 Welcome, welcome again for the another module for this particular portion of the course. So we have looked into the hydrogen energy being produced from coal and biomass and the technology behind it. So we will just look into some of the tutorial problems so that to get a hang of how to convert some of the numbers, some of the figures, efficiencies and hydrogen production numbers and all. So we will just have couple of tutorial problem and then you can practice more at your own. So first problem that we will discuss now will be the, if we are given an ultimate analysis result of the biomass sample, then we how we can calculate the empirical elemental formula. So what exactly is an element means ultimate, so what exactly is the ultimate analysis result. So ultimate analysis is done when you give any hydrocarbon or a solid or a liquid fuel to the sort of a system and it gives you how much is the mass fraction of different elements. So carbon, hydrogen, oxygen are the major components. So typically for biomass you will also have nitrogen and sulphur. So just for the keeping it a bit simple, I have not included nitrogen and sulphur in this. But if we just assume that our biomass is having only carbon, hydrogen and oxygen, then the machine or the instrument will give us the result in this form, so carbon and hydrogen and typically most of the machines which do ultimate analysis gives the oxygen as balance. So oxygen typically is not detected but indirectly whatever the numbers you get mass fractions of the different element you just subtract that from 1 you will get the oxygen mass fraction. So this carbon 45%, hydrogen 6% and oxygen 49%, whatever the number you are seeing here is actually the mass fraction and that is what the ultimate analysis of a solid fuel whether it is a coal or a biomass that is what we get through the ultimate analysis. Now our task is to calculate the empirical elemental formula. So this is something like this you will have something like C, X, H, Y, O, Z. So similar to what you see a typical hydrocarbon or any oxygenated hydrocarbon formula, but typically all the chemicals which has our standard formulas are molecules or the elements of a similar type. But here what we are looking into is when we have a heterogeneous compound like biomass or a coal which will not have any single chemical species but mixture of hundreds or even thousands of different chemical species. So in that case for a simplification for many analysis this empirical formula helps us for simple simplification we get a sort of a uniform empirical formula for a given biomass or a coal that gives us how much of moles of hydrogen or oxygen present per unit of carbon in that particular sample of wood or a coal and that we can use it for many of our studies especially when we are looking into the kinetics reactions and the output and how much of extensive reactions of water gas or boardwalk or water gas shift reaction etc they will go. So it helps us so we will just see how we get this. So we have another tutorial problem 2 we will discuss after this. So if we are looking for the solution here so let us say solution of tutorial problem 1. What exactly we are getting is we are having carbon as 45 percent, hydrogen as 6 percent and oxygen as we have got around 49 percent. So now if we put the molecular mass of or molecular mass of individual species I had it so for carbon it is 12 so we know that this we can get it from the data book hydrogen it is 1 and then oxygen it is 16 and this is in grams per mole or kg per kilo mole. Now to start with what we can give is to start with we have to assume that let us say we have 100 grams of biomass sample 100 grams of biomass sample then in 100 grams of biomass sample and all these are mass fraction so all these are mass fraction that means you have carbon as 0.45 into 100 that means you have 45 grams and for hydrogen you have 6 grams for oxygen you have 49 grams. So now what we can get is how many number of moles first we have to find out because what we are getting into is so what we have to find out we have to find out in this particular format and this is basically on the mole basis. So mole basis so we have to convert the mole basis to mass basis to mole basis. So how to do that so we have got this number of grams and we know how much is the molecular mass of each and every species. So carbon will be whatever amount that we have divided by the molecular mass of this particular species okay so 45 by 12 and it comes out to be 3.75 moles why because it is 12 gram per mole so 12 gram per mole gives us for 45 grams it gives us 3.75 moles so similarly for hydrogen it will be 6 grams divided by 1 gram per mole so it will be 6 moles okay. So now you can see you can see that