 So, good morning. So, what we are doing in the last class is about the solar radiation and estimation of solar radiation and that is one important number that we should know while designing the solar photovoltaic system. We should know how much energy is coming in on a daily basis, monthly basis, yearly basis in various regions and things like that. So, that we had discussed briefly yesterday. So, that will help us to actually find out what should be the optimal orientation of our solar collector for a given day or given month or we can find out optimal for the whole year and we should also be able to estimate the global and diffuse solar radiation for a given location. And as I said it is very important to understand this because when you are going for the big power plant it makes lot of difference if your estimation is few percent here and so what we do today in this tutorial is actually look at the some of the calculations about this solar radiation that we have done yesterday. So, to help with this tutorial is my student Sastry he will help you out. So, before we start I want to tell you about one thing that the daily solar radiation in India is how much? Average daily solar radiation in India varies from that is one number which everybody must be aware right 5 to 7 or 4 to 7 kilowatt hour per meter square per day right. So, daily average solar radiation in India varies like something like 4 to 7 kilowatt hour per meter square per day that number we know very well right. Now, whenever we characterize our solar PV cells and modules what is the number we use for the characterization? Characterization is AM 1.5 spectrum which is corresponding to 1000 watt per meter square right it is corresponding to 1000 watt per meter square. So, so basically if I have so if you need to find out if I need so this is one condition so this is one thing that is another thing if I need to find out how much energy will be generated on a daily basis if I install 1 kilowatt of module. Suppose I am installing 1 kilowatt peak you know peak indicates what is the peak power corresponding to 1000 watt per meter square right. So, if I if I say if I install 1 kilowatt peak power at any given location and I want to estimate how much energy it will generate on a daily basis. So, there are some quick ways to look at it how much energy it will generate. Now, if you look at the day if you look at one full day it starts from the morning and afternoon and evening and the day length in the in our country varies from 8 10 hours to 12 hours sometimes 13 hours right, but during the day the intensity is not 1000 watt per meter square it always keeps on changing right morning it will be very low then afternoon it will peak and then evening it will go down again. So, it is very difficult to then calculate energy right. So, see if I if I draw the plot between the intensity with respect to time the graph would look something like this right. So, this is like morning 6 o clock this may be evening 7 o clock 6 o clock depending on the season. So, overall day length is 10 hours 12 hours. Now, this 1 kilowatt peak of my module is is actually characterized when the intensity is 1000 watt per meter square here the intensity is changing from 0 to 100 to 100 500 it may be reaching to 800 or 1000 it may be going to 800 for a given particular day or above 800 right. So, if I integrate this whole curve what would I get if I integrate this curve if I do the integration of this curve what would I get I get energy right eventually I am getting watt into hours the unit of that. So, if integration of this curve will be energy and the unit of that will be watt into hours this is watt intensity is watt per meter square and this is time. So, watt hour per meter square. Now, if I integrate this and if I find out how many hours in a day the intensity is equal to 1000 watt per meter square right. So, I may find out that for within this curve for some number of hours for some number of hours intensity is equal to 1000 watt per meter square. So, suppose the integration of this whole curve the black curve is suppose the integration of the black curve is at has come to like 5400 watt hour per meter square. Now, we have because we have done over the day that is per day. So, this 5400 watt per meter square can be written as 5.4 hours into 1000 watt per meter square per day is it ok. This can be written as 5.4 times what does it mean for this particular day there are 5.4 hours during which the radiation is equivalent to 1000 watt per meter square right otherwise the day is very long it is not 5.4 hours day is about 10 hours or 12 hours, but during the if I take the average and find out the number of hours it is 5.4 hours of the time of the day the intensity will be equal to 1000 watt per meter square is that clear. So, for any if you just have the energy for the whole day it may be like 4000 watt per hour per meter square per day it may be 5000, 6000, 7000 depending on the region you are. So, if you go to Rajasthan in summer you may have 7000 watt hour per meter square 7000 watt hour per meter square per day means 7 hours of 1000 watt per meter is it clear. Now, this information I can bring back here. So, if what I what was my problem my