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We have P equals half rho A v V 1 square minus V 2 square and I also had V equals half V 1 plus V 2. So, I will substitute this here. So, we simply have 1 by 4 rho A V 1 plus V 2 into V 1 square minus V 2 square. So, we just complete this mathematical steps here. So, this is half rho A you can open the bracket here. So, this is V 1 cube minus V 1 V 2 square plus V 2, V 1 square minus V 2 cube okay. So, this is the total power that has been extracted from the wind. We also know that the original wind had an energy of half rho A V 1 cube. So, we would like to know what fraction of that original power that was there this P nought, what fraction of that P nought did the windmill manage to extract that is the point that we would like to understand. So, we are interested in using these two relationships and trying to find out what is the value of P by P nought and to use that to understand whether we are you know understanding something about what’s the best that we can do in this case. So, to do that we will simply we will just divide these two equations because that is exactly what that is, so P by P nought equals you can see here that the, you simply have a factor half here because the rho a here will remove the rho a there because they are in the numerator and denominator in both cases in there it is there both in the numerator as well as the denominator. So, the rho A will be removed and this factor 1 by 4 and 1 by 2 will combine to give you this factor 1 by 2 okay. So, that is basically what we have here. So, this is half and you have V 1 cube divided by V 1 cube, so that is 1, minus you will have V 2 square by V 1 square plus you will have V 2 by V 1 minus V 2 cube by V 1 cube. I got this simply by dividing the equation here let us say whatever this is in this page this is equation 1 and this is equation 2. So, I just got it by dividing equation 1 by equation 2 on this page. So, that’s all I have done and I have arrived at this. So, we will just for simplicity sake we would call V 2 by V 1 as some value alpha. So, that we don’t have to write you know so many terms here. So, this is simply P by P nought equals half of 1 minus alpha square plus alpha minus alpha cube where alpha is simply the ratio of V 2 by V 1 velocity far away before after the windmill divided by velocity far away before the windmill. So, this is all we have got. So in fact, all you have to do is plot P by P nought. So, now, that we have this equation let’s say I let me call this equation 3 in this page. So, I just have to make a plot of this equation P by P nought versus V 2 by V 1 for all various values of V 2 by V 1 the maximum V 2 could be I mean the minimum V 2 could be is 0, maximum V 2 could be is V 1 or we can assume something like that and then you plot it between 0 and 1 right. So, if you do that you will see that the P by P nought varies and at some point, you will have a maximum and that we will say that that is that combination of V 2 and V 1 at which maximum power is drawn. So, that’s a simple way to do it. So, you simply have to make a plot off and that is what I am going to show you also shortly P by P nought versus V 2 by V 1 which is nothing but alpha. So, if you do the plot you will see something. So, this is we can say you know as I said V 2 can be 0 or it can be equal to V 1. So, this is 1. So, if you plot it you will see some curve, something like that whatever some curve we will get and so there will be some value which is the maximum. So, that is the ratio of V 2 is and V 1 at which we can extract the maximum power from that windmill and we will get some value here that is that fraction which is the maximum fraction that we can get and hopefully you have given the way we are trying to derive this that will be our Betz limit I mean so if you have gotten this right. So, this is one way we can do it we can just plot it and come up with it or we can simply differentiate this equation because that is exactly what it is we want this maximum value. So, at that maximum value, we will simply have d P by P nought by d alpha is equal to 0. So, if you solve that equation also you will get the maximum. So, we can just solve that equation. So, d P by P nought d alpha is simply half of you will have minus 2 alpha plus 1 minus 3 alpha square. So, therefore, we simply have to set this to be 0 to get the maximum. So, that is the quadratic equation is minus 3 alpha square minus 2 alpha plus 1 is equal to 0. So, the solutions are minus B. So, it’s simple, so that’s therefore, alpha is equal to plus 2 plus or minus the square root of B square is 4 in this case minus 4 A C. So, minus 4 into A is 12 into 1 is 12, so minus 4 into minus 3 is plus 12. So, you get plus 12 here by 2 A, so which is minus 6. So, you have 2 plus or minus the square root of 16 by 6. So, this is simply we will put it right here we have, therefore, alpha is equal to 2 plus or minus 4 by 6. So, alpha can be pi minus 6 here. So, alpha can be minus 1 or it can be 1 by 3. So, if it is 2 plus 4 you have a 6 by minus 6 