Figuring out your home insulation (R-Value explained)

[porkrind]

I'm a mechanical engineer, and I recently bought a larger home. I thought I'd share with you all how to determine the average insulation value of your home.

First, get your recent heating bill. The bill will state how many "decatherms" of heat you used that month. A Therm is equal to 100,000 BTUs, so a decatherm is a million BTUs. Convert this number into BTUs per hour by multiplying decatherms by a million and dividing by 730 hours in a month. Call this number Q.

Second, estimate the number of square feet surface area your home has. Just think of it as a box, and add up all the floor, ceiling, and outer-wall area you have. Call this number A.

Third, estimate the average temperature difference in degrees Fahrenheit between the outside air temp and the inside air temp. Call this number dT.

The average thermal resistance of your house in conventional "R-Value" units will then be: Ravg = A x dT / Q. (FYI, the R-Value you see at the hardware store actually has units of ft^2*degF*hr/BTU)

Example: If you used 25 decatherms in November, that would be an average heat power, Q, of 34247 BTUs per hour. (This is about 10 kW, or 14 horsepower, just to keep your house warm).

If your house is 30 ft x 40 ft, and has two floors of 8 ft each, it has 2400 sqft of floor space, and 4640 sqft of outside surface area. So A is 4640 square feet.

Say the average temp outside between day and night is 41 degF, and you keep the inside of the house at 70 degF. This is a temp difference, dT, of 29 degF.

The average R-Value of the house is then A x dT / Q, or (4640 sqft)(29 degF)/(34247 BTU/hr), which equals 3.9.


The average R-Value you calculate is only an average. Some parts of the house will be more, some less. The trick now is to find the low R-Value areas of the house and insulate them more if possible. Usually they will be single-pane windows, and exposed masonry. Use yourself to find out which parts of the house feel cold. For reference, here are the R-Values for a square foot of some common materials:

4" thick common brick wall: 0.82
Single thickness glass window: 0.88
Double thickness glass window: 1.9
12" thick bare concrete wall or floor: 1.0
Typical sheetrock ceiling with 6" of cellulose insulation: 23
4" brick wall with 3" fiberglass insulation: 13

Notice how low single pane glass and concrete are. Chances are, this is where most of your heat is escaping. Easy things you can do are:

1. Put a sheet of clear plastic over single pane windows. This will double their insulation R-Value.
2. Lay a rug or carpet on basement floors. This will increase the R-Value from a 1 to a 4.
3. Hang a nice Persian rug or decorative comforter on a bare brick wall, such as those around fireplaces. This will also increase the R-Value from a 1 to a 4.
4. Close heat vents in unused rooms or basement rooms when not in use. This effectively doubles the R-Value of these whole rooms.
5. Install a programmable thermostat, and set it to reduce the temp in your home at night or times the home is unoccupied.
6. Don't waste your money improving the high R-Value areas of your home while you still have low R-Value areas. Think about being out in the cold yourself. It is much better to have thermal underwear on and a hat and a light jacket than it is to have a heavy coat and shorts on.
7. The US Dept of Energy divides the country up into climate zones and recommends specific wall and ceiling R-Values for each zone. For example, the coldest, zone 1, is the northeast and northern mid-west, and has a recommended ceiling R-Value of 49 and 18 for the walls.

Good luck. With a little bit of analysis, you could save some $$ on your fuel bill and help the ecosystem by burning/consuming less.
:)