Mathematical Model of Gasification of Solid Fuel

Abstract

This paper presents an innovative mathematical model of solid fuel gasification, which is not described in the available literature. The calculation of the components of the heterogeneous phase (including both solid and gaseous phases) as well as the calculation of the homogeneous phase (only gaseous components) is based on the balance of the total amounts of carbon, oxygen, hydrogen, and nitrogen entering the reactor space. Additionally, this paper introduces a new method for calculating the composition of the gaseous phase, based on reducing the heterogeneous mixture (composed of solid and gaseous phases) to a homogeneous gaseous phase. This approach to calculating the gaseous phase composition in the solid fuel gasification process has not been found by the authors in the cited literature. This paper also presents a model for calculating the heterogeneous and gaseous phases using the number of moles that participate in the assumed chemical reactions of the solid fuel gasification process. This approach to calculating the composition of the heterogeneous and gaseous phases of the solid fuel gasification process is also not represented in the cited literature. For comparison with the literature data, municipal solid waste (MSW) and cashew nut shell (Cashew Shell Char (CNSC)) were used as fuels in the calculation of gasification composition. The results of the calculation of the gaseous phase composition using the model presented in the paper show good agreement with the data from the literature. The calculation of the composition of the heterogeneous mixture during the steam gasification of MSW (α = 0.4) shows the presence of a solid phase (carbon) up to approximately 735 °C. At that temperature, the synthetic gas contains only gaseous components: CO = 33.10%, H₂ = 52.70%, CH₄ = 2.54%, CO₂ = 4.97, H₂O = 5.93% and N₂ = 0.76%. Increasing the temperature above 735 °C eliminates the solid phase from the equilibrium mixture. The literature data on solid fuel gasification generally do not consider the proportion of the solid phase (carbon) in the equilibrium mixture. To satisfy the material balance at the input and output of the gasification reactor, it is necessary to determine the proportion of the solid phase (carbon) in the equilibrium mixture. Since the proportion of the solid phase (carbon) in the heterogeneous equilibrium mixture can only be determined through measurement, the development and application of a mathematical model in engineering practice is of great importance, so this developed model can be considered a useful tool for simulating the influence of process parameters on gas characteristics.

1. Introduction

High energy consumption is primarily met by using fossil fuels (coal, oil, and natural gas), and the increasing demands for environmental conservation have led to the development of alternative fuel utilization technologies such as pyrolysis and gasification. The processes of fuel pyrolysis and gasification (of coal, biomass, municipal solid waste, and especially agricultural biomass) enable the production of high-quality gaseous fuel. This gaseous fuel, known as synthetic gas, contains valuable compounds such as $C{O}_2$, $H_2$, $CO$, $H_{2}O$, $C{H}_4$, and $N_2$. Moreover, gasification aligns with international policies on the emission of harmful gasses into the environment and supports the production of synthetic gas (Monteiro et al., 2018) [1]. The devices in which fuel gasification is carried out are called gasifiers, which have been developed over decades (Couto et al., 2015 [2] Motta et al., 2018 [3]). However, the choice of gasifiers depends on the properties of the raw material and the desired quality of the synthetic gas. In industrial practice, two main types of gasifiers are predominantly used: fixed-bed gasifiers and fluidized-bed gasifiers. Fluidized-bed gasifiers provide better mixing than fixed-bed gasifiers. A fluidized-bed gasifier can be used for various types of biomasses with high moisture content. The advantage of this type of gasifier over a fixed-bed gasifier is the uniform temperature distribution in the gasification zone, achieved during the fluidization of the pulverized biomass and air. However, the temperature must be maintained below the ash softening point to prevent its agglomeration with the fluidized biomass particles, which would disrupt the fluidization conditions (Henrich et al., 2004 [4]). Furthermore, fluidized-bed gasifiers offer simpler operation, maintenance, high efficiency, high heat, and high reaction rates (Basu, 2018 [5]; Motta et al., 2018 [3]; Safarian et al., 2019 [6]).

Gasification process models can be divided into the following: thermodynamic equilibrium models, kinetic models, and neural network models (Barauah et al., 2014 [7]). Some models use the Aspen Plus process simulator (He et al., 2012 [8]), which combines the kinetic model and the thermodynamic equilibrium model. Unlike the kinetic model, which predicts the fraction and composition of products in different zones within the reactor, the equilibrium model predicts the maximum possible yield of a specific product. The comparison of different models has shown that the thermodynamic equilibrium model is the simplest and can be used to analyze the influence of process parameters and different types of biomasses on the gasification process. Cornelius et al. [9] in their work consider a one-dimensional model of biomass gasification in a fluidized bed. The proposed model explains the influence of the hydrodynamic behavior of the fluidized bed.

The results of the model were confirmed through the experimental testing of biomass gasification. Oil shale gasification was investigated by Braun et al. [10]. N₂ and water vapor were used as gasification agents. The presented model is in agreement with the experimental data. The use of water vapor as a gasification agent during coal gasification in a fluidized-bed reactor was also investigated by Matsui et al. [11]. The modeling of biomass gasification by Babu et al. [12], coal by Macak et al. [13], and agricultural biomass by Raheem et al. [14] and Marta et al. [15] considers the influence of physical and chemical phenomena that make up pyrolysis, combustion, reduction, and drying in a mathematical form. Wheat straw gasification was investigated by Wang et al. [16]. The aim of the research was to determine the gasification process parameters for H₂ yield and tar reduction. The gasification temperature was maintained in the temperature range of 700 °C to 900 °C. A brief overview of gasification technologies used in practice is presented in a paper by Ivan et al. [17]. The paper also shows the optimal symmetry of the expanded medium and the solid fuel supplied for gasification, and the output parameters of the produced generator gas are shown. Thermochemical conversion technologies (gasification, pyrolysis, and combustion) have become known as practical technologies for municipal solid waste management (MSWM). In his dissertation, Moshi [18] presents a fixed-bed MSWM hybrid gasification (HFBG) model. The model was developed using the Aspen Plus model to improve the performance of a fixed bed gasifier. The model shows an increase in H₂ and CO of about 37.05% mole fraction in the produced gas.

The use of models based on thermodynamic equilibrium is widespread among many researchers to analyze the gasification process and is based on the following assumptions:

• All carbon content in the fuel is converted into gaseous forms of CO₄, CO₂, and CH₄, and the residence time is sufficiently long to achieve thermodynamic equilibrium of chemical reactions.
• It is assumed that the ash in the fuel (raw material) is inert in all phases of the gasification process.
• It is assumed that all product gasses behave as ideal gasses.
• Sulfur and chlorine content in the fuel are neglected.
• The chemical composition of the fuel is taken in the form of $C{H}_x{O}_y{N}_z$ and the gasification reaction using air as the agent can be written as follows:

$$\underset{feedstock}{\underbrace{C{H}_x{O}_y{N}_z+w{H}_{2}O}}+m(O_2+3.76N_2)=n_{H_2}{H}_2+n_{CO}CO+n_{C{O}_2}C{O}_2+n_{H_{2}O}{H}_{2}O+n_{C{H}_4}C{H}_4+(z/2+3.76m)N_2,$$

where

$n_{H_2},n_{CO},n_{C{O}_2},n_{H_{2}O},n_{C{H}_4},n_{N_2}$—numbers of moles of H₂, CO, CO₂, H₂O, CH₄ and N₂ in synthetic gas;

$m$—number of moles in air;

$w$—mass fraction of moisture in the fuel;

$x$—mole of hydrogen per mol of carbon;

$y$—mole of oxygen per mol of carbon;

$z$—mole of nitrogen per mol of carbon.

