So, today we will have 2 hours of lectures on combustion. So, before we start on combustion I mean this is some subject which is slightly peripheral to thermodynamics it will be used mostly when you are doing applied courses like power plant engineering or IC engines, but there are a few concepts that probably need to be reiterated. I mean many of the concepts are related to chemical reactions and there is of course the concept of adiabatic flame temperature which must be told and if this is known and reasonable knowledge is given I think it will be very useful in the future courses where power plant engineering or IC engines is taught. So before we begin combustion let us see what fuels we normally encounter and we will realize I mean I would not spend too much time on fuels. So, as engineers or as mechanical engineers we will encounter solid, liquid and gaseous fuels. So, the solid fuels are mostly used in external type of combustion engines. So, this would be you know most of the plants which are run on Rankine cycle they would use for example, coal as the main fuel of course there are lot of plants nowadays which use biomass and garbage, but coal is the most standard fuel that we use in our power plants and maybe you know some background students already have you know that coal is formed due to high pressure on plants and animals which have been buried for long and there are various forms of coal which are you know in the formation region like peat or lignite and lignite is a reasonably well formed coal that is used often and the most common coal that we use in Indian power plants is bituminous coal. One thing you can emphasize is that Indian coal is actually has a lot of big percentage of ash and the better coals are what are called as anthracites, but these are mostly used for metallurgical processes where pure coal is required otherwise in power plants we normally go for bituminous coal. So, if we come to liquids you all know that crude oil is obtained from various oil fields and this crude oil is later refined and undergoes a process called fractional distillation at different temperatures different fractions will liquefy out and finally, whatever remains in the gaseous form will probably be there at in gaseous form at room temperature and you know that most of the fuels that we encounter are going to hydrocarbon. So, coal is more or less purely carbon with some ash and other things, but most of the liquid and gaseous fuels are purely hydrocarbon and anything with more than C 5 is usually would be and above would be in liquid form. So, for example, you know what is normally called as gasoline or petrol the most common compound which is associated with gasoline is iso octane, but in general petrol consists of everything between around C 5 to C 9 a huge range and the iso octane is one of the biggest components that even thing goes in there. However, it just mixture of many compounds if you go slightly higher towards C 10 you get what is called as kerosene or sold as kerosene and if you go slightly higher towards C 12 you get what is called as diesel and if you go even higher you know the liquid starts becoming thicker and these are then no longer used in iso engines and used in furnaces and you get what is called as furnace oil you know close to C 20. And beyond this if the hydrocarbon as per molecule more than 20 carbon atoms the thing starts becoming stickier and thicker and you know at very high carbon content C 60 and all you will get what is called as star. So, these are the things that we obtain from crude oil and you know most of these are used in either iso engines or power plants. So, this probably needs to be told and finally, we have gases. So, you know anything below C 5 would be a gas. So, for example, you know butane and propane these are gases that remain after fractional distillation and butane and propane together are liquefied and sold as LPG that is liquefied petroleum gas. It is a combination of butane and propane and may be some ethane in there and when we come to methane it is mostly sold as natural gas that is because in most marshes etcetera the gas which comes out naturally after fermentation is methane and hence by default methane is called as natural gas. So, whatever is sold as natural gas is actually around 90 percent methane and the remaining could be ethane and butane and so on. So, and natural gas is used in CNG engines etcetera. So, this is roughly how the fuels would be and of course, you I think once this background is given regarding fuels we can come to the combustion of these fuels and I think here you must remind students that you know mostly it is finally a chemical reaction and they have learned chemical reactions in eleventh and twelfth standard and the first thing that we can probably tell is balancing of reactions. So, the simplest species to take normally and that is what we would start with is start with methane. Methane is C H 4 we require some amount of O 2 let me not put how much and I will get C O 2 plus H 2 O. So, this is when you can tell them to start balancing this reaction there is they know what balancing means and basically every element on the right hand side and left hand side has to be balanced and the reason is that we are not looking at nuclear reactions where atoms of one element will transmute into atoms of other elements. So, every element the atom should be balanced on either side. So, the carbon is one on the left side it is one on the right side. So, that is pretty good hydrogen you know there are four on the left side two on the right side. So, if I just add two here it will be four. Now, if I look at oxygen C O 2 consists of one O 2 and twice H 2 O will have another two O's. So, totally I require four O's on the left side. So, I will just put here too. So, this is the simplest balancing that I could think of, but here you must now tell that know whenever we burn fuel and we require some oxidant and we have just put oxygen here and of course, it must be emphasized that fuels need not be only hydrocarbon I mean rocket fuels do not consist of hydrocarbon and oxidants need not be purely oxygen I mean they can consist of other elements, but as far as we are concerned we look primarily at hydrocarbons and oxygen because this is what is used in all the heat engines that we operate and this is what is more important for us as mechanical engineers. So, oxygen is not just available directly, but it is available only in the form along with nitrogen in air and air is the most common oxygen that we will always use. So, at this point you must emphasize that all our calculations for IC engines power plant etcetera will involve the presence of nitrogen and it is very important to account for this nitrogen because it affects the net temperature of whatever we are using and at this point you can say see all these are on a molar basis here because I am saying 1 C H 4 means 1 mole of C H 4 twice O 2 would mean 2 moles of O 2, 1 C O 2 would mean 1 mole of C O 2, 2 H 2 O would mean 2 moles of H 2 O. So, in the atmosphere you know by volume there is around 21 percent oxygen and remaining 79 percent of nitrogen. So, moles and volume correspond to each other because equal number of moles occupy equal volume irrespective of the species. So, that means for 1 mole of oxygen there are 79 upon 21 moles of N 2 which roughly turns out to be 3.76 moles of nitrogen. So, hence what you must emphasize is that this reaction though is not was initially written only as a balanced chemical reaction as far as we are concerned it must be written as such C H 4 plus twice O 2 plus twice into 3.76 N 2 you see O 2 plus twice H 2 O the nitrogen does not take part in any reaction and I will just write it as twice into 3.76. So, on both sides I have added nitrogen and it is remaining unburned. Of course, you will realize that when we talk of very high temperatures we may have to account for dissociation of nitrogen or nitrogen reacting with oxygen, but we will keep it simple now and say that nitrogen does not react and we will add it on both sides and we will see that this is especially needed the nitrogen in calculating stoichiometric air fuel ratios and also the adiabatic flame temperature. So, this is what we write. So, balanced reaction we have balanced every atom of an element on either side. So, you also realize that if the same number of atoms of an element are on both sides the atoms are not going to lose any mass. So, the mass on the left and right side will also balance out. So, there is also a mass balance again this is not a nuclear reaction in a nuclear reaction you would have lost mass and you would have gained energy due to the loss of mass. Here you do not gain energy because of loss of mass, but purely chemical reaction. So, there is mass balance and mass on either side is the same. Now, you must emphasize that this does not mean that moles will balance. So, moles do not balance or need not balance as it was in some cases they will. So, for example, in the reaction that we have just written here on the left side there is 1 mole of C H 4 plus 2 moles of oxygen and plus 7.52 moles of nitrogen. So, this makes it 9.52 moles and this makes on the reactant side. If I look at the right side I see that there is 1 mole of C O 2 twice 2 moles of H 2 O and still 7.52 moles of nitrogen and I have 9.52 moles on the product side also. But if I choose a different species. So, for example, I can choose a Heptane in C 7 H 16. So, I will just write oxygen plus nitrogen giving. So, you will see that I will get 7 CO 2 plus 8 H 2 O. So, I will require 7 of O 2 here and 4 of O 2 here. So, I will have 11 O 2 here which means this is 11 times 3.76 which means I will have 11 point sorry 11 multiplied by 3.76 H 2. So, if you see the left side the reactants you will realize that there is 1 plus 11 plus 11 multiplied by 3.76. I will not do this. So, I will have to do this multiplication is just 12 plus 11 multiplied by 3.76 and products there is 7 plus 8 15 plus 11 multiplied by 3.76. So, you see that product side there is 3 moles extra because this is only 12 plus 11 multiplied by 3.76 where as it is 15. So, the number of moles will not balance that is because atoms of different elements combine and form bigger or smaller species. In this case Heptane was a very big species and that was just one mole of Heptane, but we found far more moles of carbon dioxide and water and hence the moles do not balance, but of course you will see that atoms of every element have to balance and the mass has to balance. So, as long as this is told I think we can proceed further. What we would do next is that you emphasize the fact that we will always be using 3.76 as a factor for nitrogen and this is something that they must remember now that 1 mole O 2 corresponds to 3.76 moles of N 2 in air. This is assuming air is being used and which is what we will always be using and which means that 1 mole of oxygen corresponds to 4.76 moles of air because air is composed of those 3.76 moles of nitrogen plus the 1 mole of oxygen. So, for every mole of oxygen there is 4.76 moles of air. So, this is