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Week 10 - Simulating Combustion of Natural Gas.

Aim: Simulating Combustion of Natural Gas. Objective: Part I Perform a combustion simulation on the combustor model and plot the variation of the mass fraction of the different species in the simulation using line probes at different locations of the combustor as shown in Fig. You need to plot for CO2, H2O,…

Faizan Akhtar
updated on 07 Jun 2021

Aim: Simulating Combustion of Natural Gas.

Objective:

Part I

Perform a combustion simulation on the combustor model and plot the variation of the mass fraction of the different species in the simulation using line probes at different locations of the combustor as shown in Fig. You need to plot for CO2, H2O, CH4, N2, O2, NOx emissions & Soot formation.

Part II

As you must have observed from the above simulation, the Nox and soot are getting formed at the outlet of the combustor. Such formation has harmful effects on the environment and humans. The stringent government norms also demand the least formation of Nox and soot and to satisfy those requirements, you need to check the effect of adding the water in the fuel.

In this part, you need to add the water content in the fuel from 5% to 30% by mole and observe the effect of it on the results. It is necessary to provide line plots and contours to prove your claim.

Introduction

A combustor is an area or a chamber where the combustion takes place. It can be a burner, combustion chamber, or flame holder. In a combustor, the high-pressure air is fed which is heated at constant pressure and is guided to the turbines through the nozzle-guided vanes.

A combustor must maintain stable combustion despite the velocity of the air is on the higher side. In order to do so the air mixes and ignites the fuel and to complete the combustion more air is added. The typical components of the combustor include fuel inlet, air inlet, and the walls of the combustor. The drawbacks of the combustor model include the emission of species such as carbon-di-oxide and emission of $N O_{x}$ $NO_x$ and soot formation due to incomplete combustion which is harmful to the environment. The temperature plot should be uniform, if it is not then it will lead to temperature hotspots leading to the thermal damage of the combustor.

Applications of combustion

The typical application of the combustion includes

Burning of wood or coal for household purposes.
Burning of petrol or diesel for using vehicles like the car.
Cooking purposes.
For the production of energy in thermal power plants.

Solving and Modelling approach

The geometry is loaded into the Spaceclaim.
The combustor is modeled into a cylinder. The academic license restricts the use of 512000 computational cells, making it difficult for the entire geometry to be taken for simulation.
The geometry consists of a small fuel inlet and a larger air outlet. The "Combine" option is selected from the "Prepare" ribbon and the entire geometry is combined into one.
The plane is selected from the "Design" tab and the origin is selected to create three different planes along with the geometry viz, XY, YZ, ZX.
The split option is selected and the target geometry and the target plane (XY) are selected to split the model in half.
The XZ plane is selected to split the body into one-fourth.
The 2-D geometry is selected and created as a new design.
The 2-D geometry is compatible with the academic license and the mesh refinement approach can be easily followed.
Named selections are created and the coarse mesh is selected.
The mesh is refined to reduce aspect ratio and skewness value to minimize error due to interpolation and also to increase convergence
The whole geometry is loaded into the setup.
The boundary conditions and the species models are initialized by using hybrid initialization.
The necessary monitor plots, contour plots are created to visualize the combustion inside the combustor.

Preprocessing and Solver settings

The geometry is loaded into Spaceclaim

The combine option is selected which is available in the "Design" ribbon, and the "Select Bodies to Merge" is selected and the components are merged into a single one.

The plane option is selected from the "Design" ribbon and the three plains are created as shown below

The "Split Body" is selected from the ribbon, the target body, and the XY plane is selected first, the resulting split body is shown below

The XZ plane is selected and the body is split into one-fourth as shown below

The front portion of the body is selected and is copied and pasted in a new tab with a new file name as shown below

Thus a 3-D model is converted into a 2-D model for ease in simulation. The 2-D model is loaded for meshing.

