Hello and welcome. In the last lecture, if you recall, we had looked at the implication of the gravity along with the propellant burn rate on the overall terminal performance and we had also looked at the loss of energy due to gravity and its connection with the burn rate. Now, we are ready to bring in the third force that we have considered in our earlier discussion that is the aerodynamic drag and consider its implications. So, let us begin. So, let us begin our discussion on the impact of drag. If we look at the drag as an effect which is of a reasonably smaller amount, then we find that for most missions it will essentially be an order of magnitude lower than gravity and it is treated as a tertiary effect. While in the case of gravity, we made a simplified assumption about the value of the gravity because the drag is an even smaller order effect, we can use an even simpler model to capture the effect of drag which is primarily based on an energy concept which we will look at next. So, the simplified drag model essentially is aimed at capturing the overall loss of energy that will occur during the complete trajectory through the definition of the concept of a constant average deceleration that we can introduce in the equations of motion so that they remain linear and we will try to match the overall energy that gets lost because of such a constant average deceleration against the actual energy loss which is likely to happen and it is found that such an approach gives a reasonable performance estimate. So, we again assume that our trajectory is along a straight line that is a radial line. Under that condition, we can now rewrite our equations of motion which we borrow from our earlier discussion on the equations of motion for gravity and then we just subtract one more term that is d by m on the right hand side where d is the drag and m is the mass. Now, this d by m is what I call an acceleration or with a negative sign we can even treat it as a deceleration to do drag that the vehicle will experience while moving through atmosphere and it is just a pure subtraction from the value of the thrust. Now, in this model the ad term that I have introduced is going to be treated as a constant drag acceleration term similar to a constant gravitational term that we have already introduced in the last lecture. Of course, we know that d by m is not a constant because the drag varies in a particular manner as a function of altitude and velocity while the mass varies as per the burn rate introduced. So, obviously d by m which is the actual acceleration at each time instant will vary along the trajectory. So, how does it vary and here we have something to help us which we have already seen earlier. So, let us consider the generic drag versus altitude plot which we have seen earlier as dynamic pressure versus altitude plot but now only it is scaled version. Here you will notice that the x axis which earlier was only the dynamic pressure has now been multiplied with the reference area and the drag coefficient so that this represents the drag variation as well. You will realize that the variation will remain the same except that it will get scaled to the force units rather than the pressure units because the surface area at least through the atmospheric phase of the flight will remain constant and the drag coefficient as we have discussed earlier is 1 for a bluff body assumption that we have already made. So, you will realize that the same curve can adequately represent the variation of drag with respect to altitude. Of course, in the equation that we have written down in the previous slide the drag is a function of time but you would realize is that as altitude itself is a function of time drag versus altitude is an adequate representation in the present case. Now, let us bring in the idea of the energy loss. So, as drag is a force and altitude is a distance from our basic understanding of mechanics we realize that if I calculate the area under the curve that will represent the total energy contained in the drag term because the area under the curve is nothing but the integral of capital D that is drag into dh taken over the applicable altitude. In this case as we have noted earlier also beyond 40 kilometer altitude the actual drag is negligible so that the area under the curve can adequately represent the total energy that would be lost because of drag. Now, we introduce an approximation through the red dotted triangle that is shown in the picture. Now, the reason for introducing this is not very far to see because the shape of the curve is very nearly a triangle except that it is a curved triangle. By introducing the straight lines I introduce an approximation but I have another point. If you see the triangle closely you will find that the overestimate of energy in the upper part by the triangle is to some extent compensated by the underestimate of the energy or the area by the lower part of the triangle which obviously means that in some manner the triangle can reasonably capture the overall energy that might be lost. The reason for generating this triangle is that once I create this triangle I can match this triangle with the green rectangle as the next level of energy matching. So, I can say that the area under the triangle which is same as the area under the actual curve is also the area under the rectangle and once I assume that the area under the rectangle is same as the area under the curve then the width of the rectangle is nothing but my average drag acceleration that I have used in the expression. I hope this translation is clear to you. So, once we do this then it is extremely simple to go from the actual curve to the value of average drag just by looking at certain features of the curve. So, in this case because we are drawing the triangle at the tip of the actual curve which is the maximum value of the drag by dividing this value with the instantaneous mass at that altitude it will become an instantaneous drag acceleration value which is the peak value that the system will experience. So, once I introduce this idea of the peak value of the acceleration which is nothing but the point at which the triangle has the apex then we know that the area under the triangle and the area under the rectangle will be exactly matched if the width of the rectangle is half of the peak value. This is directly from our triangle relations of area calculation which is nothing but half base into height. So, half the height is nothing but the total area as long as I match the height average value of acceleration is just the half of the peak value by matching the area and you will realize that that can represent a reasonably good approximation to a smaller order effect of drag. So, that in the initial estimates we can always get the impact of the drag for a given trajectory without too much of computational effort which would otherwise be required for solving the complete nonlinear differential equation. With this let us rework the previous example that we have seen in our gravity loss case to see what is now the order of magnitude or the impact of the drag on the terminal performance using the simplified expression that we have generated. So, let us take the same problem the burn rate of 600 pages per second and now introduce the three aerodynamic parameter necessary that is the CD naught 1 the surface area assumed to be pi meter square and one last parameter which is introduced just to kind of use the computational effort at this point is to make a stipulation that the drag will be maximum at roughly around 50 seconds which is the half of the total burn time of 100 seconds in this case. Let us assume that it would be somewhere around this the reason why this number is chosen will become clearer as we look at the solution but I just wanted to keep in mind the fact that the peak of the drag is roughly around the 10 kilometer to 12 kilometer altitude range which means that if we peg the time at around a value which results in the altitude value of around 10 to 12 kilometers then the drag calculated at that altitude will essentially represent the maximum value of drag for that particular trajectory it is an approximation but it helps us to simplify our analysis. So, let us try and determine the impact of drag. So, let us first recall the non drag values that we have already seen earlier for this case that is the final burnout velocity is 2.3 kilometers per second and the altitude is 78 