something that probably needs to be emphasized this is a number that is you know people expect that you would often remember as far as you know mechanical engineers are concerned whenever we are going to use air the 3.76 is a number that we often remember. So, once this is done what we can next go to is again a concept which is known to students which is what is called as an endothermic and exothermic reaction. So, basically if there is a reaction A plus B giving C plus D and if C A plus B is that some standard state and again you can emphasize and we will keep on emphasizing that first we will define something called as a standard state and the standard state will always be 1 atmosphere 25 degree C or 298 Kelvin. So, let us say A and B are at standard state and you they react and form C and D and C and D are brought back to standard state you will realize that to carry out the reaction at least some energy was required or some energy is liberated. Now, if energy was required to get C plus D it means that you have to provide energy and that energy is absorbed during the reaction and it now is belongs to C and D. So, that means that the energy I will put the symbol U of C plus D minus U of A plus B which is energy of C plus D minus U of A plus of products minus energy of reactants this is greater than 0. If this is greater than 0 the products are at a higher energy than the reactants and this means that the reaction needed extra energy this is what is normally called as endothermic reaction and I mean it is something that everyone is aware of, but it is just that we are emphasizing it again and vice versa if you know energy of products is less than energy of reactants then this will be an exothermic process that is because now the energy will be released out the products are at a lower energy and normally people show it like this A plus B is here and either C plus D is here or C plus D is here.  Now you realize that all reactions take place at constant pressure in fact this is what we maintain either in a combustion chamber we maintain the pressure constant or we let things happen in the atmosphere. So, the reaction is at atmospheric pressure and hence the expansion work is inbuilt. So, things are happening at constant pressure and hence what happens is that rather than working with you everyone works with H that is because the constant pressure expansion work has to be accounted for and inbuilt into whatever we are using. So, instead of using you you will see that people will use H. So, H products minus H reactants is what is called as the delta H of the reaction or the heat of the reaction in olden days. So, we will call it as the enthalpy of the reaction. So, if delta H R is negative then you know that the products had a lower energy and this is an exothermic reaction delta H R greater than 0 is an endothermic. So, this is just the reaction. Ensuring that now everyone knows we are working in enthalpy and what the heat of reaction means in this case. So, I will give just an example we will write for example, carbon in some grapitic form plus oxygen in gaseous form gives CO 2 I am not including here any time now nitrogen or anything. So, the heat of reaction here delta H R is minus 393.5 kilo joule. So, this means that the products have a lower energy than the reactants and this is an exothermic reaction and 393.5 kilo joule have come out as a result of 1 mole of carbon burning up with 1 mole of oxygen giving 1 mole of CO 2. So, the thing that must now be told is that whenever we are using fuels fuel plus oxidant better be an exothermic reaction otherwise there is no point in using it as a fuel we are using something as a fuel only because it will burn and release energy. If it is going to absorb energy then it is not a fuel maybe we are using it for something else, but if something is a fuel then it better combust or react with the oxidant and release energy. So, carbon all hydrocarbons react with oxygen and they will release energy and hence this all these set of compounds are all classified as fuels. So, fuels are anything which will react with the necessary oxidant and release energy that means they will go undergo an exothermic reaction process. So, that is the key to being called a fuel otherwise you cannot call it a fuel. So, once this is done here again I must emphasize that this delta H R is obtained when both these are evaluated at the standard state. So, you have to get this back at 1 atmosphere and 25 degree C this is initial this is also at or you should get it back to the same temperature, but all tabulation is always done at the standard state and this delta H R will be given a symbol delta H R naught that is this is the enthalpy of reaction at the standard state. If we had carried out the reaction at another temperature that is both the reactants and products were at a different temperature then sorry I mean they are different than 25 degree C, but they are equal to each other. It would have been the enthalpy of reaction at that temperature not the standard temperature, but then we will normally never tabulate such values all tabulated values are always at the standard state and it is assumed that both the reactants and products are brought down to the standard state. So, whatever enthalpy of reaction is there this is once the enthalpy of the product is calculated at the standard state. So, once we get this thing straightened out the other thing that you can tell is that you know there was there is something called as the Hess law or the Hess's law and this had a reasonable and it was it made sense to teach this earlier it just said that if I needed the enthalpy of a particular reaction or earlier it used to be called the heat of a reaction. If I did not know or if I could not get that reaction straight in a straight forward manner I could use other reaction to get the heat of reaction of an unknown reaction for example, you know C plus O 2 gives CO 2. I can burn carbon with oxygen and get carbon dioxide and you know the heat of reaction for this is easy to obtain, but if I wanted the heat of reaction for C plus O 2 giving 2 CO. So, that means this should be 2 here the heat of reaction for this is very difficult to calculate that is because if I try to burn carbon with oxygen and try to obtain CO it is very difficult even if I try my best to ensure that complete combustion does not occur because some CO 2 forms by default. So, I cannot figure out whether the energy is entirely because of reaction which leads to CO whereas, I can really try to ensure complete combustion. So, CO 2 forming is far more easier. So, in this case we can use the fact we can use Hess's law as follows C plus O 2 this is bound to be an exothermic reaction I will go lower here and I will get CO 2 and the delta H R as I said earlier was 393.5 kilo joule. I need to know this that is C plus O 2 giving I will put it as C plus O 2 giving CO plus half O 2. So, I will balance it out like this because I want to start with the same set of reactants. So, CO is formed so half O 2 is remaining. So, you can think of it in terms of moles where you say one mole of C and one mole of oxygen were there. So, one mole of CO have formed you have utilized all the CO and half mole of oxygen is remaining, but as I said you know so this is how the reaction is. I say now I have gotten this I do not know what is the heat of reaction here, but if I ensure the reaction of CO plus half O 2. So, I get hold of CO and I burn it with oxygen I will get this CO 2. So, I just do this reaction CO plus half O 2 and I will carry out this reaction and here since this is now entirely the end product is guaranteed to be CO 2. You can easily measure the heat of reaction or the enthalpy of reaction and it will be 2 minus 283.0 kilo joule. So, now I had to start with one mole of CO C and one mole of O 2 I could get CO 2. Now if I wanted this I could not do it directly, but I realize this is enthalpy that is the state function I when I subtract the enthalpy of the reactants on the product I get my enthalpy of reaction. So, here all I had to I realize is that you know if I went in steps finally the net enthalpy change should have been the same. So, if the enthalpy change here is 283 and the enthalpy change here would only the two enthalpy changes here must add up and give this enthalpy change. So, here it should be delta H R is equal to minus 110.5. So, this is what how Hess's law was used they said that if you could get to a certain set of products in steps and if you did not know the heat of reaction as they called it then of a particular step you could use the heat of reactions of other steps plus the heat of reaction of the net reaction to get the heat of reaction of a particular step. So, this is what we have done C plus O 2 I can go via two steps or I can go via single step for the single step I know the enthalpy of reaction for the two step reaction I can get the enthalpy of reaction for one of the steps. So, I will be able to get the enthalpy of reaction of the other step. Now in all this one must ensure that one is balancing the reaction everywhere and ensure that you are starting with the same number of moles and ending up with the same number of moles of products irrespective of the path you take. So, that balancing must be done and then you can go ahead. So, but then this is you know earlier this was thought over this is a very nice method to get it, but I think now probably it should be obvious because once you know that the enthalpy is the state function is built it is a property of those that gas C O plus half O 2 at one atmosphere and 25 degree C you know what is the enthalpy here you know what the enthalpy here is I know that the difference will directly give me this I know the difference will directly give me this. So, this addition giving this should be pretty obvious now and this is and now no longer people use this people what they do is they just try to tabulate the enthalpy of formation for each of this species. So, that brings us to something called as the enthalpy of formation. So, what you do is I need the enthalpy of every species at the standard state and this is what is tabulated. So, I need to somehow get this and tabulate this and this is an in fact you would have we have uploaded one sheet on the website noting down the enthalpy of certain set of species. So, what here we have to follow some ground rules and things are standardized here. So, the standards that people employ are they say that let us start with giving reference enthalpy of zero elements occurring in their natural form at standard conditions. So, if we want to assign enthalpies to all elements or sorry all species we will have to come up with some reference or some standard where we decide that you know some guys or some species will have particular enthalpy at a particular temperature we need to do this assignment and then we will find out the enthalpy of other species which are formed from those species. So, the standard or the most reference standard reference that people use is that they assign zero enthalpy to elements occurring in their natural