Wednesday, February 23, 2011

Tilt and Sun Angle

In my previous post, I calculated the solar collector area according to electric usage in KWh/day, and solar insolation is KWh/sqft/day. One factor that I did not include is the collector tilt, and it is a very important factor.

The first thing is to decide for a tilt angle. If we want to maximize energy harvesting in the winter months, then the collector tilt must be the 90° complement of the sun angle at that time of the year, at noon. This online tool is very valuable to calculate sun angle:
[ Sun Angle ]

Here are the values calculated:
November 1st, noon, sun angle = 28°
Dec 1st = 21°
Jan 1st = 20°
Feb 1st = 26°

A tilt of 70° will maximize efficiency in December and January, because its surface will be exactly perpendicular to the sun direction. This is the tilt we will choose.

To calculate how much energy can be harvested, we must convert a 0° tilt area (the base of the KWh/day solar insolation data) to a 70° tilt. The conversion is:
1 / cos(70°) = 1 / 0.35

Now we can calculate the area of a 70° tilted collector that will provide the heat needed in December:
Electric consumption in December = 75KWh/day
Sun insolation in December at 70° tilt = 0.09/0.35 KWh/sqft/day = 0.257 KWh/sqft/day
Efficiency factor = 0.6 * 0.8 ~ 50%
Energy harvested = 0.128 KWh/sqft/day

Solar collector area = (75KWh/day) / (0.128KWh/sqft/day) ~ 600 sqft.

A 400sqft collector would provide 2/3 of our need in December.
Using sun angle and KWh usage for each month of the year, we can estimate how much of the yearly electrical bill will be provided from solar energy. That will be the object of another post.

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