 In tutorial 1 we have seen very basic problems related to the steam methane reforming and other hydrogen production methods. So, these were like very preliminary knowledge about the what type of problems can be there associated with the course. Now let us with that background let us see in this second tutorial some more problems these are more detailed problems. Let us look at the problem the problem statement is that in a steam methane reforming reaction the feed gas composition is provided so these are the different species which are present in the natural gas. So, in the earlier problem we have considered only methane being present now we know that natural gas is a mixed composition with methane primarily in major constituent but then there can be other constituents as well. So, here we have taken other constituent to be ethane carbon dioxide nitrogen and let us assume that the steam to carbon ratio which is given in the process is 3. So, steam to carbon ratio is provided to be 3 and the exit gas temperature is 900 degree centigrade and pressure is 20 bar. So, different equilibrium constants are provided temperature here T represented is in degree centigrade and what we have to do is we have to calculate the exit gas composition after the two reactions first steam methane reforming and then water gas shift reaction. Now let us do this problem. So, the first reaction that will occur is SMR and the second reaction will be water gas shift. So, it is given in the problem that steam to carbon ratio is 3. So, using that if we now calculate the carbon in the natural gas so we have methane in which the carbon is 0.95 mole percent is given in ethane which is 0.03 into 2 C2H6 and also into the carbon dioxide which is 0.01. So, a total this all makes up a total of 1.02 moles. Now if carbon to steam ratio which is given to be 1 is to 3 provided that it is 1.02 mole percent of carbon. So, the steam is 3 times of this which makes it to be 3.06 mole percent for steam. Now there are two assumptions here nitrogen we know it is remaining unreacted in the process and we will assume here that the ethane remains is completely reacted in the reaction. So, to solve this problem let us first write the species what is present at the inlet and what is present at the outlet. So, methane which is given to be 0.95 let us assume that in the outlet stream the unreacted methane is X ethane C2S6 which is 0.03 mole percent we assume that it is completely reacted CO2 0.01 that we will find out N2 given is 0.01 it is not consumed in the reaction what are we have just now seen it is 3.06 what comes out at the exit that we will find out CO initially the initial mole percent was 0. Let us assume that on reaction it is Y for H2 initially it was 0 and then we will find out how much is there so these are the entries that we need to fill in the table. Now we will solve this problem using materials balance those who are from chemical engineering background they will be well versed with solving such problems but for we will be solving this this is a simple problem but little lengthier one what we will be doing is we will be doing material balance and specifically atomic balance for each of these constituents. Now if we do the material balance we start with the carbon balance what we say is whatever is entering or being generated is equal to whatever is leaving coming out. Now if we start with the carbon balance the carbon present at the inlet side is 1.02 that we have found and that is equal to if we see the two reactions again here. So now what is leaving is unreacted methane so this is what is arising from the methane side X then ethane it is completely consumed so 0 for ethane. Now we have CO which is we have represented by Y the term arising from CO and then the term arising from CO2 let the CO2 output we will find from this carbon balance. So the CO2 at the output side at the outlet can be represented as 1.02 minus X minus Y. So this is one of the entry that we can fill here 1.02 minus X minus Y. Now the second material balance we will do will be for oxygen if we do for oxygen balance again the same whatever is entering or generation equal to coming out. Now from the CO2 side it is 0.02 in the reaction which is entering with water it is associated so it is 3.06 on the product side associated with CO it is Y this is an atomic balance that we are doing it is also associated with CO2 CO2 we have just now calculated as 1.02 minus X minus Y so we will substitute for that and there are 2 moles so 1.02 minus X minus Y that is arising from the CO2 reactant what in water also that oxygen content is there. Now if with this oxygen balance we can find out the H2O which is coming out and that comes out to be 1.04 plus 2X plus Y. So in this table we have got the second entry which is that H2O coming out 1.04 plus 2X plus Y. Now the third material balance we will do for hydrogen so from methane hydrogen methane it is 0.95 in the reactant side and there are 4 hydrogen atoms to it so 0.95 into 4 this is arising from the methane in the reactant then there is ethane present on the reactant side so from ethane the mole fraction of it is 0.03 and then there are 6 atoms in ethane there are 6 atoms then it is coming from water 2 water molecules reacting to 3.06 2 hydrogen atoms which are coming from water and the amount which is leaving that is corresponding to the unreacted methane X amount of unreacted methane there are 4 hydrogen atoms in the methane so 4 acts plus there is water 2 times 1.04 plus 2X plus Y. This is hydrogen is having 2 atoms and hydrogen we have already found the content of