watch casio solar

we shall continue with the example that wehave been discussing last time; meanwhile, i could get useful information, which is adata source for indian locations - that is a http addresses given over here. i shall write it down, in case you cannotgo to this original document… hyphen s e c dot p d f. so, this has got one of the examplesthat you can see clearly on the screen; mean monthly global solar radiation on a horizontalsurface in mega joules per meter square a day, for selected locations of india, i haveshown here with something to bring in. this numbers you cannot read, but you can alwaysgo to the original source, and have a look at the data that is compiled by solar energycenter mnre indian metallurgical department.

so, this has got lots of contents which youcan have a can you focus on the screen. the sun solar radiation spectrum, and thevarious measuring instruments, and the locations where the measurements have been made andthe terminology and this is about twenty three locations that we have got, which basicallyhas got mean hourly ambient temperature and for each month, then the monthly global solarradiation and the monthly diffuse radiation, and given i have given here selected fromsrinagar which is more or less a high latitude location in india, upto trivandrum which isin down south kerala which is almost low latitude, which i have shown the latitudes here. srinagar has a latitude of 34 degrees 05 minutesnorth and of course, the corresponding longitudes

also are given over here. and whereas, trivandrumhas 8 degrees 28 minutes n. one of the ideas is not just i have taken or south, north,east, west and something middle, so that you will have a feel for the variation in thesolar radiation. if you look at trivandrum, this is the diffused radiation, this is theglobal radiation, the average is 19.45 mega joules per meter square day which is annualaverage, and it varies only between 17.38 which is a minimum i could see and goes upto 23.4 mega joules per meter square a day; that is if you statistically say that thestandard deviation is very low, and as you are going very near the equator the variationbetween january to december or their summer june to december is very small. like thatyou can make an interesting observation for

the remaining locations, and that is my ideaof showing it here. the corresponding diffuse solar radiationon horizontal surface is given; one of the idea says giving this data particularly toyou, is that you might verify the h d bar by h bar corelations for the indian locations,because this co relations were developed with mostly north american data. so, their applicabilityhas been tested, but now you have got a lot more data for the indian locations. so, youmight as well re establish or validate those corelations. so, these numbers may be usefulto us when we solve some examples for the indian locations. this was the example based upon f chart methodand a phi bar f chart method, we have been

discussing. it is at a location of latitude40 degrees n, a process heating system employing flat plate collectors facing south with aslope of 40 degrees, area 50 meter square has been installed. the collector parameters are f r u l is 2.63watts per meter square degree c, and f r tau alpha for normal; this we have done, but thisis for the sake of your continuity 0.72; and it is suppose to t minimum is 60 degree cand 12 kilo watt for 12 hours a day. of course, the other data ground reflectance is 0.2 forthe month of january h bar is 8.6 mega joules per meter square a day, and ambient temperatureis minus 5 degree c. so, this is the given data which we wanted to calculate.

what is the critical radiation level? this after a lot of preliminary calculationswe have calculated for, for example, mean declination for the month of january is minus20.9 degrees, at the end the sunset hour angle is 71.3degrees, in this case omega s dashalso will be 71.3 degrees, and we got the extra terrestrial horizontal monthly averagedaily radiation to be 14.4 mega joules per meter square day. and then the clearness indexk t bar is 0.6 - that is 8.6 upon 14.4; and the diffuse fraction h d bar by h bar we gotas 0.3 using co relation; then we got r b bar 2.32, i remember and r bar 1.91 right,all these calculations we have done in detail. and from this we proceeded to calculate rt, n, which is 0.178, once again this is a

co relation due to, and r d n is 0.164 thisis a relation due to liu and jordan; and remember that this is i t sorry i at the noon timeby h for the day or the monthly average day; this is i d by h d, in other words if youknow h d or h we can calculate i r i d. then in calculating r n we need h d by h that turned out to be 0.41,and similarly ultimately we got r n equal to 1.59. so, r bar upon r n is 1.91 by 1.59,which is about 1.20 r bar by r n is equal to 1.91 by 1.59. so, subsequently we estimated the criticalradiation level, which turned out to be 0.909 mega joules per meter square; this is fora hour. this is where i wanted to continue from here. so, this i c equal to 2.63 times60 minus minus 5 times 3600 upon 0.72 into

