10.4: Appendix D - Atomic Species Balance

Throughout this text we have made use of macroscopic mass and mole balances to solve a variety of problems with and without chemical reactions. Our solutions have been based on the application of Axiom I and Axiom II, and our attention has been focused on mass flow rates, molar rates of production owing to chemical reaction, accumulation of mass or moles, etc. In this appendix we show how problems can be solved using macroscopic atomic species balances. This approach has some advantages when carrying out calculations by hand since the number of atomic species balance equations is almost always less than the number of molecular species balance equations.

We begin our development of atomic species balance equations with Axioms I and II given by

Axiom I: $\frac{d}{dt} \int_{\mathscr{V}}c_{A} dV + \int_{\mathscr{A}}c_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = \int_{\mathscr{V}}R_{A} dV , \quad A = 1,2, ..., N \label{1}$

Axiom II $\sum_{A = 1}^{A = N}N_{JA} R_{A} = 0 , \quad J = 1, 2, ..., T \label{2}$

Here we should remember that the individual components of the chemical composition matrix, $$\left[N_{JA} \right]$$, are described by (see Sec. 6.2)

$N_{JA} = \begin{Bmatrix}\text{ number of moles of} \\ J\text{-type atoms per mole} \\ \text{ of molecular species }A\end{Bmatrix} , \quad J = 1, 2,...,T, \text{ and }A = 1, 2,...N \label{3}$

To develop an atomic species balance, we multiply Equation \ref{1} by $$N_{JA}$$ and sum over all molecular species to obtain

$\frac{d}{dt} \int_{\mathscr{V}}\sum_{A = 1}^{A = N}N_{JA} c_{A} dV + \int_{\mathscr{A}}\sum_{A = 1}^{A = N}N_{JA} c_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = \int_{\mathscr{V}}\sum_{A = 1}^{A = N}N_{JA} R_{A} dV \label{4}$

On the basis of Axiom II, we see that the last term in this result is zero and our atomic species balance takes the form

Axioms I & II: $\frac{d}{dt} \int_{\mathscr{V}}\sum_{A = 1}^{A = N}N_{JA} c_{A} dV + \int_{\mathscr{A}}\sum_{A = 1}^{A = N} N_{JA} c_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = 0 , \quad J = 1, 2,.., T \label{5}$

Here we have indicated explicitly that there are $$T$$ atomic species balance equations instead of the $$N$$ molecular species balance equations which are given by Equation \ref{1}. When $$T\leq N$$ it may be convenient to solve material balance problems using atomic species balances.

In many applications of Equation \ref{5}, diffusive transport is negligible and $$\mathbf{v}_{A}$$ can be replaced by $$\mathbf{v}$$ leading to

$\frac{d}{dt} \int_{\mathscr{V}}\sum_{A = 1}^{A = N}N_{JA} c_{A} dV + \int_{\mathscr{A}}\sum_{A = 1}^{A = N}N_{JA} c_{A} \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad J = 1, 2, ..., T \label{6}$

If the concentration is given in terms of mole fractions one can use $$c_{A} = x_{A} c$$ to express Equation \ref{6} as

$\frac{d}{dt} \int_{\mathscr{V}} \left\{ \sum_{A = 1}^{A = N}N_{JA} x_{A} \right\} c dV + \int_{\mathscr{A}} \left\{\sum_{A = 1}^{A = N}N_{JA} x_{A} \right\} c\mathbf{v} \cdot \mathbf{n} dA = 0 , \quad J = 1, 2,..., T \label{7}$

For some reacting systems these results, rather that Equation \ref{1}, \quad provide the simplest algebraic route to a solution.