form at the standard conditions. So, what does natural form mean? It means that for example, if I am considering a species oxygen and oxygen occurs as O 2 and not as O. So, O 2 at one atmosphere and 25 degree C is assigned zero enthalpy. So, you notice that the standard or references used in combustion are much different from let us say the references we used in moisture calculation where we use zero degree centigrade as the place where we wanted to set zero enthalpy, but in combustion calculation all the zeroes for naturally occurring elements are set at 25 degree C which is considered as the standard state. So, nitrogen occurring again as N 2 at these standards is assigned zero. Carbon coming in graphitic form graphitic form is considered the standard state. So, this is solid at one atmosphere and 25 degree C this is assigned zero and so on with other species that we will encounter. So, we are talking only of elements here. So, now if the elements occur in non-standard form for example, O or N then we come up with an enthalpy for these species, but we use a regular reaction to come up with an enthalpy for this. So, for example, I will just now write down I do not have the value here, but let me just tell you that I have oxygen giving 2 O. So, let us say that I have obtained the species using the standard reaction and let me see where I have the value for this. I have the value for let us say H 2. So, H 2 again here as one atmosphere and 25 degree C will be assigned zero. So, let us say that I need to form from H 2 I have to go to twice H and both of them. So, I now will find out the delta H R for this reaction and it will turn out that I will need to provide energy to actually get this reaction going and that is because I need to break the H 2 bond and form 2 hydrogen atoms. So, if I have to provide energy it is an endothermic reaction and the product will have higher energy than the reactants and the delta H R for this reaction turns out to be 436 kilo joules. So, this is a positive number and what we do now is we say that you provided 436 kilo joules for 1 mole of hydrogen, but what you actually ended up is getting 2 moles of H. So, we say we assign half the energy to each of these moles of H and we define something called as the enthalpy of formation for H or the standard enthalpy for H as 436 by 2 or 218 kilo joule per mole. So, the enthalpy of formation for H is 218 kilo joule per mole. Similarly, you will realize that for oxygen I will need to provide energy I will get 2 O. Let us say I will need to provide x amount of energy I do not have the value here right now, but if I provide x amount of energy then again I get 2 moles of O and I will assign x by 2 as the enthalpy of formation for oxygen and this is all at standard state because the heat of reaction or the enthalpy of reaction is calculated such that all species are brought back to the standard state which is 1 atmosphere and 25 degree Celsius. So, what and then let me see for example, if I now go to a compound let us say I go to H 2 O then the enthalpy of reaction then the enthalpy of formation of the species is defined on the basis of the heat of reaction for forming that species if from the elements which form the species and which should occur in that standard state. So, for example, H 2 O is formed from H 2 and O 2. So, I should consider H 2 and O 2 in their naturally occurring form at standard state I should get the reaction which forms H 2 O and then I should find out the heat of reaction necessary and using that I will get the enthalpy of formation for H 2 O. So, I will write this reaction as such H 2 gas. So, this means 1 mole of H 2 plus half mole of oxygen. So, remember I am putting half mole here we could have always written it as a balanced reaction by saying 2 H 2 and 1 O 2, but if I write this reaction like I get 1 mole of H 2 O. So, this is all 1 mole and these are at 1 atmosphere 25 degree C. So, the elements should occur in the naturally occurring form at the standard state and if I form the species that is of interest and bring it back to the standard state and I find delta H R here the delta H R will turn out to be minus 241.8 kilo joules that means there is a release of energy. So, now I see that H 2 O 1 mole of H 2 O when it is formed you know has to the heat of reaction as minus 241.8 kilo joules and since I have now assigned 0 enthalpy at standard state to both H 2 and O 2 it means that the H of products minus the H of reactants should have given me delta H R. So, now since reactants is 0 H of products is nothing but delta H R and is equal to minus 241.8 kilo joules and I say that H F for H 2 O and I put this dot here to denote standard condition is minus 241.8 kilo joule per mole. Remember that if I had written the reaction as twice H 2 plus O 2 giving me twice H 2 O in a regular balanced form that delta H R for this reaction would have been twice of this number and it would have been 483.6 minus. So, then I would have said 2 moles of H 2 O will require 483 moles or you know this many 483.6 kilo joules are liberated and you will get the same for 1 mole of H 2 O you will have minus 241. So, in this way what we do is for every species that is of interest to us we note down H F. So, for example O 2 N 2 H 2 C in graphite form these are all assigned 0. Then I will find out what is the enthalpy of formation for H 2 O C O 2 C O and any other species of interest. So, for example if you are doing high temperature calculations you will realize that nitrogen will react with oxygen to give N you know nitrogen plus oxygen let me just write here.  So, when reactions involving these are necessary you realize all of these are some kind of compounds they will not have a 0 enthalpy of reaction at standard state, but they will be formed from nitrogen and oxygen and we will list H F 0 for these species too. So, here we are talking only of relevant species to us you will realize that relevant species to us will all be formed from the compounds H C N and O because N and O come from air and almost all our fuels are hydrocarbons. If you are using some other chemistry for some other kind of fuels then you will require the relevant species there and you should have a tabulated form for all those species, but this is what is done standardly. Then you realize that you will require H F 0 for all the fuels. So, there will be H F 0 for C H 4. So, C H 4 is formed from carbon in graphitic form plus twice H 2 will give C H 4 and you will realize that there is a delta H R for this and that delta H R now you assign here to C H 4. Similarly, C 2 H 6 and all other species you can think of using C N and H M it could be alkene, alkene or it could be any of those alcohols or ketones etcetera, which would involve oxygen, but they will be formed from all of them will be formed from the standard sorry the elements in the naturally occurring form at the standard state. You find out the heat of reaction for that and accordingly assign the heat of formation for all these species. So, once this tabulation is done then at standard state you have H for every species that you are every species of interest not every general species because we are we will tabulate only for whatever is of interest every species of interest. So, once this is done then delta H R for all reactions that we know of can be easily found out. So, if we have a reaction C 2 H 6 plus O 2 giving twice C O 2 plus 3 H 2 O and I will just balance it twice O 2 and 1 and half. So, 3.5 O 2 I will at standard state I will get the H of this H of this is 0 H of this is tabulated H of this is tabulated. So, I get the H of the products I will get the H of the reactants I will just subtract this and get the H of the reaction and hence this is how I will proceed to get delta H R for reactions of interest. You will realize that this is delta H R 0 which is delta H R at standard state. Now, the one thing that we will realize is this is at standard state we need to find out the enthalpy of species at other temperatures and this is especially necessary when we need to do calculations regarding adiabatic flame temperatures etcetera. Because at that time we need to know what kind of temperatures we reach. So, though the heat of reaction is all defined at the standard state we require H for all species at higher temperatures and we realize that for every species as you start providing the energy its enthalpy will start increasing and for all species for all elements to though we started gave 0 at 25 degree C. So, let me just draw it here let us say this is temperature this is 25 degrees this is H I have given 0 here. You will realize that as temperature increases H will increase and this is what is bound to happen for every species. I have CO 2 CO 2 has a negative enthalpy of formation. So, that means at 25 degree C I start somewhere here and as I start heating it its enthalpy will go on increasing and at some point it will cross 0. It does not matter at all where it crosses 0 and this is because we have arbitrarily set the enthalpy of formation to be 0 for certain species at 25 degrees. So, at 25 degree certain species of positive enthalpy is some certain species of negative enthalpy certain species which have negative enthalpy at some higher temperature they will cross 0. So, this is really of no concern to us because finally, we will be just interested in the differences in enthalpy to figure out whether we have an exothermic or endothermic reaction and what kind of energy is there. But if we want to find out how this enthalpy increases you will realize that if I just write d H is equal to C p d t I will need to integrate I will need if I know the C p for every species then I can get the H for every species as a function of temperature. Now, it turns out that C p is a function of temperature. So, it is not a constant in fact it is going to be a constant only for noble gases. So, all monatomic gases will have C p constant, but any other species oxygen. So, let me not write oxygen, nitrogen, CO 2, H 2 O for all of these the C p itself if I plot C p it will have some shape like this it will go on increasing with temperature. So, it will not be a constant and we can more or less find this out. So, if we do statistical mechanics roughly it will turn out that diatomic gases. So, for example, monatomic gases the C v would be 3 by 2 and C p would be pi by 2, but if I go to diatomic gas I will have 6 degrees of freedom and C v will be 3 R t sorry 3 R and C p will be just 4 R it will always be C v plus R. So, this is if you if the entire diatomic gas is at very high temperature, but normally what happens is at room temperature all degrees of freedom have not yet kicked in. So, you will get a certain C p at lower temperature and as you go higher in temperature you will attain a certain C p. For example, for a diatomic gas I will start somewhere here and then I will plateau out like this. So, it will just between two values it will vary smoothly, but if I have a triatomic gas I will have far more degrees of freedom and every time an extra degree of freedom kicks in till all the degrees of freedom have you know