0.94 this 0.72 is f r tau alpha normal, butmy i c formula i need is f r tau alpha bar, which is f r tau alpha n multiplied by taualpha bar by tau alpha n. so, this is equal this tau alpha bar by tau alpha normal isequal to 0.94. i shall show the calculations for this, sothat we will have an exercise in calculating tau alpha bar also. so, if you look at itthis way the absorbed variation s bar… can you come here?fine fine. so, the absorbed radiation s bar is given by h bar r bar times tau alpha bar.so, this should be equal to the beam component h b bar r b bar tau alpha bar b plus the diffusedcomponent… d times 1 plus cos beta by 2 plus h bar into ground reflectivity into taualpha for the ground reflected radiation times

1 minus cost beta by 2.so, what we want is, tau alpha bar by tau alpha n. so, i can divide throughout withthis that should be equal to if i write h b bar as h bar minus h d bar by h bar timesr b bar upon r bar times tau alpha bar b by tau alpha n plus h d bar upon h bar timestau alpha bar d 1 plus cos beta by 2 plus rho g tau alpha bar g by tau alpha n into1 minus cos beta by 2 times 1 upon r bar, here also 1 upon r bar. this we have done,but then i wanted to say that the number 0.94 simply written you should not wonder, howyou got it how we got it. so, this is and how do we calculate tau alphabar b upon tau alpha n is the effective transmittance of product for the direct variation for whichwe need theta b bar that is the effective

angle of incidence for the direct radiation,for which we have already demonstrated that there are charts. so, from the charts i havepicked up for phi is equal to 40 degrees beta is equal to 40 degrees month of january, andgamma is equal to 0, theta b bar is equal to 41 degrees, this is approximate, becausei added from the charts. and the effective angle of incidence for the ground reflectedradiation theta g is from the brand mule and bakeman co relation 90 minus 0.5788, all theserelations are there in your, in your lectures. so, you should not be looking surprised, wherefrom these things have come. so, this will turn out to be 73.148.now you see we got a theta g greater than 60 degrees calling for some relation for thetato be valid other than the ashtre relation,

which is valid for less than 60 degrees. similarlytheta d for the diffused radiation is 59.68 minus 0.1388 beta plus 0.001497 beta squarethis will turn out to be 56.52 degrees. so, this is within that asthre limit of 60 degrees. so, what we get now is each one of them, sotau alpha bar b by tau alpha n, which will turn out to be in the relation 1 plus b 0into 1 by cos theta minus 1. so, i made the calculations, the data is not given whetherit is a one glass cover or a two glass cover - collector, we will see what difference alsoit makes. if it is two glass covers, 1 minus 0.17 by 1 by cos 41 minus 1 which is 0.9445for two glass covers. and if it is a single glass cover, i use the constant incidenceangle modifier co efficient b zero as 0.11

minus upon one upon cos 41 minus 1 this turnout to be 0.96 this is for one glass cover. if you recall the tau alpha variation withtheta, one glass cover will be a cover up above the two glass covers curve. then similarlytau alpha bar diffuse by tau alpha normal also in the less than 60 degrees zone. so,i will have 1 minus 0.17 into one upon cos 56.62 minus 1 which is 0.86183 for two glasscovers, and this will be 1 minus 0.11 into 1 upon cos 56.62 minus 1 which is about 0.91again for one glass cover. so, for the ground reflected portion, so taualpha g tau alpha n, i will use the relation proposed at iit kharagpur two times 1 plusb 0 into cos theta; this is nothing but at theta is equal to pi by 2, this will be 0,and at theta is equal to 60, it coincides

with the previous relationship. so, if youdo this and you have point this is 2 into 1 minus 0.17 into cos 73.148. so, much ofaccuracy is not needed, because there are many approximations we have been doing, thiscomes out to be 0.48; and if it is that is for two glass covers; and if it is one glasscover you have got 2 into 1 minus 0.11 into cos 73.148 which is0.516 for one glass cover.so, just lets write it down tau alpha bar b upon tau alpha n this is one glass cover,this is two glass covers, which will be 0.96 and 0.9445 tau alpha bar d by tau alpha nfor 1 glass cover 0.91 and for two glass covers 0.86183, i think. and the last one of course, is tau alpha nextone - tau alpha bar g by tau alpha n for one

