Newbie to Coal. Have a ?

Perhaps it also says something for the Underwriters Laboratories (UL) organization. They approved the unworkable wiring schematic. I wonder if the boiler they tested (giving them the benefit of the doubt that they did in fact test one before approving it to carry their UL listing) was wired that way?

Yea lsayre I will contact AHS before making such adjustments. I'll also make sure I'm around all day to be able to monitor such adjustments.
I understand it could be possible for AHS to not go through the red tape of getting "relisted". It's not like they are a fortune 500 company and probably couldn't afford to shut down operations for an extended period of time. I was just under the assumption that a simple revision of the schematic would suffice, but with regulation in the states these days...
Reasons like this I became a member of the forum... for that "insider info" that the dealer/installer didn't even know. Thanks again.

McGiever wrote:

lsayre wrote:I was told something to the effect that the boiler was actually approved (listed) by UL using the bogus schematic, and now in order to change the schematic and the manual itself they would have to jump through hoops to be re tested and listed by UL with the schematic corrected.

Poor unknowing customers have to take it in the shorts...Quite the business concept.

Another example of the lax engineering and production documentation. It's systematic of ma and pa companies. None are ISO 9000 class companies. Of the companies represented on this forum EFM impresses me the most; their record keeping is outstanding and can tell you a lot just by product serial number. That's not something AHS can do. I can unfortunately personally attest to it!

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.

jdeg wrote:

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.

Where:
HDD's = 65 degrees minus the days mean temperature

jdeg wrote:

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.

The average temperature is the high and low added together and divided by 2.

If you take this figure and subtract it from 65 which is the outside temperature at which it is assumed no inside heat will be needed you will get the number of degree days for that day.

If on a day the average temperature is 35 and you burn 60 pounds of coal, that is 30 degree days (65-35= 30) and you are burning 2 pounds of coal for each degree day.

So for your coal consumption to double it would require a day with 60 degree days which would mean an average temperature of 5 degrees, so my original statement was too quick and was wrong.

Your original statement was however quite intuitive, and this merely shows that intuition is not always the best teacher (despite our intuitive desire for it to be so). You have grasped the concept fully now.

Thanks for the clarification guys. I kind of figured there had to be some equation involved. Bear with me... I'm just at the beginning of this learning curve.

I believe I have grasped the concept. For me to determine my multiplicative factor I would need to physically weigh my daily consumption of coal and then plug it into the equation. I understand it won't be 1.33 but it shouldn't be far from that given that our situations are similar.

Agreed, I would be surprised if it was much less than 1.1 or much greater than 1.6. Less would be great though!