 Now we have to look at one more effect, one more effect has to be studied. Now the gas that you use inside the envelope, it need not be pure gas because after all it is a gas which comes from any commercial establishment, it will have some impurity. So that also is going to affect the gross and then net lift. Now we define lifting gas purity y. So in any gas container you will have the gas and air as a impurity. So this factor y is called as a purity factor. Now typically airships are not operated when this factor falls below 90%. Some companies insist 95% from safety point of view. And when you start, you will never have 100%. It will be 99.7%, 99.5% even when you have a fresh fill. And practically speaking it is a large envelope. It is very difficult to take out all the air from any gas bag. We have tried in our lab several times. Some amount of air always remains inside. Whatever section pressure etc you use, some amount of air will always remain inside. Secondly as the airship operates, this water vapor is actually a very small molecule and it also goes inside the envelope. It may not go at a very high rate but it will go inside. The gas inside will also go out slowly with time. I told you 1 liter per square meter per day is considered to be a good material property or acceptable material property. We are trying to make it lesser and lesser. So some LTA gas will go out and some moisture will come inside the envelope. So this purity will be lost over a period of time. So Y will be volume occupied by pure lifting gas upon the total lifting gas volume. So let us see. So we just assume that you are putting gas with purity Y inside. So the mass of the lifting gas will be equal to mass of the dry air which is the impurity. Now do not try to bring in wet air also in this because that is going to complicate the thing further. Then you have to do calculation inside calculation. So let us now assume that there is dry air which comes in and adds to the mass of the pure gas. So now it is also mixture. So same thing will apply V of LG, volume of lifting gas into rho of LG will be VDA into rho DA plus V PG into rho PG. Earlier we had VW here water vapor WV sorry. Now you have pure gas and air. So Anagat's law applies. So the volume fractions will add up. So volume of lifting gas will be volume occupied by the dry air plus volume occupied by the pure gas. There is at the container they have some volume total volume will be equal to the volume of each of them. So therefore volume of dry air is equal to volume of the lifting gas minus volume of the pure gas and you can then put this in the expression and then you divide by VLG. So you will get rho LG that is the density of the lifting gas will be VLG minus V PG into rho DA plus V PG rho G upon VLG. So I just copied that. Now V PG by VLG that is defined as the fraction volume of the pure gas divided by volume of the total lifting gas. Lifting gas means gas which also contains air impurity. So therefore you can replace that by 1 minus Y rho DA plus Y rho PG. Now PG is equal to rho RT you will get the same thing rho of pure gas and rho of dry air. You can relate it to the pressures, temperatures and using the same law you can relate it to pressures at the sea level standard conditions. So therefore rho LG that is density of the lifting gas. So you need this number to calculate the net static lift. That will be equal to 1 minus Y into rho 0 plus Y rho PG 0 and then the terms will come from the ratio. So now apply the same gas constant. So rho is equal to R into D0 where RD is relative density. Again you bring in the same constant relative density. So you replace rho PG 0 that is density of the pure gas under ISA conditions with relative density of the pure gas into the density at sea level condition. So the same thing you bring it down to the sea level. So you get a formula which gives you the value of density of lifting gas. Earlier we got a similar formula for air and water vapour. Now we have got a similar formula for lifting gas and air. Now air is the impurity in this case. So once again it will be inside, the value of Y will be inside. So just for your information this will be available to you obviously. So the same formula comes here. We notice that the relative density of helium is 0.1382 and hydrogen is 0.6995 compared to the air. So 1 minus will be also calculatable. So you get two simple formula for calculating the density of helium and density of the hydrogen gas. Now finally the last thing to be considered is called as super pressure and super heat. So the ambient air will not always be at the air, the temperature of the lifting gas will not remain same as ambient air. There will be some heat transfer over a period of time. So if I take a balloon with the gas inside, keep it in the atmosphere. After sometime there will be heat transfer so the air is and will also get heated up. Now how much time it takes? That is the function of the property of the envelope material. But we assume that there is some temperature increase of the gas inside that is Delta T SH. Similarly the pressure of the balloon inside is not going to be kept same as pressure outside. We have to keep slight over pressure. Why? Because we need to have the envelope tight. If I have the pressure inside same as outside and if I move the balloon the nose will bulge in. We have experienced this in one flight. If you observe carefully one of my videos shows that the nose was bulging inside. So therefore pressure inside of the gas will also be more than the ambient by an amount Delta PSP. This number is normally kept between 400 to 500 Newton per meter square Delta P it varies. Now if you give more Delta P you are stretching the membrane also the structure will also have problems. There will be more chance of tearing. If there are pinholes they will open up. This is what happened in R101. The pinholes opened up. So the formula then gets modified by the PS will become PS plus Delta PSP and TS will become TA plus Delta T SH. And with that you can calculate the lifting gas weight. Now the lifting gas weight is what? Minus the balloon. So you just put I into V and you get the lifting gas. So the same formula will now be created with the value of I. So you just have I into K V and V outside for weight of the lifting gas. Now just two simple cases before we wind up. One is assume pure lifting gas so no Y is equal to 1, no superheat, no super pressure, simple expression. ISA conditions and no superheat P will be equal to directly related to, rho will be related to the values and from there you can get the values of the gas as well as weight. Now I already mentioned to you that the water vapor contamination takes place. That molecule is very, very small. In fact water molecule is much smaller than nitrogen and oxygen so it permeates much better. Now this I will skip right now basically it is just a question of if you so basically the idea is the effect of the atmospheric air coming in is like the effect of humidity. So this is one problem which I would like you to do. Now I do not want to do it. This is the time. This is something which I would like you to note down and I will call it as a homework problem rather than a tutorial problem. So I am going to put this on Moodle page. So in order to copy it down it will come on the Moodle page. So this will take care of everything that you have seen today. Altitude, ISA plus 15, helium purity, inflation fraction, super pressure, superheat, humidity. For all these effects what is the lifting gas calculation?