 So, we have looked at inflation fraction. We found that all you need is a ratio of the pressure and temperature. We looked at atmospheric pressure. We found that if super heat is not there, there is no change in the net lift. Otherwise, there are two terms. One is the weight of the balloon air, the other is the gross lift. Both of them depend essentially on delta P. In one case, we divide by T plus super heat. In the other, we divide by only T and you have K into V as the constant parameters. So, now let us look at super pressure and super pressure alone. Now, what happens if the super pressure increases? So, first thing is when you have high super pressure, then the lifting gas is going to be compressed slightly. That is basically what super pressure is. So, what will happen is that will force the balloon air to go out. So, simply there will be lower or there will be higher i. There will be lower i. Why should it be low if you are compressing? Because the gas also is compressed. Now, more balloon air will be sucked into the balloon. So, if you maintain the envelope shape and volume because if you do not do that, if you do not do that, then the shape will alter and all our assumptions will be invalid if the volume changes. We are assuming that the envelope volume is maintained constant. Yes. Because when you apply pressure to a gas. Yeah, it is the pressure inside. You are right. Super pressure is the pressure of the gas inside, LTA gas inside. So, that is why when the pressure of the gas increases slightly, it will push out the balloon air. That is what I am saying. It will push out the balloon air. How do I explain it? See, the total volume available is the same. So, if we apply pressure on the gas, then suppose we do not allow anything to go out. So, then what will happen? This pressure of the gas is going to act on the envelope and also on the balloon air. And if we do not allow the balloon air to go out, then there will be a compression of the balloon air. And I am assuming that gas is a quantity inside the envelope which is being pressurized. So, you are right in saying that this gas under pressure will exert the pressure or convey the pressure to the surfaces inside and the balloon air. But when you take a cylinder for example, and there is gas inside, you increase the pressure. What happens to the gas? In this case, the volume is not decreasing. Point taken, point taken. So, we can say that the gas is going to exert pressure on the balloon air. And the balloon air will, there will be pressure to push the air of the balloon air out. But the balloon air is going to resist that by sucking in more air to maintain the volume. Yes. Sir, in this case, is not it so that we increase the size of the balloon air in order to increase PSP? Because when you intentionally suck in air inside the balloon air, the net volume for the LTI gas decreases and hence PSP increases and all the other. See, it depends on whether you are looking at a closed system or an open system. If you look at an open system, then we consider that the balloon air is in contact with the atmosphere always. In that case, we can say yes, the air is being sucked and the pressure is being applied to the gas inside. When I say super pressure, I basically mean that when we fill the gas in the envelope, I can fill the gas in the envelope at the pressure equal to atmospheric. So, what will happen is that there will be zero differential pressure between the balloon air, between the gas inside and outside. But I fill the gas with a higher pressure and then I lock it or I close the valve. That is what is called super pressure. It is not because of the pressure from the balloon air. It is because of the intentional pressure that you put. So, consider the case and there is no balloon air. Just to make matters simple. You can still have super pressure, which would be pressurized balloon. When you have a balloon air and envelope and now you fill in air at high pressure. So, the balloon air is at some volume. Let us say it is locked. This is now a closed system. Now I push the gas at more pressure or I just increase the pressure of the gas inside. That is what is being discussed here. Now if we assume that we are not crossing the pressure height, which is always the case in all our calculations, then the weight of the gas inside is going to remain the same. Where does the LT gas will remain the same? We are not putting more gas. We are putting the same gas under higher pressure. So, if we look at this. Now look at the expression for the weight of the balloon air. This is the same thing which is copied and pasted here. Ratio of the pressures and temperature 1 minus i into k into v. Here I have ignored the contribution of E. Now, so the difference between the two of them, delta WBA, it will be essentially if we keep now assuming the PS is same. We are not changing the ambient air pressure. We are just changing the super pressure inside. So therefore, it will be PSP 1 and PSP 2. So the difference will come because of only PSPs. Therefore, earlier we got PS2 minus PS1 because we ignored super pressure. Now we ignore PS1 and PS2 change. We only get delta PSP 2 minus delta PSP 1 into kv upon same T A plus T SH, delta T SH. Now the net gross lift, the net lift is basically the gross lift difference minus the balloon air weight. But the gross lift is not going to change. The gross lift changes only if, see what is gross lift? Essentially gross lift is equal to the volume of the air displaced by the airship. Now that will not change unless the location changes between density changes or the volume changes. So therefore, the gross lift change will be 0. There will only be a change in the air, only the change in the air in the balloon air because of super pressure. Earlier there was some super pressure, now there is different super pressure. So therefore, the net lift will actually be negative if we increase super pressure. That means if we increase super pressure, actually net lift is going to go down, not up, minus sign there. Is this clear why it happens? Why does it go down? Does it seem logical to you that when you increase super pressure, that means when PS2 is more than PS1, the net lift will actually be, it is very simple. You put it in the expression and see it. Because understand that balloon air is actually weight. It is not lift, balloon air is negative. Air in the balloon is not helping us. Air in the balloon is our, we are forced to put it. So what is happening here is the air is being sucked in. Therefore, the net lift will actually be reduced. So when super pressure increases, the net lift reduces. Now the other way around, suppose super pressure decreases. That means you are filling in the air, you are filling in the gas but at a lower pressure. Of course, it has to be at least equal to atmospheric pressure, otherwise that will cave inside. But you can put, you can put it 500 Newton per meter square more or 1000 Newton per meter square more. Let us say you have 1000, you made it 100. You have reduced it. So what will happen now? The LTA gas will expand to fill the envelope. It will push the balloon air out. When you push the balloon air out, actually you save on weight because now you have less air inside. So the net lift will increase. Now coming back to the same expression, lifting gas, the lift, gross lift. Now I am going to take care of the other factors. So the expression remains the same. PS minus the contribution because of humidity into Kv by Ta. Ignore humidity. So you get PS Kv by Ta. So therefore, this is the same expression that we saw last time. So I will just, there is only one difference between this and the last one. Has somebody noticed the difference? I have shown the same slide before when I talked about change in PS. Actually the title should not be PS. It should be PSP. It should be change in PSP. So this portion is the same. The difference will come when you look at the ignoring of the PSP. So we recall that the weight of the lifting gas is obtained by this long expression. Here I have put the relative density of the lifting gas. In this expression, let us estimate the error in WLG and WBA by ignoring the value of PSP. So can you figure out the expression for that? So basically what you do is you calculate WLG with PSP considered and then without taking the difference upon WLG when it is considered into 100%. So it will be, you can just try yourself what you will get. So interestingly the terms are, there is a term in the bracket which will remain kind of constant because we are not looking at change in the humidity. We are not looking at the change in the purity of the gas which is Y. We are just looking at change in this delta PSP. So what you do is you get the expression. The expression will be this whole expression on the top 1 minus 1 minus RDPG Y PS etc. etc. This whole thing will be the WLG when you include superpressure minus just put delta PSP equal to 0 and the difference of the 2 divided by the whole expression. So things like IK, VR and this whole square bracket will be there in both the expressions. They will all knock off. So what will remain is only PS plus delta PSP and then PS and similarly delta PSP by T into that for the ballooning. So how much error is created depends on the numerical values of parameters like I, K, V etc etc. So for any operating condition we can calculate the error. We will do this question later. It will affect I but what has happened is if you see the calculations there is some I and there is in the beginning and there is some I in the end I1 and I2 and by using the expression relationship between I2 and I1 which is P1 by T2 we can eliminate the effect of I. So that is what we have done.