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how to figure heating degree days
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Uni R_3
Member Posts: 299
Then you can start comparing your house's overall number next to others.
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heating degree days
How do you figure out heating degree days.
To know how much LP, should have been used on a house with a heat loss calc of 120,000 BTU's0 
I can help with the HDD part, but
you're on your own on the effeciency calculations.
You can get the Heating Degree Days (HDD) historical data here:
http://www.weather.gov/view/states.php
Select your state then select climate data and scroll down till you see the nearest station's chart showing the monthtodate HDD's under column 6A. There is a way to get more historical HDD data if you surf this site further.......
Good luck,
Dave0 
fuel use
Use this formula to get close to the amount of fuel needed:
F= (HL x 24 x DD) divided by (E x P x TD)
F =fuel usage in gallons per year
HL =heat loss calculation in BTU/hr.
DD =degree days per year in your area
24 =hours in a day
E =heating unit DOE rating
P =Heat value of fuel used
TD =Design temp of system
This should get you in the ballpark. It is meant for a home with hydronic heating and does not account for domestic hot water use. Too many variable to be exact, but I find a radiant heated home will figure out the same if you do a proper heat loss and use the radiant loss in your calculation.
0 
Al your formula
Al,
Would this formula be proper to use when comparing 12 months
NG usage with previous oversized CI boiler to a new Mod con with one of the recent three years worth of NG usage?
Did remove the old WH and installed the new IWH at the time of boiler change0 
formula
I've had this for many years and it still seems to work well if you consider the variables, but all the figures in the formula are constants, so it should work to compare. Just use the proper numbers. The only things that change is the efficiency of the unit and the value of the fuel if you are switching to another fuel source. Many older boiler were not DOE rated so that may be a problem.0 
I stand with Al on this
The efficiency is the real wildcard. The formula is fairly sound and I found it accurate to within 10%.
A trick you can use with it is, if you know your fuel consumption and degreedays for a given year and are confident in your heat loss basis, you can backcalculate to find your real efficiency. You may be surprised, pleasantly or otherwise.0 
Al fuel
Al, Can I use CCF for natural gas for F instead of gallons used? Do I just plug in the formula the 1675 CCF used?
Scott0 
Am I doing this right?
Are these reasonable numbers? Am I doing this right? I don't understand some of these things
If I put it 10 for design temp , I get 7277 gal. ... If I put it 82 degrees( 72 inside and 10 outside), I get 887 gal. ( less than a 72 degree diff. )
HL = 59000
DD = 5900  mountains western PA.
E = 0.82  Burnham V74
P = 140000  fuel oil
TD = 72
" F= (HL x 24 x DD) divided by (E x P x TD) "
Raising the heat load or degree days will raise the fuel usage which makes sense. Lowering the efficiency , heat content of the fuel, or design temp will raise the fuel usage which doesn't make sense to me. I must be using the wrong numbers.
I am getting 1010 gal. I'd like to understand and use this formula... would you please explain the units of measurement involved? Maybe give an example?
thanks0 
May I?
One thing that is missing is the Cd factor, usually about 0.6 for the 6000 DD range.
The TD is the same as the one used to calculate the heat loss so it's purpose is to cancel out that factor. It has to be the same used for the heat loss or else your heat loss has to change to that basis. Change the TD and your numbers will jumb out of accuracy.
Let me use your numbers:
59,000 x 5,900 x 24
72 x 0.82 x 140,000
= 1,110 (you are so far correct). Now, let's apply the Cd factor of 0.6:
x0.6 = 606.5 gallons per year, which seems excellent, but wait! Something else jumps into this; the rated AFUE versus the actual efficiency as applied to the system.
Now, that 0.82 is the rating on the V74 boiler. I see a DOE output of 156 MBH versus your 59,000 heat loss, a factor of 2.64 oversizing. In actual usage your efficiency will be (if experience serves) closer to 5560% given cycling and other variables.
In other words, instead of using the 0.82 as a divisor, use 0.55 or 0.60. Your oil usage would be more in the range of 904 to 986 gallons per year.
