# 7.3: Combustion Reactions

Computation of the rate of production or consumption of chemical species during combustion is an important part of chemical engineering practice. Efficient use of irreplaceable fossil energy resources is ecologically responsible and economically sound. Fuels are burned in combustion chambers using air as the source of oxygen ($$\ce{O2}$$) as illustrated in Figure $$\PageIndex{1}$$. The fuel enters the combustion chamber at Stream #1 and air is supplied via Stream #2. Since air is 79% nitrogen ($$\ce{N2}$$), and nitrogen can form NOX as part of the combustion reaction, it is good practice to use only the amount of air that is needed for the reaction. On the other hand, the need to burn the fuels completely requires excess air to displace the reaction equilibrium. Matters are complicated further by the presence of water in the air stream. In this simplified treatment of combustion, we will ignore the complexities associated with NOX in the exit stream and water vapor in the entrance stream.

## Theoretical Air

The first step in analyzing a combustion process is to determine the rate at which air must be supplied in order to achieve complete combustion. This is called the theoretical air requirement for the combustion process and it is based on the fuel composition.

Most fuels consist of many different hydrocarbons; however, in this simple example we consider the fuel to consist of a single molecule represented by C$$_m$$H$$_n$$O$$_p$$. Examples are propane ($$m = 3, n = 8, p = 0$$), carbon monoxide ($$m = 1, n = 0, p = 1$$), and methanol ($$m = 1, n = 4, p = 1$$). Total combustion is the conversion of all the fuel to $$\ce{CO2}$$ and $$\ce{H2O}$$. Our tools for this analysis are Axioms I and II as represented by Equations $$(7.1.4)$$ and $$(7.1.5)$$ with the global rate of production defined by Equation $$(7.1.3)$$. We first express the atomic matrix as

$\text{ Molecular species } \to \text{ C}_{m} \text{H}_{n} \text{O}_{p} \quad \ce{O2} \quad \ce{CO2} \quad \ce{H2O} \\ \begin{matrix} {carbon} \\ {hydrogen} \\ {oxygen} \end{matrix} \begin{bmatrix} { m} & { 0} & { 1} & { 0} \\ { n} & { 0} & { 0} & { 2} \\ { p} & { 2} & { 2} & { 1} \end{bmatrix} \label{16}$

and make use of this representation to express Axiom II in the form

$\sum_{A = 1}^{A = N}N_{JA} \mathscr{R}_{ A} = \begin{bmatrix} {m} & {0} & {1} & {0} \\ {n} & {0} & {0} & {2} \\ {p} & {2} & {2} & {1} \end{bmatrix}\begin{bmatrix} {\mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} } } \\ {\mathscr{R}_{ \ce{O2} } } \\ {\mathscr{R}_{ \ce{CO2} } } \\ {\mathscr{R}_{ \ce{H2O}} } \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{17}$

The atomic matrix can be expressed in row reduced echelon form according to

$\left[N_{JA} \right]^* = \begin{bmatrix} {1} & {0} & {0} & {\frac{2}{n} } \\ {0} & {1} & {0} & {\frac{4m+n-2p}{2n} } \\ {0} & {0} & {1} & {-\frac{2m}{n} } \end{bmatrix} \label{18}$

This indicates that the rank is three, and all the species production rates can be expressed in terms of a single pivot species that we chose as water. Use of Equation \ref{18} in Equation \ref{17} allows us to express Axiom II in the form

$\begin{bmatrix} {1} & {0} & {0} & { \frac{2}{n} } \\ {0} & {1} & {0} & {\frac{4m+n-2p}{2n} } \\ {0} & {0} & {1} & {- \frac{2m}{n} } \end{bmatrix}\begin{bmatrix} {\mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} } } \\ {\mathscr{R}_{ \ce{O2} } } \\ {\mathscr{R}_{ \ce{CO2} } } \\ {\mathscr{R}_{ \ce{H2O}} } \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{19}$

This equation can be partitioned according to the development in Example $$6.3.1$$ or the development in Sec. 6.4 in order to obtain

$\begin{bmatrix} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{bmatrix} \begin{bmatrix} {\mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} } } \\ {\mathscr{R}_{ \ce{O2} } } \\ {\mathscr{R}_{ \ce{CO2} } } \end{bmatrix} + \begin{bmatrix} { \frac{2}{n} } \\ {\frac{4m+n-2p}{2n} } \\ {- \frac{2m}{n} } \end{bmatrix} \begin{bmatrix} {\mathscr{R}_{ \ce{H2O}} } \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{20}$