Meshing

It is the process of discretizing the entire computational domain into a finite number of control volumes. These volumes are used for solving non-linear second-order PDE. The NS equations are solved in each volume using iterative procedures and the leftover after each iteration is called Residuals. The residuals plot plays an important criterion in determining whether the solution has converged or not. The residuals can never be zero but lower residuals mean the solution is reaching a more accurate level. There are various factors that play an important role in determining the convergence and one of the criteria is meshing. The coarse mesh has higher skewness and aspect ratio value. The mesh refinement approach is carried out to decrease the value of the mentioned parameters recommended by ANSYS-FLUENT. The finite volume discretization schemes are recommended by the CFD software package because it preserves the conservation property, it can be applied to the structured and unstructured mesh.

Baseline mesh

Element size: $2 m m$ $2mm$

Captured curvature: Yes

Captured proximity: Yes

Num cells across gaps: 1

Mesh metric

Number of Elements	Element Quality	Aspect Ratio	Skewness	Orthogonal Quality
$17037$ $17037$	$0.99564$ $0.99564$	$1.011$ $1.011$	$4.5526 e - 003$ $4.5526e-003$	$0.99981$ $0.99981$

Refined mesh

Element size: $1 m m$ $1mm$

Captured curvature: Yes

Captured proximity: Yes

Num cells across gaps: 2

Mesh metric

Number of Elements	Element Quality	Aspect Ratio	Skewness	Orthogonal Quality
$67345$ $67345$	$0.99731$ $0.99731$	$1.0035$ $1.0035$	$2.7491 e - 003$ $2.7491e-003$	$0.99988$ $0.99988$

Element size: $0.8 m m$ $0.8mm$

Captured curvature: Yes

Captured proximity: Yes

Num cells across gaps: 3

Mesh metric

Number of Elements	Element Quality	Aspect Ratio	Skewness	Orthogonal Quality
$105933$ $105933$	$0.99733$ $0.99733$	$1.0061$ $1.0061$	$2.5638 e - 003$ $2.5638e-003$	$0.9999$ $0.9999$

Element size: $0.7 m m$ $0.7mm$

Captured curvature: Yes

Captured proximity: Yes

Num cells across gaps: 4

Mesh metric

Number of Elements	Element Quality	Aspect Ratio	Skewness	Orthogonal Quality
$138094$ $138094$	$0.99691$ $0.99691$	$1.0038$ $1.0038$	$3.1476 e - 003$ $3.1476e-003$	$0.9998$ $0.9998$

Named selection

Air inlet (Type Velocity inlet)

Fuel inlet (Type Velocity inlet)

Side Walls (Type Wall)

Axis (Type Axis)

Walls (Type Wall)

Outlet (Type Pressure outlet)

Setup

The steady-state pressure-based( $M < 0.3$ $Mlt0.3$ ) solver was selected. The axis-symmetric option was selected which states that the rate of change of any quantities (density, temperature, and pressure) remains zero, the change is observed only in the axial direction (x-axis).

The energy equation is selected.

Calculation of Reynolds number

$R_{e} = \frac{ρ \cdot v \cdot D}{μ}$ $R_e=frac{rho*v*D}{mu}$

$R_{e} = \frac{1.225 \cdot 0.5 \cdot 0.08}{1.7894 e - 05}$ $R_e=frac{1.225*0.5*0.08}{1.7894e-05}$

$R_{e} = 2738.348$ $R_e=2738.348$ which justifies the flow is in transitional region but the Reynolds number is more close to the turbulent region.

Viscous model

The turbulence model was set to $k - ε$ $k-epsilon$ realizable, standard wall function.

