 Now let us look at those very very rare scenarios in which for some reason you are required to fly above the pressure height. So in these cases you have reached a situation where the ballonet is flush there is no air inside but still you want to go slightly higher. So how do you go higher? What is the option you have? Engine thrust. Yes. So you can start swiveling your engine. You can give vector thrust that is one option available. Suppose your engine is not having vectoring capability. Now what do you do? You still want to fly higher than needed, higher than the pressure altitude. Throw the stop you make the airship lighter that means throw away the ballast. So it is possible to do that but is there any danger with that? What is the danger? You cannot be able to recover it. You may throw water or you may throw even sand so that you do not hurt people below right. So you have thrown 500 kg or 300 kg of water or sand. So now when you come down to land you will be stuck at some altitude at which ballonet will become fully inflated and now you cannot come down unless you accumulate some weight. So yes you can do this. You can throw off some payload or you can get you can make yourself statically light but the danger is then you have to ensure that you are statically heavy to the required value when you come down. So that is not a safe thing. So is there any other way? In this case you are talking about changing the envelope volume as you go up allowing more but essentially you are changing the shape of the envelope. So you are saying that fly with an envelope which is folded or contracted and then allow it more and more expansion. But ultimately the envelope will reach some maximum expanded condition. I want to go above that what do I do? So yes the only way that you have is to throw some lifting gas no other way. So what happens in an airship is that there are automatic pressure relief valves. See the gas wants to expand because the ambient pressure is falling and the envelope cannot take it it will tear. So the best way is to allow the gas to expand by going out. So there are pressure relief valves which are automatically primed at a particular delta P and if for some reason the delta P becomes more than that the lifting gas is released. This is because if you do not do that there will be a very high super pressure and that can tear the envelope. Now typically we allow the super pressure or the delta P values to be approximately 500 Pascal's and these valves open when the delta P becomes twice that. This is a normal practice just for your information. So it is released or valve automatically. Now if you are operating at the pressure height of 1500 meters per meter altitude. You know that pressure also changes with every altitude. So per meter altitude the change in pressure is 10.3 Pascal and you can only go up to 1000 Pascal. So what it means is that in 50 meters itself the pressure will double from 500 to 1000 Pascal's. So within 50 meters of delta H the valve will open. So you might be able to fly perhaps only 50 meters higher than the pressure altitude before the gas is pumped out to reduce the envelope pressure. Now what will happen is that the lifting gas will be lost because valves are opening and the gas is going out. So let us see what happens because of that. So let us look at the scenario where the operating altitude is more than the pressure altitude. What is happening there is some fraction of the lifting gas is valve or released to the atmosphere. Now let us try to find out what is that fraction. So what we do is we define something like an imaginary inflation fraction at maximum altitude. So something like the inflation fraction is equal to 1 at pressure altitude. It is lower than 1 at lower altitudes and we go to a higher altitude you have let us say 1.02 or 1.03 is an imaginary number because it cannot be more than 1. So this imaginary fraction is equal to the original fraction which is 1 at pressure altitude plus delta inflation fraction which is equal to the ratio of the last gas. So you can put it like this. This is something like if the pressure altitude was actually higher so much more fraction would have been needed at the sea level. So IMA is now the inflation fraction at the maximum altitude which is higher than the operating or which is higher than the pressure altitude and there is this lost inflation fraction. If you can calculate the value of delta IL that will give you how much lifting gas has to be thrown out or how much lifting gas is lost. For example you can see here this example this number will help you. Suppose the operating altitude or called as the maximum altitude, suppose the maximum altitude is such that 2 percent of the inflation lifting gas is lost then the imaginary inflation fraction at the maximum altitude is equal to 1.02. One gives you pressure altitude 2 percent is lost hence that is equivalent to that height which would have got 2 percent more inflation fraction requirement at sea level. So if 90 percent inflation fraction gives you a pressure altitude of 1500 meters or yeah 1500 meters then 1560 meters would be at maybe 92 percent. So that 2 percent is the imaginary additional inflation fraction. Now there are 2 cases which arise listen very carefully this is a very interesting scenario now. You might exceed the pressure altitude only for a