ON DERIVING THE RELATIONSHIP BETWEEN FUEL COMPOSITION AND THE COMPOSITION OF COMBUSTION PRODUCTS

V. N. SMIRENSKII

The equation relating the composition of a fuel to the composition of the final combustion products is of essential importance for the analysis of combustion processes. In all the literature available on this question, this mathematical relationship is obtained as if indirectly, in the formulation of the problem of determining a fuel characteristic or of calculating carbon monoxide in the combustion products. The derivation itself is carried out as follows. As the initial equation, an expression is taken for the sum of dry gases in percentages or in volumetric units. Then, by means of a large number of all sorts of substitutions and transformations, one arrives at one or another final expression. Along the way, the relationship between the composition of the fuel and the composition of the combustion products is obtained. Those studying this question get the impression that there is no physical meaning in the given equation, and that it is obtained by applying a whole series of artificial transformations. In the course of the entire long and complicated derivation there is no logical clarity; the derivation itself becomes obscure and difficult to present and assimilate.

It seems more expedient to pose from the very beginning the problem of finding the connection between the initial composition of the fuel and the composition of the final combustion products. Expressions for the fuel characteristic or for determining carbon monoxide are obtained either in the course of the derivation or follow from the fundamental equation. From this fundamental expression a whole series of other essential relationships may also be obtained.

In the present work a more expedient method is given for deriving the expression of interest to us for solid and liquid fuels. In the first section a detailed derivation of the equation is given for the case of chemical incompleteness of combustion in the form of carbon monoxide. In the second section the results of deriving the complete equation are presented, with allowance for other possible elements of chemical incompleteness of fuel combustion. The equation in its complete form may be of essential importance for the analysis of gas-producer processes.

I. INCOMPLETE COMBUSTION OF FUEL WITH THE FORMATION OF CARBON MONOXIDE

It is known that the combustion of individual combustible elements making up a fuel is accompanied by changes in volumes. These changes in volumes are different for different combustible elements and, moreover, differ for different combustion reactions (complete or incomplete combustion).

During the combustion of 1 kg of fuel of a given composition, a change in volume will likewise occur, and the magnitude of this change will depend on which elements were included in the fuel, in what quantity, and how the combustion reactions proceeded. The volume of gases obtained as a result of fuel combustion will have a quite definite composition. Under such conditions, there is obviously the possibility of finding a relationship between the composition of the fuel

and the composition of the final products of combustion. This relationship can be established through the change in volumes.

In the case considered here, combustion of a solid fuel will involve changes in volume: 1) from the complete combustion of carbon to \(CO_2\) \((\Delta v_1)\), 2) from the incomplete combustion of carbon to \(CO\) \((\Delta v_2)\), from the combustion of sulfur to \(SO_2\) \((\Delta v_3)\), from the combustion of “free” hydrogen to \(H_2O\) \((\Delta v_4)\), and, as is conventionally assumed, from gasification of the nitrogen of the fuel to \(N_2\) \((\Delta v_5)\). If we take for consideration the volume of dry gases obtained after the combustion of \(1\ \mathrm{kg}\) of fuel—\(V_{c2}\ \mathrm{Nm^3/kg}\), and the volume of supplied air—\(V_{\mathrm{в}}\ \mathrm{Nm^3/kg}\), then, in order to obtain the relationship of interest to us, we shall have the following initial expression:

\[ V_{c2}=V_{\mathrm{в}}+\sum_{1}^{5}\Delta v\ \mathrm{Nm^3/kg} \tag{1} \]

It is necessary to find all changes in volume during the combustion of the individual elements of the fuel \(\sum_{1}^{5}\Delta v\), to express \(V_{c2}\), \(V_{\mathrm{в}}\), and \(\sum_{1}^{5}\Delta v\) through the composition of the fuel and the composition of the dry gases, and to substitute their values into equation (1). After the corresponding transformations we shall obtain an expression relating the composition of the fuel to the composition of the combustion products.

Let us adopt the following designations and quantities: \(K^p=C^p+0.368\,S_\ell^p\), \(C^p\), \(H^p\), \(S_\ell^p\), \(O^p\), \(N^p\%\)—the percentage values of the elements of the burned fuel; \(RO_2=CO_2+SO_2\), \(CO\), \(O_2\), \(N_2\%\)—the percentage composition of the dry gases; \(V_{CO}\), \(V_{CO_2}\), \(V_{SO_2}\ \mathrm{Nm^3/kg}\)—the volumes of the corresponding gases; \(V_{O_2}^{CO_2}\), \(V_{O_2}^{CO}\), \(V_{O_2}^{SO_2}\ \mathrm{Nm^3/kg}\)—the volumes of oxygen that went to form the corresponding gas.

