 Now what we will do is having learnt all these it is time now for us to figure out how much do we actually know okay. So I am going to give you a problem which I will give you as a homework problem. How many of you are able to solve it? You are able to solve it? You have all the answers? Anybody else? How many of you tried to solve the problem? Okay. So where did you get stuck? What is the equation? Yes, instead of 4.256 it is 5.256, correct. So very good. P ambient by P0 or P2 by P1 where 2 is any condition in the atmosphere, below that. P2 by P1 is equal to T2 by T1 times 5.256, correct. So that is one thing. What else? Anybody else? So only 2 people in the class have bothered to go back and check. That is a very low score and that is not a good idea. This is an elective course. You have chosen to take this course and if you do not show interest and do the course, do the things given to you then it is basically going to harm you only. So let us look at this problem today again. So please note down the problem first of all. This problem we have to solve today and right now. Do you have calculators? All of you? Okay. So please look at the calculators. So VEMV that is the envelope volume is 12,000 meter cube. Pressure altitude of 150,000 meter, what does it mean? What is meant by the term pressure altitude of 1500 meters? Someone should raise the hand and explain to me. Omkar, any idea? What is meant by pressure altitude of 1500 meters? No, no, no. That is pressure height. What you say that when the balloon is completely flush, that is pressure altitude or pressure height. This is something else. That is why I asked this question. This is not the pressure height. This is if the pressure altitude of 150 meters, what information is communicated to you? Yes. Sir, it means that the ambient pressure is equal to the standard atmosphere and temperature. Correct. That is right. So the airship is operating at some altitude at which the ambient air pressure is equal to what you see under ISA conditions at 1500 meters. This is the meaning of pressure altitude of 1500 meters. It does not mean that the airship is operating at 1500 meter. The airship may be operating at 5000 meters. But it so happens that the ambient air pressure at that altitude is equal to the ambient pressure that you see under ISA conditions at 1500 meters. Correct. That is ISA plus 15. So that means the ambient air temperature also is not ISA. It is ISA plus 15. Then you have Y. Helium here is 0.96. Then you have I equal to 0.95. Then delta PSP is 490 Pascals. Then delta TSH is 5 degree centigrade and E upon ES into 100 is equal to 50 percent. So you are supposed to estimate the gross and static net static lift. And now we can actually remove this because now we have a formulae for the Boulogne air weight. Now this is the information that we were discussing that PS by P naught that is the pressure of the ambient air at some altitude upon the pressure at sea level. It is called as delta pressure ratio of the atmosphere. And that is going to be T by T naught to the power 5.259. This formula holds only in the gradient level atmosphere which starts from sea level and stops at 11 kilometers. Now no airship normally goes beyond 11 kilometers. So we do not have to bother about heights above 11 kilometers. So you can assume this for your calculations. Has everybody copied it down? Okay, so now let us go ahead then. Let us solve it. So the first thing that we need to understand is that TS that is the ambient air temperature that will be equal to T0 minus lambda into HP. This is equal to 288 if I ignore that 0.16 minus lambda is 0.0065 6.5 degree per kilometer into 1500. So that means TS is going to be 278.5 degrees Kelvin. This is the first thing that we have to calculate. Now the value of delta that is equal to TS by T naught to the power 5.259 will be equal to 278.5 divided by 288. What is the value you get? Please solve this expression 3N. This is what you will get. So therefore what is the value of PS that will be 101325 into okay. Now the next thing is what is the value of delta P super pressure? This is 490 Pascals. What about the ambient temperature? Will it be equal to TS? TS is the standard air temperature. So what will it be? This will be equal to TS plus 15 that is 278.5 plus this is the value of TA to 93 okay. So now how much is the value of TA in centigrade? 20 point okay. So you remember there is a formula for getting ES. Do you remember the formula? That formula which we had for calculating the value of ES which uses the ambient air temperature in centigrade. So it is 18.678 plus you know minus C divided by 234.5. Do you remember this formula or not? Yes or no? So that is the case then I want the value of ES. No no 611.2 611.2. So what is the value of ES? 2375 units Pascals. Anybody else with some other number? Yeah I am getting 2382. So I will make it 2382. That is what I got in the calculations okay. So 2382 Pascals. Now what is the value of E? E will be equal to RD by 100 into ES. That will be 50 percent into so what is E? 1191 Pascal okay. Unless you solve it yourself you will not understand what is happening. Alright so let us move ahead now. We have a formula for gross static lift LG. LG is equal to ES minus 1 minus TA. So that becomes how much is PS? 84. How much is PS? 84691 minus. What about 1 minus RD? What is 1 minus RD for this right? Into 1191 divided by what is TA? Embedded temperature. Now how will you get the value of K? K is P0 rho 0 by T0 into G. So the value of K is actually known to us. This is the value of K. So you multiply that by 0.03416 times envelope volume which is 12,000 meter cube. So what is the value of 117? 117655. What is the units? No it will be newtons because G is taken care. Remember this K contains G also and do not make this mistake. It will be newtons okay. Now what is the lifting gas weight WLG? For that we have an expression that is equal to 1 minus 1 minus RD pure gas times Y PS plus delta PSP over TA plus delta T superheat times I into V ENV 1 minus 1 minus. Now what is the value of 1 minus RD? So 1 minus RD itself is 0.378 and Y is 0.95. Oh I am sorry. This is actually 0.8618 that is 1 minus RD into 0.96. Then you have 846919. How much is plus delta PSP is 490? This is 293.5 plus ISA plus 5 and this will be 0.95 into K which is 0.03416 into 12,000. This is WLG, lifting gas weight. So how much is lifting gas weight? So from the gross weight of 117655 newtons, the lifting gas weight itself is 1918 newtons okay. Now the next thing that we need to see is weight of the air in the balloonet. That will be PS plus delta PSP minus PS is 84691, delta PSP is 490 and E is 1191 divided by TA was 293.5 plus this is 5, 1 minus 95.95 times 0.03416 times 12,000. How much is WBA? Weight of air in the balloonet, 58 that is it. So now you have everything. So now LN is equal to LG minus WLG minus WBA. So this will be equal to what is the LG we got was? How much was the LG? WLG and air. So what is LN? The units are going to be newton 9. So that is the answer we are looking for. The net static lift is 92651 newton. And the gross weight, sorry the gross static lift was 117656 newton. Now you can yourself check what happens if you ignore humidity. What will be the percentage error? Okay. So with that we come to the end of today's lecture okay. In the next class we will look at effect of certain variations in the atmospheric properties on the net static lift. How it changes? Okay. Thank you.