When problems are stated in terms of species mass densities and species mass fractions, it is convenient to rearrange Equation \ref{5} in terms of these variables. To develop the forms of Eqs. \ref{6} and \ref{7} that are useful when mass flow rates are given, we begin by representing the species concentration in terms of the species density according to

$c_{A} = {\rho_{A} / MW_{A} } \label{8}$

The species density is now expressed in terms of the mass fraction and the total mass density according to

$\rho_{A} = \omega_{A} \rho \label{9}$

and when these two results are used in Equation \ref{6} we obtain

$\frac{d}{dt} \int_{\mathscr{V}}\sum_{A = 1}^{A = N}\left(\frac{N_{JA} \omega_{A} }{MW_{A} } \right) \rho dV + \int_{\mathscr{A}}\sum_{A = 1}^{A = N}\left(\frac{N_{JA} \omega_{A} }{MW_{A} } \right) \rho \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad J = 1, 2, ..., T \label{10}$

This represents the “mass” analog of Equation \ref{7}, and it will be convenient when working with problems in which mass flow rates and mass fractions are given. We should note that Equation \ref{10} actually represents a molar balance on the $$J^{th}$$ atomic species since an examination of the units will indicate that

$\int_{\mathscr{A}}\left\{\sum_{A = 1}^{A = N}\frac{N_{JA} \omega_{A} }{MW_{A} } \right\} \rho \mathbf{v} \cdot \mathbf{n} dA\to \begin{Bmatrix}\text{moles per} \\ \text{unit time}\end{Bmatrix} \label{11}$

To obtain the actual atomic species mass balance one would multiply Equation \ref{10} by the atomic mass of the $$J^{th}$$ atomic species, $$AW_{J}$$.

At steady state Equation \ref{4} represents a system of homogeneous equations in terms the net atomic species molar flow rates leaving the control volume. For there to be a nontrivial solution, the rank r of $$\left[N_{JA} \right]$$ must be less than $$N$$. Consequently there are at most r linearly independent atomic species balances (see Sec. 6.2.3 for details).

Example $$\PageIndex{1}$$: Production of Sulfuric Acid

In this example we consider the production of sulfuric acid illustrated in Figure $$\PageIndex{1}$$. The mass flow rate of the 90% sulfuric acid stream is specified as $$\dot{m}_{1} = 100 \ { lb}_{ m} /{ hr}$$, and we are asked to determine the mass flow rate of the pure sulfur trioxide stream, $$\dot{m}_{2}$$. As is often the custom with liquid systems the percentages given in Figure $$\PageIndex{1}$$ refer to mass fractions, thus we want to create a final product in which the mass fraction of sulfuric acid is 0.98.

In order to analyze this process in terms of the atomic species mass balance, we use the steady state form of Equation \ref{10} given here as

$\int_{\mathscr{A}}\sum_{A = 1}^{A = N}\left(\frac{N_{JA} \omega_{A} }{MW_{A} } \right) \rho \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad J \Rightarrow \ce{S}, \ce{H}, \ce{O} \tag{1}\label{a1}$

For the system shown in Figure $$\PageIndex{1}$$ we see that Equation \ref{a1} takes the form

$\dot{m}_{ 1} \sum_{A = 1}^{A = N}\left(\frac{N_{JA} \omega_{A} }{MW_{A} } \right)_{1} + \dot{m}_{ 2} \sum_{A = 1}^{A = N}\left(\frac{N_{JA} \omega_{A} }{MW_{A} } \right)_{2} = \dot{m}_{ 3} \sum_{A = 1}^{A = N}\left(\frac{N_{JA} \omega_{A} }{MW_{A} } \right)_{3} , \quad J \Rightarrow \ce{S}, \ce{H}, \ce{O} \tag{2}\label{a2}$

In the process under investigation there are three atomic species indicated by $$\ce{S}$$, $$\ce{H}$$ and $$\ce{O}$$, and there are three molecular species indicated by $$\ce{H2} \ce{SO4}$$, $$\ce{H2O}$$ and $$\ce{SO3}$$. Since only sulfur and oxygen appear in the stream for which we want to determine the mass flow rate, we must make use of either a sulfur balance or an oxygen balance. However, sulfur appears in only two of the

molecular species while oxygen appears in all three. Because of this a sulfur balance is preferred. At this point we recall the definition of $$N_{JA}$$ given by Equation \ref{3}

$N_{JA} = \begin{Bmatrix} \text{ number of moles of} \\ J\text{-type atoms per mole} \\ \text{ of molecular species }A\end{Bmatrix} , \quad J = 1, 2,...,T, \text{ and } A = 1, 2,...N \tag{3}\label{a3}$

\begin{align} N_{JA} = 1 , && J = { S} , && A = \ce{H2SO4} \nonumber\\ N_{JA} = 1 , && J = { S} , && A = \ce{SO3} \label{a4}\tag{4} \end{align}

