I there a relation between HDD and heat loss calc ?
Is there a standard relationship between HDD (heating degree days in C or F) for a general location and the calculated heat loss BTU/hr for a specific house ?
I used the SlantFin heat loss calc and came up with a reasonable number for my house, I think (85000 at -35F) (-35F because im considering a gshp but dont have much extra electric for strip heating.. in any case, -25F didnt make a huge diff..)..
The average HDD in F for this locale is 9500 (at the airport, closer to Lake Superior (ie, is more moderate)).. but it should be close.
Should I be running to the home depot for insulation/etc, or does that sound about reasonable ?
BTW, its a 1000sqft x2story slab-on-grade (but with a 5ft foundation also), 600sqft attached garage (i keep it at ~38f), upper story is 1977 code, bottom is 2001 code (jacked it up with a reno).. oil boiler into hydronic in-slab and an air handler to ducts upstairs.
Thanks for any info.
I came across several definitions for HDD on Google, but rather than get technical, it boils down to how cold an area is. It gives a measure of the 'coldness' of the outdoor temperature relative to a base temperature (usually 65 deg F).
Originally Posted by Dave72
Knowing this, however, will not enable you to calculate your heat loss, as every house (in your area) is different due to size, construction, insulation, windows, wall area, indoor temperature, etc. Only a proper heat calc will do this.
Obviously, your home with a given heat loss would have a greater or lesser loss in an area where average HDDs are more or less.
Heating demand is usually calculated at an 'outdoor design temperature' based on weather bureau data accumulated over years. One can expect that the outdoor temperature will be at or above the design temperature greater than 90% of the time.
Curious: Not familiar with the calculator that you used. (minus) 35F? Where do you live? What are typical winter temperatues? What is you indoor temperature?
You may very well need some improvements. If so, these should be done before a heat calc is done.
Since you've lived there since at least 2001, have you kept good records of how house much oil you use during the Jan-Feb time frame? You can work the heat loss calc backwards...
Taking Jan oil usage (based on early Feb fill up), and HDD in °C for same time period (you'll need to get this data from somewhere), but using some fake numbers:
500L / 1000HDD = 0.5L/HDD
Energy content of oil 38.2MJ/L. If design temp is -35C and you keep tstat at 20C, and assuming your system efficiency is 70%:
(53 * 0.5L/HDD * 38.2MJ/L * 70%) / 24hrs = 29.52MJ/hr
29.52 / 3.6 = 8.2kWh / hr
8.2kWh * 3414 = 28,000Btu / hr
This methodology is documented here:
Redo your heat loss, using your actual design conditions.
Don't fudge it to build in a safety factor.
The draw back to using heating degree days to calc how much heat you need is.
That at 2PM at 10°F outdoor temp, you need less heat then at 2 AM at 10°F outdoor temp.
HDDs doesn't allow for solar gain.
But wouldn't solar gain be taken into account when using the ratio of Energy Used / HDD during a fixed time period? I think I understand what you are saying in that this backwards method of sizing only indicates the average heat loss, and not peak loss. Do I have that correct?
Originally Posted by beenthere
HHD is a temp recording method only. Solar gain is not taken into account in anyway.
If you tracked your HHDs, and therms used(in a NG heating system).
You could show that on a 50HDD day you averaged .03 therms of gas. 1.5Therms.
But on a 45HHD day you averaged .04 therms of gas. 1.6Therms
The 50 HHD was an average day.
The 45 HHD day, was cloudly all day(house was losing heat at the same rate both during the day, and at night time).
So if you use HHD to try and size a new system. You can end up very short when at design temp conditions.
When doing operating cost comparisons.
You use the bin data for how many hours are spent at what temps.
You don't even use HHDs for that.
Same way with CDDs.
HDDs is used to determine when to make the next oil, or LP delievery by fuel companies.
For the most part.
Its taken that you don't need heat at 65°F.
Your home gains enogh heat during the day, that it will remain warm over night.
Throw a couple consecutive cloudy days in, and your house will start to become cool at night, because it won't gain enough heat during the day.
A bad example, but works good enough.
Take a greenhouse.
By HHD it may come out that it needs a 350,000BTU boiler.
In reality, it needs an 800,000 BTU boiler.
During the day, at 20°F out, with the sun out, it heats by solar gain only. No fuel used, until the clouds come out, or the sun goes down.
Heating Degree Days have nothing to do with calculated heat loss.
Heating Degree Days have everything to do with fuel consumption over the entire "Heating Season"
At what temperature is heat loss relevant?
Generally below 65°F.
Each building has a "balance point", where heating is required below this balance point. In residences, there are very small internal gains from appliances, lighting, equipment, etc. So the balance point is usually pretty high, maybe 55 or 60 degrees.
Large commercial buildings with large core areas may not require heat until it's 30 degrees out. In fact, it may require mechanical cooling when it's anything above that temp.
I was given a heat loss number of about 17 MBTU. I'm really really sure I'm not losing 17 MBTU when it's 65* outside...
Originally Posted by beenthere
You asked at what temp is heatloss relevant. Its relavent below 65°F.
You didn't ask at what temp your heatloss load was calculated at.
Questions ask in a vague context, will often get an incorrect, or unexpected answer.