 So, we will have a look at the two chapters of the two lectures that we covered. The first one was basically on introduction to propulsion system. So, it is an introduction. So, on that I do not want to do any numericals. We will do numericals based on level flight. So, the first question is a analytical question. You have to derive something and that question is about estimation of the velocity that is to be used for level flight. So, here is the first question. There is an aircraft with wing loading W by S. That means the aircraft weight is W and the wing area is S. It is cruising at a fixed altitude. So, that means rho is constant. And also the pilot flies it at an angle of attack at that velocity to ensure that its CL by CD is maximum. Given this condition, what should be the velocity of the aircraft to maintain the steady level flight or the cruise flight assuming a parabolic drag polar. That is CD is equal to CG0 plus KCL square. Remember that the parabolic drag polar is only an assumption. In real life the drag polar or the drag variation of the aircraft is not parabolic. Typically we see CD0 plus K1 CL square plus K2 CL. There is a K into CL term also. But we ignore it because by ignoring it you get a simple expression which can be used to derive some formulae easily. So, now the question is open to you. You have to derive the condition for V, the velocity for cruise such that CL by CD is maximum. And what is that velocity in terms of W by S in terms of CG0 and in K. So, your answer should contain a formula V for minimum CL by CD is equal to a function of W by S K and CD0. So, can you please derive the expression? So, the way to do this is first you get the condition for CL by CD maximum. What is the value of CL or V at which CL by CD is maximum? Does anybody know the answer already? What is the condition for CL by CD to be maximum? Yes please. CD is 0. That is clarified where is equal to 2 and 2 is equal to 1. Ok that is great. So, when the two drag terms are going to be equal that means when CD0 when CD0 is equal to K that is the condition. But can you derive this condition? So, to derive the condition what you will do is you will say that CD0, CD is equal to CD0 plus KCL square. Take a partial derivative with respect to CL such that CL by CD is maximum. Because you can easily derive it. So, ultimately as we heard the condition you will get will be CD0 is equal to K. Let us just save some time. Let us assume that the optimum condition for CL by CD to be maximum is CD0 is equal to K. If that is the case now it is very simple to derive the expression for velocity. So, please do that. Ultimately I should get an expression which says V is equal to a function of W by S, CD0 and K. So, whenever somebody gets the answer you just raise your hand. What is your answer? V is equal to you have to take K. I told you I want the answer in terms of only CD0, K, W by S and may be rho. That is all. I do not want to see anything else in the expression. W by S, rho, CD0 and K all of it is power 1 by 4. Then you do it. If you have something like square, square and power 4, 1 by 4 you do it make it little bit more elegant. Your answer may not be wrong but you can reform it slightly. It is actually very straight forward you have to just say lift equal to weight. Therefore, W is equal to half rho V square SCL. So, take S on the denominator on the left hand side. You get W by S is equal to half rho V square CL and CL is nothing but root of CD0 by K. So, that is it. It is a very straight forward expression. So that means if you know the drag polar of the aircraft that means if you know CD0 and K and if you know it is wing loading W by S, for a given altitude you can calculate what is the true airspeed required such that the CL by CD is maximum and if CL by CD is maximum then that is the condition corresponding to thrust required is minimum because we got it from drag required is equal to minimum. So, it is a nice elegant expression which can be very handy in calculating at what speed should you fly an aircraft so that the thrust required is minimum. Just look at the W by S which is easily available for every aircraft CD0, K and rho. Moving on, the next question is to try and get a variation of thrust available and thrust required at sea level for an actual aircraft and to do this exercise I have chosen this very beautiful aircraft called as the Fertile Republic A10 or also called as the Warthog. I say it is a beautiful aircraft but many people find it very ugly. But as I always say in war in combat we do not have a beauty contest, we have a contest of capability. So, let me present to you my friends, Fertile Republic in a war scenario. See how this aircraft can be used to create havoc. I have a question, as this aircraft is drifting away or peeling away it is throwing out some things. Can you see that? There are some white things coming out from the aircraft. Have you observed? What are these? What are these white things which the aircraft is shedding as it goes into a turn? Yes, smoke, smoke flare. What is the smoke flare for? Is the aircraft a smoker? Before going for a fight let me have a smoke. What is it for? It is not a smoke flare, there