problem is that I have 1 kilo watt peak is important because peak is corresponding to 1000 watt per meter square by breaking the total solar radiation in that of daily what I have done I have broken into 1000 into number of hours right. So, I can bring this information back here. So, now answer me this question. So, if I have 1 kilo watt peak system installed at a given location where the solar radiation availability is 5400 watt hour per meter square per day how much energy it will generate on a daily basis. Have you got this very clearly no doubt ok. So, because now my module is having 1 kilo watt peak and 1 kilo watt peak at 1000 watt per meter square and because for this particular location I have 5.4 hours of solar radiation and therefore, my module at that location will give me 5.4 kilo watt hour of energy right. So, depending on the location you can very easily like without just in seconds you can say if you are using 1 kilo watt module your energy generator will be 5.4 kilo watt hour. If somebody says I have only 500 watt module then energy generator will be 2.7 kilo watt hour per day right. So, if this is in a quick way you can correlate if that daily energy data is available. Now, if this is a daily data it can be converted into monthly data ok. So, daily data is 5.4 hour per meter square per day monthly data will be how much what will be the monthly solar radiation 5.4 into 30 how much it is. So, monthly how much it is? So, about 160 kilo watt kilo watt hour per meter square per day or what else it means? It means what? 160 hours of 1000 watt per meter square per day right. You can also be written that it is 160 hours of 1000 watt per meter square per day ok. What is the yearly data? I am sorry this is per month. What is the yearly data? So, 5.4 into 365. So, 1971 kilo watt hour per meter square per year this is per month or it is equivalent to 1971 hours of 1000 watt per meter square per year fine. So, in a very simple way you can actually estimate the energy that a PV module will generate depending on the solar radiation data ok. Now, the precaution that you have to take is this 5.4 hours of 1000 watt per meter square where it is available where whether it is available at the horizontal ground or whether it is available at a tilted surface right. Normally, when we estimate solar radiation the solar radiation are estimated for the horizontal ground, but you can what you need to do? You need to estimate the solar radiation at a tilted surface because your module is not going to be your horizontal surface your module is going to be a tilted surface. So, whether the radiation is mentioned for the horizontal ground or whether it is mentioned for the inclined surface like yesterday I have showed a table 3 data I have showed one is global on a horizontal surface, diffuse on a horizontal surface and global at a inclined surface. So, this data whatever you are using make sure it is in the same plane at which or in which your modules are installed is that point clear whatever the solar radiation data you use you must use the data for the plane in which your modules are installed ok. So, normally the global data are given for the horizontal surface make sure that you convert into the solar radiation data on the inclined surface fine. So, let us do some calculation I will hand out hand over to SAS 3 now. Hello, good morning. We will start from the different angles. So, that to calculate what would be the number of what would be the total radiation falling on a horizontal surface in terms of global radiation, diffuse radiation as well as total radiation. You are aware of these terminologies these are n is the day number of the air latitude is phi in degrees longitude is psi in degrees these parameters are fixed for a particular location elevation this is also fixed for a particular location in which is in kilometers declination angle delta we have to calculate in degrees slope of the collector beta again that is in our hands we have to fix that slope and hour angle omega s in degrees and number of sunshine hours just now you have calculated this n hours and maximum number of possible sunshine hours for a particular location is S max which is also in hours and these 3 parameters now you are going to calculate for a given for a given particular place. This is called monthly average daily global radiation Hg, Hg bar means this is the average term which is in kilowatt per meter square day actually we can convert kilo joules per meter square day into kilowatt per meter square day and this is the monthly average daily diffuse radiation Hd bar and again the total daily radiation falling on a tilted surface the total radiation is Hd and by calculating the terms like delta omega s and S max and n is available for a particular location we are able to calculate these 3 particular terms for a given location. Again I am just refreshing day number of the air and varies from 1 to 365 for example for January 1st n is equal to 1 and so on and declination angle is delta is equal to this is the formula I think these are given in the sheet was it given earlier hour angle omega s is equal to cos inverse minus tan pi tan delta del day length of the is equal to S max is equal to 2 by 15 into