number of moles of hydrogen is more than the number of moles of carbon but from the mass fraction data from mass fraction data it looks like hydrogen is a very small component it looks small in this number because it is molecular mass is less. So it gives a very wrong intuition so that is why this elemental analysis or this getting the empirical formula helps us and that is what we are now doing. So for oxygen similarly we will get 49 divided by 16 which is its molecular mass and we will get that as 3.062 moles it will be 3.062 and how many total number of moles total number of moles will come as 12.812 moles in total so this is total number of moles. Now it is very simple we know how many number of moles are there in the 100 grams of the sample and we know individually how much moles of how many moles of carbon, hydrogen and oxygen. So simply what we have to do we just have to calculate the mole fraction and mole fraction we know how much how to calculate so it is simply the total number of moles which we have and we know how many of the carbon moles. So carbon moles are 3.75 total number of moles is 12.812 so we get the mole fraction as 0.292 so 0.292 or 29% is the mole fraction so it does not have any unit it is just the fraction. Similarly for hydrogen it will be 6 by 12.812 and it comes out to be 0.468 and similarly for oxygen it will come out to be 3.062 by 12.812 it will come out to be finally 0.239. So what is the empirical formula that we get? Similarly we can just write C 0.292 hydrogen 0.468 and oxygen 0.239 but looking at this empirical formula it also looks a little bit means uncomfortable so what we have seen here C x o z looks a little bit uncomfortable because of this some fractions of number. So the easier version is we calculate for 1 mole of carbon or 1 atom of carbon how many hydrogen and how many oxygen so and simply what to do we just have to divide all this number by the fraction of carbon sorry 292 and what we get here is C we get here is CH 1.6 o 0.82. So this is what is our empirical formula for given biomass sample okay so this is simply we can write that that is CH 1.6 o 0.82. So and it looks quite different from what we have seen in the mass fraction because of the different molecular mass of the carbon hydrogen and oxygen. Similar is when we get into hydrogen, sulfur, phosphorus and many other compounds elements which are very much present in any of the biomass. So similarly similar exercise can give us that. So this is the solution for the our first total problem. So now moving to the second problem. So second problem is also little bit interesting from the perspective that when we convert any biomass through gasification process whether it is a coal or a biomass we convert how to get the output numbers how to convert it into efficiencies conversion efficiency of hydrogen and energy. So this particular exercise or this understanding we will get it from when we solve this tutorial problem okay. So let us see what is the tutorial problem. So it says a biomass gasifier is designed for conversion of 100 kg of dry biomass it is working with air as a gasifying medium. So here it is air as a gasifying medium means it is air gasification problem okay. So air gasification problem and it generates 250 meter cube of producer gas which we call as producer gas which is the product gas of the gasification process. So we have 100 kg per hour of dry biomass being fed and we are getting 250 meter cube of producer gas and this is at normal temperature and pressure okay which is your 1 bar and 25 degree centigrade or typically we also denoted as normal meter cube. So this is 250 normal meter cube of producer gas and now the gas analysis result of that producer gas. So gas analyzers give us the data and that is the volume fraction okay. So all these data of gases are the volume or mole fraction. So volume fraction and mole fraction are interchangeable. So the analyzer gives us the volume fraction. So hydrogen is 20% by volume, carbon monoxide 20%, CO2 12%, methane 2% and rest is nitrogen because we are using air as a gasifying media. So there will be bulk of nitrogen also. Now the question asks how many kgs of producer gas will be produced per hour? How many grams of hydrogen will be produced per kg biomass? And what is the hydrogen conversion efficiency? Means we have seen lot many efficiencies numbers and all. So here the hydrogen conversion efficiency means if you have hydrogen and that we see how many grams of hydrogen it was sorry. So it was 6%. So 6% means if you have 1 kg biomass it will have 60 grams of hydrogen in it okay. So out of 60 grams of hydrogen how much you are able to convert it into elemental hydrogen or the H2. So that is the H2 conversion efficiency. And