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Module 1: Solar Energy and Solar Radiation

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Hello. In this class, we are going to look at the basic idea of Solar Energy. We recognize that in today’s world a lot of people talk about solar energy and there is a lot of emphasis on solar energy, many new companies are coming up with solar energy-based products. So, we need to get a better sense of what it is that the; what is possible with this you know this arena of solar energy. And as an initial step in the direction, we will look at what we will see in this class, which is the transaction between sun and earth in terms of you know how much energy is coming from the sun. I am just titling it as the sun earth transaction. But basically, we are going to look at how much energy reaches our earth from the sun, and you know what is it in the context of our existence and our usage and demand etcetera. So, this is what we are going to look at. So, it is solar energy the sun to earth transaction. (Refer Slide Time: 01:13) So, our learning objectives for this class are to calculate the energy received by the earth from the sun. So, every day I mean during the daytime except maybe in places which are you know if you have I mean if you are near the poles and then sometime in the winter, you don’t get any sunlight, but beyond that almost any other place on earth, we are just getting sunlight every day. So, there is a lot of energy that is coming from the sun to the earth. So, we would like to calculate put some number to it, what is this energy how much is this energy that is coming to us from the sun. So, that’s one calculation that we going to do. So, we are going to do that in this class it must be I mean in some senses a straightforward calculation, but still a very important calculation because it puts a lot of things in perspective. So, that’s why we will go over it. Then we are going to compare what energy we are receiving from the sun relative to the energy that we use okay. So, we are going to compare it the energy received by the earth from the sun, with the energy usage by humankind. So, energy usage by humankind we have already seen in some of our previous classes I have mentioned it, we are using about 500 Exa joules of energy every year in our present state of usage, there are various reasons based on which this will this is likely to increase. There are also technological advances that are happening all the time which may help us you know sort of level this off in some way. So, we don’t know exactly what number it will settle, because on the one hand there is a large section of the world population that is still not as developed as some other sections of the world population, and always there is a desire for everybody to be developed it is natural. And therefore, simply based on those numbers you can expect the energy demand of the population to go up in a big way. At the same time, new technologies come which essentially help you do the same job with a lot less energy. So, simply because everybody in moves up in a lifestyle of their you know behavioural aspects and so on and in lifestyle accomplishments, it does not necessarily mean that the energy usage will also go up in the same manner. So, if we know double the population has the same kind of lifestyle, it does not mean energy will be double it maybe something else. So, that’s something that we have to keep in mind. So, these are factors that are competing with each other, the number of people who are trying to improve their lifestyle and the new technologies which help you get the same lifestyle with less energy usage. So, these are the competing factors based on which the energy demand over some time will change. So, in any case, we do have this number of 500 Exa joules barrier and that’s something we will refer to in the through this class ok. (Refer Slide Time: 04:04) So, let’s start with the sun, the surface of the sun is about 5500 degrees centigrade. Just under 6000 degrees centigrade So, you can you know that temperature is not high. In the sense we do experiments in the lab, sometimes you are doing arcing, you are doing the holding, welding arcs are there various arcs are there, there are various places in relatively commonplace situations where you are getting close to this temperature, you may not be at 6000, you may not be at 5500, but you may be getting somewhat close to this temperature. In our you know laboratories routinely we do we have furnaces which have the 1000 degree centigrade, 1500 degree centigrade and so on. If you start going past 1500 into 2000 degrees centigrade or something you will need a little bit more specialized furnaces, but when you do things like arcing and so on you may get even higher temperatures. So so, that’s the surface temperature of the sun, it is not high relatively speaking the interior is much much much much higher. So, you can see here that the core of the sun our current estimates put that temperature at in several millions of degrees centigrade. So, that’s where you know fusion reaction is going on which gives us all the energy that we receive right. So, that is in several million degrees C, the surface is about 5500 degrees C. So, people have studied the sun carefully, and they have a sense of what is the power of the sun. So, you know when you put a light bulb in your house you say you know 100-watt bulb 50-watt bulb and so on. So, if you think Sun is a bulb that is giving you light what is the power of this bulb. So, that is the number that you see here. The sun gives out 384 Yotta watts. So, again you know this is the reason why I went through that scale of quantities in one of our first classes, mainly because you see these kinds of numbers you have to have some sense of what is this number what is the significance of this number how big or how small that number