which is minus 1 and if it is 2 minus 4 you have minus 2 by minus 6 which is 1 by 3. So, now, let’s look at it. So, this is alpha is as I said V 2 by V 1, right. So, that is what is alpha. So, alpha being minus 1 means that the wind came hit the windmill and reversed back the same velocity with which it came okay. So, that seems very interesting, but that is not possible because the wind has two will hit itself right it will turn back and they still wind coming. So, this is not something that can be sustained because you are not distributing it anywhere the wind comes and it’s trying to head back the same direction in which it came we are not allowing density to change anything is going to change. So, it cannot do this. So, even though the equation gives us a minus one it is not mean it is not physically possible. So, alpha equal to 1 by 3 is the only one that seems possible. So, V 2 by V 1 equals 1 by 3 is the more meaningful result that we will have through this process and that is where we will get the maximum efficiency. So, we can substitute that back here and if you substitute that back here you will get a value. So, we will just see that in a moment. (Refer Slide Time: 43:36) So, the same equations I have put here what I just wrote down on the screen. So, the change in energy in the wind is rho AL V 1 square minus V 2 square that is what we have here. The power extracted from the wind is simply dE by dt which is this rho A small v V 1 square minus V 2 square and this V 1 the small v is half of V 1 plus V 2 and that is what has been put here and because of that half, this half here becomes 1 by 4 here. The kinetic energy in the incoming wind is this and therefore, power in the incoming wind is this value which is all that we did in our previous screen. (Refer Slide Time: 44:17) And therefore, when we do a P by P nought this is the value that you will get and that that’s how you arrive at this equation and then we set V 2 by V 1 is alpha and we arrived at this P by P nought equals one minus alpha square plus alpha minus alpha cube right. So, this is how we got this and if you look at this value of P by P nought. So, we will have this is equal to if you set alpha equal to 1 by 3 you will have 1 by 2, 1 minus you will have 1 by 9, plus 1 by 3, minus 1 by 27. This is kind of what you will get right. So, on this basis, we can look at it. So, this is equal to 27 and then we will have 27 minus 3 plus 9 minus 1. So, you will have 27 minutes 3 plus 9 minus 1. Yeah so, we will get half of this value. So, P by P nought if you do this calculation. So, we will get half off. So, if you complete this calculation it is 27 plus 9 is 36, minus 3 minus 1 so that is minus 4. So, this is simply 32 by 27 and so if you complete this you will get 16 by 27 okay. So 16 by 27 is this P by P nought and this is that value we saw in our Betz equation. So, this is 0.59. So, this comes to about 0.593. So, this is the derivation for the Betz equation we can get this 0.593 and that directly comes from you know the starting from the Bernoulli’s equation looking at the energy per unit volume essentially you know incoming what is outgoing and equating them and then are arriving at this value 0.593. So, as I said you can also do this as a plot. So, that is essentially what I have done here. (Refer Slide Time: 46:44) So, if you see here this value P by P nought I have plotted as a function of V 2 by V 1. So, you can easily do this you simply take any spreadsheet. So, I have just used excel you can use any other spreadsheet or any other you know plotting software that you have and you make a plot of various values of V 2 by V 1 as I said for us it is meaningful to look at values between 0 and 1 which means V 2 is either 0 if the wind has been fully stopped or V 2 is the same as V 1 in which case you know it is taking the I mean which is going un undisturbed by the windmill. So, if you do that and plot it as a function of P by P nought or P by P nought as a function of V 2 by V 1 you see this curve and at this curve, you see a maximum. So, this maximum as you can see is very close to 0.6, 0.593 is the value we got if you go down this, this is some 0.333. So, that is the value at which we get it you can see here I have just marked all the values of various values of alpha and for that the various values of P by P nought. So, you can see here at 0.3 we have reached a value of 0.592 and it is only marginally more than that 0.593 is the maximum that we can get. So, that is how the Betz limit comes about and it’s a calculation done in the nineteen twenties and as I said you can see various versions of this and the calculation works out as we described. (Refer Slide Time: 48:11) So, in conclusion, the Betz limit indicates that only about 59 per cent of the energy available in the wind can be captured this has some assumptions. So, there are ways in which you can think of you know going around these assumptions and there are some versions of the windmill which may not conform to this assumption. For example,