If steam is used as the agent in fuel gasification, then the gasification reaction can be written in the form

$$\underset{feedstock}{\underbrace{C{H}_x{O}_y{N}_z+w{H}_{2}O}}+\underset{steam}{\underbrace{m{H}_{2}O}}=n_{H_2}{H}_2+n_{CO}CO+n_{C{O}_2}C{O}_2+n_{H_{2}O}{H}_{2}O+n_{C{H}_4}C{H}_4$$

From the previous two equations, it can be seen that the composition of the gaseous phase (synthetic gas) is determined using the chemical formula of the fuel in the form and material balance of the components C, H, O, and N, and the balance constants of the chemical reactions that take place during the solid fuel gasification process.

This paper presents a thermodynamic model of solid fuel gasification not found in the literature, and numerical examples are compared with data from the existing literature. The composition of synthetic gas is determined using two methods:

• The calculation of gas composition based on the ratios of total quantities of carbon, oxygen, hydrogen, and nitrogen entering the reactor space.
• The calculation of gas composition based on the number of moles involved in assumed chemical reactions.

An important difference can be observed when defining the material balance used by many researchers in the reviewed literature and the material balance presented in this paper (balance material determined on the basis of 1 and 2). In addition, mathematical models 1 and 2 enable the determination of the solid phase, i.e., the presence of solid carbon, which is not present in the cited literature. Determining the solid phase during the solid fuel gasification process is particularly important due to satisfying the material balance at the inlet and outlet of the gasifier (reactor space).

The goals of this mathematical model presented in this paper are to predict the yield of synthetic gas and solid residue produced during gasification, based on the characteristics of solid fuel (technical and elemental analysis), and to estimate the influence of humidity, the gasification agent (air and water vapor), and temperature on the yield of gasification products.

2. Materials and Methods

The mathematical model of solid fuel gasification in this paper is based on the assumption that the following chemical reactions occur in the reactor during the gasification process (Na Deng, 2016) [19] (Uisung Lee, 2014) [20]:

Methane reaction:

$$C+2H_2=C{H}_4,$$

(1)

Boudouard reaction:

$$C+C{O}_2=2CO,$$

(2)

Water-gas reaction:

$$C+H_{2}O=CO+H_2,$$

(3)

By combining Equations (2) and (3), the following reaction is obtained:

Water-gas shift reaction:

$$CO+H_{2}O=C{O}_2+H_2,$$

(4)

The reactor space diagram depicting chemical reactions (1)–(4) is shown in Figure 1.

The literary data (Gumz, [21]) indicate that assumed chemical reactions (1)–(3) are used in the calculation of fuel gasification in an air stream, and Equation (4) is typically used when introducing steam during the gasification process. The equilibrium constants of chemical reactions (1)–(3) as functions of temperature can be determined using the following expressions (Gumz, [21]):

$$\log K_{p1}=-18.06361+{\displaystyle \frac{4662.80}{T}}-2.09594\cdot {10}^{-3}T+0.38620\cdot \hspace{0.17em}{10}^{-6}\hspace{0.17em}{T}^2+3.034338\hspace{0.17em}\cdot \log T,$$

(5)

$$\log K_{p2}=8.26730-{\displaystyle \frac{8820.690}{T}}-{1.20871410}^{-3}\hspace{0.17em}T+0.153734\cdot {10}^{-6}{T}^2+2.295483\hspace{0.17em}\cdot \log T,$$

(6)

$$\log K_{p_3}=-28.45778-{\displaystyle \frac{4825.986}{T}}-5.671122\cdot {10}^{-3}\hspace{0.17em}T+0.8255488\cdot {10}^{-6}\hspace{0.17em}{T}^2+14.515760\hspace{0.33em}\cdot \log T,$$

(7)

where

$K_{p_1}={\displaystyle \frac{p_{C{H}_4}}{{p_{H_2}}^2}}$—the equilibrium constant of chemical reaction (1), in Pa⁻¹;

$K_{p_2}={\displaystyle \frac{{p_{CO}}^2}{p_{C{O}_2}}}$—the equilibrium constant of chemical reaction (2), in Pa;

$K_{p_3}={\displaystyle \frac{p_{CO}\cdot p_{H_2}}{p_{H_{2}O}}}$—the equilibrium constant of chemical reaction (3), in Pa;

$T$—the absolute temperature during the considered chemical reactions, in K;

$p_{C{H}_4},\hspace{0.17em}{p}_{H_2},p_{CO},p_{C{O}_2},p_{H_{2}O}$—partial pressures of methane, hydrogen, carbon monoxide, carbon dioxide, and water vapor in the equilibrium mixture, in Pa.

The equilibrium constant of reaction (4) is

$$K_{p_4}={\displaystyle \frac{p_{C{O}_2}\cdot p_{H_2}}{p_{CO}\cdot p_{H_{2}O}}}={\displaystyle \frac{K_{p3}}{K_{p2}}},$$

(8)

2.1. Calculation of Gas Composition Based on the Ratio of the Total Amounts of Carbon, Oxygen, Hydrogen, and Nitrogen Entering the Reactor Space

2.1.1. Calculation of the Composition of a Heterogeneous Mixture

As a result of chemical reactions (1)–(3), in the equilibrium mixture (heterogeneous phase), seven components are formed, whose mole fractions are unknown $(x_C,x_{H_2},x_{C{H}_4},x_{C{O}_2},x_{CO},x_{H_{2}O},x_{N_2}).$

To determine the stated mole fractions, it is necessary to set up seven material balance equations with seven unknowns:

$$x_C+x_{H_2}+x_{C{H}_4}+x_{C{O}_2}+x_{CO}+x_{H_{2}O}+x_{N_2}=1,$$

(9)

$${\displaystyle \frac{\Sigma C}{\Sigma O_2}}={\displaystyle \frac{x_C+x_{C{H}_4}+x_{C{O}_2}+x_{CO}}{x_{C{O}_2}+0.5\hspace{0.33em}{x}_{CO}+0.5\hspace{0.33em}{x}_{H_{2}O}}}=L,$$

(10)

$${\displaystyle \frac{\Sigma C}{\Sigma H_2}}={\displaystyle \frac{x_C+x_{C{H}_4}+x_{C{O}_2}+x_{CO}}{x_{H_2}+2\hspace{0.33em}{x}_{C{H}_4}+x_{H_{2}O}}}=M,$$

(11)

$${\displaystyle \frac{\Sigma C}{\Sigma N_2}}={\displaystyle \frac{x_C+x_{C{H}_4}+x_{C{O}_2}+x_{CO}}{x_{N_2}}}={\displaystyle \frac{1}{I}},$$

(12)

$$K_{p_1}={\displaystyle \frac{p_{C{H}_4}}{{p_{H_2}}^2}}={\displaystyle \frac{1}{p}}\cdot {\displaystyle \frac{x_{C{H}_4}\cdot (1-x_C)}{{x_{H_2}}^2}},$$

(13)

$$K_{p_2}={\displaystyle \frac{{p_{CO}}^2}{p_{C{O}_2}}}=p\cdot {\displaystyle \frac{{x_{CO}}^2}{x_{C{O}_2}\cdot (1-x_C)}},$$

(14)

$$K_{p_3}={\displaystyle \frac{p_{CO}\cdot p_{H_2}}{p_{H_{2}O}}}=p\cdot {\displaystyle \frac{x_{CO}\cdot x_{H_2}}{x_{H_{2}O}\cdot (1-x_C)}},$$

(15)

where

$\sum C,\sum O_2,\sum H_2,\sum N_2$—the total number of kilomoles of carbon, oxygen, hydrogen, and nitrogen in the reactor space;

$L,\hspace{0.17em}M,\hspace{0.17em}I$—auxiliary parameters;

$p$—total pressure in the reactor space, in Pa.

