# get the hourly solar radiation?

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www.met.ie/latest/yesterday.asp

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

www.met.ie/latest/yesterday.asp

active| newest | oldestre.jrc.ec.europa.eu/pvgis/apps3/pvest.php

Very easy to use, just click on the "Daily Radiation" tab.

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

averagepower. 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.If it is the true integrated power input (assuming zero cloud cover), then the true average power delivered to a point on the surface

mustbe 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.

I must admit, I shall check my sources.

Steve

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