is. So, you can see here Yotta means 10 power 24. So, 10 power 24 is that you know the order of magnitude, that gets referred to as a Yotta and. So, 384 Yotta watts means this number. Here 384 into 10 power 24 watts. So, you can think of it as though there is this bulb which is out there which is which has 384 into 10 power 24 watts bulb, and that’s the bulb that is you know lighting our lives. It is the bulb that is running our lives through the best if you look at it scientifically right now if there is no sunlight, you would not have photosynthesis, and if you didn’t have photosynthesis the I mean the plants would die, the animals would die and the life as we know it would not be able to sustain itself. So, critical to our sustenance is the sunlight arrival of sunlight to you know to enable the photosynthesis. So, that is very critical to us. So, that’s all enabled using this light bulb that is out there sort of you can call it a light bulb, but it is the sun, which gives us 384 Yotta watts. So, I simply put that down as 3.84 into 10 power 26. So, that’s the starting point of our calculation because that is the power of the sun, and from there we want to estimate what is earth receiving. So, that is the thing that we are trying to estimate. (Refer Slide Time: 07:26) So, if you take this image here, it puts a few things into perspective and then we will see how we have to go about this calculation. So, you can see here some important parameters, the first thing is this number that we just saw 384 Yotta watts that’s the power of the sun. Then there is the sun which has a radius of a 0.7 million kilometres radius. So, 7, 00,000 kilometres is the radius. Earth is sitting at an orbit is going around the sun in an orbit of this radius roughly 149 a million kilometres is the radius of that earth’s orbit. It won’t be a perfect circle, but we can assume a guess always in all these calculations some approximation we will do. So, we will assume it is a circular orbit of 149 million kilometres. And then you have earth here which is which you see here and as you can see I mean it is its radius is 6371 kilometres that are the radius of the earth. Before I even get into more details of this image that I am just showing you, I will alert you to the fact that this image that this sketch that I have put now put together here, is not to scale okay. So, because simply because if it were to scale we would simply be able to see things in any reasonable manner on this screen, but it is not to scale and that’s something that you should keep in mind. So, quite simply, for example, this is only 0.7 million kilometres diameter right I mean sorry radius, and this is 149 million kilometres. So, technically this distance this radius of this orbit radius of earth’s orbit is approximately 200 times, 200 into the radius of the sun ok. So, earth’s orbit if you take it’s roughly 200 times the radius of the sun the image that I am showing you here does not have this circle at 200 times the size of the radius of that circle is not 200 times the radius of the sun. So, therefore, right there it is not to scale. So, that is point number 1, also again if you compare the size of the earth as shown in this image versus the size of the sun. If you see again radius of the sun that is 0.7 million kilometres, so that 700000 kilometres and earth is about 6371 kilometres, we just assume that is also about 7000 kilometres. So, you are looking at a factor of 100. So, this is earth is about 7000 the sun is about 7, 00,000. So, the radius of the sun is approximately 100 times the radius of the earth. So, if you compare a clearly both in terms of the comparison of the radius of the sun with the radius of the earth, and comparison of the radius of the sun with a radius of earth’s orbit in both those specific aspects, this sketch that I am drawing here is not to scale. So, that’s just something that you should keep in mind, I mean it is not very relevant to our calculation immediately, but you should keep that in mind when you see such images as to what is the sense of scale that is something you keep in mind right. So, now, what are we going to do? We need to find out what is the energy that is coming from the sun to the earth right. So, now, this power that is being released by the sun is going out in all directions uniformly. So, power is a watt is joules per second. So, 384 Yotta joules per second is leaving the sun and heading out in all directions uniformly ok. So, you have to see. So, this same energy is going to arrive at this orbit right this entire earth’s orbit as it if you think of it as a concentric sphere. So, if you can think of a sphere that is the size of the earth’s orbit and in the middle, there is the sun, whatever energy has been released from the sun that as it goes out will eventually go past this orbit right. So, this orbit is a sphere we are using only one circle in that sphere, which is where the earth is rotating right or revolving. So, after sitting in one circle in the sphere, but that is a sphere. So, this energy uniformly goes out through that bigger sphere. So, that entire energy is going out to this bigger sphere, but the earth is there only in one location in the sphere, it is not there in the entire sphere. It is there at a given instant of time at given second earth is located at one particular position in that sphere. And it also has a finite size small size. So, out of all the energy released from the sun per second, only the energy that is coming to the location where the earth is present at that instant that is the only energy it is in a position to absorb or it is in a position to receive right, rest of the energy goes somewhere else. So, if the earth is only now taking away say 1 per cent of the space in that entire sphere, then only one per cent of the energy will reach the earth. So, something like that. So, that’s the kind of idea that we are trying