kilometers when there is no drag and we would like to compare these values with the value of the same parameters when we introduce the drag term. So, let us now look at the analysis step by step. So, let us first calculate the velocity that this particular vehicle will reach in 50 seconds. So, it is the application of the same expression that is the expression with gravity but without drag and we find that this results in the velocity of 616 meters per second the corresponding altitude that is reached is around 13 kilometers which is not bad. You will find that by and large the drag will peak around this altitude for most of the trajectories and then we go to the atmospheric tables that are readily available and from that atmospheric table we read out the value of atmospheric density at this altitude. So, which is 0.267 kg per meter cube. Now, the next step is to calculate the drag which is nothing but half rho v square at 13 kilometer altitude into cd into the surface area. So, if you do this calculation you will find that the drag at this altitude is 159 kilo Newton. Further because we have proceeded up to 50 percent of the total burn time we have consumed half the propellant which means about 30 tons of propellant has been consumed. So, the effective mass at this point is only 50 tons. Based on these two quantities we now simply calculate the peak value of the drag curve as 3.18 meters per second square and compare this number with the gravitational acceleration which is about 9.81 it is one third of that but more importantly because we are going to use an average value it is not a constant it is only the peak value the average value just half of this this comes out to be 1.59 meters per second square. Now, you can see that this number is practically one sixth almost an order of magnitude lower than the value of the gravitational acceleration. So, our original hypothesis that the drag is a tertiary effect is also justified and with that small term we can also say that our approximation that we have introduced using the energy balance methodology is also justifiable. And with this drag I suggest that you do this exercise yourself as per the equation that we have seen earlier you will find that the velocity which was 2.30 without the drag becomes 2.14. So, this is the amount of loss about 160 meters per second is the loss in velocity due to drag. And similarly the altitude which was 78 kilometers becomes 70 kilometers about 8 kilometer loss in the altitude together you can add the potential and the kinetic energy loss together and that will effectively tell you how much of energy loss has happened over and above the gravity loss. You will find that this energy loss is significantly lower than the energy loss due to gravity for the same case because the ideal velocity is 3.264 kilometers per second that represents a particular energy compared to that the gravity case gives an energy lower by about 36 percent. You will find that this will add another 8 to 10 percent in the loss so that the total loss is likely to be around 45 percent of the complete energy. Let me now introduce an idea that we have not seen earlier. If you recall we had given an expression for gravity energy loss as a function of beta but we had not actually characterized the impact of beta directly but we had made a mention that chances are that the drag loss is likely to be higher if you burn faster. Let us try and examine that idea now to the same example and this time we take a significantly higher burn rate of 3000 kgs per second which is 5 times more than what was there in the previous case. You will immediately realize that if you are going to burn at 3000 kgs per second all the propellant will get burnt in 20 seconds itself and that the peak would be lower than what happens at 20 seconds. So, in this case we make a stipulation that the drag peak will occur somewhere around 15 seconds which is in the lower atmosphere and we are going to assume that this should again be roughly around 10 to 12 kilometers altitude but now because we have changed the burn rate our non-drag values change. We have already realized that the implication of a higher burn rate is to reduce the gravity loss. So, we get a higher burnout velocity at 3.07 kilometers per second and a significantly lower altitude at 23.4 kilometers but the total energy is significantly higher because the loss is much smaller. Now assuming that T equal to 15 seconds we calculate the altitude which is achieved at this time and that altitude turns out to be 11.5 kilometers. So, our assumption is justifiable that we are still in the same ballpark of 10 to 12 kilometers. The velocity now at this altitude please note instead of 616 meters per second that is what you saw in the previous case is now 3 times it is almost 1800 meters per second very high velocity at nearly the same altitude. So, obviously you see that your drag term which was 159 kilo Newton is practically 10 times at 1713 kilo Newton because the drag is significantly higher and more importantly because you have burned faster you have consumed lot more propellant in 15 seconds in 15 seconds you have actually consumed 45 times because you have consumed 45 tons of propellant the residual mass is only 35 tons so mass is also lower. So, your peak of acceleration is significantly higher which was earlier 3.48 is now on already 49 meters per second square that is practically 5 times your gravity which means your drag loss if you do this is going to be 5 times the loss that you are going to get because of gravity of course we can also calculate the average drag acceleration that is half of 49 and based on that we calculate the velocity and altitude and now you see that the impact of this drag is significantly higher the 3.07 kilometer per second has become 2.58 which is almost 0.5 or 500 meters per second lower and effectively you can say about 20 percent loss 15 to 20 percent loss in velocity. Similarly, you have another 5 kilometer loss in altitude which is also close to about you can say 20 percent together you will find that this is going to represent a very large energy loss that brings us to the same point that we had seen earlier but had not this particular result in front of us to make any comments. So, now let us look at it that if we burn faster we reduce the gravity loss but then we increase the drag loss similarly if we burn slower we increase the gravity loss but then we reduce the drag loss. So, is there going to be a golden mean of burn rate where we might be able to keep both the losses to their minimum value and thus make the mission an optimal one. So, this is the nice hypothesis that we need to examine and we need to do this by modeling both the losses together so that we now get what is called a combined loss and see if we can get a minimum for the combined loss and map it to a burn rate which is going to be applicable for that trajectory to say that if you burn the propellant at this rate then this is the most efficient mission with the given propellant and the lift of mass and the propellant that you can carry out. Considering the loss due to gravity and the loss due to aerodynamic drag. So, let us take the same example but this time we do not implement any burn rate. We are now going to look at the combined loss. So, let us see if we can get combined energy loss as a function of burn rate and locate the minima and determine the corresponding optimal burn rate. Just to simplify our analysis and the various algebraic steps, let us again assume that the drag profile peak is going to be around 12 kilometer altitude so that we can fix the density value and let us also use the beta in the range of 400 to 1200. Let us see if that gives us something. Now, I want to show you something on the right hand side which is the approximate solution and then I will tell you how the solution has been obtained. So, this solution is the solution for the loss of combined energy as a function of burn rate starting from 400 kg per second to about 1200 kgs per second. And you see that this is a classic concave curve inverted parabola with its minima lying very close to the 600 kg per second that we have used in our examples earlier. Which means the number that we had used was very close to the actual minima that you are going to get. The above curve has been obtained through MATLAB by writing a small code. I suggest that you can also try that exercise. I will tell you the steps involved