This immediately leads to a special case of the global pivot theorem (see Eq. $$(6.4.8)$$)

Global Pivot Theorem: $\begin{bmatrix} {\mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} } } \\ {\mathscr{R}_{ \ce{O2} } } \\ {\mathscr{R}_{ \ce{CO2} } } \end{bmatrix} = \begin{bmatrix} { -\frac{2}{n} } \\ {-\frac{4m+n-2p}{2n} } \\ {\frac{2m}{n} } \end{bmatrix} \begin{bmatrix} {\mathscr{R}_{ \ce{H2O}} } \end{bmatrix} \label{21}$

which provides the species global net rates of production in terms of the global net rate of production of water as given by

$\mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} } = - \frac{2}{n} \mathscr{R}_{ \ce{H2O}} , \quad \mathscr{R}_{ \ce{O2} } = - \frac{4m+n-2p}{2n} \mathscr{R}_{ \ce{H2O}} , \quad \mathscr{R}_{ \ce{CO2} } = - \frac{2m}{n} \mathscr{R}_{ \ce{H2O}} \label{22}$

In this particular problem, it is convenient to represent the global rate of production of fuel (which is negative) in terms of the global rate of production of oxygen (which is negative), thus we use the first two of Equations \ref{22} to obtain

$\mathscr{R}_{ \ce{O2} } = \frac{4m+n-2p}{4} \mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} } \label{23}$

At this point we have completed our analysis of Axiom II and we are ready to apply Axiom I using the control volume illustrated in Figure $$\PageIndex{1}$$. Application of the steady state form of Equation $$(7.1.5)$$ for both the fuel ($$\text{ C}_{m} \text{H}_{n} \text{O}_{p}$$) and the oxygen ($$\ce{O2}$$) leads to

Fuel: $- (\dot{M}_{\text{ C}_{m} \text{H}_{n} \text{O}_{p} } )_{1} + (\dot{M}_{\text{ C}_{m} \text{H}_{n} \text{O}_{p} } )_{3} = \mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} } \label{24}$

Oxygen: $- (\dot{M}_{\ce{O2} } )_{2} + (\dot{M}_{\ce{O2} } )_{3} = \mathscr{R}_{ \ce{O2} } \label{25}$

In order to determine the theoretical air needed for complete combustion, we assume that all the fuel and all the oxygen that enter the combustion chamber are reacted, thus there is no fuel in Stream #3 and there is no oxygen in Stream #3. Under these circumstances Equations \ref{24} and \ref{25} reduce to

Fuel (complete combustion): $- (\dot{M}_{\text{ C}_{m} \text{H}_{n} \text{O}_{p} } )_{1} = \mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} } \label{26}$

Oxygen (complete combustion): $- (\dot{M}_{\ce{O2} } )_{2} = \mathscr{R}_{ \ce{O2} } \label{27}$

At this point we represent $$\mathscr{R}_{ \ce{O2} }$$ in terms of $$\mathscr{R}_{ \text{ C}_{m} \text{H}_{n} \text{O}_{p} }$$ using Equation \ref{23} and this allows us to express the theoretical oxygen required for complete combustion as

$(\dot{M}_{\ce{O2} } )_{Theoretical} = (\dot{M}_{\ce{O2} } )_{2} = \frac{4m+n-2p}{4} (\dot{M}_{\text{ C}_{m} \text{H}_{n} \text{O}_{p} } )_{1} \label{28}$

If the fuel in Stream #1 contains $$N$$ molecular species that can be represented as $$\text{ C}_{m} \text{H}_{n} \text{O}_{p}$$, the theoretical oxygen required for complete combustion is given by

$(\dot{M}_{\ce{O2} } )_{Theoretical} = \sum_{i = 1}^{i = N}\left[\frac{4m+n-2p}{4} \right]_{i} \left[(\dot{M}_{\text{ C}_{m} \text{H}_{n} \text{O}_{p} } )_{1} \right]_{i} \label{29}$

This analysis for the theoretical oxygen can be extended to other fuels that may contain sulfur provided that the molecules in the fuel can be characterized by $$\text{ C}_{m} \text{H}_{n} \text{O}_{p} \text{S}_{q}$$. In this case the complete combustion products would be $$\ce{CO2}$$, $$\ce{H2O}$$ and $$\ce{SO2}$$.