Molecular volumes: for \(O_2\)—22.39, for \(CO\)—22.4, for \(CO_2\)—22.26, for \(SO_2\)—21.89 \(\mathrm{Nm^3/mol}\). Molecular weight for \(H_2\)—2.016 \(\mathrm{kg/mol}\). Specific weights: for \(SO_2\)—2.926, for \(N_2\)—1.251 \(\mathrm{kg/Nm^3}\).

Let us find all changes in volume due to the individual elements entering into the composition of \(1\ \mathrm{kg}\) of the fuel being burned.

1) \(C+O_2=CO_2\).

In view of the inequality of the molecular volumes of oxygen and carbon dioxide in this reaction, we obtain the change in gas volume by the amount:

\[ \Delta v_1=V_{CO_2}-V_{O_2}^{CO_2} =\frac{CO_2}{100}V_{c2} -\frac{22.39}{22.26}\frac{CO_2}{100}V_{c2} = \]

\[ =-0.006\,\frac{CO_2}{100}V_{c2}\ \mathrm{Nm^3/kg}. \]

The minus sign indicates a decrease in volume.

2) \(C+0.5\,O_2=CO\)

In view of the insignificant difference in molecular volumes for \(CO\) and \(O_2\), one may write that

\[ \Delta v_2=V_{CO}-V_{O_2}^{CO} =(1-0.5)\frac{CO}{100}V_{c2} =+0.5\,\frac{CO}{100}V_{c2}\ \frac{\mathrm{Nm^3}}{\mathrm{kg}}. \]

3) \(S+O_2=SO_2\).

In this case we obtain a decrease in volume by the amount:

\[ \Delta v_3 = V_{SO_2} - V_{O_2}^{SO_2} = \left(1 - \frac{23.39}{21.89}\right) \frac{SO_2}{100} V_{c2} = -0.023 \frac{SO_2}{100} V_{c2}\ \frac{\mathrm{nm}^3}{\mathrm{kg}} . \]

To make it possible to combine \(\Delta v_1\) and \(\Delta v_3\), we split \(\Delta v_3\) into two terms:

\[ \Delta v_3 = -0.006 \frac{SO_2}{100} V_{c2} - 0.017 \frac{SO_2}{100} V_{c2}. \]

We combine the first term with \(\Delta v_1\) and obtain
\[ -0.006 \frac{RO_2}{100} V_{c2}\ \frac{\mathrm{nm}^3}{\mathrm{kg}} . \]
In the second term we eliminate \(SO_2\), for which we proceed as follows. Since \(1\ \mathrm{nm}^3\) of \(SO_2\) contains \(1.47\ \mathrm{kg}/\mathrm{nm}^3\) of sulfur, we may write:

\[ V_{SO_2} = \frac{S^p_\ell}{100 \cdot 1.47}\ \mathrm{nm}^3/\mathrm{kg}. \]

\[ SO_2 = \frac{V_{SO_2}}{V_{c2}}\,100 = \frac{S^p_\ell}{1.47 \cdot V_{c2}}\ \%. \]

\[ -0.017 \frac{SO_2}{100} V_{c2} = -0.0115 \frac{S^p_\ell}{100}. \]

Thus:

\[ \Delta v_1 + \Delta v_3 = -0.006 \frac{RO_2}{100} V_{c2} - 0.0115 \frac{S^p_\ell}{100}\ \mathrm{nm}^3/\mathrm{kg}. \]

4) \(H_2 + 0.5 O_2 = H_2O.\)

With respect to the volume of dry gases, here there will be a decrease in volume by the amount of oxygen used for the combustion of “free” hydrogen,
\[ H^p - 0.126 O^p. \]

\[ \Delta v_4 = -0.5 \cdot 22.39 \frac{H^p - 0.126 O^p}{100 \cdot 2.016} = -5.55 \frac{H^p - 0.126 O^p}{100}\ \mathrm{nm}^3/\mathrm{kg}. \]

5) Assuming that the nitrogen of the fuel passes into the flue gases, we obtain an increase in volume by the amount:

\[ \Delta v_5 = + \frac{N^p}{100 \cdot 1.251} = +0.8 \frac{N^p}{100}\ \mathrm{nm}^3/\mathrm{kg}. \]

After substituting all values of \(\Delta v\),

\[ V_v = \frac{\frac{N_2}{100} V_{c2} - 0.8 \frac{N^p}{100}}{79}\,100 = \]

\[ = V_{c2}\frac{100 - (RO_2 + O_2 + CO)}{79} - \frac{0.8 N^p}{79}\ \mathrm{nm}^3/\mathrm{kg} \]

and

\[ V_{c2} = \frac{K^p}{0.539\,(RO_2 + CO)}\ \mathrm{nm}^3/\mathrm{kg} \]