glass cover it is 0.48, for two glass coversit is 0.516. so, if i plug it in the relation that we have written tau alpha bar by taualpha n is equal to 1 minus h d bar by h bar into r b by r bar r b bar by r bar 1.91 times0.94475 for do two glass covers plus 0.3 that is h d bar by h bar into 1 plus cos beta by2 times 0.86183 by r bar 1.91. the third component will be rho g times 1 minus cos 40 by 2 intotau alpha bar ground reflect is 0.48 by 1.91. if you do this, you will get 0.9282. and ifyou replace this numbers, this is for two glass covers. so, this is about 0.93, whichis given as 0.94, but small difference may be due to the theta b bar. but if i do withone glass cover tau alpha bar by tau alpha n is 0.9485; what i have done is, i have replaced this 9447 with 96;and this when with 0.91 and this one with

0.516. so, it will be enhanced soon. the truthmay be between 95 and 93, the average is 0. so, since we do not have that data, but remember,we now got the detail of calculating the transmittance of product or the absorbed radiation for aday. so, as i was mentioning or emphasizing this 1 plus be 0 into 1 by cos theta minus1, the incidence angle modifier co efficient relation for tau alpha by alpha n in general,depending upon the theta that you use for the direct radiation or the diffused radiationor the ground reflect radiation, it will give you the appropriate tau alpha by tau alphanormal. so, we got this now the other thing was whatis the non dimensional critical level? that is x c bar is i c by your i t noon equal towhich we have already calculated as 0.909

by r t n 0.178 times r n into 8.6. so, thisis also in mega joules, and now this is also in mega joules, so there is no problem thisnon-dimensional critical level will be 0.37. we got the non dimensional critical level,we got the critical level. then second problem i am separating it out, what is the maximumutilizability? that is we want phi bar max; that is from cleans correlation exponentialto the power a plus b r n by r bar times x c bar plus c x c bar square and the constantsa 2.943 minus 9.271 k t bar plus 4.031 k t bar square which will come out to be minus1.17. b will be 4.345 plus 8.853 k t bar minus 3.602 k t bar squared equal to minus 0.33.of course, it is time taking and but yet i am preferring to write this relations as manytimes as possible, because when you solve

in the examination or when you want to solvethe problems, the relations will be at different places in the text book or if you compileall of them, there are certain differences like i d by i h d by h h d bar by h bar h0 h 0 bar. so, there are minor differences, but then a repetition of these things maygive you an idea that if you are calculating the monthly average delay utilizability, theconstants a b c are there, which are different from the constants a and b in the correlation.and you should also have some sort of an idea that these things are related and correlatedto clearness index for some reason, whereas a and b in co relation for r t are just geometricfactors. so, if you remember the little bit of physical basis behind these correlations,you will not pick up the wrong relation. so,

this is 306, i will have a little if i donot continuously talk this is 0.704. so, phi bar max we got it of course, you canplug in these numbers as 0.51. so, once again first thing is all preliminary r r bar r nh 0 bar k t bar h d bar by h bar i c and i t noon, then from that, the critical radiationlevel, we have calculated, then the non dimensional critical level, then the maximum utilizabilitycorresponding to t minimum, based on which we have calculated i c. so, that is the sequence.third part solar load fraction met by the system. so, we need the variables y, whichis a c f r tau alpha bar h t bar and number of days in the month by the load. the loadin this case will be 12 kilo watts; that means, 12000 watts for 12 hours multiplied by secondsmultiplied by 31 actually this comes out to

be 16.1 giga joules, if you divide this by10 to the power 9, you will have 16.1giga joules. so, y will be area is 50, f r taualpha normal is 0.72 multiplied by tau alpha bar by tau alpha normal 0.94, just now wehave calculated times r bar 1.91 times 8.6 into 10 to the power 6 joules multiplied by31 by this chunk 12000 into 2 into 3600 into 31, which will come to 1.07. remember y isof the order of 1 2 3, whereas x dash will be little larger. so phi bar max y, which in a way representsthe maximum possible solar load fraction that the system can meet will be 0.51 times 1.0equal to 0.55; in other words if anybody calculates and comes out with a number larger than thisobviously, there is a calculation mistake.