in the code that is to be developed that is you take the solution because of gravity. Based on that solution generate the velocity at 12 kilometer altitude and with that velocity and the density at 12 kilometer altitude get the value of drag. And once you get the value of drag you can find out the time taken to reach that altitude multiply that time taken with beta to find out what is the residual mass as that altitude as a function of beta. It is not going to be a constant it will now be a function of beta. Based on that obtain the drag peak which is going to be a function of beta and then go back to the expression and recalculate the velocity and the altitude under the action of this drag acceleration term as a function of beta. Add this loss to the loss because of gravity and that becomes the combined loss. This is how that solution has been obtained. Of course you need to realize that this is a highly simplified analysis so obviously it is an approximate presentation and you will probably need to do a more rigorous analysis to arrive at the actual optimal rate. But the idea that I am trying to propose is that by making reasonable assumptions we can simplify the analytical procedure and still get a reasonable understanding of the physics involved in the process. So you realize that there is a possibility of arriving at an optimal solution for the burn rate by trying to adjust such that the combined energy loss is a minimum. Of course such an analysis would always be carried out even for the simplified case of a constant burn rate in a more rigorous manner towards the end of the design through complete nonlinear simulation of equations and numerical solutions. We will find that at the initial stages of design when you are trying to size the rocket and you want to understand how much loss would be there because of gravity and drag put together so that you can appropriately put the propellant and the ISP correspondingly as part of your design solution, you need a kind of a gross estimate of the burn rate. A very, very crude thumb rule which can be commonly used in such situations is that an optimal beta would generally lie close to a trajectory for which the gravity and the drag loss are nearly equal in their magnitude. This is based on the fact that variation of these two effects are broadly similar in nature, completely inverse of each other and also that when gravity is a secondary effect the drag is tertiary and when the drag is secondary the gravity is tertiary so that they kind of complement each other and that when there are nearly of the same order of magnitude the chances are that that would be the trajectory where you would have the minimum loss. Just to understand this let us go back to the previous picture. If you look at this picture you will find that the effective combined loss is of the order of around 47 percent which is slightly more for more value of beta beyond 600 kg per second. You already know that for this burn rate the loss due to gravity is about 36 percent so which means there is another 10 percent which has got added because of the drag and that when you start increasing the burn rate further why the energy loss due to gravity will reduce it will not reduce at the same rate at which the loss due to drag will increase because of the square of velocity. So you will realize that the loss because of drag will quickly climb up so that within the same ballpark it would also reach a reasonably high value and that it will overtake the loss because of gravity. So you can clearly see the increasing trend as you note the curve on the right hand side. To summarize drag is a smaller order effect which can be captured from energy consideration. Further an optimal burn rate exists that results in the most efficient mission for a given vehicle from the point of view of minimizing the combined loss. Hi so with this we have established a reasonable analytical basis for understanding the implication of various forces which are present in the trajectory solution and also a broad idea of how to estimate such a loss and find its impact on the terminal parameters. With this we conclude our discussion on the motion along a straight line and we will move over to the more realistic case of motion along a curvilinear path that we will look at in the next lecture. So bye see you in the next lecture and thank you.


Yeah, we on Boss Talk 101. And then got the nerve to say what you are due to somebody for talking. Yeah, he definitely, he was very keen to say, I set him up. And that was the reason I wanted to talk about and address it. Because that's a serious statement to say the way he said it. I said it, man. Didn't you hear the tone of his voice? Yeah, I heard it. He was real, having him say, I set him up. See, but what happened when you make that statement, you're in the heat of what you feel? Oh, his facial expression and everything. You seen it? He's in the heat of what he feels. Once the feeling go, so this is why you shouldn't rely on your feelings. Once the feeling leave, that initial feeling has other feelings that's attached to it. Whether you admit it, whether you process it or not, you have to hear Shane in the new video. I got it. Yeah, you do. Oh, Shane now. Watch this. That's what's so. They shamed now. Oh, nigga was just talking. Do you feel me? Play it back. Play it back. Play it back. Now, this is the second video where they come together. After they got the backlash. After they got the backlash. After we made the video. After we made the boss talk video, here you come. Now, watch this. Here you go. Do you want to speak on it? Who wants to speak on it? The thing is, first of all, nigga was just talking. Oh, he sounds so humble. Yeah, I don't care. I mean, the thing is, I was just talking with him. You would set up another black brother? I would set up Charleston white.  And I was like, I get the message. I hate niggas like him.  Listen, now he get the message, but Shane gets the last interview. Well, he said he didn't even, he had listened to him before. He already, but now all of a sudden, because of the backlash, he say, okay, I wouldn't set him up. He on the white boy platform, say I set a nigga up and the world saw it. And the world, and the world got him in his feelings. And the world rebuked him. And the world rebuked him. And the world rebuked him.  The white boy, man, that's some, I'm going to say this because I've got to say this. When I seen the way Adam 22, it hasn't Adam 22, no jump was say, he reminded me of Jerry Heller on straight out of Compton.  That's what we're going to do. I don't know. Again, defamation of character libel. I don't really care.   Cause Jerry Heller, after they made that diss song on that movie straight out of Compton, he was like, we got to do something. How did they do this? Cause he was trying to make them niggas up.  You know what? I'll call my friends at the JDL. They'll handle him. We'll see how much he likes that. He wanted to try to make sure he keep the control. See, see over the situation with those guys, because that's what Jerry Heller was doing.  Come on, Eric, we got to get organized.  You know what? I always knew that he was a hateful human being. He was trying to keep that control. He was embarrassed for them even being, cause they called him out. Like you, when you were speaking, you said, you said that's the boss. They have to go to their boss. And I could feel it in his spirit when he said that, when you said that he felt some type of way cause he was being exposed for really what was happening on that platform. I listened to the niggas say Google who he is and say nigga, you are no name. Adam Jumper is the motherfucker. The podcast don't say Adam 22 AD in the niggas sidekick. It says Adam 22. That's that white boy's name on there. And he got them two niggas right there to help aid him. The A in payphone. The white boy had with Stern, same thing. Nigga, your name ain't on there. I made my name big nigga.  I treat white boys bad if they like Adam 22. They don't run nothing in this country.  I don't like liberal white boys, period.   We play on them kind of white boys. The friendly white boy don't know how we really feel. We play on him. That redneck rice is white boy know how we really feel and we know how we got a better understanding with each other. A California white boy, we can't know, man.  Hey, nigga, I stand on everything I say. I stand on this shit, man. Yeah, we on Boss Talk 101.