I mean populated properly. I will keep on varying the C p till I reach a maximum here. So, this is what happens with all C p and they keep on varying as a function of temperature and what was done was that all these C p values were tabulated as a function of temperature or there were polynomial fits which were given to these. And we will use something which is called as a NASA format. So, a lot of this work was done initially for reactions and not only that for reactions of air to form N 2 O 2 etcetera you know whenever not only for burning of fuel, but whenever you know space vehicles re enter the space all kinds of reactions regarding nitrogen and oxygen will occur in the atmosphere and then hence a big database was created for the C p values of many hydrocarbons as well as many species involving oxygen and nitrogen. And if you go to the NIST website which is US government website it is called the National Institute of Standards and Technology you will get a whole lot of C p fits. So, we will use what is called as the NASA format. So, in the NASA format C p is a polynomial with 5 coefficients. So, I will write it as follows C p is equal to a 0 plus a 1 t plus a 2 t square plus a 3 t cube plus a 4 t 4 and so a 0 a 1 a 2 a 3 a 4 these are 5 coefficients. So, it is you see it is the fourth order polynomial with one constant and usually the temperature has to be input in Kelvin. So, when I put this T is T as 300 I will get the C p at 300 if I put T as 1000 I will get the C p at 1000 and so on. Now, using only a fourth order polynomial it is not so easy to get the fit in a huge range and what the range of interest which was of interest sorry I mean the temperature range that was of interest was mostly from somewhere between 0 sorry 25 degree Celsius and going up to some 3000 Kelvin or even up to 5000 Kelvin because such temperatures could be anticipated. And you will see that most of the fits are given between 300 Kelvin and around 3000 or 4000 Kelvin and it is not so easy to fit one single fourth order polynomial in this huge temperature range which is nearly of the order of 4000 Kelvin. So, what they have done is they have fitted two different ranges and usually they will tell that the ranges are. So, 298 Kelvin which corresponds to 25 degree C up to 4000 Kelvin is fit, but this is in split in two ranges. So, 298 Kelvin up to let us say 1500 Kelvin and then 1500 Kelvin up to 4000 Kelvin or some such split. So, it is not always going to be 1500 Kelvin the split temperature may be 2000 Kelvin 3000 Kelvin. In fact, I have given you the sheets and you must download them because we will use them in the afternoon and you will see in those sheets. Now, I am not going to project those sheets, but you will see that I have given sheets for the CP coefficients and let us say the coefficient for CO is given as CP upon R U this is universal gas constant. So, these are all molar specific heat and it is given as A 1.  So, they instead write it as A 1 plus A 2 t plus A 3 t squared plus A 4 t cube plus A pi t 4 and you will see that they are split between 300 and 1000 is one range and 1000 Kelvin to 5000 Kelvin is second range 2. So, in fact, for all the species that are given to you in this table CO CO 2 H 2 H 2 O N 2 the range is 300 to 1000 and 1000 to 5000, but it need not always be like this if you go to the website you will find that all kinds of different fits are given sometimes it is from 300 to 1500 and then 1500 to 3000. So, whatever is available is given in a certain format and you will realize that CP has been proper values or well established values of polynomial fits have been provided and it is a very good source for all our calculations. Now, this is CP now if I want H you require you realize that we could have just written d H is CP dt I can integrate d H which means I can integrate CP dt. Now, since CP is a function of temperature I cannot take it outside the integral and I will get in their format it will be written as a 1 t squared by 2 plus a 2 sorry I think they gave it as a 1 plus a 2 sorry I will just rub this here. So, I will just write it properly. So, if I want this integral of d H is integral of a 1 plus a 2 t plus a 3 t squared plus a 4 t cube plus a 5 t 4. So, I will get this as a 1 t plus a 2 t squared by 2 plus a 3 t cube by 3 plus a 4 t 4 by 4 plus a 5 t 5 by 5 and if I just write this as H then you will realize I will have to put a constant of integration. So, let and they call it as a 6 and you will realize that this a 6 will be will have to be some function of H f that is because if I now put t is equal to 298 Kelvin in this if I put t is 298 Kelvin. So, I will all the values have been put here. So, a 6 should be such that once I add up all these values it should give me H at 298 Kelvin which is H f 0. So, a 6 is in a way it is put in the constant of integration is put in such a way that once you put 298 Kelvin you will get the correct H at 298 Kelvin and it will depend on what the enthalpy of formation of that species is like. So, this is a 6. So, you will realize that in the table they have written it in this form H 0 by R t is equal to a 1 plus a 2 t plus a 3 t squared by 3 plus a 4 by 4 t square t cube plus a 5 by 5 t raise to 4 plus a 6 by t. So, they are just taken R and t on this side you just multiply by t and you will see that it is the same form. So, if you want to get H you first put the t there calculate the expression and multiply this R u which comes here and you will get your H at that particular temperature. Now, assuming this is an ideal gas H will be only a function of temperature. So, in their formula you will realize that they have put H 0 and again it means at standard pressure really and this is assuming one atmosphere pressure, but if the gas behaves as an ideal gas it is irrespective of pressure you should still get the same enthalpy. However, a s is not so straight forward.  So, for any simple compressible substance I could have written this t d s is d u plus p d v and of course, this will now be applicable to an ideal gas and if p is constant.  So, t d s is d H this is assuming this and d s would be just d H by t which means it is just c p d t by t which means it is just a 1 plus a 2 p plus a 3 p square plus a 4 p cube plus a 5 that means just upon t d t is there. So, if I integrate this s would be a 1 by t if I integrate it will be a 1 log t plus a 2 the t would go when I integrate again I will get back it a 3 is t square by 2 plus a 4 is t cube by 3 plus a 5 is t 5 by sorry t 4 by 4 and I will need to add a constant of integration which they call as a 7 and you will realize that they have written it in forms of s 0 let me just take a 2 minute break here.


The Shroud of Turin, the Turin Shroud, or Holy Shroud, is a length of linen that for centuries has been passionately argued to be the actual burial garment of Jesus Christ. However, predictably, these claims have not escaped the attention of those who dismiss such claims, with some going to incredible lengths to attempt to discredit its authenticity. Yet, just as William Coralus has pointed out during his investigation, along with many other individuals who have explained over the years that although much of the detail present upon the Shroud could be explained with printing methods already in existence as early as the medieval era, the methods used to date the Shroud to a much later creation than that of the crucifixion of Christ, these dating methods could just as easily, yet legitimately be thrown out, just as they did with the claims of those who argued its authenticity. Preserved since 1578 in the Royal Chapel of the Cathedral of San Giovanni Battista in Turin, Italy, home to some of the most astonishing pieces of art on this planet, it measures 14 feet 3 inches in length and 3 feet 7 inches wide. It portrays two faint images, one of the back and one of the front of a 5-foot 7-inch man. The image contains markings that allegedly correspond to the crucifixion wounds of Jesus, including thorn marks on the head, lacerations as if from flogging on the back, bruises on the shoulders, and various stains of what is presumed to be blood. Initial carbon dating of the Shroud was undertaken at great cost by individuals hell-bent on proving the Shroud was a hoax, we might add. These dates at the elation of those who funded the examination came back with an age for the images creation as less than 700 years ago. However, quote, a reality of our biosphere is that virtually everything is permeated with microbes and their byproducts. S.J. Mattingly and L.A. Garzavides of the University of Texas, San Antonio have been studying what is known as biogenic varnishes. These plastic-like coatings are produced by bacteria and fungi, much younger than the actual artifact itself, and an examination of a few linen fibers from the Shroud itself has proven that they are indeed coated with these microbes, end quote, from the archeological anomalies small artifacts by Corleis. Thus, examination of these threads will throw up many anomalous dates. Therefore, the true authenticity of the Shroud has not been debunked. What we personally feel would be a logical step to take going forward would be to date these biogenic varnishes in search of the oldest. This would then, by default, give us a minimum date of creation. It is a direction we hope is taken by those who believe in the Shroud's Christian origins. The reality is that, as of yet no one alive today knows for certain, via any scientific investigation, the true age or origin of the Shroud, regardless of those who hastily debunked the artifact. We instead find such results curious, and often wonder what drives their motive for dismissal. Regardless, all of these factors, if taken to discredit the Shroud, have instead made us even more interested in finding out the truth surrounding the Shroud's origins. It is an artifact which we undoubtedly find highly compelling.


Meanwhile, a senior advocate of Nigeria, Femi Faluna, has spoken on the issue of insecurity, saying that the present government has not ensured the security and welfare of Nigerians, and therefore should be impeached. Faluna said this on the sidelines of an event to mark World Press Freedom Day. So nowhere is safe in our country, but in a regime that has outlived its usefulness, bust, bust, that nobody can suggest that the president should resign. For them, it's a reasonable statement to ask the president to resign. But for me, I submit that by virtue of Section 143, and we didn't put it there, there's a provision for impeachment. If a president is unable to perform the functions of his office, the National Assembly shall move on him and ensure that he's neatly taken out of power through the process of impeachment. I think that our country has got to that state, because Section 14, Section 2B provides that the primary purpose of the government in Nigeria shall be the security and welfare of the people.  We're not asking for a coup. Hello, hope you enjoyed the news. Please do subscribe to our YouTube channel and don't forget to hit the notification button so you get notified about fresh news updates.


