# 4.1: ALL

In the previous chapter, we considered single-component systems for which there was a single density, , a single velocity, v, and no chemical reactions. In multicomponent systems we must deal with the density of individual species and this leads to the characterization of systems in terms of species mass densities and species molar concentrations, in addition to mass fractions and mole fractions. Not only must we characterize the composition of multicomponent systems, but we must also consider the fact that different molecular species move at different velocities. This leads us to the concept of the species velocity which plays a dominant role in the detailed study of separation and purification processes and in the analysis of chemical reactors. In this chapter we will discuss the concept of the species velocity and then illustrate how a certain class of macroscopic mass balance problems can be solved without dealing directly with this important aspect of multicomponent systems. While chemical reactions represent an essential feature of multicomponent systems, we will delay a thorough discussion of that matter until Chapter 6.

4.1 Axioms for the Mass of Multicomponent Systems

In Chapter 3, we studied the concept of conservation of mass for single-component systems, both for a body and for a control volume. The words associated with the control volume representation were

$$\left\{\begin{array}{l} {\rm time\; rate\; of\; change\; } \\ {\rm \; \; \; \; \; of\; mass\; in\; a\; \; } \\ {\rm \; \; \; control\; volume} \end{array}\right\}\quad =\quad \left\{\begin{array}{l} {\rm \; rate\; at\; which} \\ {\rm mass\; }enters{\rm \; the} \\ {\rm control\; volume} \end{array}\right\}\quad -\quad \left\{\begin{array}{l} {\rm \; rate\; at\; which} \\ {\rm mass\; }leaves{\rm \; the} \\ {\rm \; control\; volume} \end{array}\right\}$$ ([ZEqnNum921479])

and for a fixed control volume, the mathematical representation was given by

$$\frac{d}{dt} \int _{V}\rho \, dV \; \; +\; \; \int _{A}\rho v\cdot n\, dA \quad =\quad 0$$ ([ZEqnNum562507])

One must keep in mind that the use of vectors allows us to represent both the mass entering the control volume and mass leaving the control volume in terms of $$\rho v\cdot n$$. This follows from the fact that $$v\cdot n$$ is negative over surfaces where mass is entering the control volume and $$v\cdot n$$ is positive over surfaces where mass is leaving the control volume. In addition, one should remember that the control volume, $$V$$, in Eq. [ZEqnNum562507] is arbitrary and this allows us to choose the control volume to suit our needs.

Now we are ready to consider N-component systems in which chemical reactions can take place, and in this case we need to make use of the two axioms for the mass of multicomponent systems. The first axiom deals with the mass of species A, and when this species can undergo chemical reactions we need to extend Eq. to the form given by

$$\left\{\begin{array}{l} {\rm time\; rate\; of\; change} \\ {\rm of\; mass\; of\; species\; }A} \\ {{\rm in\; a\; control\; volume} \end{array}\right\}\quad =\quad \left\{\begin{array}{l} {\rm \; \; \; rate\; at\; which} \\ {\rm mass\; of\; species\; }A} \\ {{\rm \; \; \; \; }enters{\rm \; the} \\ {\rm \; control\; volume} \end{array}\right\}\; \; -\; \; \left\{\begin{array}{l} {\rm \; \; \; rate\; at\; which} \\ {\rm mass\; of\; species\; }A} \\ {{\rm \; \; \; \; \; }leaves{\rm \; the} \\ {\rm \; \; control\; volume} \end{array}\right\}\; \; +\; \; \left\{\begin{array}{l} {\rm \; net\; rate\; of\; production} \\ {\rm of\; the\; mass\; of\; species\; }A} \\ {{\rm \; \; \; \; \; \; \; \; \; \; owing\; to} \\ {{\rm \; \; \; }chemical\; reactions} \end{array}\right\}$$ ([ZEqnNum906921])

In order to develop a precise mathematical representation of this axiom, we require the following quantities:

$$\rho _{A} \quad =\quad \left\{\begin{array}{l} {\rm mass\; density} \\ {{\rm of\; \; species\; }A} \end{array}\right\}$$ ([ZEqnNum275116])

$$v_{A} \quad =\quad \left\{\begin{array}{l} {\rm velocity\; of} \\ {{\rm \; species\; }A} \end{array}\right\}$$ ()

$$r_{A} \quad =\quad \left\{\begin{array}{l} {\rm \; net\; mass\; rate\; of\; production} \\ {{\rm per\; unit\; volume\; of\; species\; }A} \\ {{\rm owing\; to\; }chemical\; reactions} \end{array}\right\}$$ ([ZEqnNum544342])

Here it is important to understand that $$r_{A}$$ represents both the creation of species A (when $$r_{A}$$ is positive) and the consumption of species A (when $$r_{A}$$ is negative). In terms of these primitive quantities, we can make use of an arbitrary fixed control volume to express Eq. as

Axiom I: $$\frac{d}{dt} \int _{V}\rho _{A} \, dV \; \; +\; \; \int _{A}\rho _{A} v_{A} \cdot n\, dA \quad =\quad \int _{V}r_{A} \, dV \; ,\quad \quad A=1,\; 2,\; ....,\; N$$ ([ZEqnNum662048])

In the volume $$V$$ the total mass produced by chemical reactions must be zero. This is our second axiom that we express in words as

$$\left\{\begin{array}{l} {\rm total\; rate\; of\; production} \\ {\rm \; \; \; \; of\; mass\; owing\; to\; } \\ {{\rm \; \; }chemical\; reactions} \end{array}\right\}\quad =\quad 0$$ ()

and in terms of the definition given by Eq. this word statement takes the form

$$\sum _{A=1}^{A=N}\; \int _{V}r_{A} \, dV \quad =\quad 0$$ ([ZEqnNum085083])

The summation over all N molecular species can be interchanged with the volume integration in this representation of the second axiom, and this allows us to express Eq. as

$$\int _{V}\sum _{A=1}^{A=N}\; r_{A} \, dV \quad =\quad 0$$ ()

Since the volume $$V$$ is arbitrary, the integrand must be zero and we extract the preferred form of the second axiom given by

Axiom II: $$\sum _{A=1}^{A=N}r_{A} \quad =\quad 0$$ ([ZEqnNum636688])

In Eqs. [ZEqnNum819624] and [ZEqnNum914545] we have used a mixed mode nomenclature to represent the chemical species, i.e., we have used both letters and numbers simultaneously. Traditionally, we use upper case Roman letters to designate various chemical species, thus the rates of production for species A, B and C are designated by $$r_{A}$$, $$r_{B}$$ and $$r_{C}$$. When dealing with systems containing N different molecular species, we allow an indicator, such as A or D or G, to take on values from 1 to N in order to produce compact forms of the two axioms given by Eqs. [ZEqnNum473475] and [ZEqnNum304328]. We could avoid this mixed mode nomenclature consisting of letters and numbers by expressing Eq. [ZEqnNum905675] in the form;

Axiom II: $$r_{A} \; \; +\; \; r_{B} \; \; +\; \; r_{C} \; \; +\; \; r_{D} \; \; +\; \; ....\; \; +\; \; r_{N} \quad =\quad 0$$ ()

however, this approach is rather cumbersome when dealing with N-component systems.

The concept that mass is neither created nor destroyed by chemical reactions (as indicated by Eq. [ZEqnNum049988]) is based on the work of Lavoisier1 who stated:

“We observe in the combustion of bodies generally four recurring phenomena which would appear to be invariable laws of nature; while these phenomena are implied in other memoirs which I have presented, I must recall them here in a few words.”

Lavoisier went on to list four phenomena associated with combustion, the third of which was given by

Third Phenomenon. In all combustion, pure air in which the combustion takes place is destroyed or decomposed and the burning body increases in weight exactly in proportion to the quantity of air destroyed or decomposed.

It is this Third Phenomenon, when extended to all reacting systems, that supports Axiom II in the form represented by Eq. [ZEqnNum085083]. The experiments that led to the Third Phenomenon were difficult to perform in the 18 $${}^{th}$$ century and those difficulties have been recounted by Toulmin2.

4.1.1 Molar concentration and molecular mass

When chemical reactions occur, it is generally more convenient to work with the molar form of Eqs. [ZEqnNum443376] and [ZEqnNum733074]. The appropriate measure of concentration is then the molar concentration defined by

$$c_{A} \quad =\quad {\rho _{A} \mathord{\left/ {\vphantom {\rho _{A} MW_{A} }} \right. \kern-\nulldelimiterspace} MW_{A} } \quad =\quad \left\{\begin{array}{l} {moles{\rm \; of\; species\; }A} \\ {\rm \; \; per\; unit\; volume} \end{array}\right\}$$ ()

while the appropriate net rate of production for species A is given by DSL. $$R_{A} \quad =\quad {r_{A} \mathord{\left/ {\vphantom {r_{A} MW_{A} }} \right. \kern-\nulldelimiterspace} MW_{A} } \quad =\quad \left\{\begin{array}{l} {\rm net\; }molar{\rm \; rate\; of\; production} \\ {\rm per\; unit\; volume\; of\; species\; }A} \\ {{\rm owing\; to\; chemical\; reactions} \end{array}\right\}$$ ()

Here $$MW_{A}$$ represents the molecular mass3 of species A that is given explicitly by

$$MW_{A} \quad =\quad \frac{{\rm kilograms\; of\; }A}{{\rm moles\; of\; }A}$$ ([ZEqnNum709487])

The numerical values of the molecular mass are obtained from the atomic masses associated with any particular molecular species, and values for both the atomic mass and the molecular mass are given in Tables A1 and A2 in Appendix A. In those tables we have represented the atomic mass and the molecular mass in terms of grams per mole, thus the definition given by Eq. [ZEqnNum709487] for water leads to

$$MW_{\rm H}_{\rm 2} {\rm O} \quad =\quad \frac{0.018015{\rm \; kg}}{\rm mol} \quad =\quad \frac{18.015{\rm \; g}}{\rm mol}$$ ()

In terms of $$c_{A}$$ and $$R_{A}$$, the two axioms given by Eqs. [ZEqnNum888263] and [ZEqnNum573841] take the form

Axiom I: $$\frac{d}{dt} \int _{V}c_{A} \, dV \; \; +\; \; \int _{A}c_{A} v_{A} \cdot n\, dA \quad =\quad \int _{V}R_{A} \, dV \; ,\quad A=1,\; 2,\; ....,\; N$$ ([ZEqnNum545075])

Axiom II: $$\sum _{A\; =\; 1}^{A\; =\; N}MW_{A} R_{A} \quad =\quad 0$$ ()

Here it is important to note that mass is conserved during chemical reactions while moles need not be conserved. For example, the decomposition of calcium carbonate (solid) to calcium oxide (solid) and carbon dioxide (gas) is described by

$$\left({\rm CaCO}_{\rm 3} \right)_{solid} \quad \to \quad \left({\rm CaO}\right)_{solid} \; \; +\; \; \left({\rm CO}_{\rm 2} \right)_{gas}$$ ()

thus one mole is consumed and two moles are produced by this chemical reaction.

One must be very careful to understand that the net molar rate of production per unit volume of species A owing to chemical reactions, $$R_{A}$$, may be the result of many different chemical reactions. For example, in the chemical production system illustrated in Figure 4-1, carbon dioxide may be produced by the Figure 4-1. Chemical production system

oxidation of carbon monoxide, by the complete combustion of methane, or by other chemical reactions taking place within the control volume illustrated in Figure 4-1. The combination of all these individual chemical reactions is represented by $$R_{\, {\rm CO}_{\rm 2} }$$. It is important to note that in Figure 4-1 we have suggested the stoichiometry of the reactions taking place while the actual chemical kinetic processes taking place may be much more complicated. The subject of stoichiometry will be discussed in detail in Chapter 6, and a brief discussion of chemical kinetics is given in Section 8.6.

4.1.2 Moving control volumes

In Sec. 3.3 we developed the macroscopic mass balance for a single component system in terms of an arbitrary moving control volume, $$V_{a} (t)$$. The speed of displacement of the surface of a moving control volume is given by $$w\cdot n$$, and the flux of species A that crosses this moving surface is given by the normal component of the relative velocity for species A, i.e., $$(v_{A} -w)\cdot n$$. On the basis of this concept, we can express the first axiom for the mass of species A in the form

Axiom I: $$\frac{d}{dt} \int _{V_{a} (t)}\rho _{A} \, dV \; \; +\; \; \int _{A_{a} (t)}\rho _{A} (v_{A} -w)\cdot n\, dA \quad =\quad \int _{V_{a} (t)}r_{A} \, dV \; ,\quad \quad A=1,\; 2,\; ....,\; N$$ ([ZEqnNum996159])

When it is convenient to work with molar quantities, we divide this result by the molecular mass of species A in order to obtain the moving control volume form given by

Axiom I: $$\frac{d}{dt} \int _{V_{a} (t)}c_{A} \, dV \; \; +\; \; \int _{A_{a} (t)}c_{A} (v_{A} -w)\cdot n\, dA \quad =\quad \int _{V_{a} (t)}R_{A} \, dV \; ,\quad \quad A\; \; =\; \; 1,\; 2,\; ....,\; N$$ ([ZEqnNum863340])

Throughout this chapter, we will restrict our studies to fixed control volumes in order to focus our attention on the new concepts associated with multicomponent systems. However, the world of chemical engineering is filled with moving, dynamic systems and the analysis of those systems will require the use of Eqs. [ZEqnNum645577] and [ZEqnNum863340].

4.2 Species Mass Density

The mass of species A per unit volume in a mixture of several components is known as the species mass density, and it is represented by $$\rho _{A}$$. The species mass density can range from zero, when no species A is present in the mixture, to the density of pure species A, when no other species are present. In order to understand what is meant by the species mass density, we consider a mixing process in which three pure species are combined to create a uniform mixture of species A, B, and C. This mixing process is illustrated in Figure 4-2 where we have indicated that three pure species are combined to create a uniform mixture having a measured volume of $$45\; cm^{3}$$. The total volume of the three pure species is $$50\; cm^{3}$$, thus there is a change of volume upon mixing as is usually the case with liquids. We denote this change of volume upon mixing by $$\Delta V_{mix}$$, and for the process illustrated in Figure 4-2 we express this quantity as

$$\Delta V_{mix} \quad =\quad V\; \; -\; \; \left(V_{A} +V_{B} +V_{C} \right)$$ ()

The densities of the pure species have been denoted by a superscript zero, thus $$\rho _{A}^{\rm o}$$ represents the mass density of pure species A. The species mass density of species A is defined by

$$\left\{\begin{array}{l} {species\; mass\; density} \\ {\rm \; \; \; of\; species\; }A} \end{array}\right\}\quad =\quad {\left({\rm mass\; of\; species\; }A\right)\mathord{\left/ {\vphantom {\left({\rm mass\; of\; species\; }A\right) \left(\begin{array}{l} {\rm volume\; in\; which\; species\; }A} \\ {{\rm \; \; \; \; \; \; \; \; is\; contained} \end{array}\right)} \right. \kern-\nulldelimiterspace} \left(\begin{array}{l} {\rm volume\; in\; which\; species\; }A} \\ {{\rm \; \; \; \; \; \; \; \; is\; contained} \end{array}\right)}$$ ()

and this definition applies to mixtures in which species A is present as well as to the case of pure species A. Figure 4-2. Mixing process

If we designate the mass of species A as $$m_{A}$$ and the volume of the uniform mixture as V, the species mass density can be expressed as

$$\rho _{A} \quad =\quad {m_{A} \mathord{\left/ {\vphantom {m_{A} V}} \right. \kern-\nulldelimiterspace} V}$$ ()

For the mixing process illustrated in Figure 4-2, we are given that the mass of species A is

$$m_{A} \quad =\quad \rho _{A}^{\rm o} \, V_{A} \quad =\quad (0.85\, g/cm^{3} )(15\, cm^{3} )\quad =\quad 12.75\; g$$ ()

and this allows us to determine the species mass density in the mixture according to

$$\rho _{A} \quad =\quad {m_{A} \mathord{\left/ {\vphantom {m_{A} V}} \right. \kern-\nulldelimiterspace} V} \quad =\quad \frac{12.75\, {\rm g}}{\rm 45}\_ \, {\rm cm}^{\rm 3} } \quad =\quad 0.283\, {\rm g}\mathord{\left/ {\vphantom {{\rm g} {\rm cm}^{{\rm 3} } \right. \kern-\nulldelimiterspace} {\rm cm}^{\rm 3} } \quad$$ ()

This type of calculation can be carried our for species B and C in order to determine $$\rho _{B}$$ and $$\rho _{C}$$.

The total mass density is simply the sum of all the species mass densities and is defined by

$$\left\{\begin{array}{l} {\rm total\; mass} \\ {\rm \; \; density} \end{array}\right\}\quad =\quad \rho \quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\rho _{A}$$ ([ZEqnNum971008])

The total mass density can be determined experimentally by measuring the mass, m, and the volume, V, of a mixture. For any a particular mixture, it is difficult to measure directly the species mass density; however, one can prepare a mixture in which the species mass densities can be determined as we have suggested in Figure 4-2. When working with molar forms, we often need the total molar concentration and this is defined by

$$\left\{\begin{array}{l} {\rm \; \; total\; molar} \\ {\rm concentration} \end{array}\right\}\quad =\quad c\quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}c_{A}$$ ([ZEqnNum863721])

4.2.1 Mass fraction and mole fraction

For solid and liquid systems it is sometimes convenient to use the mass fraction as a measure of concentration. The mass fraction of species A can be expressed in words as

$$\omega _{A} \quad =\quad \left\{\begin{array}{l} {\rm \; mass\; of\; species\; }A{\rm \; per} \\ {\rm unit\; mass\; of\; the\; mixture} \end{array}\right\}$$ ()

and in precise mathematical form we have

$$\omega _{A} \quad =\quad \frac{\rho _{A} }{\rho } \quad =\quad {\rho _{A} \mathord{\left/ {\vphantom {\rho _{A} \sum _{G\; =\; 1}^{G\; =\; N}\rho _{G} }} \right. \kern-\nulldelimiterspace} \sum _{G\; =\; 1}^{G\; =\; N}\rho _{G} }$$ ([ZEqnNum363676])

Note that the indicator, G, is often referred to as a dummy indicator since any letter would suffice to denote the summation over all species in the mixture. In this particular case, we would not want to use A as the dummy indicator since this could lead to confusion. The mole fraction is analogous to the mass fraction and is defined by

$$x_{A} \quad =\quad \frac{c_{A} }{c} \quad =\quad {c_{A} \mathord{\left/ {\vphantom {c_{A} \sum _{G\; =\; 1}^{G\; =\; N}c_{G} }} \right. \kern-\nulldelimiterspace} \sum _{G\; =\; 1}^{G\; =\; N}c_{G} }$$ ([ZEqnNum159844])

If one wishes to avoid the mixed-mode nomenclature in Eqs. [ZEqnNum363676] and [ZEqnNum887852], one must express the mass fraction as

$$\omega _{A} \quad =\quad \frac{\rho _{A} }{\rho } \quad =\quad \frac{\rho _{A} }{\rho _{A} \; \; +\; \; \rho _{B} \; \; +\; \; \rho _{C} \; \; +\; \; \rho _{D} \; \; +\; \; ....\; \; +\; \; \rho _{N} }$$ ()

while the mole fraction takes the form

$$y_{A} \quad =\quad \frac{c_{A} }{c} \quad =\quad \frac{c_{A} }{c_{A} \; \; +\; \; c_{B} \; \; +\; \; c_{C} \; \; +\; \; c_{D} \; \; +\; \; ....\; \; +\; \; c_{N} }$$ ([ZEqnNum881605])

Very often $$x_{A}$$ is used to represent mole fractions in liquid mixtures and $$y_{A}$$ to represent mole fractions in vapor mixtures, thus Eq. [ZEqnNum881605] represents the mole fraction in a vapor mixture while Eq. [ZEqnNum573667] represents the mole fraction in a liquid mixture.

EXAMPLE 4.1. Conversion of mole fractions to mass fractions

Sometimes we may be given the composition of a mixture in terms of the various mole fractions and require the mass fractions of the various constituents. To convert from $$x_{A}$$ to $$\omega _{A}$$ we proceed as follows:

$$x_{A} \quad =\quad {c_{A} \mathord{\left/ {\vphantom {c_{A} c}} \right. \kern-\nulldelimiterspace} c}$$ (l) $\label{GrindEQ__2_} c_{A} \quad =\quad x_{A} \, c$ $\label{GrindEQ__3_} \rho _{A} \quad =\quad MW_{A} \; c_{A} \quad =\quad MW_{A} \; x_{A} \; c$ $\label{GrindEQ__4_} \rho \quad =\quad \sum _{G\; =\; 1}^{G\; =\; N}\rho _{G} \quad =\quad \left(\sum _{G\; =\; 1}^{G\; =\; N}MW_{G} \; x_{G} \right)\; c$ $\label{GrindEQ__5_} \omega _{A} \quad =\quad \frac{\rho _{A} }{\rho } \quad =\quad \frac{MW_{A} \; x_{A} \; c}{\left(\sum _{G=1}^{G=N}MW_{G} \; x_{G} \right)\; c} \quad =\quad \frac{MW_{A} \; x_{A} }{\sum _{G=1}^{G=N}MW_{G} \; x_{G} }$

4.2.2 Total mass balance

Given the total density defined by Eq. , we are ready to recover the total mass balance for multicomponent, reacting systems. For a fixed control volume, this is developed by summing Eq. over all species to obtain

$$\sum _{A\; =\; 1}^{A\; =\; N}\frac{d}{dt} \int _{V}\rho _{A} \, dV \; \; +\; \; \sum _{A\; =\; 1}^{A\; =\; N}\int _{A}\rho _{A} \, v_{A} \cdot n\, dA \quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\int _{V}r_{A} \, dV$$ ()

The summation procedure can be interchanged with differentiation and integration so that this result takes the form

$$\frac{d}{dt} \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}\rho _{A} \, dV \; \; +\; \; \int _{A}\sum _{A\; =\; 1}^{A\; =\; N}\rho _{A} \, v_{A} \cdot n\, dA \quad =\quad \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}r_{A} \, dV$$ ()

On the basis of definition of the total mass density given by Eq. [ZEqnNum882543] and the axiom given by Eq. [ZEqnNum105047], this result simplifies to

$$\frac{d}{dt} \int _{V}\rho \, dV \; \; +\; \; \int _{A}\sum _{A\; =\; 1}^{A\; =\; N}\rho _{A} \, v_{A} \cdot n\, dA \quad =\quad 0$$ ([ZEqnNum536076])

At this point, we define the total mass flux according to

$$\left\{\begin{array}{l} {\rm \; \; \; total} \\ {\rm mass\; flux} \end{array}\right\}\quad =\quad \rho \, v\quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\rho _{A} v_{A}$$ ([ZEqnNum419769])

Since  is defined by Eq. , this result actually represents a definition of the velocity v that can be expressed as

$$v\quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\omega _{A} v_{A}$$ ([ZEqnNum731232])

This velocity is known as the mass average velocity and it plays a key role both in our studies of macroscopic mass balances and in subsequent studies of fluid mechanics, heat transfer, and mass transfer. Use of this definition for the mass average velocity allows us to express Eq. [ZEqnNum536076] as

$$\frac{d}{dt} \int _{V}\rho dV \; \; +\; \; \int _{A}\rho v\cdot ndA\quad =\quad 0$$ ([ZEqnNum386474])

This is identical in form to the mass balance for a fixed control volume that was presented in Chapter 3; however, this result has greater physical content than our previous result for single-component systems. In this case, the density is not the density of a single component but the sum of all the species densities as indicated by Eq. and the velocity is not the velocity of a single component but the mass average velocity defined by Eq. .

4.3 Species Velocity

In our representation of the axioms for the mass of multicomponent systems, we have introduced the concept of a species velocity indicating that individual molecular species move at their own velocities designated by $$v_{A}$$ where $$A=1,\; 2,\; ..\, N$$. In order to begin thinking about the species velocity, we consider a lump of sugar (species A) placed in the bottom of a tea cup which is very carefully filled with water (species B). If we wait long enough, the solid sugar illustrated in Figure 4-3 will dissolve and become Figure 4-3. Dissolution of sugar

uniformly distributed throughout the cup. This is a clear indication that the velocity of the sugar molecules is different from the velocity of the water molecules, i.e.,

$$v_{Sugar} \quad \ne \quad v_{Water}$$ ()

If the solution in the cup is not stirred, the velocity of the sugar molecules will be very small and the time required for the sugar to become uniformly distributed throughout the cup will be very long. We generally refer to this process as diffusion and diffusion velocities are generally very small. If we stir the liquid in the teacup, the sugar molecules will be transported away from the sugar cube by convection as we have illustrated in Figure 4-4. In this case, the sugar will become uniformly distributed throughout the cup in a relatively short time and we generally refer to this process as mixing. Mechanical mixing can accelerate the process by which the sugar becomes uniformly distributed throughout the teacup; however, a true mixture of sugar and water could never be achieved unless the velocities of the two species were different. The difference between species velocities is crucial. It is responsible for mixing, for separation and purification, and it is necessary for chemical reactions to occur. If all species velocities were equal, life on earth would cease immediately.

In addition to mixing the sugar and water as indicated in Figures 4-3 and 4-4, we can also separate the sugar and water by allowing the water to evaporate. In that case all the water in the tea cup would appear in the surrounding air and the sugar would remain in the bottom of the cup. This separation would not be possible unless the velocity of the water were different than the velocity of the sugar. While the difference between species velocities is of crucial interest to chemical engineers, there is a class of problems for which we can ignore this difference and still obtain useful results. In the following paragraphs we want to identify this class of problems. Figure 4-4. Mixing of sugar

To provide another example of the difference between species velocities and the effect of diffusion, we consider the process of absorption of $${\rm SO}_{2}$$ in a falling film of water as illustrated in Figure 4-5. The gas Figure 4-5. Absorption of sulfur dioxide

mixture entering the column consists of air, which is essentially insoluble in water, and $${\rm SO}_{2}$$, which is soluble in water. Because of the absorption of $${\rm SO}_{2}$$ in the water, the exit gas is less detrimental to the local environment. It should be intuitively appealing that the species velocities in the axial direction are constrained by

$$v_{\rm SO}_{{\rm 2} } \cdot k\quad \approx \quad v_{air} \cdot k$$ ([ZEqnNum501761])

in which k is the unit vector pointing in the z-direction. The situation for the radial components of $$v_{\rm SO}_{{\rm 2} }$$ and $$v_{air}$$ is quite different because the radial components are normal to the gas-liquid interface. Since the air is insoluble in water, the component of $$v_{air}$$ in the radial direction must be zero at the gas-liquid interface, i.e.,

$$v_{air} \cdot n\quad =\quad 0\; ,\quad {\rm at\; the\; gas-liquid\; interface}$$ ([ZEqnNum911488])

On the other hand, the sulfur dioxide is crossing the interface as it leaves the gas stream and enters the liquid stream. The radial component of $$v_{{\rm SO}_{2} }$$ must therefore be positive and we express this idea as

$$v_{{\rm SO}_{2} } \cdot n\quad >\quad 0\; ,\quad {\rm at\; the\; gas-liquid\; interface}$$ ()

It should be clear that $$v_{{\rm SO}_{2} } \cdot n$$ is a velocity associated with a diffusion process while $$v_{{\rm SO}_{2} } \cdot k$$ is a velocity associated with a convection process and that the latter is generally much, much larger than the former, i.e.,

$$\underbrace{v_{\rm SO}_{2} } \cdot k}_{{\rm convection}\quad >>\quad \underbrace{v_{\rm SO}_{2} } \cdot n}_{{\rm diffusion}$$ ()

The motion of a chemical species can result from a force applied to the fluid, i.e., a fan might be used to move the gas mixture through the tube illustrated in Figure 4-5. The motion of a chemical species can also result from a concentration gradient such as the gradient that causes the sugar to diffuse throughout the teacup illustrated in Figure 4-3. Because the motion of chemical species can be caused by both applied forces and concentration gradients, it is reasonable to decompose the species velocity into two parts: the mass average velocity and the mass diffusion velocity. We represent this decomposition as

$$v_{\rm SO}_{2} } \quad \quad =\quad \quad \underbrace{\; \; v\; \; }_{\begin{array}{l} {\rm mass\; average} \\ {\rm \; \; \; velocity} \end{array}\quad +\quad \underbrace{\; \; u_{\kern 1pt} {\rm SO}_{2} } \; \; }_{\begin{array}{l} {\rm mass\; diffusion} \\ {\rm \; \; \; \; \; velocity} \end{array}$$ ([ZEqnNum665161])

At entrances and exits, such as those illustrated in Figure 4-5, the diffusion velocity in the z-direction is usually small compared to the mass average velocity in the z-direction and Eq. can be approximated by

$$v_{{\rm SO}_{2} } \cdot k\quad =\quad v\cdot k\; ,\quad {\rm negligible\; diffusion\; velocity}$$ ([ZEqnNum707604])

In this text we will repeatedly make use of this simplification in order to solve a variety of problems without the need to predict the diffusion velocity. However, in subsequent courses mass transfer across fluid-fluid interfaces will be studied in detail, and that study will require a complete understanding of the role of the diffusion velocity for a variety of processes.

