# 4.8: Multiple Units

When more than a single unit is under consideration, some care is required in the choice of control volumes, and the two-column distillation unit illustrated in Figure $$\PageIndex{1}$$ provides an example. In that figure we have indicated that all the mass fractions in the streams entering and leaving the two-column unit are specified, i.e., the problem is over-specified and we will need to be careful in our degree of freedom analysis. In addition to the mass fractions, we are also given that the mass flow rate to the first column is $$1000 \ lb_{ m} /{ hr}$$. On the basis of this information, we want to predict

1. The mass flow rate of both overhead (or distillate) streams (streams #2 and #3).
2. The mass flow rate of the bottoms from the second column (stream #4).
3. The mass flow rate of the feed to the second column (stream #5).

We begin the process of constructing control volumes by making the primary cuts shown in Figure $$\PageIndex{1}$$. Those cuts have been made where information is given (stream #1) and information is required (streams #2, #3, #4, and #5). In order to join the primary cuts to form control volumes, we are forced to construct two control volumes such as we have shown in Figure $$\PageIndex{2}$$. We first form Control Volume I which connects three primary cuts (streams #1, #2 and #5) and encloses the first column of the two-column distillation unit.

In order to construct a control volume that joins the primary cuts of streams #3 and #4, we have two choices. One choice is to enclose the second column by joining the primary cuts of streams #3, #4 and #5, while the second choice is illustrated in Figure $$\PageIndex{3}$$. If Control Volume II were constructed so that it joined streams #3, #4 and #5, it would not be connected to the single source of the necessary information, i.e., the mass flow rate of stream #1. In that case, the information about stream #5 would cancel in the balance equations and we would not be able to determine the mass flow rate in stream #5. Since the data are given in terms of mass fractions and the mass flow rate of stream #1, the appropriate macroscopic balance is given by Equation $$(4.1.7)$$. For steady-state conditions in the absence of chemical reactions, the species mass balances are given by

$\int_{\mathscr{A}}\rho_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = 0 , \quad \int_{\mathscr{A}}\rho_{B} \mathbf{v}_{B} \cdot \mathbf{n} dA = 0 , \quad \int_{\mathscr{A}}\rho_{C} \mathbf{v}_{C} \cdot \mathbf{n} dA = 0 \label{89}$

Since convective effects will dominate at the entrances and exits of the two control volumes, we can express this result in the form

Species balances: $\int_{\mathscr{A}}\omega_{A} \rho \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad \int_{\mathscr{A}}\omega_{B} \rho \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad \int_{\mathscr{A}}\omega_{C} \rho \mathbf{v} \cdot \mathbf{n} dA = 0 \label{90}$

Here we have represented the fluxes in terms of the mass fractions since the stream compositions are given in terms of mass fractions that are constrained by

Mass fraction constraints: $(\omega_{A} )_{i} + (\omega_{B} )_{i} + (\omega_{C} )_{i} =1 , \quad i=1,2,3,4,5 \label{91}$

Before attempting to determine the flow rates in streams 2, 3, 4 and 5, we need to perform a degree of freedom analysis to be certain that the problem is well-posed. We begin the analysis with Control Volume II, and as our first step in the degree of freedom analysis we list the process variables as

## Control Volume II

Mass fractions: $(\omega_{A} )_{i} , \quad (\omega_{B} )_{i} , \quad (\omega_{C} )_{i} , \quad i=1,2,3,4 \label{92}$

Mass flow rates: $\dot{m}_{ i} , \quad i=1,2,3,4 \label{93}$

and we indicate the number of process variables explicitly as

 I. Three mole fractions in each of four streams 12 II. Four mass molar flow rates 4

For this process the generic degrees of freedom are given by

 Generic Degrees of Freedom (A) 16

Moving on to the generic specifications and constraints, we list the three molecular species balances given by

Species balances: $\int_{\mathscr{A}}\omega_{A} \rho \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad \int_{\mathscr{A}}\omega_{B} \rho \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad \int_{\mathscr{A}}\omega_{C} \rho \mathbf{v} \cdot \mathbf{n} dA = 0 \label{94}$

along with the four mass fraction constraints that apply at the streams that are cut by Control Volume II.