we have assumed that the windmill is a plane right. So, a single plane across which this velocity variation is happening some of the vertical axis windmills occupy some you know horizontal direction as well right. So, horizontal extent they have that is some places are closer to the wind that is incoming and some are away from the incoming wind. So, that changes the dynamics of this equation how the equation is set up and so strictly if you just look at this equation and directly apply this equation there it is not really a valid way to do it some variations are there and that can change how we look at the efficiency. In general, the actual efficiencies would be less than this limit because this only sets the upper limit for a certain type of wind. So, in summary in this class, we have tried to understand how the derivation for the efficiency of the windmill comes about and the idea that there is this limit which is the 0.593 or roughly 59 per cent, that’s sort of the maximum that we can extract the maximum power that we can extract from the wind that is incoming towards the windmill. So, I think that concludes our class it gives us a good background to understand what is possible with the windmill and you know what are some limits of this whole process. Thank you. Hello there. So, we have in the last few classes been discussing wind energy we will do so for another couple of classes there which includes today and we have had an overview of the wind energy process and we have looked at you know what are the geographical aspects associated with it, what is the wind speed aspect associated with it and how much of a difference that makes and we also did a lot of calculations what is the power, how does it relate to the wind speed. We also did the calculation associated with the Betz efficiency, we looked at some parameters associated with it to understand you know maybe what are sort of limits associated with it, how you should think about it, and so on. We looked at the fluid flow through the wind turbine and then came up with that Betz efficiency which is about 59 per cent. And I also told you that you know those calculations are sort of the best case calculation and, so typically efficiencies are going to be less than that maybe 30-40 per cent efficiency you are looking at its still a pretty good efficiency because you simply have to keep this out there and it generates electricity for you. I also told you that designs could be different. So, there are some assumptions on the design of the turbine which is used in the deficiency calculation and therefore, we had one you know one specific location where the turbine was located and the interaction between the wind and the turbine took place at that location. Whereas, you could have other designs where the interaction is spread out across a range of you know locations and therefore, the interaction with whatever is the forward part of the turbine could be different from what is the interaction with you know rear part of the turbine relatively speaking relative to the wind direction. And therefore, those factors would also have to be accounted for in your calculations if you want to get a much more you know the realistic or accurate idea of what is the real limit. So, there are various aspects associated like that and we should at least be aware, but we have gone through the basic calculation which is I think a very useful value to keep in mind and also the fact is that very large fraction of the windmills that are being put up there essentially conformed to this design that they are the horizontal axis and they sort of encounter the wind in one particular direction or on one particular plane and therefore, the Betz calculation applies to them perfectly fine. So, with that kind of a background that level of detail that we have developed. In today’s class, we will look at we continue to look at wind energy we will look more at the parts associated with this wind turbine and to the extent that those parts have specific applications we will also look at what are some material requirements that are being placed on those parts performance requirements which, therefore, require some material requirements. And therefore, the parts and materials combination is sort of what we will look at and we will see how that impacts the overall idea of this wind power okay. (Refer Slide Time: 03:22) So, our learning objectives for today’s class are to become aware of the various parts of a windmill, various important parts of the windmill in the most you know the common design that seems to be getting implemented these days and to also become, to that degree aware also of the various materials being used in the windmill, various materials being used in windmills of course, across various locations in the windmill. So, of course, this term windmill itself is a general term often. In fact, in the context of electricity generation, they tend to use the term wind turbine because there is a turbine associated with it, but I think in a commonplace you said we could still call it windmill. So, that’s in that context it is this term is