More people studying architecture is a good thing because it raises the awareness of what architecture can be. And that's a significant impact, has a significant impact on the culture. I mean, the overall American culture is not one that supports across the board good architecture. It doesn't hackneyed phrase making places. It doesn't do that. Students who enter schools of architecture today are entering it at a very young age. Perhaps when their total world experience and awareness is relatively narrow and they're making a decision to become a practicing architect and putting aside those studies as general ed, as liberal art studies that might actually in the end make them more contributing architects. The idea of reading a treatise on philosophy, the idea of being able to write a poem, the idea of being able to understand a particular piece of science, all those things, all those pieces that are not necessarily architecturally directed are very important in becoming a good architect. And fewer and fewer people are having that basic liberal arts general ed knowledge in the profession. This is a serious problem. There's the belief that students can't handle the complexity that multiple issues in a project might require. I don't believe that's true. And I think if the students can't handle that complexity, especially in the later years of their curriculum, they shouldn't be architects. And that may sound elitist or narrowing of the opportunity for the general public to enter the profession but I don't think it is. I think it would protect the profession and it would protect the outcome of our work to have people who are more able to do the work. And I have a concern that the people entering schools of architecture are inexperienced and unaware of all the choices that they might have in terms of careers and choosing it at a very young age with a narrow idea of what it's really about. In Italy, where thousands of people study architecture, they're studying architecture for exactly that reason to become aware of a piece of their lives, a piece of the culture that's important to them. They don't necessarily enter that study with the idea of being a practicing architect and that's a major difference in our culture. People in this country go to the School of Architecture with the intention of becoming an architect, of having that shingle, being able to call yourself an architect and be involved in the process of designing. I wish it were more like the Italians. I wish people would study architecture just to become aware of what it can be to be able to analyze it, to discuss it, to make decisions about how it should be. And that's not the case. And the other issue with that, when you study architecture in Italy or you study architecture in the Netherlands, there's this balance between ideas and how you make something. Our schools are very theoretical. We spend a lot of time churning ideas. We spend a lot of time on theory. We spend a lot of time on the philosophy of architecture and not so much about actually how you make it. And you can't do one without the other. You have to do both. Well, we're not making very good things in New York. It's New York's not alone. We're not making good things in San Francisco. We're not making good things in Philadelphia. We're not making things which are rich in character. They're two-dimensional. They don't give the inhabitants choices. They don't... In many cases, they're not even comfortable. But they're not dealing with energy issues. I am amazed that residential buildings in the United States do not have a place...  There should be a place to have a small garden. If it's a place to sit and read a paper and breathe the air. We don't do that. That should be a natural part of the discussion. There should be places along the street where, in good weather, people can be on the outside. There should be variety in architecture and planning. It's not happening. Things are getting more two-dimensional. They're getting flatter. There's a huge emphasis in schools of architecture on the skin and the theory of the building skin. It's fine if it has to do with the energy issues or the light issues or even the opening and closing according to seasons. But it's more than just the geometry of the skin. It's more than just the detailing of the skin unless it's for those purposes I just mentioned. It's not new that people have said that bad building is not architecture. I'd rather say that there's good architecture and there's bad architecture. Building is architecture. It may be that it's so discomforting that we don't want to call it that, but it is. Architecture has great variety. There are some very modest places that are wonderful and there are some very grand places that are terrible. I'm always amazed that people do buildings with consideration of the place or the outdoor space. I'm amazed that, especially in the United States, there's so little concern with what happens and what Lucan used to call the in-between space, the space between the outdoors and the indoors. We've lost that. It doesn't exist anymore. We have a slab and we push through an opening at the ground level and there's no sensitivity about the human experience of maybe slowing down and understanding that you're moving from one place to another. Our architecture is less sensitive than it could be and it's less rich than it could be. There's a traditional trio. It's still traditional and it's still the majority but it's not always equal. That's the architect, the owner and the contractor. Whether the contractor is a construction manager, whether the architect is a group of architects and engineers or whether the owner is the board, those three people working together. The architect has no legal agreement with the contractor. The owner has legal agreements with the architect and the contractor. The owner is more involved as a funder, as a final decision maker. But the owner needs direction and the owner needs guidance and that guidance can come from the contractor or the guidance from the architect and more and more for perhaps reasons of finance, reasons of timeliness, more and more contractors are making the decisions. They may select a material because you can get it now. You may select a material because it's less money than something else but that may not be the best thing to do in the long run or even in the short run. The architect's job is to find reasons and to be able to convince the owner that what they're proposing is the right thing to do. I think architects need great rhetorical skills but they need great analytical skills and they need to present all the conditions. You have a much better chance of convincing someone to do something if you can give them multiple reasons. You'll get good publicity. You do it because it's going to save energy. You're going to do it because it's going to give you the lowest maintenance over time. I care about the inspiration but it's the facts often that will convince somebody whether or not to do something. And the inspiration is important too but that often times it's hard to make a client who doesn't understand inspiration feel it. Because we have the knowledge of history and we have the knowledge of what makes a good building nobody else studies that. Nobody studies contractors and construction management. They don't study what is a good space. They don't know about light or shadow patterns. They don't know about energy and temperature. They don't think about that. They may look at different systems in terms of cost options and efficiencies which we do too but we have a broader view of it. I think we have a view of it that gives us more information about how to make decisions on those other than just an efficiency or a cost item or does it fit the... mechanically does it fit the dimensions of the space. I recently was in Vermont and I was driving around in a section of the Green Mountains and I left the Green Mountains and was in a little village and there were these beautiful new little modest houses that were smart. They had the right orientation. They had awnings and they had few windows on the north side and they were just sighted beautifully with the topography and I went up and knocked on the door and said, who's the architect who did this? And was told that there was no architect, that there was a housewright who was doing it who had just lived in Vermont all his life and understood the place and was making houses that were right for that spot. We need more architects like that. That person had a lot of knowledge. That person had a huge body of knowledge. I'm not so sure what the knowledge is that we architects have now and I'm especially unsure of if we're fostering a pattern of gaining knowledge in our schools of architecture.


So, we are welcoming here all of you to yet another session of Radiology Journal Club. This is a monthly event, a new endeavor by Indian radiologist, quite popular by now. This is the sixth session we are running and we have been doing a Radiology Journal Club since last September 2021 and we have covered important topics pertaining to neuroimaging, breast imaging, imaging of renal masses. Then last month we had a good session which was about pediatric neuro radiology. Also, they are shared on the YouTube channel of Indian radiologists and they are freely accessible to all our subscribers. So, do subscribe to our channel and you can revise all what we learned during these radiology journal clubs. We are thankful to the entire core team of Indian radiologists for these important events and keeping the education alive. Dr. Deepu Patkarsar, Dr. Sanjeev Mani, Dr. Jignesh Thakkar, Dr. Shalindra Singh and myself, Dr. Mithusha, my colleague Dr. Gauria Hoja and Mamta Patkar ma'am. So, as we all know it's difficult to keep in pace with all the upcoming advances and by these kinds of journal clubs we try to make it easier because these are something which we can on the go listen to and learn about. So, these articles are recent articles and our experts are the ones who have been using these articles as well as their day-to-day practice experience. So, bringing all of these together they can update us in quick 30 minutes about these upcoming advances and after this they also discuss about these important aspects with the help of two-three relevant cases. We are thankful to our education partners, Bayers as well as Fandom for their continued support because these are all events which are not possible without the support of these educational partners and we have shared their flyers as well as the leaflets and also they have shared a lot of educational materials that also will be shared on our channel soon. So, without further ado, we know that a lot of new upcoming webinars as well as hybrid events are coming from Indian ideologists. So, do subscribe to our channel for all these information and register for these upcoming conference. Sonobhas is March 19, 20th and now it's time to welcome the expert speaker for today Dr. Amit Chaudhary. Amit is assistant professor at Department of Ideology at Tata Memorial Hospital in Mumbai. He also worked as consultant ideologist at Nanavati Mac Super Speciality Hospital. He has completed his fellowship in neuroimaging from Nair College Mumbai and has several years of experience specifically to the field of oncorradiology. We welcome you Amit and over to you. Thank you Mithisha for your kind words. Let me dive straight into the today's journal club topic which is neuroimaging in the era of evolving WHO classification of brain tumors. So, as you all know recently WHO has revised the classification of CNS neoplasm in 2021 which builds on predominantly the molecular and genetic landscape of brain tumors which dictate their prognosis as well as management to large extent in addition to the histological features and then there can be certain imaging phenotypes which can be used to narrow the differentiates to certain molecular classification of these tumors. So, to begin with let me put across these two cases and all the participants can quickly vote for the diagnosis in the chat box. This is the first case wherein we have a plain T1 axial image, a T2 axial image and a post contrast T1 axial image and this is the next case. So, again a T2 axial image in the left bottom, a T2 flare image and a T1 post contrast image and audience can type in their answers in the chat box and we will discuss in these cases in further depth as we progress through the talk. Then this is another case, a case of a left frontal glioblastoma in our 53 years old lady who was operated and treated with chemotherapy of alkylating agent that is temozolomide. So, this is the first preoperative scan. This is the follow-up study in after about six months post treatment and when the patient has received chemotherapy and this is further follow-up. So, this is a response assessment scan and you can maybe in your minds make a response assessment and we will get back to these cases as we progress through the talk. Okay, moving on. So, the present article focuses mainly on 2016 CNS tumor WHO classification but at the same time it has included and incorporated the recommendations of the C impact now committee which was the committee who has recommended many of the guidelines for the new WHO 2022-2021 CNS tumor classification and these