omega s day length is nothing but sun is rotating for example say 360 degrees in 24 hours each degree can be counter each degree is divided 360 degrees by 24 hours gives 15 degrees so hour angle is at exactly noon hour angle is 0 and before noon it is positive and after noon it is considered as negative. So, two terms added 1 by 15 plus 1 by 15 which gives rise to 2 by 15 into omega s which is nothing but the day length this is just a simple calculation what could be the value of n for today and calculate the value of n for April 14th 2008. So, for April 14th 31 plus 28 plus 31 plus 14 so on 2008 is a leap year so we have to calculate February 29 days. So, this would come 1 number extra 105 I calculated for April 14th so 105 and 106 this is also important while calculating the global radiations this is depend as the hour angle at sunrise or sunrise which is given by omega s is equal to cos inverse minus tan pi tan delta and calculated at for due south condition where gamma is equal to 0 degrees in the cos theta I think you have the formula sheet in the cos theta tan if you put gamma is equal to 0 then we will get this condition omega s is equal to cos inverse minus tan pi tan delta and this is valued for between September 22 and March 21 and the location is northern hemisphere and if the days in consideration lies between March 21 and September 21 the formula varies and we have to take the slope in into account omega s is equal to cos inverse minus tan pi minus beta into tan delta. This is because the declination angle is negative in this region and also the motion of the sun intersects in the horizontal plane in east west line which also lies in south of east west line passing through the observer on an inclined plane. So, we have to consider this this is the hour angle corresponds to sunrise or sunset and again this is also calculated by gamma is equal to 0 and this is the minimum of two terms cos inverse minus tan pi tan delta, cos inverse minus tan pi minus beta and tan delta. We have considered modulus here because the hour angle is positive or negative, positive hour angle corresponds to sunrise and negative hour angle corresponds to sunset and inclined surface facing due north this is due south and this is due north just the pi minus beta is there and this is pi plus beta and these both terms are same and the day length is the same s max is equal to 2 by 15 into omega s where 2 by 15 into minus tan pi minus beta into tan delta. This is the problem calculate the hour angle at sunrise and sunset on June 21 and December 21 for a surface inclined at angle tan degrees and facing due south gamma is equal to 0 the surface is located in Mumbai 19 degrees 7 minutes north 72 degrees 51 minutes east this value is latitude this value is longitude latitude is represented in pi longitude is represented as psi. For our calculations only latitude is required longitude is not required. Are there any specialties for these two dates June 21 and December 21st? I will show the last class slide these two exactly dates are shown here this is 21st December and this is 21st this is the last class slide. So, is the problem clear to everybody? First we need to calculate delta particularly at two dates June 21st and December 21st and then we have to calculate the hour angle omega s t. So, you need to calculate hour angle at the sunrise once you know the hour angle at sunrise you can find out the sunrise time and same thing this will be equal to. So, this is with respect to 12 noon. So, if you know the sunrise time you can also know the sunset time delta formula I am showing delta you will have in your slides also delta slide the slide that yesterday slides you have with you right now. So, you will find it from your slides here. So, first of all to in order to calculate hour angle what are the parameters you need? You need delta and you need delta and you need phi that is it ok. So, you need you need to find out omega s sunrise hour angle and sunset hour angle you need to find out phi and delta. Phi is it given in the problem it is given because it is we are talking about Mumbai right. So, phi is given delta you need to calculate and we have to take account of the slope inclined at an angle tending this which is beta this is the collector tilt angle. So, there is a little catch here you know what is the catch for us always sunrise time is the same as when the sun rises for the panel which is inclined to some angle sunrise time is not always the same time when the sun rises why do you think that is the case? You got my point for us always sunrise time is the time when sun rises for the panel which is inclined sunrise time is not the same time when the sun rises and the reason being that if you have if you have the panel which is inclined which is inclined like this. So, at the time when sun rises it does not see the sun sun is below behind it it does not see the sun because sun is behind it ok. So, you have to add that much angle. So, for horizontal plane you do not have to worry about all this thing for horizontal plane sunrise time is the for the panel same, but if the plane is here F a module is inclined then sunrise for the module happens little later when sun comes to the horizon for that