You can backcalculate your actual efficiency using your actual oil usage, subject to the accuracy of your other factors (heat loss, actual degree days and Cd factor).
I hope this helps!
Brad"If you do not know the answer, say, "I do not know the answer", and you will be correct!"
Ernie White, my Dad0 
You Certainly May, and thank you
I have been monitoring and "sensoring" my boiler for a few months now, and I believe the efficiency of the boiler itself, regardless of the system, is closer to the 5560% figure you stated.
So, you're saying that:
1. the equation is correct as is is written in the previous post, but I should have used .55.60 instead of .82? I agree with you, but how does one know that? The sticker says .82... I don't think most people would go to the trouble I have gone, or have your experience with a particular boiler. When I try to look something up about a v 74 it almost seems like an oddball.
and ...
2. I should have applied a .6 Cd factor that applies to the HDD zone in which I live, that was missing in the previous equation
So the equation should be F = ((HL x 24 x DD) / (E x P x TD))x Cd... Where do you get the value of Cd?
Is that correct? And again.. thanks.0 
The Value of Cd...
To your questions in order:
1) Unless you measure the efficiency over time you will not know. The published 82% is an AFUE number tested in a laboratory under steadystate conditions, such which rarely exists in the real world and probably not in your situation either
2) You are correct to apply the Cd factor. Now, where to get that? It is published by ASHRAE
The attached is the handbook ("Cook Book") from LorenCook Company, a fine manufacturer of fans in Springfield, MO. See Page 64 for the chart. See what you make of it!
Cheers
Brad"If you do not know the answer, say, "I do not know the answer", and you will be correct!"
Ernie White, my Dad0 
Thank you
I believe I have found the info ... including the fuel use formula and the Cd factor. plus there's a heck of a lot of other info there as well.
Thank You0 
OffTopic
Your advice and knowledge is amazing Brad and your posts are always a healthy read.
Going offtopic a bit, is this formula translatable for an electric house? More precisely, can I conduct a similar exercise for my new electric powered geothermal heat pump?
Thanks.0 
3413 BTUs per kWh
Could you multiply your consumption in kilowatthours by 3413 and then divide by the total HDDs over the consumption period and then divide again by your square footage? It would be very nice if you could post that number in the BTU/DD/SF thread to show how geothermal stacks up  maybe mention inside temps and general construction (regular vs superinsulated etc.)0 
design temp
I have a question on design temp. How do you figure out what your design temp is? Is design temp based on the the average od the coldest day for my appropriate geographical area?
0 
Why, thank you, Patrick
Glad to please, it is nice that you said that!
For electric as UniR said, (or any fuel for that matter) you just substitute the perunit BTU's of the fuel you are using and substitute the efficiency, both in the divisor. All else remains the same.
3,413 per kWH for electric resistance; Eff. 0.98 to 1.0
140,000 per gallon for #2 Fuel Oil; Eff. to suit (0.550.85, do you feel lucky? So many older systems being oversized...)
92,500 per gallon for propane; Eff. similar to gas, a point or two less.
For Geothermal or any other heat pump for that matter it is indeed electricity but the efficiency (COP) is of course higher. Being in the divisor it brings the numbers down.
Example: 3,413 BTUH per kWH and an efficiency of 2.5 to 4.0 and sometimes more!
Remember though, the efficiency is not just nameplate it is how it is applied to the system. Gross oversizing hurts, a lot. Deadnuts sizing to load might get you close to nameplate. Modulation will get you closer than that.
This formula is not bad, is it? You just have to have a few "known" buildings under your belt with a few years fuel bills to get a feel for how accurate it is. From there your experience takes over as to relative efficiencies.
Hope this helps!
Brad"If you do not know the answer, say, "I do not know the answer", and you will be correct!"
Ernie White, my Dad0 
It depends...
Every area where weather stations are set up have several design temperatures. I will give you mine in Boston for relative comparison.
There is the "exteme record" (once in a lifetime). Boston's cold record was, IIRC, 18 below zero. (A real Weezbo summer!)
There is the "median of extreme lows" which for Boston is 1 degree F.