In order to reinforce our thoughts about the species velocity, the mass average velocity, and the mass diffusion velocity, we return to the definition of the mass average velocity given by Eq. [ZEqnNum374801] and express the z-component of v according to

$$v\cdot k\quad =\quad \omega {\kern 1pt} _{{\rm SO}_{2} } (v_{{\rm SO}_{2} } \cdot k)\; \; +\; \; \omega _{air} (v_{air} \cdot k)$$ ()

in which the mass fractions are constrained by

$$\omega _{{\kern 1pt} {\rm SO}_{2} } \; \; +\; \; \omega _{air} \quad =\quad 1$$ ()

On the basis of the approximation given by Eq. [ZEqnNum501761] we conclude that

$$v\cdot k\quad \approx \quad v_{{\rm SO}_{2} } \cdot k\quad \approx \quad v_{air} \cdot k$$ ()

However, the radial component of the species velocities is an entirely different matter. Once again we can use Eq. [ZEqnNum593885] to obtain

$$v\cdot n\quad =\quad \omega {\kern 1pt} _{{\rm SO}_{2} } (v_{{\rm SO}_{2} } \cdot n)\; \; +\; \; \omega _{air} (v_{air} \cdot n)\; ,\quad \quad {\rm at\; the\; gas-liquid\; interface}$$ ()

but on the basis of Eq. [ZEqnNum911488] this reduces to

$$v\cdot n\quad =\quad \omega {\kern 1pt} _{{\rm SO}_{2} } (v_{{\rm SO}_{2} } \cdot n)\; ,\quad \quad {\rm at\; the\; gas-liquid\; interface}$$ ()

Under these circumstances we see that

$$v\cdot n\quad \ne \quad v_{{\rm SO}_{2} } \cdot n\quad \ne \quad v_{air} \cdot n\; ,\quad \quad {\rm at\; the\; gas-liquid\; interface}$$ ()

and the type of approximation indicated by Eq. [ZEqnNum707604] is not valid. In this text, we will study a series of macroscopic mass balance problems for which Eq. [ZEqnNum566719] represents a reasonable approximation; however, one must always remember that neglect of the diffusion velocity is a very delicate matter, and it is considered further in Problem 4-10. As we have mentioned before, the difference between species velocities is responsible for separation and purification, and it absolutely is necessary in order for chemical reactions to occur.

4.4 Measures of Velocity

In the previous section we defined the mass average velocity according to

$$v\quad =\quad \sum _{B\; =\; 1}^{B\; =\; N}\omega _{B} \, v_{B} \; ,\quad {\rm mass\; average\; velocity}$$ ()

and we noted that the total mass flux vector was given by

$$\rho v\quad =\quad \sum _{B\; =\; 1}^{B\; =\; N}\rho _{B} \, v_{B} \quad =\quad \left\{\begin{array}{l} {\rm \; total\; mass} \\ {\rm flux\; vector} \end{array}\right\}$$ ()

By analogy we define the molar average velocity by

$$v^{*} \quad =\quad \sum _{B\; =\; 1}^{B\; =\; N}x_{B} \, v_{B} \; ,\quad {\rm molar\; average\; velocity}$$ ()

and it follows that the total molar flux vector is given by

$$c\; v^{*} \quad =\quad \sum _{B\; =\; 1}^{B\; =\; N}c_{B} \, v_{B} \quad =\quad \left\{\begin{array}{l} {\rm total\; molar} \\ {\rm flux\; vector} \end{array}\right\}$$ ([ZEqnNum482548])

In the previous section we used a decomposition of the species velocity of the form

$$v_{A} \quad =\quad v\; \; +\; \; u_{A}$$ ()

so that the species mass flux vector could be expressed as

$$\rho _{A} v_{A} \quad =\quad \underbrace{\; \rho _{A} \, v\; }_{\begin{array}{l} {\rm convective} \\ {\rm \; \; \; \; \; flux} \end{array}}\; \; +\; \; \underbrace{\; \rho _{A} \, u_{A} \; }_{\begin{array}{l} {\rm diffusive} \\ {\rm \; \; \; flux} \end{array}}$$ ()

In dealing with the molar flux vector one finds it convenient to express the species velocity as

$$v_{A} \quad =\quad v^{*} \; \; +\; \; u_{A}^{*}$$ ()

so that the molar flux takes the form

$$c_{A} v_{A} \quad =\quad \underbrace{\; c_{A} \, v^{*} \; }_{\begin{array}{l} {\rm convective} \\ {\rm \; \; \; \; \; flux} \end{array}}\; \; +\; \; \underbrace{\; c_{A} \, u_{A}^{*} \; }_{\begin{array}{l} {\rm diffusive} \\ {\rm \; \; \; flux} \end{array}}$$ ()

When convective transport dominates, the species velocity, the mass average velocity and the molar average velocity are all essentially equal, i.e.,

$$v_{A} \quad =\quad v\quad =\quad v^{*}$$ ([ZEqnNum467879])

This is the situation that we encounter most often in our study of material balances and we will make use of this result repeatedly to determine the flux of species A at entrances and exits. While Eq. [ZEqnNum467879] is widely used to describe velocities at entrances and exits, one must be very careful about the general use of this approximation. If Eq. [ZEqnNum848461] were true for all species under all conditions, there would be no separation, no purification, no mixing, and no chemical reactions!

Under certain circumstances we may want to use a total mole balance and this is obtained by summing Eq. [ZEqnNum545075] over all N species in order to obtain

$$\frac{d}{dt} \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}c_{A} \, dV \; \; +\; \; \int _{A}\sum _{A\; =\; 1}^{A\; =\; N}{\rm (}c_{A} \, v_{A} {\rm )} \cdot n\, dA \quad =\quad \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}R_{A} \, dV$$ ()

Use of the definitions given by Eqs. [ZEqnNum863721] and [ZEqnNum482548] allows us to write this result in the form

$$\frac{d}{dt} \int _{V}c\, dV \; \; +\; \; \int _{A}c\, v^{*} \cdot n\, dA \quad =\quad \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}R_{A} \, dV$$ ()

Here we should note that the overall net rate of production of moles need not be zero, thus the overall mole balance is more complex than the overall mass balance.

4.5 Molar Flow Rates at Entrances and Exits

Here we direct our attention to the macroscopic mole balance for a fixed control volume

$$\frac{d}{dt} \int _{V}c_{A} \, dV \; \; +\; \; \int _{A}c_{A} \, v_{A} \cdot n\, dA \quad =\quad \int _{V}R_{A} \, dV \; ,\quad \quad A=1,{\rm \; 2,\; ....\; }N$$ ([ZEqnNum839730])

with the intention of evaluating $$c_{A} \, v_{A} \cdot n$$ at entrances and exits. The area integral of the molar flux, $$c_{A} v_{A} \cdot n$$, can be represented as

$$\int _{A}c_{A} \, v_{A} \cdot n\, dA \quad =\quad \int _{A_{\rm e} }c_{A} \, v_{A} \cdot n\, dA \; \; +\; \; \int _{A_{i} }c_{A} \, v_{A} \cdot n\, dA$$ ([ZEqnNum349015])

where $$A_{\rm e}$$ represents the entrances and exits at which convection dominates and $$A_{i}$$ represents an interfacial area over which diffusive fluxes may dominate. In this text, our primary interest is the study of control volumes having entrances and exits at which convective transport is much more important than diffusive transport, and we have illustrated this type of control volume in Figure 4-6. There the entrances and exits for both the water and the air are at the top and bottom of the column, while the surface of the control volume that coincides with the liquid-solid interface represents an impermeable boundary at which $$c_{A} \, v_{A} \cdot n=0$$. For systems of this type, we express Eq. [ZEqnNum349015] as

$$\int _{A}c_{A} v_{A} \cdot n\, dA \quad =\quad \int _{A_{\rm e} }c_{A} v_{A} \cdot n\, dA \quad =\quad \int _{A_{\rm e}{\rm n}{\rm t}{\rm r}{\rm a}{\rm n}{\rm c}{\rm es} }c_{A} v_{A} \cdot n\, dA \; \; +\; \; \int _{A_{\rm exits} }c_{A} v_{A} \cdot n\, dA$$ ()

In order to simplify our discussion about the flux at entrances and exits, we direct our attention to an exit and express the molar flow rate at that exit as

$$\dot{M}_{A} \quad =\quad \int _{A_{\rm exit} }c_{A} v_{A} \cdot n\, dA$$ ()

On the basis of the discussion in Sec. 4.2, we assume that the diffusive flux is negligible ( $$v_{A} \cdot n\approx v\cdot n$$) so that the above result takes the form

$$\dot{M}_{A} \quad =\quad \int _{A_{\rm exit} }c_{A} v\cdot n\, dA$$ ([ZEqnNum713441])

It is possible that both $$c_{A}$$ and v vary across the exit and a detailed evaluation of the area integral is required in order to determine the molar flow rate of species A. In general this is not the case; however, it is very important to be aware of this possibility. Figure 4-6. Entrances and exits at which convection dominates

4.5.1 Average concentrations

In Sec. 3.2.1 we defined a volume average density and we use the same definition here for the volume average concentration given by

$$\langle c_{A} \rangle \quad =\quad \frac{1}{V} \int _{V}c_{A} \, dV \; ,\quad \quad {\rm volume\; average\; concentration}$$ ()

At entrances and exits, we often work with the “bulk concentration” or “cup mixed concentration” that was defined earlier in Sec. 3.2.1. For the concentration, $$c_{A}$$, we repeat the definition according to

$$\langle c_{A} \rangle _{b} \quad =\quad \frac{1}{Q_{exit} } \int _{A_{\, exit} }c_{A} \, v\cdot n\, dA \quad =\quad \frac{\int _{A_{\, exit} }c_{A} \, v\cdot n\, dA }{\int _{A_{\, exit} }v\cdot n\, dA }$$ ([ZEqnNum182517])

In terms of the bulk concentration the molar flow rate given by Eq. [ZEqnNum223174] can be expressed as

$$\dot{M}_{A} \quad =\quad \langle c_{A} \rangle _{b} \; Q_{exit}$$ ([ZEqnNum453862])

in which it is understood that $$\dot{M}_{A}$$ and $$\langle c_{A} \rangle _{b}$$ represent the molar flow rate and concentration at the exit. In addition to the bulk or cup-mixed concentration, one may encounter the area average concentration denoted by $$\langle c_{A} \rangle$$ and defined at an exit according to

$$\langle c_{A} \rangle \quad =\quad \frac{1}{A_{\, exit} } \int _{A_{\, exit} }c_{A} \, dA$$ ()

If the concentration is constant over $$A_{exit}$$, the area average concentration is equal to this constant value, i.e.

$$\langle c_{A} \rangle \quad =\quad c_{A} \; ,\quad when{\rm \; }c_{A} {\rm \; }is\; constant$$ ()

We often refer to this condition as a “flat” concentration profile, and for this case we have

$$\langle c_{A} \rangle _{b} \quad =\quad \langle c_{A} \rangle \quad =\quad c_{A} \; ,\quad \quad flat\; concentration\; profile$$ ()

Under these circumstances the molar flow rate takes the form

$$\dot{M}_{A} \quad =\quad c_{A} \; Q_{exit} \; ,\quad \quad flat\; concentration\; profile$$ ([ZEqnNum983745])

The conditions for which $$c_{A}$$ can be treated as a constant over an exit or an entrance are likely to occur in many practical applications.

When the flow is turbulent, there are rapid velocity fluctuations about the mean or time-averaged velocity. The velocity fluctuations tend to create uniform velocity profiles and they play a crucial role in the transport of mass orthogonal to the direction of the mean flow. The contribution of turbulent fluctuations to mass transport parallel to the direction of the mean flow can normally be neglected and we will do so in our treatment of macroscopic mass balances. In a subsequent courses on fluid mechanics and mass transfer, the influence of turbulence will be examined more carefully. In our treatment, we will make use of the reasonable approximation that the turbulent velocity profile is flat and this means that $$v\cdot n$$ is constant over $$A_{\, exit}$$. Both turbulent and laminar velocity profiles are illustrated in Figure 4-7 and there we #### Figure 4-7. Laminar and turbulent velocity profiles for flow in a tube

see that the velocity for turbulent flow is nearly constant over a major portion of the flow field. If we make the “flat velocity profile” assumption, we can express Eq. [ZEqnNum182517] as

$$\langle c_{A} \rangle _{b} \; \quad =\quad \; \frac{\int _{A_{\, exit} }c_{A} \, v\cdot n\, dA }{\int _{A_{\, exit} }v\cdot n\, dA } \; \quad =\quad \; \frac{\int _{A_{\, exit} }c_{A} \, dA }{\int _{A_{\, exit} }dA } \; \frac{v\cdot n}{v\cdot n} \; \quad =\quad \; \frac{1}{A_{\, exit} } \int _{A_{\, exit} }c_{A} \, dA \; \quad =\quad \; \langle c_{A} \rangle$$ ()

For this case, the molar flow rate at the exit takes the form

$$\dot{M}_{A} \quad =\quad \langle c_{A} \rangle Q_{exit} \; ,\quad \quad flat\; velocity\; profile$$ ([ZEqnNum294219])

To summarize, we note that Eq. [ZEqnNum453862] is an exact representation of the molar flow rate in terms of the bulk concentration and the volumetric flow rate. When the concentration profile can be approximated as flat, the molar flow rate can be represented in terms of the constant concentration and the volumetric flow rate

as indicated by Eq. [ZEqnNum230041]. When the velocity profile can be approximated as flat, the molar flow rate can be represented in terms of the area average concentration and the volumetric flow rate as indicated by Eq. [ZEqnNum272495]. If one is working with the species mass balance given by Eq. [ZEqnNum387174], the development represented by Eqs. [ZEqnNum839730] through [ZEqnNum366974] can be applied simply by replacing $$c_{A}$$ with $$\rho _{A}$$.

4.6 Alternate Flow Rates

There are a number of relations between species flow rates and total flow rates that are routinely used in solving macroscopic mass or mole balance problems provided that either the velocity profile is flat or the concentration profile is flat. For example, we can always write Eq. [ZEqnNum279695] in the form

$$\dot{M}_{A} \quad =\quad \int _{A_{exit} }c_{A} v\cdot n\, dA \quad =\quad \int _{A_{exit} }\frac{c_{A} }{c} c\, v\cdot n\, dA \quad =\quad \int _{A_{exit} }x_{A} \, c\, v\cdot n\, dA$$ ([ZEqnNum974643])

If either $$c\, v\cdot n$$ or $$x_{A}$$ is constant over the area of the exit, we can express this result as

$$\dot{M}_{A} \quad =\quad \langle x_{A} \rangle \dot{M}$$ ([ZEqnNum189643])

where $$\dot{M}$$ is the total molar flow rate defined by

$$\dot{M}\quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\dot{M}_{A}$$ ([ZEqnNum181666])

If the individual molar flow rates are known and one desires to determine the area averaged mole fraction at an entrance or an exit, it is given by

$$\langle x_{A} \rangle \quad =\quad {\dot{M}_{A} \mathord{\left/ {\vphantom {\dot{M}_{A} \sum _{B\; =\; 1}^{B\; =\; N}\dot{M}_{B} }} \right. \kern-\nulldelimiterspace} \sum _{B\; =\; 1}^{B\; =\; N}\dot{M}_{B} }$$ ()

provided that either $$c\, v\cdot n$$ or $$x_{A}$$ is constant over the area of the entrance or the exit. It will be left as an exercise for the student to show that similar relations exist between mass fractions and mass flow rates. For example, a form analogous to Eq. [ZEqnNum189643] is given by

$$\dot{m}_{A} \quad =\quad \langle \omega _{A} \rangle \, \dot{m}$$ ([ZEqnNum700852])

and the mass fraction at an entrance or an exit can be expressed as

$$\langle \omega _{A} \rangle \quad =\quad {\dot{m}_{A} \mathord{\left/ {\vphantom {\dot{m}_{A} \sum _{G\; =\; 1}^{G\; =\; N}\dot{m}_{G} }} \right. \kern-\nulldelimiterspace} \sum _{G\; =\; 1}^{G\; =\; N}\dot{m}_{G} }$$ ([ZEqnNum264342])

One must keep in mind that the results given by Eqs. [ZEqnNum779969] through [ZEqnNum264342] are only valid when either the concentration (density) is constant or the molar (mass) flux is constant over the entrance or exit.

When neither of these simplifications is valid, we express Eq. [ZEqnNum974643] as

$$\dot{M}_{A} \quad =\quad \int _{A_{\rm exit} }x_{A} \, c\, v\cdot n\, dA \quad =\quad \langle x_{A} \rangle _{b} \dot{M}$$ ()

where $$\langle x_{A} \rangle _{b}$$ is the cup mixed mole fraction of species A. The definition of the mole fraction requires that

$$\sum _{A\; =\; 1}^{A\; =\; N}x_{A} \quad =\quad 1$$ ([ZEqnNum276080])

and it will be left as an exercise for the student to show that

$$\sum _{A\; =\; 1}^{A\; =\; N}\langle x_{A} \rangle _{b} \quad =\quad 1$$ ([ZEqnNum359366])

This type of constraint on the mole fractions (and mass fractions) applies at every entrance and exit and it often represents an important equation in the set of equations that are used to solve macroscopic mass balance problems.

4.7 Species Mole/Mass Balance

In this section we examine the problem of solving the N equations represented by either Eq. [ZEqnNum694147] or Eq. [ZEqnNum506940] under steady-state conditions in the absence of chemical reactions. The distillation process illustrated in Figure 4-8 provides a simple example; however, most distillation processes are more complex than the one shown in Figure 4-8 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 1-5 and 1-6), thus understanding the simple process illustrated in Figure 4-8 is an important step in our studies. Figure 4-8 Distillation column

For this particular ternary distillation process, we are given the information listed in Table 4-1. 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 4-1, we have not specified $$x_{C}$$ since this mole fraction will be determined by Eq. [ZEqnNum276080]. 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 4-1. Specified conditions

In the previous chapter, we considered single-component systems for which there was a single density, , a single velocity, v, and no chemical reactions. In multicomponent systems we must deal with the density of individual species and this leads to the characterization of systems in terms of species mass densities and species molar concentrations, in addition to mass fractions and mole fractions. Not only must we characterize the composition of multicomponent systems, but we must also consider the fact that different molecular species move at different velocities. This leads us to the concept of the species velocity which plays a dominant role in the detailed study of separation and purification processes and in the analysis of chemical reactors. In this chapter we will discuss the concept of the species velocity and then illustrate how a certain class of macroscopic mass balance problems can be solved without dealing directly with this important aspect of multicomponent systems. While chemical reactions represent an essential feature of multicomponent systems, we will delay a thorough discussion of that matter until Chapter 6.

4.1 Axioms for the Mass of Multicomponent Systems

In Chapter 3, we studied the concept of conservation of mass for single-component systems, both for a body and for a control volume. The words associated with the control volume representation were

$$\left\{\begin{array}{l} {\rm time\; rate\; of\; change\; } \\ {\rm \; \; \; \; \; of\; mass\; in\; a\; \; } \\ {\rm \; \; \; control\; volume} \end{array}\right\}\quad =\quad \left\{\begin{array}{l} {\rm \; rate\; at\; which} \\ {\rm mass\; }enters{\rm \; the} \\ {\rm control\; volume} \end{array}\right\}\quad -\quad \left\{\begin{array}{l} {\rm \; rate\; at\; which} \\ {\rm mass\; }leaves{\rm \; the} \\ {\rm \; control\; volume} \end{array}\right\}\label{ZEqnNum921479}$$

and for a fixed control volume, the mathematical representation was given by

$$\frac{d}{dt} \int _{V}\rho \, dV \; \; +\; \; \int _{A}\rho v\cdot n\, dA \quad =\quad 0\label{ZEqnNum562507}$$

One must keep in mind that the use of vectors allows us to represent both the mass entering the control volume and mass leaving the control volume in terms of $$\rho v\cdot n$$. This follows from the fact that $$v\cdot n$$ is negative over surfaces where mass is entering the control volume and $$v\cdot n$$ is positive over surfaces where mass is leaving the control volume. In addition, one should remember that the control volume, $$V$$, in Eq. [ZEqnNum562507] is arbitrary and this allows us to choose the control volume to suit our needs.

Now we are ready to consider N-component systems in which chemical reactions can take place, and in this case we need to make use of the two axioms for the mass of multicomponent systems. The first axiom deals with the mass of species A, and when this species can undergo chemical reactions we need to extend Eq. to the form given by

$$\left\{\begin{array}{l} {\rm time\; rate\; of\; change} \\ {\rm of\; mass\; of\; species\; }A} \\ {{\rm in\; a\; control\; volume} \end{array}\right\}\quad =\quad \left\{\begin{array}{l} {\rm \; \; \; rate\; at\; which} \\ {\rm mass\; of\; species\; }A} \\ {{\rm \; \; \; \; }enters{\rm \; the} \\ {\rm \; control\; volume} \end{array}\right\}\; \; -\; \; \left\{\begin{array}{l} {\rm \; \; \; rate\; at\; which} \\ {\rm mass\; of\; species\; }A} \\ {{\rm \; \; \; \; \; }leaves{\rm \; the} \\ {\rm \; \; control\; volume} \end{array}\right\}\; \; +\; \; \left\{\begin{array}{l} {\rm \; net\; rate\; of\; production} \\ {\rm of\; the\; mass\; of\; species\; }A} \\ {{\rm \; \; \; \; \; \; \; \; \; \; owing\; to} \\ {{\rm \; \; \; }chemical\; reactions} \end{array}\right\}\label{ZEqnNum906921}$$

In order to develop a precise mathematical representation of this axiom, we require the following quantities:

$$\rho _{A} \quad =\quad \left\{\begin{array}{l} {\rm mass\; density} \\ {{\rm of\; \; species\; }A} \end{array}\right\}\label{ZEqnNum275116}$$

$$v_{A} \quad =\quad \left\{\begin{array}{l} {\rm velocity\; of} \\ {{\rm \; species\; }A} \end{array}\right\}$$ ()

$$r_{A} \quad =\quad \left\{\begin{array}{l} {\rm \; net\; mass\; rate\; of\; production} \\ {{\rm per\; unit\; volume\; of\; species\; }A} \\ {{\rm owing\; to\; }chemical\; reactions} \end{array}\right\}\label{ZEqnNum544342}$$

Here it is important to understand that $$r_{A}$$ represents both the creation of species A (when $$r_{A}$$ is positive) and the consumption of species A (when $$r_{A}$$ is negative). In terms of these primitive quantities, we can make use of an arbitrary fixed control volume to express Eq. as

Axiom I: $$\frac{d}{dt} \int _{V}\rho _{A} \, dV \; \; +\; \; \int _{A}\rho _{A} v_{A} \cdot n\, dA \quad =\quad \int _{V}r_{A} \, dV \; ,\quad \quad A=1,\; 2,\; ....,\; N\label{ZEqnNum662048}$$

In the volume $$V$$ the total mass produced by chemical reactions must be zero. This is our second axiom that we express in words as

$$\left\{\begin{array}{l} {\rm total\; rate\; of\; production} \\ {\rm \; \; \; \; of\; mass\; owing\; to\; } \\ {{\rm \; \; }chemical\; reactions} \end{array}\right\}\quad =\quad 0$$ ()

and in terms of the definition given by Eq. this word statement takes the form

$$\sum _{A=1}^{A=N}\; \int _{V}r_{A} \, dV \quad =\quad 0\label{ZEqnNum085083}$$

The summation over all N molecular species can be interchanged with the volume integration in this representation of the second axiom, and this allows us to express Eq. as

$$\int _{V}\sum _{A=1}^{A=N}\; r_{A} \, dV \quad =\quad 0$$ ()

Since the volume $$V$$ is arbitrary, the integrand must be zero and we extract the preferred form of the second axiom given by

Axiom II: $$\sum _{A=1}^{A=N}r_{A} \quad =\quad 0\label{ZEqnNum636688}$$

In Eqs. [ZEqnNum819624] and [ZEqnNum914545] we have used a mixed mode nomenclature to represent the chemical species, i.e., we have used both letters and numbers simultaneously. Traditionally, we use upper case Roman letters to designate various chemical species, thus the rates of production for species A, B and C are designated by $$r_{A}$$, $$r_{B}$$ and $$r_{C}$$. When dealing with systems containing N different molecular species, we allow an indicator, such as A or D or G, to take on values from 1 to N in order to produce compact forms of the two axioms given by Eqs. [ZEqnNum473475] and [ZEqnNum304328]. We could avoid this mixed mode nomenclature consisting of letters and numbers by expressing Eq. [ZEqnNum905675] in the form;

Axiom II: $$r_{A} \; \; +\; \; r_{B} \; \; +\; \; r_{C} \; \; +\; \; r_{D} \; \; +\; \; ....\; \; +\; \; r_{N} \quad =\quad 0$$ ()

however, this approach is rather cumbersome when dealing with N-component systems.

The concept that mass is neither created nor destroyed by chemical reactions (as indicated by Eq. [ZEqnNum049988]) is based on the work of Lavoisier1 who stated:

“We observe in the combustion of bodies generally four recurring phenomena which would appear to be invariable laws of nature; while these phenomena are implied in other memoirs which I have presented, I must recall them here in a few words.”

Lavoisier went on to list four phenomena associated with combustion, the third of which was given by

Third Phenomenon. In all combustion, pure air in which the combustion takes place is destroyed or decomposed and the burning body increases in weight exactly in proportion to the quantity of air destroyed or decomposed.

It is this Third Phenomenon, when extended to all reacting systems, that supports Axiom II in the form represented by Eq. [ZEqnNum085083]. The experiments that led to the Third Phenomenon were difficult to perform in the 18 $${}^{th}$$ century and those difficulties have been recounted by Toulmin2.

4.1.1 Molar concentration and molecular mass

When chemical reactions occur, it is generally more convenient to work with the molar form of Eqs. [ZEqnNum443376] and [ZEqnNum733074]. The appropriate measure of concentration is then the molar concentration defined by

$$c_{A} \quad =\quad {\rho _{A} \mathord{\left/ {\vphantom {\rho _{A} MW_{A} }} \right. \kern-\nulldelimiterspace} MW_{A} } \quad =\quad \left\{\begin{array}{l} {moles{\rm \; of\; species\; }A} \\ {\rm \; \; per\; unit\; volume} \end{array}\right\}$$ ()

while the appropriate net rate of production for species A is given by DSL. $$R_{A} \quad =\quad {r_{A} \mathord{\left/ {\vphantom {r_{A} MW_{A} }} \right. \kern-\nulldelimiterspace} MW_{A} } \quad =\quad \left\{\begin{array}{l} {\rm net\; }molar{\rm \; rate\; of\; production} \\ {\rm per\; unit\; volume\; of\; species\; }A} \\ {{\rm owing\; to\; chemical\; reactions} \end{array}\right\}$$ ()

Here $$MW_{A}$$ represents the molecular mass3 of species A that is given explicitly by

$$MW_{A} \quad =\quad \frac{{\rm kilograms\; of\; }A}{{\rm moles\; of\; }A}\label{ZEqnNum709487}$$

The numerical values of the molecular mass are obtained from the atomic masses associated with any particular molecular species, and values for both the atomic mass and the molecular mass are given in Tables A1 and A2 in Appendix A. In those tables we have represented the atomic mass and the molecular mass in terms of grams per mole, thus the definition given by Eq. [ZEqnNum709487] for water leads to

$$MW_{\rm H}_{\rm 2} {\rm O} \quad =\quad \frac{0.018015{\rm \; kg}}{\rm mol} \quad =\quad \frac{18.015{\rm \; g}}{\rm mol}$$ ()

In terms of $$c_{A}$$ and $$R_{A}$$, the two axioms given by Eqs. [ZEqnNum888263] and [ZEqnNum573841] take the form

Axiom I: $$\frac{d}{dt} \int _{V}c_{A} \, dV \; \; +\; \; \int _{A}c_{A} v_{A} \cdot n\, dA \quad =\quad \int _{V}R_{A} \, dV \; ,\quad A=1,\; 2,\; ....,\; N\label{ZEqnNum545075}$$

Axiom II: $$\sum _{A\; =\; 1}^{A\; =\; N}MW_{A} R_{A} \quad =\quad 0$$ ()

Here it is important to note that mass is conserved during chemical reactions while moles need not be conserved. For example, the decomposition of calcium carbonate (solid) to calcium oxide (solid) and carbon dioxide (gas) is described by

$$\left({\rm CaCO}_{\rm 3} \right)_{solid} \quad \to \quad \left({\rm CaO}\right)_{solid} \; \; +\; \; \left({\rm CO}_{\rm 2} \right)_{gas}$$ ()

thus one mole is consumed and two moles are produced by this chemical reaction.

