I have my gas bills for the past 2 years. I’m trying to calculate my actual btu usage. I am being charged by the 100 cubic foot. January of last year I was billed for 303. So that would mean that I used 30,300 cubic feet of natural gas. I have found on the inter webs that there is approximately 102,000 BTU in 1 cubic foot of gas. This would mean that I used 3,090,600,000 BTU’s that month. Which by dividing that into 31 days, gives me 99,696,774 BTUs per day. Then I divide that by 24hrs, and that gives me 4,154,032 BTU’s per hour. That seems like crazy town to me, since my radiators can only put out 130,000 BTU’s per hour. Can that old boiler possibly be that inefficient??? Or have I made mistakes in my math?
The crazy town numbers are due to the fact that your method isn't temperature dependent. You don't divide by the number of
days, but rather the HEATING DEGREE-days. You also need to correct for output BTUs using the boiler's efficiency numbers- some fraction of the input BTUs went straight up the flue. With an ancient boiler it could be as low as 50%.
Download a daily base 65F spreadsheet of data from a nearby weather station on
degreedays.net that covers the span between the meter-reading dates and add them up. (Include either the first meter reading date or the last, but not both or you'll overcount the HDD by a bit. Whittle down the units to BTUs per degree-day, then /24 to get to the BTUs per degree-hour.
That gives you a linear constant- every degree below the presumptive 65F heating/cooling balance point increase the heat load by that constant. If your 99% outside design temp is 10F, that would be 55F heating degrees x your derived BTU/degree-hour= implied heat load.
The method is all outlined with examples in
this bit o' bloggery.
If you can share your ZIP code and the meter reading dates for the period where you used 303 ccf I can run the numbers here. For a quickie example:
Last January Lexington ran about 1000 HDD65, and if you used 303ccf that's about 3.09 MMBTU (million BTU). 3.09 MMBTU/1000 HDD= 30,900 BTU/HDD or (/24=) 1288 BTU/degree-hour of source fuel energy.
Lexington's 99% outside design temp is +10F, and the heating degree-base of 65F is the presumptive balance point, so you have 65F-10F= 55F heating degrees.
That means at 10F outside you'd be burning 55F x 1288 BTU/F-hr= 70,840 BTU/hr, and that's SOURCE fuel energy.
Assuming 80% combustion efficiency the implied heat load would then be:
0.8 x 70,840 BTU/hr= 56,672 BTU/hr.
Of course some Januarys are much colder than others and you might live in a place colder than Lexington- this is just a dummy example. Run that napkin-math on your real meter reading dates and real weather data for the time period. One thing is certain- the heat load can't be higher than the source fuel BTUs that heated the place. It's unlikely that your boiler is running better than 80% or less than 50%, so running the numbers assuming 60% efficiency and 70% efficiency would put the ~1 sigma error bar bracket on the actual heat load, and using 80% should be a fairly firm upper bound.
