Calculation of daily solar insolation

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Discussion Overview

The discussion centers on the calculation of daily solar insolation for a specific latitude and date, focusing on the methodology and discrepancies between personal calculations and online calculator results. The scope includes theoretical calculations and potential practical applications.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a formula for calculating daily solar insolation and provides numerical results, but questions the accuracy of their result compared to online calculators.
  • Another participant suggests that the discrepancy might be due to not accounting for albedo in the calculations.
  • A different participant questions the specific online calculators used for verification, noting that they obtained a different result from Wolfram Alpha.
  • Some participants express uncertainty about the relevance of albedo and its implications for the calculations, indicating a lack of context for the initial question.
  • There is a suggestion that the calculations may only represent an intermediary step, and that actual insolation values would depend on site-specific conditions, weather, and time of year.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or the importance of albedo, with multiple competing views on the accuracy and context of the results presented.

Contextual Notes

Participants note that the calculations may not be suitable for general use without considering specific environmental factors, and that further context is necessary to fully understand the implications of the results.

jones123
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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!
 
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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.
 
Bandersnatch said:
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.
Hi, calculations were based on Part 4: Irradiation Calculations | ITACA:

upload_2017-7-17_13-51-47.png
 
jones123 said:
However, when I try to check the result with online calculators, they say the result should be 2.65E7 J/day.m2 ...
What calculators are these? I'm getting 3.39E7 out of wolfram alpha.
 
Bandersnatch said:
if you corrected for albedo
What albedo?
 
sophiecentaur said:
What albedo?
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.
 
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Bandersnatch said:
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.
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.
 

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