I have been working through a lot of my works in progress projects lately and I've been able to add so many new things to my closet. This wool mini skirt notwithstanding. But before I get into the details of the pattern, I have to show the journey that this fabric has been on. To help me get through my unfinished projects, I decided to let fate help pick which patterns I would be working on. I have something called decision fatigue. If the decision is too difficult or overwhelming for me, then I just give up on making a decision at all. So I wrote down all the projects on note cards and have been randomly drawing one every day and documenting the progress on my TikTok. Here's the video that I made for this wool skirt. Hi, you guys. Welcome back. It's day eight of the unfinished project challenge. Day eight of a lot. So let's get into it. Let's draw randomly which unfinished project I'm going to be working on today. Uh oh, I got two. I'll always wonder what that one was. This is work in progress two, which means it's one of the ones from the bin. Let me go grab it. Okay, this is going to be a good one. From what I remember, this is like a little cape jacket. Uh, I think the fabric is so pretty, right? I love like an unconventional plaid situation. I think it's wool. And it looks like the pattern is McCall's 7051. I do not have the pattern here. I do not have the instructions here. So I've got to go find that somewhere before I can do anything. We have a serious problem. Okay, so this is very clearly like a wool fabric. It has no stretch, no give. And this is the pattern. What was I thinking? There's no way I wanted to make a wool top. That makes no sense. And what's even more weird is that the sleeves are cut to be short sleeves. Like what? I don't know what to do. This is all scrap. So that gives me a little bit of hope that I can like salvage this and turn it into like a little coat. But I genuinely cannot figure out what I was planning on doing with this. I've been on quite a journey today, but what I ended up doing was cutting it all up into a completely different pattern. I ended up making McCall's 7438, which is this little mini skirt pattern. And guys, it is mini. It is very, very short. Like I can't bend over. This is a stand up only outfit. But I think given what I had to work with, it's going to be fine. Right? So as you can see, it was initially cut to be some sort of top, which is still so confusing to me. So the fact that I was able to turn that into this skirt makes me really happy. I'm going to be showing you all sorts of angles of the skirt and talking a lot about fit, the fabric I use, the pattern instructions, hacks and alterations I made and anything I'll do differently when I make another one in this video. Be sure to like it though, so other people will see it. Subscribe and click the bell if you want to be notified every time I post a video. Videos highlighting my makes and sewing pattern reviews go up every Monday. I already have dozens for you to watch if you need a little binge sesh. Check out the playlist in the top right corner of your screen. It opens in a new tab, so you won't lose your place here, I promise. If you're new here, I'm Lindsay. I sew all my own clothes and basically just come on here and talk about it, so I don't annoy the heck out of my friends and family. I am so glad you found me. Please be sure to introduce yourself in the comments so I can give you a formal welcome. The pattern is McCalls 7438 and it is described as a semi-fitted below waist skirt that has a contour waistband, back zipper and stitched hem. View A features an overlay, View B has front pleats with decorative buttons, View C has a center front pleat and View D has self-fringe. This pattern is out of print now, but there are some available on Etsy and if you haven't seen, fringe is everything right now. This skirt is $350 and made from faux leather, which you know isn't super expensive, so I see a View D in my future. This version though is View C with the center front pleat. Because I was just trying to salvage this work in progress and also trying to do it within a day, I did take some creative liberties. First, I cut the largest size of the pattern, which is a 22. That is one to two sizes bigger than my hips and four sizes bigger than my waist, but I did not want to fit around with getting that waistband to fit perfectly and doing the zipper on wool, especially considering I wasn't even sure that this make would work out. Consider it an extreme wearable muslin. So to get the waistband to fit, I just added two channels to the back waistband and inserted two pieces of elastic in them. It's definitely an interesting choice considering the fabric, but I don't hate it. It kind of adds to the casual nature of this skirt and fabric. I did though have the wear with all to not add elastic to the front of the waistband. I did not want any bulk in that area. My body does a good enough job of that on its own, so I get all the benefits of an elastic waist skirt, but a clean look of a flat front best of both worlds. The fabric is a wool or more likely a wool blend that I got several years ago. I can't remember where I got it from online somewhere, but I remember loving the fact that it was clearly a cool weather fabric, but with unconventional colors. I love plaids. They give a nod to a few different seasons in one. It means I can wear this skirt multiple months and it not seem out of place. If I wear it during the fall, the colors seem fall like. If I wear it during the winter, the substrate seems winter like, and so the timing is always right to wear this skirt. The pattern did not call for a lining, but since it was wool, I knew I would be uncomfortable if I did not line the whole thing. I also used a lining fabric for the waistband facing so as to not add a ton of extra bulk to that area. The waistband facing is sewn as instructed, but the skirt lining is actually attached at the waist and the hem. I just sewed it to the hem first, understitched that seam, pressed it inside the skirt, and then sandwiched that raw edge into the waistband seams. So the skirt is pretty short, which I acknowledge in the TikTok. Wearing it with tights definitely helps, but if I want to make other versions to wear without tights, I will need to adjust that. I think that adding a little bit of length all around will help, but so would a full seat adjustment in the back, so I will do that too. Also, with a little more time on my hands, I'll be able to use my fast fit system to find the perfect size for the waistband and be able to make a fitted one without an elastic casing. If you don't know, my fast fit system is a little guide that I created and I use all the time to figure out which size of a pattern to cut, factoring in design and wearing ease. It takes a lot of the guesswork out of picking a size. You can download it from my website and I have linked it in the description box for you. I've also linked my most recent pattern review that was for McCall's 8218. If you want to see what else I've been sewing lately and if you want to keep up with my unfinished project challenge, be sure to follow me on TikTok. But that is going to do it for me today. Thank you so much for watching and I will see you all very soon. Bye!