One must be very careful to understand that the net molar rate of production per unit volume of species A owing to chemical reactions, $$R_{A}$$, may be the result of many different chemical reactions. For example, in the chemical production system illustrated in Figure 4-1, carbon dioxide may be produced by the

Figure 4-1. Chemical production system

oxidation of carbon monoxide, by the complete combustion of methane, or by other chemical reactions taking place within the control volume illustrated in Figure 4-1. The combination of all these individual chemical reactions is represented by $$R_{\, {\rm CO}_{\rm 2} }$$. It is important to note that in Figure 4-1 we have suggested the stoichiometry of the reactions taking place while the actual chemical kinetic processes taking place may be much more complicated. The subject of stoichiometry will be discussed in detail in Chapter 6, and a brief discussion of chemical kinetics is given in Section 8.6.

4.1.2 Moving control volumes

In Sec. 3.3 we developed the macroscopic mass balance for a single component system in terms of an arbitrary moving control volume, $$V_{a} (t)$$. The speed of displacement of the surface of a moving control volume is given by $$w\cdot n$$, and the flux of species A that crosses this moving surface is given by the normal component of the relative velocity for species A, i.e., $$(v_{A} -w)\cdot n$$. On the basis of this concept, we can express the first axiom for the mass of species A in the form

Axiom I: $$\frac{d}{dt} \int _{V_{a} (t)}\rho _{A} \, dV \; \; +\; \; \int _{A_{a} (t)}\rho _{A} (v_{A} -w)\cdot n\, dA \quad =\quad \int _{V_{a} (t)}r_{A} \, dV \; ,\quad \quad A=1,\; 2,\; ....,\; N\label{ZEqnNum996159}$$

When it is convenient to work with molar quantities, we divide this result by the molecular mass of species A in order to obtain the moving control volume form given by

Axiom I: $$\frac{d}{dt} \int _{V_{a} (t)}c_{A} \, dV \; \; +\; \; \int _{A_{a} (t)}c_{A} (v_{A} -w)\cdot n\, dA \quad =\quad \int _{V_{a} (t)}R_{A} \, dV \; ,\quad \quad A\; \; =\; \; 1,\; 2,\; ....,\; N\label{ZEqnNum863340}$$

Throughout this chapter, we will restrict our studies to fixed control volumes in order to focus our attention on the new concepts associated with multicomponent systems. However, the world of chemical engineering is filled with moving, dynamic systems and the analysis of those systems will require the use of Eqs. [ZEqnNum645577] and [ZEqnNum863340].

4.2 Species Mass Density

The mass of species A per unit volume in a mixture of several components is known as the species mass density, and it is represented by $$\rho _{A}$$. The species mass density can range from zero, when no species A is present in the mixture, to the density of pure species A, when no other species are present. In order to understand what is meant by the species mass density, we consider a mixing process in which three pure species are combined to create a uniform mixture of species A, B, and C. This mixing process is illustrated in Figure 4-2 where we have indicated that three pure species are combined to create a uniform mixture having a measured volume of $$45\; cm^{3}$$. The total volume of the three pure species is $$50\; cm^{3}$$, thus there is a change of volume upon mixing as is usually the case with liquids. We denote this change of volume upon mixing by $$\Delta V_{mix}$$, and for the process illustrated in Figure 4-2 we express this quantity as

$$\Delta V_{mix} \quad =\quad V\; \; -\; \; \left(V_{A} +V_{B} +V_{C} \right)$$ ()

The densities of the pure species have been denoted by a superscript zero, thus $$\rho _{A}^{\rm o}$$ represents the mass density of pure species A. The species mass density of species A is defined by

$$\left\{\begin{array}{l} {species\; mass\; density} \\ {\rm \; \; \; of\; species\; }A} \end{array}\right\}\quad =\quad {\left({\rm mass\; of\; species\; }A\right)\mathord{\left/ {\vphantom {\left({\rm mass\; of\; species\; }A\right) \left(\begin{array}{l} {\rm volume\; in\; which\; species\; }A} \\ {{\rm \; \; \; \; \; \; \; \; is\; contained} \end{array}\right)} \right. \kern-\nulldelimiterspace} \left(\begin{array}{l} {\rm volume\; in\; which\; species\; }A} \\ {{\rm \; \; \; \; \; \; \; \; is\; contained} \end{array}\right)}$$ ()

and this definition applies to mixtures in which species A is present as well as to the case of pure species A.

Figure 4-2. Mixing process

If we designate the mass of species A as $$m_{A}$$ and the volume of the uniform mixture as V, the species mass density can be expressed as

$$\rho _{A} \quad =\quad {m_{A} \mathord{\left/ {\vphantom {m_{A} V}} \right. \kern-\nulldelimiterspace} V}$$ ()

For the mixing process illustrated in Figure 4-2, we are given that the mass of species A is

$$m_{A} \quad =\quad \rho _{A}^{\rm o} \, V_{A} \quad =\quad (0.85\, g/cm^{3} )(15\, cm^{3} )\quad =\quad 12.75\; g$$ ()

and this allows us to determine the species mass density in the mixture according to

$$\rho _{A} \quad =\quad {m_{A} \mathord{\left/ {\vphantom {m_{A} V}} \right. \kern-\nulldelimiterspace} V} \quad =\quad \frac{12.75\, {\rm g}}{\rm 45}\_ \, {\rm cm}^{\rm 3} } \quad =\quad 0.283\, {\rm g}\mathord{\left/ {\vphantom {{\rm g} {\rm cm}^{{\rm 3} } \right. \kern-\nulldelimiterspace} {\rm cm}^{\rm 3} } \quad$$ ()

This type of calculation can be carried our for species B and C in order to determine $$\rho _{B}$$ and $$\rho _{C}$$.

The total mass density is simply the sum of all the species mass densities and is defined by

$$\left\{\begin{array}{l} {\rm total\; mass} \\ {\rm \; \; density} \end{array}\right\}\quad =\quad \rho \quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\rho _{A}\label{ZEqnNum971008}$$

The total mass density can be determined experimentally by measuring the mass, m, and the volume, V, of a mixture. For any a particular mixture, it is difficult to measure directly the species mass density; however, one can prepare a mixture in which the species mass densities can be determined as we have suggested in Figure 4-2. When working with molar forms, we often need the total molar concentration and this is defined by

$$\left\{\begin{array}{l} {\rm \; \; total\; molar} \\ {\rm concentration} \end{array}\right\}\quad =\quad c\quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}c_{A}\label{ZEqnNum863721}$$

4.2.1 Mass fraction and mole fraction

For solid and liquid systems it is sometimes convenient to use the mass fraction as a measure of concentration. The mass fraction of species A can be expressed in words as

$$\omega _{A} \quad =\quad \left\{\begin{array}{l} {\rm \; mass\; of\; species\; }A{\rm \; per} \\ {\rm unit\; mass\; of\; the\; mixture} \end{array}\right\}$$ ()

and in precise mathematical form we have

$$\omega _{A} \quad =\quad \frac{\rho _{A} }{\rho } \quad =\quad {\rho _{A} \mathord{\left/ {\vphantom {\rho _{A} \sum _{G\; =\; 1}^{G\; =\; N}\rho _{G} }} \right. \kern-\nulldelimiterspace} \sum _{G\; =\; 1}^{G\; =\; N}\rho _{G} }\label{ZEqnNum363676}$$

Note that the indicator, G, is often referred to as a dummy indicator since any letter would suffice to denote the summation over all species in the mixture. In this particular case, we would not want to use A as the dummy indicator since this could lead to confusion. The mole fraction is analogous to the mass fraction and is defined by

$$x_{A} \quad =\quad \frac{c_{A} }{c} \quad =\quad {c_{A} \mathord{\left/ {\vphantom {c_{A} \sum _{G\; =\; 1}^{G\; =\; N}c_{G} }} \right. \kern-\nulldelimiterspace} \sum _{G\; =\; 1}^{G\; =\; N}c_{G} }\label{ZEqnNum159844}$$

If one wishes to avoid the mixed-mode nomenclature in Eqs. [ZEqnNum363676] and [ZEqnNum887852], one must express the mass fraction as

$$\omega _{A} \quad =\quad \frac{\rho _{A} }{\rho } \quad =\quad \frac{\rho _{A} }{\rho _{A} \; \; +\; \; \rho _{B} \; \; +\; \; \rho _{C} \; \; +\; \; \rho _{D} \; \; +\; \; ....\; \; +\; \; \rho _{N} }$$ ()

while the mole fraction takes the form

$$y_{A} \quad =\quad \frac{c_{A} }{c} \quad =\quad \frac{c_{A} }{c_{A} \; \; +\; \; c_{B} \; \; +\; \; c_{C} \; \; +\; \; c_{D} \; \; +\; \; ....\; \; +\; \; c_{N} }\label{ZEqnNum881605}$$

Very often $$x_{A}$$ is used to represent mole fractions in liquid mixtures and $$y_{A}$$ to represent mole fractions in vapor mixtures, thus Eq. [ZEqnNum881605] represents the mole fraction in a vapor mixture while Eq. [ZEqnNum573667] represents the mole fraction in a liquid mixture.

EXAMPLE 4.1. Conversion of mole fractions to mass fractions

Sometimes we may be given the composition of a mixture in terms of the various mole fractions and require the mass fractions of the various constituents. To convert from $$x_{A}$$ to $$\omega _{A}$$ we proceed as follows:

$$x_{A} \quad =\quad {c_{A} \mathord{\left/ {\vphantom {c_{A} c}} \right. \kern-\nulldelimiterspace} c}$$ (l) $\label{GrindEQ__2_} c_{A} \quad =\quad x_{A} \, c$ $\label{GrindEQ__3_} \rho _{A} \quad =\quad MW_{A} \; c_{A} \quad =\quad MW_{A} \; x_{A} \; c$ $\label{GrindEQ__4_} \rho \quad =\quad \sum _{G\; =\; 1}^{G\; =\; N}\rho _{G} \quad =\quad \left(\sum _{G\; =\; 1}^{G\; =\; N}MW_{G} \; x_{G} \right)\; c$ $\label{GrindEQ__5_} \omega _{A} \quad =\quad \frac{\rho _{A} }{\rho } \quad =\quad \frac{MW_{A} \; x_{A} \; c}{\left(\sum _{G=1}^{G=N}MW_{G} \; x_{G} \right)\; c} \quad =\quad \frac{MW_{A} \; x_{A} }{\sum _{G=1}^{G=N}MW_{G} \; x_{G} }$

4.2.2 Total mass balance

Given the total density defined by Eq. , we are ready to recover the total mass balance for multicomponent, reacting systems. For a fixed control volume, this is developed by summing Eq. over all species to obtain

$$\sum _{A\; =\; 1}^{A\; =\; N}\frac{d}{dt} \int _{V}\rho _{A} \, dV \; \; +\; \; \sum _{A\; =\; 1}^{A\; =\; N}\int _{A}\rho _{A} \, v_{A} \cdot n\, dA \quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\int _{V}r_{A} \, dV$$ ()

The summation procedure can be interchanged with differentiation and integration so that this result takes the form

$$\frac{d}{dt} \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}\rho _{A} \, dV \; \; +\; \; \int _{A}\sum _{A\; =\; 1}^{A\; =\; N}\rho _{A} \, v_{A} \cdot n\, dA \quad =\quad \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}r_{A} \, dV$$ ()

On the basis of definition of the total mass density given by Eq. [ZEqnNum882543] and the axiom given by Eq. [ZEqnNum105047], this result simplifies to

$$\frac{d}{dt} \int _{V}\rho \, dV \; \; +\; \; \int _{A}\sum _{A\; =\; 1}^{A\; =\; N}\rho _{A} \, v_{A} \cdot n\, dA \quad =\quad 0\label{ZEqnNum536076}$$

At this point, we define the total mass flux according to

$$\left\{\begin{array}{l} {\rm \; \; \; total} \\ {\rm mass\; flux} \end{array}\right\}\quad =\quad \rho \, v\quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\rho _{A} v_{A}\label{ZEqnNum419769}$$

Since  is defined by Eq. , this result actually represents a definition of the velocity v that can be expressed as

$$v\quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\omega _{A} v_{A}\label{ZEqnNum731232}$$

This velocity is known as the mass average velocity and it plays a key role both in our studies of macroscopic mass balances and in subsequent studies of fluid mechanics, heat transfer, and mass transfer. Use of this definition for the mass average velocity allows us to express Eq. [ZEqnNum536076] as

$$\frac{d}{dt} \int _{V}\rho dV \; \; +\; \; \int _{A}\rho v\cdot ndA\quad =\quad 0\label{ZEqnNum386474}$$

This is identical in form to the mass balance for a fixed control volume that was presented in Chapter 3; however, this result has greater physical content than our previous result for single-component systems. In this case, the density is not the density of a single component but the sum of all the species densities as indicated by Eq. and the velocity is not the velocity of a single component but the mass average velocity defined by Eq. .

4.3 Species Velocity

In our representation of the axioms for the mass of multicomponent systems, we have introduced the concept of a species velocity indicating that individual molecular species move at their own velocities designated by $$v_{A}$$ where $$A=1,\; 2,\; ..\, N$$. In order to begin thinking about the species velocity, we consider a lump of sugar (species A) placed in the bottom of a tea cup which is very carefully filled with water (species B). If we wait long enough, the solid sugar illustrated in Figure 4-3 will dissolve and become

Figure 4-3. Dissolution of sugar

uniformly distributed throughout the cup. This is a clear indication that the velocity of the sugar molecules is different from the velocity of the water molecules, i.e.,

$$v_{Sugar} \quad \ne \quad v_{Water}$$ ()

If the solution in the cup is not stirred, the velocity of the sugar molecules will be very small and the time required for the sugar to become uniformly distributed throughout the cup will be very long. We generally refer to this process as diffusion and diffusion velocities are generally very small. If we stir the liquid in the teacup, the sugar molecules will be transported away from the sugar cube by convection as we have illustrated in Figure 4-4. In this case, the sugar will become uniformly distributed throughout the cup in a relatively short time and we generally refer to this process as mixing. Mechanical mixing can accelerate the process by which the sugar becomes uniformly distributed throughout the teacup; however, a true mixture of sugar and water could never be achieved unless the velocities of the two species were different. The difference between species velocities is crucial. It is responsible for mixing, for separation and purification, and it is necessary for chemical reactions to occur. If all species velocities were equal, life on earth would cease immediately.

In addition to mixing the sugar and water as indicated in Figures 4-3 and 4-4, we can also separate the sugar and water by allowing the water to evaporate. In that case all the water in the tea cup would appear in the surrounding air and the sugar would remain in the bottom of the cup. This separation would not be possible unless the velocity of the water were different than the velocity of the sugar. While the difference between species velocities is of crucial interest to chemical engineers, there is a class of problems for which we can ignore this difference and still obtain useful results. In the following paragraphs we want to identify this class of problems.

Figure 4-4. Mixing of sugar

To provide another example of the difference between species velocities and the effect of diffusion, we consider the process of absorption of $${\rm SO}_{2}$$ in a falling film of water as illustrated in Figure 4-5. The gas

Figure 4-5. Absorption of sulfur dioxide

mixture entering the column consists of air, which is essentially insoluble in water, and $${\rm SO}_{2}$$, which is soluble in water. Because of the absorption of $${\rm SO}_{2}$$ in the water, the exit gas is less detrimental to the local environment. It should be intuitively appealing that the species velocities in the axial direction are constrained by

$$v_{\rm SO}_{{\rm 2} } \cdot k\quad \approx \quad v_{air} \cdot k\label{ZEqnNum501761}$$

in which k is the unit vector pointing in the z-direction. The situation for the radial components of $$v_{\rm SO}_{{\rm 2} }$$ and $$v_{air}$$ is quite different because the radial components are normal to the gas-liquid interface. Since the air is insoluble in water, the component of $$v_{air}$$ in the radial direction must be zero at the gas-liquid interface, i.e.,

$$v_{air} \cdot n\quad =\quad 0\; ,\quad {\rm at\; the\; gas-liquid\; interface}\label{ZEqnNum911488}$$

On the other hand, the sulfur dioxide is crossing the interface as it leaves the gas stream and enters the liquid stream. The radial component of $$v_{{\rm SO}_{2} }$$ must therefore be positive and we express this idea as

$$v_{{\rm SO}_{2} } \cdot n\quad >\quad 0\; ,\quad {\rm at\; the\; gas-liquid\; interface}$$ ()

It should be clear that $$v_{{\rm SO}_{2} } \cdot n$$ is a velocity associated with a diffusion process while $$v_{{\rm SO}_{2} } \cdot k$$ is a velocity associated with a convection process and that the latter is generally much, much larger than the former, i.e.,

$$\underbrace{v_{\rm SO}_{2} } \cdot k}_{{\rm convection}\quad >>\quad \underbrace{v_{\rm SO}_{2} } \cdot n}_{{\rm diffusion}$$ ()

The motion of a chemical species can result from a force applied to the fluid, i.e., a fan might be used to move the gas mixture through the tube illustrated in Figure 4-5. The motion of a chemical species can also result from a concentration gradient such as the gradient that causes the sugar to diffuse throughout the teacup illustrated in Figure 4-3. Because the motion of chemical species can be caused by both applied forces and concentration gradients, it is reasonable to decompose the species velocity into two parts: the mass average velocity and the mass diffusion velocity. We represent this decomposition as

$$v_{\rm SO}_{2} } \quad \quad =\quad \quad \underbrace{\; \; v\; \; }_{\begin{array}{l} {\rm mass\; average} \\ {\rm \; \; \; velocity} \end{array}\quad +\quad \underbrace{\; \; u_{\kern 1pt} {\rm SO}_{2} } \; \; }_{\begin{array}{l} {\rm mass\; diffusion} \\ {\rm \; \; \; \; \; velocity} \end{array}\label{ZEqnNum665161}$$

At entrances and exits, such as those illustrated in Figure 4-5, the diffusion velocity in the z-direction is usually small compared to the mass average velocity in the z-direction and Eq. can be approximated by

$$v_{{\rm SO}_{2} } \cdot k\quad =\quad v\cdot k\; ,\quad {\rm negligible\; diffusion\; velocity}\label{ZEqnNum707604}$$

In this text we will repeatedly make use of this simplification in order to solve a variety of problems without the need to predict the diffusion velocity. However, in subsequent courses mass transfer across fluid-fluid interfaces will be studied in detail, and that study will require a complete understanding of the role of the diffusion velocity for a variety of processes.

In order to reinforce our thoughts about the species velocity, the mass average velocity, and the mass diffusion velocity, we return to the definition of the mass average velocity given by Eq. [ZEqnNum374801] and express the z-component of v according to

$$v\cdot k\quad =\quad \omega {\kern 1pt} _{{\rm SO}_{2} } (v_{{\rm SO}_{2} } \cdot k)\; \; +\; \; \omega _{air} (v_{air} \cdot k)$$ ()

in which the mass fractions are constrained by

$$\omega _{{\kern 1pt} {\rm SO}_{2} } \; \; +\; \; \omega _{air} \quad =\quad 1$$ ()

On the basis of the approximation given by Eq. [ZEqnNum501761] we conclude that

$$v\cdot k\quad \approx \quad v_{{\rm SO}_{2} } \cdot k\quad \approx \quad v_{air} \cdot k$$ ()

However, the radial component of the species velocities is an entirely different matter. Once again we can use Eq. [ZEqnNum593885] to obtain

$$v\cdot n\quad =\quad \omega {\kern 1pt} _{{\rm SO}_{2} } (v_{{\rm SO}_{2} } \cdot n)\; \; +\; \; \omega _{air} (v_{air} \cdot n)\; ,\quad \quad {\rm at\; the\; gas-liquid\; interface}$$ ()

but on the basis of Eq. [ZEqnNum911488] this reduces to

$$v\cdot n\quad =\quad \omega {\kern 1pt} _{{\rm SO}_{2} } (v_{{\rm SO}_{2} } \cdot n)\; ,\quad \quad {\rm at\; the\; gas-liquid\; interface}$$ ()

Under these circumstances we see that

$$v\cdot n\quad \ne \quad v_{{\rm SO}_{2} } \cdot n\quad \ne \quad v_{air} \cdot n\; ,\quad \quad {\rm at\; the\; gas-liquid\; interface}$$ ()

and the type of approximation indicated by Eq. [ZEqnNum707604] is not valid. In this text, we will study a series of macroscopic mass balance problems for which Eq. [ZEqnNum566719] represents a reasonable approximation; however, one must always remember that neglect of the diffusion velocity is a very delicate matter, and it is considered further in Problem 4-10. As we have mentioned before, the difference between species velocities is responsible for separation and purification, and it absolutely is necessary in order for chemical reactions to occur.

4.4 Measures of Velocity

In the previous section we defined the mass average velocity according to

$$v\quad =\quad \sum _{B\; =\; 1}^{B\; =\; N}\omega _{B} \, v_{B} \; ,\quad {\rm mass\; average\; velocity}$$ ()

and we noted that the total mass flux vector was given by

$$\rho v\quad =\quad \sum _{B\; =\; 1}^{B\; =\; N}\rho _{B} \, v_{B} \quad =\quad \left\{\begin{array}{l} {\rm \; total\; mass} \\ {\rm flux\; vector} \end{array}\right\}$$ ()

By analogy we define the molar average velocity by

$$v^{*} \quad =\quad \sum _{B\; =\; 1}^{B\; =\; N}x_{B} \, v_{B} \; ,\quad {\rm molar\; average\; velocity}$$ ()

and it follows that the total molar flux vector is given by

$$c\; v^{*} \quad =\quad \sum _{B\; =\; 1}^{B\; =\; N}c_{B} \, v_{B} \quad =\quad \left\{\begin{array}{l} {\rm total\; molar} \\ {\rm flux\; vector} \end{array}\right\}\label{ZEqnNum482548}$$

In the previous section we used a decomposition of the species velocity of the form

$$v_{A} \quad =\quad v\; \; +\; \; u_{A}$$ ()

so that the species mass flux vector could be expressed as

$$\rho _{A} v_{A} \quad =\quad \underbrace{\; \rho _{A} \, v\; }_{\begin{array}{l} {\rm convective} \\ {\rm \; \; \; \; \; flux} \end{array}}\; \; +\; \; \underbrace{\; \rho _{A} \, u_{A} \; }_{\begin{array}{l} {\rm diffusive} \\ {\rm \; \; \; flux} \end{array}}$$ ()

In dealing with the molar flux vector one finds it convenient to express the species velocity as

$$v_{A} \quad =\quad v^{*} \; \; +\; \; u_{A}^{*}$$ ()

so that the molar flux takes the form

$$c_{A} v_{A} \quad =\quad \underbrace{\; c_{A} \, v^{*} \; }_{\begin{array}{l} {\rm convective} \\ {\rm \; \; \; \; \; flux} \end{array}}\; \; +\; \; \underbrace{\; c_{A} \, u_{A}^{*} \; }_{\begin{array}{l} {\rm diffusive} \\ {\rm \; \; \; flux} \end{array}}$$ ()

When convective transport dominates, the species velocity, the mass average velocity and the molar average velocity are all essentially equal, i.e.,

$$v_{A} \quad =\quad v\quad =\quad v^{*}\label{ZEqnNum467879}$$

This is the situation that we encounter most often in our study of material balances and we will make use of this result repeatedly to determine the flux of species A at entrances and exits. While Eq. [ZEqnNum467879] is widely used to describe velocities at entrances and exits, one must be very careful about the general use of this approximation. If Eq. [ZEqnNum848461] were true for all species under all conditions, there would be no separation, no purification, no mixing, and no chemical reactions!

Under certain circumstances we may want to use a total mole balance and this is obtained by summing Eq. [ZEqnNum545075] over all N species in order to obtain

$$\frac{d}{dt} \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}c_{A} \, dV \; \; +\; \; \int _{A}\sum _{A\; =\; 1}^{A\; =\; N}{\rm (}c_{A} \, v_{A} {\rm )} \cdot n\, dA \quad =\quad \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}R_{A} \, dV$$ ()

Use of the definitions given by Eqs. [ZEqnNum863721] and [ZEqnNum482548] allows us to write this result in the form

$$\frac{d}{dt} \int _{V}c\, dV \; \; +\; \; \int _{A}c\, v^{*} \cdot n\, dA \quad =\quad \int _{V}\sum _{A\; =\; 1}^{A\; =\; N}R_{A} \, dV$$ ()

Here we should note that the overall net rate of production of moles need not be zero, thus the overall mole balance is more complex than the overall mass balance.

4.5 Molar Flow Rates at Entrances and Exits

Here we direct our attention to the macroscopic mole balance for a fixed control volume

$$\frac{d}{dt} \int _{V}c_{A} \, dV \; \; +\; \; \int _{A}c_{A} \, v_{A} \cdot n\, dA \quad =\quad \int _{V}R_{A} \, dV \; ,\quad \quad A=1,{\rm \; 2,\; ....\; }N\label{ZEqnNum839730}$$

with the intention of evaluating $$c_{A} \, v_{A} \cdot n$$ at entrances and exits. The area integral of the molar flux, $$c_{A} v_{A} \cdot n$$, can be represented as

$$\int _{A}c_{A} \, v_{A} \cdot n\, dA \quad =\quad \int _{A_{\rm e} }c_{A} \, v_{A} \cdot n\, dA \; \; +\; \; \int _{A_{i} }c_{A} \, v_{A} \cdot n\, dA\label{ZEqnNum349015}$$

where $$A_{\rm e}$$ represents the entrances and exits at which convection dominates and $$A_{i}$$ represents an interfacial area over which diffusive fluxes may dominate. In this text, our primary interest is the study of control volumes having entrances and exits at which convective transport is much more important than diffusive transport, and we have illustrated this type of control volume in Figure 4-6. There the entrances and exits for both the water and the air are at the top and bottom of the column, while the surface of the control volume that coincides with the liquid-solid interface represents an impermeable boundary at which $$c_{A} \, v_{A} \cdot n=0$$. For systems of this type, we express Eq. [ZEqnNum349015] as

$$\int _{A}c_{A} v_{A} \cdot n\, dA \quad =\quad \int _{A_{\rm e} }c_{A} v_{A} \cdot n\, dA \quad =\quad \int _{A_{\rm e}{\rm n}{\rm t}{\rm r}{\rm a}{\rm n}{\rm c}{\rm es} }c_{A} v_{A} \cdot n\, dA \; \; +\; \; \int _{A_{\rm exits} }c_{A} v_{A} \cdot n\, dA$$ ()

In order to simplify our discussion about the flux at entrances and exits, we direct our attention to an exit and express the molar flow rate at that exit as

$$\dot{M}_{A} \quad =\quad \int _{A_{\rm exit} }c_{A} v_{A} \cdot n\, dA$$ ()

On the basis of the discussion in Sec. 4.2, we assume that the diffusive flux is negligible ( $$v_{A} \cdot n\approx v\cdot n$$) so that the above result takes the form

$$\dot{M}_{A} \quad =\quad \int _{A_{\rm exit} }c_{A} v\cdot n\, dA\label{ZEqnNum713441}$$

It is possible that both $$c_{A}$$ and v vary across the exit and a detailed evaluation of the area integral is required in order to determine the molar flow rate of species A. In general this is not the case; however, it is very important to be aware of this possibility.