to look at here. So, we are taking the same energy, it is going through the sphere that is the earth’s orbit and in that sphere wherever the earth has located the size of the earth, that much only the earth can grab. So, this idea we are going to put down in some numbers alright. So, that’s what we will see in our next couple of slides. So, if you look at the intensity of the. So, we now have the power of the sun which is 384 Yotta watts. So, we have to convert this to an intensity. Intensity means how much power per unit area. Now clearly as the as you go away from the sun as you go further and further away from the sun, this energy is distributed over larger and larger surface area okay. Therefore, the intensity which is the energy per unit area keeps decreasing, which is always the case right you are very close to a flame you feel the more intense heat from the flame, you move away and away and away you feel less heat correct? Same thing sun is there if you are very close to the sun, you feel the extremely high intensity of the sun as you move further and further away, you sense lower and lower intensity. That is got to do with this idea that the energy is getting distributed over a larger area, you are only one small fraction of the total area right. So, we are taking 384 Yotta watts, we want to see in this orbit that you see here in this orbit what is the intensity of this power, this power that is coming out this so many Yotta watts. In terms of intensity at this orbit, what is that intensity? How many this many joules per second is coming out from the sun, how many joules per meter square is there at the earth’s orbit every second. So, that’s the idea that we want to see. So, for that we have to understand; what is the area of corresponding to the sphere, that is the earth’s orbit and then you divide 384 by the area of that sphere. So, that it has now been distributed across that entire area of that sphere that’s the first calculation we want to do. (Refer Slide Time: 14:42) So, the same number here 384 into 10 power 26 that is the intensity of the sun; and if you look at the earth’s orbit which is 149 million kilometres; so 149 million kilometres, that is 149 into 10 power 6 kilometres, that is the same as 149 into power 9 meters. Right, it is kilometres becomes meters. So, that is 10 power 9 and then that is the same as 1.49 into 10 power 11 meters I just put a decimal place there. So, that’s the number that you see here. So, you have 100 and; 1.49 into 10 power 11, that’s the radius of earth’s orbit. So, the area of corresponding to the sphere of earth’s orbit is simply an area of the sphere equals 4 pi r square; where r is the radius of that sphere, and r happens to be this value this 1.9 into 10 power 11 meter right. So, the denominator is this area of that sphere, and this is the sphere corresponding to earth’s orbit. So, if you had a large sphere the size of earth’s orbit that is the area of that sphere. The suns energy that the sun is releasing every second is now distributed by the time it reaches earth’s orbit, it is distributed across this entire sphere equally that’s the point right. So, per meter square what is that energy, that is what we are trying to get; this is the energy released totally per second and this is the area. So, if you divide, if you just complete this calculation, you arrive at 100 and 1377 watts per meter square every second. So, that’s also there per second it is arriving so, but this many I am sorry the watts is already there. So, that is already joules per second. So, the intensity is in watts per meter square right per second. So, it’s already there in this watt unit. So, 1377 watts per meter square, this is the intensity of the energy that is reaching the earth’s orbit by the time it reaches earth’s orbit. Because you are further away it has dropped to this which is coming to our earth. Now earth’s orbit as I said this is a sphere the size of the earth’s orbit right, but we are only a small fraction of that orbit okay. So, if you see here, if you go back here, this is this whole sphere that you see corresponding to this dotted line it’s a huge massive sphere, that that is all around as an imaginary sphere. This is an imaginary sphere around the sun that we are looking at, and then on your slide, it is a two-dimensional circle, but you can think of it as a sphere which is you know sticking out of the surface and then going below the surface, in that the earth occupies only this region right. This small region that you see here is what the earth is occupying, and so, it will cut it will receive energy relative to this, that region there that’s the energy that it that will fall on it. It's like a big light that is there that is lighting up the room, you stand in one location in the room. So, only a small fraction of the light falls on you, the rest of the light is going everywhere else in the room right. The same idea is there, the sun is there it is putting out a huge amount of energy; the earth is standing in one location. So, only one part of the energy falls on the earth, all the rest of the energy goes all around and goes out into space it is not falling on any object near us of course, it falls on the moon, it falls on other planets etcetera, but basically, it is otherwise it is just hitting outright. So, that is the overall picture. So, now, what we are calculating is; what is the amount of energy that is going to fall on earth. So, that’s what we are doing. For that, we have found out per meter square what is the energy that is there in earth’s orbit or power that is there in earth’s orbit and then we are going to look at what is the area that is the earth, and that is all that is going to fall on the earth. So, the intensity at earth’s orbit is this value, 1377 watts per meter square. (Refer Slide Time: 18:57) Then in the earth’s orbit as I said this is the cross-sectional area of earth