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Here's how I answered a similar one earlier:

How long is a piece of string?

No really... There's a lot more to the question than how big the house is. What you really need to know is, how much ventilation air (intentional or not) moves through the house, and what are the areas and R-values of the surfaces the house ill loose heat through.

Here's an example:

Say a house is in a 5000 heating degree day climate where the coldest it gets is 0°F. it has an air circulation of .5 air changes per hour. Its 42' x 50', with 8' ceilings. It has R-40 insulation in the ceilings and R-20 in the walls. It's on a well insulated slab so we'll disregard foundation losses. It has doors and windows totaling 200 square feet, and they are R-2.

You size the furnace so that it will heat the house 98.5 percent of the time. That remaining 2.5% of the time is when it's so cold out the house won't quite stay at 68°, but oh well. Throw on a blanket and go to bed. Let's call that 98.5% temperature 5°.

So now you need to be able to overcome a 60° heat difference with those walls, doors and windows.

The roof is going to loose 1/r*60°*area per hour. R=40. That's 3150 BTU per hour. The walls are 1272 square feet (because we subtracted the window and door area). They loose 3871 BTU per hour. The windows loose 6000 BTU per hour.

The building has a volume of 2100x8. It changes half of that air per hour, and it takes .0182BTU to heat a cubic foot of air 1°F. That's 18345 BTU.

The total is around 31,400 BTU. Not bad for the coldest night of the winter.

Now run those numbers again when the building is 21'x50', two stories, with a 2' exposure uninsulated concrete foundation (R-1), R-15 ceiling (6" unfaced fiberglass with air blowing over it from soffit and ridge vents), R-8 walls (R-11 fiberglass, degraded from air infiltration), R-2 windows but 300 square feet of them, and 1.3 air changes per hour. Imagine that instead of a winter minimum temperature of 0°, the minimum is -30°, and your go below 2 1/2% of the time temperature is -20°. Call it an 8000 heating degree day climate. It'll be different.

85° temperature difference over a 1050 square foot attic, divided by R-15 = 5950 BTU/hr. The walls are 8' tall, plus rim joists, so 142' circumference times 17.5' is 2485, minus the windows is 2185 square feet. 85*2185/8=23215 BTU/hr. The windows are 85*300/2=12750 BTU/hr. The foundation cools to a bit below grade, so call it 142'*3', or 426 square feet. (Actual R-values depend on soil type and hydrology.) 426*85/1=36210 BTU/hr. The house has a volume of 21*50*17.5, or 18375 cubic feet, and is moving about 23000 cubic feet an hour through. 23000*.0182BTU/cu.ft*85°=35532 BTU. So this house needs a furnace capable of putting out close to 114,000 BTU per hour.

Or you could insulate the basement to R-10 with 2" of foam, which would cut down the losses there from 36210 to 3621. Chase air leaks, especially in the basement and attic. Caulk Plumbing and electrical penetrations. Put gasket kits on the doors and tight fitting interior storms on the windows. Change the infiltration from 1.25 to .6 air changes per hour by picking away at these leaks. That takes it down to 17055 BTU/hr. That will also make the windows R-3, and bring their losses down to 8500. Blow the attic full of cellulose to R-50 and go from 5950 to 1785. Now we have a whole house heat use of near 54000 BTU/hr.

In this 8000 heating degree day climate, that takes us from 257 million BTU/year to 122 million BTU/year. Weatherization let you get by with a smaller furnace, and saved you (in this example) 135 million BTU = about 1150 gallons of oil = about $3060 per year. Run that for a few years and you see what your weatherization budget is.

This is what an energy auditor can do for you. Find one who is certified and isn't just out to sell you a single product like windows.

Q: What size furnace for 3300 square foot house?

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