Substituting into equation (1) and making the corresponding transformations, we obtain:

\[ 21-1.005\,RO_2-O_2-0.605\,CO= \]

\[ =\frac{2.37\left(H^p-0.126\,O^p\right)+0.09\,N^p+0.005\,S^p_l}{K^p}\,(RO_2+CO). \]

Taking

\[ \beta=\frac{2.37\left(H^p-0.126\,O^p\right)+0.09\,N^p+0.005\,S^p_l}{K^p}+0.005, \tag{2} \]

we obtain the desired equation in final form:

\[ 21-\beta RO_2-(RO_2+O_2)-(0.6+\beta)\,CO=0 \tag{3} \]

The quantity \(\beta\), depending exclusively on the composition of the combustible mass of the fuel, is called the “fuel characteristic.” When the nitrogen and sulfur content in solid and liquid fuels is small, formula (2) is often simplified and given in the following form:

\[ \beta=2.37\,\frac{H^p-0.126\,O^p}{K^p}+0.005. \]

Usually no physical meaning is assigned to the fuel characteristic \(\beta\); however, the following may be noted here. It is known that with an increase in the chemical age of a fossil fuel, the percentage content of the principal component of its combustible mass—carbon—increases considerably, at the same time as hydrogen and oxygen decrease. Consequently, with an increase in the chemical age of a fossil fuel, the value of \(\beta\) must continuously decrease, since the denominator of the expression will increase while the numerator remains approximately constant. Thus, it should be recognized that the quantity \(\beta\) characterizes the chemical maturity of the fossil fuel of a given deposit.

Note: If the inequality of the molecular volumes of gases is neglected, the derivation of expression (3) is considerably simplified.

II. INCOMPLETE COMBUSTION OF FUEL WITH FORMATION OF CO, H₂, CₘHₙ AND H₂S

To simplify the derivation, it may be assumed that hydrogen and all hydrocarbon compounds in gases are particular cases of the general hydrocarbon \(C_mH_n\).

Using the same method of derivation, it is necessary to take into account that part of the hydrogen of the fuel in this case will go to the formation of \(C_mH_n\) and \(H_2S\), and part of the sulfur will go to the formation of \(H_2S\). Therefore we shall have:

\[ \Delta v_3=-0.006\,\frac{SO_2}{100}\,V_{cg} -0.012\,\frac{S^p_l}{100} +0.017\,\frac{H_2S}{100}\,V_{cg}\ \text{nm}^3/\text{kg}, \]

\[ \Delta v_4=-5.55\,\frac{H^p-0.126O^p}{100} +0.5\,\frac{H_2S+0.5n\,C_mH_n}{100}\,V_{cg}\ \text{nm}^3/\text{kg}. \]

A new term will be added:

\[ \Delta v_6=+\frac{H_2S+C_mH_n}{100}\,V_{cg}\ \text{nm}^3/\text{kg}. \]

\[ \Delta v_1,\ \Delta v_2\ \text{and}\ \Delta v_5 \]

will remain unchanged.

Expressions for the volume of air and gases:

\[ V_{\mathrm{a}}=V_{\mathrm{dg}}\frac{100-(RO_2+O_2+CO+C_mH_n+H_2S)}{79} -\frac{0.8N^p}{79}\ \mathrm{nm^3/kg}, \]

\[ V_{\mathrm{dg}}= \frac{K^p}{0.539(RO_2+CO+mC_mH_n+0.989H_2S)} \ \mathrm{nm^3/kg}. \tag{4} \]

The required expression will take the form:

\[ 21-\beta RO_2-(RO_2+O_2)-(0.6+\beta)CO +\left[0.198n-m(\beta-0.005)-0.21\right]C_mH_n+ \]

\[ +(0.206-0.989\beta)H_2S=0. \tag{5} \]

The fuel characteristic \(\beta\) has the same meaning as in equation (2).

The corresponding expressions for the case of combustion of gaseous fuels are derived in an analogous manner.

In the general case of chemical incomplete combustion of a gaseous fuel, the basic equation of the dependence of the final combustion products on the initial fuel is identical with equation (5).

The fuel characteristic in this case has the form:

\[ \beta= \frac{21-(RO_2^{m}+O_2^{m})-0.6CO^{m}} {RO_2^{m}+CO^{m}+mC_mH_n^{m}+0.989H_2S^{m}} + \]

\[ + \frac{(0.198n+0.005m-0.21)C_mH_n^{m}+0.206H_2S^{m}} {RO_2^{m}+CO^{m}+mC_mH_n^{m}+0.989H_2S^{m}}, \tag{6} \]

where \(RO_2^{m}=CO_2^{m}+SO_2^{m}\), \(CO^{m}\), etc. are the percentage composition of the gaseous fuel. As before, \(C_mH_n^{m}\) denotes all hydrocarbons of the fuel and hydrogen.