Figure 4-6. Entrances and exits at which convection dominates

4.5.1 Average concentrations

In Sec. 3.2.1 we defined a volume average density and we use the same definition here for the volume average concentration given by

$$\langle c_{A} \rangle \quad =\quad \frac{1}{V} \int _{V}c_{A} \, dV \; ,\quad \quad {\rm volume\; average\; concentration}$$ ()

At entrances and exits, we often work with the “bulk concentration” or “cup mixed concentration” that was defined earlier in Sec. 3.2.1. For the concentration, $$c_{A}$$, we repeat the definition according to

$$\langle c_{A} \rangle _{b} \quad =\quad \frac{1}{Q_{exit} } \int _{A_{\, exit} }c_{A} \, v\cdot n\, dA \quad =\quad \frac{\int _{A_{\, exit} }c_{A} \, v\cdot n\, dA }{\int _{A_{\, exit} }v\cdot n\, dA }\label{ZEqnNum182517}$$

In terms of the bulk concentration the molar flow rate given by Eq. [ZEqnNum223174] can be expressed as

$$\dot{M}_{A} \quad =\quad \langle c_{A} \rangle _{b} \; Q_{exit}\label{ZEqnNum453862}$$

in which it is understood that $$\dot{M}_{A}$$ and $$\langle c_{A} \rangle _{b}$$ represent the molar flow rate and concentration at the exit. In addition to the bulk or cup-mixed concentration, one may encounter the area average concentration denoted by $$\langle c_{A} \rangle$$ and defined at an exit according to

$$\langle c_{A} \rangle \quad =\quad \frac{1}{A_{\, exit} } \int _{A_{\, exit} }c_{A} \, dA$$ ()

If the concentration is constant over $$A_{exit}$$, the area average concentration is equal to this constant value, i.e.

$$\langle c_{A} \rangle \quad =\quad c_{A} \; ,\quad when{\rm \; }c_{A} {\rm \; }is\; constant$$ ()

We often refer to this condition as a “flat” concentration profile, and for this case we have

$$\langle c_{A} \rangle _{b} \quad =\quad \langle c_{A} \rangle \quad =\quad c_{A} \; ,\quad \quad flat\; concentration\; profile$$ ()

Under these circumstances the molar flow rate takes the form

$$\dot{M}_{A} \quad =\quad c_{A} \; Q_{exit} \; ,\quad \quad flat\; concentration\; profile\label{ZEqnNum983745}$$

The conditions for which $$c_{A}$$ can be treated as a constant over an exit or an entrance are likely to occur in many practical applications.

When the flow is turbulent, there are rapid velocity fluctuations about the mean or time-averaged velocity. The velocity fluctuations tend to create uniform velocity profiles and they play a crucial role in the transport of mass orthogonal to the direction of the mean flow. The contribution of turbulent fluctuations to mass transport parallel to the direction of the mean flow can normally be neglected and we will do so in our treatment of macroscopic mass balances. In a subsequent courses on fluid mechanics and mass transfer, the influence of turbulence will be examined more carefully. In our treatment, we will make use of the reasonable approximation that the turbulent velocity profile is flat and this means that $$v\cdot n$$ is constant over $$A_{\, exit}$$. Both turbulent and laminar velocity profiles are illustrated in Figure 4-7 and there we

Figure 4-7. Laminar and turbulent velocity profiles for flow in a tube
see that the velocity for turbulent flow is nearly constant over a major portion of the flow field. If we make the “flat velocity profile” assumption, we can express Eq. [ZEqnNum182517] as

$$\langle c_{A} \rangle _{b} \; \quad =\quad \; \frac{\int _{A_{\, exit} }c_{A} \, v\cdot n\, dA }{\int _{A_{\, exit} }v\cdot n\, dA } \; \quad =\quad \; \frac{\int _{A_{\, exit} }c_{A} \, dA }{\int _{A_{\, exit} }dA } \; \frac{v\cdot n}{v\cdot n} \; \quad =\quad \; \frac{1}{A_{\, exit} } \int _{A_{\, exit} }c_{A} \, dA \; \quad =\quad \; \langle c_{A} \rangle$$ ()

For this case, the molar flow rate at the exit takes the form

$$\dot{M}_{A} \quad =\quad \langle c_{A} \rangle Q_{exit} \; ,\quad \quad flat\; velocity\; profile\label{ZEqnNum294219}$$

To summarize, we note that Eq. [ZEqnNum453862] is an exact representation of the molar flow rate in terms of the bulk concentration and the volumetric flow rate. When the concentration profile can be approximated as flat, the molar flow rate can be represented in terms of the constant concentration and the volumetric flow rate

as indicated by Eq. [ZEqnNum230041]. When the velocity profile can be approximated as flat, the molar flow rate can be represented in terms of the area average concentration and the volumetric flow rate as indicated by Eq. [ZEqnNum272495]. If one is working with the species mass balance given by Eq. [ZEqnNum387174], the development represented by Eqs. [ZEqnNum839730] through [ZEqnNum366974] can be applied simply by replacing $$c_{A}$$ with $$\rho _{A}$$.

4.6 Alternate Flow Rates

There are a number of relations between species flow rates and total flow rates that are routinely used in solving macroscopic mass or mole balance problems provided that either the velocity profile is flat or the concentration profile is flat. For example, we can always write Eq. [ZEqnNum279695] in the form

$$\dot{M}_{A} \quad =\quad \int _{A_{exit} }c_{A} v\cdot n\, dA \quad =\quad \int _{A_{exit} }\frac{c_{A} }{c} c\, v\cdot n\, dA \quad =\quad \int _{A_{exit} }x_{A} \, c\, v\cdot n\, dA\label{ZEqnNum974643}$$

If either $$c\, v\cdot n$$ or $$x_{A}$$ is constant over the area of the exit, we can express this result as

$$\dot{M}_{A} \quad =\quad \langle x_{A} \rangle \dot{M}\label{ZEqnNum189643}$$

where $$\dot{M}$$ is the total molar flow rate defined by

$$\dot{M}\quad =\quad \sum _{A\; =\; 1}^{A\; =\; N}\dot{M}_{A}\label{ZEqnNum181666}$$

If the individual molar flow rates are known and one desires to determine the area averaged mole fraction at an entrance or an exit, it is given by

$$\langle x_{A} \rangle \quad =\quad {\dot{M}_{A} \mathord{\left/ {\vphantom {\dot{M}_{A} \sum _{B\; =\; 1}^{B\; =\; N}\dot{M}_{B} }} \right. \kern-\nulldelimiterspace} \sum _{B\; =\; 1}^{B\; =\; N}\dot{M}_{B} }$$ ()

provided that either $$c\, v\cdot n$$ or $$x_{A}$$ is constant over the area of the entrance or the exit. It will be left as an exercise for the student to show that similar relations exist between mass fractions and mass flow rates. For example, a form analogous to Eq. [ZEqnNum189643] is given by

$$\dot{m}_{A} \quad =\quad \langle \omega _{A} \rangle \, \dot{m}\label{ZEqnNum700852}$$

and the mass fraction at an entrance or an exit can be expressed as

$$\langle \omega _{A} \rangle \quad =\quad {\dot{m}_{A} \mathord{\left/ {\vphantom {\dot{m}_{A} \sum _{G\; =\; 1}^{G\; =\; N}\dot{m}_{G} }} \right. \kern-\nulldelimiterspace} \sum _{G\; =\; 1}^{G\; =\; N}\dot{m}_{G} }\label{ZEqnNum264342}$$

One must keep in mind that the results given by Eqs. [ZEqnNum779969] through [ZEqnNum264342] are only valid when either the concentration (density) is constant or the molar (mass) flux is constant over the entrance or exit.

When neither of these simplifications is valid, we express Eq. [ZEqnNum974643] as

$$\dot{M}_{A} \quad =\quad \int _{A_{\rm exit} }x_{A} \, c\, v\cdot n\, dA \quad =\quad \langle x_{A} \rangle _{b} \dot{M}$$ ()

where $$\langle x_{A} \rangle _{b}$$ is the cup mixed mole fraction of species A. The definition of the mole fraction requires that

$$\sum _{A\; =\; 1}^{A\; =\; N}x_{A} \quad =\quad 1\label{ZEqnNum276080}$$

and it will be left as an exercise for the student to show that

$$\sum _{A\; =\; 1}^{A\; =\; N}\langle x_{A} \rangle _{b} \quad =\quad 1\label{ZEqnNum359366}$$

This type of constraint on the mole fractions (and mass fractions) applies at every entrance and exit and it often represents an important equation in the set of equations that are used to solve macroscopic mass balance problems.

4.7 Species Mole/Mass Balance

In this section we examine the problem of solving the N equations represented by either Eq. [ZEqnNum694147] or Eq. [ZEqnNum506940] under steady-state conditions in the absence of chemical reactions. The distillation process illustrated in Figure 4-8 provides a simple example; however, most distillation processes are more complex than the one shown in Figure 4-8 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 1-5 and 1-6), thus understanding the simple process illustrated in Figure 4-8 is an important step in our studies.

Figure 4-8 Distillation column

For this particular ternary distillation process, we are given the information listed in Table 4-1. 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 4-1, we have not specified $$x_{C}$$ since this mole fraction will be determined by Eq. [ZEqnNum276080]. 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 4-1. Specified conditions

Stream #1    Stream #2    Stream #3
$$\dot{M}_{1} =1200{\rm \; moles/hr}$$

$$x_{A} \; \; =\; \; 0.3$$    $$x_{A} \; \; =\; \; 0.6$$    $$x_{A} \; \; =\; \; 0.1$$
$$x_{B} \; \; =\; \; 0.2$$    $$x_{B} \; \; =\; \; 0.3$$
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

SnplaceRule SnI. 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 $$v_{A} \cdot n$$ 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 $$v_{A}$$ 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 4-8, it should be obvious that we need to cut the entrance and exit streams and then join the cuts as illustrated in Figure 4-9 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 $$v_{A} \cdot n=0$$.

Figure 4-9. Control volume for distillation column

4.7.1 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: $$(x_{A} )_{i} \; ,\; \; (x_{B} )_{i} \; ,\; \; (x_{C} )_{i} \quad \quad i=1,2,3\label{ZEqnNum729411}$$

Molar flow rates: $$\dot{M}_{i} \; ,\quad i=1,2,3\label{ZEqnNum779582}$$

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: $$\int _{A}x_{A} \, cv\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}x_{B} \, cv\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}x_{C} \, cv\cdot n\, dA \; \; =\; \; 0\label{ZEqnNum156558}$$

along with the three mole fraction constraints that apply at the streams that are cut by the control volume illustrated in Figure 4-9.

Mole fraction constraints: $$(x_{A} )_{i} \; \; +\; \; (x_{B} )_{i} \; \; +\; \; (x_{C} )_{i} \quad =\quad 1\; ,\quad \quad i=1,2,3\label{ZEqnNum157867}$$

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: $$\dot{M}_{1} =1200{\rm \; moles/hr}\; ,\quad x_{A} =0.3\; ,\quad x_{B} =0.2$$ 3

II. Conditions for Stream #2: $$x_{A} \; \; =\; \; 0.6\; ,\quad x_{B} \; \; =\; \; 0.3$$ 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 Eq. [ZEqnNum838824]. If one specifies all three mole fractions in a particular stream, Eq. [ZEqnNum199001] 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 4-2 that provides a template for subsequent problems in which we have N molecular species and M streams.

Table 4-2. Degrees-of-Freedom

Stream Variables
compositions

flow rates

Generic Degrees of Freedom (A) (N x M) + M = 12
Number of Independent Balance Equations
mass/mole balance equations

Generic Constraints (B)

compositions

flow rates

Particular Specifications and Constraints (C) 6
Degrees of Freedom (A - B - C)

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.

4.7.2 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 4-1 given here as

Table 4-3. Specified and unknown conditions

Stream #1    Stream #2    Stream #3
$$\dot{M}_{1} =1200{\rm \; moles/hr}$$    ?    ?
$$x_{A} \; \; =\; \; 0.3$$    $$x_{A} \; \; =\; \; 0.6$$    $$x_{A} \; \; =\; \; 0.1$$
$$x_{B} \; \; =\; \; 0.2$$    $$x_{B} \; \; =\; \; 0.3$$    ?
?    ?    ?
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. [ZEqnNum106642]. These can be used to express Table 4-3 as follows:

Table 4-4. Unknowns to be determined

Stream #1    Stream #2    Stream #3
$$\dot{M}_{1} =1200{\rm \; moles/hr}$$    ?    ?
$$x_{A} \; \; =\; \; 0.3$$    $$x_{A} \; \; =\; \; 0.6$$    $$x_{A} \; \; =\; \; 0.1$$
$$x_{B} \; \; =\; \; 0.2$$    $$x_{B} \; \; =\; \; 0.3$$    $$x_{B} \; \; =\; \; 0.9-x_{C}$$
$$x_{c} \; \; =\; \; 0.5$$    $$x_{c} \; \; =\; \; 0.1$$    ?
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 4-4 allows us to express the balance equations given by Eqs. [ZEqnNum156558] as

Species A: $$0.6\; \dot{M}_{2} \; \; +\; \; 0.1\; \dot{M}_{3} \quad +\quad \quad 0\quad \quad \quad =\quad {\rm 360\; moles/hr}$$ ([ZEqnNum420022]a)

Species B: $$0.3\; \dot{M}_{2} \; \; +\; \; 0.9\; \dot{M}_{3} \; \; -\; \; \; \underbrace{\dot{M}_{3} \langle x_{C} \rangle _{3} }_{\begin{array}{l} {\rm bi-linear} \\ {\rm \; \; \; form} \end{array}}\quad =\quad 240{\rm \; moles/hr}$$ ([ZEqnNum420022]b)

Species C: $$0.1\; \dot{M}_{2} \; \; +\; \; \quad 0\quad +\; \; \quad \underbrace{\dot{M}_{3} \langle x_{C} \rangle _{3} }_{\begin{array}{l} {\rm bi-linear} \\ {\rm \; \; \; form} \end{array}}\quad \; \; =\quad 600{\rm \; moles/hr}$$ ([ZEqnNum967271]c)

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

$$\dot{M}_{2} \; \; =\; \; 480{\rm \; mol/hr}\; ,\quad \quad \dot{M}_{3} \; \; =\; \; 720{\rm \; mol/hr}\; ,\quad \quad \langle x_{C} \rangle _{3} \quad =\quad 0.767$$ ()

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

Table 4-5. Solution for molar flows and mole fractions

Stream #1    Stream #2    Stream #3
$$\dot{M}_{1} =1200{\rm \; mol/hr}$$    $$\dot{M}_{2} =480{\rm \; mol/hr}$$    $$\dot{M}_{3} =720{\rm \; mol/hr}$$
$$x_{A} \; \; =\; \; 0.3$$    $$x_{A} \; \; =\; \; 0.6$$    $$x_{A} \; \; =\; \; 0.100$$
$$x_{B} \; \; =\; \; 0.2$$    $$x_{B} \; \; =\; \; 0.3$$    $$x_{B} \; \; =\; \; 0.133$$
$$x_{C} \; \; =\; \; 0.5$$    $$x_{C} \; \; =\; \; 0.1$$    $$x_{C} \; \; =\; \; 0.767$$
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. [ZEqnNum424664] 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.

4.7.3 Solution of sets of equations

To illustrate a classic procedure for solving sets of algebraic equations, we direct our attention to Eqs. [ZEqnNum192373]. We begin by eliminating the term, $$\dot{M}_{3} \langle x_{C} \rangle _{3}$$, from Eq. [ZEqnNum622418]b to obtain the following pair of linear equations:

Species A: $$0.6\; \dot{M}_{2} \; \; +\; \; 0.1\; \dot{M}_{3} \quad =\quad 360{\rm \; mol/hr}$$ ([ZEqnNum777303]a)

Species B: $$0.4\; \dot{M}_{2} \; \; +\; \; 0.9\; \dot{M}_{3} \quad =\quad 840{\rm \; mol/hr}$$ ([ZEqnNum777303]b)

To solve this set of linear equations, we make use of a simple scheme known as Gaussian elimination. We begin by dividing Eq. [ZEqnNum813064]a by the coefficient 0.6 in order to obtain

$$\dot{M}_{2} \; \; +\; \; 0.1667\; \dot{M}_{3} \quad =\quad 600{\rm \; mol/hr}$$ ([ZEqnNum806894]a)

$$0.4\; \dot{M}_{2} \; \; +\; \; 0.9\; \dot{M}_{3} \quad =\quad 840{\rm \; mol/hr}$$ ([ZEqnNum806894]b)

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

$$\dot{M}_{2} \; \; +\; \; 0.1667\; \dot{M}_{3} \quad =\quad 600{\rm \; mol/hr}$$ ([ZEqnNum490788]a)

$$0.8333\; \dot{M}_{3} \quad =\quad 600{\rm \; mol/hr}$$ ([ZEqnNum490788]b)

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

$$\dot{M}_{2} \; \; +\; \; 0.1667\; \dot{M}_{3} \quad =\quad 600{\rm \; mol/hr}$$ ([ZEqnNum272303]a)

$$\dot{M}_{3} \quad =\quad 720{\rm \; mol/hr}$$ ([ZEqnNum272303]b)

At this point, we begin the procedure of “back substitution” which requires that Eq. [ZEqnNum663920]b be substituted into Eq. [ZEqnNum185945]a in order to obtain the final solution for the two molar flow rates.

$$\dot{M}_{2} \; \quad =\quad 480{\rm \; mol/hr}$$ ([ZEqnNum929408]a)

$$\dot{M}_{3} \quad =\quad 720{\rm \; mol/hr}$$ ([ZEqnNum929408]b)

The procedure leading from Eqs. [ZEqnNum938569] to the solution given by Eqs. [ZEqnNum854642] 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.

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 4-10 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 $$1000{\rm \; lb}_{\rm m} /{\rm hr}$$. 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 4-10. 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 4-11. We first form Control Volume I which

Figure 4-10. Two-column distillation unit

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 4-12. 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 Eq. [ZEqnNum323637]. For steady-state conditions in the absence of chemical reactions, the species mass balances are given by

$$\int _{A}\rho _{A} v_{A} \cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\rho _{B} v_{B} \cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\rho _{C} v_{C} \cdot n\, dA \; \; =\; \; 0$$ ()

Figure 4-11. Primary cuts for the two-column distillation unit

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: $$\int _{A}\omega _{A} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\omega _{B} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\omega _{C} \, \rho \, v\cdot n\, dA \; \; =\; \; 0$$ ()

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: $$(\omega _{A} )_{i} \; \; +\; \; (\omega _{B} )_{i} \; \; +\; \; (\omega _{C} )_{i} \quad =\quad 1\; ,\quad \quad i=1,2,3,4,5$$ ()

Figure 4-12. Control volumes for two-column distillation unit

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

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

Mass flow rates: $$\dot{m}_{\, i} \; ,\quad i=1,2,3,4$$ ()

and we indicate the number of process variables explicitly as

I. Three mole fractions in each of four streams

II. Four mass molar flow rates

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: $$\int _{A}\omega _{A} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\omega _{B} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\omega _{C} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\label{ZEqnNum647193}$$

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

Mass fraction constraints: $$(\omega _{A} )_{i} \; \; +\; \; (\omega _{B} )_{i} \; \; +\; \; (\omega _{C} )_{i} \quad =\quad 1\; ,\quad \quad i=1,2,3,4\label{ZEqnNum264060}$$

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

I. Balance equations for three molecular species

II. Mole fraction constraints for the four streams

Generic Specifications and Constraints (B)

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: $$\dot{m}_{\, 1} =1000{\rm \; lb}_{\rm m} /{\rm hr}\; ,\quad \omega _{A} \; \; =\; \; 0.5\; ,\quad \omega _{B} \; \; =\; \; 0.3$$

II. Conditions for Stream #2: $$\omega _{A} \; \; =\; \; 0.045\; ,\quad \omega _{B} \; \; =\; \; 0.091$$

III. Conditions for Stream #3: $$\omega _{A} \; \; =\; \; 0.069\; ,\quad \omega _{B} \; \; =\; \; 0.901$$

IV. Conditions for Stream #4: $$\omega _{A} \; \; =\; \; 0.955\; ,\quad \omega _{B} \; \; =\; \; 0.041$$

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 4-6. Degrees-of-Freedom

Stream Variables
compositions

flow rates

Generic Degrees of Freedom (A) (N x M) + M = 16
Number of Independent Balance Equations
mass/mole balance equations

Generic Specifications and Constraints (B)

compositions

flow rates

Particular Specifications and Constraints (C) 9
Degrees of Freedom (A - B - C)

This indicates that use of Control Volume II will lead to a well-posed problem, and we are assured that we can use Eqs. [ZEqnNum647193] and [ZEqnNum264060] 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 Eq. [ZEqnNum386474]. In terms of Eqs. [ZEqnNum696072] for Control Volume II, this approach leads to

:

species A: $$(\omega _{A} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{A} )_{3} \dot{m}_{3} \; \; +\; \; (\omega _{A} )_{4} \dot{m}_{4} \quad =\quad (\omega _{A} )_{1} \dot{m}_{1}$$ ([ZEqnNum825073]a)

species B: $$(\omega _{B} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{B} )_{3} \dot{m}_{3} \; \; +\; \; (\omega _{B} )_{4} \dot{m}_{4} \quad =\quad (\omega _{B} )_{1} \dot{m}_{1}$$ ([ZEqnNum825073]b)

Total: $$\dot{m}_{2} \quad \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \quad \; \; \dot{m}_{1}$$ ([ZEqnNum397112]c)

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. [ZEqnNum797722] through [ZEqnNum131831]. The same procedure can be used with Eqs. [ZEqnNum595841], and one begins by dividing Eq. [ZEqnNum509481]a by $$(\omega _{A} )_{2}$$ to obtain

:

species A: $$\dot{m}_{2} \; \; +\; \; \left[{(\omega _{A} )_{3} \mathord{\left/ {\vphantom {(\omega _{A} )_{3} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{3} \; \; +\; \; \left[{(\omega _{A} )_{4} \mathord{\left/ {\vphantom {(\omega _{A} )_{4} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{4} \quad =\quad \left[{(\omega _{A} )_{1} \mathord{\left/ {\vphantom {(\omega _{A} )_{1} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{1}$$ ([ZEqnNum343114]a)

species B: $$(\omega _{B} )_{2} \dot{m}_{2} \quad \; \; +\quad \; \; (\omega _{B} )_{3} \dot{m}_{3} \quad \; \; +\quad \; \; (\omega _{B} )_{4} \dot{m}_{4} \quad =\quad \quad (\omega _{B} )_{1} \dot{m}_{1}$$ ([ZEqnNum343114]b)

Total: $$\dot{m}_{2} \quad \quad \quad +\quad \quad \quad \dot{m}_{3} \quad \quad \quad +\quad \quad \quad \dot{m}_{4} \quad \quad =\quad \quad \; \; \dot{m}_{1}$$ ([ZEqnNum326641]c)

In order to eliminate $$\dot{m}_{2}$$ from Eq. [ZEqnNum461768]b, one multiplies Eq. [ZEqnNum126289]a by $$(\omega _{B} )_{2}$$ and subtracts the result from both Eq. [ZEqnNum865208]b. To eliminate $$\dot{m}_{2}$$ from Eq. [ZEqnNum282378]c, one need only subtract Eq. [ZEqnNum284549]a from Eq. [ZEqnNum643704]c, and these two operations lead to the following balance equations:

:

species A: $$\dot{m}_{2} \; \; +\; \; \left[{(\omega _{A} )_{3} \mathord{\left/ {\vphantom {(\omega _{A} )_{3} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{3} \; \; +\; \; \left[{(\omega _{A} )_{4} \mathord{\left/ {\vphantom {(\omega _{A} )_{4} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{4} \quad =\quad \left[{(\omega _{A} )_{1} \mathord{\left/ {\vphantom {(\omega _{A} )_{1} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{1}$$ ([ZEqnNum656618]a)

species B: $$\left\{.......\right\}\dot{m}_{3} \quad \quad \quad +\quad \quad \left\{.......\right\}\dot{m}_{4} \quad =\quad \quad \left\{.......\right\}\dot{m}_{1}$$ ([ZEqnNum656618]b)

Total: $$\left[......\right]\dot{m}_{3} \quad \quad \quad +\quad \quad \left[......\right]\dot{m}_{4} \; \; \quad =\quad \quad \left[......\right]\dot{m}_{1}$$ ([ZEqnNum994624]c)

Clearly the algebra is becoming quite complex, and it will become worse when we use Eq. [ZEqnNum894051]b to eliminate $$\dot{m}_{3}$$ from Eq. [ZEqnNum333495]c. Without providing the details, we continue the elimination process to obtain the solution to Eq. [ZEqnNum777374]c and this leads to the following expression for $$\dot{m}_{4}$$:

$$\dot{m}_{4} \quad \quad =\quad \quad \dot{m}_{1} \; \; \frac{\left[1-\frac{(\omega _{A} )_{1} }{(\omega _{A} )_{2} } \right]\quad -\quad \frac{\frac{(\omega _{B} )_{1} }{(\omega _{B} )_{3} } \left[1-\frac{(\omega _{B} )_{2} }{(\omega _{B} )_{1} } \frac{(\omega _{A} )_{1} }{(\omega _{A} )_{2} } \right]}{\left[1-\frac{(\omega _{B} )_{2} }{(\omega _{B} )_{3} } \frac{(\omega _{A} )_{3} }{(\omega _{A} )_{2} } \right]} \left[1-\frac{(\omega _{A} )_{3} }{(\omega _{A} )_{2} } \right]}{\left[1-\frac{(\omega _{A} )_{4} }{(\omega _{A} )_{2} } \right]\quad -\quad \frac{\frac{(\omega _{B} )_{4} }{(\omega _{B} )_{3} } \left[1-\frac{(\omega _{B} )_{2} }{(\omega _{B} )_{4} } \frac{(\omega _{A} )_{4} }{(\omega _{A} )_{2} } \right]}{\left[1-\frac{(\omega _{B} )_{2} }{(\omega _{B} )_{3} } \frac{(\omega _{A} )_{3} }{(\omega _{A} )_{2} } \right]} \left[1-\frac{(\omega _{A} )_{3} }{(\omega _{A} )_{2} } \right]}\label{ZEqnNum840659}$$

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

$$\dot{m}_{2} \quad =\quad {\rm 220\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{3} \quad =\quad {\rm 288\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{4} \quad =\quad {\rm 492\; lb}_{\rm m} /{\rm hr}$$ ()

In order to determine $$\dot{m}_{5}$$, we must make use of the balance equations for Control Volume I that are given by Eqs. [ZEqnNum959841]. These can be expressed in terms of two species balances and one total mass balance leading to

:

species A: $$-\; (\omega _{A} )_{1} \dot{m}_{1} \; \; +\; \; (\omega _{A} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{A} )_{5} \dot{m}_{5} \quad =\quad 0$$ ([ZEqnNum679073]a)

species B: $$-\; (\omega _{B} )_{1} \dot{m}_{1} \; \; +\; \; (\omega _{B} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{B} )_{5} \dot{m}_{5} \quad =\quad 0$$ ([ZEqnNum679073]b)

Total: $$-\; \; \dot{m}_{1} \quad \quad +\quad \quad \dot{m}_{2} \quad \quad +\quad \quad \dot{m}_{5} \quad =\quad 0$$ ([ZEqnNum294734]c)

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

$$\dot{m}_{5} \quad =\quad {\rm 780\; lb}_{\rm m} /{\rm hr}$$ ()

The algebraic complexity associated with the simple process represented in Figure 4-10 encourages the use of matrix methods that are described in Sec. 4.9.

4.9 Matrix Algebra

In Sec. 4.7 we examined a distillation process with the objective of determining molar flow rates, and our analysis led to a set of two equations and two unknowns given by

Species A: $$0.6\; \dot{M}_{2} \; \; +\; \; 0.1\; \dot{M}_{3} \quad =\quad 360{\rm \; mol/hr}$$ ([ZEqnNum153881]a)

Species B: $$0.4\; \dot{M}_{2} \; \; +\; \; 0.9\; \dot{M}_{3} \quad =\quad 840{\rm \; mol/hr}$$ ([ZEqnNum153881]b)

Some thought was necessary in order to set up the macroscopic mole balances for the process illustrated in Figure 4-7; however, the algebraic effort required to solve the governing macroscopic balances was trivial. In Sec. 4.8, we considered the system illustrated in Figure 4-10 and the analysis led to the following set of three equations and three unknowns:

species A: $$(\omega _{A} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{A} )_{3} \dot{m}_{3} \; \; +\; \; (\omega _{A} )_{4} \dot{m}_{4} \quad =\quad (\omega _{A} )_{1} \dot{m}_{1}$$ ([ZEqnNum596473]a)

species B: $$(\omega _{B} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{B} )_{3} \dot{m}_{3} \; \; +\; \; (\omega _{B} )_{4} \dot{m}_{4} \quad =\quad (\omega _{B} )_{1} \dot{m}_{1}$$ ([ZEqnNum596473]b)

Total: $$\dot{m}_{2} \quad \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \quad \; \; \dot{m}_{1}$$ ([ZEqnNum810328]c)

The algebraic effort required to solve these three equations for $$\dot{m}_{2}$$, $$\dot{m}_{3}$$ and $$\dot{m}_{4}$$ was considerable as one can see from the solution given by Eq. [ZEqnNum840659]. It should not be difficult to imagine that solving sets of four or five equations can become exceedingly difficult to do by hand; however, computer routines are available that can be used to solve virtually any set of equations have the form given by Eqs. [ZEqnNum239077].