it 1.04 plus 2X plus Y and that is along with the hydrogen coming out in the reaction. Now if we solve this then we can write it as and that will give us the amount of hydrogen coming out and this corresponds to 8.02 minus 8X minus 2Y so we will fill in this entry for hydrogen 8.02 minus 8X minus 2Y so now we have found the corresponding mole fraction of the species which are coming out in the process. Now our aim is here to find out what are the values of X and Y to find this so we have 2 unknowns and now we will be using the 2 known equations for finding these values of X and Y it is given in the problem that the equilibrium constant for the reforming reaction is given by exponential 24.383 minus 15415405 divided by the temperature and the temperature is also given in the problem to be 900 degree centigrade. So we can find the reforming equilibrium constant and when we substitute this T equal to 900 degree centigrade we get a value of so this exponential is exponential of 7.26 and that comes out to be 1426.957 so this is one value we have got and similarly we can get the equilibrium coefficient for shift reaction. So provided from the correlation the value for shift is exponential 2299 divided by temperature minus 2.79 when we substitute again temperature equal to 900 degree we get that value as 0.7901. So now we have 2 unknowns and we have 2 equations that we will be using how we will frame those equations we will be framing those equations according to the definition of equilibrium constant. So equilibrium constants these 2 equations these values we will be substituting in the definition for the equilibrium constants and then we will be getting. So let us do 1 by 1 let us first use for the reforming reaction the equilibrium constant so it is the partial pressure corresponding to the product side CO upon the standard pressure into the partial pressure for hydrogen which is one of the product upon the standard pressure raised to the stoichiometric coefficient. So the reaction was CH4 plus H2O giving CO plus 3H2. So the product is CO it is partial pressure, partial pressure of hydrogen and the stoichiometric coefficient raised to the stoichiometric coefficient divided by the reactant side. So the partial pressure of methane over the standard pressure, partial pressure of water divided by the standard pressure raised to the stoichiometric coefficient 1 and similarly we can do this for the shift reaction. So shift reaction CO plus H2O giving CO plus H2O so the partial pressure of the product side CO over standard pressure, partial pressure for hydrogen similarly for the reactant side CO sorry CO2 this is CO2 and for hydrogen. Now these are the two equations whose values we have already found so for this particular problem the reforming equilibrium constant we have found to be 1426.957 and for shift reaction we have found this value to be 0.7901. Now we will be substituting for the partial pressure in these equations to solve this problem. Now let us find out the partial pressures from the table which we have now filled the entries. So we have now the species known here and now we will be utilizing the Dalton's law for finding the partial pressure. So this is the species which we have got and now we can find also the mole fraction from here. So mole fraction we can calculate for each of these species let us tabulate that separately so that we can use these for the finding the x and y values. So now if we try to tabulate these again using the mole fractions then we can say that the species entering, leaving and the mole corresponding mole fraction we can find. So the methane on the reactant side 0.95 on the product side x ethane 0.03 mole percent completely reacted carbon dioxide 0.01 this was given in the problem and we have found that its content is in the outlet side is 1.02 minus x minus y. For nitrogen this remains unreacted it was 0.01 it remains 0.01 for water we have found it is 3.06 and then at the outlet side it is 1.04 plus 2x plus y. For CO carbon monoxide initially it was 0 we have assumed its concentration its mole percent to be y on the product side H2 initially 0 and we have found it to be 0.8.02 minus 8x minus 2y and that total contributes to 4.06 entering moles mole percent and then it is what is leaving 10.09 minus 6x minus y. Now we can find out the mole fraction corresponding to each of these species by dividing what we have got at the output divided by the total number of moles. So for methane it is x divided by 10.09 minus 6x minus y for ethane this remains 0 for carbon dioxide it is 1.02 minus x minus y numerator denominator is the total 10.09 minus 6x minus y for nitrogen again 10.09 minus 6x minus y 1.04 plus 2x plus y is the numerator for water divided by 10.09 minus 6x minus y same for carbon monoxide y divided by 10.09 minus 6x minus y and for hydrogen 8.02 minus 8x minus 2y divided by 10.09 minus 6x minus y. So this is the mole fraction yi is mole fraction. Now we know the relationship partial pressure is equal to mole fraction times the total pressure and total pressure is given in the problem to be 20 bar. Now substituting these individual mole fractions in the equilibrium constant in terms of the partial pressure. So partial pressure will be given by partial pressure will be for any species i which could be CO, CH4 or CO2 or water we can write the partial pressure in terms of the mole fraction for that species and the pressure total pressure which is 20 bar. So yi into 20 and that we will substitute in the using the table here. Now if we substitute for K reforming the equilibrium constant for the reforming