In the last class we had seen the performance of a turbojet we derived equations for the non-dimensional thrust as well as we had derived expressions for ISP of a turbojet. Now let us put in some typical numbers and see what is it that we get out of it and we will see some interesting facets of it. So firstly we will look at optimal expansion of flow in the nozzle here there are two conditions that we need to look at one is C level thrust or C level ISP and what happens at cruise conditions so at C level P0 is equal to 1 atmosphere or 1.1 mega Pascal's and T0 would be if we assume an IS ISA atmosphere model then it will be 288 Kelvin and M0 is 0 at C level before take at takeoff conditions and now if we assume the compression ratio pi C to be very low that is pi C to be of the order of around 4 then my tau C would be something like 1.484 and if we take TT4 to be 1000 Kelvin then this makes ? B is equal to something like 3.4 and we know the Q of the fuel this is kerosene so 42 mega joules per kg and let us take the CP of air and flow gases to be the same and let us take it to be 1 kilo joule per kg Kelvin okay. So with this conditions now you have to find out what is it at the cruise altitude of around 11 km so altitude whatever things that are going to change the things that are going to change are these two only atmospheric conditions and at cruise conditions this Mach number will also change so we need to change the only three parameters P0 would be lower and it would be somewhere around 22.6 kilo Pascal's and P0 would be this is we have taken ISC international standard atmosphere so therefore we get this now if we use this and calculate f by m.aa and ISP okay we have all the parameters here to calculate both of them or we will get this kind of a table m. is equal to 0 then I will also put V7 V0 both these in meters per second and lastly ISP in kilo Newton second per kg okay so if we take the C level performance that is at C level m0 is equal to 0 and using this we will get f by m.aa not as 1.8 and this would be something like 612 meters per second and this would be 0 and ISP would be something like this now at 11 km m0 would be 0.8 and this would reduce slightly to 1.76 okay now there are two things to notice one is that irrespective of the altitude you see that f by m.aa not as nearly constant right this happens because of two things one notice that at C level this is large the velocity differential is large okay the mass flow rate is very small velocity differential is large mass flow rate is very small so you end up getting a high thrust because of the velocity differential at a higher altitude the velocity differential is reduced but you have air coming in now at this velocity therefore what you get is a larger mass flow rate so you get a larger mass flow rate with a smaller velocity differential therefore you get nearly the same thrust or same non-dimensional thrust what happens to ISP ISP is come down which means that SFC is gone up if I were to write SFC also here SFC would be in kg per kg or this would be something like 0.8 this would be like 1.06 so we are cruising at an altitude wherein the SFC is low why do we do that I think it would be more beneficial if we were to stay on ground or close to ground and carry out our operation is not from the looks of it what we see is that the ISP has come down or SFC has increased so why do you think if this is happening or this is desirable non desirable what is the opinion you have so one say something so what happens no not really well actually the reason why we operate it at a higher altitude is primarily the drag is less at a higher altitude right you are not looking at drag of the engine per say you are looking at the drag of the entire aircraft wherein when I was discussing about turbofan engines I said G goes in for a bypass ratio of 9 and says that the engine of the nasal drag is very small compared to the overall drag and therefore they go in for a large bypass ratio right now what we are looking at is the drag of the entire aircraft goes down as altitude increases because roading road decreases as you go up in altitude right now density decreases drag decreases but your SFC will increase because what we have set in here is if you notice I did not change this part TT4 TT4 is the same at both the altitudes and if you look at ? B in this case ? B would be because T0 is lower so ? B for TT4 of equal to 1000 Kelvin would be 4.63 that is what we are saying is you need more fuel to bring the air from a lower temperature after compression to 1000 Kelvin because we have set that temperature so therefore we will find that the SFC will be higher right okay and that is the reason we see this now we can look at similar numbers for choke flow through the nozzle the only thing that changes would be I will just put them here itself now here we will assume a slightly larger compression ratio of round 12 which means that Tc would also go up it would be much well can also increase the turbine inlet temperature this means ? B will go up to the rest of the quantities remain the same here again sorry I had written 100 Kelvin here you are not pointed out it was to be 1000 Kelvin earlier and that makes it 4.63 okay and that is why we saw that change in ISP or SFC now we change it to 1200 Kelvin ? B would be 5.55 and this would mean correspondingly all these quantities will change because now we have choke flow in the nozzle the thrust consists of two parts one is the convective flux part and the pressure thrust part so this will again have two parts one point and at an altitude is what we see notice that as you go up an altitude the fraction of the pressure thrust is increasing okay it is a lower fraction here right or if you look at this number it is a lower fraction here compared to a higher fraction here right the overall thrust why does this happen why does pressure thrust part increase as you go up an altitude higher ambient pressure if you see is one-fifth so the pressure thrust part is going to increase if you go up an altitude okay and again you see the same feature here that SFC has increased and that is primarily because you need to heat the air from a lower temperature to a set temperature of 1200 here instead of 1000 earlier so therefore you will find that SFC is higher in this case okay fine okay then we have dealt with two things we looked at optimal optimally expanded flow through the nozzle and we looked at choke nozzle okay now let us look at what happens if you have the after burner on and the other method of increasing the thrust that is water methanol injection let us look at these two methods and how to derive expressions for them okay so firstly let us look at what happens if we have the after burner on I will retain these set of numbers because I want to use this to look at what happens without the after burner with the after burner conditions I will retain this numbers and we will get back to a similar table at the end of the discussion on afterburners now if we look at the TS diagram with the after burner on we will again assume ? to be one okay so you have isentropic compression through the intake as well as through the compressor then you have heat addition in the main combustor expansion through the turbine so this is 0 to so this is the TS diagram for efficiencies one and with the after burner switched on okay now what do we have to do if we have to analyze this okay firstly our expression for f by m.a0 the non-dimensional thrust would be m0 and 2 in this case 1 plus f plus I will call I will introduce a new parameter fab m7 by m0 into under root T7 by right this is the expression for non-dimensional thrust now if you put fab equal to 0 this is the expression that we had derived for a turbojet without the after burner switched on right when fab goes to 0 this is the expression that we get now we have fab where fab is nothing but m.fuel that is added in the after