The system of seven equations, Equations (9)–(15), with seven unknown quantities $(x_C,x_{H_2},x_{C{H}_4},x_{C{O}_2},x_{CO},x_{H_{2}O},x_{N_2})$ can be transformed into a system of three nonlinear equations with three unknowns:

$$F_1(x_{CO},x_{H_2},x_C)=0,F_2(x_{CO},x_{H_2},x_C)=0,F_3(x_{CO},x_{H_2},x_C)=0$$

$$a_1{x_{CO}}^2+b_1{x_{H_2}}^2+c_1{x_C}^2+d_1{x}_{CO}{x}_{H_2}+e_1{x}_{CO}{x}_C+f_1{x}_{H_2}{x}_C+g_1{x}_{CO}+h_1{x}_{H_2}+i_1{x}_C+k_1=0,$$

(16)

$$a_2{x_{CO}}^2+b_2{x_{H_2}}^2+c_2{x_C}^2+d_2{x}_{CO}{x}_{H_2}+e_2{x}_{CO}{x}_C+g_2{x}_{CO}+i_2{x}_C=0,$$

(17)

$$a_3{x_{CO}}^2+b_3{x_{H_2}}^2+c_3{x_C}^2+d_3{x}_{CO}{x}_{H_2}+e_3{x}_{CO}{x}_C+f_3{x}_{H_2}{x}_C+g_3{x}_{CO}+h_3{x}_{H_2}+i_3{x}_C=0.$$

(18)

The coefficients of the equations are as follows:

$$\begin{gathered} \begin{array}{l}{a}_1=p\cdot K_{p_3}\left(1+I\right),\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{b}_1=p\cdot K_{p_1}\cdot K_{p_2}\cdot K_{p_3}\cdot \left(1+I\right),\hspace{1em}{c}_1=-K_{p_2}\cdot K{p}_3\cdot \left(1+I\right),\hspace{1em}{d}_1=p\cdot K_{p_2},\\ e_1=-K_{p_2}\cdot K_{p_3}\cdot \left(1+I\right),\hspace{1em}{f}_1=-K_{p_2}\cdot K_{p_3},\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}{g}_1=K_{p_2}\cdot K_{p_3}\left(1+I\right),\hspace{1em}\hspace{1em}{h}_1=K_{p_2}\cdot K_{p_3},\\ \hspace{0.17em}{i}_1=K_{p_2}\cdot K_{p_3}\cdot \left(2+I\right),\hspace{1em}\hspace{1em}{k}_1=-K_{p_2}\cdot K_{p_3},\end{array} \\ \begin{array}{l}{a}_2=2\cdot p\cdot K_{p_3}\cdot \left(1-L\right),\hspace{1em}{b}_2=2\cdot p\cdot K_{p_1}\cdot K_{p_2}\cdot K_{p_3},\hspace{1em}{c}_2=-2\cdot K_{p_2}\cdot K_{p_3},\hspace{1em}{d}_2=-L\cdot p\cdot K_{p_2},\\ e_2=K_{p_2}\cdot K_{p_3}\cdot \left(L-2\right),\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{g}_2=K_{p_2}\cdot K_{p_3}\cdot \left(2-L\right),\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{i}_2=2\cdot K_{p_2}\cdot K_{p_3},\end{array} \\ \begin{array}{l}{a}_3=p\cdot K_{p_3},\hspace{1em}\hspace{1em}\hspace{0.17em}{b}_3=p\cdot K_{p_1}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\cdot K_{p_2}\cdot K_{p_3}\cdot \left(1-2M\right),\hspace{1em}\hspace{0.17em}\hspace{0.17em}{c}_3=-K_{p_2}\cdot K_{p_3},\hspace{1em}{d}_3=-M\cdot p\cdot K_{p_2},\\ e_3=-K_{p_2}\cdot K_{p_3},\hspace{1em}{f}_3=M\cdot K_{p_2}\cdot K_{p_3},\hspace{1em}{g}_3=K_{p_2}\cdot K_{p3},\hspace{1em}{h}_3=-M\cdot K_{p_2}\cdot K_{p_3},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{i}_3=K_{p_2}{K}_{p_3}.\end{array} \end{gathered}$$

The auxiliary sizes $L,\hspace{0.17em}M,\hspace{0.17em}I$ are determined using the following expressions:

Amount of carbon:

$$\Sigma C={\displaystyle \frac{C}{12}},\hspace{0.33em}\mathrm{kmolC}/\mathrm{kg}\hspace{0.17em},$$

(19)

where

$C$—mass fraction of carbon in biomass, in $\mathrm{kg}\hspace{0.17em}\mathrm{C}/\mathrm{kg}$.

Amount of oxygen:

$$\Sigma {\mathrm{O}}_2={\displaystyle \frac{1}{32}}\cdot (\mathrm{O}+{\displaystyle \frac{16}{18}}\cdot \mathrm{W}+0.23\cdot L_{actual})+{\displaystyle \frac{1}{36}}\cdot \alpha ,\hspace{0.17em}\mathrm{kmol}\hspace{0.17em}{\mathrm{O}}_2/\mathrm{kg},$$

(20)

where

O—mass fraction of oxygen in biomass, in kgO/kg;

W—mass fraction of moisture in biomass, in kgW/kg;

$\lambda$—excess air coefficient;

$L_{actual}$—the actual amount of air introduced into the gasifier, in $\mathrm{kg}L/\mathrm{kg}$;

$L_{actual}=\lambda \cdot L_{\min },\mathrm{kg}L/\mathrm{kg}$;

$L_{\min }==O_{\min }/0.23,\mathrm{kg}L/\mathrm{kg}$;

$O_{\min }=2.667\cdot C+8\cdot H+S-O,\mathrm{kg}O/\mathrm{kg}$;

$\alpha ={\displaystyle \frac{m_{H2O}}{m_{MSW}}}$—the amount of water vapor injected reduced per kilogram of fuel.

Amount of hydrogen:

$$\Sigma {\mathrm{H}}_2={\displaystyle \frac{1}{2}}\cdot (\mathrm{H}+{\displaystyle \frac{2}{18}}\cdot \mathrm{W})+{\displaystyle \frac{1}{18}}\cdot \alpha ,{\mathrm{kmolH}}_2/\mathrm{kg},$$

(21)

where

H—mass fraction of hydrogen in the fuel, in kg/kg.

Amount of nitrogen:

$$\Sigma {\mathrm{N}}_2={\displaystyle \frac{1}{28}}\cdot (\mathrm{N}+0.77\cdot L_{actual}),\hspace{0.33em}{\mathrm{kmolN}}_2/\mathrm{kg},$$

(22)

where

N—mass fraction of the nitrogen in the fuel, in kgN/kg.

The nonlinear system of equations (Equations (16)–(18)) can be solved by Newton’s iterative method.

From Equations (12)–(15), we express $x_{N_2}$,$x_{C{H}_4}$,$x_{C{O}_2}$, and $x_{H_{2}O}$ as follows:

$$\begin{gathered} x_{C{H}_4}={\displaystyle \frac{{x_{H_2}}^2\cdot p\cdot K_{p1}}{1-x_C}},\hspace{0.17em}{x}_{C{O}_2}={\displaystyle \frac{{x_{CO}}^2\cdot p}{Kp2\cdot (1-x_C)}},\hspace{0.17em}{x}_{H_{2}O}={\displaystyle \frac{x_{CO}\cdot x_{H_2}\cdot p}{K_{p3}\cdot (1-x_C)}}, \\ {x}_{N_2}=I\cdot (x_C+x_{CO}+{\displaystyle \frac{{x_{CO}}^2\cdot p}{K_{p2}\cdot (1-x_C)}}+{\displaystyle \frac{{x_{H_2}}^2\cdot p\cdot K_{p1}}{(1-x_C)}}). \end{gathered}$$

The algorithm for solving the system of Equations (16)–(18) is shown in Figure 2. The system of equations was solved in the programming language C++.