Then there are the bins or occurances at 99.6%, 99% and 97.5%, meaning it would only be colder in a given year 0.4%, 1% and 2.5% of total annual hours.
For Boston these numbers are respectively, 7, 12 and 15 F.
Most designers use the 99.4%. Many use zero because it makes the math easy
What is your location? I may have a weather packet for you. See the attached for Boston from the USAF."If you do not know the answer, say, "I do not know the answer", and you will be correct!"
Ernie White, my Dad0 
I don't get this
I am very interested in a formula such as this, but I'll be darned if I can get this to make any sense.
Oil Heating
H/(DT x 140,000 x E) x D x 24 x CD= GALLONS/YEAR
Where:
DT = Indoor Design Minus Outdoor Design Temp.
H = Building Heat Loss
D = Annual Degree Days
E = Seasonal Efficiency
CD= Correlation Factor CD vs. DegreeDays
Seems to me that by having the DT in the divisor of the formula isn't going to work.
The larger the difference... the smaller the result and it should be the other way around... as the indoor design temp goes up the fuel use goes down.
That ain't right.
the larger the difference between inside and outside design temp the more fuel you should use.
What the heck am I missing here?0 
I am in the Pittsburgh, Pa. area.
My ultimate goal is to use the above formula to compare the old W/M cast iron boiler which was oversized and over pumped (75 GPM) to a W/M ulta. As I understand it this formula will let me even out the temp. so I am close to comparing apples to apples.
Thanks
Scott0 
Pittsburgh Weather
As luck would have it, I have the Pittsburgh weather packet right here. Have fun!
Brad"If you do not know the answer, say, "I do not know the answer", and you will be correct!"
Ernie White, my Dad0 
I appreciate your trouble,
I wonder if you'd explain it a little further? It is important to me that I understand it and get it right.
If you use:
H/(DT x 140,000 x E) x D x 24 x CD= GALLONS/YEAR
Where:
DT = Indoor Design Minus Outdoor Design Temp.= 72
H = Building Heat Loss =59000
D = Annual Degree Days=6000
E = Seasonal Efficiency =60
CD= Correlation Factor =.6
You get 843 gal... What I don't get is... by having the design temps in the divisor, lowering the change in inside and outside temps will raise the fuel usage. A 50° difference will result in a 1213 gal. fuel use wher a 72 degree difference results in 843 ???0 
Don't mess with the divisors...
The issue you seek was responded to above in a previous post, that the design temperature in the denominator/divisor is the same as used to determine the heat loss so that it cancels out the "per degree difference" effect of the heat loss which is in the numerator.
If you were to monkey with the numerator you would have to recalculate your heat loss. So if you used a 50 degree delta versus 72, your heat loss would be lower and so would your fuel usage.
If you change (reduce) just the assumed design temperature, the fuel use will apparently go up because you are dividing by a smaller number.
Does that make sense?
And, by the way, none of this is any trouble if it helps you.
Brad0 
that actually occurred to me but
I can't get it to register , somehow... Thick skull, you know.
H/(DT x 140,000 x E) x D x 24 x CD= GALLONS/YEAR
if I'm reading you right this would make heat loss = (heat loss x 72)/72
59000/72= 819
What is the original 819 value or unit of measurement that was multiplied by 72, that is now cancelled out by dividing it by 72 ?
Is the 819 some sort of (I can't think of a better word) "R factor" ( maybe it's not that bad of a word after all ) of the building envelope that is multiplied by the desired temperature differential between inside and outside?.
There again the smaller the number, the smaller the gallons used; so I guess it would be a measurement of the "conductance" of heat from the inside to the outside.
How do you read the correlation factor graph in your attachment, and what is the correlation factor?
I'm not sure how the degree days fits mathematically into all this either.
Well, that oughta be about enough questions from Mr. Question Mark. I just like to understand how things fit together.
Thanks0 
OK I might also be confused. Some one tell me if I am on the right track with the theory of the above formula.