Example $$\PageIndex{1}$$: Determination of theoretical air

A fuel containing 60% $$\ce{CH4}$$, 15% $$\ce{C2H6}$$, 5% $$\ce{CO}$$ and 20% $$\ce{N2}$$ (all mole percent) is burned with air to produce a flue gas containing $$\ce{CO2}$$, $$\ce{H2O}$$, and $$\ce{N2}$$. The combustion process takes place in the unit illustrated in Figure $$\PageIndex{1}$$, and we want to determine the molar flow rate of theoretical air needed for complete combustion. The solution to this problem is given by Equation \ref{29} that takes the form

\begin{align} (\dot{M}_{\ce{O2} } )_{Theoretical} = & \sum_{i = 1}^{i = N} \left[ \frac{4m+n-2p}{4} \right]_i \left[(\dot{M}_{\text{ C}_{m} \text{H}_{n} \text{O}_{p} } )_{1} \right]_{i} \nonumber \\ = & \left(\frac{4+4}{4} \right)(\dot{M}_{\ce{CH4} } )_{1} + \left(\frac{8+6}{4} \right)(\dot{M}_{\ce{C2H6}} )_{1} + \left(\frac{4-2}{4} \right)(\dot{M}_{\ce{CO}} )_{1} \label{1a}\tag{1}\\ = & 2 (\dot{M}_{\ce{CH4} } )_{1} + \frac{7}{2} (\dot{M}_{\ce{C2H6}} )_{1} + \frac{1}{2} (\dot{M}_{\ce{CO}} )_{1} \nonumber \end{align}

If we let $$\dot{M}_{1}$$ be the total molar flow rate of Stream #1, we can express Equation \ref{1a} as

\begin{align} (\dot{M}_{\ce{O2} } )_{Theoretical} = & \left[2\left(0.60\right) + \frac{7}{2} (0.15) + \frac{1}{2} (0.05)\right]\dot{M}_{1} \nonumber \\ = & 1.75 \dot{M}_{1} \label{2a}\tag{2}\end{align}

This indicates that for every mole of feed we require 1.75 moles of oxygen for complete combustion. Taking the mole fraction of oxygen in air to be 0.21, we find the molar flow rate of air to be given by

$(\dot{M}_{ air} )_{Theoretical} = \left(\frac{1}{0.21} \right)1.75 \dot{M}_{1} = 8.33 \dot{M}_{1} \label{3a}\tag{3}$

The determination of the theoretical molar flow rate of air in Example $$\PageIndex{1}$$ was relatively simple because the complete combustion process involved a feed stream that could be described in terms of a single molecular form given by $$\text{ C}_{m} \text{H}_{n} \text{O}_{p}$$. In the next example, we consider a slightly more complex process in which the excess air is specified.

Example $$\PageIndex{2}$$: Combustion of residual synthesis gas

In Figure $$\PageIndex{2}$$ we have illustrated a synthesis gas (Stream #1) consisting of 0.4% methane ($$\ce{CH4}$$), 52.8% hydrogen ($$\ce{H2}$$), 38.3% carbon monoxide ($$\ce{CO}$$), 5.5% carbon dioxide ($$\ce{CO2}$$), 0.1% oxygen ($$\ce{O2}$$), and 2.9% nitrogen ($$\ce{N2}$$). The synthesis gas reacts with air (Stream #2) that is supplied at a rate which provides 10% excess oxygen and the composition of Stream #2 is assumed to be 79% nitrogen ($$\ce{N2}$$) and 21% oxygen ($$\ce{O2}$$).

In this example we want to determine: (I) the molar flow rate of air relative to the molar flow rate of the synthesis gas, and (II) the species molar flow rates of the components of the flue gas relative to the molar flow rate of the synthesis gas. We assume complete combustion so that all hydrocarbon species in Stream #1 are converted to carbon dioxide ($$\ce{CO2}$$) and water ($$\ce{H2O}$$). In addition, we assume that the nitrogen is inert so that no NOX appears in the exit stream. This means that the molecular species in the flue gas are $$\ce{N2}$$, $$\ce{O2}$$, $$\ce{CO2}$$ and $$\ce{H2O}$$.

