need the theory of solar enargy or any inform help me to calculate the required equpement to built psolar sour
487293
Do a search on this site..  \nregarding solar energy.
341004
i think you are right, you only need one.\n\nw = h*c/wavelength-eV\n\nboth should give the same w, about 1.9eV, \n\nwavelentgh = h*c/w
790576
You have to find basic wind speed, K1,K2,K3 factors, Design wind pressure, Gust factors, exposed area & solidity ratio, drag co-efficient.\nWind load on panel = ExpArea*Drag*Gust*Design Wind Pr.\nPanel load is distributed to nodes & you can analyse and find the axial forces (Comp & ten) treating these are truss members.\nThen design of these members is as explained in IS802.
121934
If you want complete info this is the site, http://science.howstuffworks.com/solar-cell.htm
10808
I would suggest that you consider installing two different solar devises; one for heating and one for electricity.  Water can be heated by collecting solar energy in a panel that allows sunlight to pass through a treated glass or plastic pane and shine on a black convoluted surface that avoids reflection.  The heat produced can not escapes as easily as sunlight enters.  This is the greenhouse effect that heats your car on a rather cool day.  Inside the solar panel there are tubes or pipes that circulate cooling water to carry off the heat to an insulated storage tank.  In cold regions the cooling water must contain antifreeze.  The panel shoud be placed on a roof nearly perpendicular to a line from the sun at noon durning the cold season.  In the upper USA that would be a southern exposure.  The solar panel can be used to pre-heat water that is further heated by an electric or gas waterheater.  The number of panels you will need will depend on the number of people using hot water and whether you use it as the sole source of warm water or as preheat.  Expect panels to cost hundreds of dollars, depending on whether you install them yourself. \n\nFor electrical loads, you should consider installing photovoltaic solar panels which produce direct current (DC).  This can be used for heating water or running electrical equipment and lights but heating may use up the entire capacity of the panel(s).  You may need a battery to store energy and a converter to provide the correct voltage (and AC?) as desired.  Expect voltaic solar panels to cost less as new technologies come to market.  DayStar is building a factory now to produce cheaper continuous film cells (that could cover a roof?).  \n\nThe wattage for various items you wish to power is usually included on the manufacturer's identification plate and depends on the size and model you have or wish to buy.  Good luck.
721867
Check with local energy company. good luck
152456
Well, I applaud your ingenuity but that is already being done albeit not in massive amounts.  It is also more expensive to build a house that is solar powered but people are doing it here and there and in the long run, it does save money.  All the info you need is on the net but I doubt you can build from scratch.  You'd have to purchase kits from companies that produce solar panels and generators.
779719
Try NASA's site:\nhttp://www.nasa.gov/vision/universe/solarsystem/index.html\n\nor Wikipedia links from:\nhttp://en.wikipedia.org/wiki/Planets
944167
PV=nRT\nV = nRT/P\nV = (114g/44g/mol)(0.08206L-atm/mol-K)(28+273K)/0.987atm\nV = 64.84L or 0.0648 m^3
545704
Well, get out your heat transfer text book and start calculating. You should have all the info you should need.
815466
u can visit www.solarsystem.nasa.gov
841975
u can pray to god to give u d equation to calculate d intensity of d sun with snow also
819964
On the vernal equinox, the sun is directly overhead the Earth's equator at local noon, and we are told that the insolation under those conditions is 1368 W/m^2.\n\nAt latitudes other than the equator, the solar energy per unit area is decreased by a factor of cos(L), where L is the latitude.  This is called the "projection effect".  At 40 degrees latitude, the incoming solar energy per unit area is therefore:\n\n1368 W/m^2 * cos(40 deg) = 1368 W/m^2 * 0.766 = 1048 W/m^2.\n\nI'm assuming the question is asking how long it would take to bring the water to a boil, not how long it would take to completely boil away the water.\n\nIn either case, you are missing a key bit of information necessary to answer the question:  the initial temperature of the water.  Let's assume that temperature is T.  I'll also assume that you are located as sea level, so the boiling point of the water is 100 C.\n\nThe specific heat capacity of water is 4.1855 J/(g*degC), and the density of water is ~1.00 gm/cm^3, so to bring one liter of water from temperature T to 100 C requires:\n\nE = 4.1855 J/(g*degC) * 1000 cm^3 * 1 gm/cm^3 * (100 degC - T) \nE = 4.1855*10^3 J/degC  * (100 degC - T)\n\nWe have a collector with a 1 m^2 surface area, so it supplies 1048 W = 1048 J/sec at 40 degrees latitude, all of which we are to assume goes into heating the water.\n\nWith these assumptions, it would take:\n\nt = [4.1855*10^3 J/degC  * (100 degC - T)]/(1048 J/sec)\n\nseconds to bring the water to a boil.\n\nFor instance, if T = 25 C, t = 299.54 sec ~= 5 minutes to bring the water to a boil.
1126540
I like this one,\nhttp://www.exploratorium.edu/ronh/solar_system/
437730
See http://members.aol.com/logan20/elecspon.html for the equation to use.
812469
I'll give you some data and you can work out the details.  Total mass of the atmosphere, 5.136E18 kg.  Assumed average temperature of the atmosphere, 15 C. Enthalpy of vaporization of water at 15 C, 44.4 kilojoules per mole (by interpolation).  That should be enough data to do the trick -- except that I have no idea how much electricity Ottawa uses.  Call the power company and ask..\n\nPostscript: I tried to reply to your private note, but you have to do an address verification protocol with Yahoo before it will accept e-mails posted through your avatar.\n\nPost-postcript (11/21):  Yahoo is still claiming that your e-mail address is unconfirmed.  A pox on them!
438682
sure. go to a science store or look up schematics on the web.