By: **lsayre** On: Thu. Jul. 10, 2014 6:32 pm

For my case, wherein on the coldest day of the coldest year in a decade I burned 110 lb's of coal, then for coal my homes heat loss is about 44,550 BTUH.

110/24 x 12,150 x 0.80 = 44,550 BTUH

Back when we were all electric for a full decade the most electricity we used in one year was 30,000 KWH. And now that we have the coal boiler our electricity demand is only 6,500 KWH per year. 30,000 - 6,500 = 23,500 KWH for heat. But for electricity we only heated the home to 62 degrees when we were asleep or not home, and to 68 degrees when we were awake and at home. Call it 65 degrees on average. We now keep the house at 69 degrees. So using my 2.5% more energy demand per degree of rise rule, the 23,500 KWH must be multiplied by 4 x 2.5% or 10%.

23,500 KWH for 65 degrees x 1.1 = 25,820 KWH for 69 degrees

25,820 KWH / 210 days of heating season = 123 KWH per day on average

Multiply by the "rule of 2.5X" and the estimated heat demand for the coldest single day in our decade of heating via electricity becomes 307.5 KWH

Multiply this by 3,412 BTU's per KWH, and the result is 1,049,190 BTU's for this entire 24 hour coldest day.

Then divide by 24 to get our homes heat loss estimate via electricity at 43,700 BTUH.

By comparison, 44,550 BTUH of heat loss calculated for coal and 43,700 BTUH of heat loss calculated for electricity are within ~2%.

Last edited by

lsayre on Thu. Jul. 10, 2014 6:54 pm, edited 2 times in total.