In dealing with sets of many equations, it is convenient to use the language of matrix algebra. For example, in matrix notation we would express Eqs. [ZEqnNum780531] according to

$$\left[\begin{array}{ccc} {(\omega _{A} )_{2} } & {(\omega _{A} )_{3} } & {(\omega _{A} )_{4} } \\ {(\omega _{B} )_{2} } & {(\omega _{B} )_{3} } & {(\omega _{B} )_{4} } \\ {1} & {1} & {1} \end{array}\right]\left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\quad =\quad \left[\begin{array}{c} {(\omega _{A} )_{1} \dot{m}_{1} } \\ {(\omega _{B} )_{1} \dot{m}_{1} } \\ {\dot{m}_{1} } \end{array}\right]\label{ZEqnNum350920}$$

In this representation of Eqs. [ZEqnNum925999], the $$3\times 3$$ matrix of mass fractions multiplies the $$3\times 1$$ column matrix of mass flow rates to produce a $$3\times 1$$ column matrix that is equal to the right hand side of Eq. [ZEqnNum350920]. In working with matrices, it is generally convenient to make use of a nomenclature in which subscripts are used to identify the row and column in which an element is located. We used this type of nomenclature in Chapter 2 where the $$m\times n$$ matrix A was represented by

$${\rm A}\quad =\quad \left[\begin{array}{cccc} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {....} & {....} & {......} & {....} \\ {a_{m1} } & {a_{m2} } & {......} & {a_{mn} } \end{array}\right]$$ ()

Here the first subscript identifies the row in which an element is located while the second subscript identifies the column. In Chapter 2 we discussed matrix addition and subtraction, and here we wish to discuss matrix multiplication and the matrix operation that is analogous to division. Matrix multiplication between A and B is defined only if the number of columns of A (in this case n) is equal to the number of rows of B. Given an $$m\times n$$ matrix A and an $$n\times p$$ matrix, B, the product between A and B is illustrated by the following equation:

([ZEqnNum897810])

Here we see that the elements of the ith row in matrix A multiply the elements of the jth column in matrix B to produce the element in the ith row and the jth column of the matrix C. For example, the specific element $$c_{11}$$ is given by

$$c_{11} \quad =\quad a_{11} b_{11} \; \; +\; \; a_{12} b_{21} \; \; +\; \; .......\; \; +\; \; a_{1n} b_{n1}$$ ()

while the general element $$c_{ij}$$ is given by

$$c_{ij} \quad =\quad a_{i1} b_{1j} \; \; +\; \; a_{i2} b_{2j} \; \; +\; \; .......\; \; +\; \; a_{in} b_{nj}$$ ()

In Eq. [ZEqnNum897810] we see that an $$m\times n$$ matrix can multiply an $$n\times p$$ to produce an $$m\times p$$, and we see that the matrix multiplication represented by AB is only defined when the number of columns in A is equal to the number of rows in B. The matrix multiplication illustrated in Eq. [ZEqnNum559949] conforms to this rule since there are three columns in the matrix of mass fractions and three rows in the column matrix of mass flow rates. The configuration illustrated in Eq. [ZEqnNum268548] is extremely common since it is associated with the solution of n equations for n unknowns. Our generic representation for this type of matrix equation is given by

$${\rm A}\, {\rm u}\quad =\quad {\rm b}\label{ZEqnNum309268}$$

in which A is a square matrix, u is a column matrix of unknowns, and b is a column matrix of knowns. These matrices are represented explicitly by

$${\rm A}\quad =\quad \left[\begin{array}{cccc} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {....} & {....} & {......} & {....} \\ {a_{n1} } & {a_{n2} } & {......} & {a_{nn} } \end{array}\right]\; ,\quad \quad {\rm u}\quad =\quad \left[\begin{array}{c} {u_{1} } \\ {.} \\ {.} \\ {u_{n} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {b_{1} } \\ {.} \\ {.} \\ {b_{n} } \end{array}\right]\label{ZEqnNum496328}$$

Sometimes the coefficients in A depend on the unknowns, u, and the matrix equation may be bi-linear (see Eqs. [ZEqnNum192373]) or non-linear. We represent systems of this type according to

$${\rm A}({\rm u})\, {\rm u}\quad =\quad {\rm b}$$ ()

to clearly indicate that the problem is not linear.

The transpose of a matrix is defined in the same manner as the transpose of an array that was discussed in Sec. 2.5, thus the transpose of the matrix A is constructed by interchanging the rows and columns to obtain

$${\rm A}\quad =\quad \left[\begin{array}{cccc} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {.} & {.} & {......} & {.} \\ {a_{m1} } & {a_{m2} } & {......} & {a_{mn} } \end{array}\right]\; ,\quad \quad {\rm A}^{\rm T} \quad =\quad \left[\begin{array}{cccc} {a_{11} } & {a_{21} } & {...} & {a_{m1} } \\ {a_{12} } & {a_{22} } & {...} & {a_{m2} } \\ {.} & {.} & {...} & {.} \\ {.} & {.} & {...} & {.} \\ {a_{1n} } & {a_{2n} } & {...} & {a_{mn} } \end{array}\right]$$ ()

Here it is important to note that A is an $$m\times n$$ matrix while $${\rm A}^{\rm T}$$ is an $$n\times m$$ matrix.

4.9.1 Inverse of a square matrix

In order to solve Eq. [ZEqnNum309268], one cannot “divide” by A to determine the unknown, u, since matrix division is not defined. There is, however, a related operation involving the inverse of a matrix. The inverse of a matrix, A, is another matrix, $${\rm A}^{-1}$$, such that the product of A and $${\it A}^{-1}$$ is given by

$${\rm A}\, {\rm A}^{-1} \quad =\quad {\rm I}$$ ()

in which I is the identity matrix. Identity matrices have ones in the diagonal elements and zeros in the off-diagonal elements as illustrated by the following $$4\times 4$$ matrix:

$${\rm I}\quad =\quad \left[\begin{array}{cccc} {1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} \end{array}\right]$$ ()

For the inverse of a matrix to exist, the matrix must be a square matrix, i.e., the number of rows must be equal to the number of columns. In addition, the determinant of the matrix must be different from zero.

As an example of the use of the inverse of a matrix, we consider the following set of four equations containing four unknowns:

$$\begin{array}{l} {a_{11} u_{1} \; \; +\; \; a_{12} u_{2} \; \; +\; \; a_{13} u_{3} \; \; +\; \; a_{14} u_{4} \quad =\quad b_{1} } \\ {} \\ {a_{21} u_{1} \; \; +\; \; a_{22} u_{2} \; \; +\; \; a_{23} u_{3} \; \; +\; \; a_{24} u_{4} \quad =\quad b_{2} } \\ {} \\ {a_{31} u_{1} \; \; +\; \; a_{32} u_{2} \; \; +\; \; a_{33} u_{3} \; \; +\; \; a_{34} u_{4} \quad =\quad b_{3} } \\ {} \\ {a_{41} u_{1} \; \; +\; \; a_{42} u_{2} \; \; +\; \; a_{43} u_{3} \; \; +\; \; a_{44} u_{4} \quad =\quad b_{4} } \end{array}\label{ZEqnNum291322}$$

In compact notation, we would represent these equations according to

$${\rm A}\, {\rm u}\quad =\quad {\rm b}\label{ZEqnNum729810}$$

We suppose that A is invertible and that the inverse, along with u and b, are given by

$${\rm A}^{-1} \quad =\quad \left[\begin{array}{cccc} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{array}\right]\; ,\quad \quad {\rm u}\quad =\quad \left[\begin{array}{c} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \end{array}\right]$$ ()

We can multiply Eq. [ZEqnNum729810] by $${\rm A}^{-1}$$ to obtain

$${\rm A}^{-1} {\rm A}\, {\rm u}\quad =\quad {\rm I}\, {\rm u}\quad =\quad {\rm u}\quad =\quad {\rm A}^{-1} {\rm b}$$ ()

and the details are given by

$$\left[\begin{array}{cccc} {1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} \end{array}\right]\left[\begin{array}{c} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{array}\right]\quad =\quad \left[\begin{array}{cccc} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{array}\right]\left[\begin{array}{c} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \end{array}\right]$$ ()

Carrying out the matrix multiplication on the left hand side leads to

$$\left[\begin{array}{c} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{array}\right]\quad =\quad \left[\begin{array}{cccc} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{array}\right]\left[\begin{array}{c} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \end{array}\right]$$ ()

while the more complex multiplication on the right hand side provides

$$\left[\begin{array}{c} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{array}\right]\quad =\quad \left[\begin{array}{cccc} {\bar{a}_{11} b_{1} } & {\bar{a}_{12} b_{2} } & {\bar{a}_{13} b_{3} } & {\bar{a}_{14} b_{4} } \\ {\bar{a}_{21} b_{1} } & {\bar{a}_{22} b_{2} } & {\bar{a}_{23} b_{3} } & {\bar{a}_{24} b_{4} } \\ {\bar{a}_{31} b_{1} } & {\bar{a}_{32} b_{2} } & {\bar{a}_{33} b_{3} } & {\bar{a}_{34} b_{4} } \\ {\bar{a}_{41} b_{1} } & {\bar{a}_{42} b_{2} } & {\bar{a}_{43} b_{3} } & {\bar{a}_{44} b_{4} } \end{array}\right]$$ ()

Each element of the column matrix on the left hand side is equal to the corresponding element on the right hand side and this leads to the solution for the unknowns given by

$$\begin{array}{l} {u_{1} \quad =\quad \bar{a}_{11} b_{1} \; \; +\; \; \bar{a}_{12} b_{2} \; \; +\; \; \bar{a}_{13} b_{3} \; \; +\; \; \bar{a}_{14} b_{4} } \\ {} \\ {u_{2} \quad =\quad \bar{a}_{21} b_{1} \; \; +\; \; \bar{a}_{22} b_{2} \; \; +\; \; \bar{a}_{23} b_{3} \; \; +\; \; \bar{a}_{24} b_{4} } \\ {} \\ {u_{3} \quad =\quad \bar{a}_{31} b_{1} \; \; +\; \; \bar{a}_{32} b_{2} \; \; +\; \; \bar{a}_{33} b_{3} \; \; +\; \; \bar{a}_{34} b_{4} } \\ {} \\ {u_{4} \quad =\quad \bar{a}_{41} b_{1} \; \; +\; \; \bar{a}_{42} b_{2} \; \; +\; \; \bar{a}_{43} b_{3} \; \; +\; \; \bar{a}_{44} b_{4} } \end{array}\label{ZEqnNum762280}$$

It should be clear that there are two main problems associated with the use of Eqs. [ZEqnNum291322] to obtain the solution for the unknowns given by Eq. [ZEqnNum762280]. The first problem is the correct interpretation of a physical process to arrive at the original set of equations, and the second problem is the determination of the inverse of the matrix A.

There are a variety of methods for developing the inverse for a matrix; however, sets of equations are usually solved numerically without calculating the inverse of a matrix. One of the classic methods is known as Gaussian elimination and we can illustrate this technique with the problem that was studied in Sec. 4.8. We can make use of the data provided in Figure 4.9 to express Eqs. [ZEqnNum305779] in the form

species A: $$0.045\, \dot{m}_{2} \; \; +\; \; 0.069\, \dot{m}_{3} \; \; +\; \; 0.955\, \dot{m}_{4} \quad =\quad \; 500{\rm \; lb}_{\rm m} /{\rm hr}$$ ([ZEqnNum216736]a)

species B: $$0.091\, \dot{m}_{2} \; \; +\; \; 0.901\, \dot{m}_{3} \; \; +\; \; 0.041\, \dot{m}_{4} \quad =\quad \; 300\; {\rm lb}_{\rm m} /{\rm hr}$$ ([ZEqnNum216736]b)

Total: $$\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad {\rm 1000\; lb}_{\rm m} /{\rm hr}$$ ([ZEqnNum249266]c)

and our objective is to determine the three mass flow rates, $$\dot{m}_{2}$$, $$\dot{m}_{3}$$ and $$\dot{m}_{4}$$. The solution is obtained by making use of the following three rules which are referred to as elementary row operations:

I. Any equation in the set can be modified by multiplying or dividing by a non-zero scalar without affecting the solution.

II. Any equation can be added or subtracted from the set without affecting the solution.

III. Any two equations can be interchanged without affecting the solution.

For this particular problem, it is convenient to arrange the three equations in the form

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \; \dot{m}_{3} \quad \quad +\quad \quad \; \dot{m}_{4} \quad \quad =\quad {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0.045\, \dot{m}_{2} \; \; +\; \; 0.069\, \dot{m}_{3} \; \; +\; \; 0.955\, \dot{m}_{4} \quad =\quad \; 500{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0.091\, \dot{m}_{2} \; \; +\; \; 0.901\, \dot{m}_{3} \; \; +\; \; 0.041\, \dot{m}_{4} \quad =\quad \; 300\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}\label{ZEqnNum491562}$$

We begin by eliminating the first term in the second equation. This is accomplished by multiplying the first equation by 0.045 and subtracting the result from the second equation to obtain

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad \; \; \; +\; \; 0.024\, \dot{m}_{3} \; \; +\; \; 0.910\, \dot{m}_{4} \quad =\quad \; 455{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0.091\, \dot{m}_{2} \; \; +\; \; 0.901\, \dot{m}_{3} \; \; +\; \; 0.041\, \dot{m}_{4} \quad =\quad \; 300\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}\label{ZEqnNum522104}$$

Directing our attention to the third equation, we multiply the first equation by 0.091 and subtract the result from the third equation to obtain

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \; {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad \; \; \; +\; \; 0.024\, \dot{m}_{3} \; \; +\; \; 0.910\, \dot{m}_{4} \quad =\quad \; 455{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad \; \; \; +\; \; 0.810\, \dot{m}_{3} \; \; -\; \; 0.050\, \dot{m}_{4} \quad =\quad 209\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}\label{ZEqnNum181816}$$

The second equation can be conditioned by dividing by 0.024 so that the equation set takes the form

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \; {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad \dot{m}_{3} \quad \; +\quad 37.917\, \dot{m}_{4} \quad =\quad 18958{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad \; \; \; +\; \; 0.810\, \dot{m}_{3} \; \; -\; \; 0.050\, \dot{m}_{4} \quad =\quad 209\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ()

We now multiply the second equation by $$0.810$$ and subtract the result from the third equation to obtain

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \; {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad \dot{m}_{3} \quad \; +\quad 37.917\, \dot{m}_{4} \quad =\quad 18958{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad 0\quad \; \; -\quad \; 30.763\, \dot{m}_{4} \quad =\quad -15147\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ()

and division of the third equation by $$-30.763$$ leads to

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \; {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad \dot{m}_{3} \quad \; +\quad 37.917\, \dot{m}_{4} \quad =\quad 18958.3{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad 0\quad \quad +\quad \quad \quad \; \dot{m}_{4} \quad \; =\quad 492.39\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}\label{ZEqnNum789895}$$

Having worked our way forward through this set of three equations in order to determine $$\dot{m}_{4}$$, we can work our way backward through the set to determine $$\dot{m}_{3}$$ and $$\dot{m}_{2}$$. The results are the same as we obtained earlier in Sec. 4.8 and we list the result again as

$$\dot{m}_{2} \quad =\quad {\rm 219\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{3} \quad =\quad {\rm 288\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{4} \quad =\quad {\rm 492\; lb}_{\rm m} /{\rm hr}\label{ZEqnNum468182}$$

The procedure represented by Eqs. [ZEqnNum453458] through [ZEqnNum789895] is extremely convenient for automated computation and large systems of equations can be quickly solved using a variety of software. Using systems such as MATLAB or Mathematica, problems of this type become quite simple.

4.9.2 Determination of the inverse of a square matrix

The Gaussian elimination procedure described in the previous section is closely related to the determination of the inverse of a square matrix. In order to illustrate how the inverse matrix can be calculated, we consider the coefficient matrix associated with Eq. [ZEqnNum491562] which we express as

$${\rm A}\quad =\quad \left[\begin{array}{ccc} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{array}\right]$$ ()

We can express Eq. [ZEqnNum879461] in the compact form represented by Eq. [ZEqnNum641032]

$${\rm A}\, {\rm u}\quad {\it =}\quad {\it b}\label{ZEqnNum180096}$$

in which u is the column matrix of unknown mass flow rates and b is the column matrix of known mass flow rates.

$${\rm u}\quad =\quad \left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {1000\, lb_{m} /hr} \\ {500\, lb_{m} /hr} \\ {300\, lb_{m} /hr} \end{array}\right]\quad =\quad \left[\begin{array}{c} {b_{2} } \\ {b_{3} } \\ {b_{4} } \end{array}\right]$$ ()

We can also make use of the unit matrix

$${\rm I}\quad =\quad \left[\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]$$ ()

to express Eq. [ZEqnNum180096] in the form

$${\rm A}\, {\rm u}\quad {\it =}\quad {\it I}\, {\it b}\label{ZEqnNum844576}$$

Now we wish to repeat the Gaussian elimination used in Sec. 4.9.1, but in this case we will make use of Eq. [ZEqnNum844576] rather than Eq. [ZEqnNum551051] in order to retain the terms associated with $${\rm I}\, {\rm b}$$. Multiplying the first of Eqs. [ZEqnNum504219] by -0.045 and adding the result to the second equation leads to

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \; \; \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \quad b_{2} \quad \quad +\quad \quad 0\quad +\quad \; 0\quad =\quad 1000\, {\rm lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad 0.024\, \dot{m}_{3} \quad \; +\quad 0.910\, \dot{m}_{4} \quad =\quad \begin{array}{ccc} {-0.045\, b_{2} \quad +} & {b_{3} \quad +} & {0} \end{array}\quad =\quad 455{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0.091\, \dot{m}_{2} \; \; +\; \; 0.901\, \dot{m}_{3} \; \; +\; \; 0.041\, \dot{m}_{4} \quad =\quad \begin{array}{ccc} {\quad 0\quad \quad +} & {\quad \; \; 0\quad +\; } & {b_{4} } \end{array}\quad =\quad 300\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ()

Note that this set of equations is identical to Eqs. [ZEqnNum522104] except that we have included the terms associated with $${\rm I}\, {\rm b}$$. We now proceed with the Gaussian elimination represented by Eqs. [ZEqnNum181816] through [ZEqnNum834391] to arrive at $\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \; \; \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \quad \; \; b_{2} \quad \quad \; +\quad \quad \; \; \; \; 0\quad \; \; \; \; +\quad \; \; \; 0\quad \quad \; \; =\quad \; \; 1000\, {\rm lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad \quad \; \; \dot{m}_{3} \quad \; +\quad 37.917\, \dot{m}_{4} \quad =\quad \begin{array}{ccc} {-1.875\, b_{2} \; \; \; \; +} & {41.667b_{3} \; \; \quad +} & {\; \; 0} \end{array}\quad \quad \; \; =\quad 18958.3{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \; \quad \quad \; \; 0\; \; \quad \quad +\; \; \quad \quad \; \dot{m}_{4} \quad =\quad \begin{array}{ccc} {-0.0464\, b_{2} \; \; +} & {1.0971\, b_{3} \; \; \; -} & {0.03251\, b_{4} \; } \end{array}=\quad 492.39\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$

([ZEqnNum536583])

This result is identical to Eq. [ZEqnNum838645] except for the fact that we have retained the terms in the second column matrix that are associated with $${\rm I}\, {\rm b}$$ in Eq. [ZEqnNum723983]. In Sec. 4.9.1, we used Eq. [ZEqnNum940351] to carry out a backward elimination in order to solve for the mass flow rates given by Eq. [ZEqnNum468182], and here we want to present that backward elimination explicitly in order to demonstrate that it leads to the inverse of the coefficient matrix, A. This procedure is known as the Gauss-Jordan algorithm and it consists of elementary row operations that reduce the left hand side of Eq. [ZEqnNum536583] to a diagonal form.

We begin to construct a diagonal form by multiplying the third equation by -37.917 and adding it to the second equation so that our set of equations take the form $\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \dot{m}_{3} \quad \; +\quad \quad \dot{m}_{4} \; \; \quad \quad =\quad \quad \; \; b_{2} \quad \quad \; +\quad \quad \; \; \; \; 0\quad \; \; \; \; +\quad \; \; \; 0\quad \quad \; \; =\quad \; 1000\, {\rm lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad \; \dot{m}_{3} \quad \; +\quad \quad \; 0\quad \quad \quad =\quad \begin{array}{ccc} {-0.1152\, b_{2} \; \; +\; } & {0.0677b_{3} \quad +} & {1.2326\, b_{4} } \end{array}\; \; =\quad 288.42{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \; \quad 0\; \; \quad \; +\quad \quad \dot{m}_{4} \quad \quad \; \; =\quad \begin{array}{ccc} {-0.0464\, b_{2} \; \; +} & {1.0971\, b_{3} \; \; \; -} & {0.03251\, b_{4} \; \; \; } \end{array}=\quad 492.39\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$

()

We now multiply the third equation by $$-1$$ and add the result to the first equation to obtain $\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \dot{m}_{3} \quad \; +\quad \quad 0\quad \quad \; =\quad \; \; 1.0464\, b_{2} \; \; -\; \; 1.09711\, b_{3} \; \; +\; \; 0.03251\, b_{4} \quad \; =\quad 507.61\, {\rm lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad \; \dot{m}_{3} \quad \; +\quad \quad \; 0\quad \quad =\quad \begin{array}{ccc} {-0.1152\, b_{2} \; \; +} & {0.0677b_{3} \quad +} & {1.2326\, b_{4} } \end{array}\quad \; =\quad 288.42{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \; \quad 0\; \; \quad \; +\quad \quad \dot{m}_{4} \quad \; \; =\quad \begin{array}{ccc} {-0.0464\, b_{2} \; \; +} & {1.0971\, b_{3} \; \; \; -} & {0.03251\, b_{4} \quad } \end{array}=\quad 492.39\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$

()

and finally we multiply the second equation by $$-1$$ and add the result to the first equation to obtain the desired diagonal form $\begin{array}{l} {\dot{m}_{2} \quad +\quad \; 0\quad \; \; +\quad \quad 0\quad \quad =\quad \; \; 1.1616\, b_{2} \; \; -\; \; 1.16484\, b_{3} \; \; -\; \; 1.20005\, b_{4} \quad \; =\quad 219.187{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad \dot{m}_{3} \quad +\quad \quad 0\quad \quad =\quad \begin{array}{ccc} {-0.1152\, b_{2} \; \; +} & {0.0677b_{3} \quad +} & {1.2326\, b_{4} } \end{array}\quad \; =\quad 288.42{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad 0\; \; \quad +\quad \; \; \dot{m}_{4} \quad \quad =\quad \begin{array}{ccc} {-0.0464\, b_{2} \; \; +} & {1.0971\, b_{3} \; \; \; -} & {0.03251\, b_{4} \quad } \end{array}=\quad 492.39{\rm \; lb}_{\rm m} /{\rm hr}} \end{array}$

([ZEqnNum673098])

Here we see that the unknown mass flow rates are determined by

$$\dot{m}_{2} \quad =\quad {\rm 219\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{3} \quad =\quad {\rm 288\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{4} \quad =\quad {\rm 492\; lb}_{\rm m} /{\rm hr}$$ ()

where only three significant figures have been listed since it is unlikely that the input data are accurate to more than 1%. The coefficient matrix contained in the central term of Eqs. [ZEqnNum673098] is the inverse of A and we express this result as

$${\rm A}^{-1} \quad =\quad \left[\begin{array}{ccc} {1.1616} & {-1.16484} & {-1.20005} \\ {-0.1152} & {0.0677} & {1.2326} \\ {-0.0464} & {1.0971} & {-0.0325} \end{array}\right]$$ ()

The solution procedure that led from Eq. [ZEqnNum935304] to the final answer can be expressed in compact form according to,

$${\rm A}\, {\rm u}\quad {\it =}\quad {\it b}\; ,\quad \Rightarrow \quad {\it A}^{-1} {\it A}\, {\it u}\quad {\it =}\quad {\it A}^{-1} \, {\it b}\; ,\quad \Rightarrow \quad {\it I}\, {\it u}\quad {\it =}\quad {\it A}^{-1} \, {\it b}\; ,\quad \Rightarrow \quad {\it u}\quad {\it =}\quad {\it A}^{-1} \, {\it b}$$ ()

and in terms of the details of the inverse matrix the last of these four equations can be expressed as

$$\left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\quad =\quad \left[\begin{array}{ccc} {1.1616} & {-1.16484} & {-1.20005} \\ {-0.1152} & {0.0677} & {1.2326} \\ {-0.0464} & {1.0971} & {-0.0325} \end{array}\right]\; \left[\begin{array}{c} {1000\, lb_{m} /hr} \\ {500\, lb_{m} /hr} \\ {300\, lb_{m} /hr} \end{array}\right]$$ ()

Given that computer routines are available to carry out the Gauss-Jordan algorithm, it should be clear that the solution of a set of linear mass balance equations is a routine matter.

4.10 Problems $${}^{*}$$4

Section 4.1

4-1. Use Eq. [ZEqnNum645577] to obtain a macroscopic mole balance for species A in terms of a moving control volume. Indicate how your result can be used to obtain Eq. [ZEqnNum434393].

Section 4.2

4-2. Determine the mass density, , for the mixing process illustrated in Figure 4-2.

4-3. A liquid hydrocarbon mixture was made by adding 295 kg of benzene, 289 kg of toluene and 287 kg of p-xylene. Assume there is no change of volume upon mixing, i.e., $$\Delta V_{mix} =0$$, in order to determine:

1. The species density of each species in the mixture.

2. The total mass density.

3. The mass fraction of each species.

4-4. A gas mixture contains the following quantities (per cubic meter) of carbon monoxide, carbon dioxide and hydrogen: carbon monoxide, 0.5 kmol/m $${}^{3}$$, carbon dioxide, 0.5 kmol/m $${}^{3}$$, and hydrogen, 0.6 kmol/m $${}^{3}$$. Determine the species mass density and mass fraction of each of the components in the mixture.

4-5. The species mass densities of a three-component (A, B, and C) liquid mixture are: acetone, $$\rho _{A} \; =\; 326.4{\rm \; kg/m}^{\rm 3}$$, acetic acid, $$\rho _{B} =326.4\, {\rm kg/m}^{\rm 3}$$, and ethanol, $$\rho _{C} =217.6{\rm \; kg/m}^{\rm 3}$$. Determine the following for this mixture:

1. The mass fraction of each species in the mixture.

2 The mole fraction of each species in the mixture.

3. The mass of each component required to make one cubic meter of mixture.

4-6. A mixture of gases contains one kilogram of each of the following species: methane (A), ethane (B), propane (C), carbon dioxide (D), and nitrogen (E). Calculate the following:

1. The mole fraction of each species in the mixture

2. The average molecular mass of the mixture

4-7. Two gas streams, having the flow rates and properties indicated in Table 4.7, are mixed in a pipeline. Assume perfect mixing, i.e. no change of volume upon mixing, and determine the composition of the mixed stream in moles/m $${}^{3}$$.

Table 4.7. Composition of gas streams

Stream #1    Stream #2
Mass flow rate    0.226 kg/s    0.296 kg/s
methane    0.48 kg/m $${}^{3}$$    0.16 kg/m $${}^{3}$$
ethane    0.90 kg/m $${}^{3}$$    0.60 kg/m $${}^{3}$$
propane    0.88kg/m3    0.220 kg/m $${}^{3}$$
4-8.Develop a representation for the mole fraction of species A in an N-component system in terms of the mass fractions and molecular masses of the species. Use the result to prove that the mass fractions and mole fractions in a binary system are equal when the two molecular masses are equal.

4-9. Derive the total mass balance for an arbitrary moving control volume beginning with the species mass balance given by Eq. [ZEqnNum996159].