The family of the turbulence model and its variants (Standard/RNG/Realizable) are considered one of the frequently used turbulence models and they are found in all commercial codes available. It is used for the high Reynolds number, equilibrium flows and steady-state without having a large pressure gradient. The major feature of this model is wall sensitivity. There are certain shortcomings of this model is that it can only be used for high Reynolds number, so the law of the wall must be employed to obtain the velocity"boundary condition" away from solid boundaries. In order to integrate the equation in the viscous sub-layer low Reynolds number must be used. The model is insensitive to the adverse pressure gradient so it ignores the shear stress originating from the wall and thus it delays separation. The best way to avoid all those shortcomings is to use the Realizable model which uses a reliability constraint in the equation as if all the normal stresses near the wall are positive and the correlation coefficient of the shear stress should not exceed one.

The near-wall treatment is chosen as a Standard wall function which should not be used below the Y+ limit of 11 as the solution accuracy might deteriorate in an uncontrollable manner.

Species model

The species model allows us to set the parameters related to species transport and combustion. In the species model "Species Transport" is selected. The species transport allows us to set the model related to the mixing and transport of species by solving conservation equations consisting of convection, diffusion, and the reaction terms in the equation. The "Reaction" is selected as "Volumetric" which enables the calculation of the reacting flow as volumetric. The None-Explicit Source Chemistry Solver was used. The Inlet Diffusion is selected from the option which enables the calculation of diffusion flux at the inlet. The Diffusion Energy Source is selected which includes the effect of enthalpy transport due to species diffusion in the energy equation. The number of volumetric species is displayed as 5 (CO2, H2O, CH4, N2, O2) this is the informational value and it cannot be edited. The mixture-material can be selected from the dropdown list as such the present challenge states that the mixture-material should be "methane-air", "methane-air2". The "Eddy-Dissipation" is selected as the Turbulence-Chemistry- Interaction enables us to use it for turbulent flows and computes only the mixing rate.

The chemical reaction involving combustion of methane with air is as follows'

$C H_{4} + a r (O_{2} + 3.76 N_{2}) \to a C O_{2} + b H_{2} O + c N_{2}$ $CH_4+ar(O_2+3.76N_2)to aCO_2+bH_2O+cN_2$

Thus equating left-hand side to the right hand side

$a = 1$ $a=1$

$2 b = 4$ $2b=4$

$3.76 a r = c$ $3.76ar=c$

$2 a r = 2 a + b$ $2ar=2a+b$

Therefore $a = 1$ $a=1$ , $b = 2$ $b=2$ , $a r = 2$ $ar=2$ $c = 7.52$ $c=7.52$

At the fuel inlet, the methane is injected which is equal to a 1-mole fraction or mass fraction.

The air mixture contains 2 moles of $O_{2}$ $O_2$ and $7.52$ $7.52$ moles of $N_{2}$ $N_2$ .

Mole fraction of $O_{2}$ $O_2$ $= \frac{2}{9.52}$ $=frac{2}{9.52}$ $= 0.21$ $=0.21$

Mole fraction of $N_{2} = \frac{7.52}{9.52}$ $N_2=frac{7.52}{9.52}$ $=$ $=$ $0.7899$ $0.7899$

Equivalent mass fraction of $O_{2} =$ $O_2=$ $0.23$ $0.23$

Equivalent mass fraction of $N_{2} = 0.77$ $N_2=0.77$

$N O_{x}$ $NO_x$ model

The generation of $N O_{x}$ $NO_x$ is quite low during the combustion process. In other words $N O_{x}$ $NO_x$ chemistry has less influence on temperature field and other combustion products. The $N O_{x}$ $NO_x$ model is activated after the species model has been set up. The methane is selected as fuel.

The "Thermal $N O_{x}$ $NO_x$ " and "Prompt $N O_{x}$ $NO_x$ " are selected for the above challenge. For particulate matters present in the combustion the fuel $N O_{x}$ $NO_x$ will be enabled only if the DPM option is enabled. The "Thermal $N O_{x}$ $NO_x$ " accounts for the product formed by the oxidation of atmospheric nitrogen in the presence of combustion air. The "Prompt $N O_{x}$ $NO_x$ " accounts for the reaction of molecular nitrogen with combustion flame giving rise to cyanide and hydrogen cyanide which eventually decompose to form nitrogen oxide.

