We're real similar. I had some data available from the first 3 months of 2013. I was averaging 0.6192 lbs/HDD/1K sq ft of Reading rice. House built in 1980.lsayre wrote: At 1.44 lbs. per HDD (base 65), it calculates out to 0.649 lbs. per HDD per thousand square feet (heated). House built in 1964.
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- mdrelyea
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- Lightning
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So by comparing pounds of coal/HDD/1000 sq ft you get the ultimate picture of how well your house is insulated compounded by how efficient your heating appliance is, that is also comparable to somebody else's. It keeps the apples and oranges right. Would that be correct?
And another thing.. When I was tinkering with HDDs I wondered why do we base it on 65 degrees instead of what we are heating the house up to? I think I've come up with a reason. We need to factor in other things that are adding heat to the house like lights, TVs, refrigerators running, other appliances that generate heat like dryers, dish washers ect. All that stuff is worth 5-8 degrees (my best guess). Would that be accurate also?
And another thing.. When I was tinkering with HDDs I wondered why do we base it on 65 degrees instead of what we are heating the house up to? I think I've come up with a reason. We need to factor in other things that are adding heat to the house like lights, TVs, refrigerators running, other appliances that generate heat like dryers, dish washers ect. All that stuff is worth 5-8 degrees (my best guess). Would that be accurate also?
- mdrelyea
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Yep, as long as you're using the same base for the HDD number. The waters can still get muddied by other factors like how windy somebody's location is or how many trees they have blocking the sun, etc...Lightning wrote:So by comparing pounds of coal/HDD/1000 sq ft you get the ultimate picture of how well your house is insulated compounded by how efficient your heating appliance is, that is also comparable to somebody else's. It keeps the apples and oranges right. Would that be correct?
I think 65 is just an estimate of when most people turn the heat on. It doesn't have to be 65. When comparing numbers ask what base is being used to be sure.Lightning wrote:And another thing.. When I was tinkering with HDDs I wondered why do we base it on 65 degrees instead of what we are heating the house up to? I think I've come up with a reason. We need to factor in other things that are adding heat to the house like lights, TVs, refrigerators running, other appliances that generate heat like dryers, dish washers ect. All that stuff is worth 5-8 degrees (my best guess). Would that be accurate also?
- hotblast1357
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Sooo what's hdd?
- lsayre
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Heating degree days.hotblast1357 wrote:Sooo what's hdd?
Daily HDD's = 65 - [(days high temp + days low temp)/2]
Daily HDD's = 65 - days mean temp
When your utility company sends you an estimated bill they use the total HDD's for the area for the month times your past computed average of energy units per HDD.
- hotblast1357
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So what is the formula for all this? Days high and low temps, are those inside or outside? I'm assuming inside..
- lsayre
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Outside temperatures.hotblast1357 wrote:So what is the formula for all this? Days high and low temps, are those inside or outside? I'm assuming inside..
- Lightning
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Like this...
Daily HDD's = 65 - [(days high temp outside + days low temp outside)/2]
65 - ((43 degrees + 36 degrees) divided by 2)
65 - 39.5
25.5
In this example we had 25.5 degrees that needed to be heated..
If we used 35 pounds of coal that day then we needed 1.37 pounds per degree.
(35 pounds divided by 25.5 degrees)
which is 1.37 pounds of coal per HDD
If I did that all right lol
Daily HDD's = 65 - [(days high temp outside + days low temp outside)/2]
65 - ((43 degrees + 36 degrees) divided by 2)
65 - 39.5
25.5
In this example we had 25.5 degrees that needed to be heated..
If we used 35 pounds of coal that day then we needed 1.37 pounds per degree.
(35 pounds divided by 25.5 degrees)
which is 1.37 pounds of coal per HDD
If I did that all right lol
- lsayre
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Looks right to me.Lightning wrote:If we used 35 pounds of coal that day then we needed 1.37 pounds per degree.
(35 pounds divided by 25.5 degrees)
which is 1.37 pounds of coal per HDD
If I did that all right lol
- mdrelyea
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We were then looking at pounds of coal per HDD per thousand square feet. So if you house is 1500 square feet it would be:Lightning wrote:which is 1.37 pounds of coal per HDD
1.37/(1500/1000) = 0.9133 lbs/hdd/1ksqft
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- warminmn
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This formula doesnt take into account the wind unless I miss something? Please don't tell me it takes the same amount of fuel on a windy day as a calm day? If so, come here on the prairie during a blizzard at zero or below, lol
- hotblast1357
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Ok then that makes sense, I am at 1.25 lbs/DDH. And at .625lbs/HDD/1,000SQFT
- mdrelyea
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No, it doesn't. It only accounts for outdoor temperature, pounds of coal, and square feet.warminmn wrote:This formula doesnt take into account the wind unless I miss something? Please don't tell me it takes the same amount of fuel on a windy day as a calm day? If so, come here on the prairie during a blizzard at zero or below, lol
HDDs aren't perfect. http://www.energylens.com/articles/degree-days. They also doesn't take into account how sunny it is and how that sun shining though your windows heats up the house. I'm sure there are other things I haven't thought of that aren't accounted for either. The linked article talks about some problems with HDDs. While the authors are primarily focused on electric heat, a lot of the discussion is relevant to any heat source.