Section 4.3

4-10. The species velocities, in a binary system, can be decomposed according to $\label{GrindEQ__1_} v_{A} \quad =\quad v\; \; +\; \; u_{A} \; ,\quad \quad v_{B} \quad =\quad v\; \; +\; \; u_{B}$ in which v represents the mass average velocity defined by Eq. [ZEqnNum803313]. One can use this result, along with the definition of the mass average velocity, to prove that $\label{GrindEQ__2_} v_{A} \quad =\quad v_{B} \; \; +\; \; \left(\frac{1}{1-\omega _{A} } \right)u_{A}$ This means that the approximation, $$v_{A} \approx v_{B}$$ requires the restriction $\label{GrindEQ__3_} \left|u_{A} \right|\quad <<\quad (1-\omega _{A} )\left|v_{A} \right|$ Since $$1-\omega _{A}$$ is always less than one, we can always satisfy this inequality whenever the mass diffusion velocity is small compared to the species velocity, i.e., $\label{GrindEQ__4_} \left|u_{A} \right|\quad <<\quad \left|v_{A} \right|$ For the sulfur dioxide mass transfer process illustrated in Figure 4-4, this means that the approximation $\label{GrindEQ__5_} v_{\rm SO}_{{\rm 2} } \cdot k\quad \approx \quad v_{air} \cdot k$ is valid whenever the mass diffusion velocity is restricted by $\label{GrindEQ__6_} \left|u_{{\rm SO}_{2} } \cdot k\right|\quad <<\quad \left|v_{{\rm SO}_{2} } \cdot k\right|$ In many practical cases, this restriction is satisfied and all species velocities can be approximated by the mass average velocity.

Next we direct our attention to the mass transfer process at the gas-liquid interface illustrated in Figure 4-5. If we assume that there is no mass transfer of air into or out of the liquid phase, we can prove that $\label{GrindEQ__7_} u_{\rm SO}_{2} } \cdot n\quad =\quad \left(1-\omega _{{\rm SO}_{{\rm 2} } \right)v_{{\rm SO}_{2} } \cdot n\; ,\quad \quad at\; the\; gas-liquid\; interface$ Under these circumstances, the mass diffusion velocity is never small compared to the species velocity for practical conditions. Thus the type of approximation indicated by Eq. [ZEqnNum668645] is never valid for the component of the velocity normal to the gas-liquid interface. As a simplification, we can treat the sulfur dioxide-air system as a binary system with species A representing the sulfur dioxide and species B representing the air.

(a). Use the definition of the mass average velocity given by Eq. [ZEqnNum146973] to prove Eq. 2.

(b) Use $$v_{\rm air} \cdot n\; \; =\; \; v_{B} \cdot n\; \; =\; \; 0$$ in order to prove Eq. 7.

Section 4.4

4-11. A three component liquid mixture flows in a pipe with a mass averaged velocity of $${\rm v}=0.9\, {\rm m/s}$$. The density of the mixture is $$\rho$$ = 850 kg/m $${}^{3}$$. The components of the mixture and their mole fractions are:n-pentane, $$x_{P} =0.2$$, benzene, $$x_{B} =0.3$$, and naphthalene, $$x_{N} =0.5$$. The diffusion fluxes of each component in the streamwise direction are: pentane, $$\rho _{P} \, u_{P} =1{\rm .564}\times {\rm 10}^{-6} {\rm kg/m}^{\rm 2} {\rm s}$$, benzene, $$\rho _{B} \, u_{B} =1{\rm .563}\times 1{\rm 0}^{-6} \, {\rm kg/m}^{\rm 2} {\rm s}$$, and naphthalene, $$\rho _{N} \, u_{N} =-3.127\times {\rm 10}^{-6} \, {\rm kg/m}^{\rm 2} {\rm s}$$. Determine the diffusion velocities and the species velocities of the three components. Use this result to determine the molar averaged velocity, $${\rm v}^{*}$$. Note that you must use eight significant figures in your computation.

Section 4.5

4-12. Sometimes heterogeneous chemical reactions take place at the walls of tubes in which reactive mixtures are flowing. If species A is being consumed at a tube wall because of a chemical reaction, the concentration profile may be of the form $\label{GrindEQ__1_} c_{A} {\rm (}r{\rm )}\quad =\quad c_{A}^{\rm o} \left[1-\phi \left({r\mathord{\left/ {\vphantom {r r_{\rm o} }} \right. \kern-\nulldelimiterspace} r_{\rm o} } \right)^{2} \right]$ Here r is the radial position and $$r_{\rm o}$$ is the tube radius. The parameter  depends on the net rate of production of chemical species at the wall and the molecular diffusivity, and it is bounded by $$0\le \phi \le 1$$. If  is zero, the concentration across the tube is uniform at the value $$c_{A}^{\rm o}$$. If the flow in the tube is laminar, the velocity profile is given by $\label{GrindEQ__2_} {\rm v}_{z} {\rm (}r{\rm )}\quad =\quad 2\langle {\rm v}_{z} \rangle \left[1-\left({r\mathord{\left/ {\vphantom {r r_{\rm o} }} \right. \kern-\nulldelimiterspace} r_{\rm o} } \right)^{2} \right]$ and the volumetric flow rate is $\label{GrindEQ__3_} Q\quad =\quad \langle {\rm v}_{z} \rangle \, \pi r_{\rm o}^{\rm 2}$ For this process, determine the molar flow rate of species A in terms $$c_{A}^{\rm o}$$, $$\phi$$, and $$\langle {\rm v}_{z} \rangle$$. When $$\phi =0.5$$, determine the bulk concentration, $$\langle c_{A} \rangle _{b}$$,and the area-averaged concentration, $$\langle c_{A} \rangle$$. Use these results to determine the difference between $$\dot{M}_{A}$$ and $$\langle c_{A} \rangle Q$$.

4-13. A flash unit is used to separate vapor and liquid streams from a liquid stream by lowering its pressure before it enters the flash unit. The feed stream is pure liquid water and its mass flow rate is 1000 kg/hr. Twenty percent (by weight) of the feed stream leaves the flash unit with a density $$\rho$$ = 10 kg/m $${}^{3}$$. The remainder of the feed stream leaves the flash unit as liquid water with a density $$\rho$$ = 1000 kg/m $${}^{3}$$ Determine the following:

1. The mass flow rates of the exit streams in kg/s.

2. volumetric flow rates of exit streams in m $${}^{3}$$/s.

Section 4.6

4-14. Show that Eq. [ZEqnNum692829] results from Eq. [ZEqnNum839821] when either $$c\, v\cdot n$$ or $$x_{A}$$ is constant over the area of the exit.

4-15. Use Eq. [ZEqnNum443450] to prove Eq.[ZEqnNum181666].

4-16. Derive Eq. [ZEqnNum700852] given that either $$\rho \, v\cdot n$$ or $$\omega _{A}$$ is constant over the area of the exit.

4-17. Prove Eq.[ZEqnNum359366].

Section 4.7

4-18. Determine $$\dot{M}_{3}$$ and the unknown mole fractions for the distillation process described in Sec. 4.7 subject to the following conditions:

Stream #1    Stream #2    Stream #3
$$\dot{M}_{1} =1200{\rm \; mol/hr}$$    $$\dot{M}_{2} =250{\rm \; mol/hr}$$    $$\dot{M}_{3} =?$$
$$x_{A} \; \; =\; \; 0.3$$    $$x_{A} \; \; =\; \; 0.8$$    $$x_{A} \; \; =\; \; ?$$
$$x_{B} \; \; =\; \; 0.2$$    $$x_{B} \; \; =\; \; ?$$    $$x_{B} \; \; =\; \; 0.25$$
$$x_{C} \; \; =\; \; ?$$    $$x_{C} \; \; =\; \; ?$$    $$x_{C} \; \; =\; \; ?$$
4-19. A continuous filter is used to separate a clear filtrate from alumina particles in a slurry. The slurry has 30% by weight of alumina (specific gravity of alumina = 4.5). The cake retains 5% by weight of water. For a feed stream of 1000 kg/hr, determine the following:

1. The mass flow rate of particles and water in the input stream

2. The volumetric flow rate of the inlet stream in m $${}^{3}$$/s.

3. The mass flow rate of filtrate and cake in kg/s.

4-20. A BTX unit, shown in Figure 4.20, is associated with a refinery that produces benzene, toluene, and xylenes. Stream #1 leaving the reactor-reforming unit has a volumetric flow rate of 10 m $${}^{3}$$/hr and is a

Figure 4.20. Reactor and distillation unit

mixture of benzene (A), toluene (B), and xylenes (C) with the following composition: $\langle c_{A} \rangle _{1} \quad =\quad {\rm 6,000}\; {\rm mol/m}^{\rm 3} \; ,\quad \quad \langle c_{B} \rangle _{1} \quad =\quad 2,000\, \, {\rm mol/m}^{\rm 3} \; ,\quad \quad \langle c_{C} \rangle _{1} \quad =\quad 2,000\; {\rm mol/m}^{\rm 3}$ Stream [GrindEQ__1_] is the feed to a distillation unit where the separation takes place according to the following specifications:

1. 98% of the benzene leaves with the distillate stream (stream #2).

2. 99% of the toluene in the feed leaves with the bottoms stream (stream #3)

3. 100% of the xylenes in the feed leaves with the bottoms stream (stream #3).

Assuming that the volumes of components are additive, and using the densities of pure components from Table I in the Appendix, compute the concentration and volumetric flow rate of the distillate (stream #2) and bottoms (stream #3) stream leaving the distillation unit.

4-21. A standard practice in refineries is to use a holding tank in order to mix the light naphtha output of the refinery for quality control. During the first six hours of operation of the refinery, the stream feeding the holding tank at 200 kg/min had 30% by weight of n-pentane, 40% by weight of n-hexane, 30% by weight of n-heptane. During the next 12 hours of operation the mass flow rate of the feed stream was 210 kg/min and the composition changed to 40% by weight of n-pentane, 40% by weight of n-hexane, and 20% by weight of n-heptane. Determine the following:

The average density of the feed streams

The concentration of the feed streams in moles/m $${}^{3}$$.

After 12 hours of operation, and assuming the tank was empty at the beginning, determine:

The volume of liquid in the tank in m $${}^{3}$$.

The concentration of the liquid in the tank, in mol/m $${}^{3}$$.

The partial density of the species in the tank.

4-22. A distillation column is used to separate a mixture of methanol, ethanol, and isopropyl alcohol. The feed stream, with a mass flow rate of 300 kg/hr, has the following composition:

Component    Species mass density
methanol    395.5 kg/m $${}^{3}$$
ethanol    197.3 kg/m $${}^{3}$$
isopropyl alcohol    196.5 kg/m $${}^{3}$$
Separation of this mixture of alcohol takes place according to the following specifications:

(a) 90% of the methanol in the feed leaves with the distillate stream

(b) 5% of the ethanol in the feed leaves with the distillate stream

(c) 3% of the isopropyl alcohol in the feed leaves with the distillate stream.

Assuming that the volumes of the components are additive, compute the concentration and volumetric flow rates of the distillate and bottom streams.

4-23. A mixture of ethanol (A) and water (B) is separated in a distillation column. The volumetric flow rate of the feed stream is 5 m $${}^{3}$$/hr. The concentration of ethanol in the feed is $$c_{A} =2,800\, {\rm mol/m}^{\rm 3}$$. The distillate leaves the column with a concentration of ethanol $$c_{A} =13,000\, {\rm mol/m}^{\rm 3}$$. The volumetric flow rate of distillate is one cubic meter per hour. How much ethanol is lost through the bottoms of the column, in kilograms of ethanol per hour?

4-24. A ternary mixture of benzene, ethylbenzene, and toluene is fed to a distillation column at a rate of $$10^{5}$$mol/hr. The composition of the mixture in % moles is: 74% benzene, 20% toluene, and 6% ethylbenzene. The distillate flows at a rate of $$75\times 10^{3}$$mol/hr. The composition of the distillate in % moles is 97.33 % benzene, 2% toluene, and the rest is ethylbenzene. Find the molar flow rate of the bottoms stream and the mass fractions of the three components in the distillate and bottoms stream.

4-25. A complex mixture of aromatic compounds leaves a chemical reactor and is fed to a distillation column. The mass fractions and flow rates of distillate and bottoms streams are given in Table 4.25. Compute the molar flow rate and composition, in molar fractions, of the feed stream.

Table 4.25 Flow rate and composition of distillate and bottoms streams.

(kg/hr)    $$\omegaup$$ $${}_{Benzene}$$    $$\omegaup$$ $${}_{Toluene}$$    $$\omegaup$$ $${}_{Benzaldehide}$$    $$\omegaup$$ $${}_{BenzoicAcid}$$    $$\omegaup$$ $${}_{MethylBenzoate}$$
Distillate    125    0.1    0.85    0.03    0.0    0.02
Bottoms    76    0.0    0.05    0.12    0.8    0.03
4-26. A hydrocarbon feedstock is available at a rate of $$10^{6}$$mol/hr, and consists of propane ( $$x_{A} =0.2$$), n-butane ( $$x_{B} =0.3$$), n-pentane ( $$x_{C} =0.2$$) and n-hexane ( $$x_{D} =0.3$$). The distillate contains all of the propane in the feed to the unit and 80% of the pentane fed to the unit. The mole fraction of butane in the distillate is $$y_{B} =0.4$$. The bottom stream contains all of the hexane fed to the unit. Calculate the distillate and bottoms streams flow rate and composition in terms of mole fractions.

Section 4.8

4-27. It is possible that the process illustrated in Figure 4-12 could be analyzed beginning with Control Volume I rather than beginning with Control Volume II. Begin the problem with Control Volume I and carry out a degree-of-freedom analysis to see what difficulties might be encountered.

4-28. In a glycerol plant, a 10% (mass basis) aqueous glycerin solution containing 3% NaCl is treated with butyl alcohol as illustrated in Figure 4.28. The alcohol fed to the tower contains 2% water on a mass basis. The raffinate leaving the tower contains all the original salt, 1.0% glycerin and 1.0% alcohol. The extract from the tower is sent to a distillation column. The distillate from this column is the alcohol containing 5% water.

Figure 4.28. Solvent extraction process

The bottoms from the distillation column are 25% glycerin and 75% water. The two feed streams to the extraction tower have equal mass flow rates of 1000 lbm per hour. Determine the output of glycerin in pounds per hour from the distillation column.

Section 4.9

4-29. In Sec. 4.8 the solution to the distillation problem was shown to reduce to solving the matrix equation, $${\rm A}\, {\rm m}={\rm b}$$, in which ${\rm A}\quad =\quad \left[\begin{array}{ccc} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{array}\right]\; ,\quad \quad {\rm u}\quad =\quad \left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {1000} \\ {500} \\ {300} \end{array}\right]$ Here it is understood that the mass flow rates have been made dimensionless by dividing by $${\rm lb}_{\rm m} /{\rm hr}$$. In addition to the matrix A, one can form what is known as an augmented matrix. This is designated by $${\rm A}\dot{\dot{\cdot }}{\rm b}$$ and it is constructed by adding the column of numbers in b to the matrix A in order to obtain ${\rm A}\dot{\dot{\cdot }}{\rm b}\quad =\quad \left[\begin{array}{ccccc} {1} & {1} & {1} & {.} & {1000} \\ {0.045} & {0.069} & {0.955} & {.} & {500} \\ {0.091} & {0.901} & {0.041} & {.} & {300} \end{array}\right]$ Define the following lists in Mathematica corresponding to the rows of the augmented matrix, $${\rm A}\dot{\dot{\cdot }}{\rm b}$$. $\begin{array}{l} {R1\quad =\quad \left\{\begin{array}{cccc} {1,} & {1,} & {1,} & {1000} \end{array}\right\}} \\ {R2\quad =\quad \left\{\begin{array}{cccc} {0.045,} & {0.069,} & {0.955,} & {500} \end{array}\right\}} \\ {R3\quad =\quad \left\{\begin{array}{cccc} {0.091,} & {0.901,} & {0.041,} & {300} \end{array}\right\}} \end{array}$ Write a sequence of Mathematica expressions that correspond to the elementary row operations for solving this system. The first elementary row operation that given Eq. [ZEqnNum270394] is $R2\quad =\quad (-0.045)\; R2\; \; +\; \; R1$ Show that you obtain an augmented matrix that defines Eq. [ZEqnNum938375].

4-30. In this problem you are asked to continue exploring the use of Mathematica in the analysis of the set of linear equations studied in Sec. 4.9.2, i.e., $${\rm A}\, {\rm m}={\rm b}$$ where the matrices are defined by ${\rm A}\quad =\quad \left[\begin{array}{ccc} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{array}\right]\; ,\quad \quad {\rm u}\quad =\quad \left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {1000} \\ {500} \\ {300} \end{array}\right]$ [GrindEQ__1_] Construct the augmented matrix $${\rm A}\dot{\dot{\cdot }}{\rm I}$$ according to ${\rm A}\dot{\dot{\cdot }}{\rm I}\quad =\quad \left[\begin{array}{ccccccc} {1} & {1} & {1} & {.} & {1} & {0} & {0} \\ {0.045} & {0.069} & {0.955} & {.} & {0} & {1} & {0} \\ {0.091} & {0.901} & {0.041} & {.} & {0} & {0} & {1} \end{array}\right]$ and use elementary row operations to transform this augmented matrix to the form ${\rm A}\dot{\dot{\cdot }}{\rm I}\quad =\quad \left[\begin{array}{ccccccc} {1} & {0} & {0} & {.} & {} & {} & {} \\ {0} & {1} & {0} & {.} & {} & {\it B} & {} \\ {0} & {0} & {1} & {.} & {} & {} & {} \end{array}\right]$ Show that the elements represented by B make up the matrix B having the property that $${\rm B}={\rm A}^{-1}$$. Use your result to calculate $${\rm m}={\rm A}^{-1} \, {\rm b}$$.

[GrindEQ__2_] Show that the inverse found in Part 1 satisfies $${\rm A}\, {\rm A}^{-1} ={\rm I}$$.

[GrindEQ__3_] Use Mathematica’s built-in function Inverse to find the inverse of A.

[GrindEQ__4_] Use Mathematica’s RowReduce function on the augmented matrix $${\rm A}\dot{\dot{\cdot }}{\rm b}$$ and show that from the row echelon form you can obtain the same results as in [GrindEQ__1_].

[GrindEQ__5_] Use Mathematica’s Solve function to solve $${\rm A}\, {\rm m}={\rm b}$$.

. Lavoisier, A. L. 1777,. Memoir on Combustion in General,  Mémoires de L’Academie Roayal des Sciences  592-600.↩

. Toulmin, S.E. 1957, Crucial Experiments: Priestley and Lavoisier, Journal of the History of Ideas,  18 , 205-220.↩

. See Section 2.1.1.↩

* Problems marked with the symbol  will be difficult to solve without the use of computer software.↩

Stream #2 Stream #3
$$\dot{M}_{1} =1200{\rm \; moles/hr}$$
$$x_{A} \; \; =\; \; 0.3$$ $$x_{A} \; \; =\; \; 0.6$$ $$x_{A} \; \; =\; \; 0.1$$
$$x_{B} \; \; =\; \; 0.2$$ $$x_{B} \; \; =\; \; 0.3$$

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

SnplaceRule SnI. 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 $$v_{A} \cdot n$$ 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 $$v_{A}$$ 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 4-8, it should be obvious that we need to cut the entrance and exit streams and then join the cuts as illustrated in Figure 4-9 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 $$v_{A} \cdot n=0$$. Figure 4-9. Control volume for distillation column

4.7.1 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: $$(x_{A} )_{i} \; ,\; \; (x_{B} )_{i} \; ,\; \; (x_{C} )_{i} \quad \quad i=1,2,3$$ ([ZEqnNum729411])

Molar flow rates: $$\dot{M}_{i} \; ,\quad i=1,2,3$$ ([ZEqnNum779582])

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: $$\int _{A}x_{A} \, cv\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}x_{B} \, cv\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}x_{C} \, cv\cdot n\, dA \; \; =\; \; 0$$ ([ZEqnNum156558])

along with the three mole fraction constraints that apply at the streams that are cut by the control volume illustrated in Figure 4-9.

Mole fraction constraints: $$(x_{A} )_{i} \; \; +\; \; (x_{B} )_{i} \; \; +\; \; (x_{C} )_{i} \quad =\quad 1\; ,\quad \quad i=1,2,3$$ ([ZEqnNum157867])

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: $$\dot{M}_{1} =1200{\rm \; moles/hr}\; ,\quad x_{A} =0.3\; ,\quad x_{B} =0.2$$ 3

II. Conditions for Stream #2: $$x_{A} \; \; =\; \; 0.6\; ,\quad x_{B} \; \; =\; \; 0.3$$ 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 Eq. [ZEqnNum838824]. If one specifies all three mole fractions in a particular stream, Eq. [ZEqnNum199001] 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 4-2 that provides a template for subsequent problems in which we have N molecular species and M streams.

Table 4-2. Degrees-of-Freedom

compositions

flow rates

## Number of Independent Balance Equations

mass/mole balance equations

Generic Constraints (B)

compositions

flow rates

## Particular Specifications and Constraints (C) 6

Degrees of Freedom (A - B - C)

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.

4.7.2 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 4-1 given here as

Table 4-3. Specified and unknown conditions

Stream #1 Stream #2 Stream #3
$$\dot{M}_{1} =1200{\rm \; moles/hr}$$ ? ?
$$x_{A} \; \; =\; \; 0.3$$ $$x_{A} \; \; =\; \; 0.6$$ $$x_{A} \; \; =\; \; 0.1$$
$$x_{B} \; \; =\; \; 0.2$$ $$x_{B} \; \; =\; \; 0.3$$ ?
? ? ?

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. [ZEqnNum106642]. These can be used to express Table 4-3 as follows:

Table 4-4. Unknowns to be determined

Stream #1 Stream #2 Stream #3
$$\dot{M}_{1} =1200{\rm \; moles/hr}$$ ? ?
$$x_{A} \; \; =\; \; 0.3$$ $$x_{A} \; \; =\; \; 0.6$$ $$x_{A} \; \; =\; \; 0.1$$
$$x_{B} \; \; =\; \; 0.2$$ $$x_{B} \; \; =\; \; 0.3$$ $$x_{B} \; \; =\; \; 0.9-x_{C}$$
$$x_{c} \; \; =\; \; 0.5$$ $$x_{c} \; \; =\; \; 0.1$$ ?

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 4-4 allows us to express the balance equations given by Eqs. [ZEqnNum156558] as

Species A: $$0.6\; \dot{M}_{2} \; \; +\; \; 0.1\; \dot{M}_{3} \quad +\quad \quad 0\quad \quad \quad =\quad {\rm 360\; moles/hr}$$ ([ZEqnNum420022]a)

Species B: $$0.3\; \dot{M}_{2} \; \; +\; \; 0.9\; \dot{M}_{3} \; \; -\; \; \; \underbrace{\dot{M}_{3} \langle x_{C} \rangle _{3} }_{\begin{array}{l} {\rm bi-linear} \\ {\rm \; \; \; form} \end{array}}\quad =\quad 240{\rm \; moles/hr}$$ ([ZEqnNum420022]b)

Species C: $$0.1\; \dot{M}_{2} \; \; +\; \; \quad 0\quad +\; \; \quad \underbrace{\dot{M}_{3} \langle x_{C} \rangle _{3} }_{\begin{array}{l} {\rm bi-linear} \\ {\rm \; \; \; form} \end{array}}\quad \; \; =\quad 600{\rm \; moles/hr}$$ ([ZEqnNum967271]c)

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

$$\dot{M}_{2} \; \; =\; \; 480{\rm \; mol/hr}\; ,\quad \quad \dot{M}_{3} \; \; =\; \; 720{\rm \; mol/hr}\; ,\quad \quad \langle x_{C} \rangle _{3} \quad =\quad 0.767$$ ()

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

Table 4-5. Solution for molar flows and mole fractions

Stream #1 Stream #2 Stream #3
$$\dot{M}_{1} =1200{\rm \; mol/hr}$$ $$\dot{M}_{2} =480{\rm \; mol/hr}$$ $$\dot{M}_{3} =720{\rm \; mol/hr}$$
$$x_{A} \; \; =\; \; 0.3$$ $$x_{A} \; \; =\; \; 0.6$$ $$x_{A} \; \; =\; \; 0.100$$
$$x_{B} \; \; =\; \; 0.2$$ $$x_{B} \; \; =\; \; 0.3$$ $$x_{B} \; \; =\; \; 0.133$$
$$x_{C} \; \; =\; \; 0.5$$ $$x_{C} \; \; =\; \; 0.1$$ $$x_{C} \; \; =\; \; 0.767$$

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. [ZEqnNum424664] 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.

4.7.3 Solution of sets of equations

To illustrate a classic procedure for solving sets of algebraic equations, we direct our attention to Eqs. [ZEqnNum192373]. We begin by eliminating the term, $$\dot{M}_{3} \langle x_{C} \rangle _{3}$$, from Eq. [ZEqnNum622418]b to obtain the following pair of linear equations:

Species A: $$0.6\; \dot{M}_{2} \; \; +\; \; 0.1\; \dot{M}_{3} \quad =\quad 360{\rm \; mol/hr}$$ ([ZEqnNum777303]a)

Species B: $$0.4\; \dot{M}_{2} \; \; +\; \; 0.9\; \dot{M}_{3} \quad =\quad 840{\rm \; mol/hr}$$ ([ZEqnNum777303]b)

To solve this set of linear equations, we make use of a simple scheme known as Gaussian elimination. We begin by dividing Eq. [ZEqnNum813064]a by the coefficient 0.6 in order to obtain

$$\dot{M}_{2} \; \; +\; \; 0.1667\; \dot{M}_{3} \quad =\quad 600{\rm \; mol/hr}$$ ([ZEqnNum806894]a)

$$0.4\; \dot{M}_{2} \; \; +\; \; 0.9\; \dot{M}_{3} \quad =\quad 840{\rm \; mol/hr}$$ ([ZEqnNum806894]b)

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

$$\dot{M}_{2} \; \; +\; \; 0.1667\; \dot{M}_{3} \quad =\quad 600{\rm \; mol/hr}$$ ([ZEqnNum490788]a)

$$0.8333\; \dot{M}_{3} \quad =\quad 600{\rm \; mol/hr}$$ ([ZEqnNum490788]b)

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

$$\dot{M}_{2} \; \; +\; \; 0.1667\; \dot{M}_{3} \quad =\quad 600{\rm \; mol/hr}$$ ([ZEqnNum272303]a)

$$\dot{M}_{3} \quad =\quad 720{\rm \; mol/hr}$$ ([ZEqnNum272303]b)

At this point, we begin the procedure of “back substitution” which requires that Eq. [ZEqnNum663920]b be substituted into Eq. [ZEqnNum185945]a in order to obtain the final solution for the two molar flow rates.

$$\dot{M}_{2} \; \quad =\quad 480{\rm \; mol/hr}$$ ([ZEqnNum929408]a)

$$\dot{M}_{3} \quad =\quad 720{\rm \; mol/hr}$$ ([ZEqnNum929408]b)

The procedure leading from Eqs. [ZEqnNum938569] to the solution given by Eqs. [ZEqnNum854642] 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.