then last one what is the energy conversion efficiency? And it is very simple energy conversion means if you have a solid fuel which you fed to the gasifier and you get the gaseous fuel. So how much of the energy or the chemical energy you have been able to conserve when you convert it from solid fuel to gaseous fuel through gasification process okay. And yeah so one number missing that is the calorific value of biomass is 17 mega joule per kg. So on dry basis okay. This is on dry basis. So now let us go one by one. So how many kgs of producer gas will be produced per hour? So we have given the volume fraction. So here solution for 2 okay. So here what we are given? We are given the volume fraction. So 100 kg per hour of biomass that we can denote it as BM okay is giving us 250 meter cube of producer gas okay for PG. That means 1 kg of biomass is giving us 2.5 meter cube of producer gas. Now this 2.5 meter cube of gas and our question is how many kg okay. So for this what we need to know? We need to derive the density okay. So we will find all the density first and then straight away dividing it with this volume we can get the mass of the gas. So now first we need to get the molecular mass of producer gas. Again it is a mixture of different gas species but the common if we get the molecular mass just like we calculate for air and we want to calculate for producer gas it will be nothing but the summation of all the volume fraction of all the ith species of gases, molecular mass of the ith species of the gases okay and this we can get. So what we have is X of hydrogen into molecular mass of hydrogen plus volume fraction of CO into molecular mass of CO plus volume fraction of CO2 into molecular mass of CO2 plus so we can just put it in bracket to avoid confusion. So plus volume fraction, so volume fraction is denoted by X so of methane into molecular mass of methane and plus we also have nitrogen that is molecular mass of sorry volume fraction of nitrogen into molecular mass of nitrogen and we have and we know all the numbers for this we can get it from the database if you do not have. So volume fraction of the hydrogen X hydrogen is given as 0.2 so 20% here we have to use the number not the percentage and X of CO is given as 0.2 X of hydro sorry CO2 is given as 0.12 or 12% and volume fraction of methane is given as 2% so 0.02 and rest is for the nitrogen and that is 46 okay and then you put the molecular mass of all. So finally you will get the number as 24.48 gram per mole okay so this is what you get as the molecular mass of the producer gas. So once you get the producer gas molecular mass so density how it is given we know that 1 mole of gas by Avogadro number or that law it occupies 24.4 liters of volume okay so 1 kilo mole will occupy 24.4 meter cube of volume. So when we look into density of producer gas it will come as molecular mass of PG which is this molecular mass of producer gas divided by 24.4 so this is again at the normal temperature and pressure 1 by 25 degree centigrade so this comes out to be 1.003 kg per meter cube okay. So you can just cross check it molecular mass is kilogram per kilo mole okay so here this is kg and kg we can convert that and this is nothing but your meter cube okay meter cube per kilo mole. So it comes out to be kg per meter cube that is the density of the producer gas. Yeah so what we have got is the density of the producer gas so density of the producer gas so we can get the overall mass of the producer gas produced per hour. So our question was how much is the mass of producer gas generated per hour so it is simply we have to multiply the volume flow rate multiplied by density so that was 250 meter cube multiplied by 1.003 kg per meter cube and this was meter cube per hour. So here we got 2250.75 kg per hour of flow rate of producer gas. So here we get the hour answer for the first question so answer for the first question is 250.75 kg per hour. So now coming to the second question second question says how many grams of hydrogen grams of hydrogen will be produced per kg of biomass. So here we have to see how much we have got into the this much is the mass flow rate. This we can also say that mass flow rate of producer gas and when we multiplied by the wall mass fraction of hydrogen mass fraction of hydrogen is still unknown to us but we can find out. So then what we will get will be the mass flow rate of the hydrogen or which is basically the mass of hydrogen generated per hour. So here we get the so to find out our second part of the problem is how much is the mass fraction of hydrogen that we have to find out that is generated by YH2 and this we get as mole fraction of hydrogen into molecular mass of hydrogen divided by our molecular mass of producer