Soot model

The "Soot model" is used for calculating the formation of soot at the combustion outlet. The one-step "Soot model" is selected which is the default "Soot model". The fuel is selected as $C H_{4}$ $CH_4$ and the oxidant is selected as $O_{2}$ $O_2$ .

Boundary condition

Air inlet: The ANSYS-FLUENT describes the inlet to be a velocity inlet. The inlet velocity of the air will be $0.5 m \sec^{- 1}$ $0.5msec^-1$ , the thermal boundary condition will remain as it is. In species, the mass fraction of $0.23$ $0.23$ will be computed for $O_{2}$ $O_2$ and the rest species will be zero.

Fuel inlet: The inlet velocity of the fuel will be $80 m \sec^{- 1}$ $80msec^-1$ . In species, the mass fraction of methane will be computed as 1 as discussed above`

Outlet: The ANSYS-FLUENT describes the outlet to be a pressure outlet, the gauge pressure will be $0 P a$ $0Pa$ .

Axis: The type axis is selected by ANSYS-FLUENT which indicates that the rate of change of quantities in the radial direction is zero.

Walls and side walls: The type wall is selected for the ANSYS-FLUENT. The wall is stationary and in the shear condition, no-slip is selected.

Monitor plots

The residual plot is created and the convergence limit is checked off.

The surface report is created. The area-weighted average temperature is plotted by selecting the outlet boundary zones. The temperature at the outlet, mass fraction of the soot, and the $N O_{x}$ $NO_x$ are calculated for the outlet.

Initialization

The solution is hybrid initialized by hitting $t = 0$ $t=0$ . The hybrid initialization solves the Laplace equation to evaluate the pressure and the velocity fields. The number of iteration is set to 10 and the tolerance level is set to 1e-6.

Post-processing

The line probes were created at plane-1 in the CFD post.

Line-1

Point	$x [m]$ $x[m]$	$y [m]$ $y[m]$	$z [m]$ $z[m]$
Point 1	$0.1$ $0.1$	$0$ $0$	$0$ $0$
Point 2	$0.1$ $0.1$	$0.08$ $0.08$	$0$ $0$

Line-2

Point	$x [m]$ $x[m]$	$y [m]$ $y[m]$	$z [m]$ $z[m]$
Point 1	$0.2$ $0.2$	$0$ $0$	$0$ $0$
Point 2	$0.2$ $0.2$	$0.08$ $0.08$	$0$ $0$

Line-3

Point	$x [m]$ $x[m]$	$y [m]$ $y[m]$	$z [m]$ $z[m]$
Point 1	$0.3$ $0.3$	$0$ $0$	$0$ $0$
Point 2	$0.3$ $0.3$	$0.08$ $0.08$	$0$ $0$

Line-4

Point	$x [m]$ $x[m]$	$y [m]$ $y[m]$	$z [m]$ $z[m]$
Point 1	$0.4$ $0.4$	$0$ $0$	$0$ $0$
Point 2	$0.4$ $0.4$	$0.08$ $0.08$	$0$ $0$

Line-5

Point	$x [m]$ $x[m]$	$y [m]$ $y[m]$	$z [m]$ $z[m]$
Point 1	$0.5$ $0.5$	$0$ $0$	$0$ $0$
Point 2	$0.5$ $0.5$	$0.08$ $0.08$	$0$ $0$

Line-6

Point	$x [m]$ $x[m]$	$y [m]$ $y[m]$	$z [m]$ $z[m]$
Point 1	$0.6$ $0.6$	$0$ $0$	$0$ $0$
Point 2	$0.6$ $0.6$	$0.08$ $0.08$	$0$ $0$

Part-2 simulation set up

The material is changed from methane-air to methane-air2 step. In the fuel inlet boundary condition, under the species tab, the input parameters are set for methane and water.

The output parameter is selected for the "Temperature", "Mass fraction of soot", "Mass fraction of $N O_{x}$ $NO_x$ .