Mass fraction constraints: $(\omega_{A} )_{i} + (\omega_{B} )_{i} + (\omega_{C} )_{i} =1 , \quad i=1,2,3,4 \label{95}$

This leads us to the second step in our degree of freedom analysis that we express as

 I. Balance equations for three molecular species 3 II. Mole fraction constraints for the four streams 4
 Generic Specifications and Constraints (B) 7

Our third step in the degree of freedom analysis requires that we list the particular specifications and constraints according to

 I. Conditions for Stream #1: $$\dot{m}_{ 1} =1000 \ lb_{ m} /{ hr} , \quad \omega_{A} = 0.5 , \quad \omega_{B} = 0.3$$ 3 II. Conditions for Stream #2: $$\omega_{A} = 0.045 , \quad \omega_{B} = 0.091$$ 2 III. Conditions for Stream #3: $$\omega_{A} = 0.069 , \quad \omega_{B} = 0.901$$ 2 IV. Conditions for Stream #4: $$\omega_{A} = 0.955 , \quad \omega_{B} = 0.041$$ 2

This leads us to the particular specifications and constraints indicated by

 Particular Specifications and Constraints (C) 9

and we summarize these results in the following table:

Table $$\PageIndex{1}$$: Degrees-of-Freedom

Stream Variables $$N \times M = 12$$ $$M = 4$$ $$(N \times M) + M = \mathbf{16}$$ $$N=3$$ $$M=4$$ $$N+M = \mathbf{7}$$ 8 1 0 0 9 0

This indicates that use of Control Volume II will lead to a well-posed problem, and we are assured that we can use Eqs. \ref{94} and \ref{95} to determine the mass flow rates in streams 2, 3, 4. This calculation is carried out in the following paragraphs.

Often in problems of this type, it is convenient to work with $$N-1$$ species mass balances and the total mass balance that is given by Equation $$(4.2.18)$$ In terms of Eqs. \ref{94} for Control Volume II, this approach leads to

## Control Volume II:

Species $$A$$: $(\omega_{A} )_{2} \dot{m}_{2} + (\omega_{A} )_{3} \dot{m}_{3} + (\omega_{A} )_{4} \dot{m}_{4} =(\omega_{A} )_{1} \dot{m}_{1} \label{96a}$

Species $$B$$: $(\omega_{B} )_{2} \dot{m}_{2} + (\omega_{B} )_{3} \dot{m}_{3} + (\omega_{B} )_{4} \dot{m}_{4} =(\omega_{B} )_{1} \dot{m}_{1} \label{96b}$

Total: $\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = \dot{m}_{1} \label{96c}$

Here we are confronted with three equations and three unknowns, and our problem is quite similar to that encountered in Sec. 4.7 where our study of a single distillation column led to a set of two equations and two unknowns. That problem was solved by Gaussian elimination as indicated by Eqs. $$(4.7.9-4.7.10)$$ through $$(4.7.17-4.7.18)$$. The same procedure can be used with Eqs. \ref{96a} - \ref{96c}, and one begins by dividing Equation \ref{96a} by $$(\omega_{A} )_{2}$$ to obtain

## Control Volume II:

Species $$A$$: $\dot{m}_{2} + \left[{(\omega_{A} )_{3} / (\omega_{A} )_{2} } \right]\dot{m}_{3} + \left[{(\omega_{A} )_{4} / (\omega_{A} )_{2} } \right]\dot{m}_{4} =\left[{(\omega_{A} )_{1} / (\omega_{A} )_{2} } \right]\dot{m}_{1} \label{97a}$

Species $$B$$: $(\omega_{B} )_{2} \dot{m}_{2} + (\omega_{B} )_{3} \dot{m}_{3} + (\omega_{B} )_{4} \dot{m}_{4} =(\omega_{B} )_{1} \dot{m}_{1} \label{97b}$

Total: $\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = \dot{m}_{1} \label{97c}$

In order to eliminate $$\dot{m}_{2}$$ from Equation \ref{97b}, one multiplies Equation \ref{97a} by $$(\omega_{B} )_{2}$$ and subtracts the result from both Equation \ref{97b}. To eliminate $$\dot{m}_{2}$$ from Equation \ref{97c}, one need only subtract Equation \ref{97a} from Equation \ref{97c}, and these two operations lead to the following balance equations:

## Control Volume II:

Species $$A$$: $\dot{m}_{2} + \left[{(\omega_{A} )_{3} / (\omega_{A} )_{2} } \right]\dot{m}_{3} + \left[{(\omega_{A} )_{4} / (\omega_{A} )_{2} } \right]\dot{m}_{4} =\left[{(\omega_{A} )_{1} / (\omega_{A} )_{2} } \right]\dot{m}_{1} \label{98a}$

Species $$B$$: $\left\{.......\right\}\dot{m}_{3} +\left\{.......\right\}\dot{m}_{4} =\left\{.......\right\}\dot{m}_{1} \label{98b}$

Total: $\left[......\right]\dot{m}_{3} +\left[......\right]\dot{m}_{4} =\left[......\right]\dot{m}_{1} \label{98c}$

Clearly the algebra is becoming quite complex, and it will become worse when we use Equation \ref{98b} to eliminate $$\dot{m}_{3}$$ from Equation \ref{98c}. Without providing the details, we continue the elimination process to obtain the solution to Equation \ref{98c} and this leads to the following expression for $$\dot{m}_{4}$$:

$\dot{m}_{4} =\dot{m}_{1} \frac{\left[1-\frac{(\omega_{A} )_{1} }{(\omega_{A} )_{2} } \right]-\frac{\frac{(\omega_{B} )_{1} }{(\omega_{B} )_{3} } \left[1-\frac{(\omega_{B} )_{2} }{(\omega_{B} )_{1} } \frac{(\omega_{A} )_{1} }{(\omega_{A} )_{2} } \right]}{\left[1-\frac{(\omega_{B} )_{2} }{(\omega_{B} )_{3} } \frac{(\omega_{A} )_{3} }{(\omega_{A} )_{2} } \right]} \left[1-\frac{(\omega_{A} )_{3} }{(\omega_{A} )_{2} } \right]}{\left[1-\frac{(\omega_{A} )_{4} }{(\omega_{A} )_{2} } \right]-\frac{\frac{(\omega_{B} )_{4} }{(\omega_{B} )_{3} } \left[1-\frac{(\omega_{B} )_{2} }{(\omega_{B} )_{4} } \frac{(\omega_{A} )_{4} }{(\omega_{A} )_{2} } \right]}{\left[1-\frac{(\omega_{B} )_{2} }{(\omega_{B} )_{3} } \frac{(\omega_{A} )_{3} }{(\omega_{A} )_{2} } \right]} \left[1-\frac{(\omega_{A} )_{3} }{(\omega_{A} )_{2} } \right]} \label{99}$

Equally complex expressions can be obtained for $$\dot{m}_{2}$$ and $$\dot{m}_{3}$$, and the numerical values for the three mass flow rates are given by

$\dot{m}_{2} = 220 \ lb_{ m} /{ hr} , \quad \dot{m}_{3} = 288 \ lb_{ m} /{ hr} , \quad \dot{m}_{4} = 492 \ lb_{ m} /{ hr} \label{100}$

In order to determine $$\dot{m}_{5}$$, we must make use of the balance equations for Control Volume I that are given by Eqs. \ref{94}. These can be expressed in terms of two species balances and one total mass balance leading to

## Control Volume I:

Species $$A$$: $- (\omega_{A} )_{1} \dot{m}_{1} + (\omega_{A} )_{2} \dot{m}_{2} + (\omega_{A} )_{5} \dot{m}_{5} =0 \label{101a}$

Species $$B$$: $- (\omega_{B} )_{1} \dot{m}_{1} + (\omega_{B} )_{2} \dot{m}_{2} + (\omega_{B} )_{5} \dot{m}_{5} =0 \label{101b}$

Total: $- \dot{m}_{1} +\dot{m}_{2} +\dot{m}_{5} =0 \label{101c}$

and the last of these quickly leads us to the result for $$\dot{m}_{5}$$.

$\dot{m}_{5} = 780 \ lb_{ m} /{ hr} \label{102}$

The algebraic complexity associated with the simple process represented in Figure $$\PageIndex{1}$$ encourages the use of matrix methods. We begin a study of those methods in the following section and we continue to study and to apply matrix methods throughout the remainder of the text. Our studies of stoichiometry in Chapter 6 and reaction kinetics in Chapter 9 rely heavily on matrix methods that are presented as needed. In addition, a detailed discussion of matrix methods is available in Appendix C1.