okay. So, this is the orbit you can think of this as the orbit, it is in an orbit and this is the cross-sectional area of in that orbit right. So, in that cross-sectional area, you will have the sunlight that is arriving. It is sunlight is heading off in all directions you can see here. So, sunlight heads off in all directions only some is falling on the earth. So, this part that is falling on the earth is what we are trying to calculate. So, for that first we need to calculate the cross-sectional area of earth, which is simple again the same thing it is pi r square and so that is pi, and this is the radius of the earth. So, that is 6371 kilometres. So, that is 6371 into 10 power 3 meters or 6.371 into 10 power 6 meters. So, that’s the radius of the earth in meters and then this radius is what we are plugging in here and so, we get the cross-sectional area of the earth as 1.27 into 10 power 14-meter square. So, that’s the cross-sectional area of earth, and that’s the cross-section that is receiving the sunlight. The sunlight is falling on it and so, even though I see I may have a 3-dimensional shape, essentially only the light that is corresponding to this flat version of my image is what is receiving the light. The 3-dimensional shape may add some I know some shading to the light as it falls on me, but this is the light that I am intersecting the cross-section of this flat surface is what intersects the light. So, the rest of it goes back right so that is basically what we are doing. So, the earth is spherical we are taking the cross-section and that cross-section is what is receiving the light. So, that is the cross-sectional area that we see here and so, if you and we also have the intensity of the radiation at the earth’s orbit. So, the intensity is in watts per meter square that is how diffuse the energy has become by the time it reached earth’s orbit, in that watts per meter square this many meter squares are intersecting that are in a position to absorb that energy. So, if you multiply this with that intensity, you get the total amount of power that is reaching the earth right. So, this is what we are going to do. (Refer Slide Time: 21:27) So, these are the two numbers we got, the previous slide we got this 1300 and I mean no two slides ago, we got this intensity at the earth’s orbit as 1377 watts per meter square. And then in the previous slide previous calculation, we got this in a mean area corresponding to us cross-section is 1.27 into 10 power 14 meters meter square. So, if you multiply these two, this is the amount of power that is falling on the earth. So 1.755 into 10 power 17 watts or joules per second is arriving on the earth. So, the sun is far away, it is releasing a very large amount of power. So, we had this value here which we will take from this place 384 Yotta watts, that’s the amount of power that is being released by the sun and. So, 384 Yotta watts and that was 384 into 10 power 24 watts or 3.84 into 10 power 26 watts. This is the power of the sun that is that’s being released by the sun, but it is getting diffused as you go further and further away, because it is getting distributed across a lot of places and then the earth because it has a small size out of this total large number that is being released by the sun only this much is being captured by the earth ok. So, out of this only so much is captured by the earth’s surface, because that’s all the surface of the earth if the earth were a larger planet more energy would fall on it in the same orbit. If you had earth as a larger planet more energy would fall on it or if the earth were closer to the sun, then by the time the suns energy reaches that orbit, it will still be at higher intensity because it is not; it’s still closer to the sun. So, the higher intensity will be there. So, the earth will absorb more energy for the same duration of time. So, if you get closer to the sun, you get more energy at the same amount of time. If you are further from sun relative to the earth you will again get less energy or if you are in the earth’s orbit with a larger size for the earth, you will get more energy. So, these are the different ways in which you can you know change these numbers, but we have a certain orbit for the earth, we have a certain size for the sun, we have certain power from the sun, we have a certain size for the earth. If you take all these factors into account, this is the amount of power that is arriving at the surface of the earth from the sun. So, that’s the number that we have available to us and is, therefore, of interest to us. (Refer Slide Time: 24:10) So, now if you now see the energy power, this is in watts per, this is the watts that we have here which has reached us from the sun. So, 1.755 into 10 power 17 watts is reaching the earth every second. So, that. So, every joule sorry these many joules are arriving every second, so that’s the power in watts. So, 1.755 into 10 power 17 watts is the power that is being received by the earth. And so, what is the consequence of this on a year-round basis. So, if you want to see what is the total energy received by earth per year just from the sun. So, then you have to take this power which is the amount of energy received by the earth every second right. So, it is a watt is in joules per second. So, these many joules is arriving on the surface of the earth every second. So, if you want to calculate what energy we received in the entire year, we have to multiply simply by the number of seconds in a year. So, for that, we have 60 seconds per minute, 60 minutes per hour, 24 hours in a day and 365 days in a year. So, actually this number that if you calculate, you will get some fairly large number some you know in basically, it will amount to a huge amount of seconds, that is there in every year and this seconds if you multiply by this power, you get the total amount of joules that is reaching the earth every year from the sun okay. So, if you do the calculation you arrive at this number. 