2.1.2. Calculation of the Composition of a Homogeneous (Gaseous) Mixture

In the gaseous phase, there are six components whose mole fractions should be determined $(y_{H_2},\hspace{0.17em}{y}_{C{H}_4},\hspace{0.17em}{y}_{C{O}_2},\hspace{0.17em}{y}_{CO},\hspace{0.17em}{y}_{H_{2}O},y_{N_2})$. The mole fractions of the homogeneous mixture will be determined using the mole fractions of the heterogeneous mixture. It follows from Equation (9) that

$$x_{H_2}+x_{C{H}_4}+x_{C{O}_2}+x_{CO}+x_{H_{2}O}+x_{N_2}=1-x_C,$$

(23)

$${\displaystyle \frac{x_{H_2}}{1-x_C}}+{\displaystyle \frac{x_{C{H}_4}}{1-x_C}}+{\displaystyle \frac{x_{C{O}_2}}{1-x_C}}+{\displaystyle \frac{x_{CO}}{1-x_C}}+{\displaystyle \frac{x_{H_{2}O}}{1-x_C}}+{\displaystyle \frac{x_{N_2}}{1-x_C}}=1,$$

(24)

$${\displaystyle \frac{x_{H_2}}{1-x_C}}\cdot p+{\displaystyle \frac{x_{C{H}_4}}{1-x_C}}\cdot p+{\displaystyle \frac{x_{C{O}_2}}{1-x_C}}\cdot p+{\displaystyle \frac{x_{CO}}{1-x_C}}\cdot p+{\displaystyle \frac{x_{H_{2}O}}{1-x_C}}\cdot p+{\displaystyle \frac{x_{N_2}}{1-x_C}}\cdot p=p,$$

(25)

$$p_{H_2}+p_{C{H}_4}+p_{C{O}_2}+p_{CO}+p_{H_{2}O}+p_{N_2}=p.$$

(26)

It follows from Equations (25) and (26) that

$$\begin{gathered} p_{H_2}={\displaystyle \frac{x_{H_2}}{1-x_C}}\cdot p,p_{C{H}_4}={\displaystyle \frac{x_{C{H}_4}}{1-x_C}}\cdot p,p_{C{O}_2}={\displaystyle \frac{x_{C{O}_2}}{1-x_C}}\cdot p, \\ {p}_{CO}={\displaystyle \frac{x_{CO}}{1-x_C}}\cdot p,p_{H_{2}O}={\displaystyle \frac{x_{H_{2}O}}{1-x_C}}\cdot p,p_{N_2}={\displaystyle \frac{x_{N_2}}{1-x_C}}\cdot p. \end{gathered}$$

(27)

Equation (27) determines the mole fractions in a homogeneous (gaseous) equilibrium mixture:

$$y_{H_2}={\displaystyle \frac{x_{H_2}}{1-x_C}},y_{C{H}_4}={\displaystyle \frac{x_{C{H}_4}}{1-x_C}},y_{C{O}_2}={\displaystyle \frac{x_{C{O}_2}}{1-x_C}},y_{CO}={\displaystyle \frac{x_{CO}}{1-x_C}},y_{H_{2}O}={\displaystyle \frac{x_{H_{2}O}}{1-x_C}},y_{N_2}={\displaystyle \frac{x_{N_2}}{1-x_C}},$$

(28)

where the equality holds as follows:

$$y_{H_2}+y_{C{H}_4}+y_{C{O}_2}+y_{CO}+y_{H_{2}O}+y_{N_2}=1.$$

(29)

The calculation of the composition of a homogeneous (gaseous) mixture can also be determined using a material balance, i.e., there are six components whose mole fractions should be determined $\left(y_{{\mathrm{H}}_2},y_{{\mathrm{CH}}_4},y_{{\mathrm{CO}}_2},y_{\mathrm{CO}},y_{{\mathrm{H}}_2\mathrm{O}},y_{{\mathrm{N}}_2}\right)$, so it is necessary to set six material balance equations:

$$y_{{\mathrm{H}}_2}+y_{{\mathrm{CH}}_4}+y_{{\mathrm{CO}}_2}+y_{\mathrm{CO}}+y_{{\mathrm{H}}_2\mathrm{O}}+y_{{\mathrm{N}}_2}=1,$$

(30)

$${\displaystyle \frac{\Sigma {\mathrm{H}}_2}{\Sigma {\mathrm{O}}_2}}={\displaystyle \frac{y_{{\mathrm{H}}_2}+2\hspace{0.33em}{y}_{{\mathrm{CH}}_4}+y_{{\mathrm{H}}_2\mathrm{O}}}{y_{{\mathrm{CO}}_2}+0.5\hspace{0.33em}{y}_{\mathrm{CO}}+0.5\hspace{0.33em}{y}_{{\mathrm{H}}_2\mathrm{O}}}}=J,$$

(31)

$${\displaystyle \frac{\Sigma {\mathrm{N}}_2}{\Sigma {\mathrm{O}}_2}}={\displaystyle \frac{y_{{\mathrm{N}}_2}}{y_{{\mathrm{CO}}_2}+0.5\hspace{0.33em}{\mathrm{y}}_{\mathrm{CO}}+0.5\hspace{0.33em}{y}_{{\mathrm{H}}_2\mathrm{O}}}}=K,$$

(32)

$$K_{p_1}={\displaystyle \frac{p_{{\mathrm{CH}}_4}}{p_{{\mathrm{H}}_2}^2}}={\displaystyle \frac{1}{p}}\cdot {\displaystyle \frac{y_{{\mathrm{CH}}_4}}{y_{{\mathrm{H}}_2}^2}},$$

(33)

$$K_{p_2}={\displaystyle \frac{p_{\mathrm{CO}}^2}{p_{{\mathrm{CO}}_2}}}=p\cdot {\displaystyle \frac{y_{\mathrm{CO}}^2}{y_{{\mathrm{CO}}_2}}},$$

(34)

$$K_{p_3}={\displaystyle \frac{P_{\mathrm{CO}}\cdot p_{{\mathrm{H}}_2}}{p_{{\mathrm{H}}_2\mathrm{O}}}}=p\cdot {\displaystyle \frac{y_{\mathrm{CO}}\cdot y_{{\mathrm{H}}_2}}{y_{{\mathrm{H}}_2\mathrm{O}}}}.$$

(35)

where J and K are auxiliary quantities.

The system of six equations, Equations (30)–(35), with six unknown quantities $\left(y_{{\mathrm{H}}_2},y_{{\mathrm{CH}}_4},y_{{\mathrm{CO}}_2},y_{\mathrm{CO}},y_{{\mathrm{H}}_2\mathrm{O}},y_{{\mathrm{N}}_2}\right)$ is transformed into a system of two nonlinear equations:

$${\mathrm{F}}_1\left(y_{\mathrm{CO}},y_{{\mathrm{H}}_2}\right)={0,\mathrm{F}}_2\left(y_{\mathrm{CO}},y_{{\mathrm{H}}_2}\right)=0:$$

$$a_1{y}_{\mathrm{CO}}^2+b_1{y}_{\mathrm{CO}}+c_1{y}_{\mathrm{CO}}{y}_{\mathrm{H}2}+d_1{y}_{{\mathrm{H}}_2}^2+e_1{y}_{{\mathrm{H}}_2}+f_1=0,$$

(36)

$$a_2{y}_{\mathrm{CO}}^2+b_2{y}_{\mathrm{CO}}+c_2{y}_{\mathrm{CO}}{y}_{\mathrm{H}2}+d_2{y}_{{\mathrm{H}}_2}^2+e_2{y}_{{\mathrm{H}}_2}=0.$$

(37)

The coefficients of the equations are as follows:

$$\begin{gathered} a_1=(1+K)\cdot {\displaystyle \frac{p}{K_{p_2}}},b_1=1+0.5K,c_1=\left(1+0.5\hspace{0.33em}K\right)\cdot {\displaystyle \frac{p}{K_{p_3}}},d_1=p\cdot K_{p_1},e_1=1, f_1=-1, \\ {a}_2=-J\cdot {\displaystyle \frac{p}{K_{p_2}}},b_2=-0.5J,c_2=\left(1-0.5\hspace{0.33em}J\right)\cdot {\displaystyle \frac{p}{K_{p_3}}},d_2=2\cdot p\cdot K_{p_1},e_2=1. \end{gathered}$$

The nonlinear system of Equations (36) and (37) can also be solved by Newton’s iterative method.

