmingly difficult and a computer solution would be required.