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 4-10 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 $$1000{\rm \; lb}_{\rm m} /{\rm hr}$$. 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 4-10. 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 4-11. We first form Control Volume I which Figure 4-10. Two-column distillation unit

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 4-12. 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 Eq. [ZEqnNum323637]. For steady-state conditions in the absence of chemical reactions, the species mass balances are given by

$$\int _{A}\rho _{A} v_{A} \cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\rho _{B} v_{B} \cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\rho _{C} v_{C} \cdot n\, dA \; \; =\; \; 0$$ () Figure 4-11. Primary cuts for the two-column distillation unit

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: $$\int _{A}\omega _{A} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\omega _{B} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\omega _{C} \, \rho \, v\cdot n\, dA \; \; =\; \; 0$$ ()

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: $$(\omega _{A} )_{i} \; \; +\; \; (\omega _{B} )_{i} \; \; +\; \; (\omega _{C} )_{i} \quad =\quad 1\; ,\quad \quad i=1,2,3,4,5$$ () Figure 4-12. Control volumes for two-column distillation unit

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

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

Mass flow rates: $$\dot{m}_{\, i} \; ,\quad i=1,2,3,4$$ ()

and we indicate the number of process variables explicitly as

I. Three mole fractions in each of four streams

II. Four mass molar flow rates

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: $$\int _{A}\omega _{A} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\omega _{B} \, \rho \, v\cdot n\, dA \; \; =\; \; 0\; ,\quad \quad \int _{A}\omega _{C} \, \rho \, v\cdot n\, dA \; \; =\; \; 0$$ ([ZEqnNum647193])

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

Mass fraction constraints: $$(\omega _{A} )_{i} \; \; +\; \; (\omega _{B} )_{i} \; \; +\; \; (\omega _{C} )_{i} \quad =\quad 1\; ,\quad \quad i=1,2,3,4$$ ([ZEqnNum264060])

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

I. Balance equations for three molecular species

II. Mole fraction constraints for the four streams

Generic Specifications and Constraints (B)

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: $$\dot{m}_{\, 1} =1000{\rm \; lb}_{\rm m} /{\rm hr}\; ,\quad \omega _{A} \; \; =\; \; 0.5\; ,\quad \omega _{B} \; \; =\; \; 0.3$$

II. Conditions for Stream #2: $$\omega _{A} \; \; =\; \; 0.045\; ,\quad \omega _{B} \; \; =\; \; 0.091$$

III. Conditions for Stream #3: $$\omega _{A} \; \; =\; \; 0.069\; ,\quad \omega _{B} \; \; =\; \; 0.901$$

IV. Conditions for Stream #4: $$\omega _{A} \; \; =\; \; 0.955\; ,\quad \omega _{B} \; \; =\; \; 0.041$$

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 4-6. Degrees-of-Freedom

compositions

flow rates

## Number of Independent Balance Equations

mass/mole balance equations

Generic Specifications and Constraints (B)

compositions

flow rates

## Particular Specifications and Constraints (C) 9

Degrees of Freedom (A - B - C)

This indicates that use of Control Volume II will lead to a well-posed problem, and we are assured that we can use Eqs. [ZEqnNum647193] and [ZEqnNum264060] 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 Eq. [ZEqnNum386474]. In terms of Eqs. [ZEqnNum696072] for Control Volume II, this approach leads to

:

species A: $$(\omega _{A} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{A} )_{3} \dot{m}_{3} \; \; +\; \; (\omega _{A} )_{4} \dot{m}_{4} \quad =\quad (\omega _{A} )_{1} \dot{m}_{1}$$ ([ZEqnNum825073]a)

species B: $$(\omega _{B} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{B} )_{3} \dot{m}_{3} \; \; +\; \; (\omega _{B} )_{4} \dot{m}_{4} \quad =\quad (\omega _{B} )_{1} \dot{m}_{1}$$ ([ZEqnNum825073]b)

Total: $$\dot{m}_{2} \quad \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \quad \; \; \dot{m}_{1}$$ ([ZEqnNum397112]c)

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. [ZEqnNum797722] through [ZEqnNum131831]. The same procedure can be used with Eqs. [ZEqnNum595841], and one begins by dividing Eq. [ZEqnNum509481]a by $$(\omega _{A} )_{2}$$ to obtain

:

species A: $$\dot{m}_{2} \; \; +\; \; \left[{(\omega _{A} )_{3} \mathord{\left/ {\vphantom {(\omega _{A} )_{3} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{3} \; \; +\; \; \left[{(\omega _{A} )_{4} \mathord{\left/ {\vphantom {(\omega _{A} )_{4} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{4} \quad =\quad \left[{(\omega _{A} )_{1} \mathord{\left/ {\vphantom {(\omega _{A} )_{1} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{1}$$ ([ZEqnNum343114]a)

species B: $$(\omega _{B} )_{2} \dot{m}_{2} \quad \; \; +\quad \; \; (\omega _{B} )_{3} \dot{m}_{3} \quad \; \; +\quad \; \; (\omega _{B} )_{4} \dot{m}_{4} \quad =\quad \quad (\omega _{B} )_{1} \dot{m}_{1}$$ ([ZEqnNum343114]b)

Total: $$\dot{m}_{2} \quad \quad \quad +\quad \quad \quad \dot{m}_{3} \quad \quad \quad +\quad \quad \quad \dot{m}_{4} \quad \quad =\quad \quad \; \; \dot{m}_{1}$$ ([ZEqnNum326641]c)

In order to eliminate $$\dot{m}_{2}$$ from Eq. [ZEqnNum461768]b, one multiplies Eq. [ZEqnNum126289]a by $$(\omega _{B} )_{2}$$ and subtracts the result from both Eq. [ZEqnNum865208]b. To eliminate $$\dot{m}_{2}$$ from Eq. [ZEqnNum282378]c, one need only subtract Eq. [ZEqnNum284549]a from Eq. [ZEqnNum643704]c, and these two operations lead to the following balance equations:

:

species A: $$\dot{m}_{2} \; \; +\; \; \left[{(\omega _{A} )_{3} \mathord{\left/ {\vphantom {(\omega _{A} )_{3} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{3} \; \; +\; \; \left[{(\omega _{A} )_{4} \mathord{\left/ {\vphantom {(\omega _{A} )_{4} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{4} \quad =\quad \left[{(\omega _{A} )_{1} \mathord{\left/ {\vphantom {(\omega _{A} )_{1} (\omega _{A} )_{2} }} \right. \kern-\nulldelimiterspace} (\omega _{A} )_{2} } \right]\dot{m}_{1}$$ ([ZEqnNum656618]a)

species B: $$\left\{.......\right\}\dot{m}_{3} \quad \quad \quad +\quad \quad \left\{.......\right\}\dot{m}_{4} \quad =\quad \quad \left\{.......\right\}\dot{m}_{1}$$ ([ZEqnNum656618]b)

Total: $$\left[......\right]\dot{m}_{3} \quad \quad \quad +\quad \quad \left[......\right]\dot{m}_{4} \; \; \quad =\quad \quad \left[......\right]\dot{m}_{1}$$ ([ZEqnNum994624]c)

Clearly the algebra is becoming quite complex, and it will become worse when we use Eq. [ZEqnNum894051]b to eliminate $$\dot{m}_{3}$$ from Eq. [ZEqnNum333495]c. Without providing the details, we continue the elimination process to obtain the solution to Eq. [ZEqnNum777374]c and this leads to the following expression for $$\dot{m}_{4}$$:

$$\dot{m}_{4} \quad \quad =\quad \quad \dot{m}_{1} \; \; \frac{\left[1-\frac{(\omega _{A} )_{1} }{(\omega _{A} )_{2} } \right]\quad -\quad \frac{\frac{(\omega _{B} )_{1} }{(\omega _{B} )_{3} } \left[1-\frac{(\omega _{B} )_{2} }{(\omega _{B} )_{1} } \frac{(\omega _{A} )_{1} }{(\omega _{A} )_{2} } \right]}{\left[1-\frac{(\omega _{B} )_{2} }{(\omega _{B} )_{3} } \frac{(\omega _{A} )_{3} }{(\omega _{A} )_{2} } \right]} \left[1-\frac{(\omega _{A} )_{3} }{(\omega _{A} )_{2} } \right]}{\left[1-\frac{(\omega _{A} )_{4} }{(\omega _{A} )_{2} } \right]\quad -\quad \frac{\frac{(\omega _{B} )_{4} }{(\omega _{B} )_{3} } \left[1-\frac{(\omega _{B} )_{2} }{(\omega _{B} )_{4} } \frac{(\omega _{A} )_{4} }{(\omega _{A} )_{2} } \right]}{\left[1-\frac{(\omega _{B} )_{2} }{(\omega _{B} )_{3} } \frac{(\omega _{A} )_{3} }{(\omega _{A} )_{2} } \right]} \left[1-\frac{(\omega _{A} )_{3} }{(\omega _{A} )_{2} } \right]}$$ ([ZEqnNum840659])

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

$$\dot{m}_{2} \quad =\quad {\rm 220\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{3} \quad =\quad {\rm 288\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{4} \quad =\quad {\rm 492\; lb}_{\rm m} /{\rm hr}$$ ()

In order to determine $$\dot{m}_{5}$$, we must make use of the balance equations for Control Volume I that are given by Eqs. [ZEqnNum959841]. These can be expressed in terms of two species balances and one total mass balance leading to

:

species A: $$-\; (\omega _{A} )_{1} \dot{m}_{1} \; \; +\; \; (\omega _{A} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{A} )_{5} \dot{m}_{5} \quad =\quad 0$$ ([ZEqnNum679073]a)

species B: $$-\; (\omega _{B} )_{1} \dot{m}_{1} \; \; +\; \; (\omega _{B} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{B} )_{5} \dot{m}_{5} \quad =\quad 0$$ ([ZEqnNum679073]b)

Total: $$-\; \; \dot{m}_{1} \quad \quad +\quad \quad \dot{m}_{2} \quad \quad +\quad \quad \dot{m}_{5} \quad =\quad 0$$ ([ZEqnNum294734]c)

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

$$\dot{m}_{5} \quad =\quad {\rm 780\; lb}_{\rm m} /{\rm hr}$$ ()

The algebraic complexity associated with the simple process represented in Figure 4-10 encourages the use of matrix methods that are described in Sec. 4.9.

4.9 Matrix Algebra

In Sec. 4.7 we examined a distillation process with the objective of determining molar flow rates, and our analysis led to a set of two equations and two unknowns given by

Species A: $$0.6\; \dot{M}_{2} \; \; +\; \; 0.1\; \dot{M}_{3} \quad =\quad 360{\rm \; mol/hr}$$ ([ZEqnNum153881]a)

Species B: $$0.4\; \dot{M}_{2} \; \; +\; \; 0.9\; \dot{M}_{3} \quad =\quad 840{\rm \; mol/hr}$$ ([ZEqnNum153881]b)

Some thought was necessary in order to set up the macroscopic mole balances for the process illustrated in Figure 4-7; however, the algebraic effort required to solve the governing macroscopic balances was trivial. In Sec. 4.8, we considered the system illustrated in Figure 4-10 and the analysis led to the following set of three equations and three unknowns:

species A: $$(\omega _{A} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{A} )_{3} \dot{m}_{3} \; \; +\; \; (\omega _{A} )_{4} \dot{m}_{4} \quad =\quad (\omega _{A} )_{1} \dot{m}_{1}$$ ([ZEqnNum596473]a)

species B: $$(\omega _{B} )_{2} \dot{m}_{2} \; \; +\; \; (\omega _{B} )_{3} \dot{m}_{3} \; \; +\; \; (\omega _{B} )_{4} \dot{m}_{4} \quad =\quad (\omega _{B} )_{1} \dot{m}_{1}$$ ([ZEqnNum596473]b)

Total: $$\dot{m}_{2} \quad \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \quad \; \; \dot{m}_{1}$$ ([ZEqnNum810328]c)

The algebraic effort required to solve these three equations for $$\dot{m}_{2}$$, $$\dot{m}_{3}$$ and $$\dot{m}_{4}$$ was considerable as one can see from the solution given by Eq. [ZEqnNum840659]. It should not be difficult to imagine that solving sets of four or five equations can become exceedingly difficult to do by hand; however, computer routines are available that can be used to solve virtually any set of equations have the form given by Eqs. [ZEqnNum239077].

In dealing with sets of many equations, it is convenient to use the language of matrix algebra. For example, in matrix notation we would express Eqs. [ZEqnNum780531] according to

$$\left[\begin{array}{ccc} {(\omega _{A} )_{2} } & {(\omega _{A} )_{3} } & {(\omega _{A} )_{4} } \\ {(\omega _{B} )_{2} } & {(\omega _{B} )_{3} } & {(\omega _{B} )_{4} } \\ {1} & {1} & {1} \end{array}\right]\left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\quad =\quad \left[\begin{array}{c} {(\omega _{A} )_{1} \dot{m}_{1} } \\ {(\omega _{B} )_{1} \dot{m}_{1} } \\ {\dot{m}_{1} } \end{array}\right]$$ ([ZEqnNum350920])

In this representation of Eqs. [ZEqnNum925999], the $$3\times 3$$ matrix of mass fractions multiplies the $$3\times 1$$ column matrix of mass flow rates to produce a $$3\times 1$$ column matrix that is equal to the right hand side of Eq. [ZEqnNum350920]. In working with matrices, it is generally convenient to make use of a nomenclature in which subscripts are used to identify the row and column in which an element is located. We used this type of nomenclature in Chapter 2 where the $$m\times n$$ matrix A was represented by

$${\rm A}\quad =\quad \left[\begin{array}{cccc} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {....} & {....} & {......} & {....} \\ {a_{m1} } & {a_{m2} } & {......} & {a_{mn} } \end{array}\right]$$ ()

Here the first subscript identifies the row in which an element is located while the second subscript identifies the column. In Chapter 2 we discussed matrix addition and subtraction, and here we wish to discuss matrix multiplication and the matrix operation that is analogous to division. Matrix multiplication between A and B is defined only if the number of columns of A (in this case n) is equal to the number of rows of B. Given an $$m\times n$$ matrix A and an $$n\times p$$ matrix, B, the product between A and B is illustrated by the following equation: ([ZEqnNum897810])

Here we see that the elements of the ith row in matrix A multiply the elements of the jth column in matrix B to produce the element in the ith row and the jth column of the matrix C. For example, the specific element $$c_{11}$$ is given by

$$c_{11} \quad =\quad a_{11} b_{11} \; \; +\; \; a_{12} b_{21} \; \; +\; \; .......\; \; +\; \; a_{1n} b_{n1}$$ ()

while the general element $$c_{ij}$$ is given by

$$c_{ij} \quad =\quad a_{i1} b_{1j} \; \; +\; \; a_{i2} b_{2j} \; \; +\; \; .......\; \; +\; \; a_{in} b_{nj}$$ ()

In Eq. [ZEqnNum897810] we see that an $$m\times n$$ matrix can multiply an $$n\times p$$ to produce an $$m\times p$$, and we see that the matrix multiplication represented by AB is only defined when the number of columns in A is equal to the number of rows in B. The matrix multiplication illustrated in Eq. [ZEqnNum559949] conforms to this rule since there are three columns in the matrix of mass fractions and three rows in the column matrix of mass flow rates. The configuration illustrated in Eq. [ZEqnNum268548] is extremely common since it is associated with the solution of n equations for n unknowns. Our generic representation for this type of matrix equation is given by

$${\rm A}\, {\rm u}\quad =\quad {\rm b}$$ ([ZEqnNum309268])

in which A is a square matrix, u is a column matrix of unknowns, and b is a column matrix of knowns. These matrices are represented explicitly by

$${\rm A}\quad =\quad \left[\begin{array}{cccc} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {....} & {....} & {......} & {....} \\ {a_{n1} } & {a_{n2} } & {......} & {a_{nn} } \end{array}\right]\; ,\quad \quad {\rm u}\quad =\quad \left[\begin{array}{c} {u_{1} } \\ {.} \\ {.} \\ {u_{n} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {b_{1} } \\ {.} \\ {.} \\ {b_{n} } \end{array}\right]$$ ([ZEqnNum496328])

Sometimes the coefficients in A depend on the unknowns, u, and the matrix equation may be bi-linear (see Eqs. [ZEqnNum192373]) or non-linear. We represent systems of this type according to

$${\rm A}({\rm u})\, {\rm u}\quad =\quad {\rm b}$$ ()

to clearly indicate that the problem is not linear.

The transpose of a matrix is defined in the same manner as the transpose of an array that was discussed in Sec. 2.5, thus the transpose of the matrix A is constructed by interchanging the rows and columns to obtain

$${\rm A}\quad =\quad \left[\begin{array}{cccc} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {.} & {.} & {......} & {.} \\ {a_{m1} } & {a_{m2} } & {......} & {a_{mn} } \end{array}\right]\; ,\quad \quad {\rm A}^{\rm T} \quad =\quad \left[\begin{array}{cccc} {a_{11} } & {a_{21} } & {...} & {a_{m1} } \\ {a_{12} } & {a_{22} } & {...} & {a_{m2} } \\ {.} & {.} & {...} & {.} \\ {.} & {.} & {...} & {.} \\ {a_{1n} } & {a_{2n} } & {...} & {a_{mn} } \end{array}\right]$$ ()

Here it is important to note that A is an $$m\times n$$ matrix while $${\rm A}^{\rm T}$$ is an $$n\times m$$ matrix.

4.9.1 Inverse of a square matrix

In order to solve Eq. [ZEqnNum309268], one cannot “divide” by A to determine the unknown, u, since matrix division is not defined. There is, however, a related operation involving the inverse of a matrix. The inverse of a matrix, A, is another matrix, $${\rm A}^{-1}$$, such that the product of A and $${\it A}^{-1}$$ is given by

$${\rm A}\, {\rm A}^{-1} \quad =\quad {\rm I}$$ ()

in which I is the identity matrix. Identity matrices have ones in the diagonal elements and zeros in the off-diagonal elements as illustrated by the following $$4\times 4$$ matrix:

$${\rm I}\quad =\quad \left[\begin{array}{cccc} {1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} \end{array}\right]$$ ()

For the inverse of a matrix to exist, the matrix must be a square matrix, i.e., the number of rows must be equal to the number of columns. In addition, the determinant of the matrix must be different from zero.

As an example of the use of the inverse of a matrix, we consider the following set of four equations containing four unknowns:

$$\begin{array}{l} {a_{11} u_{1} \; \; +\; \; a_{12} u_{2} \; \; +\; \; a_{13} u_{3} \; \; +\; \; a_{14} u_{4} \quad =\quad b_{1} } \\ {} \\ {a_{21} u_{1} \; \; +\; \; a_{22} u_{2} \; \; +\; \; a_{23} u_{3} \; \; +\; \; a_{24} u_{4} \quad =\quad b_{2} } \\ {} \\ {a_{31} u_{1} \; \; +\; \; a_{32} u_{2} \; \; +\; \; a_{33} u_{3} \; \; +\; \; a_{34} u_{4} \quad =\quad b_{3} } \\ {} \\ {a_{41} u_{1} \; \; +\; \; a_{42} u_{2} \; \; +\; \; a_{43} u_{3} \; \; +\; \; a_{44} u_{4} \quad =\quad b_{4} } \end{array}$$ ([ZEqnNum291322])

In compact notation, we would represent these equations according to

$${\rm A}\, {\rm u}\quad =\quad {\rm b}$$ ([ZEqnNum729810])

We suppose that A is invertible and that the inverse, along with u and b, are given by

$${\rm A}^{-1} \quad =\quad \left[\begin{array}{cccc} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{array}\right]\; ,\quad \quad {\rm u}\quad =\quad \left[\begin{array}{c} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \end{array}\right]$$ ()

We can multiply Eq. [ZEqnNum729810] by $${\rm A}^{-1}$$ to obtain

$${\rm A}^{-1} {\rm A}\, {\rm u}\quad =\quad {\rm I}\, {\rm u}\quad =\quad {\rm u}\quad =\quad {\rm A}^{-1} {\rm b}$$ ()

and the details are given by

$$\left[\begin{array}{cccc} {1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} \end{array}\right]\left[\begin{array}{c} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{array}\right]\quad =\quad \left[\begin{array}{cccc} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{array}\right]\left[\begin{array}{c} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \end{array}\right]$$ ()

Carrying out the matrix multiplication on the left hand side leads to

$$\left[\begin{array}{c} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{array}\right]\quad =\quad \left[\begin{array}{cccc} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{array}\right]\left[\begin{array}{c} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \end{array}\right]$$ ()

while the more complex multiplication on the right hand side provides

$$\left[\begin{array}{c} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{array}\right]\quad =\quad \left[\begin{array}{cccc} {\bar{a}_{11} b_{1} } & {\bar{a}_{12} b_{2} } & {\bar{a}_{13} b_{3} } & {\bar{a}_{14} b_{4} } \\ {\bar{a}_{21} b_{1} } & {\bar{a}_{22} b_{2} } & {\bar{a}_{23} b_{3} } & {\bar{a}_{24} b_{4} } \\ {\bar{a}_{31} b_{1} } & {\bar{a}_{32} b_{2} } & {\bar{a}_{33} b_{3} } & {\bar{a}_{34} b_{4} } \\ {\bar{a}_{41} b_{1} } & {\bar{a}_{42} b_{2} } & {\bar{a}_{43} b_{3} } & {\bar{a}_{44} b_{4} } \end{array}\right]$$ ()

Each element of the column matrix on the left hand side is equal to the corresponding element on the right hand side and this leads to the solution for the unknowns given by

$$\begin{array}{l} {u_{1} \quad =\quad \bar{a}_{11} b_{1} \; \; +\; \; \bar{a}_{12} b_{2} \; \; +\; \; \bar{a}_{13} b_{3} \; \; +\; \; \bar{a}_{14} b_{4} } \\ {} \\ {u_{2} \quad =\quad \bar{a}_{21} b_{1} \; \; +\; \; \bar{a}_{22} b_{2} \; \; +\; \; \bar{a}_{23} b_{3} \; \; +\; \; \bar{a}_{24} b_{4} } \\ {} \\ {u_{3} \quad =\quad \bar{a}_{31} b_{1} \; \; +\; \; \bar{a}_{32} b_{2} \; \; +\; \; \bar{a}_{33} b_{3} \; \; +\; \; \bar{a}_{34} b_{4} } \\ {} \\ {u_{4} \quad =\quad \bar{a}_{41} b_{1} \; \; +\; \; \bar{a}_{42} b_{2} \; \; +\; \; \bar{a}_{43} b_{3} \; \; +\; \; \bar{a}_{44} b_{4} } \end{array}$$ ([ZEqnNum762280])

It should be clear that there are two main problems associated with the use of Eqs. [ZEqnNum291322] to obtain the solution for the unknowns given by Eq. [ZEqnNum762280]. The first problem is the correct interpretation of a physical process to arrive at the original set of equations, and the second problem is the determination of the inverse of the matrix A.

There are a variety of methods for developing the inverse for a matrix; however, sets of equations are usually solved numerically without calculating the inverse of a matrix. One of the classic methods is known as Gaussian elimination and we can illustrate this technique with the problem that was studied in Sec. 4.8. We can make use of the data provided in Figure 4.9 to express Eqs. [ZEqnNum305779] in the form

species A: $$0.045\, \dot{m}_{2} \; \; +\; \; 0.069\, \dot{m}_{3} \; \; +\; \; 0.955\, \dot{m}_{4} \quad =\quad \; 500{\rm \; lb}_{\rm m} /{\rm hr}$$ ([ZEqnNum216736]a)

species B: $$0.091\, \dot{m}_{2} \; \; +\; \; 0.901\, \dot{m}_{3} \; \; +\; \; 0.041\, \dot{m}_{4} \quad =\quad \; 300\; {\rm lb}_{\rm m} /{\rm hr}$$ ([ZEqnNum216736]b)

Total: $$\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad {\rm 1000\; lb}_{\rm m} /{\rm hr}$$ ([ZEqnNum249266]c)

and our objective is to determine the three mass flow rates, $$\dot{m}_{2}$$, $$\dot{m}_{3}$$ and $$\dot{m}_{4}$$. The solution is obtained by making use of the following three rules which are referred to as elementary row operations:

I. Any equation in the set can be modified by multiplying or dividing by a non-zero scalar without affecting the solution.

II. Any equation can be added or subtracted from the set without affecting the solution.

III. Any two equations can be interchanged without affecting the solution.

For this particular problem, it is convenient to arrange the three equations in the form

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \; \dot{m}_{3} \quad \quad +\quad \quad \; \dot{m}_{4} \quad \quad =\quad {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0.045\, \dot{m}_{2} \; \; +\; \; 0.069\, \dot{m}_{3} \; \; +\; \; 0.955\, \dot{m}_{4} \quad =\quad \; 500{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0.091\, \dot{m}_{2} \; \; +\; \; 0.901\, \dot{m}_{3} \; \; +\; \; 0.041\, \dot{m}_{4} \quad =\quad \; 300\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ([ZEqnNum491562])

We begin by eliminating the first term in the second equation. This is accomplished by multiplying the first equation by 0.045 and subtracting the result from the second equation to obtain

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad \; \; \; +\; \; 0.024\, \dot{m}_{3} \; \; +\; \; 0.910\, \dot{m}_{4} \quad =\quad \; 455{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0.091\, \dot{m}_{2} \; \; +\; \; 0.901\, \dot{m}_{3} \; \; +\; \; 0.041\, \dot{m}_{4} \quad =\quad \; 300\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ([ZEqnNum522104])

Directing our attention to the third equation, we multiply the first equation by 0.091 and subtract the result from the third equation to obtain

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \; {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad \; \; \; +\; \; 0.024\, \dot{m}_{3} \; \; +\; \; 0.910\, \dot{m}_{4} \quad =\quad \; 455{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad \; \; \; +\; \; 0.810\, \dot{m}_{3} \; \; -\; \; 0.050\, \dot{m}_{4} \quad =\quad 209\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ([ZEqnNum181816])

The second equation can be conditioned by dividing by 0.024 so that the equation set takes the form

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \; {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad \dot{m}_{3} \quad \; +\quad 37.917\, \dot{m}_{4} \quad =\quad 18958{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad \; \; \; +\; \; 0.810\, \dot{m}_{3} \; \; -\; \; 0.050\, \dot{m}_{4} \quad =\quad 209\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ()

We now multiply the second equation by $$0.810$$ and subtract the result from the third equation to obtain

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \; {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad \dot{m}_{3} \quad \; +\quad 37.917\, \dot{m}_{4} \quad =\quad 18958{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad 0\quad \; \; -\quad \; 30.763\, \dot{m}_{4} \quad =\quad -15147\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ()

and division of the third equation by $$-30.763$$ leads to

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \; {\rm 1000\; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad \dot{m}_{3} \quad \; +\quad 37.917\, \dot{m}_{4} \quad =\quad 18958.3{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \quad +\quad \quad 0\quad \quad +\quad \quad \quad \; \dot{m}_{4} \quad \; =\quad 492.39\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ([ZEqnNum789895])

Having worked our way forward through this set of three equations in order to determine $$\dot{m}_{4}$$, we can work our way backward through the set to determine $$\dot{m}_{3}$$ and $$\dot{m}_{2}$$. The results are the same as we obtained earlier in Sec. 4.8 and we list the result again as

$$\dot{m}_{2} \quad =\quad {\rm 219\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{3} \quad =\quad {\rm 288\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{4} \quad =\quad {\rm 492\; lb}_{\rm m} /{\rm hr}$$ ([ZEqnNum468182])

The procedure represented by Eqs. [ZEqnNum453458] through [ZEqnNum789895] is extremely convenient for automated computation and large systems of equations can be quickly solved using a variety of software. Using systems such as MATLAB or Mathematica, problems of this type become quite simple.

4.9.2 Determination of the inverse of a square matrix

The Gaussian elimination procedure described in the previous section is closely related to the determination of the inverse of a square matrix. In order to illustrate how the inverse matrix can be calculated, we consider the coefficient matrix associated with Eq. [ZEqnNum491562] which we express as

$${\rm A}\quad =\quad \left[\begin{array}{ccc} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{array}\right]$$ ()

We can express Eq. [ZEqnNum879461] in the compact form represented by Eq. [ZEqnNum641032]

$${\rm A}\, {\rm u}\quad {\it =}\quad {\it b}$$ ([ZEqnNum180096])

in which u is the column matrix of unknown mass flow rates and b is the column matrix of known mass flow rates.

$${\rm u}\quad =\quad \left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {1000\, lb_{m} /hr} \\ {500\, lb_{m} /hr} \\ {300\, lb_{m} /hr} \end{array}\right]\quad =\quad \left[\begin{array}{c} {b_{2} } \\ {b_{3} } \\ {b_{4} } \end{array}\right]$$ ()

We can also make use of the unit matrix

$${\rm I}\quad =\quad \left[\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]$$ ()

to express Eq. [ZEqnNum180096] in the form

$${\rm A}\, {\rm u}\quad {\it =}\quad {\it I}\, {\it b}$$ ([ZEqnNum844576])

Now we wish to repeat the Gaussian elimination used in Sec. 4.9.1, but in this case we will make use of Eq. [ZEqnNum844576] rather than Eq. [ZEqnNum551051] in order to retain the terms associated with $${\rm I}\, {\rm b}$$. Multiplying the first of Eqs. [ZEqnNum504219] by -0.045 and adding the result to the second equation leads to

$$\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \; \; \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \quad b_{2} \quad \quad +\quad \quad 0\quad +\quad \; 0\quad =\quad 1000\, {\rm lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad 0.024\, \dot{m}_{3} \quad \; +\quad 0.910\, \dot{m}_{4} \quad =\quad \begin{array}{ccc} {-0.045\, b_{2} \quad +} & {b_{3} \quad +} & {0} \end{array}\quad =\quad 455{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0.091\, \dot{m}_{2} \; \; +\; \; 0.901\, \dot{m}_{3} \; \; +\; \; 0.041\, \dot{m}_{4} \quad =\quad \begin{array}{ccc} {\quad 0\quad \quad +} & {\quad \; \; 0\quad +\; } & {b_{4} } \end{array}\quad =\quad 300\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$$ ()

Note that this set of equations is identical to Eqs. [ZEqnNum522104] except that we have included the terms associated with $${\rm I}\, {\rm b}$$. We now proceed with the Gaussian elimination represented by Eqs. [ZEqnNum181816] through [ZEqnNum834391] to arrive at $\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \quad \; \; \dot{m}_{3} \quad \quad +\quad \quad \dot{m}_{4} \; \; \quad =\quad \quad \; \; b_{2} \quad \quad \; +\quad \quad \; \; \; \; 0\quad \; \; \; \; +\quad \; \; \; 0\quad \quad \; \; =\quad \; \; 1000\, {\rm lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad \quad \; \; \dot{m}_{3} \quad \; +\quad 37.917\, \dot{m}_{4} \quad =\quad \begin{array}{ccc} {-1.875\, b_{2} \; \; \; \; +} & {41.667b_{3} \; \; \quad +} & {\; \; 0} \end{array}\quad \quad \; \; =\quad 18958.3{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \; \quad \quad \; \; 0\; \; \quad \quad +\; \; \quad \quad \; \dot{m}_{4} \quad =\quad \begin{array}{ccc} {-0.0464\, b_{2} \; \; +} & {1.0971\, b_{3} \; \; \; -} & {0.03251\, b_{4} \; } \end{array}=\quad 492.39\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$

([ZEqnNum536583])

This result is identical to Eq. [ZEqnNum838645] except for the fact that we have retained the terms in the second column matrix that are associated with $${\rm I}\, {\rm b}$$ in Eq. [ZEqnNum723983]. In Sec. 4.9.1, we used Eq. [ZEqnNum940351] to carry out a backward elimination in order to solve for the mass flow rates given by Eq. [ZEqnNum468182], and here we want to present that backward elimination explicitly in order to demonstrate that it leads to the inverse of the coefficient matrix, A. This procedure is known as the Gauss-Jordan algorithm and it consists of elementary row operations that reduce the left hand side of Eq. [ZEqnNum536583] to a diagonal form.