much top is that is that clear to everybody ok. So, that is what he said if it is sun is in the if you are in a winter time that is not an issue right because sun is always in front of the observers. So, it sunrise time is always same, but if it is the summer time the sun will actually rise behind you or the behind the panel then it will take some time for the sun to come in the same plane and then sunrise time will be little later than the actual sunrise time. So, surface inclined at angle 10 actually makes this question little bit complicated if you make it 0 slight surface then it sunrise time is straight forward ok. So, do it what is the value of delta you are getting? The value of delta for December 21st would be minus 23.45 and here we are taking the slope beta term means we have to use this formula omega s t is equal to minimum of cos inverse and the pi value we are not supposed to take this directly this is in degrees and in minutes again we have to convert it into decimal value 19 degrees 7 minutes. So, this would be 19 plus 7 divided by 60 that would comes around the pi value is comes around. So, omega s t is equal to minimum of cos inverse minus tan pi tan delta and cos inverse minus tan pi minus beta tan delta and we have to take the minimum value of these two that would comes around plus or minus 94 degrees pi value is 19.2 degrees and delta 23.45 it is given 19 degrees 7 minutes right. So, 7 minutes 7 divided by 60 60 minutes is equal to 1 or 19 plus 7 divided by 60. So, it comes around 19.16. So, 19.2 we can take. Why minimum of 2? This value and this value are different because we are considering with slope also. So, this value and this value are different and the declination angle is negative at this case. So, our day in concentration is. So, the top equation is valid when your panel is flat lying flat on the ground horizontal and the bottom equation is valid when your panel is inclined to some angle beta. 21. So, you have calculated that you have got this value omega s t is equal to plus or minus. Ok everybody got this simple calculation right nothing it is just calculate 10 and 10 phi and tan delta. Again this would be same and the only difference is tan minus tan phi minus beta will come here it would be plus tan phi plus beta. This is for December 21st and this is for June 21st. So, let us say for June 21st the omega s t has come 94 degree right which means what time is sunrise rising what is the sunrise time. So, this is in hour this is in degree we convert into the time do it tell me quickly 94 degree minus 94 degree plus plus is the morning right that is on. So, plus 94 degree which is the morning time and minus 94 is the evening time. 94 degree is equivalent to how much what is hour 1 hour is equal to 15 degree ok. So, how much it is do not give me around numbers give me exact numbers because on everyday newspaper see you know in minutes also you see the difference of sunrise and sunset. So, what is the sunrise time 6 hours 26 minutes is cannot be sunrise time right because you have to subtract from the minus you are going opposite. So, your time sunrise time would be 5 hours 42 minutes. So, in the morning 5 o clock 42 minute they will be sunrise is that clear you are getting it. So, 94 degree is equal to 6 hours and some minutes how much minute it comes 6 hours and 16 minutes. So, now you have to subtract from 12 noon 6 you go back 6 hours 16 minutes which means 5 hours 44 minute is your sunrise time and your sun sunset time will be also 5 usually 6 hours 14 minutes in the p m. So, 6 14 is 6 14 p m is your sunset time right. So, what is your day length how long is the day. So, omega s you calculated is plus minus 94 degree 1 hour is equal to 15 degree. So, which means 94 degree is equal to 6 hours 14 minutes plus minus with 16 is it 6 hours 16 minute with respect to noon right 26 6.26 hours 6.26 hours or give me minute. So, 6 hours and how many minutes 16 minutes. So, this is now from 12 noon this way 6 hours and 16 minutes this way 6 hours and 16 minutes. So, if I go this way 6 hours 16 minute. So, I get 5 hours. So, 5 hours and 44 minute is my sunrise time and this way if I go 6 hours 16 minutes then I have 6 p m 6 hours and 16 minutes in the p m is my sunset time and the sum of the 2 12 hours 32 minutes is my day length. So, this day length is what S max the maximum number of sun hours this is S max double of the sunrise sun sunrise time. So, S this is nothing, but your S max. So, when we estimate a global solar radiation right there is a parameter called S max which is equal to day length maximum number of sunrise sun hours right. But in reality the number of sun hours will be less than the S max right because of the cloud cover during the monsoon and some other time. So, the S average will be less than the S max and yesterday I told you from so, S is the number of actual sunshine hours and S max is the maximum number of sunshine hours right. S a is the average actual number of sunshine hours it may be annual right and this number S a would be less than S max right. So, from where we will get S a our average value of S from where we will get this S max we can calculate right no problem absolutely as soon as we know the day for which we want