 Now let us look at the difference of the atmospheric pressure. Now we will look at only that situation where maybe of the same altitude the ambient pressure is not equal to the standard value. So let us recall from our previous slides. Now unless you read them up regularly you will not be able to appreciate but I cannot do much about it. So it is your job to go back home and look at these slides and get familiar. So just for you to recall the gross lift of the envelope basically we derived in the class that it is equal to the total pressure PS at the altitude minus 1 minus RDWV times E where RDWV is the relative density of the water vapor correct and E is the humidity. So how do you estimate humidity R H by 100. So the value of E is R H by 100 and this particular term on the numerator which is subtracting from PS is the effect of relative humidity. We saw that the effect was around 1.6 to 2 percent but still for completeness we want to leave the term there into KV upon TA what is the value of what is TA it is the ambient air temperature and what is K what is V. So V is the envelope volume and K is the constant which was rho 0 P0 P0 by rho 0 I think. So the value of that is 0.0 yeah 0.0 see the value of pi is 3.1416 this value is 0.03416 something like that okay. So it is a constant it is a constant K because it only involves rho 0 P0 P0 which are constant values. Now if you ignore the value of E or let us say put E equal to 0 then the numerator will have only the PS term so we will have a simple expression and now if I take 2 conditions condition 1 condition 2 or from operating condition 1 to condition 2 I want to see if the PS changes from PS 1 to PS 2 what will be the change in the gross lift or LG. So from the expression above you can always work out that if I do LG at 1 equal to PS1 minus this into KV upon TA1 and LG2 is equal to PS2 minus the immediate term times KV by TA2. Now the TA2 and TA1 are same here because the only thing changing is the pressure we are assuming that everything else remains constant only the pressure changes. So delta LG will be this is not correct it should be delta LG so the change in LG will be PS2 minus PS1 into KV by TA fine okay. Now we also want to recall that the weight of air in the balloon A it was obtained using a very similar expression as PS plus delta PSP which is super pressure minus the humidity effect upon TA plus TSH which is the super heat effect. Now we use 1 minus I because the air in the balloon A or the gas balloon A volume is 1 minus I times V okay. So just like we do the calculation for the lifting gas the same calculations we do also for the balloon A only thing is this is a complete expression which also has the super pressure and the super heat effect. Suppose we again ignore humidity because it is only 1.6% or so so you will get a simple expression by knocking off this term in the numerator and now WBA2 minus WBA1 will be the difference in the balloon A air weight. So in this expression you first put PS is equal to PS1 and then you put PS is equal to PS2 and in the first case it will be 1 minus I1 and then it will become 1 minus I2. Recall that the inflation fraction will change even though we are changing only PS but the effect of that is also to change the inflation fraction. The moment air in the balloon A is expelled or taken the inflation fraction cannot remain the same. So the difference of the 2 expressions will give you the expression for WBA2 minus WBA1. So going from a condition when the pressure is PS1 to a condition when pressure is PS2 everything else remaining same we have one expression for LG that is the delta LG lifting gross lift increment and we have one expression for the change in the balloon A air. So essentially if you look at net lift so I am just copying and pasting it here and now we can simplify this. Now how do we simplify this? If you see in the numerator you have delta PSP in both the terms and there is a minus sign in between. So if you expand this you will get PS2 minus I2 into PS2 plus delta PSP minus I2 into delta PSP and the same thing for the second term and if you then and then another is the same so you can take it common and then you can subtract the terms. So doing that so I have just done it for you I have just expanded the whole term for you okay. There are few who are looking at me and nodding your head you will really regret this it is better that you write down what I am doing because when it comes back when you come to the calculations at that time you will not have the slides in front of you. You will get the finished product which you can understand only if you have done these calculations okay it is only for ease in communication that we are using the power point otherwise we are actually supposed to do these derivations ourselves. If you wish I can do it in front of you here if it helps but it is already there on the screen so it is better that you do it. So you do it yourself and then check whether you get the same expression if not you have to check or if there is some mistake here you have to tell me. So what