and the second variable x dashed is area multipliedby f r u l times just a scaling factor 100 multiplied by the number of days or numberseconds in the month by the chunk 12000 12 3600 times 31. please remember when the loadis given in joules or mega joules make the consistent units, if it is kilo watts forcertain time, you have to convert it to joules; that is power is drawn over a period of time,which makes it the energy; this will come out to be 2.19. storage is the standard storagewhich is 350 kilo joules per degrees; this directly does not come in the calculationexcept in the r s which is the ratio of standard storage; this is a actual storage, and inthis case, it is equal to unity. so, now, solar load fraction f by the phibar f chart correlation is phi bar max y minus

0.015 times e to the power 3.85 f minus 1times 1 minus e to the power minus 0.15 x dash times r s to the power 0.76. so, thisi will try to put all the number that we know, so that it will be easy to iterate later on,because this is an implicit relation 0.55 minus 0.015 times this is not known, becausef is not known, 3.85 f remains as these times 1 minus e to the power minus 0.15 into x dashed,which is 2.19 into 1. so, this is 0.55 minus 0.0042 e to the power 3.85 f minus 1.so, what i have done is, i have set myself up a table calculate lhs and then the rhs.so, i know phi bar max y is about the maximum solar load fraction, if you forget all theselosses part; this is basically a loss and this implicit relation indirectly expressesf again in terms of f. so, this f i know should

be less than 0.55 i s u need to be 0.52 rhswill come out to be 0.523 oh good good agreement. i tried with 0.53 i got 0.5218. so, it maybe 52 or 53. and if you want to see the efficacy, you may start with 0.4, then this will betoo large, then you have to bring it down. so, it is not as difficult as it looks like,when you have this long equation, it is only just if you have the calculator particularly,if you have a programmable calculator, even if you have no idea, except that it will benice if you have the idea that f will be less than this phi bar max y, and particularlyf should be less than 1, then the iteration is not a difficult thing, it does not takemore than five minutes. now, this is a the solar load fraction wehave calculated, as per the specifications

of the system meets 12 kilo watts 12 hoursa day for the month, and the collector parameters are given for one month january. then youcan calculate that f january to f december and then capital f will be sigma f i l i uponsigma l i; i do not have the number i will let you know tomorrow if you want, but thewhole thing is repetitive; and it is certainly not very easy for hand calculation for allthe 12 months. so, a small program can be written for this and you can make 12 calculationswith the power of the computers available now, practically quite fast. four solar loadfraction will try to calculate if tank losses are included right and given the area overallloss co efficient product of the tank is 5.9 watts for degree c, and environment or thesurrounding temperature we would call t a

dashed is at 20 degrees c. so, t minimum weknow is 60 degree c. so, i will assume as a first years for the tank temperature tobe 62, so first guess. so, roughly that is t minimum plus 2 degrees. so, with this assumed temperature of the tanklosses from the tank q t will be equal to 5.9 that is u a product times 62 minus 20times 3600 joules 24 times 31 number of days, so many joules, this comes to 0.7 giga joules.so, it is not too small. so, now, the system is supposed to meet the load inclusive ofthe tank loss. so, load on the system inclusive of tank losses, which will be 16.1 giga joules plus 0.7 that comes to 16.8 gigajoules. now, my phi bar max y is proportional to this original 5.1 times original y 1.07into 16.1 by 16.8 which is 0.53 phi bar max

y in naught subscript. similarly x dashed should be enhanced equalto 2.19 into 16.1 by 16.8 this is equal to 2.10. so, again f is phi bar max y minus 0.015times e to the power 3.85 f minus 1 times 1 minus e to the power minus 1 5 x dashedinto r s to the power 0.76. so, you write it f is now 0.53 minus 0.015 e to the power3.85 f minus 1 into 1 minus e to the power minus 1.15 sorry it should be minus 0.15 timesx dashed is 2.10 times r s to the power 0.76. so, i set up myself the same table, now iwill come out by iterative process f t l equal to 0.51. now, you remember this is the solar load fraction, where there iwill just call it the load fraction including tank loss; that is our notation. so, this we said there is an iterative processto set up the iterative process average utilizability

