# I Calculation of daily solar insolation

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1. Jul 17, 2017

### jones123

Hi all,

I'm trying to calculate the daily solar insolation (integrated solar irradiances from sunrise to sunset) for a given location and day of the year. For latitude 43.2° and April 15th, I get:

H0 = (solar constant * scale factor for intra annual variation of sun-earth distance * 86400 seconds/pi) * (cos(latitude) * cos(declination) * sin(hour angle sunset) + (hour angle sunset * pi/180)*(sin(latitude)*(sin(declination))

or with numbers:

H0 = ((1368*0.99226178414418)*(86400/pi))*(cos(43.2°)*cos(9.6423°)*sin(99.1805°)+(99.1805°*pi/180)*sin(43.2°)*(sin(9.6423°)) = 3.39E7 J/day.m2

However, when I try to check the result with online calculators, they say the result should be 2.65E7 J/day.m2 ...

Can anyone please check and help me out if and what I am doing wrong? It would be of very great help!

Thank you!

2. Jul 17, 2017

### Bandersnatch

I can't say I entirely follow your calculations, but the end result difference looks in the ballpark of what you'd get if you corrected for albedo - which you don't seem to have done.

3. Jul 17, 2017

### jones123

Hi, calculations were based on Part 4: Irradiation Calculations | ITACA:

4. Jul 17, 2017

### Bandersnatch

What calculators are these? I'm getting 3.39E7 out of wolfram alpha.

5. Jul 22, 2017

### sophiecentaur

What albedo?

6. Jul 24, 2017

### Bandersnatch

Having no context for the question, I assumed the calculations were for energy available at the surface at some location, which must take into account the amount of energy reflected. The context provided later has shown that this is but an intermediary step, and such effects are taken account for later on.

7. Jul 24, 2017

### sophiecentaur

So it's not something that you could use for a general calculation. The 'real' answer must depend on the actual site, weather and time of year. A survey would be needed, I suppose.