Preferably this collector will be mounted adjacent to a large reflective surface.  A snow field would be perfect, however, dry sand, concrete and water will also work.  The quality of the reflective surface is called Albedo, which in Latin refers to its “whiteness.”

Here are some albedo figures for some common reflective surfaces:

Also, the lower the sun angle, the larger the reflective surface should be.  This is for two reasons; first, the law of reflection states:

1. The incident ray, the reflected ray and the normal to the reflection surface at the point of the incidence lie in the same plane.
2. The angle which the incident ray makes with the normal is equal to the angle which the reflected ray makes to the same normal.

Therefore, the lower the angle of the sun the further away the reflection point will be from the collector.

Secondly, the lower sun angle also means that the energy density of the sun light is much less.  A larger reflective area will aid in gathering more energy.

Solar Collector parts list:

*Not required for a passive system
**Quantities doubled for a passive system

Total, active system: $447.16
Total, passive system: $355.74

All parts except snap disk switch, PV panel, and DC fan were priced and purchased at the Home Depot.

Therefore, a passive collector needs to offset $355.74 in the first year’s use, an active collector needs to offset $447.16.  According to NYSERDA, the cost of home heating oil is currently $3.823 a gallon.  I need to save 117 gallons of fuel oil to offset the $447.16 collector cost.  Each gallon of home heating oil has 139,000 BTU. My boiler is 86 percent efficient, therefore, I get 119,540 BTU per gallon.

My solar collector needs to generate 13,986,000 BTU to save 117 gallons.

I expect the active solar collector I build to generate about 45,000 BTU per day.  The heating season lasts from October through April, or 212 days.  I expect 30 percent of those days to be too cloudy to generate significant heat from the collector.  I have 148 days of good solar resource, so 6,660,000 BTU can be expected.  That makes the payback approximately two years vs the one year original design goal.

Solar collector tools:

Ridgid MS1065LZ 10 inch miter saw

Makita 6213D 3/8 inch cordless drill

Ridgid R84001 3/8 inch cordless drill

Bosch 1587A Jig Saw

DeWalt D28110 rotary grinder

Construction details to follow in Part III

What if someone could design a solar collector that can be easily built and installed by the average do it yourselfer.  This is the idea that I had and I think I may have something.

Here are a few design benchmarks:

1. That solar system would need to be fabricated on site with standard power tools.
2. It should be constructed of material readily available at most home improvement stores and the like.
3. The system should be simple and easy to understand and troubleshoot.
4. It should be simple enough to construct that anyone with basic carpentry and metal working skills can build it and install it.
5. It should be efficient and relatively inexpensive, paying for itself in one year.

Those are the basic ideas I had and I believe I met most of them with my design.  What I was going for was something that would produce heat during the winter time and be optimized for cold snowy locations.

This solar collector is used to heat air, circulating air over a collector plate and returning it to the conditioned space.  Air heating panels are simpler to construct than water heating panels, their downside is that there is no storage capacity associated with them.  In other words, they work great when the sun is shining, but will not produce any heat at night.  They are suplimental heaters in most cases and cannot replace a central heating system.  That being the case, they can still save a significate amount of energy.

The main collector surface is made from aluminum drink cans.  The cans have the tops cut off and are stacked horizontally like this:

horizontal can solar collector

This arrangement is more work and requires more materials but it has several advantages over other designs:

1. Each can becomes a mini solar receiver similar to solar receivers used on large concentrated solar systems.
2. The array of receivers gathers energy more effectively because there is less reflected energy than an ordinary flat plate collector.  Once the energy strikes the collector surface, it is reflected down into the cans where the convex bottom aids in absorption.
3. If mounted on a vertical south facing wall in front of a reflective surface such as a snow field or dry sand, the array will gather much more solar energy due to the increased insolation area.
4. Aluminum is an excellent conductor of heat, thus the heat will move to the back of the collector plate, which will be cooled by forced air.

The main idea here is to make it simple yet effective.  Aside from the collector, a 250-300 CFM DC fan and a PV panel round out the system.  A small “snap disk” thermal fan switch turns the fan on and off depending on the collector temperature.

Part II will discuss tools and materials.  I expect the system to cost about $400-450 to build.  The most expensive item is the PV panel, which can be substituted with an AC wall transformer.

Part III will be a sysnopsis of my own system.

