Solar Hot Water System mounting and location

Perhaps one of the most important factors in installing any type of solar powered system is locating the solar energy collectors, be they photovoltaic panels, or thermal collectors in an optimum place. The worst thing you can do is locate a solar collector where it will not work correctly. You will never get your return on investment for the materials and installation. In that circumstance, it would have been better not to install a solar system at all. Fortunately, solar hot water is a little more forgiving in this regard than are photovoltaics. The panels must be located in a position that is unshaded between the hours of 9 am and 3 pm.

In the higher northern latitudes, location becomes more critical both from a perspective of the angle of the sun striking the solar panel, and the hours a panel will be a viable energy collector. In the northern hemisphere, the panels should be aligned to true south as closely as possible (and vice versa in the southern hemisphere). The reason being is this, the maximum energy transfer between the sun and the solar collector occurs when the sun’s energy is striking the panel perpendicularly. The further away the sun’s angle is from 90 degrees, the more spread out the sun’s energy is.

This is true for both azimuth (direction e.g. south, north, east, west) and elevation (angle the panel is mounted). In most cases five or ten degrees off on either the elevation or the azimuth will not make a big difference. Much beyond that and you will need to add collector surface area to make up for the reduced energy input into the panel.

The elevation angle is determined by your latitude above (or below) the equator. For example, my latitude is 42 degrees North. As a base figure, I would install my collector with a 42 degree angle. However, since my solar collector area is larger than I need in the summer time, I am going to increase that angle to make it work better in the winter months. For me, the optimum elevation angle appears to be around 48 degrees. I calculated this based on the PVwatts program from the National Renewable Energy Laboratory (NREL).

PV watts uses both the elevation angle and the insolation data for a particular location to give the panel energy output value. By this calculation, solar power should generate 100 percent or greater of my hot water for six months out of the year, 80-100 percent for three months out of the year, and 25-80 percent for the remaining three months for a total of 73 percent of my hot water usage. That is a large savings of electricity.

Caution: Lots of math theory ahead

Remember the whole A²+B²=C² thing from school? You probably told your math teacher “I’ll never use this stuff, why do I need to remember that?” Now you have a reason to use it. Along with A²+B²=C² there was also something else called Camp SOHCAHTOA which is a way to remember the triangle functions of sine, cosine and tangent. I am more of a visual person, so I prefer the unit circle. Either way, the proper trig function can be determined, then it is just a matter of plugging the information into your calculator.

Here are the knowns: the size of the panel is 10 feet by 4 feet wide. I am going to add 6 inches to the top and bottom as a fudge factor. So in the equation state above, C=11 feet. In order to build the proper support structure the values of A and B need to be found. We also know that the angle b is 48 degrees.

Welcome to Camp SOHCAHTOA:

Sine = Opposite/Hypotenues
Cosine=Adjacent/Hypotenuse
Tangent=Opposite/Adjacent

Using Camp SOHCAHTOA, which trig function can be used to find the value of A and which one to find the value of B? Since angle b is opposite side B, the sine function is used to find the length of side B. Therefore Side B=(sin)48 x 11 feet= 8.45 feet. Side A is adjacent to angle b, therefore the cosine function will be used to find the length of side A. Side A=(cosine)48 x 11 feet= 7.04 feet.

Lets test the math: A²+B²=C² so 7.04² + 8.45²=120.96 feet. In the above triangle; C=11 feet, so C²=121 feet. The square root of 120.96 is 10.99 feet, both of those answers are close enough for this application.

I will build a frame that is 9’6″ wide by 7’1″ deep by 8’5″ high to mount my solar panels on. The frame will be oriented to true south by using a topographical map, then confirming that with a sighting at solar noon. More in the next post.