Corrected the graph on solar insolation so that it is in kWh/m2/day not kW/m2/day
I newly started to learn intricacies of solar power generation, and I found your site is the best to learn about solar radiation, Insolation and PV cells etc. When I am dealing with “Calculation of Solar Isolation” I encountered with the following in consistency.
As mentioned in “Calculation of Solar Isolation” I downloaded Solar Isolation for 17 deg. Latitude with zero deg. Tilt by pressing “Download Data” for fig2. It gave me 365 das daily solar Isolation for the Latitude 17 deg. (Average for 365 days is 8.80 Kwh/m2/day). When I downloaded the solar radiation for Latitude 17 deg. using RETScreen Average Solar radiation over a period of 365 days is 5.29 Kwh/m2/day . Global Horizontal Radiation from NASA Surface meteorology and Solar Energy: Global Data Sets for Latitude 17 deg. and Longitude 78 deg. (Parameter: Insolation Incident On A Horizontal Surface (kWh/m^2/day) is 5.19 annual average. From this it shows data from RETScreen and NASA are nearly in consistence. But in your case it is a higher value. Will any body clarify why this mismatch.
I have found the data on this page incredible useful for my current project( building a solar car for world solar challenge 2015). How ever to I'm simply wondering if the data for first graph represents solar isolation for horizontal surface? In ether case I'm quite keen to know because because I intend to use the data to optimize specific features on the new solar car.
This is very nice. Happy to see this website. If we want to calculate the insolation, how would we use the "sunset" and "sunrise" formulas in the calculation. Looking forward having info.
Yeah me too, im having a hard time to know how they calculate the graph using their formulas
When you say module, which one (module) are you referring to.
I'm having a hard time. What formulas did you use to calculate the first graph? I really need the data for my project. Thanks!
I mean I really need to know the formulas used, so I can calculate the hourly solar radiation
I'm referring to the second interactive graph on this page. I'm trying to better understand all the factors affecting solar PV panel output. Under clear sky conditions, this chart is easy to use when the tilt angle is fixed at one value. But how would you do the equivalent thing with single-axis tracking, when a PV panel is located north of the equator, facing south and continuously pivoting automatically about an east-west axis to follow just the sun's elevation above the horizon? I'm trying to see better how the ratio of panel output - with single axis tracking - to the same panel's output, with fixed tilt angle, varies with latitude under clear sky conditions. For obvious reasons, when the effects of cloud cover are included - as in SAM simulations, for instance - you get a curve of this ratio versus latitude having considerable scatter, which isn't particularly helpful when explaining to my clients what's happening.
Can anyone help me answer the question that I posted on October 19th? Would much appreciate this. Thanks in advance!
It looks like there might be an issue with the I_D formula, which I think should be I_D = 1.353 x (0.7^AM)^0.678.
If that is not the case, then the chart has been incorrectly implemented, which uses the following method:
x1 = Math.pow(0.7, am);
Stot = Stot + 1.353 * Math.pow(x1, 0.678);
Many houses have been built by solar-unaware designers.
What happens if the receiving surface does not face the South?
It would be useful being capable of eveluating if the loss of energy to compare it against the cost of additional rods and structural componets aimed at aligning the panels to South.
Embedding the south-deviation second axis tilt in this page would be much appreciated.
I guess that clear-sky direct irradiation is a function of latitude, day-of-year, time-of-day, and also height. Most probably the height at see level is assumed, as there is no height input. Could you please implement also height input parameter? Or is the corresponding script available? Thanks in advance
The variable in vertical axis should be the global radiation(direct radiation+diffuse radiation ), not direct radiation. The diffuse radiation is about 10% of the direct component. Only in this way, can the numerical data be correct in the graph.
The curves show that at my latitude 68,5 N the module tilt should be 55,5 in order to get the best fit between module insolation and incident insulation. It says elsewhere here that the tilt should equal the latitude. So I am not sure where this statement come from or if this is true for a limited range of latitudes....