burner divided by m.air okay so it is the fuel a ratio in the after burner so what do you think of the value of this plus this what is the maximum of f and f plus ab that it can go to it can maximum go to stoichiometric which is what is the value it is 0.067 which is the stoichiometric value you cannot burn more than that so it goes to something like 0.067 now if you go to 0.067 I can still use the condition that f plus fab is less than 1 right with an error of around 6.7% so I will use that I will say f plus fab is very much less than 1 okay so that we can take this terms of and I will also introduce another parameter here just like we had TT4 by T0 as a control parameter we have a new control parameter that is TT6 by T0 right what is the maximum temperature in the cycle divided by T0 the minimum temperature so I will call TT6 by T0 as equal to ?ab remember we had TT4 by T0 as equal to ? burner similar to that I mean introducing a new control parameter to account for the afterburn okay so using this I need still these ratios T7 by T0 and M7 by M0 so I can get T7 by T0 as TT7 into TT6 by sorry this has to be TT6 okay fine these two cancel off and I will get T7 by T0 what is T7 by TT7 it is the ratio of static to stagnation so I can express this in terms of Mach number this would be 1 plus 1 by 1 plus ?-1 by 2 M7 square what is T7 by TT7 by TT6 this is flow through the nozzle right so flow through the nozzle we have assumed the flow to be isentropic so TT the total temperature does not change so this is 1 and TT6 by TT0 we have defined it as ?ab so I get T7 by T0 as equal to ?ab by M7 square now we need to do cascading of pressures to get the other parameter yes yes so why are we doing for the optimal because in case of urbanian optimally expanded if you use a convergent nozzle it is choked we will do both cases okay firstly we will take the optimal expanded flow then we look at what happens if the nozzle is choked the nozzle being choked does not essentially mean that P7 is equal to P0 it can be that P7 is greater than P0 even if the nozzle is choked right so we are taking a special case of choke flow wherein the exit pressure is equal to the ambient pressure now if we do cascading of pressures I get P7 by P0 is equal to okay this is what we get and the first term here is ratio again ratio of static to stagnation pressure this is flow through nozzle flow through jet pipe and then flow through turbine flow through the main combustor flow through compressor and this would be flow through intake and that is ?0 to the power of ? by ?-1 so we will put that down here and we know because we are considering optimally expanded flow what is P7 by P0 this would be equal to unity so we will use that so we can write 1 is equal to 1 by M7 square into flow through nozzles we are assuming all efficiencies to be one then flow through after burner again one then flow through compressor you get ? T then sorry flow through turbine then you have flow through combustor which is one then you have flow through compressor I see and you have flow through the intake which is one and lastly ?0 to the power of ? by ?-1 and here also sorry so I get I can rewrite these two in terms of ? T so I will get 1 is equal to 1 by 1 plus ?-1 by 2 M7 square is equal to ? T ? C ?0 okay and I know from the definition of ?0 that 1 plus ?-1 by 2 M0 square is equal to ?0 so using these two I can get my expression for M7 by M0 I can write M7 as equal to 2 by ?-1 T C ?0-1 and M0 similarly would be and we required in our expression for T7 by T0 this term so I can get that too so I can write T7 by T0 as equal to ? AB by T T C ?0-1 okay so sorry no-1 sorry so this is the expression that we have now we can substitute back and get our expression for the non-dimensional thrust which is S by M. AE0 is equal to or I- I can rewrite this part by taking this out is common I can rewrite this as okay so this is the expression that we get for the non-dimensional thrust now there is still one part missing we know that there is a turbine compressor power balance so these two are not independent variables and they are connected okay so let us get that expression also we had derived that expression for a case without the after burner on do you think it will change with the after burner being switched on because this is something that happens downstream of the turbine it does not get affected and therefore whatever we had derived earlier holds good so from we get Tt is equal to 1- ?0 by ? B into Tc-1 and if you substitute this expression back then we will get a by okay this is the overall expression that we get notice here that ? B is now parameter of very small consequence right the temperature at the end of the combustor is not such an important parameter whereas for the thrust the temperature at the end of the after burner is a more important parameter okay now let us look also at the ISP part okay if we have to calculate ISP we know from our previous expression for non-dimensional specific impulse that I need 1 by F into F by m.e0 so here this was the expression that we had without the after burner how would this change with the after burner I need to include here F plus Feb for the case with the after burner okay so with the after burner on you have this additional term here okay now how do we determine this quantity how do we determine this quantity what do we need to do what did we do to get 1 by F energy balance across the combustor now what do we need to do energy balance across the combustor plus the after burner so let us do that by taking energy after burner we get m.f plus m.fab into Q okay must be equal to m.a into 1 plus F Cp Tt4- Tt3 plus m.a into 1 plus F plus Fab okay this is the portion that gets added on in the after burner into Cp Tt6 by minus Tt5 okay this is the expression that we get now we know that we can use F is very much less than 1 and we also know that F plus Fab is also very much less than 1 so using this I can eliminate these two and therefore I will get F plus Fab that is you take it here and divide this by m.a you will get this into Q by Cp must be equal to Tt6- Tt5 plus Tt4- Tt3 okay now if we look at the TS diagram if we look at the TS diagram this is the TS diagram we have and what we have here is Tt6- Tt5 that is this- this plus Tt4- Tt3 fine now I can add this part what you will get is you will get as though it is a continuous rise from Tt3 to Tt6 okay if I add this part but we know that compressor turbine power balance Cp being the same this portion in this portion is same so you have to subtract one of them if you are adding this you have to subtract this okay and I also know that the Tt2 must be equal to Tt0 right because flow through the what do you have flow through the intake what kind of process is that it is an isentropic process because we are assuming all efficiency is to be one it is an isentropic process total temperature does not change okay so you get Tt2 is equal to Tt0 so we will use that here and do a little bit of jugglery as I said I will add- Tt5- Tt4- okay if you look at this I have not done anything this is what is this part turbine work in the turbine work in the compressor these two must be equal right so I can take this out these two are equal now I know that Tt2 by Tt0 is equal to 1 so this goes to 0 okay so by doing this manipulation I have been able to get this as Tt6- Tt0 which is nothing but you are looking at raising the temperature from here to Tt6 okay this was TS diagram so I can write now F plus FAB Q by CP is equal to Tt6- Tt0 now if I divide by T0 on both sides I get what is Tt6 by T0 this is ?ab and what is this this is ?0 so we get a new expression for we are looking for this expression 1 by F plus F plus FAB is equal to Q by CP T0 ?ab- ?0 okay and following this I can now write isp by a0 is equal to Q by CP T0 ?ab- ?0 into okay so this is our expression for isp with the after burner switched on I will stop here and continue.