are now being practiced or incorporated into the practice of neuro oncology. So, moving on to adult gliomas mainly this article focuses on adult gliomas and their molecular landscapes. So, previously the gliomas were mainly differentiated into IDH mutant and IDH wild type based on immunohistochemistry and their histological features. So, typically a glioma was high grade if there were features of microvascular proliferation, necrosis, hemorrhage etc and then there was IDH mutation which was either detected to be positive when the tumor was IDH mutant or when it wasn't detected the tumor was considered to be a wild type the later being of pure prognosis compared to the former tumor type and in the recent context of 2021 classification this is the status. So, broadly they are classified into IDH mutant and IDH wild type tumors. So, if one encounters IDH mutation positive status then one would either do immunohistochemistry to look for what is known as ATRX which is a protein of the alpha thalassemia retardation X gene and this is detected by IHC. If the result of this test is inconclusive then one can go on to do genetic testing to look for co-deletion or non-co-deletion of 1P19Q. So, these two findings are mutually exclusive and one of these findings can be considered for this diagnosis. So, if this ATRX protein is retained or 1P19Q that is the short arm of chromosome 1 or long arm of chromosome 19 these are simultaneously deleted that is co-deleted in that case the diagnosis is oligodendroglioma and this is a typical case of oligodendroglioma wherein on this CT scan axial image which is actually borrowed from radiopedia one can see that there is this linear calcification which is related to the calcification of the vessels present in the oligodendroglioma. So, these are really fine capillaries which are present in the oligodendroglioma which is really the better tumor of all the tumors which we are going to see today and has a far better survival and outcome compared to the rest of the tumors with a 10 year overall survival of approximately 80 to 90 percent and on this axial T2-HMR image we can see that this is a particularly based tumor which is hyper intense on this T2 axial image is associated with some diaryl thickening and this hyper intensity perhaps which is related to calcification. Next we go on to see whether this protein is lost if it is lost and then there is either non-co-delation of 1P19Q then that would imply that this is an IDH mutant non-co-delated tumor. If that is the case one would further go on to assess what is known as cyclin-dependent kinase 2AB again this is a marker which implies that the tumor has poor prognosis. So, even though if that tumor might be IDH mutant this patient would carry a far more dismal prognosis or outflow compared to patients without this abnormality and without the presence of micro vascular proliferation, hemorrhage or necrosis. So, if this is non homozygous deleted or not deleted at all and then there is no micro vascular proliferation or necrosis then that would be an IDH mutant astrocytoma which would be either grade 2 or grade 3 and this is a typical example of the same. So, this is an IDH mutant 1P19Q non-co-deleted astrocytoma which again is having a relatively good prognosis because there is no CDKN2AB homozygous deletion and there is no micro vascular proliferation or necrosis. Now again in the context of WHO 2021 classification of tumor there is nothing called as grade 1 neoplasm as far as astrocytomas are concerned that classification now applies only to the well circumscribed neoplasms and not to diffuse gliomas. So, diffuse gliomas would be grade 2 to grade 4. Then if this is homozygous deleted and or associated with micro vascular proliferation with or without necrosis that would imply that this is a higher grade IDH mutant neoplasm. On the wild side that is IDH wild or absence of IDH mutation there can be association with micro vascular proliferation again and there can be necrosis which would help in histological grading of this tumor as a high grade neoplasm. Then again there are other molecular markers like third mutation, EGFR amplification, combined gain of chromosome 7 and loss of chromosome 10. In the presence of these mutations the tumor would be a IDH wild type tumor and the classic example being glioblastoma and this is the tumor where in on T2 weighted image and this X-ray T1 post contrast image one can see that there is a tumor which is epicentered in the sphenium of corpus callosum showing certain areas of image with areas of necrosis within on this post contrast T1 weighted image showing thick irregular peripheral rind of enhancement. Moving on, so the article also describes the molecular landscape of medulloblastoma and ependymoma really briefly stating it as being beyond the scope of this article but just to give you a brief overview of these tumors. So WHO 2016 classified medulloblastomas predominantly into four subgroups that is WNT or wingless subgroup of medulloblastoma which had excellent premises amongst all the medulloblastomas then SHH or the sonic hijab mutated subgroup of medulloblastoma. Again in the infantile form or age group this has a better prognosis compared to the adult patients then there was the worst form of medulloblastoma which was earlier classified as group 3 or group 4 depending on molecular and genetic analysis and these usually had bad prognosis because of presentation with metastasis in the first diagnosis or visit. So the WHO 2021 has revised this classification wherein WNT medulloblastoma stage SHH medulloblastoma continues to be similar just that this has been further subclassified into TP53 mutant and TP53 wild and the group 3 and group 4 medulloblastomas have both been grouped together into non-WNT and non-SHH type of medulloblastomas. Now for the sake of this article we wouldn't delve any further into medulloblastomas or even appendymomas. Appendymomas as per WHO 2016 classifications were classified into posterior fossa B and posterior fossa A posterior fossa A appendymomas typically were encountered at the Foramina of Lushka would extend into the CP angles and PFBA appendymoma was usually present at the obex of the fourth ventricle and was associated with better prognosis. As far as WHO 2021 classification of appendymomas is concerned it classifies appendymomas based on their anatomical location into supratentorial infratentorial or posterior fossa appendymomas and spinal appendymomas. Moving on to neuroimaging evaluation of tumor phenotype on imaging. So where can imaging help us in coming to a particular diagnosis? So these can be done by based of conventional MRI features that is T1 weighted imaging, T2 flare imaging, post contrast imaging and in addition we can use advanced techniques which can act as adherence which include diffusion weighted imaging and perfusion weighted imaging. Perfusion weighted imaging can be dynamic susceptibility contrast imaging or arterial spin labeling based perfusion techniques. So again we have seen this case in the question quiz to you before and what do we have in this first case? So this was again an elderly gentleman who had present with neurological disturbances and focal neurological deficits and we have here T1 weighted axial plane or an enhanced image that's the T2 axial image and this is the T1 post contrast image and we can see that there is a large heterogeneous mass or tumor which is epicenter in the genu of the corpus callus extending into both the cerebral hemispheres across the splemium of the corpus callus sorry it's not the genu but the splemium of the corpus callus and on T2 weighted images we can see that there are certain dark or high-pointed areas which are lining the cavity or necrotic portion of the tumor which is consistent with hemorrhage within the tumor. Again we can see that these areas are also T1 bright on the unenounced image implying that there are some areas of hemorrhage within the tumor and then on this post contrast image we can see that there is thick irregular rind of enhancement and on perfusion weighted imaging one would expect to see very high perfusion values in excess of RCBV of 1.75 where one would put an ROI on the tumor, preferentially in the enhancing portion of the tumor and in the contralateral unaffected white matter and take the ratio of these two. So this was an IDH white type of tumor that is a glioblastoma. Now this is the other end of the spectrum wherein on this T2 axial image we see a homogeneously hyper intense or bright tumor which is located along the Sylvian fissure involves the insula, the temporal lobe and the frontal lobe and on flare image we see that this tumor is suppressed partially or almost near completely and has some peripheral hyper intense rim. So what we can see over here is that there is a T2 flare mismatch that is the tumor is bright on T2 weighted images and is suppressed on flare images that is doesn't continue to be as hyper intense on flare images. So this is an important sign which has been discussed in this lower grade diffuse glioma that is the T2 flare mismatch sign and this has been shown to correlate very specifically with IDH mutant 1P19Q non-co-deleted astrocytomas which are relatively lower grade tumors than grade 4 IDH mutant astrocytomas or even glioblastomas and carry a very good prognosis and on post contrast imaging these tumors show mild to no enhancement at all. Moving on to tumor grading. Now tumor grading can be done again grossly by looking at morphological features and advanced techniques like diffusion weighted imaging and perfusion weighted imaging. On diffusion weighted imaging one can look into ADC values wherein higher grade tumors can have lower ADC values likewise lymphomas tend to show very low ADC values compared to glial neoplasms and tumors like glioblastoma which are really the highest grade tumors will show elevated perfusion value with the cutoff being an RCBV of 1.75 which has been widely explored in depth by Mengele et al in their 2003 book published in AJNR. So this is an example of a glioblastoma or high grade tumor wherein on this T1 an enhanced image you can see a heterogeneous tumor located in the right cerebral hemisphere in the right parietotemporal region. This is a susceptibility weighted image which is a form of gradient image which shows certain hypointense areas or foci which are also known as foci of blooming which represents hemorrhage within the lesion. On post-confrost T1 weighted image we can again see a thick rind of enhancement which is irregular with central non-enhancing necrotic area which is suggestive of a high grade tumor like glioblastoma and on RCBV map from dynamic susceptibility contrast perfusion we can see the ratio of the tumor to contralateral unaffected white matter is 3.5 implying this is a very high gradation. Again you can see that this the second signal intensity is that of the normal parenchyma and the ones in pink or red that is this curve number one represent the signal intensity of the first pass of contrast through the glioblastoma and what we can see here that because this is a very high grade tumor this doesn't show a complete baseline recovery that is the recovery doesn't go right up to that level rather it stops to somewhere around 50 percent because of presence of leaky channels of neo angiogenesis within the tumor. Next is the benign tumor which we have seen earlier tumor with T2 flare mismatch that is a low-grade IDH mutant 1P19Q non-co-deleted astrocytoma again here the tumor is bright on T2 my apologies for not presenting the flare image over here but on T1 post-contrast the tumor shows barely any enhancement or mild peripheral enhancement and on perfusion imaging we can see that the tumor is not very hyper perfused the ratio of rcpv of the tumor to the white matter is approximately 1.2 so this is a relatively lower grade tumor then the other perfusion technique which can be used for non-invasive assessment is arterial spin labeling based perfusion. Now this technique can be really good for patients with history of renal failure or patients with contrast allergies or in patients in home contrast administration is not possible perhaps because of our difficulty in getting venous access in such case one can do arterial spin labeling based perfusion provided the sequence is available on the scanner now the base modern 1.580 utility scanners come loaded with this sequence so this is a great sequence for doing a non-contrast perfusion study and it has been shown that cpf that is the cerebral blood flow measurements or matrix in arterial spin labeling correlate really well with the rcpv metric of the dynamic susceptibility contrast perfusion and again this was the case of glioblastoma which was showing very high or elevated rcpv value on dsc profusion technique and was showing corresponding elevated cbf values on the asl map this is the t2 actual image of the same case and these are the t1 post contrast images wherein again we can appreciate the thick irregular rind of enhancement with certain foci of enhancement within the tumor which can again be a case in a with glioblastomas moving on now there can be an exception to this rule of hyper perfusion one typical such exception is oligodendroglioma why so this is because oligodendrogliomas also demonstrate very high perfusion