Letting $$J$$ in Equation \ref{a2} represent sulfur we have

$\frac{\dot{m}_{ 1} (\omega_{\ce{H2SO4}} )_{1} }{MW_{\ce{H2SO4}} } + \frac{\dot{m}_{ 2} }{MW_{\ce{SO3}} } = \frac{\dot{m}_{ 3} (\omega_{\ce{H2SO4}} )_{3} }{MW_{\ce{H2SO4}} } \tag{5}\label{a5}$

and an overall mass balance

$\dot{m}_{ 1} + \dot{m}_{ 2} = \dot{m}_{ 3} \label{a6}\tag{6}$

can be used to eliminate $$\dot{m}_{3}$$ in terms of $$\dot{m}_{1}$$ and $$\dot{m}_{2}$$. This allows us to solve for the mass flow rate of sulfur trioxide that is given by

$\dot{m}_{2} = \frac{\left[(\omega_{\ce{H2SO4}} )_{3} - (\omega_{\ce{H2SO4}} )_{1} \right] \dot{m}_{1} }{\left[\frac{MW_{\ce{H2SO4}} }{MW_{\ce{SO3}} } - (\omega_{\ce{H2SO4}} )_{3} \right]} \label{a7}\tag{7}$

In this case we have been able to obtain a solution using only a single atomic element balance along with the overall mass balance. Given this solution, we can easily compute the mass flow rate of pure sulfur trioxide to be

$\dot{m}_{2} = 32.65 \ { lb}_{ m} /{ hr} \label{a8}\tag{8}$

In the previous example we solved a problem in which there were three molecular species and three atomic species and we found that a solution could be obtained easily using an atomic species balance. When the number of atomic species is less that the number of molecular species, there is a definite computational advantage in using the atomic species balance. However, the physical description of most processes is best given in terms of molecular species and this often controls the choice of the method used to solve a particular problem.

Example $$\PageIndex{2}$$: Combustion of carbon and air

Carbon is burned with air, as illustrated in Figure $$\PageIndex{2}$$, with all the carbon oxidized to $$\ce{CO2}$$ and $$\ce{CO}$$. The ratio of carbon dioxide produced to carbon monoxide produced is 2:1. In this case

we wish to determine the flue gas composition when 50% excess air is used. The percent excess air is defined as

$\begin{Bmatrix} \text{ percentage of} \\ \text{ excess air}\end{Bmatrix} = \frac{\begin{Bmatrix} \text{ molar flow} \\ \text{ rate of oxygen} \\ \text{ entering}\end{Bmatrix} - \begin{Bmatrix} \text{ molar rate of } \\ \text{ consumption of oxygen} \\ \text{ owing to reaction}\end{Bmatrix}}{\begin{Bmatrix} \text{ molar rate of } \\ \text{ consumption of oxygen} \\ \text{ owing to reaction}\end{Bmatrix}} \times 100 \label{b1}\tag{1}$

and this definition requires a single application of Equation \ref{1} in order to incorporate the global molar rate of production of oxygen, $$\mathscr{R}_{\ce{O2}}$$, into the analysis.

Atomic Species Balances

The steady-state form of Equation \ref{7} is given by

$\int_{\mathscr{A}} \left\{\sum_{A = 1}^{A = N}N_{JA} x_{A} \right\} c \mathbf{v} \cdot \mathbf{n} dA = 0 , \quad J \Rightarrow \ce{C}, \ce{O}, \ce{N} \label{b2}\tag{2}$

and for the system illustrated in Figure $$\PageIndex{2}$$ we obtain

$\left\{\sum_{A = 1}^{A = N}N_{JA} x_{A} \right\}_{1} \dot{M}_{1} + \left\{\sum_{A = 1}^{A = N}N_{JA} x_{A} \right\}_{2} \dot{M}_{2} = \left\{\sum_{A = 1}^{A = N}N_{JA} x_{A} \right\}_{3} \dot{M}_{3} \label{b3}\tag{3}$