I thought the purpose of the formula discussed here was to allow a comparision snapshot in time with another snapshot in time. The major variable of course in the snapshot is the weather. The formula then allows one to compare a new fuel efficient burning heat making machine with an inefficient machine allowing for the variation of temp from one warmer winter to a previous colder winter.
Am on the right track here??
0 
Yes...
And then divide by your heated space's square footage allowing you can compare it other places and get an idea of the overall heating efficiency.0 
Sort of
The typical first use of the DegreeDay formula uses 30year averages to get a ballpark range of what it might cost to heat a particular structure.
Beyond that, if you want to do specific trending, you need to know the actual degree days (the weather essentially) for the period you are trying to compare.
Only then can you get a range of "BTU's per DegreeDay". Even so these numbers will vary a bit so the difference is the assumed efficiency. Without a constant string of realtime efficiency measurements, it is all a guess anyway. A logical assumption at best.
As you can imagine, the efficiency if taken over a year is one thing but if you use a monthly basis, it will vary. A conventional nonmodulating, noncondensing boiler will be more efficient in colder months and less so in warmer months. ModCons the opposite, generally.0 
my old coal boiler was fantastic
when it was 20 below zero and I could sit right there in front of it in the cellar and fire it.
Other than that, about the only time it was 72 in the house was when it was on it's way up past it or on it's way down past it.
So what this formula amounts to is... the heat required by your house ( determined in part by how cold it usually gets in your area and in part by how well your house is constructed ), divided by the amount of BTU avaliable in a gallon of fuel oil multiplied by an efficiency factor of your heating system in getting it into your house. It will not necessarily give you an accurate picture of your particular building's energy use unless you know exactly how many degree days there were in the period AND how efficient your system is AND have an accutrate heat loss anaylsis. If you DO know the exact degree days AND have an accurate heat loss analysis,AND the gallons of fuel used, you can figure the efficiency of your heating system.
I think I got it!
Why didn't you say so0 
Good explanation, thanks. Will see if my professional who installed the boiler will give me the heat loss calc.
What to do when subsituting natural gas for oil? Do you use the ccf for N/G???0 
Units
For gas use therms. CCF is ok if you know your local gas quality is 1,000 BTU's per CF. But it all gets down to therms either way.
Adjust to your actual gas content. Here in Boston it tends to be slightly higher than 1,000 BTU/CF (maybe 1030 or so) because they add butane or hook up a pipe to our state house.0 
Historical Data
> you're on your own on the effeciency
> calculations.
>
> You can get the Heating Degree
> Days (HDD) historical data
> here:
>
> http://www.weather.gov/view/states.php
> Select your state then select climate data and
> scroll down till you see the nearest station's
> chart showing the monthtodate HDD's under
> column 6A. There is a way to get more historical
> HDD data if you surf this site
> further.......
>
> Good luck,
>
> Dave
Dave
Is there an easy place/way to get the historical data  I am looking at doing a 23 year look at my system prior to going to geothermal, etc.
Thanks.0 
Weather Underground
www.wunderground.com0 
Brad,
Thank you for the cookbook. I know what I will be doing tonight ;) I have always wondered about airside information. This will be a great reference tool.
thanks again!0 
> OK I might also be confused. Some one tell me if
> I am on the right track with the theory of the
> above formula.
>
> I thought the purpose of the
> formula discussed here was to allow a comparision
> snapshot in time with another snapshot in time.
> The major variable of course in the snapshot is
> the weather. The formula then allows one to
> compare a new fuel efficient burning heat making
> machine with an inefficient machine allowing for
> the variation of temp from one warmer winter to a
> previous colder winter.
>
> Am on the right track
> here??
The problem in your theory Scott is that you cannot compare boiler efficiency between traditional boilers and modcons via AFUE. Depending on the system to which it's attached a modcon may deliver seasonal efficiency slighty higher than "the number" while conventional boilers will almost always deliver significantly lower than "the number".0 
It worked
Thanks Uni R  after some serious poking around I found the right weather station that has provided me with all the data I need.0 
Mike
I was trying to do just that. I was going to take my usage during a one year period for the former W/M CI gas boiler and compare the next year after the installed new ultra 230.
