Coalpail.com Forum

Alrighty YOU engineering types... I heard that heat-loss can be generalized by multiplying a constant we'll call K by the square of the temperature differential of the current inside/outside temps. This being the case, the function representing this generalization for my home (heat loss figured at 50K at zero degrees F outside) is f(x) = (50000/70^2)(70-x)^2. If this is accurate, one could use this to determine daily coal needs depending on the temps outside (not a windy day of course) to see if one is burning (ballpark figure) what they should be or maybe more than they should be.

Any takers on this thread?

I determined (roughly) that I'll need 6600 lbs of coal for the year including about 2000 lbs for DHW.

Generalization is correct. Heat loss is based on the characteristics of the home, such as # outside walls, windows, doors, attic above, cellar below....ect..
Sounds like a lot of variables, but not that bad to tally. Outside Temp. is the big factor, use too low of a number and you over estimate, too low under estimate, zero is a safe bet. To base a calc. on your general formula would be a wasted exercise, too many variables not calculated in.

Calculating heat loss from a home is a straight forward engineering procedure, knowing the desired indoor temperature and the outdoor temperature, and the R factor of the building material or insulation that's in between. It's exercise in spreadsheet discipline and detail. The big unknown is the heat loss due to air infiltration resulting from cracks and poor or slopping construction. These can be a significant fraction of the heat loss. The ACCA has a process called Manuel J for making the calculations. See: http://www.acca.org/
Estimating heat gain (A/C) is much more difficult. In the end the air infiltration loss is just a guess based on classes of construction quality.

Chapter 2 of this book,

http://www.hydronicpros.com/publications/index.php?id=24

covers the heat loss procedures in detail.

I forgot to mention that the heat loss on our new house was calculated by the HVAC people to be 50000 BTU an hour at zero degrees outside and 70 degrees inside. That being said, if one is has a Manual J analysis performed and the heat loss is determined as accurately as can be, the generalized function that I listed above should be able to tell you how much heat you should need (or rather amount of coal) for any given temperature day... given it's not windy. The function I used divided the heat loss (50,000) by the temperature differential - which if it's 70 degrees inside and zero degrees outside, is 70 squared - to find the constant 'K'. I haven't worked in this field... ever, but it seems to me that this would be correct.

Any comments?

Pret