The mass fraction of methane and water is changed accordingly and the behavior of the output parameters is observed.

Case-1

Baseline mesh

Element size: $2 m m$ $2mm$

Pressure-velocity coupling scheme: SIMPLE

Residual plot

Temperature

Pressure-velocity coupling scheme: SIMPLEC

Residual plot

Temperature

Pressure-velocity coupling scheme: COUPLED

Residual plot

Temperature

Comparison of all cases

Mesh	Number of elements	Pressure-velocity schemes	Temperature at the outlet	Mass fraction of $N O_{x}$ $NO_x$	Mass fraction of Soot
Baseline mesh	$17183$ $17183$	SIMPLE	$937.8879 K$ $937.8879 K$	$4.920071 e - 07$ $4.920071e-07$	$0$ $0$
		SIMPLEC	$1622.17 K$ $1622.17K$	$4.014398 e - 05$ $4.014398e-05$	$5.16668 e - 09$ $5.16668e-09$
		COUPLED	$1589.537 K$ $1589.537K$	$0.0001888244$ $0.0001888244$	$0.0003230682$ $0.0003230682$

The baseline simulation was performed to check how the simulation ran and to decide the best pressure-velocity schemes. The SIMPLE (Semi Implicit Pressure Linked Equation) as stated solves the pressure and momentum equation implicitly and the velocity equation explicitly. It solves the equation sequentially and requires more iteration for convergence, but less memory space. The mass fraction of the soot is zero and the $N O_{x}$ $NO_x$ was recorded close to zero, a rise is observed in the monitor plots at 500 iterations, and even the combustion was not visualized in the contour temperature. The SIMPLEC (Semi Implicit Pressure Linked Equation Consistent) manipulates the momentum equation in such a way to omit the less significant terms in the velocity equation, and again it is the variant of SIMPLE scheme, the mass fraction of the soot and the $N O_{x}$ $NO_x$ was recorded close to zero, therefore not reliable in the case of $N O_{x}$ $NO_x$ and soot. The COUPLED scheme as the name suggests that it solves the pressure and the velocity equation instantaneously, requires less iteration to convergence, requires more memory. The monitor plot produced by the COUPLED scheme produces a stable output and reaches a convergence at around 250 iterations. The mass fraction of soot and the $N O_{x}$ $NO_x$ is also stable. The results of the COUPLED scheme are reliable.

For the "COUPLED" scheme the gradient is set to Least Square Cell Method. The Pressure is set to the second-order scheme which utilizes the central differencing scheme at the cell centroid to compute the values and is second-order accurate. The momentum, mass fraction of species, mass fraction of $N O_{x}$ $NO_x$ , mass fraction of soot is set to the second-order upwind scheme which utilizes the values at the cell faces to compute the cell-centered solution at the cell centroid and is second-order accurate. The turbulence terms are discretized by using the first-order upwind scheme which utilizes the values at the cell faces as the cell averaged value indicating that it is first-order accurate.

Mesh refinement approach

The mesh refinement approach was carried to gradually increase the "Element Quality", "Orthogonal Quality", and to decrease the "Aspect Ratio", "Skewness" value to reduce any error arising due to interpolation. It can also increase convergence. The orthogonal quality should be $> 0.1$ $gt0.1$ and the skewness value should be $< 0.95$ $lt0.95$ .