We begin by choosing the $$J^{th}$$ atomic species to be carbon so that Equation \ref{b3} takes the form

Atomic Carbon Balance: $\dot{M}_{1} = \left[(x_{\ce{CO}} )_{3} + (x_{\ce{CO2}} )_{3} \right] \dot{M}_{3} \label{b4}\tag{4}$

and we continue with this approach to obtain the atomic oxygen and atomic nitrogen balance equations given by

Atomic Oxygen Balance: $2(x_{\ce{O2}} )_{2} \dot{M}_{2} = \left[2(x_{\ce{O2}} )_{3} + 2(x_{\ce{CO2}} )_{3} + (x_{\ce{CO}} )_{3} \right] \dot{M}_{3} \label{b5}\tag{5}$

Atomic Nitrogen Balance: $2(x_{\ce{N2}} )_{2} \dot{M}_{2} = 2(x_{\ce{N2}} )_{3} \dot{M}_{3} \label{b6}\tag{6}$

Directing our attention to the percentage of excess air defined by Equation \ref{b1}, we define the numerical excess air as

$\beta = \frac{ \left\{ (\dot{M}_{\ce{O2}} )_{2} \right\} - \left\{ -\mathscr{R}_{\ce{O2}} \right\}} {\left\{ -\mathscr{R}_{\ce{O2}} \right\}} = 0.5 \label{b7}\tag{7}$

At this point we must make use of the mole balance represented by Equation \ref{1} in order to represent the global net rate of production of oxygen as

Molecular Oxygen Balance: $\mathscr{R}_{\ce{O2}} = (\dot{M}_{\ce{O2}} )_{3} - (\dot{M}_{\ce{O2}} )_{2} \label{b8}\tag{8}$

This can be used to obtain a relation between the numerical excess air and the molar flow rates of oxygen given by

$\beta (\dot{M}_{\ce{O2}} )_{2} = \left(1+\beta \right)(\dot{M}_{\ce{O2}} )_{3} \label{b9}\tag{9}$

For use with the other constraining equations, this can be expressed in terms of mole fractions to obtain

$\beta (x_{\ce{O2}} )_{2} \dot{M}_{2} = \left(1+\beta \right)(x_{\ce{O2}} )_{3} \dot{M}_{3} \label{b10}\tag{10}$

This result, along with the atomic species balances and the input data, can be used to determine the flue gas composition

Analysis

We are given that the molar rate of production of carbon dioxide is two times the molar rate of production of carbon monoxide. We can express this information as

$\dot{M}_{\ce{CO2}} = \gamma \dot{M}_{\ce{CO}} , \quad \gamma = 2 \label{b11}\tag{11}$

and in terms of mole fractions this leads to

$(x_{\ce{CO2}} )_{3} = \gamma (x_{\ce{CO}} )_{3} \label{b12}\tag{12}$

The final equation required for the solution of this problem is the constraint on the sum of the mole fractions in Stream #3 that is given by

$(x_{\ce{O2}} )_{3} + (x_{\ce{N2}} )_{3} + (x_{\ce{CO}} )_{3} + (x_{\ce{CO2}} )_{3} = 1 \label{b13}\tag{13}$

In addition to this constraint on the mole fractions in Stream #3, we assume that the air in Stream #2 can be described by

$(x_{\ce{O2}} )_{2} = 0.21 , \quad (x_{\ce{N2}} )_{2} = 0.79 \label{b14}\tag{14}$

Use of Equation \ref{b8} in Eqs. \ref{b4} through \ref{b6} gives

Atomic Carbon Balance: $\dot{M}_{1} = (x_{\ce{CO}} )_{3} \left(1+\gamma \right)\dot{M}_{3} \label{b15a}\tag{15a}$

Atomic Oxygen Balance: $(x_{\ce{O2}} )_{2} \dot{M}_{2} = \left[(x_{\ce{O2}} )_{3} +\left(\gamma + \frac{1}{2} \right)(x_{\ce{CO}} )_{ 3} \right] \dot{M}_{3} \label{b15b}\tag{15b}$

Atomic Nitrogen Balance: $(x_{\ce{N2}} )_{2} \dot{M}_{2} = (x_{\ce{N2}} )_{3} \dot{M}_{3} \label{b15c}\tag{15c}$

At this point we note that we have eight unknowns and seven equations; however, one of the molar flow rates can be eliminated to develop a solution.