We begin to construct a diagonal form by multiplying the third equation by -37.917 and adding it to the second equation so that our set of equations take the form $\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \dot{m}_{3} \quad \; +\quad \quad \dot{m}_{4} \; \; \quad \quad =\quad \quad \; \; b_{2} \quad \quad \; +\quad \quad \; \; \; \; 0\quad \; \; \; \; +\quad \; \; \; 0\quad \quad \; \; =\quad \; 1000\, {\rm lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad \; \dot{m}_{3} \quad \; +\quad \quad \; 0\quad \quad \quad =\quad \begin{array}{ccc} {-0.1152\, b_{2} \; \; +\; } & {0.0677b_{3} \quad +} & {1.2326\, b_{4} } \end{array}\; \; =\quad 288.42{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \; \quad 0\; \; \quad \; +\quad \quad \dot{m}_{4} \quad \quad \; \; =\quad \begin{array}{ccc} {-0.0464\, b_{2} \; \; +} & {1.0971\, b_{3} \; \; \; -} & {0.03251\, b_{4} \; \; \; } \end{array}=\quad 492.39\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$

()

We now multiply the third equation by $$-1$$ and add the result to the first equation to obtain $\begin{array}{l} {\dot{m}_{2} \; \; \quad +\quad \dot{m}_{3} \quad \; +\quad \quad 0\quad \quad \; =\quad \; \; 1.0464\, b_{2} \; \; -\; \; 1.09711\, b_{3} \; \; +\; \; 0.03251\, b_{4} \quad \; =\quad 507.61\, {\rm lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad \; \dot{m}_{3} \quad \; +\quad \quad \; 0\quad \quad =\quad \begin{array}{ccc} {-0.1152\, b_{2} \; \; +} & {0.0677b_{3} \quad +} & {1.2326\, b_{4} } \end{array}\quad \; =\quad 288.42{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \; \quad 0\; \; \quad \; +\quad \quad \dot{m}_{4} \quad \; \; =\quad \begin{array}{ccc} {-0.0464\, b_{2} \; \; +} & {1.0971\, b_{3} \; \; \; -} & {0.03251\, b_{4} \quad } \end{array}=\quad 492.39\; {\rm lb}_{\rm m} /{\rm hr}} \end{array}$

()

and finally we multiply the second equation by $$-1$$ and add the result to the first equation to obtain the desired diagonal form $\begin{array}{l} {\dot{m}_{2} \quad +\quad \; 0\quad \; \; +\quad \quad 0\quad \quad =\quad \; \; 1.1616\, b_{2} \; \; -\; \; 1.16484\, b_{3} \; \; -\; \; 1.20005\, b_{4} \quad \; =\quad 219.187{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad \dot{m}_{3} \quad +\quad \quad 0\quad \quad =\quad \begin{array}{ccc} {-0.1152\, b_{2} \; \; +} & {0.0677b_{3} \quad +} & {1.2326\, b_{4} } \end{array}\quad \; =\quad 288.42{\rm \; lb}_{\rm m} /{\rm hr}} \\ {} \\ {0\quad \; \; +\; \quad 0\; \; \quad +\quad \; \; \dot{m}_{4} \quad \quad =\quad \begin{array}{ccc} {-0.0464\, b_{2} \; \; +} & {1.0971\, b_{3} \; \; \; -} & {0.03251\, b_{4} \quad } \end{array}=\quad 492.39{\rm \; lb}_{\rm m} /{\rm hr}} \end{array}$

([ZEqnNum673098])

Here we see that the unknown mass flow rates are determined by

$$\dot{m}_{2} \quad =\quad {\rm 219\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{3} \quad =\quad {\rm 288\; lb}_{\rm m} /{\rm hr}\; ,\quad \quad \dot{m}_{4} \quad =\quad {\rm 492\; lb}_{\rm m} /{\rm hr}$$ ()

where only three significant figures have been listed since it is unlikely that the input data are accurate to more than 1%. The coefficient matrix contained in the central term of Eqs. [ZEqnNum673098] is the inverse of A and we express this result as

$${\rm A}^{-1} \quad =\quad \left[\begin{array}{ccc} {1.1616} & {-1.16484} & {-1.20005} \\ {-0.1152} & {0.0677} & {1.2326} \\ {-0.0464} & {1.0971} & {-0.0325} \end{array}\right]$$ ()

The solution procedure that led from Eq. [ZEqnNum935304] to the final answer can be expressed in compact form according to,

$${\rm A}\, {\rm u}\quad {\it =}\quad {\it b}\; ,\quad \Rightarrow \quad {\it A}^{-1} {\it A}\, {\it u}\quad {\it =}\quad {\it A}^{-1} \, {\it b}\; ,\quad \Rightarrow \quad {\it I}\, {\it u}\quad {\it =}\quad {\it A}^{-1} \, {\it b}\; ,\quad \Rightarrow \quad {\it u}\quad {\it =}\quad {\it A}^{-1} \, {\it b}$$ ()

and in terms of the details of the inverse matrix the last of these four equations can be expressed as

$$\left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\quad =\quad \left[\begin{array}{ccc} {1.1616} & {-1.16484} & {-1.20005} \\ {-0.1152} & {0.0677} & {1.2326} \\ {-0.0464} & {1.0971} & {-0.0325} \end{array}\right]\; \left[\begin{array}{c} {1000\, lb_{m} /hr} \\ {500\, lb_{m} /hr} \\ {300\, lb_{m} /hr} \end{array}\right]$$ ()

Given that computer routines are available to carry out the Gauss-Jordan algorithm, it should be clear that the solution of a set of linear mass balance equations is a routine matter.

4.10 Problems $${}^{*}$$4

Section 4.1

4-1. Use Eq. [ZEqnNum645577] to obtain a macroscopic mole balance for species A in terms of a moving control volume. Indicate how your result can be used to obtain Eq. [ZEqnNum434393].

Section 4.2

4-2. Determine the mass density, , for the mixing process illustrated in Figure 4-2.

4-3. A liquid hydrocarbon mixture was made by adding 295 kg of benzene, 289 kg of toluene and 287 kg of p-xylene. Assume there is no change of volume upon mixing, i.e., $$\Delta V_{mix} =0$$, in order to determine:

1. The species density of each species in the mixture.

2. The total mass density.

3. The mass fraction of each species.

4-4. A gas mixture contains the following quantities (per cubic meter) of carbon monoxide, carbon dioxide and hydrogen: carbon monoxide, 0.5 kmol/m $${}^{3}$$, carbon dioxide, 0.5 kmol/m $${}^{3}$$, and hydrogen, 0.6 kmol/m $${}^{3}$$. Determine the species mass density and mass fraction of each of the components in the mixture.

4-5. The species mass densities of a three-component (A, B, and C) liquid mixture are: acetone, $$\rho _{A} \; =\; 326.4{\rm \; kg/m}^{\rm 3}$$, acetic acid, $$\rho _{B} =326.4\, {\rm kg/m}^{\rm 3}$$, and ethanol, $$\rho _{C} =217.6{\rm \; kg/m}^{\rm 3}$$. Determine the following for this mixture:

1. The mass fraction of each species in the mixture.

2 The mole fraction of each species in the mixture.

3. The mass of each component required to make one cubic meter of mixture.

4-6. A mixture of gases contains one kilogram of each of the following species: methane (A), ethane (B), propane (C), carbon dioxide (D), and nitrogen (E). Calculate the following:

1. The mole fraction of each species in the mixture

2. The average molecular mass of the mixture

4-7. Two gas streams, having the flow rates and properties indicated in Table 4.7, are mixed in a pipeline. Assume perfect mixing, i.e. no change of volume upon mixing, and determine the composition of the mixed stream in moles/m $${}^{3}$$.

Table 4.7. Composition of gas streams

Stream #1 Stream #2
Mass flow rate 0.226 kg/s 0.296 kg/s
methane 0.48 kg/m $${}^{3}$$ 0.16 kg/m $${}^{3}$$
ethane 0.90 kg/m $${}^{3}$$ 0.60 kg/m $${}^{3}$$
propane 0.88kg/m3 0.220 kg/m $${}^{3}$$

4-8.Develop a representation for the mole fraction of species A in an N-component system in terms of the mass fractions and molecular masses of the species. Use the result to prove that the mass fractions and mole fractions in a binary system are equal when the two molecular masses are equal.

4-9. Derive the total mass balance for an arbitrary moving control volume beginning with the species mass balance given by Eq. [ZEqnNum996159].

Section 4.3

4-10. The species velocities, in a binary system, can be decomposed according to $\label{GrindEQ__1_} v_{A} \quad =\quad v\; \; +\; \; u_{A} \; ,\quad \quad v_{B} \quad =\quad v\; \; +\; \; u_{B}$ in which v represents the mass average velocity defined by Eq. [ZEqnNum803313]. One can use this result, along with the definition of the mass average velocity, to prove that $\label{GrindEQ__2_} v_{A} \quad =\quad v_{B} \; \; +\; \; \left(\frac{1}{1-\omega _{A} } \right)u_{A}$ This means that the approximation, $$v_{A} \approx v_{B}$$ requires the restriction $\label{GrindEQ__3_} \left|u_{A} \right|\quad <<\quad (1-\omega _{A} )\left|v_{A} \right|$ Since $$1-\omega _{A}$$ is always less than one, we can always satisfy this inequality whenever the mass diffusion velocity is small compared to the species velocity, i.e., $\label{GrindEQ__4_} \left|u_{A} \right|\quad <<\quad \left|v_{A} \right|$ For the sulfur dioxide mass transfer process illustrated in Figure 4-4, this means that the approximation $\label{GrindEQ__5_} v_{\rm SO}_{{\rm 2} } \cdot k\quad \approx \quad v_{air} \cdot k$ is valid whenever the mass diffusion velocity is restricted by $\label{GrindEQ__6_} \left|u_{{\rm SO}_{2} } \cdot k\right|\quad <<\quad \left|v_{{\rm SO}_{2} } \cdot k\right|$ In many practical cases, this restriction is satisfied and all species velocities can be approximated by the mass average velocity.

Next we direct our attention to the mass transfer process at the gas-liquid interface illustrated in Figure 4-5. If we assume that there is no mass transfer of air into or out of the liquid phase, we can prove that $\label{GrindEQ__7_} u_{\rm SO}_{2} } \cdot n\quad =\quad \left(1-\omega _{{\rm SO}_{{\rm 2} } \right)v_{{\rm SO}_{2} } \cdot n\; ,\quad \quad at\; the\; gas-liquid\; interface$ Under these circumstances, the mass diffusion velocity is never small compared to the species velocity for practical conditions. Thus the type of approximation indicated by Eq. [ZEqnNum668645] is never valid for the component of the velocity normal to the gas-liquid interface. As a simplification, we can treat the sulfur dioxide-air system as a binary system with species A representing the sulfur dioxide and species B representing the air.

(a). Use the definition of the mass average velocity given by Eq. [ZEqnNum146973] to prove Eq. 2.

(b) Use $$v_{\rm air} \cdot n\; \; =\; \; v_{B} \cdot n\; \; =\; \; 0$$ in order to prove Eq. 7.

Section 4.4

4-11. A three component liquid mixture flows in a pipe with a mass averaged velocity of $${\rm v}=0.9\, {\rm m/s}$$. The density of the mixture is $$\rho$$ = 850 kg/m $${}^{3}$$. The components of the mixture and their mole fractions are:n-pentane, $$x_{P} =0.2$$, benzene, $$x_{B} =0.3$$, and naphthalene, $$x_{N} =0.5$$. The diffusion fluxes of each component in the streamwise direction are: pentane, $$\rho _{P} \, u_{P} =1{\rm .564}\times {\rm 10}^{-6} {\rm kg/m}^{\rm 2} {\rm s}$$, benzene, $$\rho _{B} \, u_{B} =1{\rm .563}\times 1{\rm 0}^{-6} \, {\rm kg/m}^{\rm 2} {\rm s}$$, and naphthalene, $$\rho _{N} \, u_{N} =-3.127\times {\rm 10}^{-6} \, {\rm kg/m}^{\rm 2} {\rm s}$$. Determine the diffusion velocities and the species velocities of the three components. Use this result to determine the molar averaged velocity, $${\rm v}^{*}$$. Note that you must use eight significant figures in your computation.

Section 4.5

4-12. Sometimes heterogeneous chemical reactions take place at the walls of tubes in which reactive mixtures are flowing. If species A is being consumed at a tube wall because of a chemical reaction, the concentration profile may be of the form $\label{GrindEQ__1_} c_{A} {\rm (}r{\rm )}\quad =\quad c_{A}^{\rm o} \left[1-\phi \left({r\mathord{\left/ {\vphantom {r r_{\rm o} }} \right. \kern-\nulldelimiterspace} r_{\rm o} } \right)^{2} \right]$ Here r is the radial position and $$r_{\rm o}$$ is the tube radius. The parameter  depends on the net rate of production of chemical species at the wall and the molecular diffusivity, and it is bounded by $$0\le \phi \le 1$$. If  is zero, the concentration across the tube is uniform at the value $$c_{A}^{\rm o}$$. If the flow in the tube is laminar, the velocity profile is given by $\label{GrindEQ__2_} {\rm v}_{z} {\rm (}r{\rm )}\quad =\quad 2\langle {\rm v}_{z} \rangle \left[1-\left({r\mathord{\left/ {\vphantom {r r_{\rm o} }} \right. \kern-\nulldelimiterspace} r_{\rm o} } \right)^{2} \right]$ and the volumetric flow rate is $\label{GrindEQ__3_} Q\quad =\quad \langle {\rm v}_{z} \rangle \, \pi r_{\rm o}^{\rm 2}$ For this process, determine the molar flow rate of species A in terms $$c_{A}^{\rm o}$$, $$\phi$$, and $$\langle {\rm v}_{z} \rangle$$. When $$\phi =0.5$$, determine the bulk concentration, $$\langle c_{A} \rangle _{b}$$,and the area-averaged concentration, $$\langle c_{A} \rangle$$. Use these results to determine the difference between $$\dot{M}_{A}$$ and $$\langle c_{A} \rangle Q$$.

4-13. A flash unit is used to separate vapor and liquid streams from a liquid stream by lowering its pressure before it enters the flash unit. The feed stream is pure liquid water and its mass flow rate is 1000 kg/hr. Twenty percent (by weight) of the feed stream leaves the flash unit with a density $$\rho$$ = 10 kg/m $${}^{3}$$. The remainder of the feed stream leaves the flash unit as liquid water with a density $$\rho$$ = 1000 kg/m $${}^{3}$$ Determine the following:

1. The mass flow rates of the exit streams in kg/s.

2. volumetric flow rates of exit streams in m $${}^{3}$$/s.

Section 4.6

4-14. Show that Eq. [ZEqnNum692829] results from Eq. [ZEqnNum839821] when either $$c\, v\cdot n$$ or $$x_{A}$$ is constant over the area of the exit.

4-15. Use Eq. [ZEqnNum443450] to prove Eq.[ZEqnNum181666].

4-16. Derive Eq. [ZEqnNum700852] given that either $$\rho \, v\cdot n$$ or $$\omega _{A}$$ is constant over the area of the exit.

4-17. Prove Eq.[ZEqnNum359366].

Section 4.7

4-18. Determine $$\dot{M}_{3}$$ and the unknown mole fractions for the distillation process described in Sec. 4.7 subject to the following conditions:

Stream #1 Stream #2 Stream #3
$$\dot{M}_{1} =1200{\rm \; mol/hr}$$ $$\dot{M}_{2} =250{\rm \; mol/hr}$$ $$\dot{M}_{3} =?$$
$$x_{A} \; \; =\; \; 0.3$$ $$x_{A} \; \; =\; \; 0.8$$ $$x_{A} \; \; =\; \; ?$$
$$x_{B} \; \; =\; \; 0.2$$ $$x_{B} \; \; =\; \; ?$$ $$x_{B} \; \; =\; \; 0.25$$
$$x_{C} \; \; =\; \; ?$$ $$x_{C} \; \; =\; \; ?$$ $$x_{C} \; \; =\; \; ?$$

4-19. A continuous filter is used to separate a clear filtrate from alumina particles in a slurry. The slurry has 30% by weight of alumina (specific gravity of alumina = 4.5). The cake retains 5% by weight of water. For a feed stream of 1000 kg/hr, determine the following:

1. The mass flow rate of particles and water in the input stream

2. The volumetric flow rate of the inlet stream in m $${}^{3}$$/s.

3. The mass flow rate of filtrate and cake in kg/s.

4-20. A BTX unit, shown in Figure 4.20, is associated with a refinery that produces benzene, toluene, and xylenes. Stream #1 leaving the reactor-reforming unit has a volumetric flow rate of 10 m $${}^{3}$$/hr and is a Figure 4.20. Reactor and distillation unit

mixture of benzene (A), toluene (B), and xylenes (C) with the following composition: $\langle c_{A} \rangle _{1} \quad =\quad {\rm 6,000}\; {\rm mol/m}^{\rm 3} \; ,\quad \quad \langle c_{B} \rangle _{1} \quad =\quad 2,000\, \, {\rm mol/m}^{\rm 3} \; ,\quad \quad \langle c_{C} \rangle _{1} \quad =\quad 2,000\; {\rm mol/m}^{\rm 3}$ Stream [GrindEQ__1_] is the feed to a distillation unit where the separation takes place according to the following specifications:

1. 98% of the benzene leaves with the distillate stream (stream #2).

2. 99% of the toluene in the feed leaves with the bottoms stream (stream #3)

3. 100% of the xylenes in the feed leaves with the bottoms stream (stream #3).

Assuming that the volumes of components are additive, and using the densities of pure components from Table I in the Appendix, compute the concentration and volumetric flow rate of the distillate (stream #2) and bottoms (stream #3) stream leaving the distillation unit.

4-21. A standard practice in refineries is to use a holding tank in order to mix the light naphtha output of the refinery for quality control. During the first six hours of operation of the refinery, the stream feeding the holding tank at 200 kg/min had 30% by weight of n-pentane, 40% by weight of n-hexane, 30% by weight of n-heptane. During the next 12 hours of operation the mass flow rate of the feed stream was 210 kg/min and the composition changed to 40% by weight of n-pentane, 40% by weight of n-hexane, and 20% by weight of n-heptane. Determine the following:

The average density of the feed streams

The concentration of the feed streams in moles/m $${}^{3}$$.

After 12 hours of operation, and assuming the tank was empty at the beginning, determine:

The volume of liquid in the tank in m $${}^{3}$$.

The concentration of the liquid in the tank, in mol/m $${}^{3}$$.

The partial density of the species in the tank.

4-22. A distillation column is used to separate a mixture of methanol, ethanol, and isopropyl alcohol. The feed stream, with a mass flow rate of 300 kg/hr, has the following composition:

Component Species mass density
methanol 395.5 kg/m $${}^{3}$$
ethanol 197.3 kg/m $${}^{3}$$
isopropyl alcohol 196.5 kg/m $${}^{3}$$

Separation of this mixture of alcohol takes place according to the following specifications:

(a) 90% of the methanol in the feed leaves with the distillate stream

(b) 5% of the ethanol in the feed leaves with the distillate stream

(c) 3% of the isopropyl alcohol in the feed leaves with the distillate stream.

Assuming that the volumes of the components are additive, compute the concentration and volumetric flow rates of the distillate and bottom streams.

4-23. A mixture of ethanol (A) and water (B) is separated in a distillation column. The volumetric flow rate of the feed stream is 5 m $${}^{3}$$/hr. The concentration of ethanol in the feed is $$c_{A} =2,800\, {\rm mol/m}^{\rm 3}$$. The distillate leaves the column with a concentration of ethanol $$c_{A} =13,000\, {\rm mol/m}^{\rm 3}$$. The volumetric flow rate of distillate is one cubic meter per hour. How much ethanol is lost through the bottoms of the column, in kilograms of ethanol per hour?

4-24. A ternary mixture of benzene, ethylbenzene, and toluene is fed to a distillation column at a rate of $$10^{5}$$mol/hr. The composition of the mixture in % moles is: 74% benzene, 20% toluene, and 6% ethylbenzene. The distillate flows at a rate of $$75\times 10^{3}$$mol/hr. The composition of the distillate in % moles is 97.33 % benzene, 2% toluene, and the rest is ethylbenzene. Find the molar flow rate of the bottoms stream and the mass fractions of the three components in the distillate and bottoms stream.

4-25. A complex mixture of aromatic compounds leaves a chemical reactor and is fed to a distillation column. The mass fractions and flow rates of distillate and bottoms streams are given in Table 4.25. Compute the molar flow rate and composition, in molar fractions, of the feed stream.

Table 4.25 Flow rate and composition of distillate and bottoms streams.

(kg/hr) $$\omegaup$$ $${}_{Benzene}$$ $$\omegaup$$ $${}_{Toluene}$$ $$\omegaup$$ $${}_{Benzaldehide}$$ $$\omegaup$$ $${}_{BenzoicAcid}$$ $$\omegaup$$ $${}_{MethylBenzoate}$$
Distillate 125 0.1 0.85 0.03 0.0 0.02
Bottoms 76 0.0 0.05 0.12 0.8 0.03

4-26. A hydrocarbon feedstock is available at a rate of $$10^{6}$$mol/hr, and consists of propane ( $$x_{A} =0.2$$), n-butane ( $$x_{B} =0.3$$), n-pentane ( $$x_{C} =0.2$$) and n-hexane ( $$x_{D} =0.3$$). The distillate contains all of the propane in the feed to the unit and 80% of the pentane fed to the unit. The mole fraction of butane in the distillate is $$y_{B} =0.4$$. The bottom stream contains all of the hexane fed to the unit. Calculate the distillate and bottoms streams flow rate and composition in terms of mole fractions.

Section 4.8

4-27. It is possible that the process illustrated in Figure 4-12 could be analyzed beginning with Control Volume I rather than beginning with Control Volume II. Begin the problem with Control Volume I and carry out a degree-of-freedom analysis to see what difficulties might be encountered.

4-28. In a glycerol plant, a 10% (mass basis) aqueous glycerin solution containing 3% NaCl is treated with butyl alcohol as illustrated in Figure 4.28. The alcohol fed to the tower contains 2% water on a mass basis. The raffinate leaving the tower contains all the original salt, 1.0% glycerin and 1.0% alcohol. The extract from the tower is sent to a distillation column. The distillate from this column is the alcohol containing 5% water. Figure 4.28. Solvent extraction process

The bottoms from the distillation column are 25% glycerin and 75% water. The two feed streams to the extraction tower have equal mass flow rates of 1000 lbm per hour. Determine the output of glycerin in pounds per hour from the distillation column.

Section 4.9

4-29. In Sec. 4.8 the solution to the distillation problem was shown to reduce to solving the matrix equation, $${\rm A}\, {\rm m}={\rm b}$$, in which ${\rm A}\quad =\quad \left[\begin{array}{ccc} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{array}\right]\; ,\quad \quad {\rm u}\quad =\quad \left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {1000} \\ {500} \\ {300} \end{array}\right]$ Here it is understood that the mass flow rates have been made dimensionless by dividing by $${\rm lb}_{\rm m} /{\rm hr}$$. In addition to the matrix A, one can form what is known as an augmented matrix. This is designated by $${\rm A}\dot{\dot{\cdot }}{\rm b}$$ and it is constructed by adding the column of numbers in b to the matrix A in order to obtain ${\rm A}\dot{\dot{\cdot }}{\rm b}\quad =\quad \left[\begin{array}{ccccc} {1} & {1} & {1} & {.} & {1000} \\ {0.045} & {0.069} & {0.955} & {.} & {500} \\ {0.091} & {0.901} & {0.041} & {.} & {300} \end{array}\right]$ Define the following lists in Mathematica corresponding to the rows of the augmented matrix, $${\rm A}\dot{\dot{\cdot }}{\rm b}$$. $\begin{array}{l} {R1\quad =\quad \left\{\begin{array}{cccc} {1,} & {1,} & {1,} & {1000} \end{array}\right\}} \\ {R2\quad =\quad \left\{\begin{array}{cccc} {0.045,} & {0.069,} & {0.955,} & {500} \end{array}\right\}} \\ {R3\quad =\quad \left\{\begin{array}{cccc} {0.091,} & {0.901,} & {0.041,} & {300} \end{array}\right\}} \end{array}$ Write a sequence of Mathematica expressions that correspond to the elementary row operations for solving this system. The first elementary row operation that given Eq. [ZEqnNum270394] is $R2\quad =\quad (-0.045)\; R2\; \; +\; \; R1$ Show that you obtain an augmented matrix that defines Eq. [ZEqnNum938375].

4-30. In this problem you are asked to continue exploring the use of Mathematica in the analysis of the set of linear equations studied in Sec. 4.9.2, i.e., $${\rm A}\, {\rm m}={\rm b}$$ where the matrices are defined by ${\rm A}\quad =\quad \left[\begin{array}{ccc} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{array}\right]\; ,\quad \quad {\rm u}\quad =\quad \left[\begin{array}{c} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{array}\right]\; ,\quad \quad {\rm b}\quad =\quad \left[\begin{array}{c} {1000} \\ {500} \\ {300} \end{array}\right]$ [GrindEQ__1_] Construct the augmented matrix $${\rm A}\dot{\dot{\cdot }}{\rm I}$$ according to ${\rm A}\dot{\dot{\cdot }}{\rm I}\quad =\quad \left[\begin{array}{ccccccc} {1} & {1} & {1} & {.} & {1} & {0} & {0} \\ {0.045} & {0.069} & {0.955} & {.} & {0} & {1} & {0} \\ {0.091} & {0.901} & {0.041} & {.} & {0} & {0} & {1} \end{array}\right]$ and use elementary row operations to transform this augmented matrix to the form ${\rm A}\dot{\dot{\cdot }}{\rm I}\quad =\quad \left[\begin{array}{ccccccc} {1} & {0} & {0} & {.} & {} & {} & {} \\ {0} & {1} & {0} & {.} & {} & {\it B} & {} \\ {0} & {0} & {1} & {.} & {} & {} & {} \end{array}\right]$ Show that the elements represented by B make up the matrix B having the property that $${\rm B}={\rm A}^{-1}$$. Use your result to calculate $${\rm m}={\rm A}^{-1} \, {\rm b}$$.

[GrindEQ__2_] Show that the inverse found in Part 1 satisfies $${\rm A}\, {\rm A}^{-1} ={\rm I}$$.

[GrindEQ__3_] Use Mathematica’s built-in function Inverse to find the inverse of A.

[GrindEQ__4_] Use Mathematica’s RowReduce function on the augmented matrix $${\rm A}\dot{\dot{\cdot }}{\rm b}$$ and show that from the row echelon form you can obtain the same results as in [GrindEQ__1_].

[GrindEQ__5_] Use Mathematica’s Solve function to solve $${\rm A}\, {\rm m}={\rm b}$$.

1. . Lavoisier, A. L. 1777,. Memoir on Combustion in General,  Mémoires de L’Academie Roayal des Sciences  592-600.

2. . Toulmin, S.E. 1957, Crucial Experiments: Priestley and Lavoisier, Journal of the History of Ideas,  18 , 205-220.

3. . See Section 2.1.1.

4. * Problems marked with the symbol  will be difficult to solve without the use of computer software.