to calculate and the location for which we want to calculate we can always calculate S max no problem right we can always calculate sunrise time sunset time and we can always calculate what is the possible maximum number of sunshine hours we can always calculate S a depends on when there is a cloud and when there is no cloud and it is very difficult to calculate that. So, meteorological stations will provide you what is the value of S a now again meteorological stations are not there everywhere there are only 13 stations in the country. So, many times you have to extrapolate and get the value of S a but this is how it works got it clear S max everybody can calculate now everybody so that two formula depending on whether the panel is tilted or not tilted you just need to find out delta and phi, phi is location and delta you can easily calculate everybody yes or no you cannot be neutral good we can go forward. Local upper end time is used to calculate the hour angle at hour angle corresponds to use it for the calculating hour angle at particular location. So, local upper end time is equal to standard time which is i s t minus 4 into psi s t d minus psi l plus e where psi is the longitude of the location s t d is the standard location standard value and psi l is the local value for a particular location this value will be given and this is common for all standard value and e is the equation of time where e is equal to 9.87 sin 2 b minus 7.53 cos b minus 1.5 sin b where b comes from n minus 81 into 360 by 364 where n is the day number of the year for January 1st the value of n is equal to 1 and so on. And for Indian standard conditions psi s t d is given as 82.5 degrees and local value for this problem can be taken from this 72.51 minutes. So, this is the problem determine the local upper end time corresponding to 14 30 hours at Mumbai these are the Mumbai latitude and longitudes on July 1st Indian standard time is based on 82.5 degrees. So, first we can calculate first particular day. So, yesterday I told you that for the sun actual time and the time we as per our watch are different right because as per our watch the noon time is the time when sun is at overhead position at 82.5 degree east longitude right that is our noon time. And therefore noon time actual noon time is different for different places as the sun moves. So, therefore you need to you need to find out the local upper end time for the actual measurement and the local upper end time can be estimated from this equation. So, the problem is given there determine the local upper end time corresponding to 14 30 hours. So, according to our watch time is 230 in the noon, but as for the sun motion what is the real time for Mumbai on July 1. We can start from b value we can calculate n value for this particular day this is July 1st. So, July 1st we can calculate n value and then we can calculate b value and then we can calculate e value. So, this is i s t is given standard time this is 82.5 degrees this is psi standard sorry this is i s t is 230 this is Indian standard time and psi s t d means standard longitude is this 82.5 degrees and local value is this 72.51 and we can substitute this e value. So, that we can get the local upper end time what is the n value for this July 1st? Yes for July 1st n value is 182 and we can calculate b value yes the value is 99.89 the b value comes around 99.89. And equation of time is local upper end time and standard time means 230 hours these two are different basically due to and this is from the meridian value and this is due to our solar e comes around minus 3.5 this term comes in minutes and this term also comes in minutes. So, this is standard value is 14 hours 30 minutes. So, we have to subtract 14 hours 30 minutes minus this so many minutes and this so many minutes again while calculating we have to convert this degrees minutes into decimal value 72 plus 51 by 60 while substituting here equation of time yes minus 3.5 what is that it local upper end time no it is already in minutes it is already in minutes 82.5 psi l is 72.5 then we have to convert it decimal value 72 plus 51 by 60 this is in decimal value this is not in decimal value. You got it you got the value of e and you got the value of b. So, if you put it there you will got how much time you need to subtract from i s t how much time how many minutes you have to subtract 42 minutes from i s t to get the local upper end time. So, 14 is this instead of 14.30 it will occur at 13.48 what does it mean? So, who is advance? Mumbai is advance. So, sun actually noon will occur earlier in Mumbai by that many minutes how many 48 minutes 42 minutes. So, what does it mean if you want to incline your panel for the noon time and if you follow the Indian standard time then you are doing wrong by 42 minutes big difference right. You have to find the local upper end time and then incline your panel getting it. So, the noon at Mumbai will not occur at the noon as per the time it was the clock it will occur earlier because a standard Indian standard time is based on 72.5 is based on 82.5 is everybody got this point very important the difference is not small it is a very big difference right and therefore, your gain or loss in energy can