where we have just now seen it is the ratio of the partial pressure for CO for hydrogen raised to the power 3. Similarly for methane reactant side the two reactants and steam. Now if we substitute these values now the standard pressure is taken as 1 bar. If we substitute all these values from the in terms of the mole fraction which is nothing but yi times the total pressure the denominator for each of these is 1. So partial pressure for CO is given by the mole fraction for CO times the total pressure that is 20. Similarly the mole fraction for hydrogen into 20 raised to the power 3 mole fraction for methane, mole fraction for hydrogen again we can take out that 20 out. So that remains is 20 to the power 2 and then the corresponding mole fractions for CO, hydrogen, methane and water. Let us substitute those values. So mole fraction for CO we can use from here y divided by 10.09 minus 6x minus y for the CO for hydrogen 8.02 minus 8x minus 2y divided by 10.09 minus 6x minus y whole raised to the power 3 stoichiometric coefficient. Denominator for methane x divided by 10.09 minus 6x minus y and the last term for water this is given by 1.04 plus 2x plus y divided by 10.09 minus 6x minus y. Now this we need to solve this is one equation that we will be getting we can further write it as we can simplify it to write this is whole cube divided by the denominator x 1.04 plus 2x plus y and 10.09 minus 6x minus y whole square. So this is another equation which we have let us write it as equation number 4. Same way we will be using for the shift reaction the equilibrium constant for the shift reaction there we have written the partial pressure for CO2 for hydrogen with the divide by the reactants partial pressure for CO and water and this value we have already calculated to be 0.7901. Now let us again substitute the value for the mole fraction. So partial pressure for species i is nothing but mole fraction times the total pressure. So that is the speech 20 which is the total pressure times the individual mole fraction. Now again we will use all these mole fraction from the table. So for CO2 the mole fraction is given by this let us substitute that. So for the CO2 the mole fraction was 1.02 minus x minus y divided by 10.09 minus 6x minus y. For hydrogen it is 8.02 minus 8x minus 2y divided by the same denominator we will substitute that 8.02 minus 6x minus y divided by 10.09 minus 6x minus y. Similarly for CO this is y divided by the same denominator 10.09 minus 6x minus y and for water that is 1.04 plus 2x plus y and the denominator is same. We also have total pressure for both the terms which gets cancelled out and that gives us 1.02 minus x minus y, 8.02 minus 6x minus 2y divided by y 1.04 plus 2x plus y and that is equal to 0.7901. So these are the two equations, the equation number 4 and equation number 5 that we need to solve simultaneously. So the two equations that we need to solve is y 8.02 minus 8x minus 2y whole cube 410.09 minus 6x minus y whole square x 1.04 plus 2x plus y and that value was equal to the reforming constant that was 1426.957. This is the equation number 4 rewritten here 1.02 minus x minus y, 8.02 minus 8x minus 2y divided by y times 1.04 plus 2x plus y and that value is equal to 0.7901. So that is again the equation number 5 being rewritten. Now we can see that these are polynomial equations that with fractions and that needs to be solved to find for the x and y. So we have two equations and two unknowns. So we need to solve that, it is little difficult to solve it. We can use excel to solve and find out, we can use solver to solve and find out and if we solve that we can find out that the value of x comes out to be 0.18335 and the value of y comes out to be 0.64025. So these can be easily solved by putting a value for x and y and then later on changing that value of x and y so that we can get very closer to these solutions. Now if we put these values of x and y which we have just now found using these equations, in that case we can write down for the different species the amount of amount entering, amount leaving and the mole fraction for the various constituents in the reaction. For methane let us say it is 0.95 that was given, if we substitute the value of x, it is 0.1833 and then we can find the later the mole fraction. For ethane that initially was 0.03 mole percent and that completely reacted so the at the output the mole fraction was 0. CO2 it was given to be 0.01 and substituting the values we have found it is 0.19639 for nitrogen this remain unreacted to 0.01 for water 3.06 we have found and then substituting the different values we get 2.04696 the values of x and y in the earlier table for CO initially it was 0, hydrogen it was initially 0. Now substituting x and y we can get it to be 0.64025 for hydrogen 5.272 and then if we do a total of it then at the exit this comes out to be 8.3496 so mole percent divided by the total number of moles will give us the mole fraction. So if we divide 0.1833 for methane with the total number of moles we get it to be 0.02195 this is the mole fraction of the unreacted methane in the outlet gas stream same for carbon dioxide it is 0.02352 that is the carbon dioxide left after the reaction for nitrogen for water 0.245 for CO 0.07668 and hydrogen 0.6315. So this is the mole fraction or this is the mole percent in the product gas this is the gas composition which was required in the problem and this is the corresponding mole fraction of each of the constituent present in the outlet gas stream. So this is one of the problem a little detailed one that we have solved today. Thank you.