• Collector supply temperature 148 degrees F
• Collector return temperature 166 degrees F
• Flow rate 5.8 GPM

To calculate how much energy I am receiving, I need to know the thermal gain of the collector array, in this case 166 degrees – 148 degrees = 18 degrees. The flow rate is 5.8 GPM, therefore we can say 18 degrees x 5.8 GPM = 104.4 degree gallons/minute. A gallon is equal to 8.33 pounds, therefore we can say 104.4 degree gallons/minute x 8.33 pounds = 869.65 degree pounds per minute. A BTU is the energy it takes to warm 1 pound of water 1 degree Fahrenheit. A degree pound is equal to a BTU; the energy being supplied by my solar system this after noon is:

• 869.65 BTU/Minute
• 52,179 BTU/hour

This equals:

• 15.3 kWh (3,413 BTU/kWh)
• 0.375 gallons #2 heating oil (139,000 BTU/Gallon)
• 0.573 gallons propane (91,000 BTU/Gallon)
• 0.512 CCF natural gas (102,000 BTU/CCF)

All of this is, of course, before system losses, so it would be safe to say that I was actually receiving about 10 kWh per hour during the afternoon. Still, not too shabby.

Now, today is an exceptional day because the solar tank was completely cold and the difference in temperature between the solar tank and the solar panel loop was making the heat exchanger operate very efficiently. Once the solar tank is warmed back up (which it should be by the end of the day, the collector array BTUs will come down a somewhat. What is nice is that every sunny day, I can expect to get between 7-14 kWh a day.

That is going to be expensive, I know.  I looked up the prices today of the panels and the grid tie inverter plus a tracking mount and it was a whopping $14,590.00.  The tracking mount is really not needed to make the system work, however, with a tracker you can expect a 25 percent increase in efficiency.  Federal and State tax incentives knock that down to around $10,500.00, but even so, that is about a ten year payback.

If the news is to be believed, those prices should start coming down soon because of added capacity in production of both the panels and the raw materials to make them.  Something like a 40 percent increase in manufacturing capacity last year alone.  I hope that is true for a number of reasons. I will compare prices in the spring and see if they have come down.  We do have a long list of other projects, but, perhaps I could take a second job at Lowes or the Home Depot to help pay for it.

One of the really cool things about photovoltaics, small wind turbines and micro hydro turbines is the distributed generation effect.  What this means is that instead of having a couple of huge generating plants with long high tension transmission lines bring the power to major population centers, there are many many small generators over a large area connected to the power grid and synchronized together.  This means less need for high tension power lines, huge power plants, and all the associated problems.  Demand can be met locally and more efficiently than the large power plant model.  Transmission of electrical power over any distance is inefficient, the longer the distance, the greater the inefficiency.  Its better for local small businesses (installers), better for the environment and in the long run, better for the home owner.

1. Use a magnetic compass, and calculate the magnetic declination into the compass azimuth
2. Use a topographical map, and site along a line to a known land mark
3. Use a solar noon calculator, then at solar noon, note the sun’s shadow on the ground.

The fastest way seems to be using a magnetic compass. I looked up the magnetic declination (the variation between magnetic north and true north) for my area on the NOAA web site. This site is easy to use, you just need to know your zip code. For my area, the declination is 13 degrees west of north, in other words, my N compass needle should be pointing at 347 degrees and the compass will be aligned on the true north/south axis. I don’t trust magnetic compasses that much as they are influenced by metal objects and electrical fields that are nearby.

To find true south using a topographical map, you have to be able to interpret the map, then find some land mark that is due south from your mounting location and use it to align your solar collectors. This method should be backed up by one of the other methods described here. To find topo maps, use www.topozone.com. They have the maps right on line, you just enter your place name and hit search and you should have a map of your area. I always resize the map to large and 1:25:000 aspect ratio. It may be necessary to pan around and find your house. Once the house has been found, click on the location to center the map on it. This will also give you the coordinates for your house. You should change the latitude and longitude coordinates to D/M/S format and write them down.

Finally, you can use solar noon to show you where true south is. Basically, solar noon is when the sun is at its highest point. In the northern hemisphere, that means the sun is located due south, and vice versa in the southern hemisphere. To use the solar noon method, you need to have two stakes and a piece of string. Drive the first stake into the ground at your panel mounting location and tie the string to it. Tie the string to the second stake as well. At the time of solar noon, align the string with the shadow of the stake first stake that was driven into the ground, and drive the second stake into the ground. The string is now along the north-south axis. You have to know the exact time of solar noon at your location. To find this information, I again referred to the NOAA website. You can either use the closest city in the drop down list, or enter your latitude and longitude from the topo map. If you do not know how to read a topo map as described above, you can find the reference coordinates for your community at the FCC website using place name and state, or at the US Census website using zip code lookup

For my site, I used methods two and three. This is the north-south axis at the mounting location for my solar panels.

That agrees within about a degree of what I extrapolated from the topographical map. This is another picture with a square aligned with the big side along the north-south axis, and the small side along the east-west axis.

Basically, the face of the solar panels will be aligned along the east-west axis. Incidentally, the magnetic compass was way off.

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.