is a better word for it. You are near the point but not exactly. Anybody else? Maybe this side somebody. Yes. Afterburner is afterburner behind the engine, this is ahead of the engine. See once again, see where is it coming from and why will it come in pulses? Afterburner will be a continuous, you see it is coming out in pulses, so it is not a pulse jet but it is coming in pulses, okay. Anybody else? This is active under missile called flares. Correct. Not any under missiles, only the heat seeking missile, correct. So an aircraft that flies low and in the enemy territory for attack is always going to be under the radar of heat seeking missiles. So these are called as magnesium flares. These are a kind of a safety or a counter measure. So to confuse the infrared heat seeking missiles, you throw out magnesium flares. It is also called as a CHAF, C-H-A-F-F, CHAF dispensers. So you create very high heat source so the missile can be misguided, okay. So it is a defense or a safety mechanism. So the aircraft has peeled off and now you are going to combat. The Air Force wants to buy the new aircraft F-35 which is very expensive. As you saw to buy 50 F-35, they have to cancel or retire 300 of these aircraft and this is very effective. It is doing a lot of good work. So therefore, Senator John McCain says I will not support this activity of encouraging F-35 and downgrading or retiring A-10. So they have got a new lease of life, they have been refurbished and they are still being used. So there are several interesting design features of this aircraft but our question today is on performance. Okay. So we will look at a very, very boring thing that is a steady level flight of Warthouse. No combat, no bombs, no missiles, okay. So we are going to look at this particular aircraft and these are some of the key characteristics, the wing reference area, the aspect ratio and zero lift drag coefficient. Notice that the value of CDO is very high, it is 0.032 whereas the value is normally of the order of 0.02 or even 0.01815 for very smooth aircraft. So this is not a very smooth looking aircraft. It has got lots of appendages which come out because it carries armament. It is meant to carry a lot of armament. So it is not a very pretty aircraft, not aerodynamically very smooth but does not matter, okay. It has extremely powerful engines and not one but two of them form reliability point of view. So two jet engines, each of them giving you approximately 40 kilo Newtons of static thrust at sea level. So 80 kilo Newtons of thrust available at sea level from two engines. The weight of the aircraft is 103,000 Newtons. Our job is two-fold. We have to plot the power required curve at sea level and also we have to plot the power available curve at sea level and hence get the value of the maximum velocity at sea level, okay. That is our job today, okay. So to do this, first thing you do is please note down these four or five important things somewhere in your notebook, SW47 square meters, AR of wing 6.5, CD0 0.032, the Oswald Efficiency Factor or E as it is called 0.87, W103047 Newton and T will be 4298 Newton. These numbers please note down, okay. So how do we do this? We look at some velocities, say 100 meter per second. Is this a realistic number for an aircraft of this type, 100 meter per second some people are saying no. What is the problem? Is it too high, yes? But maybe it can fly at this speed, okay. It is not too high, it is not beyond the expected Vmax. We will know very soon whether it can fly or not depend on whether the thrust available is equal to thrust required or power available equal to power required or more, okay. Let us choose a number 100 meter per second, okay. So dynamic pressure at sea level for an aircraft which has got a speed of 100 meter per second is half rho V square. Now this is the first place where many students make a mistake. Observe and notice carefully. You are putting density in kg per meter cube which is 1.225 at sea level. You are putting velocity in meters per second. When you put these numbers, although the density is in kg per meter cube, the answer q infinity will not come in kg per meter square, it will come in Newton per meter square. The reason for that is units have to be balanced on both sides, okay. So q is dynamic pressure, it could be kg per meter square, it could be n per meter square. But you check density is kg per meter cube, velocity is meter per second. So this is kg per meter cube into meter square per second square. When you cancel the various units, you will find that g or meter per second square remains. So therefore the answer that you will get will be in Newton per meter square. Many people make this mistake because they have taken kg per meter cube, they think the answer is in kg per meter square. But be careful about these things, okay. So 6125 n by n square is the dynamic pressure at sea level at this velocity. Now the lift coefficient Cl for level flight is W by Qs because lift has to be equal to weight. W is in Newton's 103, 047, Q is 6125, S is 47, these numbers are not going to change. W and S are going to remain the same. So therefore Cl will be 0.358. This is a very appropriate number, Cl