2.2. Calculation of Gas Composition Using the Number of Moles That Participate in the Assumed Chemical Reactions

During chemical reactions (1)–(3), the number of kilomoles of the components in the mixture after the establishment of chemical equilibrium is as follows:

Reaction C + 2 H₂ = CH₄:

$$\mathrm{Carbon}:{n_{\mathrm{C}}}^{(1)}=a_1-z_1,$$

(38)

$$\mathrm{Hydrogen}:{n_{{\mathrm{H}}_2}}^{(1)}=b_1-2\hspace{0.33em}{z}_1,$$

(39)

$$\mathrm{Methane}:{n_{{\mathrm{CH}}_4}}^{(1)}=z_1,$$

(40)

where

$a_1$—the number of kilo moles of carbon that enters reaction (1);

$b_1$—the number of kilo moles of hydrogen that enters reaction (1);

$z_1$—the number of kilo moles of methane in the mixture after equilibrium is established.

The total number of kilo moles in the mixture after establishing chemical equilibrium is as follows:

-

Heterogeneous mixture (solid and gaseous phase)

$$\begin{gathered} n{(s)}^{(1)}={n_{\mathrm{C}}}^{(1)}+{n_{{\mathrm{H}}_2}}^{(1)}+{n_{{\mathrm{CH}}_4}}^{(1)}, \\ n{(s)}^{(1)}=(a_1-z_1)+(b_1-2\hspace{0.33em}{z}_1)+z_1=a_1+b_1-2\hspace{0.33em}{z}_1, \end{gathered}$$

(41)

-

Homogeneous mixture (gas phase only)

$$\begin{gathered} n{(g)}^{(1)}={n_{{\mathrm{H}}_2}}^{(1)}+{n_{{\mathrm{CH}}_4}}^{(1)}, \\ n{(g)}^{(1)}=(b_1-2\hspace{0.33em}{z}_1)+z_1=b_1-z_1. \end{gathered}$$

(42)

Reaction C + CO₂ = 2 CO:

$$\mathrm{Carbon}:{n_{\mathrm{C}}}^{(2)}=a_2-{\displaystyle \frac{z_2}{2}},$$

(43)

$$\mathrm{Carbon}\mathrm{dioxide}:{n_{{\mathrm{CO}}_2}}^{(2)}=b_2-{\displaystyle \frac{z_2}{2}},$$

(44)

$$\mathrm{Carbon}\mathrm{monoxide}:{n_{\mathrm{CO}}}^{(2)}=z_2,$$

(45)

where

$a_2$—the number of kilo moles of carbon that enters reaction (2);

$b_2$—the number of kilo moles of carbon dioxide that enters reaction (2);

$z_2$—the number of kilo moles of carbon monoxide in the mixture after the establishment of chemical equilibrium.

The total number of kilo moles in the mixture after establishing chemical equilibrium is as follows:

-

Heterogeneous mixture (solid and gaseous phase)

$$\begin{gathered} n{(s)}^{(2)}={n_{\mathrm{C}}}^{(2)}+{n_{{\mathrm{CO}}_2}}^{(2)}+{n_{\mathrm{CO}}}^{(2)}, \\ n{(s)}^{(2)}=(a_2-{\displaystyle \frac{z_2}{2}})+(b_2-{\displaystyle \frac{z_2}{2}})+z_2=a_2+b_2, \end{gathered}$$

(46)

-

Homogeneous mixture (gas phase only)

$$\begin{gathered} n{(g)}^{(2)}={n_{{\mathrm{CO}}_2}}^{(2)}+{n_{\mathrm{CO}}}^{(2)}, \\ n{(g)}^{(2)}=(b_2-{\displaystyle \frac{z_2}{2}})+z_2=b_2+{\displaystyle \frac{z_2}{2}}. \end{gathered}$$

(47)

Reaction C + H₂O = CO + H₂:

$$\mathrm{Carbon}:{{\mathrm{n}}_{\mathrm{C}}}^{\left(3\right)}={\mathrm{a}}_3-z_3,$$

(48)

$$\mathrm{Aerated}\mathrm{water}:{{\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}}^{\left(3\right)}={\mathrm{b}}_3-z_3,$$

(49)

$$\mathrm{Carbon}\mathrm{monoxide}:{{\mathrm{n}}_{\mathrm{CO}}}^{\left(3\right)}={\mathrm{z}}_3,$$

(50)

$$\mathrm{Hydrogen}:{{\mathrm{n}}_{{\mathrm{H}}_2}}^{\left(3\right)}={\mathrm{z}}_3,$$

(51)

where

$a_3$—the number of kilo moles of carbon that enters reaction (3);

$b_3$—the number of kilo moles of aerated water that enters reaction (3);

$z_3$—the number of kilo moles of carbon monoxide (hydrogen) in the mixture after establishing chemical equilibrium.

The total number of kilo moles in the mixture after establishing chemical equilibrium is as follows:

-

Heterogeneous mixture (solid and gaseous phase)

$$\begin{gathered} n{(s)}^{(3)}={n_{\mathrm{C}}}^{(3)}+{n_{{\mathrm{H}}_2\mathrm{O}}}^{(3)}+{n_{\mathrm{CO}}}^{(3)}+{n_{{\mathrm{H}}_2}}^{(3)}, \\ n{(s)}^{(3)}=(a_3-z_3)+(b_3-z_3)+z_3+z_3=a_3+b_3, \end{gathered}$$

(52)

-

Homogeneous mixture (gas phase only)

$$\begin{gathered} n{(g)}^{(3)}={n_{{\mathrm{H}}_2\mathrm{O}}}^{(3)}+{n_{\mathrm{CO}}}^{(3)}+{n_{{\mathrm{H}}_2}}^{(3)}, \\ n{(g)}^{(3)}=(b_3-z_3)+z_3+z_3=b_3+z_3, \end{gathered}$$

(53)

During reactions (1)–(3), the total number of kilomoles in the mixture after establishing chemical equilibrium is as follows:

-

Heterogeneous mixture (solid and gaseous phase)

$$\begin{gathered} n(s)=n{(s)}^{(1)}+n{(s)}^{(2)}+n{(s)}^{(3)}+d, \\ n(s)=(a_1+b_1-2\hspace{0.33em}{z}_1)+(a_2+b_2)+(a_3+b_3)+d=a_0+b_0-2\hspace{0.33em}{z}_1+d, \end{gathered}$$

(54)

-

Homogeneous mixture (gas phase only)

$$\begin{gathered} n(g)=n{(g)}^{(1)}+n{(g)}^{(2)}+n{(g)}^{(3)}+d, \\ n(g)=(b_1-z_1)+(b_2+{\displaystyle \frac{z_2}{2}})+(b_3+z_3)+d=b_0-z_1+{\displaystyle \frac{z_2}{2}}+z_3+d, \end{gathered}$$

(55)

where

a₀ = a₁ + a₂ + a₃ = C/12 − O/32—the number of kilomoles of carbon that enters reactions (1), (2), and (3);

b₀ = b₁ + b₂ + b₃ = H/2 + O/32 + W/18—the auxiliary size;

d = N/28—the number of kilomoles of nitrogen in the mixture after establishing the chemical balance of reactions (1)–(3);

C, H, O, N, W—the mass fractions of carbon, hydrogen, oxygen, nitrogen, and moisture in solid fuel.