but in this scenario that elevated perfusion is related to the presence of network of capillaries within the oligodendroglioma deserving the presence of neo angiogenesis and related vessels in glioblastoma so again here we can see that the ratio of the relative cerebral blood volume of the tumor to counter lateral white matter is in excess of 1.75 obviously we can see it is about to the tune of six times roughly of that of the tumor in this scenario and here again we can see on this gradient image there are these linear hypo intensities or foci or lines of blooming which corresponds to the chicken wire type of calcification in the capillaries present in the oligodendroglioma so again this is an exception to that rule wherein not really every case which is hyper perfused is glioblastoma but then we can look at other ancillary features wherein if one finds sugar while calcification which can be better appreciated on CT scan image a cortical based tumor which is associated with thickening of the gyral folds and with which usually doesn't show significant neck crosses or hemorrhage in such cases a possibility of oligodendroglioma can be put for prospectively. Moving on the another exception is meningiomas in meningiomas tumor grading cannot be done on perfusion values this is because all meningiomas from lower grade to higher grade express EGFR and vascular endothelial growth factor which is responsible for higher vascularity in these tumors and hence most and in fact all the meningiomas have very high perfusion values so for assessing rates of meningiomas it is usually the growth rate of meningiomas which helps in assessment of the grade of meningiomas again the location of meningiomas to certain important structures like dural sinuses would imply that these meningiomas are relatively aggressive in nature than the typical convexity meningiomas. Moving on to tumor characterization now that is differentiating a tumor a glioblastoma or glioma from meningioma or from a lymphoma or any other intracranial tumor so we can make some attempt on imaging for characterization of tumor again this can be based on conventional MRI findings or advanced imaging techniques like diffusion perfusion or even diffusion tensor imaging. So diffusion tensor imaging though has not been discussed in this article so we will limit our discussion to just conventional imaging findings and advanced techniques like diffusion perfusion imaging. So this is an example this is again a case from the article by the authors wherein in that article there is a large tumor which is present in the anterior cranial fossa arising from the floor of the anterior cranial fossa which shows extensive calcifications and by virtue of this calcification this is a meningioma or was rather a meningioma in rare example and then this is another case a convexity based case towards the calvarium on this anenous CT scan gauge which is hyperdense and which doesn't show calcification. So when one doesn't encounter calcification one should consider a possibility of hemangioperisitoma as a differential for a meningioma again this doesn't negate or obviate a finding of meningioma because again higher grade meningiomas may not calcify vis-a-vis lower grade meningiomas which tend to calcify again diffusion weighted imaging. So diffusion weighted imaging is a very important tool which can allow for characterization of tumor typical example being lymphomas. So lymphomas by virtue of their high cellularity cause restriction of movement of water molecules in the interstitial space and we can see here on this axial T2 weighted image there is a tumor which is of intermediate signal intensity or signal intensity which is almost similar to that of the gray matter and on diffusion weighted imaging on the left side we can see that the tumor is restricting and extending across the entire corpus callosum and kind of encasing the lateral ventricles or traversing along the roof of the lateral ventricles and corpus callosum. So this was the case of lymphoma and again we can see the perfusion values. Remember the perfusion values which we saw for the case of glioblastoma they were very high there was poor recovery of the signal intensity poor return to baseline and there was a very high or elevated rcb value. Contrast that in cases of lymphoma these rcb values are relatively lower compared to glial neoplasms. So that can be a one differentiating factor in for narrowing down the differentials between a lymphoma and a glial neoplasm. Next is dynamic susceptibility contrast perfusion typically its percentage signal recovery which can be used for characterization of tumors. So this was an article which was published back in 2011 wherein they had shown that based on the percentage signal recovery that is the amount of return of this contrast bolus based on the signal intensity time curve to the baseline can be used to judge or estimate the grade of tumor or rather to characterize a tumor. So typically in a glial neoplasm this baseline recovery tends to be better more than 50 percent above the baseline towards the baseline. In a lymphoma there has been shown to be a phenomenon which is known as baseline overshoot which is considered to be characteristic of lymphoma. So just resuming as to how we can differentiate tumors based on the percentage signal recovery the glial tumors usually tend to have good percentage signal recovery that is they tend to reach towards the baseline. Then lymphomas demonstrate a characteristic feature which is known as a baseline overshoot that is they go over and above the baseline. This is thought to be because of interstitial leakage of contrast in cases of lymphomas where in T1 effects predominate over the T2 star effects. Standard metastasis these you have poor baseline recovery that is the return of signal after the passage of contrast bolus is less than 50 percent of what had of the signal dip. So that is how percentage signal recovery can be used to differentiate between tumors. Moving on now we discuss evaluation of tumor genotype. So we had discussed this feature before the T2 flare mismatch sign which is quite a specific feature for identifying a tumor genotype based on MR morphology that is a tumor which is nearly homogeneously hyper intense of bright on T2 weighted axial images is here completely suppressed on flare images with a small peripheral rim of hyper intensity on flare. So that is the T2 flare mismatch sign and then again these tumors have very minimal enhancement or even no enhancement at times and barely show any elevated for fusion values. So these are the RCBB map from dynamic susceptibility weighted contrast imaging or perfusion imaging wherein we can see that there are no elevated areas of blood flow or blood volume in the tumor and on the diffusion weighted imaging these tumors don't restrict. So usually tumors with restrict are lymphoma and then in post treatment scenario wherein there is tumor recurrence there will be increased cellularity and there might be restriction of diffusion. So this is one such tumor entity which can be prospectively diagnosed based on imaging alone. The other tumor entities described earlier was oligodendro glioma which by virtue of presence of the chicken wire calcification can be prospectively characterized as IDH mutant 1P19 cube co-deleted tumor. Then we saw the IDH mutant astrocytoma and finally glioblastoma IDH wild type which has thick peripheral irregular rind of enhancement with central necrotic areas and hemorrhage which are really the gross imaging features of histopathology which would show microvascular proliferation necrosis and hemorrhage within the tumor. Moving on now we go on to assess the post treatment changes or rather response assessment in brain tumor. So recently the response assessment criteria for brain tumor was defined by RANO group that is the response assessment in neuroancology which involves measuring the tumor in two axial dimensions and adding those axial dimensions. So this has been formulated for clinical trial and has been inculcated into clinical practice to some extent but again because of its tedious nature it hasn't really found a prime time in the clinical radiology scene. So the next step was BT RAT that is the brain tumor reporting and data systems which is akin to the conventional or classic bi RAT for breast cancer imaging or for example bi RAT and prostate imaging, lyrat and liver imaging so on and so forth. So how does one assess post treatment changes? So classically the post treatment changes can be appreciated in the early to early delayed phase after administration of radiotherapy and chemotherapy in patient with operated case of brain tumor and what one can see here is that this was an operated case of blastoma and we can see that in the early period that is after about three to four months of radiotherapy the imaging showed this irregular feathery kind of enhancement or the so-called switch is pattern of enhancement which is which can be seen frequently with radio necrosis or radiation necrosis in tumor. Again this can also be a feature of pseudo preparation which can be encountered in patients receiving hemozolomide chemotherapy and radiotherapy and on DAC contrast imaging we can see some reassuring RGB values wherein in this tumor we can see the RGB value is lower it is about 0.4. So this is an inverted map wherein the red areas are really the hypo or reduced perfusion areas and the blue areas are hyper or increased perfusion areas. So that is the perfusion of the normal brain parenchyma in the left cerebral hemisphere and that is the abnormal or the tumor areas. So this is abnormal this was normal. Moving on so this is another case a case of in fact oligodendroglioma who was operated and in the postoperative and post radiotherapy period on this follow-up study we can see that there is this irregular post contrast enhancement with some increase in flare signal intensity or almost similar flare signal intensity and then at further follow-up we can see that the patient is having on MRI the flare signal intensity which has reduced or stayed similar but the contrast enhancement has vanished near completely. So this is suggestive of response and this interim imaging was really pseudo progression or rather enhanced reactivity of the tissues to the agents which were administered that is radiotherapy and alkylatic chemotherapy agent. So this is how pseudo progression can be assessed and it can be best assessed on serial follow-up scans. There have been studies done which have tried to give a certain cutoff for differentiating a pseudo progression from real or true progression and these cutoff values of RCB vary from 0.9 to 2. So that is really a wide range implying that there is a lot of overlap between features of pseudo progression and actual or real progression in the post treatment setting of gliomas. So these methods are still under evaluation that is dynamic susceptibility contrast for response assessment are still under evaluation and haven't been formally incorporated yet into any of the guidelines be it rhino criteria or be it BTRAT imaging. Then there is this another example again from the author's article wherein we can see on this T1 axial post contrast enhancement there is this irregularly enhancing area which is seen in the setting of post radiotherapy and chemotherapy and this is arterial spin labeling map this area is showing a liberated effusion implying that there is some residual tumor. So if there were a follow-up in such a scenario compared to the pretreatment imaging then based on the enhancement findings and flare abnormality one would be able to categorize this as either post treatment change or stable findings or tumor progression but then as discussed earlier these would require serial follow-ups and this is another patient who had received Deversus map which is an anti angiogenic agent which is used in the setting of recurrent glioblastomas wherein all the chemotherapy arrangement has been completed in such scenarios Deversus map is usually used to tide over the abnormalities or symptoms arising from the mass effect of the tumor in these patients and this is usually because of tumor progression and associated peritumeral edema and what can one see in patients receiving Deversus map is that this drastically reduces any peritumeral edema and enhancement because of its direct action on the tumor vascularity and gives a dramatic symptom relief. However no studies to date have been able to prove its efficacy as far as overall survival is concerned. Again this agent doesn't really work that great in the setting of primary management of lymphomas that is glioblastomas or glial tumors. This is usually reserved for patients with recurrence and on perfusion as well one can see that there is complete suppression of perfusion in the patient who has received Deversus map. So that is how progression versus response versus pseudo response can be assessed. Now this is this can be said to be almost a response because we are not seeing any elevated CBF values in case of ASL perfusion in this scenario. The other technique which can be used for assessment of tumor progression versus response is FET-PET. This is an amino acid technique which employs a radio tracer called Fluoroethyl tyrosine. So this agent is really good for brain tumor imaging.  