So this is about energy grid scale energy In our power grids we use conventional power plants provide power and to stabilize the grids So it's in the next ten years. We're gonna shut down most of these cold fire and plants nuclear plants nuclear in Germany So seeing this with renewables is that they do not stabilize the grid The wind only is there when it is there not when you need energy same as the Sun and solar So renewables destabilize power grids So today we have three different measures the conventional plans to keep the grid stable We have demand side management and we have batteries So this demand side management the potential is pretty limited and batteries are really expensive So we came up with a new technology Which is based on the oldest energy storage principle mankind knows the Potter wheel Storing energy in a rotating body using the movement of the body to keep the energy If you want to load to charge it you accelerate it if you want to use the energy you decelerate it So this is how our Potter wheel of the 21st century looks like it's a so-called flywheel and in this containment There's a rotor made out of carbon fiber Levitated and magnetic bearings running in a high vacuum atmosphere to reduce friction and oil losses to make it as Efficient as possible So how does the competition with the battery look like while we're less expensive? We are sustainable and we have an unlimited cycle life Otherwise then the battery in your mobile phone. You can charge our system millions of times Our market is global. We start with interconnected grids from Germany and Europe Going to the US in 2020 and Asia Oceana around 2023 We think that there's a market for 40,000 flywheels globally each one worth that nice Tesla car over there So a business model We have partners which manufactures the components we need for the flywheels based on our specification We assemble the flywheels Put our software on and bring them to our customers Our team with 12 people now for founders and experienced advisory board with all the knowledge We don't have since your startup To prove that it works we've currently built a big grid scale pilot with the total power of one megawatt in England and Ireland and This is how we develop Next year end of next year will have our first paying customer bring it to serious production of the system Reach break even in 2021 and then scale up globally So if you're interested in energy in science rocket science without rockets In business to business in global markets and a proven and great working team talk to me And thank you. Thank you, Hendrik and time to move to questions So can you just tell me a bit more about where you place yourself in the network and who the customers are? Yeah, so our traditional customers would be utilities Not so traditional customers would be everybody who's running assets in the grid power generation assets and Customers we expect to see next year or in the years after Our company is focusing on energy storage installations. We see that in Ireland We already got questions from Ireland about very big installation there which offer the investment into the storage technology as a financial product to life insurance companies So we got pretty bright pretty wide bandwidth of customers There's a screaming demand for what you offer. I'm sorry. There's this. There's a huge demand for what you're offering I mean, it's a big problem. It's a holy grail. What what is this cost for someone to to buy and? Is anyone else doing something like this? Well, there's other companies doing like that There's companies in the U.S. There's companies in Germany What we offer is a different technology or when you really dive into the system, it's differently We can manufacture our systems at lower costs and our systems are modular So we can put together the system on the demands of fitting the demands of the customer Like making it really fits the application reducing his costs in investment as well as running costs. How much? how much so Currently we built a container with one megawatt of power for flywheels in it.  According to your hypothesis, so it's not what so did this is a new field So let's turn it around nobody suffers, but we can keep the lights on Someone's going to lose money when you make money most likely No, we replace easy conventional power plants and These people lose money and we provide the technologies that we can turn them really off And and given that there are other players out there doing similar stuff like this to what extent where are you in the patining process and How do you think about that? Yeah, so we got a couple of patents. We have a patent strategy Going forward for the next three years with about four patents each year and special aspects But most of the knowledge is in the system itself We're not gonna patent this because if we patented everybody else will know how we did it So now you can rip the system apart. You won't see what's the real IP in it How much money are you raising? So currently we raised two million and how are you raising now? Yeah, you're raising two million Yeah, and if you explored kind of grants and other financing options to help me up to now about 1 million euro in grants went into our company So we really took up speed and now we need more money to keep the speed and even get faster I wasn't sure if you said it's customized or you're just gonna have one kind of unit that everyone kind of customized Yeah, doesn't that hurt the economics? Not really so this is a modular system a platform So when we customize the system we use like 95% of common parts and the rest is adapted but also based on a on a Yeah platform thing like Lego putting things together And thank you, Hendrik we have to move on that was adaptive balancing


Hello, I welcome you all once again to my channel explore education. I am Dr. Rashmi Singh assistant professor department of education as it's kind of all three boys University of Allahabad and This time I'm going to discuss with you all the topic that is development of secondary education in India in post independence Prior to this I have already discussed with you the development of secondary education in India in pre independence period. Okay, let's start Bad boy, we started with the space and we came at the end in nineteen seventies In the class, your sergeant report in nineteen fourteen before that, your subcommittee report, all the things have happened to you Now, as soon as India is free, Tara Chan committee is sitting in the forties in nineteen forties In nineteen forties, the committee appointed Dr. Tara Chandi, then educational advisor, Dr. Tara Chan was a Shaq Sheik advisor Government of India and he was chairperson, that's why this committee was named Tara Chan committee There were some recommendations for him. Secondary schools must be multi-purpose to cater for the various needs of the pupils Every committee, every commission is saying that education is being educated in a way that you can do it very well So that the students can have various requirements in this committee There shall be one major examination and determination of the secondary stage in school They are saying that when the termination of the exam, i.e. the 12th board or the 12th examination, there should be a major examination That's what Tara Chan committee said And they said that the salary and the other service conditions of the teachers should be as recommended by CAVE Central advisory board of education, that's why it's called CAVE They said that the teachers' service conditions and salary should be the same as what CAVE has recommended Then you should know that Tara Chan committee gave recommendations That you should make an exclusively education commission for secondary education In the morning, your secondary education commission comes The Mudaliya commission, Lakshman