Tuesday, November 9, 2010

Furnace Sizing: Heat Loss Calculations

Since I am of the opinion that our broken down hunk of junk furnace is grossly oversized for the heating requirements of the house, I decided to do a heat loss calculation to do things properly. The existing furnace is a 130,000 BTU 92% efficiency, which I really think is in the order of half again too big, and that something like 90,000 BTU should have no issues keeping the house warm.

I found the CMHC website which has a fantastic article on replacing a furnace, including a detailed method of doing a heat loss calculation based on your fuel bills.


The only other information needed is data on heating degree days for the period you are planning to calculate, which are available at these two sites. The first link will have you search for a local weather station, and provides historical 30 year averages for the station selected. The second link provides a Canada wide average for each of the last ten years. Specific HDD values vary by location, and you should try to get both localized data, and compare this with the Canadian data, especially if you live in places that are warmer (Victoria BC) or colder (Fort McMurray AB) than the Canadian average.



Looking at our Union Gas bill covering November 2008 through February 2009 (actual consumption from Oct 21 2008 to Feb 17 2009), I then simply cut and paste the calculation portion of the text from the CMHC site, and went through adding my own numbers below:


Total gas consumption from December to March = 2620 m3

Estimating 5% of gas is used for hot water, gas consumption during the period for heating = 2620 * 0.95 = 2490 m3

Heating degree days for that period (from Environment Canada) = 2680 HDD

Heating consumption by degree day = 2490 m3/ 2680 HDD = 0.929 m3/HDD

Heating consumption (difference of -35 coldest, from 18 ambient) at 53 HDD/day = (53 HDD/day)(0.929 m3/HDD) = 49.2 m3/day

Where gas has an energy content of 37.5 MJ/m3, and the existing furnace has an efficiency of 92 per cent, then:

Heat loss at 53 HDD/day = (49.2 m3/day) (37.5 MJ/m3)(0.92) = 1697 MJ/day or (divide by 24) 70.7 MJ/h

According to the energy content of electricity, 3.6 MJ/h = 1 kW, then 70.7 MJ/h = 19.6 kW

This heat loss would require a furnace that produces an output of 19.6 kW or about 66,900 Btu/h (1 kW is approximately 3,412 Btu/h).

If we allow the CAN/CSA F280 permissible oversizing of 40 per cent, then the proper furnace sizing would be (1.4)(20,100 Btu/h) = approximately 93700 Btu/h.


The determination of the correct furnace size for a house is NOT an exact science. Ideally the furnace size that is most efficient for any house is the size that effectivly runs close to 'flat out' only on the absolute coldest day of the year. If it is too small, it will not be able to keep the house warm enough, and if too large, it has an easy time heating during extreme cold, and becomes less efficient in mild weather.

1 comment:

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