The mole fractions of the components in the mixture after the establishment of the chemical equilibrium of reactions (1)–(3) are as follows:

-

Heterogeneous mixture (solid and gaseous phase)

$$x_{\mathrm{C}}={\displaystyle \frac{{n_{\mathrm{C}}}^{(1)}+{n_{\mathrm{C}}}^{(2)}+{n_{\mathrm{C}}}^{(3)}}{n(s)}}={\displaystyle \frac{(a_1-z_1)+(a_2-\frac{z_2}{2})+(a_3-z_3)}{a_0+b_0-2\hspace{0.33em}{z}_1+d}}={\displaystyle \frac{a_0-z_1-\frac{z_2}{2}-z_3}{a_0+b_0-2\hspace{0.33em}{z}_1+d}},$$

(56)

$$x_{{\mathrm{H}}_2}={\displaystyle \frac{{n_{{\mathrm{H}}_2}}^{(1)}+{n_{{\mathrm{H}}_2}}^{(3)}}{n(s)}}={\displaystyle \frac{(b_1-2\hspace{0.33em}{z}_1)+z_3}{a_0+b_0-2\hspace{0.33em}{z}_1+d}}={\displaystyle \frac{b_1-2\hspace{0.33em}{z}_1+z_3}{a_0+b_0-2\hspace{0.33em}{z}_1+d}},$$

(57)

$$x_{{\mathrm{CH}}_4}={\displaystyle \frac{{n_{{\mathrm{CH}}_4}}^{(1)}}{n(s)}}={\displaystyle \frac{z_1}{a_0+b_0-2\hspace{0.33em}{z}_1+d}},$$

(58)

$$x_{{\mathrm{CO}}_2}={\displaystyle \frac{{n_{{\mathrm{CO}}_2}}^{(2)}}{n(s)}}={\displaystyle \frac{b_2-\frac{z_2}{2}}{a_0+b_0-2\hspace{0.33em}{z}_1+d}},$$

(59)

$$x_{\mathrm{CO}}={\displaystyle \frac{{n_{\mathrm{CO}}}^{(2)}+{n_{\mathrm{CO}}}^{(3)}}{n(s)}}={\displaystyle \frac{z_2+z_3}{a_0+b_0-2\hspace{0.33em}{z}_1+d}},$$

(60)

$$x_{{\mathrm{H}}_2\mathrm{O}}={\displaystyle \frac{{n_{{\mathrm{H}}_2\mathrm{O}}}^{(3)}}{n(s)}}={\displaystyle \frac{b_3-z_3}{a_0+b_0-2\hspace{0.33em}{z}_1+d}},$$

(61)

$$x_{{\mathrm{N}}_2}={\displaystyle \frac{d}{n(s)}}={\displaystyle \frac{d}{a_0+b_0-2\hspace{0.33em}{z}_1+d}},$$

(62)

-

Homogeneous mixture (gas phase only)

$$y_{{\mathrm{H}}_2}={\displaystyle \frac{{n_{{\mathrm{H}}_2}}^{(1)}+{n_{{\mathrm{H}}_2}}^{(3)}}{n(g)}}={\displaystyle \frac{(b_1-2\hspace{0.33em}{z}_1)+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}}={\displaystyle \frac{b_1-2\hspace{0.33em}{z}_1+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}},$$

(63)

$$y_{{\mathrm{CH}}_4}={\displaystyle \frac{{n_{{\mathrm{CH}}_4}}^{(1)}}{n(g)}}={\displaystyle \frac{z_1}{b_0-z_1+\frac{z_2}{2}+z_3+d}},$$

(64)

$$y_{{\mathrm{CO}}_2}={\displaystyle \frac{{n_{{\mathrm{CO}}_2}}^{(2)}}{n(g)}}={\displaystyle \frac{b_2-\frac{z_2}{2}}{b_0-z_1+\frac{z_2}{2}+z_3+d}},$$

(65)

$$y_{\mathrm{CO}}={\displaystyle \frac{{n_{\mathrm{CO}}}^{(2)}+{n_{\mathrm{CO}}}^{(3)}}{n(g)}}={\displaystyle \frac{z_2+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}},$$

(66)

$$y_{{\mathrm{H}}_2\mathrm{O}}={\displaystyle \frac{{n_{{\mathrm{H}}_2\mathrm{O}}}^{(3)}}{n(g)}}={\displaystyle \frac{b_3-z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}},$$

(67)

$${\mathrm{y}}_{{\mathrm{N}}_2}={\displaystyle \frac{\mathrm{d}}{\mathrm{n}\left(\mathrm{g}\right)}}={\displaystyle \frac{\mathrm{d}}{{\mathrm{b}}_0-z_1+\frac{{\mathrm{z}}_2}{2}+{\mathrm{z}}_3+\mathrm{d}}}.$$

(68)

The partial pressures of the components in a homogeneous mixture (gaseous phase) after the establishment of chemical equilibrium are as follows:

$$p_{{\mathrm{H}}_2}=y_{{\mathrm{H}}_2}\hspace{0.33em}p={\displaystyle \frac{b_1-2\hspace{0.33em}{z}_1+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}}\hspace{0.33em}p,$$

(69)

$$p_{{\mathrm{CH}}_4}=y_{{\mathrm{CH}}_4}\hspace{0.33em}p={\displaystyle \frac{z_1}{b_0-z_1+\frac{z_2}{2}+z_3+d}}\hspace{0.33em}p,$$

(70)

$$p_{{\mathrm{CO}}_2}=y_{{\mathrm{CO}}_2}\hspace{0.33em}p={\displaystyle \frac{b_2-\frac{z_2}{2}}{b_0-z_1+\frac{z_2}{2}+z_3+d}}\hspace{0.33em}p,$$

(71)

$$p_{\mathrm{CO}}=y_{\mathrm{CO}}\hspace{0.33em}p={\displaystyle \frac{z_2+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}}\hspace{0.33em}p,$$

(72)

$$p_{{\mathrm{H}}_2\mathrm{O}}=y_{{\mathrm{H}}_2\mathrm{O}}\hspace{0.33em}p={\displaystyle \frac{b_3-z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}}\hspace{0.33em}p,$$

(73)

$$p_{{\mathrm{N}}_2}=y_{{\mathrm{N}}_2}\hspace{0.33em}p={\displaystyle \frac{d}{b_0-z_1+\frac{z_2}{2}+z_3+d}}\hspace{0.33em}p,$$

(74)

where

p—total pressure in the reactor space after establishing chemical equilibrium.

