Topic: A set of Algebraic Equations with multiple variables

Application: Steady$-$state balance in a steam reforming plant

Problem Statement:

Steam reforming of methane is one way to produce clean hydrogen for the 'Hydrogen economy'. The main two reactions are given below:

$C{H}_4+H_{2}O>CO+3H_2$

$($Rxn$1)$

$CO+H_{2}O>C{O}_2+H_2$

A steam reforming of methane reactor takes in $C{H}_4$ and $H_{2}O.$ The output stream of the reactor is a mixture of unreacted $C{H}_4,$ and $H_{2}O,$ and the products $CO,C{O}_2,$ and $H_2.$ The composition out of the reformer is based mainly on chemical equilibrium. Given the temperature, the pressure, and the composition of the inlet stream, we want to know what the composition of the outlet stream at equilibrium $($steady state$).$

Figure $1$: Schematic diagram showing a reactor, which takes $C{H}_4$ and $H_{2}O$ as the input streams and produces $H_2,CO$ and $C{O}_2.$ The exit of the reactor contains also unreacted $C{H}_4$ and $H_{2}O.$

The reactor is well mixed, implying that the concentration of species at the exit of the reactor under steady state corresponds to the concentration in equilibrium inside the reactor.

To solve this problem we can use the method of minimum reactions.

The mole rate of carbon, oxygen and hydrogen entering the reactor can be deduced based on the input flows $C{H}_4$ and $H_{2}O.$ The mass balance for the total carbon, hydrogen and oxygen can be written in the following way:

$N_C=(N_{CH4}{)}_{in}$

$N_H=4(N_{CH4}{)}_{in}+2N_{H2O}$

$N_O=(N_{H2O}{)}_{in}$

Under steady state, the molar flow of reactants and products inside the reactor will be constant. An elemental mass balance on carbon, hydrogen and oxygen elements using the molar flow under stated state can be expressed in the following way:

$N_C=N_{CH4}+N_{CO}+N_{CO2}$

$N_H=4N_{CH4}+2N_{H2}+2N_{H2O}$

$N_O=N_{CO}+2N_{CO2}+N_{H2O}$

Where $N_{CH4},N_{CO},N_{CO2},N_{H2},N_{H2O}$ are the mole rates at the exit of the reactor. The set of three reactions given by Eq$1-$Eq$3$ contains five unknown parameters. We require two additional equations in order to be able to find a unique solution to the concentration. The extra two equations can be related to the equilibrium constant for each reaction. The equilibrium equation for Rxn$1$ can be expressed in the following way:

$Ln(K_1)=2Ln[\frac{P}{P_o{N}_t}]+3Ln(N_{H2})+Ln(N_{CO})-Ln(N_{H2O})-Ln(N_{CH4})$

Where $P$ is the pressure of the system, $P_o$ is the standard pressure $)=(1(atm)$ and $N_t$ is the total number of moles inside the reactor; calculated as the sum of all species present in reaction Rxn$1.$

The equilibrium equation for $Rxn2$ can be expressed in the following way:

$Ln(K_2)=Ln(N_{H2})+Ln(N_{CO2})-Ln(N_{H2O})-Ln(N_{CO})$

The equilibrium constants $(K_1$ and $K_2)$ can be calculated using the following polynomial expressions:

$K_1=\frac{1}{\text{e}\text{x}\text{p}(0\mathrm{.}2513z^4-0\mathrm{.}3665z^3-0\mathrm{.}58101z^2+27\mathrm{.}1337z-3\mathrm{.}277)}$

$K_2=$exp$(-0\mathrm{.}29353z^3+0\mathrm{.}63508z^2+4\mathrm{.}1778z+0\mathrm{.}31688)$

Where

$z=(\frac{1000}{T})-1\mathrm{.}0$

with $T$ being the temperature of the reaction $(K).$

Calculate the concentration of all the species at the exit of the reactor under steady state, for a known flow of $C{H}_4$ and $H_{2}O$ entering the reactor. Write a Matlab program that determines the equilibrium composition under a steady state in the methane steam reforming. Plot the flow of $C{H}_4,H_2,CO,C{O}_2$ and $H_{2}O$ at the exit of the reactor as a function of temperature in the range of $T=600K$ to $1200K.$ The pressure of the reactor is always $P=1$atm and the input flow of methane is $10$mole$\frac{s}{m}in$ while that of steam is $5$mole$\frac{s}{m}in.$

Just for your reference, the concentration of $C{H}_4$ at the exit of the reactor as a function of temperature looks like:

I also meed to reproduce a plot of CH$4$ exit comcentraion vs temperature