 In the last class, we looked at variation of certain key parameters of the atmosphere on the net static lift of an LTA system. And we could not complete all the 9 factors, there were 3 factors remaining. So we take those 3 factors today one by one. And then we will look at some simple tutorial problems so that the concepts can register in your mind. So the effect of variation of these 6 parameters the inflation fraction which represents the ratio of the lifting gas inside the envelope upon the total envelope volume. The inflation fraction changes when the airship gains or loses altitude because the ambient temperature and pressure will change. And with that there is a direct link between the inflation fraction. Atmospheric pressure comes directly in the numerator of the net static lift calculation. So when PS changes, there will be a change in the inflation fraction also and also in the lift. Superpressure essentially represents the pressure in the envelope which is more than the atmospheric to maintain the shape. This also we saw, you can model it as a change in pressure or change in super pressure, the formulae are very similar. Ambient temperature can change either slowly allowing the thermal equilibrium to happen between the outside air and the inside system. In that case we saw that there is no net static lift gain but if we do a rapid change in ambient temperature then the system does not get time to respond. And hence there is a change in the static lift. Superheat is the heating of the envelope beyond the atmospheric temperature due to the radiation that acts on the envelope. We saw that it can be modeled as an increase in temperature. And today we look at the remaining 3 factors, 2 factors and then we look at the effect of that on the lifting gas volume of the air. So relative humidity is the next but before we start I thought let us brush up a few things. So we will do a few problems that we will try to explain the concepts to us more clearly. So the airship has 80% inflation fraction at ISA sea level condition and it is flown to the altitude of 1000 meters ISA. So do the numerical yourself and see if you get the answers which are mentioned below. First is you fly it to a height of 1000 meters above mean sea level. How will you estimate the ambient pressure at 1000 meters above mean sea level? These are ISA conditions. So under ISA conditions the temperature at sea level is 288 degrees Kelvin and the pressure is 101325 Pascals. How do you get the temperature and pressure at the height of 1000 meters ISA? Okay, let us do it one by one. So can somebody answer this question? First of all how do you go about calculating this effect? So 90.2%. Okay. So how did you do it? You calculated the temperature at 1000 meters by subtracting 6.5 degrees from 288 and then you got the pressure using P by P naught or delta is equal to T by P naught to the power 5 point correct. And then in the formula I2 by I1 is equal to P2 by P1 T1 by T2. You have just put the same the effect of pressure only. Okay. Now let us look at the impact of temperature alone. How much is that? This is to be done in the same way 78.2% correct and now let us look at the effect of both of them. So just multiply the two factors that you got. So 90.2% is just the pressure, 78.2 is just the temperature and 88.2 which is actually 0.902 into 0.782 into 0.8 that is the effect. So we will notice that because of pressure the inflation fracture increases. Because of temperature it decreases and the net effect is an increase. So both of them work in the opposite direction. Okay. Another problem, now the airship is considered to be 10,000 meter cube envelope volume and we subject it to gradual increase in temperature from 20 degree centigrade to 30 degree centigrade and it is operating at sea level. The question is find the change in the gross lift that is delta Lg, the weight of the balloon air, change in the balloon air volume and change in the net static lift. So we are looking at changes in the gross lift, weight of the balloon air, net lift and the air volume of the air in the balloon air. So what is the formula for change in the gross lift because of slow increase in temperature? Do you recall the formula? So the ambient air pressure PS will remain the same. So basically gross lift is equal to the ambient air pressure into K into V upon T correct that is the simple formula if you neglect the effect of lifting gas purity is just the pressure into K into volume upon the temperature. So temperature is now becoming from A1 to A2. So therefore the delta gross lift will be the same ambient pressure PS which is prevalent into 1 upon T A1 minus 1 upon T A2 minus 1 upon T A1 into K into V. So V is 10,000 meter cube. So T2 is 288 plus 30, T1 is 288 plus 20, yes, yes, yes I am just adding to the, so 288 plus 16 is the C level value but here there is no question of C level you are given the value. So you just 273 plus whatever number is given. So 293.16 and 303.16 the value of K which is the constant is 0.03416 and the pressure at C level is 101325. So please calculate the number and tell me what is the change in the gross lift and also please tell me whether that change is positive or negative. By heating the lifting gas are you going to get a positive or a negative change in the gross lift? It should be negative it is nothing wrong it should be negative the gross static lift is going to decrease if you heat the or if the ambient temperature increases but the same cannot be said for the net static lift. Now the weight of the ballonet air so what about now there is a 10 degree constant super heat that means the envelope has now got a temperature 10 degrees higher than ambient because of the super heat the temperature of the gas inside is 10 degrees higher than atmospheric because of super heat. Now the weight of the ballonet air is also given by the same expression PS into K into V divided by T but now T will be the temperature plus the super heat. So it will become 303 plus 10 and 293 plus 10 everything else remaining the same. So the numerical value which was minus 3893 earlier it will come down or I should say there will be a positive addition to that so it will become less negative. So what is the ballonet air weight change delta V delta delta WBA minus 3649 yeah that seems reasonable minus 3649 Newton's. Now let us look at the net change in the static lift this will be the net change in the grass lift minus net change in the static lift. So earlier you got minus 3893 now you got minus 3649 so the difference between them is the change in the net static lift be careful about the signs. So will the net static lift increase or decrease increase or decrease why should it increase but so the decrease in lift is how much 3898 okay. So L n will be so delta G delta L G is minus 38 and from that you minus no minus minus of that correct so it will decrease by around 248 or 49 Newton's. So the net lift has decreased grass lift has decreased by a lot but now ballonet air has also been as also been taken out so the net lift is only 248 Newton's reduced okay and then what will be the weight of the ballonet air this is very simple you know the ballonet air mass you can use the density of the air at sea level conditions to calculate the value. So what is the volume occupied by the ballonet but be careful you must have maybe you have ignored one aspect that the density of the ballonet air will not remain same as sea level because there is a temperature change. So if the density of the ballonet air is considered to be the same so the density of the ballonet air is basically the pressure which is 101325 into the value K value upon the new temperature which is 298.2 plus 10 degrees so there will be a change in the density of the ballonet air. So what is the density of the ballonet air from 1.2256 it is going to change because of temperature. What density do you get for the ballonet air it will be P that is 101325 into K divided by the new T which is 10 degree higher due to super heat 1.16 okay I am getting 1.14 something yeah 1.1454 is what I am getting so what is the temperature you have taken in your calculations in the denominator. So how much is it 3.1 so it is earlier it was 298.2 plus 10 no look that temperature was 293 right and it became 309.3 so what will you take will you take 293 or will you take this one but why will you take that because that is only when it attains that temperature so why not take the mean of the two. So if you are waiting and if you are saying I wait till the whole system equalizes and the temperature conveys to the gas inside then you can say okay the temperature has become so one can either take the mean temperature or one can take the maximum temperature. So you take the maximum temperature so that is the case what is going to be the volume what is that more than the volume of the gas as the envelope of the 10,000 is the volume of the whole envelope here how can the balloon be 10,700 first thing is we stuck the units because the value of balloon air weight balloon air is in newtons and if you use density it is kg per meter cube so be careful about units first of all so there is a change in the balloon air weight of 365 Newton so first you convert that into kilograms if you want to work in kg per meter cube so the volume of the balloon air is going to be the change in the balloon air weight divided by rho into g how much is that yes correct that seems to be possible is it positive or is it negative but is it positive or negative 329 is it plus or minus negative that means the balloon air the balloon air is going to expel the air that is expected when you have superheat and when you are heating the body you expect that the balloon air will be throwing out the air okay so there are numbers here for you.