The chemical equilibrium constants of reactions (1)–(3) are as follows:

$$K_{p1}={\displaystyle \frac{p_{{\mathrm{CH}}_4}}{{p_{{\mathrm{H}}_2}}^2}}={\displaystyle \frac{\frac{z_1}{b_0-z_1+\frac{z_2}{2}+z_3+d}\hspace{0.33em}p}{{(\frac{b_1-2\hspace{0.33em}{z}_1+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}\hspace{0.33em}p)}^2}}={\displaystyle \frac{1}{p}}\hspace{0.33em}{\displaystyle \frac{z_1\hspace{0.33em}(b_0-z_1+\frac{z_2}{2}+z_3+d)}{{(b_1-2\hspace{0.33em}{z}_1+z_3)}^2}},$$

(75)

$$K_{p2}={\displaystyle \frac{{p_{\mathrm{CO}}}^2}{p_{{\mathrm{CO}}_2}}}={\displaystyle \frac{{(\frac{z_2+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}\hspace{0.33em}p)}^2}{\frac{b_2-\frac{z_2}{2}}{b_0-z_1+\frac{z_2}{2}+z_3+d}\hspace{0.33em}p}}=p\hspace{0.33em}{\displaystyle \frac{{(z_2+z_3)}^2}{(b_2-\frac{z_2}{2})\hspace{0.33em}(b_0-z_1+\frac{z_2}{2}+z_3+d)}},$$

(76)

$$\begin{array}{l}{K}_{p3}={\displaystyle \frac{p_{\mathrm{CO}}\hspace{0.33em}{p}_{{\mathrm{H}}_2}}{p_{{\mathrm{H}}_2\mathrm{O}}}}={\displaystyle \frac{\frac{z_2+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}\hspace{0.33em}p\hspace{0.33em}\frac{b_1-2\hspace{0.33em}{z}_1+z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}\hspace{0.33em}p}{\frac{b_3-z_3}{b_0-z_1+\frac{z_2}{2}+z_3+d}\hspace{0.33em}p}}=\\ p\hspace{0.33em}{\displaystyle \frac{(z_2+z_3)\hspace{0.33em}(b_1-2\hspace{0.33em}{z}_1+z_3)}{(b_3-z_3)\hspace{0.33em}(b_0-z_1+\frac{z_2}{2}+z_3+d)}}.\end{array}$$

(77)

The resulting system of Equations (75)–(77) can be reduced to a system of nonlinear equations of the form Φ₁(z₁, z₂, z₃) = 0, Φ₂(z₁, z₂, z₃) = 0, and Φ₃(z₁, z₂, z₃) = 0:

A₁·z₁² + C₁·z₃² + D₁·z_1·z₂ + E₁·z₁·z₃ + G₁·z₁ + I_1·z₃ + J₁ = 0,

(78)

B₂·z₂² + C₂·z₃² + D₂·z_1·z₂ + F₂·z₂·z₃ + G₂·z₁ + H₂·z₂ + I_2·z₃ + J₂ = 0,

(79)

C₃·z₃² + D₃·z_1·z₂ + E₃·z₁·z₃ + F₃·z₂·z₃ + G₃·z₁ + H₃·z₂ + I_3·z₃ + J₃ = 0.

(80)

The coefficients of the equations are as follows:

$$\begin{gathered} A_1=1+4\hspace{0.33em}p{K}_{p1},\hspace{0.17em}{C}_1=pK{p}_1,\hspace{0.17em}{D}_1=-{\displaystyle \frac{1}{2}},\hspace{0.17em}{E}_1=-\left(1+4\hspace{0.33em}p\hspace{0.33em}{K}_{p1}\right), \\ {G}_1=-\left(4b_{1}p{K}_{p1}+b_0+d\right),\hspace{0.17em}{I}_1=2\hspace{0.33em}{b}_1\hspace{0.33em}p\hspace{0.33em}{K}_{p1},\hspace{0.17em}{J}_1=b_1^2\hspace{0.33em}p\hspace{0.33em}{K}_{p1}, \\ {B}_2=-\left(1+{\displaystyle \frac{1}{4}}\hspace{0.33em}{\displaystyle \frac{K_{p2}}{p}}\right),\hspace{0.17em}{C}_2=-1,\hspace{0.17em}{D}_2={\displaystyle \frac{1}{2}}\hspace{0.33em}{\displaystyle \frac{K_{p2}}{p}},\hspace{0.17em}{F}_2=-\left({\displaystyle \frac{1}{2}}{\displaystyle \frac{K_{p2}}{p}}+2\right), \\ {G}_2=-b_2\hspace{0.33em}{\displaystyle \frac{K_{p2}}{p}},\hspace{0.17em}{H}_2=-{\displaystyle \frac{1}{2}}\hspace{0.33em}{\displaystyle \frac{K_{p2}}{p}}\hspace{0.33em}\left(b_0-b_2+d\right),\hspace{0.17em}{I}_2=b_2\hspace{0.33em}{\displaystyle \frac{K_{p2}}{p}}, \\ {J}_2=\left(b_2\hspace{0.33em}d+b_0\hspace{0.33em}{b}_2\right)\hspace{0.33em}{\displaystyle \frac{K_{p2}}{p}}, \\ {C}_3=-\left(1+{\displaystyle \frac{K_{p3}}{p}}\right),\hspace{0.17em}{D}_3=2,\hspace{0.17em}{E}_3=\left(2+{\displaystyle \frac{K_{p3}}{p}}\right),\hspace{0.17em}{F}_3=-\left(1+{\displaystyle \frac{1}{2}}{\displaystyle \frac{K_{p3}}{p}}\right), \\ {G}_3=-b_3\hspace{0.33em}{\displaystyle \frac{K_{p3}}{p}},\hspace{0.17em}{H}_3=\left({\displaystyle \frac{b_3}{2}}\cdot {\displaystyle \frac{K_{p3}}{p}}-b_1\right),\hspace{0.17em}{I}_3=\left(b_3-b_0-d\right)\hspace{0.33em}{\displaystyle \frac{K_{p3}}{p}}-b_1, \\ {J}_3=\left(b_3\hspace{0.33em}{b}_0+b_3\hspace{0.33em}d\right)\hspace{0.33em}{\displaystyle \frac{K_{p3}}{p}}. \end{gathered}$$

The system of nonlinear Equations (78)–(80) can be solved by Newton’s iterative method.

3. Results

The results obtained by applying the model are confirmed by the results presented in the literature [18,19,22] for different types of solid fuel such as municipal solid waste (MSW) and cashew nut shell (CNSC).

3.1. Validation of the Model

3.1.1. Comparison with Data (Na Deng, Dongyan Li, et al. 2016 [19])

The data presented in [19] were used to calculate the gasification composition. Fuel composition (MSW) is reduced to a working wet basis.

Fuel composition (reduced to a working wet basis) was calculated as follows:

$$\begin{array}{l}C=0.3192\hspace{0.33em}\hspace{0.33em}\mathrm{kg}/\mathrm{kg}\\ H=0.0471\hspace{0.17em}\hspace{0.33em}\mathrm{kg}/\mathrm{kg}\\ O=0.1581\hspace{0.17em}\hspace{0.33em}\hspace{0.33em}\mathrm{kg}/\mathrm{kg}\\ N=0.0198\hspace{0.17em}\hspace{0.33em}\mathrm{kg}/\mathrm{kg}\\ S=0.0030\hspace{0.17em}\hspace{0.33em}\mathrm{kg}/\mathrm{kg}\\ W=0.2400\hspace{0.17em}\hspace{0.33em}\mathrm{kg}/\mathrm{kg}\\ \begin{array}{l}\underset{¯}{A=0.2128\hspace{0.17em}\hspace{0.33em}\mathrm{kg}/\mathrm{kg}}\\ TOTAL=1.0000\end{array}\\ \mathrm{Feed}\hspace{0.17em}\mathrm{rate}=1.32\hspace{0.17em}\hspace{0.33em}\mathrm{kg}/\mathrm{h}\\ \mathrm{Gasification\; temperature}=500–1000°\mathrm{C}(\mathrm{interval}50°\mathrm{C})\\ \mathrm{Steam}\hspace{0.33em}\mathrm{and}\hspace{0.33em}\mathrm{reactor}\hspace{0.33em}\mathrm{pressure}=1.013\cdot {10}^5\mathrm{Pa}\\ \mathrm{The\; amount\; of\; water\; vapor\; introduced\; into\; the\; reactor}\mathrm{\alpha }=0.4.\end{array}$$

Figure 3 shows a comparison of the results of the gasification of municipal solid waste (MSW) obtained by applying the mathematical model presented in this work with the model by Na Deng et al. [19]. The results of the MSW gasification composition obtained by the mathematical model are mostly in agreement with the MSW gasification composition values presented in the literature [19]. Assuming that the amount of water vapor injected is at a ratio of 0.4, for the considered temperatures of 600 °C, 800 °C, and 1000 °C (Figure 3), with increasing temperature, the mole fractions of CO and H₂ increase, while the values of H₂O, CH₄, and N₂ decrease, which was expected. Namely, at low temperatures, CH₄ is present in the gas, but as the temperature increases, carbon is converted into CO and CH₄ is converted into H₂ by the mechanization reaction. Above 600 °C, CH₄ production drops sharply. Since N₂ does not participate in the reactions, its share in the gas remains constant. The share of N₂ obtained by the model is slightly lower at 800 °C and 1000 °C than shown in the paper [19], which is a consequence of the application of the equation.