momentary reason or momentarily as a transient or you might do it with a sustained manner. That means for a very long time you fly. So what happens is if it is a transient phenomenon which happens over a small period of time. So the process of changes in the pressure and the temperature inside the envelope is an adiabatic process there is no time for heat exchange to take place. So we assume that the temperature does not really change a lot it is an adiabatic process but if you allow it to fly for a long time then you are giving time for the air to come in equilibrium with the ambient conditions because we are giving time to it. So depending on whether it is a transient exceedance of pressure altitude or a sustained exceedance of pressure altitude you are going to get 2 different expressions because one process will be adiabatic one process will not be adiabatic. So let us see first a transient exceedance which is an adiabatic process. So the airship climbs from the pressure altitude to something like a maximum altitude in an adiabatic fashion. So it can be shown now I have not derived this expression but it can be shown that the inflation fraction in this case will be equal to the inflation fraction at the pressure altitude which is 1 into the pressure of the ambient air at the operating altitude in standard conditions upon pressure to the power minus 1 by gamma Lg where gamma Lg is the specific heat ratio for the lifting gas. So this particular expression comes from a standard adiabatic pressure relationships which you must have learned in thermodynamics if there is a P by P0 is equal to T by T cannot power 1 by 1 upon 1 plus 1 plus gamma etc. From there you can easily assume that the inflation fraction will be related like this and since IpH equal to 1 therefore Ima or the inflation fraction for the operating altitude or maximum altitude higher than the pressure altitude will be the ratio of pressures upon minus 1 by gamma Lg you can also call it as PS by sorry PSPH upon PS and the power plus 1 by gamma MLG. So this is one expression which we have to keep in mind that means the inflation fraction for a temporary exceedance will be the ratio of pressures to the power minus 1 by gamma Lg. Now Ima is basically equal to 1 plus delta Il so the delta Il can be back calculated as PSph by PSma power 1 by gamma Lg minus 1 here this minus 1 has been converted by changing the ratio. Notice here you have PSma this is in denominator here you have PSph this is in numerator so that takes care of this minus 1. So the expression that you get finally is that the inflation fraction or the additional inflation fraction is equal to the ratio of the pressures upon 1 by gamma Lg minus 1 which is the ratio at the pressure altitude. Now suppose we allow it to fly for a very long time above the pressure altitude then we will allow heat transfer to take place we will allow we will allow the gas to adiabatically cool. Initially it will be heated but then it will be cooled and it will be it will be heated to the temperature at that particular altitude. So now what will happen the gas is going to basically expand if it expands there will be further loss of lifting gas. So if you allow it to sustain you will find that the lifting gas loss will be larger we will see that in an example that you will solve soon. So for this we do not worry about the transient phenomena we look at the final expression we have we look at the formula which says that the inflation fraction at any condition 2 is equal to the density ratios upon the initial value. So therefore Ima will become sigma Sph by sigma Sma and since Iph is equal to 1 right. So this I1 will be equal to 1 this will become ph this will become ma. So once again Ima the inflation fraction required for the pressure height exceedance is 1 plus delta L. So therefore delta I1 is equal to the density ratio minus 1. Ima should be why should it be in the numerator? I am replacing yeah it depends if I replace so yeah I am replacing basically Iph by 1. So this value delta L see if you look here 2 is ma yeah so it should be reverse it should be reverse okay. So let us see let us see I think I made a mistake here I made a mistake here I think it should be sigma S1 by S2 so that is why there is a error here let us see. Now let us assume a situation where you are only exceeding by 1 percent that means 1 percent of lifting gas is lost whether you do it in a transient manner or in a sustained manner. So in the transient manner as we have just now seen the delta I1 will be equal to PS P H by PS ma to the power 1 by gamma Lg minus 1 and that value is equal to 0.01 because we have said there is a 1 percent change. So therefore PS ma or the pressure ambient pressure at the altitude will be equal to the ambient pressure at the pressure altitude upon 1.01 times gamma Lg. In the case of sustained delta L is the density ratio and therefore sigma S ma will be equal to sigma S P H upon 1.01. So with this now let us do the following let us calculate the values for transient as well as sustained exceedance for 1 percent lifting gas assume it to be helium assume to have no super pressure and no super heat. So I wanted to calculate these numbers so there are 2 expressions available what will be the value of gamma of helium 1.67 so try it out. So first thing is you should know what is the value of sigma at the