In this example, the development leading to Equation \ref{29} is applicable for the combustion of synthesis gas and that result takes the form

\begin{align} (\dot{M}_{\ce{O2} } )_{Theoretical} =& \sum_{i = 1}^{i = N} \left[ \frac{4m+n-2p}{4}\right]_i \left[(\dot{M}_{\text{ C}_{m} \text{H}_{n} \text{O}_{p} } )_{1} \right]_{i} \nonumber \\ = & \frac{4+4}{4} (\dot{M}_{\ce{CH4} } )_{1} + \frac{2}{4} (\dot{M}_{\ce{H2} } )_{1} + \frac{4-2}{4} (\dot{M}_{\ce{CO}} )_{1} \label{1b}\tag{1} \end{align}

The species molar flow rates in Stream #1 can be expressed in terms of the total molar flow rate, $$\dot{M}_{1}$$ according to

$(\dot{M}_{\ce{CH4} } )_{1} = 0.004\dot{M}_{1} , \quad (\dot{M}_{\ce{H2} } )_{1} = 0.528\dot{M}_{1} , \quad (\dot{M}_{\ce{CO}} )_{1} = 0.383\dot{M}_{1} \label{2b}\tag{2}$

and these results can be used in Equation \ref{1b} to obtain the molar flow rate of theoretical air in terms of the molar flow rate of synthesis gas.

$(\dot{M}_{\ce{O2} } )_{Theoretical} = \left[\frac{4+4}{4} \left(0.004\right) + \frac{2}{4} \left(0.528\right) + \frac{4-2}{4} \left(0.383\right)\right]\dot{M}_{1} = 0.4635\dot{M}_{1} \label{3b}\tag{3}$

Taking into account that Stream #1 contains oxygen, the condition of 10% excess oxygen requires

$(\dot{M}_{\ce{O2} } )_{1} + (\dot{M}_{\ce{O2} } )_{2} = \left(1.10\right)(\dot{M}_{\ce{O2} } )_{Theoretical} \label{4b}\tag{4}$

and this allows us to express the molar flow rate of oxygen in Stream #2 as

$(\dot{M}_{\ce{O2} } )_{2} = \left(1.10\right)(\dot{M}_{\ce{O2} } )_{Theoretical} - (y_{\ce{O2} } )_{1} \dot{M}_{1} \label{5b}\tag{5}$

Given that the mole fraction of oxygen ($$\ce{ O2}$$) in the synthesis gas (Stream #1) is 0.001, we can use this result along with Equation \ref{3b} to obtain

$(\dot{M}_{\ce{O2} } )_{2} = 0.509 \dot{M}_{1} \label{6b}\tag{6}$

We are given that the mole fraction of oxygen ($$\ce{O2}$$) in the air (Stream #2) is 0.21 and this leads to the total molar flow rate for air given by

$\dot{M}_{2} = 2.42 \dot{M}_{1} \label{7b}\tag{7}$

Knowing the molar flow rate of the air stream relative to the molar flow rate of the synthesis gas that is required for complete combustion is crucial for the proper operation of the system. Knowing how much air is required to achieve complete combustion will generally be determined experimentally, thus the ratio 2.42 will be adjusted to achieve the desired operating condition.

In addition to knowing the required flow rate of air, we also need to know the molar flow rates of the species in the flue gas since this gas will often be discharged into the atmosphere. Determination of these molar flow rates requires the application of Axioms I and II. We begin with the atomic matrix

$\text{ Molecular Species } \to \ce{CH4} \quad \ce{H2} \quad \ce{CO} \quad \ce{CO2} \quad \ce{H2O} \quad \ce{O2} \\ \begin{matrix} {carbon } \\ {hydrogen} \\ {oxygen } \end{matrix} \begin{bmatrix} { 1} & {0} & {1} & {1} & { 0} & {0 } \\ { 4} & {2} & {0} & {0} & { 2} & {0 } \\ { 0} & {0} & {1} & {2} & { 1} & {2 } \end{bmatrix} \label{8b}\tag{8}$

which leads to the following form of Axiom II:

Axiom II: $\sum_{A = 1}^{A = 6}N_{JA} \mathscr{R}_{ A} = \begin{bmatrix} {1} & {0} & {1} & {1} & {0} & {0} \\ {4} & {2} & {0} & {0} & {2} & {0} \\ {0} & {0} & {1} & {2} & {1} & {2} \end{bmatrix} \begin{bmatrix} \mathscr{R}_{ \ce{CH4} } \\ \mathscr{R}_{ \ce{H2} } \\ \mathscr{R}_{ \ce{CO} } \\ \mathscr{R}_{ \ce{CO2} } \\ \mathscr{R}_{ \ce{H2O} } \\ \mathscr{R}_{ \ce{O2} } \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{9b}\tag{9}$