Algebra

In order to determine the flue gas composition, we do not need to determine the absolute values of $$\dot{M}_{1}$$, $$\dot{M}_{2}$$, and $$\dot{M}_{3}$$ but only the relative flow rates. These are defined by

$\mathscr{M}_{ 2} = {\dot{M}_{2} / \dot{M}_{1} } \label{b16a}\tag{16a}$

$\mathscr{M}_{ 3} = {\dot{M}_{3} / \dot{M}_{1} } \label{b16b}\tag{16b}$

so that Eqs. \ref{b15a}-\ref{b15c} and Equation \ref{b10} take the form

$(x_{\ce{CO}} )_{3} \left(1 + \gamma \right) \mathscr{M}_{ 3} = 1 \label{b17a}\tag{17a}$

$(x_{\ce{O2}} )_{2} \mathscr{M}_{ 2} = \left[(x_{\ce{O2}} )_{3} + \left(\gamma + \frac{1}{2} \right)(x_{\ce{CO}} )_{3} \right] \mathscr{M}_{ 3} \label{b17b}\tag{17b}$

$(x_{\ce{N2}} )_{2} \mathscr{M}_{ 2} = (x_{\ce{N2}} )_{3} \mathscr{M}_{ 3} \label{b17c}\tag{17c}$

$\beta (x_{\ce{O2}} )_{2} \mathscr{M}_{ 2} = \left(1+\beta \right)(x_{\ce{O2}} )_{3} \mathscr{M}_{ 3} \label{b17d}\tag{17d}$

From these four equations we need to eliminate $$\mathscr{M}_{ 2}$$ and$$\mathscr{M}_{ 3}$$ in order to determine the mole fractions and thus the composition of the flue gas. Equations \ref{b17a} and \ref{b17c} can be used to obtain the following representations for $$\mathscr{M}_{ 2}$$ and$$\mathscr{M}_{ 3}$$

$\mathscr{M}_{ 3} = \frac{1}{(x_{\ce{CO}} )_{3} \left(1+\gamma \right)} \label{b18a}\tag{18a}$

$\mathscr{M}_{ 2} = \frac{(x_{\ce{N2}} )_{3} }{(x_{\ce{CO}} )_{3} \left(1+\gamma \right)(x_{\ce{N2}} )_{2} } \label{b18b}\tag{18b}$

Use of these two results in Equation \ref{b17b} leads to

$(x_{\ce{N2}} )_{3} = \underbrace{\left\{\frac{(x_{\ce{N2}} )_{2} }{(x_{\ce{O2}} )_{2} } \right\}}_{ known}(x_{\ce{O2}} )_{3} + \underbrace{\left\{ \frac{(x_{\ce{N2}} )_{2} \left(\gamma + \frac{1}{2} \right)}{(x_{\ce{O2}} )_{2} }\right\}}_{ known}(x_{\ce{CO}} )_{3} \label{b19a}\tag{19a}$

and when we eliminate $$\mathscr{M}_{ 2}$$ and$$\mathscr{M}_{ 3}$$ from Equation \ref{b17d} we find

$(x_{\ce{N2}} )_{3} = \underbrace{\left\{\frac{1+\beta }{\beta } \frac{(x_{\ce{N2}} )_{2} }{(x_{\ce{O2}} )_{2} }\right\}}_{ known}(x_{\ce{O2}} )_{3} \label{b19b}\tag{19b}$

These two equations, along with the constraint on the mole fractions given by

$(x_{\ce{O2}} )_{3} + (x_{\ce{N2}} )_{3} + \left(1+\gamma \right)(x_{\ce{CO}} )_{3} = 1 \label{b20}\tag{20}$

are sufficient to determine $$(x_{\ce{O2}} )_{3}$$, $$(x_{\ce{N2}} )_{3}$$ and $$(x_{\ce{CO}} )_{3}$$. These are given by

