franco b wrote:If the average temperature is half that at which you are burning 60 pounds then you will burn double that amount.
It's hard for me to grasp this concept. I don't understand how to divide temps in half. For example; if it is 10*F for a day and I burn 70lbs and the next day it is 5*F I should expect to burn 140lbs? That seems unrealistic and I actually hope it is. Perhaps I'm just not familiar with division of temperature.
It is unrealistic. The current reality for me (my home, burning Harmony anthracite, in my boiler) is:
Daily Coal Consumed = [(65 degrees - the days mean temperature) x 1.33] + 3
For a mean temp of 22.5 degrees, half way between 15 degrees (the days low) and 30 degrees (the days high), I get:
Daily Coal Consumed = [(65 degrees -22.5) x 1.33] + 3
Daily Coal Consumed = 59.5 lbs.
And for a a mean daily temperature only half of that, or 11.25 degrees, I get:
Daily Coal Consumed = [(65 degrees -11.25) x 1.33] + 3
Daily Coal Consumed = 74.5 lbs.
In the above equation I consider that 3 pounds of coal daily are providing for my homes DHW, and not its heat, thus the addition of 3 to my results.
For the case of heat only, my equation would be:
Daily Coal Consumed = (65 degrees - the days mean temperature) x 1.33
The only thing that will require change in order to apply this formula to someone else's home and coal and conditions is to change the multiplicative factor. My multiplicative factor is '1.33', but yours will most likely be something different. That is 1.33 times the heating degree days (HDD's) for my case.
HDD's = 65 degrees minus the days mean temperature