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i live in ireland and i am trying to get the hourly solar radiation levels in watts per meter square. i dont have the money to get a pyranometer. the link below shows the Global Solar Radiation in Joule / cm square for certain places in ireland, i was wondering how do you convert it to watts per meter squared. thanks

www.met.ie/latest/yesterday.asp

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## 9 Replies

Heygalicious (author)2010-04-18

No need for any arduous calculations, my lecturer in college used this website to do the calculations for him.
re.jrc.ec.europa.eu/pvgis/apps3/pvest.php
Very easy to use, just click on the "Daily Radiation" tab.

kelseymh (author)2010-04-16

Steve gave you the (almost) right answer, but maybe a bit too tersely.  Remember that power = energy per unit time, so watts = joules/s.

The Ireland Met table you link to gives the total energy (joules) delivered during daylight hours.  To convert that to an average power, you need to divide by the number of seconds of daylight.  So you need to know the times of sunrise and sunset, and use the number of seconds between them.

Don't divide by 86,400, since the power delivered at night is zero :-)

This will also give you just an average power.  As the sun traverses the sky, the power delivered to a given point will vary according to the azimuth angle.  If you really need to know the power at 8h00 vs. the power at 13h30, then you need to do some trig calculations yourself.

steveastrouk (author)2010-04-16

I am sure that you DON'T actually allow for daylight. Sounds odd, but I've checked the maths for this before. The hours of daylight up at latitude 53 where we are vary wildly over the year, so any assessment of average power by your method would only work on two days a year.

kelseymh (author)2010-04-16

Really?!?  Then what exactly are they quoting for daily solar flux?

If it is the true integrated power input (assuming zero cloud cover), then the true average power delivered to a point on the surface must be that integral divided by the integration time (sunrise to sunset).  If you divide by 86,400 then you end up with an average power that is biased below the true value.

Consider a trivial example.  The average of 2 and 4 is just (2+4)/2 = 3.  But the average of 2,4,0,0,0,0 is not 3, but 1.

I think where I'm confused is that you say those zeros should be included in the average, and I'm saying they shouldn't be.  I would be very grateful to find a reference for the correct answer.

steveastrouk (author)2010-04-17

The flux is just that, what you get during a day - and if its dark for 8-16 hours the numbers work out just fine, you don't get much solar energy, over a day, in the winter, and you get more, over a day in the summer.

I must admit, I shall check my sources.

Steve

steveastrouk (author)2010-04-17

I have just sent a polite email to the Irish Met office to ask them for a definitive answer !

kelseymh (author)2010-04-17

Thank you!  I don't disbelieve you; I just want to know where my intuition has led me astray.

steveastrouk (author)2010-04-16

Take the daily joules/ sq centimetre x 10,000 to get Joules/ sq metre and divide by 24 x 3600 for a mean daily figure.

Take Valentia 2230J/cm^2

== 22300000 Joules / m^2 / 24 / 3600 = 238 J/second and a joule/ second = 1W

so Valentia, 238 W/m^2

kelseymh (author)2010-04-16

Almost, but not quite :-)  The solar flux at night is, if I'm not mistaken, identically zero.  So you need to divide by the number of seconds of daylight (vaguely 43,200, but strongly dependent on date and latitude).