$(x_{\ce{O2}} )_{3} = \frac{1}{1+K_{3} +\left(1+\gamma \right)\left(K_{3} -K_{1} \right) / K_{2} } \label{b21a}\tag{21a}$

$(x_{\ce{N2}} )_{3} = \frac{K_{3} }{1+K_{3} +\left(1+\gamma \right)\left(K_{3} -K_{1} \right) / K_{2} } \label{b21b}\tag{21b}$

$(x_{\ce{CO}} )_{3} = \frac{\left(K_{3} -K_{1} \right) / K_{2} }{1+K_{3} +\left(1+\gamma \right)\left(K_{3} -K_{1} \right) / K_{2} } \label{b21c}\tag{21c}$

$(x_{\ce{CO2}} )_{3} = \gamma (x_{\ce{CO}} )_{3} \label{b21d}\tag{21d}$

where the constants $$K_{1}$$, $$K_{2}$$ and $$K_{3}$$ are given by

$K_{1} = \frac{(x_{\ce{N2}} )_{2} }{(x_{\ce{O2}} )_{2} } = 3.7619 \label{b22a}\tag{22a}$

$K_{2} = \frac{(x_{\ce{N2}} )_{2} }{(x_{\ce{O2}} )_{2} } \left(\gamma + \frac{1}{2} \right) = 9.4048 \label{b22b}\tag{22b}$

$K_{3} = \frac{(x_{\ce{N2}} )_{2} }{(x_{\ce{O2}} )_{2} } \left(\frac{1+\beta }{\beta } \right) = 11.2857 \label{b22c}\tag{22c}$

Use of these values in Eqs. \ref{b21a}-\ref{b21d} gives

$(x_{\ce{O2}} )_{3} = 0.0681 , \quad (x_{\ce{N2}} )_{3} = 0.7685 , \quad (x_{\ce{CO}} )_{3} = 0.0545 , \quad (x_{\ce{CO2}} )_{3} = 0.1080 \label{b23}\tag{23}$

The sum of the mole fractions is 0.999 indicating the presence of a small numerical error that could be diminished by carrying more significant figures. On the other hand, the input data for problems of this type is never accurate to $$\pm 0.1\%$$ and it would be misleading to develop a more accurate solution.

In this appendix we have illustrated how problems can be solved using atomic species balances instead of molecular species balances. In general, the number of atomic species balances is smaller than the number of molecular species balances, thus the use of atomic species balances leads to fewer equations. However, information is often given in terms of molecular species, such as the percent excess oxygen in Example $$\PageIndex{2}$$, and this requires the use of molecular species balances. Nevertheless, the algebraic effort in Example $$\PageIndex{2}$$ is less than one would encounter if the problem were solved using molecular species balances.

Problems

1. A stream of pure methane ($$\ce{CH4}$$) is partially burned with air in a furnace at a rate of 100 moles of methane per minute. The air is dry, the methane is in excess, and the nitrogen is inert in this particular process. The products of the reaction are illustrated in Figure C-1.

The exit gas contains a 1:1 ratio of $$\ce{H2O} : \ce{H2}$$ and a 10:1 ratio of $$\ce{CO} : \ce{CO2}$$. Assuming that all of the oxygen and 94% of the methane are consumed by the reactions, determine the flow rate and composition of the exit gas stream.

2. A fuel composed entirely of methane and nitrogen is burned with excess air. The dry flue gas composition in volume percent is: $$\ce{CO2}$$, 7.5%, $$\ce{O2}$$, 7%, and the remainder nitrogen. Determine the composition of the fuel gas and the percentage of excess air as defined by

$\begin{Bmatrix} \text{ percent of} \\ \text{ excess air}\end{Bmatrix} = \frac{\begin{pmatrix}\text{ molar flow} \\ \text{ rate of oxygen} \\ \text{ entering}\end{pmatrix} - \begin{pmatrix} \text{ molar rate of} \\ \text{ consumption of oxygen} \\ \text{ owing to reaction}\end{pmatrix}}{\begin{pmatrix}\text{ molar rate of} \\ \text{ consumption of oxygen} \\ \text{ owing to reaction}\end{pmatrix}} \times 100 \nonumber$