Using monthly insolation data, sun angle and collector tilt, it is possible to estimate how much of our home energy needs (based on utility bills) will be provided by the solar system.

The first important data is the solar energy falling on each horizontal square foot, daily average, for each month, in KWh/day/sqft (click on the table or graphic to bring full screen):

Check my earlier post for references on this data.

Sun angle, calculated for the 1rst day of each month, graphical representation, followed by monthly data:

Next we need to calculate the collector sun angle. The ideal collector sun angle is 90°. The graphic shows how the collector sun angle is calculated (example shows June data), and the following table, each monthly value, for a 70° tilt.

Formula is:

Collector Sunangle = 180° - Tilt - Sunangle

The data from the table shows that maximum efficiency (collector sun angle = 90°) is achieve during the winter months, which is the reason of the 70° tilt.

Next, the ratio between horizontal area and collector area is needed to estimate the energy per collector sqft. This ratio is C/A, as represented on the graphic.

Area Ratio = sin(Collector sunangle) / sin(Sunangle)

Here too, the high tilt angle favors the winter months.

From these data, we can calculate the KWh/day per collector sqft:

KWH/day/collector sqft = KWh/day/horizontal sqft * Area ratio.

It is interesting to see how much the variation in monthly insolation is reduced by choosing the right tilt angle (compare the table above to the first table).

Finally, the solar fraction per month, for a 400sqft collector at 48% efficiency (60% from commercial flat plates data, and 80% for the "DIY factor"):

The yearly solar fraction is estimated at 96%.

The same calculation with a 600sqft collector, yields a 100% solar fraction, while a 275sqft yields a 90% solar fraction.

The difference in solar fraction between 350sqft and 400sqft, is only 1%. The December solar fraction goes from 67% (400sqft) to 58% (350sqft).

It seems that the best size is anywhere between 350 and 400sqft.

Final data is the needed collector area per month. This data could be useful if I want to occult part of the collector to avoid overheating. I will try to use a drainback system, but even with drainback, the empty collector exposed to the sun will wear faster than a shaded collector.

Also the 100 or so sqft needed during summer are for hot water needs, so this data is helpful in dimensioning the hot water system alone.

## Tuesday, March 1, 2011

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Note to readers: there is an easier way to calculate the production of a solar collector through this website:

ReplyDeletehttp://solarelectricityhandbook.com/solar-irradiance.html

No need to go through the trigonometry I described in this post.