not the maximum, because this is sort of aoverall load fraction met by the solar energy system, which does not distinguish betweenthe load on the system as needed by the consumer as well as the load, because of the additionaltank losses to be met by the system. so, utilizability corresponding to this average utilizability,which we would just call it phi bar is that point phi 1 by 1.03 of t l by y; that is right;y is 1.03. so, this will be equal to 0.49. so, again i just wanted to make sure notehow we got this number 1.03. 1.03 is the original 1.07 into 16.1 by 16.8 original y multipliedby enhancement or depreciation factor, because of the tank loss.now this phi bar will turn out to be again we have got the you use a same constants equalto minus 1.17 and this is 0.49 already known

to us equal to e to the power a plus b r nby r bar, just be cautious, because everywhere else it is r bar by r n written; i do notknow in the text books and including i picked up same habit, but the relation is in termsof r n by r bar x c bar plus c x c bar square, where a is minus 1.17 and b is minus 0.3 andc is 0.704. so, r bar by r n again you see r bar by r n 1.2. so, if i plug it in everythingis known to me. so, x c bar i back calculate it should turn out to be 0.37. so, this isan equation set up to calculate x c bar. so, this is caused by a certain temperature difference. so, i will not go about the roundabout wayx c bar straight forward equal to 0.37 equal to 2.63 into that suggested average inlettemperature minus t a times time unit tau

alpha normal into point tau alpha bar by taualpha n; this is whole thing is i c by i t noon r t n r n times 8.6 into 10 to the power6 to make it joules. so, my t i bar minus t a equal to 68.5degrees. so, t i bar is 68.5remember t as a minus 5. so, it becomes plus 5 over here minus 5 is 63.5 degrees. so, suggestedso average 63,.5 plus 60 this is the minimum; this is the possible value comes to 61.7 closeto the guess 1 equal to 62. if you are not satisfied, you could start with 61.7 and youdo the calculation you may come out with 61.8. so, that is the f t n we got and f the solarload fraction will be f t l times 1 plus q t by l minus q t by l that formula we havegot which will be equal to 0.51 into 1 plus 0.7 by 16.1 minus 0.7 by 16.1, which willbe equal to 0.49. remember here this is the

original load on the system. this is the originalload on the system, because we defined f t l as q s plus q tank by l plus q tank. so,that we divided throughout with l the load on the system. so, in this formula it is importantto remember this is 16.1 not the 16.8 which included the tank losses. so, f again is wegot 0.49, which essentially means f reduced from 0.53 to 0.49 about 0.04 in 0.5 that isapproximately 8 percent reduction. so, tank losses will reduce the solar load fractionby about 8 percent, but that can significantly depend upon the minimum temperature at whichenergy delivery is needed. in the case of f chart it is a single unit, so the correlationtakes care of the losses etcetera in the case phi bar f chart instead of 60, if you havegot 80, it would have been much higher number

than 8 percent.so, this as we have said right from the beginning the once again to repeat and to make you understandthe logic behind it we estimate the tank loss assuming a tank temperature, we treat thetank process as a part of the load recalculate the coordinates, i mean variables x dashedand y; and use it to the phi bar f chart method to find out the load fraction. that load fractionis the, so called load fraction inclusive of the tank losses; whether the guess temperatureis correct or not, is found out by obtaining what we call the average utilizability, whichcorresponds to some sort of an average x c bar, which should have been correspondingto a average inlet temperature to the collector. so, we suggested possibilities being t minimumequal to 60 or whatever, and this t i bar

we take the average; and if this average isclose to the guess that we made in the beginning, then we will stop the iteration, otherwiseyou take the new value that is obtained and go through the procedure again. so, this howthe tank losses can be taken care of; next time we shall find out the effect of the heatexchanger, and then the effect of the heat exchanger plus tank losses; until then bye.