Friday, January 14, 2011

Coal in your stocking?

There was a day when you would have wished for such a gift.

It was this week back in 1912 that Nebraska experienced its lowest temperature on record: 39 degrees below zero. A thought for those of you who live in old houses—like those built around the turn of the century—imagine what that had to be like. No insulation in the walls, single-pane windows, and likely plenty of fresh air via the multiple air leaks that were inherent to the building process of the day. It's enough to make me want to blow up an air mattress and sleep next to my wood stove tonight.

What was it like the night of that record low? Let's do some oversimplified math to stay warm and paint the picture.

The house is a 800 sq. ft. bungalow with the nine foot ceilings that were common back then. Each exterior wall is 28.3-ish feet long, so the total exterior area of the walls is 28.3*9*4=1,018.8 sq. ft. The area of the ceiling (28.3*28.3) is roughly 800 sq. ft., as is the floor. So the total exterior area of the house (walls+ceilings+floors) is basically 2,618 sq. ft.

We'll be generous and say that the uninsulated walls, ceilings and floors have r-values of 3.

And we'll say that the inside of the house is 50 degrees while the outside is -39 degrees, meaning that there is a temperature differential of 89 degrees from one inside of the wall to the outside.

To figure out how many btu's it would take to maintain this temperature differential, you multiply the total square footage of the outside surfaces of the house by the temperature differential, and then you divided that by the r-value. So...(2,618*89)/3=77,667 btu's.  Just to keep the house at 50 degrees.  To keep it at 68 degrees would require 93,375 btu's.

Either way, good luck finding that many btu's in 1912. It was a cold, cold night!