being used okay. So, with that idea of our learning objectives, we will take a look at the windmill that of the various parts of the windmill. As I told you also that you know we saw in one of our earlier classes that you could have this vertical axis wind turbine in which case you do have still have some kind of a tower and something at the base within many of the parts up in the base itself at the ground level then, but most of the turbines that we see these days are horizontal axis turbines and for them, many of the parts are placed on top of the tower. So, this is the general you know design philosophy that is followed with most of these wind turbines and it seems to be getting followed across a wide range of manufacturers. So, that’s something that we have to contend with and that’s something that we will keep in mind as we put the diagrams together. Of course, I also pointed out that you know we had this relationship that the power is proportional to V cube. (Refer Slide Time: 05:13) So, much of the design of wind turbines you know strongly acknowledges this aspect that the power is proportional to V cube and therefore, both in terms of citing, citing of the wind turbine as well as the general you know the stature of the physical stature of the wind turbine is designed specifically to take advantage of higher wind speeds and therefore, those structures are typically pretty tall. So, that’s something that we would have to keep in mind okay. So, with that background let us put down some parts for the windmill. So, I am just going to draw a rough sketch the label some parts and then we will discuss what we are interested in looking at. Okay so, we will start by saying that we have the central part which is some kind of a hub and then from that, you have a blade. So, the blade will typically have. So, let me just clear this so, something like that, some design like that and something like that we will keep that for the moment and then little further down the other major part that, so this will be the blade. So, there is a third blade I will just put that down in the end. So, we get this diagram completed. A little bit behind it is a structure that looks something like this. So, this is a broken sort of similar to the break you would see maybe in an automobile or something. So, it’s a breaking system and then further down we have something like a gear we do have a gear. So, you will; the teeth of the gear will be there and then further down we have a generator so that links up to this gear, that’s our generator and we also have a system down here of some form here which is just I am just drawing this as a schematic some kind of a system down here which we will talk about in just a moment. And through all this, we also have a shaft. So, that’s what we would have there and yeah then we have a tower okay. So, these are our major parts and of course, we will have the third blade somewhere down there we are only seeing part of the blade here and that disappears behind somewhere there something that is your third blade. So, this is these are the major parts of our wind turbine we just a schematic of it. So, we will just go over it. So, this would be our look me just. So, this is our rotor blade, that’s the primary part of the wind turbine that is what you know interacts with the wind to get us the breeze. This central part here is referred to as the hub and from that you have a shaft that comes through some kind of a drive train that comes through of some sort comes through. What you have here this arrangement that you see here is the brake, then we have a gear gearbox and it is the one that you know enables the linking to the generator. So, the generator sits here, sits here and then we have a mechanism here called the yaw mechanism, yaw mechanism that’s what is sitting there. This tall structure here is the tower and finally, what we have here is the nacelle right. So, this is the nacelle. So, that is what we have here. So, these are the major parts of our wind turbine right. So, we will look at all of these parts we will try to understand what material restrictions are there and what requirements are there pretty much all the major parts we will look at. So, if you look at it the wind the rotor blade is what interacts with the wind, so to speak. So, that’s the first in the primary part of the windmill there is a hub which holds all the rotor blades together that’s what this hub does. So, this rotor blade is attached to the hub and then from the hub comes the drive train to which is attached to the brake. We discussed in one of our earlier classes that you know if the wind speeds start exceeding about 70 kilometres an hour then for most of the existing windmill wind turbine structures that are quite dangerous. So, it can damage the wind turbine the blades and therefore, we need a braking mechanism to stop the wind turbine if you want to stop the wind turbine for any reason, it could even also be for routine maintenance if you wish, but in any case for the these to deal with the idea of the situation that you may