5.5 into 10 power 24 joules is arriving at the earth every year from the sun. So, that is 5.5 million Exa joules. So, 10 power 18. So, 5.5 million into 10 power 18 joules per year is arriving from the sun on the earth. (Refer Slide Time: 26:15) So, we have now some two interesting numbers that we have that we can compare. We know as of today that humanity uses 500 Exa joules every year okay and from the sun we are receiving 5.5 million Exa joules every year okay. So, the sun is sending to us, 5.5 million Exa joules and we are using 500 Exa joules. So, these are two numbers that are of interest. So, if you compare these two numbers okay. So, given that we get 5.5 million Exa joules from the sun every year, it is of interest to see in how much time will we receive 500 Exa joules? Okay so, 500 Exa joules will take how much time? So, if you divide 500 Exa joules by 5.5 million Exa joules, you will get the amount of time in years that it takes for the sun to give us that much energy. So, if you do that, if you do 500 divided by 5.5 into 10 power 6, 5.5 million Exa joules they are both in Exa joules now 500 Exa joules divided by 5.5 million Exa joules, you get 9 into 10 power minus 5 years. So, 9 into 10 power minus 5 years, we receive this energy that we that humanity seems to be using every year okay. That if you convert that today’s that is 0.033 days if you convert that to hours that is 0.79 hours ok. So, as of now humanity or humankind is using 500 Exa joules every year, but right now it is using it from a wide variety of sources of energy. Most of the energy we use as we have seen typically is coming from fossil fuels is coming, a lot of it is coming from coal, and from other you know resources that are associated with nonrenewable sources of energy which are also not clean, but all told together we are using 500 extra joules of energy right. Now if you just keep that number in mind and not worry about the source at the moment, and you compare that against what we are receiving from the sun; then it turns out that every 0.79 hours the amount of energy arriving from the sun to the earth is equal to what we are using, which is this 500 Exa joules per year. (Refer Slide Time: 28:48) It also turns out that when radiation arrives at the earth from more than one location there is some amount of reflection. So, due to that reflection actually, even though that energy arrives at the top surface of the atmosphere; from the sun it is arriving at the top surface of the atmosphere some of it is getting reflected. So, it turns out about 70 per cent goes through and reaches the surface of the earth; meaning where we stand now when you stand outside on the in when you are out for a walk, and you feel the sunlight reaching you, that’s only about 70 per cent of the sunlight that arrived at the top part of the atmosphere. So, 30 per cent got lost due to reflections both at the top part and also some interior parts of the atmosphere, but 70 per cent reaches you on the ground. So, wherever you are about 70 per cent of what showed up at the top surface reached you. So, that is the calculation. So, if you take that also into account, this the amount of energy that we use per year instead of 0.79 hours, it takes a little longer because 30 per cent is getting reflected only 70 per cent is coming to us. So, if you just divide that by 0.7 it’s about one hour. So it takes a little longer for the amount of energy that is to get the same amount of energy because some of it is getting reflected right. So, if 100 per cent comes it would take 0.79 I mean hours. So, that is the way we have to look at it right. So, we get it in about one hour. So, if you see here, that’s the interesting part that you want to focus on; that the entire energy that we use in a year reaches us from the sun in just one hour. So, that is a very phenomenal piece of information that we have to appreciate and understand. I mean in many ways when we talk of you know non-conventional sources of energy, really the source of energy that would make the most difference to mankind is if we find a way to tap the solar energy effectively. Because as you can see even with today’s level of advancement with the level of energy usage that we do, every hour we are receiving energy from the sun equal to what we use what we require in the entire year. Okay so, in the entire year whatever energy we are utilizing, all that energy is being received from the sun by us every hour. So, you can imagine even if we are inefficiently tapping this energy that much room for error we have. We have so much room for error and still, we would be completely satisfied. So, in other words, the point you have to understand is, with the clean source of energy like solar energy you have an abundant amount of room for you know the satisfaction of all of the humanities needs okay. So, all of humanities needs can be met just by solar energy, if you just know how to tap it that’s all it is. It is already arriving at us we are not even doing anything; you don’t even have to drill to get this solar energy. It’s coming straight at you just stand outside it is hitting. So, that energy is already there, you just step out it is already reaching you. We simply need to know how to capture it and use it where we want. So, you have an automobile you should know how to capture it, use it an automobile, should know how to capture it use it in your house, you should know how to capture it you sit in your office. So, this is all that we have to do. If you know how to do this, we have an abundant source of clean energy which has no other ramifications because anyway it is coming on the earth, and anyway, earth has gotten used to it or wherever we are on earth, whatever this as of today it is already in equilibrium with this amount of sunlight that is reaching us.