This axiom can be expressed in row reduced echelon form according to

$\begin{bmatrix} {1} & {0} & {0} & {-1} & {-1} & {-2} \\ {0} & {1} & {0} & { 2} & { 3} & { 4} \\ {0} & {0} & {1} & { 2} & { 1} & { 2} \end{bmatrix} \begin{bmatrix} \mathscr{R}_{ \ce{CH4} } \\ \mathscr{R}_{ \ce{H2} } \\ \mathscr{R}_{ \ce{CO} } \\ \mathscr{R}_{ \ce{CO2} } \\ \mathscr{R}_{ \ce{H2O} } \\ \mathscr{R}_{ \ce{O2} } \end{bmatrix} = \begin{bmatrix} {0} \\ {0} \\ {0} \end{bmatrix} \label{10b}\tag{10}$

and the solution for the three non-pivot global rates of production is given by

Axiom II: \begin{align} \mathscr{R}_{ \ce{CH4} } = & \mathscr{R}_{ \ce{CO2} } + \mathscr{R}_{ \ce{H2O}} + 2 \mathscr{R}_{ \ce{O2} } \nonumber \\ \mathscr{R}_{ \ce{H2} } = & -2 \mathscr{R}_{ \ce{CO2} } - 3 \mathscr{R}_{ \ce{H2O}} - 4 \mathscr{R}_{ \ce{O2} } \label{11b}\tag{11} \\ \mathscr{R}_{ \ce{CO}} = & -2 \mathscr{R}_{ \ce{CO2} } - \mathscr{R}_{ \ce{H2O}} - 2 \mathscr{R}_{ \ce{O2} } \nonumber \end{align}

This completes our analysis of Axiom II, and we move on to the steady state for of Axiom I that is given by

Axiom I: $\int_{\mathscr{A}} c_{A} \mathbf{v}_{A} \cdot \mathbf{n} dA = \mathscr{R}_{ A} , \quad A = 1,2, ..., N \label{12b}\tag{12}$

Application of this result to the control volume illustrated in Figure $$\PageIndex{2}$$ leads to the general expression

$-\left(\dot{M}_{A}\right)_{1} - \left( \dot{M}_{A}\right)_{2} + \left( \dot{M}_{A} \right)_{3} = \mathscr{R}_{A}, \quad A = 1,2, \ldots, 7 \label{13b}\tag{13}$

while the individual species balances take the forms

\begin{align} \ce{CH4}: && -\left(\dot{M}_{\ce{CH4}}\right)_{1} = -0.004 \dot{M}_{1} = \mathscr{R}_{\ce{CH4}} \label{14b}\tag{14}\end{align}

\begin{align} \ce{H2}: && -\left(\dot{M}_{\ce{H2}}\right)_{1} = -0.528 \dot{M}_{1} = \mathscr{R}_{\ce{H2}} \label{15b}\tag{15}\end{align}

\begin{align} \ce{CO}: && -\left(\dot{M}_{\ce{CO}}\right)_{1} = -0.383 \dot{M}_{1} = \mathscr{R}_{\ce{CO}} \label{16b}\tag{16}\end{align}

\begin{align} \ce{CO2}: && -\left(\dot{M}_{\ce{CO2}}\right)_{1} + \left(\dot{M}_{\ce{CO2}}\right)_{3} = \mathscr{R}_{\ce{CO2}} \label{17b}\tag{17}\end{align}

\begin{align} \ce{N2}: && -\left(\dot{M}_{\ce{N2}}\right)_{1}-(\dot{M})_{2} + \left(\dot{M}_{\ce{N2}}\right)_{3} = \mathscr{R}_{\ce{N2}} \label{18b}\tag{18}\end{align}

\begin{align} \ce{O2}: && -\left(\dot{M}_{\ce{O2}}\right)_{1}-(\dot{M}_{\ce{O2}})_{2} + \left(\dot{M}_{\ce{O2}}\right)_{3} = \mathscr{R}_{\ce{O2}} \label{19b}\tag{19}\end{align}

\begin{align} \ce{H2O}: && (\dot{M}_{\ce{H2O}} )_{3} = \mathscr{R}_{ \ce{H2O}} \label{20b}\tag{20} \end{align}

