I thought it would be interesting to see how much money we saved every year with the wood stove.  There are several factors involved in this, namely, the heat value of the wood we are burning, the efficiency of the wood stove, the heat value of the oil we are burning,  the efficiency of the boiler and the cost for the fuel.

fire wood

According to Woodheating.org, (linked to by the DOE wood heating web site), black cherry contains 23,500,000 BTU/cord (assumes >20% moisture).  A cord measures 4 x 4 x 8 feet.  Of course, these numbers are approximations, but for my purposes, they will work.

My wood stove is 74% efficient, according the the manufacture’s data sheet.  Therefore, I get 23,500,000 BTU  x .74  = 17,390,000 BTU heat per cord of cherry firewood.

My Furnace is 84.5% efficient, according to the last time it was serviced.  A gallon of heating oil contains 139,000 BTU.  Therefore, I get 139,000 BTU x .845 = 117,455 BTU heat per gallon of oil.

If I burned a whole cord of wood this year, I generated 17,390,000  BTUs of heat for my house.  I therefore avoided using 17,390,000 BTU ÷ 117,455 BTU/Gallon oil = 148 Gallons oil.

I get my fire wood for free.  Heating oil costs about $2.76/gallon, so I saved 148 gallons x $2.76/per gallon = $408.63 which is a little too precise.  I’d say $408 ±5% to account for imprecise qualities of cord wood and wood moisture content.

We are almost out of fire wood for this year, on account of it being cold.  Next year when the solar heaters are attached to the side of the house, it will be interesting to see how much oil is saved.

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.

1. Finish repairing the walls.  I started fixing all of the cracks in the walls last spring.  Things are looking pretty good and I am almost done.  I have a few small cracks to fill in with .
2. Paint the walls.  Basically to cover up all the repairs and make it look better.  This will also allow me to monitor the basement and see if any new cracks develope.
3. Replace the rotted rim joist under the front door.  This requires that I finish removing the front deck, which will be done this weekend, weather permitting.
4. Seal up air leaks around sill plate.  This is important in the two additions as I know that they were jacked up and whatever seal was in place was destroyed.  I will use expanding foam to make a good seal around the entire house.
5. Insulate the ceiling between the main house and the basement.  I know that we are loosing a lot of heat to the basement every winter.  Now that oil prices are $4.40 per gallon I want to minimize that loss as much as possible.  I am looking to install unfaced R-19 insulation bats in all of the open bays.  This work should pay for itself in the first year.
6. Finish repairing the termite damage on the main support beam.  Termites are interesting creatures.  They seemed to consume just one part of one of the 2 x 6′s in the main support beam.  It does not seem to have changed the structural integrity, nor can I find any evidence of termite activity in the other two 2 x 6′s.  I have looked and drilled several inspection holes, which I will fill with epoxy. The beam work may take place over the winter months
7. Replace all of the lolly columns.  They are all looking a little rusty.  I am sure that they will last for several more years, but if I am doing the beam, I might as well replace all of the support columns as well.  I may pour concrete pads under the columns.
8. Build storage shelves and better organize things.  Since our basement work last spring, things are sort of piled up in the middle of the floor.  It would be nice to set up several large storage shelves.

The most important of all of that work right now is the insulation.  I am going to try and get our oil usage down to one tank full (275 gallons) this year.  It may be a bit of a stretch, last year we used 450 gallons, but I have some strategies.

First off, the insulation and air leak sealing.  Every year I do some work on this and every year it gets better. I think if I can get the basement sealed up, a lot of heat loss through the floor will be avoided.  It will also make thinks like the furnace and hot water heater more efficient because they won’t be sitting in a cold drafty basement.

Secondly, I am working on a solar furnace.  Right now, I think I have a unique design and I am researching existing patents.  If no one else has come up with my design, I will patent it.  I am building a few prototypes to try out.  I am working on something that will take advantage of snow cover and can be mounted on south facing vertical walls.

Thirdly, I am going to have more wood available this year to better utilize the wood stove.  It can be a pain sometimes, but it does save money and the wood supply is still free.  Free heat is good heat.  More on the woodstove and wood supply later.

What the? Holy Hand Grenades Batman!

Oh no, wait, that is not coming up through the floor, it is dripping on the floor… from the boiler expansion tank… or something right above the boiler expansion tank like the air vent. That seems odd. This unit appears to be leaking right along the top seem. They are relatively inexpensive, so I just went to the local hardware store and picked up a TACO 400 Hy-Vent for about $8.00.

Old leaking air vent.

This is the new air vent with pipe thread compound applied, ready to install.

I did not drain all the water out of the system to replace this, I simply turned off the boiler feed valve. Then unscrewed the old unit and quickly screwed the new unit in its place, using a little pipe thread compound on the air vent thread to make sure it does not leak. I did this when the boiler was cold to avoid being scalded. Some water leaked out while I was doing this, maybe 2-3 ounces or so.

The installation instructions on the box read:

Screw valve vertically into tapping by hand (DO NOT USE WRENCH) making certain that the cap is screwed down thightly, to prevent scale and dirt from rushing in to valve while filling system.

After system is filled, loosen cap slightly and allow air to be released slowly, otherwise scale and dirt might rush in causing valve to leak. If this does happen, push down valve stem then pull up lightly to dislodge scale.

To shut off valve, screw cap down thightly (sic), for normal venting, open cap on full turn, wherever there is a possibility of water damage, use a waste connector and run tubing to nearest drain.

An air separator in a hot water heating system bleeds air out of the pipes so that the system works correctly. It has an air vent on the top of it. The air vent has a little float in it and when there is enough air in the top chamber, the float drops down, opening a small valve and you hear a little “ssssst.” They are important because too much air in a heating system can cause pump cavitation, knocking, excessive corrosion of cast iron parts, and/or complete system failure.

What this means is our house uses much less heating oil than average. Our first year here, we burned nearly 800 gallons. Improvements in attic insulation, replacement windows, doors, weather stripping, programmable thermostats, heating zones and out door temperature reset, oil burner efficiency and so on have paid off.

Of course, there is more that can be done. This year I hope to finish insulating the basement and crawl spaces, seal up the sill plate, fix some other air leaks and either replace or block up the basement windows. I think blocking up makes more sense as all three of them are on the north side of the house and don’t admit much light to the basement as it is.

My goal is to reduce the oil usage by another 75 gallons next winter, bringing it down to 350-360 or so, which is doable. That would be about 0.222 gallons per Ft². Current heating oil prices are $4.05 a gallon and that is not supposed to come down. Much beyond this and we will have to start rebuilding the whole house.

Oil Savings:

Last year, we pre-purchased all of our oil for $2.61 a gallon. We purchase 450 gallons and paid $1,175.00. If we where using 800 gallons, that would have been $2,088.00. That means we had a savings of $914.00 over what we would have paid.

Without any further work, this year I project we will save $1417.50. If I get the basement insulation work done and we order less oil, 360 gallons vs. 450 gallons, that increases to $1,782.00. Much more than enough to pay for the needed building supplies to complete the job.

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.