have very strong winds you do need a braking system to stop the windmill if you desire to do. So, the braking system is there. Then there is a gearbox we will as I said the primary purpose of the gearbox is to link the rotors the rotating blades to the generator. You can ask why not directly connect to the generator. So, again, in fact, there is a reason for it we will discuss that shortly. It has got primarily to do with rpm control, but there are that the idea that you can connect directly to a generator is also not out of them; you know out of the realm of possibilities. So, that is also being considered and is also being implemented. So, we will look at that then there is this, that’s the generator that is sitting there. All of these parts you know especially the drive train, the brake, the gearbox, the generator are all hold housed in this structure called the nacelle which is you know outer covering, so to speak of this structure. And then at the at just below it you have the tower or at the bottom, you have the tower which is the tall structure which holds you know the all the top items all the items that you just saw are held on top of this tower and tower can be quite tall. I mean if you see the modern designs of wind turbines you are already looking at blades that are you know approaching say 50 meter, 60 meters in length and therefore, the tower is taller than that because at the lowest point you don’t want the turbine to be I know the blades to be close to the ground to hit something on the ground or you know to be unsafe for any such reason. Plus you also wanted little bit off the ground to ensure some steady flow of wind even at the lowest point and therefore, usually, the blade itself is the lowest point of the blade when it comes in that circle is still several tens of meters off the ground. So, you are looking at tower structures which are of the order of 100 meters, of the order of that 100 meters length is what you are looking at them I mean you can have variation it depends on your wind I mean the windmill blade, but you are looking at this kind of a number maybe 60 meter, 70 meters, 80 meters something like that you know some several tens of meters usually more than 50 meters is what we are looking at. Now, as you can imagine we when we discussed this idea of you know horizontal axis vs a vertical versus a vertical axis. So, I pointed out that you know in the horizontal axis you have the tower and the axis is like this pointing towards the wind this is the horizontal axis. You can also have a vertical axis where the axis is pointing upward. So, when we discussed horizontal axis versus vertical axis I pointed out that when you have a vertical axis and you have the blades in an oriented such the axis is vertical then orienting it concerning the wind does not matter because the wind is anyway always perpendicular to it right. So, the wind is always perpendicular to it, it doesn’t matter from which direction it comes in whether it comes from the front of the right or the left or the back. The axis is vertical the blades are rotating about that axis in this direction and therefore, it doesn’t matter which side the wind is coming from. So, there is no issue of orientation. Whereas, in the horizontal axis when you have the breeze coming from you know in a horizontal direction based on which direction the breeze is coming from, based on which direction the breeze is coming from the blades have to be oriented so that the breeze comes perpendicular to them okay. So, only then the blades can completely benefit from the movement of the wind. So, the blades have to keep getting reoriented based on the wind direction which would change season to season and typically will change even during the time of the day based on you know especially based on your location if you were closer to the coast you may have sea breeze at some point, you may have a land breeze at another point. So, you will have distinct variations in the direction of the breeze and you need to reorient your turbine. So, therefore, you this top nacelle arrangement nacelle with all the parts has to get reoriented so that the blades these specific rotor blades you can face the wind. And therefore, you have this your arrangement. So, it is an arrangement which is fixed on the tower on the tower and it attaches to that nacelle and then you know rotates that nacelle around so that you get the direction oriented appropriately. So, you need to have some wind sensors, you need to have some control system which decides how much to rotate so that it orients correctly. So, all that is also included in this overall system which even though I don’t I am not showing you the exact those parts here. So, there will be some control system which keeps takes care of all these orientation issues. So, these are the major parts. Now, we will go through most of these major parts and try to understand what are some specific restrictions associated with those parts and to that degree also some aspects associated with the material issues associated with these parts.