Refined mesh

Element size: $1 m m$ $1mm$

Pressure-velocity schemes: COUPLED

Residual plot

Temperature

Element size: $0.8 m m$ $0.8mm$

Pressure-velocity schemes: COUPLED

Residual plot

Temperature

Element size: $0.7 m m$ $0.7mm$

Pressure-velocity schemes: COUPLED

Residual plot

Temperature

Comparison of all cases

Mesh	Element size	Number of elements	Pressure-velocity schemes	Temperature at the outlet	Mass fraction of $N O_{x}$ $NO_x$	Mass fraction of soot
Refined mesh	$1 m m$ $1mm$	$67345$ $67345$	COUPLED	$1588.67 K$ $1588.67K$	$0.0001843973$ $0.0001843973$	$0.0002781303$ $0.0002781303$
	$0.8 m m$ $0.8mm$	$105933$ $105933$		$1586.987 K$ $1586.987K$	$0.0001853749$ $0.0001853749$	$0.0003461276$ $0.0003461276$
	$0.7 m m$ $0.7mm$	$138094$ $138094$		$1587.345 K$ $1587.345K$	$0.0001843004$ $0.0001843004$	$0.0003142322$ $0.0003142322$

The fewer deviations are observed in the results from element size 1mm and 7mm whereas the results from element size 1mm and 8mm are more deviated which brings us to the conclusion of using 7mm element size in a future calculation.

Results

Element size: $0.7 m m$ $0.7mm$

Mixture: Methane-air

Temperature at plane-1

Mass fraction of $C O_{2}$ $CO_2$

Mass fraction of $O_{2}$ $O_2$

Mass fraction of $N_{2}$ $N_2$

Mass fraction of $H_{2} O$ $H_2O$

Mass fraction of $C H_{4}$ $CH_4$

Mass fraction of NOx

Results

Element size: $0.7 m m$ $0.7mm$

Mixture: Methane-air2 step

Residual plot & Temperature plot

Mass fraction values

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$

$C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$

$C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$

$C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Temperature at plane-1

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $C H_{4}$ $CH_4$

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $C O$ $CO$

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $C O_{2}$ $CO_2$

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $N_{2}$ $N_2$

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $O_{2}$ $O_2$

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $H_{2} O$ $H_2O$

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $N O_{x}$ $NO_x$

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of soot

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $C H_{4}$ $CH_4$ for the given line probes (Line-1 to Line-6)

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $C O$ $CO$ for the given line probes (Line-1 to Line-6)

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $C O_{2}$ $CO_2$ for the given line probes (Line-1 to Line-6)

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $N_{2}$ $N_2$ for the given line probes (Line-1 to Line-6)

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $O_{2}$ $O_2$ for the given line probes (Line-1 to Line-6)

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $H_{2} O$ $H_2O$ for the given line probes (Line-1 to Line-6)

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of $N O_{x}$ $NO_x$ for the given line probes (Line-1 to Line-6)

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Mass fraction of soot for the given line probes (Line-1 to Line-6)

