 So we have looked at these 3 so far, we have looked at inflation fraction, we have looked at the atmospheric pressure and the super pressure. Now the next thing to look at will be the ambient air temperature, other things remaining same when the ambient air temperature changes and changes slowly, this is the first expression. So when it changes slowly, you are giving enough time for the thermal equilibrium, we know that if you give enough time, there will be a conduction. So when TA changes or the ambient air temperature changes, the temperature of the lifting gas and the balloon air, they also change but they take time to change. However, there are 2 things, one is that both helium and hydrogen which are commonly used LTA gases, they are excellent in thermal conductivity. So they will quickly help in the temperature equilibrium to be reached. And when the TA changes slowly, then the super heat that is created because of change in the ambient air temperature can be ignored because super heat basically means when there is temperature which has not been absorbed or which has not been equalized over a period of time. That is why that is what is the super heat. So we recall that the lifting, the net lift is basically given by this expression which we have seen so many times. If I put E equal to 0, it becomes PS by TA into KV. Therefore the gross lift change will be just because of TA2 and TA1. So it will be 1 by TA2 minus 1 by TA1 into KV and PS can be taken out as common. Now also recall that the balloon air weight difference, we saw it last time is PS plus delta PSP into 1 minus I2 upon the temperature at 2 minus the same thing at 1 into KV. So let us ignore delta PSP with respect to PS because PS is a large quantity and delta PSP is a small quantity. So to make things simple, because remember we are looking at changes in one parameter but others are not changing. Either they are not changing or we are ignoring the minor changes. So we knock off delta PSP in this expression. So you put delta PSP equal to 0, you will get as PS times 1 minus I2 minus PS times 1 minus I1 upon the temperature which is going to be A2 and A1 plus the superheat. Now suppose we assume that PS is constant, PSH is constant because we are looking at only change in the TA now. So other things remaining constant, we can show that I2 by I1 which was basically P2 by P1, P2 by T1, T2 by P1, T1 by, thank you. So those pressures will get cancelled out because pressures are same, PS is same, PSH is same. So the pressure terms will get cancelled out, the only thing will be the temperature terms. And now if you see with this you can say this I2, you can replace for I2 here and replace for I1 here and you will find or if you just do cross multiply the terms inside, you can show that they will cancel out and ultimately you will get a need expression which will say that the change in the balloon air is going to be only PS times 1 upon the temperature at 2 including the superheat minus 1 upon temperature at 1 including superheat times K into V. So this is only because we have ignored the superheat contribution, we are allowing it to slowly equalize. Later on we will see what happens if you do not allow this to happen when it is sudden. So I am just copying and pasting the 2 expressions from the previous slide and the net lift is the difference between the gross weight and the balloon air. So all of them are not neatly in terms of only the temperatures and the superheats. So once again if you ignore superheat what do you get, what happens if you ignore superheat? 0. So if TSH is 0, delta TSH is 0 then there is no net lift increase, there is a direct compensation. So if you change TSH slowly the net lift will not change. You allow an airship to stand. So there is an airship which is standing outside. The temperature of the ambient air changes. If this temperature is conveyed beautifully inside the system and if both the LTA gas and the gas in the balloon air and the balloon air get to the same temperature, there will be no net lift change because there will be a cancellation of the increase. And if you look at only the LTA gas, weight, only the balloon air weight there may be differences but the net lift will remain the same and that is what matters to you. So we have looked at all these factors. The next one we look at is superheat. Everything else remaining same. We just look at now the effect of superheat. So just like super pressure, the same expressions you will only change delta TSH and you will ignore changes in T. So if you have superheat, what is superheat? Temperature increase because of exposure to high temperature. So both the balloon air and the gas inside are going to expand because of the increase in the temperature. So interestingly balloon air weight will reduce because the balloon air gas will be expanded but WLG will remain the same, the weight of the lifting gas will be, gas is not being thrown out. So its weight will remain the same, weight of the balloon air will reduce but weight of the lifting gas will remain the same, outside volume remains the same, the only differences are in the temperatures. So since we are not displacing heavier or lighter air therefore gross lift will remain the same. So net lift will change. Now many people they have this impression that the net lift changes because you are pushing the air out from the balloon air. Therefore there is more volume available for the LTA gas. Same mass of LTA gas occupies more volume. So there is a density decrease and because of this the difference in the density between the ambient air and the density inside increases and hence the lift increases. This is a fallacy. This is something which many people assume. The mechanism is not this and I will show you in the next slide. So once again I will repeat the fallacy so that you can understand. You will see these in books