At this point we need to simplify the balance equations for carbon dioxide, nitrogen and oxygen. Beginning with carbon dioxide, we obtain

\begin{align} \ce{CO2} : && (\dot{M}_{\ce{CO2}} )_{3} = \mathscr{R}_{ \ce{CO2} } + 0.055\dot{M}_{1} \label{21ab}\tag{21} \end{align}

Moving on to the nitrogen balance given by Equation \ref{18b}, we make use of the conditions

\begin{align} \ce{N2} : && (\dot{M}_{ \ce{N2} } )_{1} = 0.029 \dot{M}_{1} , && (\dot{M}_{ \ce{N2} } )_{2} = 0.79 \dot{M}_{2} , && \mathscr{R}_{ \ce{N2} } = 0 \label{21bb}\tag{22} \end{align}

to express the molar flow rate in the flue gas as

\begin{align} \ce{N2} : && (\dot{M}_{ \ce{N2} } )_{3} = (\dot{M}_{ \ce{N2} } )_{1} + (\dot{M}_{ \ce{N2} } )_{2} = 0.029 \dot{M}_{1} + 0.79 \dot{M}_{2} \label{22b}\tag{23} \end{align}

Finally, the use of Equation \ref{7b} leads to the following result for the molar flow rate of nitrogen:

\begin{align} \ce{N2} :&& (\dot{M}_{ \ce{N2} } )_{3} = 1.94 \dot{M}_{1} \label{23b}\tag{24} \end{align}

Turning our attention to the oxygen balance, we note that the mole fractions in Streams #1 and #2 are specified, thus we can represent Equation \ref{19b} in the form

\begin{align} \ce{O2} : && (\dot{M}_{\ce{O2}} )_{3} = \mathscr{R}_{ \ce{O2} } + 0.001 \dot{M}_{1} + 0.21 \dot{M}_{2} \label{24b}\tag{25} \end{align}

Use of Equation \ref{7b} provides the simplified version of the oxygen balance given by

\begin{align} \ce{O2} : && (\dot{M}_{\ce{O2}} )_{3} = \mathscr{R}_{ \ce{O2} } + 0.509 \dot{M}_{1} \label{25b}\tag{26} \end{align}

In order to determine the molar flow rates of oxygen, carbon dioxide and water in the flue gas, we need to determine the global rates of production of $$\ce{O2}$$, $$\ce{CO2}$$ and $$\ce{H2O}$$. From Axiom II, in the form given by Eqs. \ref{11b}, and with the use of Eqs. \ref{14b}, \ref{15b} and \ref{16b}, we can express the global rates of production of the three pivot species as

\begin{align} - 0.004\dot{M}_{1} = & \mathscr{R}_{ \ce{CO2} } + \mathscr{R}_{ \ce{H2O}} + 2\mathscr{R}_{ \ce{O2} } \nonumber\\ - 0.528\dot{M}_{1} = & -2 \mathscr{R}_{ \ce{CO2} } - 3 \mathscr{R}_{ \ce{H2O}} - 4 \mathscr{R}_{ \ce{O2} } \label{26b}\tag{27} \\ - 0.383\dot{M}_{1} = & -2 \mathscr{R}_{ \ce{CO2} } - \mathscr{R}_{ \ce{H2O}} - 2 \mathscr{R}_{ \ce{O2} } \nonumber \end{align}

The solution of these three equations is given by

$\mathscr{R}_{ \ce{CO2} } = 0.387\dot{M}_{1} , \quad \mathscr{R}_{ \ce{H2O}} = 0.536\dot{M}_{1} , \quad \mathscr{R}_{ \ce{O2} } = -0.463 \dot{M}_{1} \label{27b}\tag{28}$

and these results can be used with Eqs. \ref{20b}, \ref{21ab} and \ref{25b} to provide the species molar flow rates for the flue gas, Stream #3.

\begin{align} (\dot{M}_{\ce{N2} } )_{3} =& 1.94 \dot{M}_{1} , && (\dot{M}_{\ce{O2}} )_{3} =& 0.046\dot{M}_{1} \nonumber \\ (\dot{M}_{\ce{CO2} } )_{3} =& 0.442 \dot{M}_{1} , && (\dot{M}_{ \ce{H2O}} )_{3} =& 0.536\dot{M}_{1} \label{28b}\tag{29} \end{align}

One should keep in mind that the solution for the species molar flow rates in the flue gas should be based on three steps. Application of Axiom I and Axiom II represent two of those steps while the third step, a degree-of-freedom analysis, was omitted.