in level flight normally tends to be approximately between 0.3 to 0.4. So that is a right ballpark between 0.3 to 0.4 is the expected value of Cl for level flight. Once you know the Cl, then you know the value of E which is 0.87 AR which is 0.5. So therefore you can get the value of CD. Once you know the value of CD, you can get the value of D drag and thrust equal to drag. So the thrust required will be W by Cl by CD. Okay, remember there was a mistake in my notes, in my notes I had shown this thing little bit differently I think. So Tr is W by Cl by CD. So that is 103, 047 by CD, Cl by CD which is 9.13 which is 11287 Newtons. The thrust required is only 11287 Newtons, the thrust available is actually 40,000 into 2, quite a lot. So we are comfortable at this speed, we have much more thrust than required. Power required will be thrust into velocity, velocity is 100 meter per second, thrust required is 11287. So the power required will be 112900, so it will be in watts, 1.129 internal time power 6 watts or 1129 kilowatts okay. So now in a graph paper or in a sketch against 100 meter per second on x axis you have to mark a point 1129 kilowatts on the y axis, simple. This is how you get the power required at a given speed for study level flight. So we have used two conditions here, lift equal to weight, thrust equal to drag. Power available is basically the thrust available into the velocity, thrust available is going to be given to you, two engines each of 40298, so therefore it will be 80596 into V infinity in watts which will be 80.596 kilowatts, I have just divided by 1000. So total power available is 80 kilowatts. So at 100 meter per second, the velocity of 100 will give you 8059 kilowatts. Required 1127 available 8059, required is only one eighth of available, so absolutely no problem okay. So now we need to repeat this for other velocities. So I will just show you what we have already obtained for one particular speed correct. So now we have to take a guess, at this speed power available is 8 times power required roughly and the power required is proportional to V cube, so we need to take a guess up to what number we can go but we do not know a priory, our job is to find out where is the maximum. So what we can do is take some rough numbers, so I have just taken numbers 130, 160, 190. So now each one of you, you have a choice of any two velocities, just write down on your notebook the number, choose any number between 130, 160, 190, 220, 250, 280, 310. You can choose some other number also but then you cannot verify, because for these I have the values, so you can verify okay. So pick up any two numbers, any two numbers, any two velocities of your choice and what you do is fill up this table for your aircraft or sorry for your operating condition. Also there is one more column to be put which is the power available. At sea level we assume that the power available is not going to change much with the velocity, power available will be changed, sorry it will be proportional to velocity. So it is a sea level, so the maximum value remains the same, just multiply by the velocity you have chosen. So put one more column where you have P available in kilowatts okay, so this is a exercise for individuals but you have a choice of two velocities, whatever number you like, the first thing you need to calculate is dynamic pressure for which density is the same but velocities are different. So see what velocities have you chosen, 310 okay, that is it, one velocity, so I said choose two velocities. So at 310 do you get power required more than power available which means that you have crossed the maximum speed. So we have one hint from Sushil, if his calculations are right which I assume will be then the maximum velocity is below 310 okay, so Sushil take 280 maybe because 310 is more than what your aircraft can handle, 310 m per second would be almost near Mach 1, Mach 1 is 330 or 340 okay, so 330 and 310 means you are almost near Mach number 1, in other words it seems that this aircraft cannot fly supersonic at sea level but please note this is not an exact calculation, this is a small example, small example for the classroom purposes, so do not take it literally that these numbers are perfectly correct. Also remember one more thing that CD equal to CD0 plus k2l square is actually a parabolic approximation, if you reach very high Mach numbers near one for example, there will be other drag terms which will come into play, you will have probably a critical Mach number much before 0.9, so therefore this particular simplistic calculation may not be applicable but nevertheless we want to use it for the classroom purposes, okay anybody else have you finished two velocities, which one have you chosen, 220 and 310 okay, so you have for three velocities now, so are you confirming to me that the aircraft can fly at 280, power approximate, so which is more, PA is more which means the maximum velocity is slightly more than 280, no more than 280, no more than 280, does anybody know what is the maximum velocity, you cannot, it will be more than 280 less than 310 unless somebody does for 300, okay, so let us see the table now, okay, look at the numbers, I