 So, let us see if we have understood this by doing a simple tutorial problem. So this problem will run now in the whole class for all the example that we see. So we assume there is an airship of 6000 meter cube and envelope volume and a ballonet of 2400 meter cube at sea level condition. So we are ignoring sea level conditions and for simplicity we are ignoring the relative humidity that means that E will be 0. We are also ignoring super heat and also ignoring the super pressure. So calculate I at sea level and I at h equal to 5000 meters. So first part of this I, I at sea level you should do it orally, what will be I at sea level? 0.6. Simple. 6000 total minus 2400 air difference will be 3600 divided by 6000 that will be the amount of the fraction of the gas so it is 0.6 okay. So we are coming to the same question that I showed you earlier. Sea level the value of I is 0.6 okay. Now I am asking you what happens to the I at h equal to 5000 meter it will be 1 but just confirm it. It should be 1 because we have just now seen that graph for sea level condition I equal to 0.6 it goes to 1.0. So please reconform this you may get 0.999 or 0.998 etc but reconform it please. So how do you confirm this? You have that formula I2 by I1. And now what you need to do is you need to calculate the pressure at 5000 meters. How do you do that? What is the pressure at sea level? 101325 Pascals okay. How do you get the ambient air pressure under ISA conditions at 5000 meters correct. So that is a simple formula which I would like you to remember and that is that there is a parameter called delta is a parameter called theta, theta is temperature ratio okay. So theta is equal to T upon T0, T0 being the sea level condition and T being the condition at any altitude. So T by T0 is the temperature ratio. So the value of T at any altitude up to 11 kilometers, 11000 meters will be 6.5 degree reduced for kilometer altitude from sea level value of 288.16 or take it as 288. So first get the value of temperature at this particular altitude that will be 288-5 into 6.5 right okay. So now you have the temperature at 5 kilometer altitude in ISA conditions. Now the pressure ratio that is P by P0 also called as delta is equal to theta or T by T0 to the power 5.235, 5.236 I think. This is only under ISA conditions. So the value of P by P0 or P2 by P1 where 2 is the 5000 feet and 1 is sea level that will be equal to T2 by T1 to the power 5.253, 5.256. So give me the value of P 5000 by P sea level, 0.5 okay, 3 digits is okay, 3 or 4 digits are sufficient okay. So now what will be I2 by I1, I2 by I1 will be equal to the pressure ratio only. Now to get I2 you just multiply I1 which is 0.6 with this ratio, no I think you are doing the opposite. Instead of dividing you are multiplying so you are getting 2, in fact you should divide them. So notice I2 by I1 is equal to P1 by P2, so I2 will be I1 into P1 by P2, so P1 by P2, P1 is the sea level, P2 is the ambient at 5000. So the number will be more than 1, P1 will be more than 1, 0.53 okay. And then if you multiply by I1 what do you get, 0.999, we expected to be nearly 1 correct. So it is correct. So that is right. ISL is equal to 0.0, not 0.4, 0.6 and I at 5K will be 0.999 okay. So this was for, now let us calculate the change in the gross lift, yeah it is correct, it is 0.6, I am going to correct it, ISL is 0.6 not 0.4 and I5K is 0.999 okay. So estimate delta LG, delta VBA and delta LN. Now delta VBA is what volume of the balloon air, not the weight. So you have to just get it from the density because it is ISA conditions, therefore density of air at ISA is 1.2256 kg per meter cube. So first calculate delta LG, then WBA, hence VBA and then delta N which is the net gross lift. So I will open the formula for you, you have to use these two expressions, first expression that you need to use is LG. So for LG this is LG. So delta PS is now 100. So delta PE means this, this is delta P, PS2 minus PS1 is 100 Pascals. So LG will be 100 into KV by TA, all you need is the value of K, V is the volume which is already given 6000 meter cube, TA is the ambient air temperature and you are talking about C level. So what is delta LG first? 30, 33 Newtons, 33 kilo Newton, no delta LG, 33 kilo Newton, so much change, it is only 100 Pascals, 101325 is the total pressure, your delta P is only 100, it should be 70 I think. Around 70 is what I expect, around 70, okay. Now let us go up and let us look at how do you get the value of volume of ballonet air. So for that you have to use this equation number 3. equation number 3 contains difference of the pressure. So that delta P is PS2 minus PS1, it is 100 divided by TA which is the ambient air temperature 288.16, delta PSH is 0, we are ignoring superheat, no the question is at C level, see the question is at C level, so you take it at C level. So WBA2, my WBA1 will be 100 into K into 6000 upon TA plus delta PSH, delta PSH is being ignored, right, no superheat change in the current calculations. So you will get the weight of the ballonet air and with the density you can get the volume. So what is the volume, change in the volume of the ballonet air, all students must bring calculators in this class, actually it is quite straight forward, why so much time. So what is WBA, delta WBA sorry, 71 watt, grams, kg, Newton, Newton, now you get it from density, from density get it in meter cube, no, 6 meter cubes, that is right, that is the correct answer, it should be approximately 5.9 to 6 meter cubes, okay. Now what about delta n, delta n will be equal to delta, delta n will be equal to delta, delta lg, so now is the time for me to correct it and make it 0.6.