 So, today we continue from the last class. Today's class will be a kind of revision of what we did and also we will look at some important aspects related to specific LTS systems. So, if you recall, what we have done last time is looked at various methods to estimate the static lift. We spent a lot of time trying to understand the effect of humidity and to figure out how the static lift changes in the presence of humidity. We also had a look at three additional terms. One was the impurity of the lifting gas. So, we used the term Y to take care of the lifting gas impurity. We also took care of two other aspects. One was the super pressure that is pressure inside the envelope more than the ambient and also the effect of super heat that is exposure to ambient air temperature due to which the gas inside gets heated. Now, let us look at now what happens because of all these factors to the weight of the air in the ballonet. So, recall that weight of the air in the ballonet can be given by the expression rho ba that is the density of the air in the ballonet times 1 minus i where i is the infraction fraction. Infraction fraction just to record our memory is the ratio of the lighter than air gas upon the total volume of the envelope. In case you do not have any ballonet then i will be equal to 1. And in case the whole envelope is filled by the ballonet i will be 0. So, rho ba density of the ballonet air 1 minus i into v nv is the volume available for the ballonet because v nv is the total volume and 1 minus i v nv is the volume occupied by the ballonet at any instance and g is the classical extension due to gravity which takes care of the units in this case because we have to be sure that the units are coming in newtons. So, also recall that density of the ambient air rho a can be assumed to be the static pressure minus the component or the loss because of the humidity that is 1 minus r d v v w v times e into t naught rho naught by p naught. Incidentally using simple expression p v is equal to r t you can work out or p is equal to rho r t. So, rho will be equal to p by t. So, you can get the expression for r from that and that is how you got this if you recall we got this t 0 rho 0 by p 0 using the standard gas equation. So, this term in the numerator in brackets is essentially the pressure which is going to be less by the magnitude 1 minus r d times e because of the humidity. Because of the density sorry because of the relative density I am sorry not humidity this is because of the density the density of LTA gas is lower than that of the air. Now, the density of the air in the balloon a that will be rho b a if we assume that the pressure inside the envelope is p l g p of lifting gas and the same pressure is transmitted to the balloon a. We are assuming equivalence of pressure between the balloon a and the envelope and if you also assume that the temperature inside the envelope of the lifting gas is t l g the same temperature is ultimately conveyed to the air in the balloon a. Then the density of the air in the balloon a also will be by the same expression except that instead of p star p s which is the ambient air pressure under standard conditions instead of that we will use p l g pressure of the system inside the envelope and for p a which is the ambient air temperature we will use t l g. So, by the same expression one can get the density of the air inside the balloon a recall that the density of the air inside the balloon a can be equal to the density of the outside it all depends on the temperatures depends on whether p a and t a p l g and p s and t l g and t a if they are matching then it will be the same value. Now in this we need to incorporate the effect of super pressure and super heat. Super pressure if you recall is the additional pressure that is acting why because we are inflating the balloon to a higher pressure than needed this delta p is if you recall intentional. So, what will happen is that in the numerator the term p l g will be added by delta p sp. Similarly the term in the denominator t l g there will be an additional term of delta t sh. So, other things will remain the same and this expression now is an expression that takes care of super pressure as well as super heat if there are any questions. So, this is basically a class to revise and to understand what we have learnt last time. So, in case there are any doubts you are most welcome to interrupt me we can go back and revisit a few equations if you feel right now the weight of the air in the balloon a you see the first line in the in the slide is the density of the air in the balloon a times 1 minus i v n v g. So, this rho b a from here if you replace in this expression you will get the full expression for the weight of the air in the balloon a which is just this term rho b a is inserted here and 1 minus i into k v n v. Recall that this k is the atmospheric constant which is nothing but the value of t 0 rho 0 p 0 and g. So, this g which you have here is also coupled here and that is k and that is the constant because t 0 is 15 degree Celsius or 288.16 degree Kelvin p 0 is 101325 degree meter square that is also a constant rho 0 is 1.256 kg per meter cube it is also a constant and g is just to take care of the units. So, these are the expressions now which will help us determine the total weight of the air in the balloon a and if you need for some explanation the density of the air inside the balloon a. So, this particular equations are basically they are called as the force form equations because they are more concerned about the weights. Now, the same expression I have copied and pasted here for continuity again the same expression that we have used before. Now, let us look at some simplified cases because not every time will you have issues like super pressure, super heat. So, let us see if we can take a simplified case. So, the first simplified case is dry atmosphere no super heat and no super pressure. So, now I want someone to help me what will happen to this equation in case we have this kind of a situation when I say dry atmosphere I mean that there is no effect of humidity or humidity is absent. So, when humidity is absent what do you get? What will change in this expression? No, one by one you have to raise your hand so that you allow me to also exercise some choice. So, who will let me understand what happens if the humidity is neglected or it is absent? Yes. So, E will be 0 that is right. So, if an E is 0 then 1 minus RDWV into E this term is going to vanish. What happens when the super pressure is not there? Then delta PSP is 0. Similarly, what happens when super heat is not there? Delta Ts is not present. So, if we do that the expression becomes very simple it becomes the row below the first one will become the term inside the bracket will completely go away and row BA will become PA by TA because now the lifting gas will ultimately come to equilibrium with the air outside. So, PLG will become PA and TLG will become TA. So, this will become PA this term is 0 T naught by TLG becomes TA and this whole expression we know is equal to row naught is equal to row A that is the density of the ambient air. So, if you have no humidity no super pressure no super heat then density of the air inside the balloon is equal to density of the air outside which is what I just few minutes ago told you that it can happen in certain cases. Similarly, WBA or the weight of the air in the balloon A will be this term will again go away and this will become PS this will become TA. So, it is just 1 minus I into K V ENV. So, by these two simple expressions you can get the values for your condition of dry atmosphere and super heat and super pressure neglected. Let us look at another situation now we are not working in any atmosphere but we are working under international standard atmosphere. So, what will happen in this case is that the row air will be equal to row S which is the standard ambient air density. So, the expressions will be similar only thing is it will not be ambient air it will be density of the ambient air under standard conditions and the pressure also. Now in this case what will happen is that you can also replace you can also replace this and you know the K is can be brought inside and with that you can get you can recover the value of row S right.