have the numbers only for power required, power available I do not have the numbers here, so check for the numbers against the velocity which you have chosen, if the numbers are not matching then please look at your calculations, so is there anybody who has a difference of opinion, so are these numbers okay, at least for your velocity, they seem to be okay, so we notice that the CL by CD is actually maximum at 100 itself amongst the numbers given as the velocity increases beyond 100, CL by CD is reducing, so it may be worthwhile to try 80 meter per second as a velocity, maybe at that velocities you may have a CL by CD higher than 9.13, we do not know, it is a matter of just calculating, okay, so this is something you can try out yourself, in fact what you can do write a small code and using that code you can calculate the value of power available and power required and then plot the graph, what will be nice if somebody can upload on Moodle, the value of p required p available for velocities from say 10 meter per second that will be too less I think, something like say 25 meter per second to the maximum value at C level, okay, so whoever wants to do it just to brush up their knowledge can do it and upload, so this is the, yes, yes there is a question. Can we assume that rag is called the maximum velocity can provide and then do it from behind? You can do that also but how do you know, you can do that to get the value but then you will get two answers, you will get a lower value and the higher value, so you can take the higher of the two, yes you can do that, if your question is just to find the velocity at which you can fly max, you do not have to do all this, you can do it directly by a single equation but I wanted to see the plot because I want to look at many more things, for example I also want to know what is the velocity to fly at which thrust required is minimum, what is the velocity required at which the power required is minimum, I want to confirm that CL by CD is actually minimum and I am more curious, I also want to plot CL 3 by 2 by CD and I want to confirm that CL 3 by 2 by CD is also maximum or minimum at a particular value, okay, so that is why I want you to do this, so here is the graph for power available versus power required for this example, for speeds below 100 we have not drawn it because we have not taken it up but you could do it and with that we get the intersection at around 295 meter per second, so by this very simplistic calculation the maximum speed of the aircraft at sea level will be 295 meter per second, actual value will be less than this because the drag will increase very rapidly, maybe this velocity is more than the critical Mach number, so therefore the drag will rise rapidly, so roughly this value can be considered, okay, let us look at what happens to the same thing at a higher altitude, so question number 3 is everything else remaining same just change the altitude of operation as 5000 meters, so earlier you had a thrust required and thrust available at sea level, now we want thrust required and available at a height of 5000 meters, so to do this we need to know 2 things, one thing we need to know is what is the density of air at 5000 meters, now this is the number which you can actually calculate because if you assume ISA conditions which we have to assume unless it is stated explicitly in the question you can assume the conditions to be ISA, okay, whenever there is a question if not mentioned you have to assume ISA, so under ISA conditions how will you find density of air at 5000 meters, what is the procedure you will follow? Sir we know that the temperature is 565 degrees per kilometer, okay under ISA conditions I agree that temperature drops at the rate of 6.5 degrees per kilometer, fine. So this whole formula just to memorize it under ISA conditions it becomes T by T naught to the power 5.258, yeah, so what is T1, what is T naught, how much T naught at sea level? 288 degree Kelvin, and density? 1.225 or 1.2256 kg per meter cube, correct, very good, so you know T naught, you know rho naught, you can get T1 by subtracting 6.5 degrees per kilometer and this subtraction is acceptable till 11 kilometers, from 1125 it remains constant, above that do not bother, it is not aeronautical, it is space, okay, so it is very simple, up to 11 kilometers reduce 6.5 degree per kilometer and T by T naught to the power 5.256 will give you the value of density, so that number comes to, you can just cross check this please quickly, 288.16 minus 6.5 into 5 upon 288.16 to the power 5.258 should be 0.7634, no, that should, that will give you the ratio and that ratio into 1.2256 should give you 1.73, just check it, right, so density of air at sea level is known, so 5 kilometers is known. Now we have to assume, you have to assume something now, we have to assume some variation of power required, velocity and temperature, so temperature we do not have to assume, we know it is the variation like this, now you may not need this, you can say I do not want to worry, I will take the value of, I will take the value of v true at some altitude and calculate the value of C L, then K, then C D, then T and then P, okay.