$C H_{4} = 1$ $CH_4=1$ $H_{2} O = 0$ $H_2O=0$ & $C H_{4} = 0.95$ $CH_4=0.95$ $H_{2} O = 0.05$ $H_2O=0.05$

$C H_{4} = 0.90$ $CH_4=0.90$ $H_{2} O = 0.1$ $H_2O=0.1$ & $C H_{4} = 0.85$ $CH_4=0.85$ $H_{2} O = 0.15$ $H_2O=0.15$

$C H_{4} = 0.80$ $CH_4=0.80$ $H_{2} O = 0.20$ $H_2O=0.20$ & $C H_{4} = 0.75$ $CH_4=0.75$ $H_{2} O = 0.25$ $H_2O=0.25$

$C H_{4} = 0.70$ $CH_4=0.70$ $H_{2} O = 0.3$ $H_2O=0.3$

Comparison of all cases for methane-air2step mixture

Water content in the fuel ( $%$ $%$ )	$C H_{4}$ $CH_4$ mass fraction	$H_{2} O$ $H_2O$ mass fraction	Mass fraction of $N O_{x}$ $NO_x$	Mass fraction of Soot	Temperature at the outlet
1	1	0	$0.00015995403$ $0.00015995403$	$0.00019230996$ $0.00019230996$	$1586.9917 K$ $1586.9917K$
5	0.95	0.05	$0.00012614165$ $0.00012614165$	$6.5777408 E - 05$ $6.5777408E-05$	$1539.2351 K$ $1539.2351K$
10	0.90	0.1	$9.792421 E - 05$ $9.792421E-05$	$2.2853644 E - 05$ $2.2853644E-05$	$1489.7656 K$ $1489.7656K$
15	0.85	0.15	$7.4061329 E - 05$ $7.4061329E-05$	$7.256628 E - 06$ $7.256628E-06$	$1439.0385 K$ $1439.0385K$
20	0.80	0.20	$5.4331403 E - 05$ $5.4331403E-05$	$2.2460482 E - 06$ $2.2460482E-06$	$1387.0244 K$ $1387.0244K$
25	0.75	0.25	$3.8479895 E - 05$ $3.8479895E-05$	$5.9262782 E - 07$ $5.9262782E-07$	$1333.6414 K$ $1333.6414K$
30	0.70	0.30	$2.616394 E - 05$ $2.616394E-05$	$1.4117863 E - 07$ $1.4117863E-07$	$1278.8117 K$ $1278.8117K$

Conclusion

The steady-state simulation was performed successfully for methane-air and methane-air2step mixture.
The pressure-based solver was selected since ( $M < 0.3$ $Mlt0.3$ ) and the COUPLED scheme was selected because the monitor plots were reliable and can be trusted.
The simulation performed by using methane-air mixture gave us insight about the temperature at the outlet, mass fraction of NOx and soot, $C H_{4}$ $CH_4$ , $N_{2}$ $N_2$ , $O_{2}$ $O_2$ , $H_{2} O$ $H_2O$ , $C O_{2}$ $CO_2$ and six line-probes were drawn on plane-1 to observe their variation. The fuel inlet comprises of 1mole of $C H_{4}$ $CH_4$ therefore the mass fraction of $C H_{4}$ $CH_4$ is higher near the inlet but reduces to zero at the outlet. The mass fraction of $C O_{2}$ $CO_2$ and $O_{2}$ $O_2$ was observed near the inlet as well as at the outlet. Pollutant NOx and soot were also observed at the outlet.
It is necessary to reduce the emission of pollutants such as NOx and soot to make the combustion process harmless for the environment, which led to simulate the combustion with methane-air2step mixture.
The second stage of simulation involves the addition of water content in the fuel side from 5% to 30% which brings us to numerous observations which are listed below.
As the mass fraction of water increases the temperature at the outlet decreases which further reduces any thermal damage from the hotspot region as such it could have damaged the region of the combustor.
The environmental laws were complied with, the increase in mass fraction of water brought down the levels of NOx and soot emission at the outlet which makes the combustion process environmentally friendly. The thermal NOx increases with an increase in temperature and with the greater concentration of $O_{2}$ $O_2$ species. The addition of water lowers the temperature, thus decreasing the concentration of thermal NOx, that's why there is no change in the mass fraction of $O_{2}$ $O_2$ for the six-line probes with an increase in mass fraction of water. The condition favourable for the prompt NOx is the rich fuel condition when the nitrogen reacts with the hydrocarbon to produce HCN and again atomic nitrogen reacts with the OH radicals to form NO. The addition of water into the fuel suppresses the thermal NOx and prompt NOx.The soot is formed as a result of incomplete combustion of hydrocarbon fuels or insufficient amount of OH radicals, as $O_{2}$ $O_2$ is being utilized for combustion. The addition of water suppresses the NOx and the soot formation because it makes the fuel less rich.
It also produces extra species which is CO but due to an increase in the mass fraction of water, the mass fraction of CO near the outlet approaches zero.
The mass fraction of $C H_{4}$ $CH_4$ decreases with an increase in mass fraction of water, at the outlet the mass fraction of $C H_{4}$ $CH_4$ approaches zero.
There is a significant decrease in the mass fraction of $C O_{2}$ $CO_2$ with an increase in the mass fraction of water.
The only limitation that could be observed during the combustion process is that the water must be treated before being utilized for combustion, and the presence of a water source near its vicinity.