also and this confused me a lot in the beginning. This is what I also believed that you have an airship and you have a balloon and now you have superheat. So the lifting gas inside is going to expand and hence it will push the balloon air out. So the volume available for the LTA gas increases, the mass remains same, density reduces. So rho A minus rho G, difference increases, net lift increases. No, we will see the mechanism is a bit different. We just copied the formula from the last time. The weight of the balloon air is the difference of the 1 by temperature terms into PS into KV and the net lift is equal to gross lift minus the change in the balloon air. So the net lift will be equal to PS, the same expression that you saw. And here what we are doing now is simplifying it. If you simplify it, you will get delta T SH2 minus delta T SH1 is a numerator upon T A plus T SH1 and T A plus T SH2 in the denominator. Now if you look at the relative value of T A and delta T SH1 and also T A and delta T SH2, you can actually ignore. Just to get the order of magnitude analysis, you can ignore. So if you do that, you can approximate this value as just the change in the superheat divided by square of the ambient air temperature. And between the perm on the middle and on the right, the difference because of this ignoring is not more than 3 percent. So if you are happy with 3 percent error in general, the number will change depending on PS and all that, but generally there is around 3 percent. So therefore, without much loss of accuracy, we can get an expression for net lift change. Now let us look at the fallacy. Let us look at why there is a mistake. So we recall that the net lift is basically because of the 3 terms are there. First term is the gross lift which is rho A into VG. What is this? This is the weight of the air displaced. So because you brought the airship from somewhere to this place, air of density rho A, ambient air density times the V and this G is just for the units. So the first term rho into VG is the gross lift. The second term minus rho LGI into VG is the weight of the lifting gas which has been put in the space created by bringing the airship, correct? You brought the airship, along with the airship you got what? You got the envelope, gondola, blah, blah, blah, all those structures plus you also got, so right now you are ignoring everything else. So you are getting in lifting gas in the envelope and you are getting air in the balloon air. So rho LGI into I is why I into V is the volume occupied by the lifting gas, into the lifting gas density rho LGI, it will subtract from the gross lift. Then 1 minus I into V is the volume occupied by the balloon air, balloon air also has some gas air inside which has got density that is rho B A, rho of balloon air. In general you cannot assume it to be same as air density, we do not know. It may be same as rho A but we do not know right now. So we give it a term as rho B A, density of the balloon air. So with this you get the net lift but this is not payload. Net lift is what is the vertical force you get, now you subtract the rate of the system blah, blah, blah then you will get the payload, is this point clear to everybody? Now what is the contribution of the lifting gas? The lifting gas contribution is the first part or the middle part sorry which is negative contribution. It is actually a weight. So it is not giving you lift, it is taking away from some lift. So that contribution is negative rho LGI times I into VG. Now as the value of TSH1 approaches TSH2, so we know that the inflation fraction depends on pressure and temperature. So as the value of TSH1 approaches TSH2, the inflation fraction I1 will approach I2 value. So now what will be the lift, what will be the gross lift difference? It will be the difference between the inflation fraction I2 and I1. So when I change TSH, there will be a change in the lifting gas density. It will become rho LGI2 from rho LGI1, therefore the inflation fraction will become I2 from I1 and the difference in the lifting gross lift, gross lift will change only because of the difference between the gas. Now rho LGI is the density of the lifting gas that is equal to mass upon volume. So therefore rho LGI1 will be mass upon I1 V and rho LGI2 will be mass upon I2 V where I1 and I2 are the inflation fractions as the two conditions 1 and 2. Mass of the gas unions constant, you are not throwing out gas. So put the value of rho LGI1 and rho LGI2 in this expression. You will get M upon I1 sorry M upon I2 V into I2, I2 I2 will cancel and you will get M upon I1 V into I1, I1 I1 will cancel. So therefore delta LGI will be 0 because they will just cancel out each other. If delta LGI is 0 then the lifting gas contribution to that delta LGI is 0. So lifting gas does not contribute anything to the net lift when there is a change in the superheat because it undergoes a change in the density and through that it simply cancels out. So therefore change of the lifting gas is irrelevant. So when people say that density of the lifting gas has reduced it is actually irrelevant because we have seen here that the density of the lifting gas, its change is not going to play any role in the net lift calculation. Now we come to the next point. What is left now? Yes. No, you are looking only at a change in TSH. See do not make a system suddenly open and suddenly closed and then expect to have the same relationships valid. We are looking at only the contribution of the lifting gas. So yes, I is changing from I1 to I2, I is changing. It was I1 earlier it became I2. But while becoming I2 the density also has accordingly, so that is what we are trying to say the change in the inflation fraction will take care of the density change. Therefore density change is not the mechanism of generating the net lift. We need to figure out why the net lift is changing.