So in such a scenario the brain tumor can be masked in the setting of FDG scan vis-a-vis an FET-PET scan which can be a good agent in case wherein the patient has contraindications to PET scan or there are certain anomalies in the tumor. The patient has a lot of hemorrhage and tumor which can cause susceptibility artifacts and can render DSC contrast perfusion imaging uninterpretable. Again ASL can be another good technique in the setting of hemorrhage within the tumor. Let me see if I can stop this scrolling through. So yeah. So this was the case of glioblastoma again on this FET-PET and a fusion of FET-PET MRI. We can see that there is this area of avid FET tracer uptake which is showing a tumor to white matter ratio in excess of 2. So that FET-PET the ability ratio of tumor to white matter of 2 is usually used as a cutoff. 2 to 2.5 is usually used as a cutoff for differentiating between treatment response vis-a-vis any residual or recurrent tumor. Moving on to structured reporting. Now what is structured reporting? So structured reporting employs certain radiology template and this is an attempt towards standardization of the reporting of any scans be it in any of the body area. Then again this provides a better communication between the radiology standard treating physician or surgeon. This serves as a very good checklist so that none of the pertinent findings are missed and hence there are fewer misdiagnosis and because of all the above features there is increased confidence in surgeons and treating physicians regarding decision for tumor receptability or further management of these cases. BT-RAT has made one such attempt towards standardization of follow-up of glial neoplasms who in the post-treatment settings and this is a template which is available on their website btrats.com which can be accessed by anyone. This is available for free of charge. We look at an algorithm as to how response assessment can be done in the setting of chemotherapy and radiotherapy being received in a patient who has been operated for brain tumor. So this is a very busy slide showing that algorithm but let's break it down and make it somewhat simpler. So let's say there is a patient of brain tumor who has been operated and who has been put on chemotherapy and is on follow-up and the patient curbs at the first follow-up and let's say that there is no prior imaging available. In such case we can put this or use this as a baseline study and categorize it as BT-RAT 0. Again this category is typically used in patients who have prior imaging but those are only pre-operative imaging. In the post-operative settings this can serve as a new baseline and it will be categorized as BT-RAT 0 and will be used as a baseline for subsequent follow-up studies. Say for example another patient who has been operated case of brain tumor is on chemotherapy has a prior imaging and with that prior imaging and the present follow-up imaging if there is no change in findings that would imply this is a BT-RAT category 2 that is there is no change in findings. So this can imply that there is stable residual disease or this can imply that there is post-treatment change wherein the disease has perhaps been taken care of or has been gone, reduced. So next category of assessment is mainly for patients who have been put on Bavacism map predominantly. In such cases if there is no enhancement or no mass effect and there is complete regression of flare signal and peritumeral edema then these will be assigned a criteria called BT-RAT 1A which would imply a good response to therapy. If the patient has been put on steroids and there is improvement in patient condition and imaging findings that is lack of enhancement or reduction in tumor enhancement or peritumeral edema this would be categorized as BT-RATs 1B. Again if there is improvement post-Bavacism map therapy with only improvement in enhancement but persistence of but persistence of flare signal abnormality that would be categorized as BT-RATs 1B. Can somebody tell me if I am audible? Yeah I am audible. And slides are visible right? Yes. Okay thank you. Then if the condition has been stable or there is improvement for subsequent follow-up studies for over a month then it will be again assigned a BT-RATs criteria of 1A. Now let us say that the patient has undergone radiotherapy and has undergone an imaging for certain reasons say for follow-up reasons or for certain worsening of features and if that imaging falls within the period of 90 days after initiation of therapy it is considered to be treatment related effect that is worsening of imaging findings which can imply increase in enhancement in the post-treatment setting and or increase in the flare abnormality along the operative bed or the treated tumor bed and this would be assigned a criteria of BT-RATs 3A. Next is worsening of imaging findings wherein this is seen in the settings of more than 90 days post-administration of chemotherapy. If in such settings there is worsening in either flare or enhancement but not both that would be assigned a category BT-RATs 3B. On the other hand if there is worsening of both flare that is increasing signal on flare imaging and increasing post-contrast enhancement in the post-treatment settings post 90 days of starting of treatment and if this increase in the signal abnormalities on either flare or rather both flare and ANH is less than 25% then this would still be progression but this would be BT-RATs category 3C in case this is the first search finding on the follow-up after 90 days post-administration of therapy. But if this is over a period of more than one studies or this is second or third follow-up and there is further progression or worsening of imaging findings that is in further increase in contrast enhancement, new onset epindymal enhancement or presence of new enhancing areas which has come up at a site distant from the radiotherapy portal in such scenarios this would be categorized as tumor progression that is BT-RATs 4. Again if this increase in signal is more than 25% compared to a previous study this would be categorized as BT-RATs 4. Let us look do one's exercise to understand how BT-RATs work. So this was a patient a case of glioblastoma 53 years old lady who has been operated has undergone surgery and this was the first baseline imaging post surgery. So at this instance since this is the first post-operative scan this will serve as a baseline so this would be categorized as BT-RATs 0 this will be BT-RATs 0. Then the patient came for follow-up after institution of chemotherapy that is themosolomide therapy and administration of radiotherapy and at this follow-up we see that the flare signal is unchanged along the treatment bed but there is some enhancement which is somewhat prominent compared to the prior study. So this patient was assigned a category of BT-RATs 3p that is findings are indeterminate of progression of disease versus response to therapy and his patient was advised a further follow-up. So this is how the categories can be assigned. As discussed earlier we have a prior imaging here so this was the BT-RATs 0 or baseline imaging and then on the subsequent follow-up imaging that is after administration of chemotherapy post 90 days period we can see that there is somewhat worsening in terms of a enhancement but stable flare findings and this was assigned BT-RATs category 3b. Now the same patient at further follow-up and what we can see here there is more than 25 percent increase in the flare signal there is obvious increase in the contrast enhancement which corresponds to more than 25 percent increase compared to this previous follow-up study or even baseline for that matter and then another feature which increases our diagnostic confidence of calling this progression is the elevated RCBV values. Compare this RCBV value ROI in the tumor with the control lateral side and we can see that this is approximately 2.4 implying that there is underlying tumor and progression in this case. So let us look at how this assessment can be made. So then if you recall we had prior imaging which barely had any enhancement or flare abnormality which was stable compared to the February scan and now we are comparing the March 2021 study with the October 2020 study. So we have previous imaging yes and the time since administration of chemotherapy or starting chemotherapy is more than 90 days we had seen that this was the response assessment category which was assigned during the October follow-up scan and on the current follow-up what do we see? We have both increase in flare signal abnormality along the treatment bed as well as enhancement within the tumor which is showing elevated perfusion. So more than 25 percent of both flare abnormality and enhancement hence this is Frank tumor progression BT RATS category 4. So this was a small exercise to understand how BT RATS assessment can be done and this completes the talk and I thank you all for listening patiently and daring with me during the power outage.


Hello everyone, this is Mr. P. On today's lesson, we're going to take a look at the first conditional. If you watch this lesson, you will understand how to use it. So, let's get started. First, let's do a quick overview. So, let's do a tense revision then. First, let's take a look at the present simple. Well, you know that the helping verb for the present simple is do or does. Remember that does is for the third person singular, and that is he, she, or it. We only use the helping verb for negative sentences or questions, like in these examples. He doesn't study hard. Does he study hard? He studies hard. For other pronouns like I, you, they, and we, we just need the verb study in this case, and then something else if you have an object. Like in my example, I study hard. So, notice the third example. He studies hard. Remember that for the third person singular, he, she, or it, you need to add es or s to the verb. In this case, we have i-e-s because the verb ends in y. So, let's take a look at the second tense we are going to use. That is the future simple. Well, it's simple and easy. The helping verb here will be will. So, we just need will plus the base of the verb, like in these examples. I will go home soon, or you won't go home for a negative sentence. Instead, for a question, will he go home soon? So, notice that the verb does not change. And will either. So, now let's take a look at when to use the first conditional. So, when do we use it? It is used when an action is likely possible to happen in the future. That's why you need the future simple. And how is it formed? Well, let's take a look at that. Well, first we need a conjunction, plus the present simple. In the second clause, we need will plus the base of the verb. Conjunctions such as if, when, as soon as, after, even if, unless. Here we have some examples. So, if he works, he will get paid. When I arrive, I will call you. As soon as she graduates, she will start to work. After he gets a job, he will buy a house. Even if he loses his job, he will have money saved up. Unless you see me, you won't believe me. So, notice after the first clause, we need a comma. If he works, comma. Second clause, he will get paid. Now, there is another way of saying the same thing. Let's take a look at that. So, we need will plus the base of the verb, plus a conjunction, plus the present simple. He will get paid if he works. I will call you when I arrive. She will start to work as soon as she graduates. He will buy a house after he gets a job. He will have money saved up, even if he loses his job. You won't believe me unless you see me. So, this is a preference. But remember, if you have the conjunction at the beginning, so in the first clause, you need a comma before the second clause. So now, let's practice. Complete the sentences with the right form of the verb. So, I will give you 5 minutes to complete it, and then I will give you the key. So, you need to post the video. Now that you are ready, let's take a look at the key. If I go out tonight, I will go to the cinema. She will move to London when she gets a job. We won't be late unless we hurry. As soon as we finish, we will go out.  Notice that I said he'll call you. This is just a choice. You can use the contracted form when you speak. When you write, it's better to use will. Write the whole helping verb. So, one last consideration. There are other ways of forming the first conditional. If you are interested in looking at other lessons about this topic, I suggest watching my other two lessons listed in my playlist. Do a search in my channel and you will find them. Oh, this is me. This is an example of another way of forming the conditional. This is called the zero conditional. If you are interested, notice. If you are interested, that's the present simple in the first clause. The second clause is I suggest watching. So, that's the present simple again. So, this is also called the zero conditional. There is no conditional. Well, I hope you enjoyed the lesson. If you have questions, you may type them under this video. Thank you for watching my lesson, obviously. And if you haven't subscribed to my channel, please do. You may like the lesson, probably, and then hit on the like button. And you may share the video, the video lesson, if you liked it. Until next time, take care. Bye-bye.