Swamy Mudaliya was a chairman He was the vice chancellor of Madras University And the commission was instructed to check the prevailing education system It was given responsibility and responsibility You can see what is the status of secondary education in RISH And you can tell us how it can be improved So the recommendation was to recognize the educational pattern of education They are saying that first you should make a pattern of education And then you know that in 1986, it became 10 plus 2 plus 3 The curriculum pattern must have elasticity and diversification They are saying that the curriculum pattern should be like this Where there should be diversity and elasticity That is, rigidness That is, every time there should be diversification Examination system should be reformed There should be multi-purpose schools at higher secondary level They are saying that you should make multi-purpose schools at higher secondary level Public schools should continue in the state government Must establish residential schools in rural places Because the rural areas are not as much educated as the children So you should make them residential So that the children stay and stay there And since English was very important It was said that you should make the mother tongue a regional language You should make the mother tongue a regional language The constant and realistic appraisal of the pupils' progress Should be made compulsory throughout the year This is what you can call the foundation of CCE That is continuous and comprehensive evaluation They all started with these recommendations They said that throughout the year You should make the mother tongue a regional language You should make the mother tongue a regional language The service conditions of the teaching personnel Must be improved by raising funds At that time, the salary of the teachers was very poor So it was said that it was improving So that the great intellect can be inspired in the teaching profession With a good salary Because the salary is not good, so no one wants to stay here State advisory boards of education should guide and advise the education department They said that the state advisory boards of education They should give their education and education to the teachers Education commission was founded in 1964 Which is known as the Kothai Commission It is popular and popular And on the secondary education, its main principles were That the complete overhaul of the educational structure and pattern The education and pattern Rejuvenate it completely Change it, it needs to be changed They said that you should give the primary stage of 7 to 8 years Give the lower secondary stage of 2 to 3 years In general education And in vocational education of 1 to 3 years And give the higher secondary stage of 2 to 3 years And in vocational education of 1 to 3 years Such education commission said And for curriculum, they said that the commission considered It is necessary to vocationalize secondary education In a large measure To raise enrollments in vocational courses To 20% of total enrollment The lower secondary stage And 50% of total enrollment in the higher secondary stage They are saying that diversify vocationalize secondary education And increase enrollment in it That means more than 20% of students were given a grade And they said that you should do this From lower secondary to 10th grade Around 20% of total enrollment Join vocational courses of 20% And join vocational courses of 50% of the 12th grade This was done by Kothari Ayod Then they talked about equalization of educational opportunities We have talked about this To achieve the objective of egalitarianism in educational scenario That in the objective of egalitarianism Equalization of education The commission said that Free lower secondary education Attempts should be made to extend free education To all needy and deserving students They said that you should try All those who are deserving Those who want to get education These students should be given free education Proper planning should be made About the location of the institutions Which institutions should be made It should be after Neogen The neighborhood school That should be in the reach The primary educational institution It was one kilometer away And the higher secondary It should be three kilometers away That's what they said Then guidance and better supervising facilities Should be made compulsory They said that Neuralization of Paramarsh And supervising facilities should be provided What will happen when you provide The child will stop there That is, it will be retained The wastage and stagnation will not happen That's why it is being improved Then comes your national education In 1986 And since we have discussed national education In totality That's why you can take the slide That what he said in the second education He talked about the NABU And about education The name of the national education Has also been made By MHRD and NCRT They made many proposals That what we will do Will bring vocationalization And why we are not able to bring vocationalization He said that there will be a reason behind it Like there is no well coordinated Management system Unemployability of vocational passouts Passing from vocational exams Pass out of it Mismatch between demand and supply Reluctance in accepting the concept By the society Provisions for professional growth and career advancement For the vocational passouts etc Urgent steps to strengthen the vocational system Are there for imperative You can see the problem of vocational education In the second education After that comes Your most innovative Most updated Most latest national education policy 2020 And what he is saying in the second education He is saying that the major change He was running for so many years He changed the structure He changed it to 5 plus 3 plus 3 plus 4 In fact 5 plus 3 plus 3 was 10 plus 2 plus 3 Last one He made a cluster of 10, 11, 12 Means 9, 10, 11, 12 He made a cluster of 4 And 3 from here 5, 3, 8, 3, 11, 4, 15 And 5, 3, 8, 3, 11, 4, 15 And it is 10 plus 2 plus 3 And it is 13 So it is two years Early childhood education The foundational stage He is saying that Make it 3 years 2 years primary school 5 years old Then in the preparatory stage You can study 3, 4, 5 Means 5 plus 3 Then in the plus 3, 6, 7, 8 And in the secondary stage 9 to 12 So it is 10 plus 2 10 plus 2 plus 3 So it is 10 plus 2 10 plus 2 means 5 plus 3 plus 3 plus 4 Means 12 He made it 5, 3, 8, 3, 4, 5 So it is 3 years old This is an additional Means in primary school The foundation is first forced Then the child is ready 3, 4, 5, 5, 6, 7, 8 And 9, 10, 11, 12 He made a cluster Means earlier the course of 9, 10 More than double the course of 11, 12 So the world is changing It should not be like this He made the whole course Equally distribution in 4 parts So that much burden is on him And in the secondary stage He said it should be 4 years old He studied multidisciplinary He studied subject-oriented, pedagogical He studied based on curricular style He studied with greater depth He studied with greater critical thinking He talked about attention to life He entered in To continue to have the option of exiting After grade 10 and reinterning in the next phase To pursue vocational or another course Is available in grades 11 to 12 Including a more specialized school If you read the national education policy You know the concept of multiple entry Multiple exit Means you can enter the courses In between Means you have made it flexible You have made it student-centered You have made it student-autonomic So this is the calculation of the year From the woods Dispatch to national education policy 20-20 So in this way I have completed this very topic So thank you and don't forget to like and subscribe My channel and school education I have run from my seat