Are you saying this is not a resonable comparison to examine effeciency between the two different boilers?
If the above discussed formula isn't correct for this assumption, how would you compare the two, regarding fuel consumption?
Scott0 
Benchmarking Energy Use
I calculate the energy use in Btu's / HDD / sq ft as outlined in the attached Home Energy article.
This standardized method can compare the heating efficiency of two different boiler systems in the same house in different years and can compare different buildings in the same or different cities.
You need the heating fuel use and the heating degree days for the period. Heating Degree Days corrects for climate differences.
Doug0 
Benchmarking Energy Use
Calculate your energy use in Btu's / HDD / sq ft  as outlined in the attached Home Energy article.
This standardized method can compare the overall heating efficiency of two different boiler systems in the same house in different years, or can compare different buildings in the same or different cities.
You need the heating fuel use and the heating degree days for the period. Heating Degree Days corrects for climate differences in different years or different locations.
Doug0 
Are you saying this is not a resonable comparison to examine effeciency between the two different boilers?
Not via AFUE when comparing a conventional boiler to a condensing/modulating boiler.
If the above discussed formula isn't correct for this assumption, how would you compare the two, regarding fuel consumption?
I did so in my system, but only after the boiler changed. My calculations were good enough for this house such that I could predict fuel use on a monthly (or longer) basis within ±10% of actual regardless of which boiler was used or the weather.
In doing so, I found that:
1) AFUE for my conventional boiler in my system was greatly overstatedabout 35%.
2) Heat loss calculation (Manual Jbased) for my home was greatly overstatedabout 45%. It is quite wellknown and understood that all heat loss calculations are significant overstatements35%  60% is the generally accepted range of overstatement.
3) Degree day calculation based on 65°F (the standard) and 70° room temperature (also the standard) tends to overstate heat loss as it does not account for occupancy and solar gains. For my house, I found that a degree day basis approximately 10°F below the average room temperature workede.g. for 65°F average room temp, I'd calculate degree days based on the difference between the outside temp and 55°F. I also found that using average hourly temps produced a better estimate of degree days than average daily temp. Each hour is given a "weight" of 1/24th its value to produce degree days. The shorter the period of time you are examining, the greater I found was the difference between the two calculation methods. On an annual basis, there was little differenceon a monthly basis there was often significant difference. In general daily average temp tended to overstate loss early in the season and understate late in the season.
Using AFUE and the "straight" heat loss calculation the conventional boiler thus appeared to be operating at or near its' stated efficiencyat least on a seasonal basis.
The condensing/modulating boiler on the other hand did not consume enough fuel to meet "straight" ManualJ losses. Not nearly enough.
Since it's not reasonable to believe that the modcon is operating at greater than 100% efficiency (gross fuel value) nor is it reasonable to assume that the modcon itself reduces heat loss by any large amount the only reasonable assumption was that heat loss was never as great as assumed and, more importantly, that the conventional boiler was never as efficient as stated.
I keep stressing "my system" because all of these numbers are system dependent. Just a few of the variables involved are:
1) Orientation of the structure.
2) Window placement and operation of any window coverings.
3) Degree of oversizing of the boilers.
4) Type and placement of emitters.
5) Outside temperature variability.
6) General control philosophy of the boilers.
7) Habits of the occupants to include any daily setback and desire for "quick" heat.
8) General operating temperatures of the system.
9) Number of "zones" and their relative usage/loss/emitter sizing/emitter type/operating temperatures to each other.
The list goes on and on and on...
Sorry, but I cannot give a "magic" formula to compare a conventional boiler in any given system/structure to a condensing/modulating boiler.
I can however say that a condensing/modulating boiler will in nearly every case use significantly less fuel than implied by the difference in AFUE. From reports here, 25% would seem to be the absolute least with 40% or so most common and 50%+ more common than the lower end.
Perhaps some day the US Dept. of Energy, ASHRAE and others will revise their standards for determining heating appliance efficiency. I would not hold my breath however as their true concern seems to be to make warmair furnaces appear to be much more efficient than they actually are when used in actual systems in the real world...0
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