various pressure altitudes. May I suggest something to do this question you will need to know the value of sigma that means you need the atmospheric tables or you have to calculate the values. So I will leave it for you to confirm when you go back home just observe that in a transient exceedance the pressure height is 135 meters exceeded in sustained it is only 100 meters. For a pressure height of 1000 meters when you are going to higher pressure altitude the exceedance for 1 percent gas is less. So a sustained exceedance because of the cooling of the gas okay so now let us go ahead. What you also want to calculate is the weight of the lifting gas. So how do you estimate the weight of the lifting gas? What you know is the percentage loss of the lifting gas that I am interested in knowing the weight. So the weight of the lifting gas will be equal to the weight of the lifting gas under the ground conditions divided by the inflation fraction at MA because under the ground condition the weight of the lifting gas is WLG when you filled it on the ground you filled the gas and you filled the ballonet. So there was some weight of the lifting gas right and the original inflation fraction was IIG at the ground level. So when you go to the maximum altitude the inflation factor has changed to IMA basically IMA is the percentage you know inflation fraction basically is an indication of how much gas is lost. So with this you can calculate the weight of the lifting gas that has been lost to reach a particular maximum altitude above the pressure altitude. Interestingly this formula does not depend on whether you do it in a transient fashion or in a sustained fashion. As long as you use the correct value of IMA the value of IMA will not be the same in both the cases because the formulae are different. So as long as you use the correct value of IMA depending on whether it is a transient loss or a sustained loss you can use the same formula to calculate the weight of the lifting gas. Therefore the weight that is lost is equal to the weight at the end minus weight in the beginning or in other words WLG will be equal to the weight at the beginning upon into 1 minus sorry 1 upon IMA minus 1. So you calculate the value of IMA using the formulae described earlier, appropriate to the operating condition transient or sustained and from there knowing the weight of lifting gas at sea level you can get the change value. Now once again I am copying and pasting an old formula about the nut static lift. So the nut static lift is equal to the gross lift at the new altitude minus the gross lift at the pressure altitude. So we are now calculating the change in the net static lift when you move from when you move from pressure altitude to a slightly higher altitude. So this will come simply by the ratio of pressure and temperature into K into V we are ignoring the effect of any humidity. This is the old formula. So therefore the delta ln change in the net lift will be the change in the gross lift minus the weight of the lifting gas. Because what has happened is the airship is lighter, is lighter because some gas has been thrown out. That is WLG. How much gas has gone out? You know from the inflation fraction from the INI calculations and the weight of the original lifting gas and LGMA and LGPH are available using standard expressions for gross lift. So in normal circumstances we also consider weight of the ballonet air but the ballonet is empty in this case. Therefore WBA has been ignored. So therefore one can say that delta N is equal to the expression PSMA upon TAMA minus PSPH upon TAPH into KV minus the weight of the lifting gas which in the previous slide was obtained as weight of the lifting gas on the ground into 1 upon IMA minus 1. So now everything is known to you in this expression. From a pressure altitude to a maximum altitude, higher than pressure altitude, the value of PSTA is known to you from the atmospheric tables. The value of K if you recall is the constant depending on the lifting gas. V is the volume, WLG is the lifting gas weight on the ground value that is also known to you from the inflation factor into the type of gas. So with this you can calculate. Now use the correct value of IMA whether it is a sustained or it is a transient increase. If you are operating under ISA conditions then things become simpler because then the value of TA will be equal to TS, the value of you know and therefore we also know that PS by T will become PS by TS and P by T is equal to sigma into P0 by T0. So when you put that in the expression it becomes much simpler, the ratios change to the density ratios and then you get P0, T0 as a constant value and K into V minus this expression okay. So now I think this formula is what we should note down because we will require this in calculation in the near future. So the change in net lift is equal to under ISA conditions the change in the density ratios so sigma SMA minus sigma SPH, density ratio at the maximum altitude minus density ratio at the pressure altitude into P0 by T0 which are constants P0 being 101325, T0 being 288.16 Kelvin, K is the constant for a particular lifting gas and V is the volume of the envelope minus WLG which is the weight of the lifting gas times 1 upon IMA minus 1 and this IMA will be obtained by calculations either PMA by PSH power 1 upon gamma or sigma ratio.