In general, the mathematical model gives slightly higher volume fractions of CO and CO₂, and slightly lower volume fractions of H₂, CH₄, H₂O, and N₂ than the data in the literature [19].

Figure 4 shows the effect of temperature on the mole fraction of carbon in a heterogeneous equilibrium mixture. It is observed in the presence of a solid phase (carbon) up to ≈735 °C where $C=0.00\%$. In synthetic gas, only gaseous components are present, whose mole fractions are $CO=33.10\%$, $H_2=52.70\%$, $C{H}_4=2.54\%$, $C{O}_2=4.97\%$, $H_{2}O=5.93\%$, and $N_2=0.76\%$. By increasing the temperature above 735 °C, there is no solid phase in the equilibrium mixture. The degree of carbon conversion decreases sharply when the temperature increases, so that at 700 °C, the degree of carbon conversion would be 96.26%. At 735 °C, the degree of carbon conversion is 100%.

3.1.2. Comparison with Data (Robert Elirasion Moshi [18])

In the numerical analysis, the mathematical model presented in this paper was compared with the results of the MSW gasification gas composition obtained by the model presented by Moshi Robert [18]. The composition of MSW (technical and elemental analysis) reduced to working mass (with moisture and ash) is shown in

Table 1. Elementary analysis of the working mass MSW.

Ultimate Analysis (with Moisture and Ash)	Mass Fraction (%)
Carbon (C)	16.26
Hydrogen (H)	1.57
Oxygen (O)	10.27
Nitrogen (N)	0.70
Sulfur (S)	0.09
MC (W)	59.79
Ash (A)	10.55

.

The coefficient of excess air (equivalent fraction of ER) is determined using the following expression:

$$ER={\displaystyle \frac{\frac{L}{{\dot{m}}_{MSW}}}{L_{\min }}}=0.75$$

(81)

where

$L=8.95kgL/h$—the amount of air introduced into the gasifier;

${\dot{m}}_{MSW}=6kg/h$—mass flow of MSW;

$L_{\min }={\displaystyle \frac{O_{\min }}{0.23}}={\displaystyle \frac{2.667\cdot C+8\cdot H+S-O}{0.23}}=1.9889\hspace{0.17em}kgL/kgMSW$;

$C,H,S,O$—mass fractions of carbon, hydrogen, sulfur, and oxygen in MSW, in kg/kg (Table 1).

Figure 5 shows the comparative results obtained by applying the model presented in this paper with the results presented by Robert Moshi [18]. The conformity of the results of mole fractions in the gas of both models is observed. From Figure 4, it can be seen that increasing the gasification temperature from 400 °C to 800 °C increases the mole fractions of CO and H₂. Above 800 °C, the values of the mole fractions of CO and H₂ practically remain unchanged. CH₄ decreases as the temperature increases due to the reformation of methane vapor as well as the oxidation of methane to H₂. By increasing the gasification temperature, the proportions of CO₂ and H₂O in the gas decrease as a result of reactions (2) and (3). The trend in the results obtained by the mathematical model presented in this paper is similar to that of the results of the models published in the literature (Deng et al., 2017) [19], (Han et al., 2017) [16].

By comparing the mathematical model with the work of M., Venkanta Ramanan et al. [22], the results obtained from the mathematical model are in good agreement with the experimental data (Figure 6). Cashew Shell Char (CNSC) was used as fuel, and the different mass fractions of moisture in the fuel were 7% and 10%, at a reaction temperature of 1373 K, an excess air coefficient of λ = 0.3, and a pressure in the reactor space of 1∙10⁵ Pa. A slight inconsistency of volume fractions in the synthetic gas was observed at the lower mass fraction of moisture in the fuel of 7%.

4. Conclusions

This paper presents two models for calculating the composition of synthetic gas during solid fuel gasification: the calculation of the gas composition based on the ratio of the total amounts of carbon, oxygen, hydrogen, and nitrogen entering the reactor space; and the calculation of the gas composition based on the number of moles participating in the assumed chemical reactions. Both equilibrium models agree with each other. In both models, the calculation of the composition of the heterogeneous mixture (the proportion of solid carbon) and the calculation of the composition of the homogeneous mixture (gas phase only) are presented. The model was tested with data from the literature, and good agreement was shown. Municipal solid waste (MSW) [18,19] and Cashew Shell Char (CNSC) [22] were used as fuel (raw material).

In the first example of the comparison of the results of the gas composition obtained by applying a mathematical model with data from the literature [19], the water vapor ratio was used as a means of gasification, and temperatures of 600 °C, 800 °C, and 1000 °C were considered. By increasing the gasification temperature, the mole fractions of CO and H₂ increase, while the values of H₂O, CH₄, and N₂ decrease, as expected. Namely, at low temperatures, CH₄ is present in the gas, but as the temperature increases, carbon is converted into CO and CH₄ is converted into H₂ by the mechanization reaction. Above 600 °C, the production of CH₄ drops sharply (Figure 3).

In a heterogeneous equilibrium mixture, the presence of a solid phase (carbon) is observed up to ≈735 °C, where the mole fraction of carbon decreases. At a given temperature, only gaseous components are present in synthetic gas whose mole fractions are CO = 33.10%, H₂ = 52.70%, CH₄ = 2.54, CO₂ = 4.97, H₂O = 5.93, and N₂ = 0.76%. By increasing the temperature above 735 °C, there is no solid phase in the equilibrium mixture (Figure 4). The degree of carbon conversion decreases sharply when the temperature increases, so that at 700 °C, the degree of carbon conversion would be 96.26%. At 735 °C, the degree of carbon conversion is 100% (Figure 4).

In Figure 5, the values of the volume fractions of CO, CO₂, H₂, CH₄, and H₂O obtained using the mathematical model are comparable to the data of the model presented in the literature [18]. Figure 6a shows a good agreement between the volume fractions of CO, CO₂, H₂, and CH₄ obtained by applying the mathematical model with the values obtained by the model shown in the literature [22]. By reducing the mass fraction of moisture in the working mass of CNSC fuel from 10% to 7%, slightly higher values of volume fractions of H₂ and CO were obtained using the mathematical model compared to the application of the model presented in the literature (Figure 6b) [22].

The presented model for calculating the composition of solid fuel gasification can be applied in industrial practice for the preliminary calculations, design, and operation of solid fuel gasifiers.

It can also be used as a guideline for solid fuel gasification studies and will provide a reliable experiment to predict the gasification process. In addition to the practical engineering significance, the mathematical model presented in this paper provides a theoretical contribution to the understanding of the mechanisms of the solid fuel gasification process.