I have done here is the terms which contain I1 and I2 I have club them together and the terms which are independent of I1 and I2 I have put them in the first bracket and the denominator is the same so it becomes it remains common okay shall we go ahead alright. Now notice that we have two terms in the numerator one of them has PS2 plus delta PSP with I2 the other one has PS1 plus delta PSP with I1. So if you recall very in the last few slides I have shown you one expression between I2 and I1 which related the pressures and temperatures okay. So what we can do is we can use that expression and by that you can actually eliminate what you will find is that this term PS2 plus delta P into I2 will be equal to PS1 plus delta PSP into I1 because I2 by I1 is equal to PS1 plus delta PSP dip upon PS2 plus delta PSP. So now this particular term has a negative sign this has a positive sign and both these terms are actually equal. So quite simply they will get knocked off. So here it is recall that I2 upon I1 is equal to PS1 plus delta PS1 upon PS2 delta PS2 delta SP2 and similarly for the temperatures and we are only assuming here that there is a change in PS everything else is not changing. So therefore if you assume that TA2 is equal to TA1 because there is no temperature change. If you assume that the super heat at 2 and 1 are same because we are not assuming any super heat change and if you also assume that the super pressure also remains the same the only thing that changes is PS1 and PS2. So when you do that you can easily see that the term on the right side here will become equal because these 2 will be equal and these 2 will be equal so they will knock off. So this will knock off and become 1 and you will get I2 by I1 is equal to PS1 plus delta PSP no need to put 1 or 2 similarly PS2 plus delta PSP and with that this I2 into this is equal to I1 into this and that if you put there you will get a very simple expression because the inflation fraction terms will simply cancel each other. So once that happens the difference in the ballonet air weight is nothing but difference in the ambient pressures divided by the temperature plus super heat times K into V. This expression is very similar to what we got for the gross lift. For the gross lift if you recall we got a similar expression that the gross lift change is equal to PS2 minus PS1 upon Ta KV. The only difference is that there is no super heat considered here and here super heat comes into play. So now the change in the net static lift and that is what our whole capsule is about. This whole chapter is about change in the static lift because of change in the parameters that is equal to the difference in the gross lift minus difference in the ballonet air weight. Is this point clear? Because this will come every time. Essentially in a system the gross lift is equal to weight of the air displaced and the net lift is weight of the gross lift minus the weight of the ballonet. Where does the weight in the ballonet? And the lifting gas weight itself. So the delta n will be the net gross lift that will be equal to the sorry the net lift and that will be equal to the difference in the gross lift minus difference in the ballonet air weight. For LG2 minus LG1 we have an expression which is PS2 minus PS1 KV upon Ta and for WBA to WBA1 we have an expression right above it that is PS2 minus PS1 Ta plus delta TSH KV. When you put them together you will get a simple expression in which you can actually take out PS2 minus PS1 common in the numerator. So in the case of lifting gas we have 1 by Ta in the case of ballonet we have 1 by Ta plus TSH. So therefore by simplification further you can have this numerator you can say Ta plus TSH minus Ta upon this and you know you can easily get this particular expression. So delta n can be easily calculated. Is there something missing here? I also observed just now that the Ta term is missing. This should actually be PS2 minus PS1 upon Ta delta TSH upon Ta plus TSH KV. So this I will be correcting when I upload this. The T term because if you take the denominators there are 2 different values. So when you do the subtraction it will become like that. So what do we learn from here? Then the change in the net lift when you simply provide higher pressure that is now where is the pressure changing? This PS change takes place where? This is in the atmosphere. So when you operate from any condition where PS is equal to PS2 compared to when you pressure was PS1 there is a change in the net lift and that lift is that changes simply the difference between the pressures times the ratio of superheat upon Ta plus TSH into K into V. Now if you assume that there is no superheat. If you assume that there is no superheat then you find that net lift change is 0 because it will subtract 1 upon Ta minus 1 upon Ta. So this is an interesting observation that if just the pressure is changed without any super pressure with no superheat then there is no change. So just by changing ambient pressure and keeping everything else constant including superheat net lift change is 0.