be also significant. So, when you want to put it horizontal do not put it for as per your watch put it as per the local upper end time. So, you need to calculate the local upper end time and then you can do it fine. It tells you that the noon in at Mumbai would occur earlier than the Indian standard time noon that what is telling it right how much earlier 42 minutes earlier 8 to 2.5 which location what is the location 8 to 2.5 8 to 2.5 is the standard which location Indian standard. Yeah. So, yeah I am sorry. So, it will actually occur later right. So, where sun rises earlier at 72 or at 82 sun rises earlier at 82. So, which means Mumbai will be slower later actually sun rises earlier at 82 longitude then it comes to 72 degree longitude. You are subtracting from that. So, which means that the I am sorry. So, noon noon will occur at Mumbai at 42 minutes later than the Indian standard time sorry about earlier thing it is it will occur later because the sunrise will actually occur at 82 degree first and then 72 degree which means when you want to put it. So, basically at Indian standard time sun will not be at the overhead position it will be little bit at towards the east still. So, do not blame sun do not blame it is already 12 and it has not come it is late today not like that because the IST is actually set for that particular time fine. Next problem. Now, we can go for the global radiation for a particular horizontal surface. So, now this is estimating the global solar radiation on a horizontal surface. Normally we estimate the monthly averaged daily global solar radiation ok. So, it is monthly averaged daily global solar radiation for a given surface. S max we you can calculate already right that we have done in the first problem S bar must be given average value of sunshine hour actual sunshine hours A and B are constant must be given H o is the monthly averaged daily extra terrestrial solar radiation outside the earth atmosphere. And if you know everything then you can find out H g bar which is the monthly averaged daily global solar radiation ok. This is for a particular location H naught is 24 by 5 into this thing. So, for a given location H naught is calculated from this and H g is the monthly average daily global radiation and A and B as the constants where S and S max are S is the maximum number of sunshine hours and average sunshine hours and S max is the maximum number of sunshine hours. So, from S 0 bar tell me what parameter you do not know 24 by pi. 24 by pi. So, you know everything and you know omega s is the sunrise hour angle phi is the latitude of a location delta is the declination angle and again omega s. A and B are the constants predicted and there are some measured values for A and B for some kind of 24 values are there and we can take the nearest approximate value for A and B. So, if the location is near to some determined value predetermined value then we can take that value as A and B and here this omega s term is in radiance. So, we have to calculate omega s and again it in degrees and we have to convert it into again it into radiance. Pi is the latitude of the location and delta we can calculate and omega s we can calculate and A and B values are given. S S bar will be also given and S max we have to calculate from 2 by 15 into omega s this is the estimate the monthly average daily global radiation on a horizontal surface at Vadodara 22 degrees not 73 degrees 10 minutes east during the month of March if the average sunshine hours per day is 9.5. So, this is S bar this latitude is given pi value is given nearest approximate value for Vadodara can be Ahmadabad and Ahmadabad data will be available. So, I am giving you the values of A and B constants A is 0.28 and B is 0.48 and representative day of the month for H naught for H naught calculation some days are approximated in a particular month because we are calculating monthly average daily global radiation. So, this day particularly represents the whole month radiation. So, that day is per March we can take it as 16. So, as we said the expression is for monthly average daily global solar radiation. So, that it is daily global solar radiation but is the average over the month. Now, people have found out that there are some day of the month if you do the calculation for that particular day it is the radiation is equal to the monthly average. So, this day for example January 17th if you calculate the global radiation daily global radiation on January 17th is that it is equal to the monthly average value of daily solar radiation. So, for each month there is one day if you do the calculation for that particular day it is equal to the monthly average and it is not surprised that these days are in the middle of the month most likely it is not surprising. So, January 17th February 16th March 16th April 15th. So, of course, when you take average it is it is for the middle of the month or more or less. We can start from calculating n value per March 16th and then delta value the n value per March 16th here we are not talking about any slope of the solar collector. So, we can directly take omega s is equal to tan inverse cos inverse minus tan pi minus tan delta. So, the delta value for March 16th would