gas. So we know all the three values and we know that this is 0.2 or 20% is the mole fraction or the volume fraction multiplied by 2 is the molecular mass of hydrogen that is diatomic hydrogen not the single atom diatomic and we know also the molecular mass of producer gas that we have just derived is 24.48 kilo mole per kg sorry kg per kilo mole. So this we get as 0.01634 and which is 1.634% so this is our mass fraction of the hydrogen and from this we can we can get how much is our hydrogen generation. So our hydrogen generation per kg sorry generated per hour is our producer gas generated per hour multiplied by the mass fraction of the hydrogen which we have just calculated. So this will be 250.75 which is the mass flow rate or production rate of producer gas per hour multiplied by 0.01634 which is the mass fraction of the hydrogen and this will be 0.097 kg of hydrogen per hour or 4097 grams of hydrogen per hour and then we have to get how much is the hydrogen produced per kg of biomass. So we have to divide this overall flow rate I mean production rate of hydrogen per hour by the biomass consumption per hour. So this will be 4097 grams of hydrogen per 100 kg per hour of biomass consumption. So this is hydrogen per hour and this is 100. So this comes out to be 40.97 or roughly 41 grams of hydrogen per hour of operation. So this is our how much is the hydrogen that we have got. So this is interesting to know and here you will get lead to the third part of our problem it's also. So our second one how many grams of will be produced per kg biomass. So this is roughly 41 grams of hydrogen per hour and then we are looking into hydrogen conversion efficiency. So hydrogen conversion efficiency how much we are having the this thing hydrogen into our biomass hydrogen into the biomass is given by this value 6% and that we have got is 60 grams. So this will give us the input of hydrogen from the system and that we can use to get the output conversion efficiency. So our third part of the problem hydrogen in biomass is 6% by mass and that is equal to 60 gram of hydrogen per kg of biomass. So this is our input and our hydrogen in product or in the output producer gas is 41 or we can just write the accurate value 40.97 grams of hydrogen per kg of biomass. So this is our input and this is our output and we know efficiency or conversion efficiency will be the output ratio of output to input. So our hydrogen conversion efficiency will be 40.97 divided by 60 and that comes out to be 0.683 or 68.3%. So this is interesting to know that 68.3% of the hydrogen that is present in the biomass is being converted into elemental hydrogen and where the rest of the hydrogen goes it goes in form of methane and H2O. So methane we have seen that it is also like 2% was in the producer gas. So that is also counted as our hydrogen that comes out in the useful form that is fuel. So hydrogen elemental as well as methane both are fuel but here H2O it gets condensed out. So you would have seen that in our input this thing gas H2O is not there. So H2O is not there because this is the composition of the cold gas where H2O has been condensed out and is taken out from the system. So but the rest of the hydrogen goes in the form of H2O that we have to keep in mind. So now coming to the last part so here we will just write the answer also here. So this we have got 68.3% is our conversion efficiency. So last part is what is the energy conversion efficiency of the gasifier. So again energy conversion efficiency and what we mean by energy conversion efficiency here is that it is what we have seen in the biomass gasification what we have seen in our biomass gasification lectures that it is a chemical conversion process. So what we are doing is we have a solid fuel that is biomass or mixture of varieties of oxidated carbohydrates and that is being converted into gaseous fuel. So how much energy is being conserved in the overall conversion process of biomass gasification and that we have to find out. So again very simple ratio of output energy to the input energy and that output energy we have to find out input energy is given as 17 mega joule per kg on the dry basis. So our calorific value of the biomass is given as 17 mega joule on the dry basis we have to find out the output sort of energy that comes out of the system. So that we will find out. This is the calorific value of producer gas into producer gas production rate in meter cube per hour to the calorific value of biomass in mega joule per kg into consumption rate of biomass consumption per hour. So here first we have to find out the calorific value of the producer gas. So calorific value of producer gas or in short we can write it also as Cv of producer gas will be is calculated as