be omega s is equal to minus cos inverse cos inverse minus tan pi tan delta. Once we calculate omega s and we can substitute in h naught and the a and b values are already given and s bar value is already given s max will be calculated from omega s which is 2 by 15 into omega s. The value of omega s here only we can use it omega s in radians here we can directly use in degrees. So, these are all degrees pi delta pi delta these are all degrees, but the individual term is in radians this is s bar 9.5 is the s bar and the day length is 2 by 15 into omega s that is s max term. This s max term is the 2 by 15 into omega s here 1 omega s in radians is around 1.55 radians 2 by 15 into omega s omega s is around 89.02 2 by 15 into 89.02 15 day length s max is the day length omega is the hour angle and s max is the day length. Go back to the expression of omega s and s max s max s max is 2 by 15 into omega s 2 by we can sub 89.02 into 2 by 15. Here we are considering the slope of the solar collector for December 21st to there are two conditions for omega s t. June 21st and December 21st this value will be used and after that December 21st to June 20 this is for September 22 and March 21st and this is for March 21 and September 22. So, we have to calculate the minimum value in between these two range. Here we are not considering the slope of the collector. So, we can take directly this value omega s is equal to minus tan pi tan delta cos inverse minus tan pi tan delta where s max is 2 by 15 into omega s. I think enough time we have spent on this. Go to the final. What is the s bar given to the problem? Go to the problem. s bar is 9.5. What is s max? s max will be greater than s bar. It is the maximum number of sunshine hours. How much is this? So, s max is 11.98. So, here just write down first. What is how much is A? What is A? A and B is given. A is 0.28, B is 0.48, s bar is 9.5 watt hours and s max s bar max 11.9 hours. H 0 bar H naught bar that is extra terrestrial solar radiation monthly average daily extra terrestrial solar radiation. Got it anybody? What is the problem? What is the problem in calculation? So, this is this is the expression n is 75, phi is 22, delta is minus 2.42 all values are given. So, the omega s has to be in radians 1.55, omega s has to be in the radians. Omega s in degree is 89.02, omega s in radians is 1.553. That much you can do? If you did not get it right here do it at home or in your train while going back. When I was studying our professor has given an assignment and there was not enough time. So, he said do it in the train and as soon as you get down post it before going home. So, he did our assignment in the train we posted it before we reached home. So, that you can do right? If you did not get it right here you have all the time to do the calculation in your train unless until you are going by flight for which we are not reimbursing. So, what we are calculating here is H g bar that is what we are calculating right? That is a monthly average daily global solar radiation. Everything else we need to fill in. H g bar is what we are calculating monthly average daily global solar radiation and H 0 bar is monthly average daily global extra terrestrial solar radiation outside the earth atmosphere. Fine? So, how much is the H 0 bar come? Anybody? And what do you think would be the unit of it? What do you think would be the unit of H 0 bar? What hour right? It is the energy we are talking about. What hour per meter square per day? It is a monthly average but daily global radiation. It is not monthly. Monthly averaged daily global radiation. Okay? Shastri just show the value. I think they all want to do it in the train. But the earlier expressions for kilojoules right? Because 3600 has been multiplied. So, it is actually converted into kilojoules but that 3600 would not have been there. It would have been kilowatt hour. So, because it has been converted into kilojoules. So, 34206 kilojoules per meter square per day is H 0. So, put the other terms and you will get H G bar. And H G bar that is the monthly average global radiation is less than expected to be less than H 0 or more than H 0 expected to be less right? Because during the while the rays going through the earth atmosphere something will get absorbed and and scattered and all. So, H G bar is going to be less than that. Some of these kind of ideas are useful in checking whether you are you are going doing the right way or not right? So, S bar has to be less than S max, S bar max. H G has to be less than H 0. So, these are the ideas are good to check whether you are doing your calculation in right direction or not. Okay? So, what you get eventually is 6.3 kilowatt hour per meter square per day at that location 6.3 kilowatt hour per meter square per day or we can also call it 6.3 hours of 1000 watt per meter square right? We can also write it as 6.3 hours of 1000 watt per meter square per day. Fine? So, I think many of you get used to this kind of calculations little longer, but it gives a lot of idea about what happens outside the earth and what happens on the earth right? In terms of the radiation. So, please do this calculation is there any other problem that is it? Okay, fine. We will stop here.