summation of the calorific calorific value of the every fuel element I multiplied by the mole fraction of that ith element in the gas species. So here it will be our Cv of hydrogen into mole fraction of hydrogen plus calorific value of CO into mole fraction of CO plus calorific value of methane multiplied by mole fraction of methane. Only three gas was hydrogen CO2 and nitrogen is not contributing to any in the any form of fuel in the producer gas. So this we have to find out and if we look into the data table to find out the what is the calorific value we will get that hydrogen has a calorific value of 10.8 mega joule per meter cube. CO has 12.6 mega joule per meter cube and our methane has 35.8 mega joule per meter cube. So if we put it in this formula we will get the overall calorific value of producer gas will be 5.396 mega joule per meter cube and if we substitute this value here we will get the conversion efficiency as 5.396 into 250 is the production rate of producer gas divided by 17 is the calorific value of biomass and 100 kg is the consumption rate. So we will get as 79.35%. So this is our conversion efficiency. So here conversion efficiency you can see it is almost 80%. So almost 80% because it is not a thermodynamic cycle. So typically for mechanical engineering student would be little bit sort of very much surprised that how much we are getting as such a high efficiency but yeah the chemical engineering student will be seeing that okay chemical conversion process 80% is a good enough of a efficiency. So here we are just converting sort of a calorific conversion, conserving the energy of hydrogen carbon in from one form in the biomass to the gaseous form of carbon monoxide, hydrogen and methane and whatever the overall heat losses and then there will be some sort of entropy losses also. So here the second law of thermodynamics plays a role in terms of how one chemical or one species getting converted into other species and how it overall changes the pressure, temperature and the overall number of moles inside the system that changes the entropy of the system and that is one of the losses. So apart from your heat losses and the hot gas that is coming out that is also heat is also not of a useful thing for us if it is not a useful that is also loss and some entropy losses. So overall 80% will be roughly conserved in the overall conversion process. So it will come out 79.35%. So here we get our answer to the final sort of a problem. So here we would like to sort of conclude our tutorial session but again we are talking about the air gasification into this thing and as we discussed in our lecture we also have oxy steam gasification gasification and this oxy steam gasification will have extra input for the energy input into the system in terms of oxygen just separation from the air it will involve energy and then with steam generation steam generation is very quite energy intensive process. So this will also be involved but what you will see the hydrogen conversion efficiency will be more than 100%. So typically in our oxy steam gasification oxy steam gasification typically will give around 100 grams of hydrogen per kg of biomass. So 100 grams that is 40 grams surplus. So our input was just 60 grams of hydrogen in the biomass and rest of the 40 grams will be coming from this steam. So this steam will contribute 40 grams of hydrogen into it. So that is how our oxy steam gasification helps in enhancing the overall hydrogen. We have not given this because it involves a little bit more complexity you can just look into how much will be the separation energy input for oxygen and for steam generation at a given temperature. So overall efficiency will be less overall energy efficiency energy efficiency is less for the oxy steam gasification because of lot of energy that goes into oxy steam generation but our hydrogen production hydrogen production is on the higher side that is the difference. And same philosophy applies for the coal systems also whether it is getting into the ultimate analysis results of the coal. Coal is also having little bit of hydrogen, little bit of oxygen. Many of the coal will have negligible amount of oxygen but hydrogen will be there in some amount and also as we discussed earlier the hydrogen which looks like 6% here by number of moles it is quite significant. And similarly for coal the coal to hydrogen is typically by the oxy steam gasification process where we get most of the hydrogen from the steam itself because the coal will have very less amount of hydrogen. Typically like in Indian coal it will not be more than 20 to 30 grams per kg of biomass less than half what you will get in the biomass. So with this tutorial we will end the today's lecture. Thank you.